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Topic review (newest first)
- anonimnystefy
- 2013-02-27 05:19:58
No problem and you're welcome!
- genericname
- 2013-02-27 02:44:16
Ah, thank you. It was base 2, sorry for the confusion.
- bobbym
- 2013-02-26 15:31:10
A very good answer!
- anonimnystefy
- 2013-02-26 13:41:10
Because then the step makes sense. We will have have to wait for the OP's answer.
And besides, the use of n and lg reminds me of comp. analysis.
- bobbym
- 2013-02-26 13:27:50
Why would the log be to the base two?
- anonimnystefy
- 2013-02-26 12:40:36
Hi genericname
lg(a/b)=lg(a)-lg(b) for any a and b for which the expression is defined.
So, lg(n/2)=lg(n)-lg(2)=lg(n)-1, assuming the logarithm is with base 2.
- bobbym
- 2013-02-26 09:10:06
I am getting:
- genericname
- 2013-02-26 07:58:06
Yeah.
- bobbym
- 2013-02-26 07:50:27
Hi;
Is this the problem?
- genericname
- 2013-02-26 07:33:05
(2*(n/2)*lg(n/2)) + n
= (n*lg(n/2))+n
= n*(lg n - 1) + n
How does (n*lg(n/2))+n simplify to n*(lg n - 1) + n? What happened to the n/2 that was inside? It has been a while since I last worked with log.