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New poster has been reminded to always show their work when starting schoolwork threads.
Summary: How do I solve the derivative of (((x^{x})^{x})^{x})...
How would I solve this derivative?
How would I solve this derivative?
For what values of ##x## is that function defined?Summary: How do I solve the derivative of (((x^{x})^{x})^{x})...
How would I solve this derivative?
View attachment 301752
They haven't mentioned that in the questionFor what values of ##x## is that function defined?
It might be important. How do you know it's even a well-defined function?They haven't mentioned that in the question
That's true. But, the function in the question is more complicated and involves an infinite sequence of exponentials.From what I 've read in various calculous books, usually the function ##x^x## is defined only for ##x>0##.
Yes, my intuition says for x>1 it is probably ##f(x)=\infty## but for ##0<x\leq1## it might be well defined.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!
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 $$f(x)=-\frac{W_n(-x\ln x)}{x\ln x}$$I'm on team this thing equals 1 when x is between 0 and 1
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##.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,
$$
\ln(y)=y\ln(x^{x})
$$
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.@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##.
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...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.
I don't disagree with Wolfram. The critical error is yours and made before you typed anything into Wolfram.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...
And what is my critical error?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.
Assuming ##y = (x^x)^y##. This is wrong.And what is my critical error?
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.And to be completely honest I wasn't aware of the multiple exponents convention either
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}##.Summary: How do I solve the derivative of (((x^{x})^{x})^{x})...
How would I solve this derivative?
View attachment 301752
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).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.