Question

(a) Write a code that uses to approximate f(), where f(x) tan-1z and a-v2. (Note (v2 Output k, h,r,e. Here k is the iteration index, h is the step size, r is the approximate value of () and e is the error i.e. erl. From a nicely displayed table of the error, estimate the order of the finite difference approximation you just coded, by repeatedly changing the value of h.

Answer 1

clc;
clear all;
f=@(x)atan(x);
x=sqrt(2);
k=1;
h=0.5;
err=0.1;
disp('________________________________________________')
disp('k h r e')
disp('________________________________________________')
while(err>1e-10)

r=(-f(x+2*h)+8*f(x+h)-8*f(x-h)+f(x-2*h))/(12*h);
err=abs(r-(1/3));

fprintf('%f %f %f %e ',k,h,r,err)
R(k)=err;
h=h/2;
k=k+1;

end

for j=1:k-2
Order(j)=log2(R(j)/R(j+1));
end
disp('Rate of Order')
Order'

1.000000 2.000000 3.000000 4.000000 5.000000 6.000000 7.000000 Rate of Order 0.500000 0.250000 0.125000 0.062500 0.031250 0.015625 0.007813 0.334128 0.333335 0.333333 0.333333 0.333333 0.333333 0.333333 7.949775e-04 1.504431e-06 5.868085e-07 4.687588e-08 3.08731le-09 1.954111e-10 1.224976e-11 ans - 9.0456e+00 1.3583e+00 3.6460e+00 3.9244e+00 3.9818e+00 3.9957e+00