Conference rusure::math

    			X = sqrt (3)
    			Y = 3X
    
    John

In general,

		( x ^ (1/(x-1)) )		    ( x ^ (x/(x-1)) )
( x ^ (x/(x-1)) )		  = ( x ^ (1/(x-1)) ) 

as is easily seen by taking the log base x of both sides.


Note that we could approach the problem by considering the expression
X^(1/X), since X^Y = Y^X is equivalent to X^(1/X) = Y^(1/Y).
The inverse of this function is multivalued (actually, it has exactly
two values) in the range X > 1, except at X = e.

For example, in .2:

1/(x-1)		x-1	x	X=x^(1/(x-1))	Y=xX	X^Y (= Y^X)
	
1		1	2	2.0		4.0	16.0
2		1/2	3/2	2.25            3.38	15.44
3		1/3	4/3	2.37		3.16	15.30
4		1/4	5/4	2.44		3.05	15.24
5		1/5	6/5	2.49		2.99	15.21
.
.
100		.01	1.01	2.70		2.73	15.15
.
.
big		little	1+	e-		e+	(e^e)+

Each X is (rational) ^ (integer) is rational; each Y is (rational) *
(rational) is rational.

Are there any others with both X and Y rational?

Can we prove that there are none?

If we let X = (1+1/b)^b and Y = (1+1/b)^(b+1), then X^Y = Y^X.

If b is an integer, then both X any Y are rational (although X^Y
is rational in only one case).  I suspect that this may be the
only general formula that gives rational solutions to X^Y = Y^X.

For example, if X = (5/4)^4 and Y = (5/4)^5, then X^Y = Y^X (!?).

					- Gilbert

P.S.  Note that the form in 512.2 gives all solutions to X^Y = Y^X,
where X and Y are positive numbers.

    There's something fishy with .3.  Here's what BASIC calculates:
--------------------------------
basic

VAX BASIC V2.4

Ready
print 2.44^3.05, 3.05^2.44
 15.1893       15.1947 
Ready
--------------------------------
    
    There are several stinkies here.  First of all, is BASIC's
    power computation THAT inaccurate that we see such discrepancy
    in the SECOND decimal place for two values that should be equal?
    
    Secondly, .3 says the answer should be 15.24.  Is BASIC
    inaccurate, or is .3 inaccurate ?  Or are one of them just plain
    wrong ?
    
    /Eric

re .6:  The "problem" is, the two numbers are only given to 3 significant
figures, so you should only expect 3 sig. figs. in the result, or 15.2 in each
of the numbers: the result given, and the 2 numbers you calculated.  If you
compute the 2 numbers to better accuracy, you will get better accuracy in the
result.

Betcha you ain't an engineer.

-Mike

    Whoa!  Wait a minute !  The numbers given were supposedly examples
    of RATIONAL solutions.  Now you tell me they're only accurate to
    3 figures ?  Why weren't they presented to INFINITE precision,
    which is generally quite easy for rational numbers.
    
    BTW, yes, I'm a senior softwarew engineer, been at DEC for 12
    years.
    
    /Eric

re .8:  "Rational" does not mean "exact".  If you think so, write out 1/7
to the last decimal place.  The numbers given in .3 for X and Y weren't given
exactly, so when you performed the Y**X and X**Y functions, the results were in
error.

I ran the same problem using FORTRAN Double Precision and got the following:

X=2.44140625
Y=3.051758125
X**Y=Y**X=15.23995052905966
X**Y-Y**X=0.0 (exactly)

Since X isn't 2.44 and y isn't 3.05, you got different results from the given
15.24 (which isn't exact either) and X**Y wasn't equal to Y**X

>    BTW, yes, I'm a senior softwarew engineer, been at DEC for 12
>    years.

Ahh, a "phony" (i.e. software) engineer!  (So am I) :-)  Really, I mentioned
engineering since I remember being drilled over and over in college getting a
BSEE not to use more digits in your final result than you know in one of your
measurements. I got this in physics and chemistry courses, but the engineering
courses made sure you knew this! That is to say that you measure the radius of
a circle is 1.3 units, but can't measure it any closer than .05 units, the
computed area should be given as 5.3, even though you may know pi to 1,000,000
decimal places.

-Mike

>    re .8:  "Rational" does not mean "exact".  If you think so, write out 1/7
>to the last decimal place.
    
    		Forget decimal place.  I'll write out 1/7 in infinite
    		precision:
    
    				1 / 7
    
    		There.  I suggest the .3 table be presented that way.
    		If 3.05 isn't supposed to be 3.05, put in the correct
    		number as a fraction.  This is math, not scientific roundoff
    		experimentation lab record observation psychology
    		mishmash.
    
/Eric

O.K.  I did it for you.  Here's a table with EXACT values for X and
    Y.
    
1/(x-1)		x-1	x	X=x^(1/(x-1))	Y=xX	X^Y (= Y^X)
	
1		1	2	2.0		4.0	16.0
2		1/2	3/2	2.25           27/8	15.44
3		1/3	4/3	64/27	      256/81	15.30
4		1/4	5/4	625/256	     3125/1024	15.24
5		1/5	6/5	7776/3125   46656/15625	15.21
.
.
.
.
big		little	1+	e-		e+	(e^e)+

    /Eric

    Re .2 and .5:
    
    Two questions:
    
    o		In .2, you've found two functions f1(x) and f2(x)
    		such that f1(x)^f2(x)=f2(x)^f1(x).  How can this be
    		derived algebraically from x^y=y^x, or is it just
    		"clever" substitution ?
    
    o		In .5 you claim that .2 gives ALL the solutions.  How
    		can we prove that some other f1, f2 does not exist
    		that also work ?
    
    /Eric

re .-1
	Note 512.2 hints why this gives all solutions.  Consider the
	function f(X) = X^(1/X).  A graph of this is roughly shaped like:

	       .
	      . .
	1    .   ................
	    .      
	0...
	 0     e

	Now consider the inverse function: given f(X), find X.  The
	problem is equivalent to finding two Xs for the f(X).  We see
	that this is possible only if 1 < f(X) < e^(1/e).  The equations
	for X and Y given in 512.2 yiled all the solutions in this range;
	no other equations are needed.

Let X = z^(1/(z-1)), and Y = z^(z/(z-1)), both rational.  Now z must
be rational, since Y/X = z.  Let z = a/b, where a and b are integers,
with gcd(a,b) = 1 (i.e., a/b is in 'lowest terms'), and a > b.

Then, X = (a/b)^(b/(a-b)) and Y = (a/b)^(a/(a-b)).  Now for X to be
rational, we need (a^b) to be an (a-b) power of some integer.  Since
gcd(b,a-b) = 1, this is equivalent to requiring that a = c^(a-b) for
some integer c.  Similarly, b = d^(a-b).

We will show that a-b=1.  Assume that a-b>=2.  Now c^(a-b)-d^(a-b),
when constrained by c>d>=1, is minimized when c=2 and d=1, so we know
that a-b = c^(a-b)-d^(a-b) >= 2^(a-b)-1.  But 2^n >= n+2 if n>=2, so
we reach the absurdity that a-b >= 2^(a-b)-1 >= (a-b)+2-1 = a-b+1.
Thus, the assumption that a-b>=2 is false, and so a-b=1.

So, X = ((b+1)/b)^(b) = (1+1/b)^b and Y = ((b+1)/b)^(b+1) = (1+1/b)^(b+1),
with integral b, completely describe rational solutions to the problem.


Finally, let us consider which of these solutions yield a rational X^Y.
We have X^Y = ((b+1)/b) ^ ((b+1)^(b+1)/b^b).  For this to be rational,
((b+1)/b) must be a b-power of some rational, or equivalently, (b+1) = k^b,
for some integer k.  But the only solutions to this are b=0 and {b=1,k=2},
and b=0 is not a valid solution, so we are left with X=2, Y=4, and X^Y=16
as the only solution with rational X, Y and X^Y.

					- Gilbert

This is slightly out of the subject:

Prove that  x**y + y**x  is never < 1 .

I saw this problem a few years ago. I don't have the solution, and I hope
I didn't mis-quote the original problem...

Enjoy,
Nitsan

re .-1:  Prove that x^y + y^x >= 1, for x and y > 0.

If x >= 1 and y >= 1, then x^y + y^x >= 2.  So we need only consider
the range 0 < x,y < 1.


The function x^y + y^x is continuous in the range x,y > 0, so its
minimum must either occur by a boundary of the region 0 < x,y < 1,
or at a local minima within that region.

We can find a local minima by setting derivatives to zero:

	d(x^y+y^x)/dx = y x^(y-1) + y^x log y = 0
	d(x^y+y^x)/dy = x y^(x-1) + x^y log x = 0
Or,
	y x^y + x y^x log y = 0
	x y^x + y x^y log x = 0
Or,
	x y^x - x y^x log x log y = 0

Thus, either x y^x = 0 (which can't happen), or log x log y = 1.


If 0 < y < 1, then as x approaches zero, x^y approaches zero (from above),
and y^x approaches 1 (from below).  Thus, we may be close here.


Has anyone gotten further?

Prove that x^y + y^x >= 1, for x and y > 0.

> If x >= 1 and y >= 1, then x^y + y^x >= 2.  So we need only consider
> the range 0 < x,y < 1.

If x >= 1 OR y >= 1, then x^y>1 OR y^x>1 hence x^y + y^x >= 1.
So we need only consider the range 0 < x,y < 1.
                         
    Your analysis left out the case where 0<x<1 and 1<=y (and vice versa).
          
    
    /Bevin

re .-1:  Of course.

Looking at the problem further, would it be acceptable to argue as follows:

If z(x,y) = x^y + y^x, then z(0,y) = 1 (except possibly at y = 0),
z(x,0) = 1 (except possibly at x = 0), z(1,y) >= 1, z(x,1) >= 1, and
z(0,0) can be (consistently!) defined as 1.

In the region 0 <= x,y <= 1 the function z(x,y) is continuous, at the
boundary z(x,y) is >=1, and this region contains no local minima
(this last part is not yet proven).

Putting this together with the fact that z(x,y) is >= 1 if x >=1 or
y >=1 gives the result: that z(x,y) >= 1 for x,y >= 0.

The function z(x, y) = x^y + y^x is not continuous at (0, 0): if you approach
this point along either axis the limit is 1, if you approach along the line x =
y the limit is 2.

However, if you hold y constant and consider z as a function of x, the resulting
function is continuous for x >= 0 and differentiable for x > 0.  The derivative
is yx^(y-1) + (ln y)y^x, which is always > 0; and the limit of z(x, y) as x
approaches 0 is 1, which proves the result.