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Help(Using a fourier series to find the sum of second series

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data

    I have found the complex fourier series corresponding to following function [tex]f(x) = x \cdot (\pi -x)[/tex] defined on the interval [tex](0,\pi)[/tex]

    where I get that [tex]f(x) = \frac{\pi^2}{12} + \sum(\frac{-cos(n\pi)+1}{2n^2} + \frac{cos(n\cdot \pi)}{n^3\cdot \pi} \cdot i) [/tex]

    Then I suppose to use that series to find the sum of series [tex]\sum_{n=1}^{\infty} \frac{sin(2n-1)x}{(2n-1)^3}[/tex] where [tex]x \in [-\pi,\pi][/tex]



    Do I show the two series converge to the same point???
     
    Last edited: Sep 27, 2009
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  3. Sep 27, 2009 #2

    LCKurtz

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    First of all, your expression for f(x) doesn't have any x in it. But that's a bit irrelevant because I think you misunderstand what you need to do anyway. You are given a function defined on (0, π). And you are asked a question about a series that has only sine terms. Since the problem suggests the function you are given and the series you need to evaluate are related, what might be the connection? What if you extended your function x(π-x) from (0,π) to (-π,0) by taking the odd extension? What kind of terms would be in the resulting Fourier Series? Have you studied half-range expansions?
     
  4. Sep 28, 2009 #3
    maybe I am silly old girl, but am I suppose to deduce that from the the interval on which f(x) is defined??

    Can you please give an example of a half range expression?
     
  5. Sep 28, 2009 #4

    LCKurtz

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    You are given a function on (0,π). Presumably you are going to find the Fourier Series of a 2π periodic function. What are you going to use for f(x) on (-π, 0)? You are going to have to integrate f(x)cos(nx) and f(x)sin(nx) from -π to π to calculate your coefficients. So what formula are you going to use for f(x) on (-π, 0)?

    Hint: What do you know about the Fourier coefficients for even and odd functions? What do you need to do with f(x) to have any chance of getting a series similar to what you are asked to sum? I'm asking about the an and bn here, not the complex expansion.
     
  6. Sep 29, 2009 #5
    If my assumption about my integral is semi correct then since it half a periode we are looking at its

    [tex]\frac{1}{\pi}(\int_{0}^{\pi} (x \cdot (\pi-x) \cdot e^{-inx}) dx[/tex]

    anyway I found out that [tex]e^{-inx} = cos(nx)-i\cdot sin(nx)[/tex] which also equals

    [tex](-1)^{-n)[/tex]

    does this integral then equal the series

    [tex]\sum_{n=0}^{\infty}(-1)^{-n} \cdot (x\cdot (\pi-x))[/tex]
     
    Last edited: Sep 29, 2009
  7. Sep 29, 2009 #6

    LCKurtz

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    No, that integral does not equal that series. I'm afraid you are very confused. I have asked you several leading questions, the answers to which might help you understand, but you seem to ignore the questions. Let's start again with that last hint:

    1. What do you know about the Fourier coefficients an and bn in the case where f(x) is an even or odd function on (-π, π)?

    2. In light of the answer to 1, what does f(x) need to be on (-π, 0) to have any chance of getting a series similar to what you are asked to sum?

    Please think about the answers to these two questions. They are critical to understanding this problem. Then we can proceed.
     
  8. Sep 29, 2009 #7
    if n is odd and defined on the interval [-n,n] then the corresponding [tex]b_k[/tex] in the fourier series (according to my textbook)

    [tex]b_k = \frac{2}{n} \int_{0}^{n} f(x) \cdot sin(\frac{k\cdot \pi \cdot x}{n}) dx[/tex]

    giving the corresponding fourier series since f(x) is odd

    [tex]\sum_{k=1}^{\infty} b_k \cdot sin(\frac{k\cdot \pi \cdot x}{n}) [/tex]

    but what I don't understand here is how is this connect to f(x) is defined on a complex vector space. I thought since it stated that f(x) is defined on a complex vector space then its the same as the corresponding fourier series is a complex fourier seres?

    have I misunderstood something here?
     
    Last edited: Sep 29, 2009
  9. Sep 29, 2009 #8

    LCKurtz

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    "if n is odd and defined on the interval [-n,n] then..."

    Not "if n is odd". Do you mean "if f(x) is odd"?

    "have I misunderstood something here?"

    Yes, I think so. You are given f(x) on (0,π), not a complex vector space. The fact that there is a complex form for the coefficients is irrelevant to the given question and is misleading you. Forget everything complex for the moment.

    You have partially answered my first question by giving the formula for bn if f(x) is odd. And what about the an in that case?

    What about question 2?

    And I just noticed your formula for bk has typos. Check it.
     
  10. Sep 29, 2009 #9
    yes off cause I meant if f(x) is odd my mistake.

    so everytime I get an assigment which states that f(x) is defined on the complex space. Then that has a trick question?

    According to my textbook if f(x) is odd then the corresponding fourier series doesn't contain and [tex]cos[/tex]. I guess thats answer there.
    According to my textbook then f is odd

    [tex]\int_{-n}^{n} f(x) dx = 0 [/tex]

    and according to theory the left interval [-n,0] cancels out the right interval [0,n] and thats reason why since f is is odd we either deal if the left or right interval and not the whole interval [-n,n]

    then to find the fourier series which corresponds to [tex]f(x) = x \cdot (\pi - x) [/tex] is and odd function then I find the

    [tex]\begin{array}{cccccc}b_k = \frac{2}{\pi} \int_{0}^{\pi} (x \cdot (\pi - x)) \cdot sin (\frac{k \cdot \pi \cdot x}{\pi}) dx = - \frac{\pi \cdot x \cdot cos(kx)}{k} + \frac{k^2 \cdot x^2-2\cdot (cos(kx)}{k^3} - \frac{2x \cdot sin(kx)}{k^2} + \frac{\pi \cdot sin(kx)}{k^2}|_{x=0}^{\pi} = \frac{2((k\cdot \pi-2)\cdot cos(k\pi)-k\cdot cos(kx) \cdot \pi}{k^2}\end{array}[/tex] ??

    trying to simplify (hard)

    [tex]cos(kx) = 1-k \sin^2(x) [/tex]

    then

    [tex]b_k = \frac{2((k\cdot \pi-2)\cdot cos(k\pi)-k\cdot (1-k\cdot sin^2(x)) \cdot \pi}{k^2}[/tex]

    not sure how this connect to the series then being

    [tex]\sum_{k=1}^{\infty} b_k \cdot sin(\frac{k \cdot \pi \cdot x}{\pi}) = \frac{k^2 \cdot \pi \cdot (sin^2 x) + 2 \cdot (k\pi-2) \cdot cos(k\pi)-k\cdot \pi) \cdot sin(kx) }{k^2}[/tex] and how is suppose to lead to find the sum of

    [tex]\sum_{n=1}^{\infty} \frac{sin(2n-1)\cdot x}{(2n-1)^3} [/tex] ??? series comparison test? Or is it something to do with convergence?

    (Have I made a mistake my b_k? Cause that doesn't look like a simply expression which look any like the other series to me??
     
    Last edited: Sep 29, 2009
  11. Sep 29, 2009 #10

    LCKurtz

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    You're getting there. Don't forget to put in your limits and simplify to get bk.
    Once you have bk simplified, you are ready to write down your Fourier Series. Then it will be time to look at what you got and what the question asked and see if you can answer it.
     
  12. Sep 29, 2009 #11

    Hi I have edited my previous post.

    With some extra details. Hope you can point me in the right direction regarding these question:)
     
  13. Sep 29, 2009 #12

    LCKurtz

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    Check your formula for bk. I think you dropped a factor of 2 and have a sign wrong. And you will be able to simplify it additionally by looking at what you get when k is even vs. when k is odd.

    Once you have bk correct and simplified you need to write your fourier series down, and put the bk you have calculated in the series.

    That needs to be done and correct. Then it will be time to look at what you have vs. the question you are trying to answer.
     
  14. Sep 29, 2009 #13
    Changed my post again hope you maybe will look at it again then you have the time???

    Is the first and the second series suppose to mirror each other??

    I hope you can give me a hint on this, because my sill textbook hasn't any references to series that mirror each other :(
     
  15. Sep 29, 2009 #14

    LCKurtz

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    It's getting hard to follow when you keep changing your post. Consider copying and pasting the relevant parts.

    Your bk's should only have k's and constants in their formula. No x's.

    And when you do get them figured out, you substitute their values in the series for bk, not on the right side of the equals like you have in the second equation from the bottom.

    There is no point in trying to figure out the rest of the problem until you have your Fourier Series correct. You need to do two things to continue:

    1. Work on your bk until you can show that its value is:

    [tex]b_k = \frac {4(1- cos(k\pi))}{\pi k^3}[/tex]

    2. Write down the series you get with these values for bk.
     
  16. Oct 4, 2009 #15
    I did the calculations have them on paper not here. But would simply like to how then I have this corresponding fourier series for f(x). How do I use this to find the sum of the second series? Is there a theorem which I need to apply?
     
  17. Oct 4, 2009 #16

    LCKurtz

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    When you get time, write down the series you get with those bk and post it here. It will (it really will, eventually, trust me) come out very much like the series you are looking for. You won't need any theorems, just your eyeballs. There are a couple of things I may need to point out to you yet. You know the old saying, "if it was easy, it wouldn't be difficult".:smile:
     
  18. Oct 5, 2009 #17
    Here is my calculations of the integral

    [tex]\frac{2}{\pi} (x \cdot (\pi -x ) \cdot sin(kx) dx[/tex]

    we set [tex]x \cdot (pi -x ), du = \pi - 2x [/tex]

    and [tex]dv = sin(kx) , v = \frac{-cos(kx)}{k}[/tex]

    thus using integration by parts

    [tex]x \cdot (pi -x ) \cdot \frac{-cos(kx)}{k} |_{0}^{\pi} - \int_{0}^{\pi} \frac{-cos(kx)}{k} \cdot (\pi - 2x) dx [/tex]

    yielding

    [tex]\int_{0}^{\pi} \frac{-cos(kx)}{k} \cdot (\pi - 2x) dx [/tex]

    (since the left side is zero!)

    choosing [tex]u =\pi - 2x, du = -2 [/tex]

    [tex]dv = \frac{-cos(kx)}{k}, v = \frac{-sin(kx)}{k^2}[/tex]

    and again by integral by parts

    [tex](\pi - 2x) \cdot \frac{-sin(kx)}{k^2}|_{0}^{\pi} - \int_{0}^{\pi} \frac{2sin(kx)}{k^2} dx[/tex]

    yielding

    [tex]\frac{sin(k\pi)\cdot \pi}{k^2} - \int_{0}^{\pi} 2 \cdot \frac{sin(kx)}{k} dx[/tex]

    finally result

    [tex]b_k = \frac{sin(k\pi)\cdot \pi}{k^2} - \int_{0}^{\pi} 2 \cdot \frac{sin(kx)}{k} dx = \frac{-2cos(kx)}{k^3}|_{x=0}^{\pi} = \frac{-2(cos(kx)-1}{k^3}[/tex]

    thus

    [tex]b_k = \frac{2}{\pi} \cdot (\frac{2cos(k \pi) + k \cdot sin(k\pi) \cdot \pi-2}{k^3}) = \frac{4 \cdot (cos(k\pi)-1)}{k^3\cdot \pi}[/tex]

    If I plug this into the formula for the odd function fourier series I get

    [tex]f(x) = \sum_{k=1}^{\infty} \frac{4 \cdot (cos(k\pi)-1)}{k^3\cdot \pi} \cdot sin(kx) [/tex]

    so how do I go about using this to find the sum of

    [tex]\sum_{n=1}^{\infty} \frac{sin(2n-1)x}{(2n-1)^3}[/tex] where [tex]x \in ]-\pi, \pi [ [/tex]
     
    Last edited: Oct 5, 2009
  19. Oct 5, 2009 #18

    LCKurtz

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    OK, good. Now write down your fourier series for f:

    f(x) = your sum with your formulas in for the bk.

    Also write out the first 6 terms.

    Oh I see you have already done that. Write out the first 6 terms now.
     
  20. Oct 5, 2009 #19

    LCKurtz

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    Also, what do you notice about (cos(kπ) - 1) when k is even or odd?
     
  21. Oct 5, 2009 #20
    okay here is the first six terms

    [tex]\frac{-8 \cdot sin(x)}{\pi} + 0 + \frac{-8 sin(3x)}{27 \pi} + 0 + \frac{-8 \cdot sin(5x)}{125 \pi} + 0 [/tex]
     
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