Fourier Series — Question 4

Question 4 – Solution

Strategy. Keep the endpoint jump in integration by parts, and remove the mean before seeking a periodic primitive.

Step 1: Compute the full coefficients. For $k=n\pi$, exponential-trigonometric antiderivatives at $0,2$ give $$\boxed{m=\frac E2,\qquad a_n=\frac E{1+k^2},\qquad b_n=-\frac{kE}{1+k^2}.}$$ The Fourier sum at a join is $(1+e^2)/2$, since its two limits are $e^2$ and $1$. The assigned join value does not enter any integral.

Step 2: Retain the boundary contribution. On the open interval, $f'=f$. Integration by parts gives $$\int_0^2 f'\cos kx=E+k\int_0^2 f\sin kx,\qquad \int_0^2 f'\sin kx=-k\int_0^2 f\cos kx.$$ Thus $a_n=E+kb_n$ and $b_n=-ka_n$, yielding the coefficients above. Dropping $E$ would give $(1+k^2)a_n=0$ and then $b_n=0$. The periodic extension has a jump, so it is not a globally smooth periodic solution of $f'=f$.

Step 3: Construct the periodic primitive. Integrating the mean-zero forcing gives, on $[0,2]$, $$\boxed{F(x)=e^x-1-\frac E2x.}$$ Both endpoint values are zero, so its periodic extension is continuous. Two such primitives differ by a constant on each open period; continuity connects those constants, and $F(0)=0$ fixes the remaining freedom.

Step 4: Check the transformed coefficients. Direct integration gives mean $M=-1$. If $A_n,B_n$ denote the coefficients of $F$, the polynomial terms have zero cosine integral and $\int_0^2x\sin kx=-2/k$. Consequently $$\boxed{A_n=\frac E{1+k^2},\qquad B_n=\frac E{k(1+k^2)}.}$$ Indeed $kB_n=a_n$ and $-kA_n=b_n$. These are coefficient identities; the derivative equation is verified by the closed form on each open period. Finally, a continuous piecewise smooth periodic primitive of $f$ would have net change $\int_0^2f=E\ne 0$, which is impossible. Subtracting the mean is essential.