Problem 26

**ThePerfectHacker** · June 10th 2007, 07:56 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** · June 13th 2007, 09:45 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** · June 13th 2007, 10:52 AM

Originally Posted by Singular

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** · June 13th 2007, 03:25 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** · June 13th 2007, 04:07 PM

Originally Posted by Rebesques

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** · June 13th 2007, 04:11 PM

I see what you mean now

But I wouldn't ask a blonde out for chess

**ThePerfectHacker** · June 18th 2007, 07:18 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** · June 19th 2007, 03:11 AM

Originally Posted by ThePerfectHacker

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

**ThePerfectHacker** · June 19th 2007, 06:01 AM

Originally Posted by topsquark

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.

**CaptainBlack** · June 19th 2007, 09:09 AM

Originally Posted by ThePerfectHacker

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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