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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg122670] Re: How to eliminate noises? A better way perhaps.
  • From: Dana DeLouis <dana01 at me.com>
  • Date: Sat, 5 Nov 2011 04:47:27 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com

> HornerForm.  In this case it is
> 
> k^2 (14 +
>     k^3 (-1296 + ....
>  k requires computing k^2,k^3,k^6

Hi.  As a side note to Richards excellent reply...
HornerForm for me pulled out a common k^3 term instead of just factoring out a k^2 term.

If I were to use this code in vb/vba or something, I like to add the following.
Since k^3 is calculated multiple times, I like to set that to a variable as well.

Series[p,{n,1/2,40}] //Normal;

% //. n-1/2->k
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

HornerForm[%,k] //.k^3->w

k^2 (w (k^6 (w (w (w (w (w (1516282028262917406720 w-113721152119718805504) 
      +3977900382788517888)-83414577743659008)+598075300380672)-2002639749120)+1451188224)-1296)+14)

In vb, I might change the k^2 to k*k, and note that the odd ball k^6 is just w*w.
Hense, a function without a lot of power terms might be something like:

g[n_] := Module[{k = n - 1/2, w}, 
   w = k*k*k; 
   k*k (w (w*w (w (w (w (w (w (w*1516282028262917406720 -113721152119718805504)+3977900382788517888)
                -83414577743659008)+598075300380672)-2002639749120)+1451188224)-1296)+14)]

If f[x] is the op's original equation, then a quick test at a few points:

Table[ { f[n], g[n] }, {n,35/100,53/100,1/100}];

Equal@@@% //Union
{True}

= = = = = = =
Dana DeLouis
$Version
8.0 for Mac OS X x86 (64-bit) (February 23, 2011)
= = = = = = =



On Nov 4, 7:01 am, Richard Fateman <fate... at cs.berkeley.edu> wrote:
> 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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