Methods of computing square roots

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

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[edit] 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{S} = e^{\frac{1}{2}\ln S}.

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

[edit] Rough estimation

Many of the methods for calculating square roots 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. One way of obtaining such an estimate for \sqrt{S} is calculating 3D, where D is the number of digits (to the left of the decimal point) in a positive real S. If S < 1, D is the negative of the number of zeros to the immediate right of the decimal point.

A better method of rough estimation is this:

If D is odd, D = 2n + 1, then use \sqrt{S} \approx 2 \cdot 10^n.
If D is even, D = 2n + 2, then use \sqrt{S} \approx 6 \cdot 10^n.

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

[edit] Babylonian method

Graph charting the use of the Babylonian method for approximating the square root of 100 (10) using start values x0=50, x0=1, and x0=-5. Note that using a negative start value yields the negative root.
Graph charting the use of the Babylonian method for approximating the square root of 100 (10) using start values x0=50, x0=1, and x0=-5. Note that using a negative start value yields the negative root.

A commonly used algorithm for approximating \sqrt S (and perhaps the best) is known as the "Babylonian method"[1] and 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{S}.
x_{n+1} = \frac{1}{2} \left(x_n + \frac{S}{x_n}\right),
\sqrt S = \lim_{n \to \infty} 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.

[edit] Example

We'll calculate \sqrt{S}, where S = 125348, to 6 significant figures. Here D, the number of digits in S, is 6. Then:

x_0 = 3^D = 3^6 = 729.000\,\!
x_1 = \frac{1}{2} \left(x_0 + \frac{S}{x_0}\right) = \frac{1}{2} \left(729.000 + \frac{125348}{729.000}\right) = 450.472
x_2 = \frac{1}{2} \left(x_1 + \frac{S}{x_1}\right) = \frac{1}{2} \left(450.472 + \frac{125348}{450.472}\right) = 364.365
x_3 = \frac{1}{2} \left(x_2 + \frac{S}{x_2}\right) = \frac{1}{2} \left(364.365 + \frac{125348}{364.365}\right) = 354.191
x_4 = \frac{1}{2} \left(x_3 + \frac{S}{x_3}\right) = \frac{1}{2} \left(354.191 + \frac{125348}{354.191}\right) = 354.045
x_5 = \frac{1}{2} \left(x_4 + \frac{S}{x_4}\right) = \frac{1}{2} \left(354.045 + \frac{125348}{354.045}\right) = 354.045

Therefore, \sqrt{125348} \approx 354.045.

[edit] Convergence

We let the relative error in xn be defined by

\epsilon_n = \frac {x_n}{\sqrt{S}} - 1

and thus

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

Then one can show that

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

and thus that

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

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

[edit] 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{S}, let N2 be the nearest perfect square to S. Then, calculate:

d = S - N^2 \,\!
P = \frac{d}{2N}
A = N + P\,\!
\sqrt{S} \approx A - \frac{P^2}{2A}

This can be also written as:

\sqrt{S} \approx \frac{N^4+6N^2S+S^2}{4N^3+4NS}
\sqrt{N^2 + d} \approx N + \frac{d}{2N} - \frac{d^2}{8N^3 + 4Nd}

[edit] Example

We'll find \sqrt{9.2345}.

N=3\,\!
d = 9.2345 - 3^2 = 0.2345\,\!
P = \frac{0.2345}{2 \cdot 3} = 0.0391
A = 3 + 0.0391 = 3.0391\,\!
\sqrt{9.2345} \approx 3.0391 - \frac{0.0391^2}{2 \cdot 3.0391} \approx 3.0388

[edit] Digit by digit calculation

This is a method to find each digit of the square root in a sequence. It is much 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. We will describe the method for the decimal system. It goes as follows:

Write the decimal expansion of the square. The numbers are laid out in a fashion similar to the long division algorithm, and the root will appear above its square. Divide the digits of the square into pairs, starting from the decimal point and going both left and right. The decimal point of the root should be placed above the decimal point of the square. One digit of the root will appear above each pair of digits of the square.

Begin at the left-most (non-zero) pair of digits of the square. Initialize the remainder to zero. Initialize p to zero.
For each iteration (pair of digits) do:

  1. Bring down the most significant (leftmost) pair of digits of the square not yet used (if all the digits have been used, use the digits 00) and append them to the remainder from the previous step, i.e. multiply the remainder by one hundred and add the two digits. This will be the current value c.
  2. Let p be the part of the root found so far, ignoring any decimal point (for the first step, p = 0). Determine the greatest digit x such that y = (20 \cdot p + x) \cdot x does not exceed c (notice 20p + x is simply twice p, with the digit x appended to the right). 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.

[edit] Examples

Find the square root of 152.2756.

          1  2. 3  4 
     \/  01 52.27 56                            

x        01                   1*1 <= 1 < 2*2                 x = 1
         01                     y = x*x = 1*1 = 1
2x       00 52                22*2 <= 52 < 23*3              x = 2
         00 44                  y = 2x*x = 22*2 = 44                      
24x         08 27             243*3 <= 827 < 244*4           x = 3       
            07 29               y = 24x*x = 243*3 = 729
246x           98 56          2464*4 <= 9856 < 2465*5        x = 4       
               98 56            y = 246x*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

x        02                  1*1 <= 2 < 2*2                 x = 1
         01                    y = x*x = 1*1 = 1
2x       01 00               24*4 <= 100 < 25*5             x = 4
         00 96                 y = 2x*x = 24*4 = 96                      
28x         04 00            281*1 <= 400 < 282*2           x = 1       
            02 81              y = 28x*x = 281*1 = 281
282x        01 19 00         2824*4 <= 11900 < 2825*5       x = 4       
            01 12 96           y = 282x*x = 2824*4 = 11296
2828x          06 04 00      28282*2 <= 60400 < 28283*3     x = 2
                             The desired precision is achieved: 
                             The square root of 2 is about 1.4142

[edit] 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.[2] 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. Calculating the Duplex: 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 ==> 2·1 = 1 
14 ==>2·1·4 = 8 
142 ==> 16+4 = 20 
14,21 ==> 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

[edit] Example 1

Consider 532 = 2809. Take the first group, 28. 25 is the nearest perfect square below 28. Since 28 > 25 = 52 we take 5 as the first digit in the root. For the divisor we take double this first digit, 2 times 5 = 10. 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. 28: minus 25: is 3:. Append 3 to the next digit 0. The dividend is 30. The divisor 10 goes into 30 just 3 times. (No reserve needed here for subsequent deductions.) Zero remainder appended to 9. Nine is the dividend. Now we have a digit to the right of the colon so we deduct the duplex. 32 = 9. Zero remainder. 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.

Find the square root of 2809. 
Set down the number in groups of two digits. 
The number of groups gives the number of digits in the root.
From the first group, 28, we obtain the divisor, 10. 
Put a colon after the first group to separate it. 
Since 28>25=52. And 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

[edit] 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.

[edit] 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.

[edit] Iterative methods for reciprocal square roots

The following are iterative methods for finding the reciprocal square root of S, \frac{1}{\sqrt{S}}. Once it has been found, we can find \sqrt{S} by simple multiplication: \sqrt{S} = \frac{1}{\sqrt{S}} \cdot S. The 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 \frac{1}{x^2} - S = 0. It converges quadratically:
x_{n+1} = \frac{x_n}{2} \cdot (3 - S \cdot x_n^2).
y_n = S \cdot x_n^2\, , \!
x_{n+1} = \frac{x_n}{8} \cdot (15 - y_n \cdot (10 - 3 \cdot y_n)).

[edit] Taylor series

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

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

[edit] Other methods

Finding \sqrt{S} 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.

[edit] Continued fraction expansion

Quadratic irrationals (numbers of the form \frac{a+\sqrt{b}}{c}, 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=\left\lfloor\sqrt{S}\right\rfloor\,\!
m_{n+1}=d_na_n-m_n\,\!
d_{n+1}=\frac{S-m_{n+1}^2}{d_n}\,\!
a_{n+1} =\left\lfloor\frac{\sqrt{S}+m_{n+1}}{d_{n+1}}\right\rfloor =\left\lfloor\frac{a_0+m_{n+1}}{d_{n+1}}\right\rfloor\!

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{S} = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3+\,\cdots}}}

[edit] Example, square root of 114 as a continued fraction

We begin with m0=0; d0=1; and a0=10.
\sqrt{114}=\frac{\sqrt{114}+0}{1}=10+\frac{\sqrt{114}-10}{1}= 10+\frac{(\sqrt{114}-10)(\sqrt{114}+10)}{\sqrt{114}+10} =10+\frac{114-100}{\sqrt{114}+10}=10+\frac{1}{\frac{\sqrt{114}+10}{14}}.
Next, m1=10; d1=14; and a1=1.
\frac{\sqrt{114}+10}{14}=1+\frac{\sqrt{114}-4}{14}=1+\frac{114-16}{14(\sqrt{114}+4)} =1+\frac{1}{\frac{\sqrt{114}+4}{7}}.
Next, m2=4; d2=7; and a2=2.
\frac{\sqrt{114}+4}{7}=2+\frac{\sqrt{114}-10}{7}=2+\frac{14}{7(\sqrt{114}+10)} =2+\frac{1}{\frac{\sqrt{114}+10}{2}}.

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

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

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

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

Now, loop back to the second equation above.

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

[edit] 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{p}{q} is a continued fraction convergent of \sqrt{S}, 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_{m+n} = p_m p_n + S \cdot q_m q_n\,\!
q_{m+n} = 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_{n+1} = p_1 p_n + S \cdot q_1 q_n\,\!
q_{n+1} = p_1 q_n + p_n q_1\,\!
  • To find the larger convergents quickly, iterate:
p_{2n} = p_n^2 + S \cdot q_n^2\,\!
q_{2n} = 2 p_n q_n\,\!
  • In either case, \frac{p_n}{q_n} is a rational approximation satisfying
\left|\frac{p_n}{q_n} - \sqrt{S}\right| < \frac{1}{q_n^2 \cdot \sqrt{S}}.

[edit] 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.

float fastsqrt(float val) {
    int tmp = *(int *)&val;
    tmp -= 1<<23; /* Remove last bit to not let it go to mantissa */
    /* tmp is now an approximation to logbase2(val) */
    tmp = tmp >> 1; /* divide by 2 */
    tmp += 1<<29; /* add 64 to exponent: (e+127)/2 =(e/2)+63, */
                  /* that represents (e/2)-64 but we want e/2 */
    return *(float *)&tmp;
}

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 some magic number that corresponds to a calculation involving an initial guess used in the Newton's approximation.

[edit] Inverse 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^{-{1\over2}} instead, was written by Greg Walsh, and implemented into the game Quake 3 by Gary Tarolli. As an approximation, it produced a relative error of less than 4%, and in computer graphics it is a very efficient way to normalize a vector.

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

[edit] See also

[edit] Notes

  1. ^ There is no direct evidence showing how the Babylonians computed square roots, although there are informed conjectures. (Square root of 2#Notes gives a summary and references.)
  2. ^ Vedic Mathematics: Sixteen Simple Mathematical Formulae from the Vedas, by Swami Sankaracarya (1880-1960), Motilal Banarsidass Indological Publishers and Booksellers, Varnasi, India, 1965; reprinted in Delhi, India, 1975, 1978. 367 pages.

[edit] External links

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