1. ## Fourier series cosine

Hello can anybody help me with this problem?
Find the even (cosine) extension of the function given in
Q. 6 as a Fourier-series. Write down, without making any calculations, the odd extension as a Fourier

This is the fuction given in question 6

f
(x) = sin (x)
0 (less than or equal to) x (less than or equal to) (pi);

2. Originally Posted by math_lete
Hello can anybody help me with this problem?

Find the even (cosine) extension of the function given in
Q. 6 as a Fourier-series. Write down, without making any calculations, the odd extension as a Fourier

This is the fuction given in question 6

f
(x) = sin (x)

0 (less than or equal to) x (less than or equal to) (pi);

$\displaystyle y_c = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos n x$ where $\displaystyle a_n = \frac{2}{\pi}\int_0^{\pi} \sin x \cos n x\,dx$

For the fourier sine series there's just one term

Spoiler:
$\displaystyle y_s = \sin x$

3. Originally Posted by math_lete
Hello can anybody help me with this problem?
Find the even (cosine) extension of the function given in
Q. 6 as a Fourier-series. Write down, without making any calculations, the odd extension as a Fourier

This is the fuction given in question 6

f
(x) = sin (x)
0 (less than or equal to) x (less than or equal to) (pi);
The "even extension would be, of course, f(x)= -sin(x) for $\displaystyle -\pi\le x\le 0$, sin(x) for [tex]0\le x\le \pi[tex]. Apply the usual formulas for Fourier coefficients to that. Simplifications: because this is an even function, the sine coefficients will be 0 and you can get the cosine coefficients by integrating from 0 to $\displaystyle \pi$ and doubling.

The "odd extension" which, as implied, you can do without computation, is just sin(x) for $\displaystyle -\pi\le x \le \pi$ and the Fourier series is just sin(x) itself- that is all coefficients of cos(x) are 0, the coefficient of sin(nx) is 0 unless n= 1 in which case it is 1.