Definitions

# Methods of computing square roots

This article presents and explains several methods which can be used to calculate square roots.

## Exponential identity

Pocket calculators typically implement good routines to compute the exponential function and the natural logarithm, and then compute the square root of S using the identity

$sqrt\left\{S\right\} = e^\left\{frac\left\{1\right\}\left\{2\right\}ln S\right\}.$

The same identity is used when computing square roots with logarithm tables or slide rules.

## Rough estimation

Many of the methods for calculating square roots of a positive real number S require an initial seed value. If the initial value is too far from the actual square root, the calculation will be slowed down. It is therefore useful to have a rough estimate, which may be very inaccurate but easy to calculate. If S ≥ 1, let D be the number of digits to the left of the decimal point. If S < 1, let D be the negative of the number of zeros to the immediate right of the decimal point. Then the rough estimation is this:
If D is odd, D = 2n + 1, then use $sqrt\left\{S\right\} approx 2 cdot 10^n.$
If D is even, D = 2n + 2, then use $sqrt\left\{S\right\} approx 6 cdot 10^n.$

Two and six are used because $sqrt\left\{sqrt\left\{1 cdot 10\right\}\right\} = sqrt\left[4\right]\left\{10\right\} approx 2 ,$ and $sqrt\left\{sqrt\left\{10 cdot 100\right\}\right\} = sqrt\left[4\right]\left\{1000\right\} approx 6 ,.$

When working in the binary numeral system (as computers do internally), an alternative method is to use $2^\left\{leftlfloor D/2rightrfloor\right\}$ (here D is the number of binary digits).

## Babylonian method

Perhaps the first algorithm used for approximating $sqrt S$ is known as the "Babylonian method", named after the Babylonians, or "Heron's method", named after the first-century Greek mathematician Heron of Alexandria who gave the first explicit description of the method. It can be derived from (but predates) Newton's method. This is a quadratically convergent algorithm, which means that the number of correct digits of the approximation roughly doubles with each iteration. It proceeds as follows:

1. Start with an arbitrary positive start value x0 (the closer to the root, the better).
2. Let xn+1 be the average of xn and S / xn (using the arithmetic mean to approximate the geometric mean).
3. Repeat steps 2 and 3, until the desired accuracy is achieved.

It can also be represented as:

$x_0 approx sqrt\left\{S\right\}.$
$x_\left\{n+1\right\} = frac\left\{1\right\}\left\{2\right\} left\left(x_n + frac\left\{S\right\}\left\{x_n\right\}right\right),$
$sqrt S = lim_\left\{n to infty\right\} x_n.$

This algorithm works equally well in the p-adic numbers, but cannot be used to identify real square roots with p-adic square roots; it is easy, for example, to construct a sequence of rational numbers by this method which converges to $+3$ in the reals, but to $-3$ in the 2-adics.

### Example

To calculate $sqrt\left\{S\right\}$, where S = 125348, to 6 significant figures, we will use the rough estimation method above to get x0. The number of digits in S is D=6=2·2+2. So, n=2 and the rough estimate is
$x_0 = 6 cdot 10^2 = 600.000 ,.$

$x_1 = frac\left\{1\right\}\left\{2\right\} left\left(x_0 + frac\left\{S\right\}\left\{x_0\right\}right\right) = frac\left\{1\right\}\left\{2\right\} left\left(600.000 + frac\left\{125348\right\}\left\{600.000\right\}right\right) = 404.457 ,.$

$x_2 = frac\left\{1\right\}\left\{2\right\} left\left(x_1 + frac\left\{S\right\}\left\{x_1\right\}right\right) = frac\left\{1\right\}\left\{2\right\} left\left(404.457 + frac\left\{125348\right\}\left\{404.457\right\}right\right) = 357.187 ,.$

$x_3 = frac\left\{1\right\}\left\{2\right\} left\left(x_2 + frac\left\{S\right\}\left\{x_2\right\}right\right) = frac\left\{1\right\}\left\{2\right\} left\left(357.187 + frac\left\{125348\right\}\left\{357.187\right\}right\right) = 354.059 ,.$

$x_4 = frac\left\{1\right\}\left\{2\right\} left\left(x_3 + frac\left\{S\right\}\left\{x_3\right\}right\right) = frac\left\{1\right\}\left\{2\right\} left\left(354.059 + frac\left\{125348\right\}\left\{354.059\right\}right\right) = 354.045 ,.$

$x_5 = frac\left\{1\right\}\left\{2\right\} left\left(x_4 + frac\left\{S\right\}\left\{x_4\right\}right\right) = frac\left\{1\right\}\left\{2\right\} left\left(354.045 + frac\left\{125348\right\}\left\{354.045\right\}right\right) = 354.045 ,.$
Therefore, $sqrt\left\{125348\right\} approx 354.045 ,.$

### Convergence

We let the relative error in xn be defined by
$epsilon_n = frac \left\{x_n\right\}\left\{sqrt\left\{S\right\}\right\} - 1$

and thus

$x_n = sqrt \left\{S\right\} cdot \left(1 + epsilon_n\right).$

Then one can show that

$epsilon_\left\{n+1\right\} = frac \left\{epsilon_n^2\right\}\left\{2 \left(1 + epsilon_n\right)\right\}$

and thus that

$0 leq epsilon_\left\{n+2\right\} leq min \left(frac \left\{epsilon_\left\{n+1\right\}^2\right\}\left\{2\right\}, frac \left\{epsilon_\left\{n+1\right\}\right\}\left\{2\right\}\right)$

and consequently that convergence is assured provided that x0 and S are both positive.

#### Worst case for convergence

If one is using the rough estimate above with the Babylonian method, then the worst cases are:
$S = 1; , x_0 = 2; , x_1 = 1.250; , epsilon_1 = 0.250 ,.$
$S = 10; , x_0 = 2; , x_1 = 3.500; , epsilon_1 < 0.107 ,.$
$S = 10; , x_0 = 6; , x_1 = 3.833; , epsilon_1 < 0.213 ,.$
$S = 100; , x_0 = 6; , x_1 = 11.333; , epsilon_1 < 0.134 ,.$

Thus in any case,

$epsilon_1 leq 2^\left\{-2\right\} ,.$
$epsilon_2 < 2^\left\{-5\right\} < 10^\left\{-1\right\} ,.$
$epsilon_3 < 2^\left\{-11\right\} < 10^\left\{-3\right\} ,.$
$epsilon_4 < 2^\left\{-23\right\} < 10^\left\{-6\right\} ,.$
$epsilon_5 < 2^\left\{-47\right\} < 10^\left\{-14\right\} ,.$
$epsilon_6 < 2^\left\{-95\right\} < 10^\left\{-28\right\} ,.$
$epsilon_7 < 2^\left\{-191\right\} < 10^\left\{-57\right\} ,.$
$epsilon_8 < 2^\left\{-383\right\} < 10^\left\{-115\right\} ,.$

Remember that round off errors will slow the convergence. One should keep at least one extra digit beyond the desired accuracy of the xn which you are calculating to minimize round off error.

## Bakhshali approximation

This is a method for finding an approximation to a square root which was described in an ancient manuscript known as the Bakhshali Manuscript. It is equivalent to two iterations of the Babylonian method beginning with N. The original presentation goes as follows: To calculate $sqrt\left\{S\right\}$, let N2 be the nearest perfect square to S. Then, calculate:

$d = S - N^2 ,!$

$P = frac\left\{d\right\}\left\{2N\right\}$

$A = N + P,!$

$sqrt\left\{S\right\} approx A - frac\left\{P^2\right\}\left\{2A\right\}$
This can be also written as:
$sqrt\left\{S\right\} approx N + frac\left\{d\right\}\left\{2N\right\} - frac\left\{d^2\right\}\left\{8N^3 + 4Nd\right\} = frac\left\{8N^4 + 8N^2 d + d^2\right\}\left\{8N^3 + 4Nd\right\}$

### Example

We'll find $sqrt\left\{9.2345\right\}.$

$N=3,!$

$d = 9.2345 - 3^2 = 0.2345,!$

$P = frac\left\{0.2345\right\}\left\{2 times 3\right\} = 0.0391$

$A = 3 + 0.0391 = 3.0391,!$

$sqrt\left\{9.2345\right\} approx 3.0391 - frac\left\{0.0391^2\right\}\left\{2 times 3.0391\right\} approx 3.0388$

## Digit by digit calculation

This is a method to find each digit of the square root in a sequence. It is slower than the Babylonian method (if you have a calculator which can divide in one operation), but it has several advantages:

• It can be easier for manual calculations.
• Every digit of the root found is known to be correct, i.e. it will not have to be changed later.
• If the square root has an expansion which terminates, the algorithm will terminate after the last digit is found. Thus, it can be used to check whether a given integer is a square number.

Napier's bones include an aid for the execution of this algorithm. The Shifting nth-root algorithm is a generalization of this method.

The algorithm works for any base, and naturally, the way it proceeds depends on the base chosen.

### Decimal (base 10)

Write the original number in decimal form. The numbers are written similar to the long division algorithm, and, as in long division, the root will be written on the line above. Now separate the digits into pairs, starting from the decimal point and going both left and right. The decimal point of the root will be above the decimal point of the square. One digit of the root will appear above each pair of digits of the square.

Beginning with the left-most pair of digits, do the following procedure for each pair:

1. Starting on the left, bring down the most significant (leftmost) pair of digits not yet used (if all the digits have been used, write "00") and write them to the right of the remainder from the previous step (on the first step, there will be no remainder). In other words, multiply the remainder by 100 and add the two digits. This will be the current value c.
2. Find p, y and x, as follows:
• Let p be the part of the root found so far, ignoring any decimal point. (For the first step, p = 0).
• Let y equal $\left(20 cdot p + x\right) cdot x$.
• Determine the greatest digit x such that $\left(20 cdot p + x\right) cdot x$ (that is, y) does not exceed c.
• Note: 20p + x is simply twice p, with the digit x appended to the right).
• Note: You can find x by guessing what c/(20·p) is and doing a trial calculation of y, then adjusting x upward or downward as necessary.
• Place the digit $x$ as the next digit of the root, i.e above the two digits of the square which you just brought down. Thus the next p will be the old p times 10 plus x.
3. Subtract y from c to form a new remainder.
4. If the remainder is zero and there are no more digits to bring down, then the algorithm has terminated. Otherwise go back to step 1 for another iteration.

#### Examples

Find the square root of 152.2756.

`          1  2. 3  4 `
`     /  01 52.27 56`
`         01                   1*1 <= 1 < 2*2                 x = 1`
`         01                     y = x*x = 1*1 = 1`
`         00 52                22*2 <= 52 < 23*3              x = 2`
`         00 44                  y = (20+x)*x = 22*2 = 44`
`            08 27             243*3 <= 827 < 244*4           x = 3`
`            07 29               y = (240+x)*x = 243*3 = 729`
`               98 56          2464*4 <= 9856 < 2465*5        x = 4`
`               98 56            y = (2460+x)*x = 2464*4 = 9856`
`               00 00          Algorithm terminates: Answer is 12.34`

Find the square root of 2.

`          1. 4  1  4  2`
`     /  02.00 00 00 00`
`         02                  1*1 <= 2 < 2*2                 x = 1`
`         01                    y = x*x = 1*1 = 1`
`         01 00               24*4 <= 100 < 25*5             x = 4`
`         00 96                 y = (20+x)*x = 24*4 = 96`
`            04 00            281*1 <= 400 < 282*2           x = 1`
`            02 81              y = (280+x)*x = 281*1 = 281`
`            01 19 00         2824*4 <= 11900 < 2825*5       x = 4`
`            01 12 96           y = (2820+x)*x = 2824*4 = 11296`
`               06 04 00      28282*2 <= 60400 < 28283*3     x = 2`
`                             The desired precision is achieved:`
`                             The square root of 2 is about 1.4142`

### Binary (base 2)

Inherent to digit-by-digit algorithms is a search and test step: find a digit, $, e$, when added to the right of a current solution $, r$', such that $,\left(r+e\right)cdot\left(r+e\right) le x$, where $, x$ is the value for which a root is desired. Expanding, we obtain $,rcdot r + 2re + ecdot e le x$. The current value of $,rcdot r$—or, usually, the remainder—can be incrementally updated efficiently when working in binary, as the value of $, e$ will be a single bit, and the operations needed to compute $,2cdot rcdot e$ and $,ecdot e$ can be replaced with faster bit shift operations. This gives rise to simple computer implementations:

`   int sqrt(int num) {`
`       int op = num;`
`       int res = 0;`
`       int one = 1 << 14; // The second-to-top bit is set: 1L<<30 for long`
`       // "one" starts at the highest power of four <= the argument.`
`       while (one > op)`
`           one >>= 2;`

`       while (one != 0) {`
`           if (op >= res + one) {`
`               op -= res + one;`
`               res += one << 1;`
}
`           res >>= 1;`
`           one >>= 2;`
}
`       return res;`
}

Faster algorithms, in binary and decimal or any other base, can be realized by using lookup tables—in effect trading more storage space for reduced run time.

## Duplex method for extracting a square root

The duplex method is a variant of the digit by digit method for calculating the square root of a whole or decimal number one digit at a time. The duplex is the square of the central digit plus double the cross-product of digits equidistant from the center. The duplex is computed from the quotient digits (square root digits) computed thus far, but after the initial digits. The duplex is subtracted from the dividend digit prior to the second subtraction for the product of the quotient digit times the divisor digit. For perfect squares the duplex and the dividend will get smaller and reach zero after a few steps. For non-perfect squares the decimal value of the square root can be calculated to any precision desired. However, as the decimal places proliferate, the duplex adjustment gets larger and longer to calculate. The duplex method follows the Vedic ideal for an algorithm, one-line, mental calculation. It is flexible in choosing the first digit group and the divisor. Small divisors are to be avoided by starting with a larger initial group.

In short, to calculate the duplex of a number, double the product of each pair of equidistant digits plus the square of the center digit (of the digits to the right of the colon).

`Number => Calculation = Duplex`
`574 ==> 2(5·4) + 72 = 89`
`406,739 ==> 2(4·9)+ 2(0·3)+ 2(6·7) = 72+0+84  = 156`
`123,456 ==> 2(1·6)+ 2(2·5)+ 2(3·4) = 12 +20 +24  = 56`
`88,900,777 ==> 2(56)+2(56)+2(63)+0+0 = 320+30 = 350`
`48329,03711 ==> 2(4·1)+2(8·1)+2(3·7)+2(2·3)+2(9·0)= 8+16+42+12+0 = 78`
In a square root calculation the quotient digit set increases incrementally for each step.
`Number => Calculation = Duplex:`
`1 ==> 12 = 1`
`14 ==>2(1·4) = 8`
`142 ==> 2(1·2) + 42 = 4 + 16 = 20`
`14,21 ==> 2(1·1) + 2(4·2) = 2 + 16 = 18`
`14213 ==> 6+8+4 = 18`
`142,135 ==> 10+24+4 = 38`
`1421356 ==> 12+40+12+1 = 65`
`1421,3562 ==> 4+48+20+6 = 78`
`142,135,623 ==> 6+16+24+10+9 = 65`
`142,1356,237 ==> 14+24+8+12+30 = 88`
`142,13562,373 ==> 6+56+12+4+36+25 = 139`

#### Example 1, by discussion

Consider the perfect square 2809. 532 = 2809. Use the duplex method to find the square root of 2,809.

• Set down the number in groups of two digits.
• We define a divisor, a dividend and a quotient to find the root.
• Given 2809. Consider the first group, 28.
• Find the nearest perfect square below that group.
• The root of that perfect square is the first digit of our root.
• Since 28 > 25 and 25 = 52, we take 5 as the first digit in the square root.
• For the divisor we take double this first digit (2 · 5), which is 10.
• Next, we set up a division framework with a colon.
• 28: 0 9 is the dividend and 5: is the quotient.
• We put a colon to the right of 28 and 5 and keep the colons lined up vertically. The duplex is calculated only on quotient digits to the right of the colon.
• We calculate the remainder. 28: minus 25: is 3:.
• We append the remainder on the left of the next digit to get the new dividend.
• Here, we append 3 to the next dividend digit 0, which makes the new dividend 30. The divisor 10 goes into 30 just 3 times. (No reserve needed here for subsequent deductions.)
• We repeat the operation.
• The zero remainder appended to 9. Nine is the next dividend.
• Now we have a digit to the right of the colon so we deduct the duplex, 32 = 9.
• Subtracting this duplex from the dividend 9, we get a zero remainder.
• Ten into zero is zero. The next root digit is zero. The next duplex is 2(3·0) = 0.
• The dividend is zero. We have an exact square root, 53.0.

#### Example 1, analysis and square root framework

`Find the square root of 2809.`
`Set down the number in groups of two digits.`
`The number of groups gives the number of whole digits in the root.`
`Put a colon after the first group, 28, to separate it.`
`From the first group, 28, we obtain the divisor, 10, since`
`28>25=52 and by doubling this first root, 2x5=10.`
`       Gross dividend:     28:  0  9. Using mental math:`
`              Divisor: 10)     3  0   Square: 10)  28:  30  9`
`    Duplex, Deduction:     25: xx 09  Square root:  5:   3. 0`
`             Dividend:         30 00`
`            Remainder:      3: 00 00`
`Square Root, Quotient:      5:  3. 0`

#### Example 2

Find the square root of 2,080,180,881. Solution by the duplex method: this ten-digit square has five digit-pairs, so it will have a five-digit square root. The first digit-pair is 20. Put the colon to the right. The nearest square below 20 is 16, whose root is 4, the first root digit. So, we use 2·4=8 for the divisor. Now we proceed with the duplex division, one digit column at a time. Prefix the remainder to the next dividend digit.

` divisor; gross dividend: 8) 20:  8   0   1   8    0   8   8   1`
`read the dividend diagonally up: 4   8   7  11   10  10   0   8`
`        minus the duplex:    16: xx  25  60  36   90 108  00  81`
`         actual dividend:      : 48  55  11  82   10  00  08  00`
`       minus the product:      : 40  48  00  72   00  00   0  00`
`               remainder:     4:  8   7  11  10   10   0   8  00`
`                quotient:     4:  5,  6   0   9.   0   0   0   0`

`Duplex calculations:`
`Quotient-digits ==> Duplex deduction.`
`5       ==> 52= 25`
`5 and 6 ==> 2(5·6) = 60`
`5,6,0   ==> 2(5·0)+62 = 36`
`5,6,0,9 ==> 2(5·9)+2(6·0) = 90`
`5,6,0,9,0 ==> 2(5·0)+2(6·9)+ 0 = 108`
`5,6,0,9,0,0 ==> 2(5·0)+2(6·0)+2(0·9) = 0`
`5,6,0,9,0,0,0 ==> 2(5·0)+2(6·0)+2(0·0)+92 = 81`
Hence the square root of 2,080,180,881 is exactly 45,609.

#### Example 3

Find the square root of two to ten places. Let us take 20,000 as the beginning group, using three digit-pairs at the start. The perfect square just below 20,000 is 141, since 1412 = 19881 < 20,000. So, the first root digits are 141 and the divisor doubled, 2 x 141 = 282. With a larger divisor the duplex will be relatively small. Hence, we can pick the multiple of the divisor without confusion.

`        Dividend: 2.0000 :    0    0     0     0     0     0    0    0`
`Diagonal;Divisor: 282)   : 119   62    40   102   162   182   75  112`
`    Minus duplex:        : xxxx   16    16    12    28    53   74   59`
` Actual dividend:  20000 : 1190  604   384  1008  1592  1767  676 1061`
`   Minus product:  19881 : 1128  564   282   846  1410  1692  564  846`
`       Remainder:    119 :   62   40   102   162   182    75  112  215`
`   Root quotient:   1.41 :    4    2     1     3     5     6    2    3`

Ten multiples of 282: 282; 564; 846; 1128; 1410; 1692; 1974; 2256; 2538; 2820.

## Iterative methods for reciprocal square roots

The following are iterative methods for finding the reciprocal square root of S which is $1/sqrt\left\{S\right\}$. Once it has been found, we can find $sqrt\left\{S\right\}$ by simple multiplication: $sqrt\left\{S\right\} = S cdot \left(1/sqrt\left\{S\right\}\right)$. These iterations involve only multiplication, and not division. They are therefore faster than the Babylonian method. However, they are not stable. If the initial value is not close to the reciprocal square root, the iterations will diverge away from it rather than converge to it. It can therefore be advantageous to perform an iteration of the Babylonian method on a rough estimate before starting to apply these methods.

• One method is found by applying Newton's method to the equation $\left(1/x^2\right) - S = 0$. It converges quadratically:

$x_\left\{n+1\right\} = frac\left\{x_n\right\}\left\{2\right\} cdot \left(3 - S cdot x_n^2\right).$

$y_n = S cdot x_n^2, , !$
$x_\left\{n+1\right\} = frac\left\{x_n\right\}\left\{8\right\} cdot \left(15 - y_n cdot \left(10 - 3 cdot y_n\right)\right).$

## Taylor series

If N is an approximation to $sqrt\left\{S\right\}$, a better approximation can be found by using the Taylor series of the square root function:

$sqrt\left\{N^2+d\right\} = sum_\left\{n=0\right\}^infty frac\left\{\left(-1\right)^\left\{n\right\}\left(2n\right)!d^n\right\}\left\{\left(1-2n\right)n!^2 4^nN^\left\{2n-1\right\}\right\} = N + frac\left\{d\right\}\left\{2N\right\} - frac\left\{d^2\right\}\left\{8N^3\right\} + frac\left\{d^3\right\}\left\{16N^5\right\} - frac\left\{5d^4\right\}\left\{128N^7\right\} + cdots$

As an iterative method, the order of convergence is equal to the number of terms used. With 2 terms, it is identical to the Babylonian method; With 3 terms, each iteration takes almost as many operations as the Bakhshali approximation, but converges more slowly. Therefore, this is not a particularly efficient way of calculation.

## Other methods

Finding $sqrt\left\{S\right\}$ is the same as solving the equation $x^2 - S = 0,!$. Therefore, any general numerical root-finding algorithm can be used. Newton's method, for example, reduces in this case to the Babylonian method. Other methods are less efficient than the ones presented above.

## Continued fraction expansion

Quadratic irrationals (numbers of the form $frac\left\{a+sqrt\left\{b\right\}\right\}\left\{c\right\}$, where a, b and c are integers), and in particular, square roots of integers, have periodic continued fractions. Sometimes we may be interested not in finding the numerical value of a square root, but rather in its continued fraction expansion. The following iterative algorithm can be used for this purpose (S is any natural number which is not a perfect square):

$m_0=0,!$

$d_0=1,!$

$a_0=leftlfloorsqrt\left\{S\right\}rightrfloor,!$

$m_\left\{n+1\right\}=d_na_n-m_n,!$

$d_\left\{n+1\right\}=frac\left\{S-m_\left\{n+1\right\}^2\right\}\left\{d_n\right\},!$

$a_\left\{n+1\right\} =leftlfloorfrac\left\{sqrt\left\{S\right\}+m_\left\{n+1\right\}\right\}\left\{d_\left\{n+1\right\}\right\}rightrfloor =leftlfloorfrac\left\{a_0+m_\left\{n+1\right\}\right\}\left\{d_\left\{n+1\right\}\right\}rightrfloor!.$

Notice that mn, dn, and an are always integers. The algorithm terminates when this triplet is the same as one encountered before. The expansion will repeat from then on. The sequence [a0; a1, a2, a3, …] is the continued fraction expansion:

$sqrt\left\{S\right\} = a_0 + cfrac\left\{1\right\}\left\{a_1 + cfrac\left\{1\right\}\left\{a_2 + cfrac\left\{1\right\}\left\{a_3+,cdots\right\}\right\}\right\}$

### Example, square root of 114 as a continued fraction

We begin with m0=0; d0=1; and a0=10 ($10^2=100<114$ and $11^2=121>114$ so 10 chosen).

$sqrt\left\{114\right\}=frac\left\{sqrt\left\{114\right\}+0\right\}\left\{1\right\}=10+frac\left\{sqrt\left\{114\right\}-10\right\}\left\{1\right\}= 10+frac\left\{\left(sqrt\left\{114\right\}-10\right)\left(sqrt\left\{114\right\}+10\right)\right\}\left\{sqrt\left\{114\right\}+10\right\}$ $=10+frac\left\{114-100\right\}\left\{sqrt\left\{114\right\}+10\right\}=10+frac\left\{1\right\}\left\{frac\left\{sqrt\left\{114\right\}+10\right\}\left\{14\right\}\right\}.$
$m_\left\{1\right\} = d_\left\{0\right\} cdot a_\left\{0\right\} - m_\left\{0\right\} = 1 cdot 10 - 0 = 10 ,.$
$d_\left\{1\right\} = frac\left\{S-m_\left\{0\right\}^2\right\}\left\{d_0\right\} = frac\left\{114-10^2\right\}\left\{1\right\} = 14 ,.$
$a_\left\{1\right\} = leftlfloor frac\left\{a_0+m_\left\{1\right\}\right\}\left\{d_\left\{1\right\}\right\} rightrfloor = leftlfloor frac\left\{10+10\right\}\left\{14\right\} rightrfloor = leftlfloor frac\left\{20\right\}\left\{14\right\} rightrfloor = 1 ,.$

So, m1=10; d1=14; and a1 = 1.
$frac\left\{sqrt\left\{114\right\}+10\right\}\left\{14\right\}=1+frac\left\{sqrt\left\{114\right\}-4\right\}\left\{14\right\}=1+frac\left\{114-16\right\}\left\{14\left(sqrt\left\{114\right\}+4\right)\right\}$ $=1+frac\left\{1\right\}\left\{frac\left\{sqrt\left\{114\right\}+4\right\}\left\{7\right\}\right\}.$
Next, m2=4; d2=7; and a2=2.
$frac\left\{sqrt\left\{114\right\}+4\right\}\left\{7\right\}=2+frac\left\{sqrt\left\{114\right\}-10\right\}\left\{7\right\}=2+frac\left\{14\right\}\left\{7\left(sqrt\left\{114\right\}+10\right)\right\}$ $=2+frac\left\{1\right\}\left\{frac\left\{sqrt\left\{114\right\}+10\right\}\left\{2\right\}\right\}.$

$frac\left\{sqrt\left\{114\right\}+10\right\}\left\{2\right\}=10+frac\left\{sqrt\left\{114\right\}-10\right\}\left\{2\right\}=10+frac\left\{14\right\}\left\{2\left(sqrt\left\{114\right\}+10\right)\right\}$ $=10+frac\left\{1\right\}\left\{frac\left\{sqrt\left\{114\right\}+10\right\}\left\{7\right\}\right\}.$

$frac\left\{sqrt\left\{114\right\}+10\right\}\left\{7\right\}=2+frac\left\{sqrt\left\{114\right\}-4\right\}\left\{7\right\}=2+frac\left\{98\right\}\left\{7\left(sqrt\left\{114\right\}+4\right)\right\}$ $=2+frac\left\{1\right\}\left\{frac\left\{sqrt\left\{114\right\}+4\right\}\left\{14\right\}\right\}.$

$frac\left\{sqrt\left\{114\right\}+4\right\}\left\{14\right\}=1+frac\left\{sqrt\left\{114\right\}-10\right\}\left\{14\right\}=1+frac\left\{14\right\}\left\{14\left(sqrt\left\{114\right\}+10\right)\right\}$ $=1+frac\left\{1\right\}\left\{frac\left\{sqrt\left\{114\right\}+10\right\}\left\{1\right\}\right\}.$

$frac\left\{sqrt\left\{114\right\}+10\right\}\left\{1\right\}=20+frac\left\{sqrt\left\{114\right\}-10\right\}\left\{1\right\}=20+frac\left\{14\right\}\left\{sqrt\left\{114\right\}+10\right\}$ $=20+frac\left\{1\right\}\left\{frac\left\{sqrt\left\{114\right\}+10\right\}\left\{14\right\}\right\}.$

Now, loop back to the second equation above.

Consequently, the continued fraction for the square root of 114 is: $sqrt\left\{114\right\} = \left[10;1,2,10,2,1,20,1,2,10,2,1,20,1,2,10,2,1,20,...\right].$

## Pell's equation

Pell's equation and its variants yield a method for efficiently finding continued fraction convergents of square roots of integers. However, it can be complicated to execute, and usually not every convergent is generated. The ideas behind the method are as follows:

• If (p, q) is a solution (where p and q are integers) to the equation $p^2 = S cdot q^2 pm 1!$, then $frac\left\{p\right\}\left\{q\right\}$ is a continued fraction convergent of $sqrt\left\{S\right\}$, and as such, is an excellent rational approximation to it.
• If (pa, qa) and (pb, qb) are solutions, then so is:

$p = p_a p_b + S cdot q_a q_b,!$
$q = p_a q_b + p_b q_a,!$

• More generally, if (p1, q1) is a solution, then it is possible to generate a sequence of solutions (pn, qn) satisfying:

$p_\left\{m+n\right\} = p_m p_n + S cdot q_m q_n,!$
$q_\left\{m+n\right\} = p_m q_n + p_n q_m,!$
The method is as follows:

• Find natural numbers (0 excluded) p1 and q1 such that $p_1^2 = S cdot q_1^2 pm 1$. This is the hard part; It can be done either by guessing, or by using fairly sophisticated techniques.

*To generate a long list of convergents, iterate:
$p_\left\{n+1\right\} = p_1 p_n + S cdot q_1 q_n,!$
$q_\left\{n+1\right\} = p_1 q_n + p_n q_1,!$
*To find the larger convergents quickly, iterate:
$p_\left\{2n\right\} = p_n^2 + S cdot q_n^2,!$
$q_\left\{2n\right\} = 2 p_n q_n,!$

• In either case, $frac\left\{p_n\right\}\left\{q_n\right\}$ is a rational approximation satisfying

$left|frac\left\{p_n\right\}\left\{q_n\right\} - sqrt\left\{S\right\}right| < frac\left\{1\right\}\left\{q_n^2 cdot sqrt\left\{S\right\}\right\}.$

## Approximations that depend on IEEE representation

On computers, a very rapid Newton's method based approximation to the square root can be obtained for floating point numbers when computers use an IEEE (or sufficiently similar) representation.

This technique is based on the fact that the IEEE floating point format approximates base-2 logarithm. For example, you can get the approximate logarithm of 32-bit single precision floating point number by translating its binary representation as an integer, scaling it by $2^\left\{23\right\}$, and removing a bias of 127. $x_\left\{integer\right\} / 2^\left\{23\right\} - 127 approx log_2\left(x\right)$

For example, 1.0 is represented by a hexadecimal number 0x3F800000, which would represent 1065353216 = $127 * 2^\left\{23\right\}$ if taken as an integer. Using the formula above you get $1065353216 / 2^\left\{23\right\} - 127 = 0$, as expected from $log_2\left(1.0\right)$. In a similar fashion you get 0.5 from 1.5(0x3FC00000).

In order to get the square root, we can divide the logarithm by 2 and convert the value back. The following program demonstrates the idea. Note that we intentionally allow the exponent's lowest bit to propagate into the mantissa.

` float fastsqrt(float val)  {`
`      union`
`      {`
`    int tmp;`
`    float val;`
} u;
`      u.val = val;`
`      u.tmp -= 1<<23; /* Remove last bit so 1.0 gives 1.0 */`
`      /* tmp is now an approximation to logbase2(val) */`
`      u.tmp >>= 1; /* divide by 2 */`
`      u.tmp += 1<<29; /* add 64 to exponent: (e+127)/2 =(e/2)+63, */`
`      /* that represents (e/2)-64 but we want e/2 */`
`      return u.val;`
}

In the above, the operations to remove last exponent bit and add the IEEE bias can be combined into a single operation. An additional adjustment can be made in the same operation to reduce the maximum relative error. So, the three operations, not including the cast, can be rewritten as:

`   tmp = (1<<29) + (tmp >> 1) - (1<<22) + m;`

Where m is a bias for adjusting the approximation errors. For example, with m = 0 you get accurate results for even powers of 2 (e.g. 1.0), but for other numbers the results will be slightly too big (e.g. you get 1.5 for 2.0 instead of 1.414... with 6% error). With m = -0x4C000 you get errors between about -3.5% and 3.5%.

If the approximation is to be used for an initial guess for Newton's method to the equation $\left(1/x^2\right) - S = 0$, you need to use a reciprocal form shown in the following section.

### Reciprocal of the square root

A variant of the above routine is included below, which can be used to compute the reciprocal of the square root, i.e. $x^\left\{-\left\{1over2\right\}\right\}$ instead, was written by Greg Walsh, and implemented into the game Quake 3 by Gary Tarolli.The integer-shift approximation produced a relative error of less than 4%, and the error dropped further to 0.15% with one iteration of Newton's method on the following line. In computer graphics it is a very efficient way to normalize a vector.

` float invSqrt(float x)`
` {`
`      float xhalf = 0.5f*x;`
`      union`
`      {`
`    float x;`
`    int i;`
} u;
`      u.x = x;`
`      u.i = 0x5f3759df - (u.i >> 1);`
`      x = u.x * (1.5f - xhalf * u.x * u.x);`
`      return x;`
}