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.