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Question

Since $sqrt2$ is irrational, is there a way to compute the first 20 digits of it?

What I have done so far

I started the first digit decimal of the $sqrt2$ by calculating iteratively so that it would not go to 3 so fast. It looks like this:

beginalign
sqrt 2 & = 1.4^2 equiv 1.96\
sqrt 2 & = 1.41^2 equiv 1.9881\
sqrt 2 & = 1.414^2 equiv 1.999396\
& ldots
endalign

First I tell whether it passes such that $1.x^2$ would be not greater than 3.

If that passes, I will add a new decimal to it. Let's say $y.$ $1.xy^2$

If that y fails, I increment $y$ by 1 and square it again.

The process will keep repeating. Unfortunately, the process takes so much time.

You can go on trying to compute the square of $1.414x$, where $x$ is a number between $0$ and $9$. The greatest number between $1.4140$ and $1.4149$ such that its square is less then $2$ is your next candidate to repeat the process. — 23 hours ago
@Gibbs I tried that so far. But the reason is that it takes more time to compute it. — 23 hours ago
Possible duplicate of 1. calculate-more-digits-of-square-root-of-2 2. is-there-any-simple-method-to-calculate-sqrt-x-without-using-logarithm 3. — 20 hours ago

J.G. 14.5k11626 · Accepted Answer · 2018-09-14 12:17:08Z

up vote
27
down vote

accepted

Calculating the square root of a number is one of the first problems tackled with numerical methods, known I think to the ancient Babylonians. The observation is that if $x,,y>0$ and $ynesqrtx$ the $y,,x/y$ will be on opposite sides of $sqrtx$, and we could try averaging them. So try $y_0=1,,y_n+1=frac12(y_n+fracxy_n)$. This is actually the Newton-Raphson method 5xum mentioned. The number of correct decimal places approximately doubles at each stage, i.e. you probably only have to go as far as $y_5$ or so.

answered 23 hours ago

J.G.

14.5k11626

9

Definitely one of the fastest methods: $$ y_0 = 1.colortan0;\ y_1 = 1.colortan5;\ y_2 = 1.41colortan666666666666666666666666666...;\ y_3 = 1.41421colortan568627450980392156862745...;\ y_4 = 1.41421356237colortan468991062629557889...;\ y_5 = 1.41421356237309504880168colortan962350...;\ cdots $$
â€“Â Oleg567
23 hours ago

6

@Oleg567 We could go even faster with post-Newton Householder methods, but the individual steps become more computationally complex. BTW the calculator you used to check that probably also used Newton-Raphson for the division.
â€“Â J.G.
23 hours ago

2

The beauty of this method is that the initial estimate can be way off and the method will converge quickly anyway. of course, making an educated guess to pick the initial estimate helps to reduce the number of iterations.
â€“Â Vasya
23 hours ago

3

Love the intuitive explanation for it!
â€“Â dbx
23 hours ago

1

@Paul Since it's Newton-Raphson it'll be about $2^n$ of them, but a more detailed answer than that can't be obtained without careful analysis of the specifics of the problem. However, if you look at how which digits have "gotten stuck", you can be confident from the shrinking error terms that they won't change. See the black digits in Oleg567's comment for an example.
â€“Â J.G.
13 hours ago

Â |Â
show 1 more comment

score 10 · Answer 2 · 2018-09-15 09:43:07Z

Here's the way I learnt to obtain decimal digit after decimal digit when I began middle school:

beginarraylcl
2&big( &colorred1.414,2dots \[1ex]
1,00&& 24times colorred4=96<100\
-96,&& 25times5=125>100\[1ex]
phantom-04,00&&281timescolorred1<400\
;:-2,81&&282times2>400\[1ex]
phantom-0119,00&&2824timescolorred4<11900\
phantom0-112,96&&2825times5>11900 \[1ex]
phantom00;604,00&&28282timescolorred2 < 60400 \
&&28283times3> 60400
endarray
&c.

Let me explain the procedure on the first two steps. It relies on a clever use of the identity $(x+y)^2=x^2+2xy+y^2$. Suppose more generally we want to find the square root of a number $a$.

We first find the greatest natural number $n$ such that $n^2le a$.

If $a$ is not a perfect square, i.e. if $n^2<a$, let $d$ be the first decimal digit of the square root. This is the greatest digit such that $;Bigl(n+frac d10Bigr)^2le a$. We'll transform this inequality into a more easy-to-use test:
beginalign
Bigl(n+frac d10Bigr)^2le a&iff frac2n10d+fracd^2100<a -n^2\
&iff (10times 2n+d)times dle (a-n^2)times 100
endalign
In practice, this means, we calculate the difference $a-n^2$ and add two 0s. Then we double $n$, add a digit d (this is the result of calculating $10times 2n+d$) and multiply what we obtain by this digit. Last, we test whether the result is less than $100(a-n^2)$, and retain the largest possible digit.

Looks interesting, can you talk us through it a bit? I don't really get it. e.g. where does 100 come from? — 9 hours ago
@goblin, There are some references for this method at math.stackexchange.com/a/538055/117057 and math.stackexchange.com/q/376365/117057 — 6 hours ago
@goblin: I've added an explanation for the first two steps. The following stepsruns along te same lines, only the first step is different. Hope this will make it clear. — 2 hours ago
@goblin You start off with 1 because 1 is the largest integer whoose square is less than 2. Then extend 1 by the next two digits, 00, to get 100. Now double the 1 just obtained and find the largest digit such that 2x times x is less than 100. — 2 hours ago

5xum 83k384148 · Answer 3 · 2018-09-14 12:13:39Z

The number $sqrt2$ is the solution to the equation $x^2-2=0$, so any method for numerically approximating the roots of an equation (such as the Newton method) will be able to approximate $sqrt2$.

Daniel Schepler 7,1181514 · Answer 4 · 2018-09-14 22:43:53Z

On a similar note to the answer by R. Romero: in the special case of taking the square root of an integer $N$, it is fairly straightforward to calculate the continued fraction representation of $sqrtN$.

In the particular case $N=2$, we have:
$$ sqrt2 = 1 + frac12 + frac12 + frac12 + ddots. $$
(This follows from the fact that if $x = sqrt2-1$, then $x = sqrt2-1 = frac1sqrt2+1 = frac12+x$.)

Now, from this we can calculate subsequent rational approximations to $sqrt2$:

$$
beginmatrix
& & 1 & 2 & 2 & 2 & 2 & 2 & cdots \
0 & 1 & 1 & 3 & 7 & 17 & 41 & 99 & cdots \
1 & 0 & 1 & 2 & 5 & 12 & 29 & 70 & cdots
endmatrix $$
So, for example $frac9970 approx 1.4142857$ whereas $sqrt2 approx 1.4142136$.

(It also happens that this procedure generates solutions to Pell's equation $a^2 - 2 b^2 = pm 1$; for example, $99^2 - 2 cdot 70^2 = 1$. The connection is: if $a^2 - 2 b^2 = pm 1$ then $a - b sqrt2 = pm frac1a + b sqrt2$; so if $a$ and $b$ are large positive integers satisfying Pell's equation, then $a - bsqrt2 approx pmfrac12a$ which implies $fracab - sqrt2 approx pmfrac12ab approx pmfrac1a^2sqrt2$.)

Is there somewhere I can read more about this, especially the connection between continued fractions and Pell's equation? — 9 hours ago

asky 1318 · Answer 5 · 2018-09-14 22:33:21Z

There's a general method that converges about as quickly as Newton-Raphson but is somewhat more general. It's based off of Continued Fractions:

Suppose you want to find the square root of $N$. Let $a+b = N$ where $b$ has an easy to calculate square root.

let $y_n+1 = sqrt b + fraca sqrt b + y_n$

$y_n+1$ converges to $sqrt N$.

Tyberius 7572621 · Answer 6 · 2018-09-15 03:29:23Z

Suppose you want to find the square root of $p$ and suppose your initial guess is $x/y$:

Let $mathbf M=beginbmatrix 1 & p \ 1 & 1 endbmatrix$ and $mathbf q=beginpmatrix x \ y endpmatrix$
Then $mathbf Mmathbf Mmathbf M...mathbf q$ gives a numerator and denominator the ratio of which converges to the square root of $p$. This gives an approximation to the square root of $2$ as fast as the other methods but with no floating point arithmetic until the final division.

Performs well for calculation tools optimized for Matrix arithmetic. This also gives you solutions for Pell's equation for $p=2$ as mentioned by Daniel Schepler.

Jaroslaw Matlak 3,948830 · Answer 7 · 2018-09-14 13:05:49Z

You can compute it manually using the algorithm:

$p=0$, $r=0$, $i=0$

Split the number into sections of two digits

Take i'th section $n_i$, let $k=100t+n_i$

Find the greatest number $x$, such that $$y=x(20p+x)leq k$$

Assign $p=10 p + x$, $i=i+1$, if the accyracy of the result is not satisfied, then return to 3.

Example:

02.00 00 00 00 00

$n_0 = 2$, $k=2$, therefore for $x=1$: $y=1$ and $p=1$

$n_1=0$, $k=100$, so for $x=4$: $y=24*4=96<100$ and $p=14$

$n_2=0$, $k=400$, so for $x=1$, $y=281*1=281<400$ and $p=141$

$n_3=0$, $k=11900$, so for $x=4$, $y=2824*4=11296<11900$ and $p=1414$

$n_4=0$, $k=60400$, so for $x=2$, $y=28282*2=56564<60400$ and $p=14142$

$n_5=0$, $k=383600$, so for $x=1$, $y=282841*1=282841<383600$ and $p=141421$

...

After all just remember to point the comma in place, where it should be, ie. after first number (it depends how many sections were there on the left side of our number), so you'll have:
$$sqrt2approx 1.41421$$

To obtain accuracy of 20 numbers after the comma, you should append 20 sections of 00 in the step 2. , ie.:

02.00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

Toby Mak 2,8651925 · Answer 8 · 2018-09-14 13:49:32Z

Using the fact that $sin fracpi4 = fracsqrt22$, then we have to find $2 sin fracpi4$.

We can approximate $sin x$ using the Taylor series to three terms:

$$sin x = x - fracx^33! + fracx^55! + O(x^6),$$

so we have:

$$sin fracpi4 approx fracpi4 - frac(pi/4)^33! + frac(pi/4)^55! .$$

If we approximate $pi$ as $frac227$, then we have $fracpi4 = frac1114$, then we have:

$$sin fracpi4 approxfrac1114 - frac(11/14)^33! + frac(11/14)^55!,$$

which when you multiply by $2$ to get $sqrt2$, gives $1.4147$, while the actual value is $1.4142$.

If we expand the Taylor series to more terms, or improve the approximation of $pi$ (such as $frac355113$), then we can get to $20$ correct digits.

DonÃ¢Â€Â™t you need pi to nearly 20 digits for this to work? — 9 hours ago

M. Wind 1,874715 · Answer 9 · 2018-09-15 05:36:50Z

Let us start with an initial guess $x$ for the square root of $2$. Then add a correction term $y$. Write down $(x+y)^2 - 2 = 0$. Solve this equation for $y$ by expanding it up to third order in the difference $(2-x^2)$. This is a straightforward calculation. Combining all contributions, the result is simple and elegant:

$$y = (x^4+12x^2+4)/(4x^3+8x)$$

For a rational initial guess $x$ the result $y$ is also rational, but far more close to the desired value. For example if we take $x = 3/2$, then $y=577/408$, which differs from the square root of 2 by a factor 1.0000015.

score 0 · Answer 10 · 2018-09-15 11:09:18Z

Newton-Rhapson is a good idea because of the convergence rate. However, I am more of a fan of using Taylor's expansions here since it is super easy to derive on the go to give fairly ok estimates in quite a reasonable time. So, the way to go to find $sqrtx$ is to find first the closest integer which approximates $sqrtx$ and call this $a$, then apply Taylor to $a^2$. Then Taylor says
$$ sqrtx approx a + (x-a^2)cdot frac12 a - (x-a^2)^2/2 cdot frac14 a^3 + cdots. $$
The thing that is nice here is that you also get bounds on the error you make. So, denote $f(x) = sqrtx$, then the error of a $n$th order approximation (i.e., going as far as $(x-a^2)^n/n! cdot f^(n)(a^2)$ in the approximation above) is given by $$ (x-a)^n+1/(n+1)! cdot f^(n+1)(xi)$$ for a certain $xi$ between $a^2$ and $x$. This can be estimated quite easily since this $f^(n+1)$ is monotone around $x$. Thus look at the boundaries of the domain of $xi$ and find the 'best' maximal value which you can calculate without a calculator.

Example for $x=2$. Apparently $1$ is the closest integer to $sqrt2$ and thus we will take $a=1$. Then, let's take a second order approximation
$$sqrt2 approx 1 + (2-1)cdot frac12 - (2-1)^2/2cdot frac14 = 1 + 0.5 - 0.125 = 1.375 $$
and the absolute error is given by
$$ E=left|(2-1)^3/3!cdot frac38 cdot xi^2sqrtxiright| = frac116 cdot frac1$$
for a certain $xi$ between $1$ and $2$. Since this is a decreasing function on $(1,2)$. The maximum is attained at $1$ and hence the error is bounded by
$$ E leq frac116 $$
which seems to be a good estimate since $E = 0.039dots$ and $1/16 = 0.0625$.

Edit As some of you noted this method 'looks' more difficult than Newton-Rhapson and the convergence is slower. The last part is obviously true and I would answer this question with: How quick do you need it to be and do you want to calculate it in your head or do you have a computer? Do you need to have a quick guess which is approximately equal to the value of $sqrt2$ or do you need a precise estimate. If you don't have a computer but pen and paper, the best method is Newton-Rhapson.

I would argue that my method is better if you don't have pen and paper or a computer and you are asked to give an estimation of $sqrt10$ on the go (especially for $sqrtx$ with $x$ big, the Taylor approximation is better since the $sqrtbullet$ function becomes more linear as $x$ grows).

I agree that my method looks way more difficult but it isn't if you get more familiar with it. Also, this method is super quick in terms of calculation time in your head and if you practice a little with it, it becomes way easier. Also, this method works particularly nice for $sqrtx$ where $x$ differs one from a perfect square because then the $(x-a^2)^n$ term will always be one.

Let's look at an example here. Suppose you need to calculate $sqrt122$, then first order approximation of my method gives
$$ sqrt122 approx 11 + frac12cdot 11. $$
It took me less than one second to find this approximation and the second order approximation works almost as quick here. You just need to add $frac-18cdot 11^3$. Please note that the error of the first order approximation here is approximately equal to $10^-4$.

If you apply Newton-Rhapson here you get the same approximation after one step if you choose $x_0=11$. The only thing is that I always forget what the exact form is of Newton-Rhapson. So when I want to apply it, I have to think about it where I could have immediately applied Taylor but I would say that is just my particular preference.

I'd say this is more difficult, less precise, and not as generally applicable as Newton-Raphson. — 21 hours ago
I would say it is less difficult since when you apply Newton-Rhapson you always have to find the exact algorithm and this method can be applied to find $sqrt2.243$ also quite quickly. — 20 hours ago
I agree with @leftaroundabout, but perhaps if you edit into your post an illustration of how this method could be used by hand to compute rad 2 to high accuracy, it would appear simpler. Right now, it looks much more difficult. — 17 hours ago
Taylor's converges much more slowly than Newton Raphson. Note the second order term starting with initial guess 1 is 1.4166.... already correct to two digits behind the decimal. You might get an additional correct digit at each step of the calculation heavy Taylor series. The accuracy doubles per step for Newton Raphson without the difficulty of calculating the Taylor coefficients. There might be ways to patch it up. There's an alternative series to the Taylor series for arctan that converges much faster than Taylor. — 9 hours ago

J.G. 14.5k11626 · Accepted Answer · 2018-09-14 12:17:08Z

up vote
27
down vote

accepted

Calculating the square root of a number is one of the first problems tackled with numerical methods, known I think to the ancient Babylonians. The observation is that if $x,,y>0$ and $ynesqrtx$ the $y,,x/y$ will be on opposite sides of $sqrtx$, and we could try averaging them. So try $y_0=1,,y_n+1=frac12(y_n+fracxy_n)$. This is actually the Newton-Raphson method 5xum mentioned. The number of correct decimal places approximately doubles at each stage, i.e. you probably only have to go as far as $y_5$ or so.

answered 23 hours ago

J.G.

14.5k11626

9

Definitely one of the fastest methods: $$ y_0 = 1.colortan0;\ y_1 = 1.colortan5;\ y_2 = 1.41colortan666666666666666666666666666...;\ y_3 = 1.41421colortan568627450980392156862745...;\ y_4 = 1.41421356237colortan468991062629557889...;\ y_5 = 1.41421356237309504880168colortan962350...;\ cdots $$
â€“Â Oleg567
23 hours ago

6

@Oleg567 We could go even faster with post-Newton Householder methods, but the individual steps become more computationally complex. BTW the calculator you used to check that probably also used Newton-Raphson for the division.
â€“Â J.G.
23 hours ago

2

The beauty of this method is that the initial estimate can be way off and the method will converge quickly anyway. of course, making an educated guess to pick the initial estimate helps to reduce the number of iterations.
â€“Â Vasya
23 hours ago

3

Love the intuitive explanation for it!
â€“Â dbx
23 hours ago

1

@Paul Since it's Newton-Raphson it'll be about $2^n$ of them, but a more detailed answer than that can't be obtained without careful analysis of the specifics of the problem. However, if you look at how which digits have "gotten stuck", you can be confident from the shrinking error terms that they won't change. See the black digits in Oleg567's comment for an example.
â€“Â J.G.
13 hours ago

Â |Â
show 1 more comment

score 10 · Answer 12 · 2018-09-15 09:43:07Z

Here's the way I learnt to obtain decimal digit after decimal digit when I began middle school:

beginarraylcl
2&big( &colorred1.414,2dots \[1ex]
1,00&& 24times colorred4=96<100\
-96,&& 25times5=125>100\[1ex]
phantom-04,00&&281timescolorred1<400\
;:-2,81&&282times2>400\[1ex]
phantom-0119,00&&2824timescolorred4<11900\
phantom0-112,96&&2825times5>11900 \[1ex]
phantom00;604,00&&28282timescolorred2 < 60400 \
&&28283times3> 60400
endarray
&c.

Let me explain the procedure on the first two steps. It relies on a clever use of the identity $(x+y)^2=x^2+2xy+y^2$. Suppose more generally we want to find the square root of a number $a$.

We first find the greatest natural number $n$ such that $n^2le a$.

If $a$ is not a perfect square, i.e. if $n^2<a$, let $d$ be the first decimal digit of the square root. This is the greatest digit such that $;Bigl(n+frac d10Bigr)^2le a$. We'll transform this inequality into a more easy-to-use test:
beginalign
Bigl(n+frac d10Bigr)^2le a&iff frac2n10d+fracd^2100<a -n^2\
&iff (10times 2n+d)times dle (a-n^2)times 100
endalign
In practice, this means, we calculate the difference $a-n^2$ and add two 0s. Then we double $n$, add a digit d (this is the result of calculating $10times 2n+d$) and multiply what we obtain by this digit. Last, we test whether the result is less than $100(a-n^2)$, and retain the largest possible digit.

Looks interesting, can you talk us through it a bit? I don't really get it. e.g. where does 100 come from? — 9 hours ago
@goblin, There are some references for this method at math.stackexchange.com/a/538055/117057 and math.stackexchange.com/q/376365/117057 — 6 hours ago
@goblin: I've added an explanation for the first two steps. The following stepsruns along te same lines, only the first step is different. Hope this will make it clear. — 2 hours ago
@goblin You start off with 1 because 1 is the largest integer whoose square is less than 2. Then extend 1 by the next two digits, 00, to get 100. Now double the 1 just obtained and find the largest digit such that 2x times x is less than 100. — 2 hours ago

5xum 83k384148 · Answer 13 · 2018-09-14 12:13:39Z

The number $sqrt2$ is the solution to the equation $x^2-2=0$, so any method for numerically approximating the roots of an equation (such as the Newton method) will be able to approximate $sqrt2$.

Daniel Schepler 7,1181514 · Answer 14 · 2018-09-14 22:43:53Z

On a similar note to the answer by R. Romero: in the special case of taking the square root of an integer $N$, it is fairly straightforward to calculate the continued fraction representation of $sqrtN$.

In the particular case $N=2$, we have:
$$ sqrt2 = 1 + frac12 + frac12 + frac12 + ddots. $$
(This follows from the fact that if $x = sqrt2-1$, then $x = sqrt2-1 = frac1sqrt2+1 = frac12+x$.)

Now, from this we can calculate subsequent rational approximations to $sqrt2$:

$$
beginmatrix
& & 1 & 2 & 2 & 2 & 2 & 2 & cdots \
0 & 1 & 1 & 3 & 7 & 17 & 41 & 99 & cdots \
1 & 0 & 1 & 2 & 5 & 12 & 29 & 70 & cdots
endmatrix $$
So, for example $frac9970 approx 1.4142857$ whereas $sqrt2 approx 1.4142136$.

(It also happens that this procedure generates solutions to Pell's equation $a^2 - 2 b^2 = pm 1$; for example, $99^2 - 2 cdot 70^2 = 1$. The connection is: if $a^2 - 2 b^2 = pm 1$ then $a - b sqrt2 = pm frac1a + b sqrt2$; so if $a$ and $b$ are large positive integers satisfying Pell's equation, then $a - bsqrt2 approx pmfrac12a$ which implies $fracab - sqrt2 approx pmfrac12ab approx pmfrac1a^2sqrt2$.)

Is there somewhere I can read more about this, especially the connection between continued fractions and Pell's equation? — 9 hours ago

asky 1318 · Answer 15 · 2018-09-14 22:33:21Z

There's a general method that converges about as quickly as Newton-Raphson but is somewhat more general. It's based off of Continued Fractions:

Suppose you want to find the square root of $N$. Let $a+b = N$ where $b$ has an easy to calculate square root.

let $y_n+1 = sqrt b + fraca sqrt b + y_n$

$y_n+1$ converges to $sqrt N$.

Tyberius 7572621 · Answer 16 · 2018-09-15 03:29:23Z

Suppose you want to find the square root of $p$ and suppose your initial guess is $x/y$:

Let $mathbf M=beginbmatrix 1 & p \ 1 & 1 endbmatrix$ and $mathbf q=beginpmatrix x \ y endpmatrix$
Then $mathbf Mmathbf Mmathbf M...mathbf q$ gives a numerator and denominator the ratio of which converges to the square root of $p$. This gives an approximation to the square root of $2$ as fast as the other methods but with no floating point arithmetic until the final division.

Performs well for calculation tools optimized for Matrix arithmetic. This also gives you solutions for Pell's equation for $p=2$ as mentioned by Daniel Schepler.

Jaroslaw Matlak 3,948830 · Answer 17 · 2018-09-14 13:05:49Z

You can compute it manually using the algorithm:

$p=0$, $r=0$, $i=0$

Split the number into sections of two digits

Take i'th section $n_i$, let $k=100t+n_i$

Find the greatest number $x$, such that $$y=x(20p+x)leq k$$

Assign $p=10 p + x$, $i=i+1$, if the accyracy of the result is not satisfied, then return to 3.

Example:

02.00 00 00 00 00

$n_0 = 2$, $k=2$, therefore for $x=1$: $y=1$ and $p=1$

$n_1=0$, $k=100$, so for $x=4$: $y=24*4=96<100$ and $p=14$

$n_2=0$, $k=400$, so for $x=1$, $y=281*1=281<400$ and $p=141$

$n_3=0$, $k=11900$, so for $x=4$, $y=2824*4=11296<11900$ and $p=1414$

$n_4=0$, $k=60400$, so for $x=2$, $y=28282*2=56564<60400$ and $p=14142$

$n_5=0$, $k=383600$, so for $x=1$, $y=282841*1=282841<383600$ and $p=141421$

...

After all just remember to point the comma in place, where it should be, ie. after first number (it depends how many sections were there on the left side of our number), so you'll have:
$$sqrt2approx 1.41421$$

To obtain accuracy of 20 numbers after the comma, you should append 20 sections of 00 in the step 2. , ie.:

02.00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

Toby Mak 2,8651925 · Answer 18 · 2018-09-14 13:49:32Z

Using the fact that $sin fracpi4 = fracsqrt22$, then we have to find $2 sin fracpi4$.

We can approximate $sin x$ using the Taylor series to three terms:

$$sin x = x - fracx^33! + fracx^55! + O(x^6),$$

so we have:

$$sin fracpi4 approx fracpi4 - frac(pi/4)^33! + frac(pi/4)^55! .$$

If we approximate $pi$ as $frac227$, then we have $fracpi4 = frac1114$, then we have:

$$sin fracpi4 approxfrac1114 - frac(11/14)^33! + frac(11/14)^55!,$$

which when you multiply by $2$ to get $sqrt2$, gives $1.4147$, while the actual value is $1.4142$.

If we expand the Taylor series to more terms, or improve the approximation of $pi$ (such as $frac355113$), then we can get to $20$ correct digits.

DonÃ¢Â€Â™t you need pi to nearly 20 digits for this to work? — 9 hours ago

M. Wind 1,874715 · Answer 19 · 2018-09-15 05:36:50Z

Let us start with an initial guess $x$ for the square root of $2$. Then add a correction term $y$. Write down $(x+y)^2 - 2 = 0$. Solve this equation for $y$ by expanding it up to third order in the difference $(2-x^2)$. This is a straightforward calculation. Combining all contributions, the result is simple and elegant:

$$y = (x^4+12x^2+4)/(4x^3+8x)$$

For a rational initial guess $x$ the result $y$ is also rational, but far more close to the desired value. For example if we take $x = 3/2$, then $y=577/408$, which differs from the square root of 2 by a factor 1.0000015.

score 0 · Answer 20 · 2018-09-15 11:09:18Z

Newton-Rhapson is a good idea because of the convergence rate. However, I am more of a fan of using Taylor's expansions here since it is super easy to derive on the go to give fairly ok estimates in quite a reasonable time. So, the way to go to find $sqrtx$ is to find first the closest integer which approximates $sqrtx$ and call this $a$, then apply Taylor to $a^2$. Then Taylor says
$$ sqrtx approx a + (x-a^2)cdot frac12 a - (x-a^2)^2/2 cdot frac14 a^3 + cdots. $$
The thing that is nice here is that you also get bounds on the error you make. So, denote $f(x) = sqrtx$, then the error of a $n$th order approximation (i.e., going as far as $(x-a^2)^n/n! cdot f^(n)(a^2)$ in the approximation above) is given by $$ (x-a)^n+1/(n+1)! cdot f^(n+1)(xi)$$ for a certain $xi$ between $a^2$ and $x$. This can be estimated quite easily since this $f^(n+1)$ is monotone around $x$. Thus look at the boundaries of the domain of $xi$ and find the 'best' maximal value which you can calculate without a calculator.

Example for $x=2$. Apparently $1$ is the closest integer to $sqrt2$ and thus we will take $a=1$. Then, let's take a second order approximation
$$sqrt2 approx 1 + (2-1)cdot frac12 - (2-1)^2/2cdot frac14 = 1 + 0.5 - 0.125 = 1.375 $$
and the absolute error is given by
$$ E=left|(2-1)^3/3!cdot frac38 cdot xi^2sqrtxiright| = frac116 cdot frac1$$
for a certain $xi$ between $1$ and $2$. Since this is a decreasing function on $(1,2)$. The maximum is attained at $1$ and hence the error is bounded by
$$ E leq frac116 $$
which seems to be a good estimate since $E = 0.039dots$ and $1/16 = 0.0625$.

Edit As some of you noted this method 'looks' more difficult than Newton-Rhapson and the convergence is slower. The last part is obviously true and I would answer this question with: How quick do you need it to be and do you want to calculate it in your head or do you have a computer? Do you need to have a quick guess which is approximately equal to the value of $sqrt2$ or do you need a precise estimate. If you don't have a computer but pen and paper, the best method is Newton-Rhapson.

I would argue that my method is better if you don't have pen and paper or a computer and you are asked to give an estimation of $sqrt10$ on the go (especially for $sqrtx$ with $x$ big, the Taylor approximation is better since the $sqrtbullet$ function becomes more linear as $x$ grows).

I agree that my method looks way more difficult but it isn't if you get more familiar with it. Also, this method is super quick in terms of calculation time in your head and if you practice a little with it, it becomes way easier. Also, this method works particularly nice for $sqrtx$ where $x$ differs one from a perfect square because then the $(x-a^2)^n$ term will always be one.

Let's look at an example here. Suppose you need to calculate $sqrt122$, then first order approximation of my method gives
$$ sqrt122 approx 11 + frac12cdot 11. $$
It took me less than one second to find this approximation and the second order approximation works almost as quick here. You just need to add $frac-18cdot 11^3$. Please note that the error of the first order approximation here is approximately equal to $10^-4$.

If you apply Newton-Rhapson here you get the same approximation after one step if you choose $x_0=11$. The only thing is that I always forget what the exact form is of Newton-Rhapson. So when I want to apply it, I have to think about it where I could have immediately applied Taylor but I would say that is just my particular preference.

I'd say this is more difficult, less precise, and not as generally applicable as Newton-Raphson. — 21 hours ago
I would say it is less difficult since when you apply Newton-Rhapson you always have to find the exact algorithm and this method can be applied to find $sqrt2.243$ also quite quickly. — 20 hours ago
I agree with @leftaroundabout, but perhaps if you edit into your post an illustration of how this method could be used by hand to compute rad 2 to high accuracy, it would appear simpler. Right now, it looks much more difficult. — 17 hours ago
Taylor's converges much more slowly than Newton Raphson. Note the second order term starting with initial guess 1 is 1.4166.... already correct to two digits behind the decimal. You might get an additional correct digit at each step of the calculation heavy Taylor series. The accuracy doubles per step for Newton Raphson without the difficulty of calculating the Taylor coefficients. There might be ways to patch it up. There's an alternative series to the Taylor series for arctan that converges much faster than Taylor. — 9 hours ago

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