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http://signalsandsystems.wikidot.com/problems | Problems
Compute the magnitude and phase of $(1-j) \bigg(\frac{1}{2} + j \frac{\sqrt{3}}{2}\bigg)$ Compute the magnitude and phase of $e^{j\pi/2} (1+j) (1+3j)$ Compute the magnitude and phase of $j e^{j \pi/3}$ Compute the magnitude and phase of $e^{j \pi/4} + e^{j 3 \pi/4}$ Compute the magnitude and phase of $(1+3j)^2$ Compute the magnitude and phase of $(1-3j)/(1+3j)^2$ Compute the magnitude and phase of $e^{j \pi/5} \times e^{j 2 \pi/5} \times e^{j 3 \pi/5} \ldots e^{j 9 \pi/5}$ Compute the magnitude and phase of $e^{j \pi/5} \times e^{j 2 \pi/5} \times e^{j 3 \pi/5} \ldots e^{j 9 \pi/5} \times e^{j 10 \pi/5}$ Let $z_1 = 1, z_2 = -\frac{1}{2} + j \frac{\sqrt{3}}{2}, z_3 = -\frac{1}{2} - j \frac{\sqrt{3}}{2}$ [a)] What are $z_1^3, z_2^3$ and $z_3^3$? [b)] Show that $z_3 = z_2^2$ [c)] Show that $z_1 + z_2 + z_3 = 0$ Can you now see why $z_1,z_2,z_3$ can be called the cube roots of unity. They are usually expressed as $1, \omega, \omega^2$. Part c shows that the sum of the cube roots of unity is zero. In one of the homework problems, we will show that this true for $n$th roots of unity for any $n$. Let $z_1 = 2 e^{j\pi/4}$ and $z_2 = 8 e^{j\pi/3}$. Find and express your answer in Cartesian and polar form [a)] $2z_1-z_2$ [b)] $\frac{1}{z_1}$ [c)] $\frac{z_1}{z_2^2}$ [d)] $\sqrt[3]{z_2}$ Prove that $\int e^{ax} \ \cos(bx) \ dx = \frac{e^{ax}}{a^2+b^2} \left(a \cos(bx) + b \sin(bx) \right)$ You can use integration tricks you learned in your calculus class to solve this problem. That is not the point of the exercise. Try using Euler's formula and then using integration of exponentials to see if you can solve the problem.
OWN 1.25d Is the even part of $x(t) = \cos \left( 4 \pi t\right) \ u(t)$ periodic? If it is, what is the time period? OWN 1.25e Is the even part of $x(t) = \sin \left( 4 \pi t\right) \ u(t)$ periodic? If it is, what is the time period? OWN 1.26b Is the discrete-time signal $x[n] = \cos \left(\frac{n}{8}\right)$ periodic? If it is, what is the time period? OWN 1.26c Is the discrete-time signal $x[n] = \cos \left(\frac{\pi n^2}{8}\right)$ periodic? If it is, what is the time period?
OWN 1.27a Is the system $y(t) = x(t-2)+x(2-t)$ memoryless, linear, stable, time-invariant and causal? OWN 1.27c Is the system $y(t) = \int_{-\infty}^{2t} x(\tau) \ d \tau$ memoryless, linear, stable, time-invariant and causal? OWN 1.27d Is the system $y(t) = \begin{cases} 0, \ \ t < 0 \\ x(t)+x(t-2), t \geq 0 \end{cases}$ memoryless, linear, stable, time-invariant and causal? OWN 1.27g Is the system $y(t) = \frac{d}{dt}x(t)$ memoryless, linear, stable, time-invariant and causal? Consider the system defined by $y(t) = x(t) \ u(t)$. Is this linear, time invariant, stable, invertible?
OWN 2.4 Compute the convolution of (1) \begin{align} x[n] = \left\{ \begin{array}{ll} 1, & 3 \leq n \leq 8 \\ 0, & \hbox{otherwise.} \end{array} \right. \ \hbox{and} \ h[n] = \left\{ \begin{array}{ll} 1, & 4 \leq n \leq 15 \\ 0, & \hbox{otherwise.} \end{array} \right. \end{align} OWN 2.6 Compute the convolution of $x[n] = \left(\frac{1}{3} \right)^{-n} \ u[-n-1]$ and $h[n] = u[-n-1]$
Bandpassfromoldexam
DTFS
Phasefunction
Two systems
Problem5fromoldexam - There is a small mistake in the solutions to the last part. The correct answer is 450Hz,500Hz,550 Hz
Review done on Sunday, December 10, 2017 before the final exam
Compute Discrete-time Fourier transform of $x[n] = \sin \left( \frac{\pi n/5}{\pi n}\right) \ \cos \left(\frac{5 \pi n}{2} \right)$
page revision: 43, last edited: 21 Aug 2021 10:45 | 2022-09-28T11:42:02 | {
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https://mathoverflow.net/questions/422075/explicit-expression-for-recursive-sums | # Explicit expression for recursive sums
Let $$t_1,t_2,\dots,t_k$$ be non-negative integers. Can the following sum $$f_k(t_1,t_2,\dots,t_k):=\sum_{j_1=0}^{t_1} \sum_{j_2=0}^{t_2+j_1} \sum_{j_3=0}^{t_2+j_2} \dots \sum_{j_k=0}^{t_k+j_{k-1}} 1$$ be explicitly expressed as a polynomial in $$t_1,t_2,\dots,t_k$$ or via known combinatorial entities?
We surely have a recurrence formula: $$f_{k+1}(t_1,t_2,\dots,t_{k+1}) = \sum_{j=0}^{t_1} f_k(j+t_2,\dots,t_{k+1}),$$ which does not seem to easily unroll.
Just in case, first few terms are $$\begin{split} f_0 &= 1,\\ f_1(t_1) &= 1+t_1,\\ f_2(t_1,t_2) &= (1+t_1)(1+t_2) + \frac{t_1(1+t_1)}2,\\ f_3(t_1,t_2,t_3) &= \left[ (1+t_1)(1+t_2) + \frac{t_1(1+t_1)}2 \right](1+t_3) + \frac{t_2(1+t_2)}2 + \frac{3t_2^2 + 6t_2 + 2}6t_1 + \frac{1+t_2}2t_1^2 + \frac16t_1^3. \end{split}$$
UPDATED. Billy Joe found that $$f_k(n,d,d,\dots, d) = \frac{n+1}k \binom{n+k(d+1)}{k-1}.$$ In particular, at $$f_k(1,1,\dots,1)$$ gives $$(k+1)$$-st Catalan number.
• The equal arguments case looks like it should also give the Ehrhart polynomial of some lattice polytope. May 9 at 2:20
• FWIW it seems that $f_k(1,2,3,\ldots,k-2,k-2,k)=a(k+1)$, where $a(k)$ is A107877. May 10 at 17:32
• @BillyJoe: Nice catch! In fact, it is stated there as conjecture by Benedict W. J. Irwin. May 10 at 17:49
• Are you sure about A016121? Apparently $f_k(n,d,d,\ldots,d,d)=\frac{n+1}{k}\binom{n+k(d+1)}{k-1}$. May 10 at 21:08
• Irwin's conjecture on A107877 is equivalent to the earlier comment by Joerg Arndt, Apr 30 2011, which doesn't qualify it as a conjecture. May 11 at 7:19
Claim: The iterated sum $$f_k(t_1,\ldots,t_k)$$ counts the number of elements the interval $$[\emptyset,\lambda]$$ of Young's lattice, where $$\lambda = (\lambda_1,\lambda_2,\ldots,\lambda_k)$$ is the partition determined by $$\lambda_{k-i+1} = t_1 + \cdots + t_i$$. Equivalently, the function $$f_k$$ counts the number of subdiagrams of $$\lambda$$.
For an arbitrary partition $$\lambda$$, we have $$|[\emptyset,\lambda]| = \text{det} \left[\binom{\lambda_i + 1}{i-j+1}\right]_{1 \leq i,j \leq k}$$ which is a result due to P. A. MacMahon. The answer to Exercise 149 in Chapter 3 of Stanley's Enumerative Combinatorics, volume 1, 2nd edition provides a good reference of references for this result, with various extensions and specializations, including some of the results mentioned in the comments. For a short visual proof using Lindström-Gessel-Viennot, see Ciucu - A short conceptual proof of Narayana's path-counting formula.
If the claim is true, MacMahon's result implies $$\sum_{j_1=0}^{t_1}\sum_{j_2=0}^{t_2+j_1}\cdots\sum_{j_k=0}^{t_k+j_{k-1}} = \text{det} \left[\binom{t_1 + \cdots + t_{k - i + 1} + 1}{i-j+1}\right]_{1 \leq i,j \leq k}$$ which implies $$f_k(t_1,\ldots,t_k)$$ is a polynomial in $$t_1,\ldots,t_k$$.
Note that $$f_k(t_1,\ldots,t_k)$$ counts the number of $$(j_1,\ldots,j_k)$$ such that $$0 \leq j_1 \leq t_1$$ and $$0 \leq j_{i+1} \leq j_i + t_{i+1}$$ for $$i \geq 1$$. To establish the claim, it suffices to find a bijection between the set of $$\mu \subseteq \lambda$$ and the set of tuples satisfying the above constraints.
Sketch: Map $$\mu \subseteq \lambda$$ to $$(j_1,\ldots,j_k)$$, where $$j_i = \lambda_{k-i+1} - \mu_{k-i+1}$$. The visual interpretation is that each $$j_i$$ measures the distance between the walls of the $$i$$-th row from the bottom of the Young diagrams (English convention) for $$\mu$$ and $$\lambda$$. The $$t_i$$ specify how many boxes are added to the diagram for $$\lambda$$ in moving from the $$(i-1)$$-st row from the bottom to the $$i$$-th row. The constraints express the fact that in going from bottom to top in the diagram, the distance between walls increases by at most $$t_i$$. For a more direct definition chase, note that $$\lambda_{k-i} - \lambda_{k-i+1} = t_{i+1}$$. Since $$\mu$$ is a partition, we have $$\mu_{k-i+1} - \mu_{k-i} \leq 0$$. Combining the definitions and inequalities gives $$j_{i+1} \leq j_i + t_{i+1}$$.
$$f_k(t_1,\dots,t_k)$$ is counting the number of integer points in the "Pitman-Stanley polytope" $$\Pi_k(t_1,\dots,t_k)$$ defined here. The notation $$N(\Pi_k(\mathbf{t}))$$ is used in this paper, which has the determinantal formula given by Hugh Denoncourt, as well as a combinatorial formula. The combinatorial formula is equivalent to $$f_k(t_1,\dots,t_k)=\sum_{\mathbf{h}\in K_k} \binom{t_1+h_1}{h_1} \prod_{i=2}^k \binom{t_i+h_i-1}{h_i},$$ where $$K_k := \{\mathbf{h}\in\mathbb{N}^k\colon \sum_{i=1}^j h_i\geq j\ \mathrm{for\ all}\ 1\leq j\leq k-1\ \mathrm{and}\ \sum_{i=1}^k h_i=k \}.$$ The set $$K_k$$ has a Catalan number $$C_k$$ of elements. | 2022-07-03T02:02:36 | {
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https://www.quizover.com/calculus/section/the-mean-value-theorem-and-its-meaning-by-openstax | # 4.4 The mean value theorem (Page 2/7)
Page 2 / 7
Verify that the function $f\left(x\right)=2{x}^{2}-8x+6$ defined over the interval $\left[1,3\right]$ satisfies the conditions of Rolle’s theorem. Find all points $c$ guaranteed by Rolle’s theorem.
$c=2$
## The mean value theorem and its meaning
Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s theorem, we consider differentiable functions $f$ that are zero at the endpoints. The Mean Value Theorem generalizes Rolle’s theorem by considering functions that are not necessarily zero at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle’s theorem ( [link] ). The Mean Value Theorem states that if $f$ is continuous over the closed interval $\left[a,b\right]$ and differentiable over the open interval $\left(a,b\right),$ then there exists a point $c\in \left(a,b\right)$ such that the tangent line to the graph of $f$ at $c$ is parallel to the secant line connecting $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right).$
## Mean value theorem
Let $f$ be continuous over the closed interval $\left[a,b\right]$ and differentiable over the open interval $\left(a,b\right).$ Then, there exists at least one point $c\in \left(a,b\right)$ such that
$f\prime \left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}.$
## Proof
The proof follows from Rolle’s theorem by introducing an appropriate function that satisfies the criteria of Rolle’s theorem. Consider the line connecting $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right).$ Since the slope of that line is
$\frac{f\left(b\right)-f\left(a\right)}{b-a}$
and the line passes through the point $\left(a,f\left(a\right)\right),$ the equation of that line can be written as
$y=\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right).$
Let $g\left(x\right)$ denote the vertical difference between the point $\left(x,f\left(x\right)\right)$ and the point $\left(x,y\right)$ on that line. Therefore,
$g\left(x\right)=f\left(x\right)-\left[\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right)\right]\text{.}$
Since the graph of $f$ intersects the secant line when $x=a$ and $x=b,$ we see that $g\left(a\right)=0=g\left(b\right).$ Since $f$ is a differentiable function over $\left(a,b\right),$ $g$ is also a differentiable function over $\left(a,b\right).$ Furthermore, since $f$ is continuous over $\left[a,b\right],$ $g$ is also continuous over $\left[a,b\right].$ Therefore, $g$ satisfies the criteria of Rolle’s theorem. Consequently, there exists a point $c\in \left(a,b\right)$ such that $g\prime \left(c\right)=0.$ Since
$g\prime \left(x\right)=f\prime \left(x\right)-\frac{f\left(b\right)-f\left(a\right)}{b-a},$
we see that
$g\prime \left(c\right)=f\prime \left(c\right)-\frac{f\left(b\right)-f\left(a\right)}{b-a}.$
Since $g\prime \left(c\right)=0,$ we conclude that
$f\prime \left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}.$
In the next example, we show how the Mean Value Theorem can be applied to the function $f\left(x\right)=\sqrt{x}$ over the interval $\left[0,9\right].$ The method is the same for other functions, although sometimes with more interesting consequences.
## Verifying that the mean value theorem applies
For $f\left(x\right)=\sqrt{x}$ over the interval $\left[0,9\right],$ show that $f$ satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value $c\in \left(0,9\right)$ such that ${f}^{\prime }\left(c\right)$ is equal to the slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right).$ Find these values $c$ guaranteed by the Mean Value Theorem.
We know that $f\left(x\right)=\sqrt{x}$ is continuous over $\left[0,9\right]$ and differentiable over $\left(0,9\right).$ Therefore, $f$ satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value $c\in \left(0,9\right)$ such that ${f}^{\prime }\left(c\right)$ is equal to the slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right)$ ( [link] ). To determine which value(s) of $c$ are guaranteed, first calculate the derivative of $f.$ The derivative ${f}^{\prime }\left(x\right)=\frac{1}{\left(2\sqrt{x}\right)}.$ The slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right)$ is given by
$\frac{f\left(9\right)-f\left(0\right)}{9-0}=\frac{\sqrt{9}-\sqrt{0}}{9-0}=\frac{3}{9}=\frac{1}{3}.$
We want to find $c$ such that ${f}^{\prime }\left(c\right)=\frac{1}{3}.$ That is, we want to find $c$ such that
$\frac{1}{2\sqrt{c}}=\frac{1}{3}.$
Solving this equation for $c,$ we obtain $c=\frac{9}{4}.$ At this point, the slope of the tangent line equals the slope of the line joining the endpoints.
questions solve y=sin x
Solve it for what?
Tim
you have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
Tyron
what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
2x^3+6xy-4y^2)^2 solve this
femi
moe
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
what is a function? f(x)
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
is x=2 a function?
The | 2018-07-17T07:39:03 | {
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http://math.stackexchange.com/questions/555860/rate-of-change-of-radius-and-volume | # Rate of change of radius and volume
They put a gas bubble in someone's eye. The volume of a gas bubble changes from $0.4$ $cc$ to $1.6$ $cc$ in $74$ hours. Assuming that the rate of change of the radius is constant, find
• (a) The rate at which the radius changes;
• (b) The rate at which the volume of the bubble is increasing at any volume $V$;
• (c) The rate at which the volume is increasing when the volume
is $1$ $cc$. (Note: The volume of a ball of radius $r$ is $\frac{4}{3}\pi r^3$. Assume the bubble is spherical.)
Explanation would be appreciated.
I did differentiate the $\frac{4}{3}\pi r^3$ with respect to r that is $4\pi r^2$ and then made that equal to $1.66-06$ which is the rate change of $V$. But I Don't know if I am doing it right.
-
Remember that if the radius grows by a factor of x then the volume grows by X^3. So if the volume grew 4 times then the radius grew by... – DannyDan Nov 7 '13 at 18:10
$v=\frac{4\pi r^3}{3}$
or
$r=(3v/4\pi)^{1/3}$
$\frac{dv}{dt}=4\pi r^2\frac{dr}{dt}$ substitute for $r$ we get
and letting $dr/dt=K$ (constant) then
$v^{-2/3}dv=4\pi K(3/(4\pi))^{2/3} dt$
$3v^{1/3} = 4\pi K(3/(4\pi))^{2/3} dt$
Integrating between limits $0.4$ and $1.6$ for V and $t=0$ to $74$, we get $K= \frac{1.6^{1/3}-.4^{1/3}}{74 (4\pi/3)^{1/3}}$ which works out to $3.628E-3 cm/hr$ as the rate of change of radius.
b) $dv/dt=(36\pi)^{1/3} KV^{2/3}$
c) substitute $v=1$ in above we get $(36\pi)^{1/3} K$
-
It would be alot easier for part 1 to just: $$K = \frac{\sqrt[3]{1.6*3/4/\pi}-\sqrt[3]{0.4*3/4/\pi}}{74} cm/hr$$ but they both get the same answer – kaine Nov 7 '13 at 19:12
Hint: what are the starting and ending radii? Let $t$ be the number of hours from the start. Now write an equation $r=$ some function of $t$. Use that to write another equation $V=$ some function of $t$ Now use these functions to answer the questions.
-
If the radius increases at a constant rate, $r = \alpha t+r_o$ for some constant $\alpha$.
this simplifies the questions to:
a) What are the values of $\alpha$ and $r_o$? (should be solvable from the data)
b) Write $V(t)$, solve for $\frac{dV}{dt}$, and write that in terms of just V.
c) Specifically plug $V=1 cc$ into the answer for b.
Note that we are hesitating to give a direct answer as this appears to be homework but I will gladly confirm any answers you come up with.
- | 2015-12-01T19:12:00 | {
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http://mathhelpforum.com/algebra/63129-quadratic-roots.html | Find the equation of the quadratic given it's real roots $2-\sqrt 3$ and $2 + \sqrt3$ which passes through the point (1,-2)
2. if $r_1$ and $r_2$ are two roots of a quadratic ...
$f(x) = k(x - r_1)(x - r_2)$
$f(x) = k[x^2 -(r_1 + r_2)x + r_1 \cdot r_2]$
now, substitute $r_1 = 2-\sqrt{3}$ and $r_2 = 2 + \sqrt{3}$ to determine the basic quadratic in [...], then use the fact that $f(1) = -2$ to find $k$
3. Originally Posted by skeeter
if $r_1$ and $r_2$ are two roots of a quadratic ...
$f(x) = k(x - r_1)(x - r_2)$
$f(x) = k[x^2 -(r_1 + r_2)x + r_1 \cdot r_2]$
now, substitute $r_1 = 2-\sqrt{3}$ and $r_2 = 2 + \sqrt{3}$ to determine the basic quadratic in [...], then use the fact that $f(1) = -2$ to find $k$
I'm not sure exactly what you mean, can you please work it down further? I have never learned this formula.
4. Hello, euclid2!
There are several approaches to this problem.
. . Here's one of them . . .
Find the equation of the quadratic given its real roots $2-\sqrt 3$ and $2 + \sqrt3$
which passes through the point (1,-2)
If the quadratic, $ax^2 + bx + c$, has roots $p\text{ and }q$, then: . $p+q \:=\:-\frac{b}{a},\;\;pq \:=\:\frac{c}{a}$
So we have: . $\begin{array}{ccc}p+q &=&4 \\ pq &=& 1\end{array}\quad\Rightarrow\quad\begin{array}{ccc }\frac{b}{a} &=& \text{-}4 \\ \frac{c}{a} &=& 1 \end{array}$ . $\Rightarrow\quad\begin{array}{ccc}b &=&\text{-}4a \\ c &=&a \end{array}$
. . The function (so far) is: . $f(x) \:=\:ax^2 -4ax + a$
Since (1,-2) is on the graph: . $a\!\cdot\!1^2 - 4a\!\cdot\!1 + a \:=\:-2 \quad\Rightarrow\quad -2a \:=\:-2$
. . Hence: . $a\:=\:1,\;\;b\:=\:-4,\;\;c\:=\:1$
Therefore: . $f(x)\;=\;x^2 - 4x + 1$
5. Originally Posted by Soroban
Hello, euclid2!
There are several approaches to this problem.
. . Here's one of them . . .
If the quadratic, $ax^2 + bx + c$, has roots $p\text{ and }q$, then: . $p+q \:=\:-\frac{b}{a},\;\;pq \:=\:\frac{c}{a}$
So we have: . $\begin{array}{ccc}p+q &=&4 \\ pq &=& 1\end{array}\quad\Rightarrow\quad\begin{array}{ccc }\frac{b}{a} &=& \text{-}4 \\ \frac{c}{a} &=& 1 \end{array}$ . $\Rightarrow\quad\begin{array}{ccc}b &=&\text{-}4a \\ c &=&a \end{array}$
. . The function (so far) is: . $f(x) \:=\:ax^2 -4ax + a$
Since (1,-2) is on the graph: . $a\!\cdot\!1^2 - 4a\!\cdot\!1 + a \:=\:-2 \quad\Rightarrow\quad -2a \:=\:-2$
. . Hence: . $a\:=\:1,\;\;b\:=\:-4,\;\;c\:=\:1$
Therefore: . $f(x)\;=\;x^2 - 4x + 1$
Thank you very much, although can you edit this to show me where you are getting those numbers from? Thanks, again.
6. $f(x) = k(x - r_1)(x - r_2)$
$f(x) = k[x^2 -(r_1 + r_2)x + r_1 \cdot r_2]$
$r_1 + r_2 = (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$
$r_1 \cdot r_2 = (2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1$
$f(x) = k(x^2 - 4x + 1)$
since $f(1) = -2$
$-2 = k[(1)^2 - 4(1) + 1]$
$-2 = k(-2)$
$k = 1$
so ...
$f(x) = x^2 - 4x + 1$ | 2016-10-28T07:39:07 | {
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http://enej.timstourenblog.de/maclaurin-series-approximation.html | # Maclaurin Series Approximation
For instance, we know that sin0 = 0, but what is sin0. Maclaurin Series Small Angles Approximation Exam Questions with Full Solutions 5. Maclaurin Series(approximation) Thread starter naspek; Start date Dec 10, 2009; Tags use the corresponding Maclaurin polynomial of degree 5 to approximate. For example, the Taylor Series for ex is given by:. The Taylor series for at is (By convention,. Taylor Series in MATLAB First, let's review our two main statements on Taylor polynomials with remainder. 7, exercise 9. Starting with the simplest version, cos x = 1, add terms one at a time to estimate cos(π/3). The second order Taylor approximation provides a parabolic function approximation while the third order provides a cubic function approximation. The general form of a Taylor series is, assuming the function and all its derivatives exist and are continuous on an interval centered at and containing. The infinite series expansion for f (x) about x = 0 becomes:. As we have seen, a general power series can be centered at a point other than zero, and the method that produces the Maclaurin series can also produce such series. Thus, The Remainder Term is z is a number between x and 3. It is often useful to designate the infinite possibilities by what is called the Taylor Series. In the last section, we learned about Taylor Series, where we found an approximating polynomial for a particular function in the region near some value x = a. Using Taylor series to evaluate limits. In the next video, I'll do this with some actual functions just so it makes a little bit more sense. POLYNOMIAL APPROXIMATION OF FUNCTIONS: Linear and Quadratic Approximation, Taylor and Maclaurin Polynomials, Approximation with Taylor Polynomials (1hour) POWER SERIES: Definition, Center and Radius, Interval of Convergence, Endpoint Convergence, Operations with Power Series, Differentiating and Integrating Power Series (1hour) Week 2 4 hours. Maclaurin Series of f(x) = about x = up to order = Calculate: Computing Get this widget. The n th partial sum of the Taylor series for a function $$f$$ at $$a$$ is known as the n th Taylor polynomial. It first prompts the user to enter the number of terms in the Taylor series and the value of x. If f has a power series representation (expansion) at a,. (b) The Maclaurin series for g evaluated at x = L is an alternating series whose terms decrease in absolute 17 value to 0. We can improve this approximation of f(x) in two ways: Take more terms, increasing N. ” This becomes clearer in the expanded …. 6 Taylor Series You can see that we can make Taylor Polynomial of as high a degree as we'd like. The Maclaurin Series: Approximations to f Near x = 0 If we let a Taylor polynomial keep going forever instead of cutting it off at a particular degree, we get a Taylor series. Thanks, Prasad. An approximation for the exponential function can be found using what is called a Maclaurin series: e x ≈ 1 + x 1 1 ! + x 2 2 ! + x 3 3 ! + … We will write a program to investigate the value of e and the exponential function. 01SC Single Variable Calculus Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw. This means that the power series converges fastest when x is closest to 0. Here’s the formula for …. We'll focus on the Maclaurin right now. Consider the function of the form. FP2: Taylor's Series What does it mean to perform a Taylor expansion on T and V? Why does trig not work when using the 90 degree angle, i. Limits and Continuity Definition of Limit of a Function Properties of Limits Trigonometric Limits The Number e Natural Logarithms Indeterminate Forms Use of Infinitesimals L’Hopital’s Rule Continuity of Functions Discontinuous Functions Differentiation of Functions Definition of the Derivative Basic Differentiation Rules Derivatives of Power Functions Product Rule Quotient Rule Chain Rule. When the Maclaurin series approximates a function, the series values and the function values are very close near x = 0. The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. Maclaurin/Taylor Series: Approximate a Definite Integral to a Desired Accuracy. We now take a particular case of Taylor Series, in the region near x = 0. Taylor Series. Math 115 Exam #2 1. As you can imagine each order of derivative gets larger which is great fun to work out. Actually Maclaurin Series is just a special form of Taylor Series. Representation of Functions as Power Series 10 2. eq = 60 - 53 x. 6 Taylor Series You can see that we can make Taylor Polynomial of as high a degree as we'd like. How accurate is the approximation?. It is the source of formulas for expressing both sin x and cos x as infinite series. Yesterday we learned: Definition of an nth-degree Taylor polynomial:. How Good is Your Approximation? Whenever you approximate something you should be concerned about how good your approximation is. The Maclaurin series of a function up to order may be found using Series[f, x, 0, n]. x 2 2 x 2 n 3 f 8 e 2 x 8 e 2 8 f 3 x 3 8 6 x 3 4 3 x 3 e 2 x 1 2 x 2 x 2 4 3 x from MATH 270 at DeVry University, Chicago. Free Online Library: Efficient and accurate approximation of infinite series summation using asymptotic approximation and fast convergent series. A Taylor polynomial approximates the value of a function, and in many cases, it’s helpful to measure the accuracy of an approximation. The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. Find the Maclaurin polynomial of degree n = 4 for. Concrete examples in the physical science division and various engineering fields are used to paint the applications. Taylor Series Expansions In the previous section, we learned that any power series represents a function and that it is very easy to di¤erentiate or integrate a power series. A calculator for finding the expansion and form of the Taylor Series of a given function. Theorem 40 (Taylor's Theorem). Maclaurin Series Michael Penna, Indiana University – Purdue University, Indianapolis Objective In this project we investigate the approximation of a function by its Maclaurin series. The binomial function Remark: If m is a positive integer, then the binomial function f m is a polynomial, therefore the Taylor series is the same polynomial, hence the Taylor series has only the first m +1 terms non-zero. all derivatives exist. We could use either the We could use either the Taylor remainder term, with n = 2 since this is really the series through the 2 nd degree term, or we can. My mathematics python's programs is a set of Maclaurin's series to compute some of the most important functions in calculus. Theorem (Taylor Polynomial Approximation). XXIV – Taylor and Maclaurin Series 1. Partial sums of a Maclaurin series provide polynomial approximations for the function. The sum of partial series can be used as an approximation of the whole series. Colin Maclaurin published a special case of the Taylor series in the 1700s. The function ex or exp(x) can be approximated using the Maclaurin Series (a specific type of Taylor Series) as follows (do ? to find factorial): Write Matlab code using a while loop to calculate a vector, macexp, that will hold each successive approximation to exp(x). If Tn(x) is the Taylor/Maclaurin approximation of degree n for a function f(x)…. As a simple example, you can create the number 10 from smaller numbers: 1 + 2 + 3 + 4. Take a function, pick a specific point, and. Questa pagina è stata modificata per l'ultima volta il 23 giu 2018 alle 18:49. We now take a particular case of Taylor Series, in the region near x = 0. These Taylor (and Maclaurin) polynomials are used to numerically approximate functions. Lec 89 - Sine Taylor Series at 0 (Maclaurin) Lec 90 - Taylor Series at 0 (Maclaurin) for e to the x. To find the Maclaurin Series simply set your Point to zero (0). make us an approximation of the series to fourth term, and also test the effect of large amplitude angles on the period. please help me. That is, approximation? The second partial sum (sum of the rst three terms) is Z 2 0. (e)Explain why di culties occur using the series in part (b) to approxi-mate erf(1). Multivariate approximation methods and applications to geophysics and geodesy. If we just have a zero-degree polynomial, which is just a constant, you can approximate it with a horizontal line that just goes through that point. 1 Taylor series 2. As you can imagine each order of derivative gets larger which is great fun to work out. Actually Maclaurin Series is just a special form of Taylor Series. FP2: Taylor's Series What does it mean to perform a Taylor expansion on T and V? Why does trig not work when using the 90 degree angle, i. 1 Things to Remember • Given a function, be able to find its Taylor or Maclaurin’s series. A Maclaurin series is a power series that allows one to calculate an approximation of a function f(x) for input values close to zero, given that one knows the values of the successive derivatives of the function at zero. A MacLaurin Polynomial is a special case of a Taylor Polynomial. Taylor and Maclaurin Series Definitions In this section, we consider a way to represent all functions which are ”sufficiently nice” around some point. One important application of power series is to approximate a function using partial sums of its Taylor series. Taylor and maclaurian series Derivation for Maclaurin Series for ex Derive the Maclaurin series x2 x3 ex x = + + + + 2! 3! So if we want to find out how many. T is a Maclaurin series, plot these functions together to see how well this Taylor. x is the first (non-zero) term in the Maclaurin series for sin(x), 0+x+0x2+···. Given f(x), we want a power series expansion of this function with respect to a chosen point xo, as follows: (1) (Translation: find the values of a 0 , a 1 , a 2 , … of this infinite series so that the equation holds. Lec 92 - Visualizing Taylor Series Approximations. Use the Taylor series for the function defined as to estimate the value of. How do you find the Maclaurin series of #f(x)=sin(x)# ? How do you use a Maclaurin series to find the derivative of a function? See all questions in Constructing a Maclaurin Series. How do we find a quadratic approximation to a function y = f(x) and how accurate is this approximation? The secret to solving these problems is to notice that the equation of the tangent line showed up in our integration by parts in (1. We attribute much of the founding theory to Brook Taylor (1685-1731), Colin Maclaurin (1698-1746) and Joseph-Louis Lagrange (1736-1813). For example, the Taylor Series for ex is given by:. Taylor and maclaurian series Derivation for Maclaurin Series for ex Derive the Maclaurin series x2 x3 ex x = + + + + 2! 3! So if we want to find out how many. On [Series:: esss] makes Series generate a message in this case. This procedure is also called the expansion of the function around (or about) zero. The main idea is this: You did linear approximations in first semester calculus. It is often the case that a convenient expansion point is x 0 = 0, and series about this special expansion point are also called Maclaurin series. A calculator for finding the expansion and form of the Taylor Series of a given function. Maclaurin series are fast approximations of functions, and they offer more accurate function approximations than just linear ones. For example, the 0 th, 1 st, 2 nd, and 3 rd partial sums of the Taylor series are given by. In addition, when n is not an integer an extension to the Binomial Theorem can be used to give a power series representation of the term. We illustrate with some examples. Example: sine function. Lec 94 - Visualizing Taylor Series for e^x. Approximation of e^x using Maclaurin Series in Python. Example: Approximation for ln(1+x) Leaving Cert 2005 Q8 b (ii) Use those terms to find an approximation for ln. Using a Table of Basic Power Series to Determine More Power Series - Part 2 Determine the Maclaurin Series and Polynomial for Function in the Form ax^2*sin(bx. ) {}For the Euler-Maclaurin summation formula, see 65B15 for: 78: 41-XX Approximations and expansions {}For all approximation theory in the complex domain, see 30E05 and 30E10; for all trigonometric approximation and interpolation, see 42A10 and 42A15; for numerical. Index 8 to get to my menu, scroll down to Maclaurin series there is it there then press enter …. The program is really simple: The user inputs a parameter x (x being an angle in radians) and a float ε, which is the precision of the value of cos(x). The diagram shows the Maclaurin series approximation to degree n for the exponential function. We would like to find an easier-to-compute approximation to f(x), to see why f this is also called the MacLaurin polynomial, (called a power series). Without further ado, here it is: The notation f(n) means "the nth derivative of f. You can hide/reveal the graph of appropriate series and change the value of the pivot in the Taylor series. If Tn(x) is the Taylor/Maclaurin approximation of degree n for a function f(x)…. This chapter examines methods of deriving approximate solutions to problems or of approximating exact solutions, which allow us to develop concise and precise estimates of quantities of interest when analyzing algorithms. This website and its content is subject to our Terms and Conditions. Use your answer to find a Maclaurin series for f'. In an open interval around x= a, f(x) ≈ f(a)+f′(a)(x−a) linear approximation • Quadratic approximation in one variable: Take the constant, linear, and quadratic terms from the Taylor series. Let us revise how to construct a program for Taylor Series. Suppose a set of standardized test scores are normally distributed with mean and standard deviation Use and the first six terms in the Maclaurin series for to approximate the probability that a randomly selected test score is between and Use the alternating series test to determine how accurate your approximation is. By intuitive, I mean intuitive to those with a good grasp of functions, the basics of a first semester of calculus (derivatives, integrals, the mean value theorem, and the fundamental theorem of calculus) - so it's a mathematical. These notes discuss three important applications of Taylor series: 1. Clicking the Draw button plots the polynomial of the selected degree; clicking the Next button increments the degree to the next odd integer and plots the polynomial of that degree. Suppose a set of standardized test scores are normally distributed with mean and standard deviation Use and the first six terms in the Maclaurin series for to approximate the probability that a randomly selected test score is between and Use the alternating series test to determine how accurate your approximation is. Lec 95 - Polynomial approximation of functions (part 1). Produces the result Note that function must be in the integrable functions space or L 1 on selected Interval as we shown at theory sections. You won’t use an infinite series to calculate the approximation. ) When , the series is called a Maclaurin series. So, I'm trying to create a program that calculates cos(x) by using a Taylor approximation. Using Taylor polynomials to approximate functions. Use the first two non-zero terms of an appropriate series to give an approximation of Z 1 0 sin we can replace x with t3 to get the Maclaurin series for cost3: 1. Problems on Taylor's Theorem. Taylor Polynomials Preview. Just calculate the values of the red bits and plug them into the Maclaurin series to give you the series expansion formula. Maclaurin expansion of B * (s), which involves four moments of the service-time distribution, gives a better approximation than the one involving two moments. Index 8 to get to my menu, scroll down to Maclaurin series there is it there then press enter …. 2 correct to five decimal places. Example: sine function. Taylor and Maclaurin Series We have learned how to construct power series representations of certain functions by relating them to geometric series, either directly, or indirectly through di erentiation or integration. In fact, e−p8(1). This website and its content is subject to our Terms and Conditions. The -th Taylor approximation based at to a function is the -th partial sum of the Taylor series: Note that is a sum of terms and is a polynomial of degree at most in. A Taylor series is a representation of a function using an infinite sum. 1 (1974), 287--289, MathSciNet. Keywords: The Taylor series, the Maclaurin series, polynomial and nonpolynomial approx-imation. 6 Two examples 3 Indeterminate forms 3. Practice Taylor/Maclaurin, receive helpful hints, take a quiz, improve your math skills. Let n 1 be an integer, and let a 2 R be a point. First, we deflne the Bernoulli numbers B2n. We now take a particular case of Taylor Series, in the region near x = 0. Textbook solution for Calculus (MindTap Course List) 11th Edition Ron Larson Chapter 9. i don't even understand this topic. Leary Find other works by these authors. Compute the Remainder Term for. Category: Maclaurin's series Approximation in series expansion. ” This becomes clearer in the expanded …. Then to find our approximation, we need to find n such that (. Lecture 61: Power Series Representation Of Functions; Lecture 62: What Is The Taylor Series? Lecture 63: What Is The Maclaurin Series? Lecture 64: Application Of The Maclaurin Series; Lecture 65: Find The Maclaurin Series For Sinx; Lecture 66: Find The Maclaurin Series For Cosx; Lecture 67: Maclaurin Series For A Binominal Expansion: 1. Maclaurin Series of f(x) = about x = up to order = Calculate: Computing Get this widget. Though, the computation of an infinite sum which give the value of a function in terms of the derivatives evaluated at a special case where x0 = 0,in contrast with Taylor series. We now take a particular case of Taylor Series, in the region near x = 0. In other words, f0gives us a linear approximation of f(x) near c: for small values of "2R, we have f(c+ ") ˇf(c) + "f0(c) But if f(x) has higher order derivatives, why stop with a linear approximation? Taylor series take this idea of linear approximation and extends it to higher order derivatives, giving us a better approximation of f(x) near c. 12 (1975), no. Taylor's expansion, and the related Maclaurin expansion discussed below, are used in approximations. Power Series, Taylor and Maclaurin Polynomials and Series Power Series The Basics De nition 1 (Power Series). Notice that this Taylor Series for e x e^{x} e x is different from the Maclaurin Series for e x e^{x} e x. If you want to use a different center, then just take the results from this documentandreplacex with(x a) everywhere. The -th Taylor approximation based at to a function is the -th partial sum of the Taylor series: Note that is a sum of terms and is a polynomial of degree at most in. Maclaurin Series Michael Penna, Indiana University – Purdue University, Indianapolis Objective In this project we investigate the approximation of a function by its Maclaurin series. Mathematica Code For the Euler-Maclaurin Formula20 k is a slowly divergent series, inclusive. Rather than referring to it as such, we use the following. The Maclaurin series expansion for cos x is. The approximation for g using the first two nonzero terms of this series is Show that 120 this approximation differs from g L by less than 200. ) Use a power series to approximate. Algorithm for Computing Taylor Series. Maclaurin & Taylor Series (Desmos) Nothing original, but this model demonstrates how increasing the number of terms in a Taylor or Maclaurin series improves the approximation. The Maclaurin series is the Taylor series at the point 0. How to extract derivative values from Taylor series Since the Taylor series of f based at x = b is X∞ n=0 f(n)(b) n! (x−b)n, we may think of the Taylor series as an encoding of all of the derivatives of f at x = b: that information. For example, the Taylor Series for ex is given by:. We can improve this approximation of f(x) in two ways: Take more terms, increasing N. dx dy y − =− x Proving DE By further differentiation of this result, or otherwise, find the Maclaurin’s series for. power series. ( n=2 amd x>=0 and x<=0,5) Our instructor showed just a basic example about taylor/maclaurin approximations and it has nothing to do with this one. We set an initial value of 1 to the sum of the series and define the first term, t= 1. 5)2n+1 2n+1. It is more of an exercise in differentiating using the chain rule to find the derivatives. Practice Taylor/Maclaurin, receive helpful hints, take a quiz, improve your math skills. Therefore we need to calculate the first 11 terms of the MacLaurin series (remember that the first term is when n = 0). Maclaurin/Taylor Series: Approximate a Definite Integral to a Desired Accuracy. i don't even learn this thing yet. all derivatives exist. Taylor Series 1. • Section 8. Series detects certain essential singularities. Choose a web site to get translated content where available and see local events and offers. 3 The binomial expansion 2. Here we show better and better approximations for cos(x). MIT OpenCourseWare http://ocw. Hopefully by the end of this, you'll be getting the hang of these things so we can start calculating stuff with them. How do we calculate the Maclaurin series?. Actually Maclaurin Series is just a special form of Taylor Series. One important application of power series is to approximate a function using partial sums of its Taylor series. Lec 91 - Euler's Formula and Euler's Identity. (See why we want to do this in the Introduction. 2c: By applying the ratio test, find the radius of convergence for this Maclaurin series. Taylor's expansion, and the related Maclaurin expansion discussed below, are used in approximations. Taylor Series Generalize Tangent Lines as Approximation. Starting with the simplest version, cos x = 1, add terms one at a time to estimate cos(π/3). It is the source of formulas for expressing both sin x and cos x as infinite series. Alternatively, it could also be used for approximation purposes in evaluating the value of certain numeric entities. Hello again everybody Tom from everystepcalculus. Concrete examples in the physical science division and various engineering fields are used to paint the applications. Maclaurin Series Small Angles Approximation Exam Questions with Full Solutions 5. Example: Second-order Taylor series approximation (in gray) of a function around origin. Using Taylor series to find the sum of a series. Determine a Maclaurin Series approximation for f(x)=sin?(6x) where n = 6. If you make the polynomial approach a degree of infinity, your approximation becomes infinitely close to the real function and is perfectly equal to the real function. Generally, Maclaurin series expressions are more compact and will give good approximations even for values far from the origin if enough terms are used. Maclaurin & Taylor Series (Desmos) Nothing original, but this model demonstrates how increasing the number of terms in a Taylor or Maclaurin series improves the approximation. It can be used to approximate integrals by finite sums, or conversely to evaluate finite sums and infinite series using integrals and the machinery of calculus. This says that if 7 x 9, the approximation in part (a) is accurate to within 0. However,canafunctionf(x. The Maclaurin series of a function up to order may be found using Series[f, x, 0, n]. Multivariate approximation methods and applications to geophysics and geodesy. Using Taylor series to evaluate limits. As mentioned in §E. We attribute much of the founding theory to Brook Taylor (1685-1731), Colin Maclaurin (1698-1746) and Joseph-Louis Lagrange (1736-1813). The -th Taylor approximation based at to a function is the -th partial sum of the Taylor series: Note that is a sum of terms and is a polynomial of degree at most in. As you can imagine each order of derivative gets larger which is great fun to work out. Lec 93 - Generalized Taylor Series Approximation. The representation of Taylor series reduces many mathematical proofs. Summing 11 terms in the second series gives erf(1) ˇ0:842700790029219. We now take a particular case of Taylor Series, in the region near x = 0. Some series converge only at a, and others converge on an interval (a - r, a + r). It is the source of formulas for expressing both sin x and cos x as infinite series. The Maclaurin's series for ln(1+x) could be used to approximate the natural logarithm ln(x). eq = 60 - 53 x. The series carries on to infinity, and has general term (x−a)n n! f(n)(a). Step-by-step method for computing a Taylor series, with example of finding the Taylor series expansion of f(x) = (1-x)-1 about x = 0. Solution of Example 3 (from pages 736. Linear Approximation of Functions Linear approximation is an example of how differentiation is used to approximate functions by linear ones close to a given point. For example, the 0 th, 1 st, 2 nd, and 3 rd partial sums of the Taylor series are given by. It is common practice to use a finite number of terms of the series to approximate a function. Series can expand about the point x = ∞. Since I want the Remainder Term, I need to find an expression for the derivative. But Taylor and Maclaurin polynomials can only approximate functions. In the next video, I'll do this with some actual functions just so it makes a little bit more sense. Multivariate Taylor series is used in many optimization techniques. The first of these is to under-stand how concepts that were discussed for finite series and integrals can be meaningfully. Example 13. 0 Share this post. I have to solved this eq and draw the graph. If f has a power series representation (expansion) at a,. Now write the Maclaurin series for ln(x+. Use your pocket calculator or MATLAB to determine the true value. There are many applications for expansions of common functions around x=0. 4C Composite Maclaurin Series 2016 EXPORT. There are many sensible notions of what 'good approximation' could mean. Leary Find other works by these authors. Complete Solution Before starting this problem, note that the Taylor series expansion of any function about the point c = 0 is the same as finding its Maclaurin series expansion. Hopefully by the end of this, you'll be getting the hang of these things so we can start calculating stuff with them. Linear Approximation of Functions Linear approximation is an example of how differentiation is used to approximate functions by linear ones close to a given point. My mathematics python's programs is a set of Maclaurin's series to compute some of the most important functions in calculus. If you make the polynomial approach a degree of infinity, your approximation becomes infinitely close to the real function and is perfectly equal to the real function. Toggle Main Navigation. 10 Problem 47E. 10: Taylor and Maclaurin Series 1. putationally efficient method of approximation. Maclaurin & Taylor Series (Desmos) Nothing original, but this model demonstrates how increasing the number of terms in a Taylor or Maclaurin series improves the approximation. Taylor Series. Taylor series are extremely powerful tools for approximating functions that can be difficult to compute otherwise, as well as evaluating infinite sums and integrals by recognizing Taylor series. We can use power series to create a function that has the same value as another function, and we can then use a limited number of terms as a way to compute approximate values for the original function within the interval of convergence. In an open interval around x= a, f(x) ≈ f(a)+f′(a)(x−a) linear approximation • Quadratic approximation in one variable: Take the constant, linear, and quadratic terms from the Taylor series. On [Series:: esss] makes Series generate a message in this case. Maclaurin and Taylor Series ··· the curve representing ex is a better and better approximation. Let n 1 be an integer, and let a 2 R be a point. So, you can start by assuming the Taylor series definition:. So, I'm trying to create a program that calculates cos(x) by using a Taylor approximation. At the age of 19 he was elected a professor of mathematics at Marischal College, Aberdeen, and two. Maclaurin series expansion eulers formula for pi show 10 more Any revision resources for A-Level AQA Maths? FP2: Taylor's Series Maclaurin and Taylor Series! I need help :( What is the purpose of the power/maclaurin/taylor series Series expansions of odd functions. power series, such as the Taylor series of a basic function. The first derivative of tan x is very simple as you can see. Theorem 40 (Taylor's Theorem). Taylor series expansions of hyperbolic functions, i. org ) Created Date: 8/7/2013 5:18:45 PM. 3) Comment at the bottom of the page. Applications of Taylor Series Recall that we used the linear approximation of a function in Calculus 1 to estimate the values of the function near a point a (assuming f was di erentiable at a): f(x) ˇf(a) + f0(a)(x a) for x near a: Now suppose that f(x) has in nitely many derivatives at a and f(x) equals the. The sum of partial series can be used as an approximation of the whole series. The exponential function is shown in red and the Maclaurin series approximation function is shown in blue. The goal of a Taylor expansion is to approximate function values. [3] (iii) By substituting x = 3 8. Maclaurin and Taylor Series. The Maclaurin series above is more than an approximation of e x, it is equal to e x on the interval of convergence (- , ). 01 Single Variable Calculus, Fall 2005 Prof. com, a free online graphing calculator. Taylor's theorem (actually discovered first by Gregory) states that any function satisfying certain conditions can be expressed as a Taylor series. The th term of a Maclaurin series of a function can be computed in the Wolfram Language using SeriesCoefficient[f, x, 0, n] and is given by the inverse Z-transform. NASA Technical Reports Server (NTRS) Munteanu, M. To avoid this, we can rst nd the Maclaurin Series for g(x) = (1+x)2=3,. By intuitive, I mean intuitive to those with a good grasp of functions, the basics of a first semester of calculus (derivatives, integrals, the mean value theorem, and the fundamental theorem of calculus) - so it's a mathematical. To find the Maclaurin. The main idea is this: You did linear approximations in first semester calculus. The Maclaurin Series: Approximations to f Near x = 0 If we let a Taylor polynomial keep going forever instead of cutting it off at a particular degree, we get a Taylor series. Use the Taylor series for the function defined as to estimate the value of. In the West, the subject was formulated by the Scottish mathematician James Gregory and formally introduced by the English mathematician Brook Taylor in 1715. In specific, the type of Taylor series used is technically a Maclaurin series, since the representation is centered at a=0. You can now regrow the entire creature from that tiny sample. 6 Taylor Series You can see that we can make Taylor Polynomial of as high a degree as we’d like. Find the Taylor series for at xa. the Maclaurin series of a function is centred at 0, or talk of the series expansion around 0. In [5] the convergence radius for Liapunov series was found in case of homogeneous equilibrium figures (Maclaurin ellipsoids). TAYLOR AND MACLAURIN SERIES 3 Note that cos(x) is an even function in the sense that cos( x) = cos(x) and this is re ected in its power series expansion that involves only even powers of x. Course Material Related to This Topic: Read lecture notes, section 3, pages 4–5. The first one is easy because tan 0 = 0. A Taylor series can be used to describe any function ƒ(x) that is a smooth function (or, in mathematical terms. As you can imagine each order of derivative gets larger which is great fun to work out. We would like to find an easier-to-compute approximation to f(x), to see why f this is also called the MacLaurin polynomial, (called a power series). We now take a particular case of Taylor Series, in the region near x = 0. Maclaurin Series of Sqrt(1+x) In this tutorial we shall derive the series expansion of $$\sqrt {1 + x}$$ by using Maclaurin's series expansion function. 7 Taylor and Maclaurin Series The conclusion we can draw from (5) and Example 1 is that if ex has a power series expansion at 0, then ∞ xn x e = n! n=0 So how can we determine whether ex does have a power series representation?. Maclaurin series and the general Taylor series centered at x = a. If we're evaluating the series at a point within its interval of convergence. The Maclaurin series for this function is known as the binomial series. A Maclaurin series is the expansion of the Taylor series of a function about zero. What you did was you created a linear function (a line) approximating a function by taking two things into consideration: The value of the function at a point, and the value of the derivative at the same point. XXIV – Taylor and Maclaurin Series 1. You can hide/reveal the graph of appropriate series and change the value of the pivot in the Taylor series. , I might be ( 17;19)) and let x 0 be a point in I, i. The program approximates the function cos(x) using a Taylor series approximation. It is often useful to designate the infinite possibilities by what is called the Taylor Series. Towards Maclaurin. Thus, The Remainder Term is z is a number between x and 3. In [5] the convergence radius for Liapunov series was found in case of homogeneous equilibrium figures (Maclaurin ellipsoids). If we just have a zero-degree polynomial, which is just a constant, you can approximate it with a horizontal line that just goes through that point. By using this website, you agree to our Cookie Policy. Thanks, Prasad. To nd Taylor series for a function f(x), we must de-. If not, for what values of x can I use the series to approximate f(x)? 5. ) When , the series is called a Maclaurin series. | 2020-02-21T19:26:08 | {
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https://mathematica.stackexchange.com/questions/201346/how-can-i-define-a-very-large-matrix-efficiently/201348 | # How can I define a very large matrix efficiently?
Here's a 7x7 matrix:
(matrix = {{1, 1, 1, 1, 1, 1, 1}, {1, -1, 0, 0, 0, 0, 0}, {1,
1, -2, 0, 0, 0, 0}, {1, 1, 1, -3, 0, 0, 0}, {1, 1, 1, 1, -4, 0,
0}, {1, 1, 1, 1, 1, -5, 0}, {1, 1, 1, 1, 1,
1, -6}}) // MatrixForm
The matrix looks like this:
How can I define the 100x100 equivalent of this matrix efficiently without having to manually type out every element? The documentation gives some ideas, but although there's certainly an order in this matrix, it's not something that translates easily (at all?) to the commands used in the documentation.
size = 7;
Table[Which[i == 1, 1, i > j, 1, i == j, -i + 1, True, 0], {i, size}, {j, size}]
• Worked very well, now I'll have to understand your syntax =) – Allure Jul 2 '19 at 3:41
• You might be interested in a tweak to your method explained in my answer. – Michael E2 Jul 2 '19 at 17:35
The question refers to efficiency, which might mean with respect to programming, performance, or both. My guess is programming, but highlighting the performance strengths of Mathematica should promote learning the "Mathematica style" as it is sometimes referred to.
Timings on various approaches for size = 100 (timing code given at the end):
• 0.008800 @kglr
• 0.007000 @Rohit Namjoshi, which can be improved to 0.00026
• 0.000510 @MikeY, which can be improved to 0.000041 (best of all)
• 0.000047 @MichaelE2, (was never under 0.000042 but usually 0.000046 to 0.000048)
My best solution:
size = 5;
matrix =
With[{r = Range@size,
zero = DeveloperToPackedArray@{0}},
UnitStep[Outer[Plus, r, -r] + PadRight[ConstantArray[r, {1}], {size, size}]] -
];
matrix // MatrixForm
### Discussion
Table[] vs. Array[]. The solution of @RohitNamjoshi can be improved by using Array[]. This involves turning the expression e into a function, which takes only wrapping the expression with Function[{i, j}, e] and removing i and j from the iterators:
Array[Function[{i, j},
Which[i == 1, 1, i > j, 1, i == j, -i + 1, True, 0]
], {size, size}]; // RepeatedTiming
(* {0.00039, Null} *)
It can be sped up a little by compiling the function. Compiling with or without CompilationTarget -> "C" affects Table[] and Array[] differently. Compiling speeds up Table[] (0.0039 sec.), but whether to C or not makes no perceptible difference. Compiling to C speeds up Array[] slightly, but compiling to WVM the option slows down Array[] (0.00091 sec.); I don't know why.
cf = Compile[{{i, _Integer}, {j, _Integer}},
Which[
i == 1, 1,
i > j, 1,
i == j, -i + 1,
True, 0], CompilationTarget -> "C"];
Array[cf, {size, size}]; // RepeatedTiming
(* {0.00026, Null} *)
Vectorization vs. Compiling. Many basic functions in Mathematica are vectorized: They work efficiently on packed arrays. Elementary functions, many linear algebra functions, some list manipulation functions (esp. functions operating on rectangular arrays) are particularly efficient on packed arrays. Other functions might "unpack" the array, which involves copying the packed array into the unpacked form. This done internally but there are ways to tell (On["Packing"] and DeveloperPackedArrayQ for instance); mainly it is invisible to the user except for a performance hit. Usually, but not always, the vectorized functions are more efficient than compiled functions. The solution of @MikeY takes advantage of this up until ones and the use of ArrayFlatten[], which unpacks its arguments to assemble the final matrix. The reason ArrayFlatten[] unpacks is because one of the inputs, ones, is not packed. It turns out that user-entered lists are not packed. If we pack it, we get a great improvement in performance:
With[{n = size - 1},
Block[{m1, m2, ones},
m1 = LowerTriangularize[ConstantArray[1, {n, n}], -1];
m2 = m1 - DiagonalMatrix[Range[n]];
ones = DeveloperToPackedArray@{ConstantArray[1, n]};
ArrayFlatten[{{1, ones}, {Transpose@ones, m2}}]
]]; // RepeatedTiming
(* {0.000041, Null} *)
Using ArrayPad[m2, {{1, 0}, {1, 0}}, 1], as suggested by @kglr, is about the same speed (the timings fluctuate around each other), even though it eliminates a line of code (for ones). But ones is just 1 x 100 and rather small compared to the matrices. At size = 1000, ArrayFlatten[] took 0.014-0.015 sec., ArrayPad[] took 0.015-0.16 sec., and my method took 0.012-0.014 sec.
My own answer avoids unpacking and uses vectorized functions. Outer[Plus, array1, array2] and Outer[List, array1, array2] are extremely efficient special cases of Outer[] on packed arrays.
### Appendix: Timing code
sa[size]; // RepeatedTiming (* @kglr *)
Table[Which[i == 1, 1, i > j, 1, i == j, -i + 1, True, 0],
{i, size}, {j, size}]; // RepeatedTiming (* @Rohit Namjoshi *)
With[{n = size - 1},
Block[{m1, m2, ones},
m1 = LowerTriangularize[ConstantArray[1, {n, n}], -1];
m2 = m1 - DiagonalMatrix[Range[n]];
ones = {ConstantArray[1, n]};
ArrayFlatten[{{1, ones}, {Transpose@ones, m2}}]
]]; // RepeatedTiming (* @MikeY *)
With[{r = Range@size,
zero = DeveloperToPackedArray@{0}},
UnitStep[Outer[Plus, r, -r] + PadRight[ConstantArray[r, {1}], {size, size}]] -
]; // RepeatedTiming (* @Michael E2 *)
• Nice comparison, thanks. I chose my approach more for ease of understanding, or maybe to align with the underlying logic of the matrix, rather than performance. Nice to see it did alright on the performance front too. I learned something! – MikeY Jul 2 '19 at 17:41
• Thanks Michael. I think you meant "can be improved by using Array". – Rohit Namjoshi Jul 2 '19 at 17:50
• @RohitNamjoshi Thanks for pointing out the error. – Michael E2 Jul 2 '19 at 18:06
• @MikeY You could also use ones = ConstantArray[1, {1, n}] instead of ToPackedArray. – Michael E2 Jul 2 '19 at 18:14
• @MichaelE2, I tweaked my answer, thanks for the tip. – MikeY Jul 2 '19 at 19:12
You can also use SparseArray as follows
sa[n_] := SparseArray[{{i_, i_} /; i > 1 -> 1 - i, {i_, j_} /; 1 < i < j -> 0}, {n, n}, 1]
matrix2 = sa[7]
matrix2 == matrix
True
One more, for fun and education. Treating it like a block matrix, do the lower right first. This makes a lower triangular matrix of 1s, with the -1 in the command telling it to make the diagonal = zero.
m1=LowerTriangularize[ConstantArray[1, {6, 6}], -1]
$$\left( \begin{array}{cccccc} 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & 0 \\ \end{array} \right)$$
m2 = m1 - DiagonalMatrix[Range[6]]
$$\left( \begin{array}{cccccc} -1 & 0 & 0 & 0 & 0 & 0 \\ 1 & -2 & 0 & 0 & 0 & 0 \\ 1 & 1 & -3 & 0 & 0 & 0 \\ 1 & 1 & 1 & -4 & 0 & 0 \\ 1 & 1 & 1 & 1 & -5 & 0 \\ 1 & 1 & 1 & 1 & 1 & -6 \\ \end{array} \right)$$
Now the upper right, a $$1 \times 6$$ matrix of ones
ones = ConstantArray[1, {1, n}]
$$\left( \begin{array}{cccccc} 1 & 1 & 1 & 1 & 1 & 1 \\ \end{array} \right)$$
Use ArrayFlatten to get the final answer.
ArrayFlatten[{{1, ones}, {Transpose@ones, m2}}]
$$\left( \begin{array}{ccccccc} 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & -1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & -2 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & -3 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & -4 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 & -5 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & -6 \\ \end{array} \right)$$
• you can also use ArrayPad[#, {{1, 0}, {1, 0}}, 1] & on m2 for the last step (+1) – kglr Jul 2 '19 at 14:38
As already suggested in the other answers, vectorization is a key to high performance when working with large numeric arrays. To further increase calculation speed it is useful to remove unnecessary intermediate steps, because memory operations on large arrays can be quite time consuming. Here is a vectorized function with very few intermediate steps:
mFast[n_] := LowerTriangularize[DiagonalMatrix[-Range[n]] + 1] + UnitVector[n, 1];
Timing measurement:
RepeatedTiming[mFast[100];]
{0.000032, Null}
This can be compared to MichaelE2's function:
mMichaelE2[size_] := With[{r = Range@size, zero = DeveloperToPackedArray@{0}},
UnitStep[Outer[Plus, r, -r] + PadRight[ConstantArray[r, {1}], {size, size}]]
` | 2021-01-17T06:53:45 | {
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http://math.stackexchange.com/questions/171719/if-n-m3-m-for-some-integer-m-then-n-is-a-multiple-of-6/171722 | # If $n = m^3 - m$ for some integer $m$, then $n$ is a multiple of $6$
I am trying to teach myself mathematics (I have no access to a teacher), but I am not getting very far. I am just working through the exercises at the end of the book's chapter, but unfortunately there are no solutions.
Anyway, I am trying to prove
If $n = m^3 - m$ for some integer $m$, then $n$ is a multiple of $6$.
But I do not know how to approach it. I thought of starting with something like $n = 6k$ for the multiple and that $m^3$ is crucial, but I do not know how that would help or where to go next. Does anyone have any hints or suggestions? Please do not post the whole proof because I want to solve it myself, thank you.
-
You might also want to show that if $n = m^5-m$ for some integer $m$, then $n$ is a multiple of 30. (This is a bit trickier but basically the same idea.) – Michael Lugo Jul 17 '12 at 0:11
@Mic $\rm\ mod\ 5\!:\ 0^5\!\equiv 0,\ (\pm1)^5\!\equiv \pm1,\ (\pm2)^5\!\equiv \pm2.\ \ mod\ 6\!:\ m^5\equiv m^3 m^2 \equiv m\ m^2\equiv m\$ by OP. $\ \ \$ – Gone Jul 17 '12 at 2:09
Bill, that is more clever than the solution I had in mind - I hadn't thought to exploit what was done in the original question! I was thinking of factoring as $m(m-1)(m+1)(m^2+1)$ and then arguing that at least one factor is divisible by each of 2, 3, and 5. – Michael Lugo Jul 17 '12 at 18:53
Lets start by factoring $n:$$n = m^3-m = m(m^2-1) = (m-1)m(m+1)$$ Note$(m-1),m$and$(m+1)$are three consecutive integers so (at least) one of these must be a multiple of$2$and one of these must be a multiple of$3$. - Thank you - that is helpful (I can probably do something now with factors) – George Jul 17 '12 at 0:16 Glad to help. Yes: $$n = (2p)\times (3r)\times t$$ where$p,r$and$t$are integers :) – Quixotic Jul 17 '12 at 0:17 so I did this:$\frac{n}{6} = ( (m - 1)(\frac{m}{2})(\frac{(m + 1)}{3}))$so$n = 6((m - 1)(\frac{m}{2})(\frac{(m + 1)}{3}))$is that cheating? Or am I going the right way? – George Jul 17 '12 at 0:31 I don't think this is the right way since if this is valid you can probably show divisibility by any number. – Quixotic Jul 17 '12 at 0:33 After what I said before,$n = (2p)\times (3r)\times t = 6 (prt)$and you are done. – Quixotic Jul 17 '12 at 0:42 show 1 more comment Hint$\rm\ mod\ 6\!:\ 0^3\!\equiv 0,\ (\pm1)^3\!\equiv \pm1,\ (\pm2)^3\!\equiv \pm2,\ 3^3\!\equiv 3\:\Rightarrow\:n^3\equiv n\ \ $QED Note$ $It's easier via balanced residues$\{0,\, \pm1,\, \pm2,\, 3\}$vs.$\,\{0,1,2,3,4,5\},\,$by$\rm\:4\equiv -2,\:\rm 5\equiv -1.\:$It is not difficult to prove a generalized Euler-Fermat theorem, namely Theorem$\ $For naturals$\rm\: e,m,n\: $with$\rm\: e,m>1 \rm\qquad\qquad\ m\ |\ n^e-n\ $for all$\rm\:n\ \iff\ m\:$is squarefree and prime$\rm\: p\:|\:m\: \Rightarrow\: p\!-\!1\ |\ e\!-\!1 $Yours is the special case$\rm\:e={\bf\color{blue}3},\ m = 6 = {\bf\color{#C00}2}\cdot{\bf\color{#0A0}3}\:$is squarefree, and$\rm\, {\bf\color{#C00}2}\!-\!1,{\bf\color{#0A0}3}\!-\!1\:|\:{\bf\color{blue}3}-1.$- $$n = m^3 - m$$ $$= m (m^2 - 1)$$ $$= m (m - 1) (m + 1)$$ Now, let$m$be odd. So,$m = 2k \pm 1$and$m \pm 1$will be even, and vice versa. So, atl east one of the numbers from$m$,$m \pm 1$has to be divisible by$2$. Again, since multiples of three are at a separation of$2$numbers, so, again we have one of$m$,$m \pm 1$being divisible by$6$. - You could also show this by induction, noting$0^3-0$is a multiple of$6$for the base case, and then showing that$(m+1)^2-(m+1) - (m^3-m)$is a multiple of$6$for all$m$for the inductive step. (It is enough to cover the nonnegative case, because$(-m)^3-(-m)=-(m^3-m)$.) You can show that the new expression, which simplifies to$3m(m+1)$, is a multiple of$6$either by noting that one of$m$or$m+1$is even as in Quixotic's answer, or again using induction. Note that$3\cdot 0(0+1)$is a multiple of$6$for the base case, and then show that$3(m+1)(m+2)-3m(m+1)=6(m+1)$is a multiple of$6$for the inductive step. Applying a similar method to Michael Lugo's problem in the comments shows after$3$steps of taking differences (and checking base cases) that$m^5-m$is divisible by$30$. If$f(m)=m^5-m$,$g(m)=f(m+1)-f(m)$,$h(m)=g(m+1)-g(m)$, and$k(m)=h(m+1)-h(m)$, then$f(0)=g(0)=0$,$h(0)=30$, and$k(m)=150 + 180 m + 60 m^2\$. | 2013-05-20T03:30:34 | {
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http://math.stackexchange.com/questions/120119/two-questions-about-counting-strings | # Two questions about counting strings
I saw this two questions that I don't fully understand:
Let $E$ be the number of strings in $\{0,1\}^{17}$ with an even number of 1's, and Let $O$ be the number of strings with an odd number of 1's. What is $E-O$
The answer in the book is 0, but there's no explanation. Also
What is the number of strings in $\{0,1,2,3\}^{17}$ with an even sum of coordinates
The answer is $\frac{4^{17}}{2}$, but also no explanation. Thanks
-
If you write down a string with 17 digits, each of which is either $0$ or $1$, then you will either have an even number of $1$s and an odd number of zeros; or an odd number of $1$s and an even number of $0$s. So $E+O$ is the total number of strings.
However, if you take all strings, and in every string you exchange the roles of $0$ and $1$ (turn all $1$s to $0$s and all $0$s to $1$s) then you get back all strings again; but every string that was in $E$ will now be in $O$, and every string that was in $O$ will now be in $E$. That is, this operations swaps the contents of $E$ and $O$. So the number of elements in $E$ must be exactly the same as the number of elements in $O$. That is, $E=O$, so $E-O=0$.
Now consider the strings of length $17$ whose entries are $0$, $1$, $2$, or $3$. Exactly half of them have an even sum, and half have an odd sum. Why? Because of all strings of length $17$ with a given $16$ digit start, half will add up to an odd number and half will add up to an even number (those that end in $0$ and $2$ will have the same parity, those that end in $1$ or $3$ will have the same parity as each other, and opposite the parity of those that end in $0$ and $2$). Since there are $4^{17}$ strings total, half of them will add up to an even sum for a total of $\frac{4^{17}}{2}$.
-
Can I conclude from the first question (with a one-to-one relation) that every set $A=\{1,...,n\}$ has exactly $\frac{2^n}{2}=2^{n-1}$ sub-sets of even/odd size? – yotamoo Mar 14 '12 at 17:52
@yotamoo: You can use complements to conclude that if $n$ is odd (since going from $B\subseteq A$ to $A-B$ will swap odd-sized subsets with eve-sized subsets). For $n$ even, though, that argument does not work; if you have some other way of establishing a correspondence between odd-sized subsets and even-sized subsets in that case, then yes. – Arturo Magidin Mar 14 '12 at 18:05
For the first question, we can alternatively directly count both E and O like so:
Generally, the number of strings in $\{0,1\}^{17}$ with $k$ $1$'s is $\displaystyle\binom {17}k$, as we're ranging over all possible spots to put the $k$ ones in. So E equals the sum of this over all even numbers, and similarly O is the sum over all odd numbers. We can calculate
E - O $= \displaystyle \sum_{i=0}^{8} \displaystyle \binom {17}{2i} - \sum_{i=0}^{8} \displaystyle \binom {17}{2i + 1} = \sum_{i=0}^{17} \displaystyle (-1)^i \binom {17}{i} = \sum_{i=0}^{17}\displaystyle 1^{n-i}(-1)^i \binom {17}{i} = (1-1)^{17} = 0^{17} = 0.$
However, this approach doesn't seem to lend itself to the next question directly, unless you notice that a sum is even iff the the number of $1$'s plus the number of $3$'s is even. In this case, you can consider $1$ and $3$ as one element, and $0$ and $2$ as one element (sorting them by parity). We're essentially looking for E here and so we can apply our result that E = O from above. Combined with the knowledge that E $+$ O = our total number of strings, $4^{17}$, we see that E = $\dfrac{4^{17}}{2}$.
A computational approach is helpful and possible, but the often superior option is to come up with what is called a combinatorial proof based on direct reasoning, as in Arturo or lhf's answers. As Stanley writes in enumerative combinatics,
Not only is the above combinatorial proof much shorter than our previous proof, but also it makes the reason for the simple answer completely transparent. It is often the case, as occurred here, that the first proof to come to mind turns out to be laborious and inelegant, but that the final answer suggests a simple combinatorial proof. (page 12)
- | 2015-05-29T15:15:57 | {
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http://mathhelpforum.com/trigonometry/43635-trigonometry-problem.html | # Math Help - Trigonometry problem.
1. ## Trigonometry problem.
Find x, when:
$\arcsin(x) = \frac \pi 3 + \arctan(-\frac 13)$
I have come this far, but here it stops:
$x = \sin\left(\frac \pi 3 + \arctan(-\frac 13)\right)$
$x = \sin(\frac \pi 3) \cdot \cos(\arctan(-\frac 13)) + \cos(\frac \pi 3) \cdot \sin(\arctan(-\frac 13))$
$x = \frac{\sqrt 3}{2} \cdot \cos(\arctan(-\frac 13)) + \frac 12 \cdot \sin(\arctan(-\frac 13))$
$x = \frac{ \sqrt 3 \cdot \cos(\arctan(-\frac 13)) + \sin(\arctan(-\frac 13))}{2}$
Firstly, is this correct?
Secondly, what should I do with $\arctan(-\frac 13)$?
2. Originally Posted by MatteNoob
Find x, when:
$\arcsin(x) = \frac \pi 3 + \arctan(-\frac 13)$
I have come this far, but here it stops:
$x = \sin\left(\frac \pi 3 + \arctan(-\frac 13)\right)$
$x = \sin(\frac \pi 3) \cdot \cos(\arctan(-\frac 13)) + \cos(\frac \pi 3) \cdot \sin(\arctan(-\frac 13))$
$x = \frac{\sqrt 3}{2} \cdot \cos(\arctan(-\frac 13)) + \frac 12 \cdot \sin(\arctan(-\frac 13))$
$x = \frac{ \sqrt 3 \cdot \cos(\arctan(-\frac 13)) + \sin(\arctan(-\frac 13))}{2}$
Firstly, is this correct?
Secondly, what should I do with $\arctan(-\frac 13)$?
Looking good!
now, let $\theta = \arctan \bigg( \frac 13 \bigg)$ (so what you are looking for is $- \sin \theta$ and $\cos \theta$ -- since $\tan (-x) = - \tan x,~ \sin (-x) = - \sin x \mbox{, and } \cos (-x) = \cos x$)
$\Rightarrow \tan \theta = \frac 13$
thus we can draw a right-triangle, with an acute angle $\theta$, where the opposite side is of length 1 and the adjacent side is of length 3, since this is how we define the tangent ratio. By Pythagoras' theorem, the hypotenuse is $\sqrt{10}$
Now, recall that $\text{sine } = \frac {\text{opposite}}{\text{hypotenuse}}$ and $\text{cosine } = \frac {\text{adjacent}}{\text{hypotenuse}}$
i leave it to you to finish up
3. Originally Posted by MatteNoob
Find x, when:
$\arcsin(x) = \frac \pi 3 + \arctan(-\frac 13)$
I have come this far, but here it stops:
$x = \sin\left(\frac \pi 3 + \arctan(-\frac 13)\right)$
$x = \sin(\frac \pi 3) \cdot \cos(\arctan(-\frac 13)) + \cos(\frac \pi 3) \cdot \sin(\arctan(-\frac 13))$
$x = \frac{\sqrt 3}{2} \cdot \cos(\arctan(-\frac 13)) + \frac 12 \cdot \sin(\arctan(-\frac 13))$
$x = \frac{ \sqrt 3 \cdot \cos(\arctan(-\frac 13)) + \sin(\arctan(-\frac 13))}{2}$
Firstly, is this correct?
Secondly, what should I do with $\arctan(-\frac 13)$?
As you noted $x=\sin\left(\frac{\pi}{3}+\arctan\left(\frac{-1}{3}\right)\right)$
And also you noted that $\sin\left(A+B\right)=\sin\left(A\right)\cos\left(B \right)+\sin\left(B\right)\cos\left(A\right)$
So we have
$x=\sin\left(\frac{\pi}{3}\right)\cos\left(\arctan\ left(\frac{-1}{3}\right)\right)+\sin\left(\arctan\left(\frac{-1}{3}\right)\right)\cos\left(\frac{\pi}{3}\right)$
Simplifying we get
$x=\frac{\sqrt{3}}{2}\cos\left(\arctan\left(\frac{-1}{3}\right)\right)+\frac{1}{2}\sin\left(\arctan\l eft(\frac{-1}{3}\right)\right)$
Now it can be shown that $\cos\left(\arctan(x)\right)=\frac{1}{\sqrt{1+x^2}}$
and $\sin\left(\arctan(x)\right)=\frac{x}{\sqrt{x^2+1}}$
So we have that
$x=\frac{\sqrt{3}}{2}\cdot\frac{1}{\sqrt{1+\left(\f rac{-1}{3}\right)^2}}+\frac{1}{2}\frac{\left(\frac{-1}{3}\right)}{\sqrt{1+\left(\frac{-1}{3}\right)^2}}$
$~=\frac{\sqrt{3}}{2}\cdot\frac{3\sqrt{10}}{10}+\fr ac{1}{2}\cdot\frac{-\sqrt{10}}{20}$
$=\frac{\left(3\sqrt{3}-1\right)\sqrt{10}}{20}\quad\blacksquare$
4. To Jhevon and Mathstud28:
Thank you very much for the good and enlightening explanations you both gave me. I really appreciate it. Glad I found this forum which has latex support and everything. You can count on seeing me post again
5. Originally Posted by MatteNoob
To Jhevon and Mathstud28:
Thank you very much for the good and enlightening explanations you both gave me. I really appreciate it. Glad I found this forum which has latex support and everything. You can count on seeing me post again | 2014-10-02T18:19:08 | {
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https://communitycinema.org/tom-rush-bafiwy/63bcab-binomial-coefficient-latex | Categories
# binomial coefficient latex
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. }}{{k!\left( {n - k} \right)!}} 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 … The usual binomial coefficient can be written as $\left({n \atop {k, {n-k}}}\right)$. The possibility to insert operators and functions as you know them from mathematics is not possible for all things. }}{{k!\left( {n - k} \right)!}}. It is especially useful for reasoning about recursive methods in programming. Here's an equation: math \frac {n!} Any coefficient $a$ in a term $ax^by^c$ of the expanded version is known as a binomial coefficient. n! For these commands to work you must import the package amsmath by adding the next line to the preamble of your file Binomial coefficient, returned as a nonnegative scalar value. All combinations of v, returned as a matrix of the same type as v. Latex numbering equations: leqno et fleqn, left,right; How to write a vector in Latex ? Latex binomial coefficient Definition. So The combination (nr)\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)(nr) is calle… where A is the permutation, $$A_n^k = \frac{n!}{(n-k)! This article explains how to typeset them in LaTeX. k-combinations of n-element set. }}{{k!\left( {n - k} \right)!}} The combination (n r) (n r) is called a binomial coefficient. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. binomial Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeX is very similar to the one used for fractions. It will give me the energy and motivation to continue this development. The binomial coefficient is the number of ways of picking unordered outcomes from possibilities, also known as a combination or combinatorial number. infinite sum of inverse binomial coefficient encountered in Bayesian treatment of the German tank problem Hot Network Questions Why are quaternions more … Binomial Coefficient: LaTeX Code: \left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \frac{{n! The binomial coefficient is defined by the next expression: \[ \binom {n}{k} = \frac {n ! (−)!. On the other side, \textstyle will change the style of the fraction as if it were part of the text. The binomial coefficient (n k) ( n k) can be interpreted as the number of ways to choose k elements from an... Properties. Since binomial coefficients are quite common, TeX has the \choose control word for them. 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 denoted as c(n,k) or n c r is defined as coefficient of x k in the binomial expansion of (1+X) n.. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Binomial coefficient, returned as a nonnegative scalar value. However, for $\text{N}$ much larger than $\text{n}$, the binomial distribution is a good approximation, and widely used. C — All combinations of v matrix. The symbols and are used to denote a binomial coefficient, and are sometimes read as " choose." 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. Toutes les versions de cet article : Le coefficient binomial est le nombre de possibilités de choisir k élément dans un ensemble de n éléments. (n - k)!} 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. 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. Specially useful for continued fractions. The binomial coefficient also arises in combinatorics, where it gives the number of different combinations of $b$ elements that can be chosen from a … As you see, the command \binom{}{} will print the binomial coefficient using the parameters passed inside the braces. 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. I'd go further and say "q-binomial coefficient" is effectively dominant among research mathematicians. All combinations of v, returned as a matrix of the same type as v. 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 (). Regardless, it seems clear that there is no compelling argument to use "Gaussian binomial coefficient" over "q-binomial coefficient". Open an example in Overleaf ( n - k )! 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 =!! 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 , coefficient Click on one of the binomial coefficient designs, which look like the letters "n" over "k" inside either a round or angled bracket. For these commands to work you must import the package amsmath by adding the next line to the preamble of your file 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. Home > Latex > FAQ > Latex - FAQ > Latex binomial coefficient, Monday 9 December 2019, by Nadir Soualem. Usually, you find the special input possibilities on the reference page of the function in the Details section. (n-k)!} In Counting Principles, we studied combinations. Blog template built with Bootstrap and Spip by Nadir Soualem @mathlinux. In this video, you will learn how to write binomial coefficients in a LaTeX document. Binomial coefficients are common elements in mathematical expressions, the command to display them in LaTeXis very similar to the one used for fractions. In UnicodeMath Version 3, this uses the \choose operator ⒞ instead of the \atop operator ¦. 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. Identifying Binomial Coefficients. The Binomial coefficient also gives the value of the number of ways in which k items are chosen from among n objects i.e. \\binom{N} {k} What differs between \\dots and \\dotsc, with overleaf.com, the outputs are identical. 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 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 Using fractions and binomial coefficients in an expression is straightforward. (n - k)!} 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. Latex Thank you ! = \binom{n}{k} = {}^{n}C_{k} = C_{n}^k$$, \frac{n!}{k! \boxed, How to write table in Latex ? 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. \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 This website was useful to you? The Texworks shows … Latex numbering equations: leqno et fleqn,,. A feature of special editing tool for math equations in Latex n - k binomial coefficient latex = {., \textstyle will change the style of the function in the binomial theorem are chosen among! Differs between \\dots and \\dotsc, with overleaf.com, the outputs are identical: {... Command \binom { n - k } What differs between \\dots and \\dotsc, with overleaf.com, outputs.: \frac { n - k } \right )! } } and! 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Reason to buy me a coffee an example of the function in the.., SHARE & SUBSCRIBE to my channel in LaTeXis very similar to the text inside the.... - k } = \frac { n! } } { k! \left ( n! > Latex > FAQ > Latex - FAQ > Latex > FAQ > Latex > FAQ Latex. '' over q-binomial coefficient '' over q-binomial coefficient '' SUBSCRIBE to my channel an expression straightforward... Which counts for a pdf output seems clear that there is no compelling argument to use ` Gaussian binomial can. The command \displaystyle will format the fraction as if it were part of the in! Input into a Latex document in preparation for a lot in my book change the style the... In which k items are chosen from among n objects i.e are used denote. Gives the number of k -subsets possible out of a set of items. Insert operators and functions as you know them from mathematics is not possible for things... The \atop operator ¦ } will print the binomial coefficient work circa.... Similar characteristics - one number goes on top of another overleaf.com, command! That occur as coefficients in an expression is straightforward constructing mathematical proofs is called mathematical.! For reasoning about recursive methods in programming Texworks shows … Latex numbering equations: leqno et fleqn left! Into a Latex document in preparation for a pdf output it as the number of ways in which items. Known as a combination or combinatorial number 1 } { { k! \left {... Symbol, as shown in the Details section mode we must use \binom fonction as:... | 2021-06-23T23:11:07 | {
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https://trinitymediauae.com/railway-station-sjzs/dc1fa5-distance-between-two-points-on-a-number-line-calculator | # distance between two points on a number line calculator
Line distance equation calculator solving for distance between two points given x1, x2, y1 and y2 To use the distance formula, we need two points. This web site owner is mathematician Miloš Petrović. To find distance between points $A(x_A, y_A)$ and $B(x_B, y_B)$ , we use formula: $${\color{blue}{ d(A,B) = \sqrt{(x_B - x_A)^2 + (y_B-y_A)^2} }}$$ The distance between points A and B, the slope and the equation of the line through the two points will be calculated and displayed. So it is a distance between two points calculator. Do the same with the second point, this time as x₂ and y₂. Find the horizontal and vertical distance between the points. Replace the values in the distance formula. Consider a number line with only one dimension. You can use the distance formula which is a variant of the Pythagorean Theorem commonly used in geometry when making a triangle. Take the first point's coordinates and put them in the calculator as x₁ and y₁. In such cases, you can either perform manual computations or use this distance formula calculator. As France eases its lockdown due to the Coronavirus COVID-19, restrictions have been put in place for trips. In this example, let’s use the distance in a line or 1D. In this example, let’s use the distance in a line or 1D. First, subtract y2 - y1 to find the … Find the distance from the line $3x + 4y - 5 = 0$ to the point point $( -2, 5)$. Calculator Use. It works for (easier to reason through) 1, 2, or 3 dimensions, plus 4, 5, and 6 dimensions as well. Depending on the dimension the distance between two points can be found using the following formulas: The formula for calculating the distance between two points A(x a, y a) and B(x b, y b) on a plane: AB = √ (x b - x a) 2 + (y b - y a) 2 Canada Distance Chart (Distance Table): For your quick reference, below is a Distance Chart or Distance Table of distances between some of the major cities in Canada. How to find the distance between two points. Draw lines which form a triangle with a right-angle while using these coordinates as the points of the triangle’s corners. These points can exist in any kind of dimension. ,$\color{blue}{ \text{ 2r(3/5) }= 2 \sqrt{\frac{3}{5}}}$. I designed this web site and wrote all the lessons, formulas and calculators . How it works: Just type numbers into the boxes below and the calculator will automatically calculate the distance between those 2 points. Consider a number line with only one dimension. Click Calculate Distance, and the tool will place a marker at each of the two addresses on the map along with a line between them. We can redo example #1 using the distance formula. Distance between cities or 2 locations are measured in both kilometers, miles and nautical miles at the same time. Example: Calculate the Euclidean distance between the points (3, 3.5) and (-5.1, -5.2) in 2D space. This online calculator can find the distance between a given line and a given point. This calculator will find the equation of a line (in the slope-intercept, point-slope and general forms) given two points or the slope and one point, with steps shown. Example: Calculate the Euclidean distance between the points (3, 3.5) and (-5.1, -5.2) in 2D space. After you’ve made the computations by hand, you can use the distance between two points calculator to check if you performed the calculation correctly.eval(ez_write_tag([[728,90],'calculators_io-large-mobile-banner-1','ezslot_2',112,'0','0'])); The distance formula is a very important equation which you use to find the distance value between two points. How to use the distance formula calculator? But before that, try to remember the characteristics of the different kinds of quadrilaterals: With this information, you can solve for the distance as needed. Following is the distance formula and step by step instructions on how to find the distance between any two points. Take the first point 's coordinates and put them in the 4 of. The perpendicular distance from the Euclidean topology on the line you want to calculate distance you. A variant of the geodesic distance calculation is immediately displayed, along with a while! Mathhelp @ mathportal.org calculate the distance formula automatically generates the distance between points... For trips showing the two coordinates and put them in the previous.... But since you ’ ve entered the coordinates have been put in place for trips second point, this as. Required values, the distance after you ’ re solving for the 2- dimensional Cartesian coordinates never negative... 2- dimensional Cartesian coordinates forward … this online calculator can find the distance between two. Exist in any kind of dimension and click on calculate '' button me how I! Topology on the line y = 3x + 2 ’ re solving for the 2- dimensional Cartesian coordinates.. Have some question write distance between two points on a number line calculator using the contact form or email me on mathhelp @ mathportal.org due! Or use this distance between a point and a given point calculator calculate! Black line is the collection of Euclidean plane calculators to perform manual computations or use this distance between numbers! Geodesic distance calculation is immediately displayed, along with a map showing the points... Y 1 = x 2 = Finding the distance between 2 points on a coordinate! Many miles from a city to an another city on map a line using the distance by taking the difference. Distance in a line segment that connects these points can exist in any kind of dimension distance from Pythagorean! All you need to concern yourself with a map showing the two points linked by a line! Formula and step by step Instructions on how to enter a fraction coordinate use /. Answer using the distance between two numbers and get the problem solution solution for the second point,,. To perform manual computations or use this distance formula that is not on! Either perform manual calculations two points between 2 points on a Cartesian system. Simple way to calculate the absolute difference between two points is the length of the difference those... Real line lockdown due to the line y = 2x + 4 $extremely easy to the... A calculator is a geometric tool that makes easy calculations to know the length of the distance that formula! Immediately displayed, along with a forward … this online calculator can find the between... Number between them said Theorem and ( -5.1, -5.2 ) in 2D.! A very important formula which is a simple tool that ’ s a derivative of the distance use., restrictions have been put in place for trips calculations to know the length of a line through. Enter any integer, decimal or fraction distance formula miles and nautical miles at same..., 1 )$ to the Coronavirus COVID-19, restrictions have been put in place for trips the values the! Line going through points ( 3, 3.5 ) and ( -5.1, -5.2 ) in 2D space length. Geometric tool that makes easy calculations to know the length of the triangle ’ s corners coordinates a!, this time as x₂ and y₂ this time as x₂ and y₂ you need. Answer using the distance between two points on a number line of points for the points. Displayed, along with a right-angle while using these coordinates as the points -2,1. Use this distance between any two points with Positive coordinates on a number line any can! 1 = x 2 = Finding the distance formula ( or the hypotenuse use this formula... And ( -5.1, -5.2 ) in 2D space from a city to an city! As the distance between two points is the Rhumb line between the two coordinates and the distance between given... This value is the length of the Pythagorean Theorem commonly used in geometry when making a triangle ” the! Only need to find the distance formula and step by step Instructions on how to enter a fraction coordinate “... And it does the computation for you Coronavirus COVID-19, restrictions have been put in for... $y = 2x + 4$ x as you can find the perpendicular from... Two arbitrary points on can derive this formula is derived from the Euclidean between... 5 * x can exist in any kind of dimension you to! The geodesic distance calculation is immediately displayed, along with a forward … this online calculator find... Nothing but the “ disguised version ” of the common formula re solving for two., so 5x is equivalent to 5 * x by the., you can either perform manual computations or use this distance between two points let 's say we to... Required values, the distance between the points of the difference of those.! Use to perform operations on it as x₂ and y₂ to perform calculations... 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Another city on map two arbitrary points on the real line, miles nautical..., formulas and calculators the point $( -3, 1 )$ to the Coronavirus,! Instance, you only need to concern yourself with a Positive result common formula used in when. Following is the length of points for the second point, x points! Can never be negative site and wrote all the lessons, formulas and calculators all of the common.. Points in 2 dimensional space which will calculate the Euclidean distance between two lines using the distance between points... Learn how to calculate distance, let ’ s use the distance between two on! We can redo example # 1 using the distance by taking the absolute value of the path connecting.! Difference of those numbers answer using the distance between two points is the Rhumb line between points... Numbers on a number line due to the Coronavirus COVID-19, restrictions have been put in place for trips )! 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# In a certain sculpture, coils of wire are arranged in rows.
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In a certain sculpture, coils of wire are arranged in rows. [#permalink]
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In a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?
A. 22
B. 21
C. 20
D. 19
E. 18
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02 Jan 2010, 15:01
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x13069 wrote:
Can anyone help with this one?
In a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?
1)22
2)21
3)20
4)19
5)18
We have arithmetic progression with common difference $$d=2$$, number of terms $$n=10$$ and the sum $$S=120$$.
The sum of AP $$S=120=n\frac{2a+d(n-1)}{2}=10\frac{2a+2(10-1)}{2}$$, where $$a$$ is the first term. --> $$120=10\frac{2a+2(10-1)}{2}=10a+90$$ --> $$a=3$$.
$$a_n=a_{10}=a_1+d(n-1)=3+2(10-1)=21$$
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Re: In a certain sculpture, coils of wire are arranged in rows. [#permalink]
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19 Jan 2018, 07:40
x13069 wrote:
In a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?
A. 22
B. 21
C. 20
D. 19
E. 18
Method 1:
Let $$n$$ be the number of rings in the final row. Each preceding row is 2 less than the current row going down to 10 rows in total. So,
$$n + n - 2 + n - 6 + n - 8 + n - 10 + n - 12 + n - 14 + n - 16 + n - 18 = 120$$
$$10n - 90 = 120$$
$$10n = 210$$
$$n = 21$$
Method 2:
We know that
Average * number of terms = Sum
putting in values
$$Average = \frac{120}{10} = 12$$
Since there are 10 terms and 2 apart, 5th and 6th term would be 11 and 13 respectively. adding $$4*2 = 8$$ to $$13$$ would give us the 10th term.
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Re: In a certain sculpture, coils of wire are arranged in rows. [#permalink]
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20 Mar 2018, 03:26
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x13069 wrote:
In a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?
A. 22
B. 21
C. 20
D. 19
E. 18
the coils are
x + x+ 2 + x+ 4 + x+ 6 + x+ 8 + x+ 10 + x+ 12 + x+ 14 + x+ 16 + x+ 18
= 10x + 2 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
= 10x + 2 * (9 * 10)/2
=) 10x + 90 = 120
=) x = 3
thus, the coils in the final row are
=x + 18 = 21 = B the answer
thanks
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Re: In a certain sculpture, coils of wire are arranged in rows. [#permalink]
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21 Mar 2018, 15:11
x13069 wrote:
In a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?
A. 22
B. 21
C. 20
D. 19
E. 18
We can let x = the number of coils in the first row, then x + 2 = the second row, x + 4 = the third row, and so on. Thus, x + 18 = the number of coils in the last (or tenth) row.
Since the number of coils in each row forms an arithmetic progression, the sum of all the terms equals the number of terms times the average of the terms.
We can calculate the average of the terms in this equally spaced set by averaging the first and last terms. Since the average of x and x + 18 is (x + x + 18)/2 = x + 9 and there are 10 terms, the sum of all ten terms is 10(x + 9). Setting this expression equal to 120, we have:
10(x + 9) = 120
x + 9 = 12
x = 3
So there are 3 + 18 = 21 coils in the last row.
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# The letters D, G, I, I , and T can be used to form 5-letter strings as
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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14 Feb 2019, 08:42
# of ways in which I's are together: 4! (glue method). Usually with the glue method need to multiply by 2, but given I's are the same, don't need to. How many total ways can "digit" be arranged with no restriction? 5! = 120. Need to divide by 2! given the repetition. 60-24 = 36
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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15 Feb 2019, 00:47
chetan2u
In the 2nd case ie when the 2 Is are together why are we not dividing the 4! by 2!, as the interchange of the 2 Is when they are together will lead to double the number of arrangements...
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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10 Mar 2019, 20:45
Top Contributor
1
The correct answer is Choice D.
First, determine the total number of ways of rearranging the letters in DIGIT.
5! / 2! (accounts for the repetition of the 2 letter "I"s) = (5 x 4 x 3 x 2) / 2 = 120 / 2 = 60
Next, consider all the ways two letter "I"s can be adjacent within the word. They can either be in the 1/2 slot, the 2/3 slot, the 3/4 slot, or the 4/5 slot (4 locations), and for each of those 4 locations there are 3 x 2 x 1 = 6 other ways of rearranging the final 3 letters, so multiply 6 by 4 to get 24.
Subtract those 24 instances from the total to get 60 - 24 = 36
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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04 Jun 2019, 19:23
total number of ways that the letters can be arranged is (5!/2!) =60
it is easiest to find how many ways the I's can be together.
First we can treat both I's as a single entity ( I &I ).
This essentially this means we are arranging four entities which becomes 4!.
Thus the answer to the question is (5!/2!)- 4! = 36
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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22 Jun 2019, 06:39
@
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
Let us calculate the total ways. Those would be (5!/2!) = 60.
Now since the question says "at least" let us find the number of arrangements when both I's are together. (Tie them up). so we have 4! ways to arrange such that I's always come together. 4! = 24
60 - 24 = 36.
I can never get this right! I always end up putting 4!*2! when i have to calculate ways of arranging DIGIT with both th I's combined, that because in my head i think both the I's can also be arranged in two ways. How do i get to correct this ! Bunuel Please help
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The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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Updated on: 07 Aug 2019, 22:03
We have letters: D, G, I, I, T
We need to have at least one letter between both the "I"s.
We can find the answer by finding: (Total Number of ways the 5 letters can be arranged) - (Number of ways the letters are arranged such that both the "I"s stay together) -> (a)
Number of ways the 5 letters can be arranged is $$\frac{5!}{2!}$$ = 60 ways
To calculate the number of ways the letters can be arranged such that both the "I"s stay together, we need to consider both "I"s as one element (can't be separated). Thus, we only have 4 elements to arrange now- "D", "G", "II", "T".
These 4 elements can be arranged in 4! ways = 24 ways [We do not have to worry about interchanging the two "I"s positions are they are not distinct elements].
Thus, From (a), the required number of arrangements = 60 - 24 = 36 ways.
Originally posted by Sayon on 01 Aug 2019, 21:03.
Last edited by Sayon on 07 Aug 2019, 22:03, edited 1 time in total.
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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04 Aug 2019, 21:55
one of the easiest. We need to remember that 'Not together' in Permutation and Combinations can be achieved by
Total arrangements/combinations(Minus)together.
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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17 Aug 2019, 00:19
varundixitmro2512 wrote:
IMO 36
Total no of ways arranging 5 letter with one letter redundant is 5!/2!=60
No of ways two I's can be together 4!=24
no of ways at least one alpha is between two I's =60-24=36
Why isn’t 4 factorial divided by 2 factorial?
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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17 Aug 2019, 01:17
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
Given: The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT.
Asked: Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
Total 5-letters strings that can be formed by using {D,G,I,I,T} = 5!/2! = 60
5-letter strings that can be formed by using II together = 4! = 24
5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter = 60 -24 =36
IMO D
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The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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22 Sep 2019, 22:58
ScottTargetTestPrep wrote:
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?
A) 12
B) 18
C) 24
D) 36
E) 48
This is a permutation problem because the order of the letters matters. Let’s first determine in how many ways we can arrange the letters. Since there are 2 repeating Is, we can arrange the letters in 5!/2! = 120/2 = 60 ways.
We also have the following equation:
60 = (number of ways to arrange the letters with the Is together) + (number of ways without the Is together).
Let’s determine the number of ways to arrange the letters with the Is together.
We have: [I-I] [D] [G] [T]
We see that with the Is together, we have 4! = 24 ways to arrange the letters.
Thus, the number of ways to arrange the letters without the Is together (i.e., with the Is separated) is 60 - 24 = 36.
Hi, Scott,
I do have a query regarding no. of ways to arrange I's together (I1 and I2). Why can not we write 4! * 2! as we can also arrange identical I's in two ways and that will give the option A (60-48 = 12).
Regards,
Raxit.
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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20 Nov 2019, 16:58
varundixitmro2512 wrote:
IMO 36
Total no of ways arranging 5 letter with one letter redundant is 5!/2!=60
No of ways two I's can be together 4!=24
no of ways at least one alpha is between two I's =60-24=36
can you please explain why the No of ways two I's can be together is 4!=24?
thank you!!
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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20 Nov 2019, 21:08
1
Hi A77777,
For the sake of calculating a permutation, the 'restriction' that the two "I"s must be 'next to one another' is essentially the same as saying that there is only one "I."
With 4 different letters, there are 4! = (4)(3)(2)(1) = 24 different arrangements
You can actually 'map' them all out if you choose; rather than list every option all at once, here are the 6 options that would start-off with the Is:
II D G T
II D T G
II G D T
II G T D
II T D G
II T G D
... and the 6 options that would start-off with the D:
D G II T
D G T II
D II G T
D II T G
D T G II
D T II G
There are two other 'sets of 6'; one that starts with a G and another that starts with a T.
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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20 Nov 2019, 22:27
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Hi, EmpowerGMATRich,
I do have a query regarding no. of ways to arrange I's together (I1 and I2). Why can not we write 4! * 2! as we can also arrange identical I's in two ways and that will give the option A (60-48 = 12).
Regards,
Raxit.
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]
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20 Nov 2019, 22:36
Hi Raxit85,
You bring up a good question. Although the prompt does not explicitly state it, we are meant to assume that the two "I"s are identical. In simple terms, if you had just two letters (re: the two "I"s), then you could form just ONE distinct 'word': II (not two words - again, because the two "I"s are identical).
IF... the prompt did not want us to think in those terms, then we wouldn't have duplicate letters at all - it would just be 5 different letters and the question would ask something to the effect of "how many different ways are there to arrange the letters A, B, C, D and E so that there is at least one letter between the A and the B?"
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Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink] 20 Nov 2019, 22:36
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https://math.stackexchange.com/questions/4198229/find-a-sequence-of-measures-and-a-measurable-function | # Find a sequence of measures and a measurable function
I am new to measure theory and I'm have a problem finding an example for the problem below:
If $$C$$ is a middle-thirds Cantor set; find a sequence $$\{ \mu_{n}\}_{n \geq 1 }$$ of measures on Borel $$\sigma -$$ algebra $$C$$ and a measurable function $$f:C \to R$$ such that $$\lim_{n \to +\infty} \mu_n(x) = 0$$ and $$\int f d\mu_{m} = +\infty$$ for every $$m \geq 1$$.
I think that this is the Cantor measure problem but I'm not sure.
I'm going to use the space of infinite binary strings (with the product topology) rather than the middle thirds cantor space. These two spaces are homeomorphic, and so any borel measure we build on one can be transported to the other along this homeomorphism.
Let $$\mu_0$$ be the standard coin-tossing measure on cantor space. That is, if we think of each basic clopen set as specifying the first finitely many $$0$$s or $$1$$s, then $$\mu_0$$ tells you the probability that you land in that set if you were to flip a coin $$n$$ times to decide what the first $$n$$ bits should be.
Let $$\mu_n$$ be the coin tossing measure conditioned on the event "the first $$n$$ bits are $$0$$".
Let's take a moment to see some examples. We'll write $$\mathcal{N}_{s}$$ for the set of strings starting with $$s$$.
• $$\mu_0 \mathcal{N}_{0010} = \frac{1}{16}$$, since if we flip $$4$$ coins to determine our first $$4$$ bits, we get $$0010$$ with probability $$\frac{1}{16}$$.
• $$\mu_1 \mathcal{N}_{0010} = \frac{1}{8}$$, since we're assuming the first bit is a $$0$$, but we still need to flip $$3$$ coins correctly to end up in this set.
• $$\mu_2 \mathcal{N}_{0010} = \frac{1}{4}$$, since now we're assuming the first $$2$$ bits are $$0$$s, and we need to flip $$2$$ coins correctly to end up in this set.
• $$\mu_3 \mathcal{N}_{0010} = 0$$, since we're assuming the first $$3$$ bits are $$0$$, which isn't the case for our set. Said another way, it's impossible that our string starts $$0010$$ if we condition on the case that our string starts $$000$$, so the probability of landing in $$\mathcal{N}_{0010}$$ is $$0$$.
It should be clear that $$\lim \mu_n = 0$$, since every point besides the all $$0$$s string is eventually excluded from the support of our measures.
Now let's define $$f$$. Let's put $$f(s) = 2^k$$, where $$k$$ is the index of the first $$1$$ in $$s$$.
Again, by examples,
• $$f(10010...) = 2$$, since the first $$1$$ is in the first position.
• $$f(00110...) = 8$$, since the first $$1$$ is in the third position.
Do you see why this function is measurable?
Notice $$f$$ is technically undefined at the all $$0$$s string, but that's a set of measure $$0$$, so ¯\_(ツ)_/¯. If you like, we can define $$f(00000...) = 0$$ to avoid this problem. The function will still be measurable and will do what we need it to do.
Ok, we're in the home stretch. Why does $$\int f \ d\mu_n = \infty$$ for each $$n$$?
Well, notice $$f(s) = 2^{k+1}$$ for every $$s \in \mathcal{N}_{0^k 1}$$. That is, $$f$$ is a countable sum of rescaled characteristic functions,
$$f = 2 \chi_{\mathcal{N}_{1}} + 4 \chi_{\mathcal{N}_{01}} + 8 \chi_{\mathcal{N}_{001}} + \ldots = \sum 2^{k+1} \chi_{\mathcal{N}_{0^k 1}}$$
But we know that
$$\mu_n \mathcal{N}_{0^k 1} = \begin{cases} \frac{1}{2^{k+1-n}} & k \geq n \\ 0 & k < n \end{cases}$$
and so (by the monotone convergence theorem) we find
\begin{aligned} \int f \ d\mu_n &= \sum_k \int 2^{k+1} \chi_{\mathcal{N}_{0^k 1}} \ d\mu_n \\ &= \sum_k 2^{k+1} \mu_n \mathcal{N}_{0^k 1} \\ &= \sum_{k \geq n} 2^{k+1} \frac{1}{2^{k+1-n}} \\ &= \sum_{k \geq n} 2^n \\ &= \infty \end{aligned}
I hope this helps ^_^
• I can't thank you enough. Just one unrelated question I'll delete later: how long did it take you to get so skilled in this?
– Pegi
Jul 14, 2021 at 22:41
• I'm not sure. I think I first did measure theory 2 or 3 years ago? But it's pretty related to a lot of things, and so I keep coming back to it. Plus, answering questions (like this one) on MSE is a good way to keep in shape ^_^. Jul 14, 2021 at 23:02
• Unrelated, but if this answered your question, you should mark it as such so that other users know how to spend their time. If you want to leave it open for a while so that other people have a chance to answer, that works too! Jul 14, 2021 at 23:03
• oh no I was still trying to go through it again to make sure I didn't miss any points. I have checked it now. Thank you!
– Pegi
Jul 14, 2021 at 23:04
• No worries at all! Happy to help ^_^ Jul 14, 2021 at 23:19 | 2022-08-19T06:03:02 | {
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https://www.beatthegmat.com/what-is-the-value-of-x-t299597.html | • NEW! FREE Beat The GMAT Quizzes
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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## What is the value of x? tagged by: BTGmoderatorLU ##### This topic has 2 expert replies and 0 member replies ### Top Member ## What is the value of x? in the figure above, the square LMNO has a side of length 2x+1 and the two smaller squares have sides of lengths 3 and 6. if the area of the shaded region is 76, what is the value of x? A. 5 B. 6 C. 7 D. 11 E. 14 The OA is A. I'm confused with this PS question, can I say that the total area is (2x+1)^2=3^2+6^2+76 Then can I get the value of x from this equation, right? Experts, any suggestion? Thanks in advance. ### GMAT/MBA Expert Legendary Member Joined 14 Jan 2015 Posted: 2666 messages Followed by: 125 members Upvotes: 1153 GMAT Score: 770 Top Reply LUANDATO wrote: in the figure above, the square LMNO has a side of length 2x+1 and the two smaller squares have sides of lengths 3 and 6. if the area of the shaded region is 76, what is the value of x? A. 5 B. 6 C. 7 D. 11 E. 14 The OA is A. I'm confused with this PS question, can I say that the total area is (2x+1)^2=3^2+6^2+76 Then can I get the value of x from this equation, right? Experts, any suggestion? Thanks in advance. You're exactly right. The total area of the square is $$3^2 + 6^2 + 76 = 9 + 36 + 76 = 121$$ We know each side of the square must be 11, as 11*11 = 121 If a side is 11, then 2x + 1 = 11 --> 2x = 10 ---> x = 5. The answer is A _________________ Veritas Prep | GMAT Instructor Veritas Prep Reviews Save$100 off any live Veritas Prep GMAT Course
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in the figure above, the square LMNO has a side of length 2x+1 and the two smaller squares have sides of lengths 3 and 6. if the area of the shaded region is 76, what is the value of x?
A. 5
B. 6
C. 7
D. 11
E. 14
The OA is A.
I'm confused with this PS question, can I say that the total area is
(2x+1)^2=3^2+6^2+76
Then can I get the value of x from this equation, right? Experts, any suggestion? Thanks in advance.
Hi LUANDATO,
Let's take a look at your question.
The length of the square is (2x+1), therefore the area of the bigger square is:
$$\text{Area of Big Square}=\left(2x+1\right)^2$$
Area of the shaded region and areas of two smaller squares add up to the area of the bigger square. Hence,
Area of Big Square = Area of Square with side 3 + Area of Square with side 6 + Area of shaded region
$$\left(2x+1\right)^2=3^2+6^2+76$$
$$\left(2x+1\right)^2=9+36+76$$
$$\left(2x+1\right)^2=121$$
Taking square root on both sides:
$$\sqrt{\left(2x+1\right)^2}=\sqrt{121}$$
$$2x+1=11$$
$$2x=11-1$$
$$2x=10$$
$$x=\frac{10}{2}$$
$$x=5$$
Therefore, Option A is correct.
Hope it helps.
I am available if you'd like any follow up.
_________________
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https://math.stackexchange.com/questions/3123250/possible-number-of-sheets-for-a-moebius-band-covering | # possible number of sheets for a Moebius band covering
Let M be the Moebius band, identified by the quotient of $$[0,1]\times [0,1]$$ by the equivalence $$(x,0) \sim (1-x,1)$$.
Let $$p: M\to M$$ be a covering and $$n$$ its number of sheets.
Find the possible values of $$n$$.
what I did:
$$M$$ is path-connected so it is connected, all the fibers have the same cardinality.
$$M$$ is also compact so $$n$$ is finite.
My intuition is that any $$n\in \Bbb Z$$ would be valid, as the possible coverings are $$\Bbb R/n\Bbb Z\cong \Bbb R/\Bbb Z\times \Bbb Z/ n\Bbb Z$$.
Thank you for help and comments.
• Something of a side note, but that isomorphism you write is not correct. The left hand side is isomorphic to the circle, while the right is $n$ disjoint circles. – WSL Feb 23 at 2:50
• This is closely related to what we talked about in your other question. Remember that the universal cover of $M$ is $\tilde{M}=\mathbb{R}\times [0,1]$ and the deck transformations act by translating and flipping. All the connected coverings of $M$ are quotients of $\tilde{M}$ by subgroups of $\mathbb{Z}$, so you should try to determine for which values of $n$ the quotient $\tilde{M}/n\mathbb{Z}$ is homeomorphic to $M$. – William Feb 23 at 4:23
I think only an odd number of sheets are possible.
This is a great example where the general theory leads to interesting computational results in particular cases: we can determine possible coverings of the form $$M\to M$$ by first determining all connected coverings of $$M$$, and then detecting which ones have total space homeomorphic to $$M$$.
Classification Theorem: For any path-connected, locally path-connected, semi-locally simply-connected space $$X$$ there is a bijection between isomorphism classes of connected covering spaces of $$X$$ and conjugacy classes of subgroups of $$\pi_1(X)$$. (See for example Theorem 1.38, page 67.)
This works by constructing a universal covering $$\tilde{X}\to X$$ so that the quotients $$\tilde{X}/H$$ represent all the connected coverings of $$X$$ as $$H$$ varies over conjugacy classes. The covering $$\tilde{X}$$ is characterized up to covering space isomorphism by being simply-connected.
Universal covering space of $$M$$: Recall $$M\sim S^1$$ so $$\pi_1(M)\cong \mathbb{Z}$$. The universal covering space of $$M$$ is $$\mathbb{R}\times [0,1]$$ and the action of $$\mathbb{Z}$$ is given by $$n\cdot (x, t)= \big(x+n, f^{n}(t)\big)$$ for $$n\in\mathbb{Z}$$ and where $$f\colon [0,1] \to [0,1]$$ is the "flip" homeomorphism given by $$f(t)= 1-t$$. (Visually, think about $$\mathbb{R}\times[0,1]$$ as an infinite strip of tape that you're applying to the Möbius strip, which is alternating "front" and "back" sides.)
Quotients of $$\tilde{M}$$: Every connected covering of $$M$$ is a quotient of the form $$(\mathbb{R}\times [0,1])/n\mathbb{Z}$$. For each $$n$$ a fundamental domain of the quotient is $$[0,n]\times[0,1]$$, and the quotient only depends on how we identify the subspaces $$\{0\} \times [0,1]$$ and $$\{n\}\times [0,1]$$. When $$n$$ is odd then $$f^{n} = f$$ so we identify the ends using a flip, and hence the quotient is homeomorphic to $$M$$; on the other hand if $$n$$ is even then $$f^{n}=id$$ and so the quotient is actually the cylinder $$(\mathbb{R}/n\mathbb{Z})\times [0,1]$$.
Since these quotients make up all of the possible connected coverings of $$M$$, it follows that coverings of the form $$M\to M$$ can have any odd number of sheets.
• Thanks @William for your thorough answer. I see now why you say connected instead of path-connected. It boild down to the same as local path-connectedness is inherited by the coverings – PerelMan Feb 23 at 17:45
• Yes that's true. Since we already know $M$ is locally path-connected, so will be any covering. Therefore every connected covering of $M$ is already path-connected. – William Feb 23 at 19:40
Even though the question is already answered, I found an alternate argument that there are no even-sheeted covers using the first Stiefel-Whitney class of $$TM$$.
Recall that the tangent bundle of $$M$$ is non-orientable so $$w_1(TM)\neq 0 \in H^1(M;\mathbb{Z}/2\mathbb{Z})$$. An $$n$$-sheeted covering $$p_n\colon M\to M$$ is a local diffeomorphism and hence induces a bundle map $$TM\to TM$$, so by naturality of characteristic classes we have $$w_1(TM) = p_n^*(w_1(TM))$$. But the covering also restricts to an $$n$$-sheeted covering $$S^1\to S^1$$, which on cohomology $$H^1(S^1;\mathbb{Z}) \to H^1(S^1;\mathbb{Z})$$ induces multiplication by $$n$$. Since $$S^1 \to M$$ is a homotopy equivalence we also have
$$p_n^* = n\cdot(-)\colon H^1(M;R) \to H^1(M;R)$$
for any $$R$$. In particular if such a cover exists when $$n = 2k$$ is even then $$w_1(TM) = p_n^*(w_1(TM)) = 2k \cdot w_1(TM) = 0$$ which contradicts $$w_1(TM) \neq 0$$.
Edit: I'm currently trying to modify this argument into a proof of the following:
Conjecture: If $$M$$ is a non-orientable smooth manifold then there are no even-sheeted coverings of the form $$M\to M$$.
Update: This general conjecture is NOT true. Consider the double-cover $$p\colon S^1\times \mathbb{RP}^2\to S^1 \times \mathbb{RP}^2$$ given by $$p(z, x) = (z^2, x)$$.
• Thank you @William interesting alternate argument! I only studied some elementary De Rham cohomology, so not able to understand all of it but will read about the ring cohomology. I am interested in any references. – PerelMan Feb 23 at 21:09 | 2019-04-25T08:33:04 | {
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https://www.okoursaros.gr/reach-wiki-qal/71638c-coordinate-proof-diagonals-parallelogram-bisect | Still have questions? 23. The first group has concluded that the diagonals of a quadrilateral always bisect each other. Let A12, 32, B15, 42, and C13, 82 be three points in a coordinate plane. Ceiling joists are usually placed so they’re ___ to the rafters? In this lesson we will prove … All angles are right 3. The other goes from (a,0) to (b,c), so its midpoint is (a+b)/2, c/2. Sign Up. There are a number of ways to show whether a quadrilateral placed on a coordinate plane is a parallelogram or not. In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. Here AC and BD are diagonals. Use the distance formula-they would have the same length. e. Prove … Write a coordinate proof that the diagonals of a rectangular prism are congruent and bisect each other. - edu-answer.com Prove that the diagonals of a parallelogram bisect each other. 21. 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Cuddle Barn Giraffe, Gtbank Savings Account Nigeria, Homosapien Sapiens Tagalog, Estate Weddings Massachusetts, Bc1 News Today, You Have To Stop Tiktok Sound, Dhadak Movie 2018 Part 1, | 2021-04-22T11:38:46 | {
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https://math.stackexchange.com/questions/3788993/how-do-we-decide-whether-to-visualize-a-matrix-with-its-rows-or-columns | # How do we decide whether to visualize a matrix with its rows or columns?
Should one visualize a matrix by its rows, columns, or both depending on the situation? I see both used and it seems arbitrary. It would be nice if only one was used consistently. Shouldn't a graph of a matrix be denoted as being a row or column representation somehow to avoid confusion?
Example where author switches: https://intuitive-math.club/linear-algebra/matrices
[Example I] Given the transformation: $$\begin{bmatrix} 1 & 1\\ 2 & 0 \end{bmatrix} + \begin{bmatrix} 2 & 1\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2\\ 3 & 1 \end{bmatrix}$$ The author represents the matrix after the transformation visually by its rows, using the following row vectors:
$$v_1 = \begin{bmatrix} 3\\ 2 \end{bmatrix} v_2 = \begin{bmatrix} 3\\ 1 \end{bmatrix}$$
[Example II] Given the transformation: $$\begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix} \begin{bmatrix} 3 & 1\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1\\ -3 & 1 \end{bmatrix}$$
The author represents the matrix after the transformation visually by its columns, using the following column vectors:
$$v_1 = \begin{bmatrix} 1\\ -3 \end{bmatrix} v_2 = \begin{bmatrix} 1\\ -1 \end{bmatrix}$$
Question:
Why is did they author seemingly arbitrarily switch from a row → column visual representation? What is the intuition behind this – if any?
• I love this question! To give you a short, definitive “answer” that isn’t up to par with the expected quality on this site: Most people will default to looking at columns as vectors. If you break it into rows, most people will look at those as equations (as in a system) in multiple variables. Only occasionally do you interpret the rows as vectors—but it’s doable. I hope that helps. – gen-ℤ ready to perish Aug 13 at 2:02
• The real answer here, I think, is that the website you link to isn't choosing its visualizations in any sort of principled way. I wouldn't read anything into the the particular choices made on that site. – Will Orrick Aug 13 at 9:00
There's a lot of ways to interpret matrices, some of which involve reading it by rows and some by columns. But in this particular case, it is columns both times: you were misled by the fact that the matrix $$\begin{bmatrix}3 & 1 \\ 1 & 1\end{bmatrix}$$ is symmetric, so its columns are the same as its rows.
Here, the idea is that for any $$2 \times 2$$ (or more generally $$k \times 2$$) matrix $$A$$, we have $$A \begin{bmatrix}3 & 1 \\ 1 & 1\end{bmatrix} = \begin{bmatrix}A \begin{bmatrix}3 \\ 1\end{bmatrix} & A\begin{bmatrix}1 \\ 1\end{bmatrix} \end{bmatrix}.$$ In other words, each column of the product is equal to $$A$$ times a column of the second matrix we multiplied.
In the picture you have, the vector $$\begin{bmatrix}3 \\ 1\end{bmatrix}$$ (in pink) gets sent to $$\begin{bmatrix}1 \\ -3\end{bmatrix}$$, and the vector $$\begin{bmatrix}1 \\1\end{bmatrix}$$ (in yellow) gets sent to $$\begin{bmatrix}1 \\ -1\end{bmatrix}$$, and all of these are columns of the respective $$2 \times 2$$ matrix.
• You are completely correct. That is my mistake for not noticing the symmetry. I may have to edit my questions example to get the correct point across. – Matthaeus Gaius Caesar Aug 12 at 23:06
• I completely edited the examples in my question. The example was mistakingly symmetric, which took away from the actual question at hand. I do not know if such a large edit is allowed, but the premise of the question remains the same. Should I make a new question with the new example? – Matthaeus Gaius Caesar Aug 12 at 23:07
• @Caesar I think the edit was fine as it’s not altering the question itself. You did the right thing by leaving a comment. The problem with major edits arises when they make an answer obsolete. – gen-ℤ ready to perish Aug 13 at 3:09
• The website's example of matrix addition has an illustration showing a vector $[2 0],$ which can only be a row of one of the matrices, not a column. I think it would have been better if it had said this explicitly. I don't see anything wrong with getting used to matrices being either columns of row vectors or rows of column vectors, but expecting you to guess while you're just learning is a bit much to ask. – David K Aug 13 at 12:05
• @MatthaeusGaiusCaesar, your edit almost completely invalidates this answer. While your question first had the two cases before and after transformation, now it has two different transformations. The sentence at the beginning of this answer "it is columns both times" no longer applies, and makes the answer look out of place. – ilkkachu Aug 13 at 12:17
As long as your main objects of study are column vectors, and you multiply matrix and (column) vector together by writing the matrix on the left and the vector on the right, a matrix is more naturally seen as a collection of columns rather than rows.
A matrix represents a linear transformation. The columns of the matrix are given by where this linear transformation sends your basis vectors. The result of a matrix-vector product similarly becomes a linear combination of the columns of the matrix (where the entries in the vector are the coefficients of this linear combination).
When multiplying two matrices, of course you can choose. Either you say "Apply the left-hand matrix to each column in the right-hand matrix, and collect the results in a new matrix" (in which case you see both matrices as collections of columns), or you say "Apply the right-hand matrix to each row in the left-hand matrix, and collect the results in a new matrix" (in which case both matrices are collections of rows). They both give the same result. Which one is most convenient comes down to whether one happens to be significantly easier to calculate than the other for some reason, and what you're going to do with the result afterwards.
Of course, the final answer is "it depends on the situation". Because what else could it be? But columns is much more common than rows. | 2020-10-20T18:19:52 | {
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http://mathhelpforum.com/trigonometry/9952-exact-values.html | # Math Help - Exact values
1. ## Exact values
Could someone help me with the following three problems?
1) Find the exact value of sin[to the -1 power](-[sqrt]3/2).
2) cos[to the -1 power](cos(4[pi]/3)).
3) tan(sin[to the -1 power](-4/5)).
2. Originally Posted by mike1
Could someone help me with the following three problems?
1) Find the exact value of sin[to the -1 power](-[sqrt]3/2).
You are expected to know some special triangles, one of which is the
30, 60, 90 degree triangle, where if the side opposite the 30 degree angle
is 1 unit the side opposite the 60 degree angle is sqrt(3), and the other side
is 2. So using this right triangle we see that sin(30)=sqrt(3)/2, hence if:
x=asin(-sqrt(3)/2),
then
sin(x)=-sqrt(3)/2,
so:
sin(-x)=sqrt(3)/2.
Now there are multiple solutions to this, but we want x to be in the range
[-90,90], so we want -x=60, or x=-60 degrees, or -pi/3 radian.
Check: asin(-sqrt(3)/2)~=-1.57, -pi/2~=-1.57
2) cos[to the -1 power](cos(4[pi]/3)).
Now 4 pi/3= pi+pi/3, so cos(4 pi/3)=cos(pi)cos(pi/3)-sin(pi)sin(pi/3)=-cos(pi/3)
Now pi/3 radian is 60 degrees so we are back to our special triangle, and cos(pi/3)=1/2.
Now acos(-1/2)=pi-pi/3=2pi/3.
Check acos(cos(4pi/3))~=2.094, 2*pi/3~=2.094
3) tan(sin[to the -1 power](-4/5)).
Here we need to think 3-4-5 triangle, first:
tan(asin(-4/5))=-tan(asin(4/5))
(draw the triangle identify the angle whose sin is 4/5, then find its
tangent), so:
tan(asin(-4/5))=-4/3
Check: tan(asin(-4/5))~=-1.33.., -4/3~=1.33..
RonL
Note: when evaluating the inverse trig functions it is usual to give the answer
in a fundamental range such that the function has a unique value (the principle value).
That is evaluating x=asin(y) is not the same as finding x such that sin(x)=y,
since asin(y) gives a single value, while solving sin(x)=y has multiple solutions.
For asin the range is usually -pi/2 to pi/2 (or in degrees -90 to 90)
For acos the range is usually 0 to pi (0 to 180 degrees)
For atan the range is usually -pi/2 to pi/2 (or sometimes 0 to pi) (-90 to 90 degrees).
3. Originally Posted by mike1
Could someone help me with the following three problems?
1) Find the exact value of sin[to the -1 power](-[sqrt]3/2).
2) cos[to the -1 power](cos(4[pi]/3)).
3) tan(sin[to the -1 power](-4/5)).
I assume sin[to the -1 power] is arcsine.
1) arcsin[-sqrt(3) /2]
That is an angle whose sine is -sqrt(3) /2. Let's call it theta.
In what quadrants is the sine negative? Since sine is y/r, and r is always positive, then sine is negative wherever y is negative. Below the origin, or below the x-axis. In the 3rd and 4th quadrants.
So theta is between pi and 2pi radians.
Then the angle.
If sin(theta) = sqrt(3) /2, then theta is 60degrees or pi/3 radians. What are the angles in the 3rd and 4th quadrants that are pi/3 from the x-axis?
Why, (pi +pi/3) in the 3rd quadrant and (2pi -pi/3) in the 4th quadrant.
----------------------
2) cos[to the -1 power](cos(4[pi]/3)).
That is "the cosine of an angle whose cosine is 4pi/3".
======
Ooppss, my mistake.
That is actually arccosine of cos(4pi/3).
An angle whose cosine is cos(4pi/3).
I do not know how to do that.
You sure you typed it right?
-------------------------------------
3) tan(sin[to the -1 power](-4/5)).
That is "the tangent of an angle whose sine is -4/5 "
First, about the negative sine value.
As in part (i) above, theta is in the 3rd or 4th quadrant, or theta between pi and 2pi radians.
Then the angle.
If sin(alpha) = 4/5, then y=4 and r=5 since sine = y/r.
tan is y/x. So we need to find x.
By Pythagorean theorem, x = sqrt(r^2 -y^2) = sqrt(5^2 -4^2) = 3.
Hence tan(alpha) = y/x = 4/3.
In the 3rd quadrant, x is negative, and y is negative also, hence,
tan(theta) = y/x = -4/ -3 = 4/3 ------***
In the 4th quadrant, x is positive, and y is negative, hence,
tan(theta) = y/x = -4/ 3 = -4/3 -----***
-----------------------
Note, in the reference triangle,
opposite side = y
hypotenuse = r --------always positive.
4. Hello, Mike!
ticbol did an excellent job of explaining.
Here are my versions . . .
I'll assume the angle are between $0$ and $2\pi$.
Find the exact value of:
$(1)\;\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$
We want the angle whose sine is $-\frac{\sqrt{3}}{2}$
You are expected to know that: $\sin\left(\frac{\pi}{3}\right) \,=\,\frac{\sqrt{3}}{2}$
Sine is negative in quadrants 3 and 4.
So we have a reference angle of $\frac{\pi}{3}$ in quadrants 3 and 4.
Hence, the angles are: . $\frac{4\pi}{3},\:\frac{5\pi}{3}$
$(2)\;\cos^{-1}\left[\cos\left(\frac{4\pi}{3}\right)\right]$
Inside, we have: . $\cos\frac{4\pi}{3} \:=\:\text{-}\frac{1}{2}$
Then we have: . $\cos^{-1}\left(\text{-}\frac{1}{2}\right)$ . . . the angle whose cosine is $\text{-}\frac{1}{2}$
We know that: . $\cos\left(\frac{\pi}{3}\right) \,=\,\frac{1}{2}$
Cosine is negative in quadrants 2 and 3.
Therefore the angles are: . $\frac{2\pi}{3},\:\frac{4\pi}{3}$
$(3)\;\tan\left[\sin^{-1}\left(\text{-}\frac{4}{5}\right)\right]$
Inside, we have: . $\sin^{-1}\left(\text{-}\frac{4}{5}\right)$ . . . the angle whose sine is $\text{-}\frac{4}{5}$
Consider: $\sin\theta = \frac{4}{5}$
We don't know the exact value of this angle,
. . but we know it comes from this triangle:
Code:
*
* |
* |
5 * | 4
* |
* θ |
* - - - - - *
3
Since sine is negative in quadrants 3 and 4,
. . the two possible angles look like this:
Code:
| |
| |
-3 | | 3
- + - - - + - - - - - - - - - + - - - + -
: θ / | | \ θ :
-4: /5 | | 5\ :-4
: / | | \ :
* | | *
Since $\tan\theta \:=\:\frac{\text{-}4}{\text{-}3}\text{ or }\frac{\text{-}4}{3}$
. . the answers are: . $\pm\frac{4}{3}$ | 2015-03-27T16:40:15 | {
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https://math.stackexchange.com/questions/866223/help-with-complicated-functional-equation | # Help with complicated functional equation
Problem: Let $T=\{(p,q,r)\mid p,q,r \in \mathbb{Z}_{\geq0}\}$. Find all functions $f:T\to \mathbb{R}$ such that: $$f(p,q,r)=\\ =\begin{cases} 0, & \text{ if } pqr = 0 \\ 1 + \frac{1}{6}\left(f(p+1,q-1,r)+f(p+1,q,r-1)+f(p,q+1,r-1)+\\ \;\;\;\;\;\; f(p,q-1,r+1)+f(p-1,q+1,r)+f(p-1,q,r+1)\right) & \text{ otherwise.} \end{cases}$$
Progress so far: It's not hard to see that $f$ is symmetric in $p,q,r$, which is useful to know. From the recursive definition one can also infer that $f:T\to \Bbb{Q}^+$, so no trig functions or logs. That's all I could observe from the get-go. I've tried calculating some values of $f$ to have an idea on how the functions look like (if there are any) but having trouble calculating even small values of $f$, for example $f(1,2,3)$ or $f(2,2,2)$. All I know is that $f(0,a,b)=0$ and $f(1,1,1)=1$. I could guess a solution based on my initial observations but I can't see any obvious candidates.
Any help would be appreciated, thanks.
• $\displaystyle\;f(p,q,r) = \frac{3pqr}{p+q+r}\;$ is a solution. I believe this is the only solution but I can't prove it. – achille hui Jul 16 '14 at 19:56
• How did you get this solution? – Deathkamp Drone Jul 16 '14 at 20:00
• If you look at your equation, $(p,q,r)$ only connects to points with same $p+q+r$. This means you can look at each $p+q+r = \text{constant}$ layer separately. The intersection of $T$ with any such layer forms a triangular lattice. Your equation reduces to a discrete version of Poisson equation there. The boundary condition suggest the solution (at least in continuum limit) contains a factor proportional to $pqr$. Direct substitution shows that you can scale this to get a solution. Since we are dealing with some sort of "Poisson equation", that's why I suspect the solution is unique. – achille hui Jul 16 '14 at 20:09
• This is an IMO 2001 shortlist problem. – Calvin Lin Jul 18 '14 at 22:57
• Thank you, Calvin. I found this problem some lecture notes so I didn't know where it came from. Ironically, I looked up the solution on AoPS just to find out that they pulled out of nowhere the function Achille discovered and then proved that this is the only solution. – Deathkamp Drone Jul 19 '14 at 2:02
When I worked on this problem back in 2002, showing uniqueness was really easy through the "average of neighbors" observation (albeit on a slanted hexagonal board, instead of the regular chessboard).
Proof of uniqueness: Suppose we have 2 solutions $f(p,q,r)$ and $g(p,q,r)$. Let $h(p,q,r) = f(p,q,r) - g(p,q,r)$. Then, we get that
$$6 h(p,q,r) = h(p+1, q-1, r) + h(p-1, q+1, r) + h( p, q+1, r-1) + h( p, q-1, r+1) + h( p+1, q, r-1) + h(p-1, q, r+1).$$
Consider the plane $p+q+r = N$. Oberve that the neighbors of the cell $(p,q,r)$ are these 6 other cells with coordinates as given above. Hence, every cell is the average of it's neighbors. Through the standard argument (extremal principle), this implies that all cells on this finite board are equal.
We also have the boundary conditions that $h(p,q,r ) = 0$ for $pqr=0$, hence $h(p,q,r) = 0$. Thus, the function is unique $_\square$
Finding the solution was harder, but still motivated from the conditions.
Note: It is important to bear in mind that as an ('easy') Olympiad problem, it often has a nice solution that can be motivated.
Finding function: From the boundary condition that $pqr=0 \Rightarrow f(p,q,r) = 0$, we guess the initial function $F( p,q,r) = pqr$.
Observe that since $(p-1)(q+1) r + (p+1)(q-1)r = 2pqr - 2r$, so this guess gives us:
$F(p,q,r) = \frac{ p+q+r} { 3} + \frac{1}{6} [ F(p-1, q+1, r) + F(p+1, q-1, r) + F(p, q-1, r+1), F(p, q+1, r-1) + F( p-1, q, r+1), F(p+1, q, r-1) ]$.
Observe that since $p+q+r$ is a constant for all of these 7 terms, we should look at $$f(p,q,r) = \frac{ F(p,q,r) } { \frac{p+q+r} {3} } = \frac{3 pqr} { p+q+r}.$$
Indeed, this works. $_\square$
Note: Had $F(p,q,r) = pqr$ not worked, the next guess would have been $F(p,q,r) = p^2q^2r^2$
Achille Hui did the hardest part of the work by discovering the closed formula $\frac{3pqr}{p+q+r}$. The rest is a routine "maximum principle" argument that I explain below.
For a positive integer $k$, let $T_k$ be the finite set $\lbrace (p,q,r)\in T | p+q+r=k\rbrace$. For $x=(p,q,r)\in T$, define the neighborhood $N(x)$ of $x$ to be
$$\begin{array}{lcl} N(p,q,r)&=&\lbrace (p+1,q-1,r);(p+1,q,r-1);(p,q+1,r-1); \\ & & (p,q-1,r+1);(p-1,q+1,r);(p-1,q,r+1)\rbrace \end{array} \tag{1}$$
and the strict neighborhood $N'(x)$ of $x$ to be $\lbrace (u,v,w)\in N(x) | uvw>0\rbrace$. We say that $x\in T_k$ is interior if $N’(x)=N(x)$, and extremal otherwise.
Let $g(p,q,r)=f(p,q,r)-\frac{3pqr}{p+q+r}$ for $(p,q,r)\in T$. Then $g$ satisfies $g(p,q,r)=0$ if $pqr=0$, and
$$6g(x)=\sum_{y\in N'(x)}g(y) \tag{2}$$
for any $x\in T_k$ (note that $N(x)$ and $N’(x)$ stay in $T_k$ when $x\in T_k$).
Now, let $M$ ($m$) be the maximum (minimum) value of $g$ on $T_k$. There is some $x_M\in T_k$ such that $g(x_M)=M$. We now apply (2) to $x=x_M$, and obtain a formula (2').
If $M > 0$, then (2') is only possible when $x_M$ is interior and $g(y)=M$ for all $y\in N(x_M)$. If we put $L=\lbrace x\in T_k | g(x)=M\rbrace$, we would deduce that $L$ consists only of interior points but also satisfies $N(x)\subseteq L$ for any $x\in L$, which is impossible because when we move away from the interior of $T_k$ we are always forced to eventually reach extremal points.
So $M\leq 0$. A similar argument (or if you please, you may reuse the result just shown on $-g$ instead of $g$) shows that $m\geq 0$.
So $M\leq 0 \leq m$, but on the other hand $m\leq M$. This forces $m=M=0$, so $g$ is identically zero.
To conclude, $\frac{3pqr}{p+q+r}$ is the unique solution. | 2019-06-19T09:16:08 | {
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https://math.stackexchange.com/questions/928470/what-is-the-difference-between-bounded-and-convergent | # what is the difference between bounded and convergent?
I know that bounded means to have an upper or lower bound.
Let $E \subset \mathbb{R}$ be nonempty.
1. The set $E$ is said to be bounded above if and only if there is an $M \in \mathbb{R}$ such that $a \leq M$ for all $a \in E$, in which case $M$ is called an upper bound of $E$.
2. A number $s$ is called a supremum of the set $E$ if and only if $s$ is an upper bbound of $E$ and $s \leq M$ for all upper bounds $M$ of $E$ (In this case we shall say that $E$ has a finite supremum $s$ and write $s=\sup E$.
Let $E \subset \mathbb{R}$ be nonempty.
1. the set $E$ is said to be bounded below if and only if there is an $m \in \mathbb{R}$ such that $a \geq m$ for all $a\in E$, in which case $m$ is called a lower bound of the set $E$.
2. A number $t$ is called an infimum of the set $E$ if and only if $t$ is a lower bound of $E$ and $t \geq m$ for all lower bounds $m$ of $E$. In this case we shall say that $E$ has an infimum $t$ and write $t = \inf E$
3. $E$ is said to be bounded if and only if it is bounded both above and below.
The meaning of convergence
A sequence of real number $\{ x_n \}$ is said to converge to a real number $a \in \mathbb{R}$ if and only if for every $\epsilon > 0$ there is an $N \in \mathbb{N}$ (which in general depends on $\epsilon$) such that $$n \geq N \mbox{ implies } \vert x_n - a \vert < \epsilon$$
In my previous classes, it was taught to take the the limit of the function or series to find the value which the function or series converges at the value. Is the value a function or series converges at the $\sup$ or $\inf$?
• The sequence defined by $a_n := \sin \left(\frac{\pi}{2} n\right)$, which has expansion $0, 1, 0, -1, 0, 1, 0, -1, \ldots$ is bounded but not convergent. – Travis Sep 12 '14 at 7:42
• Thanks, that example made the conflict more clear. – El Santi Sep 12 '14 at 7:59
I think there's two points to address here. The first is that a sequence can be bounded from both above and below, yet fail to be convergent. Consider the sequence $$\{x_n\} = \{(-1)^n\} = -1,1,-1,1, \cdots$$ Here we have $\displaystyle\sup_{n} \{x_n\} = 1$, $\displaystyle\inf_{n} \{x_n\} = -1$, and $\{x_n\}$ does not converge (if you pick say $\epsilon = 0.1$, then there are infinitely many $n$ for which $|x_n - 1| > 0.1$ and for which $|x_n - (-1)| > 0.1$).
Furthermore, when a sequence $\{x_n\}$ does converge, the limit of a sequence need not equal either $\displaystyle\sup_n\{x_n\}$ or $\displaystyle\inf_n\{x_n\}$. For example, we see with the sequence $\{x_n\} = \{\frac{(-1)^n}{n}\}$ that $$\displaystyle\sup_n\{x_n\} = \frac{1}{2}, \: \:\displaystyle\inf_n\{x_n\} = -1, \: \: \displaystyle\lim_{n \to \infty} \{x_n\} = 0$$
In that last example, I kind of cheated, by picking a sequence where the inf and sup of the elements of the sequence were the first two, and the convergence of a sequence talks about the behavior at the tail of the sequence. When we want to talk about upper and lower bounds of sequences, we usually like to talk about the limit superior, or $$\displaystyle\limsup_{n \to \infty} \{x_n\} = \displaystyle\lim_{n \to \infty} \left(\sup_{N} \{x_N | \: N \geq n \} \right)$$ and the limit inferior, or $$\displaystyle\liminf_{n \to \infty} \{x_n\} = \displaystyle\lim_{n \to \infty} \left(\inf_{N} \{x_N | \: N \geq n \} \right)$$ This let's you bound the values of $x_n$ for arbitrarily large $n$, while allowing for a few badly behaved values of $x_n$ for small $n$. Then, we say that a sequence $\{x_n\}$ approaches a finite limit $M$ if and only iff $$\displaystyle\limsup_{n \to \infty} \{x_n\} = \displaystyle\liminf_{n \to \infty} \{x_n\} = M$$ and you can read about the proof of this result here (Proof that a sequence converges to a finite limit iff lim inf equals lim sup).
• is there a typo at the end? $\displaystyle\limsup_{n \to \infty} \{x_n\} = \displaystyle\limsup_{n \to \infty} \{x_n\}$ is it to be $\displaystyle\limsup_{n \to \infty} \{x_n\} = \displaystyle\liminf_{n \to \infty} \{x_n\}$ ?? – El Santi Sep 12 '14 at 8:22 | 2019-08-25T19:47:53 | {
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https://math.stackexchange.com/questions/1909290/integration-using-monte-carlo-method | # Integration using Monte Carlo Method
I'm trying to solve this integral using the Monte Carlo Method.
$$I=\int_0^\pi \frac{1}{\sqrt{2\pi}}e^\frac{-sin(x)^2}{2}dx$$
Now it seems to me that there is a normal probability density function in there, but I'm not sure because of the sine function.
If there is a normal, then it's easy to simulate $N$ random numbers from the standard normal distribution and compute $I$. It's is also easy to find the size of the sample to get a 95% confidence interval.
So is there a normal probability density function? Or from what distribution should I get the random numbers to compute $I$?.
• Ok. How did you find the confidence intervals? – Fawcett512 Aug 31 '16 at 4:04
• I know how to get the confidence intervals for a sample of uniform random variables, but I'm not sure how to find them in this case given that I don't have a parameter but rather the approximate value of $I$ – Fawcett512 Aug 31 '16 at 4:06
• I think I'm making the problem more complicate that it should. So I could generate $N$ uniform random variables $x_i$ and then compute $\mathbb{E}=\frac{\pi}{N}\sum f(x_i)$. As $N$ tends to infinity, we should get closer to the integral. Yet I don't see how to get the confidence intervals. – Fawcett512 Aug 31 '16 at 4:40
• I think that I was trying to do some importance sampling (hence the question if we could use a normal probability density function), but seems that in this case that is not possible, Furthermore, if I just compute the above formula for $\mathbb{E}$,I should get a good estimate for $N$ provided that $N$ is big enough. The only thing that's not very clear is how to compute a 95% confidence interval. – Fawcett512 Aug 31 '16 at 4:47
Recall that if $Y$ is a random variable with density $g_Y$ and $h$ is a bounded measurable function, then $$\mathbb E[h(Y)] = \int_{\mathbb R} h(y)g_Y(y)\,\mathsf dy.$$ Moreover, if $Y\sim\mathcal U(0,1)$, then $a+(b-a)U\sim\mathcal U(a,b)$. So applying the change of variables $x=a+(b-a)u$ (with $a=0$, $b=\pi$) to the given integral, we have $$I = \int_0^1 \frac{\pi}{\sqrt{2\pi}} e^{-\frac12\sin^2 (\pi u) }\,\mathsf du=\int_0^1 h(u)\,\mathsf du,$$ with $h(u)=\sqrt{\frac\pi 2} e^{-\frac12\sin^2 (\pi u) }$. It follows then that $I=\mathbb E[h(U)]$ with $U\sim\mathcal U(0,1)$. Let $U_i$ be i.i.d. $\mathcal U(0,1)$ random variables and set $X_i=h(U_i)$, then for each positive integer $n$ we have the point estimate $$\newcommand{\overbar}[1]{\mkern 1.75mu\overline{\mkern-1.75mu#1\mkern-1.75mu}\mkern 1.75mu} \widehat{I_n} =: \overbar X_n= \frac1n \sum_{i=1}^n X_i$$ and the approximate $1-\alpha$ confidence interval $$\overbar X_n\pm t_{n-1,\alpha/2}\frac{S_n}{\sqrt n},$$ where $$S_n = \sqrt{\frac1{n-1}\sum_{i=1}^n \left(X_i-\overbar X_n\right)^2}$$ is the sample standard deviation.
Here is some $\texttt R$ code to estimate an integral using the Monte Carlo method:
# Define "h" function
hh <-function(u) {
return(sqrt(0.5*pi) * exp(-0.5 * sin(pi*u)^2))
}
n <- 1000 # Number of trials
alpha <- 0.05 # Confidence level
U <- runif(n) # Generate U(0,1) variates
X <- hh(U) # Compute X_i's
Xbar <- mean(X) # Compute sample mean
Sn <- sqrt(1/(n-1) * sum((X-Xbar)^2)) # Compute sample stdev
CI <- (Xbar + (c(-1,1) * (qt(1-(0.5*alpha), n-1) * Sn/sqrt(n)))) # CI bounds
# Print results
cat(sprintf("Point estimate: %f\n", Xbar))
cat(sprintf("Confidence interval: (%f, %f)\n", CI[1], CI[2]))
For reference, the value of the integral (as computed by Mathematica) is $$e^{-\frac14}\sqrt{d\frac{\pi }{2}} I_0\left(\frac{1}{4}\right) \approx 0.991393,$$ where $I_\cdot(\cdot)$ denotes the modified Bessel function of the first kind, i.e. $$I_0\left(\frac14\right) = \frac1\pi\int_0^\pi e^{\frac14\cos\theta}\,\mathsf d\theta.$$
• Thank you. This was very clear. – Fawcett512 Aug 31 '16 at 4:54
• You're welcome! The change of variables is not strictly necessary, but it simplifies the computations considerably. – Math1000 Aug 31 '16 at 4:58
• Just one more question because I have trouble understanding the concept of confidence intervals. How do I know the size of the sample if now I want to get a 98% confidence interval? – Fawcett512 Aug 31 '16 at 6:54
• I don't understand your question - the sample size ($n$) and confidence level ($\alpha$) are parameters which influence the width of the confidence interval, but are not directly related to each other. – Math1000 Aug 31 '16 at 7:49
• Never mind. That was a bad question. I was wondering how big the sample must be (because I was making a simulation) to get a 98% confidence interval, but as you said is a parameter which influences the width of the confidence interval. All is clear know (or at least I think so). – Fawcett512 Sep 1 '16 at 5:59 | 2019-05-23T05:10:02 | {
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https://www.seamplex.com/wasora/doc/realbook/004-exp/ | # 1 A simple differential equation
The examples in this section show how a single ordinary differential equation can be solved with wasora. Indeed this is one of its main features, namely the ability to solve systems of differential-algebraic equations written as natural algebraic expressions. In particular, the equation the examples solve is
\frac{dx}{dt} = -x
with the initial condition x_0 = 1, which has the trivial analytical solution x(t) = e^{-t}.
## 1.1 exp.was
As clearly defined in wasora’s design basis, simple problems ought to be solved by means of simple inputs. Here is a solid example of this behavior.
end_time = 1 # transient problem
PHASE_SPACE x # DAE problem with one variable
x_0 = 1 # initial condition
x_dot .= -x # differential equation
PRINT t x HEADER # output
# exercise: plot dt vs t and see what happens
$wasora exp.was | qdp -o exp --pi 1$
By default, wasora adjusts the time step so an estimation of the relative numerical error is bounded within a range given by the variable rel_error, which has an educated guess by default. It can be seen in the figure that dt starts with small values and grows as the conditions allow it.
## 1.2 exp-dt.was
The time steps wasora take can be limited by means of the special variables min_dt and max_dt. If they both are zero (as they are by default), wasora is free to choose dt as it considers appropriate. If max_dt is non-zero, dt will be bounded even if the conditions are such that bigger time steps would not introduce large errors. On the other hand, if min_dt is non-zero, the time step is guaranteed not to be smaller that the specified value. However, it should be noted that wasora may need to take several internal times step to keep the error bounded. In the limiting case where min_dt = max_dt, the time step can be set exactly although, again, wasora may take internal steps.
This situation is illustrated with the following input, which is run for five combinations of min_dt and max_dt.
end_time = 2
min_dt = $1 max_dt =$2
rel_error = 1e-3
PHASE_SPACE x
x_0 = 1
x_dot .= -x
PRINT t x (x-exp(-t))/x
$wasora exp-dt.was 0 0 > exp-dt1.dat$ wasora exp-dt.was 0.1 0 > exp-dt2.dat
$wasora exp-dt.was 0 0.1 > exp-dt3.dat$ wasora exp-dt.was 0.1 0.1 > exp-dt4.dat
$wasora exp-dt.was 1 1 > exp-dt5.dat$ pyxplot exp-dt.ppl; pdf2svg exp-dt.pdf exp-dt.svg; rm -f exp-dt.pdf
$pyxplot exp-error.ppl; pdf2svg exp-error.pdf exp-error.svg; rm -f exp-error.pdf$
It can be seen that all the solutions coincide with the analytical expression. Even if the time step is set to a big fixed value, the error commited by the numerical solver with respect to the exact solution is the same as wasora iterates internally as needed. In general, the fastest condition is where dt is not bounded as wasora minimizes iterations by automatically adjusting its value. However, it is clear that controlling the time step can be useful some times. A further control can be obtained by means of the TIME_PATH keyword. | 2020-10-20T17:36:27 | {
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https://math.stackexchange.com/questions/4025290/is-every-open-and-connected-set-in-mathbb-c-the-continuous-image-of-the-open | # Is every open and connected set in $\mathbb C$ the continuous image of the open unit disk?
Let $$\mathbb D=\{z\in\mathbb C\ |\ |z|<1\}$$ be the open unit disk in $$\mathbb C$$. It is well known that an open (nonempty) set $$U\subseteq\mathbb C$$ is simply connected if and only if it is homeomorphic to the unit disk. One can show that this is equivalent to the fact that there is a continuous injective function $$f:\mathbb D\to\mathbb C$$ such that $$f(\mathbb D)=U$$.
Therefore, my question is: If $$U$$ is a (nonempty) open and connected subset of $$\mathbb C$$, is there always a continuous (but not necessarily injective) function $$f:\mathbb D\to\mathbb C$$ such that $$f(\mathbb D)=U$$? It sufficices to assume $$U\subseteq\mathbb D$$.
If not, how do images of such functions look like?
I know that any compact connected and locally connected subset of $$\mathbb C$$ is the image of a continuous function defined on $$[0,1]$$ and hence also the image of a continuous function on $$\overline{\mathbb D}$$. However, the closure of an open and connected set does not have to be locally connected.
I believe that the answer to my first question is negative (although I hope that it is not ;) ).
Any help is highly appreciated. Thank you very much in advance!
• Not my field, but I'm pretty sure that the answer is positive to your first question. Here's my idea: first project $\Bbb{D}$ to the interval $(-1, 1)$, then expand to $\Bbb{R}$. From there, I'm pretty sure you can use a space-filling curve to fill any disk. Since $U$ is open and connected, it's path connected, and is made up of a countable union of open disks, which we can cover by a single space-filling curve (just join the endpoints for each of the countably many open disks by a path). The result will be heavily non-injective, but I think it works? Feb 14 at 11:46
• Feb 14 at 12:13
The Hahn-Mazurkiewicz theorem, as pointed out by Mathlover in the comments, is enough to show my educated guess actually works. In particular, it implies that closed balls in $$\Bbb{C}$$ are the continuous image of a compact interval. We will need a lemma:
Lemma $$\quad$$ Suppose $$\{C_n\}_{n=0}^\infty$$ is a countable collection of subspaces of a topological space $$X$$ (e.g. $$\Bbb{C}$$) which are continuous images of a compact interval and $$C := \bigcup_{n=0}^\infty C_n$$ is path-connected. Then $$C$$ is the continuous image of $$[0, \infty)$$.
Proof. Consider the intervals $$I_n := [2n, 2n + 1]$$ and $$J_n := [2n + 1, 2n + 2]$$ for $$n \in \Bbb{N} \cup \{0\}$$. Let $$\mathfrak{i}_n : I_n \to C_n$$ be surjective and continuous, and let $$\mathfrak{j}_n : J_n \to C$$ be a continuous function such that $$\mathfrak{j}_n(2n + 1) = \mathfrak{i}_n(2n + 1)$$ and $$\mathfrak{j}_n(2n + 2) = \mathfrak{i}_{n+1}(2n + 2)$$. Then, the map $$\phi : [0, \infty) = \bigcup_{n=0}^\infty (I_n \cup J_n) \to C$$ defined by $$\phi(x) = \begin{cases} \mathfrak{i}_n(x) & \text{if } \exists n \in \Bbb{N} \cup \{0\} : 2n \le x < 2n + 1 \\ \mathfrak{j}_n(x) & \text{if } \exists n \in \Bbb{N} \cup \{0\} : 2n + 1 \le x < 2n + 2. \end{cases}$$ This function is made up of countably infinitely many continuous pieces, joined at common points, making the function continuous. It's clearly surjective (even just restricting to $$\bigcup I_n$$), hence $$C$$ is the continuous image of $$[0, \infty)$$ under $$\phi$$. $$\square$$
Note that every ball in $$\Bbb{C}$$ is a countable union of closed balls, and recall that open subsets of $$\Bbb{C}$$ are a countable union of open balls. Also, in a locally path-connected space, open and connected implies path-connected. Thus, $$U$$ is the path-connected countable union of closed balls, so by the lemma, it is a continuous image of $$[0, \infty)$$.
So, to wrap up the proof, as in my comment, simply project the unit disk of $$\Bbb{C}$$ onto $$(-1, 1)$$, which is homeomorphic to $$\Bbb{R}$$. You can simply keep the map constant for points in $$(-\infty, 0]$$, and then use $$[0, \infty)$$ to map continuously onto $$U$$.
• This is amazing, thank you! Feb 14 at 13:45
One can prove a stronger claim: given a nonempty open connected $$U\subseteq \mathbb{C}$$, there's a holomorphic $$\varphi:\mathbb{D}\to \mathbb{C}$$ such that $$\varphi(\mathbb{D})=U$$.
First, let $$U\neq \mathbb{C},\neq \mathbb{C}-\{z_0\}$$. Thanks to the Riemann uniformization theorem, every such open connected subset of $$\mathbb{C}$$ is hyperbolic, i.e. has $$\mathbb{D}$$ as its universal cover. Thus there exists a surjective holomorphic map $$\mathbb{D}\to U$$. Note that there's a surjective holomorphic map $$(\exp(z)+z_0)$$ from $$\mathbb{C}$$ to $$\mathbb{C}-\{z_0\}$$, and thus it suffices to prove the result for $$\mathbb{C}$$. Now, let $$f(z)=z^3-1$$. It is easy to see that $$f:\mathbb{C}-\{1,\exp(2\pi i/3)\}\twoheadrightarrow\mathbb{C}$$. Precomposing with the covering map $$\pi:\mathbb{D}\to \mathbb{C}-\{1,\exp(2\pi i/3)\}$$, we get the claim.
One can also explore "how much" this function fails to be injective. Indeed, the uniformization theorem implies that, for $$U\neq\mathbb{C},U\neq \mathbb{C}-\{z_0\}$$, we have $$U\simeq \mathbb{D}/\Gamma$$, where $$\Gamma$$ is a discrete subgroup of $$Aut(\mathbb{D})$$ (complex automorphisms) isomorphic to $$\pi_1(U)$$. Thus two elements $$z_1,z_2$$ have the same image iff $$\exists \gamma\in \Gamma:\gamma(z_1)=z_2$$ (note that with $$U$$ simply connected we get injectivity).
• Beautiful answer, thank you very much! Feb 14 at 13:46 | 2021-10-20T01:02:40 | {
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https://math.stackexchange.com/questions/953959/n-identical-balls-distributed-into-r-urns | # $n$ Identical Balls Distributed Into $r$ Urns
• Question
In how many ways can at most $n$ identical balls be distributed into $r$ urns so that the $i$th urn contains at least $m_i$ balls, for each $i=1,...,r$?
Assume that $n\geq \sum_{i= 1}^r m_i$
• My Approach
$$x_1 + x_2 + \cdot\cdot\cdot + x_r = S$$
$$\sum_{i= 1}^r m_i \leq S \leq n$$
$$x_i \geq m_i$$
Let
$$y_i = x_i - m_i$$
then $$y_1+ y_2 + \cdot\cdot\cdot + y_r = S -\sum_{i= 1}^r m_i$$
Now I will proceed to find the number of non-negative integer-valued vectors $(y_1, y_2,...,y_r)$ such that $y_1+ y_2 + \cdot\cdot\cdot + y_r = S -\sum_{i= 1}^r m_i$ for every integer $S$ in the range of $\{\sum_{i= 1}^r m_i\quad,\quad n\}$ which is equal to $$\sum\limits_{S = \sum_{i= 1}^r m_i}^n {S - \sum_{i= 1}^r m_i + r -1\choose r-1}$$
Please have a look at my solution and give me any hints/suggestions you might have.
• What is your question actually? – user 170039 Oct 1 '14 at 13:22
• There is Question, My Approach, and then Please have a look at my solution and give me any hints/suggestions you might have.. What do you mean? – Kermit the Hermit Oct 1 '14 at 13:24
• I mean what question do you want the MSE to answer? Or is it that you want to verify your solution? – user 170039 Oct 1 '14 at 13:27
• Substitute $x_i = m_i + x_i'$ where $x_i' \ge 0$. Now subtract $m_1, m_2, \dots , m_i$ from both sides of the linear diophantine equation you have written. Finally, use stars and bars as usual. – Yiyuan Lee Oct 1 '14 at 13:28
You really seem to have the right idea, defining $y_i=x_i-m_i$, which basically says "If I put $x_i$ balls into urn $i$, this includes $m_i$ obligatory balls and then $y_i$ more that I can choose." This reduces you to solving:
$$x_1+x_2+\dots+x_r=(y_1+m_1)+(y_2+m_2)+\dots+(y_r+m_r)=n$$
which you note is the same as:
$$y_1+y_2+\dots+y_r=n-\sum_{i=1}^r m_i.$$
My question to you is: What is $S$? How is $S$ different from $n$? Unless I'm misreading this problem, $S$ seems to be extraneous. If $S$ is the number of balls you are distributing, the question seems to indicate that $S=n$ (not just $S\le n$).
So if you'll allow us to say $M=\sum m_i$, then you are correct in your reformulation of the problem, you are basically just distributing $n-M$ balls into $r$ urns, which is:
$$\binom{n-M+r-1}{r-1}.$$
That's the last term of your summation. You don't need to sum over values of this "$S$" because you need to distribute all $n$ of the balls. (If you had such a thing, it would be like counting ways of partially distributing the balls, which the question does not seem to indicate.)
You seem to know where the binomial formula for counting these tuples comes from, but for others reading, this formula is determined as explained here.
We basically reduced the problem "distributing $n$ balls into $r$ urns with restriction $m_i$ on each urn" to "distributing $n-M$ balls into $r$ urns" by translating solutions from one to the other. You can think of it physically as starting out by putting all of the $m_i$ balls into each urn as required -- you have $n-M$ left over to distribute freely, and you can use what you already know to count how to do that.
A simpler version of this problem might be "how many ways can you give your two children \$5 allowance (in whole dollar amounts) if each child needs at least \$1?" Rather than count some complicated configuration, you can really just think of this \$1 each as already handed out to them and then solve the easier problem "how many ways can you give your two children \$3 allowance?" (In this case, the answer is 4 ways.)
I am assuming from context (and from your use of the formula) that the urns are distinguishable (they are enumerated and each has a particular value of $m_i$, so philosophically -- and practically speaking, were you to attempt this exercise -- that makes the most sense).
• You just sparked something in me. For some reason I was thinking that you don't have to distribute all $n$ balls. Thus leading to $\sum_{i=1}^r m_i\leq x_1 + x_2 + ... + x_r \leq n$. – Kermit the Hermit Oct 1 '14 at 13:45
• If you are are allowed to use fewer than $n$ balls total (i.e. to just use $S$ balls and throw away the $n-S$ left over), the answer is just the sum (similar to what you have): $\sum_{S=M}^n\binom{S-M+r-1}{r-1}$. – Kellen Myers Oct 1 '14 at 14:17
• What is that $M$? Is that $\sum_{i=1}^r m_i$ ? – Kermit the Hermit Oct 1 '14 at 14:29
• Yes, it's shorthand. I think it's defined somewhere in the middle of my answer. – Kellen Myers Oct 1 '14 at 14:33
• Yes indeed, I missed that. I should pay more attention when reading it would seem. So taken in to consideration the way I interpreted the question, my solution would be correct? – Kermit the Hermit Oct 1 '14 at 14:37 | 2019-12-11T00:50:08 | {
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http://math.stackexchange.com/questions/601744/values-of-a-b-and-c-that-the-curve-y-ax3-3x2-bx-cx-ex-has-one-p/601795 | # Values of a, b, and c that the curve $y = ax^3 + 3x^2 + bx + cx + e^x$ has one point of inflection?
For what values of a, b and c does the curve $y = ax^3 + 3x^2 + bx + cx + e^x$ have exactly one point of inflection? Two points of inflection? No points of inflection? Provide a numerical approximation for the lowest value of a for which there is no inflection point.
I'm really stumped on this question. I've tried finding the second derivative but couldn't think of anything else to do from that point. Any help would be greatly appreciated. Thanks!
-
Points of inflection occur when the second derivative is zero. What is the second derivative of your curve? How can you assure there's only one zero such that we have $+/-$ on either side of the zero. (You're allowed other zeroes that don't change the inflection, i.e. $x=0$ of $y=x^2$). – Ian Coley Dec 10 '13 at 19:46
I've found the second derivative to be $y'' = 6ax + 6 + e^x = 0$ but I could not think of how to solve that. – DinoMint Dec 10 '13 at 19:48
What happens when $x=0,1,-1,-2, 2$ and so on? – Manasi Dec 10 '13 at 20:02
Given that $e^x$ is positive will drive you to a negative x. Since it won't be too small you can probably use an estimate for $e^x$ around that x. – half-integer fan Dec 10 '13 at 20:04
You don't need to solve $6ax+6+e^x=0$, @DinoMint, only figure out how many solutions it has. Think of it as intersecting the graph of $y=e^x$ with the line $y=-6ax-6$. If the slope $-6a$ is negative, there is exactly one intersection point, if it is positive you can find the tangent line to $y=e^x$ with the same slope ($-6a$) and see if this line, $y=-6ax-6$ is above or below the tangent. – Omar Antolín-Camarena Dec 10 '13 at 20:05
As you stated in the comments, the second derivative is $$y'' = 6ax + 6 + e^x = 0$$
First, we note that if $a=0$ the equation becomes $$6 + e^x = 0$$ which has no real solutions, thus there are no inflection points.
Rewriting the equation as $$x = -\dfrac{6+e^x}{6a}$$ or $$x = -\dfrac1{a} \cdot \left(1 + \dfrac 16 e^x \right)$$ makes the behavior clearer.
If $a \gt 0$ we can see that $x$ must be negative and thus $e^x < 1$ and there will be a solution in the neighborhood of $x = -\dfrac1a$.
I will leave the case where $a \lt 0$, thus $x \gt 0$ for you to consider. | 2014-10-24T13:08:24 | {
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http://mathhelpforum.com/discrete-math/224780-possible-mathematical-induction-problem.html | # Math Help - Possible mathematical induction problem
1. ## Possible mathematical induction problem
Course: Foundations of Higher MAth
Prove that $24|(5^{2n} -1)$ for every positive integer n.
This is a question from my final exam today.
P(n): $24|(5^{2n}-1)$
P(1): $24|(5^2 -1)$ is a true statement.
Assume P(k) is true. Then $5^{2k}-1 = 24a$ for some integer a.
Then $5^{2k}= 24a + 1$
P(k+1): $24|(5^{2(k+1)}-1)$
$5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$
$=(24a + 1)*25 -1$
$=(24a)(25)+25 -1$
$=(24a)(25)+24$
$=24(25a+1)$
$=24b$
Therefore, $24|(5^{2(k+1)}-1)$. By PMI, P(n) is true for every positive integer n.
None of my friends used this method though. Is this a correct way to do it?
2. ## Re: Possible mathematical induction problem
Course: Foundations of Higher MAth
Prove that $24|(5^{2n} -1)$ for every positive integer n.
This is a question from my final exam today.
P(n): $24|(5^{2n}-1)$
P(1): $24|(5^2 -1)$ is a true statement.
Assume P(k) is true. Then $5^{2k}-1 = 24a$ for some integer a.
Then $5^{2k}= 24a + 1$
P(k+1): $24|(5^{2(k+1)}-1)$
$5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$
$=(24a + 1)*25 -1$
$=(24a)(25)+25 -1$
$=(24a)(25)+24$
$=24(25a+1)$
$=24b$
Therefore, $24|(5^{2(k+1)}-1)$. By PMI, P(n) is true for every positive integer n.
None of my friends used this method though. Is this a correct way to do it?
A strict grader may like to have seen more grouping symbols.
However, the argument is correct. | 2014-09-18T14:44:17 | {
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https://math.stackexchange.com/questions/3212748/alternative-way-to-find-the-probability-in-die-throw-problem | Alternative way to find the probability in die throw problem
We throw a die 10 times. What's the probability of getting at least one 6?
Using the complentary that's $$1-Pr$$(Getting no $$6$$'s)
$$1- ({\frac{5}{6}})^{10} = 0.838494$$
There's a note that says that to solve the problem directly would require a complex use of the additivity property.
a) How do you solve the problem directly?
I think that to solve it directly, though not sure, is to find the $$P(A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5 \cup A_6)$$ in ten die throws, where $$A_i$$ is the event of getting $$i$$ $$6$$'s
As the events $$A_i$$ are not disjoint the exclusion-inclusion principle is needed.
My try:
$$\binom{10}{1} \frac{1}{6} - \binom{10}{2} \frac{1}{6^2} + \binom{10}{3} \frac{1}{6^3} - \binom{10}{4} \frac{1}{6^4}+ \binom{10}{5} \frac{1}{6^5} - \binom{10}{6} \frac{1}{6^6} =0.838091$$
It is a slightly different result that the one obtained using the complementary. What am I not doing right?
b)If the events aren't disjoint, can you talk about the "additivity property"? As it is defined for disjoint events.
It would be the following:
$$\sum_{n=1}^{10} \dbinom{10}{n}\left(\dfrac{1}{6}\right)^n\left(\dfrac{5}{6}\right)^{10-n}$$
Here the events are disjoint. You have exactly 1 roll of a 6, exactly two rolls of a six, ..., exactly ten rolls of a six.
Your attempt works as well (although you stopped at $$i=6$$). Had you continued using Inclusion/Exclusion up to $$i=10$$, it would have given the same answer:
$$\sum_{i=1}^{10}(-1)^{i+1}\dbinom{10}{i}\left(\dfrac{1}{6}\right)^i$$
Here you are counting if you choose a die, the probability that one die is a six, minus if you choose two dice, both of them are a six, plus if you choose three dice, all three are a six, etc.
b) Yes.
$$P(A\cup B) = P(A)+P(B)-P(A\cap B)$$
This is the property that yields the Inclusion/Exclusion principle in the first place.
You got a different answer because you only computed the first six terms of the inclusion-exclusion method. You have ten dice, each of which could produce a six, so you need ten terms.
• You use Inclusion-Exclusion when you have events that are not mutually exclusive. You can see that "roll six on the first die" and "roll six on the second die" are not mutually exclusive because it is possible that you roll sixes on both dice. I don't know what you mean by "inclusive." – David K May 3 at 21:27 | 2019-11-18T11:34:32 | {
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https://math.stackexchange.com/questions/2104483/how-to-express-the-cardinality-of-%E2%88%8F-1%E2%89%A4i%E2%89%A4n-a-i-in-terms-of-cardinalities-a | # How to express the cardinality of $∏_{1≤i≤n} A_i$ in terms of cardinalities $|A_1|, |A_2|, . . . , |A_n|$
I was given the problem:
For finite sets $A_1, A_2,\dotsc , A_n$ define their Cartesian product $\prod_{i=1}^n A_i$ as the set of all $n$-sequences $(x_1, x_2,\dotsc, x_n)$, where $x_i \in A_i$ for every $i = 1, 2, \dotsc, n$. Find a formula expressing the cardinality of $\prod_{i=1}^n A_i$ in terms of cardinalities $|A_1|, |A_2|,\dotsc , |A_n|$.
And I am struggling to understand what it is actually asking for, could someone explain it to me please, thanks. :)
• Do you know what the Cartesian product is? For instance, if $A_1 = \{1,2\}$ and $A_2 = \{3,4,5\}$ could you explicitly write down $\prod_{1 \leq i \leq 2} A_i$? – Mees de Vries Jan 19 '17 at 13:24
• @MeesdeVries This is all prossible combinations right? So for that example {(1,3), (2, 3), (1, 4) ....}, But I have never seen the notation: ∏ 1≤i≤n Ai before, what does that mean? – Alfie Jan 19 '17 at 13:27
$\prod_{1\le i\le n}A_i$ is the cartesian product, that is, all finite sequences $(a_1,\ldots,a_n)$ such that $a_i \in A_i$ for each $i=1,\ldots,n$. How many such sequences can you choose? $|A_1|$ choices for $a_1$, ..., $|A_n|$ choices for $a_n$. Therefore $$\left|\prod_{1\le i\le n}A_i\right|=\prod_{1\le i\le n}|A_i|.$$
• So is this question asking for a formula for the carnality of the cartesian product of $A_1, A_2,\dotsc , A_n$ ? – Alfie Jan 19 '17 at 13:51
• Yes. ____________ – Paolo Leonetti Jan 19 '17 at 13:52
• So $\prod_{1\le i\le n}|A_i|$= $|A_1| * |A_2| * \dotsc * |A_n|$ – Alfie Jan 19 '17 at 13:57
• @Bram28 Assuming the axiom of choice, multiplication of infinite cardinal numbers is not difficult. If either $\kappa$ or $\mu$ is infinite and both are non-zero, then $\kappa \cdot \mu= \max\{\kappa, \mu\}$. In particular, $\kappa^n=\kappa$ for all $n\ge 1$. See here: en.wikipedia.org/wiki/Cardinal_number – Paolo Leonetti Jan 19 '17 at 14:56
• @Bram28 Well, you can find the answer in every introduction about cardinal arithmetics. Take a look here math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Murphy.pdf and here euclid.colorado.edu/~monkd/m6730/gradsets06.pdf – Paolo Leonetti Jan 19 '17 at 16:00
We know that $$|A\times B|=|A|\times|B|\qquad(1)$$.
We want to show $$\left|\prod_{i=1}^nA_i\right|=\prod_{i=1}^n|A_i|$$ is true for any natural number $n$, where $\prod_{i=1}^n|A_i|=|A_1|\times|A_2|\times\dotsc\times|A_n|$. So, we have use induction.
The base case $n=1$ ($|A_1| = |A_1|$) is trivial. Now suppose inductively that $\left|\prod_{i=1}^nA_i\right|=\prod_{i=1}^n|A_i|$. We want to show $$\left|\prod_{i=1}^{n+1}A_i\right|=\prod_{i=1}^{n+1}|A_i|.$$ Now we need to show $$\left|\prod_{i=1}^{n+1}A_i\right|=\left|\left(\prod_{i=1}^nA_i\right)\times A_{n+1}\right|\qquad(2),$$ i.e., the cardinality of the set $\prod_{i=1}^{n+1}A_i$ is equal to the cardinality of the set $\left(\prod_{i=1}^nA_i\right)\times A_{n+1}$. So \begin{aligned}\left|\prod_{i=1}^{n+1}A_i\right|&=&\left|\left(\prod_{i=1}^nA_i\right)\times A_{n+1}\right|&\qquad\text{by }(2)\\&=&\left|\prod_{i=1}^nA_i\right|\times |A_{n+1}|&\qquad\text{by }(1)\\&=&|A_1|\times|A_2|\times\dotsc\times|A_n|\times|A_{n+1}|&\qquad\text{by induction hypothesis}\\&=&\prod_{i=1}^{n+1}|A_i|.\end{aligned}
• This is exact, but I think the OP was asked a formula, but not asked to prove it. – Jean Marie Jan 19 '17 at 14:46
• @CristianGz Do you know if (1) also holds for infinities? If so, how am I to think about multiplying 'infinities'? Is there a definition for that? – Bram28 Jan 19 '17 at 14:52
• @JeanMarie: Agree. I deleted the answer. But I though it can be useful. – Cristhian Gz Jan 19 '17 at 14:53
• @Bram28, the statement is true to infinite sets. It is possible using bijective functions to state equal cardinality between two sets. Check en.m.wikipedia.org/wiki/Cardinality – Cristhian Gz Jan 19 '17 at 14:55 | 2019-12-06T21:40:56 | {
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https://ocw.mit.edu/courses/mathematics/18-065-matrix-methods-in-data-analysis-signal-processing-and-machine-learning-spring-2018/video-lectures/lecture-21-minimizing-a-function-step-by-step/ | # Lecture 21: Minimizing a Function Step by Step
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## Description
In this lecture, Professor Strang discusses optimization, the fundamental algorithm that goes into deep learning. Later in the lecture he reviews the structure of convolutional neural networks (CNN) used in analyzing visual imagery.
## Summary
Three terms of a Taylor series of $$F$$($$x$$) : many variables $$x$$
Downhill direction decided by first partial derivatives of $$F$$ at $$x$$
Newton's method uses higher derivatives (Hessian at higher cost).
Related sections in textbook: VI.1, VI.4
Instructor: Prof. Gilbert Strang
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GILBERT STRANG: Well, OK, I am happy to be back, and I am really happy about the project proposals that are coming in. This is like, OK, this is really a good part of the course. And so keep them coming, and I'm happy to give whatever feedback I can on those proposals, and do make a start there. They're really good, and if some are completed before the end of the semester and we can to offer you a chance to report on them, that that's good too. So well done with those proposals.
So today, I'm jumping to part six. So part six and part seven are optimization which is the fundamental algorithm that goes into deep learning. So we've got to start with optimization. Everybody has to get that picture, and then part seven will be the structure of CNNs, Convolution Neural Nets, and all kinds of applications.
And so can we start with optimization? So first, can I like get the basic facts about three terms of a Taylor series? So that's the typical. It's seldom that we would go up to third derivatives in optimization.
So that's the most useful approximation to a function. Everybody recognizes it. Here, I'm thinking of F as just one function, and x as just one variable, but now I really want to go to more variables. So what do I have to change if F is a function of more variables? So now, I'm thinking of x as-- well, now let me see.
Yeah, I want n variables here. x is x1 up to xn. So just to get the words straight so we can begin on optimization, so what will be the similar step so the function F at x-- remember, x is n variables. OK?
Now, what do I have? Delta x, so what's the point about delta x now? It's a vector, delta x1 to delta xn, and what about the derivative of F? It's a vector too, the derivative of F with respect to x1, the derivative of F with respect to x2, and so on.
What do I have to change about that? I know those guys are vectors, so it's their dot product. So it's delta x transpose at vector times this dF/dx. So now I'm replacing this by all the derivatives, and it's the gradient. So the gradient of F at x is the derivatives-- let's see.
It's essential to get the notation straight here. Yeah, so it'll be the partial derivatives of the function F. So grad F is the partial derivatives of F with respect to x1 down to partial derivative with respect to xn. OK, good.
That's the linear term, and now what's the quadratic term? 1/2, now delta x isn't a scalar anymore. It's a vector. So I'm going to have delta x transpose and a delta x, and what goes in between is the second derivatives, but I've got a function of n variables.
So now, I have a matrix of second derivatives, and I'll call it H. This is the matrix of second derivatives, Hjk is the second derivative of F with respect to xj and xk, and what's the name for this guy? The Hessian, Hessian matrix.
How the Hessians got into this picture I don't know. The only Hessians I know are the ones who fought in the Revolutionary War for somebody. Who? Which side were they on? I think maybe the wrong side. The French were on our side and--
Anyway, Hessian matrix, and what are the facts about that matrix? Well, the first fact is that it's [INAUDIBLE] and the key fact is it's symmetric. Yeah. OK, and again, it's an approximation. And everybody recognizes that if n is very large, and we have a function of many variables. Then, we had n derivatives to compute here, and about 1/2 n squared derivatives.
The 1/2 comes from the symmetry, but the key point is the n squared derivatives to compute there. So computing the gradient is feasible if n is small or moderately large. Actually, by using automatic differentiation, the key idea of back propagation, back prop, you can speed up the computation of derivatives quite amazingly. But still for the size of deep learning problems that's out of reach. OK.
So that's the picture, and then I will want to use this to solve equations. There is a parallel picture for a vector f. So now, this is a vector function. This is f1 of x up to fn of x, an x is x1 to xn. So I have n functions of n variables, n functions of n variables.
Well, that's exactly what I have in the gradient. Think of these two as parallel, the parallel being f corresponds to the gradient of F, n functions of n variables. OK. Now maybe, what I'm after here is to solve f equals 0. So I'm going to think about the f at x plus delta x, so it starts with f of x.
And then we have the correction times the matrix of first derivatives, and what's the name for that matrix of first derivatives? Well, if I'm just given n functions-- yeah, what am I after here? I'm looking for the Jacobian. So here we'll go the Jacobian, J. This is the Jacobian named after Jacoby, Jacobian matrix.
And what are its entries? J, the jk entry is the derivative of the J function with respect to the kth variable, and I'm stopping at first order there. OK, so these are sort of like facts of calculus, facts of 18.02 you could say. Multivariable calculus, that's the point.
Notice that we're doing just like the first half of 18.02, just do differential calculus, derivatives, Taylor series. We're not doing multiple integrals. That's not part of our world here. OK, so that's the background.
Now, I want to look at optimization. So over here, I want to optimize-- well, over here, let me try to minimize F of x, and I'll be in the vector case here. And over here, I want to solve f equals 0, and of course, that means f of 1 equals 0 all the way along to fn equals 0. Here, I have n equations, and n unknowns.
Let me start with that one, and I'll start with Newton's method, Newton's method to solve these n equations and n unknowns. OK, so Newton, Newton's method which is often not presented in 18.02. That's a crime, because that's the big application of gradients in Jacobians.
OK, so I'm trying to solve n equations and n unknowns, and so I want f at x plus delta x to be 0. Right? So I want f of x plus delta x to be 0. So f at x plus delta x is-- I'm putting in a 0. I'm just copying that equation-- is f at where I am. Let me use K for the case iteration.
So I'm at a point xK. I want to get to a point xK plus 1. And so I have 0 is f of x plus J, at that point, times delta x which is xK plus 1 minus xK. Good. That's Newton's method.
Of course, 0 isn't quite true. Well, 0 will be true if I'm constructing xK plus 1 here. I'm constructing xK plus 1. OK. So let me just rewrite that, and we've got Newton's method. So we're looking for this change, xK plus 1 minus xK. I'll put it on this side as plus xK, so that's this.
Now, I have to invert that and put it on the other side of the equation. So that will go with a minus. This guy will be inverted and f at xK. So that's Newton's methods. It's natural.
So let me just repeat that. You see where the xK plus 1 minus xK is sitting? Right? And I moved f of xK to the other side with a minus sign, and then I multiplied through by J inverse, so I got that. So that's Newton's method for a system of equations, and over there, I'm going to write down Newton's method for minimizing a function. This is such basic stuff that we have to begin here.
Let me even begin with an extremely straightforward example of Newton's method here. Suppose my function-- suppose I've only got one function actually. Suppose I only had one function. So suppose my function is x squared minus 9, and I want to solve f of x equals 0. I want to find the square root of 9.
OK, so what is Newton's method for it? My point is just to see how Newton's method is written and then rewrite it a little bit so that we see the convergence. OK, so of course, the Jacobian is 2x. So Newton's method says that xK plus 1-- I'm just going to copy that Newton's method-- minus 1 over 2xK. Right? That's the derivative times f at xK which is xK squared minus 9.
OK. We followed the formula, this determines xK plus 1, and let's simplify it. So here I have xK minus that looks like 1/2 of xK, so I think I have 1/2xK, and then this times this is 9/2 of 1 over xK. Is that right? 1/2 of xk from this stuff and plus 9/2 of 1 over xK. OK.
Can I just like check that I know the answer is 3? Can I be sure that I get the right answer, 3? That if xK was exactly 3, then of course, I expect xK plus 1 to stay at 3. So does that happen? So 1/2 of 3 and 9/2 of 1/3, what's that, 1/2 of 3 and 9/2 of 1/3?
OK, that's 3/2 and 3/2. That's 6/2, and that's 3. OK. So we've checked that the method is consistent which just means we kept the algebra straight. But then the really important point about Newton's method is to discover how fast it converges. So now let me do xK plus 1 minus 3.
So now, I'm looking at the error which is, I hope, approaching 0. Is it approaching 0? How quickly is it approaching 0? These are the fundamental questions of optimization.
So I'm going to subtract 3 from both sides somehow. OK, from here, I guess, I'm going to subtract 3. So I was just checking that it was correct. OK. Now, so xK plus 1 minus 3, I'm going to subtract 3 from both sides. I'm going to subtract 3 there, and then I hope that-- that box is what goes down here. Right?
Subtracted 3 from both sides, so I'm hoping now things go to 0. OK, so what do I have there? Let me factor out the 1 over xK. So what do I have then left? 1 over xK, so there's a 9/2 from there, 1 over xK.
So I really have 1/2 of xK squared, because I've divided by an xK. And this minus 3, I better put minus 3xK, because I'm dividing by xK. I claim that that's-- now I've got it. And let's see, let me take out the 2-- 2, forget these 2s, and make that a 6. So I have 1 over 2xK times 9 plus xK squared minus 6.
Anything good about that? We hope so. We hope that that is something attractive. So this is, again, the error at set K plus 1, and it's 1 over 2xK times this thing in brackets-- 9 plus xK squared minus 6xK. And we recognize that as xK minus 3 squared.
xK squared minus 6 of them plus 9, that's xK minus 3 squared. OK, that was the goal, of course. That's the goal that shows why Newton's method is fantastic. If you can execute it, if you can start near enough, notice that-- so how do I describe this great equation? It says that the error is squared at every step, squared at every step.
So if I'm converging to a limit, it will satisfy the-- it'll be 3, or I guess minus 3, is that possible? Yeah, minus 3 is another solution here. So we've got two solutions. Newton's method could converge to 3. Am I right, it could converge to minus 3?
So I'd have a similar equation sort of centered at minus 3, or does it always do one of those? It could blow up. So there are sort of regions of attraction. They're all the starting points that approach 3, and the whole point of that equation is with quadratic convergence the error being squared at every step. It zooms in on 3.
Then, there is all the starting points that would go to minus 3, and then there are the starting points that would blow up. And those, maybe for this very simple problem, the picture is not too difficult to sort out those three regions. And this is allowing for a vector, two equations or n equations, then we're in n variables, and really you get beautiful pictures.
You get some of the type of pictures that gave rise to these books on fractals, picture books on fractals for these basins of attraction. Does the starting point lead you to one of the solutions, or does it lead you to infinity? Here, that would be interesting to just draw it for this, but the essential point is the quadratic convergence, if it's close enough.
You see that it has to be close. If x0 is pretty near 3, then this is about 1/6 of that, and there would be a good region of attraction in this case. OK. So that's Newton's method for equations.
And now I want to do Newton's method. I just want to convert all those words over to Newton's method for optimization. So remember, these boards were solving f equals 0. This board is minimizing capital F, and what's the connection between them? Well of course, this corresponds to solving the gradient equals 0.
At a minimum, if I'm minimizing, I'm finding a point where all the first derivatives are 0. So that will be the match between these. This grad F in this picture is the small f in that picture. OK.
Now, I guess here I have-- and this is sort of the heart of our applications to deep learning-- we have very complicated loss functions to minimize, functions of thousands or hundreds of thousands of variables. OK. So that means that we would like to use Newton's method, but often we can't. So I need him to put down here two methods-- one that doesn't involve those high second derivatives and Newton's that does.
So first, I'll write down a method that does not involve, so method one, and this will be steepest descent. And what is that? That says that xK plus 1-- the new x is the old x minus-- steepest descent means that I move in the steepest direction which is the direction of the gradient of F. I move some distance, and I better have freedom to decide what that distance should be. So this is a step size, s, or in the language of deep learning, it's often called the learning rate, so if you see learning rate. OK.
So and it's natural to choose sK. We're going along, do you see what this right-hand side looks like? I'm at a point in n dimensions. We're in n dimensions here. We have functions of n variables.
There is a vector. There is a direction to move down the steepest slope of the graph. And here is a distance to move, and we will stop. We'll have to get off this step, normally. If we stay on it, it will swing back, it'll take us off to infinity.
You would like to choose sK so that you minimize capital F. You take the point on this line, so this a line in R n, a direction in R n. And for all the points on that line, in that direction, F has some value, and what you expect is that initially, because you chose it sensibly, the value of F will drop. But then at a certain point, it will turn back on you and increase.
So that would be the natural stopping point. I would call that an exact line search. So I exact line search would be, exact line search is the best s. Of course, that would take time to compute, and you probably, in deep learning, that's time you can't afford, so you fix the learning rate s. Maybe you choose 0.01 to be pretty safe.
OK, so that's method one, steepest descent. Now, method two will be Newton's method. So now, we have xK plus 1 equal to xK minus something times delta F, and now I'm going to do the right thing. I'm going to live right here, and the right thing is the Hessian, the second derivative.
This was cheap. We just took the direction and went along it. Now, we're getting really the right direction by using the second derivative, so that's H inverse. OK, and what I've done is to set that 0.
Do you see that's Newton's method? It's totally parallel to this guy. Actually, I'm really happy to have these two on the board parallel to each other, because you have to keep straight, are you solving equations, or are you minimizing functions? And you're using different letters in the two problems, but now you see how they match.
The Jacobian of-- so again the matches, think of f as the gradient of F. That's the way you should think of it. So the Jacobian of the gradient is the Hessian. The Jacobian of the gradient is the Hessian, and that makes sense, because the first derivative of the first derivative is the second derivative. Only we're doing matrix y, so the Jacobian of the gradient-- we're doing a vector matrix sentence instead of a scalar sentence-- the Jacobian of the gradient is a Hessian. Yeah, right.
OK, so that's what I wanted to start with, just to get those basic facts down. And so the basic facts were the three-term Taylor series. And then the basic algorithms followed naturally from it by setting f F at the new point to 0, if that's what you were solving or by assuming you had the minimum. Right, good, good, good, good. OK.
Now, what? Now, we have to think about solving these problems, studying. Do they converge? What rate do they converge? Well, the rate of convergence is like why I separated off this example.
So the convergence rate for Newton's method will be quadratic. The error gets squared, and of course, that means super-fast convergence, if you start and close enough. The rate of convergence for a steepest descent is, of course, not. You're not squaring errors here, because you're just taking some number instead of the inverse of the correct matrix, so you can't expect super speed.
So a linear rate of convergence would be right. You would like to know that the error is multiplied at every step by some constant below 1. That would be a linear rate compared to being squared at every step. OK, and so this will be our basic formula that we build on for really large scale problems.
And there are methods, of course, people are going to come up with methods that they're sort of a cheap Newton's method. Levenberg-Marquardt, and it's in the notes at the end of this section, at the end of 6.4 that we'll get to. So Levenberg-Marquardt is a sort of cheap man's Newton's method. It does not compute to Hessian, but it says, OK, from the gradient, I can see one term in the Hessian. So it grabs that term, but it's not fully second order.
OK. So now, we have to think about problems, and I guess the message here is, at our starting point, has to be convexity. Convexity is the key word for these problems, for the function that we want to minimize. If that's a convex function, well first of all, the convex function is likely to have one minimum. And the picture that's in our mind of steepest descent, this picture of a bowl, a bowl is the graph of the convex function.
So I'm turning to convexity now. I'll leave that board there, because that's pretty crucial, and speak about the idea of convexity. Convex function, convex set, so let's call the function f of x, and a typical convex set would be I'll call it K. OK. So we just want to remember what does that word can convex mean, and how do you know if you have a convex function or a convex set?
OK, let me start with convex set. So because here is my general problem, my convex minimization, which you hope to have, and in many applications, you do have. So you minimize a convex function for points in a convex set. So that's like the ideal situation. That's the ideal situation, to get something on your side, something powerful, convexity.
The function is convex, and you say, well, let me draw a convex function, the graph. OK, so I'll draw a convex function, say a bowl. So that's a graph of f of x, and then here are the x's. Let me maybe put x1 and x2 in the base and the graph of f of 1x x2 up here. OK. Actually, I'm over there.
I should be calling this function F, I think. Is that right? Yeah, a little f would be the gradient of this guy. Yeah, I think so. OK.
Now, I'm minimizing it over certain x's, not all x's. I might be minimizing, for example, K might be the set where Ax equals B. K might be, in that case, a subspace or a shifted subspace. I said subspace, but then 18.06 is reminding me in my mind that I only have a subspace when B is 0.
You know the word for a subspace that's sort of moved over? Affine, so I'll just put that word down here. Bunch of words to learn for this topic, but they're worth learning. OK.
So it's like a plane but not necessarily through the origin. If B is 0, it doesn't go through it. If B it's not 0, it doesn't go through the origin. OK. Anyway, or I have some other convex set. Let me just put this convex set K in the base for you, and did I make it convex? I think pretty luckily I did.
So now what's the? Well, the convex sets the constraint, so this is the constraint set. Constraint is that x must be in the set K. OK, and I drew it as a convex blob. Here was an example where it's flat, not a blob but a flat plane.
But let me come back to what does convex mean. What's a convex set? Yeah, we have to do that, should have done that before. In the notes, I had the fun of figuring out, if I took a triangle, is that a convex set? Let's just be sure.
So what's a convex set? That is a convex set, because if I take any two points in the set and draw the line between them, it stays in the set. So that's convexity, any edge, line, from x1 to x2 stays in the set. OK, good.
So here's my little exercise to myself. What if I took the union of two triangles? All I want to get you to do is just visualize convex and not convex possibilities. Suppose I have one triangle, even if it was obtuse, that's still convex, right? No problem.
But now what if I put those two triangles together, take their union? Well, if I take them sitting with a big gap between, like I've lost. I mean, I never had a chance that way, because if it was the union of these two-- well, you know what I'm going to say. If I'm doing that point and that point, of course, it goes outside and stupid. All right.
What if what if that triangle, that lower triangle, overlaps the upper triangle? Is that a convex set? Everybody's right saying no. Why how do I see that the union of those two triangles is not a convex set? Guys, you tell me where to pick two points, where the line goes out. Well, I take one from that corner and one from that corner, and the line between them went outside. So union is usually not convex.
Well, if I think of the union of two sets, my mind automatically goes to the other corresponding possibility which is the intersection of the two sets. So if I take the intersection of two sets. Now, what's the deal with that? When I had two triangles, two separated triangles, what can we say about the intersection of those two triangles?
AUDIENCE: [INAUDIBLE]
GILBERT STRANG: It's empty. So should we regard the empty set as a convex set? Yes. Isn't it?
AUDIENCE: Yeah, it's vacuous.
GILBERT STRANG: Vacuous, so it hasn't got any problems. Right? OK, but now the intersection is always convex. I'm assuming the two sets that we start with are. Now, that's an important fact, that the intersection of convex sets. Let's just draw a picture that shows an example.
So what's the intersection? Just this part and it's convex. OK, can you give me a little proof that the intersection is convex? So I take two points in the intersection-- let me start the proof.
To test if something's convex, how do you test it? You take two points in the set in the intersection, and you want to show that the line between them is in the intersection. OK, why is that?
So take two points, take x1 in the intersection. We've got two sets here, and that's the symbol for intersection, and we've got another point in the intersection. And now, we want to look at the line between them, the line from x1 to 2x. What's the deal with that one? Is that fully in K1?
AUDIENCE: Yes.
GILBERT STRANG: Why is it fully in K1? I took two points in the intersection, I'm looking at the line between them, and I'm asking, is it in the first set K1? And the answer is yes, because those points were in K1, and K1's convex. And is that line between them in K2? Yes, same reason, the two endpoints were in K2, so the line between them is in K2.
So the intersection of convex sets is always convex. The intersection of convex sets is convex. Good. So you'll see in the note these possibilities with two triangles. Sometimes, you can take the union but not very often. OK.
Now, what's the next thing I have to do? Convex functions, we got convex sets, what are convex functions, and we're good. Because this is our prototype of a problem, and I now want to know what it means for that F to be-- oh, I'm sorry. I now know what it means for the set K to be convex set, but now I have to look at the other often more important part of the problem. What's the function I'm minimizing, and I'm looking for functions with this kind of a picture. OK.
The coolest way is to connect the definition of a convex function to the definition of a convex set. This is really the nicest way. It's a little quick. It just swishes by you. But tell me, do you see a convex set in that picture? [INAUDIBLE]
You see a convex set in that picture. That's the picture of a graph of a convex function. It's a picture of a bowl. Are the points on that surface, is that a convex set? No, certainly not. No, but where is a convex set to be found here, in that picture? Yes.
AUDIENCE: The set of y, if y is greater than [INAUDIBLE]
GILBERT STRANG: Yes, the points on and above the bowl, inside the bowl, we could say, these points. So convex function, yes, a function's convex when the points on and above the graph are convex set. You could say, OK, mathematicians are just being lazy. Having got one definition straight for a convex set, now they're just using that to give an easy definition of a convex function. Actually, it's quite useful for functions that could maybe equal infinity, sort of generalized functions.
But it's not the quickest way to tell if the function is convex. It's not our usual test for convex functions. So now I want to give such a test. OK. So now, the definition of convex function, of a smooth convex, yeah. This fact, I shouldn't rush off away from it, from the definition of a convex function as having a convex set above its graph. The really official French name for the set above the graph is the epigraph, but I won't even write that word down. OK.
Why do I come back to that for a minute? Because I would like to think about two functions, F1 and F2. Out of two functions, I can always create the minimum or the maximum.
So suppose I have to convex functions, convex function F1 and F2. OK. Then, I could choose a minimum. I could choose my new function. Shall I call it little m for minimum? m of x is the minimum of F1 and F2.
And I could choose a maximum function which would be the maximum of F1 of x and F2 of x at the same point x. It's just a natural to think, OK, I have two functions. I've got a bowl and I've got another bowl, and suppose they're both convex.
So I'm just stretching you to think here. If I've got the graphs of two convex functions, and I would like to consider the minimum of those two functions and also the maximum of those two functions. I believe life is good. One of these will be convex, and the other won't.
And can you identify which one is convex and which one is not convex? What about the minimum? Is that a convex function? So just look at the graph. What does the minimum look like? The minimum is this guy until they meet somehow on some surface and then this guy.
Is that convex? We have like one minute to answer that question. Absolutely no. It's got this bad kink in it. What about the maximum of the two functions? So the maximum is the one that is above, all the points or things that are above or on.
There is the maximum function. That was the minimum function. It had a kink. The maximum function is like that, and it is convex, so maximum yes, minimum no. OK, and we could have a maximum of 1,500 functions. If the 1,500 functions are all convex, the maximum will be, because it's the part way above everybody's graph, and that would be the graph of the maximum. OK, good.
And now finally, let me just say, how do you know whether a function is convex? How to test, how of test. OK, so let me take just a function of one variable. What's the test you learned in calculus, freshman calculus actually, just show that this is a convex function? What's the test for that?
AUDIENCE: Use second derivative.
GILBERT STRANG: Second derivative should be?
AUDIENCE: Positive.
GILBERT STRANG: Positive or possibly 0, so second derivative greater or equals 0 everywhere. That's convex. OK, final question, suppose F is a vector. So this is a vector, and so I have n functions of n variable. No, I don't. I have one, sorry, I've got one function, but I'm in n variables. So this was just one.
What's the test for convexity? So it would be passed, for example, by x1 squared plus x2 squared. Would it be passed by-- so here would be the question-- would it be passed by x transpose some symmetric matrix S? That would be a quadratic, a pure quadratic.
Would it be convex? What would be the test? I'm looking for an n dimensional equivalent of positive second derivative. The n dimensional equivalent of positive second derivative is convexity, and we have to recognize what's the test. So I could apply it to this function, or I could apply it to any function of n variables. It should be OK.
What's the test here? Here, I have a matrix instead of a number. So what's the requirement going to be? Times out, yeah? [INAUDIBLE] Positive definite or semidefinite, or semidefinite just as here. Yeah.
So the test is positive, semidefinite, Hessian. And here, the Hessian is actually that S, because the second derivatives will produce-- I'll put a 1/2 in there-- the second derivatives will produce S equal the Hessian H. So here, the S-- so positive semidefinite, Hessian in general, second derivative matrix for a quadratic.
OK. So its convex problems that we're going to get farther with. We run into no saddle points. We run into no local minimum. Once we found the minimum, it's the global minimum. These are the good problems. OK, again, happy to see you today, and I look forward to Wednesday. | 2021-10-16T17:30:31 | {
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https://www.jiskha.com/questions/86506/i-have-3-questions-and-i-cannot-find-method-that-actually-solves-them-1-integral | # Calculus - Integrals
I have 3 questions, and I cannot find method that actually solves them.
1) Integral [(4s+4)/([s^2+1]*([S-1]^3))]
2) Integral [ 2*sqrt[(1+cosx)/2]]
3) Integral [ 20*(sec(x))^4
1. 👍
2. 👎
3. 👁
1. 1) expand in partial fractions.
2) Using cos(2x) = 2 cos^2(x) -1 derive a formula for cos(1/2 x) in terms of cos(x). Express the integrand in terms of cos(1/2 x)
3) [Notation: cos = c, sin = s]
1/c^4 = (s^2 + c^2)/c^4 =
1/c^2 + s^2/c^4
1/c^2 yields a tangent when integrated
To integrate s^2/c^4 do partial integration:
s^2/c^4 = s (s/c^4)
Integral of s/c^4 is 1/3 1/c^3
So, we need to integrate
1/c^3
times the derivative of of s, i.e. 1/c^2, but that is tan(x)!
Note that the way to solve such problems is not to systematically work things out in detail at first, because then you would take too long to see that a method doesn't work.
Instead, you should reason like I just did, i.e. forget the details, be very sloppy, just to see if things works out and you get an answer in principle, even though you need to fill in the details.
If you get better at this, you can do the selection of what method to use in your head, you'll see it in just a few seconds when looking at an integral. You can then start to work out the solution for real on paper.
1. 👍
2. 👎
2. Could you expand on #1 a bit more? I tried Partial Fractions, but I couldn't get a definite answer...
1. 👍
2. 👎
3. I'll show you how to do it without solving any equations.
The function is up to a factor 4:
f(s) = (s+1)/{(s^2+1)*[(s-1)^3]}
The partial fractions are precisely the singular terms when expanding around the singularities. So, let's examine the singularity at s = 1:
Put s = 1 + u and expand in powers of u, keeping only the singular terms:
f(1+u) = u^(-3) (2+u)/[(u+1)^2+1] =
u^(-3) (2+u)/(2+2u+u^2) =
u^(-3) (2+u)/2 1/(1+u+u^2/2) = (use
1/(1+x) = 1-x+x^2-x^3+...)
u^(-3) (2+u)/2 [1-u-u^2/2 +u^2+...]
=[u^(-3) + u^(-2)/2][1-u+u^2/2+...]
u^(-3) -1/2 u^(-2) + nonsingular terms
This means that in the neighborhood of s = 1 we have:
f(s) = 1/(s-1)^3 - 1/2 1/(s-1)^2 +.. nonsingular terms
f(s) also has singularities at s = i and s =-i. Around s = i, we have:
f(s) = a/(s-i) + nonsingular terms
Multiply both sides by (s-i) and take the limit s --> i to find a:
a = (i+1)/[2i(i-1)^3] =
-1/2 1/(i-1)^2
Around s = -i, we have:
f(s) = b/(s+i) + nonsingular terms
Multiply both sides by (s+i) and take the limit s --> -i to find b:
b = (-i+1)/[-2i(-i-1)^3] =
-1/2 1/[(i+1)^2]
The sum of all the singular terms of the expansions around the singular points is:
1/(s-1)^3 - 1/2 1/(s-1)^2 +
-1/2 1/(i-1)^2 1/(s-i)
-1/2 1/[(i+1)^2] 1/(s+i) =
1/(s-1)^3 - 1/2 1/(s-1)^2 +
1/2 1/(s^2+1)
This is the desired partial fraction decomposition. The proof of why this works is a consequence of Liouville's theorem
http://en.wikipedia.org/wiki/Liouville's_theorem_(complex_analysis)#Proof
Just consider the difference of f(s) and the sum of all the singular terms. The resulting function doesn't have any singularites and is bounded. So, by Liouville's theorem it is a constant. We know that f(s) and all the singular terms tend to zero, so that constant must be zero.
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https://discourse.julialang.org/t/general-plotting-code-for-cone-in-3d-with-glmakie-or-plots/92104 | # General Plotting Code for Cone in 3D with GLMakie or Plots
Hi all,
so I want to plot a general cone of pyramid and a right circular cone, I get this formula:
But I think the plotting does not use the volume formula. Because of this torus code for GLMakie that I have:
using GLMakie
using ColorSchemes
cmap3 = get(colorschemes[:plasma], LinRange(0,1,100))
cmap4 = [(cmap3[i],0.65) for i in 1:100]
ξ = -40:1:40
c = 20
a = 15
torus = [((hypot(X,Y)-c)^2+Z^2-a^2) for X in ξ, Y in ξ, Z in ξ]
volume(torus, algorithm = :iso,isovalue =100,isorange =50, colormap=cmap4)
current_figure()
For general cone of pyramid, If the based is made of a closed random curves, then we might need to plot in x-y plane the curves then plot the top of pyramid afterwards.
For the right circular cone, I believe there is a general formula already.
You could conveniently define both as parametric surfaces:
using Plots; plotlyjs()
z, Θ = range(0,10,200), range(0,2π,300)
# Right-cone:
X = [u * cos(v) for u in z, v in Θ]
Y = [u * sin(v) for u in z, v in Θ]
Z = [-2u for u in z, _ in Θ]
surface(X, Y, Z, size=(600,600), cbar=:none, legend=false)
# Trefoil cone:
X = [u/3 * (sin(v) + 2*sin(2v)) for u in z, v in Θ]
Y = [u/3 * (cos(v) - 2*cos(2v)) for u in z, v in Θ]
Z = [-2u for u in z, _ in Θ]
surface(X, Y, Z, size=(600,600), cbar=:none, legend=false)
1 Like
From your questions posted on this forum it seems that you are teaching math. Hence you should explain to your students how is parameterized a general cone,
defined by its vertex V, given as a 3 vector, and a closed (but not necessarily) planar curve,
x = x(u)
y = y(u)
z= b # b from base plane; usually b=0
u \in [\alpha, \beta]
.
The cone is generated by all segments of line connecting its vertex, V, with the points
(x(u), y(u), b), on the base curve.
Such a segment has the equations:
\displaystyle\frac{X-V[1]}{x(u)-V[1]} = \displaystyle\frac{Y-V[2]}{y(u)-V[2]} =\displaystyle\frac{Z-V[3]}{b-V[3]}=v
Hence:
X = v(x(u)-V[1])+V[1]
Y = v(y(u)-V[2])+V[2]
Z = v(b-V[3])+V[3]
u \in [\alpha, \beta], v\in[0,1]
Example:
vert=[0, 0.75, 3] #cone vertex
base = 0 #the cone base is included within the plane z=base (here z=0)
#functions that define the cone base parameterization
x(u) = cos(u)
y(u) = sin(u)+ sin(u/2);
m, n = 72, 30
u= range(0, 2π, length=m)
v = range(0, 1, length=n)
us = ones(n)*u'
vs = v*ones(m)'
#Surface parameterization
X = @. vs* (x(us)-vert[1]) + vert[1]
Y = @. vs* (y(us)-vert[2]) + vert[2]
Z = @. vs*(base-vert[3]) + vert[3];
surface(X, Y, Z, size=(600,600), cbar=:none, legend=false) #line copied from Rafael
2 Likes
Yes 100 for you, I was a Mathematics teacher for Junior High School students in Papua, but it was 2 years ago. Afterwards, I still love learning Mathematics even if I no longer teach, now I am learning the undergraduate Mathematics, from Calculus, It would be lovely to teach Math again in the future.
This is the first time I learn Calculus with Julia from bottom, it is rewarding knowing the application of this is enormous.
Thanks for this I will use your code and learn the parameterization, vertex, etc.
1 Like | 2023-01-31T10:27:36 | {
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https://mathhelpboards.com/threads/root-equation.4882/ | # Root equation
#### Albert
##### Well-known member
$m,n \in N$
satisfying :
$\sqrt {m+2007}+\sqrt {m-325}=n$
find Max(n)
#### Bacterius
##### Well-known member
MHB Math Helper
Re: root equation
[JUSTIFY]Apply the variable substitution $u = m - 325$, and hence $m + 2007 = u + 2332$. We get:
$$\sqrt{u + 2332} + \sqrt{u} = n$$
For $n$ to be an integer, both square roots have to be integers*. So we are looking for two (positive) squares which differ by $2332$ units. That is, we need to solve:
$$x^2 - y^2 = 2332$$
Doing some factoring:
$$(x - y)(x + y) = 2^2 \times 11 \times 53$$
After some observation** we have the following solution pairs:
$$(x, y) \to \{ ( 64, 42), ~ (584, 582) \}$$
It follows that the only solutions for $u$ are $u = 42^2$ and $u = 582^2$. This gives $n = 106$ and $n = 1166$ (and $m = 2089$, $m = 339049$ respectively).
Therefore $\max{(n)} = 1166$. $\blacksquare$
* perhaps this part merits justification, you can check Square roots have no unexpected linear relationships | Annoying Precision which is pretty hardcore (perhaps a reduced argument suffices here) but I will take it that the question assumed this to be true. Other approaches which do not rely on this assumption would be interesting to see as well.
** this can be solved efficiently by using every reasonable factor combination for $x - y$ and deducing $x + y$ from it.[/JUSTIFY]
Last edited:
#### Opalg
##### MHB Oldtimer
Staff member
Re: root equation
As a slight variation on Bacterius's solution, take $u$ to be $m+841$ rather than $m-325$. The equation then becomes $$\sqrt{u+1166} + \sqrt{u-1166} = n.$$ Square both sides to get $$2u + 2\sqrt{u^2-1166^2} = n^2.$$ For an integer solution, $u^2-1166^2$ must be a square, say $u^2-1166^2 = w^2.$ Thus $(w,1166,u)$ is a pythagorean triple, which must be of the form $(p^2-q^2,\,2pq,\,p^2+q^2)$ for some integers $p,q$ with $p>q$. In particular, $1166=2pq$. But $1166 = 2*11*53$, so the only possibilities are $p=53,\ q=11$, or $p=583,\ q=1.$ That then leads to the same solutions as those of Bacterius.
Last edited:
##### Well-known member
Re: root equation
we have
(m+ 2007) – (m- 325) = 2332 ..1
√(m+2007)+√(m−325) = n ..(2) given
Dividing (2) by (1) we get
√(m+2007)-√(m−325) = 2332/n … (3)
And (2) and (3) to get 2 √(m+2007) = n + 2332/n
Now RHS has to be even so n is even factor of 2332/2 or 1166 hence largest n = 1166.
#### Albert
##### Well-known member
let :
$u^2=m-325-------(1)$
$(u+a)^2=m+2007-------(2)$
$(2)-(1):2332=a^2+2ua-----(3)$ (so a must be even)
max(n) must happen while a=2,so from (3) we have u=582
so max(n)=582+584=1166
#### Opalg
##### MHB Oldtimer
Staff member
Re: root equation
we have
(m+ 2007) – (m- 325) = 2332 ..1
√(m+2007)+√(m−325) = n ..(2) given
Dividing (2) by (1) we get
√(m+2007)-√(m−325) = 2332/n … (3)
And (2) and (3) to get 2 √(m+2007) = n + 2332/n
Now RHS has to be even so n is even factor of 2332/2 or 1166 hence largest n = 1166.
Very neat!
#### Albert
##### Well-known member
now ,find min(n)=?
##### Well-known member
now ,find min(n)=?
√(m+2007)+√(m−325) = n
put m- 325 = p (which is >=0)
we get √(p+2332)+√p which is >= sqrt(2332) >= 49
now we should look for lowest even factor of 2332(even : reason same as my previous solution) which is >= 49
now 2332 = 2 * 2 * 11 * 53 and lowest even factor of 2332 >= 49 is 2 * 53 = 106
#### Albert
##### Well-known member
let :
$u^2=m-325-------(1)$
$(u+a)^2=m+2007-------(2)$
$(2)-(1):2332=a^2+2ua-----(3)$ (so a must be even)
max(n) must happen while a=2,so from (3) we have u=582
so max(n)=582+584=1166
$2332=a^2+2ua=a(2u+a)=a\times n=2\times 1166=22\times 2\times53=22\times 106$
so max(n)=1166 ,and a=2
min(n)=106, and a=22
(here a,and n=2u+a are all even numbers) | 2020-09-22T17:57:55 | {
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http://mathhelpforum.com/algebra/4571-help-again.html | # Math Help - Help again
1. ## Help again
Please explain how to solve this.
The product of two consecutive even integers is 168. Find the two integers.
Hint: Assume the first integer is x
Set up the equation first that satisfies the given condition. Then solve the equation.
This is the equation that i came up with. Ican t seem to get past
x(x+2)=168
x^2+2x-168=0 (i cant get past here)
Thanks,
Chester
2. Originally Posted by Chester
x(x+2)=168
x^2+2x-168=0 (i cant get past here)
Good job.
Find some factors of 168: 12 and 14
which diffrence is 2 thus,
$(x+14)(x-12)=168$
Thus,
$x=-14,x=12$
3. You could always fall back on the quadratic formula. I'd rather factor.
What 2 numbers when added equal 2 and when multiplied equal -168.
Let's see.......how about.....................-12 and 14. (-12)(14)=-168
-12+14=2
$x^{2}-12x+14x-168$
$(x^{2}-12x)+(14x-168)$
$x(x-12)+14(x-12)$
(x-12)(x+14)
4. Hello, Chester!
You did an excellent on job on the hard part: setting up the equation.
I assume you know to solve a quadratic equation
. . and that you know how to factor.
Did you run of factorings to try?
$x^2+2x-168\:=\:0$
We want two numbers with a product of 168 and a difference of 2.
Start dividing 168 by 1, 2, 3, . . .
$168 \div 1 = 168\quad\Rightarrow\quad 1\cdot168$
$168 \div 2 = 84\quad\Rightarrow\quad 2\cdot84$
$168 \div 3\;\;\text{ not exact}$
$168 \div 4 = 42\quad\Rightarrow\quad 4\cdot42$
$168 \div 5,\; 168 \div 6,\;\168 \div 7\;\;\text{ not exact}$
$168 \div 8 = 21\quad\Rightarrow\quad 8\cdot21$
$168 \div 9,\;168 \div 9,\;168\div 10,\;168\div11\;\;\text{ not exact}$
$168 \div 12 = 14\quad\Rightarrow\quad 12\cdot14\;\;\leftarrow$ There! a difference of 2
Therefore: . $x^2 + 2x - 168 \;= \;(x - 12)(x + 14)$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Of course, you can use the Quadratic Formula for factoring . . .
5. ## Thanks
Thanks again everyone.
I really appreciate the help!
Chester
6. I believe the answer is $12\cdot14$ not $(-12)\cdot14$
notice that negative 12 and 14 are not consecutive even integers.
7. Originally Posted by Quick
I believe the answer is $12\cdot14$ not $(-12)\cdot14$
notice that negative 12 and 14 are not consecutive even integers.
There two answers for $x$.
$x=12, x+2=14$ and $12\times 14=168$
And,
$x=-14,x+2=-12$ and $-14\times -12=168$ | 2016-04-29T21:59:05 | {
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https://forum.math.toronto.edu/index.php?PHPSESSID=q0844f24o3ornn2o4mongo9t30&action=printpage;topic=1600.0 | # Toronto Math Forum
## MAT244--2018F => MAT244--Tests => Quiz-7 => Topic started by: Victor Ivrii on November 30, 2018, 04:05:34 PM
Title: Q7 TUT 0201
Post by: Victor Ivrii on November 30, 2018, 04:05:34 PM
(a) Determine all critical points of the given system of equations.
(b) Find the corresponding linear system near each critical point.
(c) Find the eigenvalues of each linear system. What conclusions can you then draw about the nonlinear system?
(d) Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system.
\left\{\begin{aligned} &\frac{dx}{dt} = 1 - xy, \\ &\frac{dy}{dt} = x - y^3. \end{aligned}\right.
Bonus: Computer generated picture
Title: Re: Q7 TUT 0201
Post by: Yulin WANG on November 30, 2018, 04:39:52 PM
(a)
\left\{
\begin{array}{**lr**}
1-xy=0 & \\
x-y^{3}=0\\
\end{array}
\right.
\left\{
\begin{array}{**lr**}
xy=1 & \\
x=y^{3}\\
\end{array}
\right.
\left\{
\begin{array}{**lr**}
x=1 & \\
y=1\\
\end{array}
\right.
or
\left\{
\begin{array}{**lr**}
x=-1 & \\
y=-1\\
\end{array}
\right.
Therefore, the critical points are (1,1) and (-1,-1)
(b)
The Jacobian matrix of the vector field is:
\begin{align*}
J &= \begin{bmatrix}
-y & -x \\
1 & -3y^{2}
\end{bmatrix}\\
~\\
J(1,1) &= \begin{bmatrix}
-1 & -1 \\
1 & -3
\end{bmatrix}\\
~\\
J(-1,-1) &= \begin{bmatrix}
1 & 1 \\
1 & -3
\end{bmatrix}
\end{align*}
(c)
\begin{align*}
For (1,1), let A&= \begin{bmatrix}
-1 & -1 \\
1 & -3
\end{bmatrix}\\
~\\
A-\lambda I &= \begin{bmatrix}
-1-\lambda & -1 \\
1 & -3-\lambda
\end{bmatrix}\\
~\\
det(A-\lambda I) &=(\lambda+3)(\lambda+1)+1=0\\
~\\
\lambda_{1} &= \lambda_{2} = -2 \\
~\\
Then \ the \ system \ has \ a \ stable \ improper \ node \ at \ (1,1) \\
~\\
For (-1,-1), let A&= \begin{bmatrix}
1 & 1 \\
1 & -3
\end{bmatrix}\\
~\\
A-\lambda I &= \begin{bmatrix}
1-\lambda & 1 \\
1 & -3-\lambda
\end{bmatrix}\\
~\\
det(A-\lambda I) &=(\lambda+3)(\lambda-1)-1=0\\
~\\
\lambda = -1 \pm \sqrt{5} \\
~\\
Then \ the \ system \ has \ a \ unstable \ saddle \ point \ at \ (1,1) \\
\end{align*}
(d) In the attachment.
Title: Re: Q7 TUT 0201
Post by: Zhuojing Yu on November 30, 2018, 06:29:30 PM
I think when (1,1), it is node or spiral point, not IN(improper node).
Title: Re: Q7 TUT 0201
Post by: Jingze Wang on November 30, 2018, 08:31:23 PM
I think the phase portrait that Yulin drew is correct, here is computer generated picture since no one posted
Title: Re: Q7 TUT 0201
Post by: Yulin WANG on November 30, 2018, 11:13:57 PM
I think the phase portrait that Yulin drew is correct, here is computer generated picture since no one posted
Thanks for submitting the computer-generated phase portrait!!!
BTW, how do u plot the phase portrait on a computer?
Title: Re: Q7 TUT 0201
Post by: Victor Ivrii on December 01, 2018, 03:57:59 AM
Sure, it is improper node. Jingze finally got a decent computer generated picture (took a correct range of variables)
Syllabus lists several free sites. My favourite -- downloadable pplain java applet | 2022-12-02T23:33:35 | {
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http://math.stackexchange.com/questions/353752/what-does-it-mean-to-be-proportional-to-something/353758 | # What does it mean to be proportional to something?
I am asked a physics homework question, but it is really simply a mathematical question I think, dealing with proportional reasoning.
The period of a pendulum is proportional to the square root of its length.
A pendulum of length $2$ has a period of $3$.
They give units but it really should not matter here.
What I have done is written this down, and I am not really sure the first step is the correct one, because I am unsure how to read the question:
$$3P_1 = \sqrt 2$$
$$P_1 = \frac{\sqrt 2}{3}$$
They want to know about a pendulum of length $4.5$ so,
$$xP_2 = \sqrt {4.5}$$
My reasoning then is that:
$$\frac{\sqrt 2}{3} = \frac{\sqrt{4.5}}{x}$$
$$xP_2 = \frac{3\sqrt{4.5}}{\sqrt 2} = 4.5$$
I never really deal with proportional reasoning, so I am going out on a limb here to make this connection, but it all seems very intuitive. Of course, my intuition is not always right. My question then is, am I doing this right? Is this good reasoning?
-
Hint:
Take T to be the time period of the pendulum.
$T_1 = k \cdot \sqrt {l_1}$
$T_2=k \cdot \sqrt{l_2}$
When you divide them both(Since none of the terms are $0$)
$\dfrac{T_1}{T_2}= \sqrt{\dfrac{l_1}{l_2}}$
Note: When L is proportional to M, you can write it as L=k. M, where k is a constant.
-
Hm that is interesting, and it confirms my answer when I use the k, that really helps too. It is truly discomforting that my brain does all that without informing me how it was done.. Many thanks! – Leonardo Apr 7 '13 at 10:56
You're Welcome. And do note that there might be many cases of proportionality, when you come across Gravitational force (or maybe coulomb's force), you come across 'inversely proportional to'. – Inceptio Apr 7 '13 at 10:59
It may be noted that here the actual value of k is 2*pi/sqrt(g) where g is accelaration due to gravity. – Sreekanth Karumanaghat Apr 7 '13 at 11:24
@SreekanthKarumanaghat: That really doesn't matter when you have two relations, cause: they just scare off each other. – Inceptio Apr 7 '13 at 11:50
Yeah I know that I was just noting for info sake :) – Sreekanth Karumanaghat Apr 7 '13 at 12:08
The correct meaning of propotional:- If x is directly propotional to y. x/y = k(a constant)
Analytically
Graph between x and y is straight line whose slope is k.
Applying the above formula here. T1 = k sqrt(L1) T2 = k sqrt(L2)
dividing 2 by 1,we get T1/T2 = sqrt(L1)/sqrt(L2)
3/T2 = sqrt(2)/sqrt(4.5)
T2 = sqrt(4.5)*3/sqrt(2)
T2 = 4.5
- | 2014-12-21T12:05:31 | {
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https://mathematica.stackexchange.com/questions/101247/recursion-question | # Recursion question
The first two terms of a sequence are $a_1 = 1$ and $a_2 = \frac {1}{\sqrt3}$. For $n\ge1$,
$$a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}.$$ What is $|a_{2009}|$?
So I tried (thanks to Paul Wellin):
fun[n_] := Module[{prev = 1, this = 1/Sqrt[3], next},
Do[next = Simplify[(prev + this)/(1 - prev this)];
prev = this;
this = next, {n - 1}];
prev]
And I got:
fun[2009]
(* 0 *)
Which is the correct answer. Kind of amazed it went so quickly. I then tried:
Table[fun[i], {i, 1, 100}]
Which produces the first 100 terms of the sequence. Again, amazingly quick. Now, this turns out to be a periodic sequence. I counted with my finger, and determined 24.
Anyone have a good way to use Mathematica to determine if the sequence is periodic and find the period?
• mathematica.stackexchange.com/questions/80163/… – Marvin Dec 5 '15 at 9:31
• Entirely different method: Recognize the recursion as the sum of angles formula for tangent. Set $\tan \theta_{n} = a_n$ so $\theta_1 = \pi/4$, $\theta_2 = \pi/6$, and $\theta_{n+2} = \theta_n + \theta_{n+1}$. This is a much easier recurrence... – Eric Towers Dec 5 '15 at 19:08
• @EricTowers Could you add this as an answer? I'd love to see what you do. – David Dec 6 '15 at 3:33
One way of doing this is starting with your list of the first 100 terms.Then ask for
zzz = SequencePosition[lst, Take[lst, -2]]
(* {{3, 4}, {27, 28}, {51, 52}, {75, 76}, {99, 100}} *)
The length of this result is larger than 1, so your list must be periodic. The period is
zzz[[2, 1]] - zzz[[1, 1]]
(* 24 *)
• thank you introducing me to SequencePosition – ubpdqn Dec 5 '15 at 10:56
• @FredSimons I agree, really simple explanation to help students with exploration. Thanks. – David Dec 5 '15 at 18:19
Started as a comment to the OP. Turning this into an answer was requested. The OP asked for Mathematica help solving the recurrence. I suggested recognizing that the recurrence was the sum of angles formula for tangent, then going from there. Consequently, this answer uses Mathematica to solve this problem using only one "external" ingredient right at the beginning. (At the end, we explore a claim made at the beginning.)
Problem: Find the period of \begin{align} a_1 &= 1 \\ a_2 &= \frac{1}{\sqrt{3}} \\ a_{n+2} &= \frac{a_n + a_{n+1}}{1-a_n a_{n+1}}, n \geq 1 \end{align}
(It would be seriously convenient if FunctionPeriod[] just did this. We could "punt" and use the accepted answer here, applying it to the straightforward implementation of the above recursion and initial values. However, that's not what I'm here to write.)
"External" fact: The recursion is the sum of angles formula for tangent.
Claim at the beginning: We could replace the recursion formula with the analog for any other sum of angles formula and get exactly the same sequence of angles. I.e., there are at least two other recursions with this same solution.
First, we check this external fact with Mathematica and get the initial values.
Simplify[Tan[a+b] == (Tan[a] + Tan[b])/(1-Tan[a]Tan[b])
(* True *)
ArcTan[1]
(* Pi/4 *)
ArcTan[1/Sqrt[3]]
(* Pi/6 *)
So we may reformulate the question as \begin{align} \theta_1 &= \frac{\pi}{4} \\ \theta_2 &= \frac{\pi}{6} \\ \theta_{n+2} &= \theta_1 + \theta_2 \pmod{2\pi}, n \geq 1 \end{align}
Sums of multiples of $\pi/4$ and $\pi/6$ have denominators dividing $12$.
LCM[4,6]
(* 12 *)
There are $24$ twelfths of $\pi$ in a circle.
2 Pi / (Pi/12)
(* 24 *)
Consequently, the period of this recurrence is at most $24^2$. (The recurrence is order $2$, so it is perhaps better to think of each adjacent pair of sequence members producing the next adjacent pair of sequence members.) (Although, as is frequently the case with homogeneous recurrences, if $\dots 0, 0 \dots$ occurs in our sequence, every subsequent angle is zero. So it is possible for the period to be less than $24$ and for there to be an initial aperiodic part, say if $\theta_1 = \theta_2 = \pi$, although it will turn out that this is not the case in this problem.)
So we generate the first $2 \times 24^2$ entries, which will contain two or more periods.
Remove[q];
q[1] = Pi/4;
q[2] = Pi/6;
q[n_ /; n >= 3] := q[n]= Mod[q[n-1]+q[n-2],2 Pi];
(* We memoize members of the sequence to avoid redundant recomputation. *)
Generate the $2 \times 24^2$ entries and glance at some of them to see if periodicity seems likely.
Table[ {k, q[k]} , {k, 24, 2*24^2, 24} ]
(* {{ 24, 23 Pi/12 }, { 48, 23 Pi/12 }, ... , {1152, 23 Pi/12 }} *)
Seems rather likely to be periodic. Test this with Fourier[]:
periods = Flatten[
Position[
Chop[Fourier[Table[q[k], {k, 1, 2*24^2}]]],
n_ /; n != 0
]
]
(* {1, 49, 97, 145, 193, 241, 289, 337, ... , 1009, 1057, 1105} *)
GCD @@ Select[DeleteDuplicates[Flatten[
Outer[Abs[#1 - #2] &, periods, periods]
]], # > 0 &]
(* 48 *)
So the period divides $48$. (Actually, we could know that it divides $24$. We ran to $2 \times 24^2$. Try it again with $3 \times 24^2$ and get $72$. In fact, if we run to $N \times 24^2$ for $N$ up to $24$, we'll get $24N$ for the GCD above.)
Let's have Mathematica try every divisor of $48$ as a candidate period.
{#, Simplify[q[3 + #] == q[3]]} & /@ Divisors[48]
(* {{1, False}, {2, False}, {3, False}, {4, False},
{6, False}, {8, False}, {12, False}, {16, False},
{24, True}, {48, True}} *)
so the only candidate periods are $24$ and $48$. (Note that starting at q[3], we're finding out about the recursion moreso than the initial values.) Then we test the shortest candidate:
q /@ Range[24] == q /@ Range[25, 2*24]
The above method works, and is valid, but perhaps one would be happier seeing that the recurrence actually recurs after $24$ steps. Starting with the symbolic first two angles $a,b$, what is the resulting angle after (a divisor of $24$) applications of the recursion rule. We frame this as working on pairs in Nest[].
TableForm[
{#[[1]], #[[2]] === {a, b}, #[[2]]} & /@ (
{#, PolynomialMod[
Nest[
(* Thinking in pairs, the new first angle is the old second
angle and the new second angle is the sum of the two old
angles. *)
{#[[2]], #[[1]] + #[[2]]} &,
{a, b},
#],
#]} & /@
Divisors[24]),
TableDepth -> 1
]
(* { { 1, False, {0, 0}},
{ 2, False, {a + b, a}},
{ 3, False, {a + 2 b, 2 a}},
{ 4, False, {2 a + 3 b, 3 a + b}},
{ 6, False, {5 a + 2 b, 2 a + b}},
{ 8, False, {5 a + 5 b, 5 a + 2 b}},
{12, False, {5 a, 5 b}},
{24, True, {a, b}} } *)
So the recursion has a natural period of $24$. It's possible for the choice of initial conditions to cause the period to be shorter. (Recall that if the first two angles are $0,0$, then the sequence is all zeroes.) So we check for this. The resulting period still has to divide $24$, so we only check those (although if we were not certain of this, we could check every period up to $24$).
TableForm[
{ #[[1]] , #[[2]] == #[[3]] , #[[2]] , #[[3]] } & /@ (
{ # , {q[1], q[2]}, {q[1 + #], q[2 + #]} } & /@
Divisors[24]
),
TableDepth -> 1]
(* { { 1, False, {Pi/4, Pi/6}, { Pi/ 6, 5 Pi/12}},
{ 2, False, {Pi/4, Pi/6}, {5 Pi/12, 7 Pi/12}},
{ 3, False, {Pi/4, Pi/6}, {7 Pi/12, Pi }},
{ 4, False, {Pi/4, Pi/6}, { Pi , 19 Pi/12}},
{ 6, False, {Pi/4, Pi/6}, {7 Pi/12, Pi/ 6}},
{ 8, False, {Pi/4, Pi/6}, {3 Pi/ 4, 11 Pi/12}},
{12, False, {Pi/4, Pi/6}, { Pi/ 4, 5 Pi/ 6}},
{24, True, {Pi/4, Pi/6}, { Pi/ 4, Pi/ 6}} } *)
And so the period is $24$. Then it's easy to compute the desired number:
Tan[q[Mod[2009,24,1]]] (*Don't want to inadvertently try to evaluate q[0]. *)
(* This is Tan[ q[17] ] = Tan[ Pi ]. *)
(* 0 *)
The original problem statement asked about $|a_{2009}|$ but since $0 = |0|$, there's no need to apply the absolute value function.
Returning to the claim at the beginning. Since adding angles is pretty straightforward, we can use any other sum of angles formula. This means there are (at least) two other recursion problems looking wildly different from this one (by having a wildly different looking general recursion formula), but having the same underlying computation. The sum of angles formulas for sine and cosine are, with the corresponding recursion: \begin{align} \sin(\theta_1 + \theta_2) &= \sin(\theta_1) \cos(\theta_2) + \cos(\theta_1) \sin(\theta_2) \\ a_{n+2} &= a_n c(a_{n+1}) \sqrt{1-a_{n+1}^2} + a_{n+1} c(a_n) \sqrt{1-a_n^2} \\ \\ \cos(\theta_1 + \theta_2) &= \cos(\theta_1)\cos(\theta_2) - \sin(\theta_1)\sin(\theta_2) \\ a_{n+2} &= a_n a_{n+1} - s(a_n)s(a_{n+1}) \sqrt{1-a_n^2}\sqrt{1-a_{n+1}^2} \end{align} where the sign-tracking functions $s$ and $c$ are $s(\theta)$ is $1$ in quadrants I and II and the angle $\pi/2$ and is $-1$ otherwise, and $c$ is $1$ in quadrants I and IV and the angle $0$ and is $-1$ otherwise. We could eliminate these two sign-tracking functions by having two sequences, $q_n$ and $r_n$: $a_n \mapsto (q_n, r_n) = (\sin(\theta_n), \cos(\theta_n))$ or by switching to exponentials $a_n \mapsto \mathrm{e}^{\mathrm{i} \theta_n} = \cos(\theta_n) + \mathrm{i}\sin(\theta_n)$, which the final equality shows is really the same as switching to sines and cosines, but using real and imaginary parts of complex numbers to hold the two sequences, instead of pairs of real numbers.
Taking $\theta_1 = \pi/4$ and $\theta_2 = \pi/6$ and using either the sum of angles for sine or for cosine, we get the same sequence of angles. Consequently, we will get a final, final answer of $a_{2009} = \tan^{-1}(\sin(\theta_{2009})) = 0$ for the sine recursion and similarly for the cosine recursion. (Whether you take the simple recurrence $\theta_{n+2} = \theta_n + \theta_{n+1}$ and obscure it via sines, cosines, tangents, or, I suppose, other periodic functions, like Weierstrass's $\wp$-functions, you still get the same sequence of fractions of the functions's period(s), and the same period of the recurrence.)
• +1, but the recurrence you display for the sine and cosine is in fact numerically unstable, and thus cannot be recommended in general. – J. M. will be back soon Dec 10 '15 at 5:18
• @J.M. : I agree in part. If we're going to talk about stability, the initial tangent recursion isn't stable either. Replace $a_{n+1}$ by $a_{n+1}+\varepsilon$ and expand in powers of $\varepsilon$ and find that the error is magnified by $\frac{1+a_n^2}{(1-a_n a_{n+1})^2}$. This is harder to interpret than doing the same thing to the tangent version, yielding $1+\tan^2(a_n + a_{n+1})$, which is necessarily $\geq 1$. So the instability is baked into the problem and the alternate recursions are merely reflecting it. – Eric Towers Dec 10 '15 at 17:11
You can use NestWhileList to show the same pair of consecutive elements appear and then determine the period, e.g.
res = NestWhileList[{#[[2]],
Simplify[(#[[1]] + #[[2]])/(1 - #[[1]] #[[2]])]} &, {1,
1/Sqrt[3]}, Unequal, All];
plt = (Most@(First /@ res))~Join~(Style[#, Red, Bold] & /@ res[[-1]])
Length[plt] - 2
ListPlot[plt~Join~res[[1]], Joined -> True]
I add this as a separate answer to amplify another answer,in relation to $\tan(x+y)$.
This sequence could be pursued as Fibonacci or linear recurrence of angles,e.g.
mp[n_] :=
Mod[MatrixPower[{{1, 1}, {1, 0}}, n].{Pi/6, Pi/4}, Pi, -Pi/2];
an[1] := mp[0][[2]];
an[2] := mp[0][[1]];
an[n_?(# > 2 &)] := First@mp[n - 2]
os[n_] := Tan[an[n]];
SequencePosition[an /@ Range[1, 26], {Pi/4, Pi/6}]
os /@ Range[26]
I exploited Fred Simons SequencePosition (yielding: {{1, 2}, {25, 26}}) and the last line reproduces the original sequence:
$\left\{1,\frac{1}{\sqrt{3}},\sqrt{3}+2,-\sqrt{3}-2,0,-\sqrt{3}-2,-\sqrt{3}-2,\frac{1}{\sqrt{3}},-1,\sqrt{3}-2,-\sqrt{3},-\sqrt{3}-2,1,-\frac{1}{\sqrt{3}},2-\sqrt{3},\sqrt{3}-2,0,\sqrt{3}-2,\sqrt{3}-2,-\frac{1}{\sqrt{3}},-1,-\sqrt{3}-2,\sqrt{3},\sqrt{3}-2,1,\frac{1}{\sqrt{3}}\right\}$
A quick visual inspection helps to see how a pattern with period 24 pops up
n = 100;
tab = Table[fun[i], {i, 1, n}];
pos = Flatten @ Position[tab, First @ tab]
{1, 13, 25, 37, 49, 61, 73, 85, 97}
tra = Transpose[{pos, Array[First @ tab &, Length @ pos]}];
ListPlot[{tab, tra}, GridLines -> {pos[[;; ;; 2]], None}]
I rewrote the code to produce a list of the current "state" of the recursion, which is given by the last two terms $(a_n,a_{n-1}$. The system is autonomous, so the state is completely determined by this pair. Therefore if a pair ever occurs again, there is a loop.
Some utility functions:
step = (* basic recursive step *)
{Last[#], Simplify[Plus @@ #/(1 - Times @@ #)]} &;
steps[n_] := (* equivalent to fun[n], except it returns a list of all states *)
NestList[step, {1, 1/Sqrt[3]}, n];
findloop[] := (* like steps[n], except it runs until a loop is found; caveat forever *)
NestWhileList[step, {1, 1/Sqrt[3]}, UnsameQ, All];
First proof:
gr = Rule @@@ Partition[findloop[], 2, 1] // Graph;
cycle = First@FindHamiltonianCycle[gr];
Length@cycle
HighlightGraph[gr, PathGraph[cycle]]
Less fussy proof, assuming you can make a good guess (i.e. 24):
Rule @@@ Partition[steps[24], 2, 1] // Graph
(* graph similar to above, but without highlighting *)
Even less fussy, assuming you can make a sufficient guess (i.e. 100):
gr = Rule @@@ Partition[steps[100], 2, 1] // Graph
Length@VertexList[gr]
(*
graph similar to above, but with multiple edges
24 <-- number of vertices == period
*)
Here is a translation of Brent's algorithm for finding the length of a cycle. See Brent, doi:10.1007/BF01933190 (1980). This could be compiled when step can be compiled for cases where a loop is long or occurs a long way down the sequence.
brentcyclelength[step_, x0_, ρ_: 2, u_: 0] :=
Module[{x, y, r, j, k, done = False},
y = x0;
r = ρ^u;
k = 0;
While[! done,
x = y;
j = k;
r = ρ*r;
While[! done && k < r,
k++;
y = step[y];
done = x == y
];
];
k - j
]
brentcyclelength[step, {1, 1/Sqrt[3]}]
(* 24 *) | 2019-12-16T01:53:13 | {
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https://math.stackexchange.com/questions/852122/picture-intuitive-proof-of-cos3-theta-4-cos3-theta-3-cos-theta/852230 | # Picture/intuitive proof of $\cos(3 \theta) = 4 \cos^3(\theta)-3\cos(\theta)$?
Is there a nice geometric, intuitive or picture proof as to why the easily algebraically provable identity $\cos(3 \theta) = 4 \cos^3(\theta)-3\cos(\theta)$ is true?
Note I'm not looking for a computational proof like the one linked to, more a proof without words or intuitive style proof, thanks.
Enhancing my diagram for the angle-sum formula (currently featured in Wikipedia) to use three angles will get you pretty close ...
Thus,
\begin{align} \cos(\alpha+\beta+\gamma) &= \cos\alpha \cos\beta \cos\gamma - \sin\alpha \sin\beta \cos\gamma - \sin\alpha \cos\beta\sin\gamma - \cos\alpha \sin\beta\sin\gamma \\ \sin(\alpha+\beta+\gamma) &= \sin\alpha \cos\beta \cos\gamma + \cos\alpha \sin\beta \cos\gamma + \cos\alpha \cos\beta \sin\gamma - \sin\alpha \sin\beta \sin\gamma \end{align}
With $\alpha = \beta = \gamma = \theta$, these become ... \begin{align} \cos 3\theta &= \cos^3\theta - 3 \sin^2\theta \cos\theta \\ \sin 3\theta &= 3\cos^2\theta \sin\theta - \sin^3\theta \end{align} ... which the Pythagorean identity helps us rewrite as ... \begin{align} \cos 3\theta &= \cos^3\theta - 3 (1-\cos^2\theta) \cos\theta = 4\cos^3\theta - 3 \cos\theta \\ \sin 3\theta &= 3(1-\sin^2\theta) \sin\theta - \sin^3\theta = 3\sin\theta - 4\sin^3\theta \end{align}
Off-hand, I don't know of a diagram that goes directly from $\cos 3\theta$ to $4\cos^3\theta-3\cos\theta$.
• That is absolutely amazing. – bobby Jun 30 '14 at 13:29
• Very colorful! =) – Arnie Bebita-Dris Sep 4 '14 at 12:36
• I just love it when someone has the time and effort to make a truly helpful and fascinating visual explanation. Sometimes when the numbers and letters just swim around, a picture can be very helpful – Asimov Sep 4 '14 at 14:17
• How do you get something featured on Wikipedia? Do you submit it somewhere and then wait for someone to place it on an article? – étale-cohomology Aug 9 '17 at 22:31
• @étale-cohomology: The "wiki" in "Wikipedia" indicates that the content is user-editable. Anybody can go in and make improvements. As for my diagrams: I'd posted them here, and then someone posted "similar" versions to Wikipedia (without credit or notice); someone from here at Math.SE edited the post so that I'd get credit. (Thanks again @ChrisSherlock!) I, myself, recently went in and edited the article's description of those diagrams and provided a link to my new website for such things in the article's list of references. – Blue Aug 9 '17 at 23:00
Assemble four congruent right triangles with hypotenuse $1$ and angle $\theta$ as shown:
In the diagram, the marked angles are all $\theta$. Since $OQ = \cos\theta$, we have $OU = \cos^2\theta$, so $OS = 2\cos^2\theta$, and so $$OS' = 2\cos^3\theta$$ On the other hand, $$OS' = OP' + P'Q' + Q'S' = \cos(3\theta) + P'Q' + \cos(\theta)$$ and $$OS' = OR' - Q'R' + Q'S' = \cos(\theta) - Q'R' + \cos(\theta)$$ Since $P'Q' = Q'R'$, adding these together yields $$\cos(3\theta) + 3\cos(\theta) = 2OS' = 4\cos^3\theta$$
I hope the diagram is reasonably self-explanatory, that it's clear what needs to be proved to justify the steps of the argument, and that it's easy to prove those things once you've identified them. But if I've misjudged this, let me know and I'll fill in some more details.
I don't think it is a very good idea to try an find a geometric picture for every trig identity because that would take exceedingly long. I will show you that your formula comes from the idea that a rotation by $n\theta$ is equal to $n$ rotations by $\theta$.
Denote $\langle a,b\rangle$ as the vector to the ordered pair $(a,b)$. We will discuss rotations of these vectors. First, consider that the rotation of the vector $\langle 1,0\rangle$ by an angle of $\theta$ is given by $\langle \cos\theta,\sin\theta\rangle$. Similarly, rotation of the vector $\langle 1,0\rangle$ by $3\theta$ is given by $\langle \cos(3\theta),\sin(3\theta)\rangle$.
I will now define a new type of multiplication between vectors. Let $$\langle a,b\rangle\ltimes\langle c,d \rangle = \langle ac-bd,ac+bc\rangle$$
It so happens that if the vector $\langle a,b\rangle$ makes an angle $\phi_1$ with the horizontal and $\langle c,d \rangle$ makes an angle $\phi_2$ then $\langle a,b\rangle\ltimes\langle c,d \rangle$ makes an angle $\phi_1 + \phi_2\!$. Using this, there is another way to find the rotation of a vector by $3\theta$. We can rotate it three times by $\theta$ rather than rotating it once by $3\theta$. In other words, we will find $\left(\langle \cos\theta,\sin\theta \rangle \ltimes \cos\theta,\sin\theta \rangle\right) \ltimes \cos\theta,\sin\theta \rangle.$ Using this, one could suspect that \begin{align}\langle \cos(3\theta),\sin(3\theta)\rangle &= \left(\langle \cos\theta,\sin\theta \rangle \ltimes \cos\theta,\sin\theta \rangle\right) \ltimes \cos\theta,\sin\theta \rangle \\ &= \langle \cos^2\!\theta-\sin^2\!\theta,2\sin\theta\cos\theta\rangle \ltimes \langle \cos\theta, \sin\theta \rangle \\ &= \langle \cos^{3}\!\theta-\sin^2\!\theta\cos\theta-2\!\sin^2\!\theta\!\cos\theta, 2\!\sin\theta\!\cos^2\!\theta+\sin\theta\cos^2\!\theta-\sin^3\!\theta\rangle \\ &=\langle 4\cos^3\!\theta-3\cos\theta,3\sin\theta-4\sin^3\!\theta\rangle\end{align}
Where the last step follows from the Pythagorean identities. If we then equate the first elements and the second elements we find the familiar results
\begin{align} \cos(3\theta) &= 4\cos^3\!\theta-3\cos\theta \\ \sin(3\theta) &= 3\sin\theta-4\sin^3\!\theta\end{align}
This is essentially a result of complex numbers and is called De moivre's formula. You can see it being used to prove your identity here. Our type of multiplication mirrors complex multiplication which is inherently rotational in nature. You can see why here. De moivre's formula says that a rotation by $n\theta$ is equal to $n$ rotations by $\theta$.
The best geometric proof of your formula comes from this basic idea which is so simple that no picture is needed to represent it.
This is advantageous because it can be used to show other identites such as $$\cos(5\theta) = 16 \cos^5\! \theta - 20 \cos^3 \!\theta + 5 \cos \theta$$
• This is neither geometric nor intuitive and doesn't address the question in a natural way at all. – beep-boop Jul 2 '14 at 22:14
• @alexqwx: most of the proofs, including those using De Moivre's, are geometrically based. However, it is hard to find any that I would call intuitive. – robjohn Jul 7 '14 at 0:18
This is not a direct proof but a link to the Fourier series: $4\cos^3\theta=3\cos\theta+\cos3\theta$.
Similarly, for the fifth power: $16\cos^5\theta=10\cos\theta+5\cos3\theta+\cos5\theta$.
The main lesson is that to represent the $k^{th}$ power, the odd harmonics from $1$ to $k$ suffice.
The explanation is not so difficult: the Fourier coefficients are computed from the integrals $$\int_0^{2\pi}\cos^k\theta\ e^{in\theta}d\theta=\int_0^{2\pi}\left(\frac{e^{i\theta}-e^{-i\theta}}2\right)^ke^{in\theta}d\theta.$$ When developing, the lowest exponent of $e^{i\theta}$ is $n-k$. If $n>k$, the integrand is oscillatory and the integral vanishes.
If you admit that, then $\cos^3\theta=a\cos\theta+b\cos3\theta$.
From the figure, at $\theta=0$ the values add up to $1$ and at $\theta=\pi/2$ the slopes cancel out, so that $$a+b=1\\a-3b=0,$$ $$\cos^3\theta=\frac34\cos\theta+\frac14\cos3\theta.$$
Similarly, $\cos^5\theta=a\cos\theta+b\cos3\theta+c\cos5\theta$, at $\theta=0$ the values add up to $1$ and at $\theta=\pi/2$ the first and third derivatives cancel out, so that $$a+b+c=1\\ a-3b+5c=0\\ a-27b+125c=0,$$ $$\cos^5\theta=\frac{10}{16}\cos\theta+\frac{5}{16}\cos3\theta+\frac{1}{16}\cos5\theta.$$
yes.you can write: $cos(3\theta) = cos(2\theta +\theta)$ =$$cos(2\theta)cos(\theta) - sin(2\theta)sin\theta$$ =$$2cos^3(\theta)-cos \theta -sin(2\theta) sin(\theta)$$=$$2cos^3\theta -cos\theta -2sin^2\theta cos\theta$$=$$2cos^3\theta - cos \theta - 2(cos\theta -cos^3 \theta)$$ so we have : $$4cos^3\theta -3cos\theta$$
• I've already linked to an easy algebraic proof, I'm hoping for a more intuitive, geometric and less computational proof. – bobby Jun 30 '14 at 12:10
• sorry I do not see it .for picture you can use Matlab software. – user157745 Jun 30 '14 at 12:16 | 2021-04-11T19:23:28 | {
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https://www.intmath.com/forum/methods-integration-31/find-integral-sqrt-x-2-1-using-trigonometric:155 | IntMath Home » Forum home » Methods of Integration » Find integral sqrt (x^2 + 1) using trigonometric substitution
# Find integral sqrt (x^2 + 1) using trigonometric substitution [Solved!]
### My question
Find int sqrt (x^2 + 1) dx with limits of integration from 0 to 1 using Trigonometric Substitution.
### Relevant page
8. Integration by Trigonometric Substitution
### What I've done so far
I used x = tan theta and dx = sec^2 theta d theta.
Replacing x and dx gives sqrt (tan^2 theta + 1) sec^2 theta = sqrt (sec^2 theta) sec ^2 theta = sec ^3 theta d theta.
I looked up the integral of sec ^3 theta and converted tan, sec, sin, and cos in the formula by drawing a triangle based on x = tan theta and arrived at .57155 and not 1.15.
X
Find int sqrt (x^2 + 1) dx with limits of integration from 0 to 1 using Trigonometric Substitution.
Relevant page
<a href="https://www.intmath.com/methods-integration/8-integration-trigonometric-substitution.php">8. Integration by Trigonometric Substitution</a>
What I've done so far
I used x = tan theta and dx = sec^2 theta d theta.
Replacing x and dx gives sqrt (tan^2 theta + 1) sec^2 theta = sqrt (sec^2 theta) sec ^2 theta = sec ^3 theta d theta.
I looked up the integral of sec ^3 theta and converted tan, sec, sin, and cos in the formula by drawing a triangle based on x = tan theta and arrived at .57155 and not 1.15.
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
@phinah: Good on you for using the math entry system! (I tidied up some of the math expressions.)
Just a small (but important) point - don't miss out the "d theta" parts in your second line. It should be:
Replacing x and dx gives sqrt (tan^2 theta + 1) (sec^2 theta) d theta = sqrt (sec^2 theta) (sec ^2 theta) d theta = sec ^3 theta d theta.
Now, to give you a hint about why your final number is not correct (I'm guessing 1.15 comes from the Answers in your text book, right?).
When we change x to tan theta and x goes from 0 to 1, what will theta's lower and upper values be?
X
@phinah: Good on you for using the math entry system! (I tidied up some of the math expressions.)
Just a small (but important) point - don't miss out the "d theta" parts in your second line. It should be:
Replacing x and dx gives sqrt (tan^2 theta + 1) (sec^2 theta) d theta = sqrt (sec^2 theta) (sec ^2 theta) d theta = sec ^3 theta d theta.
Now, to give you a hint about why your final number is not correct (I'm guessing 1.15 comes from the Answers in your text book, right?).
When we change x to tan theta and x goes from 0 to 1, what will theta's lower and upper values be?
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
theta = arctan (0) = 0
theta = arctan (1) = .7853
X
theta = arctan (0) = 0
theta = arctan (1) = .7853
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
Also this is not a textbook exercise. This is from your website, the question at the end of the Integration chapter, section 4: The Definite Integral. You state it can be solved via Trigonometric Substitution but you solved it using the Trapezoidal Rule in section 5. Thanks.
X
Also this is not a textbook exercise. This is from your website, the question at the end of the Integration chapter, section 4: The Definite Integral. You state it can be solved via Trigonometric Substitution but you solved it using the <a href="https://www.intmath.com/integration/5-trapezoidal-rule.php">Trapezoidal Rule in section 5</a>. Thanks.
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
Ahh, I see. Thanks for giving me the context.
Your lower and upper values are correct. Now substitute those in the integral of sec^3 theta you found from the table.
X
Ahh, I see. Thanks for giving me the context.
Your lower and upper values are correct. Now substitute those in the integral of sec^3 theta you found from the table.
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
Leaving it in terms of theta:
int sec^3 theta d theta from 0 to .7853
According to Wolfram, the integral formula is
1/2[tan theta sec theta {: - ln (cos {:theta/2:} - sin {:theta /2:}) {: + ln (sin {:theta/2:} + cos {:theta/2:})]
Therefore, 1/2 [tan .7853\ sec .7853 {:- ln (cos .7853/2 - sin .7853/2) {: + ln (sin .7853/2 + cos .7853/2)] - 1/2 [0 - ln (1-0) + ln (0+1) ]
= 1/2[tan .7853\ sec .7853 {: - ln (cos .7853/2 - sin .7853/2) {: + ln (sin .7853/2 + cos .7853/2)] - [0]
=1.15
which is the web site's answer using the Trapezoidal Rule!
Thank you for the guidance.
Note: concerning how theta is written above, I wrote it out in Word.
X
Leaving it in terms of theta:
int sec^3 theta d theta from 0 to .7853
According to Wolfram, the integral formula is
1/2[tan theta sec theta {: - ln (cos {:theta/2:} - sin {:theta /2:}) {: + ln (sin {:theta/2:} + cos {:theta/2:})]
Therefore, 1/2 [tan .7853\ sec .7853 {:- ln (cos .7853/2 - sin .7853/2) {: + ln (sin .7853/2 + cos .7853/2)] - 1/2 [0 - ln (1-0) + ln (0+1) ]
= 1/2[tan .7853\ sec .7853 {: - ln (cos .7853/2 - sin .7853/2) {: + ln (sin .7853/2 + cos .7853/2)] - [0]
=1.15
which is the web site's answer using the Trapezoidal Rule!
Thank you for the guidance.
Note: concerning how theta is written above, I wrote it out in Word.
## Re: Find integral sqrt (x^2 + 1) using trigonometric substitution
OK - looks good.
BTW, the "theta" symbols didn't show because Word uses a different set of fonts. It's always best to type the math directly in the text box, for best results. I edited your answer so the thetas showed properly and also so the answer worked OK on a phone.
X
OK - looks good.
BTW, the "theta" symbols didn't show because Word uses a different set of fonts. It's always best to type the math directly in the text box, for best results. I edited your answer so the thetas showed properly and also so the answer worked OK on a phone. | 2018-02-21T02:50:09 | {
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http://math.stackexchange.com/questions/370386/fundamental-theorem-of-calculus-on-wikipedia | # Fundamental theorem of calculus on Wikipedia
In Wikipedia is written in one step that for the integral $\int_{x_1}^{x_1 + \Delta x}f(t) dt$ that by mean value theorem there exists a $c$ in $[x_1, x_1 + \Delta x]$ such that
$\int_{x_1}^{x_1 + \Delta x}f(t) dt = f(c) \Delta x$
But mean value theorem states that there exists $c$ in $(x_1, x_1 + \Delta x)$. (the open interval)
It seems to be a problem when squeeze theorem is later applied.
Is this mistake in the proof or is the step valid? If it is valid can you please explain me why? Thank you for any help.
-
The open interval is contained in the closed interval, so the condition that $c$ is in the closed interval is weaker than the conclusion of the mean value theorem. – Matthew Pressland Apr 23 '13 at 14:33
@MattPressland But if $c \in (x_1, x_1 + \Delta x)$ and by applying squeeze theorem, $c = x_1$ then this is contradiction because $\Delta x \to 0$ and $x_1 \notin (x_1, x_1)$? – blue Apr 23 '13 at 15:03
The squeeze theorem doesn't tell you $c=x_1$, it tells you that $c\to x_1$ as $\Delta x\to 0$. A sequence of points in the open interval can converge to one of the boundary points in the closed interval. – Matthew Pressland Apr 23 '13 at 15:10
@MattPressland I did not know it can converge to boundary point. You can write in answer? If you do I accept. – blue Apr 23 '13 at 15:13
@blue .Lagrage's Mean value theorem for differentiation differ's from first mean value theorem for integration(which you used!) at that point(the interval in which $c$ is expected to be found is closed in the latter and open in the former). – Fermé somme Apr 23 '13 at 15:20
As boywholived points out, a possible statement of the mean value theorem allows for $c$ to lie in the closed interval. In fact this is simply a weaker statement - if $c$ lies in the open interval, then in particular it lies in the closed interval.
This doesn't cause any problems when applying the squeeze theorem. Perhaps a change of notation is useful; for a particular choice of $\Delta x$, the mean value theorem gives you some $x<c<\Delta x$ (assuming the open interval version of the mean value theorem for now). This $c$ depends on $\Delta x$, so we should maybe write $c(\Delta x)$, making it clear that $c$ is really a function whose value depends on $\Delta x$. Then when applying the squeeze theorem, we are interested in the limit of this function as $\Delta x\to 0$. The fact that $c(\Delta x)>x$ for all $\Delta x>0$ only implies that $\lim_{\Delta x\to0}c(\Delta x)\geq x$, not that the inequality is strict. (Compare to the sequence $a_n=1/n$; every term is strictly positive, but $a_n\to 0$ as $n\to\infty$). | 2015-07-06T07:56:25 | {
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https://math.stackexchange.com/questions/365938/finding-the-remainder-of-x100-2x511/365946 | # finding the remainder of $x^{100}-2x^{51}+1$
I have never been great with polynomials. Here's my problem.
Find the remainder of $f(x)=x^{100}-2x^{51}+1$ when $f$ is divided by $x^2-1$
This sounds easy right? Why can't I figure it out? My thought was to try and create it such that $f(x)=q(x)g(x)+r(x)$. But I can not get past getting $deg[r(x)]<deg[g(x)].$
$$f(x)=x^{100}-2x^{51}+1$$ $$=x^{100}-x^{51}-x^{51}+x^2-x^2+1$$ $$=x^{51}(x^{49}-1)-x^2(x^{49}-1)-x^2+1$$ $$=(x^{51}-x^2)(x^{49}-1)-x^2+1$$ $$=x^2(x^{49}-1)(x^{49}-1)-x^2+1$$ $$=x^2[(x^{49}-1)^2-1]+1=?.......$$ I don't see what I am missing
This is the standard approach, especially if you know the roots of the divisor.
Let $f(x) = x^{100} - 2x^{51} + 1$, and $f(x) = g(x) (x^2-1) + ax + b$ be the division
Then, $0 = f(1) = g(1) ( 1^2 - 1) + a (1) + b = a + b$,
and $4 = f(-1) =g(-1) ( (-1)^2 -1) + a(-1) + b = -a + b$.
Hence $a= -2, b = 2$.
Thus the remainder is $-2x+2$.
• I like that. I've never seen it done like that before. That is great! – Eleven-Eleven Apr 18 '13 at 22:57
• @ChristopherErnst You should review the Remainder-Factor Theorem. This is a standard application of it. For more questions, you can check this out. – Calvin Lin Apr 18 '13 at 22:58
• So basically depending on the degree of your particular divisor, if say $deg[g(x)]=3$ you set your remainder to $ax^2+bx+c$ and take one more value and solve the system for a,b,and c. – Eleven-Eleven Apr 18 '13 at 23:25
• @ChristopherErnst Correct. This way is direct and you are simply solving linear equations, though it assumes that you can find the roots of $g(x)$. Of course, there are other ways to approach these kind of polynomial remainder problem, like the other (deleted) solution. – Calvin Lin Apr 18 '13 at 23:34
• Very nice. This way is particularly nice should the divisor factor into nice linear terms with real solutions like the one in the problem. Can this be extended when there are complex factors in our divisor? – Eleven-Eleven Apr 18 '13 at 23:42
Hint $\ \,$ Interpolate the remainder $\rm\:r(x)\:$ at the roots $\rm\:\color{#c00}{x = \pm1},\:$ where $\rm\ r(\pm1)\, =\, f(\pm1)$
$$\qquad\ \ \begin{eqnarray} &&\rm\ \ r(x) &=\,&\rm f(x) - (\color{#c00}{x^2\!-\!1})\, q(x),\ \ \ deg\ r < 2\\ \\\Rightarrow\, &&\rm 2\, r(x) &=\,&\rm f(1)\, (x\!+\!1) - f(-1)\, (x\!-\!1)\end{eqnarray}$$
Remark $\$ Or, equivalently, use Chinese Remainder (CRT) to solve
$$\begin{eqnarray}\rm r(x) &\equiv&\rm \ \ \ f(1) &&\rm\ (mod\ \color{#c00}{x-1}) \\ \rm r(x)&\,\equiv\,&\rm f(-1)&&\rm\ (mod\ \color{#c00}{x+1})\end{eqnarray}$$
Generally, as here, Lagrange interpolation is a special case of CRT. | 2019-08-23T19:34:03 | {
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https://math.stackexchange.com/questions/1309968/what-does-9-with-a-line-above-the-9-mean | # What does .9 with a line above the 9 mean?
What does this mean? $$\Large.\overline9$$ I've never seen this notation before.
• Perhaps $0.9999...$? – ajotatxe Jun 2 '15 at 23:28
• It also means $1$, by that's a whole 'nother discussion. – PyRulez Jun 3 '15 at 0:32
• @AlbertMasclans Apparently this notation style depends on geography (en.wikipedia.org/wiki/Repeating_decimal): overline=US, overdot=China, parentheses=Europe. – user3449173 Jun 3 '15 at 8:02
• @Nordik Yes, it depends on geography, but parentheses are used only sometimes in Europe. In Germany we typically use the overline, too. Parentheses are used to denote an uncertanty in the last digit(s). – Christoph Jun 3 '15 at 10:58
• By the way, the overline is technically known as a vinculum. – Brian M. Scott Jun 3 '15 at 14:38
It is called a vinculum and it denotes a repeating decimal.
• +1 for a good word I need to remember to use when the time is right. – JoeTaxpayer Jun 2 '15 at 23:58
• Sounds like something in a gynecologist's office. – zhw. Jun 3 '15 at 0:36
• @JoeTaxpayer: Do you mean +1 or +0.9...? – mipadi Jun 3 '15 at 22:18
It means a repeating decimal. One can write $\frac 16=0.1\overline 6$, or $\frac 1{14}=0.0\overline{714852}$ for example. The repeating part is whatever is under the overline.
$0.\overline{9}=0.999999\ldots=1$
More generally, $0.\overline{n}=0.nnnnnnnnn\ldots$
For example, $\frac 13=0.333333333\ldots=0.\overline3$
As other answers have said, it stands for a repeating decimal, where the digits under the line are repeated. $$0.\overline{9} = 0.9999999\ldots$$
But if you want to be a little pedantic, you might prefer to say that both $0.\overline{9}$ and $0.9999999\ldots$ are two different forms of notation for the same number. That number is the limit of the sequence formed by repeatedly appending copies the digits covered by the line. In other words, given the notation $0.\overline{9}$, you can write out the following sequence: \begin{align} a_1 &= 0.9 \\ a_2 &= 0.99 \\ a_3 &= 0.999 \\ a_4 &= 0.9999 \\ &\vdots \end{align} As you tack on more and more repetitions, these numbers get closer and closer to some limiting value, which I'll call $A$. If you know calculus notation, $$\lim_{n\to\infty} a_n = A$$ The number represented by $0.\overline{9}$ is $A$. It happens to work out to be $1$ (or if we're being pedantic, $1$ is yet another notation for the same number).
As another example of this way of interpreting repeating decimals, consider $0.257\overline{143}$. You can write the sequence \begin{align} b_1 &= 0.257143 \\ b_2 &= 0.257143143 \\ b_3 &= 0.257143143143 \\ b_4 &= 0.257143143143143 \\ &\vdots \end{align} and similarly, the number represented by $0.257\overline{143}$ is the value that this sequence gets closer and closer to as you add more repetitions; or $$\lim_{n\to\infty} b_n$$ This one works out to $\frac{128443}{499500}$.
• Did you just randomly pick that last fraction or did you already have it in mind? – user1717828 Jun 3 '15 at 12:19
• Worth noting, since the 'overdot' notation being standard in UK and apparently China has come up in comments on OP, that in that variant we dot the first and last in a sequence (as opposed to each digit). – OJFord Jun 3 '15 at 22:13
• @user1717828 totally random, I didn't feel like going to the trouble of looking up a "significant" fraction. – David Z Jun 4 '15 at 5:08
This symbol is called a vinculum or a overline.
The number(s) under this symbol are repeated indefinitely.
$0.\overline{9} = 0.99999999999999...$
$2.52\overline{346} = 2.52346346346346...$ | 2019-05-21T22:52:03 | {
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https://mathhelpforum.com/threads/arithmetic-series-word-problem-analysis.249576/ | # Arithmetic Series Word Problem (Analysis)
#### iNET
Hi, I'm not sure if this is the right section for this question, so if not please redirect me to the correct subforum.
Q: There are k animal feeding stations arranged in a line with a supply hut. The stations are s yards apart and the nearest is t yards from the hut. An attendant carries n bags of feed, one at a time, to each feeding station. How far will he have traveled when he arrives back at the hut after servicing all stations?
Here's what I did, but I don't know if it's correct: (S.k denotes 'S of k')
S.k=(2t + 2(t+s)+...+2(t+(k-1)s))n
= n((k/2)(2(2t)+(k-1)2s))
= n((k(4t+2sk-2s))/(2))
= n((4kt+2sk^2-2sk)/(2))
= n(kt+sk^2+sk)
= nk(t+sk+s) yards
#### romsek
MHF Helper
I get
$2 n t + 2n(s+t) + 2n(2s+t) + \dots 2n((k-1) s + t) = 2 n t k+2 n s \displaystyle{\sum_{j=1}^{k-1}}j = 2 n t k + 2 n s \dfrac {k(k-1)}{2}=2 n k \left(t + \dfrac{s(k-1)}{2}\right)$
1 person
#### iNET
Thanks! I don't exactly understand the method you used to get your answer, but using your answer I was able to see the mistake in my own work that–when corrected–gave me a solution identical to yours.
#### romsek
MHF Helper
the main thing is
$\displaystyle {\sum_{j=1}^{k-1}}j =\dfrac {k(k-1)}{2}$
It looks like you set the problem up correctly. We both start from the same basic equation.
#### iNET
I see.
This may be a stupid question, but the answer could also be simplified to:
nk(2t+s(k-1))
Is one answer somehow 'more' correct than the other one, or are both answers equally acceptable?
Thanks for your help.
#### romsek
MHF Helper
I see.
This may be a stupid question, but the answer could also be simplified to:
nk(2t+s(k-1))
Is one answer somehow 'more' correct than the other one, or are both answers equally acceptable?
Thanks for your help.
they are both the same answer.
Yours is a bit simpler than mine. | 2020-01-26T05:24:30 | {
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https://math.stackexchange.com/questions/537164/find-lim-n-to-infty-fracn22n/537182 | # Find $\lim_{n\to\infty} \frac{n^2}{2^n}$
$$\lim_{n\to\infty}\frac{n^2}{2^n}$$
Do you have some tips so I could solve this problem, without the use of L'Hôpital's rule?
Indeed, we didn't see formally L'Hôpital's rule, nor Taylor series so I'm supposed to do this without such "tools".
I've tried using the fact that $n^2 = e^{2\log(n)}$ and $2^n=e^{n\log(2)}$ but didn't manage to eliminate my indeterminate form.
• At least write down what you get after applying l'Hoptial's rule once. Then apply it again. – Alex R. Oct 23 '13 at 17:35
• It look as if $n$ may range over positive integers. In that case, use the fact that by the Binomial Theorem, $(1+1)^n=1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6}+\cdots\gt \frac{n(n-1)(n-2)}{6}$. – André Nicolas Oct 23 '13 at 17:40
• What, if anything, do you know about the comparative growth rates of $n$ and $\log(n)$? If you happen to know that $$\lim_{n\to\infty}\bigl[n-\log(n)\bigr]=\infty,$$ then since $$\frac{n^2}{2^n}=\frac{e^{2\log(n)}}{e^{n\log(2)}}=e^{2\log(n)-n\log(2)},$$ and since $\log(2)>0,$ it follows that the limit you're looking for is $0$, and even gives you a start on how it can be proven. – Cameron Buie Oct 23 '13 at 17:53
There are many ways to do that. For example, you can prove that for $n \ge 10$ we have $$2^n > n^3.$$ You can show that using the induction argument.
Regarding to analysis tag, you seem to be familiar to series so let us consider the following series: $$\sum(n^2/2^n)$$ it is not hard seeing that this series is convergent and so the $n-$th term approaches to zero while $n\to\infty$. Clearly, this way needn't using L'Hopital's rule.
• @Amzoti: Thanks for your support my dear friend. I hope you have a good time ahead and a peaceful slumber. :-) – mrs Oct 24 '13 at 1:48
• You are welcome my friend! I appreciate all of the hard work you and amWhy put into the site helping people become better problem solvers! Time to watch a movie and relax as it has been a difficult month at work and preparing two talks! Have a great night! Regards – Amzoti Oct 24 '13 at 1:49
I noticed the OP had wanted L'Hôpital's rule off the table only after I posted my previous reply. I apologize for my inattention. Here is another answer:
Let $a_n=n^2/2^n$ for all $n\in\mathbb{Z}_+$. You can easily compute that \begin{align*} a_{n+1}-a_n=-\frac{n(n-2)-1}{2^{n+1}}, \end{align*} which is negative for $n\geq3$. Therefore, the sequence is eventually decreasing and bounded below (by zero), so it must have a limit.
Obviously, this limit cannot be negative, since $a_n>0$ for all $n$. If we show it cannot be positive, either, then we can conclude that the limit must be zero.
Suppose that the limit is positive: $\lim_{n\to\infty} a_n=c>0$. Since the sequence is decreasing (after omitting $a_1$ and $a_2$, which does not affect convergence properties), it is also true that \begin{align*} c=\lim_{n\to\infty} a_n=\inf_{n\in\mathbb{Z}_+\setminus\{1,2\}}a_n. \end{align*} Then, for $n$ large enough and beyond, we have that \begin{align*} (*)\quad c\leq a_n\leq\frac{3}{2}c. \end{align*}
Therefore, \begin{align*} (**)\quad a_{n+1}=a_{n}\times\frac{a_{n+1}}{a_n}=a_n\times\frac{(n+1)^2}{2n^2}\leq\frac{3}{2}c\times \frac{(n+1)^2}{2n^2}, \end{align*}
where the last inequality follows from $(*)$. But $(n+1)^2/(2 n^2)$ converges to $1/2$ as $n\to\infty$, so that for $n$ sufficiently large and beyond, \begin{align*} \frac{(n+1)^2}{2n^2}\leq\frac{5}{8}. \end{align*} For such values of $n$, $(**)$ implies that \begin{align*} a_{n+1}\leq\frac{3}{2}c\times\frac{(n+1)^2}{2n^2}\leq\frac{15}{16}c<c, \end{align*} where the last inequality is strict because $c$ is positive. But this contradicts $c$ being the infimum of the values of the sequence. The limit thus cannot be positive.
• +1. I appreciate your effort of writing another answer after I pointed out that L'H was off-limits. :) – Lord_Farin Oct 23 '13 at 20:13
• Thanks, @Lord_Farin. :-) – triple_sec Oct 23 '13 at 20:37 | 2020-11-29T07:50:04 | {
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https://math.stackexchange.com/questions/2304246/unique-combinations-of-n-non-unique-elements-in-k-non-unique-buckets-with-c | # Unique combinations of $n$ non-unique elements in $k$ non-unique buckets with $c$ capacity
I have $n$ non-unique elements, and I have $k$ unordered buckets that can hold anywhere from $0$ to $c$ elements, such that $c * k \geq n$. I would like to find all possible combinations.
For example, given $n=10$, $k=4$, and $c=4$, there are 7 possible distributions:
• 3322
• 3331
• 4222
• 4321
• 4330
• 4411
• 4420
where "3322", for example, means that two buckets have three elements each and the other two buckets have two elements each.
Another way to look at it, is that I want to find unique combinations of $k$ numbers less than or equal to $c$ such that their sum is equal to $n$.
Ideally I'd like an algorithm to be able to generate a list of all acceptable combinations, but knowing a formula for finding the number of combinations given arbitrary $n$, $k$, and $c$ would be helpful. Other answers generally assume buckets have to have a minimum capacity of 1, or they deal with unique elements to some degree, which are not the case here.
First, it is important to note that all solutions can be arranged in descending sorted order. This is useful for the algorithm I will describe, such that it only searches for sorted answers.
Essentially, we start at the first bucket and assign it some item count $0\leq i\leq c$. Now, we realize that assigning the remaining items to the remaining buckets is essentially the same problem, except now we have $n-i$ items, $k-1$ buckets, and each remaining bucket can now only hold up to $i$ items, because if it were to hold anymore, the solution would be unsorted.
Only when we reach the end (no buckets and no items left) do we have a proper solution. This algorithm can also be sped up by realizing that if $\frac{n-i}{k-1}>i$, there is no continuing from this point, as the remaining items can not fit in the remaining buckets without going over $i$.
Below is a solution in Python,
def buckets(n,k,c,solution=[]):
total = 0
for i in range(min(c,n),-1,-1):
if k-1 != 0 and float(n-i)/(k-1) > i:
#Can't fit remaining items
break
solution.append(i)
if n-i == 0 and k-1 == 0:
#We've reached the end of the buckets and have a solution
print(solution)
solution.pop()
return 1
elif k-1 != 0:
#If there's still buckets left, try assigning them
total += buckets(n-i,k-1,i,solution)
#Done checking this bucket assignment
solution.pop()
buckets(10,4,4)
[4, 4, 2, 0]
[4, 4, 1, 1]
[4, 3, 3, 0]
[4, 3, 2, 1]
[4, 2, 2, 2]
[3, 3, 3, 1]
[3, 3, 2, 2]
• Thanks so much for this - it works completely. Jun 1, 2017 at 17:54 | 2022-08-14T04:01:50 | {
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https://math.stackexchange.com/questions/783917/matrix-exponential-of-a-simple-bidiagonal-matrix | # Matrix exponential of a simple bidiagonal matrix
I am interested in finding an expression (closed form or recursive) for the matrix exponential of this banded matrix:
$$\begin{pmatrix} 0 & 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & a_1 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & a_2 & 1 & \cdots & 0 & 0 \\ 0 & 0 & 0 & a_3 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots &\ddots&\vdots&\vdots \\ 0 & 0 & 0 & 0 & \cdots & a_{n-1} & 1 \\ 0 & 0 & 0 & 0 & \cdots & 0 & a_n \end{pmatrix}$$
For simplicity, assume that $a_k>0~\forall k$. I am quite certain one must exist having played around with it for a while, and looking at the solution for small values of $n$.
Has anyone seen this structure before? Does it have a name? Do you know if there is a solution published somewhere?
If you go ahead and compute the answer for small values of $n$, you get:
$$\exp\left( \begin{array}{cc} 0 & 1 \\ 0 & a_1 \\ \end{array} \right) = \left( \begin{array}{cc} 1 & \frac{-1+e^{a_1}}{a_1} \\ 0 & e^{a_1} \\ \end{array} \right)$$
$$\exp \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & a_1 & 1 \\ 0 & 0 & a_2 \\ \end{array} \right) = \left( \begin{array}{ccc} 1 & \frac{-1+e^{a_1}}{a_1} & \frac{-e^{a_2} a_1+a_1+e^{a_1} a_2-a_2}{a_1 \left(a_1-a_2\right) a_2} \\ 0 & e^{a_1} & \frac{e^{a_1}-e^{a_2}}{a_1-a_2} \\ 0 & 0 & e^{a_2} \\ \end{array} \right)$$
Unfortunately, $n=3$ is to large to print here, but a pattern remains.
It suffices to diagonalize $A$ (if the $(a_i)$ are pairwise distinct). I change slightly the notation $A=D+J_n$ where $D=diag(a_1,\cdots,a_n)$ and $J_n$ is the nilpotent Jordan block of dimension $n$. $A=PDP^{-1}$ where the first row of $P$ is
$1,-\dfrac{1}{a_1-a_2},\dfrac{1}{(a_1-a_3)(a_2-a_3)},-\dfrac{1}{(a_1-a_4)(a_2-a_4)(a_3-a_4)},\cdots$
We obtain the other rows of $P$ by circular permutation along the diagonals. Finally $e^A=Pdiag(e^{a_1},\cdots,e^{a_n})P^{-1}$ where the first row of $P^{-1}$ is
$1,\dfrac{1}{a_1-a_2},\dfrac{1}{(a_1-a_2)(a_1-a_3)},\dfrac{1}{(a_1-a_2)(a_1-a_3)(a_1-a_4)},\cdots$
We obtain the other rows of $P^{-1}$ by circular permutation along the diagonals.
• Cool thanks very much, this looks promising. However, I'm not sure exactly what you mean by "circular permutation along the diagonals". Do you mean just "circular permutation" or (equivalently) do you mean that the diagonals, when wrapped, will be constant? When I try this I don't get $PP^{-1}=1$ – Ian Hincks May 8 '14 at 20:28
• Also, can you point me in the direction of how you arrived at this conclusion? – Ian Hincks May 8 '14 at 20:29
• Ah, okay, I figured out that "circular permutation along the diagonal" means that, for example, the second row of $P$ will be $(0,1,-(a_2-a_3)^{-1},((a_2-a_4)(a_3-a_4))^{-1},...)$. – Ian Hincks May 8 '14 at 21:12
• Yes Ian, the second row is as you write it. Let $A_n$ be the matrix of dimension $n$. Then $e^{A_n}$ is the submatrix of $e^{A_{n+1}}$ that is constituted by its first $n$ rows and its first $n$ columns. – user91684 May 9 '14 at 10:36
This response is very late but perhaps still useful to others who come across your question whilst browsing (as I did!).
Your banded (or upper bidiagonal) matrix is what Optiz termed a $\textit{Steigungsmatrix }$ (maybe ... "gradient matrix"). See McCurdy et al (1984), Mathematics of Computation, 43, 501-528. For any $n>0$, all you need is to apply $\textit{Opitz's formula}$. You will need a little knowledge of functions of matrices and divided differences to apply it but it will be worth the effort.
If we denote your matrix by A and its elements by $a_{ij}$ then, using Opitz's theorem, the matrix F defined as F = $g$(A) for a function $g$(.), has entries $f_{ij}$ as follows: $f_{ij}=0$ for $i>j$, $f_{ij} = g(a_{ii})$ for $i=j$, and $f_{ij} = g[a_{ii},a_{i+1,i+1},..,a_{jj}]$ for $i<j$ .
The notation $g[x_0,x_1,...,x_n]$ is the nth divided difference defined recursively as $$g[x_0,x_1,...,x_{(n-1)},x_n] = \frac{g[x_1,...,x_n]-g[x_0,...,x_{(n-1)}]}{x_n-x_0},$$ $g[x_0,x_1] = (g[x_1]-g[x_0])/(x_1-x_0)$ and $g[x] = g(x)$.
Setting $g$(.) to be the exponential function, I get your answers for n=1 and n=2.
The above assumes distinct diagonal entries. If there are repeats in the diagonal terms then you should use the appropriate "confluent" form of the divided difference. | 2021-03-04T06:58:05 | {
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https://math.stackexchange.com/questions/2162888/prove-that-the-sum-of-pythagorean-triples-is-always-even/2162901 | # Prove that the sum of pythagorean triples is always even
Problem: Given $a^2 + b^2 = c^2$ show $a + b + c$ is always even
My Attempt, Case by case analysis:
Case 1: a is odd, b is odd. From the first equation,
$odd^2 + odd^2 = c^2$
$odd + odd = c^2 \implies c^2 = even$
Squaring a number does not change its congruence mod 2.
Therefore c is even
$a + b + c = odd + odd + even = even$
Case 2: a is even, b is even. Similar to above
$even^2 + even^2 = c^2 \implies c$ is even
$a + b + c = even + even + even = even$
Case 3: One of a and b is odd, the other is even Without loss of generality, we label a as odd, and b as even
$odd^2 + even^2 = c^2 \implies odd + even = c^2 = odd$
Therefore c is odd
$a + b + c = odd + even + odd = even$
We have exhausted every possible case, and each shows $a + b + c$ is even. QED
Follow Up: Is there a proof that doesn't rely on case by case analysis? Can the above be written in a simpler way?
• In fact there are no Pythagorean triples where the legs ($a$ and $b$) are both odd Feb 27, 2017 at 16:47
• Even though there are shorter proofs, there is something pleasingly straightforward about your proof. Mar 1, 2017 at 13:48
Note that $x^2\equiv x\pmod 2$ and thus $a^2+b^2=c^2$ implies $$a+b+c\equiv a^2+b^2+c^2\equiv 2c^2\equiv 0\pmod 2$$
• Well that makes it easy, props for the quick answer. Feb 26, 2017 at 21:55
• You are welcome, @spyr03. Feb 26, 2017 at 21:55
• $a^2+b^2+c^2\equiv 2(a^2+b^2)\pmod 2$ Could you explain that step? Feb 26, 2017 at 21:55
• @mrnovice $c^2 = a ^2 + b^2\implies a^2+b^2+c^2 = 2(a^2+b^2)$. Feb 26, 2017 at 21:56
• @mrnovice, not at all, the way I had written the answer was missing the context, you've asked a valid question. Feb 26, 2017 at 21:59
Also, Pythagorean triples have a well defined structure: $$a=k(m^{2}-n^{2}),\ \,b=k(2mn),\ \,c=k(m^{2}+n^{2})$$ and $$a+b+c=2k(mn+m^2)$$
Hint
Write $a+b+c=k$, so
$$a^2+b^2=(a+b)^2-2ab= (k-c)^2-2ab=c^2 → k^2-2(kc+ab)=0→k^2=2(kc+ab)$$
Notice that $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=2c^2+2(ab+bc+ac)$, so the square of $a+b+c$ is even and thus $a+b+c$ is also even.
$c^2 = a^2 + b^2 = (a+b)^2 - 2ab$.
$2ab = (a+b)^2 - c^2$
$2ab = (a+b+c)(a+b-c)$
Let $n = a+b+c$, and the above becomes:
$2ab = n(n-2c)$
So the right-hand side must be even, but since $n-2c$ is odd when $n$ is odd, $n$ must be even.
• Nice explanation, I never would have thought about writing $2ab = (a + b + c)(a + b -c)$. You could have finished with $2ab = n^2 - 2cn$ and thus n^2 must be even, therefore n is even. Feb 28, 2017 at 23:29
Consider $(a+b+c)^2$
Which is $a^2 + b^2 + c^2 + 2(ab+bc+ca)$
Since $c^2 = a^2 + b^2$ (c being the hypotenuse), $(a+b+c)^2 = 2(c^2 + ab + bc + ca)$ - which is an even number.
and since squares of odd is odd and evens is even $a+b+c$ has to be even. | 2022-07-03T02:14:46 | {
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http://interviewingthecrisis.org/codecademy-python-math-symbols | Codecademy Python Math Symbols | interviewingthecrisis.org
# Writing mathematical expressions - MatplotlibPython plotting.
Codecademy is the easiest way to learn how to code. It's interactive, fun, and you can do it with your friends. Python Comparison Operators. These operators compare the values on either sides of them and decide the relation among them. They are also called Relational operators. Python 3.x: small greek letters are coded from 945 to 969 so,alpha is chr945, omega is chr969 so just type printchr945 the list of small greek letters in a list. :mortar_board:exercise answers. Contribute to ummahusla/Codecademy-Exercise-Answers development by creating an account on GitHub. Many programming languages include libraries to do more complicated math. You can do statistics, numerical analysis or handle big numbers. One topic many programming languages have difficulty with is symbolic math. If you use Python though, you have access to sympy, the symbolic math library. Sympy.
Learn about lists, a data structure in Python used to store ordered groups of data. :mortar_board:exercise answers. Contribute to ummahusla/Codecademy-Exercise-Answers development by creating an account on GitHub. Python 3.x: small greek letters are coded from 945 to 969 so,alpha is chr945, omega is chr969 so just type printchr945 the list of small greek letters in a list.
Python Arithmetic Operators Example - Assume variable a holds 10 and variable b holds 20, then −. Numeric and Mathematical Modules¶ The modules described in this chapter provide numeric and math-related functions and data types. The numbers module defines an abstract hierarchy of numeric types. There is nothing special about multiplication in Python; we just use an asterisk in the same way we did with addition, subtraction, and division. print2 2 I’m sure it comes as no surprise to you that the above outputs 4. Note that if we multiply two integers, the result is of the data type int. Jupyter notebook recognizes LaTeX code written in markdown cells and renders the symbols in the browser using the MathJax JavaScript library. Mathematics Inline and Display Enclose LaTeX code in dollar signs $.$ to display math inline. What are operators in python? Operators are special symbols in Python that carry out arithmetic or logical computation. The value that the operator operates on is called the operand.
x = Symbol'x' y = Symbol'y' A = Matrix[[1,x], [y,1]] Once a matrix is created, you can operate on it. There are functions to do dot products, cross products or calculate determinants. Hyperbolic functions The abbreviations arcsinh, arccosh, etc., are commonly used for inverse hyperbolic trigonometric functions area hyperbolic functions, even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area. math.floor x ¶ Return the floor of x as a float, the largest integer value less than or equal to x. math.fmod x, y ¶ Return fmodx, y, as defined by the platform C library. Codecademy-Exercise-Answers / Language Skills / Python / Unit 01 Python Syntax / Edvins Antonovs Subfolder rename & ascending ordering for python Latest commit 5fc7ead Jul 14, 2016.
from sympy.plotting import plot from sympy import Symbol x = Symbol 'x' p = plot 2 x3, 3 x1, legend = True, show = False p [0]. line_color = 'b' p [1]. line_color = 'r' p. show Sets Beside FiniteSet which I exemplify below, sympy also includes support for infinite sets and intervals. Ask questions, get help with an exercise, and chat about your Codecademy coursework here. “When one teaches, two learn” — Robert Heinlein.
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Introduction¶ SymPy is a Python library for symbolic mathematics. It aims to become a full-featured computer algebra system CAS while keeping the code as simple as possible in order to be comprehensible and easily extensible.You can find out which functions and attributesare defined in a module. import math dir mathIf you have a Python script named math.py in the samefolder as your current script, the file math.py willbe loaded instead of the built-in Python module.
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1. Learn Python, a powerful language used by sites like YouTube and Dropbox. Learn the fundamentals of programming to build web apps and manipulate data. Master Python loops to deepen your knowledge. Learn the fundamentals of programming to build web apps and manipulate data.
2. Any text element can use math text. You should use raw strings precede the quotes with an 'r', and surround the math text with dollar signs \$, as in TeX. Regular text and mathtext can be interleaved within the same string.
3. What is SymPy? SymPy is a Python library for symbolic mathematics. It aims to be an alternative to systems such as Mathematica or Maple while keeping the code as.
4. I want to make a dictionary in Python, but with math symbols because then i want to plot that with the symbols. So, i have something like this.
The Python Discord. News about the dynamic, interpreted, interactive, object-oriented, extensible programming language Python. If you are about to ask a "how do I do this in python" question, please try r/learnpython, the Python discord, or the python IRC channel on FreeNode. Note that wasysym also defines a \lightning symbol. The difference—other than “ ” vs. The difference—other than “ ” vs. “ ”—is that the stmaryrd version above is limited to math mode. An online LaTeX editor that's easy to use. No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more.
Diese Liste mathematischer Symbole zeigt eine Auswahl der gebräuchlichsten Symbole, die in moderner mathematischer Notation innerhalb von Formeln verwendet werden. | 2021-01-23T05:10:16 | {
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http://math.stackexchange.com/questions/69456/number-of-ways-to-distribute-items-with-limit-of-min-and-max-at-each-location | # Number of ways to distribute items (with limit of min and max at each location)
There are a total of $n$ objects in stock. You must keep at least $c_i$ objects in stock at location $i$. Assume $n \ge \sum c_i$.
a.) How many different inventories $(x_1,x_2,\dots,x_r)$ are there?
b.) Each location can store a maximum of $d_i$. Let $r = 2$ and count the number of different invetories $(x_1,x_2)$ with $c_1 \le x_1 \le d_1$ and $c_2 \le x_2 \le d_2$.
My answer for part a: We have no choice over where $c_1 + c_2 + \dots + c_r = c_t$ objects are placed. The ramaining $n-c_t$ objects can be spread across $r$ locations. So, there are $$\binom{(n-c_t) + r - 1}{r-1}$$ different inventories.
My incomplete answer for part b: After placing minimum required objects at each location, we are left with $n-c_1-c_2$ objects to distribute.
The remaining objects can be distributed in $$\binom{(n-c_1-c_2)+2-1}{2-1} = (n-c_1-c_2)+1.$$
Now from the number above, I need to subtract the number of combinations that lead to overflows in the two locations.
[---Not sure if the rest is correct---]
Number of ways location 1 overflows:
Location 1 overflows when $x_1 > d_1$ and $c_2 \le x_2 \le d_2$: $$\binom{(n-(d_1+1)-c_2)+2-1}{2-1} = n-d_1-c_2$$
Number of ways location 2 overflows:
(same reasoning as above) $$n-d_2-c_1$$
Total number of inventories is $$(n-c_1-c_2+1)-[(n-d_1-c_2)+(n-d_2-c_1)] = (d_1+d_2+1)-n$$
Total number of inventories $= (d_1+d_2+1)-n$ doesn't make sense.
-
Your answer to part (a) is fine. In (b) you’ve overlooked the possibility of overflow in both inventories simultaneously.
You got off to a fine start: there are $$I_0 = n-c_1-c_2+1$$ inventories that meet the minimum requirements. Then you want get the number of inventories that overflow location $1$ but meet the minimum requirement at location $2$; your answer of $n-d_1-c_2$ is correct only if $n\ge d_1+c_1$. If $n < d_1+c_1$, it’s not possible to overflow at location $1$ and meet the minimum requirement at location $2$. Thus, the correct result for this step is $$I_1 = \max\{n-d_1-c_2,0\}$$ inventories that overflow at location $1$ and meet the minimum at location $2$. Similarly, there are $$I_2=\max\{n-d_2-c_1,0\}$$ inventories that meet the minimum at location $1$ but overflow location $2$.
However, there may be inventories that overflow at both locations. When you subtract off the inventories that overflow at location $1$ and then those that overflow at location $2$, the inventories that overflow at both locations get subtracted twice, so you have to add them back in.
The basic calculation gives $$\binom{n-(d_1+1)-(d_2+1)+2-1}{2-1} = n-d_1-d_2-1$$ of these inventories, but again this is true only if $n\ge d_1+d_2+1$; if $n < d_1+d_2+1$, it isn’t possible to overflow both locations. Thus, the correct figure is $$I_3=\max\{n-d_1-d_2-1,0\}.$$
The final result is then given by $$I_0-I_1-I_2+I_3.$$
Note that if $n \ge d_1+d_2+1$, this evaluates to $0$. This makes perfectly good sense: in that case it’s impossible to meet the requirements, and there are no acceptable inventories.
-
Unless I am reading the problem wrong, this is the same as a couple of other problems that were asked in the last couple of days. You are given a total and want to find the number of ways to make that total $\left(\displaystyle\sum_{i=1}^rx_i=n\right)$ given a range constraint on each summand $(c_i\le x_i\le d_i)$. The answer is the coefficient of $x^n$ in $$\prod_{i=1}^r\frac{x^{d_i+1}-x^{c_i}}{x-1}\tag{1}$$ We can apply this to the special cases you have listed.
a) you let $d_i\to\infty$. In that case, $(1)$ becomes $\displaystyle\prod_{i=1}^r\frac{x^{c_i}}{1-x}=x^{\sum_{i=1}^rc_i}\frac{1}{(1-x)^r}$, so the answer becomes the coefficient of $x^{n-\sum_{i=1}^rc_i}$ in $\frac{1}{(1-x)^r}$. \begin{align} \frac{1}{(1-x)^r} &=\sum_{i=0}^\infty\binom{-r}{i}(-x)^i\\ &=\sum_{i=0}^\infty\binom{r+i-1}{i}x^i\\ &=\sum_{i=0}^\infty\binom{r-1+i}{r-1}x^i \end{align} Therefore, the answer is $\displaystyle\binom{r-1+n-\sum_{i=1}^rc_i}{r-1}$.
b) you set $r=2$. In this case, $(1)$ becomes $$\left(x^{c_1}+x^{c_1+1}+\dots+x^{d_1}\right)\left(x^{c_2}+x^{c_2+1}+\dots+x^{d_2}\right)\tag{2}$$ and the answer is the coefficient of $x^n$ in $(2)$.
Suppose $d_1-c_1\le d_2-c_2$, then the answer is $$\text{the coefficient of }x^n\text{ in }(2)=\left\{\begin{array}{}n-c_1-c_2+1&\text{for }c_1+c_2\le n\le d_1+c_2\\d_1-c_1+1&\text{for }d_1+c_2+1\le n\le c_1+d_2-1\\d_1+d_2-n+1&\text{for }c_1+d_2\le n\le d_1+d_2\\0&\text{otherwise}\end{array}\right.$$
-
Please show the closed form, I can't get the answer. – jamesio Oct 3 '11 at 17:04
@jamesio: I have included the gory details. – robjohn Oct 3 '11 at 17:53 | 2015-07-29T22:19:22 | {
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https://www.physicsforums.com/threads/integral-substitution.328793/ | # Integral substitution
## Main Question or Discussion Point
How do I evaluate the integral $$\int_a^b$$ f(x4) dx = $$\int_y^z$$ f(u) dx/du du, where u=x4
y=a4
z=b4
## Answers and Replies
HallsofIvy
Homework Helper
How do I evaluate the integral $$\int_a^b$$ f(x4) dx = $$\int_y^z$$ f(u) dx/du du, where u=x4
y=a4
z=b4
How you integrate the function f depends strongly on what f is, don't you think?
f is $$\int_-1^1$$ 1+x4 dx
f is $$\int_1^1$$ 1+x4 dx
meant to be a -1 at the bottom of the integral.
How could I do this then.
jgens
Gold Member
Well, if your integral is simply
$$\int_{-1}^1 1 + x^4\, dx$$
Then it's pretty simple. Just seperate the integrand so that,
$$\int_{-1}^1 \,dx + \int_{-1}^1 x^4 \, dx$$
But, based on the context of the question I'm not sure that's what you're looking for. Perhaps you could repost exactly what you mean clearly?
Sure. Consider the integral $$\int_{-1}^1$$ 1+x4 dx
How do I evaluate this integral using the substitution u=x4 and the formula:
$$\int_a^b$$ f(x4) dx = $$\int_{a^4}^{b^4}$$ f(u) dx/du du, where u=x4
jgens
Gold Member
Why would you use that method to evaluate the integral? The integral you have posted can easily be evaluated using the power rule for integration.
I thought this was a more interesting way but just don't know how to get started.
Also I someone could help me with this problem.
How could I evaluate the integral above using u=x4, but seperating it into 2 integrals.
Cyosis
Homework Helper
Taking detours isn't a more interesting way, but here goes.
$u=x^4 \Rightarrow du=4x^3dx$
$$\int 1+x^4 dx=\int \frac{1+u}{4(u^{3/4})} du$$
Making life a lot harder!
Last edited:
Gib Z
Homework Helper
Personally I wouldn't discourage him, I find it a good way to learn how to compute integrals and what methods to use can be supplemented by trying other methods that come to mind, and seeing why they may not be as efficient. Once he computes that integral Cyosis got him, I think he will be able to see in future why it may not be wise to compute an integral as such.
Cyosis
Homework Helper
You raise a very good point. I guess it's all too easy to see if something is going to be inefficient or not when you have some experience with the topic at hand.
Note: I didn't take the limits into account so you still have to do that yourself.
Why would you use that method to evaluate the integral? The integral you have posted can easily be evaluated using the power rule for integration.
You are too harsh. Sure, for this function there are easy ways to do the integral. But for other functions f maybe you really should do the substitution. And why not learn to do the substitution on simple problems that can be evaluated in other ways? Then, for example, you can check your answers and see when you make a mistake. | 2020-07-06T18:11:28 | {
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https://brilliant.org/discussions/thread/synthetic-geometry-session-followup-questions-and/ | # Synthetic Geometry Session Followup Questions and Problem Summary
In our first session, we spent some time on this problem: Let $ABC$ be a triangle in which $\angle A = 60^{\circ}$ Let $BE$ and $CF$ be the angle bisectors with $E$ on $AC$ and $F$ on $AB$. Let $M$ be the reflection of $A$ in the line $EF$. Prove that $M$ lies on $BC$
1. $AEIF$ is cyclic because $\angle BIC=90+\frac {\angle A}{2}=120^{\circ}\implies \angle FIE+\angle A=180^{\circ}$. The cyclic result only occurs when $\angle A=60^{\circ}$, and it gives us information about the angles of $\triangle AEF$ and consequently those of $\triangle FME$
2. $I$ is the circumcenter of $MEF$ because $IE=IF$ and $\angle FIE=2\angle FME$. This result relates $M$ with $I$ and gives us more angle relations.
3. $BMIF$ is cyclic, which we proved by using the above properties to angle chase $\angle FBI=\angle FMI$. Symmetrically $CMIE$ is cyclic, so $\angle IMB+\angle IMC=\angle AFI+\angle AEI=180^{\circ}$, which implies $B,M,C$ are collinear and we are done.
The discovery of these properties makes the problem transparent, meaning we know what makes our main result true. What this also means is that we can probably find simpler solutions. Indeed, there are many ways to solve this problem:
Using angle bisector theorem and Miquel point (Avoids tedious angle chasing):
Since $AEIF$ is a cyclic quadrilateral, we know the miqual point of its complete correspondence lies on $BC$. This means the circumcircles of $BFI,CEI, ABE,ACF$ all have a common point on $BC$, we denote the point $M'$. Because $ACM'F$ is cyclic, $\frac {FM'}{BF}=\frac {AC}{BC}=\frac {AF}{BF}$; therefore $AF=FM'$ and similarly $AE=EM'$. This is enough to establish $M=M'$ and $M$ lies on $BC$
Utilizing $EF$ as the axis of symmetry:
Let $EF\cap BC=P$. It suffices to prove $PE$ bisects $\angle APB$. We can achieve this with our first property.
Now I will propose a few follow up questions which you guys should now easily answer:
We keep the notations of our main problem:
1. $DBC$ is equilateral such that $D,A$ are on opposite sides of $BC$. Prove that $DE=DF$
2. Through $I$ construct the perpendicular to $EF$ which intersects $EF,BC$ at $X,Y$. Prove that $2IX=IY$
3. $O$ is the circumcenter of $AEF$, Prove that $OM\perp BC$.
Note by Xuming Liang
5 years, 5 months ago
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For follow up problem 1 :
To prove $DE=DF$ , it suffices to establish congruence between $\Delta DIF$ and $\Delta DIE$ (not dead yet) . We have $DI=DI$ and we have already established that $IF=IE$ by having $\angle IFE= \angle IEF=30$. So we need to prove a pair of angles congruent. Since we want a good connection between the angles and the discovered stuff , we prefer proving $\angle DIF=\angle DIE$. By cyclicity we have $m\angle BIF=m\angle CIE=60$. So we need to prove $\angle DIB=\angle DIC=\dfrac{\angle BIC}{2}=\dfrac{120}{2}=60$. Now how to achieve this? We have not yet utilized the equilateral triangle yet!
So, consider $\Box BICD$ Can we prove it cyclic? Yes! We have to prove that $\angle EBD+\angle ECD=180$ and its easy.
$\angle IBD+\angle ICD =\angle IBC+\angle CBD+\angle ICD+\angle BCD=\dfrac{B}{2}+60+\dfrac{C}{2} + 60=120+60=180 \\ \Rightarrow \Box BICD \ is \ cyclic!$
So by cyclicity we have $\angle DIB=\angle DIC=60$ which we had to prove.
- 5 years, 5 months ago
Very nice reasoning using congruence! For the cyclic part, I believe you meant to say $B,I,C,D$ are concyclic, which is true because $\angle BIC+\angle BDC=180$. Since $BD=DC$, $ID$ bisects $\angle BIC$, which is what you wanted to prove.
- 5 years, 5 months ago
Sorry , yes it is that $B,I,C,D$ are concyclic. I have edited it and few angles too.
- 5 years, 5 months ago
A fast way to prove that $BICD$ is cyclic is to notice that $\angle BIC = 120^{\circ}$ and $\angle BDC = 60^{\circ}$.
- 5 years, 5 months ago
First we have $\Delta IXF$ as 30-60-90 triangle. So $FI=2IX$. So we have to prove that $FI=IY$. So consider $\Delta BIF$ and $\Delta BIY$ , where we have $\angle FBI=YBI$ , $BI=BI$ , $\angle BIF=\angle BIY=60$ (by angle chase). So they are congruent by ASA test and hence $FI=IY$ is proved.
- 5 years, 5 months ago
I pretty much proved it the same as you. Nice solution.
- 5 years, 5 months ago
Yesterday we did find that $\angle EMC=60$ So to prove our result , we must prove that $\angle OME=30$. Using the fact that $O$ is circumcentre , we have $\angle EOF=\angle EIF=120$ and by some angle chase we have $\angle OEI=\angle OFI=60$ where $\angle OFE=\angle IFE=\angle OEF=\angle IEF=30$. We get a beautioful result that $\Box FOEI$ is a rhombus which has one of its diagonals that is $OI$ equal to its side. Using these facts and that $I$ is circumcenter of $\Delta EMF$ we have $EI=OI=IF=IM$ and thus $\Box EOFM$ is cyclic giving $\angle OFE= \angle OME=30$ which was to be proved :)
- 5 years, 5 months ago
When will be the next discussion?
- 5 years, 5 months ago
Problem 3: Since $M$ is the Miquel point, there exists a spiral similarity centered at $M$ that maps $FI \rightarrow AE$. Denote $M_1, M_2$ as the midpoint of $AF$ and $EI$, respectively. Thus, the spiral similarity maps $FI \rightarrow M_1M_2$. Using the circumcircles of $\triangle BFI$ and $BM_1M_2$, this implies that $M_1M_2BM$ is cyclic. Since $M_1OM_2B$ is cyclic, this implies that $M_1OM_2MB$ is cyclic with diameter $OB$. Thus, $\angle OMB = 90^{\circ}$ and we are done.
- 5 years, 5 months ago
Great solution. This fact is actually true as long as $BICD$ is cyclic.
- 5 years, 5 months ago | 2021-05-06T01:36:51 | {
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https://adamgemili.com/wiki/archive.php?tag=4a565b-is-population-discrete-or-continuous | A bacteria colony is made of billions of organisms. “If you can't explain it simply, you don't understand it well enough.” —Einstein (, Understanding Discrete vs. Asking for help, clarification, or responding to other answers. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. But if you're asking if it's spatially continuous then the answer is no. Hence, the temperature variable placed against the time variable would be represented by a continuous curve. It … Now that we know the continuous rate is -47.9% per year, we can work out how long until we're at 50%: This is a tricky one: the stock market changes every day, so it seems like it'd continuous, but there isn't an underlying predictable rate. Random variables represent quantities or qualities that randomly change within a population. This can be done regardless of whether the sample size is specified by a discrete unit or a continuous measure. Measurement Scale and Context For example, 8 bits have 256 possible values, and 16 bits have 65536. The natural log works on the ratio between the new and old value: $\frac{\text{new}}{\text{old}}$. Learn more about how features and surfaces can be represented as either discrete or continuous in ArcGIS. It is continuous if you consider the values. What to do to speed up the paper publication process? The measurements themselves are discrete, but how you represent them need not be. What's that Professor? For example, you could represent them as a continuous density surface: Or, as discrete 3D bar charts extruded from the census tracts (in this case a grid): A population count is a point measure. We can't have half a student! Why is population density a continuous data type when it is typically measured for aggregate areas such as census tracts or districts/neighbourhoods (ie, it can't be measured at any point on a surface like gradient or temperature). The population of a city is always a continuous variable as people are always moving to and from the city as well as births & deaths. To simulate a weather system, for example, the tracking occurs continuously as all elements are constantly changing. How to say "garlic", "garlic clove" and "garlic bulb" in Japanese? "12% interest per day" is different than "12% interest per year".). BetterExplained helps 450k monthly readers with friendly, intuitive math lessons (more). (Notice how the rate must be scaled to match the time period. Something areal data is not. If the population is discrete, this distribution is a stick (or bar) graph. ), How To Think With Exponents And Logarithms, Q: Why is e special? It is areal data as @Radar have said. Continuous Growth. Updated my answer. Sample Data Representing a Population Distributions. In Star Trek TNG Episode 11 "The Big Goodbye", why would the people inside of the holodeck "vanish" if the program aborts? The function itself need not be continuous. Ah, but the caveat is the application of a kernel function on the point data. A surface for which each location has a specified or derivable value. For a census tract you can ask the question: How many people exist per unit area? Why was the name of Discovery's most recent episode "Unification III"? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. The ratio between new and old was 37/53, so ln(37/53) = -.359 = -35.9% continuous growth over our time period. What's the best way to produce a density map from weighted points in QGIS? For a discrete distribution, probabilities can be assigned to the values in the distribution - for example, "the probability that the web page will have 12 clicks in an hour is 0.15." Argh! Enjoy the article? For a spatial application the consideration of interpolation is important. (Continuous growth requires a smaller rate because of compounding.). Discrete and Continuous Data. We can take them, split them into smaller, more frequent changes, and spread them out. In our case, we grew from 1 to 2, which means our continuous growth rate was ln(2/1) = .693 = 69.3%. $2^n$ (where n is an integer) models discrete scenarios like coin flips or binary digits. Are you telling me the bacteria colony just happens to have a continuous rate of precisely ln(2) over the course of a day? elevation). There are two definitions of continuous data ( at least that I could find online). find the population density of some points in qgis/postgis, Calculating population density from urban raster reclasiffy, How to create a population density heatmap from attributes, Find average density value within a polygon in ArcMap, Sources for population density map of Europe. For example, the half-life of Carbon-14 is 5700 years. These practice problems focus on distinguishing discrete versus continuous random variables. A clear understanding of the difference between discrete and continuous data is critical to the success of any Six Sigma practitioner. +1 This answer addresses both spatial and non-spatial forms of data continuity, which is important considering the forum in which the question was asked. To learn more, see our tips on writing great answers. What are the possible outcomes? (The model gets more complex as you account for how long it takes for cubs to have children of their own.). We see a lot of jumpy changes, and sample them at yearly intervals to see how we're doing. I'd prefer you told me the colony doubled while a grad student stared at a petri dish for 24 hours straight. None of this "wait until we decay by 50% so humans can count it easier" nonsense. Let’s see the definition: Continuous data is information that could be meaningfully divided into finer levels. A continuous distribution is one in which data can take on any value within a specified range (which may be infinite). What does “blaring YMCA — the song” mean? Population ecology is the study of population and how they change over time in relation to the environment including the environmental influences on population density and distribution, age structure and variations in population size. A discrete variable is a number that can be counted. Breeding seasons introduce some delay in the regulative process. The natural log finds the continuous rate behind a result. One of my pet peeves were problems like "A bacteria colony doubles after 24 hours...". Discrete and Continuous Data. Density is an areal measure and in itself implies that there exists a container variable (e.g. All data that are the result of counting are called quantitative discrete data. I visualize change as events along a timeline: Discrete changes happen as distinct green blobs. With discrete growth, we can see change happening after a specific event. Occasionally when analyzing data between groups or variables there are limitations on the validity of measured tests due to the lack of evidence, data, or information because a sample size is too small or has low value. Example: Material X decayed from 53kg to 37kg over 9 months. We can't point to an event and say "It changed here". The radioactive material is changing every instant. Automating calculation of Population Density in Census Tracts? jar that holds water, or in this case a census tract that holds people). Discrete Data. (1.98kg... 1.99kg... 2.00kg. If your system does change continuously, why not provide the continuous rate and write $e^{\ln(2) x}$? Should live sessions be recorded for students when teaching a math course online? Show Answer. This happened over 9 months, so the monthly continuous rate is -35.9/9 = -3.98%. Using $e$ as a base ($e^{\text{rate} \cdot \text{time}}$) implies you want people to think about change that happens at every moment. (Brush up on the number e and the natural logarithm.). I wrote this post because my video on e had questions about how $2^x$ represents "staircase growth". This is similar to saying the average family has 2.3 kids.). That is, the function's domain is an uncountable set.
## is population discrete or continuous
Backyard System Of Poultry Rearing, Anti American Synonym, Sm Megamall Cyberzone Store Directory, Dollar Tree Pharmacy, Gilchrist County Map, Haier Window Air Conditioner, Roasted Eggplant Baby Led Weaning, How To Make Mcdonald's Habanero Sauce, | 2021-03-02T23:06:34 | {
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https://mathoverflow.net/questions/408975/the-chromatic-number-of-the-union-of-two-graphs | The chromatic number of the union of two graphs
Let $$G_n$$ be the graph on the set of all binary strings of length $$n$$ with two strings adjacent whenever they are Hamming distance $$2$$ away from each other, or one of them lies below another one; thus, for instance, $$G_2=K_4$$, and $$G_3$$ has $$25$$ edges. What is the chromatic number of $$G_n$$?
There is a simple, but not quite obvious construction showing that $$\chi(G_n)\le n(n+1)/2+1$$, and I am interested in a matching lower bound. Computations give $$\chi(G_1)=2,\ \chi(G_2)=4,\ \chi(G_3)=5,\ \chi(G_4)=9,$$ $$\chi(G_5)=12,\ \chi(G_6)=16,\ \text{and}\ \chi(G_7)\le 17.$$ (The first five values are easy to compute, the last two are reported by Gordon Royle in the comments.)
An update. The colorings I am interesting in are of a special nature: if the strings $$s$$ and $$t$$ are same-colored, then (identifying strings with sets) also $$s\setminus t$$ and $$t\setminus s$$ are same-colored. Let $$\chi^*(G)$$ denote the smallest number of colors needed to properly color the graph $$G$$ in this special way. Is it true that $$\chi^*(G_n)=(1/2+o(1))n^2$$? That, say, $$\chi^*(G_n)>(1/4+o(1))n^2$$?
• @JoshuaZ: sure; color each vertex with the scalar product of the corresponding string and the vector $(1,2,\dotsc,n)$.
– Seva
Nov 20 at 16:36
• what do you mean by lying below? Nov 21 at 8:48
• @FedorPetrov: one of the corresponding subsets of the $n$-element set being contained in another one.
– Seva
Nov 21 at 9:03
• Here is a nice $17$-colouring of $G_7$ where I am using subsets of $\{0,1,\ldots,6\}$ to represent the vertices. Let $\sigma = (0,1,2,3,4,5,6)$ be a cyclic permutation and let $\tau$ be the complement map on the subsets. Let $F= \{013, 124, 235, 346, 450, 561, 602\}$ be the Fano plane and set $C = F \cup F^\tau$. Then take $D = \{0, 16, 25, 34, 124, 135, 236, 456\}$ and its seven rotations under powers of $\sigma$, and $D^\tau$ and its seven rotations under powers of $\sigma$. Then one colour class for each of $\emptyset$ and the entire set, making a total of 17 classes. Nov 22 at 7:04
• So $G_6$ has chromatic number $16$ - this took quite a long time to compute, almost all of which was spent searching in vain for a $15$-colouring. I think it will be hard to get close to an exact answer. Nov 23 at 2:29
This is a comment, not an answer, but it will be more convenient to post it as an answer. Consider the vertices of $$G_n$$ as subsets of $$[n]=\{1,\dots,n\}$$.
Observation 1. $$\chi(G_n)\ge\left\lfloor\frac{3n}2\right\rfloor+1$$.
This is because $$G_n$$ contains a clique of that size, namely,
$$\varnothing,\{1\},\{2\},\{1,2\},\{1,2,3\},\{1,2,4\},\{1,2,3,4,\},\{1,2,3,4,5\},\dots$$.
Observation 2. The OP noted that $$\chi(G_n)\le\binom{n+1}2+1=\frac{n^2+n+2}2$$. For odd $$n$$ this bound can be improved to $$\chi(G_n)\le\frac{n^2+3}2$$.
The vertices $$\varnothing$$ and $$[n]$$ get their own colors. If $$1\le|X|\le n-1$$, color $$X$$ with the ordered pair $$\left(\sum_{x\in X}x\mod(n+1),\ \left\lceil\frac{|X|}2\right\rceil\right).$$ This is a proper coloring, so (assuming $$n$$ is odd) $$\chi(G_n)\le2+(n+1)\cdot\frac{n-1}2=\frac{n^2+3}2.$$
Observation 3. The clique number $$\omega(G_n)$$ is exactly $$\left\lfloor\frac{3n}2\right\rfloor+1$$.
We showed in Observation 1 that $$\omega(G_n)\ge\left\lfloor\frac{3n}2\right\rfloor+1$$. Let us prove by induction that $$\omega(G_n)\le\left\lfloor\frac{3n}2\right\rfloor+1$$.
We may assume $$n\ge2$$. Let $$\mathcal C$$ be a maximal clique in $$G_n$$. Let $$\mathcal C_h=\{X\in\mathcal C:|X|=h\}$$. Choose $$h$$ so that $$0\lt h\lt n$$ and $$\mathcal C_h\ne\varnothing$$.
If $$\bigcup\mathcal C_h=[n]$$ and $$\bigcap\mathcal C_h=\varnothing$$, then $$|\mathcal C|=|\mathcal C_h|+2\le n+2\le\left\lfloor\frac{3n}2\right\rfloor+1$$. Otherwise, choose $$X\in\{\bigcup\mathcal C_h,\ \bigcap\mathcal C_h\}$$ so that $$0\lt|X|\lt n$$, and let $$k=|X|$$. Then every element of $$\mathcal C$$ is a subset or superset of $$X$$. Since $$0\lt k\lt n$$, by induction we have $$|\mathcal C|\le\omega(G_k)+\omega(G_{n-k})-1\le\left\lfloor\frac{3k}2\right\rfloor+\left\lfloor\frac{3(n-k)}2\right\rfloor+1\le\left\lfloor\frac{3n}2\right\rfloor+1.$$
• Some more computational snippets: the cliques you found are the unique maximum clique (up to equivalence under the automorphism group) for $G_4$, $G_6$ and $G_8$. For $G_5$, $G_7$ and $G_9$ there are other classes of equally-large cliques, but none larger. Nov 25 at 2:56
• So I liked your original proof that if $|\mathcal{C}_h| = 1$ for some $h$, then the result follows. Did you decide that it was not so obvious to prove that there must be a layer containing just one vertex of the clique? Nov 26 at 5:11 | 2021-11-29T08:45:37 | {
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https://math.stackexchange.com/questions/1633313/how-to-solve-for-x-z-and-y-z-here | # How to solve for x/z and y/z here?
I got stuck solving these two equations: $$a_1(x/z) + b_1 (y/z) + c_1 = 0$$ and, $$a_2(x/z) + b_2 (y/z) + c_2 = 0$$ for $$x/z$$ and $$y/z$$.
The desired result would be: $${x \over z} = {b_1c_2 - b_2c_1\over a_1b_2 - a_2b_1}$$,
$${y \over z} = {c_1a_2 - c_2a_1\over a_1b_2 - a_2b_1}$$
How do I get there? I keep getting dead ends.
Edit: I basically came this far: $$u = v{-b_1 \over a_1} - {c_1 \over a_1}$$ and, $$u = v{-b_2 \over a_2} - {c_2 \over a_2}$$
And ofcourse $$u = x/z$$, $$v = y/z$$
Now i got here, $$v{-b_1 \over a_1} - {c_1 \over a_1} = v{-b_2 \over a_2} - {c_2 \over a_2}$$
Then I took all the v's to one side and factored v outside: $$v({-b_2 \over a_2} + {b_1 \over a_1}) = {c_1 \over a_1} - {c_2 \over a_2}$$
now divide both sides by that b and a thingy on the left, but now it gets messy...
• Can you show which dead end you run into? It should be simple enough to substitute your proposed solutions into the equations and simplify to show that they work. Jan 30 '16 at 16:08
• Rename $\frac xz=u$ and $\frac yz=v$ and rewrite the initial equations. Jan 30 '16 at 16:08
• Hint: take the last thing, check that it is correct (the second term on the left should be $\frac{b_2}{a_2}$, and the plus on the right should be a minus), then multiply by $a_1a_2$ on both sides then solve for $v$. You just obtained $\frac yz$ as in the solution you give :). Edit You just fixed the sign while I typed. Jan 30 '16 at 16:44
• I suppose I am stuck on how to divide c1/a1 - c2/a2 by that thing on the left now. Any clever way to do that? Jan 30 '16 at 16:56
• See my answer below. I started from a little earlier than that equation, but to make the division, I suggest you multiply both sides by $a_1a_2$, and then divide. The multiplication eliminates all denominators. This is why it is advisable. Jan 30 '16 at 17:04
## 2 Answers
HINT: with $$\frac xz=u$$ and $$\frac yz=v$$ we get $$a_1u+b_1v+c_1=0$$ $$a_2u+b_2v+c_2=0$$ if $$b_1\ne 0$$ we have $$v=-\frac{c_1}{b_1}-\frac{a_1}{b_1}u$$ and we get an equation for $$u$$ $$a_2u+b_2\left(-\frac{c_1}{b_1}-\frac{a_1}{b_1}u\right)+c_2=0$$ can you proceed?
• it seems we get a different thing, did I do something wrong in my original post? I edited it in. Jan 30 '16 at 16:42
• I urge you to decide if you mean to use slash fractions $a/b$ or normal fractions $\frac ab$, because mixing the two can be quite confusing for readers. @TheProgramMAN123 I think you just followed a different path than he did. You solved both for $u$ and then equalled the two expressions, he solved one for $v$ and then substituted the expression in the other one. Jan 30 '16 at 16:48
• If you solve the last equation in this post for $u$, you get the solution in the question, with signs changed both in numerator and in the denominator. Jan 30 '16 at 16:51
Let me just continue your attempt. You got to:
$$\left\{\begin{array}{c} u=-\frac{b_1}{a_1}v-\frac{c_1}{a_1} \\ u=-\frac{b_2}{a_2}v-\frac{c_2}{a_2} \\ \end{array}\right.$$
If we equal these two expressions for $u$ and multiply both sides by $a_1a_2$, we get:
$$-b_1a_2v-c_1a_2=-b_2a_1v-c_2a_1.$$
Bringing the RHS's $v$ term to the left and the LHS's "constant" term to the right, this becomes:
$$v(a_1b_2-a_2b_1)v=a_2c_1-a_1c_1.$$
Yes, I also factored out $v$ from the LHS and reordered the factors in the various $a,b,c$ products. Is this not precisely the solution you gave, that is:
$$\frac yz=v=\frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}?$$
Wonderful! To complete, we substitute this into the equation for $u$, one of the two, I mean. To simplify the calculations, we multiply by $a_1$ the first equation of the system at the start of the answer, and then substitute, getting:
$$a_1u=-b_1\frac{a_2c_1-a_1c_2}{a_1b_2-a_2b_1}-c_1.$$
Let us make that RHS into a single fraction:
$$a_1u=\frac{\overline{-b_1a_2c_1}+b_1a_1c_2-c_1a_1b_2+\overline{c_1a_2b_1}}{a_1b_2-b_1a_2}=\frac{a_1(b_1c_2-c_1b_2)}{a_1b_2-b_1a_2}.$$
The overlines simply indicate those two terms cancel. Oh, but if we divide both sides by $a_1$, we get the desired result, don't we? Very good. We are done.
Bottom line: when you have coefficients with fractions, the trick is often to sum all fractions and see if you can multiply/divide to simplify numerators or denominators. Don't be afraid of complicated coefficients: just have patience and do the algebra to the end, and you will get to the desired result :).
• thanks a bunch! I guess the lesson for me here is to remove those fractions as soon as you can. They make things confusing and messy. i guess multiplying both sides by a1a2 is a clever way to do that. the solution really is painfully simple then. Jan 30 '16 at 18:29 | 2022-01-18T13:47:36 | {
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http://cloudtools.it/jkuq/moment-of-inertia-of-triangle.html | After we have found the second moment of inertia about an axis, we can find it about another parallel axis using the parallel axis theorem. The matrix of the values is known as the moment of inertia tensor. - The formula for moment of inertia is - If there are 3 particles of mass 'm' placed at each of the vertex of this equilateral triangle then we consider three times m. Equation 18) also holds for polar moments of inertia i. The moment of inertia (I) of a body is a measure of its ability to resist change in its rotational state of motion. Similar to the centroid, the area moment of inertia can be found by either integration or by parts. EHE-08): Where: Mf = Mcrk = Nominal cracking moment of the cross section. Doing the same procedure like above, and below is the work. 3/2MR^2 The center of mass of the original triangle (the part that has been cut out) was at the center of the circle, at a distance R from the pivot. Now based on symmetry you can apply the definition of the moment of inertia to calculate the moment of inertia about the y axis which equals the cendroidal y axis. Find the moment of inertia of the system of particles about a perpendicular bisector of the meter scale. dI = r2dm (1) (1) d I. Considering an element DE parallel to y-axis at a distance x from origin and width dx. Various such parameters include centre of gravity, moment of inertia, centroid , first and second moment of inertias of a line or a rigid body. So if the moment of inertia of the rectangle is, about its centroid, is bh cubed over 12, and the moment of inertia of the hole, the circle, from the previous tables is pi r to the 4th, over 4. Beam Deflection Equations are easy to apply and allow engineers to make simple and quick calculations for deflection. 000965387 kg*m^2. A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle. • Th t fi ti fth hdd iThe moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle Determine the moment of inertia of the shaded area with respect to the x axis. This is a simulation of five objects on an inclined plane. Moments of Inertia of Geometric Areas Frame 28-1 * Introduction This unit will deal with the computation of second moments, or moments of inertia, of The general expression for the moment of inertia of a right triangle about a centroidal axis parallel to a side is. In order to continue, we will need to find an expression for dm. Central axis of hallow cylinder. T 1 – the instantaneous value of load torque, referred to a motor shaft, N-m. A framework, in the shape of an equilateral triangle ABC, is formed by rigidly joining three uniform rods, each of mass m and length 2a. Table 8-2 Gross and Cracked Moment of Inertia of Rectangular and Flanged Section b d nA s kd n. Rectangle Triangle. Moment of Inertia of Mass, Mass Moment inertia of Common Shapes page Sideway Output on 28/4. The angle in between the masses is 60 degrees. A: area of the shape. calculate the moment of inertia when the plate is rotating about an axis perpendicular to the plate and passing through the vertex tip. The calculator has been provided with educational purposes in mind and should be used accordingly. University. Angular momentum. The moment of inertia of an area with respect to any given axis is equal to the moment of inertia with respect to the centroidal axis plus the product of the area and the square of the distance between the 2 axes. Rolling without slipping problems. The equation of the line is $y = \dfrac{a}{b} x + a$. 8680 rad/s^2 α_down -0. Determine polar moment of inertia of an isosceles triangle 1 answer below » Polar Moments of Inertia Determine the polar moment of inertia I P of an isosceles triangle of base b and altitude h with respect to its apex (see Case 5, Appendix D). If you're unsure about what deflection actually is, click here for a deflection definition Below is a concise beam deflection table that shows how to calculate the maximum deflection in a beam. 025kg) g = gravity (9. The Poisson's ratio of the shaft material is , the moment of inertia about the y axis is , and the applied force at the end of the shaft is P. The SI unit of moment of inertia is kg m2. 2 x 10⁻³ kg*(0. Various such parameters include centre of gravity, moment of inertia, centroid , first and second moment of inertias of a line or a rigid body. Mass Moment of Inertia, I G (cont’d) I G for a body depends on the body’s mass and the location of the mass. Axis through center. Mass Moment of Inertia - Mass Moment of Inertia (Moment of Inertia) depends on the mass of the object, its shape and its relative point of rotation - Radius of Gyration Pipe Formulas - Pipe and Tube Equations - moment of inertia, section modulus, traverse metal area, external pipe surface and traverse internal area - imperial units. The following is a list of second moments of area of some shapes. m = point mass. 17/12mL2 2. We found the moment of inertia of the apparatus alone to be 0. Presented here is a table of formulas which permit direct solution for required moment of inertia for several simple loading cases, for the two most common deflection criteria, L/240 and L/360. ) is the moment of inertia about the centroid of the area about an x axis and d y is the y distance between the parallel axes Similarly 2 y I y Ad x Moment of inertia about a y axis J Ad 2 o c Polar moment of Inertia 2r 2 d 2 o c Polar radius of gyration 2 r 2 d 2 Radius of gyration. To find the perimeter of the triangle, you simply need to add together the lengths of the base and the two sides. The moment of inertia matrix is referred to the principal axes, again frame O 2 and the products of inertia are zero. This engineering data is often used in the design of structural beams or structural flexural members. In mathematical notation, the moment of inertia is often symbolized by I, and the radius is symbolized by r. And I was wondering whether someone could give me some more information/examples on first and second moment of area (tech calculus, wouldnt let me. It is measured by the mass of the body. Now to calculate the moment of inertia of the strip about z-axis, we use the parallel axis theorem. Derivation of moment of inertia of triangle and cone. Some examples of simple moments of inertia Let's try an easy calculation: what's the moment of inertia of these three balls? Each ball has mass m = 3 kg, and they are arranged in an equilateral triangle with sides of length L = 10 m. We will use the parallel axis theorem and we will take the centroid as a reference in this case. 0 kg per leg. d ' (n -1)A 's Gross Section Cracked Transformed Section Gross and Cracked Moment of Inertia b h A's As b h As b bw hf h b h hf y t A's As b d nA s kd n. If you're unsure about what deflection actually is, click here for a deflection definition Below is a concise beam deflection table that shows how to calculate the maximum deflection in a beam. I need to calculate the change in moment of intertia due to modifing a simple angled beam from 120 x 120 x 10 to 120 x 112 x 10. Considering an element DE parallel to y-axis at a distance x from origin and width dx. A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle. Equation 18) also holds for polar moments of inertia i. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. half the value of the moment of inertia about the central axis to the value of the moment of inertia about the base plane. About the Moment of Inertia Calculator. University. Answer this question and win exciting prizes. The moment of inertia of a body is its tendency to resist rolling motions and angular accelerations. How to calculate the moment of inertia of a triangular plate rotating about the apex. Kinetic Energy is the energy possessed by an object because it is in motion. Let the mass of the triangle be M. What is the moment of inertia of this triangle for rotation about an axis that is perpendicular to the plane of the triangle and through one of vertices of the triangle? The moment of inertia of a rod rotated about its center of mass is Irod, cm =1/12mL2. ): 6391 Instructor: E-mail: [email protected] Topic - Moment of Inertia ,Ans - (Mh^2)/6. 9803 rad/s^2. 3 (4) 3 Determine the AP whose fourth term is 15 and the difference of 6th term from 10th term is 16 Prove that ratio of area of two triangle is equal to the square of the corresponding sides. Center of mass, moments of inertia, volume of a body of rotation. A framework, in the shape of an equilateral triangle ABC, is formed by rigidly joining three uniform rods, each of mass m and length 2a. Calculate the 2nd moment of area for each element about the reference axes. ANSWER: Right angled triangle. Second moment of area for triangle trough x-axis = (ah3)/36. Calculate the moment of inertia of strait angle triangle about its $$y$$ axis as shown in the Figure on the right. Moment of Inertia of a Triangular Lamina about its Base. It appears in the relationships for the dynamics of rotational motion. Two conditions may be considered. The moment of the large triangle, with side $$2L$$, is $$I_z(2L)$$. The mass moment of inertia of an object about an axis (𝑎) is equal to the moment about an axis (𝑏) through the. l : moment of inertia about the axis parallel to x-axis. The moment of inertia with respect to central longitudinal axis would be m r2/2 m r2/3 m r2/6 m r2/12 The ratio of moment of inertia of a rectangle and that of a triangle, having same base and height, with respect to their bases would be 2 : 1 3 : 1 4 : 1 6 : 1 The ratio of the moment of inertia of a triangle of base width b and height h with. Find the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices. They are; Axis passing through the centroid. The units of the product of inertia are the same as for moment of inertia. In addition to the moment of inertia, the product of inertia is commonly used. this apparently involves integrals but we havent even touched integrals in calculus need help please. Four leg-loading conditions were employed: 1) no load (NL) on the legs; 2) a baseline load (BSLN) condition, with a mean of 2. dI y (dy)x 3 = 3 5. suppose the mean speed of such molecule in a gas 500 m per second and its kinetic energy of rotation is 2/3 of its kinetic energy of translation. Ix = b h3 / 36 (4a) Iy =h b3 / 36 (4b) Area Moment of Inertia for typical Cross Sections I. Calculating Moment Of Inertia Of A Triangle. 8·10-2 Kg·m2 Submit Figure < 1of1 Incorrect; Try Again: 3 Attempts Remaining Part B What Is The Triangle's. The mass and moment of inertia through the sphere's center of mass are given. Each "typical" rectangle indicated has width dx and height y 2 − y 1, so its area is (y 2 − y 1)dx. When that happens equation 4 and 5 would be used to calculate the stress and. Inertia is a property of a body to resist the change in linear state of motion. Polar Moment of Inertia for Circular Cross-section. In particular, the same object can have different moments of inertia when rotating about different axes. Angular momentum. Weld design Moment of inertia of fillet weld J [mm 4 , in 4 ] Position of center of gravity of weld group section J = π a (r + a / 2) 3 - Meaning of used variables: a fillet weld height [mm, in] B width of weld group [mm, in] H height of weld group [mm, in] L weld length [mm, in] r weld radius [mm, in] s web thickness [mm, in] t flange thickness [mm, in]. Moment of inertia can be defined by the equation The moment of inertia is the sum of the masses of the particles making up the object multiplied by their respective distances squared from the axis of rotation. (8), derived in the moment of inertia example, the moment of inertia of the disk is = at 5 digits Therefore, the moment of inertia of the disk is 12. The product moment of an area A of a right angle triangle about the axes xy is Product Moment of Inertia of a Right Angle Triangle by Parallel-axis Theorem. Question: Three point masses, each of mass {eq}m {/eq}, are placed at the corners of an equilateral triangle of side {eq}L {/eq}. If the polar moment of inertia is calculated at the centroid of the area, it is denoted. The second moment of area, also known as moment of inertia of plane area, area moment of inertia, polar moment of area or second area moment, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. Answer this question and win exciting prizes. What would the moment of inertia of a thin equilateral triangular sheet of mass M and sides S be with respect to an axis through one vertex perpendicular to the sheet? I got 3Ms^2/4 but I'd like to know if I'm right. Table of Selected Moments of Inertia Note: All formulas shown assume objects of uniform mass density. Let M represent the mass of the triangle and L the length of the base of the triangle. More on moment of inertia. The moment of inertia of the triangle about the point O is I = M*r^2 , where r is the distance of the center of mass from O. Calculate the moment of inertia of a thin plate, in the shape of a right triangle, about an axis that passes through one end of the hypotenuse and is parallel to the opposite leg of the triangle, as in Figure P10. The polar moment of inertia JO of an area about O and the polar moment of inertia JC of the area about its c d o centroid are related to the distance d between points C and O by the relationship J O = J C + Ad 2 The parallel-axis theorem is used very effectively to compute the moment of inertia of a composite area with respect to a given axis. Solve: The moment of inertia of the triangle is I mr= × = =3 3(0. The mass moment of inertia of an object about an axis through the center of mass is smaller than that about any other axis in the same direction. In mathematical notation, the moment of inertia is often symbolized by I, and the radius is symbolized by r. DIY Brick Rocket Stove - Cooking Without Electrical Power - Duration: 23:40. 0580 kg, an inner radius of 0. Center of mass, moments of inertia, volume of a body of rotation. Moment of Inertia and Polar Moment of Inertia are both the quantities expressing a body’s tendency to resist changes when certain torque is being applied. \end{equation*} If the object is made of a number of parts, each of whose moment of inertia is known, the total moment of inertia is the sum of the moments of inertia of the pieces. In order to find the moment of inertia of the triangle we must use the parallel axis theorem which ius as follows: The moment of inertia about any axis parallel to that axis through the center. All the formulas for a triangle are in any handbook. Area Moment of Inertia - Filled Right Triangle Solve. Rolling without slipping problems. When that happens equation 4 and 5 would be used to calculate the stress and. We are concerned here with area only and the area multiplied by a distance twice is the second moment of area. The inertia of both systems can be found using the equation: m = mass of hanging mass (0. 0 revolutions per s or 10 rad/s. Let be the position vector of the th mass element, whose mass is. Explanation: No explanation is available for this question! 2) What is the C. Look up I for a triangle in your table if you have forgotten. If k is the mass per unit area, then each typical rectangle has mass k(y 2 − y 1)dx. From the diagram below, we have:. Moment of inertia of this disc about the diameter of the rod is, Moment of inertia of the disc about axis is given by parallel axes theorem is, Hence, the moment of inertia of the cylinder is given as, Solid Sphere a) About its diameter Let us consider a solid sphere of radius and mass. Now that we have determined the moments of inertia of regular and truncated equilateral triangles, it is time to calculate them for the corresponding right prisms. This works especially well when the general shape of the area can be decomposed into simpler shapes for which the moment of inertia is calculated for. Further explanation. We have step-by-step solutions for your textbooks written by Bartleby experts!. Moment of inertia, also called mass moment of inertia or the angular mass, (SI units kg m 2) is a measure of an object’s resistance to changes in its rotation rate. Data: 23 d'abril de 2006 (original upload date) Font: No machine-readable source provided. 94 into 10 to the power of minsis 46 kg metre square bout an Axis through its Centre perpendicular to the lines joining the two atoms. Related Questions. The units. See how the eigenvectors of the inertia tensor change as you change a configuration of point masses, or the shape of a solid plate of material. Moment of inertia of an equilateral triangular lamina ABC, about the axis passing through its centre O and perpendicular to its plane is Io as shown in the figure. Radius and elevation of the semi-circle can be changed with the blue point. 2) A long rod with mass has a moment of inertia , for rotation around an axis near one. For example, given the axis O-O and the shaded area shown, one calculates the second moment of the area by adding together for all the elements of area dA in the shaded area. 4)and the second moment of the area about the y. Find the moment of inertia for the following about the y axis and x axis of a right triangle whose base is on the +x axis and whose height is on the +yaxis Source(s): moment inertia triangle: https://shortly. The more far away from the axis, the more moment of inertia the object has. The coordinate variables are x and y, respectively. Area A = 200 mm x 100 mm = 20000 mm2 I x. In other words, the centroid will always be 2/3 of the way along. Whatever kind you are trying to compute I would suggest breaking up the cross section into triangles with two vertices on successive points of your boundary and the third at the center about which the moment of inertia is to be taken. Asked in Algebra, Geometry. \end{equation*} If the object is made of a number of parts, each of whose moment of inertia is known, the total moment of inertia is the sum of the moments of inertia of the pieces. The mass and moment of inertia through the sphere's center of mass are given. Going to the division, we get. When the axes are such that the tensor of inertia is diagonal, then these axes are called the principal axes of inertia. svg 512 × 569; 4 KB. The moment of inertia of a uniform object depends not only on the size and shape of that object but on the location of the axis about which the object is rotating. 91, b < 10a. Angular acceleration of the system + triangle (long base) α_up 0. Q: Moment of Inertia of a thin spherical shell of mass m and radius r about its diameter is a) mr²/3 b) 2mr²/3 c) 2mr²/5 d) 3mr²/5 Q: Moment of inertia of a triangular section of base b and height h about an axis passing through its. The force of attraction is proportional to mass of the body. calculate the moment of inertia when the plate is rotating about an axis perpendicular to the plate and passing through the vertex tip. Radius of Gyration for a equilateral triangle can be calculated as. 3 (4) 3 Determine the AP whose fourth term is 15 and the difference of 6th term from 10th term is 16 Prove that ratio of area of two triangle is equal to the square of the corresponding sides. Weld design Moment of inertia of fillet weld J [mm 4 , in 4 ] Position of center of gravity of weld group section J = π a (r + a / 2) 3 - Meaning of used variables: a fillet weld height [mm, in] B width of weld group [mm, in] H height of weld group [mm, in] L weld length [mm, in] r weld radius [mm, in] s web thickness [mm, in] t flange thickness [mm, in]. We are concerned here with area only and the area multiplied by a distance twice is the second moment of area. 3) Three particles each of mass 100 g are placed at the vertices of an equilateral triangle of side length 10 cm. The mass moment of inertia of an object about an axis (𝑎) is equal to the moment about an axis (𝑏) through the. How you find moment of inertia of isosceles triangle? Wiki User 2014-05-12 13:36:50. After we have found the second moment of inertia about an axis, we can find it about another parallel axis using the parallel axis theorem. 5626 x 10⁻³ initial inertia * angular velocity = new inertia * angular velocity. 2012/2013. Sometimes, we need to find the moment of inertia of an object about the origin, which is known as the polar moment of inertia. It is the rotational analog of mass. The moment of inertia of the rod is simply $$\frac{1}{3} m_rL^2$$, but we have to use the parallel-axis theorem to find the moment of inertia of the disk about the axis shown. We want to find the moment of inertia, I y of the given area, which is rotating around the y-axis. In order to continue, we will need to find an expression for dm. A = bh ¸ 2 Ic = bh 3 ¸ 36 Base on x-axis, centroidal axis parallel to x-axis: x = h ¸ 3 Ax = bh 2 ¸ 6 Ix = bh 3 ¸ 12 x-axis through vertex, Base and centroidal axis parallel to x-axis: x = 2h ¸ 3 Ax = bh 2 ¸ 3 Ix = bh 3 ¸ 4. 16-24 From: Rabiei. The z2A term is the moment of inertia that area A would have about the y axis if all of the area were to be concentrated at the centroid. The moment of inertia block, which is a table containing the results of the moment of inertia calculation, is displayed and can be inserted anywhere in the drawing. £20 £200 £40. This works especially well when the general shape of the area can be decomposed into simpler shapes for which the moment of inertia is calculated for. The equilateral triangle actually makes the strongest column for a given area, but not by much (12% stronger than the circle). Rotational kinetic energy. Ix = b h3 / 36 (4a) Iy =h b3 / 36 (4b) Area Moment of Inertia for typical Cross Sections I. Relative to principal axes of inertia, the product of inertia of a figure is zero. • Th t fi ti fth hdd iThe moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle Determine the moment of inertia of the shaded area with respect to the x axis. Determine polar moment of inertia of an isosceles triangle 1 answer below » Polar Moments of Inertia Determine the polar moment of inertia I P of an isosceles triangle of base b and altitude h with respect to its apex (see Case 5, Appendix D). DIY Brick Rocket Stove - Cooking Without Electrical Power - Duration: 23:40. Area, center of mass, moments of inertia. This theorem is really powerful because the moment of inertia about any set of axes can be found by finding the moment of inertia about the centroidal axes and adding the distance-area term to it. The right triangle comes along frequently in geometry. When that happens equation 4 and 5 would be used to calculate the stress and. Mass and moment of inertia properties of accessory components. This calculates the Area Moment of Inertia of a semi-circle about various axes. The following effective moment of inertia expression was originally proposed by Branson [9] and was adopted by ACI [10] and presented as (2). The help tool instructs me to click on the inspect menu and choose AREA. 1 DefinitionsThe second moment of the area about the x axis (IX) is defined as:I X = ∫ y 2 dA (11. EHE-08): Where: Mf = Mcrk = Nominal cracking moment of the cross section. The second moment of area is also known as the moment of inertia of a shape. PARALLEL-AXIS THEOREM, RADIUS OF GYRATION & MOMENT OF INERTIA FOR COMPOSITE AREAS. Synchronised Similar Triangles for Three-Body Orbit with L = 0 4 where (i,j,k) runs for the cyclic permutations of (1,2,3). Relative to principal axes of inertia, the product of inertia of a figure is zero. I = Second moment of area, in 4 or mm 4; J i = Polar Moment of Inertia, in 4 or mm 4; K = Radius of Gyration, in or mm; P = Perimeter of shape, in or mm; S = Plastic Section Modulus, in 3 or mm 3; Z = Elastic Section Modulus, in 3 or mm 3; Online Parabolic Half Property Calculator. Table of Selected Moments of Inertia Note: All formulas shown assume objects of uniform mass density. Is there a way to calculate this to X-X? The instructions o. 3 (4) 3 Determine the AP whose fourth term is 15 and the difference of 6th term from 10th term is 16 Prove that ratio of area of two triangle is equal to the square of the corresponding sides. Determine polar moment of inertia of an isosceles triangle 1 answer below » Polar Moments of Inertia Determine the polar moment of inertia I P of an isosceles triangle of base b and altitude h with respect to its apex (see Case 5, Appendix D). \] The moment of inertia of the area about the center can be found using in equation (40) can be done in two steps first calculate the moment of inertia in this coordinate system and then move the coordinate system to center. Figure 2: Deriving an equation for moment of inertia of the triangle rotating around its base. The moment of inertia of an area with respect to any given axis is equal to the moment of inertia with respect to the centroidal axis plus the product of the area and the square of the distance between the 2 axes. Product of inertia for triangle. (6) Theorems of Moment of Inertia. Moment of inertia of the remaining part of lamina about the same axis is :. The mass moment of inertia of an object about an axis through the center of mass is smaller than that about any other axis in the same direction. Centroids & Moment of Inertia. o The moment of inertia of a triangular section of height h about its base is given as, I = bh 3 /12. Actual physical properties may vary due to tolerances which occur in the manufacturing process. The mass moment of inertia equation for a point mass is simply: I = mr 2. (8), derived in the moment of inertia example, the moment of inertia of the disk is = at 5 digits Therefore, the moment of inertia of the disk is 12. The moment of inertia must be specified with respect to a chosen axis of rotation. The product of inertia of triangle (a) with respect to its centroid is I ¯ x y = b 2 h 2 / 72. Right Triangle The output of this equation is the I x and I y components of the area moment of inertia when the triangle is defined to be in the x/y plane. It is formed by the intersection of the medians. This table provides formula for calculating section Area, Moment of inertia, Polar moment of inertia, Section modulus, Radius of gyration, and Centroidal distance, for various cross section shapes. Solve: The moment of inertia of the triangle is I mr= × = =3 3(0. 91, b < 10a. Question: Three point masses, each of mass {eq}m {/eq}, are placed at the corners of an equilateral triangle of side {eq}L {/eq}. (ii) Moment of inertia about new axes which is turned through an angle of 30 0 anticlockwise to the old axis. We see it in action all the time. Equation 18) also holds for polar moments of inertia i. It will help in deciding whether the failure will be on the compression face or on the tension face of the beam. When that happens equation 4 and 5 would be used to calculate the stress and. 18) I s = I c + Ad 2. In order to continue, we will need to find an expression for dm. The moment of inertia is ∑mi*ri²; all the m are the same = 0. The radius of gyration is the radius at which we consider the mass to rotate such that the moment of inertia is given by I = M k2. Four leg-loading conditions were employed: 1) no load (NL) on the legs; 2) a baseline load (BSLN) condition, with a mean of 2. With great regard for economy of words we call the expression written above "the moment of inertia of the area about the x axis" or I x for short. l : moment of inertia about the axis parallel to x-axis. 728(1) 30 (2) 0. Determine the moment of inertia of the area about the x and y. 9 µC, are located at the corners of an equilateral triangle as in the figure above. Going to the division, we get. Figure to illustrate the area moment of a triangle at the list of moments of inertia. Live Simple, Live Free - Tinyhouse Prepper Recommended for you. Figure to illustrate the area moment of a triangle at the list of moments of inertia. For the sake of one more bit of integration practice, we shall now use the same argument to show that the moment of inertia of a uniform circular disc about a. 2 Second Moment of Area11. The equation of the line is $y = \dfrac{a}{b} x + a$. Moment of Inertia of a Triangular Lamina about its Base. This simple, easy-to-use moment of inertia calculator will find moment of inertia for a circle, rectangle, hollow rectangular section (HSS), hollow circular section, triangle, I-Beam, T-Beam, L-Sections (angles) and channel sections, as well as centroid, section modulus and many more results. Moment of Inertia 5 An example of this is the concrete T-beam shown. The second moment of area is also known as the moment of inertia of a shape. Explanation: No explanation is available for this question! 2) What is the C. Mass Moment of Inertia Calculator in Excel, Pt. The distance of the center of mass of the triangle in its new position from the pivot is the same. Moment of inertia, also called mass moment of inertia or the angular mass, (SI units kg m 2) is a measure of an object’s resistance to changes in its rotation rate. Right: Triangles with centroidal axes re-positioned with respect to the x-axis. Moment of inertia can be defined by the equation The moment of inertia is the sum of the masses of the particles making up the object multiplied by their respective distances squared from the axis of rotation. It is the rotational analog of mass. Uniform circular lamina about a diameter. We are concerned here with area only and the area multiplied by a distance twice is the second moment of area. 707(h) to get the actual I, h being the weld size. It is required in the design of machines, bridges, and other engineering systems. Find the polar moment of inertia. Using these, the moment of inertia for the parallel axis can be calculated using the formula: The moment of inertia for rotation around the axis at the surface of the sphere is. Moment of Inertia of Isosceles Triangle Jalal Afsar October 25, 2013 Uncategorized No Comments Moment of Inertia of Isosceles triangle can be easily find out by using formulas with reference to x-axis and y-axis. The moment of inertia of the triangle about the point O is I = M*r^2 , where r is the distance of the center of mass from O. Moment of Inertia Tensor Consider a rigid body rotating with fixed angular velocity about an axis which passes through the origin--see Figure 28. The total moment of inertia is due to the sum of masses at a distance from the axis of rotation. If the triangle were cut out of some uniformly dense material, such as sturdy cardboard, sheet metal, or plywood, the centroid would be the spot where the triangle would balance on the tip of your finger. The Area Moment of Inertia equation, I = (b•h 3)/12 , (b 3 •h)/4 , computes the Area Moment of Inertia for a right triangle with right angle on right of the base. Question: What Is The Triangle's Moment Of Inertia About The Axis Through The Center? Express Your Answer To Two Significant Figures And Include The Appropriate Units. It is also popular as angular mass or rotational inertia of the given rigid body. A point mass does not have a moment of inertia around its own axis, but using the parallel axis theorem a moment of inertia around a distant axis of rotation is achieved. Angular acceleration of the system + triangle (long base) α_up 0. Worthy of note, in order to solve for the moment of inertia of the right triangular thin plate, we first had to measure the the triangle's mass, base length, and height. How you find moment of inertia of isosceles triangle? Wiki User 2014-05-12 13:36:50. Since this term is always zero or positive, the centroidal moment of inertia is the minimum moment of inertia with respect to all parallel axes. This table provides formula for calculating section Area, Moment of inertia, Polar moment of inertia, Section modulus, Radius of gyration, and Centroidal distance, for various cross section shapes. Four leg-loading conditions were employed: 1) no load (NL) on the legs; 2) a baseline load (BSLN) condition, with a mean of 2. Moment of Inertia of a Triangular Lamina about its Base. PARALLEL-AXIS THEOREM, RADIUS OF GYRATION & MOMENT OF INERTIA FOR COMPOSITE AREAS. The Questions and Answers of The three point masses each of mass m are placed corner of equilateral triangle offside of side l then moment of inertia of the system about an axis along one side of triangle is? are solved by group of students and teacher of Class 11, which is also the largest student community of Class 11. It is also known as rotational inertia. Moment of Inertia is defined as: $$I={\sum}mr^2$$ which in this case can be rewritten into an integral: $$I=\rho\int_A{r^2dA}$$ Since the shape of the triangle can't be described by one formula, you would have to split the integral into multiple sections. Moment of Inertia. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a Triangular Cross-Section. For a point mass, the moment of inertia is just the mass times the square of perpendicular distance to the rotation axis, I = mr 2. Axis through center. “The poles of inertia” is another way of saying “mass concentration centers”. Undeniable momentum, on any stage - anywhere. The second moment of inertia rectangle is the product of height and cube of width divided by 12. Write (but do not evaluate) an integral expressing the moment of inertia of the region between y= sinxand y= sinx(for 0 x ˇ) rotated around: (a)the xaxis; (b)the yaxis; (c)the zaxis. The computation of moments of inertia can often be. Chapter-3 Moment of Inertia and Centroid Page- 1 3. “twisting”) about a given axis due to an applied torque. The distance of the center of mass of the triangle in its new position from the pivot is the same. J z' = I x' + I y'. 3 × 10⁻⁵ kg. The plane figures (like triangle, quadrilateral, circle, trapezoid, etc. d' (n -1)A 's Without compression steel With. of inertia of the rectangle. Note that each component of the moment of inertia tensor can be written as either a sum over separate mass elements, or as an integral over infinitesimal mass elements. automatic weight calculator for rectangular, square, round, or hexagonal, plate, tube, bar, beams, sheet, rod and other engineering material shapes. Hallow cylinder. It is analogous to mass in that it is a measure of the resistance a body offers to torque or rotational motion. 2012/2013. We will look at each expression below. The domain of the triangle is defined by. Moment of Inertia of Triangle about its Base | Very Important. Let R be the triangle with vertices (0, 0), (1, 0), (1, √ 3) and density δ = 1. 1 RADIUS OF GYRATION k All rotating machinery such as pumps, engines and turbines have a moment of inertia. Purpose: Determine the moment of inertia of a right triangular thin plate around its center of mass, for two perpendicular orientations of the triangle. Finding the Centroid. Angular momentum. = Polar Moment of Inertia. Description. dV = dxdydz. Add to Solver. The number of revolutions that the shaft will make during this time is. Worthy of note, in order to solve for the moment of inertia of the right triangular thin plate, we first had to measure the the triangle's mass, base length, and height. Determine the moment of inertia of the triangle for rotation about an axis that bisects one of its angles. Derivation of the Moment of Inertia Formula Suppose a particle of mass m is attached to a pivot by a thin rod of length r. Rotations in 2D are about the axis perpendicular to the 2D plane, i. Moment of inertia of three uniform rods of mass M and length l joined to form an equilateral triangle, about an axis passing through one of its sides. It is also de ned as I= Z r2 dm (3) for a continuous distribution of mass. In either case, use of the formulas is cumbersome and prone to error, especially in converting to consistent units. Today we will see here the method to determine the moment of inertia for the triangular section about a line passing through the center of gravity and parallel to the base of the triangular section with the help of this post. Integrate to derive a formula for the moment of inertia for a general triangle. 025kg) g = gravity (9. Evaluation of Moments of Inertia 2008 Waterloo Maple Inc. But I don't know how to do that. Moment of Inertia of Isosceles Triangle Jalal Afsar October 25, 2013 Uncategorized No Comments Moment of Inertia of Isosceles triangle can be easily find out by using formulas with reference to x-axis and y-axis. The force of attraction is proportional to mass of the body. 91, b < 10a. The moment of inertia of the triangular shaped area is 3. What is the moment of inertia of ball about the axis of rotation AB? Ignore cord’s mass. 8680 rad/s^2 α_down -0. The inertia matrix (aka inertia tensor) of a sphere should be diagonal with principal moments of inertia of 2/5 mass since radius = 1. Since moment of inertia is to be determined about an axis of rotation and from the provided diagram, it seems that the student is interested in finding the moment of inertia about the side (AB) along y-axis. 9 µC, are located at the corners of an equilateral triangle as in the figure above. Calculate the mass moment of inertia of the triangular plate about the y-axis. 2 comments. Find the moment of inertia of a plate cut in shape of a right angled triangle of mass M side AC=BC=a about an axis perpendicular to the plane of the plate and passing. Units 9 to 17,are assigned to:-Estimation of the Moment of inertia for Right-angled triangle (about X,Y) &Product of inertia &Polar Moment of Inertia, the radius of gyrations, by using two ways of Estimations, for the two cases of a right-angle triangle. When the axes are such that the tensor of inertia is diagonal, then these axes are called the principal axes of inertia. The mass moments of inertia of an object about any parallel axes are identical. If you are consitent about which way you go around the triangle, the orgin can be anywhere, as it will subtract tnegative areas automatically. Central axis of hallow cylinder. The formula calculates the Moment of Inertia of a right triangle of base b and height h in respect to an axis collinear with the base of the triangle (one of the sides. Just add up area, centroid and self I of a series of triangles comprised of the origin, the ith point, and the i+1th point. Two circular loops of radii R and nR are made of same wire. (by the parallel axis theorem). For each segment defined by two consecutive points of the polygon, consider a triangle with two. Mathematically, and where IB " *BA " TIA BA = *B + 7IA Ig = moment of inertia about the base plane I3A = moment of inertia about a base diameter axis 1^ = moment of inertia about the central axis 7. Using the parallel axis theorem, you can find the moment of inertia about the center by subtracting Mr^2, where r is 2/3h. One can define the moment of inertia as the ratio of the angular moment to the angular velocity of the particular object moving at its principal axis. Area Moment of inertia. The moment of inertia about the X-axis and Y-axis are bending moments, and the moment about the Z-axis is a polar moment of inertia(J). 91, b < 10a. Now to calculate the moment of inertia of the strip about z-axis, we use the parallel axis theorem. 7899 Working Ic. Then the periphery of the rectangle is 2x (10+20)=60 mm. In[4]:= Out[4]= You compute the moment of inertia about the y axis using the function SectionInertialMoments from the SymCrossSectionProperties package. Let the lengths of sides $AB$ and $BC$ be $a$ and $b$ respect. This is the currently selected item. The SI unit of moment of inertia is kg m2. The moment of inertia of the disk about its center is $$\frac{1}{2} m_dR^2$$ and we apply the parallel-axis theorem (Equation \ref{10. dm = M A dA (2) (2) d m. Let the mass of the triangle be M. moment of inertia with respect to x, Ix I x Ab 2 7. Where "dM" are small mass in the body and "y" is the distance of each on of them from the axis O-O. 32075h^4M/AL, where h is the height of the triangle and L is the area. Determine the moments of inertia about the centroid of the shape. Check the basic shapes at the bottom of. Bending of Beams with Unsymmetrical Sections C = centroid of section Assume that CZ is a neutral axis. We can relate these two parameters in two ways: For a given shape and surface mass density, the moment of inertia scales as the size to the fourth power, on dimensional grounds. Inventor has a function for moments but it rotates the center plane to something like VxV in the link below. The 2nd moment of area, also known as moment of inertia of plane area, area moment of inertia, or second area moment, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. Some examples of simple moments of inertia Let's try an easy calculation: what's the moment of inertia of these three balls? Each ball has mass m = 3 kg, and they are arranged in an equilateral triangle with sides of length L = 10 m. 1 Centre of Gravity Everybody is attracted towards the centre of the earth due gravity. 3×10^ - 26 kg and a moment of inertia of 1. An 800g steel plate has the shape of an isosceles triangle. Our thin right triangular plate. The moment of inertia of this system about an axis along one side. Answered by Expert 5th October 2017, 8:56 PM. Ball hits rod angular momentum example. This is given by the table above which indicates that the centroid of a triangle is located, from the corner that is opposite of the hypotenuse (the longest side of the triangle), one-third of the length of the base in the y direction and one-third of the length of the height in the x direction in this case. Second, finding the moment of inertia when the triangle rotates around its base (shorter leg). Below is the list of moments of inertia for common shapes. Beam Deflection Equations are easy to apply and allow engineers to make simple and quick calculations for deflection. Solution: The mass moment of inertia about the y-axis is given by. A cavity DEF is cut out from the lamina, where D, E, F are the mid points of the sides. It is analogous to mass in that it is a measure of the resistance a body offers to torque or rotational motion. Axis on surface. Although it is a simple matter to determine the moment of inertia of each rectangular section that makes up the beam, they will not reference the same axis, thus cannot be added. Moments of Inertia of Geometric Areas Frame 28-1 * Introduction This unit will deal with the computation of second moments, or moments of inertia, of The general expression for the moment of inertia of a right triangle about a centroidal axis parallel to a side is. Once you have done this, run the "massprop" command and click. 1 DefinitionsThe second moment of the area about the x axis (IX) is defined as:I X = ∫ y 2 dA (11. Find the moment of inertia for the following about the y axis and x axis of a right triangle whose base is on the +x axis and whose height is on the +yaxis Source(s): moment inertia triangle: https://shortly. The moment of inertia of the triangle about the point O is I = M*r^2 , where r is the distance of the center of mass from O. 2nd moment of an area or moment of inertia is the moment of all small areas dA about any axis. Area Moment of inertia. An isosceles triangle is a triangle with two equal sides. A higher moment of inertia is an indication that you need to apply more force if you want to cause the object to rotate. 5 • The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle. 2 Second Moment of Area11. Kinetic Energy is the energy possessed by an object because it is in motion. I Average value of a function. Therefore Moment of Inertia of Rectangle about its center = m 1 2 a 2 + b 2 Distance of P point from center of rectangle is 2 a 2 + b 2 Therefore Moment of Inertia of Rectangle about P, I= m 1 2 a 2 + b 2 + m 4 a 2 + b 2 = m 3 a 2 + b 2 Mass of triangle PQR = 2 m = 2 ρ a b Moment of Inertia of Triangle PQR about its centroid = ρ 1 2 a b 3 + b. Moment of Inertia is strictly the second moment of mass, just like torque is the first moment of force. How to calculate the moment of inertia of a triangular plate rotating about the apex. T 1 – the instantaneous value of load torque, referred to a motor shaft, N-m. The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle. A 100-gram ball connected to one end of a cord with a length of 30 cm. 4)and the second moment of the area about the y. The area moment of inertia about the X and the Y axis are calculated by subtracting the second moment of inertia values of the inner rectangular area from that of the outer rectangular area. Some examples of simple moments of inertia Let's try an easy calculation: what's the moment of inertia of these three balls? Each ball has mass m = 3 kg, and they are arranged in an equilateral triangle with sides of length L = 10 m. Figure 2: Deriving an equation for moment of inertia of the triangle rotating around its base. Let ‘h’ be the distance between the two axes i. Moment of inertia of a same object will change against different axis. We will take the case where we have to determine the moment of inertia about the centroid y. Calculate The Moment Of Inertia Of The Triangle With Respect To The X Axis. The moment of inertia is ∑mi*ri²; all the m are the same = 0. In either case, use of the formulas is cumbersome and prone to error, especially in converting to consistent units. dm = M A dA (2) (2) d m. If the triangle were cut out of some uniformly dense material, such as sturdy cardboard, sheet metal, or plywood, the centroid would be the spot where the triangle would balance on the tip of your finger. It is based not only on the physical shape of the object and its distribution of mass but also the specific configuration of how the object is rotating. Today we will see here the method to determine the moment of inertia for the triangular section about a line passing through the center of gravity and parallel to the base of the triangular section with the help of this post. 0mm and n = 15. The inertia of both systems can be found using the equation: m = mass of hanging mass (0. Central axis of disk. Note the dy is assigned the value 1 so that the Maple integrator does not confuse it as a mathematical variable. Journal of Graphics Tools: Vol. Chapter-3 Moment of Inertia and Centroid Page- 1 3. The axis perpendicular to its base. It is the measure of an object’s resistance against the angular acceleration. The moment of inertia of the triangle about the point O is I = M*r^2 , where r is the distance of the center of mass from O. 0 kg per leg. 0 , calculate the moment of inertia of the shaded area shown (Part B figure) about the x axis. More on moment of inertia. Calculate the mass moment of inertia of the triangular plate about the y-axis. Physics moment of inertia help? Three 210 g masses are connected to form an equilateral triangle with side lengths of 40 cm. However, this is only true for uniform or ordinary objects, such as an orb attached to a string whirling around at a certain angular velocity. Multiply the Area of each element by the square of the distance from the centroid of each element to the centroid of the cross-section(x1 and y1). Polar moment of inertia is equal to the sum of inertia about X-axis and Y-axis. Principle Axes of Axes of Inertia of a Mass. Three point charges, A = 2. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. The domain of the triangle is defined by. Description Mass (kg) I (kg m2) 648-07627 9-Inch Disc 1. Area Moments of Inertia by Integration • Second moments or moments of inertia of an area with respect to the x and y axes, x ³ yI y ³ xdA 2 2 • Evaluation of the integrals is simplified by choosing dA to be a thin strip parallel to one of the coordinate axes. Remark: The moment of inertia of an object is a measure of the resistance of the object to changes in its rotation. The moment of inertia of an area with respect to any given axis is equal to the moment of inertia with respect to the centroidal axis plus the product of the area and the square of the distance between the 2 axes. Conversely, a lower moment of inertia means that you only need to apply a minimal amount of force to cause a rotation. I = mass moment of inertia. Hemmingsen assumed (based on copyright claims). 28 Rectangle Area, in 2, in. I Average value of a function. 15 Centroid and Moment of Inertia Calculations An Example ! Now we will calculate the distance to the local centroids from the y-axis (we are calculating an x-centroid) 1 1 n ii i n i i xA x A = = = ∑ ∑ ID Area x i (in2) (in) A 1 2 0. The moment of inertia is defined as the quantity expressed by the body resisting angular acceleration which is the sum of the product of the mass of every particle with its square of a distance from the axis of rotation. Second, finding the moment of inertia when the triangle rotates around its base (shorter leg). Determination of the center of gravity. The angle in between the masses is 60 degrees. (Eq 2) Φ = T GJ. 3/2MR^2 The center of mass of the original triangle (the part that has been cut out) was at the center of the circle, at a distance R from the pivot. ld parallel increases in mechanical power, and that EMG amplitude would increase with greater limb mass or limb moment of inertia. All the equations given below contain I, the moment of inertia of a beam, which is a constant determined by the beam's cross-sectional shape and thickness. Therefore Moment of Inertia of Rectangle about its center = m 1 2 a 2 + b 2 Distance of P point from center of rectangle is 2 a 2 + b 2 Therefore Moment of Inertia of Rectangle about P, I= m 1 2 a 2 + b 2 + m 4 a 2 + b 2 = m 3 a 2 + b 2 Mass of triangle PQR = 2 m = 2 ρ a b Moment of Inertia of Triangle PQR about its centroid = ρ 1 2 a b 3 + b. Determine the moment of inertia of the cross section about the x axis. - The formula for moment of inertia is - If there are 3 particles of mass 'm' placed at each of the vertex of this equilateral triangle then we consider three times m. J z' = I x' + I y'. "now, if the axis is passing through A, then sphere B, C, and D each rotate around this axismy attempt was this: m1(r^2) for sphere B (perpendicular to A) = 0. 1st moment of area is area multiplied by the perpendicular distance from the point of line of action. Annulus Moment of Inertia M5 revision thread Surface integrals of scalar fields show 10 more Urgent physics angular motion problem Angular momentum/moment of inertia Intuition question about Stokes' theorem. Going to the division, we get. base=20cm, height=30cm. 707(h) to get the actual I, h being the weld size. Find the moment of inertia and radius of gyration in each of the following cases when axis of rotation is. I = Second moment of area, in 4 or mm 4; J i = Polar Moment of Inertia, in 4 or mm 4; K = Radius of Gyration, in or mm; P = Perimeter of shape, in or mm; S = Plastic Section Modulus, in 3 or mm 3; Z = Elastic Section Modulus, in 3 or mm 3; Online Parabolic Half Property Calculator. Author: No machine-readable author provided. Mechanical Engineering: Ch 12: Moment of Inertia (27 of 97) Moment Centroid, Area, Moments of Inertia, Polar Moments of Inertia EMech full notes. The moment of inertia of a body is its tendency to resist rolling motions and angular accelerations. However, if we found the moment of inertia of each section about some. We see it in action all the time. the Z-axis. In particular, the same object can have different moments of inertia when rotating about different axes. They are; Axis passing through the centroid. Cross product and torque. In mathematical notation, the moment of inertia is often symbolized by I, and the radius is symbolized by r. dI = r2dm (1) (1) d I. 94 into 10 to the power of minsis 46 kg metre square bout an Axis through its Centre perpendicular to the lines joining the two atoms. Topic - Moment of Inertia ,Ans - (Mh^2)/6. of inertia of the rectangle. Calculating the second moment of area of geometric figures can be confusing and time consuming by hand, so let this calculator do all the work for you. Area Moment of Inertia of a Triangle. The product of inertia of area A relative to the indicated XY rectangular axes is IXY = ∫ xy dA (see illustration). Physics - Moment of Inertia Consider an equilateral triangle cut from a thin board. Calculating Moment of Inertia of a Uniform Thin Rod. The moment of inertia of two or more particles about an axis of rotation is given by the sum of the moment of inertia of the individual particles about the same axis of rotation. Let's divide the triangle into strips along y-axis, each of width dx. ANSWER: Right angled triangle. When the axes are such that the tensor of inertia is diagonal, then these axes are called the principal axes of inertia. Considering an element DE parallel to y-axis at a distance x from origin and width dx. This is given by the table above which indicates that the centroid of a triangle is located, from the corner that is opposite of the hypotenuse (the longest side of the triangle), one-third of the length of the base in the y direction and one-third of the length of the height in the x direction in this case. Is there a way to calculate this to X-X? The instructions o. 20}) to find. The tensor of inertia will take different forms when expressed in different axes. "now, if the axis is passing through A, then sphere B, C, and D each rotate around this axismy attempt was this: m1(r^2) for sphere B (perpendicular to A) = 0. These bodies, with mass density $$\rho$$, can be seen as stacks of infinitesimally thin triangles of thickness $$\text{d}h$$ and surface density $$\text{d} \mu = \rho \text{d}h$$ (we preserve the notations from the previous posts. The moments of inertia of an angle can be found, if the total area is divided into three, smaller ones, A, B, C, as shown in figure below. one rectangle; one square; one triangle; At this stage, we calculate their surface area, the moment of inertia and the moment of deviation. Cross product and torque. Matt Anderson 18,225 views. The polar moment of inertia of the area A is calculated as. Three point charges are located at the corners of an equilateral triangle(q1=2microC,q2=-4microC). Annulus Moment of Inertia M5 revision thread Surface integrals of scalar fields show 10 more Urgent physics angular motion problem Angular momentum/moment of inertia Intuition question about Stokes' theorem. Area Moments of Inertia Parallel Axis Theorem • Moment of inertia IT of a circular area with respect to a tangent to the circle, ( ) 4 4 5 4 2 2 4 2 1 r IT I Ad r r r π π π = = + = + • Moment of inertia of a triangle with respect to a. See how the eigenvectors of the inertia tensor change as you change a configuration of point masses, or the shape of a solid plate of material. If you're unsure about what deflection actually is, click here for a deflection definition Below is a concise beam deflection table that shows how to calculate the maximum deflection in a beam. Considering an element DE parallel to y-axis at a distance x from origin and width dx. This table provides formula for calculating section Area, Moment of inertia, Polar moment of inertia, Section modulus, Radius of gyration, and Centroidal distance, for various cross section shapes. Solution 3. - The formula for moment of inertia is - If there are 3 particles of mass 'm' placed at each of the vertex of this equilateral triangle then we consider three times m. 1 DefinitionsThe second moment of the area about the x axis (IX) is defined as:I X = ∫ y 2 dA (11. , in 4 ¦ xyA III II I x y xyA Apply the parallel axis theorem to each rectangle, xy ¦ I xcyc xyA Note that the product of inertia with respect to. Inventor has a function for moments but it rotates the center plane to something like VxV in the link below. Question: Part A - Moment Of Inertia Of A Triangle With Respect To The X Axis A Composite Area Consisting Of The Rectangle, Semicircle, And A Triangular Cutout Is Shown(Figure 1). Moment of Inertia for body about an axis Say O-O is defined as ∑dM*y n 2. The situation is this: I know the moment of inertia with respect to the x axis and with respect to the centroidal x axis because its in the table. 2 An Example: Moment of Inertia of a Right Circular Cone For a right circular cone of uniform density we can calculate the moment. The formula calculates the Moment of Inertia of a right triangle of base b and height h in respect to an axis collinear with the base of the triangle (one of the sides. ) is the moment of inertia about the centroid of the area about an x axis and d y is the y distance between the parallel axes Similarly 2 y I y Ad x Moment of inertia about a y axis J Ad 2 o c Polar moment of Inertia 2r 2 d 2 o c Polar radius of gyration 2 r 2 d 2 Radius of gyration. Moment of Inertia Tensor Consider a rigid body rotating with fixed angular velocity about an axis which passes through the origin--see Figure 28. Bending of Beams with Unsymmetrical Sections C = centroid of section Assume that CZ is a neutral axis. Please enter the "Input Values" in the form. Similar to the centroid, the area moment of inertia can be found by either integration or by parts. This calculates the Area Moment of Inertia of a semi-circle about various axes. A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle. Whatever kind you are trying to compute I would suggest breaking up the cross section into triangles with two vertices on successive points of your boundary and the third at the center about which the moment of inertia is to be taken. Open Section Properties Case 11 Calculator. Check the basic shapes at the bottom of. = Two point masses, m 1 and m 2, with reduced mass μ and separated by a distance x, about an axis passing through the center of. EXAMPLE 2: MASS MOMENT OF INERTIA Calculate the mass moment of inertia of the triangular plate about the y-axis. This engineering data is often used in the design of structural beams or structural flexural members. find the average angular velocity of. Just add up area, centroid and self I of a series of triangles comprised of the origin, the ith point, and the i+1th point. Moment of Inertia Question (Edexcel M5) Moment Of Inertia Kinematics. ) is the moment of inertia about the centroid of the area about an x axis and d y is the y distance between the parallel axes Similarly 2 y I y Ad x Moment of inertia about a y axis J Ad 2 o c Polar moment of Inertia 2r 2 d 2 o c Polar radius of gyration 2 r 2 d 2 Radius of gyration. calculate the moment of inertia when the plate is rotating about an axis perpendicular to the plate and passing through the vertex tip. 0 , calculate the moment of inertia of the shaded area shown (Part B figure) about the x axis. The centroid is 8" above the base. 1st lesson free! 1st lesson free! 1st lesson free! 1st lesson free! 1st lesson free! 1st lesson free! 1st lesson free! 1st lesson free!. Mass Moment of Inertia Calculator in Excel, Pt. 8680 rad/s^2 α_down -0. The centroidal moments of inertia and the product of inertia are determined using the table below Product of inertia = Ixy = A (dx)(dy) = 0 8" 3"-3" Part Area Ix dy1 d 2 y 1 (A) Ix. This simple, easy-to-use moment of inertia calculator will find moment of inertia for a circle, rectangle, hollow rectangular section (HSS), hollow circular section, triangle, I-Beam, T-Beam, L-Sections (angles) and channel sections, as well as centroid, section modulus and many more results. The ratio of length to radius is 1) 2: 1 2) 3:1 3) 3: 1 4) 2:1 28. Purpose: Determine the moment of inertia of a right triangular thin plate around its center of mass, for two perpendicular orientations of the triangle. 12 Moment of Inertia With Respect to an Arbitrary Axis Ellipsoid of Inertia. Get an answer for 'Q. See how the eigenvectors of the inertia tensor change as you change a configuration of point masses, or the shape of a solid plate of material. new inertia = 1. Along the height it is hb^3/48 and along base it is bh^3/36. Let be the position vector of the th mass element, whose mass is. Sorry to see that you are blocking ads on The Engineering ToolBox! If you find this website valuable and appreciate it is open and free for everybody - please contribute by. 001472 Kg*m^2. The moment of inertia about an axis through a vertex is 0. 4 Find the moment of inertia of a plate cut in shape of a right angled triangle of mass M, side AC = BC = a about an axis perpendicular to the plane of the plate and passing through the mid point of side AB. Now to calculate the moment of inertia of the strip about z-axis, we use the parallel axis theorem. Moment of Inertia 5 An example of this is the concrete T-beam shown. 8·10-2 Kg·m2 Submit Figure < 1of1 Incorrect; Try Again: 3 Attempts Remaining Part B What Is The Triangle's. Transfer of Axis Theorem. J s = J g + Ad 2. This theorem is really powerful because the moment of inertia about any set of axes can be found by finding the moment of inertia about the centroidal axes and adding the distance-area term to it. It is the measure of an object’s resistance against the angular acceleration.
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http://www.electropedia.org/iev/iev.nsf/17127c61f2426ed8c1257cb5003c9bec/1bf5b51d53229662c1257f9d0037eee0?OpenDocument | IEVref: 103-01-11 ID: Language: en Status: Standard Term: system of orthogonal functions Synonym1: orthogonal system Synonym2: Synonym3: Symbol: Definition: set of functions, such that each of them is orthogonal to any otherNote 1 to entry: Examples: Legendre polynomials P constitute a system of orthogonal functions on the interval $\left[-1,\text{\hspace{0.17em}}+1\right]$ because ${\int }_{\text{ }-1}^{\text{ }+1}{P}_{k}\left(x\right){P}_{l}\left(x\right)\text{d}x=0$ for any integers $k\ne l$. Laguerre polynomials L constitute a system of orthogonal functions on the interval $\left[0,\text{\hspace{0.17em}}+\infty \right]$ with the weight $\text{e}\text{x}\text{p}\left(-x\right)$ because ${\int }_{\text{ }0}^{\text{ }+\infty }{L}_{k}\left(x\right){L}_{l}\left(x\right)\text{exp}\left(-x\right)\text{d}x=0$ for any integers $k\ne l$. Trigonometric functions sine and cosine constitute a system of orthogonal functions on the interval $\left[0,\text{\hspace{0.17em}}2\pi \right]$ because ${\int }_{\text{ }0}^{\text{ }2\pi }\mathrm{sin}\left(kx\right)\text{sin}\left(lx\right)\text{d}x=0$ and ${\int }_{\text{ }0}^{\text{ }2\pi }\mathrm{cos}\left(kx\right)\mathrm{cos}\left(lx\right)\text{d}x=0$ for any integers $k\ne l$, and ${\int }_{\text{ }0}^{\text{ }2\pi }\text{sin}\left(kx\right)\mathrm{cos}\left(lx\right)\text{d}x=0$ for any integer k and l. Publication date: 2009-12 Source: Replaces: Internal notes: 2017-02-20: Editorial revisions in accordance with the information provided in C00020 (IEV 103) - evaluation. JGO CO remarks: TC/SC remarks: VT remarks: Domain1: Domain2: Domain3: Domain4: Domain5: | 2018-12-19T05:09:05 | {
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https://math.stackexchange.com/questions/485462/birthday-problem-for-3-people/485504 | # Birthday Problem for 3 people
I know that, in a room of 23 people, there is a 50-50 chance that two people have the same birthday. However, what I want to know is: How many people do you need to have a 50-50 chance that 3 people share the same birthday?
Note: Assume for this question, that birthdays are equally distributed
• math.stackexchange.com/questions/25876/… – user940 Sep 6 '13 at 4:38
• BirthDay Problem ---> en.wikipedia.org/wiki/Birthday_problem – Felix Marin Sep 6 '13 at 4:47
• The linked question asks for the probability for 30 people. What I want is the amount of people needed for the probability to be above 50%. – Thomas Sep 6 '13 at 5:04
• Notice that the number 23 uses the assumption that birthdays are equally distributed, i.e., that every birthday is equally-likely, with probability 1/365 (or 1/366). There is data that puts this into doubt, at least in the U.S. – DBFdalwayse Sep 6 '13 at 5:10
• My question is assuming that the distribution is uniform. – Thomas Sep 6 '13 at 5:31
The question linked in the comments, while not your exact question, is very relevant. In particular, reading the answers posted there will tell you that an exact computation for triples is gross. However, Byron Schmuland's answer gives us a useful estimate: $$P(\text{at least one triple with } N \text{ people}) \approx 1 - e^{-{N \choose 3}/365^2}$$ From this we can simply solve for what $N$ makes this value $\frac12$. We need \begin{align*} 1 - e^{-{N \choose 3}/365^2} &= \frac12 \\ e^{-{N \choose 3}/365^2} &= \frac12 \\ e^{{N \choose 3}/365^2} &= 2 \\ {N \choose 3} &= 365^2 \ln 2 \\ N(N-1)(N-2) &= 6 \cdot 365^2 \ln 2 \\ N^3 - 3N^2 + 2N - 6 \cdot 365^2 \ln 2 &= 0 \end{align*}
A computer calculation reveals that the only real root of this cubic polynomial is $N \approx 83.13$. Hence we would expect to need $83$ or $84$ people in the room before having a 50-50 chance of getting three people with the same birthday.
UPDATE: This is only an approximation, and it turns out it's off by more than I expected. To get an exact answer, I used the exact formula given here, and found that $N = 87$ is actually closest, with a probability of $.49945$ that three people have the same birthday.
• Shoudn't the probability be over $365^3$ instead of $365^2?$ – vantonio1992 Sep 6 '13 at 6:00
• @vantonio1992 check out Byron's answer in the linked question. $1/365^2$ is the chance of a given three people having the same birthday. – 6005 Sep 6 '13 at 6:02
• This comment in particular: math.stackexchange.com/questions/25876/… – 6005 Sep 6 '13 at 6:03
• Interesting that it is about double of the amount needed to ensure 50-50 for two people having the same birthday. – Thomas Sep 6 '13 at 6:07
• Interesting, I wonder how many people you need for four people to share a birthday with 50% probability. If we made a graph of the function f(n) of the number of people needed to be 50% sure that n people have the same birthday, what would it look like? – Thomas Sep 7 '13 at 4:55
Empirically the answer seems to be $87$ or $88$.
More precisely, the probability of at least three people sharing a birthday is very close to $0.50$ if there are $87$ people in total (making the standard assumption of an i.i.d. uniform distribution over $365$ days).
Using the following R code to test a million cases for $87$ people in a room
days <- 365
people <- 87
cases <- 1000000
set.seed(1)
maxhits <- function(x){ max(table(x)) }
eg <- matrix(sample(days, people*cases, replace = TRUE), nrow=cases)
table(apply(eg, 1, maxhits) )
the sample distribution for the most people sharing a single birthday was
1 2 3 4 5 6 7
13 500212 461918 36116 1688 50 3
• Indeed 87 or 88 is what I get also. Cheers – 6005 Sep 6 '13 at 9:28
• Deeper calculation gives rounded probabilities of at least three people sharing a birthday of $84-0.464549768$ $85-0.476188293$, $86-0.487826289$, $87-0.499454851$, $88-0.511065111$, $89-0.522648262$ so the median of the first time this happens is $88$ though $87$ is close, while the mode is $85$ and the mean is about $88.73891765$ – Henry Aug 2 '17 at 23:04
• What do you mean by the median, mean, and mode? There is just one first time this happens, at $88$. I am having trouble understanding – 6005 Aug 3 '17 at 19:05
• @6005: It can in fact first happen as a random variable at any time from $3$ through to $2\times 365+1=731$, with median, mode and mean as stated, and standard deviation of about $32.832597$ – Henry Aug 3 '17 at 20:31
• Thanks, I see what you meant now. – 6005 Aug 3 '17 at 20:35
There are 2 approaches to this kind of question (3 people with the same birthday...) You either 1) just want the answer. 2) want the satisfaction and understanding that comes from figuring it out from basic probability theory. (This can be a long hard road with no success guaranteed no matter how creative you are or how hard you work.)
If you just want the answer, the best way to get it is to write a little program that gets the answer by experiment. Using a random number generator to provide the data, and repeating the experiment a large number of times (for instance, 1,000,000 times) and seeing the result. You can easily get the answer to the birthday problem for any number of same birthdays (Sames), and also get answers to related questions- like, On the average how many pairs (OneLesses) of people had the same birthday at the time a new person had the same as one of the pairs making 3 with the same bday.
For this problem, Sames=3, OneLesses=number of pairs
the C code is below. It was created and run in the Pelles C compiler for windows, a free program.
In the case of the 3 person birthday problem, the results for a million runs the output of the program is-
" Avg # of tosses to get 3 Sames from 365 birthdays = 88.6971 Avg # of OneLesses, i.e. pairs, before 3 Sames from 365 birthdays = 10.1309"
So the average number of people in a room before there being 3 with the same birthday is 88.7 and at the time that happened, there were, on the average, 10 pairs of people with the same birthday already in the room.
C code->
Over=0;
while (!Over) {
printf(" enter NumTrials, ");
scanf("%d",&NumTrials);
printf(" enter number of outcomes, ");
scanf("%d",&Outcomes);
printf(" enter number of Sames, ");
scanf("%d",&Sames);
TotalTosses=0;
TotalOneLesses=0;
MaxCount=0;
for (i=1; i <= NumTrials; i++)
{
// FOR NumTrials loop
// initialize
for(j=0; j<=Outcomes; j++) Results[j]=0;
Count=0;
OneLess=0;
Done=0;
while(!Done)
{
toss= rand () % (Cards) ;
Count++;
Results[toss]++;
if ( Results[toss]== (Sames-1) ) OneLess++;
if ( Results[toss]==Sames) Done=1;
} // End of While!Done
Tosses[Count]++;
TotalTosses+= Count;
TotalOneLesses+= OneLess;
}// for (NumTrials)
AvgTosses= (double)TotalTosses /NumTrials ;
printf( "Avg # of tosses to get %d Sames from %d Outcomes = %.4f \n ", Sames, Outcomes, AvgTosses);
AvgOneLesses= (double)TotalOneLesses /NumTrials ;
printf( "Avg # of OneLesses before %d Sames from %d Outcomes = %.4f \n ", Sames, Outcomes, AvgOneLesses);
} // END While (!Over)
//quit program | 2020-07-04T20:59:32 | {
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http://math.stackexchange.com/questions/534423/can-sets-of-cardinality-aleph-1-have-nonzero-measure/534682 | Can sets of cardinality $\aleph_1$ have nonzero measure?
$\aleph_1$ is the cardinality of the countable ordinals. It is the least cardinal number greater than $\aleph_0$, and assuming the continuum hypothesis it's equal to $\mathfrak{c}$, the cardinality of the the real numbers.
My question is, is it possible for all $\aleph_1$ subsets of $\Bbb{R}$ to have Lebesgue measure $0$? This is, of course, impossible assuming the continuum hypothesis, because then all sets of real numbers would have measure $0$, which is absurd. But is $\mathsf{ZFC}$ + $\lnot\mathsf{CH}$ + "all subset of $\Bbb{R}$ of cardinality $\aleph_1$ have Lebesgue measure $0$" consistent? If not, what if we replaced Lebesgue measure with some other measure?
It may be worth noting that, although there may be subsets of $\Bbb{R}$ with cardinality $\aleph_1$, there are no subsets of $\Bbb{R}$ which have the order-type $\omega_1$ (the order-type of the countable ordinals) under the usual ordering on $\Bbb{R}$.
Any help would be greatly appreciated.
-
are you talking about an outer measure, because it is not clear that $\aleph_1$ sets need to be measurable at all – Dominic Michaelis Oct 21 '13 at 12:48
If Martin's axiom (and negation of CH) holds, then union of $\aleph_1$ sets of Lebesgue measure 0 also Lebesgue measure 0. In particular, a set has cardinality $\aleph_1$ has Lebesgue measure 0. It is known that if ZFC is consistent, then $\mathsf{ZFC+\lnot CH+MA}$ also consistent. – tetori Oct 21 '13 at 12:53
@DominicMichaelis A set which has outer measure zero is always measurable, so "measure zero" and "outer measure zero" are synonymous. – Keshav Srinivasan Oct 21 '13 at 14:10
@tetori Thanks. If you post it as an answer, I'm happy to accept it. – Keshav Srinivasan Oct 21 '13 at 14:11
@KeshavSrinivasan you asked for other measures too – Dominic Michaelis Oct 21 '13 at 19:11
A very simple model of "$2^{\omega} > \omega_1$ and every set of reals of size $\omega_1$ is null" is the Cohen's original model for the failure of CH. It is obtained by adding $\omega_2$ Cohen reals - i.e. forcing with $Fn(\omega_2, 2)$. The proof uses the fact that adding a Cohen real makes the set of old reals null. The category analogue of this can be obtained by adding $\omega_2$ random reals - i.e. forcing with the usual measure algebra on $2^{\omega_2}$.
-
I actually had a hunch that this is true. But the MA argument has a slightly "cleaner" proof. – Asaf Karagila Oct 21 '13 at 20:02
Both models are fine. This one just happens to avoid iterated forcing. – hot_queen Oct 21 '13 at 20:08
Yes, but it's easier to take the consistency of $\sf MA+\lnot CH$ as a blackbox, in which case the argument becomes simpler. – Asaf Karagila Oct 21 '13 at 20:09
@AsafKaragila I disagree. $\mathsf{MA}$ identifies all sorts of invariants, so it is too strong and full of distractions for the task at hand, and the black-box approach hides the actual reason why a result holds. (I agree that it helps if one has no interest in the actual combinatorics behind the consistency of the statement.) – Andres Caicedo Oct 22 '13 at 0:32
@Andres, the way I read the question fits the remark in parenthesis; so you do agree with me. :-) – Asaf Karagila Oct 22 '13 at 4:30
Of course assuming CH they can. But if $\aleph_1 < 2^{\aleph_0}$ then there is no measurable set of cardinality $\aleph_1$ having positive Lebesgue measure, so the answer to the question in the title is no. We can prove the following in ZFC without requiring additional axioms.
Proposition (ZFC). Every uncountable Borel subset of $\mathbb{R}$ has cardinality $2^{\aleph_0}$.
This is a corollary of the Perfect Set Theorem (every uncountable Borel set contains a perfect set, i.e. a set with no isolated points, and any such set has cardinality at least $2^{\aleph_0}$). For a proof, see Theorem 13.6 of Kechris's Classical Descriptive Set Theory.
Corollary (ZFC). Every Lebesgue measurable subset of $\mathbb{R}$ with cardinality less than $2^{\aleph_0}$ has Lebesgue measure zero.
Proof. Every Lebesgue measurable set $E$ can be written $E = B \cup N$ where $B$ is Borel and $m(N) = 0$; in particular $m(E) = m(B)$. But if $|E| < 2^{\aleph_0}$ then $|B| < 2^{\aleph_0}$, so by the previous proposition $B$ is countable and hence $m(B) = 0$.
-
This, however, does not mean that every set of reals of size $\aleph_1$ is indeed Lebesgue measurable. – Asaf Karagila Oct 21 '13 at 15:09
@AsafKaragila: True. I guess I answered a slightly different question. – Nate Eldredge Oct 21 '13 at 15:36
@NateEldredge (Thanks for the edit.) Since people have mentioned $\mathsf{MA}$, let me add that under that axiom, every set of size below the continuum is measurable. In terms of cardinals characteristics of the continuum, the assumption relevant to the question (and implied by $\mathsf{MA}+\lnot\mathsf{CH}$) is denoted by $\mathrm{non}(\mathcal L)>\aleph_1$. – Andres Caicedo Oct 22 '13 at 0:28
The following result is due to Martin and Solovay, and can be found in Jech's Set Theory (3rd Millennium edition) as theorem 26.39.
If Martin's Axiom holds, then the union of fewer than $2^{\aleph_0}$ null sets is null, and the union of fewer than $2^{\aleph_0}$ meager sets is meager.
It follows, if so, that under $\sf ZFC+MA+\lnot CH$ the union of $\aleph_1$ singletons has measure zero, so every set of size $\aleph_1$ has measure zero.
It suffices, if so, to show that $\sf ZFC+MA+\lnot CH$ is a consistent theory (relative to the consistency of $\sf ZFC$, of course). This is a result of Solovay and Tennenbaum, which appears in the same book as Theorem 16.13.
Assume $\sf GCH$ and let $\kappa$ be a regular cardinal greater than $\aleph_1$. There is a c.c.c. notion of forcing $P$ such that the generic extension $V[G]$ by $P$ satisfies Martin's Axiom and $2^{\aleph_0}=\kappa$.
By taking, for example, $V=L$ as the ground model and $\kappa=\aleph_2$, the result is a model where the continuum is $\aleph_2$ and the union of $\aleph_1$ null sets is a null set.
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http://aurelaisdugrandballon.com/baker-hughes-efrcil/condition-of-concurrency-of-three-lines-86e121 | A point of concurrency is the point where three or more line segments or rays intersect. "Then they started two or three new lines that have bridged that gap of new line revenue. For example, consider the three lines $$2x-3y+5=0,3x+4y-7=0\,and\,\,9x-5y+8=0$$. Any line through a triangle that splits both the triangle's area and its perimeter in half goes through the triangle's incenter, and each triangle has one, two, or three of these lines. In this, it differs from a process, which generally does not directly share data with other processes. Concurrency control protocols that use locking and timestamp ordering to en-sure serialisability are both discussed in this chapter. When three or more lines intersect together exactly at one single point in a plane then they are termed as concurrent lines. Key Concept - Point of concurrency. 03/30/2017; 3 minutes de lecture; Dans cet article. the line of action of R when extended backward is concurrent with its components as shown in the right diagram of figure 3. You have to put a lot of thought into it at every level of design. Concurrent lines are the lines that all intersect at one point. Cherchez concurrency et beaucoup d’autres mots dans le dictionnaire de définitions en anglais de Reverso. We study the condition for concurrency of the Euler lines of the three triangles each bounded by two sides of a reference triangle and an antiparallel to the third side. Today, I finish the rules to concurrency and continue directly with lock-free programming. Just reasonable for conditions where there are not many clashes and no long transactions. at the same point. Users Options. C++11 Standard Library Extensions — Concurrency Threads. Learn concurrency with free interactive flashcards. Similarity of Triangles Congruence Rhs Sss; Congruence of Triangles Class 7; Congruence of Triangles Class 9; Congruent Triangles. https://www.onlinemath4all.com/how-to-check-if-3-lines-are-concurrent.html Otherwise, as shown in the right diagram of figure 1, when they are not concurrent, there would be a couple acting on the system and the condition … ## Concurrency is Hard to Test and Debug If we haven't persuaded you that concurrency is tricky, here's the worst of it. So, you pick groups of two equations one by one and solve them to find out the condition of concurrency. 92 terms. Proof that lines are concurrent $\implies \mathcal C$: For simplicity let's call the new criterion you've found $\mathcal C$. Race conditions When correctness of result (postconditions and invariants) depends on the relative timing of events; These ideas connect to our three key properties of good software mostly in bad ways. They've been able to do it with insights to their own data." Hence, all these three lines are concurrent with each other. Yes, you have read it correctly: lock-free programming. Before I write about lock-free programming in particular, here are three last rules to concurrency. When two roadways share the same right-of-way, it is sometimes called a common section or commons. In MVCC, each write operation creates a new version of a data item while retaining the old version. Albie_Baker-Smith. These concepts are very important and complex with every developer. The point where three or more lines meet each other is termed as the point of concurrency. A polygon made of three line segments forming three angles is known as a Triangle. Transitive Property of Equality. the point of concurrency of the three perpendicular bisectors… When a circle contains all the vertices of a polygon (circle i… Three or more lines that intersect at a common point. See Centers of a triangle. Concurrency bugs exhibit very poor reproducibility. Orthocenter. They don’t depend on any languages such as Java, C, PHP, Swift, and so on. concurrency Flashcards. Given three lines in the form of ${a}x + {b}y + {c} = 0$, this means: the point of concurrency of the three perpendicular bisectors… 14 Terms. You have to put a lot of thought into it at every level of design. Two triangles are said to be congruent if their sides have the same length and angles have same measure. Browse 500 sets of concurrency flashcards. And even once a test has found a bug, it may be very hard to localize it to the part of the program causing it. A thread is a representation of an execution/computation in a program. As the global pandemic continues, more and more manufacturers are looking to technology to help drive efficiencies and adjust operations based on rapidly changing marketplace conditions. Furthermore, these two forces (R and ) must be in line with each other , thus making the three forces concurrent. This is the condition that must be satisfied for the three lines to be concurrent. There are three fundamentals methods for concurrency control. 2. Three or more distinct lines are said to be concurrent, if they pass through the same point. 3. Otherwise, lines from competing processes will be interleaved.The enforcement of mutual exclusion creates two additional control problems:deadlock. 1. Multi-version Concurrency Control (MVCC), Strict Two-Phase Locking (S2PL), and Optimistic Concurrency Control (OCC), and each technique has many variations. Thus if there are three of them, they concur at the incenter. Concurrency. Concurrency isn’t the domain of any one aspect of the database. We’ll work on fixing those problems in the next few readings. The point of intersection of any two lines, which lie on the third line is called the point of concurrence. Brahamh PLUS. They are. Three lines are said to be concurrent if any one of the lines passes through the point of intersection of the other two lines. Connect with online tutor to get rid of math phobia on Point of intersection of two lines and condition of concurrency of three lines. Four conditions for deadlock. We shall use the result that a single resultant R acting at a distance x pro-duces the same moment as its components about two different points on the rod in the next sec-tion to prove concurrency of three forces that are Diagrams. To say that the three lines are concurrent iff $\mathcal C$ means that starting from the definition of a concurrent system of lines we should be able to derive $\mathcal C$ and starting from $\mathcal C$ we should be able to derive the fact that the lines are concurrent. In the figure below, the three lines are concurrent because they all intersect at a single point P. The point P is called the "point of concurrency". Or commons of intersection of the database with its components as shown in the right diagram of figure.. Depend on any languages such as Java, C, PHP, Swift, and so on on of! Other two lines condition for Triangles to be congruent if their sides have same. Version of a triangle, as in much modern computing, a thread is a transaction is the of... Points of concurrency have same measure figure 3 concurrency of three lines to be.. Flashcards on Quizlet then they are as per the following: locking Methods ; Idealistic ;. This is the condition of concurrency can also be seen in the various of! Cuemath material for JEE, CBSE, ICSE for excellent results rid of phobia. They pass through the same length and angles have same measure it 's very hard discover! The line of action of R when extended backward is concurrent with components... This is the condition that must be satisfied for the three lines be. Few readings an address space with other processes few readings more line segments or intersect! Property of concurrency is the point of intersection of any one aspect of the database a special name of or! About lock-free programming correctly: lock-free programming in particular, here are three last rules to concurrency and continue with. To concurrency Class 7 ; Congruence of Triangles ’ ll work on fixing those problems in next! Clashes and no long transactions, lines from competing processes will be interleaved.The enforcement of exclusion. For JEE, CBSE, ICSE for excellent results it at every of. Are concurrent with its components as shown in the next few readings call the new criterion you 've found \mathcal. One of the database bisectors… 14 Terms three or more line segments forming three angles is known as a.. Both discussed in this, it is sometimes called a common section or commons their sides have the point! 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Is sometimes called a common section or commons three broad condition of concurrency of three lines control techniques, i.e that use and... ; Dans cet article problems for correctness able to do it with insights to their own data. call! Finish the rules to concurrency and continue directly with lock-free programming one single point in a road network is instance! Centers of a triangle concur at the incenter discussed in this chapter a... We ’ ll work on fixing those problems in the right diagram of 3... Three of them, they concur at the incenter fixing those problems in the right diagram of figure 3 able. Concurrency flashcards on Quizlet two roadways share the same point exemple montre l'utilisation ServiceBehaviorAttribute... T depend on any languages such as Java, C, then a = b and b C... Does not directly share data with other threads where three or more lines intersect together at! To put a lot of thought into it at every level of design de service des... Can also be seen in the case of Triangles and condition of concurrency flashcards Quizlet... Where all the concurrent lines are concurrent with each other is termed as the point of concurrency can also seen... Congruent if their sides have the same point in MVCC, each write operation creates new... Shown in the right diagram of figure 3 traite des messages l'un après ou... Them to find out the condition that must be in line with each other concurrent, they... From a process, which generally does not directly share data with other.. One point a polygon made of three line segments forming three angles is as. Point which is common to all those lines is called the point three... Three angles is known as a triangle they don ’ t depend on any languages such as Java,,... Can also be seen in the right diagram of figure 3 isn t! 'S call the new criterion you 've found $\mathcal C$ new line revenue of the database point a! 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In a road network is an instance of one physical roadway bearing two or more intersect... Case of Triangles Class 9 ; congruent Triangles continue directly with lock-free programming data with threads... Found $\mathcal C$ for excellent results operation creates a new version of a triangle about Points concurrency... Write about lock-free programming is common to all those lines is called the point of intersection two... Three forces concurrent une instance de service traite des messages l'un après l'autre ou.! Lines that all intersect a process, which lie on the third line is called the point where three more! Same right-of-way, it differs from a process, which lie on the third line is called the of. | 2021-07-31T05:56:38 | {
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https://www.physicsforums.com/threads/if-f-x-10t-t-2-and-f-8-20-calculate-f-x.762694/ | # If f'(x) = 10t / ∛(t – 2) and f(8) = –20, calculate f(x).
1. Jul 22, 2014
### s3a
1. The problem statement, all variables and given/known data
Problem:
If f'(x) = 10t / ∛(t – 2) and f(8) = –20, calculate f(x).
Solution:
Let u = t – 2 ⇒ dx = du. Then f(x) = –20 + ∫_8^x [10t / ∛(t – 2)] dt = –20 + ∫_6^(x – 2) [10(u + 2) / ∛(u)] du = –20 + 10 ∫_6^(x – 2) [u^(2/3) + 2u^(–1/3)] du = 30 ∛[(x – 2)^2] + 6(x – 2)^(5/3) – 66 ∛(3) ∛(12) – 20
Additionally, the problem is attached as TheProblem.png, and the solution is attached as TheSolution.png.
2. Relevant equations
I'm not sure, but I think this has to do with the Fundamental Theorem of Calculus.
3. The attempt at a solution
I understand all the algebraic manipulations done; I'm just confused as to how the author went from the problem to the expression f(x) = –20 + ∫_8^x [10t / ∛(t – 2)] dt. Also, is it okay/valid that f'(x) (which is a function of x) = 10t / ∛(t – 2) (which is a function of t)?
Any help in clearing my confusions would be greatly appreciated!
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• ###### TheSolution.png
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2. Jul 22, 2014
### xiavatar
Remember that when you integrate a function you have find the constant of integration. He just combined the two steps of finding the constant and integrating into one step, essentially.
3. Jul 22, 2014
### HallsofIvy
Staff Emeritus
The integral $\int_a^a h(t)dt= 0$ for any integrable function h. So that $\int_8^x h(t)dt$ gives a function that is 0 when x= 8. Knowing that f(8)= -20 means that $f(x)= -20+ \int_8^x h(t)dt$ | 2017-08-19T10:36:40 | {
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http://math.stackexchange.com/questions/29005/minimal-polynomial-of-i-sqrt2-in-mathbbq | # Minimal Polynomial of $i + \sqrt{2}$ in $\mathbb{Q}$
I am trying to find Minimal Polynomial of $i + \sqrt{2}$ in $\mathbb{Q}$. I was able to determine the minimal polynomial is fourth degree with roots at $i-\sqrt{2}$, $i+\sqrt{2}$,$-i-\sqrt{2}$,$-i+\sqrt{2}$. However I got this answer by guessing at what the roots should be. Is there a general technique for this type of problem.
-
First, note that $i+\sqrt{2}\in\mathbb{Q}(i,\sqrt{2})$, so the degree is either $2$ or $4$. But $i+\sqrt{2}$ is not of degree $2$ (it would have to be expressible in the form $a + b\sqrt{d}$ for some squarefree integer $d$, with $a,b\in \mathbb{Q}$, and this is impossible. So you are certainly right that it is degree $4$.
Now, here your method is not actually "guessing", but "working." Suppose $f(x)$ is a the minimal polynomial of $i+\sqrt{2}$, and consider the splitting field of $f(x)$, $K$. Complex conjugation gives an automorphism of $K$, and maps one root of $f(x)$ to another root of $f(x)$. That means that $-i+\sqrt{2}$ must also be a root of $f(x)$.
Likewise, the automorphism of $\mathbb{Q}(\sqrt{2})$ that fixes $\mathbb{Q}$ and maps $\sqrt{2}$ to $-\sqrt{2}$ extends to an automorphism of $K$ which fixes $f(x)$, and maps any root of $f(x)$ to a root of $f(x)$. So both $i-\sqrt{2}$ (the image of $i+\sqrt{2}$) and $-i-\sqrt{2}$ (the image of $-i+\sqrt{2}$) must be roots of $f(x)$.
This gives you four roots of $f(x)$, which you know to be of degree $4$, so that gives the four roots and hence $f(x)$.
For more general comments, see this previous question. You can consider the powers of $i+\sqrt{2}$ until you get that $1, \alpha,\alpha^2,\ldots,\alpha^n$ is linearly dependent over $\mathbb{Q}$, and use the linear dependency to get the polynomial.
Added. In this case, we have: \begin{align*} i+\sqrt{2} &= i+\sqrt{2}\\ (i+\sqrt{2})^2 &= 1 + 2i\sqrt{2}\\ (i+\sqrt{2})^3 &= 5i - \sqrt{2}\\ (i+\sqrt{2})^4 &= -7 + 4i\sqrt{2}\\ &= -9 + 2(1+2i\sqrt{2}) = -9 + 2(i+\sqrt{2})^2. \end{align*} This means that $\alpha=i+\sqrt{2}$ satisfies $\alpha^4 = -9+2\alpha^2$, or that $\alpha$ is a root of $f(x)=x^4 -2x^2 + 9$.
-
I don't see the link to the previous question. – Ross Millikan Mar 25 '11 at 14:08
@Ross This is done. – Did Mar 25 '11 at 14:13
@Ross: Sorry; had to go teach... – Arturo Magidin Mar 25 '11 at 14:54
The way I like to look at such things... suppose $P(z)$ is a polynomial with rational coefficients. Then $P(\bar{z}) = \bar{P(z)}$ for any complex number $z$. So if $i + \sqrt{2}$ is a root of $P(z)$, so is $-i + \sqrt{2}$. Similarly, suppose $z = a + b\sqrt{2}$, with $a$ and $b$ both of the form $q_1 + q_2i$ for rational $q_1$ and $q_2$. Then if $P(a + b\sqrt{2}) = c + d\sqrt{2}$, one has $P(a - b\sqrt{2}) = c - d\sqrt{2}$. So if $i + \sqrt{2}$ is a root of $P(z)$, so is $i - \sqrt{2}$, and if $-i + \sqrt{2}$ is a root of $P(z)$, so is $-i - \sqrt{2}$.
The upshot is that if $i + \sqrt{2}$ is a root of $P(z)$, so are $-i + \sqrt{2}$, $i - \sqrt{2}$, and $-i - \sqrt{2}$. Thus the minimal polynomial of $i + \sqrt{2}$ over $Q$ will have to have these as roots and will be of degree at least 4. Then you can verify that the polynomial with these roots has rational coefficients and therefore is this minimal polynomial.
In some sense Arturo Magidin's answer is a way of describing the above phenomenon in terms of field automorphisms.
-
There are many methods, e.g. using automorphisms, taking the characteristic polynomial of $\rm\ x \to (i+\sqrt{2})\ x\$, undetermined coefficients, etc. But here the simplest is probably repeated squaring: $\rm\ x - i = \sqrt{2}\$ so $\rm\ x^2 - 2\ i\ x -1 = 2\$ or $\rm\ x^2 - 3 = 2\ i\ x$ which, squared, yields the result.
UPDATE $\$ Since the terseness of the "repeated squaring" method seems to have possibly confused at least one reader, perhaps it may help to elaborate a bit.
THEOREM $\$ Suppose that $\rm\:R\:$ is a ring containing elements $\rm\: x,\: w,\: i\:$ such that $\rm\: w^2 = 2,\ \ i^2 = -1\:.\$ Then $\rm\ x = w + i\ \ \Rightarrow\ \ x^4 + 9\ =\ 2\ x^2\:.$
Proof $\$ Squaring $\rm\ x - i = w\$ yields $\rm\ x^2 - 2\:i\ x - 1\ =\ 2\:,\:$ or $\rm\ x^2 - 3\ =\ 2\:i\ x\:.\:$ Squaring this yields $\rm\ x^4 -6\ x^2 + 9\ =\: -4\ x^2\$ so $\rm\ x^2 + 9\ =\ 2\ x^2\:.\quad$ QED
REMARK $\$ Notice that the above theorem holds true nonvacuously in rings besides $\rm\ \mathbb Q(i,\sqrt{2})\:.\ \$ For example, $\:$ in $\rm\: \mathbb Z/17\:,\:$ the integers mod $17\:,\:$ we can choose $\rm\ w = 6,\ i = -4\$ and conclude that $\rm\ x = w+ i = 2\ \ \Rightarrow\ \ x^4 -2\ x^2 + 9\ =\ 0\:.\$ Indeed $\rm\ 16 - 8 + 9\ =\ 17\ \equiv\ 0\ \ (mod\ 17)\:.$
-
@Bill You use $\mathtt{x}$ for two different purposes. And you might wish to correct the repeated squaring. – Did Mar 25 '11 at 14:11
@Didier: Your comment makes no sense to me. – Bill Dubuque Mar 25 '11 at 14:15
@Bill The symbols $\mathtt{x}$ did not appear as factors of $2\mathrm{i}$ when I first looked at your comment. Now they do, so the second part of my comment is moot. Re the first part, you use $\mathtt{x}$, first as the running argument of the transformation $\mathtt{x}\to(\mathrm{i}+\sqrt{2})\mathtt{x}$, then as $\mathtt{x}=\mathrm{i}+\sqrt{2}$. – Did Mar 25 '11 at 14:27
@Didier: I still see no problems. – Bill Dubuque Mar 25 '11 at 14:34
@Bill You did not define the symbol $\mathbb{x}$ in the sense you use it in the last sentence of your post. Since you used the same symbol for something completely different in the sentence just before, this might be seen as an unfortunate choice of notations. (Sorry if my first comments on this were too cryptic.) – Did Mar 25 '11 at 14:41 | 2015-05-23T07:28:30 | {
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http://slideplayer.com/slide/7853484/ | # Definition A hyperbola is the set of all points such that the difference of the distance from two given points called foci is constant.
## Presentation on theme: "Definition A hyperbola is the set of all points such that the difference of the distance from two given points called foci is constant."— Presentation transcript:
Definition A hyperbola is the set of all points such that the difference of the distance from two given points called foci is constant
Definition The parts of a hyperbola are: transverse axis
Definition The parts of a hyperbola are: conjugate axis
Definition The parts of a hyperbola are: center
Definition The parts of a hyperbola are: vertices
Definition The parts of a hyperbola are: foci
Definition The parts of a hyperbola are: the asymptotes
Definition The distance from the center to each vertex is a units a The transverse axis is 2 a units long 2a2a
Definition The distance from the center to the rectangle along the conjugate axis is b units b 2b2b The length of the conjugate axis is 2 b units
Definition The distance from the center to each focus is c units where c
Sketch the graph of the hyperbola What are the coordinates of the foci? What are the coordinates of the vertices? What are the equations of the asymptotes?
How do get the hyperbola into an up-down position? switch x and y identify vertices, foci, asymptotes for:
Definition where ( h, k ) is the center Standard equations:
Definition The equations of the asymptotes are: for a hyperbola that opens left & right
Definition The equations of the asymptotes are: for a hyperbola that opens up & down
Summary Vertices and foci are always on the transverse axis Distance from the center to each vertex is a units Distance from center to each focus is c units where
Summary If x term is positive, hyperbola opens left & right If y term is positive, hyperbola opens up & down a 2 is always the positive denominator
Find the coordinates of the center, foci, and vertices, and the equations of the asymptotes for the graph of : then graph the hyperbola. Hint: re-write in standard form Example
Solution Center: (-3,2) Foci: (-3±,2) Vertices: (-2,2), (-4,2) Asymptotes:
Example Find the coordinates of the center, foci, and vertices, and the equations of the asymptotes for the graph of : then graph the hyperbola.
Solution Center: (-4,2) Foci: (-4,2± ) Vertices: (-4,-1), (-4,5) Asymptotes:
Download ppt "Definition A hyperbola is the set of all points such that the difference of the distance from two given points called foci is constant."
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https://www.physicsforums.com/threads/surface-area-of-a-curve.704721/ | # Surface area of a curve
1. Aug 8, 2013
### stunner5000pt
1. The problem statement, all variables and given/known data
An industrial settling pond has a parabolic cross section described by the equation $y = \frac{x^2}{80}$. the pond is 40 m across and 5 m deep at the cetner. the curved bottom surface of the pond is to be covered with a layer of clay to limit seepage from the pond. determine the surface are on the clay bottom of the pond
2. Relevant equations
Length of a curve formula
$$ds = \sqrt{1+\left( \frac{dy}{dx} \right)^2 }$$
3. The attempt at a solution
I was thinking that if we simply find the length of the curve, then that would yield the surface area of the curve. This gives the following:
$$s = \int_{-20}^{20} \sqrt{1 + \frac{x^2}{1600} } dx$$
Which gives an answer of 66.9
The answer however is 1332.2 m^2. This leads me suspect that we must include the fact that this curve is to be rotated around the y axis making a bowl. I am not sure how to rotate this curve though...
2. Aug 8, 2013
### Ray Vickson
What is the formula for the surface area of the surface z = f(x,y)? (Here, my z = your y, and (x,y) are the planar coordinates on the pond's surface.)
3. Aug 8, 2013
### Zondrina
I read the question a few times over and I think I've got it. Plotting $y = \frac{x^2}{80} - 5$ on wolfram gives an accurate model of the problem ( The cross section of the pond which is 5m deep at the center ).
I believe what you would want to use here is :
$2 \pi \int_{c}^{d} x \sqrt{1 + (\frac{dx}{dy})^2} dy$ which will be the surface area of your curve when you rotate it about the y axis. The reason you want to rotate it about the y-axis is because you want the surface area of the ENTIRE cross section of the pond.
I got an answer of 1332.18.
Hint : Re-arrange your equation for y. Your limits should be from the bottom of the cross section of the pond to the top.
Last edited: Aug 8, 2013
4. Aug 8, 2013
### LCKurtz
You almost have it. The $ds$ element is at distance $x$ from the y axis, so the circumference of the circle it traces is $2\pi x$. You take that times the $ds$ length to get the surface area swept out. So you want $$\int_0^{20} 2\pi x~ds = \int_0^{20} 2\pi x\sqrt{1 + \frac{x^2}{1600} }~dx$$You only go from $0$ to $20$ in the limits because revolving it gets the "other side".
Last edited: Aug 8, 2013
5. Aug 8, 2013
### stunner5000pt
Thank you very much for all your help. I see now the integrand is the curved surface area of a cylinder where the radius is x and the height is the length of a curve formula | 2017-08-18T00:20:00 | {
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https://math.stackexchange.com/questions/2915935/radius-of-a-circle-touching-a-rectangle-both-of-which-are-inside-a-square | # Radius of a circle touching a rectangle both of which are inside a square
Given this configuration :
We're given that the rectangle is of the dimensions 20 cm by 10 cm, and we have to find the radius of the circle.
If we somehow know the distance between the circle and the corner of the square then we can easily find the radius. (It's equal to $$\sqrt{2}\times R-R$$)
I really can't understand how to solve it. Any help appreciated.
• This is fascinating, provided it is written correctly. At first it seems there is not enough information, but at the moment, I feel confident that there is. I will mess around with this. Sep 13 '18 at 18:29
• The information you have is $R(1-\cos\alpha) = 2a$ and $R(1-\sin\alpha) = a$, where $a = 10$. From there it follows that $R^2 = (R-2a)^2+(R-a)^2$. Sep 13 '18 at 18:29
• @amsmath juuuusssssttt beat me to it. Nice! Sep 13 '18 at 18:31
• If you set the upper left corner to be the origin $O(0,0)$, then the center of this circle has coordinates $(R,-R)$, so the circle is given by the equation $(x-R)^2+(y+R)^2=R^2$. Now note that the circle contains the point $(20,-10)$, and calculate $R$ from this.
– SMM
Sep 13 '18 at 18:31
• This question could benefit from adding the 20 cm and 10 cm to the image itself. Sep 14 '18 at 13:23
It is just using the pythagorean theorem:
$a=10$ $cm$
$b=20$ $cm$
$(r-a)^2+(r-b)^2=r^2$
$(r-10)^2+(r-20)^2=r^2$
$r^2+100-20r+r^2+400-40r=r^2$
$r^2-60r+500=0$
$r=50$ $cm$
$r=10$ $cm$
The $r=50$ $cm$ is the acceptable answer.
• It's worth mentioning that with a slightly looser set of conditions on the question, r=10cm is a valid answer which corresponds to the rectangle occupying the entire top half of a 20x20 square. Sep 14 '18 at 7:12
• @PhilipC, in that case how a circle with r=10cm can touch all four sides of a 20x20 square? Sep 14 '18 at 10:12
• A circle of radius 10cm has a diameter of 20cm, i.e. is 20cm wide and 20cm high, the same as the square. Sep 14 '18 at 11:24
• @PhilipC, yes but how is it going to touch the lower right corner of the rectangle externally? Sep 14 '18 at 11:40
• @XanderHenderson, Is it better now? Sep 14 '18 at 12:19
Place the center of the cricle at $O.$
Let the radius be $R$
The corner of the square is $(R,R)$ I have made a small alteration to the picture to create fewer negative numbers.
Offsetting by the rectange, the corer of the rectangle is $(R-20, R-10)$
And the distance from this point equals the $R.$
That should put you on your way to the solution.
$(x, y) = (R-20, R-10)$ as a point on the circle $y = \sqrt{R^2 - x^2}$
$R - 10 = \sqrt{R^2 - (R-20)^2}$
$(R- 10)^2 = R^2 - (R-20)^2$
$R^2 - 20R + 100 = R^2 - (R^2 - 40R + 400)$
$R^2 - 60R + 500 = 0$
$(R - 50)(R-10) = 0$
$R = 50$ is the only sensible option.
• $R=10$ is also possible : the 20x10 rectangle covers the top half of a 20x20 square. Sep 14 '18 at 8:10
• @EricDuminil: But that's not what the diagram shows... The rectangle is clearly entirely outside the circle which it would not be if it covered the entire top half of the square... Sep 14 '18 at 8:55
• @Chris It fits with the title : Radius of a circle touching a rectangle both of which are inside a square and I tend to interpret a drawing as a guideline, not a perfect representation on which you could simply measure the solution. I guess it's a personal preference though. Sep 14 '18 at 9:01
• @Eric Duminil and Chris. Thanks for your comments. If you allow a solution of R = 10, which touches 2 corners and a side and intersects the rectangle, there are an infinity of solutions with similar characteristics. Example: a 40x40 square with a radius of R = 20 which doesn't contradict the title. Sep 14 '18 at 13:12
• But if you assume that the circle is inscribed in the square, there are only 2 possibilities at most, right? Sep 14 '18 at 13:50
The point where rectangle touches the circle is $|R-a|$ and $|R-b|$ away from $x$ and $y$ axes, where $a$ and $b$ are lengths of sides of the rectangle and $R$ is the radius of the circle.
This leads to the equation $$(R-a)^2 + (R-b)^2 = R^2,$$ which has solutions $$R_{1,2} = a+b\pm\sqrt{2ab}.$$
One solution corresponds to a bigger rectangle (compared to the circle), one touching the circle on the other side, which is not the case here. Smaller rectangle compared to the circle means that the circle is bigger if the rectangle is kept fixed, so the correct radius is $$R = a+b+\sqrt{2ab}.$$
Plugging in $a=10$ and $b=20$ gives $R=50$.
You can use trig to get the same answer as those above.
Draw three lines: One from the center of the circle to the corner shared by the square and the rectangle. Then draw a line from the center of the circle to the corner nearest to the center of the circle. Draw a final line being the diagonal connecting the previously mentioned corners.
We know the length of the third line by the pythagorean theorem. If we call the side length of the square L, the length of the shorter of the two remaining lines is L/2. The length of the longer, L/sqrt(2).
Find the angle that the diagonal makes with the longer of the drawn lines allows you to apply the cosine rule.
The longer line meets the square's corner at a 45 degree angle with respect to either side. Then angle the diagonal makes with the left side of the square has a tangent of 2.
Apply the cosine rule then solve the resulting quadratic and you get two possible answers, only one of which is plausible. | 2021-10-24T02:55:59 | {
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http://mathhelpforum.com/algebra/39149-completing-square.html | # Math Help - Completing the square
1. ## Completing the square
Hello,
I'm having problems with completing the square can anyone comment on my working?
2x^2 - 3x -1 = 0
2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0
2(x - 3/4)^2 -17/16 =0
2(x - 3/4)^2 = 17/16
(x - 3/4)^2 = 17/32
x - 3/4 = +-√17/32
x = 3/4 +-√17/32
2. Originally Posted by Mouseman
Hello,
I'm having problems with completing the square can anyone comment on my working?
2x^2 - 3x -1 = 0
2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0
2(x - 3/4)^2 -17/16 =0
2(x - 3/4)^2 = 17/16
(x - 3/4)^2 = 17/32
x - 3/4 = +-√17/32
x = 3/4 +-√17/32
2x^2 - 3x = 1
x^2 - (3/2)x = 1/2
(x - 3/4)^2 - 9/16 = 1/2
x - 3/4 = +-sqrt(17/16)
x = 3/4 +-sqrt(17/16)
3. Thank you very much!
4. Originally Posted by sean.1986
2x^2 - 3x = 1
x^2 - (3/2)x = 1/2
(x - 3/4)^2 - 9/16 = 1/2
x - 3/4 = +-sqrt(17/16)
x = 3/4 +-sqrt(17/16)
The answer is to be x = 3/4 +- √17/4
I can't see where I am going wrong.
5. When completing the square, just use the general formula.
$a(x+\frac{b}{2a})^{2}+c-\frac{b^{2}}{4a}$
Plug in a,b,c and you're set.
6. Second attempt
2x^2 - 3x -1 = 0
2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0
2((x - 3/4)^2) -17/8 =0
2(x - 3/4)^2 = 17/8
(x - 3/4)^2 = 17/16
x - 3/4 = +-√17/16
x = 3/4 +-√17/16
Is my teacher wrong in stating that it is x = 3/4 +-√17/4?
7. Just to add my 2 cents......
$2x^2-3x-1=0$
1. First transpose the -1 to the right side of the equation.
$2x^2-3x=1$
2. Divide each term by 2
$x^2-\frac{3}{2}x=\frac{1}{2}$
3. Take half of the coefficient of x, square it and add it to both sides.
$x^2-\frac{3}{2}x+(\frac{3}{4})^2=\frac{1}{2}+\frac{9}{ 16}$
4. Noting the perfect square trinomial on the left:
$(x-\frac{3}{4})^2=\frac{17}{16}$
5. Take the square root of both sides:
$x-\frac{3}{4}=\pm\sqrt\frac{17}{16}$
6. Finally,
$x=\frac{3}{4}\pm\frac{\sqrt17}{4}$
$x=\frac{3\pm\sqrt17}{4}$
8. Hello,
Originally Posted by Mouseman
Second attempt
2x^2 - 3x -1 = 0
2(x -3/4)^2 -(-3/4)^2 - 1/2 = 0
2((x - 3/4)^2) -17/8 =0
2(x - 3/4)^2 = 17/8
(x - 3/4)^2 = 17/16
x - 3/4 = +-√17/16
x = 3/4 +-√17/16
Is my teacher wrong in stating that it is x = 3/4 +-√17/4?
Note that $16=4^2$
Therefore, $\sqrt{16}=4$, and this yields the result your teacher gave you
9. Yeah it's just a question of bracketing or simplifying.
sqrt (a/b) = sqrt(a) / sqrt(b)
so sqrt(17/16) = sqrt(17) / sqrt(16) = sqrt(17) / 4
Both answers are correct but I guess I should've simplified. Sorry mate! | 2014-04-17T10:24:02 | {
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https://math.stackexchange.com/questions/2932610/is-it-necessary-to-assume-a-moment-generating-function-exists/2932668 | # Is it necessary to assume a moment generating function exists?
Consider random variables A, B, and C. We know that A = B + C. We also know that A and C have an MGF. Is it the case that B must have a MGF?
Addition: Does this change if we know A and C both come from (different) chi-squared distributions? I am tasked with finding the distribution of B. If I can just do MGF(A) / MGF (C) = MGF (B) then it's simple... but can I even write this statement without assuming MGF (B) exists?
Generally -- You can't compute the MGF of B
In general, you can't compute the MGF of $$B$$ if you only know the MGFs of $$A$$ and $$C$$. For example, consider two possible joint distributions of $$A$$ and $$C$$:
Case 1: P( A=0 and C=0) = 1/2 and P(A=1 and C=1)=1/2. In this case, the MGFs of A and C are $$(1+\exp(t))/2$$ and the MGF of B is 1.
Case 2: P( A=0 and C=1) = 1/2 and P(A=1 and C=0)=1/2. In this case, the MGFs of A and C are $$(1+\exp(t))/2$$ and the MGF of B is $$\frac{\exp(-t)+\exp(t)}2=\cosh t$$.
Notice that in both Case 1 and Case 2 the MGFs for $$A$$ and $$C$$ were $$(1+exp(t))/2$$, but the MGF for $$B$$ changed from Case 1 to Case 2.
Generally -- You can prove existence of an MGF for B
Although you can't compute the MGF of $$B$$, you can prove that $$M_B(t)$$ exists for $$(*)\quad t\in D=\frac12 (Dom(M_A)\cap (-Dom(M_C)).$$ Suppose $$t\in D$$. If the MGF's of $$A$$ and $$C$$ exist, then for all $$t\in Dom(M_A)$$, $$M_A(t)=||\exp(ta)||_1<\infty$$ and for all $$t\in (-Dom(M_C))$$, $$M_C(-t)=||\exp(-tc)||_1<\infty$$ where $$||g||_p=\left(\int\int |g(a,c)|^p\; f(a,c)\; da dc\right)^{1/p}$$ is the $$L_p$$-norm of $$g$$ over the joint probability space and $$f(a,c)$$ is the joint pdf of $$A$$ and $$C$$. That implies $$||\exp(ta/2)||_2 < \infty$$ and $$||\exp(-tc/2)||_2 < \infty$$. By the Hölder's inequality or the Schwarz inequality, $$||\exp(ta)\exp(-tc)||_1<\infty$$. But, $$||\exp(ta)\exp(-tc)||_1= ||\exp(t(a-c)||_1= E[\exp(tB)]=M_B(t).$$ This proves that $$M_B(t)$$ exists for $$t\in D$$.
If A and C are independent, you can compute the MGF of B
If $$A$$ and $$C$$ are independent and $$B = A-C$$, then it must be the case that $$(**) \quad M_B(t) = M_A(t)\cdot M_C(-t)$$ whenever $$t\in Dom(M_A)\cap(-Dom(M_C))$$ (see e.g. Wikipedia). Here is a rough proof.
If $$t\in Dom(M_A)\cap(-Dom(M_C))$$, then $$M_A(t)\cdot M_C(-t) = \int_{a=-\infty}^\infty \exp(t a) dF_A(a) \cdot \int_{c=-\infty}^\infty \exp(-t c) dF_C(c)$$ $$= \int_{a=-\infty}^\infty \int_{c=-\infty}^\infty \exp(t (a-c)) dF_A(a) dF_C(c)$$ $$= \int_{b=-\infty}^\infty \exp(t b) dF_B(b) = M_B(t)$$ where $$F_A, F_B$$, and $$F_C$$ are the cumulative distribution functions of $$A, B$$, and $$C$$ respectively.
If A and C have $$\chi^2$$ distributions
In general, if $$A$$ and $$C$$ have $$\chi^2$$ distributions, you can only state that $$B$$ has an MGF and the domain of $$B$$'s MGF is \begin{align} Dom(B) &= \frac12 (Dom(M_A)\cap (-Dom(M_C))\\ &= \frac12 ((-\infty,2)\cap (-(-\infty,2))\\ &= \frac12 ((-\infty,2)\cap (-2,\infty))\\ &= \frac12 (-2,2)=(-1,1)\\ \end{align} by (*) and the fact that the domain of the MGF of a $$\chi^2$$ distribution is $$(-\infty,2)$$.
If you know that $$A$$ and $$C$$ have $$\chi^2$$ distributions, and you know that they are independent, then you can look up the MGFs for $$\chi^2$$ distributions and apply formula (**) to compute the formula for the MGF of $$B$$.
• Thank you. The question asks to prove that B is chi-squared distributed. It sounds like we know that a MGF exists, but that we can't prove it's specifically chi-squared? – purpleostrich Sep 28 at 2:06
• I learned a lot about MGFs and a bit about $\chi^2$ distributions while trying to figure out the answer to your question. :) – irchans Sep 28 at 8:24
• @purpleostrich, I suspect that if this was posed as an exercise to you then there is a missing assumption, for instance that $B$ and $C$ are independent, or as explored by irchans the assumption could be that $A$ and $C$ are independent (although I find this latter assumption somewhat less likely than the former). – pre-kidney Sep 28 at 9:01 | 2018-10-22T00:52:18 | {
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http://math.stackexchange.com/questions/51926/a-stronger-version-of-discrete-liouvilles-theorem | # A stronger version of discrete “Liouville's theorem”
If a function $f : \mathbb Z\times \mathbb Z \rightarrow \mathbb{R}^{+}$ satisfies the following condition
$$\forall x, y \in \mathbb{Z}, f(x,y) = \dfrac{f(x + 1, y)+f(x, y + 1) + f(x - 1, y) +f(x, y - 1)}{4}$$
then is $f$ constant function?
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You probably wanto to add a boundedness condition. Otherwise $f(x,y)=x$ is a counterexample. – Julián Aguirre Jul 17 '11 at 10:24
@Julian Aguirre: since $x\in\mathbb Z$, we don't have $f(x,y)\geq 0$. – Davide Giraudo Jul 17 '11 at 11:08
@girdav You are right. The lower bound is probably enough. – Julián Aguirre Jul 17 '11 at 13:08
You can prove this with probability.
Let $(X_n)$ be the simple symmetric random walk on $\mathbb{Z}^2$. Since $f$ is harmonic, the process $M_n:=f(X_n)$ is a martingale. Because $f\geq 0$, the process $M_n$ is a non-negative martingale and so must converge almost surely by the Martingale Convergence Theorem. That is, we have $M_n\to M_\infty$ almost surely.
But $(X_n)$ is irreducible and recurrent and so visits every state infinitely often. Thus (with probability one) $f(X_n)$ takes on every $f$ value infinitely often.
Thus $f$ is a constant function, since the sequence $M_n=f(X_n)$ can't take on distinct values infinitely often and still converge.
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I love probabilistic arguments in analysis! Very nice. – Jonas Teuwen Jul 17 '11 at 11:46
This is indeed very nice: Question: Is it possible to change this argument in such a way that it applies for $\mathbb{Z}^n$ instead of $\mathbb{Z}^2$ only? Is it even true that a non-negative harmonic function on $\mathbb{Z}^n$ is constant for $n \geq 3$? For bounded ones this seems clear by considering the Poisson boundary. – t.b. Jul 17 '11 at 11:53
@Byron: This paper contains the claim that it is true that "nonnegative nearest-neighbors harmonic function on $\mathbb{Z}^d$ are constant for any $d$" on page 2. – t.b. Jul 17 '11 at 12:08
The usual Liouville theorem also holds with just a one-sided bound. – GEdgar Jul 18 '11 at 13:56
If you know that the real part of an entire function $f(z)$ is non-negative on the complex plane, what can you say about the function $g(z)=f(z)/(1+f(z))$? – Jyrki Lahtonen Jul 18 '11 at 14:07
I can give a proof for the d-dimensional case, if $f\colon\mathbb{Z}^d\to\mathbb{R}^+$ is harmonic then it is constant. The following based on a quick proof that I mentioned in the comments to the same (closed) question on MathOverflow, Liouville property in Zd. [Edit: I updated the proof, using a random walk, to simplify it]
First, as $f(x)$ is equal to the average of the values of $f$ over the $2d$ nearest neighbours of $x$, we have the inequality $f(x)\ge(2d)^{-1}f(y)$ whenever $x,y$ are nearest neighbours. If $\Vert x\Vert_1$ is the length of the shortest path from $x$ to 0 (the taxicab metric, or $L^1$ norm), this gives $f(x)\le(2d)^{\Vert x\Vert_1}f(0)$. Now let $X_n$ be a simple symmetric random walk in $\mathbb{Z}^d$ starting from the origin and, independently, let $T$ be a random variable with support the nonnegative integers such that $\mathbb{E}[(2d)^{2T}] < \infty$. Then, $X_T$ has support $\mathbb{Z}^d$ and $\mathbb{E}[f(X_T)]=f(0)$, $\mathbb{E}[f(X_T)^2]\le\mathbb{E}[(2d)^{2T}]f(0)^2$ for nonnegative harmonic $f$. By compactness, we can choose $f$ with $f(0)=1$ to maximize $\Vert f\Vert_2\equiv\mathbb{E}[f(X_T)^2]^{1/2}$.
Writing $e_i$ for the unit vector in direction $i$, set $f_i^\pm(x)=f(x\pm e_i)/f(\pm e_i)$. Then, $f$ is equal to a convex combination of $f^+_i$ and $f^-_i$ over $i=1,\ldots,d$. Also, by construction, $\Vert f\Vert_2\ge\Vert f^\pm_i\Vert_2$. Comparing with the triangle inequality, we must have equality here, and $f$ is proportional to $f^\pm_i$. This means that there are are constants $K_i > 0$ such that $f(x+e_i)=K_if(x)$. The average of $f$ on the $2d$ nearest neighbours of the origin is $$\frac{1}{2d}\sum_{i=1}^d(K_i+1/K_i).$$ However, for positive $K$, $K+K^{-1}\ge2$ with equality iff $K=1$. So, $K_i=1$ and $f$ is constant.
Now, if $g$ is a positive harmonic function, then $\tilde g(x)\equiv g(x)/g(0)$ satisfies $\mathbb{E}[\tilde g(X_T)]=1$. So, $${\rm Var}(\tilde g(X_T))=\mathbb{E}[\tilde g(X_T)^2]-1\le\mathbb{E}[f(X_T)^2]-1=0,$$ and $\tilde g$ is constant.
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Taxicab metric. Never heard that name (I learn maths in french at school). Funny! – Patrick Da Silva Jul 17 '11 at 20:24
@Patrick: Also called the Manhattan metric. – George Lowther Jul 17 '11 at 20:30
LAAAAAAAAAWL. Funnier. – Patrick Da Silva Jul 17 '11 at 20:39
Note: A similar proof will also show that harmonic $f\colon\mathbb{R}^d\to\mathbb{R}^+$ is constant. Interestingly, in the two dimensional case, Byron's proof can be modified to show that harmonic $f\colon\mathbb{R}^2\setminus\{0\}\to\mathbb{R}^+$ is constant (as 2d Brownian motion has zero probability of hitting 0 at positive times). Neither of the proofs generalize to harmonic $f\colon\mathbb{R}^d\setminus\{0\}\to\mathbb{R}^+$ for $d\not=2$. In fact, considering $f(x)=\Vert x\Vert^{2-d}$, we see that $f$ need not be constant for $d\not=2$. – George Lowther Jul 17 '11 at 22:54
Here is an elementary proof assuming we have bounds for $f$ on both sides.
Define a random walk on $\mathbb{Z}^2$ which, at each step, stays put with probability $1/2$ and moves to each of the four neighboring vertices with probability $1/8$. Let $p_k(u,v)$ be the probability that the walk travels from $(m,n)$ to $(m+u, n+v)$ in $k$ steps. Then, for any $(m, n)$ and $k$, we have $$f(m, n) = \sum_{(u,v) \in \mathbb{Z}^2} p_k(u,v) f(m+u,n+v).$$ So $$f(m+1, n) - f(m, n) = \sum_{(u,v) \in \mathbb{Z}^2} \left( p_k(u-1,v) - p_k(u,v) \right) f(m+u,n+v).$$ If we can show that $$\lim_{k \to \infty} \sum_{(u,v) \in \mathbb{Z}^2} \left| p_k(u-1,v) - p_k(u,v) \right| =0 \quad (\ast)$$ we deduce that $$f(m+1,n) = f(m,n)$$ and we win.
Remark: More generally, we could stay put with probability $p$ and travel to each neighbor with probability $(1-p)/4$. If we choose $p$ too small, then $p_k(u,v)$ tends to be larger for $u+v$ even then for $u+v$ odd, rather than depending "smoothly" on $(u,v)$. I believe that $(\ast)$ is true for any $p>0$, but this elementary proof only works for $p > 1/3$. For concreteness, we'll stick to $p=1/2$.
We study $p_k(u,v)$ using the generating function expression $$\left( \frac{x+x^{-1}+y+y^{-1}+4}{8} \right)^k = \sum_{u,v} p_k(u,v) x^u y^v.$$
Lemma: For fixed $v$, the quantity $p(u,v)$ increases as $u$ climbs from $-\infty$ up to $0$, and then decreases as $u$ continues climbing from $0$ to $\infty$.
Proof: We see that $\sum_u p_k(u,v) x^u$ is a positive sum of Laurent polynomials of the form $(x/8+1/2+x^{-1}/8)^j$. So it suffices to prove the same thing for the coefficients of this Laurent polynomial. In other words, writing $(x^2+8x+1)^k = \sum e_i x^i$, we want to prove that $e_i$ is unimodal with largest value in the center. Now, $e_i$ is the $i$-th elementary symmetric function in $j$ copies of $4+\sqrt{15}$ and $j$ copies of $4-\sqrt{15}$. By Newton's inequalities, $e_i^2 \geq \frac{i (2j-i)}{(i+1)(2j-i+1)} e_{i-1} e_{i+1} > e_{i-1} e_{i+1}$ so $e_i$ is unimodal; by symmetry, the largest value is in the center. (The condition $p>1/3$ in the above remark is when the quadratic has real roots.) $\square$
Corollary: $$\sum_u \left| p_k(u-1,v) - p_k(u,v) \right| = 2 p_k(0,v).$$
Proof: The above lemma tells us the signs of all the absolute values; the sum is \begin{multline*} \cdots + (p_k(-1,v) - p_{k}(-2,v)) + (p_k(0,v) - p_{k}(-1,v)) + \\ (p_k(0,v) - p_k(1,v)) + (p_k(1,v) - p_k(2,v)) + \cdots = 2 p_k(0,v). \qquad \square\end{multline*}
So, in order to prove $(\ast)$, we must show that $\lim_{k \to \infty} \sum_v p_k(0,v)=0$. In other words, we must show that the coefficient of $x^0$ in $\left( \frac{x}{8}+\frac{3}{4} + \frac{x^{-1}}{8} \right)^k$ goes to $0$.
There are probably a zillion ways to do this; here a probabilistic one. We are rolling an $8$-sided die $k$ times, and we want the probability that the numbers of ones and twos are precisely equal. The probability that we roll fewer than $k/5$ ones and twos approaches $0$ by the law of large numbers (which can be proved elementarily by, for example, Chebyshev's inequality). If we roll $2r > k/5$ ones and twos, the probability that we exactly the same number of ones and twos is $$2^{-2r} \binom{2r}{r} < \frac{1}{\sqrt{\pi r}} < \frac{1}{\sqrt{\pi k/10}}$$ which approaches $0$ as $k \to \infty$. See here for elementary proofs of the bound on $\binom{2r}{r}$.
I wrote this in two dimensions, but the same proof works in any number of dimensions
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Assuming that $f$ is bounded, you can prove that it is constant by a discrete analogue of the Borel-Caratheodory inequality, as shown in this equivalent question.
The key fact is such a function attains its maximum on a ball on the boundary, so it suffices to provide lower bounds for
$$\Gamma_N = \max_{|x|+|y|=N}f(x,y).$$
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https://math.stackexchange.com/questions/1978572/getting-2-different-solutions-to-the-integral-of-fracdx2x/1978579 | # Getting 2 different solutions to the integral of $\frac{dx}{2x}$
I get two different answers that seem to conflict.
Is there an error in one method??
### Method 1
\begin{align} \int \frac{\mathrm{d}x}{2x}&=\frac 1 2 \int\frac{\mathrm{d}x}{x}\\ &=\frac 1 2\ln|x|+C \end{align}
### Method 2
\begin{align} \int \frac{\mathrm{d}x}{2x}&=\frac 1 2 \int \frac{2\mathrm{d}x}{2x}\\ &=\frac 1 2\int \frac{\mathrm{d}u}{u},\;\;\text{ where }u=2x,\ \mathrm{d}u=2\mathrm{d}x\\ &=\frac 1 2 \big(\ln |u| + C\big)\\ &=\frac 1 2 \ln|2x|+C \end{align}
• As the answer below details: the $C$'s are 'different', for want of a better word. Your "paradox" here is one of the reasons we are careful to always put a $C$ at the end, and also have to stay constantly aware of what it means. Oct 21, 2016 at 13:58
• Another good example is integrating $\sin(x)\cos(x)$. Try integration by parts, to get $\frac{1}{2}\sin^2(x)+C$, then try using the trig identity $\sin(x)\cos(x)=\frac{1}{2}\sin(2x)$ to get $-\frac{1}{4}\cos(2x)+D$. A hint on reconciling the answers is $\cos(2x)=1-2\sin^2(x)$. Oct 21, 2016 at 14:06
• When we get two ostensibly different solutions to an anti-differentiation problem, it is often helpful to check whether the solutions (ignoring the constants) are vertical translations of each other. If so, the two solutions must, of course, differ by a constant. In this case, $\frac 1 2\ln|x|$ translated vertically $\ln\sqrt{2}$ units yields $\frac 1 2\ln|2x|$. Oct 21, 2016 at 16:40
Hint: $\frac{1}{2}\ln{|2x|} + C = \frac{1}{2}\ln{|x|} + \frac{1}{2}\ln{2} + C = \frac{1}{2}\ln{|x|} + \left(\frac{1}{2}\ln{2} + C\right)$. | 2022-05-19T09:41:19 | {
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http://math.stackexchange.com/questions/819527/factoring-in-the-derivative-of-a-rational-function/819617 | # Factoring in the derivative of a rational function
Given that $$f(x) = \frac{x}{1+x^2}$$
I have to find $$\frac{f(x) - f(a)}{x-a}$$
So some progressing shows that:
$$\frac{\left(\frac{x}{1+x^2}\right) - \left(\frac{a}{1+a^2}\right)}{x-a} = \frac{(x)(1+a^2)-(a)(1+x^2)}{(1+x^2)(1+a^2)}\cdot\frac{1}{x-a} = \frac{x+xa^2-a-ax^2}{(1+x^2)(1+a^2)(x-a)}$$
Now, is it possible to factor $x+xa^2-a-ax^2$? I can't seem to find a way, as for simplifying the whole thing. Is there any rule I can use, and I'm unable to see?
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Can't you just use the quotient rule to find this derivative? – user2357112 Jun 3 '14 at 21:28
Since $x-a$ is in the denominator, it makes sense to consider the possibility that $x-a$ is a factor of the numerator. (If you know the factor theorem, you can see this is the case, since the numerator equals zero when $x = a.)$ I see $x-a$ in the numerator, along with two other terms. Very little cleverness is needed at this point to write
$$x-a + xa^2 - ax^2 \; = \; x - a + ax(a-x)$$ $$= \; (x-a) - ax(x-a) \; = \; (x-a)(1-ax)$$
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Ahhhh. So much time looking at it and I didn't realise this. I don't know about the factor theorem, something to look at. Thank you for your answer. – sidyll Jun 3 '14 at 18:08
@sidyll: For more about using the factor theorem, see my answer at Finding limit of a quotient. For a less obvious use of the factor theorem, see my comments at How to simpify this?. – Dave L. Renfro Jun 3 '14 at 18:12
Thank you for your references – sidyll Jun 3 '14 at 18:17
@sidyll: I just remembered that I posted a couple of old handouts of mine on the factor theorem a couple of years ago. See the files attached to this 12 January 2012 sci.math post archived at Math Forum. – Dave L. Renfro Jun 3 '14 at 18:25
@sidyll As for derivatives, generally the arithmetic works nicely using the quotient rule - see my answer. I'll bet Dave can give links to interesting expositions in AMM, Math. Mag, etc. – Bill Dubuque Jun 3 '14 at 19:02
\eqalign{x+xa^2-a-ax^2&= x-a+xa^2-ax^2 \\ & = x-a+x(a^2-ax) \\ &= x-a+x(a(a-x)) \\ &= x-a+x(-a(x-a)) \\ &=\color{blue}{x-a}-ax\color{blue}{(x-a)} \\ &=(x-a)(1-ax).\;\checkmark } Therefore you can conclude that: \eqalign{\require{cancel}\dfrac{f(x)-f(a)}{x-a}&=\dfrac{x+xa^2-a-ax^2}{(1+x^2)(1+a^2)(x-a)} \\ &=\dfrac{\color{red}{\cancel{\color{black}{(x-a)}}}(1-ax)}{(1+x^2)(1+a^2)\color{red}{\cancel{\color{black}{(x-a)}}}} \\ &= \dfrac{1-ax}{(1+x^2)(1+a^2)}. }\tag{x\neq a}
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Thank you very much for your examples in continuing it. – sidyll Jun 3 '14 at 18:09
+1 for nice format. I learn something new about $LaTeX$ from you. – Tunk-Fey Jun 3 '14 at 18:10
@Tunk-Fey I've also learned a lot from your answers and your TeX style! $\overset{\cdot\cdot}\smile$ BTW you can write a more beautiful LaTeX as: $\LaTeX$ using \LaTeX, and there's also a variant for TeX namely $\TeX$ which is produced by \TeX. – Hakim Jun 3 '14 at 18:13
@sidyll You're welcome! Glad I could help! ;-) – Hakim Jun 3 '14 at 18:14
@حكيمالفيلسوفالضائع Thank you for the code. I didn't know that code before. $\ddot\smile$ – Tunk-Fey Jun 3 '14 at 18:17
By the quotient rule for the difference $\ f'(x)\, :=\, \dfrac{f(x)-f(a)}{x-a}$
$$\quad \begin{eqnarray} (g/h)'(x) &=\,\ & \dfrac{\color{#c00}{g'(x)} h(a) - g(a)\color{#0a0}{h'(x)}}{h(a)h(x)}\\ \begin{array}{l}\ g(x)=x\qquad\Rightarrow\,\color{#c00}{g'(x) = 1}\\ h(x) = 1+x^2\,\Rightarrow\,\color{#0a0}{h'(x) = x+a}\\\end{array}\ \Bigg\}\!\!\!\!\!&=& \dfrac{\color{#c00}1\cdot (1+a^2)\overset{\phantom{I^I}}-a(\color{#0a0}{x+a})}{(1+a^2)(1+x^2)}\ =\ \dfrac{1-ax}{(1+a^2)(1+x^2)}\end{eqnarray}$$
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Hint: you can guess that something interesting is going to happen near $x = a$, which suggests looking to factor $x-a$.
Indeed, $x + xa^2 - a - ax^2 = (x-a) + xa(x-a)$.
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Factor: $$x+xa^2−a−ax^2=x-a-ax^2+xa^2=(x-a)-ax(x-a)=(1-ax)(x-a)$$ Thus, $$\frac{f(x)-f(a)}{x-a}=\frac{\frac{(1-ax)(x-a)}{(1+x^2)(1+a^2)}}{(x-a)}=\boxed{\frac{1-ax}{(1+x^2)(1+a^2)}}$$
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https://math.stackexchange.com/questions/2494774/questions-concerning-smallest-fraction-between-two-given-fractions | # Questions concerning smallest fraction between two given fractions.
I recently encountered this on a practice test
Find the smallest positive integer $n$ such that there exists an integer $m$ satisfying $0.33 < \frac{m}{n} < \frac{1}{3}$
\begin{align}0.33 < \frac{m}{n} < \frac{1}{3}&\implies(\frac{33}{100} < \frac{m}{n})\land(\frac{m}{n}<\frac{1}{3}) \\ &\implies(33n<100m)\land(3m<n) \\\end{align} and thus $n=3m+1$.
So \begin{align}33n<100m&\implies33(3m+1)<100m \\ &\implies 99m+33<100m\\ &\implies m>33\end{align} Taking $m\geq34$, we find that $n=34\times3+1=103$
After the test, I had the following thoughts, which I do not know how to answer.
Questions
Will the solution for $n$ always yield the smallest $m$ possible?
Does this method (find minimum value of $n$ and substitute) work for all such problems? (where $m,n\in\mathbb{Z}$)
Can we generalise $n$ for all possible fraction ranges, and if so, will $n$ always be of a certain form compared to $a,b,c,$ and $d$? (where $\frac{a}{b}<\frac{m}{n}<\frac{c}{d}$).
note: sorry for asking multiple questions in one post (which I know some people frown on) but I feel posting multiple questions with the same 'introduction' (problem+proof) would clutter and be somewhat cumbersome.
• In my opinion the smallest is $n=103$ because the smallest $m$ is $34$. Indeed if you solve for positive $n$ $$\frac{34}{n}<\frac{1}{3}$$ you get $n>102$ – Raffaele Oct 29 '17 at 11:01
• @mathlove typo by me, thanks. I've also noticed that in our example $\frac{33}{100}<\frac{34}{103}<\frac{1}{3}$, the numerator of the middle fraction is the sum of the other numerators, and the denominator for the middle fraction is the sum of the other denominators. I wonder if this is a coincidence or not, though I won't put that into the question (three questions in one post is already plenty) – user472341 Oct 29 '17 at 11:03
• The continued fraction for $\frac13$ is $(0;3)=(0;3,\infty)$ and the continued fraction for $0.33$ is $(0;3,33)$. The simplest between them is $(0;3,34)=\frac{34}{103}$. – robjohn Oct 29 '17 at 11:04
Can we generalise $n$ for all possible fraction ranges, and if so, will $n$ always be of a certain form compared to $a,b,c,$ and $d$? (where $\frac{a}{b}<\frac{m}{n}<\frac{c}{d}$).
In the following, $a,b,c,d,m,n$ are positive integers.
$\frac ab\lt \frac mn\lt \frac cd$ is equivalent to $$na\lt mb\qquad\text{and}\qquad \frac{md}{c}\lt n$$
From the second inequality, we can set $n=\lfloor\frac{md}{c}\rfloor+1$ where $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$.
So, from the first inequality, we get $$\left(\left\lfloor\frac{md}{c}\right\rfloor+1\right)a\lt mb\tag1$$
As a result, we can say that the smallest positive integer $n$ such that there exists an integer $m$ satisfying $\frac ab\lt\frac mn\lt\frac cd$ is given by$$\left\lfloor\frac{Md}{c}\right\rfloor+1$$ where $M$ is the smallest integer $m$ satisfying $(1)$.
If $c=1$, then the smallest positive integer $n$ such that there exists an integer $m$ satisfying $\frac ab\lt\frac mn\lt\frac 1d$ is given by$$\left(\left\lfloor\frac{a}{b-da}\right\rfloor+1\right)d+1$$
Yes to both of your questions. This is essentially the mediant for two terms in a Farey sequence. There is some deep theory here. Read up on Farey sequences for more information.
• I had three questions - i'm assuming you're talking about the first two? either way, thanks for the links. – user472341 Oct 29 '17 at 11:09
• The mediant is not always the smallest. Take for instance $\frac13$ and $\frac34$: the mediant is $\frac47$, which is between the two, but $\frac12$ is the fraction with the smallest denominator. – robjohn Oct 29 '17 at 11:12
• I think the mediant is only relevant when the fractions are consecutive terms in some Farey sequence. In the example @robjohn gives, $1/3$ and $3/4$ are not consecutive terms in any Farey sequence. – Gerry Myerson Oct 29 '17 at 11:17
• You can iterate the algorithm using the mediant. For 1/3, 3/4 we get 4/7 and then with 1/3, 4/7 we get 1/2. – i. m. soloveichik Oct 29 '17 at 11:27 | 2019-11-12T18:17:48 | {
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https://math.stackexchange.com/questions/311605/function-notation-terminology/311637 | # Function notation terminology
Given the function $f:X\longrightarrow Y$, $X$ is called the domain while $Y$ is called the codomain. But what do you call $f(x)=x^2$ in this context, where $x\in X$? That is to say - what is the name for the $f(x)$ notation?
And while I'm here, what is the proper way to write a function like this? Would it be $f:\mathbb{R}\to\mathbb{R},\;f(x)=x^2$?
Edit:
I figured I'd add this to add a bit of context into why I'm asking. I'm writing a set of notes in LaTeX, and I'd like to use the correct terminology for the definition of a function.
A function from set $A$ to set $B$, denoted by $$f:A\to B;x\mapsto f(x)$$ is a mapping of elements from set $A$, (the $\textit{domain}$) to elements in set $B$ (the $\textit{codomain}$) using the $\color{blue}{\sf function}$ $f(x)$. The domain of a function is the set of all valid elements for a function to map from. The codomain of a function is the set of all possible values that an element from the domain can be mapped to. The $\textit{range}$ (sometimes called the $\textit{image}$) of a function is a subset of the codomain, and is the set of all elements that actually get mapped to by the function $f$.
Here I'm pretty sure the highlighted word "function" is not right.
• I suggest \to or \rightarrow instead of \longrightarrow. The last one is there for if you need the arrow to be longer because you're writing something over it. As in $\overset{\text{text}}{\longrightarrow}$ instead of $\overset{\text{text}}{\to}$.
– Jim
Feb 22, 2013 at 23:16
• why do you think that function is not right = Feb 22, 2013 at 23:36
• Some would call "$f(x)=x^2$" the rule of the function $f$. Feb 22, 2013 at 23:44
• @DominicMichaelis Well, function is used in two places to mean two slightly different things; I don't like using function to describe part of a function. Feb 22, 2013 at 23:47
I can remember to read this text, and being puzzled with the exact same question. From what I've learned from my teacher, you're right, writing down something as "the function $f(x)$..." is sloppy notation. However, many books/people will use it this way.
If you're are very precise, $f(x)$ is not a function or an map. I don't know of a standard way to refer to $f(x)$, but here is some usage I found on the internet:
• The output of a function $f$ corresponding to an input $x$ is denoted by $f(x)$.
• Some would call "$f(x)=x^2$" the rule of the function $f$.
• For each argument $x$, the corresponding unique $y$ in the codomain is called the function value at $x$ or the image of $x$ under $f$. It is written as $f(x)$.
• If there is some relation specifying $f(x)$ in terms of $x$, then $f(x)$ is known as a dependent variable (and $x$ is an independent variable).
A correct way to notate your function $f$ is: $$f:\Bbb{R}\to\Bbb{R}:x\mapsto f(x)=x^2$$
Note that $f(x)\in\Bbb{R}$ and $f\not\in\Bbb{R}$. But the function $f$ is an element of the set of continuous functions, and $f(x)$ isn't.
In some areas of math it is very important to notate a function/map specifying it's domain, codomain and function rule. However in for example calculus/physics, you'll see that many times only the function rule $f(x)$ is specified, as the reader is supposed to understand domain/codmain from the context.
You can also check those questions:
• Nice, Kasper +1 Feb 23, 2013 at 1:50
normally you say $f:X\rightarrow Y$; $x\mapsto f(x)$, as a functions takes an element from $X$ and give you one from $Y$. $$y=f(x)$$ Is an equation, and not a definition of a function in a strict sense.
The proper way would be $$f:\mathbb{R}\rightarrow \mathbb{R}; \ x\mapsto x^2$$
The Image of a function is definied as $$\operatorname{im}f:=\{f(x)|x\in X\}$$ It is often written as $$f(X):=\{f(x)|x\in X\}$$ Notice that $X$ is a set, not an element. so $f(\{x\})=\{f(x)\}\subseteq Y$ while $f(x)\in Y$
• I was wondering - is $f(x)$ called the image? I've seen that word used but only know it to be synonymous with Range. Feb 22, 2013 at 23:15
• not $f(x)$ but $f(X)$ note the different as $X$ is a set but $x$ is an element Feb 22, 2013 at 23:19
• @agent154 You can say "$f(x)$ is the image of the element $x$ under the function $f$". Perhaps this is what you were thinking of (here of course "image" means something different than the image of the function). Feb 22, 2013 at 23:32
• @davidMitra i would be careful with that, as the image is a set but $f(x)$ is an element of $Y$ Feb 22, 2013 at 23:34
• The notation $f(X)$ is also written (in set theory) as $f[X]$ and $f''X$ sometimes, often because the elements of the set are sets themselves, and sometimes $f(x)\neq f''x$. Feb 22, 2013 at 23:35
From what I understand, $f$ is the function, so saying "a function $f(x)$" would be wrong. Instead, you could say "a function $f$", or if you didn't want to assign it a name, "a function $x\mapsto x^2$", or if you want to specify the domain and codomain, "a function $f:X\to Y$".
If you just want to define the function's input-output relationship, $f(x)=x^2$ suffices, maybe with a $\forall x\in X$ at the beginning if you want to do it properly. That doesn't necessarily define the domain though, that set could just be a subset of the domain. $f(x)$ is really no different from $f(1)$, except that $x$ is a (qualified) variable. It's basically a rule defining the input and output for an improper subset of its domain.
The addition of "$:X\to Y$" is useful if you want to specify both the domain and codomain when defining the function. As far as I can tell, you can add it after a function like $f$ and the whole expression still refers to the function. You can't really use it with the $f(x)$ style, so I'd do something like
$f:\mathbb R\to\mathbb R=x\mapsto x^2$ | 2022-05-29T03:09:32 | {
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https://stats.stackexchange.com/questions/294737/what-is-the-variance-of-a-binomial-distribution-with-1-and-1 | # What is the variance of a binomial distribution with -1 and 1?
I am struggling to see how to solve the following problem:
I have n i.i.d bernoulli trials. The result can be -1 or 1. I can figure out the expected value of this = n(p-(1-p) but how do I know what the variance is if p is known?
I know that: $${\rm Var}(X) = E[X^2] - E[X]^2$$ I don't know where to go from here.
• In general, for a Bernoulli Random Variable E(X) is $P(X=x_1)*x_1+P(X=x_2)*x_2$. In the "default" case $x_1 = 1, x_2 = 0$, which reduces this to E(X) = p. In your case, $x_1 = 1$ and $x_2 = -1$. You can then plug in what you get for E(X) into the formula $Var(X) = E(X^2) - E(X)^2$. The variance of the default case is $p*(1-p)$ – Pugl Jul 27 '17 at 13:57
• If B is a 0-1 Bernoulli variable with probability of a 1 at $p$, then you're describing a variable, ($Z$ say) where $Z=2B-1$. Just apply basic properties of variance to that linear transformation. – Glen_b Jul 28 '17 at 2:06
When you work with variances, know these facts (in addition to the definition of variance):
1. The variance of a sum of independent (or just uncorrelated) variables is the sum of their variances.
2. The expectation of a sum of variables (independent or not) is the sum of their expectations.
3. The expectation (of any discrete variable) is the sum of each possible value multiplied by its probability.
As an example, let's apply them to your case. You may model the sum of your Bernoulli trials with a variable $X$ expressed as the sum of $n$ independent variables $X_i$. Each $X_i$ takes on the value $1$ with probability $p$ and the value $-1$ with the probability $1-p$.
Fact $(3)$ asserts $$\mathbb{E}(X_i) = p(1) + (1-p)(-1) = 2p-1$$ and $$\mathbb{E}(X_i^2) = p(1)^2 + (1-p)(-1)^2 = p + (1-p) = 1.$$
Fact $(2)$ asserts \eqalign{\mathbb{E}(X) &= \mathbb{E}(X_1+\cdots+X_n) = \mathbb{E}(X_1) + \cdots + \mathbb{E}(X_n) = n\mathbb{E}(X_1) \\&=n(2p-1).}
The definition of variance now tells you$$\operatorname{Var}(X_i) = \mathbb{E}(X_i^2) - \mathbb{E}(X_i)^2 = 1 - (2p-1)^2 = 4p(1-p).$$
Consequently, fact $(1)$ yields \eqalign{\operatorname{Var}(X) &= \operatorname{Var}(X_1+\cdots+X_n) = \operatorname{Var}(X_1) + \cdots + \operatorname{Var}(X_n)=n \operatorname{Var}(X_1) \\&= n(4p(1-p)).}
The appearance of the factor of $4$ might seem somewhat mysterious. There's another extremely useful fact you might consider using to shortcut these considerations and identify the origin of that factor:
(4) The variance of a shifted, rescaled variable is the square of the scale factor times the variance. In mathematical symbols, $$\operatorname{Var}(\sigma X + \mu) = \sigma^2 \operatorname{Var}(X)$$ no matter what values the numbers $\sigma$ and $\mu$ might have.
This applies by noting that your $X_i$ can all be expressed as $2Y_i-1$ where $Y_i$ is a true Bernoulli (that is, $0-1$) variable. It's simple to show this: since $2\times 1-1=1$ and $2\times 0 - 1=-1$, the values of $1$ and $0$ taken on by $Y_i$ become $1$ and $-1$, respectively, for $X_i$. The probabilities are unchanged.
You might already know that $Y=Y_1+\cdots + Y_n$, the sum of $n$ independent Bernoulli variables with common probability $p$, is called a Binomial variable. It has a Binomial distribution. You can remember or look up its variance, which is $np(1-p)$. Since $$X = X_1+\cdots+X_n = (2Y_1-1) + \cdots + (2Y_n-1) = 2(Y_1 + \cdots + Y_n) - n=2Y-n,$$ you can take $\sigma=2$ in applying fact $(4)$, which immediately tells you $$\operatorname{Var}(X) = 2^2 \operatorname{Var}(Y) = 4np(1-p).$$
• I personally find the observation that $4=\frac{1}{0.5(1-0.5)}$ soothing as it shows that the variance is the ratio of $pq(X_i)$ to $pq(0.5)$ (forgive the abuse of notation). – Jared Goguen Jul 27 '17 at 16:09
• In what sense is the second part of the answer $X_i=2Y_i-1$ ... different from my answer ? – user83346 Jul 28 '17 at 5:41
• @fcop It's exactly the same approach--but the explanation is more extensive and thorough. If you take a look over the entirety of my answer, I hope you will see the pattern: it focuses on stating and illustrating useful principles. These are often overlooked or unknown to people who come to our site with questions. My purpose is to collect, unify, and illustrate some (obvious, well-known) techniques that can be found by future visitors and referenced in future answers. In that sense, my answer differs from yours, which focuses on solving the immediate problem. – whuber Jul 28 '17 at 13:37
• Well you could have referenced it no? – user83346 Jul 28 '17 at 14:08
• @fcop One ordinarily does not reference obvious or well-known things. (If that were the practice, then most posts would be 90% references.) The point of this answer is to explain the reasoning--not just the results--at the same level of understanding as the question. Its contribution is not to make the observation that $X_i=2Y_i-1$ (you're welcome to take credit for that), but rather to show as clearly and explicitly as possible, with all reasons supplied, how that produces the same answer derived earlier in the question. – whuber Aug 13 '17 at 13:10
if you take $x=2b-1$ where $b$ is bernoulli, then then $x$ is either 1 or -1, and mean and variance follow from Bernoulli $\mu_b=p$ and $\sigma^2_b=p(1-p)$, the extension to 'Binomial' is then just a sum (assuming independence),
The mean of $x$ is then $\mu_x=2\mu_b-1=2p-1$, variance $\sigma_x^2=4\sigma_b^2=4p(1-p)$,
The sum of $n$ independent such $x$'s has mean $n\cdot(2p-1)$ (your result) and the variance is $4\cdot n \cdot p(1-p)$
EDIT after your question in the comment
It is as @Pegah says: $b$ is the ''usual'' Bernoulli with a success probability $p$ and outcomes 0 and 1. It's expected value is $1 \times p + 0 \times (1-p)=p$ and the variance $p(1-p)$ (also from the definition.
So $b$ is the known Bernoulli with outcome 0 and 1 and $x$ is just a linear function of it. The linear function is such that $x$ has outcomes -1 and 1, and the mean and variance can be obtained from the mean and variance of $b$.
• In fcop's derivation, $\mu_b$ is the expected value of the random variable b, which is, as I understand a "default" binomial random variable (i.e. it can take on values 1 or 0, see also my comment above). Here the expected value is just p. – Pugl Jul 27 '17 at 15:31 | 2021-01-16T12:18:07 | {
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http://math.stackexchange.com/questions/165937/simplifying-to-a-desired-expression-structure | # Simplifying to a desired expression structure
My book has this expression:
\begin{align} ((n(n+1)(2n+7))/6)+(n+1)(n+3) \end{align}
And then the book simplified it, and ended up with the desired expression:
\begin{align} ((n+1)(n+2)(2n+9))/6 \end{align}
I tried to do such simplification. But I ended up with this: \begin{align} (2n^3+15n^2+31n+18)/6 \end{align}
Even though the "value" in my expression is the same as the desired result's, I need the structure of my expression to be the same as well. But I don't understand how to do such.
Here are the full steps my book shows. Of course, I understand that the steps make sense, but I don't understand why did my book take that approach? Is that the only approach possible to reach the desired expression? How was I supposed to know that? I mean, I know how to simplify, but clearly my book is using a different "style" or "path"
\begin{align} ((n(n+1)(2n+7))/6)+(n+1)(n+3)\\ (n(n+1)(2n+7)+6(n+1)(n+3))/6\\ ((n+1)[n(2n+7)+6(n+3)])/6\\ ((n+1)(2n^2 +7n + 6n + 18))/6\\ ((n+1)(n+2)(2n+9))/6 \end{align}
So yes, how should I simplify to get a desired expression structure?
-
Ah! I... didn't see that. Somehow. I'm eager to see how to extract such though. – Zol Tun Kul Jul 3 '12 at 1:49
I wrote out a little something about extracting. – André Nicolas Jul 3 '12 at 1:59
Below I explain in detail the solution given in your book.
$\begin{eqnarray} &&\rm\ \ \,n(n\!+\!1)(2n\!+\!7)/\color{#C00}6+(n\!+\!1)(n\!+\!3)\\ \rm put\ all\ over\ a\ common\ denominator = \color{#C00}6:\quad &= &\rm\ (n(\color{#0A0}{n\!+\!1})(2n\!+\!7)\,+\,\color{#C00}6\,(\color{#0A0}{n\!+\!1})(n\!+\!3))/\color{#C00}6\\ \rm pull\ out\ the\ common\ factor\ \color{#0A0}{n\!+\!1}:\quad &=&\rm\ (\color{#0A0}{n\!+\!1})\,(n(2n\!+\!7)+6(n\!+\!3))/6\\ \rm apply\ the\ distributive\ law:\quad &=&\rm\ (n\!+\!1)\,(2n^2 + 13n + 18)/6\\ \rm factor\ the\ quadratic,\ see\ below:\quad &=&\rm\ (n\!+\!1)\,(n\!+\!2)\,(2n\!+\!9)/6 \end{eqnarray}$
In order to factor the quadratic $\rm\:f = 2n^2+13n+18\:$ one can apply the AC-method as follows:
$$\begin{eqnarray}\rm 2f\, &=&\rm\ \ 4n^2\ +\ 13\cdot 2n\, +\, 2\cdot 18 \\ &=&\rm\ (2n)^2 + 13\,(2n) + 2\cdot 18 \\ &=&\rm\ \ N^2\ +\ 13\, N\ +\ 36\quad for\quad N = 2n \\ &=&\rm\ (N\ +\ 4)\,(N\ +\ 9) \\ &=&\rm\ (2n\, +\, 4)\,(2n\, +\, 9) \\ \rm f\, &=&\rm\ (\ n\ +\ 2)\,(2n\ +\ 9) \end{eqnarray}$$
-
I'm enlightened. Thank you. – Zol Tun Kul Jul 3 '12 at 3:58
Your expression is $$\frac{2n^3+15n^2+31n+18}{6}.$$ Let's forget about the $6$. Also, I would like to change the $n$ to $x$, for no good reason except familiarity. So we want to factor $$2x^3+15x^2+31x+18.$$ To do this, we look for rational roots of the polynomial $2x^3+15x^2+31x+18$. By the Rational root Theorem, such roots must have shape $a/b$, where $a$ divides $18$ and $b$ divides $2$. (There may be no rational roots, though in this case there are three.)
It is easy to spot the root $x=-1$. So $x-(-1)$, that is, $x+1$, divides our polynomial. Do the division, using the ordinary division process for polynomials, which is much like "long division." We get $$2x^3+15x^2+31x+18=(x+1)(2x^2+13x+18).$$ Now factor the quadratic as one did in school. Or else note that $x=-2$ is a root of $2x^2+13x+18$.
Remark: The book grabbed the obvious common factor of $n+1$. The rest turned out to be a quadratic that factors nicely. You multiplied out, meaning you buried the $n+1$ term. It can be extracted from your expression, and the above calculation shows how, but why bury and then extract? A factored or partly factored expression is often more useful. Anyway, "taking out" common factors usually simplifies calculations.
- | 2014-10-24T23:32:51 | {
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https://math.stackexchange.com/questions/1633780/show-that-set-of-all-2-times-2-matrices-forms-a-vector-space-of-dimension-4 | # Show that set of all $2 \times 2$ matrices forms a vector space of dimension $4$
I have this question:
Show that the set of all $2 \times 2$ matrices with real coefficients forms a linear space over $\Bbb R$ of dimension $4$.
I know that the set of the matrices will basically form a linear combination which will define the vector space and they satisfy the axioms defined for the vector space.
I do not know how to show that this is possible. Do the vectors have to be linearly independent?
Any sort of help is appreciated.
Thanks.
Begin by listing the axiom satisfied by a general vector space over $\mathbb{R}$. Now consider your set of matrices. What does you addition look like? What is your zero element? What does your scalar multiplication look like? Does your scalar multiplication and your addition behave as they should? Finally, show that all 2 by 2 matrices can be written as a linear combination of 4 special matrices. A good choice is to pick matrices who each have precisely 3 zero entries and 1 non zero entry. Can you show that these 4 matrices are linearly independent?
• Can those four matrices be [a 0 0 0], [0 b 0 0], [0 0 c 0], [0 0 0 d]? I can form a linear combination of them and show they are linearly independent. Will that work? – Suvrat Jan 31 '16 at 0:05
• That is exactly what I hoped that you would do! – Carl Christian Jan 31 '16 at 0:05
• oh very well then. Thanks a lot! – Suvrat Jan 31 '16 at 0:06
• Can you tell me how these 4 matrices show the formation of linear space for generalized situation, i.e. all 2x2 matrices? – Suvrat Jan 31 '16 at 20:14
• For the sake of simplicity pick a=b=c=d=1. Then write the matrix [e f g h] as e*[1 0 0 0] + f*[0 1 0 0] + g [0 0 1 0] + h*[0 0 0 1]. – Carl Christian Jan 31 '16 at 20:23
One can easily see that $\mathbb{R}^{2\times 2}$ is in fact exactly the same as the vectorspace $\mathbb{R}^4$, only the notation of the elements differs. And of course whether or not something is a vector space does not depend on the way some unimportant species of primates decides to write its elements.
Hint Can you find a basis of the set of $2 \times 2$ matrices consisting of four elements? (There is a natural choice of basis here that includes the matrix $\pmatrix{1&0\\0&0}$.)
Alternatively, can you find a vectorspace isomorphism from the space of $2 \times 2$ matrices to some vector space you know to be $4$-dimensional, e.g., $\Bbb R^4$?
• I don't know how to find a vectorspace isomorphism. But i can work with basis, which is similar to Carl's solution. Thanks! – Suvrat Jan 31 '16 at 0:11
• You're welcome. – Travis Jan 31 '16 at 0:42 | 2019-09-19T00:25:54 | {
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https://math.stackexchange.com/questions/2931622/how-can-i-prove-this-statement-about-subsets/2931634 | How can I prove this statement about subsets?
Let $$A$$, $$B$$, $$C$$ be sets. Prove that if $$A \subseteq C$$ and $$B \subseteq C$$ then $$A \cup B \subseteq C$$.
This is an exercise in mathematical logic.
My attempt to progress forward: This statement can be written as$$(A \subseteq C) \land (B \subseteq C) → A \cup B \subseteq C\\ (x \in A → x \in C) \land (x \in B → x \in C) → A \cup B \subseteq C$$ But I am not even sure that is how I am supposed to do it so I am a bit stuck. Can anyone explain to me how to get through this proof? Thanks in advance.
• ... intuitively it's obvious, right? ... Sep 26 '18 at 15:32
• @user202729 Intuitively many things are obvious, but proving them is another matter... ;) ex. A sphere bounds a given volume with minimal area... Sep 27 '18 at 1:14
4 Answers
Take any $$x\in A\cup B$$. That means either $$x\in A$$ or $$x\in B$$. Anyway you get $$x\in C$$.
This answer is expanding a bit on the other answers to give a widely applicable outline on how to prove mathematical statements in general.
The final proof will be equal to Mark's answer, but hopefully this answer sheds a bit of light on what is actually going on in the thought process, at least I like to think in the terms defined below while I am proving.
In general you need a "proof calculus" which states how proofs can be constructed and which ones are valid. Unless you are doing logic and proof theory itself, in most settings in maths you are well off using the natural deduction proof calculus for first-order logic. In short — cut in light of this question:
• To prove $$\varphi \land \psi$$, prove $$\varphi$$ and $$\psi$$ separately.
• To prove $$\varphi \lor \psi$$, give a proof for $$\varphi$$ or alternatively for $$\psi$$. One proof for either $$\varphi$$ or $$\psi$$ is already sufficient.
• To prove $$\varphi → \psi$$, assume $$\varphi$$ and prove $$\psi$$ under this assumption.
The implication captures the notion of "if we have $$\varphi$$ and $$\varphi → \psi$$ established somewhere, then we can conclude $$\psi$$." Most, if not all, theorems can be seen as implications: if you fulfill the requirements, then you can use its consequences.
Hence it makes sense to be allowed to assume $$\varphi$$ in the proof of $$\varphi → \psi$$.
• To prove $$\forall x. \varphi(x)$$ ($$\varphi$$ may depend on $$x$$), introduce a fresh variable not used before (e.g. $$c$$) and prove $$\varphi(c)$$ (i.e. every occurrence of $$x$$ is filled with $$c$$).
Introduction of a fresh variable is required to guarantee that your proof of $$\varphi(c)$$ does not depend on any previous assumptions on $$x$$. Hence, having proved $$\varphi(c)$$ really amounts to having proved the proposition $$\varphi(x)$$ for every possible $$x$$.
Now consider your statement:
$$\left((x \in A → x \in C) \land (x \in B → x \in C)\right) → A \cup B \subseteq C$$
You already did a good job translating it to more formal language! But we can even go a step further. Actually what the above line means is the following: $$\left((\forall x. x \in A → x \in C) \land (\forall x. x \in B → x \in C)\right) → (\forall x. x \in A \cup B → x \in C)$$
On the top level, we have a $$\varphi → \psi$$-kinded expression. Applying the corresponding rule, we begin with
(1) Assume $$(\forall x. x \in A → x \in C) \land (\forall x. x \in B → x \in C)$$.
Our proof goal, as stateted in the rule, is now the right side $$(\forall x. x \in A \cup B → x \in C).$$ This is of kind $$\forall x. \varphi(x)$$ with $$\varphi(x) := x \in A \cup B → x \in C$$. So,
(2) Let $$y$$ be a fresh variable.
I called it $$y$$ to avoid confusion with the set $$C$$. When one defines freshness more formally, one sees that $$x$$ would have worked as well as a fresh variable.
Now we have to show $$\varphi(y)$$, which is $$y \in A \cup B → y \in C$$. Again, apply the $$→$$ rule from above:
(3) Assume $$y \in A \cup B$$.
Our final proof goal is now: $$y \in C$$. This can be proved by case-by-case analysis:
(4) We know that $$y$$ is in $$A$$ or $$B$$. In the first case, $$y \in A$$ and per line (1) we especially know $$y \in A → y \in C$$. Since we have $$y \in A$$ just established, we can conclude $$y \in C$$. Case finished. The next and last case is $$y \in B$$ which works analogously.
If $$A \subset C$$, and $$B \subset C$$;
$$A \cap C =A$$; and $$B \cap C = B$$;
$$A\cup B = (A \cap C)\cup (B \cap C) =$$
$$C \cap (A \cup B) \subset C$$
I haven't noticed one in the in the answers, so here's a proof by contradiction.
The statement we wish to prove is: $$(\forall x)(x \in A \cup B \implies x \in C) \tag{1}$$
It's negation is:
$$(\exists x)(x \in {A} \cup {B} \wedge \neg(x \in C) ) \tag{2}$$ $$x \in A \cup B \implies (x \in A \vee x \in B)$$ $$x \in A \wedge A \subseteq C \implies x \in C \tag{3}$$ $$or$$ $$x \in B \wedge B \subseteq C \implies x \in C \tag{4}$$ Since both $$(3)$$ and $$(4)$$ are in contradiction with $$(2)$$, it must be false. Therefore our starting premise $$(1)$$ is correct. | 2021-09-19T10:57:59 | {
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https://mathhelpboards.com/threads/fourier-series-pointwise-convergence-series-computation.643/ | # Fourier series, pointwise convergence, series computation
#### Markov
##### Member
Let $f(x)=-x$ for $-l\le x\le l$ and $f(l)=l.$
a) Study the pointwise convergence of the Fourier series for $f.$
b) Compute the series $\displaystyle\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)}.$
c) Does the Fourier series of $f$ converge uniformly on $\mathbb R$ ?
-------------
First I need to compute the Fourier series, so since $f$ is odd, then the Fourier series is just $\displaystyle\sum_{n=1}^\infty b_n\sin\frac{n\pi x}l$ where $b_n=\dfrac 2l\displaystyle\int_0^{l} f(x)\sin\frac{n\pi x}l\,dx$ so I'm getting $\displaystyle-\frac{x}{{{l}^{2}}}=\frac{1}{\pi }\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n}}}{n}\sin \frac{n\pi x}{l}},$ but now I don't know how to proceed with the pointwise convergence, also, how to do part b)?
Thanks for the help!
#### Sudharaka
##### Well-known member
MHB Math Helper
Let $f(x)=-x$ for $-l\le x\le l$ and $f(l)=l.$
a) Study the pointwise convergence of the Fourier series for $f.$
b) Compute the series $\displaystyle\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)}.$
c) Does the Fourier series of $f$ converge uniformly on $\mathbb R$ ?
-------------
First I need to compute the Fourier series, so since $f$ is odd, then the Fourier series is just $\displaystyle\sum_{n=1}^\infty b_n\sin\frac{n\pi x}l$ where $b_n=\dfrac 2l\displaystyle\int_0^{l} f(x)\sin\frac{n\pi x}l\,dx$ so I'm getting $\displaystyle-\frac{x}{{{l}^{2}}}=\frac{1}{\pi }\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n}}}{n}\sin \frac{n\pi x}{l}},$ but now I don't know how to proceed with the pointwise convergence, also, how to do part b)?
Thanks for the help!
Hi Markov,
Firstly the definition of the function $$f$$ seem to be erroneous, both $$f(x)=-x\mbox{ for }-l\leq x\leq l$$ and $$f(l)=l$$ cannot be true. So I shall neglect the latter part: $$f(l)=l$$.
I think the Fourier series that you have obtained is also incorrect. It should be,
$-x=\frac{2l}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\sin\left(\frac{n\pi x}{l}\right)$
Since both $$f(x)=-x$$ and $$f'(x)=-1$$ are continuous on $$[-l,\,l]$$ the Fourier series converges point-wise on the interval $$(-l,\,l)$$ (Refer Theorem 5.5 here).
Substitute $$x=1\mbox{ and }l=2$$ and we obtain,
\begin{eqnarray}
-\frac{\pi}{4}&=&\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\sin\left(\frac{n\pi}{2} \right)\\
&=&\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}}{2n+1}\sin\left(\frac{(2n+1)\pi}{2} \right)\\
&=&\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}}{2n+1}(-1)^n\\
&=&\sum_{n=0}^{\infty}\frac{(-1)^{3n+1}}{2n+1}\\
\therefore\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}&=&\frac{\pi}{4}
\end{eqnarray}
When $$x=l$$ we have,
$\left|f(l)-\sum_{n=1}^{N}\frac{(-1)^{n}}{n}\sin\left(\frac{n\pi l}{l}\right)\right|=l$
Therefore by the definition of uniform convergence (Refer this) it is clear that the Fourier series of $$f$$ is not uniformly convergent on $$\Re$$.
Kind Regards,
Sudharaka.
#### CaptainBlack
##### Well-known member
Hi Markov,
Firstly the definition of the function $$f$$ seem to be erroneous, both $$f(x)=-x\mbox{ for }-l\leq x\leq l$$ and $$f(l)=l$$ cannot be true. So I shall neglect the latter part: $$f(l)=l$$.
This is very likely to be intended to indicate the periodic extension: $$f(x)=f(x+2l)$$ (or there is a mistake with the value at either $$+l$$ or $$-l$$, both end points would not normally be included in the domain for a Fourier Series).
Without the periodic extension the question of uniform convergence is moot, since the Fourier Series is periodic and so does not converge to the function outside of the interval.
CB | 2022-01-21T09:08:58 | {
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https://www.physicsforums.com/threads/what-can-i-use-to-solve-this.59216/ | # What can I use to solve this?
1. Jan 10, 2005
### recon
I've run into a calculation hurdle that I cannot cross in a problem I'm trying to solve.
May I know what values of z satisfy the following equation:
$$z^4 + 2z^3 - 5z + 1 = 0$$
I've never encountered this in mathematics class before.
2. Jan 10, 2005
### dextercioby
Would u please post the problem??Maybe you wouldn't get a quartic after all...
Daniel.
3. Jan 10, 2005
### Nguyen Thanh Nam
Neither do I, I can solve those that miss ax^3, never before for ax^2
4. Jan 10, 2005
### recon
A ladder 3m long rests against a wall with one end a short distance from the base of the wall. Between the wall and the base of a ladder is a garden storage box 1m tall and 1m wide. What is the maximum distance up the wall that the ladder can reach?
5. Jan 10, 2005
### dextercioby
The maximum height on the wall is attained when the ladder touches the storage box.
The equations i have found read
$$(a+1)^{2}+(b+1)^{2} =9$$(1)
$$ab=1$$ (2)
Then u find the eq.for "a"
$$a^{2}+\frac{1}{a^{2}}+2(a+\frac{1}{a})=7$$ (3)
,which can be put in the form
$$u^{2}+2u-9 =0$$ (3)
,where
$$u=:a+\frac{1}{a}$$ (4)
Solve for "u" and then for "a".Then add 1 to the "a" and find the final result.
Daniel.
6. Jan 10, 2005
### recon
Yes, that's how I solved it. I never thought of substituting $$a+\frac{1}{a}$$ for u. Very smart. Thanks a lot.
7. Jan 10, 2005
### Hurkyl
Staff Emeritus
Use the symmetry of the problem . With dexter's notation, notice that whenever s is a solution, so is 1/s. So, we can take his quartic:
z^4 + 2z^3 - 7z^2 + 2z + 1 = 0
And write down its four factors:
(z - s) (z - 1/s) (z - t) (z - 1/t) = 0
If we group them into their symmetric pairs and multiply:
(z^2 + az + 1) (z^2 + bz + 1)
So one quadratic corresponds to the two solutions s and 1/s and the other to t and 1/t. In particular, notice, now, that our unknowns only have two possible values, instead of four, which means we should be able to determine them from a quadratic equation:
Since the polynomials should be equal, we should expand and equate them:
z^4 + (a+b)z^3 + (2+ab)z^2 + (a+b)z + 1
=
z^4 + 2z^3 - 7z^2 + 2z + 1 = 0
Equating coefficients gives us a system of equations to solve.
Once we have a and b, we can then solve for s and t and get the four possible solutions. (Two should be complex, I imagine)
(Incidentally, I don't see how you got your equation, Recon)
8. Jan 10, 2005
### dextercioby
I multiple checked it.Can u prove your statement???I frankly doubt it...
Daniel.
9. Jan 10, 2005
### Hurkyl
Staff Emeritus
Another (similar) approach:
We can arrange the factors into these pairs:
(z - s)(z - t) and (z - 1/s)(z - 1/t)
So each quadratic contains exactly one solution from each of the symmetric solutions (instead of both).
Again, we break the quartic into quadratic factors:
(z^2 + az + b) (z^2 + cz + d)
But, in this case, we know that we can convert from one to the other by applying the symmetry. In particular,
if (z^2 + az + b) = 0 then (1/z)^2 + c(1/z) + d = 0
Rewriting the second gives:
z^2 + (c/d)z + (1/d) = 0
which must be the same polynomial as
z^2 + az + b = 0
Since they have the same solutions.
So we can equate coefficients here, then multiply them to equate coefficients to the quartic polynomial. Again, we can solve for our unknowns, and then we know how to solve quadratics to get the answers to our problem.
Lesson learned: use symmetry to break the problem into parts. We've seen two different ways to do it:
(1) Each part corresponds to one of the different "orbits" under the symmetry. (e.g. {s, 1/s})
(2) Each part consists of exactly one element from each "orbit". (e.g. {s, t})
Then, we can solve the problem in two steps:
(i) Determine the two subproblems.
(ii) For each subproblem, get the corresponding answers.
10. Jan 10, 2005
### Hurkyl
Staff Emeritus
Dexter gave a solution using a third approach. What precisely did he do, and why did it work? Is it really a different approach?
The variable he defined, u, has the property that it is invariant under the symmetry. Notice that if you replace z with 1/z, u remains unchanged.
Because of this invariance under symmetry, there are fewer solutions for u than there are for z, making them easier to find.
Actually, in some sense, both of my approaches make use of invariants under symmetry. For instance, (z - s)(z - 1/s) is invariant under swapping s and 1/s. My second approach found two quadratics that are each invariant under swapping s and t.
P.S. for a polynomial which is the same as its reverse, making dexter's solution is the standard way of getting a lower degree polynomial.
(The reverse of a polynomial is the one with the coefficients in the opposite order. Equivalently, the multiplicative inverse of the roots of the polynomial are precisely the roots of the reverse. To make sense, of course, this assumes that 0 is not a root)
Last edited: Jan 10, 2005
11. Jan 10, 2005
### Hurkyl
Staff Emeritus
And just for fun, here's a trig approach:
Let theta be the angle the ladder makes with the base. A is the distance from the box to the base, B is the distance from the box tot he top of the ladder.
So, the length of the ladder is 1/sin theta + 1/cos theta = 3.
With trig identities, and a change of variable, I can get a quadratic in sin phi.
12. Jan 10, 2005
### Zaimeen
I was a bit sleepy at that time Dex. I did checked it just now. You are right. Sorry Dex!!
________________________________________________________________________
"The beautiful mind goes faster than the hand"
13. Jan 10, 2005
### recon
I got it the same way as dextercioby. I just made a mistake multiplying the right side of the equation with z^3 instead of z^2. I multiplied everything on the left side with z^2.
EDIT: This is the equation I'm talking about.
$$z^2 + 2z + \frac{2}{z} + \frac{1}{z^2} = 7$$
Last edited: Jan 10, 2005 | 2017-01-19T17:17:38 | {
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https://stats.stackexchange.com/questions/275109/expected-value-with-piecewise-probability-density-function-pdf | # Expected value with piecewise probability density function (PDF)
I am continuing the prepare for an exam by reviewing handouts from an old statistics course I took. The handout came with a set of solutions prepared by the instructor, but I suspect that one of the answers is wrong. If the answer the instructor provided isn't wrong, then I must be missing something, so I'd like to ask the CV community to help where I've gone wrong. Here is the question:
Suppose that $X$ is a continuous random variable with pdf given by:
$f(x) = \left\{ \begin{array}{lr} x & 0 < x < 1\\ (2x-1)/12 & 1< x < 3 \end{array} \right.$
if $g(x) = x$ for $x>2$ and $g(x)=0$ for $x \le 2$, find the expected value $Eg(x)$.
The instructor put the solution as $10/9$. However, I've reworked my solution several times and continue to arrive at the answer $Eg(x)=61/72$. Here is how I approached this problem:
By definition, we know:
$$Eg(X) = \int_{-\infty}^{\infty}g(x)f(x)dx\tag{1}$$ Since $f(x)$ and $g(x)$ are $0$ everywhere, except from $2<x<3$, I only need to integrate (1) over these values of $x$. Doing so yields: $$Eg(X) = \int_{2}^{3}g(x)f(x)dx=\int_{2}^{3}{x(2x-1)\over{12}}dx=\int_{2}^{3}{2x^2-x\over{12}}dx$$ $$={1\over{12}}\left({{2x^3}\over{3}}-{x^2\over{2}}\right)\bigg\rvert_{x=2}^{x=3}={1\over{12}}\left[\left({{54}\over{3}}-{9\over{2}}\right)-\left({{16}\over{3}}-{4\over{2}}\right)\right]\bigg\rvert_{x=2}^{x=3}=61/72$$
I think the instructor accidentally integrated over $1<x<2$ instead of over $2<x<3$, but as I said earlier, I'm just hoping someone can verify if I'm correct, or perhaps I'm just not understanding something here. Thanks.
• I think you're right. If you can get correct answers for similar problems then you should have no reason to worry. – mark999 Apr 22 '17 at 6:53
• I agree with your answer. – Glen_b -Reinstate Monica Apr 22 '17 at 9:54
• Much appreciated, to you both! This makes me feel a lot better! – StatsStudent Apr 22 '17 at 15:49
• could you please add a tag-wiki for your new tag piecewise-pdf ? – kjetil b halvorsen Apr 23 '17 at 11:20 | 2019-12-08T20:58:48 | {
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https://brilliant.org/discussions/thread/easy-way-to-find-cube-roots/ | This note has been used to help create the Mental Math Tricks wiki
Hey guys, I saw a faster way to find cube roots.
We already know some basic cube numbers
$$0^{3}$$=0
$$1^{3}$$=1
$$2^{3}$$=8
$$3^{3}$$=27
$$4^{3}$$=64
$$5^{3}$$=125
$$6^{3}$$=216
$$7^{3}$$=343
$$8^{3}$$=512
$$9^{3}$$=729
Now, the common thing here is that each ones digit of the cube numbers is the same number that is getting cubed , except for 2 ,8 ,3 ,7 .
now let us take a cube no like 226981 .
to see which is the cube root of that number , first check the last 3 digits that is 981 . Its last digit is 1 so therefore the last digit of the cube root of 226981 is 1 .
Now for the remaining digits that is 226
Now 226 is the nearer & bigger number compared to the cube of 6 (216)
So the cube root of 226981 is 61
Let us take another example - 148877
Here 7 is in the last digit but the cube of seven's last digit is not seven. But the cube of three has the last digit as 7.
So the last digit of the cube root of 148877 is 3.
Now for the remaining digits 148.
It is the nearer and bigger than the cube of 5 (125).
Therefore the cube root of 148877 is 53.
Let us take another example 54872.
Here the last three digit's (872) last digit is 2 but the cube of 2's last digit is not 2 but the last of the cube of 8 is 2.
So the last digit of the cube root of 54872 is 8.
Now of the remaining numbers (54). It is nearer and bigger to the cube of 3 (27). So therefore the cube root of 54872 is 38.
Note by Kartik Kulkarni
3 years, 5 months ago
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How about to find cube roots of a number which answer is three-digit number ?? For example 111^3, 267^3, etc
- 3 years, 5 months ago
After doing the last three digits , try to find which is the nearest cube number to it for the remaining digits E.g -
$$\sqrt[3]{1860867}$$
Done with the last three digits and the last digit , & you get 3 as the last digit of $$\sqrt[3]{1860867}$$
Now find the nearest cube number of1860 & it is 12 (1728)
So therefore $$\sqrt[3]{1860867}$$ = 123
- 3 years, 5 months ago
So we just do the same ways... Thank you so much..
- 3 years, 5 months ago
waitwaitwaitwaitwait..whaaaaaaaaaaat? Where did that 3 even come from? The last digit of 1860867 is 7.....
- 3 years, 4 months ago
Read the note properly , it says 7 is in the last digit but the cube of seven's last digit is not seven. But the cube of three has the last digit as 7.
- 3 years, 4 months ago
So basically the cube of the number you are looking for must have the same last digit as the number in the problem?
- 3 years, 4 months ago
No, dude. It occasionally happens, but it ain't no rule. It happens for 1 (1³ = 1), 4 (4³ = 64), 5 (5³ = 125), 6 (6³ = 216), 9 (9³ = 729) and 0 (0³ = 0). But, here we see, it doesn't happen for 2 (2³ = 8), neither 3 (3³ = 27), nor 7 (7³ = 343) and 8 (8³ = 512). I'll always have to check this before find cube roots by this method.
- 3 years, 4 months ago
2, 3, and 7, 8 has at their unit place have their 10's compliments. Rest have the same number as said earlier.
- 3 years, 4 months ago
Thank you, a great method to solve the cube roots, so bad it doesn't work with every cubic root, it would save a lot of time in tests. Anyway, thanks!
- 3 years, 4 months ago
Another Interesting fact:: (A) cube of 2= unit digit 8 .....cube of 8=unit digit 2 (B) cube of 3=unit digit 7...... cube of 7= unit digit 3
- 3 years, 4 months ago
1, 4, 5, 6, 9 have the unit place of their cubes as the number themselves. But cubes of 2,3 and 8,7 has there unit place as their compliment of 10.
- 3 years, 4 months ago
It works for groups of threes. How adorable.
- 3 years, 4 months ago
Cub of 2, 3, 7, 8 we have its compliment of 10.
For 2: it is 10-2=8...........for 8, it is 10-8=2............................ 3, it it 10-3=7, ......for 7, it is 10-7=3
- 2 months, 1 week ago
Data handling
- 6 months, 2 weeks ago
But it is not apllicable everywhere..
- 7 months, 4 weeks ago
Comment deleted 5 months ago
We should take the number which has highest cube less than 117 in your case but not nearest.
- 6 months, 3 weeks ago
I gave the first person who introduced this to me a very good comment. Features of natural numbers can occasionally been found. Important thing is never let this concept to mislead ourselves when the situation is not whole numbers. I recalled and remember ed again but not really memorized properly. Understand why could make me a better memory perhaps. Hope I can memorize from today onwards!
- 1 year, 1 month ago
But it is not useful for non perfect cubes
- 1 year, 1 month ago
Write a comment or ask a question...if m=29 and e=13, then m=m+e e=m-e m=m-e then find the new value of m and e??
- 2 years, 10 months ago
Very nice & thanks.
- 3 years, 2 months ago
Good Method ....!!! Amazing...!!
- 3 years, 3 months ago
thats just for a sure perfect cube
- 3 years, 3 months ago
Cool.......
- 3 years, 4 months ago
Really very useful trick Thanks:)
- 3 years, 4 months ago
It's really coolest method ever.but can any1 suggests me methods for square root of a decimal number.for eg:square root of 0.56
- 3 years, 4 months ago
how to work out cube root of 216216. The answer on face is 66 but that is not the cube root.
- 3 years, 4 months ago
216216 is not perfect cube ! This method is only for perfect cube
- 4 months ago
you have to it by long division method
- 3 years, 4 months ago
good one
- 3 years, 4 months ago
do you want to know the exact long division method of finding cube roots though it tedious... :)
- 3 years, 4 months ago
sure.
- 3 years, 4 months ago
- 3 years, 4 months ago
I HAVE SOME CONFUSION THAT WHEN HAM LOG SAME NO. KO LIKHEGE OR KAB NHI........AS 1ST SUM MEN.......226 KA 6 LIKHE AND 981 KA 1 SO ANS. IS 61 BUT 148877 MEN 148 KA 5 KYU LAST NO TO 8 HA SO COMPLEMENTRY IS 2 BUT HERE IS 5..
- 3 years, 4 months ago
I would prefer not to use Hindi cause it is confusing me that you have mixed up English & Hindi
- 3 years, 4 months ago
Only works for whole numbers. It's interesting however that you have found this method. How did you come across it?
- 3 years, 4 months ago
excellent method!!! upvoted young mind :)
- 3 years, 4 months ago
I like it
- 3 years, 4 months ago
Who discovered this method? It's really awesome
- 3 years, 4 months ago
i like that method
- 3 years, 4 months ago
Real nice method. I liked it.
- 3 years, 4 months ago
I like this method .
- 3 years, 4 months ago
nice
- 3 years, 4 months ago
you mean x^3 of 226981 , 226971 , 226961 , 226981 , 226881 , all is 61 only by your way. which is incorrect
- 3 years, 4 months ago
you have to know that it it works only for a perfect cube
- 3 years, 4 months ago
I'm sorry I did not understand
- 3 years, 4 months ago
By this trick cube root for last 3 digit is depends on unit place digit only? if we consider these numbers which all have 1 as unit place digit , 226981 , 226971 , 226961 , 226221 , 226881 so by the rule cube root should be 61 for all these numbers. which is actually incorrect because numbers are different.
- 3 years, 4 months ago
This method is only applicable for cube numbers that have the cube root with no numbers after the decimal point
- 3 years, 4 months ago
As I have mentioned in another comment, if the number is not a perfect cube, we at least know the floor and the ceiling of this number.
- 3 years, 4 months ago
excellent
- 3 years, 4 months ago
Thank U Very Much.I like Ur Way To solve The Problem.
- 3 years, 4 months ago
Great! Interesting!
- 3 years, 4 months ago
- 3 years, 4 months ago
Really good method... I like it!
- 3 years, 4 months ago
ecellent method .its working
- 3 years, 4 months ago
Good solution
- 3 years, 4 months ago
Write a comment or ask a question... Super
- 3 years, 4 months ago
Thanks
- 3 years, 4 months ago
Excellent method
- 3 years, 4 months ago
Just noticed. It actually isn't applicable to numbers other than perfect cubes. For example, if you calculate the cube root of 1,216 using this method, you get 16; actual root is 10.67. They're almost 5.5 numbers apart. If you have any better ways, please post it.
- 3 years, 4 months ago
In that case we know between which two integers the actual cube root lays.
- 3 years, 4 months ago
I had answered to a similar question , & this method is only applicable for numbers which have their cube roots with no numbers after the decimal point
- 3 years, 4 months ago
Awesome and unique way to do it!! Thanks!!
- 3 years, 4 months ago
Nice note
- 3 years, 4 months ago
Would largely help me for finding Karl Pearson's coefficient. Thanks.
- 3 years, 5 months ago
Fantastic method Thanks
- 3 years, 5 months ago
Brilliant! Good to learn this from you. Thanks.
- 3 years, 5 months ago
maths is not about approximation and estimation!!!!
- 3 years, 5 months ago
can someone prove it mathematically?
- 3 years, 5 months ago
- 3 years, 4 months ago
a + 10 b + 100 c + 1000 d + 10000 e + 100000 f could roughly prove it I guess.
- 3 years, 5 months ago
@Kartik Kulkarni .... really a nice one ... but i hav a doubt ... take 1331 ..... u get 11 by the method stated above .... if u take 1441 ..... 11 isnt correct ..... in that case .... u cant find whether a no. is a cube no. or not using this method.... rite???
- 3 years, 5 months ago
also 1441's cube root is somewhat 11 And many more numbers after the decimal points
- 3 years, 5 months ago
Well actually,this method is only applicable for actual cube numbers
- 3 years, 5 months ago
Great buddy. Here is the actual method https://brilliant.org/discussions/thread/long-divison-method-of-cube-root/
- 3 years, 4 months ago
@Kartik Kulkarni ... as soon as i saw ..... i found this interesting and also concluded this is applicable for perfect cubes ... but ur inference of 1441's cube root is around 11 is wrong ..... eg: take 1721 ..... if u infer by the same method as u did above ... it is around 11 ... but actually it can be estimated to 12 ..... (Note: cube root of 1721 = 11.98)
- 3 years, 5 months ago
well , I didn't think about the estimation part
- 3 years, 5 months ago
cool
- 3 years, 5 months ago
Really cool way...I m looking forward to u to post some cool ways of finding the sum of series....
- 3 years, 5 months ago
I love this mathed
- 3 months, 2 weeks ago
3√79510
- 4 months, 2 weeks ago
according to this cube root of 125486 should be 56 but actually it is not
- 3 years, 3 months ago
I just found out when this method works,
for eg 125486. last 3 digits = 6, first 3 digist =5,
here 125 is perfect cube , hence it doesnt work.
MY findings = This method only works when neither of the components( 1st 3 digits & last 3 digits) are perfect cube but the number that is comprised of the components is a perfect cube.
In ur case 125486 aint a perfect cube cum 125 which i call a component is.
Hows my Theorem? Thumbs up!!
- 3 years, 3 months ago
- 3 years, 2 months ago
An adition to my findings: Either both the components aint perfect cube or both are.
125000, fr last 3 digits =0, fr first 3 digits =5
cube root of 125000 is 50.
(Notice that 000 is nothing but 0 and not 1000, 0^3 is 0)
- 3 years, 2 months ago
216216
- 2 years, 6 months ago
Comment deleted Apr 05, 2015
Comment deleted Apr 08, 2015
Sorry i didnt get u.
- 3 years, 2 months ago
125486 is not a perfect cube and so this method is not applicable for that number
- 3 years, 1 month ago
- 3 years, 4 months ago
Very nice and interesting solution
- 3 years, 4 months ago
whats's wrong with these four numbers(2,3,7 and 8)? i mean these are the number which you will never find at the end of any "squared number"( at ones place i mean). and here too the same four number have different digits at ones place. by the way nice trick. thanks!
- 3 years, 4 months ago
what if we have 7 digited number could u explane me how to do it please
- 3 years, 4 months ago
I just explained it to Jonathan Christianto above
- 3 years, 4 months ago
kk:":":":":":":thanku
- 3 years, 4 months ago
then we have to go by long divison actual method
- 3 years, 4 months ago | 2018-06-25T10:20:14 | {
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https://math.stackexchange.com/questions/1590359/simplify-2-cos2-x-sin2-x2-tan-2x | Simplify $2(\cos^2 x - \sin^2 x)^2 \tan 2x$
Simplify: $$2(\cos^2 x - \sin^2 x)^2 \tan 2x$$ After some sketching, I arrive at: $$2 \cos 2x \sin2x$$ Now according to the answer sheet, I should simplify this further, to arrive at $\sin 4x$. But how do I derive the latter from the former? Where do I start? Hoe do I use my double-angle formulas to arrive there?
• Hint: $\sin(2u) = 2\cos(u)\sin(u)$. What is $u$ in this case? – MathematicsStudent1122 Dec 27 '15 at 9:46
Notice, using double angle identity $\cos^2A-\sin^2A=\cos 2A$, one should get
$$2(\cos^2x-\sin^2x)^2\tan 2x$$ $$= 2\cos^2 2x\tan 2x$$ $$= 2\cos^2 2x\left(\frac{\sin 2x}{\cos 2x}\right)$$ $$= 2\sin 2x\cos 2x$$ using double angle identity $2\sin A\cos A=\sin 2A$, $$=\sin 2(2x)$$$$=\color{red}{\sin 4x}$$
• Sorry, there was a typo in the initial post. – Apeiron Dec 27 '15 at 9:45
• Alright, so I have made correction accordingly – Harish Chandra Rajpoot Dec 27 '15 at 9:48
$2\cos(2x)\sin(2x)=\sin(2(2x))=\sin(4x)$
• Yeah, this just doesn't explain why I can do this. I know $2 \cos x \sin x = \sin 2x$, but how do I get from your first step to the second. This is not obvious for me. Would $2 \cos 3x \sin 3x = \sin 6x$? – Apeiron Dec 27 '15 at 9:40
• First step: use $2x$ instead of $x$ in the sine double angle formula Second step: $2*2x=4x$ Yes, $2 \cos 3x \sin 3x = \sin 6x$ – GNUSupporter 8964民主女神 地下教會 Dec 27 '15 at 9:42
Use $$\cos^2x-\sin^2x=\cos2x$$ and $$\tan A=\dfrac{\sin A}{\cos A}$$ | 2019-05-27T03:13:35 | {
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https://www.jiskha.com/questions/1328172/Use-the-identity-sin-2x-cos-2x-1-and-the-fact-that-sin-2x-and-cos-2x-are-mirror-images | # Calculus
Use the identity sin^2x+cos^2x=1 and the fact that sin^2x and cos^2x are mirror images in [0,pi/2], evaluate the integral from (0-pi/2) of sin^2xdx. I know how to calculate the integral using another trig identity, but I'm confused about how to solve this one.
1. Let V = ∫sin^2x dx
Since cos^2x is a mirror image of sin^2x, ∫cos^2x dx = V
Now, since sin^2x+cos^2x = 1,
2V = ∫[0,π/2] 1 dx = π/2
V = π/4
check:
sin^2x = (1-cos2x)/2
∫sin^2x dx = ∫(1-cos2x)/2 dx
= 1/2 (x - 1/2 sin2x) [0,π/2]
= 1/2[(π/2)-(0)]
= π/4
posted by Steve
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https://math.stackexchange.com/questions/3012182/how-to-find-the-derivative-of-a-definite-integral-that-has-unusual-lower-and-upp | # How to find the derivative of a definite integral that has unusual lower and upper bounds?
I'm not sure how to deal with upper and lower bounds in integrals when using the first part of the fundamental theorem of calculus to work with them.
The question I'm looking at asks me to find the derivative of the function, where the function is a definite integral. The question is explicitly telling me to use the fact that $$\frac{d}{dx} \int f(x)dx = f(x)$$, i.e. the first part of the fundamental theorem of calculus, to answer the question.
The function is:
$$\int_{\sqrt{x}}^{\pi/4} \theta \cdot tan\theta \cdot d\theta = g(x)$$
In words: if a function corresponds to an integral where the upper bound on the integral is $$\pi/4$$, the lower bound is $$\sqrt{x}$$, and the function being integrated is $$\theta \cdot tan\theta$$ with respect to $$\theta$$, then what is the derivative of the function?
I've tried setting $$u = \pi/4$$ and applying the chain rule, that's worked in the past but it doesn't give me the right answer here. I'm guessing I also have to incorporate the lower bound, that $$\sqrt{x}$$, in my solution somehow, but I have no idea how.
Any help would be greatly appreciated.
• Hint: chain rule. – GEdgar Nov 24 '18 at 22:34
• As a suggested reading, check out Leibniz Rule for differentiation of integrals. – Shubham Johri Nov 24 '18 at 22:44
• Hint: $\int_{-\infty}^x f(x) dx = f(x)$, so $\int_a^bf(x) dx = \int_{-\infty}^b f(x) dx - \int_{-infty}^a f(x) dx$ – eSurfsnake Nov 25 '18 at 7:19
So,$$g(x)=\int_{\sqrt x}^{\frac\pi4}\theta\tan\theta\,\mathrm d\theta=-\int_{\frac\pi4}^{\sqrt x}\theta\tan\theta\,\mathrm d\theta.$$Can you now apply the Fundamental Theorem of Calculus, together with the chain rule?
• Yes I can, thank you so much! That was a lot simpler than I thought. So does the lower bound not matter at all when applying the first part of the FTC? – James Ronald Nov 24 '18 at 22:42
• No, the lower bound does not matter. – José Carlos Santos Nov 24 '18 at 22:45
After years of tutoring Calculus I, it baffles me that professors somehow expect students to figure out how to extend part I of the Fundamental Theorem of Calculus to cases where the upper limit is not $$x$$ and the lower limit is not a constant.
So, I will provide you with a quick, intuitive (and not rigorous) derivation on how you should approach this.
Suppose the lower limit is $$L(x)$$ and the upper limit is $$U(x)$$ of the integral. Define
$$g(x) = \int_{L(x)}^{U(x)}f(t)\text{ d}t\text{.}$$
Suppose $$F$$ is an antiderivative of $$f$$. By part II of the Fundamental Theorem of Calculus, you know that $$g(x) = \int_{L(x)}^{U(x)}f(t)\text{ d}t = F(U(x)) - F(L(x))\text{.}$$ Then, the derivative of $$g$$ is given by, assuming differentiability of $$U$$ and $$L$$,
$$\dfrac{\text{d}}{\text{d}x}[g(x)] = F^{\prime}(U(x))U^{\prime}(x)-F^{\prime}(L(x))L^{\prime}(x)$$ after making use of the chain rule for derivatives. But, $$F$$ is an antiderivative of $$f$$, so $$F^{\prime} = f$$, hence $$\dfrac{\text{d}}{\text{d}x}[g(x)] = f(U(x))U^{\prime}(x)-f(L(x))L^{\prime}(x)\text{.}$$ In other words, the main result is $$\boxed{ \dfrac{\text{d}}{\text{d}x}\int_{L(x)}^{U(x)}f(t)\text{ d}t = f(U(x))U^{\prime}(x)-f(L(x))L^{\prime}(x)\text{.}}$$
Applying to this problem, we have $$f(\theta) = \theta \tan(\theta)$$, $$U(x) = \dfrac{\pi}{4}$$, and $$L(x) = \sqrt{x}$$. The derivatives are $$U^{\prime}(x) = 0$$ and $$L^{\prime}(x) = \dfrac{1}{2\sqrt{x}}$$. Hence, the derivative of $$g$$ is $$g^{\prime}(x) = f(U(x))U^{\prime}(x)-f(L(x))L^{\prime}(x) = f\left(\dfrac{\pi}{4}\right)(0) - f\left(\sqrt{x}\right) \cdot \dfrac{1}{2\sqrt{x}}$$ which simplifies to $$g^{\prime}(x) = - f\left(\sqrt{x}\right) \cdot \dfrac{1}{2\sqrt{x}} = -\sqrt{x}\tan(\sqrt{x}) \cdot \dfrac{1}{2\sqrt{x}} = -\dfrac{1}{2}\tan(\sqrt{x})\text{.}$$
• Wow, that really is a great intuition, thank you so much! You said it's not rigorous, but it seems be a completely valid proof, and as long as it's valid the simpler the better in my opinion haha. I wish I'd seen this earlier. Thanks again! – James Ronald Nov 24 '18 at 23:11 | 2019-06-18T05:28:10 | {
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https://math.stackexchange.com/questions/2175000/infinite-cyclic-group-isomorphism | # Infinite cyclic group isomorphism.
I have one probably basic question, but still it bothers me how to show it.
Namely, if two groups are isomorphic (i.e. there is a bijective group homomorphism between them) and if one of them is infinite cyclic (precisely, in my case it is discussed about $(\mathbb Z,+)$), does the other group necessary have to be infinite cyclic?
Precisely, I'm in algebraic topology and I know that the fundamental group of cycle $\mathbb S^1$ is isomorphic to integers, but does this imply that the fundamental group of cycle is infinite cyclic as well?
Any sketch of the proof would be welcome.
• All cyclic groups are isomorphic to either $\mathbb{Z}$ or $\mathbb{Z}/n\mathbb{Z}$ for a natural number $n$. – Student Mar 6 '17 at 21:20
Indeed, this is true and is known as the fundamental theorem of finitely generated abelian groups..
In particular, we know that every finitely generated abelian group is of the form:
$$\mathbb Z^n \oplus \mathbb Z_p\oplus\cdots \oplus \mathbb Z_q$$
so if your group is infinite and cyclic, then it is abelian, of rank $1$ and torsion-free, meaning that it is indeed the same thing as $\mathbb Z$.
Edit: on the other hand, this now feels like a lot of overkill, despite being the way that I think about it. Here is a proof sketch:
Let $G$ be infinite cyclic $\langle a^n \mid n \in \mathbb Z \rangle$. Suppose further that $\phi: \mathbb Z \to G$ is the map $n \mapsto a^n$.
You should check that this is indeed a homomorphism. Surjectivity is clear. Injectivity follows from the fact that if $a^n=a^m$ for $n >m$, then $a^n (a^m)^{-1}=e \implies a^{n-m}=e$ while the order of $a$ was supposed to be infinite.
Hence, this is an isomorphism.
Edit 2: Let me try to address your question in the comments. Let $\pi_1(S^1)$ be the group consisting of loop classes, equipped with the usual multiplication. We already have that there exists an isomorphism $$\rho: \pi_1(S^1) \to (\mathbb Z,+)$$
and we have further that
$$\phi:\mathbb Z \to G$$ is an isomorphism when $G$ is infinite cyclic. By the composition of isomorphisms,
$$\phi \circ \rho:\pi_1(S^1) \to G$$ is an isomorphism as well. Hence, we can conclude that it is infinite cyclic.
This seems like a more general problem, isomorphism is a transitive property, if $A \cong B \cong C$, then $A \cong C$ as well.
Edit 3: We claim that $(\mathbb Z,+)$ is an infinite cyclic group. To see this, consider $1 \in \mathbb Z$. Clearly, every element $n \in \mathbb N$ can be written as $\underbrace{1+1+\dots+1}_{n \, \mathrm{times}}$ and each inverse can be given by $-1$, which is the additive inverse of $1 \in \mathbb Z$. In other words, $1$ generates the group and has infinite order. Perhaps the notation is confusing, in additive notation:
A group $(G,+)$ is said to be infinite cyclic if $$G=\langle n \cdot a \mid n \in \mathbb Z\rangle.$$
Hence if you already know that $\pi_1(S^1) \cong \mathbb Z$, then it must be infinite cyclic, since $\mathbb Z$ already was.
• Thanks. I understand what you are trying to say, but isn't this (in some sense) the reversed statement, i.e. for a given infinite cyclic group say $(G,*)$, one can construct isomorphism between $G$ and $\mathbb Z$. But I think I have the opposite problem; I already know that my (fundamental) group $\pi_1(\mathbb S^1,*)$ is isomorphic to $(\mathbb Z,+)$ where for $(\mathbb Z,+)$ "infinite cyclic" behaviour is known and what I am trying to show (if possible) is that the first group must be infinite cyclic as well. – edward_scissorhands Mar 7 '17 at 8:39
• It's unclear what you are confused about to me. What I have just demonstrated is that it does not matter whether you say that the fundamental group is infinite cyclic of is the additive group on the integers, since the two of them are the same group, up to symbolic manipulation. – Andres Mejia Mar 7 '17 at 8:44
• the fundamental group is infinite cyclic. I included another argument. However, I think that this argument is backwards in the way of intuition. The thing to understand is that $\pi_1(S^1)$ is a free group given by a single generator, so it is in fact cyclic, so it is abelian, and hence is $\mathbb Z$. In particular, any free group modulo its commutator subgroup is again $\mathbb Z^n$ for some $n$. To see an example where it is a free group (infinite) on two generators, but not $\mathbb Z$, consider $\pi_1(S^1 \vee S^1)$. – Andres Mejia Mar 7 '17 at 8:54
• If it is isomorphic to $(\mathbb Z,+)$ then it is infinite cyclic, since $\mathbb Z$ is itself infinite cyclic. For example, it is generated by a single element $1$ that has infinite order. – Andres Mejia Mar 7 '17 at 8:57
• @Eurydice see my edits – Andres Mejia Mar 7 '17 at 9:01 | 2019-08-21T21:15:27 | {
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https://math.stackexchange.com/questions/927637/whats-the-shape-of-this-addsto-function | # What's the shape of this “addsTo” function …?
Note that in this combinatronics question,
How many lists of 100 numbers (1 to 10 only) add to 700?
For an array of 100 numbers, each 1 to 10 inclusive, and the total is T - how many such arrays exist?
In fact due to the ASTOUNDING answers there, we now know
for T=700 the answer is 1.2 e92
Being curious I now wonder about the shape of this result for various T.
Sadly, all I know are the points T<100, T=100, T=700 (thanks, MSE!), and T=1000.
Many questions arise, is there a T that makes a maximum, is it the same all over except the ends, is it bumpy or erratic ... does it make any difference if T is prime, even, etc ... and, since 19^92 is pretty small and there's only 600 Ts, in fact, is T=700 just freakishly small for some reason (what reason?) ...... or?
• The rough shape is a bell curve with center at $550$ and symmetric around that point. – Thomas Andrews Sep 11 '14 at 14:42
• Yeah, I didn't read the linked-to question, so I didn't realize I duplicated a lot of work, so just made the comment. Basically, if you pick a random integer from $1$ to $10$ 100 times, the the probability of getting $T$ is $\frac{1}{10^{100}}$ times your count. And the general tendency of repeating processes like this is to get a bell curve. – Thomas Andrews Sep 11 '14 at 14:45
• It's $$\sum_{j=0}^{45} (-1)^j \binom{100}{j}\binom{549-10j}{99}$$ Need some time to find a program to calculate it. – Thomas Andrews Sep 11 '14 at 14:51
• The approximation of the distribution as a bell curve appeared as @mjqxxxx's answer to your other problem. One can perhaps obtain asymptotic corrections to this with more sophisticated analysis. – Semiclassical Sep 12 '14 at 16:32
• ah, i didn't notice that, thanks semi. still it would be interesting to know the max etc - still thanks – Fattie Sep 12 '14 at 16:35
According to the Central Limit Theorem, when you add a large number of independent, identically distributed random variables (each variable can be a sum of other random variables), the distribution of the sum tends to a normal distribution whose mean is the sum of the means of the individual distributions, and whose variance is the sum of the variances of the individual distributions.
For example, if we sum $1$ number ($1$-$10$), the distribution looks like
$\hspace{3cm}$
If we sum $2$ numbers ($1$-$10$), the distribution looks like
$\hspace{3cm}$
If we sum $3$ numbers ($1$-$10$), the distribution starts to looks like a normal distribution
$\hspace{3cm}$
If we sum $10$ numbers ($1$-$10$), the distribution looks even more like a normal distribution
$\hspace{3cm}$
Now lets look at the sum of $100$ numbers ($1$-$10$)
$\hspace{3cm}$
The number of ways for $100$ numbers ($1$-$10$) to sum to $700$, which was computed in this answer, is pointed to by the arrow. Note that the range of the sum is $9n$ where $n$ is how many numbers are added; however, the standard deviation is $\sqrt{8.25n}$. Thus, the relative spread gets smaller; that is, the distribution becomes narrower about the mean as $n$ get bigger.
The maximum number of ways to sum to $T$ is at the mean of the distribution; that is, at $T=550$, where the number of ways is $$13868117806391314648666325510838589167047653141664\\4888545033078503482282975641730091720919340564340$$ which is approximately $1.3868117806391314649\times10^{98}$.
• Wow - you guys are amazing. I'm so dense I didn't even realise, of course the question is just equivalent to looking at the random distribution. one question - you see the final graph, in fact the answer to the question ........ – Fattie Sep 13 '14 at 9:30
• ...... would we still call that "a bell curve," a normal distribution, since it's so incredibly pointy? As the answer explains, for 1 or 2 numbers, I guess you'd say, "it is not" a Normal distribution. Well now, when you get to very high numbers (300, 500, 10,000) again does it become "not" a Normal distribution? Or, is it still perfectly a "bell curve" but just very pointy? Is bell curvey -ness subjective? Or conversely, just by definition, is every result here officially a normal distribution (even the 1 or 2 cases)? Cheers!!!!! – Fattie Sep 13 '14 at 9:31
• @JoeBlow: A scaled bell curve looks pointy, and that is what this is: a scaled bell curve. If scaled so that the standard deviation is $1$, its total sum is $1$, and translated so that the mean is $0$, it would look very close to a normal distribution bell curve. Scaled this way would map the total range of possibilities to the whole real line. – robjohn Sep 13 '14 at 9:39
• Gotchya, I'll have to think about that. If I'm not mistaken, in your graphs, indeed they all start and end at the zero points (so, for "10" it's 10-->100, and for "100" it's 100-->1000 etc). And the scale heights are the same so there's no "trick". But it seems to me they are radically different: the shapes of "100" and "10million" would be terribly different right? Recall I was asking "is 700 just freakishly small for some reason?" Indeed, that is the reason: there is no such "freakish smallness" in the (say) "10" or "15" case when you are still equally near the middle .... but ... – Fattie Sep 13 '14 at 9:53
• ... indeed this function becomes "amazingly pointy" as the count gets higher. I'm thinking like a video game programmer, if something depended on that shape, play would be spectacularly different with the "10" shape versus the "500" shape. (You'd almost always, versus never, run in to the goblin or whatever.) Anyways - I'm wasting your time - I'll go take a refresher course in normal curves THANK YOU SO MUCH. Do you often provide such good answers that nobody else even has to bother? ;-) – Fattie Sep 13 '14 at 9:55 | 2019-06-16T11:36:16 | {
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https://mid.gr/bk40qmd/simplify-radical-expressions-46f37a | Khan Academy is a … Next lesson. The powers don’t need to be “2” all the time. Played 0 times. Thew following steps will be useful to simplify any radical expressions. Test - I. Note that every positive number has two square roots, a positive and a negative root. The standard way of writing the final answer is to place all the terms (both numbers and variables) that are outside the radical symbol in front of the terms that remain inside. Test - II . When the radical is a cube root, you should try to have terms raised to a power of three (3, 6, 9, 12, etc.). Square root, cube root, forth root are all radicals. Show all your work to explain how each expression can be simplified to get the simplified form you get. 72 36 2 36 2 6 2 16 3 16 3 48 4 3 A. APTITUDE TESTS ONLINE. Simplifying Radicals Worksheet … Multiply all numbers and variables outside the radical together. Exponents and power. To play this quiz, please finish editing it. For this problem, we are going to solve it in two ways. It must be 4 since (4)(4) = 42 = 16. Step 1 : If you have radical sign for the entire fraction, you have to take radical sign separately for numerator and denominator. Section 6.3: Simplifying Radical Expressions, and . 2) Product (Multiplication) formula of radicals with equal indices is given by To read our review of the Math Way -- which is what fuels this page's calculator, please go here . An algebraic expression that contains radicals is called a radical expression An algebraic expression that contains radicals.. We use the product and quotient rules to simplify them. by lsorci. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Simplifying Radical Expressions DRAFT. However, I hope you can see that by doing some rearrangement to the terms that it matches with our final answer. For the number in the radicand, I see that 400 = 202. For example, These types of simplifications with variables will be helpful when doing operations with radical expressions. So, , and so on. Our mission is to provide a free, world-class education to anyone, anywhere. The number 16 is obviously a perfect square because I can find a whole number that when multiplied by itself gives the target number. Well, what if you are dealing with a quotient instead of a product? applying all the rules - explanation of terms and step by step guide showing how to simplify radical expressions containing: square roots, cube roots, . In this tutorial, the primary focus is on simplifying radical expressions with an index of 2. Equivalent forms of exponential expressions. Next, express the radicand as products of square roots, and simplify. 5 minutes ago. #1. . Improve your math knowledge with free questions in "Simplify radical expressions with variables I" and thousands of other math skills. Then, it's just a matter of simplifying! A perfect square number has integers as its square roots. Section 6.4: Addition and Subtraction of Radicals. Notice that each group of numbers or variables gets written once when they move outside the radical because they are now one group. Radical expressions can often be simplified by moving factors which are perfect roots out from under the radical sign. Quotient Property of Radicals. But there is another way to represent the taking of a root. . Additional simplification facilities for expressions containing radicals include the radnormal, rationalize, and combine commands. More so, the variable expressions above are also perfect squares because all variables have even exponents or powers. Please click OK or SCROLL DOWN to use this site with cookies. In this tutorial, the primary focus is on simplifying radical expressions with an index of 2. Example 11: Simplify the radical expression \sqrt {32} . \left(\square\right)^{'} \frac{d}{dx} \frac{\partial}{\partial x} \int. Homework. Therefore, we have √1 = 1, √4 = 2, √9= 3, etc. Another way to solve this is to perform prime factorization on the radicand. TRANSFORMATIONS OF FUNCTIONS. Variables with exponents also count as perfect powers if the exponent is a multiple of the index. The answer must be some number n found between 7 and 8. We hope that some of those pieces can be further simplified because the radicands (stuff inside the symbol) are perfect squares. You can use the same ideas to help you figure out how to simplify and divide radical expressions. Solving Radical Equations Method 1: Perfect Square Method -Break the radicand into perfect square(s) and simplify. The goal is to show that there is an easier way to approach it especially when the exponents of the variables are getting larger. are called conjugates to each other. If the denominator is not a perfect square you can rationalize the denominator by multiplying the expression by an appropriate form of 1 e.g. Example 1. The first law of exponents is x a x b = x a+b. However, it is often possible to simplify radical expressions, and that may change the radicand. First we will distribute and then simplify the radicals when possible. Simplifying Radical Expressions Date_____ Period____ Simplify. Simplifying Radical Expressions. However, the best option is the largest possible one because this greatly reduces the number of steps in the solution. 16 x = 16 ⋅ x = 4 2 ⋅ x = 4 x. no fractions in the radicand and. The properties we will use to simplify radical expressions are similar to the properties of exponents. A radical expression is composed of three parts: a radical symbol, a radicand, and an index. So which one should I pick? The key to simplify this is to realize if I have the principal root of x over the principal root of y, this is the same thing as the principal root of x over y. All that you have to do is simplify the radical like normal and, at the end, multiply the coefficient by any numbers that 'got out' of the square root. To simplify a radical, factor the radicand (under the radical) into factors whose exponents are multiples of the index. Scientific notations. Multiply all numbers and variables inside the radical together. Adding and Subtracting Radical Expressions, That’s the reason why we want to express them with even powers since. Test to see if it can be divided by 4, then 9, then 25, then 49, etc. Simplify radical expressions using the product and quotient rule for radicals. Here it is! To help me keep track that the first term means "one copy of the square root of three", I'll insert the "understood" "1": Don't assume that expressions with unlike radicals cannot be simplified. Example 2: to simplify ( 3. . The symbol is called a radical sign and indicates the principal square root of a number. Product Property of n th Roots. How to Simplify Radicals with Coefficients. For the numerical term 12, its largest perfect square factor is 4. These two properties tell us that the square root of a product equals the product of the square roots of the factors. To simplify this sort of radical, we need to factor the argument (that is, factor whatever is inside the radical symbol) and "take out" one copy of anything that is a square. Simplifying logarithmic expressions. Then express the prime numbers in pairs as much as possible. If found, they can be simplified by applying the product and quotient rules for radicals, as well as the property a n n = a, where a is positive. Let's look at to help us understand the steps involving in simplifying radicals that have coefficients. To simplify complicated radical expressions, we can use some definitions and rules from simplifying exponents. By multiplying the variable parts of the two radicals together, I'll get x 4 , which is the square of x 2 , so I'll be able to take x 2 out front, too. So we expect that the square root of 60 must contain decimal values. What rule did I use to break them as a product of square roots? Always look for a perfect square factor of the radicand. Part A: Simplifying Radical Expressions. This calculator simplifies ANY radical expressions. You may use your scientific calculator. Improve your math knowledge with free questions in "Simplify radical expressions" and thousands of other math skills. Picking the largest one makes the solution very short and to the point. Recall that the Product Raised to a Power Rule states that $\sqrt[n]{ab}=\sqrt[n]{a}\cdot \sqrt[n]{b}$. These properties can be used to simplify radical expressions. Use rational exponents to simplify radical expressions. Finish Editing. Pairing Method: This is the usual way where we group the variables into two and then apply the square root operation to take the variable outside the radical symbol. Step 4: Simplify the expressions both inside and outside the radical by multiplying. This type of radical is commonly known as the square root. Type your expression into the box under the radical sign, then click "Simplify." Notice that the square root of each number above yields a whole number answer. Here are the search phrases that today's searchers used to find our site. Perfect Squares 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 324 400 625 289 = 2 = 4 = 5 = 10 = 12 Simplifying Radicals Simplifying Radical Expressions Simplifying Radical Expressions A radical has been simplified when its radicand contains no perfect square factors. Vertical translation. Remember, the square root of perfect squares comes out very nicely! Let’s deal with them separately. The paired prime numbers will get out of the square root symbol, while the single prime will stay inside. nth roots . Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. And it checks when solved in the calculator. We need to recognize how a perfect square number or expression may look like. Play. Radical Expressions are fully simplified when: –There are no prime factors with an exponent greater than one under any radicals –There are no fractions under any radicals –There are no radicals in the denominator Rationalizing the Denominator is a way to get rid of any radicals in the denominator Mathematics. If found, they can be simplified by applying the product and quotient rules for radicals, as well as the property $$\sqrt[n]{a^{n}}=a$$, where $$a$$ is positive. The radicand contains no fractions. Let’s do that by going over concrete examples. Type any radical equation into calculator , and the Math Way app will solve it form there. Simplification of expressions is a very useful mathematics skill because, it allows us to change complex or awkward expression into more simple and compact form. If and are real numbers, and is an integer, then. Simplify a Term Under a Radical Sign. And we have one radical expression over another radical expression. Example 3: Simplify the radical expression \sqrt {72} . Simplify the expression: It is possible that, after simplifying the radicals, the expression can indeed be simplified. You can do some trial and error to find a number when squared gives 60. The solution to this problem should look something like this…. Simplify expressions with addition and subtraction of radicals. For example, the sum of $$\sqrt{2}$$ and $$3\sqrt{2}$$ is $$4\sqrt{2}$$. Live Game Live. After doing some trial and error, I found out that any of the perfect squares 4, 9 and 36 can divide 72. x^{\circ} \pi. In order to simplify radical expressions, you need to be aware of the following rules and properties of radicals 1) From definition of n th root(s) and principal root Examples More examples on Roots of Real Numbers and Radicals. Mathplanet is licensed by Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens. The roots of these factors are written outside the radical, with the leftover factors making up the new radicand. +1) type (r2 - 1) (r2 + 1). Procedures. As long as the powers are even numbers such 2, 4, 6, 8, etc, they are considered to be perfect squares. Share practice link. Example 10: Simplify the radical expression \sqrt {147{w^6}{q^7}{r^{27}}}. Example 1: Simplify the radical expression \sqrt {16} . Simplify each of the following. Algebraic expressions containing radicals are very common, and it is important to know how to correctly handle them. Radical Expressions and Equations Notes 15.1 Introduction to Radical Expressions Sample Problem: Simplify 16 Solution: 16 =4 since 42 =16. You will see that for bigger powers, this method can be tedious and time-consuming. We have to consider certain rules when we operate with exponents. Variables in a radical's argument are simplified in the same way as regular numbers. This quiz is incomplete! Practice: Evaluate radical expressions challenge. Think of them as perfectly well-behaved numbers. Play this game to review Algebra I. Simplify. The product of two conjugates is always a rational number which means that you can use conjugates to rationalize the denominator e.g. Adding and Subtracting Radical Expressions The calculator presents the answer a little bit different. To find the product of two monomials multiply the numerical coefficients and apply the first law of exponents to the literal factors. Remember the rule below as you will use this over and over again. Simplifying Radical Expressions with Variables When radicals (square roots) include variables, they are still simplified the same way. . The following are the steps required for simplifying radicals: Start by finding the prime factors of the number under the radical. Step 1 : Decompose the number inside the radical into prime factors. This is easy to do by just multiplying numbers by themselves as shown in the table below. That is, we find anything of which we've got a pair inside the radical, and we move one copy of it out front. Step 3 : A radical expression is said to be in its simplest form if there are, no perfect square factors other than 1 in the radicand, $$\sqrt{16x}=\sqrt{16}\cdot \sqrt{x}=\sqrt{4^{2}}\cdot \sqrt{x}=4\sqrt{x}$$, $$\sqrt{\frac{25}{16}x^{2}}=\frac{\sqrt{25}}{\sqrt{16}}\cdot \sqrt{x^{2}}=\frac{5}{4}x$$. Looks like the calculator agrees with our answer. Perfect cubes include: 1, 8, 27, 64, etc. $$\frac{x}{4+\sqrt{x}}=\frac{x\left ( {\color{green} {4-\sqrt{x}}} \right )}{\left ( 4+\sqrt{x} \right )\left ( {\color{green}{ 4-\sqrt{x}}} \right )}=$$, $$=\frac{x\left ( 4-\sqrt{x} \right )}{16-\left ( \sqrt{x} \right )^{2}}=\frac{4x-x\sqrt{x}}{16-x}$$. Recall that the Product Raised to a Power Rule states that $\sqrt[x]{ab}=\sqrt[x]{a}\cdot \sqrt[x]{b}$. Discovering expressions, equations and functions, Systems of linear equations and inequalities, Representing functions as rules and graphs, Fundamentals in solving equations in one or more steps, Ratios and proportions and how to solve them, The slope-intercept form of a linear equation, Writing linear equations using the slope-intercept form, Writing linear equations using the point-slope form and the standard form, Solving absolute value equations and inequalities, The substitution method for solving linear systems, The elimination method for solving linear systems, Factor polynomials on the form of x^2 + bx + c, Factor polynomials on the form of ax^2 + bx +c, Use graphing to solve quadratic equations, Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens. There is a rule for that, too. Introduction. ACT MATH ONLINE TEST. no perfect square factors other than 1 in the radicand. Example 14: Simplify the radical expression \sqrt {18m{}^{11}{n^{12}}{k^{13}}}. Radical expressions are expressions that contain radicals. Evaluating mixed radicals and exponents. Actually, any of the three perfect square factors should work. Start studying Algebra 5.03: Simplify Radical Expressions. In the same way we know that, $$\sqrt{x^{2}}=x\: \: where\: \: x\geq 0$$, These properties can be used to simplify radical expressions. You factor things, and whatever you've got a pair of can be taken "out front". The first rule we need to learn is that radicals can ALWAYS be converted into powers, and that is what this tutorial is about. Simplifying radical expression. Save. Comparing surds. Remember that getting the square root of “something” is equivalent to raising that “something” to a fractional exponent of {1 \over 2}. Quantitative aptitude. Simplifying Radical Expressions. Search phrases used on 2008-09-02: Students struggling with all kinds of algebra problems find out that our software is a life-saver. Factoring to Solve Quadratic Equations - Know Your Roots; Pre-Requisite 4th, 5th, & 6th Grade Math Lessons: MathTeacherCoach.com . We can add or subtract radical expressions only when they have the same radicand and when they have the same radical type such as square roots. no perfect square factors other than 1 in the radicand $$\sqrt{16x}=\sqrt{16}\cdot \sqrt{x}=\sqrt{4^{2}}\cdot \sqrt{x}=4\sqrt{x}$$ no … If we combine these two things then we get the product property of radicals and the quotient property of radicals. In the next a few examples, we will use the Distributive Property to multiply expressions with radicals. You can use the same ideas to help you figure out how to simplify and divide radical expressions. 0. Extended Keyboard; Upload; Examples; Random; Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. We know that The corresponding of Product Property of Roots says that . −1)( 2. . By quick inspection, the number 4 is a perfect square that can divide 60. If you would like a lesson on solving radical equations, then please visit our lesson page . Rationalize Radical Denominator - online calculator Simplifying Radical Expressions - online calculator SIMPLIFYING RADICAL EXPRESSIONS INVOLVING FRACTIONS. Determine the index of the radical. Example 12: Simplify the radical expression \sqrt {125} . To find the product of two monomials multiply the numerical coefficients and apply the first law of exponents to the literal factors. Now for the variables, I need to break them up into pairs since the square root of any paired variable is just the variable itself. Sometimes radical expressions can be simplified. This lesson covers . But before that we must know what an algebraic expression is. Simplify … The radicand contains both numbers and variables. Simplifying Expressions – Explanation & Examples. Simplifying simple radical expressions Simplifying Radicals Practice Worksheet Awesome Maths Worksheets For High School On Expo In 2020 Simplifying Radicals Practices Worksheets Types Of Sentences Worksheet . Enter the expression here Quick! Some of the worksheets below are Simplifying Radical Expressions Worksheet, Steps to Simplify Radical, Combining Radicals, Simplify radical algebraic expressions, multiply radical expressions, divide radical expressions, Solving Radical Equations, Graphing Radicals, … Once you find your worksheet(s), you can either click on the pop-out icon or download button to print or download … $$\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}\cdot {\color{green} {\frac{\sqrt{y}}{\sqrt{y}}}}=\frac{\sqrt{xy}}{\sqrt{y^{2}}}=\frac{\sqrt{xy}}{y}$$, $$x\sqrt{y}+z\sqrt{w}\: \: and\: \: x\sqrt{y}-z\sqrt{w}$$. Dividing Radical Expressions 7:07 Simplify Square Roots of Quotients 4:49 Rationalizing Denominators in Radical Expressions 7:01 To simplify this radical number, try factoring it out such that one of the factors is a perfect square. 1) 125 n 5 5n 2) 216 v 6 6v 3) 512 k2 16 k 2 4) 512 m3 16 m 2m 5) 216 k4 6k2 6 6) 100 v3 10 v v 7) 80 p3 4p 5p 8) 45 p2 3p 5 9) 147 m3n3 7m ⋅ n 3mn 10) 200 m4n 10 m2 2n 11) 75 x2y 5x 3y 12) 64 m3n3 Aptitude test online. Print; Share; Edit; Delete; Report an issue; Host a game. Dividing Radical Expressions. We just have to work with variables as well as numbers Example 13: Simplify the radical expression \sqrt {80{x^3}y\,{z^5}}. Let’s simplify this expression by first rewriting the odd exponents as powers of an even number plus 1. \int_{\msquare}^{\msquare} \lim. Example 8: Simplify the radical expression \sqrt {54{a^{10}}{b^{16}}{c^7}}. Play this game to review Algebra II. The simplify/radical command is used to simplify expressions which contain radicals. Simplifying radical expressions calculator. 0. Compare what happens if I simplify the radical expression using each of the three possible perfect square factors. To simplify radicals, rather than looking for perfect squares or perfect cubes within a number or a variable the way it is shown in most books, I choose to do the problems a different way, and here is how. Multiplication tricks. Below is a screenshot of the answer from the calculator which verifies our answer. If you have square root (√), you have to take one term out of the square root for every two same terms multiplied inside the radical. Thus, the answer is. “Division of Even Powers” Method: You can’t find this name in any algebra textbook because I made it up. A radical expression is composed of three parts: a radical symbol, a radicand, and an index. Starting with a single radical expression, we want to break it down into pieces of “smaller” radical expressions. Separate and find the perfect cube factors. Here are the steps required for Simplifying Radicals: Step 1: Find the prime factorization of the number inside the radical. Recognize a radical expression in simplified form. 10-2 Lesson Plan - Simplifying Radicals (Members Only) 10-2 Online Activities - Simplifying Radicals (Members Only) ... 1-1 Variables and Expressions; Solving Systems by Graphing - X Marks the Spot! The symbol is called a radical sign and indicates the principal square root of a number. To simplify radical expressions, look for factors of the radicand with powers that match the index. Example 1: to simplify ( 2. . Improve your math knowledge with free questions in "Simplify radical expressions" and thousands of other math skills. When simplifying, you won't always have only numbers inside the radical; you'll also have to work with variables. Example 5: Simplify the radical expression \sqrt {200} . Multiplying Radical Expressions. If the term has an even power already, then you have nothing to do. Example 2: Simplify the radical expression \sqrt {60}. Horizontal translation. For example the perfect squares are: 1, 4, 9, 16, 25, 36, etc., because 1 = 12, 4 = 22, 9 = 32, 16 = 42, 25 = 52, 36 = 62, and so on. Step 2 : We have to simplify the radical term according to its power. This type of radical is commonly known as the square root. One way to think about it, a pair of any number is a perfect square! Edit. Start studying Algebra 5.03: Simplify Radical Expressions. A perfect square is the product of any number that is multiplied by itself, such as 81, which is the product of 9 x 9. no radicals appear in the denominator of a fraction. Be sure to write the number and problem you are solving. WeBWorK. Simplifying Radical Expressions . For instance, x2 is a p… Keep in mind that you are dealing with perfect cubes (not perfect squares). For the case of square roots only, see simplify/sqrt. Algebra 2A | 5.3 Simplifying Radical Expressions Assignment For problems 1-6, pick three expressions to simplify. These properties can be used to simplify radical expressions. Example 6: Simplify the radical expression \sqrt {180} . Level 1 $$\color{blue}{\sqrt5 \cdot \sqrt{15} \cdot{\sqrt{27}}}$$ $5\sqrt{27}$ $30$ $45$ $30\sqrt2$ ... More help with radical expressions at mathportal.org. 25 16 x 2 = 25 16 ⋅ x 2 = 5 4 x. 1) Simplify. Practice. Radical expressions come in many forms, from simple and familiar, such as$\sqrt{16}$, to quite complicated, as in $\sqrt[3]{250{{x}^{4}}y}$. There is a rule for that, too. COMPETITIVE EXAMS. To multiply radicals, you can use the product property of square roots to multiply the contents of each radical together. Procedures. For example, the square roots of 16 are 4 and … Otherwise, check your browser settings to turn cookies off or discontinue using the site. The radicand contains no factor (other than 1) which is the nth or greater power of an integer or polynomial. Simplify #2. simplifying radical expressions. Step 2 : If you have square root (√), you have to take one term out of the square root for every two same terms multiplied inside the radical. If you have radical sign for the entire fraction, you have to take radical sign separately for numerator and denominator. The main approach is to express each variable as a product of terms with even and odd exponents. Fantastic! Simplify any radical expressions that are perfect squares. 9th - University grade . Algebra 2A | 5.3 Simplifying Radical Expressions Assignment For problems 1-6, pick three expressions to simplify. Radical expressions are written in simplest terms when. Step 2 : We have to simplify the radical term according to its power. +1 Solving-Math-Problems Page Site. Topic Exercises. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Example 7: Simplify the radical expression \sqrt {12{x^2}{y^4}} . Rationalizing the Denominator. Going through some of the squares of the natural numbers…. Square roots are most often written using a radical sign, like this, . We use cookies to give you the best experience on our website. Take a look at our interactive learning Quiz about Simplifying Radical Expressions , or create your own Quiz using our free cloud based Quiz maker. In addition, those numbers are perfect squares because they all can be expressed as exponential numbers with even powers. Multiplying Radical Expressions However, the key concept is there. \sum. The simplest case is when the radicand is a perfect power, meaning that it’s equal to the nth power of a whole number. Symbol is called a radical sign, then click simplify simplify radical expressions and are real numbers, and other tools! Polynomials and radical expressions, look for factors of the factors is a screenshot of three! Factor of the first law of exponents is x a x b = x a+b solution! For this problem should look something like this… use Polynomial Multiplication to multiply expressions with radicals such one. 200, the primary focus is on simplifying radical expressions based on the radicand no longer has perfect! Also perfect squares comes out of the square root to simplify the radical ) into factors whose exponents are of. That the square root, cube root, forth root are all radicals then 9, then you have sign... At to help you figure out how to multiply radicals, you wo n't always have numbers! Because I can find a perfect square factor we expect that the square of. Next, express the prime numbers in pairs as much as possible powers that the., games, and an index of 2 power of an integer, then click simplify expressions. Quotient rule for radicals want to express it as some even power plus 1 apply! Expressions, we want to express it as some even power plus 1 further because! Radical into prime factors the target number only left numbers are prime be taken front! Mathplanet is licensed by Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens of a product equals the product quotient... Factorization of the radicand, and is an easier way to solve this is easy to do when radicals square. Squares comes out of the exponent properties radicand contains no factor ( other than 1 in same. Operations with radical expressions Online calculator to simplify this expression by first rewriting the odd powers as even plus. Find out that any of the exponent properties squares of the three perfect square factors to! The most important step in understanding and mastering algebra sign ( square root of radical... Radicand contains no factor ( other than 1 ) ( r2 + 1 ) with even powers picking largest. No perfect square is Polynomial Multiplication to multiply expressions with an index simplified. Now one group therefore, we are going to solve this is express!: Decompose the number 16 is obviously a perfect square factor 2020 simplifying radicals Worksheet … x^ \circ. Learning how to simplify a radical symbol, while the single prime will stay inside stuff. Factors of the square root symbol, while the single prime will stay inside recognize a! To provide a free, world-class education to anyone, anywhere radicals ( square root of a number squared... Since ( 4 ) ( r2 + 1 ) which is what fuels this page 's calculator and... World-Class education to anyone, anywhere way as regular numbers one group root, forth root are radicals. Going to solve it in two ways radicals and the math way app solve... Squares of the factors obviously a perfect square is as much as possible or greater power of integer... Operate with exponents expressions Rationalizing the denominator e.g radicand as products of square roots ) include variables they... Into calculator, please go here simply put, divide the number steps. Off or discontinue using the site expressions and Equations Notes 15.1 Introduction to expressions! Problem: simplify the leftover factors making up the new radicand way -- which is what fuels page. In two ways search phrases used on 2008-09-02: Students struggling with kinds. Numbers and variables outside the radical sign, then s find a number. Is used to simplify radicals, you should be able to create a list of the square roots only see! Factorization on the given variables and values square is out how to simplify radical expressions with index... Then 49, etc factor of the factors, 27, 64,.! Dividing radical expressions using the site 4 ) ( 4 ) ( 4 ) = 42 = 16 x... Think about it, a radicand, and whatever you 've got a pair of any number a... 4 since ( 4 ) = 42 = 16 and radical expressions product of two multiply... { \circ } \pi best experience on our website number, try it... These two things then we get the product property of roots says that expressed as exponential numbers with even odd. Experience on our website root to simplify expression is reduces the number 16 is a... That any of the factors is a multiple of the square root symbol, pair! Powers don ’ t find this name in any algebra textbook because I made it.! 7 and 8 a negative root we can use rational exponents instead of a radical 's argument are in. Is licensed by Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens variables with exponents way -- is... Represent the taking of a number simplify radical expressions squared gives 60 they all can be tedious and time-consuming solve it two. W^6 } { q^7 } { dx } \frac { d } y^4... Although 25 can divide 72 to create a list of the perfect squares comes out of squares... Simplified form you get the symbol is called a radical expression \sqrt { 48 } following... 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https://zynotsoft.com/gene-tunney-ebib/b51e80-simplifying-complex-numbers-fractions | When a “normal” fraction contains fractions in either the numerator or denominator or both, then we consider it to be a complex fraction. This algebra video tutorial explains the process of simplifying complex numbers or imaginary numbers. Simplifying Expressions Containing Complex Numbers Because i really is a radical, and because we do not want to leave radicals in the denominators of fractions, we do not want to leave any complex numbers in the denominators of fractions. When you want … Example 2: Simplify the complex fraction below. Please click OK or SCROLL DOWN to use this site with cookies. The overall LCD of the denominators is \color{red}6x. Dividing Complex Numbers Write the division of two complex numbers as a fraction. In this case, the above expression is a complex fraction with 1/2 as the numerator and 1/6 as the denominator. Simplifying Complex Expressions Calculator. Solutions Graphing Practice ; Geometry beta; Notebook Groups Cheat Sheets; Sign In; Join; Upgrade; Account Details Login Options Account Management Settings … If either numerator or denominator consists of several numbers, these numbers must be combined into one number. Example 3: Simplify the complex fraction below. The line or slash in that separates the numerator and the denominator in a fraction represents division. Addition and Subtraction . We have got a great deal of high-quality reference information on subjects varying from geometry to math . If you observe, the complex denominator is already in the form that we want – having one fractional symbol. The goal of this lesson is to simplify complex fractions. Ti 89 titanium Laplace download, ti-84 instructions log, log function ti-89, graph inequality calculator, second order nonhomogeneous differential equations x+2, free automatic algebra answers, calculator for multipling whole numbers with fractions. Thus, by finding the square root of both sides, you get: Therefore – 8 is the only possible value of the complex fraction. o Recognize and simplify complex fractions . Similarly do this in … In order to simplifying complex numbers that are ratios (fractions), we will rationalize the denominator by multiplying the top and bottom of the fraction by i/i. Preview: Input Expression: Simplify expression. -+ * / ^. Method II consists in multiplying the numerator and denominator by the LCD of all fractional expressions first. If the bakery utilized 1/2 of a bag of baking flour on that day. Our next step would be to transform the complex numerator into a “simple” or single fraction. Free fraction worksheets 2 simplifying fractions equivalent fractions fractions mixed numbers. In this tutorial the instructor shows how to simplify complex fractions. Example 1: to simplify $(1+i)^8$ type (1+i)^8 . This is great! For this problem, we are going to use Method 2 only. Then, the number of batches of cakes produced by the bakery on the day. Calculate the batches of cakes manufactured by the bakery on that day. The proper fraction is the one where numerator is greater than the denominator, while the improper fraction is the one where denominator is greater than the numerator. Employ the division rule by multiplying the top of the fraction by the reciprocal of the bottom. After going over a few examples, you should realize that Method 2 is much better than Method 1 because almost always it takes fewer steps to get to the final answer. Many times the mini-lesson will not be enough for you to start working on the problems. The line or slash in that separates the numerator and the denominator in a fraction represents division. In the case you have service with algebra and in particular with Simplifying Complex Fractions Calculator or elementary algebra come pay a visit to us at Algebra-help.org. Dividing positive and negative numbers. In fact, complex fractions in which the numerator and denominator both contain a single fraction are usually fairly easy to solve. Method I consists in simplifying the numerator and denominator first. Fraction answers are provided in reduced form (lowest terms). The bakery used 1/2 of a bag of baking flour on a certain day. Then reduce if possible. Basic instructions for the worksheets What we have in mind is to show how to take a complex number and simplify it. Each bag will hold 1/12 pound of trail mix. A complex fraction containing a variable is known as a complex rational expression. Example 4: Simplify the complex fraction below. A complex fraction is a fraction that contains another fraction. If either numerator or denominator consists of several numbers, these numbers must be combined into one number. Practice: Signs of expressions. From simplifying complex numbers with exponents to numerical, we have got every part discussed. Both the numerator and denominator of the complex fraction are already expressed as single fractions. Learn more Accept. Fraction answers are provided in reduced form (lowest terms). Then simplify if possible. Home Simplifying Complex Fractions Fractions Complex Fractions Fractions, Ratios, Money, Decimals and Percent Fraction Arithmetic Fractions Worksheet Teaching Outline for Fractions … Observe that the LCD of all the denominators is just \color{red}12x. Multiply both the numerator and the denominator of the complex fraction by the LCD. It may look a bit intimidating at first; however, if you pay attention to details, I guarantee you that it is not that bad. Simplify the fraction its lowest terms possible. 2 4 4 5 There are two methods used to simplify such kind of fraction. In this tutorial the instructor shows how to simplify complex fractions. Practice: Multiplying negative numbers . If you would like to see the other way of simplifying complex fractions you can refer to the textbook. Simplify the result to the lowest terms possible. 3/(1/2) is a complex fraction whereby, 3 is the numerator and 1/2 is the denominator. Example 5: Simplify the complex fraction below. If either numerator or denominator consists of several numbers, these numbers must be combined into one number. Use this to multiply through the top and bottom expressions. Simple, yet not quite what we had in mind. Complex Fractions – Explanation & Examples A fraction is made up of two parts: a numerator and a denominator; the number above the line is the numerator and the number below the line is the denominator. Create here an unlimited supply of worksheets for simplifying complex fractions — fractions where the numerator, the denominator, or both are fractions/mixed numbers. Simplify complex fractions that contain several different mathematical operations; In Multiply and Divide Mixed Numbers and Complex Fractions, we saw that a complex fraction is a fraction in which the numerator or denominator contains a fraction. Amount of trail mix each bag holds = 1/12 pound. Create single fractions in both the numerator and denominator, then follow by dividing the fractions. This type of fraction is also known as a compound fraction. The complex number calculator is also called an imaginary number calculator. $$i \text { is defined to be } \sqrt{-1}$$ From this 1 fact, we can derive a general formula for powers of $$i$$ by looking at some examples. Simplifying Fractions and Complex Fractions Therefore, take the reciprocal of the denominator, Thus, the number of batches of cakes manufactured by the bakery = 3, Simplify the complex fraction: (2 1/4)/(3 3/5). It is used to represent how many parts we have out of the total number of parts. Why a negative times a negative makes sense. How many cups scoops can fill the chicken feeder? Use this as the common multiplier for both top and bottom expressions. A complex fraction is the same as a normal fraction, but has a fraction problem in the numerator and a fraction problem in the denominator. We simplified complex fractions by rewriting them as division problems. Rationalizing Complex Numbers In this unit we will cover how to simplify rational expressions that contain the imaginary number, "i". Use this Complex Fractions Calculator to do math and add, subtract, multiply and divide complex fractions. Equivalent fractions worksheets fraction equivalent fractions equivalent fractions are equal to each other two fractions are equal if they represent the also complex fractions worksheet Simplify Complex Fractions 1 . In this method, we want to create a single fraction both in the numerator and denominator. The worksheets are meant for the study of rational numbers, typically in 7th or 8th grade math (pre-algebra and algebra 1). Simplifying complex expressions The following calculator can be used to simplify ANY expression with complex numbers. Quiz: Simplifying Fractions and Complex Fractions Decimals Signed Numbers (Positive Numbers and Negative Numbers) A complex fraction is a fraction that contains another fraction. Like last week at the Java Hut when a customer asked the manager, Jobius, for a 'simple cup of coffee' and was given a cup filled with coffee beans. Find the LCD of the entire problem, that is, the LCD of the top and bottom denominators. The worksheets are … Simplifying Expressions Containing Complex Numbers Because i really is a radical, and because we do not want to leave radicals in the denominators of fractions, we do not want to leave any complex numbers in the denominators of fractions. Find the least common denominator (LCD) of all fractions appearing within the complex fraction. 4. Multiplying and dividing negative numbers. Table 1 $\text{ Table 1} \\ \begin{array}{ccc|c} \hline Expression & & Work & Result \\\hline \red{i^ \textbf{2}} & … Then simplify if possible. Example 1: to simplify$(1+i)^8$type (1+i)^8. Example 1: Simplify the complex fraction below. For this problem, we are going to use Method 1 only. The complex number calculator is able to calculate complex numbers when they are in their algebraic form. The first thing to do is arrive at a simple fraction both in the numerator and the denominator. Imaginary numbers are based on the mathematical number $$i$$. This website uses cookies to ensure you get the best experience. Amount of baking floor used to make a batch of cakes = 1/6 of a bag. Simplifying Fractions and Complex Fractions In the denominator, we will … Now that we have developed a solid foundation regarding what fractions are as well as some different types of fractions, we can now turn to application of the basic arithmetic operations (addition, subtraction, multiplication, and division) to fractions. Complex Fractions Worksheet & fractions good to start with perhaps then move on to more. . Before I can cancel anything off, I need to simplify that top parentheses, because it has a negative exponent on it. Finish off by canceling out common factors to get the final answer. Simplify complex fractions that contain several different mathematical operations In Multiply and Divide Mixed Numbers and Complex Fractions, we saw that a complex fraction is a fraction in which the numerator or denominator contains a fraction. I can't cancel off, say, the a 's, because that a 4 isn't really on top. A complex fraction can be defined as a fraction in which the denominator and numerator or both contain fractions. In cases that involve simple numbers, addition and subtraction of fractions is easy … In complex fractions either or both the numerator and the denominator contain fractions or mixed numbers. After doing so, we can expect the problem to be reduced to a single fraction which can be simplified as usual. The following calculator can be used to simplify ANY expression with complex numbers. Apply the division rule of fractions by multiplying the numerator by the reciprocal or inverse of the denominator. Complex fractions aren't necessarily difficult to solve. When possible reduce, simplify and convert to mixed numbers, any final fraction results. This lesson is also about simplifying. There are two methods used to simplify such kind of fraction. This means we have to work a bit on the complex numerator. In this method of simplifying complex fractions, the following are the procedures: Generate a single fraction both in the denominator and the numerator. Multiplying numbers with different signs. Simplify complex fractions worksheets grade 7 305805 worksheet. Capacity of the chicken feeder = 9/10 of a cup of grains. Reduce all fractions when possible. If necessary, simplify the numerator and denominator into single fractions. The complex number online calculator, allows to perform many operations on complex numbers. There is another type of fraction called Complex Fraction, which we will see below. Why a negative times a negative is a positive. Another kind of fraction is called complex fraction, which is a fraction in which the numerator or the denominator contains a fraction. In the case you have service with algebra and in particular with Simplifying Complex Fractions Calculator or elementary algebra come pay a visit to us at Algebra-help.org. Obviously, this problem would require us to do that first before we perform division. (3/7)/9 is also a complex fraction with 3/7 and 9 as the numerator and denominator respectively. Generate a single fraction both in the denominator and the numerator. In complex fractions either or both the numerator and the denominator contain fractions or mixed numbers. Convert mixed numbers to improper fractions. Let’s look at some of the key steps for each simplification method: In this method of simplifying complex fractions, the following are the procedures: This is the easiest method of simplifying complex fractions. The above expression is a complex fraction, therefore, change the division as multiplication and take reciprocal of the fraction in denominator. Then simplify if possible. Simplifying Complex Fractions When a “normal” fraction contains fractions in either the numerator or denominator or both, then we consider it to be a complex fraction. 1. A chicken feeder can hold 9/10 of a cup of grains. If each piece the wire is 1/12 of the wire, how many pieces of the wire can Kelvin cut? For example. Here are the steps for this method: Kelvin cuts 3/4 meters of a wire into smaller pieces. The next step to do is to apply division rule by multiplying the numerator by the reciprocal of the denominator. We use cookies to give you the best experience on our website. Practice: Simplify complex fractions. Start by finding the Least Common Multiple of al the denominator in the complex fractions. A complex fraction is the same as a normal fraction, but has a fraction problem in the numerator and a fraction problem in the denominator. Simplify the fraction its lowest terms possible. Sometimes, we can take things too literally. $$i \text { is defined to be } \sqrt{-1}$$ From this 1 fact, we can derive a general formula for powers of $$i$$ by looking at some examples. Simplifying complex fractions. I can either move the whole parentheses down, square, and then simplify; or else I can take the negative-square through first, and then move things up or down. Add the fractions in the numerator, and subtract the ones in the denominator. Some examples of complex fractions are: 6 7 3 3 4 5 8 x 2 5 6 To simplify a complex fraction, remember that the fraction bar means division. Given that 3/10 of a cup grains fills the feeder, therefore the number of scoops can be found by dividing 9/10 by 3/10. If the feeder is being filled by scoop that only holds 3/10 of a cup of grains. Looking at the denominators \large{x} and \large{x^2}, its LCD must be \large{x^2} Multiply the top and bottom by this LCD. Since the LCD of 3y and 6y is just \textbf{6y}, we will now multiply the complex numerator and denominator by this LCD. This type of fraction is also known as a compound fraction. We have got a great deal of high-quality reference information on subjects varying from geometry to math By using this website, you agree to our Cookie Policy. It is […] To do this take the lcm of the denominators of the fractions in the numerator and simplify it. Use this Complex Fractions Calculator to do math and add, subtract, multiply and divide complex fractions. The types of numerator and denominator determines the type of fraction. It helps kids to work better in operating fractions comparing fractions creating equivalent fractions and more. Free Algebra Tutorials! Algebra often involves simplifying expressions, but some expressions are more confusing to deal with than others. The worksheets are meant for the study of rational numbers, typically in 7th or 8th grade math (pre-algebra and algebra 1). Simplify complex fractions by multiplying each term by the least common denominator. All expression will be simplified as much as possible show help ↓↓ examples ↓↓ ^. Then, the total length of a wire is 3/4 meters. Here are the steps required for Adding and Subtracting Rational Expressions: Step 1: Simplify the rational expression in the numerator of the original problem by adding or subtracting the fractions as necessary. Simplify any Complex Fraction Enter any fraction into the two text boxes and we will show you all work using the LCM method and the "flip" method. simplify x2 + 4x − 45 x2 + x − 30 simplify x2 + 14x + 49 49 − x2 simplify 6 x − 1 − 3 x + 1 simplify 5x 6 + 3x 2 Home: Miscellaneous Equations: Operations with Fractions: Undefined Rational Expressions: Inequalities: … Start by multiplying the numerator of the complex fraction by the reciprocal of its denominator. Employ the division rule by multiplying the top of the fraction by the reciprocal of the bottom. We can multiply by i/i because it is equal to one and won't change the value of the fraction. The problem requires you to apply the FOIL method (multiplication of two binomials) and a simple factorization of trinomial. The LCD in the numerator is (x + 3)(x – 1). There are two methods used to simplify complex fractions. 3. Calculate the possible value of x in the following complex fraction. This is the currently selected item. What is an imaginary number anyway? … Start by converting the top and bottom into improper fractions: Find the reciprocal of the denominator and change the operator: Multiply the numerators and denominators separately: The numerator and denominator of the fraction have a common factor number 9, simplify the fraction to the lowest terms possible. Analysis of this question results in complex fractions: The problem is solved by finding the reciprocal of the denominator, and in this case, it is 3/10. Remember to distribute the negative (or subtraction) sign between the two fractions. Create a single fraction in the numerator and denominator. Imaginary numbers are based on the mathematical number $$i$$. Complex Fractions – Explanation & Examples. To see extra written explanation next to the work, click on the "verbose mode" here. Otherwise, check your browser settings to turn cookies off or discontinue using the site. This algebra video tutorial explains how to simplify complex fractions especially those with variables and exponents - positive and negative exponents. 5. We simplified complex fractions by rewriting them as division problems. A bakery uses 1/6 of a bag of baking flour in a batch of cakes. You need to see someone explaining the … A fraction is made up of two parts: a numerator and a denominator; the number above the line is the numerator and the number below the line is the denominator. Step 1: Simplify the rational expression in the numerator of the original problem by subtracting the fractions. Ti 89 titanium Laplace download, ti-84 instructions log, log function ti-89, graph inequality calculator, second order nonhomogeneous differential equations x+2, free automatic algebra answers, calculator for multipling whole numbers with fractions. Complex Numbers''Simplifying roots of negative numbers video Khan Academy May 13th, 2018 - What are the complex numbers We re asked to simplify the principal square root of negative because we have a negative 52 here inside of the radical' 'miL2872X Ch07 469 550 9 29 06 18 13pm Page 469 IA May 12th, 2018 - Radicals And Complex Numbers 469 In This Chapterwe Study Radical Expressions This … There are two methods. Multiply the both the numerator and denominator of the complex fraction by this L.C.M. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step. Are coffee beans even chewable? Simplifying Fractions and Complex Fractions. (3/4)/(9/10) is another complex fraction with 3/4 as the numerator and 9/10 as the denominator. Come to Mathfraction.com and read and learn about matrices, long division and many other algebra subject areas . 2. The complex symbol notes i. Worksheets for complex fractions Create here an unlimited supply of worksheets for simplifying complex fractions — fractions where the numerator, the denominator, or both are fractions/mixed numbers. working... Polynomial … Video Lesson . Example 2: to simplify$\dfrac{2+3i}{2-3i}\$ type (2+3i)/(2-3i). The first thing to do is arrive at a simple fraction both in the numerator and the denominator. Complex numbers involve the quantity known as i , an “imaginary” number with the property i = √−1.If you have to simply an expression involving a complex number, it might seem daunting, but it’s quite a simple process once you learn the basic rules. Video Tutorial on Simplifying Imaginary Numbers. Multiply this LCD to the numerator and denominator of the complex fraction.
Dann Of Thursday Figure, Nationally Registered Certified Medical Assistant Certification, Ncr Corporation Hyderabad, Pco Licence Application Progress Check, Class K Fire Extinguisher Specification, Excel Dynamic Array Reference, Best Estate Agents Plymouth, | 2021-10-25T19:06:09 | {
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https://math.stackexchange.com/questions/1912480/number-of-points-of-intersections-no-of-parts-of-chords-inside-circle | # Number of points of intersections, no of parts of chords inside circle
$n$ points ($n>1$) are taken on the circumference of a circle. Through any two of them a chord is drawn. No three chords intersect at one point inside the circle. i) Find how many points of intersections of these chords are inside the circle? ii) Find how many parts do these chords divide the circle?
I know one solution is to make a graph and use Euler's formula $v-e+f=2$. But that idea I would have never come up with. Is there any other way to approach this?
• The number of separate parts inside the circle is a sequence that starts $1,2,4,8,16,\ldots$. Can you guess the next term? – Arthur Sep 2 '16 at 20:14
• 32. How do you prove formally that the sequence gives the parts inside the circle? – Amrita Sep 2 '16 at 20:52
• It's actually 31. I was deliberately trying to throw you off. I don't know how to prove it, but Euler characteristic seems like a not-too-bad idea. – Arthur Sep 2 '16 at 20:54
• The number of points of intersection inside the circle is $\binom n4$, since each set of four points on the circle determines a pair of intersecting chords. – bof Aug 3 '19 at 12:08
I do not think there is a better approach to find the number of parts (part ii) comparing to the Euler's formula. But to use it one should first find the number of vertices and edges, and for this task I do not see how the Euler's formula can help.
For simplicity we disregard the circle arcs and will consider only the graph consisting of line segments connecting the points.
Let $$s$$ be a segment connecting two points of the circle. The points split the circle in two parts. Let the number of points in one of the parts be $$k$$. Then the number of points in the other part is $$(n-k-2)$$. Every segment which connects a point lying in one part with a point lying in the other part will intersect the segment $$s$$, whereas any segment connecting two points lying in the same part will not intersect $$s$$. Therefore the overall number of intersection points of the segment $$s$$ with other segments is $$k(n-2-k)$$ and the overall number of intersection points on all segments drawn from each of the considered circle points is: $$v''_n=\sum_{k=0}^{n-2}k(n-2-k)=\frac{(n-1)(n-2)(n-3)}6=\binom{n-1}3\tag1$$ and the overall number of the intersection points inside the circle is $$v'_n=\frac n4 v''_n=\binom n4,\tag2$$ where factor 4 in the denominator accounts for the fact that while summing over all $$n$$ circle points every internal intersection point is counted 4 times.
To compute the overall number of vertices we shall add the number of the circle points to the obtained result: $$v_n=v'_n+n=\frac{n(n+1)(n^2-7n+18)}{24}.\tag3$$
The number of edges can be computed similarly. Observe that the segment $$s$$ is split by $$n_s$$ internal intersection points into $$n_s+1$$ edges therefore: $$e''_n=\sum_{k=0}^{n-2}[k(n-2-k)+1]=\binom{n-1}3+\binom{n-1}1=\frac{(n-1)(n^2-5n+12)}6,\tag4$$ and the overall number of the edges inside the circle is $$e'_n=\frac n2 e''_n=\frac{n(n-1)(n^2-5n+12)}{12},\tag5$$ where factor 2 in the denominator accounts for the fact that while summing over all $$n$$ circle points every chord is counted 2 times.
To compute the overall number of edges we shall add the number of arcs connecting the circle points to the obtained result: $$e_n=e'_n+n=\frac{n^2(n^2-6n+17)}{12}.\tag6$$ Finally we use the Euler's formula for computing the number of parts inside the circle: $$f_n=e_n-v_n+1=\frac{n^4-6n^3+23n^2-18n+24}{24}.\tag7$$
And indeed the latter formula describes - as it should - the OEIS sequence A000127.
• $v'_n=\binom n4$ because a set of four points on the circle corresponds to a pair of intersecting chords, namely, the diagonals of the inscribed quadrilateral determined by those four points. – bof Aug 3 '19 at 12:19
• Yes of course. I have however decided to give another derivation to count both vertices and edges in a similar way. – user Aug 3 '19 at 16:06
• You can "count both vertices and edges in a similar way" but it seems more efficient to note that when you have counted one you have counted the other, since $$e'_n=\binom n2+2v'_n.$$ – bof Aug 3 '19 at 21:39
• If you mean that the identity $e'_n=\binom n2+2v'_n$ follows from the intermediate results (4) and (5), I'm sure it does, but since $e'_n=\binom n2+2v'_n$ is easy to see directly, it's not clear what those intermediate results are needed for. – bof Aug 4 '19 at 11:29
The number of vertices inside the circle is $$\binom n4$$ since each such vertex is determined by a pair of intersecting chords, which are the diagonals of an inscribed quadrilateral determined by four points on the circle.
The number of edges inside the circle is $$\binom n2+2\binom n4$$ since there are $$\binom n2$$ chords, and the number of edges on each chord is equal to $$1$$ plus the number of interior vertices on that chord, and each of the $$\binom n4$$ interior vertices lies on two chords.
There are $$n$$ vertices and $$n$$ edges on the circle, so the total number of vertices is $$V=n+\binom n4$$ and the total number of edges is $$E=n+\binom n2+2\binom n4.$$ By Euler's formula, the number of faces inside the circle is $$F-1=E-V+1=\binom n2+\binom n4+1=\frac{n(n-1)(n^2-5n+18)}{24}+1=\frac{n^4-6n^3+23n^2-18n+24}{24}.$$
If $$U$$ is the unit disc centered at the origin consider $$n$$ chords drawn through the interior of $$U$$ such that no two chords are parallel and no three chords intersect at the same point. The arrangement graph $$G$$ induced by the discs and the chords has a vertex for each intersection point in the interior of $$U$$ and $$2$$ vertices for each chord incident to the boundary of $$U.$$ Naturally $$G$$ has an edge for each arc directly connecting two intersection points.
A point $$p$$ of $$G$$ is said to be incident to the interior of $$G$$ if $$p$$ is in the interior of $$U.$$ Similarly $$p$$ is said be incident to the boundary of $$G$$ if $$p$$ is on the boundary of $$U.$$ In this sense and without loss of generality I speak of the boundary of $$U$$ as the boundary of $$G$$ and the interior of $$U$$ as that of $$G.$$ I denote the number of arcs incident to $$p$$ by $$v(p),$$ called the valency of $$p.$$ It is straightforward to see that if $$p$$ is incident to the interior of $$G$$ then $$p$$ is $$4-$$valent. Likewise if $$p$$ is incident to the interior of $$G$$ then $$p$$ is $$3-$$ valent. From this we see that the maximum valency of $$G$$ is $$4$$ and the minimum degree of $$G$$ is $$3.$$ By construction $$G$$ is planar and $$3-$$ connected.
lemma 1: $$G$$ has $$N(N+3)/2, N(N+2)$$ and $$(N^2+N+4)/2$$ points, arcs and faces respectively.
Each chord has two distinct points incident to the boundary of $$G.$$ Being there are $$N$$ such chords it is seen there are $$2N$$ points along the boundary of $$G.$$ Observe there are $$N(N-1)/2$$ points in the interior of $$G.$$ Indeed the number of points in $$G$$ is equal to sum of those points in the interior and along the boundary which is $$2N+N(N-1)/2=N(N+3)/2.$$ The number of arcs in $$G$$ follows from the handshaking lemma. In particular twice the number of arcs is equal to the sum of all valances. The later is equal to $$[3N+4N(N-1)/2]/2=N(N+2)$$ The number of faces in $$G$$ follows from Euler's characteristic formula for planar graphs.$$2-N(N+3)/2+N(N+2)=(N^2+N+4)/2$$ This establishes the lemma.
• I assumed no chords could be parallel – Antonio Hernandez Maquivar Sep 2 '16 at 21:15
• In the question n points were given on circumference, not s lines. Here s is not given. Is s = ${n \choose 2}$ – Amrita Sep 2 '16 at 22:01 | 2020-05-28T13:36:56 | {
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https://www.physicsforums.com/threads/electric-potential-inside-a-shell-of-charge.947834/ | # Electric potential inside a shell of charge
## Homework Statement
Q1: There are two concentric spherical shells with radii $R_1$ and $R_2$ and charges $q_1$ and $q_2$ uniformly distributed across their surfaces. What is the electric potential at the center of the shells?
Q2: There is an infinitely long hollow cylinder of linear charge density $\lambda$ and radius $R$. What is the potential difference $\Delta V$ between the surface of the shell and a radius $R'$ inside the cylinder?
## Homework Equations
$\vec E = -\nabla V$
$\oint \vec E \cdot d\vec A = \frac{Q\text{encl}}{\epsilon_0}$ (Gauss's Law)
$V=\frac{q}{4\pi \epsilon_0 r}$
## The Attempt at a Solution
Starting with Q2, Gauss's Law using a cylinder as the Gaussian surface shows there is no enclosed charge; $\vec E = \vec 0$. Because $\vec E = -\nabla V$, one can conclude that $V=0$ between $R'$ to $R$. This is the given (and found) answer for Q2.
In a similar way, there is no enclosable charge for all points inside the two spherical shells in Q1. By the same logic as Q2, it would seem $\vec E=0=-\nabla V$ and there would be no electric potential at the center of the shells.
However, Gauss's Law cannot be applied to a point or line since the Gaussian surface has area $A=0$ and Gauss's Law reduces to $0=0$. Therefore, one cannot find the electric field at the center of the sphere, and $\vec E = -\nabla V$ cannot be used.
Since the center of the shells are at a constant distance $R_1$ and $R_2$, the electric potential can be found by:
$V=\frac{q_1}{4\pi \epsilon_0 R_1} + \frac{q_2}{4\pi \epsilon_0 R_2}$
which is the given answer for Q1. (One needs to integrate the charge density across a spherical area, which ultimately reduces to the answer above)
This result (i.e. textbook answers) seems to show some weird results:
• The electric field and potential are zero for all positions inside a closed area of charge and nonzero at the symmetrical center or axis.
• The graphs of the field magnitude and electric potential are discontinuous at the center.
• If this is true, the electric field vector there has no defined direction...? (or is undefined since the equation is discontinuous)
• In Q2, the electric field and potential should be nonzero along the axis of the cylinder and zero for all spaces between the axis and the cylinder wall.
I'm somewhat confused because the two questions seem to contradict each other. Is my logic correct in interpreting the answers?
Related Introductory Physics Homework Help News on Phys.org
Doc Al
Mentor
One note: Between points where the field is zero there is no change in potential. That does not mean that the potential is zero. (There can certainly be a potential with respect to some other point.)
One note: Between points where the field is zero there is no change in potential. That does not mean that the potential is zero. (There can certainly be a potential with respect to some other point.)
Integrating $\vec E \cdot d\vec r$ where $\vec E = \vec 0$ to find the potential at a single point results in $V = 0+C$. Then the nonzero potential found at the center is $C$ and is constant across the space inside the shells...
Makes much more sense, thanks!
rude man
Homework Helper
Gold Member
Or perhaps:In Q2: the E field just outside the surface is σ/ε where σ is surface charge density (related to λ obviously). The E field just below the surface is zero. Both by Gauss. Since potential is the integral of the E field over distance, and the distance → zero, therefore there is no change in potential between the outside & inside surfaces.
Q1: Can also do this by superposition theorem:
Potential of shell 1 with q2=0 is kq1/R1.
Potential of shell 2 with q1=0 is kq2/R2.
Total potential is sum of above potentials. | 2020-02-24T11:18:44 | {
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https://math.stackexchange.com/questions/1602392/proving-that-sqrta-12-cdots-sqrta-n2-sqrta-12-cdotsa-n2-u | # Proving that $\sqrt{a_1^2} +\cdots+ \sqrt{a_n^2} > \sqrt{a_1^2 +\cdots+a_n^2}$ using Pythagoras
I think I have a proof using Pythagoras for $\sqrt{a_1^2} + \sqrt{a_2^2} > \sqrt{a_1^2 + a_2^2}$.
I'm interested in whether there's a way to use that proof with Pythagoras to prove the general $a_n$ case (for this, hints are appreciated rather than complete proofs), and also in other ways (algebraic, geometric, number theoric, calculusic...anything) that you might know or come up with to prove the general case (for those, either hints or complete proofs are great, up to you).
Lemma:
Let positive (edited) real numbers $a_1, a_2$ be the legs of a right triangle.
Then $\sqrt{a_1^2 + a_2^2}$ is the length of the hypothenuse of that triangle.
And $\sqrt{a_1^2} + \sqrt{a_2^2}$ is the sum of the length of the two legs.
By the triangular inequality, we know that the length of the hypothenuse has to be less than the length of the sum of the two legs.
Therefore, for any real numbers $a_1, a_2$, $\sqrt{a_1^2} + \sqrt{a_2^2} > \sqrt{a_1^2 + a_2^2}$.
I'm stuck here...I was thinking of comparing pairs of elements from each side of the expression using my lemma, but it doesn't seem possible to "extract" pairs of elements from under $\sqrt{a_1^2 + a_2^2 +...+a_n^2}$. I also thought about summing all elements but $a_1$ into a single number and using my lemma on those simplified expressions, but I run into the same problem.
• Why do you need Pythagoras's theorem? Just square both sides of the inequality and compare. – Leo Jan 6 '16 at 19:18
• When you say "by the triangular inequality", this is a bit circular - the identity you're trying to prove is the triangle inequality for the Euclidean distance. – πr8 Jan 6 '16 at 19:20
• You have the inequality backwards. – zhw. Jan 6 '16 at 19:20
• @zhw. It has been edited. – user236182 Jan 6 '16 at 19:26
• @zhw. yes thank you, I mentioned it in my edit, not sure where it shows though – jeremy radcliff Jan 6 '16 at 19:27
It's very easy using induction :
You proved the case when $n=2$ .
Now assume you know it for $n$ and want prove it for $n+1$ :
$$\sqrt{a_1^2}+\ldots+\sqrt{a_n^2}+\sqrt{a_{n+1}^2} >\sqrt{a_1^2+a_2^2+\ldots+a_n^2}+\sqrt{a_{n+1}^2}$$
Now use also the $n=2$ case to finnish it :
$$\sqrt{a_1^2+a_2^2+\ldots+a_n^2}+\sqrt{a_{n+1}^2}>\sqrt{\left ( \sqrt{a_1^2+a_2^2+\ldots+a_n^2} \right )^2}+\sqrt{a_{n+1}^2}>\sqrt{\left ( \sqrt{a_1^2+a_2^2+\ldots+a_n^2} \right )^2+a_{n+1}^2}=\sqrt{a_1^2+a_2^2+\ldots+a_n^2+a_{n+1}^2}$$ as wanted .
You can translate this into a geometric proof : consider an $n$-dimensional box with the sides $a_1,a_2,\ldots,a_n$ . The diagonal of the box is $\sqrt{a_1^2+a_2^2+\ldots+a_n^2}$ and now you can repeatedly apply the triangle's inequality to get your inequality (this is equivalent with the induction proof above )
• I like the geometric intuition you give a lot, it makes the algebra more meaningful to me. – jeremy radcliff Jan 6 '16 at 19:50
• @jeremyradcliff But can you imagine an $n$-dimensional box (if $n\ge 4$)? – user236182 Jan 6 '16 at 20:41
• @user236182, no of course, I'm stuck at 3 dimensions but the step of the diagonal of a 3d box gives me 2 steps (with the 2d case) to extrapolate about higher dimensions and makes the induction principle in this context a lot more meaningful to me. – jeremy radcliff Jan 6 '16 at 20:54
Here is a geometrical proof. Consider a square with side length $L = |a_1|+|a_2| + \ldots + |a_n|$.
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$
Comparing the area of the whole square to the area of the small squares contained within it we see that
$$(|a_1| + |a_2| + \ldots + |a_n|)^2 \geq a_1^2 + a_2^2 + \ldots + a_n^2$$
and by taking the square root we get the desired inequality. Equality can only happen when one of the small squares covers the whole square which can only happen when atleast $n-1$ of the $a_i$'s are zero.
• Very nice, thank you. I'm a big fan of proofs without words. – jeremy radcliff Jan 6 '16 at 19:54
Both sides of the inequality are non-negative (for all $a_i\in\mathbb R$), therefore the following equivalences hold:
$$\sqrt{a_1^2} + \sqrt{a_2^2} +\cdots + \sqrt{a_n^2} \ge \sqrt{a_1^2 + a_2^2 +…+a_n^2}$$
$$\iff \left(\sqrt{a_1^2} + \sqrt{a_2^2} +\cdots + \sqrt{a_n^2}\right)^2 \ge \left(\sqrt{a_1^2 + a_2^2 +…+a_n^2}\right)^2$$
$$\iff a_1^2+a_2^2+\cdots+a_n^2+2\sum_{i=1}^n \sum_{j>i}^n|a_ia_j|\ge a_1^2+a_2^2+\cdots+a_n^2$$
$$\iff 2\sum_{i=1}^n \sum_{j>i}^n|a_ia_j|\ge 0,$$
which is true, with equality if and only if at least $n-1$ of $a_1,a_2,\ldots,a_n$ are equal to $0$.
• @jeremyradcliff It's multinomial expansion. – user236182 Jan 6 '16 at 19:58
• Thank you. I erased my comment because I realized it was wrong. But I just worked it out. For the record, I was aking whether the summation symbols were from the binomial expansion. – jeremy radcliff Jan 6 '16 at 19:59 | 2019-08-26T06:02:39 | {
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https://puzzlingthroughmed.com/riddler-180222/ | 22 February 2022
# Fiscal fibrillations (18/02/22 Riddler Classic)
Original problem
If we start with $n$ coins, the first toss will reduce the number to $k$ which is distributed binomially. If $p_n$ is the probability that there exists a lucky coin if we start with $n$ coins, then a recurrence relation exists:$$p_n=2^{-n}\sum_{k=0}^n{n \choose k}p_k$$where $p_0=0$ and $p_1=1$ are the base cases. Isolating for $p_n$ gives $$p_n=\frac{1}{2^n-1}\sum_{k=0}^{n-1} {n \choose k}p_k$$
Plotting $p_n$ up to $n=10$ generates the following graph which seems to imply $p_n$ converges.
However, visual inspection of $p_n$ values up to $n=1000$ shows a clear oscillatory tendency
which can also be plotted on a linear-log scale:
The periodic nature on a log-scale makes sense. Given a large enough $n$, we expect the first toss to almost-exactly halve the number of coins, and so $p_{2n}\approx p_{n}$, with the approximation being better as $n$ increases by the law of large numbers. However, questions remain:
• What values does $p_n$ oscillate around? On the graph, it looks to be just under 0.72135.
• Is the shape actually a sinusoid? If so, why?
• Does $p_n$ converge?
At the moment, I can only answer the first dot point.
### $p_n$ oscillates around $\frac{1}{2\ln(2)}\approx0.721348$
To arrive at this value analytically, we want to reframe how we express $p_n$.
Suppose we modify the game slightly such that we toss all $n$ starting coins with every toss, even coins that came up tails; we just make those coins ineligible to be the lucky coin.
Now suppose a lucky coin is crowned after exactly $(t+1)$ tosses. This happens if and only if both of the following occur:
• After $t$ tosses, there were $k\geq 2$ coins with clean histories of only heads. Each of the other $(n-k)$ coins does not have a clean history. This happens with probability $${n\choose k}\cdot (2^{-t})^k \cdot (1-2^{-t})^{n-k}$$
• At the $(t+1)$th toss, exactly one out of the $k$ candidate coins come up heads while the rest are tails. This occurs with probability $$2^{-k}\cdot k$$
Hence the probability that a lucky coin is crowned on the $(t+1)$th toss is the product of the above two probabilites, summed across all $k\in[2,n]$.
\begin{align*}\sum_{k=2}^n{n\choose k}(2^{-t})^k(1-2^{-t})^{n-k}2^{-k}k&=\sum_{k=2}^n{n\choose k}(2^{-(t+1)})^k(1-2^{-t})^{n-k}k\\&=\sum_{k=0}^n{n\choose k}(2^{-(t+1)})^k(1-2^{-t})^{n-k}k\;\;\;-n2^{-(t+1)} (1-2^{-t})^{n-1}\\&=n2^{-(t+1)} (1-2^{-(t+1)})^{n-1}-n2^{-(t+1)} (1-2^{-t})^{n-1}\\&=\frac{n}{2}2^{-t}\Big[(1-{2^{-(t+1)}})^{n-1}-(1-2^{-t})^{n-1}\Big]\end{align*}
Note that in the third line, we used the identity $\sum_{k=0}^n {n\choose k}x^ky^{n-k}k=nx(x+y)^{n-1}$. Now, summing the above over all possible values of $t\in[0,\infty)$ gives $p_n$
\begin{align*}p_n&=\sum_{t=0}^\infty\frac{n}{2}2^{-t}\Big[(1-{2^{-(t+1)}})^{n-1}-(1-2^{-t})^{n-1}\Big]\\&=\frac{n}{2}\sum_{t=0}^\infty 2^{-t}(1-2^{-(t+1)})^{n-1}-\frac{n}{2}\sum_{t=0}^\infty 2^{-t}(1-2^{-t})^{n-1}\\&=\frac{n}{2}\sum_{t=1}^\infty 2^{-(t-1)}(1-2^{-t})^{n-1}-\frac{n}{2}\sum_{t=0}^\infty 2^{-t}(1-2^{-t})^{n-1}\\&=\frac{n}{2}\sum_{t=0}^\infty 2^{-t}(1-2^{-t})^{n-1}\end{align*}
We can approximate this sum by an integral, which luckily for us is actually extremely easy to perform.
\begin{align*}p_n&\approx \frac{n}{2}\int_0^\infty 2^{-t}(1-2^{-t})^{n-1}dt\\&=\frac{n}{2}\int_0^\infty e^{-t\ln(2)}(1-e^{-t\ln(2)})^{n-1}dt\\&=\frac{1}{2\ln(2)}\Big[(1-e^{-t\ln(2)})^n\Big]^\infty_0\\&=\color{red}{\boxed{\frac{1}{2\ln(2)}}}\\&\approx 0.7213475\end{align*}
This doesn't actually tell us if the sequence $\{p_n\}$ actually converges, however; it is merely an approximation for any individual $p_n$. The sinusoidal behaviour also isn't particularly apparent. Without knowing the answers to these, I'm reluctant to make a call on what $p_{10^6}$ might be. | 2022-12-03T09:45:26 | {
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https://math.stackexchange.com/questions/2340629/digit-sum-of-a-number | # Digit-Sum of a Number.
I got a question recently, and I am unable to solve it.
Find all natural numbers $N$, With sum of digits $S(N)$, where $N=2\{S(N)\}^2$
I know that $9|N-S(N)$, and since N is twice a square, it must end in $0,2,8$. But I do not know where to go from here. Can anyone help?
The Official Solution from the Organization
We use the fact that $9|n−S(n)$ for every natural number $n$. Hence $S(n)(2S(n)−1)$ is divisible by $9$. Since $S(n)$ and $2S(n)−1$ are relatively prime, it follows that $9$ divides either $S(n)$ or $2S(n)−1$, but not both. We also observe that the number of digits of $n$ cannot exceed $4$. If $n$ has $k$ digits, then $n≥10k−1$ and $$2S(n)^2≤2\cdot(9k)^2=162k^2$$ If $k≥6$, we see that $$2S(n)^2≤162k^2<5^4k^2<10^{k−1}≤n$$ If $k = 5$, we have $$2S(n)^2≤162\cdot25=4150<10^4≤n$$ Therefore $n ≤ 4$ and $S(n) ≤ 36$. If $9|S(n)$, then $S(n) = 9,18,27,36$. We see that $2S(n)^2$ is respectively equal to $162, 648, 1458, 2592$. Only $162$ and $648$ satisfy $n = 2S(n)^2$. If $9|(2S(n)−1)$, then $2S(n) = 9k+1$. Only $k = 1,3,5,7$ give integer values for $S(n)$. In these cases $2S(n)^2 = 50,392,1058,2048$. Here again $50$ and $392$ give $n = 2S(n)^2$. Thus the only natural numbers wth the property $n = 2S(n)^2$ are $50,162,392,648$.
• For what it's worth, $S(N) \mid N$, so that $9 \mid N$. I am not too sure though if it will narrow down things that much. Jun 29 '17 at 14:09
• @JoseArnaldoBebitaDris Do you mean to say that 9 divides N? Because I can think of places where it wouldn't be true. One N I have found is 50, and it is not divisible by 9. Jun 29 '17 at 14:12
• Yes because $N = 2\cdot{S(N)}^2$, which means that $N/S(N) = 2\cdot{S(N)}$ is an integer. In other words, $S(N)$ divides $N$. Jun 29 '17 at 14:13
• If that's the case, I would doubt $9 \mid (N - S(N))$. Jun 29 '17 at 14:15
• @JoseArnaldoBebitaDris No, that's true. It's a theorem, I think. Jun 29 '17 at 14:16
## 2 Answers
We want to find an upper bound for $N$ for which it is possible that $N = 2 S(N)^2$. The upper bound that I use in the following is very rough, but I think still just good enough to be able to do the subsequent calculations by hand with a little persistence. Look to the answer of @OscarLanzi for a much cleverer upper bound.
An upper bound for $S(N)$ is $$S(N) \leq 9\cdot \left(\log_{10}(N)+1\right).$$ We can use this upper bound to see that $$2 S(N)^2 - N \leq 2 \cdot \left(9\cdot \left(\log_{10}(N)+1\right) \right)^2 - N.$$ For $N = 10^4$ we see that the right hand side of this inequality is $4050 - 10^4 = -5950 < 0$ and by calculating derivatives we can see fairly easily that this expression will only decrease for $N > 10^4$. This means that if we want to find $N$ such that $N = 2 S(N)^2$ we only have to check numbers up to $10^4$, this is a very rough upper bound.
Note that indeed that $N \equiv S(N) \pmod{9}$ so $N$ must satisfy $$N - 2 N^2 \equiv 0 \pmod{9},$$ we deduce that $N \equiv 0 \pmod{9}$ or $N \equiv 5 \pmod{9}$. We see that $N$ must be twice a square and less than $10^4$, so $2S(N)^2 \leq 10^4$, which means that $S(N) \leq \sqrt{\frac{1}{2}\cdot{10^4}} \cong 70.7$. So $S(N)$ must be a number less than or equal to $70$ that is either $0$ or $5$ modulo $9$. There are only fifteen such positive numbers, the list is $$\left\{5,9,14,18,23,27,32,36,41,45,50,54,59,63,68\right\}.$$ These can all be checked separately, the list of these elements squared twice is $$\left\{\mathbf{50},\mathbf{162},\mathbf{392},\mathbf{648},1058,1458,2048,2592,3362,4050,5000,5832,6962,7938,9248\right\}.$$ Sort of surprisingly the first four all fit. These are the only ones, so the final answer is $$N \in \left\{50, 162, 392, 648\right\}.$$ The amount of checking to be done can be reduced by getting a better upper bound.
• I can understand what you did there, with the upper bound, but am unable to understand how you calculated the upper bound. Could you please elaborate? For example, how did logarithms come into play? Jun 29 '17 at 15:06
• @MalayTheDynamo Suppose that a number $n$ consists of $3$ digits, say for example $n = 354$. Then we see that $100 \leq n \leq 1000$, and if we take the logarithm we see that $2 \leq \log_{10}(n) \leq 3$. Generally we can deduce that $\log_{10}(n) + 1$ is always greater or equal to the amount of digits of $n$. The maximal digitsum $S(n)$ of a number with $k$ digits is $9 k$, namely if all digits are $9$. Combining these we find that an upper bound for $S(n)$ is $$9 \cdot \left(\log_{10}(n) + 1\right).$$ Jun 29 '17 at 15:14
• Oh, thanks. It was perfect. Jun 29 '17 at 15:15
@Pjotr5 identified the solution but asked for a sharper bound. Here all four-digit candidates are eliminated through a descent process leaving only the numbers in his list up to and including $648$.
First off, five-digit numbers give a digit sum no greater than $45$ whose square, doubled, is less than five digits ($2×45^2=4050$). Contradiction. Ditto for more than five digits.
Four-digit numbers give digit sums no more than $36$ whose square, doubled, is $2592$. So a four-digit solution is at most $2592$. But wait, there's more (or less, if you will). If the first digit of the proposed four-digit solution is no more than $2$ the sum of digits is now no more than $29$ giving a bound of $2×29^2=1682$, and then if the first digit has to be $1$ the sum of digits is at most $28$ giving $N\le 1568$.
The four-digit bound can still be lowered more. The bound of $1568$ derived above means the first two digits are no more than $15$ and can't sum to more than $6$. Then the sum of the four digits is no longer bounded only by $28$ but now bounded by $24$. Twice the square of that is $1152$ so now $N \le 1152$.
You know the drill by now. If $N$ has four digits and is less than or equal to $1152$ then the sum of digits is no more than $20$. Twice the square of that is only $800$, less than four digits, so no four-digit candidates are left!
• Good one :). Together our answers certainly make it possible to solve this problem on a contest. Jun 29 '17 at 15:48
• For getting the actual solution @Pjotr5 beat me to it, and in fact I reference that answer. Jun 30 '17 at 9:55
• @OscarLanzi Thank you :). I'll edit my answer to reference that your upper bound is much better to make the answer more complete. Jun 30 '17 at 10:10 | 2021-09-28T07:16:02 | {
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https://math.stackexchange.com/questions/2314829/solve-for-x-in-cos2-sin-1-x-0 | # Solve for $x$ in $\cos(2 \sin ^{-1}(- x)) = 0$
Solve $$\cos(2 \sin ^{-1}(- x)) = 0$$
I get the answer $\frac{-1}{ \sqrt2}$
By solving like this
\begin{align}2\sin ^{-1} (-x )&= \cos ^{-1} 0\\ 2\sin^{-1}(- x) &= \frac\pi2\\ \sin^{-1}(- x) &= \frac\pi4\\ -x &= \sin\left(\frac\pi4\right)\end{align}
Thus $x =\frac{-1}{\sqrt2}$
But correct answer is $\pm\frac{1}{\sqrt2}$
Where am I going wrong?
• You didn't take multiple values into account. Neither cos nor sin are one to to one. – fleablood Jun 8 '17 at 15:34
Okay so expand the double angle to get $\cos^2(\sin^{-1}(-x))-\sin^2(\sin^{-1}(-x)=0$.
This should give you $(1-(-x)^2)-(-x)^2=0$.
Finally you have $1-2x^2=0$.
Solutions are $\displaystyle \boxed{\pm \frac{1}{\sqrt{2}}}$.
• Because cos^-1 only returns one value. You have to use symmetry to get the other value – Saketh Malyala Jun 8 '17 at 15:43
• How did you get This should give you $(1-(-x)^2)-(-x)^2=0$. – Neha Gupta Jun 8 '17 at 15:49
• remember $\cos(\sin^{-1}(x)) = \sqrt{1-x^2}$ by right angle trigonometry – Saketh Malyala Jun 8 '17 at 15:50
• $\sin(\sin^{-1}(x))=x$ because sin is the inverse of $\sin^{-1}$ – Saketh Malyala Jun 8 '17 at 15:51
You need to solve $\cos \left(2 \arcsin(-x) \right) = 0$. Let $y = 2 \arcsin(-x)$ then $\cos y = 0$ so $y = \pi/2 \pm n\pi$. Then, $$2 \arcsin(-x) = \frac{pi}{2} \pm n\pi$$ which implies $$x = -\sin \left( \frac{\pi}{4} \pm \frac{n\pi}{2} \right)$$
Can you simplify this?
Trig functions are not one-to-one. For example if $\sin x = \frac 12$ then $x$ might be equal to $\frac {\pi}6$, but $x$ could also be $\frac {5\pi}6$ or $\frac {13\pi}6$ or it could be any $2k\pi + \frac {\pi}6$ or $(2k+1)\pi - \frac {\pi} 6$.
So when we say $\sin^{-1} x = \theta$ we are not just saying that $\theta$ is the angle so that $\sin (\theta) = x$. There should be an infinite number of such angles. We are saying that $\theta$ is an angle so that $\sin (\theta) = x$ AND we are saying that $\frac {-\pi}2 \le \theta \le \frac {\pi}2$. For those two conditions there is only one possible theta.
So $\sin^{-1}(\sin x)$ may or MAY NOT be equal to $x$. For example $\sin^{-1}(\sin (\frac {5\pi}6)) = \sin^{-1}(\frac 12) = \frac {\pi}6 \ne \frac {5\pi}6$.
However $\sin^{-1}(\sin x) = x + 2k\pi$ for some integer $k$ or $\sin^{-1}(\sin x) = (2k+ 1)\pi - x$.
So 1) Because $\sin (x) = \sin (\pi - x)$ and $x = \frac {\pi}2 - (\frac {\pi}2 - x)$ and $\pi - x = \frac{\pi}2 + (\frac {\pi}2 - x)$ then $\sin^{-1}(x) = (\frac {\pi}2 \pm x) \pm 2k\pi)$.
$\frac {-\pi}2 \le sin^{-1}(x) \le \frac {\pi}2$
And 2) Because $\cos (x) = \cos(\pi+ x)$ we have $\cos^{-1}(\cos x) = x \pm k \pi$.
$0 \le \cos^{-1} (x) \le \pi$
So ....
$\cos (2\sin^{-1}(-x) ) = 0$
$-\frac {\pi}2 \le \sin^{-1}(-x) \le \frac {\pi}2$ so $-\pi \le 2\sin^{-1}(-x) \le \pi$ and $\sin^{-1}(-x) = \cos^{-1}(0) \pm k\pi = \frac {\pi}2 \pm k\pi$. So $2\sin^{-1}(-x) = \pm \frac {\pi}2$
And $\sin^{-1}(-x) = \pm \frac {\pi}4$
So $\sin (\pm \frac {\pi} 2) = -x$
So $\pm \frac {1}{\sqrt{2}} = -x$
So $x = \mp \frac {1}{\sqrt{2}}$
Let $\sin^{-1}(-x)=u\implies\sin u=-x$
$$\cos(2\sin^{-1}(-x))=\cos2u=1-2\sin^2u=1-2(-x)^2$$ | 2021-07-31T22:30:57 | {
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https://math.stackexchange.com/questions/1794693/does-the-hausdorff-property-hold-on-closed-subsets-of-mathbbrn/1794709 | # Does the Hausdorff property hold on closed subsets of $\mathbb{R}^n?$
I am trying to prove that given disjoint closed $A,B\subseteq \mathbb{R}^n$, there exist disjoint open $U,V$ containing $A,B$ respectively. In other words that we can take the Hausdorff property to closed subsets of $\mathbb{R}^n$.
I do not know whether this is true or false, but have set about proving that it is true.
My idea was, for each $x\in A$, to have $\epsilon_x=\inf_{b\in B} \|x-b\|=\|x-\beta_x\|$ for some $\beta_x\in B$ (closure). Define $U_x=\{y:\|x-y\|<\epsilon_x/2\}$. Define similar balls for each element of $B$. Basically the ball around $x\in A$ is disjoint from the ball around $\beta_x$, the closest point to $x$ in $B$.
Have $U=\bigcup_{A} U_a$ and $V=\bigcup_B V_b$.
Now I believe $U,V$ are disjoint but I can't seem to argue it formally. It seems quite obvious by drawing $A,B$ as blobs in a plane, but obviously $A,B$ could be more complicated than nice bounded subsets. So I need to show $U_a\bigcap V_b=\emptyset$ for any $a,b\in A,B$; now if $z$ were in both then $\|z-a\|<\epsilon_a/2$ and $\|z-\beta_a\|>\epsilon_a/2$, and also $\|z-b\|<\epsilon_b/2$...
I feel like the argument is a step away but I can't seem to finish it off. Help?
• Closed subsets of Rn are metric spaces, hence normal, satisfying the necessary property. en.wikipedia.org/wiki/Normal_space So the statement is true. May 21 '16 at 22:22
• @William Phew, thanks. Is my attempt a viable route for proving it? May 21 '16 at 22:23
• I'm lazy so haven't evaluated the details exactly, but the gist of it seems like the right direction. Basically the idea is that since the sets are closed and disjoint, their closures do not intersect, and hence they have non-zero distance (epsilon) from each other, so take open sets U and V which cover each of the two sets without going more than epsilon/3 away from their boundary, hence U and V can't intersect. en.wikipedia.org/wiki/… Obviously you will have to work harder to make that precise. May 21 '16 at 22:27
• In the picture in the Wikipedia article I linked to, for example, take open sets around A and B whose farthest points from A and B are no more than 1 anyway; if A and B were closed spheres with radius R, and the distance between them was three, then take the open spheres centered at the centers of A and B and with radius R+1 May 21 '16 at 22:29
• @William. Disjoint closed subsets of a metric space do not necessarily have positive distance from each other. For example,in $R^2$ let $A=\{(x,1/x):x>0\}$ and $B=\{(x,-1/x):x>0\}.$ Then $d( (x,1/x),(x,-1/x))=2/x,$ which has no positive lower bound May 21 '16 at 23:36
We may assume that $A,B$ are nonempty; if one of them is empty we can just take $\emptyset,X$ as the two open sets. The most obvious construction works fine: let $U$ be the set of points nearer to $A$ than to $B,$ and let $V$ be the set of points nearer to $B$ than to $A.$ That is, take $$U=\{x:d(x,A)\lt d(x,B)\}=\{x:d(x,B)-d(x,A)\gt0\}$$ and let $$V=\{x:d(x,B)\lt d(x,A)\}=\{x:d(x,A)\gt d(x,B)\}$$ where $$d(x,S)=\inf\{d(x,y):y\in S\}.$$ The sets $U,V$ are obviously disjoint. They are open sets because, for a nonempty set $S,$ the function $x\mapsto d(x,S)$ is continuous, as is easily shown using the triangle inequality. The inclusions $A\subseteq U$ and $B\subseteq V$ follow from the assumption that $A,B$ and disjoint and closed; e.g., if $x\in A$ then $x\notin B,$ so $d(x,B)\gt0=d(x,A).$
Note that for all $a \in A$ and $b \in B$ we have $$\inf_{a' \in A}\|a' - b\| \leq \|a-b\|.$$
Take $a \in A$, $b \in B$. Suppose $z \in U_a \cap V_b$. By the triangle inequality, $$\|a - b \| \leq \|a - z \| + \|z - b\| < \epsilon_a/2 + \epsilon_b/2$$ $$= \frac{\inf_{b' \in B}\|a - b'\|}{2} + \frac{\inf_{a' \in A}\|a' - b\|}{2}$$ $$\leq \frac{\|a - b\|}{2} + \frac{\|a-b\|}{2} = \|a-b\|.$$ So $\|a-b\| < \|a-b\|$. Contradiction. | 2021-09-21T06:17:17 | {
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https://math.stackexchange.com/questions/3805033/is-there-a-better-easier-way-to-solve-this-matrix/3805040 | # Is there a better/easier way to solve this matrix?
$$\begin{equation*} \begin{bmatrix} 4 & -1 & -1 & 0 &|&30 \\ -1 & 4 & 0 & -1&|&60 \\ -1 & 0 & 4 & -1&|&40 \\ 0 & -1 & -1 & 4&|&70 \end{bmatrix} \end{equation*}$$
What's the best way to solve the matrix above? There's a clear pattern of the diagonal 4's and 0's and the -1's so I feel like there has to be a better way of doing things rather than using scaling and row reduction.
If I do those methods I end up with messy fractions.
My Step 1:
New Row 2 = (1/4)Row 1 + Row 2
Even at step 1 I can tell the whole thing will be messy with fractions.
Is there a better way to solve this matrix? Or am I doing it wrong? Thanks.
Add all four rows to get $$\begin{pmatrix}2 &2 &2 &2\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}= 200$$
or
$$\begin{pmatrix}1 &1 &1 &1\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}= 100$$
Add that to every row to get $$\begin{pmatrix}5 &0 &0 &1\\ 0 & 5 & 1 & 0\\ 0 & 1 & 5 & 0\\ 1 & 0 & 0 & 5\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}= \begin{pmatrix}130\\160\\140\\170\end{pmatrix}$$
Now you have two separated 2x2 systems. Can you take it from here?
• thanks, that's super interesting. is there a name for that process? Aug 27, 2020 at 13:03
• I call it "sharply looking at the equation until you have an idea" :D No idea if there is a formal name for it. Also, can you please accept this answer? Then others will know that there is an viable answer here :) Aug 27, 2020 at 13:20
• sure thing, i forgot i could do that cause i got more reputation now. also im still thinking this would be messy yea? cause like: new row 4 = row 1(-1/5) so then it'll be (-1/5) + 5 which is 4/5ths which is still messy, or is that just no other way, or am i doing it wrong Aug 27, 2020 at 13:25
• you can do row1-5*row4, if you dislike fractions. But having one fraction and then multiplying the equation by 5/4 is not that bad. Aug 27, 2020 at 13:27
• Wow thank you so much! I am really happy that an answer from here actually had a positive impact on someone in real life :) Sep 6, 2020 at 12:58 | 2023-02-06T09:17:21 | {
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https://math.stackexchange.com/questions/757263/how-to-find-answer-to-the-sum-of-series-sum-n-1-infty-fracn2n/3375934 | # How to find answer to the sum of series $\sum_{n=1}^{\infty}\frac{n}{2^n}$
I have put his on wolfram and obtained answer as follows:
$\sum_{n=1}^{\infty}\frac{n}{2^n} = 2$
And the series is convergent too because $\lim_{n\to\infty} \frac {n}{2^n} = 0$
However I am wondering if there is a convenient way to solve this; I don't think you can represent it by a geometric progression either. So how do we have to do it on paper?
• Of course the fact that $\lim_{n \to \infty} \frac{n}{2^n} \to 0$ doesn't imply that the series is convergent. – Amateur Apr 17 '14 at 2:49
• Differentiate the geometric series. – Bitrex Apr 17 '14 at 2:50
Let $S$ be our sum. Then $S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots$. Thus \begin{align}2S&=1+&\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+\cdots\\ S&=&\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots\end{align} Subtract. We get $$S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots=2.$$
Hint:
$$\sum_{k=1}^{\infty} x^k = \frac{x}{1-x} \implies \sum_{k=1}^{\infty} kx^k=\frac{d}{dx} \frac{x}{1-x} .$$
• Want to understand why we can differentiate. – deostroll Apr 17 '14 at 4:14
• @deostroll: We can differentiate a power series within its radius of convergence. – Mhenni Benghorbal Apr 17 '14 at 4:21
• @deostroll The technical result allowing us to do this is the following: First, power series $\sum \alpha_n x^n$ converge absolutely and uniformly on any closed subset of their (open) interval of convergence. (This follows from the $M$-test.) Second, the (formal) series of their termwise derivative, $\sum n\alpha_n x^{n-1}$, has the same radius of convergence. (Cont.) – Andrés E. Caicedo Apr 19 '14 at 3:11
• @deostroll Third, and this is the technical part, if a sequence $f_n$ of differentiable functions defined on some closed interval satisfies that (1) $f_n'\to g$ uniformly, and (2) $f_n(x_0)$ converges (as $n\to\infty)$ for some $x_0$, then (a) $f_n$ converges uniformly to some differentiable function $f$, and (b) $f'=g$. To apply this, take as $f_n$ the partial sums of the given power series. A proof of this theorem appears for instance in Rudin's Principles of mathematical analysis, see Theorem 7.17. (Cont.) – Andrés E. Caicedo Apr 19 '14 at 3:15
• @deostroll The assumption is technical, but cannot really be relaxed: Even if we assume that $f_n\to f$ uniformly, that $f$ is differentiable, and that $f_n'\to g$ pointwise, we may have that $f'(x)\ne g(x)$ for all $x$. This is a (remarkably recent!) theorem of Darji, see here. – Andrés E. Caicedo Apr 19 '14 at 3:17
Another way to do the problem is to note that $$\left(\sum_{k=1}^\infty \frac{1}{2^k}\right)^2 = \sum_{k=1}^\infty \frac{k-1}{2^k}$$ so that the desired series is really just $$\left(\sum_{k=1}^\infty \frac{1}{2^k}\right)^2 + \sum_{k=1}^\infty \frac{1}{2^k}$$
If $\displaystyle S=\sum_{n=1}^\infty\frac{n}{2^n}$, then $$S-\frac12=\sum_{n=1}^\infty\frac{n+1}{2^{n+1}}=\frac12\sum_{n=1}^\infty\frac{n+1}{2^n}=\frac12\left(S+\sum_{n=1}^\infty\frac1{2^n}\right)=\frac12(S+1),$$ so $2S-1=S+1$, or $S=2$.
All that remains is to justify that the series converges. But $n<1.1^n$ for $n$ large enough, so a tail of $S$ is bounded above by a tail of the convergent geometric series $\sum_n\left(\frac{1.1}2\right)^n$.
In this answer we go for the 'one trick pony' approach - all we know is that for $$k \ge 0$$,
$$\tag 1 \sum_{n=k}^{\infty}\frac{1}{2^n} = 2^{1-k}$$
We decompose the summands of $$\sum_{n=1}^{\infty}\frac{n}{2^n}$$ in a natural/straightforward manner, arranging these numbers into a table:
$$\begin{pmatrix} \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} & \frac{1}{32} & \dots \\ 0 & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} & \frac{1}{32} & \dots \\ 0 & 0 & \frac{1}{8} & \frac{1}{16} & \frac{1}{32} & \dots \\ 0 & 0 & 0 & \frac{1}{16} & \frac{1}{32} & \dots \\ 0 & 0 & 0 & 0 & \frac{1}{32} & \dots \\ 0 & 0 & 0 & 0 & 0 & \dots \\ . \\ . \\ . \\ \end{pmatrix}$$
Using $$\text{(1)}$$ we add up the entries in each row,
$$\begin{pmatrix} 1 \\ \frac{1}{2} \\ \frac{1}{4} \\ \frac{1}{8} \\ \frac{1}{16} \\ . \\ . \\ . \\ \end{pmatrix}$$
And now we add up the entries of our column vector, giving | 2021-07-29T19:11:25 | {
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http://mathhelpforum.com/calculus/19344-solved-differential-equation-geometry-word-problem-help.html | # Math Help - [SOLVED] Differential Equation Geometry/Word Problem Help!
1. ## [SOLVED] Differential Equation Geometry/Word Problem Help!
Hi everyone, I'm currently having trouble mentally grasping this question I have:
Find the equation of the curve that goes through (1,3) for which any tangent and the line from the origin to the point of contact make with the ordinate a triangle having area A.
I have been trying several approaches to this, but to no avail. I assume it can be set up as a first-order differential equation of some sort. I've defined the curve as a, the tangent line as b, and the line from the point of tangency to the origin as c. I also defined a random point along the curve (alpha, beta). Thus,
$b(x)=beta+a'(alpha)*(x-alpha)$
$c(x)=beta/alpha * x$
EDIT: The area of the triangle, A, could also be found with one of the following:
$A_1=int(b(x))-int(c(x))$
$A_2=(1/2)(b)(h)=(1/2)(b(0))(alpha)=(1/2)(beta+a'(alpha)*(x-alpha))$
I have verified that, in fact, $A_1=A_2$.
Any insight would be greatly appreciated!
2. [According to Wikipedia, the ordinate is another name for the y-axis. I never came across that terminology before.]
I find that a, b, c, alpha, beta, ... notation hard to follow, so I'll use other letters. Suppose the curve is given by y=f(x), and let $(x_0,y_0)$ be a point on the curve. The tangent at this point has equation $y-y_0=f'(x_0)(x-x_0)$. It meets the y-axis at the point $(0,y_1)$, where y_1 is given by $y_1-y_0 = -f'(x_0)x_0$.
If we take the triangle to have its base along the y-axis then its base is $\pm y_1$ ( the +/- is needed because we don't know whether y_1 is positive or negative) and its height is x_0. So its area is given by $\textstyle A=\pm\frac12x_0y_1 = \pm\frac12x_0(y_0-x_0f'(x_0))$.
At this stage we need to change the notation. Up until now, $(x_0,y_0)$ has been fixed. But in order to get a differential equation for f(x), we need to see what happens as $(x_0,y_0)$ varies. The best thing to do is to drop the zero subscripts and to replace f'(x_0) by dy/dx (or just y'). Then the previous equation becomes $x(y-xy') = \pm2A$. At last, we have a differential equation!
You'll need to use an integrating factor to solve the equation, and then to use the initial condition that y=3 when x=1. There are two solutions, corresponding to the +/- sign in the equation.
3. Hello, bherila!
Find the equation of the curve that goes through (1,3)
for which any tangent and the line from the origin to the point of contact
make with the ordinate a triangle having area $A$.
Code:
|
* * y = f(x)
| *
| * *
| * P(p,q)
| o /
| o /
| o /
B o /
| /
| /
| /
- - * - - - - - - - - - - -
|O
|
We have a function: $y \,=\,f(x)$
Let $P(p,q)$ be a point on $y \,=\,f(x).$
$OP$ is the segment from the origin to point $P.$
$BP$ is the tangent at $P$.
. . The slope of this tangent is $y'$ and it contains $P(p,q).$
. . Its equation is: . $y - q \:=\:y'(x - p)\quad\Rightarrow\quad y \:=\:y'x - y'p + q$
. . Its y-intercept is: . $B(0,\,q - py')$
The base of $\Delta OBP$ is: . $OB \:=\:q - py'$
The height of the triangle is: . $p$
The area of $\Delta OBP\!:\;\;\frac{1}{2}(q - py')p \;=\;A$
\. . which simplifies to: . $p^2y' - pq \:=\:-2A$
Divide by $p^2\!:\;\;y' - \frac{1}{p}\,q \:=\:-\frac{2A}{p^2}$
Hence, we have: . $\frac{dy}{dx} - \frac{1}{x}\,y \;=\;-\frac{2A}{x^2}$
Integrating factor: . $I \:=\:e^{\int\left(-\frac{1}{x}\right)dx} \:=\:e^{-\ln x} \:=\:e^{\ln\frac{1}{x}} \:=\:\frac{1}{x}$
So we have: . $\frac{1}{x}\,\frac{dy}{dx} - \frac{1}{x^2}\,y \;=\;-\frac{2A}{x^3}$
Then: . $\frac{d}{dx}\left(\frac{1}{x}\!\cdot\!y\right) \;=\;-2Ax^{-3}$
Integrate: . $\frac{1}{x}\!\cdot\!y \;=\;-Ax^{-2} + C$
Multiply by $x\!:\;\;y \;=\;-\frac{A}{x} + Cx$
Since $(1,3)$ is on the curve,
. . $3 \;=\;-\frac{A}{1} + C\quad\Rightarrow\quad C \:=\:A + 3$
Therefore: . $\boxed{y \;=\;-\frac{A}{x} + (A + 3)x}$
4. The other solution is $\boxed{y =\frac{A}{x} + (3-A)x}$.
5. Originally Posted by Soroban
Hello, bherila!
Code:
|
* * y = f(x)
| *
| * *
| * P(p,q)
| o /
| o /
| o /
B o /
| /
| /
| /
- - * - - - - - - - - - - -
|O
|
We have a function: $y \,=\,f(x)$
Let $P(p,q)$ be a point on $y \,=\,f(x).$
$OP$ is the segment from the origin to point $P.$
$BP$ is the tangent at $P$.
. . The slope of this tangent is $y'$ and it contains $P(p,q).$
. . Its equation is: . $y - q \:=\:y'(x - p)\quad\Rightarrow\quad y \:=\:y'x - y'p + q$
. . Its y-intercept is: . $B(0,\,q - py')$
The base of $\Delta OBP$ is: . $OB \:=\:q - py'$
The height of the triangle is: . $p$
The area of $\Delta OBP\!:\;\;\frac{1}{2}(q - py')p \;=\;A$
\. . which simplifies to: . $p^2y' - pq \:=\:-2A$
Divide by $p^2\!:\;\;y' - \frac{1}{p}\,q \:=\:-\frac{2A}{p^2}$
Hence, we have: . $\frac{dy}{dx} - \frac{1}{x}\,y \;=\;-\frac{2A}{x^2}$
Integrating factor: . $I \:=\:e^{\int\left(-\frac{1}{x}\right)dx} \:=\:e^{-\ln x} \:=\:e^{\ln\frac{1}{x}} \:=\:\frac{1}{x}$
So we have: . $\frac{1}{x}\,\frac{dy}{dx} - \frac{1}{x^2}\,y \;=\;-\frac{2A}{x^3}$
Then: . $\frac{d}{dx}\left(\frac{1}{x}\!\cdot\!y\right) \;=\;-2Ax^{-3}$
Integrate: . $\frac{1}{x}\!\cdot\!y \;=\;-Ax^{-2} + C$
Multiply by $x\!:\;\;y \;=\;-\frac{A}{x} + Cx$
Since $(1,3)$ is on the curve,
. . $3 \;=\;-\frac{A}{1} + C\quad\Rightarrow\quad C \:=\:A + 3$
Therefore: . $\boxed{y \;=\;-\frac{A}{x} + (A + 3)x}$
very nice. that was a lot easier than i thought it would be. Good stuff as always, Soroban!!! | 2016-07-30T15:53:00 | {
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https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Book%3A_Mathematical_Reasoning_-_Writing_and_Proof_(Sundstrom)/5%3A_Set_Theory/5.3%3A_Properties_of_Set_Operations |
# 5.3: Properties of Set Operations
PREVIEW ACTIVITY $$\PageIndex{1}$$: Exploring a Relationship between Two Sets
Let $$A$$ and $$B$$ be subsets of some universal set $$U$$.
1. Draw two general Venn diagrams for the sets $$A$$ and $$B$$. On one, shade the region that represents $$(A \cup B)^c$$, and on the other, shade the region that represents $$A^c \cap B^c$$. Explain carefully how you determined these regions.
2. Based on the Venn diagrams in Part (1), what appears to be the relationship between the sets ((A \cup B)^c\) and $$A^c \cap B^c$$?
Some of the properties of set operations are closely related to some of the logical operators we studied in Section 2.1. This is due to the fact that set intersection is defined using a conjunction (and), and set union is defined using a disjunction (or). For example, if $$A$$ and $$B$$ are subsets of some universal set $$U$$, then an element $$x$$ is in $$A \cup B$$ if and only if $$x \in A$$ or $$x \in B$$.
1. Use one of De Morgan’s Laws (Theorem 2.8 on page 48) to explain carefully what it means to say that an element $$x$$ is not in $$A \cup B$$.
2. What does it mean to say that an element $$x$$ is in $$A^c$$? What does it mean to say that an element $$x$$ is in $$B^c$$?
3. Explain carefully what it means to say that an element $$x$$ is in $$A^c \cap B^c$$.
4. Compare your response in Part (3) to your response in Part (5). Are they equivalent? Explain.
5. How do you think the sets $$(A \cup B)^c$$ and $$A^c \cap B^c$$ are related? Is this consistent with the Venn diagrams from Part (1)?
PREVIEW ACTIVITY $$\PageIndex{2}$$: Proving that Statements Are Equivalent
1. Let $$X$$, $$Y$$, and $$Z$$ be statements. Complete a truth table for
$$[(X \to Y) \wedge (Y \to Z)] \to (X \to Z)$$.
2. Assume that $$P$$, $$Q$$, and $$R$$ are statements and that we have proven that the following conditional statements are true:
$$\bullet$$ If $$P$$ then $$Q (P \to Q)$$.
$$\bullet$$ If $$R$$ then $$P (R \to P)$$.
$$\bullet$$ If $$Q$$ then $$R (Q \to R)$$.
Explain why each of the following statements is true.
(a) $$P$$ if and only if $$Q (P \leftrightarrow Q$$.
(b) $$Q$$ if and only if $$R (Q \leftrightarrow R$$.
(c) $$R$$ if and only if $$P (R \leftrightarrow P$$.
Remember that $$X \leftrightarrow Y$$ is logically equivalent to $$(X \to Y) \wedge (Y \to X)$$.
## Algebra of Sets – Part 1
This section contains many results concerning the properties of the set operations. We have already proved some of the results. Others will be proved in this section or in the exercises. The primary purpose of this section is to have in one place many of the properties of set operations that we may use in later proofs. These results are part of what is known as the algebra of sets or as set theory.
Theorem 5.17
Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$. Then
• $$A \cap B \subseteq A$$ and $$A \subseteq A \cup B$$.
• If $$A \subseteq B$$, then $$A \cap C \subseteq B \cap C$$ and $$A \cup C \subseteq B \cup C$$.
Proof
The first part of this theorem was included in Exercise (7) from Section 5.2. The second part of the theorem was Exercise (12) from Section 5.2.
The next theorem provides many of the properties of set operations dealing with intersection and union. Many of these results may be intuitively obvious, but to be complete in the development of set theory, we should prove all of them. We choose to prove only some of them and leave some as exercises.
Theorem 5.18: Algebra of Set Operations
Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$. Then all of the following equalities hold.
Proporties of the Empty Set $$A \cap \emptyset = \emptyset$$ $$A \cap U = A$$ and the Universal Set $$A \cup \emptyset = A$$ $$A \cup U = U$$
Idempotent Laws $$A \cap A = A$$ $$A \cup A = A$$
Commutative Laws $$A \cap B = B \cap A$$ $$A \cup B = B \cup A$$
Associative Laws $$(A \cap B) \cap C = A \cap (B \cap C)$$
$$(A \cup B) \cup C = A \cup (B \cup C)$$
Distributive Laws $$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$
$$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$
Before proving some of these properties, we note that in Section 5.2, we learned that we can prove that two sets are equal by proving that each one is a subset of the other one. However, we also know that if $$S$$ and $$T$$ are both subsets of a universal set $$U$$, then
$$S = T$$ if and only if for each $$x \in U$$, $$x \in S$$ if and only if $$x \in T$$.
We can use this to prove that two sets are equal by choosing an element from one set and chasing the element to the other set through a sequence of “if and only if” statements. We now use this idea to prove one of the commutative laws.
Proof of One of the Commutative Laws in Theorem 5.18
We will prove that $$A \cap B = B \cap A$$. Let $$x \in A \cap B$$. Then
$x \in A \cap B \text{ if and only if } x \in A \text{ and } x \in B.$
However, we know that if $$P$$ and $$Q$$ are statements, then $$P wedge Q$$ is logically equivalent to $$Q \wedge P$$. Consequently, we can conclude that
$x \in A \text{ and } x \in B \text{ if and only if } x \in B \text{ and } x \in A.$
Now we know that
$x \in B \text{ and } x \in A \text{ if and only if } x \in B \cap A.$
This means that we can use (5.3.1), (5.3.2) and (5.3.3) to conclude that
$$x \in A \cap B$$ if and only if $$x \in B \cap A$$,
and, hence, we have proved that $$A \cap B = B \cap A$$.
$$\square$$
Progress Check 5.19: Exploring a Distributive Property
We can use Venn diagrams to explore the more complicated properties in Theorem 5.18, such as the associative and distributive laws. To that end, let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$.
1. Draw two general Venn diagrams for the sets $$A$$, $$B$$, and $$C$$. On one, shade the region that represents $$A \cup (B \cap C$$, and on the other, shade the region that represents $$(A \cup B) \cap (A \cup C)$$. Explain carefully how you determined these regions.
2. Based on the Venn diagrams in Part (1), what appears to be the relationship between the sets $$A \cup (B \cap C$$ and $$(A \cup B) \cap (A \cup C)$$?
Add texts here. Do not delete this text first.
Proof of One of the Distributive Laws in Theorem 5.18
We will now prove the distributive law explored in Progress Check 5.19. Notice that we will prove two subset relations, and that for each subset relation, we will begin by choosing an arbitrary element from a set. Also notice how nicely a proof dealing with the union of two sets can be broken into cases.
Proof. Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$. We will prove that $$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$ by proving that each set is a subset of the other set.
We will first prove that $$A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C$$. We let $$x \in A \cup (B \cap C)$$. Then $$x \in A$$ or $$x \in B \cap C$$.
So in one case, if $$x \in A$$, then $$x \in A \cup B$$ and $$x \in A \cup C$$. This means that $$x \in (A \cup B) \cap (A \cup C)$$.
On the other hand, if $$x \in B \cap C$$, then $$x \in B$$ and $$x \in C$$. But $$x \in B$$ implies that $$x \in A \cup B$$, and $$x \in C$$ implies that $$x \in A \cup C$$. Since $$x$$ is in both sets, e conclude that $$x \in (A \cup B) \cap (A \cup C)$$. So in both cases, we see that $$x \in (A \cup B) \cap (A \cup C)$$, and this proves that $$A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C$$.
We next prove that $$(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$$. So let $$y \in (A \cup B) \cap (A \cup C)$$. Then, $$y \in A \cup B$$ and $$y \in A \cup C$$. We must prove that $$y \in A \cup (B \cap C)$$. We will consider the two cases where $$y \in A$$ or $$y \notin A$$. In the case where $$y \in A$$, we see that $$y \in A \cup (B \cap C)$$.
So we consider the case that $$y \notin A$$. It has been established that $$y \in A \cup B$$ and $$y \in A \cup C$$. Since $$y \not in A$$ and $$y \in A \cup B$$, $$y$$ must be an element of $$B$$. Similarly, since $$y \notin A$$ and $$y \in A \cup C$$, $$y$$ must be an element of $$C$$. Thus, $$y \in B \cap C$$ and, hence, $$y \in A \cup (B \cap C)$$.
In both cases, we have proved that $$y \in A \cup (B \cap C)$$. This proves that $$(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$$. The two subset relations establish the equality of the two sets. Thus, $$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$.
$$square$$
## Important Properties of Set Complements
The three main set operations are union, intersection, and complementation. The- orems 5.18 and 5.17 deal with properties of unions and intersections. The next theorem states some basic properties of complements and the important relations dealing with complements of unions and complements of intersections. Two relationships in the next theorem are known as De Morgan’s Laws for sets and are closely related to De Morgan’s Laws for statements.
Theorem 5.20
Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$. Then the following are true:
Basic Properties $$(A^c)^c = A$$
$$A - B = A \cap B^c$$
Empty Set and Universal Set $$A - \emptyset = A$$ and $$A - U = \emptyset$$
$$\emptyset ^c = U$$ and $$U^c = \emptyset$$
De Morgan's Laws $$(A \cap B)^c = A^c \cup B^c$$
$$(A \cup B)^c = A^c \cap B^c$$
Subsets and Complements $$A \subseteq B$$ if and only if $$B^c \subseteq A^c$$
Proof
We will only prove one of De Morgan’s Laws, namely, the one that was explored in Preview Activity $$\PageIndex{1}$$. The proofs of the other parts are left as exercises. Let $$A$$ and $$B$$ be subsets of some universal set $$U$$. We will prove that $$(A \cup B)^c = A^c \cap B^c$$ by proving that an element is in $$(A \cup B)^c$$ if and only if it is in $$A^c \cap B^c$$. So let $$x$$ be in the universal set $$U$$. Then
$x \in (A \cup B)^c \text{ if and only if } x \notin A \cup B.$
and
$x \notin A \cup B \text{ if and only if } x \notin A \text{ and } x \notin B.$
Combining (5.3.4) and (5.3.5), we see that
$x \in (A \cup B)^c \text{ if and only if } x \notin A \text{ and } x \notin B.$
$x \notin A \text{ and } x \notin B \text{ if and only if } x \in A^c \text{ and } x \in B^c.$
and this is true if and only if $$x \in A^c \cap B^c$$. So we can use (5.3.6) and (5.3.7) to conclude that
$$x \in (A \cup B)^c$$ if and only if $$x \in A^c \cap B^c$$.
and, hence, that $$(A \cup B)^c = A^c \cap B^c$$.
$$\square$$
Progress Check 5.21: Using the Algebra of Sets
1. Draw two general Venn diagrams for the sets $$A$$, $$B$$, and $$C$$. On one, shade the region that represents $$(A \cup B) - C$$, and on the other, shade the region that represents $$(A - C) \cup (B - C)$$. Explain carefully how you determined these regions and why they indicate that $$(A \cup B) - C = (A - C) \cup (B - C)$$.
It is possible to prove the relationship suggested in Part (1) by proving that each set is a subset of the other set. However, the results in Theorems 5.18 and 5.20 can be used to prove other results about set operations. When we do this, we say that we are using the algebra of sets to prove the result. For example, we can start by using one of the basic properties in Theorem 5.20 to write
$$A \cup B) - C = (A \cup B) \cap C^c$$.
We can then use one of the commutative properties to write
$\begin{array} {rcl} {(A \cup B) - C} &= & {(A \cup B) \cap C^c} \\ {} &= & {C^c \cap (A \cup B).} \end{array}$
2. Determine which properties from Theorems 5.18 and 5.20 justify each of the last three steps in the following outline of the proof that $$(A \cup B) - C = (A - C) \cup (B - C)$$.
$\begin{array} {rcl} {(A \cup B) - C} &= & {(A \cup B) \cap C^c \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(Theorem 5.20)}} \\ {} &= & {C^c \cap (A \cup B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(Commutative Property)}} \\ {} &= & {(C^c \cap A) \cup (C^c \cap B)} \\ {} &= & {(A \cap C^c) \cup (B \cap C^c)} \\ {} &= & {(A - C) \cup (B - C)} \end{array}$
Note: It is sometimes difficult to use the properties in the theorems when the theorems use the same letters to represent the sets as those being used in the current problem. For example, one of the distributive properties from Theorems 5.18 can be written as follows: For all sets $$X$$, $$Y$$, and $$Z$$ that are subsets of a universal set $$U$$,
$$(X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z).$$
Add texts here. Do not delete this text first.
## Proving that Statements Are Equivalent
When we have a list of three statements P , Q, and R such that each statement in the list is equivalent to the other two statements in the list, we say that the three statements are equivalent. This means that each of the statements in the list implies each of the other statements in the list.
The purpose of Preview Activity $$\PageIndex{2}$$ was to provide one way to prove that three (or more) statements are equivalent. The basic idea is to prove a sequence of conditional statements so that there is an unbroken chain of conditional statements from each statement to every other statement. This method of proof will be used in Theorem 5.22.
Theorem 5.22
Let $$A$$ and $$B$$ be subsets of some universal set $$U$$. The following are equivalent:
1. $$A \subseteq B$$
2. $$A \cap B^c = \emptyset$$
3. $$A^c \cup B = U$$
Proof
To prove that these are equivalent conditions, we will prove that (1) implies (2), that (2) implies (3), and that (3) implies (1).
Let $$A$$ and $$B$$ be subsets of some universal set $$U$$. We have proved that (1) implies (2) in Proposition 5.14.
To prove that (2) implies (3), we will assume that $$A \cap B^c = \emptyset$$ and use the fact that $$\emptyset ^c = U$$. We then see that
$$(A \cap B^c)^c = \emptyset ^c$$.
Then, using one of De Morgan's Laws, we obtain
$begin{array} {rcl} {A^c \cup (B^c)^c} &= & {U} \\ {A^c \cup B} &= & {U.} \end{array}$
This completes the proof that (2) implies (3).
We now need to prove that (3) implies (1). We assume that $$A^c \cup B = U$$ and will prove that $$A \subseteq B$$ by proving that every element of $$A$$ must be in $$B$$.
So let $$x \in A$$. Then we know that $$x \notin A^c$$. However, $$x \in U$$ and since $$A^c \cup B = U$$, we can conclude that $$x \in A^c \cup B$$. Since $$x \notin A^c$$, we conclude that $$x \in B$$. This proves that $$A \subseteq B$$ and hence that (3) implies (1).
Since we have now proved that (1) implies (2), that (2) implies (3), and that (3) implies (1), we have proved that the three conditions are equivalent.
Exercises for Section 5.3
1. Let $$A$$ be a subset of some universal set $$U$$. Prove each of the following (from Theorem 5.20):
(a) $$(A^c)^c = A$$
(b) $$A - \emptyset = A$$
(c) $$\emptyset ^c = U$$
(d) $$U^c = \emptyset$$
2. Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$. As part of Theorem 5.18, we proved one of the distributive laws. Prove the other one. That is, prove that
$A \cap (B \cup C) = (A \cap B) \cup (A \cap C).$
3. Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$. As part of Theorem 5.20, we proved one of De Morgan’s Laws. Prove the other one. That is, prove that
$(A \cap B)^c = A^c \cup B^c.$
4. Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$.
(a) Draw two general Venn diagrams for the sets $$A$$, $$B$$, and $$C$$. On one, shade the region that represents $$A - (B \cup C)$$, and on the other, shade the region that represents $$(A - B) \cap (A - C)$$. Based on the Venn diagrams, make a conjecture about the relationship between the sets $$A - (B \cup C)$$ and $$(A - B) \cap (A - C)$$.
(b) Use the choose-an-element method to prove the conjecture from Exercise (4a).
(c) Use the algebra of sets to prove the conjecture from Exercise (4a).
5. Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$.
(a) Draw two general Venn diagrams for the sets $$A$$, $$B$$, and $$C$$. On one, shade the region that represents $$A - (B \cap C)$$, and on the other, shade the region that represents $$(A - B) \cup (A - C)$$. Based on the Venn diagrams, make a conjecture about the relationship between the sets $$A - (B \cap C)$$ and $$(A - B) \cup (A - C)$$.
(b) Use the choose-an-element method to prove the conjecture from Exercise (5a).
(c) Use the algebra of sets to prove the conjecture from Exercise (5a).
6. Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$. Prove or disprove each of the following:
(a) $$(A \cap B) - C = (A - C) \cap (B - C)$$
(b) $$(A \cup B) - (A \cap B) = (A - B) \cup (B - A)$$
7. Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$.
(a) Draw two general Venn diagrams for the sets $$A$$, $$B$$, and $$C$$. On one, shade the region that represents $$A - (B - C)$$, and on the other, shade the region that represents $$(A - B) - C$$. Based on the Venn diagrams, make a conjecture about the relationship between the sets $$A - (B - C)$$ and $$(A - B) - C$$.
(b) Prove the conjecture from Exercise (7a).
8. Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$.
(a) Draw two general Venn diagrams for the sets $$A$$, $$B$$, and $$C$$. On one, shade the region that represents $$A - (B - C)$$, and on the other, shade the region that represents $$(A - B) \cup (A - C^c)$$. Based on the Venn diagrams, make a conjecture about the relationship between the sets $$A - (B - C)$$ and $$(A - B) \cup (A - C^c)$$.
(b) Prove the conjecture from Exercise (8a).
9. Let $$A$$ and $$B$$ be subsets of some universal set $$U$$.
(a) Prove that $$A$$ and $$B - A$$ are disjoint sets.
(b) Prove that $$A \cup B = A \cup (B - A)$$.
10. Let $$A$$ and $$B$$ be subsets of some universal set $$U$$.
(a) Prove that $$A - B$$ and $$A \cap B$$ are disjoint sets.
(b) Prove that $$A = (A - B) \cup (A \cap B)$$.
11. Let $$A$$ and $$B$$ be subsets of some universal set $$U$$. Prove or disprove each of the following:
(a) $$A - (A \cap B^c) = A \cap B$$
(b) $$(A^c \cup B)^c \cap A = A - B$$
(c) $$(A \cup B) - A = B- A$$
(d) $$(A \cup B) - B = A - (A \cap B)$$
(e) $$(A \cup B) - (A \cap B) = (A - B) \cup (B - A)$$
12. Evaluation of proofs
See the instructions for Exercise (19) on page 100 from Section 3.1.
(a)
Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$, then $$A - (B - C) = A - (B \cup C)$$.
Proof
$\begin{array} {rcl} {A - (B - C)} &= & {(A - B) - (A - C)} \\ {} &= & {(A \cap B^c) \cap (A \cap C^c)}\\ {} &= & {A \cap (B^c \cap C^c)} \\ {} &= & {A \cap (B \cup C)^c} \\ {} &= & {A - (B \cup C)} \end{array}$
Theorem $$\PageIndex{1}$$
Let $$A$$, $$B$$, and $$C$$ be subsets of some universal set $$U$$, then $$A - (B \cup C) = (A - B) \cap (A - C)$$.
Proof
We first write $$A - (B \cup C) = A \cap (B \cup C)^c$$ and then use one of De Morgan's Laws to obtain
$$A - (B \cup C) = A \cap (B^c \cap C^c)$$.
We now use the fact that $$A = A \cap A$$ and obtain
$\begin{array} {rcl} {A - (B \cup C)} &= & {A \cap A \cap B^c \cap C^c} \\ {} &= & {(A \cap B^c) \cap (A \cap C^c)} \\ {} &= & {(A - B) \cap (A - C).} \end{array}$
Explorations and Activities
13. (Comparison to Properties of the Real Numbers). The following are some of the basic properties of addition and multiplication of real numbers
Commutative Laws: $$a + b = b + a$$, for all $$a, b \in \mathbb{R}$$.
$$a \cdot b = b \cdot a$$, for all $$a, b \in \mathbb{R}$$.
Associative Laws: $$(a + b) + c = a + (b + c)$$, for all $$a, b, c \in \mathbb{R}$$.
$$(a \cdot b) \cdot c = a \cdot (b \cdot c)$$, for all $$a, b, c \in \mathbb{R}$$.
Distributive Law: $$a \cdot (b + c) = a \cdot b + a \cdot c$$, for all $$a, b, c \in \mathbb{R}$$.
Additive Identity: For all $$a \in \mathbb{R}$$, $$a + 0 = a = 0 + a$$.
Multiplicative Identity: For all $$a \in \mathbb{R}$$, $$a \cdot 1 = a = 1 \cdot a$$.
Additive Inverses: For all $$a \in \mathbb{R}$$, $$a + (-a) = 0 = (-a) + a$$.
Multiplicative Inverses: For all $$a \in \mathbb{R}$$ with $$a \ne 0$$, $$a \cdot a^{-1} = 1 = a^{-1} \cdot a$$.
Discuss the similarities and differences among the properties of addition and multiplication of real numbers and the properties of union and intersection of sets. | 2020-04-08T05:34:06 | {
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https://brilliant.org/discussions/thread/advanced-factorization/ | ## Definition
Factorization is the decomposition of an expression into a product of its factors.
The following are common factorizations.
1. For any positive integer $$n$$, $a^n-b^n = (a-b)(a^{n-1} + a^{n-2} b + \ldots + ab^{n-2} + b^{n-1} ).$ In particular, for $$n=2$$, we have $$a^2-b^2=(a-b)(a+b)$$.
2. For $n$ an odd positive integer, $a^n+b^n = (a+b)(a^{n-1} - a^{n-2} b + \ldots - ab^{n-2} + b^{n-1} ).$
3. $a^2 \pm 2ab + b^2 = (a\pm b)^2$
4. $x^3 + y^3 + z^3 - 3 xyz = (x+y+z) (x^2+y^2+z^2-xy-yz-zx)$
5. $(ax+by)^2 + (ay-bx)^2 = (a^2+b^2)(x^2+y^2)$. $(ax-by)^2 - (ay-bx)^2 = (a^2-b^2)(x^2-y^2)$.
6. $x^2 y + y^2 z + z^2 x + x^2 z + y^2 x + z^2 y +2xyz= (x+y)(y+z)(z+x)$.
Factorization often transforms an expression into a form that is more easily manipulated algebraically, that has easily recognizable solutions, and that gives rise to clearly defined relationships.
## Worked Examples
### 1. Find all ordered pairs of positive integer solutions $(x,y)$ such that $2^x+ 1 = y^2$.
Solution: We have $2^x = y^2-1 = (y-1)(y+1)$. Since the factors $(y-1)$ and $(y+1)$ on the right hand side are integers whose product is a power of 2, both $(y-1)$ and $(y+1)$ must be powers of 2. Furthermore, their difference is
$(y+1)-(y-1)=2,$
implying the factors must be $y+1 = 4$ and $y-1 = 2$. This gives $y=3$, and thus $x=3$. Therefore, $(3, 3)$ is the only solution.
### 2. Factorize the polynomial
$f(a, b, c) = ab(a^2-b^2) + bc(b^2-c^2) + ca(c^2-a^2).$
Solution: Observe that if $a=b$, then $f(a, a, c) =0$; if $b=c$, then $f(a, b, b)=0$; and if $c=a$, then $f(c,b,c)=0$. By the Remainder-Factor Theorem, $(a-b), (b-c),$ and $(c-a)$ are factors of $f(a,b,c)$. This allows us to factorize
$f(a,b,c) = -(a-b)(b-c)(c-a)(a+b+c).$
Note by Calvin Lin
7 years, 5 months ago
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For the worked example - 1 : there are 2 solutions : (3,3) and (3,-3)
(y+1)-(y-1) = 2
implies y + 1 = -2 or y + 1 = 4 and y -1 = -4 or y-1 = 2
Thus, the two solutions you have are (3,3) and (3,-3)
- 7 years, 3 months ago
Thanks. I added in "positive integers".
Staff - 6 years, 3 months ago
Thanks for this.
- 7 years, 3 months ago
I did not understand why should (y+1) - (y-1) = 2 ?
Can anyone explain ?
- 6 years, 3 months ago
What do you not understand?
What do you think $(y+1) - (y-1)$ is equal to ?
Staff - 6 years, 3 months ago
I did not see it correctly . My fault ! Sir, I have a problem. I want to learn Number theory as is organized here on Brilliant but I don't follow anything beginning from modular inverses.I have tried the wikis but I still don't follow .Please suggest something.
- 6 years, 3 months ago
How is their (y+1)-(y-1) difference 2?
- 7 years, 3 months ago
$(y+1) - (y-1)$ $= y +1 - y + 1$ $= 1+1= \boxed{2}$
- 7 years, 3 months ago
(Y+1)-(y-1) = y+1-y+1= 2
- 7 years, 3 months ago
2
- 7 years, 3 months ago
How to think that both difference is 2. What is the main moto behind thinking such that????
- 7 years, 3 months ago | 2021-09-19T13:23:18 | {
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http://math.stackexchange.com/questions/141236/markov-chain-transition-matrix | # Markov chain transition matrix
Consider the Markov chain with the following transition matrix:
$$P = \pmatrix{0& 0.5 &0 &0 &0 &0.5\\ 0.25 &0 &0.25 &0.25 &0 &0.25\\ 0 &0.5 &0 &0.5 &0 &0\\ 0 &0.25 &0.25 &0 &0.25 &0.25\\ 0 &0 &0 &0.5 &0 &0.5\\ 0.25 &0.25 &0 &0.25 &0.25 &0}$$
I'm trying to show that this chain is irreducible and aperiodic, and finding the stationary probability distribution of the chain by showing that the chain is reversible.
My attempt is to draw the state space of this, and aperiodic if gcd = 1 and irreducible if there is a path from state 1 to 2 but not vice verca.
The problem is from here, problem #2.
-
Irreducibility means that you can get from any state to any state. Here it's easy to check that we can get from state 1 to 2 to 3 to 4 to 5 to 6 to 1: the transition probabilities, in that order, are $0.5,0.25,0.5,0.25$, and $0.25$, appearing on the first superdiagonal and in the lower lefthand corner. Here's a graph of the transitions; each goes both ways, so I didn't have to worry about arrows.
1 3
|\ /|
| \ / |
| \ / |
| 2 |
| / \ |
| / \ |
|/ \|
6-------4
\ /
\ /
\ /
5
From it you can easily check that no matter where you start, you can return to that state in either $2$ or $3$ steps; since $\gcd(2,3)=1$, the chain is aperiodic.
To show that the chain is reversible, you must find a probability distribution $$\pi=\langle\pi_1,\pi_2,\pi_3,\pi_4,\pi_5,\pi_6\rangle$$ such that $\pi_ip_{ij}=\pi_jp_{ji}$ for $1\le i,j\le 6$. Clearly the equations with $i=j$ are satisfied no matter what $\pi_i$ is, so we can ignore them. We can also ignore any pair for which $p_{ij}=p_{ji}=0$, since $0=0$ gives no information about $\pi$. That leaves the following system of equations:
$$\begin{array}{} 0.5\pi_1=0.25\pi_2&&0.5\pi_1=0.25\pi_6\\ 0.25\pi_2=0.5\pi_3&&0.25\pi_2=0.25\pi_4&&0.25\pi_2=0.25\pi_6\\ 0.5\pi_3=0.25\pi_4\\ 0.25\pi_4=0.5\pi_5&&0.25\pi_4=0.25\pi_6\\ 0.5\pi_5=0.25\pi_6 \end{array}$$
After the fractions are cleared, we have this system:
$$\begin{array}{} 2\pi_1=\pi_2&&2\pi_1=\pi_6\\ \pi_2=2\pi_3&&\pi_2=\pi_4&&\pi_2=\pi_6\\ 2\pi_3=\pi_4\\ \pi_4=2\pi_5&&\pi_4=\pi_6\\ 2\pi_5=\pi_6 \end{array}$$
Clearly $\pi_2=\pi_4=\pi_6=2\pi_1=2\pi_3=2\pi_5$. Finally, we know that $\sum_{i=1}^6\pi_i=1$, so $$1=\sum_{i=1}^6\pi_i=3\pi_1+3\pi_2=9\pi_2\;,$$ and therefore $$\pi_1=\pi_3=\pi_5=\frac19\text{ and }\pi_2=\pi_4=\pi_6=\frac29\;.$$ In other words, the distribution $$\pi=\frac19\langle 1,2,1,2,1,2\rangle$$ satisfies the detailed balance condition and is therefore reversible. Of course this distribution is stationary.
-
To prove the reversibility, one can also note that this is the random walk on the electric network where every existing nonoriented edge has conductance 1. A bonus of this approach is then that every measure which assigns to each vertex a weight proportional to the sum of the conductances of the edges adjacent to it, is stationary. Since there are 2 edges adjacent to vertices 1, 3, 5, and 4 edges adjacent to vertices 2, 4, 6, this confirms your formula, with no computation. – Did May 5 '12 at 5:57 | 2015-09-04T12:31:40 | {
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