MATLAB 数值微积分

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学习笔记：郭彥甫 (Yan-Fu Kuo), 台大生機系 , MATLAB教學 - 10數值微積分

Representing Polynomials in MATLAB

• Polynomials were represented as row vectors
• for example,consider the equation $f(x)=x^3-2x-5$
• To enter this polynomial into MATLAB, usep = [1 0 -2 -5]

Values of Polynomials:polyval()

• Plot the polynomial $f(x)=9x^3-5x^2+3x+7$
• for $-2\leq x \leq5$.

a = [9,-5,3,7];
x=-2:0.01:5;
f = polyval(a,x);
plot(x,f,'LineWidth',2);
xlabel('x');
ylabel('f(x)');
set(gca,'FontSize',14)

Polynomial Differentiation（多项式微分）:polyder()

• Given $f(x)=5x^4-2x^2+1$
  
  1. What is the derivative(导数) of the function$f'(x)$?
  2. What is the derivative of the function value of $f'(7)$?

>> p = [5 0 -2 0 1];
>> polyder(p) 
ans =     20     0    -4     0
>> polyval(polyder(p),7)
ans = 6832

Exercise

• Plot the polynomial$f(x)=(20x^3-7x^2+5x+10)(4x^2+12x-3)$ and its derivative for $-2 \leq x \leq1$.

• Hint:conv()

  p1 = [20 -7 5 10];
  p2 = [4 12 -3];
  a = conv(p1,p2);
  x=-2:0.01:1;
  hold on
  plot(x,f,'b--','linewidth',2);
  plot(x,polyval(polyder(a),x),'r','linewidth',2);
  hold off
  % plot(x,f,'b--',x,polyval(polyder(a),x),'r','linewidth',2);
  xlabel('x')
  ylabel('y')
  legend('f(x)',"f'(x)");

  

Polynomial Intergration(积分):polyint()

• for a polynomial
  

  $$f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$$ the integration is $$\int f(x)=\frac{1}{n+1}a_nx^{n+1}+\frac{1}{n}a_{n-1}x^n+...+a_0x+k$$

• Given $f(x)=5x^4-2x^2+1$

  1. What is the integral of the function $\int {f(x)}$ with a constant of 3?

     >> p = [5 0 -2 0 1];
     >> polyint(p,3)
     ans =    1.0000         0   -0.6667         0    1.0000    3.0000

  2. What is the derivative of the function value of $\int {f(7)}$ ?

     >> polyval(polyint(p,3),7))
     ans = 1.6588e+04

Numerical Differentiation (数值微分)：diff()

• The simplest method: finite difference approximation(有限差分近似)

• Calculating a secant line(割线) in the vicinty(附近) of $x_0$

  $$f'(x_0)=\lim _{h\rightarrow0}\frac{f(x_0+h)-f(x_0)}{h}$$ where $h$ represents a small change in $x$.

• diff() : calculates the differences between adjacent(附近的) elements of a vector

  >> x = [1 2 5 2 1];
  >> diff(x)
  ans =  1     3    -3    -1

• Exercise : obtain the slope(斜率) of a line between 2 points (1,5) and (2,7)

  x = [1 2];y=[5 7]
  slope = diff(y)./diff(x)

• Given $f(x)=sin(x)$, find $f'(x_0)$ at $x_0 = \pi/2$ using $h=0.1$

  $f'(x_0)=\lim_{h\rightarrow0}\frac{f(x_0+h)-f(x_0)}{h}$

  >> x0 = pi/2
  x0 = 1.5708
  >> h=0.1
  h = 0.1000
  >> x = [x0 x0+h]
  x = 1.5708    1.6708
  >> y = [sin(x0) sin(x0+h)]
  y = 1.0000    0.9950
  >> m = diff(y)./diff(x)
  m = -0.0500

Exercise

• Given $f(x)=sin(x)$, write a script to find the error of $f'(x)$ at $x_0=\pi/2$ using various $h$

• We have already known $f'(\pi/2) = 0$

  x0 = pi/2;
  for h=[0.1 0.01 0.001 0.0001 0.00001 0.000001]
    x=[x0,x0+h];
    y=[sin(x0),sin(x0+h)];
    slop=diff(y)./diff(x);
    W=['h=',num2str(h),' error=',num2str(slop)];
    disp(W)
  end

How to Find the $f'$ over An Interval(区间) $[0,2\pi]$?

• In the previous example, $x_0 = \pi / 2$.

• Strategy:

  1. Create an array in the interval$[0,2\pi]$
  2. The step is the $h$.
  3. Calculate the $f'$ at these points

• For example:

  h = 0.5;
  x = 0:h:2*pi;

• Find $sin'(x)$ over $x = [0,2\pi]$.

  h = 0.5;
  x = 0:h:2*pi;
  y = sin(x);
  m = diff(y)./diff(x);  % m = f'(x)

Various Step Size

• The derivatives of $f(x) = sin(x)$ caculated using various $h$ values

  g = colormap(lines);hold on;
  for i=1:4
    x = 0:power(10, -i):pi;
    y = sin(x); m = diff(y)./diff(x);
    plot(x(1:end-1),m,'Color',g(i,:));
  end
  hold off;
  set(gca,'Xlim',[0,pi/2]); set(gca,'YLim',[0,1.2]); 
  set(gca,'FontSize',18); set(gca,'FontName','symbol');
  set(gca,'XTick',0:pi/4:pi/2);
  set(gca,'XTickLabel',{'0','p/4','p/2'});  % xtick是刻度（小竖线）；xticklabel 刻度值（竖线下面的数值）
  h = legend('h=0.1','h=0.01','h=0.001','h=0.0001');
  set(h,'FontName','Times New Roman'); box on; 

  

Exercise

• Given $f(x) = e^{-x}sin(x^2/2)$ ,plot the approximate derivatives $f'$ of $h=0.1,0.01$, and $0.001$

  g = colormap(lines);hold on;
  for i=1:3
    x = 0:power(10,-i):2*pi;
    y = exp(-x).*sin(x.^2./2);
    m = diff(y)./diff(x);
    plot(x(1:end-1),m,'Color',g(i,:));
  end
  hold off;
  set(gca,'Xlim',[0,pi*2]); set(gca,'YLim',[-0.3,0.3]);
  set(gca,'FontSize',18); set(gca,'FontName','symbol');
  set(gca,'XTick',0:pi/2:pi*2);
  set(gca,'XTickLabel',{'0','pi/2','pi','3/2pi','2pi'});
  set(gca,'YTick',-0.2:0.1:0.2);
  set(gca,'YTickLabel',{'-0.2','-0.1','0','0.1','0.2'});
  h = legend('h=0.1','h=0.01','h=0.001');
  set(h,'FontName','Times New Roman'); box on;

  

Second and Third Derivatives(二阶、三阶导)

• The second derivative $f''$ and third derivative $f'''$ can be obtained using similar approaches

• Given $f(x) = x^3$ , plot $f'$ and $f''$ for$-2\leq x \leq2$.

  x = -2:0.005:2;
  y = x.^3;
  m = diff(y)./diff(x);
  m2 = diff(m)./diff(x(1:end-1));
  plot(x,y,x(1:end-1),m,x(1:end-2),m2);
  xlabel('x','FontSize',18);
  ylabel('y','FontSize',18);
  legend('f(x) = x^3',"f''(x)","f''''(x)");
  set(gca,'FontSize',18);

Numerical Integration(数值积分)

• Calculating the numerical value of a definite integral

  $s = \int_a^b f(x)d(x)\approx \sum_{i=0}^nf(x_i)\int_a^bL_i(x)dx$

• Quadrature method — approximating the integral by using a finite set of points(正交方法 - 使用有限的点集近似积分)

Numerical Quadrature Rules(数值求积分准则)

• Basic quadrature rules:

  1. Midpoint rule(zeroth-order approximation)(中点准则（零阶近似），使用矩形预测面积大小)

     

  2. Trapezoid rule(first-order approximation)（梯形准则（一阶近似），使用梯形预测面积大小）

     

Midpoint Rule Using sum()

• Example:

  $A = \int_0^24x^3dx = x^4|^2_0 = (2)^4-(0)^4 = 16$

  >> h = 0.05;
  >> x = 0:h:2;  % x is a vector
  >> midpoint = (x(1:end-1)+x(2:end))./2;  % midpoint is a vector
  >> y = 4*midpoint.^3;  % y is a vector
  >> s = sum(h*y)
  s =15.9950

Trapezoid Rule Using trapz()

• Example:

  $A = \int_0^24x^3dx = x^4|^2_0 = (2)^4-(0)^4 = 16$

  >> h = 0.05;
  >> x = 0:h:2;
  >> y = 4*x.^3;
  >> s = h*trapz(y)
  16.0100
  
  % the forth step above equals
  trapezoid = (y(1:end-1)+y(2:end))/2;
  s = h*sum(trapezoid)

Second-order Rule:1/3 Simpson’s

• Example:

  $A = \int_0^24x^3dx = x^4|^2_0 = (2)^4-(0)^4 = 16$

>> h = 0.05;
>> x = 0:h:2;
>> y = 4*x.^3;
>> s = h/3*(y(1)+2*sum(y(3:2:end-2))+4*sum(y(2:2:end))+y(end))
s = 16

Comparison

• Midpoint rule — zeroth-order
• Trapezoid rule — first-order
• Simpson’s rule — second-order
• higher order of approximation results to a higher accuracy (逼近的次数越高越精准）

Review of Function Handles(@)

• A handle is a pointer to a function

• Can be used to pass function to other functions

• For example, the input of the following function is another function:

  function [y] = xy_plot(input,x)
    % xy_plot receives the handle of a function and plots that
    % function of x
    y = input(x);
    plot(x,y,'r--');
    xlabel('x');
    ylabel('function(x)');
  end

• When we need to use the function xy_plot ,use xy_plot(@sin,0:0.01:2*pi);

Numerical Intergration:integral()

• Numerical integration on a function from using global adaptive quadrature and default error tolerances.

• Example:

  $\int_0^2\frac{1}{x^3-2x-5}$

  >> y = @(x) 1./(x.^3-2*x-5);
  >> integral(y,0.2)
  ans = -0.4605

  $\int_0^2sin(x)$

  y = @(x) sin(x);
  integral(y,0,2)

Double and Triple Integrals(双重和三重积分)

• Example:

  $f(x,y) = \int_0^\pi\int_\pi^{2\pi} ( y\cdot {sin(x)} + x \cdot {cos(y)} )dx dy$

  f = @(x,y) y.*sin(x)+x.*cos(y);
  integral2(f,pi,2*pi,0,pi)

• Example:

  $f(x,y) = \int_{-1}^1\int_0^1\int{_0 ^\pi} (y\cdot {sin(x)}+z \cdot {cos(y)} )dx dy dz$

  f = @(x,y,z) y.*sin(x)+z.*cos(y)
  integral3(f,0,pi,0,1,-1,1)