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Full Version: What changes were made for 2.023 of Android Free42?
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I just got the Android update to 2.023 but do not see any info on the Free42 web site yet on the changes...
Found my answer on the Google Play site....
I feel o raised to the 0 power still should be equal to 1...here is a video that shows why I say that... https://www.youtube.com/watch?time_conti...0_mi8ngNnM
You made improvements on the SOLVE function for version 2.023...If the video I linked above convinces you that 0 to the 0 pwer should be 1, then you should make another "improvement" to the original HP 42S and implement back 0^0=1. I think improvements in accuracy in this case is warrented of that video convinces you that 0^0 is 1...
Wait, I take it back...The latest version of Mathematica says 0 raised to the 0 power is "indeterminate"...So ignore the video I posted here...I go with Mathematica on the issue...Keep it as is...
Looks like RPL machines likes 0^0=1...go28 and droid48sx gave this answer...

Go41 displays DATA ERROR

My free42 for android also gave 1, maybe it's not updated yet.

Go97 gives error

15C android gives error

Android Voyage 200 gives undefined

This is very interesting...

I am not a mathematician (yet), so may I ask opinions from Gerson, Joe, Ed, Dieter, and others about this subject?

Thanks

Edited: Real 71B gives warning 0^0 and afterwards displays 1
Real 50g gives an interrogation signal "?"
Real 48gx confirms 1
Real Casio FX-880p gives S/N ERROR
(02-10-2019 02:06 AM)Jlouis Wrote: [ -> ]Looks like RPL machines likes 0^0=1...go28 and droid48sx gave this answer...

Go41 displays DATA ERROR

My free42 for android also gave 1, maybe it's not updated yet.

Go97 gives error

15C android gives error

Android Voyage 200 gives undefined

This is very interesting...

I am not a mathematician (yet), so may I ask opinions from Gerson, Joe, Ed, Dieter, and others about this subject?

Thanks

Edited: Real 71B gives warning 0^0 and afterwards displays 1
Real 50g gives an interrogation signal "?"
Real 48gx confirms 1
Real Casio FX-880p gives S/N ERROR
If you manually check for Android updates it should update to 2.023...When you do it will show the new error message...
(02-10-2019 03:04 AM)zeno333 Wrote: [ -> ]If you manually check for Android updates it should update to 2.023...When you do it will show the new error message...

Thanks, I did the update and it shows the error now.
The source archive contains a HISTORY file with the changelog:

Code:
2019-02-09: release 2.0.23 * SOLVE now tries harder when secant extrapolation gets stuck due to the secant   being excessively steep. * 0^0 now returns Invalid Data, not 1, for all combinations of real and complex   arguments. The rationale for returning 1 was mathematically questionable, and   Invalid Data is what the real HP-42S returns. * iPhone version: added basic iPhone X support.

If you are interested in the actual code changes, you can have a look at my Free42 release mirror repository on GitHub (core/gtk only because I had no other history when I started collecting the releases in a repository).
That information is readily available on Thomas Okken's website at this address:

http://thomasokken.com/free42/history.html
I was more interested in know why some HP, like 28s and 48 calculators shows 1 as an answer.
(02-10-2019 02:06 AM)Jlouis Wrote: [ -> ]Real 50g gives an interrogation signal "?"

It returns "?" in exact mode and "1" in approximate mode.
(02-10-2019 02:06 AM)Jlouis Wrote: [ -> ]<snip>
This is very interesting...

I am not a mathematician (yet), so may I ask opinions from Gerson, Joe, Ed, Dieter, and others about this subject?
I am not a mathematician either! I think that it is correct to say that $$0^0$$ is undefined. Consider the function
$$f(x)=x^{-1/\ln(x)}.$$
As $$x\to 0^+$$ both $$x$$ and the power it is raised to tend to zero, so it is tempting to write $$f(0)$$ as $$0^0$$. But for all positive values of $$x$$, $$f(x)$$ is equal to $$1/e$$, so
$$\lim_{x\to 0^+}f(x)={1\over e}.$$
So to give a value to the expression $$0^0$$ it is necessary to know how the limit was reached. Because of this the expression itself is indeterminate.

Nigel (UK)
(02-10-2019 10:36 AM)grsbanks Wrote: [ -> ]That information is readily available on Thomas Okken's website at this address:

http://thomasokken.com/free42/history.html

It was not there yet when I made my posting asking about it...I was put up there later than usual..
(02-10-2019 08:35 AM)SammysHP Wrote: [ -> ]The source archive contains a HISTORY file with the changelog:

Code:
2019-02-09: release 2.0.23 * SOLVE now tries harder when secant extrapolation gets stuck due to the secant   being excessively steep. * 0^0 now returns Invalid Data, not 1, for all combinations of real and complex   arguments. The rationale for returning 1 was mathematically questionable, and   Invalid Data is what the real HP-42S returns. * iPhone version: added basic iPhone X support.

If you are interested in the actual code changes, you can have a look at my Free42 release mirror repository on GitHub (core/gtk only because I had no other history when I started collecting the releases in a repository).

That was added afrter I asked my question...
Heh. Since it always takes a while for the Android and iOS versions to appear in their respective on-line stores, I decided to upload them before updating my own web site for a change. And guess what, this time the Android version appeared in the Play Store INSTANTLY. Oh well!

You still could have checked the release notes in the Play Store though. Or waited a few minutes for my web site to get updated.
(02-10-2019 08:17 PM)Nigel (UK) Wrote: [ -> ]
(02-10-2019 02:06 AM)Jlouis Wrote: [ -> ]<snip>
This is very interesting...

I am not a mathematician (yet), so may I ask opinions from Gerson, Joe, Ed, Dieter, and others about this subject?
I am not a mathematician either! I think that it is correct to say that $$0^0$$ is undefined. Consider the function
$$f(x)=x^{-1/\ln(x)}.$$
As $$x\to 0^+$$ both $$x$$ and the power it is raised to tend to zero, so it is tempting to write $$f(0)$$ as $$0^0$$. But for all positive values of $$x$$, $$f(x)$$ is equal to $$1/e$$, so
$$\lim_{x\to 0^+}f(x)={1\over e}.$$
So to give a value to the expression $$0^0$$ it is necessary to know how the limit was reached. Because of this the expression itself is indeterminate.

Nigel (UK)

Thanks Nigel
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