lost at fourier coefficient

t_n_k

Joined Mar 6, 2009
5,455
Keep in mind that

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

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

t_n_k

Joined Mar 6, 2009
5,455
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.
 

Thread Starter

stupid

Joined Oct 18, 2009
81
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

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.
 

Georacer

Joined Nov 25, 2009
5,182
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?
 

Thread Starter

stupid

Joined Oct 18, 2009
81
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:

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?
 

t_n_k

Joined Mar 6, 2009
5,455
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:
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.
 

Thread Starter

stupid

Joined Oct 18, 2009
81
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

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.
 

Georacer

Joined Nov 25, 2009
5,182
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?
 

t_n_k

Joined Mar 6, 2009
5,455
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
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.
 
Top