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Is there any kind of known pattern to $\sqrt 2$ in base 2? Is there any classification categories for decimal digits of numbers that for example would put $\sqrt 2, \sqrt 3 \cdots \sqrt n$ into separate category than $\sqrt[3] 2, \sqrt[3] 3 \cdots \sqrt[3] n$

Or given a decimal expansion with arbitrary precision, or definition of digit in the nth place (not necessarily decimal expansion) to get a probability of which class of numbers the number is likely to be? e.g. transcedental or algebraic of degree k?

jimjim
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  • If there was one on base $2$, then there would be one on base $10$ as well, making $\sqrt2$ rational. – barak manos Nov 09 '15 at 09:47
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    @barakmanos : The question did not ask for a "repeating pattern", e.g. liouville's number has a simple pattern and it is transcendental. – jimjim Nov 09 '15 at 09:50
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    I believe it remains unknown whether $\sqrt {2} $ is "normal" in any base i.e. for each natural number $n$ having uniform distribution of all $n$-tuples of digits in that base. – Simon Nov 09 '15 at 10:00
  • @Arjang The answer depends sensitively on what you will accept as a 'pattern', then. – preferred_anon Nov 09 '15 at 10:07
  • @DanielLittlewood : what are some possible examples that could be considered a pattern? At this stage looking for any in's into what subject in mathematics deals with anything related to this. Simon mentioned number being "normal", I think that is a start. – jimjim Nov 09 '15 at 10:10
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    The wikipedia page is pretty interesting, and points out that it is unknown if $\sqrt 2$ is normal. In fact, we have never proven that a number is normal, except for numbers which were constructed to be normal. https://en.wikipedia.org/wiki/Normal_number – pancini Nov 09 '15 at 10:14
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    See this :http://community.wolfram.com/web/community/groups/-/m/t/1063480 – Mariusz Iwaniuk Apr 17 '17 at 08:03
  • see this related question:

    https://math.stackexchange.com/q/5004475/870236

     – Math Admiral Nov 30 '24 at 15:49

2 Answers2

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One expression for the square root of 2 is given by $\sqrt{2} = \sum_{k=0}^{\infty} (2k+1) \binom{2k}{k} 2^{-3k -1} $ which is rather useful for computing the digits exactly. It also means that any pattern in the binary digits of $\sqrt{2}$ should be somewhat related to the the digits of the central binomial coefficients. There's probably a combinatorial interpretation of this.

Incidentally, $(2k+1) \binom{2k}{k}$ is asymptotic to $\frac{2^{2k + 1}}{\sqrt{\pi}}$ so the number of terms affecting the kth digit is linear in k. Also, the greatest power of 2 dividing the central binomial coefficient is known as Gould's sequence, i,e, $\binom{2k}{k} = 2^{2^{h(k)}}$ where $h(k)$ is the hamming weight of k.

saolof
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there is a method to calculate decimal places of root2, probably that could help find a pattern with it. Although this is not a very obvious one, it probably could (very loosely) help to find a pattern. The method works a bit like dividing, so you have to calculate the next decimal place, then the next. It does not work on approximations by the way.

http://www.murderousmaths.co.uk/books/sqroot.htm

On this link you will see a few methods, but scroll down past the cyan shortcut box and you will see the one shown by Frank La Fontaine.

I'm sorry that I can't show an actual pattern, but hopefully this helps.

ruby duby
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