Problem 26

ThePerfectHacker · #1 $Old$ June 10th, 2007, 09:56 PM

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.

Singular · #2 $Old$ June 13th, 2007, 11:45 AM

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"

ThePerfectHacker · #3 $Old$ June 13th, 2007, 12:52 PM

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".

Rebesques · #4 $Old$ June 13th, 2007, 05:25 PM

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

ThePerfectHacker · #5 $Old$ June 13th, 2007, 06:07 PM

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).

Rebesques · #6 $Old$ June 13th, 2007, 06:11 PM

I see what you mean now

But I wouldn't ask a blonde out for chess

ThePerfectHacker · #7 $Old$ June 18th, 2007, 09:18 PM

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.

topsquark · #8 $Old$ June 19th, 2007, 05:11 AM

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