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chess64
Birch & Swinnerton Dyer
Offline Joined: 20 Feb 2005 Posts: 4872
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Week 4 Solution
The statement of Germany 1996 #2 is the contrapositive of this problem, and a nice solution to that is given here . Some people also used the contrapositive technique for this problem.
Posted: Sun May 21, 2006 5:43 am
seesaw
Hodge Conjecture
Offline Joined: 09 Mar 2006 Posts: 53
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I will take over the proof contest, but I want someone else to help me (as I am not always available.) Are there any volunteers?
Anyway:
Week 5
Prove that, for all values of , (provided that and are defined),
(PM me your proof)
Posted: Sun May 21, 2006 6:41 am
darkdevil
P versus NP
Offline Joined: 21 Sep 2005 Posts: 25 Location: localhost
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now you are the one who gets the solutions? not chess64 anymore?
Posted: Sun May 21, 2006 10:04 am
darkdevil
P versus NP
Offline Joined: 21 Sep 2005 Posts: 25 Location: localhost
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and btw.....what is ??
Posted: Sun May 21, 2006 10:09 am
seesaw
Hodge Conjecture
Offline Joined: 09 Mar 2006 Posts: 53
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darkdevil wrote:
now you are the one who gets the solutions? not chess64 anymore?
Yes, I am the one who gets the solutions (for now at least).
darkdevil wrote:
and btw.....what is ??
is .
Posted: Sun May 21, 2006 10:16 am
seesaw
Hodge Conjecture
Offline Joined: 09 Mar 2006 Posts: 53
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Scoreboard
Last edited by seesaw on Sun Jun 04, 2006 2:51 pm; edited 3 times in total
Posted: Sun May 21, 2006 1:31 pm
drunner2007
Navier-Stokes Equations
Offline Joined: 23 Mar 2005 Posts: 1117 Location: United States
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Remark
any new updates/remarks?
_________________ It is never too late to be what you could have been...
The only way to predict the future is to make it.
Posted: Thu May 25, 2006 5:33 am
seesaw
Hodge Conjecture
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I might not be available for the next few days, so I am posting week 6 today.
Week 5 Solution
matt276eagles wrote:
Since the maximum of the function is , we have for all values of . Using the double angle identity for sine, we obtain .
Dividing by , which is positive for all values of such that and are defined, we have . By the definitions of tangent and secant, we have , which is the desired result.
I would like to remind everybody to try to write proofs that are as clear as possible.
Posted: Fri May 26, 2006 11:38 am
seesaw
Hodge Conjecture
Offline Joined: 09 Mar 2006 Posts: 53
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I think that after Week 6 I may start again from week 1 and have proof contests in 6-week cycles. Any feedback on that idea?
Also, does anyone want to volunteer to help out with the proof contest or to be a backup?
Week 6
Let , , and be fixed natural numbers such that, for every positive integer , there is a triangle whose sides have length , , and . Prove that these triangles are not scalene.
Posted: Fri May 26, 2006 11:42 am
seesaw
Hodge Conjecture
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Week 6 Solution
Jiang wrote:
Let and . For there to be a triangle for every positive integer n, must be true for all n. Dividing the inequality by , we get . Because , as n gets larger, also gets larger without limit. While because , as n gets larger, approaches more and more to 0. So the inequality would eventually prove to be false. If either two or all of the sides are same, it is obviouse that they will always be the same for any n. A scalene triangle must have all of its sides in different lengths, so it would not be possible.
I am sorry to say that due to certain circumstances, I am no longer able to run the Proof Contest. If there are any volunteers to take over the contest please PM me.
Posted: Sun Jun 04, 2006 2:54 pm
darkdevil
P versus NP
Offline Joined: 21 Sep 2005 Posts: 25 Location: localhost
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Week 7
Prove that has only one real root
Prove that if is a root and then
Posted: Tue Jun 06, 2006 5:48 am
1=2
Birch & Swinnerton Dyer
Offline Joined: 06 Jan 2006 Posts: 5825 Location: One Brownie Point!
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Since week 7 has no solution posted yet, and it is overdue, I would like to turn this into a proof marathon.
Here's an easy one.
Prove that
_________________
Hmm... apparently quoting Disney movies gets me muted on FTW.
Posted: Mon Jul 17, 2006 6:02 am
lotrgreengrapes7926
Navier-Stokes Equations
Offline Joined: 31 Mar 2006 Posts: 1855
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nutz_for2.718281828 wrote:
Since week 7 has no solution posted yet, and it is overdue, I would like to turn this into a proof marathon.
Here's an easy one.
Prove that
Click the smiley.
Divide by 3. Then, AM-GM, as long as
.
_________________
Posted: Mon Jul 17, 2006 6:35 am
SplashD
Birch & Swinnerton Dyer
Offline Joined: 09 Apr 2006 Posts: 2762 Location: Je ne sais pas
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Does AM-GM work for more than two numbers?
_________________ "You don't win no silver, you only lose the gold." -David Glen Eisley
Posted: Mon Jul 17, 2006 6:37 am
lotrgreengrapes7926
Navier-Stokes Equations
Offline Joined: 31 Mar 2006 Posts: 1855
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SplashD wrote:
Does AM-GM work for more than two numbers?
Yes.
OK, this post is not too short anymore.
_________________
Posted: Mon Jul 17, 2006 6:38 am
SplashD
Birch & Swinnerton Dyer
Offline Joined: 09 Apr 2006 Posts: 2762 Location: Je ne sais pas
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IF you divide by 3, wouldn't you still have a two on the RHS?
_________________ "You don't win no silver, you only lose the gold." -David Glen Eisley
Posted: Mon Jul 17, 2006 6:40 am
lotrgreengrapes7926
Navier-Stokes Equations
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SplashD wrote:
IF you divide by 3, wouldn't you still have a two on the RHS?
OK, I meant divide by 6, to get Click to reveal hidden content .
From AM-GM,
. Since,
for all
,
must be true.
And nutzfore, you have to state .
_________________
Posted: Mon Jul 17, 2006 6:44 am
matt276eagles
Navier-Stokes Equations
Offline Joined: 08 Jun 2005 Posts: 1235 Location: Princeton, NJ
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nutz_for2.718281828 wrote:
Since week 7 has no solution posted yet, and it is overdue, I would like to turn this into a proof marathon.
Here's an easy one.
Prove that
Huh? For and , the inequality gives .
Posted: Mon Jul 17, 2006 9:53 am
SplashD
Birch & Swinnerton Dyer
Offline Joined: 09 Apr 2006 Posts: 2762 Location: Je ne sais pas
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matt276eagles wrote:
nutz_for2.718281828 wrote:
Since week 7 has no solution posted yet, and it is overdue, I would like to turn this into a proof marathon.
Here's an easy one.
Prove that
Huh? For and , the inequality gives .
lotrgreengrapes said
_________________ "You don't win no silver, you only lose the gold." -David Glen Eisley
Posted: Mon Jul 17, 2006 9:58 am
amcavoy
Yang-Mills Theory
Offline Joined: 21 Mar 2005 Posts: 922
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Here's another easy one:
For , prove the AM-GM inequality. That is, prove that
Posted: Mon Jul 17, 2006 10:24 am
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