Fourier Cosine Series — Question 4

Question 4 – Solution

Strategy. Remove the endpoint-slope contribution while retaining the residual’s generally nonzero mean.

Step 1: Compute the raw data. For $k=n\pi$, direct integration gives $$\boxed{\text{mean}=e-1,\qquad a_n=\frac{2(e(-1)^n-1)}{1+k^2}.}$$ Every raw finite cosine sum has zero endpoint derivatives, disagreeing with $f'(0)=1$ and $f'(1)=e$.

Step 2: Cancel the slope contribution. The derivative $\ell'=1+(e-1)x$ matches both target endpoint slopes. Two integrations by parts give its positive-frequency coefficient $2(e(-1)^n-1)/k^2$; the constant second derivative has zero cosine integral. Subtracting yields $$\boxed{m_r=\frac 56(e-1)-\frac 12,\qquad c_n=-\frac{2(e(-1)^n-1)}{k^2(1+k^2)}.}$$ The residual has zero endpoint derivatives, so the leading slope term cancels and fourth-power decay remains. Its mean must still be included.

Step 3: Certify the function error. The bound $|c_n|\leq 2(1+e)/(\pi^4n^4)$ gives uniform absolute convergence. The continuous piecewise smooth even extension identifies the sum with $r$. Consequently $$\boxed{\|e^x-A_N\|_\infty\leq\frac{2(1+e)}{\pi^4} \int_N^\infty t^{-4}dt=\frac{2(1+e)}{3\pi^4N^3}.}$$ The choice $\boxed{N=7}$ makes this bound less than $10^{-4}$.

Step 4: Verify the derivative behavior. Derivative coefficients are bounded by a constant times $n^{-3}$, so that series converges uniformly. Together with convergence at one point, this justifies differentiation. Every residual partial sum has zero endpoint slopes; adding $\ell$ gives $A_N'(0)=1,A_N'(1)=e$ exactly. Slope matching does not force exact endpoint values at finite $N$; those are controlled by the function-error bound and become exact in the limit.