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Approximating Square Roots and Cube Roots

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Approximating Square Roots and Cube Roots Simple formulas … Approximating Square Roots and Cube Roots feature Do they work? The idea of equivalent geometric forms is used in this study to devise simple formulas to estimate the square root and the cube root of an arbitrary positive number. The resulting formulas are easy to use and they don’t take much time to calculate. They give excellent estimates which compare very favourably with those given by a pocket calculator. Introductory Remarks Ali Ibrahim Hussen Historically the Babylonians were the �irst to devise an iterative method for computing square roots of numbers using rational operations that are easy to carry out ([1], [2]). The ancient Chinese method (200 BCE), commented on by Liu Hui in the third century CE, is similar in procedure to the long division method used in schools even today ([2]). The Greek mathematician Heron of �lexandria, who gave the �irst explicit description of the Babylonians iterative method, also devised a method for cube root calculation in the �irst century CE ([�]). In ��� CE, �ryabhata (Indian mathematician and astronomer) gave a method for computing cube roots of numbers of arbitrary size ([5]). In the sixteenth century, Isaac Newton devised an iterative method used to calculate square and cube roots of arbitrary numbers. The method used in this article is different from the methods mentioned above. It is based on the idea of changing a geometric Keywords: Square root, cube root, estimation, accuracy, geometric mean, harmonic mean, inequality, square, cube 11 At Right Angles | Vol. 4, No. 1, March 2015 Vol. 4, No. 1, March 2015 | At Right Angles 11 �igure or form into an equivalent ‘regular’ one numbers are close to one another, then the AM is We can do better that this by taking and estimate quite close to the GM. To our surprise, with equal area or volume, e.g. changing a quite close to the GM. . Then our estimate for √ is: we �ind that thissecond idea algorithmworks very well. rectangle into a square, or a rectangular cuboid Geometrically this may be viewed as follows. 96 푎 So here is our for approximating into a cube. � 푎 � � � Assume that . Within the rectangle we 2 2 the square root of aAlgorithm given positive II number . The idea was used by ancient Indian engineers draw a square on the shorter side , then slice the 96 Step 1: who worked with designing and constructing leftover portion into two equal halves and stack ≈ 6 62 Write as ×where and are temples (600 BCE). TheySulba-sutras used a practical method them (one each) on to two adjacent sides of the TheSecond error approach now is about to approximating2%. square Step 2: for changing a rectangle into a square. The square (Figure� 2). The resulting shape is a square reasonably close to each other. method appears in the by of side � ( minus a small piece in the corner, roots. � Step 3: � Compute the AM and the HM of and . Baudhayana (Sulba-sutras means the ‘Rule of the which is a square of side | |. If is close to , In the study of statistics we encounter Chord’) ([6]). The idea is used also by David W. then this portion will be very small. So if we Compute the average of the AM and HM severalarithmetic different mean kinds of meansgeometric de�ined for mean any Henderson in his attempt to solve quadratic and ignore the little square, the larger square may be given pair of positive numbers and .Wehave computed in Step 2. That is, compute: First approach to approximating square roots. harmonic mean cubic equations geometrically ([7]). In our case taken to be the one we seek. the (AM), the �irst algorithm 푎 2 an algebraic analysis is performed instead of a (GM) and the (HM), among � 푎 � geometric one. 2 2 푎 The above reasoning gives us our many others (not listed here). These are de�ined 1. Estimation of Square Roots for approximating the square root of a given as follows: This is our estimate for √. positive number . Algorithm I 2 This algorithm gives us extremely good results — Step 1: AM (푎 GM √ HM Let be the given number whose square root is 2 푎 far better than we would expect! The following Example 3. required. (Here, of course, is positive.) We start Write as where and are close �ach of these has its signi�icance and uses. �f examples illustrate this. by expressing as a product , with as close to toStep each 2: other. particular interest to us is the fact that if , Let 6. Let us take 24 and as possible (the closer the better). The task of � then: � 25. Then our estimate for √6 is: computing √√ may be expressed Then our estimate of √ is ( . HM < GM < AM geometrically as: Construct a square whose area always 24푎25 2×24×25 2 The closer and are to each other, the better � 푎 � �245 푎 � is equal to that of a rectangle with dimensions These inequalities are never violated. The HM and 2 2 24푎25 2 49 will be our estimate. (Figure 1). AM lie on opposite sides of the GM. 4 ≈244949795 This algorithm gives us fairly good results For example, if 2 and , then 96 a √ab provided we are wise in our initial choice of Compare this with the true value: Example 1. numbers. The following examples illustrate this. AM (2 푎 5 GM √2×4 √6 244949742 …. The estimate is correct to 2 Example 4. b Let . Let us take and 2×2× seven decimal places. HM 2 Figure 1. � . Then our estimate for √ is 2푎 Let . Let us take and � ( . This may be compared with and, of course, 2<4<5. . Then our estimate for √ is: the actual value which is roughly 9 (an error geometric mean ofExample about 0 2.0%). 푎 2× × 9 6 Recall that in our approach we start by selecting � 푎 � � 푎 � positive numbers and such that . We 2 2 푎 2 6 9 Now note that √ is the (GM) of Let 0. Let us take and arithmetic mean wish to compute the GM of and . But as the GM 72 the quantities and . We may try to approximate � . Then our estimate for √0 is ≈ 622 is dif�icult to compute, we choose to compute the 22 the GM by the (AM) of and . It � (0 9 ≈ 7. This may be AM instead. We could also choose to compute the is well known that the AM exceeds the GM compared with the actual value which is roughly This estimate is correct to four decimal places. HM (this too is an easy computation). Whichever (strictly) if the numbers are unequal. But if the (an error of about 0%). (TheExample true 5. value is 6227766 ….) b (a + b)/2 one we choose to compute will then be our estimate for the GM. over-estimate Let 2. Let us take 45 92 and 245 49. Then our estimate for If we use the AM, we always get an a under-estimate √2 is: (a + b)/2 for the GM. And if we use the HM, we always get 92푎49 2×92×49 Empty square an . � 푎 � 2 2 92푎49 of side 1 a b both 2 Now an interesting idea strikes us: why not | − | 6 72 54 compute the AM and HM, and then take their � 푎 � ≈44725955 Figure 2. 2 6 6 592 average? Their respective errors may then just cancel each other, and we may just get an This estimate is correct to eight decimal places. 1312 At Right Angles | Vol. 4, No. 1, March 2015 Vol. 4, No. 1, March 2015 | At Right Angles 1312 2 At Right Angles ∣ Vol. 4, No. 1, March 2015 Vol. 4, No. 1, March 2015 ∣ At Right Angles 3 We can do better that this by taking and estimate quite close to the GM. To our surprise, . Then our estimate for √ is: we �ind that thissecond idea algorithmworks very well. 96 푎 So here is our for approximating � 푎 � � � 2 2 the square root of aAlgorithm given positive II number . 96 Step 1: ≈ 6 62 Write as ×where and are TheSecond error approach now is about to approximating2%. square reasonablyStep 2: close to each other. roots. Step 3: Compute the AM and the HM of and . In the study of statistics we encounter Compute the average of the AM and HM severalarithmetic different mean kinds of meansgeometric de�ined for mean any computed in Step 2. That is, compute: given pair of positiveharmonic numbers mean and .Wehave the (AM), the 푎 2 (GM) and the (HM), among � 푎 � 2 2 푎 many others (not listed here). These are de�ined as follows: This is our estimate for √. 2 AM (푎 GM √ HM This algorithm gives us extremely good results — 2 푎 far better than we would expect! The following �ach of these has its signi�icance and uses. �f examplesExample 3. illustrate this. particular interest to us is the fact that if , Let 6. Let us take 24 and then: 25. Then our estimate for √6 is: HM < GM < AM always 24푎25 2×24×25 2 � 푎 � �245 푎 � These inequalities are never violated. The HM and 2 2 24푎25 2 49 AM lie on opposite sides of the GM. 4 ≈244949795 For example, if 2 and , then 96 Compare this with the true value: AM (2 푎 5 GM √2×4 6 244949742 …. The estimate is correct to 2 √ Example 4. 2×2× seven decimal places. HM 2 2푎 Let . Let us take and and, of course, 2<4<5. . Then our estimate for √ is: 푎 2× × 9 6 Recall that in our approach we start by selecting � 푎 � � 푎 � positive numbers and such that .
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