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Can (1-cos(npi))=(-1)^(n+1)?

  1. Dec 9, 2011 #1
    1. The problem statement, all variables and given/known data

    To find the Fourier series for

    f(x)=
    0, -2<x<0
    x, 0≤x<1
    1, 1≤x<2

    2. Relevant equations

    f(x)=a_0/2+Ʃ(from n=0 to ∞) (a_n*cos(npix/p) + b_n*sin(npix/p))

    3. The attempt at a solution

    So p=2, interval=[-2,2]

    a_0=3/4,

    a_n=(2/n^2pi^2)*(cos(npi/2)-1),

    Here is my problem:

    did the exercise twice and keep getting:

    b_n=(2/n^2pi^2)(sin(npi/2)+(npi/2)*(1-cos(npi))

    I have the solution:

    b_n=(2/n^2pi^2)(sin(npi/2)+(npi/2)*(-1)^(n+1))

    I know cos(npi)=(-1)^n

    and I've been told sin(npi)=(-1)^(n+1)

    My steps:

    1) b_n=1/2[∫(from 0 to 1) x*sin(npix/2)dx + ∫(from 1 to 2) sin(npix/2)dx]

    2) integration by parts for 1st integral and subsequent integration:

    → b_n=1/2[-(2x/npi)cos(npix/2)|(0 to 1) + (4/n^2pi^2)sin(npix/2)|(0 to 1) - (2/npi)cos(npix/2)|(1 to 2)]

    3) I am left with:

    b_n=1/2[(2/npi) + (4/n^2pi^2)sin(npi/2) - (2/npi)cos(npi)

    after regrouping is where (1-cos(npi)) comes from.

    Thus original question is:

    Can (1-cosnpi)=(-1)^(n+1)? or am I mistaken somewhere?

    Thank you
     
    Last edited: Dec 9, 2011
  2. jcsd
  3. Dec 9, 2011 #2

    HallsofIvy

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    I don't know where you got "[itex]sin(n\pi)= (-1)^{n+1}= 1- cos(n\pi)[/itex]". None of those equalities is correct.

    [itex]sin(n\pi)= 0[/itex] for all n, not a power of -1. [itex]cos(n\pi)= -1[/itex] if n is odd, 1 if n is even so [itex]cos(n\pi)= (-1)^n[/itex] and [itex]1+ cos(n\pi)[/itex] is 2 if n is even, 0 if n is odd.
     
  4. Dec 9, 2011 #3
    As I said, I "know cos(npi)=(-1)^n and I was told sin(npi)=(-1)^(n+1)" so I believe you that it is wrong. I imagine that that someone wanted to say sin(npi/2)=(-1)^(n+1)???

    As for the "equalities", I was wondering, not stating, seeing as the equations "looked" similar.

    Therefore, if all this is wrong, would it be possible for you to help me figure out where I went wrong?

    I think it safe to say I will remove that "P.S." from the question so as not to confuse...
     
    Last edited: Dec 9, 2011
  5. Dec 9, 2011 #4

    LCKurtz

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    Why would you imagine that? Try it with n = 0 or 2.
     
  6. Dec 9, 2011 #5

    LCKurtz

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    Your an is correct but I think your a0 should be -5/4.
    I didn't check your remaining steps, but what about the integral from -2 to 0?
     
    Last edited: Dec 9, 2011
  7. Dec 9, 2011 #6
    The reason I "imagine" that is because I had some sort of logic in my head that said that if
    sin(npi/2)=something, when n is an odd number than n+1 was gonna give me an odd number...
    I see as I try to explain that it doesn't always work. But I had "sin(npi)=(-1)^(n+1) in my class notes which is a mistake and probably didn't help my understanding...

    The thing is that in my school book. It keeps telling me things like "=(-1)^(n+1).
    And a similar problem to my initial question just popped up again:

    Expand f(x)=x, -2<x<2
    x = Odd f(x)
    Thus
    b_n = ∫ (0 to 2) x*sin(npix/2)dx

    When I do this, I get (-4/npi)cos(npi)

    and it gives me (4/npi)(-1)^(n+1)

    as in original post, I was getting: (1-cos(npi)) and the answer was (-1)^(n+1)

    So is it when there is a negative in front that I have to do ^(n+1)???

    Could someone please respond to the actual question about where (-1)^(N+1) comes from...
     
    Last edited: Dec 9, 2011
  8. Dec 9, 2011 #7
    @ LCKurtz
    The answer in the book is 3/8 for a_0/2
    And the -2 to 0 disappears 'cause the function is 0 on that interval
    P.S. Thank you for trying to help me with the real question
     
  9. Dec 9, 2011 #8

    vela

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    This is right.

    This is wrong. You're making another error somewhere, which is causing you to mistakenly think that [itex]1-\cos n\pi = (-1)^{n+1}[/itex] must hold. You need to turn this around. The latter equality is obviously wrong, which means you must have made a mistake somewhere, so you should be looking for it.

    It seems a bit strange that you're struggling with basic algebra when you're learning about Fourier series. There's nothing mysterious going on here:
    [tex]-(-1)^n = (-1)(-1)^n = (-1)^{n+1}[/tex]
     
  10. Dec 9, 2011 #9
    My brain works in VERY mysterious ways, but thank you. That does clear things up.
    I am working hard on trying to figure out where my mistake is.
    I get confused when sometimes the answer is (-1)^(n+1)
    and in other cases like:

    b_n=(2/pi)∫(0 to pi)sin(nx)dx

    I get (2/npi)(1-cos(npi))and the answer in that case is (2/pi)(1-(-1)^n)/n

    instead of (2/pi)(1+(-1)^(n+1))/n
     
    Last edited: Dec 9, 2011
  11. Dec 9, 2011 #10

    LCKurtz

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    For some reason I had written the function as -2 on that interval. Guess my eyeball fell on the -2 instead of the 0.
     
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