Brahmagupta, the Kuttaka and Bhaskara II PDF

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The history of Mathematics. Those are the mathematician....

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Indian Mathematics https://www.storyofmathematics.com/indian.html Despite developing quite independently of Chinese (and probably also of Babylonian mathematics), some very advanced mathematical discoveries were made at a very early time in India. Mantras from the early Vedic period (before 1000 BCE) invoke powers of ten from a hundred all the way up to a trillion and provide evidence of the use of arithmetic operations such as addition, subtraction, multiplication, fractions, squares, cubes, and roots. A 4th Century CE Sanskrit text reports Buddha enumerating numbers up to 10 53, as well as describing six more numbering systems over and above these, leading to a number equivalent to 10 421. Given that there are an estimated 1080 atoms in the whole universe, this is as close to infinity as any in the ancient world came. It also describes a series of iterations in decreasing size, in order to demonstrate the size of an atom, which comes remarkably close to the actual size of a carbon atom (about 70 trillionths of a metre).

The Indians were also responsible for another hugely important development in mathematics. The earliest recorded usage of a circle character for the number zero is usually attributed to a 9th Century engraving in a temple in Gwalior in central India. But the brilliant conceptual leap to include zero as a number in its own right (rather than merely as a placeholder, a blank or empty space within a number, as it had been treated until that time) is usually credited to the 7th Century Indian mathematicians

Brahmagupta The great 7th Century Indian mathematician and astronomer Brahmagupta wrote some important works on both mathematics and astronomy. He was from the state of Rajasthan of northwest India (he is often referred to as Bhillamalacarya, the teacher from Bhillamala), and later became the head of the astronomical observatory at Ujjain in central India. Most of his works are composed in elliptic verse, a common practice in Indian mathematics at the time, and consequently have something of a poetic ring to them. https://www.storyofmathematics.com/indian_brahmagupta.html The Kuttaka

Brahmagupta gave a clearer explanation than Aryabhata had done of a method of solving what we call linear Diophantine equations, that is, equations of the form ax = by + c, where a, b, and c are given integers, and x and yunknown integers to be found. He applied this technique to computations involving astronomy and the calendar. We shall illustrate the method with such a computation, not one taken from Brahamgupta's work, but entirely in the spirit of that work. It is well known that 19 solar years are almost exactly equal to 235 lunar months. Given that the moon was full on January 30, 2010, what is the next year in which it will be full on February 5? If we choose one 235th of a solar year as a unit of time T, so that T ≈ 1.55 days, or 37 hours, 18 minutes, then one year is 235 T and according to the fundamental relation, one month is 19 T. Since our unit of time T is about a day and a half, the period from January 30 to February 5, which is six days, amounts to 4 T, approximately. Thus we would like to find an integer number of years y and an integer number of months x such that

That is, we want x lunar months to exceed y solar years by 4T. Canceling T, we see that we need to solve 19x = 235y + 4. There are infinitely many solutions if there are any at all, since if ( x0, y0) is a solution, so is (x0 + 235k, y0 + 19k) for any integer k whatever. Conversely, any two solutions (x0, y0) and ( x1, y1) will differ by (235 k, 19k) for some k. Thus the problem is to find one solution. One way to do this is by trial and error: Just look at multiples of 235 until you find one that leaves a remainder of 15 when divided by 19 (since 15 + 4 = 19). Thus you begin with

Continuing in this way, you eventually get to 13 × 235 = 19 × 160 + 15 = 19 × 161 − 4, so that 19 × 161 = 13 × 235 + 4. Thus, we can take x = 161, y = 13. In particular, the year will be 2010 + 13 = 2023. (This is correct!) This method of finding a year on which the Moon will be full on a particular date is remarkably accurate, considering that the time period T is actually about 37 hours, and hence not exactly a day and a

half. When it goes wrong in a short-term prediction, the moon will be full a day later or earlier in the predicted year. Thus, the solution of linear Diophantine equations is not difficult. The only disadvantage to the method used above is the tedious trial-and-error procedure of getting one solution. That is where the method called the kuttaka(pulverizer) comes in. This technique shortens the labor of finding the first solution by a considerable amount, especially when the coefficients a, b, and c are large. Here are the steps you follow: 1. First, be sure the equation is written ax = by + c, where a and b are positive, and b > a. In other words, the constant term c needs to be on the same side of the equation as the larger coefficient, and the two coefficients must have the same sign. If they don't, replace y by a new variable z = − y, and then they will have the same sign. You can then multiply the equation by −1 if necessary to get them both positive. The constant term c may be positive or negative. (This “normalizing” is not absolutely essential, but experience shows that one has to be very careful when executing the kuttaka. The experienced user can handle variants in the method, but the beginner had better follow rigid rules.) 2. Second, perform the Euclidean algorithm procedure to find the greatest common divisor d of a and b. If it is larger than 1, then the expression ax − by can only be a multiple of that greatest common divisor, so if c is not a multiple of it, there are no solutions, and you are finished. 3. If d divides c, take all of the quotients—except the last one, which yields a remainder of 0 — and write them in a column. To illustrate with the equation 19 x = 235y + 4, which we considered above, we have the column

4. Next, augment that column with two more numbers at the bottom. The first one is c/d if the number of quotients is even and −c/d if it is odd. In our case, we have an even number of quotients, and so we adjoin 4 at the bottom. The second additional number, which forms the bottom row of the array, is always 0. Thus we get the following column:

5. Now operate on this column, at each stage modifying the bottom entry and the entry two rows above it, as follows: The entry two rows above the bottom gets replaced by its product with the number below it, plus the number below that. Thus in this example, the first thing to do is to replace the 2 in the third row (counting the bottom row as row 1) by 2 × 4 + 0 = 8. The second part of the procedure is to erase the bottom number. Repeating this procedure until there are only two rows left yields

We should now have a solution, and indeed we do: x = 396, y = 32. It is not the smallest solution, however. We get a smaller one by subtracting 235 from x and 19 from y, yielding x = 161, y = 13. This procedure needs to be practiced on some simple equations, such as 3 x = 23y + 1 and 17x = 11y − 5, before the details will fall into place. The number of errors that can creep into this procedure is rather large. If the answer you get doesn't check when you put the values of x and y back into the equation, look for the following possible mistakes: 1. a and b must both be positive and their greatest common divisor d must also divide c if the equation is to have any solutions. 2. When the equation is written ax = by + c, you must have b > a. 3. Do not include the last quotient from the Euclidean algorithm in the column. 4. Adjoin c/d to the column of quotients (ignore the remainders in this algorithm), if you have an even number of quotients. (If c/d is negative, leave it negative in this case.) If the number of quotients is odd, adjoin −c/d. (If c/dis negative, make it positive in this case.) These are the commonest sources of errors when carrying out this procedure. But of course, you also have to do the divisions with remainder carefully, avoiding computational errors. Bhaskara II

Labeled as one the “greatest mathematicians of medieval India”, the 12th-Century mathematician Bhaskara II wrote many books containing mathematical and astronomical feats which would not be discovered elsewhere for another 500 years. https://www.storyofmathematics.com/bhaskara-II

Algebra in the Works of Bhaskara II

The Lilavati of Bhaskara II contains a collection of problems in algebra, which are sometimes stated as though they were intended purely for amusement. For example, One pair out of a flock of geese remained sporting in the water, and saw seven times the half of the square-root of the flock proceeding to the shore, tired of the diversion. Tell me, dear girl, the number of the flock. Like countless other unrealistic algebra problems that have appeared in textbooks over the centuries, this story is a way of posing to the student a specific quadratic equation, namely , whose solution is x = 16. The Vija Ganita (Algebra) As mentioned in Chapter 19, Bhaskara II advertised his Algebra as an object of intellectual contemplation. We may agree that it fits this description. The problems, however, are just as fanciful as in the Lilavati. For example, the rule for solving quadratic equations is applied in the Vija Ganita (p. 212 of the Colebrooke translation) to find the number of arrows x that Arjuna (hero of the Mahabharata) had in his quiver, given that he shot them all, using to deflect the arrows of his antagonist, to kill his antagonist's horse, six to kill the antagonist himself, three to demolish his antagonist's weapons and shield, and one to decapitate him. In other words, . Combinatorics Bhaskara gives a thorough treatment of permutations and combinations, which already had a long history in India. He describes combinatorial formulas such as

by saying Let the figures from one upward, differing by one, put in the inverse order, be divided by the same in the direct order; and let the subsequent be multiplied by the preceding and the next following by the foregoing. The several results are the changes by ones, twos, threes, etc. He illustrates this principle by asking how many possible combinations of stressed and unstressed syllables there are in a six-syllable verse. His solution is as follows: The figures from 1 to 6 are set down, and the statement of them, in direct and inverse order is

The results are: changes with one long syllable, 6; with two 15; with three, 20; with four, 15, with five, 6; with all long, 1. Bhaskara assures the reader that the same method can be used to find the permutations of all varieties of meter. He then goes on to develop some variants of this problem, for example, A number has 5 digits and the sum of the digits is 13. If zero is not a digit, find the total number of possible numbers.

To solve this problem, you have to consider the possibility of two distinct digits (for example, 91111, 52222, 13333, 55111, 22333), three distinct digits (for example 82111, 73111) and count all the possible rearrangements of the digits. Bhaskara reports that the initial syllables of the names for colors “have been selected by venerable teachers for names of values of unknown quantities, for the purpose of reckoning therewith.” He proceeds to give the rules for manipulating expressions involving such quantities; for example, the rule that we would write as (− x − 1) + (2x − 8) = x − 9 is written

where the dots indicate negative quantities. The syllable ya is the first syllable of the word for black, and ru is the first syllable of the word for species. Bhaskara gives the rule that we express as the quadratic formula for solving a quadratic equation by radicals, then goes on to give a criterion for a quadratic equation to have two (positive) roots. He also says (pp. 207–208 of the Colebrooke translation) that “if the solution cannot be found in this way, as in the case of cubic or quartic equations, it must be found by the solver's own ingenuity.” That ingenuity includes some work that would nowadays be regarded as highly inventive, not to say suspect; for example (p. 214), Bhaskara's solution of the equation

Bhaskara warns that multiplying by zero does not make the product zero, since further operations are to be performed. Then he simply cancels the zeros, saying that, since the multiplier and divisor are both zero, the expression is unaltered. The result is the equation we would write as . Bhaskara clears the denominator and writes the equivalent of 9 x2 + 12x = 60. Even if the multiplication by zero is interpreted as multiplication by a nonzero expression that is tending to zero, as a modern mathematician would like to do, this cancelation is not allowed, since the first term in the numerator is a higher-order infinitesimal than the second. Bhaskara is handling 0 here as if it were 1. Granting that operation, he does correctly deduce, by completing the square (adding 4 to each side), that x = 2. Bhaskara says in the Vija Ganita that a nonzero number divided by zero gives an infinite quotient. This fraction , of which the denominator is cipher, is termed an infinite quantity. In this quantity consisting of that which has cipher for its divisor, there is no alteration, though many be inserted or extracted; as no change takes place in the infinite and immutable GOD [Vishnu], at the period of the destruction or creation of worlds, though numerous orders of beings are absorbed or put forth. By the time of Bhaskara, the distinction between a rational and an irrational square root was well known. The Sanskrit word for an irrational root is carani, according to the commentator Krishna (Plofker, 2009, p. 145), who defines it as a number, “the root of which is required but cannot be found without residue.” Bhaskara gives rules such as and .

Geometry in the Works of Bhaskara II In his work Siddhanta Siromani (Crest Jewel of the Siddhantas), written in 1150, Bhaskara tackled the extremely difficult problem of finding the area of a sphere. As we have seen (Section 7.2 of Chapter 7), the Egyptians had deduced correctly that the area of a hemisphere is twice the area of its circular base, and (Section 14.2 of Chapter 14) Archimedes had proved rigorously that the surface of a sphere is four times the equatorial disk it contains. In order to achieve that result, Archimedes had to make use of the method of exhaustion, which can be seen as an anticipation of integral calculus. Something similar can be said about Bhaskara's approach, which was numerical and based on Aryabhata's trigonometry, in contrast to the metric-free approach used by Archimedes. The discussion we are about to give is based on the exposition of this result given by Plofker (2009, pp. 196–201). As was stated in the previous chapter, in constructing his table of sine differences, Aryabhata I chose 225' of arc as the constant difference, dividing the arc of one quadrant of a circle of radius 3438' into 24 equal pieces. Bhaskara II started from that point, dividing a complete great circle of a sphere—which we can think of as the equator—into 96 equal pieces. Each of these pieces is regarded as one unit of length. He then imagined the lines of longitude drawn through these 96 points running from pole to pole, thereby partitioning the sphere into 96 mutually congruent sectors. In each sector, he then imagined the circles of latitude drawn, dividing each quadrant of a line of longitude into 24 equal arcs between the pole and the equator, 48 between the two poles. Thus the surface of the sphere was partitioned into 96 × 48 = 4608 regions, 192 of which (those having a vertex at one of the poles) are curvilinear triangles, and the other 4416 of which are curvilinear trapezoids. A set of 20 of these trapezoids lying just above the equator is shown in Fig. 21.1. Since they are very small, one can imagine that they actually are planar triangles and trapezoids. For a typical trapezoid whose upper and lower edges are at co-latitudes ( k − 1) × 225' and k × 225'—these are the distances along a line of longitude from the pole—the lengths of these edges are proportional to the radii of those circles of latitude. In terms of the unit of length (1') chosen by Aryabhata, the radius r of the circle at co-latitude k × 225' is sin ( k × 225'); that is, it is given in the second column of the table of sines displayed in the previous chapter. The length of the arc of that circle inside the sector is 225'. However, these are minutes of arc on the circle of latitude, not on the sphere. A minute of arc on a circle of radius r is minutes of arc on a greatcircle of a sphere of radius R. Thus, the portion of the circle of latitude of radius r inside each sector has length minutes of spherical arc, where R = 3438. Since Bhaskara's unit of length is 225 of Aryabhata's units, we need to divide by 225. Altogether then, the length of that arc inside a given sector at co-latitude k × 225' is . The area of each trapezoid (again, treated as if it were a plane trapezoid) is numerically equal to the average of these lengths for the upper and lower edges, since each trapezoid has altitude equal to one unit. All we have to do then is sum up the areas of the 4416 trapezoids and the 192 triangles in order to find the area of the sphere. This is done most easily by finding the area of a half-sector and doubling it. To find the area of each triangle or trapezoid between the pole and the equator in a given sector, one has only to take the average of the lengths of the opposite sides (counting the “side” at the pole as having length 0), multiply by the altitude, and then add up the results. Thus, we need to find

Bhaskara's “polygonal” method of getting the area of a sphere.

Bhaskara saw how this sum could be rewritten to eliminate the 2 in each denominator, except for the very last term. He evaluated it by adding up all of the sines in the table and then subtracting half of the last one. This was a simple exercise in arithmetic, and we noted in the previous chapter that the sines in the table add up to 54, 233. Therefore, the area of a half-sector is numerically

(The number 1719 is half of the sine of a 90° arc.) A full sector is then twice this amount, or 30.54916. Bhaskara observed that this is, within the limits of precision, precisely the diameter of the sphere, since . Thus, it seems that if a sphere is partitioned into sectors of unit opening, the area of each sector in square units is numerically equal to the length of the diameter. Since the total area of the sphere is 96 times the area of this sector—that is, it is this area times the number of units of length in the circumference, Bhaskara concluded (correctly) that the area of a sphere [in square units] is equal to its diameter times its circumference. Bhaskara would have had to construct a finer table of sines in order to test the result with a smaller unit of length (by partitioning the sphere into more than 4608 regions). As a practical matter, since the actual diameter is about 0.01 units larger than the value he used, while he had the circumference correct, he would have gotten a numerical value for the area that is too small by 1%. In our terms, his unit of area was

, and the numerical approximation that he used for the area was

where

The accurate value of cn would be 4. Bhaskara's procedure amounts to taking n = 24. By direct computation, we get c24 ≈ 3.96023, which is, as already noted, 1% too small. That Bhaskara understood the principle of infinitesimal approximation is shown by another of his results, in which he says that the difference between two successive sines in the table, that is, sin ((k + 1) × 225') − sin (k × 225'), is 225 cos (k × 225')/R (where R = 3438). This result seems to prefigure the infinitesimal relation that calculus books write as

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