Mathematical Induction Proofs and Examples, Study notes of Calculus

Download Mathematical Induction Proofs and Examples and more Study notes Calculus in PDF only on Docsity! Lecture Note #3 (Mathematical Induction) Exercises ========================================================================================== 3) Strong form of Mathematical Induction #1. [Example 5.4.2, p. 270] Define a sequence s0, s1, s2,.. as follows s0 = 0, s1 = 4, sk = 6sk-1 – 5sk-2 for all integers k >= 2. Actually the whole proof is shown in the textbook (p. 270-271). You can see it there. But since I went over this in the class, I type what I wrote (or close to it) here. a) Find the first four terms • s0 = 0 • s1 = 4 • s2 = 6*s1 – 5*s0 = 6*4 – 5*0 = 24 • s3 = 6*s2 – 5*s1 = 6*24 – 5*4 = 144 – 20 = 124 b) We are given that sn = 5n – 1. Prove true. Proof: By induction on n. Let P(n) = 5n – 1 (for all n >= 0). 1) Basis step (n = 0 and n = 1): When n = 0, s0 = 0 (as given) and P(0) = 50 – 1 = 1 – 1 = 0. When n = 1, s0 = 4 (as given) and P(1) = 51 – 1 = 5 - 1 = 4. Therefore, s0 = P(0) and s1 = P(1) … (A) 2) Inductive step: [Inductive hypothesis] Assume si = P(i) for all integer 0 <= i <= k, where k >= 1, that is, si = 6si-1 – 5si-2 = P(i) = 5i - 1. [Inductive statement] Show sk+1 = P(k+1), that is, sk+1 = 6sk – 5sk-1 = P(k+1) = 5k+1 - 1. 𝐿𝐿𝐿𝐿𝐿𝐿 = 𝑠𝑠𝑘𝑘+1 = 6 ∙ 𝑠𝑠𝑘𝑘 − 5 ∙ 𝑠𝑠𝑘𝑘−1 = 6 ∙ [5𝑘𝑘 − 1] − 5 ∙ [5𝑘𝑘−1 − 1] = 6 ∙ 5𝑘𝑘 − 6 − 5 ∙ 5𝑘𝑘−1 + 5 = 6 ∙ 5𝑘𝑘 − 1 − 5𝑘𝑘 = 5 ∙ 5𝑘𝑘 − 1 = 5𝑘𝑘+1 − 1 = 𝑃𝑃(𝑘𝑘 + 1) = 𝑅𝑅𝐿𝐿𝐿𝐿 Therefore, ek+1 = P(k+1) … (B) By (A) and (B), we can conclude that the statement is true (for all integers n >= 0). QED. #2. Write the Strong Mathematical Induction version of the problem given earlier, “For all integer n >= 4, n cents can be obtained by using 2-cent and 5-cent coins.” Note the basis steps should prove P(4) and P(5). Proof: 1) Basis step (n = 4 and n = 5): When n = 4, we can make 4 cents by two 2-cent coins. When n = 5, we can make 5 cents by one 5-cent coin. … (A) 2) Inductive step: [Inductive hypothesis] Assume n >= 6, and k cents can be obtained using 2-cent and 5-cent coins (only), for all k, 4 <= k <= n. By the inductive hypothesis, we can make n-2 cents (by taking out one 2-cent coin). So we can add a 2-cent coin to make n cents. … (B) By (A) and (B), we can conclude that the statement is true (for all integers n >= 1). QED. #3. [Section 5.4, Exercise #5, p. 277] Suppose that e0, e1, e2,… is a sequence defined as follows. e0 = 12, e1 = 29 ek = 5ek-1 – 6ek-2 for all integers k >= 2. Prove that en = 5*3n + 7*2n for all integer n >= 0. Use Strong Mathematical Induction. Actually the whole proof is shown at the back of the textbook (p. A-38). You can see it there. But since I went over this in the class, I type what I wrote (or close to it) here. Proof: Let P(n) = 5*3n + 7*2n for all integer n >= 0. We show the equivalency between the sequence en and P(n), that is, en = 5en-1 – 6en-2= P(n) = 5*3n + 7*2n. 3) Basis step (n = 0 and n = 1): When n = 0, e0 = 12 (as given) and P(0) = 5*30 + 7*20 = 5 + 7 = 12. When n = 1, e0 = 29 (as given) and P(1) = 5*31 + 7*21 = 15 + 14 = 29. Therefore, e0 = P(0) and e1 = P(1) … (A) 4) Inductive step: [Inductive hypothesis] Assume ei = P(i) for all integer 0 <= i <= k, where k >= 1, that is, ei = 5ei-1 – 6ei-2 = P(i) = 5*3i + 7*2i . [Inductive statement] Show ek+1 = P(k+1), that is, ek+1 = 5ek – 6ek-1 = P(k+1) = 5*3k+1 + 7*2k+1 . 𝐿𝐿𝐿𝐿𝐿𝐿 = 𝑒𝑒𝑘𝑘+1 = 5 ∙ 𝑒𝑒𝑘𝑘 − 6 ∙ 𝑒𝑒𝑘𝑘−1