UnkleRhaukus
  • UnkleRhaukus
Evaluate the sum at \(x=\tfrac\pi2\) \[S(x) =\frac{1}2+\frac2\pi\sum\limits_{n=1,3,5...}^\infty\frac{\sin({n x})}n\\=\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{\sin({(2r-1) x})}{2r-1}\\ \]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
UnkleRhaukus
  • UnkleRhaukus
\[S\left(\tfrac\pi2\right) =\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{\sin(\pi r-\tfrac\pi2)}{2r-1}\\ =\frac{1}2+\frac2\pi\sum\limits_{r=0}^\infty\frac{(-1)^{r-1}}{2r-1}\]
anonymous
  • anonymous
looks to me a series of log(1+x) and log(1-x)
UnkleRhaukus
  • UnkleRhaukus
how

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
also r should start in 1 and not 0
UnkleRhaukus
  • UnkleRhaukus
hmm
anonymous
  • anonymous
see this http://www.wolframalpha.com/input/?i=series+log%281%2Bx%29
UnkleRhaukus
  • UnkleRhaukus
but that has even terms in it
anonymous
  • anonymous
eliminate using series for log (1-x) try it if it not works use ix instead of x
UnkleRhaukus
  • UnkleRhaukus
im not sure how to do that
anonymous
  • anonymous
see that \[\frac{1}{2}\frac{\log(1+x)}{\log(1-x)}=x+\frac{x^3}{3}+\frac{x^5}{5}...\] Replace x by ix
anonymous
  • anonymous
now do you see it
anonymous
  • anonymous
Signs will now alternate
UnkleRhaukus
  • UnkleRhaukus
is there another way to do this?
UnkleRhaukus
  • UnkleRhaukus
@mukushla
experimentX
  • experimentX
|dw:1355650436133:dw|
UnkleRhaukus
  • UnkleRhaukus
ln 2 ?
experimentX
  • experimentX
let me check for generating function
experimentX
  • experimentX
probably the one by @NotSoBright works just put x=i and get the sum of series ... it should work out.
experimentX
  • experimentX
\[ {1 \over 1 - x } = 1 + x + x^2 + ....\\ {1 \over 1 +x} = 1 - x + x^2 - ... \\ ------\text{ Adding }------\\ {1 \over 1 - x} + {1 \over 1 +x} = {2 \over 1 -x^2}= 2( 1 +x^2 + x^4 + ...) \\ \text{Integrating both sides} \\ {1 \over 2} \log \left( 1 + x \over 1- x\right) = x+\frac{x^3}{3}+\frac{x^5}{5}... \\ \text{Replace x by i you get} \\ \log 2 - \frac{\log \sqrt 2} 2 + i\frac \pi 2 = i \left( 1 - {1 \over 3} + \frac 15 - ... \right) \] comparing imaginary terms, you get that sum is pi/2
experimentX
  • experimentX
woops sorry ... \[ \log 2 - \frac{\log \sqrt 2} 2 + i\frac \pi 4 = i \left( 1 - {1 \over 3} + \frac 15 - ... \right) \] should have been pi/4
experimentX
  • experimentX
|dw:1355652797759:dw| putting that value you would get 1/2 + 2/pi * pi/4 = 1/2 + 1/2 = 1
experimentX
  • experimentX
this is definitely expansion of what i wrote earlier. If you have Mathematica you can try this code Plot[1/2 + 2/Pi *Sum[ Sin[(2 n - 1) x]/(2 n - 1), {n, 1, 30}], {x, 0, 2 Pi}]
UnkleRhaukus
  • UnkleRhaukus
do you think that is the answer they want for question 3.?
experimentX
  • experimentX
well ... i think there are lots of series that uses the convergence of Fourier series to find the value of series. like ... for example: \( \sum {1 \over n^2} , \sum {1 \over n^4} \) .. i think it is other way ... probably to show that the summation of series and the value from Fourier are equal.
experimentX
  • experimentX
lol .. it seems there are quite a lot of questions!! I would put the values in the function ..
UnkleRhaukus
  • UnkleRhaukus
i think your both getting the right result, i just dont follow the method, and id be surprised if this is the method they are looking for
experimentX
  • experimentX
i guess not, it's quite long
UnkleRhaukus
  • UnkleRhaukus
can i just assume that we know the sum of the alternating harmonic series is ln2, (in the context of Fourier series)
experimentX
  • experimentX
yes you can!! it's pretty well known
UnkleRhaukus
  • UnkleRhaukus
how do i get the \(\pi/4\) bit?
experimentX
  • experimentX
Do you know complex logarithm? \[ \log (1+i) = \log \sqrt 2 + i{\pi \over 4}\]
UnkleRhaukus
  • UnkleRhaukus
i have not studied complex logarithms , and they are not in the book the question has come from
experimentX
  • experimentX
this shouldn't be difficult \[ \large z = |z|e^{i \arg(z)} \\ \log(z) = \log |z| + i\arg(z)\] as you see z = 1+i, |z| = \sqrt(2) and arg(z) = pi/4
UnkleRhaukus
  • UnkleRhaukus
i think i had the completely idea wrong idea of what this question was asking, because in question 1 i know S(1/2) so..
UnkleRhaukus
  • UnkleRhaukus
\[\begin{align*} % S(x) S(x) &=\frac{1}2+\frac2\pi\sum\limits_{r=1}^\infty\frac{\sin\big((2r-1)x\big)}{2r-1}\\ \\ \\% S(π/2) S\left(\tfrac\pi2\right) &=\frac{1}2+\frac2\pi\sum\limits_{r=1}^\infty\frac{\sin(\pi r-\tfrac\pi2)}{2r-1}\\ 1&=\frac{1}2+\frac2\pi\sum\limits_{r=1}^\infty\frac{(-1)^{r-1}}{2r-1}\\ \frac\pi4&=\sum\limits_{r=1}^\infty\frac{(-1)^{r-1}}{2r-1}\\ \\ \end{align*}\]
UnkleRhaukus
  • UnkleRhaukus
1 Attachment

Looking for something else?

Not the answer you are looking for? Search for more explanations.