Fourier Series of an Odd Function and a Series Identity: 04-BS-5, December 2014, Q2

Question 2: Fourier Series of an Odd Function and a Series Identity (14+6 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $f(x)$, period $p=2$ (half-period $l=1$): $f(x)=x(1+x)$ on $(-1,0)$, $f(x)=x(1-x)$ on $(0,1)$.

Find. (A) The full Fourier series of $f(x)$. (B) The value of $\sum_{n=1}^{\infty}(-1)^{n-1}/(2n-1)^3$, using the series at a convenient point.

f(x): odd, period-2 parabolic zig-zag (two periods shown)

Approach. Check symmetry first: $f(-x)=-f(x)$ for every $x$ in $(0,1)$, so $f$ is odd and its Fourier series contains sine terms only ($a_0=a_n=0$). Compute $b_n$ by integrating by parts twice, then evaluate the series at $x=\tfrac12$ — a point where every sine factor collapses to $\pm1$ — to isolate the target sum.

Confirm odd symmetry. For $0\lt x\lt1$, $-x\in(-1,0)$ so $f(-x)=(-x)(1+(-x))=-x(1-x)=-f(x)$. Hence $f$ is odd, $a_0=0$, $a_n=0$ for all $n$, and (with half-period $l=1$) $$f(x)=\sum_{n=1}^{\infty}b_n\sin(n\pi x),\qquad b_n=\dfrac{2}{l}\int_0^lf(x)\sin\!\left(\dfrac{n\pi x}{l}\right)dx=2\int_0^1x(1-x)\sin(n\pi x)\,dx$$
Integrate by parts twice. With $u=x-x^2$, $dv=\sin(n\pi x)dx$, two applications of integration by parts (the boundary terms vanish at $x=0,1$ since $u(0)=u(1)=0$) give $$b_n=2\left[\dfrac{2\big(1-(-1)^n\big)}{n^3\pi^3}\right]=\dfrac{4\big(1-(-1)^n\big)}{n^3\pi^3}$$ so $b_n=0$ when $n$ is even, and for $n$ odd, $(-1)^n=-1$ gives $$\boxed{b_n=\dfrac{8}{n^3\pi^3}\quad(n\text{ odd}),\qquad b_n=0\quad(n\text{ even})}$$
Write the series over odd indices $n=2k+1$. $$\boxed{f(x)=\dfrac{8}{\pi^3}\sum_{k=0}^{\infty}\dfrac{\sin\big((2k+1)\pi x\big)}{(2k+1)^3}}$$
(B) Evaluate at $x=\tfrac12$. Directly, $f(\tfrac12)=\tfrac12(1-\tfrac12)=\tfrac14$. From the series, $\sin\!\big((2k+1)\pi/2\big)=(-1)^k$, so $$\dfrac14=\dfrac{8}{\pi^3}\sum_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)^3}\ \Longrightarrow\ \sum_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)^3}=\dfrac{\pi^3}{32}$$ Re-indexing $n=k+1$ (so $2k+1=2n-1$ and $(-1)^k=(-1)^{n-1}$) reproduces the requested identity exactly: $$\boxed{\dfrac{\pi^3}{32}=\sum_{n=1}^{\infty}\dfrac{(-1)^{(n-1)}}{(2n-1)^3}}$$ (numerically both sides equal $0.968946\ldots$).

Fourier coefficients and the resulting identity
Quantity	Result
a₀, a_n	0 (f is odd)
b_n, n even	0
b_n, n odd	8/(n³π³)
Σ(−1)ⁿ⁻¹/(2n−1)³	π³/32 ≈ 0.968946

Topic: Fourier sine series of an odd periodic function and series summation via pointwise evaluation.

Key relations: for an odd $f$ of period $2l$, $f(x)=\sum b_n\sin(n\pi x/l)$ with $b_n=\tfrac2l\int_0^lf(x)\sin(n\pi x/l)dx$; evaluating the resulting series at a special point where $\sin(n\pi x/l)$ takes simple values ($\pm1$ or $0$) converts an infinite Fourier series into a closed-form numerical series identity.

Why this works: Dirichlet’s theorem guarantees pointwise convergence of the Fourier series to $f(x)$ at every point where $f$ is continuous (true here at $x=\tfrac12$), so substituting $x=\tfrac12$ is legitimate and directly ties the series to the elementary value $f(\tfrac12)=\tfrac14$.

Common pitfall: choosing an evaluation point where the series does not simplify (e.g. an irrational multiple of the period), or forgetting the sign pattern $(-1)^k$ introduced by $\sin((2k+1)\pi/2)$, which is what turns the sum into an alternating series rather than $\sum1/(2n-1)^3$.

Source: Kreyszig, Advanced Engineering Mathematics, Fourier Series (Ch. 11).

Question 2 of 7: Fourier Series of an Odd Function and a Series Identity

Question 2: Fourier Series of an Odd Function and a Series Identity (14+6 marks)