Fourier Transform formulas \begin{align*} \hat{u}(k) & = \int _{-\infty }^{\infty } u(x) e^{-i k x} \,dx \\ u(x) & = \frac{1}{2\pi }\int _{-\infty }^\infty \hat{u}(k) e^{i k x} \,dk \\ \hat{u}(k) & = \hat{G}(k) \hat{h}(k)\\ u(x) = g(x) h(x) &\Longleftrightarrow \hat{g}(k) \ast \hat{h}(k) = \hat{u}(k) = \int _{-\infty }^\infty \hat{g}(k-\tau ) \hat{h}(\tau ) \,d\tau \\ \hat{u}(k) = \hat{g}(k) \hat{h}(k) &\Longleftrightarrow g(x) \ast h(x) = u(x) = \int _{-\infty }^\infty g(x-\tau ) h(\tau ) \,d\tau \\ e^{i x d}u(x) & \Longleftrightarrow \hat{u}(k-d)\\ u(x-a) & \Longleftrightarrow e^{-i a k} \hat{u}(k)\\ \frac{du}{dx} & \Longleftrightarrow i k \hat{u}(k)\\ \int _a^x u(x) \,dx & \Longleftrightarrow \frac{ \hat{u}(k)}{i k} + c \delta (k) \end{align*}
Fourier Series formulas \begin{align*} f(x) &= a_0 + \sum \limits _{k=1}^\infty \left ( a_k \cos (k x) + b_k \sin (k x) \right ) \\ a_0 &= \frac{1}{2 \pi } \int _{-\pi }^\pi f(x) \, dx \\ a_k &= \frac{1}{\pi } \int _{-\pi }^\pi \cos (k x) f(x) \, dx = c_k + c_{_k}\\ b_k &= \frac{1}{\pi } \int _{-\pi }^\pi \sin (k x) f(x) \, dx = i(c_k - c_{-k}) \\ f(x) &= \sum \limits _{-\infty }^\infty c_k e^{i k x} \\ c_k &= \frac{1}{2 \pi } \int _{-\pi }^\pi f(x) e^{-i k x} \, dx \\ \text{square wave} &\stackrel{\text{Fourier series}}{\longleftrightarrow } \sum _{k=1,3,5,\dots }^{\infty } \frac{4}{\pi k} \sin (kx) \\ f(x) &\stackrel{\text{Fourier series}}{\longleftrightarrow } \sum _{-\infty }^\infty c_k e^{ikx} \\ f(x) &\stackrel{\text{Fourier transform}}{\longleftrightarrow } \frac{1}{2\pi } \int _{-\infty }^{\infty } \hat{f}(\omega ) e^{i\omega x} \hspace{2mm} \text{where} \hspace{2mm} \omega =\frac{2\pi }{T}k\\ \delta (x) &\stackrel{\text{Fourier series}}{\longleftrightarrow } \frac{1}{2\pi }+ \frac{1}{\pi } \left (\cos x+\cos 2x+\cos 3x + \dots \right ) \\ x &\stackrel{\text{Fourier series}}{\longleftrightarrow } 2 \left (\frac{\sin x}{1} - \frac{\sin 2x}{2} +\frac{\sin 3x}{3} - \dots \right )\\ x &\stackrel{\text{Fourier series}}{\longleftrightarrow } 2 \left (\frac{\sin x}{1} - \frac{\sin 2x}{2} +\frac{\sin 3x}{3} - \dots \right )\\ |\sin (x)| &\stackrel{\text{Fourier series}}{\longleftrightarrow } \begin{cases} a_0 &= \frac{2}{\pi }\\ a_k &= \begin{cases} 0, & \text{k odd}.\\ \frac{4}{\pi } \left ( \frac{1}{1-k^2} \right ), & \text{k even}. \end{cases} \end{cases}\\ \hat{H}(k) = \hat{f}(k) \hat{g}(k) &\stackrel{F^{-1}}{\longleftrightarrow } H(x) = f(x) \circledast g(x) = \int _{-infty}^{\infty } f(\tau ) g(x-\tau ) \,d\tau \\ u(x) &= f(x) \circledast g(x) \\ \hat{u}(k) &= \hat{f}(k) \hat{g}(k) \\ \end{align*}
1.
When period \(T\) is not \(2 \pi \) replace \(k\) by \(\frac{2 \pi }{T} k\) in all formulas for Fourier series.
2.
Plancherel formula \(2\pi \int _{-\infty }^\infty |f(x)|^2 \, dx = \int _{-\infty }^\infty | \hat{f}(k) |^2 \,dk\)
3.
Parseval’s formula \( \int _{-\pi }^{\pi } |f(x)|^2 \,dx = 2\pi \sum \limits _{k=1}^\infty |c_k|^2\)
4.
Parseval’s formula again \(2\pi a_0^2+\pi \left (a_1^2+b_1^2+a_2^2+b_2^2+\dots \right ) = \int _{-\infty }^\infty f^2(x) \,dx\)
5.
Inner products \( 2\pi \int _{-\infty }^\infty f(x)\bar{g}(x) \, dx = \int _{-\infty }^\infty \hat{f}(k) \bar{\hat{g}}(k) \, dk\)
6.
integration by parts \(\int u v' = [uv]- \int u' v\) so pick the one that is easy to differentiate for \(u\) and the one that
is easy to integrate for \(v\).
7.
properties of odd and even functions Let \(o,e\) be odd and even functions, then \(e+e=e, o+o=o, e \times e=e, o \times o=e, o \times e=o, \frac{e}{e}=e, \frac{e}{o}=o\)
8.
trig identities \begin{equation*} \begin{aligned} \sin ^2(x) &= \frac{1}{2}-\frac{1}{2}\cos (2x)\\ \cos ^2(x) &= \frac{1}{2}+\frac{1}{2}\cos (2x)\\ \sin ^3(x) &= \frac{3}{4}\sin (x) - \frac{1}{4}\sin (3x)\\ \cos ^3(x) &= \frac{3}{4}\cos (x) -\frac{1}{2}\cos (2x)\\ \sin (2x) &= 2\sin (x)\cos (x)\\ \cos (2x) &= \cos ^2(x)-\sin ^2(x)\\ &=1-2\sin ^2(x)\\ &=2\cos ^2(x)-1\\ \tan (2x) &=\frac{2\tan (x)}{1-\tan ^2(x)}\\ \sin (A \pm B) &= \sin (A)\cos (B)\pm \cos (A)\cos (B)\\ \cos (A \pm B) &= \cos (A)\sin (B)\mp \sin (A)\sin (B)\\ \int \cos ^n(x) \,dx &= \frac{\cos ^{n-1}(x) \sin (x)}{n}+ \frac{n-1}{n} \int \cos ^{n-2} \,dx\\ &= \frac{1}{2} \cos x \sin x + \frac{x}{2} \hspace{5mm} \text{n even}\\ &= \frac{1}{3} \cos ^2x \sin x + \frac{2}{3} \sin x \hspace{5mm} \text{n odd}\\ \int \sin ^n(x) \,dx &= \frac{-\sin ^{n-1}(x) \cos (x)}{n}+ \frac{n-1}{n} \int \sin ^{n-2} \,dx\\ &= \frac{-1}{2} \sin x \cos x + \frac{x}{2} \hspace{5mm} \text{n even}\\ &= \frac{-1}{3} \sin ^2x \cos x - \frac{2}{3} \cos x \hspace{5mm} \text{n odd}\\ \int x^n e^{ax} \,dx &= \frac{1}{a} \left ( x^n e^{ax} - n \int x^{n-1} e^{ax} \,dx \right )\\ \end{aligned} \end{equation*}
9.
exp/trig \begin{equation*} \begin{aligned} \sin (x) &= \frac{e^{i\theta }-e^{-i\theta }}{2i}\\ \cos (x) &= \frac{e^{i\theta }+e^{-i\theta }}{2}\\ re^{i\theta } &= r \left ( \cos (\theta ) + i \sin (\theta ) \right ) \\ \ln (r e^{i\theta }) &= \ln (r) + i\theta + 2 k \pi i \\ F(e^{-\frac{x^2}{2}}) &=\int _{-\infty }^{\infty } e^{-\frac{x^2}{2}} e^{-i k x} \,dx = e^{-\frac{k^2}{2} \sqrt{2\pi }}\\ F(e^{-{x^2}}) &= e^{-\frac{k^2}{4} \sqrt{\pi }}\\ \int _0^\infty e^{-x^2} \,dx &= \frac{\sqrt{\pi }}{2} \\ \int _{-\infty }^\infty e^{-x^2} \,dx &= \sqrt{\pi } \\ \end{aligned} \end{equation*}
Laplace
- 1.
- To find solution to Laplace on disk, or radius \(r\), use polar. The solution is \begin{equation*} \begin{aligned} u(r,\theta ) = a_0 &+ a_1 r \cos (\theta )+b_1 r \sin (\theta )\\ &+ a_2 r^2 \cos (2 \theta )+b_2 r^2 \sin (2 \theta ) \dots \\ \end{aligned} \end{equation*} Where the \(a_k\) and \(b_k\) found from finding Fourier series of \(u(r,\theta )\) evaluated at boundary as normally done.
- 2.
- solution inside the circle is \begin{equation*} u(r,\theta )=\frac{1}{2\pi } = \int _{-\pi }^{\pi } \frac{1-r^2}{1+r^2-2r \cos (\theta -\zeta ) \,d\zeta } \end{equation*} If we are given \(u_0=\delta \) at point on circle as boundary conditions, use the above formula, much easier.
misc. items
- 1.
- The function that minimizes \(\int _a^b \frac{1}{2} (u'(x))^2-f u(x) \,dx\) is the solution of \( u''(x)=f\)
- 2.
- every function is made up of odd/even parts \begin{equation*} \begin{aligned} f_{\text{odd part}} &= \frac{f(x)-f(-x)}{2} \\ f_{\text{even part}} &= \frac{f(x)+f(-x)}{2} \end{aligned} \end{equation*}
