This is the question I am to solve:

Given the function $f(x)=\ln(3x+1)$, compute approximations to $f'(0)$ using the centered 3-point formula:


Use Matlab to generate a plot of the absolute error as a function of the step-size h for the centered 3-point formula approximation to f′(0). At what values of h is the error minimized, that is, what step-sizes yield the maximum accuracy for this problem? (The magnitude of this number will be quite small.)

I have written the following MATLAB .m script file to generate the matrix of values $x, f(x), f'(x),$ and $\approx f'(x)$. However, I cannot figure out how to evaluate the error at more than one value for $h$ at a time, if there is a way.

h = 0.1;

x = -0.3:h:0.3;

f = @(x)(log(3.*x+1));

fp = @(x)((3)./(3.*x+1));

N = length(x);

A = zeros(N,4);

for i=1:N

for j = 1:3

   A(i,1) = x(i);

   A(i,2) = f(A(i,1));

   A(i,3) = fp(A(i,1));



for i=2:N-1

for j = 4

   A(i,j) = (A(i+1,j-2)-A(i-1,j-2))./(2.*h);



mdpt = ceil(N/2);

pt = A(mdpt,4);

err = abs(pt-fp(0));


up vote 0 down vote accepted

There are probably a few ways to answer this question. Perhaps you could generate the approximations of $f'(0)$ using the following code:

x0 = 0;

n = 1000;

h = logspace(-16, -1, n); % Generates n numbers logarithmically spaced between 10^(-16) and 10^(-1)

fp_approx = (f(x0 + h) - f(x0 - h)) ./ (2*h); % The approximations of f'(x_0)

Then compute the error as

fp_exact = fp(x0)*ones(size(fp_approx));

err = abs(fp_approx - fp_exact);

You could take $n$ much larger (say, $n = 10^6$), then find the minimum $h$ using min:

[min_err, idx] = min(err);


  • Thank you for the help. I took away my for loops and basically pasted what you had suggested into my script file, so I have my assigned functions followed by the pasted material. To complete the question, would adding the line "plot(err,h)" be correct? If so, the plot generated is Which shows both a blue line at the bottom and one going toward the top middle. – Khori Wood Oct 22 '13 at 21:44
  • Glad to help. I get that plot as well when I type plot(err, h). Since you are interested in the absolute error as a function of h, you might want to swap the order of h and err: plot(h, err). One caveat though -- because of the very small numbers involved (recall that some of the numbers are on the order of $10^{-16}$), the plot doesn't show the graph of the absolute error very clearly. Instead you could use a log-log plot, loglog(h, err), which shows the absolute error as a function of $h$ much more clearly (on a log-log scale). – Kyle Oct 23 '13 at 2:23

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.