Need help with this Calculus problem!

Question

5. The integrals defined in line (3) are meaningful numbers for any integrable function f. Treating a given function as if it were a combination of periodic functions opens the door to Fourier analysis, a powerful tool with many applications in science, engineering, and mathematics. Present exact calculations where they are requested below, but use suitable software to produce the corresponding plots.

f(x)=b_{1}sin(x)+b_{2}sin(2x)+...=\sum_{k=1}^{\infty}b_{k}sin(kx).

sum = sigma sign

(a) Let f(x)=1 for 0<=x<=pi. Find a formula for B_{n}(f) valid for every integer n>=1

(b) (1 mark) Make three plots showing y=f(x) and y=S_{N}(x) on the same axes, where

\[S_{N}(x)=\sum_{n=1}^{N}B_{n}(f)\sin(nx).\]

Use N=3 for the first plot, N=5 for the second, and N=11 for the third.

(c) (1 mark) Let (g(x)=x for 0<=x<=pi. Find a formula for B_{n}(g) valid for every integer n>=1

(d) (1 mark) Make three plots showing y=g(x) and y={S}_{N}(x) on the same axes, where

{S}_{N}(x)=\sum_{n=1}^{N}B_{n}(g)\sin(nx).}

Use N=3 for the first plot, N=5 for the second, and N=11 for the third.

Answer 1

(a) For a function $f(x) = 1$ defined on $[0, \pi]$ we can compute the Fourier sin series coefficients using the formula:

$$B_n(f) = \frac{2}{L}\int_0^L f(x)\sin{\left(\frac{n\pi x}{L}\right)}dx$$

For $f(x) = 1$, $[0, \pi]\Rightarrow L = \pi - 0 = \pi$, we get

\begin{align*}

B_n(1) &= \frac{2}{\pi}\int_0^\pi (1)\sin{\left(\frac{n\pi x}{\pi}\right)}dx, \\

&= \frac{2}{\pi}\int_0^\pi\sin{nx}dx, \\

&= \frac{2}{\pi}\left.\left[-\frac{1}{n}\cos{nx}\right]\right|_0^\pi, \\

&= \frac{2}{n\pi}\left[-\cos{n\pi} + \cos{0}\right], \\

&= \boxed{\frac{2}{n\pi}\left[1-\cos{n\pi}\right].}

\end{align*}

(b)

# Define B_n using the latest provided expression

def B_n_latest_provided(n):

return 2 / (n * np.pi) * (1 - np.cos(n * np.pi))

# Function to calculate S_N(x) using the latest B_n formula

def S_N_latest_provided(N, x_values):

sum_sn = np.zeros_like(x_values)

for n in range(1, N + 1):

B_n = B_n_latest_provided(n)

sum_sn += B_n * np.sin(n * x_values)

return sum_sn

# Plot settings for the latest B_n

fig, axes = plt.subplots(3, 1, figsize=(10, 15))

for i, N in enumerate(N_values):

# Calculate S_N(x) for the current N using the latest B_n

SN_x_values_latest_provided = S_N_latest_provided(N, x_values)

# Plot f(x) and S_N(x) using the latest B_n

axes[i].plot(x_values, f_x_values, label='f(x) = 1')

axes[i].plot(x_values, SN_x_values_latest_provided, label=f'S_{N}(x) using latest B_n')

axes[i].set_title(f'N = {N}')

axes[i].legend()

plt.tight_layout()

plt.show()

(c) For a function $g(x) = x$ defined on $[0, \pi]$ we can compute the Fourier sin series coefficients using the formula:

$$B_n(g) = \frac{2}{L}\int_0^L g(x)\sin{\left(\frac{n\pi x}{L}\right)}dx$$

For $g(x) = x$, $[0, \pi]\Rightarrow L = \pi - 0 = \pi$, we get

\begin{align*}

B_n(1) &= \frac{2}{\pi}\int_0^\pi (x)\sin{\left(\frac{n\pi x}{\pi}\right)}dx, \\

&= \frac{2}{\pi}\int_0^\pi x\sin{nx}dx, \\

&= \frac{2}{\pi}\left.\left[\frac{\sin{nx}-nx\cos{nx}}{n^2}\right]\right|_0^\pi, \\

&= \boxed{\frac{2}{\pi}\left[\frac{\sin{n\pi} - n\pi\cos{n\pi}}{n^2}\right].}

\end{align*}

(d)

# Define B_n using the new provided expression for g(x)

def B_n_for_g(n):

return 2 / np.pi * (np.sin(n * np.pi) - n * np.pi * np.cos(n * np.pi)) / n**2

# Function to calculate S_N(x) using the new B_n formula for g(x)

def S_N_for_g(N, x_values):

sum_sn = np.zeros_like(x_values)

for n in range(1, N + 1):

B_n = B_n_for_g(n)

sum_sn += B_n * np.sin(n * x_values)

return sum_sn

# Plot settings for the new B_n for g(x)

fig, axes = plt.subplots(3, 1, figsize=(10, 15))

for i, N in enumerate(N_values):

# Calculate S_N(x) for the current N using the new B_n for g(x)

SN_x_values_for_g = S_N_for_g(N, x_values)

# Plot g(x) and S_N(x) using the new B_n for g(x)

axes[i].plot(x_values, g_x_values, label='g(x) = x')

axes[i].plot(x_values, SN_x_values_for_g, label=f'S_{N}(x) using new B_n for g(x)')

axes[i].set_title(f'N = {N}')

axes[i].legend()

plt.tight_layout()

plt.show()

Hope this helps!