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Calculus shapes

Ordinary derivatives I already have as primes. A second-order linear ODE is the first equation that looks like physics.

𝑦′′+𝑝(𝑥)𝑦′+𝑞(𝑥)𝑦=𝑔(𝑥){𝑦}^{\prime\prime }+𝑝(𝑥){𝑦}^{\prime }+𝑞(𝑥)𝑦=𝑔(𝑥)
(3.44)

Constant coefficients, the characteristic polynomial sitting beside it:

𝑦′′−3𝑦′+2𝑦=0{𝑦}^{\prime\prime }-3{𝑦}^{\prime }+2𝑦=0
(3.45)
𝑟2−3𝑟+2=0{𝑟}^{2}-3𝑟+2=0
(3.46)

An Euler equation, so I remember the 𝑥𝑥 in the coefficients is not a typo:

𝑥2𝑦′′+𝑥𝑦′−𝑦=0{𝑥}^{2}{𝑦}^{\prime\prime }+𝑥{𝑦}^{\prime }-𝑦=0
(3.47)

Cauchy–Euler with a first derivative I cannot drop:

𝑥2𝑦′′+5𝑥𝑦′+4𝑦=0{𝑥}^{2}{𝑦}^{\prime\prime }+5𝑥{𝑦}^{\prime }+4𝑦=0
(3.48)

Bessel, because a 1𝑥\frac{1}{𝑥} coefficient next to 𝑦′{𝑦}^{\prime } is where fractions in ODEs go to die:

𝑥2𝑦′′+𝑥𝑦′+(𝑥2−𝑛2)𝑦=0{𝑥}^{2}{𝑦}^{\prime\prime }+𝑥{𝑦}^{\prime }+({𝑥}^{2}-{𝑛}^{2})𝑦=0
(3.49)

Legendre:

(1−𝑥2)𝑦′′−2𝑥𝑦′+𝑛(𝑛+1)𝑦=0(1-{𝑥}^{2}){𝑦}^{\prime\prime }-2𝑥{𝑦}^{\prime }+𝑛(𝑛+1)𝑦=0
(3.50)

Sturm–Liouville, so a weight and an eigenvalue share a line:

(𝑝(𝑥)𝑦′)′+𝑞(𝑥)𝑦+λ𝑤(𝑥)𝑦=0{(𝑝(𝑥){𝑦}^{\prime })}^{\prime }+𝑞(𝑥)𝑦+\lambda 𝑤(𝑥)𝑦=0
(3.51)

First-order linear, integrating factor and all:

𝑦′+𝑃(𝑥)𝑦=𝑄(𝑥){𝑦}^{\prime }+𝑃(𝑥)𝑦=𝑄(𝑥)
(3.52)

Separable, written as differentials because that is how I will actually compute it:

d𝑦d𝑥=𝑔(𝑥)h(𝑦)\frac{\mathrm{d}𝑦}{\mathrm{d}𝑥}=𝑔(𝑥)h(𝑦)
(3.53)
d𝑦h(𝑦)=𝑔(𝑥) d𝑥\frac{\mathrm{d}𝑦}{h(𝑦)}=𝑔(𝑥)\, \mathrm{d}𝑥
(3.54)

Exactness, 𝑀𝑀 and 𝑁𝑁 on one line:

𝑀(𝑥,𝑦) d𝑥+𝑁(𝑥,𝑦) d𝑦=0𝑀(𝑥,𝑦)\, \mathrm{d}𝑥+𝑁(𝑥,𝑦)\, \mathrm{d}𝑦=0
(3.55)
∂𝑀∂𝑦=∂𝑁∂𝑥\frac{\partial 𝑀}{\partial 𝑦}=\frac{\partial 𝑁}{\partial 𝑥}
(3.56)

The Euler–Lagrange equation, because a calculus of variations line is still an ODE:

∂𝐿∂𝑞−dd𝑡(∂𝐿∂𝑞˙)=0\frac{\partial 𝐿}{\partial 𝑞}-\frac{\mathrm{d}}{\mathrm{d}𝑡}\left(\frac{\partial 𝐿}{\partial \dot{𝑞}}\right)=0
(3.57)

Hamilton’s equations as a pair:

𝑞˙=∂𝐻∂𝑝𝑝˙=−∂𝐻∂𝑞\begin{aligned}\dot{𝑞} & =\frac{\partial 𝐻}{\partial 𝑝} \\ \dot{𝑝} & =-\frac{\partial 𝐻}{\partial 𝑞}\end{aligned}
(3.58)

Schrodinger, so a second spatial derivative and a Ψ\Psi have to coexist with Planck’s constant:

𝑖ℏ∂Ψ∂𝑡=−ℏ22𝑚∇2Ψ+𝑉Ψ𝑖\hbar \frac{\partial \Psi }{\partial 𝑡}=-\frac{{\hbar }^{2}}{2𝑚}{\nabla }^{2}\Psi +𝑉\Psi
(3.59)

Partials for a function of two variables. Heat, wave, Laplace — the three I will actually meet.

∂𝑢∂𝑡=𝑘∂2𝑢∂𝑥2\frac{\partial 𝑢}{\partial 𝑡}=𝑘\frac{{\partial }^{2}𝑢}{\partial {𝑥}^{2}}
(3.60)
∂2𝑢∂𝑡2=𝑐2∂2𝑢∂𝑥2\frac{{\partial }^{2}𝑢}{\partial {𝑡}^{2}}={𝑐}^{2}\frac{{\partial }^{2}𝑢}{\partial {𝑥}^{2}}
(3.61)
∂2𝑢∂𝑥2+∂2𝑢∂𝑦2=0\frac{{\partial }^{2}𝑢}{\partial {𝑥}^{2}}+\frac{{\partial }^{2}𝑢}{\partial {𝑦}^{2}}=0
(3.62)

Laplace in polar, because a 1𝑟\frac{1}{𝑟} in front of a derivative is not optional:

∂2𝑢∂𝑟2+(1𝑟)∂𝑢∂𝑟+(1𝑟2)∂2𝑢∂θ2=0\frac{{\partial }^{2}𝑢}{\partial {𝑟}^{2}}+\left(\frac{1}{𝑟}\right)\frac{\partial 𝑢}{\partial 𝑟}+\left(\frac{1}{{𝑟}^{2}}\right)\frac{{\partial }^{2}𝑢}{\partial {\theta }^{2}}=0
(3.63)

The wave equation in two space dimensions:

∂2𝑢∂𝑡2=𝑐2(∂2𝑢∂𝑥2+∂2𝑢∂𝑦2)\frac{{\partial }^{2}𝑢}{\partial {𝑡}^{2}}={𝑐}^{2}\left(\frac{{\partial }^{2}𝑢}{\partial {𝑥}^{2}}+\frac{{\partial }^{2}𝑢}{\partial {𝑦}^{2}}\right)
(3.64)

Transport, then Burgers, then a viscous regularisation. Three one-line PDEs, not one.

∂𝑢∂𝑡+𝑐∂𝑢∂𝑥=0\frac{\partial 𝑢}{\partial 𝑡}+𝑐\frac{\partial 𝑢}{\partial 𝑥}=0
(3.65)
∂𝑢∂𝑡+𝑢∂𝑢∂𝑥=0\frac{\partial 𝑢}{\partial 𝑡}+𝑢\frac{\partial 𝑢}{\partial 𝑥}=0
(3.66)
∂𝑢∂𝑡+𝑢∂𝑢∂𝑥=ν∂2𝑢∂𝑥2\frac{\partial 𝑢}{\partial 𝑡}+𝑢\frac{\partial 𝑢}{\partial 𝑥}=\nu \frac{{\partial }^{2}𝑢}{\partial {𝑥}^{2}}
(3.67)

A first-order PDE with a directional derivative:

𝑎(𝑥,𝑦)∂𝑢∂𝑥+𝑏(𝑥,𝑦)∂𝑢∂𝑦=𝑐(𝑥,𝑦,𝑢)𝑎(𝑥,𝑦)\frac{\partial 𝑢}{\partial 𝑥}+𝑏(𝑥,𝑦)\frac{\partial 𝑢}{\partial 𝑦}=𝑐(𝑥,𝑦,𝑢)
(3.68)

Clairaut: mixed partials commute when they exist and are continuous. I need both orders written, not a slogan.

∂2𝑢∂𝑥∂𝑦=∂2𝑢∂𝑦∂𝑥\frac{{\partial }^{2}𝑢}{\partial 𝑥\partial 𝑦}=\frac{{\partial }^{2}𝑢}{\partial 𝑦\partial 𝑥}
(3.69)

Gradient, divergence, curl, Laplacian. If nabla becomes a letter, the rest of vector calculus is noise.

∇𝑓=(∂𝑓∂𝑥∂𝑓∂𝑦∂𝑓∂𝑧)\nabla 𝑓=\begin{pmatrix}\frac{\partial 𝑓}{\partial 𝑥} \\ \frac{\partial 𝑓}{\partial 𝑦} \\ \frac{\partial 𝑓}{\partial 𝑧}\end{pmatrix}
(3.70)
∇⋅𝐹=∂𝐹1∂𝑥+∂𝐹2∂𝑦+∂𝐹3∂𝑧\nabla \cdot 𝐹=\frac{\partial {𝐹}_{1}}{\partial 𝑥}+\frac{\partial {𝐹}_{2}}{\partial 𝑦}+\frac{\partial {𝐹}_{3}}{\partial 𝑧}
(3.71)
∇×𝐹=(∂𝐹3∂𝑦−∂𝐹2∂𝑧∂𝐹1∂𝑧−∂𝐹3∂𝑥∂𝐹2∂𝑥−∂𝐹1∂𝑦)\nabla \times 𝐹=\begin{pmatrix}\frac{\partial {𝐹}_{3}}{\partial 𝑦}-\frac{\partial {𝐹}_{2}}{\partial 𝑧} \\ \frac{\partial {𝐹}_{1}}{\partial 𝑧}-\frac{\partial {𝐹}_{3}}{\partial 𝑥} \\ \frac{\partial {𝐹}_{2}}{\partial 𝑥}-\frac{\partial {𝐹}_{1}}{\partial 𝑦}\end{pmatrix}
(3.72)
Δ𝑢=∇2𝑢=∂2𝑢∂𝑥2+∂2𝑢∂𝑦2+∂2𝑢∂𝑧2\Delta 𝑢={\nabla }^{2}𝑢=\frac{{\partial }^{2}𝑢}{\partial {𝑥}^{2}}+\frac{{\partial }^{2}𝑢}{\partial {𝑦}^{2}}+\frac{{\partial }^{2}𝑢}{\partial {𝑧}^{2}}
(3.73)

Integrals. The differential d𝑥\mathrm{d}𝑥 is part of the notation, not a factor I invented.

∫𝑎𝑏𝑓(𝑥) d𝑥{\int }_{𝑎}^{𝑏}𝑓(𝑥)\, \mathrm{d}𝑥
(3.74)
∬𝐷𝑓(𝑥,𝑦) d𝑥 d𝑦{\iint }_{𝐷}𝑓(𝑥,𝑦)\, \mathrm{d}𝑥\, \mathrm{d}𝑦
(3.75)
∭𝑉𝑓(𝑥,𝑦,𝑧) d𝑥 d𝑦 d𝑧{\iiint }_{𝑉}𝑓(𝑥,𝑦,𝑧)\, \mathrm{d}𝑥\, \mathrm{d}𝑦\, \mathrm{d}𝑧
(3.76)

A contour integral, because a closed path is not a dummy index:

∮𝐶𝑓(𝑧) d𝑧=2π𝑖∑𝑘Res(𝑓,𝑧𝑘){\oint }_{𝐶}𝑓(𝑧)\, \mathrm{d}𝑧=2\pi 𝑖\sum _{𝑘}\text{Res}(𝑓,{𝑧}_{𝑘})
(3.77)

Line and surface integrals, the statements I will actually quote:

∮∂Ω𝑃 d𝑥+𝑄 d𝑦=∬Ω(∂𝑄∂𝑥−∂𝑃∂𝑦) d𝐴{\oint }_{\partial \Omega }𝑃\, \mathrm{d}𝑥+𝑄\, \mathrm{d}𝑦={\iint }_{\Omega }\left(\frac{\partial 𝑄}{\partial 𝑥}-\frac{\partial 𝑃}{\partial 𝑦}\right)\, \mathrm{d}𝐴
(3.78)
∬∂Ω𝐹⋅𝑛 d𝑆=∭Ω(∇⋅𝐹) d𝑉{\iint }_{\partial \Omega }𝐹\cdot 𝑛\, \mathrm{d}𝑆={\iiint }_{\Omega }(\nabla \cdot 𝐹)\, \mathrm{d}𝑉
(3.79)
∮∂Σ𝐹⋅ d𝑟=∬Σ(∇×𝐹)⋅𝑛 d𝑆{\oint }_{\partial \Sigma }𝐹\cdot \, \mathrm{d}𝑟={\iint }_{\Sigma }(\nabla \times 𝐹)\cdot 𝑛\, \mathrm{d}𝑆
(3.80)
∑𝑛=1∞1𝑛2=π26\sum _{𝑛=1}^{\infty }\frac{1}{{𝑛}^{2}}=\frac{{\pi }^{2}}{6}
(3.81)
∏𝑘=1𝑛(1+𝑥𝑘)\prod _{𝑘=1}^{𝑛}(1+{𝑥}_{𝑘})
(3.82)

An infinite product I actually meet:

sin⁡(π𝑥)π𝑥=∏𝑛=1∞(1−𝑥2𝑛2)\frac{\sin (\pi 𝑥)}{\pi 𝑥}=\prod _{𝑛=1}^{\infty }\left(1-\frac{{𝑥}^{2}}{{𝑛}^{2}}\right)
(3.83)

Nested radicals and a cube root, because Chrome already lied to me once about 𝑥\sqrt{𝑥}.

𝑥3+𝑥2+1\sqrt[3]{𝑥}+\sqrt{\sqrt{{𝑥}^{2}+1}}
(3.84)

A fourth root over a nested square root, not the same picture twice:

𝑎+𝑏4𝑎+𝑏3\frac{\sqrt[4]{𝑎+\sqrt{𝑏}}}{\sqrt{𝑎+\sqrt[3]{𝑏}}}
(3.85)

Hats and tildes for Fourier, a bold vector, floor and ceiling, a norm:

𝑓^(ξ)=∫−∞∞𝑓(𝑥)𝑒−2π𝑖𝑥ξ d𝑥\hat{𝑓}(\xi )={\int }_{-\infty }^{\infty }𝑓(𝑥){𝑒}^{-2\pi 𝑖𝑥\xi }\, \mathrm{d}𝑥
(3.86)
𝑦~=𝑦−𝑦mean\tilde{𝑦}=𝑦-{𝑦}_{\text{mean}}
(3.87)
⌊𝑥⌋≤𝑥<⌈𝑥⌉\lfloor 𝑥\rfloor \leq 𝑥<\lceil 𝑥\rceil
(3.88)
∥𝐴𝑣∥≤∥𝐴∥ ∥𝑣∥\|𝐴𝑣\|\leq \|𝐴\|\, \|𝑣\|
(3.89)

Cauchy–Schwarz, Holder, triangle. Three inequalities, three shapes.

∣⟨𝑢,𝑣⟩∣≤∥𝑢∥ ∥𝑣∥|\langle 𝑢,𝑣\rangle |\leq \|𝑢\|\, \|𝑣\|
(3.90)
∥𝑓𝑔∥1≤∥𝑓∥𝑝 ∥𝑔∥𝑞\|𝑓𝑔{\|}_{1}\leq \|𝑓{\|}_{𝑝}\, \|𝑔{\|}_{𝑞}
(3.91)
∥𝑥+𝑦∥≤∥𝑥∥+∥𝑦∥\|𝑥+𝑦\|\leq \|𝑥\|+\|𝑦\|
(3.92)

Evaluation bar, continued fraction, underbrace as a reminder of what a clump is:

𝐹(𝑥)∣𝑥=01=𝐹(1)−𝐹(0)𝐹(𝑥){|}_{𝑥=0}^{1}=𝐹(1)-𝐹(0)
(3.93)
1+12+13+14+151+\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{5}}}}
(3.94)
𝑥+𝑦+𝑧⏟three terms\underbrace{𝑥+𝑦+𝑧}_{\text{three terms}}
(3.95)

A Taylor polynomial with a Lagrange remainder, which mixes sum, factorial, and an evaluation I have not justified yet:

𝑓(𝑥)=∑𝑘=0𝑛𝑓(𝑘)(𝑎)𝑘!(𝑥−𝑎)𝑘+𝑓(𝑛+1)(𝑐)(𝑛+1)!(𝑥−𝑎)𝑛+1𝑓(𝑥)=\sum _{𝑘=0}^{𝑛}\frac{{𝑓}^{(𝑘)}(𝑎)}{𝑘!}{(𝑥-𝑎)}^{𝑘}+\frac{{𝑓}^{(𝑛+1)}(𝑐)}{(𝑛+1)!}{(𝑥-𝑎)}^{𝑛+1}
(3.96)

Taylor in two variables, because a double sum is not a single sum with a fake index:

𝑓(𝑥,𝑦)=∑𝑖=0𝑛∑𝑗=0𝑛−𝑖(1𝑖!𝑗!)∂𝑖+𝑗𝑓∂𝑥𝑖∂𝑦𝑗(𝑎,𝑏)(𝑥−𝑎)𝑖(𝑦−𝑏)𝑗+𝑅𝑛𝑓(𝑥,𝑦)=\sum _{𝑖=0}^{𝑛}\sum _{𝑗=0}^{𝑛-𝑖}\left(\frac{1}{𝑖!𝑗!}\right)\frac{{\partial }^{𝑖+𝑗}𝑓}{\partial {𝑥}^{𝑖}\partial {𝑦}^{𝑗}}(𝑎,𝑏){(𝑥-𝑎)}^{𝑖}{(𝑦-𝑏)}^{𝑗}+{𝑅}_{𝑛}
(3.97)