Q13Mathematics I

Question

Find a Fourier series for the function $f (x) = x^{2}$ in the interval $- π < x < π$ and deduce the following:

\frac{1}{1 ^{2}} - \frac{1}{2 ^{2}} + \frac{1}{3 ^{2}} - \frac{1}{4 ^{2}} + \dots = \frac{π ^{2}}{12}

Answer 1

The goal is to mathematically determine the Fourier series expansion of the continuous function $f (x) = x^{2}$ over the perfectly symmetric interval $[- π, π]$ and utilize it to prove an infinite sum identity.

Step 1: Exploiting Mathematical Symmetry

The function $f (x) = x^{2}$ is strictly an even function because $f (- x) = (- x)^{2} = x^{2} = f (x)$ . In Fourier analysis, any perfectly even function will contain absolutely zero sine terms because sine functions are odd. Therefore, without any complex integration, we can declare all $b_{n} = 0$ .

Step 2: Calculating the Cosine Coefficients

We calculate the DC component, $a_{0}$ , using symmetry to integrate from $0$ to $π$ instead of $- π$ to $π$ :

a_{0} = \frac{2}{π} \int_{0}^{π} x^{2} d x = \frac{2}{π} [\frac{x ^{3}}{3}]_{0}^{π} = \frac{2 π ^{2}}{3}

Next, we systematically evaluate the harmonic cosine coefficients $a_{n}$ :

a_{n} = \frac{2}{π} \int_{0}^{π} x^{2} cos (n x) d x

This requires repeated integration by parts (or tabular integration). Integrating $x^{2} cos (n x)$ yields:

a_{n} = \frac{2}{π} [x^{2} \frac{sin ( n x )}{n} + 2 x \frac{cos ( n x )}{n ^{2}} - 2 \frac{sin ( n x )}{n ^{3}}]_{0}^{π}

Because $sin (nπ) = 0$ and $sin (0) = 0$ , the first and third terms completely vanish. Only the middle term survives:

a_{n} = \frac{2}{π} (2 π \frac{cos ( nπ )}{n ^{2}} - 0) = \frac{4}{n ^{2}} cos (nπ)

Since $cos (nπ)$ mathematically equals $(- 1)^{n}$ for all integers $n$ , we get $a_{n} = \frac{4 ( - 1 ) ^{n}}{n ^{2}}$ .

The rigorously derived Fourier series is therefore:

f (x) = \frac{a _{0}}{2} + n = 1 \sum \infty a_{n} cos (n x) = \frac{π ^{2}}{3} + 4 n = 1 \sum \infty \frac{( - 1 ) ^{n}}{n ^{2}} cos (n x)

Step 3: Deductive Evaluation at $x = 0$

To prove the required identity, we must evaluate this infinite function series precisely at $x = 0$ . At $x = 0$ , the original function $f (0) = 0^{2} = 0$ . Furthermore, $cos (n \cdot 0) = cos (0) = 1$ for all terms in the infinite sum.

0 = \frac{π ^{2}}{3} + 4 n = 1 \sum \infty \frac{( - 1 ) ^{n}}{n ^{2}}

We shift the constant term to isolate the summation:

- \frac{π ^{2}}{3} = 4 (\frac{- 1}{1 ^{2}} + \frac{1}{2 ^{2}} - \frac{1}{3 ^{2}} + \frac{1}{4 ^{2}} - \dots)

We divide strictly by $- 4$ to resolve the alternating signs and complete the mathematical proof:

\frac{π ^{2}}{12} = \frac{1}{1 ^{2}} - \frac{1}{2 ^{2}} + \frac{1}{3 ^{2}} - \frac{1}{4 ^{2}} + \dots

Q13Mathematics I

Question

Answer

Step 1: Exploiting Mathematical Symmetry

Step 2: Calculating the Cosine Coefficients

Step 3: Deductive Evaluation at x=0

Step 3: Deductive Evaluation at $x = 0$