User Name Remember Me? Password

 Math Events Math Events, Competitions, Meetups - Local, Regional, State, National, International

 July 21st, 2009, 02:30 AM #1 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 1998 Putnam Mathematical Competition Okay, since hardly no one is posting on Math Olympiads section anymore, I will post some interesting ones here more often. Here's an interesting Putnam problem: B-1. Find the minimum value of for . For a challenge, try not to look at the solutions on the Putnam site or anything similar (unless if you completely give up). July 21st, 2009, 02:32 AM #2 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Re: 1998 Putnam Mathematical Competition Okay. First thing that comes to my mind is to substitute , simply. Does that help you guys? July 21st, 2009, 07:56 AM #3 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: 1998 Putnam Mathematical Competition u^3 = x^3 + 3x + 3/x + 1/x^3, so x^3 + 1/x^3 = u^3 - 3u. u^6 = x^6 + 6x^4 + 15x^2 + 20 + 15/x^2 + 6/x^4 + 1/x^6 u^4 = x^4 + 4x^2 + 6 + 4/x^2 + 1/x^4 u^2 = x^2 + 2 + 1/x^2 u^6 - 6u^4 = x^6 - 9x^2 - 16 - 9/x^2 + 1/x^6 u^6 - 6u^4 + 9u^2 - 2 = x^6 + 1/x^6 So the numerator is u^6 - u^6 + 6u^4 - 9u^2 + 2 - 2 = 6u^4 - 9u^2 and the denominator is u^3 + u^3 - 3u = 2u^3 - 3u so the fraction simplifies to Now you need to take the derivative wrt x... July 21st, 2009, 02:48 PM #4 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Re: 1998 Putnam Mathematical Competition Great job on the solution, CRGreathouse! Since now we have the equation, all we need to do is take the derivative wrt , and obey the inequality of . Putnam isn't too hard compared to IMO, as long we take a good time to look at the problem and look for different situations and good solutions, and also using past mathematical experiences and knowledge. Tags 1998, competition, mathematical, putnam Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Solarmew New Users 5 December 6th, 2012 11:23 AM jcmc2112 Math Events 3 December 5th, 2010 08:02 AM johnny Math Events 1 June 8th, 2010 12:23 PM Cat Math Events 8 July 13th, 2009 05:34 AM johnny Calculus 3 July 31st, 2007 04:29 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      