Matlab实现——Differentiation Using Limits

difflim.m

%Program 6.1 (Differentiation Using Limits). To approximate %f'(x) numerically by generating the sequencef'(x)≈Dk=%[f(x+(10^-k)*h)-f(x-(10^-k)*h]/[2*(10-k)*h]  for k=0,… ,n%until |Dn+1-Dn| > |Dn-Dn-1| or |Dn-Dn-1| < tolerance, %which is an attempt to find the best approximation f’(x)≈Dn.function  [L,n]=difflim(f,x,toler)%Input  - f is the function input as a string 'f '%       - x is the differentiation point%       - toler is the desired tolerance%Output - L=[H' D' E']: %                    H is the vector of step sizes%                    D is the vector of approximate derivatives%                    E is the vector of error bounds%         - n is the coordinate of the "best approximation"max1=15;h=1;H(1)=h;D(1)=(feval(f,x+h)-feval(f,x-h))/(2*h);E(1)=0;R(1)=0;for n=1:2   h=h/10;   H(n+1)=h;   D(n+1)=(feval(f,x+h)-feval(f,x-h))/(2*h);   E(n+1)=abs(D(n+1)-D(n));   R(n+1)=2*E(n+1)/(abs(D(n+1))+abs(D(n))+eps);endn=2;while((E(n)>E(n+1))&(R(n)>toler))&n<max1   h=h/10;   H(n+2)=h;   D(n+2)=(feval(f,x+h)-feval(f,x-h))/(2*h);   E(n+2)=abs(D(n+2)-D(n+1));   R(n+2)=2*E(n+2)/(abs(D(n+2))+abs(D(n+1))+eps);   n=n+1;endn=length(D)-1;L=[H' D' E'];

fun.m

function f=fun(x)f=sin(cos(1/x));

untitled.m

x=1/sqrt(2);toler=10e-13;f='fun';[L,n]=difflim(f,x,toler)