Question

in the fourier series expansion of function f(x)= x (-π,π) fourier coeffecent bn is given by​

Answer 1

Answer:

f(x)=π23+4 ∑n=1+∞(−1)nn2 cos(nx)

Step-by-step explanation:

Calculate Fourier series of f(x)=x2 , x∈ [−π,π] and determine module and phase spectrum

f(x)=a02+∑n=1+∞an cos(nx) + bn sin(nx)

a0=1π∫π−πx2 dx=23π2

an=1π∫π−πx2 cos(nx) dx=1π 2(π2n2−2)sin(nπ)+4πncos(nπ)n3=4(−1)nn2

bn=0∀n≥1

because f is even

f(x)=π23+4 ∑n=1+∞(−1)nn2 cos(nx)

mark brainliest thanks