# Fourier Fun

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### Fourier Series

Allows us to expand any periodic funciton on the range $(-L,L)$ in terms of sinusoidal functions that are periodic on that interval

$f(x)=\sum _{ n=0 }^{ \infty }{ A_{ n }\cos (\frac { n\pi /*x }{ L } ) } +\sum _{ n=0 }^{ \infty }{ B_{ n }\sin (\frac { n\pi x }{ L } ) }$

• Recall Euler’s formula: $e^z = e^{s+it} = e^s e^{it} = e^{s}(\cos(t)+i\sin(t))$
• Euler’s identity gives $e^{i\pi} + 1 = 0$ (i.e., $e^{i/pi}=-1$)
• This is a consequence of Euler’s formula:
• $e^{ix} = \cos x + i \sin x$ ⇒ let $x = \pi$ $e^{i\pi} = \cos(\pi) + i\sin(\pi) = -1 + i(0) = -1$
• Since the sines and cosines can be combined into a complex exponential, we can use this equivalency to simplify $f(x)$ into a single term:

$f(x) = \sum_{n=-\infty}^{\infty}{a_n e^{in\pi x/L}}$

where $a_n=\frac{1}{2L} \int _{ -L}^{L}{f(x)e^{-in\pi x/L}dx}$

### Review of sine and cosine:

#### Cosine Function

• Even function ⇒ $\cos(-x) = cos(x)$ ⇒ symmetric
• $\int _{-\pi}^{\pi}{\cos(\theta)d\theta} = 0$

#### Sine Function

• Odd function ⇒ $\sin(-x) = -\sin(x)$ 1 ⇒ antisymmetric
• $\int _{-\pi}^{\pi}{\sin(\theta)d\theta} = 0$

### The Fourier Transform

Note: a majority of these equations I got from the first reference  I listed

Let $g(t)$ be a function of time and $G(\omega)$ be a function of frequency

• Aside: $\omega \equiv 2\pi \upsilon$ where $\omega =$ angular frequency and $\upsilon =$ oscillation frequency
• Then, the FT of $g(t)$ (if it exists) is …

$G(\omega) = \mathcal{F} \{g(t) \} = \sqrt { \frac { |b| }{ (2\pi )^{ 1-a } } } \int _{ -\infty }^{ \infty }{g(t)e^{ib\omega t}dt}$

$g(t) = \mathcal{F}^{-1} \{g(t) \} = \mathcal{F} \{G(\omega ) \} = \sqrt { \frac { |b| }{ (2\pi )^{ 1+a } } } \int _{ -\infty }^{ \infty }{G(\omega)e^{-ib\omega t}d\omega}$

#### Some common parameter choices

• Physics and Mathematica default: $a = 0, b =1$

$G(\omega) = \sqrt { \frac { |1| }{ (2\pi )^{ 1-0 } } } \int _{ -\infty }^{ \infty }{g(t)e^{i(1)\omega t}dt} = \sqrt { \frac { 1 }{ (2\pi ) } } \int _{ -\infty }^{ \infty }{g(t)e^{i\omega t}dt}$

$G(\omega) = \sqrt { \frac { 1 }{ (2\pi ) } } \int _{ -\infty }^{ \infty }{g(t)e^{i\omega t}dt}$

• Pure mathematics and systems engineering: $a=1, b =-1$

$G(\omega) = \sqrt { \frac { |-1| }{ (2\pi )^{ 1-1 } } } \int _{ -\infty }^{ \infty }{g(t)e^{i(-1)\omega t}dt} = \sqrt { \frac { 1 }{ (2\pi )^0 } } \int _{ -\infty }^{ \infty }{g(t)e^{-i\omega t}dt} = \int _{ -\infty }^{ \infty }{g(t)e^{-i\omega t}dt}$

$G(\omega) = \int _{ -\infty }^{ \infty }{g(t)e^{-i\omega t}dt}$

• Classical physics: $a=-1, b=1$

$G(\omega) = \sqrt { \frac { |1| }{ (2\pi )^{ 1-(-1) } } } \int _{ -\infty }^{ \infty }{g(t)e^{i(1)\omega t}dt} = \sqrt { \frac { 1 }{ (2\pi )^2 } } \int _{ -\infty }^{ \infty }{g(t)e^{i\omega t}dt} = \frac{1}{2\pi }\int _{ -\infty }^{ \infty }{g(t)e^{-i\omega t}dt}$

$G(\omega) = \frac{1}{2\pi }\int _{ -\infty }^{ \infty }{g(t)e^{-i\omega t}dt}$

• Signal processing: $a = 0, b = -2\pi$

$G(\omega) = \sqrt { \frac { |-2\pi| }{ (2\pi )^{ 1-(0) } } } \int _{ -\infty }^{ \infty }{g(t)e^{i(-2\pi)\omega t}dt} = sqrt { \frac { 2\pi }{ 2\pi} } \int _{ -\infty }^{ \infty }{g(t)e^{-i2\pi \omega t}dt} = \int _{ -\infty }^{ \infty }{g(t)e^{-i2\pi \omega t}dt}$

$G(\omega) = \int _{ -\infty }^{ \infty }{g(t)e^{-i2\pi \omega t}dt}$

### Fourier Transform in Quantum Mechanics

Note: most of this material comes from the second link listed under my sources.

#### Conjugate pairs

• A conjugate pair is a pair of variables that are related to one another via the FT.
• Two conjugate pairs that exist in nature are:
1. Time ($t$) and frequency ($\upsilon$):
• $G(\upsilon) = \mathcal{F} \{ g(t) \} = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}{e^{-i\upsilon t}g(t)dt}$
• $g(t) = \mathcal{F} \{ G(\upsilon) \} = \mathcal{F}^{-1} \{ g(t) \} = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}{e^{i\upsilon t}G(\upsilon) d\upsilon }$
2. Position ($x$) and momentum ($\rho$):
• $\phi(\rho) = \mathcal{F} \{ \Psi(x) \} = \frac{1}{\sqrt{2\pi \hbar}} \int_{-\infty}^{\infty}{e^{-i\rho x/ \hbar} \Psi (x)dx}$
• $\Psi(x) = \mathcal{F} \{ \phi (\rho) \} = \mathcal{F}^{-1} \{ \Phi(x) \} = \frac{1}{\sqrt{2\pi \hbar}} \int_{-\infty}^{\infty}{e^{i\rho x/ \hbar} \phi (\rho) d\rho}$

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