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Task:

Let $f(x)=\begin{cases}\frac{1}{\pi}x-2 \quad \text{ for } 2\pi\leq x < 3\pi \\ 4-\frac{1}{\pi}x \quad \text{ for } 3\pi \leq x < 4\pi\end{cases}$, $f:\mathbb{R}\to\mathbb{R}$

and $f(x+2\pi)=f(x)$ for all $x\in\mathbb{N}$. How do I create a Fourier-Series for that function?

Solution/Problem:

The function is periodical for every $2\pi$. That's why $T=2\pi$. $$a_k=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cdot \cos(k\cdot x)dx,$$ $k=0,1,2,3,\dots$

$$b_k=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cdot \sin(k\cdot x) dx,$$ $k=1,2,3,\dots$ and

$$F_n(x)=\frac{a_0}{2}+\sum\limits_{k=1}^{n}\left(a_k\cos(k\cdot x)+b_k\cdot \sin(k\cdot x)\right)$$ is the Taylor-Polynomial, the Taylor-Series is $$F_\infty(x)=\frac{a_0}{2}+\sum\limits_{k=1}^{\infty}(a_k\cos(k\cdot x)+b_k\cdot \sin(k\cdot x)).$$

Because the function is periodical, we can change the integration-limits to $2\pi$, $4\pi$ or is this wrong? Afterwards, I've computed the integrals for $a_k$ and $b_k$ with $k$ arbitrary except for $a_0$, which needs to be computed too.

$$a_k=\frac{1}{\pi}\left(\displaystyle\int\limits_{2\pi}^{3\pi}\frac{1}{\pi}x-2\cdot \cos(kx)dx+\int\limits_{3\pi}^{4\pi}4-\frac{1}{\pi}x\cdot \cos(kx)dx\right)$$

$$a_k=\frac{2\sin(2\pi k)+\sin(3\pi k)-4\sin(4\pi k)}{\pi k}+\frac{13}{2}$$ for $k>0$ and $a_0=1$.

(I've left out the integration-steps, because that takes too mucht time to type in $\LaTeX$)

The computation for $b_k$ is similar except that we need to insert the other function and multiply by $\sin(kx)$.

$$b_k=\frac{\sin(3\pi k)-\sin(4\pi k)}{\pi^2k^2}+\frac{-2\cos(2\pi k)-\cos(3\pi k)+4\cos(4\pi k)}{\pi k}+\frac{13}{2}$$

and therefore, the Taylor-Polynomial is $$F_n(x)=\frac{1}{2}+\sum\limits_{n=1}^{n}(a_k\cdot \cos(kx)+b_k\cdot \sin(kx))dx$$ and the Series:

$$F_\infty(x)=\frac{1}{2}+\sum\limits_{n=1}^{\infty}(a_k\cdot \cos(kx)+b_k\cdot \sin(kx))dx,$$

which is a kinda weird solution and I've probably made some mistakes while computing. Can you tell me what I've done wrong and how to do it right?

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The problem is unfortunately in the part you skipped: the integrals, here is the result you should arrive to

\begin{eqnarray} a_0 &=& 1 \\ a_k &=& \frac{1}{k^2\pi^2}[-\cos 2\pi k + 2 \cos 3\pi k - \cos 4\pi k] = \frac{2}{k^2\pi^2}(-1 + (-1)^k) ~~~ k = 1, 2, \cdots \\ b_k &=& 0 \end{eqnarray}

This is a plot that shows convergence for different number of terms in the series $N$

enter image description here


EDIT Code to generate the previous plot

import matplotlib.pyplot as plt
import numpy as np


# fourier series
def fseries(x, nmax = 10):

    k = np.arange(1, nmax + 1)
    s = (-2 + 2 * (-1)**k) * np.cos(x * k) / (k * np.pi)**2
    return 0.5 + sum(s)

# original function
def f(x):

    x = x[(x > 2 * np.pi) & (x < 4 * np.pi)]
    y = np.zeros_like(x)

    i = x < 3 * np.pi
    y[i] = x[i] / np.pi - 2

    i = x > 3 * np.pi
    y[i] = 4 - x[i] / np.pi

    return x, y

# axis
ax = plt.gca()
ax.set_xlabel(r'$x/\pi$')
ax.set_ylabel(r'$f(x)$')


# plot fourier series
x = np.linspace(0, 5 * np.pi, num = 500)
for n in [1, 5, 20]:
    y = np.array([fseries(k, nmax = n) for k in x])
    plt.plot(x / np.pi, y, lw = 1, label = r'$N = {n}$'.format(n = n))

# plot original function
x1 = np.linspace(0, 5 * np.pi, num = 500)
x1, y1 = f(x1)

plt.plot(x1 / np.pi, y1, lw = 3, label = r'$f(x)$')
ax.legend(frameon = False)
plt.show()
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  • $\begingroup$ Thank you, I found my error: I forgot the brackets because we're multiplying $f(x)$ with $\cos(kx)$. $\endgroup$ – Doesbaddel Dec 11 '19 at 21:07
  • $\begingroup$ How did you do the plot? it looks pretty beautiful! $\endgroup$ – Doesbaddel Dec 12 '19 at 6:40
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    $\begingroup$ @Doesbaddel It's python, I'd be happy to include some MWV of the script if you're interested $\endgroup$ – caverac Dec 12 '19 at 10:34
  • $\begingroup$ Yeah, sure! I would like to see it. $\endgroup$ – Doesbaddel Dec 12 '19 at 10:44
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    $\begingroup$ @Doesbaddel Done! $\endgroup$ – caverac Dec 12 '19 at 18:34

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