How do I compute the derivative of (x^x)^(x^x)^(x^x)....

[L_ucifer]

Summary: How do I solve the derivative of (((x^{x})^{x})^{x})...

How would I solve this derivative?

 

[Delta2]

Hmm, not sure if this will help but if we set $f(x)$ this function then I *think* the following equation holds
$$f(x)=(x^x)^{f(x)}$$

Take the derivative of this equation in LHS and RHS and see what you get. You will form a functional equation for $y=f'(x)$ which you might be able to solve for y.

 

[PeroK]

Summary: How do I solve the derivative of (((x^{x})^{x})^{x})...

How would I solve this derivative?
View attachment 301752

For what values of $x$ is that function defined?

 

[L_ucifer]

For what values of $x$ is that function defined?

They haven't mentioned that in the question

 

[PeroK]

They haven't mentioned that in the question

It might be important. How do you know it's even a well-defined function?

 

[Delta2]

From what I 've read in various calculous books, usually the function $x^x$ is defined only for $x>0$.

 

[PeroK]

From what I 've read in various calculous books, usually the function $x^x$ is defined only for $x>0$.

That's true. But, the function in the question is more complicated and involves an infinite sequence of exponentials.

Its domain (and range) may be more constrained!

 

[Delta2]

That's true. But, the function in the question is more complicated and involves an infinite sequence of exponentials.

Its domain (and range) may be more constrained!

Yes, my intuition says for x>1 it is probably $f(x)=\infty$ but for $0<x\leq1$ it might be well defined.

 

[Office_Shredder]

I'm on team this thing equals 1 when x is between 0 and 1

 

[Delta2]

Using the method outlined in post #2 I was able to prove (given that the function and its derivative are well defined) that $$f'(x)(\frac{1}{f(x)}-x\ln x)=f(x)(\ln x+1)\Rightarrow f'(x)=f^2(x)\frac{\ln x+1}{1-f(x)x\ln x}$$ .

So what do you think, is the starting equation claimed at #2 valid? If its valid though it can't be $f(x)=1,0<x<1$. It is $f(x)=1$ only for $x=1$.

 

[Delta2]

I'm on team this thing equals 1 when x is between 0 and 1

Intuitively it seems so, however I think it is one of the cases where our intuition misleads us. I did a minor research using wolfram, and it seems that wolfram says this function is equal to $$f(x)=-\frac{W_n(-x\ln x)}{x\ln x}$$

where $$W_n(z)$$ the analytic continuation of the product log function.

TBH never heard of $W_n(z)$ during my undergraduate studies, neither during my master studies.

https://www.wolframalpha.com/input?i=y=(x^x)^y

 

[Fred Wright]

We evaluate the derivative in terms of the Lambert W function. We have,
$$
y=(x^x)^{{(x^x)}^{...}}
$$
$$
y=(x^{x})^{y}
$$
Taking the logarithm,
$$
\ln(y)=y\ln(x^{x})
$$
$$
\ln(y)=e^{\ln(y)}\ln(x^{x})
$$
and so,
$$
-\ln(y)e^{-\ln(y)}=-x\ln(x)
$$
We consign both sides of the equation to the Lambert W function,
$$
W(-\ln(y)e^{-\ln(y)})=W(-x\ln(x))
$$
The Lambert W function obeys,
$$
W(z)e^{W(z)}=z
$$
and we have,
$$
\ln(y)=-W(-x\ln(x))
$$
Exponentiation yields
$$
y=e^{-W(-x\ln(x))}
$$
and from the property of the Lambert W function,
$$
e^{-W(-x\ln(x))}=\frac{W(-x\ln(x))}{-x\ln(x)}
$$
we find,
$$
y=\frac{W(-x\ln(x))}{-x\ln(x)}
$$
For brevity of notation let,
$$
f(x)=-x\ln(x)
$$
$$
\frac{df(x)}{dx}=-(1+\ln(x))
$$
Taking the derivative,
$$
\frac{dy}{dx}=\frac{(1+\ln(x))W(f)}{f^2}+\frac{\frac{dW(f)}{dx}}{f}
$$
$$
\frac{dW(f)}{dx}=\frac{dW(f)}{df}\frac{df}{dx}
$$
By implicit differentiation,
$$
\frac{dW(f)}{df}=\frac{W(f)}{f(1+W(f))}
$$
and thus,
$$
\frac{dy}{dx}=\frac{(1+\ln(x))W(f)}{f^2}-\frac{(1+\ln(x))W(f)}{f^2(1+W(f))}
$$
$$
=\frac{(1+\ln(x))W^2 (-x\ln(x))}{(x\ln(x))^2 (1+W(-x\ln(x)))}

$$
which holds for $x\ln(x) \ne \frac{1}{e}$.

 

[Delta2]

@Fred Wright to be honest I am not familiar with the W Lambert function. Is it the same as the analytic continuation of the product log function ($W_n(z)$ see post 11 and associated wolfram link). If it is the same I think your final result coincide with mine. Just put $$f(x)=-\frac {W(-x\ln x)}{x\ln x}$$ and I think we have the same result (by f(x) I mean the initial function with the infinite $x^x$.

 

[PeroK]

We evaluate the derivative in terms of the Lambert W function. We have,
$$
y=(x^x)^{{(x^x)}^{...}}
$$
$$
y=(x^{x})^{y}
$$
Taking the logarithm,
$$
\ln(y)=y\ln(x^{x})
$$

If we take $x = \frac 1 2$, then$$ \frac{\ln(y)}{y} = -\frac 1 2 \ln(2) \approx -0.3466$$This gives $$y \approx 0.767$$Which is wrong because with $x = \frac 1 2$ we have $z = (x^x) \approx 0.707$ and $z^z \approx 0.783$. Note that, as indicated above $y = 1$ for $0 \le x \le 1$.

Equally, we could have $$\ln(y) = y\ln(z^z)$$where z = $x^x$, which contradicts $$\ln(y) = y\ln(x^x)$$

 

[PeroK]

@Fred Wright to be honest I am not familiar with the W Lambert function. Is it the same as the analytic continuation of the product log function ($W_n(z)$ see post 11 and associated wolfram link). If it is the same I think your final result coincide with mine. Just put $$f(x)=-\frac {W(-x\ln x)}{x\ln x}$$ and I think we have the same result (by f(x) I mean the initial function with the infinite $x^x$.

You can be sceptical all you like, but there are obvious contradictions in what you are doing. It's not rigorous, not valid and simply wrong.

 

[Delta2]

You can be sceptical all you like, but there are obvious contradictions in what you are doing. It's not rigorous, not valid and simply wrong.

This looks like a perfect self critique for you Mr Perok. I can tell you got guts though, you disagree not only with me and fredwright but with Wolfram too...

 

[PeroK]

This looks like a perfect self critique for you Mr Perok. I can tell you got guts though, you disagree not only with me and fredwright but with Wolfram too...

I don't disagree with Wolfram. The critical error is yours and made before you typed anything into Wolfram.

You don't need guts when you have a modicum of mathematical ability.

 

[Delta2]

I don't disagree with Wolfram. The critical error is yours and made before you typed anything into Wolfram.

You don't need guts when you have a modicum of mathematical ability.

And what is my critical error?

All you do is some approximate calculations, the results are approximately close and still you claim we are wrong.

 

[PeroK]

And what is my critical error?

Assuming $y = (x^x)^y$. This is wrong.

 

[berkeman]

Thread closed temporarily. Please take this disagreement to PMs, gentlemen, and include me in the PM conversation. Once you have arrived at a consensus, we can re-open this thread to finish sorting it out. Thanks.

 

[berkeman]

Update -- after a fruitful thread in the Advisor forum, things seem to have been resolved. I'm reopening this thread so that the participants can tie this thread off.

 

[PeroK]

It depends on how we interpret multiple exponents. The convention, which I didn't know, is to go right to left (or top to bottom). For example, if there are no brackets, then the convention is $x^{(x^x)}$, rather than as I expected $(x^x)^x$.

With this convention, we don't have a constant function and the calculations involving the Lambert function are correct.

Apologies to all concerned.

 

[Delta2]

Well ok, @PeroK except from being an excellent scientist, has the courage and honesty to admit his mistake when he really understands he has done one.
And to be completely honest I wasn't aware of the multiple exponents convention either

 

[PeroK]

That said, the assumption of convergence for the right-to-left limit does indeed appear to be false. And $f(x)$ is not well-defined for all $x \in (0, 1)$.

 

[Mark44]

And to be completely honest I wasn't aware of the multiple exponents convention either

The key to this is understanding the associativity of operators, or how they group. This is an important concept in programming, but something not discussed very much in mathematics courses, at least in my experience. Something I posted in an advisors-only forum section was that different tools can give different results for repeated exponents. For example, Excel evaluates 2^2^3 as if it were written (2^2)^3, producing 64. That is, the associativity is left-to-right. OTOH, Python and other programming languages that have an exponentiation operator evaluate 2**2**3 as if it were written 2**(2**3), which produces 256. Here the associativity is right-to-left.

 

[PhDeezNutz]

This thread topic is way out of my wheelhouse but I have to commend @PeroK, @Delta2 and the moderation staff for keeping things civil and coming to a consensus. This is a testament to the quality of PF.

 

[Orodruin]

Summary: How do I solve the derivative of (((x^{x})^{x})^{x})...

How would I solve this derivative?
View attachment 301752

To be a bit of a nitpicker here … the formula in text does not match that of the image. The text would indeed be the limit of $x^{x^n}$ whereas the image with standard interpretation of nested exponentials (working top-down) would not be. I think most people would not interpret $e^{x^2}$ as $(e^{x})^2=e^{2x}$.

The confusion is therefore understandable as the problem depends on which version you read - image or text.

 

[Prof B]

To be a bit of a nitpicker here … the formula in text does not match that of the image. The text would indeed be the limit of $x^{x^n}$ whereas the image with standard interpretation of nested exponentials (working top-down) would not be. I think most people would not interpret $e^{x^2}$ as $(e^{x})^2=e^{2x}$.

The confusion is therefore understandable as the problem depends on which version you read - image or text.

There is nothing nitty about it. Many problems in these forums are stated ambiguously or inconsistently, many OPs are cavalier about the issue, and many of the readers play white knight to the cavalier OPs. In this particular example, the question in the title is ambiguous, the question in the summary asks for the derivative of x^(x^infinity), and the question in the picture asks for the derivative of a function that satisfies f(x) = (x^x)^f(x).

 

[Prof B]

The derivative can be found pretty easily without any knowledge of the W() function. Using the formula in the image, we get $y = (x^x)^y$. By laws of exponents, $y = x^{xy}$. Take logarithms of both sides to get $\ln{y} = xy\ln{x} = yx\ln{x}$. Differentiate with respect to x to get
$$ y'/y = y'x\ln{x} + y\ln{x} + y$$
Solve the linear equation for y' to get
$$ y' = \frac{y\ln{x}+y}{1/y - x\ln{x}} $$

 

[Delta2]

@Prof B I think you almost revealed the complete solution to the OP...

Remember the rules here in PF are that we don't do that, we just have to guide the OPs using hints and other methods in order to make them think and find the answer for themselves!