Trying to study the error of FDM for a second order derivative versus step size I calculated the coefficients and validated them, but the output has errors for small step sizes.

The function in question is

$$ f(x) = e^{\sin(x)} $$

With a derivative of $$ f''(x) = \left(\cos^2(x) - \sin(x)\right)e^{\sin(x)} $$

The calculated FDM

$$\frac{\partial^{(2)}f}{\partial x^{(2)}}\approx -\frac{5}{2h^2}f(x) - \frac{1}{12h^2}f(x-2h) + \frac{4}{3h^2}f(x-h) + \frac{4}{3h^2}f(x+h) - \frac{1}{12h^2}f(x+2h)$$

And my Octave implementation:

stps = 1e-7:1e-6:3e-5;
outs = [];
x = pi/2;
for h = stps
  coefficients = [-5/2 -1/12 4/3 4/3 -1/12]./h^2;
  steps = [0 -2 -1 1 2].*h;

  outs(end+1) = sum(coefficients .* exp(sin(x + steps))); 

plot(stps,outs, "linewidth",1.5 , stps, ones(size(stps)).* (cos(x)^2 - sin(x))*exp(sin(x)) , "linewidth",1.5)

Which produces the following plot: FDM output plot

Are those rounding errors caused by the nature of floating point numbers and operations or something else?

  • 1
    $\begingroup$ Here's what's happening: math.stackexchange.com/questions/2213240/… $\endgroup$
    – user14717
    Commented Feb 2, 2018 at 15:10
  • $\begingroup$ Try 'format long' to get more precision, then you'll see this happen at $h\approx 10^-18$ rather than $10^-7$. $\endgroup$
    – user14717
    Commented Feb 2, 2018 at 15:12
  • 1
    $\begingroup$ @user14717 format long only affects the terminal output, not the internal accuracy -- it makes no difference here. $\endgroup$ Commented Feb 2, 2018 at 17:57
  • 1
    $\begingroup$ It's normal, but it's easier to see what's happening if you plot $\log|f''_{\mathrm{estimate}}-f''_{\text{true}}|$ vs $\log(\text{stepsize})$ instead, and make sure to include large stepsizes too (up to $\sim 1$). $\endgroup$
    – Kirill
    Commented Feb 2, 2018 at 22:09
  • 1
    $\begingroup$ @user1471 There is none, apart from the Symbolic Toolbox's vpa (which is not a silver bullet). $\endgroup$ Commented Feb 2, 2018 at 22:54

1 Answer 1


Summarizing and formulating the answer:

(part of it was already given in the comments by user14717, Christian Clason, and Kirill)

While performing numerical differentiation using finite differences, one would observe two sources of error: truncation and rounding. Truncation error (for simplicity, what order terms are truncated from the Taylor series expansion) will be determined by a discretization scheme that is being used and will decrease with $h$. Unfortunately, that will be eventually stopped by an increase in the rounding error. Therefore, there usually exists an optimal discretization step $h$ when the total error (the combined effect of Taylor series truncation and rounding in finite-precision arithmetic) is minimal. Certainly, this value depends on the discretization scheme, function under-approximation, chosen point, etc. So, one would expect to see a plot that is similar to the following one:

relative error vs step size

Here, I plot the relative error in the approximation of the second derivative using your scheme vs analytical derivative as a function of the step size $h$. Since you are using a standard five-point stencil for a second-order derivative, the error is expected to drop as $\mathcal O(h^4)$, until you hit $h\approx5\cdot 10^{-3}$. At this point, the rounding error starts dominating and we see the oscillations in the relative error together with its gradual increase. However, using this scheme you are able to get 10 significant digits from your derivative pretty safely using double precision, which I would say is a very reasonable result.

If interested, you might invest your time adjusting your discretization scheme, increased precision (vpa), etc.

A couple of minor notes:

  • loglog plots are usually very useful when studying refinements, errors, timings, etc.
  • a line on the plot usually implies continuity of the data; in this particular case, where you are using finite differences at discrete points, markers are preferred, in my opinion.

Slightly modified Octave code:

x = pi/2;
stps = logspace(-8,0,80);
numpoints = size(stps,2);
outs = zeros(numpoints,1);
analytic = ones(numpoints,1)*(cos(x)^2 - sin(x))*exp(sin(x));
for i=1:numpoints
  coefficients = [-5/2 -1/12 4/3 4/3 -1/12]./h^2;
  steps = [0 -2 -1 1 2].*h;
  outs(i) = sum(coefficients .* exp(sin(x + steps))); 

loglog(stps',abs(outs-analytic)./abs(analytic),"kx");hold on;
xlim([1E-8,1E-0]);   ylim([1E-12,1E+1]);
xlabel("h");         ylabel("Rel. error: |num-an|/|an|");
  • 1
    $\begingroup$ In some cases, it may be interesting to use the "complex step" approach, which can provide almost machine accuracy results. $\endgroup$
    – Laurent90
    Commented Jun 3, 2022 at 16:28
  • $\begingroup$ @Laurent90 yep, complex-step technique is a very powerful instrument where it is applicable. $\endgroup$
    – Anton Menshov
    Commented Jun 3, 2022 at 16:31

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