Induction Proof $T(n) = 9T(n/3) + n^2$

Question

How can I prove that the reccurence

T(n) = 9T(n/3) + n²

leads to T(n) = O(n² log(n)) using the substitution method and a proof by induction? I’m not allowed to use the Master Theorem.

Using induction and assuming T(n) ≤ cn² log n, I got to the following point:

T(n) = 9 * T(n/3) + n²

≤ 9c ( n² / 9 + log(n/3)) +n²

= cn² + 9c log(n/3) + n²

Thank you.

What are the policies for the class that you’re taking? Are you allowed to ask questions on Stack Overflow? — templatetypedef, May 07 '20 at 16:05
Can you show your intermediate work that got you to the inequality you came up with above? — templatetypedef, May 07 '20 at 16:46
Sure. T(n) = 9 * T(n/3) + n^2 <= 9c ( n^2/9 + log(n/3)) +n^2 = cn^2 + 9c * log(n/3) + n^2 — Joshua, May 07 '20 at 18:04

score 1 · Answer 1 · answered May 07 '20 at 18:10

1

I think you've made a math error in your substitution. If we assume that T(n) ≤ cn² log n, then we'd get

T(n) = 9T(n / 3) + n²

≤ 9(c(n / 3)² log(n / 3)) + n²

= 9((1 / 9)cn² log (n / 3)) + n²

= cn² log(n / 3) + n²

You're very close to having things complete at this point. As a hint, suppose that the logarithm is a base-3 logarithm. What happens if you then use properties of logarithms to simplify cn² log(n / 3)?

answered May 07 '20 at 18:10

templatetypedef

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Thank you! Then I have cn^2 (log(n) - 1) + n^2 = cn^2 log(n) + (1 - c)log(n). Can I say that c has to be greater than 1 one? because then cn^2 log(n) + (1 - c)log(n) would be smaller than cn^2 log(n) and I would be done. – Joshua May 08 '20 at 13:20
btw How do I write math content in Stackoverflow? – Joshua May 08 '20 at 13:22
Alas, it’s hard. I’m using the HTML tags ^{and _{for that.}} – templatetypedef May 08 '20 at 15:19
Check your math. You’re not distributing the terms in the multiplication correctly. – templatetypedef May 08 '20 at 15:45
And now can I assume that c is >= 1, so that im done? – Joshua May 09 '20 at 21:14

Induction Proof $T(n) = 9T(n/3) + n^2$

1 Answers1