Find the trigonometric Fourier series for the function f(x): [-T/2, T/2] - R given by the expression: f(x) = ² FS(2)=sinhr|1+2 +[₁ FS(x) = FS(x) = FS(x) sinh r sin T 70 sinh T T [ 21 12 2 ∞ (-1)" n² +1 + Σn=1 + ΣΩ + Ex-1 (-1)" n²+1 (-1)" n+1 (cos 2nx + n sin 2nx) 2na)]. (-1)" n²+1 (cos 2nx - n sin 2nx) nx)]. (cos 2nx - (cos 2nx - n sin 2nx) sin 2nx)

Consider f(x) = sin(x²). Is the function even, odd or neither, and which conse- quence does the answer have for its Fourier series?

= 1) The function f(x) periodic on the interval [0, 2л] has complex Fourier series f(x): Σ(1/n²) einx where the sum over n goes from - infinity to infinity. Convert this to cosine and sine Fourier Series by finding the values of A's and B's in the expression Ao + ΣAn cos(nx) + Σ Bn sin(nx) where each sum goes from 1 to infinity. Hint: consider the n and -n term together in the complex Fourier Series or use Euler's identity.

Find the trigonometric Fourier series for the function f(x): [-T/2, π/2] → R given by the expression: ƒ(2) - {0 = O о O O cos 2x if x = [-π/2, 0] 0 if x = (0, π/2] O ∞ FS(x) = -2 cos(2x) + 1 n=1 FS(x) = −cos(2x) + Σ2 FS(x) = −sin(2x) + Σn=2 FS(x) = cos(2x) + Σn=0 FS(x) = cos(2x) + Ex-1 - n² cos² (n²-1)π 2n cos² n cos² (=) 2 n cos² (n²-1)T (+) 2 2(n²-1)π NE (+) (n.²-1) 2n cos² * ( =) 2 (n²+1)π -sin(2nx). -sin(2nx). -sin(nx). sin(2nx). -sin(2nx).