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Consider the function, $$f(x)=x^{x^{x^{x^{.^{.^{.{^.}}}}}}}$$While differentiating it with respect to $x$ (provided that $f(x)$ evaluates to a finite value), we were made to write $f(x)=x^{f(x)}$.

My question is: Why we can't write it as $f(x)=(f(x))^x$ and then differentiate implicitly? Both the ways of writing mean the same thing.

FFjet
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Vansh Saini
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1 Answers1

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You have to be a bit careful here. The point is that, for each $x$, $f(x)$ is defined as $f(x) = \lim_{n\to \infty} p_n(x)$, where $p_n$ means applying $n$ times "$x$ to the power", e.g, $p_2(x) = x^{x^x}$. (Note that is is not the same as $(x^x)^x$!) Then it follows that

$$ f(x) = \lim_{n\to \infty} p_n(x) = \lim_{n\to \infty} p_{n+1}(x) = \lim_{n\to \infty} x^{p_{n}(x)} = x^{\lim_{n\to \infty} p_{n}(x)} = x^{f(x)}. $$

We cannot do a similar sequence of equalities to arrive at your expression $f(x) = f(x)^x$.

Inzinity
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  • Many thanks. I wanted to know will the ordered pair $(x,f(x))$ satisfy the equality $f(x)={f(x)}^x$ and if it satisfies it then this form ought to be correct. – Vansh Saini Sep 13 '19 at 04:47
  • @VanshSaini Fixing $f(x)=y$, the only solution to $y = y^x$ is $x=1$ unless $y=1$ or $y=0$. Hence the only function $f$ defined on an open subset of $\mathbb R$ satisfying $f(x)=f(x)^x$ is the function $f(x)=1$. – Inzinity Sep 13 '19 at 04:50
  • Got it.................. – Vansh Saini Sep 13 '19 at 04:58