hobz Posted March 9, 2008 Share Posted March 9, 2008 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? Link to comment Share on other sites More sharing options...
D H Posted March 9, 2008 Share Posted March 9, 2008 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 ... Link to comment Share on other sites More sharing options...
hobz Posted March 9, 2008 Author Share Posted March 9, 2008 Yeah ok, that was kinda stupid. Reused a function from a sampled function, didn't notice the 2*pi. Thanks for clearing it up! Link to comment Share on other sites More sharing options...
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