let n - 2i = 2m ie 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 = 0 ie 2i = 0 ie n-2m = 0 ie m=n/2
therefore the summation is
SUM { lg((2m) } where m = 1 to n/2
now the variable is not important in the summation hence substituting i for
m we get :
SUM { lg(2i) } where i = 1 to n/2 which is your
expression on the left side.
what say ?
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
>
>
> >
>
>
--
Ciao,
Ajinkya
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