Solving x^y=y^x

[sasa]

This guy have quite interesting samples to play with...
Is it possible to solve this one on Prime?

[Albert Chan]

solve(subst(x^y=y^x, y=t*x), x) ==> [ t^(1/(t-1)) ]

So, [x, y] = [ t^(1/(t-1)), t^(t/(t-1)) ]

But, I may have discovered a XCas bug ...

solve([x^y=y^x, y=t*x], x) ==> *crash*

[Dieter]

(10-24-2018 06:06 PM)sasa Wrote:  This guy have quite interesting samples to play with...
Is it possible to solve this one on Prime?

I didn't watch the video, but this equation can be directly solved by means of the Lambert W function.
The WP34s, for example, handles this easily:

LN
RCL/ L
CHS
Wp
RCL/ L

Enter x, get y and vice versa.

Replace Wp with Wm (the other branch of Lambert's W function) and get the second solution. For x=e both solutions are equal (x=y=e)

X=2 => Y=2
X=3 => Y=2,47805268...
X=4 => Y=2

So, does the Prime feature the W function?

Dieter

[sasa]

(10-24-2018 07:21 PM)Albert Chan Wrote:  solve(subst(x^y=y^x, y=t*x), x) ==> [ t^(1/(t-1)) ]

Yes, but without substitution hint, it seems it cannot conclude that is needed.
Perhaps with another function?

[John Keith]

(10-24-2018 07:37 PM)Dieter Wrote:  So, does the Prime feature the W function?

No, but it should IMHO.

[sasa]

(10-24-2018 07:37 PM)Dieter Wrote:  I didn't watch the video...

Two limitations from the video (few seconds): x<>y and x,y where real numbers.

[Albert Chan]

(10-24-2018 07:37 PM)Dieter Wrote:  The WP34s, for example, handles this easily:

LN
RCL/ L
CHS
Wp
RCL/ L

Enter x, get y and vice versa.

Thanks, Dieter. I was curious how above get derived ...

Found a very close solution for infinite tetration c = x^c, or ln(c)/c = ln(x).
From the video (~ 7:00), using basic algebra, he got c = W(-ln(x)) / (-ln(x))

For x^y = y^x, ln(y)/y = ln(x)/x, so just substitute above: c=y, ln(x)=ln(x)/x
--> y = W(-ln(x)/x) / (-ln(x)/x)

[ijabbott]

(10-24-2018 09:58 PM)Albert Chan Wrote:  Found a very close solution for infinite tetration c = x^c, or ln(c)/c = ln(x).
From the video (~ 7:00), using basic algebra, he got c = W(-ln(x)) / (-ln(x))

For x^y = y^x, ln(y)/y = ln(x)/x, so just substitute above: c=y, ln(x)=ln(x)/x
--> y = W(-ln(x)/x) / (-ln(x)/x)

I like his series of videos on geometric algebra - something that isn't currently handled very well by our favourite, hand-held, symbolic calculators.

[Erwin]

(10-24-2018 10:35 PM)ijabbott Wrote:  

(10-24-2018 09:58 PM)Albert Chan Wrote:  Found a very close solution for infinite tetration c = x^c, or ln(c)/c = ln(x).
From the video (~ 7:00), using basic algebra, he got c = W(-ln(x)) / (-ln(x))

For x^y = y^x, ln(y)/y = ln(x)/x, so just substitute above: c=y, ln(x)=ln(x)/x
--> y = W(-ln(x)/x) / (-ln(x)/x)

I like his series of videos on geometric algebra - something that isn't currently handled very well by our favourite, hand-held, symbolic calculators.

Hallo,

The question in the video for Infinite Tetration was the point for me to put out my old calculator. So I can only show a solution for Infinite Tetration on the HP71b to see if the number converges.
Define the W-Funktion y=x*exp(x) and put the formula on minute 12:10 in the keyboard. Looks like like this:

Definiton of the W-function in BASIC
10 DEF FNW(X) = FNROOT(0,10,FVAR*EXP(FVAR)-X)

and then his example with 1.4 on the keyboard calling there defined W-function:
FNW((-LN(1.4)))/(-LN(1.4))
Result: 1.88666330624

and then his example with SQR(2) on the keyboard calling there defined W-function:
FNW((-LN(SQR(2)))/(-LN(SQR(2)))
Result: 1.99999999998 not exactly 2 but this seems is a limitation on the more then 30 year old calculator.

The possibility for defining a function and using it straight from the keyboard was a revolution in this times.

best regards
Erwin

[Albert Chan]

(10-26-2018 04:54 PM)Erwin Wrote:  Definiton of the W-function in BASIC
10 DEF FNW(X) = FNROOT(0,10,FVAR*EXP(FVAR)-X)

and then his example with 1.4 on the keyboard calling there defined W-function:
FNW((-LN(1.4)))/(-LN(1.4))
Result: 1.88666330624

and then his example with SQR(2) on the keyboard calling there defined W-function:
FNW((-LN(SQR(2)))/(-LN(SQR(2)))
Result: 1.99999999998 not exactly 2 but this seems is a limitation on the more then 30 year old calculator.

Unless W function is built-in, it might be easier solve directly for infinite tetration.
So, for any x, solve for y, such that x^y = y

Below solved with my Casio FX-115MS solver, using x as initial guess for y:

For x=1.4, y = 1.886663306
For x= √2, y = 2
For x=1.5, y does not converge

Last x > e^(1/e), thus diverged. see Exponential Reiterated

[CyberAngel]

(10-24-2018 08:23 PM)John Keith Wrote:  

(10-24-2018 07:37 PM)Dieter Wrote:  So, does the Prime feature the W function?

No, but it should IMHO.

If it's not in KhiCAS then it's not coming.
Completeness compared to other CAS
demands a full implementation of Lambert W
including (i)laplace, solve, int, diff eq's, etc.
I hope there are no GPL problems...

[Erwin]

(10-28-2018 08:21 PM)Albert Chan Wrote:  Unless W function is built-in, it might be easier solve directly for infinite tetration.[...]

Of course you are right - its much easier in your case. The HP71b has the possibility to define systemwide new functions and use it in Calculator and BASIC mode (of course FORTH too) - that was the reason why I grab my calculator to try it :-)

[Tim Wessman]

(10-28-2018 08:51 PM)Erwin Wrote:  possibility to define systemwide new functions and use it in Calculator and BASIC mode (of course FORTH too) - that was the reason why I grab my calculator to try it :-)

Prime does the same thing - either through a program or a simple define a function interface. The issue purely lies in tying that into the inner workings of the CAS to be able to recognize when it should use that function in simplifications, integrals, etc. They really are two separate things.