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

[Stavros M. <mac...@gm...>]

I wonder if anyone is using NaN payloads to add, say, high-precision
numbers or numerical differentiation or intervals arithmetic?
Similar to using it for symbolic calculation but should be simpler....

On Thu, Feb 29, 2024 at 2:48 PM Richard Fateman <fa...@gm...> wrote:

> Typically, yes the Newton iteration would be terminated either
> when the result is deemed close enough/ but not always.
>
> A really fast sqrt program might start with pretty good initial
> guess and an in-line expansion of 2 iterations.
>
> See https://en.algorithmica.org/hpc/arithmetic/rsqrt/  for a fun
> exposition of
> how to compute sqrt(1/x) using 1 Newton step. Maybe 2.
>
> There is the well-known potential to understand (or debug or optimize)
> programs by
> "symbolic execution".  This has something of that flavor.
>
>  The novelty presented by using NaNs is that you could
> try to run and better understand a large, opaque, proprietary numerical
> program by
> inserting some NaN(s) in the input and look at the output.
>
>  I am unaware of anyone doing this for real, or
> even an implemented framework by which one could try to do this.
> RJF
>
>
>
>
> On Thu, Feb 29, 2024 at 11:30 AM Stavros Macrakis <mac...@gm...>
> wrote:
>
>> 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...>
>> wrote:
>>
>>> In the context of taking a numerical program and reusing it
>>> with 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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>>