I had made the substituition but I don't get to go back to 2i.
n-2i=2m
Then
SUM[lg(2m)]= SUM [lg2+lgm]=SUM[lg2]+SUM[lgm]= SUM [1] +SUM[lgm]
OBS: SUM = SUMMATION.
I stop here. I dont know how to continue. Can you help me.
Thanks in advance
Allysson
Ajinkya Kale escreveu:
substitute
let n - 2i = 2mie 2i = n-2m
hence
SUM { lg (n-2i) } = SUM { lg (2m) }
no the limits
upper limit = i = n/2-1 ie 2i = n-2 ie n-2m = n-2 ie m=1
lower limit = i = 0ie 2i = 0 ie n-2m = 0ie m=n/2
therefore the summation is
SUM { lg((2m) }
substitute n-2i = 2m in firsst equationyou will get the left one on
reducing the terms in terms of m.
On 10/19/07, Allysson Costa [EMAIL PROTECTED] wrote:
Anyone can give a explanation how I get the equation below true?
Why lg(n-2i) became lg(2i)?
Thanks in advance.
Allysson
