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My calculus book says that with: $$a=x^{x^{x^{.^{.^{.}}}}}$$ (exponent tower goes on forever), then: $$x=a^\frac{1}{a}$$

I tried it out with $a=3$ so $x=3^\frac{1}{3}$ and then ran a python program to test it. I did: $$(3^\frac{1}{3})^{(3^\frac{1}{3})^{(3^\frac{1}{3})^{.^{.^{.}}}}}$$

Between 500 runs and 10.000 runs the answer stayed: 2.4780526802882967

Maybe I don't appreciate enough what infinity means, but why isn't it closer to $3$ after 10.000 runs? Is the answer $x=a^\frac{1}{a}$ really correct? Will this really finally be 3 if it would run infinitely?

For reference, this was the little python program:

x=3**(1/3)
y=x
for i in range(0,100000):
    y=x**y
    print(y)
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    $\begingroup$ What exactly does your calculus book say? I somehow doubt that the infinite exponent tower is defined for $\lvert x\rvert > 1$. $\endgroup$ – Clement C. Jan 31 '16 at 18:10
  • $\begingroup$ @ClementC.: You would think so, wouldn't you? But your (and my) intuition is faulty $-$ the sequence of finite exponent towers is bounded above by $a$. $\endgroup$ – TonyK Jan 31 '16 at 18:38
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    $\begingroup$ @ClementC. Such power tower only converges for $e^{-e} \leq x \leq e^{\frac1e}$. If it converges, then it must statisfy $x^a=a$ and hence $x=a^{\frac1a}$. $\endgroup$ – wythagoras Jan 31 '16 at 18:38
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    $\begingroup$ My bad. But then, if I understood correctly this and the answer below (I feel I am missing something)-- the issue is that the sequence defined by $x_1 = x = 3^{1/3}$ and $x_{n+1} = x_n^x$ does converge, but not to the tower in question? @wythagoras (i.e., it converges to one of the two values $\ell$ such that $\frac{\ln \ell}{\ell} = \frac{\ln 3}{3}$, but "sadly" not the right one.) $\endgroup$ – Clement C. Jan 31 '16 at 18:55
  • $\begingroup$ @wythagoras So basically the issue is that the function doesn't converge for $x=3^\frac{1}{3}$? $\endgroup$ – GambitSquared Jan 31 '16 at 20:26
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You are confusing the direct and converse implication.

The book doesn't say that if we set $x=a^{\frac{1}{a}}$ the sequence converges.

It says that if the series converges to $b$, then you have $x=b^{\frac{1}{b}}$, where $b$ is the limit.

If you apply the theorem in this case, it tells you that your $x$ is equal to the limit to the limit, not that

If everything worked well in your programming, this means that you must have $$3^{\frac{1}{3}}= 2.4780526802882967^{\frac{1}{ 2.4780526802882967}} $$

Note that in general, a number $x$ could be written in more than one way as $a^\frac{1}{a}$, but the tower can only converge to one $a$. This means that if you start with another one, the limit will be different.

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$$y_{n+1}=x^{y_n}$$ $$\ln y_{n+1}=y_n \ln x=y_n (\frac{1}{3}\ln 3)$$ What you are getting is $y_{n+1}=y_n=2.4780526802882967$,

$$\frac{\ln y_{n}}{y_n}= \frac{1}{3}\ln 3$$

Which should have been true only for $y_n =3$ if $\frac{\ln x}{x}$ was a one to one function except that it's not.Plot

The above plot shows the function $\frac{\ln x}{x}$ .Observe that the line $y=\frac{\ln 3}{3}$ intercepts the graph at 2 points- $x=3$ and $x=2.4780526802882967$.

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