50g - Exponent of zero

[Claudio L.]

(09-29-2017 01:24 AM)Mike (Stgt) Wrote:  This said, the result XPON(0) = inf looks very reasonable for me, it has nothing to do with the internal representation of 0 and its exponent in your HP-50G.

Ciao.....Mike

But it does have to do with the internal representation, since XPON returns exactly the exponent stored on every real number (except zero as it happens on the 50g).
Also, your argument uses continuous functions. Let's not forget that XPON is not continuous, but discrete, as it only exists for the numbers representable by the machine. It exists for 0.999999999999 and 1.0 but is undefined in between.
If we choose to ignore this granularity and treat it like continuous, then the approach to 0 with an error e:
limit XPON(0+e) = -499 because e would be the smallest number representable
limit XPON(0-e)=-499 as well

So why -Inf? If both limits coincide then this is an avoidable discontinuity and its value should be -499 (now I'm confused...)

Actually, somebody might make the argument that XPON's domain excludes the whole range:
-1e-499<x<+1e-499

So it's not only undefined at 0, it is undefined for every real number that is not representable by the machine in that range. In that case you can't even take a limit when it approaches zero, because the function only exists in isolated "dots", is discrete.

But if this is the case, then XPON is mostly a practical function to retrieve the stored exponent, since it's defined only for machine representable numbers. In that case why not define it with a 0 for x=0. After all, 0 is a special number because its mantissa cannot be represented by 1<M<10, therefore why not make a special definition for a special number?

From this practical point of view, -Inf sounds just as made up as 0, because it's a theoretical number obtained from limits, but the actual function XPON is not defined in the vicinity of zero, so how do you determine that the result is -Inf? It seems Undefined would be more appropriate.