Heron Of Alexandria-math Contributions-square Root Method And Area Of A Triangle 6k5i55

HERON OF ALEXANDRIA --Math contributions. N K Srinivasan Ph D Introduction This article is intended for middle school and high school students and math teachers---to introduce some of the math works of this great mathematician and inventor of the Greek period. Heron lived in Alexandria ,a city in Egypt, which was under the Greek rulers and later a part of Roman empire. His name is also spelled as 'Hero' ,due to some translators using different texts. His period is roughly between 10 AD and 70 AD. His exact year of birth or year of death is not known. Alexandria was a prosperous sea-port and a center of learning at that time. There was a 'museum' or temple precinct, with the famous Alexandria Library inside, in which scholars gathered to do research , writing and giving lectures in an Academy. This library, a treasure trove of great knowledge of the Greeks and others, was burned up and the last damage was during Ootaman empire.



Heron was a prolific inventor and engineer , besides being a mathematician and physicist. He is classed with such geniuses as Pythagoras, Leonardo da vinci and even Thomas Edison of our times. Before taking up his math works, I will give a short, partial list of his inventions, some of which were practical devices and some fascinating "toys". 1 His most famous device is a small steam rotor with steam going out of nozzles from a spherical bulb. This device is considered as the first steam engine. 2 He made a water operated device to open and close temple doors. 3 He made a coin vending machine--to dispense holy water when a coin is dropped into the pan of a balance. 4 He developed syringes. 5 He built land surveying equipment , called 'diopter' like a theodolite. 6 He made a musical drum operated by wind power to make thunder-like sound. 7 He built an odometer to measure the distances traveled by horse drawn carts.



8 He is credited with studies on laws of mirror reflection and several optical devices with mirrors. 9 He developed a puppet theater using weights and pulleys. Math works Now we shall turn to his mathematical discoveries given in his book " Metrika". He wrote several books which are like lecture notes. Possibly 'Metrika' is a compilation of other's works as well. Be that as it may, the following math formulae are attributed to Heron by historians. 1 Area of a triangle His formula for area of a triangle is ,perhaps, most widely known and is mentioned in almost all geometry books.​ Heron's formula : area = sqrt[ s (s-a) (s-b) (s-c)] where s is the semi-perimeter ; s = (a+b+c)/2 This is a very useful formula for area of a triangle because it does not require a knowledge of any angle ,but only the length of the three sides.



The proof or derivation of this formula is pretty long and tedious; you will find a proof in the book: "William Dunham: Journey through Genius." Note that this formula can fail if one of the sides is equal to semi-perimeter s. In that case,alter the length of the side slightly to a lower value. Much later, Brahma Gupta , an Indian mathematician and astronomer of 6th century, derived a similar formula: an area formula for a cyclic quadrilateral : area = sqrt[(s-a)(s-b)(s-c)(s-d)] [which reduces to Heron's formula when a=0.] A cyclic quadrilateral is a quadrilateral circumscribed by a circle. It is a moot point whether Brahma Gupta knew Heron's formula. 2Formula for square root and cube root ,using iteration. Heron gave a simple, but powerful formula to calculate square root of any number N by an iterative procedure. Start with an approximate root x(0) ;



Then x(1) = (1/2)[x(0) + N/x(0)] Now you plough back x(1) on the right side and get the next approximation x(2). Repeat the 'iterations'. Usually 3 or 4 iterations are sufficient to get the square root of N to four or five decimal place accuracy. Example: To find square root of N = 2. Let us begin with x(0) = 1. Then x(1) = (1/2) [ 1 + 2/1] = 1.5 x(2) = (1/2) [ 1.5 + 2/1.5] = 1.4166 x(3) = (1/2) [ 1.4166 + 2/1.4166] = (1/2) [1.4166 + 1.41183] =1.414215 We can stop here; we already have accuracy to fourth decimal place. My Casio calculator gives: sqrt(2)=1.414213 Calculators often use the same or similar 'algorithm' to get square roots. Heron's formula, at once simple and of powerful nature, may appear strange. It takes the arithmatic mean of x and N/x as the next



approximation. Let the actual root be x'. If x is an underestimate of the square root, then N/x will be an overestimate of the square root. Therefore Heron takes the arithmetic mean of x and N/x by adding them and dividing by two. This is a qualitative explanation for Heron's formula. This formula is also called "Babylonian method" by some authors. [ Newton-Raphson method: This formula can be easily derived from Newton_Raphson formula , which uses Calculus.If you are not familiar with calculus, you can skip this section. To find the sqrt(N) , we write the formula: f(x) = x2 -N = 0 Newton -Raphson iterative formula : x(n+1) = x(n) - f(x)/f'(x) where f'(x) is the derivative or differential coefficient of f(x); Here, f'(x) = 2x So , N-R formula becomes: x(n+1) = x(n) - (x(n)2 -N)/2x(n)



x(n+1) = x(n) - x(n)/2 + N/ 2x(n) = x(n)/2 + N/2x(n) = (1/2) [ x(n) + N/x(n)] This is the same formula as Heron's! Newton_Raphson method has the power of 'second order convergence' or "quadratic convergence' which means that the accuracy improves by two decimal places with each iteration. So, you need only two iterations for 4 -place accuracy if the initial root is close to the actual one. Therefore it is not surprising that Heron's formula is very powerful. 3 Heron's Cube root formula To find the cube root of N, find two numbers a, and b, such that a3 < N < b3. Let d = N - a3 and D = b3 - N Cube root of N = C = a + [bd /(bd+aD)] (b-a) Example : Find the cube root of 30 . Let a = 3 b= 4



d = 30 - 27 = 3 D = 64 - 30 = 34 Then C = 3 + 12/114 = 3.10526 My Casio calculator gives: 3.1068 [ There are several simpler formulae for cube roots based on secant method for solving such problems. The present author uses a simpler formula based on linear interpolation: C = a+ (d/d+D)[b-a]. The procedure can be iterated with next approximation.] Applying Newton-Raphson method, we can get the following iteration formula: x' = (1/3) [ 2x + N/x2 ] For N =30, start with x =3 x' = (1/3) [ 6 + 30/9] = 3.1111 x'' = (1/3) [ 6.2222 + 30/9.679] = (1/3) [ 6.2222 + 3.099494] = 3. 1072385 This formula is the one widely used for cube roots, though the initial guess should be close to the actual root.



The Casio calculator gives: Cube root of 30 = 3.1072289. Note that Heron's formula for cube root is not derived from N _R method. Heron's method starts with two numbers which are either greater or lesser than the root and makes approximation to the root. Halley's Method You may recall the name of the famous astronomer Edmond Halley (1678-1715) ,known for the comet "Halley's comet" which visits us once in 76 years. Halley extended the Newton-Raphson method by taking the second derivative of the function f(x): f"(x). His formula was later modified by Bailey and is very powerful for solving such equations [f(x)= 0] and for finding square roots and cube roots. His iterative formula ,starting with initial root x is as follows: x' = x - (2f(x)f'(x))/ [ 2 f(x)2 -f(x)f" (x)] This iteration has 'cubic convergence' ---which means that the error gets reduced to three decimal places with each iteration.



Example :To find cube root of 30, f(x) = x3 -30 =0, then f'(x) = 3x2 and f"(x) = 6x Let us take x = 3.0 as the initial root. f'(x) =27 f"(x) = 18 Plugging in , we get: x' = 3.1071428 The Casio calculator gives : 3.106880

The error is : 0.0003 , in just one iteration.! Note: If you have learned Taylor series expansion for a function f(x) near x=a, you can easily derive Newton-Raphson formula and Halley's formula. Try this! -------------------------------------------------------- 4 Heronic mean Heron developed a formula for mean value using both arithmetic mean and geometric mean. If a and b are two numbers,then Heronic mean is as follows: Heronic mean H = (1/3) [ a + sqrt(ab) + b ] If a=3 and b= 5, then H = (1/3) ( 3 + 2.828 + 5 ) = 3.6095 This is a weighted average of arithmetic mean and geometric mean:



H = (2/3)4 + (1/3)2.828 = 3.6093 The motivation for using this mean is not clear and Heron must have found some application . Summary Heron's contributions are wide in scope and was far reaching in his time . His inventions and engineering skills must have been very powerful in transforming the society . His book 'metrica' was found only in 1896.While several methods for extracting square roots of numbers exixt, Heron's method or algorithm is the most popular and rapid method. References 1 William Dunham -- Journey through genius 2 An article by J J O'Connor and E F Robertson [Univ of St Andrews] 3 T L Heath -- A history of Greek Mathematics-- 2 vols Cambridge U Press. 4 N K Srinivasan--Numerical methods for engineers--CBS publishers, New Delhi. 5 "Mathworld" websites ----------------------------------------------------