Given $S = \Sigma a_k + b_k$ and $I = \int^{\pi}_{-\pi} f^2 dx$, determine if $S=I $ . Function $f=\|cos x|$

Question

So, the first part of the task was to find the Fourier series for the function $f=\left|\cos x\right|$ and now I have to tell whether $\sum a_k + b_k = \int^{\pi}_{-\pi} f^2 \,dx$. The coefficients $a_k, b_k$ are the coefficients for Fourier series.

I computed Fourier series representation of the function: $f(x) = -\frac{1}{2\pi}$. For some reason I got a negative coefficient $a_0 = \frac{1}{2\pi}\int^{-\pi}{\pi} \left|\cos x\right| = \frac{1}{2\pi}\int^{-\pi}_{\pi} \left|\cos x\right|\,dx = -\frac{1}{2\pi}\int_{-\pi}^{-\frac{\pi}{2}} \cos x \,dx + \frac{1}{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos x \,dx- \frac{1}{2\pi}\int^{\pi}_{\frac{\pi}{2}}\cos x \,dx = -\frac{1}{\pi}$. The middle term cancels out so I'm left with: $$-\frac{1}{2\pi}\int_{-\pi}^{-\frac{\pi}{2}}\cos x \,dx- \frac{1}{2\pi}\int^{\pi}_{\frac{\pi}{2}}\cos x \,dx = \frac{1}{2\pi}\left[\sin x \right]^{-\frac{\pi}{2}}_{-\pi}+\frac{1}{2\pi}\left[\sin x \right]^{\pi}_{\frac{\pi}{2}}= -\frac{1}{2\pi}-\frac{1}{2\pi}= -\frac{1}{\pi}$$ I know this cannot be true as $f(x)$ should be positive and all the rest of the coefficients for Fourier series cancel out. Anyway in the result I get that Fourier representation for the given function is: $f(x) = -\frac{1}{2\pi}$ Then I calculate the given integral $$I: \int^{\pi}_{-\pi} f^2 \,dx = \int^{\pi}_{-\pi}\cos^2 x \,dx =\pi$$ But I am not sure about the sum $\sum a_k +b_k$. Is it just a sum of all the coefficients, in this case: $a_0 + a_n + b_n = -\frac{1}{\pi}$? So, I get that $S \neq I$, but I know I am wrong about the $a_0$ coefficient.

Answer 1

$\int_{-\pi}^{\pi} \cos^2{x} \; dx$ is not equal to $-\frac{1}{3}\sin^3{x} \big\rvert_{-\pi}^{\pi}$, you should actually use the power reduction formula: $\cos^2{x}=\frac{1}{2}\left(1+\cos{2x}\right)$. So the integral is equal to $\frac{x}{2}+\frac{1}{4}\sin{2x} \big\rvert_{-\pi}^{\pi}=\pi$. Just differentiate $-\frac{1}{3}\sin^3{x}$ with respect to $x$ to see that the integrand is not equivalent.