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Fourier sine series integration

  1. Dec 24, 2014 #1
    1. The problem statement, all variables and given/known data
    The question is to get Fourier sine series of e^-x =f(x) on 0<x<1

    2. Relevant equations

    Bn = 2/L ∫ (e^-x) * sin(nπx/L) over the limits 1 to 0, where L = 1

    f(x) = summation of Bn*sin(nπx/L)

    3. The attempt at a solution
    So I integrated ∫ by part integration

    so I took u = sin(nπx) and v = -e^-x and u' = nπ*cos(nπx)

    so equation was

    (-e^-x)*sin(nπx) - ∫-(e^-x)*nπ*(cos(nπx))

    then again I did part integration where u = nπ*cos(nπx), v = e^-x and u' = -n^2*π^2*sin(nπx)

    so I got

    (e^-x)*nπ*(cos(nπx)) + ∫(e^-x)*n^2*π^2*sin(nπx)

    ∫(e^-x)*sin(nπx) = (-e^-x)*sin(nπx) - (e^-x)*nπ*(cos(nπx)) - ∫(e^-x)*n^2*π^2*sin(nπx)

    Now taking ∫(e^-x)*sin(nπx) = L i can write the above equation as

    L = (-e^-x)*sin(nπx) - (e^-x)*nπ*(cos(nπx)) - n^2*π^2 L

    So L = (-e^-x)*sin(nπx) - (e^-x)*nπ*(cos(nπx))/(1+n^2*π^2)

    Can someone please confirm if my integration is correct? Many thanks. I will only put the limits after someone confirms the integration is correct.
     
  2. jcsd
  3. Dec 24, 2014 #2

    LCKurtz

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    It looks OK except you dropped the ##2##.
     
  4. Dec 24, 2014 #3
    What 2 are you talking about? I was not asking if the Bn is correct in the post, just wanted to know if the integration is correct. Can you confirm if there is any other 2 I need to worry about?
     
  5. Dec 24, 2014 #4

    LCKurtz

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    That 2.
     
  6. Dec 24, 2014 #5
    Okay so after doing the integration part by putting the limits I get nπ*(1-((e^-1)*(-1)^n))/(1+n^2*π^2). Is that correct? Keep in mind that 2 of Bn I am still keeping it out as its a constant.
     
  7. Dec 24, 2014 #6
    This equation is the relationship that one uses for Bn when one is expanding f(x) as a so-called "half wave" Fourier sine series (that I was referring to in another thread)

    I confirm that the integration is correct. And I evaluated the integral with the functions that you used for u and v interchanged, so I'm sure that this result is correct. But, you can always confirm that the integration is correct simply by differentiating it.

    However, I would not have used the symbol L for the integral, since you already used that symbol earlier to represent the length of the x interval.

    Chet
     
  8. Dec 24, 2014 #7
    Yes, I confirm this result too.
     
  9. Dec 25, 2014 #8
    Thanks. So after multiplying it by 2 I get the Bn, however in the question now its asking me to show that the fourier sine series of ## f(x) = e^-x ## is equal to ## \frac {1} {1+π^2} ## -## \frac{3} {1+9π^2} ## + ## \frac {5} {1+25π^2} ## ...

    How is this possible...? I thought f(x) would just be Bn*sin(nπx). when I did that for n=1, My Bn numerator was like 6.65 when I replaced the n in the final integral by 1. Also, as its getting multiplied by sin(nπx) shouldn't all terms be just zero?
     
  10. Dec 25, 2014 #9
    Are they asking for the value of the series at some specific value of x?

    Chet
     
  11. Dec 25, 2014 #10
    Nope nothing specific, like I stated in the question at the top 0<x<1. this is the only information given about x value...

    its actually asking me to find the fourier sine series of F(x) = ## e^{-x} ## and then use the result of this to show

    ## \frac {1} {1+π^2}- \frac {3} {1+9π^2} + \frac {5} {1+25π^2} \ ##... = ## \sum_{k=0}^\infty \frac {(2k+1)(-1)^k} {1+(2k+1)^2 (π^2)} \ ## = ## \frac {\sqrt {e}} {2π(e+1)} \ ##

    I can see the relationship between the summation and the terms in the series...but don't know how I can actually get those terms in the first place...The summation represents the Bn I suppose and the Bn I got from the integration doesn't match this...
     
  12. Dec 25, 2014 #11
    I can see how they got this result from your solution. You need to evaluate both sides of the equation for the specific case of x = 1/2.

    Chet
     
  13. Dec 25, 2014 #12
    I calculated Bn for n =1 and then multiplied with sin(nπx) with x = 0.5, the answer comes 0.79 which is not the first term of the series...how do we even get the exponential format from this?
     
  14. Dec 25, 2014 #13
    You're supposed to derive this result algebraically. Write out your expressions for B1, B2, B3, B4, and B5 algebraically. What is e^-x when x =1/2?
     
  15. Dec 25, 2014 #14

    2 ( ## \frac {π (e^{-1} +1) } {1+π^2} \ ## - ## \frac {3π (e^{-1} +1) } {1+9π^2} \ ## + ## \frac {5π (e^{-1} +1) } {1+25π^2} \ ##) = ## \sqrt{e} \ ##

    The 2 is from the Bn formula so now taking ## \ 2π(e^{-1} +1) \ ## common and moving it to other side I get

    ## \frac {1} {1+π^2} \ ## - ## \frac {3} {1+9π^2} \ ## + ## \frac {5} {1+25π^2} \ ## = ## \frac { \sqrt{e} } { \ 2π(e^{-1} +1) } \ ##

    ## \frac {1} { \frac {1} {e^{-1}}} \ ## = e so

    ## \frac {1} {1+π^2} \ ## - ## \frac {3} {1+9π^2} \ ## + ## \frac {5} {1+25π^2} \ ## = ## \frac { \sqrt{e} } { \ 2π(e+1) } \ ##

    Seems like its correct...Sorry I took so long to understand this...Thanks alot . anyways can you please help with that heat equation. I think I am almost near to the solution to that as well...
     
  16. Dec 25, 2014 #15
    Nicely done. For the heat transfer problem, you do the exact same thing you did on this problem to get the Fourier Series. You can even use the exact same half wave equation to get Bn.

    Chet
     
  17. Jan 6, 2015 #16
    Hi I just realized when x =1/2 its not ## \sqrt {e} \ ## but its ## \frac {1} { sqrt{e} } \ ## as its ## \ e^{-x} \ ##. So then what I did can't be correct right? If I take x =-1/2 then my signs becomes opposite then the signs of the terms in the series...What do I do? Please help...
     
  18. Jan 6, 2015 #17
    You need to check your algebra. I solved the problem and got the desired result, and so did you too earlier. Find out what you're doing wrong now.

    Chet
     
  19. Jan 6, 2015 #18
    But that was wrong! I got the correct result by writing ## \ e^{-x} \ ## when x = 1/2 as ## \sqrt{e} \ ##. That's the wrong part. You told it was correct earlier but I think you also didn't realise that I was actually doing it wrong....How do I get the desired result by doing it the correct way now
     
  20. Jan 6, 2015 #19
    I didn't just look over what you did and agree to it. I obtained the result on my own even before you presented your results. My results agreed with the relationship that was given.

    Chet
     
  21. Jan 6, 2015 #20
    So you got it by taking x as 0.5? Because if you do that then sqrt e just remains in the denominator instead of being on the numerator
     
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