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Analysis and transforms

Fourier series, not just the transform. A sum from −∞-\infty with a complex exponential is a different table than a definite integral.

𝑓(𝑥)=∑𝑛=−∞∞𝑐𝑛𝑒2π𝑖𝑛𝑥𝑓(𝑥)=\sum _{𝑛=-\infty }^{\infty }{𝑐}_{𝑛}{𝑒}^{2\pi 𝑖𝑛𝑥}
(3.132)
𝑐𝑛=∫01𝑓(𝑥)𝑒−2π𝑖𝑛𝑥 d𝑥{𝑐}_{𝑛}={\int }_{0}^{1}𝑓(𝑥){𝑒}^{-2\pi 𝑖𝑛𝑥}\, \mathrm{d}𝑥
(3.133)

Inversion of the transform, so the hat has to come off:

𝑓(𝑥)=∫−∞∞𝑓^(ξ)𝑒2π𝑖𝑥ξ dξ𝑓(𝑥)={\int }_{-\infty }^{\infty }\hat{𝑓}(\xi ){𝑒}^{2\pi 𝑖𝑥\xi }\, \mathrm{d}\xi
(3.134)

Parseval:

∫−∞∞∣𝑓(𝑥)∣2 d𝑥=∫−∞∞∣𝑓^(ξ)∣2 dξ{\int }_{-\infty }^{\infty }{|𝑓(𝑥)|}^{2}\, \mathrm{d}𝑥={\int }_{-\infty }^{\infty }{\left|\hat{𝑓}(\xi )\right|}^{2}\, \mathrm{d}\xi
(3.135)

Convolution, an asterisk that is not multiplication:

(𝑓∗𝑔)(𝑥)=∫−∞∞𝑓(𝑥−𝑦)𝑔(𝑦) d𝑦(𝑓\ast 𝑔)(𝑥)={\int }_{-\infty }^{\infty }𝑓(𝑥-𝑦)𝑔(𝑦)\, \mathrm{d}𝑦
(3.136)

Laplace transform:

(L𝑓)(𝑠)=∫0∞𝑒−𝑠𝑡𝑓(𝑡) d𝑡(\mathcal{L}𝑓)(𝑠)={\int }_{0}^{\infty }{𝑒}^{-𝑠𝑡}𝑓(𝑡)\, \mathrm{d}𝑡
(3.137)

The Gamma function as an integral, and as a functional equation:

Γ(𝑧)=∫0∞𝑡𝑧−1𝑒−𝑡 d𝑡\Gamma (𝑧)={\int }_{0}^{\infty }{𝑡}^{𝑧-1}{𝑒}^{-𝑡}\, \mathrm{d}𝑡
(3.138)
Γ(𝑧+1)=𝑧Γ(𝑧)\Gamma (𝑧+1)=𝑧\Gamma (𝑧)
(3.139)

Riemann zeta, then Euler’s product. If the exponent −𝑠-𝑠 falls off 𝑛𝑛, this is broken.

ζ(𝑠)=∑𝑛=1∞𝑛−𝑠\zeta (𝑠)=\sum _{𝑛=1}^{\infty }{𝑛}^{-𝑠}
(3.140)
ζ(𝑠)=∏𝑝  prime(1−𝑝−𝑠)−1\zeta (𝑠)=\prod _{𝑝\, \text{ prime}}{(1-{𝑝}^{-𝑠})}^{-1}
(3.141)

A generating function:

𝐺(𝑥)=∑𝑛=0∞𝑎𝑛𝑥𝑛𝐺(𝑥)=\sum _{𝑛=0}^{\infty }{𝑎}_{𝑛}{𝑥}^{𝑛}
(3.142)

Exponential generating function, so 𝑛!𝑛! sits in the denominator of a summand:

𝐸(𝑥)=∑𝑛=0∞𝑎𝑛𝑥𝑛𝑛!𝐸(𝑥)=\sum _{𝑛=0}^{\infty }{𝑎}_{𝑛}\frac{{𝑥}^{𝑛}}{𝑛!}
(3.143)

Geometric series, finite and infinite:

∑𝑘=0𝑛𝑟𝑘=1−𝑟𝑛+11−𝑟\sum _{𝑘=0}^{𝑛}{𝑟}^{𝑘}=\frac{1-{𝑟}^{𝑛+1}}{1-𝑟}
(3.144)
∑𝑘=0∞𝑟𝑘=11−𝑟,∣𝑟∣<1\sum _{𝑘=0}^{\infty }{𝑟}^{𝑘}=\frac{1}{1-𝑟},\quad |𝑟|<1
(3.145)

A hypergeometric term I will not expand by hand:

2𝐹1(𝑎,𝑏;𝑐;𝑧)=∑𝑛=0∞(𝑎)𝑛(𝑏)𝑛(𝑐)𝑛𝑧𝑛𝑛!{}_{2}𝐹_{1}(𝑎,𝑏;𝑐;𝑧)=\sum _{𝑛=0}^{\infty }\frac{{(𝑎)}_{𝑛}{(𝑏)}_{𝑛}}{{(𝑐)}_{𝑛}}\frac{{𝑧}^{𝑛}}{𝑛!}
(3.146)

Residue at a simple pole:

Res(𝑓,𝑎)=lim⁡𝑧→𝑎(𝑧−𝑎)𝑓(𝑧)\text{Res}(𝑓,𝑎)=\lim _{𝑧\to 𝑎}(𝑧-𝑎)𝑓(𝑧)
(3.147)

Cauchy’s integral formula, an 𝑛𝑛-th derivative on the left and a power in the denominator:

𝑓(𝑛)(𝑎)=𝑛!2π𝑖∮𝐶𝑓(𝑧)(𝑧−𝑎)𝑛+1 d𝑧{𝑓}^{(𝑛)}(𝑎)=\frac{𝑛!}{2\pi 𝑖}{\oint }_{𝐶}\frac{𝑓(𝑧)}{{(𝑧-𝑎)}^{𝑛+1}}\, \mathrm{d}𝑧
(3.148)

A commutator and a Poisson bracket, so a square bracket is not a matrix fence:

[𝐴,𝐵]=𝐴𝐵−𝐵𝐴[𝐴,𝐵]=𝐴𝐵-𝐵𝐴
(3.149)
{𝐹,𝐻}=∂𝐹∂𝑞∂𝐻∂𝑝−∂𝐹∂𝑝∂𝐻∂𝑞\{𝐹,𝐻\}=\frac{\partial 𝐹}{\partial 𝑞}\frac{\partial 𝐻}{\partial 𝑝}-\frac{\partial 𝐹}{\partial 𝑝}\frac{\partial 𝐻}{\partial 𝑞}
(3.150)

Tensor product and a wedge of differentials:

𝑈⊗𝑉𝑈\otimes 𝑉
(3.151)
ω=𝑓 d𝑥+𝑔 d𝑦+h d𝑥∧ d𝑦\omega =𝑓\, \mathrm{d}𝑥+𝑔\, \mathrm{d}𝑦+h\, \mathrm{d}𝑥\wedge \, \mathrm{d}𝑦
(3.152)
dω=∂𝑓∂𝑥 d𝑥+∂𝑔∂𝑦 d𝑦\mathrm{d}\omega =\frac{\partial 𝑓}{\partial 𝑥}\, \mathrm{d}𝑥+\frac{\partial 𝑔}{\partial 𝑦}\, \mathrm{d}𝑦
(3.153)

Bra-ket, because a pipe in the middle is not a determinant:

⟨φ ∣ ψ⟩\langle \varphi \, |\, \psi \rangle
(3.154)
∣ψ⟩=∑𝑛𝑐𝑛∣𝑛⟩|\psi \rangle =\sum _{𝑛}{𝑐}_{𝑛}|𝑛\rangle
(3.155)

Expectation and variance, operators with a subscript that is a measure:

E[𝑋]=∫Ω𝑋 d𝑃\mathbb{E}[𝑋]={\int }_{\Omega }𝑋\, \mathrm{d}𝑃
(3.156)
Var(𝑋)=E[(𝑋−E[𝑋])2]\text{Var}(𝑋)=\mathbb{E}[{(𝑋-\mathbb{E}[𝑋])}^{2}]
(3.157)

A matrix exponential as a series, because a power of a matrix is not a power of a scalar:

𝑒𝑡𝐴=∑𝑘=0∞(𝑡𝐴)𝑘𝑘!{𝑒}^{𝑡𝐴}=\sum _{𝑘=0}^{\infty }\frac{{(𝑡𝐴)}^{𝑘}}{𝑘!}
(3.158)

Variation of parameters for a second-order equation, a pair of integrals I want to see simultaneously:

𝑦𝑝=𝑢1𝑦1+𝑢2𝑦2,𝑢1′𝑦1+𝑢2′𝑦2=0𝑢1′𝑦1′+𝑢2′𝑦2′=𝑔\begin{aligned}{𝑦}_{𝑝}={𝑢}_{1}{𝑦}_{1}+{𝑢}_{2}{𝑦}_{2},\quad {𝑢}_{{1}^{\prime }}{𝑦}_{1}+{𝑢}_{{2}^{\prime }}{𝑦}_{2} & =0 \\ {𝑢}_{{1}^{\prime }}{𝑦}_{{1}^{\prime }}+{𝑢}_{{2}^{\prime }}{𝑦}_{{2}^{\prime }} & =𝑔\end{aligned}
(3.159)

The implicit function theorem’s derivative, a fraction of partials:

d𝑦d𝑥=−∂𝐹∂𝑥∂𝐹∂𝑦\frac{\mathrm{d}𝑦}{\mathrm{d}𝑥}=-\frac{\frac{\partial 𝐹}{\partial 𝑥}}{\frac{\partial 𝐹}{\partial 𝑦}}
(3.160)

Multivariable chain rule, Jacobian times a velocity:

dd𝑡𝑓(γ(𝑡))=∇𝑓(γ(𝑡))⋅γ′(𝑡)\frac{\mathrm{d}}{\mathrm{d}𝑡}𝑓(\gamma (𝑡))=\nabla 𝑓(\gamma (𝑡))\cdot {\gamma }^{\prime }(𝑡)
(3.161)