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## 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 show that
$\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$
# Solutions
2. Since
$\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
$\sin^4 x = \frac{{( e^{ix}-e^{-ix} )}^4}{16}$
$= \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$.
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 $f_0(x) = \frac{1}{\sqrt{2\pi}}$,
The Fourier coefficients for the above functions are:
$a_{-4} = a_{4} = \sqrt{2\pi} \times 1/16$, $a_{-2} = a_{2} = - \sqrt{2\pi} \times 4/16$, $a_0 = \sqrt{2\pi} \times 6/16$
Since $a_m = < f_m, f >$ and setting $f(x) = \sin^4(x)$,
$\int_0^{2\pi} \sin^4(x) dx = <1, f> = \sqrt{2\pi} \times < f_0, f >$
$= \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$
3. Since
$\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
$\sin^2 x = \frac{e^{i 2x}+e^{-i 2x}-2}{4}$
$a_m = \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \sin^2(x) e^{-imx} dx$
$= \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{i 2x}+e^{-i 2x}-2}{4} e^{-imx} dx$
$= \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{-i (m-2)x}+e^{-i (m+2)x}-2e^{-imx}}{4} dx$.
Because $\int_0^{2\pi} e^{inx} dx = 2\pi$ for $n = 0$ and $\int_0^{2\pi} e^{inx} dx = 0$ for $n \neq 0$,
$m = 2 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \sqrt{2\pi}/4$,
$m = 0 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi \times (-2) = - \sqrt{2\pi}/2$,
$m = -2 : a_m = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \sqrt{2\pi}/4$,
Then,
$\sum |a_n|^2 = {(\sqrt{2\pi}/4)}^2 + {(- \sqrt{2\pi}/2)}^2 + {(\sqrt{2\pi}/4)}^2 = \frac{3 \pi}{4}$.
And, it was shown in Prob 2 that $\int_0^{2\pi} \sin^4(x) dx = \frac{3 \pi}{4}$.
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, $\{-2,-1,0,1,2\} \mapsto \{4,2,1,3,5\}$.
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