Single Variable
Derivative
A function of a real variable \(y=f(x)\) is differentiable at a point \(a\) of its domain, if its domain contains an open interval \(I\) containing \(a\), the limit:
\(L=\lim_{h\rightarrow0}\frac{f(a+h)-f(a)}{h}\)
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
\(f(x)=\sum_{i=0}^\infty\frac{f^{i}(x_0)}{i!}(x-x_0)^i\)
\(e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+o(x^3)\)
Indefinite Integral
\(\int f(x)dx=F(x)+C\)
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 \(\int f(x)dx=\Phi(x)+C\)
\(\int_a^bf(x)dx=\Phi(b)-\Phi(a)\)
Multivariable
Double Integral
How to calculate?
- find the domain of each variable
- step by step, each step for a variable
z.B. \(\sigma\): y=cx, x=a, x=b, Ox. c>0, b>a>0, \(\iint_\sigma(x+y)d\sigma\)
evidently, \(a\leq x\leq b\), \(0\leq y\leq cx\)
\(\iint_\sigma(x+y)d\sigma\)
\(=\int_a^bdx\int_0^{cx}(x+y)dy\)
\(=\int_a^b[xy+\frac{y^2}{2}]_0^{cx}dx\)
\(=\int_a^b(\frac{2c+c^2}{2})x^2dx\)
\(=\frac{1}{6}(2c+c^2)(b^3-a^3)\)
If introduce new prarms u,v s.t. \(x=x(u,v)\), \(y=y(u,v)\)
then \(\iint_D f(x,y)dxdy=\iint_{D'}f[x(u,v),y(u,v)]|\frac{\partial(x,y)}{\partial(u,v)}|dudv\)
Jacobian \(J=|\frac{\partial(x,y)}{\partial(u,v)}|=x'_uy'_v-x'_vy'_u\)
Line Integral
\(\int_Cf(x,y)ds=\int_a^bf(x(t),y(t))\sqrt{x'(t)^2+y'(t)^2}dt\)
z.B. \(L\) is a part of unit circle in the first quadrant (第一象限), \(\int_Lxyds\)
Parametric equation of \(L\) is: \(x=\cos t\), \(y=\sin t\), \(0\leq t\leq\frac{\pi}{2}\)
Then \(\int_Lxyds=\int_0^{\frac{\pi}{2}}\cos t\sin t\sqrt{(-\sin t)^2+\cos^2t}dt\)
\(=\int_0^{\frac{\pi}{2}}\cos t\sin tdt=\frac{1}{2}\)
or
\(y=\sqrt{1-x^2}\), \(0\leq x\leq 1\)
Then \(\int_Lxyds=\int_0^1x\sqrt{1-x^2}\times\sqrt{1+\frac{x^2}{1-x^2}}dx\)
\(=\int_0^1xdx=\frac{1}{2}\)
Numerical Series
Convergence 收敛
Divergence 发散
p- series \(\sum_n\frac{1}{n^p}\): when p>1, convergent; when p<=1, divergent.
z.B. \(\sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)}\)
\(a_n=\frac{1}{n(n+1)(n+2)}=\frac{n+2-n}{2n(n+1)(n+2)}\)
\(=\frac{1}{2n(n+1)}-\frac{1}{2(n+1)(n+2)}=b_n-b_{n+1}\)
\(\lim_{n\rightarrow\infty}b_n=\lim_{n\rightarrow\infty}\frac{1}{2n(n+1)}=0=b\)
\(\therefore \sum_{n=1}^\infty a_n=b_1-b=\frac{1}{4}\)