Taylor expansion
- Taylor expansion: f(x)≈f(x0)+1!f′(x0)(x−x0)+2!f′′(x0)(x−x0)2+...
Single-variable calculus
Derivatives:
dxdxn=nxn−1
dxdax=axln(a)
dxdln(x)=1/x
dxdtan(x)=sec2(x)
dxdcot(x)=−csc2(x)
dxdsec(x)=sec(x)tan(x)
dxdcsc(x)=−csc(x)cot(x)
∫tan=ln∥sec∥
∫cot=ln∥sin∥
∫sec=ln∥sec+tan∥
∫csc=ln∥csc−cot∥
∫a2−u2du=sin−1(au)
∫uu2−a2du=a1sec−1(au)
∫a2+u2du=a1tan−1(au)
Continuous: left limit = right limit = value
Differentiable: continuous and no sharp points / asymptotes
L'Hospital's - for indeterminate forms: (g(x)f(x))′=g′(x)f′(x)
Integration by parts: ∫udv=uv−∫duv, LIATE
Expansions:
ex=∑n!xn
sin(x)=∑0∞(2n+1)!(−1)nx2n+1
cos(x)=∑0∞(2n)!(−1)nx2n
Geometric Sum: a1st1−r1−rn+1
Multivariable calculus
- Polar: r,θ,z
- Spherical: ρ,θ,ϕ
- Clairut's Thm: Conservative function fxy=fyx
- Lagrangian - solves minimize f subject to g = c
- solution will always be tangent to f
- ∇f=λ∇g - gives us n constraints
- remember g = c is a constraint too
- to do this efficiently, define the Lagrangian L(x,λ)=f−λ⋅g
- taking deriv wrt λ and setting = 0 enforces g = c
- taking deriv wrt other variables and setting = 0 enforces other conditions
- therefore final eq just becomes ∇L=0