Question

Let $f(x) = \begin{cases} h(x), & 0<x<C \\-h(-x), & -C<x<0 \end{cases}$ and $f(x+2C)=f(x) \; \forall x \in \mathbb{R}$. If the Fourier series of $f(x) = \sum_{n=0}^\infty \left(a_n \cos\frac{n\pi x}{C} + b_n \sin\frac{n\pi x}{C}\right)$ then $\sum_{n=0}^\infty a_n b_n =$

• A. $\left(\int_{-C}^{C} h(x)\cos\frac{n\pi x}{C}dx\right) \left(\int_{-C}^{C} h(-x)\sin\frac{n\pi x}{C}dx\right)$
• B. $\int_{-C}^{C} f(x)\cos\frac{n\pi x}{C}dx$
• C. $2 \int_{0}^{C} h(x)\sin\frac{n\pi x}{C}dx$
• D. $2 \int_{-C}^{0} -h(-x)\sin\frac{n\pi x}{C}dx$

Hint:

• Determine if $f(x)$ is even or odd. $f(-x) = \begin{cases} h(-x), & 0<-x<C \implies -C<x<0
  -h(-(-x)), & -C<-x<0 \implies 0<x<C \end{cases}$ So $f(-x) = \begin{cases} -h(x), & 0<x<C
  h(-x), & -C<x<0 \end{cases}$. Comparing with $f(x)$: For $0<x<C$, $f(x)=h(x)$, so $f(-x)=-f(x)$. For $-C<x<0$, $f(x)=-h(-x)$, so $f(-x) = -f(x)$. Thus $f(x)$ is an odd function.
• For an odd function, Fourier coefficients $a_n = 0$ for all $n \ge 0$.
• Fourier coefficients $b_n$ are generally non-zero for an odd function.
• The sum $\sum_{n=0}^\infty a_n b_n = \sum_{n=0}^\infty (0 \cdot b_n) = 0$.
• Evaluate the given options. Option (b) is $\int_{-C}^{C} f(x)\cos\frac{n\pi x}{C}dx$. Since $f(x)$ is odd and $\cos$ is even, $f(x)\cos(\cdot)$ is odd. The integral of an odd function over $[-C, C]$ is 0. This matches the sum's value.

Answer 1

The Correct Option is B

Given a function \( f(x) \) defined as:

\[ f(x) = \begin{cases} h(x), & 0<x<C \\ -h(-x), & -C<x<0 \end{cases} \]

and it satisfies the periodic condition \( f(x + 2C) = f(x) \) for all real \( x \).

The Fourier series expansion of \( f(x) \) is given by:

\[ f(x) = \sum_{n=0}^\infty \left(a_n \cos \frac{n \pi x}{C} + b_n \sin \frac{n \pi x}{C} \right) \]

We are asked to find the value of the series:

\[ \sum_{n=1}^\infty a_n b_n \]

Step 1: Properties of \( f(x) \)

• Note that \( f(x) \) is an odd function, since: \[ f(-x) = \begin{cases} h(-x), & 0<-x<C \implies -C<x<0 \\ -h(x), & -C<-x<0 \implies 0<x<C \end{cases} \] which means \( f(-x) = -f(x) \).
• Because \( f(x) \) is odd and has period \( 2C \), its Fourier series contains only sine terms; the cosine coefficients \( a_n = 0 \) for \( n \geq 1 \).

Step 2: Since \( a_n = 0 \) for \( n \geq 1 \), the product \( a_n b_n = 0 \) for all \( n \geq 1 \).

Step 3: Therefore,

\[ \sum_{n=1}^\infty a_n b_n = 0 \]

Step 4: Checking the options, the one that matches this is:

\[ \int_{-C}^C f(x) \cos \frac{n \pi x}{C} \, dx \]

because this integral is zero for an odd function multiplied by an even function (cosine), and the sum of the product \( a_n b_n \) also evaluates to zero.

Hence, the correct answer is option 2.