Re: [EXTERNAL] Re: 1-button calcula tor ~ "all elementary fns from a single operator"

OK, I downloaded the "EML compiler" from here:

https://github.com/VA00/SymbolicRegressionPackage/tree/master

The EML compiler requires python3 & numpy (and probably other stuff that I might already have installed).

I then did a trivial script to convert from Mathematica expressions to Maxima expressions.

EML requires that log(0)=minf, so I defined mylog(x):=if (x=0) then minf else log(x), and used mylog(x) instead of log(x) in the definition of eml.

However, I now require that %e^minf = 0.

What is the magic in Maxima to make this happen?  I did "? minf", but that didn't provide any help.

-----Original Message-----
From: Henry Baker <[email protected]>
Sent: May 1, 2026 11:23 AM
To: Przemek Klosowski via Maxima-discuss <[email protected]>
Subject: Re: [Maxima-discuss] [EXTERNAL] Re: 1-button calculator ~ "all elementary fns from a single operator"

Obviously, Google was confused.

Q: If z = exp(x)-log(y), then x = log(z + log(y)) might look a tad better ? Is this log(z+log(y)) function universal in the same sense as eml() ?

Q: if we have a 1BC expression for f(x)=y, can we then trivially compute x=f^-1(y) ?

Q: The 1BC paper seems to want to rely on functions defined over the reals; I suspect that functions defined over the complex numbers might give additional 1BC results that might be prettier ? Perhaps the constant %pi*%i might work to force things into the complex plane ?

Q: I've always had a fondness for asinh(x), as it is bijective onto the reals, and has many of the same properties/characteristics as "gradual underflow" floating point numbers. I'm wondering if it could be part of a universal 1BC function ?

-----Original Message-----
From: Przemek Klosowski via Maxima-discuss
Sent: May 1, 2026 9:24 AM
To:
Subject: Re: [Maxima-discuss] [EXTERNAL] Re: 1-button calculator ~ "all elementary fns from a single operator"

> BTW, Google just told me that the obvious differential equation for > eml(x,y) is
>
> dy/dx = y*exp(x)

(%i1) eq:'diff(y,x)=y*exp(x);
dy x
(%o1) ── = %e y
dx
(%i2) ode2(eq,y,x);
x
%e
(%o2) y = %e %c

what am I missing?



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