HP 48GX Collector’s Edition (merged) - please don't start two threads

[carey]

(11-19-2023 08:18 PM)M0R33z Wrote:  Maybe adding also some of useful applications to the ROM

alg48v42, erable, Meta Kernel, ...

I think that was called the HP-49g :)

[johnb]

As I've said elsewhere, if Moravia or SwissMicros were to release a Collector's Edition of any of the 48 series, it would likely take entirely new tooling of the enclosure and the mechanicals... at which point they (and we) would all be better served by a Collector's Edition 50gII.

To appeal to the greatest number of people, it would need to have 50g functionality, more memory, a faster processor, and better/more connectivity. Since a lot of people hate the keyboard and/or physical packaging of the 50g, the 50gII would probably look and feel more like a 48-series. It might even have the old-school classic beveled keys-on-hinges. A bonus nicety would be a return to the old HP color scheme.

A manufacturer probably wouldn't want to go much further beyond that (such as various alternate modes of operatoin, etc.) because development costs would be enormous just for what's described here. However, it would probably cost near nothing to make the platform Open Source, so that the adventurous could hack their own custom operating modes or OS.

[Also, like I've said elsewhere, if anyone builds this, sign me up for at least two of them, maybe more... one daily driver, one backup, one to keep NIB to sell when someone else's daily driver finally quits...]

[avsebastian]

0 ^ 0 = 1 ????

the Same on hp 50g.

I think it is a very serious mistake

[Liam Hays]

I can dream of an HP 51G collector's edition, with a re-created RPL OS (no Saturn emulation), a nice classic keyboard and chassis, and a SHARP memory display like the DM42 (low lag, low power).

They would never do this, but what if HP manufactured a calculator and didn't make software for it? Then, the community™️ could step in and make the software on their own.

Also, 0^0=1 is on purpose: https://www.hpcalc.org/hp48/docs/faq/48faq-5.html#ss5.2

[EugeneNine]

One of my projects has been to re-create the 48 series. Should be able to 3dprint the plastics, use a modern MCU using the code from x48 for the emulation. Then start making upgrades and additions,

[lvt]

(11-20-2023 07:47 PM)carey Wrote:  

(11-19-2023 08:18 PM)M0R33z Wrote:  Maybe adding also some of useful applications to the ROM

alg48v42, erable, Meta Kernel, ...

I think that was called the HP-49g

More like a 48.5G to me

[Steve Simpkin]

(11-21-2023 02:19 AM)Liam Hays Wrote:  I can dream of an HP 51G collector's edition, with a re-created RPL OS (no Saturn emulation), a nice classic keyboard and chassis, and a SHARP memory display like the DM42 (low lag, low power).

They would never do this, but what if HP manufactured a calculator and didn't make software for it? Then, the community™️ could step in and make the software on their own.

Also, 0^0=1 is on purpose: https://www.hpcalc.org/hp48/docs/faq/48faq-5.html#ss5.2

c3d’s project, DB48X, approaches some of your goals. It is a new implementation of RPL for the DM42.
https://www.hpmuseum.org/forum/thread-20157.html

[brouhaha]

If HP were ever to offer a new RPL calculator, it seems likely that it would be a derivative of the 50G, since that is the most recent RPL calculator, and already runs on a modern processor architecture (ARM), though the specific processor chip in the 50G is discontinued.

However, as far as I've ever heard, there is little enthusiasm among the HP engineers who are qualified to do such a thing. The absolute minimum it would take to make that happen would be convincing Moravia and Royal that it would sell well enough to justify a fairly major engineering investment, and even that wouldn't guarantee that it would happen.

(my opinions only, I don't work for HP, Moravia, or Royal)

[vaklaff]

RPL users, you might want to take a look at Christophe de Dinechin's DB48X. IMHO it's a more promising way than trying to talk Moravia into a new RPL machine.

[c3d]

(11-21-2023 08:10 AM)vaklaff Wrote:  RPL users, you might want to take a look at Christophe de Dinechin's DB48X. IMHO it's a more promising way than trying to talk Moravia into a new RPL machine.

Thank you! Just got 0.5.0 out. We are entering the "alpha" stage now, i.e. I think it's usable by a general public.

[Allen]

I understand from some previous discussion time out of mind that the source code for the HP 48 Series was lost at some point in the late 2000's?

[Joe Horn]

(11-21-2023 12:38 AM)avsebastian Wrote:  0 ^ 0 = 1 ????

the Same on hp 50g.

I think it is a very serious mistake

The simplest and clearest explanation I've ever seen is based on the definition of powers, namely, "a^b" means "start with the multiplicative identity (1), then multiply by a, b times." Example: "2^3" means "start with 1 then multiply by 2, three times." Therefore, "0^0" means "start with 1, then multiply by 0, zero times." The result, of course, is 1.

The USUAL definition of powers, "a^b means start with a, and multiply by a, b times" is incorrect, since that yields a^(b+1). Instead, the process always begins with the multiplicative identity, just as products always start with the additive identity.

[FLISZT]

'0 ^ 0' ⇒ the 50g returns " ? " in exact mode.

[Hans S.]

(11-21-2023 12:38 AM)avsebastian Wrote:  0 ^ 0 = 1 ????

au contraire; 0^0 is exactly 1, as per definition (apart from the pointless question mark inflation).

Hans

[Voldemar]

(11-21-2023 01:55 PM)FLISZT Wrote:  '0 ^ 0' ⇒ the 50g returns " ? " in exact mode.

0^0 undefined

  Untitled.gif (Size: 1.29 KB / Downloads: 233)
HP 50g in Approx mode 0.^0.=­1.

[Johnh]

On the question of what 0^0 equals, obviously it's a conundrum, but I'm sure proper mathematicians are clear on what it should be. But here's my take, and the answer is 1:

If you say that 0^0 should be the limit of x^x as x approaches zero, then you can easily see the result heading towards 1 ie:

0.1^0.1 = 0.79433
0.01^0.01 = 0.95499
0.001^0.001 = 0.99312
0.0001^0.0001 = 0.99908
0.00001^0.00001 = 0.99988 etc

And barring a typo above, then I think all calculators will agree if considered that way.

[brouhaha]

(11-21-2023 08:30 PM)Johnh Wrote:  On the question of what 0^0 equals, obviously it's a conundrum, but I'm sure proper mathematicians are clear on what it should be. But here's my take, and the answer is 1:

If you say that 0^0 should be the limit of x^x as x approaches zero, then you can easily see the result heading towards 1 ie:.

And on the other hand, you can just as easily look at it as the limit of 0^x, as x approaches 0 from the right:

0^3 = 0
0^2 = 0
0^1 = 0
0^0 = ?

You can trivially construct cases that approach 0^0 from other directions, and yield other limiting values. This is why 0^0 is generally considered to be undefined.

There are good arguments for considering 0^0 to be1, but when doing so one should still be aware that it is actually a discontinuity, and that considering it to be 1 is an adopted convention and not literally "the true value" of 0^0.

[avsebastian]

(11-21-2023 08:45 PM)brouhaha Wrote:  

(11-21-2023 08:30 PM)Johnh Wrote:  On the question of what 0^0 equals, obviously it's a conundrum, but I'm sure proper mathematicians are clear on what it should be. But here's my take, and the answer is 1:

If you say that 0^0 should be the limit of x^x as x approaches zero, then you can easily see the result heading towards 1 ie:.

And on the other hand, you can just as easily look at it as the limit of 0^x, as x approaches 0 from the right:

0^3 = 0
0^2 = 0
0^1 = 0
0^0 = ?

You can trivially construct cases that approach 0^0 from other directions, and yield other limiting values. This is why 0^0 is generally considered to be undefined.

There are good arguments for considering 0^0 to be1, but when doing so one should still be aware that it is actually a discontinuity, and that considering it to be 1 is an adopted convention and not literally "the true value" of 0^0.

yeah

0 ^ 0 = "nothing" ^ "nothing" = 1 ??? ilogical

If 0 / 0 = error, is not the same?

[Johnh]

My point is, with two 'Zeros' in 0^0, I was exploring the limit when both 'Zeros' are given equal weight, as they approach 0. The actual answer is undefined, but I was curious to know if x^x would approach 0 or 1.

[Gil]

If 0^0 is or were to be defined as 1, as for x^0, with x≠0, or with the argument of the limit x^x, with x —> 0+,
I can't help consider, as a dummy, x^0=x^(a-a),
with 0 = a-a for any a,
then x^(a-a) = x^a/x^a=(x/x)^a = 1^a... =1.

But stop! x is =0
—> with x/x, we have "0/0"...

But of course I am no mathematician.

A between (interesting, but perhaps not satisfactory) solution of the HP50G is indeef
0.^0. —> 1. (or 0.^0 —>1. or 0^0. —> 1., with the latter result not so clear for me),
with the dot that could be understood as "an approaching value",
and 0^0 in exact mode with the HP50G that results in '?'.