Question

If $f(x) = x^3$, $0 \le x \le 4$, $f(x+4) = f(x)$ $\forall x \in \mathbb{R}$ and the Fourier series of $f(x)$ is $f(x) = \sum_{n=0}^{\infty} \left(a_n \cos \frac{n\pi x}{2} + b_n \sin \frac{n\pi x}{2}\right)$, then $a_0 =$}

• A. \(8\)
• B. \(32\)
• C. \(16\)
• D. \(24\)

Hint:

Always double check whether a Fourier question writes the constant term as $\frac{a_0}{2}$ or simply as part of a summation $\sum a_n \cos(\dots)$. If written inside the summation directly, $a_0$ represents the exact absolute mathematical average $\frac{1}{T}\int f(x)dx$ instead of $\frac{2}{T}\int f(x)dx$.

Answer 1

The Correct Option is C

Concept: The standard representation of a Fourier series for a function with period $T = 2L$ defined on an interval $[c, c+2L]$ is given by: \[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos \frac{n\pi x}{L} + b_n \sin \frac{n\pi x}{L} \right) \] However, notice carefully that in the problem description, the summation begins from $n=0$ with no separate constant term divided by 2: \[ f(x) = \sum_{n=0}^{\infty} \left(a_n \cos \frac{n\pi x}{2} + b_n \sin \frac{n\pi x}{2}\right) \] Evaluating the term inside the summation specifically for $n=0$: \[ a_0 \cos(0) + b_0 \sin(0) = a_0 \cdot 1 + 0 = a_0 \] Thus, the continuous definition of the DC value (average value) under this specific notation requires setting up the average value over the period: \[ a_0 = \frac{1}{T}\int_{0}^{T} f(x)\,dx \] Here, the period $T = 4$, as indicated by the translation condition $f(x+4) = f(x)$.

Step 1: Set up the definitive integral.
Given $f(x) = x^3$ over the primary period interval $[0, 4]$, the formula for this non-standard definition coefficient $a_0$ is: \[ a_0 = \frac{1}{4} \int_{0}^{4} x^3 \, dx \]

Step 2: Perform the integration.
Using the standard power rule of calculus ($\int x^n dx = \frac{x^{n+1}}{n+1}$): \[ \int_{0}^{4} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{4} \] Substituting the limits of integration: \[ = \left( \frac{4^4}{4} \right) - \left( \frac{0^4}{4} \right) = \frac{256}{4} - 0 = 64 \]

Step 3: Scale by the pre-integral coefficient.
Multiply the integrated value by the fraction outside: \[ a_0 = \frac{1}{4} \times 64 = 16 \] This gives $a_0 = 16$, which matches Option (C).