Question

find the series of f(x) =x² in (0.2π)​

Answer 1

Step-by-step explanation:

Since this an even function, all of the Fourier coefficients for the sine terms are zero, bn=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=1 .

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

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))

to simplify the integral:

an=2π∫π0xsin(x)cos(nx)dx

The resulting Fourier series is:

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