1. ## Find derivative!

The task is defined the following way:
"We have the function $f(x) = x^{x^{x^{\ldots}}}$
Find it's derivative!"

There is no more additional information about f(x) so, obviously we have to find its natural domain, and the set where it is differetiable.

I am pretty sure that such set, where f(x) is differetiable is the interval (0, 1] (not sure if it is the maximal).

I have found a formula, for the derivative, but it contains $x^{x^{x^{\ldots}}}$ terms (founding the inverse function leads to that formula). The problem is: can you find the finite formula for the derivative (does it exist???). Or maybe you can find the derivative written as an infinite series?

And in the end, one thing. Please, do not confuse the function with $\left(\left(\left(x^x\right)^x\right)^x\right)^{.. .}$ like I did in the first look. It is the function defined that way:

$
f_1 (x) = x
$

$
f_{n+1} (x) = x^{f_n (x)}
$

$
f (x) = \lim_{n \rightarrow \infty} f_n (x)
$

2. Have you tried taking the natural logarithm (f(x) = exp(ln(f(x))) to use the power property of the logarithm? Try to do it for the first few cases, can you detect a pattern and extend it to the general case?

3. It seems to be connected to the Ackermann Function
problem is that your function is an "infinite" ackermann function thus I do not know if this is what you are looking for.

4. Originally Posted by TD!
Have you tried taking the natural logarithm (f(x) = exp(ln(f(x))) to use the power property of the logarithm? Try to do it for the first few cases, can you detect a pattern and extend it to the general case?
Frankly speaking I haven't tried taking the natural logarithm of this function, and calculate it's derivative (I doubt if it will lead me to the solution, the formulas will go more and more complicated). Maybe I should.

There is also another problem with this attitude.... Shouldn't be the sequence of functions uniformly convergent to have the property that the limit of the derivatives is the derivative of the limit? I can prove that it is convergent on the (0, 1] interval, but is it uniformly convergent (I belive yes!)?

But there is another thing. The function can be described this way:

$
y = x^y
$

And from that, you can determine the inverse function:
$
x = e^{\frac{\ln y}{y}}
$

From that I can prove uniquiness (of course I should prove convergence first)...

Originally Posted by ThePerfectHacker
It seems to be connected to the Ackermann Function
problem is that your function is an "infinite" ackermann function thus I do not know if this is what you are looking for.
Hm... Ackermann function is a discreete thus it has no derivative. But, maybe it is some trail...

5. Start out with what you wrote.

$y=x^y$

Now take the natural log of both sides, and differentiate.

$ln(y)=yln(x)$

6. ## maybe next step:)

so...
I wonder if this method is valid , please write what do you think about it:
$
lny= ylnx
x=e^(lny)/y
$

$
dx/dy = (1/y^2 - (lny)/y^2)*e^(ln/y) =$

$
= ((1-lny)*e^(lny/y))/y^2
$

so, now let's find dy/dx
$
dy/dx = y^2/((1-lny)*e^(lny/y)) =
$

finally =
$(x^x^x^x^....)^2/((1-lnx^x^x^x....)*x)

$

*). xxxxx what you can see means x power x power x power x.....

What do you think?????

7. Originally Posted by Jameson
Start out with what you wrote.

$y=x^y$

Now take the natural log of both sides, and differentiate.

$ln(y)=yln(x)$

Yes, and you will get what harold84 did.

$
f'(x) = \frac{(x^{x^{x^{\ldots}}})^2}{x \left(1-\ln \left( (x^{x^{x^{\ldots}}} \right) \right) }
$

But, look at my first post. The problem is not if you can find SOME formula for the f(x) derivative. You are to find FINITE formula for the derivative, or prove that there is no FINITE formula for that (if you will find it as a infinite series, or infinite product I will also be happy ).

8. ## Hmm...

Yes, as I can see noone is able to solve it. If I find a solution I will post it here.

9. I have this thought: isn't $f(x) = x^{x^{x^{\ldots}}}$ function equal to
$f\begin{cases} \rightarrow \infty \mbox{ for x}>1 \\ \mbox{1 for x}\in (0,1> \\ \mbox{for x}<0 \mbox{ does not exist} \end{cases}$
Function is differentiable in interval $x\in(0,1>$, where $f'(x)=0$
Well, that's my point, what do you think of it?

Do you think, in this case the function is differentable in interval $x\in (1,\infty)$(it's somehow constant here).

10. How did you find this ooold topic :P ... um.... nevermind

I guess you are wrong because

$
\left (\left (\left( x^x \right)^x \right)^x \right)^{...} \neq x^{x^{x^{...}}}
$

$x^{x^{x^{...}}}$ is a function f(x) which satisfies equation

$x^{f(x)} = f(x)$

Hence for $0 < x < 1$ we cannot have f(x) = 1

11. I wonder where I've found this old topic myself now...

$\left (\left (\left( x^x \right)^x \right)^x \right)^{...} \neq x^{x^{x^{...}}}$explaines everything.

12. That's a power tower: $h(z)=z^{z^{\ddots}}$

My dots are going in the wrong direction but you know what I mean. Note that:

$z^h=h$

$e^{h\ln(z)}=h$

$\ln(z)e^{h\ln(z)}=h\ln(z)$

$\ln(z)=h\ln(z)e^{-h\ln(z)}$

$-\ln(z)=-h\ln(z)e^{-h\ln(z)}$

At this point it's in the form $ge^g$ which allows us to take the Lambert-W function of both sides:

$h(z)=-\frac{W\big[-ln(z)\big]}{ln(z)};\quad e^{-e}\leq z\leq e^{1/e}$

We can now take it's derivative although I don't know how to derive the derivative for the Lambert-W function:

$\frac{d}{dz} W(z)=\frac{W[z]}{z(1+W[z])}$

Thus using the chain rule:

$h'(z)=\frac{W\big[-\ln(z)\big]}{z\ln^2(z)}-\frac{W\big[-\ln(z)\big]}{z\ln^2(z)\bigg(1+W\big[-\ln(z)\big]\bigg)}$

Check out Mathworld on power towers.

13. Originally Posted by shawsend
At this point it's in the form $ge^g$ which allows us to take the Lambert-W function of both sides:

$h(z)=-\frac{W\big[-ln(z)\big]}{ln(z)};\quad e^{-e}\leq z\leq e^{1/e}$
exp(-exp(1)) = 0.0660 (approx.)

But it seems to me, that z can be every number greater than 0 less than 1.

It is obvious (convergence) that $0 < z < 1$, but do the conditions should be more strict?

Starting with equation
$
-h \ln(z) e^{-h \ln z} = - \ln z
$

We see that it has to be $-h \ln z \in [-1, \infty)$ to get the correct value on the LHS (when we use Lambert W function). However we can see that $-h \ln z > 0$ since $h > 0$ and $\ln z < 0$ (assumptions on z)

And the second problem is on RHS, $- \ln z$ must be in the domain of W. Hence $- \ln z \in \left[-\frac{1}{e}, \infty \right)$ (which is obviously true).

Thus condition on z is: $z \in (0, 1)$

Originally Posted by shawsend
We can now take it's derivative although I don't know how to derive the derivative for the Lambert-W function:

$\frac{d}{dz} W(z)=\frac{W[z]}{z(1+W[z])}$
It's the consequence of inverse function derivative theorem
$(f^{-1})'(x_0) = \frac{1}{f'(f^{-1}(x_0))}$

Having $f(t) = t e^t$, we get $W = f^{-1}$

$f'(t) = (t+1)e^t$

Thus
$W'(z) = \frac{1}{f'(W(z))} = \frac{1}{(W(z)+1) e^{W(z)}} = \frac{W(z)}{(W(z)+1)W(z) e^{W(z)}}
$

We get
$W'(z) = \frac{W(z)}{z(W(z)+1)}$

Since $W(z) e^{W(z)} = z$

-----------------

To sum up... It looks like we finally have some solution, at least we could express this derivative by some special function (which is satisfying I suppose). Hence probably there is no finite solution for this derivative (its existence would imply finite formula for Lambert W function, which I suppose does not exist).

14. Originally Posted by albi
(which is satisfying I suppose).
End special function discrimination: equal rights for special functions.

Thanks also for that derivation above for the derivative!