Chapter 28Definition of Fourier Series and Worked Examples

Example 28.1: Fourier series of $f(x)=x$ (period 2π)

Find the Fourier series of the function obtained by extending $f(x)=x\ (-\pi \lt x\le\pi)$ with period $2\pi$ (Figure 28.1).

Solution. We use the formulas of Theorem 28.1. First we find $a_0$. Since $f(x)=x$ is an odd function ($f(-x)=-f(x)$), its integral over the symmetric interval $[-\pi,\pi]$ is

$$ a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}x\,dx=0 $$

(when an odd function is integrated over an interval symmetric about the origin, the part with $x \gt 0$ and the part with $x \lt 0$ cancel each other exactly, giving $0$). Next we find $a_k$ ($k\ge1$). Since $x\cos kx$ is also an odd function (odd function × even function = odd function), for the same reason

$$ a_k=\frac1\pi\int_{-\pi}^{\pi}x\cos kx\,dx=0 $$

Finally we find $b_k$. Since $x\sin kx$ is an even function (odd function × odd function = even function), we can use $\int_{-\pi}^{\pi}=2\int_{0}^{\pi}$. By integration by parts (taking $u=x,\ dv=\sin kx\,dx$ gives $du=dx,\ v=-\frac1k\cos kx$),

$$ \int_0^{\pi}x\sin kx\,dx=\Bigl[-\frac{x}{k}\cos kx\Bigr]_0^{\pi}+\frac1k\int_0^{\pi}\cos kx\,dx=-\frac{\pi}{k}\cos k\pi+\frac1k\Bigl[\frac1k\sin kx\Bigr]_0^{\pi}=-\frac{\pi(-1)^k}{k}+0 $$

(we used $\cos k\pi=(-1)^k$ and $\sin k\pi=0$ (where $k$ is an integer)). Therefore

$$ b_k=\frac1\pi\cdot2\times\left(-\frac{\pi(-1)^k}{k}\right)=\frac{2(-1)^{k+1}}{k} $$

From the above, the Fourier series is

$$ f(x)=\sum_{k=1}^{\infty}\frac{2(-1)^{k+1}}{k}\sin kx=2\sin x-\sin2x+\frac23\sin3x-\frac12\sin4x+\cdots $$

(verified with SymPy: $a_0=a_k=0$ and $b_k=2(-1)^{k+1}/k$ for $k=1,\ldots,5$). The reason that $a_0=a_k=0$ is that $f(x)=x$ is an odd function—in general, no $\cos$ terms at all appear in the Fourier series of an odd function. We will prove this fact in general in Section 28.3.

Example 28.2: Fourier series of $f(x)=x$ (period 4)

Find the Fourier series of the function obtained by extending $f(x)=x\ (-2 \lt x\le2)$ with period 4 ($L=2$, Figure 28.2).

Solution. We use the formulas of Theorem 28.2 (with $L=2$). Since $f(x)=x$ is an odd function, as in Example 28.1, $a_0=0$ and $a_k=0\ (k\ge1)$ hold for exactly the same reason (the integral of an odd function over a symmetric interval). For $b_k$, we use the fact that $x\sin\frac{k\pi}{2}x$ is an even function and compute by integration by parts ($u=x,\ dv=\sin\frac{k\pi}{2}x\,dx$ and $v=-\frac{2}{k\pi}\cos\frac{k\pi}{2}x$).

$$ b_k=\frac12\int_{-2}^{2}x\sin\frac{k\pi}{2}x\,dx=\int_0^2 x\sin\frac{k\pi}{2}x\,dx =\left[-\frac{2x}{k\pi}\cos\frac{k\pi}{2}x\right]_0^2+\frac{2}{k\pi}\int_0^2\cos\frac{k\pi}{2}x\,dx $$

The second term on the right-hand side is $\left[\frac{4}{k^2\pi^2}\sin\frac{k\pi}{2}x\right]_0^2=\frac{4}{k^2\pi^2}\sin k\pi=0$ (since $k$ is an integer, $\sin k\pi=0$). The first term is $-\frac{4}{k\pi}\cos k\pi=-\frac{4(-1)^k}{k\pi}$ at $x=2$ and $0$ at $x=0$, so

$$ b_k=-\frac{4(-1)^k}{k\pi}=\frac{4(-1)^{k+1}}{k\pi} $$

Hence

$$ f(x)=\sum_{k=1}^{\infty}\frac{4(-1)^{k+1}}{k\pi}\sin\frac{k\pi}{2}x $$

(verified with SymPy: $a_0=a_k=0$ and $b_k=4(-1)^{k+1}/(\pi k)$ for $k=1,\ldots,4$). In general, a similar computation shows that for the same form $f(x)=x\ (-L \lt x\le L)$ we get $b_k=\dfrac{2L(-1)^{k+1}}{k\pi}$ (let us check that substituting $L=\pi$ returns exactly $b_k=2(-1)^{k+1}/k$ of Example 28.1).

Example 28.3: Fourier cosine series of $f(x)=x^2$ (period 2π)

Find the Fourier series of the function obtained by extending $f(x)=x^2\ (-\pi \lt x\le\pi)$ with period $2\pi$ (Figure 28.3).

Solution. $f(x)=x^2$ is an even function ($f(-x)=(-x)^2=x^2=f(x)$), so by Theorem 28.3(I) it is a Fourier cosine series, and $b_k=0$. For $a_0$ we have

$$ a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}x^2\,dx=\frac{1}{2\pi}\left[\frac{x^3}{3}\right]_{-\pi}^{\pi}=\frac{1}{2\pi}\cdot\frac{2\pi^3}{3}=\frac{\pi^2}{3} $$

For $a_k\ (k\ge1)$, since the integrand $x^2\cos kx$ is an even function, we write $\int_{-\pi}^\pi=2\int_0^\pi$ and compute by integration by parts, repeated twice. With $u=x^2,\ dv=\cos kx\,dx$ we have $du=2x\,dx,\ v=\frac1k\sin kx$, so

$$ \int_0^{\pi}x^2\cos kx\,dx=\left[\frac{x^2}{k}\sin kx\right]_0^{\pi}-\frac2k\int_0^{\pi}x\sin kx\,dx $$

The first term is zero because $\sin k\pi=0$. The second term, $\int_0^\pi x\sin kx\,dx$, has exactly the same form as (an intermediate step in) the computation of Example 28.1, and equals $-\frac{\pi(-1)^k}{k}$. Therefore

$$ \int_0^{\pi}x^2\cos kx\,dx=0-\frac2k\left(-\frac{\pi(-1)^k}{k}\right)=\frac{2\pi(-1)^k}{k^2} $$ $$ a_k=\frac1\pi\cdot2\times\frac{2\pi(-1)^k}{k^2}=\frac{4(-1)^k}{k^2} $$

Hence

$$ f(x)=\frac{\pi^2}{3}+\sum_{k=1}^{\infty}\frac{4(-1)^k}{k^2}\cos kx=\frac{\pi^2}{3}-4\cos x+\cos2x-\frac49\cos3x+\cdots $$

(verified with SymPy: $a_0=\pi^2/3$ and $a_k=4(-1)^k/k^2$ for $k=1,\ldots,4$). Incidentally, if we formally substitute $x=\pi$ ($f(\pi)=\pi^2$) into this expression, then since $\cos k\pi=(-1)^k$ we get $\pi^2=\frac{\pi^2}{3}+\sum_{k=1}^\infty\frac{4}{k^2}$, that is, $\sum_{k=1}^\infty\frac1{k^2}=\frac{\pi^2}{6}$, the answer to the famous “Basel problem”—however, the grounds for substituting into the series here (the theory of pointwise convergence), even though this is not a point of discontinuity on the boundary, will be studied in the next chapter, so here we only present it as an interesting by-product.

Example 28.4: Complex Fourier series of $f(x)=x$ (period 2π)

For the function $f(x)=x\ (-\pi \lt x\le\pi)$ (period 2π) of Example 28.1, find the complex Fourier coefficients $C_k$ directly, and verify that they are consistent with the result $a_k=0,\ b_k=2(-1)^{k+1}/k$ of Example 28.1.

Solution. By Theorem 28.4 ($L=\pi$), for $k\ne0$ we have

$$ C_k=\frac{1}{2\pi}\int_{-\pi}^{\pi}x\,e^{-ikx}\,dx $$

Integration by parts ($u=x,\ dv=e^{-ikx}dx$, $v=-\frac{1}{ik}e^{-ikx}=\frac{i}{k}e^{-ikx}$) gives

$$ \int_{-\pi}^{\pi}x\,e^{-ikx}\,dx=\left[\frac{ix}{k}e^{-ikx}\right]_{-\pi}^{\pi}-\frac{i}{k}\int_{-\pi}^{\pi}e^{-ikx}\,dx $$

The second term on the right-hand side vanishes when $k$ is a nonzero integer: $\int_{-\pi}^{\pi}e^{-ikx}dx=\left[\frac{i}{k}e^{-ikx}\right]_{-\pi}^{\pi}=\frac{i}{k}\bigl(e^{-ik\pi}-e^{ik\pi}\bigr)=\frac{i}{k}\bigl((-1)^k-(-1)^k\bigr)=0$ (we used $e^{\pm ik\pi}=\cos k\pi\pm i\sin k\pi=(-1)^k$). The first term is

$$ \left[\frac{ix}{k}e^{-ikx}\right]_{-\pi}^{\pi}=\frac{i\pi}{k}e^{-ik\pi}-\frac{i(-\pi)}{k}e^{ik\pi}=\frac{i\pi}{k}(-1)^k+\frac{i\pi}{k}(-1)^k=\frac{2i\pi(-1)^k}{k} $$

Therefore

$$ C_k=\frac{1}{2\pi}\cdot\frac{2i\pi(-1)^k}{k}=\frac{i(-1)^k}{k}\qquad(k\ne0) $$

Also, $C_0=a_0=0$ (the value found in Example 28.1). (Verified with SymPy: $C_k=i(-1)^k/k$ for $k=\pm1,\pm2,\pm3$.) Let us see that this corresponds to $a_k,b_k$ of Example 28.1. By the formula in the “Math Note” box,

$$ C_k=\frac{a_k-ib_k}{2}=\frac{0-i\cdot\frac{2(-1)^{k+1}}{k}}{2}=\frac{-i(-1)^{k+1}}{k}=\frac{i(-1)^k}{k} $$

This agrees completely with the directly computed $C_k$. With concrete numerical values, we have confirmed that the only difference is whether we view the same Fourier series of the same function in its real form ($\cos,\sin$) or in its complex form ($e^{ikx}$).

Example 28.5: Fourier sine series of a square wave (Example No. 1 of the original notes)

Let $f(x)=\begin{cases}-\dfrac{\pi}{4} & (-\pi \lt x\le0)\\[4pt]\dfrac{\pi}{4} & (0 \lt x\le\pi)\end{cases}$ (period 2π, Figure 28.4). Expand $f(x)$ in a Fourier series.

Solution. Let $f(x)=\dfrac{a_0}{2}+\displaystyle\sum_{k=1}^{\infty}(a_k\cos kx+b_k\sin kx)$.

[i] The coefficients $a_k\ (k=0,1,2,\ldots)$.

$$ a_k=\frac1\pi\int_{-\pi}^{\pi}f(x)\cos kx\,dx=\frac1\pi\int_0^{\pi}\frac{\pi}{4}\cos kx\,dx+\frac1\pi\int_{-\pi}^0\left(-\frac{\pi}{4}\right)\cos kx\,dx $$

(For $k=0$ as well, the same expression can be used, with $\cos0x=1$.) Carrying out the integrations, we get

$$ =\frac14\left[\frac1k\sin kx\right]_0^{\pi}-\frac14\left[\frac1k\sin kx\right]_{-\pi}^0=\frac14\cdot\frac1k(\sin k\pi-0)-\frac14\cdot\frac1k(0-\sin(-k\pi))=0 $$

(We used $\sin k\pi=0,\ \sin(-k\pi)=0$. For $k=0$, both sides are $0$ from the start.) Hence $a_k=0$ (for all $k=0,1,2,\ldots$).

[ii] The coefficients $b_k\ (k=1,2,3,\ldots)$.

$$ b_k=\frac1\pi\int_{-\pi}^{\pi}f(x)\sin kx\,dx=\frac1\pi\int_0^{\pi}\frac{\pi}{4}\sin kx\,dx+\frac1\pi\int_{-\pi}^0\left(-\frac{\pi}{4}\right)\sin kx\,dx $$ $$ =\frac14\left[-\frac1k\cos kx\right]_0^{\pi}-\frac14\left[-\frac1k\cos kx\right]_{-\pi}^0 $$

The first term is $-\frac1{4k}(\cos k\pi-\cos0)=-\frac1{4k}\bigl((-1)^k-1\bigr)=\frac{1-(-1)^k}{4k}$. The second term is $-\frac1{4k}(\cos0-\cos(-k\pi))=-\frac1{4k}\bigl(1-(-1)^k\bigr)=-\frac{1-(-1)^k}{4k}$. Since this is a subtraction, we must be careful with the sign:

$$ b_k=\frac{1-(-1)^k}{4k}-\left(-\frac{1-(-1)^k}{4k}\right)=\frac{1-(-1)^k}{2k} $$

From [i] and [ii],

$$ f(x)=\sum_{k=1}^{\infty}\frac{1-(-1)^k}{2k}\sin kx $$

This is the result (verified with SymPy: $a_k=0$ and $b_k=(1-(-1)^k)/(2k)$ for $k=1,\ldots,5$).

Discussion (connection with π). The coefficient $\dfrac{1-(-1)^k}{2k}$ vanishes when $k$ is even, because $1-(-1)^k=0$; only when $k$ is odd does $\dfrac{1-(-1)}{2k}=\dfrac1k$ remain. Therefore

$$ f(x)=\sin x+\frac13\sin3x+\frac15\sin5x+\cdots $$

Now let us substitute $x=\dfrac{\pi}{2}$. From the graph (Figure 28.4) it is clear that $f\!\left(\frac{\pi}{2}\right)=\frac{\pi}{4}$. On the other hand, since $\sin\frac{\pi}{2}=\sin\frac{5\pi}{2}=\sin\frac{9\pi}{2}=\cdots=1$ and $\sin\frac{3\pi}{2}=\sin\frac{7\pi}{2}=\cdots=-1$, we can derive

$$ \frac{\pi}{4}=f\!\left(\frac\pi2\right)=1-\frac13+\frac15-\frac17+\cdots=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{2k-1} $$

which is the Leibniz series for π—a surprisingly simple expression of the number $\pi$ as an infinite series. Fourier series are not merely “a tool for decomposing waves”; they also show us $\pi$ making an appearance in unexpected places.

This series, however, converges very slowly. Computing the partial sums $S_N=\sum_{k=1}^{N}\frac{(-1)^{k-1}}{2k-1}$ numerically, we obtain

($\pi/4=0.785398\ldots$). Each time the number of terms is multiplied by 10, the error becomes roughly one-tenth—in other words, the error decreases roughly in inverse proportion to the number of terms, which is a very slow convergence. This phenomenon, and how the series behaves near a jump discontinuity (the Gibbs phenomenon), will be checked numerically in Exercise 28.4.

Example 28.6: Fourier cosine series of a triangle wave (Example No. 2 of the original notes)

Let $f(x)=\begin{cases}-x & (-\pi \lt x\le0)\\ x & (0 \lt x\le\pi)\end{cases}$ (that is, $f(x)=|x|$, period 2π, Figure 28.5). Expand $f(x)$ in a Fourier series.

Solution. Let $f(x)=\dfrac{a_0}{2}+\displaystyle\sum_{k=1}^{\infty}(a_k\cos kx+b_k\sin kx)$.

[i] The coefficients $a_k\ (k=1,2,\ldots)$. We use the indefinite integral formula $\int x\cos kx\,dx=\frac1kx\sin kx+\frac1{k^2}\cos kx$ from the “Math Note” above.

$$ a_k=\frac1\pi\int_{-\pi}^{\pi}f(x)\cos kx\,dx=\frac1\pi\left(\int_0^{\pi}x\cos kx\,dx-\int_{-\pi}^0 x\cos kx\,dx\right) $$ $$ =\frac1\pi\left[\frac1kx\sin kx+\frac1{k^2}\cos kx\right]_0^{\pi}-\frac1\pi\left[\frac1kx\sin kx+\frac1{k^2}\cos kx\right]_{-\pi}^0 $$

In each square bracket, the term $\frac1kx\sin kx$ vanishes at both the upper and the lower limit, since $\sin k\pi=\sin0=0$ (for $x=0,\pm\pi$ we always have $\sin kx=0$). Only the terms $\frac1{k^2}\cos kx$ remain, so

$$ =\frac1\pi\left\{\frac{(-1)^k}{k^2}-\frac1{k^2}\right\}-\frac1\pi\left\{\frac1{k^2}-\frac{(-1)^k}{k^2}\right\}=\frac{2\bigl\{(-1)^k-1\bigr\}}{\pi k^2}\qquad(k=1,2,\ldots) $$

For $a_0$, we substitute $k=0$ ($\cos0x=1$) into the formula for $a_k$:

$$ a_0=\frac1\pi\int_{-\pi}^{\pi}f(x)\,dx=\frac1\pi\left(\int_0^{\pi}x\,dx+\int_{-\pi}^0(-x)\,dx\right)=\frac1\pi\left(\frac{\pi^2}{2}+\frac{\pi^2}{2}\right)=\pi $$

[ii] The coefficients $b_k\ (k=1,2,\ldots)$. Since $f(x)=|x|$ is an even function, Theorem 28.3(I) tells us that we should have $b_k=0$ from the outset. This can also be confirmed by actual computation. Using the indefinite integral formula $\int x\sin kx\,dx=-\frac1kx\cos kx+\frac1{k^2}\sin kx$, we get

$$ b_k=\frac1\pi\left(\int_0^{\pi}x\sin kx\,dx-\int_{-\pi}^0 x\sin kx\,dx\right)=\frac1\pi\left[-\frac{x}{k}\cos kx+\frac1{k^2}\sin kx\right]_0^{\pi}-\frac1\pi\left[-\frac{x}{k}\cos kx+\frac1{k^2}\sin kx\right]_{-\pi}^0 $$

In both square brackets, the term $\frac1{k^2}\sin kx$ vanishes, because $\sin kx=0$ at $x=0,\pm\pi$. Evaluating the remaining terms $-\frac{x}k\cos kx$, the first term is $-\frac{\pi}{k}(-1)^k-0=-\frac{\pi(-1)^k}{k}$ and the second term is $0-\left(-\frac{-\pi}{k}\cos(-k\pi)\right)=0-\frac{\pi(-1)^k}{k}\cdot(-1)$; simplifying, we obtain

$$ b_k=\frac1\pi\left(-\frac{\pi(-1)^k}{k}\right)-\frac1\pi\left(-\frac{\pi(-1)^k}{k}\right)=0 $$

Thus we confirm once again that $b_k=0$. From [i] and [ii],

$$ f(x)=\frac{\pi}{2}+\sum_{k=1}^{\infty}\frac{2\bigl\{(-1)^k-1\bigr\}}{\pi k^2}\cos kx=\frac{\pi}{2}-\frac{4}{\pi}\cos x-\frac{4}{9\pi}\cos3x-\frac{4}{25\pi}\cos5x-\cdots $$

This is the result (note that for even $k$ we have $(-1)^k-1=0$, so the coefficient vanishes). (Verified with SymPy: $a_0=\pi$, $a_k=2\{(-1)^k-1\}/(\pi k^2)$, and $b_k=0$ for $k=1,\ldots,5$.)

Example 28.7: Sawtooth wave (Example No. 3 of the original notes, computed in two ways, real and complex)

For $f(x)=\dfrac{x}{2}\ (-4 \lt x \lt 4)$ (period 8, $L=4$, Figure 28.6), expand $f(x)$ in a Fourier series.

Solution (real Fourier series). Since $f(x)=x/2$ is an odd function, by Theorem 28.3(II) it becomes a Fourier sine series ($a_k=0$).

$$ f(x)=\sum_{k=1}^{\infty}b_k\sin\frac{k\pi}{4}x\qquad\text{(we assume this form).} $$

Using the indefinite integral formula $\int x\sin\frac{k\pi}{4}x\,dx=-\frac{4}{k\pi}x\cos\frac{k\pi}{4}x+\left(\frac{4}{k\pi}\right)^2\sin\frac{k\pi}{4}x$ (this is $\int x\sin ax\,dx=-\frac{x}{a}\cos ax+\frac1{a^2}\sin ax$ with $a=\frac{k\pi}{4}$ substituted), we obtain

$$ b_k=\frac14\int_{-4}^{4}f(x)\sin\frac{k\pi}{4}x\,dx=\frac18\int_{-4}^{4}x\sin\frac{k\pi}{4}x\,dx=\frac18\left[-\frac{4}{k\pi}x\cos\frac{k\pi}{4}x+\left(\frac{4}{k\pi}\right)^2\sin\frac{k\pi}{4}x\right]_{-4}^{4} $$

At $x=\pm4$ we have $\frac{k\pi}{4}x=\pm k\pi$, so $\sin(\pm k\pi)=0$ and the second term vanishes at both the upper and the lower limit. Evaluating the first term, at $x=4$ it is $-\frac{4}{k\pi}\cdot4\cos k\pi=-\frac{16}{k\pi}(-1)^k$, and at $x=-4$ it is $-\frac{4}{k\pi}\cdot(-4)\cos(-k\pi)=\frac{16}{k\pi}(-1)^k$. Subtracting,

$$ b_k=\frac18\left(-\frac{16(-1)^k}{k\pi}-\frac{16(-1)^k}{k\pi}\right)=\frac18\cdot\left(-\frac{32(-1)^k}{k\pi}\right)=-\frac{4(-1)^k}{k\pi}=\frac{4(-1)^{k+1}}{k\pi} $$

Therefore

$$ f(x)=\sum_{k=1}^{\infty}\frac{4(-1)^{k+1}}{k\pi}\sin\frac{k\pi}{4}x $$

holds (verified with SymPy: $b_k=4(-1)^{k+1}/(\pi k)$ for $k=1,\ldots,4$).

Solution (complex Fourier series). Let us also find the Fourier series of the same function in complex form, using Theorem 28.4 ($L=4$). Setting $f(x)=\sum_{k=-\infty}^{\infty}C_ke^{i\frac{k\pi}{4}x}$, we have

$$ C_k=\frac{1}{2\cdot4}\int_{-4}^{4}f(x)\,e^{-i\frac{k\pi}{4}x}\,dx=\frac18\int_{-4}^{4}\frac{x}{2}\,e^{-i\frac{k\pi}{4}x}\,dx=\frac1{16}\int_{-4}^{4}x\,e^{-i\frac{k\pi}{4}x}\,dx $$

(We substituted $f(x)=x/2$ and absorbed the factor $\frac12$ of $x/2$ into the coefficient, using $\frac18\times\frac12=\frac1{16}$.)

By integration by parts ($u=x,\ dv=e^{-i\frac{k\pi}{4}x}dx$, $v=\frac{4}{ik\pi}\cdot(-1)e^{-i\frac{k\pi}{4}x}=i\frac{4}{k\pi}e^{-i\frac{k\pi}{4}x}$),

$$ \int x\,e^{-i\frac{k\pi}{4}x}\,dx=i\frac{4}{k\pi}x\,e^{-i\frac{k\pi}{4}x}-\int i\frac{4}{k\pi}e^{-i\frac{k\pi}{4}x}\,dx=i\frac{4}{k\pi}x\,e^{-i\frac{k\pi}{4}x}+\left(\frac{4}{k\pi}\right)^2 e^{-i\frac{k\pi}{4}x} $$

(In the last integral, applying the same replacement once more reverses the sign and gives $+\left(\frac4{k\pi}\right)^2e^{-i\frac{k\pi}4x}$.) Hence

$$ C_k=\frac1{16}\int_{-4}^{4}x\,e^{-i\frac{k\pi}{4}x}\,dx=\frac{1}{16}\left[i\frac{4}{k\pi}x\,e^{-i\frac{k\pi}{4}x}+\left(\frac{4}{k\pi}\right)^2e^{-i\frac{k\pi}{4}x}\right]_{-4}^{4} $$

For the term $\left(\frac4{k\pi}\right)^2e^{-i\frac{k\pi}4x}$, the value at $x=4$ is $\left(\frac4{k\pi}\right)^2e^{-ik\pi}=\left(\frac4{k\pi}\right)^2(-1)^k$, and at $x=-4$ it is the same value $\left(\frac4{k\pi}\right)^2(-1)^k$ (since $e^{ik\pi}=(-1)^k$ as well), so it cancels exactly when we subtract. Only the term $i\frac4{k\pi}xe^{-i\frac{k\pi}4x}$ remains, and

$$ C_k=\frac1{16}\left(i\frac{16}{k\pi}(-1)^k-\left(-i\frac{16}{k\pi}(-1)^k\right)\right)=\frac1{16}\cdot i\frac{32}{k\pi}(-1)^k=\frac{2i(-1)^k}{k\pi} $$

(At $x=4$: $i\frac4{k\pi}\cdot4\cdot(-1)^k=i\frac{16}{k\pi}(-1)^k$; at $x=-4$: $i\frac4{k\pi}\cdot(-4)\cdot(-1)^k=-i\frac{16}{k\pi}(-1)^k$; the difference is $2i\frac{16}{k\pi}(-1)^k$. Multiplying this by $1/16$ gives the expression above.) (Verified with SymPy: $C_k=2i(-1)^k/(\pi k)$ for $k=\pm1,\pm2,\pm3$.)

Let us check whether this agrees with the result for the real Fourier series. By the formulas in the “Math Note” box, $a_k=C_k+C_{-k}$ and $b_k=i(C_k-C_{-k})$. Since $C_{-k}=\frac{2i(-1)^{-k}}{-k\pi}=-\frac{2i(-1)^k}{k\pi}$ (because $(-1)^{-k}=(-1)^k$),

$$ a_k=\frac{2i(-1)^k}{k\pi}-\frac{2i(-1)^k}{k\pi}=0,\qquad b_k=i\left(\frac{2i(-1)^k}{k\pi}-\left(-\frac{2i(-1)^k}{k\pi}\right)\right)=i\cdot\frac{4i(-1)^k}{k\pi}=-\frac{4(-1)^k}{k\pi}=\frac{4(-1)^{k+1}}{k\pi} $$

This agrees completely with the result of the real Fourier series, $a_k=0,\ b_k=4(-1)^{k+1}/(k\pi)$. We have confirmed that the real computation and the complex computation arrive at the same answer.

Exercise 28.1: Computing a complex Fourier series

Find the complex Fourier series of the function obtained by extending $f(x)=e^x\ (-\pi \lt x\le\pi)$ with period 2π.

Hint: Substitute directly into the formula of Theorem 28.4 ($L=\pi$) and compute an integral of the form $\int e^{(1-ik)x}dx$. Use $e^{\pm ik\pi}=(-1)^k$. At the end, making the denominator real (multiply by $1+ik$ to clear $1-ik$) gives a clean form.

→ Solution

Exercise 28.2: Fourier cosine series of period 2L

Find the Fourier cosine series of the function obtained by extending $f(x)=|x|\ (-3 \lt x\le3)$ with period 6 ($L=3$).

Hint: Noting that $f$ is an even function, we see from Theorem 28.3(I) that $b_k=0$ from the start. For $a_0,a_k$, apply integration by parts, as in Example 28.6, to the formulas of Theorem 28.2.

→ Solution

Exercise 28.3: Proof that only cosine terms appear in the Fourier series of an even function

Let $f(x)$ be an even function of period $2L$ ($f(-x)=f(x)$). Prove directly that $b_k=\frac1L\int_{-L}^{L}f(x)\sin\frac{k\pi}{L}x\,dx=0$ (for all $k$) by splitting the integral into $[-L,0]$ and $[0,L]$ and applying the substitution $x\to-x$ to the first half.

Hint: In the “Intuition” box of Section 28.3 we said intuitively that “since the integrand is an odd function, the integral over a symmetric interval is 0.” Here we show that intuition properly, at the level of computation, by actually splitting the integral in two and making a substitution.

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Exercise 28.4: Numerical experiment on the Fourier series of the square wave (Gibbs phenomenon)

For the partial sums $S_N(x)=\sum_{k=1}^{N}\frac{1-(-1)^k}{2k}\sin kx$ of the Fourier sine series of the square wave in Example 28.5, compute them in Python for $N=5,20,100$, and check numerically (a) how the value at the point of continuity $x=\pi/2$ approaches $\pi/4$, and (b) how an “overshoot” that does not disappear even as $N$ is increased occurs near the point of discontinuity $x=0$.

Hint: The phenomenon in (b) is called the Gibbs phenomenon. It is known that the size of the overshoot does not approach $0$ as $N\to\infty$ but remains at about 9% of the size of the jump (about 1.179 times, expressed as a ratio).

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Exercise 28.5: Working backward from complex coefficients to real coefficients (checked with the sawtooth wave)

From the complex Fourier coefficients $C_k=\dfrac{2i(-1)^k}{k\pi}\ (k\ne0)$ of the sawtooth wave obtained in Example 28.7, recompute the real Fourier coefficients $a_k,b_k$ yourself using the formulas $a_k=C_k+C_{-k}$ and $b_k=i(C_k-C_{-k})$, and verify that they agree with the result $a_k=0,\ b_k=\frac{4(-1)^{k+1}}{k\pi}$ for the real Fourier series in the main text of Example 28.7.

Hint: This backward computation has already been carried out once in the main text, but retrace it yourself by hand, keeping in mind that $(-1)^{-k}=(-1)^k$.

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Exercise 28.6: Fourier sine series by odd extension

Extend $f(x)=x(\pi-x)\ (0 \lt x \lt \pi)$ to the interval $(-\pi,0)$ by $f(x)=-f(-x)$, that is, as an odd function, and then periodically with period $2\pi$, and expand the result in a Fourier sine series.

Hint: Since this is an odd extension, $a_k=0$, and it suffices to compute $b_k=\frac2\pi\int_0^{\pi}x(\pi-x)\sin kx\,dx$ using integration by parts twice (the pattern of the two-fold integration by parts in Example 28.3 is a useful reference).

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