Aproximating a messy product ...

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2 total messages Started by Levent Aydin Mon, 04 Jan 1993 23:25

#1012

Author: Levent Aydin
Date: Mon, 04 Jan 1993 23:25

25 lines
1555 bytes

	Hello everyone;
	I need to approximate a function given in terms of a product with a
simple function that I can deal with.The expression is as follows:

               n              /            \  
              __             |    x-m_i     | 
        g(x)= || 0.5*{ 1+ erf| -------------| }
              i=1            |  sqrt(2)*s_i | 
                              \            /  

	Here, m_i and s_i are two known sequences of non-zero real numbers.You 
can assume that the sequence m_i is a descending sequence and the value of s_i is
directly proportional to the value of m_i for each i.Also it is known that the
sequence m_i is such that it has only couple values which are significantly
larger than the rest of them.
	So far all I could figure was that the product is determined by the terms
with large values of m_i.I think that the product can be approximated by a
function similar to one of those terms, i.e. 0.5*{1+erf[(x-mo)/(sqrt(2)*so)]}.
The question is what are the optimum values for 'mo' and 'so'.
	Any input will be greatly appretiated.Thank you very much...


							L  E  V  E  N  T
P.S: The function erf(x) is the standard error function.

Re: Aproximating a messy product ...

#1142

Author: Richard Simard
Date: Mon, 11 Jan 1993 19:50

45 lines
2301 bytes

In article <1iah0fINNjhj@solar.usc.edu> laydin@solar.usc.edu (Levent Aydin) writes:
>
>	I need to approximate a function given in terms of a product with a
>simple function that I can deal with.The expression is as follows:
>
>               n              /            \  
>              __             |    x-m_i     | 
>        g(x)= || 0.5*{ 1+ erf| -------------| }
>              i=1            |  sqrt(2)*s_i | 
>                              \            /  
>
************************************************************

I don't think it would be a good idea to approximate by a
single term of the same form , whatever m_0 and s_0.

It would be best to separate the 2 factors with m_i much
larger than the others. Consider the function erf(y). It is
a monotonously increasing func; it is practically 0 from 
- infinity to 0 - 3 standard deviations, and  practically 1 
from 0 +
3 stand. dev. to + infinity; inbetween, it increases 
smoothtly.
Thus, you could approximate one factor  in the above
product by the following
approx.:

f(x) = 0.5 , for x = - infinity to m_i - 3 stand. dev.
     = 1 , for x = + infinity to m_i + 3 stand. dev.
     = a cubic polynomial with inflexion at m_i in-between.
 
Also you could approximate the n-2 factors with the m_i close
together by the function h(x) where

h(x) 	= 0.5^(n-2) , for x= - infinity to min{m_i - a}
     	= 1 , for x = max{m_i + b} to +infinity
	= Ae^(px), in-between 

the parameter a is around 1 to 3 standard dev of  in min m_i
the parameter b is between 1 to 3 standard dev. from max m_i
and A and p are chosen so the straight line px + c1 represent
well enough the log of the product of the (n-2) factors.

--
From:  simardr@iro.umontreal.ca

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