Question

The Fourier series expansion is given as: 
\[ \frac{a_{0}}{2} + a_{1}\cos x + a_{2}\cos 2x + \cdots + b_{1}\sin x + b_{2}\sin 2x + \cdots \] 

For the function \[ f(x) = x + \frac{x^{2}}{4}, \quad -\pi \le x \le \pi \] 
The value of \(a_0\) and \(b_2\) are:

• A. $a_{0}=\frac{3\pi^{2}}{5},b_{2}=-2$
• B. $a_{o}=\frac{\pi^{2}}{6},b_{2}=+1$
• C. $a_{o}=\frac{\pi^{2}}{6},b_{2}=-1$
• D. $a_{0}=\frac{5\pi^{2}}{3},b_{2}=+2$

Hint:

When the interval is symmetric $[-\pi, \pi]$, always check if the function is Odd or Even. Integral of an Odd function over this interval is always 0, saving you significant calculation time.

Answer 1

The Correct Option is C

Concept: The Fourier coefficients for a function $f(x)$ on the interval $[-\pi, \pi]$ are calculated using: $a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx$ and $b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) sin(nx) dx$.

Step 1: Calculate $a_0$.
$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} (x + \frac{x^2}{4}) dx$. Note: $\int_{-\pi}^{\pi} x dx = 0$ (odd function over symmetric interval). $a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{x^2}{4} dx = \frac{1}{4\pi} [\frac{x^3}{3}]_{-\pi}^{\pi} = \frac{1}{4\pi} (\frac{\pi^3}{3} - \frac{-\pi^3}{3}) = \frac{1}{4\pi} \cdot \frac{2\pi^3}{3} = \frac{\pi^2}{6}$.

Step 2: Calculate $b_2$.
$b_2 = \frac{1}{\pi} \int_{-\pi}^{\pi} (x + \frac{x^2}{4}) sin(2x) dx$.
Split the integral: $b_2 = \frac{1}{\pi} [\int_{-\pi}^{\pi} x sin(2x) dx + \int_{-\pi}^{\pi} \frac{x^2}{4} sin(2x) dx]$.
Note: $\frac{x^2}{4} sin(2x)$ is (Even $\times$ Odd) = Odd, so its integral is 0. $b_2 = \frac{1}{\pi} \int_{-\pi}^{\pi} x sin(2x) dx$. Using integration by parts: $= \frac{1}{\pi} [x(\frac{-cos 2x}{2}) - (1)(\frac{-sin 2x}{4})]_{-\pi}^{\pi}$ $= \frac{1}{\pi} [(\frac{-\pi cos 2\pi}{2}) - (\frac{-(-\pi) cos (-2\pi)}{2})] = \frac{1}{\pi} [-\frac{\pi}{2} - \frac{\pi}{2}] = \frac{-\pi}{\pi} = -1$.

Step 3: Result.
$a_0 = \frac{\pi^2}{6}$ and $b_2 = -1$.