Question

Find the Fourier series expansion of f(x) = sin x; in (- 1,1) Hence deduce that 1/1^ 2 + 1/3^ 2 +....= pi^ 2/3​

Answer 1

Since this an even function, all of the Fourier coefficients for the sine terms are zero, bn=0bn=0.

The constant term at the beginning of the Fourier series, which represents the mean value of the function over its defined periodic interval, is:

1π∫π0xsin(x)dx=11π∫0πxsin⁡(x)dx=1.

To get the Fourier coefficients for the cosine terms, anan, use the trigonometry identity:

sin(x)cos(nx)=12(sin(x−nx)+sin(x+nx))sin⁡(x)cos⁡(nx)=12(sin⁡(x−nx)+sin⁡(x+nx))

or

sin(x)cos(nx)=12(sin(−(n−1)x)+sin((n+1)x))sin⁡(x)cos⁡(nx)=12(sin⁡(−(n−1)x)+sin⁡((n+1)x))

to simplify the integral:

an=2π∫π0xsin(x)cos(nx)dxan=2π∫0πxsin⁡(x)cos⁡(nx)dx

The resulting Fourier series is:

$f(x)=1−12cos(x)−2∑∞n=2(−1)n(n2−1)cos(nx)$