The problem of the week of purdue university seems a little boring on last few problems. Here is another website from The Mathematics Department of the Hong Kong University of Science and Technology
http://www.math.ust.hk/excalibur/, which is more challenge.
Monday, November 28, 2011
Wednesday, November 9, 2011
Problem 11 fall 2011
The problem 11 of POW is quite interesting.
The system can be simplified.
Let $x_i$ represent the ith man, $d(x_i, x_j)$ is the distance between ith and jth man. Because all men shoot at the same time, so if $d(x_i, x_j)< \min \{ d(x_i,x_k), d(x_j,x_k)\}$, then $x_i, x_j$ die at the same time. So the each pair of $x_i, x_j$ which isolate far away from the others will be both dead at the same time, that means these kinds of pair men can be canceled from the whole. Beside these, there is another kind of situation. That is $d(x_i,x_{i+1})$<$d(x_{i+1},x_{i+2})$<$d(x_{i+2},x_{i+3})$ like a chain or like the domino. Here the first two men will shoot each other, only the last man will survive in a chain.
The question is, if there are two chain remain, what will happen? All last one of each chain will remain alive.
If n is odd number, simplified the whole, then get some chains. Then all the last of chain will remain alive. Since n is odd number, so there at least one chain in the whole (may be need detail discussion).
If n is even number, all men are paired isolated, then n men die without any survival.
The system can be simplified.
Let $x_i$ represent the ith man, $d(x_i, x_j)$ is the distance between ith and jth man. Because all men shoot at the same time, so if $d(x_i, x_j)< \min \{ d(x_i,x_k), d(x_j,x_k)\}$, then $x_i, x_j$ die at the same time. So the each pair of $x_i, x_j$ which isolate far away from the others will be both dead at the same time, that means these kinds of pair men can be canceled from the whole. Beside these, there is another kind of situation. That is $d(x_i,x_{i+1})$<$d(x_{i+1},x_{i+2})$<$d(x_{i+2},x_{i+3})$ like a chain or like the domino. Here the first two men will shoot each other, only the last man will survive in a chain.
The question is, if there are two chain remain, what will happen? All last one of each chain will remain alive.
If n is odd number, simplified the whole, then get some chains. Then all the last of chain will remain alive. Since n is odd number, so there at least one chain in the whole (may be need detail discussion).
If n is even number, all men are paired isolated, then n men die without any survival.
Thursday, November 3, 2011
Problem 10 fall 2011
Problem 9 is a little hard for me to solve, because I did not touch the series problem for at least 4 years. So I do problem 10 first.
There are lots of examples can be list here like this one. Let $a=\sqrt{3}, b=\log_{\sqrt{3}}2$, then
$$a^b=\sqrt{3} ^\log_{\sqrt{3}}2$$ $$ =2 $$
And it is easy know that $\sqrt{3}$ is a irrational number. The following to prove $\log_{\sqrt{3}}2$ is also a irrational number.
If it is not, then $\log_{\sqrt{3}}2 = \frac{p}{q}$, where $(p,q)=1, p,q \in \mathbb{N}$.
Therefore $\sqrt{3}^{p/q}=2$, which can be write as $\sqrt{3}^p=2^q$. Square both side, we get $3^q=4^p$. Since $p,q$ are both nature number ($>0$), but R.H.S. is an odd number, L.H.S is an even number. Then the assumtion faile. Then $\log_{\sqrt{3}}2$ is an irrational number.
There are lots of examples can be list here like this one. Let $a=\sqrt{3}, b=\log_{\sqrt{3}}2$, then
$$a^b=\sqrt{3} ^\log_{\sqrt{3}}2$$ $$ =2 $$
And it is easy know that $\sqrt{3}$ is a irrational number. The following to prove $\log_{\sqrt{3}}2$ is also a irrational number.
If it is not, then $\log_{\sqrt{3}}2 = \frac{p}{q}$, where $(p,q)=1, p,q \in \mathbb{N}$.
Therefore $\sqrt{3}^{p/q}=2$, which can be write as $\sqrt{3}^p=2^q$. Square both side, we get $3^q=4^p$. Since $p,q$ are both nature number ($>0$), but R.H.S. is an odd number, L.H.S is an even number. Then the assumtion faile. Then $\log_{\sqrt{3}}2$ is an irrational number.
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