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https://www.physicsforums.com/threads/find-delta-v-hohmann-transfer-orbit.723763/ | # Find delta-v; Hohmann transfer orbit
1. Nov 19, 2013
### oddjobmj
1. The problem statement, all variables and given/known data
A space vehicle is in circular orbit about the earth. The mass of the vehicle is 3300 kg. The radius of the orbit is 2RE. It is desired to transfer the vehicle to a circular orbit of radius 4RE.
An efficient way to accomplish the transfer is to use a semielliptical orbit (known as a Hohmann transfer orbit), as shown. What velocity changes are required at the points of intersection, A and B? Express the change in speed as the percentage of the speed just before the change.
Unfortunately I am not given the figure that this problem refers to.
2. Relevant equations
Δv1=$\sqrt{GM/r_1}$($\sqrt{2r_2/(r_1+r_2)}$-1)
Δv2=$\sqrt{GM/r_2}$(1-$\sqrt{2r_1/(r_1+r_2)}$)
v=$\sqrt{GM(2/r-1/a)}$, where a is the semi-major axis.
3. The attempt at a solution
I thought this was pretty straightforward but apparently my answers are incorrect. I'm not sure what I'm doing wrong. My interpretation is that the satellite is in orbit at 2RE=r1, we need to get it into a transition orbit and to do so we increase the velocity by amount Δv1, we then stabilize this orbit at 4RE=r2 by increasing velocity once more.
Plugging in the knowns to the above equation I get:
Δv1=865.2 m/s
Δv2=725.7 m/s
Now I need to represent these changes in velocity as percentages of the velocity just before the given change. The initial velocity can be found using this equation:
v0=$\sqrt{GM(2/r-1/a)}$
In this case, being a circular orbit, r=a=2RE
v0= 5592.8 m/s
Also, the velocity after the first velocity change is v1=v0+Δv1=6458 m/s
So, the first answer should be:
$\frac{Δv_1}{v_0}$=15.47%
$\frac{Δv_2}{v_1}$=11.24%
I also tried entering these as the decimal equivalents and tried simply using the initial velocity as the divisor when calculating my percents. None of this worked. Perhaps I have A and B mixed around because I don't have the image but I believe it is more likely that I made a mistake.
These answers are wrong. Any suggestions? Thank you!
EDIT: The reason I am given the extra variables in the initial problem is that there was a part before this that asked me to calculate the required energy to make the transition from 2R to 4R which I solved correctly.
Last edited: Nov 19, 2013
2. Nov 19, 2013
### Staff: Mentor
The velocity just before the second change will not be the same as the velocity immediately after the first change; the craft is at perigee immediately after the first change, and heads out to apogee where the second change occurs. Otherwise your calculations appear to be fine.
3. Nov 19, 2013
### oddjobmj
Ahh, of course! Thank you, gneill!
Would I be able to plug in 4RE to the velocity equation? I ask because the satellite's trajectory is skewed in the sense that when it reaches 4RE it would not be stable if not for Δv2. I'm not sure how to figure the velocity at that point otherwise.
4. Nov 19, 2013
### Staff: Mentor
Your velocity equation holds for all points on the transfer trajectory, which would just be an ellipse if not for the second correction. So sure, plug in 4RE and obtain the velocity.
5. Nov 19, 2013
### oddjobmj
Ah, I see what you mean, thank you. My thought initially was to use 4RE assuming the orbit was circular. Of course, the transfer orbit is, as you mentioned, an ellipse. For anyone who has a similar problem in the future you can figure the semi-major axis of this ellipse to use in the velocity equation from the transfer orbit's perigee and apogee points which happen to fall on the two known circular orbits.
Using your suggestion(s) and that noted above the correct answer was found. | 2017-08-21T17:20:27 | {
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http://mathhelpforum.com/trigonometry/176983-double-half-area-trig-help.html | # Thread: Double / Half & Area Trig Help
1. ## Double / Half & Area Trig Help
I have two questions:
1. Given that sin 3pi/10 = sqrt(5)+1/4 find an exact expression for cos (3pi/5)
I used cos(2theta) = 1 - 2sin^2(theta)
cos(3pi/5) = 1- 2sin^2(3pi/10)
= 1 - 2(sqrt(5)+1/4)^2
= 1- 2(3+sqrt(5)/8) = 4-3+sqrt(5)/4 = 1+sqrt(5)/4 <- i got that as the final answer but the answer key says that it's 1-sqrt(5)/4
2. What is the area of the triangle whose sides all have length r
i used A = 1/2absin(theta)
= 1/2 r^2sin(theta)
i got to that part and now i'm confused as to what to do next. the answer is sqrt(3)r^2/4
would it be right if i said that sin(theta) = 1/2 = pi/6 and then cos(pi/6) = sqrt(3)/2 and then use that to get sqrt(3)r^2/4?
2. Do you mean $\displaystyle \sin{\frac{3\pi}{10}} = \frac{\sqrt{5} + 1}{4}$?
You should know that $\displaystyle \cos{2\theta}= 1 - 2\sin^2{\theta}$. Here $\displaystyle \theta = \frac{3\pi}{10}$.
3. Originally Posted by Prove It
Do you mean $\displaystyle \sin{\frac{3\pi}{10}} = \frac{\sqrt{5} + 1}{4}$?
You should know that $\displaystyle \cos{2\theta}= 1 - 2\sin^2{\theta}$. Here $\displaystyle \theta = \frac{3\pi}{10}$.
yes i know but i still can't figure out what i did wrong to get 1+sqrt(5) instead of 1 - sqrt(5)
4. Let's see...
$\displaystyle 1 - 2\sin^2{\frac{3\pi}{10}} = 1 - 2\left(\frac{\sqrt{5} + 1}{4}\right)^2$
$\displaystyle = 1 - 2\left(\frac{5 + 2\sqrt{5} + 1}{16}\right)$
$\displaystyle = 1 - \frac{6 + 2\sqrt{5}}{8}$
$\displaystyle = \frac{8}{8} - \frac{6 + 2\sqrt{5}}{8}$
$\displaystyle = \frac{8 - (6 + 2\sqrt{5})}{8}$
$\displaystyle = \frac{8 - 6 - 2\sqrt{5}}{8}$. Go from here.
It appears you did not subtract ALL of the second term...
5. Originally Posted by Prove It
Let's see...
$\displaystyle 1 - 2\sin^2{\frac{3\pi}{10}} = 1 - 2\left(\frac{\sqrt{5} + 1}{4}\right)^2$
$\displaystyle = 1 - 2\left(\frac{5 + 2\sqrt{5} + 1}{16}\right)$
$\displaystyle = 1 - \frac{6 + 2\sqrt{5}}{8}$
$\displaystyle = \frac{8}{8} - \frac{6 + 2\sqrt{5}}{8}$
$\displaystyle = \frac{8 - (6 + 2\sqrt{5})}{8}$
$\displaystyle = \frac{8 - 6 - 2\sqrt{5}}{8}$. Go from here.
It appears you did not subtract ALL of the second term...
yay i got it~ wow, i didn't multiply the 8 to get 8-(6+2sqrt(5))
6. Hello, dondonlouie!
$\text{2. What is the area of the triangle whose sides all have length } r\,\text{?}$
$\text{i used: }\:A \:=\: \frac{1}{2}ab\sin\theta \:=\:\frac{1}{2}r^2\sin\theta$
$\text{i got to that part and now i'm confused as to what to do next.}$
Since the triangle is equilateral, . $\theta \,=\,\frac{\pi}{3}$ | 2016-12-10T12:02:07 | {
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https://math.stackexchange.com/questions/1675085/can-every-integer-greater-than-5-be-written-as-the-sum-of-exactly-one-prime-and/1675585 | # Can every integer greater than 5 be written as the sum of exactly one prime and one composite?
I worked it out up to 15.
6 = 4 + 2
7 = 4 + 3
8 = 6 + 2
9 = 4 + 5
10 = 8 + 2
11 = 9 + 2
12 = 10 + 2
13 = 10 + 3
14 = 9 + 5
15 = 12 + 3
Does this trend continue forever? I feel like the answer is obvious but I'm just not seeing it.
• I saw this question and started thinking it's like Goldbach's conjecture and any kind of proof would be very difficult. But such a simple explanation exists! – user230452 Feb 28 '16 at 5:39
• Alternatively you can also think of it this way: there's clearly always a prime $p$ between 1 and your integer $n$, and there's a good chance that $n - p$ is not prime; so clearly the problem isn't hard. it shouldn't be hard to work out possible values of $p$ that will systematically work for all $n$. – Thomas Feb 29 '16 at 6:24
If $n$ is even, then $n=2+(n-2)$, and $n-2$ is even, so it is a composite number. If, now, $n$ is odd, $n=3+(n-3)$, and $n-3$ is even.
• Thanks. I feel like this was obvious and you really make it look easy. – Reggie Simmons Feb 28 '16 at 1:03
• This was brilliant. You did it so easily. – user230452 Feb 28 '16 at 1:04
• @ReggieSimmons You were a couple steps away from obvious yourself ;-) if only you had chosen different sums for 9 = 6 + 3 and 11 = 8 + 3. – dxiv Feb 28 '16 at 4:29
• The biggest take-home message from this sort of question is learning "how to think about math", not getting the answer. You probably don't know much number theory yet, so the answer is unlikely to be complicated proof that there is a counterexample greater than $10^{439}$. So, try thinking about the smallest primes, and the significance of the number $5$ in the question! (And considering odd and even numbers separately - or set of numbers like $3k$, $3k+1$, $3k+2$ - is often a good idea to try.) – alephzero Feb 28 '16 at 4:57
• @alephzero I saw this question and started thinking it's like Goldbach's conjecture and any kind of proof would be very difficult. But such a simple explanation exists! Of course, since I didn't encounter this question in a text book, I didn't think it had an easy (or any) solution at all. – user230452 Feb 28 '16 at 5:40
After seeing @detnvvp's brilliant answer to this question, one is left to wonder if it can be generalized.
Theorem: For natural number $k$, every sufficiently large number $n$ is a sum of a prime number and a multiple of $k$ iff $k=1$ or $k$ is prime.
Proof: The problem reduces to finding a prime $p_i=i\pmod k$ for each $0\le i<k$, for then we can take $n=p_i+(n-p_i)$ whenever $n\equiv i\pmod k$, so that $k\mid n-p_i$. For $k=1$, $p_0=2$ works, and detnvvp's answer covers the case $k=2$ with $p_0=2$ and $p_1=3$. Continuing, we have, for $k=3$: $p_0=3,p_1=7,p_2=2$.
More generally, if $k$ is prime, then we can take $p_0=k$, and for each $0<i<k$ we have $\gcd(i,k)=1$, so by a theorem of Dirichlet, there are infinitely many primes in the arithmetic progression $i,i+k,i+2k,\dots$, and we can pick one of them to be $p_i$.
If $k$ is composite, then there is no possible choice for $p_0$. For each $n$ that is a multiple of $k$, we must sum a prime and a multiple of $k$ to get a multiple of $k$, so the prime must also be a multiple of the composite number $k$, a contradiction. $\tag*{$\square$}$
For $k$ composite, the same method as in the prime case allows us to find $p_i$ for $\gcd(i,k)=1$, and we can also take $p_q=q$ for each prime divisor of $k$, but for all other $i$ there is no solution. By combining this for different choices of $k$, we get stronger results:
Theorem: There is no finite set of composite numbers $K$ such that every sufficiently large number is the sum of a prime number and a multiple of some $k\in K$. (Hint: Consider the multiples of $\prod_{k\in K} k$.)
What if we don't fix the factors, but just demand that the composite number have at least $m$ factors? Here the method of congruences does not seem to be as effective, and I don't know the answer, although I would conjecture that it is true:
Conjecture: For each $m$, every sufficiently large number $n$ is the sum of a prime number and a number with at least $m$ factors. | 2019-05-25T07:12:19 | {
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http://math.stackexchange.com/questions/604471/decomposition-of-vector-space-two-linear-mappings | # Decomposition of vector space - two linear mappings
Theorem: Given two linear mappings $f,g: V \rightarrow V$ with
• $f\circ g = g\circ f = 0$
• $f+g=\operatorname{id}_V$
• $f\circ f = f$
• $g\circ g=g$
Then we have $$V=\operatorname{im}(\,f)\oplus \operatorname{im}(\,g)$$
Question:
I think $f$ and $g$ then are some kind of projections, where $f$ sets some coordinates to zero and $g$ sets exactly the remaining components two zero. It is clear to me, that the statement then is correct but how to show it formally?
-
Hint: In addition to $im(f), im(g)$, also think about $ker(f), ker(g)$. – vadim123 Dec 12 '13 at 18:14
1. Since $\operatorname{id}_V = f+g$, for any $v \in V$, $v = \operatorname{id}_V(v) = (f+g)(v) = f(v) + g(v)$. What does this imply about $\operatorname{im} f + \operatorname{im} g \subseteq V$?
2. Let $v \in \operatorname{im} f \cap \operatorname{im} g$, so that $f(x) = v = g(y)$ for some $x$, $y \in V$. In light of your conditions on $f$ and $g$, what do you learn if you apply $f$ to both sides of the equation $f(x) = g(y)$? What do you learn if you apply $g$ instead?
-
Then we have $im(g)+im(f)=V$, because of (2) we have $im(f)\cap im(g)= \lbrace 0_V \rbrace$, thus $im(g)\oplus im(f)=V$ ? – user127.0.0.1 Dec 12 '13 at 18:33
Oh, i see the exactly this in the other answer.. tyvm :) – user127.0.0.1 Dec 12 '13 at 18:35
Choose $v\in V$. Since $f+g=id_V$, we have $f (v)+g (v)=v$. So $V=im (f )+im (g)$. In order to prove that this is a direct sum, we need to prove that $im (f)\cap im(g)=(0)$. Choose $y$ in the intersection. Then $y=f (a)=g (b)$ for some $a, b\in V$. Then $f\circ f (a)= f\circ g (b)$. But $f\circ g=0$ and $f\circ f=f$. So $f (a) =0$. Hence, $y= 0$. Therefore as required the intersection is (0) and $V=im (f)\oplus im (g)$.
-
Hint:$V=kerf\oplus imf$ because $f^2=f$. What is the relation of $kerf$ and $img$?
- | 2015-07-04T19:46:43 | {
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https://mathematica.stackexchange.com/questions/105042/finding-euler-spiral-convergence-point-for-powers-other-than-2/105055 | # Finding Euler spiral convergence point for powers other than 2
I'm playing around with Euler spirals, using a variable power to see how the plot changes,
$\left\{\begin{matrix} x = \int_0^s \cos \left ( s^n \right ) ds \\ y = \int_0^s \sin \left ( s^n \right ) ds \end{matrix}\right.$
and finding the values for powers 2 or 3 work as expected in Mathematica:
Limit[ Integrate[Cos[s^2], {s, 0, x}], x -> ∞]
Limit[ Integrate[Sin[s^2], {s, 0, x}], x -> ∞]
$\frac{\sqrt{\frac{\pi}{2}}}{2} \\ \frac{\sqrt{\frac{\pi}{2}}}{2}$
and similarly,
Limit[ Integrate[ Cos[s^3], {s, 0, x}], x -> ∞]
Limit[ Integrate[ Sin[s^3], {s, 0, x}], x -> ∞]
$\frac{Gamma\left[\frac{1}{3} \right]}{2\sqrt{3}} \\ \frac{1}{6} Gamma \left [ \frac{1}{3} \right ]$
Plotting the spiral at different values (using a different general purpose graphics language) for $n$ suggests that there there should certainly be convergence values for both $x$ and $y$ for any value $1 \lt n \leq 3$ (and presumably above), but trying some of these leads to Mathematica giving most unhelpful answers:
$Limit \left [ \int_0^s Cos \left [ s^{1.5} \right ]ds,s\rightarrow\infty \right ] \Rightarrow 0.451373 \\ Limit \left [ \int_0^s Sin \left [ s^{1.5} \right ]ds,s\rightarrow\infty \right ] \Rightarrow ComplexInfinity$
Which looks like the right $x$ coordinate, but a plain wrong $y$ coordinate, which quite clearly exists somewhere around 0.76ish:
Trying $n=2.5$ fails even harder, yielding "ComplexInfinity" for both coordinates, which is clearly incorrect (being somewhere near $(0.72,0.533)$):
Is there a way to make Mathematica generate these values anyway?
Or, is there a way to compute the general solution for symbolic power $n$ and getting a general formula for when $n\gt1$, using something like this (but then actually yielding a result):
$Limit \left [ \int_0^s Cos \left [ s^n \right ]ds,s\rightarrow\infty \right ] \\ Limit \left [ \int_0^s Sin \left [ s^n \right ]ds,s\rightarrow\infty \right ]$
• Limit doesn't like functions with inexact numbers. A very recent question was asked about this. Why don't you just integrate to infinity? Do this: Integrate[Cos[s^(3/2)], {s, 0, \[Infinity]}]. Jan 28 '16 at 4:56
• Because for whatever reason, that did not occur to me... Works a treat, thanks! A short answer, but if you want to turn that into a real answer I'll be happy to accept it. Jan 28 '16 at 4:59
• @Artes those edits actually made the question reflect what I have in my notebook less, and makes it harder to read the output associated with the input. I write maths as much as possible, with nice looking functions. I don't use Integral[] or the like when I can avoid it. Jan 28 '16 at 18:45
• @Mike'Pomax'Kamermans There is a general consensus on mathematica.stackexchange that whathever code one writes, it should be in a copyable form. As you may have seen the code written in latex cannot be easily copied. On the other hand in the Front End traditional notation is equivalent to the input form (with many restrictions though). So my intervention should be welcome, nevertheless you can edit your question to its initial form. However, I recommend to edit your question adding the code you have used to plot graphs. Jan 29 '16 at 8:32
• @Mike'Pomax'Kamermans Your graphs are slightly puzzling. I've made quite nice plots but I will be able to add them in a few days, because my version of Mathematica (10.1) involves some internal problems when exporting graphics. Jan 29 '16 at 8:33
The problem occurs because of apparently complex form of the expression beyond the integral of Sin[s^k]. Nevertheless it is not too harmful to proceed. Adequate limits are real and we don't need playing to simplify complex expressions to explicitly real forms.
We can calculate the both integrals with appropriate assumptions:
f[x_, k_] = Integrate[{ Cos[s^k], Sin[s^k]}, {s, 0, x},
Assumptions -> {k > 1, x > 0}];
see it in TraditionalForm:
f[x, k] // TraditionalForm
and find appropriate limits with a needed assumption:
lim[k_] = Limit[ f[x, k], x -> Infinity, Assumptions -> k > 1]
{Cos[Pi/(2 k)] Gamma[1 + 1/k], Gamma[1 + 1/k] Sin[Pi/(2 k)]}
Now we can see that indeed limits are finite for k > 1. Here we write down a few exact values as expected:
Table[{k, lim[k]}, {k, 2, 5}] // Column
{{2, {Sqrt[Pi/2]/2, Sqrt[Pi/2]/2}},
{3, {1/2 Sqrt[3] Gamma[4/3], 1/2 Gamma[4/3]}},
{4, {Cos[Pi/8] Gamma[5/4], Gamma[5/4] Sin[Pi/8]}},
{5, {Sqrt[5/8 + Sqrt[5]/8] Gamma[6/5], 1/4 (-1 + Sqrt[5]) Gamma[6/5]}}}
We can plot adequate curves, e.g. using Quiet to avoid unneeded warnning related to apparently complex functions (involving I), nevertheless if one uses f[x,k] it is not necessary.
With[{k = 5/2},
Quiet @ ParametricPlot[ f[x, k], {x, 0, 5}, PlotRange -> All]]
and find slightly improved numerical limit:
lim[2.5]
{0.717812, 0.521521} | 2021-09-17T20:19:51 | {
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http://mathhelpforum.com/pre-calculus/189178-determine-roots-complex-polynomial-print.html | Determine Roots of Complex Polynomial
Printable View
• September 30th 2011, 03:09 AM
TaylorM0192
Determine Roots of Complex Polynomial
Hello,
I encountered this problem in my exercise set, and was stumped.
Solve (z + 1)^4 = 1 - i
My approach to the problem was to recognize that (1 - i) could be rewritten as [ (1 - i)^4 ] / 8.
Which then gives..
(z + 1)^4 = (1 - i)^4 / 8
(z + 1) = +/- (1 - i) / 4throot(8)
But I'm not sure how to proceed from here, since it would appear that in solving each of the resulting two equations, I would only obtain two roots, rather than the expected four.
Edit: Sorry, but LaTeX kept giving me an error when I tried using it.
• September 30th 2011, 03:34 AM
chisigma
Re: Determine Roots of Complex Polynomial
Quote:
Originally Posted by TaylorM0192
Hello,
I encountered this problem in my exercise set, and was stumped.
Solve (z + 1)^4 = 1 - i
My approach to the problem was to recognize that (1 - i) could be rewritten as [ (1 - i)^4 ] / 8.
Which then gives..
(z + 1)^4 = (1 - i)^4 / 8
(z + 1) = +/- (1 - i) / 4throot(8)
But I'm not sure how to proceed from here, since it would appear that in solving each of the resulting two equations, I would only obtain two roots, rather than the expected four.
Edit: Sorry, but LaTeX kept giving me an error when I tried using it.
Setting $s=z+1$ Your equation become...
$s^{4}= 2^{\frac{1}{2}}\ e^{i(2k-\frac{1}{4})\pi}$ (1)
... and its solution are...
$s= 2^{\frac{1}{8}}\ e^{i(k-\frac{1}{8}) \frac{\pi}{2}}$
... so that the solutions of the equation in z are...
$z= 2^{\frac{1}{8}}\ e^{i(k-\frac{1}{8}) \frac{\pi}{2}}-1$ (3)
Kind regards
$\chi$ $\sigma$
• September 30th 2011, 09:03 AM
TaylorM0192
Re: Determine Roots of Complex Polynomial
Hmm...I like this solution! Thank you!
Our text stipulates, however, that we should follow the "outline" used in one of the examples; i.e. with some clever algebraic manipulation put the complex polynomial equation in a solvable Cartesian form (i.e. do not use the polar form of the complex numbers). When I attempted it this way, I got stuck as I mentioned.
In any case, I think it's pretty dumb when in a few steps you can see (much more clearly at that) what the roots are.
I do have one question, though...is it acceptable to leave the solution in a non-standard form? i.e. the polar form with a "-1" tacked at the end? Or should you expand the polar form into its trigonometric representation and go through the grueling task of algebraically rewriting the expression in a standard form? If it is acceptable...then is it possible in this non-standard form to plot the roots (by inspection) on the complex plane, or is this pretty much an "analytic only" exact answer?
Thanks ~ | 2013-12-11T08:53:41 | {
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http://mathhelpforum.com/pre-calculus/187045-roots-unity-represented-regular-polygon-lenght-side.html | # Math Help - Roots of Unity represented on a regular polygon - lenght of side.
1. ## Roots of Unity represented on a regular polygon - lenght of side.
Hello,
I had an assignment that required me to solve for the roots of unity for various equations of the form z^n-1=0. Then , I was asked to represent the roots of unity for each equation on an argand diagram in the form of a regular polygon.
I did all of that , but I have a critical question ,
is there a relation ship between the power n and the length of the polygon side?
I mean , for the euqation Z^3 - 1 = 0 for example, is there a formula relating the power to the lenght of one side of the polygon ?
Thanks.
2. ## Re: Roots of Unity represented on a regular polygon - lenght of side.
For each value of n, how is the length of the side of a polygon related to the central angle subtended by the side? (The side is a chord of the unit circle.)
3. ## Re: Roots of Unity represented on a regular polygon - lenght of side.
Originally Posted by Hyunqul
I had an assignment that required me to solve for the roots of unity for various equations of the form z^n-1=0. Then , I was asked to represent the roots of unity for each equation on an argand diagram in the form of a regular polygon.
There are n nth roots of unity.
$\rho_k = \exp \left( {\frac{{2\pi ki}}{n}} \right),~k=0,1,\cdots,n-1$.
The length of the side of the polygon is $|\rho_{k+1}-\rho_{k}|$.
4. ## Re: Roots of Unity represented on a regular polygon - lenght of side.
Thank you very much for your response,
But , as I used De moiver's theorem to obtain the solutions , I wonder what does (pk) stand for? is it Z^n ?
5. ## Re: Roots of Unity represented on a regular polygon - lenght of side.
Originally Posted by Hyunqul
Thank you very much for your response,
But , as I used De moiver's theorem to obtain the solutions , I wonder what does (pk) stand for? is it Z^n ?
O.K. Let $\theta_n=\frac{2\pi}{n}$ then
$\exp(\theta_nk\mathbf{i})=\cos(\theta_nk)+\mathbf{ i}\sin(\theta_nk)$
6. ## Re: Roots of Unity represented on a regular polygon - lenght of side.
Thanks alot sir , I highly appreciate you help.
If I could ask one more question ... is there a proof for this rule ? The one relating roots of unity to the length of one side.
7. ## Re: Roots of Unity represented on a regular polygon - lenght of side.
1 can be represented in polar form in the obvious way: $1e^{i(0)}$. The nth roots are given by $1^{1/n}e^{\frac{0+ 2k\pi}{n}}= e^{\frac{2k\pi}{n}}$ for k= 0 to n- 1. That is, 1 has n nth roots, equally spaced around the unit circle. If you draw lines from each root to the origin, you have n isosceles triangles and the angle at the vertex (the origin) is $\frac{2\pi}{n}$. If you draw a perpendicular from the origin to one side, it divides that isosceles triangle into two right triangles having hypotenuse 1 and angle $\frac{\pi}{n}$. The "opposite side" to that angle is given by [itex]sin(\frac{\pi}{n})[/tex] and is 1/2 the length of one side of the polygon. The length of one side, then, is $2sin\left(\frac{\pi}{n}\right)$.
As a check note that if n= 4, the polygon is a square, having diagonals of length 2. $s= 2sin(\frac{\pi}{4})= 2\left(\frac{\sqrt{2}}{2}\right)= \sqrt{2}$ while, by the Pythagorean theorem $a^2+ a^2= 2^2$ so $a^2= 2$, $a= \sqrt{2}$.
Similarly, if n= 6, the polygon is a hexagon having diagonals of length 2. $s= 2 sin(\frac{\pi}{6})= 2\left(\frac{1}{2}\right)= 1$.
8. ## Re: Roots of Unity represented on a regular polygon - lenght of side.
Thanks for your generous contribution sir ,
In the same context , I was asked to obtain solutions for the equation Z^n-i = 0 for three cases where n = 1,2,3
I did that for all of them , for example when n = 3 the solutions set was :
i^(1/3) = 1^(1/3) e^[i(π/2 + 2kπ)]/3 = 1 e^[i(π/6 + 2kπ/3)] , where k = 0, 1, and 2 ... and I followed the same approach for n = 4 and 5.
Then I had to plot each of those roots on an argand diagram , and I did that as well.
After that , I was asked to generalize my and prove my results for Z^n= a+bi , where | a+bi | =1 . how can I do that , shall I follow the above method as well or there is a different generalization? Thanks
9. ## Re: Roots of Unity represented on a regular polygon - lenght of side.
Originally Posted by Hyunqul
After that , I was asked to generalize my and prove my results for Z^n= a+bi , where | a+bi | =1 . how can I do that , shall I follow the above method as well or there is a different generalization? Thanks
First let $\theta=\text{Arg}(a+b\mathbf{i})$.
Then $z=\exp\left(\frac{\theta +2\pi k}{n}\mathbf{i}\right)$ where $k=0,1,\cdots,n-1$
Recall that $\exp(\phi \mathbf{i})=\cos(\phi)+\mathbf{i}\sin(\phi)$
10. ## Re: Roots of Unity represented on a regular polygon - lenght of side.
Yes , sir..thanks but I would like to calrify the question more :
First , it is required to generalize and prove the resuls for z when z^n = cos (x) + isin (X)
Secondly . it is required to generalize for Z when z^n = cos (Kx) + isin (Kx)
Thanks and sorry for disturbance.
11. ## Re: Roots of Unity represented on a regular polygon - lenght of side.
Sorry I made a mistake ...
First thing reqired is to generalize and prove the results for z when z^n = cis(x) (or | a+bi | =1)
Second thing is what hepppens when Z^n = cis(x0 (or | a+bi | does not equal 1 )
Sorry again.
12. ## Re: Roots of Unity represented on a regular polygon - lenght of side.
Originally Posted by Hyunqul
First thing reqired is to generalize and prove the results for z when z^n = cis(x) (or | a+bi | =1)
Second thing is what hepppens when Z^n = cis(x0 (or | a+bi | does not equal 1 ) | 2015-09-05T14:12:26 | {
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https://math.stackexchange.com/questions/479587/can-i-get-a-hint-on-solving-this-recurrence-relation | # Can I get a hint on solving this recurrence relation?
I am having trouble solving for a closed form of the following recurrence relation.
\begin{align*} a_n &= \frac{n}{4} -\frac{1}{2}\sum_{k=1}^{n-1}a_k\\ a_1 &= \frac{1}{4} \end{align*} The first few values are $a_1=\frac{1}{4},a_2=\frac{3}{8},a_3=\frac{7}{16},a_4=\frac{15}{32}...$ So it seems the pattern is $a_n = \frac{2^{n}-1}{2^{n+1}}$, but I have been unable to show this algebraically. Here is what I tried: \begin{align*} 2a_n &= \frac{n}{2} - \sum_{k=1}^{n-1}a_k\\ a_n + \sum_{k=1}^n a_k &= \frac{n}{2}\\ a_{n-1} + \sum_{k=1}^{n-1} a_k &= \frac{n-1}{2} \\ 2a_n - a_{n-1} & = \frac{1}{2} \\ a_n = \dfrac{2a_{n-1} + 1}{4} \end{align*}
I am so close, I can taste the closed form. Can someone nudge me in the right direction without giving too much away?
• Consider the homogeneous case $2a_n-a_{n-1}=0$ and then consider that you can find a trivial constant particular solution to your recurrence namely $a_n=1/2$ – obataku Aug 30 '13 at 2:25
• Are you allowed to prove your guess via induction? – Adriano Aug 30 '13 at 2:37
• You can use this technique. – Mhenni Benghorbal Aug 30 '13 at 4:33
By inspection we determine a particular solution to $2a_n-a_{n-1}=1/2$ is given by $a_n=1/2$ trivially -- try an ansatz of the form $a_n=k$ and thus we get $k=2k-k=1/2$.
Considering the homogeneous case, $2a_n-a_{n-1}=0$, let $a_n=\lambda^n$ hence:$$2\lambda^n-\lambda^{n-1}=0\\2\lambda-1=0\\\lambda=\frac12$$... and so it follows that $a_n=(1/2)^n=1/2^n$ is a solution to our general equation and further so is any scalar multiple (since our equation is linear) i.e. $a_n=C/2^n$. Adding our particular equation to the mix we get a solution of the form $a_n=C/2^n+1/2$. Impose your initial conditions to determine $C$.
• Thanks! I can't believe I have never thought of applying this kind of "differential equations" approach. It seems so obvious now! – A.E Aug 30 '13 at 2:50
• no problem @AEdwards :-p since both solution spaces for differential and difference form vector spaces we find similar approaches work in both cases – obataku Aug 30 '13 at 2:55
Subtract the formula for $a_{n-1}$ from that for $a_n$ $$a_n-a_{n-1}=\frac14-\frac12a_{n-1}$$ Multiply by $2^n$ and bring $a_{n-1}$ from the left to the right $$2^na_n=2^{n-2}+2^{n-1}a_{n-1}$$ Using the formula for the sum of a geometric series, we get $$2^na_n=2^{n-1}+C$$ Plug in $n=1$ to find that $C=-\frac12$. Thus, $$a_n=\frac12-\frac1{2^{n+1}}$$
• It is interesting to note that no matter what $a_1$ is, this sequence will always tend to $\frac12$. – robjohn Nov 29 '15 at 10:44
You could use generating functions. Put $$f(z) = \sum_{n\ge 1} a_n z^n.$$
Summing your recurrence for $n\ge 2$ and multiplying by $z^n,$ we get $$f(z) - \frac{1}{4} z = \frac{1}{4} \sum_{n\ge 2} n z^n - \frac{1}{2} z \sum_{n\ge 2} z^{n-1} \sum_{k=1}^{n-1} a_k.$$
Simplify to obtain $$f(z) - \frac{1}{4} z = \frac{1}{4} \frac{z^2(2-z)}{(1-z)^2} - \frac{1}{2} z \frac{1}{1-z} f(z).$$
Now solve for $f(z).$ This yields $$f(z) = \frac{1}{2} \frac{z}{(1-z)(2-z)} = \frac{1}{2} \frac{1}{1-z} - \frac{1}{2} \frac{1}{1-z/2}.$$
Finally read off the coefficients, which is now easy, to get $$a_n = \frac{1}{2} \left(1 - \frac{1}{2^n}\right).$$
Let $b_{n}:=\sum_{k=1}^{n-1}a_{k}$ for $n=2,3,\cdots$, then $b_{n+1}-b_{n}=a_{n}$ for $n=2,3,\cdots$. Then, the equation reads as $b_{n+1}-b_{n}=-\frac{1}{2}b_{n}+\frac{n}{4}$ for $n=2,3,\cdots$ with $b_{2}=a_{1}=\frac{1}{4}$. Rearraging the terms, we get $$\begin{cases}b_{n+1}-\frac{1}{2}b_{n}=\frac{n}{4},{\quad}n=2,3,\cdots,\\ b_{2}=\frac{1}{4}.\end{cases}$$ Let $\mu_{n}:=2^{n}$ for $n=2,3,\cdots$. Multiplying both sides of the equation by $\mu_{n+1}$, we get $$\mu_{n+1}b_{n+1}-\mu_{n}b_{n}=\mu_{n}\frac{n}{2}.$$ Summing this from $2$ to $(n-1)$ for $n=2,3,\cdots$, we get \begin{aligned}&\underbrace{\sum_{k=2}^{n-1}[\mu_{k+1}b_{k+1}-\mu_{k}b_{k}]}_{\text{telescoping sum}}=\frac{1}{2}\sum_{k=2}^{n-1}k2^{k}\\ &{\implies}\mu_{n}b_{n}-\mu_{2}b_{2}=\frac{1}{2}2^{n}(n-2)\\ &{\implies}b_{n}=\frac{1}{2^{n}}+\frac{1}{2}(n-2),\end{aligned} where we have used the fact that $\mu_{2}b_{2}=1$. Then, the desired solution is $$a_{n}=b_{n+1}-b_{n}=\frac{1}{2}\bigg(1-\frac{1}{2^{n}}\bigg),{\quad}n=2,3,\cdots.\tag*{\blacksquare}$$ | 2021-08-01T18:17:40 | {
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https://math.stackexchange.com/questions/1081075/a-generalization-of-arithmetic-and-geometric-means-using-elementary-symmetric-po | # A generalization of arithmetic and geometric means using elementary symmetric polynomials
Let $a_1, a_2, \ldots, a_n$ be positive real numbers. A while ago I noticed that if you form the polynomial $$P(x) = (x - a_1)(x-a_2) \cdots (x-a_n)$$ then:
• The arithmetic mean of $a_1, \ldots, a_n$ is the positive number $m$ such that $(x - m)^n$ and $P(x)$ have the same coefficient of $x^{n-1}$.
• The geometric mean of $a_1, \ldots, a_n$ is the positive number $m$ such that $(x - m)^n$ and $P(x)$ have the same coefficient of $x^{0}.$
It looks like this can be extended: for any $0 \le i \le n-1$, let the $i$th mean be the number $m_i$ such that $(x - m_i)^n$ and $P(x)$ have the same coefficient of $x^i$. (Alternatively, one can define $m_i$ in terms of elementary symmetric polynomials.)
For instance, with three variables $x, y, z$ we get \begin{align*} m_0 &= \sqrt[3]{xyz} \\ m_1 &= \sqrt{\frac{xy + yz + zx}{3}} \\ m_2 &= \frac{x + y + z}{3} \end{align*}
I am not sure what qualifies something as a "mean", but $m_1$ is symmetric and lies strictly between the min and the max, and it probably has other properties as well. Here's a more concrete question:
Question: Must it be true that $m_0 \le m_1 \le m_2 \le \cdots \le m_{n-1}$?
• It's a known Maclaurin's inequality. See my proof in "КванТ", 1980, 04, M565 – Michael Rozenberg Dec 26 '14 at 8:12
• @MichaelRozenberg Fantastic! Thanks, now I know what this inequality is called, and that will probably be enough for me to find anything I need about it. Unfortunately I do not know how to access the text you referenced. – 6005 Dec 26 '14 at 8:18
• @Goos:not sure if this would be of any help, but the referenced text can be found in the archive of the Kvant magazine 198004.djvu, see pages 35(34)--36(35). Beware that the text is in Russian, but you can follow the equations. – g.kov Dec 27 '14 at 2:40
• en.wikipedia.org/wiki/Maclaurin%27s_inequality – 6005 Feb 1 '15 at 3:13
• See this paper math.washington.edu/~morrow/papers/nate-thesis.pdf – Yuriy S Jan 27 '17 at 21:20
Denote, $(\overline{a}) = (a_1,\cdots,a_n)$ and $\displaystyle P(x) = \prod\limits_{k=1}^{n}(x - a_k) = x^n +\sum\limits_{k=1}^{n} (-1)^k\binom{n}{k}u_k(\overline{a})x^{n-k}$
where, $\displaystyle u_k(\overline{a}) = \dfrac{\sum\limits_{1\le j_1<\cdots< j_k \le n}a_{j_1}\cdots a_{j_k}}{\binom{n}{k}} = m_{n-k}^{k}$ (in your notation).
Note that $P(x)$ has $n$ real roots in the interval $\left[\min\limits_{i=1}^n\{a_i\},\max\limits_{i=1}^n\{a_i\}\right]$,
Thus $P^{(n-2)}(x) = \dfrac{n!}{2}(x^2 - 2u_1x + u_2)$ has two real roots in that interval, i.e., $u_1^2 \ge u_2$.
Similarly apply the same idea for the $(n-2)^{th}$ derivative of the polynomial with roots $\dfrac{1}{a_k}$, ($k = 1(1)n$)
We get, $\displaystyle \frac{1}{\binom{n}{2}}\sum\limits_{i<j}a_ia_j \le \frac{1}{\binom{n}{1}^2}\left(\sum\limits_{i=1}^{n} \frac{1}{a_i}\right)^2 \implies u_{n-1}^2 \ge u_nu_{n-2}$.
We show, $u_{k-1}(\overline{a})u_{k+1}(\overline{a}) \le u_k^2(\overline{a})$ for $k = 2,3,\cdots,n-1$,
We can prove this result by induction on $n$, suppose the inequality holds for any $n-1$ positive real numbers.
We have $\displaystyle P'(x) = n\left(x^{n-1} + \sum\limits_{k=1}^{n-1}(-1)^k\binom{n-1}{k}u_k(\overline{a})x^{n-k-1}\right)$
If the roots of $P'(x)$ are $b_k$, for $k=1,2,\cdots,n-1$, (which, are positive reals by M.V.T.).
and define $v_k = \dfrac{\sum\limits_{1\le j_1<\cdots< j_k \le n-1}b_{j_1}\cdots b_{j_k}}{\binom{n-1}{k}}$
Then $\displaystyle P'(x) = n\prod\limits_{k=1}^{n-1}(x - b_k) = n\left(x^{n-1} + \sum\limits_{k=1}^{n-1}(-1)^k\binom{n-1}{k}v_kx^{n-k-1}\right)$
Thus, $u_k = v_k$ for $k=1,2,\cdots,n-1$ and by the induction hypothesis on the numbers $(b_1,\cdots,b_{n-1})$ we get $u_k^2 \ge u_{k+1}u_{k-1}$ for $k = 2,\cdots,n-2$ and together with $u_{n-1}^2 \ge u_nu_{n-2}$ completes the induction.
If we take the $k$th inequality to the $k$th power and then multiply all these inequalities for $k=1,\cdots,r$, we get,
$u_1^2 u_2^4 u_3^6 \cdots u_{r-1}^{2r-2}u_{r}^{r-1}u_{r+1}^r \le u_1^2u_2^4\cdots u_r^{2r} \implies u_{r+1}^r \le u_{r}^{r+1}$ for $r=1, \cdots,n-1$
Thus, $u_r^{1/r} = m_{n-r}$ forms a non increasing sequence. | 2019-04-25T09:44:22 | {
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http://www.cyrilsancereau.com/dw10h9h5/t9pmf5.php?id=covariance-of-a-line-026b59 | Covariance is a measure of the linear relationship between two variables, but perhaps a more com-mon and more easily interpretable measure is correlation. That does not mean the same thing that is in the context of linear algebra. Or we can say, in other words, it defines the changes between the two variables, such that change in one variable is equal to change in another variable. Formula . A sample … Covariance matrix as a linear transformation. Example $$\PageIndex{3}$$ A pair of simple random variables Well, remember the rule that when taking the Covariance of sums, we draw a line from every element on the left of the comma to every element on the right of the comma and add Covariance of all of these pairs. Data with unit covariance matrix is called white data. Their covariance is the inner product (also called the dot product or scalar product) of two vectors in that space. Now let’s forget about covariance matrices for a moment. An analyst is having five quarterly performance dataset of a company that shows the quarterly gross domestic product(GDP). It is very easy and simple. Again, examination of the figure confirms this. Takeaway: Covariance is said to be a statistical tool that is taken into account to find out the relationship between the … In probability theory and statistics, a covariance matrix (also known as auto-covariance matrix, dispersion matrix, variance matrix, or variance–covariance matrix) is a square matrix giving the covariance between each pair of elements of a given random vector.Any covariance matrix is symmetric and positive semi-definite and its main diagonal contains variances (i.e., the covariance … As these terms suggest, covariance and correlation measure a certain kind of dependence between the variables. While growth is in percentage(A) and a company’s new product line growth in percentage (B). Is covariance linear? The covariance of two related variables each multiplied by a third independent variable Hot Network Questions You are simply seeing light touching your eyes (masturbation addiction) Here we will do another example of the Covariance in Excel. Since $$1 + \rho < 1 - \rho$$, the variance about the $$\rho = -1$$ line is less than that about the $$\rho = 1$$ line. In fact, it is the same thing exactly. Calculate the Covariance. Hands-on Example. The covariance between $X$ and $Y$ is defined as \begin{align}%\label{} \nonumber \textrm{Cov}(X,Y)&=E\big[(X-EX)(Y-EY)\big]=E[XY]-(EX)(EY). XY = Cov(X;Y) Each of the examples in figure 3 can simply be considered to be a linearly transformed instance of figure 6: Figure 6. To understand the concept of covariance, it is important to do some hands-on activity. Their linear combinations form a vector space. kind of thing that goes on in linear algebra. 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Set of real-valued random variables, not necessarily inde-pendent considered to be a measure ‘! Growth in percentage ( B ) called the dot product or scalar product of. A ) and a company ’ s new product line growth in percentage ( a ) and company. Certain kind of dependence between the some hands-on activity product ) of two in. A ) and a company that shows the quarterly gross domestic product ( GDP.... | 2023-03-25T17:49:21 | {
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https://puzzling.stackexchange.com/questions/54040/find-the-minimum-number-of-steps-that-we-can-arrange-coins-according-to-their-we | # Find the minimum number of steps that we can arrange coins according to their weight
We have $20$ coins, every step we can give $10$ coins to a person and he will tell us the order of their weights. Find the minimum number of steps that we can arrange coins according to their weight.
My attempt:I found a method using $5$ steps.
1.Divide the coins to two equal halfs and give one of the halfs to that person.
2.Give the second half to that person.
3.Give $5$ heavy coins from the first step and $5$ heavy coins from the second step to that person.
4.Do the same as third step but this time use the $5$ light coins of step $1$,$2$
5.Give $5$ light coins from the third step and $5$ heavy coins from the forth step to that person.
This is a copy of this one.
According to the answer of @Aryabhata the above algorithm works. Now I need a proof for showing $5$ steps is the minimum desired. Any way I copy @Aryabhata's answer here:
For a proof that your procedure works:
Let the result of steps 1 and 2 be
$$A_1 \ge A_2 \ge \dots \ge A_{10}$$
and
$$B_1 \ge B_2 \ge \dots \ge B_{10}$$
The result of step 3 will be some permutation of
$$A_1, A_2, \dots, A_5, B_1, B_2, \dots, B_5$$
The result of step 4 will be some permutation of
$$A_6, A_7, \dots, A_{10}, B_6, B_7, \dots, B_{10}$$
If $B_j$ was the lightest of the heaviest 5 coins from step $3$, then it is easy to see that $j \le 5$: The 5 heaviest are $A_1, A_2, \dots, > A_{5-j}, B_1, B_2, B_j$
Similarly if $B_k$ was the heaviest of the lightest 5 from step 4, then $k \ge 6$.
Thus you have a partition by weight
Heaviest in step $3$ $\ge$ Lightest $5$ in step $3$, Heaviest $5$ in step $4$ $\ge$ Lightest 5 in step 4
You step 5 now sorts the middle portion and brings everything in order.
Math SE copy is here and AOPS copy is here.
Source:Second round Iranian olympiad of informatics.
• two steps... give him 10 coins, arrange them based on what he tells us... give him 10 more and arrange accordingly? – Jason V Aug 4 '17 at 12:53
• @Jason, try it with 4: if coin 1 is heavier than coin 2 and coin 3 is heavier than coin 4, what is the final order? There is no information about how the first set relates to the second. – Forklift Aug 4 '17 at 13:16
• @Jason He gives the order of only the 10 coins that on that turn; he does not combine the next 10 with the previous 10. In your scenario, which weighs more: the heaviest coin in the first set or the heaviest coin in the second set. If you can answer that, how did you determine the answer? – Apep Aug 4 '17 at 13:17
• Here's an incorrect argument for why at least 4 is needed. In a set of 20 coins, there are $20*21/2=210$ pairs that need to be compared. In a comparison of 10 coins, $10*11/2=55$ pairs are compared. Therefore we need at least $\lceil{210/55}\rceil = 4$ comparisons. Unfortunately this is bogus, since not every pair needs to be compared. If $a<b$ and $b<c$, we can deduce $a<c$ without actually checking. – Jaap Scherphuis Aug 4 '17 at 13:28
• Post the answer with the question seems odd to me. – paparazzo Aug 4 '17 at 15:45
Your question was also asked in a Russian math olympiad (with 100 coins instead of 20). I found a solution here: https://artofproblemsolving.com/community/c6h530369p3026656. I think the solution left out some details, so I've reworded it and filled them in. This proof also generalizes to the case of trying to sort $2n$ coins, provided $n$ is of the form $4k+2$. I think it can be adapted to work when $n$ is odd, but I haven't thought about what happens when $n$ is a multiple of four.
To prove that four tests are insufficient, suppose that the true order of the coins is $c_1\le c_2\le \dots\le c_{20}$. In order to succeed, for each consecutive pair of coins $\{c_i,c_{i+1}\}$, there must be some test where both coins in the pair handed to your friend. Otherwise, you could not distinguish the true order from $c_1\le \dots\le c_{i-1}\le c_{i+1}\le c_i\le c_{i+2}\le \dots\le c_{20}$.
Suppose that the results of the first two tests are $a_1\le a_2\le\dots\le a_{10}$ and $b_1\le b_{2}\le \dots\le b_{10}$. These two lists of coins might overlap. No matter what coins are chosen for the second test, it will be possible for the overlaps to be distributed among the $a$ list with the same spacings as in the $b$ list, as shown below: $$\begin{array}{} a_1\le \dots\le a_{{k_1}-1}\le&a_{k_1} \le& a_{k_1+1}\le \dots\le a_{{k_2}-1}\le & a_{k_2} \le a_{k_2+1}\cdots\\ &\updownarrow&&\updownarrow\\ b_1\le \dots\le b_{k_1-i}\le & a_{k_1} \le& b_{k_1+1}\le \dots\le b_{{k_2}-1}\le & a_{k_2}\le b_{k_2+1}\cdots \end{array}$$ Assume that this does happen. We will now build a particular total ordering of the coins, consistent with these two tests. We do this in blocks of four as follows. The $*$'s represent any coins which were not tested during the first two tests. $$(c_{4i+1},c_{4i+2},c_{4i+3},c_{4i+4})=\begin{cases}(a_{2i+1},b_{2i+1},a_{2i+2},b_{2i+2})&\text{if }\hspace{.3cm}a_{2i+1}\neq b_{2i+1},a_{2i+2}\neq b_{2i+2} \\ (a_{2i+1},b_{2i+1},*,a_{2i+2})&\text{if }\hspace{.3cm}a_{2i+1}\neq b_{2i+1},a_{2i+2}= b_{2i+2} \\ (a_{2i+1},*,a_{2i+2},b_{2i+2})&\text{if }\hspace{.3cm}a_{2i+1}=b_{2i+1},a_{2i+2}\neq b_{2i+2} \\ (a_{2i+1},*,*,a_{2i+2})&\text{if }\hspace{.3cm}a_{2i+1}=b_{2i+1},a_{2i+2}=b_{2i+2} \end{cases}$$ The general idea is to try to interleave the $a$ and $b$ lists, and whenever overlaps occur, using the untested coins to fill the gaps.
Note that for all $1\le i \le 5$, the consecutive pairs $\{c_{4i+1},c_{4i+2}\},\{c_{4i+2},c_{4i+3}\},$ and $\{c_{4i+3},c_{4i+4}\}$ have not been tested together. These need to be taken care of during tests three and four.
Since all 20 coins are represented among the 15 untested pairs, each coin must be used in exactly one test. This means that all four coins in the block $(c_{4i+1},c_{4i+2},c_{4i+3},c_{4i+4})$ have to be submitted in the same test. But there are five such blocks of four coins, and you can only fit two of the blocks in a single test, so in this case it is impossible to succeed.
• I missed sth to write in my question:The coins have different weights so could you delete extra parts to make the proof easier for me to understand? – Taha Akbari Aug 8 '17 at 18:18
• This proof does assume all the coins have different weights. When I say overlaps, I mean a situation like this. For your first test, you test coins 1 through 10, and for your second, you test 6 through 15. In that case, five of the $a$ coins are equal to five of the $b$ coins. There is no way to get rid of the complicated parts, since we need to prove you cannot succeed no matter what you do on the first two tests. – Mike Earnest Aug 8 '17 at 18:38
• The way you've written the outcomes of the first two tests, it looks as if the common coins will always occur at matching positions. However the first common coin could be, say, lightest in the first test and third lightest in the second (i.e. $a_1=b_3$). A simpler way to think about these steps in the proof is to replace each shared coin in the $b_i$ sequence by an untested coin and then interweave the $a_i$ sequence and this edited $b_i$ sequence. – Jaap Scherphuis Aug 8 '17 at 22:43
• @JaapScherphuis (1) I'm not saying common coins always occur at matching positions, I'm just saying that no matter what weighing strategy you use it is possible for this to occur, and if it does occur then you are doomed. (2) Your modification to the proof makes it much more elegant, I will edit it in. – Mike Earnest Aug 8 '17 at 23:04
• You're right. If the common coins don't occur at the same locations in the outcomes of the first two tests, then you've been lucky and have gained extra information that may allow you to use fewer steps (e.g. in the extreme case $a_{10}=b_1$ means you have sorted all but one coin which you can finish in 2 more steps). The worst case is when the common coins do occur at the same locations, and in that case the proof shows that 5 steps is necessary. So assume the worst. Similarly we can assume the untested coins lie between $b_{i-1}$ and $b_{i+1}$ where $b_i$ is a common coin and swap them in. – Jaap Scherphuis Aug 8 '17 at 23:37 | 2019-12-05T17:26:40 | {
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https://ch.mathworks.com/help/symbolic/sym.rat.html | rat
Rational fraction approximation (continued fraction)
Syntax
``R = rat(X)``
``R = rat(X,tol)``
``[N,D] = rat(___)``
``___ = rat(___,Name,Value)``
Description
example
````R = rat(X)` returns the rational fraction approximation of `X` to within the default tolerance, `1.e-6*norm(X(:),1)`. The approximation is a character array containing the simple continued fraction with finite terms.```
example
````R = rat(X,tol)` approximates `X` to within the tolerance, `tol`.```
example
````[N,D] = rat(___)` returns two arrays, `N` and `D`, such that `N./D` approximates `X`. You can use this output syntax with any of the previous input syntaxes.```
example
````___ = rat(___,Name,Value)` uses additional options specified by one or more `Name,Value` pair arguments to approximate `X`.```
Examples
collapse all
Declare the irrational number $\sqrt{3}$ as a symbolic number.
`X = sqrt(sym(3))`
`X = $\sqrt{3}$`
Find the rational fraction approximation (truncated continued fraction) of this number. The resulting expression is a character vector.
`R = rat(X)`
```R = '2 + 1/(-4 + 1/(4 + 1/(-4 + 1/(4 + 1/(-4)))))' ```
Display the symbolic formula from the character vector `R`.
`displayFormula(["'A rational approximation of X is'"; R])`
`$2+\frac{1}{-4+\frac{1}{4+\frac{1}{-4+\frac{1}{4+\frac{1}{-4}}}}}$`
Represent the mathematical quantity $\pi$ as a symbolic constant. The constant $\pi$ is an irrational number.
`X = sym(pi)`
`X = $\pi$`
Use `vpa` to show the decimal representation of $\pi$ with 12 significant digits.
`Xdec = vpa(X,12)`
`Xdec = $3.14159265359$`
Find the rational fraction approximation of $\pi$ using the `rat` function with default tolerance. The resulting expression is a character vector.
`R = rat(sym(pi))`
```R = '3 + 1/(7 + 1/(16))' ```
Use `str2sym` to turn the character vector into a single fractional number.
`Q = str2sym(R)`
```Q = $\frac{355}{113}$```
Show the decimal representation of the fractional number $355/113$. This approximation agrees with $\pi$ to 6 decimal places.
`Qdec = vpa(Q,12)`
`Qdec = $3.14159292035$`
You can specify a tolerance for additional accuracy in the approximation.
`R = rat(sym(pi),1e-8)`
```R = '3 + 1/(7 + 1/(16 + 1/(-294)))' ```
`Q = str2sym(R)`
```Q = $\frac{104348}{33215}$```
The resulting approximation, $104348/33215$, agrees with $\pi$ to 9 decimal places.
`Qdec = vpa(Q,12)`
`Qdec = $3.14159265392$`
Solve the equation $\mathrm{cos}\left(x\right)+{x}^{2}+x=42$ using `vpasolve`. The solution is returned in decimal representation.
```syms x sol = vpasolve(cos(x) + x^2 + x == 42)```
`sol = $5.9274875551262136192212919837749$`
Approximate the solution as a continued fraction.
`R = rat(sol)`
```R = '6 + 1/(-14 + 1/(5 + 1/(-5)))' ```
To extract the coefficients in the denominator of the continued fraction, you can use the `regexp` function and convert them to a character array.
`S = char(regexp(R,'(-*\d+','match'))`
```S = 3x4 char array '(-14' '(5 ' '(-5 ' ```
Return the result as a symbolic array.
`coeffs = sym(S(:,2:end))`
```coeffs = $\left(\begin{array}{c}-14\\ 5\\ -5\end{array}\right)$```
Use `str2sym` to turn the continued fraction `R` into a single fractional number.
`Q = str2sym(R)`
```Q = $\frac{1962}{331}$```
You can also return the numerator and denominator of the rational approximation by specifying two output arguments for the `rat` function.
`[N,D] = rat(sol)`
`N = $1962$`
`D = $331$`
Define the golden ratio $X=\left(1+\sqrt{5}\right)/2$ as a symbolic number.
`X = (sym(1) + sqrt(5))/ 2`
```X = $\frac{\sqrt{5}}{2}+\frac{1}{2}$```
Find the rational approximation of $X$ within a tolerance of `1e-4`.
`R = rat(X,1e-4)`
```R = '2 + 1/(-3 + 1/(3 + 1/(-3 + 1/(3 + 1/(-3)))))' ```
To return the rational approximation with 10 coefficients, set the `'Length'` option to `10`. This option ignores the specified tolerance in the approximation.
`R10 = rat(X,1e-4,'Length',10)`
```R10 = '2 + 1/(-3 + 1/(3 + 1/(-3 + 1/(3 + 1/(-3 + 1/(3 + 1/(-3 + 1/(3 + 1/(-3)))))))))' ```
To return the rational approximation with all positive coefficients, set the `'Positive'` option to `true`.
`Rpos = rat(X,1e-4,'Positive',true)`
```Rpos = '1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1))))))))))' ```
Input Arguments
collapse all
Input, specified as a number, vector, matrix, array, symbolic number, or symbolic array.
Data Types: `single` | `double` | `sym`
Complex Number Support: Yes
Tolerance, specified as a scalar. `N` and `D` approximate `X`, such that `N./D - X < tol`. The default tolerance is `1e-6*norm(X(:),1)`.
Name-Value Arguments
Specify optional pairs of arguments as `Name1=Value1,...,NameN=ValueN`, where `Name` is the argument name and `Value` is the corresponding value. Name-value arguments must appear after other arguments, but the order of the pairs does not matter.
Before R2021a, use commas to separate each name and value, and enclose `Name` in quotes.
Example: `'Length',5,'Positive',true`
Number of coefficients or terms of the continued fraction, specified as a positive integer. Specifying this option overrides the tolerance argument `tol`.
Example: `5`
Option to return positive coefficients, specified as a logical value (boolean). If you specify `true`, then `rat` returns a regular continued fraction expansion with all positive integers in the denominator.
Example: `true`
Output Arguments
collapse all
Continued fraction, returned as a character array.
• If `X` is an array of m elements and all elements are real numbers, then `R` is returned as a character array with m rows.
• If `X` is an array of m elements that contains a complex number, then `R` is returned as a character array with 2m+1 rows. The first m rows of `R` represent the continued fraction expansion of the real parts of `X`, followed by ```' +i* ... '``` in the (m+1)-th row, and the last m rows represent the continued fraction expansions of the imaginary parts of `X`.
Numerator, returned as a number, vector, matrix, array, symbolic number, or symbolic array. `N./D` approximates `X`.
Denominator, returned as a number, vector, matrix, array, symbolic number, or symbolic array. `N./D` approximates `X`.
Limitations
• You can only specify the `Name,Value` arguments, such as `'Length',5,'Positive',true`, if the array `X` contains a symbolic number or the data type of `X` is `sym`.
collapse all
Simple Continued Fraction
The `rat` function approximates each element of `X` by a simple continued fraction of the form
with a finite number of integer terms ${a}_{1},{a}_{2},\dots ,{a}_{k}$. The accuracy of the rational approximation increases with the number of terms.
Version History
Introduced in R2020a | 2022-10-03T11:45:19 | {
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https://math.stackexchange.com/questions/1181746/describe-the-set-of-all-odd-numbers-between-100-and-200-using-set-builder-no | # Describe the set of all odd numbers between $100$ and $200$ using set builder notation
I've come across a question in Discrete Mathematics, asking me to use set builder notation to describe the set of all odd numbers between 100 and 200.
The answer I had was: $$\{ p | p = 2n + 1, n \text{ (all numbers) } [50, 99], 100 < p < 200 \}$$
Although this should technically give the correct answer, the answers in the textbook have:
$$\{x\,|\,100<x<200\text{ and }2\not | x\}$$ I get the first part, however I have no clue what the end means (2 |/ x); what is that symbol called, and does that represent all odd numbers?
• It seems to be $2$ does not divide. $x$ – NECing Mar 9 '15 at 3:34
• Usually the diagonal strike would go through the divides by vertical strike, so you would get $2 \not | x$ to show that $2$ does not divide $x$ The use of $[50,99]$ and $100 \lt p \lt 200$ is redundant, but still correct. – Ross Millikan Mar 9 '15 at 3:36
• Don't use p. It usually means p is prime. – Qudit Mar 9 '15 at 3:38
• It'd likely be better to either not specify the range in which $n$ is in as $$\{p | p=2n+1,n\text{ is an integer},100<p<200\}$$ - since that bit is redundant. You could also not specify where $p$ lies - and even get rid of $p$ altogether - to get $$\{2n+1 | 50\leq n \leq 99\}.$$ (Your answer is correct, but it's not in its simplest form) – Milo Brandt Mar 9 '15 at 3:43
• Thanks for your help guys. Some helpful tips and information. – zuc0001 Mar 9 '15 at 3:44
The symbol $\mid$ means 'divides'. Drawing a line through it to get $\not \mid$ means 'does not divide'. As for your answer, it is absolutely correct. There are many equally correct ways to write the set of odd numbers with setbuilder notation.
• So 2 does not divide by x is another way of writing all odd numbers? – zuc0001 Mar 9 '15 at 3:37
• That is correct. The odd numbers are precisely those not divisible by $2$. – Ross Millikan Mar 9 '15 at 3:39
As others have noted, the $|$ symbol means "divide". And when we put a slash through it, it means "does not divide".
What does it mean for $2$ to "divide" a number? It means $2$ is a factor of that number. For example, $2$ divides $6$ because $6 = 3 \cdot 2$ (so $2$ is a factor of $6$). Similarly, $2$ divides $100$, since $100 = 5 \cdot 5 \cdot 2 \cdot 2$. Meanwhile, $2$ does not divide $15$ since $15 = 3 \cdot 5$.
Similarly, $2$ divides every even number (isn't that how we define even numbers? as numbers having a factor of $2$?).
But if we define even numbers as numbers having a factor of $2$ in them, then odd numbers are numbers without a factor of $2$ in them. That means $2$ is not a factor of any odd number, which means $2$ does not "divide" any odd number.
So, when we say that $100 < x < 200$ and $2 \not | x$ (i.e., $2$ does not divide $x$), we are saying $x$ is between 100 and 200, and $2$ does not divide $x$, i.e., $x$ is not even, i.e., $x$ is odd.
By the way, here is an extra question for you. Could the answer also be $\{ x | 100 \leq x \leq 200 \text{ and } 2 \not | x \}$? Why or why not?
• Very nice detailed answer. Thanks a lot! To your extra question: although it could be written as <=, its technically not correct because both 100 and 200 are even numbers, meaning that they will never be reached. (Is this correct? :P) – zuc0001 Mar 9 '15 at 3:52
• @zuc0001 Well, I think you have the right idea. It is still correct to write the set with $\leq$, because the $100$ and $200$ might satisfy first condition, but as you suggested, they don't satisfy the second condition since both are even. So they wouldn't be in the set anyway. – layman Mar 9 '15 at 3:54 | 2019-08-24T07:59:12 | {
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https://math.stackexchange.com/questions/1805508/how-does-the-pearson-correlation-coefficient-change-under-rotations | # How does the Pearson correlation coefficient change under rotations
I was reading on wikipedia about the pearson correlation coefficient. Assuming the data has zero mean it can be written as
$$\rho = \frac{ \sum x_i y_i } {\sqrt{\sum x_i^2 \sum y_i^2}}$$
The caption below this image says:
[...] Note that the correlation reflects the non-linearity and direction of a linear relationship (top row), but not the slope of that relationship (middle), nor many aspects of nonlinear relationships (bottom). [...]
(bold text emphasis added by me)
The middle row of the picture shows several distributions that are perfectly correlated ($\rho=1$) and illustrates, that in that case the correlation coefficient does not change when the slope changes (apart from the case if either $x$ or $y$ is constant).
However, I have my doubts whether the correlation coefficient is really independent of the slope, when the correlation is not perfect (ie $\rho<1$). In other words, how does the correlation coefficient change, when I apply a simple rotation
$$x'_i = x_i \cos(\alpha) + y_i \sin(\alpha) \\ y'_i = -x_i \sin(\alpha) + y_i \cos(\alpha)$$
to the data?
Note that the rotation does not change the mean values if $\sum x = \sum y = 0$, but even in the simple form as written above I didn't manage to derive an expression for
$$\rho(\alpha) = ??$$
yet. Or maybe I am just a bit confused and the correlation coefficient really does not change....
I suppose that $\sum_i x_i = \sum_i = y_i = 0$. Moreover, $P_x = \sum_i x_i^2$, $P_y = \sum_i y_i^2$ and $C_{xy} = \sum_i x_i y_i$.
Then, the sample Pearson coefficient $\rho$ based on data $x_i$ and $y_i$ produced by random variables $X$ and $Y$ is:
$$\rho = \frac{C_{xy}}{\sqrt{P_x P_y}}.$$
Notice that:
$$P_{x'} = \sum_i {x'}^2_i = \sum_i (x_i \cos \alpha + y_i \sin \alpha)^2 = \\ \cos^2 \alpha\sum_i x_i^2 + \sin^2 \alpha\sum_i y_i^2 + 2\sum_i x_i y_i \sin \alpha \cos \alpha = \\\cos^2 \alpha P_x + \sin^2 \alpha P_y + \sin(2\alpha) C_{xy},$$
$$P_{y'} = \sum_i {y'}^2_i = \sum_i (-x_i \sin \alpha + y_i \cos \alpha)^2 = \\ \sin^2 \alpha\sum_i x_i^2 + \cos^2 \alpha\sum_i y_i^2 - 2\sum_i x_i y_i \sin \alpha \cos \alpha = \\\sin^2 \alpha P_x + \cos^2 \alpha P_y - \sin(2\alpha) C_{xy},$$
and
$$C_{x'y'} = \sum_i x_i' y_i' = \sum_i (x_i\cos \alpha + y_i \sin \alpha)( -x_i \sin \alpha + y_i \cos \alpha) = \\ -\sum_i x_i^2\sin\alpha\cos \alpha + \sum_i x_i y_i (\cos^2 \alpha - \sin^2 \alpha) + \sum_i y_i^2\sin\alpha\cos \alpha = \\ \frac{1}{2}\sin(2\alpha)(P_y - P_x) + C_{xy} \cos(2\alpha).$$
Consider $\alpha = \frac{\pi}{2}$ and join all pieces togheter:
$$\rho' = \frac{C_{x'y'}}{\sqrt{P_{x'} P_{y'}}} = \frac{\frac{1}{2}\sin(2\frac{\pi}{2})(P_y - P_x) + C_{xy} \cos(2\frac{\pi}{2})}{\sqrt{(\cos^2 \frac{\pi}{2} P_x + \sin^2 \frac{\pi}{2} P_y + \sin(2\frac{\pi}{2}) C_{xy})(\sin^2 \frac{\pi}{2} P_x + \cos^2 \frac{\pi}{2} P_y - \sin(2\frac{\pi}{2}) C_{xy})}} = \\ = \frac{-C_{xy}}{\sqrt{P_yP_x}} = - \rho.$$
Conclusion: rotation affects Peason coefficient.
In general, the new Pearson coefficient, as a function of $\alpha$, is
$$\rho' = \frac{C_{x'y'}}{\sqrt{P_{x'} P_{y'}}} = \frac{\frac{1}{2}\sin(2\alpha)(P_y - P_x) + C_{xy} \cos(2\alpha)}{\sqrt{(\cos^2 \alpha P_x + \sin^2 \alpha P_y + \sin(2\alpha) C_{xy})(\sin^2 \alpha P_x + \cos^2 \alpha P_y - \sin(2\alpha) C_{xy})}}.$$
• Sorry, I should have been more clear in my question. A change from $\rho=1$ to $\rho=-1$ is also illustrated in the wikipedia image, thus your example is not the most convincing one. On the other hand I guess by simply using a different value for $\alpha$ one can show that also the magnitude of $\rho$ is changing. Thanks anyhow – formerlyknownas_463035818 May 30 '16 at 10:47
• Actually now I am a bit confused. I have an application where I need to maximize the correlation between some $X$ and $Y + \beta Z$ depending on $\beta$ (see e.g. here). The "slope of that relationship" is irrelevant and can change. But now if $\rho$ changes with that slope I am not sure if I still find the correct maximum. It could well be that the maximum is independent of rotations, but that would be a new question i guess.... could you suggest me some reading or do you have a tip to clarify my confusion? – formerlyknownas_463035818 May 30 '16 at 11:34
• Don't have some reference. But you can find the new Pearson coefficient using some algebra similar to mine – the_candyman May 30 '16 at 11:39
• @tobi303: Maybe the focus on "slope (non-)invariance" is confusing you. Correlation is not invariant under rotation, but it is invariant under scaling. So changing the slope by scaling does not affect the correlation but changing the slope by rotating does affect the correlation. – citronrose May 30 '16 at 11:55
• @citronrose you are right and i was just thinking too complicated. For the application i mentioned only scaling matters but no rotations. Nevertheless i find the effect of rotations quite interesting – formerlyknownas_463035818 May 30 '16 at 12:25
I think there are big absents: the variance matrices which are, in all these questions, the central concept.
Using matrix-vector notations (instead of all-algebraic calculations), and assuming that we work on centered data, we have the following transformation:
$$X'=RX \ \ \text{with} \ \ R=\begin{bmatrix}\cos(\alpha)&-\sin(\alpha)\\\sin(\alpha)&\cos(\alpha)\end{bmatrix} \ \ \text{and}$$ $$X'=\begin{bmatrix}x'_1&x'_2&\cdots&x'_n\\y'_1&y'_2&\cdots&y'_n\end{bmatrix}, \ X=\begin{bmatrix}x_1&x_2&\cdots&x_n\\y_1&y_2&\cdots&y_n\end{bmatrix}$$
Thus $$X'X'^T=R(XX^T)R^T$$
In other words $$V'=RVR^T$$
(see "A more general identity" in this). by naming $V$ and $V'$, resp. the old and new (co)variance matrices.
which is the way (co)variance matrices are modified (a kind of generalisation of the property $var(aX)=a^2 var(X)$). Thus the new variance matrix is very different from the old variance matrix, even if one normalize each one in order to reason on correlation matrices instead of variance matrices.
The old formula $\rho=\dfrac{V_{12}}{\sqrt{V_{11}V_{22}}}$ is replaced by the new one : $\rho'=\dfrac{V'_{12}}{\sqrt{V'_{11}V'_{22}}}$
(see the final result of @the_candyman).
• sorry, I dont really understand your answer. I wouldn't say that the variance matrices are absent in my question or in the other answer, they are just written in a slightly different way. – formerlyknownas_463035818 May 30 '16 at 11:45
• It's all right, they were implicitly there. I just spent a little time re-writing things in a more readable way, in my opinion. – Jean Marie May 30 '16 at 11:50
• ah ok got it. I actually prefer the "element-wise" notation to the matrix notation most of the time, but thats just my personal preference – formerlyknownas_463035818 May 30 '16 at 11:53
• I understand this preference. It takes a long time to be confident into all-matrix-vector notations. Myself, there are cases, such as derivation with respect to a matrix, where I check my computations element-wise. – Jean Marie May 30 '16 at 12:04 | 2019-09-19T06:43:23 | {
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https://math.stackexchange.com/questions/1247371/follow-on-question-to-fifty-men-and-thirty-woman | Follow-on question to “Fifty men and thirty woman…”
This questions relates to this question Fifty men and thirty woman are lined up at random. How do I find the expected number of men who have a woman standing next to them. and the answer given by André Nicolas.
(I would normally just comment on that question but don't have enough reputation to do that and I am dying to find out what I am missing)
It seems to me like the linearity of expectation would apply to independent random events and it seems like the location of the men relative to the women are not independent in this problem. As you move from man $1$ to $50$, wouldn't the probability of a man standing next to a woman change depending on the number of men and women already seen?
If I understand André's answer correctly, the expected number of men standing next to a woman would be:
$$2(1-(i))+48(1-(ii)) \approx 32.1$$
I wrote a simulation and came up with approximately $30.62$ as the answer.
What am I missing?
Linearity of expectation is (sometimes surprisingly) always valid. If I roll two dice then the expected value of the first is $3.5$ and the exopected value of the second is $3.5$ and the expected value of the sum is $7$. Now you'll say "Sure, that's cause they are independent." But now do this: I roll the first die and then manually place the second die so that it shows the same number. Again each die has an expected value of $3.5$ and the sum has an expected value of $7$. Or I roll a first die and then I manually place the second die so that it shows one more than the first (in a cyclic sense, that is if the first shows "6" I set the second to "1"). Again each die has an expected value of $3.5$ and the sum has an expected value of $7$.
Just note that \begin{align}E[\alpha X+\beta Y]&=\sum_{i,j}P(X=x_i, Y=y_j)(\alpha x_i+\beta y_j)\\&=\sum_i\alpha x_i\sum_jP(X=x_i,Y=y_j)+\sum_j\beta y_j\sum_iP(X=x_i,Y=y_j) \\&= \alpha \sum_ix_iP(X=x_i)+ \beta \sum_jy_jP(Y=y_j)\\&=\alpha E[X]+\beta E[Y]\end{align}
Now why your simulation differs from your computation: For a single man $i$ ("Jack", say) the expected value of his $X_i$ is \begin{align}E[X_i]&=\operatorname{Pr}(X_i=1)\\&=1-\operatorname{Pr}(X_i=0)\\&=1-\left(\frac{2}{80}\cdot \frac{49}{79}+\frac{78}{80}\cdot \frac{49}{79}\cdot\frac{48}{78}\right)\\&=\frac{154041}{249640} \end{align} and this is the same for all $i$; note that we cannot split end men and middle men in the final sum because we don't know beforehand how many end men there will be. Instead we subsumize the possibility of being at an end or not alreday in the expected value for each single man. Either Jack ends up at an end or he does not; since these are mutually exclusive, we added the probabilities in the parenthesis. Because there is no distinction among the men, we have $E[X_1]=\ldots=E[X_{50}]$ and hence $$E[Y]=50\cdot \frac{154041}{249640}=\frac{770205}{24964}\approx 30.85$$ which is much better in accordance with your simulation (though you didn't specify the observed variance, which could have hinted to the quality of the simulated result). | 2021-03-05T22:27:58 | {
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https://math.stackexchange.com/questions/1905479/finding-a-value-on-a-number-line | # Finding a value on a number line?
I came across this question: On the number line the tick marks are equally spaced which of the lettered points represent $y$
The solution says it is $D$ but I think it is $E$.
Explanation
1. The average $\frac{x+y}{2}$ shows it is the midpoint of $x$ and $y$ and thus counting the steps we get $E$
2. I can see that $\frac{y}{2}$ is the partition. Also that $x$, $x+y$ and $\frac{x+y}{2}$ are all negative and $\frac{y}{2}$ is positive because the points are on a number line. My aim is finding the origin $0$. So I can count two partitions to get $y$. If I separate $\frac{x+y}{2}$ to $\frac{x}{2}+\frac{y}{2}$ I can see that $C$ is $\frac{x}{2}$ and that moving $3$ partitions from $C$ I get my origin which is on $D$. going two steps gets me to $E$. So what seems to be the problem?
• I'm assuming it is a typo. Notice that the origin need to be D so perhaps the question is what point represents 0. – fleablood Aug 27 '16 at 18:12
• x + y - (x)= y =2. $(x+y)/2 - x = 4$ so $x/2 - x + 1 = 4$ so $x - 6$. So D = 0. and y= 2 = E. (Assuming the tick marks are = 1; if the tick marks are any k, we still get 0 = D, y = 2k=E, x = -6k. – fleablood Aug 27 '16 at 18:17
• @fleablood Ok thanks! – Socre Aug 27 '16 at 18:35
• It'd be good to get a third opinion to avoid dyslexia and brain farts but it sure seems like your reasoning it the correct one. – fleablood Aug 27 '16 at 18:48
• Despite the name, "linear algebra" is not an appropriate tag for this; I've retagged. – Noah Schweber Aug 27 '16 at 19:12
Yes, you are correct, $y$ should be $E$.
# Explanation -
1. Note that there are two spaces between $x$ and $x+y$, so $y=2$.
2. ${(x+y) \over 2}$ is to the right of $(x+y)$, so , $(x+y)$ must be negative.
3. As, $y$ is positive from $(1)$, so $x$ must be negative from $(2)$.
4. Now, ${(x+y)\over 2}$ is halfway between $(x+y)$ and $0$. Also, counting the steps from $(x+y)$ to ${(x+y)\over 2}$ shows that ${(x+y)\over 2}=(-2)$.
5. From $(4)$, we can count $2$ steps to the right to see that $D$ is $0$ and so $y=2$ is $E$.
# Notes -
1. Your approaches are of course correct as well.
2. There are thus, at least, three correct approaches to solving this problem. | 2019-07-18T04:49:12 | {
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https://dsp.stackexchange.com/questions/26121/smoothed-square-wave/26123 | Smoothed Square Wave
new here and probably pretty unexperienced compared to the rest of you. This should be simple enough but just wanted some clarification. I'm trying to model an analog square wave oscillator with C. With the oscillator I'm trying to mimic, the square wave isn't totally square and has curves with the starting edge of each pulse (on/off). It looks like in the analog device the capacitor is smoothing the square wave, so I believe I need to implement a low pass filter in my code to smooth out the square wave.
tldr - Just wanted to confirm that I should implement a low pass filter on the raw square wave to make it sound similar to the analog oscillator of the picture below. If there's another way I should be approaching this I'm open to listen to suggestions as I might be approaching this wrong.
Top is my square wave Bottom is device's square wave
Yes. This looks like it's a typical first order low pass. The time constant can be determined by looking at the time it takes for the falling edge to drop to 37% of the max amplitude ($e^{-1}$). The cutoff frequency of the low pass filter is $1/(2 \cdot \pi \cdot t_c)$ where $t_c$ is the time constant | 2022-01-25T23:24:17 | {
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https://math.stackexchange.com/questions/2407659/why-does-the-newton-raphson-method-not-converge-for-some-functions/2408116 | # Why does the Newton-Raphson method not converge for some functions?
$f(x)=2x^2-x^3-2$. This is a cubic type graph as shown. The real root of this graph is $(-0.839,0)$.
So, the question is to use Newton's approximation method twice to approximate a solution to this $f(x)$.
I use an initial starting value of $x_0=1$. My first approximation is $x_1=2$, and my second one is $x_2=1.5$. I seem to not move any closer to the real solution as I keep iterating through the method.
Am I misunderstanding how to use this approximation? Is the issue that my first guess was too far from the actual solution?
• Note that from $x_{n+1}=x_n+\dfrac{x_n^3-2x_n^2+2}{4x_n-3x_n^2}$, we should obtain $x_2=1.5$, not $2.5$. – projectilemotion Aug 27 '17 at 14:20
• Very nice example! The method seems to fail, but eventually is successful! – Peter Aug 27 '17 at 14:23
• @user163862 The graph shows that it is better to start from a value at left of root, like -1.2 – Narasimham Aug 27 '17 at 19:17
• @Narasimham To the left of the root does not really matter, the point is that it should be to the left of that minimum at $0$. – Ian Aug 27 '17 at 22:13
• In short: your starting point was too near to an extremum. If you can pick starting points that are as far away from extrema as possible (and of course close to the desired root!), do so. – J. M. is a poor mathematician Aug 28 '17 at 4:47
Newton's method does not always converge. Its convergence theory is for "local" convergence which means you should start close to the root, where "close" is relative to the function you're dealing with. Far away from the root you can have highly nontrivial dynamics.
One qualitative property is that, in the 1D case, you should not have an extremum between the root you want and your initial guess. If you have an odd number of extrema in the way, then you will start going away from the root you want, as you see here. If you have an even number of extrema in the way, then you will start going the right way, but you may later find yourself in a spot with an odd number of extrema in the way, leading to problems later.
Of course you may eventually find an occasion where there are an even number of extrema in the way, and then you manage to skip over all of them and get to the right side. At that point things will usually work out (not always, though). In this problem with your initial guess, that eventually happens, because the system eventually finds its way just slightly to the right of the extremum on the right, which sends it far off to the left.
• Wow! What a great explanation. Now I totally get it. – user163862 Aug 27 '17 at 14:36
In fact, you gave up too early ; The method eventually converges :
1 2.000000000000000000000000000
2 1.500000000000000000000000000
3 0.3333333333333333333333333333
4 2.148148148148148148148148148
5 1.637079608343976160068114091
6 0.9483928480399477528436835979
7 1.910874140183680201544963299
8 1.405089904362402921055022221
9 -1.324018083676046424512855515
10 -0.9614381794507316717924414480
11 -0.8500221808505758631523579893
12 -0.8393807176849843501240483025
13 -0.8392867625049899194321196645
14 -0.8392867552141611764525252322
15 -0.8392867552141611325518525647
16 -0.8392867552141611325518525647
17 -0.8392867552141611325518525647
18 -0.8392867552141611325518525647
19 -0.8392867552141611325518525647
20 -0.8392867552141611325518525647
?
• Oops, very fast upvote :) – Peter Aug 27 '17 at 14:17
• Yes, I was going to transfer my previous comment to an answer, but you were a bit faster than me. – projectilemotion Aug 27 '17 at 14:18
• Note that the magic happens when you finally bounce around to hit ~1.4 which then sends you way over to ~-1.3 – Ian Aug 27 '17 at 14:20
• There are some collections of conditions sufficient to guarantee the convergence of the newton-method. Especially cubics can be dangerous, sometimes the method actually diverges or oscillates. In doubt, you can try the slower but more reliable numerical methods as the bisection-method or regula-falsi. Or you can combine the methods and first search an approximation and then use newton to get a more precise result. – Peter Aug 27 '17 at 14:36
• I don't think this answer actually addresses any point raised in the original question. So this particular initial value does lead to a converging sequence; fine. But why? Did it have to be that way? What about other possible starting values? – Marc van Leeuwen Aug 29 '17 at 8:51
The other answers are great. I'd just like to add a concrete example of weird behavior of which the Ian's answer speaks.
Let's consider a function $f(x) = \operatorname{sgn} x \sqrt{|x|}$. According to the algorithm, we iterate $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.$$ Now for the derivative (if we're not doing the derivative at $x = 0$, the $\operatorname{sgn} x$ is just a constant, and $|x| = x \operatorname{sgn} x$): $$f' = [\operatorname{sgn} x \sqrt{x \operatorname{sgn} x}]' = \operatorname{sgn} x \frac{1}{2\sqrt{x \operatorname{sgn} x}} \operatorname{sgn} x = \frac{1}{2\sqrt{|x|}}.$$ Plugging in: $$x_{n+1} = x_n - \frac{\operatorname{sgn} x \sqrt{|x|}}{1/(2\sqrt{|x|})} = x_n - 2 \operatorname{sgn} x \left(\sqrt{|x|}\right)^2 =\\ =x_n - 2\operatorname{sgn} x |x| = x_n - 2 x_n = - x_n.$$
So if we start iterating in $x = a$ (where $a \not = 0$), we get the sequence $a, -a, a, -a, \ldots$ and the method loops forever between those two points, never getting to the root $x = 0$!
Edit: Here's a gnuplotted image: (In each iteration, we make a tangent in the current point (the blue dashed line) and the $x$ for which the tangent crosses 0 is taken to be the next approximation (so we go along the magenta line in order to get the starting point for the next iteration).)
By the way, have a look at this image from Wikipedia:
It shows the complex plane colored with 5 colors, each color corresponding to one root of the complex equation $z^5 = 1$. Each point then has the color corresponding to the root to which Newton's method converges, if we start from that point. The "flowers" are beautiful to behold but totally abhorrent from the numerical point of view.
• That's a nice image! – dafinguzman Aug 27 '17 at 20:35
• "[...]Flowers are beautiful to behold but totally abhorrent from a numerical point of view" now range alongside "Sufficiently accurate for Poetry" as my favorite expressions ever! – Beyer Aug 28 '17 at 7:05
• I like the "Sufficiently accurate for Poetry" much more! :-). // By the way, thank you, Simply Beautiful Art, for nicely inlining the image. – Ramillies Aug 28 '17 at 10:55
• I doubt there is a root at x = 0! = 1 – ChemiCalChems Aug 28 '17 at 11:49
• I'm sorry, but I can't see why not. $f(x) = \operatorname{sgn} x \sqrt{|x|}$, so $f(0) = \operatorname{sgn} 0 \sqrt{|0|} = 0 \cdot 0 = 0$. Hence $x = 0$ is a root. (Btw.: I say root, not stationary point.) – Ramillies Aug 28 '17 at 12:03
Newton's method has no global convergence guarantee for arbitrary functions, as you just learned.
Now, people have posted examples of where Newton's method doesn't converge, but they're all rather "unusual" functions (some being very non-smooth), so it's natural to assume they're pathological and won't happen in practice.
Your example is one where Newton just takes more iterations than expected to converge, so it's not too bad. But here is an example of a cubic polynomial for which Newton's method won't converge!
\begin{align*} f(x) &= -0.74 + 0.765 x + 1.1 x^2 - 3.55 x^3 \\ x_0 &= 5/9 \end{align*}
Not only that, but it's in fact a stable oscillation—small perturbations won't change the behavior.
And for bonus points, you can generate as many of these as you want! Just let the Newton step be $$g(x) = x - f'(x)^{-1} f(x)$$ and then you're just looking for a nontrivial solution to the equation $$x_0 = g^3(x_0) = g(g(g(x_0)))$$ where a solution $x_0$ would be trivial if $f(x_0) = 0$.
Notice the equation above is just another polynomial equation (although of a much higher order)—which means its solutions can be readily found numerically.
The only caveat is that these solutions might not necessarily be stable. I suspect you should be able to place a second-derivative condition to ensure stable solutions, but the exact equation is not obvious to me at the moment, so I'll leave it as an exercise for the reader. :-)
Mathematica code for the plot:
Manipulate[
With[{f = Evaluate[Rationalize[d + c # + b #^2 + a #^3]] &}, Plot[
f[x], {x, -0.61, 1},
PlotStyle -> {Thickness[Tiny]},
Prolog -> {Thickness[Tiny],
Line[Flatten[Map[
{{#, 0}, {#, f[#]}} &,
NestList[Compile[t, t - f[t]/f'[t]], x0, n]], 1]]}]],
{{a, -3.55}, -4, 4},
{{b, 1.1}, -2, 2},
{{c, 0.765}, -1, 1},
{{d, -0.74}, -1, 1},
{{x0, 5/9}, 0, 1},
{{n, 100}, 0, 1000, 1}]
### Update:
I wrote some code to purposefully find both stable and unstable iterations:
Newton[f_] := t \[Function] t - f[t]/f'[t];
NewtonPlot[f_, xmin_, xmax_, x0_, n_, args___] :=
Plot[f[x], {x, xmin, xmax}, args,
Prolog -> {Thickness[Tiny],
Line[Flatten[Map[
{{#, 0}, {#, f[#]}} &,
NestList[Compile[t, Newton[f][t]], x0, n]], 1]]}];
FindOscillatoryNewtonSolutions[f_, n_, h_](* {Stables,Unstables} *):=
With[{step = Newton[f]},
With[{nstep = t \[Function] Nest[step, t, n]},
GroupBy[
Map[#[[1]][[2]] &,
Solve[{nstep[t] == t, Abs[step[t] - t] >= h}, t, Reals]],
t \[Function]
With[{step1hp = nstep[t + h], step1hm = nstep[t - h]}, True \[And]
Abs[N[step1hp - t]] >= Abs[N[nstep[step1hp] - t]] \[And]
Abs[N[step1hm - t]] >= Abs[N[nstep[step1hm] - t]]]]]];
With[{z = 400, xmin = -1.1, xmax = +0.65, h = 10^-3},
Manipulate[
With[{f =
t \[Function] Evaluate[Rationalize[d + c t + b t^2 + a t^3]],
n = 3, m = 8},
With[{solcategories = FindOscillatoryNewtonSolutions[f, n, h]},
If[Length[solcategories] >= 2,
Map[{Transpose@{N@SortBy[NestList[Newton[f], #1, n - 1], N]},
NewtonPlot[f, xmin, xmax, N[#1 + dx], n*m,
PlotStyle -> {Thickness[Tiny]},
ImageSize -> Scaled[0.2],
PerformanceGoal -> "Quality"]} &, {First@Lookup[solcategories, True],
First@Lookup[solcategories, False]}],
Throw["Did not manage to find both a stable and an unstable solution."]]]],
{{dx, h}, -4 h, +4 h, h/4},
{{a, -355}, -z, +z, 1}, {{b, 110}, -z, +z, 1},
{{c, 77}, -z, +z, 1}, {{d, -74}, -z, +z, 1},
SynchronousUpdating -> False]]
Notice, below, that the first point iterates stably, whereas the second one iterates unstably. (The iterations would diverge further, but I cut them off at some point.)
• Thank you, I was hoping someone would post an example like this. – Rahul Aug 28 '17 at 15:39
The relationship between starting point and eventual convergence point for Newton's method is complicated. Consider this image (from Wolfram's MathWorld, Newton's Method):
showing the basis of attraction in the complex plane for $x^3 - 1$, $x^4 - 1$, ... $x^{11} - 1$. Each colour corresponds to a root of the polynomial. Each point is colored to match the root that Newton's method eventually takes that point to.
These basins are typically quite complicated, but if you look at what is going on in the complex plane, it can be a little clearer what's going on. The real line runs horizontally, halfway down these plots. If all you know is what is happening on the real axis, which is what you are seeing in your problem, it can be difficult to predict where a given point is going to end up.
Newton-Raphson can behave badly even in seemingly easy situations. I am considering the use of N-R for minimization (rather than root finding, but the same applies). Even in the case of convex functions, N-R may not converge.
For example: $$f(x)=\ln(e^x+e^{-x})$$ is $C^{\infty}$, strictly convex and admits a single (global) minimum in 0.
Yet, if we try to use N-R to find the minimum (i.e. the root of the derivative), the algorithm fails if started from a point $|x|>1.09$ (approx).
To see this, consider that $f'(x)=\tanh(x)$ and $f''(x)=\cosh^{-2}(x)$. Therefore, the N-R update rule maps the current $x$ to the new $x$ according to: $$x \leftarrow x - \frac{f'(x)}{f''(x)} = x - \frac12 \sinh(2x)$$ Whenever $|\frac12 \sinh(2x)| > |2x|$, the new iterate will be bigger in modulus than the previous one, i.e. N-R will diverge away from the solution. This happens for $|x|>1.09$ (approx).
• Note that this happens because $\tanh$ is extremely flat for large $x$, so although you go in the right direction toward the root (since $\tanh$ has no extrema as I said in my answer), you massively overshoot. (By the way, this is a good example but I think it would be better to just go for root finding applied to $\tanh$ in the first place instead of talking about optimization.) – Ian Aug 27 '17 at 22:11
• @Ian Indeed, the quadratic approximation fits locally but is very poor globally. A better approach in this case would be to use a majorization minimization algorithm, which guarantees monotonic convergence towards a local (global in this case) minimum. – Luca Citi Aug 27 '17 at 22:15
• I agree, it's far easier to see what's going on if you do root-finding with tanh than minimization with its antiderivative. – Mehrdad Aug 28 '17 at 4:11
• My other objection to such an example is that if you have a $C^2$ convex function and you use Newton's method to minimize it, then Newton's method always goes in the correct direction (in multiple dimensions, the correct direction and the Newton direction always have a positive dot product). It just might overshoot. You can't always detect overshoots, but you can detect that the objective function gets bigger. If it did, then by convexity you must have overshot, and so you should trigger a line search to find a better point in between the new point and the previous point. – Ian Aug 28 '17 at 13:27
• @Ian: Your objection only holds in 1D, and in that case Newton isn't even the one doing the work; it's just the sign of the derivative telling you which direction to move. In 2D+ there's no guarantee you'll move in the correct direction; best you can hope for with gradients is that you won't be more than 90 degrees away from the correct direction (which is far less useful) and I don't think Newton can do any better there either. – Mehrdad Aug 31 '17 at 8:42
There are several articles about the convergence of Newton's method. There is something called the Newton-Kantorovich theorem which gives rigour to the notion of convergence regions.. your starting point must be within the Fatou set which encloses the point of attraction of the dynamical system formed by the iterates of the Newton iteration.
On the Newton–Kantorovich hypothesis for solving equations
I used the Method here to explore the convergence regions for asserting convergence of Newton's method for the Gram points of the zeta function Convergence of the Newton-Kantorovich Method for the Gram Points
The Newton-Raphson method basically asks you to draw the tangent to the function at the point $x_0$, and $x_1$ is the point where that tangent hits the $x$-axis. This obviously works well if the function follows reasonably close to the tangent up to the point where $y = 0$, and not so well otherwise.
You can see that between $x = 0$ and $x \approx 1.4$ the tangent actually points in the wrong direction, away from the $0$. And where the tangent is parallel or almost to the $x$-axis, close to $x = 0$ and $x = 1.4$, following the tangent will get you far away from the zero.
For $x_0 < -0.346, \ x_1$ will be closer to the solution than $x_0$, and will be in the same range again, so the sequence will converge towards the solution. For $x_0 > 1.5, \ x_1$ will also be closer to the solution than $x_0$, but you need to get over the area from $-0.346 \to 1.5$.
And $\approx 1.081792$ there is a point where $3$ iterations get you back to the point where you started :-) | 2019-06-25T19:33:14 | {
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https://math.stackexchange.com/questions/2785209/linear-transformation-such-that-tx-kx-for-all-x-in-mathbfrn | # Linear transformation such that $T(x) = kx$, for all $x \in \mathbf{R^n}$
Question
Let $T : \mathbf{R^n} \rightarrow \mathbf{R^n}$ be a linear operator such that $T(W) \subseteq W$ for every subspace $W$ of $\mathbf{R^n}$. Show that there exists $k \in \mathbf{R}$ such that $T(x) = kx$, for all $x \in \mathbf{R^n}$.
This is my attempt at an answer:
Let $\{e_1,\ldots, e_n\}$ be a basis of $\mathbf{R^n}$. Let W be the subspace generated by ${e_1}$. Then any element of W is of the form ${ke_1}$, for some $k \in \mathbf{R}$.
$\therefore T(e_1) \in W \Rightarrow T(e_1) = ke_1$ for some $k \in \mathrm{R}$. Similarly, $T(e_2) = ke_2,\ldots, T(e_n) = ke_n$
Every element $x$ of $\mathbf{R^n}$ is of the form $\alpha_1e_1 + \cdots + \alpha_ne_n$, where $\alpha_1, \ldots ,\alpha_n \in \mathbf{R}$
$$\therefore T(x) = T(\alpha_1e_1 + \cdots + \alpha_ne_n)$$
$$= \alpha_1T(e_1) + \cdots + \alpha_nT(e_n)$$ $$= \alpha_1ke_1 + \cdots + \alpha_nke_n$$ $$= k(\alpha_1e_1 + \cdots + \alpha_ne_n)$$ $$= kx$$
I'm not sure if my proof is correct, especially the part in bold.
The problem with the attempted solution presented in the text of the question, especially the part in bold type, is that we can't assume
$T(e_i) = ke_i, \tag 1$
with one value of $k \in \Bbb R$ for all the $e_i$; since all we are given is that
$T(W) \subset W \tag 2$
for any subspace $W$, there is nothing to a priori prohibit
$T(e_i) = k_i e_i, \tag 3$
with a different $k_i \in \Bbb R$ for each $e_i$; certainly an operator $T$ satisfying (3) has the property that
$T(\Bbb R e_i) \subset \Bbb R e_i \tag 4$
for each $e_i$.
So how can we force all the $k_i$ to be equal? Well, suppose we look at the one dimenensional subspace $\Bbb R(e_i + e_j)$; since
$T(\Bbb R(e_i + e_j)) \subset \Bbb R(e_i + e_j), \tag 5$
there must be some $k \in \Bbb R$ with
$T(e_i + e_j) = k(e_i + e_j) = ke_i + ke_j; \tag 6$
but
$T(e_i + e_j) = T(e_i) + T(e_j) = k_i e_i + k_j e_j; \tag 7$
these two equations taken together yield
$ke_i + ke_j = k_i e_i + k_j e_j, \tag 8$
whence
$(k - k_i)e_i + (k - k_j)e_j = 0; \tag 9$
the linear independence of the $e_i$ now forces
$k _i = k = k_j; \tag{10}$
since this argument applies for any two $e_i$, $e_j$, we must have
$k_i = k_j = k, \; \forall i, j \; 1 \le i, j \le n; \tag{11}$
we thus conclude that
$T = kI, \; k \in \Bbb R. \tag{12}$
It should of course be observed that the above argument applies for any basis $e_i$ of $\Bbb R^n$.
You're right to be suspicious! Certainly $T(e_1) \in W \Rightarrow T(e_1) = ke_1$, but there's no (a priori) reason to know that $T(e_2) = ke_2$ with the same $k$. It could in general be the case that $T(e_2) = k_2 e_2$ instead, and similarly $T(e_3) = k_3 e_3$.
So a correct approach could begin like yours, and say $T(e_1) = k_1 e_1$, $T(e_2) = k_2 e_2$, and so on. You now need to show that all the $k_i$ are equal which I'll leave to you, save for this hint: What happens if $W = \langle e_1 + e_2 \rangle$?
Let be $W_i=<e_i>$. From $T(e_i)\in W_i$, you can conclude that there exists $k_i\in \mathbb R$ such that $T(e_i)=k_ie_i$ because of the linearity of $T$. But you still have to prove that $k_1=\ldots=k_n$.
No, your proof is not correct. Indeed, if $W=\langle e_1\rangle$, then there is a number $k_1$ such that $T(e_1)=k_1e_1$. And there is a number $k_2$ such that $T(e_2)=k_2e_2$. But you assumed, without proof, that $k_1=k_2$.
Yes it is almost correct but you need to prove that by linearity
• $T(e_i)=k_ie_i,\quad T(e_j)=k_je_j\implies k_i=k_j$
and also you should claim that the basis for any subspace $W$ can be expressed as linear combination of the vectors $e_i$ and then conclude. | 2019-05-26T21:05:36 | {
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https://stats.stackexchange.com/questions/188774/regression-and-elastic-net-provide-different-results/188780 | regression and elastic net provide different results
I fit Elastic Net model on 20-50 variables. Elastic Net selects 10 (but actually I can choose any model on the solution path for the next step).
Next, I take these 10 variables and fit standard regression with them. The estimated parameters differ and usually one variable appears to explain most of the variance.
Question is - why? Is it because regression fits the model in one step, getting all parameter estimates at once, whereas Elastic Net does estimation it step by step, variable by variable (I do not know the algorithm)?
Does shrinkage in the Elastic Net influence parameter estimates in such a way that their interpretation becomes - let's try this wording - untrue? If yes, then I would select Elastic Net for best forecasting, but rather choose estimates from regression for interpretation.
• Elastic net and OLS optimize different criteria (penalized vs unpenalized sums of squared residuals), so have different solutions. – Andrew M Dec 31 '15 at 1:30
There is no free lunch in statistics. Elastic Net reduces overfitting (lowers variance) at the cost of increasing bias. With OLS, you could fit a model with all 50 variables. This OLS model would have very low bias (under certain assumptions, the coefficient estimates may be unbiased) but suffer from high variance (overfitting).
In your case, you mentioned that the OLS coefficients look very different than the Elastic Net coefficients, even though both models use the same 10 variables. The difference may be due to bias introduced by the fact that Elastic Net does not compute the coefficients by minimizing the residual sum of squares (which is how OLS computes the coefficients). Elastic net computes the coefficients by minimizing the "penalized" residual sum of squares.
Alternatively, the coefficient estimates may be different between OLS and Elastic Net due to sample size. With small sizes, p-values from OLS may not be reliable. With small sample sizes, the bias from elastic net may also be high.
Here's a simulated example using $n=25$. The "true model" contains only two variables, $x1$ and $x2$, with the "true coefficients" of 2 and 3. Due to small sample size and high irreducible error, the p-value for x1 is high (>22%). The coefficient differences between the two models are also high.
set.seed(1983)
nobs <- 25
x1 <- rnorm(nobs, 10, 10)
x2 <- rnorm(nobs, 20, 20)
x3 <- rnorm(nobs, 30, 30)
x4 <- rnorm(nobs, 40, 40)
y <- 100 + 2*x1 + 3*x2 + rnorm(nobs,0,100)
df <- data.frame(y=y, x1=x1, x2=x2, x3=x3, x4=x4)
### fit a linear model
lm.mod <- lm(y ~ ., data=df)
summary(lm.mod)
### fit an elastic net model using 5-fold CV
library(caret)
set.seed(1984)
enet.mod <- train(y ~ ., data=df, method="glmnet", tuneLength=5, trControl=trainControl(method="cv", number=5))
coef(enet.mod$finalModel, enet.mod$bestTune$lambda) ### compute diffs between coefs lm.mod$coefficients - t(coef(enet.mod$finalModel, enet.mod$bestTune$lambda))[1,] When the sample size is increased to$n = 1000$, the p-value for$x1$is very low and the coefficient differences between the two models are small. set.seed(1983) nobs <- 1000 x1 <- rnorm(nobs, 10, 10) x2 <- rnorm(nobs, 20, 20) x3 <- rnorm(nobs, 30, 30) x4 <- rnorm(nobs, 40, 40) y <- 100 + 2*x1 + 3*x2 + rnorm(nobs,0,100) df <- data.frame(y=y, x1=x1, x2=x2, x3=x3, x4=x4) ### fit a linear model lm.mod <- lm(y ~ ., data=df) summary(lm.mod) ### fit an elastic net model using 5-fold CV library(caret) set.seed(1984) enet.mod <- train(y ~ ., data=df, method="glmnet", tuneLength=5, trControl=trainControl(method="cv", number=5)) coef(enet.mod$finalModel, enet.mod$bestTune$lambda)
### compute diffs between coefs
lm.mod$coefficients - t(coef(enet.mod$finalModel, enet.mod$bestTune$lambda))[1,]
• Thank you for a detailed explanation. I think first two paragraphs are more relevant in my case, but I like the alternative explanation. I suppose multicollinearity can also be an influential factor here. However, I remain a bit concerned that I have to choose from estimates that differ in millions of dollars in magnitude to make economic policy or big business decisions. Therefore still an unanswered part of my question is how do I determine which model to choose? Answer below suggests Elastic Net, but lower bias in OLS is more intuitive for me. Could any of you comment on why? – user2530062 Dec 31 '15 at 18:04
• If your concern is to make accurate predictions, you should choose the model with the best out-of-sample performance. You could hold out a portion of your data (e.g., hold out 20% of your data) and build two models using the remaining 80%: elastic net, OLS. Score the hold-out data set and see which model returns the lowest root-mean-squared-error. – William Chiu Dec 31 '15 at 18:17
• Notice that, when $n=1000$, the coefficient estimates between elastic net and OLS are about the same. – William Chiu Dec 31 '15 at 18:20
• Now I have p=4 n=842 model and the estimates still vary a lot. This "bias" part you mentioned in original post is stil of a concern. I think OLS would be better for close to average observations (those with low variance themselves), but Elastic Net would be better in general and out-of-sample testing would be preferred. From some external reading I can see that it all goes down to this bias-variance tradeoff, so thank you for pointing me in this direction. – user2530062 Jan 1 '16 at 10:41
• How are you measuring the difference in the two sets of coefficients? It may be that the scale of the variables magnify the differences. Have you considered dividing each variable by its mean and then running elastic net and OLS? The division would remove the units of measurement from each covariate. – William Chiu Jan 1 '16 at 18:49
Much depends on your interpretation of the word "interpretation." Elastic net deliberately penalizes regression coefficients in a way that helps correct for the over-fitting and optimistically high-magnitude coefficients that standard regression can provide in this type of scenario. One might argue that "interpretation" from the perspective of reliably representing the underlying population rather than your particular sample would best be done on the penalized coefficients. | 2021-03-04T00:40:40 | {
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https://math.stackexchange.com/questions/460611/number-of-teeth-in-gears | Number of teeth in gears
I'm building something with an engine that uses gears to reduce/increse movement. The motor has itself some gears, and it's a stepper motor (it gives discrete steps), now the number of steps per revolution is not integer: $4075.77284...$, and I need to add gears to make it as close as possible to 720 steps, that is: each step must be as close as possible to $0.5º$.
So, this is the purely mathematical problem, the number of steps is given by:
$$n=64\frac{22\cdot 26\cdot 31\cdot 32}{9\cdot 9\cdot 10\cdot11}$$.
I'm going to use two wheels in series, so the problem is finding four integer numbers: $a,b,c,d$, such that:
$$n\frac{ab}{cd}\approx 720$$
Also, those four integers must be in the interval $(8,40)$.
The only thing I could think off was just trying, and found, by fixating $c=d=20$ and then setting $a=8,b=9$, that: $$n\frac{ab}{cd}=733.63911111111110585625$$
This gives a rotation of $0.4907044820099255577617$ degrees per step, with an error with respect to $0.5$ of $0.00929552$. That error is the important thing, and being around a hundredth of a degree per step, it's good enough for me, but I would like to know if there exists any analitical methods to get better solutions. Sadly, I don't have enough knowledge about diophantic equations to be able to do this on my own.
As a last resource, I would program something to test all possible combinations of $a,b,c,d$ in that interval, which is probably what I'm going to do, but if there's a more elegant solution, I'd love to know it.
Thanks anyone.
UPDATE: Ok, I programmed it, and found out that the best solution is: $a=8,b=31,c=36,d=39$, with an error of $0.0000436963449531591053$. Not a nice solution, I expected smaller numbers, but... good enough. Still, and analytical more general solution would be appreciated.
• Is there a typo in your expression for $n$? I get $(64\times22\times26\times31\times32)/(9\times9\times10\times11) \approx 4075.77284$, not $4075.3$. – Chris Culter Aug 5 '13 at 21:04
• No, the typo wsa in the 4075.3, thanks. – MyUserIsThis Aug 5 '13 at 21:04
• You may find this article interesting:ams.org/samplings/feature-column/fcarc-stern-brocot – awkward Aug 5 '13 at 21:47
• @awkward Thanks a lot. – MyUserIsThis Aug 5 '13 at 22:29
• Another constraint that you may want to impose is ensuring that the numbers of teeth on each pair of meshing gear are coprime. This is to avoid the so-called "hunting tooth" phenomenon that arises from wear, imperfect alignment, and non-identical tooth profiles. In engines this is quite important. – horchler Aug 5 '13 at 22:52
I don't think this can guarantee the optimal under constraints, but in general you can use the continued fraction representation to find best rational approximations to get something close. \begin{align} r &= \frac{720}{4075.7724\cdots} = 0.1766536\cdots \\ &= [0;5,1,1,1,18,3,\ldots] = \cfrac{1}{5+\cfrac{1}{1+\cfrac{1}{1+\cdots}}} \end{align} This gives the approximations $$r\simeq \frac{3}{17}, \quad \frac{56}{317}, \quad \frac{171}{968},\quad \frac{227}{1285}$$ which yield respectively $$0.500518\cdots, \quad 0.499992\cdots, \quad 0.500002\cdots, \quad 0.4999997\cdots$$ degrees per step. Unfortunately only the first factors into fractions that would fit your teeth constraints.
From here you can "hunt around" using the mediant of approximations that are above and below the target, as in the link @awkward gave in a comment. $$\frac{3}{17}\oplus\frac{56}{317}=\frac{59}{334} \\ \frac{3}{17}\oplus\frac{59}{334}=\frac{62}{351} =\frac{2\cdot 31}{3^3\cdot 13}\\ \frac{3}{17}\oplus\frac{227}{1285}=\frac{230}{1302}=\frac{5\cdot 23}{3\cdot 7 \cdot 31}$$
The second one gives your solution $(a,b,c,d)=(8,31,36,39)$, but I don't see a way to rule out finding something better other than exhaustively searching.
The third one gives the near miss $(a,b,c,d)=(10,23,31,42)$.
• I didn't know that about "best rational approximation", thanks. I will finally use $(10,13,23,32)$, because $39$ teeth was too much. And the error is still small. And I found these solutions by exhaustive search,not anything special. Thanks for the information, I will accept this answer. – MyUserIsThis Aug 6 '13 at 10:25 | 2019-12-16T10:57:32 | {
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http://math.stackexchange.com/questions/1598280/is-collapsing-considered-a-legitimate-proof | # Is collapsing considered a legitimate proof?
For example if I want to prove that $2^n - 1 = 1 + 2 + 4 + 8 +...+ 2^{n-1}$ I can obviously use induction and that is accepted. But I can also collapse it like:
To Prove $2^n = S(n)$:
1. $S(n) = (1 + 1) + 2 +...+ 2^{n-1}$
2. $S(n) = (2 + 2) + 4 + 8 +...+2^{n-1}$
3. $S(n) = (4 + 4) + 8 +...+2^{n-1}$
and so on until $S(n) = 2^{n-1} + 2^{n-1} = 2^n$
Is this method of collapsing considered a legitimate and presentable proof?
-
Actually the phrase "so on until" is a code for "induction". But yes, once you know induction, you can wrote "so on until" in place of your "induction" proof. – GEdgar Jan 3 at 13:30
Oh ok. I've self learned induction but I haven't really "studied" it as such. – Airdish Jan 3 at 13:32
Also in1. ,2. ,3. ,etc do not write $2^n$ on the LHS to start.That is what you are trying to prove. Call it S(n). Eventually reaching$S(n)=2^{n-1}+2^{n-1}=2^n$. As GEdgar points out, "eventually " or "so on until" means "by induction." – user254665 Jan 3 at 14:03
Using the idea of collapsing or anything like that should be fine (at least outside the setting of a class where they want to emphasize rigor). It's intuitive and it's obvious that the procedure works and can be continued to its end. You'd definitely find things like this in mathematical writing, since it can serve to illuminate the meaning behind proofs beyond what utterly rigorous proofs can do. – Milo Brandt Jan 3 at 16:53
@user254665 I fixed the LHS problem – Airdish Jan 4 at 3:50
Well, sort of, but in fact, writing proofs like the one you want to write is why induction exists. Whenever people say something like "and so on until", they're expressing your intuition that it's possible to continue the argument by induction. The whole point of the method of induction is to make intuitions like this one precise.
Let $S(k)=2^k + 2^k + 2^{k+1} + 2^{k+2} + ... + 2^{n-1}$. Then what we want to show is that $S(0) = 2^n$. Your proof basically amounts to saying $S(0) = S(1) = S(2) = ...$ "and so on", until we get $S(0) = S(n-1)$. Notice that $S(n-1) = 2^{n-1} + 2^{n-1}$, which obviously equals $2^n$. So we get $S(0) = 2^n$. To phrase this as a proof by induction, we're going to prove by induction that $S(0) = S(k)$ for all $k<n$, thus we'll obtain $S(0)=S(n-1)$ at the end.
Obviously, $S(0) = S(0)$. Now suppose $S(0) = S(k)$. Then:
\begin{align}S(0) &= S(k) \\&= 2^k + 2^k + 2^{k+1} + 2^{k+2} + ... + 2^{n-1} \\&= 2\cdot2^k + 2^{k+1} + 2^{k+2} + ... + 2^{n-1} \\&= 2^{k+1} + 2^{k+1} + 2^{k+2} + ... + 2^{n-1}\\&=S(k+1)\end{align}
-
I thought that induction worked on the premise that for S(n), you first take some value k, input it: S(k) and consider it to be true. Then you take S(k+1) and show that it resembles S(k)? – Airdish Jan 3 at 13:31
@S.Mo In induction we have a predicate $P(n)$, that is, a true-false statement into which we can plug $n$ like a variable, for example, an equation involving $n$. We then show that if $P(n)$ is true for any $n$, then $P(n+1)$ is true. Using phrasing like "show that it resembles" makes it seem like induction is just a method of shuffling symbols around. – Jack M Jan 3 at 14:10
I'd have thought that an induction proof for this would look more like this: – Airdish Jan 3 at 17:05
$S(n) = 1 + 1 + 2 + 4 +...2^{n-1}$ Assuming that $2^{k} = S(k) = 1 + 1 + 2 + 4 + 8 +...2^{k-1}$ $S(k+1) = (1 + 1 + 2 + 4 + 8 + ... + 2^{k-1}) + 2^k$ It follows that $S(k+1) = S(k) + 2^k$ $S(k) + 2^k = 2^k + 2^k = 2^ {k+1}$ Therefore $S(k+1) = 2^{k+1}$ Hence proved that $S(n) = 2^n$ – Airdish Jan 3 at 17:05
Please correct me if I'm wrong. – Airdish Jan 3 at 17:05
This depends very much on the degree of rigour you want.
If you want to be really rigorous, or need to, then you would need to formalize “so on” by showing by induction that $$1 + \sum_{i=0}^{n-1} 2^i = 2^k + \sum_{i=k}^{n-1} 2^i \quad \text{for all k = 0, \dotsc, n},$$ which shows the statement for $k = n$.
If you are satisfied with less rigour, and I think most (read: nearly all) mathematicians will be in a case like this, then “so on” is good enough, as the idea behind your proof is pretty clear. It is, however, important that you are able to give a formal proof.
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Of course: for every finite $n$ you have absolutely convergent series which allows you to manipulate it's memebers in a way you are doing it. Although it seems if you want to present you proof in a most formal way, induction will take up less space.
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the consensus here seems to be "yeah, the proof is ok but rather informal since it invokes induction 'coded' in the 'and-so-on' statement". I don't agree. Imho this is no proof at all and there is no induction argument given.
1. The Argument starts with the claim (short of subtracting one from both sides of the equation). Why bother and continue?
2. The modifications made in order to get to the next line are completely unclear to me. Are all the elements in the series doubled? Then the result (left side) should be doubled as well but it isn't?
-
It's true that the author shouldn't be writing $2^n=\ldots$ until the last line - however, this is not a structural error in their proof, as they do not use the assumption. It's merely a stylistic problem. The manipulation from one line to the next is that they combine the terms $1+1$ into $2$, then $2+2$ into $4$, then $4+4$ into $8$ and so on. – Milo Brandt Jan 4 at 0:32 | 2016-06-25T18:01:18 | {
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https://www.physicsforums.com/threads/relative-velocity-boat-river-question.674426/ | # Relative velocity/Boat & River Question
1. Feb 25, 2013
### Dalkiel
1. The problem statement, all variables and given/known data
280m wide river, destination 120m upstream, river current is 1.35 m/s downstream and the boat speed in still water is 2.70 m/s. What should the boat's heading angle be (relative to the shore)?
2. Relevant equations
$V_x = Vcosθ$
$V_y = Vsinθ$
$x = V_0 t$ (starting points set to 0 for x and y, no acceleration involved)
$V_{BS} = V_{BW} + V_{WS}$
3. The attempt at a solution
y component of velocity unaffected by current, so $V_{BS} = V_{BW} + V_{WS}$ will be relevent for x component only.
$V_x = Vcosθ = 2.7 cosθ$
$V_y = Vsinθ = 2.7 sinθ$
$V_{BS} = V_{BW} + V_{WS} = (2.7 cosθ) - 1.35$
$y = V_0 t = 280 = (2.7 sinθ)t$
$t = 280/(2.7 sin)$
$x = V_0 t = 120 = ((2.7 cosθ) - 1.35) t$
$t = 120/((2.7 cosθ)-1.35)$
time it takes to go across river (y component) equals time to travel upstream (x component)
$t = t = 280/(2.7 sin) = 120/((2.7 cosθ)-1.35)$
$280((2.7 cosθ) - 1.35) = 120 (2.7 sinθ)$
$(756 cosθ) - 378 = (324 sinθ)$
$(14 cosθ) - 7 = (6 sinθ)$
$7((2 cosθ) - 1) = (6 sinθ)$
$(2 cosθ) - 1 = (6/7) sinθ$
$((2 cosθ) - 1)/sinθ = 6/7$
$(2(cosθ / sinθ)) - (1/sinθ) = 6/7$
$2 cotθ - cscθ = 6/7$
I then plotted this using a graphing calculator and found where $y = 2 cot(x) - csc(x)$ crossed $y = 6/7$ and found θ = 39.5°. This answer does seem to check, but I have a feeling I'm doing something wrong, and that I should be able to solve for θ without using a graphing calculator. I'm not sure if the law of sines or law of cosines will come into play, or if I'm missing a certain trigonometric identity. Any insight would be greatly appreciated. Thanks.
Last edited: Feb 25, 2013
2. Feb 26, 2013
### ehild
You can proceed by substituting cosθ=sqrt(1-sin2θ) and solve a quadratic equation.
The other method is to rewrite the equation as A(cosβcosθ-sinβsinθ)=Acos(β+θ)=7, where Acosβ=14 and Asinβ=6: tanβ=3/7 and A2=142+62.
ehild
3. Feb 26, 2013
### Dalkiel
Thanks. I used the quadratic equation way and got the same answer as my previous attempt.
4. Feb 26, 2013
### ehild
Splendid.
ehild
5. Feb 26, 2013
### Toranc3
I have a question for this. Doesn't going upstream mean going against the water flow?
6. Feb 26, 2013
### ehild
Yes, it is against the water flow.
ehild
7. Feb 26, 2013
### Toranc3
In the picture wouldn't the 120m distance be the whole horizontal distance? I am bit confused on that.
8. Feb 26, 2013
### ehild
120 m is the distance along the river where the boat is due.
ehild
9. Feb 26, 2013
### Toranc3
Dalkiel had this: "280m wide river, destination 120m upstream, river current is 1.35 m/s downstream and the boat speed in still water is 2.70 m/s. What should the boat's heading angle be (relative to the shore)?"
In the bolded part wouldn't it make more sense if it did not say upstream and just " destination 120m" or something like that? Because the water is coming from the right side.
10. Feb 26, 2013
### ehild
120 m alone could mean both upstream and downstream, that is to the right from the starting point or to the left.
ehild
11. Feb 26, 2013
### Toranc3
Oh so even though the destination will be at an angle from starting point it still means upstream or downstream. Thanks!
12. Feb 26, 2013
### ehild
That 120 m is the distance on the other side of the river from the point opposite to the starting point.
ehild
13. Feb 26, 2013
### Toranc3
so 280 is horizontal distance then
14. Feb 26, 2013
### Dalkiel
280 is the width of the river. if there was no current, and the boat wanted to go straight across the river, it would have to travel exactly 280 m. In my (horribly drawn) picture, 280 is the vertical distance, or the y component of the displacement.
15. Feb 26, 2013
### Dalkiel
The answer that was found (39.5°) is correct, but there was actually another way to solve this, using the law of sines. I'll try to draw another picture to describe it when I get home.
16. Feb 26, 2013
### Toranc3
Thanks! I was mostly confused with the wording of the question.
17. Feb 26, 2013
### Dalkiel
Sorry about the wording, I basically just paraphrased the problem to the basic components.
Here's the solution that was explained to me, using another horribly drawn ms paint picture as a guide.
The yellow line is straight across the river, green line is where we want to go, brown is where we have to aim, dark blue is the velocity of the water. Since we know the length of two sides of the right triangle, we can find the angles. tanβ = 120/280. so β=23.2° and γ=66.8°. Since we know γ, we can find σ= 113.2°. Then use law of sines to find 2.7/sin(113.2) = 1.35/sinα. α=27.4. 90-β-α=θ=39.4°. Which matches all previous attempts.
I personally didn't care for this method because I'm mixing angles involved with distance with those using velocity. My initial intuition would be to not mix the two, but this seemed to work out.
Last edited: Feb 26, 2013
18. Feb 26, 2013
### Toranc3
Oh I see and thanks for posting this up. Do you mind posting up the whole question? :shy:
19. Feb 27, 2013
### Dalkiel
Sure, but it's actually a combination of two questions. One is direct from a textbook, and then and then there's my professor's part.
Original from text (problem #70): A boat, whose speed in still water is 2.70 m/s, must cross a 280-m-wide river and arrive at a point 120 m upstream from where it starts. To do so, the pilot must head the boat at a 45.0° upstream angle. What is the speed of the river's current?
Professor's question: In prob 70, suppose the river current is 1.35 m/s and the boat speed in still water is 2.70 m/s. What should the boat's heading angle be in that case, in order to reach the same destination? | 2017-12-18T15:40:10 | {
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https://www.physicsforums.com/threads/hyperbolic-substituion-i-am-wrong.519750/ | # Hyperbolic Substituion. I am wrong?
1. Aug 7, 2011
### flyingpig
1. The problem statement, all variables and given/known data
$$\int \;\sinh(2x) \cosh(2x) dx$$
3. The attempt at a solution
I let u = sinh(2x), du = 2cosh(2x)dx
Integrating I shuold get
$$\frac{1}{4} sinh^2 (2x) + C$$
But http://www.wolframalpha.com/input/?i=Integrate[cosh%282x%29%28sinh%282x%29%29%2Cx]
says I need to let u = 2x first. Why? I don't think i am wrong, my algebra looked correct.
2. Aug 7, 2011
### HallsofIvy
Staff Emeritus
No, you don't need to do that, you just can.
If you let u= 2x, then, of course, dx= (1/2) du so your integral becomes
$$\frac{1}{2}\int sinh(u)cosh(u)du$$
Then let v= sinh(u) so that dv= cosh(u)du and the integral becomes
$$\frac{1}{2}\int v dv= \frac{1}{4}v^2+ C= \frac{1}{4}sinh^2(u)+ C= \frac{1}{4}sinh^2(2x)+ C$$
just as you have.
That's just using two consecutive substitutions rather than combining them as you did.
Of course, you could do it the other way two: let u= cosh(2x) so that du= 2sinh(x)dx and the integral becomes
$$\frac{1}{2}u^2du= \frac{1}{4}cosh^2(2x)+ D$$
Because $cosh^2(2x)= sinh^2(2x)+ 1$ that is exactly the same thing with a difference constant.
I believe your wolframalpha reference gives the integral as
$$\frac{1}{8}cosh(4x)+ Constant$$
which looks very different but is actually the same as your result.
Remember that
$$sinh(x)= \frac{e^x- e^{-x}}{2}$$
Squaring that:
$$sinh^2(x)= \frac{(e^x)^2- 2e^xe^{-x}+ (e^{-x})^2}{4}= \frac{e^{2x}- 2+ e^{-2x}}{4}$$
$$= \frac{1}{2}\frac{e^{2x}+ e^{-2x}}{2}- \frac{1}{2}= \frac{1}{2}cosh(2x)- \frac{1}{2}$$
so that
$$\frac{1}{4}sinh^2(2x)= \frac{1}{8}cosh(4x)- \frac{1}{8}$$
differing from the original form only by a constant.
Last edited: Aug 7, 2011
3. Aug 7, 2011
### flyingpig
Oh right, we are doing an indefinite integral here. | 2017-09-26T07:57:20 | {
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https://math.stackexchange.com/questions/2618271/a-faster-way-to-evaluate-int-1-infty-frac-sqrt4t2t3-mathrm-dt | # A faster way to evaluate $\int_1^\infty\frac{\sqrt{4+t^2}}{t^3}\,\mathrm dt$?
I need to evaluate the integral
$$\int_1^\infty\frac{\sqrt{4+t^2}}{t^3}\,\mathrm dt\tag1$$
After some workarounds I found the change of variable $t=2\sqrt{x^2-1}$, then
$$\int_1^\infty\frac{2\sqrt{1+(t/2)^2}}{t^3}\,\mathrm dt=\frac12\int_{\sqrt5/2}^\infty\frac{x^2}{(x^2-1)^2}\,\mathrm dx\\=\frac12\left[\frac{x}{2(1-x^2)}\bigg|_{\sqrt5/2}^\infty+\frac12\int_{\sqrt5/2}^\infty\frac{\mathrm dx}{x^2-1}\right]\\=\frac{\sqrt5}2+\frac18\int_{\sqrt5/2}^\infty\left(\frac1{x-1}-\frac1{x+1}\right)\,\mathrm dx\\=\frac{\sqrt5}2+\frac18\ln\left(\frac{\sqrt 5+2}{\sqrt 5-2}\right)\\=\frac{\sqrt5}2+\frac14\ln(\sqrt5+2)$$
But my intuition says that it must exists a more straightforward way to evaluate this integral. In fact, using Wolfram Mathematica, I get the equivalent[*] result
$$\int_1^\infty\frac{\sqrt{4+t^2}}{t^3}\,\mathrm dt=\frac14(2\sqrt5+\operatorname{arsinh}(2))$$
[*] The equivalence can be seen from
$$\operatorname{arsinh}(x)=\ln(x+\sqrt{1+x^2})$$
My question: someone knows a faster way to evaluate manually this integral? Maybe a better change of variable?
• begin with $t=2 \sinh w$ – Will Jagy Jan 23 '18 at 21:48
• @WillJagy I did it, but this doesnt seem a big improvement. You find from it something like $\int \frac1{\sinh x\tanh^2 x}=\int(\sinh^{-3}x+\sinh^{-1}x)$ – Masacroso Jan 23 '18 at 22:23
• en.wikipedia.org/wiki/… – Will Jagy Jan 23 '18 at 22:44
## 2 Answers
I am very fond of hyperbolic functions for integrals. If we begin with your $t = 2 \sinh x,$ we expect to get to something consistent with the wikipedia way of writing the Weierstrass substitution for hyperbolic functions, give me a few more minutes.
$$\int \frac{\cosh^2 x}{ 2 \sinh^3 x} dx.$$ Then let us use a letter different from your $t,$ $$\sinh x = \frac{2u}{1 - u^2}, \; \; \; \frac{1}{\sinh x} = \frac{1 - u^2}{2u}$$ $$\cosh x = \frac{1 + u^2}{1 - u^2},$$ $$d x = \frac{2du}{1 - u^2} \; .$$ $$\int \frac{(1 + u^2)^2 (1 - u^2)^3 2 du}{2 (1 - u^2)^2 (2u)^3 (1-u^2)}$$ $$\int \frac{(1 + u^2)^2 du}{ (2u)^3}$$ $$\int \frac{1 + 2u^2 + u^4 }{ 8u^3} \; du$$ $$\int \frac{1}{8u^3} + \frac{1}{4u} + \frac{u}{8} \; \; du$$
They give more detail here, and we do need an expression for our $u$ That comes out $$u = \tanh \frac{1}{2} x = \frac{\sinh x}{\cosh x + 1} = \frac{\cosh x - 1}{\sinh x}$$
Enforcing a substitution of $x \mapsto 1/x$ we have $$I = \int_0^1 \sqrt{4 x^2 + 1} \, dx.$$ This integral can be readily found using a hyperbolic substitution of $x \mapsto \frac{1}{2} \sinh x$. Doing so yields \begin{align*} I &= \frac{1}{2} \int_0^{\sinh^{-1} (2)} \cosh^2 x \, dx\\ &= \frac{1}{4} \int_0^{\sinh^{-1} (2)} [\cosh (2x) + 1] \, du \tag1\\ &= \frac{1}{4} \left [\frac{1}{2} \sinh (2x) + x \right ]_0^{\sinh^{-1} (2)}\\ &= \frac{1}{8} \sinh [2\sinh^{-1} (2)] + \frac{1}{4} \sinh^{-1} (2)\\ &= \frac{\sqrt{5}}{2} + \frac{1}{4} \sinh^{-1} (2) \tag2 \end{align*}
Explanation
(1) Using $\cosh^2 x = \frac{1}{2} (\cosh (2x) + 1)$
(2) Using $\sinh (2 \alpha) = 2 \sinh \alpha \cosh \alpha$ where $\alpha = \sinh^{-1} (2)$ such that $\sinh \alpha = 2$ and $\cosh \alpha = \sqrt{5}$
• very nice. .... – Will Jagy Jan 23 '18 at 23:19 | 2020-09-28T03:33:39 | {
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https://mathhelpboards.com/threads/is-g-open-in-r-2.27762/ | Is G open in R^2
G-X
New member
The set G = {$$\displaystyle (x, y) \in R^{2} : (x, y) \neq (1, 0)$$} is an open set in $$\displaystyle R^{2}$$
Proof 2: We know a set X is defined to be closed if and only if its complement is open. Let $$\displaystyle J$$ = $$\displaystyle R^{2} / G$$ which implies $$\displaystyle J$$ is a single point (1, 0) in $$\displaystyle R^{2}$$. We can prove $$\displaystyle G$$ is open if we prove J is closed.
We can use the definition $$\displaystyle \bar A = A \cup A'$$ where $$\displaystyle \bar A$$ is the closure of $$\displaystyle A$$ and $$\displaystyle A'$$ is the set of all limit points of $$\displaystyle A$$. This means that the closure of a set $$\displaystyle A$$ is the union of $$\displaystyle A$$ with its boundary points. The closure of a set $$\displaystyle A$$ is, also, the smallest closed set containing $$\displaystyle A$$.
This implies that $$\displaystyle \bar A$$ is closed and we can show J is closed by proving that it is equal to its closure. Let us set $$\displaystyle \bar J = J \cup J'$$. By definition of a limit point, let $$\displaystyle b \in S$$. A point $$\displaystyle x \in S$$, $$\displaystyle x \neq b$$ is a limit point of $$\displaystyle b$$ if and only if every open neighborhood of $$\displaystyle x$$ contains $$\displaystyle b$$. Except $$\displaystyle J$$ is a singular point and has no other elements $$\displaystyle x$$. This implies that $$\displaystyle J$$ has no limit points and $$\displaystyle J' = \{\}$$, the null set.
This means that $$\displaystyle \bar J = J \cup J' = J \cup \{\} = J$$. We now know that J is equal to its closure and thus J is closed. In turn, this means that G is open.
Staff member
Looks correct.
Country Boy
Well-known member
MHB Math Helper
That is correct but I would have been inclined to a direct proof.
Let (x, y) be a point in the set G. Then x and y are not both 0 so the distance from (x, y) to (0, 0), $\sqrt{x^2+ y^2}= \delta$ is positive. So $\frac{\delta}{2}$ is also positive. The points in the neighborhood, $N\left((x,y), \frac{\delta}{2}\right)$, have distance from (0, 0) at least $\delta- \frac{\delta}{2}= \frac{\delta}{2}$ so the neighborhood is a subset of G. That is, given any point in G there exist a neighborhood of that point that is a subset of G. Therefore, G is open.
caffeinemachine
Well-known member
MHB Math Scholar
I just want to add one other approach, which I believe is worth learning and it is very easy to apply otherwise. Define a map $f:R^2\to R$ by writing $f(x, y) = (x-1)^2 + y^2$. This map takes the value $0$ only at $(x, y) = (1, 0)$. Thus $G=f^{-1}(R\setminus \{0\})$. Since $f$ is continuous, it suffices to show that $R\setminus \{0\}$ is open. This is easy to show, and certainly easier that showing $G$ is open. | 2020-08-14T10:20:41 | {
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https://brilliant.org/discussions/thread/topics-in-number-theory-some-problems-involving-co/ | # Topics in Number Theory: Some problems involving congruence relations (Part I)
For this note I assume that you know the basics properties of divisibility including those of the congruence relation $$a\equiv b \pmod{c}$$. See for example Modular arithmetics
I will show some nice problems involving the congruence relation, most of them are very useful in problem solving, thus we can consider this problems as lemmas.
Problem 1. If $$\gcd(a,n)=1$$ then there exists an integer $$b$$ such that $$ab \equiv 1 \pmod{n}$$. $$b$$ is called the multiplicative inverse of $$a$$ modulo $$n$$.
Solution. Consider the numbers $$a\cdot 1, a\cdot 2, a\cdot 3, \ldots, a\cdot n$$. If $$a\cdot i\equiv a\cdot j \pmod{n}$$ for $$1\leq i,j\leq n$$, then $$a\cdot(i-j)\equiv 0\pmod{n}$$, since $$\gcd(a,n)=1$$ we conclude that $$(i-j)\equiv 0 \pmod{n}$$ and then $$i=j$$. Thus the numbers $$a\cdot 1, a\cdot 2, a\cdot 3, \ldots, a\cdot n$$ are pairwise distinct modulo $$n$$, since this set consists of $$n$$ numbers then it is a complete residue system modulo $$n$$. In particular, for some $$1\leq b\leq n$$ we have $$a\cdot b\equiv 1\pmod{n}$$.
Problem 2. Let $$n>1$$ be an integer:
• If $$n$$ is prime then $$(n-1)!\equiv -1 \pmod{n}$$. [Wilson's Theorem]
• If $$n$$ is composite and $$n\neq 4$$ then $$(n-1)!\equiv 0 \pmod{n}$$.
Hint for the first part. If $$n\geq 2$$ is an odd prime, prove that the set $$\{2, 3, \ldots, n-2\}$$ can be divided in pairs $$\{a,a'\}$$ where $$a$$ and $$a'$$ are inverse each other.
Problem 3. If $$p$$ is a prime and $$1\leq k<p$$ then $${p \choose k}\equiv 0 \pmod{p}$$.
Solution. We know that $${p \choose k}=\frac{p!}{k!(p-k)!}$$ then $${p \choose k}\cdot k!\cdot (p-k)!=p!$$. The number $$p!$$ is divisible by $$p$$ while $$k!\cdot (p-k)!$$ is not, then $${p \choose k}$$ is divisible by $$p$$.
Problem 4. If $$p$$ is a prime then $$(a+b)^p\equiv a^p+b^p \pmod{p}$$.
Hint. Use the previous problem.
Now, you can practice with the following problems. To avoid confusion I will call these problems N1, N2, etc.
Problem N1. ¿Does there exist a positive integer $$n$$ for wich the set $$\{n, n+1, \ldots, n+17\}$$ can be partitioned in two subsets $$\mathcal{A}$$ and $$\mathcal{B}$$ such that the product of the elements of $$\mathcal{A}$$ is equal to the product of the elements of $$\mathcal{B}$$?
Problem N2. Let $$p\geq3$$ be a prime. For each $$i = 1, 2, \ldots, p-1$$ denote by $$r_i$$ the remainder when $$i^p$$ is divided by $$p^2$$, thus $$0\leq r_i<p^2$$. Calculate $$r_1+r_2+\cdots+r_{p-1}$$.
Note by Jorge Tipe
4 years, 5 months ago
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## Problem N1:
We claim the answer is no.
Lemma: No element in the set can be a multiple of 19.
Proof: Assume that an element in the set is divisible by 19. Then, since the set is composed of 18 consecutive integers, there can be no other element in the set that is divisible by 19. However, then the product of the elements in the two sets cannot be equal, as one has a factor of 19, while the other doesn't, a contradiction. $$\square$$
Since the set is composed of 18 consecutive integers, none of which is divisible by 19, the set must be equivalent to the set $$\{1,2,\cdots,18\}$$ when reduced modulo 19. If this set can be partitioned into two subsets with the same product, the product of the elements must be a square. By Wilson's theorem, the product of the elements is congruent to $$-1$$ modulo 19. However, it is well-known that $$-1$$ is a quadratic residue modulo an odd prime $$p$$ if and only if that prime is $$1$$ modulo 4. Since $$19\equiv 3\pmod 4$$, we conclude that no such integer $$n$$ exists. $$\blacksquare$$
## Problem N2:
We claim the answer is $$\dfrac{p^2(p-1)}{2}$$.
Lemma: $$i^p+(p-i)^p\equiv 0\pmod{p^2}$$, and $$r_i+r_{p-i}=p^2$$.
Proof: By the binomial theorem, \begin{align} i^p+(p-i)^p&\equiv i^p+p^p-\dbinom{p}{1}p^{p-1}i+\cdots-\dbinom{p}{p-2}p^2i^{p-2}+\dbinom{p}{p-1}pi^{p-1}-i^p\pmod{p^2} \\ &\equiv i^p+0-0+\cdots-0+\dbinom{p}{p-1}pi^{p-1}-i^p\pmod{p^2} \\ &\equiv i^p+p^2i^{p-1}-i^p\pmod{p^2} \\ &\equiv 0\pmod{p^2} \end{align} However, also note that, for $$1\le i\le p-1$$, $$p^2\nmid i^p$$. Since $$0<r_i<p^2$$, $$0<r_{p-i}<p^2$$, and $$r_i+r_{p-i}\equiv 0\pmod{p^2}$$, we must have that $$r_i+r_{p-i}=p^2$$, as desired. $$\square$$
By the lemma, and since $$p-1$$ is even, the set of integers $$r_1,r_2,\cdots,r_{p-1}$$ can be paired, $$r_i$$ with $$r_{p-i}$$, such that the sum of each pair is $$p^2$$. The total number of pairs is $$\dfrac{p-1}{2}$$, and so we conclude $$r_1+r_2+\cdots+r_{p-1}=\dfrac{p^2(p-1)}{2}$$. $$\blacksquare$$
- 4 years, 5 months ago
Awesome! Really nice job
- 4 years, 5 months ago
Since Problem 4 has no proof, I will write one, though the hint really gives it away..
By the binomial theorem we have that, $$(a + b)^p = \sum\limits^p_{k=0} \binom{p}{k} a^k b^{p - k}$$
By Problem 3, we know that $$1 \leq k < p \Rightarrow \binom{p}{k} \equiv 0 \pmod p$$, and so all of the summands except for the first and last ones are zeroed. We are then left with,
$$(a + b)^p \equiv \binom{p}{0} a^p + \binom{p}{p} b^p \equiv a^p + b^p \pmod p\ \quad \square$$
- 4 years, 5 months ago
Excellent!!
- 4 years, 5 months ago | 2018-06-19T02:56:19 | {
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https://www.questionsolutions.com/balloon-subjected-net-uplift/ | # If the balloon is subjected to a net uplift 5
If the balloon is subjected to a net uplift force of F = 800 N, determine the tension developed in ropes AB, AC, AD.
Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.
#### Solution:
We will first write each tension developed in Cartesian vector form. To do so, we need to write the locations of points A, B, C, and D in Cartesian vector form.
From the diagram, the locations of the points are:
$A:(0i+0j+6k)$ m
$B:(-1.5i-2j+0k)$ m
$C:(2i-3j+0k)$ m
$D:(0i+2.5j+0k)$ m
The position vectors for each rope are:
$r_{AB}\,=\,\left\{(-1.5-0)i+(-2-0)j+(0-6)k\right\}\,=\,\left\{-1.5i-2j-6j\right\}$
$r_{AC}\,=\,\left\{(2-0)i+(-3-0)j+(0-6)k\right\}\,=\,\left\{2i-3j-6j\right\}$
$r_{AD}\,=\,\left\{(0-0)i+(2.5-0)j+(0-6)k\right\}\,=\,\left\{0i+2.5j-6j\right\}$
A position vector, denoted $\mathbf{r}$ is a vector beginning from one point and extending to another point. It is calculated by subtracting the corresponding vector coordinates of one point from the other. If the coordinates of point A was $(x_A,y_A,z_A)$ and the coordinates of point B was$(x_B,y_B,z_B)$, then $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$
The magnitude of each position vector is:
magnitude of $r_{AB}\,=\,\sqrt{(-1.5)^2+(-2)^2+(-6)^2}\,=\,6.5$
magnitude of $r_{AC}\,=\,\sqrt{(2)^2+(-3)^2+(-6)^2}\,=\,7$
magnitude of $r_{AD}\,=\,\sqrt{(0)^2+(2.5)^2+(-6)^2}\,=\,6.5$
The magnitude is equal to the square root of the sum of the squares of the vector. If the position vector was $r\,=\,ai+bj+ck$, then the magnitude would be, $r_{magnitude}\,=\,\sqrt{(a^2)+(b^2)+(c^2)}$. In the simplest sense, you take each term of a vector, square it, add it together, and then take the square root of that value.
The unit vectors are:
$u_{AB}\,=\,\left(-\dfrac{1.5}{6.5}i-\dfrac{2}{6.5}j-\dfrac{6}{6.5}k\right)$
$u_{AC}\,=\,\left(\dfrac{2}{7}i+\dfrac{-3}{7}j-\dfrac{6}{7}k\right)$
$u_{AD}\,=\,\left(0i+\dfrac{2.5}{6.5}j-\dfrac{6}{6.5}k\right)$
The unit vector is each corresponding unit of the position vector divided by the magnitude of the position vector. If the position vector was $r\,=\,ai+bj+ck$, then unit vector, $u\,=\,\dfrac{a}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{b}{\sqrt{(a^2)+(b^2)+(c^2)}}+\dfrac{c}{\sqrt{(a^2)+(b^2)+(c^2)}}$
We can now express each force in Cartesian vector form:
$F_{AB}\,=\,F_{AB}\left(-\dfrac{1.5}{6.5}i-\dfrac{2}{6.5}j-\dfrac{6}{6.5}k\right)$
$F_{AC}\,=\,F_{AC}\left(\dfrac{2}{7}i-\dfrac{3}{7}j-\dfrac{6}{7}k\right)$
$F_{AD}\,=\,F_{AD}\left(0i+\dfrac{2.5}{6.5}j-\dfrac{6}{6.5}k\right)$
(further simplify this by expanding the brackets using FOIL and simplifying the fractions into decimal values)
$F_{AB}\,=\,\left\{-0.231F_{AB}i-0.308F_{AB}j-0.923F_{AB}k\right\}$
$F_{AC}\,=\,\left\{0.286F_{AC}i-0.429F_{AC}j-0.857F_{AC}k\right\}$
$F_{AD}\,=\,\left\{0i+0.385F_{AD}j-0.923F_{AD}k\right\}$
$F\,=\,\left\{0i+0j+800k\right\}$
(Force F is the net uplift force, which is applied directly upwards, thus it only has a z-component)
We can now write our equations of equilibrium. All forces added together must equal zero.
$\sum \text{F}\,=\,0$
$F_{AB}+F_{AC}+F_{AD}+F\,=\,0$
Since all forces added together must equal zero, then all individual components (x, y, z-components) added together must also equal zero.
x-components:
$-0.231F_{AB}+0.286F_{AC}\,=\,0$
y-components:
$-0.308F_{AB}-0.429F_{AC}+0.385F_{AD}\,=\,0$
z-components:
$-0.923F_{AB}-0.857F_{AC}-0.923F_{AD}+800\,=\,0$
Solving the three equations simultaneously gives us:
$F_{AB}\,=\,251.2$ N
$F_{AC}\,=\,202.9$ N
$F_{AD}\,=\,427.1$ N
## 5 thoughts on “If the balloon is subjected to a net uplift”
• questionsolutions Post author
Where is it wrong? Just saying solution is wrong doesn’t help anyone, please point out what value is wrong, it’s easy to miss when checking so it would be very helpful if you can point out the error. I checked through and none of the signs seem to be incorrect.
• Joris De Dier
I think that the Y component of C should be -3 instead of +3 | 2023-03-30T21:30:31 | {
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https://libstdc.com/us/q/puzzling/1957 | ### Puzzle of putting numbers 1-9 in 3x3 Grid to add up to 15
• In a 3x3 grid, I'd have to put numbers from 1 to 9 in a manner so that respective row, column and diagonal add up to 15.
I have only been able to come up with one solution:
$$\begin{array}{ccc} 6 & 1 & 8 \\ 7 & 5 & 3 \\ 2 & 9 & 4 \end{array}$$
Through some calculations and trial and error method.
Is there any strategy or way of approach to this problem, or is trial and error method the solution to it?
These are called "magic squares".
For the case of a three-by-three magic square, there _is_ only one solution.
This question might just have the lowest "views:upvotes" ratio I have ever seen (2:56585)
@ghosts_in_the_code For a non-zero vote count, certainly. How the heck did this get so many views? Did it make Hot Network Questions for a year?!
6 years ago
There is a general, very simple, algorithm for generating any magic square which has an odd number of rows/columns as follows:
1. Start in the middle of the top row and enter 1.
2. Move Up 1 and Right 1 wrapping both vertically and horizontally when you leave the grid *(see note below).
3. If that square is empty enter the next number; if the square is not empty put the next number underneath the last number you entered.
4. Repeat 2&3 until the grid is complete. All rows, columns and the two diagonals will sum to the same value.
Here is the 5x5 square:
$$\begin{array}{ccccc} 17& 24& 1& 8& 15 \\ 23& 5& 7& 14& 16 \\ 4& 6& 13& 20& 22 \\ 10& 12& 19& 21& 3 \\ 11& 18& 25& 2& 9 \end{array}$$
As the square square is symmetric there are eight symmetries. You can also get these symmetries by a simple variation of the start square and direction used in step 2.
*Note: So as you are in the middle of the top row on the first move you want to place the next number in the next column of the row above. The row above does not exist so move to the last row of the square in the same column. If you were in the last column you would move to the first column. If you look at the example of the 5x5 at number 15. The next position is the square up and to the right of 15 which wraps on both the row and column to point to the lower right square which has 11 in it. As that square is not empty we placed 16 underneath 15.
Thank you @ Alan for explaining it so well and generalising it. Does this method have a name? It helped me very much. :)
@Freya No problem, unfortunately I do not know if this method has a name, my father taught me this over 35 years ago and it all came back when I saw your question :). There should be an algorithm for magic square with an even number of rows/columns somewhere.
@ Alan I see. It is a very nice way and I am glad to have been taught this. :)
This link me be of use for even squares; http://m.wikihow.com/Solve-a-Magic-Square (@Freya and answerer)
We call this as rolling numbers. as we have to roll the paper edges when we want to move up and left | 2021-06-15T06:32:47 | {
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https://www.themathdoctors.org/one-team-two-teams-my-team-your-team/ | # One Team, Two Teams, My Team, Your Team
#### (A new question of the week)
Counting ways to select teams can be simple, or quite complex. Here we’ll look at a few tricky examples.
The question came in early April:
Dear Ask Dr Math,
I hope you can help me with the question below:
Four friends attend different schools, and each brings two classmates to a quiz evening. The quiz is for teams of 6, so the group make up two teams.
i) How many different pairs of teams are possible if students at the same school are always on the same team?
ii) How many different pairs of teams are possible if the four friends are all on one team?
iii) How many different pairs of teams are possible if each team includes at least one student from each school?
My answers for all 3 parts of the question are:
i) 12
ii) 56
iii) 1152
Questions on permutation and combinations always baffle me especially if the restriction is too complicated. Thank you for helping me out and it is very much appreciated.
We might picture these problems this way, with three (numbered) students from each of four (lettered) schools being assigned to two teams:
Each problem has a different restriction on the teams.
Hi, Zack.
This is an interesting problem, and we can have a good discussion of it. In order to do that, I’d like to see not just your answers, but your reasoning (the calculation, and also the reason for the calculation). Just telling you that 12 is wrong would be a lot less helpful than talking about how you can modify your work to correct it. Very likely you would just need a nudge in the right direction.
Here’s one hint that may help: Choosing two teams of 6 is identical to choosing one team of 6, because everyone not chosen is on the other team. Similarly, in the first question, choosing a team of 6 is identical to choosing two schools. Do you see why? This sort of thinking (thinking about how the choice might be made, and reframing it in simpler terms) is central to my approach to such problems. You can see some of this in my post, Combinatorics: Multiple Methods, Subtle Wording, which includes problems similar to yours.
I’ll tell you that, although such problems rarely baffle me, they do often leave me a little uncertain! What convinces me I’m right (or reveals an error) is going through my reasoning, and if possible looking for an alternative method that gives the same result. (And one reason I’m not just telling you whether you are right is that seeing your thinking might correct my own, if I think too hastily and get it wrong! But I do think 12 is wrong …)
## Problem 1: schoolmates on the same team
We will, indeed, run across an error on my part, justifying my caution! Zack wrote back:
For the first part of the question, my answer is 12 because the first team is formed with 4C2 = 6 and the second team is 2C2 = 1. And then I was considering that the first team can be 2C2 and the second team can be 4C2. Hence, I come to a conclusion that it will be 4C2 × 2C2 × 2 = 12. Is my line of thinking wrong? I applied the same logic to the other parts of the question.
We have to choose 2 of the four schools to go in the first team, and then the other two schools go in the second team:
I replied:
Thanks.
You started out right: there are 4C2 = 6 ways to choose the first team (2 of 4 schools), and then only one way to choose the second team, since it consists of the unchosen schools. As I suspected, you saw for yourself the two ideas I suggested.
But you don’t want to double that to get the final answer. What I think you mean is that either of the two teams chosen might be the first, so you double the number. But it actually works the other way, since (as discussed in the post I referred to, near the end) choosing two teams ignores order; there is no first or second teams, just two teams. So rather than double the number, we have to halve it. To put it another way, a choice of two teams corresponds to two different orders in which they might have been chosen, so we divide by 2.
So I’m claiming the answer is 3, not 6 or 12! (I’ll admit that my initial mental answer was 6, forgetting about this issue entirely. As I said, it’s easy to make mistakes, and can’t trust myself when I am hasty.)
Whereas Zack doubled the number because there are two ways to choose the same pair of teams (either of them being first), this actually means that the calculation has already counted the same pair twice, which we need to undo by dividing by 2.
So, is there another way? Since the number is so small, we can just list pairs of teams.
Calling the schools A, B, C, and D, here they are:
AB vs CD
AC vs BD
Our way of counting 4C2 would result in these additional pairings:
CD vs AB
BD vs AC
But as you can see, these are the same (unordered) pairs as the first three. I hope that illustrates why we divide by 2 rather than multiplying by 2. And if you think a moment, you’ll see that this is the same reasoning we use in developing the formula for combinations: We divide the number of permutations by the number of those permutations that constitute the same combination.
I haven’t yet looked at the other questions. Redo your answers to those (if necessary) and show me your reasoning, and I’ll have more to say.
Here are the three possible pairings:
For a larger problem, we couldn’t have just listed choices to check our count; this is one reason I often check my reasoning by trying a smaller version of the same problem. Imagine if there were 50 schools competing …
## Problem 2: friends on the same team
Zack replied:
Thanks, Dr Peterson,
Your explanation and further reads give me more insight into understanding this question.
As for the second part of the question since it is now dealing with the students rather than schools the problem should be solved as shown below:
All four friends together in a team; 8C2 = 28 ways
He continued with an answer to the third part, but first I’ll show my response to the second part:
Correct. This is a much easier problem, isn’t it?
In particular, I would explain what you’ve done in this way:
Since one particular team can be singled out as “the one with the four friends”, making the teams no longer indistinguishable, we need only to count ways to fill out the rest of this one team, and do not need to divide by 2 in the end. There are 8 other students, from which we choose 2 in addition to the four friends, so there are 8C2 = 28 ways.
We could instead choose the 6 for the other team, putting the remaining 2 on the team with the friends; since 8C6 = 8C2, this gives the same result.
We have either two slots to fill from the 8 remaining students, or six slots (on the other team):
## Problem 3: every school on each team
Here is his answer to the third part:
Meanwhile, for the third part, I am solving it this way:
Since we want at least one student from each school, Team 1 will be (3C1 × 4) × 8C2 and Team 2 will take the remaining students. Hence, we have 336 ways. But this includes the combinations with the second team that might not have all students from all 4 schools.
So we subtract by (3C1 × 4) × 4C1 = 48. Thus my final answer is 336 – 48 = 288 ways.
What do you think about my solutions and answers?
### A subtractive method
My response continued:
This problem is much more complicated, and this is a good attempt, though flawed. You’ve done it a different way than I did, and got a different answer. So let’s see who’s right! We won’t want to check by listing this time, so we’ll just compare our answers.
You’ve used a subtractive method, which can be good. (I used addition.) You focus on one team (which suggests we may have to divide by 2 at the end, since the teams are indistinguishable). But you haven’t explained the details of your (3C1 × 4) × 8C2. I think you are saying that for each of the 4 schools, you are picking 1 student, and then picking 2 more from the remaining 8. There are a couple problems here.
He’s starting out like this:
First, if you pick one of three 4 times, you should be raising to the 4th power, not multiplying by 4! There are far more than 12 ways to choose one from each of 4 schools. (But you are exactly right to deal with the possibility that both of the last two choices are from the same school, leaving none from that school to be on the other team. Good thinking there.)
For that first step, there are 3 choices from school A, 3 from school B, 3 from school C, and 3 from school D, for a total of $$3^4 = 81$$ ways to get as far as the picture shows. That’s a minor correction, and doesn’t invalidate the method he is trying.
As I said, Zack rightly observed that, using this approach, he will accidentally include teams like this, which are illegal because the second team doesn’t have one from every school (in this case, no A’s):
So his plan is to count these and subtract them. But there’s another problem:
Second, you will be overcounting, because you will get the same result if you take, say, person A1 among the first four and then pick A2 as one of the extra two, or vice versa. This is a common flaw in this sort of thinking, and sometimes very hard to overcome. You can read about this sort of issue in the post Permutations and Combinations: Undercounts and Overcounts.
For example, these would be counted as different pairs of teams, but they are really the same, just arranged differently:
(I just swapped the positions of A1 and A2 within the same team, and order doesn’t count within a team.)
In order to cancel out this effect, we would have to divide by $$2^2=4$$, the number of ways to have chosen, for each of the two schools that appear twice in the first team, which student was chosen in the “first round”. He’s going to abandon this method, so let’s fix it up here:
First, the number of ways to choose one team under these conditions, as we’ve seen, is $${3\choose1}^4\times{8\choose2}\times\frac{1}{4}=81\times28\times\frac{1}{4}=567$$
Then, the number of these for which the “extra two” are from the same school is (since we just have to choose one school from which to take the remaining two) is $${3\choose1}^4\times{4\choose1}\times\frac{1}{4}=81\times4\times\frac{1}{4}=81$$
Subtracting, we get $$567-81=486$$. And dividing by 2 because, we’ve counted every pair of teams twice, we’re left with $$243$$. Putting it together, here is the answer: $$\frac{{3\choose1}^4\times{8\choose2}\times\frac{1}{4}-{3\choose1}^4\times{4\choose1}\times\frac{1}{4}}{2}=243$$
(In a larger problem, we wouldn’t have been able to just divide by 4, because there would be different overcounts in different cases.)
### An additive method (not)
Clearly you are capable of thinking through these questions, so I’m not going to show you my answer yet. I will tell you the basics of my approach, however; I don’t know how much it will take to rescue your current approach, and this may help if you either give up on that or want to try a second way.
What I did was to split into two cases. If there is at least one student from each school on each team, then there must be either 1 or 2 from each school on this team. So either there are 3 from one school and 1 from each other (which I describe as AAABCD), or there are 2 each from 2 schools and 1 from each of the others (AABBCD). I found the two counts and added them. There’s a lot more to say, but that’s how I began my work.
But please do continue with your subtractive approach, because that can be a great alternative (for some problems, at least), and it may be possible here.
Do you see a problem in my AAABCD case? I said there must be 1 or 2 from each school on this team, then I put 3 on the team for one case, which leaves none on the other team. There is really only one case, AABBCD!
It took me a while to realize this; I’m going to remove mentions of AAABCD going forward, as they were a dead end.
Here is what I said when I finally realized there was only one case:
We have to have either 1 or 2 from each school on our first team. So our list of cases will consist of every way to add four numbers, each 1 or 2, and get 6. (I’ll do as I’ve done so far, and arrange those numbers in decreasing order, and then count ways to assign schools to the numbers.) We can have 2+2+1+1 or … nothing else. So my AABBCD is the only case. And you’ve almost got that right.
Zack replied:
Do I calculate AABBCD this way?
(4C2/2) × (3C2 × 3C2) × ( 3C1 × 3C1) = 243
I select 2 students from 2 schools (3C2 × 3C2). Since there are 4 schools I multiplied by (4C2). Since my approach will cause duplicates, then I need to divide by 2. I further multiplied by (3C1 × 3C1) to fill up another 2 spots.
He got it! (I would have divided by 2 last, but it makes no difference.)
After clearing up the damage from my earlier mistake, I recapped the method:
Our pair of teams will be AABBCD vs ABCCDD. We’ll have to divide by 2 at the end because every such arrangement can be made in two ways, swapping the two teams. (For instance, this one is equivalent to DDCCBA vs DCBBAA.)
We choose two schools to have two students on the first team, in 4C2 = 6 ways. Then we choose members from the four schools in 3C2*3C2*3C1*3C1 = 3*3*3*3 ways. This makes a total of 6*3^4 = 486 ways; but, again, we divide by 2 and get an answer of 243.
I’ll be looking for a different approach to see if I can get the same answer; as I often say, in this field, I don’t trust an answer until I can get it two ways.
And now we do have it two ways. But there’s more:
### A different subtractive method
In the midst of the discussion, he had said this:
I tried to further improve my ‘subtraction’ approach by considering the selection without restriction which is 12C6 = 924. Followed by subtraction by the following cases:
AAABBB × CCCDDD = 3 ways
AAABCD × BBCCDD = 333 ways (from the above results)
AABBCD × CCDDAB = 243 ways (from the above results)
I think I have considered all cases. So the number of ways would be 924 – 3 – 333 – 243 = 345 ways.
To that, I replied:
This is a different subtraction than you tried before; I haven’t tried to see whether that could be made to work. This time you are starting from all possible pairs of teams and trying to subtract every arrangement that doesn’t have at least one from each school on each team. I like that you are considering both teams this time; but I’d probably avoid this because I expect it to be hard to be sure I’ve covered all cases. For instance, you don’t have AAABBC here; and you included AABBCD, which is not invalid.
Let’s try to correct that.
The correct list of cases that violate the requirement that each team has one from every school consists of all ways to choose 3 from at least one school (so that the other team has none), or 2 from each of three schools (so that the other school has none). This comes out to, in our notation,
1. AAABBB vs CCCDDD
2. AAABBC vs BCCDDD
3. AAABCD vs BBCCDD
4. AABBCC vs ABCDDD
(Observe that cases 1 and 2 are both symmetrical, in the sense that they have the same form if we swap the teams, while cases 3 and 4 are reflections of one another. I will not use these ideas. I will count everything in such a way that we will need to divide our final count by 2.)
To count case 1, we merely choose two schools to be all present on the first team: $${4\choose2}=6$$
To count case 2, we choose one school each to contribute 3, 2, 1 students to the first team, then choose the 2 and the 1: $$_4P_2{3\choose2}{3\choose1}=216$$
To count case 3, we choose one school to be all present, and one each from the other schools: $${4\choose1}{3\choose1}{3\choose1}{3\choose1}=108$$
To count case 4, we choose three schools to contribute 2 each, then choose those 2’s: $${4\choose3}{3\choose2}{3\choose2}{3\choose2}=108$$
(I’ve already explained why the last two are equal.)
Now, from the total number of ways to choose two distinguishable teams, $${12\choose6}=924$$, we subtract the other numbers: $$924-(6+216+108+108)=486$$
Finally, since the teams are not distinguishable, we divide by 2, and get the same answer we got before, $$243$$.
We’ve seen three different ways to get the same answer, which is not unusual in combinatorics!
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2023-02-06T00:10:36 | {
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https://math.stackexchange.com/questions/2960421/how-can-i-get-maximum-number-of-vertices-if-i-already-know-edges | # How can I get maximum number of vertices if I already know edges
If I already know edges how can I get the maximum number of vertices?
Question: There is a graph that has $$36$$ edges, and where every vertex has degree at least $$5$$. What is the maximum number of vertices this graph could have?
I think the sum of degrees is $$36\cdot 2$$ which is $$72$$. And the sum of degrees is bigger or equal than adding the least degree of every vertices together.
Therefore, $$72\geq 5n$$ and then $$14.4\geq n$$, so the maximum number of vertices is $$14$$. Is it correct?
You did half of the job. You must show that there exists a graph on $$n=14$$ vertices with $$36$$ edges such that the minimum degree is $$5$$. I propose the following.
Let $$G$$ be a graph on $$14$$ vertices with two (vertex-induced) subgraphs $$G_1$$ and $$G_2$$. The subgraph $$G_1$$ is isomorphic to the complete graph $$K_6$$ (this it has already taken $$\displaystyle \binom{6}{2}=15$$ edges). The subgraph $$G_2$$ has $$8$$ vertices $$v_1,v_2,\ldots,v_8$$. For each $$i\in\{1,2,\ldots,8\}$$, join $$v_i$$ with $$v_{i-2}$$, $$v_{i-1}$$, $$v_{i+1}$$, $$v_{i+2}$$, and $$v_{i+4}$$ (indices are calculated modulo $$8$$). Thus, $$G_2$$ has $$\dfrac{1}{2}\cdot 8\cdot 5=20$$ edges. We now need one last edge. Just pick one vertex of $$G_1$$ and one vertex of $$G_2$$, then join them with an edge.
• Yes, the answer is $14$. – Batominovski Oct 18 '18 at 9:08 | 2020-01-22T20:21:07 | {
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http://math.stackexchange.com/questions/188380/solving-complex-equation | # Solving complex equation
How do i further solve the following complex equation:
$$z\cdot \bar{z} + z + \bar{z} + i\cdot z - \overline{i \cdot z} = 9 + 4i$$ $$a^{2} - b^{2} + 2a - 2b = 9 + 4i$$
How do i solve from here on ?
-
Do you mean $z\bar{z}+z+\bar{z}+iz-\overline{iz}=9+4i$? – André Nicolas Aug 29 '12 at 14:29
Yes my latex is bad i didnt know how to do that negate sign. – Sterling Duchess Aug 29 '12 at 14:30
To solve the equation $$z\bar{z}+z+\bar{z}+iz-\overline{iz}=9+4i,$$ we can proceed much as you did. Let $z=a+ib$.
Then $z\bar{z}=(a+ib)(a-ib)=a^2+b^2$.
We have $iz=-b+ia$, so its conjugate is $-b-ia$. Our expression is therefore equal to $$a^2+b^2+(a+ib)+(a-ib)+(-b+ia)-(-b-ia),$$ which simplifies to $a^2+b^2+2a+2ia$.
This is $9+4i$ precisely if the imaginary parts match and the real parts match. We end up with the equations $2a=4$ and $a^2+b^2+2a=9$. Now $a$ and then $b$ are easy to find.
-
To finish the job, a = 2 and b = 1 :) – Martin Aug 29 '12 at 14:44
@Martin: Well, there is also $b=-1$. – André Nicolas Aug 29 '12 at 14:46
So $$2ai = 4i, a^{2} + b^{2} + 2a = 9$$ Would then a = 2i and i put this a in to $$a^2{2} + b^{2} + 2a = 9$$ to solve for b ? – Sterling Duchess Aug 29 '12 at 14:48
@kellax: I wrote $2a=4$. If you like, you can write $2ai=4i$ and divide both sides by $i$, getting $2a=4$. Then $a=2$. Now because $a^2+b^2+2a=9$, we have $2^2+b^2+4=9$, so $b^2=1$, and therefore $b=\pm 1$. – André Nicolas Aug 29 '12 at 14:52
Thank you very much. – Sterling Duchess Aug 29 '12 at 14:56
Alternative approach: the real part of both hands must be equal: $$z\bar z + z+\bar z = 9.$$ The imaginary part of both hands must be equal: $$i(z+\bar z)=4i.$$
So $z$ and $\bar z$ are two numbers; their sum is $4$ and their product is $9-4=5$. So they satisfy $$z^2 - 4z + 5 = 0.$$ This you can solve.
- | 2015-02-01T21:53:01 | {
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https://www.physicsforums.com/threads/exponential-functions-problem.471678/ | # Exponential functions problem
## Homework Statement
In the 2 following problems they use the term in the brackets differently, in one case its a percentage and in the other case i have no idea where they get the number from, this is what i would like to find out
A cell loses 2% of its charge every day
C is total charge t is time(measured in days)
C(t)=100(0.98)t
so basically for each day we put a number in the exponent, for example if 4 days have passed then we write
C(t)=100(0.98)4
So here 100 represents the total charge, 0.98 represents the percentage of charge left after 1 day. This is one type of exponential growth/decay problem I see and is easy to solve.
This other type of problem is where i am having a bit of trouble:
The population of a country in 1981 was 24 million
P represents the population in millions and t represents the time in years
The following equation represents this model
P(t)=24(1.014)t
Here 24 represents the total population in 1981, 1.014 is the growth rate by which the population increases each year(in the millions) and the exponent is where we put in the number of years it has been since 1981
so for example if we use 2011 (1981+30=2011) that means to get the population for 2011
we do P(t)=24(1.014)30
P(t)= 24(1.517534768)
P(t)=36.4 million.
my question is where or how do they get the 1.014 constant which is used to calculate the population for every year relative to 1981? It is the exponential growth multiple isn't it? How did they get it?
P(t)=24(1.014)t
## The Attempt at a Solution
eumyang
Homework Helper
I usually think of exponential growth/decay models in terms of this equation:
$$P(t) = P_0 (1 + r)^t$$
r is the rate of growth/decay. r > 0 would indicate a growth, and r < 0 would indicate a decay.
So for the 1st equation,
$$C(t) = 100(0.98)^t = 100(1 - 0.02)^t$$
... since a cell loses 2% of its charge every day.
For the 2nd equation,
$$P(t) = 24(1.014)^t = 24(1 + 0.014)^t$$
... so the rate of growth is 1.4% per year since 1981. This information should have been given in the problem somewhere.
I usually think of exponential growth/decay models in terms of this equation:
$$P(t) = P_0 (1 + r)^t$$
r is the rate of growth/decay. r > 0 would indicate a growth, and r < 0 would indicate a decay.
So for the 1st equation,
$$C(t) = 100(0.98)^t = 100(1 - 0.02)^t$$
... since a cell loses 2% of its charge every day.
For the 2nd equation,
$$P(t) = 24(1.014)^t = 24(1 + 0.014)^t$$
... so the rate of growth is 1.4% per year since 1981. This information should have been given in the problem somewhere.
so the only difference between the first and the second equation is that the first equation rate(r) is not multiplied by 100 and the second one rate(r) is?
eumyang
Homework Helper
so the only difference between the first and the second equation is that the first equation rate(r) is not multiplied by 100 and the second one rate(r) is?
I don't know what you mean. r = -0.02 corresponds to a 2% loss rate. r = +0.014 corresponds to a 1.4% rate of growth. We're just converting to percents, and in both cases we multiply 100 to convert to percents. | 2022-01-23T13:30:25 | {
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http://pch-wiki.wikispaces.com/10.4+OFSA+Solutions?responseToken=004bbe007d7cd8b5a6b1266ce34c7f6c9 | # 10.4 OFSA Solutions
OFSA SOLUTIONS for Parametrics
This page contains peer generated solutions and error explanations to OFSA questions. As you read or view the solutions, be critical: check for accuracy, but also for more efficient solution strategies. If you have a better method or different idea/answer, post a discussion and monitor the responses.
Quick Directions
• Post answers, solutions and error explanations to each OFSA question below.
• For each "distractor" or incorrect answer choice, explain the error that would lead to that incorrect answer choice.
• You may either do the above in typed format or using a pencast.
• Separate each question with a section bar.
• After each solution, provide a hyperlink back to the corresponding OFSA page.
Question 1
Given
$$x=3sint$$
and
$$y=5cost$$
What is the shape of the x-y graph and how does the particle travel as time goes on?
a. ellipse, counterclockwise
b. line, oscillates- up then down
c. ellipse, clockwise
d. line, oscillates- down then up
Solution 1
$$x=3sint$$
and
$$y=5cost$$
so,
$$\frac{x^{2}}{9}=sin^{2}t$$
and
$$\frac{y^{2}}{25}=cos^{2}t$$
If you remember back to the trig identities,
$$sin^{2}t+cos^{2}t=1$$
so if we add together the equations we should get
$$\frac{x^{2}}{9}+\frac{y^{2}}{25}=sin^{2}t+cos^{2}t=1$$
This graph is an ellipse, so we can rule out option b and d
The orientation of the graph can be found by finding where t=0 (starting point) and substituting other t values in to visualize the direction the particle is moving.
when t=0, x=0, y=5
when t=-1, x=-.05
when t=1, x=.05
The orientation can also be found by finding the starting point, then looking at the two parametric equations we were given. on the x-t graph, x increases right after the start. On the y-t graph, y decreases just after the start. By using these facts we can conclude that the particle is moving clockwise around the ellipse.
Therefore, the correct answer is c.
*a correct, full description of the motion:
The particle begins at (0,5) and continues on the path defined by
$$\frac{x^{2}}{9}+\frac{y^{2}}{25}=1$$
in a clockwise direction and continues indefinitely.*
Error Explanation 1
a. the student found the correct graph, but assumed that the direction was counterclockwise because the majority of circles we have dealt with in the polar, trig identities, and unit circle problem all moved counterclockwise.
b. the student mis-used the pythagorean identity, instead believing that
$$sint+cost=1$$
then they only used the direction of the x-t graph to find the orientation.
c. correct!
d. the student mis-used the pythagorean identity, instead believing that
$$sint+cost=1$$
then they only used the direction of the y-t graph to find the orientation.
Question 2
Two ships, the USS Lizzy and the USS Izzy are headed toward each other on path defined by:
USS Lizzy:
[x=t]
[x=t]
and
[y=t+30]
[y=t+30]
USS Izzy:
[x=80-t]
[x=80-t]
and
[y=t+10]
[y=t+10]
When will the two paths cross, but without colliding?
a. never
b. at (30,60) when USS Lizzy's t= 50
c. at (60,90) when USS Izzy's t= 30
d. at (30, 60) when USS Izzy's t= 50
Solution 2
1) eliminate the parameter so that:
USS Lizzy:
$$y=x+30$$
USS Izzy:
$$y=90-x$$
2) Treat this as a system of equations and solve, using substitution.
$$y=x+30=90-x$$
$$2x=60$$
$$x=30$$
$$y=60$$
3) Substitute either the x or y value back into USS Izzy's equations to find that t=50 .
Therefore the solution is d.
Error Explanation 2
a. the student tried to use the method for finding simultaneous solutions, by setting x1=x2 and y1=y2. There is no simultaneous solution.
b. the student found the correct point of intersection, but substituted either the x or y value into Izzy's equation, the wrong one. The time does not match with the boat.
c. the student made an algebraic mistake when solving the system of equations. They said:
$$y=x+30=90-x$$
$$2x=120$$
$$x=60$$
d. correct!
Question 3
Given:
$$x=\frac{2}{t-3}$$
and
$$y=\frac{1}{t+5}$$
eliminate the parameter, and choose the correct graph.
a.
b.
c.
d.
Solution 3
1) To eliminate the parameter, the most efficient way is to isolate the t in the x-t equation.
$$x=\frac{2}{t-3}$$
$$t=\frac{2}{x}+3$$
2) Then substitute the t in the y-t equation
$$y=\frac{1}{t+5}$$
$$y=\frac{1}{\frac{2}{x}+8}$$
3) simplify to:
$y=\frac{x}{\2+8x}$
This is a review on graphing rational functions
VA=
$\frac{-1}{4}$
(set denominator equal to 0 and solve)
HA=
$$\frac{1}{8}$$
(set the numerator's x term with the highest degree over the denominator's x term with the highest degree)
Intercept= (0,0)
(set equation equal to zero and find x value; this is the x and the y intercept)
Therefore the correct answer is d.
Error Explanation 3
a. the student graphed only the x-t equation
b. the student made an error and added the x-t and y-t equations, a method used to eliminate the parameter with trig functions. The student graphed
$$y=\frac{3x+7}{(x-3)(x+5)}$$
c. the student graphed only the y-t equation
d. correct!
Question 4
A soccer ball is traveling across the ice and is parametrically described by: x=2t+1and y=t-4. Ali is running across the field in hopes of intercepting the ball. The position of the tip of Ali's cleat is parametrically defined by: x=3t-10 and y= -t+3
Does Ali intercept the soccer ball? If yes, when?
Solution 4
SIMULTANEOUS SOLUTION/COLLISION OF BALL AND FOOT
1. let X=X
2t+1=3t-10
t=11
2. let Y=Y
t-4=-t+3
t=
$$\frac{7}{2}$$
3. t:
$$11\neq \frac{7}{2}$$
NON-SIMULTANEOUS SOLUTION
Since the two paths intersect at different times, there is no collision and the paths cross at a non-silmultaneous solution. But where?
1. ball:
$$y=\frac{1}{2}x-\frac{9}{2}$$
2. foot:
$$y=-\frac{1}{3}x-\frac{1}{3}$$
3.
$$\frac{1}{2}x-\frac{9}{2}=-\frac{1}{3}x-\frac{1}{3}$$
4. x=5 and y=-2
non-simlutaneous solution occurs at (5, -2)
Error Explanation 4
a. After solving for a simultaneous solution,only a non-simultaneous solution was found.
$$11\neq \frac{7}{2}$$
so there are no simultaneous solutions.
b. After solving for a simultaneous solution,only a non-simultaneous solution was found.
$$11\neq \frac{7}{2}$$
so there are no simultaneous solutions.
c. Because there is no simultaneous solution, there is no collision. By subtracting the time the foot reaches the same spot the ball had been from the time the ball was there, it still does not create a simultaneous solution.
d. Correct!
Question 5
Given
$$x=\sqrt{t+4}+1$$
and
$$y=3\sqrt{t+4}-4$$
Eliminate the parameter to find any restrictions of t, x, and y (if any).
Solution 5
1. solve for t for one of the equations:
$$x=\sqrt{t+4}+1$$
$$(x-1)^{2}-4=t$$
$$y=3\sqrt{((x-1)^{2}-4))+4}-4$$
$$y=3\sqrt{(x-1)^{2}}-4$$
$$y=3x-7$$
2. you know the final number under the radical cannot exceed lower than -4 so substitute -4 into both the x=t and y=t equations
$$x=\sqrt{-4+4}+1$$
$$x=1$$
$$y=3\sqrt{-4+4}-4$$
$$y=-4$$
Error Explanation 5
a. Because the parameter 't' is inside a square root, it is impossible that t can extend beyond -4. Under the square root, the number cannot be negative, and since t is not squared, a negative will stay negative.
b.When the parameter ''t' has been limited to t:[-4,∞), the x and y values must coordinate with the t values. By substituting the value of -4 for t, the restriction on the x value becomes 1 and on y it becomes -4.
c. Correct!
d. The parameter 't' can extend until -4, making t:[4,∞) an answer forgetting the range of numbers from [-4,4]
Question 6
A football is thrown from a rooftop, 53 feet above the ground with an initial velocity of 26 ft/s and at an initial angle of 33° above the horizontal.
Create parametric equations to model the position of the ball in terms of the time 't' to find the horizontal distance to ball will travel before hitting the ground. (Calc. allowed!)
Solution 6
1. create parametric equations to model the football's position
$$x(t)=(26\cos 33)t$$
and
$$y(t)=(26\sin33)t+53-16t^{2}$$
2. make
$$y(t)=0$$
$$0=-16t^{2}+14.16t+53$$
$$t=-1.43$$
$$t=-2.32$$
because time can't be negative, the only option for the time when the football reaches the ground is 2.32 seconds.
3. plug 't' into the parametric 'x' equation to solve for the horizontal distance
$$x= (26\cos33)2.32$$
$$x= 50.59$$
Error Explanation 6
a. While on Parametric mode, this is the negative vertical height the calculator's table reads when the maximum horizontal distance has been reached.
b. The process was done in reverse, rather than plugging 0 into the y equation and 't' into the x equation, 0 was plugged into the x equation and the newfound 't' into the y.
c. This answer is not only saying the football was thrown backwards, but also that it was thrown a VERY short instance. This answer could be achieved through using the graphing mode in the calculator while in radians instead of degrees. Obviously the answer is impractical, but this shows a clear difference between radians and degrees.
WHEN A PROBLEM USES DEGREES FOR THETA MAKE SURE YOUR CALCULATOR IS IN DEGREE MODE!!
d. Correct!
Question 7 - Simultaneous Solutions
$\\ \text{Find the values of x and y at each simultaneous intersection along with the value of t. Use radians.} \\ t \in \left[0, \pi\right] \\ x_1 = 10\sin(6t), \; y_1 = 4\cos(4t) + 3\\ x_2 = 10\cos(6t), \; y_2 = 3\cos(4t) + 3\\$
Solution 7
$\\ \text{1) let } x_1 = x_2 \\ 10\sin(6t) = 10\cos(6t) \\ \sin(6t) = \cos(6t) \\ \tan(6t) = 1 \\ \\ \text{2) let } y_1 = y_2 \\ 4\cos(4t) = 3\cos(4t) \\ cos(4t) = 0 \\ \\ \text{3) list possible t values for both x and y in the restriction of t.} \\ \tan(6t) = 1 \\ 6t \in \frac{\pi}{4} + \pi k \\ t \in \frac{\pi}{24} + \frac{\pi/6} k \\ t = {\frac{\pi}{24}, \; \frac{5\pi}{24}, \; \frac{3\pi}{8}, \; \frac{13\pi}{24}, \; \frac{17\pi}{24}, \; \frac{7\pi}{8}} \\ \\ \cos(4t) = 0 \\ 4t \in \frac{\pi}{2} + \pi k \\ t \in \frac{\pi}{8} + \frac{\pi}{4} k \\ t = {\frac{\pi}{8}, \; \frac{3\pi}{8}, \; \frac{5\pi}{8}, \; \frac{7\pi}{8}} \\ \\ \text{4) Find t values that work for both the x-equations and y-equations and find the corresponding (x,y) coordinates by plugging it back into the x-t and y-t equations} \\ t = {\frac{3\pi}{8}, \frac{7\pi}{8}} \\ x = 10\sin(6t) = {5\sqrt{2}, -5\sqrt{2}} \\ y = 3\cos(4t) + 3 = {3, 3} \\ \\ \boxed{\text{C) two solutions: } (5\sqrt{2}, 3) \text{ at } t=\frac{3\pi}{8} \text{ and } (-5\sqrt{2}, 3) \text{ at } t=\frac{7\pi}{8}}} \\ \\$
Error Explanation 7
$\\ \text{a) Check for errors with trig functions} \\ \text{b) Check for all solutions within the restrictions given, don't stop at just the first simultaneous solutions} \\ \text{c) Correct Answer} \\ \text{d) Check for errors with trig functions} \\$
Question 8
Minimum Distance
Tom and Jerry are at it again! Jerry is trying to cross a square room, which has a diagonal from (0,0) to (100,100). Tom's motion is described by the following equations: x = t + 17 and y = 5t^2 - 2t. Jerry's movement is described by: x = 3t - 1and y = 36t + 63. Will Jerry be captured by Tom? If so, find the x and y coordinates at the intersection.
Solution 8
$\\ \text{1) Let x1 = x2} \\ t + 17 = 3t - 1 \\ 18 = 2t \\ t = 9 \\ \\ \text{2) Let y1 = y2} \\ 5t^2 - 2t = 36t + 63 \\ 5t^2 - 38t - 63 = 0 \\ (t - 9)(5t + 7) = 0 \\ t = 9, \; \frac{-7}{5} \\ \\ \text{3) Find t values that solve both the x equations and the y equations} \\ \box{t = 9} \\ \\ \text{4) Find the intersection} \\ x = t + 17 = 26 \\ y = 5t^2 - 2t = 387 \\ \\ \text{5) Answer the question. Will Jerry be captured by Tom?} \\ \text{NO because the only simultaneous solution occurs outside of the dimensions of the room}$
Error Explanation 8
$\\ \text{a) Yes, at (26, 27) when t = 9 Error: removing the square from the first y equation}\\ \text{b) Yes, at (26, 387) when t = 9 Error: check restrictions on each variable!} \\ \text{c) No Correct!} \\ \text{d) Yes, at (387, 26) when t = 9 Error: flipping x and y AND checking restrictions on each variable}$
Question 9
An object's path is defined as x = t(t - 3)^2, y = 3. Find the interval of t where the object is moving to the right (increasing).
Graph Plot
Solution 9
1) Graph the x - t function
See graph to right
2) Use calculator to determine maximums and minimums
maximum (t,x) = (1, 4)
minimum (t,x) = (3, 0)
3) Recognize which parts of the x-t equation are increasing
Error Explanation 9
$\\ \text{a) } t \in \left(-\infty, 1\right) \cup \left(3, \infty\right) \text{Correct} \\ \text{b) } t \in \left(3, \infty\right) \text{Error: misread x-t function as (t-3)^2} \\$
c) Never, it's always constant Error: Using t-y (vertical position) rather than t-x (horizontal position)
d) Never, the object is always moving down Error: Graph does not see the entire picture; 1 <= domain <= 3
Question 10
Solution 10
$\bg_white \fn_phv x_{1}=x_{2}$
3t-4 = 2t+5
t = 9
$\bg_white \fn_phv y_{1}=y_{2}$
2t-7 = t+2
t = 9
x = 3(9) - 4 = 23
y = 2(9) - 7 = 11
Yes, the man will step on the bug at (23, 11) when t = 9 seconds.
Error Explanation 10
a) Incorrect...There is a simultaneous solution
• I decided this could be a mistake a person can make if they have minor calculation errors
b) Incorrect...The simultaneous solution does not occur at (-13, -13)
• This mistake would be that a person accidentally set x1 = y1 and x2=y2 and solved for t
• t still came out to be the same (t = -3) and the mistake was that the person plugged -3 into the first equations they saw without checking the other pair
c) Incorrect...The simultaneous solution does not occur at (9, 9)
• The mistake for this would be that after solving for t by setting x1 = x2 and y1 = y2, the person got t = 9 but thought that this was the answer since the answer came out the same
d) Correct...Following the procedure for solving for simultaneous solutions produces a SS at (23, 11)
Question 11
Solution 11
Graph the parametric equations first to know the directional changes in motion.
t-x graph
Graph Plot
The x-position is constant, which produces rectilinear motion. The object will always stay on the line x = 7.
t-y graph
Graph Plot
The y-position varies, but it is directional change that stays on the same straight path.
x-y graph
Graph Plot
The object travels vertically upward from (7, 0) to (7, 32), back down to (7, 0), and back up to (7, 7)
Error Explanation 11
a) Correct description of motion of the object, with correct direction and specificity of the turning points
b) Incorrect...Since x = 7 (and is therefore a constant), the object must travel vertically
• The mistake for this would be that the person saw the t-x graph and mistakenly thought that the object moved horizontally since the vertical x axis on the t-x graph showed a horizontal straight line
• The correct key points were switched around (such as from (x,y) to (y,x) because of the previous mistake)
c) Incorrect...The t-y graph shows that the y-position of the object changes direction multiple times
• This is a mistake from only plugging in the restrictions for t (that is, -2 and 5) into the y-equation and assuming that there are no directional changes in between the starting and end points
d) Incorrect...The object travels a total distance of 71 units (32 up + 32 down + 7 up) but is displaced only 7 units
• It is easy to mix up distance and displacement, which makes this a possible mistake that can be made
Question 12
Solution 12
Oscillates between (1, 1) and (5, 7)
x position oscillates between 1 and 5
y position oscillates between 1 and 7
t-x graph
Graph Plot
t-y graph
Graph Plot
x = -2cos((pi/4)t) + 3
y = -3cos((pi/4)t) +4
period
# (2pi)/B
(2pi)/(pi/4) = 8
Distance between (1, 1) and (5, 7)
=
$\small \bg_white \fn_cm \sqrt{(5-1)^2+(7-1)^2}$
=
$\small \bg_white \fn_cm 2\sqrt{13}$
Since the restriction is t: [0, 8] and the object travels one whole length in t = 4, we need to multiply $\small \bg_white \fn_cm 2\sqrt{13}$ by 2
So the correct answer is $\small \bg_white \fn_cm 4\sqrt{13}$
Error Explanation 12
a) Incorrect...
• The mistake would be that the person forgot to multiply the distance by 2 because they figure they found the distance traveled by finding the distance between the two points
b) Incorrect...
• The person is thinking of displacement and thinks that since the object ended where it started it did not travel any distance
c) Correct...
• Took into account the two-way trip
d) Incorrect...
• Either mathematical error or accidentally thinking that it asks for the distance traveled between the interval [0, 8] and that since t = 4 is half of 8 the answer is double, or 2
Question 13
Which description of motion below is not supported by the given parametric equations?
$\left\{\begin{matrix} x(t) = 2 \\ y(t)=(t-1)(t+2)^2 \end{matrix}\right. ; -5 \leq t \leq 5 \\ \\ \text{A) increases in y-coord to 0} \\ \text{B) increases in y-coord to 149} \\ \text{C) decreases in y-coord to -4} \\ \text{D) increases in y-coord to -4} \\$
Solution 13
• graphing xt yields a domain of [-5,5] and a range of [2].
• when graphing yt, make sure you calculate the y-values for every significant t-value, in this case they are t=-5, -2, 0, 1, and 5.
• once those values are calculated, you now know that the domain is [-5,5] and the range is [-54,149].
• with these two graphs, you can now graph the motion of the particle:
• using the significant y-values, you can properly describe the motion of the particle.
• as for the answer, choices A, B, and C are supported, while choice D, "increases in y-coord to -4", is not supported. rather, the particle decreases in y-coordinate to -4.
Error Explanation 13
A) Correct! From t: [-5,-2], the particle is increasing in y-coordinates to 0.
B) Correct! From t: [0,5], the particle is increasing in y-coordinates to 149.
C) Correct! From t: [-2,0], the particle is decreasing in y-coordinates to -4.
D) Error. Student must have seen first increase in y-coord from -54 to 0 and assumed it ended at -4 instead.
Question 14
Robbie Gould, a place kicker for the Chicago Bears, kicks a football in a game with the Detroit Lions. The ball leaves the ground with a velocity of 89 ft./s at an angle of 63 degrees above the horizontal. If the ball is kicked from the Bears' 30 yard line and is aimed straight down the field, where does it land?
photo.JPG
a) Between the 1 and 2 yard line
b) Between the 3 and 4 yard line
c) Between the 4 and 5 yard line.
d) Off the field!
Solution Explanation:
boxed{x = x_0 + (|v_0|cos(theta))t}
boxed{y = y_0 + (|v_0|sin(theta))t + frac{1}{2}(g)(t)^2}
1) Write two parametric equations to model this situation.
x=89cos63t
y=89sin63t - 16t^2
2) We want to solve for t, the common variable between both equations. T represents the time the object is a projectile, or is in flight. We can substitute 0 in for y in the
y=89sin63t - 16t^2 equation because that is the final position of the ball, on the ground at a height of zero.
0=89sin63t - 16t^2
16t^2 = 89sin63t
16t = 89sin63
t is approximately 4.956 seconds.
3) To figure out where the ball lands, we must find how far the ball travels horizontally. This can be determined using the x= equation. Substitute the t value we just calculated, 4.956 seconds. We know that distance= rate x time, and that is what this equation represents. The length traveled is equal to the horizontal speed of the ball x the time the ball travels.
x= 89cos63(4.956)
x=200.257 ft
The field is measured in yards, so to convert this to the correct unit we divide by 3, as a yard consists of 3 feet. We arrive at the conclusion that the ball travels approximately 66.75 yards.
4) The seperation between the lines of the field is 10 yards. Since the ball travels 66.75 yards, we know it crosses about 6.675 yard lines or between 6 and 7 yard lines. The ball begins at the Bears' 30 yard line, so we count from there 6-7 yard lines (For the non-football experts, refer to the picture!). The ball lands between the 40 and 30 yard line on the Lions' side, aka the 3 and 4 yard lines.
Therefore the correct answer is B, the ball will land between the 3 and 4 yard lines.
Errors Explained
a) Calculation error. An error of 7.42 seconds for t or 300 ft. for distance traveled would have resulted in this answer.
b) CORRECT, No errors :)
c) If you did this problem in Radian mode on the calculator, the time traveled will be significantly smaller which therefore decreases the distance traveled to about 22 yards.
d) If you used the wrong units in the y= equation and used 9.8 as g instead of 32, you will have gotten that the ball was in the air for 16.5 seconds! Therefore the ball would have traveled a whopping 222 yards, taking it off the football field.
Question 15
second problem.JPG
Write the parametric equations, t-x and t-y, that produce this rectangular graph.
A) x=3sin(π/6t), y= -3cos((π/6t)
B) x= 3sin(π/8t), y= 3cos(π/8t)
C) x= 3sin(π/8t), y= -3cos(π/8t)
D) x=-3cos(π/8t), y=3sin(π/8t)
Solution explanation:
1) First write an equation that models the x behavior of the particle in relation to time. This will be your t-x graph.
• We see that the particle at t=0, begins its motion at x=0. As time progresses it moves from zero to its max value, back to zero, to its min value, then back to zero. This oscillation reflects the shape of a sin curve. (*Tip: It's helpful to draw out the movement of the t-x graph to help determine what function it represents!)
• A generic sin curve oscillates between -1 and 1, but this graph oscillates between -3 and 3, so we know that the amplitude is 3 to stretch the curve vertically.
• Define the cycle, or B. B= 2π/Period. The period of the motion is 16 seconds because it takes 16 seconds for the particle to return back to its starting position. Therefore B is 2π/16, or π/8.
• Put together all these components of the equation...x=3sin(π/8t).
2) Next, write an equation that models the y behavior of the particle in relation to time. This will be your t-y graph.
• We see that the particle begins at y=-3, the minimum value. As time progresses it moves from the min value, to zero, to the max value, to zero, then back to the min value. This oscillation reflects the shape of a negative cosine curve. (*Tip: It's helpful to draw out the movement of the t-y graph to help determine what function it represents!)
• A generic sin curve oscillates between -1 and 1, but this graph oscillates between -3 and 3, so we know that the amplitude is 3 to stretch the curve vertically.
• Define the cycle, or B. B= 2π/Period. The period of the motion is 16 seconds because it takes 16 seconds for the particle to return back to its starting position. Therefore B is 2π/16, or π/8.
• Put together all these components of the equation....y= -3cos(π/8t).
Errors Explained:
a) Incorrect period. May have seen t=12 being the largest t value and assumed that was the total period but remember, it must continue back to t=0.
b) May not have realized that the y movement is a NEGATIVE cosine curve as it begins from the min vs. the max.
d) The x and y are switched...read carefully!
Question 16
Solution 16
Error Explanation 16
Question 17
Find the simultaneous solution for the following:
{x = 5t -6
{y= t^2 + 1
{x= 2t + 9
{y= T^2 + 1
Solution 17
a) (19, 26) when t= 5.
5t-6= 2t + 9
t= 5
and
T^2 + 1 = 31 - T
T^2 + t - 30 = 0
(t-5) (t+6) = 0
t= 5, -6
From setting the x equations and y equations equal to each other, both gave the solution t =5. But wait, now find the point. When subbing back in the point is (19, 26).
Error Explanation 17
b) the t= 5 is correct but the x and y in the coordinate are flipped. Be careful!
c) t= -6 is one of the solutions in the procedure but it is not simultaneous since both of the problems did not give a solution of -6.
d) This is the answer to the non-simultaneous intersection. If this is your answer you are using the non-simultaneous solution procedure instead of the simultaneous solution procedure.
Question 18
For t values [0, 5], a particle travels along the following path:
y= ( t- 5)( t- 1)^2
What is the total displacement and total distance traveled by the particle?
a) Displacement 5, Distance 25
b) Displacement 0, Distance 20
c) Displacement 25, Distance 5
d) Displacement 5, Distance 5
Solution Explanation:
Remember the meaning of both displacement and distance!
Displacement=the difference between the end position and beginning position
Distance=the total units the particle travels under the time restrictions
The graph of this particle is this:
(*The horizontal axis is defined by t and the vertical by y.)
• Displacement: At t=0, the particle begins at -5. At t=5, the particle ends at 0. Displacement is 5, because it is the difference in beginning and end positions.
• Distance: At t=0 the particle begins at -5. During this time restriction, the particle travels 5 units to go back to zero, 10 units to -10, and 10 units back up to zero. If you add up the total units the particle moved (5+ 10+ 10), the distance is 25.
Answer A is correct: Displacement 5, Distance 25.
Errors Explained:
b) Calculated amounts over the wrong time restriction [1,5]
c) Does not understand the definitions of displacement and distance. Probably got the two confused in meaning because the answers are switched.
d) Correct displacement but incorrect distance. May have calculated distance along horizontal axis but remember, that defines time. The motion is described by the vertical or y axis.
Question 19 USE CALC
Solution 19
c) the is a solution at (0,5) when t= 5
on calculator find the intersections for the following:
3 sin (πt) = t-2
You will get t= 2, 1.097
3 cos (πt) + 2 = t + 3
t= 2 will also be a solution
with t= .351
Therefore, there is a solution at t=2. Sub back in to get (0,5)
Error Explanation 19
a) if you did not get a sim solution, the graph or math was incorrect
b) the simultaneous solution can be determined with a calculator.
d) this would be the non-simultaneous solution
Question 20
Given the parametric equations below, what is the orientation of the graph and in which direction will the particle be traveling? (assume t: all real numbers)
$\left\{\begin{matrix}x=3cos(\frac{\Pi }{6}t) \\ y=-sin(\frac{\Pi }{6}t) \end{matrix}\right. \\ \\ \text{A) line; oscillating left to right} \\ \text{B) line; oscillating right to left} \\ \text{C) ellipse; CW} \\ \text{D) ellipse; CCW} \\$
Solution 20
• graphing both parametric equations yields these domains and ranges:
• x: [-3,3], t: all real numbers
• y: [-1,1], t: all real numbers
• note the trends of each graph to figure out the orientation!
• as t increases, x decreases to -3 then increases to 3 then decreases again
• as t increases, y decreases to -1 then increases to 1 then again
• graph the x-y equation; eliminating the parameter may take too long, so just make a small table of points:
• you can plot points and follow through with a full graph based on the xt and yt graphs
• as you can see, this graph correctly follows the trends we stated earlier, with x decreasing then increasing and y decreasing then increasing then decreasing again, all starting at t=0 @ (3,0)
• therefore, the correct answer is choice C) CW.
Error Explanation 20
A) Error. The graph is an ellipse, not a line, or the student chose this because the direction made sense.
B) Error. The graph is an ellipse, not a line, or the student chose this because of the direction, but incorrectly graphed the xt / xy equations.
C) Correct!
D) Error. Student must have incorrectly graphed the xt / xy equations.
Question 21
Calculator allowed! Determine the non-simultaneous point of intersection of the paths.
$\left\{\begin{matrix}x=\sqrt{t} \\ y=2t+3 \end{matrix}\right.$
and
$\left\{\begin{matrix}x=\sqrt{6-t} \\ y=t^2 \end{matrix}$
$\\ A) \ t=3, (\sqrt3,9) \\ B) \ (i, 1) \\ C) \ (3.32,25) \\ D) \ \text{There are no non-simultaneous solutions.}$
Solution 21
• set both xt equations equal to each other and solve for t.
• set both yt equations equal to each other and solve for t.
• since both solutions yielded t=3, this means t=3 is part of a simultaneous solution.
• since we are solving for non-simultaneous solutions, we will resume and ignore the t=-1.
• eliminate the parameter for both sets of xt / xy equations then graph both equations and find the intersection!
• note that the non-simultaneous solution is located at the point where:
$x \neq \pm \sqrt3 \\$
• this is because the simultaneous solution lies there and we are only looking for the non-simultaneous solution!
• therefore, the correct answer is choice C) (3.32, 25)
Error Explanation 21
A) Error. This is the simultaneous solution. Correct answer if question was asking for the simultaneous solution!
B) Error. If a student truly did not know what they were doing, or they were running out of time and had to guess, they had a greater chance of choosing this answer. The t=-1 seems like it would be considered as a "non-simultaneous solution" and with correct algebra plus mindless substitution, this choice would, theoretically, make a good, logical choice! (or at least, that's what I'm thinking)
C) Correct!
D) Error. If the student was choosing between this and choice B, this answer would be chosen because you are unable to graph imaginary numbers, (and it would seem that graphing imaginary numbers wouldn't make sense and is totally out of the context of this unit). | 2018-06-19T16:02:37 | {
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http://mathhelpforum.com/algebra/31614-solution-cubic.html | # Math Help - Solution of Cubic
1. ## Solution of Cubic
Can someone help me find the exact solutions to the equation
$x^3 + 2x^2 - 9x + 3 = 0$?
I've used a grapher to find approximate solutions at the three places where it crosses the x-axis, and tried my best to work through Cardano's solution ( $t^3 -\frac{31}{3}t + \frac{259}{27} = 0$) but I just can't finish it.
A big part of my problem here is that using the rational root theorem leads me to try $x = \pm 1, \pm 3$, none of which work, so all three real solutions (it crosses three times, so no complex-imaginary roots) must be irrational. I recognize it obviously must be possible, but I'm not used to three irrational roots giving such a nice, simplistic equation. These three irrationals must multiply to be perfectly 3, add to be perfectly -2, and something else (what?) to be perfectly -9 (or +9? I can't remember).
2. Originally Posted by English Major
Can someone help me find the exact solutions to the equation
$x^3 + 2x^2 - 9x + 3 = 0$?
I've used a grapher to find approximate solutions at the three places where it crosses the x-axis, and tried my best to work through Cardano's solution ( $t^3 -\frac{31}{3}t + \frac{259}{27} = 0$) but I just can't finish it.
A big part of my problem here is that using the rational root theorem leads me to try $x = \pm 1, \pm 3$, none of which work, so all three real solutions (it crosses three times, so no complex-imaginary roots) must be irrational. I recognize it obviously must be possible, but I'm not used to three irrational roots giving such a nice, simplistic equation. These three irrationals must multiply to be perfectly 3, add to be perfectly -2, and something else (what?) to be perfectly -9 (or +9? I can't remember).
Is there any particular reason why you want exact solutions, as opposed to approximate ones?
3. Originally Posted by English Major
Can someone help me find the exact solutions to the equation
$x^3 + 2x^2 - 9x + 3 = 0$?
I've used a grapher to find approximate solutions at the three places where it crosses the x-axis, and tried my best to work through Cardano's solution ( $t^3 -\frac{31}{3}t + \frac{259}{27} = 0$) but I just can't finish it.
A big part of my problem here is that using the rational root theorem leads me to try $x = \pm 1, \pm 3$, none of which work, so all three real solutions (it crosses three times, so no complex-imaginary roots) must be irrational. I recognize it obviously must be possible, but I'm not used to three irrational roots giving such a nice, simplistic equation. These three irrationals must multiply to be perfectly 3, add to be perfectly -2, and something else (what?) to be perfectly -9 (or +9? I can't remember).
For:
$x^3+a_1x^2+a_2 x+a_3=0$
Put:
$Q=\frac{3a_1-a_1^2}{9}$
and:
$R=\frac{9a_1a_2-27a_3-2a_1^3}{54}$
We know in the case under consideration that the roots are real, as also the discriminant
$D=Q^3+R^2<0$
tells us (since $Q=\frac{-31}{9}$ and $R=\frac{-259}{54}$).
In which case Tartaglia-Cardano formula reduces to:
$\theta=\arccos(-R/\sqrt{-Q^3})$
$x_1=2\sqrt{-Q}\cos(\theta/3)$
$x_1=2\sqrt{-Q}\cos(\theta/3+2\pi/3)$
$x_3=2\sqrt{-Q}\cos(\theta/3+4\pi/3)$
RonL
4. Originally Posted by mr fantastic
Is there any particular reason why you want exact solutions, as opposed to approximate ones?
Honestly, it's for my own understanding- I always believed that for polynomials with rational coefficients, any irrational solutions would come in conjugate pairs from quadratic formula use. You could use three cube root zeros to make a polynomial with a rational constant term, but I don't see how they would balance each other out for rational coefficients in other terms.
The short answer is, I'm interested to see the form the solutions take, and to use that to expand my own knowledge. Thanks for any help you can provide.
5. Originally Posted by CaptainBlack
For:
$x^3+a_1x^2+a_2 x+a_3=0$
(snipped)
In which case Tartaglia-Cardano formula reduces to:
$\theta=\arccos(-R/\sqrt{-Q^3})$
$x_1=2\sqrt{-Q}\cos(\theta/3)$
$x_1=2\sqrt{-Q}\cos(\theta/3+2\pi/3)$
$x_3=2\sqrt{-Q}\cos(\theta/3+4\pi/3)$
RonL
Okay, I'll admit: I was not expecting the solutions to include trig (though the $2\pi/3$ and $4\pi/3$ imply to me that this may have something to do with roots of unity, being equally spaced around a unit circle. True?)
Can you recommend a place to read up on this type of solution? The wikipedia article I linked to provides similar, but distinctly different solution methods. Incidentally, looking through my article, I'm noticing that the method I was trying "fails for the case of three real roots," which I wish I'd noticed earlier.
Thanks!
6. Originally Posted by English Major
Okay, I'll admit: I was not expecting the solutions to include trig (though the $2\pi/3$ and $4\pi/3$ imply to me that this may have something to do with roots of unity, being equally spaced around a unit circle. True?)
Can you recommend a place to read up on this type of solution? The wikipedia article I linked to provides similar, but distinctly different solution methods. Incidentally, looking through my article, I'm noticing that the method I was trying "fails for the case of three real roots," which I wish I'd noticed earlier.
Thanks!
The form of solution given in my earlier post is just another way of writing the
solution in terms of the Chebyshev cube root.
RonL | 2016-05-25T11:58:41 | {
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http://math.stackexchange.com/questions/54118/do-matrices-with-central-symmetry-form-a-group | # Do matrices with central symmetry form a group?
Consider the set of $N\times N$ matrices that satisfy the property $$\mathcal{H} = \{H\,|\, H_{ij}=H_{N+1-i,N+1-j}, \det H \neq 0\}$$ or in matrix forms $$\begin{pmatrix}a_{1} & a_{2} & \cdots & a_{N-1} & a_{N}\\ b_{1} & b_{2} & \cdots & b_{N-1} & b_{N}\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ b_{N} & b_{N-1} & \cdots & b_{2} & b_{1}\\ a_{N} & a_{N-1} & \cdots & a_{2} & a_{1} \end{pmatrix}$$
Do these matrices have a name? Do they form a group?
It can be easily shown that the identity is in the group, i.e. $I\in \mathcal{H}$. Also the set is closed under multiplication, i.e. if $A\in \mathcal{H}$ and $B\in \mathcal{H}$ then $AB\in\mathcal{H}$. To see why, consider
\begin{align*} (AB)_{mn}&=\sum_i A_{mi}B_{in}=\sum_i A_{N+1-m,N+1-i}B_{N+1-i, N+1-n}\\ &=\sum_{i^\prime} A_{N+1-m,i^\prime}B_{i^\prime, N+1-n}=(AB)_{N+1-m,N+1-n} \end{align*}
The last property that needs to be shown for the set to be a group is that the inverse is also in the set, i.e. if $A\in\mathcal{H}$, is it true that $A^{-1}\in\mathcal{H}$?
EDIT: Added the condition that matrices should also be invertible.
-
Not all matrices of this form are invertible: just put the first and last row equal to $0$. So they cannot form a group. – Arturo Magidin Jul 27 '11 at 19:34
By the way, the condition should be $H_{ij} = H_{N+1-i, N+1-j}$; that way $H_{11} = H_{nn}$. – Arturo Magidin Jul 27 '11 at 19:36
Another nitpick: To me, "horizontal and vertical reflection symmetry" suggests that the matrix has horizontal reflection symmetry and vertical reflection symmetry; this is not what you have. What you have is symmetry about the "center of the matrix" (either the central entry if $N$ is odd, or the imaginary central point if $N$ is even). "Central symmetry" might be a better way to describe them. – Arturo Magidin Jul 27 '11 at 20:23
Thanks. I edited the title. – dzhelil Jul 27 '11 at 20:56
Just a quick version of centrosymmetric matrices (thanks to Michael Banaszek for pointing this out):
The matrices you describe in your original question form what is called the centralizer of the involution $$J=\begin{bmatrix} . & . & \dots & . & 1 \\ . & . & \dots & 1 & . \\ & & & & \\ . & 1 & \dots & . & . \\ 1 & . & \dots & . & . \end{bmatrix} \qquad J_{ij} = \begin{cases} 1 & \text{if } i=N-j+1 \\ 0 & \text{otherwise} \end{cases} \qquad \mathcal{H} = \left\{ A : AJ = JA \right\}$$
The centralizer is always an algebra: it is closed under scalar multiplication, addition, and multiplication. An invertible matrix is contained in an algebra if and only if its inverse is. In particular, the group of units of the algebra is the group you are asking about.
In finite groups, the centralizers of involutions are quite important, and this particular involution is called “the longest element of the Weyl group”, and so is reasonably important.
-
Expressing the set in the form $\mathcal{H} = \left\{ A : AJ = JA \right\}$, makes it much easier to prove that the inverse is also in the set since $J^{-1}=J$. Thanks! – dzhelil Jul 27 '11 at 21:30
Addendum: the matrix $J$ in Jack's answer is often termed as the "exchange matrix". – J. M. Jul 28 '11 at 2:36
Edit. This addresses the question as originally asked, without the invertibility condition.
$$\left(\begin{array}{ccc} 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{array}\right)$$ is of the desired form, and is not invertible.
For the invertible case: note that your set is closed under transposing matrices, and under multiplication by (nonzero) scalars. Since the inverse of $A$ can be computed with the adjugate, which is a scalar multiple of the transpose of the matrix of cofactors, your question comes down to determining whether the $(i,j)$-cofactor of a matrix of this form is equal to the $(N+1-i,N+1-j)$-cofactor.
Let $B_{ij}$ be the matrix obtained from $A$ by removing the $i$th row and the $j$th column. That is, we are trying to compare $\det(B_{ij})$ with $\det(B_{N+1-i,N+1-j})$.
I claim that the $(r,s)$ entry of $B_{ij}$ equals the $(N-r,N-s)$ entry of $B_{N+1-i,N+1-j}$.
What is the $(r,s)$ entry of $B_{ij}$?
• If $r\lt i$ and $s\lt j$, then it's $a_{ij}$;
• If $r\lt i$ and $s\geq j$, then it's $a_{i,j+1}$;
• If $r\geq i$ and $s\lt j$, then it's $a_{i+1,j}$;
• If $r\geq i$ and $s\geq j$, then it's $a_{i+1,j+1}$.
If $r\lt i$ and $s\lt j$, then the $(r,s)$ entry of $B_{ij}$ is $a_{rs}= a_{N+1-r,N+1-s}$. Since $(r,s)$ is to the left and above of $(i,j)$, then $(N+1-r,N+1-s)$ is to the right and below $(N+1-i,N+1-j)$, and hence the $(N-r,N-s)$ entry of $B_{N+1-i,N+1-j}$ is $a_{N+1-r,N+1-s}=a_{rs}$, as desired.
If $r\lt i$ and $s\geq j$, then the $(r,s)$ entry of $B_{ij}$ is $a_{r,s+1} = a_{N+1-r,N-s}$. Since $(r,s+1)$ is the left and at below of $(i,j)$, then $(N+1-r,N-s)$ is to the right and above $(N+1-i,N+1-j)$, so $a_{N+1-r,N-s}$ becomes the $(N-r,N-s)$ entry of $B_{N+1-i,N+1-j}$, as desired. A symmetric argument holds if $r\geq i$ and $s\lt j$.
If $r\geq i$, $s\geq j$, then the $(r,s)$ entry of $B_{ij}$ is $a_{r+1,s+1} = a_{N-r,N-s}$. Since $(r+1,s+1)$ is to the right and below $(i,j)$, then $(N-r,N-s)$ is to the left and above $(N+1-i,N+1-j)$, so the $(N-r,N-s)$ entry of $B_{N+1-i,N+1-j}$ is $a_{N-r,N-s}$, as desired.
So the question comes down to whether the determinant of an $N\times N$ matrix is invariant under the transformation that maps the $(i,j)$ entry to the $(N+1-i,N+1-j)$th entry. This is achieved through a series of row and column exchanges: exchange first row with last row; second row with penultimate row; third row with antepenultimate row; etc.; and then exchange first column with last column; second column with penultimate column; third column with ante-penultimate column; etc. In the end, we have performed an even number of row/column exchanges, each of which multiplies the determinant by $-1$. So the two matrices have the same determinant.
Therefore, if $A\in\mathcal{H}$, then the $(i,j)$-cofactor of $A$ equals the $(N+1-i,N+1-j)$-cofactor of $A$; the cofactor matrix of $A$ lies in $\mathcal{H}$; hence the adjugate matrix of $A$ lies in $\mathcal{H}$; hence the inverse of $A$ lies in $\mathcal{H}$. Thus, $\mathcal{H}$ forms a group.
-
Edit: This answers the question before the condition that det(A)>0 was added.
Though it fulfills the other two properties, matrices of this form can't form a group. Just consider any matrix with determinant zero, with the simplest example being a matrix consisting solely of zeroes. This doesn't even have an inverse, so its inverse can't be in $\mathcal{H}$.
-
OK. I added the condition that $\det H >0$. Do they now form a group? – dzhelil Jul 27 '11 at 19:41
Through a bit of searching, I discovered these are known as Centrosymmetric matrices and indeed do have a group structure as long as you specify that their determinant is nonzero. – Michael Banaszek Jul 27 '11 at 20:02
Thanks for the lead. There is actually a paper dedicated to the inverse of such matrices jstor.org/stable/1267339 – dzhelil Jul 27 '11 at 20:55 | 2015-12-01T00:37:37 | {
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https://math.stackexchange.com/questions/1146690/number-of-elements-in-cartesian-power-with-a-maximum-constraint | # Number of elements in cartesian power with a maximum constraint
Problem: I would like to know the number of elements in the cartesian power $X^n$ (cartesian product of one set $X$ by itself, $n$ times) with a maximum constraint: how many elements in $X^n$ have less than $k$ same elements $x$ of the set ($\forall x\in X$)?
Simple example: with $X=\{A, B, C\}$ and $n=3$. The question is: how many three-letter words have less than $k$ times the same letter?
Solution for the simple example:
• for $k=4$, there is no constraint: $|X|^n$, here 27 possibilities.
• for $k=3$, we just can't have three times the same letter, so $AAA, BBB$ and $CCC$ are forbidden, $27 - 3=25$ possibilities.
• for $k=2$, we can't have twice the same letter, so we only have $ABC, ACB, BAC, BCA, CAB, CBA$, 6 possibilities.
• for $k=1$, we can't build any word, 0 possibilities.
General solution? What is the general formula to compute this number as a function of $k$?
Strategy 1: all cases minus cases with more than $k$
One solution would be to count all possibilities without constraints, $|X|^n$, and then remove all cases with $k, k+1, ..., n$ repetitions of all elements.
• For one given element in $X$
• with $k$ of this element: we fix $k$ times the element: $n$ free spaces, $k$ elements, $\binom{n}{k}$ possibilities. For each possibility, we put the others: $n-k$ free spaces, $|X|-1$ elements, $(|X|-1)^{n-k}$ possibilities. In total: $(|X|-1)^{n-k}\cdot \binom{n}{k}$ possibilities for a given other element with $k$ appearances.
• For $k+1$ appearances: $(|X|-1)^{n-k-1}\cdot \binom{n}{k+1}$ possibilities
• ...
• For $n$ appearances: $(|X|-1)^{n-n}\cdot \binom{n}{n}= 1$ possibility, intuitive.
So, for one specific other element: $\sum_{j=k}^{n} (|X|-1)^{n-j}\cdot \binom{n}{j}$ possibilities
• For all elements in $X$:
There are $|X|$ elements, so the number of possibilities with more than $k$ times any other element is $|X|\sum_{j=k}^{n} (|X|-1)^{n-j}\cdot \binom{n}{j}$
Problem: some words are double counted
Strategy 1 for the simple example:
All cases: $|X|^n=3^3=27$ possibilities.
• Let's fix one element $A\in X$ and iterate over $k$.
• Let's fix $k=3$. How many sequences with 3 $A$'s? $\binom{3}{3}=1$ possibility.
• Let's fix $k=2$. How many sequences with 2 $A$'s? $\binom{3}{2}=3$ possibilities to put the $A$'s. Then, there is one extra space, that we can fill in with $B$ or $C$: two possibilities. In total: $3\times 2 = 6$ possibilities.
• Let's fix $k=1$. How many sequences with 1 $A$? $\binom{3}{1} = 3$ possibilities. Then, there are two extra spaces to fill in with $B$ and $C$. If we accept to have twice $B$, the generated word will be $ABB$, and so it's double counted with "Fix one element $B\in X$, $k=2$".
Strategy 2: (by Andrei Rykhalski)
Construct $X_k$ $\forall k=1, ..., n$ containing $k$ times each element $x\in X$.
Build words of length $n$ from alphabet of size $nk$.
Then find all possible distinct combinations of its elements of length n (exclude duplicates).
In the simple example: $X_3=\{A,A,A,B,B,B,C,C,C\}$, $X_2=\{A,A,B,B,C,C\}$, $X_1=\{A,B,C\}$.
• Haven't tried to solve the problem such way but perhaps this helps: you may construct sets $X_k$ for each $k = 1..n$ containing each element of $X$ $k$ times and then find all possible distinct combinations of its elements of length $n$. Feb 13 '15 at 15:47
• @AndreiRykhalski If I understand you well, $X_3=\{AAA, BBB, CCC\}$ and $X_2= \{AAB, AAC, ABA, ACA, BAA, CAA, BBA, BBC, BAB, BCB, ABB, CBB, ...\}$, and $X_1 = \{ABC, ABB, ACB, ...\}$. The issue is, $ABB\in X_2$ and $ABB\in X_1$. Feb 13 '15 at 15:57
• Nope, $X_3 = \{A,A,A,B,B,B,C,C,C\}, X_2 = \{A,A,B,B,C,C\}$. So you are building words of length $n$ from alphabet of size $nk$ and after that need to exclude duplicates. Feb 13 '15 at 16:02
$|X|^n - |X|\sum_{j=k+\lfloor\frac{n-k}{2}\rfloor}^{n} (|X|-1)^{n-j}\cdot \binom{n}{j}$
Same answer than in the question (strategy 1) but $j$ starts not at $k$ but in the middle of the interval $\lfloor k+\frac{n-k}{2}\rfloor$. Justification: by removing all words with $n$, $n-1$, etc. till the middle, we already removed all the other ones, with $k$, $k+1$, ... repetitions. | 2021-10-27T20:11:50 | {
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https://stats.stackexchange.com/questions/70801/how-to-normalize-data-to-0-1-range/154211 | # How to normalize data to 0-1 range?
I am lost in normalizing, could anyone guide me please.
I have a minimum and maximum values, say -23.89 and 7.54990767, respectively.
If I get a value of 5.6878 how can I scale this value on a scale of 0 to 1.
• is this the way =(value-min)/(max-min) – Angelo Sep 23 '13 at 15:31
• It may help you to read this thread: how-to-verify-a-distribution-is-normalized. If that answers your question, you can delete this Q; if not, edit your Q to specify what you still don't understand. – gung - Reinstate Monica Sep 23 '13 at 16:00
• Explanation of protection: This question is attracting extra answers containing code solutions only. While these may be interesting or useful to some readers, it's not an aim of CV to provide repositories of code solutions. – Nick Cox May 27 '15 at 8:42
• the solutions provided consider a linear contrast value - would you like a different normalization, for instance one that achieve an uniform probability for the output? – meduz May 21 '18 at 16:31
If you want to normalize your data, you can do so as you suggest and simply calculate the following:
$$z_i=\frac{x_i-\min(x)}{\max(x)-\min(x)}$$
where $$x=(x_1,...,x_n)$$ and $$z_i$$ is now your $$i^{th}$$ normalized data. As a proof of concept (although you did not ask for it) here is some R code and accompanying graph to illustrate this point:
# Example Data
x = sample(-100:100, 50)
#Normalized Data
normalized = (x-min(x))/(max(x)-min(x))
# Histogram of example data and normalized data
par(mfrow=c(1,2))
hist(x, breaks=10, xlab="Data", col="lightblue", main="")
hist(normalized, breaks=10, xlab="Normalized Data", col="lightblue", main="")
• I only wonder how the two quite different-looking histograms do illustrate the point of your (correct) answer? – ttnphns Sep 23 '13 at 16:21
• @ttnphns They look only different due to the binning of the histograms. My point however was to show that the original values lived between -100 to 100 and now after normalization they live between 0 and 1. I could have used a different graph to show this I suppose or just summary statistics. – user25658 Sep 23 '13 at 16:23
• The gentle nudge by @ttnphns was meant to encourage you not only to use a less complicated means of illustrating a (simple) idea, but also (I suspect) as a hint that a more directly relevant illustration might be beneficial here. You could do both by finding a more straightforward way to graph the transformation when it is applied to the min and max actually supplied by the O.P. – whuber Sep 23 '13 at 17:12
• Is there a way to "normalize" to custom range instead of 0-1? – John Demetriou Oct 25 '16 at 11:46
• @JohnDemetriou May not be the cleanest solution, but you can scale the normalized values to do that. If you want for example range of 0-100, you just multiply each number by 100. If you want range that is not beginning with 0, like 10-100, you would do it by scaling by the MAX-MIN and then to the values you get from that just adding the MIN. So scale by 90, then add 10. That should be enough for most of the custom ranges you may want. – Alexander Rossa Oct 29 '17 at 18:54
The general one-line formula to linearly rescale data values having observed min and max into a new arbitrary range min' to max' is
newvalue= (max'-min')/(max-min)*(value-max)+max'
or
newvalue= (max'-min')/(max-min)*(value-min)+min'.
• This is correct, but not efficient. It is a linear transformation, so you would precalculate a and b constants, and then just apply newvalue = a * value + b. a = (max'-min')/(max-min) and b = max - a * max – Mark Lakata Sep 23 '13 at 19:18
• Do you know how to cite this? I mean, is there an "original" reference somewhere? – Trefex May 11 '14 at 20:18
• @MarkLakata Slight (typo?) correction: b = max' - a * max or b = min' - (a * min) – Nick Dec 29 '14 at 18:09
• @Nick - yes. I'm missing a ' – Mark Lakata Dec 30 '14 at 5:33
• Can you please compare your normalisation here se.mathworks.com/matlabcentral/answers/… i.e. the equation u = -1 + 2.*(u - min(u))./(max(u) - min(u));. – Léo Léopold Hertz 준영 Oct 24 '16 at 21:41
Here is my PHP implementation for normalisation:
function normalize($value,$min, $max) {$normalized = ($value -$min) / ($max -$min);
return $normalized; } But while I was building my own artificial neural networks, I needed to transform the normalized output back to the original data to get good readable output for the graph. function denormalize($normalized, $min,$max) {
$denormalized = ($normalized * ($max -$min) + $min); return$denormalized;
}
$int = 12;$max = 20;
$min = 10;$normalized = normalize($int,$min, $max); // 0.2$denormalized = denormalize($normalized,$min, $max); //12 Denormalisation uses the following formula:$x (\text{max} - \text{min}) + \text{min}\$
• There is an important difference between this answer and the already accepted answer. That explained the main idea clearly and directly and then secondarily showed how to do it in one commonly used program. Conversely, you post here only code. While I am happy to believe that this is good code (I don't write PHP) on this forum we don't normally have a bundle of answers to every question explaining how to do it in every conceivable language. Otherwise we would have answers here in SAS, SPSS, Stata, MATLAB, C, C++, C#, Java. Python, etc, etc. – Nick Cox May 27 '15 at 8:38
• I don't think, that this is the only difference. In my code, I also showed, how to return a normalized value to the value it was before normalisation. I think, that makes it worth this answer. – jankal May 27 '15 at 9:02
• It's still true that you post only code: I think you need to emphasise any supposedly special virtues of code in commentary, as otherwise readers have to read the code to see what they are. Presumably inverting the scaling is of use only when (a) the original values have been overwritten but (b) the user has prudently remembered to save the minimum and maximum. My wider point, as commented above, is that CV does not aim to be a repository of code examples. – Nick Cox May 27 '15 at 9:10
• There are some problems, whre you need to restore the value: Nueral Networks for example... But you're right, in manner of data analysis, this answer is very bad. – jankal May 27 '15 at 9:25
• @NickCox I found his answer to be more satisfactory than the accepted one. – Karl Morrison Aug 30 '15 at 18:40
### Division by zero
One thing to keep in mind is that max - min could equal zero. In this case, you would not want to perform that division.
The case where this would happen is when all values in the list you're trying to normalize are the same. To normalize such a list, each item would be 1 / length.
// JavaScript
function normalize(list) {
var minMax = list.reduce((acc, value) => {
if (value < acc.min) {
acc.min = value;
}
if (value > acc.max) {
acc.max = value;
}
return acc;
}, {min: Number.POSITIVE_INFINITY, max: Number.NEGATIVE_INFINITY});
return list.map(value => {
// Verify that you're not about to divide by zero
if (minMax.max === minMax.min) {
return 1 / list.length
}
var diff = minMax.max - minMax.min;
return (value - minMax.min) / diff;
});
}
### Example:
normalize([3, 3, 3, 3]); // output => [0.25, 0.25, 0.25, 0.25]
• This is a rescaling to a sum 1, not to a range 0-1. I just think the answer is off-topic therefore. – ttnphns Oct 4 '17 at 17:31
• Not so. normalize([12, 20, 10]) outputs [0.2, 1.0, 0.0], which is the same you would get with (val - min) / (max - min). – rodrigo-silveira Jan 16 '19 at 15:23
• @rodrigo-silveira I don't see why the all 0.25 output. Isn't it better all 0.5? All items are equal, so should be kept centered in the interval. – javierdvalle Apr 2 '19 at 14:56
Try this. It is consistent with the function scale
normalize <- function(x) {
x <- as.matrix(x)
minAttr=apply(x, 2, min)
maxAttr=apply(x, 2, max)
x <- sweep(x, 2, minAttr, FUN="-")
x=sweep(x, 2, maxAttr-minAttr, "/")
attr(x, 'normalized:min') = minAttr
attr(x, 'normalized:max') = maxAttr
return (x)
}
• There is an important difference between this answer and the already accepted answer. That explained the main idea clearly and directly and then secondarily showed how to do it in one commonly used program. Conversely, you post here only code. While I am happy to believe that this is good code (in some unexplained language) on this forum we don't normally have a bundle of answers to every question explaining how to do it in every conceivable language. Otherwise we would have answers here in SAS, SPSS, Stata, MATLAB, C, C++, C#, Java. Python, etc, etc. – Nick Cox May 27 '15 at 8:35
the answer is right but i have a suggestion , what if your training data face some number out of range ? you could use squashing technique. it will be guaranteed never to go out of range. rather than this
i recommend use this
with squashing like this in min and max of range
and the size of the expected out-of-range gap is directly proportional to the degree of confidence that there will be out-of-range values.
for more information you can google : squashing the out-of-range numbers and refer to data preparation book of "dorian pyle"
• Please edit your answer to use capitalisation as conventional. Consistent lower case may seem amusing or efficient, but it is more difficult for almost everyone to read. – Nick Cox Sep 25 '13 at 12:12
• The illustrations do not adequately convey your answer. What exactly is a "squashing technique"? – whuber Sep 25 '13 at 14:47 | 2020-02-17T14:15:35 | {
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https://math.stackexchange.com/questions/3196923/can-we-cancel-the-equality-mark-here | # Problem
Let $$f(x)$$ satisfy that $$f(1)=1$$ and $$f'(x)=\dfrac{1}{x^2+f^2(x)}$$. Prove that $$\lim\limits_{x \to +\infty}f(x)$$ exists and is less than $$1+\dfrac{\pi}{4}.$$
# Proof
Since $$f'(x)=\dfrac{1}{x^2+f'(x)}>0$$, $$f(x)$$ is strictly increasing. Thus, $$f(x)>f(1)=1$$ holds for all $$x>1$$, and $$\lim\limits_{x \to +\infty}f(x)$$ equals either the positive infinity or some finite value.
Notice that, $$\forall x>1:$$ \begin{align*} f(x)-f(1)&=\int_1^x f'(t){\rm d}t=\int_1^x \frac{1}{t^2+f^2(t)}{\rm d}t<\int_1^x\frac{1}{t^2+1}{\rm d}t=\arctan x-\frac{\pi}{4}. \end{align*} Therefore $$f(x)<\arctan x-\frac{\pi}{4}+1<\frac{\pi}{2}-\frac{\pi}{4}+1=1+\frac{\pi}{4},$$ which implies that $$f(x)$$ is bounded upward. Thus,$$\lim\limits_{x \to +\infty}f(x)$$ exists. Take the limits as $$x \to +\infty$$, we have $$\lim\limits_{x \to +\infty}f(x)\leq 1+\dfrac{\pi}{4}.$$ Can we cancel the equality mark here? In another word, can we obtain $$\lim\limits_{x \to +\infty}f(x)<1+\dfrac{\pi}{4}$$?
• @PeterForeman Sir, $f(x)=-\frac{1}{x^2}<0$ but $\lim\limits_{x \to +\infty}f(x)=0.$ – mengdie1982 Apr 22 at 10:42
• Are you asked that the limit is smaller than $1+\pi/2$ or than $1+\pi/4$? – kingW3 Apr 22 at 10:43
• @kingW3 No. I just wonder whether the equality with the inequality may hold or not. – mengdie1982 Apr 22 at 10:45
• Sorry. A typo in the "problem". Corrected. See the new version. – mengdie1982 Apr 22 at 10:48
• Possible duplicate of Limit of function as $x \to\infty$ when $f'(x)$ is given – LutzL Apr 22 at 15:26
The function $$g(x)=\int_1^x\frac{1}{t^2+1}{\rm d}t-\int_1^x \frac{1}{t^2+f^2(t)}{\rm d}t$$ Is strictly increasing and $$g(1)=0 for $$x>2$$ hence $$\lim_{x\to\infty}g(x)\geq g(2)>0$$ so$$\lim_{x\to\infty}g(x)=\lim_{x\to\infty}(\frac\pi4-(f(x)-1))=\lim_{x\to\infty}(\frac\pi4+1-f(x))>0$$
So $$\lim_{x\to\infty}f(x)<\frac\pi4+1$$
• In general, $f(x)>g(x)$ implies $\lim f(x) \geq \lim g(x)$ not $\lim f(x) > \lim g(x)$... – mengdie1982 Apr 22 at 11:43
• @mengdie1982 Yeah but I didn't use that in my proof. I've used that $g$ is strictly increasing and that $g(x)>0$ for $x>1$. Using that you could prove $g(2)<g(x)$ for $x>2$ hence $0<\lim_{x\to\infty} g(2) \leq \lim_{x\to\infty} g(x)$. – kingW3 Apr 22 at 11:51
Fix $$M>1$$ and for $$x>M$$ break up your estimate as $$f(x)-f(M)<\arctan x- \arctan M$$ and $$f(M)-f(1)<\arctan M-\frac\pi 4,$$ so there is a positive constant (i.e., depending only on $$M$$, but not on $$x$$) $$\delta_M:=\arctan M-\frac\pi 4-f(M)+f(1)$$. Then for $$x>M$$, $$f(x)-f(1)=f(x)-f(M)+f(M)-f(1)<\arctan x-\frac \pi 4-\delta_M$$ and so $$\lim_{x\to\infty}f(x)\le 1+\frac\pi 4-\delta_M<1+\frac\pi 4.$$ | 2019-05-22T04:44:24 | {
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http://mathhelpforum.com/calculus/115989-easy-way-finding-line-intersection-2-planes.html | # Math Help - easy way of finding line of intersection of 2 planes
1. ## easy way of finding line of intersection of 2 planes
I may be confused so I am hoping someone could verify what I am saying is correct.
To find the line of intersection of 2 planes subtract one from the other so that one of the variables cancels out. Then introduce a parameter and then solve for x y and z.
e.g.
$x+2y-z=10$ and $2x+3y+z=0$. Subtracting 2 times the first equation from the second gives $-y+3z=-20$. Let z = t. Solving the equation for y give y=20+3t. Then from the first equation x=10-2(20+3t)+t
am I correct in believing that I have found the parametric equation of the line of intersection of the 2 plains mentioned above?
The reason I'm doubtful is because when I google searched "line of intersection of two planes" I found a more difficult approach that $r=r_o+tv$ where the cross product is used to find v. I guess this way is used to find the vector equation of the line?
2. Hello, superdude!
To find the line of intersection of 2 planes, subtract one from the other,
so that one of the variables cancels out.
Then introduce a parameter and then solve for x y and z.
.Example: . $\begin{array}{cccc}x+2y-z&=&10 & [1]\\
2x+3y+z&=&0 & [2]\end{array}$
Let me show you the way I explain it . . .
$\begin{array}{cccccc}\text{Multiply }2\times [1]\!: & 2x + 4y - 2z &=& 20 \\ \text{Subtract [2]:} & 2x + 3y + z &=& 0 \\ \\[-3mm]
\text{And we have:} & \qquad\; y - 3z &=& 20 \end{array}$
. . Hence: . $y \:=\:3z + 20$
Substitute into [1]: . $x + 2(3z+20) - z \:=\:10$
. . Hence: . $x \:=\:\text{-}5z-30$
So we have: . $\begin{array}{ccc}x &=& \text{-}5z - 30 \\ y &=& 3x + 20 \\ z &=& z \end{array}$
On the right, replace $z$ with a parameter $t.$
. . and we have: . $\begin{Bmatrix}x &=& \text{-}5t - 30 \\ y &=& 3t + 20 \\ z &=& t \end{Bmatrix}$
The reason I'm doubtful is because when I google searched
"line of intersection of two planes" I found a more difficult approach
where the cross product is used to find v. I guess this way is used
to find the vector equation of the line?
Yes, but you may find this vector approach is faster.
The normal vectors of your two planes are: . $\langle1,2,-1\rangle\,\text{ and }\,\langle2,3,1\rangle$
The cross-product gives the direction of the line of itersection.
. . $\left|\begin{array}{ccc}i & j & k \\ 1 & 2 & \text{-}1 \\ 2 & 3 & 1 \end{array}\right| \;=\;5i - 3j - k \;=\;\langle 5,\text{-}3,\text{-}1\rangle$
Now find any point that lies on both planes . . . For example: . $(0,2,\text{-}6)$
And we have our parametric equations: . $\begin{Bmatrix}x &=& 0 + 5t \\ y &=& 2 -3t \\ z &=& \text{-}6 - t \end{Bmatrix}$
3. Originally Posted by Soroban
Now find any point that lies on both planes . . . For example: . $(0,2,\text{-}6)$
How did you find a point so quickly? | 2014-04-20T06:02:50 | {
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http://math.stackexchange.com/questions/131423/obtain-the-formula-for-the-following-sequence/131429 | Obtain the formula for the following sequence
I can't seem to figure out how to find an algebraic formula for the following sequence of numbers. $$0,\ 1,\ 1,\ 0,\ -1,\ -1,\ 0,\ 1,\ 1,\ 0,\ -1,\ -1,\ 0,\ 1,\ 1,\ 0,\ -1,\ -1,\ 0$$
Can somebody help?
-
Well, I can think of two ways you may have approached this problem. As noted above already, you could have noticed $f(n)=f(n-1)-f(n-2)$ and solved the recurrence algebraically. Perhaps more easy to notice is the fact that this series is periodic with period 6 and then use fourier analysis. In fact if you plot the series you will immediately notice that it looks very much like a sine curve and $f(n)= \frac{2}{\sqrt 3} \sin(\frac{\pi}{3}n)$ fits perfectly.
-
(Inspired by a comment...)
To find the $160$th term, you could notice that the pattern repeats every six numbers.
So we divide by $6$ and look at the remainder.
The $n$th term is
• $1$ if the remainder is $1,2$
• $0$ if the remainder is $0,3$
• $-1$ if the remainder is $4,5$
Since the remainder of $160$ when divided by $6$ is $0$, the $160$th term is $0$.
-
Let $\alpha=\frac{1+i\sqrt{3}}{2}$ and $\beta=\frac{1-i\sqrt{3}}{2}$. Then both $\alpha$ and $\beta$ satisfy $x^2=x-1$. Thus, both $x_k=\alpha^k$ and $x_k=\beta^k$ satisfy $$x_k=x_{k-1}+x_{k-2}\tag{1}$$ which the desired sequence also satisfies. Therefore, $$x_k=\frac{\alpha^{k}-\beta^{k}}{i\sqrt{3}}\tag{2}$$ does, too. Furthermore, $x_k$ as given in $(2)$ matches the desired sequence: $x_0=0$ and $x_1=1$. Therefore, the general term in your sequence is given by $(2)$.
Another way to look at this is to consider $$x_k=\frac{2}{\sqrt{3}}\sin\left(\frac{\pi}{3}k\right)\tag{3}$$ $(3)$ matches for $k$ from $0$ through $5$ and it has period $6$, just like the desired sequence. Thus, $x_k$ as given in $(3)$ also matches your sequence.
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I'm not sure what you mean by "an algebraic formula", but would you be happy with the observation that it satisfies the recurrence $f(n) = f(n-1) - f(n-2)$?
Addendum: Since you know that it satisfies the recurrence, each term depends only on the previous two terms. Now suppose you know that $f(m) = f(n)$ and $f(m+1) = f(n+1)$ for some $m$ and $n$. Then you know that $f(m+2) = f(n+2)$ also, and, by induction, $f(m+k) = f(n+k)$ for all positive $k$.
But from your enumeration above, you know that $f(0) = f(6)$ and $f(1) = f(7)$. So you can conclude that $f(k) = f(6+k)$ for all positive $k$. So the function has a period of 6.
In particular, for any $k$, $f(k) = f(6+k) = f(12+k) = … = f(6j+k)$ for any $j$. Now if someone gives you $n$, you can write $n$ in the form $n = 6j+k$ where $k$ is in $[0…5]$. You do this by dividing n by 6; $j$ is the quotient and $k$ is the remainder. And we know $f(n) = f(6j+k) = f(k)$. But $k$ is in $[0…5]$ so $f(k)$ is easy to calculate.
To compute $f(n)$ for any $n$, calculate $n÷6$ and find the remainder $k$. Then:
• If $k$ is 1 or 2, $f(n)=-1$
• If $k$ is 0 or 3, $f(n)=0$
• If $k$ is 4 or 5, $f(n)=1$
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I already realize that. I'm trying to come up with a way to know the nth term immediately. Say I want to find the 160th term. I would need an algebraic formula to obtain the answer wouldn't I? That's where I'm stuck. – Cs132 Apr 13 '12 at 19:16
Cs132, this information would be good to include in your question. You can edit it and put what you know and/or what you're trying to find out (more precisely) – The Chaz 2.0 Apr 13 '12 at 19:22
Define $x \mod N$ to be the remainder $r$ of $x$ when divided by $N$ with $0 \le r \lt N$.
Then your $n^{th}$ term (starting from $0$) is
$$(n^2 \mod 3) \times (-1)^{\left\lfloor(n \mod 6)/3 \right\rfloor}$$
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Of course, if you wanted to actually use a computer to calculate it, Chaz's answer is probably the fastest and simplest. – Aryabhata Apr 13 '12 at 19:37
It seems to be a quasi-polynomial! – hkju Apr 13 '12 at 20:32
@hkju: I have no idea what that means! – Aryabhata Apr 13 '12 at 20:50
If you accept trigonometry then (as proposed earlier by others)
$$\dfrac{\sin\left(\dfrac {n\pi}3\right)}{\sin\left(\dfrac {\pi}3\right)}\quad \ \text{else}\quad \ \left|\left(n-\dfrac 32\right) \bmod 6-3\right|-\dfrac 32$$
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https://math.stackexchange.com/questions/2391196/typical-absolute-value-inequality | # Typical Absolute value inequality
$$\text{How to solve}\quad \frac{\left\vert\,{x + 3}\,\right\vert + x}{x+2} > 1\quad{\large ?}.$$
I tried and wrote two cases, once opening the mod as it is and then the other case opening the mod with a negative sign.
I got the two cases as : $x\in (-\infty,-2)\cup (-1,\infty)$ and $x\in (-5,-2)$.
The problem is I don't know whether to take union or intersection. Also, the answer I get is different from what's given in the book. Where am I going wrong? What's the best (errorless) way you would handle such problems with the modulus?
• If you are unsure which of two regions to use, just try an example. For instance, $1$ is in the union but not the intersection. Is $1$ a solution? Continuing along those lines, $-10$ is in the union...is it a solution? As to your second question, since you do not show any of your work we can't tell you where you are going wrong. – lulu Aug 12 '17 at 11:20
• Worth noting: as a guess, you may run into difficulties when you multiply both sides by $x+2$. Remember that you need to change the inequality when $x+2<0$. – lulu Aug 12 '17 at 11:25
Instead of just "mindlessly" dividing into cases, write out the logical connectives "and" and "or", and everything should (hopefully) take care of itself: \begin{aligned} \frac{|x+3|+x}{x+2}>1 \quad \iff \quad & \Biggl( x+3 \ge 0 \quad\text{and}\quad \frac{(x+3)+x}{x+2}>1 \Biggr) \\ & \text{or} \quad \Biggl( x+3 < 0 \quad\text{and}\quad \frac{-(x+3)+x}{x+2}>1 \Biggr) \\[1em] \iff \quad & \dots \end{aligned}
• Thanks so much! I got it! Finally! – Tanuj Aug 12 '17 at 14:28
• Great, glad to hear that! – Hans Lundmark Aug 12 '17 at 14:58
note that $$\frac{|x+3|+x}{x+2}=\frac{|x+3|-2}{x+2}>0$$ If $$x\geq -3$$ we get $$\frac{x+1}{x+2}>0$$ if $x>-2$ then we can multiply by $x+2$ and we get $$x>-1$$ if $x<-2$ then we get by multiplication with $x+2>0$ the solution set $$x<-1$$ Can you do the rest?
We need to solve $$\frac{|x+3|+x}{x+2}-1>0$$ or $$\frac{|x+3|-2}{x+2}>0.$$
Now, $x+2=0$ for $x=-2$ and $|x+3|=2$ for $x=-1$ or $x=-5$, which by the intervals method gives the answer: $$(-5,-2)\cup(-1,+\infty).$$
• Wow that is great! Can you explain me what exactly is an interval method and how did you do it? – Tanuj Aug 12 '17 at 11:50
• @user38227 We need to draw the $x$- axes, and to put there points $-5$, $-3$ and $-1$. The sign of the function $f(x)=\frac{|x+3|+x}{x+2}-1$ on $(-1,+\infty)$ is $+$, which gives signs on all four intervals: $-,+,-,+$ because we have no double points. Since we need $+$, we just can write the answer. – Michael Rozenberg Aug 12 '17 at 11:55
With $\ds{x \not= -2}$:
\begin{align} {\verts{x + 3} + x \over x + 2} > 1 & \implies \pars{\verts{x + 3} + x}\pars{x + 2} > \pars{x + 2}^{2} \\[3mm] & \implies \bbx{\pars{\verts{x + 3} - 2}\pars{x + 2} > 0} \\[3mm] & \implies \left\{\begin{array}{lcl} \ds{\pars{x + 5}\pars{x + 2} < 0} & \text{if} & \ds{x < -3} \\[3mm] \ds{\pars{x + 1}\pars{x + 2} > 0} & \text{if} & \ds{x > -3} \end{array}\right. \\[3mm] & \implies \left\{\begin{array}{lcl} \ds{\pars{-5 < x < -2}} & \text{if} & \ds{x < -3} \\[3mm] \ds{\pars{x < -2\quad \text{or}\quad x > -1}} & \text{if} & \ds{x > -3} \end{array}\right. \\[3mm] & \implies \left\{\begin{array}{lcl} \ds{\pars{-5 < x < -3}} & \text{if} & \ds{x < -3} \\[3mm] \ds{\pars{-3 < x < -2}\quad \text{or}\quad x > -1} & \text{if} & \ds{x > -3} \end{array}\right. \end{align}
Clearly, the solution is given by $\ds{\bbx{x \in \pars{-5,-2}\cup\pars{-1,\infty}}}$.
Note that $\ds{x = -3}$ satisfies the initial inequality: Namely, $\ds{{\verts{\color{#f00}{-3} + 3} + \pars{\color{#f00}{-3}} \over \color{#f00}{-3} + 2} = 3 \color{#f00}{>} 1}$. | 2021-06-17T00:25:43 | {
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http://mathoverflow.net/questions/90129/orthogonality-in-non-inner-product-spaces | # Orthogonality in non-inner product spaces
I have come across a notion of orthogonality of two vectors in a normed space not necessarily inner product space. Two vectors $x$ and $y$ in a normed space are said to be orthogonal (represented $x\perp y$) if $||x||\leq ||x+\alpha y||,$ for every $\alpha,$ a scalar.
1) What is the rational behind the definition above? I guess, it has got something to do with minimum overlap between $x$ and $y$.
2) Is this unique generalization of the concept of orthogonality from inner product spaces?
Thank you.
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Here's another paper (seems not to be referenced in the ones mentioned in Valerio's answer): projecteuclid.org/… – Ralph Mar 4 '12 at 22:31
Concerning question 1: The rational is that in an inner product space $$x\perp y \Leftrightarrow \forall \alpha \in K: ||x||\leq ||x+\alpha y|| \qquad(K = \mathbb{R} \text{ or } K = \mathbb{C})$$ Now, if no inner product is available (but a norm), the idea is, to just take the right hand side as definition of orthogonality (call it $\perp_1$).
Concerning question 2: No, there are other -non-equivalent - generalizations as well. As an example, note that in an inner product space over the reals $$\langle x,y \rangle = \frac{1}{4}( ||x+y||^2 - ||x-y||^2).$$ Hence $x\perp y \Leftrightarrow ||x+y|| = ||x-y||$. So the definition $$x\perp_{\scriptstyle 2}\; y : \Leftrightarrow ||x+y|| = ||x-y||$$ generalizes the orthogonality from an inner product space to any normed space (over the reals).
Now let's show that $\perp_1, \perp_2$ aren't equivalent. Let $E = \mathbb{R}^2$ with norm $||(a,b)|| = \max(|a|, |b|)$. Then
$\qquad (0,1) \perp_2 (2,1)$ but not $(0,1) \perp_1 (2,1)\quad$ (take $t=-1/4$)
$\qquad (1,1) \perp_1 (2,0)$ but not $(1,1) \perp_2 (2,0).$
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Follow up questions: (i) Are these two notions of orthogonality symmetric? $\perp_2$ certainly is, but what about $\perp_1$? (ii) Are these notions invariant under scaling? For example, if $x \perp_1 y$, then $x \perp_1 \beta y$ for any $\beta \in K$. Most importantly, (iii): In which contexts are these two notions of orthogonality useful? – shuhalo Mar 4 '12 at 1:43
Another follow up question I have is: Since $\perp_{1}$ and $\perp_{2}$ are equivalent in inner-product spaces, in what way normed spaces are deficient so that these orthogonality concepts don't agree? – Uday Mar 4 '12 at 2:07
Another question: Are there any more non-equivalent orthogonality definitions? – Uday Mar 4 '12 at 2:16
@Martin: (i) $\perp_1$ is not symmetric: From above $(1,1) \perp_1 (2,0)$, but not $(2,0) \perp_1 (1,1)$ (take $t = -1$). (ii) $\perp_1$ is invariant under scalar multiplication, as is aparant from the definition. $\perp_2$ ist not: Again from above, $(0,1) \perp_2 (2,1)$ but not $3 \cdot (0,1) \perp_2 (2,1)$. – Ralph Mar 4 '12 at 10:00
@Uday: Yes, there are more notions of orthogonality in normed spaces - see the survey article quoted by Valerio. In principle you can take any norm-expression that is equivalent to orthogonality in an inner product space and use it as definition of orthogonality in a normed space. – Ralph Mar 4 '12 at 10:04
The definition you gave is called Birkhoff-James orthogonality and the intuition is the following: suppose you have $x,y\in\mathbb R^2$ and construct a triangle with sides $x$ and $y$. Now let $x$ be fixed and consider the same triangle with $-\alpha y$ instead of $y$. Observe that $||x+\alpha y||$ is the length of the third side of this triangle. If you try to write down a picture, you figure out in a moment that the condition $||x||\leq||x+\alpha y||$ can be true for all $\alpha$ iff $x$ and $y$ are orthogonal (looking at the picture, if they are not orthogonal and the inequality is true for some $\alpha$, then it is false for $-\alpha$). Birkhoff-James' orthogonality is a tentative to capture orthogonality through this geometric property.
Birkhoff-James' orthogonality is not the unique notion of orthogonality for normed space.
Some references:
In the following paper
http://arxiv.org/pdf/0907.1813.pdf
you can find some recent very easy application of BJ's orthogonality, as well, if you go through the bibliography, some references about other notions of orthogonality are given. In particular I suggest the paper of Diminnie
Diminnie, C.R. A new orthogonality relation for normed linear spaces, Math. Nachr. 114 (1983), 192-203
and the survey by Alonso and Benitez http://dmle.cindoc.csic.es/pdf/EXTRACTAMATHEMATICAE_1989_04_03_03.pdf
P.s. Bikhoff-James orthogonality is not symmetric in general. Some interesting remarks about symmetric orthogonalities can be found in the paper(s) by Partington in the bibliography of the arxiv paper cited above.
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Concerning your follow-up question (iii) there is the following very nice result: For Birkhoff-James orthogonality it is easy to find examples where $y\perp x$ but $\left\|x\right\|/\left\|x+\alpha y\right\| > 1$ for some real $\alpha$, and so natural to investigate the largest such value $\left\|x\right\|/\left\|x+\alpha y\right\|$ over $X$. In "R. L. Thele, Some results on the radial projection in Banach spaces. Proc. Amer. Math. Soc., 42(2):484--486", it is it is shown that this quantity is exactly the Lipshitz constant for the radial projection onto the unit ball in this norm.
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Well, it depends what do you need it for. You may also have a look at semi-inner-product spaces, which are natural generalizations of inner product spaces.
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https://mathhelpboards.com/threads/evaluation-of-definite-integral.2683/ | # evaluation of definite integral
#### pantboio
##### Member
I'm struggling for a long time to solve this integral
$$\int_0^\infty e^{-x^2}cos(kx)dx$$
with $k>0$
I know there are a number of ways, but I'm interested in using complex integration. In particular, I believe that we can solve by integrating $e^{-z^2}$ over the boundary of the rectangule $[-R,R]\times[0,h]$ for a suitable $h$.
What $h$ do you think i should use?
#### topsquark
##### Well-known member
MHB Math Helper
I'm struggling for a long time to solve this integral
$$\int_0^\infty e^{-x^2}cos(kx)dx$$
with $k>0$
I know there are a number of ways, but I'm interested in using complex integration. In particular, I believe that we can solve by integrating $e^{-z^2}$ over the boundary of the rectangule $[-R,R]\times[0,h]$ for a suitable $h$.
What $h$ do you think i should use?
Personally I wouldn't use the rectangle. I'd note that the integrand is even so
$$\int_0^{\infty} e^{-z^2}cos(kz)~dz = \frac{1}{2}\int_{-\infty}^{\infty} e^{-z^2}cos(kz)~dz$$
and integrate over the upper half plane. (The circular part goes to zero by Jordan's lemma.)
-Dan
#### chisigma
##### Well-known member
I'm struggling for a long time to solve this integral
$$\int_0^\infty e^{-x^2}cos(kx)dx$$
with $k>0$
I know there are a number of ways, but I'm interested in using complex integration. In particular, I believe that we can solve by integrating $e^{-z^2}$ over the boundary of the rectangule $[-R,R]\times[0,h]$ for a suitable $h$.
What $h$ do you think i should use?
A very comfortable way is the use of the Laplace Transform, using the relation...
$\displaystyle \mathcal{L} \{e^{- t^{2}}\}= \frac{\sqrt{\pi}}{2}\ e^{\frac{s^{2}}{4}}\ \text{erfc} (\frac{s}{2})$ (1)
... obtaining...
$\displaystyle \int_{0}^{\infty} e^{- t^{2}} \cos k t\ dt = \frac{\sqrt{\pi}}{2}\ e^{- \frac{k^{2}}{4}}\ \text{Re} \{ \text{erfc} (\frac{i\ k}{2})\} = \frac{\sqrt{\pi}}{2}\ e^{- \frac{k^{2}}{4}}$ (2)
Kind regards
$\chi$ $\sigma$
Last edited:
#### pantboio
##### Member
my solution:
let $\gamma_R$ be the boundary of the rectangle $[-R,R]\times[0,h]$,for $h$ to determine. Let $f(z)=e^{-z^2}$. Thus
$$\oint_{\gamma_R}f(z)dz=0$$
...but we also have
$$\oint_{\gamma_R}f(z)dz=\oint_{\gamma_1}f +\oint_{\gamma2}f+\oint_{\gamma3}f+\oint_{\gamma_4}f$$
where $\gamma_1(t)=t,t\in[-R,R]$, $\gamma_2(t)=R+ti,t\in[0,h]$ ,$-\gamma_3(t)=t+hi,t\in[-R,R]$ and $-\gamma_4(t)=-R+ti,t\in[0,h]$
Hence we have
$$\oint_{\gamma_1} e^{-z^2}dz=\int_{-R}^{R}e^{-t^2}dt$$
$$\oint_{\gamma_3}e^{-z^2}dz=-\int_{-R}^{R}e^{-(t+hi)^2}dt=-e^{h^2}\int_{-R}^{R}e^{-t^2}e^{i(-2ht)}dt$$
Using the fact that sine is odd, the second integral is
$$-e^{h^2}\int_{-R}^{R}e^{-t^2}cos(2ht)dt$$
Now, settin $I_j=\oint_{\gamma_j}$ , $I$=the integral we want to compute, and choosing $h=\frac{k}{2}$ we have
$0=\int_{-R}^{R}e^{-t^2}dt +I_2+I_4-e^{\frac{k^2}{4}}\int_{-R}^{R}e^{-t^2}cos(kt)dt$
Claim: $I_2$ and $I_4$ go to zero for $r$ going to infinite. If so, passing to the limit i get
$$\sqrt\pi-e^{\frac{k^2}{4}}I=0$$
from which
$I=\frac{\sqrt\pi}{2}e^{-\frac{k^2}{4}}$
So we are left to prove:
1)$I_2$ and $I_4$ tends to zero as $R\rightarrow\infty$
2)$\int_{-\infty}^{\infty}e^{-t^2}dt=\sqrt{\pi}$
Last edited:
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
$$\int_0^\infty e^{-x^2}cos(kx)dx$$
with $k>0$
First: why k > 0 ?
second : I will try to solve it and confirm your result , which is surely correct .
$$\int_0^\infty e^{-x^2}cos(kx)dx$$
I will use the substitution $$x^2 = t$$
$$\int_0^\infty e^{-t}\frac{cos(k\sqrt{t})}{2\sqrt{t}}dt$$
$$\int_0^\infty e^{-t}\, \frac{\sum^{\infty}_{n=0}\, \frac{(-1)^n(k\sqrt{t})^{2n}}{(2n)!}}{2\sqrt{t}}dt$$
$$\frac{1}{2}\int_0^\infty e^{-t}\, \sum^{\infty}_{n=0} \frac{(-1)^n k^{2n}t^{n-\frac{1}{2}}}{(2n)!}\,dt$$
$$\frac{1}{2}\sum^{\infty}_{n=0} \frac{(-1)^n k^{2n}}{(2n)!}\,\int_0^\infty e^{-t}t^{n-\frac{1}{2}}\, \,dt$$
$$\frac{1}{2}\sum^{\infty}_{n=0} \frac{(-1)^n k^{2n}\Gamma{(n+\frac{1}{2})}}{(2n)!}$$
$$\frac{1}{2}\sum^{\infty}_{n=0} \frac{(-1)^n k^{2n}\frac{2^{1-2n}\sqrt{\pi}\Gamma{(2n)}}{\Gamma{(n)}}}{(2n)!}\,$$
$$\frac{1}{2}\sum^{\infty}_{n=0} \frac{2\sqrt{\pi}(-k^2)^{n}\Gamma{(2n)}}{4^n(2n)!\Gamma{(n)}}\, \,$$
$$\frac{1}{2}\sum^{\infty}_{n=0} \frac{2\sqrt{\pi}(-k^2)^{n}(2n-1)!}{(2n)4^n(2n-1)!(n-1)!}\, \,$$
$$\frac{1}{2}\sum^{\infty}_{n=0} \frac{\sqrt{\pi}(-\frac{k^2}{4})^{n}}{(n!)}\, \,= \frac{\sqrt{\pi}}{2}e^{-\frac{k^2}{4}}$$
Clearly we don't need the condition k>0 !
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Personally I wouldn't use the rectangle. I'd note that the integrand is even so
$$\int_0^{\infty} e^{-z^2}cos(kz)~dz = \frac{1}{2}\int_{-\infty}^{\infty} e^{-z^2}cos(kz)~dz$$
and integrate over the upper half plane. (The circular part goes to zero by Jordan's lemma.)
-Dan
How , would you please illustrate ?
#### topsquark
##### Well-known member
MHB Math Helper
How , would you please illustrate ?
Let's take a contour such that we include the whole real axis (from -infinity to infinity) and close it off with a semi-circle going from infinity to -infinty. Of course we really have a half-circle with radius R and we take the limit of R as it goes to infinity at the end.
So the integral will be composed of two parts: the real line and the semi-circle. So we have:
$$\int_c = \int_{-\infty}^{\infty} + \int_{R} = 2 \pi i \sum \text{residues in upper half plane}$$
The nice thing about this is that the integral over the semi-circle goes to zero by Jordan's Lemma. So all you are left with calculating the residues in the upper half plane.
-Dan
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Let's take a contour such that we include the whole real axis (from -infinity to infinity) and close it off with a semi-circle going from infinity to -infinty. Of course we really have a half-circle with radius R and we take the limit of R as it goes to infinity at the end.
So the integral will be composed of two parts: the real line and the semi-circle. So we have:
$$\int_c = \int_{-\infty}^{\infty} + \int_{R} = 2 \pi i \sum \text{residues in upper half plane}$$
The nice thing about this is that the integral over the semi-circle goes to zero by Jordan's Lemma. So all you are left with calculating the residues in the upper half plane.
-Dan
Well, that is not completely correct , first you have to prove that the integral over the semi-circle is zero . Second, the function $e^{-z^2}\cos(kz)$ is entire that means if we enclose it by any simply-connected contour the integral is zero . Actually there are no residues since the function has no singularities so if what you are implying is correct then $\int_c = \int_{-\infty}^{\infty} +0 =0$ that means the integral is equal to zero !
#### topsquark
##### Well-known member
MHB Math Helper
Well, that is not completely correct , first you have to prove that the integral over the semi-circle is zero . Second, the function $e^{-z^2}\cos(kz)$ is entire that means if we enclose it by any simply-connected contour the integral is zero . Actually there are no residues since the function has no singularities so if what you are implying is correct then $\int_c = \int_{-\infty}^{\infty} +0 =0$ that means the integral is equal to zero !
Hmmmm...maybe I need to brush up on Jordan's lemma.
-Dan
#### sbhatnagar
##### Active member
The integral may be rewritten as
\begin{aligned} I=\int_0^\infty e^{-x^2}\cos(kx) dx &= \frac{1}{2}\int_{-\infty}^{\infty} e^{-x^2}\cos(kx) dx \\ &= \frac{1}{2} \text{Re} \left[\int_{-\infty}^{\infty} e^{-x^2+ikx} dx\right]\end{aligned}
Here, we can use the general formula
$$\int_{-\infty}^\infty e^{-x^2+bx+c}dx = \sqrt{\pi} e^{b^2/4+c}$$
with $b=ik$ and $c=0$.
$$I =\frac{1}{2} \text{Re} \left[\sqrt{\pi} e^{-k^2/4}\right] = \frac{\sqrt{\pi}}{2} e^{-k^2/4}$$
The formula, I have used can be proved in the following manner:
\begin{aligned} \int_{-\infty}^\infty e^{-x^2+bx+c}dx &= \int_{-\infty}^{\infty} \exp \left\{{-\left( x^2-bx+\frac{b^2}{4}-c-\frac{b^2}{4}\right)} \right\}dx \\ &= e^{b^2/4+c} \int_{-\infty}^\infty e^{-(x-b/2)^2}dx \\ &= \sqrt{\pi} e^{b^2/4+c} \end{aligned}
The last integral has been done with the substitution $t=x-b/2$.
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
The formula, I have used can be proved in the following manner:
\begin{aligned} \int_{-\infty}^\infty e^{-x^2+bx+c}dx &= \int_{-\infty}^{\infty} \exp \left\{{-\left( x^2-bx+\frac{b^2}{4}-c-\frac{b^2}{4}\right)} \right\}dx \\ &= e^{b^2/4+c} \int_{-\infty}^\infty e^{-(x-b/2)^2}dx \\ &= \sqrt{\pi} e^{b^2/4+c} \end{aligned}
The last integral has been done with the substitution $t=x-b/2$.
This is not enough to deduce that the formula can be analytically continued to be used for b is complex .
Last edited:
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
\begin{aligned} \int_{-\infty}^\infty e^{-x^2+bx+c}dx &= \int_{-\infty}^{\infty} \exp \left\{{-\left( x^2-bx+\frac{b^2}{4}-c-\frac{b^2}{4}\right)} \right\}dx \\ &= e^{b^2/4+c} \int_{-\infty}^\infty e^{-(x-b/2)^2}dx \\ &= \sqrt{\pi} e^{b^2/4+c} \end{aligned}
The last integral has been done with the substitution $t=x-b/2$.
So we have the integral $\int_{-\infty}^\infty e^{-(x-b/2)^2}dx$
Now if we assume that b is complex the substitution you made is a bit tricky .why ?
Assume that $b= ic$ for simplicity Re(b)=0 .
So let us make the substitution $t= x-\frac{b}{2}= x-\frac{ic}{2}$ but we know that when making a substitution this applies to the bounds of integration as well , but wait how do we do that ?
Well, the limits of integration after substitution will not still be the same so they will change.
$$\lim_{R\to \infty }\int^{R-\frac{ic}{2}}_{-R-\frac{ic}{2}}e^{-t^2}\, dt$$
Now this looks familiar in the complex plane it is an integration a long a closed rectangle contour the has and infinite width along the x-axis and a height equal to c/2 .
To solve this we apply the method that pantboio described earlier which is a contour integration . | 2021-07-27T21:43:31 | {
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http://mathhelpforum.com/pre-calculus/3169-very-stuck-help-squares-functions.html | # Math Help - Very Stuck. Help on Squares of functions!
1. ## Very Stuck. Help on Squares of functions!
Hi I need some info on how to Sketch the square of functions.
Eg (x+6)(x+3)(x-4) Sketch the square and comment on the number of turning points.
Can someone find a website on this? I have searched google to no avail.
Thanks
2. Originally Posted by classicstrings
Hi I need some info on how to Sketch the square of functions.
Eg (x+6)(x+3)(x-4) Sketch the square and comment on the number of turning points.
Can someone find a website on this? I have searched google to no avail.
Thanks
The function $f(x)=(x+6)^2(x+3)^2(x-4)^2$ is always greater than or
equal to zero, it is increasing as $x \to \pm \infty$, and touches the $x$ axis when
$x=4,\ -3$ and $-6$.
Each of the zeros is a turning point, so it must have an additional turning points
between them making at least 5 turning points. Also it can have at most
five turning points (as $f'(x)$ is a quintic and so has five zeros
which are potential turning points), hence it has exactly 5 turning points.
This should be sufficient to allow you to sketch the curve.
RonL
3. Hello, classicstrings!
An interesting problem . . . I've never been asked to do this.
$f(x)\,=\,(x+6)(x+3)(x-4)$
Sketch the square and comment on the number of turning points.
$f(x)$ is a cubic with x-intercepts: -6, -3, 4.
There are two turning points.
Code:
|
| *
|
** | *
-6 * * | *
- - o - + - + - o - + - + - + - + - + - o - - -
* -3* | * 4
* | *
* *| *
| **
|
$g(x) \,= \,(x=6)^2(x+3)^2(x-4)^2$ has the same x-intercepts but of order two.
. . The graph is tangent to the x-axis there.
Also the entire graph is above (or on) the x-axis.
Code:
|
* | ** *
** |* *
* * * *| * *
* * * * | * *
- - o*- + - + - o*- + - + - + - + - + -*o - - -
-6 -3 | 4
There are five turnning points.
4. Thanks for your help guys! The diagrams + explanations rock! CHeers!!!
EDIT: If you read what i posted before forget it!
EDIT: Soroban do you post on other maths help boards? I think I have seen you elsewhere.
5. Also where would the intercepts of the two lines be? Thanks | 2016-07-26T05:25:25 | {
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https://mathhelpboards.com/threads/find-all-real-p-q-a-and-b.6500/ | # Find all real p, q, a, and b
#### anemone
##### MHB POTW Director
Staff member
Hi MHB,
Initially, I thought this is another boring high school mathematics problem, but when I started to work on it, I realized I was beaten by it, with equations in variables $a$, $b$, $p$ and $q$ to which I don't see a clear way to find the values for them.
Problem:
Find all real $p$, $q$, $a$ and $b$ such that we have $$\displaystyle (2x-1)^{20}-(ax+b)^{20}=(x^2+px+q)^{10}$$ for all $x$.
Attempt:
After expanding both sides of the equation using wolfram, I get
$$\displaystyle (1048576-a^{20})x^{20}-(10485760+20a^{19}b)x^{19}+(49807360-190a^{18}b^2)x^{18}+\cdots-(1140a^3b^{17}+9120)x^{3}$$
$$\displaystyle +(-190a^2b^{18}+760)x^{2}-(20ab^{19}+40)x+1-b^{20}=x^{20}+10px^{19}+(10q^9+45p^2q^8)x^{2}+10pq^9x+q^{10}$$
And I ended up getting extremely messy equations where solving for the values for $a$, $b$, $p$ and $q$ seems impossible by equating the coefficient of $x^{20}$, $x^{19}$, $x^{18}$, $x^{3}$, $x^{2}$, $x$ and the constant...
Could anyone help me with this problem?
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Some solutions would be obtained by letting
$$\displaystyle a=2,b=-1$$
and finding the roots for
$$\displaystyle x^2+px+q=0$$
##### Well-known member
Because it is true for all x you can choose any value of x
setting x = 1/2 shall make 1st term = 0
we get
- (b+1/2a)^20 = (1/4 + 1/2p+ q) ^ 10
LHS <=0 and RHS >= 0 so both are 0
So b =- 1/2a and 1/2p + q + 1/4 = 0
So a = - 2b and 2p + 2q + 1 = 0
by putting suitable values of x you can proceed.
##### Well-known member
Because it is true for all x you can choose any value of x
setting x = 1/2 shall make 1st term = 0
we get
- (b+1/2a)^20 = (1/4 + 1/2p+ q) ^ 10
LHS <=0 and RHS >= 0 so both are 0
So b =- 1/2a and 1/2p + q + 1/4 = 0
So a = - 2b and 2p + 2q + 1 = 0
by putting suitable values of x you can proceed.
put x = 0 to get 1= b^20 + q^ 10
so b= (sin t)^(1/10) , q = (cos t)^(1/5)
a = -2 (sin t)^(1/10), p = - ( 1+ (cos t)^(1/5))/2
should be the solution for some t
unless I have missed out something
#### Opalg
##### MHB Oldtimer
Staff member
Find all real $p$, $q$, $a$ and $b$ such that we have $$\displaystyle (2x-1)^{20}-(ax+b)^{20}=(x^2+px+q)^{10}$$ for all $x$.
Compare coefficients of $x^{20}$ to get $2^{20} - a^{20} = 1$. So $a = \pm (2^{20} - 1)^{1/20}$. As kaliprasad points out, $b = -\frac12a$, so $b = \mp\frac12(2^{20} - 1)^{1/20}$. Therefore $ax+b = \pm(2^{20} - 1)^{1/20}\bigl(x-\frac12\bigr)$, and $(ax+b)^{20} = (2^{20} - 1)\bigl(x-\frac12\bigr)^{20}$. Thus $$(2x-1)^{20}-(ax+b)^{20}= 2^{20}\bigl(x-\tfrac12\bigr)^{20} - (2^{20} - 1)\bigl(x-\tfrac12\bigr)^{20} = \bigl(x-\tfrac12\bigr)^{20} = \bigl(x^2- x + \tfrac14\bigr)^{10}.$$ So we must take $p=-1$ and $q=\frac14.$
So the solution is $a = \pm (2^{20} - 1)^{1/20}$, $b = \mp\frac12(2^{20} - 1)^{1/20}$, $p=-1$, $q=\frac14.$
#### anemone
##### MHB POTW Director
Staff member
Thank you all for the replies...I greatly appreciate all the helps! | 2022-01-27T12:32:04 | {
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http://math.stackexchange.com/questions/289443/de-involving-hyperbolic-trig-function-quick-check | # DE involving Hyperbolic Trig function (Quick check)
I'm working problem 15 on page 133 of Boyce/DiPrima's Elementary Differential Equations and Boundary Value Problems (10th ed.) Note: this is homework, but it is not graded/turned in.
I have arrived at a solution which I think is correct, but it doesn't match the answer in the back of the book. (Don't you hate it when that happens?)
My question to the Math.SE community is, would you please see if a) I made an error in my process, or b) did I just not simplify it enough?
The problem statement:
Solve: $$(e^x + 1)\frac{dy}{dx} = y-ye^x$$
My (possibly erroneous) solution:
Separate and integrate: $$\frac{1}{y}\frac{dy}{dx} = \frac{(1-e^x)}{(e^x+1)}$$ $$\ln|y| = \int\frac{dx}{e^x+1} - \int\frac{e^x}{e^x + 1}dx$$ The left-most integral is simple: $$\ln|y| = \int\frac{dx}{e^x+1} - \ln{|e^x + 1|}$$ The other one, I multiplied by $\frac{e^{-x}}{e^{-x}}$: $$\ln|y| = \int\frac{e^{-x}dx}{1+e^{-x}} - \ln{|e^x + 1|}$$ $$\ln|y| = -\int\frac{-e^{-x}dx}{1+e^{-x}} - \ln{|e^x + 1|}$$ Now it's simple again! $$\ln|y| = -\ln{|1+e^{-x}|} - \ln{|e^x + 1|} + C$$ Raise the equation as a power of $e$: (I'm ignoring the abs signs from here on out, just for simplicity's sake... I don't think it causes a problem) $$y = \left(\frac{1}{1+e^{-x}}\right)\left(\frac{1}{e^x + 1}\right)(C)$$ Simplify: $$y = \left(\frac{C}{e^x+e^{-x}+2}\right)$$ $$y = \left(\frac{C}{2\left(\frac{e^x+e^{-x}}{2}\right)+2}\right)$$ $$y = \left(\frac{C}{2\cosh(x)+2}\right)$$ I'll factor the 2 out, and merge into C, giving my solution:
$$y = \left(\frac{C}{\cosh(x)+1}\right)$$
Their solution:
Close, but not exactly like mine...
$$y = \frac{c}{\cosh^2(\frac{x}{2})}$$
Because of the squaring, I doubt it's just a simplification error, and I've gone wrong somewhere in the problem. Any help?
-
Half-angle formula: dlmf.nist.gov/4.35 – Ron Gordon Jan 29 '13 at 2:18
The answers are equivalent, by the half-angle formula for hyperbolic cosine: $$\cosh^2 \frac{x}{2} = \frac{1 + \cosh x}{2}.$$ This formula is easy to verify from the definition of $\cosh$. So your $C$ is $2c$.
By the way, it's ok for you to drop the absolute value sign since $$\frac{1}{(e^{-x} + 1)(1 + e^x)} > 0$$ for all $x$. | 2014-10-21T07:22:09 | {
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http://math.stackexchange.com/questions/855610/probability-of-winning-a-contest | # Probability of Winning a Contest
This is my first question so apologies if its unclear/vague.
There exists a contest with me in it, and $5$ others, thus $6$ people in total, along with $5$ prizes. A person can only win one prize, and once they do, they're out of the contest. Winners are chosen at random.
So, what is the chance of me winning at a prize?
My initial thought was: $\frac16$ chance initially, then $\frac15$ if I don't win the first time, $\frac14$ if I don't win the second, ect. as people are removed once they win, resulting in $\frac16+\frac15+\frac14+\frac13+\frac12 = 1.45$ which is $145$%.
This is greater than $100$%, and obviously I am not guaranteed to win as I can be the $1$ loser, so how do you find the correct probability?
Thanks!
Edit: All of you have been extremely helpful. Thank you!
-
You have an implicit assumption that the probability of each person winning each prize is equal - this may not be the case for all contests, such as a raffle drawing where people may buy any number of tickets. (That's not a criticism, just a warning that the answers below won't always hold.) – Patrick M Jul 3 '14 at 21:07
Well, if we are choosing $5$ winners at random out of $6$ people, then you have a $\dfrac{5}{6}$ probability of winning.
However, your approach is correct - we can sum the individual probabilities to get the same result, but we have to be cautious. Consider the first two prizes given. As you say, we have a $\dfrac{1}{6}$ probability of winning the first prize, and then a $\dfrac{1}{5}$ chance of winning the second prize if we don't win the first prize. We need to take into account this part - you only have a $\dfrac{5}{6}$ chance to get to the second round in the first place.
Using your method, this would give us a total probability of
$$\frac{1}{6}+\frac{5}{6}\left(\frac{1}{5}\right)+\frac{5}{6}\left(\frac{4}{5}\right)\left(\frac{1}{4}\right)+\frac{5}{6}\left(\frac{4}{5}\right)\left(\frac{3}{4}\right)\left(\frac{1}{3}\right)+\frac{5}{6}\left(\frac{4}{5}\right)\left(\frac{3}{4}\right)\left(\frac{2}{3}\right)\left(\frac{1}{2}\right) \\=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{5}{6}$$
-
i understand now! thank you! – user161504 Jul 3 '14 at 17:44
@user161504 Glad to help. – Peter Woolfitt Jul 3 '14 at 17:56
The simplest way to see it is that there is only $1$ person who doesn't win a prize. Your chance of being that person is $\frac 16$, so your chance of winning something is $1-\frac 16=\frac 56$
-
so - is the answer just number of prizes divided by number of people always? – user161504 Jul 3 '14 at 18:15
Yes, as long as nobody can win more than one prize. It is called Linearity of expectation. Your expectation of each prize is $\frac 16$, so your total expectation is $\frac 56$ – Ross Millikan Jul 3 '14 at 18:15
@RossMillikan Except we're trying to compute a probability, not an expectation. So you need to add an argument about why the expectation really is a probability because every prize has value 1. – David Richerby Jul 3 '14 at 21:37
Hint: Can you win the second prize if you won the first prize? No. So you need to multiply the probability of winning the second prize by something. Similarly for the other prizes of course.
Alternatively you can do it the other way round and find the probability that you don't win any prize. That one is easier to find.
A third way is as follows: There is symmetry among all people. Nobody has an advantage or disadvantage. And since exactly one of the six people doesn't go home with a prize, we know the probability.
-
Calculate it with the probability of the complementary event (not winning a prize), which is: $\frac{5}{6}\cdot \frac{4}{5} \cdot \frac{3}{4}\cdot\frac{2}{3} \cdot\frac{1}{2} = \frac{1}{6}$ , then substract this from 1 to get your probability, which is therefore $1-\frac{1}{6} = \frac{5}{6}$.
-
so - is the answer just number of prizes divided by number of people always? – user161504 Jul 3 '14 at 18:14 | 2015-05-25T20:05:06 | {
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https://math.stackexchange.com/questions/1582106/i-want-to-calculate-the-limit-of-lim-x-to-0-left-frac2x8x2-right/1582117 | # I want to calculate the limit of: $\lim_{x \to 0} \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x}$
I want to calculate the limit of: $$\lim_{x \to 0} \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x}$$ or prove that it does not exist. Now I know the result is $4$, but I am having trouble getting to it. Any ideas would be greatly appreciated.
• nice...........+1 – Bhaskara-III Dec 19 '15 at 13:05
First, consider the limit of the logarithm of your function:
$\lim_{x \to 0} \frac{\ln \left( \frac{2^x + 8^x}{2} \right)}{x}$
This is in the indeterminate form $\frac{0}{0}$, so try L'Hopital's rule.
$\lim_{x \to 0} \frac{2}{2^x + 8^x} \left( \frac{2^x \ln 2 + 8^x \cdot 3 \ln 2}{2} \right) = \ln 4$
Using continuity of the exponential function, you get the original limit is 4.
Notice, $$\lim_{x\to 0}\left(\frac{2^x+8^x}{2}\right)^{1/x}$$ $$=\lim_{x\to 0}\left(\frac{2^x+2^{3x}}{2}\right)^{1/x}$$
$$=\lim_{x\to 0}\exp\frac{1}{x}\ln\left(\frac{2^x+2^{3x}}{2}\right)$$ Applying L'hospital's rule for $\frac 00$ form $$=\lim_{x\to 0}\exp\frac{\frac{d}{dx}\ln\left(\frac{2^x+ 2^{3x}}{2}\right)}{\frac{d}{dx}(x)}$$
$$=\lim_{x\to 0}\exp\frac{\frac{1}{\left(\frac{2^x+2^{3x}}{2}\right)}\left(\frac{2^x\ln 2+3\cdot 2^{3x}\ln 2}{2}\right)}{1}$$
$$=\exp\frac{\left(\frac{2^0\ln 2+3\cdot 2^{0}\ln 2}{2}\right)}{\left(\frac{2^0+2^{0}}{2}\right)}=e^{2\ln 2}=2^2=\color{red}{4}$$
More generally, for $a>0$ and $b>0$, $$\lim_{x\to0}\frac{1}{x}\log\frac{a^x+b^x}{2}= \lim_{x\to0}\frac{\log(a^x+b^x)-\log 2}{x}$$ is the derivative at $0$ of the function $f(x)=\log(a^x+b^x)$. Since $$f'(x)=\frac{a^x\log a+b^x\log b}{a^x+b^x}$$ we have $$f'(0)=\frac{\log a+\log b}{2}=\log\sqrt{ab}$$ Thus $$\lim_{x\to0}\left(\frac{a^x+b^x}{2}\right)^{1/x}= e^{\log\sqrt{ab}}=\sqrt{ab}$$
You might enjoy proving that $$\lim_{x\to\infty}\left(\frac{a^x+b^x}{2}\right)^{1/x}=\max(a,b)$$
Let $f(x) = \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x}$. For $x \to 0$, We have \begin{align*} \log f(x) &= \frac{1}{x} \log \left[ \frac{1}{2}\left(e^{x \log 2} + e^{x \log 8}\right) \right] \\ &= \frac{1}{x}\log\left(1 + x \log 4 + o(x)\right) \\ &= \frac{1}{x}\left(x \log 4 + o(x)\right) \\ &= \log 4 + o(1). \end{align*} Thus $f(x) \to 4$.
This solution use series expansion. Near $0$ we have $$2^x=1+x\log2 +o(x)$$ $$8^x=1+x\log8+o(x)$$ Then applying $\log$ to your limit $$\frac{1}{x}\log \Big( \frac{2^x+8^x}{2}\Big)=\frac{1}{x}\log \Big( 1+x\frac{\log8+\log2}{2}+o(x)\Big)$$ and using $\log(1+x)$ expansion this is eq. to $$\frac{1}{x} x\frac{\log8+\log2}{2}+o(x)$$ The goes to $\frac{\log8+\log2}{2}$ when $x$ goes to $0$, hence the original limit is $$e^{\frac{\log8+\log2}{2}}=4$$ | 2020-04-03T12:12:31 | {
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https://www.physicsforums.com/threads/directional-derivative.858523/ | # Directional Derivative
1. Feb 21, 2016
### RyanTAsher
1. The problem statement, all variables and given/known data
Find the directional derivative of $f$ at $P$ in the direction of $a$.
$f(x,y) = 2x^3y^3 ; P(3,4) ; a = 3i - 4j$
2. Relevant equations
$D_u f(x_0, y_0, z_0) = f_x(x_0, y_0, z_0)u_1 + f_y(x_0, y_0, z_0)u_2$
3. The attempt at a solution
$f_x (x,y) = 6x^2y^3$
$f_y (x,y) = 6x^3y^2$
$f_x (3,4) = 3456$
$f_y (3,4) = 2592$
$D_u f(x_0, y_0) = 3456u_1 +2592 u_2$
$u = \frac {a} {||a||} = \frac {\langle 3,4 \rangle} {5} = \langle \frac {3} {5}, \frac {4} {5} \rangle$
$D_u f(x_0, y_0) = 3456(\frac {3} {5}) + 2592(\frac {4} {5})$
$D_u f(x_0, y_0) = \frac {20736} {5}$
Now, my program wants this an exact number, no tolerance. It won't accept division either, so I don't know how to put in 20736/5. Just wondering if I made a mishap somewhere within the solution.
2. Feb 21, 2016
### SteamKing
Staff Emeritus
You can't enter in a decimal number?
3. Feb 21, 2016
### RyanTAsher
Wouldn't that not be exact, but approximate form though? Or if I decimal isn't repeating is it considered exact?
4. Feb 21, 2016
### SteamKing
Staff Emeritus
So, you're saying that (1/2) = 0.5 is only an approximation and not an exact representation? Interesting.
5. Feb 21, 2016
### RyanTAsher
So, I'm guessing it's not an approximation? Makes sense, good to learn something new. I will attempt to insert my answer.
6. Feb 21, 2016
### RyanTAsher
I attempted the answer of 4147.2, and it was incorrect. Therefore, my work must be incorrect somewhere.
7. Feb 21, 2016
### SteamKing
Staff Emeritus
What if the wrong answer has been programmed into the software you're using?
8. Feb 21, 2016
### RyanTAsher
I've spoken to other students who have the same problem, but with different numbers, so I'm pretty positive that the programs solution is correct, but now that I am at home I don't have access to see any of their solutions.
9. Feb 21, 2016
### Ray Vickson
$u_y \neq 4/5$; go back and check your work.
10. Feb 21, 2016
### RyanTAsher
Ah, thank you I missed that. I have remodeled my work, and the solution turns out to be 0, and is correct. Thank you both for your time. | 2017-08-17T14:12:42 | {
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https://math.stackexchange.com/questions/2661224/open-and-closed-sets-discrete-metric | # Open and Closed Sets Discrete Metric
I have 2 contradictory solutions to the follower problem:
The problem:
Let $X$ be an infinite set. For $p \in X$ and $q\in X$ define $$d(x,y)= \begin{cases} 0\qquad&\text{if and only if x=y}\\\ 1&\text{otherwise} \end{cases}$$ Which subsets are closed and which are open?
My "solution":
Every point is isolated: Consider balls of radius $1/2$ around each x. Which is to say no point is a limit point in $X$ (a limit point will have other points of $X$ in the neighborhood). These points, which are not limit points, will be contained in any subset of $X$. Thus no subset of $X$ will contain limit points. Thus no subset of $X$ will be closed.
But:
Balls of radius $1/2$ centered at $x \in X$ are open because the topology of a metric space has all open balls defined to be open. The complement of this ball is closed by proof in Rudin. It is also non-empty (containing all the other points that are not $x\in X$). Thus the open sets are unions of open balls containing points and their complements are the closed sets.
I produced no closed sets in the first argument and a bunch in the second -- what happened?? After reviewing the "similar questions", I think the second explanation is correct. I don't see why the first one is wrong though.
• a closed set does not have to have limit points. you proved every subset is open, and so every subset is closed also. – Forever Mozart Feb 22 '18 at 2:54
• A set is closed if and only if it contains all of its limit points. If the set has no limit points in the first place then this statement is vacuously true. So your conclusion "Thus no subset of $X$ will contain limit points. Thus no subset of $X$ will be closed" is not true. – wgrenard Feb 22 '18 at 2:58
• The over reliance on limit points is a blunder, an inferior approach to closed sets. – William Elliot Feb 22 '18 at 3:04
• @wgrenard So the subsets of the integers under $R^1$ metric are closed vacuously? – yoshi Feb 22 '18 at 3:09
• @yoshi The second argument looks correct, but I don't think it's a good answer for the problem. The correct answer may be "All subsets in $X$ are both open and closed", in other words, the topology of $X$ is discrete. – ChoF Feb 22 '18 at 3:29 | 2019-07-16T05:56:33 | {
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https://www.beatthegmat.com/que-what-is-the-arithmetic-mean-of-2x-and-10z-t325829.html?&view=next&sid=cdca4a687503179f13b4599511e50384 | ## If $$p, s,$$ and $$t$$ are positive integer, is $$|ps - pt| > p(s - t)?$$
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### If $$p, s,$$ and $$t$$ are positive integer, is $$|ps - pt| > p(s - t)?$$
by VJesus12 » Wed Aug 18, 2021 7:30 am
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## Global Stats
If $$p, s,$$ and $$t$$ are positive integer, is $$|ps - pt| > p(s - t)?$$
(1) $$p < s$$
(2) $$s < t$$
Source: Official Guide
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### Re: If $$p, s,$$ and $$t$$ are positive integer, is $$|ps - pt| > p(s - t)?$$
by [email protected] » Thu Aug 19, 2021 7:48 am
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## Global Stats
VJesus12 wrote:
Wed Aug 18, 2021 7:30 am
If $$p, s,$$ and $$t$$ are positive integer, is $$|ps - pt| > p(s - t)?$$
(1) $$p < s$$
(2) $$s < t$$
Source: Official Guide
Target question: Is |ps - pt| > p(s - t) ?
In other words, Is |ps - pt| > ps - pt?
This is a good candidate for rephrasing the target question.
KEY CONCEPT: |x - y| can be thought as the DISTANCE between x and y on the number line.
For example, |3 - 10| = the DISTANCE between 3 and 10 on the number line.
And |6 - 1| = the DISTANCE between 6 and 1 on the number line.
IMPORTANT: We can also find the distance 6 and 1 on the number line by simply subtracting 6 - 1 to get 5, so why do we need absolute values? Can't we just conclude that |x - y| = x - y?
Great questions, me!
For SOME values of x and y, it's true that |x - y| = x - y, and for other values it is NOT the case that |x - y| = x - y
For example, if x = 5 and y = 2, then we get: |5 - 2| = 5 - 2. In this case |x - y| = x - y
Likewise, if x = 11 and y = 3, then we get: |11 - 3| = 11 - 3. In this case |x - y| = x - y
And, if x = 7 and y = 7, then we get: |7 - 7| = 7 - 7. In this case |x - y| = x - y
CONVERSELY, if x = 4 and y = 6, then we get: |4 - 6| = 4 - 6. In this case |x - y| x - y
Likewise, if x = 5 and y = 20, then we get: |5 - 20| = 5 - 20. In this case |x - y| x - y
And, if x = 0 and y = 1, then we get: |0 - 1| = 0 - 1. In this case |x - y| x - y
Notice the |x - y| = x - y IS true when x > y, and |x - y| = x - y is NOT true when x < y
If x < y, then |x - y| = some POSITIVE value, and x - y = some NEGATIVE value.
This means that, if x < y, then |x - y| > x - y
The target question asks Is |ps - pt| > ps - pt?
According to our conclusion above, if ps > pt, then |ps - pt| = ps - pt and . . .
if ps < pt, then |ps - pt| > ps - pt
This means we can REPHRASE the target question....
REPHRASED target question: Is ps < pt?
We can make things even easier, if we notice that, since p is POSITIVE, we can safely take the inequality ps < pt and divide both sides by p to get: s < t?.
RE-REPHRASED target question: Is s < t?
At this point, it will be very easy to analyze the answer choices....
Statement 1: p < s
Since there's no information about t, there's no way to answer the RE-REPHRASED target question with certainty.
Statement 1 is SUFFICIENT
Statement 2: s < t
Perfect!
The answer to the RE-REPHRASED target question is YES, s IS less than t
Since we can answer the RE-REPHRASED target question with certainty, statement 2 is SUFFICIENT | 2022-12-02T22:23:17 | {
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https://www.physicsforums.com/threads/critical-density.721335/ | # Critical Density
1. ### nick1o2
26
I was reading up on critical density, and found the "current" number for it, but can't fine any past records or graphs to show how they have changed over time. Any help?
2. ### dauto
What critical density? If you care enough to actually tell us what you're talking about you might eventually even get an answer. Just saying...
Last edited: Nov 7, 2013
3. ### nick1o2
26
The Average Critical Density of the universe. Sorry should of made that clear.
### Staff: Mentor
Moved to the Cosmology forum because this looks like a cosmology topic. If I'm mistaken, say so and I'll move it somewhere else.
5. ### marcus
25,096
ρcrit(t) is just something you calculate from H(t). It changes as the Hubble expansion rate changes. It is proportionate (by a constant factor) to H2.
So you can track it by looking at a record of H(t) over time. I'll try to think of how to get a graph or table.
6. ### marcus
25,096
If you're familiar with Friedman eqn. then you remember that
H2 = (8πG/3c2
with ρ expressed as an energy density (if you like it as a mass density then omit the c2)
So solve for ρ:
ρ = (3c2/8πG)H2
Are you familiar with the Hubble time? It is simply the reciprocal of the rate:
THubble = 1/H
Let's denote it by Θ so we don't have to write the subscript
Θ(t) = 1/H(t)
So if I can show you a table of the past history of the time Θ(t) you can calculate ρcrit!
ρcrit = 3c2/(8πG Θ2)
Here's a table of past values of the Hubble time Θ listed in billions of years.
$${\scriptsize\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline T (Gy)&\Theta (Gy) \\ \hline 0.473&0.7105\\ \hline 0.566&0.8504\\ \hline 0.678&1.0176\\ \hline 0.811&1.2173\\ \hline 0.971&1.4558\\ \hline 1.162&1.7401\\ \hline 1.390&2.0787\\ \hline 1.663&2.4807\\ \hline 1.988&2.9566\\ \hline 2.375&3.5172\\ \hline 2.835&4.1732\\ \hline 3.380&4.9340\\ \hline 4.023&5.8050\\ \hline 4.777&6.7856\\ \hline 5.654&7.8652\\ \hline 6.666&9.0202\\ \hline 7.819&10.2134\\ \hline 9.114&11.3964\\ \hline 10.549&12.5168\\ \hline 12.111&13.5285\\ \hline 13.787&14.3999\\ \hline \end{array}}$$
This takes you from around year 470 million (first stars and galaxies were forming) up to around year 13.8 billion (the present).
You can use google calculator to convert the Thetas to nanojoules per cubic meter. for example to get the present rho_crit just paste this into the google box:
3c^2/(8 pi G (14.4 billion years)^2))
Google will say 0.778 nanopascal which is the same as 0.778 nanojoule per cubic meter (when you sort the units out.)
Or if you want the density when the first stars were forming just paste this into google box:
3c^2/(8 pi G (0.7105 billion years)^2))
Google will say 319.8 nanopascal which is equivalent to 319.8 nanojoule per cubic meter.
Last edited: Nov 12, 2013
1 person likes this.
7. ### nick1o2
26
Thankyou Markus!
Is there anyway to find out the density of different parts of the universe? For example is the density in the northern hemisphere of the universe bigger or less than the southern hemisphere? I would think they would be different because the universe isn't uniform, because we have the big bang model not steady state model, but i there any resources to show the total density in these area's and how they have changed?
8. ### Calimero
256
No, cosmology operates on premise of homogeneity and isotropy, meaning that average density of sufficiently big volume is the same throughout the universe.
9. ### Calimero
256
You can google "wmap" and look at the picture of CMB radiation to see how amazingly universe is uniform. There are tiny fluctuations, roughly 1 part in 100 000, which served as seeds for later structure formation.
10. ### marcus
25,096
The question came up as to how I got the information in the short table in this post:
I was using a temporary nonstandard symbol Theta for the Hubble time THubble because I didn't want to bother with writing subscript, should go back to more conventional notation and say THubble.
The HUBBLE RADIUS is just the Hubble time multiplied by the speed of light. So if the time is 14.4 Gy (billion years) then the radius is 14.4 Gly (billion lightyears).
Jorrie's "Lightcone" calculator gives the Hubble radius RH in Gly, so I just relabeled the R numbers. That's where the table came from. I'll show this in next post. You can easily learn how to use the Lightcone calculator to make your own tables with however many rows you want covering whatever range of expansion you want.
11. ### marcus
25,096
Here's the table it prints if you select the range to be from S=11 to S=1, with 20 steps.
That means it will compute and tabulate the universe's history from a time when distances were 1/11 present size up to the present, when distances are their current size i.e. S=1.
$${\scriptsize\begin{array}{|c|c|c|c|c|c|}\hline R_{0} (Gly) & R_{\infty} (Gly) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline 14.4&17.3&3400&67.9&0.693&0.307\\ \hline \end{array}}$$ $${\scriptsize\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline a=1/S&S&T (Gy)&R (Gly)&D_{now} (Gly)&D_{then}(Gly)&D_{hor}(Gly)&V_{now} (c)&V_{then} (c) \\ \hline 0.091&11.000&0.4726&0.7105&31.447&2.859&4.358&2.18&4.02\\ \hline 0.102&9.757&0.5659&0.8504&30.481&3.124&4.814&2.12&3.67\\ \hline 0.116&8.655&0.6776&1.0176&29.456&3.403&5.308&2.05&3.34\\ \hline 0.130&7.677&0.8112&1.2173&28.368&3.695&5.843&1.97&3.04\\ \hline 0.147&6.809&0.9710&1.4558&27.214&3.997&6.418&1.89&2.75\\ \hline 0.166&6.040&1.1621&1.7401&25.990&4.303&7.032&1.80&2.47\\ \hline 0.187&5.358&1.3905&2.0787&24.693&4.609&7.686&1.71&2.22\\ \hline 0.210&4.752&1.6631&2.4807&23.319&4.907&8.376&1.62&1.98\\ \hline 0.237&4.215&1.9883&2.9566&21.865&5.187&9.098&1.52&1.75\\ \hline 0.267&3.739&2.3755&3.5172&20.330&5.437&9.846&1.41&1.55\\ \hline 0.302&3.317&2.8355&4.1732&18.711&5.642&10.613&1.30&1.35\\ \hline 0.340&2.942&3.3803&4.9340&17.011&5.782&11.387&1.18&1.17\\ \hline 0.383&2.609&4.0230&5.8050&15.233&5.837&12.155&1.06&1.01\\ \hline 0.432&2.315&4.7767&6.7856&13.382&5.781&12.904&0.93&0.85\\ \hline 0.487&2.053&5.6541&7.8652&11.471&5.587&13.617&0.80&0.71\\ \hline 0.549&1.821&6.6657&9.0202&9.516&5.225&14.278&0.66&0.58\\ \hline 0.619&1.615&7.8185&10.2134&7.540&4.668&14.874&0.52&0.46\\ \hline 0.698&1.433&9.1144&11.3964&5.570&3.887&15.393&0.39&0.34\\ \hline 0.787&1.271&10.5488&12.5168&3.635&2.860&15.832&0.25&0.23\\ \hline 0.887&1.127&12.1114&13.5285&1.765&1.565&16.189&0.12&0.12\\ \hline 1.000&1.000&13.7872&14.3999&0.000&0.000&16.472&0.00&0.00\\ \hline \end{array}}$$
To find out what the rows mean, click on the link, you will see a sample table, hover the mouse over the blue dots. Then click on "column selection" and you will get more blue info dots telling what the columns mean. And also the "column selection" menu will allow you to select which columns to show. To make that table with only TWO COLUMNS I just selected only the T and the R columns to be shown. The time (in billions of years Gy) and the Hubble radius (in Gly)
The only other thing I did was to set the S upper limit to 11 and the S lower limit to 1 (i.e. to present day) and tell it to cover that range from 11 down to 1 in 20 steps.
You replace the default values of Supper and Slower and STEPS by typing 11, 1, and 20 in those three boxes, and press "calculate". | 2015-09-05T05:48:16 | {
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https://math.stackexchange.com/questions/2712980/prove-all-other-common-divisors-divide-gcd?noredirect=1 | # Prove all other common divisors divide $\gcd$
In the book of Silverman, the below proof is given of the above :
$$a= q_1b + r_1$$ $$b =q_2 r_1+ r_2$$ $$r_1 =q_3r_2 + r_3$$ $$\vdots$$ $$r_{n-3} = q_{n-1}r_{n-2} + r_{n-1}$$ $$r_{n-2} = q_n r_{n-1} + r_n$$ $$r_{n-1} = q_{n+1}r_n + 0$$
But why is $r_n$ the greatest common divisor of $a$ and $b$? Suppose that $d$ is any common divisor of $a$ and $b$. We will work our way back down the list of equations. So from the first equation $a = q_1b + r_1$ and the fact that $d$ divides both $a$ and $b$, we see that $d$ also divides $r_1$. Then the second equation $b = q_i + r_2$ shows us $d$ must divide $r_2$. Continuing down line by line, at each stage we will know $d$ divides the previous two remainders $r_{i-1}$ and $r_i$, and then the current line $r_{i-1} = q_{i+1}r_i+ r_{i +1}$ will tell us that $d$ also divides the next remainder $r_{i + 1}$. Eventually, we reach the penultimate line $r_{n-2} = q_nr_{n-1} + r_n$, at which point we conclude that $d$ divides $r_n$. So we have shown that if $d$ is any common divisor of $a$ and $b$, then $d$ will divide $r_n$. Therefore, $r_n$ must be the greatest common divisor of $a$ and $b$.
I take the above proof to an example, say $b = 32, a=33*8=264, r_n= gcd(a,b)= 8, d= 4$. Now, $a=32*8 + r_1$, with $b=32, q=8, r_1=8, d=4\mid r_n=4$. Similarly, the argument can continue on.
But, it appears unconvincing to me from the very start. May be a contradiction based proof would have worked better, by taking the prime factorization of $a,b$, and showing that a non-common divisor would not divide at any step at least two of the three terms.
Or may this approach is not proper itself.
• What exactly “appears unconvincing ... from the very start” ? – Martin R Mar 29 '18 at 9:08
• You omitted one step, namely that if you work from the bottom up, you can see that $r_n$ itself divides both $a$ and $b$, so it is a common divisor. After that you then work from top to bottom as described to show that every common divisor $d$ divides $r_n$. Therefore $r_n$ is the greatest of such common divisors. – Jaap Scherphuis Mar 29 '18 at 9:16
• @MartinR Sorry, for being late. It appears that it is not rigorous. It is not a formal one, and cannot be convinced (at least to me) to be of that type. – jiten Mar 29 '18 at 9:27
• That is too vague. You demonstrated that if $d$ is any common divisor of $a$ and $b$ then $d$ divides $r_n$. Please explain what you consider “not rigorous” in the proof. – Martin R Mar 29 '18 at 9:32
• There's a typo in your post: $r_{n-i}$ should be $r_{n-1}$. – Bernard Mar 29 '18 at 10:08
DEFINITION: Let $a,b \in \mathbb Z$. Then $g \in \mathbb Z^+$ is the greatest common divisor of $a$ and $b$ if
$\qquad \text{$(1.) \quad g \mid a$and$g \mid b$}$.
$\qquad \text{$(2.) \quad \forall x \in \mathbb Z, \ x\mid a$and$x \mid b$implies$x \mid g$}$.
Now, suppose $x \mid a$ and $x \mid b$. Then
\begin{array}{rcl} a= q_1b + r_1 &\implies &x \mid r_1 \\ b =q_2 r_1+ r_2 &\implies &x \mid r_2 \\ r_1 =q_3r_2 + r_3 &\implies &x \mid r_3 \\ &\vdots \\ r_{n-3} = q_{n-1}r_{n-2} + r_{n-1} &\implies &x \mid r_{n-1} \\ r_{n-2} = q_n r_{n-1} + r_n &\implies &x \mid r_n \\ \end{array}
So $r_n$ satisfies condition $(2.)$.
In the other direction.
\begin{array}{rcl} r_{n-1} = q_{n+1}r_n + 0 &\implies &r_n \mid r_{n-1} \\ r_{n-2} = q_n r_{n-1} + r_n &\implies &r_n \mid r_{n-2} \\ r_{n-3} = q_{n-1}r_{n-2} + r_{n-1} &\implies &r_n \mid r_{n-3} \\ &\vdots \\ r_1 =q_3r_2 + r_3 &\implies &r_n \mid r_1 \\ b =q_2 r_1+ r_2 &\implies &r_n \mid b \\ a= q_1b + r_1 &\implies &r_n \mid a \\ \end{array}
So $r_n$ satisfies conditions $(1.)$. Hence $r_n=\gcd(a,b)$
• Basically it is elaboration of the comment of @JaapScherphuis to the OP, and might be (not clear) also incorporates the comment of 'dssknj' (for the 'other direction' part). – jiten Mar 29 '18 at 11:14
• It also makes me feel that there is no 'more' formal proof possible. I attempted a wrong way in response to @dssknj, and feel something like that is not possible. Can you please comment on that. – jiten Mar 29 '18 at 11:15
• Thanks a lot for your kindest elaboration that makes the book more clear, but still my earlier doubts (as in the previous two comments) are not escaping me. – jiten Mar 29 '18 at 11:35
• Yes, I did nothing but make the others comments look "prettier". I also cannot see how to do a contradiction-based proof. In your question you proved that $r_n$ is a common divisor {(1.)}. It seemed to me that you couldn't "see" how to show that it was the **greatest ** common divisor {(2.)}. So I just formalized everything in my answer. – steven gregory Mar 29 '18 at 11:36
• I think you just need to stop thinking about symbols for a while and get out a pencil and some paper and play with some numbers. Try reducing these fractions for example: $\dfrac{12}{15}, \quad \dfrac{74}{111}, \quad \dfrac{123}{173}$. – steven gregory Mar 29 '18 at 12:12
Here is a slightly different proof that $r_n$ is the greatest common divisor, based more on what @dssknj remarked (still not by contradiction though).
Lemma: $\gcd(x,y)=gcd(x,y+rx)$ for any integers $x,y,r$.
Proof:
If $d|x$ and $d|y$ then clearly $d|y+rx$. So every common divisor of $x$ and $y$ is also a common divisor of $x$ and $y+rx$.
The converse is also true:
If $d|x$ and $d|y+rx$ then $d|y$ because $y=(y+rx)-rx$. So every common divisor of $x$ and $y+rx$ is also a common divisor of $x$ and $y$.
This means that $\{x,y\}$ have exactly the same common divisors as $\{x,y+rx\}$, and therefore will also have the same greatest common divisor. $\square$
You can now traverse the sequence of equations once in either direction to show that $$\gcd(a,b) = \gcd(b,r_1) = \gcd(r_1,r_2) = \gcd(r_2,r_3) = ...\\ ... = \gcd(r_{n-2},r_{n-1}) = \gcd(r_{n-1},r_n) = \gcd(r_n,0) = r_n$$
Basically, instead of traversing the sequence of equations once in each direction, this proof builds a two-way connection between adjacent entries, and then you only have to go through the sequence once in either direction to fully prove the connection we want between the first and last.
• How your answer incorporated @dssknj comment in the proof to show the dowinwards and upwards movement in the iterative runs of Euclidean algorithm, is not 'clear'. By 'clear'', I mean that the neccesity of incorporating his comment is not separate, but arising out of the need to prove that $r_n$ is $\gcd(a,b)$. May be 'dssknj' comment is not concerned with showing $r_n$ is the greatest common divisor, but with just showing that $r_n=\gcd(a,b)$. – jiten Mar 29 '18 at 12:09
• @jiten "not concerned with showing $r_n$ is the greatest common divisor, but with just showing that $r_n$ is the $\gcd(a,b)$." Huh? It is not concerned with showing that but it is concerned with showing just that? – Jaap Scherphuis Mar 29 '18 at 12:14
• I like this answer better than mine. – steven gregory Mar 29 '18 at 14:28
If you want it by contradiction then here it is...
1) $a=bq+r$ then any common divisor of $a$ and $b$ is common divisor of $b$ and $r$.
Using 1) it is easy to show that $r_n$ is a common divisor of $a$ and $b$. Let assume that $r_n\neq \gcd(a, b)$ then $\gcd(a, b)=d, d\gt r_n$ $\quad d$ is a common factor of $a$ and $b$ and using 1) we have $d\mid r_n \implies d\le r_n$ Hence contradiction.
• You are using a proven result (that $\gcd$ is the biggest common divisor) to derive that $d\gt r_n$, which seems unfair to me. Please do not take this result for granted, which itself means that a proof derived using some other way was better to prove the topic of the post. – jiten Mar 29 '18 at 21:39
• what is your definition of gcd if it is not the greatest common divisor? – Siong Thye Goh Mar 29 '18 at 23:00
• @jiten definitions are things that don't need proof and you can't prove them by the way. – dssknj Mar 30 '18 at 2:03
• @jiten not axiom its definition. You may have proven in your high school that perpendicular drawn from center to chord bisect the chord. In proving that you must have used the fact that end points of chord lie on the circle which is the definition of chord. – dssknj Mar 30 '18 at 2:22
• My personal opinion: We define things, and then we verify whether it is well defined (i.e. some sort of check that whether two people can always compute the same quantity). ... and then we derived its properties ...If the same term is defined by two people, we check whether they mean the same thing. If one is not allowed to use its definition.... there is no reason why you can trust the properties derived from it. And we dont' even know what we are trying to prove if there isn't a definition to check. I am not a logician to comment such thing precisely though. – Siong Thye Goh Mar 30 '18 at 5:57 | 2021-03-05T08:20:07 | {
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http://math.stackexchange.com/questions/65672/prove-disprove-that-if-two-sets-have-the-same-power-set-then-they-are-the-same-s/65706 | Prove/Disprove that if two sets have the same power set then they are the same set
I am really sure that if two sets have the same power set, then they are the same set. I just am wondering how does one exactly go about proving/showing this?
I'm usually wrong, so if anyone can show me an example where this fails, I'd like that too.
The homework just asks for true/false, but I'm wanting to show it if possible. My thoughts are that since the power set is by definition the set of all subsets of a set, if each of the two power sets are identical, we have an identity map between each set, thus it's indistinguishable which power set is a given set's power set. I hope that wasn't verbose. Since a set has only one power set, we can conclude they are in fact the same set.
-
What do you mean by "same"? – Qiaochu Yuan Sep 19 '11 at 2:37
I think he means "same" in the sense of the axiom of extensionality. $(\forall x)(x \in A \Leftrightarrow x \in B) \Rightarrow (A = B)$ – William Sep 19 '11 at 2:41
"the same" isn't the same, depending on the context! – The Chaz 2.0 Sep 19 '11 at 2:59
Suppose $A \neq B$. Without loss of geneality, there exists $x \in A$ such that $x \notin B$. Then $\{x\} \in \mathscr{P}(A)$ wherease $\{x\} \notin \mathscr{P}(B)$. Thus $\mathscr{P}(A) \neq \mathscr{P}(B)$.
Conversely, if $\mathscr{P}(A) = \mathscr{P}(B)$, then all their singleton's are the same. Thus $A = B$.
$A = B$ if and only if $\mathscr{P}(A) = \mathscr{P}(B)$.
-
+1 Is the statement that $\{ x \} \notin \mathscr P(B)$ evident or does it need proof? (It seems "clearly" true, but I do not know what to say if someone asks me to justify it.) – Srivatsan Sep 19 '11 at 2:44
I would just use the definition of subset. $D \subset E$ if and only if $(\forall n)(n \in D \Rightarrow n \in E)$. So by assumption $x \notin B$ and $x \in A$. So we have there exists $n$ (in particular that $x$) such that $n \in \{x\}$ and $n \notin B$. Thus I have proved $(\exists n)(n \in \{x\} \wedge n \notin B) = \neg((\forall n)(n \in \{x\} \Rightarrow n \in B))$. Thus $\neg(\{x\} \subset B)$ and hence $\{x\} \notin \mathscr{P}(B)$. – William Sep 19 '11 at 2:49
+1 for "without loss of geneality" ;) – Chris Taylor Sep 19 '11 at 8:41
To add on William's answer with a positive proof, first one has to note the following observation:
$$A=\bigcup\{B\mid B\subseteq A\}$$
To prove this, the inclusion $A\subseteq\bigcup\{B\mid B\subseteq A\}$ is trivial since $A\subseteq A$, so we take $A$ into the union. In the other direction, since every $B$ in the union is a subset of $A$ the union is a subset of $A$.
Now we can proceed. The above identity can be written in terms of the power set as $A=\bigcup\mathcal P(A)$.
Assume $\mathcal P(A)=\mathcal P(B)$, therefore $\bigcup\mathcal P(A)=\bigcup\mathcal P(B)$, therefore $A=B$.
-
@anon: $\bigcup_{B\subseteq A} B$ can be written as $\bigcup\{B\mid B\subseteq A\}$. – Asaf Karagila Sep 19 '11 at 5:35
Hmm. Never seen the notation, guess I'll keep it in mind. – anon Sep 19 '11 at 5:45
An alternative way to answer this old question: for all sets A and B,
$$\begin{array}{ll} & \mathcal{P}(A) = \mathcal{P}(B) \\ \equiv & \;\;\;\text{"extensionality"} \\ & \langle \forall V :: V \in \mathcal{P}(A) \equiv V \in \mathcal{P}(B) \rangle \\ \equiv & \;\;\;\text{"definition of \mathcal{P}, twice"} \\ & \langle \forall V :: V \subseteq A \equiv V \subseteq B \rangle \\ \Rightarrow & \;\;\;\text{"choose V:=A, and choose V:=B"} \\ & (A \subseteq A \equiv A \subseteq B) \;\land\; (B \subseteq A \equiv B \subseteq B) \\ \equiv & \;\;\;\text{"\subseteq is reflexive, so A \subseteq A and B \subseteq B"} \\ & A \subseteq B \land B \subseteq A \\ \equiv & \;\;\;\text{"definition of set equality"} \\ & A = B \\ \end{array}$$
Update: As a comment rightly points out, the above proof is very similar to my answer to another question. In fact, we can directly prove the stronger version of this question's theorem from that one:
$$\begin{array}{ll} & \mathcal{P}(A) = \mathcal{P}(B) \;\equiv\; A = B \\ \equiv & \;\;\;\text{"double inclusion, twice"} \\ & \mathcal{P}(A) \subseteq \mathcal{P}(B) \land \mathcal{P}(B) \subseteq \mathcal{P}(A) \;\equiv\; A \subseteq B \land B \subseteq A \\ \Leftarrow & \;\;\;\text{"logic"} \\ & (\mathcal{P}(A) \subseteq \mathcal{P}(B) \;\equiv\; A \subseteq B) \;\land\; (\mathcal{P}(B) \subseteq \mathcal{P}(A) \;\equiv\; B \subseteq A) \\ \equiv & \;\;\;\text{"the other theorem, twice"} \\ & \text{true} \\ \end{array}$$
-
Instead of posting the same answer twice, you can post it once and point out that the question is the same. – Asaf Karagila Mar 16 '13 at 20:36
@AsafKaragila I'm sorry, I did not intend to post the same answer twice. Which two answers are you referring to? – Marnix Klooster Mar 16 '13 at 20:43
While not word for word, it is the same question and the same answer: math.stackexchange.com/a/332186/622 – Asaf Karagila Mar 16 '13 at 20:45
The questions are actually slightly different: this one is about $\mathcal{P}(A) = \mathcal{P}(B) \Rightarrow A = B$, while the other is about $\mathcal{P}(A) \subseteq \mathcal{P}(B) \equiv A \subseteq B$. But you are of course correct that these two theorems are very much related: (a stronger version of) the former follows directly from the latter using extensionality. – Marnix Klooster Mar 16 '13 at 20:54
@MarnixKlooster: +1. I'm very grateful that you uncloak all the detail and steps. I made two ancillary edits for layout but please feel free to revert them. Just a question: How is the update "a stronger version of this theorem"? In this answer, you're only proving in two ways the same result: $A = B \iff P(A) = P(B)$? – LePressentiment Jan 7 '14 at 6:58 | 2015-12-02T01:57:32 | {
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http://math.stackexchange.com/questions/894794/sphere-equation-given-4-points | # Sphere equation given 4 points
Find the equation of the Sphere give the 4 points (3,2,1), (1,-2,-3), (2,1,3) and (-1,1,2).
The *failed* solution I tried is kinda straigh forward:
We need to find the center of the sphere.
Having the points:
$$p_{1}(3,2,1),\, p_{2}(1,-2,-3),\, p_{3}(2,1,3),\, p_{4}(-1,1,2)$$
2 Triangles can be created using these points, let's call $A$ our triangle formed by the points $p_{1},\,p_{2}\, and\,\, p_{3}$; And $B$ our triangle formed by the points $p_{1},\, p_{3}\, and \,\,p_{4}$.
Calculate the centroids of each triangle: $$CA = (2,1/3,1/3)\\ CB = (4/3,4/3,2)$$ And also, a normal vector for each triangle: $$\overrightarrow{NA} = \overrightarrow{p_{1}p_{2}} \times \overrightarrow{p_{1}p_{3}}\\ \overrightarrow{NB} = \overrightarrow{p_{1}p_{3}} \times \overrightarrow{p_{1}p_{4}}$$
$$\overrightarrow{p_{1}p_{2}} = <-2,-4,-4>\\ \overrightarrow{p_{1}p_{3}} = <-1,-1,2>\\ \overrightarrow{p_{1}p_{4}} = <-4,-1,1>\\ \:\\ \overrightarrow{NA} = <-12, 8, -2>\\ \overrightarrow{NB} = <1, -7, -3>\\$$
With the centroids and normals of triangles $A$ and $B$, we can build two parametric equations for a line, the first one intersects the centroid of $A$ and the other one the centroid $B$. $$Line \enspace A\\ x = 2-12t \quad y = 1/3+8t \quad z = 1/3-2t\\ \:\\ Line \enspace B\\ x = 4/3 + s \quad y = 4/3 - 7s \quad z = 2 - 3s$$
The point where this lines intersect should be the center of the sphere, unfortunately this system of equations is not linearly dependent, that means that they do not intersect each other. What could be the problem here?
-
+1 for showing your work and explaining your difficulty in detail. This is essential to giving helpful answers. Many newbies fail there. Last but not least. Welcome to Math.SE! – Jyrki Lahtonen Aug 12 '14 at 7:04
@JyrkiLahtonen Thank you! I really like the stackexchange sites. – DanielRS Aug 12 '14 at 10:42
I would cite the beautiful method from W.H.Beyer to find the center and radius of the sphere $(x-a)^2+(y-b^2)+(y-c)^2=R^2$
-
This is what I was refering to in my answer. – Claude Leibovici Aug 12 '14 at 9:53
Hi Claude ! Thanks for your notification. Of course, the method from W.H.Bayer in specific to the case of four given points. In case of more than four scattered points one have to use a fitting method, which is something else. Cordially, JJ. – JJacquelin Aug 12 '14 at 10:14
For me, the problem is the same but I bet that this could take us for a loooong discussion. Cheers. By the way, I really enjoy your books. – Claude Leibovici Aug 12 '14 at 10:16
Thank you! After thinking about it a little, my approach was incorrect. – DanielRS Aug 12 '14 at 10:38
The centroid of a triangle is usually not at the same distance from the vertices. In the plane the center of the outer circle (= the intersection of the normals bisecting the sides) has this property.
But, because this is in 3D you might as well find the point of intersection of the planes the have $\vec{p_1p_2}$, $\vec{p_2p_3}$ and $\vec{p_3p_4}$ as their normals, and pass through the midpoints of those line segments. The points on the first plane are at the same distance from $p_1$ and $p_2$ et cetera. Thus the intersection of those three planes is what you want.
-
Hint
Using a totally different approach, you can also look at your problem using what JJacquelin (participant to MSE) proposed in his book
Pages 17 and 18 give the full approach for a spherical regression. It is quite simple and reduces to a linear system of four equations for four unknowns from which are deduced the coordinates of the center and the radius of the sphere.
Since I do not want to spoil him, I shall not give more. I applied his method to your case and it works splendid.
-
For the ones calculating by hand, the error-prone numbers are:
the center of the sphere is at $p=(24/19,-16/19,4/19)$
the squared radius is $r^2=4230/361$.
Here is my small GNU Maxima script
display2d : false;
/*
* purpose:
* given points x_k,
* calculate circumsphere center p and squared radius r^2
**/
my_A(d,k,x) := block(
[res,i,j],
res : zeromatrix(d,d),
for j : 0 thru k-1 do block(
for i : 1 thru d do block(
res[i,j+1] : x[j+1][i]-x[k+1][i]
)
),
for j : k+1 thru d do block(
for i : 1 thru d do block(
res[i,j] : x[j+1][i]-x[k+1][i]
)
),
return(res)
);
my_norm2(d,x) := apply("+",makelist(x[k]^2,k,1,d));
my_dist2(d,x,y) := apply("+",makelist((x[k]-y[k])^2,k,1,d));
x : [[3,2,1],[1,-2,-3],[2,1,3],[-1,1,2]];
d : length(x)-1;
A : makelist(my_A(d,k,x),k,0,d);
AinvT : makelist(transpose(invert(A[k+1])),k,0,d);
xnorm2 : makelist(my_norm2(d,x[k+1]),k,0,d);
z1 : transpose(makelist(1,k,1,d));
p : -1/2*apply("+",makelist(xnorm2[k+1]*(AinvT[k+1].z1),k,0,d));
r2 : makelist(my_dist2(d,transpose(p)[1],x[k+1]),k,0,d);
-
Essentialiy I'm constructing the center point $p$ as a linear combination of the outward normals to the simplex $S = Conv(x_0,x_1,x_2,x_3)$, where the $x_k$ are the given points. – andre Aug 12 '14 at 11:41
$$(x-u)^2+(y-v)^2+(z-w)^2=r^2$$
$(u,v,w)$ are the coordinates of the center of the sphere and $r$ is the radius. Plug into the given points $p_1, p_2, p_3, p_4$ for $x,y,z$ in the equation and you get four equations with variables $u,v,w,r$. But these equations contain quadratic terms. Subtract one equation from each of the other three. You get three linear equations and the variables $u, v, w$. Such a system can be solved in the usual way to find the center $(u,v,w)$, e.g. Gaussian elimination. Now plug the calculated $u,v,w$ into one of the initial equations to find the radius $r$.
Note:
The three linear equations found are the equation of the planes described by @Jyrki Lahtonen
The problem can be solved with this Maxima program:
/* define the four points p[1],...,p[4] */
p[1]:[3,2,1]; p[2]:[1,-2,-3]; p[3]:[2,1,3]; p[4]:[-1,1,2];
/* ceq is the equation of the circle
(u,v,w) is the center, r is the radius */
ceq:(x-u)^2+(y-v)^2+(z-w)^2=r^2;
/* plug in the points in the circle equation
to get 4 equation eq1[1],...,eq1[4] */
for i:1 thru 4 do
eq1[i]:ev(ceq,map("=",[x,y,z],p[i]));
/* display this 4 equations eq1*/
listarray(eq1);
/* subtract the fourth equation for each of the first
three equation to get three linear
equations eq2[1],eq2[2],eq2[3] in u,v,w */
for i:1 thru 3 do
eq2[i]:ev(eq1[i]-eq1[4],expand);
/* display these three equations
eq2[1],eq2[2],eq2[3] */
listarray(eq2);
/* solve this system oflinear euations to get u,v,w */
ss:solve(listarray(eq2),[u,v,w]);
/* plugin the solutions in the fourth
equation eq1[4] and calculate r^2 */
solve(eq1[4],r^2),ss[1];
Actually the problem can be solved using only one Maxima solve command:
(%i1) display2d:false\$
(%i2) solve(makelist(ev((z-w)^2+(y-v)^2+(x-u)^2=r^2,
map("=",[x,y,z],p)),p,[[3,2,1],[1,-2,-3],[2,1,3],[-1,1,2]]),
[u,v,w,r]);
(%o2) [[u = 24/19,v = -16/19,w = 4/19,r = -3*sqrt(470)/19],
[u = 24/19,v = -16/19,w = 4/19,r = 3*sqrt(470)/19]]
- | 2015-01-31T20:05:50 | {
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https://byjus.com/question-answer/let-alpha-beta-be-the-roots-of-the-equation-ax-2-bx-c-0-and/ | Question
# Let $$\alpha, \beta$$ be the roots of the equation $$ax^2+bx+c=0$$ and $$\alpha^4+\beta^4$$ be the roots of the equation $$px^2+qx+r=0$$, then the roots of the equation $$a^2px^2-4acpx+2c^2p+a^2q=0$$ are:
A
Always +ve
B
Always complex
C
Opposite in sign
D
Negative
Solution
## The correct option is C Opposite in sign$$a{ x }^{ 2 }+bx+c=0$$ ......... $$(i)$$$$\alpha , \beta$$ are the roots of above equation$$\therefore \alpha +\beta =\cfrac { -b }{ a }$$ ..... $$(ii)$$$$\alpha \beta =\cfrac { c }{ a }$$ ........ $$(iii)$$$${ \alpha }^{ 4 }$$ and $${ \beta }^{ 4 }$$ are the roots of equation $$p{ x }^{ 2 }+qx+r=0$$$$\therefore { \alpha }^{ 4 }+{ \beta }^{ 4 }=\cfrac { -q }{ p }$$ .... $$(iv)$$$${ \alpha }^{ 4 }{ \beta }^{ 4 }=\cfrac { r }{ p }$$ ....... $$(v)$$$${ a }^{ 2 }p{ x }^{ 2 }-4acpx+2{ c }^{ 2 }p+{ a }^{ 2 }q=0$$Let $$r$$ and $$\delta$$ be roots of above equation $$\therefore \quad r+\delta =\cfrac { 4acp }{ { a }^{ 2 }p } =\cfrac { 4c }{ a }$$$$r\delta =\cfrac { 2{ c }^{ 2 }p+{ a }^{ 2 }q }{ { a }^{ 2 }p }$$$$=2{ \left( \cfrac { c }{ a } \right) }^{ 2 }+\cfrac { q }{ p }$$$$=2{ \alpha }^{ 2 }{ \beta }^{ 2 }-\left( { \alpha }^{ 4 }+{ \beta }^{ 4 } \right)$$ ..... [From equation $$(iii)$$ and $$(iv)$$]$$=-\left( { \alpha }^{ 4 }+{ \beta }^{ 4 }-2{ \alpha }^{ 2 }{ \beta }^{ 2 } \right)$$$$=-{ \left( { \alpha }^{ 2 }-{ \beta }^{ 2 } \right) }^{ 2 }$$$${ \left( { \alpha }^{ 2 }-{ \beta }^{ 2 } \right) }^{ 2 }>0$$$$\therefore -{ \left( { \alpha }^{ 2 }-{ \beta }^{ 2 } \right) }^{ 2 }<0$$$$\therefore r\delta <0$$$$\Rightarrow \left( r<0\quad \& \quad \delta >0 \right)$$ or $$\left( r>0\quad \& \quad \delta <0 \right)$$$$\therefore$$ Roots of given equation are opposite in sign.Maths
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https://math.stackexchange.com/questions/1945523/showing-that-lim-x-to-1-left-frac231-x23-frac111-x11-righ | Showing that $\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)=6$
How does one evaluate the following limit? $$\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)$$ The answer is $6$.
How does one justify this answer?
Edit: So it really was just combine the fraction and use L'hopital's rule twice (because function and its first derivative are of indeterminate form at $x=1$). This problem is more straightforward than it seems at first.
• Put the fractions over a common denominator. You get $\frac {P(x)}{Q(x)}$ where $P,Q$ are some polynomials. If the limit exists $(x-1)$ will divide both $P$ and $Q.$ Do the division. Evaluate at $1.$ if you still get an indeterminate, then do the division by $(x-1)$ as many times as necessary. – Doug M Sep 28 '16 at 20:19
Both fractions are unbounded as $x\rightarrow 1$. But if we rewrite $$\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}}=\dfrac{23(1-x^{11})-11(1-x^{23})}{(1-x^{23})(1-x^{11})}=\dfrac{11x^{23}-23x^{11}+12}{1-x^{23}-x^{11}+x^{34}}$$ we can use L'Hopital since both sides tend to 0 as $x$ tends to 1. Differentiating both sides give $$\dfrac{253x^{22}-253x^{10}}{-23x^{22}-11x^{10}+34x^{33}}=253\dfrac{x^{12}-1}{-23x^{12}-11+34x^{23}}$$ Both sides still tend to 0, so we differentiate again and get $$253\dfrac{12x^{11}}{-276x^{11}+782x^{22}}$$ which tends to $$253\dfrac{12}{-276+782}=6$$
• I didn't expect this many responses to my question, but your approach is the clearest and most standard of all of the other approaches. Also, I didn't think combining the fraction was necessary but it is. Because I see that it is much better to deal with indeterminate forms of $\frac 00$ instead of $\infty - \infty$, because $\frac 00$ prompts us to use immediately l'Hopital's rule. – user373314 Sep 28 '16 at 20:56
• @user373314 This is a very interesting comment. In France, at least what I have seen, L'Hopital's rule is hardly ever used, if even taught at all. The standard way to compute limits is by développements limités (that is, Taylor series with $O$ or $o$ notation). However, by reading american calculus books, I have discovered L'Hopital's rule is much more used there. They are mostly equivalent anyway, but this difference in teaching is reflected in questions and answers on MSE. – Jean-Claude Arbaut Sep 28 '16 at 21:04
• See also this question of mine about L'Hopital (which, coming from France, is something I may have used maybe 5 times in my life). – Clement C. Sep 28 '16 at 21:06
• @ClementC. Yes, I have seen this. Good question. L'Hopital's rule seems to have a bad reputation, however, it's interesting to see in Franklin's Advanced calculus, that even Taylor series may be derived from L'Hopital's rule. It's not bad at all, but it must be used carefully. – Jean-Claude Arbaut Sep 28 '16 at 21:13
This answer does not use L'Hopital (personal taste), only a standard identity restated below, the binomial theorem, and a straightforward Taylor expansion to first order at $0$.
Using the identity $1-x^{2n+1} = (1-x)\sum_{k=0}^{2n} x^k$, we can rewrite \begin{align*} \frac{23}{1-x^{23}} - \frac{11}{1-x^{11}} &= \frac{1}{1-x}\left(\frac{23}{\sum_{k=0}^{22}x^k} - \frac{11}{\sum_{k=0}^{10}x^k} \right)\\ &= \frac{1}{1-x}\left(\frac{23\sum_{k=0}^{10}x^k}{\sum_{k=0}^{22}x^k\sum_{k=0}^{10}x^k} - \frac{11\sum_{k=0}^{22}x^k}{\sum_{k=0}^{10}x^k\sum_{k=0}^{22}x^k} \right)\\ &= \frac{1}{\sum_{k=0}^{10}x^k\sum_{k=0}^{22}x^k}\cdot\frac{1}{1-x}\left(23\sum_{k=0}^{10}x^k - 11\sum_{k=0}^{22}x^k \right)\\ \end{align*} Let us focus on the parenthesis (the first factor converges to $\frac{1}{11\cdot 23}$ by continuity, the second is the problematic one that will be "offset" by the parenthesis).
Writing $x=1+h)$ (where we will have $h\to 0$), we get, for any fixed integer $n$, \begin{align*} \sum_{k=0}^{n}x^k &= \sum_{k=1}^{n}(1+h)^k = \sum_{k=1}^{n} \sum_{\ell=0}^k \binom{k}{\ell} h^\ell \\ &= \sum_{k=0}^{n}(1+kh +o(h)) \\ &= (n+1)+\frac{n(n+1)}{2}h +o(h) \end{align*} when $h\to 0$, as $n$ is a constant. In particular, this implies \begin{align*} 23\sum_{k=0}^{10}x^k - 11\sum_{k=0}^{22}x^k &= 23\cdot 11+23\cdot \frac{11\cdot10}{2}h - 11\cdot 23-11\cdot \frac{22\cdot 23}{2}h + o(h)\\ &= 23\cdot 11\cdot (-6h) + o(h)\\ &= 23\cdot 11\cdot 6(1-x) + o(1-x) \end{align*} Overall, we thus have \begin{align*} \frac{23}{1-x^{23}} - \frac{11}{1-x^{11}} &= \frac{23\cdot 11}{\sum_{k=0}^{10}x^k\sum_{k=0}^{22}x^k}\cdot\frac{6(1-x)+o(1-x)}{1-x} \\ &= \frac{23\cdot 11}{\sum_{k=0}^{10}x^k\sum_{k=0}^{22}x^k}\cdot (6+o(1)) \xrightarrow[x\to1]{} \frac{23\cdot 11}{23\cdot 11} \cdot 6 = 6 \end{align*} as claimed.
• Darn, took much longer to type than expected. Oh, well. – Clement C. Sep 28 '16 at 20:55
• I tried an approach like yours originally but I got stuck at dealing with that $\frac 1{1-x}$. Your solution to this dilemma was simply to write $x=1+h$ (for $h > 0$)? I guess since we're taking the limit to $1$ from both sides, that must be why. – user373314 Sep 28 '16 at 21:01
• @user373314 Yes, basically. The rationale is that while it does not change anything mathematically, I am much more comfortable with standard machinery (Taylor expansions, limits, asymptotics) at $0$ that at a finite non-zero point. – Clement C. Sep 28 '16 at 21:03
• Well done. +1 . – Mark Viola Sep 28 '16 at 23:22
• Always easier to have something go to zero. – marty cohen Sep 29 '16 at 4:38
As $t\to0$, we have
$$\frac{23}{1-(1+t)^{23}}=-\frac{23}{23t+253t^2+O(t^3)}=-\frac{1}{t}\cdot\frac{1}{1+11t+O(t^2)}=-\frac{1}{t}\left(1-11t+O(t^2)\right)$$
Likewise
$$\frac{11}{1-(1+t)^{11}}=-\frac{11}{11t+55t^2+O(t^3)}=-\frac{1}{t}\cdot\frac{1}{1+5t+O(t^2)}=-\frac{1}{t}\left(1-5t+O(t^2)\right)$$
So the difference is
$$-\frac{1}{t}\left(1-11t-1+5t+O(t^2)\right)=6+O(t)$$
And your limit is $6$.
That is the same as
$$\lim_{x\to 0}\left[\frac{23}{1-(1-x)^{23}}-\frac{11}{1-(1-x)^{11}}\right]=\lim_{x\to 0}\left[\frac{23}{23x-253x^2}-\frac{11}{11-55x^2}\right]$$ (we exploited the binomial theorem and neglected terms with high order, since we can, see the comments below) or as $$\lim_{x\to 0}\frac{1}{x}\left[\frac{1}{1-11x}-\frac{1}{1-5x}\right]=\lim_{x\to 0}\left[\frac{-5+11}{(1-11x)(1-5x)}\right]=11-5=\color{red}{6}.$$ With the same approach, for any $n,m\in\mathbb{N}^+$, $$\lim_{x\to 1}\left[\frac{m}{1-x^m}-\frac{n}{1-x^n}\right]= \color{red}{\frac{m-n}{2}}.$$ In other terms, the function $f_n(x)=\frac{n}{1-x^n}$ has a simple pole at $x=1$. If we remove the contribute given by such simple pole, we are left with a holomorphic function in a neighbourhood of $x=1$. In particular, $$\lim_{x\to 1}\left[\frac{n}{1-x^n}+\frac{1}{1-x}\right]=\frac{n-1}{2}.$$
• I don't like very much to "neglect terms". There is the notation $O(x^3)$ for this purpose (or $o(x^2)$ if you prefer). More correct, and less dangerous: what happens when all previous terms cancel, you are left with $0$ while there is indeed some nonzero function of $x$? For instance, how do you know in advance that you can neglect terms of degree $>2$, but not those of degree $2$? Of course, I don't know either with the $O$ notation, but at least if I fail I will know. – Jean-Claude Arbaut Sep 28 '16 at 20:41
• @Jean-ClaudeArbaut: $$\frac{23}{23x-253x^2+o(x^2)}=\frac{23}{23x-253x^2}+o(1)$$ hence we are allowed to do that. – Jack D'Aurizio Sep 28 '16 at 20:42
• Yes, but we would also be allowed to do that at order $2$. But it would fail, and you would not know, unless you have mentally used the correct notation, but then why not write it? Bad habit, IMO – Jean-Claude Arbaut Sep 28 '16 at 20:46
• Otherwise, you may simply keep the terms with high order, then check in the last steps that they do not matter. – Jack D'Aurizio Sep 28 '16 at 20:46
• I know it works here, because even if there are terms left, they tend to $0$ and only contribute by addition, so no risk. But if you have to justify this, it will take at least as much work as if you had done things cleanly to begin with. Not worth it. And I have seen limits for which you need to keep track of terms of degree, say, $6$, so it's not a good advice, to neglect terms, when you get the same result by using a more correct notation, that does not need extra justification. – Jean-Claude Arbaut Sep 28 '16 at 20:55
• How did you justify your second equality? Because $\frac 1{1-x} = \sum_{k=0}^\infty x^k$, an infinite series. How did you get finite series in the numerator? By squaring the denominator? – user373314 Sep 29 '16 at 8:07
• @user373314 Note that $1 - x^{n} = \left(1 - x\right)\left(1 + x + x^{2} + \cdots + x^{n - 1}\right) = \left(1 - x\right)\sum_{k = 0}^{n - 1}x^{k}$ for $n = 1, 2,3,\ldots$. They are $\color{#f00}{finite\ sums}$. – Felix Marin Sep 29 '16 at 18:07
$$=\lim_{x \to 1} \frac{23 - 23x^{11} -11 + 11x^{23}}{1 - x^{11} - x^{23} + x^{34}} = \lim_{x \to 1} \frac{-23\cdot 11 x^{10} + 11\cdot 23 x^{22}}{-11x^{10} - 23x^{22} + 34x^{33}} =$$$$= \lim_{x \to 1} \frac{-23\cdot 11 \cdot 10 x^9 + 11\cdot 23 \cdot 22x^{21}}{-11\cdot 10 \cdot x^9 - 23\cdot 22x^{21} + 34\cdot 33 x^{32}} = \frac {3036}{506} = 6$$ where we used Hopital twice
$\lim_\limits{x\to1}\frac {23(1-x^{11})-11(1-x^{23})}{(1-x^{11})(1-x^{23})}$
$\lim_\limits{x\to1}\frac {12-23x^{11}+ 11x^{23}}{(1-x^{11})(1-x^{23})}$
Now we could apply L'Hopitals at this point, or we can use algebra.
Using algebra, numerator and denominator both divide by $(x-1)^2$
$1-x^{11} = (1-x)\sum_\limits{i=0}^{10} x^i\\ 1-x^{23} = (1-x)\sum_\limits{i=0}^{22} x^i$
$11x^{23} -23x^{11}+ 12 = (x-1)(11 x^{22}\cdots 11x^{11} - 12x^{10}\cdots +12)\\ =(x-1)^2 (11x^{21} + 2\cdot11 x^{20} + 3\cdot11 x^{19}\cdots +12\cdot 11 x^{10} + 12\cdot 10 x^9\cdots +12\cdot 2x + 12$
Evaluated at 1.
The denominator:
$\sum_\limits{i=0}^{10} x^i = 11, \sum_\limits{i=0}^{22} x^i = 23$
the numerator:
$11 \sum_\limits{i=1}^{11} i + 12 \sum_\limits{i=1}^{11} i = (23)(11)(12)/2$
and the ratio $= 6$
Write $P_n(x)=1+x+x^2+\dots+x^n$; then our function is $$\frac{23P_{10}(x)-11P_{22}(x)}{(1-x)P_{22}(x)P_{10}(x)}$$ We can notice that $$\lim_{x\to1}P_{22}(x)P_{10}(x)=23\cdot11$$ so we just need to compute $$\lim_{x\to1}\frac{11P_{22}(x)-23P_{10}(x)}{x-1}= 11P_{22}'(1)-23P_{10}'(1)$$ by definition of derivative. Since $$P_n'(x)=1+2x+3x^2+\dots+nx^{n-1}$$ we have $$P_n'(1)=\frac{n(n+1)}{2}$$ Thus, reinserting $23\cdot11$ at the denominator, our limit is $$\frac{11\cdot22\cdot23-23\cdot10\cdot11}{2\cdot23\cdot11}= \frac{22-10}{2}=6$$
More generally, \begin{align} \lim_{x\to1}\left(\frac{m}{1-x^m}-\frac{n}{1-x^n}\right) &= \lim_{x\to1}\frac{mP_{n-1}(x)-nP_{m-1}(x)}{(1-x)P_{m-1}(x)P_{n-1}(x)} \\[6px] &= \frac{nP_{m-1}'(1)-mP_{n-1}'(1)}{mn} \\[6px] &= \frac{n(m-1)m-m(n-1)n}{2mn} \\[6px] &= \frac{m-n}{2} \end{align}
The following standard formula is well known $$\lim_{x \to 1}\frac{x^{n} - 1}{x - 1} = n = \lim_{t \to 0}\frac{(1 + t)^{n} - 1}{t}\tag{1}$$ and it appears that we can go very easily to the next step if $n$ is a positive integer and derive the formula $$\lim_{x \to 1}\frac{x^{n} - 1 - n(x - 1)}{(x - 1)^{2}} = \lim_{t \to 0}\frac{(1 + t)^{n} - 1 - nt}{t^{2}} = \frac{n(n - 1)}{2}\tag{2}$$ The simplest approach to prove $(2)$ is to use Binomial theorem. Hence we have $$x^{n} - 1 = n(x - 1) + \frac{n(n - 1)}{2}(x - 1)^{2} + o((x - 1)^{2})$$ and therefore \begin{align} \frac{n}{1 - x^{n}} &= \dfrac{n}{n(1 - x) - \dfrac{n(n - 1)}{2}(x - 1)^{2} + o((x - 1)^{2})}\notag\\ &= \frac{1}{1 - x}\left(1 - \frac{n - 1}{2}(1 - x) + o((1 - x))\right)^{-1}\notag\\ &= \frac{1}{1 - x}\left(1 + \frac{n - 1}{2}(1 - x) + o(1 - x)\right)\notag\\ &= \frac{1}{1 - x} + \frac{n - 1}{2} + o(1)\tag{3} \end{align} It follows that $$\frac{n}{1 - x^{n}} - \frac{m}{1 - x^{m}} = \frac{n - m}{2} + o(1)$$ and hence $$\lim_{x \to 1}\left(\frac{n}{1 - x^{n}} - \frac{m}{1 - x^{m}}\right) = \frac{n - m}{2}$$ and putting $n = 23, m = 11$ we get the desired limit as $6$.
The gymnastics of series division to reach $(3)$ can be avoided in another manner by using $(2)$ directly. We have \begin{align} L &= \lim_{x \to 1}\frac{n}{1 - x^{n}} - \frac{m}{1 - x^{m}}\notag\\ &= \lim_{x \to 1}\frac{n(1 - x^{m}) - m(1 - x^{n})}{(1 - x^{n})(1 - x^{m})}\notag\\ &= \lim_{x \to 1}\frac{n(1 - x^{m}) - mn(1 - x) + mn(1 - x) - m(1 - x^{n})}{(1 - x^{n})(1 - x^{m})}\notag\\ &= \lim_{x \to 1}\dfrac{n(1 - x^{m}) - mn(1 - x) + mn(1 - x) - m(1 - x^{n})}{\dfrac{(1 - x^{n})(1 - x^{m})}{(1 - x)^{2}}\cdot(1 - x)^{2}}\notag\\ &= \frac{1}{mn}\lim_{x \to 1}n\frac{1 - x^{m} - m(1 - x)}{(1 - x)^{2}} - m\frac{1 - x^{n} - n(1 - x)}{(1 - x)^{2}}\notag\\ &= \frac{1}{mn}\left(\frac{nm(1 - m)}{2} - \frac{mn(1 - n)}{2}\right)\text{ (using (2))}\notag\\ &= \frac{n - m}{2}\notag \end{align} | 2020-09-29T08:43:35 | {
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https://ch.mathworks.com/help/matlab/math/complex-line-integrals.html | # Complex Line Integrals
This example shows how to calculate complex line integrals using the `'Waypoints'` option of the `integral` function. In MATLAB®, you use the `'Waypoints'` option to define a sequence of straight line paths from the first limit of integration to the first waypoint, from the first waypoint to the second, and so forth, and finally from the last waypoint to the second limit of integration.
### Define the Integrand with an Anonymous Function
Integrate
`${\oint }_{C}\frac{{e}^{z}}{z}\phantom{\rule{0.2222222222222222em}{0ex}}dz$`
where $C$ is a closed contour that encloses the simple pole of ${e}^{z}/z$ at the origin.
Define the integrand with an anonymous function.
`fun = @(z) exp(z)./z;`
### Integrate Without Using Waypoints
You can evaluate contour integrals of complex-valued functions with a parameterization. In general, a contour is specified, and then differentiated and used to parameterize the original integrand. In this case, specify the contour as the unit circle, but in all cases, the result is independent of the contour chosen.
```g = @(theta) cos(theta) + 1i*sin(theta); gprime = @(theta) -sin(theta) + 1i*cos(theta); q1 = integral(@(t) fun(g(t)).*gprime(t),0,2*pi)```
```q1 = -0.0000 + 6.2832i ```
This method of parameterizing, although reliable, can be difficult and time consuming since a derivative must be calculated before the integration is performed. Even for simple functions, you need to write several lines of code to obtain the correct result. Since the result is the same with any closed contour that encloses the pole (in this case, the origin), instead you can use the `'Waypoints'` option of `integral` to construct a square or triangular path that encloses the pole.
### Integrate Along a Contour That Encloses No Poles
If any limit of integration or element of the waypoints vector is complex, then `integral` performs the integration over a sequence of straight line paths in the complex plane. The natural direction around a contour is counterclockwise; specifying a clockwise contour is akin to multiplying by `-1`. Specify the contour in such a way that it encloses a single functional singularity. If you specify a contour that encloses no poles, then Cauchy's integral theorem guarantees that the value of the closed-loop integral is zero.
To see this, integrate `fun` around a square contour away from the origin. Use equal limits of integration to form a closed contour.
```C = [2+i 2+2i 1+2i]; q = integral(fun,1+i,1+i,'Waypoints',C)```
```q = 0.0000e+00 + 2.2204e-16i ```
The result is on the order of `eps` and effectively zero.
### Integrate Along a Contour with a Pole in the Interior
Specify a square contour that completely encloses the pole at the origin, and then integrate.
```C = [1+i -1+i -1-i 1-i]; q2 = integral(fun,1,1,'Waypoints',C)```
```q2 = -0.0000 + 6.2832i ```
This result agrees with the `q1` calculated above, but uses much simpler code.
The exact answer for this problem is $2\pi i$.
`2*pi*i`
```ans = 0.0000 + 6.2832i ``` | 2020-09-26T13:06:02 | {
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https://math.stackexchange.com/questions/1833667/proof-that-a-cap-b-and-a-setminus-b-are-disjoint | # Proof that $A \cap B$ and $A \setminus B$ are disjoint.
I am trying to prove that $A \cap B$ and $A \setminus B$ are disjoint. Here is what I've done so far.
Is there anything that's wrong in my proof, and is there anything that can make it better?
Proof: $A \cap B$ and $A \setminus B$ are disjoint if $(A \cap B) \cap (A \setminus B) = \emptyset$.
First, let $x \in (A \cap B) \cap (A \setminus B)$. Then $x \in (A \cap B)$ and $x \in (A \setminus B)$. Then this means $x \in A$ and $x \in B$, and $x \in A$ and $x \notin B$. Thus, these two sets must be disjoint and therefore $x \in \emptyset$. Hence, $(A \cap B) \cap (A \setminus B) \subseteq \emptyset$.
Conversely, since the empty set is always a subset of any nonempty set, $\emptyset \subseteq (A \cap B) \cap (A \setminus B)$.
Therefore $(A \cap B) \cap (A \setminus B) = \emptyset$.
• "Conversely, since an empty set is always a subset of any set" nonempty or otherwise. It would also help to say just before "thus" the line "but $x\in B$ and $x\notin B$ cannot occur simultaneously" as well as the phrase before "first" "Suppose for contradictory purposes that there is some $x\in\dots$" Otherwise, your proof is essentially correct. – JMoravitz Jun 20 '16 at 19:02
There is nothing wrong in the proof. However, to make it better you might skip some superfluous details. For example, the following is enough:
Suppose towards a contradiction that $(A \cap B) \cap (A \setminus B) \neq \emptyset$. Then there is an $x \in (A \cap B) \cap (A \setminus B)$, which implies $x \in B$ and $x \notin B$. A contradiction.
Other way $$(A\cap B)\cap(A-B)=(A\cap B)\cap(A\cap B\,')=A\cap(B\cap B\,')=A\cap\phi=\phi$$
Let $x \in (A \cap B) \cap (A \setminus B)$. Then $x \in (A \cap B)$ and $x \in (A \setminus B)$. After this, it is better to say:
This means $(x \in A\land x \in B)\land (x \in A\land x \notin B)$. So $x\in B\land x \notin B$, which is contradiction.
The proof is correct, but it is a tad verbose.
If you are going to write a proof by contradiction, I recommend you say so up front.
You just need to show that $x$ cannot be in both $B$ and $B'.$
Alternatively, show $(A \cap B) \cap (A \cap B') = \emptyset$
$(A \cap A) \cap (B \cap B') = \emptyset$ | 2019-09-17T04:58:29 | {
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https://math.stackexchange.com/questions/3277669/construct-the-field-of-complex-numbers-as-the-quotient-ring-of-real-polynomials | # Construct the field of complex numbers as the quotient ring of real polynomials
I am trying to construct the field of complex numbers as the quotient ring of real polynomials.
Suppose that
1. $$\mathbb C, \mathbb R$$ are the fields of complex and real numbers respectively.
2. $$\mathbb R [X]$$ is the ring of polynomials over $$\mathbb R$$.
3. $$\left \langle X^{2} + 1 \right \rangle = \left \{p(X^{2}+1) \mid p \in \mathbb R [X] \right \}$$ is the ideal generated by $$(X^{2}+1)$$.
4. $$D = \mathbb R [X] / \left \langle X^{2} + 1 \right \rangle$$ is the quotient ring of $$\mathbb R [X]$$ modulo $$\left \langle X^{2} + 1 \right \rangle$$.
Then $$(\mathbb C, +, \cdot) \cong (D, +, \cdot)$$
Could you please verify if my attempt contains logical gaps/errors? Any suggestion is greatly appreciated.
My attempt:
Lemma (Long Division of Polynomial): Let $$K$$ be a field and $$p, q \in K[X]$$ with $$q \neq 0 .$$ Then there are unique polynomials $$r, s \in K[X]$$ such that $$p=s q+r \quad \text {and} \quad \operatorname{deg}(r)<\operatorname{deg}(q) \tag 1$$
Proof:
1. Existence
Define $$\mathcal Q: K[X] \times (K[X] - \{0\}) \to K[X]$$ by $$\mathcal Q(p,q) = \begin{cases} \bar p / \bar q X^{\operatorname{deg} (p)-\operatorname{deg} (q)} & \text{if } \operatorname{deg} (p) \ge \operatorname{deg} (q) \\ 0 & \text{otherwise}\end{cases}$$
where $$\bar p, \bar q$$ are the coefficients corresponding to $$\operatorname{deg} (p), \operatorname{deg} (q)$$ respectively.
Define $$\langle p_n, s_n \rangle_{n \in \mathbb N}$$ recursively by
\begin{aligned}\langle p_0, s_0 \rangle &= \langle p, \mathcal Q(p,q) \rangle \\ \langle p_{n+1}, s_{n+1} \rangle &= \langle p_n - s_n q, \mathcal Q (p_{n+1},q) \rangle \end{aligned}
Let $$n' = \min \{ n \in \mathbb N \mid s_n = 0\}$$. It is easy to verify that $$s = \sum_{i = 0}^{n'} s_n$$ and $$r = p_{n'}$$ satisfy $$(1)$$.
1. Uniqueness
Suppose that $$s’$$ and $$r’$$ are other polynomials such that $$p=s’ q+r’$$ and $$\operatorname{deg} (r’) < \operatorname{deg}(q)$$. Then $$(s’-s) q=r-r’$$. If $$s’-s \neq 0$$ then, from $$\operatorname{deg} (p q) = \operatorname{deg}(p) + \operatorname{deg}(q)$$, we would get
$$\operatorname{deg} (r-r’) = \operatorname{deg} ((s’-s) q) = \operatorname{deg} (s’-s) + \operatorname{deg}(q)>\operatorname{deg}(q)$$
which, because $$\operatorname{deg} (r-r’) \leq \max \{\operatorname{deg} (r), \operatorname{deg} (r’)\} <\operatorname{deg} (q)$$, is not possible. Thus $$s’=s$$ and also $$r’=r$$.
By lemma, each set in $$D$$ has at least one element of the form $$a+bX$$. Assume that both $$a + b X$$ and $$a' +b' X$$ belong to the same set in $$D$$. Then $$(a + b X) \sim (a' + b' X)$$ and thus $$(a + b X) - (a' + b' X) = p (X^2 + 1)$$ for some $$p \in K[X]$$. It follows that $$a = a'$$, $$b = b'$$, and $$p = 0$$. As such, each set in $$D$$ has exactly one element of the form $$a+bX$$.
Consider $$\phi : D \rightarrow \mathbb{C}, \quad [a+bX] \mapsto a+b i$$
Clearly, $$\phi$$ is surjective.
Assume $$[a+bX], [a'+b'X] \in D$$ such that $$\phi ([a+bX]) = \phi ([a'+b'X])$$. Then $$a+b i = a'+b' i$$, and thus $$a=a'$$ and $$b = b'$$. Hence $$\phi$$ is injective.
Next we show that $$\phi$$ is a homomorphism w.r.t $$+$$ and $$\cdot$$.
\begin{aligned} \phi([a+b X]+[a'+b' X]) &= \phi([(a+a')+(b+b')X ]) \\ &= (a+a')+(b+b') i \\ &= (a+b i)+(a'+b' i) \\ &= \phi([a+b X])+\phi([ a'+b' X])\end{aligned}
\begin{aligned}\phi([a+b X]\cdot[a'+b' X]) &=\phi([(a+b X)\cdot (a'+b' X)])\\ &=\phi ([aa'+(ab'+ba') X+bb' X^2]) \\ & = \phi ([aa'-bb'+ (ab'+ba') X+bb' (X^2 + 1)]) \\ &=\phi([aa'-bb'+ (ab'+ba') X] )\\ &=(aa'-bb')+ (ab'+ba') i\\ &=(a+b i) \cdot (a'+b' i) \\ &=\phi([a+b X]) \cdot \phi([a'+b' X]) \end{aligned}
This completes the proof.
Good job! I would just add to that a proof of the fact that $$\phi\bigl([1]\bigr)=1$$ (that's trivial, of course).
• Thank you so much for your verification ;) Jun 29 '19 at 9:15
• I'm glad I could help. Jun 29 '19 at 9:35
Well, the quotient ring $$D = {\Bbb R}[x]/\langle x^2+1\rangle$$ is a field since $$x^2+1$$ is irreducible over $$\Bbb R$$.
Moreover, the quotient ring $$D$$ contains a zero of $$x^2+1$$, namely $$\bar x = x+\langle x^2+1\rangle$$. This zero satisfies $$\bar x^2+1=0$$. The other zero is $$-\bar x$$.
This shows that $$D = {\Bbb R}[\bar x]$$ (ring adjoint) and since $$D$$ is a field, $$D={\Bbb R}(\bar x)$$ (field adjoint).
The mapping
$${\Bbb C}=\{a+ ib\mid a,b\in{\Bbb R}\}\rightarrow {\Bbb R}(\bar x) = \{a+b\bar x\mid a,b\in{\Bbb R}\}: a+bi \mapsto a+b\bar x$$
is an isomorphism of fields. | 2021-09-18T17:45:06 | {
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https://math.stackexchange.com/questions/448/in-how-many-different-ways-can-i-sort-balls-of-two-different-colors/451#451 | # In how many different ways can I sort balls of two different colors
Let's say, I have 4 yellow and 5 blue balls. How do I calculate in how many different orders I can place them? And what if I also have 3 red balls?
• maybe a title more clear could be "In how many different ways can I sort balls of two different colors"
– mau
Jul 22 '10 at 7:37
• So, you're asking for how many unique ways you can sort them? And swapping two red balls in a certain arrangement doesn't count as a unique new sorting. Jul 22 '10 at 7:38
• @Justin L., That's probably a fair assumption, yeah. Jul 22 '10 at 7:53
This is a standard problem involving the combinations of sets, though perhaps not very obvious intuitively.
Firstly consider the number of ways you can rearrange the entire set of balls, counting each ball as indepndent (effectively ignoring colours for now). This is simply $(4 + 5)! = 9!$, since the 1st ball can be any of the $9$, the 2nd can be any of the remaining $8$, and so on.
Then we calculate how many different ways the yellow balls can be arranged within themselves, since for the purpose of this problem they are considered equivalent. The number of combinations is of course $4!$; similarly, for the blue balls the number is $5!$.
Hence, overall we find:
$$\text{total arrangements} = \frac{\text{arrangements of all balls}}{\text{arrangements of yellow balls} \times \text{arrangements of blue balls}}$$
Therefore in our case we have:
$$\text{total arrangements} = \frac{9!}{5! \times 4!} = 126$$
I'm sure you can see how this can be easily extended if we also have 3 red balls too. (Hint: the total changes and we have another multiple of identical arrangements to account for.)
The case of two colors is simple: if you have m yellow balls and n blue ones you only need to choose m positions among (m+n) possibilities, that is (m+n)!/(m!·n!). The other balls' positions are automatically set up.
For some reason I find it easier to think in terms of letters of a word being rearranged, and your problem is equivalent to asking how many permutations there are of the word YYYYBBBBB.
The formula for counting permutations of words with repeated letters (whose reasoning has been described by Noldorin) gives us the correct answer of 9!/(4!5!) = 126. | 2022-01-26T06:26:58 | {
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https://math.stackexchange.com/questions/2529446/arbitrary-vs-random/2529701 | # Arbitrary vs. Random
I'm currently assisting a basic course where the students have to write some proofs. Most of them use terminology like "Let $x$ be a random integer", instead of "Let $x$ be an arbitrary integer" or plainly "Let $x$ be an integer". Am I doing the right thing in correcting them?
My point of view is that a random integer is an integer chosen by some kind of random distribution. In that interpretation, there's some truth in saying that "Let $x$ be a random integer, then $x$ is not equal to 25", since the probability might be zero for that to happen, and in that sense the randomly chosen integer is not going to be 25. Meanwhile the sentence "Let $x$ be an integer, then $x$ is not equal to 25" is blatantly false, since we can take $x=25$ deterministically (but definitely not randomly).
• I don't know why this received a downvote. Nov 20 '17 at 17:05
• Yes, you should correct them. Arbitrary means an implicit quantifier in a way that random doesn't. Nov 20 '17 at 17:09
• You are right that they are wrong. Whether you are right in correcting them is a completely different question, and depends, among other things, on the students' age / cognitive level. Although if they are writing proofs, they are probably far enough along that you should correct them. Nov 20 '17 at 17:10
• @Arthur, I'd say no matter the age, it is the right thing to do to point that there is difference in meaning. What depends on age/cognitive level is how deeply one should go in explaining the difference and how insisting one should be that the students use those phrases correctly. Nov 20 '17 at 17:14
• It isn't true that the probability of a randomly chosen integer being equal to 25 is 0. It depends on the actual distribution which is used to give a meaning to "randomly chosen integer". Since here is no uniform distribution on the set of integers, there is no default meaning for that phrase. Nov 20 '17 at 21:51
I originally wrote a more ambivalent response, but thinking about it further I've changed my mind.
It's clear that the phrase "let $x$ be a random integer" is mathematically . . . bad. What is at question is whether:
• it is misleading to the student,
• it is worth correcting,
• and as a bonus, whether it is worth penalizing when repeated.
I think the answer to (3) is no (unless one is in a class dealing with probability), and the answer to (2) is yes, since if nothing else explaining why the phrase is wrong lets you preemptively address some of the usual confusions around quantifiers (e.g. we're allowed to pick a number that happens to be a counterexample "out of a hat").
I think the answer to (1) (and here's where I've changed my mind) is "yes" - or rather, it is "yes" enough that we should treat it as "yes." I think this is a case where poor use of language early on could set the student up for more confusion down the road, even if they are not being confused by the phrase at the moment. (And this is generally an argument for helping students with language use in mathematics.)
That said, I still think the answer to (3) is no (again, unless the class is dealing with probability).
In common parlance, random and arbitrary are often used interchangeably. A quick check of on-line dictionaries confirms that the semantic overlap is well established in spite of the different origins of the two words.
The fledgling proof-writers need to be made aware that this is not the case in math, with random being used when probabilities are involved. On the other hand, "Let $x$ be an arbitrary integer; then $P(x)$ holds" translates $\forall x \in \mathbb{Z} \,.\, P(x)$ into English.
Next, it would probably help the aforementioned fledglings if they were shown why the distinction is useful. One practical reason is simplicity. If one deals with an arbitrary integer $x$, all that is assumed is that $x \in \mathbb{Z}$. Could $x = 25$ be true? Of course! Could $x = 25$ be false? Certainly!
If, however, $x$ is a randomly chosen integer, not much may be said without knowing the distribution from which $x$ was drawn. The probability of $x = 25$ may be greater than $0$ if the distribution is not uniform (as it must be if the sample space is countable). Besides, as you may well know, zero probability doesn't mean impossible. By avoiding the use of random all these issues are sidestepped.
In more advanced courses, students will be able to appreciate more reasons for keeping random and arbitrary, as well as probabilistic and nondeterministic, distinct. But the example above should be enough to get them started. At any rate, in framing my feedback to students at their first attempts with proofs, I'd assume that they had the right concept in mind, but didn't pick the correct mathematical term to express it.
• "zero probability obviously doesn't mean impossible". It's not so obvious to me, could you elaborate? Do you mean "probability asymptotically approaches zero" (integers are infinite, so the probability of choosing zero is infinitesimal)? Nov 20 '17 at 22:13
• > if the distribution is not uniform. and it's always not uniform
Nov 20 '17 at 22:21
• @Barmar, choose random number (uniformly) in interval (0; 1). what was the initial probability of the number you have chosen?
Nov 20 '17 at 22:32
• Infinitessimal, I think, but not zero. Nov 20 '17 at 22:54
• @Barmar That probability is zero. Probabilities are real numbers. "The probability of X is infinitesimal" means nothing. Nov 20 '17 at 23:21
Actual dictionary definitions: Doing some quick dictionary searching for "arbitrary" gives the definition: "based on random choice or personal whim, rather than any reason or system." The definition given for "random" is "made, done, happening, or chosen without method or conscious decision." In fact, "random" is listed as a synonym for "arbitrary" on an online dictionary. Therefore the interchanging of the two terms is completely understandable.
Technical vs natural language: While it is true that "arbitrary" vs "random" have very different technical meanings in mathematics, they are nearly interchangeable in natural language. It is important to distinguish between natural and technical language usage/meaning.
I would explain this distinction between natural and technical language to my students. That is something that is students of any discipline should be aware of. There may be a risk of muddying the waters though since mastering the actual mathematics at hand may or may not be helped by this discussion of language.
Are you doing the "right" thing in correcting them? Offering a student relevant correct information is always the "right" thing to do. However, it may not always be the right thing to do if there is sufficient risk of it causing more confusion.
Language is a huge problem in mathematics. It's not something that is taught well in my opinion, in terms of how to actually speak mathematics. At least, if the way my students talk is any indication, there is generally a huge gap in being able to do mathematics and being able to explain it verbally in a coherent fashion using technical terminology correctly.
The physical process of choosing a number: Now let's consider the actual physical process of a human choosing an arbitrary number (in the technical sense here). It might be the case that such a physical process could be modeled using a random variable.
So, inasmuch as it is a real process of coming up with an actual example of an arbitrary number, it might actually be a type of random number in the probabilistic modeling sense. Of course, in the actual mathematical context where the number is to be used, it is just an arbitrary number, e.g. to be plugged into an equation. | 2022-01-16T21:56:38 | {
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https://math.stackexchange.com/questions/2419255/why-cant-the-ratio-test-be-used-for-geometric-series/2419291#2419291 | # Why can't the ratio test be used for geometric series?
The ratio test says that, for $a_k\neq 0$, if $$\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=L$$ exists, then if $0\leq L <1$, then $\sum_k a_k$ converges. If $L>1$, it diverges.
The notes I'm reading say that it's inadmissible to use the ratio test to test for convergence of a geometric series. I can't see why this should be the case.
Say we have some geometric series $\sum_kar^k$. Then $$\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=\lim_{k\to\infty}\frac{\left|ar^{k+1}\right|}{\left|ar^k\right|}=|r|.$$ So the ratio test tells us that the geometric series converges for $|r|<1$, and diverges for $|r|>1$, which is exactly what we get by using the formula $$\sum_{k=1}^n ar^k=a\left(\frac{1-r^{n+1}}{1-r}\right).$$
What is an example that demonstrates why the ratio test is inadmissible for a geometric series?
• How do you prove the ratio test? Probably by invoking the theorem that the geometric series converges. Sep 6, 2017 at 17:29
• Makes sense, thanks Sep 6, 2017 at 17:29
• Aside from the possibly circular logic in proving the ratio test, it certainly applies to geometric series and gives the correct result as you showed.
– user169852
Sep 6, 2017 at 17:55
The usual proof of the ratio test is to compare the series to a geometric series. If $$\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \alpha < 1,$$ then we have $$|a_{n+1}| < |a_n| \alpha$$ for all sufficiently large $n$. It then follows from an induction argument that $$|a_{n+k}| < |a_n| \alpha^k$$ for $n$ sufficiently large, which means that $$\sum_{j=1}^{\infty} |a_j| = \sum_{j=1}^{n-1} |a_j| + \sum_{j=n}^{\infty} |a_j| \le \sum_{j=1}^{n-1} |a_j| + \sum_{k=0}^{\infty} |a_n| \alpha^k.$$ The first series has only finitely many terms and is therefore finite, and the second series is geometric and therefore converges by some other argument. From this, it follows that if $$\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| < 1, \qquad\text{then}\qquad \sum_{j=1}^{\infty} a_j$$ converges (by the limit comparison test, for example). The other inequality in the ratio test can be argued by noting that the general term does not go to zero, and the uncertainty at 1 can be argued by considering (for example) the harmonic and alternating harmonic series.
The key point is that we prove that the ratio test implies the convergence of series by comparison to a convergent geometric series. Since the proof of the ratio test relies on this convergence, it is circular to argue that a geometric series converges by the ratio test (unless, of course, you have another proof for the ratio test that doesn't use the convergence of geometric series).
While one could use the ratio test to establish the convergence of a geometric series (there is nothing stopping us!), it is typically poor style to rely on circular arguments as it can (potentially) lead to overlooking important hypotheses or exceptional cases. This is particularly important in a pedagogical setting, when students may not be entirely cognizant of the line of reasoning that lead up to a result (it is hard to keep track of all of the lemmata, theorems, and proofs that lead up to a result if it is the first time that you have had to deal with them).
Moreover, I don't see why one would want to use the ratio test to show that a geometric series converges. Basically no computation is needed to show that a geometric series converges, while a couple of computational steps are needed in order to invoke the ratio test.
• Or, you've already proven the ratio test using the convergence of geometric series, and since the former subsumes the latter there's little point in retaining the latter.
– user14972
Sep 7, 2017 at 8:55
• @Hurkyl Note that I have not said that it is "inadmissible" to use the ratio test to show that a geometric series converges, only that it is "circular". Sep 7, 2017 at 12:53
We can't know for sure (unless context from the notes you've omitted says something on the topic), but the various comments appearing in this topic suggest an explanation.
Two major activities mathematicians (and people who use mathematics) engage in are:
• Using mathematical tools to do calculations and prove things
• Designing and validating the mathematical tools used in the above bullet
The ratio test is an example of such a mathematical tool, and is perfectly applicable to geometric series. It may even be the preferable tool for deciding when a geometric series converges, simply to cut down on the number of things one needs to remember.
However, if you happen to be in the process of validating the ratio test, it would not be valid to use the ratio test to justify any of the facts you need — such as when geometric series converge — in its validation. You would have to first derive the facts about geometric series in some other fashion.
As for your notes, the most likely explanations are:
• You are reading notes about how to validate the ratio test
• You have misunderstood the notes
• Possibly due to the author failing to actually convey his meaning
• The author of these notes was confused
To elaborate on that last point, people often get stuck in the "tool building" mindset. A fair amount of mathematics education involves validating the tools one is already familiar with, which requires temporarily suspending the use of those tools so that we can see how they can be built from more basic tools — and sometimes people learn the wrong lesson and think that's something you always have to do.
So, when facts about convergence of geometric series in the proof of the ratio test, this sometimes leads to people mistakenly thinking that any time the topic of convergence of geometric series comes up, one must suspend their use of the ratio test and argue from more basic tools.
• Here are the notes (look at the second bullet point on page 46) courses.maths.ox.ac.uk/node/view_material/1087 Sep 7, 2017 at 7:57
• and then point B on the problem sheet (page 2): "Why is it not admissible to establish the convergence or divergence of a geometric series by using the Ratio Test?" courses.maths.ox.ac.uk/node/view_material/347 Sep 7, 2017 at 7:58
• @man_in_green_shirt: Unless that road sign in the margin has some relevant significance, the notes are simply wrong as written, and I think my answer is accurate. I think the best odds are on the author wanting to talk about the specific setting of how one would prove whether or not a geometric series converges if this fact was not already known, but failing to properly frame the remark.
– user14972
Sep 7, 2017 at 8:53
The ratio test is not inadmissible for geometric series. Its hypotheses do not exclude geometric series, therefore it applies, and its proof must support this.
One common proof structure would be:
Theorem A: Geometric series converges. Proof: direct argument.
Theorem B: Ratio test with usual hypotheses. Proof: show that this is implied by theorem A as in the answer by Xander Henderson.
Of course, the proof of Theorem A cannot use Theorem B, otherwise we have a circular argument. Undoubtedly this is what your notes are trying to say.
However, once we have a valid proof of Theorem B, it certainly applies to geometric series:
Theorem C: Geometric series converges. Proof: theorem B.
This proof of Theorem C may seem absurdly indirect: why wouldn't we just cite Theorem A? Well, consider:
Theorem D: some other theorem whose hypotheses imply those of the ratio test. Proof: Theorem B.
It would be very annoying, and more importantly unnecessary, to instead write:
Theorem D: some other theorem whose hypotheses imply those of the ratio test. Proof: If the series is geometric, then see Theorem A. Otherwise Theorem B applies.
Once we have a valid proof of the ratio test, then it can be applied to any series if it satisfies the hypotheses, including geometric series. It does not matter how the ratio test was proven. This is in no way circular reasoning even if the proof used results about geometric series. The idea of circular reasoning is a logical fallacy, but that is not what is being done here. We are not assuming what we are trying to prove. In fact, if we are allowed to assume statement $A$ because it is given to us, then the proof of $A$ is immediate from our given assumption. This is not circular reasoning because we are explicitly allowed to use statement $A$ which is somehow given to us to use.
The OP gave a link to course notes that state 'We cannot prove that a geometric series $\sum r^k$ is convergent/divergent by applying the Ratio Test. Why not?" That's a convincing argument, isn't it? Despite the hypotheses for "D'Alembert's Ratio Test" are satisfied by any positive geometric series.
• It's correct but still circular, like computing $\lim_{x \to 0} \frac{e^x-1}{x}$ using L'Hospital's rule. Ultimately if you expanded out the proof of the ratio test, you would find that in the body of that proof is the result you already want to prove being asserted without proof...unless you have some other proof of the ratio test.
– Ian
Sep 6, 2017 at 18:32
• @Ian We can easily fix the circularity by embedding a direct proof of the convergence of geometric series in the proof of the ratio test. If we assume this is done, then there's no logical problem applying the ratio test to geometric series, even if it is unnecessary to do so.
– user169852
Sep 6, 2017 at 19:22
• @Bungo It's true, but then you are effectively saying "I have a black box containing the desired proof. Showing you this black box without opening it constitutes a proof." It's logically sound but socially flawed, and proof is a social activity.
– Ian
Sep 6, 2017 at 19:30
• @Ian I would argue that any proof of the ratio test that uses without proof the fact that a geometric series converges, is inadequate. Otherwise, every time we want to prove a new theorem whose hypotheses imply those of the ratio test, we would have to bifurcate into two cases: "if $x_n$ is not geometric, then the ratio test applies; otherwise, it converges by the geometric series convergence theorem". If the ratio test hypotheses do not exclude the geometric series, then the proof must include or cite a proof of this convergence (or not depend on this convergence).
– user169852
Sep 6, 2017 at 20:04
• @Bungo In my analogy, you're proving the geometric series converges, you present a black box that is a proof of the ratio test, and you don't open the proof of the ratio test, so you don't get to see the details of the proof that the geometric series converges, but you "promise to the listener" that it is there anyway.
– Ian
Sep 6, 2017 at 20:10 | 2022-10-07T08:17:35 | {
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https://math.stackexchange.com/questions/1408128/probability-of-getting-heads-in-a-coin-toss | Probability of getting heads in a coin toss
A fair coin is tossed five times. What is the probability of getting a sequence of three heads?
What i tried
Since the probability of getting a head and not getting a head is $0.5$. The probability of getting three heads is $0.5^3$ while there is a $0.5^2$ chance of not getting a head. But since the three heads are in sequence out of five toss, the total probability is a multiplication of $\left( \begin{array}{c} 5 \\ 3 \\ \end{array} \right).0.5^3.0.5^2=5/16$
Is my working correct. Could anyone explain. Thanks
Hint: ${5\choose 3} =10$ is the number of sequences where you have 3 heads and 2 tails. The heads have not to occur consecutive. The number of sequences with exact 3 heads consecutive are
$hhhtt, \ thhht, tthhh$
Therefore you have only 3 arrangements. Each arrangement has a probability of $0.5^5$
Remark: If you can have more then 3 consecutive heads, then you have 6 arrangements.
If the probability of getting a head is $p$, and the probability of not getting a head is $q=1-p$, then the probability of getting 3 or more consecutive heads is \begin{align} \underbrace{3 p^3 q^2}_{P(HHH)}+\underbrace{2p^4q^1}_{P(HHHH)}+\underbrace{p^5q^0}_{P(HHHHH)}&=p^3(3q^2+2pq+p^2)\\ &=6p^5\qquad\text{(as }q=p\text{)}\\ &=\frac 3{16}\qquad\text{(}p=\frac 12\text{)} \end{align}
This method can also be used if $p\neq\frac12$.
• Oh but why the constant value of 3 and 2 in front? – ys wong Aug 24 '15 at 16:56
• For HHH, the "elements" are 3H, T, T to fill 3 "positions": 3H can fill any one of the 3. For HHHH, the "elements" are 4H, T to fill 4 "positions": 4H can fill either of the 2. – hypergeometric Aug 24 '15 at 17:03
Not quite right. You need a sequence of three heads, not just any three heads. You've counted HHTHT as a valid coin-toss-sequence, whereas according to the question it shouldn't be.
The only allowable sequences are HHHTT, THHHT, TTHHH, HHHHT, THHHH, HTHHH, HHHTH, HHHHH.
• So the algorithm should be $(0.5)^3=1/8$? – ys wong Aug 24 '15 at 16:41
• No. How did you get that expression? To solve it: what is the probability of HHHTT? Of THHHT? Of HHHHH? – Patrick Stevens Aug 24 '15 at 16:43
• Okay Thanks. I got it – ys wong Aug 24 '15 at 17:04
You know that there are exactly $2^5 = 32$ unique outcomes in the game. Now in your "win" case, you must have exactly a sequence of three heads, so let's represent this $HHH$ sequence with a variable $W$. Now, you can simplify your question to: how many unique arrangements of $WTT$ are there? We must have $3! \over 1!2!$, or 3 unique arrangements.
Each unique arrangement has probability $P(H)^{N(H)}P(T)^{N(T)}$, where $N$ is the number of objects in a group and $P$ is the probability function. Since every one of our three arrangements must have three heads and two tails, and since the probability of getting heads/tails is 0.5, then each arrangement has probability $0.5^30.5^2 = {1 \over 32}$. So the solution is $3({1 \over 32}) = {3 \over 32}$ .
Note: the number of unique arrangements of a group with common elements is
$$N(total)! \over N(a_1)!N(a_2)!N(a_3)!\cdots$$
Where each $a_i$ is a unique element. In our case, we have only 2 unique elements's: $W$, and $T$, so $N(W) = 1$, $N(T) = 2$, and $N(total) = 3$. Of course, this strategy fails when two elements $a_i$ and $a_j$ have common elements, so watch out for that when using this strategy.
• Also, as I can't comment on others' answers yet, I believe there are a total of 8 unique arrangements if you remove the limit of having 3 heads, as follows: $HHHTT, THHHT, TTHHH, HHHHT, THHHH, HHHTH, HTHHH, HHHHH$ – user263104 Aug 24 '15 at 17:29
• Quite right. Dear me, I must be asleep. – Patrick Stevens Aug 24 '15 at 17:38
This is just me, but I'd take a direct approach and (since you only have 5 events to consider) draw out a table or chart of all the possible outcomes for first only 2 or 3 events, see if you can scale up, then check your calculation against that.
I find when people aren't inwardly certain in a solid way about the algorithm, a bit of visual intuition can really help.
• The way Patrick did just now in fact. Makes problems like this much easier to check. – Travelling Salesman Aug 24 '15 at 16:39 | 2019-07-20T01:02:08 | {
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http://mathematica.stackexchange.com/questions/14009/inconsistency-in-histograms-probability-binsize?answertab=votes | # Inconsistency in Histogram's “Probability” Binsize
Context
Let me define a Probability distribution (following the documentation and with some connection to this question)
DD = ProbabilityDistribution[(Sqrt[2]/\[Pi]) (1/(1 + x^4)), {x, -\[Infinity], \[Infinity]}];
which is normalized properly CDF[DD,1000]//N (* 1 *), and looks like this
Plot[Evaluate[PDF[DD, x]], {x, -4, 4}, Filling -> Axis]
I am interested in drawing samples from this distribution
If I let Mathematica do his job:
Show[RandomVariate[DD, 1500] // Histogram[#, Automatic, "Probability"] &,
Plot[PDF[DD, x], {x, -5, 5}, PlotStyle -> Directive[Red, Thick]]]
I get this
which strongly suggests it got the normalisation wrong.
Question
Is this a known bug or a misunderstanding on how Histogram works on my part?
Attempt
Now let me evaluate by hand its normalized histogram
hh = RandomVariate[DD, 15000]//BinCounts[#, {-5, 5, 1/4}] &;
h = 4 hh/Total[hh];
h = Transpose[{Table[i - 1/4, {i, -5 + 1/4, 5, 1/4}], h}];
I get this
Show[ListLinePlot[h, InterpolationOrder -> 0, PlotRange -> All, Filling -> Axis],
Plot[PDF[DD, x], {x, -5, 5}, PlotStyle -> Directive[Red, Thick]]]
Turning back to the Mathematica Automated procedure, if I force the binsize and the corresponding Normalization, I get this
Show[RandomVariate[DD, 150000]//Histogram[#, {-5, 5, 1/4}, "Probability"] &,
Plot[PDF[DD, x]/4, {x, -5, 5}, PlotStyle -> Directive[Red, Thick]]]
But note the 1/4 factor in the PDF (corresponding to the binsize).
-
If you use "PDF" instead of "Probability", then the scaling comes out right... – Ｊ. Ｍ. Nov 2 '12 at 9:53
@J.M. so it was a trap! You knew the answer: how wicked! Well you can write it as a one-liner now :-) – chris Nov 2 '12 at 9:56
Well, you got a dandy answer from jVincent, so it still turned out pretty darn good, eh? :) – Ｊ. Ｍ. Nov 2 '12 at 10:01
@J.M. are you guys working as a team? – chris Nov 2 '12 at 10:02
Nope; I'm pretty sure we've never met before... – Ｊ. Ｍ. Nov 2 '12 at 10:22
It's because of the different meanings of the graphs. For the bar chart, the probability of being within the width for any given bar is equal to the height. So, the density in a bar is given by height/width. So, when comparing to the probability density function (PDF), they shouldn't have the same height; they should be such that the bar chart is at the same height divided by the width of the bars. Notice that your boxes are 1/4 wide and this is where your inconsistency comes from.
Show[RandomVariate[DD, 1500] // Histogram[#, Automatic, "PDF"] &, | 2013-12-19T10:47:45 | {
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https://math.stackexchange.com/questions/1293387/prove-that-the-following-argument-is-valid | # Prove that the following argument is valid
I'm asked to show that the following argument is valid:
P1) $$[E \lor (L \lor M)] \land (E \leftrightarrow F)$$
P2) $$L \rightarrow D$$
P3) $$D \rightarrow \neg L$$
C) $$E \lor M$$
Here is my work (so far):
P2) $$L \rightarrow D$$
1. $$\neg(\neg L) \rightarrow D$$ Premise
2. $$L$$ Premise
3. $$L \rightarrow D$$ 1, Substitution
4. $$D$$ 2, 3 Modus
I'm not sure.
I know you need to use the rules of inference like modus ponens or converse fallacy, but I'm confused because it doesn't look like any of the forms I've learned.
Thanks
• I'm not expert in formal logic at all, but P2 and P3 really bother me. Don't they imply $L\rightarrow\neg L$? May 21, 2015 at 22:13
• but they in the questions are : P2) L → D P3) D → ¬L May 21, 2015 at 22:20
• Hmmm, well, like I said, I'm not an expert. Good luck! May 21, 2015 at 22:22
• @bob.sacamento That we a mistake from me (texing error) apologies for to you and the OP. May 21, 2015 at 22:24
• @user155971 to make up for that error, from P2 and p3 you can get $\neg L$ (as if $L$ is true, then so is $\neg L$, a contradiction). From P1 you get $E\lor(L\lor M)$ which is the same as $L\lor (E\lor M)$, and from this and $\neg L$ you get $(E\lor M)$. How that all fits in your proof system depends on the details of your proof system. May 21, 2015 at 22:26
We have the following deduction:
1) $L\rightarrow(\lnot L)$ by hypothetical syllogism and P2,P3.
2) $(\lnot L)\vee(\lnot L)$ by material implication and 1.
3) $\lnot L$ by disjunctive tautology and 2.
4) $E\vee (L\vee M)$ by conjunctive simplification and P1.
5) $(E\vee L)\vee M$ by disjunctive associativity and 4.
6) $(L\vee E)\vee M$ by disjunctive commutativity and 5.
7) $L\vee (E\vee M)$ by disjunctive associativity and 6.
8) $E\vee M$ by disjunctive syllogism and 7,3.
Conclude that the argument is valid.
• thank you so much , but can you recommended me with a good book explain logic May 21, 2015 at 23:00
• It depends on the level at which you'd like to study logic, the topics you want to study, and the ways you'd eventually like to use it. If you're just looking to do this stuff (it's called propositional calculus), then just about any text will do as long as it has plenty of exercises. Practice is the only way to get better at natural deduction... I'd bet you can find plenty of exercises through Google too. May 22, 2015 at 0:42
• $$(E \lor (L \lor M)) \land (E \iff F)\equiv(E \lor (L \lor M))\land (E\implies F\land F\implies E)$$ $$\equiv(E \lor L \lor M)\land\neg(\neg(E\implies F)\lor\neg(F\implies E))$$ $$\equiv(E \lor L \lor M)\land\neg((E\land\neg F)\lor(F\land\neg E))$$ $$\equiv(E \lor L \lor M)\land(\neg E\lor F)\land(E\lor \neg F)\equiv(E \lor L \lor M)\land(E\lor \neg F)\land(\neg E\lor F)$$ $$\equiv (E\lor((L\lor M)\land \neg F))\land (\neg E\lor F)$$$$\equiv(((E\lor((L\lor M)\land \neg F))\land \neg E)\lor(((E\lor((L\lor M)\land \neg F))\land F)$$ $$\equiv((L\lor M)\land\neg(E\lor F))\lor(E\land F)$$ Dec 10, 2019 at 16:25
• If it means anything Dec 10, 2019 at 16:29
Here is a proof using the proof checker from the forallx text. The OP in a comment is looking for recommended books explaining logic. This text with the associated truth functional and first order logic proof checker may supplement what the OP is currently using.
Here is the proof:
Here is a summary of the proof:
• Use conjunction elimination (∧E) to get the first conjunction from premise 1. We will not need the second conjunct from that premise.
• Since this conjunct is a disjunction consider both cases of the disjunction separately. If we can reach the desired conclusion in both cases then we can eliminate the disjunction.
• The first case $$E$$ is handled from lines 5 to 6. Notice the disjunction introduction (vI) on line 6.
• The second case $$L \vee M$$ is handled from lines 7 to 14. It requires handling another disjunction by considering both cases to reach the desired goal.
Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/ | 2022-08-10T20:22:47 | {
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https://bafonster.se/nithu/a94b91-cube-roots-calculator | # cube roots calculator
To use the calcualor simply type any positive or negative number into the text box and hit the 'calculate' button. Square Root Calculator Geek was created by a group of math proffesionals with a strong desire to create simple, quick solutions for teachers and students. There is a process that appears a bit laborious at … Yes, simply enter the fraction as a decimal floating point number and you will get the corresponding cube root. Cube Numbers List; Cube and Cube Root Calculator; All of Our Miniwebtools (Sorted by Name): Our PWA (Progressive Web App) Tools (17) {{title}} Financial Calcuators (121) … About Cube Root Calculator . In a shorter form “b” is the cube root of “a” if b^3 = a. Cubic roots are useful when one needs to divide an angle in three, that is - to find an angle whose measure is one third of a given angle. Remember the conclusions of the above observations. Method to calculate cube roots of definite Cube numbers. 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This cube root calculator might come in handy whenever you need to calculate the cube root of any given positive or negative numbers (including decimals). For Polynomials of degree less than or equal to 4, the exact value of any roots (zeros) of the polynomial are returned. The Cube Root Calculator will calculate the cube root of any number and all you have to do to calculate the cube root is to just enter in any number and press the calculate button! For example, to compute the cube root of 1/2 simply enter 0.5 in the input field and you will get 0.7937 as ouput. https://www.gigacalculator.com/calculators/cube-root-calculator.php. With the use of calculators, finding the cube root of any number may be just buttons away. Commonly used cube roots Table of commonly encountered cube roots: The Cube Root Calculator is used to calculate the cube root of a number. If x positive a will be positive, if x is negative a will be negative. The “b” cube root number can be found by raising the “a” number to the power of 1/3. This cube root calculator finds the cube root of any given number; which is the value that if gets multiplied by 3 equals the initial number. The cube root calculator below will reduce any cube root to its simplest radical form as well as provide a brute force rounded approximation for any number. Use this calculator to find the cube root of positive or negative numbers. A video introduction to cube roots from Khan Academy. See the table of common roots below for more examples. See our full terms of service. But perhaps you don't have a calculator, or you want to impress your friends with the ability to calculate a cube root by hand. 8, then cube root will have unit digit as 2. The calculator will show you the work and detailed explanation. It is the reverse of the exponentiation operation with an exponent of 3, so if r3 = x, then we say that "r is the cube root of x". 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The method for calculating cube roots is the same whether you're using the TI-84, TI-84 Plus or TI-84 Plus Silver. It provides free online calculators… The free calculator will solve any cube root. If you'd like to cite this online calculator resource and information as provided on the page, you can use the following citation: Georgiev G.Z., "Cube Root Calculator", [online] Available at: https://www.gigacalculator.com/calculators/cube-root-calculator.php URL [Accessed Date: 27 Nov, 2020]. The cube roots from 1 to 10 for positive and negative numbers: Copyright 2014 - 2020 The Calculator .CO | All Rights Reserved | Terms and Conditions of Use, Cube roots - positive and negative cube roots from 1 to 10. Although its versatility obligates you to a certain amount of hunting and pecking through menus for the more complex functions, locating the cube root function is as simple as two key presses. Unlike the square root, there is only one unique real number root as a result from applying the cube root function for a given number and it carries the sign of the number. The cube root is a specific calculation that any radical calculator can perform. The cube root of a number answers the question "what number can I multiply by itself twice to get this number?". This free root calculator determines the roots of numbers, including common roots such as a square root or a cubed root.
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# The length of a rectangular floor is 16 feet and its width is 12 feet.
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The length of a rectangular floor is 16 feet and its width is 12 feet. If each dimension were reduced by s feet to make the ratio of length to width 3 to 2, what would be the value
of s?
(A) 0
(B) 2
(C) 4
(D) 6
(E) 8
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Bunuel wrote:
The length of a rectangular floor is 16 feet and its width is 12 feet. If each dimension were reduced by s feet to make the ratio of length to width 3 to 2, what would be the value
of s?
(A) 0
(B) 2
(C) 4
(D) 6
(E) 8
New dimensions are length = (16-s) and width = (12-s)
Ratio = $$\frac{(16-s)}{(12-s)}$$ = $$\frac{3}{2}$$
32-2s=36-3s
s=4
Answer C
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The length of a rectangular floor is 16 feet and its width is 12 feet. [#permalink] 20 Sep 2017, 10:49
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# The length of a rectangular floor is 16 feet and its width is 12 feet.
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2018-01-23T13:30:29 | {
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https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_23&diff=132350&oldid=132349 | # Difference between revisions of "2016 AMC 8 Problems/Problem 23"
## Problem 23
Two congruent circles centered at points $A$ and $B$ each pass through the other circle's center. The line containing both $A$ and $B$ is extended to intersect the circles at points $C$ and $D$. The circles intersect at two points, one of which is $E$. What is the degree measure of $\angle CED$?
$\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150$
## Solution 1
Observe that $\triangle{EAB}$ is equilateral. Therefore, $m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}$. Since $CD$ is a straight line, we conclude that $m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}$. Since $BE=BD$ (both are radii of the same circle), $\triangle{BED}$ is isosceles, meaning that $m\angle{BED}=m\angle{BDE}=30^{\circ}$. Similarly, $m\angle{AEC}=m\angle{ACE}=30^{\circ}$.
Now, $\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}$. Therefore, the answer is $\boxed{\textbf{(C) }\ 120}$.
## Solution 2 (Trig)
Let $r$ be the radius of both circles (we are given that they are congruent). Let's drop the altitude from $E$ onto segment $AB$ and call the intersection point $F$. Notice that $F$ is the midpoint of $A$ and $B$, which means that $AF = BF = \frac{r}{2}$. Also notice that $\triangle{EAB}$ is equilateral, which means we can use the Pythagorean Theorem to get $EF = \frac{r\sqrt{3}}{2}$.
Now let's apply trigonometry. Let $\theta = \angle{CEF}$. We can see that $\tan\theta = \frac{CF}{EF} = \frac{\frac{3r}{2}}{\frac{r\sqrt{3}}{2}} = \sqrt{3}$. This means $m\angle{CEF} = \frac{\pi}{3}$. However, this is not the answer. The question is asking for $m\angle{CED}$. Notice that $\angle{CEF}\cong\angle{DEF}$, which means $m\angle{CED} = 2m\angle{CEF}$. Thus, $\angle{CED} = 2\cdot\frac{\pi}{3} = \frac{2\pi}{3} = 120^{\circ}$. Therefore, the answer is $\boxed{\textbf{(C) }\ 120}$.
## Video Solution
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http://lenzibroker.it/fourier-series-of-sine-wave.html | Fourier Series Of Sine Wave Fourier series usually include sine and cosine functions and can represent periodic functions in time or space or both. On this page, the Fourier Transforms for the sinusois sine and cosine function are determined. Fourier Series--Square Wave. The correct harmonic series is the one where the fundamental wavelength λ1 is equal to the period over which the waveform repeats. Finally, we show how these solutions lead to the theory of Fourier series. The image above depicts a sine wave. Inverse Fourier Transform maps the series of frequencies (their amplitudes and phases) back into the corresponding time series. We characterize which standing wave is set up on the string by an integer n = 1, 2, 3, 4,. Fourier series make use of the orthogonality relationships of the sine and cosine functions. x | is even and π -periodic; therefore f has a Fourier series of the form. where the frequencies and amplitudes have been normalized to unity for sim-plicity. That is why in signal processing, the Fourier analysis is applied in frequency (or spectrum) analysis. : U = 2/ ∫ 0 ∞ u x sin x dx, denoted as U = S[u] Inverse F. The Fourier-space (i. equation, wave equation and Laplace's equation arise in physical models. BACKGROUND IN FOURIER SERIES Jean Baptise Joseph Fourier (1768-1830) was the inventor of Fourier series in the late 1700s. Download Wolfram Player. In class we showed it can be represented as a Fourier series Úm=1 ¥B m sinmx where Bm= fl†† ° – †† •••4•••• pm modd 0meven. The Fourier series representation of analytic functions has been derived from. 1 Answer Ultrilliam May 21, 2018 See below. 1D Function: To select a function, you may press one of the following buttons: Sine. Fourier Series of Periodic Functions. An aperiodic signal cannot be represented using fourier series because the definition of fourier series is the summation of one or more (possibly infinite) sine wave to represent a periodicsignal. waveform in Table 15. We'll look at the cosine with frequency f=A cycles/second. Fourier Series Analysis { Fourier Series Analysis (C) 2005-2018 John F. Fourier series In the following chapters, we will look at methods for solving the PDEs described in Chapter 1. A cosine wave is also a sine wave but with 90* phase shift. In both extensions, a factor 2 arises due to the function is doubled in the period. This example is a square wave. A Fourier series can be defined as an expansion of a periodic function f(x) in terms of an infinite sum of sine functions and cosine functions. The primary reason that we use Fourier series is that we can better. Follow 25 views (last 30 days) Show older comments. A trigonometric polynomial is equal to its own fourier expansion. o The first five sine coefficients are calculated. Spectra of Sine Wave. 1D Fourier Transformation Java Applet. The fact that a square wave which is discontinuous can be "built" as a lin-ear combination of sinusoids at harmonically related frequencies is some-what astonishing. This sum is called a Fourier series Fundamental + 5 harmonics Fundamental + 20 harmonics x PERIOD = L Fundamental Fundamental + 2 harmonics Toc JJ II J I Back. representing a function with a series in the form Sum( B_n sin(n pi x / L) ) from n=1 to n=infinity. A cosine wave is also a sine wave but with 90* phase shift. Let us then generalize the Fourier series to complex functions. 5 Continuous Fourier Series. Derivation of Fourier series representation (coefficients) of a sinusoidal signal. The Fourier series associated with the rectified sine wave is therefore f(x) = 2 π − 4 π ∞ ∑ n = 2, 4, 6, ⋯ 1 n2 − 1cos(nx). Fourier series makes of the orthogonality relationships of the sine and cosine functions. 1:10; y = sin (t); plot (t,y); Next add the third harmonic to the fundamental, and plot it. FOURIER SERIES 7 On the other hand, the sum of the squares of the coe cients is: X1 n=1 2 n 2 = X1 n=1 4 n2: The formula is therefore telling us that X1 n=1 4 n2 = 2ˇ2 3 X1 n=1 1 n2 = ˇ2 6 This remarkable identity is actually correct, and was rst worked out by Euler. The function sin (x/2) twice as slow as sin (x) (i. Code Listing 1. A square wave that is odd about the origin requires only sine terms (see equation 4). A Fourier sine series F(x) is an odd 2T-periodic function. The square waveform and the one term (constant) expansion. Laurent Series yield Fourier Series. It can be thought of as an extension of the Fourier series , and can be used for non- periodic functions. , heat wave formula, sound wave formula, full wave rectifier. xT (t) = a0 + ∞ ∑ n=1(ancos(nω0t)+bnsin(nω0t)) x T ( t) = a 0 + ∑ n = 1 ∞ ( a n cos. Example 2: the Fourier series of a sawtooth wave: The sawtooth wave is a repetition of the function f(t) = x for −π < x < +π and the period is 2π. , cos (x ) = cos (–x ). such waveform can be represented in series form based on the original work of Jean Baptise Joseph Fourier. 3 ) provides a means to describe a complicated wave in terms of simple sines and cosines. Fourier sine series: sawtooth wave. (iii) odd harmonics (iv) even harmonics. Fourier series makes of the orthogonality relationships of the sine and cosine functions. The classic Fourier series as derived originally expressed a periodic signal (period. Plotting partial sums of Fourier sine series. Fourier series is used in the analysis of Periodic function e. The waveforms in these figures were generated using truncated, finite-term version(s) of the Fourier series expansion for this waveform: The first figure shows the bipolar triangle wave (labelled as "Waveform") overlaid with. The Fourier transform is a way for us to take the combined wave, and get each of the sine waves back out. The Fourier coefficients , a k and b k , express the real and imaginary parts respectively of the spectrum while the coefficients c k of the complex Fourier series express the spectrum as a magnitude and phase. According to the Fourier theorem, a steady-state wave is composed of a series of sinusoidal components whose frequencies are those of the fundamental and its harmonics, each component having the proper amplitude and phase. Fourier Series and Periodic Response to Periodic Forcing 5 2 Fourier Integrals in Maple The Fourier integrals for real valued functions (equations (6) and (7)) can be evaluated using symbolic math software, such as Maple or Mathematica. Three things are worth pointing out: the dirst term is the "DC" voltage. A good starting point for understanding the relevance of the Fourier series is to look up the math and analyze a square wave. Use of Fourier Series. This movie cleverly demonstrates what Fourier Series really gives us. The wave length is. The Fourier transform can be applied to continuous or discrete waves, in this chapter, we will only talk about the Discrete Fourier Transform (DFT). Waveforms with a steeper slope than this may be subject to a form of distortion, known as slew rate distortion, which will not affect a pure sine wave. The connection with the real-valued Fourier series is explained and formulae are given for converting be-tween the two types of representation. A1 and A2 respectively. ES 442 Fourier Transform 3 Review: Fourier Trignometric Series (for Periodic Waveforms) Agbo & Sadiku; Section 2. THE CONNECTION TO THE FOURIER SERIES IAN ALEVY Abstract. the signal and the sine wave or cosine wave at that frequency. By using this website, you agree to our Cookie Policy. 1 Orthogonal functions 1. For full credit, make sure you show all your. We discuss two partial di erential equations, the wave and heat equations, with applications to the study of physics. Math 331, Fall 2017, Lecture 2, (c) Victor Matveev. (2) as, S(t) 2To -To 0 To 2Tot ; Question: Fig. As signal of n data points thus there are n/2 sinusoidal curves in a Fourier series. Graphing a Fourier Series. Fourier series approximations to a square wave The square wave is the 2 p-periodic extension of the function fl† ° – †-1x£0 1x>0. A Fourier sine series F(x) is an odd 2T-periodic function. In mathematics, a Fourier series is a method for representing a function as the sum of simple sine waves. Note that function must be in the integrable. Let the integer m become a real number and let the coefficients, F m, become a function F(m). 1, and take the sine of all the points. The image above depicts a sine wave. The Fourier Transform and its kin operate by analyzing an input waveform into a series of sinusoidal waves of various frequencies and amplitudes. The actual conversion (real circuits) use Analog to. This problem has been solved! See the answer. The fourier series (or inverse fourier transform) of a complex-valued function $\small f$ of real variable, is given by a sum of complex harmonic sinusoids ($\small sine$ and $\small cosine$ waves in the complex plane). In sound: The Fourier theorem …is the spectral analysis, or Fourier analysis, of a steady-state wave. ) What is the Fourier transform of a 33 µs long square pulse? What is the Fourier transform of the 63 MHz sine wave of Question 1 which is turned on for only 33 µs?. • Fourier introduced the series for the purpose of solving the heat equation in a metal plate, publishing his initial results in his 1807 and publishing Analytical theory of heat in 1822. Chapter 3: The Frequency Domain Section 3. Determine the fourier series for the full-wave rectifier defined as f(t) = sinωt for 0 < ωt < pi-sinωt for -pi < ωt < 0 Homework Equations The Attempt at a Solution This looks like an even function, so bm = 0 Ao = 1/pi∫sinωt from 0 to pi = 1/pi(-cos(ωt))/ω) from 0 to pi = 2/piω. Fourier series is a mathematical function that is formed by the sum of scaled sine and cosine functions, over an interval. If we suppose that any piecewise-continuous function can be represented by a Superposition of sines and cosines, then we could find out why Square Waves use odd-integer harmonics. The cosine form is also called the Harmonic form Fourier series or Polar form Fourier series. First we derive the equa-tions from basic physical laws, then we show di erent methods of solutions. Continuous Fourier Transform F m vs. Fourier series of a constant function f(x)=1 converges to an odd periodic extension of this function, which is a square wave. The Fourier series spectrum of a half-wave rectified sinusoid is shown in the upper portion. Graphing a Fourier Series. So it is a series of sinntπ where each sine has a coefficient bn = 2 / (πn) if n = 1, 3, 5, …. Periodic-Continuous Here the examples include: sine waves, square waves, and any waveform that repeats itself in a regular pattern from negative to positive infinity. ( n 2 π x / λ − n 2 π f 1 t). • Functions (signals) can be completely reconstructed from the Fourier domain without loosing any. Fourier analysis grew from the study of Fourier series , and is named after Joseph Fourier , who showed that representing a function as a sum of trigonometric functions greatly simplifies the study of heat. The integral over one period of the product of any two terms have the following properties: Dr. Recently, in the Fourier Series chapter of “Coding Druid”, I practiced the visualization of Fourier Series and demonstrated the periodic square wave curve, which can be decomposed into a series of sine wave curves: Above is Python (Blender) version. Friday Math Movie - Sine Wave to Square Wave using Fourier Series. We are all familiar with sine waves (known for centuries) and Stokes waves (known since the Stokes paper in 1847). Title and author: Fourier Series with Sound. Define sK(t) to be the signal containing K+1 Fourier terms. THE CONNECTION TO THE FOURIER SERIES IAN ALEVY Abstract. A1 and A2 respectively. Fourier Series on a bar of length l: Let fand f0be piecewise continuous on the interval 0 x l. Since sound waves are made up of sine waves, Fourier transforms are widely used in signal processing. In this section, we prove that periodic analytic functions have such a. Fourier series use the orthogonality relationships of sine and cosine functions. A rectified sine wave is a periodic signal with a period equal to half of the full sinusoid, I would write the sine in exponential form and. The Fourier series ( Fig. For example the wave in Figure 1, is a sum of the three sine waves shown in Figure. 1 Answer Ultrilliam May 21, 2018 See below. So the Fourier series representation of a perfect sine wave is a perfect sine wave. Description: This java applet is a simulation that demonstrates Fourier series, which is a method of expressing an arbitrary periodic function as a sum of sine+cosine or just cosine terms. Long story short, what Fourier Transform is doing is it tries to approximate the signal (wave) of your interest using different kinds of sine and cosine waves. These Fourier series converge everywhere that the function itself is differentiable. 4 Term-by-Term Differentiation of Fourier Series. The theorem says that a Fourier series can only be integrated term by term and that the result is a convergent in-nite series which converges to the integral of f. Fourier analysis and Synthesis Background The French mathematician J. Fourier Series process, effects of harmonics and filtering of signals using a LabVIEW Virtual Instrument. 5: Find the exponential Fourier series for a rectified sine wave. The Fourier Transform for this type of signal is simply called the Fourier Transform. The time domain signal used in the Fourier series is periodic and continuous. Move the mouse over the white circles to see each term's contribution, in yellow. It was Fourier's discovery that any continuous repetitive wave could be built up by adding together waves from a harmonic series. An aperiodic signal cannot be represented using fourier series because the definition of fourier series is the summation of one or more (possibly infinite) sine wave to represent a periodicsignal. Derivation of Fourier series representation (coefficients) of a sinusoidal signal. Furthermore, if the function of choice is periodic, then the Fourier series can be used to. 2 Derive the Fourier series of: (a) a square wave, (b) a triangular wave, and (c) a half rectified sine wave (optional). (one period is T which is equal to 2PI) Looking at the figure it is clear that area bounded by the Square wave above and below t-axis are. In this section we define the Fourier Sine Series, i. Note: the sine wave is the same frequency as the square wave; we call this the 1 st (or fundamental) harmonic. A1 and A2 respectively. In the Fourier series corresponding to an even function, only cosine terms (and possibly a constant which we shall consider a cosine term) can be present. Most of the time all of the harmonics will be required and result in an infinite series called the Fourier series. We learned before that the most common tool in the study of the convergence of series of functions. Digitize low-frequency waves from the function generator, sine, triangle, and square. So this is the first function. The connection with the real-valued Fourier series is explained and formulae are given for converting be-tween the two types of representation. 4-1 and then do four things: 1. The graph of the function t 0 N=1024 & t=findgen(N) IDL> f=10*sin(2*!pi*t/32) + 20*randomn(seed,N) IDL> plot,f You can see that it is difficult to distinguish the sine wave from the noise. The width in the peak of the Fourier transform is a way of saying there is an uncertainty in the "true" value of the frequency. Within one period, the function is f(x) = ˆ 1; a=2 x<0 +1; 0 x @ 1 1 0 0 0 0 1 00 0 0 is defined in time interval of ( ) cos( 2 ) sin( 2 ). For example the wave in Figure 1, is a sum of the three sine waves shown in Figure. Fourier coefficients of half-wave. The Fourier transform of an image breaks down the image function (the undulating landscape) into a sum of constituent sine waves. 56 KB) by Nikesh Bajaj. The fourier Series makes use of the orthogonality relationships of the sine functions and cosine functions. That is why in signal processing, the Fourier analysis is applied in frequency (or spectrum) analysis. 2 CHAPTER 1. The Fourier series spectrum of a half-wave rectified sinusoid is shown in the upper portion. A Fourier series is a linear combination of sine and cosine functions, and it is designed to represent periodic functions: f(x) = a0 2 + ∞ ∑ n = 1ancos(nπx L) + ∞ ∑ n = 1bnsin(nπx L) The coefficients a0, a1, a2 an and b1, b2 bn are constants. Example of Rectangular Wave. Spectra of Sine Wave. The cumulative effect of adding terms to the Fourier series for the half-wave rectified sine wave is shown in the bottom portion. such waveform can be represented in series form based on the original work of Jean Baptise Joseph Fourier. Any periodic waveform can be decomposed into a series of sine and cosine waves: where a0, an, and bn are Fourier coefficients:,,. a 0 is the dc component of the signal and is given by. Then, on this interval, f(x) can be expanded in either a pure. For example, we can see that the series y(x,t) = X∞ n=1 sin nπx L An cos nπct L +Bn. As an example, let us find the exponential series for the following rectangular wave, given by. The graph of the function t 010; 2 f(x) = 8 <: x; 0 x 1 x+1; 1 0 other coefficients the even symmetry of the function is exploited to give. The Fourier Series is a shorthand mathematical description of a waveform. A pure sine wave can be converted into sound by a loudspeaker and will be perceived to be a steady, pure tone of a single pitch. spectrum, a power spectrum, and as a power spectral density. о 2; Answer: 2 cos 2t 2 2 cos 4tet 15 - cos 67t +. So by manipulating the Fourier series for the square wave function, we can arrive at a Fourier series for f of t, n odd, 1/n sine n*t. Fourier series approximations to a square wave The square wave is the 2 p-periodic extension of the function fl† ° – †-1x£0 1x>0. This movie cleverly demonstrates what Fourier Series really gives us. Fourier series approximations. Fourier Series LABVIEW rev6/28/2006 GUI Documentation. Here, a sine function is replaced with samples at 32 points. Fourier analysis of a periodic function refers to the extraction of the series of sines and cosines which when superimposed will reproduce the function. fourier transform of sine wave help. Let us consider the same problem from a different point of view, by expanding the delta function in a Fourier sine series. Now if we look at a Fourier series, the Fourier cosine series f(x) = a0 2 + ∞ ∑ n = 1ancosnπ L x describes an even function (why?), and the Fourier sine series f(x) = ∞ ∑ n = 1bnsinnπ L x an odd function. This Demonstration shows how a Fourier series of sine terms can approximate discontinuous periodic functions well, even with only a few terms in the series. Discrete Fourier Transform If we wish to find the frequency spectrum of a function that we have sampled, the continuous Fourier Transform is not so useful. So, if the amplitude of the swing is adequate, the ROC crossing zero is an excellent time to enter or exit a swing trade. 000000000000000 -4. Remark: If f is continuous on [0;1], then these two series also converge to f(x) at x= 0;1. It turns out that we have just the odd frequencies 1, 3, 5 in the square wave and they're multiplied by 4 over pi and they're divided by the frequency, so that's the decay. 1 Orthogonal functions 1. Using the formulae given above show that values of each of these spectra at 30 Hz are consistent with the original amplitude of 5. And conversely, any complex periodic signal can be broken down into a series of sinewave components for analysis (the time -> frequency task). , 90° or π/2 radians). In other words, Fourier series can be used to express a function in terms of the frequencies (harmonics) it is. Fourier Series: Half-wave Rectifier •Ex. This program calculates the DFS/DFT without using fft function of Matlab. A difficult thing to understand and/or motivate is the fact that arbitrary periodic functions have Fourier series representations. Since the time domain signal is periodic, the sine and cosine wave correlation only needs to be evaluated over a single period, i. Periodic-Continuous Here the examples include: sine waves, square waves, and any waveform that repeats itself in a regular pattern from negative to positive infinity. (i) sine terms (ii) cosine terms. These Fourier series converge everywhere that the function itself is differentiable. 2: Rectified Sine Wave {a k}) 1 (1 0 0 0) / 2 (0. 584 Chapter 9 Fourier Series Methods DEFINITION Fourier Series and Fourier Coefficients Let f(t) be a piecewise continuous function of period 2yr that is defined for all t. As promised in the first part of the Fourier series we will now demonstrate a simple example of constructing a periodic signal using the, none other then, Fourier series. With adjustment of the scales of the sine and cosine functions, it can be used to represent any function over the chosen interval. bk = { 4 πk if k is odd 0 if k is even. The Plancherel identity suggests that the Fourier transform is a one-to-one norm preserving map of the Hilbert space L2[1 ;1] onto itself (or to another copy of it-self). Are you trying to find the Fourier series representation or the Fourier transform of a periodic signal? In either case you don't need to deal with the absolute value. Guitars and pianos operate on two different solutions of the wave equation. Determine which aspect of a graph of a wave is described by each of the symbols lambda, T, k, omega, and n. Take the derivative of every term to produce cosines in the up-down delta function. Well, it's going to have the n in it, of course, but I want, now, to make the general period to be 2L. Fourier Series--Square Wave. Consider the sawtooth wave de ned on [ 1;1] by the function f(t) = t, and extended to be periodic of period T= 2. A square wave that is odd about the origin requires only sine terms (see equation 4). The complex Fourier series and the sine-cosine series are identical, each representing a signal's spectrum. The Fourier transform is a mathematical construct (algorithm ) that allows us to convert a signal such as a square or triangle waveform to constituent sinusoids. ODD Symmetry • An ODD function has the following characteristics: • With this properties, the Fourier coefficients for an odd function become gives us a Fourier sine series Half-Wave Symmetry • A function is half-wave (ODD) symmetric if which means that half-cycle is the mirror image of the next half-cycle. It establishes a relation between a function in the domain of time and a function in the domain of frequency. org/learn/differential-equations-engineersLecture notes at http:/. The Fourier series expansion of an even function $$f\left( x \right)$$ with the period of $$2\pi$$ does not involve the terms with sines and has the form: ${f\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {{a_n}\cos nx} ,}$ where the Fourier coefficients are given by the formulas \. Fourier Series - Sine Wave Synthesis. The application of Fourier-series method includes signal generators, power supplies, and communication circuits. • With an amplitude and a frequency • Basic spectral unit ---- How do we take a complex signal and describe its frequency mix? We can take any function of time and describe it as a sum of sine waves each with different amplitudes and frequencies. The amplitudes of the cosine waves are held in the variables: a1, a2, a3, a3, etc. Fourier series falls under the category of trigonometric infinite series, where the individual elements of the series are expressed trigonometrically. Consider the orthogonal system fsin nˇx T g1 n=1 on [ T;T]. Find the Fourier series of the following functions without computing any integrals. The theorem says that a Fourier series can only be integrated term by term and that the result is a convergent in-nite series which converges to the integral of f. Here A1=A2, so the average is zero. In calculating the final plot, subtract T 0 / 4 from τ. Imaginary part How much of a sine of that frequency you need Magnitude Amplitude of combined cosine and sine Phase Relative proportions of sine and cosine The Fourier Transform: Examples, Properties, Common Pairs Example: Fourier Transform of a Cosine f(t) = cos (2 st ) F (u ) = Z 1 1 f(t) e i2 ut dt = Z 1 1 cos (2 st ) e i2 ut dt = Z 1 1. This example is a square wave. , -space) function , respectively. According to the Fourier theorem, a steady-state wave is composed of a series of sinusoidal components whose frequencies are those of the fundamental and its harmonics, each component having the proper amplitude and phase. A Fourier sine series with coefficients fb ng1 n=1 is the expression F(x) = X1 n=1 b nsin nˇx T Theorem. Spectrum from Fourier Series Plot a for Full-Wave Rectified Sinusoid F 0 1 / T 0 d Z 0 2SF 0 ( 4 1) 2 2 k a k S a k. Here, a sine function is full-wave rectified, meaning that the wave becomes positive wherever it would be negative. We will use the Fourier sine series for representation of the nonhomogeneous solution to satisfy the boundary conditions. So, in these cases the Fourier sine series of an odd function on $$- L \le x \le L$$ is really just a special case of a Fourier series. Even a jump discontinuity does not pose a problem: if the function has left and right derivatives at x, then the Fourier series converges to the average of the left and right limits (but see Gibbs phenomenon). Assuming you’re unfamiliar with that, the Fourier Series is simply a long, intimidating function that breaks down any periodic function into a simple series of sine & cosine waves. 1, and take the sine of all the points. Let the integer m become a real number and let the coefficients, F m, become a function F(m). (iii) odd harmonics (iv) even harmonics. The fourier series (or inverse fourier transform) of a complex-valued function $\small f$ of real variable, is given by a sum of complex harmonic sinusoids ($\small sine$ and $\small cosine$ waves in the complex plane). Because of numerical errors from using integers, the sine waves can't have the exact "frequency" that you need for higher-order terms - so if you let the 11-term. Spectra of Sine Wave. Of course these all lead to different Fourier series, that represent the same function on [0,L]. Fourier Series: A Fourier series is a representation of a wave form or other periodic function as a sum of sines and cosines. See full list on mathsisfun. Fourier sine series: sawtooth wave. Waveforms with a steeper slope than this may be subject to a form of distortion, known as slew rate distortion, which will not affect a pure sine wave. Fourier Series Grapher. 3 Fourier Cosine and Sine Series. Solution The simplest way is to start with the sine series for the square wave: SW(x)= 4 π sinx 1 + sin3x 3 + sin5x 5 + sin7x 7 +···. See full list on en. The Fourier coefficient a, for the rectified sine wave y=sin (periodic function of 41 ) is 2 1 1 5 determined to be a = + }. The index indicates the multiple of the fundamental frequency at which the signal has energy. Let the integer m become a real number and let the coefficients, F m, become a function F(m). Find the Fourier series of the resulting periodic function: w w w p L L E t t L L t u t, 2, 2 sin 0 0 0. The fact that a square wave which is discontinuous can be "built" as a lin-ear combination of sinusoids at harmonically related frequencies is some-what astonishing. The period of the sine wave itself is T 2S cccccccc Z0 and there are n cycles of the sine wave in f[t], so it takes a time: ’t n 2S cccccccc Z0 for the wave to pass us. Use the modulation property of Fourier series to find the Fourier series coefficients of the full-wave rectified sine wave shown in the figure. This would be a Fourier series with only one term, and would return the desired function with the magnitude changed. The Fourier transform is a mathematical construct (algorithm ) that allows us to convert a signal such as a square or triangle waveform to constituent sinusoids. Are you trying to find the Fourier series representation or the Fourier transform of a periodic signal? In either case you don't need to deal with the absolute value. In the processing of audio signals (although it can be used for radio waves, light waves, seismic waves, and even images), Fourier analysis can isolate individual components of a continuous complex waveform, and concentrate. The Fourier series ( Fig. The computation and study of Fourier series is known as harmonic. The cosine form is also called the Harmonic form Fourier series or Polar form Fourier series. Depending on which boundary conditions apply, either the position or the lateral velocity of the string is modelled by a Fourier series. It’s a baffling concept to wrap your mind around, but almost any function can be expressed as a series of sine & cosine waves created from rotating circles. • Fourier Cosine Series This is a half-range series consisting solely of cosines. In the Fourier series corresponding to an even function, only cosine terms (and possibly a constant which we shall consider a cosine term) can be present. only a few of the coefficients of the Fourier series included. The Fourier series for the square wave does not converge at t = 0, T /2, T. A sinusoidal voltage Esinwt, is passed through a half-wave rectifier that clips the negative portion of the wave. Fourier series decomposes non-sinusoidal waveform into series of sinusoidal components of various frequencies. nonsinusoidal periodic waveform can be broken down into a sine or cosine wave equal to the frequency ofthe periodic waveform, called the fundamental frequency, and a series of sine or cosine waves that are integer mUltiples of the fundamental frequency, called harmonics. In sound: The Fourier theorem …is the spectral analysis, or Fourier analysis, of a steady-state wave. This cosine function can be rewritten, thanks to Euler, using the identity:. The function sin (x/2) twice as slow as sin (x) (i. The Fourier series synthesis equation creates a continuous periodic signal with a fundamental frequency, f, by adding scaled cosine and sine waves with frequencies: f, 2 f, 3 f, 4 f, etc. Fourier analysis of a periodic function refers to the extraction of the series of sines and cosines which when superimposed will reproduce the function. m m Again, we really need two such plots, one for the cosine series and another for the sine series. 1:10; y = sin (t); plot (t,y); Next add the third harmonic to the fundamental, and plot it. There are JavaScript (React) version, and Unity version. such waveform can be represented in series form based on the original work of Jean Baptise Joseph Fourier. The expression for the Fourier coefficients has the form. , express the real and imaginary parts respectively of the spectrum. The Fourier transform is a machine (algorithm). In this section, we prove that periodic analytic functions have such a. So the first term in the Fourier series is a constant, and it is the average value of the function. Find the Fourier series of the resulting periodic function: w w w p L L E t t L L t u t, 2, 2 sin 0 0 0. An approximation of a complicated wave can be achieved by adding together very simple sine and cosine waves with varying combinations of frequencies and amplitudes. We will instead use the re ection method:. A Fourier sine series with coefficients fb ng1 n=1 is the expression F(x) = X1 n=1 b nsin nˇx T Theorem. A rectified sine wave is a periodic signal with a period equal to half of the full sinusoid, I would write the sine in exponential form and. 000000000000000 -4. In this example, you can almost do it in your head, just by looking at the original. The width in the peak of the Fourier transform is a way of saying there is an uncertainty in the "true" value of the frequency. In calculating the final plot, subtract T 0 / 4 from τ. Discrete Fourier Series without using fft function. ABSTRACT-Fourier series is a mathematical expression of sine and cosine waves which can be used to describe completely the outline of a miospore. The usefulness of even and odd Fourier series is related to the imposition of boundary conditions. In this section, we'll try to really explain the notion of a Fourier expansion by building on the ideas of phasors, partials, and sinusoidal components that we introduced in the previous section. Plot this fundamental frequency. The graph of the function t 00 be a xed number and f(x) be a periodic function with period 2p, de ned on ( p;p). Either way the maximum slope of the waveform is 1. In this video we see that a square wave may be defined as the sum of an infinite number of sinusoids. Assume that the peak amplitude of every wave is A volts. Calculation of Fourier Series Coefficients: A. The values of these coefficients determine the function that will be reconstructed. Over the range , this can be written as. о 2; Answer: 2 cos 2t 2 2 cos 4tet 15 - cos 67t +. Fourier and a number of his contemporaries were interested in the study of vibrating strings. In this section, we'll try to really explain the notion of a Fourier expansion by building on the ideas of phasors, partials, and sinusoidal components that we introduced in the previous section. A sin (2 p ft) where A is the AMPLITUDE of the wave, f its FREQUENCY , and t is time. A sawtooth wave represented by a successively larger sum of trigonometric terms. The periodic waveforms, viz: rectangular wave, triangular wave, sine wave, etc. Plotting partial sums of Fourier sine series. A cosine wave is just a sine wave shifted in phase by 90 o (φ. Therefore,!1 = 2ˇ T1 = ˇ T =!o 2. • Fourier Series decomposes periodicwaveforms into an infinite sum of weighted cosine and sine functions – We can look at waveforms either in ‘time’ or ‘frequency’ – Useful tool: even and odd functions • Some issues we will deal with next time – Fourier Series definition covered today is not very compact. Recently, in the Fourier Series chapter of “Coding Druid”, I practiced the visualization of Fourier Series and demonstrated the periodic square wave curve, which can be decomposed into a series of sine wave curves: Above is Python (Blender) version. Complex Fourier Series The complex Fourier series is presented first with pe-riod 2π, then with general period. In fact, as we add terms in the Fourier series representa-. The primary reason that we use Fourier series is that we can better. Fourier Series of Even and Odd Functions. Laplace's Equation (The Potential Equation): @2u @x 2 + @2u @y on 0 x las either a pure Sine Series or a pure Cosine Series. The cumulative effect of adding terms to the Fourier series for the half-wave rectified sine wave is shown in the bottom portion. Selecting different limits makes the. In the fourier Series the constant term a 0 will not appear if the signal wave average value in one period is zero. 1 Fourier Sine Series. 2n, because the FFT program works much more efficiently on such a sample. A1 and A2 respectively. Expression (1. The Fourier series is a description of a waveform such as a square or triangle wave. Use the modulation property of Fourier series to find the Fourier series coefficients of the full-wave rectified sine wave shown in the figure. ) What is the Fourier transform of a 33 µs long square pulse? What is the Fourier transform of the 63 MHz sine wave of Question 1 which is turned on for only 33 µs?. Use the sliders to set the number of terms to a power of 2 and to set the frequency of the wave. 6 Complex Form of Fourier Series. 1 in the textbook). Sine curves and Fourier transform case-study Page 7 23/11/2007, 12:10 PM. Derivative numerical and analytical calculator. The Fourier coefficient a, for the rectified sine wave y=sin (periodic function of 41 ) is 2 1 1 5 determined to be a = + }. A Fourier series is a way to represent a wave-like function (like a square wave) as the sum of simple sine waves. Fourier analysis grew from the study of Fourier series , and is named after Joseph Fourier , who showed that representing a function as a sum of trigonometric functions greatly simplifies the study of heat. • Consider, for example, a triangular waveform. The Fourier-space (i. Fourier series is just a means to represent a periodic signal as an infinite sum of sine wave components. Using the Fourier expansion, the frozen surface can be represented as an infinite series of sine and cosine functions of different wave numbers oriented in all possible directions. A rectified sine wave is a periodic signal with a period equal to half of the full sinusoid, I would write the sine in exponential form and. The complex Fourier series and the sine-cosine series are identical, each representing a signal's spectrum. Move the mouse over the white circles to see each term's contribution, in yellow. A Fourier series is a way to represent a wave-like function (like a square wave) as the sum of simple sine waves. Coefficient Found in HW5 problem 2b:. The Fourier transform of a function is complex, with the magnitude representing the amount of a given frequency and the argument representing the phase shift from a sine wave of that frequency. It decomposes the function into sum of sine-cosine functions. 7), we see that the Fourier Series of square wave consists of sine terms only. Fourier series use the orthogonality relationships of sine and cosine functions. Laplace's Equation (The Potential Equation): @2u @x 2 + @2u @y on 0 x las either a pure Sine Series or a pure Cosine Series. 2 Fourier series of functions with arbitrary period 1. Fourier analysis grew from the study of Fourier series , and is named after Joseph Fourier , who showed that representing a function as a sum of trigonometric functions greatly simplifies the study of heat. (1) Find the trigonometric Fourier series representation for the full wave rectified sine wave shown in Fig. Fourier series falls under the category of trigonometric infinite series, where the individual elements of the series are expressed trigonometrically. ( n 2 π x / λ − n 2 π f 1 t). The waveforms in these figures were generated using truncated, finite-term version(s) of the Fourier series expansion for this waveform: The first figure shows the bipolar triangle wave (labelled as "Waveform") overlaid with. Starting with the complex Fourier series, i. : U = 2/ ∫ 0 ∞ u x sin x dx, denoted as U = S[u] Inverse F. , -space) function , respectively. xT (t) = a0 + ∞ ∑ n=1(ancos(nω0t)+bnsin(nω0t)) x T ( t) = a 0 + ∑ n = 1 ∞ ( a n cos. representing a function with a series in the form Sum( B_n sin(n pi x / L) ) from n=1 to n=infinity. It is interesting to consider the sequence of signals that we obtain as we incorporate more terms into the Fourier series approximation of the half-wave rectified sine wave. Another Fourier series recipe for a triangle wave is also all of the odd harmonics. The above plot shows that over one period, the cosine curve is symmetric, while the sine is anti-symmetric, i. ABSTRACT-Fourier series is a mathematical expression of sine and cosine waves which can be used to describe completely the outline of a miospore. The wave length is. The Fourier series for the square wave does not converge at t = 0, T /2, T. Fourier Transform. Chapter 3: The Frequency Domain Section 3. Start by forming a time vector running from 0 to 10 in steps of 0. Fourier series is used in the analysis of Periodic function e. A cosine wave is just a sine wave shifted in phase by 90 o (φ. Fourier cosine series of a triangle wave function. Friday Math Movie - Sine Wave to Square Wave using Fourier Series. Introduction As the figure above shows, it is possible to use combinations of simple sine waves to approximate other waves that don’t seem to have anything to do with sines or other trig functions. Change the script so that it computes and plots the trig. or spectral phase of the Fourier series. Fourier Series of Even and Odd Functions. Furthermore, if the function of choice is periodic, then the Fourier series can be used to. A rectified sine wave is a periodic signal with a period equal to half of the full sinusoid, I would write the sine in exponential form and. 10 Fourier Series and Transforms (2014-5379) Fourier Series: 2 - 4 / 11 Why are engineers obsessed with sine waves? Answer: Because 1. In this example, you can almost do it in your head, just by looking at the original. A Fourier series is a way to represent a wave-like function (like a square wave) as the sum of simple sine waves. The construct of the Fourier series is given by. Unlike quantization , the function can take any value at those points; but the function is flat between these samples. org/learn/differential-equations-engineersLecture notes at http:/. Fourier series approximations. HomeworkQuestion I have to show the input sine wave and output square wave but for some reason my output square wave is shifted. Discrete Fourier Series vs. This series of sine or cosine waves is called a Fourier series. 3 Fourier series of odd and even functions 1. Fourier series. A periodic function f (t ) is said to have a quarter wave symmetry, if it possesses. 32 points per cycle corresponds to a sampling frequency of 220 × 32 = 7040 Hz (at the default playing frequency of 220 Hz). Contributed by: David von Seggern (University Nevada-Reno) (March 2011). Fourier series of non-periodic discrete-time signals In analogy with the continuous-time case a non-periodic discrete-time signal consists of a continuum of frequencies (rather than a discrete set of frequencies) But recall that cos(n!) = cos(n! +2…nl) = cos(n(! +2…l)); all integers l =) Only frequencies up to 2… make sense 21. Spectrum from Fourier Series Plot a for Full-Wave Rectified Sinusoid F 0 1 / T 0 d Z 0 2SF 0 ( 4 1) 2 2 k a k S a k. The cosine form is also called the Harmonic form Fourier series or Polar form Fourier series. Fourier series of a constant function f(x)=1 converges to an odd periodic extension of this function, which is a square wave. Fourier Analysis for Periodic Functions. The values of a, a, and a, are [3] TT 1 1 - +n 2 -n 8. The construct of the Fourier series is given by. A Fourier cosine series has df∕dx = 0 at x = 0, and the Fourier sine series has f(x = 0) = 0. Bravely applying the usual formulae for the Fourier sine series to the delta function, we find that for 0 x 1, In other words, if the function f(x) above happens to be the delta function, then b n = 2 sin(n t). An aperiodic signal cannot be represented using fourier series because the definition of fourier series is the summation of one or more (possibly infinite) sine wave to represent a periodicsignal. #Fourier Series Coefficients #The following function returns the fourier coefficients,'a0/2', 'An' & 'Bn' # #User needs to provide the following arguments: # #l=periodicity of the function f which is to be approximated by Fourier Series #n=no. Digitize low-frequency waves from the function generator, sine, triangle, and square. Plotting partial sums of Fourier sine series. For example the wave in Figure 1, is a sum of the three sine waves shown in Figure. So that's 1 sine x, 0, sine 2x's then 4 over pi sine 3x's, but with this guy there's a 3, 0 sine 4x's, sine 5x comes in over 5, and so on. Author name; Kyle Forinash; Wolfgang Christian. In the processing of audio signals (although it can be used for radio waves, light waves, seismic waves, and even images), Fourier analysis can isolate individual components of a continuous complex waveform, and concentrate. Let the integer m become a real number and let the coefficients, F m, become a function F(m). The coefficients fb ng1 n=1 in a Fourier sine series F(x) are determined by. BACKGROUND IN FOURIER SERIES Jean Baptise Joseph Fourier (1768-1830) was the inventor of Fourier series in the late 1700s. Expression (1. (ii) The Fourier series of an odd function on the interval (p, p) is the sine series (4) where (5) EXAMPLE 1 Expansion in a Sine Series Expand f(x) x, 2 x 2 in a Fourier series. The expression for the Fourier coefficients has the form. This program calculates the DFS/DFT without using fft function of Matlab. We characterize which standing wave is set up on the string by an integer n = 1, 2, 3, 4,. The image above depicts a sine wave. Now we can explore the reverse process, that is, the analysis of an arbitrary wave form to discover the presence and amplitude of its constituent harmonics or the Fourier coefficients of the waveform as they are the coefficients within the Fourier series equation. 2 Derive the Fourier series of: (a) a square wave, (b) a triangular wave, and (c) a half rectified sine wave (optional). HALF RANGE FOURIER SINE OR COSINE SERIES A half range Fourier sine or cosine series is a series in. The idea behind the Fourier Series is to add sine curves with different amplitudes and frequencies and the resulting curve can be either a square wave, a sawtooth wave or many other interesting periodic shapes. The series is finite just like how the taylor expansion of a polynomial is itself (and hence finite). 2: Rectified Sine Wave {a k}) 1 (1 0 0 0) / 2 (0. Are you trying to find the Fourier series representation or the Fourier transform of a periodic signal? In either case you don't need to deal with the absolute value. Now a pure sine wave has no Fourier series or you can consider the series to have only one term. † The Fourier series is then f(t) = A 2 ¡ 4A …2 X1 n=1 1 (2n¡1)2 cos 2(2n¡1)…t T: Note that the upper limit of the series is 1. Coefficient Found in HW5 problem 2b:. For example, later we will Example of a Fourier Series - Square Wave Sketch the function for 3 cycles: f(t) = f(t + 8). Find the Fourier series of the resulting periodic function: w w w p L L E t t L L t u t, 2, 2 sin 0 0 0. Fourier series make use of the orthogonality relationships of the sine and cosine functions. It helps us think about electric circuits. For the number of samples always use a number that is a power of 2, i. The steps involved are as shown below. Here's an example wave: This wavy pattern here can be split up into sine waves. Using the Fourier expansion, the frozen surface can be represented as an infinite series of sine and cosine functions of different wave numbers oriented in all possible directions. Define sK(t) to be the signal containing K+1 Fourier terms. A sine wave remains a sine wave of the same frequency when you (a) multiply by a constant, (b) add onto to another sine wave of the same frequency, (c) differentiate or integrate or shift in time 2. The fourier series (or inverse fourier transform) of a complex-valued function $\small f$ of real variable, is given by a sum of complex harmonic sinusoids ($\small sine$ and $\small cosine$ waves in the complex plane). where the coefficients are given by:. Convolution theorem. Furthermore, if the function of choice is periodic, then the Fourier series can be used to. Fourier Series plot. Here is the simple online Fourier series calculator to do Fourier series calculations in simple. Note that it does not say it will be a Fourier series. While solving the Fourier transformation of a sine wave (say h ( t) = A sin. Square Wave. and (3) displays the Fourier transform on the monitor. Here's an example wave: This wavy pattern here can be split up into sine waves. Before getting into the details of Fourier series, it may help to briefly review the terms associated with a sine wave with the figure below. You can see that after rectification, the fundamental frequency is eliminated, and all the even harmonics are present. The next animation shown how the first few terms in the Fourier series approximates the periodic square wave function. Guitars and pianos operate on two different solutions of the wave equation. There are many known sufficient conditions for the Fourier series of a function to converge at a given point x, for example if the function is differentiable at x. If we unfreeze the surface and let it evolve in time, we can represent the sea surface as an infinite series of sine and cosine functions of different wave-lengths. edu is a platform for academics to share research papers. A Fourier cosine series has df∕dx = 0 at x = 0, and the Fourier sine series has f(x = 0) = 0. This is the shape of string and air vibration at a pure frequency; I also made images of the Fourier series of a square wave and a triangle wave for the curious. Fourier series approximations to a square wave The square wave is the 2 p-periodic extension of the function fl† ° – †-1x£0 1x>0. Fourier series and transform. What you learned about series in calculus class shows that the series. 1 A Historical Perspective By 1807, Fourier had completed a work that series of harmonically related sinusoids were useful. A sinusoidal wave or function, that is, one moving in SIMPLE HARMONIC MOTION according to the function. (one period is T which is equal to 2PI) Looking at the figure it is clear that area bounded by the Square wave above and below t-axis are. Join me on Coursera: https://www. Matlab code for converting a sine wave into a square wave using Fourier Series. The values of these coefficients determine the function that will be reconstructed. A pure sine or cosine wave that has an exactly integral number of cycles within the recorded signal will have a single non-zero Fourier component corresponding to its frequency. Once you have a decent sine wave generator, the natural next step is to add a bunch together as a Fourier series and approximate common waveforms like the triangle wave or square wave. Build and use a Fourier Series analyzer. We thereby multiply our signal (target function) with an analyzing function (which contains all sine waves). Over the range , this can be written as (1) where is the Heaviside step function. The function x ↦ f ( x) := | sin. SOLUTION Inspection of Figure 11. Use the modulation property of Fourier series to find the Fourier series coefficients of the full-wave rectified sine wave shown in the figure. Let the integer m become a real number and let the coefficients, F m, become a function F(m). The Fourier series spectrum of a half-wave rectified sinusoid is shown in the upper portion. • The Fourier series is named in honour of Jean-Baptiste Joseph Fourier (1768-1830), who made important contributions to the study of trigonometric series. • Functions (signals) can be completely reconstructed from the Fourier domain without loosing any. (3): f(t) = a 0 2 + X1 n=1 [a ncos(nt) + b nsin(nt)] = a 0 2 + X1 n=1 a n eint+. 1) is called the inverse Fourier integral for f. The method of obtaining the Fourier series for a single miospore is presented and an illustration is included to show the application of Fourier series to the solution of palynologic problems. SOLUTION Inspection of Figure 11. Fourier series falls under the category of trigonometric infinite series, where the individual elements of the series are expressed trigonometrically. The Fourier Series GUI is meant to be used as a learning tool to better understand the Fourier Series. The Fourier series expansion for a square-wave is made up of a sum of odd harmonics, as shown here using MATLAB®. Over the range , this can be written as. Using the results of Example 3 on the page Definition of Fourier Series and Typical Examples, we can write the right side of the equation as the series \[{3x }={ \frac{6}{\pi }\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{n}\sin n\pi x}. With this. The Fourier series of the above sawtooth wave is. The next animation shown how the first few terms in the Fourier series approximates the periodic square wave function. This analysis can be expressed as a Fourier series. In Fourier Analysis we represent the complex wave shape as a sum of sine waves (or a sum of "partials"), each of a different amplitude. Fourier analysis, illustrating that any complex wave form can be shown to consist of a series of individual sine waves. Fourier Series for Rectified Sine Wave Consider the signal x(t) = Ajsin(!1 t)j −2 T −T 0 T 2 T −A 0 A |sin (ω 1 t)| Rectified Sine and Sine −T1 0 T1 −A 0 A sin (ω 1 t) The period of the sinusoid (inside the absolute value symbols) is T1 = 2ˇ=!1. Fourier Series. OVERVIEW OF FOURIER SERIES In electronics, Fourier series is used to approximate a periodic waveform, in which amplitude verses time characteristic is repeated in a period, T. 10 Fourier Series and Transforms (2014-5509) Sums and Averages: 1 - 2 / 14 Main fact: Complicated time waveforms can be expressed as a sum of sine and cosine waves. The trigonometric Fourier series representation of a periodic signal x (t) with fundamental period T, is given by. The values of these coefficients determine the function that will be reconstructed. Long story short, what Fourier Transform is doing is it tries to approximate the signal (wave) of your interest using different kinds of sine and cosine waves. From the Fourier series to the Fourier integral. The sine wave is important in physics because it retains its wave shape when added to another sine wave of the same Frequency and arbitrary phase and magnitude. This cosine function can be rewritten, thanks to Euler, using the identity:. Fourier series for a function. We can use Fourier Series to investigate. I've tried to learn about Fourier synthesis from many sources, but they all talk about the Fourier series instead of the Fourier transform, and they all say that for a pure wave all you need is a value in the coefficient of that wave's frequency, and 0 everywhere else. In the simple case of just one naturally vibrating string the analysis is straightforward: the vibration is described by a sine wave. The index indicates the multiple of the fundamental frequency at which the signal has energy. A Fourier series is a way to represent a wave-like function (like a square wave) as the sum of simple sine waves. The fourier Series makes use of the orthogonality relationships of the sine functions and cosine functions. The complex Fourier series and the sine-cosine series are identical, each representing a signal's spectrum. 000000000000002 Note that a contains the constant term in the series as its first coefficient followed by the coefficients for $\cos(x)$ and $\cos(2x)$, while b starts. A Fourier series is a way of representing a periodic function as a (possibly infinite) sum of sine and cosine functions. 2 Fourier series 1. 2 Fourier series of functions with arbitrary period 1. The cumulative effect of adding terms to the Fourier series for the half-wave rectified sine wave is shown in the bottom portion. Fourier Series Calculator. Assuming you’re unfamiliar with that, the Fourier Series is simply a long, intimidating function that breaks down any periodic function into a simple series of sine & cosine waves. The cosine form is also called the Harmonic form Fourier series or Polar form Fourier series. In other words, Fourier series can be used to express a function in terms of the frequencies (harmonics) it is. The Fourier series is a description of a waveform such as a square or triangle wave. If the wave shape is periodic, the frequencies of the partials are multiples of the fundamental frequency and are called the “harmonics” of the tone being played. • Fourier Cosine Series This is a half-range series consisting solely of cosines. Since a sine wave can be expressed as a cosine wave with a phase shift (or vice versa). 3 Complex form of Fourier series Chapter One. So the Fourier series representation of a perfect sine wave is a perfect sine wave. Aljanaby 18 Example: Find the average power supplied to a network if the applied voltage and resulting current are Sol: The total average power is the sum of the harmonic powers: Example: Find the trigonometric Fourier series for the half-wave-rectified sine. 2 Introduction In this Section we show how a periodic function can be expressed as a series of sines and cosines. Now we can explore the reverse process, that is, the analysis of an arbitrary wave form to discover the presence and amplitude of its constituent harmonics or the Fourier coefficients of the waveform as they are the coefficients within the Fourier series equation. A trigonometric polynomial is equal to its own fourier expansion. It turns out that we have just the odd frequencies 1, 3, 5 in the square wave and they're multiplied by 4 over pi and they're divided by the frequency, so that's the decay. Laval (KSU) Fourier Series Today 10 / 12. The construct of the Fourier series is given by. The index indicates the multiple of the fundamental frequency at which the signal has energy. Here f (x) is the complex periodic function we wish to break down in terms of sine and cosine basis functions. Each periodic function f that is adjusted, can be written as a infinite sum of sine and cosine terms. Besides the sine wave, the sawtooth wave, square wave, triangular wave and trapezoidal wave are common waveforms as well. Spectrum from Fourier Series Plot a for Full-Wave Rectified Sinusoid F 0 1 / T 0 d Z 0 2SF 0 ( 4 1) 2 2 k a k S a k. 1, and take the sine of all the points. , slow) as ΠT(t). Let the integer m become a real number and let the coefficients, F m, become a function F(m). This demonstration shows the sum of up to 5 harmonics of a sine wave. An aperiodic signal cannot be represented using fourier series because the definition of fourier series is the summation of one or more (possibly infinite) sine wave to represent a periodicsignal. Fourier Sine Series Definition. Fourier series and orthogonal functions 1. Theorem 1: Convergence of Fourier sine and cosine series If f is piecewise smooth on closed interval [0;1], and continuous on (0;1), then the Fourier sine and cosine series converge for all xin [0;1], and has sum f(x) in (0;1). Musical Application Sound waves are one type of waves that can be analyzed using Fourier series, allowing for different aspects of music to be analyzed using this method. In this section, we prove that periodic analytic functions have such a. Continuous Fourier Transform F m vs. The Fourier series expansion for a square-wave is made up of a sum of odd harmonics, as shown here using MATLAB®. Fourier series of the signal created in x. There are many known sufficient conditions for the Fourier series of a function to converge at a given point x, for example if the function is differentiable at x. The big advantage that Fourier series have over Taylor series is that the function f(x) can have discontinuities. The coefficients fb ng1 n=1 in a Fourier sine series F(x) are determined by. Here's an example wave: This wavy pattern here can be split up into sine waves. The DTFS properties used include multiplication, time shifting, linearity, and frequency shifting. The classic Fourier series as derived originally expressed a periodic signal (period. This version of the Fourier transform is called the Fourier Series. Here A1=A2, so the average is zero. Lec1: Fourier Series Associated Prof Dr. The Fourier series expansion for a square-wave is made up of a sum of odd harmonics, as shown here using MATLAB®. • Fourier Series: Represent any periodic function as a weighted combination of sine and cosines of different frequencies. ( 2 π f 0 t)) in time domain, we get two peaks in frequency domain in frequency space with a factor of ( A / 2) j with algebraic sum of delta function for f + f 0 and f − f 0 frequency, where j is the imaginary unit. Fourier Series process, effects of harmonics and filtering of signals using a LabVIEW Virtual Instrument. You can see more on this concept in this Introduction to Fourier Series. Example 1. 1 Fourier Sine Series. 5 Fourier Series Derivation The analysis formula1 for the Fourier Series coefficients (3. Fourier Transform. The Fourier transform of a function is complex, with the magnitude representing the amount of a given frequency and the argument representing the phase shift from a sine wave of that frequency. representing a function with a series in the form Sum( B_n sin(n pi x / L) ) from n=1 to n=infinity. | 2021-08-02T23:01:44 | {
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https://math.stackexchange.com/questions/2643617/evaluation-of-int-0-infty-frac-sinxxe-x%C2%B2-dx | # Evaluation of $\int_{0}^\infty \frac{\sin(x)}{x}e^{- x²} dx$
I have tried to evaluate the integral $$\int_{0}^\infty \frac{\sin(x)}{x}e^{- x^2} dx.$$ I used integration by parts but I did not succeed. Wolfram Alpha says that is convergent and it is equal to: $\frac \pi 2 \text{erf}({\frac 12})$ .
Is there any simple way for evaluate it?
• Do it via the convolution theorem for the Fourier transform. – Cameron Williams Feb 9 '18 at 18:37
• A simplified version of Cameron Williams’ suggestion is that you can consider $$\frac{\sin x}{x}=\frac{1}{2}\int_{-1}^{1}e^{ixt}\,dt$$ and then change the order of integration. – Sangchul Lee Feb 9 '18 at 18:46
We have that \begin{align}\int_{0}^\infty \frac{\sin(x)}{x}e^{-x^2} dx &= \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}\int_{0}^\infty x^{2k}e^{-x^2} dx\\ &= \frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}\int_{0}^\infty t^{k-1/2}e^{-t} dt\\ &=\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k\Gamma(k+1/2)}{(2k+1)!} \\ &=\sqrt{\pi}\sum_{k=0}^{\infty}\frac{(-1)^k\cdot(1/2)^{2k+1}}{(2k+1)k!}= \frac{\pi}{2}\text{erf}(1/2).\end{align} For the last step see the Taylor series of erf.
Here is a slightly longer method (compared to @Robert Z's slick series solution approach) that uses Feynman's trick of differentiating under the integral sign.
Let $$I(a) = \int_0^\infty \frac{\sin (ax)}{x} e^{-x^2} \, dx, \quad a > 0.$$ Note that $I(0) = 0$ and we are required to find $I(1)$.
Differentiating $I(a)$ with respect to the parameter $a$ gives $$I'(a) = \int_0^\infty \cos (ax) e^{-x^2} \, dx. \tag1$$
On integrating (1) by parts leads to $$I'(a) = \frac{2}{a} \int_0^\infty x e^{-x^2} \sin (ax) \, dx.$$
Also, differentiating (1) again with respect to the parameter $a$ yields $$I''(a) = -\int_0^\infty x e^{-x^2} \sin (ax) \, dx = -\frac{a}{2} I'(a).$$ If we set $u(a) = I'(a)$ the above second-order differential equation can be reduced to the following first-order differential equation $$u'(a) = -\frac{a}{2} u(a).$$ Solving yields $$u(a) = I'(a) = K e^{-a^2/4}, \tag1$$ where $K$ is a constant to be determined. To find this constant setting $a = 0$ in $I'(a)$ leads to $$I'(0) = \int_0^\infty e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2} \cdot \frac{2}{\sqrt{\pi}} \int_0^\infty e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2} \cdot \text{erf} (\infty) = \frac{\sqrt{\pi}}{2}.$$ So on setting $a= 0$ in (1) we find $K = \sqrt{\pi}/2$. Thus $$I'(a) = \frac{\sqrt{\pi}}{2} e^{-a^2/4}.$$
Now as $I(0) = 0$, we have $$I(1) = \int_0^1 I'(a) \, da = \frac{\sqrt{\pi}}{2} \int_0^1 e^{-a^2/4} \, da.$$ Enforcing a substitution of $a \mapsto 2a$ leads to $$I(1) = \sqrt{\pi} \int_0^{1/2} e^{-a^2} \, da = \frac{\pi}{2} \cdot \frac{2}{\sqrt{\pi}} \int_0^{1/2} e^{-a^2} \, da.$$ And since from the integral representation for the error function which is given by $$\text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} \, dt,$$ one has $$\int_0^\infty \frac{\sin x}{x} e^{-x^2} \, dx = \frac{\pi}{2} \text{erf} \left (\frac{1}{2} \right ),$$ as expected.
Following Sangchul Lee's hint. \begin{align} I:=\int^\infty_0 \frac{\sin(x)}{x}e^{-x^2}\,dx=\frac{1}{2}\int^\infty_{-\infty}\frac{\sin(x)}{x}e^{-x^2}\,dx=\frac{1}{4}\int^\infty_{-\infty}\int^1_{-1}e^{ixt}e^{-x^2}\,dt\,dx \end{align} Since $|e^{ixt}e^{-x^2}|=e^{-x^2}$ the double integral is clearly absolutely convergent so by Tonelli-Fubini we can interchange the integration order to obtain: \begin{align} I=\frac{1}{4}\int^{1}_{-1}\int^{\infty}_{-\infty}e^{ixt}e^{-x^2}\,dx\,dt=\frac{1}{4}\int^{1}_{-1}\int^{\infty}_{-\infty}e^{-\frac{t^2}{4}-(x-it/2)^2}\,dx\,dt \end{align} where we have completed the square on the last integral. Now by an easy contour integration on a rectangle we get \begin{align} \int^\infty_{-\infty} e^{ixt}e^{-x^2}\,dx=\sqrt[]{\pi}e^{-t^2/4} \end{align} So with substitutation and parity (P) we can conclude: \begin{align} I=\frac{\sqrt[]{\pi}}{4}\int^1_{-1}e^{-t^2/4}\,dt\stackrel{\color{red}{u=t/2}}{=}\frac{\sqrt[]{\pi}}{2}\int^{1/2}_{-1/2}e^{-u^2}\,du\stackrel{\color{red}{P}}{=}\frac{\pi}{2}\cdot\frac{2}{\sqrt[]{\pi}}\int^{1/2}_0e^{-u^2}\,du=\frac{\pi}{2}\operatorname{erf}\left(\frac 12\right) \end{align} | 2021-06-24T08:43:46 | {
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https://math.stackexchange.com/questions/1920684/how-to-evaluate-a-limit-of-the-indeterminate-form-0-00/1920697 | # How to evaluate a limit of the indeterminate form $(0/0)^0$
How to find the $\lim_{n \to \infty} \left(\dfrac{(n+1)(n+2)\cdots(n+2n)}{n^{2n}}\right)^{1/n}$? I know how to find it for the indeterminate form of $1^{\infty}$ by converting it into $0/0$ form, but this cannot be converted into any known indeterminate form: $(0/0)^0$ . Can we convert it into an integral and then try to solve at as infinity is involved?. May someone help? Also please don't use any theorems or formula for limits except perhaps L'Hospital rule which I know. If you do please provide its proof as well.
• see a technique for similar question math.stackexchange.com/a/1913608/72031 – Paramanand Singh Sep 9 '16 at 20:08
• I know, I'm not new here. – Matt Sep 12 '16 at 17:16
• @RaghavSingal Just assumed you had forgotten -- this does happen. If not, are you waiting for more details or a more complete answer? – Clement C. Sep 12 '16 at 17:37
First "trick:" convert to the exponential form.
$$\left(\frac{\prod_{k=1}^{2n}(n+k)}{n^{2n}}\right)^{\frac{1}{n}} =\exp\left({\frac{1}{n}\ln\left(\frac{\prod_{i=1}^{2n}(n+k)}{n^{2n}}\right)}\right)$$ Now, let us focus on the exponent: $$\frac{1}{n}\ln\left(\frac{\prod_{k=1}^{2n}(n+k)}{n^{2n}}\right) = \frac{1}{n}\ln\left(\prod_{k=1}^{2n}\frac{n+k}{n}\right) = \frac{1}{n}\ln\left(\prod_{k=1}^{2n}\left(1+\frac{k}{n}\right)\right) = \frac{1}{n}\sum_{k=1}^{2n}\ln\left(1+\frac{k}{n}\right)$$ At that point, it starts to really look like a Riemann sum, so let us massage it a little bit more: $$\frac{1}{n}\ln\left(\frac{\prod_{k=1}^{2n}(n+k)}{n^{2n}}\right) = \frac{1}{n}\sum_{k=1}^{2n}\ln\left(1+\frac{k}{n}\right) = \frac{2}{2n}\sum_{k=1}^{2n}\ln\left(1+2\frac{k}{2n}\right) = 2\cdot\frac{1}{2n}\sum_{k=1}^{2n}f\!\left(\frac{k}{2n}\right)$$ for $f\colon [0,1]\to \mathbb{R}$ defined by $f(x)=\ln(1+2x)$.
We have$^{(\dagger)}$ $$\frac{1}{2n}\sum_{k=1}^{2n}f\!\left(\frac{k}{2n}\right)\xrightarrow[n\to\infty]{} \int_0^1 f = \frac{1}{2}(3\ln 3 -2)$$ and by continuity of the exponential your limit will be $$\left(\frac{\prod_{k=1}^{2n}(n+k)}{n^{2n}}\right)^{\frac{1}{n}} \xrightarrow[n\to\infty]{} e^{2\int_0^1 f} = e^{3\ln 3 -2)} = \frac{27}{e^2}.$$
$(\dagger)$ Here, we use the following theorem, which essentially follows from the definition of Riemann integration:
Theorem. (Riemann sums converge to the integral.) Let $f\colon [a,b]\to\mathbb{R}$ be a continuous function. Then $$\frac{1}{n}\sum_{k=1}^n f\left(a+k\frac{b-a}{n}\right) \xrightarrow[n\to\infty]{} \int_a^b f$$ and $$\frac{1}{n}\sum_{k=0}^{n-1} f\left(a+k\frac{b-a}{n}\right) \xrightarrow[n\to\infty]{} \int_a^b f.$$
This is a particular case of a slightly more general theorem ($f$ only needs to be Riemann integrable, and here we took a regular subdivision of $[a,b]$ in intervals of the same length $\frac{b-a}{n}$ instead of an arbitrary subdivision.)
In our case, $a=0$ and $b=1$, and we take a subdivision with $m\stackrel{\rm def}{=} 2n$ points: $$\frac{1}{m}\sum_{k=1}^m f\left(\frac{k}{m}\right) \xrightarrow[m\to\infty]{} \int_0^1 f.$$
• How $\frac{n+k}{n^{2n}}$ becomes $\frac{n+k}{n}$? – Von Neumann Sep 9 '16 at 18:23
• @FourierTransform It did not. $\frac{1}{n^{2n}} \prod_{k=1}^{2n} (n+k)$ became $\prod_{k=1}^{2n} \frac{n+k}{n}$. – Clement C. Sep 9 '16 at 18:24
• Oh! Sorry. I read wrong. Great! – Von Neumann Sep 9 '16 at 18:25
• I know this comment doesn't really add much, but I'm just wondering how you came up with this whole process. It seems so genius to me that I can only conclude that it took hours. – Polygon Sep 9 '16 at 18:28
• Whenever I see something of the form $a_n^{b_n}$, my first instinct is that I don't understand powers, so I rewrite it $e^{b_n \ln a_n}$ (or $2^{b_n \log_2 a_n}$ if it's computer science). Then, after that, I don't understand products: if there are products inside the logarithm, I "convert" them into a sum of logarithms instead. (The rationale being, I know of more results on sum and series than on products). And a sum of the form $\frac{1}{n}\sum_k g(\frac{k}{n})$ begs for being recognized as a Riemann sum -- it does not always work, but it's worth trying for the 90% cases where it does. – Clement C. Sep 9 '16 at 18:32
Maybe you can use the exponential form.
Forgive me for what I'm going to write, I know it's bad formalism but it may be effective:
$$\left(\frac{0}{0}\right)^0 = e^{0\cdot \ln\left(\frac{0}{0}\right)}$$
Then you may use hospital for the limit inside the logarithm and proceed...
$$\text{let } y = \left(\dfrac{(n+1)(n+2)\cdots(n+2n)}{n^{2n}}\right)^{1/n}$$
$$\implies\log_e y = {1\over n}\log_e \left(\dfrac{(3n)!}{n! \times n^{2n}}\right)$$
$$\implies e^{{1\over n}\log_e \left(\dfrac{(3n)!}{n! \times n^{2n}}\right)} = y$$
$$\text{let } z = \log_e \left(\dfrac{(3n)!}{n! \times n^{2n}}\right)$$
$$\implies z = \log_e (3n)!- \log_e n! - 2n\log_e n$$
$$\color{red}{\implies z = 3n\log_e (3n) - 3n - (n\log_e n - n) - 2n\log_e n}$$
$$\implies z = 3n\log_e (n) + 3n\log_e 3 - 2n - n\log_e n - 2n\log_e n$$
$$\implies z = n(3\log_e 3 - 2)$$
$$\text{substituting the value of z in y}$$
$$y= e^{{1\over n}z}$$
$$\implies y= e^{3\log_e 3 - 2}$$
$$\text{Finally finding the limit}$$ $$\implies \lim_{n \to \infty} y = \lim_{n \to \infty} e^{3\log_e 3 - 2} = e^{3\log_e 3 - 2} = {e^{3\log_e 3}\over e^2} = {27\over e^2}$$
$$\color{red}{RED} \leftarrow \text{Stirling approximation}$$
• I am getting $$e^{{1\over n}\log_e \left(\dfrac{(3n)!}{n! \times n^{2n}}\right)} = y$$, any idea how should i proceed ? – A---B Sep 9 '16 at 18:50
• @A---B Stirling approximation for example! – Von Neumann Sep 9 '16 at 18:51
• Then you should complete you answer. – A---B Sep 9 '16 at 18:55
• @A---B Nope. I just gave a Hint, since the question was about the indeterminate form. – Von Neumann Sep 9 '16 at 18:56
• hey, I got the answer. i will more than happy to edit it inside your answer. maybe that downvote will turn into upvote. – A---B Sep 9 '16 at 19:36
HINT:
Let $A=\left(\dfrac{(n+1)(n+2)\cdots(n+2n)}{n^{2n}}\right)^{1/n}$
$$\implies\ln A=\dfrac1n\sum_{r=1}^{2n}\ln\left(1+\dfrac rn\right)$$
$$\implies\ln A=2\cdot\dfrac1m\sum_{r=1}^m\ln\left(1+2\cdot\dfrac rm\right)$$
$$\text{As }\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$
$$\implies\ln A=2\int_0^1\ln(1+2x)\ dx$$
Now integrate by parts,
$$\int\ln(1+2x)\ dx=\ln(1+2x)\int dx-\int\left(\dfrac{d\{\ln(1+2x)\}}{dx}\cdot\int dx\right)dx$$
$$=x\ln(1+2x)-\int\dfrac x{1+2x}dx$$
Now for $\int\dfrac x{1+2x}dx,$ set $1+2x=y$ to ultimately find that
$$\ln A=2\left(\dfrac32\ln(1+2)-1\right)=3\ln 3-2=\ln\dfrac{3^3}{e^2}\text{ as }\ln(e)=1$$
• should it not be ..$ln\left(1+\dfrac{r}{2n}\right)$? – G Cab Sep 9 '16 at 18:51
• @GCab, No, Please find the updated answer. – lab bhattacharjee Sep 10 '16 at 1:58
• @right, so .. one mis-attention per each .. – G Cab Sep 10 '16 at 16:30 | 2019-08-22T10:11:33 | {
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https://flex.phys.tohoku.ac.jp/texi/calc/calc_18.html | ### RPN Calculations and the Stack
The central component of an RPN calculator is the stack. A calculator stack is like a stack of dishes. New dishes (numbers) are added at the top of the stack, and numbers are normally only removed from the top of the stack.
In an operation like 2+3, the 2 and 3 are called the operands and the + is the operator. In an RPN calculator you always enter the operands first, then the operator. Each time you type a number, Calc adds or pushes it onto the top of the Stack. When you press an operator key like +, Calc pops the appropriate number of operands from the stack and pushes back the result.
Thus we could add the numbers 2 and 3 in an RPN calculator by typing: 2 RET 3 RET +. (The RET key, Return, corresponds to the ENTER key on traditional RPN calculators.) Try this now if you wish; type M-# c to switch into the Calc window (you can type M-# c again or M-# o to switch back to the Tutorial window). The first four keystrokes "push" the numbers 2 and 3 onto the stack. The + key "pops" the top two numbers from the stack, adds them, and pushes the result (5) back onto the stack. Here's how the stack will look at various points throughout the calculation:
. 1: 2 2: 2 1: 5 .
. 1: 3 .
.
M-# c 2 RET 3 RET + DEL
The .' symbol is a marker that represents the top of the stack. Note that the "top" of the stack is really shown at the bottom of the Stack window. This may seem backwards, but it turns out to be less distracting in regular use.
The numbers 1:' and 2:' on the left are stack level numbers. Old RPN calculators always had four stack levels called x, y, z, and t. Calc's stack can grow as large as you like, so it uses numbers instead of letters. Some stack-manipulation commands accept a numeric argument that says which stack level to work on. Normal commands like + always work on the top few levels of the stack.
The Stack buffer is just an Emacs buffer, and you can move around in it using the regular Emacs motion commands. But no matter where the cursor is, even if you have scrolled the .' marker out of view, most Calc commands always move the cursor back down to level 1 before doing anything. It is possible to move the .' marker upwards through the stack, temporarily "hiding" some numbers from commands like +. This is called stack truncation and we will not cover it in this tutorial; see section Truncating the Stack, if you are interested.
You don't really need the second RET in 2 RET 3 RET +. That's because if you type any operator name or other non-numeric key when you are entering a number, the Calculator automatically enters that number and then does the requested command. Thus 2 RET 3 + will work just as well.
Examples in this tutorial will often omit RET even when the stack displays shown would only happen if you did press RET:
1: 2 2: 2 1: 5
. 1: 3 .
.
2 RET 3 +
Here, after pressing 3 the stack would really show 1: 2' with Calc: 3' in the minibuffer. In these situations, you can press the optional RET to see the stack as the figure shows.
(*) Exercise 1. (This tutorial will include exercises at various points. Try them if you wish. Answers to all the exercises are located at the end of the Tutorial chapter. Each exercise will include a cross-reference to its particular answer. If you are reading with the Emacs Info system, press f and the exercise number to go to the answer, then the letter l to return to where you were.)
Here's the first exercise: What will the keystrokes 1 RET 2 RET 3 RET 4 + * - compute? (*' is the symbol for multiplication.) Figure it out by hand, then try it with Calc to see if you're right. See section RPN Tutorial Exercise 1. (*)
(*) Exercise 2. Compute @c{$(2\times4) + (7\times9.4) + {5\over4}$} 2*4 + 7*9.5 + 5/4 using the stack. See section RPN Tutorial Exercise 2. (*)
The DEL key is called Backspace on some keyboards. It is whatever key you would use to correct a simple typing error when regularly using Emacs. The DEL key pops and throws away the top value on the stack. (You can still get that value back from the Trail if you should need it later on.) There are many places in this tutorial where we assume you have used DEL to erase the results of the previous example at the beginning of a new example. In the few places where it is really important to use DEL to clear away old results, the text will remind you to do so.
(It won't hurt to let things accumulate on the stack, except that whenever you give a display-mode-changing command Calc will have to spend a long time reformatting such a large stack.)
Since the - key is also an operator (it subtracts the top two stack elements), how does one enter a negative number? Calc uses the _ (underscore) key to act like the minus sign in a number. So, typing -5 RET won't work because the - key will try to do a subtraction, but _5 RET works just fine.
You can also press n, which means "change sign." It changes the number at the top of the stack (or the number being entered) from positive to negative or vice-versa: 5 n RET.
If you press RET when you're not entering a number, the effect is to duplicate the top number on the stack. Consider this calculation:
1: 3 2: 3 1: 9 2: 9 1: 81
. 1: 3 . 1: 9 .
. .
3 RET RET * RET *
(Of course, an easier way to do this would be 3 RET 4 ^, to raise 3 to the fourth power.)
The space-bar key (denoted SPC here) performs the same function as RET; you could replace all three occurrences of RET in the above example with SPC and the effect would be the same.
Another stack manipulation key is TAB. This exchanges the top two stack entries. Suppose you have computed 2 RET 3 + to get 5, and then you realize what you really wanted to compute was 20 / (2+3).
1: 5 2: 5 2: 20 1: 4
. 1: 20 1: 5 .
. .
2 RET 3 + 20 TAB /
Planning ahead, the calculation would have gone like this:
1: 20 2: 20 3: 20 2: 20 1: 4
. 1: 2 2: 2 1: 5 .
. 1: 3 .
.
20 RET 2 RET 3 + /
A related stack command is M-TAB (hold META and type TAB). It rotates the top three elements of the stack upward, bringing the object in level 3 to the top.
1: 10 2: 10 3: 10 3: 20 3: 30
. 1: 20 2: 20 2: 30 2: 10
. 1: 30 1: 10 1: 20
. . .
10 RET 20 RET 30 RET M-TAB M-TAB
(*) Exercise 3. Suppose the numbers 10, 20, and 30 are on the stack. Figure out how to add one to the number in level 2 without affecting the rest of the stack. Also figure out how to add one to the number in level 3. See section RPN Tutorial Exercise 3. (*)
Operations like +, -, *, /, and ^ pop two arguments from the stack and push a result. Operations like n and Q (square root) pop a single number and push the result. You can think of them as simply operating on the top element of the stack.
1: 3 1: 9 2: 9 1: 25 1: 5
. . 1: 16 . .
.
3 RET RET * 4 RET RET * + Q
(Note that capital Q means to hold down the Shift key while typing q. Remember, plain unshifted q is the Quit command.)
Here we've used the Pythagorean Theorem to determine the hypotenuse of a right triangle. Calc actually has a built-in command for that called f h, but let's suppose we can't remember the necessary keystrokes. We can still enter it by its full name using M-x notation:
1: 3 2: 3 1: 5
. 1: 4 .
.
3 RET 4 RET M-x calc-hypot
All Calculator commands begin with the word calc-'. Since it gets tiring to type this, Calc provides an x key which is just like the regular Emacs M-x key except that it types the calc-' prefix for you:
1: 3 2: 3 1: 5
. 1: 4 .
.
3 RET 4 RET x hypot
What happens if you take the square root of a negative number?
1: 4 1: -4 1: (0, 2)
. . .
4 RET n Q
The notation (a, b) represents a complex number. Complex numbers are more traditionally written @c{$a + b i$} a + b i; Calc can display in this format, too, but for now we'll stick to the (a, b) notation.
If you don't know how complex numbers work, you can safely ignore this feature. Complex numbers only arise from operations that would be errors in a calculator that didn't have complex numbers. (For example, taking the square root or logarithm of a negative number produces a complex result.)
Complex numbers are entered in the notation shown. The ( and , and ) keys manipulate "incomplete complex numbers."
1: ( ... 2: ( ... 1: (2, ... 1: (2, ... 1: (2, 3)
. 1: 2 . 3 .
. .
( 2 , 3 )
You can perform calculations while entering parts of incomplete objects. However, an incomplete object cannot actually participate in a calculation:
1: ( ... 2: ( ... 3: ( ... 1: ( ... 1: ( ...
. 1: 2 2: 2 5 5
. 1: 3 . .
.
(error)
( 2 RET 3 + +
Adding 5 to an incomplete object makes no sense, so the last command produces an error message and leaves the stack the same.
Incomplete objects can't participate in arithmetic, but they can be moved around by the regular stack commands.
2: 2 3: 2 3: 3 1: ( ... 1: (2, 3)
1: 3 2: 3 2: ( ... 2 .
. 1: ( ... 1: 2 3
. . .
2 RET 3 RET ( M-TAB M-TAB )
Note that the , (comma) key did not have to be used here. When you press ) all the stack entries between the incomplete entry and the top are collected, so there's never really a reason to use the comma. It's up to you.
(*) Exercise 4. To enter the complex number (2, 3), your friend Joe typed ( 2 , SPC 3 ). What happened? (Joe thought of a clever way to correct his mistake in only two keystrokes, but it didn't quite work. Try it to find out why.) See section RPN Tutorial Exercise 4. (*)
Vectors are entered the same way as complex numbers, but with square brackets in place of parentheses. We'll meet vectors again later in the tutorial.
Any Emacs command can be given a numeric prefix argument by typing a series of META-digits beforehand. If META is awkward for you, you can instead type C-u followed by the necessary digits. Numeric prefix arguments can be negative, as in M-- M-3 M-5 or C-u - 3 5. Calc commands use numeric prefix arguments in a variety of ways. For example, a numeric prefix on the + operator adds any number of stack entries at once:
1: 10 2: 10 3: 10 3: 10 1: 60
. 1: 20 2: 20 2: 20 .
. 1: 30 1: 30
. .
10 RET 20 RET 30 RET C-u 3 +
For stack manipulation commands like RET, a positive numeric prefix argument operates on the top n stack entries at once. A negative argument operates on the entry in level n only. An argument of zero operates on the entire stack. In this example, we copy the second-to-top element of the stack:
1: 10 2: 10 3: 10 3: 10 4: 10
. 1: 20 2: 20 2: 20 3: 20
. 1: 30 1: 30 2: 30
. . 1: 20
.
10 RET 20 RET 30 RET C-u -2 RET
Another common idiom is M-0 DEL, which clears the stack. (The M-0 numeric prefix tells DEL to operate on the entire stack.) | 2020-07-16T18:28:37 | {
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https://www.physicsforums.com/threads/can-greens-theorem-disagree-with-itself-sometimes.821302/ | # Can Green's Theorem disagree with itself sometimes?
Tags:
1. Jun 29, 2015
### kostoglotov
1. The problem statement, all variables and given/known data
Firstly, I was seeking any clarification on whether I've made any mistakes. Secondly, further insight into Green's Theorem, if my working is all good.
Regarding the vector field $\vec{F} = \frac{1}{x^2+y^2} \langle -y,x,0 \rangle$ I decided to test Green's Theorem out.
It would seem that for this vector field $\oint_C \vec{F}\cdot d\vec{r} \neq \int\int_D (\nabla \times \vec{F})\cdot \hat{k} \ dA$
2. Relevant equations
3. The attempt at a solution
So $curl \ \vec{F} = \vec{0}$ so $\int\int_D (\vec{0})\cdot \hat{k} \ dA = 0$ regardless of our path C or domain D.
So let's consider the unit circle as our path
$C: \ x^2+y^2=1$ for $\oint_C \vec{F} \cdot d\vec{r}$
Standard polar form conversion $\vec{r}(t) = \langle cos(t), sin(t), 0 \rangle$
So $\oint_C \vec{F} \cdot d\vec{r} = \int_0^{2\pi}\vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) dt = 2\pi$
So Green's Theorem doesn't work for all vector fields, even if it's a simple enclosed region...is this due to a discontinuity at (x,y) = (0,0)?
2. Jun 29, 2015
### Orodruin
Staff Emeritus
The point is that the computation of curl F is only valid away from r = 0, where the vector field is singular. If you include the appropriate delta function at r = 0, you will find that Green's theorem holds.
3. Jun 29, 2015
### HallsofIvy
Staff Emeritus
No one (including Green) has ever claimed that Green's theorem works for all vector fields. A correct statement of Green's theorem is
"If L and M have continuous partial derivatives inside a simple closed curve, then $$\oint_C (Ldx+ Mdy)= \int\int_D\left(\frac{\partial M}{\partial x}- \frac{\partial L}{\partial y}\right) dxdy$$.
Your example does not have "continuous partial derivatives" at the origin. (And if you try to exclude (0, 0) by adding a small circle around it as part of the boundary, the boundary is no longer "simple".)
4. Jun 29, 2015
### kostoglotov
... :) So what does all that mean?
5. Jun 29, 2015
### kostoglotov
Does this mean a region that does not include the origin will be fine for Green's Theorem?
6. Jun 29, 2015
### Orodruin
Staff Emeritus
This seems too restrictive. It should be straight forward to generalise it to arbitrary distributions.
7. Jun 30, 2015
### HallsofIvy
Staff Emeritus
Yes, if you are willing to go to "homotopies", "homologies", and "algebraic topology".
8. Jun 30, 2015
### Zondrina
Green's theorem as you know it has been proven when the region $D$ is simple (type I or type II):
$$\oint_C \vec F \cdot d \vec r = \iint_D Q_x - P_y \space dA$$
The vector field is given as:
$$\vec F = \frac{-y \hat i + x \hat j}{x^2 + y^2}$$
For the given vector field, there is an issue at the origin because the vector field is undefined at the origin and it doesn't have continuous partial derivatives there either. Any attempt to apply Green's theorem directly will fail for any region containing the origin.
This poses a problem for the general version of Green's theorem, but Green's theorem can be extended to apply to regions that are not simple, i.e the region will have a hole somewhere. We do this by dividing the larger region into two simpler regions where we can apply Green's theorem and superimpose the results.
So taking the curve as the positively oriented unit circle $C_1: x^2 + y^2 = 1$, it encloses the region $D: x^2 + y^2 \leq 1$. This is going to cause a problem because it contains the origin, and as mentioned before, we can't do that for this particular vector field.
By enclosing the origin in a negatively oriented, smaller circle $C_2: x^2 + y^2 = r, \space 0 < r < 1$, we can extend Green's theorem to show that:
$$\oint_C \vec F \cdot d \vec r = 2 \pi$$
Where $C = C_1 \cup C_2$.
We do this by dividing $D$ into two simple regions $D'$ and $D''$ such that $D = D' \cup D''$ and we apply Green's theorem to the two simple regions like so:
$$\iint_D Q_x - P_y \space dA = \iint_{D'} Q_x - P_y \space dA + \iint_{D''} Q_x - P_y \space dA = \oint_{\partial D'} Pdx + Qdy + \oint_{\partial D''} Pdx + Qdy$$
The line integrals are along common boundary lines and are opposite in direction, so they cancel and we get:
$$\oint_{\partial D'} Pdx + Qdy + \oint_{\partial D''} Pdx + Qdy = \oint_{C_1} P dx + Q dy + \oint_{C_2} P dx + Q dy = \oint_C P dx + Q dy$$
This is Green's theorem as we know it because we have shown:
$$\iint_D Q_x - P_y \space dA = \oint_C P dx + Q dy$$
Even though we divided the region $D$ into two simpler regions before applying Green's theorem. This allows you to get around that pesky origin problem.
So for the problem at hand, we have specifically shown:
$$\oint_{C_1} \vec F \cdot d \vec r = \oint_{C_2} \vec F \cdot d \vec r = \int_0^{2 \pi} d \theta = 2 \pi$$
9. Jul 10, 2015
### Remixex
You're trying to use Green's Theorem without the field being of C1 class, it's not continuously differentiable.
As the previous poster mentioned you need to use the second form of Green's theorem if you wish to use it at all in this problem, you'll end up integrating by definition regardless, just in a smaller circle | 2017-11-19T01:56:45 | {
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https://mathematica.stackexchange.com/questions/94363/spacing-out-random-walks-so-they-dont-overlap | # Spacing out random walks so they don't overlap
From an external simulation program, I have lots of particle tracks which follow random walks, all of which start at or near the origin $(0,0)$. This means that when I plot the walks in Mathematica, it looks a bit of a mess because all the tracks overlap with each other.
For example:
randomwalks = Table[
Accumulate@RandomVariate[StableDistribution[1, 1.7, 0, 0, 2], {512, 2}],
{i, 3}];
ListLinePlot[randomwalks, PlotRange -> Full, PlotTheme -> "Minimal",
Axes -> None, Frame -> True, AspectRatio -> 1];
This gives the image on the left - the fact that all the walks start near the origin means that they overlap and it's hard to see what is going on.
What I'd like to plot instead is something like the image on the right, where the random walks have been manually translated to stop them overlapping.
Is there a way to achieve this translation automatically with Mathematica?
The direction of each jump in the random walk is important, so I want to keep this in the visualization (hence translations only and no rotations).
Note also the choice of a Lévy-stable distribution for the random walk rather than the normal distribution. This typically gives walks that are more spread-out with longer jumps (due to the fat distribution tails). In turn this leads to a more diverse spread of shapes to try and fit together, and means arranging the shapes in a simple grid (e.g. via Column[Graphics /@ Line /@ randomwalks] is not always as successful as I'd like.
My first thought was to calculate the bounding rectangle for each track, and then use the code in An algorithm to space out overlapping rectangles? to space out those bounding rectangles. However, this might leave large regions of whitespace due to the nature of the random walks, so perhaps a bounding polygon might be better to give a tighter arrangement of the tracks.
Given I have the track data itself, is there perhaps a way to do it without using the bounding rectangles? As suggested in a comment, perhaps using the convex hull of each path and then packing the resulting polygons?
regions = ConvexHullMesh[#] &/@ randomwalks
centroids = RegionCentroid[#] &/@ regions
Once we have the centroids, one approach would be to minimize the Euclidean distance between each centroid whilst ensuring the regions don't overlap - but that would need a function such as RegionOverlapQ[] or similar.
• One idea: Take the convex hull of each path, and then your problem is reduced to packing convex polygons. – Joseph O'Rourke Sep 10 '15 at 11:23
• Caveat: Optimal packing of convex polygons into a rectangular box is NP-hard, so you will be forced to use heuristics. (Mma has ConvexHull[ ].) – Joseph O'Rourke Sep 10 '15 at 11:34
• Why not just Column[Graphics /@ Line /@ randomwalks] or something? – Kuba Sep 10 '15 at 11:41
• @Kuba because I fancied a challenge? :-) – dr.blochwave Sep 10 '15 at 11:43
• OMG My code there is mess – Dr. belisarius Sep 10 '15 at 14:21
One approach is to rasterize the lines separately and use correlation to measure overlap. The lines are positioned one by one, as close to the centre as possible without overlapping any previously positioned lines. Heike used this in her answer to the word cloud question and I borrowed the idea here.
Here's a quick & dirty implementation:
randomwalks =
Table[Accumulate@
RandomVariate[StableDistribution[1, 1.7, 0, 0, 2], {512, 2}], {i, 15}];
r = 2.05 Max@Abs@randomwalks;
bin[walk_] := Binarize@ColorNegate@Erosion[
Rasterize[Graphics[Line@walk, PlotRange -> r], ImageSize -> 500],
DiskMatrix[3]]
iter[im1_, im2_] := Module[{corr, minx, miny},
{minx, miny} = 250 - PixelValuePositions[corr, Min@ImageData@corr][[1]];
Sow[-{minx, miny} r/250];
offsets = Reap[Fold[iter, bin /@ randomwalks]][[-1, 1]] ~Prepend~ {0, 0};
ListLinePlot[
Transpose[(Transpose[randomwalks, {1, 3, 2}] + offsets), {1, 3, 2}],
PlotRange -> Full, PlotTheme -> "Minimal", Axes -> None,
Frame -> True, AspectRatio -> 1]
• Very much different from mine, but I am confident there are many different approaches on this problem (as already outlined on the question). – kirma Sep 12 '15 at 15:41
• This one has ended up being more appropriate for my needs, although @kirma's answer is a very interesting idea nonetheless! – dr.blochwave Sep 21 '15 at 12:29
This solution is primarily a proof of concept; its run-time complexity makes it impractical for large amounts of items. Large amount means more than six, in this case.
Nonetheless, the idea is the following: individual plots are converted to polynominoes to be laid out on a rectangular grid. Every possible alignment on a finite grid is encoded to a Boolean equation representing legal layouts (every piece is present on exactly one position, and the pieces don't overlap), and a satisfying layout instance is searched for this equation. Result is baked back to datasets as offsets, which can be plotted now without overlaps.
ClearAll[layOutDatasets];
layOutDatasets[datasets_List, numinstances_Integer,
resolution_Integer, maxdim_List] :=
Module[
{normalizeDatasets, generateTile, generateAlignments,
generateOrientations, atMostOne, exactlyOne},
normalizeDatasets[
dss_List] := ((With[{offset =
First@CoordinateBoundingBox@#}, # - offset & /@ #]) & /@
dss)/Max[Max@Abs[Subtract @@ CoordinateBoundingBox@#] & /@ dss];
generateTile[ds_List] :=
ImageData[
ImageCrop@
Binarize[ImageResize[#, resolution + 2], {0.99, 1}]] /. {0 ->
True, 1 -> False} &@
ImageReflect@
Rasterize[
Style[Graphics[{Thickness[1/(2 resolution)], Line@ds},
PlotRange -> {{-1/resolution,
1 + 1/resolution}, {-1/resolution, 1 + 1/resolution}}],
Antialiasing -> False], ImageSize -> 8 (resolution + 2)];
generateAlignments[patt_List] :=
With[{maxoff = maxdim - Dimensions@patt},
Flatten[Table[{i, j}, {i, 0, maxoff[[1]]}, {j, 0, maxoff[[2]]}],
1]];
generateOrientations[patt_List] :=
False] & /@ generateAlignments[patt];
atMostOne[v_List] := BooleanCountingFunction[{0, 1}, Length@v] @@ v;
exactlyOne[v_List] := BooleanCountingFunction[{1}, Length@v] @@ v;
normdatasets = normalizeDatasets[datasets];
pieces = generateTile /@ normdatasets;
placemasks = generateOrientations[#] & /@ pieces;
With[{vars =
varmaps =
Flatten@MapIndexed[
c[#2[[1]]][#2[[2]]] -> Reverse@#1/resolution &,
generateAlignments[#] & /@ pieces, {2}]},
sols =
SatisfiabilityInstances[
And @@ (atMostOne /@
Flatten[
Transpose[
Flatten[
MapIndexed[#1 && c[#2[[1]]][#2[[2]]] &,
placemasks, {4}], 1], {3, 1, 2}], 1]) &&
And @@ MapIndexed[
With[{constants = c@First@#2 /@ Range@Length@#1},
vars, numinstances];
offsols = (Pick[vars, #, True] /. varmaps) & /@ sols;
Function[{set, off}, # + off & /@ set], {normdatasets, #}] & /@
offsols]]];
An example of use:
ListLinePlot[#, Axes -> False, AspectRatio -> Automatic] &@
First@layOutDatasets[
Table[Accumulate@
RandomVariate[StableDistribution[1, 1.7, 0, 0, 2], {512, 2}], {i,
5}], 1, 5, {15, 15}]
Another run:
EDIT: Note that there is no reason this method would be limited to a rectangular grid, regular alignments or even specific rotations. Generating more varied alternatives is always an option (although it increases size of the search space, which increases time complexity exponentially), and generating random alignments instead of regular lattice alignment is also an option. Hexagonal lattice would be quite interesting, but "rasterizing" graphics to such a grid is somewhat cumbersome.
• This looks really cool - I'll give it a go later on and report back properly! – dr.blochwave Sep 12 '15 at 11:29
• @blochwave It's more than a bit unpolished and has hacks all around - but I decided to dump it here anyway, because polishing might have never happened... The core idea is pretty simple, though. – kirma Sep 12 '15 at 11:33
• you never know, community action might help polish it! – dr.blochwave Sep 12 '15 at 11:36 | 2020-02-18T21:17:21 | {
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https://visiondatatechs.com/czqd01kx/minimum-spanning-tree-cut-property-624be0 | And it is called "spanning" since all vertices are included. In this tutorial, we’ve discussed cut property in a minimum spanning tree. A minimum spanning tree is a special kind of tree that minimizes the lengths (or “weights”) of the edges of the tree. A set of k-smallest spanning trees is a subset of k spanning trees (out of all possible spanning trees) such that no spanning tree outside the subset has smaller weight. Whether the problem can be solved deterministically for a general graph in linear time by a comparison-based algorithm remains an open question. MST of G is always a spanning tree. A path in the maximum spanning tree is the widest path in the graph between its two endpoints: among all possible paths, it maximizes the weight of the minimum-weight edge. In this way, the weight of and would be . ( F Now let’s define a cut of : The cut divided the graph into two subgraphs and . If the graph is dense (i.e. A spanning tree is one reaching all the vertices: V0 = V. In the rest of this discussion we will equate tree T with it’s set of edges E 0. A second algorithm is Prim's algorithm, which was invented by Vojtěch Jarník in 1930 and rediscovered by Prim in 1957 and Dijkstra in 1959. Shortest path algorithms like Prim’s algorithm and Kruskal’s algorithm use the cut property to construct a minimum spanning tree. , which is less than: Both minimum and maximum cut exist in a weighted connected graph. What is the point of the “respect” requirement in cut property of minimum spanning tree? ζ {\displaystyle 2^{r \choose 2}\cdot r^{2^{(r^{2}+2)}}\cdot (r^{2}+1)!} ( Theorem 1: Let G=(V,E) be a undirected, connected, weighted graph. .[2]. {\displaystyle F'(0)>0} Prim’s Algorithm. [1], If the weights are positive, then a minimum spanning tree is in fact a minimum-cost subgraph connecting all vertices, since subgraphs containing cycles necessarily have more total weight. [citation needed]. Prim's and Kruskal's algorithm both produce the minimum spanning tree. If they belong to the same tree, we discard such edge; otherwise we add it to T and merge u and v. The correctness of Kruskal’s algorithm can be proved by induction and cut-property of minimum spanning tree 2. According to the cut property, the total cost of the tree will be the same for these algorithms, but is it possible that these two algorithms give different MST with the same total cost, given that we choose it in alphabetic order when faced with multiple choices. Maximum spanning trees find applications in parsing algorithms for natural languages[43] Therefore if we include the edge , then it won’t be a minimum spanning tree. ∖ Property. (AKA bottleneck shortest path tree) Cut Property: Suppose S and T partition V such that 1. ) You can kind of intuit this for our example. The high level overview of all the articles on the site. And what we need to prove is that X with e added 3 is also a part of some possibly different minimum spanning three. n In all of the algorithms below, m is the number of edges in the graph and n is the number of vertices. It means the weight of the edge should be greater than the edge . roads), then there would be a graph containing the points (e.g. 4.3 Minimum Spanning Trees. that e belongs to an MST T1. MST algorithms rely on the cut property. 3 A minimum spanning tree (MST) of an edge-weighted graph is a spanning tree whose weight (the sum of the weights of its edges) is minimum. Then deleting e will break T1 into two subtrees with the two ends of e in different subtrees. Children of the node correspond to the definition of the pair s find out in the MST ( )! Subgraph of G is a spanning tree is said to be minimalif the is... Indeed, this algorithm has the smallest weight among those that cross $u$ and $V-U.! Must be in all of the DT contains a comparison between two edges, which also takes O m. Will the cut minimum spanning tree cut property: the cut set is or no '' identity weight on our graph find!, it is possible to solve the problem can also be used to describe financial markets minimum spanning tree cut property e is a. Endpoint is in one graph and creates two graphs. [ 5 ] [ 39 ] minimum spanning tree cut property 8 ] to... Which joins and be of higher weight than therefore is a spanning tree... Weight edge across a cut edge be approached in a minimum spanning tree from a G. Any edges, e.g property states that a minimum spanning tree of a graph G= ( V ; )... These definitions here be less than the previous one ] [ 39 ] [ 40 ] note! Different parts of partition set into two subgraphs: Next is the highest weighted edge edge be... Algorithm executes a number of steps and among which is connected, undirected graph minimum spanning tree cut property edge weights be.... And e joins two vertices from different parts of partition, the minimum spanning is... More sets here is roughly proportional to its length that cut property states that any minimum weight edge crossing cut. Algorithm use the optimal decision trees a crossing edge must be part of the MST should be of. A number of phases sets of vertices a distinct weight then there would be light... Minimum cost edge e of a graph G = ( V, '... Of using the decision trees linear number of processors it is easy see! Edited on 18 December 2020, at 16:35 since all vertices are included see an example edge the. Cost, representing the least expensive path for laying the cable only along certain paths ( e.g been. Describe an algorithm due to Kruskal Substructure • greedy Choice property • Prim s... Here, we ’ ll create a cycle is minimized, over spanning trees, there are two edges e.g! That divides a graph G with positive edge weights, find a cut is the of. Which means that must have been a tree minimum spanning tree cut property G is a tree! ( MST ), but a MBST is not a minimum spanning tree of connected! Must have been a tree to start with to find an MST since all vertices are.!, the total weight of every other spanning tree. ) few use for. Let G= ( V ; e ) be a undirected, weighted graph set is both minimum and cut! Cut and replace it with the minimum spanning tree. ) to see that edge. That this problem is unrelated to the same but there are multiple spanning.., the number of edges and thus the same cardinality ( namely, ) the edges which joins and must! Weighted perfect matching but then, minimum spanning tree cut property the trimming procedure did not remove any edges e.g. Property states that a cut is in the tree. ) and then construct the minimum tree. The graph 's size efficiently$ X\cup \ { e\ } $is part of some different! Is valid for all other minimum spanning tree. ) connect and among which the... Many others other endpoint is in the MST contains no cycles z?.. Problem, multi-terminal minimum cut will disconnect the graph minimum weighted edge in the MST scientist! Only one, unique minimum spanning tree with weight greater than or equal the. Ll also demonstrate how to find an MST for the minimum weight edge the identity on. By successively selecting edges to include in, it disconnects the graph into two or more sets the definition the. Presented the correctness of cut and replace it with the minimum spanning tree e. Vertices not yet included you can kind of intuit this for our example property which it! Same cardinality ( namely, ) ( log log n ( log log ). When constructing a minimum spanning tree. ) we can conclude that the edge is a can. Expensive path for laying the cable only along certain paths ( e.g cable! Two spanning trees have the same weight, ( V, e ' ) is list... ) 3 ) won ’ T be part of some minimum spanning tree of. [ 5 ] [ ]! That X with e added 3 is also included in any MST algorithm is at,. Vertex from: Again, when we remove from, it is possible to e ciently zoom in the. Conclude that the edge is a subgraph T that is:... cut property, ” we can define efficient! A number of edges that connect and among which is the number potential. Is O ( m log n ) 3 ) graph if and disconnects graph! Vertex if there are two popular variants of a minimum spanning tree. ) the... Cross the cut property Definition 3 with a linear number of steps efficient external storage sorting algorithms and graph... Prim 's algorithm, which also takes O ( m log n ).! To be minimalif the sum is minimized, over spanning trees can also be in... The Question is presented as follows: Prove the following cut property, the crossing edge must be and. Has a distinct weight then there will only be one with the smallest weight among all the edges whose endpoint... Mst is necessarily a MST is necessarily a MST V such that.! Not in the MST as follows: Prove the following cut property, ” we conclude! By assuming the edge should be part of the edge between X y. Idea: assume not, then this edge is the point of the spanning tree )! Ve discussed cut property Definition 3 focused on distance from the cut property to construct a minimum spanning tree.... Another graph … minimum spanning tree ( MST ) is a part of the MST is necessarily a is... Edge-Unweighted every spanning tree. ) we check whether endpoints u and V belongs to cut. Distance from the cut property is valid for all minimum spanning trees satisfy a very important which. Weights or costs with each edge has a distinct weight then there would be interesting here to see the! Not yet included, you will understand the spanning tree. ) in another graph that all edges in! Is identical to the definition of the pair sets,, and cut edge edge any. A light edge that crosses the cut set, cut vertex if there exists edge! Node of the minimum weight edge crossing the cut set contains the vertices remove... Minimum sum of edge weights ( connected ) a set of edges connects. Edge-Weighted graph is a spanning tree. ) internal node of the cut property construct... Many times, each for a solution Otakar Borůvka in 1926 ( see Borůvka 's algorithm except. The reverse of Kruskal 's algorithm: assume not, then any spanning tree has n − 1 edges minimum! Joins and valid for all other minimum spanning tree. ) processors it provably... Its weight, which is the minimum spanning tree ( MST ) is the same number potential. Except for the step of using the decision trees steps in the tree. ) following... Using the decision trees to find a min weight is in the MST be! E of a graph G= ( V, e ) is a list minimum spanning tree cut property edges of phase! Discussed cut property property which makes it possible to solve the problem is the same minimum weight... Which joins and if the minimum spanning tree given determines T since is! Find out in the Next section$ and $V-U$ expensive path for laying the.. Graph $G$ smallest edge crossing the cut property ) with edge weights is small... The soft heap, an approximate priority queue and e joins two vertices from different parts of.! Connected, i.e two algorithms that … Prove the following cut property to construct minimum..., there can be defined as a partition that divides a graph containing the points ( e.g ve taken and. The identity weight on our graph, partitions the vertex set of edges and thus the same weight executes! To describe financial markets Next section minimalif the sum is minimized, over spanning trees also! The node correspond to the two ends of e in different subtrees cut the! Small as possible the maximum sum of weights of all the articles on the.. Algorithms that … Prove the following cut property in a cross the cut property showed! Like Kruskal ’ s algorithm and Kruskal ’ s algorithm • Kruskal ’ s find out in the weight! Node of the MST assume all edges weights are unique and then construct the MST simplify the proof an. O ( m log n ) time the removal of the minimum tree! Are unique be connected and acyclic of weights of the edge between X and y larger than edge... One MST if they share the same weight cut set contains the vertices ''! Us now describe an algorithm due to Kruskal u $and$ V-U \$,... To maintain two sets, minimum spanning tree cut property and e joins two vertices from different parts partition...
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http://math.stackexchange.com/questions/132625/n-choose-k-leq-left-fracen-k-rightk | ${n \choose k} \leq \left(\frac{en}{ k}\right)^k$
This is from page 3 of http://www.math.ucsd.edu/~phorn/math261/9_26_notes.pdf.
Copying the relevant segment:
Stirling’s approximation tells us $\sqrt{2\pi n} (n/e)^n \leq n! \leq e^{1/12n} \sqrt{2\pi n} (n/e)^n$. In particular we can use this to say that $${n \choose k} \leq \left(\frac{en}{ k}\right)^k$$
I tried the tactic of combining bounds from $n!$, $k!$ and $(n-k)!$ and it didn't work. How does this bound follow from stirling's approximation?
-
A related question: math.stackexchange.com/q/132519/7266 – Fabian Apr 16 '12 at 18:51
First of all, note that $n!/(n-k)! \le n^k$. Use Stirling only for $k!$.
${n \choose k} \le \frac{n^k}{k!} \le \frac{n^k}{(\sqrt{2\pi k}(k/e)^k)} \le \frac{n^k}{(k/e)^k} = (\frac{en}{k})^k$
-
Everything is right except that your inequalities are all pointing backwards. Other than that, good answer! – David Speyer Apr 16 '12 at 18:54
thanks, just noticed that. – Wonder Apr 16 '12 at 18:54
This might be useless, but the inequality you're using (namely $k! \ge k^k e^{-k}$) has a very elementary proof (without need for the full Stirling) : $$e^k = \sum_{i = 0}^{\infty} \frac{k^i}{i!} \ge \frac{k^k}{k!}$$ – Joel Cohen Apr 16 '12 at 21:11
Great, that is very nice. Thanks for pointing it out. – Wonder Apr 17 '12 at 2:40
\begin{align*} \binom{n}k&=\frac{n!}{k!(n-k)!}\\ &\le\frac{e^{1/12n} \sqrt{2\pi n} (n/e)^n}{\sqrt{2\pi k}(k/e)^k\sqrt{2\pi(n-k)}((n-k)/e)^{n-k}}\\ &=\frac{e^{1/12n}\sqrt{n}}{\sqrt{2\pi k(n-k)}}\left(\frac{n/e}{k/e}\right)^k\left(\frac{n/e}{(n-k)/e}\right)^{n-k}\\ &\le\frac{e^{1/12n}\sqrt{n}}{\sqrt{2\pi k(n-k)}}\left(\frac{n}{k/e}\right)^k\\ &\le\frac{e^{1/12n}\sqrt{n}}{\sqrt{2\pi(n-1)}}\left(\frac{en}k\right)^k\\ &\le\left(\frac{en}k\right)^k \end{align*}
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Isn't $({n/e \over (n-k)/e})^{n-k} \gt 1$ ? – adamG Feb 23 '13 at 12:15
@adamG: It’s $\left(1+\frac{k}{n-k}\right)^{n-k}\le e^k$. – Brian M. Scott Feb 23 '13 at 12:22
Thanks Brian for the clarification! – adamG Feb 23 '13 at 13:13
@adamG: My pleasure! (Over the years I’ve been hung up often enough over such things.) – Brian M. Scott Feb 23 '13 at 13:16
I found a different proof of this fact avoiding Stirling.
Note $f(x) = (\frac{ex}{k})^k$ is a $C^2$ strictly convex function.
So $f(x) + f'(x)h < f(x+h)$ for $0 < h$
In particular, letting $h = 1$ we get
$$f'(x-1)+ f(x-1) < f(x)$$ $$(\frac{e(x-1)}{k})^{k-1}e + (\frac{e(x-1)}{k})^k< (\frac{ex}{k})^k$$ Noting that $(\frac{k}{k-1})^{k-1} < e$ since the ratio limits to $e$ from below. Substituting in the LHS for the second $e$, we get $$(\frac{e(x-1)}{k})^{k-1}(\frac{k}{k-1})^{k-1} + (\frac{e(x-1)}{k})^k< (\frac{ex}{k})^k$$ $$(\frac{e(x-1)}{k-1})^{k-1} + (\frac{e(x-1)}{k})^k< (\frac{ex}{k})^k$$
Now the result follows by induction on $n+k$ for $k\lt n$, and Pascal's formula.
$\binom {n-1}{k} + \binom {n-1}{k-1} = \binom {n}{k}$
$\binom {n-1}{k} \le (\frac{e(n-1)}{k})^k$ and $\binom {n-1}{k-1} \le (\frac{e(n-1)}{k-1})^{k-1}$ imply $\binom{n}{k} \le (\frac{en}{k})^k$
The base case $\binom{n}{n}$ and $\binom{n}{0}$ and $\binom{n}{1}$ are trivial so we can avoid the technicality where $k=1$ and $k=0$.
- | 2015-07-31T17:31:09 | {
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https://www.mathworks.com/help/matlab/ref/istril.html?nocookie=true | Accelerating the pace of engineering and science
istril
Determine if matrix is lower triangular
Description
example
tf = istril(A) returns logical 1 (true) if A is a lower triangular matrix; otherwise, it returns logical 0 (false).
Examples
expand all
Test Lower Triangular Matrix
Create a 5-by-5 matrix.
`D = tril(magic(5))`
```D =
17 0 0 0 0
23 5 0 0 0
4 6 13 0 0
10 12 19 21 0
11 18 25 2 9```
Test D to see if it is lower triangular.
`istril(D)`
```ans =
1```
The result is logical 1 (true) because all elements above the main diagonal are zero.
Test Matrix of Zeros
Create a 5-by-5 matrix of zeros.
`Z = zeros(5);`
Test Z to see if it is lower triangular.
`istril(Z)`
```ans =
1```
The result is logical 1 (true) because a lower triangular matrix can have any number of zeros on its main diagonal.
Input Arguments
expand all
A — Input arraynumeric array
Input array, specified as a numeric array. istril returns logical 0 (false) if A has more than two dimensions.
Data Types: single | double
Complex Number Support: Yes
expand all
Lower Triangular Matrix
A matrix is lower triangular if all elements above the main diagonal are zero. Any number of the elements on the main diagonal can also be zero.
For example, the matrix
$A=\left(\begin{array}{cccc}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& 0\\ -1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& 0\\ -2& -2& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& 0\\ -3& -3& -3& 1\end{array}\right)$
is lower triangular. A diagonal matrix is both upper and lower triangular.
Tips
• Use the tril function to produce lower triangular matrices for which istril returns logical 1 (true).
• The functions isdiag, istriu, and istril are special cases of the function isbanded, which can perform all of the same tests with suitably defined upper and lower bandwidths. For example, istril(A) == isbanded(A,size(A,1),0). | 2015-03-02T01:06:24 | {
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https://getreal.life/jskvvg/aaa1f3-number-of-bijective-functions-from-set-a-to-set-b | ## number of bijective functions from set a to set b
Example: The function f(x) = x 2 from the set of positive real numbers to positive real numbers is both injective and surjective. So #A=#B means there is a bijection from A to B. Bijections and inverse functions. To prove a formula of the form a = b a = b a = b, the idea is to pick a set S S S with a a a elements and a set T T T with b b b elements, and to construct a bijection between S S S and T T T.. toppr. A function is said to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. A bijective function has no unpaired elements and satisfies both injective (one-to-one) and surjective (onto) mapping of a set P to a set Q. De nition (Function). = 24. Answered By . How satisfied are … In mathematics, a bijective function or bijection is a function f : ... Cardinality is the number of elements in a set. This will help us to improve better. Related Questions to study. A function f : A -> B is called one – one function if distinct elements of A have distinct images in B. Let f : A ----> B be a function. Ivanova (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. If the number of bijective functions from a set A to set B is 120 , then n (A) + n (B) is equal to (1) 8 (3) 12 (4) 16. The number of surjections between the same sets is $k! Upvote(24) How satisfied are you with the answer? One way to think of functions Functions are easily thought of as a way of matching up numbers from one set with numbers of another. or own an. A function on a set involves running the function on every element of the set A, each one producing some result in the set B. 6. Then the second element can not be mapped to the same element of set A, hence, there are 3 choices in set B for the second element of set A. 8. }\] The notation $$\exists! Set A has 3 elements and the set B has 4 elements. This article was adapted from an original article by O.A. Its inverse, the exponential function, if defined with the set of real numbers as the domain, is not surjective (as its range is the set of positive real numbers). x \in A\; \text{such that}\;}\kern0pt{y = f\left( x \right). For Enquiry. More specifically, if g(x) is a bijective function, and if we set the correspondence g(a i) = b i for all a i in R, then we may define the inverse to be the function g-1 (x) such that g-1 (b i) = a i. D. 6. How many functions exist between the set \{1,2\} and [1,2,...,n]? So, for the first run, every element of A gets mapped to an element in B. f : R → R, f(x) = x 2 is not surjective since we cannot find a real number whose square is negative. Determine whether the function is injective, surjective, or bijective, and specify its range. Set Theory Index . Power Set; Power Set Maker . It means that every element “b” in the codomain B, there is exactly one element “a” in the domain A. such that f(a) = b. A common proof technique in combinatorics, number theory, and other fields is the use of bijections to show that two expressions are equal. x$$ means that there exists exactly one element $$x.$$ Figure 3. Sep 30,2020 - The number of bijective functions from the set A to itself when A constrains 106 elements isa)106!b)2106c)106d)(106)2Correct answer is option 'A'. B. B. The cardinality of A={X,Y,Z,W} is 4. The set A of inputs is the domain and the set B of possible outputs is the codomain. Below is a visual description of Definition 12.4. The number of bijective functions from set A to itself when there are n elements in the set is equal to n! The set A has 4 elements and the Set B has 5 elements then the number of injective mappings that can be defined from A to B is. The function f is called as one to one and onto or a bijective function, if f is both a one to one and an onto function. One to One and Onto or Bijective Function. Functions . Injective, Surjective, and Bijective Functions. MEDIUM. Number of functions from one set to another: Let X and Y are two sets having m and n elements respectively. Therefore, each element of X has ‘n’ elements to be chosen from. Class 12,NDA, IIT JEE, GATE. Education Franchise × Contact Us. The element f(x) is called the image of x. }$ . A function f: A → B is bijective or one-to-one correspondent if and only if f is both injective and surjective. Similarly there are 2 choices in set B for the third element of set A. combinatorics functions discrete-mathematics. Take this example, mapping a 2 element set A, to a 3 element set B. (a) We define a function f from A to A as follows: f(x) is obtained from x by exchanging the first and fourth digits in their positions (for example, f(1220)=0221). C. 1 2. Prove that a function f: R → R defined by f(x) = 2x – 3 is a bijective function. Let A be a set of cardinal k, and B a set of cardinal n. The number of injective applications between A and B is equal to the partial permutation: [math]\frac{n!}{(n-k)! The natural logarithm function ln : (0,+∞) → R is a surjective and even bijective (mapping from the set of positive real numbers to the set of all real numbers). The number of non-bijective mappings possible from A = {1, 2, 3} to B = {4, 5} is. Answer. Onto function could be explained by considering two sets, Set A and Set B, which consist of elements. Any ideas to get me going? Business Enquiry (North) 8356912811. Business … explain how we can find number of bijective functions from set a to set b if n a n b - Mathematics - TopperLearning.com | 7ymh71aa. Answer: c Explaination: (c), total injective mappings/functions = 4 P 3 = 4! | EduRev JEE Question is disucussed on EduRev Study Group by 198 JEE Students. How many of them are injective? toppr. Bijective / One-to-one Correspondent. Answer. The term for the surjective function was introduced by Nicolas Bourbaki. Answered By . Misc 10 (Introduction)Find the number of all onto functions from the set {1, 2, 3, … , n} to itself.Taking set {1, 2, 3}Since f is onto, all elements of {1, 2, 3} have unique pre-image.Total number of one-one function = 3 × 2 × 1 = 6Misc 10Find the number of all onto functio Bijective. For understanding the basics of functions, you can refer this: Classes (Injective, surjective, Bijective) of Functions. To prove there exists a bijection between to sets X and Y, there are 2 ways: 1. find an explicit bijection between the two sets and prove it is bijective (prove it is injective and surjective) 2. Let A, B be given sets. Sets and Venn Diagrams; Introduction To Sets; Set Calculator; Intervals; Set Builder Notation; Set of All Points (Locus) Common Number Sets; Closure; Real Number Properties . Thus, bijective functions satisfy injective as well as surjective function properties and have both conditions to be true. In essence, injective means that unequal elements in A always get sent to unequal elements in B. Surjective means that every element of B has an arrow pointing to it, that is, it equals f(a) for some a in the domain of f. Then, the total number of injective functions from A onto itself is _____. A bijection (or bijective function or one-to-one correspondence) is a function giving an exact pairing of the elements of two sets. This can be written as #A=4.:60. A bijective function is one that is both ... there exists a bijection between X and Y if and only if both X and Y have the same number of elements. A function f from A to B is a rule which assigns to each element x 2A a unique element f(x) 2B. A function f (from set A to B) is bijective if, for every y in B, there is exactly one x in A such that f(x) = y. Alternatively, f is bijective if it is a one-to-one correspondence between those sets, in other words both injective and surjective. 1 answer. Hence f (n 1 ) = f (n 2 ) ⇒ n 1 = n 2 Here Domain is N but range is set of all odd number − {1, 3} Hence f (n) is injective or one-to-one function. Get Instant Solutions, 24x7. The notion of a function is fundamentally important in practically all areas of mathematics, so we must review some basic definitions regarding functions. Academic Partner. More clearly, f maps distinct elements of A into distinct images in B and every element in B is an image of some element in A. Become our. Watch Queue Queue answr. I don't really know where to start. Need assistance? In a function from X to Y, every element of X must be mapped to an element of Y. Thanks! This video is unavailable. Identity Function. Then the number of injective functions that can be defined from set A to set B is (a) 144 (b) 12 (c) 24 (d) 64. If for every element of B, there is at least one or more than one element matching with A, then the function is said to be onto function or surjective function. If the function satisfies this condition, then it is known as one-to-one correspondence. asked Aug 28, 2018 in Mathematics by AsutoshSahni (52.5k points) relations and functions; class-12; 0 votes. The function f(x) = x+3, for example, is just a way of saying that I'm matching up the number 1 with the number 4, the number 2 with the number 5, etc. Now put the value of n and m and you can easily calculate all the three values. The words mapping or just map are synonyms for function. A function $$f$$ from set $$A$$ to set $$B$$ is called bijective (one-to-one and onto) if for every $$y$$ in the codomain $$B$$ there is exactly one element $$x$$ in the domain $$A:$$ \[{\forall y \in B:\;\exists! An identity function maps every element of a set to itself. 10:00 AM to 7:00 PM IST all days. Definition: Set A has the same cardinality as set B, denoted |A| = |B|, if there is a bijection from A to B – For finite sets, cardinality is the number of elements – There is a bijection from n-element set A to {1, 2, 3, …, n} Following Ernie Croot's slides 9. Answer From A → B we cannot form any bijective functions because n (a) = n (b) So, total no of non bijective functions possible = n (b) n (a) = 2 3 = 8 (nothing but total no functions possible) Prev Question Next Question. EASY. Onto Function A function f : A -> B is said to be onto function if the range of f is equal to the co-domain of f. The question becomes, how many different mappings, all using every element of the set A, can we come up with? Answer/Explanation. Can you explain this answer? To define the injective functions from set A to set B, we can map the first element of set A to any of the 4 elements of set B. I tried summing the Binomial coefficient, but it repeats sets. D. neither one-one nor onto. By definition, two sets A and B have the same cardinality if there is a bijection between the sets. A. Functions: Let A be the set of numbers of length 4 made by using digits 0,1,2. A ⊂ B. share | cite | improve this question | follow | edited Jun 12 '20 at 10:38. Contact. Here it is not possible to calculate bijective as given information regarding set does not full fill the criteria for the bijection. If X and Y have different numbers of elements, no bijection between them exists. f (n) = 2 n + 3 is a linear function. Set Symbols . A different example would be the absolute value function which matches both -4 and +4 to the number +4. 1800-212-7858 / 9372462318. Contact us on below numbers. Problem. What is a Function? Element set B, which consist of elements, no bijection between the sets using every of! To itself when there are 2 choices in set B this can be written #. One set to another: Let X and Y are two sets, a. 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http://math.stackexchange.com/questions/169128/is-there-a-sequence-in-0-1-such-that-the-product-of-all-its-terms-is-frac/169170 | # Is there a sequence in $(0,1)$ such that the product of all its terms is $\frac{1}{2}$?
I am trying to construct a sequence $\{x_{n}\} \in (0,1)$ such that such that the product of all its terms is $\frac{1}{2}$.
Please can I have any clue to solve my problem?
Thanks.
-
how long does your sequence have to be? If it's finite then the problem is trivial. – Robert Mastragostino Jul 10 '12 at 17:47
@RobertMastragostino: I want the sequence to be infinite, – Hassan Muhammad Jul 10 '12 at 17:55
If you take any sequence $a_1,a_2,a_3,\ldots$ whose sum is $\log_b (1/2)$, then $b^{a_1}, b^{a_2}, b^{a_3},\ldots$ is a sequence whose product is $1/2$.
Later note: Notice that $\frac 1 2 + \frac 1 4 + \frac 1 8 + \frac 1 {16} + \cdots = 1$. If you multiply every term by $\log_b \frac 1 2$ then you get a series whose sum is $\log_b \frac 1 2$.
Still later note: If $b>1$, then $\log_b(1/2)<0$, and $b^{a_n}$ will be in $(0,1)$ if $a_n<0$.
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The terms of the sequence must be in the open interval $(0,1)$ – Hassan Muhammad Jul 10 '12 at 18:39
@Hassan Take $b=1/2$ for instance. – alex.jordan Jul 10 '12 at 19:17
You can write that product for $b=1/2$ in an intuitive (but not very compact) form: $\sqrt{1/2}\sqrt{\sqrt{1/2}}\sqrt{\sqrt{\sqrt{1/2}}}\cdots$. The sequence of factors however has a very nice recursive definition: $x_1=\sqrt{1/2}$, $x_{n+1}=\sqrt{x_n}$ – celtschk Jul 10 '12 at 19:27
@HassanMuhammad : If $b>1$, then $\log_b(1/2)<0$, and $b^{a_n}$ will be in $(0,1)$ if $a_n<0$. – Michael Hardy Jul 10 '12 at 22:03
@MichaelHardy: Thanks for your answer (+1) – Hassan Muhammad Jul 11 '12 at 7:31
show 1 more comment
Try to find a sequence such that $\frac{n+1}{2n}=\prod_{j=2}^nx_j$ (it will do the job). We have $x_2=3/4$ and $$x_{n+1}=\frac{\prod_{j=2}^{n+1}x_j}{\prod_{j=2}^nx_j}=\frac{n+2}{2(n+1)}\frac{2n}{n+1}=\frac{n(n+2)}{(n+1)^2}=\frac{n^2+2n}{(n+1)^2}<\frac{n^2+2n\color{red}{+1}}{(n+1)^2}=1.$$ So $x_n=\frac{n^2-1}{n^2}$ does the job.
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I don't think $x_{n}=\frac{n^2-1}{n^2}$ will do the job because if $n=1$, $x_{n}=0$ which is not in $(0,1)$, and of course $\prod_{n=1}^{\infty}=0$ which is not my goal. – Hassan Muhammad Jul 10 '12 at 18:25
@HassanMuhammad: I believe the product starts at $n=2$. – user26872 Jul 10 '12 at 18:26
@HassanMuhammad Just shift the sequence if you want it to start at $1$ or $0$, but it doesn't matter. – Davide Giraudo Jul 10 '12 at 18:33
@oen: But that is not my question. I wrote the product of all its term, so n can be 1 – Hassan Muhammad Jul 10 '12 at 18:35
@HassanMuhammad: As Davide Giraudo mentions, you can just shift the sequence. Let $m = n-1$. – user26872 Jul 10 '12 at 18:38
$$x_n=2^{-2^{-n}}\qquad (n\geqslant1)$$
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How about this telescoping product? Let $a_n=2-1/n$ and then let $x_n=a_n/a_{n+1}$. Then $$\prod_{k=1}^n x_k = \left(\frac{a_1}{a_2}\right)\left(\frac{a_2}{a_3}\right)\cdots\left(\frac{a_n}{a_{n+1}}\right)=\frac{a_1}{a_{n+1}}=\frac{n+1}{2n+1}$$ It's easy to verify that $0<x_n<1$ and the partial products obviously tend to $1/2$.
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To see that the terms are in the desired range, it helps to write a formula for $x_n$ in terms of $n$. – Michael Lugo Jul 10 '12 at 21:30
I like the telescoping product, too. – JL344 Jul 11 '12 at 5:17
Take any decreasing sequence $(\pi_k)_{k\ge0}$ such that $$\pi_0=1;\quad \forall k\gt0,\pi_k \lt \pi_{k-1};\quad\lim_{k\to\infty}\pi_k=1/2.$$ We simply set $(\pi_k)_{k\ge1}$ as a sequence of partial products, $$\pi_k=\prod_{n=1}^k x_n,\text{ where }x_n=\pi_n/\pi_{n-1},$$ guaranteeing that $$\prod_{n=1}^\infty x_n=1/2.$$
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Are you sure that for a natural number n, $n\gt 1$, $x_{n}$ is in $(0,1)$? – Hassan Muhammad Jul 10 '12 at 19:09
As long as the sequence $\pi_n$ is positive and strictly decreasing. – JL344 Jul 10 '12 at 19:32
Recall either of Euler's two famous expressions for $\sin x$: $$\sin x=x\prod_{n=1}^\infty \cos\left(\frac{x}{2^n}\right),$$ or $$\sin x=x\prod_{n=1}^\infty\left(1-\frac{x^2}{\pi^2n^2}\right).$$
Now let $x=\dfrac{\pi}{6}$.
Or else use the following formula of Viète $$\frac{2}{\pi}=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2}\cdots,$$ and multiply both sides by $\dfrac{\pi}{4}$.
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This also gives $$\prod_{n\ge2}\left(1-\frac{1}{n^2}\right)=\frac{1}{2}.$$ (Though it is telescoping and thus easier..) – anon Jul 10 '12 at 23:09
$$\frac{1}{2}=(e^{\frac{1}{2}}-1)\prod_{k=1}^{\infty}\left(\frac{2}{e^{2^{-(k+1)}}+1} \right)$$ | 2014-07-10T00:19:27 | {
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https://math.stackexchange.com/questions/1098526/if-the-probability-of-the-union-of-two-independent-equally-likely-events-is-1 | # If the probability of the union of two independent equally likely events is $1$, what is the probability of each?
$E_1$ and $E_2$ are events in a probability space satisfying the following constraints:
• $\operatorname{Pr}(E_1)=\operatorname{Pr}(E_2)$
• $\operatorname{Pr}(E_1\cup E_2)=1$
• $E_1$ and $E_2$ are independent.
The probability of $E_1$ is ...(a) $0$, (b) $1/4$, (c) $1/2$, (d) $1$
I think the answer should be (d) 1
Reasoning.
1. E1 and E2 are equally likely.
2. Sum of their probability is 1. This is possible if and only if both of their probabilities are either 1(edit: if the events are independent) or 0.5( edit: if these are dependent and exhaustive events)
3. E1, E2 are independent events. This implies that they both doesn't belong to same same space.
Hence E1 and E2 are certain events with probability 1.
• See math notation guide. I replaced the image with text and formulas. – user147263 Jan 10 '15 at 5:36
• @Fundamental: Wow!!! Thanks. I will try to use these symbols from next time. :) – Prabhakar Jan 10 '15 at 11:03
Your reasoning for (2) is in error. Imagine a 3-sided die. The probability of $A=\{1,2\}$ occurring and $B=\{2,3\}$ occurring is equal and their union is a certain event.
You can get the answer quickly by elimination:
$$1=P(E_1 \cup E_2) < P(E_1)+P(E_2) = 2P(E_1) = 2P(E_2)$$
The strict less-than comes from being independent. So $P(E_1),P(E_2)$ are both larger than $1/2$
You can explicitly prove this, too
$$1=P(E_1 \cup E_2) = P(E_1)+P(E_2)-P(E_1)P(E_2) = 2P(E_1) - P(E_1)^2$$
$$\iff P(E_1)^2 - 2P(E_1)+1=(P(E_1)-1)^2=0$$
• in the example of a 3 sided die, the occurrences of any 2 events is either dependent or mutually exclusive. As per the 3rd point in the image, the events must be independent. So, I don't think we can consider this example for this question. – Prabhakar Jan 10 '15 at 5:01
• But your reasoning for $(2)$ without having considered their independence was in error. You didnt state that until already concluding it must be $1/2$ or $1$. You would have been marked off for this in a solution. – David Peterson Jan 10 '15 at 5:03
• Thanks. I missed it. I just updated the point 2. – Prabhakar Jan 10 '15 at 5:13
• : according to you which option is the correct answer? – Prabhakar Jan 10 '15 at 5:17
• If $(x-1)^2=0$ then $x$ must be.... – David Peterson Jan 10 '15 at 5:18 | 2019-11-17T15:21:49 | {
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https://www.freemathhelp.com/forum/threads/combination-question.114317/ | # Combination Question
#### Swazination
##### New member
How many combinations can I make if I have 4 letters A,G,C,T and I want to make groups of 5. Yes, you can repeat the letters.
#### Denis
##### Senior Member
From AAAAA to TGCAA
or (if replaced by digits 1 to 4):
from 1111 to 43211
Yes?
#### ksdhart2
##### Full Member
Combinations/Probability problems of this sort almost always come down to pattern recognition. In fact, learning how to find a pattern and extrapolate what comes next will help you greatly in nearly every aspect of mathematics. So let's try using those skills and see what we can come up with for this problem. Let's start with a much simpler version and build up as we go along. Suppose we had two letters (A and G) and one slot. How many combinations would there be? Well, obviously the answer is two. That's not very interesting nor does it seem helpful just yet, but we'll keep in mind all the same.
Now suppose we had two letters and two slots. How many combinations would there be then? Thinking about it, we can force the first slot to be A, then there's two possibilities for the second slot. Similarly, if we force the first slot to be G, we have two additional combinations. So we have two possibilities for the first slot, and for each of those two possibilities, we have two possibilities for the second slot. That gives us a total of 2 * 2 = 4 = 22 combinations.
Now suppose we had two letters and three slots. How many combinations would there be then? I see that we can arbitrarily fix the first two slots and then let the third one vary. Since we already know that there's four possible ways to arrange the first two slots, and for each of those four possibilities, there's two choices for the third slot, that gives us a total of 4 * 2 = 8 = 23 combinations. Are you seeing a pattern? How many combinations would there be with four slots? Five slots? n slots?
Now let's suppose we had three letters (A, G, and C) and two slots. By the exact same logic as before, we have three choices for the first slot, and for each of those three choices, we have three choices for the second slot. That gives us a total of 3 * 3 = 9 = 32 combinations. How many combinations would there be with three letters and three slots? Four slots? n slots? By this point, you should definitely be starting to see the bigger pattern and know the full answer.
#### pka
##### Elite Member
How many combinations can I make if I have 4 letters A,G,C,T and I want to make groups of 5. Yes, you can repeat the letters.
In the post you use the term combinations which do not allow for repetitions . What it seems you want is a multi-set.
NOTE that order does not matter.
If we want the number of ways to put $$\displaystyle \bf{N}$$ identical items (in this case five choices) into $$\displaystyle \bf{k}$$ distinct cells (in this case four letters) is
$$\displaystyle \dbinom{N+k-1}{N}$$
(in this case $$\displaystyle \dbinom{5+4-1}{5}=\dfrac{8!}{5!\cdot 3!})$$
If you mean that order does matter then it is simply $$\displaystyle \bf 4^5$$
Last edited: | 2019-03-19T08:00:12 | {
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https://math.stackexchange.com/questions/1498406/finding-number-of-integers-divisible-by-2-3-or-4-using-inclusion-exclusion-prin | # Finding number of integers divisible by 2, 3 or 4 using inclusion-exclusion principle.
I want to find number of integers from 1 to 19 (both included) which are divisible by 2 or 3 or 4. Lets denote it by N. So counting and enumerating them gives N = 12. Integers are 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16 and 18.
I thought of applying inclusion-exclusion principle for finding N. Here is how I proceeded:
Lets denote the followings:
N1 = (Number of integers which are divisible by 2) + (Number of integers which are divisible by 3) + (Number of integers which are divisible by 4)
N2 = (Number of integers which are divisible by 2*3 = 6) + (Number of integers which are divisible by 3*4 = 12) + (Number of integers which are divisible by 2*4 = 8)
N3 = (Number of integers which are divisible by 2*3*4 = 24)
Then using inclusion-exclusion principle we have: N = N1 - N2 + N3
Finding N1, N2 and N3 (denoting Greatest Integer Function using floor function):
N1 = $\lfloor \frac{19}{2} \rfloor + \lfloor \frac{19}{3} \rfloor + \lfloor \frac{19}{4} \rfloor = 9 + 6 + 4 = 19$
N2 = $\lfloor \frac{19}{6} \rfloor + \lfloor \frac{19}{12} \rfloor + \lfloor \frac{19}{8} \rfloor = 3 + 1 + 2 = 6$
N3 = $\lfloor \frac{19}{24} \rfloor$ = 0
Therefore, N = N1 - N2 + N3 = 19 - 6 + 0 = 13.
From here, I am getting N = 13 which is wrong as I counted them manually above (N = 12).
Can anyone point out the mistake I am making here and suggest the correct way of doing this using inclusion-exclusion principle?
• Hint: If an integer is divisible by 2 and divisible by 4, is it divisible by 8? – Element118 Oct 26 '15 at 13:22
• Indeed, each (Number of integers which are divisible by $km$) should be replaced by (Number of integers which are divisible by $k$ or by $m$), likewise for (Number of integers which are divisible by $kmn$). – Did Oct 26 '15 at 13:28
So $N_1=19$, $N_2=\lfloor 19/6 \rfloor+\lfloor 19/4 \rfloor+\lfloor 19/12 \rfloor=3+4+1=8$ and $N_3=\lfloor 19/12 \rfloor=1$ which gives you your $12$. | 2019-09-20T09:40:40 | {
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https://math.stackexchange.com/questions/2731637/is-the-numerator-of-sum-k-0n-1k-binomnk-frac12k1-always-a | # Is the numerator of $\sum_{k=0}^{n}{(-1)^k\binom{n}{k}\frac{1}{2k+1}}$ always a power of $2$ in lowest terms?
Is the numerator of $$\sum_{k=0}^{n}{(-1)^k\binom{n}{k}\frac{1}{2k+1}}$$ always a power of $2$ in lowest terms, and if so, why? Is there a combinatorial or probabilistic proof of this?
• How much numerical evidence for this do you have? Apr 10, 2018 at 22:38
• This is equal to the product of all even numbers up to $2n$ divided by the product of all odd numbers up to $2n+1$. So the question is why is there enough downstairs to cancel all the odds from upstairs. Apr 10, 2018 at 22:45
• @HenningMakholm Through about $n=50$. Put $\sqrt{\frac{1-x}{x}}\arctan\left(\sqrt{\frac{x}{1-x}}\right)$ into Mathematica (or Wolfram Alpha) and check its Taylor series for yourself. Apr 10, 2018 at 22:46
• @ArnaudMortier Ah, so $2^n n!\frac{2^n n!}{(2n+1)!}=\frac{2^{2n}}{(2n+1)\binom{2n}{n}}$. Yeah, that explains it. Apr 10, 2018 at 22:59
• I edited my answer to add a combinatorial proof of the crucial identity. Apr 11, 2018 at 0:26
Proved in edit at the end: this sum is equal to the fraction $${\prod_{i=1}^n 2i\over\prod_{i=1}^{n} 2i+1}$$
Now: $\prod_{i=1}^n 2i=2^n n!$
and $\prod_{i=1}^{n} {2i+1}={(2n+1)!\over \prod_{i=1}^n 2i}={(2n+1)!\over 2^n n!}$
Therefore your sum is equal to $${(2^n n!)^2\over (2n+1)!}={2^{2n}\over(2n+1){2n \choose n}}$$
Where the result is now clear.
Edit: combinatorial proof of the identity ${\prod_{i=1}^n 2i\over\prod_{i=1}^{n} 2i+1}$.
Define the sum $$S_{n,m}=\sum_{k=0}^{n}{(-1)^k\binom{n}{k}\frac{1}{2k+m}}$$ Claim: for any $n,m$, one has $$S_{n,m}={\prod_{i=1}^{n}2i\over \prod_{i=0}^{n}2i+m }$$
The claim implies the desired identity when $m=1$.
Proof of the claim:
The identity ${n+1\choose k}={n\choose k}+{n\choose k-1} \$ implies immediately that $$S_{n+1,m}=S_{n,m}-S_{n,m+2}$$
Moreover, the claim holds when $n=0$, where $S_{0,m}=\frac1m$
The result follows now from an immediate induction on $n$.
• This is nice, but I think I had something more bijective in mind. Like an application of the sieve method. Say, we consider the set $\{0,\pm1, \pm2,\dots,\pm n\}$ and the properties $P_i=$ something happens at $\pm i$, $i=1,\dots,n$. Then see where we have exactly $0$ properties, i.e. something happens only at $0$. Apr 11, 2018 at 0:58
• @AlexanderBurstein Interesting. I'll try to think in this direction too. Apr 11, 2018 at 1:01
Consider the function $$F(q)=\sum_{k=0}^n (-1)^k {n\choose k} q^{2k}=(1-q^2)^n$$ If we denote the sum you want to calculate by $\phi=\sum_{k=0}^n (-1)^k {n\choose k}\frac{1}{2k+1}$, then $$\phi = \int_0^1 F(q)dq = \int_0^1 (1-q^2)^n dq=\frac{(2n)!!}{(2n+1)!!}$$ where $k!!=k(k-2)(k-4)\cdots$. This simplifies to $$\phi=\frac{2^{2n} (n!)^2}{(2n+1)!}$$ Now note that $${2n \choose n}=\frac{(2n)!}{(n!)^2}\in \mathbb{Z}$$ so $(n!)^2$ divides $(2n)!$ which in turn means it divides $(2n+1)!$. So the numerator of $\phi$ is necessarily a power of $2$ in lowest terms.
• Thanks! That explains it. But what would be a combinatorial/probabilistic proof of this identity? Kind of curious about that still. Although, that’s probably deducible from the answer somehow. Apr 10, 2018 at 23:00
• If you insist on the combinatorial/probabilistic proof, let me think about it a bit. I'm sure there is one. Apr 10, 2018 at 23:06
• @Hamed (+1) I got one, if you are interested :) Apr 11, 2018 at 0:30
• It's too bad I can't accept both answers, this one is very good, too. Apr 11, 2018 at 1:00 | 2022-08-15T10:54:11 | {
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https://www.chebfun.org/examples/approx/RationalInterp.html | A rational function is a quotient of two polynomials. Many numerical algorithms make use of rational interpolants and approximants, including well-known methods for acceleration of convergence of sequences and series like the Aitken delta-squared formula and the epsilon and eta algorithms [2].
In Chebfun, since functions are defined on intervals, a natural form of rational interpolation involves sample points in an interval. In particular, if $f$ is a function defined on $[a,b]$ and $m$ and $n$ are nonnegative integers, then one may ask for a rational function of type $(m,n)$ that interpolates $f$ at the $N+1$ Chebyshev points in $[a,b]$, where $N = m+n$. By "type $(m,n)$", we mean representable as a quotient $p(x)/q(x)$ with $p$ of degree at most $m$ and $q$ of degree at most $n$. If $N>m+n$, then the problem is overdetermined one may still ask for a rational approximant to $f$ but now based on some kind of least-squares fit.
Here are three examples taken from Chapter 26 of [4]; some of our wording is also copied from there. The first is a problem that can be solved analytically. Suppose we seek a type $(1,1)$ rational function $r$ with $r(-1)=1+\varepsilon$, $r(0)=1$, and $r(1)=1+2\varepsilon$, where $\varepsilon$ is a parameter. The exact solution is
x = chebfun('x');
r = @(ep) 1 + (4/3)*ep*x./(x-(1/3));
Here are plots for $\varepsilon=0.1$ and $0.001$:
MS = 'markersize'; LW = 'linewidth'; FS = 'fontsize'; clf
for j = 1:2
ep = 10^(1-2*j);
subplot(2,1,j), plot(r(ep),LW,1.6), axis([-1 1 0 3])
hold on, plot([-1 0 1],[1+ep 1 1+2*ep],'.k',MS,20)
text(-.8,2.3,['ep = ' num2str(ep)],FS,12)
end
We see that in this particular case, the interpolant could hardly be considered to be an good approximation over the interval $[-1,1]$. It has a pole at $x=1/3$ that gets weaker as $\varepsilon$ decreases to zero (the residue is $4\varepsilon/3$), but never goes away so long as $\varepsilon$ is nonzero. One can think of this as a nearly-cancelling pole-zero pair: the nearby zero is at $1/(3-4\varepsilon))$.
So we see that rational interpolation can be tricky! "Spurious" pole-zero pairs of rational approximations may appear in unexpected places, leading to approximants that lack the properties one might like. This happens in exact arithmetic, and it happens even more often in finite-precision arithmetic, where rounding errors may generate additional pole-zero pairs that shouldn't be there in principle.
Chebfun has a command ratinterp for computing rational interpolants, which uses Chebyshev points by default but can also be applied with arbitrary points. We could have generated the above pictures with the commands
fj = [1+ep 1 1+2*ep];
[p,q] = ratinterp(fj,1,1);
r = p./q;
Here is our second example. Define
f = cos(exp(x));
and suppose we want to construct rational interpolants of type $(n,n)$ to $f$ based on samples at $2n+1$ Chebyshev points in $[-1,1]$. Here is a table of the maximum errors obtained by ratinterp for $n=1,2,3,4,5,6$:
disp(' (n,n) Error ')
for n = 1:6
[p,q] = ratinterp(f,n,n);
err = norm(f-p./q,inf);
fprintf(' (%1d,%1d) %7.2e\n',n,n,err)
end
(n,n) Error
(1,1) 2.46e-01
(2,2) 7.32e-03
(3,3) Inf
(4,4) 6.11e-06
(5,5) 4.16e-07
(6,6) 6.19e-09
We seem to have very fast convergence, but what has gone wrong with the type $(3,3)$ approximant? The first plot below reveals that the problem is another spurious pole. In such cases, one can often get better results by calling ratinterp with a value of $N$ specified bigger than the default of $m+n$. In this case the code out a certain kind of linear least-squares fit, and we can see in the lower plot that the choice $N=15$ eliminates the spurious pole for this problem. We use a magenta color as a symbol for a robust computation.
clf
subplot(2,1,1)
[p,q] = ratinterp(f,3,3);
hold off, plot(p./q,LW,1.6), hold on
xx = chebpts(7); plot(xx,f(xx),'.k',MS,16)
title(['Type (3,3) rational interpolant to cos(e^x) in 7 Chebyshev points'])
subplot(2,1,2)
[p,q] = ratinterp(f,3,3,15);
hold off, plot(p./q,'m',LW,1.6), hold on
xx = chebpts(16); plot(xx,f(xx),'.k',MS,16)
title(['Type (3,3) least-squares approximant to cos(e^x) in 16 Chebyshev points'])
For our final example, suppose we want the interpolant of type $(8,8)$ to $\exp(x)$ in $17$ Chebyshev points. A computation with ratinterp is very successful, as shown by the first plot below. However, this success is a tribute to the robustness features of ratinterp when called in its default mode. By default, ratinterp uses an approximation procedure that is more robust than pure interpolation, based on the singular value decomposition, which often manages to eliminate spurious poles introduced by rounding errors. We can turn off this robustness by repeating the computation with a tolerance of $0$ rather than the default $10^{-14}$, and that is what is done in the second panel of the plot. The plot looks the same! In fact it isn't though; there's a spurious pole-zero pair that has gone undected. We can see it like this:
f = exp(x);
[p,q] = ratinterp(f,8,8);
clf, subplot(2,1,1), plot(p./q,'m',LW,1.6), hold on
xx = chebpts(17); plot(xx,exp(xx),'.k',MS,16)
tol = 0;
[p,q] = ratinterp(f,8,8,[],[],tol);
subplot(2,1,2), plot(p./q,LW,1.6), hold on
xx = chebpts(17); plot(xx,exp(xx),'.k',MS,16)
format long
spurious_zeros = roots(p)
spurious_poles = roots(q)
separation = spurious_poles - spurious_zeros
spurious_zeros =
-0.705459053873448
0.374674058140232
spurious_poles =
-0.705459053873448
0.374674058140232
separation =
1.0e-15 *
0
-0.499600361081320
The computation of the first panel, by contrast, had no spurious poles. To achieve this, ratinterp actually returned an approximation of type less than the allowed $(8,8)$, as we can see like this:
[p,q,rh,mu,nu] = ratinterp(f,8,8);
degree_of_p = mu
spurious_zeros = roots(p)
degree_of_q = nu
spurious_poles = roots(q)
degree_of_p =
8
spurious_zeros =
Empty matrix: 0-by-1
degree_of_q =
4
spurious_poles =
Empty matrix: 0-by-1
The interpolation algorithm of ratinterp, and the code itself, originate with [3]. The generalization to least-squares approximations and robust approximants comes from [1], though that paper discusses just roots of unity on the unit circle, not Chebyshev points in $[-1,1]$.
### References
1. P. Gonnet, R. Pachon and L. N. Trefethen, Robust rational interpolation and least-squares, Electronic Transactions of Numerical Analysis, 38 (2011), 146-167.
2. Journal of Computational and Applied Mathematics special issue on interpolation and extrapolation, volume 122, numbers 1-2, October 2000.
3. R. Pachon, P. Gonnet and J. van Deun, Fast and stable rational interpolation in roots of unity and Chebyshev points, SIAM Journal on Numerical Analysis, 50 (2012), 1713-1734.
4. L. N. Trefethen, Approximation Theory and Approximation Practice, SIAM, 2013. | 2021-09-22T11:25:04 | {
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https://web2.0calc.com/questions/a-ball-is-thrown-upward-what-is-the-intial-vertical-speed-the-accleration-of-gravity-is-9-8m-s-2-and-max-height-is-6-5m-neglect-air-resis | +0
# a ball is thrown upward. what is the intial vertical speed? the accleration of gravity is 9.8m/s^2 and max height is 6.5m. neglect air resis
0
2073
10
a ball is thrown upward. what is the intial vertical speed? the accleration of gravity is 9.8m/s^2 and max height is 6.5m. neglect air resistance
Guest Sep 16, 2014
#3
+26750
+8
I think I'd just use v2 = u2 + 2as where v = final velocity (= 0), a = gravitational acceleration (-9.8m/s2), s = distance (= 6.5m)
0 = u2 - 2*9.8*6.5
$${\mathtt{u}} = {\sqrt{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{6.5}}}} \Rightarrow {\mathtt{u}} = {\mathtt{11.287\: \!160\: \!847\: \!617\: \!969\: \!5}}$$
initial velocity ≈ 11.3 m/s
Alan Sep 16, 2014
#1
+2353
+5
Hi there!
There are two formula's you need here.
At the start, the ball only has kinetic energy (speed).
This kinetic energy can be calculated as
$$E_k = \frac{1}{2}mv^2$$
where Ek is the kinetic energy, m is the mass and v is the speed.
At 6.5 meters all the kinetic energy has transformed into potential energy (since we neglect air resistance) which has the following formula;
$$E_p = mgh$$
Where Ep is the potential energy, m is the mass, g is acceleration of gravity and h the height.
Therefore if we equate $$E_p = E_k$$
We have
$$\frac{1}{2}mv^2 = mgh$$
Which we van rewrite to
$$\begin{array}{lcl} \frac{1}{2}v^2 = gh \mbox{ (divide both sides by m)}\\ v^2 = 2gh\\ v = \sqrt{2gh}\\ \mbox{ fill in g = 9.8 and h = 6.5}\\ v = \sqrt{2*9.8*6.5} \approx 11m/s \end{array}$$
So the speed at the start was approximately 11 meters per second.
Reinout
reinout-g Sep 16, 2014
#2
+92781
+8
Ok I might do it by calculus
$$\\\ddot y=-9.8\\ \dot y=-9.8t+v\\ y=-4.9t^2+vt\\ \mbox{max height when }\doty=0\qquad(y=6.5m)\\ -9.8t+v=0\\ v=9.8t\\\\ y=-4.9t^2+vt\\ 6.5=-4.9t^2+9.8t^2\\ 4.9t^2-6.5=0\\ t=\sqrt{\frac{6.5}{4.9}}\\\\ \mbox{Initial Velocity}=9.8\times\sqrt{\frac{6.5}{4.9}}=11.287\;\;m/s\\\\$$
Melody Sep 16, 2014
#3
+26750
+8
I think I'd just use v2 = u2 + 2as where v = final velocity (= 0), a = gravitational acceleration (-9.8m/s2), s = distance (= 6.5m)
0 = u2 - 2*9.8*6.5
$${\mathtt{u}} = {\sqrt{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{6.5}}}} \Rightarrow {\mathtt{u}} = {\mathtt{11.287\: \!160\: \!847\: \!617\: \!969\: \!5}}$$
initial velocity ≈ 11.3 m/s
Alan Sep 16, 2014
#4
+2353
0
I'm not sure whether this is an English expression, but in Dutch we have the expression;
'there are several ways that lead to Rome'.
I think that it pretty much sums up these answers.
I didn't know the equation Alan used.
Melody's approach is one I hadn't directly thought of, but I think I like it better than sticking your nose into physics formulas
reinout-g Sep 16, 2014
#5
+92781
0
In NSW, Australia schools.
If you are doing physics you have to use the physics formulas.
If you are doing mathematics you have to use calculus.
Melody Sep 16, 2014
#6
+26750
0
There is another (rather unpleasant) phrase: "There is more than one way to skin a cat"
The kinematics equation I used is derived directly from calculus - see my last post at http://web2.0calc.com/questions/two-students-on-a-balcony-18-8m-above-the-street-one-student-throws-a-ball-vertically-downward-13-1-m-s-at-the-same-instant-the-onther-st_1#r6
Alan Sep 16, 2014
#7
+2353
0
Ai, that sounds awefully unpleasant.
Several of our expressions sound very strange when you literally translate them such as;
it is a truth as a cow!
or
I can do this with two finger up my nose
or
That's as correct as a bus
So given 'all craziness on a stick' are there more english expressions that are funny or strange?
reinout-g Sep 16, 2014
#8
+26750
+3
I think the "skin a cat" expression probably has more to do with the "cat-o-nine-tails" (a whip used to keep recalcitrant sailors in order in days of yore) than cute furry animals!
I must be mad, but I really like the expression "I can do this with two fingers up my nose". From now on I shall use it whenever I can!!
Alan Sep 16, 2014
#9
+92781
+3
I think that "There is more than one way to skin a cat.", must be a common phrase in many counties.
Here's another. "He is as crooked as a dog's hind leg" Crooked means dishonest.
Melody Sep 16, 2014
#10
+92781
0
I liked that expression too Alan. It may also make its way into my vocabulary !
Melody Sep 16, 2014 | 2018-07-19T19:28:44 | {
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http://bfqw.shakemymag.it/moment-of-inertia-cylinder.html | # Moment Of Inertia Cylinder
Area Moment of Inertia - Imperial units. Proposed Subject usage: Mathematics / Physics (A/AS level), Sports Science (Degree Yr 1/2) Introduction Moment of inertia of an object is an indication of the level of force that has to be applied in order to set the object, or keep the object, in motion about a defined axis of rotation. The moment of inertia of a solid cylinder is equal to one half of the mass multiplied by the square of the radius. where O-O is the axis around which one is evaluating the mass moment of inertia, and r is the perpendicular distance between the mass and the axis O-O. The moment of inertia for a flywheel may be calculated using the general equation for rotational inertia of a rigid body as shown below. Moment of inertia definition, the sum of the products of the mass and the square of the perpendicular distance to the axis of rotation of each particle in a body rotating about an axis. Here's a solid cylinder. Now using the standard result for the moment of inertia of a square lamina about an axis perpendicular to the plane of the lamina and through the centre, we have the MI of the element as $$\frac 16\delta m y^2=\frac 23\rho y^4\delta x$$ So the MI is given by $$\frac 23\frac {3M}{a^2H}\int_0^H\frac{a^4x^4}{16H^4}dx=\frac{1}{40}Ma^2$$. o The moment of inertia of a thin disc of mass m and radius r about an axis passing through its C. 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. What exactly is the area moment of inertia (also called the second moment of area)? It is a geometrical property of any area. their moments of inertia. We know MOI about the axis passing through center of solid cylinder along the length is MR (square)/2 Now cut the cylinder into two equal halves MOI about COM will become MOI about diameter of semicircular cylinder then use Parrallel axis theorem. Moment of Inertia Moment of Inertia depends on the choice of axis of rotation. How to use Moment of Inertia Converter Select the unit to convert from in the input units list. 5 m, and q = 30°. In the preceding section, we defined the moment of inertia but did not show how to calculate it. Literal definition : Hollow cylinder (drilled) rotating about an external axis parallel to the axis of the cylinder itself. (I can see the cylindirical coordiates would be. I am computing the $\hat{I}$ - moment of inertia tensor - of a cylinder with height 2h and radius R, about its axis of symmetry at the point of its centre of mass. Polar Moment of Inertia vs. In the problem we are required to find moment of inertia about transverse (perpendicular) axis passing through its center. Find moment of inertia of a uniform hollow cylinder Home Problems and Answers Classical Mechanics Find moment of inertia of a uniform hollow cylinder We know that the moment of inertia for hoop with radius R is mR2. So in particular, I've got for you a cylinder. (Use any variable or symbol stated above as necessary. Perform the following analysis to determine the moment of inertia of the platter. It suggests that to turn the shaft at an angle, more torque is required, which means more polar moment of inertia is required. $$I_{xx} = \int_m \left( y^2 + z^2 \right) \, dm$$ Where $I_{xx}$ is the moment of inertia of a continuous body about the $x$ axis in a. It is the special "area" used in calculating stress in a beam cross-section during BENDING. 37 m about an axis parallel to the center-of-mass axis and passing through the edge fo the cylinder. AP Physics Practice Test: Rotation, Angular Momentum ©2011, Richard White www. What exactly is the area moment of inertia (also called the second moment of area)? It is a geometrical property of any area. In order to calculate the moment of inertia geometrically, the shapes of the objects must be identified. Ask Question Calculate the moment of inertia of the cylinder defined below when the cylinder is rotated. I {\displaystyle I} is the moment of inertia of the flywheel about its axis of symmetry. Hi, I would like to find a way in AutoCAD to calculate the (momet of inertia). The moment of inertia of a composite body is the sum of the moments of inertia of the parts of the body, the same axis of reference being used for each part. 2 Torque & Moments of Inertia and repeat the experiment to determine the moment of inertia of the ring or bar. " Or : The product mass and the square of the perpendicular distance from the axis of rotation is known as moment of inertia. The moment of inertia relative to centroidal axis x-x, can be found by application of the Parallel Axes Theorem (see below). It is the sum of the mass of each particle in the body with the square of its. Mass moments of inertia have units of dimension ML 2 ([mass] × [length] 2 ). A second cylinder; this one having a moment of inertia of I2 and initially not rotating, drops onto the first cylinder. The moment of inertia is a measure of the resistance of an object to changes in its rotational motion, just as mass is a measure of the tendency of an object to resist changes in its linear motion. Moment of Inertia. A homogeneous solid cylinder of mass m, length L, and radius R rotates about an axis through point P, which is parallel to the cylinder axis. Application of the parallel axis theorem allows a determination of the moment of inertia about many other axes. face is half that of a hollow ring having the same mass and the same radius. Unfortunately most rotating bodies do not have the mass concentrated at one radius and the moment of inertia is not calculated as easily as this. The Moment of Inertia of the cylinder I2 = MR2. Also called "Moment of Inertia". J : Moment of inertia to motor (kg m2) J M: Moment of inertia of motor (kg m2) J G1: Moment of inertia of gear 1 (kg m2) J G2: Moment of inertia of gear 2 (kg m2) Js : Moment of inertia of screw shaft (kg m2) m : Mass of transfer material (kg) (Note) Moment of inertia of cylindrical components Where m : Mass of cylinder (kg). Bending moment refers to the algebraic sum of all moments located between a cross section and one end of a structural member; a bending moment that bends the beam convex downward is positive, and one that bends it convex upward is negative. Bookmark Moment. (1) is the generalization to extended bodies of the definition for a single mass point. The density is then (1) and the moment of inertia tensor is (2) (3) (4). 16x10 5 mm 4 = 41. To do this you will first find the total moment of inertia of the system, which represents the combination of the disk's moment of inertia, I 1, and the moment of inertia of the ring or bar, I 2. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1. Starting from rest, the mass now moves a distance 0. Every rigid object has a de nite moment of inertia about a particular axis of rotation. Basic Stress Equations Dr. Determine the Moment of Inertia. If ø63 is used within a pressure range from 0. 25 cm and a mass of 750. The following other wikis use this file: Usage on ca. png 101 × 123; 3 KB. which is the sum of all the elemental particles masses multiplied by their distance from the rotational axis squared. 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. Moment of Inertia--Rod The moment of inertia of a thin rod of length h , mass M , and cross-sectional area A can be computed as the limiting case of the moment of inertia of a cylinder as the radius , so the tensor becomes. Rotational Inertia • Rotational Inertia (or “Moment of Inertia”) depends on the mass if the spinning object and where that mass is located • I = Σ mr 2 (units kg m 2) 13 Inertia Rods • Two rods have equal mass and length. Note: These section properties are calculated with respect to the major axis only. Where m is the mass of the flywheel (kg), r is the radius of gyration (m) and k is an inertial constant to account for the shape of the flywheel. 2nd MOMENT of AREA. The Area Moment Of Inertia of a beams cross-sectional area measures the beams ability to resist bending. In this section, we show how to calculate the moment of inertia for several standard types of objects, as well as how to use known moments of inertia to find the moment of inertia for a shifted axis or for a compound object. The moment of inertia of is given by: Where we have: m: mass R: radius ( from the axis O to the object ) The following is a list of moment of inertia for some common homogeneous objects, where M stands for mass and the red line is the axis the objects rotating about. I mean, it’s a pretty basic quantity, right? But think about it. Literal definition : Hollow cylinder (drilled) rotating about an external axis parallel to the axis of the cylinder itself. The total moment of inertia is due to the sum of masses at a distance from the axis of rotation. Knowing the area moment of inertia is a critical part of being able to calculate stress on a beam. The unit of dimension of the second moment of area is length to fourth power, L 4 , and should not be confused with the mass moment of inertia. I am computing the $\hat{I}$ - moment of inertia tensor - of a cylinder with height 2h and radius R, about its axis of symmetry at the point of its centre of mass. Moment of Inertia Introduction: Moment of inertia of an object is an indication of the level of force that has to be applied in order to set the object, or keep the object, in motion about a defined axis of rotation. If the cylinder is rotating around a horizontal axis (like a baton), then the water certainly contributes to the moment of inertia. as far as i can tell its giving the LEAST and the GREATEST as i & j. To do this you will first find the total moment of inertia of the system, which represents the combination of the disk's moment of inertia, I 1, and the moment of inertia of the ring or bar, I 2. The following other wikis use this file: Usage on ca. Mathematically, we describe the effect size and shape have on rotation with something called an object's moment of inertia, abbreviated I. For each component (wheels, brakes, etc. There are several different materials involved in these parts. The smallest Moment of Inertia about any axis passes throught the centroid. Mass moment of inertia and area moment of inertia both are called as moment of inertia, hence sometimes confusing. It's I = (MR^2)/2 Where M is the total mass and R is the radius of the cylinder. 8 Solid sphere rotating about the central axis. It's a constant density cylinder. 本作品采用知识共享 cc0 1. Note: These section properties are calculated with respect to the major axis only. The equation specifying the proportionality is a rotational version of Newton’s second law: ˝ = I (1). Determine the coefficient of MR^2 and any other terms that may be there. It represents how difficult it overcomed to change its angular motion about that axis. Where m is the mass of the flywheel (kg), r is the radius of gyration (m) and k is an inertial constant to account for the shape of the flywheel. The larger the Moment of Inertia the less the beam will bend. and inside diameter is 4 in. For basic shapes there are tables that contain area moment of inertia equations which can be viewed below. Be advised that the "moment of inertia" encountered in Statics is not the same as the moment of inertia used in Dynamics. Mechanical Tips By Er Saurav Sahgal Moment Of Inertia. (b) The skater with arms extended is. Mass is a measure of inertia, the tendency of an object to resist changes in its motion. But instead of limits from to, the limit is set from zero to. Let ! be its density. 41 kg and radius 2. A Hollow Cylinder With Radius R And Wall Thickness T The. The process involves adding up the moments of infinitesmally thin cylindrical shells. Area Moment of Inertia or Moment of Inertia for an Area - also known as Second Moment of Area - I, is a property of shape that is used to predict deflection, bending and stress in beams. Selecting of a model Select models by applying the moment of inertia and rotation time which have been found to the charts below. 01 18-Jun-2003 1. 00 m, at which point the end of the rope is moving at 6. The mass of an oxygen atom is 2. Moment of inertia, also known as rotational inertia, is analogous to the inertia of linear motion. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1. 1, December 23, 1997 Page 5 Make a series of measurements of I, the moment of inertia of the rigid body, with the masses m1 and m2 placed an equal distance r (r1 = r2) from the axis of rotation. Knowing the area moment of inertia is a critical part of being able to calculate stress on a beam. Similarly, an inertia moment (or inertia torque) is defined in rotary motion as a function of the mass moment of inertia and the angular acceleration (the second time derivative of rotation angle)—see Table 2. It is the rotational analogue to mass. List of moments of inertia From Wikipedia, the free encyclopedia The following is a list of moments of inertia. It is necessary to specify a moment of inertia with respect to an axis of rotation. Because the object consists of two uniform shapes (a hollow cylinder or ring and a solid cylinder or disk) the following equations for uniform objects can be used. ) Determine the moment of inertia about an axis a length L units to the left of the left mass. Enter the value to convert from into the input box on the left. Derivation of the moment of inertia of a hollow/solid cylinder. Rotational Kinetic Energy And Moment Of Inertia Pwiki. The units of the area moment of inertia are meters to the fourth power (m^4). Introduction: The moment of inertia depends in general about which axis the object is rotated. (I can see the cylindirical coordiates would be. The moment of inertia of the threshing cylinder is dependent on the power of the motor. We expect this position vector to precess about the axis of rotation (which is parallel to) with angular velocity. Moment of Inertia (I) Calculator Common Shapes Rotational Inertia Calculator getcalc. Investigation 1: The Moment of Inertia Goals: • To study how two objects having the same mass can have dramatically different "resistances" to changes in rotational velocity (i. asked by Wolves on June 27, 2013; physics. Moment of Inertia (I) - It is the moment of the moment or second moment of mass or area of body. Moment of inertia equation. Consider the moment of inertia of the hollow cylinder of Example 5 as being that of a solid cylinder of radius b minus that of a solid cylinder of radius a, with both cylinders having the same length t. For basic shapes there are tables that contain area moment of inertia equations which can be viewed below. rigid bar A rigid bar with a mass M and length L is free to rotate about a frictionless hinge at a wall. What Is Moment Of Inertia And How To Calculate It For A Rod. Media in category "Moments of inertia" The following 152 files are in this category, out of 152 total. So what I'd like you to do is, for the cylinder, I'd like you to compute its moment of inertia around its central axis. Wallace Bending Moment "x" Bending Moment z x y z x y M x σ σ M y "y" Bending Moment σ = σ ⋅ = M y ⋅ I and M x x x y y where: M x and M y are moments about indicated axes y and x are perpendicular from indicated axes Ix and Iy are moments of inertia about indicated axes Moments of Inertia: h c b D I R b. Polar Moment of Inertia is utilized to calculate. There are several different materials involved in these parts. Thin walled cylinder rotating about the central axis. 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. 37 m about an axis parallel to the center-of-mass axis and passing through the edge fo the cylinder. The moment of inertia of a hollow circular cylinder of any length is given by the expression shown. Unless a rigid body is rotating around its center of mass, the mass moment of inertia of a system will not be the same as the mass moment of inertia of the rigid body itself. The moment of inertia of an object is a calculated measure for a rigid body that is undergoing rotational motion around a fixed axis: that is to say, it measures how difficult it would be to change an object's current rotational speed. Find a formula by keywords Calculate own formula Loading Calculate too:. Problem #2: A cylinder having a moment of inertia I=(1/2)MR2 rolls down an incline as pictured below: Calculate the angular velocity of the cylinder when it reaches the bottom of the inclined plane. The cardboard tube, in contrast to the can, is hollow. The moment of inertia essentially describes an object's resistance to rotational motion in response to a torque. It is just like how mass determines the force needed for a desired acceleration. Find Moment of Inertia of a Solid Cylinder Calculator at CalcTown. I started with some simple drawings of the four shapes for which I want to calculate mass moment of inertia: solid cylinder, hollow cylinder, disk, and a block. Problem 817 Determine the moment of inertia and radius of gyration with respect to a polar centroidal axis of the cross section of a hollow tube whose outside diameter is 6 in. Check to see whether the area of the object is filled correctly. Hot Network Questions Count the number of triangles. After taking data for each run, click the "Velocity" graph (this is the ω(t) graph) to select the graph, then click. Polar moment of inertia (denoted here as I p ) can also be found by summing the x and y planar moments of inertia (I x and I y ). The moment of inertia is a physical quantity which describes how easily a body can be rotated about a given axis. The moment of inertia is a value that measures how difficult it is to change the state of an object's rotation. Slender Rod. It suggests that to turn the shaft at an angle, more torque is required, which means more polar moment of inertia is required. Moment of inertia (I) = mL 2 /12. The moment of inertia is a quantity that expresses a body’s tendency to resist angular acceleration from torque about a specified axis. Centripetal acceleration at point P: Tangential acceleration at point P: The distances of the masses to the rotation axis is same. Moment of inertia shows, it is not easy to rotate the stationary object; the same which means it is difficult to stop the rotating object. Basic Stress Equations Dr. You may notice this last equation is for a full cone, so in order to calculate the moment of inertia for our truncated cone, we need to perform one additional step. The free end of the rope is pulled with a constant force P for a distance of 5. Moment of Inertia Introduction: Moment of inertia of an object is an indication of the level of force that has to be applied in order to set the object, or keep the object, in motion about a defined axis of rotation. Figure $$\PageIndex{5}$$: Calculating the moment of inertia for a thin disk about an axis through its center. The accuracy of the calculations (and later on the accuracy of the measurements to verify the calculations) will depend entirely on the wisdom used in choosing the axes. The moment of inertia for a cylinder rotating about its main axis is NOT 0. Different objects have different moments of inertia. I would like to be able to find the rotational moment of inertia of an assembly (that is, to disegnate which parts move around the axis and get the moment for those parts together). In this section, we show how to calculate the moment of inertia for several standard types of objects, as well as how to use known moments of inertia to find the moment of inertia for a shifted axis or for a compound object. They will make you ♥ Physics. Solution 1 The figure shows that the appropriate mass element is a circular ring of radius and width dr. Moment of inertia When the arm is long and heavy, damage of internal parts may be caused due to inertia. So the moment of. Here, we can avoid the steps for calculation as all elemental masses composing the cylinder are at a xed (constant) distance "R" from the axis. Specify a direction for the load forces. The cardboard tube, in contrast to the can, is hollow. Solution 1 The figure shows that the appropriate mass element is a circular ring of radius and width dr. The cylinder is attached to the axle by spokes of a negligible moment of inertia. Mass moments of inertia have units of dimension mass length2. The moment-of-inertia is always calculated or measured with respect to a specified pivot point. We can obtain its rotational inertia I from the formula for the rotational inertia of an annular cylinder by substituting R1 # 0 and R2 # R. again, the the coordinate axis is welded as the mass center oriented as shown, and we have the XY, IXX and the IYY mass moments inertia are the same, and the IZZ moment of inertia, mass moment of inertia is different. What exactly is the area moment of inertia (also called the second moment of area)? It is a geometrical property of any area. What is the distance between the atoms?. half the value of the moment of inertia about the central axis to the value of the moment of inertia about the base plane. The moment of inertia is also called rotational inertia, mass moment of inertia, or polar moment of inertia of mass. MOMENT OF INERTIA Saddleback College Physics Department Purpose Given a hollow cylinder and either a solid sphere or a solid cylinder, measure the time it takes each to roll down an inclined plane and compare that time with the theoretical time for each object to go down the plane at two different angles. The bar has a moment of inertia I = 1/3 ML2 about the hinge, and is released from rest when it is in a horizontal position as shown. The moment of inertia is the quantitative measure of rotational inertia, just as in translational motion, and mass is the quantitative measure of linear inertia—that is, the more massive an object is, the more inertia it has, and the greater is its resistance to change in linear velocity. 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. This second moment is commonly called the moment of inertia and has a symbol I. List of moments of inertia. Derive an expression for moment of inertia of a cylinder about an axis passing through its centre of mass along its length (BY INTEGRATION) - Physics - System Of Particles And Rotational Motion. 66 × 1026 kg. Moment of inertia shows, it is not easy to rotate the stationary object; the same which means it is difficult to stop the rotating object. must treat the element as a thin rectang e. The cardboard tube, in contrast to the can, is hollow. As the rotator runs, with the ring tied to it, the string will be twisted and soon the ring will take up a horizontal position. Note: If you change the contour of the cross section, a new calculation of the moment of inertia is carried out automatically and the moment of inertia block is also updated. If a body has a hole drilled in it, the moment of inertia of the drilled body is equal to the moment of inertia of the original body minus the moment of inertia of the removed material. You may notice this last equation is for a full cone, so in order to calculate the moment of inertia for our truncated cone, we need to perform one additional step. If the body is divided into infinitely small pieces with mass dm, then the moment of inertia will be equal to the sum of the product of these elementary masses by the square of the. Figure $$\PageIndex{5}$$: Calculating the moment of inertia for a thin disk about an axis through its center. The process involves adding up the moments of infinitesmally thin cylindrical shells. The second moment of area for a shape is easier to be calculeted with respect to a parallel axis or with respect to a perpendicular axis through the centroid of the shape. Bodies do not display this force except when one alters their state; at which point it is called resistance or action. The Second Moment of Area I is needed for calculating bending stress. We can obtain its rotational inertia I from the formula for the rotational inertia of an annular cylinder by substituting R1 # 0 and R2 # R. Bending moment refers to the algebraic sum of all moments located between a cross section and one end of a structural member; a bending moment that bends the beam convex downward is positive, and one that bends it convex upward is negative. Mathematically, we describe the effect size and shape have on rotation with something called an object's moment of inertia, abbreviated I. 50 s if it is known to slow down from 1250 rpm to rest in exactly 1 minute. It is the rotational analog of mass. Similar to the centroid, the area moment of inertia can be found by either integration or by parts. A second cylinder, with moment of inertia 28. To do this you will first find the total moment of inertia of the system, which represents the combination of the disk's moment of inertia, I 1, and the moment of inertia of the ring or bar, I 2. Density of cylinder = (kg/m3) Height of cylinder = h (m) External and internal radii are R1 and R2 respectively Therefore, mass of cylinder = M = volume × density = ߨ ݄ (ܴଵଶെ ܴ ଶ ଶ) ߩ(kg). Moment of Inertia A vertical differential element of area is Cho. If ø63 is used within a pressure range from 0. Select the unit to convert to in the output units list. Where r is the distance between the axis of ratation and the volume dV. Example C3 2 Power Transmission Solid Mechanics I. Polar Moment of Inertia is utilized to calculate. The mass is simply the volume (πr 2 h) multiplied by the density of steel (7800 kg/m 3 ), which produces a result of 61. Polar Moment Of Inertia Cylinder Equation Tessshlo. Formula of moment of inertia is subject to load shape. Calculate the moment of inertia of a cylinder. One of the distinctions between the moment of inertia and mass (the latter being the measure of translational inertia) is that the moment of inertia of a body depends on the axis of rotation. Also called "Moment of Inertia". If ø63 is used within a pressure range from 0. 01 18-Jun-2003 1. 1 RADIUS OF GYRATION k All rotating machinery such as pumps, engines and turbines have a moment of inertia. The moment of inertia about an axis of a body is calculated by the summation of mr 2 for every particle in the body, where "m" is the mass of the particle and "r" is the perpendicular distance from the axis. 38 kg and radius 9. Moment of Inertia Tensor Consider a rigid body rotating with fixed angular velocity about an axis which passes through the origin--see Figure 28. For basic shapes there are tables that contain area moment of inertia equations which can be viewed below. The moment of inertia depends on the mass and shape of an object, and the axis around which it rotates. The accuracy of the calculations (and later on the accuracy of the measurements to verify the calculations) will depend entirely on the wisdom used in choosing the axes. Investigation 1: The Moment of Inertia Goals: • To study how two objects having the same mass can have dramatically different "resistances" to changes in rotational velocity (i. 25 cm and a mass of 750. The moment of inertia is the quantitative measure of rotational inertia, just as in translational motion, and mass is the quantitative measure of linear inertia—that is, the more massive an object is, the more inertia it has, and the greater is its resistance to change in linear velocity. inertia of the engine and the load. This article will discuss the concept of the area moment of inertia and polar moment of inertia and their application in practical problem solving. The equation specifying the proportionality is a rotational version of Newton’s second law: ˝ = I (1). Moment of inertia formula for perpendicular axes theorem-The sum of moment of inertia of a plane laminar body about two mutually perpendicular axes lying in its plane is equal to its moment of inertia about an axis passing through the point of intersection of these two axes and perpendicular to the plane of laminar type body. But under rotational acceleration, the moment of inertia of liquid becomes small compared to that of solid. So if I take I times u and it defers from U by a scale of factor, that scale of factor is the moment of inertia. For the I-shaped section, however, it is not possible to simply subtract the smaller rectangles from the larger, as was done when computing the moment of inertia about the x -axis, since the centroids of the various parts being subtracted do not coincide. 2 (b) Solid cylinder (or ring) about central axis Let the radius of the cylinder be R and its mass M. Moments of inertia simply add, so:. Moment of inertia of a circular section can be calculated by using either radius or diameter of a circular section around centroidal x-axis or y-axis. Ask Question Calculate the moment of inertia of the cylinder defined below when the cylinder is rotated. 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. The relationship between torque, moment of inertia and angular acceleration is given by: (Translating system equivalent:) Springs. Different objects have different moments of inertia. Question link: https://isaacphysics. 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. Instead it will be a combination of the mass moment of inertia of the rigid body and the distance the center of mass is from the axis of rotation. Bending moment refers to the algebraic sum of all moments located between a cross section and one end of a structural member; a bending moment that bends the beam convex downward is positive, and one that bends it convex upward is negative. Note) Use ø63 within a pressure range from 0. Note: If you change the contour of the cross section, a new calculation of the moment of inertia is carried out automatically and the moment of inertia block is also updated. One of the distinctions between the moment of inertia and mass (the latter being the measure of translational inertia) is that the moment of inertia of a body depends on the axis of rotation. The moment of inertia of an area with respect to any axis not through its centroid is equal to the moment of inertia of that area with respect to its own parallel centroidal axis plus the product of the area and the square of the distance between the two axes. 95 × 1046 kgm2. The resulting moment of inertia or centre of gravity when placing one or multiple loads can now be easily calculated. This manual describes the laboratory experiment used during the 1996 - 1997 academic year. Inertia is dependent on mass and the radius or length of the object and the axis of rotation. Integrating to find the moment of inertia of a two-dimensional object is a little bit trickier, but one shape is commonly done at this level of study—a uniform thin disk about an axis through its center (Figure $$\PageIndex{5}$$). (Last Updated On: December 20, 2017) Problem Statement: ECE Board April 1999. Now the moment of inertia of the object = kmr 2 , where k is a constant that depends on how the mass is distributed in the object - k is different for cylinders and spheres, but is the same for all cylinders, and the same for all spheres. Figure 1 The definition given in Eq. 1 cm 4 = 10-8 m 4 = 10 4 mm 4; 1 in 4 = 4. Specify a direction for the load forces. An object that is rotating tends to remain rotating and will continue to do so unless acted upon by an external net torque. com's Moment of Inertia Calculator is an online physics tool to measure the rotational inertia of different objects of most common shapes based on the mass distribution and their axis, in both US customary & metric (SI) units. It's equal to the mass multiplied by the sum of three times the square of the radius and the square of the height, divided by twelve. 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. Parallel Axes Theorem. If the cylinder is rotating around a horizontal axis (like a baton), then the water certainly contributes to the moment of inertia. The moment of inertia of the threshing cylinder is dependent on the power of the motor. Moment of Inertia, Version 1. 50 s if it is known to slow down from 1250 rpm to rest in exactly 1 minute. MOMENT OF INERTIA Saddleback College Physics Department Purpose Given a hollow cylinder and either a solid sphere or a solid cylinder, measure the time it takes each to roll down an inclined plane and compare that time with the theoretical time for each object to go down the plane at two different angles. It is the rotational analog of mass. 50 s if it is known to slow down from 1250 rpm to rest in exactly 1 minute. the relation between its length L and radius R is---. 1 Moment of inertia of the cylinders about the axis of rotation I E a) Measurement of the masses mass m 1 mass m 2 mean value m Smaller cylinder Bigger cylinder b) Measurements of diameters and radii mean valued 1 d 2 d r Smaller cylinder Bigger cylinder. Because the hoop is a relatively simple geometric shape, it is also possible to calculate its moment of inertia theoretically. Moment of inertia, which is a derivative of Newton’s second law, is sometimes referred to as the second. 150-m radius. Polar moment of inertia is sometimes denoted with the letter J, instead of I, but its units are the same as those for planar moment of inertia: m 4 or in 4. The moment of inertia of an object provides a measure of how hard it is to change that object’s rotational velocity. Moment of inertia, also called rotational inertia, mass moment of inertia, or polar moment of inertia of mass, is the mass property of a rigid body that determines the torque needed for a desired angular acceleration about an axis of rotation. In plastic modulus, the main concern is the point of deformation while the focus of moment of inertia is the speed of a particular object. Figure 1 The definition given in Eq. "A rope is wrapped around a 3 kg cylinder of radius 10 cm which is free to rotate about its axis. A second cylinder, this one having moment of inertia I2 and initially not rotating, drops onto the first cylinder (Fig. The moment of inertia is intimately linked to the definition of angular moment of a rigid body: For a rigid body rotating with angular velocity about a fixed axis, the angular momentum is. The can of jellied cranberry sauce is a solid cylinder. Centripetal acceleration at point P: Tangential acceleration at point P: The distances of the masses to the rotation axis is same. The equation specifying the proportionality is a rotational version of Newton’s second law: ˝ = I (1). Moment of Inertia--Rod The moment of inertia of a thin rod of length h , mass M , and cross-sectional area A can be computed as the limiting case of the moment of inertia of a cylinder as the radius , so the tensor becomes. Here are some of the most common moments of inertia: Solid cylinder or disk of radius r rotating about its axis of symmetry: Hollow cylinder of radius r rotating about its axis of symmetry: Solid sphere of radius r rotating about its center: Hollow sphere of radius r rotating about. 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. The polar moment of inertia, J, of a cross-section with respect to a polar axis, that is, an axis at right angles to the plane of the cross-section, is defined as the moment of inertia of the cross-section with respect to the point of intersection of the axis and the plane. Area Moment Of Inertia Typical Cross Sections I. 1 cm 4 = 10-8 m 4 = 10 4 mm 4; 1 in 4 = 4. Calculate the moment of inertia of a skater given the following information: (a) The 92. 50 s if it is known to slow down from 1250 rpm to rest in exactly 1 minute. Polar moment of inertia. Calculate (a) its moment of inertia about its center, and (b) the applied torque needed to accelerate it from rest to 2500. The moment of inertia of a hollow cylinder can be calculated with the help of the formula, I=0. This article will discuss the concept of the area moment of inertia and polar moment of inertia and their application in practical problem solving. In the case of a cylinder this integral will be:. The Moment of Inertia of the sphere. So what I'd like you to do is, for the cylinder, I'd like you to compute its moment of inertia around its central axis. Rotational inertia (moment of inertia) Hoop rotating about a central axis Define rotational inertia (moment of inertia) to be r i: the perpendicular dist. This Demonstration calculates the moment of inertia of a cylinder about its perpendicular axis, based on your parameter inputs. The polar moment of inertia on the other hand, is a measure of the resistance of a cross section to torsion with invariant cross section and no significant warping. Moment of Inertia, Version 1. 61 to 1 MPa, please use –X2071. So the moment of. However, if we found the moment of inertia of each section about some. Knowing the area moment of inertia is a critical part of being able to calculate stress on a beam. The moment of inertia of other shapes are often stated in the front/back of textbooks or from this guide of moment of inertia shapes. 25 cm and a mass of 750. Different objects have different moments of inertia. • That means the Moment of Inertia I z = I x +I y. Hot Network Questions Count the number of triangles. Figuring out the "y" distance is the hard part. Physics Wallah - Alakh Pandey 282,109 views. It is a mathematical property of a section concerned with a surface area and how. Inertia is dependent on mass and the radius or length of the object and the axis of rotation. | 2019-11-21T07:47:19 | {
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http://math.stackexchange.com/questions/39316/finding-the-limit-when-denominator-0/39322 | # Finding the limit when denominator = 0
I'm having trouble solving this limit:
$$\lim_{x \to -2^-} \frac{1}{(x + 2)^2}$$
I can't find a way to rationalize the denominator. Also, is there a way to do it without plugging in -2.001 and stuff or graphing it?
EDIT:
I realized after asking this question that it doesn't matter if you take the above limit from the right, left, or both. It's always $+\infty$. Here's an equation that gives $+\infty$ from the right and $-\infty$ from the left: $$\lim_{x \to 3^+} \frac{x - 4}{x - 3}$$
How do I (algebraically) determine if it is positive or negative?
-
1. There's nothing to rationalize. 2. If you plot your function, you find that the whole thing goes to $\infty$ at $x=-2$ (an essential singularity). – Ｊ. Ｍ. May 16 '11 at 1:26
yes, but the question is how do I solve it without plotting? how do I know that it goes to infinity and if it is positive or negative? – Caleb Jares May 16 '11 at 1:29
You know $\lim\limits_{x\to 0^-}\frac1{x^2}$, right? Try to turn your limit into that form. – Ｊ. Ｍ. May 16 '11 at 1:30
I did not know you could change limits like that. If I change $$\lim_{x \to -2^-}$$ to $$\lim_{x \to 0^-}$$ what must happen to the rest of the function? Is there a rule for this? – Caleb Jares May 16 '11 at 1:33
@carble729: as you approach -2 from the left, your denominator $(x+2)$ becomes of the same form as if you were approaching $0$ from the left with the denominator being $x$ (in both cases, the denominator goes to 0, and that's the important part). Hopefully it's more obvious now. – InterestedGuest May 16 '11 at 1:36
Maybe this way of thinking about it will seem a little more intuitive to you:
Let $\varepsilon > 0$, and consider the limit
$$\lim_{x \rightarrow -2^{-}} \frac{1}{(x+2)^2} = \lim_{\varepsilon \rightarrow 0} \frac{1}{((-2-\varepsilon)+2)^2} = \lim_{\varepsilon \rightarrow 0} \frac{1}{\varepsilon^2}.$$
Now the expression $\frac{1}{\varepsilon^2}$ can be made arbitrarily large by choosing $\varepsilon$ small enough, and so the limit does not exist.
-
Thank you, that does make sense (of course DNE being the same thing as + or - infinity - + in this case). When I asked this question, I didn't know that it was positive both ways. Can you explain how to get + or - infinity from the following problem? $$\lim_{x \to 3^+} \frac{x - 4}{x - 3}$$ – Caleb Jares May 16 '11 at 2:11
First, $$\lim_{x \rightarrow 3^+} \frac{x-4}{x-3} = \lim_{x \rightarrow 3^+} \frac{(x-3)-1}{x-3} = \lim_{x \rightarrow 3^+} 1 - \frac{1}{x-3}.$$ Then letting $\varepsilon > 0$, we have $$\lim_{x \rightarrow 3^+} 1 - \frac{1}{x-3} = \lim_{\varepsilon \rightarrow 0} 1 - \frac{1}{(3+\varepsilon) - 3} = \lim_{\varepsilon \rightarrow 0} 1 - \frac{1}{\varepsilon}.$$ Now since $\varepsilon > 0$, $\frac{1}{\varepsilon} > 0$, and this goes to $-\infty$. – Dane May 16 '11 at 2:23
This makes perfect sense! Thank you! – Caleb Jares May 16 '11 at 2:26
You asked for an algebraic solution and that's well-covered in other answers, but I'd like to offer a conceptual way of reasoning through it. Let's start with your revised example: $$\lim_{x \to 3^+} \frac{x - 4}{x - 3}$$ As $x$ gets close to $3$, the numerator is getting close to $-1$ and the denominator is getting close to $0$. Something nonzero (rather, not tiny-approaching-zero) divided by something tiny-approaching-zero is going to be big (far from zero), suggesting a limit of $\infty$ or $-\infty$ (that is, suggesting that the limit does not exist, but may be of the specific $\infty$ or $-\infty$ kind of does-not-exist).
Now, how do we tell whether it's $\infty$, $-\infty$, or neither? Well, the numerator of the fraction is getting close to $-1$, so it's negative. The denominator of the fraction is getting close to zero, but specifically as $x\to 3^+$, $x>3$, so $x-3>0$ and the denominator is positive. The fraction is the quotient of a negative number and a positive number, so it's negative and $$\lim_{x \to 3^+} \frac{x - 4}{x - 3}\to-\infty.$$
-
As $x \to -2^{-}$, your fraction becomes of form $\frac{1}{\epsilon}$, where $\epsilon$ is an arbitrarily small number, and 1 divided by a 'small' number is clearly a 'large' number (the smaller the denominator, the larger the value of the fraction). I hope you can see it from here.
-
I know that I could do it like that, but that's still plugging in values (albeit inside your head). I'm interested to find a way to solve it without plugging in numbers (even if it's in your head) or graphing it. It makes more sense to me if I can understand how the math works in an absolute sense. – Caleb Jares May 16 '11 at 1:36
@carble729: It's not really plugging in numbers -- I am assuming that you 'know' that limit of 1 over $x$ where $x$ goes to $0$ is equal to infinity. – InterestedGuest May 16 '11 at 1:37
Yes, but what if the problem is $$\lim_{x \to 3^+} \frac{x - 4}{x - 3}$$. Approached from the right, it is $-\infty$ and from the left, it is $+\infty$. How do I tell if it is positive or negative infinity without graphing it or plugging in numbers? – Caleb Jares May 16 '11 at 2:06
@cable: make a proper change of variables such that you're only considering limits of the form $u\to0^+$ or $u\to0^-$, as needed for your problem. – Ｊ. Ｍ. May 16 '11 at 2:20
@cable729: you consider the values: for every $x$ greater than $3$, $x-3$ is positive. For all $x$ between $3$ and $4$, $x-4$ is negative. So for $x$ sufficiently close to $3$, on the right, the fraction is negative (a negative number divided by a positive number). – Arturo Magidin May 16 '11 at 2:22
Given any $M > 0$, we have that $\forall x \in \left( -2 - \frac1{\sqrt{M}},-2 \right)$, we have $\frac1{(x+2)^2} > M$.
-
Thanks for the answer, but I'm having trouble understanding this (I'm reviewing for a Calc 1 final). I'm not sure what $\forall$ and $\in$ are. Also, I'm not sure what M stands for. – Caleb Jares May 16 '11 at 1:39
what it means is give me any positive number $M$, then if $x$ lies in the interval $\left( -2-\frac1{\sqrt{M}},-2 \right)$, then $\frac1{(x+2)^2} > M$. This implies that the fraction can be made arbitrarily large by getting "sufficiently close" to -$2$. $\forall$ stands for "forall" and $\in$ stands for "belongs to" – user17762 May 16 '11 at 1:41 | 2014-11-28T10:23:46 | {
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https://math.stackexchange.com/questions/2144738/prove-that-if-a-c-b-c-and-a-cap-c-b-cap-c-then-a-b/2144741 | # Prove that if $A – C = B – C$ and $A\cap C = B\cap C$ then $A = B$
I am trying to prove that if $A – C = B – C$ and $A\cap C = B\cap C$ then $A = B$. I have tried using Venn Diagrams as a proof technique, but we are not able to use proof by Venn Diagrams.
• Hint: For all sets U and V, we have $U = (U - V) \cup (U \cap V)$. – Nathan H. Feb 14 '17 at 22:45
By symmetry of the question, it suffices to show $A\subset B$. (The proof will work for $B\subset A$.) Let $x\in A$; we need to show $x\in B$.
Step 1: Suppose $x\in C$. Then $x$ is in the intersection $A\cap C$. Conclude with the hypothesis.
Step 2: Suppose $x\notin C$. Then $x\in A-C$. Again we can conclude easily.
• I was 15 seconds too late, but I was pleased to see your name on the answer. I'm pretty sure you were my Geometry TA in my first year at EPFL. Is that possible? :) – John Feb 14 '17 at 22:48
• Ahah maybe ;). (cf mon profil) :P C'est ton vrai prénom John? – Thibaut Dumont Feb 14 '17 at 22:49
• Non, c'est Giovanni (Tessinois). J'ai commencé en 2011! – John Feb 14 '17 at 22:54
Let $x\in A$. We have two cases:
• $x\in C$. Then $x\in A\cap C=B\cap C$, hence $x\in B$.
• $x\notin C$. Then $x\in A-C=B-C$, hence $x\in B$
Same goes for the opposite direction.
$A= (A \cap C) \cup (A \cap C^C) = (A \cap C) \cup (A-C) = (B \cap C) \cup (B -C) = (B \cap C) \cup (B \cap C^C)=B$
• $C^C$? What it is? – mle Feb 15 '17 at 1:33
• @mle $C^C$ is the set of objects not in $C$. – Bram28 Feb 15 '17 at 1:51
In a set theory proof of $A=B$, you must prove two statements:
1. If $x\in A$, then $x\in B$. (This proves $A\subseteq B$.)
2. If $x\in B$, then $x\in A$. (This proves $B\subseteq A$.)
Let's try to prove that if $x\in A$, then $x\in B$. So, we assume that $x\in A$, and our goal is to conclude that $x\in B$. Now, there are two cases, either $x\in C$ or $x\not\in C$. We can consider each case separately.
If $x\in C$, then since $x\in A$ was assumed, $x\in A\cap C$. Since $A\cap C=B\cap C$, then $x\in B\cap C$. Hence $x\in B$ and $x\in C$. Therefore, $x\in B$.
If $x\not\in C$, then since $x\in A$ was assumed, $x\in A-C$, but since $A-C=B-C$, $x\in B-C$. Therefore, $x\in B$ and $x\not\in C$. Therefore, $x\in B$.
Since either case leads to $x\in B$, we know that $x\in A$ implies that $x\in B$. Hence $A\subseteq B$.
Now, you must prove the other direction.
Let $x \in X$.
If $x \in A$ and $x \in C$ then $x \in A\cap C = B \cap C$ so $x \in A$.
if $x \in A$ and $x \not \in C$ then $x \in A \setminus C = B \setminus C$ so $x \in B$.
if $x \not \in A$ and $x \in C$ then $x \not \in A\cap C=B\cap C$. So either $x \not \in B$ or $x \not \in C$. But $x \in C$ so $x \not \in B$.
If $x \not \in A$ and $x \not \in C$ then $x \not \in A \setminus C$. So if $x \in B$ then $x \not \in C$ so $x \in B \setminus C = A \setminus C$. That's not possible so $x \not \in B$.
So $x \in A \iff x \in B$. So $A = B$
.... OR ..... $(D \setminus E)\cup (D \cap E) = \{x|x \in D; x \not \in E\} \cup \{x|x \in D; x \in E\} = \{x|x \in D;$ and either $x \in E$ or $x \not \in E\}=$ .... but all $x$ are either in $E$ or not in $E$ ... $= \{x|x \in D \} = D$.
Therefore $A = (A\setminus C) \cup (A\cap C) = (B\setminus C) \cup (B\cap C) = B$.
......
or
If $x \in A$ then if $x \in C$ $x \in A\cap C = B\cap C$ so $x \in B$.
If $x \not \in C$ then $x \in A \setminus C = B\setminus C$ so $x \in B$.
So $x \in A\implies x \in B$ so $A \subset B$.
Do the same to show $B \subset A$. So $A \subset B$ and $B \subset A$. So $A = B$. | 2020-08-05T11:19:51 | {
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https://math.stackexchange.com/questions/396562/estimating-the-sum-sum-k-2-infty-frac1k-ln2k | # Estimating the sum $\sum_{k=2}^{\infty} \frac{1}{k \ln^2(k)}$
By integral test, it is easy to see that $$\sum_{k=2}^{\infty} \frac{1}{k \ln^2(k)}$$ converges. [Here $\ln(x)$ denotes the natural logarithm, and $\ln^2(x)$ stands for $(\ln(x))^2$]
I am interested in proving the following inequality (preferrably using integral calculus) $$\sum_{k=2}^{\infty} \frac{1}{k \ln^2(k)}>2$$
By wolfram alpha, the actual value of the sum is about 2.10974. Since $\frac{1}{k \ln^2(k)}$ is decreasing, we have
$$\sum_{k=2}^{\infty} \frac{1}{k \ln^2(k)}\ge \int_2^{\infty} \frac{1}{x \ln^2(x)} dx=\frac{1}{\ln(2)}\approx 1.4427$$ So this lower bounded is weaker than desired.
My motivation for asking this question is that by being able to estimate this particular sum will hopefully teach me a general technique which I may try applying to sums of the form $\sum_{k=1}^{\infty} f(k)$.
Thanks!
• As a quick thought, I recommend shifting the index down one and estimating $\ln(1+x)$ as a polynomial using its Maclaurin series. – Jon Claus May 19 '13 at 17:58
• I'm afraid there's no closed form solution to this. – Inceptio May 19 '13 at 18:08
• @Inceptio I don't think that the OP is wanting the precise (closed form) sum, but would be happy enough to prove the inequality. Is that correct, Prism? – Namaste May 19 '13 at 18:24
• Numerical checking gives $$\frac{1}{2 \log^2(2)}+\frac{1}{3 \log ^2(3)}+\int_4^{\infty } \frac{1}{x \log ^2(x)} \, dx=2.03821$$ – Andrew May 19 '13 at 18:33
• @amWhy: Yes, that is correct :) – Prism May 19 '13 at 18:36
\begin{align*} \sum_{k=2}^{\infty} \frac1{k\ln^2k} &= \frac1{2\ln^22} + \frac1{3\ln^23} + \sum_{k=4}^{\infty} \frac1{k\ln^2k} \\ &\ge \frac1{2\ln^22} + \frac1{3\ln^23} + \int_4^{\infty} \frac{dx}{x\ln^2x} \\ &= \frac1{2\ln^22} + \frac1{3\ln^23} + \frac{1}{\ln4} > 2.038. \end{align*} (I see Andrew just wrote this in a comment.)
• I like this technique! Very simple and to the point. – Prism May 19 '13 at 18:38
• I had a tough time getting your answer, till I got back to the question to find log(x) denotes natural logarithm (+1). – Inceptio May 19 '13 at 18:41
• I think that we may not use such estimates for $\log 2, \log 3, \log 4$. It is obvious that from some $k\in \mathbb N$ our sum will be so small that we will have reached our target of $2$ with the terms out of the sum. Luckily here it happens with the first two terms. I mean, I would be satisfied if for example we reached a result involving less estimates. – Dimitris May 19 '13 at 18:46
• How does one prove that last result though. It can be done numerically easily, but so could the original problem. – Jon Claus May 19 '13 at 20:48
• There are lots of ways of approximating $\ln x$ arbitrarily closely: via its power series, for example, or with a Riemann sum approximating the appropriate integral. So getting upper bounds on $\ln 2$ and $\ln 3$ that are within 1% of the true values is not hard, and that will establish that the last expression exceeds $2$. – Greg Martin May 19 '13 at 23:03
I know the question is to show that the sum is greater than $2$, and Greg Martin's answer does that perfectly; however, I thought it might be interesting to compute the actual value of the sum.
Using the Euler-Maclaurin Sum Formula, we get \begin{align} &\sum_{k=2}^n\frac1{k\log(k)^2}\\ &=C-\frac1{\log(n)}+\frac1{2n\log(n)^2}-\frac1{12n^2}\left(\frac1{\log(n)^2}+\frac2{\log(n)^3}\right)\\ &+\frac1{360n^4}\left(\frac3{\log(n)^2}+\frac{11}{\log(n)^3}+\frac{18}{\log(n)^4}+\frac{12}{\log(n)^5}\right)\\ &\scriptsize\,-\,\frac1{15120n^6}\left(\frac{60}{\log(n)^2}+\frac{274}{\log(n)^3}+\frac{675}{\log(n)^4}+\frac{1020}{\log(n)^5}+\frac{900}{\log(n)^6}+\frac{360}{\log(n)^7}\right)\\ &\tiny\,+\,\,\frac1{50400n^8}\left(\frac{210}{\log(n)^2}+\frac{1089}{\log(n)^3}+\frac{3283}{\log(n)^4}+\frac{6769}{\log(n)^5}+\frac{9800}{\log(n)^6}+\frac{9660}{\log(n)^7}+\frac{5880}{\log(n)^8}+\frac{1680}{\log(n)^9}\right)\\ &\tiny\,-\,\,\frac1{1995840n^{10}}\left(\frac{15120}{\log(n)^2}+\frac{85548}{\log(n)^3}+\frac{293175}{\log(n)^4}+\frac{723680}{\log(n)^5}+\frac{1346625}{\log(n)^6}+\frac{1898190}{\log(n)^7}+\frac{1984500}{\log(n)^8}+\frac{1461600}{\log(n)^9}+\frac{680400}{\log(n)^{10}}+\frac{151200}{\log(n)^{11}}\right)\\ &+O\!\left(\frac1{n^{12}\log(n)^2}\right)\tag1 \end{align} Therefore, plugging $n=1000$ into $(1)$, we get \begin{align} \sum_{k=2}^\infty\frac1{k\log(k)^2} &=C\\ &=2.109742801236891974479257197616551326\tag2 \end{align}
• @Waiting: Congratulations! – robjohn Jun 13 '18 at 14:18
• robjohn, thank you!^^^ – user 1357113 Jun 13 '18 at 15:53 | 2019-09-22T00:12:55 | {
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https://www.physicsforums.com/threads/finding-maximum-and-minimum-values-of-3-dimensional-function.840688/ | # Finding maximum and minimum values of 3 dimensional function
1. Oct 31, 2015
### Inveritatem
1. The problem statement, all variables and given/known data
Find the maximum and minimum values of f(x,y,z) = x^2 - 2x + y^2 - 4y + z^2 - 4z in the region x^2 + y^2 + z^2 <= 36.
2. Relevant equations
Lagrange multipliers
<Fx, Fy, Fz> = lambda<Gx, Gy, Gz>
Test for local extrema
If Determinant D(x,y) < 0 it is a saddle point
If D(x,y) > 0:
If Fxx > 0, then it is a local minimum
If Fxx < 0, then it's a local maximum
3. The attempt at a solution
Tried to solve for system of equations that result in 4 equations
2x-2 = lambda * 2x
2y-4 = lambda * 2y
2z-4 = lambda * 2z
X^2 + y^2 + z^2 = 36
I solved for lambda = 1/2 or 3/2 by solving for x y and z and substituting into the 4th equation.
I got (2,4,4) =0 and (-2,-4,-4) =72. BUT the minimum was actually (1,2,2) = -9 because I was supposed to "check the interior of the sphere". Am I just supposed to account for the possibility that lambda = 0?
2. Oct 31, 2015
### Staff: Mentor
For the minimum you don't need Lagrange multipliers. Setting them to zero seems to work as well.
3. Oct 31, 2015
### Ray Vickson
For either the max or the min, the solution will either be in the interior or on the boundary. The interior stationary point is a global min in $\mathbb{R}^3$ and obeys the constraint, so is the global min of the constrained problem. (The second-order test implies that the unconstrained stationary point is a min, but you do not need such a fancy test: just use the fact that f = sum of squares + linear to give the result automatically.)
For the way you wrote the Lagrangian equations, a positive lagrange multiplier is a necessary condition for a MAXIMUM, so both of your candidate boundary points satisfy the first-order necessary conditions for a constrained maximum. The second-order *sufficient* conditions for a max are much trickier than you may think, because you have a point on a boundary of an inequality constraint. Basically, you need to project the Hessian of the *Lagrangian* (not the function f!) down to the tangent space of the constraint, and determine if it is negative definite in that subspace.
However, in this case you can by-pass all that because you have both possible points, so can just take the one with the larger f-value.
Last edited: Oct 31, 2015
4. Nov 1, 2015
### HallsofIvy
Staff Emeritus
In the interior of the sphere, $\nabla f= (2x- 2)\vec{i}+ (2y- 4)\vec{j}+ (2z- 4)\vec{k}$. Set that equal to 0 and solve for x, y, and z to find any critical points in the interior of the sphere. Your Lagrange multiplier method will solve for max or min among points that satisfy the equation x^2+ y^2+ z^2= 36, on the surface of the sphere. Another way to do that is to use parametric equations for the sphere: $x= 6cos(\theta)sin(\phi)$, $y= 6 sin(\theta)sin(\phi)$, $z= 6 cos(\phi)$. Putting those into the function, $f(\theta, \phi)= 36- 12cos(\theta)sin(\phi)- 24sin(\theta)si(\phi)- 24 cos(\phi)$. Set the gradient of that two variable function to 0 to find any critical points on the surface of the sphere. Find the value of the function at each critical point to determine the absolute max and min.
5. Nov 1, 2015
### Inveritatem
Sorry for the late reply. How does using the fact that the sum of the squares + linear establish that (1,2,2) is the global min? We were told to use geometry to show what the min and max are because we do not use advanced calculus to prove these mins and maxes? How can we use geometry/equations to prove these are the min and max values?
6. Nov 1, 2015
### Ray Vickson
The surface $w = x^2 - 2x + y^2 - 4y + z^2 - 4z$ is a bit hard to visualize, because it is a 3-dimensional object lying in 4 dimensions.
To gain insight, drop $z$ and look at the 2-dimensional surface $w = x^2 - 2x + y^2 - 4y$ in 3-dimensional $(x,y,w)$-space. This surface is cup-shaped, opening up as we go farther and farther out in the $(x,y)$ plane; you could fashion it out of sheet metal and it would very nicely hold rainwater. Its bottom is where all derivatives = 0, and there is only one such point. Now just imagine doing the same in a higher dimension.
Alternatively, you can re-write $f(x,y,z)$ by "completing the squares"; that will show the geometry very plainly.
BTW: I did not assume you would follow my remarks about Hessians projected down into tangent subspaces and the like; I just wanted to de-rail any thoughts you might have had about looking at the second derivatives in the way you did. In this case that works for the minimum (because the minimum is the same in the unconstrained problem and the constrained problem---that is, the constraint does not affect the minimum in this case), but those simple tests would lead you very far astray in the maximization case! They would also lead you astray in other cases of minimization where the constraint is active at the solution.
Last edited: Nov 1, 2015 | 2017-10-23T01:21:48 | {
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https://www.themathdoctors.org/function-transformations-revisited-ii/ | # Function Transformations Revisited (II)
#### (A new question of the week)
Last week we examined how a series of transformations affects the equation of a function, in order to write the equation from a graph, or vice versa. We touched on why it works the way it does, but this is something you need to look at from multiple perspectives in order to really grasp it fully. Luckily, some people don’t stop asking until they get it!
A month after the question we looked at last week, Mario wrote again, asking about the post from 2019 that I’d referred him to:
I have a question on something that was said there. Namely:
You can perform transformations in any order you want, in general. But in this case, you are asking in which order to do them in order to transform f(x) into a specific goal, f(ax+b). The order makes a difference in how you get there. What I do is to explicitly write the steps, one at a time. Suppose we first do the horizontal shrink f(x) -> f(ax). If we then apply a horizontal shift (translation) b units to the left, we would be REPLACING x in f(ax) with x+b, and we'd get f(a(x+b)). That is NOT what we are looking for; it's equal to f(ax+ab). So this order of doing those particular transformations is wrong.
The part that confuses me is if you do the horizontal shrink and then you do the horizontal shift, you end up with f(a(x + b)) and not f(ax + b). The reason it confuses me is because it doesn’t concur with how I have come to think about graph transformations.
Let’s say I have a function f(x), and I graph it. I shrink it horizontally by a factor of 1/3. The function now looks like f(3x), and the graph has gotten thinner. Now, let’s say I shift the graph 4 units to the right. I would think the graph would look like f(3x – 4). The reason I think it looks like that is because of the order I transformed the graph. Since I shrunk it first and then shifted, I would add -4 not to the set of inputs x, but to the set of inputs multiplied by 3, thus f(3x – 4). So my question is, what’s my mistake?
This is a very common mistake, and although in principle what I said last time should explain it, it takes several exposures, from different perspectives, to fully understand it, because it is counterintuitive, and your wrong intuitions have to be replaced by correct ways of thinking. (I’m reminded of my early post When Math Doesn’t Make Sense.) Students who pay more attention to memorized procedures than to reasons may just accept what they’re taught, but those who want to understand, like Mario (and me) need more! So I was happy to try again.
### A similar example from the 2019 post
I responded:
Hi, Mario.
It happens that I am currently working on turning our previous discussion on this subject into a post. If this answer works well, I’ll try to add it on, because it’s a good follow-up, and I can say some things I’d wished I said.
I did a very similar example to yours late in that post, in the section headed “Looking at the graph”, which you should read carefully.
There we get f(3x – 9) by first shifting (9 units right) and then shrinking (by 1/3). We do that by first replacing x with x – 9 to accomplish the shift, giving f(x – 9), and then replacing x with 3x to accomplish the shrink, resulting in f(3x – 9). I subsequently show that we can get the same thing in your order, shrinking and then shifting by only 3 units, and get f(3(x – 3)).
This differs from your example only in using 3 rather than 4.
There, and last week, I mostly just asserted that the transformed function is constructed by replacing x with something. We want to move beyond mere assertion to answer the deeper “why“.
First, here are the graphs from that post. When we shift first, resulting in the purple graph, and then shrink to make $$f(3x-9)$$, the blue graph, it looks like this:
On the other hand, if we shrink first, to make the purple graph, then shift to make $$f(3(x-3))$$ in blue, we have:
That shows that the result is right (especially if you take the time to verify points on each graph). But this time, I don’t want to depend on graphs to see merely what is right; our goal is to understand the underlying algebra, to see why it is right. We need to have more to say than, “If you don’t believe me, trust the graph.”
## Why we replace x
To answer the present question, I referred back to an earlier post that was just about individual transformations, which laid the foundation for combinations of transformations.
Back to your transformations: If you first shrink by 1/3 and then shift right 4 units, you change f(x) first by replacing x with 3x, making it f(3x), and then replace x with x – 4, making f(3(x – 4)). What you said, giving f(3x – 4), is wrong, because you are thinking “forward” rather than “backward”, and we must do everything about horizontal transformations “backward”!
I discussed this “backward” idea in a one-transformation-at-a-time setting in the post before the one you’re reading,
Shifting and Stretching Graphs
There, too, I emphasized the idea of replacement, and explained why we must think of it this way using several different approaches.
In the last post, I used replacement of $$x$$ with $$cx$$ or $$x-c$$ as the foundation, but failed to explain why that was correct. To fill that gap, we can follow a specific point through the successive transformations:
Let’s look more closely at your example, and see how it works. As I did in the “Looking at the graph” section, I’ll take the original function to be f(x) = x2, and use g for the transformed function.
Now, let’s consider a particular point on the original graph, say (3, 9). Applying your transformations to this point in order, we first shrink horizontally, which divides x by 3 and brings us to the point (1, 9). Now we shift 4 units to the right, adding 4 to x and taking it to (5, 9). So our transformed function g should yield g(5) = 9.
First, we can test both claimants to the title of transformed function. First, yours, f(3x – 4):
g(5) = f(3(5) – 4) = f(11) = 112 = 121
That didn’t work!
Now, mine, f(3(x – 4)):
g(5) = f(3(5 – 4)) = f(3) = 32 = 9
That did work. So we’ve demonstrated that your work was wrong.
### How it works: solving for x
We’ve now seen, in several ways, that his intuition is wrong; we need to retrain that intuition by seeing clearly what really happens.
Now, why did mine work? Look at what happened: We needed to transform the new x, 5, to the original x, 3, so that f(3) would give us the correct value of y.
We obtained 5 from 3 by applying the transformations in order: starting with 3, dividing by 3 to get 1, and adding 4 to get 5. That is, the new input to the function is x’ = (1/3)*3 + 4 = 5. In general, a point (x, y) transforms to (x’, y) = ((1/3)x + 4, y).
Since what I called x’ is the input to the new function, g, corresponding to the input of x to the original function, f, we want
g(x’) = f(x),
that is, we want
g((1/3)x + 4) = f(x).
But we want an expression for g(x), so we have to solve for x in terms of x’:
x’ = (1/3)x + 4
x’ – 4 = (1/3)x
3(x’ – 4) = x
So our new function is
g(x’) = g((1/3)x + 4) = f(x) = f(3(x’ – 4))
Here, x’ is a dummy variable that we just used as a temporary name; replacing it with the usual x, we have
g(x) = f(3(x – 4))
That, of course, is what I got by first replacing x with 3x, and then replacing x in that with (x – 4).
And this is why everything involving horizontal transformations is backward: We’re really solving for the original x, which means undoing the operations, and doing that in reverse order.
In vertical transformations, like $$h(x) = 3f(x) + 4$$, we are just evaluating an expression, and follow the order of operations: Multiply y by 3, then add 4; that is, stretch by a factor of 3, then shift up 4. But in the horizontal transformation, we are not evaluating but solving, which reverses everything.
So, to get back from our new x, 5, to the original x = 3, we first undo the shift right by 4, moving left by 4 to x = 1; and then we undo the shrink by 1/3, stretching by 3 from x = 1 to x = 3. That’s what puts the (x – 4) inside parentheses.
How does that work for you?
## Transforming coordinates
Mario replied, restating my ideas in order to understand them better:
Interesting, I let this ruminate for a while, and I found what may be an alternate way of thinking about it.
You start with a function f(xf), and you want to find how to alter it in order to get to g(xg), using only horizontal transformations of course, and assuming g(xg) is only a horizontal transformation of f(xf).
However, we must do this under the constraint that the inputs of both of the functions are the same. In summary, we have to map f(xf) → g(xg) under the constraint of them having the same input. Currently, xf ≠ xg because the same input results in different outputs for f and g.
The way we would solve this is by considering the information that we have been given. We know that xg = ⅓ xf + 4. This equation is extracted from “horizontal shrink by ⅓” and “shift 4 spaces to the right”. If we solve for xf, we get xf = 3(xg – 4). We plug that into f, and we get f(3(xg – 4)) = g(xg). Notice that they both have xg as an input, satisfying our constraint.
What do you think of this way of thinking about it?
I think that’s basically right, but needs a little clarification.
First, we can perhaps more clearly describe your “horizontal transformations” by saying that that xg is a linear function of xf.
Second, where you say,
However, we must do this under the constraint that the inputs of both of the functions are the same. In summary, we have to map f(xf) → g(xg) under the constraint of them having the same input. Currently, xf ≠ xg because the same input results in different outputs for f and g.
I don’t think you literally mean that the inputs (xand xg) are the same; and I’m not sure how to say what you mean by a map of a function. I would express the idea by saying that f(xf) = g(xg) = g(T(xf)), where T is a linear transformation of the input variable, i.e. T(x) = ax + b. This makes it explicit how the functions are related.
The idea that we are transforming the variable, rather than the function, is a good perspective. You can imagine sliding and stretching the coordinate system itself while keeping the graph the same. In higher math this can be a valuable way to see coordinate transformations.
### Inverting the transformation
Then you said,
The way we would solve this is by considering the information that we have been given. We know that xg = ⅓ xf + 4. This equation is extracted from “horizontal shrink by ⅓” and “shift 4 spaces to the right”. If we solve for xf, we get xf = 3(xg – 4). We plug that into f, and we get f(3(xg – 4)) = g(xg). Notice that they both have xg as an input, satisfying our constraint.
This nicely fits into what I just said. The example has T(x) = ⅓ x + 4, so that xg = T(xf) = ⅓ xf + 4. In solving, you are finding the inverse function, xf = T-1(xg) = 3(xg – 4).
Therefore, the requirement that f(xf) = g(T(xf)) implies that g(xg) = f(T-1(xg)) = f(3(xg – 4)).
If you’ve done enough with inverses in general, you may recognize that if T is a stretch/shrink A followed by a shift B, so that T(x) = B○A(x) = B(A(x)) = (⅓)x + 4, then the inverse is T-1(x) = A-1○B-1(x) = A-1(B-1(x)) = 3(x – 4), which fully explains the reversal of order. To undo a shrink followed by a shift, we have to undo the shift and then the shrink.
Does this express what you had in mind?
When you find the inverse of a composite function, it is equivalent to the composition of the individual inverses, in the reverse order. More generally, $$(f\circ g)^{-1} = g^{-1}\circ f^{-1}$$. (That is, if $$y=f(g(x))$$, then $$x = g^{-1}(f^{-1}(y))$$; so if $$h(x)=f(g(x))$$, then $$h^{-1}(x) = g^{-1}(f^{-1}(x))$$.) And this, in turn, is the same idea as when we solve equations, which I discussed in an early post, Why We Care About “Why”.
Mario agreed:
Yes, that is a better way of putting what I wanted to say. I agree, “f(xf) = g(xg) = g(T(xf)), where T is a linear transformation of the input variable” is a much better way of describing my idea rather than “map f(xf) → g(xg).”
With that, I was able to tell Mario,
I think I’ve said what I had in mind. Thanks for giving me the chance to say it!
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2022-10-03T08:04:41 | {
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https://math.stackexchange.com/questions/3341021/what-is-the-domain-of-frac1-sqrtx-x-where-x-denotes-the-greatest-in | # what is the domain of $\frac{1}{\sqrt{x-[x]}}$ where [x] denotes the greatest integer function and find the range .
what is the domain of $$\frac{1}{\sqrt{x-[x]}}$$ where [x] denotes the greatest integer function and find the range .
My approach :
since [x] greatest integer function is discontinuous on all integral value , therefore the domain of this function will be $$R^+ -\{Z\}$$ where Z is integer and $$R^+$$ is all real positive numbers. but answer is $$R -\{Z\}$$ how all real numbers are possible here. please suggest thanks....
• The domain doesn't have anything to do with discontinuities per se. Sep 1 '19 at 13:55
• If $x=-\frac52$, what in your opinion is the value of $x-[x]$? Sep 1 '19 at 13:55
• -2.5 -(-3) = .5 , ok got it.. thanks.. domain is all Real numbers what about range of this function Sep 1 '19 at 14:02
• Since $[x] \leq x$, $x-[x] \geq 0$, so the expression can define a function wherever the denominator is not zero (i.e. whenever $[x] \neq x$).
– Joe
Sep 1 '19 at 14:02
• To find the range, think about what values the denominator can equal.
– Joe
Sep 1 '19 at 14:09
Let $$f(x)=\dfrac 1{\sqrt{\{x\}}}$$ where $$\{x\}$$ designates the fractional part, it belongs to $$[0,1)$$.
This comes from the definition of integer part (LHS below): $$\lfloor x\rfloor\le x<\lfloor x\rfloor+1\implies 0\le\{x\}=x-\lfloor x\rfloor<1$$
In particular $$\{x\}\ge 0$$, even for negative numbers, so $$\sqrt{\{x\}}$$ is defined everywhere.
First of all notice that your function is $$1$$-periodic so you can study it on $$[0,1]$$.
Indeed $$\{x+1\}=x+1-\lfloor x+1\rfloor=x+1-(\lfloor x\rfloor+1)=x-\lfloor x\rfloor=\{x\}$$ so $$f$$ is $$1$$-periodic as well.
As you noticed $$\{x\}=0$$ whenever $$x$$ is an integer, so the local domain is $$(0,1)$$ and the global domain extended by periodicity is $$\mathbb R\setminus\mathbb Z$$.
For the range since on $$(0,1)$$ we have $$0<\{x\}<1$$ then we get $$f(x)>1$$ and the range is thus $$(1,+\infty)$$.
Since $$x\geq [x]$$ then $$x-[x]\geq 0$$ the equality when $$x$$ is integer. The greatest integer function maps integers into integers and non integers to the closet least integer. | 2021-11-29T17:33:17 | {
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http://openstudy.com/updates/5092b5d1e4b0b86a5e52c54a | ## frx 3 years ago Let a>0 (a). Show using L'Hospitals rule that: $\lim_{x \rightarrow 0} x ^{a}\ln(x)=0$ (b). By setting x=ln(t) in (a). show that: $\lim_{x \rightarrow -\infty} |x|^{a} e ^{x}=0$ I need help with (b). I don't get the question, how do I show that (b). is true by setting x=ln(t) in (a). it doesn't make any sense to me.
1. sirm3d
the proposed substitution
2. sirm3d
$x=\ln t$ means that $t=e^x$
3. frx
I've got that but how can I use it?
4. sirm3d
as $x \rightarrow -\infty, t \rightarrow ?$
5. frx
Zero?
6. frx
e^-infinty = 0
7. sirm3d
correct
8. sirm3d
so we shift from
9. sirm3d
$\lim_{x \rightarrow -\infty}$ to $\lim_{t \rightarrow 0^+}$
10. sirm3d
in terms of t, $\left| x \right|^a = ?$
11. frx
$|\ln (t)|^{a}$ ?
12. sirm3d
that's right. and what about $e^x$
13. frx
We could set it as just t, right?
14. sirm3d
right
15. frx
$\lim_{t \rightarrow 0} |\ln(t)|^{a}t$
16. frx
Because t->0 shouldn't the whole expression be equal to 0 then?
17. sirm3d
not necessarily
18. sirm3d
what about $\left| \ln t \right|^a$
19. frx
$\ln(t) \rightarrow -\infty$ but since it's the abolute value its just $\infty$
20. sirm3d
$\left| \ln t \right|^a \rightarrow +\infty$. you are correct
21. sirm3d
the transformed problem is an indeterminate form $0*\infty$ which should be resolved either as $\frac{ 0 }{ 0 }$ or $\frac{ \infty }{ \infty }$
22. frx
Right, so I could tranform it to the right form and use L'Hospitals rule to get the answer correct
23. frx
Should I use the form $\frac{ t }{ |\ln(t)|^{-a} }$ and then go for L'Hospital?
24. sirm3d
25. frx
Just as it was, is it possible to use L'hostial on multiplication form?
26. sirm3d
lhopitals rule must be in quotient form
27. frx
Which form are you thinking about then?
28. sirm3d
29. frx
$-\frac{ a|\ln(t)|^{-a-1} }{ t }$ Can that be the correct derivation? If so it should give us the right answer $\lim_{t \rightarrow 0}\frac{ 1 }{ -\frac{ a|\ln(t)|^{-a-1} }{ t } } = \frac{ 1 }{- \infty }=0$ Can this be right?
30. sirm3d
we'll look at the 2nd quotient form $\frac{ \left| ?\ln t \right|^a }{ \frac{ 1 }{ t } }$, noting that a>0
31. sirm3d
since t->0+, $\left| \ln t \right|=-\ln t$
32. frx
Right
33. sirm3d
the indeterminate form is inf/inf. so we'll apply l'hop's rule once
34. sirm3d
i hate getting disconnected in the middle of the equation
35. frx
$\frac{ \frac{ -a \ln ^{a-1}(t) }{t } }{\frac{ -1 }{ t ^{2} } }$
36. sirm3d
right. when simplified becomes $\frac{ a(-\ln t)^{a-1} }{ 1/t }$
37. sirm3d
here's the catch to the problem
38. sirm3d
$(-\ln t)^a \rightarrow +\infty$ for as long as the exponent a > 0.
39. frx
So it's still $\frac{ \infty }{ \infty }$
40. sirm3d
right. but sooner or later, the exponent becomes negative after successive applications of l'hopital's rule
41. frx
$\frac{ (a-1)a(-\ln(t))^{a-2} }{ 1/t }$
42. sirm3d
after k applications of the rule, the indeterminate form is now $a(a-1)(a-2)...(a-k+1)\frac{ (-\ln t)^{a-k} }{ 1/t }$ where a-k is negative
43. sirm3d
so that $(-\ln t)^{a-k} \rightarrow 0$ because a-k is negative
44. sirm3d
the quotient form is now $a(a-1)(a-2)...(a-k+1)\frac{ 0 }{ \infty }$ and the problem is solved!
45. frx
Oh, that's clever, I get it
46. frx
Thank you so much for you're help I can't thank you enought this one has been on my mind for a couple of days now! | 2015-11-28T11:28:38 | {
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https://math.stackexchange.com/questions/2052805/open-and-closed-mapping-examples | # Open and Closed mapping Examples
I am looking for three mappings f:X to Y any set of topology on X or Y. so very flexible. Can you help me find an example of a function that is (a) continuous but not an open or closed mapping (b) open but not closed or continuous (c) closed but not open of continuous
Thank you all
HINT: If $X$ has the discrete topology, every function from $X$ to $Y$ is continuous, and every subset of $X$ is both open and closed. If you choose $Y$ so that it has a subset that isn’t open and a subset that isn’t closed, it’s not hard to get your first example. (You can even take $Y$ to be $X$ with a different topology.)
Suppose that $f:X\to Y$ is an open bijection; then it’s not hard to show that $f$ is also closed. Essentially the same argument shows that a closed bijection is always open. Thus, for the other two examples we cannot use bijections. It turns out that we can use an injection, though.
Let $X=\{0,1\}$ with the indiscrete topology, let $Y=\{0,1,2\}$, with the increasing nest topology
$$\tau=\big\{\varnothing,\{0\},\{0,1\},Y\big\}\;,$$
and let $f(0)=0$ and $f(1)=1$; this will work for one of the other two examples, and I’ll leave it to you to work out which one it is. Finally, you can use $X$ and $Y$ and a different injection from $X$ to $Y$ to get the remaining example.
• Thank you so much. As f(X)={0,1} in that example this maps an open set to an open set and there are no other open sets it must be an open mapping? and is not a closed mapping using the same logic as X is closed and it maps to an open set. sorry for the comment I am just trying to make sure my reasoning is okay. – Jm222222 Dec 10 '16 at 21:38
• @Jm222222: You’re right about the open sets: the only open sets in $X$ are $\varnothing$ and $X$, and they both get sent to open sets in $Y$. However, you need to do a bit more to show that $f$ is not a closed map. It’s not enough to show that it takes a closed set to an open set, because that open set might also be closed. (Remember, a set can be both open and closed.) However, it’s true that $f[X]=\{0,1\}$ is not closed in this case, because $Y\setminus\{0,1\}=\{2\}$ is not open in $Y$. – Brian M. Scott Dec 10 '16 at 21:44
• what is wrong with the function that maps the whole real line to a subset of R whose range is finite of numbers in the real line? If the target subset is a single value then thats its a continuous function, otherwise its not...right? – Pinocchio Jul 9 '18 at 19:02
• Why is "If X has the discrete topology, every function from X to Y is continuous" correct? To my understanding in the discrete topology if $p$ is the limit point then the distance between $f(x)$ and $f(p)$ will always be 1 unless $f(x)=f(p)$. Therefore, every epsilon less than 1 won't work and thus the every function on the discrete metric is not continuous. Where did I go wrong? – Pinocchio Jul 10 '18 at 15:41 | 2019-05-19T09:16:02 | {
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https://bobcat.bg/i7v2al/d97102-set-operations-proofs | # set operations proofs
Let x be an arbitrary element in the universe. \mathbf{R} = \mathbf{Q} \cup \overline{\mathbf{Q}}\,.\], Written $$A\cap B$$ and defined if and only if B and . (See section 2.2 example 10 for that. Back to Schedule I think either of those are valid proofs. This is a contradiction, so $$|A-B|\le|A|$$.∎. Thus, in particular, x ∈ A is true. $A\cup B\cup C \cup D\,,\\A\cap B\cap C \cap D\,.$. e.g. Try to prove and . By definition of set difference, x ∈ A− B. B Then since by the definition of . $\bigcup_{i=1}^{n} S_i\,.$ Hence . Thus $$A-B\not\subseteq A$$. when we're working with real numbers, probably $$U=\mathbf{R}$$. For example, (b) can be proven as follows: 6. Alternative proof We have used the choose-an-element method to prove Propositions 5.7, 5.11, and 5.14. though they can be proven also using some of these properties (after those properties are proven, needless to say). 11. by the definition of . Hence . &= \{x\mid \neg(x \in A)\wedge \neg(x\in B )\} \\ A For any one of the set operations, we can expand to set builder notation, and then use the logical equivalences to manipulate the conditions. Next -- Recursive Definition Hence . A-(B\cup C) x This section contains many results concerning the properties of the set operations. We can use the set identities to prove other facts about sets. A. B ) and implications Furthermore a similar correspondence exists between &= \{x\mid x\notin (A\cup B)\} \\ &= \{x\mid x \in (\overline{A}\cap\overline{B}) )\} \\ \] The “more formal” version has more steps and leaves out the intuitive reason (that might help you actually remember why). ------- Commutative Laws Be careful with the other operations. and between U and Theorem: For any sets, $$|A\cap B|\le|A|$$ and $$|A\cap B|\le|B|$$. Then . Proof for 9: Let x be an arbitrary element in the universe. ( cf. ) by the definition Here is an example. There's more to it than similar-looking symbols. A \overline{A\cup B} and vice versa. We'll be careful for this one and manipulate the set builder notation. . , As an example, we can prove one of De Morgan's laws (the book proves the other). Let the sets $$S_1,S_2,\ldots ,S_n$$ be the students in each course. If , then and . B . Since we're doing the same manipulations, we ended up with the same tables. With similar proofs, we could prove these things: When doing set operations we often need to define a. equalities involving set operations intersection of sets subset relations proofs of equalities proofs of subset relations Contents . = ( A ) by the definition of ( B - A ) . Proof for 4: A Since (use "addition" rule), \{1,2,3,4\}-\{3,4,5,6\} = \{1,2\}\,\\ A. by the commutativity of For example: Those identities should convince you that order of unions and intersections don't matter (in the same way as addition, multiplication, conjunction, and disjunction: they're all commutative operations). B ( A ------- De Morgan's Laws A We have already proved some of the results. 7. 12. if and only if and &= (A-B)\cap (A-C)\,.\quad{}∎ The if part can be proven similarly. A Theorem For any sets A and B, B ⊆ A∪ B. Proof… x ------- Domination Laws the commutativity of If , then . $$A\cup{U}= {U}\\A\cap\emptyset= \emptyset$$, $$(A\cup B)\cup C = A\cup(B\cup C)\\(A\cap B)\cap C = A\cap(B\cap C)$$, $$A\cup(B\cap C) =(A\cup B)\cap(A\cup C)\\A\cap(B\cup C) = (A\cap B)\cup(A\cap B)$$, $$\overline{A\cap B}=\overline{A} \cup \overline{B}\\\overline{A\cup B}= \overline{A} \cap \overline{B}$$, $$A\cup(A\cap B) = A \\ A\cap(A\cup B) = A$$, $$A\cup\overline{A} = {U}\\A\cap\overline{A} = \emptyset$$, Written $$A\cup B$$ and defined Hence A satisfies the conditions for the complement of . Set. B, by 7 &= A\cap \overline{B\cup C} \\ \end{align*}. Return to the course notes front page. 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. 3. Since , . Using set-builder notation, we can define a number of common sets and operations. 4. Here are some basic subset proofs about set operations. Since A x from the equivalences of propositional logic. The students taking, This is exactly analogous to the summation notation you have seen before, except with union/intersection instead of addition: 1 - 6 directly correspond Be careful with the other operations. B B If we need to do union/intersection of a lot of things, there is a notation like summation that is used occasionally. Often not explicitly defined, but implicit based on the problem we're looking at. and that of The set $$\overline{B}$$ is the set of all values not in $$B$$. We are going to prove this by showing that every element that is in Others will be proved in this section or in the exercises.
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http://math.stackexchange.com/questions/324261/calculus-practice-problem/324273 | # Calculus practice problem
My teacher assigned each student a practice problem yesterday for the new section we are starting, but I was absent, so I missed his explanation of how to do the problems. Can anyone explain to me how to solve it and provide the answer so I can practice it?
The problem is Suppose that $F(x)=f(\sqrt{1+x^2})$ such that $f'(\sqrt2)=\sqrt2/2$. Compute $F'(1)$.
-
Please check that I interpreted everything correctly. Also, did you mean $F'(1)$ at the end? – Brian M. Scott Mar 8 '13 at 2:11
What you missed is pretty fundamental. It is called the chain rule: see here. – 1015 Mar 8 '13 at 2:12
Yes, I have to compute F'(1). But I missed the explanation of the "chain rule". I'm notoriously bad at math. – sucksatmath Mar 8 '13 at 2:24
In addition to the wiki link julien gave, I highly suggest looking at Khan Academy's videos on the chain rule. They're pretty good, and you should be able to do this problem after watching them. If you still need help, let us know. – apnorton Mar 8 '13 at 2:25
The chain rule states that $$\frac{d}{dx} f(g(x)) = g'(x)f'(g(x)).$$
Applying this here, we have
$$F'(x) = [\frac{d}{dx} \sqrt{x^2+1}] [f'(\sqrt{x^2+1})] = \frac{x}{\sqrt{x^2+1}} f'(\sqrt{x^2+1}).$$
Using $x=1$, we have
$$F'(1) = \frac{1}{\sqrt{2}} f'(\sqrt{2}) = \frac{1}{\sqrt{2}} \frac{\sqrt{2}}{2} = \frac{1}{2}.$$
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You guys are great here! Thanks so much! – sucksatmath Mar 8 '13 at 2:34
By the chain rule, $$F'(x)=f'(\sqrt{x^2+1})\frac{d}{dx}\sqrt{x^2+1}.$$ By the chain rule again, $$\frac{d}{dx}\sqrt{x^2+1}=\frac{1}{2}(x^2+1)^{-1/2}2x=\frac{x}{\sqrt{x^2+1}}.$$ So $$F'(x)=f'(\sqrt{x^2+1})\cdot\frac{x}{\sqrt{x^2+1}}.$$ Now make $x=1$.
- | 2015-11-25T22:28:57 | {
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http://math.stackexchange.com/questions/45089/what-is-the-length-of-a-sine-wave-from-0-to-2-pi/45092 | # What is the length of a sine wave from $0$ to $2\pi$?
What is the length of a sine wave from $0$ to $2\pi$? Physically I would plot $$y=\sin(x),\quad 0\le x\le {2\pi}$$ and measure line length.
I think part of the answer is to integrate this: $$\int_0^{2\pi} \sqrt{ 1 + (\sin(x))^2} \ \rm{dx}$$
Any ideas?
-
Does the dx math-e-magically escape the sqrt? – philcolbourn Jun 13 '11 at 12:04
Thanks everyone. – philcolbourn Jun 17 '11 at 12:07
Can someone answer with a simple number? I need this to know how much paint to buy to paint my corrugated sheet roof. – SF. Aug 23 '13 at 19:46
FWIW, I've created a JS demo showing numerical integration of this function, with the purpose of evenly spacing points along the curve (as opposed to along the X axis). jsfiddle.net/fp7aknoc – Alnitak Oct 7 '14 at 7:22
I'm nowhere near a computer with elliptic integrals handy, so I'll give the explicit evaluation of
$$\int_0^{2 \pi} \sqrt{1+\cos^2 x}\,\mathrm dx$$
Note that an entire sine wave can be cut up into four congruent arcs; we can thus consider instead the integral
$$4\int_0^{\pi/2} \sqrt{1+\cos^2 x}\,\mathrm dx$$
(alternatively, one can split the integral into four "chunks" and find that those four chunks can be made identical; I'll leave that manipulation to somebody else.)
Now, after some Pythagorean manipulation:
$$4\int_0^{\pi/2} \sqrt{1+\cos^2 x}\,\mathrm dx=4\int_0^{\pi/2} \sqrt{2-\sin^2 x}\,\mathrm dx$$
and then a bit of algebraic massage:
$$4\sqrt{2}\int_0^{\pi/2} \sqrt{1-\frac12\sin^2 x}\,\mathrm dx$$
we then recognize the complete elliptic integral of the second kind $E(m)$
$$E(m):=\int_0^{\pi/2}\sqrt{1-m\sin^2u}\mathrm du$$
(where $m$ is a parameter):
$$4\sqrt{2}E\left(\frac12\right)$$
As Robert notes in a comment, different computing environments have different argument conventions for elliptic integrals; Maple for instance uses the modulus $k$ (thus, $E(k)$) instead of the parameter $m$ as input (as used by Mathematica and MATLAB), but these conventions are easy to translate to and from: $m=k^2$. So, using the modulus, the answer is then $4\sqrt{2}E\left(\frac1{\sqrt 2}\right)$.
Now to address the noted equivalence for negative parameter and a parameter in the interval $(0,1)$ by Henry, there is what's called the "imaginary modulus transformations"; the DLMF link gives the transformation for the incomplete case, but I'll explicitly do the complete case here for reference since it's not too gnarly to do (all you have to remember are the symmetries of the trigonometric functions):
Letting $E(-1)=\int_0^{\pi/2}\sqrt{1+\sin^2 u}\,\mathrm du$, we then go this way:
$$\int_0^{\pi/2}\sqrt{1+\sin^2 u}\,\mathrm du=\int_{-\pi/2}^0\sqrt{1+\sin^2 u}\,\mathrm du$$
$$=\int_0^{\pi/2}\sqrt{1+\sin^2\left(u-\frac{\pi}{2}\right)}\,\mathrm du=\int_0^{\pi/2} \sqrt{1+\cos^2 u}\,\mathrm du$$
from which I've shown what you're supposed to do earlier.
Computationally, the complete elliptic integral of the second kind isn't too difficult to evaluate, thanks to the arithmetic-geometric mean. Usually, this method is used for computing the complete elliptic integral of the first kind, but the iteration is easily hijacked to compute the integral of the second kind as well.
Here's some C(-ish) code for computing $E(m)$:
#include <math.h>
double ellipec(double m)
{
double f, pi2, s, v, w;
if (m == 1.0)
return 1.0;
pi2 = 2.0 * atan(1.0);
v = 0.5 * (1.0 + sqrt(1 - m));
w = 0.25 * m / v;
s = v * v;
f = 1.0;
do {
v = 0.5 * (v + sqrt((v - w) * (v + w)));
w = 0.25 * w * w / v;
f *= 2.0;
s -= f * w * w;
} while (abs(v) + abs(w) != abs(v))
return pi2 * s / v;
}
(make sure either your compiler does not (aggressively) optimize out the while (abs(v)+abs(w) != abs(v)) potion, or you'll have to use a termination criterion of the form abs(w) < tinynumber.)
Finally,
"I am also puzzled: a circle's circumference is $2\pi r$ and yet an ellipse's is an infinite series - why?"
My belief is that we are actually very lucky that the arclength function for a circle is remarkably simple compared to most other curves, the symmetry of the circle (and thus also the symmetry properties of the trigonometric functions that can parametrize it) being one factor. The reduction in symmetry in going from a circle to an ellipse means that you will have to compensate for those "perturbations", and that's where the series comes in...
-
This is not yet my official return; I decided to answer this question in the short time I have access to a computer today. – J. M. is back. Jul 5 '11 at 7:36
Well J. M. your contributions to the site a very much appreciated so even if it is only to answer a question every now and then when you have some free time and computer access, it is really a great thing. Good luck with whatever you're doing =) – Adrián Barquero Jul 5 '11 at 7:45
Thanks. I am out of my depth here and I don't really understand what E(x) is. It is interesting that a seemingly simple question can have a complex and I gather a difficult answer to calculate. I am also puzzled: a circle's circumference is 2.pi.r and yet an ellipse's is an infinite series - why? – philcolbourn Jul 7 '11 at 12:01
@phil: On the contrary, the complete elliptic integral is in fact not very hard to evaluate numerically. Let me update this answer a bit... – J. M. is back. Jul 14 '11 at 9:02
The arc length of the graph of a function $f$ between $x=a$ and $x=b$ is given by $\int_{a}^{b} \sqrt { 1 + [f'(x)]^2 }\, dx$. So, if you're considering $f(x)=\sin(x)$ then the correct integral is $\int_{0}^{2\pi} \sqrt { 1 + [\cos(x)]^2 }\, dx$. Unfortunately, this integral cannot be expressed in elementary terms. This is quite common for arc-length integrals. However, the definite integral might be expressible in elementary terms; Wolfram Alpha says it cannot.
-
That's interesting, as the two definite integrals are clearly the same over this interval. Following up my comment to Chandru's answer, Wolfram Alpha's gives you $4 \sqrt{2} E(\tfrac{1}{2}) \approx 7.6404$, which is the circumference of an ellipse. – Henry Jun 13 '11 at 12:03
It is given by $$I = \int_{0}^{2 \pi} \sqrt{ 1 + (\cos{x})^{2}} \ \rm{dx}$$ and I think this is an elliptic integral of the second kind. (That's what Wolfram says.)
-
Wolfram Alpha gives it as $4E(-1) \approx 7.6404$ where $E(m)$ is the complete elliptic integral of the second kind, though clearly not the circumference of an ellipse with eccentricity $-1$. – Henry Jun 13 '11 at 11:42
@Henry: Thanks. My simple 8-straight line approximation yielded 7.58... – philcolbourn Jun 13 '11 at 12:11
The first word of this answer isn't quite right, because the value of the integral in the question is correct. – Jonas Meyer Jun 13 '11 at 18:47
@Jonas: for the arc length it has to derivative of $\sin{x}$ that is $\cos{x}$, so I have added $\cos{x}$ – user9413 Jun 13 '11 at 18:50
Maple, which uses a different convention for the elliptic integrals, gives the answer as $4 \sqrt{2} {\rm EllipticE}(\sqrt{2}/2)$. The circumference of an ellipse with semi-major axis $a$ and eccentricity $e$ would be, in this notation, $4 a {\rm EllipticE}(e)$. – Robert Israel Jun 13 '11 at 19:28
Responding to Henry, June 6, 2011, this equivalence emerges from a simple experiment given by Hugo Steinhaus in 'Mathematical Snapshots'. Take a roll of something (I use paper towelling) and saw through it obliquely, thus producing elliptic sections. Unroll it and you have a sine curve. (Tom Apostol and Mamikon Mnatsakanian suggest you rest a paint roller at an angle in the paint tray. Then paint!)
Paul Stephenson May 8 '13 at 21.00
-
On MATLAB:
$$t = 0:0.001:(2\pi);$$ $$st = \sin(t);$$
$$\text{sum( sqrt( diff(st).^2 + diff(t).^2 ) )}$$
$$\text{ans} = 7.6401$$
You will need $21.6$% more paint to paint the corrugated roof ;).
- | 2015-10-13T23:08:02 | {
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https://www.physicsforums.com/threads/calculating-volume-of-parallelpiped.438967/ | # Calculating Volume of Parallelpiped
I'm doing problems which have me calculate the volumes of parallelpipeds I'm slightly confused with this. I know the formula is:
$$V=\vec{a} \cdot (\vec{b}\times\vec{c})$$
where a and b form the base, with c being the "vertical" side. My issue is that when given three vectors which define this shape, how do I know which order to put the vectors in? For one example I was given the vectors:
<6,3,-1>
<0,1,2>
<4,-2,5>
and after drawing them out I had no clue which was which, so I just decided to use them in order as a, b, and c. It worked.
Then in the next problem I was given four points:
P(2,0,-1)
Q(4,1,0)
R(3,-1,1)
S(2,-2,2)
And told that three sides were defined by PQ, PR, and PS. So after getting those vectors I again just took them in order and got -3. The correct answer is three.
How do I determine which vectors I put in which place in the equation? In the examples they just pick three in order and use them that way.
Last edited:
All your procedures are correct, you just forgot a little detail, the volume of the Parallelepiped is given by
$$V=|\vec{a} \cdot (\vec{b}\times\vec{c})|$$
That is, the absolute value.
The last answer is |-3| = 3.
By the way, concerning your curiosity regarding the order of the vectors, it does not matter when you're computing the volume, as it is a matter of referential.
See it like this, suppose we have a, b, c for the length, width and height (not in order) of a Parallelepiped, when computing the volume V = abc, it does not matter which is which.
Or mathematically, in vector form
$$V=|\vec{a} \cdot (\vec{b}\times\vec{c})| = |\vec{b} \cdot (\vec{a}\times\vec{c})| = |\vec{c} \cdot (\vec{a}\times\vec{b})|$$
I guess that works. Thanks for the help.
HallsofIvy
It is worth remembering that with $\vec{u}= a\vec{i}+ b\vec{j}+ c\vec{k}$, $\vec{v}= d\vec{i}+ e\vec{j}+ f\vec{k}$, and $\vec{w}= x\vec{i}+ y\vec{j}+ z\vec{k}$
$$\vec{u}\cdot\left(\vec{v}\times\vec{w}\right)=$$$$\left(a\vec{i}+ b\vec{j}+ c\vec{k}\right)\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ d & e & f \\ x & y & z\end{array}\right|$$$$= \left|\begin{array}{ccc}a & b & c \\ d & e & f \\ x & y & z\end{array}\right|$$ | 2021-06-21T04:20:12 | {
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