Giac/Xcas updates ported to HP Prime?

[EdDereDdE]

Hello everybody,

I'd just like to know if the evolution of Giac/Xcas will be reflected in Primes' updates?
Giac/Xcas is currently at 1.2.3 (2017), whereas Prime (the "real" one) is at 1.1.2-11.

Any information about the long term strategy please?

Thanks,
EdDereDdE from Germany

P.S. my first HP calculator was the HP-28C. After decades of HP-48G I got my Prime only recently. And i'm new to this forum, maybe my question was already answered elsewhere?

[Tim Wessman]

Yes. Right now it is 1.2.X (don't remember what off the top of my head)

The primary reason for things to be "out of sync" has to do with translations being out, testing not having the bandwidth to look at new features or issues, or documentation not being ready.

For example, the last major giac version bump was near the end of a long development cycle and dropping it in with ~1 month till release is simply not a good idea.

[EdDereDdE]

Thank you Tim for the info.
I'm well aware of development cycle issues, as I'm a developer myself since decades.
So, I suppose we will get long term updates and impovements over time.
Good to know.

B.t.w., what the 0^0 (Zero ^ Zero) "undef" error?
per Definition it should be 1 (as HP-48G and others "know").

[grsbanks]

(02-14-2017 10:10 PM)EdDereDdE Wrote:  B.t.w., what the 0^0 (Zero ^ Zero) "undef" error?
per Definition it should be 1 (as HP-48G and others "know").

Interestingly enough, the HP 50g returns a question mark to the stack if you try and evaluate 0⁰.
   
Yes, this was done with Emu48 but that uses the original ROM of the 50g so it's reasonable to assume that you'll get the same result with a real 50g.

[EdDereDdE]

This indicates, that the error was commited sometime after the HP-48G.
My real calculators (except the Prime) and the Android Apps return 0^0=1...
Same as the Windows 10 standard calculator in science mode...

Here a list of (Android) Apps i use:
Droid48 https://play.google.com/store/apps/detai....x48&hl=de
Droid48sx https://play.google.com/store/apps/detai...48sx&hl=de
Ti-59 https://play.google.com/store/apps/detai...ti5x&hl=de

A special mention goes to Handycalc:
https://play.google.com/store/apps/detai...calc&hl=de

And, naturally, the HP Prime app

[grsbanks]

FWIW, Free42 also evaluates 0⁰ as 1. I don't know about the original HP-42S (Free42 does not use the HP-42S ROM).

[Simone Cerica]

0^0 is an indeterminate form.

https://www.wolframalpha.com/input/?i=0%5E0

[grsbanks]

Indeed it is.

How about this? https://www.math.hmc.edu/funfacts/ffiles....3-5.shtml

[John Keith]

(02-15-2017 08:18 AM)grsbanks Wrote:  Interestingly enough, the HP 50g returns a question mark to the stack if you try and evaluate 0⁰.

Yes, this was done with Emu48 but that uses the original ROM of the 50g so it's reasonable to assume that you'll get the same result with a real 50g.

The physical HP50g also returns the question mark. I doubt it is a bug, rather a more modern definition of 0^0.

John

[toml_12953]

(02-15-2017 09:36 AM)Simone Cerica Wrote:  0^0 is an indeterminate form.

https://www.wolframalpha.com/input/?i=0%5E0

But it is defined as 1.

"This is highly debated. Some believe it should be defined as 1 while others think it is 0, and some believe it is undefined. There are good mathematical arguments for each, and perhaps it is most correctly considered indeterminate.
Despite this, the mathematical community is in favor of defining zero to the zero power as 1 though, at least for most purposes."

https://medium.com/i-math/the-zero-power....dyf9rnl2d

Tom L

[parisse]

0^0 is undefined. If you try to define it as 1, then some limit computations will fail.

[compsystems]

(02-13-2017 09:32 PM)Tim Wessman Wrote:  Yes. Right now it is 1.2.X (don't remember what off the top of my head)

The primary reason for things to be "out of sync" has to do with translations being out, testing not having the bandwidth to look at new features or issues, or documentation not being ready.

For example, the last major giac version bump was near the end of a long development cycle and dropping it in with ~1 month till release is simply not a good idea.

Another option is to incorporate some commands or functions that users need them, for example xCAS has EVALB, but that returns TRUE OR FALSE and not 1/0 because step by step programs that I have shown is more didactic

Sometimes the CAS returns true / false other times 1/0, I would like to see uniformity ie always true / false, the following code throws in the form of a string, but that argument is not valid as test type value

PHP Code:

#cas
  evalBool( expr1 ):=
  begin
   return( ifte( subst(expr1, '=','==' ), "true", "false") );
  end;
#end 


About incomplete documentation, TIM in a previous thread, mentions the idea that users could collaborate with the documentation and translation, as this would improve the handling and use of the calculator.

[retoa]

a^0=1 for a<>0 (1)

0^n=0 for n<>0 (2)

0^0=? that gives 1 if you apply (1) and 0 if you apply (2), both of which are correct, so 0^0 is undefined.
You can not set it as 1 per default, it would be in contradiction with (2) , nor you can set it to 0, as it would be in contradiction with (1).

Easy but interesting math...

[Joe Horn]

(02-21-2017 01:59 PM)retoa Wrote:  a^0=1 for a<>0 (1)

0^n=0 for n<>0 (2)

0^0=? that gives 1 if you apply (1) and 0 if you apply (2), both of which are correct, so 0^0 is undefined.
You can not set it as 1 per default, it would be in contradiction with (2) , nor you can set it to 0, as it would be in contradiction with (1).

Easy but interesting math...

I disagree with that reasoning, and here's why. I teach my students that "Y to the X power" means, "Start with an accumulator of 1, then multiply the accumulator by Y, X times." So 3^4 means start with an accumulator of 1, then multiply the accumulator by 3, 4 times.

0^n (where n<>0) means "Start with 1, then multiply the accumulator by 0, n times." The result will always be 0, as you said.

n^0 (where n<>0) means "Start with 1, then multiply the accumulator by n, zero times." The result will be 1, as you said.

0^0 means "Start with 1, then multiply the accumulator by 0, 0 times." The result is 1. And that's why 0^0=1, QED.

One objection to the above is, "But why do you start with 1? Isn't that kinda arbitrary?" There are two answers to this. (a) Starting with any other number gets the wrong answer for non-zero values of X and Y. (b) It's a perfect parallel to the definition of multiplication, which says "X times Y" means "Start with an accumulator of 0 (the additive identity), then add X to the accumulator Y times." Just as multiplication is defined as repeated addition, starting with the additive identity (0), so too powers are defined as repeated multiplication, starting with the multiplicative identity (1). And that's why 0.^0. returns 1 in RPL.

Another objection is, "But 2^3 doesn't mean 1*2*2*2; it simply means 2*2*2." To which I reply, So you want to define Y^X as "Start with Y, then multiply by Y, X-1 times"? No good; that leaves Y^0 undefined, unlike my definition above.

Although arguments by authority should not be as convincing as arguments by logic, here are a few just to illustrate that I'm not alone in my opinion.

"Some textbooks leave the quantity 0^0 undefined, because the functions x^0 and 0^x have different limiting values when x decreases to 0. But this is a mistake. We must define x^0=1 for all x, if the binomial theorem is to be valid when x = 0 , y = 0 , and/or x = -y . The theorem is too important to be arbitrarily restricted! By contrast, the function 0^x is quite unimportant."
-- Concrete Mathematics p.162 (R. Graham, D. Knuth, O. Patashnik)

"The number of mappings from the empty set to the empty set is obviously 1, but mathematically it's 0^0. Therefore 0^0=1." -- sci.math FAQ

"0^0=1" -- Leonhard Euler

"Zero raised to the zero power is one. Why? Because mathematicians said so. No really, it’s true." -- http://www.askamathematician.com/2010/12...-disagree/

[parisse]

I disagree, because x^y is defined as exp(y*ln(x)) if y is not an integer.

[grsbanks]

That, too, is undefined because y*ln(x) is the undefined "0 x ∞" when y=0 and x=0.

[Han]

(02-21-2017 02:42 PM)Joe Horn Wrote:  "Zero raised to the zero power is one. Why? Because mathematicians said so. No really, it’s true." -- http://www.askamathematician.com/2010/12...-disagree/

http://www.askamathematician.com/2010/12...-disagree/

From the link is a more important conclusion (IMO):

Quote:There are some further reasons why using 0^0 = 1 is preferable, but they boil down to that choice being more useful than the alternative choices, leading to simpler theorems, or feeling more “natural” to mathematicians. The choice is not “right”, it is merely nice.

I actually think the "Calculus Teacher" response explains it most clearly why 0^0 is indeterminate (from continuity point of view). The "Mathematician" response simply says that 0^0 = 1 is a choice of convenience (and not a matter of correctness).

[EdDereDdE]

Maybe we can consider it as a bit like in In quantum computing the phrase "cat state" often refers to the special entanglement of qubits wherein the qubits are in an equal superposition of all being 0 and all being 1.

I'm not good at explaining exactly what I meean, you get my point maybe.

[Han]

(02-21-2017 09:15 PM)EdDereDdE Wrote:  I'm not good at explaining exactly what I meean, you get my point maybe.

Is that a Heisenberg reference? :-)

[EdDereDdE]

(02-21-2017 09:17 PM)Han Wrote:  Is that a Heisenberg reference? :-)

Hehe might be