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Re: How to eliminate noises? A better way perhaps.

  • To: mathgroup at smc.vnet.net
  • Subject: [mg122632] Re: How to eliminate noises? A better way perhaps.
  • From: Richard Fateman <fateman at cs.berkeley.edu>
  • Date: Fri, 4 Nov 2011 05:59:20 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201111021121.GAA03503@smc.vnet.net> <j8tl5t$f3t$1@smc.vnet.net>

On 11/3/2011 1:59 AM, Bob Hanlon wrote:
> Use higher precision.
.. so say you all...
Well, that's the brute force method, and one reason to recommend it is 
it doesn't require much thought.

On the other hand, the person asking the question has an apparently 
large expression which he knows has a double zero at n=1/2  and he wants
to know the behavior of the expression from 0.35 to 0.53.  Namely, 
around that zero.

Let us call that expression p.  A non-brute force, but mostly automatic 
method is to note that (since p is a polynomial of degree 29) it is 
EXACTLY equal to s, where

s = Normal[Series[p, {n, 1/2, 29}]]

A plot of s can be done in ordinary float arithmetic and looks to the 
eye just like the plot of p, done with high working precision. Unique to 
computer algebra systems, it is also possible to look at the structure 
of s, and note that expanded around n=1/2 it is mighty sparse.

Let k = n-1/2  .  The expression being plotted is then

14 k^2 - 1296 k^5 + 1451188224 k^11 - 2002639749120 k^14 +
  598075300380672 k^17 - 83414577743659008 k^20 +
  3977900382788517888 k^23 - 113721152119718805504 k^26 +
  1516282028262917406720 k^29


Sometimes a better way to format expressions for evaluation is to use
HornerForm.  In this case it is


k^2 (14 +
    k^3 (-1296 + ....

Evaluating this expression at a value for k requires computing
k^2,k^3,k^6,and some other arithmetic for a total of about 18 arithmetic 
operations, which can be done in double-float.
The real win here is that you can show how much more insight you might 
get from using computer algebra.

RJF



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