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Few questions (convolutions, 'mapping', linspace)

From: Przemek Klosowski
Subject: Few questions (convolutions, 'mapping', linspace)
Date: Mon, 16 May 1994 13:18:06 -0400


I have a couple of questions that came in my recent using of Octave; I wonder
if anyone could help:

 - I want to convolute the theoretical function S(Q,w) with the experimental 
   resolution r(w), to compare the resulting Sr(Q,w) with experimental data. 
   The formula is:
      Sr(Q,w) =   |  S(Q,w-v) r(v) dv 

   I was hoping to use the quad("fun",-w0,w0), where fun would be defined
   as 'function y=fun(x); y=S(Q,w-x)*r(x); endfunction'. This didn't quite
   work, as I need to access Q,w from within that function. I can't pass
   them as parameters of fun() because quad() expects only one-parameter fun;
   I couldn't figure the way to pass them as globals either.

 - I often find myself defining a function (y=fun(x)), and hoping to be able
   to 'map' (in the LISP sense) the vectors/matrices with this function, for 
   instance to plot it. The naive approach, y=f(x) for a vector/matrix 'x', 
   does not result in 

         y = [f(x(1)),  f(x(2)),  f(x(3)) ... f(x(N))],

   which is fine because the arithmetic operators in the function definition 
   have meaning for matrices as well. For vector 'x' I can always do 

         for i=1:length(x); y(i)=f(x(i)); endfor

   but I wonder if there is a neater, general way. I suppose I could be defining
   my functions to do the explicitly recognize vector arguments, but
   that still leaves the question whether I have to write explicit 'for' loops
   or is there a better way.
 - I noticed that the linspace operator is fairly slow: linspace(-1,1, 10000) 
   25 seconds on a 150MHz r4400, while the a=(0:.0001:1); is very quick, and 
   FFT of the resulting matrix is only few seconds. What is the advantage of 
   over the range operator, and why is it so slow?


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