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Induction and Recursion.

Induction and Recursion. Recursive definition. Explicit definition. Factorial. Geometric progression. Power of relation on a set. The recursive definition of a function makes reference to earlier versions of itself. The main connection between recursion and induction is that

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Induction and Recursion.

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  1. Induction and Recursion. Recursive definition Explicit definition • Factorial • Geometric progression • Power of relation on a set

  2. The recursive definition of a function makes reference • to earlier versions of itself. • The main connection between recursion and induction is that • objects are defined by means of a natural sequence. • Induction is usually the best (possibly the only) way • to prove results about recursively defined objects.

  3. How to find a closed form for a recursively defined function? In general there is no ready to use recipe. Some simple cases are • Linear function of integer n: g1(n) = an+b g1(n+1) = a(n+1)+b= g1(n) + a • Quadratic function of integer n: g2(n) = an2+bn+c g2(n+1) = a(n+1)2+b(n+1)+c = g2(n) +2an+(a+b)

  4. ……..definition of composition …..by IH ….association of composition ……..definition of composition Example. Prove that for all positive integers m and n, Proof. Let m be arbitrary positive integer and then prove by induction on n1. 1) Basis. n=1: (by recursive definition of Rm+1) 2) IH: Assume that for some k 1 we have IS: We need to prove that

  5. Proof. Denote . We need to prove that S satisfies the definition of transitive closure, i. e. i) S includes R, R S ii) S is transitive iii) S is the smallest relation satisfying i) and ii). Transitive closures again. Theorem. The transitiveclosure of R is ii) assume(x, y) S and (y, z) S to prove that (x, z) S. (x, y) S implies that (x, y) Rn, n is some integer, nZ+. (y, z) S implies that (y, z) Rm, m is some integer , nZ+. (x, z) Rno Rm= Rn+m, by the previous Theorem. (x, z) S , since Rn+mS

  6. By the definition of S as , it suffices to prove that for any nZ+ (Rn T) We are going to prove it by induction on n 1. iii) Suppose T is any relation, satisfying i) RT and ii) T is transitive. and prove that S  T . 1) Basis: when n=1, R1=RT by assumption. 2) IH: Assume that for n=k, where k is some integer, k 1, Rk T . IS: Need to prove Rk+1T Suppose (x, y) Rk+1=Rko R (by definition of composition). It implies that there exists some zA, that (x, z)Rk and (z, y)R. . (x, z)Rk implies (x, z)T , because Rk T by IH. (z, y)R implies (z, y)T , because RT by assumption. (x, z)T and (z, y)T imply (x, y)T by transitive property of T. We proved by induction that for any n 1, Rn T.

  7. , n  2 Fibonacci Numbers The sequence named forFibonacci(1202) is introduced in terms of rabbits. Suppose that every month a breeding pair of rabbits produce a pair of offspring. The offspring in turn start breeding two months later, and so on. Denote Fn the number of pairs in month n. Suppose you buy a pair of rabbits in month 1 (F0=0, F1=1) then you still have one pair in month 2 (F2=1), but in month 3 they start breeding, F3=2. The sequence begins 0, 1, 1, 2, 3, 5, 8, 13, 21,… You can observe, that in a month n you have all pairs from the previous month (n1), plus offspring pairs for all pairs from month (n2), i. e. :

  8. Let’s prove the closed formula for the sum for all integers n  0 ? 1) Basis. Take n=0 and check that By the recurrent formula for n =2 we have so F21=0 and F0=0. So, Fibonacci numbers can be defined recursively: These numbers have numerous applications in CS, mathematics, theory of games. Proof by induction on n  0.

  9. 2) IH: Assume that the summation formula is correct for n=k, where k is some integer k  0, i. e. IS: We need to prove that …………by IH ……..by recurrence formula for (k+3) >1

  10. Strong Induction IP 2) n 0 [P (n) P (n+1)] 2) n 0 [(m n P (m)) P (n+1)] • Sometimes we need stronger assumption to prove IS. • It happens when to prove IS for (n +1) you need to refer not only • to the previous n , but to smaller numbers. Strong Induction Principle: Let AN denote a subset that satisfies the following two properties: 1) 0A; 2) if 0, 1, …kA, then k+1A. Then A=N. So, Strong Induction differs from Induction Principle in step 2):

  11. As we proceed n  n +1, we can not derive case n +1 from the case n (n+1)5 (n+1)4 n +1 n Example. Prove that every amount of postage of 12 cents or more can be formed using only 4 cent and 5 cent stamps. We are going to prove by induction on n  12, that any postage n = 5i + 4j, where integers i, j 0. We can infer n+1= 5i + 4j either from (n+1)5= 5(i 1)+ 4j or from (n+1)4= 5i + 4(j 1) We need to prove an extended basis here, that is to prove the basis on more then one point.

  12. Proof. 1) Basis (extended): check that we can form postage of 12, 13, 14, and 15 cents using only 4c and 5c stamps. 12=43; 13=5+ 42; 14=52+4; 15=53. 2) IH: Assume that any postage 12, … k, where k 15 can be made using 4c and 5c stamps (strong induction) IS Prove that conjecture is true for the postage k +1. By IH any postage greater or equal then 12c and less or equal k can be formed using 4c and 5c stamps. In particular it is true for k 3. Together with one more 4c stamp it gives k +1 postage.

  13. We already did this Theorem. The following three principles are equivalent: (a) the Induction Principle; (b) the Strong Induction Principle; and (c) the Well-Ordering principle. Proof. It suffices to prove (a)(b), (b)(c), and (c)(a).

  14. Proof of (a)(b): Let AN and A satisfies the two properties: (1) 0A; and (2) if 0, 1, …kA then k+1A. We need to prove A=N. To make use of Induction Principle we define the auxiliary set C={m| mN and 0, 1, …mA}. Thus, C  A N (3). Since 0A by (1), 0C by the definition of C. If kC, that is, 0, 1, …kA, then by the property (2) of the set A, k+1A, which implies that k+1C, because 0, 1, …k, k+1A. Thus, set Csatisfies the properties of the Induction Principle, so C=N. Therefore A=N because of (3).

  15. Proof of (b)(c): Assume BN and B (4). Suppose the Well-Ordering Principle is false, i.e. suppose B does not have the smallest element (5). Prove that this leads to contradiction to the Strong Induction Principle. Let A=NB (6). By (5) 0B, thus 0A by (6). If the numbers 0, 1, …kA, then these numbers do not belong to B. Then k+1 does not belong to B, because otherwise it would be the smallest element of B contradicting to (5). Thus, k+1A and by the Strong Induction Principle A=N, which implies that B = by (6), contradicting to (4). So, (c) is proved.

  16. Example. Prove that for all n 5. n Fn+1 0 1 1 1 1 1.5 2 2 2.25 3 3 3.375 4 5 5.0625 5 8 7.59375 6 13 11.390625 7 21 17.0859375 Basis

  17. Fk+2= Fk+1+ Fk > (3/2)k + (3/2)k1 = (3/2)k 1((3/2)+1) = (3/2)k 1 (5/2) = (3/2)k 1 (10/4) > (3/2)k 1 (9/4) = (3/2)k 1(3/2)2 = (3/2)k+1 ……recursive definition …….IH for n=k and n=k1 ……algebra Proof by induction on n 5 that Fn+1>(3/2)n 1) Basis (extended) n=5: F6=8>(3/2)5 = 7.59375 n=6: F7= 13>(3/2)6 = 11.390625 2) IH : Assume that for all n=5, …k, where k is some integer k 6 we have Fn+1>(3/2)n (strong induction). IS : We need to prove inequality Fn+1>(3/2)n for n = k +1, i. e. Fk+2>(3/2)k+1

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