MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Convolution Integral


Hi,
if you have problems with dummy variables, it makes sense to  work with 
pure functions, e.g. Function[x,expression(x)] or expression(#) (note 
the default variable #)&. E.g.
conv[f1_, f2_] := Module[{u}, Evaluate[Integrate[f1[u] f2[# - 
u],{u,0,#}] ] &]
the "Evaluate" is necessary because (..)& does not evaluate its 
arguments. Now, try e.g. f=conv[Cos, Sin[# + .1] &], this gives another 
pure function that can be evaluated by e.g.: f[0.3]
Daniel

Mr Ajit Sen wrote:
> Dear Mathgroup,
> 
> Could anyone please help me with the following?
> 
>  I'd like to find the convolution of 2 arbitrary
> functions, f(t) and g(t) in the Laplace transform
> sense,i.e., 
> 
> convolve[f[t],g[t]]=Integrate[f[u]*g[t-u],{u,0,t}]
> 
> Thus, I'd like convolve[Sin[t],Exp[-t]] to return
> 
>    (Exp[-t]-Cos[t]+Sin[t])/2 .
> 
> My several attempts with function definitions such as
> 
>   convolve[f_,g_]:=Integrate[f[u]*g[t-u],{u,0,t}]
> 
>  
> convolve[f[t_],g[t_]]:=Integrate[f[u]*g[t-u],{u,0,t}]
> 
>   convolve[f_,g_][t_]:=Integrate[f[u]*g[t-u],{u,0,t}]
> 
> all failed (because of the dummy u ? )
> 
> Thanks in advance.
> Sen.
> 
> 
> 
> 	
> 	
> 		
> ___________________________________________________________ 
> Yahoo! Messenger - NEW crystal clear PC to PC calling worldwide with voicemail http://uk.messenger.yahoo.com
> 


  • Prev by Date: Re: Nonautonomous ODEs
  • Next by Date: Re: Convolution Integral
  • Previous by thread: Convolution Integral
  • Next by thread: Re: Convolution Integral