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In questions like, find the derivative of $f(x)=x^{x^{x^{x^{x^{.^{.^{.}}}}}}}$, how can we formally show that $y=x^y$?

We use this technique for all type of iterations, e.g. $y=\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$, we say $y= \sqrt{6+y}$ and solve the quadratic equation. Intuitively they seem to use the fact that $\infty +1=\infty$, that is the expression has an infinite number of terms so adding or deleting one term won't change the expression. But This logic is quite informal or we can say non-rigorous.

Can we somehow show this formally, e.g. by using the $\epsilon - \delta$ definition of limit or something like that ?

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  • $\begingroup$ $$x^y=x^{\left(x^{x^{x^{x^{x^{.^{.^{.}}}}}}}\right)}=x^{x^{x^{x^{x^{.^{.^{.}}}}}}} = y$$ It's not $\infty$ if it converges, so I don't see how it's non-rigorous. $\endgroup$ – user85798 May 31 '14 at 6:44
  • $\begingroup$ @Oliver I mean let $y'= {^n}a$, now $n \to \infty$ is y, that is $y=\lim\limits_{n \to \infty} y'$. $\endgroup$ – user103816 May 31 '14 at 6:47
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Let $a_n$ be a tower of n $x$, e.g. $a_1=x$, $a_2=x^x$, for some fixed $x$. Suppose the sequence $(a_n)$ converges to a limit $L$. Then, the sequence $(a_{n+1})$ also converges to $L$. But also, $a_{n+1}=f(a_{n})$ for any $n$, where $f(y)=x^y$. Since $f$ is a continuous function in $y$ for any $x>0$, $(a_{n+1})$ converges to $f(L)$. By the uniqueness of limits $L=f(L)$.

Edit We can also proof that the limit exists. Set $b_n=\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$, where there are n $6$. Then $(b_n)$ is an increasing sequence, and bounded above by e.g. $\sqrt{12}$. This follows from induction: if $b_n<\sqrt{12}$, then $b_{n+1}=\sqrt{6+b_n}<\sqrt{6+\sqrt{12}}<\sqrt{6+6}=\sqrt{12}$. Then, by standard results (http://en.wikipedia.org/wiki/Monotone_convergence_theorem), $(b_n)$ converges.

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  • $\begingroup$ Can we do this without presuming that the limit exists. E.g. in $y=\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$ we don't know whether the limit exists or not. So can we prove the same without assuming a priory of existence of the limit? $\endgroup$ – user103816 May 31 '14 at 10:20
  • $\begingroup$ So for x = 1.1, what you call a tower of n converges for n going to infinity, because of the calculations you show for the nested square root sequence? $\endgroup$ – gnometorule Jun 2 '14 at 19:48
  • $\begingroup$ I haven't proven that the tower converges. I proved that $(b_n)$ converges, since the OP asked me to in the comments. $\endgroup$ – Sarastro Jun 2 '14 at 22:48
  • $\begingroup$ I asked Wolfram Alpha, and it happens the tower converges for 0<x<=e^(1/e). $\endgroup$ – Brilliand Jun 6 '14 at 22:42
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In addition to the answer of Sarastro :

$x^{x^{x^{x^{x^{.^{.^{.}}}}}}}= -\frac{W(-ln(x))}{ln(x)}$

where $W(X)$ is the Lambert W function which is real for $X\ge-1/e$ . Hense $-ln(x)\ge-1/e$ which implies $0<x\le e^{1/e}$

The maximum value of $x$ for convergence is $x=e^{1/e}$ where $W(-1/e)=-1$ and the limit is $-\frac{-1}{ln(e^{1/e})}=e$. This is consistent with $y=x^y$ when $x=e^{1/e}$ and $y=e$

Properties of the Lambert W function can be found in : http://mathworld.wolfram.com/LambertW-Function.html

and page 13 in http://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function and, what is more, page 12 the antiderivative of the function $f(x)=x^{x^{x^{x^{x^{.^{.^{.}}}}}}}$ formally expressed as an integral of LambertW or as another special function : the generalized Sophomores dream function.

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For $x>0$, define $y_0:=x$, and let $y_{n+1}:=x^{y_n}$ ($n=0,1,...$). Let $y=\lim_{n\rightarrow\infty}y_n$ where this limit exists, whence $y=x^y$. Then, taking logarithms, we have $\ln y=y\ln x$. Differentiating gives $$\frac{1}{y}=\frac1y\frac{\mathrm dy}{\mathrm dx}\ln x+\frac yx$$or$$\frac{\mathrm dy}{\mathrm dx}=\frac{1}{\ln x}\left(1-\frac{y^2}{x}\right).$$

The limit exists when $y=x^y$ has a solution, namely for all $x$ in the interval $(0\;,\,\mathrm e^{1/\mathrm e}]$. To see this, it is best to start with an analysis of the graph of $y=x^y$. We can isolate $x$ as $$x=\exp\frac{\ln y}{y}.$$Let us temporarily reverse the roles of $x$ and $y$ so that we are in familiar notational territory. That is, consider the graph of $$y=\exp\frac{\ln x}{x}.$$ The exponential function is increasing everywhere; so let us look at $x\mapsto (\ln x)/x$. This has a derivative$$\frac{1-\ln x}{x^2}.$$This quantity is positive in the interval $(0\;,\,\mathrm e)$, and so $x\mapsto (\ln x)/x$ is an increasing function in the interval. The compound function $$f:x\mapsto\exp\frac{\ln x}{x}$$is therefore also increasing in $(0\;,\,\mathrm e]$. Hence $f$ has an inverse$f^{-1}:x\mapsto f^{-1}(x)$, where $f^{-1}(x)$ is the unique solution $y\in(0\;,\,\mathrm e]$ to our original equation $y=x^y$. In other words, the equation $y=x^y$ has a unique solution $y$ for each $x\in(0\;,\,\mathrm e^{1/\mathrm e}]$.

It remains to prove that $y_n\rightarrow y$ as $n$ increases. We consider the three cases (1) $x=1$, (2) $0<x<1$, and (3) $1<x\leqslant\mathrm e^{1/\mathrm e}$ separately. Case 1 is trivial; so we look at case 2 now.

Assume that $0<x<1$. We will show that $0<y_{2n}<y_{2n+2}<y<1$ and $0<y<y_{2n+3}<y_{2n+1}<1$ for $n=0,1,...$. The proof is by induction. Initially we have $$0<y<1,$$ $$0<y_0<y<1,$$ $$0<y_0<y<y_1<1,$$ $$0<y_0<y_2<y<y_1<1,$$ $$0<y_0<y_2<y<y_1<1,$$ $$0<y_0<y_2<y<y_3<y_1<1.$$ Here each line is obtained by raising $x$ to the power of the corresponding terms in the previous line, in reverse order, and completing the line with an initial $0$. Now suppose that we have got as far as proving $0<y_{2k}<y_{2k+2}<y<1$ and $0<y<y_{2k+3}<y_{2k+1}<1$ for some $k$. Then applying to this last sequence the same procedure as we did initially, and omitting $y_0$, gives $$0<y_{2k+2}<y_{2k+4}<y<1,$$ while applying this procedure again to the last sequence yields $$0<y<y_{2k+5}<y_{2k+3}<1.$$ This completes the inductive step. We have shown that the even-indexed terms are bounded above by $y$ and strictly increasing: so they converge to a limit, $u$, say. By continuity of the functions involved, this limit satisfies $u=x^u$, and hence $u=y$ by the previously established uniqueness. Similarly the odd-indexed terms converge to $y$ too.

The proof for $1<x\leqslant\mathrm e^{1/\mathrm e}$ (case 3) is simpler. We show first that $1<y_n<y_{n+1}<y\leqslant\mathrm e$ ($n=0,1,...$), by induction. Initially we have $$1<y_0<y_1<y\leqslant\mathrm e.$$ Now suppose that we have reached proving $1<y_{k-1}<y_k<y\leqslant\mathrm e$ for some $k$. Then $$\frac{\ln y_{k+1}}{\ln y_k}=\frac{y_k\ln x}{y_{k-1}\ln x}=\frac{y_k}{y_{k-1}}>1,$$ whence $1<y_k<y_{k+1}<y\leqslant\mathrm e$, completing the inductive step. Thus, for this range of $x$, the $y_n$ form an increasing upper-bounded sequence, which therefore has a limit. As we reasoned before, this limit is $y$, the unique solution to $y=x^y$ in $(0\;,\,\mathrm e]$.

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  • $\begingroup$ We need to first prove that the limit exists. Can you give me some reference where it is shown that the limit actually exists. $\endgroup$ – user103816 May 31 '14 at 10:25
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    $\begingroup$ @user31782: I have added a proof to my answer. There may well be neater proofs somewhere on the internet, but I can't locate any. $\endgroup$ – John Bentin Jun 1 '14 at 11:33
  • $\begingroup$ Thankyou. This is pretty much what I was looking for. For some reason(s) I'm not yet accepting any answer. But I'll accept it later. The reason is that I lack some background; once I learn things completely I will accept the answer. $\endgroup$ – user103816 Jun 1 '14 at 11:48
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I think a better approach in the given case is to begin to rewrite the formula to that form in which we want to approach the result. I'd write the sequence of iterates $$ a_1=x, \\ a_2=\phantom a _x x, \\ a_3=\phantom a _{ \phantom a _x x} x , \\ a_4=\phantom a_{\phantom a _{ \phantom a _x x} x } x , \\ ...,\\ a_\infty=\phantom a_{ \phantom a_{\phantom a _{ \phantom a _ \ldots x} x } x } x $$ which indicates, that we begin with a "given" value $x$ (or even better $a_0=1$) and proceed by applying the value $x$ as an exponentiation-base for the intermediate value - and can then observe whether we get convergence in the partial evaluations or not. If we get convergence, then we can apply the formula $y=x^y$ (where $y=a_\infty$ in the above notation)

For the infinite expression involving roots (or also: for infinite periodic continued fractions) we do not have immediately the correct notation in that style; but for the continued fractions-problem there is this notational scheme using matrix-multiplication: beginning with a known(!) value, proceeding always on known values until a reasonable intermediate result occurs and do then the final division of the two gotten numbers. Then, proceeding some steps further, one can observe whether the process converges or not.
For the "continued root" - problem I think I've solved this one time earlier using the concept of Carleman-matrices which helped in a similar way like that matrix-method for continued fractions, but don't have it at hand at the moment.

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