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5 Numrical Analysis

Errors

Assume that r is the root of function f(x)=0f(x)=0.

  • Forward Error: rxa|r-x_a|
  • Backward Error: f(xa)f(x_a)
  • error magnification factor=relative forward errorrelative backward error\text{error magnification factor} = \frac{\text{relative forward error}}{\text{relative backward error}}

Basic Arithmetic

Evaluating a polynomial

Direct form:O(n2n^2)

P(x)=a0+a1x+a2x2+...+anxnP(x)=a_0+a_1x+a_2x^2+...+a_nx^n

Horner's method:O(n)

P(x)=a0+x(a1+x(a2+x(...(an1+anx)))P(x)=a_0+x(a_1+x(a_2+x*(...(a_{n-1}+a_nx)))

Example:求二进制幂O(n)->O(logn\log n)

an=ap(2)a^n=a^{p(2)}=abm2m+...+bi2i+b0=a^{b_m2^m+...+b_i2^i+b_0}

a2p+bi=a2paa^{2p+b_i}=a^{2p}*a or a2pa^{2p}

Multiplication

Calculate x=yzx=yz

we could devide y and z into 2 parts of n2\frac n2 bits

yab
zcd

Then follow the 4 equations, we only need 2 n2\frac n2 bits muliplications to calculate v and w

{u=(a+b)(c+d)v=acw=bdx=v2n+(uvw)2n2+w\begin{cases} u = (a+b)(c+d) \\ v = ac \\ w = bd \\ x= v2^n + (u - v - w)2^{\frac n2} + w \end{cases}

For u, we could proof that we just need 1 n2\frac n2 bits muliplication + some addition and shift

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T(n)=3T(n/2)+lnT(n)=O(nlog23)T(n) = 3T(n/2) + ln \Rightarrow T(n) = O(n^{\log_2^3})

Solving Equations

Newton's Method

x0=initial guessx_0=initial\ guess

xi+1=xif(xi)f(xi)x_{i+1}=x_i-\frac{f(x_i)}{f'(x_i)}

Convex Optimization

Convex combination: (x,y){αx+βyα+β=1,α,β0}(x,y)\rightarrow\{\alpha x + \beta y|\alpha + \beta =1, \alpha,\beta \geq 0\}

Convex set S: (x1,...,xk)S(x1, . . . , xk) \in S, then convex combination of the points x1, . . . , xk should \in S.

A function f : RnRR^n \rightarrow R is convex if x,y\forall x, y and α[0,1]\forall \alpha \in [0, 1]

f(αx+(1α)y)αf(x)+(1α)f(y)f(\alpha x + (1-\alpha) y) \leq \alpha f(x) + (1-\alpha) f(y)

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Convex Hull

Given n points in plane, S={(xi,yi)i=1,2,...,n}S = \{(x_i, y_i)| i = 1,2, ... , n\} , Convex Hull(CH(S)): smallest polygon containing all points in S.

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Finding upper tangent, same for lower tangent

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