History of Mathematics Spring 2015 Homework 3 Solutions

Transcription

1 History of Mathematics Spring 015 Homework Solutions Due: Friday, February 6, Show that the transformation x z (a /na 0 ) converts the equation of degree n into an equation in z that lacks the (n 1)-st degree term. a n x n + a x + + a 1 x + a 0 0 There was a typo here; the transformation should have been x z (a / ). Fun Fact: If you discover a typo, or thing there is a typo, you can tell/ask me about it before handing in the homework. Don t be shy. We have to use the binomial formula. But I ll give two solutions, one with more words than the other. Solution 1. (Wordy) It is clear that if we replace x by z (a / ) and expand all the powers of z (a / ), and collect equal powers of z, we will get a polynomial in z. Since the highest power of z in (z (a / )) k is z k, all powers of z will be n. The only place where a power z n can appear is on expanding (z (a / )) n so that the equation in z will begin with a n z n, followed by lower powers of z. A power of z to n 1 appears when expanding (z (a / )) n and (z (a / )), and nowhere else. In (z (a / )) n it will appear as ( n a ) z a z so that on multiplying by a n we get a z ; in (z (a / )) it appears as z so that on multiplication by the coefficient a we get a z, so that on collecting powers of z they all cancel. Solution. Just formulas. Notice how the change in order of summation is done. 0 a n x n + a x + + a 1 x + a 0 where k k0 0 a k a ) k z for 0, 1,..., n. If n we have b n b k b a k x k k0 0 k k kn a k a ) k ( n 1 n 1 k0 ( a k z a ) k k0 a k a ) k z a k a ) k a k a ) k ( n n a a ) 0 ( n + n 1. Solve by the Cardano-Tartaglia formula/method x 6x 16. Then show that a k 0 k b z, ( + ) and ( ) 81 0 and see that the root given by the formula is 6 in disguise. 0 a ) k z a n a ) 0 a n ; more relevantly, a n a ) 1 0.

2 The solution will be x u v where u, v are found by solving uv 6, u v 16. Solving for v, we get v 1 u ; using this in u v 16 and rearranging we get Solving for u, we get u 16 ± u 6 16u ± ± 0. It is a fact, and easy to verify, that choosing to + or sign in the expression for u makes no difference in the final solution one obtains, so I ll choose + and have u in the form Then v 961 u u (81 0 ) ( )( ) Thus the Cardano Tartaglia solution is As suggested by the exercise we check that x ) 961( (81 0 ). ( ) ( + ) ( ) + ( ) + ( ) ( ) + ( ) , ( ) ( ) ( ) + ( ) ( ) ( ) This implies (very sort of) that u , v , so that x u v ( + ) ( + ) 6.. Solve the following equations by the Cardano-Tartaglia formula/method. Show work! (Before applying the method, you may have to depress some of the equations. How to depress is explained in the previous exercise.) (a) x 6x + 18x 4 0. To depress substitute x z +. The equation in z is z +6z 4. For the Cardano-Tartaglia solution we have to find u, v so uv 6, u v 4, thus v /u, u 6 4u 8 0. Solving for u, u ±. Selecting the + sign we get v u 4 +. Then z u v The solution to the original equation is x + +. (b) x 10x After some horrible computations, I get the solution in the form x

3 (c) x x 6 0. The Cardano-Tartaglia solution works out to x , 9 9 which is no other than in disguise. 4. François Viète ( ) was the most important French mathematician of the sixteenth century. He was a lawyer, member of parliament and when he got a chance, a mathematician. He also did cryptography and there is a story that he successfully deciphered a Spanish code, which helped France to gain some advantage in its war against Spain. So convinced was king Philip II of Spain that the code was undecipherable, that he complained to the Pope that the French were using magic against Spain, contrary to the practice of the Christian Church. He wrote important works on algebra, geometry, and most notably trigonometry. But he is specially remembered as a pioneer in the development of symbolic notation. Some notation had already been developed before Viète, but he was the first to use the same letter for a quantity and its powers. He used the symbols + and as we do, but had no symbol for equality. We explore here his method for solving cubic equations. To solve x + px q, Viète writes the equation in the form x + ax b (so a p/, b q/). Then he looks for a solution x of the form x a y y. Show that this results into the following equation for y: y6 + by a. Use this method to solve x 6x 16. If one substitutes x a y y into y6 + by a one gets b multiplying by y we get y 6 + by a. ( ) ( ) a a y y + a y y a y a y y + a y y y + a a a ay y y y ; The equation x 6x 16 is, of course, the same as in exercise ; Viète s equation for y is exactly the same as Cardano-Tartagkia s equation for u, namely y y Using what we did above, y and Viéte s solution looks like 1 x It is, of course, Viète derived the formula sin x + sin y sin x + y cos x y from the diagram below, wherein the angles x DOA and y COD appear as central angles of a unit circle. Fill in the details of the following skecth of Viète s proof: sin x + sin y AB + CD AE AC cos x y sin x + y cos x y.

4 4 I begin with some angle chasing. Because EC is parallel tood, we have ECO COD y.triangle COA is isosceles, its angle at O is x + y, so the angles at C, A equal (π x y)/. It follows that ACD will satisfy ACD π ( ECO + OCA) π ( y + π x y ) y x. Because AE and CD are parallel, we also have Notice also that by basic trigonometry EAC ACD y x. AE AC cos( EAC) AC cos y x AC cos x y. Now sin x AB, sin y CD. since CD BE, we have AB + CD AB + BE AE. Thus sin x + sin y AE AC cos y x. It remains to prove that AC sin(x + y). For this we can use a very basic and nice geometry result (which is not even hard to prove). I mean the law of sines. Let r be the radius of the circumcircle of a ABC (the circle into which the triangle is inscribed) of sides a, b, c. Then a sin A b sin B c sin C r. (Here a, b, c are the sides opposite to the vertices A, B, C, respectively.) To apply this to our case I will add a point B somewhere on the circle to the left of the segment AC:

5 1 HOMEWORK RULES 5 Because angles AOC and AB C subtend the same arc, the first being central, the second from the circle, it is another basic geometry fact that AB C 1 ( AOB) (x + y)/. Applying the law of sines now to triangle AB C, which is inscribed in a circle of radius 1, we get We are done. 1 Homework Rules AC sin( B ) x + y, ; i.e., AC sin. Homework should be well presented. No torn pages, very little if any crossed out material. If ruled paper is used, obey the lines. If the paper isn t ruled, write straight. Avoid double columns. Don t try to cram in the maximum amount of writing into the minimum amount of space. If your writing is hard to read, hand in a typed version. (How do you know if your writing is hard to read? Ask a friend or family member to read it.) If your homework extends over more than one page, the pages should be stapled together. If you don t do it yourself, I ll be happy to staple them for you, and charge you 5 to 10 points for the staple. You are university students, on the way to becoming university graduates. One expects a university graduate to be able to write clearly, to express him or herself correctly. I will expect you to be articulate in exams and homework, and to write in complete sentences. A sentence, among other things, must contain a verb. Here is an example: is not a sentence. On the other hand, 5 x 5 is a sentence; the verb is implicit in the equal symbol; x 5 reads: x equals 5. Punctuation should be used reasonably well. For example, you should know the difference between Call me Ishmael (the first line of the classic American novel Moby Dick) and Call me, Ishmael. Or, to give another example, the difference between Sharks don t swim here, and Sharks; don t swim here. Your final answers in exams or homework should be easy to spot. I should not have to hunt for it. The work you present should be your own. Ideally, you did it all by yourself. But discussing homework with a fellow student or friend is OK, as long as it reduces to a discussion, and not to you copying verbatim what someone else did. Going to the internet and consulting with some anonymous gremlin (Clegg ex Cramster, for example) should be an absolute no-no. But, in last instance, if you get your answers from somewhere else, be it a friend or an internet location, at the very least make sure you understand the answer and try not to be too obvious. That is, don t make it too easy for me to realize that it wasn t your work.