# Problem 26

• Jun 10th 2007, 07:56 PM
ThePerfectHacker
Problem 26
This one is not so bad.

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

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

*)And the condition that $\displaystyle f(x) \not \equiv 0$ 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 $\displaystyle -1$ and other to be $\displaystyle \infty$. 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.
• Jun 13th 2007, 09:45 AM
Singular
Question number 2

4 moves, if the black player is very dumb :D

Example :

White Black
1. e2-e3 b7-b6
2. f1-c4 b8-a6
3. d1-h5 b6-b5
4. h5xf7 "checkmate"
• Jun 13th 2007, 10:52 AM
ThePerfectHacker
Quote:

Originally Posted by Singular
Question number 2

4 moves, if the black player is very dumb :D

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". :)
• Jun 13th 2007, 03:25 PM
Rebesques
I think its three moves, using queen+bishop. But if my opponent blocks this, I have lost the game already :o I totally suck at chess :o
• Jun 13th 2007, 04:07 PM
ThePerfectHacker
Quote:

Originally Posted by Rebesques
I think its three moves, using queen+bishop. But if my opponent blocks this, I have lost the game already :o I totally suck at chess :o

In fact it can be done in fewer moves!!! (If you are playing against a blonde).
• Jun 13th 2007, 04:11 PM
Rebesques
I see what you mean now :D But I wouldn't ask a blonde out for chess :p
• Jun 18th 2007, 07:18 PM
ThePerfectHacker
1)I have found this problem in a book. Say $\displaystyle f(x) = a_0+a_1x+...+a_nx^n$. We make the following observations. Since $\displaystyle f(x) \geq 0$ it must mean that $\displaystyle n$ is even and $\displaystyle a_n>0$. Since $\displaystyle g(x) = f(x)+...+f^{(n)}(x)$ it means that $\displaystyle g(x)$ is a polynomial of even degree and the leading coefficient is $\displaystyle a_n>0$.
Therefore, $\displaystyle \lim_{x\to \infty} g(x) = \lim_{x\to -\infty} g(x) = + \infty$. This tells us that $\displaystyle g(x)$ must have a minimum value (since it is a continous function). Say $\displaystyle c$ is the point where $\displaystyle g(x)$ is mimimal. Then it means that $\displaystyle g'(c) = 0$. But we know that $\displaystyle g'(x) = f'(x) + f''(x) + ... +f^{(n)}(x) + f^{(n+1)}(x) = g(x) - f(x)$ because $\displaystyle f^{(n+1)}(x) = 0$ since the degree of $\displaystyle f(x)$ is $\displaystyle n$. Thus, by what just stated we have that $\displaystyle g'(c) = g(c) - f(c) = 0$. Thus, $\displaystyle g(c) = f(c) \geq 0$ since $\displaystyle f(x)\geq 0$ by hypothesis. So if $\displaystyle x$ is any real number then $\displaystyle g(x) \geq g(c) \geq 0$ for $\displaystyle g(c)$ 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.
• Jun 19th 2007, 03:11 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
1)I have found this problem in a book. Say $\displaystyle f(x) = a_0+a_1x+...+a_nx^n$. We make the following observations. Since $\displaystyle f(x) \geq 0$ it must mean that $\displaystyle n$ is even and $\displaystyle a_n>0$. Since $\displaystyle g(x) = f(x)+...+f^{(n)}(x)$ it means that $\displaystyle g(x)$ is a polynomial of even degree and the leading coefficient is $\displaystyle a_n>0$.
Therefore, $\displaystyle \lim_{x\to \infty} g(x) = \lim_{x\to -\infty} g(x) = + \infty$. This tells us that $\displaystyle g(x)$ must have a minimum value (since it is a continous function). Say $\displaystyle c$ is the point where $\displaystyle g(x)$ is mimimal. Then it means that $\displaystyle g'(c) = 0$. But we know that $\displaystyle g'(x) = f'(x) + f''(x) + ... +f^{(n)}(x) + f^{(n+1)}(x) = g(x) - f(x)$ because $\displaystyle f^{(n+1)}(x) = 0$ since the degree of $\displaystyle f(x)$ is $\displaystyle n$. Thus, by what just stated we have that $\displaystyle g'(c) = g(c) - f(c) = 0$. Thus, $\displaystyle g(c) = f(c) \geq 0$ since $\displaystyle f(x)\geq 0$ by hypothesis. So if $\displaystyle x$ is any real number then $\displaystyle g(x) \geq g(c) \geq 0$ for $\displaystyle g(c)$ 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
• Jun 19th 2007, 06:01 AM
ThePerfectHacker
Quote:

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. :mad: I had a chance doing that 4 move checkmate many times, but never this one.
• Jun 19th 2007, 09:09 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
So it has a name. I never had a chance to do it on anyone. :mad: I had a chance doing that 4 move checkmate many times, but never this one.

Which also has a name "scholar's mate"

RonL