Lecture 1 Maths for Computer Science. Denis TRYSTRAM Lecture notes MoSIG1. sept. 2017

Transcription

1 Lecture 1 Maths for Computer Science Denis TRYSTRAM Lecture notes MoSIG1 sept / 21

2 Context The main idea of this preliminary lecture is to show how to obtain some results in Mathematics (in the field of combinatorics, counting, basic number theory). We will show how to handle simple and less easy results with very basic tools that do not require any sophisticated background. 2 / 21

3 Brief overview of proving techniques Induction Contradiction contradictio in contrarium Combinatoric proofs Bijections between sets Pigeon holes Proofs by computers (for instance the four colors conjecture, which is no more a conjecture). A nice way for proving results on integers is to represent them by sets of items (bullets, squares, numbers, intervals) and to use the Fubini s principle 1. Geometric proofs 1 explanation is coming soon... 3 / 21

4 Content 1 A preliminary example 2 Basic summations (triangular numbers) 3 Sum of odd numbers 4 / 21

5 A preliminary example An old (simple) geometrical proof This first example has been provided by Al Khwarizmi in the XIIth century. It is a nice illustration of what a geometrical proof is. Solve the equation x x = / 21

6 A preliminary example An old (simple) geometrical proof The solution of the equation x x = 39 is determined by means of the surfaces of elementary pieces. We first represent the left hand side x x. x 5/2 x 2 x 6 / 21

7 A preliminary example First example (2) The surface of the cross depicted below is equal to the right hand side. Adding the 4 little squares in the border leads to a total surface of = 64, which is the square of 8. 25/ / 21

8 A preliminary example First example (2) The surface of the cross depicted below is equal to the right hand side. Adding the 4 little squares in the border leads to a total surface of = 64, which is the square of 8. 25/ We finally deduce x: = 3. 7 / 21

9 A preliminary example Remark This method is incomplete, since there exists another root for this equation (x =-13). However, as at the time of Al Khawariszmi, Maths were mainly used for solving geometrical problems of the daily life, negative roots were not considered... 8 / 21

10 A preliminary example Another simple way to prove The informal idea is to establish a one-to-one correspondence between elements of a set (integers). Fubini s principle 2 : Enumerate the elements of a set by two different methods, one leading to an evidence. 2 Guido Fubini / 21

11 A preliminary example Content 1 A preliminary example 2 Basic summations (triangular numbers) 3 Sum of odd numbers 10 / 21

12 Basic summations (triangular numbers) Triangular numbers Definition: Triangular numbers are defined as the sum of the n first integers: n = n k=1 k. There exist many proofs for this result, the simplest one is obtained in writing this sum forward and backward and gathering the terms two by two as follows: 2.Δ n = n + n + n = (n+1) + (n+1) + + (n+1)! 2 n is n times n + 1, thus, n = (n+1).n 2 11 / 21

13 Basic summations (triangular numbers) Another way of looking at this process (1) Use the Fubini s principle. 12 / 21

14 Basic summations (triangular numbers) Another way of looking at this process (1) Use the Fubini s principle. n is represented by piles of bullets arranged as a triangle. Putting two copies up side down gives the n by n + 1 rectangle. n" n&1" n&2" 2" 1" 1" 2" 3" n" 12 / 21

15 Basic summations (triangular numbers) Another way of looking at this process (2) The following figure proves the same result by using a geometric (contiguous) argument instead of bullets. n! 1! 2! 3! n! n! 13 / 21

16 Basic summations (triangular numbers) Another way of looking at this process (2) The following figure proves the same result by using a geometric (contiguous) argument instead of bullets. n! 1! 2! 3! n! n! The sum is represented by a tesselation of boxes of size 1 by 1. The global result is determined by the surface of half the square ( n2 2 ) plus n times half of the surface of the unit square in the diagonal. Thus, n2 2 + n. 1 2 = (n+1).n 2 13 / 21

17 Basic summations (triangular numbers) Sum of two consecutive triangular numbers An interesting question is to compute n + n 1. Computing the first ranks leads us to an evidence: = 1, then, 4, 9, 16, 25, 36,... It is natural to guess n = n 2, which is easy provable by induction (or alternatively, using the expression n + n 1 = n + 2 n 1 since n = n + n 1 ). 14 / 21

18 Basic summations (triangular numbers) Sum of two consecutive triangular numbers An interesting question is to compute n + n 1. Computing the first ranks leads us to an evidence: = 1, then, 4, 9, 16, 25, 36,... It is natural to guess n = n 2, which is easy provable by induction (or alternatively, using the expression n + n 1 = n + 2 n 1 since n = n + n 1 ). This result can be directly obtained using a geometric pattern: n$ n"1$ Δ n $ 14 / 21

19 Basic summations (triangular numbers) Content 1 A preliminary example 2 Basic summations (triangular numbers) 3 Sum of odd numbers 15 / 21

20 Sum of odd numbers Sum of odds Determine the sum of the first odd integers, denoted by S n = n 1 k=0 (2k + 1). This result may again be established by using Fubini s principle. The bullets depict the consecutive odd numbers. The arrangement of the bullets gives two ways for counting. 16 / 21

21 Sum of odd numbers S n = n 2 17 / 21

22 Sum of odd numbers Rearranging the figure as a butterfly gives an insight of another expression of n + n 1. Δ n! The analytic expression is as follows: Δ n + Δ n-1 = n n-1 = (2n-1)! 18 / 21

23 Sum of odd numbers We can also imagine an alternative construction which uses four copies of S n that exactly correspond to an 2n by 2n square as depicted in the figure. 2n#1% 2n#1% 19 / 21

24 Sum of odd numbers We can also imagine an alternative construction which uses four copies of S n that exactly correspond to an 2n by 2n square as depicted in the figure. 2n#1% 2n#1% This leads to 4.S n = (2n) 2, thus S n = n / 21

25 Sum of odd numbers Generalization (1) Both previous examples of triangular and sum of odd numbers are special cases of arithmetic progressions: starting at p 1 = a, p n = p n 1 + b for n > 2. The figure below shows that the sum of the n first elements of an arithmetic progression is equal to n.a + n 1.b. a b 20 / 21

26 Sum of odd numbers Generalization (1) Both previous examples of triangular and sum of odd numbers are special cases of arithmetic progressions: starting at p 1 = a, p n = p n 1 + b for n > 2. The figure below shows that the sum of the n first elements of an arithmetic progression is equal to n.a + n 1.b. a b For instance, a = 1 and b = 2 for the sum of the first n odd numbers. We have S n = n + 2. n 1 = n + n(n 1) = n / 21

27 Sum of odd numbers Generalization (2) Another interesting case is for a = 1 and b = 4 (1, 5, 9, 13, 17,...). The sum is equal to n(2n 1) = n + 4 n 1. = n + 3 n 1 which is also a triangular number ( 2n 1 ). Δ n Δ n-1 Δ n-1 Δ n-1 21 / 21