Fourier Series — Question 1

Question 1 – Solution

Strategy. Use parity to separate coefficients, then use orthogonal projection to measure the combined approximation.

Step 1: Separate the two symmetries. The even part is $x^2$ and the odd part is $x$. Odd integrands give zero in the cross terms. Integration by parts gives $$\boxed{a_0=\frac{2\pi^2}{3},\qquad a_n=\frac{4(-1)^n}{n^2},\qquad b_n=\frac{2(-1)^{n+1}}n.}$$ For example, $\int_0^\pi x\sin nx=-\pi(-1)^n/n$ and $\int_0^\pi x^2\cos nx=2\pi(-1)^n/n^2$.

Step 2: Identify the periodic join. Inside the base interval the sum is $x+x^2$. At the join, the left limit is $\pi^2+\pi$ and the right limit is $\pi^2-\pi$, so the sum is $\pi^2$. Periodicity identifies the location of the join, not its two one-sided limits. The assigned value equals this average; the target still has a jump.

Step 3: Prove the projection claim. Orthogonality gives $$\boxed{p(x)=\pi^2/3-4\cos x+2\sin x.}$$ The residual $f-p$ is orthogonal to the three basis functions. For any other $q=A+B\cos x+C\sin x$, $$\|f-q\|_2^2=\|f-p\|_2^2+ 2\pi(A-\pi^2/3)^2+\pi(B+4)^2+\pi(C-2)^2.$$ Thus $p$ is the unique minimizer in this span.

Step 4: Compute the error independently. The odd cross term integrates to zero, giving $\|f\|_2^2=2\pi^5/5+2\pi^3/3$, while $\|p\|_2^2=2\pi^5/9+20\pi$. Consequently $$\boxed{\min\|f-q\|_2^2=\frac{8\pi^5}{45}+\frac{2\pi^3}{3}-20\pi.}$$ Changing isolated endpoint values changes neither integrals nor coefficients. It may change whether the assigned values equal the series sums.