# lost at fourier coefficient

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

1. ### stupid Thread Starter Active Member

Oct 18, 2009
81
0
hi,
i m stuck at the ...part for both an & bn..
how to proceed thence?

thanks
atupid

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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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
783
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
5,151
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.

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
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
5,151
1,266
You said that the answer for bn is
$
b_{\small{n}}=[2\cdot(-1)^{\small{n+1}}]/n \Leftrightarrow\\
b_{\small{n}}=[(-1)\cdot2\cdot(-1)\cdot(-1)^{\small{n}}]/n \Leftrightarrow\\
b_{\small{n}}=-[2\cdot(-1)^{\small{n}}]/n
$

does that shed any light?

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
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.