2
$\begingroup$

I've been told that a function $f$ can be written by the Fourier series:

$$f(x) = 1 + \frac{1}{2}\sin(x)+ \sum_{n=1}^\infty \frac{1}{2^n}\cos(nx)$$

The way I get it is that a Fourier series is written by: $$ f(x) = \frac{1}{2}a_0 + \sum_{n=1}^\infty \big(a_n\cos(nx)+b_n\sin(nx) \big) $$

And that a function then will be even if $b_n = 0$, and odd if $a_n = 0$. I would say that $a_n = \frac{1}{2^n}$ so that $a_0 = 1$. So I don't quite get the part before the sum that just equals 1 (I thought it should be ½). I understand know that $b_1 = ½$ but if i'm suppose to write $a_n$ and $b_n$ on complex form i have been told that one has to do this: $c_0 = ½ * a_0, c_n = ½ *(a_n-i*b_n) , c_(-n) = ½ *(a_n + i*b_n)$

But I don't know how to do this, if I only know what $b_1$ is ?

$\endgroup$
  • $\begingroup$ $b_1 = \frac12$ for your series. $\endgroup$ – mrf Nov 11 '15 at 9:57
  • $\begingroup$ The function is neither, since it involves both sine terms and cosine terms $\endgroup$ – JVV Nov 11 '15 at 9:58
  • $\begingroup$ that would make sense - but then why does ½ *a_0 not equal ½ ?? $\endgroup$ – Linda Nov 11 '15 at 10:00
  • $\begingroup$ Why should $½ *a_0$ equal ½ ? $\endgroup$ – NeerajKumar Nov 11 '15 at 10:07
  • $\begingroup$ The part before the sum should correspond to $½ *a_0$, and I would say that $a_0 = 1$ $\endgroup$ – Linda Nov 11 '15 at 12:23

Your Answer

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

Browse other questions tagged or ask your own question.