Fourier Cosine Series — Question 6

Question 6 – Solution

Strategy. Use an average with a nonnegative kernel and unit total mass to preserve the range of the data.

Step 1: Exhibit a negative partial sum. The mean is $1/2$ and $\boxed{a_n=2\sin(n\pi/2)/(\pi n)}$. Thus $C_1=1/2+(2/\pi)\cos x$ has minimum $\boxed{1/2-2/\pi<0}$ at $\pi$, despite $0\leq f\leq 1$.

Step 2: Compute the weights and kernel. Mode $n$ appears in $N+1-n$ of the partial sums, giving $$\sigma_N=\frac 12+\sum_{n=1}^N\left(1-\frac n{N+1}\right)a_n\cos nx.$$ Counting pairs with index difference $n$ in the squared modulus gives $K_N=1+2\sum_{n=1}^N(1-n/(N+1))\cos nt$. The squared-modulus definition proves $K_N\geq 0$, and integrating the finite expansion gives $\int_{-\pi}^{\pi}K_N=2\pi$.

Step 3: Prove preservation of the range. Insert the finite kernel expansion into the convolution. The average integral of $F(x-t)\cos nt$ is $a_n\cos nx/2$; after shifting variables, its sine component vanishes by evenness of $F$. The constant term gives mean $1/2$. This proves the formula. Because $0\leq F\leq 1$ and the kernel is nonnegative with mass $2\pi$, $$\boxed{0\leq\sigma_N(x)\leq 1.}$$

Step 4: Identify the limits and tradeoff. Ordinary partial sums tend to one left of the jump, zero right of it, and $1/2$ at the jump. The extension is continuous at zero and $\pi$, with values one and zero. Arithmetic means of a convergent sequence have the same limit, so $\sigma_N$ has exactly these limits. Averaging damps the higher modes and prevents overshoot, but spreads the finite transition instead of fitting the edge as sharply.