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Let $n\neq m^q$ and $round(\sqrt[q]{n})=m$ for some $\{m,n,q\}\in\mathbb{N}$. Then, is there any way to find the $n^r$th digit of $\sqrt[q]{n}$, for $r\in\mathbb{N}$? I once saw something about someone claiming to know the $2^{2020}$th digit of $\sqrt2$, so I'm curious now. However, I myself have no idea how to ever find the digit.

Zuter_242
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You can find a method for calculating every digit of $\sqrt[q]{n}$ at this link:

https://en.wikipedia.org/wiki/Nth_root#Computing_principal_roots

Lee Mosher
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  • Yes, but wouldn't this take too long, especially for very large values of $r$? – Zuter_242 May 11 '21 at 14:38
  • "Too long" is rather vague. One would expect the computation time to increase as $r$ increases no matter what method is used. – Lee Mosher May 11 '21 at 14:40
  • You might enjoy studying some numerical analysis, together with the calculus that underlies it. For example, the theory of Taylor polynomials gives useful error formulas such as the Lagrange remainder term and the integral remainder term. This gives general methods for computing decimal values of fairly arbitrary functions which are correct to an arbitrarily given number of digits. – Lee Mosher May 11 '21 at 14:44
  • Apologies for vagueness. However, even if $r$ exceeded only 32, $2^r$ would exceed 4 billion. Even counting to 4 billion would take a longer amount of time than the longest human lifespan (according to the source below). In conclusion I doubt it would be possible to compute the 4 billionth digit of $\sqrt2$ without a computer. (https://www.infoplease.com/askeds/counting-billion#:~:text=Dividing%20the%20hours%20by%2024,46%20minutes%2C%20and%2040%20seconds.) – Zuter_242 May 11 '21 at 14:51