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

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

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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