Fourier Sine Series — Question 7

Question 7 – Solution

Strategy. Remove the nonzero boundary values before asking a sine series for uniform approximation.

Step 1: Compute the raw coefficients. For $k=n\pi$, an antiderivative is $e^x(\sin kx-k\cos kx)/(1+k^2)$. Evaluation gives $$\boxed{b_n=\frac{2k(1-(-1)^ne)}{1+k^2}.}$$ The odd periodic extension has opposite one-sided values at each endpoint, so its jump averages are zero. Every raw sine partial sum is also zero there.

Step 2: Cancel the leading boundary contribution. Integration of the linear interpolant gives $b_n^{\ell}=2(1-(-1)^ne)/k$. Therefore $$\boxed{c_n=b_n-b_n^{\ell}= -\frac{2(1-(-1)^ne)}{k(1+k^2)}.}$$ The raw coefficients are of order $1/n$, while $c_n$ are of order $1/n^3$. The residual has zero values at both endpoints; the leading boundary term cancels.

Step 3: Establish a uniform certificate. Since $|c_n|\leq 2(1+e)/(\pi^3n^3)$, the residual series converges absolutely and uniformly. Its continuous piecewise smooth odd extension identifies the sum with $r$ throughout the closed interval. Thus $$\boxed{\|e^x-A_N\|_\infty\leq\frac{1+e}{\pi^3N^2}.}$$ For $N=12$, the bound is below $10^{-3}$, so twelve modes suffice.

Step 4: Interpret the endpoint comparison. For the raw sum $S_N$, $f-S_N$ equals one at zero and $e$ at one. For $A_N$, both endpoint errors are exactly zero, and the entire error has the small bound above. The graph shows these two errors on the same scale. The corrected approximant includes the non-sine function $\ell$; it is not a pure sine sum and therefore is not constrained to vanish at the endpoints.