Consider the following recurrence relation: G(n) G(n-1) + 2n-1 (a) Calculate G(0), G(1), G(2), G(3), G(4), and G(5). G(0) = 1 G(1) 2 G(2) = 5 G(3) 10 G(4) 17 G(5) = 26 if n = 0 if n > 0. (b) Analyze the sequences of differences. (Enter your answers as a comma-separated list of values ordered with respect to increasing n.) first differences: 1,3,5,7,9 second differences: 1,2,5,10,17,26 X
Consider the following recurrence relation: G(n) G(n-1) + 2n-1 (a) Calculate G(0), G(1), G(2), G(3), G(4), and G(5). G(0) = 1 G(1) 2 G(2) = 5 G(3) 10 G(4) 17 G(5) = 26 if n = 0 if n > 0. (b) Analyze the sequences of differences. (Enter your answers as a comma-separated list of values ordered with respect to increasing n.) first differences: 1,3,5,7,9 second differences: 1,2,5,10,17,26 X
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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Question
need help with second diff.
Transcribed Image Text:Consider the following recurrence relation:
if n = 0
G(n) = G(n-1) + 2n - 1 if n > 0.
(a) Calculate G(0), G(1), G(2), G(3), G(4), and G(5).
G(0) = 1
G(1)
= 2
G(2) 5
G(3)
= 10
G(4) 17
G(5)= 26
second differences:
(b) Analyze the sequences of differences. (Enter your answers as a comma-separated list of values ordered with respect to increasing n.)
first differences:
1,3,5,7,9
✓
✔
✔
✔
1,2,5,10,17,26
What does this suggest about the closed form solution?
This suggests that the closed form will have degree 2
X
(c) Find a good candidate for a closed-form solution.
P(n) = n²+1
(d) Prove that your candidate solution is the correct closed-form solution.
(Induction on n.) Let f(n) = n²+1
2
=
Base Case: If n = 0, the recurrence relation says that G(0) = 1, and the formula says that f(0) = 0² + 1
Inductive Hypothesis: Suppose as inductive hypothesis that G(k-1) = (k − 1)² +1
4²
Inductive Step: Using the recurrence relation,
G(K) = G(k-1) + 2k1, by the second part of the recurrence relation
k-1
+ 1 + 2k - 1, by inductive hypothesis
- 2k + 1 + 1 + 2k - 1,
for some k > 0.
= 1
, so they match.
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