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I want to find the $100th$ digit after decimal in $\sqrt 2$.

$\sqrt 2 = 1.41421356237...$

for example $7th$ digit after decimal is $5$. Is there any general method for such problems? I thought it might use the representation of real numbers as a cauchy sequence, but was not able to proceed further.

Thanks in advance!

José Carlos Santos
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Raghukul Raman
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  • have you tried approximation using a fraction ? –  Sep 22 '17 at 12:26
  • what kind of approximation?? – Raghukul Raman Sep 22 '17 at 12:31
  • using fractions ... –  Sep 22 '17 at 12:34
  • With a general method, you mean a method that allows to calculate the $10^{20}$ th digit as well ? I heard that there is a formula for $\pi$ in base $2$ that can calculate specified digits without calculating $\pi$ itself (but still with some effort), but I do not think that such a formula is known for $\sqrt{2}$ and base $10$. You can of course calculate $\pi$ arbitary precise, for example with Heron's method or a continued fraction method, but this is not what you want, is it ? – Peter Sep 22 '17 at 12:35
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    But using a fraction is similar to computation, isn't it? I thought of using cauchy sequence to represent $\sqrt 2$ $A_1 = 1, A_{n+1} = \frac{A_n + 2/A_{n}}{2}$ this converges to $\sqrt 2$ – Raghukul Raman Sep 22 '17 at 12:39
  • @RaghukulRaman In principle, yes. – Peter Sep 22 '17 at 12:39
  • @RaghukulRaman I could give you the first $100$ digits of $\sqrt{2}$ after the decimal point (The last is $7$), but I do not think that this is the intent of the question. – Peter Sep 22 '17 at 12:44
  • @Peter yes any arbitrary digit. and for $\pi$, yes there are some methods that I heard of, but I want the digits of $\sqrt 2$ – Raghukul Raman Sep 22 '17 at 12:45
  • @Peter do you have some method of computing first $100$ digits of $\sqrt 2$? – Raghukul Raman Sep 22 '17 at 12:46
  • As said, I do not think any formula is known. There is an onn-line applet for the first $2\cdot 10^9$ digits from $\sqrt{2}$, would this be interesting ? – Peter Sep 22 '17 at 12:47
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    @RaghukulRaman PARI/GP allows calculations with high precision, Wolfram alpha (online available) as well – Peter Sep 22 '17 at 12:48
  • If you want to do it by hand, you could use the Taylor series for $\sqrt{x}$. – J126 Sep 22 '17 at 12:48
  • @JoeJohnson126 how can we get the $100th $ digit? terms of taylor's series could be non terminating(eg. $1/3 = 0.3333...$) so no precise answer could be formulated. – Raghukul Raman Sep 22 '17 at 12:53
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    @RaghukulRaman In theory you can do this, if you take enogh terms, the $100$-th digit will be correct. But in practice, the calculation (by hand) well be awful. $100$ digits is really much for hand-calculation! And still this is no method for an arbitary digit. $1000$ digits would be a truly enormous task. – Peter Sep 22 '17 at 13:00

3 Answers3

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Yes, there is a general method for this problem; basically, there is a way to modify the "long division" algorithm to give you square roots instead of quotients. A detailed explanation and worked example can be found here:

https://xlinux.nist.gov/dads/HTML/squareRoot.html

An alternative technique: suppose we want the first few digits of the square root of two. Take the integer 2000000000000. (Note that there must be an even number of zeros.) There is some integer n such that when squared, it is less than 2000000000000, and (n+1) squared is greater. You can find that integer via binary search in about 40 guesses; the digits that n and n+1 have in common are definitely the digits of the square root of two. Want more digits? Add more zeros. Each additional pair of zeros will increase the binary search length by a small amount.

Eric Lippert
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This could be of help to you in your pursuit. Good luck!

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Every CAS can do this

$$1.4142135623730950488016887242096980785696718\\75376948073176679737990732478462107038850387534327641573$$

Dr. Sonnhard Graubner
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