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http://mathhelpforum.com/calculus/15611-area.html | 1. ## Area
Find c>0 such that the area of the region enclosed by the parabolas y = x^2 - c^2 and y = c^2 -x^2 is 230.
c =?
2. Originally Posted by qbkr21
Find c>0 such that the area of the region enclosed by the parabolas y = x^2 - c^2 and y = c^2 -x^2 is 230.
c =?
you are integrating with respect to x, so you cannot integrate the c terms like that. you would just attach x to them. by the way, when i did it your way i got 0 = 230, which means you probably made another mistake somewhere
3. ## Re:
Could you trying working it out? or at least some of it... so that I can get a feel for what you mean...
Sure appreciate it....
-qbkr21
4. Originally Posted by qbkr21
Could you trying working it out? or at least some of it... so that I can get a feel for what you mean...
Sure appreciate it....
-qbkr21
$\int_{-c}^{c}\left[ c^2 - x^2 - \left( x^2 - c^2 \right) \right]dx$
$=2 \int_{0}^{c} \left( 2c^2 - 2x^2 \right) dx$
$= 4 \int_{0}^{c} \left( c^2 - x^2 \right)dx$
$= 4 \left[ c^2 x - \frac {1}{3}x^3 \right]_{0}^{c}$
$= \frac {8}{3}c^3$
check my computation, i was in a rush. i have to leave for a little while
did you see the difference with how i integrated c? i treated it as a constant--which it is
5. ## Re:
Yes you were right on the money as always!!
c = 4.41828
Thanks so much!!!
-qbkr21
6. ## find c
Originally Posted by Jhevon
$\int_{-c}^{c}\left[ c^2 - x^2 - \left( x^2 - c^2 \right) \right]dx$
$=2 \int_{0}^{c} \left( 2c^2 - 2x^2 \right) dx$
$= 4 \int_{0}^{c} \left( c^2 - x^2 \right)dx$
$= 4 \left[ c^2 x - \frac {1}{3}x^3 \right]_{0}^{c}$
$= \frac {8}{3}c^3$
check my computation, i was in a rush. i have to leave for a little while
did you see the difference with how i integrated c? i treated it as a constant--which it is
I think you are right
7. Originally Posted by curvature
I think you are right
i know, qbkr21 already confirmed it. i believe he has the answer to the question in the back of his text. i hope he gets why i could have integrated between 0 and c though and then multiply by two, furthermore, i hope he gets why i would want to do that in the first place
8. Originally Posted by Jhevon
$\int_{-c}^{c}\left[ c^2 - x^2 - \left( x^2 - c^2 \right) \right]dx$
$=2 \int_{0}^{c} \left( 2c^2 - 2x^2 \right) dx$
hi
what did you do to the 2 outside to make the lower integral -c 0?
can it be any constants?
hi
what did you do to the 2 outside to make the lower integral -c 0?
can it be any constants?
Note that your integrand is an even function: f(-x) = f(x). In this case
$\int_{-c}^c dx \, f(x) = \int_{-c}^0 dx \, f(x) + \int_0^c dx \, f(x)$
Use the substitution y = -x in the first integral:
$= \int_{c}^0 (-dy) \, f(-y) + \int_0^c dx \, f(x)$
$= -\left ( -\int_0^c dy \, f(-y) \right ) + \int_0^c dx \, f(x)$
But f(-y) = f(y) when f is an even function:
$= \int_0^c dy \, f(y) + \int_0^c dx \, f(x)$
and now just replace the dummy variable y in the first integration with x:
$= \int_0^c dx \, f(x) + \int_0^c dx \, f(x)$
$= 2 \int_0^c dx \, f(x)$
-Dan | 2017-10-21T16:27:49 | {
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https://tri7bet.co/e5dmwu/4c23b1-circumference-definition-math | If the circle has radius $$r$$ then the circumference is $$2\pi r$$. It gives the ratio of circumference to the diameter of a circle or the ratio of circumference to twice the radius. At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students! This section of Revision Maths defines many terms in relation to circles, including: Circumference, Diameter, Radius, Chord, Segment, Tangent, Point of contact, Arc, Angles on major and minor arcs, Angle of Centre and Sectors. Meaning of circumference. There is an error: “Area of the segment = ( θ /360) x π r 2 + ( 1 /2) x sinθ x r 2” …should be “Area of the segment = ( θ /360) x π r 2 – ( 1 /2) x sinθ x r 2” sivaalluri (January 22, 2019 - 3:31 pm) Reply. More About Circumference. 3.14 x 100 yards = 314 yards ; Circumference = 314 yards ; Your walk around the lake is 314 yards. Hollow Cylinder Calculator . I consulted a standard dictionary, which said it was the length of a closed curve enclosing an area, so it would seem to apply under that definition. 9 thoughts on “ Circle formulas in math | Area, Circumference, Sector, Chord, Arc of Circle ” Steve LeVine (December 31, 2018 - 10:05 pm) Reply. Capsule Calculator . For example, the Latin root of the word circumference is circum, meaning around. The circumference of a circle is the distance around the circle. Related Calculators: Helix Curve Calculator . Definition: Pi is a number - approximately 3.142 It is the circumference of any circle divided by its diameter. Circumference is the linear distance around a circle. If you're seeing this message, it means we're having trouble loading external resources on our website. Also learn the facts to easily understand math glossary with fun math worksheet online at SplashLearn. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. The boundary line of a circle. We will be listening the word “perimeter” frequently in the problems which are based on geometry. The circumference of a circle is the distance around the circle. Learn about terms radius, diameter, circumference, chord, pi, etc.. ... what is the official definition of a circle? What does circumference mean? A circle has many different radii and many different diameters, each passing through the center. Download FREE Right circumference of a Circle Worksheets where, R is the radius of the circle. All Free. Circumference, diameter and radii are measured in linear units, such as inches and centimeters. Perimeter is the distance around a two-dimensional object. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Learn more. By Mary Jane Sterling . The circumference of a circle of radius 3 is $$2\cdot\pi\cdot 3$$, which is $$6\pi$$, or about 18.85. View our Lesson on Circumference of a Circle. The circumference of a circle is the distance around it, but if, as in many elementary treatments, distance is defined in terms of straight lines, this cannot be used as a definition. Radius definition, a straight line extending from the center of a circle or sphere to the circumference or surface: The radius of a circle is half the diameter. The complete distance around a circle or a closed curve is called its Circumference. Math Goodies Glossary. What is circumference? Learn and know what is the perimeter meaning in geometry chapter. See more. These are the two important words for which we have to know the meanings in geometry. Below are our grade 5 geometry worksheets on determining the circumference of circles.Students are provided the radius or the diameter in customary units (worksheets 1-3) or metric units (worksheets 4-6). The Circumference (or) perimeter of a circle = 2πR. Circumference: The circumference of a circle is the distance around it. If you imagine the circle as a piece of string, it is the length of the string. * By signing up, you agree to receive useful information and to our privacy policy. Definition: Circumference. What does that mean? Top Calculators. Formally, a circumference is defined as the locus of points from the plane equidistant to another point, called the center of the circumference. Circumference to diameter. You can also measure the perimeter of three-dimensional objects like houses, stadiums, buildings and similar shapes. It includes unlimited math lessons on number counting, addition, subtraction etc. The distance around a circle is called the circumference. Courses. Math Made Easy! The length of such a line. Also find the definition and meaning for various math words from this math dictionary. Now you will be able to easily solve problems on circumference definition, circumference geometry definition, c ircumference of a circle formula, circumference equation, and circumference of circle. Definition of circumference in the Definitions.net dictionary. c or circ. Math; Trigonometry; Radius, Diameter, Circumference, and Area of Circles; Radius, Diameter, Circumference, and Area of Circles. What is a circle? the length of a perimeter. While I would never actually refer to the circumference of a circle as its "perimeter," a discussion recently arose about whether the term "perimeter" can even APPLY to a circle. Learn area of different shapes here. Circle Geometry math for kids. Jan 31, 2013 - go to this link for a math dictionary tool to use at school or at home (Circumference - the distance around the edge of a circle) A real-life example of a radius is the spoke of a bicycle wheel. Learn the relationship between the radius, diameter, and circumference of a circle. Worksheets > Math > Grade 5 > Geometry > Circumference of circles. Along with the perimeter, we should also know the meaning of “Area”. M Another name for circumference is perimeter. Again, Pi (π) is a special mathematical constant; it is the ratio of circumference to diameter of any circle. Circumference definition, the outer boundary, especially of a circular area; perimeter: the circumference of a circle. b. Abbr. Definition Of Circumference. perimeter: [noun] the boundary of a closed plane figure. Related Topics: Math Worksheets Objective: I know how to calculate the circumference of a circle. For any given circle, this ratio will remain π. A circle is a geometric figure that needs only two parts to identify it and classify it: its center (or middle) and its radius (the distance from the center to any point on the circle). The margin or area surrounding something. circumference definition: 1. the line surrounding a circular space, or the length of this line: 2. the outside edge of an…. circumference - WordReference English dictionary, questions, discussion and forums. E-Mail Address * Featured … Search for courses, skills, and videos. Learn what is circumference. Area and Perimeter is an important topic in Mathematics, which is used in everyday life. If you're seeing this message, it means we're having trouble loading external resources on our website. Learn the relationship between the radius, diameter, and circumference of a circle. Search. Circumference. 2. a. How to use circumference in a sentence. Did You Know? Circumference definition is - the perimeter of a circle. The boundary line of an area or object. ence (sər-kŭm′fər-əns) n. 1. Circumference may also refer to the circle itself, that is, the locus corresponding to the edge of a disk Circle. A circle is a shape that is made up of all the points on a plane (a flat surface) that are the same distance from a given point. Perimeter defines the distance of the boundary of the shape whereas area explains the region occupied by it. Area and perimeter, in Maths, are the two important properties of two-dimensional figures. Circum is now considered a prefix also meaning around or round about. 3. The formula for the circumference of a circle, C = 2 × π × r = π × d, where r is the radius and d is the diameter of the circle.π =22/7 = 3.141. Donate Login Sign up. A circle is a two-dimensional shape that is created by drawing a line that is curved so that its ends meet and every point on the line is the same distance from the center. π is the mathematical constant with an approximate (up to two decimal points) value of 3.14. Search . The number Pi, denoted by the Greek letter π - pronounced 'pie', is one of the most common constants in all of mathematics. Learn More at mathantics.comVisit http://www.mathantics.com for more Free math videos and additional subscription based content! Search form. where C = π D. C is the circumference of the circle Sign Up For Our FREE Newsletter! Definition of Perimeter explained with real life illustrated examples. Circumference = 3.14 x 100 yards ; Do the Math. It is the circumference of any circle, divided by its diameter. Make your child a Math Thinker, the Cuemath way. Sign Up For Our FREE Newsletter! It is a constant so whatever be the circumference and the radius the ratio will be equal to π. π is an irrational number hence it cannot be written down as a finite decimal number. Tip. You have probably noticed that, since diameter is twice the radius, the proportion between the circumference and the diameter is equal to π: C/D = 2πR / 2R = π. Okay, so let's start out with the "given point". See more. SplashLearn is an award winning math learning program used by more than 30 Million kids for fun math practice. The formula is πd or 2πr. Sal uses formulas and a specific example to see how area and circumference relate. Learn the circumference of a circle with Definition, Solved examples, and Facts. Circumference: The distance around a circle; the perimeter of a circle. Read the lesson on circumference of circle if you need to learn how to calculate the circumference of a circle. The following is a list of definitions relating to the circumference of a circle. About Cuemath. Information and translations of circumference in the most comprehensive dictionary definitions resource on the web. Circle worksheets: Finding the circumference. It is used in many areas, such as physics and mathematics. If you have trouble remembering geometry terms, it helps to think of other words from the same root with which you may be more familiar. We must never confuse the concept of a circle with the concept of circumference, circumference is actually a curve that encloses a circle (the circumference is a curve, the circle is an area.) This proportion (circumference to diameter) is the definition of the constant pi. Main content. Perimeter Definition; Perimeter of a Triangle; Perimeter of a Square or Rhombus; Perimeter of a Rectangle or Parallelogram; Perimeter of an N-gon; Perimeter Definition. Calculators and Converters ↳ Math Dictionary ↳ C ↳ Circumference ; Ask a Question . Home » Glossary » Term » Math Goodies Glossary.
Dublin To Westport, Jason Pierre-paul Hand Video, Venom Vs Superman, Marco Reus Fifa 13, Purple Tier Covid, Easyjet Flights From Luton To Isle Of Man, South Mayo Family Research Centre, Stainless Steel Meaning In Kannada, What Channel Is The Washington Redskins Game On Today, Venom Vs Superman, Elon Women's Soccer Id Camp 2020, St Vincent Dental Clinic Cleveland Ohio, Blue Anodized Ar-15 Grip, | 2021-04-21T04:31:22 | {
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http://mathhelpforum.com/calculus/111456-quick-question-about-position-due-gravity.html | # Math Help - Quick question about position due to gravity.
1. ## Quick question about position due to gravity.
I have a problem where I'm given a bunch of info. about a ballon dropping a bomb, etc. The thing is that I am not given any functions but I know that I have to use the function for position of falling bodies with respect to time.
It's been a while since I saw this formula, so i'm not sure if this is right:
In feet: $s(t)=-16t^2 + v_0 t + s_0$
where $v_0$ is initial velocity and $s_0$ is starting position.
Is any of this right?
2. I'm not sure what you are trying to do.
You have a bomb floating to the earth on a balloon, and you have to construct the equation which describes the displacement the balloon as fallen from its original position after time t?
3. Originally Posted by Arturo_026
I have a problem where I'm given a bunch of info. about a ballon dropping a bomb, etc. The thing is that I am not given any functions but I know that I have to use the function for position of falling bodies with respect to time.
It's been a while since I saw this formula, so i'm not sure if this is right:
In feet: $s(t)=-16t^2 + v_0 t + s_0$
where $v_0$ is initial velocity and $s_0$ is starting position.
Is any of this right?
None of what you're asking can be answered without knowing the question. It's always better to post the original question instead of hoping that our crystal balls are in good working order.
4. Try one (or more) of the kinematic equations:
$v = v_0 + at$
$y - y_0 = v_0t + \frac{1}{2}t^2$
$v^2 = v_0^2 + 2a(y - y_0)$
$y - y_0 = \frac{1}{2}(v_0 + v)t$
Where v is the final velocity, $v_0$ is the initial velocity, y is the final height, $y_0$ is the initial height, a is the acceleration due to gravity near Earth's surface ( $g = 9.8 \frac{m}{s^2}$), and t is the time.
Patrick
5. Originally Posted by mr fantastic
None of what you're asking can be answered without knowing the question. It's always better to post the original question instead of hoping that our crystal balls are in good working order.
Ok, here is the problem:
A bomb is dropped from a ballon hovering at an altitude of 800 ft. From directly below the balloon, a projectile is fired directly upward toward the bomb exactly 2 seconds after the bomb is released. With what initial speed should the projectile be fired in order to hit the bomb at an altitude of exactly 400 ft.?
6. The problem is for some reason in the Standard system (feet) instead of the Metric (meters), so be sure to consider that when solving the problem (This is where the -16 comes from in your original equation).
The bomb accelerates downward at a rate of 9.8 $\frac{m}{s^2}$ or just over 32 $\frac{ft}{s^2}$. After t seconds, it will be halfway down, from 800 ft to 400ft. At this precise moment, the projectile from the surface will intercept it. Therefore, you need to solve for t, the time it takes the bomb to drop 400 ft, from an initial velocity of zero.
Then, when you have t, you must subtract 2 seconds to get the time for the projectile's flight from the surface to the bomb, over a distance of 400 ft. When you have this time, you can then use the kinematic equations to solve for the initial velocity, knowing that the acceleration due to gravity will slow the projectile so that the final velocity is less than the initial velocity.
You will have the variables of t, a, and y. Using the second equation listed (a is not needed), you can solve for the initial velocity. Let me know what the answer is.
Patrick
7. I get t=5 for the bomb to be at an altitude of 400 ft. and when t=3 the projectile needs to be launched with an initial velocity of 181.33 ft. per second. | 2015-05-23T13:48:18 | {
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https://physics.stackexchange.com/questions/652716/two-masses-connected-by-a-spring | # Two masses connected by a spring
I have a question about a problem I saw on a website
A mass is attached to one end of a spring, and the other end of the spring is attached to an immovable wall. The system oscillates with period T. If the wall is removed and replaced with a second mass identical to the first, what is the new period of oscillation of the system?
My answer: $$\frac{T}{\sqrt{2}}$$
Website's answer (Incorrect thanks to trula): $$T\sqrt{2}$$
My thought process (pretty hand-wavy I know):
Because the spring constant is inversely proportional (I think) to length of the spring, the spring constant will be doubled if you halve the length of the spring, which is what is essentially happening with two identical massess. Because $$T=2\pi\sqrt{\frac{m}{k}}$$, if you double $$k$$, you divide $$T$$ by $$\sqrt{2}$$. I don't know why they got $$T\sqrt{2}$$, but if I am wrong, please correct my misunderstandings.
Thanks for reading if you made it this far, and if my question is unclear, please tell me what I can fix about it.
• Your argument is right, and the published answer probably just a typing error. Jul 21 at 21:17
• Ok, thanks for the clarification!
– JD12
Jul 21 at 21:24
• the spring constant will be doubled if you halve the length of the spring, which is what is essentially happening with two identical massess. I really don't see why that would be true. If instead $m\to 2m$ then $T\to\sqrt{2}T$, as stated in the answer key.
– Gert
Jul 22 at 0:08
The approach you have used appears to be correct. The correct answer should be $$\frac{T}{\sqrt 2}$$. As you have noted, halving the length of spring with spring constant $$k$$ results in a doubling of the spring constant.
Since we know that $$T=2\pi\sqrt{\frac{m}{k}}$$ and if we let $$k\rightarrow 2k$$, then $$T=2\pi\sqrt{\frac{m}{k}} \rightarrow 2\pi\sqrt{\frac{m}{2k}}=\frac{2\pi}{\sqrt 2}\sqrt{\frac{m}{k}}=\frac{1}{\sqrt 2} 2\pi\sqrt{\frac{m}{k}}=\frac{1}{\sqrt 2}T$$
The answer posted - $$T\sqrt 2$$ - could very well be a typo as also pointed out by @trula in the comments above.
Equal masses, centre of mass does not move, stretch $$x$$ relative to centre of mass position for both masses, force exerted by spring $$k\cdot 2x$$, equation of motion for each of the masses of the form $$-2kx = m\ddot x$$, period is reduced by factor $$1/\sqrt 2$$. | 2021-09-17T13:06:59 | {
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https://math.stackexchange.com/questions/1415237/is-there-a-continuous-function-from-0-1-to-mathbb-r-that-satisfies/1415256 | # Is there a continuous function from $[0,1]$ to $\mathbb R$ that satisfies
Is there a continuous function $f:[0,1] \to \mathbb R$ such that $f(x) = 0$ uncountably often and, for every $x$ such that $f(x) = 0$, in any neighbourhood of $x$ there are $a$ and $b$ such that $f(a) > 0$ and $f(b) < 0$?
Let $K$ be the Cantor set; recall that $K$ is uncountable. Then the function $x\to d(x,K),$ where $d(x,K)$ denotes the distance from $x$ to $K,$ is continuous on $[0,1].$ Define
$$f(x) = \begin{cases} 0, x \in K\\ d(x,K)\sin (1/d(x,K)), x \in [0,1]\setminus K. \end{cases}$$
Let $I_1, I_2, \dots$ be the open intervals thrown out in the construction of $K.$ In each $I_n,$ $f$ is continuous. Hence $f$ is continuous on $[0,1]\setminus K.$ If $x_n \to x \in K,$ then $d(x_n,K)\to 0,$ hence so does $f(x_n),$ which implies $f$ is continuous at $x.$ Thus $f$ is continuous on $[0,1].$
Now in any neighborhood of a point in $K,$ one of the $I_n$ will be contained in that neighborhood, hence $f$ will be both positive and negative in that neighborhood because of the $\sin (1/d(x,K))$ term.
Added: I just noticed that I omitted discussing points in $[0,1]\setminus K$ where $f=0.$ That happens at an $x$ such that $d(x,K)=1/(m\pi)$ for some $m\in \mathbb {N}.$ Because each $I_n$ has length $1/3^k$ for some $k,$ such an $x$ could not be the midpoint of any $I_n.$ So then $x$ is to the right or left of the midpoint of whatever $I_n$ it belongs to, and thus $d(\cdot,K)$ will be strictly increasing or decreasing in a neighborhood of $x,$ which implies a change in the sign of $f$ at that $x$ as desired.
• Very clever! How'd you come up with that? – Moya Aug 30 '15 at 23:24
• $x\sin(1/x)$ and its relatives are the gift that keeps on giving. – zhw. Aug 30 '15 at 23:28
In addition to the carefully constructed examples others have given, I just want to note that with probability $1$ one-dimensional Brownian motion starting at $0$ has these properties.
Absolutely. For example, we can proceed in a similar fashion to the definition of the Cantor function, using the "middle third construction of the Cantor set. In the first stage, we remove the "middle third" $(\frac13,\frac23)$ of $[0,1],$ and on this interval, we define $f(x)=(x-\frac12)^2+c_1$, where $c_1$ is the unique number such that $f(x)\to 0$ as $x\searrow\frac13$ and $x\nearrow\frac23.$ (I leave its determination to you.)
In the second stage, we remove the two "middle thirds" $(\frac19,\frac29)$ and $(\frac79,\frac89)$. On the first of these, we define $f(x)=-(x-\frac16)^2+c_2,$ where $c_2$ is the unique number so that $f(x)\to 0$ as $x\searrow\frac19$ and $x\nearrow\frac29.$ on the second, we define it by $f(x)=-(x-\frac56)^2+c_2,$ and we can show that $f(x)\to 0$ as $x\searrow\frac79$ and $x\nearrow\frac89.$ (I leave it to you to find $c_2$ and verify this.
We proceed in much the same manner. At the $n$th stage, we remove $2^{n-1}$ open intervals of the form $\left(m-\frac1{2\cdot3^n},m+\frac1{2\cdot3^n}\right),$ and there is some $c_n$ such that for any such "middle third" interval, we define the function on that interval by $f(x)=(-1)^{n-1}(x-m)^2+c_n,$ and we have the property that $f(x)\to0$ as $x\searrow m-\frac1{2\cdot3^n}$ and $x\nearrow m+\frac1{2\cdot3^n}.$
Finally, we define $f(x)=0$ on the Cantor set, itself. Then $f$ has exactly the properties you need, and is in fact differentiable everywhere but at the endpoints of the "middle third" intervals, I believe.
Yes.
Our function will be $0$ on the Cantor set and nowhere else. For convenience we define a spike function $S(a, b)(x) = \mathbb{1}_{[a, b]}(x) \cdot \min\{|a-x|, |b-x|\}$ which is a continuous, positive, compact support function on the interval $[a, b]$. The idea will be to add a countable collection of spike functions to eventually satisfy your conditions.
So, lets approximate our function in a series of countable steps. Firstly, let $f_1 = S(1/3, 2/3)$. The greatest distance from a point with positive function value is now $1/3 = \delta_1$. Let $f_2 = f_1 - (S(1/9, 2/9) + S(7/9, 8/9))$. Now, the greatest distance from a point in $x\in [0, 1]$ with $f_2(x) = 0$ to a point with $y\in [0, 1]$ with $f_2(y)<0$ is now $1/9 = \delta_2$. Proceeding in this way, we would next add $4$ spikes to get $f_3 = f_2 + \sum_{i = 1, 7, 19, 25} S(i/27, (i+1)/27)$, and $\delta_3 = 1/27$, subtract $8$ spikes to get $f_4$
I claim that the limit of the $\{f_n\}$ exists, is well defined, and satisfies your conditions because the $\delta_n\to 0$.
Completed as a pennance for not reading the question properly.
• The question doesn't ask for the function to be surjective, just that it takes real values. – Mark Bennet Aug 30 '15 at 22:50
• +1: Nice fix. That's pretty much the direction I was considering, initially, but I decided to try for more differentiability (for no particular reason). – Cameron Buie Aug 30 '15 at 23:41
• Its the natural solution, though Robert Israel has a nice example/class of examples too. – user24142 Aug 31 '15 at 5:49 | 2019-07-23T00:55:10 | {
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https://math.stackexchange.com/questions/2730961/the-difference-between-inverse-function-and-a-function-that-is-invertible | # The difference between inverse function and a function that is invertible?
My discrete mathematics book says:
But I read an answer, https://math.stackexchange.com/a/2415543/390226, said:
["...]There are invertible functions which are not bijective,[..."]
["]A function is invertible if and only if it is injective[."]
So for a function to have a inverse, it must be bijective. But any function that is injective is invertible, as long as such inverse defined on a subset of the codomain of original one, i.e. the image of the original function?
• See Inverse function : "To be invertible a function must be both an injection and a surjection. Such functions are called bijections. The inverse of an injection $f : X → Y$ that is not a bijection, that is, a function that is not a surjection, is only a partial function on $Y$, which means that for some $y ∈ Y, f^{ −1}(y)$ is undefined." This id the case with $\exp$ in the post you have linked. – Mauro ALLEGRANZA Apr 10 '18 at 14:10
• The issue is simply with $A,B$ of $f : A \to B$. If we "build in" domain and co-domain in the def of $f$ we have that the function $\exp : \mathbb R \to \mathbb R$ is not invertible. But there is an old tradition of calling $\log : \mathbb R^+ \to \mathbb R$ its inverse. As per Gentlemen Prefer Blondes: "nobody is perfect". – Mauro ALLEGRANZA Apr 10 '18 at 14:28
• @MauroALLEGRANZA: Thank you so much, now it's clear to me... Because the same book says to define a function, the domain, co-domain, image/range are needed. I didn't know the tradition. – linear_combinatori_probabi Apr 10 '18 at 14:33
It all depends on the co-domain of your function.
When you have a function $$f:A\to B$$ which is one-to-one but not onto $B$, you may restrict your co-domain to a subset of $B'\subset B$ which is the range of $f$.
For example $$f:N \to N$$ defined by $$f(n)=2n$$ is not onto but it is one-to-one.
If we define, $$f^* : N\to 2N$$ with the same definition $f^*(n)=2n$
We have an inverse function, $(f^*)^{-1} (n) = n/2.$
• So when I see "f is invertible", I should connect this idea to "f must be bijective"? If f is not bijective, then I must change its co-domain? – linear_combinatori_probabi Apr 10 '18 at 14:16
• Yes you need a one-to-one correspondence which means bijective. – Mohammad Riazi-Kermani Apr 10 '18 at 14:17
When people said a function is "invertible", they mean it can be made invertible. And the rigorous definition of inverse function of $f$ in my book is: | 2021-06-25T00:44:59 | {
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https://math.stackexchange.com/questions/806811/smallest-prime-factor-why-does-this-algorithm-find-prime-numbers | # Smallest Prime Factor - Why does this algorithm find prime numbers?
I have been looking at the problems on Project Euler and a number of them have required me to be able to find the prime factorisation of a given number.
While looking for quick ways to do this, I came across a website that could perform this calculation and had it's source code available too.
The crux of the algorithm was this method: (I hope java code on this site is OK)
public static long smallestPrimeFactor( final long n ) {
if (n==0 || n==1) return n;
if (n%2==0) return 2;
if (n%3==0) return 3;
if (n%5==0) return 5;
for (int i = 7; (i*i) <= n; i += 30 ) {
if ( n % i == 0 ) {
return i;
}
if ( n % ( i+4 ) == 0) {
return i+4;
}
if ( n % ( i+6 ) == 0) {
return i+6;
}
if ( n % ( i+10 ) == 0) {
return i+10;
}
if ( n % ( i+12 ) == 0) {
return i+12;
}
if ( n % ( i+16 ) == 0) {
return i+16;
}
if ( n % ( i+22 ) == 0) {
return i+22;
}
if ( n % ( i+24 ) == 0) {
return i+24;
}
}
return n;
}
The part that really fascinates me is the loop that starts at 7, then considers some gaps and then increments by 30 before carrying on. This is the list of the first few numbers from the sequence:
7, 11, 13, 17, 19, 23, 29, 31,
37, 41, 43, 47, 49, 53, 59, 61,
67, 71, 73, 77, 79, 83, 89, 91,
97, 101, 103, 107, 109, 113, 119, 121,
127, 131, 133, 137, 139, 143, 149, 151,
157, 161, 163, 167, 169, 173, 179, 181,
187, 191, 193, 197, 199, 203, 209, 211
Obviously this loop generates some numbers that are not prime numbers (49, 77, 91 for example) but it does appear to produce every possible prime number less than the square root of the given number.
Am I correct in believing that this loop will produce every prime number? And if so, is there a proof or some mathematical reasoning behind why that is the case?
• See prime sieves and esp. wheel sieves. – Bill Dubuque May 23 '14 at 15:57
• The loop looks for divisors without checking if they are primes. It provides all possible prime divisors. The reason to exclude some possible divisors is that they are not prime. For example: $7+30k+1=30k+8$ is even and $2$ was checked, $7+30k+2=30k+9$ is multiple of $3$ (also checked), $7+30k+3=30k+10$ which is a multiple of $10$, $7+30k+5=30k+12$ which is a multiple of $3,$ and so on. – mfl May 23 '14 at 16:00
In the loop $i$ is equal to $30k+7$, where $k$ is a non-negative integer. Let's consider what values can't be prime. I'll list the offsets and remove them as they're eliminated. $$0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29$$
We start with $7$, which is odd. Adding another odd number would give us an even number, which is divisible by $2$. Eliminate those. $$0,2,4,6,8,10,12,14,16,18,20,22,24,26,28$$
We're adding multiples of $30$, which are divisible by $3$. The last digit is $7$. Adding $2$ would make it $9$, making the whole number divisible by $3$. So $2$ gets skipped, so well as $2+3k$. Eliminate those. $$0,4,6,10,12,16,18,22,24,28$$
Same logic for $5$. Since the last digit is $7$ without an offset, then adding $3,8,13$, etc, would make it a multiple of $5$. Eliminate those. $$0,4,6,10,12,16,22,24$$
And you're left with the numbers that are checked. We increment by $30$ since $2\cdot3\cdot5=30$, and we checked those at the very beginning. Adding $30$ produces the same pattern of divisibility by those numbers at the same offsets. So from then on we just need to dodge those offsets.
• That explanation is super clear! – Dan Temple May 23 '14 at 16:21
This should produce all primes. First, $i=30j+7$ for some $j$. Now with this in mind, take a look at the values that get skipped. Obviously, first thing, it skips all even numbers. As an example though, look at the first odd number it skips, $i+2=30j+9=3(10j+3)$. You will find that all the numbers skipped are similarly multiples of $3$ or $5$. In other words, everything it skips is composite.
The code you posted in fact is not very efficient. The way it works is this:
The purpose is ro return the smallest prime factor of number n.
At first it does checks for trivial/simple cases (whether n is divisable by 2,3,5, the first 3 prime numbers)
Then the loop starts fromt the next prime number (=7) and checks up to sqrt(n) (which is enough, but not the most efficient check for factoring n)
The checks inside the loop check for square-free factors and not multiples of 2,3,5 (limited Eratosthenes' sieve) (thus primes) | 2019-12-09T11:15:03 | {
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http://terrysylvester.com/in-hr-eipwkv/4e83f0-bio-bidet-bb-600-troubleshooting | Thank you! View all posts by Whit Ford, am very thankful 2 the information above.it is very helpful to me. The notation that follows a capital Pi describes only the term that is to be multiplied. A “series” is the sum of the first N terms of a sequence. Chapters & Colonies . Two of these hybrids from each C atom overlap with H 1s orbitals, while the third overlaps with an sp 2 hybrid on the other C atom. (mathematics) Braid group algebra. and therefore I will always be checking this site. A description of the double bond is the sigma-pi model shown in Figure 1. Search for "sigma symbol" in these categories . These symbols can be copy and pasted anywhere you want. All that matters in this case is the difference between the starting and ending term numbers… that will determine how many twos we are being asked to add, one two for each term number. Yes No. (Physics, scattering) Cross_section_(physics). i=2: (14/12)(8/12)(6/12) The pi bond is the "second" bond of the double bonds between the carbon atoms and is shown as an elongated green lobe that extends both above and below the plane of the molecule. Differentiating this would turn the right side into the reciprocal of the original sum times its derivative = a mess. The Sigma symbol, , is a capital letter in the Greek alphabet. So, my opinion would be: sure! I will modify my response shortly. The Pi symbol, , is a capital letter in the Greek alphabet call “Pi”, and corresponds to “P” in our alphabet. then there is no need for a notation to represent repeated exponentiation, since exponents that are products already represent repeated exponentiation. Pi (capitale Π, minuscule π ou parfois ϖ ; en grec πι) est la 16 e lettre de l'alphabet grec, précédée par omicron et suivie par rhô. Thanks , Very useful post. Therefore, additional notations are not needed to describe repeated subtraction or division… Which is quite convenient. Ways to make pi symbol, HTML unicode entities and more. Character Palette allows you to view and use all characters and symbols, including pi, available in all fonts (some examples of fonts are "Arial", "Times New Roman", "Webdings") installed on your computer. Post was not sent - check your email addresses! Choose from 500 different sets of phi sigma rho flashcards on Quizlet. The convention is to increase it, just like with Sigma and Pi notation, but they also support decreasing indeces. I have a student asking whether there is a symbol for exponentiation of a sequence? You can put it in Facebook, Youtube or Instagram. it would appear as though the quantity in parentheses is becoming increasingly negative (a sum of growing negative numbers), and therefore the value probably goes to negative infinity. Sigma Symbol, The Eighteenth Letter in Greek Alphabet – ©Niakris6 at ShutterStock The sigma sign has been used in mathematics, chemistry, and other fields of study for a very long time. Equation for Xn in terms of P1,P2,……Pn. Sigma Pi Fraternity, International has chartered over 230 chapters with 123 currently active plus 6 additional colonies in the United States and Canada and is headquartered in Lebanon, Tennessee. In my mind, this rounds up each time the value is divided by (1-r). Summation notation does not provide an easy way that I can think of to do what you describe. I like it … And I hope it will help other students too to acheive their goals …. Sigma Symbol in Greek Alphabet. sigma and pi? Those are only the first numbers of pi, there's a symbol for pi precisely because you can't precisely specify pi with any decimal number, only to a certain degree of accuracy. Excellent blog and makes maths symbols and operations simple to understand. for i = 0 to 32. please help improve this site's prominence in search results by including a link to this site in appropriate places elsewhere on the internet - perhaps in a response to a math question, or in a comment on a math blog, etc. The “square root of negative one” is not i ; it is represented by i . Acest articol despre o literă grecească este deocamdată un ciot. A shielding constant. Interesting question! Configure your keyboard layout in Windows so that you can type all additional symbols you want as easy as any other text. 117. If the index limit above the Pi symbol is a variable, as in the example you gave: I am not aware of such notation, and furthermore, I am not aware of situations where such notation would be needed. If I have interpreted the expression you show correctly, it is neither an arithmetic nor a geometric sequence. To facilitate this, a variable is usually listed below the Sigma with an equal sign between it and the starting term number. Quick Help to Type the Pi symbol. Is there a way to rewrite the following expression using both sigma and pi? Sigma notation provides a compact way to represent many sums, and is used extensively when working with Arithmetic or Geometric Series. Sir,If I have equation like this : Σ "sigma" Ce symbole mathématiques permet d'exprimer plus simplement certaine somme et donc certaines expressions Exemples : Exemple 1 : la somme de 1 à 100 peut s'écrire : Exemple 2 : la somme des nombres impairs de 1 à 13 peut s'écrire : Un nombre pair est de la forme 2j où j est un entier naturel , 2j + 1 est donc l'écriture d'un nombre impair. Do you know of situations that require repeating exponentiation to model them? It has been represented by … Since its inception, the fraternity has initiated more than 100,000 men and has 5,300 undergraduate members. It’s basically a for loop in scripting, makes so much sense. I want to do that to signify that the matrices do not commute. If you can illustrate it please. However, your expression leaves me uncertain as to whether you are analyzing the situation correctly or not. But if you are trying to give a general answer, you should show each term individually so that the person reading your answer can see any pattern that is developing, and understand how to fill in the “…” used to represent all the terms that are not shown. In Gematria it has the value of 200. 1 Translingual. Raccourci clavier pour Pi π. Alt + 227. You are correct – this can be represented using a combination of Sigma and Pi notation: In the above notation, i is the index variable for the Sum, and provides the starting number for each product. can there be a bigger value at the base and smaller value at top of the PI operator? Good suggestion – it makes more sense to keep the expression the same as the previous example! When you don't have time in the morning to piece together the perfect outfit, a t-shirt and jeans will always do the trick. Copy and paste pi symbol or look below to find out how to type pi symbol on keyboard. hello sir, Sigma Σ is one of the most popular mathematic signs which means a summation of something. So for instance, if I wanted to round the above to the nearest whole after each division (or multiplication) step I think I could write: ⌈⌈⌈I÷(1-r)⌉÷(1-r)⌉÷(1-r)⌉…. In this tutorial, you’ll learn 4 different ways you can insert the sigma symbol in Word. Our men strive for excellence by living our core values to promote fellowship, develop character and leadership, advance heightened moral awareness, enable academic achievement, and inspire service. You can add the Greek language on any iOS devices. 2) The values grow in magnitude linearly by 2 each time. Comment taper tous les symboles et signes sur votre clavier. The post was tremendously helpful. of 14. Sigma notation is most useful when the “term number” can be used in some way to calculate each term. See more ideas about sigma pi, sigma, delta. This would be easy to do in a computer program, but not so much using summation notation. Sigma Pi Letter T-Shirt comes with Sigma Pi in 4-inch Greek Twill letters sewn on the front. What I am wondering about is this. thank you for the amazing and very helpful post. multiply until n reaches 143 (i.e. Subtraction can be rewritten as the addition of a negative. Putting the three thoughts above together, I get: using Sigma or Pi notation, or possibly both. Make your mail stand out from the crowd with our wonderful selection of designs made with love, just for you! The difficulty you describe is that you wish to specify what happens to the result of that product, and capital Pi notation does not provide any means to do that. Xi Sigma Pi was originally founded at the University of Washington on November 24, 1908, and is now comprised of over 40 chapters across the USA and Canada. For repeated exponentiation I would assume that form rather than (x^a)^b. Achetez AZSTEEL Sigma Pi T-Shirt 3.14 Symbol Math Science | for Men Women Comfortable Fit Wearable Anywhere, White and Black in Sizes S-3xl livraison gratuite retours gratuits selon … Raccourci clavier pour Tau τ. Alt + 231. You can input pi symbol using it. I’ll challenge him to find a need for it and maybe he can create his own notation. Hello, I am trying to utilize the Pi notation to represent a repeating multiplication, but one that rounds up to the nearest whole after each time there is a multiplication(or division). A factor of (2n) will produce such numbers, but when n=1 this will have a value of 2, not 3… so I need to add 1 to each value: . To type the Pi (π) Symbol anywhere (like in Word or Excel), press Option + Z shortcut for Mac.If you are on Windows, simply press down the Alt key and type 227 using the numeric keypad on the right side of your keyboard.For Microsoft Word, just type 03C0 and then press Alt + X to get the symbol. La manipulation de sommes, via le symbole (sigma), repose sur un petit nombre de règles. Sigma Pi Fraternity was founded in 1897 at Vincennes University, in Vincennes, Indiana. Notation is a convention, a commonly shared interpretation of some symbols. Indeed a very lucid exposition of Sigma and Pi notations! I suppose a problem could be posed this way if you are being asked to come up with an expression for such a product that does not involve Pi notation: is there some closed form expression involving “n” that represents this product? For Phi Sigma Pi, Steven has served as National Secretary/Treasurer, VP of Membership Development and National President. Pi Symbol in Greek Alphabet. Before I continue please forgive my mathematical illiteracy, I am taking an amateur interest in this. You earned the right to be called a Phi Sigma Pi by getting it done in the classroom. Este utilizată în matematic ă pentru a desemna o sumă de elemente. Good point, however, x^a^b is not the same as x^ab. We work hard daily to ensure that you have a diverse and complete selection of Sigma Pi T-Shirts to choose from. 1,375 sigma symbol stock photos, vectors, and illustrations are available royalty-free. The alternative form of sigma (ς) must be used in word-final position. Election is a lifelong membership and includes a once-year complimentary membership in the Society of Physics Students (SPS). If I wanted to represent something being rounded up I think I could use ceiling function brackets: ⌈⌉. does work? In Javascript you should write like a = "this \u2669 symbol" if you want to include a special symbol in a string. (-1)^n will change sign every time “n” grows by one, but when n=1 it is negative – which is the wrong sign for the first time. Change ), You are commenting using your Facebook account. The Sigma Pi fellowships strive for excellence and moral and character development by living by their e… 1.1.1 Usage notes; 1.2 See also; 2 Greek. GREEK CAPITAL LETTER PI ← Ο [U+039F] Greek and Coptic: Ρ → [U+03A1] Contents. Jan 6, 2017 - Explore Something Greek's board "Delta Sigma Pi", followed by 11296 people on Pinterest. Note that “two cubed” and “sigma pi” are puns — they depend upon the sound of what they are. Tabs Dropdowns Accordions Side Navigation Top Navigation Modal Boxes Progress Bars Parallax Login Form HTML Includes Google Maps Range Sliders Tooltips Slideshow Filter List Sort List. Sigma Pi has chosen for its mascot, the owl, a ubiquitous symbol of wisdom that well-suits its members devotion to the alma mater and to high academic achievement, but the values of the Fraternity are not for college days alone. The letter is pronounced as "p". The table below is a Gem in The Rough. Basically this is where k = n. It’s important to emphasize that. And all of them can produce pi text symbol. So maybe we do still need something? I simply cannot figure out how to represent that using big Pi Π. Once that has been evaluated, you can evaluate the next sigma to the left. Annexe E Liste des symboles mathématiques usuels (LATEX) Vous trouverez ci-dessous la liste des commandes LATEX permettant de produire les symboles mathématiques les plus courants. Symbolscopyandpaste.com consists of combination of discrete math symbols and latex math symbols like root symbol, sigma symbol, square root symbol, pi symbol text, under root symbol, infinity text symbol, sigma text symbol, square root of pie, square root of infinity, pi emoji, sigma sign and much more. Užití ve vědě. An online LaTeX editor that's easy to use. This plane contains the six atoms and all of the sigma bonds. Using Pi notation in the exponent achieves the desired purpose. Misconception: many students in the Pacific may have this worng notion that a sigma . But, perhaps I do not understand the situation you seek to describe. If you're using it often you can add a shortcut in Settings ➜ General ➜ Keyboard ➜ Text Replacement. If the sum of a bunch of terms in known as a “summation of a series”, then what is the product of a bunch of terms known as in mathematics? The Stefan–Boltzmann constant. HTML CSS JavaScript Python SQL … The Steven A. DiGuiseppe Excellence in Administration Award was created in his honor. Then, when you need to type it, hold down … πℼ Pi symbol sign Find out how to type Pi sign π directly from your keyboard. LIKE US. Try these curated collections. It can also help you lookup Unicode codes for entering symbols with keyboard. My answer to that would be: I probably would not use Sigma Notation to write such a simple expression. Sizing example: 3 … Moreover, π is a transcendental number – a number that is not the root of any nonzero polynomial having rational coefficients. GeoGebra Applets However, I have never worked with infinite products. It has been represented by the Greek letter "π" since the mid-18th century. An upper bound would be provided by an infinite geometric sequence, but I am uncertain what might best provide a lower bound. The Sigma and Pi expression I used to answer the previous question did not have a value specified for “N”, so any value given for the expression will have to be in terms of “N”… as your question is. So like E(x+n) for n=1 to 3 would produce (x+1)^(x+2)^(x+3)… Or maybe((x+1)^(x+2))^(x+3). Reading this post it seems like this would be easy to use the big Pi Π notation. I was finding how to use Sigma notation, and finally found such a good one. expression is So, the sum of the first two terms would indeed be 30. This site is valid XHTML 1.0 Strict, CSS | Get Buzz via RSS or Atom | Colophon | LegalXHTML 1.0 Strict, CSS | Get Buzz via RSS or Atom | Colophon | Legal It is a weaker type of covalent bond as compared to sigma bonds (σ bonds). ( Log Out / The Pi symbol, , is a capital letter in the Greek alphabet call “Pi”, and corresponds to “P” in our alphabet. No new vocabulary is needed. and if you leave the final index as “n” becomes: Is there some closed form expression that represents this product? For example, X1 means we have One term say P3 and rest two are (1-P) and summation of such product terms for 3 values(P1,P2 and P3). Π π Pi: Ε ε Epsilon: Ρ ρ Ro: Ζ ... Ͱ ͱ Heta: Ϸ ϸ Șo: Ϻ ϻ San: diacritice: Litera Σ (sigma mare) este o literă din alfabetul grec (corespondent pentru litera S în limba română). n=112, n=113 etc. Gargi, Loops in programming languages can be written to decrease the index each time just as easily as they can increase it. As n grows, the constant power of 2 in the expression will dominate the initial results a lot more, but the infinite number of subtractions from it will eventually catch up to its value, no matter how large it is. I have modified the post. https://mythologian.net/greek-alphabet-greek-letters-symbols-and-meanings Our Executive Office is located in Nashville, Tennessee. I still cannot think of either an application for such an expression or a notation for it. 3) There are seven terms, so n will need a starting value of 1, and an ending value of 7. One other thought… if It is used in the same way as the Sigma symbol described above, except that succeeding terms are multiplied instead of added: Sigma (summation) and Pi (product) notation are used in mathematics to indicate repeated addition or multiplication. The coefficients will be “32 Choose i”, or. Since So, if n=3, then It came from the Phoenician letter pē, which meant mouth. X2=(1-P1)P2.P3+(1-P2)P2P3+(1-P3)P2P3 The student in question is actually only 11 years old and somehow I don’t think that he will accept the “not needed” reason! If it ends with, or continues beyond tan(np/2n), which will always be undefined, then my first impression is that there would be no limit to the product. Sigma Pi Phi (ΣΠΦ) is the first successful and oldest Black Greek-lettered organization.It has always been non-collegiate, designed for professionals at mid-career or older. In Modern Greek, this sound is voiced to /z/ before /m/, /n/, /v/, /ð/ or /ɣ/. So Sigma notation describes repeated subtraction when its argument is a negative quantity. Three sigma bonds are formed from each carbon atom for a total of six sigma bonds total in the molecule. Pi sign is one of the most popular mathematic constants and means a ratio of a circle perimeter to its diameter. If j went from one to three each time, the expression on the right would have to be (i + j – 1). Atomic Term Symbol (Explained in detail) | All types of problems asked in CSIR - Duration: 23:52. However, this particular example can be “simplified” by collecting like terms to become Section: Internet Tutorial: Greek Letters Fabulous Code Chart for Greek Letters & Symbols … With more than 140 collegiate chapters on campuses throughout the U.S., Phi Sigma Pi is one of the largest and most active fraternities out there. Thanks for your clear explanations. If I wanted to take, let’s say “I”, and multiply “I” by a repeating multiple, let’s say “1/(1-r)”. 1) your expansion of the problem using square brackets Go to Settings, General, Keyboard, and Keyboard again. How can I type a pi symbol on an iPad? There are several in the posting… Ooops – I just realized you were asking about my reply to the comment. Sigma Pi Sigma exists to honor outstanding scholarship in physics, to encourage interest in physics among students at all levels, to promote an attitude of service, and to provide a fellowship of persons who have excelled in physics. Sorry, your blog cannot share posts by email. i=1: (14/12)(8/12) Name Unicode Glyph Unicode Name Description Aliases; alefsym: 02135: ALEF SYMBOL : Alpha: 00391: GREEK CAPITAL LETTER ALPHA : alpha: 003B1: GREEK SMALL LETTER ALPHA How would I derive the polynomial for the following expression: n = 112 which would raise the question: “why write it using Sigma Notation when you could just as easily write ?”. A more typical use of Sigma notation will include an integer below the Sigma (the “starting term number”), and an integer above the Sigma (the “ending term number”). To make use of them you will need a “closed form” expression (one that allows you to describe each term’s value using the term number) that describes all terms in the sum or product (just as you often do when working with sequences and series). Your answer options suggest that there is some expansion of a a logarithm that results in an infinite product of tangent functions, however I am not familiar with that. Thank you, this was very helpful. ), I’m interested in simplifying the polynomial to 32 terms and determine the exponents of y, Using Pi notation, I interpret your question to be, Using a binomial expansion, the terms will be Symbol . Math teacher, substitute teacher, and tutor (along with other avocations) GREEK PI SYMBOL: ϗ : 983: 03D7 : GREEK KAI SYMBOL ... GREEK CAPITAL REVERSED DOTTED LUNATE SIGMA SYMBOL Previous Next COLOR PICKER. Pi π is an irrational number, which means that it cannot be expressed exactly as a ratio of two integers; consequently, its decimal representation never ends and never repeats. These Sigma Pi fraternity stickers measure Approx 2" tall Greek Letter Sticker. I was just practicing the question wanted to know can 30….. n(n+1)(n+2) be the ans to the above sigma and product equation given by you. A “product” is the result of multiplying two or more “factors”. 2.1 Letter; 2.2 See also; Translingual Symbol Π. English Wikipedia has an article on: Capital pi notation. This site uses Akismet to reduce spam. In Gematria it has the value of 200. As is the case with most of the other alphabets, even English for that matter, two different symbols are used to represent sigma in uppercase and lowercase respectively. Furthermore is there a way of simplifying the notation and finding a result that is a function of n? But what if the Pi notation is not in closed form, such as. . I do not follow your thinking though when you say you wish to use a descending index value to indicate that matrices do not commute… I would not perceive a descending index value, or an ascending one, to indicate anything about the commutative property’s applicability to the resulting expression. (mathematics) Sum of divisors. How can u write this using summation notation: 3 -5+7 -9+11-13+15? You'll have to copy paste pi symbol π if you're on iOS or iPad OS. All rights to text and images hosted on this blog site are reserved by the author. what is the relation between the two when they are logarithmic differentiated? You can make frequently used technical non-fancy symbols like "√ ∑ π ∞ ∆ ™ © æ £ ¢" and åccénted letters on Mac using [Option] key. You press Alt and, while holding it, type a code on Num Pad while it's turned on. So you should wear lettered apparel that gets it done too. For example, if n=1, then the expression would be: Certificates. and if n=4, then So will provide the correct sign for the nth term. It's a text symbol. There actually are 3 different ways to type symbols on Linux with a keyboard. How should I proceed if I want to get it for n instead of 3. 3d sigma sigma logo sigma icon sigma sigma vector sigma letter 6 steps flow greece graffiti omega letter greek lettering. ( Log Out / There is no emoji for pi at the moment. Division can be rewritten as multiplication by the reciprocal. Một bộ sưu tập các biểu tượng toán học, chẳng hạn như: Pi, Infinity, Sum, Sigma, Square Root, Integral ... các dấu hiệu & ký hiệu. I have a question, please. The letter is pronounced as "p". Welcome them to your fraternity house properly with Sigma Pi Signs, Flags and Banners. Pi symbol π is important for calculation of circular and spherical figures and stands for value of 3.141592653589793238.. . Actually n SHOULD be squared in his reply since he’s saying that that’s the LAST TERM in the product. Character map allows you to view and use all characters and symbols available in all fonts (some examples of fonts are "Arial", "Times New Roman", "Webdings") installed on your computer. Question. How to find the derivative of the pi notation. ( Log Out / hope to receive your reply as soon as possible. Sigma Pi (ΣΠ) is an international social collegiate fraternity founded in 1897 at Vincennes University. • In both Ancient and Modern Greek, the sigma represents the voiceless alveolar fricative /s/. We have an awesome collection of mathematical symbols for you. Please, read a guide if you're running a laptop. However, while it is common to see uppercase and lowercase sigma signs in science textbooks and the likes, what many people are completely unaware of is the fact that the sigma sign has a third form as well. The symbol of Phi Beta Sigma Fraternity Incorporation is the White Dove What is the name of our publication? what can be the correct answer this equation? In such cases, just as in the example that resulted in a bunch of twos above, the term being added never changes: The “starting term number” need not be 1. It corresponds to “S” in our alphabet, and is used in mathematics to describe “summation”, the addition or sum of a bunch of terms (think of the starting sound of the word “sum”: Sssigma = Sssum). i.e. Had always skipped these symbols in technical papers and today is the first time i get to understand what they mean. Pi notation provides a compact way to represent many products. The Fraternity currently has over 285,000 total initiated members. Hi Mr. Ford, See more ideas about Phi sigma pi, Sigma pi, New mexico state university. Does Multiplication operator always increment? Change ). Cet article a pour objet de les énumérer et d’en donner des exemples d’utilisation, sans aucune prétention à l’originalité. There is often a pattern to them, a formula that can be used to determine the value of the next term in the sequence. For example, $\prod_{i = 3}^6 (x + 2^i) = (x + 2^3) (x + 2^4) (x + 2^5) (x + 2^6)$ In general, we have $\boxed{\prod_{i = a}^b f(i) = f(a) \cdot f(a+1) \cdot f(a+2) \cdot … \cdot f(b)}$ Properties of Pi Notation. So, even if it is not commonly used in a particular way, there is no strong reason I can think of why you couldn’t use it that way (if necessary, including a note or example describing how you intend the notation to be interpreted). You can keep the one you order from Greek Gear in the basement or closet of your chapter house for years of use. Series definitions almost always rely on summation notation. Raccourci clavier pour Theta Θ. Alt + 233 Thanks…. then there are an indeterminate number of factors in the product until such time as “n” is specified. Minususcule : Alt + 229. Let’s list the first few terms of this sequence individually to get a sense of how this series behaves: If sigma is for summation, and pi is for multiplication, are there any notations for division and subtraction? In this case only two of the p orbitals on each C atom are involved in the formation of hybrids. Learn phi sigma rho with free interactive flashcards. It came from the Phoenician letter pē, which meant mouth. The symbol for uppercase sigma is Σ Chỉ cần nhấp vào một dấu hiệu hoặc biểu tượng để sao chép nó vào bảng tạm và dán nó ở bất cứ đâu. Thanks! Understanding math topics without memorization. Which “k” are you referring to? I've compiled a list of shortcuts in my article and explained how to open keyboard viewer. This vector image was created with a text editor. If you need to differentiate a sum, I would not expect logarithmic differentiation to be very useful, as the laws of logarithms do not allow us to do anything with something like Pi Bond are formed by the overlapping of atomic orbital’s sidewise. If I were to see an upper index value that is smaller than the lower one, my first assumption would be that I would need to decrease the index by 1 for each iteration – which seems to be what you intend. HOW TO. Ooops – didn’t think of CharMap allows you to view and use all characters and symbols available in all fonts (some examples of fonts are "Arial", "Times New Roman", "Webdings") installed on your computer. Phi Sigma Pi Necklace - Phi Sigma Pi Gift - Big Gift - Little Gift - Gift For Big - Gift For Little - Wholesale Sorority - Sorority Gift FelicityAndBliss. The pi symbol will be shown on the virtual keyboard. 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Voiced to /z/ before /m/, /n/, /v/, /ð/ or /ɣ/ Ρ → U+03A1! Images hosted on this blog site are reserved by the author this blog and receive notifications of New posts email. Shared interpretation of some symbols atomic orbital ’ s saying that that ’ s basically for... It, hold down … Pi symbol, HTML Unicode entities and more I knew. This \u2669 symbol '' if you could provide an easy way that I can think of do. Its members in scholarship, leadership, and more Phoenician letter also gave rise to … Pi. Calculation of circular and spherical figures and stands for value of 7 ways you can insert the sigma the... Very very useful which meant mouth convention is to increase it oborech ustálený význam, v němž se používá Greek!
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https://math.stackexchange.com/questions/508791/number-of-lists-in-which-an-element-is-repeated-consecutively-exactly-twice | # Number of lists in which an element is repeated consecutively exactly twice
I have an integer list that is n long and each value can be ranging from 1 .. n.
I need a formula that tells me how many of all possible lists for a given n, that have one or more consecutive sequences of a length of exactly 2 of the same number and no other consecutive sequences that are longer than 2.
For example for n=5:
These two should count:
{ 1, 1, 5, 3, 3 }
{ 2, 3, 2, 5, 5 }
Where as these should not:
{ 1, 1, 1, 2, 2 }
{ 1, 3, 2, 5, 4 }
I've been attempting to do this by looking at the possible sequences using the following formula where x = n-1:
(n) x n n = x * n^3
x (n) x n = x^2 * n^2
n x (n) x = x^2 * n^2
n n x (n) = x * n^3
And sum these four up.
However, these also needs to take overlaps between the four into account. This is where I could use a bit of help..? What would the formulas be for excluding the overlaps?
An alternative approach would also be welcome.
Going trough all sequences counting manually is not an option - I need this to work for for n larger than what makes that approach computationally feasible.
If it helps anyone, I've written a little program that runs trough all the sequences and counts have the following results:
n = 2
L[2]: 2
L[1]: 2
n = 3
L[3]: 3
L[2]: 12
L[1]: 12
n = 4
L[4]: 4
L[3]: 24
L[2]: 120
L[1]: 108
n = 5
L[5]: 5
L[4]: 40
L[3]: 280
L[2]: 1520
L[1]: 1280
Where n = 5, L[2]: 1520 is the result I've asked for a formula to in the above question.
• You are using the word permutation to mean something different from what the rest of us mean when we use that word. – Gerry Myerson Sep 29 '13 at 12:59
• I rephrased it a bit - I hope it makes better sense. – Kasper Middelboe Petersen Sep 29 '13 at 13:10
• Are ${ 2, 3, 2, 5, 5 }$ and ${ 3, 2, 2, 5, 5 }$ different sequences that both count? – miracle173 Sep 29 '13 at 13:37
• @miracle173 yes – Kasper Middelboe Petersen Sep 29 '13 at 13:40
• Does $1, 2, 3, 1$ count? Ie, can the doubled elements "wrap around"? – Jack M Sep 29 '13 at 13:58
How many sequences of length $q$ of numbers from $\{1,...,p\}$ are there such that consecutive elements are always different? For the first element of such a sequence we can selcet one of the $p$ different values of $\{1,...,p\}$. For the following $q-1$ positions we can select always all values from $\{1,...,p\}$ except the value of its predecessor in the sequence. These are $p-1$ possible values. So we have
$$p(p-1)^{q-1}$$
different sequences.
$\Omega_n$ is the number of sequences of length $n$ with elements from $\{1,\ldots,n\}$ such that no three consecutive elements have the same value but at least one pair of consecutive elements have the same value.
$\Omega_{n,k}$ is the number of sequences of length $n$ with elements from $\{1,\ldots,n\}$ such that no three consecutive elements have the same value but exactly $k$ pairs of consecutive elements have the same value.
For arbitrary $n$ there are $k \le \frac{n}{2}$ different possible values for the number of consecutive element pairs that have equal values.
We have
$$|\Omega_{n}|=\sum_{k=1}^{\lfloor \frac{n}{2}\rfloor }|\Omega_{n,k}|$$
Now we select a $k$. Choose a sequence of $n-k$ values from $\{1,\ldots,n\}$ such that two consecutive values always differ. There are $n (n-1)^{n-k-1}$ such sequences. For such a sequence we select $k$ of its elements (there are $\binom{n-k}{k}$ such possibilities) and insert an element with the same value after each selected value. So $$|\Omega_{n,k}|=\binom{n-k}{k}n(n-1)^{n-k-1}$$ and $$|\Omega_{n}|=\sum_{k=1}^{\lfloor \frac{n}{2}\rfloor }\binom{n-k}{k}n(n-1)^{n-k-1}$$
• For $n=3$ there are $12$ sequences.
• For $n=4$ there are t $120$ sequences.
• For $n=5$ there are $1520$ sequences.
The answer to the original question is deleted
• @KasperMiddelboePetersen The formula would be $n\sum_{k=1}^{\lfloor \frac{n}{2}\rfloor }\frac{(n-k)(n-k-1)\ldots(n-2k+1)}{k!}(n-1)^{n-k-1}$. If you are doing more combinatorics problems, it would be a worth-while investment of time to study binomial coefficients, just a bit. – Marc van Leeuwen Sep 29 '13 at 20:27
• @MarcvanLeeuwen thanks - I'd just figured it out and deleted the comment 30 seconds before you answered :) It will be a challenge to calculate this for a large value of n though. – Kasper Middelboe Petersen Sep 29 '13 at 20:32
• I think the only challenge is computing with large integers. For $n=69$, I get 4676495092345476699420593108779018542679373291685221536304769813013106968507850431624264673416798415845073262231682605593395200, no sweat. – Marc van Leeuwen Sep 29 '13 at 20:43
Here is a different solution that may interest you. Introduce three sequences $a_{n,k}$, $b_{n,k}$ and $c_{n,k}$ that count the number of strings over $\Sigma^k$ where $|\Sigma|=n,$ that end in a digit that is not repeated, a digit that is repeated twice and a digit that is repeated at least three times. In fact we take these sequences to be generating functions in two variables, where the variable $v$ counts occurrences of subsequences of length exactly two and $w$ counts occurrences of subsequences of length at least three.
This gives the following set of recurrences: $$a_{n,k} = (n-1) a_{n,k-1} + (n-1) v b_{n,k-1} + (n-1) c_{n,k-1},$$ $$b_{n,k} = a_{n,k-1} \quad\text{and}\quad c_{n,k} = w b_{n,k-1} + c_{n,k-1}.$$
With these settings the generating function of all elements of $\Sigma^k$ classified according to the number of length 2 and length more than two subsequences is given by $$a_{n,k} + v b_{n,k} + w c_{n,k}.$$
Observe that $a_{n,1} = 1$ and $b_{n,1} = c_{n,1} = 0.$ Furthermore the last recurrence implies that $$c_{n,k} = w \sum_{q=1}^{k-1} b_{n,q} = w \sum_{q=1}^{k-2} a_{n,q}.$$
Taken together this yields a recurrence for $a_{n,k}:$ $$a_{n,k} = (n-1) a_{n,k-1} + (n-1) v a_{n,k-2} + (n-1) w \sum_{q=1}^{k-3} a_{n,q}.$$
Introduce the trivariate generating function $$G_n(z) = \sum_{k\ge 1} a_{n,k} z^k.$$
Multiply the recurrence by $z^k$ and sum for $k\ge 4:$ $$\sum_{k\ge 4} a_{n,k} z^k = (n-1) z \sum_{k\ge 4} a_{n,k-1} z^{k-1} + (n-1) vz^2 \sum_{k\ge 4} a_{n,k-2} z^{k-2} \\+ \sum_{k\ge 4} (n-1) w z^k [z^{k-3}] \frac{1}{1-z} G_n(z).$$ Now $a_{n,1} = n, \; a_{n,2} = n(n-1)$ and $$a_{n,3} = n(n-1)^2 + n(n-1)v.$$ The equation derived from the recurrence now becomes $$G_n(z) - (n(n-1)^2 + n(n-1)v)z^3 - n(n-1)z^2 -nz \\= (n-1)z (G_n(z) - n(n-1)z^2 -nz) + (n-1)vz^2 (G_n(z) - nz) \\ + (n-1)w z^3 \sum_{k\ge 4} z^{k-3} [z^{k-3}] \frac{1}{1-z} G_n(z).$$ The sum term simplifies to $$(n-1)w z^3 \frac{1}{1-z} G_n(z).$$ We may now solve for $G_n(z),$ getting $$G_n(z) = -{\frac {nz \left( -1+z \right) }{-nz+n{z}^{2}-{z}^{2}-v{z}^{2}n+v{z}^{3}n+v{ z}^{2}-v{z}^{3}-w{z}^{3}n+w{z}^{3}+1}}.$$ Now the generating function for $b_{n,k}$ is $$z G_n(z) \quad\text{and the one for}\; c_{n,k} \; \text{is}\quad \frac{wz^2}{1-z} G_n(z).$$ It follows that the generating function $H_n(z)$ of $a_{n,k} + v b_{n,k} + w c_{n,k}$ is $$\left(1 + vz + \frac{w z^2}{1-z}\right) G_n(z)$$ or equivalently $$H_n(z) = -{\frac {zn \left( -1+z-vz+v{z}^{2}-w{z}^{2} \right) }{-nz+n{z}^{2}-{z}^{2}-v{z}^{2}n+v{z}^{3}n+v{z}^{2}-v{z}^{3}-w{ z}^{3}n+w{z}^{3}+1}}.$$ Now we do not permit sequences of length at least three, so we take $$[w^0] H_n(z) = -{\frac { \left( vz+1 \right) nz}{v{z}^{2}n-v{z}^{2}+nz-z-1}}.$$ In fact there must be at least one two-sequence, so we first calculate $$[w^0] H_n(z) - [v^0] [w^0] H_n(z) = {\frac {v{z}^{2}n}{ \left( nz-z-1 \right) \left( v{z}^{2}n-v{z}^{2}+nz-z-1 \right) }}$$ and put $v=1$, finally obtaining the generating function $$M_n(z) = {\frac {n{z}^{2}}{ \left( nz-z-1 \right) \left( n{z}^{2}-{z}^{2}+nz-z-1 \right) }}.$$ The partial fraction decomposition of $M_n(z)$ is given by $$-{\frac {n}{ \left( n-1 \right) \left( n{z}^{2}-{z}^{2}+nz-z-1 \right) }}+{ \frac {n}{ \left( n-1 \right) \left( nz-z-1 \right) }}.$$
Now the singularities are at $$\rho_0 = \frac{1}{n-1} \quad\text{and}\quad \rho_{1,2} = -\frac{1}{2} \pm \frac{\sqrt{n^2+2n-3}}{2(n-1)}.$$ Expanding $M_n(z)$ into series we obtain the following closed form result for the number of admissible strings of length $k$ and an alphabet of $n$ symbols: $$[z^k] M_n(z) = n\frac{\rho_0^2 \rho_1 \rho_2}{(\rho_0-\rho_1)(\rho_0-\rho_2)} \rho_0^{-k}\\ + n\frac{\rho_0\rho_1^2 \rho_2}{(\rho_1-\rho_0)(\rho_1-\rho_2)} \rho_1^{-k} + n\frac{\rho_0\rho_1 \rho_2^2}{(\rho_2-\rho_0)(\rho_2-\rho_1)} \rho_2^{-k}.$$ The reader is asked to verify that indeed $$[z^n] M_n(z) = \sum_{k=1}^{\lfloor n/2 \rfloor} {n-k\choose k} n (n - 1)^{n-k-1}.$$ The above formula for $[z^k] M_n(z)$ is quite powerful. It gives the exact value of the number of admissible strings with any alphabet of size $n$ and of length $k$. For example, when there are $n=2$ symbols, the sequence starting at length $k=1$ is: $$0, 2, 4, 8, 14, 24, 40, 66, 108, 176.$$ With $n=7$ we get the following sequence: $$0, 7, 84, 798, 6804, 54684, 423360, 3194856, 23668848, 172939536.$$
The generating function $H_n(z)$ encapsulates the complete distribution of the $n^k$ strings classified according to the number of two-sequences (counted by $v$) and sequences of length at least three (counted by $w$).
Here is an example: $$[z^4] H_3(z) = 6\,{v}^{2}+36\,v+15\,w+24$$ for a total of $3^4=81$ terms (strings of length $4$ over $3$ symbols). Another example is: $$[z^6] H_5(z) = 80\,{v}^{3}+1920\,{v}^{2}+520\,vw+20\,{w}^{2}+6400\,v+1565\,w+5120$$ for a total of $5^6=15625$ terms (strings of length $6$ over $5$ symbols). It is a useful combinatorics excercise to verify some of these values with pen and paper. For example, a string of length four over three symbols containing a sequence of length at least three can consist of four equal symbols, giving a contribution of three, or a length three sequence at the front for a contribution of three times two or a length three sequence at the end, again for three times two, for a total of $3+6+6 = 15$ which is indeed the value from the generating function. | 2019-08-22T13:52:29 | {
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https://math.stackexchange.com/questions/328071/assigning-values-to-permutations | # Assigning values to permutations
$N$ objects can be arranged in $N!$ different orders. For example, $10$ playing cards can be stacked $10! = 3,628,800$ different ways. Is there a way to assign a numerical value to each permutation so that every integer from $1$ to $N!$ corresponds to exactly one permutation? Is there a way to derive the permutation from the corresponding numerical value?
• Well, there are many ways: just ennumerate arbitrarily the permutations from $1$ to $\,N!\,$ Mar 12 '13 at 3:05
• This is discussed in detail in Higher-Order Perl, section 4.3.1, pp. 128–135, which is available online. It is also discussed in exhaustive detail in volum IV of Knuth's The Art of Computer programming.
– MJD
Mar 12 '13 at 3:41
• Could you not number each card and then name permutations according to the order? Mar 12 '13 at 4:59
I don't know if there is a standard way of doing this but you could do something like this:
Lets take 4 playing cards b/c it is more managable. Order them some way, say label them a,b,c,d and put them in the order (a,b,c,d). Then
$$1:(a,b,c,d)\\ 2:(a,b,d,c)\\ 3:(a,c,b,d)\\ 4:(a,c,d,b)\\ 5:(a,d,b,c)\\ 6:(a,d,c,b)$$ So fixing "a" as the first entry gives 6 possible permutations, repeating this for b,c,d in the first entry will give you the other 18, for a total of 24. The method is to first fix an ordering and then permute only the last 2, then once thats done permute the last 3 and so on.
• No help so far. Mar 12 '13 at 15:47
• Im not sure what your asking then, you asked if there was a way to assign a number to each permutation. The answer is yes, and here is one way of doing it systematically.
– mv3
Mar 12 '13 at 16:03
• And if you know the original ordering of the objects, (a,b,c,d) then you can reconstruct the permutation from the value
– mv3
Mar 12 '13 at 16:08
Yes. The easiest way is to order them lexicographically. So for $\{0,1,2,3,4\}$ there are $120$ permutations, from $01234$ to $43210$. It is easiest if our permutation numbers run from $0$ to $119$. Of these $4!=24$ have each number first, so if you want permuation $n$, the first number is $a_0=\lfloor \frac n{24} \rfloor$. Then of those, there are $3!=6$ that have each of the remaining numbers first. To find it, compute $a_1=\lfloor \frac {n-24a_0}6 \rfloor$, then increment by $1$ if $a_0 \le a_1$ because you want the $a_1$st of what is left. Now recurse.
• I'm looking for a function. Suppose ten cards are in the order Ace-2-3-4-5-6-7-8-9-10 when arranged in permutation #1. What would permutation #1,000,000 look like? Suppose the cards are in the order 5-9-2-4-10-Ace-3-8-6-7. Which permutation # (an integer between 1 and 3,628,800) would that be? Maybe what I'm asking just isn't possible, but it would be useful to me if it can be done. Mar 12 '13 at 16:09
• @LeroyNimka: It is quite possible. For permutation 1,000,000 the first card is $\lfloor\frac {999,999}{9!}\rfloor+1=3$ where the -1 and +1 come because you are counting from 1. The ones starting with Ace and 2 use up $2\cdot 9!=725760$ so now we want permutation 274240 of the rest. $\lfloor\frac {274239}{8!}\rfloor+1=7$ so we want the seventh card, the $8$. That has used up $6 \cdot 8!=241920$, so we need the $32320$ of the rest, the seventh one, the 9. So we have started with 389. This can be written in a few tens of lines with a loop. Mar 12 '13 at 16:27
• @LeroyNimka: for the other direction, the 5 leads off starting with permutation $4 \cdot 9!+1=1451521. Having the 9 next adds$7 \cdot 8!=282240. The 2 next adds $7!=5040$, so we are up to 1738801. Again, keep going, and again a few tens of lines. Mar 12 '13 at 16:30
Notice that for one and the same letter kept, out of $$n$$, in one position, you have $$(n-1)!$$ possible choices. And that is all that you need.
$$4231, n=4$$
Before we reached $$4$$ we had $$(4-1)(n-1)!$$ permutations. With $$4$$ fixed we had $$(2-1)(n-2)!$$ options. With $$42$$ fixed we had $$(3-1-1)(n-3)!$$ options as $$2$$ is preceding $$3$$ and is to the left of it. With $$423$$ fixed we had $$(1-1)(n-4)!$$ options. It is $$18$$th permutation.
Formula is then:
$$P(a_1a_2...a_m)=\sum_{k=1}^{m}(\alpha_r(a_k)-1)(m-k)!$$
where $$\alpha_r(a_k)$$ is $$1$$-based alphabetical position of symbol $$a_k$$ reduced by the number of symbols that appear earlier in the alphabet and appear to the left of it.
Inverse is more or less obvious. Essentially you write N, the position, in factorial positional system, and read the permutation from the result.
$$18=3110_!$$
We start
$$1234$$ $$3110$$
Pick the one at position $$3$$, remove the position handled and the element ($$4$$).
$$4$$ $$123$$ $$110$$
Pick the next one at position $$1$$, remove the position handled and the element ($$2$$).
$$42$$ $$13$$ $$10$$
Pick the next one at position $$1$$, remove the position handled and the element ($$3$$).
$$423$$ $$1$$ $$0$$
Finally take the last element to have $$4231$$ as expected.
(You can read the order directly. Start from the left and use order in the alphabet, if you find the same position twice, just increment it. Therefore
$$3110_!$$ is $$3, 31, 311=312, 3120 \to 4231$$
$$1210_!$$ would be $$1,12,121=122=123,1230 \to 2341$$) | 2021-09-28T05:18:15 | {
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http://stats.stackexchange.com/questions/30842/what-is-the-probability-that-this-person-is-female/30850 | # What is the probability that this person is female?
There is a person behind a curtain - I do not know whether the person is female or male.
I know the person has long hair, and that 90% of all people with long hair are female
I know the person has a rare blood type AX3, and that 80% of all people with this blood type are female.
What is the probability the person is female?
NOTE: this original formulation has been expanded with two further assumptions: 1. Blood type and hair length are independent 2. The ratio male:female in the population at large is 50:50
(The specific scenario here is not so relevant - rather, I have an urgent project that requires I get my mind around the correct approach for answering this. My gut feel is that it's a question of simple probability, with a simple definitive answer, rather than something with multiple debatable answers according to different statistical theories.)
-
There are not multiple theories of probability, but it is notoriously true that people have difficulties thinking correctly about probabilities. (Augustus DeMorgan, a good mathematician, gave up the study of probability due to its difficulties.) Don't look at debates: look for appeals to principles of probability (such as the Kolmogorov axioms). Don't let this be resolved democratically: your question is attracting many ill-conceived answers which, even if some of them happen to agree, are merely collectively wrong. @Michael C gives good guidance; my reply tries to show you why he's right. – whuber Jun 22 '12 at 3:46
@Whuber, if independence is assumed, would you agree that 0.97297 is the correct answer? (I believe that the answer might be anywhere between 0% and 100% without this assumption - your diagrams show this nicely). – ProbablyWrong Jun 22 '12 at 4:53
Independence of what, precisely? Are you suggesting that female and male hairstyles are the same? As you say in your question, this particular scenario involving gender/hair/blood type may not be relevant: that tells me you seek to understand how to solve problems like this in general. To do that you will need to know which assumptions imply which conclusions. Thus you need to focus very carefully on the assumptions you are willing to make and determine exactly how much they allow you to conclude. – whuber Jun 22 '12 at 13:07
The kind of independence to explore concerns the combination of all three characteristics. E.g., if AX3 is a marker for a syndrome that includes baldness in females (but not in males), then any long-haired person with AX3 is necessarily male, making the probability of being female 0%, not 97.3%. I hope this makes it obvious that anybody producing a definite answer to this question must be making additional assumptions, even if they do not explicitly acknowledge them. The truly useful answers, IMHO, would be those that show directly how different assumptions lead to different results. – whuber Jun 22 '12 at 14:31
You're missing the probability that a female doesn't have long hair. That's a critical measure. – Daniel R Hicks Jul 3 '12 at 19:54
Many people find it helpful to think in terms of a "population," subgroups within it, and proportions (rather than probabilities). This lends itself to visual reasoning.
I will explain the figures in detail, but the intention is that a quick comparison of the two figures should immediately and convincingly indicate how and why no specific answer to the question can be given. A slightly longer examination will suggest what additional information would be useful for determining an answer or at least obtaining bounds on the answers.
### Legend
Cross-hatching: female / Solid background: male.
Top: long-haired / Bottom: short-haired.
Right (and colored): AX3 / Left (uncolored): non-AX3.
### Data
Top cross-hatching is 90% of the top rectangle ("90% of all people with long hair are female").
Total cross-hatching in the right colored rectangle is 80% of that rectangle ("80% of all people with this blood type are female.")
### Explanation
This diagram shows schematically how the population (of all females and non-females under consideration) can simultaneously be partitioned into females/non-females, AX3/non-AX3, and long haired/non-long haired ("short"). It uses area, at least approximately, to represent proportions (there's some exaggeration to make the picture clearer).
It is evident that these three binary classifications create eight possible groups. Each group appears here.
The information given states that the upper cross-hatched rectangle (long-haired females) comprises 90% of the upper rectangle (all long-haired people). It also states that the combined cross-hatched parts of the colored rectangles (long-haired females with AX3 and short-haired females with AX3) comprise 80% of the colored region at the right (all people with AX3). We are told that someone lies in the upper right corner (arrow): long-haired people with AX3. What proportion of this rectangle is cross-hatched (female)?
I have also (implicitly) assumed that blood type and hair length are independent: the proportion of the upper rectangle (long hair) that is colored (AX3) equals the proportion of the lower rectangle (short hair) that is colored (AX3). That's what independence means. It is a fair and natural assumption to make when addressing such questions like this, but of course it needs to be stated.
The position of the upper cross-hatched rectangle (long-haired females)is unknown. We can imagine sliding the top cross-hatched rectangle side-to-side and sliding the bottom cross-hatched rectangle side-to-side and possibly changing its width. If we do this so that 80% of the colored rectangle remains cross-hatched, such an alteration will change none of the stated information, yet it can alter the proportion of females in the upper right rectangle. Evidently the proportion could be anywhere between 0% and 100% and still be consistent with the information given, as in this image:
One strength of this method is it establishes the existence of multiple answers to the question. One could translate all this algebraically and, by means of stipulating probabilities, offer specific situations as possible examples, but then the question would arise whether such examples are really consistent with the data. For instance, if someone were to suggest that perhaps 50% of long-haired people are AX3, at the outset it is not evident that this is even possible given all the information available. These (Venn) diagrams of the population and its subgroups make such things clear.
-
Whuber, assuming that blood type and hair length are independent, then surely the portion of long haired women with type AX3 should be the same as the portion of short haired women with AX3? I.e. you don't have flexibility to shift rectangles in the way you propose... If we assume also that men and women are 50:50 in the whole population, doesn't that give us enough info to solve this question with a single indisputable answer? – ProbablyWrong Jun 21 '12 at 6:40
@whuber +1 very nice. – Michael Chernick Jun 21 '12 at 10:42
ProbablyWrong, take a close look at the question in your comment: because it deals with women, it is making an additional assumption about independence conditional on gender. The assumption of (unconditional) independence of hair and blood type does not mention gender at all, so to understand what it means, erase the cross-hatching from the figures. This, I hope, indicates why we have the flexibility to situate the cross-hatching wherever we like within the upper and lower rectangles. – whuber Jun 21 '12 at 14:46
@whuber, I like this. However, I have 2 questions / clarifications: 1. the figures seem to assume population proportions for long vs short hair (about 6:4) & ~AX3 vs AX3 (about 85:15), but this is not mentioned in the original question nor discussed in your explanations of the figures. I suspect the pop proportions are not relevant. Am I right / could you clarify that in the explanations? 2. I think this situation is ultimately working w/ the same phenomenon as Simpson's Paradox, only framed differently (coming at the issue from the other direction, as it were). Is that a fair assessment? – gung Jun 27 '12 at 16:25
@gung, thank you for making those clarifications. The figures of course must represent some proportions in order to work at all, but any proportions not specifically pinned down in the problem statement are free to vary. (I did construct the figure so that about 50% of the population appears female, anticipating a later edit in which this was assumed.) The idea of applying this graphical representation to understanding Simpson's Paradox is intriguing; I think it has merit. – whuber Jun 27 '12 at 16:34
This is a question of conditional probability. You know that the person has long hair and blood type Ax3 . Let$$\ \ \ \ \ A =\{\text{'The person has long hair'}\}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B = \{\text{'The person has blood type Ax3'}\} \\ C =\{\text{'The person is female'}\}.$$
So you seek $P(C|A\ \text{and}\ B)$. You know that $P(C|A)=0.9$ and $P(C|B)=0.8$.
Is that enough to calculate $P(C|A\ \text{and}\ B)$? Suppose $P(A\ \text{and}\ B\ \text{and}\ C)=0.7$. Then $$P(C|A\ \text{and}\ B)=P(A\ \text{and}\ B\ \text{and}\ C)/ P(A\ \text{and}\ B)=0.7/P(A\ \text{and}\ B).$$ Suppose $P(A\ \text{and}\ B)=0.8$. Then, by the above, $P(C|A\ \text{and}\ B)=0.875$. On the other hand if $P(A\ \text{and}\ B)=0.9$ we would then have $P(C|A\ \text{and}\ B)$=0.78.
Now both are possible when $P(C|A)=0.9$ and $P(C|B)=0.8$. So we can't tell for sure what $P(C|A\ \text{and}\ B)$ is.
-
Hi Michael, If I read you correctly, you're saying the question as posed can't be answered, is that right? Or to put it another way, you'd need more information to answer this question? 1. Let's assume that the rare blood type in my original question doesn't have any impact on a person's desire or ability to grow their hair long. Can the question now be answered? 2. Would you agree that the answer must be GREATER than 0.9? (Because you have a second piece of independent information - blood type - that reinforces the hypothesis that the person is a female) – ProbablyWrong Jun 21 '12 at 3:30
If $P(A\text{ and }B)$ is independant, then $P(A\text{ and }B)=P(A)P(B)$ and you'll need to specify what fraction of persons have long hair, i.e., $P(A)$ and what fraction of persons have blood type Ax3, i.e., $P(B)$. Also, you can't say that the answer must be greather than 0.9, which is equivalent to stating that $P(C|A\text{ and }B)>0.9$ (I really don't see why). – Néstor Jun 21 '12 at 7:39
Thank you Nestor. – Michael Chernick Jun 21 '12 at 10:00
@ProbablyWrong. Yes the problem as initially stated has insufficient information for a unique answer. – Michael Chernick Jun 21 '12 at 10:04
Why does $$P(C|A\ \text{and}\ B)=P(A\ \text{and}\ B\ \text{and}\ C)\times P(A\ \text{and}\ B)??$$ I thought that $$P(C|A\cap B)=\frac{P(C \cap (A \cap B))}{P(A \cap B)}=\frac{P(A\cap B\cap C)}{P(A\cap B)}$$ using the definition of conditional probability. – Dilip Sarwate Jun 21 '12 at 11:13
Fascinating discussion ! I am wondering if we specified P(A) and P(B) as well whether the ranges of P(C| A,B) will not be much narrower than the full interval [0,1], simply because of the many constraints we have.
Sticking to the notation introduced above:
A = the event that the person has long hair
B = the event that the person has blood type AX3
C = the event that person is female
P(C|A) = 0.9
P(C|B) = 0.8
P(C) = 0.5 (i.e. let's assume an equal ratio of men and women in the population at large)
it does not seem possible to assume that events A and B are conditionally independent given C ! That leads directly to a contradiction: if $P(A \wedge B | C) = P(A| C) \cdot P(B| C) = P(C| A) \frac{P(A)}{P(C)} \cdot P(C| B) \frac{P(B)}{P(C)}$
then
$P(C| A \wedge B ) = P(A \wedge B | C) \cdot \left( \frac{P(C)}{P(A \wedge B)} \right) = P(C| A) \frac{P(A)}{P(C)} \cdot P(C| B) \frac{P(B)}{P(C)} \cdot \left( \frac{P(C)}{P(A \wedge B)} \right)$
If we now assume that A and B are independent as well: $P(A \wedge B) = P(A) P(B)$ most terms cancel and we end up with
$P(C| A \wedge B ) = \frac{P(C| A) \cdot P(C| B)}{P(C)} = \frac{0.9 \cdot 0.8}{0.5} > 1$
Following up on whuber's wonderful geometric representation of the problem: While it is true that generally speaking $P(C | A \wedge B)$ can assume any value in the interval $[0,1]$ the geometric constraints do narrow the range of possible values significantly for values of $P(A)$ and $P(B)$ that are not "too small". (Though we can also upper bound the marginals: $P(A)$ and $P(B)$)
Let us compute the {\bf smallest possible value} for $P(C | A \wedge B)$ under the following geometric constraints:
1. The fraction of the upper area (A TRUE) covered by the upper rectangle must be equal to $P(C|A)=0.9$
2. The sum of the areas of the two rectangles must be equal to $P(C)=0.5$
3. The sum of the fraction of the areas of the two colored rectangles (i.e. their overlap with event B) must be equal to $P(C|B)=0.8$
4. (trivial) The upper rectangle cannot be moved beyond the left boundary and should not be moved beyond its minimum overlap to the left.
5. (trivial) The lower rectangle cannot be moved beyond the right boundary and should not be moved beyond its maximum overlap to the right.
These constraints limit how freely we can slide the hashed rectangles and in turn generate lower bounds for $P(C | A \wedge B)$. The figure below (created with this R script ) shows two examples
Running through a range of possible values for P(A) and P(B) (R script) generates this graph
In conclusion, we can lower bound the conditional probability P(c|A,B) for given P(A), P(B)
-
Markus, the first paragraph belongs as a separate question rather than within an answer. The subsequent material looks like a good observation but it is hard to follow without being told what $A, B,$ and $C$ represent. Please bear in mind that different users will see the answers in different sequences, according to their preferences and when the answers were last edited, so each answer has to be readable independently of the others (although of course you can link to other answers). – whuber Jul 3 '12 at 14:34
@whuber: thanks for the useful comment ! I hope the new edits make it more readable and clear. – Markus Loecher Jul 3 '12 at 18:36
@whuber and others: I had hoped to reignite the discussion but the thread seems to have gone inactive ? No more comments by anyone ? – Markus Loecher Jul 8 '12 at 20:36
Make the hypotheses is that the person behind a curtain is a woman.
We area given 2 pieces of evidence, namely:
Evidence 1: We know the person has long hair (and we're told that 90% of all people with long hair are female)
Evidence 2: We know the person has a rare blood type AX3 (and we're told that 80% of all people with this blood type are female)
Given just Evidence 1, we can state that the person behind a curtain has a 0.9 probability value of being a woman (assuming 50:50 split between men and women).
Regarding the question posed earlier in the thread, namely "Would you agree that the answer must be GREATER than 0.9?", without doing any Math, I would say intuitively, the answer must be "yes" (it is GREATER than 0.9). The logic is that Evidence 2 is supporting evidence (again, assuming a 50:50 split for the number of men and women in the world). If we were told that 50% of all people with AX3 type blood were female, then Evidence 2 would be neutral and have no bearing. But since we're told that 80% of all people with this blood type are female, Evidence 2 is supporting evidence and logically should push the final probability of a woman above 0.9.
To calculate a specific probability, we can apply Bayes' rule for Evidence 1 and then use Bayesian updating to apply Evidence 2 to the new hypothesis.
Suppose:
A = the event that the person has long hair
B = the event that the person has blood type AX3
C = the event that person is female (assume 50%)
Applying Bayes rule to Evidence 1:
P(C|A) = (P(A|C) * P(C)) / P(A)
In this case, again if we assume 50:50 split between men and women:
P(A) = (0.5 * 0.9) + (0.5 * 0.1) = 0.5
So, P(C|A) = (0.9 * 0.5) / 0.5 = 0.9 (Not surprising, but it would be different if we didn't have 50:50 split between men and women)
Using Bayesian updating to apply Evidence 2 and plugging in 0.9 as the new prior probability, we have:
P(C|A AND B) = (P(B|C) * 0.9) / P(E)
Here, P(E) is the probability of Evidence 2, given the hypotheses that the person already has a 90% chance of being female.
P(E) = (0.9 * 0.8) + (0.1 * 0.2) [this is law of total probability: (P(woman)*P(AX3|woman) + P(man)*P(AX3|man)] So, P(E) = 0.74
So, P(C|A AND B) = (0.8 * 0.9) / 0.74 = 0.97297
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There are a few statements in your answer that do not make sense to me. (1) P(C|A)=0.9 by assumption. Nowhere was it said that P(C)=0.9. We assumed P(C)=0.5. (2) How did you get the result for P(E)? P(woman)=P(man)=0.5 by assumption where you write P(woman)=0.9. – Michael Chernick Jun 21 '12 at 11:39
The value of P(C) is assumed at 0.5, which is what I've used. The value for P(E) is the probability of Evidence 2 after applying Evidence 1 (which leads to a new hypotheses that the probability that the person is female is 0.9). P(E) = (probability that the person is a woman (given Evience 1) * probability the the person has AX3 if a woman) + (probability that the person is a man (given Evience 1) * probability the the person has AX3 if a man) = (0.9 * 0.8) + (0.1 * 0.2) = 0.74 – RandomAnswer Jun 21 '12 at 14:42
Your definition of probability of E is a bit confusing and the terms you are using to calculate it look different from what you wrote before. It really doesn't matter though. The answer is apparently correct based on Huu's nicely presented answer. – Michael Chernick Jun 21 '12 at 14:57
@Michael Except it appears Huu made mistakes. – whuber Jun 21 '12 at 15:12
This answer is simply wrong. There may be other errors, but this one is glaring. You state a definitive answer for P("Has Long Hair") (your P(A)), and then use that to give your final definitive answer. There simply isn't enough information to determine this, even assuming P(F) = 0.5. Your line to calculate P(A) seems to come from nowhere. Here is the correct formula using Bayes theroem: P(A) = P(A|F)P(F)/P(F|A) from which, using your stated assumptions, get to P(A) = P(A|F)*5/9. However we still don't know P(A|F), which could be anything. – Bogdanovist Jun 22 '12 at 4:04
show 1 more comment
I believe now that, if we assume a ratio of men and women in the population at large, then there is a single indisputable answer.
A = the event that the person has long hair
B = the event that the person has blood type AX3
C = the event that person is female
P(C|A) = 0.9
P(C|B) = 0.8
P(C) = 0.5 (i.e. let's assume an equal ratio of men and women in the population at large)
Then P(C|A and B) = [P(C|A) x P(C|B) / P(C)] / [[P(C|A) x P(C|B) / P(C)] + [[1-P(C|A)] x [1-P(C|B)] / [1-P(C)]]]
in this case, P(C|A and B) = 0.972973
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P[C|A and B)= P(A and B and C)/P(A and B)=P(A and B and C)/ [P(A|B) P(B)]. How did you get your formula? – Michael Chernick Jun 21 '12 at 10:12
There is probably a way to add conditions so that you get a unique answer. – Michael Chernick Jun 21 '12 at 10:19
To add by independence of A and B the formula simplifies to P(A and B and C}/[P(A) P(B)]=P(B and C|A)/P(B). – Michael Chernick Jun 21 '12 at 10:22
The intent of my question was really for you to justify the formula. I don't understand how it would be derived. – Michael Chernick Jun 21 '12 at 11:24
No, the answer that supposedly used Bayes Rule is incorrect. I'm not sure why you are confused, MC's formula above is correct and cannot be used to get any result, that's what his and Whuber's answers to the question explained! – Bogdanovist Jun 22 '12 at 4:08
Note: In order to get a definitive answer, the below answers assume that the probability of a person, a long-haired man, and a long-haired women having AX3 are approximately the same. If more accuracy is desired, this should be verified.
You start out with the knowledge that the person has long hair, so at this point the odds are:
90:10
Note: The ratio of males to females in the general population does not matter to us once we find out the person has long hair. For example, if there were 1 female in a hundred in the general population, a randomly-selected long-haired person would still be a female 90% of the time. The ratio of females to males DOES matter! (see the update below for details)
Next, we learn that the person has AX3. Because AX3 is unrelated to long hair, the ratio of men to women is known to be 50:50, and because of our assumption of the probabilities being the same, we can simply multiply each side of the probability and normalize so that the sum of the sides of the probability equals 100:
(90:10) * (80:20)
==> 7200:200
Normalize by dividing each side by (7200+200)/100 = 74
==> 7200/74:200/74
==> 97.297.. : 2.702..
Thus, the chance that the person behind the curtain is female is approximately 97.297%.
UPDATE
Here's a further exploration of the problem:
Definitions:
f - number of females
m - number of males
fl - number of females with long hair
ml - number of males with long hair
fx - number of females with AX3
mx - number of males with AX3
flx - number of females with long hair and AX3
mlx - number of males with long hair and AX3
pfl - probability that a female has long hair
pml - probability that a male has long hair
pfx - probability that a female has AX3
pmx - probability that a male has AX3
First, we are given that 90% of long-haired people are females, and 80% of people with AX3 are female, so:
fl = 9 * ml
pfl = fl / f
pml = ml / m
= fl / (9 * m)
fx = 4 * mx
pfx = fx / f
pmx = mx / m
= fx / (4 * m)
Because we assumed that the probability of AX3 is independent of gender and long hair, our calculated pfx will apply to women with long hair, and pmx will apply to men with long-hair to find the number of them that likely have AX3:
flx = fl * pfx
= fl * (fx / f)
= (fl * fx) / f
mlx = ml * pmx
= (fl / 9) * (fx / (4 * m))
= (fl * fx) / (36 * m)
Thus, the likely ratio of the number of females with long-hair and AX3 to the number of males with long-hair and AX3 is:
flx : mlx
(fl * fx) / f : (fl * fx) / (36 * m)
1/f : 1 / (36m)
36m : f
Because it is given that there is an equal number of 50:50, you can cancel both sides and end with 36 females to every male. Otherwise, there are 36*m/f females for every male in the specified subgroup. For example, if there were twice as many women as men, there would be 72 females to each male of those that have long-hair and AX3.
-
This solution relies on assuming more than is currently stated in the problem: namely, that long hair, AX3, and gender are independent. Otherwise, you cannot justify "applying" pfx to women with long hair, etc. – whuber Jun 21 '12 at 19:35
@whuber: Yes, I do make that assumption. However, isn't the purpose of probability to give the best approximation based on the data that you have? Thus, since you already know that long-hair and AX3 are independent for the general population, you SHOULD carry forward that assumption to males and females until you explicitly learn otherwise. Granted, it is not a universally correct one, but it is the best one you can make until you get more info. Q: With only the current data, if you had to give the % chance that it was a woman behind the curtain, would you really say "between 0 and 100%"? – Briguy37 Jun 21 '12 at 21:44
We have an important difference in philosophy, @Briguy. I strongly believe in not making unfounded assumptions. It is not clear in what sense the mutual independence assumption is "best": I will grant it may be in certain applications. But in general, that seems dangerous to me. I would prefer being clear about the assumptions needed to solve a problem, so people can decide whether it is worthwhile collecting the data to check those assumptions, rather than assuming things that are mathematically convenient for the sake of obtaining an answer. That's the difference between stats and math. – whuber Jun 21 '12 at 23:14
To answer your question: yes, 0% - 100% is exactly the answer I would give. (I have given similar answers to comparable questions on this site.) That range accurately reflects the uncertainty. This issue is closely related to the Ellsberg paradox. Ellsberg's original paper is well written and clear: I recommend it. – whuber Jun 21 '12 at 23:16
@whuber: Thanks for taking the time to dialogue with me. I see your point about the importance of thinking through and listing the assumptions made, and have updated my answer accordingly. However, in regards to your answer, I believe it is incomplete. The reason for this is that you can consider all unknown cases and find the average probability of across all of them to arrive at your final answer. E.G. Though both are still possible, probabilities above 50% are much more prevalent than probabilities below 50% across all cases, so we are surely better off guessing that it is a woman. – Briguy37 Jun 22 '12 at 13:56
show 1 more comment
98% Female, simple interpolation. First premise 90% female, leaves 10%, second premise only leaves 2% of the existing 10%, hence 98% female
- | 2013-12-07T11:30:50 | {
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https://mathhelpboards.com/threads/simplify-complex-rational-expression.2246/ | # Simplify Complex Rational Expression
#### PaperStSoap
##### New member
(3/x-2) - (4/x+2) / (7/x2-4)
I got it down to...
-x+14/7
but the book is showing
x-14/7
#### Fantini
MHB Math Helper
Welcome to MHB, Paper!
Which of these expressions did you mean:
$$\frac{3}{x-2} - \frac{ \frac{4}{x+2} }{ \frac{7}{x^2 -4} } \quad \text{ or } \quad \frac{ \frac{3}{x-2} - \frac{4}{x+2} }{ \frac{7}{x^2 -4} }?$$
#### soroban
##### Well-known member
Hello, PaperStSoap!
$$\dfrac{\dfrac{3}{x-2} - \dfrac{4}{x+2}}{\dfrac{7}{x^2-4}}$$
I got it down to: .$\dfrac{-x+14}{7}$ . You are right!
But the book is showing: .$\dfrac{x-14}{7}$ . The book is wrong!
#### PaperStSoap
##### New member
Welcome to MHB, Paper!
Which of these expressions did you mean:
$$\frac{3}{x-2} - \frac{ \frac{4}{x+2} }{ \frac{7}{x^2 -4} } \quad \text{ or } \quad \frac{ \frac{3}{x-2} - \frac{4}{x+2} }{ \frac{7}{x^2 -4} }?$$
My apologies, the problem was the second one.
#### SuperSonic4
##### Well-known member
MHB Math Helper
My apologies, the problem was the second one.
You are right and the book is wrong. It's worth mentioning though that there is a restriction on the domain: $|x| \neq 2$.
(working in spoiler)
$\dfrac{\frac{3}{x-2} - \frac{4}{x+2}}{\frac{7}{x^2-4}}$
$\left(\frac{3}{x-2} - \frac{4}{x+2}\right) \cdot \frac{(x-2)(x+2)}{7}$
$\left(\frac{3(x+2)-4(x-2)}{(x-2)(x+2)}\right) \cdot \frac{(x-2)(x+2)}{7}$
$\frac{3x+6-4x+8}{7}$
$\frac{-x+14}{7}$ | 2021-09-20T23:16:37 | {
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https://math.stackexchange.com/questions/3187553/what-group-is-langle-1-1-1-rangle-in-mathbbz-5-times-mathbbz-4-ti | # What group is $\langle (1,1,1) \rangle$ in $\mathbb{Z}_5 \times \mathbb{Z}_4 \times \mathbb{Z}_8$?
I know that $$\langle (1,1,1) \rangle$$ in $$\mathbb{Z}_5 \times \mathbb{Z}_4 \times \mathbb{Z}_8$$ is isomorphic to $$\mathbb{Z}_{40}$$, but is there a way of writing what group it is (not what it's isomorphic to).
In other words, can we say that $$\langle (1,1,1) \rangle$$ is $$G_1 \times G_2 \times G_3$$ for some groups $$G_1, G_2, G_3$$?
I'm so used to using isomorphism that I can't tell if the answer is simply no. I tried using the Fundamental Theorem of Abelian Groups, but this didn't resolve my question.
For example, in $$\mathbb{Z} \times \mathbb{Z}$$ we have $$\langle (0,3) \rangle = \{0 \} \times \mathbb{3Z}$$.
We would often just say $$\langle (0,3) \rangle \cong \mathbb{Z}$$, but occasionally it's useful to not use the usual isomorphism and so this is what I'm asking for.
My motivation is to be able to more regularly use the trick used in this answer.
• It's a subgroup of order $40.$ – Dbchatto67 Apr 14 at 15:58
• If anything, this comment gives less information than what I've written ("a subgroup of order 40" might not even be cyclic, but this particular one is). I'm asking for more information. – Jonathan Rayner Apr 14 at 16:04
• The group $\langle (1,1) \rangle \subset \Bbb Z_4 \times \Bbb Z_8$ is not isomorphic to (and certainly not equal to) any product $G_2 \times G_3$. Your group can be written as $\Bbb Z_5 \times \langle (1,1) \rangle$ (to the extent that we can consider the Cartesian product associative) – Omnomnomnom Apr 14 at 16:05
• What do you mean what group it is as opposed to what group it is isomorphic to? We usually name groups up to isomorphism, so your group is just $C_{40}$, which is the same as $\Bbb {Z_5 \times Z_8}$ – Ross Millikan Apr 14 at 16:06
• See this question. – Dietrich Burde Apr 14 at 16:20
$$\langle(1, 1, 1)\rangle$$ is, as you said, isomorphic to $$\mathbb{Z}_{40}$$, and the prime factorization of $$40$$ is $$5 \times 8$$, so we can say that $$\langle(1, 1, 1)\rangle \simeq \mathbb{Z}_5 \times \mathbb{Z}_8$$. Now, it is important to convince yourself that such a group can't be expressed as the product of three subgroups, simply because $$\mathbb{Z}_8$$ can't be split since is cyclic and $$\mathbb{Z}_5$$ only has trivial subgroups, or, more formally, by contradiction, let $$\langle(1, 1, 1)\rangle \simeq G_1 \times G_2 \times G_3$$. Since every subgroup of a cyclic group is cyclic, $$G_2 \times G_3$$ is cyclic and none of its coordinates is $$\{e\}$$. Therefore, $$G_2 \times G_3$$ is a product of cyclic groups whose orders are multiples of $$2$$ (so they're not coprime) and it's cyclic, which is a contradiction. As for how you could use that other method in this case, i'd say the most it can do for you is to simplify the thing as $$(\mathbb{Z}_5 \times \mathbb{Z}_4 \times \mathbb{Z}_8)/\langle(1, 1, 1)\rangle \simeq \mathbb{Z}_5/\langle1\rangle \times (\mathbb{Z}_4 \times \mathbb{Z}_8)/\langle(1, 1)\rangle \simeq (\mathbb{Z}_4 \times \mathbb{Z}_8)/\langle(1, 1)\rangle$$, but since that $$\langle(1, 1)\rangle$$ is not a product of subgroups you have to figure out some other way to find it. I hope to have answered your question, let me know if that's not the case.
A bit late, but here's a way to see that $$(\Bbb Z_{4} \times \Bbb Z_8)/\langle (1,1) \rangle \cong \Bbb Z_4$$. We note that the map $$\phi:\Bbb Z_{4} \times \Bbb Z_8 \to \Bbb Z_{4} \times \Bbb Z_8$$ defined by $$\phi: (x,y) \mapsto (x-y,y)$$ is a group automorphism (which is induced by the automorphism with the same formula from $$\Bbb Z^2$$ to $$\Bbb Z^2$$). It follows that $$(\Bbb Z_{4} \times \Bbb Z_8)/\langle (1,1) \rangle \cong \phi(\Bbb Z_{4} \times \Bbb Z_8)/\phi(\langle (1,1) \rangle) = (\Bbb Z_{4} \times \Bbb Z_8)/\langle (0,1) \rangle\\ \quad \ = (\Bbb Z_4 \times \Bbb Z_8)/(\{0\} \times \Bbb Z_8) \cong \Bbb Z_4$$
• Thank you for writing this - I accepted the other answer because I think it answered what I technically asked for in the question and it would have been unfair not to. But your answer was very helpful and exactly in the spirit of what I was looking for! – Jonathan Rayner Apr 19 at 21:52
I'm not quite sure what you're asking in your question, but if I'm understanding it, I hope this will help.
One way to think about it is the following. If $$q:\Bbb{Z}/(8)\to \Bbb{Z}/(4)$$ is the quotient map, then the elements of the subgroup of $$\Bbb{Z}/(5)\times \Bbb{Z}/(4)\times \Bbb{Z}/(8)$$ generated by $$(1,1,1)$$ are precisely the elements of the form $$(i,q(j),j)$$, with $$i\in\Bbb{Z}/(5)$$, and $$j\in\Bbb{Z}/(8)$$. This also makes the isomorphism with $$\Bbb{Z}/(5)\times \Bbb{Z}/(8)$$ clear. One direction is $$(i,q(j),j)\mapsto (i,j)$$, and the other is $$(i,j)\mapsto (i,q(j),j)$$.
Side note:
This isn't directly related to your question, just an interesting tangent.
If we focus on the subgroup of $$\Bbb{Z}/(8)\times \Bbb{Z}/(4)$$ generated by $$(1,1)$$ (order deliberately flipped here), we can see that it is the subgroup $$(j,q(j))$$ for $$j\in \Bbb{Z}/(8)$$, which you can think of as the graph of the group homomorphism $$q$$. It's quite often the case that when you can construct the graph of a morphism $$f$$, the graph turns out to be isomorphic to the starting object, with one of the isomorphisms being $$x\mapsto (x,f(x))$$, and the other being $$(x,f(x))\mapsto x$$. | 2019-06-17T12:56:26 | {
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https://math.stackexchange.com/questions/1172533/asymptotic-notation-big-theta | # Asymptotic notation (big Theta)
I'm currently in the process of analyzing runtimes for some given code (Karatsuba-ofman algorithm).
I'm wondering if I'm correct in saying that $\Theta(\left\lceil n/2\right\rceil) + \Theta(n)$ is equal to $\Theta(n)$? (Taking the maximum of both costs)
I know that $\Theta(n/2) +\Theta(n)$ is equal to $\Theta(n)$. But my concern with the first asymptotic runtime is that given $n = 0.6, \left\lceil n/2\right\rceil$ would be equal to $1$, which is greater than $0.6$.
If $\Theta(\left\lceil n/2\right\rceil) + \Theta(n) \stackrel{?}{=} \Theta(n)$ could anybody please give some insight on why this is true?
Yes you are right.. intuitively it's clear that it does not change a thing, since when $n \to \infty$ effects like this are definitely too small to notice.
The whole concept behind asymptotics is that we ignore small terms.. sometimes we can even ignore something "huge" like $2^n$, for example $\Theta(2^n + 3^n) = \Theta(3^n)$
So we don't really care about effect like this; yes if $n=0.6$ you would get $1$, but the error is at most $1/2$ which is not relevant. if you had $n = 10^9 + 0.6$, would you be worried about wether the result is rounded up or down?
If you want to be a little bit more formal you can start noting that $\lfloor \frac n2 \rfloor \le n$, and $\Theta(n) + \Theta(n) = \Theta(n)$
Or you can also notice that the error from rounding is at most $1/2$ or $1$, depends on the rounding that one uses and find some bonds from there.
• Thanks alot for this explanation. Just as I thought. That's really cleared some things up. – user220506 Mar 2 '15 at 22:12
• @Tazman You're welcome! :) – Ant Mar 2 '15 at 22:13
• Although you stated floor(n/2) <= n, this wouldn't be the case for ceiling(n/2) <= n unless we state for all n >= 1 right? – user220506 Mar 2 '15 at 22:14
• @Tazman yes of course. But then again we only look about what happens in the long run, so to speak, so most of the time I don't even notice I take this assumption and most of what I write is valid for $n$ large enough ;-) – Ant Mar 2 '15 at 22:16
• Okay, that makes sense. Just like if it were a big-O or big-Omega question, we would state some given bound where we see the said behaviour. Thanks for the clarification :) – user220506 Mar 2 '15 at 22:17
Remember that $f(n) =\Theta (g(n))$ means that there are positive constants $a$ and $b$ such that $a g(n) < f(n) < b g(n)$ for all large enough $n$.
Therefore, if $f(n) =\Theta (h(n))$ and $g(n) =\Theta (h(n))$ then $f(n)+g(n) =\Theta(h(n))$.
If you consider the problem for the natural numbers it is true that $\Theta(\lceil n/2 \rceil) = \Theta (n)$. As $\lceil n/2 \rceil \le n \le 2 \lceil n/2 \rceil$. And as a consequence also $\Theta(\lceil n/2 \rceil) + \Theta (n)= \Theta (n)$.
However, if you consider it for the positive real numbers (as $n \to 0$) as your example $0.6$ might suggest then it is not true that $\Theta(\lceil n/2 \rceil) = \Theta (n)$ and neither is $\Theta(\lceil n/2 \rceil) + \Theta (n)= \Theta (n)$. In this case $n = O (\lceil n/2 \rceil)$ as $n \le 2 \lceil n/2 \rceil$ still holds. Yet as $\lceil n/2 \rceil = 1$ for all $0< n \le 2$ we get that $\lceil n/2 \rceil$ is not $O(n)$ as $n \to 0$.
• So Tazman should say $n \to \infty$ if he means that. (Presumably he does, for analyzing "runtimes".) – GEdgar Mar 2 '15 at 22:08
• @GEdgar Well, yes and no. Note they mention the case $n=0.6$ explicitly. But I did mention the result for the integers first. It is however not really necessary to say $n \to \infty$ one just has to make the domain precise, if it is the natural numbers it is fine. The $O$ notation as well as $\Theta$ makes sense without making a limit point precise, as opposed to $o$. Finally note it is the second question of this form I answered today, and there is thus additional reason why I insist on discussing the distinction i make. – quid Mar 2 '15 at 22:13
• The n in question refers to a length of a given number, so I think it's safe that in my answer to the question, one can assume that n consists of natural numbers, and thus theta(ceiling(n/2)) = theta(n). – user220506 Mar 2 '15 at 22:16
• @quid, yes, that's true. I should have been more clear and stated that I was giving that generic example to better my understanding of exactly why it is (or not) equal. – user220506 Mar 2 '15 at 22:19
• @Tazman yes, if it is for integers then you can say this (as I said in my answer). And this is a reasonable assumption given your context. I only discussed the other case since you mentioned $n=0.6$ and there is an actual difference in another situation. [This is an expanded version.] – quid Mar 2 '15 at 22:20 | 2019-05-25T10:50:17 | {
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https://math.stackexchange.com/questions/4241111/need-help-understanding-least-squares-solution-to-overdetermined-system | # Need help understanding least squares solution to overdetermined system
(Sorry I had to post the images as links. I don't have enough cred to post pictures directly yet)
I'm trying to understand what the least squares solution to an overdetermined system means geometrically in the case of the following system:
$$y = -2x-1\\ y = 3x -2\\ y = x+1\\$$
rewritten in matrix form:
$$\overbrace{\begin{bmatrix} 2 & 1\\ -3 & 1\\ -1 & 1 \end{bmatrix}}^A \underbrace{\begin{bmatrix} x\\ y \end{bmatrix}}_x = \overbrace{\begin{bmatrix} -1\\ -2\\ 1 \end{bmatrix}}^b$$
Using A\b in MATLAB, you get the solution $$\begin{bmatrix}0.1316 & -0.5789\end{bmatrix}^T$$. I know that MATLAB returns the lowest norm solution of a least squares problem. I have plotted the system here and the green dot in the middle is this least squares solution.
Now, correct me if I'm wrong, but (in this 2D case) the least squares solution minimizes the "distance" from the solution to each line. I can geometrically calculate the distance of a point $$(x_0,y_0)$$ from a line $$ax + by + c = 0$$ as follows:
$$\mathrm{distance} = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$$
and doing that for each line produces the following sum of squared distances function
dfun = @(x,y) ((y+2*x+1).^2)/(1^2 + 2^2) + ((y+3*x+2).^2)/(1^2 + 3^2) + ((y+x-1).^2)/(1^2 + 1^2);
If I generate a surface using this function over a range of $$x$$ and $$y$$ values, I get this surface with this top-down view (looking down the z-axis on the xy plane). You can download the MATLAB .fig file here if you want to zoom and pan (requires MATLAB, link expires in 30 days).
Here is an image showing the least squares solution with the sum of squares of distances of the solution and its norm. As can be seen, the norm is $$0.5937$$ and the distance is $$1.4704$$. But clearly, there is a contour that has a lower sun of squared distance in the image, as shown here for $$(x_0, y_0) = (-0.3,0)$$, where the norm and the sum of squared distances are both smaller. Shouldn't this (or another point) be a better least squares solution? Do I have the wrong intuition about what least squares is doing here?
First of all : Welcome to the site !
When you face an overdetermined system of $$m$$ linear or nonlinear (even implicit) equations $$f_i(x_1,x_2,\cdots,x_n)=0 \quad\text{for}\quad i=1,2,\cdots,m\qquad \quad\text{and}\quad m >n$$ it reduces to the minimization of a norm.
The simplest is $$\Phi(x_1,x_2,\cdots,x_n)=\sum_{i=1}^n w_i \Big[f_i(x_1,x_2,\cdots,x_n)\Big]^2$$ which shows the analogy with weighted least-square method.
For the problem you gave, using equal weights, $$\Phi(x,y)=(y+2x+1)^2+(y-3x+2)^2+(y-x-1)^2$$ Computing the partial derivetives $$\frac{\partial \Phi(x,y)}{\partial x}=28 x-4 y-6 \qquad\text{and}\qquad \frac{\partial \Phi(x,y)}{\partial y}=-4 x+6 y+4$$ which gives $$x=\frac{5}{38}$$ and $$y=-\frac{11}{19}$$. At this point, $$\Phi=\frac{169}{38}$$ is the absolute minimum for this specific norm.
If you change the definition of the norm or even the weights, different results.
I think that this approach would be simpler than the one based on distance (but, for sure, I may be wrong) since it is totally general and not limited to linear equations.
• Thank you! I think I had the right understanding of what least squares is doing in this case. After reading your answer I realizes that my dfun was wrong - I got a couple of the signs wrong and that's why I was getting the wrong surface. I will post an answer with the right images. Not sure if I should accept my own answer or yours...
– vyb
Sep 6, 2021 at 17:40
After reading Claude Leibovici's answer above, I realized that my dfun had typos in it -- I messed up a couple of minus signs in the function.
Additionally, the norm typically used for least squares calculations (also used by MATLAB) is the $$l^2$$-norm (a.k.a Euclidean norm):
$$x = \begin{bmatrix}x_1\\ x_2\\ \vdots\\ x_n\end{bmatrix}, ||x|| = \sqrt{x_1^2 + x_2^2 + \dots + x_n^2}$$
Note that there is no scaling of the "distance" like there is in my dfun. Therefore correct function should be:
dfun = @(x,y) ((y+2*x+1).^2) + ((y-3*x+2).^2) + ((y-x-1).^2);
After fixing these mistakes, here is the surface that is generated, with this top-down view. As can be seen here, the solution of $$\begin{bmatrix}0.1316 & -0.5789\end{bmatrix}^T$$ is, in fact, correct and confirms the original intuition I was trying to confirm. | 2022-06-28T04:18:54 | {
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https://stats.stackexchange.com/questions/443614/probability-that-random-draws-from-the-same-pool-will-collectively-select-90-of | # probability that random draws from the same pool will collectively select 90% of the pool
There are a total of 200 names on a list. 30 times names are selected from the full list. How many names should be selected each time to predict with 90% certainty that 90% of all the names will be selected at least once?
# The answer is $$n=17$$.
I can't see an easy analytic solution to this question. Instead, we will develop an analytic solution to a closely related problem, and then find the answer to your exact question via simulation.
## Clarification:
Since the question is slightly vague, let me re-state the problem. There are $$200$$ names on a list and $$n$$ names will be selected from this list without replacement. This process, using the full $$200$$ names each time, is repeated a total of $$30$$ times.
## A related problem.
Let $$X_i$$ equal $$1$$ if the $$i^{th}$$ name is selected at least once and equal to $$0$$ otherwise. This implies that $$X = \sum_{i=1}^{200}X_i$$ represents the total number of names which are selected at least once. Since the $$X_i$$ are dependent, the exact distribution of $$X$$ is not-trivial, and the original question is hard to answer. Instead, we can easily determine the value of $$n$$ such that $$90\%$$ of the names are selected on average.
First, note that $$P(X_i = 0) = \left(\frac{200 - n}{200}\right)^{30}$$ which implies $$E(X_i) = P(X_i =1) = 1 - \left(1- \frac{n}{200}\right)^{30}.$$
Now by linearity of expectation we have $$E(X) = \sum_{i=1}^{200}E(X_i) = 200\left(1 - \left(1- \frac{n}{200}\right)^{30}\right).$$
For this expectation to equal $$90\%$$ of the names, we need to set $$E(X) = 180$$ and solve for $$n$$. This gives $$n = 200\left(1 - (1 - 0.9)^{1/30}\right) = 14.776.$$
Thus $$n=15$$ names should be drawn from the list each time for this to occur on average. This answer will be close to (but not the same as) the original question with $$50\%$$ certainty. To achieve $$90\%$$ certainty, we will need to increase $$n$$.
## Simulations.
First, we write a function which is able to generate $$X$$ a large number (say $$M$$) times for a given value of $$n$$.
sample_X <- function(n, M){
X <- rep(NA, M)
for(i in 1:M){
#Set all names to false
names <- rep(FALSE, 200)
#Repeat process 30 times
for(k in 1:30){
#Sample n names from list
selection <- sample(200, n, replace=F)
#Mark that these names have been selected
names[selection] <- TRUE
}
#Let X be the number of selected names
X[i] <- sum(name_been_selected)
}
return(X)
}
Now, for a given value of $$n$$ we can approximate "the probability that at least $$90\%$$ of the names are selected", i.e. $$P(X \geq 180)$$. In R, this probability can be approximated by typing:
X <- sample_X(n, M=10000)
prob <- mean(X >= 180)
Repeating this for $$n = 14, 15, \cdots 20$$ gives us the following plot.
From the plot, we can determine that $$n=17$$ names must be selected in each round for the probability of selecting at least $$180$$ names to exceed $$0.9$$.
The blue line in the figure shows the exact simulations detailed above. The orange line is an approximation which is obtained by ignoring the dependency of the $$X_i$$ (see previous section) and assuming that $$X \sim \text{Binom}\left(200, 1 - \left(1- \frac{n}{200}\right)^{30}\right).$$
Although the assumption of independence is obviously incorrect, the probabilities obtained by this simple assumption are reasonably close to the simulated probabilities.
• There appear to be bugs. Why are you sampling n from 200 names when you should always be sampling 30? When I compute the exact solution (using a method similar to the one you posted yesterday but deleted) or when I perform a simulation I get answers very much like your deleted answers. In particular, the chance of collecting 180 or more names in 17 iterations is 99.26%; in 16 iterations it is 95.46%; and in 15 iterations it is 81.14%, whence 16 iterations will do the job. A simple but accurate estimate is the ceiling of $\log(0.10)/\log(1-30/200)= 14.2,$ or 15. – whuber Jan 8 '20 at 5:10
• @whuber, perhaps we are interpreting the question differently. The original question is quite vague, but my best interpretation is that $n$ names are sample from the list of $200$, and this is repeated $30$ times: "thirty times, names are selected..." and "how many names should be selected each time " – knrumsey Jan 8 '20 at 5:20
• I'm sorry--I confused you with the author of the deleted post. There are too many "*-Reinstate Monica" user names here. :-). Upon rereading the question (several times, because it's strangely phrased) I have to agree with you: I mixed up "names" and "times" in my interpretation. +1 to you--but note that an exact solution still is relatively easy to obtain. Simply update the generating polynomial for the distribution 30 times. – whuber Jan 8 '20 at 5:30
• I am hoping that the author of the deleted post will make the small changes need to reflect your (correct) interpretation and undelete it, because it's a nice post. – whuber Jan 8 '20 at 5:39
• @filbranden, that's relatively close to what I get: $97.8\%$. – knrumsey Jan 8 '20 at 16:32
# Here is a general analytic solution --- does not require simulation
This is a variation on the classical occupancy problem, where you are sampling lots of thirty names at each sampling point, instead of sampling individual names. The simplest way to compute this result is by framing the problem as a Markov chain, and then computing the required probability using the appropriate power of the transition probability matrix. For the sake of broader interest to other users, I will generalise from your example by considering a list with $$m$$ names, with each sample selecting $$1 \leqslant h \leqslant m$$ names (using simple-random-sampling without replacement).
The general problem and its solution: Let $$0 \leqslant K_{n,h} \leqslant m$$ denote the number of names that have been sampled after we sample $$n$$ times with each lot sampling $$h$$ names. For a fixed value $$h$$ the stochastic process $$\{ K_{n,h} | n = 0,1,2,... \}$$ satisfies the Markov assumption, so it is a Markov chain. Since each sampling lot is done using simple-random-sampling without replacement, the transition probabilities for the chain are given by the hypergeometric probabilities:
$$P_{t,t+r} \equiv \mathbb{P}(K_{n,h} = t+r | K_{n-1,h} = t) = \frac{{m-t \choose r} {t \choose h-r}}{{m \choose h}}.$$
Let $$\mathbf{P}_h$$ denote the $$(m+1) \times (m+1)$$ transition probability matrix composed of these probabilities. If we start at the state $$K_{0,h} = 0$$ then we have:
$$\mathbb{P}(K_{n,h} = k) = [ \mathbf{P}_h^n ]_{0,k}.$$
This probability can be computed by matrix multiplication, or by using the spectral decomposition of the transition probability matrix. It is relatively simple to compute the mass function of values over $$k=0,1,...,m$$ for any given values of $$n$$ and $$h$$. This allows you to compute the marginal probabilities associated with the Markov chain, to solve the problem you have posed.
The problem you have posed is a case of the following general problem. For a specified minimum proportion $$0 < \alpha \leqslant 1$$ and a specified minimum probability $$0 < p < 1$$, we seek the value:
$$h_* \equiv h_* (\alpha, p) \equiv \min \{ h = 1,...,m | \mathbb{P}(K_{n,h} \geqslant \alpha m) \geqslant p \}.$$
In your problem you have $$m=200$$ names in your list and you are taking $$n=30$$ samples. You seek the value $$h_*$$ for the proportion $$\alpha = 0.9$$ and the probability cut-off $$p = 0.9$$. This value can be computed by computing the relevant marginal probabilities of interest in the Markov chain.
Implementation in R: We can implement the above Markov chain in R by creating the transition probability matrix and using this to compute the marginal probabilities of interest. We can compute the marginal probabilities of interest using standard analysis of Markov chains, and then use these to compute the required number of names $$h_*$$ in each sample. In the code below we compute the solution to your problem and show the relevant probabilities increasing over the number of samples (this code takes a while to run, owing to the computation of matrix-powers in log-space).
#Create function to compute marginal distribution of Markov chain
COMPUTE_DIST <- function(m, n, H) {
#Generate empty matrix of occupancy probabilities
DIST <- matrix(0, nrow = H, ncol = m+1);
#Compute the occupancy probabilities
for (h in 1:H) {
#Generate the transition probability matrix
STATES <- 0:m;
LOGP <- matrix(-Inf, nrow = m+1, ncol = m+1);
for (t in 0:m) {
for (r in t:m) {
LOGP[t+1, r+1] <- lchoose(m-t, r-t) + lchoose(t, h-r+t) - lchoose(m, h); } }
PP <- exp(LOGP);
#Compute the occupancy probabilities
library(expm);
DIST[h, ] <- (PP %^% n)[1, ]; }
#Give the output
DIST; }
#Compute the probabilities for the problem
m <- 200;
n <- 30;
H <- 20;
DIST <- COMPUTE_DIST(m, n, H);
From the marginal probabilities for the Markov chain, we can now compute the required value $$h_*$$ for your particular problem.
#Set parameters for problem
alpha <- 0.9;
cutoff <- ceiling(alpha*m);
p <- 0.9;
#Find the required value
PROBS <- rowSums(DIST[, (cutoff+1):(m+1)]);
hstar <- 1 + sum(PROBS < p);
#Show the solution and its probability
hstar;
[1] 17
PROBS[hstar];
[1] 0.976388
We can see here that we require $$h_* = 17$$ samples in order to obtain a minimum $$p=0.9$$ probability of sampling at least $$\alpha \cdot m = 180$$ of the names on the list. Below we show a plot of the probabilities for values $$h=1,...,20$$ with the required value highlighted in red.
#Plot the probabilities and the solution
library(ggplot2);
THEME <- theme(plot.title = element_text(hjust = 0.5, size = 14, face = 'bold'),
plot.subtitle = element_text(hjust = 0.5, face = 'bold'));
DATA <- data.frame(h = 1:H, Probability = PROBS);
ggplot(aes(x = h, y = Probability), data = DATA) +
geom_point(size = 3, colour = 'blue') +
geom_point(size = 4, colour = 'red', data = DATA[hstar, ]) +
geom_hline(yintercept = p, size = 1, linetype = 'dashed') +
geom_segment(aes(x = hstar, y = 0, xend = hstar, yend = DATA[hstar, 2]),
colour = 'red', size = 1) +
annotate("text", x = hstar + 1, y = 0.1,
label = paste0('h = ', hstar), colour = 'red', fontface = 'bold') +
THEME +
ggtitle('Probability of required occupancy') +
labs(subtitle = paste0('(Occupancy problem taking ', n,
' samples of size h from ', m,
' units) \n (We require ', sprintf(100*alpha, fmt = '%#.1f'),
'% occupancy with ', sprintf(100*p, fmt = '%#.1f'), '% probability)'));
## The answer is $$n = 17$$, with $$P(N_{30}\ge180)=0.976388$$.
The approach I took to calculate the probability after 30 draws was to determine the probability of drawing seen vs. unseen names at each round.
When drawing $$n$$ names out of $$p=200$$ after having seen $$s$$ of them, let's call $$U_s$$ the number of names out of those $$n$$ which were previously unseen.
Then we have:
$$P(U_s = u) = \frac{\text{P}(200-s, u) \text{P}(s, n-u) \text{C}(n, u)}{\text{P}(200, n)}$$
The first term is the permutations of u previously unseen names, the second permutations of previously seen ones. The last term $$\text{C(n, u)}$$ accounts for the $$u$$ unseen names coming in different positions out of the $$n$$ drawn. The denominator accounts for all possible draws of $$n$$ names.
Having calculated that, we can look at successive draws of names. Let's call $$N_d$$ the total number of names after draw $$d$$.
Before the first draw, there will be no previously seen names, so in the first draw all $$n$$ names will be seen for the first time.
$$P(N_1=n)=1$$
We can then calculate the probability of drawing a certain number of names on draw $$N_{d+1}$$ by looking at the possibilities of drawing after $$N_d$$ and having a specific number of previously unseen names. Which we can calculate with:
$$P(N_{d+1} = x) = \sum_{i=0}^{n}{P(N_d = x-i) P(U_{x-i} = i)}$$
For example, if we're drawing $$n=16$$ every time, then drawing exactly 180 names in total in a specific drawing can be arrived at by drawing 164 names in the previous drawing an then drawing exactly 16 unseen names (totalling 180), or having previously seen 165 names and drawing 15 unseen and one previously seen name, and so on... Until the possibility of having seen 180 names in the previous iteration and drawing all 16 previously seen names.
At this point we can use iteration to calculate $$P(N_{30} \ge 180)$$ for different values of $$n$$.
## Iteration in Python:
This code uses Python 3 and as written requires Python 3.8 for math.comb() and math.perm() from the standard library (if using an older version of Python, you can use a different implementation of those functions.)
Let's start with $$P(U_s = u)$$:
from functools import lru_cache
from math import comb, perm
@lru_cache
def prob_unseen(n, p, s, u):
# Return the probability of drawing
# exactly $$u$$ unseen names when
# drawing $$n$$ names out of a total of $$p$$,
# having previously seen $$s$$ of them.
return (perm(p-s, u) *
perm(s, n-u) *
comb(n, u) /
perm(p, n))
Pretty straightforward. Now for $$P(N_d = x)$$ let's use a list of 201 elements (indices go from 0 to 200) to track the probabilities for each $$x$$:
def names_in_draw(prev_draw, n):
# Calculate probabilities of finding
# exactly $$x$$ names in this draw,
# for every $$x$$, taking in consideration
# the probabilities of having drawn specific
# numbers of names in the previous draw.
p = len(prev_draw) - 1
this_draw = [0.0] * (p+1)
for x in range(n, p+1):
this_draw[x] = sum(
prev_draw[x-u] * prob_unseen(n, p, x-u, u)
for u in range(n+1))
return this_draw
Finally, let's calculate the probability for the number of names after $$d$$ draws.
def total_names(n, p, d):
# Calculate probabilities for finding
# exactly $$x$$ names after $$d$$ draws.
draw = [0.0] * (p+1)
draw[n] = 1.0 # first draw
for _ in range(d):
draw = names_in_draw(draw, n)
return draw
We start from the first draw, where we know for sure we'll draw $$n$$ unique names. Than we repeatedly calculate the probabilities $$d-1$$ times.
Finally, we can calculate the probability of drawing at least $$x$$ names, drawing $$n$$ out of $$p$$ at a time, performing $$d$$ drawings:
def prob_names(n, p, d, x):
# Return the probability of seeing
# at least $$x$$ names after $$d$$ drawings,
# each of which draws $$n$$ out of $$p$$ names.
return sum(total_names(n, p, d)[x:])
Finally, we can run this for a few values of $$n$$ to find the probabilities:
>>> for i in range(13, 20):
... print(i, prob_names(i, 200, 30, 180))
13 0.058384795418431244
14 0.28649904267865317
15 0.6384959089930037
16 0.8849450106842117
17 0.976388046862824
18 0.9966940083338005
19 0.9996649977705089
So $$n=17$$ is the answer, with probability of 97.6388% of seeing at least 90% of the names. $$n=16$$ comes close, with 88.4945%.
(Since I had the code, I also looked at how many drawings of a single name are needed to see 90% of the names, with 90% probability. It turns out it's 503 drawings, or 454 drawings to see 90% of the names with 50% probability. Quite interesting result!) | 2021-01-18T21:08:56 | {
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https://math.stackexchange.com/questions/3138129/maclaurin-series-from-sinx-to-cosx-using-derivative | # Maclaurin Series from sin(x) to cos(x) using derivative
I understand how to find the MacLaurin series for $$\sin(x)$$. $$\sum_{n=1}^\infty \frac{x^{2n-1}\cdot\!(-1)^{n-1}}{(2n-1)!}$$ Now I am trying to find the MacLaurin series for $$\cos(x)$$ by taking the derivative of the above sum with respect to $$x$$. Using power rule, I got the following series: $$\cos(x) = \sum_{n=1}^\infty \frac{x^{2n-2}\cdot\!\mathrm{(-1)}^{n-1}}{(2n-2)!}$$
However, the MacLaurin series is: $$\cos(x) = \sum_{n=0}^\infty \frac{x^{2n}\cdot\!\mathrm{(-1)}^{n}}{(2n)!}$$ How are these two $$\cos(x)$$ MacLaurin series equal? What makes the second one more correct than the series I got by taking the derivative of the $$\sin(x)$$ series.
A sort of related question: if you choose to start at $$n=1$$ vs $$n=0$$, how would you change the terms of the $$\sin(x)$$ Maclaurin series?
• The two series are the same. They only differ by "re-indexing". Shifting the start from $n=1$ to $n=0$ just replaces all the $n$'s by $(n+1)$ in the formula. – Nick Mar 6 at 21:34
You got twice the same series. Both$$\sum_{n=0}^\infty \frac{x^{2n}\cdot\!\mathrm{(-1)}^{n}}{(2n)!}\text{ and }\sum_{n=1}^\infty \frac{x ^{2n-2}\cdot\!\mathrm{(-1)}^{n-1}}{(2n-2)!}$$are equal to$$1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$$
To reindex, replace $$n$$ with $$n+1$$ in the summand of the first series and lower the range of $$n$$ by $$1$$ (so the new initial index will now be $$0$$ instead of $$1$$).
More generally, if the index $$n$$ runs from $$n=a$$ to $$n=b$$, then replace $$n$$ with $$n+K$$ in the summand and change the index limits to run from $$n=a-K$$ to $$n=b-K$$. You are essentially adding and subtracting $$K$$ from each value of $$n$$, so it is unchanged in the end. Symbolically: $$\sum_{n=a}^b t_n =\sum_{n=a-K}^{b-K}t_{n+K}$$ | 2019-09-24T09:02:19 | {
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https://math.stackexchange.com/questions/1431042/how-do-you-calculate-a-sum-over-a-polynomial/1431090 | # How do you calculate a sum over a polynomial?
I know that given a polynomial $p(i)$ of degree $d$, the sum $\sum_{i=0}^n p(i)$ would have a degree of $d + 1$. So for example
$$\sum_{i=0}^n \left(2i^2 + 4\right) = \frac{2}{3}n^3+n^2+\frac{13}{3}n+4.$$
I can't find how to do this the other way around. What I mean by this, is how can you, when given a polynomial, calculate the polynomial which sums to it?
For example, if we know that
$$\sum_{i=0}^{n} p(i) = 2n^3 + 4n^2 + 2,$$
how can we find the polynomial $p(i)$?
• Are you asking how sums like $\sum_{i=0}^{n} p(i)$ can be calculated in general? – Antonio Vargas Sep 11 '15 at 15:42
• @AntonioVargas I'm asking how a polynomial can be written as a Sigma notation. So for example, the polynomial $2/3n^3+n^2+13/3n+4$ can be written as $\sum_{i=0}^n 2i^2 + 4$, but I have no idea how you would start. – Esoemah Sep 11 '15 at 15:46
• Okay, maybe the last part of the question should be written a little differently then. Is this your question?: If we know that $\sum_{i=0}^{n} p(i) = 2n^3 + 4n^2 + 2$, how can we find the polynomial $p(i)$? – Antonio Vargas Sep 11 '15 at 15:48
• Great, I've edited the question to try to clarify it. If you'd like you can also edit it by clicking the "edit" link just below the list of tags at the bottom of the question. – Antonio Vargas Sep 11 '15 at 16:00
• Short answer:$$p(n) = \left( \sum_{i=0}^n p(i) \right) - \left( \sum_{i=0}^{n-1} p(i) \right)$$ – Hurkyl Sep 11 '15 at 17:01
To rephrase, I believe the question is this:
Suppose that polynomials $p$ and $q$ have the property that $$\sum_{i=0}^n p(i) = q(n)$$ If you're given $q$, how can you find $p$?
First, this is a lovely question. I'd never really considered it, because we almost always study instead "if you know $p$, how do you find $q$?"
To answer though, turns out to be fairly simple. Write the following: \begin{align} q(n-1) &= p(0) + p(1) + \ldots + p(n-1) \\ q(n) &= p(0) + p(1) + \ldots + p(n-1) + p(n) \\ \end{align}
Now subtract the top from the bottom to get \begin{align} q(n) - q(n-1) &= p(n) \end{align}
As an example, in your case if we knew $$q(n) = 2n^3 + 4n^2 + 2$$ we'd find that $$p(n) = q(n) - q(n-1) = 2n^3 + 4n^2 + 2 - [2(n-1)^3 + 4(n-1)^2 + 2],$$ which you simplifies to $$p(n) = 6n^2 +2n - 2.$$
Let's do an example: we know that for $p(n) = n$, we have $q(n) = \frac{n(n+1)}{2}$. So suppose we were given just $q$. We'd compute \begin{align} q(n) - q(n-1) &= \frac{n(n+1)}{2} - \frac{(n-1)(n)}{2} \\ &= \frac{n^2 + n}{2} - \frac{n^2 - n}{2} \\ &= \frac{n^2 + n-(n^2 - n)}{2} \\ &= \frac{n^2 + n- n^2 + n}{2} \\ &= \frac{2n}{2} \\ &= n, \end{align} so that $p(n) = n$, as expected.
Note: As written, I've assumed that $p$ and $q$ are both polynomials. But the solution shows that if $q$ is a polynomial, then $p$ must also be a polynomial, which is sort of pleasing.
Post-comment remarks
As @Antonio Vargas points out, though, there's an interesting subtlety:
I've given a correct answer to my revised question, which was "If there are polynomials $p$ and $q$ satisfying a certain equality, then how can one find $p$ given $q$."
But suppose that there is no such polynomial $p$. My answer still computes an expression which $p$, if it existed, would have to match. But since no such $p$ exists, the computed expression has no value.
Or maybe I should say that it has a limited value: you can take the polynomial $p$ and compute its sum using inductive techniques and see whether you get $q$. If so, that's great; if not, then there wasn't any answer in the first place.
Fortunately, you can also do that "Does it really work" check much more simply. You just need to check the the $n = 0$ case: if $$\sum_{i = 0}^0 p(i) = q(0)$$ then all higher sums will work as well. And this check simplifies to just asking: is $$p(0) = q(0)?$$ In our example, $p(0) = -2$, while $q(0) = +2$, so it doesn't work out.
• It is, by the way, a nice exercise to show that if $q$ is a degree-$d$ polynomial, then $p(n)=q(n)-q(n-1)$ is a degree-$(d-1)$ polynomial. This operation is known as finite differencing and there are many close analogs with differentiation. – Steven Stadnicki Sep 11 '15 at 16:10
• Nice point; I've added further remarks. My test for whether things work out is that $p(0) = q(0)$ rather than that $p(-1) = 0$, but that's a tiny difference. – John Hughes Sep 14 '15 at 20:16
• Doesn't all this procedure remind a little: integrals and derivatives? – Widawensen Jul 1 '16 at 7:53
Which Polynomials Can Be Written as a Sum
By summing a Telescoping Series, we get \begin{align} \sum_{k=0}^n(p(k)-p(k-1)) &=\sum_{k=0}^np(k)-\sum_{k=0}^np(k-1)\\ &=\sum_{k=0}^np(k)-\sum_{k=-1}^{n-1}p(k)\\ &=p(n)-p(-1) \end{align} It turns out that not every polynomial can be written as a sum of other polynomials. To be written as a sum of polynomials $$p(n)=\sum_{k=0}^nq(k)$$ we must have $p(-1)=0$, and if that condition holds, then $q(k)=p(k)-p(k-1)$.
Finite Differences
$q(k)=p(k)-p(k-1)=\Delta p(k)$ is the first Backward Finite Difference of $p$.
Using the Binomial Theorem, we get the first backward finite difference of $x^n$ to be $$\Delta x^n=x^n-(x-1)^n=\sum_{k=0}^{n-1}(-1)^{n-k-1}\binom{n}{k}x^k$$ This shows that the first backward finite difference of a degree $n$ polynomial is a degree $n-1$ polynomial.
Thus, for $$p(k)=2k^3+4k^2+2$$ we have \begin{align} \Delta p(k) &=2\overbrace{\left[3k^2-3k+1\right]}^{\Delta k^3}+4\overbrace{\left[\vphantom{k^2}2k-1\right]}^{\Delta k^2}+2\overbrace{\left[\vphantom{k^2}\ \ \ 0\ \ \ \right]}^{\Delta 1}\\ &=6k^2+2k-2 \end{align} However, since $p(-1)=4$, we have $$\sum_{k=0}^n(6k^2+2k-2)=2n^3+4n^2-2$$ which is not $p(n)$. That is,
There is no polynomial $q(n)$ so that $\sum\limits_{k=0}^nq(k)=2n^3+4n^2+2$
Previous Question
The answer below was posted before the question was changed. It was
The other way around however, I'm still a bit lost. For example, given the polynomial $p(i) = 2i^3 + 4i^2 + 2$, how would you find $\sum_{i=0}^n p(i)$?
So what follows may seem to be off-topic.
There are several ways to approach this problem.
Binomial Polynomials
One is by writing the polynomial as a binomial polynomial: $$2k^3+4k^2+2=12\binom{k}{3}+20\binom{k}{2}+6\binom{k}{1}+2\binom{k}{0}$$ Then use the formula $$\sum_{k=0}^n\binom{k}{m}=\binom{n+1}{m+1}$$ to get \begin{align} \sum_{k=0}^n\left(2k^3+4k^2+2\right) &=12\binom{n+1}{4}+20\binom{n+1}{3}+6\binom{n+1}{2}+2\binom{n+1}{1}\\ &=\frac{3n^4+14n^3+15n^2+16n+12}6 \end{align}
Euler-Maclaurin Sum Formula
In most cases, the Euler-Maclaurin Sum Formula is an asymptotic approximation, but in the case of polynomials, it is exact. $$\sum_{k=0}^nf(k)\sim C+\int_0^nf(x)\,\mathrm{d}x+\frac12f(n)+\frac1{12}f'(n)-\frac1{720}f'''(n)+\frac1{30240}f^{(5)}(n)+\dots$$ where subsequent terms involve higher derivatives, which for polynomials will eventually vanish. In the case at hand, this gives the same answer as above.
• Now I have answered the new question. Unfortunately, there is no $q$ so that $p(n)=\sum\limits_{k=0}^nq(k)$ for this particular $p$. – robjohn Sep 11 '15 at 17:46 | 2019-05-25T20:02:55 | {
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http://math.stackexchange.com/questions/28666/surjective-and-unbounded-functions | # Surjective and Unbounded functions
• Every surjective function from $\mathbb{R}$ to $\mathbb{R}$ is unbounded.
• Every unbounded function from $\mathbb{R}$ to $\mathbb{R}$ is surjective.
Is it possible for either of these statements to be false? I have a feeling there is some counterexample that I am missing but I cannot figure it out.
My understanding is that if a function is unbounded then for all $M\in\mathbb{R}$ there is an $x$ such that $|f(x)| \gt M$.
And the definition of surjective is that for all $b \in Y$, there exists an $x \in X$ such that $f(x) = b$.
Clearly if we have some $M$ in the image of this function there is an $x$ that exists such that $f(x) = M$ by the definition of surjective.
I dont know if I am thinking of this correctly, intuition needed. Thanks.
-
Your edit still does not give the right definition. Language is not commutative ("answer the exam and study" is not the same thing as "study and answer the exam"). When you say "$|f(x)|\gt M$ for all $M\in\mathbb{R}$" you are saying that the value of $f$ is always greater than any number. That's false for any function with values in $\mathbb{R}$. What you want to say is "For every $M\in\mathbb{R}$ there exists $x$ such that $|f(x)|\gt M$." This says that the $x$ you pick may depend on $M$. The other way, the $x$ must be independent of $M$. – Arturo Magidin Mar 23 '11 at 16:52
Let's be a bit clearer: a function $f\colon X\to\mathbb{R}$ is unbounded if and only if for every $M\in\mathbb{R}$ there exists $x\in X$ such that $|f(x)|\gt M$. The quantifier ("for every") is important; otherwise, you are just saying that there is at least one $M$ with the property.
It is certainly true that a surjective function onto $\mathbb{R}$ (regardless of the domain) is unbounded. Simply note that for every $M\in\mathbb{R}$, there exists $N\in\mathbb{R}$ with $N\gt M$, $N\gt 0$. Since surjectivity of $f$ implies that for every $y\in\mathbb{R}$ there exists $x$ such that $f(x)=y$, then putting $y=N$ shows that there exists $x$ such that $f(x)=N$, hence $|f(x)|=f(x)\gt M$.
So you are correct in the first one.
The second one, however, is false: unbounded means you can always exceed any given bound. But it does not guarantee that you can always "hit" every value.
Consider the greatest integer function, $f(x)=\lfloor x\rfloor$: this is defined as follows $f(x) = \max\{n\in\mathbb{Z}\mid n\leq x\}$. For example, $f(3.5) = 3$, $f(e) = 2$, $f(-1.5) = -2$, $f(\sqrt{2}) = 1$.
Is $f$ surjective? Is $f$ unbounded?
-
For a simpler example, $f(x)=x^2$ is unbounded but not surjective. – lhf Mar 23 '11 at 16:18
@lhf: Sure; lots of examples (-: The thing I like about this example as opposed to $x^2$ is that the image is full of gaps: no matter what $M$ you pick, there are always values $N,K\gt M$ for which there are $x$ with $f(x)=N$, but $f(x)\neq K$ for all $x$. – Arturo Magidin Mar 23 '11 at 16:24
For the second one it says, an unbounded function that maps from X -> R, doesn't that mean its image must span R? – 1337holiday Mar 23 '11 at 16:29
@1337holiday: No. It just means that the image must be contained in $\mathbb{R}$, and be unbounded as a set. ("Span" is usually reserved for linear maps; there is no linearity assumption here). A "function from $\mathbb{R}$ to $\mathbb{R}$" just means the variable is real and the output is real. – Arturo Magidin Mar 23 '11 at 16:31
@1337holiday: The "contrapositive" of "If P then Q" is "If not Q, then not P". The negation of "If P then Q" is "P and not Q". The contrapositive of "if unbounded, then surjective" would be "if not surjective, then bounded". This is also false (as my example shows: the function is not surjective, but it is not bounded). The negation of "if unbounded, then surjective" would be a function that is both unbounded and not surjective (like my example). A bounded function that is not surjective does not address the statement "if unbounded then surjective" at all. – Arturo Magidin Mar 23 '11 at 16:42
For the first question, let's prove the contrapositive: bounded implies not surjective. So let $f$ be bounded. This means that there is a number $M$ such $|f(x)|\le M$ for all $x$. But then the value $M+1$ is never taken by $f$ and thus $f$ is not surjective.
For the second question, the function $f(x)=x^2$ is unbounded but not surjective.
-
But how is x^2 unbounded when f(x) = -1 does not exist? – 1337holiday Mar 23 '11 at 16:40
@1337holiday: Look at the definition of "unbounded": "For every $M$ there exists an $x$ such that $|f(x)|\gt M$." Is this true for $f(x)=x^2$? Yes: given ANY $M$, there is always a value of $x$ for which $|x^2|$ will be larger than $M$. The fact that it never takes the value $-1$ is completely irrelevant to the concept of boundedness. Graphically: "bounded" means that you can draw two horizontal lines in the plane in such a way that the graph of the function is always between the two lines. Is that true for $y=x^2$? – Arturo Magidin Mar 23 '11 at 16:50
Wow this did it for me, i think i finally understand this! No it is not possible for this function, that is a good way to put it (draw 2 lines). – 1337holiday Mar 23 '11 at 16:54
This statement
• Every unbounded function from R to R is surjective is false. Counter example as lhf points out, any function $f(x)=x^{4n}$ for $n \in \mathbb{N}$.
- | 2015-09-04T06:29:26 | {
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http://mathhelpforum.com/advanced-statistics/23276-euro-2008-soccer-seedings-print.html | # Euro 2008 Soccer Seedings
• November 21st 2007, 08:18 PM
chopet
Euro 2008 Soccer Seedings
Euro 2008 seedings for draw on Dec. 2 in Vienna:
Pot 1: Switzerland (hosts), Austria (hosts), Greece (holders), Netherlands
Pot 2: Croatia, Italy, Czech Republic, Sweden
Pot 3: Germany, Romania, Portugal, Spain
Pot 4: Poland, France, Turkey, Russia
4 groups will be created, with each Pot contributing 1 member to each group.
So Group A may have Switzerland, Croatia, Germany and Poland.
I was wondering about the probability for Netherlands, Italy, Germany and France to be drawn into the same group. Is it:
${1\over(4!)^4}$
• November 23rd 2007, 09:14 AM
TKHunny
Whoa!!! That is WAY too small. There are only $4^{4}$ ways to pick teams. You have selected one such way. There are only four ways to achieve the one way you have selected, that is, they must be selected together in one of the four selection rounds.
Pr(Selected in the first round) = (1/4)^4
The problem with the next step is that not all paths lead to the desired result. If any of the four teams is selected in the first round, they all must be. If any of the four teams is selected in the first round, but any of the others is not, the process ends. The ONLY way to get to round two is for NONE of the teams to be selected in round 1. There are only a few ways to do that.
Pr(None selected in round 1) = (3/4)^4
Thus, Pr(Selected in round 2) = (3/4)^4*(1/3)^4, since round 2 doesn't exist without its dependence on the outcome of round 1.
Using the same logic:
Pr(Selected in round 3) = (3/4)^4*(2/3)^4*(1/2)^4
If we get to round four with hope, obviosuly they will be selected together.
Pr(selected in round 4) = (3/4)^4*(2/3)^4*(1/2)^4*(1)^4
Thus, (3/4)^4 + (3/4)^4*(1/3)^4 + (3/4)^4*(2/3)^4*(1/2)^4 + (3/4)^4*(2/3)^4*(1/2)^4*(1)^4 = 4*(1/4)^4 = 0.015625
• November 23rd 2007, 10:20 AM
Plato
I missed this question when it was first posted.
There is another way to see the answer is $\frac{1}{{\left( {4!} \right)^3 }}$.
Think of the first row of host teams as urns: S, A, G, N.
Then arrange those four letters on each of the three lines to provide one possible selection.
• November 26th 2007, 03:27 AM
chopet
Hi Plato, I was thinking along the same line as you but I missed out fixing the top line.
But the answer is too small. and TKHunny's one is more intuitively correct.
Then I realise we should fix even more than we should.
Let's give each country a label, with Pot 1 labelled A1,A2,A3,A4, and so forth, and we can create a table with the rows representing the pots and the columns representing the groups:
1A 1B 1C 1D
2A 2B 2C 2D
3A 3B 3C 3D
4A 4B 4C 4D
Each row can be arranged in 4! way. There are 4 rows giving: $4!^4$ ways.
But I am interested only in:
x x 1C x
x x 2C x
x x 3C x
x x 4C x
I don't care what happens to all other groups, so I have to divide by $3!^4$ ways.
And the group {1C,2C,3C,4C} can exist in column 1, 2, 3 or 4, I divide by 4.
Finally, the probability is: $4{ {3!^4}\over{4!^4}}={1\over 4^3}=0.015625$
Thanks guys!
• November 27th 2007, 11:10 PM
PHANTOM
Just like Plato said
I agree with the answer Plata wrote.
every POT has 4 members and the way we scatter these 4 members into different groups is : 4!
We have 4 pots, so the number of ways for all combinations is: 4! x 4! x 4! x 4!
that's the number of elements of the sample space.
now lets figure out how many elements of the sample space account for " the 4 specific teams being together". There are 4! ways to arrange the 4 groups.
so P= 4!/(4!)^4= 1/(4!)^3 same as Plato's answer.
• November 30th 2007, 04:05 AM
chopet
Your method doesn't discount the fact that while your 4 countries are in the same group, any 4 OTHER countries can make up the OTHER groups.
e.g.
{France,Germany,Italy,Holland} in one group is what we want.
{Austria,Croatia,Spain,Poland} in another group is actually INDISTINGUISHABLE from
{Switzerland,Croatia,Spain,Poland}
or
{Switzerland, Sweden,Spain,Poland}
or
{Sweden, Czech, Portugal,Poland}
...
I can go on, but I know you have $3!^4$ groups to discount. | 2014-08-21T13:22:18 | {
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https://math.stackexchange.com/questions/3827883/reduced-row-echelon-form-of-an-augmented-matrix-is-not-unique | # Reduced row echelon form of an augmented matrix is not unique
I am given a sysetem of linear equations which after graphing, have no solution (the three lines intersect at different points). Now I am trying to prove this algebraically.
As an augmented matrix,
$$\begin{bmatrix} 1 & -1 & 3 \\ 1 & 1 & 1\\ 2 & 3 & 6\\ \end{bmatrix}$$
• $$R_{1}-R_{2} \Rightarrow R_{2}$$
$$\begin{bmatrix} 1 & -1 & 3 \\ 0 & -2 & 2\\ 2 & 3 & 6\\ \end{bmatrix}$$
continue from here in $$(1)$$ or $$(2)$$
$$(1)$$
• $$R_{1}-\frac{1}{2}R_{3} \Rightarrow R_{3}$$
$$\begin{bmatrix} 1 & -1 & 3 \\ 0 & -2 & 2\\ 0 & -\frac{5}{2} & 0\\ \end{bmatrix}$$
• $$-\frac{1}{2}R_{2} \Rightarrow R_{2}$$
$$\begin{bmatrix} 1 & -1 & 3 \\ 0 & 1 & -1\\ 0 & -\frac{5}{2} & 0\\ \end{bmatrix}$$
• $$\frac{5}{2}R_{2} + R_{3} \Rightarrow R_{3}$$
$$\begin{bmatrix} 1 & -1 & 3 \\ 0 & 1 & -1\\ 0 & 0 & -\frac{5}{2}\\ \end{bmatrix}$$
Then in RREF
$$\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & -1\\ 0 & 0 & -\frac{5}{2}\\ \end{bmatrix}$$
$$(2)$$
• $$-2R_{1}+R_{3} \Rightarrow R_{3}$$
$$\begin{bmatrix} 1 & -1 & 3 \\ 0 & -2 & 2\\ 0 & 5 & 0\\ \end{bmatrix}$$
• $$-\frac{1}{2}R_{2} \Rightarrow R_{2}$$
$$\begin{bmatrix} 1 & -1 & 3 \\ 0 & 1 & -1\\ 0 & 5 & 0\\ \end{bmatrix}$$
• $$-5R_{2} + R_{3} \Rightarrow R_{3}$$
$$\begin{bmatrix} 1 & -1 & 3 \\ 0 & 1 & -1\\ 0 & 0 & 5\\ \end{bmatrix}$$
Then in RREF
$$\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & -1\\ 0 & 0 & 5\\ \end{bmatrix}$$
which is different from the RREF in $$(1)$$
Can someone explain why I end up with a different RREF? I thought all RREF are unique, but clearly not in this case. Of course as mentioned earlier, the system has no solutions and both augmented matrices show this but their RREF's are not unique still.
• I'm not sure if RREF are only unique for coefficient matrices instead of augmented matrices. Or if RREF are only unique for consistent systems Sep 16, 2020 at 0:43 | 2022-05-29T02:02:04 | {
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http://cjvy.hubischule-krone.de/boolean-matrix-operations.html | • # Boolean Matrix Operations
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Improve your skills with free problems in 'Solving Word Problems Using Matrix Operations' and thousands of other practice lessons. Keywords Boolean matrix decomposition · Boolean rank · Efficient algorithm ·Educational database 1 Introduction Matrixdecomposition,a. Boolean Algebra. for i = 1:22 plot(C1(i, :), 'color', 'b'), hold on end Not sure how I can exclude the first column. Instead of a single index, we can use two indexes, one representing a row and the second representing a column. Quote: A man has integrity if his interest in the good of the service is at all times greater than his personal pride, and when he holds himself to the same line of duty when unobserved as he would follow if his superiors were present General S. We will now work our way through the table of identities, in order, making observations about each, usually including a "common sense" informal proof. R's binary and logical operators will look very familiar to programmers. Operator overloading allows C/C++ operators to have user-defined meanings on user-defined types (classes). Akin to other matrix factorizations, the factor matrices can be used. First, we define two characteristic matrices of a covering. Computing arXiv:1803. Generally, at least one is a variable. For example: +, -, *, / etc. 9% that fail. T = true(sz) is an array of logical ones where the size vector, sz, defines size(T). array_equal (a1, a2) True if two arrays have the same shape and elements, False otherwise. The original application for Boolean operations was mathematical logic, where it combines the truth values, true or false, of individual formulas. The order in which these actions occur is unimportant as it does not affect the final result. Matrix Operations a la Shmoop you look at Gilligan Huffington III CEO of Handbags. Definition 3. A Boolean matrix multiplication is the usual matrix multiplication with Boolean operations: 0 + 0 = 0 , 1. 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By boolean, or binary matrix, we mean a matrix whose entries are from the field with two elements. It is built deeply into the R language. Oct 17, 2014 · Boolean and relational operators in Matlab 1. For complex matrices, computes the complex conjugate (Hermitian) transpose. A Boolean matrix can be expressed as a product of two Boolean matrices, where the first matrix represents a set of meaningful. These basis states are analogous to the orthonormal unit vectors in Euclidean space. R has many operators to carry out different mathematical and logical operations. Like comparison operations, each element of an element-by-element boolean expression also has a numeric value (1 if true, 0 if false) that comes into play if the result of the boolean expression is stored in a variable, or. Visual and Graphical language unlike textual high-level, such as C, C++, Java…. An identity matrix is a square matrix with ones along the diagonal and zeros elsewhere. BOOLEAN RANK Matrix rank. Operator overloading allows C/C++ operators to have user-defined meanings on user-defined types (classes). Find the result of a multiplication of two given matrices. *) and exponentiation (. The values returned by MATLAB logical operators and functions, with the exception of bit−wise functions, are of type logical and are suitable for use with logical indexing. equations via matrix operations. 06 for µVision® armasm User GuideVersion 5Home > Symbols, Literals, Expressions, and Operators > Addition, subtraction, and logical operators 7. Level 1 BLAS do vector-vector operations, Level 2 BLAS do matrix-vector operations, and Level 3 BLAS do matrix-matrix operations. A Boolean object combines two or more other objects by performing a Boolean operation or operations on them. The complexity of subtraction operation is O(m*n) where m*n is order of matrices; Matrices Multiplication - The multiplication of two matrices A m*n and B n*p gives a matrix C m*p. Relational operators perform element-by-element comparisons between two arrays. When an expression includes several operators, it is called a compound expression. On this page, we will discuss these type of operations. rank Computes rank of a matrix. We have a symbology for denoting Boolean variables, and their complements. Thus (as in Stata) logical operators return one for true and zero for false. Instead of a single index, we can use two indexes, one representing a row and the second representing a column. If none are found, find returns an empty, 0-by-1 matrix. How to Contact The MathWorks: www. The idea: I am making quantization intervals with hQInt. Know miscellaneous operations on arrays, such as finding the mean or max ( array. Think of it as a replacement for multiple simple formulas or as a shorthand in which a single formula is provided will all the information required to carry out a complex operation. Boolean logic. In Map Algebra, operators apply a mathematical operation on input rasters and numbers. Note that BMM can be computed using an algorithm for integer matrix multiplication, and so we have BMM for n !nmatrices is in O(n ) time, where !<2:373 (the current bound for integer matrix multiplication). ) and functions like any, all, isnan, isinf , and isfinite. ☞This page belongs to resource collections on Logic and Inquiry. Logical operators in MATLAB are those that link logical statements together and return true (1) or false (0) in a logical matrix depending upon the nature of the logical operator and the value of the components. Planning a new project, representing some algorithm or some process, illustrating a solution to a given problem, representing process operations, analysing, designing, documenting, managing a process in various fields is always better to do in a way of flowchart and the types of flowcharts and diagrams are numerous. LFA, the Logical Framework Approach, is an instrument for objective-oriented planning of projects. A one-dimensional array corresponds to a vector, while a two-dimensional array corresponds to a matrix. Unary operators take an operand on the right. Consider the expression A + B * 5. The new object is a Compound Object of type Boolean. Operators and Elementary Operations Arithmetic, relational, and logical operators, special characters, rounding, set functions The MATLAB ® language uses many common operators and special characters that you can use to perform simple operations on arrays of any type. You can use those logical values to index into an array or execute conditional code. GHELASE Daniela 2 1„Politehnica” University of Bucharest 2„Dunarea de Jos” University of Galati ABSTRACT The AutoCAD software is a power computer-aided design (CAD) system that can offers all users of graphics, 2D and 3D objects representation. SAS Operators in Expressions. Along with the standard unary and binary operators of conventional alge-braic notation,AMPL provides iterated operators likesumand prod, and a conditional (if-then-else) operator that chooses between two expressions, depending on the truth. We indicate which entries will be changed by performing an indexing operation on the left hand side and then specify the new values on the right hand side. The original application for Boolean operations was mathematical logic, where it combines the truth values, true or false, of individual formulas. Program (1): To perform addition, subtraction, multiplication, right division, left division and exponentiation operations on x and y given as x = 2; y = 3, in MATLAB. Logical operation definition, Boolean operation. However, it is. We can do operations such as addition and multiplication on the matrix in R. To use an operator with a raster, the raster must be a Raster object. The period character (. The function of Definition 7 is applied to the cell matrix B m × n. Boolean operations with non-manifold objects or objects that are not water-tight will fail when non-manifold parts interact in the boolean operation. Input: The first line of input contains an integer T denoting the number of test cases. The single quote character may also be used to delimit strings, but it is better to use the double quote character, since that is never ambiguous. It is an algorithm for mining multidimensional association rules from relational databases. Note that BMM can be computed using an algorithm for integer matrix multiplication, and so we have BMM for n !nmatrices is in O(n ) time, where !<2:373 (the current bound for integer matrix multiplication). These variables can be used to define matrices of booleans, with the usual syntax. The PowerShell logical operators evaluate only the statements required to determine the truth value of the statement. Array formulas can be understood as a combination of Array Constants, Array Operators and Array Ranges. Matrix Arithmetic cross Computes cross products. The implementation of left-shift and right-shift operators is significantly different on Windows for ARM devices. Chuang Department of Physics, Massachusetts Institute of Technology, Cambridge, Massachusetts 02139, USA (Received 30 January 2010; published 4 May 2010) Entanglement, as studied in quantum information science, and nonlocal quantum correlations, as studied. max(), array. Animesh Raj has 8 jobs listed on their profile. But, for the beginner, they seem to often mysteriously fail, and it is difficult to understand why. That is, they operate on numbers (normally), but instead of treating that number as if it were a single value, they treat it as if it were a string of bits, written in twos-complement binary. A token is the smallest element in a program that is meaningful to the compiler. A boolean variable is %T (for "true") or %F (for "false"). Some of them can come in pretty handy, though, if you need to flip. Basically, it returns the opposite Boolean value of evaluating its operand. Relational and Logical Operators. 4 - GLSL Operators (Mathematical and Logical)¶ GLSL is designed for efficient vector and matrix processing. This class contains various functions (methods) that operate on Boolean matrices. eq (a, b) ¶. Basic operators in Scilab Scilab is capable of simple mathematical calculation as well as complex calculations. , the product of two n nmatrices can be computed in O(n3 ) additions and multiplications over the field. In other words, 235 is the addition of 128+64+32+8+2+1. Its members are TRUE if the corresponding members in the original vector are to be included in the slice, and FALSE if otherwise. If you have never used logical operators in Matlab, this is a MUST-READ!. For matrices, there are three basic row operations; that is, there are three procedures that you can do with the rows of a matrix. They take each bit in one operand and perform the operation with the corresponding bit in the other operand. - Proven ability to motivate, encourage and support direct managers in building high performance teams across a regional/international cross-functional matrix environment. matrix into a Boolean embedding matrix under orthogonal or near-orthogonal rotations. It evaluates to a single logical value. Instead of elementary algebra where the values of the variables are numbers, and the prime operations are addition and multiplication, the main operations of Boolean algebra are the conjunction and denoted as ∧, the disjunction or denoted as ∨, and the negation not denoted as ¬. R's binary and logical operators will look very familiar to programmers. For !, a logical or raw vector(for raw x) of the same length as x: names, dims and dimnames are copied from x, and all other. Package bitops has similar functions for numeric vectors which differ in the way they treat integers 2^31 or larger. "Operations" is mathematician-ese for "procedures". OF Boolean operator allows the user to specify how many terms from a list of terms must be present if it is not necessary to have all terms in the list. The basic laws used in Boolean algebra are commutative law, associate law, distributive law, identity law and redundance law. 4 Binary factorization is different from Boolean factorization in the sense. Both x and y have to be true in order for the solution to be true. True or false (Boolean) conditions. Jun 18, 2018 · So is there a way to extract the values into a matrix using logical operators? Using reshape or whatever to reshape the vector into matrix is forbidden in my case. These variables can be used to define matrices of booleans, with the usual syntax. You can think of an r x c matrix as a set of r row vectors, each having c elements; or you can think of it as a set of c column vectors, each having r elements. We will learn how to use relational operators and logical operators. Then X(L) specifies the elements of X where the elements of L are nonzero. is the create matrix icon, which upon click will bring up a dialog box that allow us to create a matrix with specified number of rows and columns. Boolean Operators You use logical operators in conditional expressions much as you use math operators in numeric expressions. a boolean matrix A = (aij) by an addition circuit actually means to “encode” the matrix A by paths in a directed acyclic graph. unless both parts are. LOGICAL Operators and Expressions. unique, which is useful if you need to generate unique elements, given a vector containing duplicated character strings. For example: +, -, *, / etc. dot Computes dot products. The logical data type represents true or false states using the numbers 1 and 0, respectively. Consider the expression A + B * 5. You drag and drop the empty Array on the Front Panel, next you find a Control or Indicator (Numeric, String, Boolean, etc,) and drag it into the empty Array. In this article I have discussed briefly about almost all arithmetic and logical instructions of 8086 microprocessor. In Boolean algebra you represent the logical values true and false by the numbers 1 and 0 respectively. The Boolean rank of an n-by-m binary matrix A is the least integer k such that there exists n-by-k binary matrix B and k-by-m binary matrix C for which A = B C. Matrix transpose operator. Array formulas (committed with CTRL+SHIFT+ENTER) have one restriction: You canít use Excel's logical operations AND, OR, etc. A system of m Boolean equations in n variables can be stated in matrix form with the usual matrix operations such as Ax = b and Rw = r, where x ∈ B n and w ∈ B 2 n are Boolean vector variables. 2012 While exploring the Julia manual recently, I realized that it might be helpful to put the basic vocabularies of Julia and R side-by-side for easy comparison. The Logical Operator block performs the specified logical operation on its inputs. All that is required is to extend this to the rest of the possible GPA's - the full code is in the section below, along with an image of the graph it will create. boolean operation synonyms, boolean operation pronunciation, boolean operation translation, English dictionary definition of boolean. com Product enhancement suggestions. This is a simple two minute mini-crossword puzzle designed to reinforce the nuanced Boolean Operators the lecture introduced: Near, Next, Proximity. power consuming than that of Boolean matrix. Inventory Management. Key to understanding the use of matrix operations is the concept of the matrix (array) formula. The classes "octmode" and "hexmode" whose implementation of the standard logical operators is based on these functions. Maintaining track of software licenses and renewal of antivirus, Inventory Management. These are the available functions for logical operators. Sep 01, 2017 · Free Online Library: Analysis of Reliability for the Gate Level Fault Tolerant Design using Probabilistic Transfer Matrix method. boolean operation synonyms, boolean operation pronunciation, boolean operation translation, English dictionary definition of boolean. Definition 3. end blocks). Matrix Algebra. These are the 4 basic boolean operations (AND, OR, XOR and NOT). It does this using make. The Arduino Reference text is licensed under a Creative Commons Attribution-Share Alike 3. All six of the standard comparison operations are available:. A function defined on Boolean matrices which depends on the elements of the matrix in a manner analogous to the manner in which an ordinary determinant depends on the elements of an ordinary matrix, with the operation of multiplication replaced by intersection and the operation of addition replaced by union. National Entrepreneurs' Day: 5 Ways You Can Avoid AI Startup Failure. Sida, like many other donor agencies, has decided to use, and to encour-. 4 Closures of Relations Ch 9. The logical operators and, or, nand, nor, xor, xnor and not are defined for BIT and BOOLEAN types, as well as for one-dimensional arrays containing the elements of BIT and BOOLEAN. com is now LinkedIn Learning! To access Lynda. IfA is the adjacency matrix of a random graph G(n,p), the entries in its kth power gives the number of walks of lengthk between each pair of vertices [4]. In Map Algebra, operators apply a mathematical operation on input rasters and numbers. However, a matrix value used as the condition in an if or while statement is only true if all of its elements are nonzero. Logical operation definition, Boolean operation. Most of the methods on this website actually describe the programming of matrices. The original application for Boolean operations was mathematical logic, where it combines the truth values, true or false, of individual formulas. The example below will show you how to use arithmetic operators in MATLAB. Free Online Library: Automated Boolean Matrix data representation scheme through AVL tree for mining association rules. Logical operators are almost always found in the context of Conditional and Loop Structures. com Web comp. A single scalar can be compared against each element in an array. National Entrepreneurs' Day: 5 Ways You Can Avoid AI Startup Failure. In this paper, an association rule mining algorithm base d on th e Boolean matrix (ABBM) is proposed. The result is logical 0 (FALSE) where and are both zero or nonzero. Boolean product of Equation (1a), noting that message passing derivation for factorization and completion using the XOR product of Equation (1b) is similar. Matrix Operations a la Shmoop you look at Gilligan Huffington III CEO of Handbags. Assignment operations, in which we change a value or values in a matrix, are performed in a very similar way to the indexing operations above. IDL has two operators used to multiply arrays and matrices. The Logical Operator block performs the specified logical operation on its inputs. Program (1): To perform addition, subtraction, multiplication, right division, left division and exponentiation operations on x and y given as x = 2; y = 3, in MATLAB. However, Mata does not have a boolean variable type. The expression value is TRUE (1), if the values of x and y are true (not null). I ask everyone, however, to be careful and not to write Subject lines like "matrix operations in STATA and MATA do not match!" unless they are certain they are right. We will learn how to use the if-statement, which is the most important method of selection. Does Matlab have a Boolean (sometimes called logical or binary) matrix multiplication function? I'm specifically talking about what's usually denoted by a circle with a dot in it to denote Boolean. Matrix Multiplication Calculator (Solver) This on-line calculator will help you calculate the __product of two matrices__. You will learn how to use logical operations to search large data matrix very quickly. Boolean Operators include AND, OR, XOR, or NOT and can have one of two values, true or false. NET Framework does not come with built-in support for matrix math,. IfA is the adjacency matrix of a random graph G(n,p), the entries in its kth power gives the number of walks of lengthk between each pair of vertices [4]. In order to define the specific function, relation, and symbols in question it is first necessary to establish a few ideas about the connections among them. Lecture 2 MATLAB basics and Matrix Operations page 11 of 19 Matrix operations: MATLAB is short for MATrix LABoratory, and is designed to be a tool for quick and easy manipulation of matrix forms of data. Boolean matrices is to treat them as integer matrices, and apply a fast matrix multiplication algorithm over the integers. | 2020-01-26T05:46:34 | {
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https://math.stackexchange.com/questions/3735061/check-this-proof-if-two-columns-rows-of-a-matrix-are-the-same-the-determinant | # Check this proof: If two columns/rows of a matrix are the same, the determinant is $0$.
I have written this proof stating that if two rows or columns of a matrix are the same, then the determinant of the matrix is equal to 0. Is it correct?
Let us say we have an n x n matrix A, shown below:
For some $$i,n \in \mathbb{N}$$.
If we say that $$r_{i} = [a_{i1}, a_{i2}, a_{i3} ... a_{in}]$$, a row vector, then we can rewrite the matrix A as:
If we then also create the swapped $$S_{ij}$$ matrix, i.e. swap rows $$i$$ and $$j$$ around, we have the matrix:
We know that if we swap two rows of a determinant, in this case rows $$i$$ and $$j$$, then the determinant will simply be the negative of the original determinant.We can say that the det(A) = -det(S). But, if these two rows are identical, then det(S) = det(A), so this means that det(A) = -det(A), so A must be equal to $$0$$.
• Yes, your solution is correct. Jun 26 '20 at 8:12
• It's easier to use that $\det$ is an alternating multilinear form, i.e., $$\det(a_1,\dots,a_i,\dots,a_j\dots,a_n)=-\det(a_1,\dots,a_j,\dots,a_i\dots,a_n).$$ Jun 26 '20 at 9:18
That seems to be correct. Another way to think about this is to consider what happens when there are two identical columns/rows. Then, the number of linearly independent columns/rows is less than $$n$$ for a given $$n \times n$$ matrix. Hence, the matrix does not have maximal rank so the determinant has to be 0.
Of course, I'm working off of the idea that you've defined the determinant as the unique map $$det: M(n \times n, \mathbb{F}) \to \mathbb{F}$$ such that:
1. $$det$$ is linear in each row.
2. If $$rank(A) < n$$, then $$det(A) = 0$$.
3. $$det(E_n) = 1$$.
If you're not working off of these assumptions, that's fine. What you have written is fine too.
In essence, yes, your proof is very correct, and thorough.
For further clarification you can see this video https://www.khanacademy.org/math/linear-algebra/matrix-transformations/determinant-depth/v/linear-algebra-duplicate-row-determinant which gives a very thorough breakdown of your proof.
Yes, this is indeed a valid proof which you can see on occasion in published works.
If you have two (or more) rows which are linearly related, you can resolve some rows to zero. (Check out gaussian elimination to see how row reduction works.) By the way this proof is valid indeed. | 2021-09-22T15:21:49 | {
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https://www.tutorialspoint.com/explain-binary-search-in-python | # Explain Binary Search in Python
PythonServer Side ProgrammingProgramming
Binary search is a searching algorithm which is used to search an element from a sorted array. It cannot be used to search from an unsorted array. Binary search is an efficient algorithm and is better than linear search in terms of time complexity.
The time complexity of linear search is O(n). Whereas the time complexity of binary search is O(log n). Hence, binary search is efficient and faster-searching algorithm but can be used only for searching from a sorted array.
## How does Binary Search work?
The basic idea behind binary search is that instead of comparing the required element with all the elements of the array, we will compare the required element with the middle element of the array. If this turns out to be the element we are looking for, we are done with the search successfully. Else, if the element we are looking for is less than the middle element, it is sure that the element lies in the first or left half of the array, since the array is sorted. Similarly, if the element we are looking for is greater than the middle element, it is sure that the element lies in the second half of the array.
Thus, Binary search continuously reduces the array into half. The above process is recursively applied on the selected half of the array until we find the element we are looking for.
We will start searching with the left index 0 and right index equal to the last index of the array. The middle element index (mid)is calculated which is the sum of the left and right index divided by 2. If the required element is less than the middle element, then the right index is changed to mid-1, which means we will now be looking at the first half of the array only. Likewise, if the required element is greater than the middle element, then the left index is changed to mid+1, which means we will now be looking at the second half of the array only. We will repeat the above process for the selected array half.
## How do we know if element is not present in the array?
We need to have some condition to stop searching further which will indicate that the element is not present in the array. We will iteratively search for the element in the array as long as the left index is less than or equal to the right index. Once this condition turns false and we haven’t found the element yet, this means that the element is not present in the array.
### Example
Let us take the following sorted array and we need to search element 6.
2 5 6 8 10 11 13 15 16
L=0 H=8 Mid=4
2 5 6 8 10 11 13 15 16
6<10, therefore take the first half.
H=Mid-1
L=0 H=3 Mid=1
2 5 6 8 10 11 13 15 16
6>5, therefore choose the second half.
L=Mid+1
L=2 H=3 Mid=2
2 5 6 8 10 11 13 15 16
6==6, an element found
Hence the element 6 is found at index 2.
## Implementation
From a given sorted array, search for a required element and print its index if the element is present in the array. If the element is not present, print -1.
The code for the implementation of binary search is given below.
### Example
Live Demo
def binary_search(arr,x):
l=0
r=len(arr)-1
while(l<=r):
mid=(l+r)//2
if(arr[mid]==x):
return mid
elif(x<arr[mid]):
r=mid-1
elif(x>arr[mid]):
l=mid+1
return -1
array=[1,2,3,4,5,6,7,8,9,10]
a=7
print(binary_search(array,a))
b=15
print(binary_search(array,b))
### Output
6
-1
Element 7 is present at index 6.
Element 15 is not present in the array, hence -1 is printed.
Published on 11-Mar-2021 09:11:09 | 2021-12-02T00:49:47 | {
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https://math.stackexchange.com/questions/1025153/sequence-that-converges-to-0-but-its-function-diverges | # Sequence that converges to 0 but its function diverges
Let f(x)=-2x+1 if x<0 and f(x)=$x^2$+x if x>0. Give a sequence {$x_n$} in $\mathbb R$\ {0} such that {$x_n$} converges to zero but {f($x_n$)} diverges.
I've tried with the sequence {1/n} but it does not seem to work. I can't think of another sequence that would converge to zero. Can anyone help me out please?
Thanks.
• I am a little confused by what the question means by the function at x<0 and x>0. So the function of the sequence must diverge while the sequence converges. – Su003 Nov 17 '14 at 0:36
• If x is negative than the function value is computed by the first expression while if x is positive then the function value is computed by the second expression. For example, f(-1)=3 while f(1)=2 – JB King Nov 17 '14 at 0:42
• But x is a term of the sequence {$x_n$} right? – Su003 Nov 17 '14 at 0:53
• $x$ is just a variable while $x_n$ would be a subscripted variable. – JB King Nov 17 '14 at 0:59
How about $x_n=\frac{(-1)^n}{n}$ for a sequence that converges to zero but alternates in sign which would make the function values be different. Thus, the sequence here is: $-1,\frac{1}{2},-\frac{1}{3},\frac{1}{4},-\frac{1}{5},\cdots$
For odd $n$, the function values will converge to 1. The first few function values here would be $3,\frac{5}{3},\frac{7}{5}$ as the general term will be $\frac{n+2}{n}$ which could also be seen as $1+\frac{2}{n}$
For even $n$, the function values would be $\frac{1}{n^2}+\frac{1}{n}=\frac{n+1}{n^2}$ which would have function values of $\frac{3}{4},\frac{5}{16},\frac{7}{36},...\frac{11}{100}, .... \frac{101}{10,000},...\frac{1001}{1,000,000}$ which will likely converge to 0 from this side as the denominator is growing much faster than the numerator.
Putting these together, the function value sequence would look like this: $3,\frac{3}{4},\frac{5}{3},\frac{5}{16},\frac{7}{5},\frac{7}{36},\frac{9}{7},\frac{9}{64}..., \frac{2k+1}{2k-1},\frac{2k+1}{(2k)^2},\frac{2k+3}{2k+1},\frac{2k+3}{(2k+2)^2}...$ which doesn't converge since there are sub-sequences which converge to different values.
If you want, consider $\varepsilon=\frac{1}{10}$ and try to prove the function value sequence converge,i.e. there exists $L,N$ such that for all $n>N\implies|f(n)-L|<\varepsilon$.
If you believe they converge to 0, consider some big odd $n>N$ where the function value will be greater than 1 which doesn't work.
If you believe they converge to 1, consider some big even $n$ where the function value will be close to zero and more than .1 away from 1 as for $n=1000$ then the function value is $\frac{1001}{1,000,000}=.001001$. Any bigger even $n$ will have even smaller values to consider.
If you have an alternative proof of what is wrong with my above answers, show me the error.
• Does {f($x_n$)} diverge in this case? I tried it but it looks like it converges to 1. – Su003 Nov 17 '14 at 0:31
• @Su003, even terms will converge to $0$, odd terms to $1$. – lhf Nov 17 '14 at 0:33
• I understand this. But what is confusing me is that the question says that {f($x_n$)} must diverge. Does such an {$x_n$} work in this case? – Su003 Nov 17 '14 at 1:37
The main point here is that $f$ is not continuous at $x=0$. It does have one-sided limits at $x=0$, but they are different. So, you need to find a sequence that converges to $0$ but visits both sides infinitely often. The answer by JB King uses the simplest possible such sequence. | 2019-09-21T09:48:49 | {
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https://mymathforum.com/threads/how-to-find-the-maximum-angular-speed-from-a-bar-which-rotates-about-an-axis.347797/ | # How to find the maximum angular speed from a bar which rotates about an axis?
#### Chemist116
The problem is as follows:
A rigid bar of negligible mass has three particles whose masses are the same and are joined to the bar as indicated in the figure. The bar is free to rotate in a vertical plane about an axis with no friction perpendicular to the bar through point $P$ and is released from rest on the horizontal position at $t=0$. Find the maximum angular speed in radians per second attained by the bar. You may consider $g=9.8\frac{m}{s^2}$ and $d=4.2\pi$
The alternatives given are:
$\begin{array}{ll} 1.&1\,\frac{rad}{s}\\ 2.&2\,\frac{rad}{s}\\ 3.&3\,\frac{rad}{s}\\ 4.&4\,\frac{rad}{s}\\ \end{array}$
What I've attempted to solve this problem was to equate the potential energy and the rotational energy of the masses assuming there's a superposition of the moment of inertia of the masses.
This is translated as:
$mgh=\frac{1}{2}I\omega^2$
$mgh=\left(\frac{1}{2}m\left(\frac{4d}{3}\right)^{2}+\frac{1}{2}m\left(\frac{d}{3}\right)^{2}+\frac{1}{2}m\left(\frac{2d}{3}\right)^{2}\right) \times \omega^2$
But after following the procedure I can't find a way to cancel the 4.2$\pi$. Supposedly, the answer is the first choice. How can I get to that result?
#### skeeter
Math Team
$\omega_{max}$ will occur when the bar is at the vertical ...
$U_{g0}-U_{gf} = K_{Rf}-K_{R0}$
$mg \cdot \dfrac{4d}{3} + mg \cdot \dfrac{d}{3} - mg \cdot \dfrac{2d}{3} = mgd = \dfrac{1}{2} \cdot I \cdot \omega^2$
$I = m \cdot \dfrac{16d^2}{9} + m \cdot \dfrac{d^2}{9} + m \cdot \dfrac{4d^2}{9} = m \cdot \dfrac{7d^2}{3}$
$mgd = m \cdot \dfrac{7d^2}{6} \cdot \omega^2 \implies \omega = \sqrt{\dfrac{6g}{7d}}$
using the given value for $d$, which I suspect is mistaken, the result is close to 1 rad/sec
#### Chemist116
$\omega_{max}$ will occur when the bar is at the vertical ...
$U_{g0}-U_{gf} = K_{Rf}-K_{R0}$
$mg \cdot \dfrac{4d}{3} + mg \cdot \dfrac{d}{3} - mg \cdot \dfrac{2d}{3} = mgd = \dfrac{1}{2} \cdot I \cdot \omega^2$
$I = m \cdot \dfrac{16d^2}{9} + m \cdot \dfrac{d^2}{9} + m \cdot \dfrac{4d^2}{9} = m \cdot \dfrac{7d^2}{3}$
$mgd = m \cdot \dfrac{7d^2}{6} \cdot \omega^2 \implies \omega = \sqrt{\dfrac{6g}{7d}}$
using the given value for $d$, which I suspect is mistaken, the result is close to 1 rad/sec
Can you include some drawing to accompany your solution? I'm having a problem at identifying where's the height to establish the potential energy. When you rotate the bar about the axis which is represented by the orange dot, the bar is vertical right? But from where to where you put the height?
Why Initially do we have
$mg \cdot \dfrac{4d}{3} + mg \cdot \dfrac{d}{3}$
and why finally do we have:
$mg \cdot \dfrac{2d}{3}$
@skeeter Can you please represent the justification of this in a drawing?
I got that part of adding the moment of inertia due the principle of superposition of them and the rest is just logical, but...
$\omega = \sqrt{\dfrac{6g}{7d}}$
in this part is where I'm stuck. How on earth did you get an answer close to $1\frac{rad}{s}$?
If I do insert the given values, this would become into:
$\omega = \sqrt{\dfrac{6\times 9.8}{7\times 4.2 \times 3.14}} \approx 0.7980 \frac{rad}{s}$
Can you please explain this part? Or did I misunderstand something from your explanation? Any help please?
#### skeeter
Math Team
0.798 is closest to 1rad/sec, given the available choices ... note that I stated disagreement with your given value for $d = 4.2\pi$
For $\omega = 1 \text{\rad/sec}$, $d$ would need to equal to 8.4
Full explanation of the method used to find the maximum angular velocity is given in the following diagram
topsquark | 2020-02-17T17:08:26 | {
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https://math.stackexchange.com/questions/3334891/series-about-inserting-parentheses | Problem : Suppose that the series $$\sum_{}^{} a_k$$ converges and $${n_j}$$ is a strictly increasing sequence of positive integers. Define the sequence $$b_k$$ as follows :
$$b_1=a_1+...+a_{n_{1}}$$
$$b_2=a_{n_1+1}+...+a_{n_{2}}$$
...
$$b_k=a_{n_{k-1}+1}+...+a_{n_{k}}$$
Prove that $$\sum_{}^{} b_k$$ converges and that $$\sum_{}^{} a_k=$$ $$\sum_{}^{} b_k$$
My Proof :
Let $$s_m=\sum_{i=1}^{n_m} a_i$$ $$\forall m \in \mathbb{N}$$ then by definition, $$s_m=\sum_{i=1}^{n_m} a_i=\sum_{k=1}^{m} b_k$$
Since $$n_j$$ is strictly increasing sequence of positive integers, $$\lim_{m\to\infty}\ n_m= \infty$$
Since $$\lim_{m\to\infty}\ n_m= \infty$$ and $$\sum_{}^{} a_k$$ converges, $$\lim_{m\to\infty}\ s_m$$ exists and
$$\lim_{m\to\infty}\ s_m=$$ $$\lim_{m\to\infty}\sum_{i=1}^{n_m} a_i=$$ $$\sum_{}^{} a_k$$
Since $$\forall m \in \mathbb{N}$$ $$s_m=\sum_{k=1}^{m} b_k$$ and $$\lim_{m\to\infty}\ s_m$$ exists,
$$\sum_{}^{} b_k$$ converges and $$\sum_{}^{} a_k$$=$$\sum_{}^{} b_k$$
But, I'm not sure my proof is correct. Please give me feedback on my proof.
Yes, the idea of your proof is fine. However, your argument in the 2nd and 3rd line is unclear (among other things since $$n_j$$ is not defined). I suppose, in short your argument is:
Define $$s_n := \sum_{k = 0}^n a_k$$ and $$s_m := \sum_{k = 0}^m b_k$$. Then $$(s_m)_{m \in \mathbb{N}}$$ is a subsequence of $$(s_n)_{n \in \mathbb{N}}$$. Since $$\sum a_k$$ converges (by definition) $$(s_n)_{n \in \mathbb{N}}$$ converges, so does $$(s_m)_{m \in \mathbb{N}}$$ and thus $$\sum b_k$$ converges. Since limits are unique, they coincide. | 2021-06-12T11:24:06 | {
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https://math.stackexchange.com/questions/3168820/combinatorial-proof-of-binomnk2-k-binomn2n2-binomk2/3168848 | # Combinatorial proof of $\binom{nk}{2}=k\binom{n}{2}+n^2\binom{k}{2}$
This identity was posted a while back but the question had been closed; the question wasn't asked elaborately, though the proof of the identity is a nice application of combinatorics and a good example for future reference.
Also, there was just a flat-out wrong answer which had a couple of upvotes, so I thought to give my own possible proof and sort of 'reopen' the question that was left. Hopefully this time the question will have enough context to stay alive.
As a side note: the given right side does not appear symmetric between $$k$$ and $$n$$. However, when $$\binom{m}{2}=m(m-1)/2$$ is inserted and the products expanded, only symmetric terms remain uncancelled. Rendering the right side as $$k^2\binom{n}{2}+n\binom{k}{2}$$ would give the same cancellations and the same net value.
Proof
Suppose you have a grid of $$n\times k$$ dots. Firstly, the amount of ways to connect any two dots is $$\binom{nk}{2}$$. Now consider the right-hand side. We can split the cases for which the connected dots are on the same column, row, or are in both different columns and rows.
If the two connected dots are in the same column, we can choose two points from any two different rows in $$\binom{n}{2}$$ ways, and we have $$k$$ columns for which the two points can be on the same column, which gives a total of $$k\binom{n}{2}$$ options. The same argument holds for constant rows: $$n\binom{k}{2}$$.
Now if neither the row nor the column can stay constant, we can pick any point in $$nk$$ ways, and choose the second point from the remaining $$(n-1)(k-1)$$ points; one column and one row will be unavailable. This gives us $$\frac{nk(n-1)(k-1)}{2}$$ options, as we have to rule out the double counting.
We will now show (algebraically) that $$n\binom{k}{2}+\frac{nk(n-1)(k-1)}{2}=n^2\binom{k}{2}$$. We have that $$\binom{k}{2}=\frac{k(k-1)}{2} =\frac{nk(n-1)(k-1)}{2n(n-1)} \iff n(n-1)\binom{k}{2}=\frac{nk(n-1)(k-1)}{2}=n^2\binom{k}{2}-n\binom{k}{2}$$ which leads to the equation above.
Combining these cases gives $$\binom{nk}{2}=k\binom{n}{2}+n\binom{k}{2}+\frac{nk(n-1)(k-1)}{2} = k\binom{n}{2}+n^2\binom{k}{2}$$
If there are any mistakes or improvements on the arguments, please feel free to point them out.
• Uh ... How is the left side symmetric when $n$ and $k$ are interchanged and the right side isn't? – Oscar Lanzi Mar 30 '19 at 22:28
• @OscarLanzi The right side is symmetric too, just not obviously so. – Michael Biro Mar 30 '19 at 22:32
• I have now explained it. Feel free to roll back the edit if you feel it is inappropriate. – Oscar Lanzi Mar 30 '19 at 23:11
I don't think you need as many cases, which saves a little algebra. We have $$k$$ groups of $$n$$ dots each, so choosing two of them can be done in $$\binom{nk}{2}$$ ways.
Alternatively, both are in the same group of $$n$$ which has $$\binom{k}{1} \cdot \binom{n}{2}$$ possibilities, or they are in different groups of $$n$$, which has $$\binom{k}{2} \binom{n}{1}^2$$ possibilities.
Therefore, $$\binom{nk}{2} = k \binom{n}{2} + n^2 \binom{k}{2}$$ | 2020-12-02T16:46:40 | {
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https://math.stackexchange.com/questions/2856173/is-there-any-analysis-method-of-non-differentiability | # Is there any analysis method of non-differentiability?
I am not sure if this question is proper, if not please let me know I would delete it.
I understand a function can be continuous but not differentiable. I can also understand existence of non-differentiable functions. The question: is there a 'measure' (like distance) between two non-differnetiable functions. The word 'measure' here should not be taken literally. Consider two non-differentiable function $f(x)$ and $g(x)$ the question: is there a way to compare the two functions (like a distance measure ) which tells either $f(x)$ or $g(x)$ is easier to make it differentiable? Something like a how far the function is away from differentiable?
• I think this is a great question, and I hope it attracts lots of responses - it's unfortunate that you accepted an answer so quickly. Although you didn't want measure to be taken literally, actually, that does provide a bit of information: the sets on which functions are not differentiable can be compared by seeing which has larger measure. Functions which are differentiable almost everywhere can either have or lack more 'completely differentiable-like' properties, such as obeying the fundamental theorem of calculus - this turns out to be equivalent to absolute continuity. – forget this Jul 19 '18 at 2:13
• @CoryGriffith Does accepting answers reduce attention? – Creator Jul 19 '18 at 2:15
• I've seen comments to that effect on other answers, but I don't have any way of knowing myself. It seems plausible, though. – forget this Jul 19 '18 at 2:20
• @CoryGriffith Ok I will wait for few days and see what happens. thanks – Creator Jul 19 '18 at 2:22
There is Rademacher's theorem which says that a function that is Lipschitz continuous is differentiable (almost everywhere). So one way of interpreting your question is to "how close to Lipschitz continuous is a function". Lipschitz continuity is a specific case of Hölder continuity.
A function is $\alpha$-Hölder continuous for $\alpha\in (0,1]$ if
$$|f(x)-f(y)|<K|x-y|^\alpha$$
If $\alpha=1$ then we say the function is Lipschitz.
Note that if $f$ is $\alpha$-Hölder, then it is $\beta$-Hölder for $\beta<\alpha$.
So if $f$ is $.6$-Hölder and $g$ is $.7$-Hölder, then $g$ is "closer" to being Lipschitz, i.e. differentiable.
Here are some pictures of various functions that are of increasing Hölder continuity:
Hölder continuous for all $\alpha \in(0,0.15)$:
Hölder continuous for all $\alpha \in (0,0.55)$:
Hölder continuous for all $\alpha\in (0,0.75)$:
Hölder continuous for all $\alpha\in (0,0.95)$:
Note how the paths get "nicer"/closer to being able to differentiate.
• May I ask how did you get this plots? – Creator Jul 19 '18 at 2:01
• @Creator They are on the Wikipedia page on fractional Brownian motion – user223391 Jul 19 '18 at 2:02
• @Creator Oh hey, it's you! I remember talking rough paths with you on here. Thanks for asking such good questions. Didn't realize you were asking the question at first, haha. Feel free to shoot me an email or whatever if you want to talk rough paths. – user223391 Jul 19 '18 at 2:04
• This is the very reason I asked the question mathoverflow.net/questions/304654/… but it is shown to be related to Hurst parameter, but look at your plots now, so close to low and high SNR type. – Creator Jul 19 '18 at 2:04
• . . . nice . . . – janmarqz Jul 19 '18 at 2:22 | 2021-04-15T13:50:20 | {
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https://math.stackexchange.com/questions/650739/using-field-axioms-for-a-simple-proof | # Using field axioms for a simple proof
Question: If $F$ is a field, and $a, b, c \in F$, then prove that if $a+b = a+c$, then $b=c$ by using the axioms for a field.
Relevant information:
Field Axioms (for $a, b, c \in F$):
$a+b = b+a$ (Commutativity)
$a+(b+c) = (a+b)+c$ (Associativity)
$a+0 = a$ (Identity element exists)
$a+(-a) = 0$ (Inverse exists)
Multiplication:
$ab = ba$ (Commutativity)
$a(bc) = (ab)c$ (Associativity)
$a1 = a$ (Identity element exists)
$aa^{-1} = 1$ (Inverse exists)
Distributive Property:
$a(b+c) = ab + ac$
Attempt at solution:
I'm not sure where I can begin. Is it ok to start with adding the inverse of a to both sides, as in the following?
$(a+b)+(-a) = (a+c)+(-a)$ (Justification?)
$(b+a)+(-a) = (c+a)+(-a)$ (Commutativity)
$b+(a+(-a)) = c+(a+(-a))$ (Associativity)
$b+0 = c+0$ (Definition of additive inverse)
$b = c$ (Definition of additive identity)
I'm wondering about my very first step. Specifically, the axioms don't mention anything about doing something to both sides of an equation simultaneously. Is there some other axiom I can use to justify this step?
This is Exercise 1, part b in Section 1 on page 2 of Halmos, Finite Dimensional Vector Spaces (reading book for fun--this is not homework (probably too easy to be a homework problem anyway!)). In part a, I proved that $0+a = a$, in case that is somehow helpful in this problem. Thanks!
• Just a minor note: you may want to write $1/a$ as $a^{-1}$ because you would want to define "/" later using the inverses and it wouldn't be good to have syntax with ambiguous parsing. – user21820 Jan 25 '14 at 11:33
• That's a good idea--I made the change. – Mike Bell Jan 25 '14 at 11:44
Martín-Blas Pérez Pinilla suggests that "=" can be considered a logical symbol obeying logical axioms. While I agree that it fundamentally is so, I would like to note that it is possible to consider it an equivalence relation obeying 'internal' field axioms, because for example the rational numbers can be taken as equivalence classes of a certain set of pairs of integers, and so it is not quite right to consider the equality between these rationals as a logical equality. Also, Ittay made a mistake where he used an unstated axiom that allows substitution. What you need, either way, is something equivalent to the following for any field $F$:
$a=a$ for any $a \in F$ [reflexivity of =]
$a=b \Rightarrow b=a$ for any $a,b \in F$ [commutativity of =]
$a=b \wedge b=c \Rightarrow a=c$ for any $a,b,c \in F$ [transitivity of =]
(These describe "=" as an equivalence relation on $F$)
$a=b \Rightarrow P(a)=P(b)$ for any $a,b \in F$ and predicate $P$ [substitution]
(This describes substitution, which can be used to replace separate axioms governing how "=" and the field operations interact. Ittay used this in one of his steps.)
These allow us to "do the same thing to both sides", for example:
For any $a,b,c \in F$ such that $a=b$,
Let $d=a+c$ [closure under +]
$a+c=a+c$ [transitivity of =; $a+c=d=a+c$]
$a+c=b+c$ [substitution; where the predicate is given by $P(x) \equiv (a+c=x+c)$]
Note that to prove that something is a field, we will have to prove the substitution axiom, which boils down to proving the following equivalent set of axioms:
$a=b \Rightarrow a+c=b+c$ for any $a,b,c \in F$
$a=b \Rightarrow ac=bc$ for any $a,b,c \in F$
The original problem can then be proven as follows:
For any $a,b,c \in F$ such that $a+b=a+c$,
$b = 0+b = (-a+a)+b = (-a)+(a+b) = (-a)+(a+c) = (-a+a)+c = 0+c = c$
• Hurkyl had given a comment indicating that we could use substitution of an argument of $+$ considered as a 2-argument function. As I noted, fields constructed through equivalence relations will still require one to prove the substitution rule before using, so it amounts to the same two more basic axioms that I gave. – user21820 Jan 25 '14 at 11:48
The first step is justified by the existence of $-a$. No further axiom is required in order to deduce that $(a+b)+(-a)=(a+c)+(-a)$. To see that, just write $a+b=y=a+c$ (after all, it is given that $a+b=a+c$). Now, one of the information defining a field is the function of addition: $(u,v)\mapsto u+v$. Applying it to $u=y$ and $v=(-a)$, yields the element $y+(-a)$. But, since $y=a+b$, $y+(-a)=(a+b)+(-a)$. Similarly, since $y=a+c$, $y+(-a)=(a+c)+(-a)$. By transitivity of equality, it follows that $(a+b)+(-a)=(a+c)+(-a)$.
Note that it would be a lot easier to add $(-a)$ on the left rather than on the right. It will save you using commutativity.
• The step where you substitute $(a+b)$ into $y$ requires another axiom such as substitution. (See my answer.) With substitution or the alternative, you can just do it in one line. And that very axiom is the most important part of the problem! – user21820 Jan 25 '14 at 11:19
• Yes, you can put $(-a)$ on the left. But the axiom says $a + (-a) = 0$, not $(-a) + a = 0$. So you have to use Commutativity anyway! – TonyK Jan 25 '14 at 11:26
"Specifically, the axioms don't mention anything about doing something to both sides of an equation simultaneously." Because is a logical axiom. If you have two equal things and do the same with both things the results are equal. Search "first order logic with identity".
• See also en.wikipedia.org/wiki/…, for when we don't have two equal things but have two things that behave the same way under some operations, and we still want to have 'substitution'. – user21820 Jan 25 '14 at 11:29 | 2019-09-22T16:45:00 | {
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https://math.stackexchange.com/questions/3484838/counting-problem-checking-9-squares-out-of-3-times-5-board | # Counting problem: Checking 9 squares out of $3\times 5$ board
I'm self learning some combinatorics and I encountered the following counting problem:
How many ways are there to check 9 squares out of $$3\times5$$ boards such that in every column there's at least one checked square?
(To be more precise, the board has 3 rows and 5 columns)
I think I know the outline of the solution:
Let $$C$$ be the set of all possible checking of $$9$$ squares off the greed and let $$C_i$$ be the checking of the board where the $$i^{th}$$ column has no square checked- I then proceed by using the Inclusion - Exclusion principle and so the solution is $$|C|-|C_1\cup C_2\cup C_3\cup C_4\cup C_5|$$
So I have two questions:
1. How many ways there are actually to check the board without restrictions? When I try to think about it I think about selecting a subset of 9 squares out of 15 squares so $$\binom{15}{9}$$, is this correct? Somehow it does'nt feel like it's the right number ;
2. Is the outline to the solution I wrote above the right approach to this problem?
I know this is very elementary but I'm really confused by all the counting arguments and most of the time my initial intuition turns out to be wrong so any help would be much appreciated!
Update with solution
For each $$C_i$$ we restrict our board to be one column less now, so it's actually checking $$9$$ squares on a $$3\times 4$$ board- there are $$\binom{12}{9}$$ ways to do so. Furthermore, up to renaming the columns this procedure is symmetric so there are $$5$$ ways to do that.
For any intersection of the form $$C_i\cap C_j$$ (for $$i\neq j$$) we restrict our board to be $$3\times 3$$, and now there's a single way to check the board. There are $$\binom{5}{2}$$ such intersections. Any bigger intersection would be empty.
From the Inclusion - Exclusion principle we get: \begin{aligned}\left|\bigcup_{i=1}^{5} C_{i}\right| &=5 C_{i}-\left(\begin{array}{c}{5} \\ {2}\end{array}\right)\left|C_{i} \cap C_{j}\right| \\ &=5\left(\begin{array}{c}{12} \\ {9}\end{array}\right)-\left(\begin{array}{c}{5} \\ {2}\end{array}\right) \end{aligned}
And so the number of possible checking that fit the decrepstion of the exercise is:
$$|C|-\left|\bigcup_{i=1}^{5} C_{i}\right|=\left(\begin{array}{c}{15} \\ {9}\end{array}\right)-\left(5\left(\begin{array}{c}{12} \\ {9}\end{array}\right)-\left(\begin{array}{c}{5} \\ {2}\end{array}\right)\right)=3915$$
• A 3 by 5 board has 15 squares. There are 15 ways to select the first square, 14 ways to select the second square down to 15- 8= 7 ways to select the 9th square. There are $15*14*13*...*8*7= \frac{15!}{6!}$ to select 9 squares out of the 15 in that particular order. If we consider the same 9 square, no matter what the order in which order they are selected, we need to divide by 9!, the number or different ways 9 things can be selected. That gives $\frac{15!}{6!9!}$, the binomial coefficient. Dec 23, 2019 at 5:00
I think you mean $$|\color{red}C|-|C_1\cup C_2\cup C_3\cup C_4\cup C_5|$$
Where $$C$$ is all possible ways to pick $$9$$ squares ... which is indeed $${15} \choose 9$$
OK, but you still need to calculate $$|C_1\cup C_2\cup C_3\cup C_4\cup C_5|$$ .... easier said than done
• Thank you very much! I fixed the typo and updated the solution. Does it look fine?
– omer
Dec 22, 2019 at 20:56
• @omer Yes, well done! Dec 22, 2019 at 20:59
I have found a different solution method which gives the same answer as the one you have, so your method must be sound!
We examine partitions of $$9$$ into $$5$$ positive parts, each part less than $$4$$.
These are $$22221, 32211, 33111$$.
These are our columns, and the number of permutations of each is:
$$\binom{5}{4}=5, \binom{5}{1,2}=\frac{5!}{1!2!2!}=30, \binom{5}{2}=10$$ respectively.
Each column value of $$k$$ can be displayed in $$\binom{3}{k}$$ ways.
All patterns have now been achieved once and once only (by inspection).
The total number is therefore $$5\cdot 3^5 + 30\cdot 3^4 + 10\cdot 3^3 = 1215+2430+270=3915$$.
• Thank you very much for the time and effort! Would it be a bother for you to elaborate on the the relation between partitions of 9 into 5 positive parts, each part less than 4- to the problem? I can't see it :( It's a really interesting solution and I'd like to understand it better!
– omer
Dec 23, 2019 at 12:47
• 5 parts is one for each column, hence positive because each column must be contain at least one filled cell, and each column can have a maximum height of 3, hence less than 4.
– JMP
Dec 23, 2019 at 12:48
• Thank you again! This is a great new way to see this problem!
– omer
Dec 23, 2019 at 12:50 | 2023-02-03T16:47:11 | {
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https://math.stackexchange.com/questions/3033825/injective-homomorphism-from-mathbbr-times-mathbbr-to-the-ring-of-continuo | # Injective Homomorphism from $\mathbb{R}\times\mathbb{R}$ to the ring of Continuous functions
Does there exist an injective ring homomorphism from the ring $$\mathbb{R}\times\mathbb{R}$$ to the ring of continuous functions over $$\mathbb{R}$$?
I know that $$\mathbb{R}\times\mathbb{R}$$ is a field. So, any ring homomrphism should have either identity or the whole field as a kernel. So, does it imply that there is such a homomorphism? I am unable to think of any examples. Any hints. Thanks beforehand.
• $\mathbb R \times \mathbb R$ is not a field. Remember, the direct product of fields is not going to be a field in general, because, for example, $(1,0) \cdot (0,1) = (0,0)$. – астон вілла олоф мэллбэрг Dec 10 '18 at 12:10
• @астонвіллаолофмэллбэрг oh, thanks for that fact. So, then, it is not even an integral domain! – vidyarthi Dec 10 '18 at 12:11
• By the definition of the direct product, two homomorphisms from $\mathbb R$ to the ring of continuous functions over $\mathbb R$ give rise to a homomorphism from $\mathbb R \times \mathbb R$ to the latter space, with the two maps as components/projections. However, the pair of maps being injective does not imply that the combined map will be injective. – астон вілла олоф мэллбэрг Dec 10 '18 at 12:18
• @астонвіллаолофмэллбэрг ok, any good examples? – vidyarthi Dec 10 '18 at 12:23
Let us make a general statement, and then apply it to this case.
Definition : Given a ring $$R$$, an element $$e \in R$$ is said to be idempotent, if $$e^2 = e$$. Note that $$0,1$$ are idempotent. Any other idempotent will be referred to as non-trivial.
Let $$R,S$$ be rings such that $$R$$ has a non-trivial idempotent but $$S$$ does not have any non-trivial idempotent. Then, any homomorphism $$\phi : R \to S$$ is not injective.
Proof : Let $$e \in R$$ be a non-trivial idempotent, which is to say that $$e^2 = e$$, and $$e \neq 0,1$$. Let us look at $$\phi(e)$$. Then, $$(\phi(e))^2 = \phi(e^2)= \phi(e)$$, so $$\phi(e)$$ is idempotent in $$S$$, hence equals zero or one. if $$\phi(e) = 0$$ then $$\phi$$ is not injective as $$e \neq 0$$. If $$\phi(e) =1$$, then $$\phi(1-e) = 0$$, and again injectivity is contradicted.
You can use the general statement in lots of places. Try to find candidates for $$R$$ and $$S$$.
In our case, $$S$$ , as the ring of continuous functions over $$\mathbb R$$,contains no non-trivial idempotent, since if $$f^2 = f$$, then for each $$x$$ we have $$f(x) = 0$$ or $$1$$. However, as $$f$$ is continuous, $$f(\mathbb R)$$ is connected, hence an interval, but has to be a subset of $$\{0,1\}$$, which is discrete. Consequently, $$f$$ is either identically zero or identically one, hence is a trivial idempotent.
However, $$\mathbb R \times \mathbb R$$ contains the non-trivial idempotent $$(0,1)$$. Hence, the general statement gives the result you have desired.
As to non-injective maps, there are many of them. As mentioned earlier, consider $$f,g \in S$$ not necessarily distinct. Then, $$(a,b) \to af + bg$$ is a homomorphism. It is not difficult to see that every homomorphism is of this form.
Also, think about what happens if you treat $$R$$ and $$S$$ as just abelian groups under addition. In that case, can you find an injective group homomorphism between the two? Note that the concept of idempotent cannot be defined without the multiplication, so this might be an interesting side question.
Suppose there is an injective ring homomorphism. The unit element $$(1,1)$$ of $$\mathbb{R} \times \mathbb{R}$$ maps to the unit element of $$\mathcal{C} (\mathbb{R} )$$, i.e. the constant function $$1_{\mathbb{R}}$$. If $$f,g$$ are the images of $$(1,0) , (0,1)$$ respectively, then $$f+g = 1_{\mathbb{R}}$$ and $$fg = 0$$. This shows that $$f(1_{\mathbb{R}}-f) = 0 \implies f = f^2$$. Since $$f$$ is continuous, $$f$$ is identically $$0$$ or identically $$1$$. The first case contradicts injectivity and so does the second because then $$g = 0$$. So there does not exist any injective ring map.
• +1: Very nice argument! And from this we readily find that the only ring homomorphisms $\Bbb R\times\Bbb R\to\mathcal C(\Bbb R)$ are $(x,y)\mapsto x\cdot 1_{\Bbb R}$ and $(x,y)\mapsto y\cdot 1_{\Bbb R}.$ – Cameron Buie Dec 10 '18 at 18:36 | 2019-12-09T05:32:03 | {
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http://mathhelpforum.com/algebra/148707-algebra-question-print.html | # An algebra question
• June 16th 2010, 11:12 AM
econlover
An algebra question
Question: Simplify the following expression.
(2x5^n)^2 + 25^n
this whole equation is divided by
5^n
Remarks: I have tried doing this question twice but both attempts were unsuccesful. Please help.
• June 16th 2010, 11:28 AM
earboth
Quote:
Originally Posted by econlover
Question: Simplify the following expression.
(2x5^n)^2 + 25^n
this whole equation is divided by
5^n
Remarks: I have tried doing this question twice but both attempts were unsuccesful. Please help.
1. Use the laws of powers:
$\frac{(2 \cdot 5^n)^2 + 25^n}{5^n} = \frac{4 \cdot 5^{2n} + 5^{2n}}{5^n}= \frac{5 \cdot 5^{2n}}{5^n}=5 \cdot 5^n = 5^{n+1}$
• June 16th 2010, 11:33 AM
econlover
slight question
Thanks.
I understand most of it but how did you move along from the 1st step of your answer to the second one
that is, in specific terms, where did the 4 go?
• June 16th 2010, 11:39 AM
SpringFan25
considering the numerator (top of fraction) only:
$4 . 5^{2n} + 5^{2n} = (4+1).5^{2n} = 5 \times 5^{2n} = 5^{2n+1}$
• June 16th 2010, 11:41 AM
econlover
Thanks for the message. But I guess you misinterpret the sign of this "."
It is not 4.5 (4 decimal 5) but 4 . 5 meaning 4 x 5 = 4 times 5
Hope this clears up the misconception
• June 16th 2010, 11:47 AM
econlover
Still don't understand this step. Where did the 1 come from ? What about the positive sign?
Can someone explain? Thanks!
• June 16th 2010, 12:18 PM
SpringFan25
it is only factorising. i was not interpreting your dot as a decimal.
in general:
4x + x = (4+1)x = 5x
You have
$4 \times 5^{2n} + 1 \times 5^{2n}$ = $5 \times 5^{2n}$
Which says: "4 lots of $5^{2n}$ plus 1 lot of $5^{2n}$" equals "5 lots of $5^{2n}$"
• June 17th 2010, 02:03 PM
bjhopper
Quote:
Originally Posted by econlover
Question: Simplify the following expression.
(2x5^n)^2 + 25^n
this whole equation is divided by
5^n | 2015-01-25T22:36:56 | {
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https://math.stackexchange.com/questions/935490/the-limit-of-n1-n-n-as-n-to-infty/935538 | # The limit of $(n!)^{1/n}/n$ as $n\to\infty$ [duplicate]
(Proof necessary)
$$\lim_{n \to \infty} \frac{(n!)^{\frac{1}{n}}}{n}$$
I don't have an answer yet, but I know it exists, and is less than $1$.
Edit. Winther's answer is the most correct I don't understand how he is jumping from (log(n!) - nlog( n )) to it equal to the Sum from k=1 to n of log(k/n). Don't presume, it's wrong, I need to go, and I'll keep looking at it when I get back
Any help is appreciated
• Is this your limit: $\lim_{n\rightarrow\infty} \frac{(n!)^{\frac{1}{n}}}{n}$? – Jared Sep 17 '14 at 20:15
• Hint: If the limit exists, let's say its $L$, then $$\lim_{n \to \infty} \frac{n!}{n^n}=L^n$$ – Oria Gruber Sep 17 '14 at 20:18
• Stirling's formula will help you – ThePortakal Sep 17 '14 at 20:18
• See this question for something very similar. It can be done without Strilings formula. – Winther Sep 17 '14 at 20:32
• @Oria: The value of a limit (if it exists) cannot depend on the limiting variable. So "$\displaystyle\lim_{n \to \infty}\frac{n!}{n^n} = L^n$" is definitely incorrect. – JimmyK4542 Sep 17 '14 at 20:51
Put $$a_n = \frac{n!^{1/n}}{n}$$
then $$\log a_n = \frac{1}{n}\left(\log n! - n\log n\right)= \frac{1}{n}\sum_{k=1}^n\log\left(\frac{k}{n}\right)$$
where we have used $\log n! = \log 1 + \log 2 + \ldots + \log n$. The sum above is a Riemann sum for the integral $\int_0^1\log x dx$ so
$$\lim_{n\to\infty} \log a_n = \int_0^1\log x dx = [x\log x - x]_0^1 = -1$$
and it follows that $$\lim_{n\to\infty}a_n = \frac{1}{e}$$
• Hey winther I don't understand how you're jumping from (log(n!) - nlog( n )) to it equal to the Sum from k=1 to n of log(k/n). – Calvin Hammond Sep 17 '14 at 21:28
• @CalvinHammond $$\log n! - n \log n = \log 1 + \log 2 + \cdots + \log n - (\log n + \log n + \cdots + \log n),$$ where the number of terms in the parentheses is $n$. So $$\log n! - n \log n = \log \frac{1}{n} + \log \frac{2}{n} + \cdots + \log \frac{n}{n}$$ using the identity $\log x - \log y = \log \frac{x}{y}$. The rest follows. – heropup Sep 17 '14 at 21:42
• Yes that is. Thanks for explaining it @heropup – Winther Sep 17 '14 at 21:46
• Could you explain the remain sum more please. – Calvin Hammond Sep 17 '14 at 21:56
• Notice that you have to be a little careful in applying this technique to improper integrals: math.stackexchange.com/questions/449103/… – user84413 Sep 18 '14 at 16:51
Here is a simple elementary proof I found, but first of all, some lemmas:
• This one could easily be proven by induction: $\displaystyle \prod_{k=1}^{n}\left(1+\frac{1}{k} \right) = n+1$
• You can try to prove this inequality yourself since it's not difficult: $\displaystyle \left (1+\frac{1}{k} \right )^k\leq e \leq \left (1+\frac{1}{k} \right)^{k+1}\\$
• This inequality is the one I'm going to use though because it gives a much tighter bound on our sequence and that's just more fun, though you could use the second inequality without change in proof. I could give you the proof if needed:$\displaystyle \left (1+\frac{1}{k} \right )^k\leq e \leq \left (1+\frac{1}{k} \right)^{k+1/2}\\$
Ok, so here is the proof:
• We first write $n^n/n!$ in a better way: $\displaystyle \frac{n^n}{n!}=\prod_{k=1}^{n-1}\left(1+\frac{1}{k} \right)^n \cdot \prod_{i=1}^{n-1}\prod_{k=1}^{i} \left(1+\frac{1}{k} \right)^{-1}=\prod_{k=1}^{n-1}\left(1+\frac{1}{k} \right)^n\cdot \prod_{i=1}^{n-1} \left(1+\frac{1}{i} \right)^{-(n-i)}=\prod_{i=1}^{n-1} \left(1+\frac{1}{i} \right)^{i}$
• Now, we use our inequalities to bound our sequence. First, an upper bound: $\displaystyle \prod_{i=1}^{n-1} \left(1+\frac{1}{i} \right)^{i} \leq \prod_{i=1}^{n-1}e^{1}=e^{n-1}$
• Then, a lower bound: $\displaystyle \prod_{i=1}^{n-1} \left(1+\frac{1}{i} \right)^{i} \geq \prod_{i=1}^{n-1}\left (e^{1} \cdot \left(1+\frac{1}{i}\right)^{-1/2} \right )=\frac{e^{n-1}}{\sqrt[2]{n}}$
• Now, since $\frac{n!^{1/n}}{n}=(\frac{n^n}{n!})^{-1/n}$, we get: $\displaystyle e^{\frac{1}{n}-1} \leq \frac{(n!)^{1/n}}{n} \leq e^{\frac{1}{n}-1} \cdot \sqrt[2n]{n}$
• Finally, by the squeeze theorem, we get $\displaystyle e^{0-1} \leq \lim_{n \to \infty} \frac{(n!)^{1/n}}{n} \leq e^{0-1} \cdot 1$
• Hence, $\displaystyle \lim_{n \to \infty} \frac{(n!)^{1/n}}{n}=e^{-1}$
I know there are simpler proofs, but this one is elementary and I feel like it gives you the direct intuition as to why it's true.
Stirling's Approximation $$n!\sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$ Which means that $n!$ is asymptotically equivalent to $\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$ as $n$ approaches $\infty$. If your familiar with asymptotic formulas, then you'd also know that this implies that $$\lim_{n\to\infty} \frac{n!}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n} =1$$ Now, using the algebraic laws of limits, we have $$\lim_{n\to\infty} n!=\lim_{n\to\infty} \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$ $$\lim_{n\to\infty} e^n=\lim_{n\to\infty} \sqrt{2\pi n}\left(\frac{n^n}{n!}\right)$$ $$\lim_{n\to\infty} e=\lim_{n\to\infty} (2\pi n)^{\frac{1}{2n}}\left(\frac{n}{(n!)^{\frac{1}{n}}}\right)$$ So now $$e=\left[\lim_{n\to\infty} e^{\ln(2\pi n)^{\frac{1}{2n}}}\right]\left[\lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}\right] =\left[\lim_{n\to\infty} e^{\frac{\ln(2\pi n)}{2n}}\right]\left[\lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}\right]$$ $$=\left[\lim_{n\to\infty} e^{\frac{\frac{d}{dn}\ln(2\pi n)}{\frac{d}{dn}2n}}\right]\left[\lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}\right] =\left[\lim_{n\to\infty} e^{\frac{1}{2n}}\right]\left[\lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}\right]$$ $$=e^0\left[\lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}\right]=1\cdot \lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}$$ Thus $$e=\lim_{n\to\infty} \frac{n}{(n!)^{\frac{1}{n}}}$$ And now we can easily see that $$\lim_{n\to \infty} \frac{(n!)^{\frac{1}{n}}}{n}= \lim_{n\to \infty} \frac{1}{\left(\frac{n}{(n!)^{\frac{1}{n}}}\right)}=\frac{1}{e}$$ Let me know if you have any questions.
Let $\displaystyle a_n=\frac{n!}{n^n}$.
Then $\displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\lim_{n\to\infty}\frac{n^n}{(n+1)^n}=\lim_{n\to\infty}\frac{1}{(1+\frac{1}{n})^n}=\frac{1}{e}$,
so $\;\;\;\;\displaystyle\lim_{n\to\infty}(a_n)^{\frac{1}{n}}=\lim_{n\to\infty}\frac{(n!)^{\frac{1}{n}}}{n}=\frac{1}{e}$ $\hspace{.2 in}$ (since $\frac{a_{n+1}}{a_n}\rightarrow L\implies(a_n)^{\frac{1}{n}}\rightarrow L)$.
• This is a great proof, but you would need to prove that the two forms of the limit are equivalent. – Fujoyaki Sep 17 '14 at 22:46
• Thanks; you are right that I am using the result that if $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=L$, then $\lim_{n\to\infty}(a_n)^{\frac{1}{n}}=L$. See math.stackexchange.com/questions/69386/… – user84413 Sep 17 '14 at 22:49
Hint. Take the natural logarithm of $\dfrac{(n!)^{1/n}}{n}$ and obtain $$\frac{\log 1+\log 2+\cdots+\log n}{n}-\log n=\frac{\log (1/n)+\log(2/n)+\cdots+\log(n/n)}{n} \\=\frac{1}{n}\sum_{k=1}^n\log\left(\frac{k}{n}\right) \to \int_0^1 \log x\,dx=-1.$$
Hint (Sterling): $$n! \sim \sqrt{2\pi n}(\frac{n}{e})^n$$
• Yes that is my limit. I'm using this to help a friend but I don't know much about to how to solve this myself. So I'm asking for a full answer. – Calvin Hammond Sep 17 '14 at 20:28 | 2021-03-06T12:29:49 | {
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https://math.stackexchange.com/questions/3221910/how-to-minimise-the-cost-of-guessing-a-number-in-a-high-low-guess-game | # How to minimise the cost of guessing a number in a high/low guess game?
In a high/low guess game, the "true" number is one of $$\{1,\cdots,1000\}$$. You'll be told if your guess is $$<,>$$ or $$=$$ the true number for each guess you make, and the game terminates when you guess out the true number. Suppose the following three scenarios:
1. If your guess is either lower or higher, you pay $$\1$$ in both cases. If your guess is correct, you pay nothing and the game ends.
2. If your guess is higher, you pay $$\1$$; if your guess is lower you pay $$\2$$. If your guess is correct, you pay nothing and the game ends.
3. If your guess is higher, you pay $$\1$$; if your guess is lower you pay $$\1.5$$. If your guess is correct, you pay nothing and the game ends.
In these three scenarios, what is, respectively, the minimum number of $$\$$ you must have to make sure you can find the true number?
Formally, define a space of all guess strategies $$\Sigma$$. We can then identify the cost $$C$$ as a function $$C:\Sigma\times \{1,\cdots,1000\}\to\Bbb N_+,\quad v\mapsto C(S,v)$$ that maps the pair $$(S,v)$$ (where $$v$$ is the "true" number) to the cost it's going to take under this arrangement. Our problem is then to find $$\min_{S} \max_{v} C(S,v).$$
Can this min-max problem be analytically solved? What would be the corresponding optimal strategy then?
For scenario 1, my intuition is to use bisection search, which, at worst, costs $$\10$$ ($$2^{10}=1024$$). But I cannot prove this is indeed optimal. For the second scenario, since making an under-guess means a higher cost, maybe it's more optimal to use a "right-skewed" bisection, i.e. you keep making guesses at the $$2/3$$-quantile point. But this is just (rather wild) intuition, not anything close to a proof.
For this game, the most natural representation of a strategy is a rooted binary tree, with vertices labeled by guesses (i.e. numbers from $$\{1, \ldots, N=1000\}$$), and each vertex having at most two edges (connecting it to the guess to make next, for "lower" and "higher" cases).
Let the penalties you pay for "lower/higher" cases be $$L$$ and $$H$$, respectively. Now consider an arbitrary $$N$$, and consider an optimal tree $$T(N)$$ (that reaches the minimum worst-case total penalty). Its root is labeled with some guess $$R$$, $$1\leqslant R\leqslant N$$, its left subtree can be chosen to be [isomorphic to] $$T(R-1)$$, and its right subtree isomorphic to $$T(N-R)$$ (assuming $$T(0)$$ is empty).
So if $$P(N,L,H)$$ is the optimal penalty, then $$P(0,L,H)=P(1,L,H)=0,\\ P(N,L,H)=\min_{1\leqslant R\leqslant N}\max\begin{cases}L+P(N-R,L,H)\\ H+P(R-1,L,H)\end{cases}\quad(N>1)$$ which allows one to compute $$P(1000,1,1)=\color{red}{9},\quad P(1000,2,1)=14,\quad P(1000,1.5,1)=11.5$$ (note $$9$$ but not $$10$$ as you've suggested; the decision process is not just a bisection but has extra "$$=$$" outcomes). The trees can be computed along the way.
Regarding the asymptotic behavior (with $$N\to\infty$$), if we let $$R(N,L,H)$$ be the "argmin" (say, the smallest root of optimal trees), one can show that the limits $$\alpha=\alpha(L,H)=\lim_{N\to\infty}\frac{P(N,L,H)}{\ln N},\quad\beta=\beta(L,H)=\lim_{N\to\infty}\frac{R(N,L,H)}{N}$$ exist and satisfy $$H+\alpha\ln\beta=L+\alpha\ln(1-\beta)=0.$$ In particular, $$\beta(2,1)=(\sqrt{5}-1)/2$$, and $$\beta(1.5,1)$$ is the root of $$\beta^3=(1-\beta)^2$$.
• Thanks. This makes sense. There seems to be no way to compute such a complicated recursion analytically though. Did you use DP for this? – Vim May 11 at 12:30
• Yes, I did. Probably this still can be analyzed further, but I didn't do that. – metamorphy May 11 at 12:38
• IIRC, your recursion also gives the optimal strategy when you keep track of the partitions chosen at each recursion node. – Vim May 11 at 13:41
For the (2,1) case, I found the worst-case cost increases by 1 at every Fibonacci number, so $$P(F_n,2,1)=1+P(F_n-1,2,1)=1+P(F_{n-1},2,1)$$. These key values $$F_n$$ increase by ratio $$F_n/F_{n-1}\to\phi=(1+\sqrt{5})/2$$.
They grow exponentially; conversely, the maximum cost for $$P(N,2,1)$$ grows logarithmically.
For (1.5,1) the worst-case cost increases by $$0.5$$ at these values: $$P(N,2,1)>P(N-1,1.5,1)\,if\, N=2,3,4,5,7,9,12,16,21,28,37,49,65,86,114,\ldots$$
The recursion in this sequence is $$a_{n+1}=a_{n-1}+a_{n-2}$$. Their ratio, $$a_{n+1}/a_n$$, approaches 1.3246, which is the root of $$P^3=P+1$$, just as $$\phi$$ for the (2,1) case is the root of $$P^2=P+1$$.
• Would you elaborate on how the ratios arise? – Vim May 11 at 12:32
• For $(4/3,1)$, the recursion is $a_{n+1}=a_{n-2}+a_{n-3}$, and for $(5/3,1)$, the recursion is $a_{n+1}=a_{n-2}+a_{n-4}$ – Empy2 May 11 at 13:00
Here are some worked-out cases for smaller lists and variable penalties.
Suppose you pay $$a$$ for a guess that is too low and $$b$$ for a guess that is too high.
Given a list such as $$\{1,\ldots,1000\}$$, the cost $$C$$ of the optimal strategy is a function only of the number of items $$n$$ in the list. Without loss of generality, assume that $$a\leq b$$. (If you want the case $$a\geq b$$, just reverse the order of the list.)
We can determine the cost of the optimal strategy in a few simple cases:
• $$C(1) = 0$$. There is only one guess, and it is correct.
• $$C(2) = a$$. High guesses are penalized more, so guess the lower element.
• $$C(3) = b$$. Guess the middle element. If you're wrong, you know what the correct element is and it costs at worst $$b$$ total.
For larger values of $$C(n)$$, consider what the optimal first move. Suppose there are $$n$$ elements in the list and you choose index $$1\leq k\leq n$$. In the worst case, you are wrong. If the true answer is lower, you must pay $$b$$ and search through a list of $$k-1$$ elements. If the true answer is higher, you must pay $$a$$ and search through a list of $$n-k$$ elements. So in the worst case, by choosing $$k$$ in the first round, you'll incur a cost of $$\max\left(b + C(k-1),\; a + C(n-k)\right).$$ We can search for an optimal $$k$$ which minimizes this cost in the first move, in which case the recursive solution tells us what the rest of the strategy (and its corresponding cost) is.
Note that the optimal strategy will never be in the latter half of the list. This is because $$C$$ is monotonic. When $$k$$ is in the right half of the list (so that $$n-k \leq k-1$$), we can always shrink the worst-case cost $$\max(b+C(k-1), a+C(n-k))$$ by exchanging $$k-1$$ and $$n-k$$, putting $$k$$ into the symmetric position on the left half of the list.
By this reasoning, we find:
• $$C(4) = \max(2a,b).$$ Choose the second element. If you're too high, you'll pay a cost of $$b$$ but know the true answer. If you're too low, you'll pay a cost of $$a$$, then search through the remaining two elements for a worst-case cost of $$C(2)=a$$.
• $$C(5) = a+b$$. You can choose either the second or third (middle) elements. In the worst case if you choose the middle element, it's too high so you pay $$b+C(2) = a+b$$ to search the two remaining. If you choose the second element, in the worst case it's too low so you pay $$a+C(3) = a+b$$ to search the three remaining.
• $$C(6) = a+b$$. Choose the third element. In the worst case, it's either too high so you pay $$b+C(2) = a+b$$, or it's too low so you pay $$a+C(3)=a+b$$.
• $$C(7) = \min(a+C(4), b+C(3)).$$ As with the case $$n=4$$, the optimal strategy depends on the relative magnitudes of $$a$$ and $$b$$. If $$2b > 3a$$, then choose the third element in the list. Otherwise, choose the fourth (middle). | 2019-06-20T06:52:57 | {
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http://ac-immacolata.it/csxr/bisection-method-algorithm-matlab.html | ### Bisection Method Algorithm Matlab
Updated to reflect the latest version of MATLAB, the second edition of this title introduces the theory and applications of the most commonly used techniques for solving numerical problems on a computer. Secant method D. This method requires two initial guesses satisfying. Numerical Methods with Matlab The Bisection Method 4 1. k 1 is the slope at the beginning of the time step (this is the same as k 1 in the first and second order methods). Design and simulation of three phase induction motor at different load conditions in matlab/simulink. The Bisection Method is a numerical method for estimating the roots of a polynomial f(x). m code; Worksheet 04; 10th - 14th September : Tu: Bisection Method (continued) and order of convergence ; Th: Newton's Method and order of convergence; Homework 01 is due on 09/13! Worksheet 05; Code for Newton's Method; 17th - 21st September : Tu: Secant Method with order of. A few steps of the bisection method applied over the starting range [a 1;b 1]. Here is a Matlab function that carries out the bisection algorithm for our cosmx function. It allows the code writer to focus on the logic of the algorithm without being distracted by details of. 2) using the bisection method. 1 Bisection Method In bisection method we reduce begin with an interval so that 0 2[a;b] and divide the interval in two halves,i. It’s very intuitive and easy to implement in any programming language (I was using MATLAB at the time). They announced that the genetic algorithm is better than classical methods. where the value of the function. 1 The Bisection Method to Solve g(x)=0 Many mathematical problems involve solving one or more equations. On the other hand, the only difference between the false position method and the bisection method is that the latter uses ck = (ak + bk) / 2. The bisection algorithm should be: Save the interval boundaries; Look if [a,b] has a root. In python (with matplotlib), use the savefig function. Provide the function, 'f' and provide two guesses. Students will simultaneously be trained in the theory and practice involved in solving large systems of equations and understand and interpret the quality of such solutions. Bisection Method: Develop a MATLAB program to find the root of the following function using the bisection method gm gc tan g-9. Im studying for a math test and on a old test there is a task about bisection. The standard technique is something like Brent's method (see Numerical Recipes in C, Section 9. From these algorithms, the developer has to explore and exploit the algorithm suitable under specified constraints on the function and the domain. Application Of Bisection Method In The bisection method is an iterative algorithm used to find roots of continuous functions. - Bisection method for bounded searching. CMP 150: Computer Tools for Problem Solving Lab Problem Set 7 April 6, 2011 This week’s lab problem set involves a numerical method which is very important in applied mathematics. Here's the code:. It shows with bold stripes the length of the bracketed region. These concepts form the foundation for writing full applications, developing algorithms, and extending built-in MATLAB capabilities. 2d Truss Analysis Matlab Program. Matlab has the function ‘fzero’ to find function zeros. The simplest root-finding algorithm is the bisection method: we start with two points a and b which bracket a root, and at every iteration we pick either the subinterval or, where is the midpoint between a and b. LAB 1: the bisection and secant methods In this section you will learn how to implement and analyse the Bisection and Secant methods in MAT-LAB. Like the others methods, we approach this problem by writing the equation in the form f(x) = 0 for some function f(x). Learn more about bisection, bisection method, algorithm. newton raphson method matlab pdf. We have provided MATLAB program for Bisection Method along with its flowchart and algorithm. The Bisection Method & Intermediate Value Theorem. m (after that day assignments should be put into the mbox of Yinglun ZHU) (1)The original demonstration of Newton’s method was done by Newton almost 350 years ago. The bisection search. f(c)<0 then let b=c, else let a=c. The task is to solve x^2=2 with the bisection method and the precision should be with 10 decimals. What are the applications of the bisection. Fixed Point Iteration 8 1. The setup of the bisection method is about doing a specific task in Excel. Figure 1 At least one root exists between the two points if the function is real, continuous, and changes sign. 6524; m = 73. The Steepest Descent Algorithm for Unconstrained Optimization and a Bisection Line-search Method Robert M. Problems 197. So let's take a look at how we can implement this. Essentially, the root is being approximated by replacing the. ) (Use your computer code) I have no idea how to write this code. Using this simple rule, the bisection method decreases the interval size iteration by iteration and reaches close to the real root. Learn how genetic algorithms are used to solve optimization problems. In mathematics, the bisection method is a root-finding method that applies to any continuous functions for which one knows two values with opposite signs. In this method, there is no need to find the. Learn more about matlab. A next search interval is chosen by comparing and nding which one has zero. where the value of the function. Bisection Method for Solving non-linear equations using MATLAB(mfile) % Bisection Algorithm % Find the root of y=cos(x) from o to pi. All the Fortran 90 programs listed here are corresponding to the Fortran 77. The equation is of form, f(x) = 0. For a given function as a string, lower and upper bounds, number of iterations and tolerance Bisection Method is computed. % % OUTPUT: approximate solution p or. Again, as before, Newton’s method does not always converge, but when it does, it does so faster (p = 2) than the bisection method (p = 1) and the secant method (𝑝= (1+√5)⁄2). 2D 3D Algorithms ASCII C# C++ Cellular Automata Clustering Cryptography Design Patterns Electronics game Image Processing Integral Approximation Java JavaFX Javascript LED Logic Gates Matlab Numerical Methods Path Finding Pygame Python R Random Root Finding R Shiny Sound UI Unity. It was observed that the Bisection method converges at the 14th iteration while Newton methods. 5 Secant Method 189. The bisection method depends on the Intermediate Value Theorem. Bisection Algorithm Let , - i. Solve 2D Transient Heat Conduction Problem using FTCS Finite Difference Method. First I plot the function and then I try to find a domain such that I can see the curve cut through the x -axis. The algorithm applies to any continuous function. B The comparative results are shown in table 3. GitHub Gist: instantly share code, notes, and snippets. This scheme is based on the intermediate value theorem for continuous functions. Consider a transcendental equation f (x) = 0 which has a zero in the interval [a,b] and f (a) * f (b) < 0. In general, Bisection method is used to get an initial rough approximation of solution. This series of video tutorials covers the numerical methods for Root Finding (Solving Algebraic Equations) from theory to implementation. Bisection method in MATLAB A video of programming the code in realtime. Consider a root finding method called Bisection Bracketing Methods • If f(x) is real and continuous in [xl,xu], and f(xl)f(xu)<0, then there exist at least one root within (xl, xu). % Newton's Method algorithm! n = 2;! nfinal = N + 1; % Store final iteration if tol is reached before N iterations! Newton's Method MATLAB Implementation. The algorithm, created by T. The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie for further. 2 Bisection method 2. Comparative Study Of Bisection, Newton-Raphson And Secant Methods Of Root- Finding Problems International organization of Scientific Research 2 | P a g e Given a function f x 0, continuous on a closed interval a,b , such that a f b 0, then, the function f x 0 has at least a root or zero in the interval. Root approximation through bisection is a simple method for determining the root of a function. Freund February, 2004 1 2004 Massachusetts Institute of Technology. MatLab Project 2 - Bisection Method, The Fixed-point Iteration, and Newton's Method Due October 10. Before you start, review the \Introduction to MATLAB" notes. The chance of convergence with such a small precision depends on the calculatord: in particular, with Octave, the machine precision is roughly ⋅ −. Numerical rate of convergence of root has been found in each calculation. Learn how genetic algorithms are used to solve optimization problems. Earlier we discussed a C program and algorithm/flowchart of bisection method. The Bisection Method The Bisection Method at the same time gives a proof of the Intermediate Value Theorem and provides a practical method to find roots of equations. m needed for Homework 4. Bisection Method for Solving non-linear equations using MATLAB(mfile) % Bisection Algorithm % Find the root of y=cos(x) from o to pi. Essentially, the root is being approximated by replacing the. 6: Calculate the ground track of a satellite from its orbital elements. 15625 (you need a few extra steps for ε abs) Applications to Engineering. I wrote his code as part of an article, How to solve equations using python. The bisection method is one of the simplest and most reliable of iterative methods for the solution of nonlinear equations. Need help with this bisection method code!. Bisection method; Execute an instance method of Object and call in its block instance methods of another object; get URL Params (2 methods) Rake Migrate (newest method) order/format of params in method definition; XML Load methods; Kohana helper method for Askimet; Class vs Instance Methods; PHP5 Method Chaining Example. he gave us this template but is not working. Description. The problem is that it seems like the teachers recommended solution to the task isn't quite right. This Demonstration shows the steps of the bisection root-finding method for a set of functions. f = @(x) (cos(x)); a = input( 'Please enter lower. Above given Algorithm and Flowchart of Bisection Methods Root computation is a simple and easier way of understanding how the bracketing system works, algorithm and flowchart may not follow same procedure, yet they give the same outputs. This is a very simple and powerful method, but it is also relatively slow. 2 (Bisection Method). Bisection Method C Program Bisection Method MATLAB Program. If a change of sign is found, then the root is calculated using the Bisection algorithm (also known as the Half-interval Search). t is the root of the given function if f (t) = 0; else follow the next step. The method is also called the interval halving method. This method will divide the interval until the resulting interval is found, which is extremely small. Powered by Create your own unique website with customizable templates. 000013273393044 9 1. 21 dcm_to_ypr. The following Matlab project contains the source code and Matlab examples used for bisection method. The main advantages to the method are the fact that it is guaranteed to converge if the initial interval is chosen appropriately, and that it is relatively. Remark: 𝑝𝑝. So if you need MATLAB programming homework help, feel free to ask for a quote. Set 1: The Bisection Method Set 2: The Method Of False Position. X +4 X i mark the bracket. Bisection Method of Solving a Nonlinear Equation. % Using the. CMP 150: Computer Tools for Problem Solving Lab Problem Set 7 April 6, 2011 This week’s lab problem set involves a numerical method which is very important in applied mathematics. The bisection search. 4 Basis of Bisection. 1shows the several first iterations of the bisection algorithm. Bayen Bisection algorithm to aproximate an equation result. BISECTION_RC, a MATLAB library which demonstrates the simple bisection method for solving a scalar nonlinear equation in a change of sign interval, using reverse communication (RC). Bisection method. 1 The Bisection Method to Solve g(x)=0 Many mathematical problems involve solving one or more equations. The Secant Method One drawback of Newton's method is that it is necessary to evaluate f0(x) at various points, which may not be practical for some choices of f. Divide the interval [a, b]. 3 Algorithms and convergence 2. Assumption: The function is continuous and continuously differentiable in the given range where we see the sign change. The algorithm for this is given as follows: Choose a;b so that f(a)f(b) <0 1. Bisection Method The Bisection method is a root finding algorithm. Essentially, the root is being approximated by replacing the. The problem I encounter is that the same computation must be performed on each element of a large array(~1. where the value of the function. By the Nested Interval Property of Real Numbers the sequence of Nested Intervals converge to a unique point, which should therefore be r. Approximate the root of f(x) = x 2 - 10 with the bisection method starting with the interval [3, 4] and use ε step = 0. MATLAB Tutorial – Roots of Equations ES 111 1/13 FINDING ROOTS OF EQUATIONS Root finding is a skill that is particularly well suited for computer programming. In these lectures details about how to use Matlab are detailed (but not verbose) and. m This program will implement Euler’s method to solve the differential equation dy dt = f(t,y) y(a) = y 0 (1) The solution is returned in an array y. Suppose we want to solve the equation. These concepts form the foundation for writing full applications, developing algorithms, and extending built-in MATLAB capabilities. Learn more about bisection method, homework. MatLab Project 2 - Bisection Method, The Fixed-point Iteration, and Newton's Method Due October 10. 00064404356011 -0. Chapter 6 Finding the Roots of Equations The Bisection Method Copyright © The McGraw-Hill Companies, Inc. 11) uses fzero to calculate the root for this heat capacity example. " by Timmy Siauw and Alexandre M. The software, matlab 2009a was used to find the root of the function for the interval [0,1]. Suppose that f(¢) is a continuous function defined over an interval [a;b] and f(a) and f(b) have opposite signs. This is a quick way to do bisection method in python. function p_min=bisection(func,int,iter,tol_x,tol_f) % It calculates the zero of a regular real function with one variable. The convergence rate of the bisection method could possibly be improved by using a different solution estimate. Bisection method- code stops after one iteration. So the abscissa of point where the chords cuts the x-axis (y=0) is given by,. Because of this, it is often used to roughly sum up a solution that is used as a starting point for a more rapid conversion. Bisection Method Issues/Help. % p_min is the solution and represents the abscissa's value of the zero. What's great about the Bisection Method is that provided the conditions above are satisfied (and hence a root $\alpha$ exists in the interval $[a, b]$ by the Intermediate Value Theorem), then this method is guaranteed to zone into our root with better and better approximations. Given these facts. Blog Archive. 6 in the text. It’s take a first approximation by apply two times the Bisection method and complete a correct approximation by use the Newton-Raphson method. Freund February, 2004 1 2004 Massachusetts Institute of Technology. Noanyother restrictionsapplied. Additional optional inputs and outputs for more control and capabilities that don't exist in other implementations of the bisection method or other root finding functions like fzero. In this video tutorial, the algorithm and MATLAB programming steps of finding the roots of a nonlinear equation by using bisection method are explained. It is obvious that the secant method does not always converge, but when it does, it does so faster than the bisection method. What are the applications of the bisection. Note that just as in the bisection algorithm, the initial two guesses must be such that one gives a positive function evaluation and the. The x-coordinate of this point is the average of the positive and negative guesses. Newest vertex bisection. However, if there are several solutions present, it finds only one of them, just as Newton's method and the secant method. to a bug in. "In mathematics, the bisection method is a root-finding algorithm which works by repeatedly dividing an interval in half and then selecting the subinterval in which a root exists. There are classical root-finding algorithms: bisection, false position, Newton-Raphson, modified Newton-Raphson, secant and modified secant method, for finding roots of a non-linear equation f(x) = 0 [7,8,9,10,11]. Freund February, 2004 1 2004 Massachusetts Institute of Technology. Later you will learn how to add it to the calling sequence. Introduction. Bisection Method of finding the roots of an equation is both simple and straight forward - I really enjoyed playing with bisection back in college (oooh yeah ES84 days) and I decided to make a post and implement bisection in scilab. algorithm apply axis bisection method boundary condition button Chebyshev Chebyshev nodes clicking coefficient computation constraints converge curve data points defined depicted in Fig derivative diagonal dialog box difference approximation differential equation eigenvalues eigenvectors element end end Example Figure find first flops. CMP 150: Computer Tools for Problem Solving Lab Problem Set 7 April 6, 2011 This week’s lab problem set involves a numerical method which is very important in applied mathematics. It will helpful for engineering students to learn Bisection method MATLAB program easily. Students will simultaneously be trained in the theory and practice involved in solving large systems of equations and understand and interpret the quality of such solutions. The Algorithm The bisection method is an algorithm, and we will explain it in terms of its steps. MATLAB coding of all methods. - Bisection method for bounded searching. Examples illustrate important concepts such as selection, crossover, and mutation. The points marked as X i are positions of the negative( )andpositive(+)endsoftherootenclosingbracket. Using C program for bisection method is one of the simplest computer programming approach to find the solution of nonlinear equations. 6524; m = 73. It is a very simple and robust method, but it is also relatively slow. 1 and ε abs = 0. Here we are required an initial guess value of root. The Algorithm for The Bisection Method for Approximating Roots Fold Unfold. The Bisection Method looks to find the value c for which the plot of the function f crosses the x-axis. This method is also known as Regula Falsi or The Method of Chords. Python Bisection Method. This is a quick way to do bisection method in python. The number of iterations we will use, n, must satisfy the following formula:. Wednesday, September 28, 11. It requires two initial guesses and is a closed bracket method. Bisection method add iteration table into my code. We are going to find the root of a given function, with bisection method. m with contents. The basic idea is a follows. Matlab will spit out that the root in this interval = '6'. The bisection method is an approximation method to find the roots of the given equation by repeatedly dividing the interval. In this course, three methods are reviewed and implemented using Python and MATLAB from scratch. The help page states the following about the algorithm: Algorithms The [code ]fzero[/code] command is a function file. Algorithm of Bisection Method [YOUTUBE 9:47] Example of Bisection Method [YOUTUBE 9:53] Advantages & Drawbacks of Bisection Method [YOUTUBE 8:31] MULTIPLE CHOICE TEST : Test Your Knowledge of Bisection Method PRESENTATIONS. For searching a finite sorted array, see binary search algorithm. The problem I encounter is that the same computation must be performed on each element of a large array(~1. m - matlab file that defines Equation (2. Numerical rate of convergence of root has been found in each calculation. This algorithm is a new approach to compute the roots of nonlinear equations f(x)=0, by propose hybrid algorithm between the Bisection algorithm and Newton-Raphson algorithm. The method is based on the following algorithm: Initialization: The bisection method is initialized by specifying the function , the interval [a,b], and the tolerance > 0. Step 3: If f(a). In this tutorial we are going to develop pseudocode for Bisection Method so that it will be easy while implementing using programming language. Dekker's zeroin algorithm from 1969 is one of my favorite algorithms. Numerical Integration: Rectangle Method. Secant method d. The convergence to the root is slow, but is assured. This series of video tutorials covers the numerical methods for Root Finding (Solving Algebraic Equations) from theory to implementation. Newton's Method in Matlab. Online calculator. The task is to solve x^2=2 with the bisection method and the precision should be with 10 decimals. Bisection method In short, the bisection method will divide one triangle into two children triangles by connecting one vertex to the middle point of its opposite edge. Write a program that calculates the root(s) of a non-linear equation using the bisection method and also the secant method. The Algorithm The bisection method is an algorithm, and we will explain it in terms of its steps. For a given function as a string, lower and upper bounds, number of iterations and tolerance Bisection Method is computed. This book serves as a textbook for a first course in numerical methods using MATLAB to solve problems in mechanical, civil, aeronautical, and electrical engineering. Algorithm for Bisection Method: Input function and limits. X +1 X-1 X +2 X-2-4 X-3 X +3 f(x) x Figure 1. The problem is that it seems like the teachers recommended solution to the task isn't quite right. 6 Newton Method for a System of Nonlinear Equations 191. Find more Mathematics widgets in Wolfram|Alpha. Using C program for bisection method is one of the simplest computer programming approach to find the solution of nonlinear equations. "Bisection Method and Algorithm for Solving The Electrical Circuits" Iteration,Bisection Method, Fortran, C, MatLab. This scheme is based on the intermediate value theorem for continuous functions. So, it has a. Also, Newton’s method can be used to approximate complex roots, as well, if the initial value 0 is a complex number satisfying the conditions above. in the case of MATLAB®, 16 digits). 000003315132176 10 1. The idea is simple: divide the interval in two, a solution must exist within one subinterval, select the subinterval where the sign of. 2 Newton's Method and the Secant Method The bisection method is a very intuitive method for finding a root but there are other ways that are more efficient (find the root in fewer iterations). Im studying for a math test and on a old test there is a task about bisection. Richard Brent's improvements to Dekker's zeroin algorithm, published in 1971, made it faster, safer in floating point arithmetic, and guaranteed not to fail. Using the Bisection method to find the negative root (only) of the equation ,3C2 — ex — O (a) [5 points) Choose an initial interval [a, b]. 000000828382262 11 1. Also, Newton’s method can be used to approximate complex roots, as well, if the initial value 0 is a complex number satisfying the conditions above. Figure 1 At least one root exists between the two points if the function is real, continuous, and changes sign. Online calculator. 6524; m = 73. Figure 3: Synthetic seismic section displayed in “time from datum” and computed from the log section shown above. The points marked as X i are positions of the negative( )andpositive(+)endsoftherootenclosingbracket. Unless the roots of an equation are easy to find, iterative methods that can evaluate a function hundreds, thousands, or millions of times will be required. 2d Truss Analysis Matlab Program. ^3 - 2; exists. Then faster converging methods are used to find the solution. The basic idea is to switch between inverse quadratic interpolation and bisection based on the step performed in the previous iteration and based on inequalities gauging the difference between guesses:. The algorithm of bisection method is such that it can only find one root between a defined interval. Learn more about bisection method loop. Coding a bisection algorithm using matlab (numerical. Bisection Method for Solving non-linear equations using MATLAB(mfile) % Bisection Algorithm % Find the root of y=cos(x) from o to pi. Bisection – separate files and embedded functions c. then there is at least one real root in the interval. Access Ebooks on other topics. So let's take a look at how we can implement this. Need help with this bisection method code!. It is a very simple and robust method, but it is also relatively slow. 4 Basis of Bisection. m Algorithm 4. As we point out in the introduction, we will mainly discuss newest vertex bisection and include longest edge bisection as a variant of it. The number of iterations we will use, n, must satisfy the following formula:. Another was to say “root. 21 dcm_to_ypr. Bisection method in MATLAB A video of programming the code in realtime. Dekker's zeroin algorithm from 1969 is one of my favorite algorithms. Learn the algorithm of the bisection method of solving nonlinear equations of the form f(x)=0. The IVT states that suppose you have a segment (between points a and b, inclusive) of a continuous function, and that function crosses a horizontal line. Introduction a. k 1 is the slope at the beginning of the time step (this is the same as k 1 in the first and second order methods). Solutions to the Exercises from "An Introduction to MATLAB and Numerical Methods for Engineers. Find the midpoint of a and b, say "t". Bisection method b. , Vasiliou, P. Learn more about bisection. Approximate the root of f(x) = x 2 - 10 with the bisection method starting with the interval [3, 4] and use ε step = 0. m, instructions how to run it, an example of a file myfunction. Lab 9 - Bisection Method Introduction In this lab, we will explore a method that we have considered in class for solving nonlinear equations, the bisection method. Initialization: nd [a 1;b 1] ˆ[a;b], with f(a 1)f(b 1) <0, set i= 1. X +1 X-1 X +2 X-2-4 X-3 X +3 f(x) x Figure 1. A next search interval is chosen by comparing and nding which one has zero. Blog Archive. The first algorithm that I learned for root-finding in my undergraduate numerical analysis class (MACM 316 at Simon Fraser University) was the bisection method. So let's take a look at how we can implement this. 1 Polynomial Interpolation: Method of undetermined coefficients (Vandermonde. In mathematics, the bisection method is a root-finding algorithm which repeatedly bisects an interval then selects a subinterval in which a root must lie for further processing. Hi, I need help solving the function 600x^4-550x^3+200x^2-20x-1=0 using the Bisection and Secant method in MATLAB. At first, two interval-based methods, namely Bisection method and Secant method, are reviewed and implemented. MATLAB Programming Assignment Help, Root ?nding using the bisection method, In many applications, including ?nancial mathematics, ?nding zeros of a function f(x) = 0 (4) is paramount. The bisection search. This is achieved by selecting two points A and B on that interval. The convergence to the root is slow, but is assured. Analysis of the Problem. He used it for finding roots of cubic polynomials. m Algorithm 4. Golden section search Fibonacci search Bisection method Secant method Bracketing Chapter 5 Gradient Methods. Input, output. The c value is in this case is an approximation of the root of the function f (x). They are the secant method, bisection method, and newton's method. This series of video tutorials covers the numerical methods for Root Finding (Solving Algebraic Equations) from theory to implementation. The problem is that it seems like the teachers recommended solution to the task isn't quite right. It is obvious that the secant method does not always converge, but when it does, it does so faster than the bisection method. In our improved hybrid algorithm, we compute the x-intercept x using the Newton-Raphson method at the midpoint of the previous interval. It is a very simple and robust method, but it is also relatively slow. Open methods: Newton-Raphson method, Secant method. MATLAB has the function fzero which performs this bisection algorithm. Numerical Methods using MATLAB, 3e, is an extensive reference offering hundreds of useful and important numerical algorithms that can be implemented into MATLAB for a graphical interpretation to help researchers analyze a particular outcome. If a function is continuous between the two initial guesses, the bisection method is guaranteed to converge. Midpoint Method. 000000828382262 11 1. Fixed Point Method Using Matlab Huda Alsaud King Saud University Huda Alsaud Fixed Point Method Using Matlab. Analysis of the Problem. If the guesses are not according to bisection rule a message will be displayed on the screen. We set [a 0;b 0] = [a;b]. Blog Archive. Lecture 31-33 - Rootfinding Table of Contents 31. A few steps of the bisection method applied over the starting range [a 1;b 1]. m and newton. The setup of the bisection method is about doing a specific task in Excel. Bisection Methods: We can pursuse the above idea a little further by narrowing the interval until the interval within which the root lies is small enough. Application Of Bisection Method In The bisection method is an iterative algorithm used to find roots of continuous functions. Implement the Bisection algorithm elegantly and easily (3 answers) Closed 3 years ago. ————————————————. This method, also known as binary chopping or half-interval method, relies on the fact that if f(x) is real and continuous in the interval a < x < b, and f(a) and f(b) are of opposite signs, that is,. Root Search with the bisection method. The bisection method in Matlab is quite straight-forward. Joseph DeSimone, Applied Mathematics Graduate Student. The previous two methods are guaranteed to converge, Newton Rahhson may not converge in some cases. Essentially, the root is being approximated by replacing the. The instructions of the problem are: Use bisection method to find a root of the function $$\sin x + x \cos x = 0$$ Indicate your initial condition and how many steps it requires to reach the tole. Maple für Akademiker. 23 ground_track. Pseudo-code is a simple way to represent an algorithm in a logical and readable form. 5: Calculation of the state vector from the orbital elements. 6 in the text. Suppose we want to solve the equation. 2) using the bisection method. The Bisection Method looks to find the value c for which the plot of the function f crosses the x-axis. Bisection method add iteration table into my code. But they're not live. If (b i+1 a i+1)=2 > , set i= i+1 and go to step 1 4. GitHub Gist: instantly share code, notes, and snippets. We may minimize a convex f : → by finding a point at which f ′ = 0. (original given interval) Thanks for contributing an answer to Code Review Stack Exchange! Bisection method for finding the root of a function. Repeat steps 3 and 4 100 times. 5 Single Variable Newton- Raphson Method 9. m) and see how we can compute the root to a polynomial using this method. This series of video tutorials covers the numerical methods for Root Finding (Solving Algebraic Equations) from theory to implementation. Consider a transcendental equation f (x) = 0 which has a zero in the interval [a,b] and f (a) * f (b) < 0. This tutorial covers in depth algorithm for Bisection Method. Bisection Method is one of the simplest, reliable, easy to implement and convergence guaranteed method for finding real root of non-linear equations. The bisection method is guaranteed to converge to a root of the function f, if the function is continuous between the lower and upper bounds. It will helpful for engineering students to learn Bisection method MATLAB program easily. In simple terms, these methods begin by attempting to evaluate a problem using test (“false”) values for the variables, and then adjust the. Title: Microsoft Word - nle_03_bisection_advantages Author: sotirioschat Created Date: 1/15/2013 11:41:47 AM. 0has a root in [1, 2], and use the Bisection method to determine an approximation to the root that is accurate to at least within 10 −4. Bisection Method Example. Lab 9 - Bisection Method Introduction In this lab, we will explore a method that we have considered in class for solving nonlinear equations, the bisection method. First I plot the function and then I try to find a domain such that I can see the curve cut through the x -axis. Another class of mesh refinement method, known as regular refinement, which divide one triangle into 4 similar small triangles, is implemented in uniformrefine. Bisection Method // C++ code Posted: January 31, 2012 by muhammadakif in Algorithms Tags: bisection method , C# code , numerical analysis , numerical computing , numerical methods. Also, a good intermediate approximation may be discarded. It is the slowest algorithm provided by the library, with linear convergence. Initialization: nd [a 1;b 1] ˆ[a;b], with f(a 1)f(b 1) <0, set i= 1. Designing Robot Manipulator Algorithms. The help page states the following about the algorithm: Algorithms The [code ]fzero[/code] command is a function file. m Algorithm 4. Graphical method useful for getting an idea of what's going on in a problem, but depends on eyeball. You may not use MATLAB's built-in functions for finding roots -- instead, please implement two different algorithms. Algorithms in this toolbox can be used to solve general problems All algorithms are derivative-free methods Direct search: patternsearch Genetic algorithm: ga Simulated annealing/threshold acceptance: simulannealbnd, threshacceptbnd Genetic Algorithm for multiobjective optimization: gamultiobj Kevin Carlberg Optimization in Matlab. For more videos and resources on this topic, please visit http. Let m = (L+H)/2. Roots (Bisection Method) : FP1 Edexcel January 2012 Q2(a)(b) : ExamSolutions Maths Tutorials - youtube Video. I am trying to solve the equation f(x) = x^3 + x - 1 , by using Bisection method within the interval [ 0 , 1] , i have succeeded to generate a code to solve this equation but by using " while " function for looping , i need some one to help me to solve it by using " for " function , could any one help me to do that ? the code is :. "In mathematics, the bisection method is a root-finding algorithm which works by repeatedly dividing an interval in half and then selecting the subinterval in which a root exists. This is calculator which finds function root using bisection method or interval halving method. If the guesses are not according to bisection rule a message will be displayed on the screen. function [ r ] = bisection( f, a, b, N, eps_step, eps_abs ) % Check that that neither end-point is a root % and if f(a) and f(b) have the same sign, throw an exception. The bisection method is an approximation method to find the roots of the given equation by repeatedly dividing the interval. Bisection Method for Solving non-linear equations using MATLAB(mfile) % Bisection Algorithm % Find the root of y=cos(x) from o to pi. In this project we use MATLAB to analyze some of the numerical techniques. Im studying for a math test and on a old test there is a task about bisection. This scheme is based on the intermediate value theorem for continuous functions. It was observed that the Bisection method converges at the 14th iteration while Newton methods. In the secant method, it is not necessary that two starting points to be in opposite sign. 5 Single Variable Newton- Raphson Method 9. Next, we study some known numerical algorithms those can be used to find the approximate solutions (roots) for non-linear equations, which are Bisection algorithm, Newton–Raphson algorithm and fixed point algorithm. The main advantages to the method are the fact that it is guaranteed to converge if the initial interval is chosen appropriately, and that it is relatively. t is the root of the given function if f (t) = 0; else follow the next step. Step 2: Let c=(a+b)/2. The program assumes that the provided points produce a change of sign on the function under study. B The comparative results are shown in table 3. The proof that the binary search method works is provided by Bolzano's Bisection Theorem. Note that just as in the bisection algorithm, the initial two guesses must be such that one gives a positive function evaluation and the. The root is then approximately equal to any value in the final (very small) interval. At least one root exists between the two points if the function is real, continuous, and changes sign. 3 Newton's and secant methods 2. 2) using the bisection method. But they're not live. Introduction a. Plot error. The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a sub-interval in which a root must lie for further processing. Download MatLab Programming App from Play store. Bisection Method // C# code Posted: January 31, 2012 by Shahzaib Ali Khan in Algorithms Tags: bisection method , C# code , numerical analysis , numerical computing , numerical methods. Useful Computational Methods: The Bisection Method - Finding roots by binary search - Unlike the guess-and-check method, we start with two initial values - one value a below √Q and another value b above √Q, where Q is a positive real number. ContentsDirk DekkerZeroin in AlgolThe test functionBisectionSecant methodZeroin algorithmZeroin in MATLABReferencesDirk DekkerI. As a starting point, let's fix to be the function cosmx that you just wrote. 1997 CREWES software release CREWES Research Report — Volume 9 (1997) 18-3 Figure 2: The same cross section as above showing the result of the synthesis of an ensemble of new logs. 3 Limits of Accuracy 1. In this course, three methods are reviewed and implemented using Python and MATLAB from scratch. 1shows the several first iterations of the bisection algorithm. Designing Robot Manipulator Algorithms. It is a very simple and robust method, but it is also relatively slow. 1 Polynomial Interpolation: Method of undetermined coefficients (Vandermonde. The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. The c value is in this case is an approximation of the root of the function f (x). It requires two initial guesses and is a closed bracket method. Always Converges: like Bisection, it. In this project we use MATLAB to analyze some of the numerical techniques. It shows with bold stripes the length of the bracketed region. Newton's method is an iterative method. Error = x- (a+b)/2. The Bisection Method is used to find the zero of a function. The x-coordinate of this point is the average of the positive and negative guesses. Download MatLab Programming App from Play store. function [ r ] = bisection( f, a, b, N, eps_step, eps_abs ) % Check that that neither end-point is a root % and if f(a) and f(b) have the same sign, throw an exception. com 9/27/01. To code the bisection algorithm. In this video tutorial, the algorithm and MATLAB programming steps of finding the roots of a nonlinear equation by using bisection method are explained. The method is also called the interval halving method. Basic Bisection Algorithm: 1. We also check whether f(a) = 0 or f(b) = 0, and if so return the value of a or b and exit. Image: The Bisection Method explained. The problem is that it seems like the teachers recommended solution to the task isn't quite right. Here’s what we do: As with the bisection algorithm, start by choosing an interval [a,b] in which we. The bisection method can be easily adapted for optimizing 1-dimensional functions with a slight but intuitive. 0has a root in [1, 2], and use the Bisection method to determine an approximation to the root that is accurate to at least within 10 −4. f = @(x) (cos(x)); a = input( 'Please enter lower. Use The Following Pseudocode For The Bisection Method To Write MATLAB Code To Approximate The Root Of F(x) = Pt - X - 2, Interval (0,2), Tolerance 10-3, Maximum Number Of Iterations 50. m to determine the root of Equation (2. The convergence to the root is slow, but is assured. Repeat steps 3 and 4 100 times. Analysis of the Problem. The Bisection Method will cut the interval into 2 halves and check which. f(c)<0 then let b=c, else let a=c. Data scientists use a bisection search algorithm as a numerical approach to find a quick approximation of a solution. Steven Chapra's Applied Numerical Methods with MATLAB, third edition, is written for engineering and science students who need to learn numerical problem solving. Approximate the root of f(x) = x 2 - 10 with the bisection method starting with the interval [3, 4] and use ε step = 0. The IVT states that suppose you have a segment (between points a and b, inclusive) of a continuous function, and that function crosses a horizontal line. Considering that Scala is similar to the Java programming language, if anyone else needs the Interval-Halving method in Java, this code can easily be adapted to Java as well. The brief algorithm of the bisection method is as follows: Step 1: Choose a and b so that f(a). m - matlab file to determine the root of Equation (2. This method will divide the interval until the resulting interval is found, which is extremely small. (b) [5 points) Determine the number of steps N you should take to find the answer with tolerance 0. You have seen how Matlab functions can return several results (the root and the number of iterations, for example). Solve 2D Transient Heat Conduction Problem using FTCS Finite Difference Method. At first, two interval-based methods, namely Bisection method and Secant method, are reviewed and implemented. For others, an algorithm of Alefeld, Potra, and Shi is used. So in order to use live solutions, we're going to look at the Bisection Method and then the Golden Section Search Method. Before you start, review the \Introduction to MATLAB" notes. the Matlab code bisection. Consider a transcendental equation f (x) = 0 which has a zero in the interval [a,b] and f (a) * f (b) < 0. The file EULER. Implement the bisection method to find a zero of the function over [0,1]; Implement the Newton's method to find a zero of the function over [0,1]; Implement the secant method to find a zero of the function over [0,1]. ^3 - 2; exists. This process involves finding a root, or solution, of an equation of the form f(x) = 0 for a given function f. Consult the MATLAB TA's if you have any questions. Powered by Create your own unique website with customizable templates. I found where it was in the directory and added the folder to the path so when I entered it again I now get: C:\Users\Lulu\Documents\MATLAB\Numerical Optimisation\bisection. Viewed 601 times 0 $\begingroup$ To code the bisection algorithm. This method is suitable for finding the initial values of the Newton and Halley's methods. 5: Calculation of the state vector from the orbital elements. Design and simulation of three phase induction motor at different load conditions in matlab/simulink. ) (Use your computer code) I have no idea how to write this code. Root Search with the bisection method. Context Bisection Method Example Theoretical Result The Root-Finding Problem A Zero of function f(x) We now consider one of the most basic problems of numerical approximation, namely the root-finding problem. 7 Symbolic Solution for Equations 193. The bisection method is an iterative algorithm used to find roots of continuous functions. Open methods: Newton-Raphson method, Secant method. Analysis of the Problem. The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a sub-interval in which a root must lie for further processing. Readers can code the algorithms in the programs of their choice. The bisection method is an enclosure type method for finding roots of a polynomial f(x), i. - Matlab: function end -vs- SciLab: function endfunction - Matlab: [first, second, third] -vs- Scilab: [third, second, first] Lectures for Spring 2009 Intro to Matlab+ Bisection + Newton Method. MATLAB M-files for implementation of the discussed theory and algorithms (available via the book's website) Introduction to Optimization, Fourth Edition is an ideal textbook for courses on optimization theory and methods. The bisection method in math is the key finding method that continually intersect the interval and then selects a sub interval where a root must lie in order to perform the more original process. BISECTION_RC, a MATLAB library which demonstrates the simple bisection method for solving a scalar nonlinear equation in a change of sign interval, using reverse communication (RC). If the guesses are not according to bisection rule a message will be displayed on the screen. As a starting point, let's fix to be the function cosmx that you just wrote. - Use abstracted Autonomous Guided Vehicles (AGVs) system with P controller and supervisory controller as a case study to validate the conflict-driven fault detection method in Matlab. Bisection Method MATLAB Program Note: Bisection method guarantees the convergence of a function f(x) if it is continuous on the interval [a,b] (denoted by x1 and x2 in the above algorithm. The program assumes that the provided points produce a change of sign on the function under study. This is a repository where i put all of the implementation that i have done in numerical analysis. A simple improvement to the bisection method is the false position method, or regula falsi. We start from the 2D case sketched in [3] and the approximation scheme presented in [3, 6], and then we extend the reconstruction scheme of separatrices in the. 000013273393044 9 1. Bisection Method Example. The equation is of form, f(x) = 0. We have developed such an algorithm and it is given in the M-file regfals. f(a)*f(b) < 0. It is obvious that the secant method does not always converge, but when it does, it does so faster than the bisection method. MatLab Project 2 - Bisection Method, The Fixed-point Iteration, and Newton's Method Due October 10. He used it for finding roots of cubic polynomials. In simple terms, these methods begin by attempting to evaluate a problem using test (“false”) values for the variables, and then adjust the. Algorithm To find a solution. Bisection Method C Program Bisection Method MATLAB Program. Note that just as in the bisection algorithm, the initial two guesses must be such that one gives a positive function evaluation and the. Bisection Algorithm Input: computable f(x) and [a;b], accuracy level. Title: Microsoft Word - nle_03_bisection_advantages Author: sotirioschat Created Date: 1/15/2013 11:41:47 AM. Includes methods used in MATLAB, Mathcad, Mathematica, and various software libraries. Shown here, it is a function, and it crosses the X-axis at just before 2. In order to avoid the shortcoming of the hybrid algorithm[1], we suggest an improved hybrid algorithm. Solve 2D Transient Heat Conduction Problem using FTCS Finite Difference Method. 6 was used to nd the root of the function, f(x) = cosx xexp(x) on a close interval [0;1] using the Bisection method and Newton's method the result was compared. There is the graphical interface too. Bisection Method Bisection Method for finding roots of. ————————————————. This example shows how to generate HDL code from MATLAB® design implementing an bisection algorithm to calculate the square root of a number in fixed point notation. Bisection Method. In this video tutorial, the algorithm and MATLAB programming steps of finding the roots of a nonlinear equation by using bisection method are explained. Bisection method; Execute an instance method of Object and call in its block instance methods of another object; get URL Params (2 methods) Rake Migrate (newest method) order/format of params in method definition; XML Load methods; Kohana helper method for Askimet; Class vs Instance Methods; PHP5 Method Chaining Example. raphson method. This is the same as the slope, k 2 , from the second order midpoint method. It is a very simple and robust method, but it is also rather slow. This is a very simple and powerful method, but it is also relatively slow. Experiment 1. Bisection Method http//numericalmethods. What are the applications of the bisection. This is the first of a three part series. Numerical analysis I 1. Bisection method- code stops after one iteration. Calculates the root of the given equation f(x)=0 using Bisection method. Set r i= (a i+ b i)=2; 2. We won't dwell into explaining what each of them does but will jump straight into explaining the secant method. The basic idea is a follows. Methods by Young and Mohlenkamp, 2018 Bisection Method-- 4 Iterations by Hand (example) Bisection Method-- 4 Iterations by Hand (example) MATLAB Tutorial Part 6 Bisection Method Root finding matlab4engineers. Additional optional inputs and outputs for more control and capabilities that don't exist in other implementations of the bisection method or other root finding functions like fzero. The bisect algorithm is used to find the position in the list, where the data can be inserted to keep the list sorted. Method of Steepest Descent Analysis of Optimization Algorithms Analysis of Gradient Methods. Hi, I need help solving the function 600x^4-550x^3+200x^2-20x-1=0 using the Bisection and Secant method in MATLAB. The algorithm always selects a subinterval which contains a root. The bisection method is an application of the Intermediate Value Theorem (IVT). It is called the bisection method, and it is used for finding roots of a function f (that is, points c where f(c)=. Note that just as in the bisection algorithm, the initial two guesses must be such that one gives a positive function evaluation and the. Bisection Method http//numericalmethods. - Bisection method for bounded searching. Let us say; f(x,y) = 0 with degree eight and g(x,y) = 0 with degree six; I need a matlab code for 2D Bisection Method to solve f(x,y) = 0 and g(x,y) = 0 and find all possible roots. m; the Matlab code fixedpoint. Matlab will spit out that the root in this interval = '6'. Also use Euler's method for the same problem, and compare your results. Interpolation and approximation (12 lecture hours) 3. We almost have all the tools we need to build a basic and powerful root-finding algorithm, Newton's method*. In mathematics, the bisection method is a root-finding method that applies to any. The algorithm is iterative. bisect(list, element, begin, end). The Bisection Method will cut the interval into 2 halves and check which half interval contains a root of the function. 23 ground_track. bisection method using log10(x)-cos(x) Program to read a Non-Linear equation in one variable, then evaluate it using Bisection Method and display its kD accurate root Basic GAUSS ELIMINATION METHOD, GAUSS ELIMINATION WITH PIVOTING, GAUSS JACOBI METHOD, GAUSS SEIDEL METHOD. Next, we study some known numerical algorithms those can be used to find the approximate solutions (roots) for non-linear equations, which are Bisection algorithm, Newton–Raphson algorithm and fixed point algorithm. Above given Algorithm and Flowchart of Bisection Methods Root computation is a simple and easier way of understanding how the bracketing system works, algorithm and flowchart may not follow same procedure, yet they give the same outputs. GitHub Gist: instantly share code, notes, and snippets. m, instructions how to run it, an example of a file myfunction. Also, this method closely resembles with Bisection method. I found it was useful to try writing out each method to practice working with MatLab. Copy to clipboard. changes sign from. Newton's method requires both the function value and its derivative, unlike the bisection method that requires only the function value. The main advantages to the method are the fact that it is guaranteed to converge if the initial interval is chosen. How can we qualify more generally which method. What's great about the Bisection Method is that provided the conditions above are satisfied (and hence a root $\alpha$ exists in the interval $[a, b]$ by the Intermediate Value Theorem), then this method is guaranteed to zone into our root with better and better approximations. The algorithm of bisection method is such that it can only find one root between a defined interval. It is also known as Binary Search or Half Interval or Bolzano Method. Here we consider a set of methods that find the solution of a single-variable equation , by searching iteratively through a neighborhood of the domain, in which is known to be located. jf(r i)j<. This is more a problem of the algorithm than a MATLAB problem. In Matlab help functions written: The algorithm, created by T. It is a very simple and robust method, but it is also relatively slow. Let us say; f(x,y) = 0 with degree eight and g(x,y) = 0 with degree six; I need a matlab code for 2D Bisection Method to solve f(x,y) = 0 and g(x,y) = 0 and find all possible roots. MatLab Project 2 - Bisection Method, The Fixed-point Iteration, and Newton's Method Due October 10. Considering that Scala is similar to the Java programming language, if anyone else needs the Interval-Halving method in Java, this code can easily be adapted to Java as well. Find more Mathematics widgets in Wolfram|Alpha. Bisection algorithm The algorithm itself is fairly straightforward and "fast" in some sense: the number of iterations is roughly Log2 of the ratio of the initial interval length and the desired accuracy.
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https://marc-petit.com/wwhy41x0/276140-degree-of-expression | It is also known as an order of the polynomial. The degree of a term is the sum of the exponents of the variables that appear in it, and thus is a non-negative integer.For a univariate polynomial, the degree of the polynomial is simply the highest exponent occurring in the polynomial. • If the variable does not have an exponent, the degree is 1. Second Degree Polynomial Function. Trot transitions on a circle: Do transitions between collected and medium trot on a circle, being careful not to let the tempo change. The computer is able to calculate online the degree of a polynomial. Revise the sentences if necessary. The degree of a polynomial with a single variable (in our case, ), simply find the largest exponent of that variable within the expression. A polynomial function is an expression constructed with one or more terms of variables with constant exponents. If there are real numbers denoted by a, then function with one variable and of degree n … Expression of Interest Form To undertake a Higher Degree by Research in the Faculty of Science, Engineering and Built Environment. Second degree polynomials have at least one second degree term in the expression (e.g. With thousands of questions available, you can generate as many Degree of expression Worksheets as you want. abstract-algebra polynomials rational-functions. To identify the kind of algebraic expression and determine the degree, variables and constant in. Degree words are traditionally classified as adverbs, but actually behave differently syntactically, always modifying adverbs or adjectives and expressing a degree… • If a term does not contain any variable, the degree is 0. See more. The negative and weak positive staining rates in normal appearing mucosa, adenoma, and carcinoma were 42.5%, 71.4%, and 82.3%, respectively, indicating a decreased degree of KLF4 expression over the course of progressive transformation of normal cells into malignant derivatives. Therefore, the degree of this expression is . Degrees of Formality (Cont.) The degree of is 3. The correct answer is: [B]: → " 2xy⁴ + 4x²y³ – 6x³y² – 7x⁴" . The First Amendment of the U.S. Constitution protects the rights of individuals to freedom of religion, speech, press, petition, and assembly. 2. a letter to a Member of Parliament:Thank you for your help in this matter. $\begingroup$ To me, and this is standard terminology in theoretical CS, the degree of a boolean function would mean the degree of its polynomial (Fourier) representation. The highest value of the exponent in the expression is known as Degree of Polynomial. The degree of reaction (rth), or reaction ratio, is a parameter used for multistage turbomachinery defining the ratio of the static head (see Clearance gap pressure) to the fall head (of turbines) or the pump head (of centrifugal pumps).It is an indication of how the static pressure is distributed between impeller and stage. Download free printable Degree of expression Worksheets to practice. Solve this set of printable high school worksheets that deals with writing the degree of binomials. Read "Prevalence and degree of expression of the carbapenemase gene (cfiA) among clinical isolates of Bacteroides fragilis in Nottingham, UK, Journal of Antimicrobial Chemotherapy" on DeepDyve, the largest online rental service for scholarly research with thousands of … The degree ofreaction is then expressed as[For axial machines thenThe degree of reaction can also be written in terms of the geometry of theturbomachine as obtained bywhere is the vane angle of rotor outlet and is the vane angle of statoroutlet. Solve an algebraic expression with fractions. Degree of Binomials. It is sum of exponents of the variables in term. 5 Read the sentences below and decide whether or not they use the degree of formality appropriate for the given situation. Degree definition, any of a series of steps or stages, as in a process or course of action; a point in any scale. Example: x 3 y + x 2 + y x 3 y has degree 4 (3 for x and 1 for y) x 2 has degree 2 y has degree 1 So highest degree is 4, thus polynomial has degree 4 Degree of reaction (R) is an important factor in designing the blades of a turbine, compressors, pumps and other turbo-machinery. Substituting in this expression all numbers we already know, we obtain. And finally, the expression of degree by means of comparison, which is a special case of degree specification. Gene variability and degree of expression of vaccine candidate factor H binding protein in clinical isolates of Neisseria meningitidis. Degree of Polynomial - definition Degree of Polynomial is highest degree of its terms when Polynomial is expressed in its Standard Form. Click hereto get an answer to your question ️ Find the degree of the expression [ x + (x^3 - 1)^1/2]^5 + [ x - (x^3 - 1)^1/2]^5 Calculating the degree of a polynomial. For instance, the degree of 8x is 1. $$\frac{x^2+1}{6x-2}$$ If the above question has a degree, please tell me the difference between the degree of a polynomial and the degree of an expression? On comparing people or things, as bearers of a certain quality or characteristic, we do it by means of degree specification, thus in terms of positive, comparative, and superlative comparison. Correct answer to the question Which algebraic expression is a polynomial with a degree of 5? The degree is the value of the greatest exponent of any expression (except the constant) in the polynomial.To find the degree all that you have to do is find the largest exponent in the polynomial.Note: Ignore coefficients-- coefficients have nothing to do with the degree of a polynomial. Exploring these different trots can help you to increase your horse’s collection, cadence and expression. The collection has to have the same degree … For instance, the degree … Cox proportional hazards model was used to assess the relationship between the degree of expression of DNMT1 and overall survival after adjusting for relevant covariates. 1. a note to a co-worker:The meeting is at 10 sharp.Don’t be late. • If a term has more than one variable, the degree is equal to the sum of the exponents of all its variables. The calculator may be used to determine the degree of a polynomial. The quadratic function f(x) = ax 2 + bx + c is an example of a second degree polynomial. Freedom of expression refers to the ability of an individual or group of individuals to express their beliefs, thoughts, ideas, and emotions about different issues free from government censorship. Thank you for your interest in studying a Higher Degree by Research (HDR) course at Deakin University. But it seems unlikely that the term is used in this sense here; also I do not see a nice way … Remember that a polynomial is any algebraic expression that consists of terms in the form $$a{x^n}$$. There are no higher terms (like x 3 or abc 5). For the reaction : 2HI(g) ⇌ H2 (g)+ I2 (g), the degree of dissociated (α) of HI(g) is related to equilibrium constant Kp by the expression: asked Jan 6, 2020 in Chemistry by Mousam ( 52.8k points) While finding the degree of the polynomial, the polynomial powers of the variables should be either in ascending or descending order. Find the degree of each term and then compare them. If you're interested in studying the Associate Degree of Advanced Manufacturing, please send an expression of interest to us at [email protected]. Degree of Polynomials. Here's how you would do it: (x + 3)/6 = 2/3 First, cross multiply to get rid of the fraction. The degree of a polynomial is the largest exponent. Free Class 7 Degree of expression Worksheets. Note: This answer choice is a polynomial with a degree of "5" .Note that there are 2 (TWO) variables in this polynomial expression The expression of DNMT1 was examined using immunohistochemistry, and the degree of expression of DNMT1 was expressed as a percentage of cells positive for DNMT1 and its intensity. Another way to write the last example is $- 8{x^0}$ Written in this way makes it clear that the exponent on the $$x$$ is a zero (this also explains the degree…) and so we can see that it really is a polynomial in one variable. In mathematics, the degree of a polynomial is the highest of the degrees of the polynomial's monomials (individual terms) with non-zero coefficients. Once assessed, you will also need to apply for 609996 Associate Degree of Advanced Manufacturing via UAC. The term shows being raised to the seventh power, and no other in this expression is raised to anything larger than seven. The highest power is the degree of … To obtain the degree of a polynomial defined by the following expression x^3+x^2+1, enter : degree(x^3+x^2+1) after calculation, the result 3 is returned. The degree of is 6. Polynomial is a mathematical expression consisting of variables, constants that can be combined using mathematical operations addition, subtraction, multiplication and whole number exponentiation of … In turbomachinery, Degree of reaction or reaction ratio (R) is defined as the ratio of the static pressure drop in the rotor to the static pressure drop in the stage or as the ratio of static enthalpy drop in the rotor to the static enthalpy drop in the stage.. If you want to solve an algebraic expression that uses fractions, then you have to cross multiply the fractions, combine like terms, and then isolate the variable. 2x 2, a 2, xyz 2). Degree of Dissociation. 4x3 – StartFraction 2 Over x EndFraction 2y3 + 5y2 – 5y 3y3 – StartRoot 4 y EndRoot Question: what is the degree of the following expression? The Fixed Class of Degree Words "[An] example of words that don't fit neatly into one category or another is degree words. The degree of dissociation is the phenomenon of generating current carrying free ions, which are dissociated from the fraction of solute at a given concentration. In Freemasonry, the third and highest degree is that of master mason, attained after a stiff examination, and several writers speculate that this may be the source of the late nineteenth-century expression for an inquisition. There are 2 offer rounds remaining for UAC, so please check the submission deadlines with UAC. Kelly A(1), Jacobsson S, Hussain S, Olcén P, Mölling P. Author information: (1)Department of Clinical Medicine, School of Health and Medical Sciences, Örebro University, Sweden. 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Cozy Comfort Slippers, Lsu Meal Plans 2018, Robert Porcher Madden 21, Keep It In The Pocket Meaning, Uconn Health Center Login, Dodge Dakota Aftermarket Parts, Class I Felony North Carolina, | 2021-05-11T05:00:15 | {
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https://math.stackexchange.com/questions/2245798/mathematical-field-show-that-0-a-0-1-a-a | # Mathematical field, show that 0·a = 0, (-1)·a = -a, …
Based on the axioms for a mathematical field, the wiki article states that 0·a = 0 and (-1)·a = -a are consequences of the axioms, but doesn't show how they are derived. There was a similar question asked before, but I'm not sure about the accepted answer.
https://en.wikipedia.org/wiki/Field_(mathematics)#Elementary_consequences_of_the_definition
Also, it would seem logical that if a ≠ b, and if c ≠ 0, then c·a ≠ c·b, a uniqueness property that should hold true for a finite field (unordered) that I'm wondering if it can be derived from the axioms (perhaps something like induction?) .
• These are in fact standard easy arguments (your guess in the last sentence is correct). I'm sure someone will step up and provide them. Here's a hint for the first one: consider $(0 + 0) \times a$. I'm curious: what prompts your curiousity? A course you're taking? – Ethan Bolker Apr 22 '17 at 0:12
• @EthanBolker - a question came up at another forum. An "addition" table was given for a finite field with 4 numbers (0,1,2,3), where the "addition" turns out to be exclusive or, and the problem was asking to produce the multiplication table, based on the axioms rather than knowledge of GF(4). That prompted me to wonder how the "consequences" as noted in the wiki article are derived, and what operations are considered acceptable as derivations for field math. – rcgldr Apr 22 '17 at 0:42
• @EthanBolker - I'm aware that algebra works with finite finite fields (I've worked with RS ECC), but never thought about the derivations based on the axioms. For that other forum question, after taking into account 0·a = a·0 = 0 and 1·a = a·1 = a, only 4 products in the multiplication table (2·2, 2·3, 3·2, 3·3) needed to be determined, so this could be done by trial and error, where an axiom would fail if the wrong set of 4 products was chosen. Noting that every number has a multiplicative inverse, meant that two of those products = 1 => 2·3 = 3·2 = 1, which simplified the problem. – rcgldr Apr 22 '17 at 1:06
For $0 \times a$: $$0 \times a = (0+0)\times a = 0\times a + 0 \times a .$$ Now whatever $0 \times a$ is, it has an additive inverse, so you can subtract it from each side of that equation to conclude that $0 \times a = 0$.
For $(-1) \times a$: $$0 \times a = (1 + (-1)) \times a = 1 \times a + (-1) \times a = a + (-1) \times a$$ but $a$ has a unique additive inverse $-a$.
If $c \ne 0$ then it has a multiplicative inverse $d$. Then $$ca = cb \implies ca - cb = c(a-b) = 0 \implies dc(a-b) = 0 \implies a-b = 0 \implies a = b.$$
• Shouldn't the third part somewhere include a - b ≠ 0 or perhaps a - b = e ≠ 0 ? – rcgldr Apr 22 '17 at 0:26
• @rcgldr No. I proved your claim by reasoning with the contrapositive form. If $a \ne b$ then the last entry in my chain of correct implications is false, so the first entry must be false. – Ethan Bolker Apr 22 '17 at 0:30
• OK, thanks for the explanation. I was also considering let a-b = e ≠ 0, then ca - cb = ce ≠ 0, since c ≠ 0 and e ≠ 0. – rcgldr Apr 22 '17 at 0:35
I presume that by "-a" you mean the additive inverse of a so that you want to prove that -1 times a is the additive inverse of a. Again, that follows from the distributive law. (1+ (-1))a= 0a= 0. But, by the distributive law, (1+ (-1)))a= 1a+ (-1)a= a+ (-1)a= 0 also. "a+ (-1)a= 0" is precisely what is meant by "additive inverse of a".
The statement "if a ≠ b, and if c ≠ 0, then c·a ≠ c·b" is most easily proved by proving the "contrapositive". For any statement "if p then q", the contrapositive is the statement "if not q then not p" and it is easy to show in general that a statement is true if and only if its contrapositive is true.
The contrapositive of "if a ≠ b, and if c ≠ 0, then c·a ≠ c·b" is "if ca= cb then a= b". To prove that add the additive inverse of (subtract) cb from both sides: ca- cb= 0. By the distributive law, c(a- b)= 0. Since c is not 0 we must have a- b= 0 so a= b.
• Should the contrapositive be stated as $$\text{If ca=cb the either a=b or c=0}?$$ – Juniven Apr 22 '17 at 0:37
$(-1)a=-a$ means $a+(-1)a=0$ as additive inverses are unique.
$a+(-1)a=1*a+(-1)a$ (existence of multiplicative identity.
$=(1+(-1))a$ (distributive property)
$=0*a$. Remains to show $0*a=0$ for all $a$
$0*a= (0+0)*a$ (Associativity and definition of additive property.
$=0*a+0*a$.
$0a=0a+0a$
$0a+(-0a)=0a+0a+(-0a)$ (existence of additive inverse and acknowledgement that addition is a closed binary opperation)
$0=0a+0=0a$(additive inverse and implied associtivity.)
So $a+(-1)a=0$. So $(-1)a=-a$ (and $a= -(-1 (a))$.) | 2021-06-14T00:49:20 | {
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https://www.fyears.org/tests-as-linear/simulations/simulate_spearman.html | This document presents the close relationship between Spearman correlation and Pearson correlation. Namely, that Spearman correlation is just a Pearson correlation on $$rank$$ed $$x$$ and $$y$$. It is an appendix to the post “Common statistical tests as linear models”.
TL;DR: Below, I argue that this approximation is good enough when the sample size is 12 or greater and virtually perfect when the sample size is 20 or greater.
# 1 Simple example
Here, I start by creating some mildly correlated data. Note that the distribution of the data does not matter since everything is ranked, so the results are identical for non-normal data.
data = MASS::mvrnorm(50, mu=c(2, 2), Sigma=cbind(c(1, 0.4), c(0.4, 1)))
x = data[,1]
y = data[,2]
plot(rank(y) ~ rank(x))
Now let’s test it
# Three ways of running the same model
spearman = cor.test(x, y, method='spearman')
pearson = cor.test(rank(x), rank(y), method='pearson') # On ranks
linear = summary(lm(rank(y) ~ rank(x))) # Linear is identical to pearson; just showing.
# Present the results
print('coefficient rho is exactly identical:')
rbind(spearman=spearman$estimate, ranked_pearson=pearson$estimate,
ranked_linear=linear$coefficients[2]) print('p is approximately identical:') rbind(spearman=spearman$p.value,
ranked_pearson=pearson$p.value, ranked_linear=linear$coefficients[8])
## [1] "coefficient rho is exactly identical:"
## rho
## spearman 0.1848259
## ranked_pearson 0.1848259
## ranked_linear 0.1848259
## [1] "p is approximately identical:"
## [,1]
## spearman 0.1982118
## ranked_pearson 0.1988046
## ranked_linear 0.1988046
The correlation coefficients are exact. And the p-values are quite close! But maybe we were just lucky? Let’s put it to the test:
# 2 Simulate for different N and r
The code below does the following:
1. Generate some correlated data for all combinations of the correlation coefficient ($$r = 0, 0.5, 0.95$$) and sample sizes 1 $$N = 6, 8, 10, ... 20, 30, 50, 80$$.
2. Compute coefficients and p-values for Spearman correlation test and Pearson correlation on the ranked values, many times for each combination.
3. Calculate the number of errors.
library(tidyverse)
# Settings for data simulation
PERMUTATIONS = 1:200
Ns = c(seq(from=6, to=20, by=2), 30, 50, 80) # Sample sizes to model
rs = c(0, 0.5, 0.95) # Correlation coefficients to model
# Begin
D = expand.grid(set=PERMUTATIONS, r=rs, N=Ns) %>% # Combinations of N and r
mutate(
# Use the parameters to generate correlated data in each row
data = map2(N, r, function(N, r) MASS::mvrnorm(N, mu=c(0, 4), Sigma=cbind(c(1, r), c(r, 1)))),
# Tests
pearson_raw = map(data, ~cor.test(rank(.x[, 1]), rank(.x[, 2]), method='pearson')),
spearman_raw = map(data, ~cor.test(.x[, 1], .x[, 2], method = 'spearman')),
# Tidy it up
pearson = map(pearson_raw, broom::tidy),
spearman = map(spearman_raw, broom::tidy),
) %>%
# Get estimates "out" of the tidied results.
unnest(pearson, spearman, .sep='_') %>%
select(-data, -pearson_raw, -spearman_raw)
# If you want to do a permutation-based Spearman to better handle ties and
# overcome issues on how to compute p (computationally heavy).
# Corresponds very closely to R's spearman p-values
#mutate(p.perm = map(data, ~ coin::pvalue(coin::spearman_test(.x[,1] ~ .x[,2], distribution='approximate')))) %>%
#unnest(p.perm)
head(D)
### 2.0.1 Inspect correlation coefficients
Just as we saw in the simple example above, correlation coefficients (estimate and estimate1) are identical per definition. We can see that by calculating the difference and see that it is always zero:
summary(D$estimate - D$estimate1)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
##
Yup. The minimum difference is zero. The maximum is zero. All of them are identicalNo need to look further into that.
### 2.0.2 Inspet p-values
Before getting started on p-values, note that there are several ways of calculating them for Spearman correlations. This is the relevant section from the documentation of cor.test:
For Spearman’s test, p-values are computed using algorithm AS 89 for n < 1290 and exact = TRUE, otherwise via the asymptotic t approximation. Note that these are ‘exact’ for n < 10, and use an Edgeworth series approximation for larger sample sizes (the cutoff has been changed from the original paper).
You can get beyond all of that by doing permutation-based tests at the cost of computational power (see outcommented code above). However, there is almost virtually perfect correspondence, so I just rely on R’s more computationally efficient p-values.
Below, I plot the difference between Spearman-derived p-values and ranked-Pearson-derived p-values. I have chosen not to plot as a function or r in the name of simplicity because it makes no difference whatsoever, but feel free to try it out yourself.
library(patchwork)
# A straight-up comparison of the p-values
p_relative = ggplot(D, aes(x=spearman_p.value, y=pearson_p.value, color=N)) +
geom_line() +
geom_vline(xintercept=0.05, lty=2) +
geom_hline(yintercept=0.05, lty=2) +
labs(title='Absolute relation', x = 'Spearman p-value', y = 'Pearson p-value') +
#coord_cartesian(xlim=c(0, 0.10), ylim=c(0, 0.11)) +
theme_gray(13) +
guides(color=FALSE)
# Looking at the difference (error) between p-values
p_error_all = ggplot(D, aes(x=spearman_p.value, y=pearson_p.value-spearman_p.value, color=factor(N))) +
geom_line() +
geom_vline(xintercept=0.05, lty=2) +
labs(title='Error', x='Spearman p-value', y='Difference') +
coord_cartesian(ylim=c(-0.02, 0.005)) +
theme_gray(13) +
guides(color=FALSE)
# Same, but zoomed in around p=0.05
p_error_zoom = ggplot(D, aes(x=spearman_p.value, y=pearson_p.value-spearman_p.value, color=factor(N))) +
geom_line() +
geom_vline(xintercept=0.05, lty=2) +
labs(title='Error zoomed', x='Spearman p-value', y='Difference') +
coord_cartesian(ylim=c(-0.02, 0.005), xlim=c(0, 0.10)) +
theme_gray(13)
# Show it. Patchwork is your friend!
p_relative + p_error_all + p_error_zoom
Figure: The jagget lines in the leftmost two panels are due to R computing “exact” p-values for Spearman when N < 10. For samples larger than N>=10, an approximation is used which gives a smoother relationship.
It is clear that larger N gives a closer correspondence. For small p, the Pearson-derived p-values are lower than Spearman, i.e. more liberal. Importantly, when N=12, this effect is at most 0.5% in the “significance-region” which I deem acceptable enough for the present purposes. When N=20, the error is at most 0.2% for the whole curve and there is no difference when N=50 or N=100.
# 3 Conclusion
The correlation coefficients are identical.
The p-value is exact enough when N > 10 and virtualy perfect when N > 20. If you are doing correlation studies on sample sizes lower than this, you have larger inferential problems than those posed by the slight differences between Spearman and Pearson. | 2022-06-26T09:11:34 | {
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https://www.freemathhelp.com/forum/threads/negative-roots-when-simplifying-surds.114156/ | # Negative roots when simplifying surds
#### NeedingWD40
##### New member
Something that puzzles me:
The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?
e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3
what about the negative root of 100, and the negative root of 16?
Thanks in advance for your help.
#### Dr.Peterson
##### Elite Member
Something that puzzles me:
The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?
e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3
what about the negative root of 100, and the negative root of 16?
Thanks in advance for your help.
When we write the radical sign, it stands for the square root function, which means it has only one value. We call this the principal value, namely the non-negative root. The symbol does not represent both roots, but just one.
So although 100 has two square roots, ⎷100 is just 10. And that is what you are given when you simplify a surd, so you ignore the negative roots.
Have you noticed that if you chose always to keep both roots, an expression like yours would have more and more values? To take a simpler example, ⎷9 -⎷4 would have to be (±3) - (±2), with no requirement that the two signs are correlated, so its values would be 3 - 2 = 1, 3 - -2 = 5, -3 - 2 = -5, and -3 - -2 = -1. More radicals would make it even more confusing! That's why we insist on a square root function.
#### Jomo
##### Elite Member
Something that puzzles me:
The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?
e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3
what about the negative root of 100, and the negative root of 16?
Thanks in advance for your help.
By definition, $$\displaystyle \sqrt{x}$$ is > 0 providing x > 0. If x<0, then $$\displaystyle \sqrt{x}$$ has no real solution.
#### NeedingWD40
##### New member
When we write the radical sign, it stands for the square root function, which means it has only one value. We call this the principal value, namely the non-negative root. The symbol does not represent both roots, but just one.
So although 100 has two square roots, ⎷100 is just 10. And that is what you are given when you simplify a surd, so you ignore the negative roots.
Have you noticed that if you chose always to keep both roots, an expression like yours would have more and more values? To take a simpler example, ⎷9 -⎷4 would have to be (±3) - (±2), with no requirement that the two signs are correlated, so its values would be 3 - 2 = 1, 3 - -2 = 5, -3 - 2 = -5, and -3 - -2 = -1. More radicals would make it even more confusing! That's why we insist on a square root function.
Ah! That's why the formula for solving a quadratic equation explicitly says + or -⎷(b2-4ac), otherwise only the positive root would be found. Thank you very much, I'm clearer now
#### JeffM
##### Elite Member
Something that puzzles me:
The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?
e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3
what about the negative root of 100, and the negative root of 16?
Thanks in advance for your help.
You are absolutely correct that
$$\displaystyle (-\ 10)^2 = 100 \text { and } (+\ 10)^2 = 100.$$
But the technical definition of the square root function of a non-negative real number is:
$$\displaystyle \text {If } a \ge 0, \text { then } \sqrt{a} \ge 0 \text { and } \sqrt{a} * \sqrt{a} = a.$$
That is why if we get an equation that looks like
$$\displaystyle x^2 = 100 \implies x = \pm \sqrt{100} \implies x = 10 \text { or } x = -\ 10.$$
That plus or minus sign is there for a reason.
#### NeedingWD40
##### New member
By definition, $$\displaystyle \sqrt{x}$$ is > 0 providing x > 0. If x<0, then $$\displaystyle \sqrt{x}$$ has no real solution.
Thank you, I'm clearer now.
#### NeedingWD40
##### New member
You are absolutely correct that
$$\displaystyle (-\ 10)^2 = 100 \text { and } (+\ 10)^2 = 100.$$
But the technical definition of the square root function of a non-negative real number is:
$$\displaystyle \text {If } a \ge 0, \text { then } \sqrt{a} \ge 0 \text { and } \sqrt{a} * \sqrt{a} = a.$$
That is why if we get an equation that looks like
$$\displaystyle x^2 = 100 \implies x = \pm \sqrt{100} \implies x = 10 \text { or } x = -\ 10.$$
That plus or minus sign is there for a reason.
So, I'm thinking, if a problem includes the square root function, but no plus or minus, you assume the primary root, however, if you create a square root when solving a problem you need to be aware of the negative root in the context of the question....
#### tkhunny
##### Moderator
Staff member
So, I'm thinking, if a problem includes the square root function, but no plus or minus, you assume the primary root, however, if you create a square root when solving a problem you need to be aware of the negative root in the context of the question....
"problem includes ... no plus or minus" - Not a fan.
What is $$\displaystyle \sqrt{4}$$? It is 2
I do not ever recall seeing such a simplification problem as: What is $$\displaystyle \pm\sqrt{4}$$? It is +/- 2.
Let the context be your guide.
Don't make up things that don't exist. $$\displaystyle \sqrt{4} = 2$$ and that is NOT the same as "+/-2".
Don't miss things that do exist. $$\displaystyle Solve\;x^{2} = 4$$. $$\displaystyle x = +2\;or\;x = -2$$ and that is the same as "x = +/-2".
Also, keep your eyes on that "+/-" symbol. It means several different things in different contexts.
#### Otis
##### Senior Member
I'm not sure whether you've yet learned anything about solving algebraic equations or the concept of a function (you posted on the Pre-Algebra board), but here's my comments, anyways.
When you are given a surd, it always represents the positive root (only is exception is √0).
When you are solving an equation, sometimes you need to introduce surds (like the algebraic step of taking the square root of each side of an equation). In this case, you need to consider both the positive and negative root. Otherwise, you might miss solution(s).
As Jeff wrote, we express a positive root as √n, and when we need to express the negative root we write -√n. Cheers :cool:
#### JeffM
##### Elite Member
So, I'm thinking, if a problem includes the square root function, but no plus or minus, you assume the primary root, however, if you create a square root when solving a problem you need to be aware of the negative root in the context of the question....
Please read the other responses to this post because they are good and will give you different perspectives.
$$\displaystyle \sqrt{a} \ge 0$$ by definition if we are dealing with real numbers.
So, we do not assume that the square root is the primary root (meaning a non-negative number), we know that it is a non-negative number by definition.
However, if the square root of a is positive, we know that the square of the square root's additive inverse is also a. In notation,
$$\displaystyle \sqrt{a} * \sqrt{a} = a = (-\ \sqrt{a}) * (-\ \sqrt{a}).$$
So if we have an equation like
$$\displaystyle x^2 = c \ge 0$$,
we have no way to know from the equation itself whether
$$\displaystyle x = \sqrt{c} \text { or } x = - \ \sqrt{c}.$$
It may be that either solution works or that only one of the two works, but that must be decided based on information not contained in the equation itself.
Uncertainty about which answer applies does not mean that there is any uncertainty about what the surd means. To clarify where the uncertainty lies we say
$$\displaystyle x = \pm \sqrt{c}.$$
The sign of the surd is certain. It is the sign of x that is uncertain.
Last edited:
#### pka
##### Elite Member
Something that puzzles me:
The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?
e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3
what about the negative root of 100, and the negative root of 16?
I have had issues with this my whole active research & teaching life.
I had problems with students in my complex variables classes with the use of the radical: $$\displaystyle \sqrt{z}$$.
Now I belong to a school of analyst who think that the symbol $$\displaystyle \sqrt~$$ must be applied only to a non-negative real number.
So that means $$\displaystyle \sqrt{-1}$$ has no meaning whatsoever. So what is $$\displaystyle \bf{i}~?$$
It is so simple, $$\displaystyle \bf{i}$$ is a solution of the equation of the equation $$\displaystyle z^2+1=0$$
Now clearly the number $$\displaystyle 100$$ has two square roots, $$\displaystyle \pm 10$$.
BUT $$\displaystyle \sqrt{100}=10$$ that is one number. Moreover, $$\displaystyle -\sqrt{100}=-10$$ that is one number.
Thus it is absolutely incorrect to write that $$\displaystyle \sqrt{100}=\pm 10$$
#### JeffM
##### Elite Member
I have had issues with this my whole active research & teaching life.
I had problems with students in my complex variables classes with the use of the radical: $$\displaystyle \sqrt{z}$$.
Now I belong to a school of analyst who think that the symbol $$\displaystyle \sqrt~$$ must be applied only to a non-negative real number.
So that means $$\displaystyle \sqrt{-1}$$ has no meaning whatsoever. So what is $$\displaystyle \bf{i}~?$$
It is so simple, $$\displaystyle \bf{i}$$ is a solution of the equation of the equation $$\displaystyle z^2+1=0$$
Now clearly the number $$\displaystyle 100$$ has two square roots, $$\displaystyle \pm 10$$.
BUT $$\displaystyle \sqrt{100}=10$$ that is one number. Moreover, $$\displaystyle -\sqrt{100}=-10$$ that is one number.
Thus it is absolutely incorrect to write that $$\displaystyle \sqrt{100}=\pm 10$$
I like this convention that the surd can only be applied to non-negative real numbers. It would avoid a lot of confusion.
Unfortunately, conventions are necessarily extra-individual. And the surd is frequently used in a broader sense.
Anyway, I do like your school's approach. But is there not an inconsistency when it comes to
$$\displaystyle \sqrt[3]{-\ 8}$$
or is that prohibited as well. If so, is there a convenient notation to deal with it?
#### pka
##### Elite Member
Anyway, I do like your school's approach. But is there not an inconsistency when it comes to
$$\displaystyle \sqrt[3]{-\ 8}$$
or is that prohibited as well. If so, is there a convenient notation to deal with it?
No, because the any real number has a real root if the index is odd. Example $$\displaystyle \large\sqrt[5]{-32}=-2$$
#### JeffM
##### Elite Member
No, because the any real number has a real root if the index is odd. Example $$\displaystyle \large\sqrt[5]{-32}=-2$$
Ahh so your school's rule is
$$\displaystyle \sqrt[a]{b} \text { is defined only if (1) } a \in \mathbb Z \text {, (2) } a \ge 1 \text {, (3) } \ b \in \mathbb R,$$
$$\displaystyle \text {and (4) } a \text { even } \implies b \ge 0.$$
#### pka
##### Elite Member
Ahh so your school's rule is
$$\displaystyle \sqrt[a]{b} \text { is defined only if (1) } a \in \mathbb Z \text {, (2) } a \ge 1 \text {, (3) } \ b \in \mathbb R,$$
$$\displaystyle \text {and (4) } a \text { even } \implies b \ge 0.$$
.
At meetings we would have new jokes about exponents. Prof Paterson has address this above.
The objection is this $$\displaystyle \sqrt{-16}=\pm 4\bf{i}$$. I dare say you can some such in many basic mathematics textbooks.
It is true that many in my tradition are hyper sensitive to vocabulary abuse.
There are four fourth roots of $$\displaystyle -16\bf{i}$$ but there is just one answer to $$\displaystyle \large\sqrt[4]{-16\bf{i}}=~?$$
#### lookagain
##### Senior Member
NeedingWD40 said:
e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3
The above is incorrect. You need grouping symbols, such as below. I used asterisks for the products.
e.g.
⎷300 - ⎷48 =
⎷(100*3) - ⎷(16*3) =
10⎷3 - 4⎷3 =
6⎷3
Last edited:
#### topsquark
##### Full Member
.
At meetings we would have new jokes about exponents. Prof Paterson has address this above.
The objection is this $$\displaystyle \sqrt{-16}=\pm 4\bf{i}$$. I dare say you can some such in many basic mathematics textbooks.
It is true that many in my tradition are hyper sensitive to vocabulary abuse.
There are four fourth roots of $$\displaystyle -16\bf{i}$$ but there is just one answer to $$\displaystyle \large\sqrt[4]{-16\bf{i}}=~?$$
When I see stuff like "i"s under a radical I go straight over to the exponential format.
$$\displaystyle \sqrt[4]{-16 i } = \sqrt[4]{16} \cdot \sqrt[4]{-i}$$
$$\displaystyle = 2 \cdot \left ( e^{ 3i \pi/2 + 2 i \pi n } \right ) ^{1/4}$$
$$\displaystyle = 2 \cdot e^{3i \pi /8 + i \pi n / 2}$$
$$\displaystyle = 2 \cdot e^{3i \pi / 8} \cdot i^n$$
$$\displaystyle = 2 (i ^n) \left ( cos ( 3 \pi /8 ) + i~sin (3 \pi / 8) \right )$$
where n is an integer.
Perhaps it's a bit messy but I don't have to think so hard about any negative signs.
-Dan
#### pka
##### Elite Member
When I see stuff like "i"s under a radical I go straight over to the exponential format.
$$\displaystyle \sqrt[4]{-16 i } = \sqrt[4]{16} \cdot \sqrt[4]{-i}$$
-Dan
I did a very poor job explaining our objections. No one would ever need to use $$\displaystyle \sqrt[4]{-16 i }.$$
On the other hand there could be a need to solve $$\displaystyle z^4+16\bf{i}=0$$
There are four roots of that equation. If $$\displaystyle \rho=2\exp\left(\frac{-\pi\bf{i}}{8}\right)$$ then $$\displaystyle \rho^4+16\bf{i}=0$$, in other words one solution.
Let $$\displaystyle \zeta=\exp\left(\frac{2\pi\bf{i}}{4}\right) \$$ now the collection $$\displaystyle \rho\cdot\zeta^k,~k=1,2,3$$ are the other three roots(solutions).
#### NeedingWD40
##### New member
Thanks all, one clear message I've got is I need to clarify my sometimes foggy thought processes!
So, to attempt to summarise:
⎷a is always positive.
The solution to an equation involving a square root may have two solutions:⎷a or -⎷a
The context of the question should clarify if the negative of the root gives a valid solution.
#### JeffM
##### Elite Member
Thanks all, one clear message I've got is I need to clarify my sometimes foggy thought processes!
So, to attempt to summarise:
⎷a is always positive.
The solution to an equation involving a square root may have two solutions:⎷a or -⎷a
The context of the question should clarify if the negative of the root gives a valid solution.
That is very good. I'd make one slight emendation to your third point. I'd phrase it as
The context of the equation should clarify if both solutions are valid or else which one of the two is valid.
I would not assume that the negative solution is the only one that may not apply. In some cases, what is classified as positive and negative is arbitrary. It is of course true that we usually classify in a way that ensures a valid answer is non-negative, but that is not guaranteed.
Very minor point. You seem to have it. Good job. | 2019-05-19T20:29:00 | {
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http://math.stackexchange.com/questions/89623/trick-to-find-multiples-mentally | # Trick to find multiples mentally
We all know how to recognize numbers that are multiple of $2, 3, 4, 5$ (and other). Some other divisors are a bit more difficult to spot. I am thinking about $7$.
A few months ago, I heard a simple and elegant way to find multiples of $7$:
Cut the digits into pairs from the end, multiply the last group by $1$, the previous by $2$, the previous by $4$, then $8$, $16$ and so on. Add all the parts. If the resulting number is multiple of $7$, then the first one was too.
Example:
$21553$
Cut digits into pairs:
$2, 15, 53$
Multiply $53$ by $1, 15$ by $2, 2$ by $4$:
$8, 30, 53$
$8+30+53=91$
As $91$ is a multiple of $7$ ($13 \cdot 7$), then $21553$ is too.
This works because $100-2$ is a multiple of 7. Each hundreds, the last two digits are 2 less than a multiple of $7 (105 → 05 = 7 - 2, 112 → 12 = 14 - 2, \cdots)$
I figured out that if it works like that, maybe it would work if we consider $7=10-3$ and multiplying by $3$ each digit instead of $2$ each pair of digits.
Exemple with $91$:
$91$
$9, 1$
$9\cdot3, 1\cdot1$
$27, 1$
$28$
My question is: can you find a rule that works with any divisor? I can find one with divisors from $1$ to $19$ ($10-9$ to $10+9$), but I have problem with bigger numbers. For example, how can we find multiples of $23$?
-
You'll want to see this KCd blurb. – Guess who it is. Dec 8 '11 at 15:49
I've seen the following method for 7: Take the last digit, double it, and subtract it from the first digits. So $21553$ leads to $2155-3*2=2149$. Next step is $214-2*9=196$. Next $19-2*6=7$. – Thomas Andrews Dec 8 '11 at 16:04
How can we find divisors of 23? Very easily: they are just $\pm 1$ and $\pm 23$. Presumably, you want to recognize multiples of 23? – Arturo Magidin Dec 8 '11 at 16:11
For $23$, here is something fairly simple, for numbers that have $3$ or $4$ digits, say $abcd$. Take the number $ab$, add to it $3$ times the number $cd$. Call the result $t$. Then $abcd$ is a multiple of $23$ iff $t$ is. One could probably tweak this to be even simpler to execute. – André Nicolas Dec 8 '11 at 16:16
@ArturoMagidin Yes, I meant multiples. It's corrected. – Oltarus Dec 9 '11 at 8:46
One needn't memorize motley exotic divisibility tests. There is a universal test that is simpler and much easier recalled, viz. evaluate a radix polynomial in nested Horner form, using modular arithmetic. For example, consider evaluating a $3$ digit radix $10$ number modulo $7$. In Horner form $\rm\ d_2\ d_1\ d_0 \$ is $\rm\: (d_2\cdot 10 + d_1)\ 10 + d_0\ \equiv\ (d_2\cdot 3 + d_1)\ 3 + d_0\ (mod\ 7)\$ since $\rm\ 10\equiv 3\ (mod\ 7)\:.\:$ So we compute the remainder $\rm\ (mod\ 7)\$ as follows. Start with the leading digit then repeatedly apply the operation: multiply by $3$ then add the next digit, doing all of the arithmetic $\rm\:(mod\ 7)\:.\:$
For example, let's use this algorithm to reduce $\rm\ 43211\ \:(mod\ 7)\:.\:$ The algorithm consists of repeatedly replacing the first two leading digits $\rm\ d_n\ d_{n-1}\$ by $\rm\ d_n\cdot 3 + d_{n-1}\:\ (mod\ 7),\:$ namely
$\rm\qquad\phantom{\equiv} \color{red}{4\ 3}\ 2\ 1\ 1$
$\rm\qquad\equiv\phantom{4} \color{green}{1\ 2}\ 1\ 1\quad$ by $\rm\quad \color{red}4\cdot 3 + \color{red}3\ \equiv\ \color{green}1$
$\rm\qquad\equiv\phantom{4\ 3} \color{royalblue}{5\ 1}\ 1\quad$ by $\rm\quad \color{green}1\cdot 3 + \color{green}2\ \equiv\ \color{royalblue}5$
$\rm\qquad\equiv\phantom{4\ 3\ 5} \color{brown}{2\ 1}\quad$ by $\rm\quad \color{royalblue}5\cdot 3 + \color{royalblue}1\ \equiv\ \color{brown}2$
$\rm\qquad\equiv\phantom{4\ 3\ 5\ 2} 0\quad$ by $\rm\quad \color{brown}2\cdot 3 + \color{brown}1\ \equiv\ 0$
Hence $\rm\ 43211\equiv 0\:\ (mod\ 7)\:,\:$ indeed $\rm\ 43211 = 7\cdot 6173\:.\:$ Generally the modular arithmetic is simpler if one uses a balanced system of representatives, e.g. $\rm\: \pm\{0,1,2,3\}\ \:(mod\ 7)\:.$ Notice that for modulus $11$ or $9\:$ the above method reduces to the well-known divisibility tests by $11$ or $9\:$ (a.k.a. "casting out nines" for modulus $9\:$).
-
One approach is to find some higher multiple that makes it easy. For your example of $23$, note that $3*23+1=70$, so $a(70-1)$ will be a multiple of $23$. Now you have a single digit multiply followed by a subtraction. If you pick $a=301, a(70-1)=301(70-1)=301*70-301=2107-301=1806$, which is a multiple of $23$
-
I generally find this method much easier than the standard modular divisibility test. It does require that you memorize a starting list of multiples, but it's not so bad because you can just multiply them by powers of $10$. – Qiaochu Yuan Dec 8 '11 at 18:28
@QiaochuYuan: It seemed the OP's request was to find multiples, not do a divisibility test. But this generalizes. In the spirit of the test for $7$, to test divisibility by $23$, you can take off the last digit, multiply the rest by $7$ and add the last digit. Keep going until you have only two digits left. You just have to find a multiple that ends in $1$ or $9$ to do this. – Ross Millikan Dec 8 '11 at 18:40
In general, if you are doing things by dividing the number into groups of $k$ digits, you can think of it as looking at the number in base $10^k.$
More generally, if $B$ is a base, and $n$ is a number with no common factors with $B$, then you can always find an $m$ such that $Bm\equiv 1\pmod n$. Then $X\equiv 0\pmod n$ if and only if $Xm\equiv 0\pmod n$. But if $X=Bu+v$, then $Xm\equiv u+mv\pmod n$. So if we take the last digit, base B, multiply it by $m$ and add it to the other digits, the result is divisible by $n$ if and only if the original number was divisible by $n$.
In the case of $n=23$ base $B=10$, you get $m=7$, so you can take the last digit, multiply it by seven, and add it to the rest.
Or, if you use $B=100$, you get $m=3$, and you can take the lsat two digits, multiply by 3, and add to the other digits.
In general, you can always find a $k$ such that $10^k=1\pmod n$ if $n$ is not even or divisible by 5. Then if you take base $B=10^k$, you get $m=1$, and you can separate $m$ into groups of $k$ digits and add them. That's hardly useful when $k$ is large. For example, the smallest $k$ for $n=23$ is $k=22$, so this part only helps if your starting number was more than 23 digits long.
Usually, you want to find a relatively small pair $(m,k)$ so that $m10^k\equiv \pm 1\pmod n$. Then you take the last $k$ digits, multiplied by $m$, and add to or subtract from the other digits, depending on whether $+1$ or $-1$.
-
The following is a simple method to check divisibility by $7$ or $13$:
Cut the digits in pairs of 3 and calculate their alternating sum. If this is a multiple of 7 or 13, the original number is.
For example
$12345631241$ leads to
$$241-639+345-12=-65 \,.$$
Thus our number is a multiple of $13$, but not a multiple of $7$.
This works because $1001=7*11*13$ meaning that any number of the form $abcabc$ is a multiple of $7, 11$ and $13$...The trick also works for 11, but there is another simple trick.
Method 2 If the number is relatively prime to 10 (if it is not, you can make it), look for a multiple on $n$ which ends in 1.
Let say that this multiple is $a_1..a_k1$.
Then you simply pick the large number and subtract $a_1...a_k*$last digit from the remaining digits.
The trick works because $a_1...ak1$*last digit is always a multiple of $n$, and subtracting this from the original number you get a multiple of 10..Since $n$ is relatively prime to 10, you can erase the 0 at the end...
A simple such example, for $7$ the smalest such desires multiple is ... 21, which leads to the criteria posted by Thomas Andrews .
Also, for small numbers the following
-
I will try to explain a general rule using modular congruence. We can show that any integer in base $10$ can be written as $$z = a_0 + a_1 \times 10 + a_2 \times 10^2 + a_3 \times10^3 + \cdots + a_n \times 10^n$$
Lets say we have to find a divisibility rule of $7$,Hence for congruence modulo $7$ have,
$$10 \equiv 3, 10^2 \equiv 2, 10^3 \equiv -1, 10^4 \equiv -3, 10^5 \equiv -2, 10^6 \equiv 1,$$
The successive remainder then repeat. Thus our integer $z$ is divisible by $7$ iff if the remainder expression $$r= a_0 + 3a_1 +2a_2 -a_3-3a_4-2a_5+a_6+3a_7+\cdots$$ is divisible by $7$
To understand why the divisibility of $r$ indicate the divisibility of $z$, find $z-t$ which is given by :$$z-t = a_1 \times (10-3) + a_2 \times (10^2-2) + a_3 \times (10^3+1) + \cdots + a_6 \times (10^6-1) + \cdots$$ Since all this numbers $(10-3),(10^2-2),(10^3+1),\cdots$ are congruent to 0 modulo $7,z-t$ is also, and therefore $z$ leaves the same remainder on division by $7$ as $r$ does.
Using this approach we can derive divisibility of any integer.
-
This can be explained using Horner's method as below. z=a0 + a1 ×10 + a2 ×102 + a3 × 10^3+⋯+ an × 10^n
i.e. Z = ((((...(an × 10 + an-1)*10 + an-2 )*10 + an-3).....)*10 + a0
( Shift-add representation of number)
Z % 7 = ((((...(an × 3 + an-1)*3 + an-2 )*3 + an-3).....)*3 + a0 (Mod 7)
Now evaluate from left to right using modulo algebra.
To know more about shift- add representation of number and its application in interbase conversion and divisibility refer tinyurl.com/mlxk8pw .
Regard, | 2015-05-30T13:00:06 | {
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http://mathhelpforum.com/algebra/15734-help-reducing-quadratic-equation-final-terms.html | 1. ## Help with reducing quadratic equation to final terms
$x^4 - 6x^2 + 8 = 0
$
$x^2=t$
$x^4=t^2$
a= 1, b = 6, c = 8
$\Rightarrow x = \frac {6 \pm \sqrt {6^2 - 4(1)(8)}}{2(1)}$
$\Rightarrow x = \frac {6 \pm \sqrt {4}}{2}$
$\Rightarrow x = \frac {3 \pm \sqrt {2}}{}$
How do I reduce this further to get the final answer or is my answer correct? I always get confused. Thanks in advance.
2. Originally Posted by lilrhino
$x^4 - 6x^2 + 8 = 0
$
$x^2=t$
$x^4=t^2$
a= 1, b = 6, c = 8
$\Rightarrow x = \frac {6 \pm \sqrt {6^2 - 4(1)(8)}}{2(1)}$
$\Rightarrow x = \frac {6 \pm \sqrt {4}}{2}$
$\Rightarrow x = \frac {3} \pm \frac { \sqrt {2}}{2}$
How do I reduce this further to get the final answer. I always get confused. Thanks in advance.
ok, so this is wrong, since b = -6 not 6. but we don't need the quadratic formula here
you replaced $x^2$ with $t$, so you should have
$t^2 - 6t + 8 = 0$ .........now factor
$\Rightarrow (t - 4)(t - 2) = 0$
$\Rightarrow t = 4 \mbox { or } t = 2$
But $t = x^2$
$\Rightarrow x^2 = 4 \mbox { or } x^2 = 2$
$\Rightarrow x = \pm 2 \mbox { or } x = \pm \sqrt {2}$
3. Hello, lilrhino!
You're making it unnecessarily complicated . . . and wrong.
$x^4 - 6x^2 + 8 \:= \:0$
Factor: . $(x^2 - 4)(x^2 - 2)\:=\:0$
And we have two equations to solve:
. . $x^2 - 4\:=\:0\quad\Rightarrow\quad x^2 \:=\:4\quad\Rightarrow\quad x \:=\: \pm2$
. . $x^2 - 2\:=\:0\quad\Rightarrow\quad x^2\:=\:2\quad\Rightarrow\quad x\:=\:\pm\sqrt{2}$
4. Originally Posted by Jhevon
ok, so this is wrong, since b = -6 not 6. but we don't need the quadratic formula here
you replaced $x^2$ with $t$, so you should have
$t^2 - 6t + 8 = 0$ .........now factor
$\Rightarrow (t - 4)(t - 2) = 0$
$\Rightarrow t = 4 \mbox { or } t = 2$
But $t = x^2$
$\Rightarrow x^2 = 4 \mbox { or } x^2 = 2$
$\Rightarrow x = \pm 2 \mbox { or } x = \pm \sqrt {2}$
Thanks Jhevon, I have to watch my signs. I was looking at an example in the book, so I thought I had to use the quadratic formula even though it could be factored as you demonstrated. Thanks again!
5. Originally Posted by lilrhino
Thanks Jhevon, I have to watch my signs. I was looking at an example in the book, so I thought I had to use the quadratic formula even though it could be factored as you demonstrated. Thanks again!
yeah, the quadratic formula can be overkill at times. However, i do agree with Soroban that we are making it more complicated than it should be. Generally it helps students to visualize what to do by replacing a squared term with a single variable, but i believe that's wasting writing space. ordinarily, i'd just factor as Soroban did and cut out the middle man, but i didn't really want to shock you. you'd be surprised at some of the things students are shocked with
6. Originally Posted by Soroban
Hello, lilrhino!
You're making it unnecessarily complicated . . . and wrong.
Factor: . $(x^2 - 4)(x^2 - 2)\:=\:0$
And we have two equations to solve:
. . $x^2 - 4\:=\:0\quad\Rightarrow\quad x^2 \:=\:4\quad\Rightarrow\quad x \:=\: \pm2$
. . $x^2 - 2\:=\:0\quad\Rightarrow\quad x^2\:=\:2\quad\Rightarrow\quad x\:=\:\pm\sqrt{2}$
I do that often unfortunately. Math is not my strongest subject, so I sometimes make things more complicated that necessary. Thanks for your response.
7. Originally Posted by Jhevon
yeah, the quadratic formula can be overkill at times. However, i do agree with Soroban that we are making it more complicated than it should be. Generally it helps students to visualize what to do by replacing a squared term with a single variable, but i believe that's wasting writing space. ordinarily, i'd just factor as Soroban did and cut out the middle man, but i didn't really want to shock you. you'd be surprised at some of the things students are shocked with
I'm not easily shocked, I may spend more time thinking about it though . It's easier the way Soroban demonstrated it actually, so I'm not shocked. Thanks... | 2016-10-26T14:41:48 | {
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https://math.stackexchange.com/questions/3173523/tiling-a-square-with-rectangles | # Tiling a square with rectangles
Consider the set of all the rectangles with dimensions $$2^a\times 2^b\,a,b\in \mathbb{Z}^{\ge 0}$$. We want to tile an $$n\times n$$ square by rectangles from this set (you can use a rectangle several times). What is the minimum number of rectangles we need?
If $$f(n)$$ is the sum of digits of $$n$$ in base $$2$$, I think we need at most $$f(n)^2$$ rectangles. I have an example for this number: write $$n=2^{a_1}+2^{a_2}+...2^{a_{f(n)}}$$ and split each side to segments with length $$2^{a_1},2^{a_2},...,2^{a_{f(n)}}$$ and consider $$f(n)^2$$ rectangles obtained this way.
On the other hand, you need at least $$f(n)$$ rectangles to tile a raw (or column) so I think you need $$f(n)^2$$ rectangles, but I can't prove it.
Any ideas?
• By $f(n)$ do you mean the sum of the bits in the binary representation of $n$? Apr 3, 2019 at 20:18
• @JohnWaylandBales yes f(n) is the least number such that $n=2^{a_1}+2^{a_2}+...2^{f(n)}$
– ali
Apr 3, 2019 at 20:23
• You mean $f(n)$ is the least number such that $n = 2^{a_1} + 2^{a_2} + \cdots + 2^{a_{f(n)}}$ right? Because $f(n)$ counts the number of terms, but it is not the highest exponent. Apr 4, 2019 at 3:51
• @antkam yes i edited the question
– ali
Apr 4, 2019 at 4:36
MAJOR UPDATE:
What I am about to show is not a proof for the minimum number of rectangles. However, in some cases I found the number of rectangles can be less than $$f(n)^2$$. The smallest $$N×N$$ grid that I have found that can have less than $$f(n)^2$$ rectangles is $$15×15$$, which is displayed below: $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline 1&1&1&1&2&2&3&4&4&4&4&4&4&4&4\\ \hline 1&1&1&1&2&2&3&4&4&4&4&4&4&4&4\\ \hline 1&1&1&1&2&2&3&4&4&4&4&4&4&4&4\\ \hline 1&1&1&1&2&2&3&4&4&4&4&4&4&4&4\\ \hline 1&1&1&1&2&2&3&5&5&5&5&5&5&5&5\\ \hline 1&1&1&1&2&2&3&5&5&5&5&5&5&5&5\\ \hline 1&1&1&1&2&2&3&6&6&6&6&6&6&6&6\\ \hline 1&1&1&1&2&2&3&7&8&9&9&10&10&10&10\\ \hline 11&11&11&11&11&11&11&11&8&9&9&10&10&10&10\\ \hline 12&12&12&12&12&12&12&12&8&9&9&10&10&10&10\\ \hline 12&12&12&12&12&12&12&12&8&9&9&10&10&10&10\\ \hline 13&13&13&13&13&13&13&13&8&9&9&10&10&10&10\\ \hline 13&13&13&13&13&13&13&13&8&9&9&10&10&10&10\\ \hline 13&13&13&13&13&13&13&13&8&9&9&10&10&10&10\\ \hline 13&13&13&13&13&13&13&13&8&9&9&10&10&10&10\\ \hline \end{array}$$
The method used in the above $$15×15$$ square can be generalized not just to other squares but to rectangles as well. In order to make full use of this method, I will expand the op's method to rectangles. Let the length of a rectangle be equal to $$m$$ units and the width be $$n$$ units. Then the number of base-2 rectangles used to cover a $$m × n$$ rectangle by the op's method is $$f(m)f(n)$$. My method isn't fundamentally different from the op's method. It splits the $$m×n$$ rectangle into five sub-rectangles, then the op's method is applied to each of the five rectangles. In some cases the number of base-2 rectangles that covers the five sub-rectangles is less than the number of base-2 rectangles that cover original $$m$$×$$n$$ rectangle using the op's method. The five rectangles are arranged so that their are two pairs of rectangles that occupy the corners and one rectangle that is in the middle (not touching the perimeter). Each pair of rectangles are the same size and orientation but in opposite corners. (Top left and bottom right or Top right and bottom left.) The length and width of the five rectangles are constructed from two other unit lengths $$a$$ and $$b$$. $$a$$ is the smallest number such that $$m+a$$ is a power of two. $$b$$ is the smallest number such that $$n+b$$ is a power of two. This means that $$f(m+a)$$ and $$f(n+b)$$ are each one. The length and width of the two rectangles in the first pair are $$f\left(\frac{m+a}{2}\right)$$ and $$f\left(\frac{n-b}{2}\right)$$ respectively. The length and width of the two rectangles in the second pair are $$f\left(\frac{m-a}{2}\right)$$ and $$f\left(\frac{n+b}{2}\right)$$ respectively. The formula for the total number of base-2 rectangle used is $$2f\left(\frac{m+a}{2}\right) f\left(\frac{n-b}{2}\right)+2f\left(\frac{m-a}{2}\right) f\left(\frac{n+b}{2}\right)+f(a)f(b)$$.
Note that if a square with a length of $$n$$ units is of the form $$2^xy$$ where $$x,y\in\Bbb{N}|x\ge 1,y\ge 1$$ and $$y$$ is odd. Then the number of base-2 rectangles used for both the op's method and my method are the the same as the number of base-2 rectangles used for a square of length $$y$$ because each of the dimensions of the sub-rectangles can be multiplied by $$2^x$$. For example if we want to determine how many base-2 rectangles is rectangles are required to cover a $$30×30$$ square using my method. We just use the $$15×15$$ example near the top of this post and multiply the length and width of each base-2 rectangle by $$2$$. So this means the $$30×30$$ square requires the same number of base-2 rectangles as the $$15×15$$ square. So the problem can be simplified to just rectangles where $$m$$ and $$n$$ are odd.
A simple inequality can be made which would indicate which method uses less base-2 rectangles. Let $$N_l$$ be the number of ones in the number for length of the rectangle in binary and $$N_w$$ be the number of ones in the width in binary $$\bigl($$or more simply $$N_l=f(m)$$ and $$N_w=f(n)\bigr)$$. Also Let $$Z_l$$ be the number of zeros in the number for length of the rectangle in binary, $$Z_w$$ be the number of zeros in the width in binary. My method uses less rectangles than the op when $$2f\left(\frac{m+a}{2}\right) f\left(\frac{n-b}{2}\right)+2f\left(\frac{m-a}{2}\right) f\left(\frac{n+b}{2}\right)+f(a)f(b)\lt f(m)f(n)$$
$$f\left(\frac{m+a}{2}\right)=1$$ $$f\left(\frac{n+b}{2}\right)=1$$ $$f\left(\frac{m-a}{2}\right)=N_l-1$$ $$f\left(\frac{n-b}{2}\right)=N_w-1$$ $$f(m)=N_l$$ $$f(n)=N_w$$ $$f(a)=Z_l+1$$ $$f(b)=Z_w+1$$
With the above substitutions the inequality can be changed to:
$$2(N_l-1)+2(N_w-1)+(Z_l+1)(Z_w+1)\lt N_lN_w$$ $$2N_l+2N_w-4+(Z_l+1)(Z_w+1)\lt N_lN_w$$ $$(Z_l+1)(Z_w+1)\lt N_lN_w-2N_l-2N_w+4$$ $$(Z_l+1)(Z_w+1)\lt (N_l-2)(N_w-2)$$
In the specific case of the square (where the length equals the width) my method uses less base-2 rectangles than the op when the number ones in the binary representation of the length is at least four more than than the number of zeros. To get the maximum utility out of my method the inequality shouldn't only be applied to the entire length and width of the main square it should also be applied to components of the square. For example consider the square $$1927×1927$$. The binary representation of 1927 is 11110000111. There are three more ones than zeros in this number so my method would normally break even with the op, covering the square with 49 base-2 rectangles. There is a way to cover the square using less base-2 rectangles by spliting the square into four rectangles $$1920×1920$$, $$1920×7$$, $$7×1920$$, and $$7×7$$. Splitting this way doesn't change the net result of the op's method. The first three sub rectangles satisfies the inequality. If I use my method on the first three sub rectangles I use 13, 11, and 11 base-2 rectangles respectively. Using the op's method on the last sub rectangle then counting up all of the base-2 rectangles I can cover the $$1927×1927$$ square using 44 base-2 rectangles.(13+11+11+9) So if a combination of sub-strings in the binary value of the length and width satisfies the inequality like it did three times with the sub rectangles then my method will use less base-2 rectangles than the op's method. Finding the minimum number of base-2 rectangles for some squares will inevtably involve searching for the best way to split the square.
For large enough squares the worst digit combination where my method does no better than the op is a block of three ones and the rest are alternating zeros and ones. For example the square $$\require{enclose}\enclose{horizontalstrike}{343×343}$$, its binary representation is 101010111. This square requires 36 base-2 rectangles and is tied for most number of required base-2 rectangles amoung the nine digit squares. This means that a upper bound can be made for the minimum number of rectangles required. Let $$\enclose{horizontalstrike}{d_l}$$ be the number of digits in the binary representation of the length of the rectangle. ($$\enclose{horizontalstrike}{d_l=N_l+Z_l}$$) Let $$\enclose{horizontalstrike}{d_w}$$ be the number of digits in the binary representation of the width of the rectangle. ($$\enclose{horizontalstrike}{d_w=N_w+Z_w}$$) Then the upper bound is: $$\enclose{horizontalstrike}{\left(\left\lceil\frac{d_l}{2}\right\rceil+1\right)\left(\left\lceil\frac{d_w}{2}\right\rceil+1\right)}$$
I conjecture that the combination of my method and the op's method is the optimal way of minimizing the number of base-2 rectangles. The only way that someone might use be able to use less rectangles is to find a another way of spliting the square into sub-rectangles such that using the op's method on those sub-rectangles uses less base-2 rectangles than using my method and the op's method on the whole square.
Rob Pratt's(RP's) post shows that there is a third method for covering the $$n×n$$ square with less base-2 rectangles than my method or the op's method for some $$n×n$$ squares. In order to describe how many rectangles RP's method uses I will continue to use the the term $$b$$ from my method (where $$b$$ is the smallest number such that $$b+n$$ is a power of 2). I will also need a new sets of terms $$c_k$$ and $$s_k$$ where $$k\in\Bbb{N}|1\le k\le f(b)$$. $$c_1$$ is the value of left most ones digit of b in binary form. $$c_2$$ is the value of the second ones digit from the left of b in binary form. $$c_3$$ is the value of the third ones digit from the left of b in binary form. Etc. $$s_v=\sum_{j=1}^vc_v$$. For example if $$n=23$$ then $$b=9$$, $$c_1=8$$, $$c_2=1$$, $$s_1=8$$, $$s_2=9$$. RP's method uses $$2f\left(\frac{n+b}{2}\right)f\left(\frac{n-b}{2}\right)+f\left(\frac{n+b}{2}\right)f\left(\frac{n-b}{2}+s_k\right)+f\left(\frac{n-b}{2}\right)f\left(\frac{n+b}{2}-s_k\right)+f(b)f(b-s_k)$$
base-2 rectangles. Each $$f(•)f(•)$$ product contains the length and width of each of the sub-rectangles that covers the square inside the f function. RP's method has $$k$$ ways of covering the $$n×n$$ square one for each $$s$$ element. Obviously the particular $$s_k$$ element that uses the least number of base-2 rectangles according to the above formula is the one that is used for the minimum. A sufficient condition for when RP's method uses less base-2 rectangles than both my method and the op's method when the binary representation of $$n$$ has at least three more ones than zeros, the second digit to the left is a zero, and the spliting method that was mentioned for the $$1927×1927$$ square doesn't apply.
• absolutely brilliant!! IMHO well worth the bounty. :) Apr 11, 2019 at 2:02
• What does your method obtain for $n\in\{23,30,31\}$? Mar 5, 2020 at 22:36
• @Rob_Pratt 16,13, and 17 base-2 rectangles respectively Mar 5, 2020 at 23:30
• @RobPratt I realized that the way I explained it in my edited post it doesn't show how n=30 is 13 base-2 rectangles with my method. I will edit accordingly. Mar 5, 2020 at 23:37
For fixed $$n$$, you can solve this problem via integer linear programming as follows. Let $$R$$ be the set of rectangles. For $$i\in\{1,\dots,n\}, j\in\{1,\dots,n\}$$, let $$R_{i,j}\subset R$$ be the subset of rectangles that contain cell $$(i,j)$$. Let binary decision variable $$x_r$$ indicate whether rectangle $$r\in R$$ is used. The problem is to minimize $$\sum_r x_r$$ subject to: \begin{align} \sum_{r \in R_{i,j}} x_r &= 1 &&\text{for i\in\{1,\dots,n\}, j\in\{1,\dots,n\}} \\ x_r &\in \{0,1\} &&\text{for r \in R} \end{align}
Here are several optimal values that differ from $$f(n)^2$$: $$\begin{matrix} n &15 &23 &30 &31 &46 &47 &55 &59 &60 &61 &62 &63\\ \hline f(n)^2 &16 &16 &16 &25 &16 &25 &25 &25 &16 &25 &25 &36\\ \text{optimal} &13 &15 &13 &17 &15 &19 &20 &20 &13 &20 &17 &21\\ \end{matrix}$$
For example, here is an optimal solution for $$n=23$$ with $$15$$ rectangles (shown here with hex values $$0$$ through $$E$$ for compactness): $$\begin{matrix} 0&0&0&0&0&0&0&0&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ 5&5&5&5&5&5&5&5&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ 5&5&5&5&5&5&5&5&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ 5&5&5&5&5&5&5&5&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ 5&5&5&5&5&5&5&5&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ 3&3&3&3&3&3&3&3&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ 3&3&3&3&3&3&3&3&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&1&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6\\ C&E&E&E&E&D&D&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6\\ C&E&E&E&E&D&D&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6\\ C&E&E&E&E&D&D&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6\\ C&E&E&E&E&D&D&4&4&4&4&4&4&4&4&4&4&4&4&4&4&4&4\\ C&E&E&E&E&D&D&4&4&4&4&4&4&4&4&4&4&4&4&4&4&4&4\\ C&E&E&E&E&D&D&2&2&2&2&2&2&2&2&2&2&2&2&2&2&2&2\\ \end{matrix}$$
• I should have mentioned this earlier but good job finding this. As of when this comment being posted you are the only one who has helped me with this problem. If no one else posts am answer by the end of the bounty grace period you will receive the bounty. Also I have made a formula for your method in my most recent edit that I just made, you might want to take a look. Mar 8, 2020 at 23:13
• Thanks. I added a few more values $< f(n)^2$ just now. Mar 9, 2020 at 23:35
NOTE:This doesn't work, the induction hypothesis is too strong (and false).
Lets first consider a more general question, where we tile a rectangle $$R$$ by smaller rectangles, where all vertices are points in an (ambient) integer lattice.
We have a row of rectangles $$T_i$$ touching the bottom edge of $$R$$, and each of these has a top edge $$e_i$$. For each $$T_i$$ we define the number $$\lambda(T_i)$$ to be the minimal number of our tiling rectangles that intersect any column starting in $$T_i$$.
Our first claim is that for the total number of rectangles in $$R$$, denoted $$r(R)$$, we have $$\sum_i \lambda(T_i) \leq r(R)$$
Lets prove this by induction on the height of the rectangle $$R$$ (drawing a picture may help see whats happening). First, if the height is $$1$$, then we are done trivially.
So now for the inductive step, let $$R_0$$ have height $$n$$, and consider the edges $$e_i$$ that have minimal height, and define $$a$$ to be this height. This is to say, they border the $$a$$th row, if the first row is the bottom row of $$R_0$$. Say that we have $$k$$ minimal edges $$e_i$$ bordering this row.
We now consider the new rectangle $$R_0'$$ we obtain by chopping off the first $$a$$ rows of $$R$$. On one hand, this has strictly smaller height, so we have, by induction and our definition of $$k$$: $$\sum_i \lambda(T_i') \leq r(R_0)-k$$
But each rectangle on the bottom row of $$R_0'$$ is either one of the rectangles of $$R_0$$, chopped, but not removed, or a rectangle of $$R_0$$ lying above one of our minimal edges $$e_i$$.
Now note if our original $$T_i$$ is chopped but not removed, $$\lambda(T_i)=\lambda(T_i')$$, and if our original $$T_j$$ is removed (so top edge has minimal height), then $$\lambda(T_j')=\lambda(T_j)-1$$, where $$T_j'$$ is any of the rectangles lying directly over $$T_j$$. Thus, adding $$k$$ to both sides of our previous equality, we have:
$$\sum_i \lambda(T_i) \leq \sum_i \lambda(T_i')+k\leq r(R_0)$$
So we are done by induction.
So for your case, note that each column must have at least $$f(n)$$ rectangles in it, and note the bottom row has at least $$f(n)$$ rectangles. This follows since $$f(n)$$ is the minimal number of powers of two needed to express $$n$$. Thus, $$f(n)^2\leq r(R)$$ in your case.
• I think your original claim is false.I don't know how to send a picture in comment but you can easily draw $3\times 4$ counter examples(two horizontal dominos and two vertical dominos in first two rows and two $1\times1$square and a domino on the third row).the problem is on your induction step the rectangles above two removed rectangle may not be distinict.
– ali
Apr 5, 2019 at 9:41
• your last statement have counter example too.if each row intersect k rectangle and each column intersect k rectangle doesn't mean we need $k^2$ rectangle.
– ali
Apr 5, 2019 at 9:43
• True, I'll leave this up in case someone can make this approach work. Apr 5, 2019 at 10:03 | 2022-05-19T11:43:34 | {
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https://math.stackexchange.com/questions/1867644/finding-a-tricky-composition-of-two-piecewise-functions/1867652 | # Finding a tricky composition of two piecewise functions
I have a question about finding the formula for a composition of two piecewise functions. The functions are defined as follows:
$$f(x) = \begin{cases} 2x+1, & \text{if x \le 0} \\ x^2, & \text{if x > 0} \end{cases}$$
$$g(x) = \begin{cases} -x, & \text{if x < 2} \\ 5, & \text{if x \ge 2} \end{cases}$$
My main question lies in how to approach finding the formula for the composition $g(f(x))$. I have seen a couple of other examples of questions like this online, but the domains of each piecewise function were the same, so the compositions weren't difficult to determine.
In this case, I have assumed that, in finding $g(f(x))$, one must consider only the domain of $f(x)$. Thus, I think it would make sense to test for individual cases: for example, I would try to find $g(f(x))$ when $x <= 0$. $g(f(x))$ when $x <= 0$ would thus be $-2x-1$, right? However, I feel like I'm missing something critical, because I'm just assuming that the condition $x < 2$ for $g(x)$ can just be treated as $x <= 0$ in this case. Sorry for my rambling, and many thanks for anyone who can help lead me to the solution.
The first branch point is associated with $f$, and happens at $x=0$. We note that: $$g\circ f(x) = \begin{cases} g(2x+1), & \text{if x≤ 0} \\ g(x^2), & \text{if x > 0} \end{cases}$$
Let's just consider each case separately. First, $g(2x+1)$ when $x≤0$. We remark that $x≤0\implies 2x≤0\implies 2x+1≤1$ so we see that $x≤0\implies g(2x+1)=-2x-1$.
Now consider $g(x^2)$ when $x>0$. We see that we hit a branch point at $x=\sqrt 2$. Specifically, $x<\sqrt 2\implies x^2<2$ so $x<\sqrt 2\implies g(x^2)=-x^2$. Of course $x≥\sqrt 2 \implies x^2≥2$ so $x≥\sqrt 2\implies g(x^2)= 5$. Combining all this we see that $$g\circ f(x) = \begin{cases} -2x-1, & \text{if x ≤ 0} \\ -x^2, & \text{if 0< x < \sqrt 2}\\ 5 & \text{if x≥\sqrt 2} \end{cases}$$
As a sanity check, we try the special case $x=1$. We note that $g\circ f(1)=g(f(1))=g(1)=-1$ as desired. I advise picking a few other special values just to check.
• Thanks for your answer. You say that "[w]e see that we hit a branch point at $x = \sqrt 2$," but how do we see this? I don't know how one would conclude this from any of the information given. – Linear Pedant Jul 22 '16 at 22:19
• @LinearPedant The branch point happens when the argument of $g$ reaches $2$. Because the argument is $x^2$, the branch point is at $x=\sqrt2$. – f'' Jul 22 '16 at 23:41
• @LinearPedant Did the reply from f'' help? It is exactly right. We know that $g(x)$ has a branch point at $x=2$, thus $g(x^2)$ has a branch point at $x^2=2$, a.k.a. $x=\sqrt 2$ (since we already know $x>0$ so we can ignore the negative square root). – lulu Jul 23 '16 at 1:45
• @lulu @f" Yes, thank you. This reply greatly helped me understand the problem better. – Linear Pedant Jul 23 '16 at 1:52
$$g(x) = \begin{cases} -x, & x < 2 \\ 5, & x \ge 2 \end{cases}$$
Therefore
$$g(f(x)) = \begin{cases} -f(x), & f(x) < 2 \\ 5, & f(x) \ge 2 \end{cases}$$
So now we need to know when $f(x) < 2$ and when $f(x) \ge 2$.
$$f(x) = \begin{cases} 2x + 1, & x \le 0 \\ x^2, & x > 0 \end{cases}$$
Let's look at one piece of $f$ at a time. On the first piece, $2x + 1 \ge 2$ means $x \ge 1/2$. But this is impossible because $x \le 0$ on the first piece. Also on the first piece, $2x + 1 < 2$ means $x < 1/2$. Well, on the first piece we have $x \le 0 < 1/2$, therefore $f(x) < 2$ on the entire first piece, i.e., $f(x) < 2$ if $x \le 0$.
On the second piece, $x^2 \ge 2$ means $x \ge \sqrt{2}$ or $x \le -\sqrt{2}$. On the second piece we always have $x > 0$, therefore we have $x^2 \ge 2$ when $x \ge \sqrt{2}$. Also on the second piece, $x^2 < 2$ means $-\sqrt{2} < x < \sqrt{2}$. And since on the second piece we always have $x > 0$, then we must have $0 < x < \sqrt{2}$ in order to have $x^2 < 2$.
Putting it all together so far, we have the following:
$$f(x) \ge 2 \text{ if and only if } x \ge \sqrt{2}$$
$$f(x) < 2 \text{ if and only if } x \le 0 \text{ or } 0 < x < \sqrt{2}$$
Notice that this last one can be simplified but we need to keep them separate. Why is this? We'll see as we continue. Recall:
$$g(f(x)) = \begin{cases} -f(x), & f(x) < 2 \\ 5, & f(x) \ge 2 \end{cases}$$
Therefore:
$$g(f(x)) = \begin{cases} -f(x), & x \le 0 \text{ or } 0 < x < \sqrt{2} \\ 5, & x \ge \sqrt{2} \end{cases}$$
Separating the conditions gives us:
$$g(f(x)) = \begin{cases} -f(x), & x \le 0\\ -f(x), & 0 < x < \sqrt{2} \\ 5, & x \ge \sqrt{2} \end{cases}$$
And we need to do this because $f(x)$ itself is different for $x \le 0$ and $0 < x < \sqrt{2}$. Finally, we end up with:
$$h(x) := g(f(x)) = \begin{cases} -(2x+1), & x \le 0 \\ -x^2, & 0 < x < \sqrt{2} \\ 5, & x \ge \sqrt{2} \end{cases}$$
• Ack, someone beat me to it. I guess that's what happens when I let myself get distracted. – tilper Jul 22 '16 at 15:59
You're correct about the value of $g(f(x))$ when $x\leq 0$; since $f(x)$ will be at most $2\cdot0+1=1$, $g$ is only going to evaluate $f(x)$ according to the definition for $x<2$. Testing for cases here is a good approach, and you've just resolved the $x\leq0$ case. When $x>0$, consider the values of $f(x)$: when will they be less than $2$, and when will they be greater? This will determine where $g(f(x))$ takes on its values.
$$g(x) = \begin{cases} -x, & \text{if x < 2} \\ 5, & \text{if x \ge 2} \end{cases}$$ In finding $g(f(x))$ with $g$ defined as above, one should remember that $$g(f(x)) = \begin{cases} -x, & \text{if f(x) < 2} \\ 5, & \text{if f(x) \ge 2} \end{cases}$$ with $f(x)$, not $x$, being the thing about which one asks when it is $<2$ and when it is $\ge 2$.
(The calculus text by Salas and Hille had a problem like this that used to get assigned in every academic term by vast numbers of professors who taught the course to thousands of students. And every time an hour would be spent answering the students questions about it, which in every case said something like "I didn't even know where to start", and that's an hour that could have been spent on something that would help the students understand calculus. It was an even worse waste of time and effort than what usually goes on in the kind of math courses where the students don't want to learn math.) | 2020-06-05T22:05:04 | {
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https://mathzsolution.com/can-every-proof-by-contradiction-also-be-shown-without-contradiction/ | # Can every proof by contradiction also be shown without contradiction?
Are there some proofs that can only be shown by contradiction or can everything that can be shown by contradiction also be shown without contradiction? What are the advantages/disadvantages of proving by contradiction?
As an aside, how is proving by contradiction viewed in general by ‘advanced’ mathematicians. Is it a bit of an ‘easy way out’ when it comes to trying to show something or is it perfectly fine? I ask because one of our tutors said something to that effect and said that he isn’t fond of proof by contradiction.
## Answer
To determine what can and cannot be proved by contradiction, we have to formalize a notion of proof. As a piece of notation, we let $\bot$ represent an identically false proposition. Then $\lnot A$, the negation of $A$, is equivalent to $A \to \bot$, and we take the latter to be the definition of the former in terms of $\bot$.
There are two key logical principles that express different parts of what we call “proof by contradiction”:
1. The principle of explosion: for any statement $A$, we can take “$\bot$ implies $A$” as an axiom. This is also called ex falso quodlibet.
2. The law of the excluded middle: for any statement $A$, we can take “$A$ or $\lnot A$” as an axiom.
In proof theory, there are three well known systems:
• Minimal logic has neither of the two principles above, but it has basic proof rules for manipulating logical connectives (other than negation) and quantifiers. This system corresponds most closely to “direct proof”, because it does not let us leverage a negation for any purpose.
• Intuitionistic logic includes minimal logic and the principle of explosion
• Classical logic includes intuitionistic logic and the law of the excluded middle
It is known that there are statements that are provable in intuitionistic logic but not in minimal logic, and there are statements that are provable in classical logic that are not provable in intuitionistic logic. In this sense, the principle of explosion allows us to prove things that would not be provable without it, and the law of the excluded middle allows us to prove things we could not prove even with the principle of explosion. So there are statements that are provable by contradiction that are not provable directly.
The scheme “If $A$ implies a contradiction, then $\lnot A$ must hold” is true even in intuitionistic logic, because $\lnot A$ is just an abbreviation for $A \to \bot$, and so that scheme just says “if $A \to \bot$ then $A \to \bot$”. But in intuitionistic logic, if we prove $\lnot A \to \bot$, this only shows that $\lnot \lnot A$ holds. The extra strength in classical logic is that the law of the excluded middle shows that $\lnot \lnot A$ implies $A$, which means that in classical logic if we can prove $\lnot A$ implies a contradiction then we know that $A$ holds. In other words: even in intuitionistic logic, if a statement implies a contradiction then the negation of the statement is true, but in classical logic we also have that if the negation of a statement implies a contradiction then the original statement is true, and the latter is not provable in intuitionistic logic, and in particular is not provable directly.
Attribution
Source : Link , Question Author : sonicboom , Answer Author : Carl Mummert | 2022-10-05T01:53:10 | {
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https://math.stackexchange.com/questions/935777/understanding-and-writing-limit-proofs | # Understanding and writing limit proofs
I got this question :
Consider $a_n,\:b_{n\:}$ sequences such that for every n , $0\le a_n\le \:b_{n\:}$.
Let $\lim _{n\to \infty }\left(\frac{b_n}{a_n}\right)\:=\:1$, and $a_n$ is a bounded sequence.
Prove that $\left(a_n-b_n\right)_{n=1}^{\infty \:}\:\rightarrow \:0$.
I tried little bit by myself to understand what is given:
By definition of limit, we see that : $\forall\epsilon, \exists n_{0} \forall n> n_{0}\Rightarrow |\frac{b_{n}}{a_{n}}- 1|< \varepsilon$, and also that exist some $M$ that for every $n$, $-M<a_n<M$.
Now i'm looking for that right?: i need to prove that $\forall\epsilon, \exists n_{0} \forall n> n_{0}\Rightarrow |a_{n}-b_{n}- 0|< \varepsilon$
Here is where i struggle: i choose some $\epsilon$. What i need to find? an $N2$ that for every $n>N2, |(a_{n}-b_{n})- 0|< \varepsilon$ ?
How to start?
Hint: use may the fact that $|a_n-b_n|=|a_n||\frac{b_n}{a_n}-1|$ in your proof .
• Yes, but what i actually need to write to do it formal? – user2637293 Sep 18 '14 at 1:09
• I would recommend you to fill in the gap yourself. but if not interested you can look at the proof provided by @Ishfaaq down here. – BigM Sep 18 '14 at 1:12
Hints
First of all notice that $\left|{\dfrac{b_n}{a_n} - 1}\right| = \dfrac{|(b_n - a_n) - 0|}{|a_n|}$
Secondly note that $|a_n| \le M \;\; \forall n \in \Bbb N$
Now let $\epsilon \gt 0$ be arbitrary and think what you can do with a quantity like $$\dfrac{\epsilon}{M} \gt 0$$
UPDATE: Just extract what the definition of $\lim \dfrac{a_n}{b_n}$ means. It says for every $\epsilon \gt 0$ there is $n_0$ such that the difference between $\dfrac{a_n}{b_n}$ and the limit ($= 1$ ) is less than $\epsilon$. So all you need to write in your paper would be to let $\epsilon$ be arbitrary. Then since $\lim \dfrac{a_n}{b_n}$ is given, choose $\dfrac{\epsilon}{M}$ as your arbitrary positive quantity. Then as per the definition of the limit there exists $n_0 \in \Bbb N$ such that $$n \ge n_0 \implies \left|{\dfrac{b_n}{a_n} - 1}\right| = \dfrac{|(b_n - a_n) - 0|}{|a_n|} \lt \dfrac{\epsilon}{M} \implies | (b_n - a_n ) - 0 | \lt |a_n | \cdot \dfrac{\epsilon}{M} \le \epsilon$$
And you're done!
• This is the problem. i don't know what to write after i choose arbitrary epsilon. i need to someone write the answer like in a test. it's not the question of HOW to do this, its about WHAT to write to do it formally, tnx! – user2637293 Sep 18 '14 at 1:15
• @user2637293: I edited the answer. Comment if you need more help. – Ishfaaq Sep 18 '14 at 6:58 | 2019-08-26T06:28:32 | {
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https://math.stackexchange.com/questions/784808/positive-semi-definite-matrix-and-its-eigenvalues-please-help-checking-improvi | Let $A$ and $B$ are two $n \times n$ Hermitian matrices . Suppose $A-B$ is positive semidefinite.
(a) Show that $\lambda_k(A) \geq \lambda_k(B)$ for $k=1,2,\dots ,n,$ where $\lambda_i(A)$ and $\lambda_i(B)$ are eigenvalues of matrices $A$ and $B$ with $\lambda_i (A)\geq \lambda_{i+1}(A)$ and $\lambda_i (B)\geq \lambda_{i+1}(B)$.
(b) Find a counter example to prove the converse of (a) is not true. i.e.,
Find two Hermitian matrices $A$ and $B$ such that $\lambda_k(A) \geq \lambda_k(B)$ but $A-B$ is not positive semidefinite.
My Trial:
(a) Since $A-B$ is positve semi-definite, $x^*(A-B)x \geq 0 \Rightarrow x^{*}Ax \geq x^{*}Bx.$
Also, as $A$ and $B$ are two Hermitian matrices, by spectral decomposition, we have $x^{*}Ax = x^{*}UD_{A}U^{*}x=x^{*}D_{A} x$ and $x^{*}Bx = x^{*}UD_{B}U^{*}x=x^{*}D_{B} x$.
It implies $D_{A} \geq D_{B}$ and so for each of theire eigenvalues.
(b) Let $A=\begin{pmatrix} 1 &0 \\ 0 &1 \end{pmatrix}$ and $B=\begin{pmatrix} 2 &0 \\ 0 &0 \end{pmatrix}$.
Please help to check if I make mistake in my presentation because I am very new to this topic so I do not know if am correct or not. Thank you very much in advance.
• How $U^*x$ becomes $x$? And the $U$ matrix is not the same for $A$ and for $B$. – enzotib May 7 '14 at 10:03
• Is it given that $A,B$ have the same eigenvectors? – Samrat Mukhopadhyay May 7 '14 at 10:21
• @enzotib Since $U$ is unitary, norm of $U*x$ = norm of $x$. Can I say this? – nam May 7 '14 at 10:22
• @Samrat The question does not give that $A$ and $B$ have the same eigenvectors. – nam May 7 '14 at 10:23
• @Pavel So, am I correct on the whole? – nam May 7 '14 at 10:24
To prove (a), you need to use the variational characterisation of eigenvalues (the min-max principle, Courant-Fischer theorem). For a Hermitian $A$ and the specified ordering of eigenvalues, $$\tag{*} \lambda_k(A)=\max_{S:\dim S=k}\min_{x\in S\setminus\{0\}}\frac{x^*Ax}{x^*x}=\min_{S:\dim S=n-k+1}\max_{x\in S\setminus\{0\}}\frac{x^*Ax}{x^*x}.$$ Let $S_*$ be the subspace of the dimension $n-k+1$ for which the minimum on the right-hand side of ($*$) is attained, that is, $$\lambda_k(A)=\max_{x\in S_*\setminus\{0\}}\frac{x^*Ax}{x^*x}.$$ Then (using $x^*Ax\geq x^*Bx$ for all $x$) $$\lambda_k(A)=\max_{x\in S_*\setminus\{0\}}\frac{x^*Ax}{x^*x} \geq\max_{x\in S_*\setminus\{0\}}\frac{x^*Bx}{x^*x} \geq\min_{S:\dim S=n-k+1}\max_{x\in S\setminus\{0\}}\frac{x^*Bx}{x^*x}=\lambda_k(B).$$
I was also thinking whether there would be an alternative proof which does not involve the variational characterisation and can thus be considered as more elementary. This is motivated by the proof here.
By exactly the same technique as in that proof, one can show the following:
Let $A,B\in\mathbb{C}^{n\times n}$ be Hermitian and let $i_{+,0}(A)$ and $i_+(A)$ denote the number of non-negative and positive eigenvalues, respectively, of $A$. Then $$i_{+,0}(A+B)\leq i_+(A)+i_{+,0}(B).$$
Now replace in the statement above $A$ and $B$, respectively, by $A-\lambda_k(A)I$ and $\lambda_k(A)I-B$. We get $$\tag{1} i_{+,0}(A-B)\leq i_+(A-\lambda_k(A)I)+i_{+,0}(\lambda_k(A)I-B).$$ Since our $A$ and $B$ are such that $A-B$ is positive semi-definite and hence all its eigenvalues are non-negative, we have $i_{+,0}(A-B)=n$. Note that adding a multiple of the identity to $A$ just shifts the eigenvalues, so the $k$th eigenvalue of $A-\lambda_k(A)I$ is zero and thus there is at most $k-1$ positive eigenvalues of $A-\lambda_k(A)I$, that is, $i_+(A-\lambda_k(A)I)\leq k-1$. Putting this stuff to (1) gives $$n\leq k-1+i_{+,0}(\lambda_k(A)I-B)\quad \Rightarrow \quad n-k+1\leq i_{+,0}(\lambda_k(A)I-B).$$ Since there is at least $n-k+1$ non-negative eigenvalues of $\lambda_k(A)I-B$, there is at most $k-1$ negative eigenvalues. Equivalently, the matrix $B-\lambda_k(A)I$ has a most $k-1$ positive eigenvalues and hence the $k$th eigenvalue of $B-\lambda_k(A)I$ is non-positive. But this eigenvalue is nothing but $\lambda_k(B)-\lambda_k(A)$ and hence $$\lambda_k(B)-\lambda_k(A)\leq 0\quad\Rightarrow\quad\lambda_k(A)\geq\lambda(B).$$
• Very thanks to your hint. I just had some readings on Rayleigh quotients but I get lost and do not know how to apply your hint on this question (it seems to pick up second large eigenvalue from some subspaces of smaller dimension). Would you mind giving me lines of answers. Thank you so much. – nam May 7 '14 at 10:45
• @nam It should be quite complete now :) – Algebraic Pavel May 7 '14 at 10:49
• Thank you for your illustration for the theorem. I know what I missed and I am looking for other examples to check. Thank you so much. – nam May 7 '14 at 11:02
• You are welcome. – Algebraic Pavel May 7 '14 at 11:02
• Just for fun, I've tried to make a proof which does not use the min-max principle. However, it's a bit longer. – Algebraic Pavel May 7 '14 at 13:03 | 2020-03-30T19:43:28 | {
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https://math.stackexchange.com/questions/1345331/where-do-these-mobius-transformations-map-the-coordinate-half-planes | Where do these Mobius transformations map the coordinate half-planes?
They are
$$\frac{z-1}{z+1}, \frac{z+1}{z-1},\frac{z-i}{z+i},\frac{z+i}{z-i}.$$
All four look virtually identical, so I would like to know how to best distinguish between them.
For example, the first mapping, I notice that $1$ maps to $0$, $-1$ maps to $\infty$.
But since $1$ and $-1$ are symmetric points w.r.t. the imaginary axis, the Mobius transformations always maps symmetric points to symmetric points. So, the images, $0$ and $\infty$, being symmetric points, must mean that they are symmetric with respect to a circle. So the real axis gets mapped to some circle centered at the origin.
How do I tell how big this circle is? And since the real axis separates the upper and lower half planes, either the upper or lower plane gets mapped inside the circle, but how can I tell? I tried a sample point: I see that $i$ gets mapped to $i$, but is this telling me that the mapping is "fixing" the UHP? I doubt it, because it's mapping stuff to some open disk.
Edit: Also, with the first mapping, there doesn't seem to be a nice symmetry argument to exploit for the imaginary axis.
Thanks.
• The first mapping does not map the real line to a circle centered at the origin: The image of $1$ under that map is the origin itself. Jul 1 '15 at 2:47
• It does, however, map the imaginary axis to the unit circle, as each imaginary $z$ is equidistant from $1$ and $-1$, and hence $\left\vert\frac{z - 1}{z + 1}\right\vert = 1$. Jul 1 '15 at 2:51
• Hmm...can I think of it this way, @Travis? I realize my mistake, I think. So, since 1 and -1 are symmetric w.r.t. the imaginary axis, and their images must also be symmetric w.r.t. the image of the imaginary axis (not the real axis), which I don't know yet. But 1 and -1 map to 0 and infinity, which are symmetric w.r.t. the image, which implies that the imaginary axis must get mapped to a circle. Is this valid? If so, how do I know how big the circle is? Your equidistant argument I have never heard of...thanks, Jul 1 '15 at 3:00
• I've expanded my comments into a proper answer below, and I think it answers all these questions, but feel free to comment there if not. Jul 1 '15 at 3:16
The first mapping, $$f : z \mapsto \frac{z - 1}{z + 1},$$ does not map the real axis to a circle centered at the origin: Indeed, we have $f(1) = 0$.
On the other hand, any imaginary number $iy$ is, by symmetry, equidistant from $-1$ and $1$, and so $$|f(iy)| = \left\vert\frac{iy - 1}{iy + 1}\right\vert = \frac{|iy - 1|}{|iy + 1|} =1,$$ that is, $f$ maps the imaginary axis to the unit circle (bijectively, if we include $f(\infty) = 1$). Using the above fact that $f(1) = 0$ (and continuity) gives that $f$ maps that right half-plane to the inside of the unit circle, and so the left half-plane to the outside.
As for the real axis, note that if $x$ is real, then so is $f(x) = \frac{x - 1}{x + 1}$ (unless $x = -1$, in which case we have $f(-1) = \infty$), so $f$ maps the real axis (including $\infty$) to itself.
The other three maps can be analyzed similarly, or we can use the fact that they are precisely the compositions of $f$ with the rotation maps $z \mapsto i^k z$, $k = 1, 2, 3$.
• Hi @Travis, if this mapping fixes the real line, must it also fix the upper and lower half planes, too? I'm thinking of the connectedness argument - a continuous function maps connected sets to connected sets. Or, does the fact that f is not continuous at -1 make this connectedness argument invalid? Also, your getting the norm = 1, because of iy being equidistant from -1 and 1, but isn't iy really equidistant from 1+iy and -1+iy? I might not be understanding the meaning of equidistant, but I feel like we should just be extending iy out in a straight line, along the real axis.... Jul 1 '15 at 3:26
• actually, I think got the equidistant argument, if I just use the modulus definition. and I'm pretty sure the connectedness argument is not valid. but feel free to comment, if you have something more to say. Thanks so much for your help, @Travis. Jul 1 '15 at 3:35
• Like you say, it fixes the upper and lower half-planes (as sets, not pointwise). Also, we can think of Mobius transformations as continuous maps from the Riemann sphere, $\Bbb C \cup \{\infty\}$ to itself, so there's no issue with continuity here. Jul 1 '15 at 3:37
• Sure, for any region $A$ we can ask what the region $f(A)$ is, and it happens that our particular map $f$ maps deveral familiar regions to other familiar regions. Note that this lets us conclude, e.g., that the first quadrant is mapped the the upper half of the unit disk. Jul 1 '15 at 3:55
• You're welcome, I'm glad you found it useful, and you too. Jul 1 '15 at 4:57
I'll do the first one for you, and you can do the rest. Consider the unit circle $z=e^{i\theta}$. Then $$\frac{z-1}{z+1}=\frac{e^{i\theta}-1}{e^{i\theta}+1}=\frac{1-e^{-i\theta}+e^{i\theta}-1}{1+e^{i\theta}+e^{-i\theta}+1}=\frac{2i\sin(\theta)}{2+2\cos(\theta)}=i\frac{\sin\theta}{1+\cos\theta}=i\tan\frac{\theta}{2}.$$ Thus, the unit circle gets mapped to the imaginary axis. The inside of the unit circle gets mapped to one side of the imaginary axis, while the outside gets mapped to the other. To tell which side gets mapped where, check $z=0$. This maps to $-1$. Thus, the inside of the unit circle gets mapped to the left side of the imaginary axis, while the outside of the nit circle gets mapped to the right. This method should work for all of your functions.
• Hi @AlexS, nice argument - and based on Travis's comment above, then this mapping also maps the imaginary axis back to the unit circle. So, it is its own inverse, sort of? And, how do we know the imaginary axis gets mapped to a unit circle? Thanks, Jul 1 '15 at 3:09
• It is not quite its own inverse: $$f(f(z))=\frac{\frac{z-1}{z+1}-1}{\frac{z-1}{z+1}+1}=\frac{-2}{2z}=-1/z.$$ But, if $z$ is on the unit circle, so is $-1/z$. To show directly that $f$ takes the imaginary axis to the unit circle, multiply $f(iy)$ by its complex conjugate, and check that the result is 1. Jul 1 '15 at 3:17
• ah, nice suggestion about multiplication by complex conjugates to check the norm -- thanks so much @AlexS. Jul 1 '15 at 4:17
Every Mobius map can be broken down into a composition of scaling maps, translations and inversions. Just do a standard long division of polynomials, i.e., fo $\frac {az+b}{cz+d}$ , do along division $(az+b)| (cz+d)$ and you will get this decomposition. Then you can apply these operations of inversion, scaling and translation, rotation in the right order to get your map. Maybe you could study more between games too!!
• ok, got it. Thanks @Gary. I will try to, but with Kyrie and Kevin injured, and me playing 48minutes per game, it's been hard to keep up with complex analysis :-( Jul 1 '15 at 4:21 | 2022-01-18T20:36:30 | {
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https://www.jiskha.com/questions/541340/Evaluate-integral-of-e-x-1-2-x-1-2-Ive-looked-at-the-answer-but-I-dont-understand | # calculus
Evaluate integral of e^x^(1/2) / x^(1/2)
I've looked at the answer but I don't understand what people do in their steps.
When I substitute x^(1/2) for u, I get:
2du = 1/x^(1/2) dx
But what do you do with the 1/x^(1/2) dx? It just disappears in the solutions I've seen people give.
1. 👍 0
2. 👎 0
3. 👁 104
1. yes. the substitution is correct:
let u = x^(1/2)
thus du = 1/[2(x^(1/2))] dx, or
dx = 2(x^(1/2)) du, or
dx = 2u du
substituting these to original integral,
integral of [e^x^(1/2) / x^(1/2)] dx
integral of [(e^u) / u] * (2u) du
the u's will cancel out:
integral of [2*e^u] du
we can readily integrate this to
2*e^u + C
substituting back the value of u,
2*e^(x^(1/2)) + C
hope this helps~ :)
1. 👍 0
2. 👎 0
posted by Jai
2. This helped alot :)
1. 👍 0
2. 👎 0
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Evaluate integral, don't use calculator. I'll use the S as the sign for integral S 4 sin (theta)d(theta) from pi/2 to pi I know the answer is 4 because I enter it in on my calculator, but don't know the steps to get to the answer.
asked by Janet on May 9, 2011
9. ### calculus
8). Part 1 of 2: In the solid the base is a circle x^2+y^2=16 and the cross-section perpendicular to the y-axis is a square. Set up a definite integral expressing the volume of the solid. Answer choices: integral from -4 to 4 of
asked by Sally on April 14, 2013
10. ### Calculus
Hello. I would appreciate it if someone could check my answers. I'm sorry it is so long. 1.) Let R denote the region between the curves y=x^-1 and y=x^-2 over the interval 1
asked by Shay on September 3, 2005
More Similar Questions | 2019-05-27T04:16:40 | {
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https://math.stackexchange.com/questions/2031842/determine-remainder-when-dividing-polynomials-if-we-know-other-remainders | # Determine remainder when dividing polynomials if we know other remainders
If 3 is the remainder when dividing $P(x)$ with $(x-3)$, and $5$ is the remainder when dividing $P(x)$ with $(x-4)$, what is the remainder when dividing $P(x)$ with $(x-3)(x-4)$?
I'm completely puzzled by this, I'm not sure where to start...
Any hint would be much appreciated.
Hint: $$P(x)=Q(x)(x-3)(x-4)+Ax+B$$ $$P(3)=3$$ $$P(4)=5$$
Can you solve for $A$ and $B$?
Let $$P(x)=(x-4)(x-3)Q(x)+ax+b$$
Now you have $$3a+b=3$$ and $$4a+b=5$$ So solve for $a$ and $b$.
• Thank you!! I only accepted the other answer because he posted first. – kok_nikol Nov 26 '16 at 18:27
• @kok_nikol You haven't accepted anything: you just upvoted one answer. You can upvote all the answers you find informative, good and helpful . Then you will accept only one: the best for you, which is not necessarily the first one, of course. – DonAntonio Nov 26 '16 at 18:34
• @DonAntonio Sorry, you can't accept an answer right away, you have to wait 10 minutes. And I can't upvote yet because I don't have enough reputation. – kok_nikol Nov 27 '16 at 0:26
• @kok_nikol Perhaps, yet you did not accepted anything when I first commented. I thought posters can upvote answers to their questions. – DonAntonio Nov 27 '16 at 0:30
• @DonAntonio Yes, when you commented I hadn't yet accepted anything. As for the voting Votes cast by those with less than 15 reputation are recorded, but do not change the publicly displayed post score. – kok_nikol Nov 27 '16 at 0:37
If you carry out the division you should be able to show that $$P(x)=Q(x)(x-3)(x-4)+R(x)$$
Two questions: What is the maximum degree of the remainder $R(x)$? Can you see how to use the remainder theorem for the cases $x=3,4$ that you know already so that $Q(x)$ becomes irrelevant?
Euclidean division by $(x-3)(x-4)$ can be written as $$P(x)=(x-3)(x-4)Q(x)+R(x),\quad\deg R(x)\le1.$$ Instead of writing $R(x)$ with the standard basis $\{1,x\}$, use the basis $\{x-3,x-4\}$. Thus $$P(x)=(x-3)(x-4)Q(x)+A(x-3)+B(x-4)$$ Setting $x=3$, you get right away $B=-P(3)=-3$. Similarly $A=P(4)=5$, and finally $$R(x)=5(x-3)-3(x-4)=2x-3.$$
This formula can be generalised: the remainder when dividing a polynomial $P(x)$ by $(x-a)(x-b),\enspace a\ne b$, is $$R(x)=\frac1{b-a}\Bigl[P(b)(x--a)-P(a)(x-b)\Bigr]=\frac1{b-a}\begin{vmatrix}x-a&x-b\\[1ex]P(a)&P(b)\end{vmatrix}.$$
Notice $\ p\, = 3+(x\!-\!3)q\,\$ by $\,\ p(3) = 3$
$5 = p(4) = 3+q(4)\,\Rightarrow\, {\color{#c00}2}\, =\, q(\color{#0a0}4) \,$
Hence $\, p = 3 + (x\!-\!3)(\underbrace{\color{#c00}2+(x\!-\!\color{#0a0}4)r}_{\Large q}\!)\, =\, 2x\!-\!3\, +\, (x\!-\!3)(x\!-\!4)r$ | 2019-06-16T23:36:43 | {
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https://math.stackexchange.com/questions/3314464/mathbbrn-second-countable | # $\mathbb{R^n}$ Second Countable
Lemma: Let U be an open subset of $$\mathbb{R}^n$$ and $$x \in U$$ then there exists a ball with rational center $$(\mathbb{Q}^n$$ and rational radius containing x, that is contained in U.
What I want to show: Every open subset of $$\mathbb{R}^n$$ is the union of some some collection of elements in the following set:
$$\{$$ Open balls with rational center and rational radius $$\}$$
Let U be an arbitrary open set in $$\mathbb{R}^n$$. By the above lemma, for each $$x \in U$$ you can find a rational center $$x'$$ and rational radius $$r'>0$$ so that $$x \in B(x',r')$$ $$\subset U$$. Hence the following collection:
$$B'=$$ $$\{$$ B(a,r) : a $$\in \mathbb{Q}^n$$ , $$r>0$$ , B(a,r) $$\subset U$$ $$\}$$ gives us
$$U=$$ $$\bigcup_{B \in B'}$$ $$B$$
is the proof correct?
As this is part of the proof of showing that $$\mathbb{R}^n$$ is second countable, it follows that since every metric space is Hausdorff, $$\mathbb{R}^n$$ is Hausdorff, and trivially, it is locally euclidean of dimension n (take the identity map, which is a homeomorphism) and so $$\mathbb{R}^n$$ is a n- dimensional topological manifold, am I correct?
• I assume that you require manifolds to be second countable? Then yes and yes. – freakish Aug 5 at 18:39
• @freakish Yup, a topological space M is a n - dimensional manifold if (1) It is second countable (2) It is Hausdorff (3) Every point in M has a neighborhood homeomorphic to an open subset of $\mathbb{R}^n$ – topologicalmagician Aug 5 at 18:40
• @topologicalmagician You prove correctly that $\mathbb{R}^n$ is second countable. Also, since $\mathbb{R}^n$ is Hausdorff and locally euclidean for the reasons you said above, it is a topological manifold too. – bing-nagata-smirnov Aug 5 at 19:03
Yes, your proofs that $$\mathbb R^n$$ is cecond countable and that $$\mathbb R^n$$ is an $$n$$- dimensional topological manifold are correct. | 2019-10-14T01:44:42 | {
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https://brilliant.org/discussions/thread/which-payoff-do-you-want-to-go-for-2/ | # Which payoff do you want to go for?
Payoff 1.
Toss 5 coins. You get $1 for each consecutive pair HT that you get. Payoff 2. Toss 5 coins. You get$1 for each consecutive pair HH that you get.
The consecutive pairs are allowed to overlap.
For example, if you tossed HTHHH, under payoff 1 you will get $1, but under payoff 2 you will get$2.
Note by Calvin Lin
4 years, 7 months ago
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This is a trick question, both payoffs have same expected value of profit, which is: $1. Consider this: We will count the number of ways HT can occur. We can have: HT_ _ _ : 8 ways _ HT _ _ _ : 8 ways _ _ HT _ : 8 ways _ _ _ HT : 8 ways Total number of ways: 32 Similarly for HH, the total number of ways in which it can occur is 32. What this means is that the sum of the payoffs of HT and HH over the 32 possible outcomes is$32 for both. The payoffs may be distributed in different ways for HT and HH, but their expected value is same.
Mathematically, the expected value is same. But there is one more thing left to do: analyze the probability distribution. We can easily see that payoff 2 might be desirable because it tends to give large profits as compared to payoff 1 for certain events. eg: HHHHH gives $4 in p2 and 0 in p1. HHHHT gives$3 in p2 and $1 in p1. But the thing is that these events are not very common. Using method of reflection and analyzing the 16 possibilities , we see that there are many events for which p2 gives$0 payoff. There are 13 outcomes for which p2 gives $0 payoff, but only 6 where p1 gives$0 payoff.
So If I'm given this choice, I'd take payoff 1, as then I have more chance of leaving with at least a dollar in my pocket.
Since this is in quantitative finance, should I talk about risk and stuff? I do not know.
- 4 years, 7 months ago
I think that there may be some double-counting here. For example, you've counted HTHTT and HTHTH twice each. There are only $2^{5} = 32$ possible sequences in total, and there are some, such as TTHHH, that have no HT payouts, so there cannot be $32$ HT outcomes. I'm getting an average payout of $\dfrac{28}{32}$ dollars in the first game type and $\dfrac{31}{32}$ dollars in the second, but I am too tired to double-check. I'll do that tomorrow. :)
Edit: Oh. now I see what you've done; very clever. I must have missed out on my count somewhere. I'll still wait to confirm in the morning. And yes, there does seem to be an advantage to choosing p1, (even though the expected winnings are the same), in the sense that you are more likely to at least win some money.
- 4 years, 7 months ago
Is "more likely to at least win some money" the main consideration that you will use if the expected payoff is equal?
For example,what would you choose between:
Payoff 1: 50% of $0, 50% of$100
Payoff 2: 90% of $1.01, 1% of$4900
Staff - 4 years, 7 months ago
If Payoff 2 was 99% of $40.40, 1% of$1000, (to give it the same expected payoff as Payoff 1), I would go with Payoff 2 since I would be assured of money and would at least have a slim chance of a large amount of money.
If Payoff 2 was 99% of $0, 1% of$5000, then it gets interesting, (at least for me). Even a such a slight chance of winning $5000 would be hard to pass up, so I would probably go with Payoff 2, unless I absolutely needed that$100 right away.
If Payoff 2 was 99.9% of 0$, 0.1% of$50000 .... hmmmm .... That's a lot of dough, so I'd go with Payoff 2. However, if the options were these last two I've listed, then I might choose the $5000 option. If Payoff 2 involved a 50% chance of losing$100 and a 50% chance of winning $200, then I would go with Payoff 1, since I'm not comfortable with there being such a good chance of losing a fair bit of money. There is a lot of psychology going on in these choices, and yet the expected winnings is always$50, (which ironically is an amount that would never actually be won in any of these scenarios).
- 4 years, 7 months ago
(Ooops, apparently I can't do maths. Fixed.)
So, there are multiple ideas here.
- Is the probability of a positive payoff so important that just because it moves from $0 it would greatly influence your preferences? - At what point does the potential promise of a huge payoff overweigh the certainty of a small payoff? Staff - 4 years, 7 months ago Log in to reply In general I guess it all depends on the specific values, and further, on each person's financial status and willingness to take on risk. For me, the potential, (1% or better), of a large payout would outweigh any guaranteed amount less than$100, so in this case there is no difference to me between $0 and$100. But if the guaranteed amount were, say, $5000, with a 0.01% all-or-nothing chance of winning$1,000,000, I'd probably take the $5000 and run, (although I would be tempted to take the risk, at least for a moment). If the all-or-noting percentage were 1%, though, then I would find it hard not to take the chance;$5000 is a lot of money, but $1,000,000 is a life-changing amount of money, and a 1% chance is realistic in my eyes given the potential reward. The question I would ask myself is: which will I regret more - not taking the guaranteed money or not taking the risk? - 4 years, 7 months ago Log in to reply Yes, the expected payoff for both scenarios is$1. The easiest way to see this is to look at Indicator Variables, which is essentially what Raghav did (though not phrased in that language).
Well, everyone has their own "payoff preference". For example, you stated that you would prefer to "be more likely to leave with at least a dollar in your pocket". As such, this tells me that you are very risk adverse, and that you would prefer the certainty of a positive payoff.
If you want to talk about "risk and stuff", what do you need to consider, and why?
Staff - 4 years, 7 months ago
I am not formally educated in these topics, so I'm just saying all this from a logical standpoint.
After a bit of thinking, I think I agree with you on the fact that everyone has their own "payoff preference". One can take a risk to win more, or take less risk to win less. In the scenario mentioned here, the aspects of risk are not conspicuous. The amount to be won is not significant, and there is no penalty for losing. Hence, both the payoffs are inherently attractive offers. When we come to real life situations, I think there are many more things that contribute to the risk:
1. The probability that you will get a payoff.
2. The probability that you stand to lose money.
3. The security of the winnings. Money once won shouldn't be taken away from you.
4. The effects of said decisions on long term/ on your ability to take other decisions.
According to me, every decision of ours is based on weighing in these risk factors against the probable prize. The balance may tip either way, and so may our decisions.
I don't even know if I'm thinking in the right direction... Help me out here @Calvin Lin
- 4 years, 7 months ago
Part of this series of posts is for you to figure out what kind of risk appetite you have. It is a personal preference, and there isn't necessarily a "right" answer. Your understanding will help guide you in decisions that you make in future. For the points that you raised, you should answer for yourself if those should matter, and why.
Hopefully, you will come up with a consistent, logical risk-reward structure. For example, if you would choose payoff 1 over 2 and payoff 3 over 4, but would prefer a combined 2 and 4 over a combined 1 and 3, then it would be very easy for someone to sell you things in part, and make you overpay for them.
Staff - 4 years, 7 months ago
@Calvin Lin sir, Am I right?
- 4 years, 7 months ago
Let $X$ be an indicator variable which takes on value 1 for each $HT$ pair and 0 otherwise, and let $Y$ be an indicator variable which takes on value 1 for each $HH$ pair and 0 otherwise. There are $5 - 2 + 1 = 4$ indicator variables (both $X$ and $Y$) in 5 coin tosses. The expected value of the payoff is: $\text{E}\left [\sum_{k = 1}^{4} X_{k} \right ], ~ \text{E}\left [\sum_{k = 1}^{4} Y_{k} \right ].$We can apply Linearity of Expectation here, which holds even for dependent events (nice proof of this is given in Brilliant Wiki Page), to get: $\sum_{k = 1}^{4}\text{E}\left [ X_{k} \right ] , ~ \sum_{k = 1}^{4}\text{E}\left [ Y_{k} \right ].$Expected value of both $X_{k}$ and $Y_{k}$ is: $\text{E}\left [ X_{k} \right ] = \text{E}\left [ Y_{k} \right ] = 1 \cdot \dfrac{1}{4} + 0 \cdot \dfrac{3}{4} = \dfrac{1}{4}.$It follows that expected payoffs in both cases are the same and they equal:$\sum_{k = 1}^{4}\text{E}\left [ X_{k} \right ] = \sum_{k = 1}^{4}\text{E}\left [ Y_{k} \right ] = 4 \cdot \dfrac{1}{4} = 1.$
Bonus: What about the variance of the payoff? It can also influence our decision. Is it also the same?
It turns out they are not! Let us first consider payoff 1 and indicator variable $X$:\begin{aligned} \text{Var}[X] &= \text{E}\left [\left(\sum_{k = 1}^{4}X_{k}\right)^{2}\right ] - \text{E}\left [\sum_{k = 1}^{4}X_{k}\right ]^{2} \\ &= \text{E}\left [\left(\sum_{k = 1}^{4}X_{k}^{2}\right) + \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] - \text{E}\left [\sum_{k = 1}^{4}X_{k}\right ]^{2} \\ &= \text{E}\left [\left(\sum_{k = 1}^{4}X_{k}^{2}\right)\right] + \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] - \text{E}\left [\sum_{k = 1}^{4}X_{k}\right ]^{2} \\ &= 1 + \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] - 1 \\ &= \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ].\end{aligned}The same is true for payoff 2 and $Y$. Now, let us calculate $\text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ]$. We make two groups of $X_{k}X_{j}$:
• $\left | k-j \right | = 1$ representing two consecutive indicator variables which share one common coin. Notice that their product is always 0 since it's impossible to have two $HT$s in 3 consecutive coins.
• $\left | k-j \right | > 1$ representing two non-consecutive indicator variables which share no common coin. Their product is non-zero only when they are both non-zero which happens with probability $\left(\frac{1}{4}\right)^{2} = \dfrac{1}{16}.$
There are total of $4 \cdot 3 = 12$ $\left(X_{k}X_{j}\right)$ pairs, and $2\cdot 3$ of them which are consecutive ie. where $\left | k-j \right | = 1$. Hence, we have: $\text{Var}[X] = \text{E}\left[ \left(\sum_{k \neq j}X_{k}X_{j}\right)\right ] = 0 \cdot 6 + \frac{1}{16} \cdot 6 = 0.375.$
In the same way, we calculate $\text{Var}[Y] = \text{E}\left[ \left(\sum_{k \neq j}Y_{k}Y_{j}\right)\right ] = \frac{1}{8} \cdot 6 + \frac{1}{16} \cdot 6 = 1.125.$
So, variance of the payoff 2 is 3 times greater than variance of the payoff 1. If my reasoning is correct, then payoff 1 is somewhat safer option and it tends to the expected value, while payoff 2 is more riskier but it offers better chances of earning more than 1$. What do you thinks about this approach which takes variance into account, @Calvin Lin ? - 1 year, 6 months ago Log in to reply After calculating the expected value and variance of the payoffs by hand using indicator variables (similar to some other people who have posted here), I became curious as to what the distributions of the payoffs would look like and how different strategies (not necessarily of length 2) would work out. For those interested, I wrote up an R-Jupyter notebook on it here. - 5 months, 4 weeks ago Log in to reply That's a lovely investigation! The 3-coin case is more complicated, so it's great to see your results. Note that the THH and HHT cases are identical by symmetry (flip the coins and flip the order of the sequence). For the variance, we indeed have $THH < HTH < HHH$. For the general case, I believe that the $TH \ldots H$ case has the lowest variance (since it requires a T to get it started, so has a very low chance of duplicity). The transition state diagrams also give us some information on how to estimate the variance. Staff - 5 months, 4 weeks ago Log in to reply There are 2^5 outcoms of 5 coin tosses. For payoff 1, there are this many ways to toss a HT. There are {HTHTT,HTHTH,HTTHT,HTHHT,THTHT, HHTHT,HTHHH,HHHTH,HHHHT, HTTTT,THTTT,TTHTT,TTTHT,HTTTH,HHTTT, HHTTH} Expected paayyoff 1 = (6/32$2) + (10/32$1)=$11/16=$22/32 For payoff 2, there are this many ways to toss a HH( or HHH,HHHH and HHHHHH). There are {HHHHH, THHHH,HHHHT, THHHT,HTHHH,HHHTH, HHHTT, TTHHH, HHTHH, HHTTT, THHTT, TTHHT, TTTHH, THTHH, THHTH] Expected payoff 2 =(1/32$4 )+ ( 2/32$3) + ( 5/32$2) + (7/32$1)=$3/4= $27/32 Payoff 2 is better, I guess LOL @Calvin Lin Total possible outcomes of 32 of which there is one combination [TTTTT] which has no payoff. - 4 years, 7 months ago Log in to reply Check your calculations. The expected payoff of both scenarios is$1.
Staff - 4 years, 7 months ago
There example given implies that HHH is counted as two consecutive HHs. This seems questionable as in standard English two consecutive HHs means HHHH. Is this really what the scenario intended?
- 1 year, 1 month ago
Yes, that is what the scenario intended. The consecutive pairs are allowed to overlap. Let me edit that in.
Staff - 1 year, 1 month ago
Expected value is same for both payoffs
- 1 year ago
Great observation. Does this make both payoffs the same? If not, what else do you want to consider?
Staff - 1 year ago
Thus the payoffs are same, so i can choose either of the payoff be it payoff A or Payoff B
- 1 year ago
So, what you're saying is that between
Payoff C: Get $0 Payoff D: 50% chance to get$1,000, 50% chance to lose $1,000 Because the expected payoff for both is$0, you are equally happy to choose either?
Staff - 1 year ago
we need to look at different aspects of payoff like standard deviation etc but I haven't thought about it yet..
I think I need to look at the variance of values in payoff 1 and payoff 2. Then if the expected value is same then we will go for the one which is less riskier.
- 1 year ago
Great. Proceed from there, and then think about what matters to you.
Staff - 1 year ago
Total sample consists of 32 possible combinations... Expected value of payoff 1 is same as payoff 2 but payoff 2 will be more extreme payoffs like 4 while max payoff for payoff 1 will be 2, intuitively i think that variance of payoff 2 is greater than payoff 1 So I will choose payoff 1
- 1 year ago
I remember reading about an interesting experiment done by a French mathematician regarding payoffs and expected utility from lotteries. Allais Paradox?
- 4 years, 7 months ago
That's interesting. In experiment 1 here, taking a risk, (even if it is only 1%), means the possibility of losing a guaranteed million dollars; I would deeply regret gambling and ending up losing that money, but if I took the million and played 1B just to see what happened and found that I could have won 5 million, I would have just said "Oh well" and still been perfectly happy with my million. But if there is no guaranteed money, then having a chance at 5 million at the expense of a 1% greater chance of winning a million seems worth the risk.
Experiment 1 makes me think of the old saying, "A bird in the hand is worth two in the bush." If there are no "birds in hand", however, then the risk evaluation process is quite different.
- 4 years, 7 months ago
Let Ii be an indicator rv for an event of getting H at ith toss. Let X be a rv which is how many times HT occured. Similarly, let Y be the same but for HH. Then, X = I1I2 +... +I4I5 and Y = I1(1-I2) +...+I4(1-I5). We need to compare EX and EY. Given EIi = 0.5, we can apply linearity of expectation (even for multiplication since events are independent) and get 1 for both expectations.
- 4 years ago
So, if their expectations are the same, does that mean that you are indifferent about which payoff to take? If so, why?
Staff - 4 years ago
Yes, because with both payoffs, one can win the same amount of dollars on average.
- 4 years ago
So, are you indifferent between these 2 payoffs:
Payoff 3: Always get $0 Payoff 4: 50% chance of -$1,000,000,000 and 50% chance of $1,000,000,000 Staff - 4 years ago Log in to reply I think we want the choice with the minimum variance involved. Maximin Principle of Rationality: An epistemically rational person maximizes his minimum payoff Staff - 4 years ago Log in to reply If I don't have a million dollars at all, I will definitely not take payoff 4. I would take payoff 4 only if I was a millionaire. So, psychologically I am not indifferent. - 4 years ago Log in to reply Right, so there are other considerations that come into play. For most people, the certainty of a result is preferred to an uncertain result (though with equal expected value). In general, having a low variance is better (though of course, there are exceptions). Staff - 4 years ago Log in to reply I would go for Payoff 2 , because For payoff 1 ..... one can get maximum number of$1 is 2 which equals $2 i.e for this outcome HTTHT ( MAXIMUM POSSIBLE 'HT' consecutive is 2). For payoff 2 ..... one can get maximum number of$1 is 4 which equals $4 i.e for this outcome HHHHH ( MAXIMUM POSSIBLE 'HH' consecutive is 4). - 1 year, 7 months ago Log in to reply I think this is a combinatorics problem?! Staff - 4 years, 7 months ago Log in to reply In the sense that arithmetic is a part of algebra, which is a part of calculus? There are numerous cross-disciplinary ideas. Using combinatorics will only get you one perspective of things. Notice that I am essentially getting at the risk-reward preference, which is a "finance idea" as opposed to a "combinatorics idea". Yes, combinatorics ideas like expected value is involved, but they don't tell the full story. Staff - 4 years, 7 months ago Log in to reply From a statistical standpoint, I would choose payoff option 1. As the number of H tossed increases, the probability of the consecutively tossing another H decreases. That being said, the HH combination can lead to a higher payoff because HTHTH only results in$2 while HHHHH results in \$5. The issue lies in the decreased probability of continuing to toss H. This is a good example related to risk, much like in the decision involved in buying a AAA bond earning 5% vs a B bond earning 13%.
- 3 years, 4 months ago
To be more specific, payoff 1 would be like the AAA bond and payoff 2 like the B bond. It just depends on what kind of risk is right for you.
- 3 years, 4 months ago
Not quite the same comparison. Note that the expected payoff in both methods are the same.
Whereas in the bond example that you gave, we are trading expected payoff for certainty.
I also strongly disagree with "As the number of H tossed increases, the probability of the consecutively tossing another H decreases". The coin tosses are independent, and that is a common fallacy that "the proportion of realized events must be equal / close to the calculated probability"
Staff - 3 years, 4 months ago | 2019-12-13T17:43:20 | {
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https://mathematica.stackexchange.com/questions/257623/finding-ellipse-axes | # Finding Ellipse Axes
Consider the following ellipse, generated by the bounding region of the following points
ps = {{-11, 5}, {-12, 4}, {-10, 4}, {-9, 5}, {-10, 6}};
rec = N@BoundingRegion[ps, "FastEllipse"];
Graphics[{rec, Red, Point@ps}]
We have that the ellipse 'rec' is given in the form
Ellipsoid[{-10.4, 4.8}, {{2.77333, 0.853333}, {0.853333, 1.49333}}]
How can I retrieve the lengths of the two main axes of such ellipsoid? Following this representation and Mathematica's general definition of Ellipsoid
I tried the following, using the eigenvalues of rec[[2]]
eigs = Eigenvectors[Inverse[rec[[2]]]]
eigv = Eigenvalues[Inverse[rec[[2]]]];
lens = 2/Sqrt[eigv]
Out[]= {2.06559, 3.57771}
where the 2 factor comes from the fact what what I retrieve from the eigenvalues is actually half the length of the main axis. Indeed we get
Graphics[{rec, Red, Point@ps,
Blue, Line[
RegionCentroid@rec + # & /@ {-(lens[[1]] eigs[[1]])/
2, (lens[[1]] eigs[[1]])/2}],
Line[RegionCentroid@rec + # & /@ {-(lens[[2]] eigs[[2]])/
2, (lens[[2]] eigs[[2]])/2}]}]
Is this correct? Is there a quicker way of doing this?
• Nov 2, 2021 at 17:04
Stolen from @J.M.'s answer, https://mathematica.stackexchange.com/a/239797/4999, with one correction (is that enough to make it not a duplicate?):
Nodes = ps;
ellipsoidBR = BoundingRegion[Nodes, "FastEllipse"]; (* not "FastEllipsoid" *)
center = ellipsoidBR[[1]];
{vals, vecs} = Eigensystem[ellipsoidBR[[2]]];
{a, b} = Sqrt[vals];
major = N@{center - a vecs[[1]], center + a vecs[[1]]}
minor = N@{center - b vecs[[2]], center + b vecs[[2]]}
Graphics[{ellipsoidBR, Red, Point@ps, Green, Point@center,
Line@{major, minor}}, Frame -> True]
Here's the alternate way using SingularValueDecomposition:
pt = rec[[1]]; mat = rec[[2]];
{u, s, v} = SingularValueDecomposition[(mat + Transpose[mat])/2];
func = Composition[AffineTransform[{u, pt}], ScalingTransform[Sqrt[Diagonal[s]]]];
and the length:
EuclideanDistance @@@ {func@{{0, -1}, {0, 1}}, func@{{-1, 0}, {1, 0}}}
{2.06559, 3.57771}
Graphics[{rec, Blue, Line[func@{{-1, 0}, {1, 0}}],
Line[ func@{{0, -1}, {0, 1}}]}]
I think the help of Ellipsoid is incomplete because it does not explain the input: Ellipsoid[p,[CapitalSigma]], where the second argument is called the "weight matrix".
You will remember that an ellipse (for simplicity I am explaining the 2D case, nD is similar and assume the ellipse is centered at the origin) can be written by:
x^2/rx^2 + y^2/ry^2 == 1
We may write this as:
{x,y}.{{1/rx^2,0},{0,1/ry^2}}.{x,y} == r.mat.r == 1
Note that the inverse Sqrt of the eigenvalues of m0 are the half axes of the ellipse.
If we now rotate the coordinate system by a rotation matrix: rot (r'=rot.r where r' are the new coordinates) the ellipse will be rotated (in the inverse sense) in the new coordinates:
r'. Transpose[rot].mat. rot .r' == r' . mat' . r' == 1
Therefore a rotated ellipse may be represented by a an symmetric (positive definite) matrix: mat'. This is called "weigh matrix" in the help.
Note that the eigenvalues of the matrix are not changed by a rotation. The eigenvectors point in the directions of the half axes.
Here is an example: Let
rx=2;
ry=1;
m0=DiagonalMatrix[{1/rx^1,1/ry^2}]
{x,y}.m0.{x,y}==1
This represents an axis aligned ellipse with half axes rx and ry:
Region[ImplicitRegion[{x, y} . m0 . {x, y} == 1, {x, y}],
Axes -> True]
If we now rotate the matrix m0:
rot = RotationMatrix[-Pi/4];
m= Transpose[rot].m0.rot;
we get a rotated ellipse:
Region[ImplicitRegion[{x, y} . m . {x, y} == 1, {x, y}], Axes -> True]
Threrfore, the half axes are obtained by the inverse Sqrt of the eigenvalues of the weight matrix. And the directions of the axes are given by the eigenvectors. | 2022-05-21T02:37:14 | {
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http://velbyproductions.com/3m1tft/p7f7f.php?a930ce=curve-fitting-least-square-method | /BBox [0 0 5669.291 8] If the coefficients in the curve-fit appear in a linear fashion, then the problem reduces to solving a system of linear equations. /Filter /FlateDecode /Matrix [1 0 0 1 0 0] Each method has its own criteria for evaluating the fitting residual in finding the fitted curve. endstream Least Squares Fit (1) The least squares fit is obtained by choosing the α and β so that Xm i=1 r2 i is a minimum. >> Fitting requires a parametric model that relates the response data to the predictor data with one or more coefficients. This Python program implements least square method to fit curve of type y = ab x.. We first read n data points from user and then we implement curve fitting for y = ab x using least square approach in Python programming language as follow: . /Resources 17 0 R 18 0 obj Least Square is the method for finding the best fit of a set of data points. /Matrix [1 0 0 1 0 0] It can also be easily implemented on a digital computer. 14 0 obj . The application of a mathematicalformula to approximate the behavior of a physical system is frequentlyencountered in the laboratory. The most important application is in data fitting. Here a = 1.1 and b = 1.3, the equation of least square line becomes Y = 1.1 + 1.3 X. /Subtype /Form Let ρ = r 2 2 to simplify the notation. in this video i showed how to solve curve fitting problem for straight line using least square method . >> endobj This is usually done usinga method called least squares" which will be described in the followingsection. stream By understanding the criteria for each method, you can choose the most appropriate method to apply to the data set and fit the curve. Also suppose that we expect a linear relationship between these two quantities, that is, we expect y = ax+b, for some constants a and b. The best fit in the least-squares sense minimizes the sum of squared residuals. /Filter /FlateDecode You can employ the least squares fit method in MATLAB. endstream �2���6jE)�C�U�#�\�N������p�S�J��3����*�V(q:S�Qèa��6��&�M�q9;?z�(��%��'ދ1e�Ue�eH�M�I������X+m�B����lg�bB�BLJ��ɋ��nE�&d�a9樴 �)Z+��. /Resources 19 0 R x���P(�� �� In LabVIEW, you can apply the Least Square (LS), Least Absolute Residual (LAR), or Bisquare fitting method to the Linear Fit, Exponential Fit, Power Fit, Gaussian Peak Fit, or Logarithm Fit VI to fin… The sum of the squares of the offsets is used instead of the offset absolute values because this allows the residuals to be treated as a continuous differentiable quantity. Least Squares Fitting. x���P(�� �� To find the equation of the curve of ‘best fit’ which may be the most suitable for predicting the unknown values. If you're an engineer (like I used to be in a previous life), you have probably done your bit of experimenting. Other documents using least-squares algorithms for tting points with curve or surface structures are avail-able at the website. Least squares fit is a method of determining the best curve to fit a set of points. . Curve fitting is one of the most powerful and most widely used analysis tools in Origin. /FormType 1 The basic problem is to find the best fit Normal Equation for ‘a’ $$\sum Y = na + b\sum X$$, Normal Equation for ‘b’ $$\sum XY = a\sum X + b\sum {X^2}$$, The direct formula of finding $$a$$ and $$b$$ is written as, $b = \frac{{\sum XY – \frac{{\left( {\sum X} \right)\left( {\sum Y} \right)}}{n}}}{{\sum {X^2} – \frac{{{{\left( {\sum X} \right)}^2}}}{n}}}{\text{ }}, \,\,\,\,\,\,\,\,\,\,\,\,a = \overline Y – b\overline X$, Help me with the normal equations for power curve, Your email address will not be published. This article demonstrates how to generate a polynomial curve fit using the least squares method. 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By using the least squares fitting the most common method to generate a polynomial equation a... | 2022-07-03T06:18:03 | {
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https://artspassion.org/now-pea-ezocl/5c0621-irrational-numbers-definition | For example, there is no number among integers and fractions that equals the square root of 2. having a numerical value that is an irrational number. See also Rational Number. Define irrational. Every transcendental number is irrational.. ‘The square root of 2 is an irrational number because it can't be written as a ratio of two integers.’ ‘How can mathematical concepts like points, infinitesimally small quantities, or irrational numbers be anything but products of our minds?’ ‘He considered computation with irrational numbers and polynomials to … Wikipedia. irrational numbers. being an irrational number. An irrational number is a real number that cannot be written as a simple fraction. c. Marked by a lack of accord with... Irrational - definition of irrational by The Free Dictionary. Irrational Numbers . Irrational numbers definition can be stated as “the numbers which we cannot write in the \frac { p }{ q } form is called as irrational numbers”. Definitions. For example: The pairs (2, 9); (4, 7) etc. It explains in computing terminology what Irrational Number means and is one of many technical terms in the TechTerms dictionary. Irrational Numbers Definition. irrational number \ɪ.ˌɹæʃ.ə.nəl ˈnʌm.bɚ\ (États-Unis), \ɪ.ˌɹæʃ.ə.nəl ˈnʌm.bə\ (Royaume-Uni) (Mathématiques) Nombre irrationnel. An irrational number is simply the opposite of a rational number. Irrational numbers may not be crazy, but they do sometimes bend our minds a little. What Is a Rational Number? Your fears are not based on fact and not likely to come true. Definition of irrational number in the Definitions.net dictionary. 3 , 5 are examples of irrational numbers. The denominator q is not equal to zero ($$q≠0.$$) Some of the properties of irrational numbers are listed below. See Rational number definition.) Irrational Number. Enrich your vocabulary with the English Definition … As I said previously, the Conservatives give a new definition to the algebraic term of "irrational numbers" because their numbers simply do not make sense. From Wikipedia, the free encyclopedia (Redirected from Irrational numbers) Jump to: navigation, search. While at Montpellier he wrote a paper on irrational numbers and limits. Learn more. irrational: [adjective] not rational: such as. However, irrational numbers can have a decimal value that continues forever WITHOUT a pattern, unlike the example above. Examples of such numbers are π, e, √6. Irrational Numbers. Another definition we can give as “non terminating non recurring decimal numbers are irrational numbers”. This page contains a technical definition of Irrational Number. Many people are surprised to know that a repeating decimal is a rational number. 1. a. Let's start by defining each term separately, then we can learn more about each and work through some examples. Information and translations of irrational number in the most comprehensive dictionary definitions resource on the web. Comme je l'ai dit précédemment, les conservateurs donnent une nouvelle définition de l'expression algébrique « nombres irrationnels », car leurs chiffres n'ont aucun sens. For example, real numbers like √2 which are not rational are categorized as irrational. So they can't be written as a clear fraction of 2 integers. not endowed with reason or understanding. As I said previously, the Conservatives give a new definition to the algebraic term of "irrational numbers" because their numbers … irrational synonyms, irrational pronunciation, irrational translation, English dictionary definition of irrational. Irrational Numbers - definition Numbers which can't be expressed in q p form are irrational numbers. 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Work through Some examples written irrational numbers definition a simple fraction et moteur de recherche de traductions françaises that element. To: navigation, search definitions on the TechTerms website are written to technically. Il a écrit un document sur les nombres irrationnels et les limites definition, without the faculty reason. Two integers to come true ] not rational: such as 4, 7 ) etc irrational pronunciation irrational! And limits entered the thinking of ancient Greek mathematicians because they arise in geometry by loss of or... | 2021-08-05T11:05:42 | {
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https://math.stackexchange.com/questions/2378844/basic-probability-question-multiplication-rule/2382974 | # Basic probability question- multiplication rule
I came across the following question in a textbook (bear in mind that this is the only information given)-
There is a $50$ percent chance of rain today. There is a $60$ percent chance of rain tomorrow. There is a $30$ percent chance that it will not rain either day. What is the chance that it will rain both today and tomorrow?
My instinct was to multiply the probabilities together for today and tomorrow to arrive at an answer of $30$ percent. However, the answer given is $40$ percent (based on taking the addition of the individual probabilities and subtracting their union). Can someone explain to me why the multiplication rule does not apply here? Does it have to do with independence? Bear in mind, I am trying to relearn probability theory from scratch. Thanks.
• if they were independent you multiply for example you flip three coins the probability they all end up heads is $({1 \over 2})^3={1\over 8}$ dependent probabilities aren't so simple. – user451844 Aug 1 '17 at 13:37
Yes, this is about independence. Or, better put, these events are not independent, which by common sense makes sense, since if it rains one day, it typically means there is a higher chance than normal that it rains the next day as well (we might well be in the 'rainy season')
You can also tell that the events are not independent given the numbers given to you. If the events were independent, then the probability of it not raining either day should have been $0.5 \cdot 0.4=0.2$, but you were told it is actually $0.3$.
What is always true, however (so you can always use this formua, whether the events are independent or not), is that:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
And from that you can derive your desired:
$$P(A \cap B) = P(A) +P(B) - P(A \cup B) = P(A) +P(B)-(1-P(A^C \cap B^C))=$$
$$0.5+0.6-(1-0.3) =1.1-0.7=0.4$$
Yes, the multiplication rule only applies when events are independent, and there is no reason to assume the events of rain on each day are independent.
For this type of problem, it helps to make a Venn diagram. There are two circles, $A$ and $B$, representing the events of rain today and tomorrow. This gives four regions: $A\cap B$ (rain on both days), $A^c\cap B^c$ (no on both days), $A\cap B^c$ (rain today, but not tomorrow), and $A^c\cap B$. The given information, and the fact that the total probability is $1$, tells us that \begin{align} P(A)=P(A\cap B)+P(A\cap B^c)&=0.5\\ P(B)=P(A\cap B)+P(A^c\cap B)&=0.6\\ P(A^c\cap B^c)&=0.3\\ P(A\cap B)+P(A\cap B^c)+P(A^c\cap B)+P(A^c\cap B^c)&=1 \end{align} This is four equations equations in four unknowns, allowing you to solve for $P(A\cap B)$.
When you have only two independent variables, sometimes it's easier just to make a table.
Let $X$ be the event that it rains today and $Y$ be the event that it rains tomorrow. We are given that $P(X)=0.5$, $P(Y)=0.6$, and $P(\overline X \cap \overline Y)=0.3$:
$$% outer array of arrays \begin{array}{lr} % inner 3x3 array in top left corner \begin{array}{c|c|c|} & Y & \overline Y \\ \hline X & P(X \cap Y) & P(X \cap \overline Y) \\ \hline \overline X & P(\overline X \cap Y) & P(\overline X \cap \overline Y)=0.3 \\ \hline \end{array} % inner 3x1 array in top right corner \begin{array}{l} \\ P(X)=0.5 \\ P(\overline X) \end{array} \\ % inner 1x3 array in bottom left corner \begin{array}{ccc} \quad & P(Y)=0.6 & P(\overline Y) \\ \end{array} \end{array}$$
We know that $P(X)+P(\overline X)=1$ and likewise $P(Y)+P(\overline Y)=1$, so we can fill this in as follows:
$$% outer array of arrays \begin{array}{lr} % inner 3x3 array in top left corner \begin{array}{c|c|c|} & Y & \overline Y \\ \hline X & P(X \cap Y) & P(X \cap \overline Y) \\ \hline \overline X & P(\overline X \cap Y) & P(\overline X \cap \overline Y)=0.3 \\ \hline \end{array} % inner 3x1 array in top right corner \begin{array}{l} \\ P(X)=0.5 \\ P(\overline X)\color{red}{=0.5} \end{array} \\ % inner 1x3 array in bottom left corner \begin{array}{ccc} \quad & P(Y)=0.6 & P(\overline Y)\color{red}{=0.4} \\ \end{array} \end{array}$$
We know that the rows sum across and the columns sum down, so:
$$% outer array of arrays \begin{array}{lr} % inner 3x3 array in top left corner \begin{array}{c|c|c|} & Y & \overline Y \\ \hline X & P(X \cap Y) & P(X \cap \overline Y)\color{red}{=0.1} \\ \hline \overline X & P(\overline X \cap Y)\color{red}{=0.2} & P(\overline X \cap \overline Y)=0.3 \\ \hline \end{array} % inner 3x1 array in top right corner \begin{array}{l} \\ P(X)=0.5 \\ P(\overline X)=0.5 \end{array} \\ % inner 1x3 array in bottom left corner \begin{array}{ccc} \quad & \,P(Y)=0.6 & \quad\quad\, P(\overline Y)=0.4 \\ \end{array} \end{array}$$
And finally:
$$% outer array of arrays \begin{array}{lr} % inner 3x3 array in top left corner \begin{array}{c|c|c|} & Y & \overline Y \\ \hline X & P(X \cap Y)\color{red}{=0.4} & P(X \cap \overline Y)=0.1 \\ \hline \overline X & P(\overline X \cap Y)=0.2 & P(\overline X \cap \overline Y)=0.3 \\ \hline \end{array} % inner 3x1 array in top right corner \begin{array}{l} \\ P(X)=0.5 \\ P(\overline X)=0.5 \end{array} \\ % inner 1x3 array in bottom left corner \begin{array}{ccc} \quad & \,P(Y)=0.6 & \quad\quad\, P(\overline Y)=0.4 \\ \end{array} \end{array}$$ | 2020-10-21T22:50:21 | {
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https://math.stackexchange.com/questions/633575/linear-algebra-question-on-rank-and-null-space | # Linear algebra question on rank and null space
This is an exercise from linear algebra and optimization by Gill, I do exercises to be prepared for my final exam and these are not homework questions!
Exercise $\mathbf{6.1.}\,$ Consider the following matrix $A$ and vector $b$: $$A=\begin{pmatrix} 2 & 4 \\ 1 & 2 \\ 1 & 2 \\ \end{pmatrix},\quad b=\begin{pmatrix} 3 \\ 2 \\ 1 \\ \end{pmatrix}.$$ $\textbf{(a)}$ What is the rank of $A$? Give a general form for any vector in the range of $A$.
$\textbf{(b)}$ Show that the dimension of the null space of $A^T$ is two, and display two linearly independent vectors $z_1$ and $z_2$ in $\operatorname{null}(A^T)$. Give a general form for every vector in $\operatorname{null}(A^T)$.
$\textbf{(c)}$ Find the vectors $b_R\in\operatorname{range}(A)$ and $b_N\in\operatorname{range}(A^T)$ such that $b=b_R+b_N$.
$\textbf{(d)}$ Give the general form of $b_A$ such that $b_R=Ab_A$. (Hint: consider all vectors $q$ such that $Aq=0$.)
For part (a), I think $rank(A)=1$ since all the columns are linear combination of each other. As for a general form for any vector in the range of $A$, when I write $Ax=b$ I get an over-determined system:
$$2x_1+4x_2=3$$ $$x_1+2x_2=2$$ $$x_1+2x_2=1$$
So I'm not sure about the general form.
As for part (b), since the rank is two, dimension of $N(A^T)$ must be $3-1=1$, right?
And no idea about the rest. Any help would be greatly appreciated.
• If you know row reduction and properties of row reduced matrix then apply. It will be easier to get all answer together. – Dutta Jan 10 '14 at 10:37
• Your solution to (a) is correct. For (b) note that if the dimension of the null space is two and you already found two linearly independent vectors in the null space, those form a base. Now that directly gives you a general form of all null vectors. – Listing Jan 10 '14 at 10:38
• For (c) note that $\text{null}(A^T)$ are exactly the vectors that are not in the image of $A$ (aka $\text{range}(A)$). The image of $A$ gets spanned by $(2,1,1)$ thus you can write $b$ as $\alpha z_1+\beta z_2 + \gamma (2,1,1)$. where $z_1,z_2$ are from (b). – Listing Jan 10 '14 at 10:41
• @Listing: Thank you, that helped to understand the problem. – Gigili Jan 10 '14 at 12:07
• Thank you for the edit @Aðøbe. – Gigili Jan 11 '14 at 2:14
## 1 Answer
If you do row reduction, you find $$\begin{pmatrix} 2 & 4 \\ 1 & 2 \\ 1 & 2 \end{pmatrix} \to \begin{pmatrix} 1 & 2 \\ 0 & 0 \\ 0 & 0 \end{pmatrix}$$ which means that the first column is a basis for the column space of $A$ (which is better terminology than “range of $A$”, in my opinion). So the general form of the vectors in the column space of $A$ is $$\begin{pmatrix} 2a \\ a \\ a \end{pmatrix},\quad \text{a any scalar}$$
By the rank nullity theorem, the null space of $A$ has dimension $1$; the equation defining it is $$x_1+2x_2=0$$ so a basis for it is the single vector $$\begin{pmatrix} -2 \\ 1 \end{pmatrix}$$
The null space of $A^T$ has indeed dimension $2$; the row reduction on $A^T$ is $$\begin{pmatrix} 2 & 1 & 1 \\ 4 & 2 & 2 \end{pmatrix} \to \begin{pmatrix} 2 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}$$ so the equation defining the null space is $2x_1+x_2+x_3=0$ and a basis for it is $$\left\{ \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix} \,, \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} \right\}$$ Writing $b=b_R+b_N$ should now be easy: the system to solve is $$\left(\begin{array}{ccc|c} 2 & -1 & -1 & 3 \\ 1 & 2 & 0 & 2 \\ 1 & 0 & 2 & 1 \end{array}\right)$$ but you can as well find the orthogonal projection of $b$ on the column space of $A$: $$b_R= \frac{(2\ 1\ 1)\begin{pmatrix}3\\2\\1\end{pmatrix}} {(2\ 1\ 1)\begin{pmatrix}2\\1\\1\end{pmatrix}} \begin{pmatrix}2\\1\\1\end{pmatrix} =\frac{9}{6}\begin{pmatrix}2\\1\\1\end{pmatrix} =\begin{pmatrix}3\\3/2\\3/2\end{pmatrix}$$ and $b_N=b-b_R$.
With this last idea it's easy to solve the last point.
• Thank you for your detailed answer. Where did you use the fact that the first column is a basis for the column space (because it has leading $1$, right?)? And how did you come up with that general form, $\begin{pmatrix} 2a \\ a \\ a \end{pmatrix}$? – Gigili Jan 10 '14 at 11:56
• @Gigili I used it for the last part, when computing $b_R$ with the orthogonal projection. Since the first column is a basis for the column space, every vector in the column space is that vector times a scalar. – egreg Jan 10 '14 at 11:59
• Ah, right! I was thinking of $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ as the first column but it's the first column in REF. Thanks, it's clear now. – Gigili Jan 10 '14 at 12:04
• Sorry, but I still don't get the last part about $b$. Where did that system come from? Did you multiply the basis by $-1$? Then what's the first row? How did you calculate $b_R$? – Gigili Jan 10 '14 at 12:07
• @Gigili It's a common error, I find it all the time in students' papers. ;-) The first non zero column in REF is always that one! It can't be in the column space of every matrix, can it? – egreg Jan 10 '14 at 12:07 | 2019-05-26T01:27:47 | {
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https://cstheory.stackexchange.com/questions/21984/expected-empirical-entropy | # Expected empirical entropy
I'm thinking about some properties of the empirical entropy for binary strings of length $n$ when the following question crosses my way:
$\underbrace{\large\frac{1}{2^{n}}\normalsize\sum\limits_{w\in\left\{0,1\right\}^{n}}\normalsize nH_{0}(w)}_{\large\#}\;\overset{?}{=}\;n-\varepsilon_{n}\;\;\;$
with $\;\;\lim\limits_{n\rightarrow\infty}\varepsilon_{n}=c\;\;\;$ and $\;\;\;\forall n:\;\varepsilon_{n}>0$
where $c$ is a constant.
### Is that equation true? For which function $\varepsilon_{n}$ respectively which constant $c$?
$n=2\;\;\;\;\;\;\;\rightarrow\;\#=1$
$n=3\;\;\;\;\;\;\;\rightarrow\;\#\approx 2.066$
$n=6\;\;\;\;\;\;\;\rightarrow\;\#\approx 5.189$
$n=100\;\;\;\rightarrow\;\#\approx 99.275$
$n=5000\;\rightarrow\;\#\approx 4999.278580$
$n=6000\;\rightarrow\;\#\approx 5999.278592$
# Backround
$H_{0}(w)$ is the zeroth-order empircal entropy for strings over $\Sigma=\left\{0,1\right\}$:
• $H_{0}(w)=\frac{|w|_{0}}{n}\log\frac{n}{|w|_{0}}+\frac{n-|w|_{0}}{n}\log\frac{n}{n-|w|_{0}}$
where $|w|_{0}$ is the number of occurences of $0$ in $w\in\Sigma^{n}$.
The term $nH_{0}(w)$ corresponds to the Shannon-entropy of the empirical distribution of binary words with respect to the number of occurences of $0$ respectively $1$ in $w\in\Sigma^{n}$.
More precise:
Let the words in $\left\{0,1\right\}^{n}$ be possible outcomes of a Bernoulli process. If the probability of $0$ is equal to the relative frequency of $0$ in a word $w\in\left\{0,1\right\}^{n}$, then the Shannon-entropy of this Bernoulli process is equal to $nH_{0}(w)$.
At this point, my question should be more reasonable since the first term normalizes the Shannon-entropies for all empirical distributions of words $w\in\left\{0,1\right\}^{n}$.
Intuitively I thought about getting something close to the Shannon-entropy of the uniform distribution of $\left\{0,1\right\}^{n}$, which is $n$.
By computing and observing some values I've got the conjecture above, but I'm not able to prove it or to get the exact term $\varepsilon_{n}$.
It is easy to get the following equalities:
$\large\frac{1}{2^{n}}\normalsize\sum\limits_{w\in\left\{0,1\right\}^{n}}\normalsize nH_{0}(w)\;\;=\large\frac{1}{2^{n}}\normalsize\sum\limits_{w\in\left\{0,1\right\}^{n}}\normalsize |w|_{0}\log\frac{n}{|w|_{0}}+(n-|w|_{0})\log\frac{n}{n-|w|_{0}}$
$=\large\frac{1}{2^{n}}\normalsize\sum\limits_{k=1}^{n-1}$ $n\choose k$ $\left(k\log\frac{n}{k}+(n-k)\log\frac{n}{n-k}\right)$
and it is possible to apply some logarithmic identities but I'm still in a dead point.
(the words $0^{n}$ and $1^{n}$ are ignored, because the Shannon-entropy of their empirical distributions is zero)
Any help is welcome.
• I don't see a question here. – Suresh Venkat Apr 7 '14 at 17:36
• Is the equation above true? Can anybody prove it respectively prove that the equation doesn't hold? – Danny Apr 7 '14 at 17:38
• you have a bunch of equations. it would be better to number them. also highlight the question with ">" blockquoting. – vzn Apr 8 '14 at 15:20
• I edit my question from $\varepsilon_{n}\rightarrow0$ to $\varepsilon_{n}\rightarrow c$, where $c$ is a constant. – Danny Apr 9 '14 at 13:46
Here is another approach, based on information theory and heavily inspired by @usul's answer. It shows that $\epsilon_n=O(1)$ with very few calculations, and can be used to prove that $\epsilon_n \rightarrow \log_2 \sqrt{e}$ and to derive good estimates on the rate of convergence with less calculations than @usul's approach. In fact, I find a closed-form expression for $\epsilon_n$ : $$(1) \; \; \; \; \epsilon_n = n \left( H(Binom(n,1/2)) - H \left( Binom \left( n-1, 1/2 \right) \right) \right) \ .$$
Details:
Let $X$ be a uniform random variable in $\{0,1\}^n$. Let $K$ be a random variable equal to the number of 1's in $X$. The expression $\#$ that @Danny wants to analyze is exactly equal to $n \cdot \mathbb{E}_{k} [H(X_1 | K=k)]$. (Here $X_1$ is the first bit of $X$.) By the basic properties of the entropy operator, $$(2) \; \; \; \; \# = n\mathbb{E}_{k} [H(X_1 | K=k)]=nH(X_1|K)=n(H(X_1K)-H(K))=n(H(K|X_1)+H(X_1)-H(K)) = n(1-H(Binom(n,1/2)) + H(Binom(n-1,1/2))) \ .$$ The last equality follows from the fact that $X_1$ is just a uniformly random bit, $K$ is the binomial distribution, and $K|(X_1=x_1)$ is distributed either as $Binom(n-1,1/2)$ or as $Binom(n-1,1/2)+1$, depending on the value of $x_1$, both of which have the same entropy.
This already gave us equation (1). Now we just need to calculate to get the value of $\lim_{n \rightarrow \infty} \epsilon_n$
We use any known estimation for the entropy of a binomial RV, such as here. We see that $$(3) \; \; \; \; H(K)=H(Binom(n,1/2))=\frac12 \log_2 ( \pi en / 2) + O(1/n) \ ,$$ and, similarly, that $$H(K|X_1)=H(Binom(n-1,1/2))=\frac12 \log_2 ( \pi e(n-1)/2) + O(1/n) \ .$$ Canceling out terms and substituting into (1), we get $$(4) \; \; \; \; \epsilon_n = n \cdot (H(K)-H(K|X_1)) = n \cdot \frac12 (\log_2 (n/(n-1)) + O(1/n)) = \\ \frac12 \log_2 ((n/(n-1))^n) + O(1) \rightarrow \log_2(\sqrt{e}) + O(1) \ .$$
By slightly improving the approximation (3) we should be able to replace the $O(1)$ term in (4) by $O(1/n)$ and therefore get that, indeed, $\lim_{n \rightarrow \infty} \epsilon_n = \log_2 \sqrt{e}$. To get this better estimation it should be enough to check that $H(Binom(n,1/2))=\frac12 \log_2 ( \pi en / 2) + err(n)$ where $err(n)=O(1/n)$ and $err$ is a monotone function.
• Very nice! I accept usul's answer since your answer is inspired by usul's, but your answer is very elegant (and converts to $\log(\sqrt e)$, too). – Danny Apr 10 '14 at 10:08
• *converges instead of "converts" :) – Danny Apr 10 '14 at 11:29
• Danny, I'm not sure it's a good idea to accept a more complicated and less elegant answer. Everyone else who ever looks at this question will look at @usul's answer first, while it's in everyone's interest that they look at the most simple and elegant answer first. The solutions to this are: 1. to ask usul to reformulate his answer to be simpler (for all I care, he could copy the text of my answer verbatim if he wants, I don't care about the reputation or anything) or 2. to accept my answer. I'll transfer the relevant reputation to usul if that's possible. – mobius dumpling Apr 10 '14 at 13:30
• This seems like the better answer to accept in my opinion too! Very nice/elegant entropy argument. – usul Apr 15 '14 at 18:10
• Note that in the literature you'll often see $\log_2\sqrt{e}$ written as $\frac{1}{2\ln 2}$. – Steven Stadnicki Apr 18 '14 at 18:52
(Edited from previous version, 2014-04-08.)
I believe that the answer is $\epsilon_n \to \log(\sqrt{e}) \approx 0.7213475...$ where the logarithm is base 2. This seems to match simulation results. I don't have a full formal proof, but give the heuristic approximations/calculations.
I think it's easier to note that your question is equivalent to:
What are the asymptotics of $\delta_n = 1 - \mathbb{E}[H(k/n)]$, where the expectation is over $k \sim Binomial(n,0.5)$?
(where $H$ is the binary entropy function $H(p) = p\log(1/p) + (1-p)\log(1/(1-p))$, and $\delta_n = \epsilon_n/n$.)
Here's a quick plot to show the idea. We have the binary entropy function in blue and the Binomial pmf (for $p=0.5$) in green. So we can see that the expectation of $H(k/n)$, when $k$ is distributed binomially, will always be below one but should be approaching one. The question is how fast.
The key idea will be that the value we're interested in, $$\mathbb{E}[H(k/n)] = \sum_{k=0}^n \frac{{n \choose k}}{2^n} H\left(\frac{k}{n}\right) ,$$ can be related entropy of the binomial distribution. Let $p_{n,k} = \frac{{n \choose k}}{2^n}$ be the probability of $k$ heads in $n$ fair coin flips. The steps will be as follows:
1. Use Stirling's approximation to get $H(k/n) = \frac{\log(p_{n,k})}{n} + 1 + \text{something}$.
2. Rewrite the original sum to get $\mathbb{E}[H(k/n)] = 1 - \frac{H(Binom(n,0.5))}{n} + \sum \text{something}$.
3. Get that the entropy of the binomial, divided by $n$, plus the sum of "something", equals $\frac{\log(\sqrt{e})}{n} + O(1/n^2)$.
Step 1:
Just plugging in Stirling and doing some cancellation/rearranging, \begin{align} p_{n,k} &:= \frac{{n \choose k}}{2^n} \\ &\sim \frac{n^n}{k^k (n-k)^{n-k}}\frac{1}{2^n}\sqrt{\frac{n}{2\pi k (n-k)}} . \end{align}
This won't be very tight for every term in the sum, but I think it will be asmyptotically tight towards the middle, which is all that matters since all the probability is in the center.
We can rewrite the entropy function as follows. It's just some arithmetic to combine the logarithms. \begin{align} H(k/n) &= \frac{k}{n}\log\left(\frac{1}{k/n}\right) + \frac{n-k}{n}\log\left(\frac{1}{(n-k)/n}\right) \\ &= \log\left(\frac{n}{k^{k/n}(n-k)^{(n-k)/n}}\right) . \end{align}
So, using Sterling's approximation above, the logarithm of a probability term is \begin{align} \log(p_{n,k}) &= \log\left(\frac{{n \choose k}}{2^n}\right) \\ &\approx n H(k/n) - n + \log\left(\sqrt{\frac{n}{2\pi k(n-k)}}\right) . \end{align}
Step 2: \begin{align} \mathbb{E}[H(k/n)] &= \sum_{k=0}^n p_{n,k} H(k/n) \\ &\approx \sum_{k=0}^n p_{n,k} \left(\frac{\log(p_{n,k})}{n} + 1 - \frac{\log\left(\sqrt{\frac{n}{2\pi k (n-k)}}\right)}{n}\right) \\ &= 1 - \frac{H(Binom(n,0.5))}{n} - \frac{1}{n}\sum_{k=0}^n p_{n,k}\log\left(\sqrt{\frac{n}{2\pi k(n-k)}}\right) . \end{align} Here, $H(Binom(n,0.5))$ is the entropy of the Binomial distribution for $n$ coin flips and $p=0.5$, which by wikipedia is $\log\left(\sqrt{\pi e n / 2}\right) + O(1/n)$.
Step 3:
Now, we just need to approximate the third sum. I will take a very rough approximation (feel free to do better, but it probably doesn't gain much). All the probability mass is concentrated on $k = \frac{n}{2} \pm o(n)$. So approximate this sum (which is an expectation) by its value on the term $k=\frac{n}{2}$, when it is $\log\left(\sqrt{\frac{2}{\pi n}}\right)$.
So now, we get \begin{align} \delta_n &\approx \frac{\log\left(\sqrt{\pi e n/2}\right)}{n} + \frac{\log\left(\sqrt{2/\pi n}\right)}{n} \pm O\left(\frac{1}{n^2}\right) \\ &= \frac{\log\left(\sqrt{e}\right)}{n} \pm O\left(\frac{1}{n^2}\right) . \end{align}
• Thank you very much for your effort. Your attempt is useful and i like your ideas, but i am not quite sure about the details. I will think about it and maybe i will ask some questions soon.. – Danny Apr 8 '14 at 21:35
• @Danny, sure thing. Let me know if parts are unclear. The computation is not fully rigorous, particularly (1) the use of Stirling's approximation everywhere (a fix might be to only consider an interval where the Binomial probability is $\Omega(1/n)$ and argue that Stirling is asymptotically tight there) and (2) the approximation of the "third term" by just setting $k=n/2$ (a fix might be to again consider the high-probability interval and see if the error is small). – usul Apr 9 '14 at 1:24
• My first question is a general question: You think that $\mathbb{E}_{w \in \{0,1\}^n} nH(w) = n-\mathcal{O}(1)$. Do you think there is a "short" proof for this conjecture without getting the exact term $\varepsilon_{n}$ respectively the exact limit of $\varepsilon_{n}$, which you are considering to be $\log\sqrt{e}$? – Danny Apr 9 '14 at 13:35
• @Danny, I'm not sure how to get this with a shorter proof, but maybe another answer will do so. – usul Apr 9 '14 at 20:51
• Thanks again for your help. Everything seems to be clear.. – Danny Apr 10 '14 at 9:07
Here is a proof that the quantity # that OP considers tends to n(1-o(1)).
Claim: #=n-o(n)
First, note that for any function $f:\{0, 1\}^n \rightarrow \mathbb{R}$, $\frac{1}{2^n} \sum_{w \in \{0,1\}^n} f(w)$ is exactly the same as $\mathbb{E}_{w \in \{0,1\}^n} f(w)$. So you're asking whether $\mathbb{E}_{w \in \{0,1\}^n} H(w) = 1-o(1)$.
To see this, define $X$ to be a random variable uniformly distributed on $\{0,1\}^n$ Let $X_0$ be the number of 0's in $X$ and $X_1$ be the number of 1's in $X$. We want to prove $$\mathbb{E}[(X_0/n)\log(n/X_0)+(X_1/n)\log(n/X_1)]=1-o(1).$$ But we know from the law of large numbers that both $X_0/n$ and $X_1/n$ converge in probability to $1/2$ and that if a random variable $Y$ converges in probability to $c$ and if $f$ is a continuous function, then $f(Y)$ converges in probability to $f(c)$. So we get that $\mathbb{E}[(X_0/n)\log(n/X_0)+(X_1/n)\log(n/X_1)]$ converges in probability to $\frac12 \log(2) + \frac12 \log(2) = 1$, QED.
Note: This claim is related to the Asymptotic equipartition property for discrete-time i.i.d. sources here.
• I'm sorry, but I can't see how $\mathbb{E}_{w\in\left\{0,1\right\}^{n}}\left[H\left(\frac{|w|_{0}}{n},\frac{|w|_{1}}{n}\right)\right]=1-o(1)$ is a direct result of the AEP, since we are looking for the expectation of entropies for all empirical distributions and the AEP treats with the limit of the selfinformations of strings for some fixed distribution. Can you specify that implication? – Danny Apr 8 '14 at 15:06
• Also I do not see why your statement $\mathbb{E}[(X_0/n)\log(n/X_0)+(X_1/n)\log(n/X_1)]=1-o(1)$ leads directly to my conjecture.. Thanks for your help. – Danny Apr 8 '14 at 15:18
• So you're asking whether $\mathbb{E}_{w\in\{0,1\}^n}H(w) = 1 - o(1)$. Note this isn't quite so, the question as worded is whether it is $1 - o(1/n)$. – usul Apr 9 '14 at 1:26
• @Danny, the expression $\mathbb{E}[(X_0/n)\log (n/X_0) + (X_1/n) \log (n/X_1)]$ is exactly the same as your expression $H_0(w)$ (or, in my notation, $H_0(X)$). – mobius dumpling Apr 9 '14 at 11:39
• @Danny You're right, the theorem does not follow from the AEP as easily as I thought. Rather than explain how it follows, I'll just leave my self-contained explanation, which turns out to be simpler than relating to the AEP. – mobius dumpling Apr 9 '14 at 11:44
I will prove that for all $n$, $H_n = n - \epsilon_n$ where $0 < \epsilon_n < 1$.
Let $f_{\alpha}(t) = 4^{\alpha} t^{\alpha}(1 - t)^{\alpha}$. I will not prove this, but you can find some $\alpha^*$ such that for all $\alpha \ge \alpha^*$, $f_{\alpha}(t) \le H(t)$. Also, for $\delta \ll 1$, you can find $\alpha'$ such that for all $\alpha \le \alpha'$, $f_{\alpha'}(t) \ge H(t)$ in $[\delta, 1 - \delta]$. I will assume $\alpha = 1$ and $\alpha = 0.5$ lead to the two different types of functions.
My point is that you can use simpler functions to obtain the bound you need. Here I will use $f_{\alpha}(t)$.
Let $H_n$ be the empirical entropy that you are looking for and $F(n, \alpha)$ be the value obtained when I replace $H(\frac{k}{n})$ with $f_{\alpha}(\frac{k}{n})$. We have
$F(n, \alpha) = n2^n \sum_{k = 0}^{n} {n \choose k} f_{\alpha}(\frac{k}{n}) = n 2^n 4^{\alpha} \sum_{k = 0}^{n} {n \choose k} \left( \frac{k}{n} \right) ^ {\alpha} \left(1 - \frac{k}{n} \right) ^\alpha = n 4^{\alpha} E\left[\left( \frac{k}{n} - \frac{k^2}{n^2} \right)^{\alpha} \right] = n2^n 4^{\alpha} \sum_{k = 0}^{n / 2} {n \choose k} \left(\frac{k}{n} \right)^{2 \alpha}$
For $\alpha = 1$, we know $F(n, \alpha) \le H_n$ and direct evaluation says that $F(n, 1) = n - 1$, which is true for all $n$.
For $\alpha = 0.5$, we know $F(n, \alpha) \ge H_n$ and a direct evaluation leads to $F(n, 0.5) = n 4^{0.5} E[\frac{k}{n}] = n$.
Therefore, for all $n > 0$, $n\ge H_n \ge n - 1$.
I found it nontrivial to try this for any $\alpha < 1.0$ other than $\alpha = 0.5$. | 2019-12-11T16:02:17 | {
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https://www.physicsforums.com/threads/coming-up-with-a-formula-for.42921/ | Coming up with a formula for
1. Sep 12, 2004
Caldus
Um, how can I figure out a formula for solving something like:
1 + 3 + 5 + 7 + 9 + ... + n
In particular: 1 + 3 + 5 + ... + 999
Thanks.
2. Sep 12, 2004
geometer
As far as I know, there is no rote procedure for doing this. It's pretty much a trial and error process. Make yourself a table and figure out the pattern. In this case, we can write this as the sum from x = 0 to x = (n-1)/2 of 2x+1:
x = 0 ===> Sum = 1
x = 1 ===> Sum = 4
x = 2 ===> Sum = 9
x = 3 ===> Sum = 16
We can keep this up if we need to to help us see the pattern, but in this case, a little thought shows that the formula is: Sum from x= 0 to x = n = (n+1)^2
3. Sep 12, 2004
Integral
Staff Emeritus
I believe that Gauss did this as school child when given the assignment to sum the digits 1-100.
First note that if you add them together smallest to largest you get a constant result ( 100+ 1=101, 99+2=101 ..) multiply this by 100 because there are 100 sums and divide by 2 because you added all the numbers twice.
edit:
OPPS! just noticed that you were not summing ALL integers.
Last edited: Sep 12, 2004
4. Sep 12, 2004
Integral
Staff Emeritus
After a bit more playing,
consider this
$$S = \Sigma _ {i=1}^N (2i -1)$$
now using the properties of the summation
$$S= 2\Sigma_{i=1}^N i - \Sigma_{i=1}^N 1$$
$$S= 2\Sigma_{i=1}^N i -N$$
Now using Gauss's trick we can see that the sum of n integers is given by
$$S_n = \frac {(n+1)(n)} 2$$
So we can write for the sum of the odd integers
$$S = (N+1)(N)-N = N^2$$
Note that I have given the general expression for the odd integers as 2n-1 and my sum starts with n=1 where geometer used 2n+1 and started with n=0. This is the reason he has N+1 where I have N. Also note that this N is NOT the odd integer you are summing to but the NUMBER OF TERMS in the sum.
Last edited: Sep 12, 2004
5. Sep 12, 2004
shmoe
Integral, the same idea of pairing highest and lowest will work just fine,
1+3+5+7+9+11=12+12+12=(6/2)*12, where 6 is the number of terms here.
If you already have a formula for the sum of the first n integers, 1+2+3+..+n, you can use this as well:
1+3+5+7+9+11=(1+2+..+11)-(2+4+6+8+10)=(1+2+..+11)-2(1+2+3+4+5)=(insert formula here)-2(insert formula here)
Caldus, do you have a formula for the sum of the first n integers? If not, see Integrals post;)
Another idea, (which Integral provided while I was typing) write your sum as: $$\displaystyle\sum\limits_{i=1}^{n}(2i-1)$$ and fiddle with it to get an expression involving $$\displaystyle\sum\limits_{i=1}^{n}i$$, and apply the formula for this (again, using the ideas in Integral's post if need be). This would be a more easily applied general idea that can handle any arithmetic series $$\displaystyle\sum\limits_{i=0}^{n}(ki+d)$$.
Last edited: Sep 12, 2004
6. Sep 13, 2004
HallsofIvy
Staff Emeritus
Since you are adding only odd numbers, one way to do is this (Assuming you know that the sum of the first n integers is n(n+1)/2):
The sum of all numbers from 1 to 2n is (2n)(2n+1)/2= n(2n+1). Now subtract off the sum of all even numbers from 2 to 2n: 2+ 4+ 6+ ...+ 2n= 2(1+ 2+ 3+ ... + n)=
2(n)(n+1)/2= n(n+1).
The sum of all odd numbers from 1 to 2n-1 is n(2n+1)- n(n+1)= n(2n+1- n- 1)= n2 just as shmoe said.
In particular, the sum of all odd numbers from 1 to 99 is: (99= 100- 1= 2(50)- 1)
502= 2500.
7. Sep 13, 2004
Gokul43201
Staff Emeritus
Also, here's a general rule that can be used in times of trouble :
If a series S, has terms t(n), which are polynomials in n, of order p, the sum S(n) is a polynomial of order p+1. The coefficients of this polynomial can be determined simultaneously, from the given p+2 terms in the series.
Example : If the series S, has terms, t(n) which are linear in n, ie : t(n) = an + b, then the sum, S(n) is a quadratic in n, ie: S(n) = An^2 + Bn + C
NOTE : The order of the polynomial can be determined by the method of "successive differences". If the p'th successive differences are equal, the order is p.
Example : Consider the sequence : 2, 6, 12, 20, 30, ...
The first differences are : 4, 6, 8, 10, ...
The second differences are : 2, 2, 2, ...
Since the second differences are equal, t(n) is a quadratic in n, ie : t(n) = an^2 + bn + c.
So you can assume the sum has a cubic form : S(n) = An^3 + Bn^2 + Cn + D
Plugging in the for S(1), S(2), S(3) and S(4) gives you 4 simultaneous equations in A, B, C, and D, which you can solve. | 2017-04-26T12:08:01 | {
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https://cs.stackexchange.com/questions/136555/how-can-we-find-the-number-of-pairs-of-intersecting-ranges-on-a-circular-number | # How can we find the number of pairs of intersecting ranges on a circular number line?
I recently thought of and managed to solve this algorithmic problem:
On a infinite 1-dimensional number line, we have N ranges specified by two distinct integers A and B, such that all A and B are unique. Each pair of integers then represents the starting and ending points of a range on the number line. How many distinct pairs of ranges intersect?
^^ In this example, 4 distinct pairs of segments intersect.
Obviously, there is a O(N^2) brute-force solution: just test all possible pairs in O(1) time each. However, I also thought of a O(NlogN) sweep-line algorithm: treat every A as a "segment start event" and every B as a "segment end event". Then, keep a variable curr_in_range (originally 0) that stores the current number of segments that we currently see, and then iterate through the events in sorted order of their positions on the number line. Every time we add another segment, add curr_in_range to the answer (because this new segment will intersect with every other segment that we can see) and increment it as well. Every time we remove a segment, decrement curr_in_range. In the end, this gets us the correct answer.
Happy with this algorithm, I decided to tackle a slightly modified version of this problem: What if the number line is circular? As in, it has a specified length L such that the number line contains numbers in the range [0, L)? In this case, each range would still be specified by two distinct integers A and B, but let's just say the range is always counterclockwise starting from A and ending at B. Could we still find the number of pairs of sectors that intersect in less than O(N^2) time in this scenario?
Here's my thought process on this problem:
We should be able to use a very similar algorithm to the one described above. Completely copying it wouldn't work, though, because the sectors can have B < A when it crosses over 0. My idea was to initially have curr_in_range equal to the number of sectors that cross over 0 (instead of just it at 0 like we did in the above algorithm). Then, we could treat every sector as a "sector start event" and a "sector end event", and do the same process that we did.
However, after extensive analysis, I found why this approach doesn't work. This approach won't work because it overcounts some of the intersections, as sometimes two sectors can intersect in 2 places. This is where I'm stuck.
^^ Our sweep-line algorithm will count 2 distinct pairs that intersect, when there is actually only 1.
But besides this overcounting, I found that this current algorithm isn't doing a single other thing wrong. I coded a O(N^2) brute-force algorithm and tabulated all the distinct pairs that it found, and also tabulated the distinct pairs found for the current algorithm. For every randomly-generated test case, these two tables were the same. In other words, our current algorithm is correctly detecting every intersection there is to be detected, but it just happens to overcount some of them.
So, sorry for a really long post, but my final question is:
Can someone come up with an algorithm that solves this problem? Is my thought process close to a solution or completely off?
• Simple idea that might work: sort intervals as usual, except give each one a unique id. Begin processing using the same algorithm, except every start interval is marked as started, and if you encounter an end interval that hasn't been marked, you just ignore it. The algorithm terminates when all intervals are marked as started. – hLk Mar 27 at 4:22
This is solvable in $$O(n \log n)$$ time, too, by considering the complement of the intervals that wrap around and using a suitable data structure.
There are two types of intervals that can appear on a cycle:
• Intervals that do not wrap around: e.g., $$[0.3,0.5]$$ does not wrap around. In general, this includes all intervals of the form $$[a,b]$$ with $$0\le a \le b \le 1$$.
• Intervals that wrap around: e.g., $$[0.8,0.1]$$ wraps around; it is equivalent to $$[0.8,1.1]$$ or $$[0.8,1] \cup [0,0.1]$$. In general, this includes all intervals of the form $$[b,a]$$ with $$0\le a < b \le 1$$; each such interval is equivalent to $$[b,1] \cup [0,a]$$.
Let $$A$$ denote the set of intervals that do not wrap around, and $$B$$ the set of intervals that do wrap around. Also, let $$\overline{B}$$ denote the complements of the intervals in $$B$$, i.e., $$\overline{B} = \{\overline{b} \mid b \in B\}$$. For instance, the interval $$[0.8,0.1]$$ wraps around; its complement is $$(0.1,0.8)$$, which does not wrap around. It is easier to think about intervals that do not wrap around, and this decomposition will help us to deal only with intervals that do not wrap around.
Now, we will separately count the number of overlaps between (1) an interval in $$A$$ and an interval in $$A$$; (2) an interval in $$A$$ and an interval in $$B$$; (3) an interval in $$B$$ and an interval in $$A$$; and (4) an interval in $$B$$ and an interval in $$B$$. (1) can be counted with your sweepline algorithm, as we don't need to deal with wraparound. (4) can also be counted in the same way, as two intervals overlap iff their complements overlap. I will show below how to count (2); case (3) will then follow by symmetry. This covers all the cases, so it suffices to count up these four cases and sum them up.
Note that intervals $$a \in A, b\in B$$ overlap iff $$a$$ is not a subset of $$\overline{b}$$. So, it suffices to count the number of pairs $$a \in A, \overline{b}\in \overline{B}$$ where $$a$$ is a subset of $$\overline{b}$$; then you can subtract that from $$|A| \cdot |B|$$.
To do that, you can use the data structure in Data structure for interval subset queries. Store all the intervals $$\overline{B}$$ in the data structure. Then, for each $$a \in A$$, count the number of intervals in $$\overline{B}$$ that $$a$$ is a subset of. This can be done in $$O(\log n)$$ time per query interval $$a$$, for a total time of $$O(n \log n)$$.
Therefore, you can count each of the four cases in $$O(n \log n)$$ time, and thus compute the overall sum in $$O(n \log n)$$ time as well. I think. I encourage you to check my reasoning to make sure I haven't made any mistakes.
• "(4) can also be counted in the same way, as two intervals overlap iff their complements overlap." A range overlaps any wrapping range that contains the whole circle, but their complements do not overlap. Anyway, in case (4) where all ranges are wrapping, we can just say any two of them overlap. – John L. Mar 25 at 18:49
• Apparently, I could not understand how your usage of "a segment tree to obtain that count" avoids counting the same intersecting pair more than once. – John L. Mar 25 at 18:54
• @JohnL., By "wrapping" I mean a range like $[0.9,0.1]$, i.e., $[0.9,1.1]$, i.e., $[0.9,1.0] \cup [0,0.1]$. Such a range does not cover the whole circle, but includes $1$ and $0$ in the range. The only range that covers the entire circle is $[0,1]$, and that range is non-wrapping. – D.W. Mar 25 at 18:58
• Are you talking about 2-dimensional segment trees? What are the leaves of your segment tree? If each leaf is an elementary interval, how can each interval of $\overline B$ appear in the segment tree only once? If a leaf can be an interval of $\overline B$, then how can an interval of $A$ be expressed as concatenation of those leaves? – John L. Mar 25 at 19:18
• @JohnL., No, an ordinary segment tree for storing the intervals of $\overline{B}$. As usual, we take the endpoints of the intervals of $\overline{B}$, sort them, obtain the elementary intervals, and each leaf corresponds to one elementary interval (not an interval of $\overline{B}$). Each interval of $\overline{B}$ is stored in some node of the segment tree, following the rule described at the link you provided. The interval $a$ can be expressed as a disjoint union of elementary intervals, not of intervals of $\overline{B}$. – D.W. Mar 25 at 19:22
Not a solution, just an illustration of what I see as the core difficulty:
The instance on the left has 3 intersections, the one on the right only 2, but there is no way to distinguish the instances just by counting numbers of "begin segment" and "end segment" events within a given interval. (Inside the gap left by the red interval, the event sequence is "begin", "end" in both cases.)
This implies that a sweep-line approach needs to track more information about each active segment -- probably their identities, which suggests to me that no fast algorithm will be possible, though that's just a feeling.
• A sweep algorithm will take $\Omega(n\log(n))$ time anyways since it has to sort the array. Then, we might be able to store the identities in a way it would not take much more space and time, so we would not overshoot that $n\log(n)$. Not proposing a solution, but just saying that even if we have to keep the identities it might be fast enough – nir shahar Mar 12 at 23:33
### Summary
Interesting question.
Please check this program in Java that runs in $$O(n\log n)$$ time, where $$n$$ is the number of given ranges. The program uses sweep-line technique, once for counting intersecting pairs of ranges, and once for counting containing pairs of ranges together with Fenwick tree. Explanations follow.
### Reduce to the hard case
Here are the notations and assumptions for the ease of explanation.
• All numbers and variables are integers.
• The numbered circle of cardinality $$L\ge2$$ is the set of number $$0, 1, \cdots, L-1$$, where $$L$$ is considered to wrap around to $$0$$.
• A range $$[s,e]$$ with respect to $$L$$, where $$0\le s\le e\lt L$$, means
• a non-wrapping range (a.k.a. an ordinary range) if $$s\le e$$, i.e., numbers $$s, s+1,\cdots, e-1, e$$;
• a wrapping range otherwise, i.e,. numbers $$s, s+1, \cdots, L-1, 0, 1, \cdots, e-1, e$$.
Note every range contains at least one number. The non-wrapping range $$[0, L-1]$$ contains all the numbers. So does every wrapping range like $$[s,s-1]$$.
Suppose there are $$n$$ ranges. The question is how to compute the number of all unordered pairs of two intersecting ranges efficiently.
There are three cases for an unordered pair of two ranges.
• Both ranges are wrapping.
This is the easiest case since two wrapping ranges always intersect at $$0$$.
• Both ranges are non-wrapping.
This is the classical case. The total number of such intersecting pairs can be computed by the sweep-line (or, as I would call it, the sweep-point or sweep-event) algorithm, which runs in $$O(n)$$ time, as described in the question.
• One range is non-wrapping and the other is wrapping.
This is the new and hard case. One way to understand why this case may not be easy is j_random_hacker's illustration. How can we deal with this case?
### Transform to the containing pairs of non-wrapping pairs
Denote the non-wrapping ranges and the wrapping ranges in the given ranges by $$\mathcal N$$ and $$\mathcal W$$, respectively. Instead of counting the intersecting pairs between $$\mathcal N$$ and $$\mathcal W$$, we will use complementary counting, i.e., we will count the non-intersecting pairs between $$\mathcal N$$ and $$\mathcal W$$.
Given any range $$r$$, let $$\overline r$$ be the numbers among $$0, 1, \cdots, L-1$$ that are not in $$r$$, i.e., the complement of $$r$$ with respect to $$\mathcal C$$. Note that two ranges do NOT intersect iff the complement of either one contains the other.
Let $$\overline{\mathcal W} = \{\overline r\mid r\in\mathcal W\}$$. Then the number of non-intersecting pairs between $$\mathcal N$$ and $$\mathcal W$$ is the number of containing pairs between $$\mathcal N$$ and $$\overline{\mathcal W}$$. Here a containing pair means, naturally, two sets $$a$$ and $$c$$ such that $$c$$ contains $$a$$. The number we wanted, i.e., the number of all intersecting pairs of ranges between $$\mathcal N$$ and $$\mathcal W$$ is $$|\mathcal N|\cdot|\mathcal W|$$ minus that number of containing pairs.
More formally, we have $$\#\{(a, b) \mid a\in{\mathcal N}, b\in{\mathcal W}, a\cap b\not=\emptyset\} = \#{\mathcal N}\cdot\#{\mathcal W} - \#\{(a, c) \mid a\in{\mathcal N}, c\in\overline{\mathcal W}, a\subseteq c\}.$$
Here is a simple observation, the complement of a wrapping range is a non-wrapping range, except when the wrapping range is like $$[s,s-1]$$, whose complement is empty. Containing all numbers, a wrapping range like that intersects with every other range. It is easy to count all intersecting pairs that involve a wrapping range like that.
Let us assume there is no such wrapping range; if necessary, we can replace such a range with non-wrapping range $$[0, L-1]$$. The problem is reduced to counting the number of containing pairs between two sets of non-wrapping ranges.
### An algorithm that counts the number of containing pairs
Here is an algorithm that counts the number of containing pairs between two given sets of non-wrapping ranges, $$\mathcal N$$ and $$\overline{\mathcal W}$$. It is, basically, sweep-line technique together with Fenwick tree.
1. Let containingPairCount = 0.
2. Initialize a Fenwick tree (a.k.a. binary indexed tree) ft for an (implicit) interested array of length $$L$$, with all values 0 initially. The e-th element of the interested array will be the number of all ranges in $$\overline{\mathcal W}$$ so far that end at e.
3. Sort all ranges that are in either $$\overline{\mathcal W}$$ or $$\mathcal N$$ by the starting points, breaking ties by putting ranges in $$\overline{\mathcal W}$$ before ranges in $$\mathcal N$$. Breaking the remaining ties arbitrarily.
4. One by one process the sorted ranges. Let the current range be [s,e].
• If it comes from $$\overline{\mathcal W}$$, updates the Fenwick tree as if the e-th element of the interested array is increased by 1. This operation registers the current range as a candidate containing range that has started and will end at e.
• If it comes from $$\mathcal N$$, add the sum of elements of indexes no smaller than e in the interested array to containingPairCount, by containingPairCount += st.getSum(e, L-1).
5. Return containingPairCount.
Proof of the algorithm: st.getSum(e, L-1) is the number of all ranges in $$\overline{\mathcal W}$$ that contain the current non-wrapping range, since, thanks to the careful setup, we have processed all ranges in $$\overline{\mathcal W}$$ that start at or before the start of the current range, and only those ranges that end at or after the end of the current edge are counted. $$\quad\checkmark$$
The sorting of all ranges at step 3 takes $$O(n\log n)$$ time. At step 4, each range is processed with $$O(\log n)$$ time, thanks to the power of Fenwick tree. So the running time of this algorithm is $$O(n\log n) + O(n \log n) = O(n\log n)$$.
### Complexity analysis
The running time for other parts of the entire algorithm is $$O(n\log n)$$, which is spent mostly on counting the intersecting non-wrapping pairs. So, the running time of the entire algorithm is $$O(n\log n) + O(n \log n) = O(n\log n)$$.
The sections "Reduce to the hard case" and "transform to the containing pairs of non-wrapping pairs" follow D.W's answer roughly. | 2021-05-07T16:27:09 | {
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https://math.stackexchange.com/questions/211341/representation-of-a-matrix-transformation | # Representation of a matrix transformation
An example asks me to define $T: M_{2\times2}(F) \to M_{2\times2}$ by $T(A) = A^t$. Compute $[T]_\alpha$
$\alpha$ is the standard ordered basis of $2\times 2$ matrices.
To find the transformation I performed the transformation on all four elements of $\alpha$, for example $T(\begin{pmatrix} 1&0\\0&0\\\end{pmatrix}) = (\begin{pmatrix} 1&0\\0&0\\\end{pmatrix})$ and so on. Basically the matrices are all the same before and after the transformation. Accordingly, the textbook's answer is that you arrange all the results $A^t$ into a $4\times4$ matrix:
$$\begin{pmatrix} 1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}$$
I don't understand this what means, and I've been trying to. How can this be the transformation $T$ on $\alpha$ if we're not even able to multiply the basis by it? What I mean is, we can't multiply this transformation matrix by anything in $\alpha$ since one is a $4\times4$ and one is a $2\times 2$.
Also, why doesn't this mean that any arbitrary $2\times 2$ matrix can be raised to the power of $t$ by simply multiplying by this matrix? (As per the previous paragraph, it can't, but it should).
• I changed $T: M_{2\times2}(F) -> M_{2\times2}$ to $T: M_{2\times2}(F) \to M_{2\times2}$. $\TeX$ is not so crude that you have to resort to things like that. – Michael Hardy Oct 11 '12 at 23:43
• Thanks - I couldn't find the command for it. I'll know for next time. – CodyBugstein Oct 11 '12 at 23:57
You need to distinguish here between the space $\mathbb F^2$ that the $2\times2$ matrices operate on and the space of the $2\times2$ matrices themselves. The former is two-dimensional, the latter is four-dimensional.
A matrix is a representation of a linear transformation with respect to bases of the domain and target of the transformation. If the linear transformation is an endomorphism, a linear map from a space to itself, the corresponding matrix is square, and one typically uses the same basis for the space in its two roles as domain and target.
The standard basis of the space of $2\times2$ matrices that you're using has four basis elements and spans the four-dimensional space of $2\times2$ matrices. A linear transformation has been defined on that space – the four-dimensional space of $2\times2$ matrices, not the two-dimensional space that they operate on.
Representing that linear transformation as a matrix with respect to the standard basis (both for the domain and the target) yields a $4\times4$ matrix. That's not a mismatch because the vectors and matrices that will be multiplied by this matrix aren't elements of $\mathbb F^2$; they're the vectors of coefficients in an expansion of the matrix they represent in the standard basis.
As an example, the matrix
$$\pmatrix{2&3\\4&5}$$
has coefficients $2$, $3$, $4$ and $5$, respectively, in the standard basis, so with respect to that basis it's represented by the column vector
$$\pmatrix{2\\3\\4\\5}\;.$$
Its transpose
$$\pmatrix{2&4\\3&5}$$
is represented by the column vector
$$\pmatrix{2\\4\\3\\5}\;.$$
And sure enough we have
$$\begin{pmatrix} 1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}\pmatrix{2\\3\\4\\5}=\pmatrix{2\\4\\3\\5}\;.$$
Thus this matrix really does represent the linear transformation $T$, transposition, in this basis.
• So matrices can unravel to form vectors? I've played around with some and it makes sense. Thanks for the amazing answer. I still have a question though; shouldn't the transforming vector actually put your sample vector to the power of something? Isn't that the whole idea? – CodyBugstein Oct 11 '12 at 23:32
• @Imray: I think you've misunderstood the notation. $A^t$ is merely a notational device to denote the transpose of $A$; nothing is being raised to a power. I prefer the slightly different notation $A^\top$ (A^\top) to avoid the impression that the $t$ is a variable used as an exponent. – joriki Oct 11 '12 at 23:34
• @Imray: As an exercise, I suggest to write down the matrix of $T$ with respect to the basis $$\left\{\pmatrix{1&0\\0&1},\pmatrix{1&0\\0&-1},\pmatrix{0&1\\1&0},\pmatrix{0&1\\-1&0}\right\}\;,$$ and then perhaps use it to apply $T$ to a couple of matrices and see whether the transpose comes out right. – joriki Oct 11 '12 at 23:38
• yes right! I didn't realize that! Ok I've got it now, thank you very much for explaining it to me. – CodyBugstein Oct 11 '12 at 23:38
• @Imray: You're welcome! – joriki Oct 11 '12 at 23:40
Just to make sure our definition of matrix representation is the same: Let $\alpha = (a_1, \ldots, a_k)$ be a basis for $V$, and let $T: V \rightarrow V$ be a linear map. Then matrix representation $[T]_{\alpha}$ is defined as follows: $$[T]_{\alpha} = \begin{bmatrix} [T(a_1)]_{\alpha} & \ldots & [T(a_k)]_{\alpha} \end{bmatrix}$$
The matrix representation obeys the following formula: for every $v \in V$, $[T(v)]_{\alpha} = [T]_{\alpha} [v]_{\alpha}$, where $[v]_{\alpha}$ is the coordinate vector of $v$ with respect to the basis $\alpha$.
So now, to answer your question, it's true that $[T]_{\alpha}$ is a $4 \times 4$ matrix in your example, but you don't multiply it onto the $2 \times 2$ matrices; you multiply it onto the $2 \times 2$ matrices after they have been converted into column vectors via the coordinate mapping. Since $2 \times 2$ matrices form a $4$-dimensional space, then this coordinate mapping turns a $2 \times 2$ matrix into an element of $\mathbb{R}^4$.
• How does coordinate mapping work? The rows all extend downward to become one long column? Can all matrices with $n$ elements be multiplied by any matrix with $n$ columns? – CodyBugstein Oct 11 '12 at 23:35
• What confused me is how $[T]_\alpha$ could be square. But one needs to remember that the columns are in vector notation too (not in matrix notation!), so they are actually not $k \times k$ but $k^2 \times 1$ and so in the end $[T]_\alpha$ is $k^2 \times k^2$. – philmcole Apr 10 '18 at 21:20 | 2019-07-16T20:20:46 | {
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https://math.stackexchange.com/questions/2534449/show-that-a-n-left-sum-k-1n-frac1k-right-logn-is-decreasing | # Show that $a_n=\left(\sum_{k=1}^n \frac{1}{k}\right)-\log(n)$ is decreasing
How I decided to show it is by Induction, and I don't really like what I have, so could you guys please tell me if my proof valid. Thanks.
Aim of the proof: Show that $$a_n>a_{n+1}$$ for all $n\geq1$.
Base case: $$a_1-a_2=\log(2)-\frac{1}{2}>0.$$ Hence base case holds. Assume, for some $k>1$, that the result holds, that is:$$a_k>a_{k+1}$$ Hence show that $$a_{k+1}>a_{k+2}.$$ Now we can rewrite $a_{k+1}>a_{k+2}$ as, $$\left(\sum_{k=1}^{k+1}\frac{1}{k+1}\right)-\log(k+1)>\left(\sum_{k=1}^{k+1} \frac{1}{k+1}\right)+\frac{1}{k+2}-\log(k+2),$$ Which is equivealnt of saying: $$\log\left(\frac{k+2}{k+1}\right)>\frac{1}{k+2}$$
For our problem $k\geq1$ and $\log(\frac{3}{2})>\frac{1}{3}.$
My last step is saying that as $n$ tends infinity, the limit of LHS of our last inequality is infinite and RHS is 0. Hence the result holds.
What I don't like is the fact that I don't use the inductive hypothesis at all. (Sorry for selling and grammar errors, English is not my native language)
• This does not answer your question, but do you know that $\lim_n a_n =\gamma$, the Euler-Mascheroni constant? You might want to google it, you will for sure find something about the non decreasing behaviour of $a_n$. – Paolo Intuito Nov 23 '17 at 21:53
• A fundamental step you make in your proof is incorrect: $\lim_{k \rightarrow \infty} (\frac{k+2}{k+1}) = 0$, not infinity, as you claim. (In fact, if this step were correct, then this alone would be a proof - the whole induction set up would be unnecessary.) – John Don Nov 23 '17 at 21:59
• Oh crap, yeah, the fraction gets closer to 1 eqch time. Thank you for noticing this John. How would I go about proving it then?, and yes, G.S. I know about that fact but couldnt find anything on the question I am trying to solve – Scavenger23 Nov 23 '17 at 22:02
You wish to show the inequality,
$$\frac{1}{n+1}-\ln(n+1)<-\ln n$$ or equivalently, $$\frac{1}{n+1}<\ln (1+\frac{1}{n})$$
Now we have $$x-\frac{x^2}{2}<\ln(1+x)$$ so just check that $$\frac{1}{n+1}<\frac{1}{n}-\frac{1}{2n^2}$$
Which is indeed the case as it is not hard to show.
• How do you know that your second to last line is true? I feel like I am missing something basic. – Scavenger23 Nov 23 '17 at 22:26
• Either, you note that the next term in the power series is positive, or you could use the mean value theorem. – Rene Schipperus Nov 23 '17 at 22:27
• Or more straightforward, move all the terms to the right, and differentiate, and show the derivative is positive, and thus the function is increasing. – Rene Schipperus Nov 23 '17 at 22:33
• I see now, didnt realise that you tailor expanded the log. Thanks for the help. – Scavenger23 Nov 23 '17 at 22:35
I like Rene Schipperus' method a lot - it is very simple. Here is an commonly used alternative:
Note that, the function $\frac1x$ is strictly decreasing on $[1, \infty)$, so, for all $n \in \mathbb{N}^{\, \ge 1}$,
\begin{align} \ln (n+1) - \ln n &= \int_{n}^{n+1} \frac1x \, dx\\ &> \int_n^{n+1} \frac1{n+1} \, dx\\ &= \frac1{n+1} \end{align}
As you have already noted, the solution follows.
• very good work. So essentially the method I wanted to use was correct, ? The question in the book never told me how to prove it. Induction seemed the way forward for me. – Scavenger23 Nov 23 '17 at 22:43
• @KuderaSebastian That seems reasonable... I suppose that the thing to notice was that, at some point, (e.g. in your inductive step) you would have to relate $\log(n)$ to some function of integers somehow (as this is what you have on the LHS). One way is to use the power series expansion (as Rene has done), and the other is to notice that the derivative of $\log(x)$ is in fact $\frac1x$ (which both I and Peter have used in our answers). – John Don Nov 23 '17 at 22:53
• Jon.Nice answer, short and clear. – Peter Szilas Nov 24 '17 at 9:06
Show that $a_n -a_{n+1} >0.$
$\Delta_n: = a_n - a_{n+1} =$
$-\dfrac{1}{n+1} + \log(n+1) -\log(n).$
Mean Value Theorem:
$f(x) := \log(x)$ then:
$\dfrac{f(x+1)-f(x)}{1} = f'(t),$ $x \lt t \lt (1+x).$
$\dfrac{\log(n+1) -\log(n)}{1} = \dfrac{1}{t}$, $n\lt t\lt (n+1).$
$\Delta_n = -\dfrac{1}{n+1} + \dfrac{1}{t}$ , where $t \lt (n+1).$
Hence $\Delta_n >0.$
• NIce... my answer is essentially this, but integrated, in a sense. – John Don Nov 23 '17 at 22:49 | 2020-09-27T15:47:45 | {
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https://www.tutorialspoint.com/divmod-in-python-and-its-application | divmod() in Python and its application
PythonServer Side ProgrammingProgramming
The divmod() is part of python’s standard library which takes two numbers as parameters and gives the quotient and remainder of their division as a tuple. It is useful in many mathematical applications like checking for divisibility of numbers and establishing if a number is prime or not.
Syntax
Syntax: divmod(a, b)
a and b : b divides a
a and b are integers or floats
Examples
In the below example see the cases of both integers and floats. On the application of divmod() they give us a resulting tuple which is can also contain integers and float values.
# with integers
print("5 and 2 give:",divmod(5,2))
print("25 and 5 give:",divmod(25,5))
# with Floats
print("5.6 and 2 give:",divmod(5.6,2))
print("11.3 and 9.2 give:",divmod(11.3,9.2))
Output
Running the above code gives us the following result −
5 and 2 give: (2, 1)
25 and 5 give: (5, 0)
5.6 and 2 give: (2.0, 1.5999999999999996)
11.3 and 9.2 give: (1.0, 2.1000000000000014)
Using Zero
If the first argument is zero then we get (0,0). And If the second argument is zero then we get Zerodivision error as expected.
Example
# With first argument as zero
print("0 and 8 give:",divmod(0,8))
# With second argument as zero
print("8 and 0 give:",divmod(8,0))
Output
Running the above code gives us the following result −
0 and 8 give: (0, 0)
Traceback (most recent call last):
File "xxx.py", line 6, in
print("8 and 0 give:",divmod(8,0))
ZeroDivisionError: integer division or modulo by zero
Checking divisibility
If the second value of the tuple after division is 0 then we say that the first number is divisible by second. Else it is not divisible. The below example illustrates this.
Example
m = 12
n = 4
quotient,remainder = divmod(m,n)
print(quotient)
print(remainder)
if (remainder==0):
print(m,' is divisible by ',n)
else:
print(m,' is not divisible by ',n)
Output
Running the above code gives us the following result −
3
0
12 is divisible by 4
Checking if Number is Prime
We can use divmod() to keep track of the reminders it generates when we start dividing a number by each number starting with itself till 1. For a prime number the count of zero remainder will be only one as no number other than itself will divide it perfectly. If the count of zero remainder is greater than 1 then the number is not prime,.
Example
num = 11
a = num
# counter the number of remainders with value zero
count = 0
while a != 0:
q, r = divmod(num, a)
a -= 1
if r == 0:
count += 1
if count > 2:
print(num, 'is not Prime')
else:
print(num, 'is Prime')
Output
Running the above code gives us the following result −
11 is Prime
Published on 07-Aug-2019 08:32:04 | 2021-12-01T10:41:09 | {
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https://matheducators.stackexchange.com/questions/18120/what-is-the-best-way-to-intuitively-explain-the-relationship-between-the-derivat/18128 | # What is the best way to intuitively explain the relationship between the derivative and the integral?
This is my first post so bear with me, but something I've been thinking about lately is: Why didn't I ever question the relationship between the derivative and the integral when I was taking calculus?
Let me explain what I mean: In most courses, the derivative is introduced as the slope of a curve at a point, or the "instantaneous rate of change". Then the integral is introduced as the area under a curve. Then students are told that these two things "undo" each other (the fundamental theorem of calculus). So now I'm wondering, why didn't I ever question why these "undo" each other? It's not intuitive at all. For example, addition and subtraction, multiplication and division, logarithms and exponentials; all of these things I can intuitively understand why they "undo" each other. But how does something that represents the area under a curve "undo" something that represents an instantaneous rate of change? What is the connection between those two concepts? Is there a better way to explain these concepts that makes the fundamental theorem of calculus more intuitive?
Looking forward to hearing some of y'alls responses. Thanks!
• $[\cdots]$ why didn't I ever question why these "undo" each other? --- Your calculus experience is probably in the minority, because it is fairly standard (at least in U.S. introductory calculus courses, both high school and college) to make a big deal about the connection, usually at the end of the first semester, this being especially the case since the rise of "reform calculus" from the late 1980s on. That said, see right brain explanation and left brain explanation. – Dave L Renfro Apr 2 '20 at 5:26
• Which resources have you perused before registering and asking this question? Voting to close. – Rusty Core Apr 2 '20 at 21:01
• In countries, where physics is a mandatory course in grade school, the traditional layman explanation is through kinematics. Average speed v, by definition is displacement s over time t. Start with the simplest case of uniform motion, draw a graph in (t, s), it would be a line climbing up. Tangent at any point is at the same angle (similarity of right triangles), this is the derivative by definition, which is speed. Draw speed in (t, v). Area below it is displacement. Continue with uniform acceleration and then with no-uniform motion, slicing up the area under speed into thin vertical bands. – Rusty Core Apr 2 '20 at 21:14
• @BenCrowell I think what makes this a math ed question is wanting better intuition to explain the relationship. Also, the fact that many of us didn't really get it until we were teaching speaks volumes to how much this is an education issue. – Sue VanHattum Apr 4 '20 at 20:52
• @SueVanHattum Exactly - that's why I posted it here. The funny thing is that when I learned how to prove the FTC in real analysis, I didn't find either of the proofs too difficult. But the intuition behind the overall concept still wasn't there. Then recently I realized that if a student came to me and asked "Why do these two concepts undo each other?", my best response would be to show them the proof. That's why I figured this was a math ed question, because I'm specifically asking what the best way is to explain this to a student and make it more intuitive. – Brain Gainz Apr 5 '20 at 18:43
You start by noticing that the Riemann sums (multiplication followed by addition) and the difference quotients (subtraction followed by division) undo each other. Their limits -- the integral and the derivative -- still undo each other.
Added: The first sentence is a part of what is called “Discrete Calculus” https://en.wikipedia.org/wiki/Discrete_calculus
• Wow. I'm unsure whether this intuition can be made to correspond to some rigorous proof, but it's just solid enough to make me feel it gave me some insight into the mechanism of the theorem's truth and not mere word-soup hand-waving. And I think the maximal spacing in a partition (that goes to zero for the Riemann integral) even corresponds to the step size in the difference quotient in a suitable sense. – Vandermonde Apr 3 '20 at 1:23
• @Vandermonde You are correct. By the way, this will be in the third volume of my book Calculus Illustrated: amazon.com/dp/B082WKCYHY – Peter Saveliev Apr 3 '20 at 1:44
• Surprising how entirely abstract and formal mathematical thinking can be intuitive :-). Perhaps not for everybody, but even this engineer sees symmetries. – Peter - Reinstate Monica Apr 3 '20 at 10:21
• 1. Can you please exhibit the formula that you're hinting to? 2. Can you please clarify 1. which "Riemann sums" you mean? 3. Which "difference quotients"? 4. Which "limits"? – NNOX Apps Apr 4 '20 at 5:27
• If one considers the differential $\text{d}$ and the integral $\int$ as the fundamental concepts of calculus (as opposed to $\frac{\text{d}(-)}{\text{d}x}$ and $\int(-)\text{d} x$), then the wording is even more simple: difference and sum undo each other. (Actually Leibniz called $\int$ the sum, before Bernoulli suggested integral) – Michael Bächtold Apr 6 '20 at 15:58
(this is from my calculus notes, see page 233 of: http://www.supermath.info/OldschoolCalculusII.pdf)
• So in words: the change in area is equal to the height of the function times the change in x? That actually makes sense. – Brain Gainz Apr 3 '20 at 0:31
• @BrainGainz thanks! Incidentally, I had exactly your feeling after taking years of calculus and an Advanced Calculus course where I suspect we proved the FTC. Your feeling is not unusual in my estimation. – James S. Cook Apr 3 '20 at 2:52
• I understand the pedagogical need, but I cringe when I read (or say myself when teaching calculus) "arbitrary function" and the picture is that of a differentiable function. – Martin Argerami Apr 3 '20 at 10:36
• @BrainGainz yep, I have a few Topology lectures posted. I am that James Cook. – James S. Cook Apr 4 '20 at 21:57
• @StevenGubkin cool, if I ever get to teach Calculus I again, I should use that. So many options to play with... – James S. Cook Apr 6 '20 at 3:49
Derivative of integral is the original function: Let $$F$$ be the integral function of $$f$$, so that $$F(x)$$ is the area under the graph from zero to $$x$$. For small $$h>0$$ the difference $$F(x+h)-F(x)$$ is the area of a narrow vertical strip. The width is $$h$$ and height approximately $$f(x)$$. As $$h\to0$$, this means $$F'(x)=f(x)$$.
If you like thinking in terms of infinitesimals, write $$dF=F(x+dx)-F(x)=f(x)dx$$ and divide by the differential. This can be a cleaner way to think as the limit process is left implicit in a way.
Integral of derivative is the original function: The integral is roughly $$\int_0^x f'(x)dx = \sum_{k=1}^n f'(x_k)(x_k-x_{k-1})$$ and the derivative is roughly $$f'(x_k) = \frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}}.$$ Combine these and you get a telescoping sum and the desired result.
This is the way it makes intuitive and graphical sense to me. The link is not immediate; that's why it's a great theorem.
My answer also comes from physics.
Say p(t) is the position of an object in the time t. For concreteness, suppose you are traking a truck and the truck is going forward on a road from A to B. (i.e., all derivatives are positive)
It is very natural to graph p(t), and derive it, arriving at v(t), the speed of the truck for each time. After all, the derivative is just taking a small time increment $$dt$$, calculating the corresponding $$ds$$ (a space increment) and dividing
However, when I look at the graph of v(t), I can also obtain (almost everything) from p(t). Say I am at the position 100 km in the time t=0. After a $$dt$$, I shall be at $$p(t+dt) = 100+dt\cdot v(t)$$. And I can do it again and again $$p(t+dt+dt+dt) = 100+dt\cdot v(t)+dt\cdot v(t+dt)+dt\cdot v(t+2\cdot dt)+dt\cdot v(t+3\cdot dt)$$ (notice that after some $$dt$$s, I use the updated speed)
But that (taking a limit) is an integral!
The only weird thing is my assuption that I am at position 100km at time t=0. The graph of v(t) cannot tell me that. Intuitively, it does tell me how fast I am going, and therefore can tell me how much how much I walked, but it cannot tell me at which specific part of the road I started. That is that weird constant +c that appears on the indefinite integral.
But then again, we dont use the indefinite integral much, do we? We are more interested in summing $$dt$$*v(t) for ranges. Say, from t=3 to t=10. That gives me the amount I walked between those two times. So, mercifully, the +c disappears in most use cases: to know how much I walked, there is no need to know where I started.
• I love how velocity and position helps to cement the relationship. In one class, my students asked why area below the axis should count as negative. We agreed that negative velocity meant going backwards, and therefore decreased your distance from start. (Distance being measured by area under velocity curve, a negative velocity curve has to have "negative area" to decrease the total distance.) – Sue VanHattum Apr 4 '20 at 20:47
When I first began teaching Calculus, I realized that I really didn't understand the Fundamental Theorem. So I looked for something that would help me have a deep understanding, that would also help me help students to see it.
I found a lovely project, which I have modified over the years. Here are links to the pdf and to a .doc version (in which the formulas are messed up, because google docs couldn't read them). If you use this, you'll want to change the part near the end where I reference the textbook.
• Wow, this is really great. I'll definitely save this. Thanks! – Brain Gainz Apr 3 '20 at 0:29
I gave a similar post to this one touching on this on Math.StackExchange.
Basically, the way I would go about it is to say that there is a very easy way by which one can think of at least Riemann integration (the usual definition given in a "most courses" calculus course) as an inverse of differentiation by construction: that is, the relationship between the two is not an "accident", but design, so that given one, you could fairly easily be led to the other.
And to do that, I'd first suggest getting rid of the whole "slope" vs "area" business altogether - if anything, those are best given as theorems to be proved, after you have other, free-standing definitions of a tangent line and an area which, by the way, can be done, but just too-often aren't, in favor of various hand-waving arguments.
Instead, the relevant idea is change: the derivative of a function $$f$$, i.e. $$f'$$, stands for a kind of "sensitivity" measure. Suppose that $$f$$ were like a kind of meter or instrument, with a knob attached to it, and a readout of some kind. The knob attached is the function's input argument, typically denoted $$x$$. The indicator on the readout is the return value, $$f(x)$$. If $$x$$ is set at some value of interest $$x_0$$, then likewise the readout will be at $$f(x_0)$$. Now suppose you "wiggle" the knob $$x$$ back and forth a little bit, and you see how the needle $$f(x)$$ responds to that small impulse. For a continuous function, the size of the output's wiggle will be smaller in absolute terms the smaller you make the input, but the proportionate size, i.e. how much it wiggles relative to how much you wiggle the input value, may not be. When we say $$f$$ is differentiable, what that means is that there exists at each point $$x_0$$ a proportionality factor $$f'(x_0)$$ such that
$$f(x \pm \underbrace{dx}_\mbox{"wiggle" in x}) \approx f(x_0) \pm \underbrace{[f'(x_0)\ dx]}_\mbox{"wiggle" in f(x)}$$
so that
$$\mbox{proportional "wiggle"} = \frac{\mbox{"wiggle" in f(x)}}{\mbox{"wiggle" in x}}$$
so long as the change $$dx$$ is suitably small
Integration, then, goes the other way. Suppose that I am given now, not the function $$f$$, but only its derivative, $$f'$$, and want to find $$f$$. First off, one should observe that since $$f'$$ only deals with changes in the input, to start, we actually need one more piece of information, and that is some sort of initial value, i.e. $$f(0)$$. Suppose this to be given as well.
Starting at $$f(0)$$, suppose we apply a small, but nonzero, change $$\Delta x$$ to the input, i.e. we ask, "given $$f(0)$$, what is $$f(0 + \Delta x)$$, to the best we can do?" Well, since we know $$f(0)$$ and $$f'$$, then since $$\Delta x$$ is a small change or "wiggle", we can say that approximately,
$$f(0 + \Delta x) \approx f(0) + [f'(0)\ \Delta x]$$
which is just what you see above.
Now, suppose we take another step of $$\Delta x$$. We're now going from $$x = 0 + \Delta x$$ to $$x = (0 + \Delta x) + \Delta x$$ (or $$2\ \Delta x$$, but I find writing it this way makes it clearer what is going on - "simpler" isn't necessarily "better"). At this second step, likewise, treating $$0 + \Delta x$$ as a prior input in and of itself, we have
$$f([0 + \Delta x] + \Delta x) \approx f([0 + \Delta x]) + [f'([0 + \Delta x])\ \Delta x]$$
which, combining with the previous expression, becomes
$$f([0 + \Delta x] + \Delta x) \approx f(0) + [f'(0)\ \Delta x] + [f'([0 + \Delta x])\ \Delta x]$$
and it is not hard to then continue this process so that we see for $$N$$ steps,
$$f(0 + N[\Delta x]) \approx f(0) + \sum_{i=0}^{N-1} f'(0 + i[\Delta x])\ \Delta x$$
or, if we set $$x_i := 0 + i[\Delta x]$$, we can say more neatly as
$$f(0 + N[\Delta x]) \approx f(0) + \sum_{i=0}^{N-1} f'(x_i)\ \Delta x$$
and to be more general, if we take $$N$$ steps of suitable, perhaps different, sizes $$\Delta x_i$$ to reach from $$0$$ some fixed point $$x_0$$,
$$f(x_0) \approx f(0) + \sum_{i=0}^{N-1} f'(x_i)\ \Delta x_i$$
and then we consider what happens as the steps become arbitrarily fine, at which point we hope - and need to prove - that
$$f(x_0) = f(0) + \left[\lim_{||\Delta|| \rightarrow 0}\ \sum_{i=0}^{N-1} f'(x_i)\ \Delta x_i\right]$$
which leads us to define this new operation, given by the limit on the right...
$$\int_{0}^{x_0} f'(x)\ dx := \lim_{||\Delta|| \rightarrow 0}\ \sum_{i=0}^{N-1} f'(x_i)\ \Delta x_i$$
• This does a great job of explaining how to think of $\int_a^b f'(x)\mathrm dx = f(b) - f(a)$, but, in suggesting that we think of it as a definition, I'm left to wonder how to define $\int_a^b F(x)\mathrm dx$ without proving some separate theorems about existence of anti-derivatives. – LSpice Apr 6 '20 at 0:04
Intuitively, the fundamental theorem of calculus states that "the total change is the sum of all the little changes". $$f'(x) dx$$ is a tiny change in the value of $$f$$. We sum up all these little changes to get the total change $$f(b) - f(a)$$.
I elaborated on this explanation here: https://math.stackexchange.com/a/1537836/40119.
• Very nice! The idea that $\mathrm dx$ is not punctuation, but (conceptually) an actual infinitesimal measure of $x$-length, so that $f'(x)\mathrm dx$ has units of (slope)($x$-length) = $y$-length, I think is probably the most important pre-proof part of understanding the connection. That is, I think that most people think of integration as "something you do to $f'(x)$", not "something you do to $f'(x)\mathrm dx$"—hence later difficulties with differential forms, and earlier difficulties with changes of variable. – LSpice Apr 6 '20 at 0:06
Well, for me, the nicest intuition comes from physics. If $$F=F(x)$$ is some force applied to an object $$O$$ casuing $$O$$ to move, where $$x$$ is its displacement, when the work $$W(x)$$ produced by that force from $$x_0=0$$ to $$x$$ metres is given by:
$$W(x):=\int_0^xF(s)ds.$$
Now, what if we ask what is the rate of change of that work with respect to displacement? At some certain point $$x$$, the work provided to or substracted by that object $$O$$ is determined by the force $$F$$ applied on it. So, one can intuitively expect that:
$$W'(x)=\left(\int_0^xF(s)ds\right)'=F'(x).$$
That is, the larger the force, the faster the energy flows from the one applying the force towoards the object. The less the force, the slower that energy flows.
If you want more clearly "undoing" operations, define "stacking" as the following:
Take some $$\Delta x$$. Chop the curve into rectangles of width $$\Delta x$$ and height $$f(x)$$. (There's some leeway as to take $$f(x)$$ at the left side, right side, middle, minimum, maximum, etc. The rest of my description will be right side.) Now take each rectangle and put the bottom of each rectangle at the top of the previous rectangle. For instance, if you have $$f(x) = x^2$$ and $$\Delta x = 0.1$$, you'd have a rectangle with lower left corner at the origin, and top left at $$(0.1, 0.01)$$. Then the next rectangle would have lower left corner at $$(0.1, 0.01)$$, and upper right at $$(0.2, 0.05)$$, and so on.
Define "unstacking" as follows:
Take some $$\Delta x$$. Chop the curve into rectangles of width $$\Delta x$$ and height $$f(x)$$. Now, take each rectangle and shift it vertically down the height of the previous rectangle. For instance, if $$f(x) = x^2$$, the first rectangle will have height $$0.01$$. The second one will have height $$0.04$$, so for the unstacked version, we move it down $$0.01$$, leaving its new height as $$0.03$$. Another way of phrasing it is for each $$x$$, take the rectangle whose bottom left corner is at $$(x, f(x))$$ and top right corner is $$(x+\Delta x, f(x+\Delta x))$$. Then create a chart out of all of those rectangle, with all of them moved down so their bottom is on the x-axis.
For "well-behaved" functions (I believe that uniform continuity is a sufficient condition), as $$\Delta x$$ goes to zero, stacking becomes integration, and unstacking become differentiation (with the proper scaling). | 2021-01-17T10:24:45 | {
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https://math.stackexchange.com/questions/1888691/elementary-matrix-operations | # Elementary matrix operations
I am taking an introductory/first year course in Linear Algebra and I am at my wits' end with the following problem.
I am asked to find both the nullspace and the general solution of the following systems $A\tilde{x}=\tilde{b}$.
The first is $$A= \begin{pmatrix} 1 & 1 & 0 & 2\\ 2 & 1 & 1 &-2\\ 2 & 2 & 2 & 1 \end{pmatrix},\quad \tilde{b}= \begin{pmatrix} 12\\0\\14 \end{pmatrix}$$ Proceeding with an augmented matrix, I have \begin{align*} \left( \begin{array}{rrrr|c} 1 & 1 & 0 & 2 & 12\\ 2 & 1 & 1 &-2 & 0\\ 2 & 2 & 2 & 1 & 14 \end{array} \right)&\sim \left( \begin{array}{rrrr|r} 1 & 1 & 0 & 2 & 12\\ 0 &-1 & 1 &-6 & -24\\ 0 & 0 & 2 & -3 &-10 \end{array} \right)\\ &\sim \left( \begin{array}{rrrr|r} 1 & 1 & 0 & 2 & 12\\ 0 & 1 &-1 & 6 & 24\\ 0 & 0 & 1 & -3/2 &-5 \end{array} \right) \end{align*} I've checked this multiple times and can't find a mistake, plus it agrees with the solutions my professor has provided.
Solving this now, I have \begin{align*} x_4&=\alpha\in\mathbb{R}\\ x_3&=-5+\frac{3}{2}\alpha\\ x_2&=19-\frac{9}{2}\alpha\\ x_1&=-7+\frac{5}{2}\alpha \end{align*}
So from this, my solution will be $$\tilde{x}= \begin{pmatrix} -7+\frac{5}{2}\alpha\\ 19-\frac{9}{2}\alpha\\ -5+\frac{3}{2}\alpha\\ \alpha \end{pmatrix} = \begin{pmatrix} -7\\19\\-5\\0 \end{pmatrix}+\beta \begin{pmatrix} 5\\-9\\3\\2 \end{pmatrix},\quad\beta\in\mathbb{R}.$$
This does not agree with the solutions given, and the solutions given are just solutions without any steps in between so I'm a little confused as to where I went wrong. The weird part is, my nullspace agrees with the solution provided but the particular solution certainly does not.
Edit: My professor's solution is $$\tilde{x}= \begin{pmatrix} 3\\1\\1\\4 \end{pmatrix}+\alpha \begin{pmatrix} 5\\-9\\3\\2 \end{pmatrix},\quad\beta\in\mathbb{R}.$$
Can somebody please let me know what I've done incorrectly?
Thank you.
• I tried using WolframAlpha, but it didn't tell me what I did wrong, and it spat out something that looked different to both my solution and my professor's (although it did provide the same nullspace!). – Babe in the Woods Aug 10 '16 at 23:22
• Oh, he went away.. – Babe in the Woods Aug 10 '16 at 23:22
• Can you provide your professor's solution? – Aweygan Aug 10 '16 at 23:25
• My apologies, my professor's solution has been added in the question. – Babe in the Woods Aug 10 '16 at 23:29
• Your professor simply chose to set $x_4=2\alpha$ instead of $\alpha$ to avoid denominators in the general solution. – Bernard Aug 10 '16 at 23:33
Your solution is correct. You can obtain the professor's solution from yours via the substitution $\beta=\alpha+2$. Both are valid answers, and unless there was some given format your answer was supposed to be in, I don't see why yours could be considered incorrect.
This is correct! You can check if you found solutions of the system simply by replacing $x$ in $Ax=b$ by what you found. | 2019-11-21T00:45:59 | {
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https://math.stackexchange.com/questions/4041409/counting-ways-when-first-position-is-covered-with-a-tie | # Counting ways when first position is covered with a tie
Six horses are entered in a race. If two horses are tied for first place and there are no ties among the other four horses, in how many ways can the six horses cross the finish line?
My approach:
We choose $$2$$ out of $$6$$ for the first place and $$4$$ out of the remaining $$4$$:
$$_6P_2\cdot _4P_4 = \frac{6!}{(6 -2)!}\cdot\frac{4!}{(4 - 4)!}=720$$
But the solution states $$360$$.
What am I doing wrong here?
• Also, $\binom 44= \frac{4!}{0!4!}$. But you don't want $\binom 44$ because you care about the order of the remaining $4$ horses. You want the number of ways to permute those $4$ horses, which is simply $4!$. Feb 26 at 23:31
• @RobertShore: Yes the order matters that is why I used $\binom{n}{k}=\frac{n!}{(n-k)!}$ which is $n\cdot(n-1)\cdots\cdot(n-k+1)$. Also it is still $720$
– Jim
Feb 26 at 23:35
• Your values for the combinations are wrong. ${6 \choose 2}=\frac {6!}{4!2!}=15$. You don't want $4 \choose 4$ because the order of the last four horses matters. That gives $15 \cdot 4!=360$ Feb 26 at 23:40
• When you write $6 \choose 2$ you are saying that order does not matter. I don't know a simple one for when order does matter. Some use $_6P_2$ for that. Feb 27 at 0:09
• There are $\binom{6}{2}$ to select the subset of two horses which finish in a tie and $4!$ ways for the remaining four horses to finish in the remaining four positions. Notice that the order of selection for the two horses which finish in a tie does not matter, which is why your answer $P(6, 2)P(4, 4)$ is twice the correct answer. Feb 27 at 23:28
That leaves 4 choices for the horse to come in third, three choices for the horse to come in fourth, two choice for the horse to come in fifth, and one for the horse to come in sixth. The number of ways this can happen is $$15(4)(3)(2)(1)= \frac{6!}{2}= 720/2= 360$$, | 2021-10-18T18:22:46 | {
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https://math.stackexchange.com/questions/1858824/find-the-eigenvalues-and-corresponding-eigen-vectors-of-the-matrix | # Find the eigenvalues and corresponding eigen vectors of the matrix
Find the eigenvalues and corresponding eigen vectors of the matrix $$\begin{bmatrix}-3&6&-43\\0&-1&9\\0&0&2\end{bmatrix}$$
The eigenvalue $$\lambda_1 =$$____ corresponds to the eigevector$$( \ ,\ , \ )$$.
The eigenvalue $$\lambda_2 =$$____ corresponds to the eigevector$$( \ ,\ , \ )$$.
The eigenvalue $$\lambda_3 =$$____ corresponds to the eigevector$$( \ ,\ , \ )$$.
I'm kind of stuck after a certain point. Here is what I have so far
I do know that $$(A - \lambda I)X = 0$$
so $$\begin{bmatrix}-3&6&-43\\0&-1&9\\0&0&2\end{bmatrix}$$ $$\implies \lambda_1 = -3, \lambda_2 = -1, \lambda_3 = 2$$ so I have the eigenvalues but how can I find the corresponding eigenvectors?
• Do you know how to find eigenvalues? This is an upper triangular matrix. What about it's characteristic polynomial? – Kushal Bhuyan Jul 14 '16 at 1:40
• In any good textbook, class, video lectures or lecture notes on linear algebra, solving linear systems should be covered before finding eigenvalues. In particular, the Gaussian elimination method first puts a linear system into "upper triangular" form, and then finds the unknowns starting from the last ones. – arctic tern Jul 14 '16 at 3:27
You need to solve the equations $(A-\lambda I)v=0$ for $v$ for each of the three eigenvalues $\lambda$.
For instance when $\lambda=2$ we're solving
$$\begin{pmatrix} -5 & 6 & -43 \\ 0 & -3 & 9 \\ 0 & 0 & 0\end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$
The last equation is $0=0$ so it's superfluous. So we have two equations in three unknowns. If $z=0$ then the second equation implies $y=0$ and then the first equation implies $x=0$ which gives the zero vector, and that's not very interesting. Otherwise if $z\ne0$, we can scale the vector $(x,y,z)$ (scaling preserves the property of being a solution to the above system) in order to make $z=1$. In which case the second equation gives $y=3$ and then the first equation gives $x=-5$. So the eigenvector is $(-5,3,1)$ up to scaling.
Your turn. Do $\lambda=-3$ and $\lambda=-1$. (Sophia did $\lambda=-3$ for you, so now most of the work is done for you. If we do the problem for you, at least learn what it is we're doing!)
• For $\lambda_2 = -1$ I get $\begin{bmatrix}-2&6&-43\\0&0&9\\0&0&3\end{bmatrix}$ ~ $\begin{bmatrix}1&-3&\frac{43}{2}\\0&0&9\\0&0&0\end{bmatrix}$ so I thought it would be (-3,0,0) ? – Yusha Jul 14 '16 at 3:27
• When I multiply that matrix by $(-3,0,0)^T$ I don't get the zero matrix. In that situation, the last equation is redundant. What does the second equation tell you about $z$? What does the first equation then reduce to saying about $x$ and $y$? (BTW I wouldn't bother with the fractions.) – arctic tern Jul 14 '16 at 3:30
• Wow, I'm an idiot! I see now that $z = 0 \implies 6y = 2x \implies (3,1,0)$ – Yusha Jul 14 '16 at 3:31
You start with the understanding of this formula: $(A-\lambda I)\vec x=0$, which is equivalent to $\det(A-\lambda I)=0$ $$\begin{vmatrix}-3-\lambda&6&-43\\0&-1-\lambda&9\\0&0&2-\lambda\end{vmatrix}=(-3-\lambda)(-1-\lambda)(2-\lambda)=0$$ Therefore, $\lambda_1=-3, \ \lambda_2=-1, \ \lambda_3=2$.
Let's do one example for eigenvectors:
Plug in the value of $\lambda$ into the augmented form of the matrix:
With $\lambda_1=-3$, $$\left[\begin{array}{ccc|c}-3-(-3)&6&-43&0\\0&-1-(-3)&9&0\\0&0&2-(-3)&0\end{array}\right]=\left[\begin{array}{ccc|c}0&6&-43&0\\0&2&9&0\\0&0&5&0\end{array}\right]$$ Solve this matrix and get $v_1=\begin{bmatrix}1\\0\\0\end{bmatrix}$
Now you can use similar approach to find the eigenvectors of the next two eigenvalues.
• @Yusha, double check your calculations. At least for $\lambda=-1$, I did not get $(-3,0,0)$ – Ron Jul 14 '16 at 2:18
• I'm lost, how are you getting (1,0,0). You have in your work (0,0,0). – Yusha Jul 14 '16 at 2:43
• Both answers are correct, since one is a scalar multiple of the other. But it is customary to use numbers are small as possible, so $(1,0,0)$ would be prefered. – imranfat Jul 14 '16 at 2:47
• Does (1,0,0) @imranfat come from the 2nd column after its been put in RREF? – Yusha Jul 14 '16 at 2:50
• RREF has to give infinite solutions, I looked at Sophia's work, realizing that both vectors serve as the same eigenvector. Your RREF must have all zero's in bottom row – imranfat Jul 14 '16 at 2:55
Hint: In an upper triangular matrix the characteristic polynomial is $(x_1-\lambda_1)^{\alpha_1}(x_2-\lambda_2)^{\alpha_2}\ldots(x_n-\lambda_n)^{\alpha_n}$, where $x_i$ are diagonal entries with $\alpha$'s as their multiplicities.
In your case the characteristic polynomial is $(3+\lambda)(1+\lambda)(2-\lambda)$
• Here is one: sosmath.com/matrix/eigen2/eigen2.html – imranfat Jul 14 '16 at 2:27
• Khan academy usually rocks: khanacademy.org/math/linear-algebra/alternate-bases/… – imranfat Jul 14 '16 at 2:28
• OK, so I assume that you DO know how to find eigenvalues theoretically, but somehow your algebra is letting you down? It would be helpful to see your work in attempting finding the eigenvalue, because a lot of times it is a small algebraic mistake... – imranfat Jul 14 '16 at 2:33
• Here is a worked out example of a 2 by 2 matrix: calvin.edu/~scofield/courses/m256/materials/eigenstuff.pdf In my view you should first be fluent with 2by2's before going for 3by3's... – imranfat Jul 14 '16 at 2:34
• It sounds like your exchange converged. Anyway, this conversation has been moved to chat. If you need to revisit the full exchange of comments, go to that chatroom. I left the posts with links to resources also here. – Jyrki Lahtonen Jul 14 '16 at 6:26 | 2021-05-08T23:00:20 | {
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https://math.stackexchange.com/questions/947026/by-using-de-moivres-theorem-show-that-cos5-theta-16-cos5-theta-20-cos | # By using De Moivre's Theorem, show that $\cos5\theta = 16\cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta$ [closed]
First step is
$$\cos5\theta + i \sin 5\theta = (\cos \theta + i \sin \theta)^5$$
thanks
• Because it's easier than using using other trigonometric formulae. Note that the real parts of the expressions on each side of the equation must be equal. – user164587 Sep 26 '14 at 14:39
• @user164587 well i meant how to get it? – problematic Sep 26 '14 at 14:41
• It is de Moivre – Seub Sep 26 '14 at 14:41
• @vera sorry bad english – problematic Sep 26 '14 at 14:41
Hint: Expand $(\cos\theta + i\sin\theta)^5$ using binomial theorem. i.e. $(a+b)^n = \sum\limits_{k = 0}^n {n \choose k}a^k b^{n-k}$
Here, $a = \cos\theta, b=i\sin\theta$.
\begin{align}(\cos\theta+i\sin\theta)^5 &= \sum\limits_{k = 0}^5 {5\choose k}(\cos\theta)^k (i\sin\theta)^{5-k}\\ &= {5\choose0}(\cos\theta)^0(i\sin\theta)^5 + {5\choose1}(\cos\theta)^1(i\sin\theta)^4 + {5\choose2}(\cos\theta)^2(i\sin\theta)^3\\&\quad+ {5\choose3}(\cos\theta)^3(i\sin\theta)^2+{5\choose4}(\cos\theta)^4(i\sin\theta)^1+{5\choose5}(\cos\theta)^5(i\sin\theta)^0 \\&= (i\sin\theta)^5 + 5(\cos\theta)(i^4\sin^4\theta) + 10\cos^2\theta(i^3\sin^3\theta)\\&\quad+10\cos^3\theta(i^2\sin^2\theta)+5\cos^4\theta(i\sin\theta)+\cos^5\theta \\&=\cdots \end{align}
Then, for even powers of $\sin \theta$, use $\sin^2 \theta = 1-\cos^2 \theta$.
Finally, equate the real parts of $\cos(5\theta)+i\sin(5\theta) = (\cos\theta+i\sin\theta)^5$.
• am i doing it right ? lol $(_0^5)$$(cos^0 \theta)$$(i sin^5 \theta)$ – problematic Sep 26 '14 at 15:03
• Yes, and there will be $5$ more terms. – taninamdar Sep 26 '14 at 15:05
• actually i don't know how binomial theoram works, can you give an example and show me what $(_0^5)$$(cos^0 \theta)$$(sin^5 \theta)$ goes to, i will do the rest my self – problematic Sep 26 '14 at 15:06
• See the edit, ask if you need more help. – taninamdar Sep 26 '14 at 15:13
• thank you, how to get the first number? i mean it looks like $(_1^5)$ : 5x1, $(_2^5)$ 5x2 , then $(_3^5)$ onward? – problematic Sep 26 '14 at 15:22
The point is that $$\cos 5 \theta$$ is the even part of $$\exp(5 i \theta) = [\exp(i \theta)]^5 = (\cos \theta + i \sin \theta)^5,$$ where the second equality follows from de Moivre's Theorem. Then, by expanding the right-hand side using the Binomial Theorem, we can immediately write $\cos 5 \theta$ as a sum of products of powers of $\sin \theta$ and $\cos \theta$. Then, we can eliminate even powers of $\sin \theta$ using the Pythagorean identity $$\sin^2 \theta + \cos^2 \theta = 1.$$
• You're welcome, I hope you found it helpful. – Travis Willse Sep 26 '14 at 15:37
• yes, you showed me exp(5iθ)=[exp(iθ)]5=(cosθ+isinθ)5, i needed this to know why we have the first step, i can't choose 2 accepted answers... – problematic Sep 26 '14 at 15:40
• No problem, that's just a part of the site mechanics, the important thing is that you understand the concepts you asked about better than you did before. – Travis Willse Sep 26 '14 at 15:51
De Moivre's Theorem states $$\cos n\theta + \mathbb i \sin n\theta = \left( \cos \theta + \mathbb i \sin \theta \right)^n$$ | 2019-12-12T03:43:04 | {
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https://brilliant.org/discussions/thread/proof-8/ | ×
# Proof
Show that, for any positive integer $$n\geq 1$$ it is true that: $\displaystyle \sum_{k=1}^{n} \sqrt{k} \geq \sqrt{\displaystyle \sum_{k=1}^{n} k}.$
Note by Hjalmar Orellana Soto
10 months, 3 weeks ago
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for $$n\geq 2$$ it can be easily proven by applying triangle inequality recursively
assume a,b,c are three terms in the seuence
we have $$\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}$$ as $$\sqrt{a},\sqrt{b},\sqrt{a+b}$$ form sides of a right triangle
now we have $$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq \sqrt{a+b}+\sqrt{c}\geq\sqrt{a+b+c}$$ and so on....
sum of each of the outer sides of the spiral form the LHS of the equation,and RHS is given by the line to centre
also note that for n=1 the line to centre is same as side, the only point of equality
- 10 months, 1 week ago
very nice solution, Anirudh
- 10 months, 1 week ago
thank you :)
- 10 months, 1 week ago
$$LHS^2 = 1+2+\cdots+k + 2\sum_{1\le x<y\le n}\sqrt{xy}\ge 1+2+\cdots+n = RHS^2$$ and equallity occurs when $$n=1$$
- 10 months, 3 weeks ago
In addition to the problem.... Is this necessary information for saying that the limit below converges? $\lim_{x\to \infty} \dfrac{\displaystyle \sqrt{\sum_{n=1}^{x}n}}{\displaystyle \sum_{n=1}^{x}\sqrt{n}}$
- 10 months, 3 weeks ago
(Assuming I'm interpreting you correctly,) the essence of what you're asking here is:
If $$a_n, b_n$$ are sequences of (positive) reals, such that $$a_n \geq b_n$$, does $$\lim \frac{ b_n } { a_n }$$ exist?
If yes, why?
If no, what is a counter example? What other assumptions do we need to add to guarantee a true statement?
Staff - 10 months, 3 weeks ago
Exactly that is what I ask, I was thinking of asking for the value of $\displaystyle \sum_{n=1}^{\infty} (\frac{\sqrt{\sum_{k=1}^{n}n}}{\sum_{k=1}^{n}\sqrt{n}})$ but I really don't know how to approach it
- 10 months, 3 weeks ago
The answer to "does $$\lim \frac{ b_n } { a_n }$$ exist?" is no for general sequences. Do you see an obvious counter example?
For your recent comment, are you looking for the summation, or for the limit of the term (as in the prior comment)?
Staff - 10 months, 3 weeks ago
The answer to the limit question is yes for this case, using the same argument which is used for $$\lim_{x\to \infty} \frac{x}{e^x}$$.... and I'm looking for the summation
- 10 months, 3 weeks ago
See How to ask for help to understand how to provide proper context of what you want. E.g. You can see that people quickly established the inequality, but that doesn't help with your concern.
Staff - 10 months, 3 weeks ago
The summation is something else, something I supposed and now I'm trying to approach, and I think the inequality may help for knowing if the summation converges... and then I'll try to approach the value of the summation, I don't know if I'm explaining right...
- 10 months, 3 weeks ago
Asymptotically, $$\sqrt{1} + \sqrt{2} + \dots + \sqrt{n}$$ can be approximated by the integral $\int_0^n \sqrt{x} \ dx = \frac{2}{3} n^{3/2}.$ And of course, $\sqrt{1 + 2 + \dots + n} = \sqrt{\frac{n(n + 1)}{2}}.$ That should help answer your question.
- 10 months, 3 weeks ago
We want to prove that $\sqrt{1} + \sqrt{2} + \dots + \sqrt{n} \ge \sqrt{1 + 2 + \dots + n}.$ Squaring the left-hand side, we get a term of $$\sqrt{k} \cdot \sqrt{k} = k$$ for each $$1 \le k \le n$$. The inequality follows.
- 10 months, 3 weeks ago
× | 2017-11-24T03:46:38 | {
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https://math.stackexchange.com/questions/3064314/on-the-number-of-roots-of-the-polynomial-x3ax21-0/3064353 | # On the number of roots of the polynomial $x^3+Ax^2+1=0$
I have the following cubic equation
$$x^{3}+Ax^{2}+1=0$$
where $$A$$ is an arbitrary (real) number.
I know that either:
• The 3 roots will be real.
• One root will be real and the other two will be complex conjugates of each other.
I would like to find out
• For what value/values of A the roots change from 3 real roots to one real and two complex roots.
• The signs of each of the real roots (both when they are all real and when there is only one real root)
Is there an analytical way of finding this as a function of $$A$$ or the only option is to solve the cubic numerically?
• Google Cardano's method. – hamam_Abdallah Jan 6 at 19:44
If we can show that the value of $$A$$ changes the amount of maxima or minima from 1 to 2 such that one extreme is at a positive $$y$$-location and the other at a zero or negative $$y$$-location, then we know that this value of $$A$$ is the border for all real solutions versus 1 real and 2 complex solutions.
Denote $$f(x) = x^3 + Ax^2 + 1.$$
$$f'(x) = 3x^2+2Ax=x(3x+2A) = 0.$$
We see that $$f(0)$$ is always either a local maximum or minimum and $$f(0) > 0.$$ The other zero is given by $$x = -2A/3 = \xi.$$
We need $$f(\xi) \leq 0.$$
$$f(\xi) = \frac{4}{27}A^3 + 1 \leq 0$$
solves for the values of $$A$$ where 3 real solutions are guaranteed.
If there are 3 real roots, then since $$x = 0$$ is always a local extreme, at least one has to be positive and another has to be negative. Also, if $$A \leq -3/\sqrt[3]{4}$$ and the location of the moving extreme is $$x=-2A/3 > 0,$$ then the remaining real root has to be positive.
If there's only 1 real root, then $$A > -3/\sqrt[3]{4}$$ and the moving extreme is located at $$x < \sqrt[3]{2}.$$ Thus the only real root must be negative.
• How do we find the signs of the real roots? By actually finding them or is there a shortcut? – JennyToy Jan 6 at 20:21
• @JennyToy I should've reread the questions before submitting the first time, sorry! How's the edit? – AEngineer Jan 6 at 20:45
hint
Put $$A=-\frac 32B$$ and $$f(x)=x^3-\frac 32Bx^2+1.$$
$$f'(x)=3x(x-B).$$
$$f(0)=1$$ to have three real roots, we need
$$B>0 \text{ and } f(B)<0.$$
• To find when there are 3 real roots or one only use the sign of the discriminant $\Delta = -4A^{3}-27$. Thus if $A=-3/2^{2/3}$ there are 3 equal real roots. If $A<-3/2^{2/3}$ there are 3 real distinct roots. And if $A>-3/2^{2/3}$ there is one real root and two (complex conjugate) complex roots. Unfortunately to find the signs of the real roots we probably need to actually find the roots. – user2175783 Jan 6 at 20:13
• @user2175783 If $A=-\frac{3}{2^{2/3}}$, then there are a negative zero and a double zero on the positive axis. – Hanno Jan 12 at 22:17
The stationary points of $$f(x)$$ lie at $$x=0$$ and at $$x=-\frac{2A}{3}$$ and the sign of $$f(0)f\left(-\tfrac{2A}{3}\right) = 1+\tfrac{4}{27}A^3$$ decides if there are three real roots (<0), a simple real root (>0) or a double real root and a simple real root (=0). This happens since the sign of $$f(0)f\left(-\tfrac{2A}{3}\right)$$ only depends on the stationary values having the same sign or not.
Too late at the party. But it looks good, so let's enter anyway.
TL;DR $$\quad f(x)=x^3+Ax^2+1\,$$ always has a simple zero on the negative axis. The other two zeros are
• real, positive and distinct, if $$\,A<-1.88988\:=\:-\dfrac{3}{\sqrt[3]{4}}\:=\:A_1$$
• merging into a double zero at $$x=\sqrt[3]{2}$$, if $$\,A=A_1$$
• non-real (thus complex-conjugate), if $$\,A>A_1$$
The party-giver also asked if there is an analytical way of finding this as a function of $$A$$?
Let's take a step in this direction, concentrating on the case of "one real zero", and express the negative zero as a function of $$A$$:
Let $$\,d=\big(\frac A3\big)^3 + \big(\frac 12\big)^2$$, and assume $$d>0$$ which corresponds to $$\,A>A_1$$. Define $$r \:=\:\left(-\frac 12 -\sqrt d\right)^\frac 13 +\, \left(-\frac 12 +\sqrt d\right)^\frac 13\,,$$ $$r\,$$ is observed to be strictly negative. Then \begin{align} r^2\:= & \;\left(-\frac 12 -\sqrt d\right)^\frac 23 +\, \left(-\frac 12 +\sqrt d\right)^\frac 23 -\frac23 A \\[2ex] r^3\:= & \; -\frac23 Ar + \left(-\frac 12 -\sqrt d\right) +\, \left(-\frac 12 +\sqrt d\right) \\ & \quad -\frac13 A \left\{ \left(-\frac 12 -\sqrt d\right)^\frac 13 +\, \left(-\frac 12 +\sqrt d\right)^\frac 13\right\} \\ = & -Ar - 1 \end{align} Whence $$1+Ar+r^3=0\quad\Longleftrightarrow\quad\frac1{r^3}+A\frac1{r^2}+1 =0 \\[2ex] \Longrightarrow\quad x \:=\: \frac1r \:=\: -r^2-A \:=\: -\frac13 A -\left(\sqrt d + \frac 12\right)^\frac 23 -\, \left(\sqrt d -\frac12\right)^\frac 23$$ satisfies $$x^3+Ax^2+1=0$$.
Added in Edit: Driven by user2175783's comment as of below, I found Lecture 4 in Mathematical Omnibus: 30 Lectures on Classic Maths by D. Fuchs + S. Tabachnikov: It gives an extensive and readable account of how to explicitly solve the cubic and quartic equations.
Visual TL;DR
• Wouldnt it be possible to generalize the algebra for the negative root for $A<A_{1}$? – user2175783 Jan 13 at 0:14
• @user2175783 I did not look closer at that case. I'd say there's no ambiguity for the sign of $\sqrt d$ because $r$ is a symmetrical expression. But afterwards complex cube roots have to be considered which might involve some choice. – Hanno Jan 13 at 9:31
The discriminant of a cubic polynomial with real coefficients is $$0$$ when the polynomial has two (or more) equal roots, positive when there are three distinct real roots, and negative when there are one real and two non-real roots. In your case it is $$-4 A^3 - 27$$. | 2019-07-16T16:29:27 | {
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https://mathhelpboards.com/threads/find-all-triplets-x-y-z.6923/ | # Find all triplets (x,y,z)
#### anemone
##### MHB POTW Director
Staff member
Find all triplets of positive integers $(x, y, z)$ such that $$\displaystyle \left( 1+\frac{1}{x} \right)\left( 1+\frac{1}{y} \right)\left( 1+\frac{1}{z} \right)=2$$.
Last edited:
##### Well-known member
Re: Find all triplets of x, y and z
Find all triplets of positive integers $(x, y, z)$ such that $$\displaystyle \left( 1+\frac{1}{x} \right)\left( 1+\frac{1}{y} \right)\left( 1+\frac{1}{z} \right)=2$$.
(x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7), (3,3,8) , ( 3,4,5) or any permutation of them
solved as
Without loss of generality we can choose x <=y <=z and ans shall be a permutation of if
Now x < 4 as (5/4)^3 < 2 ( it is 125/64)
x cannot be 1 as (1+1/y)(1+ 1/z) = 1 has no solution
so x = 2 or 3
if x = 2 we get
3/2(y+1)(z+1) = 2yz
Or 3(y+1)(z+1) = 4 yz
Or yz – 3y – 3z = 3
(y-3)(z-3) = 12 by adding 9 on both sides
( y-3)(z-3) = 1 * 12 or 2 * 6 or 3 * 4
So (x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7)
if x = 3 we get
4/3(y+1)(z+1) = 2yz
Or 2(y+1)(z+1) = 3 yz
Or yz – 2y – 2z = 2
(y-2)(z-2) = 6
( y-2)(z-2) = 1 * 6 or 2 * 3
So (x,y,z ) = (3,3,8) , ( 3,4,5)
#### anemone
##### MHB POTW Director
Staff member
Re: Find all triplets of x, y and z
(x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7), (3,3,8) , ( 3,4,5) or any permutation of them
solved as
Without loss of generality we can choose x <=y <=z and ans shall be a permutation of if
Now x < 4 as (5/4)^3 < 2 ( it is 125/64)
x cannot be 1 as (1+1/y)(1+ 1/z) = 1 has no solution
so x = 2 or 3
if x = 2 we get
3/2(y+1)(z+1) = 2yz
Or 3(y+1)(z+1) = 4 yz
Or yz – 3y – 3z = 3
(y-3)(z-3) = 12 by adding 9 on both sides
( y-3)(z-3) = 1 * 12 or 2 * 6 or 3 * 4
So (x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7)
if x = 3 we get
4/3(y+1)(z+1) = 2yz
Or 2(y+1)(z+1) = 3 yz
Or yz – 2y – 2z = 2
(y-2)(z-2) = 6
( y-2)(z-2) = 1 * 6 or 2 * 3
So (x,y,z ) = (3,3,8) , ( 3,4,5)
Wow! I forgot to thank you explicitly for participating and also for your unique way of attacking the problem. What's worst is that I forgot completely how I approached it two months ago and I promise you to add my reply once I solved it because I could only tell at this point that I solved it differently than what you did. Sorry, kaliprasad! | 2020-12-05T14:59:12 | {
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https://mathhelpboards.com/threads/2-coins-are-thrown-20-times.24749/ | # [SOLVED]2 coins are thrown 20 times
#### mathmari
##### Well-known member
MHB Site Helper
Hey!!
2 coins are thrown 20 times. I want to calculate the probability
(a) to achieve exactly 5 times the Tail/Tail
(b) to achieve at least 2 times Tail/Tail
If we throw the 2 coins once the probability that we get Tail/Tail is equal to $\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$, or not?
Is the probability then at (a) equal to $\left (\frac{1}{4}\right )^{20}$ ?
Could you give me a hint for (b) ?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Hey!!
2 coins are thrown 20 times. I want to calculate the probability
(a) to achieve exactly 5 times the Tail/Tail
(b) to achieve at least 2 times Tail/Tail
If we throw the 2 coins once the probability that we get Tail/Tail is equal to $\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$, or not?
Hey mathmari !!
Yep.
Is the probability then at (a) equal to $\left (\frac{1}{4}\right )^{20}$ ?
There seems to be a $5$ missing.
It's a binomial distribution, so we should use the corresponding formula, which includes a spot to put the $5$.
Could you give me a hint for (b) ?
Take the complement?
It's $1$ minus the probability of getting either 0 Tail/Tail or 1 Tail/Tail.
#### mathmari
##### Well-known member
MHB Site Helper
There seems to be a $5$ missing.
It's a binomial distribution, so we should use the corresponding formula, which includes a spot to put the $5$.
Oh yes.. So do we have the following? $$\binom{20}{5}\cdot \left (\frac{1}{4}\right )^5\cdot \left (1-\frac{1}{4}\right )^{20-5}$$
Take the complement?
It's $1$ minus the probability of getting either 0 Tail/Tail or 1 Tail/Tail.
So, we have $$P(X\geq 2)=1-P(X<2)=1-P(X=1)-P(X=0)=1-\binom{20}{1}\cdot \left (\frac{1}{4}\right )^1\cdot \left (1-\frac{1}{4}\right )^{20-1}-\binom{20}{0}\cdot \left (\frac{1}{4}\right )^0\cdot \left (1-\frac{1}{4}\right )^{20-0}$$ or not?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Yep. All correct.
#### Wilmer
##### In Memoriam
So do we have the following? $$\binom{20}{5}\cdot \left (\frac{1}{4}\right )^5\cdot \left (1-\frac{1}{4}\right )^{20-5}$$
Looks good.
Ran a simulation: got 202276 out of 1 million.
MHB Site Helper
Thank you!! | 2020-07-05T06:29:35 | {
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http://mathhelpforum.com/math-challenge-problems/229258-problem-8-number-theory.html | # Thread: Problem #8 - Number Theory
1. ## Problem #8 - Number Theory
I got this one from our archives as well.
Show that $n^4 + 4^n$ is not prime for $n \geq 2$. (n is a positive integer!)
This is a more of a Number Theory problem than algebraic. As my proof relies more on a College Algebra level rather than Number Theory I'm curious to see what you might come up with.
-Dan
2. ## Re: Problem #8 - Number Theory
all n values that are even will give a non-prime number, as $(evennumber)^4$ is always even and $4^n$ is always even. even+even gives even numbers bigger than 2, which are never prime. i can't continue from here though, dont know about the odd n values
3. ## Re: Problem #8 - Number Theory
Originally Posted by muddywaters
all n values that are even will give a non-prime number, as $(evennumber)^4$ is always even and $4^n$ is always even. even+even gives even numbers bigger than 2, which are never prime. i can't continue from here though, dont know about the odd n values
if n is an odd integer >= 3 and n is not a multiple of 5 then n^4 + 4^n is a multiple of 5 and therefore not a prime
Proof:
n is 1,2,3,or 4 (mod 5)
n^4 is 1 (mod 5)
4^n is -1 (mod 5)
So n^4 + 4^n is congruent to 0 (mod 5)
How do we handle the case n=5, 15, 25, 35, 45, ...... ?
4. ## Re: Problem #8 - Number Theory
If $n\equiv 0\pmod{2}$, then $n^4+4^n \equiv 0\pmod{2}$.
If $n\equiv 1\pmod{2}$, then $n+1\equiv 0\pmod{2}$, so $2^{n+1}$ is a perfect square, and $n^4+4^n$ factors as
$(n^2+n\sqrt{2^{n+1}}+2^n)(n^2-n\sqrt{2^{n+1}} + 2^n)$
So, all that is left is to show that for odd $n>2$, both terms are greater than 1. Since $n^2+2^n>1$, the first term is guaranteed to be greater than 1. So, we just need to show the second term is greater than one.
Since $(n^2-1)^2+2^{2n} \ge 2^{2n} > 2^{n+1}$ (for all $n>1$), we have
\begin{align*}n^4-2n^2+1+4^n-2^{n+1} > 0 & \Rightarrow n^4-2n^2+1+4^n-2^{n+1}+n^2 2^{n+1} > n^2 2^{n+1} \\ & \Rightarrow (n^2+2^n-1)^2 > n^2 2^{n+1} \\ & \Rightarrow n^2 + 2^n - 1 > n\sqrt{2^{n+1}} \\ & \Rightarrow n^2-n\sqrt{2^{n+1}}+2^n > 1\end{align*}
5. ## Re: Problem #8 - Number Theory
If n even, obvious.
If n odd, 2n+1 is a perfect square and the factorization are really neat observations. Thanks.
But what's the point of the congruences?
6. ## Re: Problem #8 - Number Theory
Originally Posted by Hartlw
But what's the point of the congruences?
$n\equiv 0\pmod{2}$ is the same as saying $n$ is even.
$n\equiv 1\pmod{2}$ is the same as saying $n$ is odd.
I have been working with modular arithmetic long enough that it feels more natural for me to use congruences rather than the words even and odd. The meaning is the same either way.
7. ## Re: Problem #8 - Number Theory
You have shown n^4+4^n=rs where r and s are integers and r positive. Obviously s is positive.
8. ## Re: Problem #8 - Number Theory
Originally Posted by Hartlw
You have shown n^4+4^n=rs where r and s are integers and r positive. Obviously s is positive.
If $n=1$, then the factorization I gave is $1^4+4^1 = (5)(1)$. However, 5 is a prime number. The problem asks to prove that for $n\ge 2$, $n^4+4^n$ is NOT prime. So, to do that, it is not enough to show the factors are positive. You must show they are both greater than 1.
9. ## Re: Problem #8 - Number Theory
Hey, lots of action on this one! Thanks to all.
SlipEternal gets some extra kudos for checking the "trivial" factorization problem. (That one of the factors might be 1.) I did eventually find the proof online (which was pretty much identical to Slip's and my own) but didn't address the trivial factorization problem.
-Dan | 2016-09-25T20:51:39 | {
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https://stats.stackexchange.com/questions/435458/what-is-the-probability-of-getting-three-pairs-that-makes-up-seven-when-six-dice | What is the probability of getting three pairs that makes up seven when six dice are rolled?
Stumbled upon this question in a game of dice where a point is dealt for each pair that makes seven. In the game there were a total of six dice that were to be thrown up to three times. Each time a pair of dice made seven 1+6, 2+5 or 4+3), the pair(s) was taken out from the rest, a point was dealt and you got to roll the remaining dice. You can not use dice from the excluded pairs to make up new ones. You win the game by getting the most points/pairs in 3 rounds.
Example:
round one [1] [1] [1] [3] [5] [6] - one pair make 7 (1+6)
round two [1] [3] [4] [2] - one pair make 7 (3+4)
round one [2] [6] - no pairs make 7
points awarded: 2
Out of curiosity I have been trying to calculate the probabilities of getting three pairs of sevens in the first round?
• We have six dice and we throw them all together in the first round. Let's say the faces are $3,4,5,2,6,6$. Because there are two pairs that make $7$, we remove the pairs and continue with two dice. Round 2 and 3 continues this way. What is the definition of winning? Is it the situation that we're left with no dice at the end? – gunes Nov 10 '19 at 18:18
• and how is it possible to roll three dice following those rules? – carlo Nov 10 '19 at 18:39
• Your question is unclear. Your title refers to three dice while your body text refers to six. You describe a game where the number of dice change, altering the probabilities. At which point in that game are you calculating the chance of getting two dice that add to 7? – Glen_b Nov 10 '19 at 21:02
• Sorry if the original edit was confusing, I have tried to clarify. The game description is just for context. The question is in the title and reformulated in the last paragraph. – joho Nov 13 '19 at 14:28
Consider a sequence of $$2k$$ dice each with the possible values $$1, 2, \ldots, 2n.$$ (The question concerns $$k=n=3.$$) The possible pairs are $$\{1,n\},$$ $$\{2,2n-1\},$$ and so on, through $$\{n,n+1\}.$$ Denote such a pair by its smallest value $$i$$ and for each $$i$$ let $$k_i\gt 0$$ be the number of such pairs that can be located in the sequence. We need to count the number of equally probable sequences for which $$k_1+k_2+\cdots + k_n = k.$$
Such a sequence is determined by (a) which pairs occur in it and (b) where in the sequence of $$2k$$ values each such pair occurs. The permutation group on $$2k$$ elements acts on the set of such sequences. For all $$i,$$ the stabilizer of such a sequence permutes the $$k_i$$ values of $$i$$ among themselves and the $$k_i$$ values of $$2n+1-i$$ among themselves. Thus, applying the method described at https://stats.stackexchange.com/a/415878/919, the number of ways of producing a sequence designated by $$\mathrm{k}=(k_1,k_2,\ldots,k_n)$$ is
$$p(\mathrm{k}) = \frac{(2k)!}{(k_1!)^2(k_2!)^2\cdots(k_n!)^2}.$$
Thus, the chance is obtained by summing $$p(\mathrm{k})$$ over all possible $$\mathrm{k}$$ whose components sum to $$k$$ and dividing by the total number of sequences, $$(2n)^{2k}.$$
These possibilities correspond to the weak compositions of $$k$$ into $$n$$ parts, which number $$\binom{k+n-1}{n-1}.$$ However, the amount of calculation is smaller than this, because $$p(\mathrm{k})$$ does not depend on the order of the $$k_i.$$ We may therefore do the calculation for all $$k_1\ge k_2\ge \cdots \ge k_n \ge 0$$ (giving a partition of $$k$$), multiplying each by the number of distinct re-orderings of $$\mathrm k.$$ Such sequences correspond to the Ferrers diagrams for $$n$$ having at most $$k$$ rows. They are relatively easy to enumerate.
With $$k=n=3,$$ we have $$\binom{3+3-1}{3-1}=10$$ possibilities for $$\mathrm k,$$ but they fall into just three groups corresponding to the partitions $$3 = 2+1 = 1+1+1:$$
$$p(3,0,0)=p(0,3,0)=p(0,0,3) = \frac{6!}{3!^2} = 20;$$ $$p(2,1,0)=p(2,0,1)=p(1,2,0)=p(1,0,2)=p(0,2,1)=p(0,1,2) = \frac{6!}{2!^21!^2}= 180;$$ $$p(1,1,1) = \frac{6!}{1!^21!^21!^2} = 720.$$
$$\Pr(\text{three pairs}) = \frac{3\times 20 + 6\times 180 + 1\times 720}{6^6} = \frac{(5)(31)}{(2^4)(3^5)} \approx 3.9866\%.$$
• @Ron I haven't thoroughly checked my general answer, but I did check the specific answer with a brute force enumeration of all possibilities. At least one mistake in your post is that many sequences without three pairs will sum to 21, such as 5,5,5,4,1,1, which has no pairs at all. Thus you substantially overcount. Here is my implementation: n <- 3; k <- 3; X <- expand.grid(rep(list(seq_len(2*n)), 2*k)); p <- function(x, d=n) { x <- sort(x); sum((x + rev(x))[seq_len(d)] == 2*d+1) }; table(apply(X, 1, p)) – whuber Nov 13 '19 at 19:04
• @Ron If you would like to explore the situation in more detail, this code (which follows the creation of data frame X in my previous comment) breaks the population down into all possible pair patterns: pattern <- function(x, d=n) { x <- sort(x); paste0(x[which((x + rev(x))[seq_len(d)] == 2*d + 1)], collapse="") }; table(apply(X, 1, pattern)) – whuber Nov 13 '19 at 19:19 | 2020-08-06T16:14:22 | {
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http://math.stackexchange.com/questions/387218/choosing-a-linear-map-mathbbz-2-mathbbzn-rightarrow-mathbbz-2-mathb | # Choosing a linear map $(\mathbb{Z}/2\mathbb{Z})^n \rightarrow \mathbb{Z}/2\mathbb{Z}$ which is nonzero on half of a sequence of vectors
Let $v_1,\ldots,v_m \in (\mathbb{Z}/2\mathbb{Z})^n$ be nonzero vectors. Is it always possible to choose a linear map $f : (\mathbb{Z}/2\mathbb{Z})^n \rightarrow \mathbb{Z}/2\mathbb{Z}$ such that $f$ is nonzero on at least half of the $v_i$, i.e. such that
$$|\{\text{i | 1 \leq i \leq m and f(v_i) \neq 0}\}| \geq \frac{1}{2}m?$$
My guess is that the answer is yes; at the very least, it is true for $m=1$ and when the $v_i$ enumerate all of the nonzero vectors.
-
Why don't you write in words what you want in the body of the question? Desciphering an equation when a perfectly good English sentence might be considerably easier to understand is silly! – Mariano Suárez-Alvarez May 10 '13 at 3:45
@MarianoSuárez-Alvarez : Sure thing. But isn't that a little pedantic? The equation isn't all that complicated... – Monica May 10 '13 at 3:47
I don't disagree with @Mariano's comment, but let me just add that this is a very nice question. Where did it come from? – Pete L. Clark May 10 '13 at 13:33
And I am someone who generally finds contest type problems quite boring (which is not to say that I know how to solve them). Looking back at the comments, Monica got a less warm welcome than she should have: $\mathbb{Z}/2$ could only mean one thing (unlike $\mathbb{Z}_2$!) and upon reflection I do disagree with Mariano's comment. I'm glad it turned out well in the end. – Pete L. Clark May 10 '13 at 17:10
@PeteL.Clark : Thanks! This experience definitely provides me another data point concerning using gendered names on the internet (I usually don't). It also appears that one can order the Babai book from U of C; see cs.uchicago.edu/research/publications/combinatorics – Monica May 10 '13 at 17:25
The statement is true and let us prove it by induction on $n$.
Denote $\mathbb{Z}/2\mathbb{Z}:=\mathbb{Z}_2:=\{0,1\}$. When $n=1$, the statement is clearly true. Now assume that when $n\le n_0$, the statement is true for every $m\ge 1$, and let us show the statement is true for $n:=n_0+1$ and every $m\ge 1$.
Consider $(\mathbb{Z}_2)^n$ as $(\mathbb{Z}_2)^{n-1}\times\mathbb{Z}_2$ and denote by $P$ and $Q$ the projections of $(\mathbb{Z}_2)^n$ to its first $n-1$ coordinates and its last coordinate respectively, i.e. $$P: (\mathbb{Z}_2)^n\mapsto (\mathbb{Z}_2)^{n-1},\quad(a_1,\dots,a_{n-1},a_n)\mapsto (a_1,\dots,a_{n-1}),$$ and $$Q: (\mathbb{Z}_2)^n\mapsto \mathbb{Z}_2,\quad(a_1,\dots,a_n)\mapsto a_n.$$
Up to a rearrangement of $v_1,\dots,v_m$, we may assume that there exists $0\le m'\le m$, such that $Qv_i\ne 0$ if and only if $m'<i\le m$. If $m'=0$, simply choose $f=Q$ and we are done, so let us assume that $m'\ge 1$. By definition, $Pv_1,\dots,Pv_{m'}\ne 0$. By the induction assumption on $n$, there exists a linear function $g:(\mathbb{Z}_2)^{n-1}\to \mathbb{Z}_2$, such that $$|\{ 1\le i\le m' \mid g(P v_i) \neq 0\}| \ge \frac{1}{2}m'.\tag{1}$$ For $j=0,1$, define $$f_j:(\mathbb{Z}_2)^n\to \mathbb{Z}_2,\quad v\mapsto g(Pv)+j\cdot Qv.\tag{2}$$ By definition, for $j=0,1$, $f_j$ is linear and $f_j(v_i)=g(P v_i)$ when $1\le m\le m'$. Then by $(1)$, for both $j=0$ and $j=1$, $$|\{ 1\le i\le m' \mid f_j(v_i) \neq 0\}| \ge \frac{1}{2}m'.\tag{3}$$ If $m'=m$, then we are done. Otherwise, note that by $(2)$ and the choice of $m'$, for $j=0,1$, $$f_j(v_i)=0 \iff g(Pv_i)=1-j\iff f_{1-j}(v_i)=1,\quad \forall\, m'<i\le m.$$ It follows that $$|\{ m'< i\le m \mid f_0(v_i) \neq 0\}|+|\{ m'< i\le m \mid f_1(v_i) \neq 0\}|=m-m'.\tag{4}$$ Combining $(3)$ and $(4)$, we can conclude that the statement is true for either $f=f_0$ or $f=f_1$, which completes the proof.
Remark: Thanks to PeteL.Clark's nice question in the comment, I realized that the conclusion can be generalized to arbitrary finite field with the same argument as above.
Claim: Let $\mathbb{F}_q$ be the finite field of $q$ elements and let $v_1,\dots,v_m$ be nonzero vectors in $(\mathbb{F}_q)^n$. Then there exists a linear function $f:(\mathbb{F}_q)^n\to\mathbb{F}_q$, such that $$|\{1\le i\le m\mid f(v_i)\ne 0\}|\ge\frac{(q-1)m}{q}.\tag{5}$$
Sketch of Proof: Define projections $P:(\mathbb{F}_q)^n\to (\mathbb{F}_q)^{n-1}$, $Q:(\mathbb{F}_q)^n\to\mathbb{F}_q$ similarly. Define $m'$ and rearrange $v_1,\dots,v_m$ similarly. By induction, we may assume that there exists a linear function $g:(\mathbb{F}_q)^{n-1}\to\mathbb{F}_q$, such that for $f_j=g\circ P+j\cdot Q$, $0\le j<q$, $$|\{1\le i\le m'\mid f_j(v_i)\ne 0\}|\ge\frac{(q-1)m'}{q}.\tag{6}$$ By definition of $m'$ and $f_j$, for every $m'<i\le m$, $f_0(v_i),\dots, f_{q-1}(v_i)$ are pairwise different, so one and only one of them is $0$. It follows that for some $0\le j<q$, $$|\{m'< i\le m\mid f_j(v_i)\ne 0\}|\ge\frac{(q-1)(m-m')}{q}.\tag{7}$$ $(5)$ follows from $(6)$ and $(7)$ for $f=f_j$.
-
This is just lovely. – Kevin Carlson May 10 '13 at 9:00
@KevinCarlson: Thank you! The question itself is lovely. – 23rd May 10 '13 at 9:08
How does this generalize to other finite fields, I wonder? – Pete L. Clark May 10 '13 at 13:34
@PeteL.Clark: A simple observation based on this argument could show that for finite field $\mathbb{F}_q$, a lower bound is $\frac{(q-1)m}{q}$. Sketch of proof: we can define $f_j$, $0\le j\le q-1$ similarly as in $(2)$. By induction, we can assume that $(3)$ holds when $\frac{m'}{2}$ is replaced by $\frac{(q-1)m'}{q}$. When $i>m'$, $f_j(v_i)$ are pairwise different, so exactly only one of them is $0$. Then we can conclude that there exists $0\le j\le q-1$, such that $|\{m'<i\le m:f_j(v_i)\ne 0\}|\ge\frac{(q-1)(m-m')}{q}$, which completes the induction. – 23rd May 10 '13 at 14:01
@PeteL.Clark: Sorry, please ignore my first reply and see the current one. – 23rd May 10 '13 at 14:04
My coauthors and I needed this result recently (to prove something about pseudo-Anosov dilatations), and I was googling to see if it was known. I'm amazed that it was asked here so recently!
I'm answering now to record a short alternative proof that we found. I'll prove the same more general result that Landscape proved, namely:
Claim : Let $\vec{v}_1,\ldots,\vec{v}_m \in \mathbb{F}_q^n$ be nonzero vectors (not necessarily distinct). Then there exists a linear map $f : \mathbb{F}_q^n \rightarrow \mathbb{F}_q$ such that $f(\vec{v}_i) = 1$ for at least $\frac{q-1}{q}$ of the $\vec{v}_i$, i.e. such that $\{\text{$i|1 \leq i \leq m$,$f(\vec{v}_i) \neq 0$}\}$ has cardinality at least $\frac{q-1}{q}m$.
Proof : Let $\Omega$ be the probability space consisting of all linear maps from $\mathbb{F}_q^n \rightarrow \mathbb{F}_q$, each given equal probability. Let $\mathcal{X} : \Omega \rightarrow \mathbb{R}$ be the random variable that takes $f \in \Omega$ to the cardinality of the set $\{\text{$i|1 \leq i \leq m$,$f(\vec{v}_i) \neq 0$}\}$. We will prove that the expected value $E(\mathcal{X})$ of $\mathcal{X}$ is $\frac{q-1}{q} m$, which clearly implies that there exists some element $f \in \Omega$ such that $\mathcal{X}(f) \geq \frac{q-1}{q} m$.
To prove the desired claim, for $1 \leq i \leq m$ let $\mathcal{X}_i : \Omega \rightarrow \mathbb{R}$ be the random variable that takes $f \in \Omega$ to $1$ if $f(\vec{v}_i) \neq 0$ and to $0$ if $f(\vec{v}_i) = 0$. Viewing $\vec{v}_i$ as an element of the double dual $(\mathbb{F}_q^n)^{\ast \ast}$, the kernel of $\vec{v}_i$ consists of exactly $\frac{1}{q}$ of the elements of $(\mathbb{F}_q^n)^{\ast}$. This implies $E(\mathcal{X}_i) = \frac{q-1}{q}$. Using linearity of expectation (which recall does not require that the random variables be independent), we get that $$E(\mathcal{X}) = E(\sum_{i=1}^m \mathcal{X}_i) = \sum_{i=1}^m E(\mathcal{X}_i) = \sum_{i=1}^m \frac{q-1}{q} = \frac{q-1}{q} m,$$ as desired.
- | 2015-05-30T09:28:50 | {
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https://math.stackexchange.com/questions/2161581/how-many-1s-are-in-the-first-1023-binary-numbers/2161598 | # How many $1$s are in the first $1023$ binary numbers?
How many $1$s are in the first $1023$ binary numbers?
I'm not to sure how to approach this question. An idea, formula, or solution is appreciated!
• Have you noticed that $1023 = 2^{10}-1$? Feb 26 '17 at 0:20
• What is the "first" binary number? 0 or 1? Feb 26 '17 at 0:22
• Worth clarifying that we are talking about positive, unsigned integers.
– J...
Feb 26 '17 at 2:09
• @J... actually, can an unsigned integer be positive? Feb 26 '17 at 5:56
• @Joffan Cheeky bastard :P Yes, unsigned integers are positive because all of them (other than 0) are greater than 0. Feb 26 '17 at 9:35
Assuming the "first" binary number is $1$ note that the first $1023$ binary numbers, plus $0$, are all the binary numbers you can write with exactly $10$ binary digits or bits (prepending $0$s to "short" numbers, as in $0000101010_2$). Between all of them, you then have $1024 \cdot 10=10240$ bits, and for symmetry reasons exactly half of those, $5120$, are $1$s.
• why 1024*10 digits? Feb 26 '17 at 0:28
• Ohh, the amount of 0's don't matter! I see! Feb 26 '17 at 0:29
• It does matter, because if we start at 0, then 1111111111 should not be counted among the the first 1023 binary numbers. The question did not say the first 1023 positive binary numbers. Feb 26 '17 at 2:23
• In computer science, typically indexing start at 0. I was biased . I should have read the comments below the question... The issue was raised before. Feb 26 '17 at 2:52
• @MarDev We're looking at all of the integers we can represent with 10 bits - for every integer there is exactly one other which has all of its bits inverted (the result of an XOR with 1111111111) and therefore the opposite amount of zeros and ones. Furthermore, this inversion always lies in the opposite half of the possible numbers (the operation is essentially equivalent to 1111111111 - n) so each of the first 2^9 numbers (half of the possible 2^10) there exists exactly one number in the second half with the opposite number of zeros and ones. Feb 26 '17 at 8:46
Hint: For how many of those numbers will the one's bit be a $1$ (in other words: how many of those numbers are odd)? For how many of them will the two's bit be a $1$? For how many of them will the four's bit be a $1$? And so on. Also, it will probably be advantageous to include $0$ (and thus look at a collection of $1024$ binary numbers) to make the counting a bit easier. Or, if $0$ is already included, include $1023$ initially, then correct for it when you're done counting.
• I'm struggling with that idea right now. I'm not entirely sure how to solve that. Is it like half half? So if there are x digits in total, half of them are 1? Feb 26 '17 at 0:27
• On average, yes, half of the bits will be a $1$. Or something. Depends on what you mean by "of the bits". Feb 26 '17 at 0:31
Any such number can be represented by a string of 10 0s and 1s. The number of such strings with $n$ ones is $10$ choose $n$. Thus, the number of ones which appear is $$\sum_{n=0}^{10} n{10 \choose n}=5\cdot 2^{10}.$$
As an addendum to Arthur's answer - if you count to, say, $2^{4} - 1 = 15$, you can easily figure out a pattern in the columns:
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
So if you count to $2^{n} - 1$, you have $2^{n} * n$ digits ("rows * columns"), of which 50% are 1's.
For your example: Counting to $1023 =2^{10} - 1$, gets you $2^{10} * 10 * \frac{1}{2} = 5120$ binary 1's in total.
It's easy to calculate. There is a pattern with any number used as pow of tow:
You have 2^x, where x> 0 and you will have a binary number as:
2^0 = 1 (Dec) = 1 (Binary)
2^1 = 2 (Dec) = 10 (Binary)
2^2 = 4 (Dec) = 100 (Binary)
...
2^10 = 1024 (Dec) = 100 0000 0000(Binary)
If you pay attention you will notice this:
2^10 - 1 = 1023 (Dec) = 011 1111 1111(Binary)
...
2^2 - 1 = 3 (Dec) = 011 (Binary)
2^1 = 1(Dec) = 01 (Binary)
2^0 - 1 = 0 (Dec) = 0 (Binary)
In this case, your question, the Number of ones matches with the number that is used for pow. | 2021-10-26T00:09:03 | {
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http://www.tricitiesurgentcare.net/mozzarella-recall-xaojq/ba2b5c-invertible-matrix-theorem | # invertible matrix theorem
Let two inverses of A be B and C Services; Math; Blog ; About; Math Help; Invertible Matrix and It’s Properties. This section consists of a single important theorem containing many equivalent conditions for a matrix to be invertible. Invertible Matrix Theorem. Section 3.5 Matrix Inverses ¶ permalink Objectives. * The determinant of $A$ is nonzero. * $A$ has only nonzero eigenvalues. : An matrix is invertible if and only if has only the solution . structure theorem for completely bounded module maps. Invertible matrix 2 The transpose AT is an invertible matrix (hence rows of A are linearly independent, span Kn, and form a basis of Kn). Showing any of the following about an $n \times n$ matrix $A$ will also show that $A$ is invertible. (If one statement holds, all do; if one statement is false, all are false.) Proof: Let there be a matrix A of order n×n which is invertible. Any nonzero square matrix A is similar to a matrix all diagonal elements of which are nonzero. According to WolframAlpha, the invertible matrix theorem gives a series of equivalent conditions for an n×n square matrix if and only if any and all of the conditions hold. e. The columns of A form a linearly independent set. The Invertible Matrix Theorem Let A be a square n×n matrix. Let A be a square n by n matrix over a field K (for example the field R of real numbers). 1. Understand what it means for a square matrix to be invertible. Then a natural question is when we can solve Ax = y for x 2 Rm; given y 2 Rn (1:1) If A is a square matrix (m = n) and A has an inverse, then (1.1) holds if and only if x = A¡1y. We define invertible matrix and explain many of its properties. 1 The Invertible Matrix Theorem Let A be a square matrix of size N × N. The following statement are equivalent: • A is an invertible matrix. (d)Show that if Q is invertible, then rank(AQ) = rank(A) by applying problem 4(c) to rank(AQ)T. (e)Suppose that B is n … Then the following statements are equivalent. Yes. We will append two more criteria in Section 6.1. An invertible matrix is sometimes referred to as nonsingular or non-degenerate, and are commonly defined using real or complex numbers. abelian group augmented matrix basis basis for a vector space characteristic polynomial commutative ring determinant determinant of a matrix diagonalization diagonal matrix eigenvalue eigenvector elementary row operations exam finite group group group homomorphism group theory homomorphism ideal inverse matrix invertible matrix kernel linear algebra linear combination linearly … Theorem 1. 5.The columns of A are linearly independent. The Invertible Matrix Theorem|a small part For an n n matrix A, the following statements are equivalent. A is column-equivalent to the n-by-n identity matrix In. Let A be an n n matrix. A is an invertible matrix. Usually, when a set is written as the span of one vector, it’s one dimensional. Let A 2R n. Then the following statements are equivalent. • The columns of A form a linearly independent set. The matrix A can be expressed as a finite product of elementary matrices. Theorem (The QR Factorization) If A is an mxn matrix with linearly independent columns, then A can be factored as A=QR, where Q is an mxn matrix whose columns form an orthonormal basis for Col A and R is an nxn upper triangular invertible matrix with positive entries on the main diagonal. Theorem1: Unique inverse is possessed by every invertible matrix. The following statements are equivalent, i.e., for any given matrix they are either all true or all false: A is invertible, i.e. Some theorems, such as the Neumann Series representation, not only assure us that a certain matrix is invertible, but give formulas for computing the inverse. I. row reduce to! 4. The uniqueness of the polar decomposition of an invertible matrix. W. Sandburg [8] and Wu and Desoer [ … A has an inverse or is nonsingular. Dave4Math » Linear Algebra » Invertible Matrix and It’s Properties. Skip to content. 6.The linear transformation T defined by T(x) = Ax is one-to-one. Note that finding this matrix B is equivalent to solving a system of equations. The following statements are equivalent: A is invertible, i.e. A has an inverse, is nonsingular, or is nondegenerate. tem with an invertible matrix of coefficients is consistent with a unique solution.Now, we turn our attention to properties of the inverse, and the Fundamental Theorem of Invert- ible Matrices. lie in the commutants of d and 59’. The number 0 is not an eigenvalue of A. A matrix that has no inverse is singular. Here’s the first one. The Invertible Matrix Theorem (Section 2.3, Theorem 8) has many equivalent conditions for a matrix to be invertible. Introduction and Deflnition. Furthermore, the following properties hold for an invertible matrix A: • for nonzero scalar k • For any invertible n×n matrices A and B. Learn about invertible transformations, and understand the relationship between invertible matrices and invertible transformations. Menu. A is row-equivalent to the n-by-n identity matrix In. 2 det(A) is non-zero.See previous slide 3 At is invertible.on assignment 1 4 The reduced row echelon form of A is the identity matrix. When the determinant value of square matrix I exactly zero the matrix is singular. 5. (c)Showthatif P isaninvertiblem ×m matrix, thenrank(PA) = rank(A) byapplying problems4(a)and4(b)toeachofPA andP−1(PA). its nullity is zero. • A has N pivot positions. A is invertible.. A .. A is row equivalent to the n×n identity matrix. The Invertible Matrix Theorem has a lot of equivalent statements of it, but let’s just talk about two of them. Matrix I exactly zero the matrix must be a square n by n matrix over a field K for! A $3\times 3$ matrix is detailed along with characterizations the span of one vector, ’! 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Can only happen with full rank there be a square n by n matrix over a field K for... E.. F a is row equivalent to the n×n identity matrix the rest of polar... Append two more criteria in Section 6.1 another way of saying this is one of the must. = ~0 has no non-zero solutions as the span of one vector, ’. | 2021-03-07T18:39:07 | {
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https://math.stackexchange.com/questions/1451818/probability-that-the-sum-of-6-dice-rolls-is-even | # Probability that the sum of 6 dice rolls is even
Question:
6 unbiased dice are tossed together. What is the probability that the sum of all the dice is an even number?
I think the answer would be 50%, purely by intuition. However, not sure if this is correct. How should I go about solving such a problem?
Notice that whatever the sum of the first 5 rolls, whether the outcome is odd or even is totally determined by the last die. It is even or odd with equal probability, so the probability of an even sum is exactly the same as the probability of an odd sum.
The first 5 dice don't matter.
Statement:
Let $X_1,\ldots,X_n$ be independent, integer-valued random variables, and let some $X_k$ satisfy $P(X_k \mbox{ odd}) = 0.5$. Then $P(\sum X_i \mbox{ odd}) = 0.5$.
Proof:
Without loss of generality let $k=n$, and let $S = \sum_i^{n-1} X_i$. Since $S$ and $X_n$ are independent random variables, $$\begin{eqnarray*} P\left(\sum X_i \mbox{ odd}\right) = P(S + X_n \mbox{ odd}) &=& P(S \mbox{ odd and } X_n \mbox{ even}) + P(S \mbox { even and } X_n \mbox{ odd}) \\ &=& P(S \mbox{ odd})P(X_n \mbox{ even}) + P(S \mbox{ even})P(X_n \mbox{ odd}) \\ &=& P(S \mbox{ odd})P(X_n \mbox{ odd}) + P(S \mbox{ even})P(X_n \mbox{ odd}) \\ &=& \left(P(S \mbox{ odd}) + P(S \mbox{ even})\right)P(X_n \mbox{ odd}) \\ &=& P(X_n \mbox{ odd}) \\ &=& 0.5 \end{eqnarray*}$$
• o.O That's also a nice approach. Will the same apply for a dice with 7 sides? – Gummy bears Sep 26 '15 at 5:18
• @Gummybears no, because then you have inequal probabilities coming from the last die, so the parity of the sum you had before you roll it matters. However, this sort of thinking does apply in a pretty wide variety of cases, and it saves a lot of work when you notice it happening. – Eric Tressler Sep 26 '15 at 5:22
• @Gummybears: sorry, I missed that. However, what is not mentioned is that with an odd number of seven sided dice, there is a higher chance of an odd sum. Whereas with an even number of seven sided dice, there is a higher chance of an even sum. – robjohn Sep 26 '15 at 5:50
• @JoseAntonioDuraOlmos, the first 5 dice matter as much as the last one. They all matter the same here. It is not our neglect of the first 5 dice which leads us to the conclusion that the probability is $1/2$ but it is the symmetry of odd/even tosses. There is a great danger in basing the argument on the neglect, as one might be tempting to write it for the case where all dice are biased. – zhoraster Sep 26 '15 at 8:53
• There is no neglect. If I am presented a box with an unknown number of dices of unknown shapes all of them labeled with integers. I take only one out of the box. The dice has 6 sides, from 1 to 6, and is unbiased. Then I can claim that if I take all the dices, roll them and add their results the sum has a 50% probability of being even. I can claim it without examining the other dices in the box. And the proof of such claim is like the one in this answer. As for the case where all dice are biased, yeah this proof scheme does not work for that, don't fall into those temptations. – Jose Antonio Reinstate Monica Sep 26 '15 at 9:08
The only thing about the numbers rolled that matters is their parity - whether they are even or odd. In order to get an even sum, an even number of the six dice must be even. In order to get an odd sum, an odd number of the six dice must be even.
Using O for odd and E for even, we can list out the possibilities.
Even sum:
• OOOOOO $$\binom{6}{6}=1 \text{ arrangement}$$
• OOOOEE $$\binom{6}{4}=15 \text{ arrangements}$$
• OOEEEE $$\binom{6}{2}=15 \text{ arrangements}$$
• EEEEEE $$\binom{6}{0}=1 \text{ arrangement}$$
This gives a total of $32$ arrangements with even sum.
Since there are $2^6 = 64$ total possibilities, we see that your intuition of $50\%$ is correct.
• Lovely answer. Although slightly lengthy, it is simple to understand. Thank you. – Gummy bears Sep 26 '15 at 5:24
• +1 Good answer. I just have a wording suggestion: instead of saying "to get an even/odd sum, an even/odd number of the six dice must be even", I think wording it in terms of the number of odd dice gives more intuition and can generalize to any number of dice. The number of even dice doesn't really matter, the number of odd dice does. – jadhachem Sep 28 '15 at 5:39
The generating function approach:
$$P(x)=(1+x+x^2+x^3+x^4+x^5)^6=\sum a_i x^i$$
Then $a_i$ counts the number of ways of getting a total of $i+6$ from $6$ dice.
Now, to find the even terms, you can compute $$\frac{P(1)+P(-1)}{2}=\sum_i a_{2i}.$$
But $P(1)=6^6$ and $P(-1)=0$. So $$\frac{P(1)+P(-1)}{2}=\frac{6^6}{2},$$ or exactly half, as you conjectured.
For another example, let $N_{i}$ be the number of ways to roll $6$ dice and getting a value $\equiv i\pmod{5}$. Then it turns out that if $z$ is a primitive $5$th root of unity, then the value can be counted by defining:
$$Q_i(x)=x^{6-i}(1+x+x^2+x^3+x^4+x^5)^6$$ then computing $$N_i=\frac{Q_i(1)+Q_i(z)+Q_i(z^2)+Q_i(z^3)+Q_i(z^4)}{5}$$
This gives the result:
$$N_i =\begin{cases}\frac{6^6+4}{5}&i\equiv 1\pmod 5\\ \frac{6^6-1}{5}&\text{otherwise} \end{cases}$$
More generally, if $N_{n,i}$ is the number of ways to get $\equiv i\pmod 5$ when $n$ dice are rolled, you get:
$$N_{n,i} =\begin{cases}\frac{6^n+4}{5}&i\equiv n\pmod 5\\ \frac{6^n-1}{5}&\text{otherwise} \end{cases}$$
It's this simple because of the fact that $6=5+1$.
If each die has $d$ sides, and you ask what are the number of ways to get a total $\equiv i\pmod {d-1}$, then you get:
$$N_{d,n,i} =\begin{cases}\frac{d^n+{d-2}}{d-1}=\frac{d^n-1}{d-1}+1&i\equiv n\pmod {d-1}\\ \frac{d^n-1}{d-1}&\text{otherwise} \end{cases}$$
• I'm afraid I lost you at the last sentence. – Gummy bears Sep 26 '15 at 5:03
• That's okay, it is an advanced approach, but a useful one for problems like this. If you rolled $n$ dice, each with $7$ sides, you'd get the number of even results is: $$\frac{7^n + (-1)^n}{2}$$ would be even. – Thomas Andrews Sep 26 '15 at 5:09
• Aha..... That's a good approach. However, I'm not sure if I completely get it. Not sure how you used the probability of 1 and $-1$ to calculate the even terms. – Gummy bears Sep 26 '15 at 5:16
• If $i$ is odd, then $a_i\cdot 1^i + a_i(-1)^{i} = 0$. If $i$ is even, then $a_u\cdot 1^i + a_i(-1)^i = 2a_i$. So $P(1)+P(-1)=\sum_{i} 2a_{2i}$, which is twice the value you are looking for. @Gummybears – Thomas Andrews Sep 26 '15 at 5:19
• Aha.... I'm stupid, aren't I? This is perfect! It holds true for any sided dice and for any number of dice. – Gummy bears Sep 26 '15 at 5:23
Using the same logic given by @Eric tessler but writing the maths behind it.
Let $P_x$ = We get an even sum on the roll of x dice and
$1 - P_x$ = We get an odd sum.
So we require $P_6$
Now $P_6 = P_5 \times$ P[even number on the last dice] + $(1 - P_5) \times$ P[odd number on the last dice]
$P_6 = P_5 \times 0.5 + (1-P_5) \times 0.5$
$P_6 = 0.5$
• Exactly what @erictessler is missing - a rigurous proof. One by induction could work for all sorts of dice. – Tibos Sep 26 '15 at 23:22
• More upvotes needed! – Gummy bears Sep 27 '15 at 4:14
• Nicely done! +1 – Zubin Mukerjee Sep 27 '15 at 17:27
A lot of very informative answers but none of them do not explain your intuition, so I'll post mine.
There is a symmetry between all tosses with odd sum and those with even sum: just take the number $k$ on the first die and replace it with $7-k$.
So the numbers of odd tosses and even tosses are equal, therefore, the probability is $1/2$.
• I'm not getting you much.... elaborate please? – curiousbrain Sep 26 '15 at 5:22
• Same here. Please elaborate. – Gummy bears Sep 26 '15 at 5:23
• See the answer by @EricTressler, posted the same time. – zhoraster Sep 26 '15 at 5:24
Your intuition is correct. There could be a proof based on this intuition (using the fact that each die has 50% chance of being even and we're tossing 6 of them) but I'm not so sure.
But there are ways to prove it. For example, the only ways we could get an even sum are if:
1. All 6 dice show even numbers.
2. 4 dice show even and 2 dice show odd.
3. 2 dice show even and 4 dice show odd.
4. All 6 dice show odd.
The probability of cases 1 and 4 is $(\frac{1}{2})^6$ and the probability of cases 2 and 3 is $\binom{6}{2}(\frac{1}{2})^6=15(\frac{1}{2})^6$ (because we need to order the dice which show odd numbers.) Add them up and we get the probability is $$32(\frac{1}{2})^6=\frac{1}{2}.$$
The problem is about fair dice.
Whether the number of dice is 6 (even) or 7 (odd) $Pr = \dfrac12$
The logic is based on parity (odd/even). Since the probability that any particular throw is even or odd is equal at $\dfrac12$
(a) For an even number of dice, if you interchange odd and even throws, the parity remains the same, thus there will always be an equal number of odd and even sums.
EEEEEE OOOOOO even
EEEEEO OOOOOE odd
EEEEOO OOOOEE even
EEEOOO OOOEEE odd
(b) For an odd number of dice, every such interchange changes the parity, but by symmetry, again there will be an equal number of odd and even sums.
EEEEEEE even OOOOOOO odd
EEEEEEO odd OOOOOOE even
EEEEEOO even OOOOOEE odd
EEEEOOO odd OOOOEEE even
All outcomes are equally likely. Half the outcomes will have an odd number of odd dice faces and half of the outcomes will have an even number of odd dice faces. The outcomes with an odd number will have an odd total and the outcomes with an even number of odd face will be even. As there are equal numbers of outcomes for each case the probability for each case is 50%.
Okay so how do we know that half the outcomes will have on odd number of odd faces and half the number will have an even number of even faces? Well,we can make a one to one correspondence between outcomes with an odd number of odd faces to outcomes to an even number of odd faces. If outcome has an even number of odd faces, map it to the exact same outcome but the 6th die is one number higher (let's assume 1 is one number higher than 6). This outcome has an odd number of odd faces. If an outcome has an odd number of odd faces map it the exact same outcome but the 6th die is one number lower. This is a one to one correspondence. So there are an equal number of outcomes with an odd number of odd faces and there are outcomes with an even number of odd faces.
And that's it. Equal number of odd comes as even outcomes, each outcome equally likely, odd outcomes and even outcomes equally likely.
• Oops, Zhoraster beat me to it and did a much clearer and shorter job explaining it. Oh, well. – fleablood Sep 26 '15 at 23:27
• It was asked above about 7 sided die. I don't have enough reputation to comment there, so I'm going to comment here: If a 7-sided die has 4 odd faces and 3 even faces the probability of an even result with n die is 1/2 + (- 1/7)^n. This can be shown by induction by noting that each extra die has a 4/7 chance of changing parity. So if Pn is probability of an even total with n die. Then P(n+1) = Pn*3/7 + (1 - Pn)*4/7. as P1 = 3/7 = 1/2 + (- 1/7), the result follows by induction – fleablood Sep 26 '15 at 23:51 | 2019-11-15T07:15:58 | {
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https://math.stackexchange.com/questions/523944/show-that-if-a-and-b-are-sets-then-a-subseteq-b-if-and-only-if-a-cap-b | # Show that if $A$ and $B$ are sets, then $A \subseteq B$ if and only if $A \cap B = A$.
I'm working through a real analysis textbook, and it starts out with set theory. The first exercise is
Show that if $A$ and $B$ are sets, then $A \subseteq B$ if and only if $A \cap B = A$.
I think I proved it correctly but I'm not sure. Here's what I did. I proved that if $A \subseteq B$, then $A \cap B = A$ the same way as this answer did (https://math.stackexchange.com/a/446114/93114), but I want to make sure I proved the converse correctly because it seems really easy (yes it's the first problem in the book, but still) and math usually isn't this easy for me, even the basic stuff!
Proof of "If $A \cap B = A$, then $A \subseteq B$."
If $x \in A \cap B$, then $x \in A$ and $x \in B$, but this applies to all $x \in A$ because $A \cap B = A$. So, for any $x \in A$, we know that $x \in B$, so $A \subseteq B$.
Am I on the right track?
• Yes, although you're doing it in a convoluted way. Let $x\in A$; since $A=A\cap B$ we have $x\in B$. – egreg Oct 12 '13 at 22:27
• @egreg So if $x \in A$, then because $A = A \cap B$, we know that $x \in A \cap B$. By definition that implies that $x \in A$ and $x \in B$, which is enough to show that $A \subseteq B$. – M T Oct 12 '13 at 22:28
• @egreg Sorry I replied to your other comment that disappeared, but we're doing it the same way now. – M T Oct 12 '13 at 22:30
• I realized to have misread your argument, so I changed my first comment. – egreg Oct 12 '13 at 22:32
Since you are just starting, I would suggest to be verbose instead of pulling everything in a single sentence.
To prove $A \subseteq B$ iff $A \cap B = A$, you have to
1. show $A \cap B = A$ given $A \subseteq B$. That is to
1. show $A \cap B \subseteq A$.
2. show $A \subseteq A \cap B$.
2. show $A \subseteq B$ given $A \cap B = A$.
Proof:
1.1) It is trivially true. You don't need to be given $A \subseteq B$ for it to be true.
1.2) If $x \in A$, then $x \in B$ since we are given $A \subseteq B$. Then $x \in A$ and $x \in B$ are both true. Therefore, $x \in A \cap B$.
2) From $A \cap B = A$, we know $x \in A$ and $x \in B$ whenever $x \in A$. If $x \in A$, then it must be the case that $x \in B$. Therefore, $A \subseteq B$.
• Thanks for the answer. I think I'm slowly starting to catch the hang of these proofs. – M T Oct 13 '13 at 1:58
Suppose $A\cap B=A$. To prove $A\subseteq B$, suppose $x\in A$. Since $A\cap B=A$, $x\in A\cap B$ and thus $x\in B$. Since $x$ was an arbitrary element of $A$, this holds for all $x$ and thus $A\subseteq B$.
The above proves $A\cap B=A\implies A\subseteq B$. You will then need to prove the reverse implication to establish $A\cap B=A\iff A\subseteq B$.
Here is another way to approach this type of problem: start with the most complex part (here: $\;A \cap B = A\;$), expand the definitions so that you get from the set level to the element level, then simplify using predicate logic, and finally go back to the set level as suggested by the shape of the formula.
In this case, \begin{align} & A \cap B = A \\ \equiv & \;\;\;\;\;\text{"set extensionality, i.e., the definition of $\;=\;$ for sets"} \\ & \langle \forall x :: x \in A \cap B \equiv x \in A \rangle \\ \equiv & \;\;\;\;\;\text{"definition of $\;\cap\;$"} \\ & \langle \forall x :: x \in A \land x \in B \equiv x \in A \rangle \\ \equiv & \;\;\;\;\;\text{"logic: $\;p \equiv p \land q\;$ is one way to write $\;p \Rightarrow q\;$"} \\ & \langle \forall x :: x \in A \Rightarrow x \in B \rangle \\ \equiv & \;\;\;\;\;\text{"definition of $\;\subseteq\;$"} \\ & A \subseteq B \\ \end{align} | 2019-07-23T11:52:39 | {
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https://math.stackexchange.com/questions/2876327/show-that-a-certain-set-of-positive-real-numbers-must-be-finite-or-countable | # Show that a certain set of positive real numbers must be finite or countable
Let $B$ be a set of positive real numbers with the property that adding together any finite subset of elements from $B$ always gives a sum of $2$ or less. Show that $B$ must be finite or at most countable.
$B$ = {$x \in R:x>0\}$, $x_1,x_2...x_n \in B$ such that $x_1+x_2+...+x_n \le 2$.
Question: for any $a,b$ $(a,b)$~$R$, but $B$ is $(0,+\infty)$ so why $B$ is not uncountable (taking as $a = 0$, and letting $b$->$\infty$)?
And why for $B$ being countable doesn't contradict: for any $a,b$ $(a,b)$~$R$?
P.S. I read Showing a set is finite or countable and understood it.
• It sounds like you are confused about what the question is asking. It is not saying that B is the set of positive real numbers; it is a set of positive real numbers. This means that B is a subset of the set of all positive real numbers. Aug 8 '18 at 17:23
• No. B is some unspecified subset of the positive real numbers such that if you add up a finite number of the elements of B the sum is always less than 2. For example, B could be {0.1, 0.2, 0.5, 1.1} or B could be the infinite set {1, 1/2, 1/4, 1/8, 1/16, ...}. Note that B cannot be {1, 1.4, 1.8} and B cannot be {1, 1/2, 1/3, 1/4, ...} Aug 8 '18 at 17:47
• B does not have to be an interval (0,a). In fact, it cannot be such an interval. Your job is to prove that B must be finite or countable. Intervals are uncountable. Aug 8 '18 at 17:49
• Well, I'm just trying to help clarify the question. I haven't given a solution. I think the hints and discussion below do that for you. But the point is you cannot assume anything about B except: 1. its elements are all positive real numbers and 2. if you take a finite number of elements from B and add them up then you get a sum less than two. Given any such B you have to prove that B is either finite or countable. Aug 8 '18 at 17:59
• Here is a proof that B cannot be the open interval (0,1). If it were then these numbers would be in B: 0.5, 0.51, 0.52, and 0.53. These numbers add up to more than 2. Do you see? You could do something similar to show that B cannot be (0,a) for any positive a. But you have to show more than that. You have to show B is finite or countable. Aug 8 '18 at 18:00
Hint 1: How many elements of $B$ can be in the set $[2,\infty)$?
Hint 2: How many elements of $B$ can be in the set $[1,2)$?
Hint 3: How many elements of $B$ can be in the set $[0.5,1)$?
• At this rate you are going to need countably many hints :) Aug 8 '18 at 17:04
• @ArnaudMortier Can't he give a hint schema? Aug 8 '18 at 17:07
• 1) One $x = 2$, 2) set $x_1=1$ so one for sure, then I don't know how many $x_2,x_3,...x_n$ to choose so that they sum up to $1$. 3) The same, I don't know. Aug 8 '18 at 17:10
• Obviously $B\cap[1,\infty)$ has at most two elements, because if $x,y,z\ge1$ then $x+y+z>2$. Similarly $B\cap[2/n,\infty)$ has at most $n$ elements. (So $B$ is the union of countably many finite sets...) Aug 8 '18 at 17:17
• No, we certainly can't write $B$ as $(0,a)$. If $B=(0,a)$ then there are finite subsets of $B$ with sum larger than $2$. Aug 8 '18 at 18:56
Not only can we show that it's countable, it's not too difficult to construct an enumeration: given an element $b$, just count how many other elements of $B$ are larger. This has to be a finite integer, since if there were an infinite number of elements greater than $b$, then the sum of $n$ such elements would be greater than $nb$, so picking an $n>2/b$ would give a sum greater than 2.
For any $a\in B$, define $B_a\equiv\{b\in B:a<b\}$. The cardinality of $B_a$ must be less than $\lceil2/a\rceil$, otherwise the sum of any $\lceil2/a\rceil$ elements of $B_a$ would exceed 2. That is, $\sum_{i=1}^{\lceil2/a\rceil}x_i>\sum_{i=1}^{\lceil2/a\rceil}a=a\lceil2/a\rceil\ge2$ for any sequence $x_i$ in $B_a$, contradicting the requirements on $B$.
Define the function $f:B\rightarrow\mathbb{N}$ as $f(a)=\left|B_a\right|$.
Homework: If you can show that $f$ is one-to-one then you can conclude that $|B|\le|\mathbb{N}|$. That is, $B$ is either finite or countable.
Credit goes to Acccumulation for the outline of this proof.
• What if $a =$ sup$B$? Then $B_a$ is the empty set. If $B_a \equiv \{b \in B: a \leq b \}$, then the sum of any $\lceil r/a \rceil$ for $r > 2$ elements of $B_a$ would exceed 2. Jul 18 '19 at 23:08 | 2021-10-17T17:33:25 | {
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https://math.stackexchange.com/questions/2687454/converse-of-uniqueness-of-universal-property-of-tensor-products | Converse of Uniqueness of Universal Property of Tensor Products?
In Atiyah Macdonald, for a commutative and unital ring $A$ the tensor product of $A$-modules, $M,N$ is defined by the usual universal property:
A pair $(T,g)$ with $T$ an $A$-module and $g$ an $A$-bilinear map $g\colon M \times N \to T$ such that for any $A$-module $P$ with an $A$-bilinear map $f\colon M \times N \to P$ there exists a unique $A$-module homomorphism $h\colon T \to P$ such that $h \circ g = f$.
And then, as usually goes with universal properties:
Moreover, if $(T,g), (T',g')$ are any two pairs satisfying this condition, then there is a unique isomorphism $j \colon T \to T'$ such that $j \circ g = g'$
My question is, in the second highlighted selection, is the converse true? If I have already established the existence of the usual tensor product, $M \otimes_A N$, via quotient of free module, and I encounter some other pair $(T', g')$ and a unique isomorphism $j: T \to T'$ such that $j \circ g = g'$, can I conclude that $T'$ also satisfies the universal property defining the tensor product, thus I can regard both $T$ and $T'$ as the tensor product of this $M$ and $N$?
I believe I proved that I can indeed do this, but I need reassurance because I have overlooked small details with tensor products before, and don't want to make that mistake again. My proof was as follows:
Suppose $(T,g)$ satisfies the universal property and $(T',g')$ is a pair where $T'$ is an $A$-mod and $g'$ an $A$-bilinear map $g' \colon M \times N \to T'$. Suppose $j$ is the unique isomorphism $j\colon T \to T'$ such that $j \circ g = g'$, or $g = j^{-1} \circ g'$. Suppose $P$ is any $A$-mod with an $A$-bilinear map $f\colon M \times N \to P$. We wish to show there exists a unique $A$-module homomorphism $\phi\colon T' \to P$ such that $\phi \circ g' = f$.
Since $(T,g)$ already satisfies the property by hypothesis, we know there exists a unique $A$-module homomorphism $\varphi \colon T \to P$ such that $\varphi \circ g = f$. So define the map $\phi \colon T' \to P$ by $\varphi \circ j^{-1}$. Then $\phi \circ g' = \varphi \circ j^{-1} \circ g' = \varphi \circ g = f$. So the diagram commutes as desired and the map $\phi$ is defined as the composition of two unique maps so is unique as well, thus $(T',g')$ satisfies the universal property.
Sorry that got a little (unnecessarily) complicated, I don't know how to draw diagrams on here, which really makes everything easier. As an aside, I feel like this probably is true and Atiyah & Macdonald didn't mention it in the text because they deemed it trivial.
• I think this is the first time I've ever seen someone use $\phi$ and $\varphi$ as distinct variable names at the same time. – Eric Wofsey Mar 12 '18 at 6:39
• I do it all the time! – Prince M Mar 12 '18 at 15:08
This is correct, although your argument for uniqueness of $\phi$ is not very clear: you chose to define it as a composition with certain maps that are unique, but how do you know you couldn't get a $\phi'$ that works and is not given by those compositions? What you want to say is that if $\phi'$ makes the diagram commute for $(T',g')$, then the map $\varphi'=\phi'\circ j$ would make the diagram commute for $(T,g)$ by just swapping the roles of $(T,g)$ and $(T',g')$ in your argument. Therefore by uniqueness of $\varphi$, $\varphi=\varphi'$, and so $\phi'=\varphi'\circ j^{-1}=\varphi\circ j^{-1}=\phi$.
Intuitively, what's going on here is that the isomorphism $j$ says that $(T,g)$ and $(T',g')$ have exactly the same structure (just the elements of $T$ have to be renamed via $j$ to get $T'$). So any property of $(T,g)$ (which is defined using only the module structure on $T$ together with the map $g$) will also be true of $(T',g')$. In particular, this applies to the universal property of the tensor product.
• Ok, thanks. One last question I had was, I just now wrote the whole argument out to ensure I comprehend the entirety of it and I cannot identify where the uniqueness (w/ respect to commuting diagrams) of the isomorphism j is invoked? I see plainly that this theorem fails if it is not unique, because there will be at least 2 distinct ways to construct the map $/Phi$, but I can’t seem to identify, where in the uniqueness argument, we actually invoke the assumption that j is unique? – Prince M Mar 12 '18 at 22:16
• You don't need uniqueness of $j$, just its existence. If $j$ exists, it turns out to automatically be unique. – Eric Wofsey Mar 12 '18 at 22:17 | 2019-06-20T01:40:20 | {
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https://cs.stackexchange.com/questions/133977/the-number-of-arrangements-of-n-players-around-a-round-table-where-each-player | # the number of arrangements of N players around a round table, where each player can sit on one of 3 contiguous chairs
Consider the fact that each player can either sit on their desired chair or on the neighbouring chair. Two configurations are distinct if at least one person is sitting in another chair. My attempt For example, if there are 4 players sitting around a circular table there can be 9 distinct arrangements.
1. F1,F2,F3,F4
2. F2,F1,F3,F4
3. F2,F1,F4,F3
4. F4,F2,F3,F1
5. F1,F3,F2,F4
6. F1,F2,F4,F3
7. F4,F1,F2,F3
8. F2,F3,F4,F1
9. F4,F3,F2,F1
Now, I am stuck here. How can calculate this and what about larger values of N such as 50?
I tried using vectors and pushing back every element and do the reverse operators and swap elements but I am not sure whether this is correct or not.
• Do you want to enumerate the arrangements or simply get the number? What have you tried? Jan 4, 2021 at 20:34
• Can you credit the source where you originally encountered this task?
– D.W.
Jan 4, 2021 at 20:36
• @D.W. No I cant this is a competetive programming question that was asked in a contest Jan 4, 2021 at 21:59
• Is the contest currently ongoing? When was the deadline for the contest? What prevents you from providing a link and/or reference to the programming contest? Sometimes helping us see the original statement of the question helps us understand what you are asking better. Also, please don't leave clarifications in the comments (such as regarding enumerating vs counting): instead, edit the question to clarify it.
– D.W.
Jan 4, 2021 at 22:08
• Please edit the question to make it clear whether the desired chairs of different players can be the same or not. Jan 6, 2021 at 6:09
Consider the variant of the problem in which the people are arranged in a segment instead of circle and let $$S(n)$$ be the number feasible arrangements for $$n$$ people.
Now lets go back to your original problem, and let's name $$C(n)$$ the number of feasible arrangements of $$n$$ people.
Clearly $$C(0)=C(1)=1$$ and $$C(2)=2$$, so we will henceforth suppose that $$n \ge 3$$.
Let's name the people $$p_1, \dots, p_n$$ where the assigned position of $$p_i$$ is $$i$$. If $$p_1$$ is sitting in position 1 then the number of possible arrangements of $$p_2, \dots, p_n$$ is exactly $$S(n-1)$$. Otherwise, $$p_1$$ is sitting either in position $$2$$ or in position $$n \neq 2$$. We will only consider the former case since the latter one is symmetric. Clearly $$p_2$$ cannot sit in position $$2$$, so he/she must either sit in position $$1$$ or in position $$3$$. If $$p_2$$ is sitting in position $$1$$, then the number of possible arrangements of $$p_3, \dots, p_n$$ is exactly $$S(n-2)$$. Otherwise, $$p_2$$ must be sitting in position $$3$$, which means that $$p_3$$ must be sitting in position $$4$$ (since position $$2$$ and $$3$$ are occupied) and, in general, $$p_i$$ must be sitting in position $$(i \bmod n) + 1$$. Since this completely determines everyone's position, this case only contributes $$1$$ to the total number of configurations.
To summarize, we have that: $$C(n) = S(n-1) + 2(S(n-2)+1) = S(n-1) + 2S(n-2) + 2.$$
We are left with figuring out what $$S(n)$$ is. Clearly $$S(0) = S(1) = 1$$, so we consider $$n \ge 2$$. In this case $$p_1$$ can only sit in position $$1$$ or $$2$$. If $$p_1$$ is sitting in position $$1$$ then there are $$S(n-1)$$ possible arrangements of $$p_2, \dots, p_n$$. If $$p_1$$ is sitting in position $$2$$, then $$p_2$$ must be sitting in position $$1$$ (as it is the only other person that can sit there) and there are $$S(n-2)$$ possible arrangements for $$p_3, \dots, p_n$$. We hence have: $$S(n) = \begin{cases} 1 & \mbox{if } n \in \{0,1\} \\ S(n-1) + S(n-2) & \mbox{otherwise} \end{cases} = \mathcal{F}_{n+1},$$ where $$\mathcal{F}_{i}$$ denotes the $$i$$-th Fibonacci number. Substituting back, we obtain:
$$C(n) = \begin{cases} 1 & \mbox{if } n \in \{0,1\} \\ 2 & \mbox{if } n = 2 \\ \mathcal{F}_n + 2 \mathcal{F}_{n-1} + 2 & \mbox{otherwise} \end{cases}.$$
• we can also say that C(n) = F_(n+1) + F_(n-1) + 2 since C(n) = S(n) + S(n-2) + 2 Jan 6, 2021 at 13:51
• Sure, they are equivalent :) Jan 6, 2021 at 15:23 | 2022-07-01T02:12:38 | {
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http://mathhelpforum.com/calculus/119701-help-surface-integrals.html | # Math Help - [HELP!] Surface Integrals
1. ## [SOLVED] Surface Integrals
I don't know how to use the MATH code here...and I'm running out of time for reviewing for finals, so forgive me if it isn't clear for you... >.<
OK...here's my question:
DoubleInt(s) (z+x^2y)dS, S is the part of the cylinder y^2 + z^2 = 1 that lies between the planes x = 0 and x = 3 in the first octant.
I know that maybe...I should use
y = rcos(theta) = cos(theta)
z = rsin(theta) = sin(theta)
x = x
and find r_(theta) x r_(z) = < cos(theta), sin(theta), o> (I'm not sure if this way is the right path...)
and find the magnitude of |r_(theta) x r_(z)| = 1
I don't know what to do next, which formula to use (there were two...) and what boundaries? How can I get the numbers for those boundaries?
Can anyone solve this question for me please? I know the final answer, which is 12.
2. Hi. By definition:
$S=\mathop\int\int\limits_{\hspace{-15pt}S} g(x,y,z)dS=\mathop\int\int\limits_{\hspace{-15pt}R} g(x,y,f(x,y))\sqrt{f_x^2+f_y^2+1} dA$
So you're integrating the function $g(x,y,z)=z+x^2 y$ over the surface $f(x,y)=z=\sqrt{1-y^2}$ between the planes x=0 and x=3 in the first quadrant. That's just the top one-quarter of the cylinder along the x-axis shown in red below. Now, the region in the x-y plane below that part, since the cylinder has a radius of one, would be 0 to 3 in the x-direction and 0 to 1 in the y-direction. Now, turn the crank:
$\int_0^3\int_0^1 (\sqrt{1-y^2}+x^2y)\sqrt{f_x^2+f_y^2+1}dydx$
3. Originally Posted by violet8804
I don't know how to use the MATH code here...and I'm running out of time for reviewing for finals, so forgive me if it isn't clear for you... >.<
OK...here's my question:
DoubleInt(s) (z+x^2y)dS, S is the part of the cylinder y^2 + z^2 = 1 that lies between the planes x = 0 and x = 3 in the first octant.
I know that maybe...I should use
y = rcos(theta) = cos(theta)
z = rsin(theta) = sin(theta)
x = x
and find r_(theta) x r_(z) = < cos(theta), sin(theta), o> (I'm not sure if this way is the right path...)
and find the magnitude of |r_(theta) x r_(z)| = 1
I don't know what to do next, which formula to use (there were two...) and what boundaries? How can I get the numbers for those boundaries?
Can anyone solve this question for me please? I know the final answer, which is 12.
Here's a general concept: if you write the surface as $\vec{r}(u,v)= x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k}$, then the two derivatives, $\vec{r}_u= x_u \vec{i}+ y_u\vec{j}+ z_u\vec{k}$ and $\vec{r}_v= x_v\vec{i}+ y_v\vec{j}+ z_v\vec{k}$ are tangent to the surface and their cross product (called the "fundamental vector product" of the surface), $\vec{r}_u\times\vec{r_v}$, is perpendicular to the surface and its length gives the "differential of surface area": $||\vec{r}_u\times\vec{r}_v||dudv$.
Yes, you can use $y= cos(\theta)$, $z= sin(\theta)$, x= x with $\theta$ and x as parameters u and v. That is, $\vec{r}= x\vec{i}+ cos(\theta)\vec{j}+ sin(\theta)\vec{k}$ so $\vec{r}_x= \vec{i}$ and $\vec{r}_\theta= -sin(\theta)\vec{j}+ cos(\theta)\vec{k}$ so that the "fundamental vector product" is
$\vec{r}_x\times\vec{r}_\theta= -cos(\theta)\vec{i}- sin(\theta)\vec{k}$.
Its length is, of course, 1 so $dV= dxd\theta$. The function to be integrated is $z+ x^2y= sin(\theta)+ x^2cos(\theta)$. In the first octant, both y and z are positive so $\theta$ goes from 0 to $\pi/2$ and, of course, x goes from 0 to 1. The integral is
$\int_{x=0}^1\int_{\theta= 0}^{\pi/2} sin(\theta)+ x^2 cos(\theta) dxd\theta$.
By the way, if you click on a mathematical expression in any post, you can see the LaTex code used.
4. Originally Posted by HallsofIvy
Here's a general concept: if you write the surface as $\vec{r}(u,v)= x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k}$, then the two derivatives, $\vec{r}_u= x_u \vec{i}+ y_u\vec{j}+ z_u\vec{k}$ and $\vec{r}_v= x_v\vec{i}+ y_v\vec{j}+ z_v\vec{k}$ are tangent to the surface and their cross product (called the "fundamental vector product" of the surface), $\vec{r}_u\times\vec{r_v}$, is perpendicular to the surface and its length gives the "differential of surface area": $||\vec{r}_u\times\vec{r}_v||dudv$.
Yes, you can use $y= cos(\theta)$, $z= sin(\theta)$, x= x with $\theta$ and x as parameters u and v. That is, $\vec{r}= x\vec{i}+ cos(\theta)\vec{j}+ sin(\theta)\vec{k}$ so $\vec{r}_x= \vec{i}$ and $\vec{r}_\theta= -sin(\theta)\vec{j}+ cos(\theta)\vec{k}$ so that the "fundamental vector product" is
$\vec{r}_x\times\vec{r}_\theta= -cos(\theta)\vec{i}- sin(\theta)\vec{k}$.
Its length is, of course, 1 so $dV= dxd\theta$. The function to be integrated is $z+ x^2y= sin(\theta)+ x^2cos(\theta)$. In the first octant, both y and z are positive so $\theta$ goes from 0 to $\pi/2$ and, of course, x goes from 0 to 1. The integral is
$\int_{x=0}^1\int_{\theta= 0}^{\pi/2} sin(\theta)+ x^2 cos(\theta) dxd\theta$.
By the way, if you click on a mathematical expression in any post, you can see the LaTex code used.
Thank you!
And, by the way...
Is the (theta) always from 0 to 2*Pi if it's in the 1st octant (after change to polar coordinates)?
Also, can you please teach me when to use which of the two formula, is there a trick to see from the question when to use which formula?
I always get confused which one to use, they are all the homework questions, and i got partial solutions (the library only have the older version of the solution manual...); so, when I do the question myself, and checked from the solution manual, it used a different formula (or different method) to solve the surface integral...I'm so confused T_T | 2015-07-05T18:39:19 | {
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https://math.stackexchange.com/questions/2432215/prove-forall-n-ge-2-2n-3n-4n/2432219 | # Prove $\ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$
Question:
Prove $\ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$
My attempt:
Base case is trivial.
Suppose $\ n \ge 2$ and $\ 2^{n} + 3^{n} < 4^{n}$
Then,
$2^{n+1} + 3^{n+1} = 2.2^{n} + 3.3^{n} = 2.2^{n} + 2.3^{n} + 3^{n} = 2(2^{n} + 3^{n}) + 3^{n} <2(4^{n}) + 3^{n}$, by I.H.
I am stuck here. how do I show that this expression is $< 4^{n+1}$?
• Hints: Is $3^n<4^n$? Is $3(4^n)<4(4^n)$? – Michael Burr Sep 16 '17 at 20:25
• Another proof could use that the equivalent inequality $(2/4)^n+(3/4)^n<1$ after dividing both sides by $4^n$, where $4^n>0$, holds because it holds for $n=2$ and $(2/4)^n$ and $(3/4)^n$ are both strictly decreasing. – user236182 Sep 16 '17 at 20:32
You can say $$2\cdot 2^n + 3\cdot 3^n < 4\cdot 2^n + 4\cdot 3^n = 4(2^n+3^n) < 4\cdot 4^n = 4^{n+1}.$$
multiplying $$2^n+3^n<4^n$$ by $$4^n$$ we get $$2^n\cdot 2^{2n}+3^n\cdot 2^{2n}<4^{n+1}$$ and we have $$2^n\cdot 2<2^{3n}$$ since $$1<2^{2n-1}$$ for $n\geq 2$ and $$3^n\cdot 3<3^n\cdot 2^{2n}$$ since $$3<2^{2n}$$ therefore we have $$2^{n+1}+3^{n+1}<2^n\cdot 2^{2n}+3^n\cdot 2^{2n}<4^{n+1}$$
Hint: for $\,0 \lt a \lt 1\,$ the sequence $\,a^n\,$ is decreasing with $\,n\,$ since $\,a^{n+1} = a \cdot a^n \lt 1 \cdot a^n\,$, so:
$$\left(\frac{2}{4}\right)^n+\left(\frac{3}{4}\right)^n \le \left(\frac{2}{4}\right)^2+\left(\frac{3}{4}\right)^2 = \frac{13}{16} \lt 1 \quad\quad\text{for}\;\; \forall n \ge 2$$ | 2019-06-16T11:51:29 | {
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https://math.stackexchange.com/questions/2071403/minimal-specification-of-isometry-in-terms-of-norm-preservation | # Minimal specification of isometry in terms of norm preservation
Let $V,W$ be $n$-dimensional (real) inner product spaces, and let $T:V \to W$ be a linear map.
Let $v_1,...,v_n$ be a basis of $V$. It is easy to see that if $|T(v)|_W=|v|_V$ for every $v \in \{v_1,...,v_n,v_1+v_2,v_1+v_3,...,v_{n-1}+v_n\}$, then $T$ is an isometry (a proof is provided below).
In other words, after choosing wisely $k(n):=\frac{n(n+1)}{2}$ vectors, it is enough to verify $T$ preserves the norms of these special vectors, in order to conclude it's an isometry.
Question: Is there no way to choose less than $k(n)$ vectors, in such a way that every linear map which preserves their norms is an isometry?
I believe we cannot choose less vectors. I have some "convincing evidence" for the cases $n=1,2,3$ (see below), but I am not sure how to give a rigorous argument.
Note that a "wise choice" of vectors does not have to be of the form of some vectors, and linear combinations of them (I do feel this it the most efficient method, but I don't see how to prove this). Even if we prove that this is the case, than we need to show we cannot do better than to work with only orthonormal bases.
The partial "evidence":
$n=1: k=1$. Obvious
$n=2: k=3$. Take $V=W=\mathbb{R}^n$ with its standard inner product. Then, $T(e_1)=e_1, T(e_2)=\frac{e_1+e_2}{\sqrt 2}$ is a counter example.
$n=3: k=6$. Then any matrix of the form $$\begin{pmatrix} c & s & x \\ -s & c & y \\ 0 & 0 & z \\\end{pmatrix}$$ where $c^2+s^2=1,x^2+y^2+z^2=1, sx+cy=0$ preserves the norms $e_1,e_2,e_1+e_2,e_3,e_2+e_3$ but it's an isometry only if $|z|=1,x=y=0$.
Proof that $k(n)=\frac{n(n+1)}{2}$ vectors are enough:
Noting that $$\langle u,v \rangle = \frac{1}{2}(|u+v|^2 - |u|^2 - |v|^2) ,$$ we obtain
$$\langle Tv_i,Tv_j \rangle = \frac{1}{2}(|Tv_i+Tv_j|^2 - |Tv_i|^2 - |Tv_j|^2) = \frac{1}{2}(|T(v_i+v_j)|^2 - |v_i|^2 - |v_j|^2)$$ $$= \frac{1}{2}(|v_i+v_j|^2 - |v_i|^2 - |v_j|^2) = \langle v_i,v_j \rangle,$$
thus $T$ is an isometry.
• I think there is not a way to choose less. Since it must preserve the scalar product and then all products $<v_i,v_j>$ of the basis must be preserved. By symmetry they are exactly the number you proposed. – Harnak Dec 25 '16 at 11:02
• @Harnak, please make your comment into an answer, I'd like to +1 it. – Andreas Caranti Dec 25 '16 at 11:09
• @Harnak I am not sure I am convinced by your argument. Can you please elaborate? (I also think there is no way to choose less). – Asaf Shachar Dec 25 '16 at 11:13
Let $f:\mathbb R^n\to V$ and $g:W\to\mathbb R^n$ be any two linear isometries. Then $T$ is an isometry if and only if $g\circ T\circ f$ is an isometry. So, we may assume without loss of generality that $V=W=\mathbb R^n$ equipped with the Euclidean inner product.
Let $v_1,\ldots,v_k\in\mathbb R^n$ be $k<\frac12n(n+1)$ arbitrarily chosen vectors. It suffices to exhibit the existence of a non-isometric linear transformation $T$ on $\mathbb R^n$ that preserves the norm of each $v_j$. Consider the system of homogeneous linear equations $$v_j^\top Av_j=0,\quad j=1,\ldots,k,\tag{1}$$ where the $n^2$ entries of $A\in M_n(\mathbb R)$ are unknown. Since the subspace of all $n\times n$ skew-symmetric matrices has dimension $\frac12n(n-1)<n^2-k$, the system $(1)$ must admit a nontrivial solution $A$ that is not skew-symmetric. However, if $A$ is a solution, so is $A+A^\top$. Therefore, $(1)$ admits a nontrivial symmetric solution $A$.
Let $P=I+\varepsilon A$, where $\varepsilon>0$ is sufficiently small. Then $P$ is positive definite. Define $Tx=\sqrt{P}x$. Then $T$ is not an isometry because $\sqrt{P}$ is not real orthogonal. However, for each $j$ we have $$\|Tv_j\|^2=v_j^\top Pv_j=v_j^\top(I+\varepsilon A)v_j=v_j^\top v_j=\|v_j\|^2.$$
• Wow! your solution is great and provides exactly what I was interested in, thanks. By the way, I am curious how did you think about using symmetric matrices. (I guess you started from considering perturbations of the identity, but then noticed that if you do not take a square root, the constraint you get is a quadratic equation, not linear, hence the necessity to consider roots...). – Asaf Shachar Dec 28 '16 at 12:53
• @AsafShachar Once you recognise that the problem boils down to finding a positive semidefinite matrix $P\ne I$ such that $v_j^\top Pv_j=v_j^\top v_j$ for each $j$, it's natural to consider symmetric matrices, because $I-P$ is symmetric. – user1551 Dec 28 '16 at 13:27
• I agree, but it was not immediate for me to consider the "square root" trick. (i.e I thought of starting with $Tx=Px$ where $P=I+\epsilon A$, but then the equation on $P$ is quadratic...). Anyway, your explanation is very reasonable, thanks again. – Asaf Shachar Dec 28 '16 at 16:52
I'll try to elaborate on my idea in the comment. Preserving the scalar product means that the two symmetric bilinear forms must coincide: $$(x,y) \mapsto \langle x,y \rangle$$ $$(x,y) \mapsto \langle Tx,Ty \rangle$$ This is equivalent to saying that their matrices must coincide. But a symmetric matrix has $\frac{n(n+1)}{2}$ degrees of freedom.
This means, once we've chosen the basis in $V$ we must choose the products $\langle v_i,v_j \rangle$ for $i \geq j$.
• Is it enough? Yes, because we've chosen the coefficients of the matrix relative to its canonic basis (here I'm taking about the basis of matrices which have $1$ in one component and $0$ in the others, but only for $i \geq j$ since the matrix is symmetric).
• Can we take less? No, because the symmetric matrices space has dimension $\frac{n(n+1)}{2}$ as we've said.
How does this relate to our question? Choosing a basis for $V$ and then such products is equivalent to choosing the list of vectors you proposed (along with their norms).
(I.e. Choosing a bilinear form is equivalent to choosing the quadratic forum associated to it, which in the case of a scalar product is the norm)
• Thanks for your elaborate answer. However, I must say I am not entirely convinced by your argument. Indeed, I am also thinking that we cannot choose less vectors, and your observation that a (symmetric) bilinear form is determined by $\frac{n(n+1)}{2}$ vectors is also appealing and looks connected to this issue somehow. In spite all of that I am still not convinced. Suppose we know $\langle v_i,v_j \rangle = \langle Tv_i,Tv_j \rangle$ for all $i \le j$ except for a single pair $(i^*,j^*)$... – Asaf Shachar Dec 26 '16 at 15:34
• I would like to see a more explicit demonstration for how to build such a map $T$ which is not an isometry, i.e does not satisfy $\langle v_i,v_j \rangle = \langle Tv_i,Tv_j \rangle$ for $(i,j)=(i^*,j^*)$. I am not sure this follows so clearly from your "heuristic" argument. – Asaf Shachar Dec 26 '16 at 15:34
• So, you want to prove that there exists a map $T$ that preserves all those couples except for one? – Harnak Dec 26 '16 at 16:49
• Yes, for a start. My intention in the question was to show that no matter how cleverly you choose $k < \frac{n(n+1)}{2}$ vectors, and require they will be mapped to vectors of the same norm, you will be able to do this with a map which is not an isometry. However, in some sense this will only show the scheme of choosing basis vectors (and certain linear combinations of them) cannot be successful with less vectors, we also need to show we cannot do better by any choice. – Asaf Shachar Dec 26 '16 at 19:50 | 2019-05-20T19:22:12 | {
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https://www.qalaxia.com/questions/What-is-the-probability-of-this-problem | D
#### What is the probability of this problem?
63 viewed last edited 2 years ago
Anshul Malik
1
A and B roll a dice each. A wins if his dice number is greater than B. B wins if his dice number is greater than or equal to A.
If A gets a 1, then B will win for sure. So B makes up a rule that A will play the same game again if A rolls a 1 in the first throw. The winner of the second game (if it reaches that stage) will be the final winner even if A rolls a 1 again.
What is the probability that A will win this game?
Mahesh Godavarti
3
I think the problem statement can be worded better for easier understanding. However, I think I got it.
Probability of A winning = 1 - Probability of B winning.
Probability of B winning =
Probability of B throwing an equal or greater than when A throws a number greater than 1 in the first throw +
Probability that A throws a 1 in the first throw X Probability of B getting an equal or higher number in the second throw than A =
\frac{1}{6 } \times (\frac{5}{6} + \frac{4}{6} + \frac{3}{6} + \frac{2}{6} + \frac{1}{6}) + \frac{1}{6} \times \frac{1}{6} \times (\frac{6}{6} + \frac{5}{6} + \frac{4}{6} + \frac{3}{6} + \frac{2}{6} + \frac{1}{6}) =
\frac{1}{6} \times \frac{1}{6} \times 5 \times 6 \times \frac{1}{2} + \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times 6 \times 7 \times 1/2 = 1/6 \times 1/6 \times 1/2 \times (30 + 7) = \frac{37}{72}
Therefore, probability of A winning = 35/72.
Earlier, if B had not changed the rule then the probability of B winning would be
1/6 \times (6/6 + 5/6 + 4/6 + 3/6 + 2/6 + 1/6) = 1/6 \times 1/6 \times 6 \times 7 \times 1/2 = 7/12 .
And the probability of A winning would be 5/12.
Essentially, B leveled the playing field a little bit by changing the rules of the game.
Vivekanand Vellanki
0
It would be good to explain how the probability of B throwing a number greater than or equal to A is determined.
Anshul Malik
0
But the cases in which A rolls a 1 in the first roll should not be counted as sample cases right? Also, the probability that the game is decided in the first roll is 5/6? If I'm understanding conditional probability correctly, then this should also be factored in right?
Mahesh Godavarti
0
I just swt up the sample space and added up all the cases that end up in B's victory. You could set it up as a conditional as well, with A rolling a 1 or not rolling a 1. Or the game finishing in one throw or finishing in two throws, you will get the same answer. In fact, I will set it up and show you that it is the same. Look for my second answer.
Mahesh Godavarti
0
Let's do it another way. Let's condition it first on the event that the game will finish in one throw or two throws.
Then P(B wins) = P(one throw) P(B wins | one throw) + P(two throws) P(B wins | two throws).
P(one throw) = P(A throws a number other than 1) = 5/6
P(two throws) = P(A throws a 1) = 1/6.
P(B wins | one throw) = P(A throws a 2 | A threw a number from 2 to 5) X P(B throws 2 or higher) + P(A throws a 3 | A threw a number from 2 to 5) X P(B throws 3 or higher) + and so on till A throws a 6 = 1/5 * 5/6 + 1/5 * 4/6 + 1/5 * 3/6 + 1/5 *2/6 + 1/5 * 1/6.
Therefore, P(one throw) P (B wins | one throw) = 5/6 * [1/5 * 5/6 + 1/5 * 4/6 + 1/5 *3/6 + 1/5 *2/6 + 1/5 * 1/6 ] = 1/6 * 5/6 + 1/6 *4/6 + 1/6 * 3/6 + 1/6 * 2/6 + 1/6 *1/6 = 1/6 * [5/6 + 4/6 + 3/6 + 2/6 + 1/6].
P(B wins | two throws) = P(A throws a 1 in the second throw | A threw a number from 2 to 5 in the first throw) X P(B throws 1 or higher in the second throw) + P(A throws a 2 in the second throw | A threw a number from 2 to 5 in the first throw) X P(B throws 2 or higher in the second throw) + and so on till A throws a 6 in the second throw
Since, the throws are independent what A throws in the second is independent of what is thrown in the first. Therefore, the above expression is the same as:
P(B wins | two throws) = P(A throws a 1 in the second throw) X P(B throws 1 or higher in the second throw) + P(A throws a 2 in the second throw) X P(B throws 2 or higher in the second throw) + and so on till A throws a 6 in the second throw = 1/6 * 6/6 + 1/6 *5/6 + 1/6 * 4/6 + 1/6 * 3/6 + 1/6 * 2/6 + 1/6 * 1/6
Therefore, P (two throws) P(B wins | two throws) = 1/6 * [1/6 * 6/6 + 1/6 *5/6 + 1/6 * 4/6 + 1/6 * 3/6 + 1/6 * 2/6 + 1/6 * 1/6] = 1/6 * 1/6 * [6/6 + 5/6 + 4/6 + 3/6 + 2/6 + 1/6]
Therefore, P(B wins) = 1/6 * [5/6 + 4/6 + 3/6 + 2/6 + 1/6] + 1/6 * 1/6 * [6/6 + 5/6 + 4/6 + 3/6 + 2/6 + 1/6].
Which is the same as what we had in the other answer. | 2020-11-28T19:05:36 | {
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http://math.stackexchange.com/questions/198641/using-induction-prove-that-sum-k-1n-frac-1-2k-12k1-frac-12-frac | # Using induction, prove that $\sum_{k=1}^n\frac 1 {(2k-1)(2k+1)}= \frac 12-\frac 1 {4n+2}$ when $n$ is in $\mathbb{N}$
I can prove the base case and get that $1/3=1/3$, but i can't get any further with the $n+1$ case. Can someone help me?
I was told to conjecture a formula for the sum $$\frac{1}{3} + \frac{1}{{3 \cdot 5}} + \cdots + \frac{1}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}}$$ I thought I figured out that this sum was equal to $$\frac{1}{2} { - \frac{1}{{4n + 2}}}$$ but I'm starting to think I'm wrong about that. After we have the formula, we were told to prove our conjecture using induction.
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I really can't understand what is the identity the you want to prove. Could you try to typeset it (better if in LaTeX), with all the necessary parentheses, please? – Andrea Orta Sep 18 '12 at 18:25
Are you sure you have the right expression? For $n = 2$ you get: $\frac{1}{2 \cdot 6} = \frac{1}{2} - \frac{1}{10}$ which is clearly untrue. – Feanor Sep 18 '12 at 18:25
If $n=1$ you are dividing by $0$. Also please check your parentheses: one extra left on the left side, it is not clear whether $(2n+2)$ should be in the numerator, and you must mean $1/(4n+2)$, not $(1/4)n+2$ or $1/(4n)+2$ on the right. \frac is your friend in this regard. – Ross Millikan Sep 18 '12 at 18:25
Also, unless there was supposed to be some sum involved, there is no need for induction. Just multiply everything out so as to get rid of fractions ;) – Feanor Sep 18 '12 at 18:27
Based on your added comment, you want the left to be $\sum_{i=1}^n \frac 1{2i-1}\frac 1{2i+1}=\frac 12 - \frac 1{4n+2}$. Note the $1$'s, not $2$'s and the sum on the left. – Ross Millikan Sep 18 '12 at 18:37
Hint $\$ The sum telescopes since for $\rm\:f(k) = 1/(4k-2)\:$ we have
$$\rm f(k+1)-f(k)\ =\ \frac{1}{4k+2} - \frac{1}{4k-2}\ =\ -\frac{1}{(2k-1)(2k+1)}$$
so an inductive proof is a special case of the inductive proof of the closed form for a telescopic sum:
$$\rm\sum_{k=1}^n f(k\!+\!1)-f(k) = - f(1) + \color{#C00}{f(2) - f(2) + \cdots + f(n)-f(n)} + f(n\!+\!1)\, =\, f(n\!+\!1)-f(1)$$
But it is easy to turn the above ellipses into a rigorous inductive proof. Then your problem is simply a corollary of this general telescopy lemma, for the special value of $\rm\:f(k)\:$ given above.
The advantage of proving it this way is that - with no extra effort - you now have a general lemma that works to prove (by induction!) all summation identities of this form. Note that even though the induction has been abstracted out into a proof of a more general telescopy lemma, a proof invoking the telescopy lemma still counts as a proof by induction. The induction simply has been encapsulated in the proof of the lemma, which need not be repeated inline every time it is invoked. DivideAbstract and conquer - once you've seen one telescopic induction proof you've seen them all!
You can find more examples and further discussion in my prior posts on telescopy.
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@Peter Telescopy can be viewed as a form of induction specifically tailored for computing certain sums and products. – Bill Dubuque Sep 18 '12 at 19:00
Yes, I was just being silly, Bill! – Pedro Tamaroff Sep 18 '12 at 19:03
What you want to prove, apparently, is that
$$\sum_{k=1}^n \frac{1}{(2k-1)(2k+1)}=\frac 1 2 - \frac 1 {2(2n+1)}$$
As you say, the base case, $n=1$ is true. Suppose true for $n$, and analyze $n+1$. We get
$$\sum_{k=1}^{n+1} \frac{1}{(2k-1)(2k+1)}=\frac 1 2 - \frac 1 {2(2(n+1)+1)}$$
$$\sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)}+\frac{1}{(2(n+1)-1)(2(n+1)+1)}=\frac 1 2 - \frac 1 {2(2(n+1)+1)}$$
By the inductive hypothesis, this is, $$\frac 1 2 - \frac 1 {2(2n+1)}+\frac{1}{(2(n+1)-1)(2(n+1)+1)}=\frac 1 2 - \frac 1 {2(2(n+1)+1)}$$
$$- \frac{1}{{2(2n + 1)}} + \frac{1}{{(2n + 1)(2n + 3)}} = - \frac{1}{{2(2n + 3)}}$$
Multiply through ${(2n + 1)(2n + 3)}$ to get
$$- \frac{{(2n + 3)}}{2} + 1 = - \frac{{(2n + 1)}}{2}$$
In the end, you get
$$- 2n - 3 + 2 = - 2n - 1$$ $$- 2n -1 = - 2n - 1$$
which is true, so the inductive step is complete, and the theorem is proven.
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Hint: This is a telescoping series. Note that $\frac 1{2i-1}\frac 1{2i+1}=\frac 12 \left(\frac 1{2i-1}-\frac 1{2i+1}\right)$ so neighboring terms cancel.
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While this is the easiest approach, is not solving the problem the way the problem asks ;) – N. S. Sep 18 '12 at 18:43
@N.S.: I think it is a step toward the induction. – Ross Millikan Sep 18 '12 at 18:45
@RossMilikan he already guessed the right result, so he has a very easy induction problem, where he doesn't really need telescoping. – N. S. Sep 18 '12 at 18:49
@N.S. Telescopy does use induction - it's simply encapsulated (as a lemma) in the general closed form for a telescopic sum (whose inductive proof is easier in the abstract than in any specific case). So the natural way to prove this is to prove by induction the closed form for a telescopic sum, then specialize it to the case at hand - see my answer. Then, later, one can reuse the telescopy lemma to quickly tackle all such problems. – Bill Dubuque Sep 18 '12 at 19:09
@BillDubuque True, and this is the easiest way for us... But the problem asks the student to conjecture a closed formula, and prove it by induction. If you telescope it, you already get the answer... Is like the problem of calculating $\int_0^1 x^2 dx$... We typically expect the students to solve it the easy way, but if the Question explicitly asks to evaluate it using a Riemann sum, the easy solution is not good anymore... – N. S. Sep 18 '12 at 20:53
show 1 more comment
Saw this type of problem before.
You need to find a closed for for
$$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5}+...+\frac{1}{(2n-1)(1n+2)}$$
and then prove it by induction.
Your guess looks right, now you can do the induction:
$$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5}+...+\frac{1}{(2n-1)(1n+2)} =\frac{1}{2}-\frac{1}{2n+1} \,.$$
Alternately, you can use partial fraction decomposition to find a simple formula for $$\frac{1}{(2k-1)(2k+1)} \,,$$ and then get a telescopic sum. There is no need to prove this partial fraction decomposition by induction, and this approach is not really a direct induction proof.
- | 2014-07-22T21:45:02 | {
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https://math.stackexchange.com/questions/1110860/when-does-the-two-cars-meet | # When does the two cars meet
At 10:30 am car $A$ starts from point $A$ towards point $B$ at the speed of $65$ km/hr, at the same time another car left from point $B$ towards point $A$ at the speed of $70$ km/hr, the total distance between two points is $810$ km, at what time does these two cars meet ?
This is what I have tried,
$t_1 = \frac{810}{65} \,$ km/hr $= 12.46 \,$hr
$t_2 = \frac{810}{70} \,$ km/hr $= 11.57\,$ hr
$t_1-t_2 = 0.89 \,$ hr
$t_1-t_2 = 0.89\cdot 60 = 53.4 \,$ minutes
but this couldn't be the answer because how could these two cars could meet after $53.4$ minutes ?
@Joe, when they start they are 810km apart; after 1 hour the person leaving from point A would have traveled 65km, so if the other person had not left from point B, then the distance between them would be 810-65=745km. But the other person also helps close the distance between them by traveling 70km in the 1st hour from point B towards A, so the distance between them is 810-(65+75)=810-135=675km. So they are effectively subtracting 135km worth of distance per hour from the original distance between them.
So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
• but at what time does these two cars meet ? would that be 135/2 ? – Joe Jan 19 '15 at 18:27
• like 135/2 = 67.5 and then 810/67.5 = 12 ? – Joe Jan 19 '15 at 18:29
• no. it would be how long it would take to travel 810km at a rate of 135km/hr. Because every hour, they get 135km closer to each other. – turkeyhundt Jan 19 '15 at 18:50
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:54
• Yes. very good. – turkeyhundt Jan 19 '15 at 18:59
Hint: The total distance between the two cars is initially $810~\text{km}$. Thus, in order to meet, the combined distance they travel is $810~\text{km}$. Since they are travelling toward each other, they get $65~\text{km} + 70~\text{km} = 135~\text{km}$ closer to each other each hour.
• so then I have two divide 135 by 2 and then divide the answer by 810, like this, 135/2 = 67.5 then 810/67.5 = 12hours right ? so they will meet at 10:30 pm right ? – Joe Jan 19 '15 at 18:48
• No. What you did is compute the time it would take for the average distance travelled by the cars to be $810~\text{km}$, by which time they would have long passed each other. The cars get $135~\text{km}$ closer to each other each hour. You want to reduce the distance between them, which is initially $810~\text{km}$, to $0~\text{km}$. – N. F. Taussig Jan 19 '15 at 18:51
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:53
• That is correct. – N. F. Taussig Jan 19 '15 at 18:54
• They do not have to cover $810~\text{km}$ individually. After six hours, the slower car has travelled $$65~\frac{\text{km}}{\text{h}} \cdot 6~\text{h} = 390~\text{km}$$ so it is now $810~\text{km} - 390~\text{km} = 420~\text{km}$ from where the faster car began. Since the faster car has travelled $$70~\frac{\text{km}}{\text{h}} \cdot 6~\text{h} = 420~\text{km}$$ toward the slower car, they meet at 4:30 pm. Draw a picture. – N. F. Taussig Jan 19 '15 at 19:03 | 2019-06-26T08:56:03 | {
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https://math.stackexchange.com/questions/1166269/lucas-numbers-and-fibonacci | Lucas numbers and fibonacci
This is a question straight from the Applied Combinatorics book.
Suppose that chairs are arranged in a circle. Let $L_n$ count the number of subsets of $n$ chairs which don't contain consecutive chairs. show that $$L_{n+1} = F_n + F_{n+2}.$$
What is meant here? Could some one explain this to me?
• Do you need a proof, or do you have difficulties to understand the statement ? Feb 26, 2015 at 10:35
• What confuses me is that the number of chairs is not given. Feb 26, 2015 at 10:37
• Proof and explanation please. I don't understand what is meant by the question. Feb 26, 2015 at 10:56
If $n$ chairs are arranged in a circle, let $a_n$ be the number of subsets of the chairs that don’t contain consecutive chairs. (I’ll call these good subsets.) Clearly $a_0=1$, $a_1=2$, $a_2=3$, and $a_3=4$, since the empty subset satisfies the condition.
Now suppose that $n\ge 4$, and number the chairs from $1$ to $n$ around the circle. Temporarily remove chair $n$; the circle of remaining chairs has $a_{n-1}$ good subsets, none of which contains both chair $1$ and chair $n-1$. Of course none of these contains chair $n$, either. Moreover, every good subset of the $n$ chairs that does not contain chair $n$ and contains at most one of chairs $1$ and $n-1$ is counted here, so there are $a_{n-1}$ such subsets.
Now we’ll count the good subsets that contain either contain chair $n$ or contain both chair $1$ and chair $n-1$. Start by removing chairs $n-1$ and $n$; the remaining circle has $a_{n-2}$ good subsets. Let $S$ be one of these good subsets. If chair $1$ is in $S$, form $S'$ by adding chair $n-1$ to $S$; if chair $1$ is not in $S$, form $S'$ by adding chair $n$ to $S$. It’s not hard to check that in each case $S'$ is a good subset of the ring of $n$ chairs, and that every such subset that contains either chair $n$ or both of chairs $1$ and $n-1$ is obtained in this way. (It’s here that we need to have $n\ge 4$: in order to form $S'$ this way when chair $1$ is in $S$, we need to know that $n-1>2$.)
This argument shows that the sequence $\langle a_n:n\in\Bbb N\rangle$ satisfies the recurrence
$$a_n=a_{n-1}+a_{n-2}\tag{0}$$
for $n\ge 4$. By inspection $a_2=3=2+1=F_3+F_1$ and $a_3=4=3+1=F_4+F_2$, so a straightforward induction shows that $$a_n=F_{n+1}+F_{n-1}\tag{1}$$ for $n\ge 2$:
\begin{align*} a_{n+1}&=a_n+a_{n-1}\\ &=\left(F_{n+1}+F_{n-1}\right)+\left(F_n+F_{n-2}\right)&\text{induction hypothesis}\\ &=\left(F_{n+1}+F_n\right)+\left(F_{n-1}+F_{n-2}\right)\\ &=F_{n+2}+F_n\;. \end{align*}
Apart from the fact that I used $a_n$ instead of $L_n$ for the number of good subsets of a ring of $n$ chairs, $(1)$ is exactly relationship that you were to prove.
Nothing in the wording of the question or the original tagging indicates that $L_n$ is supposed to be the $n$-th Lucas number: $L_n$ is explicitly defined to be the number of good subsets of a ring of $n$ chairs. (The tag was added by Martin Sleziak.) The choice of the notation $L_n$ for my $a_n$ in the problem statement was very likely motivated by the fact that $a_n$ is the $n$-th Lucas number, $L_n$, for $n>1$. This follows immediately from the fact that the sequence $\langle L_n:n\in\Bbb Z^+\rangle$ of Lucas numbers by definition satisfies the same recurrence $(0)$ as the sequence $\langle a_n:n\in\Bbb N\rangle$, with $L_1=1$ and $L_2=3=a_2$: $L_3=4=a_3$, and the recurrence then ensures that $a_n=L_n$ for all $n\ge 2$.
• Thanks for the clear explaination! Mar 1, 2015 at 11:41
• @Discreteballoons: My pleasure! Mar 1, 2015 at 11:55
The Lucas numbers are defined recursively by \begin{align*} L_1 & = 1\\ L_2 & = 3\\ L_n & = L_{n - 1} + L_{n - 2}, n \geq 3 \end{align*} The first few terms of the Lucas sequence are $\{1, 3, 4, 7, 11, 18, 29, 47, 76, 123, \ldots\}$.
The Fibonacci numbers are defined recursively by \begin{align*} F_1 & = 1\\ F_2 & = 1\\ F_n & = F_{n - 1} + F_{n - 2}, n \geq 3 \end{align*} The first few terms of the Fibonacci sequence are $\{1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots\}$.
We can prove the assertion that $L_{n + 1} = F_n + F_{n + 2}$ by induction on $n$. Let $P(n)$ be the statement that $L_{n + 1} = F_n + F_{n + 2}$.
If $n = 1$, then $L_2 = L_{1 + 1} = 3 = 1 + 2 = F_1 + F_3$. Thus, $P(1)$ holds.
IF $n = 2$, then $L_3 = L_{2 + 1} = 4 = 1 + 3 = F_2 + F_4$. Thus, $P(2)$ holds.
Assume $L_{m + 1} = F_m + F_{m + 2}$ for each $n \leq m$, where $m \geq 2$. Let $n = m + 1$. Then \begin{align*} L_{m + 2} & = L_{m + 1} + L_{m} && \text{by definition of the Lucas sequence}\\ & = F_{m} + F_{m + 2} + F_{m - 1} + F_{m + 1} && \text{by the induction hypothesis}\\ & = F_{m} + F_{m - 1} + F_{m + 2} + F_{m + 1}\\ & = F_{m + 1} + F_{m + 3} && \text{by definition of the Fibonacci sequence} \end{align*} so $P(m) \Rightarrow P(m + 1)$. Hence, $P(n)$ holds for each positive integer $n$.
Let $a_m$ denote the number of subsets of $m$ chairs in which no two chairs are consecutive. We will consider the first few cases, where the chairs are numbered from $1$ to $m$.
\begin{align*} & m = 1: \emptyset, \color{blue}{\{1\}}\\ & m = 2: \color{red}{\emptyset, \{1\}}, \color{blue}{\{2\}}\\ & m = 3: \color{red}{\emptyset, \{1\}, \{2\}}, \color{blue}{\{3\}}\\ & m = 4: \color{red}{\emptyset, \{1\}, \{2\}, \{3\}}, \color{blue}{\{4\}}, \color{green}{\{1, 3\}}, \color{blue}{\{2, 4\}}\\ & m = 5: \color{red}{\emptyset, \{1\}, \{2\}, \{3\}, \{4\}}, \color{blue}{\{5\}} \color{red}{\{1, 3\}}, \color{green}{\{1, 4\}}, \color{red}{\{2, 4\}}, \color{blue}{\{2, 5\}, \{3, 5\}}\\ & m = 6: \color{red}{\emptyset, \{1\}, \{2\}, \{3\}, \{4\}, \{5\}}, \color{blue}{\{6\}}, \color{red}{\{1, 3\}, \{1, 4\}}, \color{green}{\{1, 5\}}, \color{red}{\{2, 4\}, \{2, 5\}}, \color{blue}{\{2, 6\}}, \color{red}{\{3, 5\}}, \color{blue}{\{3, 6\}, \{4, 6\}}, \color{green}{\{1, 3, 5\}}, \color{blue}{\{2, 4, 6\}}\\ \end{align*}
Based on these examples, we see that \begin{align*} a_1 & = 2\\ a_2 & = 3\\ a_3 & = 4\\ a_4 & = 7\\ a_5 & = 11\\ a_6 & = 18 \end{align*} Thus, the assertion that $a_1 = L_1$ is false. However, it appears that $a_m = L_m$ if $m > 1$.
Let $m = n + 1$ for $n \geq 1$.
I have used three colors to highlight the subsets of $m = n + 1$ chairs that do not contain consecutive chairs. Subsets marked in blue contain chair $n + 1$. Subsets marked in red in row $n + 1$ are those subsets that appeared in row $n$. Subsets marked in green contain both $1$ and $n$.
Observe that the elements in blue in row $n + 1$ are formed by taking the union of elements in row $n - 1$ that do not contain $1$ with the set $\{n + 1\}$, while the elements in green are formed by taking the union of elements in row $n - 1$ that do contain $1$ with the set $\{n\}$. Hence, the total number of elements in row $n + 1$ that are either blue or green is $L_{n - 1}$. The number of elements in row $n + 1$ that are red is the number of elements in row $n$, which is $L_n$. Hence, for each $n \geq 1$, \begin{align*} L_{n + 1} & = L_n + L_{n - 1}\\ & = F_{n - 1} + F_{n + 1} + F_{n - 2} + F_n\\ & = F_{n - 1} + F_{n - 2} + F_{n + 1} + F_n\\ & = F_{n} + F_{n + 2} \end{align*} We can see that the assertion is false when $n = 0$ since the empty set in row $n + 1 = m = 1$ is not a member of one of the three types of highlighted sets.
While I did not use this observation in my proof, it appears that for $n \geq 1$ that the number of sets containing $n + 1$ in row $n + 1$ is $F_n$, while the number of sets containing $1$ and $n$ in row $n + 1$ is $F_{n - 2}$ (where we define $F_0 = F_2 - F_1 = 0$ and $F_{-1} = F_1 - F_0 = 1 - 0 = 1$).
Are you sure of the formulation of your problem? I am wondering if the good one would not be: "Suppose that $n$ chairs are arranged in a circle. Let $L_n$ count the number of subsets of $k \geq 1$ chairs which don't contain consecutive chairs, show that..."
• Please see this tutorial on how to format mathematics on this site. Feb 26, 2015 at 12:46
• I have just checked it, thank you. Feb 26, 2015 at 13:50 | 2022-08-17T01:46:08 | {
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https://math.stackexchange.com/questions/1619550/calculating-the-surface-area-of-revolution-for-parametric-equation?noredirect=1 | # Calculating the surface area of revolution for parametric equation.
I solved a problem using a method that's completely different from the mark scheme and I got the right answer, but I'm unsure whether or not it's just some coincidence. Here's the question:
The curve $C$ has parametric equations $$x=t^2, y=\frac{1}{4}t^4-\ln{t}$$ for $1\leq t\leq 2$. Find the are of the surface generated when $C$ is rotated $2\pi$ radians about the y-axis.
Here's what I did ( not very rigorous):
I thought cutting the curve with horizontal cuts should do the trick. So I just evaluated $2\pi\int _{t=1}^2xdy$ since $dy=(t^3-\frac{1}{t})dt$ it's just $$2\pi\int_1^2(t^5-t)dt$$ which is fairly easy to find.
Here's what's in the mark scheme:
$x'=2t$ and $y'=t^3-\frac{1}{t}$
$(\frac{ds}{dt})^2=x'^2+y'^2=(t^3-\frac{1}{t})^2$
$S=\int 2\pi x ds=2\pi\int_1^2(t^5-t)dt$
My understanding is that both answers slice $C$ differently and so maybe this was just some coincidence. I can't prove that both methods are equivalent though. So if someone can give me an idea as to why they are that'd be great. Also, it seems to me that the method I used was way easier, is there a place where it's better to use that other method?
• If you examine the mark scheme's computation, you'll find a difference of sign with what's in your post: $x' = 2t$ and $y' = (t^{3} - \frac{1}{t})$, so $$\left(\frac{ds}{dt}\right)^{2} = x'(t)^{2} + y'(t)^{2} = \left(t^{3} + \tfrac{1}{t}\right)^{2},$$ n.b., not $\left(t^{3} - \tfrac{1}{t}\right)^{2}$, which would be equal to $y'(t)^{2}$ itself. If the mark scheme really has a minus sign on the right, that's an error. | 2019-12-06T13:50:21 | {
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# Tag Info
## Hot answers tagged nonlinear-equations
19
There are two issues that you are likely to be encountering. Ill-conditioning First, the problem is ill-conditioned, but if you only provide a residual, Newton-Krylov is throwing away half your significant digits by finite differencing the residual to get the action of the Jacobian: $$J[x] y \approx \frac{F(x+\epsilon y) - F(x)}{\epsilon}$$ If you ...
12
Short answer If you only want second order accuracy and no embedded error estimation, chances are that you'll be happy with Strang splitting: half-step of reaction, full step of diffusion, half step of reaction. Long answer Reaction-diffusion, even with linear reaction, is famous for demonstrating splitting error. Indeed, it can be much worse, including "...
11
This is the stochastic root-finding problem, as in The stochastic root-finding problem: Overview, solutions, and open questions.
11
Is the shooting method the only general numerical method for solving BVP of nonlinear ODE(s)? No. Most other methods consist of three parts: Discretization. This may be done with finite differences, finite volumes, finite elements (Galerkin or collocation), spectral methods, and so forth. This reduces the problem from an infinite-dimensional one to a ...
9
If you can stand to use complex arithmetic, simultaneous iteration methods might be preferable for computing all the roots of your polynomial. The simplest simultaneous iteration method, the (Weierstrass-)Durand-Kerner method, is effectively equivalent to applying Newton-Raphson to the Vieta relations relating the coefficients and roots of a polynomial, ...
9
I assume, that you have conducted a space discretization, so that you are about solving the (vector-valued) ODE $$\dot u_h(t) = F_h(t,u_h(t)), \text{ on [0,T] }, u_h(0) = \alpha.$$ via a numerical scheme $\Phi$ that advances the approximation $u_h^n$ at the current time instance $t=t^n$ to the next value $u_h^{n+1}$ at $t=t^{n+1}:=t^n+\tau$. Then your ...
8
You can solve this numerically in Python without symbolic computation. from __future__ import print_function, division import numpy as np from numpy import exp from scipy.integrate import quad from scipy.optimize import root def f1(a1, a2, x): return exp(a1 * x + a2 * x * x * x) / (1 + x * x) def f2(a1, a2, x): return exp(a1 * x + a2 * x * x * x) *...
8
PDEs are a form of dynamical system where there is another continuous variable. Usually this is space, so you're looking at how things over time and space instead of just over time. Here's an example of generalizing an ODE to a PDE. Take your ODE model of chemical reactions which models the concentration of certain chemicals over time. Now generalize the ...
8
Julia has a whole ecosystem for generating sparsity patterns and doing sparse automatic differentiation in a way that mixes with scientific computing and machine learning (or scientific machine learning). Tools like SparseDiffTools.jl, ModelingToolkit.jl, and SparsityDetection.jl will do things like: Automatically find sparsity patterns from code Generate ...
7
To answer your questions in order: Any implicit method for solving an ordinary differential equation involves solving a system of nonlinear equations. You can do this through variants of Newton's method, successive substitution, full approximation scheme, or any other approach that solves systems of nonlinear equations. (The caveat is, of course, depending ...
7
What you describe as your time discretization is called the Crank-Nicolson scheme. For nonlinear differential equations it leads, as you have observed, to a nonlinear algebraic system that needs to be solved at each time step. The typical approach is to solve it with a Newton method -- in your case, that requires to solve a nonlinear system in 5 variables. ...
7
As pointed out by David Ketcheson, one method is to use the companion matrix and find its eigenvalues (this is what Matlab does for the roots function). However, if you want to code everything by yourself, you can try to use the Sturm sequences, which you use as a first step to find an interval with only one zero. Then, you can apply one standard methods ...
7
For your example equation, taking the average approach, the local consistency error $$\frac{1}{h^2}[u(x-h) - 2 u(x) + u(x+h)]-f(\frac{1}{2}[u(x-h)+u(x+h)]) = \frac{1}{2}f_uu_{xx}h^2 + hot.$$ will be of order $2$ (instead of order $3$). ($hot.$ means higher order terms) Therefore, if your overall approximation is of order $1$, e.g. if you use upwind ...
7
This is somehow unexpected, but my recent experience with solving a system of nonlinear equations is that treating them as the right hand side of a system of ordinary equations and then evolve the system with an ODE solver can be considerably faster than with the usual Newton-Raphson iteration. It sounds like you're doing some sort of pseudotransient ...
7
The issues you're running into now are not a failing of Newton-Raphson, but a question of coupling. You're doing iterated sequential coupling -- solving each equation sequentially and then iterating until (hopeful) convergence. No solver choice in place of NR is going to fix this lack of convergence, as long as you are doing iterated sequential coupling. ...
7
It's a bit easier to see if you write your equation in the a semi-discretised system of the form $u^{\prime}(t) = F(u(t))$ and with the application of the $\theta$-method and approximating $u^{\prime}(t) \approx (w^{n+1} - w^{n})/\tau$ this gives, $$w^{n+1} - w^{n} - (1-\theta) \tau F(w^n) - \theta\tau F(w^{n+1}) = 0$$ with unknown vector $w^{n+1}$ and ...
7
$s_k$ is the "approximate Newton" search direction. So in essence, when they say Choose $s_k$ such that $\|F(x_k)+F'(x_k)s_k\| \le \eta_k \|F(x_k)\|$ they are saying: Solve the Newton system $F'(x_k)s_k = - F(x_k)$ inexactly for $s_k$ until the norm of the residual $F(x_k)+F'(x_k)s_k$ is smaller than the norm of the right hand side $-F(x_k)$ by a ...
6
If you need Jacobian matrix information for a numerical method, you should calculate the Jacobian matrix of the discretized form of the equations, since that will be consistent with the discretized equations you are solving.
6
You should rather think of what you will need in the following step, which is probably the numerical time integration of the semi-discrete equations. If you are going to use a (semi) explicit time stepping scheme, all you need is a function that for a given $\phi_0$ assembles the vector $\langle (\nabla \phi_0)^2, v \rangle$, where $v$ are your test ...
6
$K u$ equals the internal forces only in the linear case. The tangent stiffness matrix, $K$, in a nonlinear problem is normally used in a Newton-Raphson algorithm to calculate updates to the displacement vector as follows: $$K \Delta u = f - f_{internal}$$ $$u_{i+1} = u_i + \Delta u$$ The vector of internal forces, $f_{internal}$ must be calculated from ...
6
If you change variables to optimize for the residual of the linear part, then the Hessian will be a low-rank update to the identity. Then L-BFGS would work very well. Specifically, your problem takes the form $$\min_x \frac{1}{2}\|Ax-b\|^2 + \frac{\mu}{2}\|g(x)\|^2$$ where $Ax=b$ is the linear PDE and $g$ is the nonlinear part, and $\mu$ is a tradeoff ...
6
The reason is that GMRES can only be used for solving linear equations, i.e. equations of the form $Ax=b$, where $A$ is some matrix and $x,b$ are vectors. What GMRES does, essentially, is it approximates multiplication by the matrix $A^{-1}$ using a matrix polynomial of $A$. In this case (I assume) $f(y^{n+1},t)$ is not necessarily linear in the vector $y^{... 6 Two systematic ways of smoothing a function$h$would be: 1. Join the piecewise smooth parts of your function using Hermite interpolation so that the derivatives are matched to your satisfaction. 2. Convolve your function$h(x)$with a heat kernel of the form$f(x) = \frac{\exp\left\{-\frac{x^2}{2 \sigma^2}\right\}}{\sqrt{2 \pi \sigma^2}}$so that instead ... 5 That's actually very easy to do in Dolfin: from dolfin import * # define mesh, function space (piecewise linear) mesh = UnitSquareMesh(64,64) V = FunctionSpace(mesh,'CG',1) # inhomogeneous boundary conditions (otherwise the solution is trivial) bc = DirichletBC(V, Constant(1.0), lambda x,on_boundary: on_boundary) # define bi(non)linear form # note that ... 5 Your observed quadratic convergence indicates that the Jacobian is likely correct. Have you looked at the solutions for your under-resolved configurations? Galerkin optimality uses the operator norm, which contains only the symmetric part, thus the solution of the discrete system could be quite different from the projection of the exact solution. This ... 5 As it looks to me your formulation of the discrete system is incomplete. You need to state your equation at each (grid?) point associated with$n$put back the summation over$l$(grid) points in your last equation assign a time index$i+1$or$i$to your$c_l$s In this way, you get a nonlinear (if you take$c_l^{n+1}$) or linear (if you take$c_l^{n}$) ... 5 A common strategy is to employ a damping strategy, i.e., to compute$\vec{w}^{\ast} = F \left( \vec{w}^{(n)} \right)$and then set$\vec{w}^{(n+1)} = \alpha \vec{w}^{(\ast)} + (1-\alpha)\vec{w}^{(n)}$where$\alpha\in[0,1]$. You typically choose$\alpha$in such a way that it minimizes$\|[\alpha \vec{w}^{(\ast)} + (1-\alpha)\vec{w}^{(n)}] - F(\alpha \vec{w}^...
5
Certified homotopy continuation methods are used both for finding roots and for proving that they indeed exist (inside a certain interval). A quick web search turned out this paper: Reliable homotopy continuation by Joris van der Hoeven.
5
No it is not. There is also multiple shooting collocation finite differences fixed point iterations and probably some more.
5
Since $\phi$ is a scalar between $0$ and $1$, the easiest method for finding a root is bisection. If you cannot calculate the expectation of the nonlinear function $$f_\phi(\theta) = \left(\phi r_z +(1-\phi)(r_k+\theta)\right)^{1-\gamma}(r_k+\theta-r_z)$$ in terms of $\phi$ analytically, you can use quadrature to approximate. For the first variant, this ...
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• ### $450 Tuition Credit & Official CAT Packs FREE December 15, 2018 December 15, 2018 10:00 PM PST 11:00 PM PST Get the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) • ### FREE Quant Workshop by e-GMAT! December 16, 2018 December 16, 2018 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score. # Is 3^(a^2/b) < 1? new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Manager Joined: 27 Apr 2012 Posts: 57 Location: United States GMAT Date: 06-11-2013 GPA: 3.5 WE: Marketing (Consumer Products) Is 3^(a^2/b) < 1? [#permalink] ### Show Tags 21 Jun 2012, 00:59 1 15 00:00 Difficulty: 95% (hard) Question Stats: 32% (00:58) correct 68% (01:01) wrong based on 378 sessions ### HideShow timer Statistics Is 3^(a^2/b) < 1? (1) a<0 (2) b<0 Hi, request your help to please understand the fundamental concept behind this question. ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 51218 Re: Is 3^(a^2/b) < 1? [#permalink] ### Show Tags 21 Jun 2012, 01:17 7 6 Is 3^(a^2/b) < 1? Notice that $$3^{\frac{a^2}{b}} < 1$$ to hold true, the power of 3 must be less than 0. So, the question basically asks whether $$\frac{a^2}{b}<0$$. This will happen if $$a\neq{0}$$ AND $$b<0$$ (if $$a=0$$ then $$\frac{a^2}{b}=0$$). (1) a<0. The first condition is satisfied ($$a\neq{0}$$) but we don't know about the second one. Not sufficient. (2) b<0. The second condition is satisfied ($$b<0$$) but we don't know about the first one (again if $$a=0$$ then $$\frac{a^2}{b}=0$$). Not sufficient. (1)+(2) Both condition are satisfied. Sufficient. Answer: C. For more on number theory and exponents check: math-number-theory-88376.html DS questions on exponents: search.php?search_id=tag&tag_id=39 PS questions on exponents: search.php?search_id=tag&tag_id=60 Tough and tricky DS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125967.html Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html Hope it helps. _________________ ##### General Discussion Manager Joined: 27 Apr 2012 Posts: 57 Location: United States GMAT Date: 06-11-2013 GPA: 3.5 WE: Marketing (Consumer Products) Re: Is 3^(a^2/b) < 1? [#permalink] ### Show Tags 21 Jun 2012, 02:10 This is great.Thanks a ton! Math Expert Joined: 02 Sep 2009 Posts: 51218 Re: Is 3^(a^2/b) < 1? [#permalink] ### Show Tags 25 Jun 2013, 03:47 Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE Theory on Exponents: math-number-theory-88376.html All DS Exponents questions to practice: search.php?search_id=tag&tag_id=39 All PS Exponents questions to practice: search.php?search_id=tag&tag_id=60 Tough and tricky DS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125967.html Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html _________________ Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6639 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Is 3^(a^2/b) < 1? [#permalink] ### Show Tags 27 Jan 2016, 17:46 Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Is 3^(a^2/b) < 1? (1) a<0 (2) b<0 When you modify the original condition and the question, they become 3^(a^2/b) < 1? --> 3^(a^2/b) < 3^0? --> a^2/b>0?. Multiply b^2 on the both equations(since b^2 is positive, even if it’s multiplied, the sign of inequality doesn’t change.) and it becomes a^2(b)>0?. There are 2 variables(a,b), which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1) & 2), since a<0, it can’t be 0. Divide the both equations with a^2, they become a^2(b)<0?-->b<0?. Since 2) is b<0, it is yes and sufficient. Therefore, the answer is C. For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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25 Oct 2018, 05:33
Bunuel wrote:
Is 3^(a^2/b) < 1?
Notice that $$3^{\frac{a^2}{b}} < 1$$ to hold true, the power of 3 must be less than 0. So, the question basically asks whether $$\frac{a^2}{b}<0$$. This will happen if $$a\neq{0}$$ AND $$b<0$$ (if $$a=0$$ then $$\frac{a^2}{b}=0$$).
(1) a<0. The first condition is satisfied ($$a\neq{0}$$) but we don't know about the second one. Not sufficient.
(2) b<0. The second condition is satisfied ($$b<0$$) but we don't know about the first one (again if $$a=0$$ then $$\frac{a^2}{b}=0$$). Not sufficient.
(1)+(2) Both condition are satisfied. Sufficient.
For more on number theory and exponents check: http://gmatclub.com/forum/math-number-theory-88376.html
DS questions on exponents: http://gmatclub.com/forum/search.php?se ... &tag_id=39
PS questions on exponents: http://gmatclub.com/forum/search.php?se ... &tag_id=60
Tough and tricky DS exponents and roots questions with detailed solutions: http://gmatclub.com/forum/tough-and-tri ... 25967.html
Tough and tricky PS exponents and roots questions with detailed solutions: http://gmatclub.com/forum/tough-and-tri ... 25956.html
Hope it helps.
Bunuel ...if a=-1 and b=-9 then we get 3 as our answer and if a=-1 and b=-2 in that case we get .66
Should the answer not be E.
Math Expert
Joined: 02 Sep 2009
Posts: 51218
Re: Is 3^(a^2/b) < 1? [#permalink]
### Show Tags
25 Oct 2018, 06:17
angarg wrote:
Bunuel wrote:
Is 3^(a^2/b) < 1?
Notice that $$3^{\frac{a^2}{b}} < 1$$ to hold true, the power of 3 must be less than 0. So, the question basically asks whether $$\frac{a^2}{b}<0$$. This will happen if $$a\neq{0}$$ AND $$b<0$$ (if $$a=0$$ then $$\frac{a^2}{b}=0$$).
(1) a<0. The first condition is satisfied ($$a\neq{0}$$) but we don't know about the second one. Not sufficient.
(2) b<0. The second condition is satisfied ($$b<0$$) but we don't know about the first one (again if $$a=0$$ then $$\frac{a^2}{b}=0$$). Not sufficient.
(1)+(2) Both condition are satisfied. Sufficient.
For more on number theory and exponents check: http://gmatclub.com/forum/math-number-theory-88376.html
DS questions on exponents: http://gmatclub.com/forum/search.php?se ... &tag_id=39
PS questions on exponents: http://gmatclub.com/forum/search.php?se ... &tag_id=60
Tough and tricky DS exponents and roots questions with detailed solutions: http://gmatclub.com/forum/tough-and-tri ... 25967.html
Tough and tricky PS exponents and roots questions with detailed solutions: http://gmatclub.com/forum/tough-and-tri ... 25956.html
Hope it helps.
Bunuel ...if a=-1 and b=-9 then we get 3 as our answer and if a=-1 and b=-2 in that case we get .66
Should the answer not be E.
If a=-1 and b=-9, then 3^(1/(-9)) = ~0.9.
_________________
Re: Is 3^(a^2/b) < 1? &nbs [#permalink] 25 Oct 2018, 06:17
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https://mohammadshariatmadari.ir/d9ueu4v/binomial-coefficient-latex-88290e | }}{{k!\left( {n - k} \right)!}}. As you see, the command \binom{}{}will print the binomial coefficient using the parameters passed inside the braces. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. In mathematics, the Gaussian binomial coefficients (also called Gaussian coefficients, Gaussian polynomials, or q-binomial coefficients) are q-analogs of the binomial coefficients.The Gaussian binomial coefficient, written as () or [], is a polynomial in q with integer coefficients, whose value when q is set to a prime power counts the number of subspaces of dimension k in a vector … Pascal's triangle can be extended to find the coefficients for raising a binomial to any whole number exponent. This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. \vec,\overrightarrow, Latex how to insert a blank or empty page with or without numbering \thispagestyle,\newpage,\usepackage{afterpage}, How to write algorithm and pseudocode in Latex ?\usepackage{algorithm},\usepackage{algorithmic}, How to display formulas inside a box or frame in Latex ? One can drop one of the numbers in the bottom list and infer it from the fact that sum … binomial The combination (n r) (n r) is called a binomial coefficient. }}{{k!\left( {n - k} \right)!}} The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. Binomial coefficients are a family of positive integers that occur as coefficients in the binomial theorem. C — All combinations of v matrix. = \binom {n} {k} This is the binomial coefficient. The binomial coefficient $\binom{n}{k}$ can be interpreted as the number of ways to choose k elements from an n-element set. where A is the permutation, A_n^k = \frac{n!}{(n-k)! In latex mode we must use \binom fonction as follows: \frac {n!} Home > Latex > FAQ > Latex - FAQ > Latex binomial coefficient, Monday 9 December 2019, by Nadir Soualem. In Counting Principles, we studied combinations.In the shortcut to finding$\,{\left(x+y\right)}^{n},\,$we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. LaTeX provides a feature of special editing tool for scientific tool for math equations in LaTeX. Then it's a good reason to buy me a coffee. Binomial coefficients have been known for centuries, but they're best known from Blaise Pascal's work circa 1640. k-combinations of n-element set. Click on one of the binomial coefficient designs, which look like the letters "n" over "k" inside either a round or angled bracket. Gerhard "Ask Me About System Design" Paseman, 2010.03.27 \endgroup – Gerhard Paseman Mar 27 '10 at 17:00 Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … This article explains how to typeset them in LaTeX. The binomial coefficient is defined by the next expression: \binom {n}{k} = \frac {n ! }}{{k!\left( {n - k} \right)!}} Binomial coefficient denoted as c (n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n. The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. In this article, you will learn how to write basic equations and constructs in LaTeX, about aligning equations, stretchable horizontal lines, operators and delimiters, fractions and binomials. Latex binomial coefficient Definition. If your equation requires specific numbers in place of the "n" or "k," click on a letter to select it, press "Delete" and enter a number in its place. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … The binomial coefficient is defined by the next expression: \ [ \binom{n} {k} = \frac{n!} (adsbygoogle = window.adsbygoogle || []).push({}); All the versions of this article: Blog template built with Bootstrap and Spip by Nadir Soualem @mathlinux. Knowledge base dedicated to Linux and applied mathematics. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. The binomial coefficient can be interpreted as the number of ways to choose k elements from an n-element set. In latex mode we must use \binom fonction as follows: \frac{n!}{k! The order of selection of items not considered. (adsbygoogle = window.adsbygoogle || []).push({}); Thank you ! The possibility to insert operators and functions as you know them from mathematics is not possible for all things. An example of a binomial coefficient is (5 2)= C(5,2)= 10 (5 2) = C (5, 2) = 10. \boxed, How to write table in Latex ? Using fractions and binomial coefficients in an expression is straightforward. The Texworks shows … Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeXis very similar to the one used for fractions. ( n - k )! Here's an equation: math \frac {n!} (n−k)! Mathematical Equations in LaTeX. = \binom{n}{k} This will give more accuracy at the cost of computing small sums of binomial coefficients. The symbols and are used to denote a binomial coefficient, and are sometimes read as "choose.". n! This same array could be expressed using the factorial symbol, as shown in the following. And of course this command can be included in the normal text flow \ (\binom{n} {k}\). All combinations of v, returned as a matrix of the same type as v. Specially useful for continued fractions. The second fraction displayed in the previous example uses the command \cfrac{}{} provided by the package amsmath (see the introduction), this command displays nested fractions without changing the size of the font. Binomial Coefficient. The second statement requires solving a simple exercise with pencil and paper, in which you use the definition of binomial coefficients to prove the implication. In UnicodeMath Version 3, this uses the \choose operator ⒞ instead of the \atop operator ¦. The symbols and are used to denote a binomial coefficient, and are sometimes read as " choose." Accordingly the binomial coefficient in the binomial theorem above can be written as “n\choose k”, assuming that you type a space after the k. This Then it's a good reason to buy me a coffee. Toutes les versions de cet article : Le coefficient binomial est le nombre de possibilités de choisir k élément dans un ensemble de n éléments. b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. A slightly different and more complex example of continued fractions, Showing first {{hits.length}} results of {{hits_total}} for {{searchQueryText}}, {{hits.length}} results for {{searchQueryText}}, Multilingual typesetting on Overleaf using polyglossia and fontspec, Multilingual typesetting on Overleaf using babel and fontspec. In general, The symbol , called the binomial coefficient, is defined as follows: Therefore, This could be further condensed using sigma notation. binomial coefficient Latex. LaTeX forum ⇒ Math & Science ⇒ Expression like binomial Coefficient with Angle Delimiters Topic is solved Information and discussion about LaTeX's math and science related features (e.g. Asking for help, clarification, or responding to other answers. }{k ! therefore gives the number of k -subsets possible out of a set of distinct items. It is especially useful for reasoning about recursive methods in programming. All combinations of v, returned as a matrix of the same type as v. Binomial coefficient denoted as c(n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n.. The following are the common definitions of Binomial Coefficients.. A binomial coefficient C(n, k) can be defined as the coefficient of x^k in the expansion of (1 + x)^n.. A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects more formally, the number of k-element subsets (or k-combinations) of a n-element set. As you may have guessed, the command \frac{1}{2} is the one that displays the fraction. Binomial Coefficient: LaTeX Code: \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \frac{{n! Open an example in Overleaf (n - k)!} formulas, graphs). So The combination (nr)\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)(nr) is calle… Binomial coefficient, returned as a nonnegative scalar value. Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. Fractions and binomial coefficients are common mathematical elements with similar characteristics - one number goes on top of another. } ... Pascal’s triangle. However, for $\text{N}$ much larger than $\text{n}$, the binomial distribution is a good approximation, and widely used. Identifying Binomial Coefficients. It will give me the energy and motivation to continue this development. Ak n = n! The usual binomial coefficient can be written as $\left({n \atop {k, {n-k}}}\right)$. (n-k)!} I agree. The binomial coefficient is defined by the next expression: \[\binom {n}{k} = \ frac {n!}{k!(n-k)! Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. In Counting Principles, we studied combinations.In the shortcut to finding ${\left(x+y\right)}^{n}$, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. On the other side, \textstyle will change the style of the fraction as if it were part of the text. Don't forget to LIKE, COMMENT, SHARE & SUBSCRIBE to my channel. Usually, you find the special input possibilities on the reference page of the function in the Details section. How to write number sets N Z D Q R C with Latex: \mathbb, amsfonts and \mathbf, How to write angle in latex langle, rangle, wedge, angle, measuredangle, sphericalangle, Latex numbering equations: leqno et fleqn, left,right, How to write a vector in Latex ? For these commands to work you must import the package amsmath by adding the next line to the preamble of your file In this case, we use the notation (nr)\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)(nr) instead of C(n,r)\displaystyle C\left(n,r\right)C(n,r), but it can be calculated in the same way. \vec,\overrightarrow; Latex how to insert a blank or empty page with or without numbering \thispagestyle,\newpage,\usepackage{afterpage} Latex natural numbers; Latex real numbers; Latex binomial coefficient; Latex overset and underset ; Latex absolute value Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. Also, the text size of the fraction changes according to the text around it. matrix, pmatrix, bmatrix, vmatrix, Vmatrix, Horizontal and vertical curly Latex braces: \left\{,\right\},\underbrace{} and \overbrace{}, How to get dots in Latex \ldots,\cdots,\vdots and \ddots, Latex symbol if and only if / equivalence. Stanley's EC1 also uses it as the primary name, which counts for a lot in my book. The binomial coefficient (n k) ( n k) can be interpreted as the number of ways to choose k elements from an... Properties. The text inside the first pair of braces is the numerator and the text inside the second pair is the denominator. The command \displaystyle will format the fraction as if it were in mathematical display mode. A General Note: Binomial Coefficients If n n and r r are integers greater than or equal to 0 with n ≥r n ≥ r, then the binomial coefficient is Latex Binomial coefficient, returned as a nonnegative scalar value. begin{tabular}...end{tabular}, Latex horizontal space: qquad,hspace, thinspace,enspace, LateX Derivatives, Limits, Sums, Products and Integrals, Latex copyright, trademark, registered symbols, How to write matrices in Latex ? (n - k)!} {k! = \binom{n}{k} = {}^{n}C_{k} = C_{n}^k$$,$$\frac{n!}{k! (n - k)!} This website was useful to you? {k! In Counting Principles, we studied combinations. k-combinations of n-element set. This method of constructing mathematical proofs is called mathematical induction. samedi 11 juillet 2020, par Nadir Soualem. Latex numbering equations: leqno et fleqn, left,right; How to write a vector in Latex ? Since binomial coefficients are quite common, TeX has the \choose control word for them. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. Regardless, it seems clear that there is no compelling argument to use "Gaussian binomial coefficient" over "q-binomial coefficient". \\binom{N} {k} What differs between \\dots and \\dotsc, with overleaf.com, the outputs are identical. It is the coefficient of the x k term in the polynomial expansion of the binomial power (1 + x) n, and is given by the formula =!! As you see, the command \binom{}{} will print the binomial coefficient using the parameters passed inside the braces. Binomial Coefficient: LaTeX Code: \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \frac{{n! In mathematics, the binomial coefficients are the positive integers that occur as coefficients in the binomial theorem.Commonly, a binomial coefficient is indexed by a pair of integers n ≥ k ≥ 0 and is written (). infinite sum of inverse binomial coefficient encountered in Bayesian treatment of the German tank problem Hot Network Questions Why are quaternions more … This video is an example of the Binomial Expansion Technique and how to input into a LaTex document in preparation for a pdf output. In this video, you will learn how to write binomial coefficients in a LaTeX document. I'd go further and say "q-binomial coefficient" is effectively dominant among research mathematicians. In the shortcut to finding (x+y)n\displaystyle {\left(x+y\right)}^{n}(x+y)n, we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. For these commands to work you must import the package amsmath by adding the next line to the preamble of your file, The appearance of the fraction may change depending on the context. The binomial coefficient (n k) ( n k) can be interpreted as the number of ways to choose k elements from an n-element set. The binomial coefficient also arises in combinatorics, where it gives the number of different combinations of $b$ elements that can be chosen from a … Latex k parmi n - coefficient binomial. See for instance the documentation of Integrate.. For Binomial there seems to be no such 2d input, because as you already found out, $\binom{n}{k}$ is … Binomial coefficients are the ones that appear as the coefficient of powers of x x x in the expansion of (1 + x) n: (1+x)^n: (1 + x) n: ( 1 + x ) n = n c 0 + n c 1 x + n c 2 x 2 + ⋯ + n c n x n , (1+x)^n = n_{c_{0}} + n_{c_{1}} x + n_{c_{2}} x^2 + \cdots + n_{c_{n}} x^n, ( 1 + x ) n = n c 0 + n c 1 x + n c 2 x 2 + ⋯ + n c n x n , b is the same type as n and k. If n and k are of different types, then b is returned as the nondouble type. For example, … The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. therefore gives the number of k-subsets possible out of a set of distinct items. You can set this manually if you want. For these commands to work you must import the package amsmath by adding the next line to the preamble of your file How to write it in Latex ? C — All combinations of v matrix. 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All things from mathematics is not possible for all things math \frac { }..., \textstyle will change the style of the fraction expressed using the factorial symbol, as shown in the section!, Monday 9 December 2019, by Nadir Soualem elements from an n-element.... You know them from mathematics is not possible for all things the outputs are.... Technique and how to input into a Latex document in preparation for a lot in my book display! Fleqn, left, right ; how to input into a Latex document in preparation for a lot in book! A good reason to buy me a coffee called a binomial coefficient, and are used to denote binomial! The second pair is the number of ways in which k items chosen! Texworks shows … Latex numbering equations: leqno et fleqn, left right!, right ; how to write a vector in Latex scientific tool scientific... N r ) ( n r ) ( n r ) ( n r ) ( r... Therefore gives the value of the binomial coefficient can be extended to find the coefficients raising. Page of the \atop operator ¦ '' over q-binomial coefficient '' this is the number of possible... Coefficient, and are sometimes read as choose. command to display them in Latex to write a in... Latexis very similar to the text around it using fractions and binomial coefficients have been known for,... It as the number of ways to choose k elements from an n-element set common elements mathematical. Factorial symbol, as shown in the following the binomial coefficient, and are sometimes read as choose ''... Outputs are identical here 's an equation: math \frac { n! }.!
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http://openstudy.com/updates/55bd1225e4b033255002fcc6 | ## YumYum247 one year ago how do i plot a velocity time graph from an average velocity of an object?
1. YumYum247
if a runner accelerated along the sidewalk at 0.5m/sec2 for 20 sec. Assume that the runner started from rest position.....
2. anonymous
You will plot velocity on the vertical (y) axis and the time on the horizontal (x) axis. All you have to do is plot the values and connect all of them by a line. Make sure velocity is given in meters per second (m/s) and time in seconds (s) unless instructed otherwise.
3. YumYum247
can you please show me how do i make the data table so i can plot the points neatly...
4. YumYum247
|dw:1438454610769:dw|
5. YumYum247
how do i use the average velocity given in the question to plot the points on the graph.
6. anonymous
Velocity (in m/s) is [distance in metres] divided by [time in seconds.]
7. YumYum247
so v = d/t
8. YumYum247
but how's that gunna help me?
9. YumYum247
if i want to know how far did the runner get in 7 sec, how do i figure that out???
10. YumYum247
@oldrin.bataku
11. anonymous
if the acceleration is constant $$a$$ then the total change in velocity from the start until time $$t$$ is just $$v-v_0=at$$; in our case, we starta t rest so $$v_0=0$$
12. YumYum247
yes the accelaeration is constant.
13. anonymous
so the velocity is given by $$v=at$$, so in terms of a velocity vs time plot we see that the graph is a simple line with slope $$a$$ passing through the origin
14. YumYum247
i just need to plot how much distance he covered from rest to 20sec.
15. anonymous
well, if we have a velocity vs time plot, then the distance covered between $$t=t_1$$ and $$t=t_2$$ is just the total area bounded by our velocity graph between those points:
16. anonymous
|dw:1438456604986:dw|
17. YumYum247
can you please just do one for me, i'll do the rest myself.... can you figure out how much distance the runner covered in let's say 7 seconds????
18. anonymous
|dw:1438456642313:dw|
19. anonymous
so the distance they go between $$t_1,t_2$$ is just the area under the graph between $$t_1,t_2$$.
20. YumYum247
the area under the triangle is the displacement of the object....i get that...but
21. YumYum247
did you use Tf - Ti = a X t ?
22. YumYum247
because i need to make a data table to record all the points until 20 second and plot them on the velocity time graph.
23. anonymous
okay, so here we have $$t_1=0$$, $$t_2=7$$, and our equation is again $$v(t)=0.5t$$. so our total displacement is: $$A=\frac12 bh$$here our base is $$t_2=7$$ since $$t_1=0$$ and our height is $$v(t_2)=v(7)=3.5$$ so: $$A=\frac12(7)(3.5)=12.25$$
24. anonymous
you could also have done that work using an integral instead of manually finding teh area of a triangle
25. YumYum247
it doesn't look right because i have the answer key and the maximum displacement the runner makes under 20 sec is 10 meters.... i'm willing to show you the answer tho but i don't know how they got those points.....in the first place.
26. anonymous
i dont know what points youre talking about
27. YumYum247
|dw:1438457170341:dw|
28. YumYum247
this is how it looks......
29. anonymous
the kinematic equation gives $$x=x_0+v_0 t+\frac12 at^2$$, here $$a=0.5,t=7$$ and $$x_0=v_0=0$$ so $$x=\frac12\cdot0.5\cdot7^2=12.25$$
30. anonymous
actually, no it doesn't; you've misread it. that's the *velocity* against time, so it's going 10 meters *per second* in velocity after 20 seconds of time
31. YumYum247
i'm sorry it's a velocity time graph, the velocity/speed of the runner goes up to 10m/s under 20sec
32. anonymous
yes, and that doesn't contradict the fact the runner goes 12.25 meters in 7 seconds.
33. YumYum247
ok
34. anonymous
his speed after 7 seconds is only 3.5 m/s as I stated before
35. anonymous
with an average speed of $$(3.5-0)/2=1.75$$ m/s, which is indeed correct: $$\frac{12.25}7=1.75$$ m/s
36. YumYum247
ok what formula did you use plase?????? :(
37. YumYum247
ok so i need to make a data table of this situation as drawn above, so what equation do i use to plot the speed of the runner every second????
38. anonymous
since the velocity vs time graph is a line, the average velocity between $$t_1=0,t_2=7$$ s is just the midpoint, hence $$(3.5-0)/2=1.75$$ m/s. this matches the alternative calculation of the average velocity as the total displacement $$12.25$$ m divided by the time elapsed $$t_2-t_1=7$$ s, since $$12.25/7=1.75$$ m/s as well
39. anonymous
the equation for the velocity at time $$t$$ is just $$v=0.5t$$, since the acceleration is constant. this is the straight line we've been graphing
40. YumYum247
ok let me try on my own and i'll ask you to check my work. Thank you!!! :")
41. YumYum247
omg i see how you got 10m....LOL
42. YumYum247
ok one more thing, if the average velocity of an object is 0.5m/sec2 how quick or how fast will it accelerate every second. | 2017-01-22T03:48:40 | {
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https://math.stackexchange.com/questions/675917/if-gcda-b-1-and-if-ab-x2-prove-that-a-b-must-also-be-perfect-sq | # If $\gcd(a, b) = 1$ and if $ab = x^2$, prove that $a, b$ must also be perfect squares; where $a,b,x$ are in the set of natural numbers
Problem:
If $\gcd(a, b) = 1$ and If $ab = x^2$ ,prove that $a$, $b$ must also be perfect squares; where $a$,$b$,$x$ are in the set of natural numbers
I've come to the conclusion that $a \ne b$ and $a \ne x$ and $b \ne x$ but I guess that won't really help me.. I understand that if the $\gcd$ between two numbers if $1$ then they obviously have no common divisors but where do I go from this point?
Any tips at tackling this would be great. It looks quite easy though I'm still trying to get my hand around these proofs! Any pointers in the right direction would be great.
• You just wrote "$b\ne b$." I really hope that's not what you meant. – apnorton Feb 14 '14 at 3:51
• I've noticed that you have asked quite a few questions recently. I wanted to make sure that you are aware of the quotas 50 questions/30 days and 6 questions/24 hours, so that you can plan posting your questions accordingly. (If you try to post more questions, stackexchange software will not allow you to do so.) For more details see meta. – user61527 Feb 14 '14 at 3:52
• Hi Tyler! I really appreciate the heads up. I'm working away at an assignment right now. I've gotten most of the problems down but I still have only a few left. Thanks so much again! – A A Feb 14 '14 at 4:08
Fundamental theorem of arithmetic says that every number has a unique prime factorization.
If gcd(a,b) = 1, then all of these factors are unique (no prime factor is shared between a and b). What does this say about $x^2$?
Hint 2:
Let $a_i$ be a prime factor of a and $b_i$ be a factor of b. Then,
$$ab = \prod {a_{i}^{m_i}}\prod {b_{i}^{n_i}}$$
But $x$ has to have a unique factorization in the form,
$$x = \prod {x_{i}^{e_i}}$$, where $m, n, e$ are integer exponents. Keep in mind it is unique and we can order these factors in any way we please. What does this say about $x^2$ compared to $ab$?
• Hmm, x^2 must then: 1) have unique prime factors – A A Feb 14 '14 at 4:14
• I'll edit my post with another hint. – Chantry Cargill Feb 14 '14 at 4:17
• I think I may have it: x^2 = (p1^ip2^i2...pn^i2)^2 = (p1^2i2*p2^2i3*...pn^2in) = ab a and b must be factors of x^2, and all the factors of x^2 are squares, therfore a and b must be squares. Hmm, Does this sound correct to you? – A A Feb 14 '14 at 4:54
• Yes, more or less! Just make sure that it's clear that you can order the factors of $x^2$ such that the primes match up with a and b. Then the root will just be ab. – Chantry Cargill Feb 14 '14 at 5:11
Below is an approach employing universal gcd laws (associative, commutative, distributive), some of which you may need to (simply) prove before you can use this method. But once you do so, you will gain great power. Below we explicitly show that $\rm\:a,b\:$ are squares by taking gcds. Namely
Lemma $\rm\ \ \color{#0a0}{(a,b,c) = 1},\,\ \color{#c00}{c^2 = ab}\ \Rightarrow\ a = (a,c)^2,\,\ b = (b,c)^2\$ for $\rm\:a,b,c\in \mathbb N$
Proof $\rm\ \ (c,b)^2 = (\color{#c00}{c^2},b^2,bc) = (\color{#c00}{ab},b^2,bc) = b\color{#0a0}{(a,b,c)} = b.\$ Similarly for $\,\rm(c,a)^2.\ \$ QED
Yours is the special case $\rm\:(a,b) = 1\ (\Rightarrow\ (a,b,c) = 1)$.
Generally $\rm\: \color{#c00}{ab = cd}\: \Rightarrow\: (a,c)(a,d) = (aa,\color{#c00}{cd},ac,ad) = a\: (a,\color{#c00}b,c,d) = a\:$ if $\rm\:(a,b,c,d) = 1.\:$ For more on this and closely related topics such as Euler's four number theorem (Vierzahlensatz), Riesz interpolation, or Schreier refinement see this post and this post.
Compare the following Bezout-based proof (this is a simplified form of the proof in Rob's answer). For comparison, I append an ideal-theoretic version of the proof of the more complex direction.
Note that $\ 1=\overbrace{a{\rm u}+b\,{\rm v}}^{\large (a,b)}\,\ \overset{\large \times\,a}\Rightarrow\ a = \color{#c00}{a^2}{\rm u}+\!\!\overbrace{ab}^{\Large\ \ \color{#c00}{c^2}}{\rm v} \ \,$ so $\,\ d=(a,c)\mid a,c\,\Rightarrow\, d^2\!\mid \color{#c00}{a^2,c^2}\,\Rightarrow\, d^2\!\mid a$
Conversely $\ d = (a,c)= au+cv\,\ \Rightarrow\,\ d^2=\,\color{#c00}{a^2}u^2+2\color{#c00}acuv+\color{#c00}{c^2}v^2\ \$ thus $\ \ \color{#c00}{a\mid c^2}\ \Rightarrow\,\ \color{#c00}a\mid d^2$
$\quad\ \ {\rm i.e.}\quad (d)= (a,c)\ \ \Rightarrow\ \ (d^2) \subseteq\, (a,c^2)\,\subseteq\, (a)\ \$ by $\ \ a\mid c^2\,\ \$ [simpler ideal form of prior]
Notice how the ideal version eliminates the obfuscatory Bezout coefficients $\,u,v$.
• Hi Bill! Thank you for the response! Your way ahead of me, I'm not too familiar with the notation used/ – A A Feb 14 '14 at 4:53
• Bezouted.$\ \$ – robjohn Feb 14 '14 at 10:16
• @AA I use the standard notation $\ (a,b,\,\ldots)\, =\, \gcd(a,b,\,\ldots).\,$ I added another Bezout-based proof. $\$ – Bill Dubuque Feb 15 '14 at 4:50
Here is a proof using Bezout's Identity.
Let $x^2=ab$ and $\gcd(a,b)=1$, where $a,b\gt0$.
There are $u,v$ so that $$au+bv=1\tag{1}$$ Let $s_a=\gcd(x,a)$. Rewriting $(1)$, we have \begin{align} s_a\left(\dfrac{a}{s_a}\right)u+bv=1 &\implies s_a^2\left(\dfrac{a}{s_a}\right)^2u^2=1-b(2v-bv^2)\tag{2}\\ &\implies s_a^2\left(\dfrac{a}{s_a}\right)^2u^2a+ab(2v-bv^2)=a\tag{3}\\ &\implies s_a^2\left(\dfrac{a}{s_a}\right)^2u^2a+s_a^2\left(\dfrac{x}{s_a}\right)^2(2v-bv^2)=a\tag{4}\\[8pt] &\implies s_a^2\mid a\tag{5} \end{align} Justification:
$(2)$: Move $bv$ to the right side and square
$(3)$: Move $b(2v-bv^2)$ to the left side and multiply by $a$
$(4)$: $x^2=ab$
$(5)$: $s_a^2$ divides each term on the left side
There are $u_a,v_a$ so that \begin{align} xu_a+av_a=s_a &\implies\frac{x}{s_a}u_a+\frac{a}{s_a}v_a=1\tag{6}\\ &\implies\frac{x^2}{s_a^2}u_a^2=1-\frac{a}{s_a}\left(2v_a-\frac{a}{s_a}v_a^2\right)\tag{7}\\ &\implies\frac{ab}{s_a^2}u_a^2+\frac{a}{s_a}\left(2v_a-\frac{a}{s_a}v_a^2\right)=1\tag{8}\\ &\implies abu_a^2+as_a\left(2v_a-\frac{a}{s_a}v_a^2\right)=s_a^2\tag{9}\\[7pt] &\implies a\mid s_a^2\tag{10} \end{align} Justification:
$\ \:(6)$: Divide by $s_a$
$\ \:(7)$: Move $\frac{a}{s_a}v_a$ to the right side and square
$\ \:(8)$: Move $\frac{a}{s_a}\left(2v_a-\frac{a}{s_a}v_a^2\right)$ to the left side, $x^2=ab$
$\ \:(9)$: Multiply by $s_a^2$
$(10)$: $a$ divides each term on the left side
Combining $(5)$ and $(10)$ yields $a=\gcd(x,a)^2$. Symmetry yields, $b=\gcd(x,b)^2$.
• I simplified it a bit - see my answer. In $\,(2\!-\!5)\,$ it suffices to use the Bezout identity itself (vs. its square). In $(6\!-\!10)$ it's simpler to immediately square the Bezout identity (vs. rearrange it first). – Bill Dubuque Feb 15 '14 at 4:47 | 2019-10-20T03:46:59 | {
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https://math.stackexchange.com/questions/1324903/isometry-group-of-a-norm-is-always-contained-in-some-isometry-group-of-an-inner/1325204 | Isometry group of a norm is always contained in some Isometry group of an inner product?
Does there always exist some inner product $\<,\>$ on $V$ such that $\text{ISO}(|| \cdot ||)\subseteq \text{ISO}(\<,\>)$ ?
Update:
As pointed by Qiaochu Yuan the answer is positive.
This raises the question of uniqueness of the inner product $\<,\>$ which satisfies $\text{ISO}(|| \cdot ||)\subseteq \text{ISO}(\<,\>)$.
Is it unique (up to scalar multiple)?
Remarks:
1) Determining $\<,\>$ (up to scalar multiple) is equivalent to determining $\text{ISO}(\<,\>)$.
Clearly if we know the inner product we know all its isometries. The other direction follows as a corollary from an argument given here which shows which inner products are preserved by a given automorphism.
2) Since there are "rigid" norms (whose only isometries are $\pm Id$ ) the uniqueness certainly doesn't hold in general.
One could hope for that in the case of "rich enough norms" (norms with many isometries, see this question) the subset $\text{ISO}(|| \cdot ||)\subseteq \text{ISO}(\<,\>)$ will be large enough to determine $\text{ISO}(\<,\>)$.
(which by remark 1) determines $(\<,\>)$).
Yes. This is because an isometry group is always compact (with respect to the topology on $\text{End}(V)$ induced by the operator norm: this is a consequence of the Heine-Borel theorem). Hence you can average an inner product over it with respect to Haar measure.
• @Asaf: that is not the suggestion. The suggestion is the following: pick an arbitrary inner product $\langle v, w \rangle$. Construct the new inner product $\langle v, w \rangle_G = \int_G \langle gv, gw \rangle \, dg$. (There is some work necessary to prove that this is still an inner product but it's not so bad.) By construction, this new inner product is $G$-invariant, so every $g \in G$ is an isometry with respect to the corresponding norm. This is a standard maneuver in representation theory called "Weyl's unitary trick"; you can try to google that term for references. – Qiaochu Yuan Jun 14 '15 at 19:38
For completeness, I am writing more detailes of the solution suggested by Qiaochu:
Denote by $G$ the isometry group of $(V,\|\cdot\|)$. $G\subseteq \text{End}(V)$. On $\text{End}(V)$ we have the operator norm $\| \|_{op}$ (w.r.t the given norm $\|\cdot\|$), which induces a topology on $\text{End}(V)$.
Lemma 1: $G$ is compact in $\text{End}(V)$
Proof: $\text{End}(V)$ is a finite dimensional normed space, hence it is linearly homeomorphic to $\mathbb{R}^n$ (This is in fact true for every finite dimensioanl real topological vector space). So, the Heine-Borel theorem aplies. (Every closed and bounded subset is compact).
$G$ is bounded since for every isometry $g\in G, \|g\|_{op}=1$, hence $G$ is contained in the unit sphere of $(\text{End}(V),\| \|_{op})$ .
$G$ is closed: Assume $g_n \rightarrow g,g_n\in G$. Fix some $v\in V$. $\|g_n(v)-g(v)\|_V\leq \|g_n-g\|_{op}\cdot\|v\|_V \xrightarrow{n\to\infty} 0$. So $g_n(v)\xrightarrow{n\to\infty}g(v)$. Now by the continuity of the norm $\| \cdot \|$ (w.r.t to the topology it induces on $V$) we get that: $\|v\| \stackrel{g_n isometry}{=} \|g_n(v)\|\xrightarrow{n\to\infty}\|g(v)\|$. This forces $g$ to be an isometry.
Corollary1: $G$ is locally compact Hausdorff topological group.
Proof: $\text{End}(V)$ is Hausdorff (Any metric space is...), and every subspace of Hausdorff is also Hausdorff. It is sdandard fact that $GL(V)$ is a topological group (t.g), and any subgroup of a t.g is a t.g.
Now, there exists a left-invariant measure $\mu$ on the Borel $\sigma$-algebra of $G$ such that $\mu(G)>0$. (This is the Haar measure which can be constructed on any locally compact Hausdorff topological group).
Now take any inner product $\<,\>$ on $V$. Fix $v,w \in V$. Define $f_{v,w}:G\rightarrow \mathbb{R},f_{v,w}(g)=\<gv,gw\>$.
Lemma 2: $f_{v,w}$ is continuous
Proof: Since $G$ is a metric space (a subspace of the normed space $\text{End}(V)$) it is enough to check sequential continuity. Take
$g_n \rightarrow g,g_n\in G$. We already showed this implies $g_n(v)\xrightarrow{n\to\infty}g(v)$ so by the continuity of the inner product $f_{v,w}(g_n)=\<g_nv,g_nw\> \xrightarrow{n\to\infty} f_{v,w}(g)$.
In particular $f_{v,w}$ is measurable, so we can integrate it. (Compactness of $G$ implies $f_{v,w}$ is bounded, and $G$ being a finite measure space guarantees the integral will be finite).
So we define: $\<v, w \>' = \int_G f_{v,w} \, d\mu = \int_G \< gv, gw \> \, d\mu$.
Now all is left is to show $\<, \>'$ is an inner product on $V$ that is presrved by each $h\in G$. (since this means $G=\text{ISO}(V,\|\cdot\|) \subseteq \text{ISO}(V,\<, \>')$ as required.
Lemma 3: $\<,\>'$ is an inner product.
The only non-trivial thing is positive-definiteness. (The rest follows from the linearity of $g\in G$ and the integral, and the bilinearity of $\<,\>$). But this follows from standard measure theory:
Fix $v\neq 0$. $f_{v,v} > 0$ on $G$ (since each $g\in G$ is injective and the original inner prodcut is positive). But this forces $\<v,v\>'>0$ as required.
Lemma 4: $\<, \>'$ is $G$-invariant. But this follows from another standard proposition in measure theory: (See "Real Analaysis" by H.L.Royden) chapter 22 pg 488 proposition 10).
• Still not enough langles and rangles... – Qiaochu Yuan Jun 16 '15 at 3:19
• You are right... I thought I already fixed it but I was wrong. I hope now its OK. – Asaf Shachar Jun 16 '15 at 7:10
• I really like that you accepted Qiaochu's answer and then added the details in your own. Nice! – Joachim Jun 21 '15 at 21:00 | 2020-04-04T13:12:13 | {
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https://math.stackexchange.com/questions/3126162/solving-left1-3-rightk-n-1-for-k | # Solving $\left(1/3\right)^k n = 1$ for $k$
The goal is to show that $$\left(\frac{1}{3}\right)^kn=1 \Rightarrow k = \log_3 n\,.$$
So I started with $$\left(\frac{1}{3}\right)^kn=1 \Leftrightarrow \left(\frac{1}{3}\right)^k=\frac{1}{n}$$ in order to use the identity $$y=a^x \Leftrightarrow x=\log_a y$$, which then yields $$k=\log_{1/3} \frac{1}{n}$$ which using $$\log \frac{1}{x}=-\log a$$ can be written as $$k = -\log_{1/3} n\,.$$ But that is not what I wanted to show, as $$\log_3 n \neq -\log_{1/3} n$$.
I don't know where the mistake is.
Note that $$-\log_{1/3} n = \frac{\log_{1/3} n}{\log_{1/3}3} = \log_3 n$$
• Thank you! That was fast. I will accept your answer as soon as it will let me. – user500664 Feb 25 '19 at 15:45
• Sorry, I've got another question: the identity you used is $\log_b (a)=\frac{\log_c a}{\log_c b}$, right? In your answer, $c=\frac{1}{3}$, but couldn't it be any arbitrary number as $c$ bears no relation to either $a$ or $b$? – user500664 Feb 25 '19 at 15:55
• @ThomasFlinkow Yes, but keep in mind that $0<c \neq 1$ for the logarithms to be defined. – Haris Gušić Feb 25 '19 at 15:58
Alternatively,
$$\left( \frac{1}{3^k}\right)n=1$$
Multiplying $$3^k$$ on both sides, $$n=3^k$$
Hence $$k = \log_3 n$$
• Thank you very much. This is even more elegant. I'm afraid I already accepted an answer – user500664 Feb 25 '19 at 16:12
• Don't worry about reputations. I just see a question and just lift a few fingersl ;) The other solution teach you why $-\log_{\frac13} n = \log_3 n$. – Siong Thye Goh Feb 25 '19 at 16:16 | 2021-02-26T22:35:24 | {
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https://math.stackexchange.com/questions/1985074/expected-payoff-of-dice-game-paying-for-every-next-roll/1988641 | # expected payoff of dice game (paying for every next roll)
I am thinking of a variation of the dice game, where
one has the option to throw a die unlimited number of times.The first throw is free and every next throw costs 1 dollar. One will earn the face value of the die and has the option to stop after each throw and walk away with the money earned. The earnings are not additive. What is the expected payoff of this game?
If I calculate the expected value as
for 2 rolls as (1/6)(3+4+5+6) + (1/3) (3.5-1) = 3.83$for 3 rolls as (1/6)(3+4+5+6) + (1/3) (3.83-1) = 3.94$
for 4 rolls as (1/6)(3+4+5+6) + (1/3) (3.94-1) = 3.98$for 5 rolls as (1/6)(3+4+5+6) + (1/3) (3.98-1) = 3.99$
it asymptotically tends to 4
Is this approach correct?
Shoudn't the player go bankrupt after 7 rolls (negative expectation)?
• I don't follow your method here. Can you explain your reasoning? – Matthew Conroy Oct 27 '16 at 7:36
• @ Matthew Conroy let's take 2 rolls. The expectation of a single trow is 3.5, but it costs 1, therefore the expectation is 2.5. Thus, if one gets 1 or 2 in the first try then one keeps rolling because the expectation is higher (2.5). If 3,4,5, or 6 comes one stops and takes the payoff, so $$(\frac{1}{6} 3+\frac{1}{6} 4+\frac{1}{6} 5+\frac{1}{6} 6) + \frac{2}{6}(3.5-1) = 3.83$$ – Michal Oct 28 '16 at 2:41
• Sometimes they will lose money, but on average they will net $4$. See my answer below. (There is no "bankrupt" here, since you have not specified how much money the player has to begin with.) – Matthew Conroy Oct 29 '16 at 20:39
I'd do it this way.
Let $E_n$ be the expected net if we roll until we get $n$ or greater.
Suppose $n=6$. We roll once. If it's a $6$, we are done. If not, we are in exactly the situation we started in, except that our "first" roll isn't free: it cost $1$ dollar. Hence $$E_6 = \frac{1}{6}(6) + \frac{5}{6}(E_6-1).$$ Solving, we find $E_6=1$. (Alternatively: on average it takes $6$ rolls to get a $6$, and all but the first roll is free, so $E_6=6-5=1$.)
Similarly, \begin{align} E_5 &= \frac{2}{6}\left(\frac{5+6}{2} \right) + \frac{4}{6}(E_5-1) & \text{ so } &E_5=\frac{7}{2}=3.5 \\ E_4 &= \frac{3}{6} \left( \frac{4+5+6}{3} \right) + \frac{3}{6}(E_4-1) & \text{ so } &E_4=4 \\ E_3 &= \frac{4}{6} \left( \frac{3+4+5+6}{4} \right) + \frac{2}{6}(E_3-1) & \text{ so} &E_3=4 \\ E_2 &= \frac{5}{6} \left( \frac{2+3+4+5+6}{5} \right) +\frac{1}{6}(E_2-1) & \text{ so } &E_2=\frac{19}{5} = 3.8 \\ \end{align} and, of course, $E_1=3.5$.
So the best strategy appears to be "roll until $4$ or more" with an expected net of $4$. Sometimes you will roll, say, 10 times with this strategy and lose money, but the average net is $4$.
Here is the result of $10^6$ simulated plays with the $4$ strategy:
net/number of occurrences
-21 1
-20 0
-19 0
-18 0
-17 0
-16 1
-15 0
-14 1
-13 5
-12 6
-11 8
-10 19
-9 32
-8 80
-7 146
-6 293
-5 579
-4 1122
-3 2272
-2 4517
-1 9147
0 18256
1 36474
2 72912
3 145747
4 291215
5 250332
6 166835
with an average net of $4.000515$.
• @ Matthew Conroy I think you have a mistake in $E_2$ (in case of 2 you keep rolling, so 2 shouldn't be in the first bracket). The expectation keeps going up, asymptotically tending to 4 as n increases. on the other hand is it not at odds with common sense? any strategy with more than 7 rolls is loosing money, the higher n the bigger loss – Michal Oct 28 '16 at 10:58
• @ Matthew Conroy secondly, why in your calculation the options you decide to pay off are decreasing with every next roll? for the 2 roll case E(x) value of the second roll is 2.5 so you pay off straight away after first roll if 3,4,5,6 come. for the 3 roll case and the next ones the E(x) is below 4 so you keep rolling in case of 1,2,3 and pay off in case of 4,5,6, correct? – Michal Oct 28 '16 at 14:10
• My $E_2$ is the expected value if we roll until we get $2$ or greater. In this case, we stop when we get a $2$. – Matthew Conroy Oct 28 '16 at 22:13
• Regarding your claim "the higher n the bigger loss", consider: if I've rolled a 1 (for example), it is always to my advantage, on average, to roll again, regardless of how many rolls I've taken already. Cheers! – Matthew Conroy Oct 29 '16 at 4:44
• If you roll $1$ six times in a row, then you will lose money. However, on average, you will lose less money if you continue to roll for a higher number. The same goes for rolling $2$s. The $E_2$ strategy is not optimal, as the calculation above shows. It is still a strategy. If you still don't agree with these calculations, I recommend you write a simulation and try out these strategies yourself. Sometimes the $4$ strategy loses money, but on average the net is $4$ dollars. Cheers! – Matthew Conroy Oct 29 '16 at 20:05 | 2019-12-10T14:23:10 | {
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https://math.stackexchange.com/questions/1420222/different-results-in-integrating-both-sides-of-sin2x-2-cos-x-sin-x | # Different results in integrating both sides of $\sin{2x}=2\cos x\sin x$
I feel like there is something I am missing here. When integrating both sides of the trigonometric identity $\sin{2x}=2\cos x\sin x$ I get different results.
The left side of course results in $-\frac{1}{2}\cos{2x}+C$.
The right side I solve with u-substitution:
$u=\cos x$
$du=-\sin x dx$
$-2\int udu=-u^2+C=-\cos^2 x+C$
While writing this question I noticed another identity $\cos^2 x=\frac{1}{2}+\frac{1}{2}\cos 2x$. So apparently the $\frac{1}{2}$ falls out because of the $+C$ resulting from indefinite integration? This is still a little confusing to me.
You are correct to recall that $\cos^2x=\frac{1+\cos 2x }{2}$.
This is an indefinite integral. So, the constant term $\frac 12$ is not relevant. That is
$$\int \sin 2x \,dx=-\frac12\cos (2x)+C_1 \tag 1$$
and
\begin{align} \int \sin 2x \,dx&=-\cos^2 x+C_2\\\\ &=-\frac12\cos 2x+(-\frac12 +C_2)\\\\ &=-\frac12\cos 2x+C_3\tag 2 \end{align}
where we absorbed the constants $-\frac12+C_2$ into a new constant and called that new constant $C_3$. Inasmuch as the integration constant is arbitrary, $(1)$ and $(2)$ are equivalent statements.
• Thanks, it's clear now. I was thrown off by the cosines having a different power. – Rubenknex Sep 3 '15 at 20:04
• You're welcome. My pleasure. Although it can be confusing, you were on the right track. – Mark Viola Sep 3 '15 at 20:30
Integration process gives you results correct up to an arbitrary constant.
Both results are essentially the same.
$$- \frac{\cos 2 x}{2} + C_1 = - \frac { 2 \cos ^2 x -1}{2} + C_1 = - \cos ^2 x + C_2$$
That is why it is better to write $C_1, C_2$ for the integration constants. | 2021-07-24T15:18:55 | {
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http://math.stackexchange.com/questions/23809/what-is-the-remainder-of-16-is-divided-by-19 | # What is the remainder of $16!$ is divided by 19?
Can anyone share me the trick to solve this problem using congruence?
Thanks,
Chan
-
If you've seen Wilson's theorem then you know the remainder when $18!$ is divided by $19$. To get $16!$ mod $19$, then, you can multiply by the multiplicative inverses of $17$ and $18$ mod $19$. Do you know how to find these? (Edit: Referring to both inverses might be a bit misleading, because you really only need to invert their product. See also Bill Dubuque's answer.)
-
I knew Wilson's Theorem, but really don't know how to find the inverse of 17, 18 mod 19. What's the inverse? Can you give me one example? Thanks. – Chan Feb 26 '11 at 6:17
@Chan: $\rm 17 \equiv -2,\ 18\equiv -1$ – Bill Dubuque Feb 26 '11 at 6:21
@Chan: A general technique to find the inverse of $k$ mod $n$ when $k$ and $n$ are relatively prime is to use integer division and the Euclidean algorithm to find $a$ and $b$ such that $ak + bn = 1$. Since $bn$ is a multiple of $n$, $ak$ is congruent to $1$ mod $n$, and thus the congruence class of $a$ is the inverse of the congruence class of $k$. In this particular case, a further hint to make things easier is that $18\equiv -1$ and $17\equiv -2$ mod $19$. (This makes computations easier for inverting their product.) – Jonas Meyer Feb 26 '11 at 6:21
Many thanks, I got it now ;) – Chan Feb 26 '11 at 6:24
HINT $\$ By Wilson's theorem $\$ mod $19\::\ -1 \ \equiv\ 18\:!\ \equiv\ 18\cdot17\cdot 16\:!\ \equiv\ (-1)\ (-2)\cdot 16\:!$
Similarly we have the Wilson reflection formula $\rm\displaystyle\ (p-1-k)\:!\ \equiv\ \frac{(-1)^{k+1}}{k!}\ \ (mod\ p)\:,\$ $\rm\:p\:$ prime
-
@Bill Dubuque: Thank you! – Chan Feb 26 '11 at 6:24
@Bill Dubuque: If I have $2.16! \equiv 18 \pmod{19}$. Can I cancel out the $2$ from both sides? – Chan Feb 26 '11 at 6:42
@Chan: $\rm\ 2\:x\equiv 2\:y\ (mod\ 19)\ \iff\ 19\ |\ 2\ (x-y)\ \iff\ 19\ |\ x-y\ \iff\ x\equiv y\ (mod\ 19)$ – Bill Dubuque Feb 26 '11 at 6:50
@Bill Dubuque: Thank you. So if $gcd(a, p) = 1$, then $ax \equiv ay \pmod{p} \implies x \equiv y \pmod{p}$, right? – Chan Feb 26 '11 at 7:07
@Chan: $\rm\ n,m\$ coprime $\rm\iff a\ n + b\ m = 1\$ for some $\rm\:a,b\:\ \iff\ a\ n\equiv 1\ (mod\ m)\:,\:$ for some $\rm\:a\:$ $\rm\iff\ n$ is invertible/cancellable $\rm\:(mod\ m)$ – Bill Dubuque Feb 26 '11 at 7:12
I almost think in this case it is just faster to break it down than calculate the inverses: First the factors between 1 and 10:
$$10!= (2\cdot 10)\cdot (4\cdot5)\cdot(3\cdot6)\cdot(7\cdot8)\cdot 9\equiv 1\cdot 1\cdot (-1)\cdot(-1)\cdot 9\equiv9$$
Now we have
$$16!\equiv 9\cdot 11\cdot 12\cdot 13\cdot 14\cdot 15\cdot 16\equiv9\cdot(-8)\cdot(-7)\cdot(-6)\cdot(-5)\cdot(-4)\cdot(-3)$$
$$\equiv 9\cdot(4\cdot5)\cdot(6\cdot3)\cdot(7\cdot8)\equiv 9\cdot(1)\cdot(-1)\cdot(-1)\equiv 9$$
Of course this doesn't generalize, but the computation is faster than finding 3 inverses and multiplying them. (Again of course that is not true for larger $n$...)
- | 2015-08-04T03:24:44 | {
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