The first sum is - lg n + lg (n-2) + lg (n-4) + .... + lg 4 + lg 2 The second sum is - lg 2 + lg 4 + ... + lg (n-2) + lg n
Both are same. To derive second from first, let 2m=(n-2i). Now adjust the limits. On 10/18/07, Allysson Costa <[EMAIL PROTECTED]> wrote: > > > Dear friends, > > I'm begginer at algorithm theory. Then, I'm solving some recurrences. > > How can the equation below be true? > > (2 summation of lg(n-2i) from i=0 to [n/2 -1] ) > > equal to > > (2 summation of lg(2i) from i=1 to [n/2]) > > How can I change lg(n-2i) in lg(2i)? > > Thanks in advance. > > Allysson > > > > --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---