lost at fourier coefficient

Discussion in 'Homework Help' started by stupid, Jun 11, 2010.

  1. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    0
    hi,
    i m stuck at the ...part for both an & bn..
    how to proceed thence?

    thanks
    atupid
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    782
    Keep in mind that

    cos(-n\pi)=cos(n\pi)=-1  \ for \ n  \ odd \ integer

    cos(-n\pi)=cos(n\pi)=+1 \ for \ n  \ even \ integer
     
  3. stupid

    Thread Starter Active Member

    Oct 18, 2009
    81
    0
    tnk,
    the answer seems to be as follow;
    for Bn = [2(-1)^n+1]/n

    how to obtain the n+1?

    thanks
    stupid
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I think you will find that ...

    B_{n}=-\frac{2cos(n\pi)}{n} \ for \ n=1,2,3,...

    So

    B_{1}=+\frac{2}{1}

    B_{2}=-\frac{2}{2}

    B_{3}=+\frac{2}{3}

    B_{4}=-\frac{2}{4}

    etc.
     
  5. stupid

    Thread Starter Active Member

    Oct 18, 2009
    81
    0
    tnk,
    i have been given the answer Bn = [2(-1)^n+1]/n

    im wondering how the power of n+1 came about not how to compute it as u had shown..

    thanks
    stupid

     
  6. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    Well, let's break it down:
    cos(n\cdot\pi)=
    =1\ \ \   for\  n=0
    =-1\  for\  n=1
    =1\ \ \    for\  n=2
    =-1\  for\  n=3
    etc...
    Now let's take a look at:
    (-1)^n=
    =1\ \   \ for\  n=0
    =-1\  for\  n=1
    =1\ \   \ for\  n=2
    =-1\  for\  n=3

    Do you see that they are actually the same series?
     
  7. stupid

    Thread Starter Active Member

    Oct 18, 2009
    81
    0
    the power in question is n+1 not merely n
    how did the +1 came about?

    i m wondering what could be going wrong with my simple asking?
    all i wanted to know is how did the power of n+1 come about, not how to compute it.
    :confused:

     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The (-1)^(n+1) part simply ensures that with n odd, the sign of the harmonic amplitude is positive and with n even, the sign is negative. It's just a neat way of writing the solution without expressly including the cosine function.
     
  9. stupid

    Thread Starter Active Member

    Oct 18, 2009
    81
    0
    let me try one more time...
    some of other answers have (-1)^n
    others have (-1)^n+1

    why some have "+1" while others no?

    regards,
    stupid

     
  10. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    You said that the answer for bn is
    <br />
b_{\small{n}}=[2\cdot(-1)^{\small{n+1}}]/n \Leftrightarrow\\<br />
b_{\small{n}}=[(-1)\cdot2\cdot(-1)\cdot(-1)^{\small{n}}]/n \Leftrightarrow\\<br />
b_{\small{n}}=-[2\cdot(-1)^{\small{n}}]/n<br />
    does that shed any light?
     
  11. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    First of all (-1)^n+1 could be interpreted as

    [(-1)^n]+1

    rather than (-1)^(n+1). I think the latter is probably what is intended.

    To elaborate on your issue ....

    Suppose you have the case of (-1)^n

    If n is odd then (-1)^n is -1

    If n is even then (-1)^n is +1

    Suppose you have the case of (-1)^(n+1)

    If n is odd then (-1)^(n+1) is +1

    If n is even then (-1)^(n+1) is -1

    What you have is a pair of simple expressions which alternate the sign of successive terms of an infinite series and in which the required nth term sign convention [being +ve or -ve] is to occur for the case of n being odd or even.
     
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