Single Variable

Derivative

A function of a real variable y=f(x)y=f(x) is differentiable at a point aa of its domain, if its domain contains an open interval II containing aa, the limit:

L=lim⁡h→0f(a+h)−f(a)hL=\lim_{h\rightarrow0}\frac{f(a+h)-f(a)}{h}

L’Hôpital’s Rule

Used to find the limit value of an expression

The limits of the numerator (分子) and denominator (分母) are both 0 or infinite.

Taylor’s Formular

f(x)=∑i=0∞fi(x0)i!(x−x0)if(x)=\sum_{i=0}^\infty\frac{f^{i}(x_0)}{i!}(x-x_0)^i

ex=1+x+x22!+x33!+o(x3)e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+o(x^3)

Indefinite Integral

∫f(x)dx=F(x)+C\int f(x)dx=F(x)+C

Definite Integral

The definition of definite integral is based on finding the area of a curved-sided trapezoid (曲边梯形). Therefore, the definition of definite integral is used to find the area, i.e. to get a number.

Newton Leibniz Formula

Given ∫f(x)dx=Φ(x)+C\int f(x)dx=\Phi(x)+C

∫abf(x)dx=Φ(b)−Φ(a)\int_a^bf(x)dx=\Phi(b)-\Phi(a)

Multivariable

Double Integral

How to calculate?

  1. find the domain of each variable
  2. step by step, each step for a variable

z.B. σ\sigma: y=cx, x=a, x=b, Ox. c>0, b>a>0, ∬σ(x+y)dσ\iint_\sigma(x+y)d\sigma

evidently, a≤x≤ba\leq x\leq b, 0≤y≤cx0\leq y\leq cx

∬σ(x+y)dσ\iint_\sigma(x+y)d\sigma

=∫abdx∫0cx(x+y)dy=\int_a^bdx\int_0^{cx}(x+y)dy

=∫ab[xy+y22]0cxdx=\int_a^b[xy+\frac{y^2}{2}]_0^{cx}dx

=∫ab(2c+c22)x2dx=\int_a^b(\frac{2c+c^2}{2})x^2dx

=16(2c+c2)(b3−a3)=\frac{1}{6}(2c+c^2)(b^3-a^3)

If introduce new prarms u,v s.t. x=x(u,v)x=x(u,v), y=y(u,v)y=y(u,v)

then ∬Df(x,y)dxdy=∬D′f[x(u,v),y(u,v)]∣∂(x,y)∂(u,v)∣dudv\iint_D f(x,y)dxdy=\iint_{D'}f[x(u,v),y(u,v)]|\frac{\partial(x,y)}{\partial(u,v)}|dudv

Jacobian J=∣∂(x,y)∂(u,v)∣=xu′yv′−xv′yu′J=|\frac{\partial(x,y)}{\partial(u,v)}|=x'_uy'_v-x'_vy'_u

Line Integral

∫Cf(x,y)ds=∫abf(x(t),y(t))x′(t)2+y′(t)2dt\int_Cf(x,y)ds=\int_a^bf(x(t),y(t))\sqrt{x'(t)^2+y'(t)^2}dt

z.B. LL is a part of unit circle in the first quadrant (第一象限), ∫Lxyds\int_Lxyds

Parametric equation of LL is: x=cos⁡tx=\cos t, y=sin⁡ty=\sin t, 0≤t≤π20\leq t\leq\frac{\pi}{2}

Then ∫Lxyds=∫0π2cos⁡tsin⁡t(−sin⁡t)2+cos⁡2tdt\int_Lxyds=\int_0^{\frac{\pi}{2}}\cos t\sin t\sqrt{(-\sin t)^2+\cos^2t}dt

=∫0π2cos⁡tsin⁡tdt=12=\int_0^{\frac{\pi}{2}}\cos t\sin tdt=\frac{1}{2}

or

y=1−x2y=\sqrt{1-x^2}, 0≤x≤10\leq x\leq 1

Then ∫Lxyds=∫01x1−x2×1+x21−x2dx\int_Lxyds=\int_0^1x\sqrt{1-x^2}\times\sqrt{1+\frac{x^2}{1-x^2}}dx

=∫01xdx=12=\int_0^1xdx=\frac{1}{2}

Numerical Series

Convergence 收敛

Divergence 发散

p- series ∑n1np\sum_n\frac{1}{n^p}: when p>1, convergent; when p<=1, divergent.

z.B. ∑n=1∞1n(n+1)(n+2)\sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)}

an=1n(n+1)(n+2)=n+2−n2n(n+1)(n+2)a_n=\frac{1}{n(n+1)(n+2)}=\frac{n+2-n}{2n(n+1)(n+2)}

=12n(n+1)−12(n+1)(n+2)=bn−bn+1=\frac{1}{2n(n+1)}-\frac{1}{2(n+1)(n+2)}=b_n-b_{n+1}

lim⁡n→∞bn=lim⁡n→∞12n(n+1)=0=b\lim_{n\rightarrow\infty}b_n=\lim_{n\rightarrow\infty}\frac{1}{2n(n+1)}=0=b

∴∑n=1∞an=b1−b=14\therefore \sum_{n=1}^\infty a_n=b_1-b=\frac{1}{4}