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 𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 and estimate quite close to the GM. To our surprise, with equal area or volume, e.g. changing a quite close to the GM. 𝑏𝑏𝑏𝑏 𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏𝑎𝑎𝑎𝑎𝑎𝑎. Then our estimate for √𝑎𝑎𝑎𝑎 is: we �ind that thissecond idea algorithmworks very well. rectangle into a square, or a rectangular cuboid Geometrically this may be viewed as follows. 𝑎𝑎 𝑎𝑎𝑎𝑎 𝑎𝑎 9𝑎𝑎6𝑎𝑎 + 𝑎𝑎𝑎𝑎 So here is our for approximating into a cube. �𝑎𝑎𝑎𝑎𝑎𝑎 푎 �𝑎𝑎 � � Assume that 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎. Within the 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎rectangle we 2 𝑎𝑎𝑎𝑎𝑎𝑎 2 𝑎𝑎𝑎𝑎𝑎𝑎 the square root of aAlgorithm given positive II number 𝑛𝑛𝑛𝑛. The idea was used by ancient Indian engineers draw a square on the shorter side 𝑎𝑎𝑎𝑎, then slice the 𝑎𝑎9𝑎𝑎6𝑎𝑎 Step 1: who worked with designing and constructing leftover portion into two equal halves and stack 𝑎𝑎 ≈ 𝑎𝑎𝑎𝑎𝑎𝑎6𝑎𝑎𝑎𝑎 6𝑎𝑎2 Write 𝑛𝑛𝑛𝑛 as 𝑎𝑎𝑎𝑎 ×𝑏𝑏𝑏𝑏where 𝑎𝑎𝑎𝑎 and 𝑏𝑏𝑏𝑏 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 approximating𝑏𝑏𝑏𝑏𝑏𝑏2%. 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 � (𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 minus a small piece in the corner, roots. � Step 3: � Compute the AM and the HM of 𝑎𝑎𝑎𝑎 and 𝑏𝑏𝑏𝑏. Baudhayana (Sulba-sutras means the ‘Rule of the which is a square of side |𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎|. If 𝑎𝑎𝑎𝑎 is close to 𝑎𝑎𝑎𝑎, 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 𝑎𝑎𝑎𝑎 and 𝑏𝑏𝑏𝑏.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 𝑎𝑎 𝑎𝑎𝑎𝑎 +𝑏𝑏𝑏𝑏 2𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 an algebraic analysis is performed instead of a (GM) and the (HM), among � + �𝑎𝑎 geometric one. 2 2 𝑎𝑎𝑎𝑎 +𝑏𝑏𝑏𝑏 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 √𝑛𝑛𝑛𝑛. positive number 𝑛𝑛𝑛𝑛. Algorithm I 𝑎𝑎 2𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 This algorithm gives us extremely good results — Step 1: AM 𝑎𝑎 (𝑎𝑎𝑎𝑎+𝑏𝑏𝑏𝑏𝑎𝑎𝑎𝑎 GM 𝑎𝑎 √𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏𝑎𝑎 HM 𝑎𝑎 𝑎𝑎 Let 𝑛𝑛𝑛𝑛 be the given number whose square root is 2 𝑎𝑎𝑎𝑎 +𝑏𝑏𝑏𝑏 far better than we would expect! The following Example 3. required. (Here, of course, 𝑛𝑛𝑛𝑛 is positive.) We start Write 𝑛𝑛𝑛𝑛 as 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎where 𝑎𝑎𝑎𝑎 and 𝑎𝑎𝑎𝑎 are close �ach of these has its signi�icance and uses. �f examples illustrate this. by expressing 𝑛𝑛𝑛𝑛 as a product 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎, with 𝑎𝑎𝑎𝑎 as close to toStep each 2: other. particular interest to us is the fact that if 𝑎𝑎𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏, Let 𝑛𝑛𝑛𝑛 𝑎𝑎�. Let us take 𝑎𝑎𝑎𝑎𝑎𝑎2𝑎𝑎4 and 𝑎𝑎𝑎𝑎 as possible (the closer the better). The task of � then: � 𝑏𝑏𝑏𝑏 𝑎𝑎 �𝑎𝑎�. Then our estimate for √6 is: computing √𝑛𝑛𝑛𝑛𝑛𝑛√𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 may be expressed Then our estimate of √𝑛𝑛𝑛𝑛 is (𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎. HM < GM < AM𝑎𝑎 geometrically as: Construct a square whose area always 𝑎𝑎 2𝑎𝑎4+2𝑎𝑎5 2×2𝑎𝑎4×2𝑎𝑎5 𝑎𝑎 𝑎𝑎2 The closer 𝑎𝑎𝑎𝑎 and 𝑎𝑎𝑎𝑎 are to each other, the better � + �𝑎𝑎 �2𝑎𝑎𝑎 푎 � is equal to that of a rectangle with dimensions These inequalities are never violated. The HM and 2 2 2𝑎𝑎4+2𝑎𝑎5 2 4𝑎𝑎9 will be our estimate. 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎(Figure 1). AM lie on opposite sides of the GM. 4𝑏𝑏𝑏𝑏𝑏𝑏 𝑎𝑎 ≈2𝑎𝑎4494𝑏𝑏9795𝑎𝑎 This algorithm gives us fairly good results For example, if 𝑎𝑎𝑎𝑎𝑎𝑎2 and 𝑏𝑏𝑏𝑏 𝑎𝑎 𝑏𝑏, then 𝑎𝑎96𝑏𝑏 a √ab provided we are wise in our initial choice of Compare this with the true value: Example 1. 𝑎𝑎 numbers. The following examples illustrate this. AM 𝑎𝑎 (2 + 𝑏𝑏𝑎𝑎 𝑎𝑎 5𝑎𝑎 GM 𝑎𝑎 √2×𝑏𝑏𝑏𝑏4𝑎𝑎 √6 𝑎𝑎 2𝑎𝑎4494𝑏𝑏9742 …. The estimate is correct to 2 Example 4. b Let 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛. Let us take 𝑎𝑎𝑎𝑎 𝑛𝑛 𝑎𝑎𝑎𝑎𝑎𝑎 and 2×2×𝑏𝑏 seven decimal places. HM 𝑎𝑎 𝑎𝑎 𝑎𝑎𝑎𝑎2𝑎𝑎 Figure 1. 𝑎𝑎𝑎𝑎� 𝑎𝑎 𝑎𝑎𝑎𝑎𝑏𝑏. Then our estimate for √𝑛𝑛 is 2+𝑏𝑏 Let 𝑛𝑛𝑛𝑛 𝑎𝑎 𝑎𝑎 . Let us take 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎and � (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎 𝑎𝑎 𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎. This may be compared with and, of course, 𝑎𝑎𝑎𝑎2<4<5. 𝑏𝑏𝑏𝑏 𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏𝑎𝑎. Then our estimate for √𝑎𝑎𝑎𝑎 is: the actual value which is roughly 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎9𝑏𝑏 (an error geometric mean ofExample about 0 2.𝑎𝑎0𝑎𝑎%). 𝑎𝑎 𝑎𝑎 + 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 2×𝑎𝑎 × 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎 𝑎𝑎9 6𝑏𝑏 Recall that in our approach we start by selecting � + �𝑎𝑎 � + � positive numbers 𝑎𝑎𝑎𝑎 and 𝑏𝑏𝑏𝑏 such that 𝑛𝑛𝑛𝑛 𝑎𝑎 𝑎𝑎𝑎𝑎 . We 2 2 𝑎𝑎 + 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 2 6 𝑎𝑎9 Now note that √𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 is the (GM) of Let 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛0. Let us take 𝑎𝑎𝑎𝑎 𝑛𝑛 𝑎𝑎 and arithmetic mean wish to compute the GM of 𝑎𝑎𝑎𝑎 and 𝑏𝑏𝑏𝑏. But as the GM 72𝑎𝑎 the quantities 𝑎𝑎𝑎𝑎 and 𝑎𝑎𝑎𝑎. We may try to approximate 𝑎𝑎𝑎𝑎� 𝑎𝑎𝑎𝑎𝑎𝑏𝑏𝑎𝑎. Then our estimate for √𝑛𝑛0 is 𝑎𝑎 ≈ 𝑎𝑎𝑎𝑎𝑎𝑎622𝑏𝑏𝑏𝑏 is dif�icult to compute, we choose to compute the 22𝑏𝑏 the GM by the (AM) of 𝑎𝑎𝑎𝑎 and 𝑎𝑎𝑎𝑎. It � (𝑎𝑎𝑎𝑎𝑎𝑎0𝑏𝑏𝑎𝑎𝑎𝑎 𝑛𝑛9𝑏𝑏𝑏𝑏 ≈ 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎7. 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 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 (an error of about 0𝑎𝑎𝑎𝑎𝑎𝑎%). (TheExample true 5. value is 𝑎𝑎𝑎𝑎𝑎𝑎6227766𝑏𝑏 ….) b (a + b)/2 one we choose to compute will then be our estimate for the GM. over-estimate Let 𝑛𝑛𝑛𝑛 𝑎𝑎 𝑎𝑎�. Let us take 𝑎𝑎𝑎𝑎𝑎𝑎4𝑎𝑎5 𝑎𝑎 9𝑏𝑏� and 𝑏𝑏𝑏𝑏 𝑎𝑎�𝑏𝑏𝑏𝑏4𝑎𝑎� 𝑎𝑎 �𝑏𝑏𝑏𝑏9. Then our estimate for If we use the AM, we always get an a under-estimate √2𝑏𝑏 is: (a + b)/2 for the GM. And if we use the HM, we always get 𝑎𝑎 9𝑏𝑏2+4𝑏𝑏𝑏𝑏9 2×9𝑏𝑏2×4𝑏𝑏𝑏𝑏9 Empty square an . � + � 2 2 9𝑏𝑏2+4𝑏𝑏𝑏𝑏9 of side 1 a b both 2 Now an interesting idea strikes us: why not | − | 𝑎𝑎 𝑎𝑎6𝑎𝑎 72𝑏𝑏 5𝑎𝑎𝑎𝑎4𝑎𝑎 compute the AM and HM, and then take their 𝑎𝑎 � + �𝑎𝑎 ≈4𝑎𝑎472𝑎𝑎𝑎𝑎5955𝑎𝑎 Figure 2. 2 𝑎𝑎6 𝑎𝑎6𝑎𝑎 𝑎𝑎𝑎𝑎592 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 𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 and estimate quite close to the GM. To our surprise, 𝑏𝑏𝑏𝑏 𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏𝑎𝑎𝑎𝑎𝑎𝑎. Then our estimate for √𝑎𝑎𝑎𝑎 is: we �ind that thissecond idea algorithmworks very well.

𝑎𝑎 𝑎𝑎𝑎𝑎 𝑎𝑎 9𝑎𝑎6𝑎𝑎 + 𝑎𝑎𝑎𝑎 So here is our for approximating �𝑎𝑎𝑎𝑎𝑎𝑎 푎 �𝑎𝑎 � � 2 𝑎𝑎𝑎𝑎𝑎𝑎 2 𝑎𝑎𝑎𝑎𝑎𝑎 the square root of aAlgorithm given positive II number 𝑛𝑛𝑛𝑛. 𝑎𝑎9𝑎𝑎6𝑎𝑎 Step 1: 𝑎𝑎 ≈ 𝑎𝑎𝑎𝑎𝑎𝑎6𝑎𝑎𝑎𝑎 6𝑎𝑎2 Write 𝑛𝑛𝑛𝑛 as 𝑎𝑎𝑎𝑎 ×𝑏𝑏𝑏𝑏where 𝑎𝑎𝑎𝑎 and 𝑏𝑏𝑏𝑏 are TheSecond error approach now is about to approximating𝑏𝑏𝑏𝑏𝑏𝑏2%. square reasonablyStep 2: close to each other. roots. Step 3: Compute the AM and the HM of 𝑎𝑎𝑎𝑎 and 𝑏𝑏𝑏𝑏. 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 mean𝑎𝑎𝑎𝑎 and 𝑏𝑏𝑏𝑏.Wehave the (AM), the 𝑎𝑎 𝑎𝑎𝑎𝑎 +𝑏𝑏𝑏𝑏 2𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 (GM) and the (HM), among � + �𝑎𝑎 2 2 𝑎𝑎𝑎𝑎 +𝑏𝑏𝑏𝑏 many others (not listed here). These are de�ined as follows: This is our estimate for √𝑛𝑛𝑛𝑛. 𝑎𝑎 2𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏 AM 𝑎𝑎 (𝑎𝑎𝑎𝑎+𝑏𝑏𝑏𝑏𝑎𝑎𝑎𝑎 GM 𝑎𝑎 √𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏𝑎𝑎 HM 𝑎𝑎 𝑎𝑎 This algorithm gives us extremely good results — 2 𝑎𝑎𝑎𝑎 +𝑏𝑏𝑏𝑏 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 𝑎𝑎𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏, Let 𝑛𝑛𝑛𝑛 𝑎𝑎�. Let us take 𝑎𝑎𝑎𝑎𝑎𝑎2𝑎𝑎4 and then: 𝑏𝑏𝑏𝑏 𝑎𝑎 �𝑎𝑎�. Then our estimate for √6 is: HM < GM < AM𝑎𝑎 always 𝑎𝑎 2𝑎𝑎4+2𝑎𝑎5 2×2𝑎𝑎4×2𝑎𝑎5 𝑎𝑎 𝑎𝑎2 � + �𝑎𝑎 �2𝑎𝑎𝑎 푎 � These inequalities are never violated. The HM and 2 2 2𝑎𝑎4+2𝑎𝑎5 2 4𝑎𝑎9 AM lie on opposite sides of the GM. 4𝑏𝑏𝑏𝑏𝑏𝑏 𝑎𝑎 ≈2𝑎𝑎4494𝑏𝑏9795𝑎𝑎 For example, if 𝑎𝑎𝑎𝑎𝑎𝑎2 and 𝑏𝑏𝑏𝑏 𝑎𝑎 𝑏𝑏, then 𝑎𝑎96𝑏𝑏 𝑎𝑎 Compare this with the true value: AM 𝑎𝑎 (2 + 𝑏𝑏𝑎𝑎 𝑎𝑎 5𝑎𝑎 GM 𝑎𝑎 √2×𝑏𝑏𝑏𝑏4𝑎𝑎 6 𝑎𝑎 2𝑎𝑎4494𝑏𝑏9742 …. The estimate is correct to 2 √ Example 4. 2×2×𝑏𝑏 seven decimal places. HM 𝑎𝑎 𝑎𝑎 𝑎𝑎𝑎𝑎2𝑎𝑎 2+𝑏𝑏 Let 𝑛𝑛𝑛𝑛 𝑎𝑎 𝑎𝑎 . Let us take 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎and and, of course, 𝑎𝑎𝑎𝑎2<4<5. 𝑏𝑏𝑏𝑏 𝑎𝑎𝑎𝑎𝑏𝑏𝑏𝑏𝑎𝑎. Then our estimate for √𝑎𝑎𝑎𝑎 is: 𝑎𝑎 𝑎𝑎 + 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 2×𝑎𝑎 × 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎 𝑎𝑎9 6𝑏𝑏 Recall that in our approach we start by selecting � + �𝑎𝑎 � + � positive numbers 𝑎𝑎𝑎𝑎 and 𝑏𝑏𝑏𝑏 such that 𝑛𝑛𝑛𝑛 𝑎𝑎 𝑎𝑎𝑎𝑎 . We 2 2 𝑎𝑎 + 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 2 6 𝑎𝑎9 wish to compute the GM of 𝑎𝑎𝑎𝑎 and 𝑏𝑏𝑏𝑏. But as the GM 72𝑎𝑎 𝑎𝑎 ≈ 𝑎𝑎𝑎𝑎𝑎𝑎622𝑏𝑏𝑏𝑏 is dif�icult to compute, we choose to compute the 22𝑏𝑏 AM instead. We could also choose to compute the This estimate is correct to four decimal places. HM (this too is an easy computation). Whichever (TheExample true 5. value is 𝑎𝑎𝑎𝑎𝑎𝑎6227766𝑏𝑏 ….) one we choose to compute will then be our estimate for the GM. over-estimate Let 𝑛𝑛𝑛𝑛 𝑎𝑎 𝑎𝑎�. Let us take 𝑎𝑎𝑎𝑎𝑎𝑎4𝑎𝑎5 𝑎𝑎 9𝑏𝑏� and 𝑏𝑏𝑏𝑏 𝑎𝑎�𝑏𝑏𝑏𝑏4𝑎𝑎� 𝑎𝑎 �𝑏𝑏𝑏𝑏9. Then our estimate for If weunder-estimate use the AM, we always get an √2𝑏𝑏 is: for the GM. And if we use the HM, we always get 𝑎𝑎 9𝑏𝑏2+4𝑏𝑏𝑏𝑏9 2×9𝑏𝑏2×4𝑏𝑏𝑏𝑏9 an . � + � both 2 2 9𝑏𝑏2+4𝑏𝑏𝑏𝑏9 Now an interesting idea strikes us: why not 𝑎𝑎 𝑎𝑎6𝑎𝑎 72𝑏𝑏 5𝑎𝑎𝑎𝑎4𝑎𝑎 compute the AM and HM, and then take their 𝑎𝑎 � + �𝑎𝑎 ≈4𝑎𝑎472𝑎𝑎𝑎𝑎5955𝑎𝑎 2 𝑎𝑎6 𝑎𝑎6𝑎𝑎 𝑎𝑎𝑎𝑎592 average? Their respective errors may then just cancel each other, and we may just get an This estimate is correct to eight decimal places.

13 At Right Angles | Vol. 4, No. 1, March 2015 Vol. 4, No. 1, March 2015 | At Right Angles 13 Vol. 4, No. 1, March 2015 ∣ At Right Angles 3 2. Estimation of Cube Roots

�sing the quadratic formula we �ind that the roots Using the quadratic formula �e �ind that the roots Here is the actual cube root, computed using a of this equation are: calculator: Let 𝑛𝑛𝑛𝑛 be the given number whose cube root is of this equation are: ��� required. We start by expressing 𝑛𝑛𝑛𝑛 as a product 13 √251277 √𝑛𝑛 𝑥𝑥𝑥𝑥𝑥𝑥 ± . 10 ≈ 2.154434690. 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎, with 𝑎𝑎𝑎𝑎, 𝑎𝑎𝑎𝑎 and 𝑎𝑎𝑎𝑎 as close as possible to each 𝑥𝑥𝑥𝑥 =1𝑥𝑥 . 12 468 3 other (the closer the better; here, of course, We must use the positive root. So our estimate for Our estimate is correct to six decimal places. We must use the positive root. So our estimate for 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎�/� ). The�/� task of computing the cube root of 10 is: the cube root of 𝑛𝑛 is: 𝑛𝑛𝑛𝑛 = (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 may be expressed as: Compute 13 √251277 the geometric mean (GM) of 𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎. This task may + ≈ 2.154434558. √𝑛𝑛 12 468 be given a geometric form as follows: Construct a 1 𝑎𝑎 𝑛𝑛 1.81𝑛𝑛5. 3 References cube whose volume is equal to that of a cuboid Civilization before Greece and Rome with dimensions 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎. Here is the actual cube root, computed using a The Nine Chapters on the Mathematical Art: Companion and Commentary [1] Saggs H W F. , Yale University Press. calculator: �/� As we did with the square root, we start with the 𝑛𝑛 𝑛𝑛 1.8171. [2] Crossly, John W.A C. History Lun Anthony. of Greek Mathematics , Oxford University Press. arithmetic mean (AM) of 𝑎𝑎𝑎𝑎𝑎𝑎� 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 and then see how Heron’s Formula for Cube Root Our estimate is correct to two decimal places. [3] Heath, Thomas. , Vol 2. Oxford: Clarendon Press. we can ‘improve’ it. Let 𝑑𝑑𝑑𝑑 = � (𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 be the Aryabhatiya [4] Smyly J Gilbart. , Hermatena (Trinity College, Dublin) 19 (42). AM. It is well known that the AM exceeds the GM if We can do better by writing 𝑛𝑛 =2𝑎𝑎 2 𝑎𝑎 3/2. Then Baudhayana Sulba-sutras [5] Apte, Mohan. Geometric(Marathi). Solution Rajhans of Quadratic Publications and Cubic (Pune, Equations India). the numbers are unequal. So let us consider we have: [6] Prakash. , Bombay, 1968. subtracting a suitably small quantity ℎ from 𝑑𝑑𝑑𝑑. We � 1 3 11 [7] Henderson, David W. , Department Of Mathematics, Cornell must �ind ℎ so that (𝑑𝑑𝑑𝑑 𝑑𝑑 ℎ𝑎𝑎 =𝑛𝑛𝑛𝑛. Write 𝑥𝑥𝑥𝑥 =𝑑𝑑𝑑𝑑𝑑𝑑�, 𝑑𝑑𝑑𝑑 = �2𝑎𝑎�𝑎𝑎 �= 𝑎𝑎 University, Ithaca, NY. 3 2 𝑛𝑛 so 𝑥𝑥𝑥𝑥 is going to be our estimate for the desired � cube root. We argue as follows: 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑛𝑛𝑛𝑛 35 = . � � � � 3𝑑𝑑𝑑𝑑 1188 𝑛𝑛𝑛𝑛 =𝑑𝑑𝑑𝑑 𝑑𝑑 3𝑑𝑑𝑑𝑑 ℎ 𝑎𝑎 3𝑑𝑑𝑑𝑑ℎ 𝑑𝑑 ℎ 𝑎𝑎 � � � Hence the equation we must solve is: ∴𝑑𝑑𝑑𝑑 𝑑𝑑 𝑛𝑛𝑛𝑛𝑛𝑛3𝑑𝑑𝑑𝑑 ℎ 𝑑𝑑 3𝑑𝑑𝑑𝑑ℎ ALI IBRAHIM HUSSEN has a B.Sc. in Irrigation and Agricultural Machinery from the University of Mosul, Iraq, � 11 35 and a M.Sc in Civil Engg. (Hydraulic) from the University of Newcastle-Upon-Tyne, UK. He has worked for the 𝑥𝑥𝑥𝑥� 𝑑𝑑 𝑑𝑑𝑑𝑑𝑑� . (we drop the ℎ term since ℎ 𝑛𝑛𝑛𝑛)𝑎𝑎 𝑛𝑛 1188 Scientific Research Council in Baghdad, and for the Civil Engg. Dept., University of Tikrit, Iraq. He now works � as a teacher of Mathematics. Ali Hussen may be contacted at [email protected]. ∴𝑑𝑑𝑑𝑑 𝑑𝑑 𝑛𝑛𝑛𝑛𝑛𝑛3𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑 𝑑𝑑𝑑𝑑� � �sing the quadratic formula we �ind that the roots 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑛𝑛𝑛𝑛 of this equation are: ∴ 𝑛𝑛 𝑥𝑥𝑥𝑥(𝑑𝑑𝑑𝑑 𝑑𝑑 𝑥𝑥𝑥𝑥𝑎𝑎 3𝑑𝑑𝑑𝑑 11 √127149 (since 𝑑𝑑𝑑𝑑𝑑𝑑ℎ=𝑥𝑥𝑥𝑥and ℎ=𝑑𝑑𝑑𝑑𝑑𝑑𝑥𝑥𝑥𝑥). 𝑥𝑥𝑥𝑥 = 𝑥𝑥 . 12 39𝑛𝑛

In the last line we change thequadratic approximation equation sign We must use the positive root. So our estimate for to an equality sign and solve the resulting the cube root of 𝑛𝑛 is: equation — which is now a , 11 √127149 not a cubic equation, and therefore easy to solve 𝑥𝑥𝑥𝑥 = 𝑎𝑎 𝑛𝑛 1.81712𝑎𝑎1𝑛𝑛2. — for 𝑥𝑥𝑥𝑥. The answer we get will be our estimate 12 39𝑛𝑛 forExample the cube 6. root of 𝑛𝑛𝑛𝑛. ThisExample is accurate 7. to six decimal places.

Let us estimate the cube root of 𝑛𝑛𝑛𝑛 = 𝑛𝑛. Let us estimate the cube root of 1𝑎𝑎. Write 𝑛𝑛 =1𝑎𝑎 2 𝑎𝑎 3. Our initial estimate for the Write 1𝑎𝑎 =2𝑎𝑎 2 𝑎𝑎 5/2. Our initial estimate for the cube root of 1𝑎𝑎 is the AM of 2𝑎𝑎 2𝑎𝑎 5/2, i.e., cube� root of 𝑛𝑛 is the AM of 1𝑎𝑎 2𝑎𝑎 3, i.e., � � 𝑑𝑑𝑑𝑑 = � (2 𝑎𝑎 2 𝑎𝑎 5/2𝑎𝑎 = 13/𝑛𝑛. The value of 𝑑𝑑𝑑𝑑 =� (1 𝑎𝑎 2 𝑎𝑎 3𝑎𝑎 =2. The value of � �𝑑𝑑𝑑𝑑 𝑑𝑑 𝑑𝑑𝑑𝑑푑 𝑑�𝑑𝑑𝑑𝑑𝑑𝑑 is: �𝑑𝑑𝑑𝑑 𝑑𝑑 𝑑𝑑𝑑𝑑푑 𝑑�𝑑𝑑𝑑𝑑𝑑𝑑 is: � � 2 𝑑𝑑𝑑𝑑 1 (13/𝑛𝑛𝑛𝑛 𝑑𝑑 1𝑎𝑎 37 = = 3 𝑎𝑎 2 3 13/2 14𝑎𝑎4

Hence the equation we must solve is: Hence the equation we must solve is: 1 13 37 𝑥𝑥𝑥𝑥 (2 𝑑𝑑 𝑥𝑥𝑥𝑥𝑎𝑎 = . 𝑥𝑥𝑥𝑥� 𝑑𝑑 𝑑𝑑𝑑𝑑𝑑� . 3 𝑛𝑛 14𝑎𝑎4 1514 At Right Angles | Vol. 4, No. 1, March 2015 Vol. 4, No. 1, March 2015 | At Right Angles 1514 4 At Right Angles ∣ Vol. 4, No. 1, March 2015 Vol. 4, No. 1, March 2015 ∣ At Right Angles 5 Using the quadratic formula �e �ind that the roots Here is the actual cube root, computed using a of this equation are: calculator: 13 √251277 ��� 𝑥𝑥𝑥𝑥𝑥𝑥 ± . 10 ≈ 2.154434690. 12 468 We must use the positive root. So our estimate for Our estimate is correct to six decimal places. the cube root of 10 is: 13 √251277 + ≈ 2.154434558. 12 468 References Civilization before Greece and Rome The Nine Chapters on the Mathematical Art: Companion and Commentary [1] Saggs H W F. , Yale University Press. [2] Crossly, John W.A C. History Lun Anthony. of Greek Mathematics , Oxford University Press.Heron’s Formula for Cube Root [3] Heath, Thomas.Aryabhatiya , Vol 2. Oxford: Clarendon Press. [4] Smyly J Gilbart.Baudhayana Sulba-sutras , Hermatena (Trinity College, Dublin) 19 (42). [5] Apte, Mohan. Geometric(Marathi). Solution Rajhans of Quadratic Publications and Cubic (Pune, Equations India). [6] Prakash. , Bombay, 1968. [7] Henderson, David W. , Department Of Mathematics, Cornell University, Ithaca, NY.

ALI IBRAHIM HUSSEN has a B.Sc. in Irrigation and Agricultural Machinery from the University of Mosul, Iraq, and a M.Sc in Civil Engg. (Hydraulic) from the University of Newcastle-Upon-Tyne, UK. He has worked for the Scientific Research Council in Baghdad, and for the Civil Engg. Dept., University of Tikrit, Iraq. He now works as a teacher of Mathematics. Ali Hussen may be contacted at [email protected].

We retrieved this snippet from http://www.futilitycloset.com/2014/11/04/all-right-then-2/:

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15 At Right Angles | Vol. 4, No. 1, March 2015 Vol. 4, No. 1, March 2015 | At Right Angles 15 Vol. 4, No. 1, March 2015 ∣ At Right Angles 5