Comments on: GREEN CHICKEN http://www.ephblog.com/2009/11/09/green-chicken/ All Things Eph Thu, 18 Mar 2010 06:49:06 +0000 http://wordpress.org/?v=abc hourly 1 By: Dick Swart http://www.ephblog.com/2009/11/09/green-chicken/#comment-75286 Dick Swart Fri, 13 Nov 2009 23:25:19 +0000 http://www.ephblog.com/?p=24246#comment-75286 In the hubbub of Mr Moore, a phrase has been used by David to capture and provide a rationalization for the continuing back-and-forth with some vehemence, denial, reversal of fields, and the burning of straw men (a penny for the old Guy) to arrive at conclusions. The phrase is "iterate to agreement". I believe I catch the sense in which David applies the phrase: If we argue long enough, and you keep proving me wrong, it isn't that I'm a bad person, it just means we are coming closer and closer to a correct answer that may not be what I was positive it was, but means that I helped you arrive at your answer. Now certainly mathiness is next to Godliness, and Kepler's Equation has been proven through iteration and agile software seem to include iteration in the defining process. Can you who laugh in the face of the Green Chicken questions, develop a proof for the semantic process posited by David? Beat Amherst! In the hubbub of Mr Moore, a phrase has been used by David to capture and provide a rationalization for the continuing back-and-forth with some vehemence, denial, reversal of fields, and the burning of straw men (a penny for the old Guy) to arrive at conclusions.

The phrase is “iterate to agreement”. I believe I catch the sense in which David applies the phrase: If we argue long enough, and you keep proving me wrong, it isn’t that I’m a bad person, it just means we are coming closer and closer to a correct answer that may not be what I was positive it was, but means that I helped you arrive at your answer.

Now certainly mathiness is next to Godliness, and Kepler’s Equation has been proven through iteration and agile software seem to include iteration in the defining process.

Can you who laugh in the face of the Green Chicken questions, develop a proof for the semantic process posited by David?

Beat Amherst!

]]> By: dm '10 http://www.ephblog.com/2009/11/09/green-chicken/#comment-74581 dm '10 Tue, 10 Nov 2009 03:43:39 +0000 http://www.ephblog.com/?p=24246#comment-74581 David: If you take the constraint inequality x^4+4 <= 5x^2 and rearrange it, you get x^4-5x^2+4<=0 which can be written (x^2-4)(x^2-1)<=0. So the roots are -2,-1,1, and 2 and by testing a few values you can see that the inequality holds from -2 to -1 and from 1 to 2. David: If you take the constraint inequality x^4+4 <= 5x^2 and rearrange it, you get x^4-5x^2+4<=0 which can be written (x^2-4)(x^2-1)<=0. So the roots are -2,-1,1, and 2 and by testing a few values you can see that the inequality holds from -2 to -1 and from 1 to 2.

]]> By: David http://www.ephblog.com/2009/11/09/green-chicken/#comment-74552 David Tue, 10 Nov 2009 01:16:22 +0000 http://www.ephblog.com/?p=24246#comment-74552 Whoops. This is wrong because I don't account for the fact that averages don't care about order, so, the combination A and B is the same as the combination B and A. So, I need something like: [latex]100\choose{2}[/latex] [latex]\frac{100!}{98!2!} = 100*99/2 = 4950[/latex] Whoops. This is wrong because I don’t account for the fact that averages don’t care about order, so, the combination A and B is the same as the combination B and A. So, I need something like:


100\choose{2}


\frac{100!}{98!2!} = 100*99/2 = 4950

]]> By: David http://www.ephblog.com/2009/11/09/green-chicken/#comment-74549 David Mon, 09 Nov 2009 17:32:18 +0000 http://www.ephblog.com/?p=24246#comment-74549 4) Let me try b first. You want to have every possible pair of numbers produce a unique average. The number of different pairs is just [latex]100 * 99 = 9900[/latex] It is easy to have a set of numbers that produces this many pairs, just ensure that no two pairs sum to the same total. There are plenty of sets that have this feature. Easiest (for me) ensure that the numbers are of radically different sizes). So, how about: [latex]10^n[/latex] for [latex]n = {1, \dots, 100}[/latex] 4) Let me try b first.

You want to have every possible pair of numbers produce a unique average. The number of different pairs is just


100 * 99 = 9900

It is easy to have a set of numbers that produces this many pairs, just ensure that no two pairs sum to the same total. There are plenty of sets that have this feature. Easiest (for me) ensure that the numbers are of radically different sizes). So, how about:


10^n
for
n = {1, \dots, 100}

]]> By: David http://www.ephblog.com/2009/11/09/green-chicken/#comment-74547 David Mon, 09 Nov 2009 17:25:46 +0000 http://www.ephblog.com/?p=24246#comment-74547 dm: Thanks! I did forget to check the endpoints. I realize that this should be obvious, but how did you determine that the endpoints are 2 and -2? (My 8th grader daughter is just starting to do math like this, so I need to get a clue!) dm: Thanks! I did forget to check the endpoints. I realize that this should be obvious, but how did you determine that the endpoints are 2 and -2? (My 8th grader daughter is just starting to do math like this, so I need to get a clue!)

]]> By: dm '10 http://www.ephblog.com/2009/11/09/green-chicken/#comment-74545 dm '10 Mon, 09 Nov 2009 17:20:06 +0000 http://www.ephblog.com/?p=24246#comment-74545 David: setting derivative to zero gives you the points where the function "turns around" (i.e. local maxima and minima), but to find a maximum you also need to check the edges of the domain in case there's some point at which the function keeps increasing without turning around. In this case the allowed answers are [-2,-1] and [1, 2], so you'd need to check 2 and -2 in addition to 1 and -1. It turns out that the maximum value occurs at f(2)=22. Anyway, you're right that that particular problem isn't terribly difficult - it's all fairly straightforward calculation and just remembering to check all the possibilities. Some of the others have more interesting tricks. David: setting derivative to zero gives you the points where the function “turns around” (i.e. local maxima and minima), but to find a maximum you also need to check the edges of the domain in case there’s some point at which the function keeps increasing without turning around. In this case the allowed answers are [-2,-1] and [1, 2], so you’d need to check 2 and -2 in addition to 1 and -1. It turns out that the maximum value occurs at f(2)=22.

Anyway, you’re right that that particular problem isn’t terribly difficult – it’s all fairly straightforward calculation and just remembering to check all the possibilities. Some of the others have more interesting tricks.

]]> By: Ronit http://www.ephblog.com/2009/11/09/green-chicken/#comment-74544 Ronit Mon, 09 Nov 2009 17:17:33 +0000 http://www.ephblog.com/?p=24246#comment-74544 I was a miserable failure at math, but I did enable latex on ephblog a while ago when we did <a href="http://www.ephblog.com/2009/08/06/play-with-tiles/" rel="nofollow">this thread</a>... so feel free to post your work here. syntax: <strong>[latex]</strong>e^{\i \pi} + 1 = 0<strong>[/latex]</strong> will display as... [latex]e^{\i \pi} + 1 = 0[/latex] I was a miserable failure at math, but I did enable latex on ephblog a while ago when we did this thread… so feel free to post your work here.

syntax:
[latex]e^{\i \pi} + 1 = 0[/latex]
will display as…

e^{\i \pi} + 1 = 0

]]> By: David http://www.ephblog.com/2009/11/09/green-chicken/#comment-74543 David Mon, 09 Nov 2009 17:12:10 +0000 http://www.ephblog.com/?p=24246#comment-74543 It would be the <b>geekiest EphBlog thread ever</b> if we started to work this out together. So, let's go! 3) This can't be as simple as X = -1, can it? It has been a long time since I took calculus, but don't you just take the derivative and set it equal to zero? That gives you X as either 1 or -1. Both work in the restricted set of allowed answers, so then you just plug in and see which yields the biggest answer. Surely the problem is much harder then that . . . It would be the geekiest EphBlog thread ever if we started to work this out together. So, let’s go!

3) This can’t be as simple as X = -1, can it? It has been a long time since I took calculus, but don’t you just take the derivative and set it equal to zero? That gives you X as either 1 or -1. Both work in the restricted set of allowed answers, so then you just plug in and see which yields the biggest answer.

Surely the problem is much harder then that . . .

]]> By: nuts http://www.ephblog.com/2009/11/09/green-chicken/#comment-74541 nuts Mon, 09 Nov 2009 17:03:06 +0000 http://www.ephblog.com/?p=24246#comment-74541 Ronit here's an <a href="http://i279.photobucket.com/albums/kk143/nfsagan/Green-Chicken-contest-2009.jpg" rel="nofollow">image</a> of the PDF if you want to post it. </a> Ronit here’s an image of the PDF if you want to post it.

]]> By: JeffZ http://www.ephblog.com/2009/11/09/green-chicken/#comment-74540 JeffZ Mon, 09 Nov 2009 16:43:23 +0000 http://www.ephblog.com/?p=24246#comment-74540 Good stuff. The Green Chicken certainly belongs on any list of the quirkiest / coolest Eph traditions (Mountain Day, The Walk, Trivia, etc. ...). Good stuff. The Green Chicken certainly belongs on any list of the quirkiest / coolest Eph traditions (Mountain Day, The Walk, Trivia, etc. …).

]]> By: Diana http://www.ephblog.com/2009/11/09/green-chicken/#comment-74539 Diana Mon, 09 Nov 2009 16:35:08 +0000 http://www.ephblog.com/?p=24246#comment-74539 Steve Miller is also a professor. Steve Miller is also a professor.

]]> By: David http://www.ephblog.com/2009/11/09/green-chicken/#comment-74538 David Mon, 09 Nov 2009 16:20:58 +0000 http://www.ephblog.com/?p=24246#comment-74538 1) Shout out to one of my former interns (Hai!) for participating in this contest, although, to judge by the pictures, he was there in 2008 (when Williams lost) and not in 2009 . . . 2) All good Ephs who want to avoid any "uncomfortable learning" today should not discuss the gender/racial breakdown of the participants . . . 1) Shout out to one of my former interns (Hai!) for participating in this contest, although, to judge by the pictures, he was there in 2008 (when Williams lost) and not in 2009 . . .

2) All good Ephs who want to avoid any “uncomfortable learning” today should not discuss the gender/racial breakdown of the participants . . .

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