Q20Mathematics I

Question

Find half range cosine series for the function $f (x) = 2 x - 1, 0 < x < 1$ . Hence deduce that:

\frac{π ^{2}}{8} = \frac{1}{1 ^{2}} + \frac{1}{3 ^{2}} + \frac{1}{5 ^{2}} + \dots

Answer 1

The mathematical objective is to express the simple non-periodic function $f (x) = 2 x - 1$ , defined strictly on the half-interval $x \in (0, 1)$ , exclusively as an infinite sum of cosine waves. To achieve this, we theoretically extend the function into the negative domain as a perfectly even function. By definition, an even function's Fourier series will automatically possess absolutely zero sine terms ( $b_{n} = 0$ for all $n$ ).

Step 1: Calculating the DC Component ( $a_{0}$ )

The formal half-range Fourier cosine series formula over an interval $(0, l)$ strictly dictates that the constant term is calculated as:

a_{0} = \frac{2}{l} \int_{0}^{l} f (x) d x

For this problem, the fundamental interval length $l$ is precisely $1$ . We substitute this and our function into the integral:

a_{0} = 2 \int_{0}^{1} (2 x - 1) d x

We execute the basic polynomial integration strictly:

a_{0} = 2 [x^{2} - x]_{0}^{1} = 2 ((1^{2} - 1) - (0 - 0)) = 2 (0) = 0

The complete absence of a DC offset makes mathematical sense because the area of the function exactly above the x-axis perfectly cancels the area exactly below it on the interval $(0, 1)$ .

Step 2: Calculating the Harmonic Coefficients ( $a_{n}$ )

The general formula for the cosine coefficients is rigorously defined as:

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

Substituting our specific $f (x)$ and setting $l = 1$ yields a complex integral requiring integration by parts:

a_{n} = 2 \int_{0}^{1} (2 x - 1) cos (nπ x) d x

We meticulously apply the integration by parts formula $\int u d v = uv - \int v d u$ by setting $u = 2 x - 1$ and $d v = cos (nπ x) d x$ :

a_{n} = 2 [(2 x - 1) \frac{sin ( nπ x )}{nπ} - \int 2 \frac{sin ( nπ x )}{nπ} d x]_{0}^{1}

We now rigidly integrate the remaining sine term:

a_{n} = 2 [(2 x - 1) \frac{sin ( nπ x )}{nπ} - 2 (- \frac{cos ( nπ x )}{n ^{2} π ^{2}})]_{0}^{1}

a_{n} = 2 [(2 x - 1) \frac{sin ( nπ x )}{nπ} + \frac{2 cos ( nπ x )}{n ^{2} π ^{2}}]_{0}^{1}

We systematically evaluate this exact mathematical expression at the rigorous boundaries $1$ and $0$ . A critical realization is that $sin (nπ) = 0$ for all integer values of $n$ , and $sin (0) = 0$ . Therefore, all sine terms universally vanish.

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

a_{n} = 2 (\frac{2 ( - 1 ) ^{n}}{n ^{2} π ^{2}} - \frac{2 ( 1 )}{n ^{2} π ^{2}}) = \frac{4}{n ^{2} π ^{2}} ((- 1)^{n} - 1)

We must strictly analyze this resulting term for different parities of $n$ : - When $n$ is an even integer ( $2, 4, 6...$ ), $(- 1)^{n} = 1$ . The parenthesis becomes $(1 - 1) = 0$ . All even harmonics absolutely disappear. - When $n$ is an odd integer ( $1, 3, 5...$ ), $(- 1)^{n} = - 1$ . The parenthesis becomes $(- 1 - 1) = - 2$ . Thus, $a_{n} = \frac{4}{n ^{2} π ^{2}} (- 2) = \frac{- 8}{n ^{2} π ^{2}}$ .

The perfectly derived half-range Fourier series consists solely of odd-harmonic cosine waves:

f (x) = n = 1, 3, 5... \sum \infty \frac{- 8}{n ^{2} π ^{2}} cos (nπ x)

Step 3: Deductive Proof of the Infinite Series

To deductively prove the given identity, we must selectively evaluate the derived Fourier series at a specific coordinate point. The most mathematically fruitful point is the boundary $x = 0$ .

At $x = 0$ , the original function evaluates perfectly to $f (0) = 2 (0) - 1 = - 1$ . Furthermore, the cosine term within the sum evaluates perfectly to $cos (0) = 1$ for all values of $n$ .

We substitute these exact rigid values back into our Fourier series equation:

- 1 = n = 1, 3, 5... \sum \infty \frac{- 8}{n ^{2} π ^{2}} \cdot (1)

We can factor out the absolute constants $\frac{- 8}{π ^{2}}$ entirely from the infinite summation:

- 1 = \frac{- 8}{π ^{2}} (\frac{1}{1 ^{2}} + \frac{1}{3 ^{2}} + \frac{1}{5 ^{2}} + \dots)

To finalize the mathematical proof, we simply multiply both sides of the exact equation by $\frac{- π ^{2}}{8}$ , which immediately isolates the summation series and confirms the identity flawlessly:

\frac{π ^{2}}{8} = \frac{1}{1 ^{2}} + \frac{1}{3 ^{2}} + \frac{1}{5 ^{2}} + \dots

Q20Mathematics I

Question

Answer

Step 1: Calculating the DC Component (a0​)

Step 2: Calculating the Harmonic Coefficients (an​)

Step 3: Deductive Proof of the Infinite Series

Step 1: Calculating the DC Component ( $a_{0}$ )

Step 2: Calculating the Harmonic Coefficients ( $a_{n}$ )