Problem 27

ThePerfectHacker · #1 $Old$ June 18th, 2007, 09:29 PM

1)In a middle of a battlefield there is an odd number of Soviet soldiers ( $n>1$ ). Each one is standing a different distance away from anyone else. Also each soldier has a Colt Python .357 Magnum Revolver.* The soldiers decide to play a game. Each one is going to shot his nearest oppenent. At a signal each Soviet shots instantenously at his target. Show that there will be one Soviet still standing alive.
(Assume that the Soviets had nothing to drink and hence they have perfect aim). What happens if there is an even number?

2)Let $(s_n)$ be a convergent sequence with $\lim \ s_n = s$ . Show that the sequence $\left( 1 + \frac{s_n}{n} \right)^n$ is also convergent and furthermore, $\lim \ \left( 1+\frac{s_n}{n} \right)^n = e^s$ .

*)Look how beautiful that gun is!

red_dog · #2 $Old$ June 24th, 2007, 12:37 AM

2) If $(s_n)$ is convergent then $(s_n)$ is bounded and $\lim_{n\to\infty}\frac{1}{n}=0$ , so $\lim_{n\to\infty}\frac{s_n}{n}=0$ .
Then $\displaystyle \lim_{n\to\infty}\left(1+\frac{s_n}{n}\right)^n=\lim_{n\to\infty}\left[\left(1+\frac{s_n}{n}\right)^{\frac{n}{s_n}}\right]^{\frac{s_n}{n}\cdot n}=e^{\lim_{n\to\infty}s_n}=e^s$ .

galactus · #3 $Old$ June 24th, 2007, 08:47 AM

Quote:

*)Look how beautiful that gun is!

I know, I own one.

ThePerfectHacker · #4 $Old$ June 24th, 2007, 10:33 AM

Quote:

Originally Posted by galactus $View Post$

I know, I own one.

Wow!

But if I had to chose to own guns I would chose:
1)M4A1 Carbine
2)M1 Garand
3)Maschinengewehr 42 (Another classic. If only the good guys had this instead of the .30 Cal the war might have been different

)

I have an interest in World War 2 weapons because a lot of the computer games I play and played have these. Here is a nice list.

I think the Germans did have the best and the coolest weapons. I also love the British weapons. The Americans were okay. But the Soviets were a complete embarassement

.

---
One of my best friends is a Redneck (which is why I make fun of Rednecks). And he has a Magum .357 as well as a Magnum .44 and a Remington Shotgun. One of these days we will have some fun

.

Jhevon · #5 $Old$ June 24th, 2007, 11:29 AM

Quote:

Originally Posted by ThePerfectHacker $View Post$

Wow!

But if I had to chose to own guns I would chose:
1)M4A1 Carbine
2)M1 Garand
3)Maschinengewehr 42 (Another classic. If only the good guys had this instead of the .30 Cal the war might have been different

)

I have an interest in World War 2 weapons because a lot of the computer games I play and played have these. Here is a nice list.

I think the Germans did have the best and the coolest weapons. I also love the British weapons. The Americans were okay. But the Soviets were a complete embarassement

.

---
One of my best friends is a Redneck (which is why I make fun of Rednecks). And he has a Magum .357 as well as a Magnum .44 and a Remington Shotgun. One of these days we will have some fun

.

Those are rifles and machine guns, why would you want to own such guns? Someone after you?

ThePerfectHacker · #6 $Old$ June 24th, 2007, 10:24 PM

1)This was a competition problem, though it was not posed exactly as I posted it. We will argue by induction, first for $n=3$ this is clearly true. Now say it is true for $2n-1$ and we want to show it is true for $2n+1$ people. Out of all the distances amoing the soldier consider the minimal distance. This distance is determined by two soldiers, call them $S_1$ and $S_2$ . By the problem $S_1$ must shoot toward $S_2$ and $S_2$ must shoot toward $S_1$ . Now there are two possibilities. 1)Somebody shoots at them (those two soldiers). 2)Somebody does not shoot at them. If #1 then one of those soldiers is hit twice, so it is impossible for all the soldiers to be hit because the number of people exceeds the number of bullets. And if #2then we have $2n-1$ people shooting amongst themselves. Hence it is as if those two soldiers have nothing to do with them. And hence by induction someone remains alive.

2)I saw this amazing sequence problem on a forum. He is how I would do it:
Infinite series is the answer here.
Since $(s_n)$ is convergent then $|s_n|\leq A$ for some $A>0$ . Choose $N \in \mathbb{N}$ so large that $\frac{A}{N} < 1$ . Then for all $n\geq N$ we have:
$\left( 1 + \frac{s_n}{n} \right)^n = e^{n \ln \left(1+\frac{s_n}{n} \right)} = e^{ s_n -\frac{s^2_n}{n}+...}$ .
Now,
$\lim \left( s_n - \frac{s_n^2}{n} - ... \right) = s_n - 0 + 0 ... = s_n$ (by Uniform Convergence) since $\frac{|s_n|}{n^m} \leq \frac{A}{n^m} \to 0$ .
Since $e^x$ is continous by definition we see that this converges to $e^s$ .

Jhevon · #7 $Old$ June 24th, 2007, 10:37 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

1)This was a competition problem, though it was not posed exactly as I posted it. We will argue by induction, first for $n=3$ this is clearly true. Now say it is true for $2n-1$ and we want to show it is true for $2n+1$ people. Out of all the distances amoing the soldier consider the minimal distance. This distance is determined by two soldiers, call them $S_1$ and $S_2$ . By the problem $S_1$ must shoot toward $S_2$ and $S_2$ must shoot toward $S_1$ . Now there are two possibilities. 1)Somebody shoots at them (those two soldiers). 2)Somebody does not shoot at them. If #1 then one of those soldiers is hit twice, so it is impossible for all the soldiers to be hit because the number of people exceeds the number of bullets. And if #2then we have $2n-1$ people shooting amongst themselves. Hence it is as if those two soldiers have nothing to do with them. And hence by induction someone remains alive.

2)I saw this amazing sequence problem on a forum. He is how I would do it:
Infinite series is the answer here.
Since $(s_n)$ is convergent then $|s_n|\leq A$ for some $A>0$ . Choose $N \in \mathbb{N}$ so large that $\frac{A}{N} < 1$ . Then for all $n\geq N$ we have:
$\left( 1 + \frac{s_n}{n} \right)^n = e^{n \ln \left(1+\frac{s_n}{n} \right)} = e^{ s_n -\frac{s^2_n}{n}+...}$ .
Now,
$\lim \left( s_n - \frac{s_n^2}{n} - ... \right) = s_n - 0 + 0 ... = s_n$ (by Uniform Convergence) since $\frac{|s_n|}{n^m} \leq \frac{A}{n^m} \to 0$ .
Since $e^x$ is continous by definition we see that this converges to $e^s$ .

what about red_dog's answer to problem 2? Is it valid?

ThePerfectHacker · #8 $Old$ June 25th, 2007, 08:35 AM

Quote:

Originally Posted by Jhevon $View Post$

what about red_dog's answer to problem 2? Is it valid?

I really hope he would be formal (like I was).