QUESTION 10An algorithm has time complexity f(n)%3Dn*+200x+10000.Of is O(1)Of is O(n?)Of is O(n*)Of is O(n°)

Answered: .2 An algorithm has time complexity…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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6. In this problem, you may use the fact (which we will prove in Chapter 6) that an integer n is not divisible by 3 if and only if there exists an integer k such that n = 3k +1 or n = 3k + 2. (a) Prove that for all integers n, if 3 | n2, then 3 | n. (b) Prove that for all integers i and j, if 3 | (i2 + j?), then 3 | i and 3 | j.
3 . Use the baby-step giant-step algorithm to find h, k such that 2^h ≡ 7 (mod 53), 2^k ≡ 9 (mod 53).
5. (a) Use the Binomial Theorem to expand (a + b)* (b) Recall the multinomial theorem says that n! (ri + x2 + +)" Σ %3D n! x n2! x..x ng! where the sum is over all solutions to n1 +. . +ng n, and each n; 2 0. Use the multinomial theorem to expand the following: (a + b+ c)* This is tedious but doable.
3.4. A sequence s,, S2, S3, ... is defined recursively as follows: Sk = 5Sk-1 + (Sk-2)² for all integers k 2 3 S1 = 4 S2 = 8 Use (strong) mathematical induction to prove that s, is divisible by 4 for all integers n > 1.
Prove this Adams-Bashforth method using polynomial interpolation and prove it's second order accurate in time. At p" + |
4. n! (read "n factorial") is the product of the first n positive integers, that is 2! = 2 1, 3! = 3 2 1, etc. (a) How many zeroes are in the rightmost places (digits) of the large number 17! ? (b) Find the smallest natural number n such that n! ends in exactly 37 zeroes.
3. Let F be the n-th Fibonacci number, defined recursively by Fo= 0, F₁ = 1 and Fn Fn-1+ Fn-2 for n ≥2. Prove the following by induction (or strong induction): (a) Let + = ¹+√5. For all n ≥ 1, Fn = 2 (b) For all n ≥ 1, ₁₁ F² = FnFn+1. =
= The Fibonacci numbers are a sequence numbers defined by F₁ = 1, F₂ = Fn = Fn-1 + Fn-2 for n ≥ 3. 1, and
Solve this now in one hour and take a thumb up plz
Sort 6, 2, 5, 4, 3 in increasing order using the following algorithm sort(a(1),..., a(n): real numbers with n ≥ 2) for i:=1 to n-1 for j:=1 to n-i if a(j) > a(j+1) then interchange a(j) and a(j+1) {a(1),...,a(n) is in increasing order} Which of the following steps do not attained during the procedure? 1. 2, 5, 6, 4, 3 II. 2, 6, 4, 5, 3 III. 2, 4, 3, 5, 6 IV. 2, 6, 5, 4, 3 V. 2, 4, 5, 6, 3 II and V I and II III and IV only V

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.2 An algorithm has time complexity f(n)3n*+200x²+10000. Of is O(1) Ofis O(n?) Of is O(n*) f is O(n°)

.2 An algorithm has time complexity f(n)3n+200x²+10000. Of is O(1) Ofis O(n?) Of is O(n) f is O(n°)