Fourier Cosine Series — Question 1

Question 1 – Solution

Strategy. Separate the mean convention from positive-frequency normalization, and distinguish function convergence from derivative convergence.

Step 1: Compute the mean and coefficients. The zero-frequency integral gives $a_0=\pi$, so the mean term is $\pi/2$. Integration by parts for $n\geq 1$ gives $$\int_0^\pi x\cos nx\,dx=\frac{(-1)^n-1}{n^2},\qquad \boxed{a_n=\frac{2((-1)^n-1)}{\pi n^2}.}$$ Even modes vanish; odd coefficients are $-4/(\pi n^2)$. Using $a_0$ in place of $a_0/2$ adds the erroneous constant $\pi/2$.

Step 2: Identify the extension and its limits. The extension is $|x|$ on $[-\pi,\pi]$, repeated with period $2\pi$. It is continuous, with corners at multiples of $\pi$. At zero its two limits are zero; at $\pi$ both are $\pi$. Therefore $$\boxed{x=\frac\pi 2-\frac 4\pi\sum_{j=0}^\infty \frac{\cos((2j+1)x)}{(2j+1)^2},\qquad 0\leq x\leq\pi.}$$

Step 3: Bound the uniform error. The bound $|a_n|\leq 4/(\pi n^2)$ gives absolute uniform convergence by the Weierstrass test. The Fourier convergence theorem identifies the sum with $f$, including the endpoints. For $N\geq 1$, $$\boxed{\|f-C_N\|_\infty\leq\frac 4\pi\sum_{n>N}n^{-2}\leq\frac 4{\pi N}.}$$

Step 4: Reject an unjustified derivative limit. Every finite derivative is a sine sum, so $C_N'(0)=C_N'(\pi)=0$. The target has one-sided endpoint derivatives one. Thus $\boxed{\sup_{[0,\pi]}|C_N'-1|\geq 1}$ for every $N$. Uniform convergence of functions alone does not permit differentiating the limit or passing endpoint derivative conditions to it.