When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Furthermore, please do not "hide" any portion of the solution. Please use LaTeX for posting solutions. Thanks.

We can prove the assertion with $n$ instead of $2002$ by induction on $n$:

Assume it’s proven for $n-1$. If $N\le 2^{n-1}$, then pick $n-1$ elements of the $n$-element set, and perform the construction on those, choosing all the white sets from within the chosen $n-1$-element set. If, on the other hand, $N>2^{n-1}$, then interchange the roles of black and white, and do the same as above, since then the number of  black sets would have to be $2^n-N<2^{n-1}$.

is this a plausible solution? I came up with it myself and I think there are some errors in it:

First let the set $ S$ be the set of the first 2002 natural numbers. Then define "special" subsets to be subsets that contain an increasing series of consecutive numbers up to 2002, each subset containing all the elements of the previous one:

{1}, {1, 2}, {1, 2, 3}, ..., {1, 2, 3, ...., 2002}

let the set $ K$ contain all such subsets. Next, define an "awesome subset pair" to be a pair of subsets whose union is in the set $ K$. Now consider parity. For the case when N is even and greater than 3(since the case where N= 0, 1, 2, and 3 are trivial), coloring an awesome subset pair plus its union and the null subset white gets you the smallest such even number of sets N, for which N is greater than 3. A subset in the set $ K$ may have many such "awesome subset pairs" , therefore, for relatively small values of N(e.g. N=4), you use a relatively "small"(as in a subset with little elements) in the set $ K$ and take its "awesome subset pairs".

for N=4:
{1,2}
{}
{1}
{2}

In this way you can keep picking subsets and coloring them white from the set $ K$ and their "awesome subset pairs" for increasing values of N. After the point where N is too large even for all the "awesome subset pairs" and all the subsets in the set $ K$ to hold, we simply use the same technique on the other subsets. First, define a subset to be "weird" if its elements are not consecutive in any way. (e.g. {1, 5, 178, 696}) Next define a series of finite sets $ A_{1}, A_{2}, ..., A_{i}$, for some number i, which each contain subsets which are defined to be "weirdly special", which means that each subset is "weird" and each subset after the first one contains all the elements of the previous one. Let these $ A_i$ sets contain the unions/combinations of the remaining subsets(all of which are "weird"). Then the "awesome subset pair" approach can once again be implemented and now the conditions can be filled for any value of N.

[b]2002 USAMO Problem 1.[/b] Let $ S$ be a set with $ 2002$ elements, and let $ N$ be an integer with $ 0 \leq N \leq 2^{2002}$. Prove that it is possible to color every subset of $ S$ either black or white so that the following conditions hold:

(a) the union of any two white subsets is white;
(b) the union of any two black subsets is black;
(c) there are exactly $ N$ white subsets.

[b]Official Solution.[/b]

We prove that this can be done for any $ n$-element set, where $ n$ is a positive integer. Let $ S_{n}\equal{}\left \{ 1,2,\ldots,n \right \}$ and an integer $ N$ with $ 0\leq N\leq 2^{n}$.

We induct on $ n$. The base case $ n\equal{}1$ is trivial. Assume that the desired coloring can be done to the subsets of set $ S_{n}\equal{}\left \{ 1,2,\ldots,n \right \}$ and the integer $ N_{n}$ with $ 0\leq N\leq 2^{n}$. We show that there is a desired coloring for set $ S_{n\plus{}1}\equal{}\left \{ 1,2,\ldots,n,n\plus{}1 \right \}$ and integer $ N$ with $ 0\leq N_{n\plus{}1}\leq 2^{n\plus{}1}$. We consider the following cases.

[b]Case (i).[/b] $ 0\leq N_{n\plus{}1}\leq 2^{n}$. Applying the induction hypothesis to $ S_{n}$ and $ N_{n}\equal{}N_{n\plus{}1}$, we get a coloring of all subsets of $ S_{n}$ satisfying the given conditions (a), (b) and (c). All uncolored subset of $ S_{n\plus{}1}$ contains element $ n\plus{}1$, we color all of them blue. It is not hard to see that this coloring of all subsets of $ S_{n\plus{}1}$ satisfying the given conditions (a), (b) and (c).

[b]Case (ii).[/b] $ 2^{n}\plus{}1\leq N_{n\plus{}1}\leq 2^{n\plus{}1}$. Applying the induction hypothesis to $ S_{n}$ and $ N_{n}\equal{}2^{n\plus{}1} \minus{} N_{n\plus{}1}$, we get a coloring of all subsets of $ S_{n}$ satisfying the given conditions (a), (b) and (c). All uncolored subset of $ S_{n\plus{}1}$ contains element $ n\plus{}1$, we color all of them blue. Finally, we switch the color of each subset: if it is blue now, we recolor it red; if it is red now, we recolor it blue. It is not hard to see that this coloring of all subsets of $ S_{n\plus{}1}$ satisfying the given conditions (a), (b) and (c).

Thus, our induction is complete.  :lol:

The above solution is nice, but the solution randomly uses the colors blue and red.
Here is a solution that is different from all the other solutions presented:
[hide="Solution"]
If $N = 0$, we can color every subset black. Similarly if $N = 2^{2002}$, we color every subset white.
Other wise we write the binary form of $N$, and the $1$'s be at the position $x_i$ for $0 \le i \le n$. We color a subset based on the largest element within the subset. If the largest element is an $x_i$, then we color it white. Otherwise, we color the subset black. Notice that this satisfies the first two properties. Also, this means that the total number of white subsets is the sum of $2^{x_i}$ for all $i$ because we simply choose a subset with element $x_i$ and elements from $(0,..,x_i-1)$. However, this is just $N$ because of the creation of $x_i$ from the binary representation of $N$.
[/hide]

Wow quantumbyte that is an absolutely amazing proof. Did you come up with it? How long did it take you? I knew it wasn't worth reading any of the other solutions.

So I fond the following while doing this problem and I was wondering whether anyone could help me rigorize it:
[hide="my work"]
Consider all the white subsets of such a coloring: there must exist one with a maximal size, $n$. (otherwise, if there were two with this size, they could be combined to form another, contradicting the maximality).
Now, notice that this maximal set can be broken into two disjoint subsets in $2^{n-1}-1$ ways.
[hide="example"]
Let the subset be $\{a,b,c,d\}$.  Then, it can be broken up as:
$\{a\}\;\;\;\{b,c,d\}$
$\{b\}\;\;\;\{a,c,d\}$
$\{c\}\;\;\;\{a,b,d\}$
$\{d\}\;\;\;\{a,b,c\}$
$\{a,b\}\;\;\;\{c,d\}$
$\{a,c\}\;\;\;\{b,d\}$
$\{a,d\}\;\;\;\{b,c\}$
for a total of $2^{4-1}-1=7$ different ways.  This can be proven with binomial summations (essentially $\sum_{i=1}^{\lfloor\frac{n}{2}\rfloor}{\dbinom{n}{i}}$).
[/hide]
Now, I noticed that, in order for the conditions 1 and 2 to hold, we must have that, of each of the aforementioned pairs of subsets, we must have that atleast one is white.  Finally, I noted, for the cases $n=4,5$ that there are choices of these subsets such that we can get exactly $2^{n-1}$ white subsets all the way to $2^n-1$ white subsets.

If I was able to rigorously prove that such choices existed for all $n$, $6\le n\le2002$, this would finish the problem.
[/hide]

If this is unclear, please let me know and I'll try to fix it :)

The solution of @quantumbyte is perfect :)

I will give a proof too which is essentially the same as the official solution.
But since I have problems with the phrasing, I would like suggestions :)

[hide="Solution"]
Define $T_n=(1,2, \cdots n)$ and $F_n$ the set of all it's subsets.
Call a colouring [i]'valid'[/i] of the subsets of $F_n$ is it satisfies the given conditions.
If there exists a [i]valid[/i] colouring (satisfying the given properties) for $n$ element subset for $N=N$, call the set of white coloured subsets $W_{n,N}$

[b]Claim:
[/b]For any $n \in Z^+$ and $0 \le N \le 2^n$, $W_{n,N}$ that is a [i]valid[/i] colouring exists.

[b]Proof:
[/b]I will prove this by induction on $n$.
The base case is trivial.
Now let the result be true for $n=k$, that is a [i]valid[/i] colouring exists for $0 \le N \le 2^k$
For $n=k+1$, by definition, $F_k \subset F_{k+1}$
If $0 \le N \le 2^k$
Then $W_{k+1,N}=W_{k,n}$
Let if possible there exist $A,B \in W_{k,n}$ such that $A \cup B \in F_{k+1}-W_{k+1,N}$
$\implies A \cup B \in F_k - W_{k,n}$ or $ A \cup B \in  F_{k+1}-F_k$
The first case is not possible by our induction hypothesis and the second is not possible since $k+1 \in  F_{k+1}-F_k$ but $k+1$ is not an element of $A \cup B$
Hence the result here

Due to symmetry in white and black.
For $2^k < N \le 2^{k+1}$
Set $W_{k+1,N}=F_{k+1}-W_{k+1,2^{k+1}-N}$

Hence the induction hypothesis is complete and the claim is true.

$\implies n=2002$ also satisfies the conditions and thus we are done :)
[/hide]

Pretty easy but very nice  :D
[hide=Solution]
Replace 2002 with any positive integer $k$. We will prove the desired result by induction on $k$. Our base case is trivial.

Now, suppose the conclusion holds for some $k>1$. Take our valid coloring of the $2^k$ subsets of $S$ where $0 \le N \le 2^k$ subsets are white. Add in the subsets with the $k+1$-st element by coloring each of the $2^k$ subsets containing the $k+1$-st element black. This means there are at least $2^k+1$ black subsets. Observe all white subsets do not contain the $k+1$-st element so conditions (a) and (c) hold by the inductive hypothesis. If two black subsets both do not contain the $k+1$-st elements then condition (b) holds by the inductive hypothesis again. If at least one of the black subsets contains the $k+1$-st element then the union of the subsets does too, but since all subsets with the $k+1$-st element are colored black condition (b) holds true. This proves the desired claim whenever $0 \le N \le 2^k$.

If $2^k+1 \le N \le 2^{k+1}$, then simply swap the colorings of the subsets from the above case, so all black subsets become white and all white subsets become black. Condition (c) clearly holds true. Condition (b) still holds since all black subsets do not have the $k+1$-st element so we may use the inductive hypothesis. Condition (a) holds for all pairs of white subsets without the $k+1$-st element by similar reasoning. If at least one white subset has the $k+1$-st element then the union does too, but all subsets with the $k+1$-st element are white so (a) is true. This completes the induction and we are done.
[/hide]

Here's my solution: WLOG assume that $S=\{1,2, \dots ,2002\}$. We'll prove a more general result as given below:-

[b]GENERALISATION:[/b] Let $S(m)=\{1,2, \dots ,m\}$ be a set with $m$ elements, and let $N$ be an integer with $0 \leq N \leq 2^m$. Prove that it is possible to color every subset of $S$ either black or white so that the following conditions hold:
[list][*] the union of any two white subsets is white;
[*] the union of any two black subsets is black;
[*] there are exactly $N$ white subsets. [/list]

[b]PROOF:[/b] We proceed by induction on $m$. The base cases $m=1$ and $m=2$ are trivially true. So assume that the assertion is true for some $m>2$. We'll show that it is true for $m+1$ also. We use the notation $C(m,N)$ for a coloring which satisfies the given conditions for some $m,N$. Note that the null set $\{\phi \}$ does not affect the problem conditions $1$ and $2$, so we can use this set to increase or decrease the number of white or black colored subsets by one. Our construction will use this fact. Now, according to our inductive hypothesis, $C(m,N)$ exists for all $0 \leq N \leq 2^m$. Call the set $A'=\{a_1,a_2, \dots ,a_r,m+1\}$ the [i]inverse[/i] of a subset $A=\{a_1,a_2, \dots ,a_r\}$ of $S(m)$. We make two cases as follows-

[u]CASE-1[/u] $[$The null set is colored white in $C(m,N)]$
Color the [i]inverse[/i] of a subset of $S(m)$ by the same color as the subset. Then the number of white subsets become $2N$, and we get the coloring $C(m+1,2N)$. Also, if we first discolor the null set and then color it black then we get a new coloring, namely $C(m+1,2N-1)$.

[u]CASE-1[/u] $[$The null set is colored black in $C(m,N)]$
Color the [i]inverse[/i] of a subset of $S(m)$ by the same color as the subset. Then the number of white subsets become $2N$, and we get the coloring $C(m+1,2N)$. Also, if we first discolor the null set and then color it white then we get a new coloring, namely $C(m+1,2N+1)$.

As mentioned before, the color of the null set doesn't affect problem conditions $1$ and $2$, and so we can color the null set accordingly, and easily get $C(m+1,2N)$ and $C(m+1,2N-1)$ for all $0 \leq N \leq 2^m$, which shows that $S(m+1)$ also satisfies the given assertion. Hence, done. $\blacksquare$

WLOG $S={1,2,\dots,2002}$. Define a "subset pyramid" of a set as follows: in the first row, it contains all of the one element subsets, in the second row, it contains all of the two element subsets and so forth such that the kth row contains all of the k element subsets. An example for ${1,2,3,4}$ is shown: $$1,2,3,4$$ $$12,13,14,23,24,34$$ $$123,234,124,134$$ $$1234$$ Clearly the union of any two sets is either in the same row or further down the subset pyramid. Also, define a subset pyramid as "inferior" if it has less rows than another subset pyramid.
Now, I present an algorithm to construct the white and black sets of subsets. Let's start with $S$ as black. The main claim is that if we take an inferior subset pyramid of another subset pyramid, then we can switch the color and keep the conditions of the problem. This is easy to see: say, for example, we take away the subset pyramid of ${1,2,\dots m}$ from the subset pyramid of $S$, where $m<2002$. We can see that the property of the subset pyramid that the union of two sets are at the same row or further down still holds even after removing the inferior pyramid. Thus, we can take out inferior subset pyramids and switch colors from white to black as much as we need while still satisfying the constraints of the problem.
In summary, our algorithm is that we remove an inferior pyramid from $S$, make it white, then take another inferior subset pyramid from that one and make it black, and then with that subset pyramid, we can take an inferior pyramid and make it white and so forth. Let the first inferior pyramid have $a_1$ elements, the next one have $a_2$, and so forth all the way to $a_i$ such that $$0<a_i<\cdots<a_2<a_1\leq 2002$$ We have $$N=\sum_{k=1}^{i} (-1)^k*2^{a_k}$$ From experimentation with binary representation of base ten numbers, it is easy to see that all $N$ are attainable (each switch in color either adds/replaces with ones or zeroes to the binary representation, so we can handpick each $a_k$ to get our value of $N$). Since we have shown all $N$ are attainable while satisfying the constraints, we are done.
Remark: I am really sorry for the lack in clarity of the solution, I am still at a beginning level in trying to put down my ideas on paper, so feel free to ask questions about the solution

Here's a simple constructive solution.
Write each subset as a binary string where a $1$ in the one's place represents the first number being in the subset and a $0$ in the one's place represents the first number not being in the subset, and similarly for the other digits.  Call the function converting a subset to a number in this way $f$.
I claim that coloring the subsets $S$ in which
$$f(S)\oplus N\le N$$
white and the others black is a valid coloring, where $\oplus$ is the Binary Exclusive-Or operator.
This colors exactly $N$ subsets white because The XOR function is reversible.  ($f(S)\oplus N\oplus N = f(S)$)

It can be seen quite easily that
$$f(S_1 \cup S_2)\oplus N = (f(S_1) || f(S_2)) \oplus N$$
where $||$ is the bitwise or operator.  Now, notice that if both
$f(S_1)\oplus N \le N$ and $f(S_2) \oplus N \le N$ then $$(f(S_1) || f(S_2)) \oplus N \le N$$ and if both $f(S_1)\oplus N > N$ and $f(S_2) \oplus N > N$ then $$(f(S_1) || f(S_2)) \oplus N > N\blacksquare$$

Of course, further elaboration must be made on the two points above.

EDIT:  Apparently, $f(S)\oplus N \le N$ is synonymous with what quantumbyte posted, that the maximum element in the subset is in the set that $N$ represents!

Replace 2002 with $n$. We induct on $n$, $n=0$ being trivial. If $N\le 2^{n-1}$, we claim we can use the construction from $n-1$ to get $N$ white subsets, and color the rest black. We are adding the $n$'th element to the construction for $n-1$, call it $x$.

The union of any subsets that were originally black is black.
The union of a black subset not containing $x$, $S_1$ and a black subset containing $x$, $S_2$, is
\[ S_1\cup S_2 = (S_1\cup (S_2\setminus \{x\}))\cup \{x\}, \]
which is black. The union of two black subset containing $x$, $S_1$ and $S_2$, is
\[ S_1\cup S_2 = ((S_1\setminus \{x\}) \cup (S_2\setminus\{x\}) )\cup \{x\}\]
which is black, so this construction works. If $N>2^{n-1}$, we can use the same construction as before, but switching black and white, and replacing $N$ with $2^n-N$.

Solution: it's true for $N=0$, so for $N>0$, let $m,k\in \mathbb{N}_0$ st. $N=2^m+k, k<2^m$\\If $k=0,$  color all subsets of $\{1,2,...,m+1\}$ that contains number $m+1$ white. then color other subsets of $\{1,2,...,m+1\}$ which is not taken and have the smallest size until $k=2^m-1$. This algorithm should satisfy all the constraints.

We will generalize $2002$ with $n,$ and induct on $n.$ The base case $n=1$ is trivial. Assume that we can do the task for a $n,$ we will prove that we can then do the task for $n+1.$ For $n+1,$ we have $2^n$ new subsets, other than the previous $2^n$ subsets. For $N= 1, \ldots, 2^n,$ we can just use the induction hypothesis to color the subsets. When $N=2^n + 1,$ what we do is we use symmetry. So, for $2^n+1,$ we color $2^n - 1$ subsets which we previously colored white for the case when $N=2^n-1,$ black; yielding $2^n +1$ white subsets meeting the condition. Essentially, we color $2^n-k$ subsets black(which is obviously possible, ie just make the whites black) where $1 \leq k \leq 2^n$ for $N=2^n + k.$ Therefore, our induction is complete and the conclusion follows.

We will show a more general claim, with every instance of $2002$ in the problem statement replaced with $n$. To show this more general claim, we use induction.

The base case $n = 1$ is trivial.

For our inductive step, assume that the claim is true for $n$. We will now show the claim for $n+1$, which is equivalent to adding an element $e$ to the set $S$ (we let the set $S$ plus element $e$ be $S_2$). We would like to show that for all $N_2$ from $0$ to $2^{n+1}$, inclusive, there exists a valid coloring of the subsets of $S_2$. Then by inductive hypothesis, we know that for each $N$ from $0$ to $2^n$, inclusive, there exists a coloring of the subsets of $S$ that satisfies the conditions.

Take an arbitrary value of $N$. Then let $T$ be the subsets we color white. Consider the set $U$ of subsets, formed by all the subsets in $T$, as well as all the subsets in $T$ plus the element $e$. Then it's easy to see that the union of any two sets in $U$ lies in $U$. Therefore, we have a construction for $N_2 = 2N$, for any arbitrary $N$ from $0$ to $2^n$, inclusive, if we color all the subsets in $U$ white.

Now, for odd numbers, we do the following:

[list]
[*] if the empty set is in $U$, we remove it from $U$, and

[*] if the empty set is not in $U$, we add it to $U$.
[/list]
In the first case, we have a construction for $N_2 = 2N-1$, and in the second case, we have a construction for $N_2 = 2N+1$. Moreover, if $N = 2^n$, the empty set has to be in $U$, so $N_2$ never goes out of bounds. This is a valid construction for odd numbers, so our inductive step is complete. $\square$

[hide=Solution]We claim that such a coloring exists for all positive integers $n.$ We proceed by induction. Assume if we don't color a set white, we color it black. Let $S_n=\{1,2,...,n\}.$

We claim that such a coloring exists for all positive integers $n.$ We proceed by induction, letting $S_n=\{1,2,...,n\}.$

[i]Base Case $n=1.$[/i] If $N=0,$ we color none of the subsets white. If $N=1,$ we color the empty set white.

[i]Inductive Step.[/i] Call a subset $P\in S_{k+1}=\{1,2,...,k,k+1\}$ [i]old[/i] if $P\in S_k$ and [i]new[/i] if $P\not\in S_k.$

[b]Case 1. $0\le N\le 2^k.$[/b] We color all the old subsets white as in $n=k.$ This coloring satisfies all the conditions by the inductive hypothesis.

[b]Case 2. $2^k+1\le N\le 2^{k+1},$[/b] We know $N=2^k+c$ where $1\le c\le 2^k.$ We color all the old subsets as in $n=k$ and $N=c.$ We color all the new subsets white.

By the inductive hypothesis, the union of two old, white subsets is old and white, and the union of two old, black subsets is old and black. Also, the union of two new, white subsets is new and white, and there are no new, black subsets. Thus, our coloring satisfies (a) and (b). There are $c$ old, white subsets and $2^k$ new, white subsets, so (c) is also satisfied.[/hide]

We will prove the more general case where $|S_n|=n$ for some $n\geq 0$, so that we may use induction. WLOG let the set $S_n$ be $\{1, 2, \dots, n\}$.

The base case $n=0$ is trivial. Suppose this is true for $S_{n-1}$; we show it is true for $S_n$.
Switch black and white for a subset with $N$ white subsets to get a set with $2^n-N$ white subsets. So, we only need to show this is true for $0\leq N\leq 2^{n-1}$. To achieve this, we may take some coloring in $S_{n-1}$ with $N$ white subsets, and color the missing $2^{n-1}$ subsets containing $n$ black. It is easy to see that this is valid, so we are done. $\blacksquare$

[quote=MithsApprentice]Let $S$ be a set with $n$ elements, and let $N$ be an integer with $0 \leq N \leq 2^{n}$. Prove that it is possible to color every subset of $S$ either black or white so that the following conditions hold:

(a) the union of any two white subsets is white;
(b) the union of any two black subsets is black;
(c) there are exactly $N$ white subsets.[/quote]

Induction on $n$;when there are $n+1$ elements in $S$ now.Use the same set as in the $k$th step and colour all remaining subsets black.All these extra "black" subsets are of the form $$\mathcal{M}\cdot n^x +\{n \} \;\;\;\;\;\; x \in \{0,1\}$$,clearly this works since $$\mathcal{M}\cdot n^x+n \cup \mathcal{M}=\mathcal{M}.\blacksquare$$

Let $\mathcal{S}$ have $m$ elements and $\mathcal{P}$ be the set of its subsets. By inducting on $m$ we can show that coloring is possible for any $n \leq |\mathcal{P}|$. If $m = 1$ we color both subsets black for $n = 0$ , the empty set white for $n = 1$ and both subsets white for $n = 2$. Suppose now that a coloring is possible for $m$ (and any $n$). Consider a set $\mathcal{S}$ with $m + 1$ elements. Let $b$ be any element of $\mathcal{S}$.  For $n\leq 2^m$ , use induction to color just $n$ subsets of $S - {b}$ white and color black all subsets of $S$ which include $b$. Then the union of two white subsets is still a subset of $S - {b}$ and hence (by assumption) white. The union of two black subsets of $S - {b}$ is black for the same reason. If one black subset includes b, then so does the union, which must therefore be black. For $n > 2^m$, we have $2^{m+1} - n < 2^m$ , so we can find a coloring for $2^{m+1} - n$ and then swap the colors.

We induct on $2^n$, the base case is easy to see. Let the problem be true for $0 \leq N \leq 2^k$. For $N \geq2^k$ the coloring of the induction hypothesis works and for values of $N$ greater than this we just switch white and black.

We will prove this by induction on the number of elements. With 1 element it is trivial. Assume the claim is true for n elements. Either N<2^n or N>2^n.
If N<2^n, consider an element "x" and color all subsets containing x black. Then consider the set of elements that aren't x. This problem is equivalent to splitting those into a white and black category with N white subsets. If N>2^n, use a similar argument but replace black with white. Thus when n=2002 the statement is true.

[quote=2002 AMO/1]Let $S$ be a set with 2002 elements, and let $N$ be an integer with $0 \leq N \leq 2^{2002}$. Prove that it is possible to color every subset of $S$ either black or white so that the following conditions hold:

(a) the union of any two white subsets is white;
(b) the union of any two black subsets is black;
(c) there are exactly $N$ white subsets.[/quote]

We will prove this with induction on $m$, the number of elements (in the problem, $m=2002$).
Base case. $m=0$.
This is trivial, as we can color the empty set either black or white (as desired).
Inductive step:
Assume it is true for $m=M$. Then, we can generate the following construction for $m=M+1$:
If $0 \leq N \leq 2^M$, choose one element and color all subsets with that element black. Color all other elements in accordance with a construction for $m=M$ (with $N$ white subsets).
If $2^M \leq N \leq 2^{M+1}$, choose one element and color all subsets with that element white. Color all other elements in accordance with a construction for $m=M$ (with $N-2^M$ white subsets).

Replace $2002$ with $k\ge 1$. We induct on $k$.

The result is trivial at $k=1$: for $N=0$, color no subsets white, for $N=1$, color $\{\}$ white, and for $N=2$, color all subsets white.

Suppose $k>1$. Let $s$ be an element of $S$ and $S\setminus \{s\} = S'$. If $N\le 2^{k-1}$, use the induction hypothesis to make exactly $N$ white subsets of $S'$, then color every subset of $S$ including $s$ black. If $N>2^{k-1}$, flip the coloring scheme for $N' = 2^k-N \le 2^{k-1}$. We are done.

Generalize:
[quote=Rephrased Problem]Let $S$ be a set with $n$ elements, and let $N$ be an integer with $0\le N\le2^n$. Prove that it is possible to color every subset of $S$ either black or white so that the following conditions hold:

$(\text a)$ the union of any two white subsets is white;
$(\text b)$ the union of any two black subsets is black;
$(\text c)$ there are exactly $N$ white subsets.[/quote]
We induct on $n$.
The base case $n=1$ is true for $N=0$ with all subsets being colored black, and for $N=1$ it's true for only the empty set being white.

Suppose it's true for $n$. By switching white subsets with black subsets it suffices to show (for the induction step) that when $S$ has $n+1$ elements, the statement holds for $0\le N\le2^n$. Let $s\in S$ and let $W$ be the set of white subsets of $S\setminus\{s\}$ and let $B$ be the set of black subsets of $S\setminus\{s\}$ for this value of $N$. Then define $W'=\{w\cup\{s\}\mid w\in W\}$ and $B'=\{b\cup\{s\}\mid b\in B\}$. Color the elements of $W'\cup B'\cup B$ black and the elements of $W$ white. This satisfies $(\text a)$ and $(\text c)$. Now we show that it satisfies $(\text b)$. Let $x,y\in W'\cup B'\cup B$. If either of $x,y$ is in $W'\cup B'$, then $s\in x\cup y$ so $x\cup y\notin W$. If $x,y\in B$ then $x\cup y\in B$ too.

[hide = Solution] Call a subset [i]white[/i] if $a \in S' \subseteq S$ and [i]black[/i] if $a \not \in S' \subseteq S$. Then obviously if $a \in A_1 \cap A_2 \implies a \in A_1 \cup A_2$ and if $a \not \in B_1 \cap  B_2 \implies a \not \in B_1 \cup B_2$. Now we show that this is possible for all $0 \leq N \leq 2^n$.

Assume this is true for all $k \leq n - 1$. (The base case is trivial). For the $0 \leq N \leq 2^n$, we have the cases when $0 \leq N \leq 2^{n - 1}$ is satisfied. If $2^{n -1} + 1 \leq N \leq 2^n$, we may then consider coloring the subsets black rather than white. We then have that for $ 2^{n - 1} + 1 \leq N \leq 2^n$ that we can indeed color $2^n - N$ subsets black (due to the induction hypothesis), and hence we can color the remaining $N$ subsets white, and we are done. $\blacksquare$ [/hide]

We use induction, and it clearly works for $|S|=1$. Assume that it is true for $|S|=k$ elements with the set of subsets
		$\mathcal{P}(S)$. We now add an element $m$ to $S$ and order $\mathcal{P}(S)$ such that the first $2^k$ elements don't contain $m$, while the last $2^k$ do. By the induction assumption, we may may color any number of $0\le N\le 2^k$ of the first $2^k$ elements white, while we color all the other elements black, and the union of those $N$ white subsets can never form a black subset containing $m$. Also clearly, the union of any black subset stays black firstly by the induction assumption and secondly because for the last $2^k$ subsets the union of two of them will always contain an $m$ and will thus never be white. By symmetry, we could also color all the white subsets black and vice verse, so we can achieve all $0\le N\le 2^k$ number of white subsets such that the conditions hold true.

[hide = probably wrong sol]It suffices to show that for all reals $a,$ there exists a set $T$ of subsets of $\{1,2,\dots,a\}$ such that the union of any two subsets in $T$ is also in $T.$ We will prove this by induction. Clearly, this is true for all $a\leq 2^0.$ Now, suppose that this is true for $a=b.$ This means that it is true for all $a\leq 2^b.$ Without the element $b+1,$ suppose that there are $r$ elements in $T.$ With the element $b+1,$ suppose that there are $s$ elements in $T.$ We are given that $r,s\leq 2^b$ and we want to prove that $r+s$ takes on all values $\leq 2^{b+1}.$ This is clearly true. We conclude that the following conditions can be satisfied for all $0\leq N\leq 2^{2002},$ as desired.
[/hide]

We prove a stronger statement: Let $S$ be a set with $q$ elements, $\{1,2,\dots, q\}$, and let $N$ be an integer with $0 \leq N \leq 2^q.$ It is possible to color every subset of $S$ either black or white so that the three conditions hold.

We will proceed with induction on $q.$ When $q=1$, it is trivial. Assume that it is true for $q=k-1.$ Now, let $q=k.$ If $N\le 2^{k-1}$ then color all subsets containing $k$ black. Note that none of these sets can break the condition (a), because for two sets to have a union that contains $k$, one of the sets must contain $k$, and that set must be colored black. Thus, by the induction hypothesis, we are done, since no matter what $N$ is, there must be a way to color the subsets not containing $k$ so that there are $N$ white subsets.

If $N>2^{k-1}$ note that switch the color of every subset and conditions will still be satisfied. The induction step is thus proved, and along with it the assertion.

We induct on $n$ to prove the statement for general $n$; the base case is obvious. Now, first suppose $N \leq 2^{n-1}$. Pick some $n-1$-element subset of $S$, and color its elements such that there are $N$ white subsets according to the conditions given, using the inductive hypothesis. Next, color all subsets containing $N$ black. It is easy to check that this works. For $N > 2^{n-1}$, we can simply re-color every white set black and every black set white because the colors are symmetric.

Replace $2002$ with any positive integer $n$; we then claim that it is possible to color every subset of a set $S_n$ with $n$ elements, say $\{1, 2, \cdots, n\}$ black and white for any $0 \le N \le 2^n$ to satisfy the problem conditions (this obviously finishes the problem). We will prove this with induction, with the base case of $n = 1$ being obvious.

Now, for the inductive step, assume that the claim holds for $n = k$; we will then show it holds for $n = k + 1$ as well. First, if $0 \le N \le 2^k$, then we can take the same construction as we did for $S_k$ from the inductive hypothesis, and color any supersets of $\{ k + 1 \}$ black. If $2^k + 1 \le N \le 2^{k + 1}$, then take the construction from $S_k$ where we had to color $2^{k + 1} - N$ subsets white, and color any supersets of $\{k + 1\}$ black. Then, swap the color of every subset; this gives us $N$ white subsets of $S_{k + 1}$ satisfying the problem conditions, and hence the claim is proven.

Let $A(k)$ denote the assertion of the problem statement where $2002$ is replaced with the positive integer $k$. We will show that $A(k)$ is true for all positive integers $k$ via induction on $k$.

For a base case, consider $A(1)$. WLOG let $S = \{0\}$. For $N = 0$, let there be no white sets and the rest black, for $N = 1$, let $\emptyset$ be white and $\{0\}$ black, and for $N = 2$, let $\emptyset$ and $\{0\}$ be white and none black. Thus $A(1)$ is true.

Now, assume $A(k)$ is true for some fixed $k\ge 1$. We will show that this implies $A(k+1)$ is true as well. Let $T$ be a fixed $k$ element subset of $S$ and let $u$ be the element contaned in $S$ but not $T$. First suppose $N\le 2^k$. Then, since $A(k)$ is true, we can color $N$ subsets of $T$ white and the rest black while satisfying conditions (a) and (b). Then, color all subsets of $S$ containing $u$ black. This coloring clearly satisfies (a) and (b). Now, suppose $N > 2^k$. Then, $2^k - (2^{k+1} - N) < 2^k$, so since $A(k)$ is true, we can color $2^k - (2^{k + 1} - N)$ subsets of $T$ white and the rest black while satisfying conditions (a) and (b). Also color all subsets of $S$ containing $u$ white as well. Similarly to before, this coloring satisfies (a) and (b), and furthermore we get $2^k - (2^{k+1} - N) + 2^k = N$ white subsets, as needed. Hence, $A(k+1)$ is true, so the induction is complete.

It follows that $A(2002)$ is true, so we are done.

If $N = 0$ or $2^n$ for some $0\leq n\leq2002$, then the construction is trivial.

Let $N = 2^n + k$ with $0<k<2^{n}$. We can then construct $N$ white sets as follows.

Let the largest white set be $\{x_1, x_2, \dots, x_{n+1}\}$. Let all subsets of $\{x_1, x_2, \dots, x_n\}$ excluding the empty set be colored white. Then, there are $2^n$ white sets. There exists a maximal $m$ such that \[k = \sum_{i=1}^{m} {n \choose i} + l\] with $0\leq l < {n \choose m+1}$. Then color all sets where $i\leq m$ elements of $\{x_1, \dots, x_n\}$ are removed white. Also color the first $l$ sets where $m+1$ elements are removed white. For example, if $k=n+1$, $\{x_2, x_3, \dots,x_n, x_{n+1}\}, \{x_1, x_3, \dots, x_n, x_{n+1} \}, \dots, \{x_1, x_2,\dots, x_{n-1}, x_{n+1}\}$ and $\{x_3, x_4, \dots, x_{n+1}\}$ are colored white.

This construction ensures exactly $N$ subsets are white. Additionally, any union of two white sets either falls into the first $2^n$ subsets or the latter $k$ subsets, i.e. is white. Since all white sets consist of the union of a subset of $\{x_1, x_2, \dots, x_n\}$ and $\{x_{n+1}\}$, they cannot be written as the union of two black sets.

Thus, a construction exists for all $N$ and we are done.

This is garbage wording but here we go.
Switch $2002$ with $k=n$ we will prove by induction that if the problem works for $k$ than the problem works for $k=n+1$.
Suppose we have constructions for $k=n$.
If we switch to $k=n+1$ and $N \leq 2^{n}$ we can choose an element color all subsets that contain that element black, for the other subsets repeat a construction.
If $N > 2^{n}$ we can make use of a construction for $N'= N-2^n$
Choose an element color the subsets with the element in accordance to the $N'$ construction. Then color the rest of the subsets white.
Since this problem works for $k=1$ we are finished.

We will prove this result for all numbers $n$ using induction.

For $n=1$, it is clearly possible so this covers our base case.

Now, we proceed with the inductive step on $n=k+1$. We have two cases:

[list]
    [*]If $N \le 2^k$, then we take a coloring of subsets of $\{1,2,\dots,k\}$ with $N$ white sets and color every subset containing $k+1$ black.
    [*]If $N > 2^k$, then we take a coloring of subsets of $\{1,2,\dots,k\}$ with $N-2^k$ white sets and color every subset containing $k+1$ white.
[/list]

Hence, the inductive step is done and we have proved the desired result. $\square$


[hide]
$\log_{10} 2 = \frac a b$ with a,b integers.
$2 = \sqrt[b]{10^a}$
$2^b = 10^a$
$2^b = 2^a \cdot 5^a$
Not possible unless $a = b = 0$, however $\frac a b = \frac 0 0$ is indeterminate.
Contradiction.
[/hide]

[hide="Generalization"]The same idea can be used to prove that $\log_{10}n$ for any $n\in\mathbb{N}$ is either integer (if $n$ is a power of $10$) or irrational (in all the other cases).[/hide]


$ AB \cdot AD = 320$.
\begin{eqnarray*} \frac {AB \cdot AX}{2} + \frac {DZ \cdot DX}{2} & = & \frac {AB \cdot (AD/2)}{2} + \frac {(AB/2) \cdot (AD/2)}{2} \\
& = & \frac {3}{8} \cdot AB \cdot AD \\
& = & \frac {3}{8} \cdot 320 \\
& = & \boxed{120} \end{eqnarray*}

$ AB \cdot AD = 320$.
\begin{eqnarray*}
\frac {(AB+DZ) \cdot AD}{2} - \frac {AD \cdot ((3/4) \cdot XY)}{2}
& = & \frac {(AB+(AB/2)) \cdot AD}{2} - \frac {AD \cdot ((3/4) \cdot AB)}{2} \\
& = & \frac {3}{8} \cdot AB \cdot AD \\
& = & \frac {3}{8} \cdot 320 \\
& = & \boxed{120}
\end{eqnarray*}


function
LaTeX
algebra
domain
topology
real analysis
real analysis unsolved

Suppose $ I\equal{}[a,b]$ and let $ f: I\minus{}>R$ and $ g: I\minus{}>R$ be continuous functions on $ I$. Also let the set $ E\equal{} \{x \in I: f(x)\equal{}g(x)\}$ have the property that if $ (x_n) \subseteq E$ and $ x_n \minus{}> x_0$, then $ x_0 \in E$.

So then $ f(x_n) \equal{} g(x_n)$ for all $ n \in N$.  Well then that implies $ \lim(f(x_n))\equal{}\lim(g(x_n))$.  And that $ f(x_0)\equal{}g(x_0)$

Which means $ x_0 \in E$

However, I'm a little concerned about this because I didn't use properties of a closed bounded interval.  IS there something wrong with my logic to solving this problem.

maybe you should say that since I=[a,b] any $ x_n \to x_o$ must converge inside of I? besides that im convinced

$ \text{\LaTeX}$ tip: use \to or \rightarrow to get $ f: I\to\mathbb{R}$ or $ f: I\rightarrow\mathbb{R}$ (and that was \mathbb{R} instead of $ R$.) It looks better than trying to simulate an arrow with ->.

Let $ h(x)=f(x)-g(x).$ Then $ h(x)$ is a continuous function, and $ E=\{x\in E: h(x)=0\}=h^{-1}(\{0\}).$ The inverse image of a closed set, which means that $ E$ is closed relative to $ I,$ no matter what kind of set $ I$ is. So the fact that $ I$ was closed and compact is essentially irrelevant to the problem

$ I\equal{}(a,b)$  and the statement is false...

$ f(x)\equal{}g(x)\equal{}x$
$ x_n \equal{} a\plus{}\frac{1}{n} \to x_0\equal{}a \notin I \to x_0 \notin E\subset I$

^I noticed that when you I realized what your post meant.

kenn4000: That gets into the whole notion of "relatively closed." If we say that the domain of $ f$ is $ (a,b)$ then we can in effect take the attitude that nothing exists outside of that domain. Under that notion, your example of $ x_n\equal{}a\plus{}\frac1n$ is a sequence that has no limit, and as such does not invalidate the statement.

i guess youre the professor, you know best =(


I'm trying to gather information about summer math REUs.  Does anyone have recommendations for well-run programs?

I think we have three good [url=http://www.math.tamu.edu/research/undergrad/REU/]REU programs[/url], in particular "Algebraic Methods in Computational Biology". I worked with the "Matrix Analysis" program for the last two years: it's probably not as strong, but still fun. However, I don't plan to be in College Station this summer. :P

University of Wisconsin-Madison has not bad ones. Usually they are run by Ken Ono, so the topics revolve around modular forms and similar stuff. You may check about the 2008 program [url=http://www.math.wisc.edu/~ono/reu08.html]here[/url].


Got this cryptology problem here:

I have three (3) vaults which is locked by four (4) numbers.
Intricately engraved in each vault are these numbers:

Vault #1
0260059300820968 0264035402640593 0992W0968 18690593067805930264

Vault #2
1024072311230822 0822082205221123 1024072311230822 032111230822

Vault #3
28597377 058664 28597377 91860059

Each set of number (there are four sets) represents one number (hence there are 4 numbers per vault).
What are those four (4) numbers which un-locks the vaults?

By the way, the letter "W" in Vault #1 is not a typo.

Could you please help me out with the solution to the cryptology problem above (the answer and the process to get those numbers)?

Thanks a lot guys.

(this is my second posting of this problem... i hope i am not spamming... if i am, please just delet, thanks!!!)

i think the "W" is a clue...

since it was indicated in the instruction that it is not a typographical error

@jhollenbeck
thanks for trying dude... i sure do hope you could help...

[quote="probability1.01"]Are the four numbers you want to end up with each single digit?  I mean you could just take those numbers as they are to be the numbers required to open the vaults :P[/quote]

to answer your question yes...
the problem gives me four sets of numbers:
say:
   12345  23456 34567 45678
and the problem is to find out what single digit number translates to each set:
say:
   12345 = x
   23456 = y
   34567 = z
   45678 = a

it could be a simple substitution to find the single digit, but i still don't get it...  :|

hey guys, thanks for the help... please continue rith your replies, though i hope to find the solution by tomorrow...  :huh:

[quote="mathmanman"]If the "w" is not a typo, could this be in another base?[/quote]

Probably not since if you go to a base larger than 10, the single digits would still be the single digits.
If the base is less than 10, there wouldn't be 9's or W's.  And there probably wouldn't be 0's out in front either.

I don't really have any ideas though.

[quote="mathmanman"]If the "w" is not a typo, could this be in another base?[/quote]

that's possible... i am usually shot from work, that's why i tried to find if there's a forum in the web that could help me out with my assignments (personal studies) in cryptology...

here are some of the solution in a previous decryption problems in the same genre:
[img]http://img.photobucket.com/albums/v618/limreb/codes.jpg[/img]

(the above is based on a substitution of patterns to get the answer)

(the above is based on the keypads on a cellular phone)

I think it might have something to do with sequences of 3 digits, by just looking at it you can see there are quite a few repeats of these.

That's true... after noticing that, I was originally thinking that maybe each three-numer sequence stands for a letter, but that can't be right; if you look at the second set of numbers in the first vault, there are only five numbers between the two "026" sets. Hmm...

[quote="CrystalClear"]by the way... the four character password are all numeric...

i'll try the sequencing decode, but the "W" on the first vault got me stumped...

your help guys would be most appreciated... (the first three days, i thought no one would help)[/quote]

Well I mean you can always spell out a number or something.

Yes.
We should analyze it a bit, for example, which sequences are repeating? if we assume it's a substitution, it could represent spaces, or "e".

i guess the first thing that needs be done is to determine what
the letter "W" is doing in the third set of the of the first "password"...

since it is not a typographical error, then it must be a clue?
an indicator to a solution? i tried looking as ascii values and
there doesn't seem to be any significance to that letter...

secondly, there are repeating consecutive values in the number sets...
but what do we do with these, should we remove it?

i have never really had experience with these kinds of puzzles but here is what i have found...

I believe the numbers for the first vault are: 0 9 2 7

My reasoning follows:

Let every 4 numbers represent a letter.  The W tells us that the number spelled out by that series of numbers has 3 letters and contains a W in the middle.  The only single digit number satisfying this condition is TWO.  Let the set of numbers 0992 represent T and 0968 represent O.

Since i must go i will explain the rest briefly.  We see that the first number must be zero using the fact that 0968 represents O.  Using information from here we decode the last number as 7 and finally the third as 9.

This seems right to me, but i did it quickly and i do not have time to solve the rest....

My only qualm is that vault 3 has a series of 6 digits, which would contradict my reasoning that every 4 numbers represents a letter.  If that is not a typo, my idea is trashed

;)  oh well

I don't know, I was under the impression that all three vaults had different codes.

ooh well that would make more sense, probability1.01... :oops:

I guess I'll wait to see what CrystalClear says about my answer for the first vault before looking at the others

hi guys, these are the answers to my cryptology problem:

[b]Vault #1: the letter "w" was a clue... 0992W0968
0992 = t
0968 = o
no other number from 0 - 9 has a letter w when you spell it out...

i used country codes based on the member states of the United Nations...
no member state has a name that starts with the letter "w" and "x"...
(please correct me if am wrong)

Vault #3:
a = 1
b = 2
c = 3
...
z = 26

divide numbers 1 - 26 by the vaule of "pi" and get the first 2 digits to the right of the decimal point.

Vault #2: i used dates based on the zodiac signs (Aries - Sagittarius) to replace numbers 1 - 9[/b]

@arp
thanks!!! your solution was very logical for vault #1...


$ 26\equal{}5\plus{}6\plus{}7\plus{}8$ and $ 63\equal{}20\plus{}21\plus{}22$.  How many of the integers $ \{1, 2, 3, ... , 100 \}$ can be expressed as the sum of two or more consecutive positive integers?

[quote="IW@IT"]The number in the form $ 2^n$ can't be expressed as the sum of two or more consecutive positive integers
So, there are $ 100 \minus{} 6 \equal{} 94$ numbers[/quote]

Could you please explain why?  Also, the answer [hide="is"]93[/hide].

[hide="No, he forgot 1."]Trivial case: $ 1 \equal{} 1 \plus{} 0$.

By contradiction: let $ 0 \leq k < n$ denote the upper and lower limits of a series of consecutive numbers such that its sum is of the form $ 2^{m}$ (by problem statement, $ n \minus{} k > 1$). The sum of this series can be expressed as

$ \frac{1}{2}[n^{2} \plus{} n \minus{} k^{2} \minus{} k] \equal{} \frac{1}{2}[(n \minus{} k)(n \plus{} k \plus{} 1)] \equal{} 2^{m}$.

If we make the substitution $ z \equal{} n \minus{} k$, and multiply by two on both sides, we will find that

$ 2^{m \plus{} 1} \equal{} (z)(z \plus{} 2k \plus{} 1)$,

but this is impossible, since $ z$ and $ z \plus{} 2k \plus{} 1 \equiv z \plus{} 1$ have different parities (exactly one of them must be odd, but the equation can only be true if both of them are powers of two). Thus, we can discard all powers of two, of which there are $ 6$ in $ (1, 100]$ (Technically, $ 1$ is a power of two, but we already counted it).

The other numbers [i]can[/i] be expressed as desired: odd numbers are necessarily of the form $ n \plus{} [n \plus{} 1]$, and the remaining even numbers can always be expressed as

$ [n \minus{} (2^{m} \minus{} 1)] \plus{} [n \minus{} (2^{m} \minus{} 2)] \plus{} \dots \plus{} [n \plus{} (2^{m} \minus{} 1)] \plus{} [n \plus{} 2^{m}] \equal{} 2^{m}(2n \plus{} 1)$,

where $ 2^{m}$ is the highest power of two that is a factor of the number.

$ \therefore 100 \minus{} (6 \plus{} 1) \equal{} \boxed{93}$[/hide]

[quote="metafor"]even numbers can always be expressed as

$ [n \minus{} (2^{m} \minus{} 1)] \plus{} [n \minus{} (2^{m} \minus{} 2)] \plus{} \dots \plus{} [n \plus{} (2^{m} \minus{} 1)] \plus{} [n \plus{} 2^{m}] \equal{} 2^{m}(2n \plus{} 1)$,

where $ 2^{m}$ is the highest power of two that is a factor of the number.[/quote]  You might want to rethink this expansion for 48 (among several others) ;)

Oops. :P

In the first case, $ 2^{m} \leq n$; when $ 2^{m} > n$, they can be expressed as $ [2^m \minus{} n] \plus{} [2^m \minus{} (n \minus{} 1)] \plus{} \dots \plus{} [2^m \plus{} (n \minus{} 1)] \plus{} [2^m \plus{} n] \equal{} 2^m(2n \plus{} 1)$. Please tell me if I forgot something else.


function
limit
logarithms
calculus
calculus computations

Find pointwise limits of functions when $x \in (0, 1)$:
1.  $f_{n}(x) = nxe^{-nx^{2}}$
2. $f_{n}(x) = n^{2}x^{n}(1-x)$

Edited: The first one converges pointwise to 0 because the exponential function dominates. The second one is converging pointwise to zero, right? If so is infinity times zero always equal  zero?

[quote="akech"]Find pointwise limits of functions when $x \in (0, 1)$:
1.  $f_{n}(x) = nxe^{-nx^{2}}$
[/quote]
Use the root test,

$|nxe^{-nx^{2}}|^{1/n}= n^{1/n}\cdot x^{1/n}\cdot e^{-x^{2}}|$
Now,
$n^{1/n}\to 1$ and $x^{1/n}\to 1$ and $e^{-x^{2}}<1$
So the root test limit is <1, so it converges to zero.
[quote]
 If so is infinity times zero always equal  zero?[/quote]

No

$\lim \frac{1}{n}\cdot n^{2}=+\infty$
$\lim \frac{1}{n^{2}}\cdot n = 0$

And $\lim \frac{1}{n}\cdot cn=c$ for any $c$.

The statement $n^{\frac{1}{n}}$ approaches 1 eventually needs a proof right or is the proof so trivial so we do not need it?
What is the statement that $\frac{1}{n}* cn$ approaches c for all c supposed to be answering in this post. Can't you just crossout n since n is positive leaving you with a constant which we know stays a constant all of the time?

[quote="akech"]The statement $n^{\frac{1}{n}}$ approaches 1 eventually needs a proof right or is the proof so trivial so we do not need it?
[/quote]

The proof is not trivial.

I will use the following beautiful theorem.

[b]Theorem:[/b] Let $\{s_{n}\}$ be a sequence of non-zero terms. Then,

$\liminf \left| \frac{s_{n+1}}{s_{n}}\right| \leq \liminf \left| s_{n}\right|^{1/n}\leq \limsup \left| s_{n}\right|^{1/n}\leq \limsup \left| \frac{s_{n+1}}{s_{n}}\right|$

[b]Corollary:[/b] If $\lim \left| \frac{s_{n+1}}{s_{n}}\right|$ is defined (finite or infinite) then all the liminf's and limsup's are equal. So this limit is equal to $\lim |s_{n}|^{1/n}$ which also must exist.

[b]Theorem:[/b] $n^{1/n}\to 1$

[b]Proof:[/b] Define the sequence of non-zero term $s_{n}=n$.

Then, $\lim \left| \frac{n+1}{n}\right| = 1$ by corollary we have that $\lim n^{1/n}$ must exist and furthermore equal to 1.

[quote="akech"]What is the statement that $\frac{1}{n}* cn$ approaches c for all c supposed to be answering in this post.[/quote]
It's answering the question, "If $\lim_{x\to x_{0}}f(x) = 0$ and $\lim_{x \to x_{0}}g(x) = \infty$, is it true that $\lim_{x \to x_{0}}f(x)g(x) = 0$?"  That question, which you didn't ask, is the sensible way to interpret the question you did ask, namely, "is infinity times zero always equal to zero?"

Also, the easiest way to realize that $\lim_{n \to \infty}n^{\frac{1}{n}}= 1$ is by noting that $\lim_{n \to \infty}\frac{\ln n}{n}= 0$.


search
number theory solved
number theory

Let $\tau (n)$ denote the number of positive divisors of positive integres $n$. Show that the equation $\tau(an) = n$ has no solutions in $n$ for infinitely many positive integers $a$

IMO 2004 Shortlist, problem N1. Search button.

Rushil     , you are post alot ofproblem
  you should give them in 1 post

It is sort of a rule of the forum (I read in an announcement by Valentin) that each problem should be posted as a different topic to avoid confudion and misunderstanding!

Rushil is right. Please continue the discussion about this problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34224]here[/url].


This a nice one: find all pairs of natural numbers a and b s.t. ab+a+b/a^2+b^2+1.

I can't uderstate what does it mean.
Please, write clearly (more detailed)

he probably means that he wants all pairs of positive integers a and b for which the raport

[u] ab+a+b [/u]
a2+b2+1

is also a positive integer.

Oh, there are LOTS of other solns!  :D Actually, I have a solution for this one, I should have posted it at the 'Proposed problems' section. I was really tired when I posted it.. Sorry. Bye!

I have found folowing solutions
a=k^2
b=(k+1)^2
k - positive integer

Are there other solutions?

(0,1),(1,0),(1,1),(k 2 ,(k+1) 2 ),((k+1) 2 ,k 2 ) (k belongs to positive integer)

PS: here in my country,{natural numbers}={0} \cup {positive integers}.

These are all the solns. Can you prove these are the only ones?

Denote k=(a2+b2+1)/(ab+a+b).  (1)
Due to (1) we obtain
a2-k(b+1)a+(b2-kb+1)=0   (2)
Denote discriminant (2) as D=k2(b+1)2-4b2+4kb-4

If a=b then it is obvious a=b=1.

Suppose WLOG (how does it decode?) a>b.

We claim k=2. Since (2) has integer solution a then D is perfect square.
Suppose k\geq b+1 ==> k2(b+1)2 \geq D < (k(b+1)+1)2 ==> D=k2(b+1)2 k=b+1 and b=1, k=2, a=4.
Suppose k\leq b. In that case a'=(b2-kb+1)/a is a second positive integer root of (2) and a'b then (b,a') is solution and b>a'  ==> through some steps we must obtain pair (a'',b'') with b''<k ==> k=2.

Now if k=2 then D=16b ==> b=n2 ==> according to (2) and assumption a>b we obtain a=b+1+0.5\sqrt D ==> a=(n+1)2

So all solutions
(1,1);(n2,(n+1)2);((n+1)2,n2)

That's hat I did, but I had a hint: I found the problem in a book at the chapter concerning Fermat descendence, so this gave me a clue.


vector
symmetry
superior algebra
superior algebra solved

hi all,
I've a simple question but I can't solve it: prove that if E is a vector space and B1 and B2 are two basis of E then there exists a bijection between B1 and B2. I think we have to use Zorn lemmaand Cantor-Bernstein but I got nowhere  :(

What is basis in abstract vector space? Do you mean normalized space? Or Hammel basis?

A basis in a general vector space is simply an independent  set spanning the entire space (and a set of elements is called independent if it has no finite linearly dependent subset).

yes, it is what I meant. So I think there exists a bijection but how could we prove that ?

Here is what I think works, but it is very hard to explain it. Take $ B_1=(e_i)_{i\in I}, B_2=(f_j)_{j\in J}$. Now, for each i from I there exists a finite set $ A_i$ subset of J such that $ e_i=\sum_{j\in A_i}{s_j f_j}$ for some scalars $ s_j$. We will prove that in fact the union of all A_i covers J. Indeed, if not, then there is $ j_0$ in J that is not in any $ A_i$. Write that $ f_{j_0}$ is a linear combination of $ e_t$ and you will find a linear dependent family formed by vectors $ e_i$. Impossible. Thus, J is covered by a family of finite sets, the cardinal of this family being the cardinal of I. Thus, J has at most cardinal as J. By symmetry, conclude. Hope it's correct.

If $E$ is a finite-dimensional space, then statement is true.

Suppose $E$ is an infinite-dimensional space.
Consider subsets $B_{1n}$ s.t. each $x\in B_{1n}$ can be expressed as linear combination of at most $n$ elements of $B_2$. It is clear $\bigcup_{n}B_{1n}=B_1$.
On the other hand, for each $n$-tuple $(y_1,...,y_n)\in B_2^n$ there are at most $n$ distinct $x_1,...,x_n\in B_1$ which can be expressed only using $y_1$, ..., $y_n$. It follows that $|B_{1n}|\leq|nB_2^n|=|B_2|$.
So we conclude $|B_1|=\left|\bigcup_{n}B_{1n}\right|\leq |B_2\cup B_2\cup...|=|B_2|$.


Helo
Proof this:
$ n\mathbb N\cap m\mathbb N\equal{}(m.n)\mathbb N$ if n and m are primes numbers.

We'll do this by showing each side is a subset of the other. I'm changing $ m$ and $ n$ to $ p$ and $ q$ to make it clearer. Also, you need the hypothesis $ p \neq q$.

Suppose $ x \in (pq) \mathbb{N}$. Then $ x$ is a positive integral multiple of $ p$ and $ q$. Thus $ x \in p\mathbb{N}$ and $ x \in q\mathbb{N}$. So $ x$ is in the intersection of them.

Suppose $ x$ is in $ p\mathbb{N}$ and $ q\mathbb{N}$. If $ a \mid c$, $ b \mid c$, and $ (a,b) \equal{} 1$, then we have $ ab \mid c$, so $ x$ is a multiple of $ pq$ and is in $ (pq)\mathbb{N}$.


integration
calculus
calculus computations

I am most certainly wrong.
$ \dfrac{2}{3}x^6 \minus{} \dfrac{15}{4}x^4 \plus{} \dfrac{4}{3}x^3 \plus{} \dfrac{9}{2}x^2 \plus{} 8x \plus{} ln(4x^4 \minus{} 4x^3 \minus{} 3x^2 \plus{} 2x \plus{} 1)\plus{}C$
After all I AM only in Algebra I. :blush:

I plan to play with this integral over my spring vacation.  My friend told me it took him 15 pages of algebra to compute.


How many digits do you know by memory? Be honest now...

224 :P

EDIT: Hmmm, I put a 200+ option, but I don't see it, so I'll just do "100-200"
EDIT2: Well, I tried to fix it, but its even more screwed up now...

I used to know 80+ but my memory sagged from not reciting it too often so it just dropped down to 40....don't worry I plan to memorize another 100 by freshman year!  :D

55...

Unless you're going to memorize 100000 digits and hold a world record, it's pointless.

Oh and there's no poll (if you intended one to be there).

[quote="yagaron"]that's will mathematicians don't have girlfriends

they spend all day trying to memorize digits of pi

huauhauhau[/quote]

Well, go look at Ubemaya's blog... You will find that that first sentence isn't true if you go look at his blog.

I know 100 digits of pi, but it's pretty useless unless you want to impress some people (most of 'em probably won't believe you, anyway).

I try to time myself how long it takes me to type those useless digits:
3.14159265358979323846264338327950288419716939937510528097494459230781640628620899862803482513421170679
Eh, that was horrible: 25 seconds.

I probably messed up somewhere there anyways...

you commit an error

before 3421170679  have a 1 :)

or is on purpose?

how knows

a question

suppose you know n digits of pi.

can you square pi ?

multiply it by 2?

because, if you can do that, it's usefull, you is better than a calculator,

if not, you is worst than a paper

There was supposed to be a poll, but it got really screwed up and messed, so i just killed it.

So there's no poll, but u can still say whatever...

The first 15  :D they seem so rhythmic... (3.14)(1592)(6535)(8979)

:maybe: I'll never win a pi recitation contest  :(

I don't understand why people are so obsessed with memorizing so many digits.
I read a book that said 39 digits of $\pi$ are enough to calculate the circumference of the universe to the accuracy of the diameter of a hydrogen atom.

... but that doesn't make sense, how do you calculate the circumference of the universe?
Anyways, you get the idea.

Also, a lot of people think people who like math just memorize $\pi$ all day, which is probably part of the reason not many people like math.

just make music

[quote]f[n_] := Play[Cos[2*Pi*220*2^((n + 3)/12)*t], {t, 0, 1/8}, DisplayFunction -> Identity]
RealDigits[N[Pi, 100]][[1]];
Show[f /@ %]; [/quote]

There are savants that remember things like that... or that memorize pi by visualizing abstract landscapes and stuff...

lol, you cant set it to t=1/4 instead and then people can sing along to that... experts can do it on 1/6  :rotfl:

mozart can also memorize all digits of digits of pi seting $t = \frac{1}{oo}$

but he died because of that :(  poor guy

[quote="xscapezaer"]That's not necessarily true. Math ppl don't necessarily just sit around and memorize digits of $\pi$. It's just bored math losers like yours truly, with nothing better to do, that do stuff like this...

Btw, no offense to anyone who posted here, not saying anyone's a bored math loser...[/quote]

I know math ppl don't sit around memorizing $\pi$, but, first, a some ppl do, and second, its totally pointless.

pie is good.  mmm...

pi is also good, but not as tasty.

I know 13  digits.  Lucky 13!   :lol:   My mum was born on Fri. 13

3.141592653?

I know someone who knows 34, and someone who knows 64

Oh :blush: , really? Where did I get 6000? NVM

[quote="i_like_pie"][quote="Nexmus"]There's this Chinese guy who holds the world record, with like 6000 digits in 1 hour.[/quote]
The world record is 43000 digits, although it took him 24 hours, 4 min to recite them all.

I think the most digits per hour is the runner-up to the world record: 40000 digits in 5 hours, 21 min.[/quote]

I'm not sure how up-to-date this is but http://en.wikipedia.org/wiki/Akira_Haraguchi

[quote="i_like_pie"]I've seen that, but if you read the entire article, you will see:

[quote="Wikipedia"]Despite Haraguchi's efforts and detailed documentation, the Guinness World Records have not yet accepted any of his records set, and accredit instead the record to another Japanese man, Hiroyuki Goto, who managed to recite Pi by 42,195 decimal places in 1995.[/quote]

Once (if?) they accept Haraguchi's records, it will blast the old one by almost 60,000 digits.[/quote]

wow, i only can do 64 digits

I can only do the first 15 because they have this kind of rhythm to it, like A M Y said :D but I know this 7th grader at my school who memorized 952 digits of pi.  Actually, he recited more than 1100, but he made a mistake, so only the first 952 count.

To tell you the truth, what is the point of memorizing pi? It's not like when you're calculating the circumference of the circle, you'll actually multiply every single little digit...  :huh:

It's not as useless as a lot of people think. I earned a free slice of pizza for being able to recite pi, even though I only knew about 180 digits, and I totally screwed up and didn't even finish.

Wow.. Where do you guys find so many digits? My math teacher has a poster and that's where I got my measly 17 digits...

3.14159265358979323

How does memorizing pi help you solve math?   :huuh:
You can just use 3.14.
All I know is 3.1415926.   :oops:

here is a document i made listing lots of pi's digits


LaTeX
number theory unsolved
number theory

Find all positive integer $n$ such that the equation $x^3+y^3+z^3=n \cdot x^2 \cdot y^2 \cdot z^2$ has positive integer solutions.

let's assume that $x \geq y \geq z$, thus $3x^3 \geq x^3+y^3+z^3=nx^2y^2z^2$, so
$x \geq \frac{ny^2z^2}{3}$. Because $x^3+y^3+z^3=nx^2y^2z^2$ thus $x^2|y^3+z^3$, so
$2y^3 \geq y^3+z^3 \geq x^2 \geq \frac{n^2y^4z^4}{9}$.
If $z>1$ then $y \leq \frac{18}{16n^2}$ so $y=1$ but $y \geq z$ - contradiction.
If $z=1$ then $y^3+1 \geq \frac{n^2y^4}{9}$, so $9+\frac{9}{y^3} \geq n^2y$.
-If $y=1$ then $x^3+2=nx^2$ so the only solution is $(x,y,z,n)=(1,1,1,3)$ $(x^2|2=>x=1)$ so $n=3$ satisfies.
-If $y>1$ then $10 \geq n^2y$ so
a) $n=1$ then for instance $(x,y,z,n)=(3,2,1,1)$ is a solution so $n=1$ also satisfies
b) $n=2$ then $y=2$ or $y=1$ then there are no solutions-easy to check.
c) $n=3$ and $y=1$ that gives $(1,1,1,3)$ what we had before.
(q.e.d.)
Answer : $n=1,3$

Actually, isn't this infinite descent? After reducing to one of x,y,z being 1, then solve n and get n=1 and n=3 only.  :)

at last the sith has won with the latex code... :)


AMC USA(J)MO USAMO AIME Harvard college MIT

Looking around at some of our talented members on AoPS, I have no doubt that this year, it will be tough to qualify for USAMO. But how tough? Hopefully, it won't be much more than last year, or I have no chance to qualify (I am a junior).
Also: predict the AIME floor. Doesn't apply to me, though.

It depends a lot on the difficulty of the tests also, not just the calibur of competitors.

I'm predicting 220-240.  Dunno about floor...don't really care too much either :P

I'm thinking not that much. Maybe at most 10 points? It still has more factors than just AOPS which determines the index.

Yea, I don't think AoPS raises it that much.
Students getting smarter will eventually lead to harder problems from the AMC people.
Also, USAMO qualification is basically top 250 people, so it doesn't really matter about the index :D

[quote="krustyteklown"]Students getting smarter will eventually lead to harder problems from the AMC people.[/quote]

I hope my efforts will lead to this effect. I would be happy to see the various qualifying scores will come back down a bit. I was notified that 2006 exams were prepared a long time ago, so there is a 0% chance I have inspired problems on them - but I'll be checking the AIME A & B closely in 2K7! ;)

Don't worry, I'm sure Thomas and I will resort to writing hideously difficult AIME problems while we're getting dominated during Math 55, so I predict a massive drop in index as soon as they start accepting our problems!

I thought 55 = harvard, Mildorf = MIT? (dunno about Albert) *confused*

It wasn't unusually easy. 2 good problems were able to be solved, ie give the 3 digit answer from "assumptions" I'd say 9 is a pretty safe score to get into USAMO, at least its what I'm aiming for :D

the whole thing was unusually easy (can't talk about the TST, but besides that ... eh)

I would say that 9 is definitely safe for 10th graders, since 9 is the highest floor that I've seen so far. Unless it continues to get easier or people get smarter and they don't make it harder, I doubt it will go above that next year.

[quote="beta"]What's a "safe" AIME score for 10th graders?

Also was 2005 AMC/AIME unusually easy, or did people just improve(through AoPS, )[/quote]

I'm thinking it was unsually easy, because  was able to solve number 12, even when I can't solve some of the number 12's on the 1980's AIME's  :blush:

Where can I find a copy of AIME 2005? I don't think it's on Kalva. :(

Well I'm aiming for 12 on AIME either way, I just hope that it will be a safe score for USAMO.

[quote="qvc122"]Haitao Mao qualified for USAMO and also made to MOP.[/quote]

I know. But how did he do on AIME? Maybe like 11 or so?

Haitao, I believe he got a 9. But a decent 14 I think on the USAMO. We talked, but I'm not 100% sure. He is also the founder of the phrase - ORZNORZ...

Just to clarify.... if the 2006 AIME is the [b]same [/b]difficulty as the 2005 AIME, should a 240 Index qualify..... or will AoPS make it slightly higher?
Because I'm averaging 8 on AIMEs and a 125 on AMC 12's (I'm a junior).
I need to get 130/11, which is not too farfetched........

How much will AoPS raise MOP cutoff(for blue)? Not that I have any chance but it's always good to shoot high.

I really don't think the floor will rise to 10.  Last year's AIME wasn't exceptionally easy, but 9 is still one of the highest cutoffs that we've had with a cutoff of 233.  So, a cutoff of over 240 is unlikely.  By comparison, 2004's cutoff was 210 and the floor was 7.

I say AIME floor = 8, USAMO index 217-225.

You can easily screw up on AIME by just simply being careless.

[quote="1234567890"]So, do you think last year was an exception, or is the general trend that the index will increase?[/quote]

It's hard to say, but either way I am hoping for a hard AIME (not impossible, but hard like something in the late 1990's). But the point is that AIME you can solve a lot of problems but it's really easy to make a careless mistake somewhere, you just have to hope to be lucky not making any(or practice a lot).

I'm personally hoping for a cutoff between 215-220 because I see me topping out at 8 this year haha.

[quote="1234567890"]So, do you think last year was an exception, or is the general trend that the index will increase?[/quote]

I don't think we will ever have a long term trend of increasing floor and cutoff scores, because, as mentioned before, a few years of high scores will prompt tougher questions.

I think he might mean going to MOP, but I'm not sure. If not, it should be re-worded as "Is a score of 7-10 high enough to qualify for USAMO as a 7th grader?" ;)

I think the MOP deal is ONLY for freshmen, in order to qual for MOP as a 7th/8th grader, you need to score top 30 on the USAMO. (at least that's how it was in 2004)

[quote="joml88"][quote="PenguinIntegral"]Is a 7-10 USAMO qualifying for a 7th grader?[/quote]

Depends on the year.  The highest floor has been is a 9.  Who knows what will happen though.  They could give a really easy (relatively speaking) AIME and the floor goes to 10.  Or it can be like 1994 and then the floor would be much lower.[/quote]

Yes, then I believe that if I was taking the AMC tests this year, I could go to the USAMO in seventh grade. Not any higher though.


[quote="kimby_102"][quote="FieryHydra"]The sequence $a_{0}, a_{1},a_{2},...$ satisfies
$a_{m+n}+a_{m-n}=2(a_{m}+a_{n})$
for all nonnegative integers $m$ and $n$ with $m\geq n$. If $a_{1}=1$, determine $a_{2006}$.[/quote]
i think this is very interesting...
[hide]
after some experiment or "got my hands dirty" (as ACoPS puts), i found that $a_{0}=0$, $a_{1}=1$, $a_{2}=2^{2}$, $a_{3}=3^{2}$, etc...
so the answer is $2006^{2}$..
[/hide][/quote]
Very good! Can you now complete the proof?
[hide]Induction.[/hide]


Just thinking.. it would be pretty hard to inscribe a [b]non-convex[/b] quadrilateral inside a circle, wouldn't it? :)

Some convex quadrilateral can't be inscribed in a circle (for example a really long one), so I think the problem can just say "quadrilateral ABCD..."
Course, I could be wrong. At least what TripleM says is true.

[quote="JBL"][color=cyan]Nope -- there's the case where ACBD is a convex quadrilateral, which makes ABCD a concave quadrilateral which might be able to be inscribed in a circle.[/color][/quote]
How can that be? would it stil be a quadrilateral? (I tried it with a couple of quadrilaterals, but I get a triangle with two lines extended, making something like a fish.)

Wait, is this considered a quadrilateral?

[quote="JBL"][color=cyan]Nope -- there's the case where ACBD is a convex quadrilateral, which makes ABCD a concave quadrilateral which might be able to be inscribed in a circle.  Basically, it's just the way that he makes sure you subtract the two diagonals and not two of the sides.[/color][/quote]

Ohh true. I never really thought of that as being a quadrilateral though. OK, now time to try the problem.

[color=cyan]Well, yes, two of its edges would cross each other.  But you can still treat such a figure as a quadrilateral and do many quadrilateraly things to it.  I mean, it's four points and four edges.  Just in an unusual position.

I promise this is the case that the wording of the problem was meant to rule out.[/color]

I'm more concerned about the absolute values. I haven't worked on this problem, but it seems to me that it has to be greater than zero by the triangle inequality.

[color=cyan]About the problem itself, I thought it might be possible to reduce the problem to 4 quarters: if in a triangle AC + CB - BA < R, where R is the circumradius, we'll be done by taking each of the four triangles ABC, ABD, ACD, and BCD and adding the results together.  I don't of course know if this is true, but I thought it might be the way to go.[/color]

yeah, I thought of that too. However, I'm not sure how to get R in formulas. I would have to use the law of sines/the area formula, and I don't think either one would work out nicely.

a little thought (definitely not a proof cuz it's incomplete):

[hide]The >0 part is self explanatory by triangle inequality. For <2, if you let AC be the diameter, ( length =2), to test for the greatest possibility, and move B infinitely close to A, that would make AB=0 and BC=2. Doing the same to the other side, DA=0 and DC=2, this way, DB=0. AB, 0 + BC, 2, +CD, 2 +DA, 0 - AC, 2 - BD, 0= 2. Since moving a point infinitely to another point would annul the concept of triangle, the equation has to be <2.[/hide]

[color=cyan]But you still have to show that doing that will give you a maximal value.  That is, you have to show that by moving two vertices closer and closer together, you increase the value of the expression we want to show is less than 2.  That doesn't look easy.

By the way, I should note that my earlier idea sucks, because what I wanted to prove isn't true.  For example, consider the case of an equilateral triangle.[/color]

[quote="JBL"][color=black]But you still have to show that doing that will give you a maximal value.  That is, you have to show that by moving two vertices closer and closer together, you increase the value of the expression we want to show is less than 2.  That doesn't look easy.[/color][/quote]
trying to work on that

This problem has been unanswered long enough now.

A solution can be found at http://www.kalva.demon.co.uk/short/soln/sh852.html


A circle $ \Gamma$ is inscribed in a quadrilateral $ ABCD$. If $ \angle A\equal{}\angle B\equal{}120^{\circ}, \angle D\equal{}90^{\circ}$ and $ BC\equal{}1$, find, with proof, the length of $ AD$.

Let $ M, \ N, \ P, \ Q$ be the points where the sides $ AB, \ BC, \ CD, \ DA$ touch the circle.

Then $ DQ\equal{}r, \ AQ\equal{}AM\equal{}MB\equal{}BN\equal{}\frac{r\sqrt{3}}{3}\Rightarrow AD\equal{}\frac{3\plus{}\sqrt{3}}{3}\cdot r$

$ NC\equal{}\frac{r}{\tan 15^{\circ}}\equal{}r(2\plus{}\sqrt{3})$

$ 1\equal{}BC\equal{}BN\plus{}NC\Rightarrow r\equal{}\frac{3}{6\plus{}4\sqrt{3}}$

$ AD\equal{}\frac{\sqrt{3}\minus{}1}{2}$


We know that Mathcounts is right around the corner and we're all excited! What are you goals? As for me, I don't really know. What games are you gonna take? Do you think you're gonna make countdown (in chapter, states, nats)?

major goal: make nats. a perfect at chapters would be nice too =).

First in chapter (3rd last year) with a 44+ (likely), win countdown (possible).
Top 10 in Texas (45th last year) with 36+ (doable if I work hard enough and don't make careless errors).

ahh, texas. tough state. michigan has a ton of competition, but our populations not as massive as yours, so...its easier. good luck though! =]

lol..I've got the puny Sunflower State.  But there are lots of good people here, atleast this year.  I would like to get to nats, but the probabiliity i make it is $ \frac{1}{\text{infinity}}$

Goals:

School: 46
Chapter: 42+
State: 35+ (maybe make nats?)
National (if I make it): 25+

Ach, my county/state is pretty good, so I will probably die but here are my goals.

Chapter: 40+
State: 30+
Nats(basically make it, but if I do): 20+

NikeBalla You live in Michigan?

I would like to possibly make countdown at chapter. Most of the good students in my state live in my chapter so the states is just like chapters all over again. Nats possibly... Setting my expectations high.

my goal is to make state. since all members of the top two teams always make it to cd(william, etc.)I'll have to make the top ten to make state. get a 35 and I think i'm in.

[quote="jjx1"]william[/quote]

OMG! JJX1! You revealed somebody's name! You're not supposed to do that!

Goal: Don't fail.

100th view!

anyway, I have no clue about my chapter but I think I might be able to make state. I really suck with speed because I can solve a problem but it takes me longer than the next person.

Um, here are my goals: (i've already got 1st place School round)

Chapter: 1st place individual (hopefully score a 46), 2nd place countdown (I'VE GOTTA AIM HIGH!! :D ), 4th place team (to opt for my whole team to go to state. This year I've got FOUR people on my team! YEAH!)

State: 2nd place individual, 4th place countdown, 10th place team. Most importantly, MAKE NATIONALS!!!!! :D

Nationals: Top 30ish (?), possibly make countdown if I practice more? :maybe:

Yeah....

[quote="echen"]NikeBalla You live in Michigan?

[/quote]

yeah, looking at your username, i think i might know who you are. i got a perfect on amc 8, im in math III, im doing decently in mandelbrot...

do you go to country day? okay, im pming you...

I want $ \frac{44}{46}$ on chapter, which is 2 wrong on sprint for dumb errors...would say my school sprint/target score, but that's not allowed.

If I can get $ \frac{41}{46}$ at state, I'll be borderline for nationals, california's really strong this year.  Miller and Harker have their best teams in awhile, and james4l, pythag011 will be tough to beat...

Another goal: first at chapter countdown, top 3 state countdown.

Does CA have official or unofficial countdown?

Unofficial, unfortunately.  It's because there are two competitions, one for North California and one for South California, so we can't really have official countdown. :(

If countdown was official, my chances would be MUCH higher...

Yeah, uh, considering I came first at chapter last year and I was awful, I think I'll get first at chapter. My goal for chapter is to somehow get 1st written, 1st countdown, and first team. Heh, our countdown is bracket, so it'll be annoying.

Hm got fourth at state last year, so I'm aiming for first. Also, considering I got 35 and first got 36, and I was very bad back then, it's pretty possible. I'm also going to try to make top 4 countdown. Dunno about team.

Heh, I'm trying to aim for countdown at Nats, but I'm too lazy to actually work really hard. Maybe I should fix that...

Make school team (duh)

Make Chapter (that will be so hard)

Make State (hopefully)

Make Nats (in 50 years)

[quote="funtwo"]i'm aiming for 44+ at states, but i might not make top 10 with that[/quote]
Oh really? Texas is really such a hard state...? :o

How would an extremely hard test minimize flukes? Do you mean that there would be less ties and less stupid mistakes?

Pretty much. It would have a better chance of giving accurate rankings.

[quote="AIME15"]While we're hoping for things, I hope that the format is changed to 40 questions, 20 minutes :)[/quote]

NOOOOOOOOOOOOOOOOOOO!  I would epically fail then.  Don't even give them that idea!

[quote="math154"]I'm hoping for an extremely hard state test this year, so flukes will be minimized, but I don't think that will happen, given the difficulty of last year's test. :([/quote]

meaning last year was...easy...or hard?

sorry, i dont remember, i was quite (STUPID, IDIOTIC, FAILISH, RETARDED) last year.

[quote="mathluver2.718"]To make MC Nats this year. (I hope :maybe: )
it would also be nice to get at least a 44 @ Chapter and State.[/quote]
Same here! But I guess I only need a 38-40 for first place in my state.

My 6th grade goals are

43+ Chapter, make state
40+ State, mkae Nats (I actually have a shot, I beat Fantasylover 4 out of 5 times)
39+ Nats (yeah right...)

Realistically, 30+ Nats

There are also a lot of coaches who would like to see a difficult test.  It tends to more accurately identify the top students and teams.  Last year was difficult.  It seemed that at least one part of every test last year was hard.  Good luck to those of you getting ready for the Feb 7th Chapter tests.

Thanks!

I am praying for a resonably hard test, because my strengths are in hard problems, not computation and stuff.

[quote="xpmath"]Also, considering I got 35 and first got 36, and I was very bad back then, it's pretty possible.[/quote]

You just insulted almost every state.

Right now I am wishing that I didn't skip 5th grade (is Orlando fun?), but it's fun to be an assistant coach. Sort of. Kinda. Not really.

N-E-ways, good luck to everybody competing!

to be nice: if you work hard? is that implying that i cant? haha

yeah, it sucks you cant even come to chapters...meh. at least you have next year.

meh. im excited for mathcounts (too bad mine is on the 28th...sucks.)

[quote="nikeballa96"][quote="math154"]Relatively speaking, it was a killer (compared to many previous years' tests).[/quote]

oh. maybe that explains why i failed so badly. does anyone who knows my intelligence level think its possible for me to get a 40+? another goal: first countdown, first written at chapter. my chapters BEASTLY (same as echen), so if i can get top 2 at chapter, i have a relatively good chance of making nats.

argh, this is the first time i've realized how truly badly i want to make nats.[/quote]

A 40+ on what? Nats?

If Nats, certainly possible. Just work a little more. I almost got 40+ and you can beast me on FTW.

Just compare yourself with those at ICAE. If you are top ten their, you will most likely be top 10 at chapter and state. Also DCDS invitational is coming up so see how you do there.

ROFL who is this weird kid you speak of named "hirsh jain." is he smart?

and athunder, echen goes to icae. think of dcd kids... think of who's username would be e.chen. E*** CHEN. omg look at the michigan private forum.

anyways, bogtro, i was talking about states. see, im stupid.

Last Year (I think)
School-28 (5th)
Chapter-38 (3rd)
State-33 (2nd)
Nats-20 (140-149)

This Year
School-already took

Goals
Chapter- 44 or 2nd
State- 35 or 4th and make nats
Nats- 24 or top half of all

I got really lucky all of last year, so my goals this year are not that much higher.
GL All!!!!!

Uhh... I won't be able to participate in MathCounts this year, because I live in a different country, and my school doesn't do it.


inequalities
calculus
derivative
real analysis
real analysis unsolved

Prove that if $ x>0$ and $ a\ge 2$ , then $ x^a\plus{}\frac {1}{x^a}\minus{}2\le \left(x\plus{}\frac 1x\right)^a\minus{}2^a$ .

Rewrite the inequality and then use the usual calculus tricks. If no one gets it I will write something down.

Notice that \[ \left\{ x\plus{}\frac{1}{x}, 1,1\right\} \succ \left\{ x,\frac{1}{x},2\right\}\]
Let $ f(x)\equal{}x^a$; then $ f''(x)\equal{}a(a\minus{}1)x^{a\minus{}2}$ is positive as $ a\geq2$. Thus by Karamata's,
\[ f(x\plus{}1/x) \plus{} f(1) \plus{} f(1)\geq f(x) \plus{} f(1/x) \plus{} f(2) \]
which is what we were trying to prove.

We are asked to show $ (x^a  \minus{} 1)^2  \le (x^2  \plus{} 1)^a  \minus{} (2x)^a$. Use a derivative test.


http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=229747
is here you will find the directions
please join
if i do not respond to your shout, please pm me

[quote="7h3.D3m0n.117"]I thought we agreed that there would be no more Hurt and Heals?  :wink:[/quote]

I think this is a method to avoid your rule by making it in his blog instead of the fun factory forum....

[quote="7h3.D3m0n.117"]I thought we agreed that there would be no more Hurt and Heals?  :wink:[/quote]
Who is this we?  :wink:

yes, in fact this is performed solely on my blog except this advertisement and yes it is to create spam (but on my blog)

Alright if it's on your blog then I have no problem with it.

Although you posted this in the [i]Games[/i] forum instead of the Fun Factory. (But it got moved so...)


vector
function
inequalities
logarithms
topology
Functional Analysis
triangle inequality

[quote="harazi"]A very nice fact: any metric space is isometric to a closed set of a vector normed space.[/quote]

Let $(X,d)$ be a metric space. The plan is this:

(1) Show that every metric space can be embedded isometrically into a normed space;

(2) Take an ambedding such as in (1) and tweak it slightly so as to make the elements of the metric space embed as a linearly independent set;

(3) Apply (2) to the completion $\overline X$ of $X$ with respect to the metric $d$, embedding $\overline X$ into a normed space $(E,\|\cdot\|)$ as a linearly independent set. $X$ then coicides with the intersection between $\overline X$ (which is closed in $E$, being complete) and the linear subspace $L$ of $E$ spanned by $X$. Hence, $X$ is closed in $(L,\|\cdot\|)$.

I'll just outline a proof for (1) and (2). The metric space we're working with will still be called $(X,d)$.

(1) We can embed $X$ in a Banach space $\mathcal C(S,\mathbb R)$ of bounded real functions on a set $S$ (to be constructed), with the $\sup$ norm. First fix a point $p\in X$, whose image will be the origin of the space. In general, for a point $x\in X$, let's denote by $(x_{s})_{s\in S}$ its image in $\mathcal C(S,\mathbb R)$. Hence $p_{s}=0,\ \forall s\in S$. Now, in order to construct $S$, for every pair $x,y\in X$ of points in $X$, put an element $s$ in $S$. Choose $|x_{s}|\le d(x,p),\ |y_{s}|\le d(y,p)$ such that $|x_{s}-y_{s}|=d(x,y)$ (possible by the triangle inequality), and, using an argument similar to the one used in proving the Hahn-Banach Theorem (extension for one point at a time plus Zorn's Lemma), find a real $z_{s}$ for every $z\in X$ such that $|z_{s}-t_{s}|\le d(z,t),\ \forall z,t\in X\ (*)$. This will solve (1).

(2) Take the construction from (1), and modify it as follows: for each $n$-tuple $x^{1},\ldots,x^{n}$ of points in $X$ and each $n$-tuple $a_{1},\ldots,a_{n}$ of reals, not all zero, add another point $s$ to $S$, find reals $x^{i}_{s},\ i=\overline{1,n}$ less than $1$ in absolute value such that $\sum a_{i}x^{i}_{s}\ne 0$, and then find, for each $z\in X$, a real $z_{s}$ less than $1$ in absolute value such that $(*)$ holds. The proof that this is indeed possible is along the same lines as the one sketched in the previous paragraph. With this enlarged set $S$, the functions $(x_{s})_{s}$ will be bounded on $S$, linearly independent, and $\sup_{s\in S}|x_{s}-y_{s}|=d(x,y),\ \forall x,y\in X$, i.e. $x\mapsto (x_{s})_{s}$ is an isometric embedding, as desired.

[quote="Kent Merryfield"]I'm trying to imagine the embedding of some connected graph (of no special structure), presumably into $\mathbb{R}^{n}$ for some large-enough $n.$

Well, $\mathbb{R}^{n}$ but probably not with the Euclidean norm - the $1$ norm looks more promising.[/quote]

Every finite metric space embeds isometrically into $\mathbb{R}^{N}$ equipped with the norm $\|\cdot\|_{\infty}$. However, for every fixed $p < \infty$, there exists a finite metric space (or a graph on $n$ vertices equipped with the shortest path metric) on $n$ points s.t. every its embedding into $({\mathbb R}^{N}, \|\cdot\|_{p})$ has distortion at least $\Omega(\log n)$.


An interesting fact about Pythagorean triples was shown to me by a friend - if the triple is (a,b,c), then a^2 = b + c. Im not sure how to prove this fact yet, but if it is true for all triples, the triplet depends only on the value of b and the triplets all fall into a general equation or formula. ive been trying to do this, but each time i do it i end up with something wrong. could neone derive the formula?

It's false. For example, 8, 15, 17 is a Pythagorean triple, but 8^2 =64, whereas 15+17=32. However, if c-b=1, then a^2+b^2=c^2, or a^2=c^2-b^2=(c-b)(c+b)=b+c. But in general a^2=(c-b)(b+c), so that only works when c-b=1.

Not sure I agree (or I don't understand the terms) :

if a + b = c then a = (b+c)(b-c)

So a^2 = b+c is only true for (3,4,5)

Sorry ComplexZeta I didn't see your answer.

My favorite result about Pythagorean triple is the following :

if the triple is (a,b,c), then 60 | a*b*c

If you want a general formula for pythagorean triples: see below

[hide]

a=x^2-y^2, b=2xy, c=x^2+y^2 gives all possible triples. They are 'primitive' (ie it isn't a multiple of any other one) then you choose x and y so they are coprime and one is even, one is odd.

[/hide]

That doesn't generate all of them. The general form is [hide]x(y^2-z^2), 2xyz, x(y^2+z^2), where y>z.[/hide]

No, it does generate them all. You've added in an x to get all multiples, however other combinations can get these multiples anyway. For example:
6,8,10 has x=3, y=1.

Interesting that you should mention this, since this topic was brought up yesterday or the day before, I believe:

[quote="mathfanatic"]I'll just discuss a specific case here:

Take any positive, odd integer x.
Square it.
Divide it by 2. It should come out to being a.5 (that is, an integer plus 1/2)
Take a and a+1.

The triplet (x, a, a+1) is a perfect square. [/quote]

Since a + (a+1) = 2a+1 = x^2, this is true.

All pythagorean triples that can be generated using mathfanatic's method satisfy your condition that a^2 = b + c.  But it doesn't apply to things like 3-4-5, which can't be generating using the specific method from above.[/quote]

Why do you say 3-4-5 can't be generated that way? 3^2/2=4.5, so (3, 4, 5) is a Pythagorean triple.

Yeah, 3-4-5 works, just anything without the last 2 differing by 1 wouldn't. For example, using my formula above, 20,21,29.

Erp.

How is it that, somehow by random luck, I pick a bad one?  I mean a good one...one that [i]does[/i] work...  Anyways, here's one that doesn't work: 28-45-51...

Of course, any of the multiples of triples don't work either.  For instance: 6-8-10 is no good.

Oy...I need more sleep.  And ComplexZeta passed mathfanatic again.  This could get interesting...


geometry
parallelogram
rhombus
geometry unsolved

:?: Help please :?:
Rhombuses $ ABCD$ and $ A_1B_1C_1D_1$ are equal. Side $ BC$ intersects sides $ B_1C_1$ and $ C_1D_1$  at points $ M$ and $ N$ respectively. Side $ AD$ intersects sides $ A_1B_1$ and $ A_1D_1$ at points $ Q$ and $ P$ respectively. Let $ O$ be intersection of lines $ MP$ and $ QN$. Find $ m(\angle A_1B_1C_1)$, if $ m(\angle QOP) \equal{} \frac{1}{2}m(\angle B_1C_1D_1)$ .

Parallelogram formed by the lines $ A_1B_1, BC, C_1D_1, DA$ has all maltitudes equal, it is a rhombus, $ NQ$ is one of its diagonals, $ NQ$ bisects the angle $ \angle D_1NM$ internally and the $ \angle MNC_1$ externally. Likewise, parallelogram formed by the lines $ B_1C_1, BC, D_1A_1, DA$ is a rhombus, $ MP$ is one of its diagonals, $ MP$ bisects the $ \angle NMB_1$ internally and the $ \angle C_1MN$ externally $ \Longrightarrow\ O$ is the excenter of the $ \triangle C_1MN$ in the angle $ \angle NC_1M$ $ \Longrightarrow$ $ \alpha \equal{} \angle MON \equal{} 90^\circ \minus{} \frac{\angle NC_1M}{2} \equal{} 90^\circ \minus{} \alpha,$ $ 2\alpha \equal{} 90^\circ,$ $ \angle A_1B_1C_1 \equal{} 180^\circ \minus{} \angle B_1C_1D_1 \equal{} 90^\circ.$


Here is my problems in MIC (Mathsvn Inequality Contest)  :)
[b]Problem 3 (Round 1)[/b]
Given $ a, b, c \geq\ 0$. Prove that:

$ \frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \leq\ \frac {(a \plus{} b)(a \plus{} b \plus{} 2c)}{(3a \plus{} 3b \plus{} 2c)^2} \leq\ \frac {1}{8}$

[b]Problem 5 (Round 1)[/b]
Given A, B, C are three angles of a acute-angled triangle.Prove that:

$ (\frac {cos{\frac {A}{2}}}{tgA})^2 \plus{} (\frac {cos{\frac {B}{2}}}{tgB})^2 \plus{} (\frac {cos{\frac {C}{2}}}{tgC})^2 \plus{} \frac {3}{4}. \frac {sinA \plus{} sinB \plus{} sinC}{tgA \plus{} tgB \plus{} tgC} \geq\ \frac {9}{8}$

[b]Problem 1 (Round 2)[/b]
Given $ a, b, c \geq\ 0$ satisfy $ a \plus{} 2b \plus{} 3c \equal{} 4.$

Prove that: $ (a^2b \plus{} b^2c \plus{} c^2a \plus{} abc)(ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} abc) \leq\ 8$

[b]Problem 3 (Round 3)[/b]
Given the reals $ a, b, c$ satisfy $ ab \plus{} bc \plus{} ca \geq\ 11.$
Prove that: $ \sqrt [3]{a^2 \plus{} 3} \plus{} \frac {7}{5\sqrt [3]{14}} .\sqrt [3]{b^2 \plus{} 3} \plus{} \frac {\sqrt [3]{9}}{5}. \sqrt [3]{c^2 \plus{} 3} \geq\ \frac {23}{5\sqrt [3]{2}}$

And here is another problems in MIC Shortlist:
[b]1)[/b] Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3$.
Prove that: $ 3 \geq\ \frac {a}{a \plus{} (b \minus{} c)^2} \plus{} \frac {b}{b \plus{} (c \minus{} a)^2} \plus{} \frac {c}{c \plus{} (a \minus{} b)^2} \geq\ \frac {4}{5}$

[b]2)[/b] Given $ a, b, c \geq\ 0$. Prove that: $ \frac {\sqrt {ab}}{ ab \plus{} c^2} \plus{} \frac {\sqrt {bc}}{ bc \plus{} a^2} \plus{} \frac {\sqrt {ca}}{ ca \plus{} b^2} \geq\ \frac {9(a^3 \plus{} b^3 \plus{} c^3)}{2(a \plus{} b \plus{} c)^4}$

Dear my friends, we hope that we'll receive the comments about our problems and about MIC :)
Also see the test here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=258595
http://www.mathlinks.ro/Forum/viewtopic.php?t=256950

[quote="nguoivn"]And here is another problems in MIC Shortlist:
[b]1)[/b] Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3$.
Prove that: $ 3 \geq\ \frac {a}{a \plus{} (b \minus{} c)^2} \plus{} \frac {b}{b \plus{} (c \minus{} a)^2} \plus{} \frac {c}{c \plus{} (a \minus{} b)^2} \geq\ \frac {4}{5}$

[/quote]
 We have : $ a \leq a \plus{} (b \minus{} c)^2 ,b \leq b \plus{} (c \minus{} a)^2$ and $ c \leq c \plus{} (a \minus{} b)^2$.
Hence $ 3 \geq\ \frac {a}{a \plus{} (b \minus{} c)^2} \plus{} \frac {b}{b \plus{} (c \minus{} a)^2} \plus{} \frac {c}{c \plus{} (a \minus{} b)^2}$

The other hand, let be $ c \equal{} Min \left(a,b,c \right)$
$ LHS \geq \frac {a}{a \plus{} b^2} \plus{} \frac {b}{b \plus{} a^2} \geq \frac {(a \plus{} b)^2}{a^2 \plus{} b^2 \plus{} 3ab} \geq \frac {9}{9 \plus{} ab} \geq \frac {4}{5}$
Because of $ ab \leq \frac {9}{4}$.
Equality occurs iff $ a \equal{} b \equal{} \frac {3}{2}$ and $ c \equal{} 0$.

[quote="nguoivn"][quote="Hong Quy"]
$ \frac {(a \plus{} b)^2}{a^2 \plus{} b^2 \plus{} 3ab} \geq \frac {9}{9 \plus{} ab}$
Because of $ ab \leq \frac {9}{4}$.
[/quote]
We can expand and obtain: $ ab(a \plus{} b)^2 \geq\ 9ab$
But $ ab(a \plus{} b)^2 \leq\ 9ab$. So, your proof was wrong  :maybe:
Besides:
[quote="nguoivn"][b]Problem 3 (Round 1)[/b]
Given $ a, b, c \geq\ 0$. Prove that:

$ \frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \leq\ \frac {(a \plus{} b)(a \plus{} b \plus{} 2c)}{(3a \plus{} 3b \plus{} 2c)^2} \leq\ \frac {1}{8}$

[/quote]

That's so interesting when we received about 20 proofs different with another for it from competitors :lol:[/quote]
 No,this's mean $ \frac {(a \plus{} b)^2}{a^2 \plus{} b^2 \plus{} 3ab} \equal{}\frac {9}{9 \plus{} ab}$ , because of $ a\plus{}b\equal{}3$ and $ a^2\plus{}b^2\plus{}3ab\equal{}(a\plus{}b)^2\plus{}ab\equal{}9\plus{}ab$

\
[quote="nguoivn"][b]Problem 3 (Round 1)[/b]
Given $ a, b, c \geq\ 0$. Prove that:

$ \frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \leq\ \frac {(a \plus{} b)(a \plus{} b \plus{} 2c)}{(3a \plus{} 3b \plus{} 2c)^2} \leq\ \frac {1}{8}$

[/quote]

[quote="Hong Quy"] No,this's mean $ \frac {(a \plus{} b)^2}{a^2 \plus{} b^2 \plus{} 3ab} \equal{} \frac {9}{9 \plus{} ab}$ , because of $ a \plus{} b \equal{} 3$ and $ a^2 \plus{} b^2 \plus{} 3ab \equal{} (a \plus{} b)^2 \plus{} ab \equal{} 9 \plus{} ab$[/quote]
Hi, notice that: $ a\plus{}b\plus{}c\equal{}3$ => $ a\plus{}b \leq\ 3$. Can you suppose that $ a\plus{}b\equal{}3?$  :huh:

[quote="nguoivn"][quote="Hong Quy"] No,this's mean $ \frac {(a \plus{} b)^2}{a^2 \plus{} b^2 \plus{} 3ab} \equal{} \frac {9}{9 \plus{} ab}$ , because of $ a \plus{} b \equal{} 3$ and $ a^2 \plus{} b^2 \plus{} 3ab \equal{} (a \plus{} b)^2 \plus{} ab \equal{} 9 \plus{} ab$[/quote]
Hi, notice that: $ a \plus{} b \plus{} c \equal{} 3$ => $ a \plus{} b \leq\ 3$. Can you suppose that $ a \plus{} b \equal{} 3?$  :huh:[/quote]
But $ f(a,b,c) \geq f(a,b,0)$
Hence, we have: $ a\plus{}b\equal{}3$  and must prove that : $ \frac{a}{a\plus{}b^2}\plus{}\frac{b}{b\plus{}a^2} \geq \frac{4}{5}$

[quote="Hong Quy"]But $ f(a,b,c) \geq f(a,b,0)$
Hence, we have: $ a \plus{} b \equal{} 3$  and must prove that : $ \frac {a}{a \plus{} b^2} \plus{} \frac {b}{b \plus{} a^2} \geq \frac {4}{5}$[/quote]

Oh, sorry, my friend. Now, I understand your proof  :oops:  (indeed, it's similar to my proof)
And nice your proof for problem 3  :)


Sequence $ a_n $is decreasing and tends to zero. Prove that if
$\sum_{n=1}^{\infty}a_n=+\infty$ then $\sum_{n=1}^{\infty} min(a_n,\frac{1}{n})=+\infty;$

Firstly we show that if a serie $\sum_{n\ge0}x_n\to \infty$  then $\sum_{n\ge0}\min(x_n,1)\to\infty$.
If $(x_n)_n\ge 1$  for a finite number of $n$-s then there exists an indice $n'$ s.t. $x_n<1\forall n>n'$ so we have $\sum_{n\ge0}\min(x_n,1)=\sum_{i=1}^{n'}\min(x_i,1)+\sum_{n>n'}x_n$ and the last sum we know that goes to $\infty$ when $n\to\infty$.
If $(x_n)<1$ for a finite number of $n$-s then there is an indice $n''$ s.t. $x_n>1 \forall n>n''$ so the sum $\sum_{n\ge0}\min(x_n,1)=\sum_{i=1}^{n''}\min(x_i,1)+\sum_{n>n''}1$ which obviously goes to $\infty$ when $n\to\infty$.
Now from Cauchy condensation criterion we get that the serie $\sum_{n\ge0}a_n$ has the same nature with the serie $\sum_{n\ge0}2^na_{2^n}$, so $\sum_{n\ge0}2^na_{2^n}\to\infty$.
Using the same criterion for the serie $\sum_{n\ge0}\min(a_n,\frac1n)$ we get that it has the same nature with $\sum_{n\ge0}2^n\min(a_{2^n},\frac1{2^n})=\sum_{n\ge0}\min(2^na_{2^n},1)\to\infty$ (because of the fact proved above).

Thanks great Kuba, it is almost my solution.
I wonder whether there is solution non-using the Cauchy condensation criterion ?


The charge Q of an electrified insulated metal sphere is reduced from 20 to 16 units in 10 minutes.  Find the coefficient $\mu$ of leakage, if $Q=Q_{o}\times e^{-\mu t}$; $Q_o$ being the initial charge and t being in seconds.  Hence find the time taken by half the charge to leak away.

My book says the answer is $\mu=0.00037$ minutes, yet if you try to plug the variables in and take the natural logarithms you'll get $\mu=0.022314355$.  I don't get what's wrong.

P.S.  If I get it wrong with a minor calculation mistake, plz don't blame me...blame these guys:  :oops: and :blush:.

:blush:

I did get it wrong with a minor calculation mistake.  :rotfl:

[quote="ROLEX4life"]That wasn't even a physics problem. That was just a simple intermediate algebra problem.[/quote]

It was a physics problem. This is [b]Introductory Physics[/b], therefore according to the stated rules the physics problems presented here [u]are[/u] intermediate algebra problems.

$Q = Q_{0}\cdot e^{-\mu t}$

$\Rightarrow \log (Q) = \log (Q_{0})-\mu t$

$\Rightarrow \log (16) = \log (20)-\mu (\text{ 10 minutes})$

$\Rightarrow \mu =  \frac{(\log 16-\log 20)}{-\text{10 minutes}}$

$\Rightarrow \mu =\frac{(2.7726-2.9957)}{-\text{10 minutes}}= 0.0223 \text{ per minute}$.

$\therefore \mu = \frac{0.0223}{\text{min}}\times\frac{1 min}{60 seconds}= 0.000372 \text{ per second}$.