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Numerical differentiation

hobz
in Analysis and Calculus

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I have written the following MATLAB code

h = 10e-12;
f = @(x)cos(2*pi*x);

x = 0:.01:pi;

fd = (f(x+h)-f(x-h))/(2*h);

hold on
plot(x,f(x))
plot(x,fd, 'r')

 

The result is quite surprising.

fd, which is the derivative of f evaluated numerically, has an amplitude roughly 6 times larger than what the real derivative is.

Can anyone explain how this comes to happen?

The "real derivative" of [math]f(x)=cos(2\pi x)[/math] with respect to x is [math]-2\pi \sin(2\pi x)[/math]. 2*pi is a bit over six ...

  • Author

Yeah ok, that was kinda stupid. Reused a function from a sampled function, didn't notice the 2*pi. Thanks for clearing it up!

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