Question

If $f(x)=e^{-x}$ in $(-1,1)$ and the Fourier series of $f(x)$ is given by \[ f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos nx+b_n\sin nx), \] then the Fourier coefficient $a_0$ is:

• A. 1
• B. 0
• C. $2\cosh 1$
• D. $2\sinh 1$

Hint:

Use identities: \[ \sinh x=\frac{e^x-e^{-x}}{2} \] to simplify Fourier integrals.

Answer 1

The Correct Option is D

Concept: For a function defined on $(-L,L)$, the Fourier coefficient $a_0$ is given by: \[ a_0=\frac{1}{L}\int_{-L}^{L} f(x)\,dx \] Here $L=1$, so: \[ a_0=\int_{-1}^{1} e^{-x}\,dx \]

Step 1: Write the integral.
\[ a_0=\int_{-1}^{1} e^{-x}\,dx \]

Step 2: Integrate.
\[ \int e^{-x}dx=-e^{-x} \] So, \[ a_0=\left[-e^{-x}\right]_{-1}^{1} \]

Step 3: Apply limits.
\[ a_0=-(e^{-1})-(-e^{1}) \] \[ a_0=e-e^{-1} \]

Step 4: Rewrite in hyperbolic form.
\[ e-e^{-1}=2\sinh 1 \] \[ \Rightarrow a_0=2\sinh 1 \]