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(Geometric Progression asymptotics) Let f(n) = Σi-1 b² for some constant b>0. Then show that : ● • f(n) = 0(1) if b < 1 f(n) = O(n) if b = 1 • f(n)= (bn) if b > 1 In order to solve this problem, it will be useful to recall that for a geometric progression with first term a a common ratio r the sum of the first n terms is a(r"-1)

(Geometric Progression asymptotics) Let f(n) = Σi-1 b² for some constant b>0. Then show that : ● • f(n) = 0(1) if b < 1 f(n) = O(n) if b = 1 • f(n)= (bn) if b > 1 In order to solve this problem, it will be useful to recall that for a geometric progression with first term a a common ratio r the sum of the first n terms is a(r"-1)

Advanced Engineering Mathematics
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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Solve the problem with short explanations following the hints(2nd image)
gn). 4. (Geometric Progression asymptotics) Let f(n) = 1 b² for some constant b>0. Then show that : f(n)= (1) if b < 1 f(n) = O(n) if b = 1 f(n) = (bn) if b > 1 M In order to solve this problem, it will be useful to recall that for a geometric progression with first term a and common ratio r the sum of the first n terms is a(r"-1) r-1 1: -1. f+L £11 Lie
Transcribed Image Text:gn). 4. (Geometric Progression asymptotics) Let f(n) = 1 b² for some constant b>0. Then show that : f(n)= (1) if b < 1 f(n) = O(n) if b = 1 f(n) = (bn) if b > 1 M In order to solve this problem, it will be useful to recall that for a geometric progression with first term a and common ratio r the sum of the first n terms is a(r"-1) r-1 1: -1. f+L £11 Lie
Very good answer given above - basically you have the summation in terms of b and n (call this g(b, n))by substituting the parameters of the geometric series. For each of the cases you need to then give upper and lower bounds for this function. When b < 1, you need to give constants 2₁, 22 such that 2₁ ≤ g(b, n) ≤ 22. Similarly, when b = 1 you need to give constants y₁, y2 such that y₁n ≤ g(b,n) ≤ y₂n. Finally, when b < 1, you need to give constants x1, x2 such that x₁bn ≤ g(b,n) ≤ x₂bn.
Transcribed Image Text:Very good answer given above - basically you have the summation in terms of b and n (call this g(b, n))by substituting the parameters of the geometric series. For each of the cases you need to then give upper and lower bounds for this function. When b < 1, you need to give constants 2₁, 22 such that 2₁ ≤ g(b, n) ≤ 22. Similarly, when b = 1 you need to give constants y₁, y2 such that y₁n ≤ g(b,n) ≤ y₂n. Finally, when b < 1, you need to give constants x1, x2 such that x₁bn ≤ g(b,n) ≤ x₂bn.
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