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How would I calculate the next digit of the decimal representation of square root of two?

1.4142135...

Say I want to calculate the digit after 5... how would I do that?

I am not allowed to use a calculator.

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    $\begingroup$ Newton's method. Given an approximation $x_n$, the next approximation given by the method is $$ x_{n+1}=\frac{x_n}2+\frac1{x_n}. $$ The number of accurate digits increases dramatically (essentially doubling with each iteration). It is all rational arithmetic, so (in principle, at least), doable without a calculator. $\endgroup$ – Andrés E. Caicedo Sep 1 '14 at 18:30
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    $\begingroup$ Fly By Night: I have been doing the "guess method" to find square roots by hand... just taking me some time, and I was wondering if there was a quicker method. $\endgroup$ – Jin Sep 1 '14 at 18:33
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    $\begingroup$ For instance, if you start with $x_n=1.4142135$, then $x_{n+1}=1.4142135623730964\dots$ while $\sqrt2=1.4142135623730950\dots$ $\endgroup$ – Andrés E. Caicedo Sep 1 '14 at 18:33
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    $\begingroup$ Another approach is to follow the usual algorithm, which you probably are already familiar with. It of course has the disadvantage of being much slower than Newton's method, since you really are just computing the digits one by one. $\endgroup$ – Andrés E. Caicedo Sep 1 '14 at 18:35
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    $\begingroup$ @AndresCaicedo: Ha! The very same link! $\endgroup$ – TonyK Sep 1 '14 at 18:35
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There is a very nice method, by hand, that is seldom taught nowadays.

You want to take the square root of $2$ to $n$ digits, so we will write $2$ followed by $2n$ zeros. Notice, $2n$. Write them by pairs, like this for $5$ digits:

2 00 00 00 00 00

Now, if your number has several digits, you write them by pairs too (starting from the right), for example for the square root of 20451, to five digits, you would write

2 04 51 00 00 00 00 00

From now on, we continue with the square root of $2$, but the method works or any number. Actually, we are really computing the square root of any integer, truncated to the lower integer: since we add $2n$ zeros, it's like computing the square root of $2\times10^{2n}=10^n\sqrt{2}$, and by truncating you get $n$ decimals of $\sqrt{2}$.

This is the general setting, not let's see the method.


So you write your number, with additional zeros

2 00 00 00 00 00

You take mentally the integer (truncated) square root of the last part on the left. Here it's the integer square root of $2$, so it's $1$ (it's the floor of $1.414...$). If you know your table of squares up to $10^2=100$, it's enough to find this.

You will write this one on the right, like for a division

2 00 00 00 00 00    | 1

Now we enter the "general step" (we will repeat this one for each pair of zeros.

You subtract $1^2$ from $2$, and write the rest ($1=2-1^2$) under the preceding, plus one pair of zeros:

2 00 00 00 00 00    | 1
1 00

On the right, there is the "current square root", we will complete it by adding digits.

Now, find the highest possible digit ($0$ to $9$) such that the operation $2d \times d$ is not greater than $100$. $2d$ denote the number obtained by concatenating twice the current square root, to the chosen digit.

For example, you try $20\times0=0$, $21\times1=21$, $22\times2=44$, $23\times3=46$, $24\times4=96$, $25\times5=125$. Stop! Too large, so the next digit is actually a $4$.

The rest is $100-96=4$.

So we update the current square root on the right (chosen digit $4$), and the current rest (also $4$), and write down two more zeros:

2 00 00 00 00 00    | 14
1 00
   4 00

This was the first step, and we will repeat exactly the same several times:

Step 2

Twice the current root $14\times2=28$, and try $28d\times d$ so that it's not greater than 400. Since $282\times2>400$, the next digit is a $1$, and we subtract $281\times1=281$ from $400$ ($400-281=119$), and update both the rest and the current root (next digit $1$).

2 00 00 00 00 00    | 141
1 00
   4 00
   1 19 00

Step 3

Twice the current root, so $282$, and try $282d\times d$ so that it's not greater than $11900$. Since $2824\times4=11296$ and $2825\times5=14125$, too large, the next digit is a $4$.

The next rest is $11900-11296=604$.

2 00 00 00 00 00    | 1414
1 00
   4 00
   1 19 00
      6 04 00

Step 4

Twice the current root, so $2828$, and we try $2828d\times d$ so that it's not greater than $60400$. The good choice is $28282\times 2=56564$, and $60400-56564=3836$ is the next rest, and next digit is $2$:

2 00 00 00 00 00    | 14142
1 00
   4 00
   1 19 00
      6 04 00
        38 36 00

Step 5

Twice the current root is $28284$, so we try $28284d\times d$ so that it's not greater than $383600$. The good choice is $1$: $282841\times1=282841$ is the largest possible, and $383600-282841=100759$.

2 00 00 00 00 00    | 141421
1 00
   4 00
   1 19 00
      6 04 00
        38 36 00
        10 07 59

Hence, we have the last rest, and the updated square root to $5$ digits, and we have the equation:

$$2 00 00 00 00 00=141421^2+10 07 59$$

We could of course continue, adding as many pairs of zeros as we wish. With two more pairs, you will get the next digit you are after.


In the usual setting (at least what I was taught by my father, who learned it in high school in the sixties, in France), you write the successive products under the square root like this:

2 00 00 00 00 00    | 141421
1 00                |------------------
   4 00             | 24×4=96
   1 19 00          | 281×1=281
      6 04 00       | 2824×4=11296
        38 36 00    | 28282×2=56564
        10 07 59    | 282841×1=282841
                    |

That way, it's relatively compact, yet it's easy to do the subtractions.

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \root{2}&=\root{288 \over 144}={1 \over 12}\root{289 - 1} ={17 \over 12}\root{1 - {1 \over 289}} ={17 \over 12}\sum_{n = 0}^{\infty}{1/2 \choose n}\, {\pars{-1}^{n} \over 289^{n}} \end{align}

$$ \begin{array}{rcl} {17 \over 12} &\approx& \color{#66f}{1.41}\color{#c00000}{6666666666\ldots} \\ {17 \over 12}\,\pars{1 - {1 \over 578}} &\approx& \color{#66f}{1.41421}\color{#c00000}{5686274\ldots} \\ \underbrace{{17 \over 12}\,\pars{1 - {1 \over 578} - {1 \over 668168}}} _{\ds{=\ \color{#00f}{{222337 \over 157216}}}} &\approx& \color{#66f}{1.414213\ {\Large\underbrace{5}_{\uparrow}}}\ 6\color{#c00000}{6049\ldots} \end{array} $$

$$ \color{#c00000}{\large\color{#66f}{\root{2}}\approx{222337 \over 157216}\approx \color{#66f}{1.414213}\color{#00f}{\LARGE 5}\color{#66f}{6}60492570731\ldots} $$

$$ \mbox{Notice that}\quad \root{2} = {222337 \over 157216} + {\cal R}\quad \mbox{where}\quad {\cal R} \equiv {17 \over 12}\sum_{n = 3}^{\infty}{1/2 \choose n}\, {\pars{-1}^{n} \over 289^{n}} $$

Also, \begin{align} \verts{\cal R}&< {17 \over 12}\sum_{n = 3}^{\infty} \verts{\Gamma\pars{3/2} \over \Gamma\pars{n + 1}\Gamma\pars{3/2 - n}}\, {1 \over 289^{n}} \\[3mm]&={17 \over 12}\Gamma\pars{3/2}\sum_{n = 3}^{\infty} \verts{\Gamma\pars{n - 1/2}\sin\pars{\pi\bracks{n - 1/2}} \over \pi\,\Gamma\pars{n + 1}}\,{1 \over 289^{n}} \\[3mm]&<{17 \over 24\root{\pi}}\sum_{n = 3}^{\infty} {\Gamma\pars{n - 1/2} \over \Gamma\pars{n + 1}}\,{1 \over 289^{n}} <{17 \over 24\root{\pi}}\,{\Gamma\pars{5/2} \over \Gamma\pars{4}} \sum_{n = 3}^{\infty}{1 \over 289^{n}} \\[3mm]&={17 \over 192}\,{1 \over 24054048}\approx 3.6809\times 10^{-9} \end{align}

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  • $\begingroup$ @Jean-ClaudeArbaut I'll do it later. Right now it's too late over here. Thanks. $\endgroup$ – Felix Marin Sep 2 '14 at 9:20

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