Exercise: What is the Fourier transform of the function $f(x)=0$ ?

The Fourier transform of the function $f(x)=0$ is also zero. Since the function $f(x)$ is identically zero, its Fourier transform will also be zero at all frequencies. In other words, all the frequency components of the function $f(x)$ are absent, resulting in a zero Fourier transform.

(a) By expanding both sides as Fourier Sin Series, show that the solution to the equation
$$\frac{d^2 y}{d x^2}+y=2 x$$
with boundary conditions $y(x=0)=0, y(x=1)=0$ is
$$y(x)=\frac{4}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n\left(1-n^2 \pi^2\right)} \sin (n \pi x) .$$
(b) Show that the r.m.s. value of $y(x)$ is
$$\sqrt{\left\langle y^2(x)\right\rangle}=\frac{4}{\pi} \sqrt{\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^2\left(1-n^2 \pi^2\right)^2}}$$

(a) To find the solution to the given differential equation, we expand both sides as a Fourier sine series. The general form of a Fourier sine series is given by:
$$y(x) = \sum_{n=1}^{\infty} B_n \sin(n \pi x)$$
where $B_n$ are the Fourier coefficients.

Taking the second derivative of $y(x)$ with respect to $x$, we have:
$$\frac{d^2 y}{d x^2} = -\sum_{n=1}^{\infty} B_n (n \pi)^2 \sin(n \pi x)$$

Substituting the expressions for $\frac{d^2 y}{d x^2}$ and $y$ into the differential equation, we get:
$$-\sum_{n=1}^{\infty} B_n (n \pi)^2 \sin(n \pi x) + \sum_{n=1}^{\infty} B_n \sin(n \pi x) = 2x$$

Matching the coefficients of the same sine functions on both sides, we obtain the following equation for the Fourier coefficients $B_n$:
$$-(n \pi)^2 B_n + B_n = 2 \delta_{n,1}$$
where $\delta_{n,1}$ is the Kronecker delta.

Simplifying the equation, we find:
$$B_n = \frac{4 (-1)^{n+1}}{n (1 – n^2 \pi^2)}$$

Therefore, the solution to the given differential equation with the specified boundary conditions is:
$$y(x) = \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n (1 – n^2 \pi^2)} \sin(n \pi x)$$

(b) The root mean square (r.m.s.) value of $y(x)$ can be calculated as follows:
$$\sqrt{\left\langle y^2(x)\right\rangle} = \sqrt{\frac{1}{1-0} \int_{0}^{1} y^2(x) dx}$$

Substituting the expression for $y(x)$ and performing the integration, we obtain:
$$\sqrt{\left\langle y^2(x)\right\rangle} = \sqrt{\frac{4}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{n^2 (1 – n^2 \pi^2)^2}}$$

Therefore, the r.m.s. value of $y(x)$ is given by the equation:
$$\sqrt{\left\langle y^2(x)\right\rangle} = \frac{4}{\pi} \sqrt{\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^2 (1 – n^2 \pi^2)^2}}$$

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