Re: Numerical differentiation, recursion of functions not working

[Martin Helm]

Am Montag, 29. März 2010 20:56:05 schrieb Martin Helm:
> > > Stefan
> 
> You can (just for fun) do some lisp like higher order function trickery,
>  for example
> 
> function df = deriv(f)
>   df = @(x) (f(x+1e-5)-f(x-1e-5))/2e-5;
> endfunction
> 
> df = deriv(@sin)
> 
> => df =
> @(x) (f (x + 1e-5) - f (x - 1e-5)) / 2e-5
> 
> 
> d2f = deriv(df)
> 
> => d2f =
> @(x) (f (x + 1e-5) - f (x - 1e-5)) / 2e-5
> 
> This seems to look the same but is not   :-)
> 
> df(0)
> => 1.00000
> 
> d2f(0)
> => 0
> 
> as it should be!
> 
> Think about it (the trick is I am using closures here by using anonymous
> functions - or more lispy: lambda expressions)
> 
> - mh
> 

Go even one step further and define

function dfh = df_h ( f, h)
  dfh = @(x) (f (x+h) - f (x-h))/2/h;
endfunction

D = @(f) df_h (f, 1e-5) # partial evaluation to get fixed stepsize

=> D =
@(f) df_h (f, 1e-5)

d2sin = D(D(@sin)) 

=> d2sin =
@(x) (f (x + h) - f (x - h)) / 2 / h

d2sin ([0:0.2:pi])

ans =

   0.00000  -0.19867  -0.38942  -0.56464  -0.71736  -0.84147  -0.93204  
-0.98545  -0.99957  -0.97385  -0.90930  -0.80850  -0.67546  -0.51550  -0.33499  
-0.14112

This works of course not only for builtin functions.

- mh