Unsolvable Exponential: Solving x = y^y

[soandos]

Is it possible to solve x = y^y for y?

 

[CRGreathouse]

There's a closed form in Lambert's W, or you could use Newton's method. (Actually, depending on how fast you can evaluate logarithms and exponentials, the secant method is probably faster.)

 

[SW VandeCarr]

I'm sure I'm missing something here, but isn't the solution simply y equals the yth root of x?.

 

[mathman]

I'm sure I'm missing something here, but isn't the solution simply y equals the yth root of x?.

Since y is unknown, the yth root of x is unknown.

 

[soandos]

Just some clarification:

is there a way to get an exact value, or just a numeric approximation?
also, i do not really understand the Lambert W function. the link just says that it is the inverse function, but not how.

also, how is the following true (or is it false?):
Sqrt[x]^Sqrt[x]==1/x^Sqrt[x]

 

[WarPhalange]

also, how is the following true (or is it false?):
Sqrt[x]^Sqrt[x]==1/x^Sqrt[x]

False. 1/x = x^(-1), so you'd have [x^(1/2)]^[x^(1/2)] = [x^(-1)]^[x^(1/2)]

Also, a quick numerical check can show you it's false. Take 4, for example. Square root 4 = 2, so you have 2^2 = (1/4)^2, so that's false.

 

[CRGreathouse]

is there a way to get an exact value, or just a numeric approximation?

Is there a way to get an exact value for $\sqrt 2$, or just a numeric approximation? Because Newton's method is used for square roots as well.

This is a philosophical question more than a mathematical one, I suspect.

 

[Irrational]

*erroneous post*

 

[soandos]

to get an exact answer, i think the only real way is with a continuous fraction. could someone please explain to me exactly what Newtons method is?

 

[CRGreathouse]

http://www.google.com/search?q=Newton's method

 

[soandos]

I am not sure that i understand. if the root is negative, then it will never work, as the derivatives of the function are x^x (1 + \text {Log}[x]) for the first derivative, and x^{-1 + x} + x^x (1 + \text {Log}[x])^2 for the second derivative. i believe that both of these are non-continuous at any negative. is this right, or am is making a mistake?

 

[CRGreathouse]

I am not sure that i understand. if the root is negative, then it will never work, as the derivatives of the function are x^x (1 + \text {Log}[x]) for the first derivative, and x^{-1 + x} + x^x (1 + \text {Log}[x])^2 for the second derivative. i believe that both of these are non-continuous at any negative. is this right, or am is making a mistake?

You didn't sy you were looking at x < 0. You'll need to use the secant method (as I recommended in my first post!) or something similar instead of Newton in that case. Otherwise you could use Newton's method or Halley's method to compute W and get the answer that way.

I generally recommend Newton's method only because it is what is taught. The secant method is often better. Of course you can always fall back onto bisection if you have trouble... or don't want to use something complex like regula falsi.

 

[soandos]

i am not sure that i understand. i thought from what i read that the secant method has the same kind of problem that Newton's method does, namely that if there is no continuous second derivative, then there is no guarantee that it will converge.
also, from what i understand of Halley's function, is that it is dependent on Newtons method, so for -x it will also fail. in fact, i believe that all of these methods are for continuous functions only.
is there a non-iterative way of doing this, or at least one that will work for negative x?

 

[CRGreathouse]

i am not sure that i understand. i thought from what i read that the secant method has the same kind of problem that Newton's method does, namely that if there is no continuous second derivative, then there is no guarantee that it will converge.
also, from what i understand of Halley's function, is that it is dependent on Newtons method, so for -x it will also fail.

I suggested Halley's method for solving W, not x^x.

in fact, i believe that all of these methods are for continuous functions only.

Your function is continuous, what's the problem?

 

[soandos]

i thought that it is not continious at all negative x.

 

[CRGreathouse]

i thought that it is not continious at all negative x.

Well, just take absolute values and find the magnitude, then see if a solution is possible with a sign change.

 

[soandos]

how could that work if the magnitude of the absolute value is a different expresion? even with a sign change, the absolute value is changing more than that.

 

[soandos]

also, how could someone compute W using halleys mehtod?

 

[CRGreathouse]

also, how could someone compute W using halleys mehtod?

http://keithbriggs.info/W-ology.html

 

[soandos]

thanks. but what about the previous question?

 

[CRGreathouse]

thanks. but what about the previous question?

I've given lots of ways to solve the problem. Pick one and go with it.

 

[JANm]

Let us try the transformation: Switch (x,y), and after the slolution of:

y=x^x then switch(x,y) again.

is that a good start for the Newton(-Raphson) method?

 

[soandos]

that is the whole problem to begin with, finding the solution.

 

[JANm]

Beste Soandos,

y=x^x=exp(x*ln(x)), dy/dx=d(x*ln(x))/dx*exp(x*ln(x)) by the chain rule.

d(x*ln(x)/dx = ln(x)+x/x by summation rule, under the condition: x<>0

dy/dx=(ln(x)+1)*x^x under the condition: x<>0.

Fill this in in the Newton method and solve.

greetings

 

[soandos]

and how does this help with negative x?
it is just what we had before.

 

[JANm]

Sorry added in Yesterday, for negative x the ln(x) becomes complex but is calcuable.

 

[soandos]

as you can see from earlier posts, this has arelady been dealth with. Newtons does not work where the derivative is non-continious. see earlier posts.

 

[JANm]

OK. If it is true that x^x can be solved than x^-x can also be solved.
For at least all x even there is then the solution x^x=x^-x.
This is interesting for I always wondered what rationals are even.
Compare this to the fact that gamma(x) is a extension of the factorial x! for a unwhole number.

Greetings

 

[soandos]

not what i said. i meant (-|x|)^(-|x|)==x^x. that is the case that i was talking about. in that case, there is no continuous derivative. how is gamma related?

 

[JANm]

Gamma(n+1)=1*2*3...(n-1)*n=n!
FI gamma(3/2)=pi^(1/2)/2=(1/2)!=0,8862269254,
The recurrence equation n!=n*(n-1)! becomes for gamma:
.........gamma(n+1)=n *gamma(n).

Now about solving x^x, with x<0. Not all are solvable at the same tim but for even x...

3/2 is odd, for this value (-3/2)^(-3/2)<>(3/2)^(-3/2)
I suppose q=2/3 is even (not sure!), and (-2/3)^(-2/3)=(2/3)^(-2/3),
but 2, 4, 10354 are even, so (-10354)^-10354=10354^10354, and for even negative values the "exponential is solvable" I meant.

greetings

 

[soandos]

and for negative irrational?

 

[JANm]

Yes the negative irrationals, so the algebraic numbers, which are solutions of polynomal equations and the trancendentals like -e and -pi will be the numbers for which the discontinuity of x^x arises. Whether one can say of irrationals that they are even or not I don't know.

 

[soandos]

I think (based on a a mathematica Table[If[Re[x^x]==x^x,x,0],{x,-200,-.01,.01}]
but am not sure that the only time that it will be continuous is if the number is an integer.
Can someone find a counter-example or prove this?

 

[Ed Aboud]

Where would an equation like this arise?

 

[CRGreathouse]

I think (based on a a mathematica Table[If[Re[x^x]==x^x,x,0],{x,-200,-.01,.01}]
but am not sure that the only time that it will be continuous is if the number is an integer.
Can someone find a counter-example or prove this?

I really don't know what you mean. *What* will be continuous when *what* is an integer?

1. $$\Re(x^x)-x^x$$ is continuous exactly at the integers
2. $$[\Re(x^x)=x^x]$$ is continuous exactly at the integers, where [] is the Iverson Bracket
3. $$\Re(x^x)=x^x$$ iff $x\in\mathbb{Z}$
4. $$\Re(x^x)=x^x$$ implies $x\in\mathbb{Z}$
5. Something else...?