Question

Fourier series! Find the series of cosines
f(x) = 1, 0 ≤ x < π/2
0, π/2 < x ≤ π

Answer 1

To me this says what is the value of x when cos(x) = 1 on a range of 0 to π/2

Answer 2

and what is the value of x when cos(x) = 0 on a range of π/2 to π
Do you have a unit circle?

Answer 3

not

Answer 4

My thinking is not right?

Answer 5

yes but how to solve?

Answer 6

Oh, well you can use cos^-1(cos(x)) = cos^-1(0) this works for both of them.
side note: cos^-1 is equal to arccos
and arccos(cos(x)) = x

Answer 7

I know I'm a bit late to the game, but @Arfney, what you're suggesting has little to do with the problem at hand. What we want to find is the trigonometric series expansion of the given piecewise function.

Answer 8

Before I begin, the formulas were taken from this link: http://www.thefouriertransform.com/series/coefficients.php

\[f(x)=\begin{cases}1&\text{for }0\le x<\dfrac{\pi}{2}\\0&\text{for }\dfrac{\pi}{2}< x\le\pi\end{cases}\]
\[f(x)\sim a_0+\sum_{n=1}^\infty a_n\cos 2nx+\sum_{n=1}^\infty b_n\sin2nx\]
where
\[\begin{align*}a_0&=\frac{1}{\pi}\int_0^\pi f(x)~dx\\
a_n&=\frac{2}{\pi}\int_0^\pi f(x)\cos 2nx~dx\\
b_n&=\frac{2}{\pi}\int_0^\pi f(x)\sin 2nx~dx
\end{align*}\]
\[\begin{align*}a_0&=\frac{1}{\pi}\int_0^{\pi/2} dx\\
&=\frac{1}{2}\\\\
a_n&=\frac{2}{\pi}\int_0^{\pi/2} \cos 2nx~dx\\
&=\frac{2}{\pi}\frac{1}{2n} \bigg[\sin 2nx\bigg]_0^{\pi/2}\\
&=\frac{1}{n\pi}\sin n\pi\\
&=0\\\\
b_n&=\frac{2}{\pi}\int_0^{\pi/2} \sin 2nx~dx\\
&=\frac{2}{\pi}\left(-\frac{1}{2n}\right)\bigg[\cos 2nx\bigg]_0^{\pi/2}\\
&=-\frac{1}{n\pi}(\cos n\pi-1)\\
&=-\frac{1}{n\pi}((-1)^n-1)\\
&=\frac{1-(-1)^n}{n\pi}\\
\end{align*}\]
So you have
\[f(x)\sim\frac{1}{2}+\sum_{n=1}^\infty \frac{1-(-1)^n}{n\pi}\sin(nx)\]
Note that \((-1)^n=\begin{cases}-1&\text{for odd} n\\1&\text{for even }n\end{cases}\). Using this, we can split up the sine series as follows:
\[\sum_{n=1}^\infty \frac{1-(-1)^n}{n\pi}\sin(2nx)
=
\sum_{k=1}^\infty \frac{1-(-1)^{2k}}{2k\pi}\sin(2(2k)x)+\sum_{k=0}^\infty \frac{1-(-1)^{2k+1}}{(2k+1)\pi}\sin(2(2k+1)x)\\
\cdots=\sum_{k=1}^\infty \frac{1-1}{2k\pi}\sin(4kx)+\sum_{k=0}^\infty \frac{1-(-1)}{(2k+1)\pi}\sin(2(2k+1)x)\\
\cdots=\frac{2}{\pi}\sum_{k=0}^\infty \frac{1}{(2k+1)}\sin(2(2k+1)x)\]
Finally,
\[f(x)\sim\frac{1}{2}+\frac{2}{\pi}\sum_{k=0}^\infty \frac{1}{(2k+1)}\sin(2(2k+1)x)\]
The final product:
http://www.wolframalpha.com/input/?i=%281%2F2%29%2B%282%2Fpi%29*Sum%5B1%2F%282k%2B1%29*Sin%5B2%282k%2B1%29x%5D%2C%7Bk%2C0%2C10%7D%5D

Answer 9

@SithsAndGiggles. I am really sure that I did not understand it back when I answered, and I did not do my research =/. I do kinda remember Fourier Series now, but I think you covered everything in the problem. Thanks for letting me know about this.