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The task is: Expand $f(x) = x$ in a cosines serie.

My approach was to use the trigonometry fourier formula

$$ c_0 + \sum_{k=1}^{\infty } a_k \cos (k \Omega t) + b_k \sin (k \Omega t) $$

set $b_k = 0$ and calculate my $a_k$

$$ a_k = \frac{2}{l} \int_0^l \cos \left( \frac{k \pi t}{l} \right) f(t) dt$$

however I get

$$ \frac{2 t \sin \left( \frac{k \pi t}{l} \right)}{k \pi} + \frac{2 l \cos ( k \pi )}{(k \pi)^2} - \frac{2l}{k \pi} $$

The answer should be

$$ \frac{l}{2} - \frac{4 l}{ \pi ^2} \sum_{k=0}^{\infty} \frac{(-1)^{k+1} }{k} \sin (\frac{k \pi x}{l}) $$

So I believe that I'm have a totally wrong approach here. I would appreciate any help to put me in the right direction.

Hippalectryon
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iveqy
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  • The answer you give is the Fourier series of $f(x) = x$ on $[-l,l]$ not $[0,l]$. On what interval do you mean? It makes a difference! – Winther Oct 02 '14 at 21:16
  • The question does not specify an interval. I guess it would make more sense to use -l to l, however I still don't get rid of the sin and cos functions in my $a_k$ – iveqy Oct 02 '14 at 21:25
  • If you use $-l$ to $l$, then because $x$ is an odd function, cosine is an even function, and the product of even and odd functions is odd, the cosine terms in your Fourier series will drop out, leaving you with only sine terms. – Geoff Oxberry Oct 02 '14 at 21:28

1 Answers1

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Firstly, we need to know the interval which $f(x)=x$ is defined on. Let's assume it's defined on $(0,L)$. We also notice that $f$ is an odd function.

Since we want to approach $f$ through Fourier cosine series, we have to expand it in an even way on $(-L,L)$.

Thus, we define:

$g(x)=\begin{cases} -x,& -L <x<0 \\\\ x, &0<x<L \end{cases} $

Since $g(x)$ is even on $(-L,L)\implies b_k=0,\, \forall k\ge 0$.

$a_0=\displaystyle \dfrac 1 {2L}\cdot \int_{-L}^Lg(x)\,dx =\frac 1 L \cdot \int_0^L x\, dx=\frac L 2 $

$a_k\begin{array}[t]{l}=\displaystyle \dfrac 2 {2L}\cdot \int_{-L}^Lg(x)\cdot \cos\left(\frac {2k \pi x}{2L}\right)\,dx =\frac 1 L \cdot \int_{-L}^L g(x)\cdot\cos \left(\dfrac {k \pi x} {L}\right)\, dx\\\\=\displaystyle\frac 2 L \cdot \int_0^L g(x)\cdot \cos\left(\dfrac {k \pi x} {L}\right)\, dx=\frac 2 L \cdot \int_0^L x\cdot \cos\left(\dfrac {k \pi x} {L}\right)\, dx \\\\ =\displaystyle \frac{ 2\bigg(-1+(-1)^k\bigg)L}{k^2\pi^2} =\begin{cases}\dfrac{-4L}{k^2\pi^2}, & k: \text{ odd integer} \\\\ 0, & k: \text{ even integer} \end{cases} \end{array} $

So, the Fourier cosine series of $g$ on $(-L,L)$ is:

$$a_0+\sum_{k=1}^{\infty} a_k \cdot \cos\left(\dfrac {k \pi x} {L}\right)$$

$$\dfrac L 2 - \dfrac {4L} {\pi^2}\sum_{k=1,3,5\ldots}^{\infty} \frac 1 {k^2} \cdot \cos\left(\frac {k \pi x}{L}\right) $$

If you restrict $g$ on $(0,L)$ then you have the Fourier cosine series of $f$.

thanasissdr
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