my goal is to find the Fourier series of f on the given interval:

$$f(x) = \begin{cases} 0, & \text{if } -\pi < x < 0 \\ \sin(x), & \text{if } 0 \le x < \pi \end{cases}$$

I know there aren't any $\sin(nx)$ in the series but i'm having trouble with the $\cos(nx)$. this is my integral:

$$\frac1\pi\int_0^\pi \sin(x)\cos(nx)=\frac{\cos(nx)+1}{\pi(1-n^2)}=\frac{(-1)^n+1}{\pi(1-n^2)}$$

normaly i simply take the result and put it in the formula yet i don't know how to bring the result to this form:

$$\sum_{i=2}^\infty (\text{something}) \cos(nx)$$

please not it starts from 2 (i never encountered a task which it starts from 2).... any ideas what shall i do next?

  • $\begingroup$ You should edit your first question instead of creating a new essentially identical one. $\endgroup$ – mrf Aug 14 '13 at 21:54
  • $\begingroup$ i tagged the first to be deleted... why is that important...? i could also reconnect with proxy and new account and you wouldnt know its me. i just need some help i'm not here to stay... if you aren't willing to help it's oki... $\endgroup$ – jack Aug 14 '13 at 21:57
  • $\begingroup$ How does a person learn to construct arrays in TeX without learning that one can write $\le$ instead of $<=$? ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 14 '13 at 21:59
  • 2
    $\begingroup$ it took me a lot of time to find the article that explain how to use the tags correctly.. unfortunately <= was not explained there. sorry guys but this site isn't really newbie friendly.. anyhow can we stay focus on the question and laugh on my writing skills later...? $\endgroup$ – jack Aug 14 '13 at 22:04
  • 1
    $\begingroup$ again many thanks... i just wish that math books authors would simply explain the materiel with full detail examples instead of just focusing on theory. i don't know why there are so many exercises that are "left for the reader"... as the author says.. $\endgroup$ – jack Aug 14 '13 at 22:58

If: $$f(x)=\begin{cases} 0, & \text{if } -\pi < x < 0 \\ \sin(x), & \text{if } 0 \le x < \pi \end{cases}$$ then we find the coefficients as follows \begin{aligned} a_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}\left(\begin{cases} 0, & \text{if } -\pi < x < 0 \\ \sin(x), & \text{if } 0 \le x < \pi \end{cases}\right)\cos(nx){dx},\\ &=\frac{1}{\pi}\int_{0}^{\pi}\sin(x)\cos(nx){dx},\\ &=-{\frac {\cos \left( \pi \,n \right) +1}{\pi \, \left( {n}^{2}-1 \right) }}=-{\frac {(-1)^n +1}{\pi \, \left( {n}^{2}-1 \right) }},\\ =&\begin{cases} 0,& \text{if } n \text{ is odd} \\ {-\frac {2}{\pi \, \left( n^2-1\right) }}, & \text{if } n \text{ is even} \end{cases} \end{aligned} if $n=1$ in the second from last line we would take the limit and see that it is zero; alternatively we could plug $n=1$ in the integral and see that that is zero, but either way we don't need any odd $n$ they are all zero.

Now we check the $b_n$: \begin{aligned} b_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}\left(\begin{cases} 0, & \text{if } -\pi < x < 0 \\ \sin(x), & \text{if } 0 \le x < \pi \end{cases}\right)\sin(nx){dx},\\ &=\frac{1}{\pi}\int_{0}^{\pi}\sin(x)\sin(nx){dx},\\ &=-{\frac {\sin \left( \pi \,n \right) }{\pi \, \left( {n}^{2}-1 \right) }},\\ =&\begin{cases} 0,& \text{if } n\ne 1 \\ -1/2 & \text{if } n=1 \end{cases} \end{aligned} so we have to be careful there, we either take the limit when $n\rightarrow1$ or put $n=1$ in the integral but we do need one of the $sin$ terms. We then have: \begin{aligned} f(x)&=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n \cos(nx)+b_n\sin(nx),\\ &=\frac{1}{\pi}+\frac{\sin(x)}{2}-\sum_{n=1}^{\infty}{\frac {2}{\pi \, \left( 4n^2-1\right) }} \cos(2nx). \end{aligned} enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.