Chapter 1. The Genesis of Fourier Analysis

Exercises

1. If $z = x + iy$ is a complex number with $x, y \in \mathbb{R}$, we define

$$|z| = (x^2 + y^2)^{1/2}$$

and call this quantity the modulus or absolute value of $z$.
$\quad \text{(a)}$ What is the geometric interpretation of $|z|$?
Solution
The geometric interpretation of $|z|$ is the length of the vector $z$.

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$\quad \text{(b)}$ Show that if $|z| = 0$, then $z = 0$.
Solution
If $|z| = 0$ then by definition $0 = (x^2 + y^2)^{1/2} \implies 0 = x^2 + y^2 \implies x = y = 0$. $\square$

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$\quad \text{(c)}$ Show that if $\lambda \in \R$, then $|\lambda z| = |\lambda||z|$, where $|\lambda|$ denotes the standard absolute value of a real number.
Solution
Calculate that $\lambda z = \lambda x + i \lambda y$, so $|\lambda z| = ((\lambda x)^2 + (\lambda y)^2 )^{1/2} = |\lambda||z|$. $\square$

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$\quad \text{(d)}$ If $z_1$ and $z_2$ are two complex numbers, prove that

$$|z_1z_2| = |z_1||z_2| \quad \text{and} \quad |z_1 + z_2| \leq |z_1| + |z_2|.$$

Solution
Let $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$. Calculate that $z_1 \cdot z_2 = (x_1 + iy_1) \cdot (x_2 + iy_2) = x_1x_2 + ix_1y_2 + iy_1x_2 - y_1y_2 = (x_1x_2 - y_1y_2) + i(x_1y_2 + y_1x_2)$. So $|z_1z_2|^2 = |(x_1x_2 - y_1y_2) + i(x_1y_2 + y_1x_2)|^2 = (x_1x_2-y_1y_2)^2 + (x_1y_2 + y_1x_2)^2 = x_1^2x_2^2 - 2x_1x_2y_1y_2 + y_1^2y_2^2 + x_1^2y_2^2 + 2x_1x_2y_1y_2 + y_1^2x_2^2 = x_1^2x_2^2 + y_1^2y_2^2 + x_1^2y_2^2 + y_1^2x_2^2$.
Calculate separately that $(|z_1||z_2|)^2 = ((x_1^2 + y_1^2)^{1/2} \cdot (x_2^2+y_2^2)^{1/2})^2 = (x_1^2 + y_1^2)(x_2^2+y_2^2) = x_1^2x_2^2 + x_1^2y_2^2 + y_1^2x_2^2 + y_1^2y_2^2$.
Therefore $|z_1z_2|^2 = (|z_1||z_2|)^2$. Since both $|z_1z_2|$ and $|z_1||z_2|$ are non-negative we can conclude that$|z_1z_2| = |z_1||z_2|$.
For the inequality we need to show that

$$|z_1 + z_2| \leq |z_1| + |z_2| \\ \iff |(x_1 + x_2) + i(y_1+y_2)| \leq |x_1 + iy_1| + |x_2 + iy_2| \\ \iff \sqrt{(x_1 + x_2)^2 + (y_1 + y_2)^2 } \leq \sqrt{x_1 ^ 2 + y_1^2} + \sqrt{x_2^2 + y_2^2} \\ \iff (x_1 + x_2)^2 + (y_1+y_2)^2 \leq (x_1^2 + y_1^2) + 2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)} + (x_2^2 + y_2^2) \\ \iff 2x_1x_2 + 2y_1y_2 \leq 2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)} \\ \iff x_1x_2 + y_1y_2 \leq \sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)} \\ \iff x_1^2x_2^2 + 2x_1x_2y_1y_2 +y_1^2y_2^2 \leq (x_1^2+y_1^2)(x_2^2+y_2^2) \\ \iff x_1^2x_2^2 + 2x_1x_2y_1y_2 +y_1^2y_2^2 \leq x_1^2x_2^2 + x_1^2y_2^2 + y_1^2x_2^2 + y_1^2y_2^2 \\ \iff 2x_1x_2y_1y_2 \leq x_1^2y_2^2 + y_1^2x_2^2 \\ \iff 0 \leq x_1^2y_2^2 - 2x_1x_2y_1y_2 + y_1^2x_2^2\\ \iff 0 \leq (x_1y_2 - y_1x_2)^2 \\ \square$$

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$\quad \text{(e)}$ Show that if $z \neq 0$, then $|1/z| = 1/|z|$.
Solution
Using the previous result

$$|1/z||z| = |1/z \cdot z| = |1| = 1$$

Dividing both sides by $|z|$ (which we know isn't zero thanks to $\text{(b)}$ ) gives the result. $\square$

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2. If $z = x + iy$ is a complex number with $x, y \in \R$, we define the complex conjugate of $z$ by

$$\overline{z} = x - iy.$$

$\quad \text{(a)}$ What is the geometric interpretation of $\overline{z}$?
Solution
The geometric interpetation of $\overline{z}$ is a reflection of the vector $z$ over the $x$-axis.

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$\quad \text{(b)}$ Show that $|z|^2 = z\overline{z}$.
Solution

$$z\overline{z} = (x + iy)(x - iy) = x^2 + y^2 = |z|^2 \quad \square$$

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$\quad \text{(c)}$ Prove that if $z$ belongs to the unit circle, then $1/z = \overline{z}$.
Solution
If $z$ belongs to the unit circle that means that $|z| = 1$. Using part $\text{(b)}$ we get that $z\overline{z} = |z|^2 = 1$. Diving both sides by $z$ gives the result. $\square$

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3. A sequence of complex numbers $\{w_n\}_{n=1}^{\infty}$ is said to converge if there exists $w \in \mathbb{C}$ such that

$$\lim_{n \to \infty} |w_n - w| = 0,$$

and we say that $w$ is a limit of the sequence.
$\quad \text{(a)}$ Show that a converging sequence of complex numbers has a unique limit.
Solution
Let $x, y$ be limits of $\{w_n\}_{n=1}^\infty$

$$0 = \lim_{n \to \infty}|w_n - x| = \lim_{n \to \infty}|w_n - y| \\$$

So

$$\begin{aligned} |x - y| &= \lim_{n \to \infty} |x - y| = \lim_{n \to \infty} |x - w_n + w_n - y| \\ &\leq \lim_{n \to \infty} |x - w_n| + |w_n - y| \quad \text{(using 1 (d))} \\ &= \lim_{n \to \infty} |w_n - x| + |w_n - y| \\ &= \lim_{n \to \infty}|w_n - x| + \lim_{n \to \infty}|w_n - y| \\ &= 0 + 0 = 0 \\ \end{aligned}$$

This shows that $|x - y| \leq 0$, we also have $|x - y| \geq 0$ by the definition of $|\cdot|$ so we have $|x - y| = 0$. Using $\text{1 (b)}$ we get $x - y = 0 \implies x = y$ so any two limits of a converging sequence of complex numbers are equal, hence this limit is unique. $\square$

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The sequence $\{w_n\}_{n=1}^\infty$ is said to be a Cauchy sequence if for every $\epsilon > 0$ there exists a positive integer $N$ such that

$$|w_n - w_m| < \epsilon \quad \text{whenever } n, m > N.$$

$\quad \text{(b)}$ Prove that a sequence of complex numbers converges if and only if it is a Cauchy sequence. [Hint: A similar theorem exists for the convergence of a sequence of real numbers. Why does it carry over to sequences of complex numbers?]
Solution

$$\lim_{n \to \infty} |w_n - w| = 0 \iff \lim_{n \to \infty} |x_n + iy_n - (x + iy)| = 0 \\ \iff \lim_{n \to \infty} |x_n - x + i(y_n + y)| = 0 \\ \iff \lim_{n \to \infty} |x_n - x| = 0 \text{ and } \lim_{n \to \infty} |y_n - y| = 0 \\ \iff |{x_n}_a - {x_n}_b| < \epsilon \quad \text{ whenever } a, b > N_1 \text{ and } \\ |{y_n}_a - {y_n}_b| < \epsilon \quad \text{ whenever } a, b > N_2$$

$$|w_n - w_m| < \epsilon \iff |x_n + iy_n - (x_m + iy_m)| < \epsilon \\ \iff |x_n - x_m + i(y_n - y_m)| < \epsilon \\ \iff |x_n - x_m| < \epsilon \text{ and } |y_n - y_m| < \epsilon \\$$

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A series $\sum_{n=1}^\infty z_n$ of complex numbers is said to converge if the sequence formed by the partial sums

$$S_N = \sum_{n=1}^N z_n$$

converges. Let $\{a_n\}_{n=1}^\infty$ be a sequence of non-negative real numbers such that the series $\sum_{n}a_n$ converges.
$\quad \text{(c)}$ Show that if $\{z_n\}_{n=1}^\infty$ is a sequence of complex numbers satisfying $|z_n| \leq a_n$ for all $n$, then the series $\sum_{n} z_n$ converges. [Hint: Use the Cauchy criterion.]
Solution
Start with the Cauchy criterion for $\sum_{n}a_n$. For any $\epsilon > 0$ there exists $n$ such that for all $N, M > K$ we have

$$\epsilon > |A_N - A_M| = \left|\sum_{n=1}^N a_n - \sum_{m=1}^M a_m \right| = \left|\sum_{j=\text{min}(N, M)}^{\text{max}(N, M)} a_j\right|$$

Finish with the Cauchy criterion for $\sum_{n}z_n$

$$|S_N - S_M| = \left|\sum_{n=1}^N z_n - \sum_{m=1}^M z_m\right| = \left|\sum_{j=\text{min}(N, M)}^{\text{max}(N, M)} z_j\right| \\ \leq \sum_{j=\text{min}(N, M)}^{\text{max}(N, M)} |z_j| \leq \sum_{j=\text{min}(N, M)}^{\text{max}(N, M)} a_j \leq \left|\sum_{j=\text{min}(N, M)}^{\text{max}(N, M)} a_j\right| < \epsilon \quad \square$$

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4. For $z \in \Complex$, we define the complex exponential by

$$e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}.$$

$\quad \text{(a)}$ Prove that the above definition makes sense, by showing that the series converges for every complex number $z$. Moreover, show that the convergence is uniform on every bounded subset of $\Complex$.
Solution
First we use 3. $\text{(c)}$ to show that the series converges. Let $z \in \Complex$, then we have a sequence of non-negative real numbers $\frac{|z|^n}{n!}$ whose sum converges to $e^{|z|}$. Since $\left|\frac{z^n}{n!}\right| \leq \frac{|z|^n}{n!}$ we are done.
We go on to show that the convergence is uniform on bounded subsets. Let $S \subset C$ be bounded with radius $R>0$, so for each $z \in S$ we have $|z| < R$. Let $\epsilon > 0$. Regardless of what $z$ we choose from $S$

$$\left| \sum_{n=0}^N \frac{z^n}{n!}-e^z \right| = \left| \sum_{n=N}^\infty \frac{z^n}{n!} \right| \leq \sum_{n=N}^\infty \left|\frac{z^n}{n!}\right| \leq \sum_{n=N}^\infty \frac{|z|^n}{n!} \leq \sum_{n=N}^\infty \frac{R^n}{n!}$$

So choose $N$ large enough so that we have

$$\epsilon > \left|\sum_{n=0}^N \frac{R^n}{n!} - e^R\right| = \left| \sum_{n=N}^\infty \frac{R^n}{n!} \right| = \sum_{n=N}^\infty \frac{R^n}{n!}.$$

Note that the above is valid because of the convergence for the exponential series of positive reals. Combining inequalities yields the desired result

$$\left| \sum_{n=0}^N \frac{z^n}{n!}-e^z \right| < \epsilon. \quad \square$$

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$\quad \text{(b)}$ If $z_1, z_2$ are two complex numbers, prove that $e^{z_1}e^{z_2} = e^{z_1 + z_2}$. [Hint: Use the binomial theorem to expand $(z_1 + z_2)^n$, as well as the formula for the binomial coefficients.]
Solution
This problem had stumped me for a while. I ended up taking a break from this problem, learning some combinatorics, and coming back to it. Upon doing so I realized that there is a very easy combinatorical solution to this problem using the basic method of "double-counting". Take a look at the following infinite list of terms.

$$\begin{array}{ccccc} 1 \\[4pt] a & b \\[4pt] \frac{a^2}{2} & ab & \frac{b^2}{2} \\[4pt] \frac{a^3}{6} & \frac{a^2b}{2} & \frac{ab^2}{2} & \frac{b^3}{6} \\[4pt] \vdots & \vdots & \vdots & \vdots & \vdots \end{array}$$

Here the term at the $n + 1$th row and the $k + 1$th column is given by the formula $\frac{a^{n-k}b^{k}}{(n -k)!k!}$, where we always have $k \leq n$, so the sum of all of these terms is $\sum_{n=0}^{\infty}\sum_{k=0}^{n} \frac{a^{n-k}b^k}{(n-k)!k!}$.

The key insight of the double-counting method is that there are two different ways of adding up all of the terms- you can add all the rows or add all of the columns- but the result is the same.

If we add all the rows we get

$$\sum_{n=0}^{\infty}\sum_{k=0}^{n} \frac{a^{n-k}b^k}{(n-k)!k!} \\ = 1 + (a + b) + \left(\frac{a^2}{2} + ab + \frac{b^2}{2}\right) + \left( \frac{a^3}{6} + \frac{a^2b}{2} + \frac{ab^2}{2} + \frac{b^3}{6} \right) + \cdots \\ = \frac{(a + b)^0}{0!} + \frac{(a + b)^1}{1!} + \frac{(a + b)^2}{2!} + \frac{(a + b)^3}{3!} + \cdots \\ = \sum_{n=0}^{\infty} \frac{(a + b)^n}{n!} = e^{a + b}$$

If we add all the columns we get

$$\begin{aligned} &\sum_{n=0}^{\infty}\sum_{k=0}^{n} \frac{a^{n-k}b^k}{(n-k)!k!} \\ &= \left(1 + a + \frac{a^2}{2} + \frac{a^3}{6} + \cdots \right) + \left( b + ab + \frac{a^2b}{2} + \cdots \right) \\ &+ \left( \frac{b^2}{2} + \frac{ab^2}{2} + \cdots \right) + \left(\frac{b^3}{6} + \cdots\right) + \cdots \\ &= \sum_{n=0}^{\infty} \frac{a^n}{n!} + b\sum_{n=0}^{\infty}\frac{a^n}{n!} + \frac{b^2}{2}\sum_{n=0}^{\infty}\frac{a^n}{n!} + \frac{b^3}{6}\sum_{n=0}^{\infty} \frac{a^n}{n!} + \cdots \\ &= \sum_{n=0}^{\infty}\frac{a^n}{n!} \left( 1 + b + \frac{b^2}{2} + \frac{b^3}{6} + \cdots \right) \\ &= \sum_{n=0}^{\infty}\frac{a^n}{n!} \cdot \sum_{k=0}^{\infty}\frac{b^k}{k!} = e^a \cdot e^b \end{aligned}$$

This is the heart and soul of the proof, but it is a little bit hand-wavey. I hope that the above gives the reader a clear intuition behind the proof. In order to turn this intuition into more rigorous mathematics, we use the binomial expansion

$$(a + b)^n = \sum_{k=0}^{n} \frac{n!a^{n-k}b^k}{(n-k)!k!}$$

to calculate that

$$\sum_{n=0}^{\infty}\sum_{k=0}^{n} \frac{a^{n-k}b^k}{(n-k)!k!} = \sum_{n=0}^{\infty} \frac{(a+b)^n}{n!} = e^{a + b}$$

and combinatorically rearrange the series to switch from adding the rows to adding the columns via the identity

$$\sum_{n=0}^{\infty}\sum_{k=0}^{n} f(n,k) = \sum_{k=0}^{\infty}\sum_{n=k}^{\infty} f(n,k)$$

which holds because both the LHS and RHS are summing over each $k, n \in \N$ such that $k \leq n$. Plugging in $f(n,k)=\frac{a^{n-k}b^{k}}{(n-k)!k!}$ to the identity yields

$$\sum_{n=0}^{\infty}\sum_{k=0}^{n} \frac{a^{n-k}b^k}{(n-k)!k!} = \sum_{k=0}^{\infty}\sum_{n=k}^{\infty} \frac{a^{n-k}b^k}{(n-k)!k!} \\ = \sum_{k=0}^{\infty} \frac{b^k}{k!} \sum_{n=k}^{\infty} \frac{a^{n-k}}{(n-k)!} = \sum_{k=0}^{\infty} \frac{b^k}{k!} \sum_{n=0}^{\infty} \frac{a^n}{n!} = e^a \cdot e^b \qquad \square$$

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$\quad \text{(c)}$ Show that if $z$ is purely imaginary, that is, $z = iy$ with $y \in \R$, then

$$e^{iy} = \cos y + i \sin y.$$

This is Euler's identity. [Hint: Use power series.]
Solution
The general idea is

$$e^{iy} = \sum_{n=0}^{\infty}\frac{(iy)^n}{n!} = 1 + (iy) + \frac{(iy)^2}{2} + \frac{(iy)^3}{6} + \cdots \\ = 1 + iy - \frac{y^2}{2} -\frac{iy^3}{6} + \cdots \\ = (1 - \frac{y^2}{2} + \cdots) + (y - \frac{y^3}{6} + \cdots)i \\ = \cos y + i \sin y$$

More rigorously,

$$e^{iy} = \sum_{n=0}^{\infty}\frac{(iy)^n}{n!} = \sum_{n=0 \text{, } n \text { is even}}^{\infty}\frac{(iy)^n}{n!} + \sum_{n=0 \text{, } n \text{ is odd}}^{\infty}\frac{(iy)^n}{n!} \\ = \sum_{n=0}^{\infty} \frac{(iy)^{2n}}{(2n)!} + \sum_{n=0}^{\infty} \frac{(iy)^{2n+1}}{(2n+1)!} \\ = \sum_{n=0}^{\infty} \frac{i^{2n}(y)^{2n}}{(2n)!} + \sum_{n=0}^{\infty} \frac{i^{2n+1}(y)^{2n+1}}{(2n+1)!} \\ = \sum_{n=0}^{\infty} \frac{i^{2n}(y)^{2n}}{(2n)!} + i\sum_{n=0}^{\infty} \frac{i^{2n}(y)^{2n+1}}{(2n+1)!} \\ = \sum_{n=0}^{\infty} \frac{(-1)^{n}(y)^{2n}}{(2n)!} + i\sum_{n=0}^{\infty} \frac{(-1)^{n}(y)^{2n+1}}{(2n+1)!} \\ = \cos y + i \sin y \quad \square$$

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$\quad \text{(d)}$ More generally,

$$e^{x + iy} = e^x(\cos y + i \sin y)$$

whenever $x, y \in \R$, and show that

$$\left| e^{x + iy} \right| = e^{x}$$

Solution
Combining parts $\text{(b)}$ and $\text{(c)}$ gives

$$e^{x + iy} = e^x \cdot e^{iy} = e^x ( \cos y + i \sin y)$$

and we can calculate that

$$|e^{x + iy}| = |e^x (\cos y + i \sin y)| = |e^x| |\cos y + \sin y| = e^x \quad \square$$

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$\quad \text{(e)}$ Prove that $e^z = 1$ if and only if $z = 2 \pi k i$ for some integer $k$.
Solution
Let $z = x + iy$ with $x, y \in \R$

$$1 = e^z = e^{x+iy} = e^x(\cos y + i \sin y) \\ \iff e^x \cos y = 1 \text{ and } \sin y \cdot e^x = 0$$

Remember that $x$ and $y$ are real so

$$\sin y \cdot e^x = 0 \iff \sin y = 0 \iff y = 2 \pi k \text{, for some integer } k$$

We can now plug in $y$ to the other equation to find $x$

$$e^x \cos y = 1 \iff e^x \cos 2\pi k = 1 \iff e^x = 1 \iff x = 0$$

So in conclusion

$$e^z = 1 \iff y = 2 \pi k \text{ and } x = 0 \iff z = 2\pi ki \quad \square$$

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$\quad \text{(f)}$ Show that every complex number $z = x + iy$ can be written in the form

$$z = r e^{i \theta},$$

where $r$ is unique and in the range $0 \leq r < \infty$, and $\theta \in \R$ is unique up to an integer multiple of $2 \pi$. Check that

$$r = |z| \quad \text{and} \quad \theta = \arctan(y/x)$$

whenever these formulas make sense.
Solution

Start by dividing out the radius $r = |z| = \sqrt{x^2 + y^2}$ from $z$,

$$z = x + iy \\ \frac{z}{r} = \frac{x}{r} + i \frac{y}{r} \\$$

Multiply by $r$ but don't cancel on the right hand side,
$$z = r \left( \frac{x}{r} + i \frac{y}{r} \right)$$

Next, consider the following right triangle.

By the diagram and basic trig we have $\cos \theta = \frac{x}{r}$, $\sin \theta \ = \frac{y}{r}$, and $\theta = \arctan (y/x)$. Plugging in,

$$z = r \cdot ( \cos \theta + i \sin \theta) = re^{i\theta} \quad \square$$

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$\quad \text{(g)}$ In particular, $i = e^{i \pi / 2}$. What is the geometric meaning of multiplying a complex number by $i$? Or by $e^{i \theta}$ for any $\theta \in \R$
Solution
Multiplying by $i$ is rotating $\pi / 2$ radians, in general multipling by $e^{i \theta}$ is rotating by $\theta$ radians.

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$\quad \text{(h)}$ Given $\theta \in \R$, show that

$$\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2} \quad \text{and} \quad \sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}.$$

There are also called Euler's indentities.
Solution

$$e^{i \theta} = \cos \theta + i \sin \theta \\ e^{-i \theta} = e^{i \cdot (-\theta)} = \cos (-\theta) + i \sin (-\theta) = \cos \theta - i \sin \theta \\ e^{i \theta} + e^{-i\theta} = (\cos \theta + i \sin \theta) + (\cos \theta - i \sin \theta) = 2 \cos \theta \\ \cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2} \\ e^{i \theta} - e^{-i \theta} = (\cos \theta + i \sin \theta) - (\cos \theta - i \sin \theta) = 2i \sin \theta \\ \sin \theta = \frac{e^{i \theta} - e^{-i\theta}}{2i} \quad \square$$

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$\quad \text{(i)}$ Use the complex exponential to derive trigonometric identities such as

$$\cos (\theta + \phi) = \cos \theta \cos \phi - \sin \theta \sin \phi,$$

and then show that

$$2 \sin \theta \sin \phi = \cos(\theta - \phi) - \cos(\theta + \phi), \\ 2 \sin \theta \cos \phi = \sin(\theta + \phi) + \sin(\theta - \phi).$$

This calculation connects the solution given by d'Alember in terms of traveling waves and the solution in terms of superposition of standing waves.
Solution
First calculate that

$$\begin{aligned} &2 \cos(\theta + \phi)\\ &= e^{i(\theta + \phi)} + e^{-i(\theta + \phi)} = e^{i\theta}e^{i\phi} + e^{-i\theta}e^{-i\phi} \\ &= (\cos \theta + i \sin \theta) (\cos \phi + i \sin \phi) + (\cos \theta - i \sin \theta)(\cos \phi - i \sin \phi) \\ &= (\cos\theta \cos\phi + i \cos\theta \sin\phi + i \sin \theta \cos \phi - \sin \theta \sin \phi) \\ &+ (\cos \theta \cos \phi - i \cos \theta \sin \phi - i \sin \theta \cos \phi - \sin \theta \sin \phi) \\ &= 2 \cos \theta \cos \phi - 2 \sin \theta \sin \phi \quad \square \end{aligned}$$

Using this formula

$$\begin{aligned} &\cos(\theta - \phi) - \cos(\theta + \phi) \\ &= \cos(\theta + (-\phi)) - \cos(\theta + \phi) \\ &= (\cos \theta \cos (- \phi) - \sin \theta \sin (-\phi)) - (\cos \theta \cos \phi - \sin \theta \sin \phi) \\ &= \cos \theta \cos \phi + \sin \theta \sin \phi - \cos \theta \cos \phi + \sin \theta \sin \phi \\ &= 2 \sin \theta \sin \phi \quad \square \end{aligned}$$

Now let's calculate sine's angle addition formula

$$\begin{aligned} &2i \sin(\theta + \phi) \\ &= e^{i (\theta + \phi)} - e^{-i (\theta + \phi)} = e^{i\theta}e^{i\phi} - e^{i(-\theta)}e^{i(-\phi)} \\ &= (\cos \theta \cos \phi + i \cos \theta \sin \phi + i \sin \theta \cos \phi - \sin \theta \sin \phi) \\ &- (\cos \theta \cos \phi - i \cos \theta \sin \phi - i \sin \theta \cos \phi - \sin \theta \sin \phi) \\ &= 2i \cos \theta \sin \phi + 2i\sin \theta \cos \phi \\ \end{aligned} \\ \sin(\theta + \phi) = \cos \theta \sin \phi + \sin \theta \cos \phi$$

We can now solve the last identity

$$\begin{aligned} &\sin(\theta + \phi) + \sin(\theta - \phi) \\ &= (\cos \theta \sin \phi + \sin \theta \cos \phi) + (\cos \theta \sin (-\phi) + \sin \theta \cos (-\phi)) \\ &= (\cos \theta \sin \phi + \sin \theta \cos \phi) + (-\cos \theta \sin \phi + \sin \theta \cos \phi) \\ &= 2\sin \theta \cos \phi \quad \square \end{aligned}$$

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5. Verify that $f(x) = e^{inx}$ is periodic with period $2 \pi$ and that

$$\frac{1}{2\pi} \int_{-\pi}^{\pi} e^{inx} \, dx = \begin{cases} 1 \quad \text{if } n = 0, \\ 0 \quad \text{if } n \neq 0. \end{cases}$$

Use this fact to prove that if $n, m \geq 1$ we have

$$\frac{1}{\pi} \int_{-\pi}^{\pi} \cos nx \cos mx \, dx = \begin{cases} 0 \quad \text{if } n \neq m, \\ 1 \quad \text{if } n = m, \end{cases}$$

and similarly

$$\frac{1}{\pi} \int_{-\pi}^{\pi} \sin nx \sin mx \, dx = \begin{cases} 0 \quad \text{if } n \neq m, \\ 1 \quad \text{if } n = m. \end{cases}$$

Finally, show that

$$\int_{-\pi}^{\pi} \sin nx \cos mx \, dx = 0 \quad \text{for any } n, m.$$

[Hint: Calculate $e^{inx}e^{-imx} + e^{inx}e^{imx}$ and $e^{inx}e^{-imx} - e^{inx}e^{imx}$.]

Solution

First we need to show that $f(x) = e^{inx}$ is periodic with period $2 \pi$. Plug in and use $4(\text{b})$ to get

$$\tag{1} f(x + 2\pi) = e^{in(x + 2\pi)} = e^{inx + 2\pi in} = f(x)e^{2\pi in}.$$

By Euler's identity $\left(4 (\text{c})\right)$

$$e^{2\pi i n} = \cos(2\pi n) + i \sin(2\pi n) = 1,$$

plugging this into $(1)$ gives

$$f(x + 2 \pi) = f(x),$$

so by definition $f$ is periodic with period $2 \pi$.

Next we need to show that

$$\tag{2} \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{inx} \, dx = \begin{cases} 1 \quad \text{if } n = 0, \\ 0 \quad \text{if } n \neq 0. \end{cases}$$

Calculate that

$$\int_{-\pi}^{\pi} e^{inx} \, dx = \int_{-\pi}^{\pi} \cos(nx) + i \sin(nx) \, dx \\ = \int_{-\pi}^{\pi} \cos(nx) \, dx + i \int_{-\pi}^{\pi} \sin(nx) \, dx$$

If $n \neq 0$ then

$$= \left. \frac{\sin(nx)}{n} \right|_{-\pi}^{\pi} - i \left. \frac{\cos(nx)}{n} \right|_{-\pi}^{\pi} \\[4px] = \frac{\sin(n\pi) - \sin(-n\pi)}{n} - i \frac{\cos(n\pi) - \cos(-n\pi)}{n} = 0,$$

the last equality following from $\sin(n\pi) = 0$ and $\cos(x) = \cos(-x)$.
The case of $n=0$ is trivial because $e^{inx} = 1$.

Next we need to show that if $n, m \geq 1$ then

$$\frac{1}{\pi} \int_{-\pi}^{\pi} \cos nx \cos mx \, dx = \begin{cases} 0 \quad \text{if } n \neq m \\ 1 \quad \text{if } n = m. \end{cases}$$

Using the hint,

$$e^{inx}e^{-imx} + e^{inx}e^{imx} \\ = (\cos(nx) + i \sin(nx))(\cos(-mx) + i \sin(-mx)) \\ + (\cos(nx) + i\sin(nx))(\cos(mx) + i \sin(mx)) \\ = (\cos nx \cos mx - i\cos nx \sin mx + i\sin nx\cos mx + \sin nx \sin mx ) \\ + ( \cos nx \cos mx + i \cos nx \sin mx + i \sin nx \cos mx - \sin nx \sin mx ) \\ = 2\cos nx \cos mx + 2i \sin nx \cos mx,$$

and similarly

$$e^{inx}e^{-imx} - e^{inx}e^{imx} = 2 \sin nx \sin mx - 2i \cos nx \sin mx.$$

Integrate both sides to get

$$\int_{-\pi}^{\pi} e^{i(n-m)x} \, dx + \int_{-\pi}^{\pi} e^{i(n+m)x} \, dx \\ = 2\int_{-\pi}^{\pi} \cos nx \cos mx \, dx + 2i\int_{-\pi}^{\pi} \sin nx \cos mx \, dx,$$

and similarly

$$\int_{-\pi}^{\pi} e^{i(n-m)x} \, dx - \int_{-\pi}^{\pi} e^{i(n+m)x} \, dx \\ = 2\int_{-\pi}^{\pi} \sin nx \sin mx \, dx - 2i\int_{-\pi}^{\pi} \cos nx \sin mx \, dx.$$

If $n = m$ then of course $n - m = 0$, so we can plug into $(2)$ to get

$$2 \pi = 2\int_{-\pi}^{\pi} \cos nx \cos mx \, dx + 2i\int_{-\pi}^{\pi} \sin nx \cos mx \, dx$$

and similarly

$$2 \pi = 2\int_{-\pi}^{\pi} \sin nx \sin mx \, dx - 2i\int_{-\pi}^{\pi} \cos nx \sin mx \, dx.$$

If $n \neq m$ then we get

$$0 = 2\int_{-\pi}^{\pi} \cos nx \cos mx \, dx + 2i\int_{-\pi}^{\pi} \sin nx \cos mx \, dx,$$

and similarly

$$0 = 2\int_{-\pi}^{\pi} \sin nx \sin mx \, dx - 2i\int_{-\pi}^{\pi} \cos nx \sin mx \, dx.$$

The result follows from equating real and imaginary parts. $\quad \square$

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6. Prove that if $f$ is a twice continuously differentiable function on $\R$ which is a solution of the equation

$$f''(t) + c^2f(t) = 0,$$

then there exist constants $a$ and $b$ such that

$$f(t) = a \cos ct + b \sin ct.$$

This can be done by differentiating the two functions $g(t) = f(t) \cos ct - c^{-1}f'(t)\sin ct$ and $h(t) = f(t) \sin ct + c^{-1}f'(t)\cos ct$.

Solution

Let's start by following the hint. First we differentiate $g(t)$,

$$g'(t) = \frac{d}{dt} \left[ f(t)\cos ct \right] - c^{-1}\frac{d}{dt} \left[ f'(t) \sin ct \right] \\[4px] = \left(f'(t)\cos ct + f(t) \cdot -c\sin ct\right) - c^{-1}\left(f''(t)\sin ct + f'(t) \cdot c \cos ct\right) \\ = \left(f'(t)\cos ct - c f(t) \sin ct \right) - c^{-1} \left( -c^2 f(t) \sin ct + cf'(t) \cos ct \right) \\ = f'(t) \cos ct - c f(t) \sin ct + c f(t) \sin ct -f'(t) \cos ct = 0.$$

Then we differentiate $h(t)$,

$$h'(t) = \frac{d}{dt} \left[ f(t) \sin ct \right] + c^{-1} \frac{d}{dt} \left[ f'(t) \cos ct \right] \\ = \left(f'(t) \sin ct + f(t) \cdot c \cdot \cos ct \right) + c^{-1}\left( f''(t) \cos ct + f'(t) \cdot (-c) \cdot \sin ct \right) \\ = \left(f'(t) \sin ct + f(t) \cdot c \cdot \cos ct \right) + c^{-1}\left( -c^2 f(t) \cos ct + f'(t) \cdot (-c) \cdot \sin ct \right) \\ = f'(t) \sin ct + f(t) \cdot c \cdot \cos ct - c f(t) \cos ct - f'(t) \cdot \sin ct = 0. \\$$

Therefore there exist constants $a$ and $b$ such that

$$g(t) = a, \quad h(t) = b \\ f(t) \cos ct - c^{-1}f'(t)\sin ct = a, \quad f(t) \sin ct + c^{-1}f'(t)\cos ct = b \\[4px] c^{-1}f'(t) \sin ct = f(t) \cos ct - a, \quad f(t) = \frac{b - c^{-1}f'(t)\cos ct}{\sin ct}$$

Plug the right into the left

$$c^{-1}f'(t)\sin ct = \frac{b - c^{-1}f'(t)\cos ct}{\sin ct}\cos ct - a \\[4px] c^{-1}f'(t)\sin^2 ct = b\cos ct - c^{-1}f'(t) \cos^2 ct - a \sin ct \\ c^{-1}f'(t)\sin^2 ct + c^{-1}f'(t) \cos^2 ct = b\cos ct - a \sin ct \\ c^{-1}f'(t) (\sin^2 ct + \cos^2 ct) = b\cos ct - a \sin ct \\ c^{-1}f'(t) = b\cos ct - a \sin ct \\ f'(t) = cb\cos ct - ca \sin ct \\ \int_{0}^{t} f'(\tau) \, d\tau = b \int_{0}^{t}c \cos c\tau \, d\tau - a \int_{0}^{t} c \sin c \tau \, d\tau \\ f(t) - f(0) = b\left.\sin c\tau \right|_{0}^{t} + \left.a \cos c \tau \right|_{0}^{t} \\ f(t) - f(0) = b \sin ct + a \cos ct - a$$

It suffices to show that

$$f(0) = a$$

We know that

$$f\left(\frac{\pi}{2c}\right) = \frac{b - c^{-1}f'(t)\cos \frac{\pi}{2}}{\sin \frac{\pi}{2}} = b$$

We can plug this in to get

$$f\left(\frac{\pi}{2c}\right) - f(0) = b \sin \frac{\pi}{2} + a \cos \frac{\pi}{2} - a = b-a \\ f(0) = a \quad \square$$

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7. Show that if $a$ and $b$ are real, then one can write

$$a \cos ct + b \sin ct = A \cos (ct - \phi),$$

where $A = \sqrt{a^2 + b^2}$, and $\phi$ is chosen so that

$$\cos \phi = \frac{a}{\sqrt{a^2+b^2}} \quad \text{and} \quad \sin \phi = \frac{b}{\sqrt{a^2+b^2}}.$$

Solution

First use the cosine angle addition formula from $4(i)$ to calculate that

$$\begin{aligned} \cos (ct - \phi) &= \cos(ct) \cos (-\phi) - \sin(ct) \sin(-\phi) \\ &= \cos ct \cos \phi + \sin ct \sin \phi. \end{aligned}$$

Using the following right triangle

we can choose $\phi$ such that $\cos \phi = \frac{a}{\sqrt{a^2+b^2}}$ and $\sin \phi = \frac{b}{\sqrt{a^2+b^2}}$. Plugging in,

$$\cos(ct - \phi) = \frac{a}{\sqrt{a^2+b^2}} \cos ct + \frac{b}{\sqrt{a^2+b^2}} \sin ct$$

Multiply both sides by $A = \sqrt{a^2+b^2}$ to get the desired result,

$$A\cos(ct - \phi) = a \cos ct + b \sin ct. \quad \square$$

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8. Suppose $F$ is a function on $(a,b)$ with two continuous derivatives. Show that whenever $x$ and $x + h$ belong to $(a,b)$, one may write

$$F(x+h) = F(x) + hF'(x) + \frac{h^2}{2}F''(x) + h^2\varphi (h),$$

where $\varphi(h) \to 0$ as $h \to 0$.
Deduce that

$$\frac{F(x+h) + F(x-h) - 2F(x)}{h^2} \to F''(x) \quad \text{as } h \to 0.$$

Hint: This is simply a Taylor expansion. It may be obtained by noting that

$$F(x+h) - F(x) = \int_{x}^{x+h} F'(y) \, dy,$$

and then writing $F'(y) = F'(x) + (y-x)F''(x) + (y-x)\psi(y-x)$, where $\psi(h) \to 0$ as $h \to 0$.

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9. In the case of the plucked string, use the formula for the Fourier sine coefficients to show that

$$A_m = \frac{2h}{m^2} \frac{\sin mp}{p(\pi - p)}.$$

For what position of $p$ are the second, fourth, $\dots$ harmonics missing? For what position of $p$ are the third, sixth, $\dots$ harmonics missing?

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10. Show that the expression of the Laplacian

$$\triangle = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$$

is given in polar coordinates by the formula

$$\triangle = \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}.$$

Also, prove that

$$\left|\frac{\partial u}{\partial x}\right|^2 + \left| \frac{\partial u}{\partial y} \right|^2 = \left| \frac{\partial u}{\partial r} \right|^2 + \frac{1}{r^2} \left| \frac{\partial u}{\partial \theta} \right|^2 .$$

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11. Show that if $n \in \Z$ the only solutions of the differential equation

$$r^2 F''(r) + rF'(r) - n^2F(r) = 0,$$

which are twice differentiable when $r > 0$, are given by linear combinations of $r^n$ and $r^{-n}$ when $n \neq 0$, and $1$ and $\log r$ when $n = 0$.
Hint: If $F$ solves the equation, write $F(r) = g(r)r^n$, find the equation satisfied by $g$, and conclude that $rg'(r) + 2ng(r) = c$ where $c$ is a constant.

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Problem

1. Consider the Dirichlet problem illustrated above.
More precisely, we look for a solution of the steady-state heat equation $\triangle u = 0$ in the rectangle $R = \{(x,y): 0 \leq x \leq \pi, 0 \leq y \leq 1 \}$ that vanishes on the vertical sides of $R$, and so that

$$u(x,0) = f_0(x) \quad \text{and} \quad u(x,1) = f_1(x),$$

where $f_0$ and $f_1$ are the initial data which fix the temperature distribution on the horizontal sides of the rectangle.

Use separation of variables to show that if $f_0$ and $f_1$ have Fourier expansions

$$f_0(x) = \sum_{k=1}^{\infty} A_k \sin kx \quad \text{and} \quad f_1(x) = \sum_{k=1}^{\infty} B_k \sin kx,$$

then

$$u(x,y) = \sum_{k=1}^{\infty} \left( \frac{\sinh k(1-y)}{\sinh k}A_k + \frac{\sinh ky}{\sinh k}B_k \right) \sin kx.$$

We recall the definitions of the hyperbolic sine and cosine functions:

$$\sinh x = \frac{e^x - e^{-x}}{2} \quad \text{and} \quad \cosh x = \frac{e^x + e^{-x}}{2}.$$

Compare this result with the solution of the Dirichlet problem in the strip obtained in Problem 3, Chapter 5.

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