## Countability

1. Group the following sets according to their cardinality:

    a. $\mathbb{N} = \{ 1,2,3,4,\dots \}$
    - $\mathbb{Z} = \{ \dots, -2, -1,0,1,2, \dots \}$
    - $\mathbb{N} \times \mathbb{N}$
    - $\mathbb{Q}$ = Set of all fractions $\frac{n}{m}$ where $n,m \in \mathbb{Z}$
    - $\mathbb{R}$
    - The open interval $(0,1)$
    - The closed interval $[0,1]$
    - $2^{\mathbb{N}}$ = Set of all subsets of $\mathbb{N}$.
    - $2^{\mathbb{R}}$ = Set of all subsets of $\mathbb{R}$.
    - $\mathbb{R}^{\mathbb{R}}$ = Set of all functions from $\mathbb{R}$ to itself.

Cook up other examples and post them on the wiki!

2. Let $X$ be any set.  Show that the cardinality of $2^{X}$ is larger than the cardinality of $X$.  
(Hint: Let $f: X \to 2^X$ be a bijection.  Consider the set of all elements $x \in X$ such that $x$ is not an element of $f(x)$.)


## Fourier Series


1. Compute the Fourier Series of the following functions.  Do both the exponential and sin/cos expansions.
    a. $f(x) = \sin^3(3x)\cos^2(4x)$
    - $g(x) = x(x-2\pi)$  
      (Hint: Use integration by parts)

2. Show that  
$$\int_0^{2\pi} \sin^4(x) dx = \frac{3 \pi}{4}$$  
(Hint: write out the exponential fourier expansion of $\sin^4(x)$.)

3. Compute the exponential Fourier coefficients of $\sin^2(x)$:  
$$a_n = \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \sin^2(x) e^{-inx} dx$$  
and use this to verify that  
$$\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$$  

# Solutions

2. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$,

$$\begin{array}{ccl} \sin^4 x &=& \frac{{( e^{ix}-e^{-ix} )}^4}{16} \\
  &=& \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}. \end{array} $$

If we express any periodic function $f(x)$ as 

$$f(x) = \sum a_n f_n(x),$$

where $f_n(x) = \frac{e^{inx}}{\sqrt{2\pi}}$ and $f_0(x) = \frac{1}{\sqrt{2\pi}}$, then the Fourier coefficients for the above functions are:

$$\begin{array}{rcl} a_{-4} = a_{4} &=& \sqrt{2\pi} \times 1/16, \\
 a_{-2} = a_{2} &=& - \sqrt{2\pi} \times 4/16 \\
 a_0 &=& \sqrt{2\pi} \times 6/16 \end{array} $$


Since $a_m = < f_m, f >$ and setting $f(x) = \sin^4(x)$,

$$ \begin{array}{ccl} \int_0^{2\pi} \sin^4(x) dx = <1, f> &=& \sqrt{2\pi} \times < f_0, f > \\
&=& \sqrt{2\pi} \times a_0 \\ 
 &=& \frac{3 \pi}{4} \end{array} $$


3. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$, we have $\sin^2 x = \frac{e^{i 2x}+e^{-i 2x}-2}{4}$.  Therefore,

$$ \begin{array}{ccl} a_m &=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \sin^2(x) e^{-imx} dx \\
  &=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{i 2x}+e^{-i 2x}-2}{4} e^{-imx} dx \\
  &=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{-i (m-2)x}+e^{-i (m+2)x}-2e^{-imx}}{4} dx.  \end{array} $$

Because $\int_0^{2\pi} e^{inx} dx = 2\pi$ for $n = 0$ and $\int_0^{2\pi} e^{inx} dx = 0$ for $n \neq 0$, we have

$$a_2 = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \sqrt{2\ pi}/4,$$

$$a_0 = \frac{1}{4 \sqrt{2\pi}} \times 2\pi \times (-2) = - \frac{\sqrt{2\pi}}{2},$$

and

$$a_{-2} = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \frac{\sqrt{2\pi}}{4}.$$

It follows that

$$\sum |a_n|^2 = {(\sqrt{2\pi}/4)}^2 + {(- \sqrt{2\pi}/2)}^2 + {(\sqrt{2\pi}/4)}^2 = \frac{3 \pi}{4}$$.

It was shown in Prob 2 that $\int_0^{2\pi} \sin^4(x) dx = \frac{3 \pi}{4}$, so therefore, 

$$\int_0^{2\pi} \sin^4(x) dx = \sum |a_n|^2 = \frac{3 \pi}{4}$$



## Cardinality

Cardinality of the natural numbers (countable): $\mathbf{N}$,$\mathbf{Z}$  
Cardinality of the real numbers (continuum): $\mathbf{R}$

Proofs:  
- $\mathbf{Z}=\mathbf{N}$ under the bijection $n \mapsto 2n+1$ for nonnegative $n$ and $n \mapsto 2|n|$ for negative $n$. For example, $\{-2,-1,0,1,2\} \mapsto \{4,2,1,3,5\}$.