Algorithms for approximating $\sqrt{2}$

Question

Well, "Solving" is the wrong term since I am speaking about irrational numbers. I just don't know which word is the correct word... So that can be part $1$ of my question... what is the correct word since you obviously can't "solve" an irrational number because it goes forever.

Part $2$ (my real question) are there algorithms for figuring out the answer to a problem like the square root of $2$ other than guess-and-checking your way to infinity? Again, I'm obviously not asking for an algorithm to give me the never ending answer because that's crazy... but for example if I wanted to know what the $15^{th}$ decimal place of the square root of $2$ was, is there an algorithm for that?

Thank you! (I'm new here and know nothing about how to format math questions so any help or links would be appreciated as well, thanks!)

If you could help me figure out better tags to use in my question that would be nice too :) — Albert Renshaw, Jan 31 '13 at 06:17
What do you mean by 'solving' $\sqrt{2}$? I'd say that $\sqrt{2} = \sqrt{2}$. If you mean show that $\sqrt{2}$ is the root of some equation (say $f(x) = x^2-2$), then simply substitute in and show that the algebra works out. Or do you mean calculating rational approximation / decimal approximation? — Calvin Lin, Jan 31 '13 at 06:20
Are you looking for methods of computing square roots? There is a simple long-division-like method that gives you correct digits one by one. — , Jan 31 '13 at 06:31
^Brilliant! I don't have enough reputation to up-vote you but this is what I was looking for as well, thank you :) — Albert Renshaw, Jan 31 '13 at 06:46
What you're asking is how to compute decimal digits of $\sqrt{2}$. — , Feb 01 '13 at 04:35

Rustyn · Accepted Answer · 2013-01-31T07:27:37.890

20

You can use newton's method to compute the digits of $\sqrt{(2)}$:
Let: $$ f(x) = x^2 -2 $$ Define the iteration: $$ x_0 = 1\\ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} $$ This will converge to $\sqrt{2}$ quadratically.

If you want to compute other square roots:

Consider:
$$g(x) = x^2 - a$$

Which has the iterants: $$ x_{n+1}=\frac{1}{2}\left(x_n+\frac{a}{x_n}\right) $$ As mentioned below.

There's also what's called the continued fraction expansion of an algebraic number. You can use a finite continued fraction expansion.

As an example: $$ x_0 = 1 \\ x_1 = \frac{1}{2}\left(x_0 + \frac{2}{x_0}\right) =\frac{1}{2}\left( \large \textbf{1} + \frac{2}{ \large \mathbf{1}}\right) = \frac{3}{2}\\ x_2 = \frac{1}{2}\left(x_1 + \frac{2}{x_1}\right) = \frac{1}{2}\left( \large \mathbf{\frac{3}{2}} + \frac{2}{ \large \mathbf{\frac{3}{2}}}\right), \text{ etc. } $$

Added

Since we are using Newton's method, and you are wondering why it converges to the root of $f(x)$,

Note the following:
$\textbf{Theorem} $: Suppose that the function $f$ has a zero at $\alpha$, i.e., $f(\alpha) = 0$

If $f$ is continuously differentiable and its derivative is nonzero at $\alpha$, then there exists a neighborhood of $\alpha$ such that for all starting values $x_0$ in that neighborhood, the sequence ${x_n}$ will converge to $\alpha$.

So if we choose our starting guess appropriately, Newton's method always converges to the root of the equation if $f$ has these properties .

edited Jan 31 '13 at 07:27

answered Jan 31 '13 at 06:20

Rustyn

8,407

In the case of the square root of $a$, this simplifies to $x_{n+1}=\frac{1}{2}\left(x_n+\frac{a}{x_n}\right)$. – André Nicolas Jan 31 '13 at 06:23
Are you saying 1+(1-(1^2-2)/(21))+((1-(1^2-2)/(21))-(1^3-2)/(31))+(((1-(1^2-2)/(21))-(1-(1^3-2)/(31)))-(4^2-2)/(41))...
Sorry it's been a LONG time since I've worked with iterations and what-not I just want to make sure I take the sum of each n value in your defined iteration. (I'm not familiar with Newton's Method)
– Albert Renshaw Jan 31 '13 at 06:34
@AlbertRenshaw No, you don't sum them up. – Rustyn Jan 31 '13 at 06:36
What do I do with them? Or are you saying when I reach infinity that alone will be the answer? – Albert Renshaw Jan 31 '13 at 06:36
@AlbertRenshaw For example, if $x_0=1$, $x_1 = \frac{1}{2}(1+\frac{2}{1}) = \frac{3}{2}$, $x_2 = \frac{1}{2}(\frac{3}{2}+\frac{2}{1.5})$, etc. You see that if you keep doing this process indefinitely, you will obtain $\sqrt{(2)}$. – Rustyn Jan 31 '13 at 06:36
so x_∞=sqrt(2)? – Albert Renshaw Jan 31 '13 at 06:44
Can you offer any intuition as to why this happens? Or has no-one logically understood it yet? – Albert Renshaw Jan 31 '13 at 06:45
@AlbertRenshaw Yes, you see Newton's method always converges to the root of $f(x) = x^2 - 2$, which happens to be $\sqrt{(2)}$ for an appropriate starting guess. There is a proof for the convergence of newton's method out there somewhere give me a second. – Rustyn Jan 31 '13 at 06:51
Very often, the Newton Method is awesomely efficient. Roughly speaking, the number of correct digits doubles with each iteration. Try it for $\sqrt{2}$ on your calculator, starting say at $1.5$. In three iterations, you will get $\sqrt{2}$ to calculator accuracy. – André Nicolas Jan 31 '13 at 08:22
1

@AlbertRenshaw have a look at the wikipedia page (http://en.wikipedia.org/wiki/Newton's_method). There is an animation that should give you a strong intuition for how it works. – in_mathematica_we_trust Jan 31 '13 at 08:33
1

I posted a really neat alternative to Newton's method for computing square-roots efficiently at the bottom of this page: http://math.stackexchange.com/questions/287213/fast-square-roots/287453#287453. – Peder Jan 31 '13 at 15:16
If we start with the initial value $x_0$=-1 the it will converge to -$\sqrt{2}$. So we have to use bisection method for one or two iterations then we can switch over to newtons method. – Kumar Feb 01 '13 at 06:44
@AnjaliVijaya See the theorem, we choose an appropriate neighborhood. – Rustyn Feb 01 '13 at 07:00

score 13 · Answer 2 · edited Apr 13 '17 at 12:21

13

A related problem. Another way to go is the Taylor series. Derive the Taylor series of the function $\sqrt{x}$ at the point $x=1$

$$ \sqrt{x} = 1+{\frac {1}{2}} \left( x-1 \right) -{\frac {1}{8}} \left( x-1 \right) ^{2}+{\frac {1}{16}} \left( x-1 \right)^{3}-{\frac{5}{128} } \left( x-1 \right)^{4}+O\left( \left( x-1 \right) ^{5} \right). $$

If you plug in $x=2$, you get an approximate value for the $\sqrt{2}\sim 1.398437500$. Increasing the number of terms in the series improves the approximation.

Added: We can write the Taylor series of $\sqrt{x}$ explicitly by finding the $n$th derivative of $\sqrt{x}$ as

$$ \sqrt{x} = \sum _{n=0}^{\infty }\frac{\sqrt {\pi }}{2}\,{\frac {{a}^{\frac{1}{2}-n} \left( x-a\right)^{n}}{\Gamma\left( \frac{3}{2}-n \right)n! }}.$$

Substituting $a=1$ in the above formula gives the Taylor series at the point $a=1$:

$$\sqrt{x} = \sum _{n=0}^{\infty }\frac{\sqrt {\pi }}{2}\,{\frac { \left( x-1\right)^{n}}{\Gamma\left( \frac{3}{2} - n \right)n! }}.$$

Putting $x=2$ in the above equation, we have:

$$\sqrt{2} = \sum _{n=0}^{\infty }\,{\frac {\sqrt{\pi}}{2\,\Gamma\left( \frac{3}{2} - n \right)n! }}. $$

edited Apr 13 '17 at 12:21

Community

1

answered Jan 31 '13 at 06:59

Mhenni Benghorbal

47,431

Very nice indeed! – Albert Renshaw Feb 01 '13 at 03:02
@AlbertRenshaw: You are welcome. – Mhenni Benghorbal Feb 01 '13 at 03:23
4

Using taylor with $\sqrt{1+1}$ converges extremely slowly since we are so far away from the starting point of our iteration. Eg at the very edge of divergence in fact! It is much faster to use that $$ \sqrt{2} = \frac{7}{10}\sqrt{1 - \frac{1}{50}} $$ and now use the taylorexpansion with $1/50$ instead of $1$. – N3buchadnezzar Feb 16 '14 at 15:57
2

Your formula is wrong: I guess you mean: sqrt(2) = 10/7 sqrt(1-1/50). – Nov 24 '19 at 22:21

score 11 · Answer 3 · answered Jan 31 '13 at 15:31

You can also compute square roots using continued fractions. For example for $\sqrt{2}$ you have $$ \sqrt{2}=1+(\sqrt{2}-1)=1+\frac{(\sqrt{2}-1)(\sqrt{2}+1)}{\sqrt{2}+1}=1+\frac{1}{\sqrt{2}+1} $$ where $1$ is the integer part of $\sqrt{2}$. Then repeat the process for $\sqrt{2}+1$ whose integer part is $2$: $$ \sqrt{2}+1=2+(\sqrt{2}-1)=2+\frac{(\sqrt{2}-1)(\sqrt{2}+1)}{\sqrt{2}+1}=2+\frac{1}{\sqrt{2}+1} $$ therefore by repeating the process we have $$ \sqrt{2}=1+\frac{1}{2+\frac{1}{\sqrt{2}+1}}=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\cdots}}}} $$

I like this peder, thank you for elaborating on this for him. I mentioned this in my answer but your motivating example deserves a +1. — Rustyn, Jan 31 '13 at 19:43

score 4 · Answer 4 · edited Apr 10 '19 at 08:57

Following Rystyn's answer: it is nice to write down the decimals to understand how good the convergence is in Newton's method:

1.000000000000000000000000000000000000000000000000000000000000000000000  
1.500000000000000000000000000000000000000000000000000000000000000000000  
1.416666666666666666666666666666666666666666666666666666666666666666666  
1.414215686274509803921568627450980392156862745098039215686274509803921   
1.414213562374689910626295578890134910116559622115744044584905019200054  
1.414213562373095048801689623502530243614981925776197428498289498623195  
1.414213562373095048801688724209698078569671875377234001561013133113265  
1.414213562373095048801688724209698078569671875376948073176679737990732  
1.414213562373095048801688724209698078569671875376948073176679737990732 
1.414213562373095048801688724209698078569671875376948073176679737990732

Felix Marin · Answer 5 · 2014-02-17T20:11:48.243

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ It's best to evaluate $\sqrt{2} = 2\sqrt{1\over 2}$ and the guess $3/4$ for $\sqrt{1 \over 2}$: It yields $38$ exact decimal places in $5$ iterations !!!. $$ x_{n + 1} = \half\,\pars{x_{n} + {1 \over 2x_{n}}}\quad\mbox{with}\,\ n \geq 0\,,\quad x_{0} = {3 \over 4}\ \mbox{and}\ \root{2} = 2\lim_{n \to \infty}x_{n} $$

1.500000000000000000000000000000000000000 -> 2.250000000000000000000000000000000000000

1.416666666666666666666666666666666666667 -> 2.006944444444444444444444444444444444444

1.414215686274509803921568627450980392157 -> 2.000006007304882737408688965782391387928

1.414213562374689910626295578890134910117 -> 2.000000000004510950444942772099280764361

1.414213562373095048801689623502530243615 -> 2.000000000000000000000002543584239585437

1.414213562373095048801688724209698078570 -> 2.000000000000000000000000000000000000000

1.414213562373095048801688724209698078570 -> 2.000000000000000000000000000000000000000

1.414213562373095048801688724209698078570 -> 2.000000000000000000000000000000000000000

1.414213562373095048801688724209698078570 -> 2.000000000000000000000000000000000000000

1.414213562373095048801688724209698078570 -> 2.000000000000000000000000000000000000000

Algorithms for approximating $\sqrt{2}$

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