1. ## Problem 37

1)Let $\displaystyle A,B\in \mathbb{R}^+$. Define $\displaystyle a_0=A$, $\displaystyle a_1=B$ and $\displaystyle a_n = a_{n-1}+a_{n-2} \mbox{ for }n\geq 2$. Find the radius of convergence of $\displaystyle \sum_{n=0}^{\infty}a_nx^n$.

The next two problems are for the younger kids, so give them a chance.

2)Let $\displaystyle S$ be a set (non-empty) of finite terms. A "partition" is breaking the set into two sets (non-empty) so that together they have the all the elements of $\displaystyle S$ but none of eachother. For example, let $\displaystyle S=\{1,2,3,4,5\}$ then $\displaystyle \{1,2,3\} \mbox{ and }\{4,5\}$ are partitions. But $\displaystyle \{1,2,3,4\} \mbox{ and }\{4,5\}$ are not. Say that $\displaystyle S$ has $\displaystyle n$ elements. How many different paritions are there in terms of $\displaystyle n$?

3)Given an $\displaystyle 8\times 8$ checkerboard what is the maximum number of checkers which can be placed so that no two are adjacent. Prove your answer. "Adjacent" means either horizontally or veritcally next to eachother not diagnolly.

2. #1) the initial values of $\displaystyle a_0$ and $\displaystyle a_1$ do not matter. you just need to notice that $\displaystyle \lim_{n\to\infty}\frac{a_n}{a_{n-1}}=\phi=\frac{1+\sqrt{5}}{2}$
so to find the radius we just use the ratio test and get

$\displaystyle \lim_{n\to\infty}\left|\frac{x^na_n}{x^{n-1}a_{n-1}}\right|=\lim_{n\to\infty}\left|x\left(1+\frac{a _{n-2}}{a_{n-1}}\right)\right|=|x|\left(1+\frac{1}{\phi}\right) <1$

so $\displaystyle |x|<\frac{1}{1+\frac{1}{\phi}}$ which makes the radius of convergence $\displaystyle \frac{1}{1+\frac{1}{\phi}}=\frac{\phi}{1+\phi}$

are the partitions $\displaystyle \{1,2,3\}\cup\{4,5\}$ and $\displaystyle \{4,5\}\cup\{1,2,3\}$ considered to be different? this will affect the solution. i however assume that they are not.

4. Originally Posted by putnam120
are the partitions $\displaystyle \{1,2,3\}\cup\{4,5\}$ and $\displaystyle \{4,5\}\cup\{1,2,3\}$ considered to be different? this will affect the solution. i however assume that they are not.
No they are not considered different maokid7, give the younger kids a chance.

For #1 it would be nice if you can give a full proof.

5. well ill start by proving my first assertion $\displaystyle \lim_{n\to\infty}\frac{a_n}{a_{n-1}}=\phi$

Let $\displaystyle L=\lim_{n\to\infty}\frac{a_n}{a_{n-1}}$. Then after expanding $\displaystyle a_n$ by the recursion we have

$\displaystyle L=\lim_{n\to\infty}\frac{a_{n-1}+a_{n-2}}{a_{n-1}}=\lim_{n\to\infty}1+\frac{a_{n-2}}{a_{n-1}}$. Notice that the second term is just the reciprocal of the original limit so we solve

$\displaystyle L=1+\frac1L\Longrightarrow L^2-L-1=0\Longrightarrow L=\frac{1\pm\sqrt{5}}{2}$. We choose the positive solution because all the terms in the sequence are obviously positive and thus a negative ratio is not possible.

6. Originally Posted by putnam120
well ill start by proving my first assertion $\displaystyle \lim_{n\to\infty}\frac{a_n}{a_{n-1}}=\phi$

Let $\displaystyle L=\lim_{n\to\infty}\frac{a_n}{a_{n-1}}$. Then after expanding $\displaystyle a_n$ by the recursion we have

$\displaystyle L=\lim_{n\to\infty}\frac{a_{n-1}+a_{n-2}}{a_{n-1}}=\lim_{n\to\infty}1+\frac{a_{n-2}}{a_{n-1}}$. Notice that the second term is just the reciprocal of the original limit so we solve

$\displaystyle L=1+\frac1L\Longrightarrow L^2-L-1=0\Longrightarrow L=\frac{1\pm\sqrt{5}}{2}$. We choose the positive solution because all the terms in the sequence are obviously positive and thus a negative ratio is not possible.
But you are assuming the limit exists. First show that it exists.

7. First we show that the sequence is bounded.

$\displaystyle \left|\frac{a_n}{a_{n-1}}\right|\Longrightarrow 1<\left|1+\frac{a_{n-2}}{a_{n-1}}\right|<2$ since each term in the sequence is obviously larger than the previous term.

now define $\displaystyle R_n=\frac{a_n}{a_{n-1}}$ and look at the ratio $\displaystyle \frac{R_{n+1}}{R_n}=\frac{a_{n+1}a_{n-1}}{a_n^2}$ which is just $\displaystyle \frac{a_n^2-1}{a_n^2}<1$. (easy induction proof ill leave the verification to the reader, but if this were for some competition i would include the proof ) so the sequence is monotonically decreasing.

and every bounded monotonic sequence converges.

8. Originally Posted by ThePerfectHacker
3)Given an $\displaystyle 8\times 8$ checkerboard what is the maximum number of checkers which can be placed so that no two are adjacent. Prove your answer. "Adjacent" means either horizontally or veritcally next to eachother not diagnolly.
That simply means that all the pieces should either be on the white squares or black squares only. Seeing as the amount of black squares is equal to the amount of white squares, then that means the maximum amount of pieces which can be placed is: $\displaystyle \frac{1}{2} (8 \times 8) = 32 \ pieces$

9. All these three problems are my own I hope you liked them.
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Maokid7 solved the first problem the trick here is to use the ratio test for evaluating radius of convergence for infinite series. The link I gave show that no matter what two possitive numbers are chosen thier ratio is the divine proportion. So the radius is $\displaystyle 1/\phi$.
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The second problem requires the knowledge combinations formula. Rieview them!

Now for a set $\displaystyle S$ we can constrct a partion in the following way. For simplicity let us say $\displaystyle S=\{ a,b,c \}$ and show how we can construct partitions. Let $\displaystyle P_1$ and $\displaystyle P_2$ be two partitions. So we cannot have $\displaystyle P_1 = \{ a,b,c\}$ then that would mean $\displaystyle P_2 = \{ \}$ (empty set) which is against the rules of the problem. Similarly $\displaystyle P_2 \not = \{a,b,c\}$. However, let us focus on $\displaystyle P_1$ only. Because once we know $\displaystyle P_1$ then $\displaystyle P_2$ is completely determined. For example, if $\displaystyle P_1 = \{ a \}$ then $\displaystyle P_2 = \{ b,c\}$, this is what I mean by being completely determined. So let us count the number of different ways of choosing elements for $\displaystyle P_1$, because again once I know that we know what $\displaystyle P_2$ is. We cannot chose all the elements at once as explained above. But we can chose $\displaystyle 2$ elements for $\displaystyle P_1$ the number of ways this can be done is $\displaystyle {3\choose 2} = 3$. And these are:
$\displaystyle P_1 = \{ a , b\} \mbox{ and }P_2=\{c\}$
$\displaystyle P_1=\{ b, c\} \mbox{ and }P_2 = \{a \}$
$\displaystyle P_1=\{ a, c\}\mbox{ and }P_3 = \{b \}$.
Now let us count the number of ways choosing $\displaystyle 1$ element for $\displaystyle P_1$ there are $\displaystyle {3\choose 1}=3$ ways. And these are:
$\displaystyle P_1 = \{a\} \mbox{ and }P_2=\{b,c\}$
$\displaystyle P_2 = \{b\} \mbox{ and }P_2 =\{a,c\}$
$\displaystyle P_3 = \{c\} \mbox{ and }P_3 = \{a,b\}$
So in total there are $\displaystyle {3\choose 1}+{3\choose 2} = 6$ ways of creating ordered partitions. What do I mean by "ordered" I mean we make a distinction between which partition is first, $\displaystyle P_1$, and which partition is second, $\displaystyle P_2$. The important realization is that I overcounted the unordered partitions twice. Because these play symettric roles in this problem. Thus, the answer should be divided by two to give us $\displaystyle 3$.

Okay, since we understand this easy example let us generalize it for large values $\displaystyle n$ where $\displaystyle n$ is the number of elements in $\displaystyle S$. If you follow everything what I did you will agree that the number of ordered partitions is:
$\displaystyle {n\choose 1}+{n\choose 2}+...+{n\choose {n-1}}$.
And the reason why I start with $\displaystyle 1$ and go up to $\displaystyle n-1$ is because as I explained otherwise if we use all the elements in one partition then the remaining partition must be the empty set which we do not want. Do you realize that sum? Do you know the fabulous identity that,
$\displaystyle {n\choose 0}+{n\choose 1}+...+{n\choose n} = 2^n$.
But, $\displaystyle {n\choose 0} = {n\choose n} = 1$.
Thus, the total number of ordered partitions is,
$\displaystyle {n\choose 1}+...+{n\choose {n-1}} = 2^n - 2$.
Divide this number by two to get unordered partitions which is,
$\displaystyle \boxed{ 2^{n-1} - 1 }$.
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The last problem is easy but I posted it to come up with an elegant proof. Hopefully you will learn from this solution. We will use the "pigeonhole principle". It says given $\displaystyle n$ pigeonholes and $\displaystyle m$ holes, if we place these pigeons into the holes there there exists at least one hole containing $\displaystyle 2$ pigeons. Think about it, a very trivial observation, but an important one. So this is what we will do (you will see why). Give each tile in the $\displaystyle 8\times 8$ board a number. Let the upper left one be $\displaystyle 1$ to its right $\displaystyle 2$ to its right $\displaystyle 3$ and go on up to $\displaystyle 8$. Then $\displaystyle 9$ is directly below $\displaystyle 1$ and go until to the right until $\displaystyle 18$. So cover up the entire board with these $\displaystyle 64$ numbers. Now define "blocks". Block $\displaystyle \#1$ is the pair of $\displaystyle 1,2$ tiles. Block $\displaystyle \#2$ is the pair of $\displaystyle 3,4$ tiles. And so on. So there are $\displaystyle 32$ different blocks. Okay, I argue that if I place $\displaystyle 33$ peices on the checkerboard then two of them must be adjancet. To see this note that we have $\displaystyle 32$ blocks. So two of the peices end up in the same block since $\displaystyle 33>32$. This means those two pieces must be adjacent to each other. This shows $\displaystyle 33$ is too much. Now show it is possible to have $\displaystyle 32$ non-adjacent peices which will show $\displaystyle 32$ is the maximum number.

10. Just one thing TPH... I don't know about the American and English schools, but here in SA we don't do partitions and the combinatorics formula in school. So that's a completely unknown section of mathematics for someone like me. I'll probably only do those things next year when I get to university.

But thanks for the elementary probs, they're cool

11. Originally Posted by janvdl
Just one thing TPH... I don't know about the American and English schools, but here in SA we don't do partitions and the combinatorics formula in school. So that's a completely unknown section of mathematics for someone like me. I'll probably only do those things next year when I get to university.
That is a such a bad excuse. If you really like math you do it on your own. And you learn stuff by yourself not taught in school.

12. Originally Posted by ThePerfectHacker
That is a such a bad excuse. If you really like math you do it on your own. And you learn stuff by yourself not taught in school.
Look I would have studied it if i had the time... I write my finals in two weeks...

There's no time for doing unnecessary work at this time.

13. I would have to agree with TPH on this. I have found that most of the mathematics I have actually learned has been from outside the classroom even at college. If you are truly interested in something you would want to go beyond what the class covers and try to obtain a deeper understanding.
Also you tend to retain the information better if you had to go out on your own and "discover" it.

14. Originally Posted by putnam120
If you are truly interested in something you would want to go beyond what the class covers and try to obtain a deeper understanding...
I've done that with Calculus, but I DONT have the time to go deeper into anything now.

I'm two weeks away from writing the most important exam of my life.

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