Re: [Maxima-discuss] putting a symbol into a numeric formula, a use for a NaN?

[Henry B. <hb...@pi...>]

Re symbols/NaNs:

A long time ago, I was attempting to derive Lisp-style
bounds declarations at compile time using 'forward' and
'backward' type inferencing.

For the heck of it, I tried my type inferencer out on a
Newton sqrt iteration, and it dutifully tried to do the
computation at compile time, except using Lisp-style 'bounds'
as a data type.  Since the Lisp standard requires *rationals*
for declaring numeric bounds, I quickly got bounds with enormous
denominators.

I never tried using range arithmetic with floating point bounds
as approximations to the rational bounds, but perhaps that might
be the *best* use for FP range arithmetic in Lisp !  So long as
the FP bounds are *always* looser than the rational bounds, then
the FP bounds should still be useful, and the FP type inference
calculations will be far more efficient than the rational type
inference calculations were.

BTW, I also worked out a system using Henrici *circular arithmetic*
for implementing continued fractions for sqrt's; perhaps not the
fastest way to compute sqrt approximations, but maybe the most
elegant?

-----Original Message-----
From: Stavros Macrakis <mac...@gm...>
Sent: Feb 29, 2024 11:30 AM
To: Richard Fateman <fa...@gm...>
Cc: <max...@li...> <max...@li...>
Subject: Re: [Maxima-discuss] putting a symbol into a numeric formula, a use for a NaN?

Won't the Newton iteration typically be wrapped in a while clause? How will the termination condition be evaluated?


On Thu, Feb 29, 2024 at 12:32 AM Richard Fateman <fa...@gm... (mailto:fa...@gm...)> wrote:
In the context of taking a numerical program and reusing itwith partial symbolic input, here's another example..
 
Here is one step of a newton iteration:  x[n+1] = x[n] - f(x[n])/df(x[n])... 
newtstep(f,x,val):= block([d:diff(f,x)], val-  subst(val, x,f/d));
 
how does this work?  val is a guess... here, a guess the root of x^2-9...
 
newtstep(x^2-9,x, 5.0);  --> returns 3.4
newtstep(x^2-9,x, 3.4);  --> returns 3.023529411764706
 newtstep(x^2-9,x, 3.023529411764706); --> returns 3.00009155413138
   you get the idea.
 
Now try putting an expression in there for the guess.. [ could use a NaN perhaps]
newtstep(x^2-9,x,3-eps)$ /* could put 5-eps or anything else...
  newtstep(x^2-9,x,%)$  
  newtstep(x^2-9,x,%)$    --> 
    -((eps^8-24*eps^7+504*eps^6-6048*eps^5+45360*eps^4-217728*eps^3+653184*eps^2-1119744*eps+839808)/(8*eps^7-168*eps^6+2016*eps^5-15120*eps^4+72576*eps^3-217728*eps^2+373248*eps-279936))
 
whew!   try to understand this by using Taylor expansion..
 
  taylor(%,eps,0, 9) -->  3+eps^8/279936+eps^9/209952
 
The newtstep program could instead be some large mysterious
 iterative numerical program in a language that normally does not
allow symbols. But we might fool it by using NaNs as substitutes for symbolic
expressions. 
If you had this all hooked up via trap handling 
you might get some insight as to how a program works..
 
RJF
 
 


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