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#1 2013-02-26 07:33:05

genericname
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Log question

(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.

#2 2013-02-26 07:50:27

bobbym
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Re: Log question

Hi;

Is this the problem?

Last edited by bobbym (2013-02-26 07:51:47)


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#3 2013-02-26 07:58:06

genericname
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Re: Log question

Yeah.

#4 2013-02-26 09:10:06

bobbym
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Re: Log question

I am getting:

Last edited by bobbym (2013-02-26 09:10:52)


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#5 2013-02-26 12:40:36

anonimnystefy
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Re: Log question

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.


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

#6 2013-02-26 13:27:50

bobbym
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Re: Log question

Why would the log be to the base two?


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#7 2013-02-26 13:41:10

anonimnystefy
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Re: Log question

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.

Last edited by anonimnystefy (2013-02-26 13:42:40)


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

#8 2013-02-26 15:31:10

bobbym
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Re: Log question

A very good answer!


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#9 2013-02-27 02:44:16

genericname
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Re: Log question

Ah, thank you. It was base 2, sorry for the confusion.

#10 2013-02-27 05:19:58

anonimnystefy
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Re: Log question

No problem and you're welcome!


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

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