1、的近似值与精确值的绝对误差wuq,wuh和的精确值yx如下 4.46550187104484wuq =2.99953806706506 0.23701636931441 0.02299427519787 0.00229231810861wuh =1.50335165262053 0.22121835003744 0.02283621026072 0.002290737434248.2.2 中心差商公式求导及其MATLAB程序利用精度为的三点公式计算的近似值和误差估计的MATLAB主程序function n,xi,yx,wuc=sandian(h,xi,fi,M)n=length(fi); yx
2、=zeros(1,n); wuc=zeros(1,n); x1= xi(1); x2= xi(2); x3= xi(3);y1=fi(1); y2=fi(2); y3=fi(3); xn= xi(n); xn1= xi(n-1); xn2= xi(n-2); yn=fi(n); yn1=fi(n-1); yn2=fi(n-2);for k=2:n-1yx(1)=(-3*y1+4*y2-y3)/(2*h); yx(n)=(yn2-4*yn1+3*yn)/(2*h);yx(2)=( fi(3)- fi(1)/(2*h); yx(k)=( fi(k+1)- fi(k-1)./(2*h);wuc(1)
3、=abs(h.2.*M./3); wuc(n)=abs(h.2.*M./3);wuc(2:n-1)=abs(-h.2.*M./6);endfunction x,yxj, wuc=sandian3(h,xi,fi,M)yxj=zeros(1,3); wuc=zeros(1,3);x2= xi(2); y1=fi(1);for t=1:3s(t)=(2*t-5)*y1-4*(t-2)*y2+(2*t-3)*y3)/(2*h); x=xi; y=s(t); yxj(t)=y;if t=2wuc(2)=abs(-h.2*M/6);elsewuc(1:2:3)=abs(h.2*M/3);例 8.2.3
4、设已给出的数据表85:表85x1.000 0 1.100 0 1.200 0 1.300 0 1.400 0 1.500 0 1.600 0 f(x)0.250 0 0.226 8 0.206 6 0.189 0 0.173 6 0.160 0 0.147 9 M= 0.750 2,试用三点公式计算下列问题:(1)当h=0.1时,在x=1.000 0,1.100 0,1.200 0,1.300 0,1.400 0,1.500 0,1.600 0处的一阶导数的近似值,并估计误差;(2)当h=0.2时,在x=1.000 0,1.200 0, 1.400 0,1.600 0处的一阶导数的近似值,并估
5、计误差;(3)当h=0.3时,在x=1.000 0,1.300 0 ,1.600 0处的一阶导数的近似值,并估计误差;(4) 表85中的数据是函数在相应点的数值,将(1)至(3)计算的一阶导数的近似值与的一阶导数值比较,并求出它们的绝对误差.解 (1)保存M文件sandian.m,sandian3.m;(2)在MATLAB工作窗口输入如下程序 syms x,y=1/(1+x)2); yx=diff(y,x,1),yx3=diff(y,x,3),运行后将屏幕显示的结果为yx = yx3 =-2/(1+x)3 -24/(1+x)5(3)在MATLAB工作窗口输入如下程序h=0.1; xi=1.00
6、00:h:1.6000;fi=0.2500 0.2268 0.2066 0.1890 0.1736 0.1600 0.1479;x=1:0.001:1.6; yx3 =-24./(1+x).5; M= max(abs(yx3);n1,x1,yx1,wuc1=sandian(h,xi,fi,M)yxj1=-2./(1+xi).3,wuyxj1=abs(yxj1- yx1)h=0.2; fi=0.2500 0.2066 0.1736 0.1479;n2,x2,yx2,wuc2=sandian(h,xi,fi,M)yxj2=-2./(1+xi).3,wuyxj2=abs(yxj2- yx2)h=0.
7、3; fi=0.2500 0.1890 0.1479; M=max(abs(yx3);x3,yx3, wuc3=sandian3(h,xi,fi,M)yxj3=-2./(1+xi).3,wuyxj3=abs(yxj3- yx3)或 h1=0.1,x=1.0000,1.1000,1.2000,1.3000,1.4000,1.5000,1.6000;f=0.2500,0.2268,0.2066,0.1890,0.1736,0.1600,0.1479;xi=x(1:3);f11=f(1: M= 0.7502;x11,yxj11,wuc11=sandian3(h1,xi,f11,M), xi= x(4
8、:6);f12=f(4:x12,yxj12,wuc12=sandian3(h1,xi,f12,M), xi=x(5:7);f13=f(5: x13,yxj13,wuc13=sandian3(h1,xi,f13,M), h2=0.2, xi= x(1:5);f21= f(1:x21,yxj21,wuc21=sandian3(h2,xi,f21,M),xi= x(2:f22=f(2:x22,yxj22,wuc22=sandian3(h2,xi,f22,M),xi= x(3:f23=f(3:x23,yxj23,wuc23=sandian3(h2,xi,f23,M), h3=0.3, xi= x(1:
9、3:f31= f(1: x31,yxj31,wuc31=sandian3(h3,xi,f31,M),将运行的结果(略).8.2.3 理查森外推法求导及其MATLAB程序(一)一般形式的理查森外推法及其MATLAB程序利用理查森外推法计算的近似值和误差估计的MATLAB程序function Dy,dy,jdw,n=diffext1(fun,x0,jdwc,max1)h=1;j=1; n=1;jdW=1;xdW=1; x1=x0+h;x2=x0-h;Dy(1,1)=(feval(fun,x1)- feval(fun,x2)/(2*h);while(jdWjdwc)&(jwu)&(kmax1) )h
10、=h/(10(k) ; H(k+1)=h; x1=x0+H(k+1); x2=x0-H(k+1);Dy(k+1)=(feval(fun,x1)- feval(fun,x2)./(2* H(k+1);W(k+1)=abs(Dy(k+1)- Dy(k);k=k+1;n=length(Dy(1:k)-2;Dy= Dy(2:k); W=W(2: H=H(2:例 8.2.7 设.取初始步长h0=0.2,精度wu=0.00001, 用变步长的中心差商公式计算的近似值和误差估计,与精确值4.46550187104484比较.解 输入程序wu=0.00001;max1=100;h0=0.2;n,H,Dy,W=
11、 difflim(fun,x0,h0,wu,max1)jdwc=4.46550187104484- Dy运行后屏幕显示迭代次数n, 变步长数组H,用变步长的中心差商公式计算的近似值Dy,误差估计值W,jdwc,精确值与近似值Dy的差如下n = 2H =.020* 4.43395716561354 4.46549870972618 4.46550187410688W = 2.48353880661895 0.03154154411264 0.00000316438070jdwc =.0315*8.2.4 牛顿(Newton)多项式求导及其MATLAB程序利用牛顿插值多项式求导的MATLAB主程序
12、function df,A,P=diffnew(X,Y)n=length(X);A=Y;for j=2:n for i=n:-1:j A(i)=(A(i)- A(i-1)/(X(i)-X(i-j+1); end endx0=X(1);df=A(2);chsh=1;m=length(A)-1; for k=2:m chsh=chsh*(x0-X(k); df=df+chsh*(A(k+1); P=poly2sym(A);例8.2.9 根据下表给定的一组数据(X ,Y)写出 (8.30)式的具体形式及其精度和名称,并用它计算的近似值,取4位小数计算.X3.135 2 3.335 2 3.535 2
13、 3.735 2 3.935 2 4.135 2 4.335 2 4.535 2 Y0.126 6 -0.060 2 -0.603 2 -0.998 0 -0.119 4 0.995 3 -0.654 2 0.158 1 解 因为表中所给数据(X ,Y)是等差数列,公差为h=0.2, 即x0=3.135 2, x1= x0+h, x2= x0+2h, x3= x0+3h, x4= x0+4h, x5= x0+5h, x6= x0+6h, x7= x0+7h.输入程序X=3.1352,3.3352,3.5352,3.7352,3.9352,4.1352,4.3352,4.5352 ;Y=0.1266,-0.0602,-0.6032,-0.9980,-0.1194,0.9953,-0.6542 0.1581;df,A,P=diffnew(X,Y)的近似值df和阶牛顿多项式P及其系数向量A如下df = -0.2428A =0.1266 -0.9340 -4.4525 10.5083 16.1667 -72.4818 64.7309 108.6155P=633/5000*x7-467/500*x6-1781/400*x5+1478916440133901/140737488355328*x4+4550512123488957/281474976710656*x3-27833/384*x
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