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...?

 

[JANm]

The real part of x^x= Re(x^x) = x^x - Im(x^x).
Function: If Re(x^x)=x^x then function = x else 0?

 

[soandos]

sorry i was really unclear.
what i meant was that i think that Im[x^x]=0 iff x is an integer, and if not, then it is some other value (assuming that domain of x in this case is x<0)
i thought that this is pretty much intuitive as you are taking a non-odd root of a negative number.
the table was just to check the assumptions.
with respect to

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

i think that all of them are true, and about the iverson bracket, which i think can be strengthened by saying that it is equal to 1 exactly at the integers.

 

[CRGreathouse]

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}

i think that all of them are true, and about the iverson bracket, which i think can be strengthened by saying that it is equal to 1 exactly at the integers.

I think that all four are false.

 

[soandos]

how could that be? is there any other way that one can take a non-odd root and end up with a number of the same sign if the domain is restricted to negative numbers?

 

[CRGreathouse]

how could that be? is there any other way that one can take a non-odd root and end up with a number of the same sign if the domain is restricted to negative numbers?

You didn't say that the domain was restricted to negative numbers.

On the negative numbers, I'd think that the function in #2 would be discontinuous precisely at the integers.

For #3 and #4, what about noninteger rational numbers with odd denominators and even numerators?

I'm not entirely sure about #1 on the negative reals, but it might be true. The others look false even on the negative reals.

 

[soandos]

To be honest, i am not sure, and that is why i ran the mathimatica table. the only numbers that showed up were the integers. even with odd denominators and even numerators, there are still irrational.can't prove it, just going off mathematica here, in fact is there a possibility that it is wrong? as a test case it says that (-1)^(1/3) has an imaginary part, though to the best of my knowledge, it is entirely real. could it be that because there is more than one solution to a given expression that it just picks the imaginary ones?

 

[CRGreathouse]

can't prove it, just going off mathematica here, in fact is there a possibility that it is wrong?

It's hard to imagine a numerical calculation of an irrational number being right.

For me, (-1)^(1/3) = -1 since (-1)^3 = -1 and I take the principal root.

 

[JANm]

' Made a graph of y=abs(x)^x.
I thought that it would be much more jumpy. It looks like some third degree with a relative maximum at (-,3; 1,4) and a relative minimum at (0,2;0,8). The difference with the third degree is the horizontal assymptote y=0 for small x.
So for positive y there can be three, two or one solution.
If there is a solution for negative x, x has to be a even rational.
For uneven rationals there are solutions with y>-rel max_y=-1,4 approximately.

 

[soandos]

I plotted the same thing that you did, and it only intersected the x-axis in one place, but
ln(abs(x)^x) gives the intersections with the x-axis at -1,1 and with point symmetry at the origin. i am not sure, but i think that the original (with out the log) did not have this symmetry (just by looking at the graphs). how is that possible?

 

[JANm]

I plotted the same thing that you did, and it only intersected the x-axis in one place, but
ln(abs(x)^x) gives the intersections with the x-axis at -1,1 and with point symmetry at the origin. i am not sure, but i think that the original (with out the log) did not have this symmetry (just by looking at the graphs). how is that possible?

How can this intersection take place? I have an horizontal assymptote at the line y=0.

From the derivative =0 I calculated the x-values of the local/relative extrema.
x_relmax=-e^-1 and x_relmin=e^-1. Should I call it local or relative I am not sure of...

I also found a formula closer to the end result:

y=sgn(x)*abs(x)^x.

 

[soandos]

that gets an entirely differnt plot, where the negative is very messed up

 

[JANm]

Dear Soandos
Don't know why you say that. This function is well defined. We still need swap (x,y) after it is well understood, and that would give two valuedness for x as a function of y. My mathlab gave a good drawing exept from 0 to 1/e it didnot understand the heaviside jump it had to make. We have an exponential function of a positive value and multiply it with the sign function to make it negative for x<0. What the value is at x=0 we still have to decide, but more important this solution has values which work and other values that don't work for the asked function in the start of this thread: x =y^y .
Assumed was that x negative had no solutions at all, but I claim that for y=odd rational this function is well defined and that the sign thing is not important there...
A uneven root of a negative number gives a negative number so to say...
So after the former function is properly inverted (by selecting an area of one valuedness)
we have to submit all the values where y is not a uneven rational.

Measure theory says that rational numbers are countable (so certainly the odd ones) and therefore a null set from all the possible values arosen. Around each odd rational a interval can be taken in such a manner that the sum of all the intervals converges and moreover can be made as small a one wishes.

A null set seems small but another fact is that rational lie close to any number on the value-line. It needs to be proven that that also counts for odd rationals only, but I see no problem in that proof. So let's call that the thesis of JANm. This means that for any given negative number x and a presision factor delta there can be found an y (odd rational) with abs(x-y^y) < delta.

That is as far as we got thusfar...
greetings

 

[remix220]

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

y must represent a number therefor it cannot be multiplied or divided, no?
I'm bad in math!
I could see where x = y, but I don't see where you get a answer from exponent, if you don't know what y represents. Hmmmm...
Where did you come a crossed this problem?

http://www.simonsingh.net/images/465.jpg

 

[soandos]

I do not understand your first two lines.
logs can be used to reduce exponential eqations, like 4^x=250 (x=Log[4]/Log[250]), and the
x th roots of both sides are not taken.
the first time that I saw something like it was solving for the period in a compound interest forumla.

 

[rscosa]

Hi!
it is impossible to find a explicit expression for the solutions of you equation because that means that you can solve the integral x^x which is impossible to express in tems of elementary functions (read wikipedia about this topic) so the answer is: You only can get numerical approximation of you solution if it exists.