Homework 2 from lecture 11 to lecture 20

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1 Homework 2 from lecture 11 to lecture 20 June 14, 2016 Lecture Take a look at Apollonius Conics in the appendices. 2. UseCalculus toproveapropertyinapollonius book: LetC beapointonahyperbola. Let CB be the perpendicular from that point to the diameter. Let G and H be the intersections of the diameter with the curve, and choose A on the diameter, or the diameter extended, so that AH : AG = BH : BG. Then AC will be tangent to the curve at C. As a special case with modern notation, we consider a hyperbola given by the equation x 2 y2 = 1 with the intersection points with the real axis H = (3,0) and G = ( 3,0) Let C = ( 4, 4 7 ) be a point in the hyperbola and B := ( 4,0). If A = (x 3 0,0) is a point such that the ratio of line segments AH = BH holds, we need to prove that the AG BG line passing through C and A is a tangent line of this hyperbola and need to find out x 0. Lecture Archimedes in his book Measurement of a Circle presented, without a word of justification, the inequality < 3 < (1) As an explanation of the probable steps leading to the right-hand bound, show first that < ( 1 52 )2 =

2 2 and then 1 3 = < 15 1 ( 26 1 ) = Obtain the left-hand bound in a similar manner by replacing 1 52 by To find approximate value for π, Archimedes consider the inscribed and circumscribed regular polygons ofacircle withdiameter 1. Let us denoteby p n andp n theperimeters of the inscribed and circumscribed regular polygons of n sides. Notice the perimeter of the circle is π. Then we have p 6 < p 12 < p 24 < p 48 < p 96 <... < p n < π < P n <... < P 96 < P 48 < P 24 < P 12 < P 6 with the following properties: P 2n = 2p np n p n +P n, p 2n = p n P 2n. When n = 6, we know p 6 = 3 and P 6 = 2 3. By the inequality in (1), we obtain 3 = p 6 < π < P 6 = 2 3 < , i.e., we obtain approximate inequalities for π: < π < When n = 12, by using p 12 and P 12, find new approximate inequalities for π. 1 Lecture [Example] (from Diophantus s work) Find four numbers such that when any three of them are added together, their sum is one of four given numbers. Say the given number sums are 20,22,24 and 27. Solution: Let x be the sum of all four numbers. Then the numbers are just x 20,x 22,x 24 and x It follows that x = (x 20)+(x 22)+(x 24)+(x 27) 1 Archimedes calculated for p 96 and P 96 together with (1) to obtain 2 If x 1 +x 2 +x 3 = 20, then x 4 = x < π < 31, i.e., < π <

3 3 which is solved by x = 31. Then the required numbers are 11,9,7 and 4. Here is a homework problem: Find three numbers such that when any two of them are added, the sum is one of three given numbers. Say the given sums are 40, 60 and [Example] (from Diophantus s Book II, Problem 8) Divide a given square number, say 16, into the sum of two squares. Solution: By using the modern notation, let one of the required squares be x 2. Then 16 x 2 must be equal to a square. Assume that this square is (2x 4) 2 where x is the unknown. By solving the equation 16 x 2 = (2x 4) 2, we find that the two numbers are x = and 2x 4 =. 5 5 Here is a homework problem: squares. Divide the square number 36 into the sum of two Lecture Read Hypatia from Internet. Provide a short biography for the female Greek mathematician Hypatia. Lecture Use the method discussed on the pages of the notes to solve the following system of linear equations (negative numbers are allowed): { x+y = 200, 300x+500y = [Example](from Mathematical Treatise in Nine Sections) Find a root of the equation x 2 71,824 = 0, by carrying out the following steps: (a) Take 200 as an initial approximation and reduce the roots by 200 through the transformation y = x 200. (b) With 60 as an approximation to the roots of the transformed equation, make a second substitution z = y 60.

4 4 (c) By trial, find an integral root z of the third equation and use it to obtain the desired root x. By letting y = x 200 and z = y 600, we get the equation z z 4224 = 0. Trial shows that z = 8 is a solution. By using the above method to find a root of the equation x = 0. Lecture Solve the following problem from the Bakhshālī manuscript: 3 One person goes 4 yojanas a day. When he has proceeded for seven days, the second person, whose speed is 9 yojanas a day, departs. In how many days will the second person over take the first? 2. The sixth-century Hindu mathematician Āryabhata had the following procedure for finding the area of a circle: Half the circumference multiple by half the diameter is the area of a circle. How accurate is this rule? Lecture The following problem is from the Algebra of Abû Kâmil ( ): Find a number such that if 7 is added to it and the sum multiplied by the root 3 times the number, then the result is 10 times the number. In other words, to solve (x+7) 3x = 10x. Hint: put x = 1 3 y2. 2. Prove: 18 8 = 2. Lecture Read the history of the first university in the western world: University of Bologna from the appendix. 2. One of the problems from the Maasei Hoshev (The Art of Calculation, 1321): A barrel has various holes. The first hole empties the full barrel in 3 days; the second hole empties the full barrel in 5 days; another hole empties the full barrel in 20 hours; and another hole empties the full barrel in 12 hours. If all the holes are opened together, how much time will it take to empty the barrel? 3 This is a mathematical manuscript written on birch bark which was found near the village of Bakhshali in It is notable for being the oldest extant manuscript in Indian mathematics. Hoernle thought that the manuscript was from the 9th century, but the original was from 3rd or 4th century.

5 5 3. Prove the result of Oresme ( ): becomes infinite. 2 3 n Lecture Read Fibonacci sequence in the appendices. 2. Show that the sum of the squares of the first n Fibonacci numbers is given by the formula F 2 1 +F F2 n = F nf n+1. Hint: F 2 n = F n(f n+1 F n 1 ) = F n F n+1 F n F n It can be established that each positive integer equals to a sum of Fibonacci numbers, none taken more than once. For example, 5 = F 3 + F 4, 6 = F 1 + F 3 + F 4, 7 = F 1 +F 2 +F 3 +F 4. Write the integers 50 and 75 in this manner. Lecture This problem is from the work of Piero della Francesca ( ). A fountain has two basins. One above and one below, each of which has three outlets. The first outlet of the top basin fills the lower basin in two hours, the second in three hours, and the third in four hours. When all three upper outlets are shut, the first outlet of the lower basin empties it in three hours, the second in four hours, and the third in five hours. If all the outlets are opened, how long will it take for the lower basin to fill? 2. This problem is also from the work of Piero della Francesca ( ). Of three workmen, the second and third can complete a job in 10 days. The first and third can do it in 12 days, while the first and second can do it in 15 days. In how many days can each of them do the job alone?