Thread: 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);

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);

where

For the fourier sine series there's just one term

Spoiler:

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 , 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 and doubling.

The "odd extension" which, as implied, you can do without computation, is just sin(x) for 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.