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Math Help - Problem 26

  1. #1
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    Problem 26

    This one is not so bad.

    1)Let f(x) be an n-th degree* polynomial function such that f(x)\geq 0. Define the function g(x) as:
    g(x) = f(x) + f'(x) + f''(x) + ... + f^{(n)}(x). Show that g(x)\geq 0.

    2)What is the least number of moves that a player can make to give a checkmate?

    *)And the condition that f(x) \not \equiv 0 because the degree of a zero polynomial is not defined. The degree of a constant non-zero polynomial is defined to be zero. However, some authors in field theory differ on their defintions of the degree of the zero polynomial. Some define it to be -1 and other to be \infty. The way I learned it the zero polynomial had an undefined degree. This is why I make such a comment just in case you spotted the mistake in my first sentence.
    Last edited by ThePerfectHacker; June 11th 2007 at 11:57 AM.
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    Junior Member Singular's Avatar
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    Question number 2

    4 moves, if the black player is very dumb

    Example :

    White Black
    1. e2-e3 b7-b6
    2. f1-c4 b8-a6
    3. d1-h5 b6-b5
    4. h5xf7 "checkmate"
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    Quote Originally Posted by Singular View Post
    Question number 2

    4 moves, if the black player is very dumb

    Example :

    White Black
    1. e2-e3 b7-b6
    2. f1-c4 b8-a6
    3. d1-h5 b6-b5
    4. h5xf7 "checkmate"
    I am sorry, but that answer is not correct.
    Apparently the player can be even "dummer".
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    Super Member Rebesques's Avatar
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    I think its three moves, using queen+bishop. But if my opponent blocks this, I have lost the game already I totally suck at chess
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    Quote Originally Posted by Rebesques View Post
    I think its three moves, using queen+bishop. But if my opponent blocks this, I have lost the game already I totally suck at chess
    In fact it can be done in fewer moves!!! (If you are playing against a blonde).
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    Super Member Rebesques's Avatar
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    I see what you mean now But I wouldn't ask a blonde out for chess
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    1)I have found this problem in a book. Say f(x) = a_0+a_1x+...+a_nx^n. We make the following observations. Since f(x) \geq 0 it must mean that n is even and a_n>0. Since g(x) = f(x)+...+f^{(n)}(x) it means that g(x) is a polynomial of even degree and the leading coefficient is a_n>0.
    Therefore, \lim_{x\to \infty} g(x) = \lim_{x\to -\infty} g(x) = + \infty. This tells us that g(x) must have a minimum value (since it is a continous function). Say c is the point where g(x) is mimimal. Then it means that g'(c) = 0. But we know that g'(x) = f'(x) + f''(x) + ... +f^{(n)}(x) + f^{(n+1)}(x) = g(x) - f(x) because f^{(n+1)}(x) = 0 since the degree of f(x) is n. Thus, by what just stated we have that g'(c) = g(c) - f(c) = 0. Thus, g(c) = f(c) \geq 0 since f(x)\geq 0 by hypothesis. So if x is any real number then g(x) \geq g(c) \geq 0 for g(c) is the smallest value of the function.

    2)A long time ago somebody challenged me to find the shortest checkmate. The following is my solution. The strange think is that it is played by black rather than white!
    The are several version by the idea is the same.
    WHITE] Play Queens Knight any way.
    BLACK] Plays Kings Pawns
    WHITE] Plays Sicilian Defense on Kings Side (Move Bishop Pawn)
    BLACK] Plays a Checkmate with a Queen.
    So the Black player wins in just two moves.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    1)I have found this problem in a book. Say f(x) = a_0+a_1x+...+a_nx^n. We make the following observations. Since f(x) \geq 0 it must mean that n is even and a_n>0. Since g(x) = f(x)+...+f^{(n)}(x) it means that g(x) is a polynomial of even degree and the leading coefficient is a_n>0.
    Therefore, \lim_{x\to \infty} g(x) = \lim_{x\to -\infty} g(x) = + \infty. This tells us that g(x) must have a minimum value (since it is a continous function). Say c is the point where g(x) is mimimal. Then it means that g'(c) = 0. But we know that g'(x) = f'(x) + f''(x) + ... +f^{(n)}(x) + f^{(n+1)}(x) = g(x) - f(x) because f^{(n+1)}(x) = 0 since the degree of f(x) is n. Thus, by what just stated we have that g'(c) = g(c) - f(c) = 0. Thus, g(c) = f(c) \geq 0 since f(x)\geq 0 by hypothesis. So if x is any real number then g(x) \geq g(c) \geq 0 for g(c) is the smallest value of the function.

    2)A long time ago somebody challenged me to find the shortest checkmate. The following is my solution. The strange think is that it is played by black rather than white!
    The are several version by the idea is the same.
    WHITE] Play Queens Knight any way.
    BLACK] Plays Kings Pawns
    WHITE] Plays Sicilian Defense on Kings Side (Move Bishop Pawn)
    BLACK] Plays a Checkmate with a Queen.
    So the Black player wins in just two moves.
    This is just essentially the "Fool's Mate."

    -Dan
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    Quote Originally Posted by topsquark View Post
    This is just essentially the "Fool's Mate."
    So it has a name. I never had a chance to do it on anyone. I had a chance doing that 4 move checkmate many times, but never this one.
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker View Post
    So it has a name. I never had a chance to do it on anyone. I had a chance doing that 4 move checkmate many times, but never this one.
    Which also has a name "scholar's mate"

    RonL
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