My Math Forum  

Go Back   My Math Forum > College Math Forum > Complex Analysis

Complex Analysis Complex Analysis Math Forum


Reply
 
LinkBack Thread Tools Display Modes
April 9th, 2015, 10:35 AM   #1
Newbie
 
Joined: Oct 2013

Posts: 3
Thanks: 0

Recurrence sequence - proof

Hey, can you help me with this problem?
Let m be positive integer and let's define a recurrence sequence:
a(0)=a(1)=1,
a(n+2)=a(n+1)+a(n)*e^(2πi*(n+1)/m).
Prove that:
a(2m)=a(m)+1
greggor is offline  
 
April 9th, 2015, 03:30 PM   #2
Global Moderator
 
Joined: Dec 2006

Posts: 18,034
Thanks: 1393

Did you intend "m" to appear in the equation a(n+2) = a(n+1) + a(n)*e^(2πi*(n+1)/m)?
skipjack is offline  
April 10th, 2015, 03:02 AM   #3
Newbie
 
Joined: Oct 2013

Posts: 3
Thanks: 0

Yes, the sequence is dependent on the fixed m>=1
greggor is offline  
Reply

  My Math Forum > College Math Forum > Complex Analysis

Tags
proof, recurrence, sequence



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Recurrence Relation of a Sequence JSimmonds49 Algebra 1 August 18th, 2013 05:54 PM
proof that sequence -> -infinity PrototypePHX Real Analysis 3 July 9th, 2013 12:39 PM
Proof of a theorem about recurrence relations ilikecereal Applied Math 1 February 28th, 2011 01:39 PM
Proof of recurrence sequence convergence natkoza Real Analysis 2 December 6th, 2010 01:20 PM
proof recurrence riotsandravess Number Theory 2 November 22nd, 2010 09:49 PM





Copyright © 2017 My Math Forum. All rights reserved.