Re: [Axiom-developer] Re: bootstrap Meta/Boot

[M. Edward (Ed) Borasky]

C Y wrote:
>> Incidentally, I "do" have Dick Pountain's "Object-Oriented Forth",
>> which is probably as good a reference for implementing other
>> languages on a Forth base as you're likely to find anywhere. If
>> you're in the mood to re-invent a wheel or two, I'd recommend
>> starting there.
> 
> Thanks!  I see it's on Amazon for $35 - is that a good price?

New or used? If you're looking for used, try "addall.com" -- they hit
all of the used booksellers.


> I dunno - despite the availability of CASs for years now, a lot of
> folks see to be reluctant to give up their HP calculators ;-).  I
> wonder why someone doesn't re-create the HP-48 - that seems to be a
> sweet spot in calculator history.

As far as I know from a CAS perspective, the current HP-49 is upward
compatible at the RPL level from the 48, though not at the assembly
level. As far as re-creating the HP-48, I haven't heard anyone else ask
for that, but there *is* a mighty band of folks who would like to see
the HP-15C (scientific equivalent of the HP-12) re-created.

I sort of collect calculators -- I have a TI Voyage 200, TI 89 Platinum,
HP 49 and an HP-12 Platinum at the moment. Somewhere I have a TI SR-52
as well from long ago. The one I carry around in my pocket is the HP-12,
and the one I use at my desk is the HP-49. And both of the TIs have a
bug in their differentiator -- they get the derivative of 1/(1 - u^m)
wrt u as some ghastly expression involving natural logs and hyperbolic
sines, while Axiom (and everybody else) gives it as

               m - 1
            m u
   (1)  ---------------
          m 2     m
        (u )  - 2u  + 1
                                                     Type: Expression
Integer

wxMaxima:

(%i1) 1/(1-u^m);
(%o1) 1/(1-u^m)
(%i2) diff(%o1, u);
(%o2) (m*u^(m-1))/(1-u^m)^2

I can't do it on the HP-49 -- it's at work. But the last time I did it,
it got the same answer.

Just on the off chance you're wondering, "u" is utilization, "m" is the
number of processors, and the expression is the approximate relative
response time (stretch factor) for a processor-bound job. That is, if
one such job takes one second on a uniprocessor with zero utilization
(otherwise idle machine), a large collection of such jobs takes on the
average 1/(1-u^m) seconds each on an m-processor machine with a
utilization of u. And the derivative is of course simply how fast the
stretch factor grows as utilization increases.

Hopefully this won't be on the SAT if students use a TI calculator. ;)