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It is an open problem to decide whether $\pi^{\pi^{\pi^\pi}}$ is an integer or not (though it is obviously not), as mentioned in this video by Matt Parker.

The integer part of this number have almost $10^{18}$ digits (WolframAlpha estimates it as $10^{10^{17.8236}}$). For comparison, we only know $10^{14}$ digits of $\pi$ as of this day. If this doesn't seem completely out of reach to modern computing (as we 'only' require $10$ thousand times more computational power), then there is the problem of precision. This raises the following question:

How many digits of $\pi$ must we know in order to determine the value of $\pi^{\pi^{\pi^\pi}}$ up to the first decimal place?

A comment below notices this question might not be well posed if $\pi^{\pi^{\pi^\pi}}$ happens to be very close to an integer. In this sense, it is more precise to ask instead "How many digits of $\pi$ must we know in order to determine the value of $\pi^{\pi^{\pi^\pi}}$ with an error inferior to $\varepsilon$" for some small value of $\varepsilon$, say $0.01$.

Actually, I want to ask a more general question. Given some function $f$ and a target value $x$ , how good of an approximation is $f(u)$ to $f(x)$ in terms of how good of an approximation is $u$ to $x$? The answer should also depend on $f$ and $x$, naturally.

I know a certain 'niceness' from the function $f$ is required for this to be an answerable question, though I'm not sure what would be an appropriate definition of 'nice'. I feel it should be something more restrictive than 'continuous' but no more than 'smooth'.

I feel differentials have a part to take in this analysis, as one has $$|f(x)-f(u)|\approx |f'(x)| \cdot |x-u|$$ but this is not an inequality, just another approximation. For this to be useful we must know how good of an approximation this is.

If I'm being too vague with my request, I apologize. What I'm looking for is to be able to answer questions such as the one above; to have a function, say $f(x) :\equiv x^{x^{x^x}}$, and determine, for instance, how many digits from a specific input, say $\pi$, do I need in order to get a sufficiently good output.

Alma Arjuna
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    It's possible that no amount of precision suffices. It could be that it is an integer (say $n$), and so even if we knew $|\pi-x|<\delta$ with $\delta$ arbitrarily small, and $d$ arbitrarily large, $z^{z^{z^z}}$ with the approximation $x-\delta$ would be slightly less than $n$, while the function with $x+\delta$ would be larger. – Varun Vejalla Dec 09 '24 at 22:05
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    @VarunVejalla this is a good observation! I should've asked for a small error, say of $0.01$, instead of a specific digit. This would give me the digit I'm looking for except for the case where the output is an integer. – Alma Arjuna Dec 09 '24 at 22:53
  • See also https://math.stackexchange.com/questions/13050/eee79-and-ultrafinitism and https://math.stackexchange.com/questions/13054/how-to-show-eee79-is-not-an-integer and https://math.stackexchange.com/questions/390708/proof-that-left-pi-pi-right-pi-pi-and-now-pi-left-pi-pi-pi and https://math.stackexchange.com/questions/4043917/relaxation-of-the-pi-pi-pi-pi-problem and probably several others, here and on MathOverflow. – Gerry Myerson Dec 09 '24 at 23:50
  • For your more general question, see Lipschitz continuity. – mr_e_man Dec 10 '24 at 04:17
  • @mr_e_man Lipschitz continuity is far too restrictive for what I'm interested in investigating. – Alma Arjuna Dec 10 '24 at 04:33
  • It's less restrictive than "smooth" or even "continuously differentiable". Lipschitz continuity gives an inequality $$|f(x)-f(u)|\leq C\cdot|x-u|,$$ not an approximation. ($C$ can be gotten by bounding $|f'|$, if $f'$ exists.) What else did you want? – mr_e_man Dec 10 '24 at 04:44

2 Answers2

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By direct calculation, with about $30$ digits in $\pi$, I get

$$\pi^{\pi^{\pi^\pi}}\approx9.080222\cdot10^{666262452970848503};$$

so, $7$ digits in the result. Then with $60$ digits in $\pi$, I get

$$9.080222455390617769723931713284287746;$$

so, $37$ digits. Then with $120$ digits in $\pi$, I get $97$ digits in the result. Then with $1200$ digits in $\pi$, I get $1177$ digits in the result.

So it looks like it's just subtracting $23$ digits. In that case, to tell whether the result is an integer, we'd need at least this many digits in $\pi$:

$$666262452970848503\\+2\\-23\\=666262452970848482.$$

(The $+2$ is for the leftmost digit $9$ and the desired digit after the decimal point.)

It's just about the same huge number. In other words, this answer is essentially right.

I am using interval arithmetic to automatically take care of precision and rounding issues.

mr_e_man
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  • Of course we can get the $23$ (or something it's close to) by taking the derivative of $f(x)=x^{x^{x^x}}$: $$f'(\pi)\approx f(\pi)\cdot10^{20}$$ – mr_e_man Dec 10 '24 at 02:46
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Finding the first decimal digit of $f(x)=x^{x^{x^x}}$ is equivalent to finding the fractional part of $g(x)=\log_{10}f(x)=x^{x^x}\log_{10}{x}$ to some reasonable precision... specifically, enough precision to determine where it falls in the list $\left[ 0, \log_{10}2\approx 0.301, \ldots \log_{10}9 \approx 0.954\right]$. When $x=\pi$, the log in question is $$ g(\pi)=\pi^{\pi^\pi}\log_{10}\pi \approx 6.663 \times 10^{17}. $$ Now, if you were to introduce a small error $\delta$ to $\pi$, this would change, and the size of the change would depend on the derivative of $g(x)$ at $\pi$: $$ g(\pi + \delta) - g(\pi) \approx g'(\pi)\cdot\delta\approx 10^{20}\cdot\delta $$ (where I estimated the derivative numerically, but one could certainly find the exact expression and approximate that instead). Since we need a decent estimate for the fractional part of $g(\pi)$, we just need our error in the value of $\pi$ to be such that $g'(\pi) \cdot \delta < 1$; so $\delta < 10^{-20}$ or so should suffice. In other words, you should actually only need $20$-odd digits of $\pi$ to determine the first digit of $f(\pi)$ and its exact decimal length (assuming the next digits don't happen to be, say, $0000$ or $9999$).

mjqxxxx
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    The question is about the first digit after the decimal point. – mr_e_man Dec 10 '24 at 00:13
  • Or is it? I'm not sure. – mr_e_man Dec 10 '24 at 00:23
  • @mr_e_man you're right, it is about the first digit after the decimal point. – Alma Arjuna Dec 10 '24 at 00:25
  • If you apply a similar approach to $f(x + \delta)$ so that its error is roughly $\Delta$, you get $\delta \approx \frac{\Delta}{f(x)f'(x)}$ which is a fairly horrible expression but with $x = \pi$ and $\Delta = 1$ gives an incredibly small value - the denominator includes $f(x) \times \ln f(x) \times \ln \ln f(x) \times \ln \ln \ln f(x)$ with some other terms. – ConMan Dec 10 '24 at 00:37