Find Fourier Expansion Coefficient for $A\sin^2(ax)$

This solution details finding the coefficient of the complex exponential term $e^{ikx}$ within the Fourier expansion of the function $u(x) = A\sin^2(ax)$, specifically for the case where $k = -2a$. We will use trigonometric identities to simplify the function and express it in terms of complex exponentials.

Trigonometric Simplification

First, rewrite $u(x)$ using the identity $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$:

$u(x) = A \left( \frac{1 - \cos(2ax)}{2} \right)$

$u(x) = \frac{A}{2} - \frac{A}{2}\cos(2ax)$

Expressing in Complex Exponentials

Next, use the identity $\cos(\phi) = \frac{e^{i\phi} + e^{-i\phi}}{2}$ to convert the cosine term:

$\cos(2ax) = \frac{e^{i(2ax)} + e^{-i(2ax)}}{2}$

Substitute this back into the expression for $u(x)$:

$u(x) = \frac{A}{2} - \frac{A}{2} \left( \frac{e^{i(2ax)} + e^{-i(2ax)}}{2} \right)$

Simplify the expression:

$u(x) = \frac{A}{2} - \frac{A}{4} e^{i(2ax)} - \frac{A}{4} e^{-i(2ax)}$

This can be written in the general form $u(x) = \sum c_k e^{ikx}$ as:

$u(x) = \frac{A}{2}e^{i(0)x} - \frac{A}{4}e^{i(2a)x} - \frac{A}{4}e^{i(-2a)x}$

Identifying the Specific Coefficient

The Fourier expansion represents the function as a sum of terms $c_k e^{ikx}$. We are looking for the coefficient corresponding to $k = -2a$. In the derived expression:

$u(x) = \frac{A}{2}e^{i(0)x} - \frac{A}{4}e^{i(2a)x} - \frac{A}{4}e^{i(-2a)x}$

The term with $k = -2a$ is $- \frac{A}{4}e^{i(-2a)x}$.

Therefore, the coefficient of $e^{ikx}$ for $k = -2a$ is $ - \frac{A}{4}$.