Fourier Cosine series

[willmoore21]

This is more of an understanding of a question than an actual question, I'm not 100% sure what it's asking on the second line.


The question is:

Expand f(t) = sin(t) where 0 < t < pi as a Fourier cosine series. (easy enough can probably do this)

It then says

Hence determine the value of the series: (here is the equation, but I can't get it to work in Latex because I don't know how to put in summations and things like infinity).

I can most likely find the fourier series of sin(t), but is the question asking me to get the equation that is stated as the value of the series. In other words am I trying to end up with the equation given?

Thanks

 

[galactus]

One can use Fourier series to evaluate series such as \(\displaystyle \displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{{\pi}^{2}}{6}\).

What is the series you're supposed to evaluate?.

 

[willmoore21]

One can use Fourier series to evaluate series
such as \(\displaystyle \displaystyle
\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{{\pi}^{2}}{6}\)
What is the series you're supposed to evaluate?.

It's this one:
\(\displaystyle \displaystyle \sum_{n=1}^{\infty}\frac{1}{{(2n)}^{2}{-1}}=\frac{1}{{2}^{2}{-1}}+\frac{1}{{4}^{2}{-1}}+\frac{1}{{6}^{2}{-1}}+....\)

Got latex to work after some copy and paste of yours . . .

 

[willmoore21]

Now that I have latex to work here is a post of the whole question, but I don't understand what it is asking me when it says "hence determine the value of the series"

The question is:

Expand f(t) = sin(t) where 0 < t < pi as a Fourier cosine series.

Hence determine the value of the series:

\(\displaystyle \displaystyle \sum_{n=1}^{\infty}\frac{1}{{(2n)}^{2}{-1}}=\frac{1}{{2}^{2}{-1}}+\frac{1}{{4}^{2}{-1}}+\frac{1}{{6}^{2}{-1}}+....\)

 

[galactus]

There are easier ways to sum this than using Fourier series.

\(\displaystyle \displaystyle \sum_{n=1}^{\infty}\frac{1}{4n^{2}-1}=\frac{1}{2}\sum_{n=1}^{\infty}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]\)

This series 'telescopes'.

In doing so, we notice all cancels except the first and last terms.

\(\displaystyle \displaystyle\lim_{n\to\infty}\frac{1}{2}\left[1-\frac{1}{2n+1}\right]=\frac{1}{2}\)


But, if you have to use a Fourier Cosine series, then \(\displaystyle a_{0}=\displaystyle \frac{2}{\pi}\int_{0}^{\pi}sin^{2}(x)dx=\frac{4}{\pi}\)

\(\displaystyle a_{n}=\displaystyle \frac{2}{\pi}\int_{0}^{\pi}sin(x)cos(nx)dx=\frac{-4}{\pi (n^{2}-1)}\) if n is even and 0 if n is odd.

\(\displaystyle sin(x)=\frac{2}{\pi}+\displaystyle \sum_{n=1}^{\infty}a_{n}cos(nx)\)


If n is even, \(\displaystyle \displaystyle \frac{2}{\pi}-\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{1}{(2n)^{2}-1}=0\)

\(\displaystyle \displaystyle -\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{1}{(2n)^{2}-1}=\frac{-2}{\pi}\)

\(\displaystyle \displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n)^{2}-1}=\frac{-2}{\pi}\cdot \frac{-\pi}{4}=\frac{1}{2}\)

 

[willmoore21]

Galactacus, where you have stated a₀, you have integrated sin(x) which would indeed equal 4/pi, but you've wrote sin^2(x), which would equal 1. I take it you meant to write the integral of sin(x)?


Thanks

EDIT:

Also, surely an=0 when even and vice versa?

 

[willmoore21]

Galactacus, where you have stated a₀, you have integrated sin(x) which would indeed equal 4/pi, but you've wrote sin^2(x), which would equal 1. I take it you meant to write the integral of sin(x)?


Thanks

EDIT:

Also, surely an=0 when even and vice versa?

I had it figured out after a little while, just messed up some minus signs, which means an=0 when even instead of the other way around.

Thanks Galatacus.