{"history":[],"article":{"id":"dd4b9cc1-6175-4a67-9957-d452d9a60ae1","title":"Methods of computing square roots","slug":"methods-of-computing-square-roots","content":":::summary\n:::\n\n# \n**Methods of computing [square root](https://bliptext.com/articles/square-root)s** are [algorithms](https://bliptext.com/articles/algorithm) for approximating the non-negative [square root](https://bliptext.com/articles/square-root) \\sqrt{S} of a positive [real number](https://bliptext.com/articles/real-number) S. Since all [square root](https://bliptext.com/articles/square-root)s of [natural numbers](https://bliptext.com/articles/natural-number), other than of [perfect squares](https://bliptext.com/articles/square-number), are [irrational](https://bliptext.com/articles/irrational-number),\nsquare roots can usually only be computed to some finite precision: these methods typically construct a series of increasingly accurate [approximations](https://bliptext.com/articles/numerical-approximation).\nMost square root computation methods are iterative: after choosing a suitable [initial estimate](https://bliptext.com/articles/) of \\sqrt{S}, an iterative refinement is performed until some termination criterion is met.\nOne refinement scheme is [Heron's method](https://bliptext.com/articles/), a special case of [Newton's method](https://bliptext.com/articles/newton-s-method).\nIf division is much more costly than multiplication, it may be preferable to compute the [inverse square root](https://bliptext.com/articles/) instead.\nOther methods are available to compute the square root [digit by digit](https://bliptext.com/articles/), or using [Taylor series](https://bliptext.com/articles/).\nRational approximations of square roots may be calculated using [continued fraction expansions](https://bliptext.com/articles/).\nThe method employed depends on the needed accuracy, and the available tools and computational power.\nThe methods may be roughly classified as those suitable for mental calculation, those usually requiring at least paper and pencil, and those which are implemented as programs to be executed on a digital electronic computer or other computing device.\nAlgorithms may take into account convergence (how many iterations are required to achieve a specified precision), computational complexity of individual operations (i.e. division) or iterations, and error propagation (the accuracy of the final result).\nA few methods like paper-and-pencil synthetic division and series expansion, do not require a starting value.\nIn some applications, an [integer square root](https://bliptext.com/articles/integer-square-root) is required, which is the square root rounded or truncated to the nearest integer (a modified procedure may be employed in this case).\n\n# History\nProcedures for finding square roots (particularly the [square root of 2](https://bliptext.com/articles/square-root-of-2)) have been known since at least the period of ancient Babylon in the 17th century BCE.\n[Babylonian mathematicians](https://bliptext.com/articles/babylonian-mathematics) calculated the square root of 2 to three sexagesimal \"digits\" after the 1, but it is not known exactly how.\nThey knew how to approximate a hypotenuse using\n(giving for example for the diagonal of a gate whose height is rods and whose width is rods) and they may have used a similar approach for finding the approximation of \\sqrt 2.\n[Heron's method](https://bliptext.com/articles/heron-s-method) from first century Egypt was the first ascertainable algorithm for computing square root.\nModern analytic methods began to be developed after introduction of the [Arabic numeral](https://bliptext.com/articles/arabic-numeral) system to western Europe in the early Renaissance.\nToday, nearly all computing devices have a fast and accurate square root function, either as a programming language construct, a compiler intrinsic or library function, or as a hardware operator, based on one of the described procedures.\n\n# Initial estimate\nMany iterative square root algorithms require an initial [seed value](https://bliptext.com/articles/seed-value).\nThe seed must be a non-zero positive number; it should be between 1 and S, the number whose square root is desired, because the square root must be in that range.\nIf the seed is far away from the root, the algorithm will require more iterations.\nIf one initializes with x_0=1 (or S), then approximately iterations will be wasted just getting the order of magnitude of the root.\nIt is therefore useful to have a rough estimate, which may have limited accuracy but is easy to calculate.\nIn general, the better the initial estimate, the faster the convergence.\nFor Newton's method (also called Babylonian or Heron's method), a seed somewhat larger than the root will converge slightly faster than a seed somewhat smaller than the root.\nIn general, an estimate is pursuant to an arbitrary interval known to contain the root (such as [x_0,S/x_0]).\nThe estimate is a specific value of a functional approximation to over the interval.\nObtaining a better estimate involves either obtaining tighter bounds on the interval, or finding a better functional approximation to f(x).\nThe latter usually means using a higher order polynomial in the approximation, though not all approximations are polynomial.\nCommon methods of estimating include scalar, linear, hyperbolic and logarithmic.\nA decimal base is usually used for mental or paper-and-pencil estimating.\nA binary base is more suitable for computer estimates.\nIn estimating, the exponent and [mantissa](https://bliptext.com/articles/mantissa-logarithm) are usually treated separately, as the number would be expressed in scientific notation.\n\n## Decimal estimates\nTypically the number S is expressed in [scientific notation](https://bliptext.com/articles/scientific-notation) as where 1\\leq a<100 and n is an integer, and the range of possible square roots is where.\n\n### Scalar estimates\nScalar methods divide the range into intervals, and the estimate in each interval is represented by a single scalar number.\nIf the range is considered as a single interval, the arithmetic mean (5.5) or geometric mean times 10^n are plausible estimates.\nThe absolute and relative error for these will differ.\nIn general, a single scalar will be very inaccurate.\nBetter estimates divide the range into two or more intervals, but scalar estimates have inherently low accuracy.\nFor two intervals, divided geometrically, the square root can be estimated as\nThis estimate has maximum absolute error of 4\\cdot10^n at a = 100, and maximum relative error of 100% at a = 1.\nFor example, for S = 125348 factored as, the estimate is.\n, an absolute error of 246 and relative error of almost 70%.\n\n### Linear estimates\nA better estimate, and the standard method used, is a linear approximation to the function y = x^2 over a small arc.\nIf, as above, powers of the base are factored out of the number S and the interval reduced to [1,100], a secant line spanning the arc, or a tangent line somewhere along the arc may be used as the approximation, but a least-squares regression line intersecting the arc will be more accurate.\nA least-squares regression line minimizes the average difference between the estimate and the value of the function.\nIts equation is y=8.7x-10.\nReordering,.\nRounding the coefficients for ease of computation,\nThat is the best estimate on average that can be achieved with a single piece linear approximation of the function y=x2 in the interval [1,100].\nIt has a maximum absolute error of 1.2 at a=100, and maximum relative error of 30% at S=1 and 10.\nTo divide by 10, subtract one from the exponent of a, or figuratively move the decimal point one digit to the left.\nFor this formulation, any additive constant 1 plus a small increment will make a satisfactory estimate so remembering the exact number isn't a burden.\nThe approximation (rounded or not) using a single line spanning the range [1,100] is less than one significant digit of precision; the relative error is greater than 1/22, so less than 2 bits of information are provided.\nThe accuracy is severely limited because the range is two orders of magnitude, quite large for this kind of estimation.\nA much better estimate can be obtained by a piece-wise linear approximation: multiple line segments, each approximating some subarc of the original.\nThe more line segments used, the better the approximation.\nThe most common way is to use tangent lines; the critical choices are how to divide the arc and where to place the tangent points.\nAn efficacious way to divide the arc from y = 1 to y = 100 is geometrically: for two intervals, the bounds of the intervals are the square root of the bounds of the original interval, 1×100, i.e. [1,2√100] and [2√100,100].\nFor three intervals, the bounds are the cube roots of 100: [1,3√100], [3√100,(3√100)2], and [(3√100)2,100], etc.\nFor two intervals, 2√100 = 10, a very convenient number.\nTangent lines are easy to derive, and are located at x = √1*√10 and x = √10*√10.\nTheir equations are: and.\nInverting, the square roots are: and.\nThus for :\nThe maximum absolute errors occur at the high points of the intervals, at a=10 and 100, and are 0.54 and 1.7 respectively.\nThe maximum relative errors are at the endpoints of the intervals, at a=1, 10 and 100, and are 17% in both cases.\n17% or 0.17 is larger than 1/10, so the method yields less than a decimal digit of accuracy.\n\n### Hyperbolic estimates\nIn some cases, hyperbolic estimates may be efficacious, because a hyperbola is also a convex curve and may lie along an arc of Y = x2 better than a line.\nHyperbolic estimates are more computationally complex, because they necessarily require a floating division.\nA near-optimal hyperbolic approximation to x2 on the interval [1,100] is y=190/(10-x)-20.\nTransposing, the square root is x = -190/(y+20)+10.\nThus for :\nThe floating division need be accurate to only one decimal digit, because the estimate overall is only that accurate, and can be done mentally.\nA hyperbolic estimate is better on average than scalar or linear estimates.\nIt has maximum absolute error of 1.58 at 100 and maximum relative error of 16.0% at 10.\nFor the worst case at a=10, the estimate is 3.67.\nIf one starts with 10 and applies Newton-Raphson iterations straight away, two iterations will be required, yielding 3.66, before the accuracy of the hyperbolic estimate is exceeded.\nFor a more typical case like 75, the hyperbolic estimate is 8.00, and 5 Newton-Raphson iterations starting at 75 would be required to obtain a more accurate result.\n\n### Arithmetic estimates\nA method analogous to piece-wise linear approximation but using only arithmetic instead of algebraic equations, uses the multiplication tables in reverse: the square root of a number between 1 and 100 is between 1 and 10, so if we know 25 is a perfect square (5 × 5), and 36 is a perfect square (6 × 6), then the square root of a number greater than or equal to 25 but less than 36, begins with a 5.\nSimilarly for numbers between other squares.\nThis method will yield a correct first digit, but it is not accurate to one digit: the first digit of the square root of 35 for example, is 5, but the square root of 35 is almost 6.\nA better way is to the divide the range into intervals halfway between the squares.\nSo any number between 25 and halfway to 36, which is 30.5, estimate 5; any number greater than 30.5 up to 36, estimate 6.\nThe procedure only requires a little arithmetic to find a boundary number in the middle of two products from the multiplication table.\nHere is a reference table of those boundaries:\nThe final operation is to multiply the estimate k by the power of ten divided by 2, so for ,\nThe method implicitly yields one significant digit of accuracy, since it rounds to the best first digit.\nThe method can be extended 3 significant digits in most cases, by interpolating between the nearest squares bounding the operand.\nIf, then \\sqrt{a} is approximately k plus a fraction, the difference between a and k2 divided by the difference between the two squares:\nwhere\nThe final operation, as above, is to multiply the result by the power of ten divided by 2;\nk is a decimal digit and R is a fraction that must be converted to decimal.\nIt usually has only a single digit in the numerator, and one or two digits in the denominator, so the conversion to decimal can be done mentally.\nExample: find the square root of 75.\n75 = 75 × 102·0\n, so a is 75 and n is 0.\nFrom the multiplication tables, the square root of the mantissa must be 8 point something because 8 × 8 is 64, but 9 × 9 is 81, too big, so k is 8; something is the decimal representation of R. The fraction R is 75 - k2 = 11, the numerator, and 81 - k2 = 17, the denominator.\n11/17 is a little less than 12/18, which is 2/3s or .67, so guess .66 (it's ok to guess here, the error is very small).\nSo the estimate is\n8 + .66 = 8.66\n. √75 to three significant digits is 8.66, so the estimate is good to 3 significant digits.\nNot all such estimates using this method will be so accurate, but they will be close.\n\n## Binary estimates\nWhen working in the [binary numeral system](https://bliptext.com/articles/binary-numeral-system) (as computers do internally), by expressing S as where, the square root can be estimated as\nwhich is the least-squares regression line to 3 significant digit coefficients.\n\\sqrt{a} has maximum absolute error of 0.0408 at a=2, and maximum relative error of 3.0% at a=1.\nA computationally convenient rounded estimate (because the coefficients are powers of 2) is:\nwhich has maximum absolute error of 0.086 at 2 and maximum relative error of 6.1% at\na = 0.5\nand\na = 2.0\n.\nFor, the binary approximation gives.\n, so the estimate has an absolute error of 19 and relative error of 5.3%.\nThe relative error is a little less than 1/24, so the estimate is good to 4+ bits.\nAn estimate for a good to 8 bits can be obtained by table lookup on the high 8 bits of a, remembering that the high bit is implicit in most floating point representations, and the bottom bit of the 8 should be rounded.\nThe table is 256 bytes of precomputed 8-bit square root values.\nFor example, for the index 111011012 representing 1.851562510, the entry is 101011102 representing 1.35937510, the square root of 1.851562510 to 8 bit precision (2+ decimal digits).\n\n# Heron's method\nThe first explicit [algorithm](https://bliptext.com/articles/algorithm) for approximating is known as **Heron's method**, after the first-century Greek mathematician [Hero of Alexandria](https://bliptext.com/articles/hero-of-alexandria) who described the method in his [AD 60](https://bliptext.com/articles/ad-60) work [Metrica](https://bliptext.com/articles/hero-of-alexandria).\nThis method is also called the **Babylonian method** (not to be confused with the **Babylonian method** for approximating hypotenuses), although there is no evidence that the method was known to Babylonians.\nGiven a positive real number S, let\nx0 > 0\nbe any positive [initial estimate](https://bliptext.com/articles/).\nHeron's method consists in iteratively computing\nuntil the desired accuracy is achieved.\nThe sequence defined by this equation converges to\nThis is equivalent to using [Newton's method](https://bliptext.com/articles/newton-s-method) to solve x^2 - S = 0.\nThis algorithm is [quadratically convergent](https://bliptext.com/articles/quadratic-convergence): the number of correct digits of x_n roughly doubles with each iteration.\n\n## Derivation\nThe basic idea is that if \\ x\\ is an overestimate to the square root of a non-negative real number \\ S\\ then will be an underestimate, and vice versa, so the average of these two numbers may reasonably be expected to provide a better approximation (though the formal proof of that assertion depends on the [inequality of arithmetic and geometric means](https://bliptext.com/articles/inequality-of-arithmetic-and-geometric-means) that shows this average is always an overestimate of the square root, as noted in the article on [square roots](https://bliptext.com/articles/square-root), thus assuring convergence).\nMore precisely, if \\ x\\ is our initial guess of and is the error in our estimate such that then we can expand the binomial as:\nand solve for the error term\nTherefore, we can compensate for the error and update our old estimate as Since the computed error was not exact, this is not the actual answer, but becomes our new guess to use in the next round of correction.\nThe process of updating is iterated until desired accuracy is obtained.\nThis algorithm works equally well in the [p-adic numbers](https://bliptext.com/articles/p-adic-number), but cannot be used to identify real square roots with p-adic square roots; one can, for example, construct a sequence of rational numbers by this method that converges to\n+3\nin the reals, but to\n−3\nin the 2-adics.\n\n## Example\nTo calculate for to six significant figures, use the rough estimation method above to get\nTherefore to three decimal places.\n\n## Convergence\n[[File:Babylonian_method_graphs.svg|thumb|300px|link=|Semilog graphs comparing the speed of convergence of Heron's method to find the square root of\n100\nfor different initial guesses.\nNegative guesses converge to the negative root, positive guesses to the positive root.\nNote that values closer to the root converge faster, and all approximations are overestimates.\nIn [ the SVG file,] hover over a graph to display its points.]]\nSuppose that Then for any natural number Let the [relative error](https://bliptext.com/articles/relative-error) in \\ x_n\\ be defined by\nand thus\nThen it can be shown that\nAnd thus that\nand consequently that convergence is assured, and [quadratic](https://bliptext.com/articles/quadratic-convergence).\n\n### Worst case for convergence\nIf using the rough estimate above with the Babylonian method, then the least accurate cases in ascending order are as follows:\nThus in any case,\nRounding errors will slow the convergence.\nIt is recommended to keep at least one extra digit beyond the desired accuracy of the \\ x_n\\ being calculated, to avoid significant round-off error.\n\n# Bakhshali method\nThis method for finding an approximation to a square root was described in an ancient [Indian](https://bliptext.com/articles/languages-of-india) manuscript, called the [Bakhshali manuscript](https://bliptext.com/articles/bakhshali-manuscript).\nIt is equivalent to two iterations of the [Babylonian method](https://bliptext.com/articles/) beginning with x0.\nThus, the algorithm is quartically convergent, which means that the number of correct digits of the approximation roughly quadruples with each iteration.\nThe original presentation, using modern notation, is as follows: To calculate \\sqrt{S}, let x_0^2 be the initial approximation to S. Then, successively iterate as:\nThis can be used to construct a rational approximation to the square root by beginning with an integer.\nIf x_0 = N is an integer chosen so N^2 is close to S, and d = S - N^2 is the difference whose absolute value is minimized, then the first iteration can be written as:\nThe Bakhshali method can be generalized to the computation of an arbitrary root, including fractional roots.\n\n## Example\nUsing the same example as given with the Babylonian method, let S = 125348.\nThen, the first iteration gives\nLikewise the second iteration gives\n\n# Digit-by-digit calculation\nThis is a method to find each digit of the square root in a sequence.\nThis method is based on the [binomial theorem](https://bliptext.com/articles/binomial-theorem) and basically an inverse algorithm solving.\nIt is slower than the Babylonian method, but it has several advantages:\nDisadvantages are:\n[Napier's bones](https://bliptext.com/articles/napier-s-bones) include an aid for the execution of this algorithm.\nThe shifting\nn\nth root algorithm is a generalization of this method.\n\n## Basic principle\nFirst, consider the case of finding the square root of a number\nS\n, that is the square of a base-10 two-digit number\nXY\n, where\nX\nis the tens digit and\nY\nis the units digit.\nSpecifically:\nS\nwill consist of 3 or 4 decimal digits.\nNow to start the digit-by-digit algorithm, we split the digits of\nS\nin two groups of two digits, starting from the right.\nThis means that the first group will be of 1 or 2 digits.\nThen we determine the value of\nX\nas the largest digit such that\nX2\nis less than or equal to the first group.\nWe then compute the difference between the first group and\nX2\nand start the second iteration by concatenating the second group to it.\nThis is equivalent to subtracting 100X^2 from\nS\n, and we're left with.\nWe divide\nS'\nby 10, then divide it by\n2X\nand keep the integer part to try and guess\nY\n. We concatenate\n2X\nwith the tentative\nY\nand multiply it by\nY\n. If our guess is correct, this is equivalent to computing: and so the remainder, that is the difference between\nS'\nand the result, is zero; if the result is higher than\nS'\n, we lower our guess by 1 and try again until the remainder is 0.\nSince this is a simple case where the answer is a perfect square root\nXY\n, the algorithm stops here.\nThe same idea can be extended to any arbitrary square root computation next.\nSuppose we are able to find the square root of\nS\nby expressing it as a sum of n positive numbers such that\nBy repeatedly applying the basic identity\nthe right-hand-side term can be expanded as\nThis expression allows us to find the square root by sequentially guessing the values of a_is.\nSuppose that the numbers have already been guessed, then the m-th term of the right-hand-side of above summation is given by where is the approximate square root found so far.\nNow each new guess a_m should satisfy the recursion\nwhere X_{m} is the sum of all the terms after Y_{m}, i.e. the remainder, such that X_m \\geq 0 for all with initialization X_0 = S. When X_n = 0, the exact square root has been found; if not, then the sum of the a_is gives a suitable approximation of the square root, with X_n being the approximation error.\nFor example, in the decimal number system we have\nwhere 10^{n-i} are place holders and the coefficients.\nAt any m-th stage of the square root calculation, the approximate root found so far, P_{m-1} and the summation term Y_m are given by\nHere since the place value of Y_m is an even power of 10, we only need to work with the pair of most significant digits of the remainder X_{m-1}, whose first term is Y_{m}, at any m-th stage.\nThe section below codifies this procedure.\nIt is obvious that a similar method can be used to compute the square root in number systems other than the decimal number system.\nFor instance, finding the digit-by-digit square root in the binary number system is quite efficient since the value of a_i is searched from a smaller set of binary digits {0,1}.\nThis makes the computation faster since at each stage the value of Y_m is either Y_m = 0 for a_m = 0 or for a_m = 1.\nThe fact that we have only two possible options for a_m also makes the process of deciding the value of a_m at m-th stage of calculation easier.\nThis is because we only need to check if for a_m = 1.\nIf this condition is satisfied, then we take a_m = 1; if not then a_m = 0.\nAlso, the fact that multiplication by 2 is done by left bit-shifts helps in the computation.\n\n## Decimal (base 10)\nWrite the original number in decimal form.\nThe numbers are written similar to the [long division](https://bliptext.com/articles/long-division) algorithm, and, as in [long division](https://bliptext.com/articles/long-division), the root will be written on the line above.\nNow separate the digits into pairs, starting from the decimal point and going both left and right.\nThe decimal point of the root will be above the decimal point of the square.\nOne digit of the root will appear above each pair of digits of the square.\nBeginning with the left-most pair of digits, do the following procedure for each pair:\n\n### Examples\n**Find the square root of 152.2756.**\n1 2. 3 4 / \\/ 01 52.27 56 01 1*1 <= 1 < 2*2 x=1\n01 y = x*x = 1*1 = 1\n00 52 22*2 <= 52 < 23*3 x=2\n00 44 y = (20+x)*x = 22*2 = 44\n08 27 243*3 <= 827 < 244*4 x=3\n07 29 y = (240+x)*x = 243*3 = 729\n98 56 2464*4 <= 9856 < 2465*5 x=4\n98 56 y = (2460+x)*x = 2464*4 = 9856\n00 00 Algorithm terminates: Answer=12.34\n\n## Binary numeral system (base 2)\nThis section uses the formalism from [the digit-by-digit calculation section above](https://bliptext.com/articles/), with the slight variation that we let, with each a_m = 2^m or a_m = 0.\nWe iterate all 2^m, from 2^n down to 2^0, and build up an approximate solution, the sum of all a_i for which we have determined the value.\nTo determine if a_m equals 2^m or 0, we let.\nIf (i.e. the square of our approximate solution including 2^m does not exceed the target square) then a_m = 2^m, otherwise a_m = 0 and.\nTo avoid squaring P_m in each step, we store the difference and incrementally update it by setting with.\nInitially, we set for the largest n with.\nAs an extra optimization, we store and (2^m)^2, the two terms of Y_m in case that a_m is nonzero, in separate variables c_m, d_m:\nc_m and d_m can be efficiently updated in each step:\nNote that:\nwhich is the final result returned in the function below.\nAn implementation of this algorithm in C:\nFaster 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](https://bliptext.com/articles/space-time-tradeoff).\n\n# Exponential identity\n[Pocket calculators](https://bliptext.com/articles/calculator) typically implement good routines to compute the [exponential function](https://bliptext.com/articles/exponential-function) and the [natural logarithm](https://bliptext.com/articles/natural-logarithm), and then compute the square root of S using the identity found using the properties of logarithms and exponentials :\nThe denominator in the fraction corresponds to the nth root.\nIn the case above the denominator is 2, hence the equation specifies that the square root is to be found.\nThe same identity is used when computing square roots with [logarithm tables](https://bliptext.com/articles/logarithm-table) or [slide rules](https://bliptext.com/articles/slide-rule).\n\n# A two-variable iterative method\nThis method is applicable for finding the square root of and converges best for S \\approx 1.\nThis, however, is no real limitation for a computer-based calculation, as in base 2 floating-point and fixed-point representations, it is trivial to multiply S \\,\\!\nby an integer power of 4, and therefore \\sqrt{S} by the corresponding power of 2, by changing the exponent or by shifting, respectively.\nTherefore, S \\,\\!\ncan be moved to the range.\nMoreover, the following method does not employ general divisions, but only additions, subtractions, multiplications, and divisions by powers of two, which are again trivial to implement.\nA disadvantage of the method is that numerical errors accumulate, in contrast to single variable iterative methods such as the Babylonian one.\nThe initialization step of this method is\nwhile the iterative steps read\nThen, (while c_n \\to 0).\nThe convergence of c_n \\,\\!, and therefore also of a_n \\,\\!, is quadratic.\nThe proof of the method is rather easy.\nFirst, rewrite the iterative definition of c_n as\nThen it is straightforward to prove by induction that\nand therefore the convergence of a_n \\,\\!\nto the desired result \\sqrt{S} is ensured by the convergence of c_n \\,\\!\nto 0, which in turn follows from.\nThis method was developed around 1950 by [M. V. Wilkes](https://bliptext.com/articles/maurice-wilkes), [D. J. Wheeler](https://bliptext.com/articles/david-wheeler-computer-scientist) and [S. Gill](https://bliptext.com/articles/stanley-gill) for use on [EDSAC](https://bliptext.com/articles/electronic-delay-storage-automatic-calculator), one of the first electronic computers.\nThe method was later generalized, allowing the computation of non-square roots.\n\n# Iterative methods for reciprocal square roots\nThe following are iterative methods for finding the reciprocal square root of S which is 1/\\sqrt{S}.\nOnce it has been found, find \\sqrt{S} by simple multiplication:.\nThese iterations involve only multiplication, and not division.\nThey are therefore faster than the [Babylonian method](https://bliptext.com/articles/methods-of-computing-square-roots).\nHowever, they are not stable.\nIf the initial value is not close to the reciprocal square root, the iterations will diverge away from it rather than converge to it.\nIt can therefore be advantageous to perform an iteration of the Babylonian method on a rough estimate before starting to apply these methods.\n\n## Goldschmidt's algorithm\nGoldschmidt's algorithm is an extension of [Goldschmidt division](https://bliptext.com/articles/goldschmidt-division), named after Robert Elliot Goldschmidt, which can be used to calculate square roots.\nSome computers use Goldschmidt's algorithm to simultaneously calculate \\sqrt{S} and.\nGoldschmidt's algorithm finds \\sqrt{S} faster than Newton-Raphson iteration on a computer with a [fused multiply–add](https://bliptext.com/articles/fused-multiply-add) instruction and either a pipelined floating-point unit or two independent floating-point units.\nThe first way of writing Goldschmidt's algorithm begins\nand iterates\nuntil b_i is sufficiently close to 1, or a fixed number of iterations.\nThe iterations converge to and\nNote that it is possible to omit either x_n and y_n from the computation, and if both are desired then x_n = S y_n may be used at the end rather than computing it through in each iteration.\nA second form, using [fused multiply-add](https://bliptext.com/articles/fused-multiply-add) operations, begins\nand iterates\nuntil r_i is sufficiently close to 0, or a fixed number of iterations.\nThis converges to and\n\n# Taylor series\nIf N is an approximation to \\sqrt{S}, a better approximation can be found by using the [Taylor series](https://bliptext.com/articles/taylor-series) of the [square root](https://bliptext.com/articles/square-root) function:\nAs an iterative method, the [order of convergence](https://bliptext.com/articles/order-of-convergence) is equal to the number of terms used.\nWith two terms, it is identical to the [Babylonian method](https://bliptext.com/articles/).\nWith three terms, each iteration takes almost as many operations as the [Bakhshali approximation](https://bliptext.com/articles/), but converges more slowly.\nTherefore, this is not a particularly efficient way of calculation.\nTo maximize the rate of convergence, choose N so that is as small as possible.\n\n# Continued fraction expansion\nThe [continued fraction](https://bliptext.com/articles/continued-fraction) representation of a real number can be used instead of its decimal or binary expansion and this representation has the property that the square root of any rational number (which is not already a perfect square) has a periodic, repeating expansion, similar to how rational numbers have repeating expansions in the decimal notation system.\n[Quadratic irrationals](https://bliptext.com/articles/quadratic-irrational) (numbers of the form, where a, b and c are integers), and in particular, square roots of integers, have [periodic continued fractions](https://bliptext.com/articles/periodic-continued-fraction).\nSometimes what is desired is finding not the numerical value of a square root, but rather its [continued fraction](https://bliptext.com/articles/continued-fraction) expansion, and hence its rational approximation.\nLet S be the positive number for which we are required to find the square root.\nThen assuming a to be a number that serves as an initial guess and r to be the remainder term, we can write Since we have, we can express the square root of S as\nBy applying this expression for \\sqrt{S} to the denominator term of the fraction, we have\n

\nCompact notation\nThe numerator/denominator expansion for continued fractions (see left) is cumbersome to write as well as to embed in text formatting systems.\nSo mathematicians have devised several alternative notations, like\nWhen r = 1 throughout, an even more compact notation is:\nFor repeating continued fractions (which all square roots of non-perfect squares do), the repetend is represented only once, with an overline to signify a non-terminating repetition of the overlined part:\nFor √2, the value of a is 1, so its representation is:\nProceeding this way, we get a [generalized continued fraction](https://bliptext.com/articles/generalized-continued-fraction) for the square root as\nThe first step to evaluating such a fraction to obtain a root is to do numerical substitutions for the root of the number desired, and number of denominators selected.\nFor example, in canonical form, r is 1 and for √2, a is 1, so the numerical continued fraction for 3 denominators is:\nStep 2 is to reduce the continued fraction from the bottom up, one denominator at a time, to yield a rational fraction whose numerator and denominator are integers.\nThe reduction proceeds thus (taking the first three denominators):\nFinally (step 3), divide the numerator by the denominator of the rational fraction to obtain the approximate value of the root:\nrounded to three digits of precision.\nThe actual value of √2 is 1.41 to three significant digits.\nThe relative error is 0.17%, so the rational fraction is good to almost three digits of precision.\nTaking more denominators gives successively better approximations: four denominators yields the fraction, good to almost 4 digits of precision, etc.\nThe following are examples of square roots, their simple continued fractions, and their first terms — called convergents — up to and including denominator 99:\nIn general, the larger the denominator of a rational fraction, the better the approximation.\nIt can also be shown that truncating a continued fraction yields a rational fraction that is the best approximation to the root of any fraction with denominator less than or equal to the denominator of that fraction — e.g., no fraction with a denominator less than or equal to 70 is as good an approximation to √2 as 99/70.\n\n# Approximations that depend on the floating point representation\nA number is represented in a [floating point](https://bliptext.com/articles/floating-point) format as m\\times b^p which is also called [scientific notation](https://bliptext.com/articles/scientific-notation).\nIts square root is and similar formulae would apply for cube roots and logarithms.\nOn the face of it, this is no improvement in simplicity, but suppose that only an approximation is required: then just b^{p/2} is good to an order of magnitude.\nNext, recognise that some powers, p, will be odd, thus for 3141.59 = 3.14159 rather than deal with fractional powers of the base, multiply the mantissa by the base and subtract one from the power to make it even.\nThe adjusted representation will become the equivalent of 31.4159 so that the square root will be √31.4159.\nIf the integer part of the adjusted mantissa is taken, there can only be the values 1 to 99, and that could be used as an index into a table of 99 pre-computed square roots to complete the estimate.\nA computer using base sixteen would require a larger table, but one using base two would require only three entries: the possible bits of the integer part of the adjusted mantissa are 01 (the power being even so there was no shift, remembering that a [normalised](https://bliptext.com/articles/normalized-number) floating point number always has a non-zero high-order digit) or if the power was odd, 10 or 11, these being the first two bits of the original mantissa.\nThus, 6.25 = 110.01 in binary, normalised to 1.1001 × 22 an even power so the paired bits of the mantissa are 01, while .625 = 0.101 in binary normalises to 1.01 × 2−1 an odd power so the adjustment is to 10.1 × 2−2 and the paired bits are 10.\nNotice that the low order bit of the power is echoed in the high order bit of the pairwise mantissa.\nAn even power has its low-order bit zero and the adjusted mantissa will start with 0, whereas for an odd power that bit is one and the adjusted mantissa will start with 1.\nThus, when the power is halved, it is as if its low order bit is shifted out to become the first bit of the pairwise mantissa.\nA table with only three entries could be enlarged by incorporating additional bits of the mantissa.\nHowever, with computers, rather than calculate an interpolation into a table, it is often better to find some simpler calculation giving equivalent results.\nEverything now depends on the exact details of the format of the representation, plus what operations are available to access and manipulate the parts of the number.\nFor example, [Fortran](https://bliptext.com/articles/fortran) offers an function to obtain the power.\nEffort expended in devising a good initial approximation is to be recouped by thereby avoiding the additional iterations of the refinement process that would have been needed for a poor approximation.\nSince these are few (one iteration requires a divide, an add, and a halving) the constraint is severe.\nMany computers follow the [IEEE](https://bliptext.com/articles/ieee-floating-point-standard) (or sufficiently similar) representation, and a very rapid approximation to the square root can be obtained for starting Newton's method.\nThe technique that follows is based on the fact that the floating point format (in base two) approximates the base-2 logarithm.\nThat is\nSo for a 32-bit single precision floating point number in IEEE format (where notably, the power has a [bias](https://bliptext.com/articles/exponent-bias) of 127 added for the represented form) you can get the approximate logarithm by interpreting its binary representation as a 32-bit integer, scaling it by 2^{-23}, and removing a [bias](https://bliptext.com/articles/exponent-bias) of 127, i.e.\nFor example, 1.0 is represented by a [hexadecimal](https://bliptext.com/articles/hexadecimal) number 0x3F800000, which would represent if taken as an integer.\nUsing the formula above you get, as expected from \\log_2(1.0).\nIn a similar fashion you get 0.5 from 1.5 (0x3FC00000).\nTo get the square root, divide the logarithm by 2 and convert the value back.\nThe following program demonstrates the idea.\nThe exponent's lowest bit is intentionally allowed to propagate into the mantissa.\nOne way to justify the steps in this program is to assume b is the exponent bias and n is the number of explicitly stored bits in the mantissa and then show that\nThe three mathematical operations forming the core of the above function can be expressed in a single line.\nAn additional adjustment can be added to reduce the maximum relative error.\nSo, the three operations, not including the cast, can be rewritten as\nwhere a is a bias for adjusting the approximation errors.\nFor example, with a = 0 the results are accurate for even powers of 2 (e.g. 1.0), but for other numbers the results will be slightly too big (e.g. 1.5 for 2.0 instead of 1.414... with 6% error).\nWith a = −0x4B0D2, the maximum relative error is minimized to ±3.5%.\nIf the approximation is to be used for an initial guess for [Newton's method](https://bliptext.com/articles/newton-s-method) to the equation, then the reciprocal form shown in the following section is preferred.\n# [Methods of computing square roots | upload.wikimedia.org/wikipedia/commons/7/7e/Log2approx.png]\n\n## Reciprocal of the square root\nA variant of the above routine is included below, which can be used to compute the [reciprocal](https://bliptext.com/articles/multiplicative-inverse) of the square root, i.e., x^{-1/2} instead, was written by Greg Walsh.\nThe 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](https://bliptext.com/articles/newton-s-method) on the following line.\nIn computer graphics it is a very efficient way to normalize a vector.\nSome VLSI hardware implements inverse square root using a second degree polynomial estimation followed by a [Goldschmidt iteration](https://bliptext.com/articles/division-algorithm).\n\n# Negative or complex square\nIf S < 0, then its principal square root is\nIf S = a+bi where a and b are real and b ≠ 0, then its principal square root is\nThis can be verified by squaring the root.\nHere\nis the [modulus](https://bliptext.com/articles/absolute-value) of S.\nThe principal square root of a [complex number](https://bliptext.com/articles/complex-number) is defined to be the root with the non-negative real part.","current_revision":null,"created_by":"kACnuQmH5k9Nr5VNVEpVeg7XpvUmMctD","created_at":"2025-01-25T22:22:38.646Z","updated_at":"2025-01-25T22:22:38.646Z","search_vector":"'comput':3 'method':1 'root':5 'squar':4","search_content":" \n \n\n# \n**Methods of computing [square root](https://bliptext.com/articles/square-root)s** are [algorithms](https://bliptext.com/articles/algorithm) for approximating the non-negative [square root](https://bliptext.com/articles/square-root) \\sqrt{S} of a positive [real number](https://bliptext.com/articles/real-number) S. Since all [square root](https://bliptext.com/articles/square-root)s of [natural numbers](https://bliptext.com/articles/natural-number), other than of [perfect squares](https://bliptext.com/articles/square-number), are [irrational](https://bliptext.com/articles/irrational-number),\nsquare roots can usually only be computed to some finite precision: these methods typically construct a series of increasingly accurate [approximations](https://bliptext.com/articles/numerical-approximation).\nMost square root computation methods are iterative: after choosing a suitable [initial estimate](https://bliptext.com/articles/) of \\sqrt{S}, an iterative refinement is performed until some termination criterion is met.\nOne refinement scheme is [Heron's method](https://bliptext.com/articles/), a special case of [Newton's method](https://bliptext.com/articles/newton-s-method).\nIf division is much more costly than multiplication, it may be preferable to compute the [inverse square root](https://bliptext.com/articles/) instead.\nOther methods are available to compute the square root [digit by digit](https://bliptext.com/articles/), or using [Taylor series](https://bliptext.com/articles/).\nRational approximations of square roots may be calculated using [continued fraction expansions](https://bliptext.com/articles/).\nThe method employed depends on the needed accuracy, and the available tools and computational power.\nThe methods may be roughly classified as those suitable for mental calculation, those usually requiring at least paper and pencil, and those which are implemented as programs to be executed on a digital electronic computer or other computing device.\nAlgorithms may take into account convergence (how many iterations are required to achieve a specified precision), computational complexity of individual operations (i.e. division) or iterations, and error propagation (the accuracy of the final result).\nA few methods like paper-and-pencil synthetic division and series expansion, do not require a starting value.\nIn some applications, an [integer square root](https://bliptext.com/articles/integer-square-root) is required, which is the square root rounded or truncated to the nearest integer (a modified procedure may be employed in this case).\n\n# History\nProcedures for finding square roots (particularly the [square root of 2](https://bliptext.com/articles/square-root-of-2)) have been known since at least the period of ancient Babylon in the 17th century BCE.\n[Babylonian mathematicians](https://bliptext.com/articles/babylonian-mathematics) calculated the square root of 2 to three sexagesimal \"digits\" after the 1, but it is not known exactly how.\nThey knew how to approximate a hypotenuse using\n(giving for example for the diagonal of a gate whose height is rods and whose width is rods) and they may have used a similar approach for finding the approximation of \\sqrt 2.\n[Heron's method](https://bliptext.com/articles/heron-s-method) from first century Egypt was the first ascertainable algorithm for computing square root.\nModern analytic methods began to be developed after introduction of the [Arabic numeral](https://bliptext.com/articles/arabic-numeral) system to western Europe in the early Renaissance.\nToday, nearly all computing devices have a fast and accurate square root function, either as a programming language construct, a compiler intrinsic or library function, or as a hardware operator, based on one of the described procedures.\n\n# Initial estimate\nMany iterative square root algorithms require an initial [seed value](https://bliptext.com/articles/seed-value).\nThe seed must be a non-zero positive number; it should be between 1 and S, the number whose square root is desired, because the square root must be in that range.\nIf the seed is far away from the root, the algorithm will require more iterations.\nIf one initializes with x_0=1 (or S), then approximately iterations will be wasted just getting the order of magnitude of the root.\nIt is therefore useful to have a rough estimate, which may have limited accuracy but is easy to calculate.\nIn general, the better the initial estimate, the faster the convergence.\nFor Newton's method (also called Babylonian or Heron's method), a seed somewhat larger than the root will converge slightly faster than a seed somewhat smaller than the root.\nIn general, an estimate is pursuant to an arbitrary interval known to contain the root (such as [x_0,S/x_0]).\nThe estimate is a specific value of a functional approximation to over the interval.\nObtaining a better estimate involves either obtaining tighter bounds on the interval, or finding a better functional approximation to f(x).\nThe latter usually means using a higher order polynomial in the approximation, though not all approximations are polynomial.\nCommon methods of estimating include scalar, linear, hyperbolic and logarithmic.\nA decimal base is usually used for mental or paper-and-pencil estimating.\nA binary base is more suitable for computer estimates.\nIn estimating, the exponent and [mantissa](https://bliptext.com/articles/mantissa-logarithm) are usually treated separately, as the number would be expressed in scientific notation.\n\n## Decimal estimates\nTypically the number S is expressed in [scientific notation](https://bliptext.com/articles/scientific-notation) as where 1\\leq a 0\nbe any positive [initial estimate](https://bliptext.com/articles/).\nHeron's method consists in iteratively computing\nuntil the desired accuracy is achieved.\nThe sequence defined by this equation converges to\nThis is equivalent to using [Newton's method](https://bliptext.com/articles/newton-s-method) to solve x^2 - S = 0.\nThis algorithm is [quadratically convergent](https://bliptext.com/articles/quadratic-convergence): the number of correct digits of x_n roughly doubles with each iteration.\n\n## Derivation\nThe basic idea is that if \\ x\\ is an overestimate to the square root of a non-negative real number \\ S\\ then will be an underestimate, and vice versa, so the average of these two numbers may reasonably be expected to provide a better approximation (though the formal proof of that assertion depends on the [inequality of arithmetic and geometric means](https://bliptext.com/articles/inequality-of-arithmetic-and-geometric-means) that shows this average is always an overestimate of the square root, as noted in the article on [square roots](https://bliptext.com/articles/square-root), thus assuring convergence).\nMore precisely, if \\ x\\ is our initial guess of and is the error in our estimate such that then we can expand the binomial as:\nand solve for the error term\nTherefore, we can compensate for the error and update our old estimate as Since the computed error was not exact, this is not the actual answer, but becomes our new guess to use in the next round of correction.\nThe process of updating is iterated until desired accuracy is obtained.\nThis algorithm works equally well in the [p-adic numbers](https://bliptext.com/articles/p-adic-number), but cannot be used to identify real square roots with p-adic square roots; one can, for example, construct a sequence of rational numbers by this method that converges to\n+3\nin the reals, but to\n−3\nin the 2-adics.\n\n## Example\nTo calculate for to six significant figures, use the rough estimation method above to get\nTherefore to three decimal places.\n\n## Convergence\n[[File:Babylonian_method_graphs.svg|thumb|300px|link=|Semilog graphs comparing the speed of convergence of Heron's method to find the square root of\n100\nfor different initial guesses.\nNegative guesses converge to the negative root, positive guesses to the positive root.\nNote that values closer to the root converge faster, and all approximations are overestimates.\nIn [ the SVG file,] hover over a graph to display its points.]]\nSuppose that Then for any natural number Let the [relative error](https://bliptext.com/articles/relative-error) in \\ x_n\\ be defined by\nand thus\nThen it can be shown that\nAnd thus that\nand consequently that convergence is assured, and [quadratic](https://bliptext.com/articles/quadratic-convergence).\n\n### Worst case for convergence\nIf using the rough estimate above with the Babylonian method, then the least accurate cases in ascending order are as follows:\nThus in any case,\nRounding errors will slow the convergence.\nIt is recommended to keep at least one extra digit beyond the desired accuracy of the \\ x_n\\ being calculated, to avoid significant round-off error.\n\n# Bakhshali method\nThis method for finding an approximation to a square root was described in an ancient [Indian](https://bliptext.com/articles/languages-of-india) manuscript, called the [Bakhshali manuscript](https://bliptext.com/articles/bakhshali-manuscript).\nIt is equivalent to two iterations of the [Babylonian method](https://bliptext.com/articles/) beginning with x0.\nThus, the algorithm is quartically convergent, which means that the number of correct digits of the approximation roughly quadruples with each iteration.\nThe original presentation, using modern notation, is as follows: To calculate \\sqrt{S}, let x_0^2 be the initial approximation to S. Then, successively iterate as:\nThis can be used to construct a rational approximation to the square root by beginning with an integer.\nIf x_0 = N is an integer chosen so N^2 is close to S, and d = S - N^2 is the difference whose absolute value is minimized, then the first iteration can be written as:\nThe Bakhshali method can be generalized to the computation of an arbitrary root, including fractional roots.\n\n## Example\nUsing the same example as given with the Babylonian method, let S = 125348.\nThen, the first iteration gives\nLikewise the second iteration gives\n\n# Digit-by-digit calculation\nThis is a method to find each digit of the square root in a sequence.\nThis method is based on the [binomial theorem](https://bliptext.com/articles/binomial-theorem) and basically an inverse algorithm solving.\nIt is slower than the Babylonian method, but it has several advantages:\nDisadvantages are:\n[Napier's bones](https://bliptext.com/articles/napier-s-bones) include an aid for the execution of this algorithm.\nThe shifting\nn\nth root algorithm is a generalization of this method.\n\n## Basic principle\nFirst, consider the case of finding the square root of a number\nS\n, that is the square of a base-10 two-digit number\nXY\n, where\nX\nis the tens digit and\nY\nis the units digit.\nSpecifically:\nS\nwill consist of 3 or 4 decimal digits.\nNow to start the digit-by-digit algorithm, we split the digits of\nS\nin two groups of two digits, starting from the right.\nThis means that the first group will be of 1 or 2 digits.\nThen we determine the value of\nX\nas the largest digit such that\nX2\nis less than or equal to the first group.\nWe then compute the difference between the first group and\nX2\nand start the second iteration by concatenating the second group to it.\nThis is equivalent to subtracting 100X^2 from\nS\n, and we're left with.\nWe divide\nS'\nby 10, then divide it by\n2X\nand keep the integer part to try and guess\nY\n. We concatenate\n2X\nwith the tentative\nY\nand multiply it by\nY\n. If our guess is correct, this is equivalent to computing: and so the remainder, that is the difference between\nS'\nand the result, is zero; if the result is higher than\nS'\n, we lower our guess by 1 and try again until the remainder is 0.\nSince this is a simple case where the answer is a perfect square root\nXY\n, the algorithm stops here.\nThe same idea can be extended to any arbitrary square root computation next.\nSuppose we are able to find the square root of\nS\nby expressing it as a sum of n positive numbers such that\nBy repeatedly applying the basic identity\nthe right-hand-side term can be expanded as\nThis expression allows us to find the square root by sequentially guessing the values of a_is.\nSuppose that the numbers have already been guessed, then the m-th term of the right-hand-side of above summation is given by where is the approximate square root found so far.\nNow each new guess a_m should satisfy the recursion\nwhere X_{m} is the sum of all the terms after Y_{m}, i.e. the remainder, such that X_m \\geq 0 for all with initialization X_0 = S. When X_n = 0, the exact square root has been found; if not, then the sum of the a_is gives a suitable approximation of the square root, with X_n being the approximation error.\nFor example, in the decimal number system we have\nwhere 10^{n-i} are place holders and the coefficients.\nAt any m-th stage of the square root calculation, the approximate root found so far, P_{m-1} and the summation term Y_m are given by\nHere since the place value of Y_m is an even power of 10, we only need to work with the pair of most significant digits of the remainder X_{m-1}, whose first term is Y_{m}, at any m-th stage.\nThe section below codifies this procedure.\nIt is obvious that a similar method can be used to compute the square root in number systems other than the decimal number system.\nFor instance, finding the digit-by-digit square root in the binary number system is quite efficient since the value of a_i is searched from a smaller set of binary digits {0,1}.\nThis makes the computation faster since at each stage the value of Y_m is either Y_m = 0 for a_m = 0 or for a_m = 1.\nThe fact that we have only two possible options for a_m also makes the process of deciding the value of a_m at m-th stage of calculation easier.\nThis is because we only need to check if for a_m = 1.\nIf this condition is satisfied, then we take a_m = 1; if not then a_m = 0.\nAlso, the fact that multiplication by 2 is done by left bit-shifts helps in the computation.\n\n## Decimal (base 10)\nWrite the original number in decimal form.\nThe numbers are written similar to the [long division](https://bliptext.com/articles/long-division) algorithm, and, as in [long division](https://bliptext.com/articles/long-division), the root will be written on the line above.\nNow separate the digits into pairs, starting from the decimal point and going both left and right.\nThe decimal point of the root will be above the decimal point of the square.\nOne digit of the root will appear above each pair of digits of the square.\nBeginning with the left-most pair of digits, do the following procedure for each pair:\n\n### Examples\n**Find the square root of 152.2756.**\n1 2. 3 4 / \\/ 01 52.27 56 01 1*1 \nCompact notation\nThe numerator/denominator expansion for continued fractions (see left) is cumbersome to write as well as to embed in text formatting systems.\nSo mathematicians have devised several alternative notations, like\nWhen r = 1 throughout, an even more compact notation is:\nFor repeating continued fractions (which all square roots of non-perfect squares do), the repetend is represented only once, with an overline to signify a non-terminating repetition of the overlined part:\nFor √2, the value of a is 1, so its representation is:\nProceeding this way, we get a [generalized continued fraction](https://bliptext.com/articles/generalized-continued-fraction) for the square root as\nThe first step to evaluating such a fraction to obtain a root is to do numerical substitutions for the root of the number desired, and number of denominators selected.\nFor example, in canonical form, r is 1 and for √2, a is 1, so the numerical continued fraction for 3 denominators is:\nStep 2 is to reduce the continued fraction from the bottom up, one denominator at a time, to yield a rational fraction whose numerator and denominator are integers.\nThe reduction proceeds thus (taking the first three denominators):\nFinally (step 3), divide the numerator by the denominator of the rational fraction to obtain the approximate value of the root:\nrounded to three digits of precision.\nThe actual value of √2 is 1.41 to three significant digits.\nThe relative error is 0.17%, so the rational fraction is good to almost three digits of precision.\nTaking more denominators gives successively better approximations: four denominators yields the fraction, good to almost 4 digits of precision, etc.\nThe following are examples of square roots, their simple continued fractions, and their first terms — called convergents — up to and including denominator 99:\nIn general, the larger the denominator of a rational fraction, the better the approximation.\nIt can also be shown that truncating a continued fraction yields a rational fraction that is the best approximation to the root of any fraction with denominator less than or equal to the denominator of that fraction — e.g., no fraction with a denominator less than or equal to 70 is as good an approximation to √2 as 99/70.\n\n# Approximations that depend on the floating point representation\nA number is represented in a [floating point](https://bliptext.com/articles/floating-point) format as m\\times b^p which is also called [scientific notation](https://bliptext.com/articles/scientific-notation).\nIts square root is and similar formulae would apply for cube roots and logarithms.\nOn the face of it, this is no improvement in simplicity, but suppose that only an approximation is required: then just b^{p/2} is good to an order of magnitude.\nNext, recognise that some powers, p, will be odd, thus for 3141.59 = 3.14159 rather than deal with fractional powers of the base, multiply the mantissa by the base and subtract one from the power to make it even.\nThe adjusted representation will become the equivalent of 31.4159 so that the square root will be √31.4159.\nIf the integer part of the adjusted mantissa is taken, there can only be the values 1 to 99, and that could be used as an index into a table of 99 pre-computed square roots to complete the estimate.\nA computer using base sixteen would require a larger table, but one using base two would require only three entries: the possible bits of the integer part of the adjusted mantissa are 01 (the power being even so there was no shift, remembering that a [normalised](https://bliptext.com/articles/normalized-number) floating point number always has a non-zero high-order digit) or if the power was odd, 10 or 11, these being the first two bits of the original mantissa.\nThus, 6.25 = 110.01 in binary, normalised to 1.1001 × 22 an even power so the paired bits of the mantissa are 01, while .625 = 0.101 in binary normalises to 1.01 × 2−1 an odd power so the adjustment is to 10.1 × 2−2 and the paired bits are 10.\nNotice that the low order bit of the power is echoed in the high order bit of the pairwise mantissa.\nAn even power has its low-order bit zero and the adjusted mantissa will start with 0, whereas for an odd power that bit is one and the adjusted mantissa will start with 1.\nThus, when the power is halved, it is as if its low order bit is shifted out to become the first bit of the pairwise mantissa.\nA table with only three entries could be enlarged by incorporating additional bits of the mantissa.\nHowever, with computers, rather than calculate an interpolation into a table, it is often better to find some simpler calculation giving equivalent results.\nEverything now depends on the exact details of the format of the representation, plus what operations are available to access and manipulate the parts of the number.\nFor example, [Fortran](https://bliptext.com/articles/fortran) offers an function to obtain the power.\nEffort expended in devising a good initial approximation is to be recouped by thereby avoiding the additional iterations of the refinement process that would have been needed for a poor approximation.\nSince these are few (one iteration requires a divide, an add, and a halving) the constraint is severe.\nMany computers follow the [IEEE](https://bliptext.com/articles/ieee-floating-point-standard) (or sufficiently similar) representation, and a very rapid approximation to the square root can be obtained for starting Newton's method.\nThe technique that follows is based on the fact that the floating point format (in base two) approximates the base-2 logarithm.\nThat is\nSo for a 32-bit single precision floating point number in IEEE format (where notably, the power has a [bias](https://bliptext.com/articles/exponent-bias) of 127 added for the represented form) you can get the approximate logarithm by interpreting its binary representation as a 32-bit integer, scaling it by 2^{-23}, and removing a [bias](https://bliptext.com/articles/exponent-bias) of 127, i.e.\nFor example, 1.0 is represented by a [hexadecimal](https://bliptext.com/articles/hexadecimal) number 0x3F800000, which would represent if taken as an integer.\nUsing the formula above you get, as expected from \\log_2(1.0).\nIn a similar fashion you get 0.5 from 1.5 (0x3FC00000).\nTo get the square root, divide the logarithm by 2 and convert the value back.\nThe following program demonstrates the idea.\nThe exponent's lowest bit is intentionally allowed to propagate into the mantissa.\nOne way to justify the steps in this program is to assume b is the exponent bias and n is the number of explicitly stored bits in the mantissa and then show that\nThe three mathematical operations forming the core of the above function can be expressed in a single line.\nAn additional adjustment can be added to reduce the maximum relative error.\nSo, the three operations, not including the cast, can be rewritten as\nwhere a is a bias for adjusting the approximation errors.\nFor example, with a = 0 the results are accurate for even powers of 2 (e.g. 1.0), but for other numbers the results will be slightly too big (e.g. 1.5 for 2.0 instead of 1.414... with 6% error).\nWith a = −0x4B0D2, the maximum relative error is minimized to ±3.5%.\nIf the approximation is to be used for an initial guess for [Newton's method](https://bliptext.com/articles/newton-s-method) to the equation, then the reciprocal form shown in the following section is preferred.\n# [Methods of computing square roots | upload.wikimedia.org/wikipedia/commons/7/7e/Log2approx.png]\n\n## Reciprocal of the square root\nA variant of the above routine is included below, which can be used to compute the [reciprocal](https://bliptext.com/articles/multiplicative-inverse) of the square root, i.e., x^{-1/2} instead, was written by Greg Walsh.\nThe 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](https://bliptext.com/articles/newton-s-method) on the following line.\nIn computer graphics it is a very efficient way to normalize a vector.\nSome VLSI hardware implements inverse square root using a second degree polynomial estimation followed by a [Goldschmidt iteration](https://bliptext.com/articles/division-algorithm).\n\n# Negative or complex square\nIf S < 0, then its principal square root is\nIf S = a+bi where a and b are real and b ≠ 0, then its principal square root is\nThis can be verified by squaring the root.\nHere\nis the [modulus](https://bliptext.com/articles/absolute-value) of S.\nThe principal square root of a [complex number](https://bliptext.com/articles/complex-number) is defined to be the root with the non-negative real part.","revisionCount":0},"hasMore":false,"nextCursor":null}