Magic-League.com Forum Index Magic-League.com
Forums of Magic-League: Free Online tcg playing; casual or tournament play.
 
 FAQFAQ   SearchSearch   MemberlistMemberlist   UsergroupsUsergroups   RegisterRegister 
 ProfileProfile   Log in to check your private messagesLog in to check your private messages   Log inLog in 

A Math Problem!



 
Reply to topic    Magic-League.com Forum Index -> Other - Non-Magic
Author Message
coolcreep



Joined: 18 Feb 2006
Posts: 588

PostPosted: Wed Nov 19, 2008 9:33 pm    Post subject: A Math Problem! Reply with quote

Ok here is the math problem:


An infinite sequence a0, a1, a2, a3... satisfies

a(m-n) + a(m+n) = a(2m) + a(2n)

for all integers with m>=n>=0

Prove that all integers in the sequence are equal.
Back to top
Onyx



Joined: 16 Oct 2004
Posts: 11

PostPosted: Wed Nov 19, 2008 9:35 pm    Post subject: Reply with quote

www.dumb.com
Back to top
Amaterasu



Joined: 14 Sep 2006
Posts: 74

PostPosted: Wed Nov 19, 2008 9:49 pm    Post subject: Reply with quote

...if n can = 0...then, damn it...it's 0...which makes the problem rather easy :/
Back to top
aldaryn



Joined: 02 Sep 2004
Posts: 501

PostPosted: Wed Nov 19, 2008 10:02 pm    Post subject: Reply with quote

The original problem isn't clear... (m-n), (m+n), (2m), and (2n) are subscripts, right? I think that is where Amaterasu got confused, thinking it was normal multiplication.

Anyway, are you trying to prove that the actual integers IN the sequence are equal, i.e. a0=a1=a2=....? Your wording is not very clear.
Back to top
blamin8or



Joined: 31 Aug 2008
Posts: 17

PostPosted: Thu Nov 20, 2008 2:48 am    Post subject: Reply with quote

aldaryn wrote:
are you trying to prove that the actual integers IN the sequence are equal, i.e. a0=a1=a2=....? Your wording is not very clear.


coolcreep wrote:
Prove that all integers in the sequence are equal.


lol

i think the key is that all integers m>=n. Setting m=n if the series is divergent it breaks and if it is convergent it breaks so it must be neither.
Back to top
coolcreep



Joined: 18 Feb 2006
Posts: 588

PostPosted: Thu Nov 20, 2008 12:30 pm    Post subject: Reply with quote

I used a proof that was basically the same one that Evenkael used, first showing that a_0=a_2m, then showing that a_0=a_(2m-1). The problem with the proof by contradiction is that you do not justify changing a_(m+n), a_2m and a_2n all to k. By doing so you are assuming that all terms excluding a_i are equal, which is too narrow an assumption.
Back to top
Display posts from previous:   
Reply to topic    Magic-League.com Forum Index -> Other - Non-Magic All times are GMT - 7 Hours
Page 1 of 1

 


Powered by phpBB © 2001, 2005 phpBB Group

All content on this page may not be reproduced without consent of Magic-League Directors.
Magic the Gathering is TM and copyright Wizards of the Coast, Inc, a subsidiary of Hasbro, Inc. All rights reserved.


About Us | Contact Us | Privacy Policy