inequalities
Sequence
algebra
IMO Shortlist

The sequence $c_{0}, c_{1}, . . . , c_{n}, . . .$ is defined by $c_{0}= 1, c_{1}= 0$, and $c_{n+2}= c_{n+1}+c_{n}$ for $n \geq 0$. Consider the set $S$ of ordered pairs $(x, y)$ for which there is a finite set $J$ of positive integers such that $x=\textstyle\sum_{j \in J}{c_{j}}$, $y=\textstyle\sum_{j \in J}{c_{j-1}}$. Prove that there exist real numbers $\alpha$, $\beta$, and $M$ with the following property: An ordered pair of nonnegative integers $(x, y)$ satisfies the inequality \[m < \alpha x+\beta y < M\] if and only if $(x, y) \in S$.

[i]Remark:[/i] A sum over the elements of the empty set is assumed to be $0$.

Anyone has a solution for this problem ? I can find the solution for all algebra problems in the shortlist except this problem..  :(

I think that I didn't understand the problem,
you want we showing that S is limited?
I think that if we take $ (x_{n},y_{n})=(\sum_{i=1}^{n}c_{i},\sum_{i=1}^{n}c_{i-1})$
$ \exists n\in\mathbb{N}: \ min(x_{n},y_{n})=min(\sum_{i=1}^{n}c_{i},\sum_{i=1}^{n}c_{i-1})>M$
then we have $ (x_{n},y_{n})\in S$ and $ \alpha x_{n}+\beta y_{n}>M$

I'm sorry aviateurpilot, I don't really understand your post.  I worked on this problem a couple weeks ago for a couple days and couldn't hammer together a full proof.  I have good reason to believe that $ (x,y)\in S$ if and only if $ -\gamma-1<x-\gamma y<\gamma$ where $ \gamma$ is the positive real such that $ \gamma^{2}=\gamma+1$, but the proof always eludes me.  I'll give it another try, but if anyone can solve this I would love to see it too.

Okay, I think that I have obtained a solution.  If anyone sees errors in this, let me know so I can try to fix them, and also I would love to see a shorter way to do this problem if there is one.

I claim that $ (x,y)\in S$ if and only if $ -1-\gamma<x-\gamma y<\gamma$ where $ \gamma$ is the positive root of $ x^{2}-x-1$.

First, we need an important fact about Fibonacci numbers.  Let $ f_{0}=0,f_{1}=1$ and $ f_{n+2}=f_{n+1}+f_{n}$ for all $ n\ge 0$.  Then we have for all $ k\ge 0$

$ f_{k+1}-\gamma f_{k}=\frac{1}{2\gamma-1}(\gamma^{k+1}-(1-\gamma)^{k+1})-\frac{\gamma}{2\gamma-1}(\gamma^{k}-(1-\gamma)^{k})$
$ =\frac{1}{2\gamma-1}(\gamma (1-\gamma)^{k}-(1-\gamma)^{k+1})=(1-\gamma)^{k}$.

Knowing this, we can now proceed.  First suppose $ (x,y)\in S$.  Then we have some set $ J$ as mentioned in the problem, and so

$ x-\gamma y=\sum_{j\in J}(c_{j}-\gamma c_{j-1})=\sum_{j\in J}(1-\gamma)^{j-1}$ for the moment as long as we ignore the fact if $ 1\in J$.  If we let $ p$ be the greatest odd element in $ J$, we can deduce that
$ x-\gamma y\le (1-\gamma)^{0}+(1-\gamma)^{2}+\cdots+(1-\gamma)^{p-1}<\frac{1}{1-(1-\gamma)^{2}}=\gamma$.

Similiarly, if we account that $ 1$ can be in $ J$, we have $ x-\gamma y>(0-\gamma)+(1-\gamma)+(1-\gamma)^{3}+\cdots>-\gamma+\frac{1-\gamma}{1-(1-\gamma)^{2}}=-1-\gamma$.  This concludes the first direction.

To prove the other direction, suppose the two integers $ x,y$ satisfy $ -1-\gamma<x-\gamma y<\gamma$.  Now, for any positive integer $ x$, consider the following algorithm:  Pick the greatest Fibonacci number less than or equal to $ x$, subtract it, and then repeat until we arrive at zero.  In this way, we can derive a set $ J$ such that $ \sum_{j\in J}c_{j}=x$ and we can guarantee that $ 1$ or $ 2$ is not in $ J$ (this means that there is at most one $ 1$ in the sum).

Now, for any $ i\ge 4$, if $ i-2$ and $ i-1$ do not belong to $ J$ and $ i$ does, we can remove $ i$ from the set and add $ i-2$ and $ i-1$ to the set.  In this manner, we start with the smallest value in $ J$ and try to achieve a configuration with precisely one $ 1$ if we do not have it already.  Suppose that there is precisely one $ 1$ in some partition of $ x$, i.e., $ min{J}=3$.  Then for the corresponding $ y$ induced from this $ J$, we have $ x-\gamma y=\sum_{j\in J}c_{j}-\gamma c_{j-1}$ from which we can deduce the two bounds:

$ x-\gamma y\le (1-\gamma)+(1-\gamma)^{2}+(1-\gamma)^{4}+\cdots<0$ and

$ x-\gamma y\ge (1-\gamma)+(1-\gamma)^{3}+\cdots>-1$.

By moving adding $ 1$ to $ J$, we can increase $ y$ by 1 without changing $ x$, and so $ -\gamma-1<x-\gamma y<-\gamma$, thus giving us a new value in the interval.  Also, before adding $ 1$ to $ J$, if we just remove $ 3$ and add $ 2$, we do not change $ x$ but we decrease $ y$ by 1.  Thus $ \gamma-1<x-\gamma y<\gamma$ which also gives us a new value.  But since the interval is of length $ 2\gamma+1$, there can only be at most $ 3$ values in this interval.  But since they are all accounted for, we can finish this case.

So all we have left is the case when any partition of $ x$ does not contain exactly one $ 1$.  In this case, with the same strategy above we can find that $ x-\gamma y<\gamma-1$.  However, to derive a lower bound, we must notice that the only way for $ x$ to not contain exactly one $ 1$ is if the smallest element in $ J$ is even (because of the type of "move" described above).  Hence for some $ t\in\mathbb{N}$ we have $ x-\gamma y>(1-\gamma)^{2t}+(1-\gamma)^{2t+1}+(1-\gamma)^{2t+3}+\cdots>0$.  But then using this bound, we know that there cannot be a pair $ (x,y)$ in the region with a value greater than this one, and moreover, only one value with a different $ y$ can fit below this interval.  But to derive this value, simply just add $ 1$ to the set $ J$, and from here we obtain the desired $ y$.  Because there are no more possibilites for $ y$, we are finished.

O strongly believe this problem(and therefore zgorkster's solution) is incorrect. The IF part is pretty intuitive once one notices the fact that $c_{n}=F_{n-1}$ and achieves the bound
$-\alpha -1<x-\alpha *y<\alpha$. But then you realoze that for every $y$ we have at least two integers $x$ satisfying the above inequality, but Zeckendorf's representation tells us that for every y only one x is possible

Altough there's one correction that would make the problem correct: if instead he asked that the statement is true for all pairs $(x,y)$ of POSITIVE INTEGERS, because then the lower bound would be simply $-1<x-y\alpha<\alpha\iff y\in (\frac{x}{\alpha}-1,\frac{x}{\alpha}+\frac{1}{\alpha}$. But then it wouldn't make sense to redefine the sequence $c_{n}$ by shifitng Fibonacci's indices if you weren't gonna use $c_{0}=F_{-1}=1$
What a mess

First let $\phi_1>\phi_2$ be the roots to $\phi^2-\phi-1=0$. That is, $\phi_1=\frac{1+\sqrt{5}}{2}$ and $\phi_2=\frac{1-\sqrt{5}}{2}$.
We claim that $m=-1$, $M=\phi_1$, $\alpha=\phi_2$ and $\beta=1$ works.

It is well-known that for all $n\geq 0$, $$c_n=\frac{\phi_1^{n-1}-\phi_2^{n-1}}{\sqrt{5}}$$
Hence,
$$c_n\phi_2+c_{n-1}=\phi_2\left(\frac{\phi_1^{n-1}-\phi_2^{n-1}}{\sqrt{5}}\right)+\frac{\phi_1^{n-2}-\phi_2^{n-2}}{\sqrt{5}}=\phi_2^{n-1}$$

Let the pairs in $S$ be $(a_J, b_J)$ with $J\in\mathbb{N}$, with $a_J=\sum_{j\in J} c_j$ and $b_J=\sum_{j\in J} c_{j-1}$.

We first show that $\phi_2 a_J+b_J\in (-1,\phi_1)$.
$$\phi_2 a_J+b_J=\sum_{j\in J} (c_j \phi_2+c_{j-1})=\sum_{j\in J} \phi_2^{j-1}$$
Since $-1<\phi<0$, this sum is bounded by the following estimate:
$$\sum_{k=0}^{\infty} \phi_2^{2k+1}<\sum_{j\in J} \phi_2^{j-1}<\sum_{k=0}^{\infty} \phi_2^{2k}$$
Note that $\sum_{k=0}^{\infty} \phi_2^{2k+1}=\frac{\phi_2}{1-\phi_2^2}=-1$ and $\sum_{k=0}^{\infty} \phi_2^{2k}=\frac{1}{1-\phi_2^2}=-\frac{1}{\phi_2}=\phi_1$
Hence for all $(x,y)\in S$, we have $\alpha x+\beta y\in (m,M)$.

Next we show the other direction.
[b]Claim:[/b] For all $(x,y)\in \mathbb{N}^2$ with $-1<\phi_2 x+y<\phi_2$, there exists $J\subset\mathbb{N}$ such that $\phi_2 x+y=\sum_{j\in J} \phi_2^{j-1}$.
[i]Proof:[/i] For $x=y=0$, choose $J=\emptyset$. Suppose at least one of $x$ and $y$ is nonzero. Then
$$\phi_2 x+y=\phi_2^{i_1}+\phi_2^{i_2}+\cdots+\phi_2^{i_k}$$ for some sequence $i_1\leq i_2\leq \cdots \leq i_k$. This sequence exists, as we can choose $i_1=i_2=\cdots=i_y=0$, and $i_{y+1}=i_{y+2}=\cdots=i_{y+x}=1$.

From all such sequences, consider those with minimum length $k$. From these, consider those with minimum $i_1$. Then consider those with minimum $i_2$, then minimum $i_3$, and so on. We end up with a sequence $j_1\leq j_2\leq \cdots \leq j_k$. It suffices to show that all these terms are different.
Suppose otherwise, i.e. there exists $r$ such that $j_r=j_{r+1}$.

Case I: $j_r=j_{r+1}\geq 2$.
Then noting that $2\phi_2^2=1+\phi_2+\phi_2^3-\phi_2=1+\phi_2^3$, we can replace $(j_r,j_{r+1})$ with $(j_r-2,j_r+1)$. This contradicts the minimality of elements in the sequence.

Case II: $j_r=j_{r+1}=0$.
Note that $j_t\neq 1$ for all $t$, as $1+\phi_2=\phi_2^2$ and we can replace the $(0,1)$ with $2$. This contradicts the minimality of sequence length.
In fact, we cannot have two elements in the sequence differing by $1$. Hence,
$$\phi_2 x+y=\sum_{r=1}^k \phi_2 j_r>2+\phi_2^3+\phi_2^5+\cdots=2-\phi_2^2=\phi_1$$
Contradiction.

Case III: $j_r=j_{r+1}=1$.
Note that $j_t\neq 0$ or $2$ for all $t$. Following the same logic as Case II, we cannot have two elements in the sequence differing by $1$.
$$\phi_2 x+y<2\phi_2+\phi_2^4+\phi_2^6+\cdots=2\phi_2-\phi_2^3=1$$
Contradiction.

Hence the lemma is proven. Now note that for all pairs of $(x,y)$ with $-1<\phi_2x+y<\phi_1$, there exists $J_{x,y}\subset \mathbb{N}$ such that $\sum_{j\in J_{x,y}}\phi_2^{j-1}=\phi_2 x+y$. Then $\phi_2 x+y=\phi_2 a_{J_{x,y}}+b_{J_{x,y}}$. Since $\phi_2$ is irrational, it follows that $a_{J_{x,y}}=x$ and $b_{J_{x,y}}=y$. Hence we are done.

Solved with [b]anser[/b].

Let $\phi = \frac{1+\sqrt{5}}{2}$, satisfying $\phi^2 - \phi - 1 = 0$.

We'll show $$-\phi^2 < x - \phi y < \phi$$ for all $(x, y) \in S$. First by solving the recurrence we get $c_n = -\frac{1 - \sqrt{5}}{2\sqrt{5}}\cdot \phi^n + \frac{1 + \sqrt{5}}{2\sqrt{5}}\cdot \left(-\frac{1}{\phi}\right)^n$. Then we have $$c_{n + 1} - \phi c_n = -\phi \cdot \left(-\frac{1}{\phi}\right)^n.$$ If $(x, y) \in S$, then $x - y$ is the sum of this over distinct values of $n \geq 0$. This means $x - \phi y$ is between $$-\phi\cdot \left(1 + \frac{1}{\phi^2} + \cdots\right) = -\phi \cdot \frac{\phi^2}{\phi^2 - 1} = -\phi^2$$ when the sum has only positive terms, and $$-\phi \cdot -\frac{1}{\phi}\cdot \left(1 + \frac{1}{\phi^2} + \cdots\right) = \phi$$ when the sum has only negative terms.

Now suppose that $-\phi^2 < x-\phi y < \phi$, or $\frac{x}{\phi} - 1 < y < \frac{x}{\phi} + \phi$. Let the Fibonacci representation of $x$ be the unique way to express $x$ as a sum of nonadjacent Fibonacci numbers. The claim is that the smallest term in the Fibonacci representation of $x$ is $F_n$ for $n$ even iff $(\frac{x}{\phi} - 1, \frac{x}{\phi} + \phi)$ contains 3 integers, that is, $2-\phi < \left\{\frac{x}{\phi}\right\} < 1$.

Since $F_n = \frac{1}{\sqrt{5}}\phi^n - \frac{1}{\sqrt{5}}\frac{1}{(-\phi)^n}$,
\[\frac{F_n}{\phi} = \frac{1}{\sqrt{5}}\phi^{n-1} + \frac{1}{\sqrt{5}}\frac{1}{(-\phi)^{n+1}} = f_{n-1} + \frac{1}{\sqrt{5}}\left(\frac{1}{(-\phi)^{n-1}} + \frac{1}{(-\phi)^{n+1}}\right).\]
So $\left\{\frac{x}{\phi}\right\} = \sum_i \frac{1}{\sqrt{5}}\left(\frac{1}{(-\phi)^{i-1}} + \frac{1}{(-\phi)^{i+1}}\right)$, where the sum is over all $i$ s.t. $F_i$ is in the Fibonacci representation of $x$.

The maximum and minimum possible values (both unattainable) of this sum in some order are given by
\[\frac{1}{\sqrt{5}}\left[\frac{1}{(-\phi)^{n-1}} + 2\left(\frac{1}{(-\phi)^{n+1}} + \frac{1}{(-\phi)^{n+3}} + \cdots\right)\right]= \frac{1}{\sqrt{5}}\cdot\frac{1}{(-\phi)^{n-1}}(3\phi - 4)\]
and
\[\frac{1}{\sqrt{5}}\left[\frac{1}{(-\phi)^{n-1}} + \frac{1}{(-\phi)^{n+1}} + \frac{1}{(-\phi)^{n+2}} + 2\left( \frac{1}{(-\phi)^{n+4}} + \cdots\right)\right] = \frac{1}{\sqrt{5}}\cdot\frac{1}{(-\phi)^{n-1}}(2\phi -1).\]

When $n$ is even, $\frac{1}{\sqrt{5}}\cdot\frac{1}{(-\phi)^{n-1}}(3\phi - 4) < 0$, and $\frac{1}{\sqrt{5}}\cdot\frac{1}{(-\phi)^{n-1}}(2\phi -1)\ge \frac{1}{\sqrt{5}}\cdot\frac{1}{-\phi}(2\phi - 1) = -\phi + 1$, so $2-\phi < \{\frac{x}{\phi}\} < 1$.

When $n$ is odd (note $n\ge 3$), then $\frac{1}{\sqrt{5}}\cdot\frac{1}{(-\phi)^{n-1}}(3\phi - 4) > 0$, and $\frac{1}{\sqrt{5}}\cdot \frac{1}{(-\phi)^{n-1}}(2\phi -1) \leq \frac{1}{\sqrt{5}\cdot \phi^2}\cdot (2\phi - 1) = 2 - \phi$, so the $0 < \{\frac{x}{\phi}\} < 2-\phi$.

We can always obtain 2 possibilities for $y$, depending on whether or not we include $c_1 = 0$ in the sum for $x$. If the smallest term in the Fibonacci representation of $x$ is $F_n$ for $n$ even, then by "breaking down" $F_n$, we may express $x$ as a sum of 1 and some larger Fibonacci numbers. Then there are 3 possibilities for $y$ by considering whether we choose the 1 in $x$ to be $c_3$, $c_2$, or $c_3 + c_1$.

For typesetting purposes let $a = \phi = \frac{\sqrt{5}+1}{2}$. We claim that a pair of nonnegative integers $(x,y)$ is in $S$ iff $$-1-a < x-ay < a.$$

First, we prove that this is necessary. Note that for positive $n$, $c_{n+1}$ is the $n$th Fibbonaci number and is therefore equal to $\frac{a^n - (-a)^{-n}}{\sqrt{5}}$. Therefore, for $n>2$, $c_n-ac_{n-1}$ can be expressed as $$\frac{1}{\sqrt{5}} ((a^{n-1} - (-a)^{-(n-1)})-a(a^{n-2} - (-a)^{-(n-2)}) = \frac{1}{\sqrt{5}}(a(-a)^{-(n-2)}-(-a)^{-(n-1)}).$$

To get a lower bound for $x-ay$, take all positive integers $j$ where $c_j-ac_{j-1}$ is negative, and add the $c_j-ac_{j-1}$ value for each $j$. Notice that for $j\ge 2$ this quantity is only negative when $j$ is odd, and when $j=1$ the quantity is negative, so these are the terms that we need to add up. Note that when $j\ge 2$ and $j$ is odd we have $c_j-ac_{j-1} = \frac{1}{\sqrt{5}}(-a^{-j+3}-a^{-j+1}).$ As this is a geometric sequence, it's sum is $\frac{1}{\sqrt{5}}(-1-a^{-2})\cdot \frac{1}{1-a^{-2}} = -1$ after some computation. Considering $j=1$, we add another $0-a\cdot 1 = -a$, making a lower bound for $x-ay$ equal to $-1-a$.

Similarly, to get an upper bound for $x-ay$, take all positive integers $j$ where $c_j-ac_{j-1}$ is negative, and add the $c_j-ac_{j-1}$ value for each $j$. Notice that for $j\ge 2$ this quantity is only positive when $j$ is even, and we don't have to account for $j=1$. Note that when $j$ is even we have $c_j-ac_{j-1} = \frac{1}{\sqrt{5}}(a^{-j+3}+a^{-j+1}).$ As this is a geometric sequence, it's sum is $\frac{1}{\sqrt{5}}(a+a^{-1})\cdot \frac{1}{1-a^{-2}} = a$ using an earlier computation. Thus an upper bound for $x-ay$ is $a$. This proves necessity.

Now we prove sufficiency. If we take a particular set of $c_i$ that sum to $x$, this gives us a corresponding $y$ that works. Note that for a particular $x$ there are either two or three values of $y$ that could work. Let $Z$ be the Zeckendorf representation of $x$ ($x$ written as a sum of nonconsecutive Fibonacci numbers). If there are two possible values of $y$, then noting that we can take the $c_i$ corresponding to $Z$, or take the $c_i$ corresponding to $z$ in addition to $c_1$ gives us two values of $y$ that work. The $x$ that have three possible values of $y$ that could work can be solved by the following claim:

[b][color=#f00]Claim:[/color][/b] If $x$ has three possible values of $y$, then the smallest element of $Z$ is equal to $F_k$, where $k$ is even. ($F_0 = 0, F_1 =1$)
[i]Proof.[/i] It suffices to prove the contrapositive: when the lowest element of $Z$ is equal to $F_k$ for an odd $k>1$, there are only two possible values of $y$ appearing from that $x$. Define two numbers to be equivalent modulo $a$ if they differ by a multiple of $a$. Then it is easy to show that we need to prove $x$ is equivalent to a number between $0$ and $a-1$ modulo $a$ (exclusive).

For every Fibonacci number $F_i$ in $Z$, consider adjusting to $F_i - aF_{i-1}$. It is not hard to show the adjustment is negative if $i$ is even and positive if $i$ is odd. The sum of all these adjusted terms must clearly be equivalent to $x$ modulo $a$. The adjusted term for $F_k$ must be equal to $$\frac{1}{\sqrt{5}}(a^{-k+2}+a^{-k})$$ by earlier work. Now we place upper and lower bounds to the sum of the remaining adjusted terms.

An lower bound can be formed by taking all possible adjusted terms coming from even $F_i$. Because this is the Zeckendorf representation $F_{k+1}$ cannot be included, thus the minimum possible value is $\frac{1}{\sqrt{5}}(-a^{-j+2}-a^{-j})$ summed over all even values of $j$ at least $k+3$, which is equal to $\frac{1}{\sqrt{5}}(-a^{-k-1}-a^{-k-3})\cdot\frac{1}{1-a^{-2}} = \frac{1}{\sqrt{5}}(-a^{-k}-a^{-k-2}).$

An upper bound can be formed by taking all possible adjusted terms coming from odd $F_i$ greater than $k$. This is equal to $\frac{1}{\sqrt{5}}(a^{-j+2}+a^{-j})$ summed over all even $j$ at least $k+2$, which is equal to $\frac{1}{\sqrt{5}}(a^{-k}+a^{-k-2})\cdot\frac{1}{1-a^{-2}} = \frac{1}{\sqrt{5}}(a^{-k+1}+a^{-k-1}).$

Therefore, the sum of all adjusted terms must be positive, and it must be less than $\frac{1}{\sqrt{5}}(a^{-k+2}+a^{-k+1}+a^{-k}+a^{-k-1}) \le \frac{1}{\sqrt{5}}(a^{-1}+a^{-2}+a^{-3}+a^{-4}) = a-1$, proving the claim. $\blacksquare$

Now take the Zeckendorf representation of $x$ and replace $F_k$ by $F_{k-1} + F_{k-3} + F_3 + F_2$. Substitute the appropriate $c_i$ (with $i>2$), noting that we use $c_3$. This allows us to represent one $y$. Now replacing $c_3$ with $c_2$ gives us a way to represent a lower $y$, and adding $c_1$ allows us to represent a higher $y$. Therefore, we can represent three values of $y$, finishing the problem.

Set $\phi$ to be the golden ratio. Observe that
\[ (1 - \phi)^n = c_{n+2} - c_{n+1}\phi; \]
Indeed, verify that $\frac{1}{1 - \phi} = -\phi$ since this implies $\phi^2 = \phi + 1$, and then induction suffices.

We claim that $(x,y) \in S \iff -1 - \phi < x - \phi y < \phi$. Indeed, defining $x$ and $y$ for a fixed set, we have
\[ x - y\phi = \sum_{s \in S} (1 - \phi)^{s - 1}. \]
On the one hand,
\[ \sum_{s \in S} (1 - \phi)^{s - 1} \ge (1-\phi)^{-1} + (1 - \phi)^{-3} + \cdots = \frac{1}{(1-\phi)(1 - (1 - \phi)^2)} = \frac{1}{(1 - \phi)(\phi - 1) } = -1-\phi, \]
on the other hand
\[ \sum_{s \in S} (1 - \phi)^{s - 1} \le (1 - \phi)^0 + (1 - \phi)^2 + \cdots = \phi.\]
This proves necessity.

Suppose $\phi > x - \phi y > 0.$ Then for each positive $x$, there is a unique $y$ which satisfies the inequality. Particularly,
\[ \left \lfloor \frac{x}{\phi} \right \rfloor = y. \]

We will create some process which never terminates, allowing us to generate solutions.

Note that for $x=0$, $y=0$, we can set $x = 0$.
Whenever we add $c_2$, $x$ increments by $1$, and $y$ increments by $0$. Whenever we add $c_3$, $x$ increments by $1$, and $y$ increments by $1$.

Now when $x$ increments by $1$, $y$ will increment by at most $1$. Note that $y$ increments at most twice in a row, and $y$ stays fixed at most once in a row.

At any given time, we take the two largest terms with adjacent indices and merge them, i.e. $c_n + c_{n+1} = c_{n+2}$. We will show this process never generates a term with a coefficient of $2$.

We show that $c_n$ for $n > 2$ never doubles up. Firstly, $c_3$ never doubles up, since before any $c_3$ placement we are forced to put a $c_2$, so the first $c_3$ is absorbed by either $c_2$ or $c_4$. If there is a leftover $c_3$, the first $c_3$ gets turned into $c_4$, and then $c_3$ merges with $c_4$ to $c_5$. So no issues here.

Similar arguments hold for $n > 3$.

Now for $c_2$, the only issue is if we have $c_2 + c_4$, and we place a singular $c_3$.
We will show this never happens.
Indeed suppose that it did. The only way for this to occur is to start with no terms with index less than $5$, place $c_2$, place a singular $c_3$, place a $c_2$, and then place a singular $c_3$ again, and place $c_2$. This means that $x$ increments by $5$ and $y$ increments by $2$. But
\[ \left \lfloor \frac{x + 5}{\phi} \right \rfloor + \left \lfloor \frac{x}{\phi} \right \rfloor \ge 3. \]
This is a contradiction.

Suppose $-1 < x - \phi y < 0$. Then for each fixed $y$, there is a unique $x$ which satisfies the inequality. This is, of course,
\[ -1 + \phi y < x < \phi y \implies x = \lfloor \phi y \rfloor.  \]

Now whenever $y$ increments by $1$, $x$ increments by either $1$ or $2$.
Again note we can set $y$ and $x$ to have no terms for now.
If we add $c_2$ to $y$, we have $x$ incrementing by $1$. If we add $c_3$ to $y$, we have $x$ incrementing by $2$.
Now $x$ can only increment by $2$ at most twice in a row, since $3\phi < 5$. $x$ can only increment by $1$ once at a time since $2\phi > 3$
We apply the same process; the proofs for $c_n$ for $n > 2$ are analogous. It suffices to show that having 3 $c_2$ placements and 2 $c_3$ placements (in a row) is impossible; what this means is we have $x$ incrementing by $7$ and $y$ incrementing by $5$.
However,
\[ \lfloor \phi y + 5\phi \rfloor - \lfloor \phi y \rfloor \ge 8, \]
since $5\phi > 8$.

For the last case $-1 - \phi < x - \phi y < - \phi$, we let $x$ determine a fixed $y$. Note $(0,1)$ can be attained by adding $c_0$ to $y$, and we are done by the first process.

Whew.


abstract algebra
superior algebra
superior algebra solved

let $ I$ be a finitely-generated ideal in a ring $ A$ with $ I \equal{} I^2$. why is $ I$ of the form $ eA$ for some idempotent $ e$?

it's enough to show that $ I$ is principal (for if $ I\equal{}cA$, then $ c\equal{}dc^2$ for some $ d$, and we can take $ e\equal{}dc$).

after having found out that elementary calculations don't work, I've found a solution using nakayama's lemma. a bit confusing since this exercise was posed in a context with a weaker nakayama lemma (qing liu, ex. 1.2.4)

$ I$ is a finitely-generated $ A$-module such that $ I*I \equal{} I$. nakayama implies the existence of some $ a \in I$ such that $ I(1 \minus{} a) \equal{} 0$. in particular, $ I$ is generated by $ a$ and $ a$ is idempotent.

See my reply to Colin McLarty on Jan 4, 2005 in "[HM] Question on ideals" here
http://mathforum.org/kb/thread.jspa?threadID=1111328&messageID=3648504


Hi everyone out there in the states,
anyone else doing All-State orchestra, band, chorus or jazz ensemble out there?

I'm a violinist, in the Massachusetts All-State orchestra. We're playing Stravinski's L'oiseau de Feu (Firebird Suite) this year (the festival is March 16-18 for me).

What about you?

For those of you who don't know what All-states are, its a program organized by public school music educators across the country. Each year, students from public high schools go to auditions for the ensemble they wish to play in; the first 'level' you audition for are the district level ensembles. If you do well at the district audition, you recieve a recommendation for the all state auditions (and I've heard that if you do extremely well at those auditions, they have national level orchestras every couple of years, although having gotten into all states twice, i havn't personally heard of anything beyond it). Once you are accepted into an ensemble, you have a few weeks to prepare the music before you go to the music festival, rehearse until you drop dead for two days straight, and have a concert. Although rehearsals can get pretty tedious, its still a fun experience overall, and a great way to meet other talented musicians from your area.

[quote="Halluci"]Hi everyone out there in the states,
anyone else doing All-State orchestra, band, chorus or jazz ensemble out there?

I'm a violinist, in the Massachusetts All-State orchestra. We're playing Stravinski's L'oiseau de Feu (Firebird Suite) this year (the festival is March 16-18 for me).

What about you?[/quote]
Hey Halluci,

In Canada, it works somewhat differently (we don't call them all-state bands, obviously).  Our provincial honour bands are run by each provincial band association, and Alberta's band is run once a year in October.  It has pretty much the same format as yours - intensive rehearsals for several days, and then a performance at the end.  The Alberta Band Association (ABA), as well as those of Saskatchewan, British Columbia, Manitoba, Ontario, etc., are all tied closely with the Canadian Band Association, which has administrative ties to each of its member organizations.

The Canadian Band Association (CBA) also runs its own band, called the National Youth Band (NYB).  Like the Alberta Honour Band (AHB), it runs once a year, but it holds nationwide auditions.  It usually takes place in May, and lasts for about a week.  I didn't get into this year's band, but I'm going to audition for the 2007 band.

We also have a National Youth Orchestra (NYO), but it is a separate organization from the CBA.  They both receive funding from the government and local sponsers.  The NYO is held every summer, and has pretty much the same format as the NYB.

Personally, I play, and have played, in a variety of ensembles.  I participated in the AHB last October, which was hosted in Red Deer.  At our concert, we played the following works:

- The Fisher Who Died in his Bed; John Herberman
- Puzta; Jan Van der Roost
- Lord of the Rings Suite: I. Gandalf; Johan de Meij
- Enigma Variations: IX. Nimrod (for concert band); Edward Elgar
- Galop; Dmitri Shostakovich

I also play 1st alto in an ensemble called the "Littlebirds", which is a big band run by the Edmonton Jazz Society, a non-profit organization devoted to maintaining legitimate jazz in Edmonton.  It is an auditioned high school big band which accepts members from the Edmonton metropolitan area - some of the members live pretty far away.  We have 4 gigs a year, 2 of which we've already played.  The next one is scheduled for March, and the final one for May.

This summer, I will also be joining the University of Alberta's summer concert band.  Anyone who has a competent proficiency level on their instrument can join; in addition to undergraduate and prospecting university students, people from community bands regularly participate in it.

And as for my school bands... meh, they're pretty inconsequential.  I'm almost done with that anyway.

I was in South Carolina's All-state band from 2001 to 2004 as a percussionist.  There's another pretty good program that  BOA does called the Honor Band of America, which I participated in in 2003 and 2004.  Some of my favorite experiences in band were in those ensembles, especially in 2003 HBOA and 2004 All-state.

I was in the Missouri All-state Orchestra this year as a violinist. We played:

Bernstein's Overture to Candide
Tchaikovsky's Romeo and Juliet Overture
A commissioned work by Michael Mclean, Scherzo Fantastique.

It was really an amazing experience, as I'm sure anyone else who's been to one of these things can relate to. I've actually played a good chunk of the Firebird Suite before in my city's civic orchestra, and I enjoyed it a lot. Definitely a very interesting piece, as well as very challenging. What did you think of it?

I went this year as a percussionist in Washington; we played:
Hector Berlioz's Roman Carnival
Chaubrier's España Rhapsody
Commissioned work by David Ott

Forgot to mention, I was in All-state Orchestra. My brother was in All-state band, and they played:
March Slave, by Tchaikovsky
among others.

did you know that the texas average all region honors band is as good as the other state bands?
my region played english dances set II
Did you know that the music the texas all-state band plays is specifically made for texas all-state band since they are so good?
I dont think so, so why don't you go on youtube and search texas all-state band and listen to slalom and Nitor

how dramatic...

I've been in NJ All-State Orchestra as a violinist, and last year we did Capriccio Italien by Tchaikovsky, Carneval by Dvorak, and El Salon Mexico by Copland.

Did anybody on these forums go to MENC All Eastern Orchestra last March? (Frenergy, Shostakovich 5)

[quote="Halluci"]For those of you who don't know what All-states are, its a program organized by public school music educators across the country. Each year, students from public high schools go to auditions for the ensemble they wish to play in; the first 'level' you audition for are the district level ensembles. If you do well at the district audition, you recieve a recommendation for the all state auditions (and I've heard that if you do extremely well at those auditions, they have national level orchestras every couple of years, although having gotten into all states twice, i havn't personally heard of anything beyond it). Once you are accepted into an ensemble, you have a few weeks to prepare the music before you go to the music festival, rehearse until you drop dead for two days straight, and have a concert. Although rehearsals can get pretty tedious, its still a fun experience overall, and a great way to meet other talented musicians from your area.[/quote]

It works slightly differently for us in California. We just apply straight to the All-State Orchestra instead of working through the District level first.

Also, since CA is such a big state, we break it up, so we also have an All-Southern Orchestra program (don't know if there's one for Norcal) and then we also have All-State.

I've made it to All-Southern for the past 3 years as a violinist, and it's a great experience!
Last year, I think we played some Rachmaninoff symphony or something. This year, we had a bit more variety, playing Slava by Bernstein, Beethoven's Egmont, and a Tchaikovsky Symphony


There was a question on a test (duke math meet) to find the largest prime factor of (3^12)-1.

I have no idea on how to even start this problem. Could someone please tell me how to solve this problem, and provide a more generalized method of solving any problems of this type? Thanks ;)

Now try the following slightly harder question: find the three prime factors of $2^{22}+1$.

Let's see,

2^22+1 is in the form of x^22+1. I might be able to express this as x^22+1-1+1, so it equals x^22-1+2, so that in turn equals (x-1)(x^21+x^20+x^19+...+x+1)+2. As for how to proceed with this, I have no clue. However, I am sure there must be a more efficient way. What do you suggest ComplexZeta?


limit
algebra
polynomial
real analysis
real analysis unsolved

Hi!

I have some confusiong about solving
$\frac{s+5}{(s+1)(s^{2}+1)}=\frac{A}{s+1}+\frac{Bs+C}{s^{2}+1}$

One could simply solve this by partial fraction decomposition.
So far so god, but it's offen more easy to use limits.
So
$A=\lim_{x\to-1}\frac{s+5}{s^{2}+1}=\frac{6}{2}=3$

this is not [b]OK[/b] if one solves by partial fraction decomposition...
Since $A=-B=2 ,C=3$
as for next limit it would go to $s^{2}+1=0 \to s=+-i$
taking limit $s\to i$ one gets 0 in denominator...

could some one please explain limit method for solving this type of problems

oh sorry.. this was an aquard misstake.. but what about case were one get 0 in denominator?

well doing the limit as s approaches i doesnt make sense in achieving a solution

[quote="mlok"]I think it does: equating residues of both sides at $\pm i$, you would find $B$ and $C$. But instead of using residues, it's better to multiply by common denominator and stuff.[/quote]
why it's is better? by limmit method (if you can it good(!)) you can find $B,C ...Z$ allmost instantly (well in nice limits anyway).
but using by partial fraction decomposition can take long time and realy nasty calculations whn you for ex have power of 5 or more...

[quote="greengrass"]When you solve, i.e.,

$\frac{1}{(x-a)(x-b)}\, = \, \frac{A}{x-a}+\frac{B}{x-b}\Rightarrow 1 \, = \, A(x-b)+B(x-a)$

you assumed $x \not= a,b$. So, chossing $x \, =\, a, b$ to calc $A,B$ is not strictly correct. Doing that you are actualy working with limits.[/quote]
You don't actually need to argue with limits. If two polynomials with coefficients in a field are evaluated as equal in infinitely many different points, the polynomials must be equal in every point. This follows from the factor theorem.

[quote="matumba"]why it's is better? [/quote]
I meant it's better to multiply by the common denominator before setting $s=\pm i$. This avoids words like "limit" or "residue" - you just compare the values of two polynomials at certain points.


a+b+c=1
xa+yb+zc = 1
x^2a+y^2b+z^2c = 1

What is the value of x^3a+y^3b+z^3c in terms of x y z?

[hide]Aren't they all (x, y, and z) 1?

The others (a, b, and c) can be any three numbers that add up to 1.[/hide]

x+y+z=1
ax+by+cz = 1
a^2x+b^2y+c^2z = 1

What is the value of a^3x+b^3y+c^3z in terms of a b c?

Hopefully its more undstandable now then

That's the same thing, isn't it? Do you mean $a^{2}x+b^{2}y+c^{2}z = 1$ or $a^{2x}+b^{2y}+c^{2z}=1$?

oh sorry about that.
The first part with the a^2 *x etc

if we have s_k=ax^k+by^k+cz^k, and we are given s_0, s_1, s_2,

we want s_3

you should learn about linear recurrences

it can be shown that

$s_{k}-(x+y+z)s_{k-1}+(xy+yz+zx)s_{k-2}-(xyz)s_{k-3}=0$

for $k\ge 3$

sooo...we have $s_{3}=x+y+z-xy-yz-zx+xyz$

[quote="aahmadi"][quote="Altheman"]if we have s_k=ax^k+by^k+cz^k, and we are given s_0, s_1, s_2,

we want s_3

you should learn about linear recurrences[/quote]
Do any of the AOPS books discuss these? If not, do you know of any good sources on the topic?[/quote]

this is kind of in aops...but not really...this is similar to newton's sums

[quote="aahmadi"][quote="Altheman"]if we have s_k=ax^k+by^k+cz^k, and we are given s_0, s_1, s_2,

we want s_3

you should learn about linear recurrences[/quote]
Do any of the AOPS books discuss these? If not, do you know of any good sources on the topic?[/quote]

I found a good source, with problems, solutions, and general methods:

http://me.in-berlin.de/~rws/rgf.pdf


limit
real analysis
real analysis unsolved

Hello,

Does anybody have an idea of what is the most general situation on which you would have a that

\[\lim_{\lambda\rightarrow+\infty}\sum_{n=1}^\infty a_nf(c_n\lambda) = \sum_{n=1}^\infty a_n \lim_{\lambda\rightarrow+\infty}f(c_n\lambda)\]

Asuming that $\lim_{\lambda\rightarrow+\infty}f(\lambda)$, $\lim_{\lambda\rightarrow-\infty}f(\lambda)$ both exist and $f(0)$ is defined.

I know that if the series is absolutely convergent, then I can easely do it, but what about when the series is not absolutely convergent. What do I have to know in order to be able to interchange such limits?

Unfortunately, as far as I know, the only general theorem that allows you to exchange the limits in the expression $\lim_{\lambda\to\infty}\lim_{n\to\infty}S(n,\lambda)$ is the one that requires uniform convergence in $n$ or in $\lambda$. It immediately takes care of the case when $\sum|a_k|<+\infty$ and $f$ is bounded but doesn't yield much more than that. So, however sad it is, it seems that there is no really universal criterion and you'll have to prove the possibility to exchange the limits individually for each particular case. As usual, I'll be happy if someone proves me wrong on this issue.  :)


Is it  possible to  set the numbers $ 1,2,3...9$ as $ a(1) , a(2), a(3),....a(9)$ such that the numbers $ \mid a(1) \minus{} 1 \mid, \mid a(2) \minus{} 2 \mid,\mid a(3) \minus{} 3 \mid...\mid a(9) \minus{} 9 \mid$ are all different, and the same with the number $ 1,2,3,...9$?

[quote="Johan Gunardi"][hide]No, because
\[ |a(1) - 1| + |a(2) - 2| + \cdots + |a(9) - 9| = \pm1\pm1\pm2\pm2\pm\cdots\pm9\pm9\le( - 1 - 1 - 2 - 2 - 3 - 3 - 4 - 4 - 5) + (5 + 6 + 6 + 7 + 7 + 8 + 8 + 9 + 9) = 40 < 1 + 2 + \ldots + 9.
\]
[/hide][/quote]

Dear Gunardi see this link.Sorry for the language of the website, but, it says that  The program [b]Turbo Pascal [/b] found 96 solutions

http://garciacapitan.auna.com/problemas/diferencias/index.htm

I misunderstood your problem. I thought that $ \{|a(1)\minus{}1|,|a(2)\minus{}2|,\ldots,|a(9)\minus{}9|\}\equal{}\{1,2,\ldots,9\}$.

If they are different but not necessarily equal $ 1,2,\ldots,9$, I think the problem is quite easy.


since $a$ is positive, plug in $\frac{-b}{2a}$ for $x$, and get your answer

[quote="MCrawford"]The least value of the function $ax^2 + bx + c$ ($a > 0$) is:
\[ \text{(A)}\ -\frac{b}{a} \qquad \text{(B)}\ -\frac{b}{2a} \qquad \text{(C)}\ b^2 - 4ac \qquad \text{(D)}\ \frac{4ac-b^2}{4a} \qquad \text{(E) none of these} \][/quote]

[hide="Answer"] (D) [/hide]

[hide="Hint"]
$ax^2 + bx + c = a(x^2 + \frac {b}{a} x + \frac{c}{a})$
Then complete the square.
[/hide]

[quote="amirhtlusa"][quote="AntonioMainenti"][quote="mel"]Whoops--kunny is right. I forgot that the question was asking for the minimum value of the function, not the minimum value for x.  :blush:[/quote]Hmm, I don't think there is a minimum value for $x$.[/quote]
what????
since $a$ is positive, it means that the function is concave up, so there must be a minimum. the answer is $D$.[/quote]Notice that he said minimum value of $x$, not $f(x)$.

[quote="eryaman"][/quote][quote="amirhtlusa"][quote="AntonioMainenti"][quote="mel"]Whoops--kunny is right. I forgot that the question was asking for the minimum value of the function, not the minimum value for x.  :blush:[/quote]Hmm, I don't think there is a minimum value for $x$.[/quote]
what????
since $a$ is positive, it means that the function is concave up, so there must be a minimum. the answer is $D$.[/quote][quote="eryaman"]Notice that he said minimum value of $x$, not $f(x)$.[/quote]
but the question is asking for the minimum value of the function!! ;)

[quote="amirhtlusa"][/quote][quote="eryaman"][/quote][quote="amirhtlusa"][quote="AntonioMainenti"][quote="mel"]Whoops--kunny is right. I forgot that the question was asking for the minimum value of the function, not the minimum value for x.  :blush:[/quote]Hmm, I don't think there is a minimum value for $x$.[/quote]
what????
since $a$ is positive, it means that the function is concave up, so there must be a minimum. the answer is $D$.[/quote][quote="eryaman"]Notice that he said minimum value of $x$, not $f(x)$.[/quote][quote="amirhtlusa"]
but the question is asking for the minimum value of the function!! ;)[/quote]...but the person I quoted said "the minimum value for $x$"...  ;)

[quote="mel"]lol. This has got to be the most times I've EVER been quoted.

Ok...I meant "I forgot that the question was asking for the minimum value of the function, not the value of x at which the function is a minimum."

Note to self: must be more careful... :P[/quote]
yes please :lol:
by the way the $x$ value at min is $\frac {-b}{2a}$, you were right, but you just didn't read the quesion well!! :lol:

[quote="random_mathematician"]its the axis of symmetry, ie. $-b/2a$

i.e. B[/quote] ;) .  Read the question, closely, again.

not again, please don't start it over ;) , $\frac {-b}{2a}$ is the $x$-intersept at the minimum, the question is asking for the value of the $f(\frac{-b}{2a})$

lol! At least I'm not the only one... :P

Is it true that great minds think alike?


trigonometry
inequalities
inequalities proposed

If ABC is a triangle with the angles A, B, C less than 90 degrees, find the maximum of the expression:
\[ E \equal{} \frac {\cos A\cos B\cos C}{\sin ^{2} A\sin ^{2} B\sin ^{2} C} .
\]

perform the substitution $ (A,B,C)\equal{}\left(90\minus{}\frac{A}{2}, 90\minus{}\frac{B}{2}, 90\minus{}\frac{C}{2}\right)$. recall that $ \sin \frac{A}{2}\equal{}\sqrt{\dfrac{yz}{(x\plus{}y)(x\plus{}z)}}$ and $ \cos \frac{A}{2}\equal{}\sqrt{\dfrac{x(x\plus{}y\plus{}z)}{(x\plus{}y)(x\plus{}z)}}$, where $ (x,y,z)\equal{}(s\minus{}a,s\minus{}b,s\minus{}c)$.

then, we have that

$ \dfrac{\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{\cos^2\frac{A}{2}\cos^2\frac{B}{2}\cos^2\frac{C}{2}}\equal{}\dfrac{(x\plus{}y)(y\plus{}z)(z\plus{}x)}{(x\plus{}y\plus{}z)^3}$

finally, it follows easily that the maximum value is $ \frac{8}{27}$.

Thank you for this nice and ingenious solution. The inequality geometrical meann <or = than arithmethical mean is then used to get that 8/27.

Jean


[quote="combuster"]How to solve X?

X(G-I) = (BL)^2 + X + BC

Thnx![/quote]
My friends : :gathering:  :gathering:  :gathering: [color=red]X,G,I,B,L,C [/color]? Why you do you give  question !
 Let $ S \in S^{ \plus{} \plus{} }_{n}(R)$ Slove equation with $ X$ unknown : $ SX \plus{} XS \equal{} 2S^{2009}$


[quote="ch1n353ch3s54a1l"]In $\triangle ABC$, cevians $AE$, $BF$, and $CG$ intersect at $D$. Prove that $\frac{[ADB]}{[BDC]}=\frac{AF}{FC}$.

I thought this might be easy for intermediate, so I posted it here.  :ninja:[/quote]

[hide]$\frac{[ADB]}{[BDC]}=\frac{[AFB]}{[BFC]}$ because of shared altitudes and bases

and

$\frac{[AFB]}{[BFC]}=\frac{AF}{FC}$ because of shared altitude.[/hide]


At least how many stars should be drawn in a table 4 x 4 such that after eliminating arbitrary two columns and arbitrary two rows at least one star will remain in the table?

(prove your answer)

[quote="peeta"]
(prove your answer)[/quote]

;)

ok, ok ,i will do it as soon as possible :D

maybe you should post the configuration also...

(as I thought it was 8...)

the problem is from russian$1996$ the $7$th problem...

i will talk about it later...after posting my solution...

dear peter,im sure about it ;)

the right form of it in russian $1961$ is :prove that we can draw $7$stars in it.

Well, it seems you're right ashegh..
But, could you show us your solution please?

The Russian problem he's talking about is :
[quote]Given a table 4x4.
a) Find, how 7 stars can be put in its fields in such a way, that
erasing of two arbitrary lines and two columns will always leave at
least one of the stars.
b)Prove that if there are less than 7 stars, you can always find two
columns and two rows, that if you erase them, no star will remain in
the table.[/quote]

 :)

i was looking for a solution as well ... and i finally found one:

$\begin{array}{|c|c|c|c|} \hline & & & \star\\\hline \star & \star & &\\\hline \star & & \star &\\\hline &\star &\star &\\ \hline \end{array}$

dear mathman, i dont have any program which can draw the diagram...

and i think i can show u the solution by using some latex...

but ,suppose we want to put the stars in the squares inwich after eliminating arbitrary two columns and arbitrary two

rows at least one star will remain in the table.

then share the table to two color ,black and white,like the squares in chess.

then if we put the$7$ stars in each $7$squares of $8$squares wich has the same color,by eliminaing each two

arbitrary rows and each two arbitrary columns ,there exists a star wich is remaining in table.

actually,the tings i said,arent the solution.but i wanted to say drawing the stars in the table ,isnt special.

So, a solution with 7 is possible:
[quote="peeta"]$\begin{array}{|c|c|c|c|} \hline & & & \star\\\hline \star & \star & &\\\hline \star & & \star &\\\hline &\star &\star &\\ \hline \end{array}$[/quote]

And a solution with 6 is trivially impossible: by the pigeonhole principle we can remove at least 4 stars by removing 2 columns, and now there are only 2 stars left so we remove their 2 rows and we got an empty field.

AAAHH! That's [b]exactly[/b] what I wanted to write Peter VDD!  (I thought about it during my philosophy test  :P )

Oh, and, don't worry, I am not angry at all ;)

[quote="mathmanman"]Oh, and, don't worry, I am not angry at all ;)[/quote]Of course not, why would you be angry to find the right solution? :P

Well, I originally said so because of the "ahh" in caps + exclamation marks :)

But, yeah, you're right ;)


Let $x;y;z$ are positive real numbers satisfying  $xyz=1$ and $k$ is a number satisfying $k \geq 8$.Prove that:
$\frac{1}{\sqrt{1+kx}}+\frac{1}{\sqrt{1+ky}}+\frac{1}{\sqrt{1+kz}}\geq \frac{3}{\sqrt{1+k}}$

If you can show that,
$\frac{3}{\sqrt{1+kx}+\sqrt{1+ky}+\sqrt{1+kz}}\geq \frac{1}{\sqrt{1+k}}$

Then you can easily make a problem by AM-HM inequality.

I just do not see a way to show that.

[quote]
$\frac{3}{\sqrt{1+kx}+\sqrt{1+ky}+\sqrt{1+kz}}\geq\frac{1}{\sqrt{1+k}}$
[/quote]
It's not true for $x,y\rightarrow 0$.
Actually there is another strong one:
1)forall $x,y,z\in\mathbb{R_{+}}$ and $\lambda\geq\frac{5}{4}$ there is
$\sum\frac{1}{\sqrt{1+\lambda x}}\geq\frac{3}{\sqrt{1+\lambda}}$
2)forall $x,y,z\in\mathbb{R_{+}}$ and $0<\lambda<\frac{5}{4}$ there is
$\sum\frac{1}{\sqrt{1+\lambda x}}>2$
and 2) is very easy by 1). But 1)... Oh so terrible

[quote="scpajmb"]Actually there is another strong one:
1)forall $x,y,z\in\mathbb{R_{+}}$ and $\lambda\geq\frac{5}{4}$ there is
$\sum\frac{1}{\sqrt{1+\lambda x}}\geq\frac{3}{\sqrt{1+\lambda}}$
2)forall $x,y,z\in\mathbb{R_{+}}$ and $0<\lambda<\frac{5}{4}$ there is
$\sum\frac{1}{\sqrt{1+\lambda x}}>2$
and 2) is very easy by 1). But 1)... Oh so terrible[/quote]
It's not true.

Actually
1)forall $x,y,z\in\mathbb{R_{+}},xyz=1$ and $\lambda \ge 8$
 there is
$\sum\frac{1}{\sqrt{1+\lambda x}}\geq\frac{3}{\sqrt{1+\lambda}}$
2)forall $x,y,z\in\mathbb{R_{+}},xyz=1$ and $0 < \lambda < 8$
 there is
$\sum\frac{1}{\sqrt{1+\lambda x}}>1$
3)forall $x,y,z\in\mathbb{R_{+}},xyz=1$ and $0 < \lambda \le \frac{5}{4}$ there is
$\sum\frac{1}{\sqrt{1+\lambda x}}\le\frac{3}{\sqrt{1+\lambda}}$
4)forall $x,y,z\in\mathbb{R_{+}},xyz=1$ and $\lambda>\frac{5}{4}$ there is
$\sum\frac{1}{\sqrt{1+\lambda x}}<2$.[/quote]
How do you prove these inequalities?

[quote="shalex"][quote]
$\frac{3}{\sqrt{1+kx}+\sqrt{1+ky}+\sqrt{1+kz}}\geq\frac{1}{\sqrt{1+k}}$
[/quote]
It's not true for $x,y\rightarrow 0$.
Actually there is another strong one:
1)forall $x,y,z\in\mathbb{R_{+}}$ that $xyz=1$ and $\lambda\geq\frac{5}{4}$ there is
$\sum\frac{1}{\sqrt{1+\lambda x}}\geq\frac{3}{\sqrt{1+\lambda}}$
2)forall $x,y,z\in\mathbb{R_{+}}$ that $xyz=1$ and $0<\lambda<\frac{5}{4}$ there is
$\sum\frac{1}{\sqrt{1+\lambda x}}>2$
and 2) is very easy by 1). But 1)... Oh so terrible[/quote]

I'm sorry for making some mistakes which have been corrected


function AMC AIME USA(J)MO USAMO ARML MATHCOUNTS

[u]Rules:[/u]
I will pose a question.  You try to give me the most popular answers.  If the answer you gave isn't up there, don't be surprised.  It just didn't make our survey.
You will get points based on how many people out of 100 gave the response you give.  So if 36 people said an answer you gave, you get 36 points.  If it's not on the board, you get a strike.  After three strikes, we throw out the question.
And I was unable to gather up 100 people to vote, so I just played with the numbers a bit.

So... let's play!

[b]We surveyed 100 people, top 6 answers on the board.
NAME SOMETHING THAT YOU CAN DO WITH MOST GRAPHING CALCULATORS, BUT NOT WITH SCIENTIFIC CALCULATORS.[/b]
$\boxed{<1>}$
$\boxed{<2>}$
$\boxed{<3>}$
$\boxed{<4>}$
$\boxed{<5>}$
$\boxed{<6>}$

Do we post the answers in this thread or PM them to you?

Graphing Differentials?  Number one answer!

[b]We surveyed 100 people, top 6 answers on the board.
NAME SOMETHING THAT YOU CAN DO WITH MOST GRAPHING CALCULATORS, BUT NOT WITH SCIENTIFIC CALCULATORS.[/b]
$\boxed{\text{GRAPH FUNCTIONS}}\boxed{47}$
$\boxed{<2>}$
$\boxed{<3>}$
$\boxed{<4>}$
$\boxed{<5>}$
$\boxed{<6>}$

[b]nutz_for2.718281828: 47[/b]

[quote="lasagnaman"]generate random numbers![/quote]

I can do that with my TI 30X-IIS, so that shouldn't be up there... :)

I would say program/use graphics, or something like that.

Generate random numbers? *err* First strike.
Using Apps such as games! *ding* Okay, number five.
Let's see programming! *ding* Number two.

[b]We surveyed 100 people, top 6 answers on the board.
NAME SOMETHING THAT YOU CAN DO WITH MOST GRAPHING CALCULATORS, BUT NOT WITH SCIENTIFIC CALCULATORS.[/b]
$\boxed{\text{GRAPH FUNCTIONS}}\boxed{47}$
$\boxed{\text{WRITE PROGRAMS}}\boxed{23}$
$\boxed{<3>}$
$\boxed{<4>}$
$\boxed{\text{PLAY GAMES}}\boxed{6}$
$\boxed{<6>}$
Strikes: 1

[b]b-flat: 23
Ignite168: 6[/b]
nutz_for2.718281828: 47

Aren't you going to keep track of which people got strikes?

Solve Equations

The only penalty for getting a strike is not scoring any points.  So no, I will not be keeping track of how many strikes each person got.
All three of those are wrong.  Let's see the other answers:

[b]We surveyed 100 people, top 6 answers on the board.
NAME SOMETHING THAT YOU CAN DO WITH MOST GRAPHING CALCULATORS, BUT NOT WITH SCIENTIFIC CALCULATORS.[/b]
$\boxed{\text{GRAPH FUNCTIONS}}\boxed{47}$
$\boxed{\text{WRITE PROGRAMS}}\boxed{23}$
$\boxed{\text{FACTOR}}\boxed{11}$
$\boxed{\text{DRAW PICTURES}}\boxed{7}$
$\boxed{\text{PLAY GAMES}}\boxed{6}$
$\boxed{\text{WASTE TIME}}\boxed{3}$

Okay, let's move on.
[b]We surveyed 100 people, top 6 answers on the board.
NAME A WIDELY KNOWN MATHEMATICS COMPETITION.[/b]
$\boxed{<1>}$
$\boxed{<2>}$
$\boxed{<3>}$
$\boxed{<4>}$
$\boxed{<5>}$
$\boxed{<6>}$
Note that some responses that I got while I was running the survey were very general.  Just to let you know, I am looking for specific answers.  If it's too general, I will tell you.  It will not count as a strike, but you won't get any points.

b-flat: 23
Ignite168: 6
nutz_for2.718281828: 47

The AMC Series which includes AMC, AIME, USAMO, etc., if that's too general, than USAMO.

I would say ARML and then, Mandelbraught (not spelled correctly, but whatever).

Mathcounts, I would have posted earlier, but I never view this topic anymore! WOOHOO! I have a score of 47!

[quote="lingomaniac88"]Mathcounts is #2.  nutz_for2.718281828 duplicated an answer, so we can't give him credit.  No strike, though.
Yes, "the AMC series" is too general.  But USAMO is up there, so you'll get the points.
ARML? Number 4.
AMC is number one.
IMO is down at the bottom.
Putnam and Balkan MO, however, well, neither of them made the survey.
Just one answer remains...

[b]We surveyed 100 people, top 6 answers on the board.
NAME A WIDELY KNOWN MATHEMATICS COMPETITION.[/b]
$\boxed{\text{AMC}}\boxed{36}$
$\boxed{\text{MATHCOUNTS}}\boxed{29}$
$\boxed{<3>}$
$\boxed{\text{ARML}}\boxed{7}$
$\boxed{\text{USAMO}}\boxed{5}$
$\boxed{\text{IMO}}\boxed{3}$
Strikes: 2

[b]b-flat: 30
daine: 29
Ignite168: 11[/b]
nutz_for2.718281828: 47
[b]ragnarok23: 3
ShoeFactory: 36[/b][/quote]
AIME?


I'm not sure if people who applied were notified, but the returning alumni just received an e-mail saying that the American Airlines flight certificates will not be available this year. I hope this doesn't prevent anyone from coming to Mathcamp this year.  :(

People who applied were not yet notified of whether or not they're in (at least, I wasn't). I would imagine that they will try to help out those in need.

Aww nuts, that's not good. Is anyone certain that there would be some sort of compensation?

We're going to keep this in mind for financial aid to people from Canada.  Within the US, you can usually get fares that are even cheaper than the flight certificates, or at least pretty close, if you shop around a bit, so hopefully prices won't be too bad.  If it turns out to be a real problem then we'll certainly address it more closely.


Two numbers are called twin primes if (a) they are both primes and (b) they are 2 numbers apart from each other e.g. 5 and 7.  No one has yet proven if there is an infinite number of twin primes or not.

Here's my challenge:

Define triple primes as three numbers which are (a) prime and (b) are 2 numbers apart from each other e.g. 3, 5 and 7
1.  Prove that 3, 5, and 7 is the only possible set of triple primes.
2.  Now from your proof explain why 3, 5, and 7 are triple primes.

[hide]Let (p, p+2, p+4) represent a set of triple primes. (p+2) mod 3 :equiv: p mod 3 +2, and (p+4) mod 3  :equiv: p mod 3 +1, so one of the three is divisible by 3.

3, 5, 7 is a triplet of primes because 3= 3*1 and therefore is a prime.[/hide]

[hide]let the three numbers be x, x+2 and x+4

we know that x must be odd because only one even number is a prime (2). Now, at least one of the three numbers must be divisible by three. if x/3 left a remainder of 1, then x+2 must be divisible by 3 because x-1 would be divisible by three making x+2 divisible by three. if x/3 left a remainder of 2, then x+4 must be divisible by three because x+1 would be divisible by three and thus, x+4 must also be divisible by 3. of 3 consecutive odd or even numbers, one of them must be divisible by 3.

3,5,7 is the only set of triple primes because of the previous rule. since at least one of the numbers must be divisible by 3, the only way that number can be prime is if it is 3, because a prime's only factors are itself and 1. thus, x=3 and the other 2 numbers are 5 and 7.[/hide]

[hide]

Theorem.  {3, 5, 7} is the only set of triple primes.

Lemma.  For every three consecutive odd numbers, one is divisible by 3.

Proof of Lemma.  Let our odd nunbers be o, o+2, and o+4.  In mod 3, they are o, o+2, and o+1.  If o is congruent to 0 mod three, then o is divisible by 3.  If o is congruent to 1 mod three, then o+2, which is one of our odd numbers, is divisible by 3.  Finally if o is congruent to 2 mod three, o+4   :equiv: o+1 :equiv: 0 mod 3.

Proof of Theorem.  By our lemma, the only set of triple primes must contain a prime multiple of three.  The only prime multiple of three is 3, which is also the smallest prime.  Thus our only set is 3, 5, and 7.[/hide]

1. [hide]Any prime number other than 3 is either 1 more than a multiple of 3 or 2 more than a multiple of 3. If a 3x+1 (where x is any integer) prime is the lowest number in a set of triple primes, the middle number is 3x+3, a multiple of 3. If a 3x+2 prime is the lowest number in a set of triple primes, the greatest number is 3x+6, a multiple of 3. The only way to avoid either the middle or greatest number being a multiple of three is to make the smallest number a multiple of 3. The only prime number that's a multiple of 3 is 3 itself, and therefore 3, 3+2, 3+4 is the only set of triple primes.[/hide]

2. [hide]3 is the only prime number that is a multiple of 3 and can therefore avoid what happens to 3x+1 and 3x+2 primes. So there is no problem using it as the lowest term of a set of triple primes.[/hide]

[quote="joml88"]No one has yet proven if there is an infinite number of twin primes or not.[/quote]

The Twin Prime Conjecture may be close to being proven. http://mathworld.wolfram.com/news/2004-06-09/twinprimes/

Nice problem by the way.

Hi all,

I am new to this group. I saw this problem, and thought to give it a try.

Every prime number greater than 6 is of the form (6k+1) or (6k-1). So if you have a triple prime set, and suppose two of them are (6k+1) and (6k-1) for some k, then the third number that can be a member of this triple is either (6k+3) or (6k-3), both of which are not primes.

(3,5,7) is the only such triple.

Sambit


This is a very very simple standard exercise that has me dumbfounded:

We have two lines that we are first supposed to show they intersect each other in a point. Then determine the plane that includes both these lines on "normal form" (Ax + By + Cz + D = 0).

Now, since the question doesn't say that we have an orthonormalized coordinate system, I simply set up the equation for the plane on parametric form and solved it to get the plane on normal form. However, I also checked what would have happened if I had just taken the vector product of the two lines' direction vectors (assuming it was a ON-system), and then assuming that the vector product (A, B, C) could be used as coefficients in "Ax + By + Cz".

For some reason this works. I see two problems in the calculation: 1) I am calculating the vector product using a formula that only works in positively oriented orthonormalized coordinate systems. 2) If I truly get a normal vector in 1) there would then be another error since the identification between coefficients and the vector-coordinates only works in ON-systems.

If you think about it, this will work every time since the coordinate system could just as well be orthonormal, and then you would get the same two equations using both methods. If you change the coordinate system, the equation will not change using the first method. However, I'd like some more insight: Why does it work, looking at definitions and whatnot of vector and scalar product?

It is good to think about stuff like that

suppose you have $(a,b,c)$ and $(d,e,f)$ as vectors through the origin defining a line through the origin( we will make it easy for now on us) and we would like to have the plane containing both of them

essentially you want to find p q r such that $px+qy+rz=0$ becomes the equation right
essentially you want to solve the homogenous system of linear equations
$pa+qb+rc=0$
$pd+qe+rf=0$

Well there is a general theorem for solving systems of n homogenous equations in n variables, by taking all  possible(n-1) determinants
you can easily check this is true, and you will find that the given coefficents are exactly what you would have find when taking cross products

If the coordinate system isn't orthonormal, that cross-product isn't the normal vector to the plane, but it's still there in the equation. Everything is the same except the interpretation; the vectors involved are just vectors, with no special angles between them.

That "cross product" still satisfies the equations fredbel posted; those just aren't dot products anymore.


smiley count bug for :?: -see image

The smiley counts for 2.

AoPS skin, Firefox, Vista

It's not a bug. It's just that some smilies just have extra codes making them count as 2: Like for example: the question mark, blush, and some others just count for 2.

Actually it's a bug, but not one that I will try to correct anytime soon :)

Actually, that happened to me too, and it still doesn't work

[quote="Valentin Vornicu"]but not one that I will try to correct anytime soon :)[/quote]
It's probably not going to be fixed soon (though that was from a year ago).

oh...some of them don't work for me..wait what!?!
not gonna be fixed soon?? :(


analytic geometry
graphing lines
slope
geometry
geometric transformation
homothety
ratio

There are given several equilateral triangles in a plane,such that each triangle can be obtained by another with parallel moving.Also known that any two triangles have a common point. Prove that there exist three points,such that each triangle contains at least one of them.

It's a piercing number type of problem.
You may find something relevant here :
http://www.combinatorics.org/Volume_15/PDF/v15i1r27.pdf

Pierre.

Let's say the slopes of the edges of the triangle are $ a$, $ b$, $ c$.  For any triangle $ T$, the line containing the slope $ a$ edge of $ T$ splits the plane into two regions.  Let $ H_a(T)$ be the region not containing $ T$.  Let $ S_a$ be the set of triangles $ T$ such that there exist triangles $ U,V$ such that $ U \cap V \subset H_a(T)$.  Define $ S_b, S_c$ similarly.

First, I claim that no triangle $ T$ is in $ S_a$, $ S_b$, and $ S_c$.  For any triangle $ U$, the intersection of $ T$ and $ U$ is homothetic to $ T$ and $ U$; call the homothety ratio $ r(T,U)$.  Suppose $ U_{ab} \cap U_{ac} \subset H_a(T)$, $ U_{ba} \cap U_{bc} \subset H_b(T)$, and $ U_{ca} \cap U_{cb} \subset H_c(T)$, with $ U_{ab}$ closer to the slope $ b$ edge of $ T$ than $ U_{ac}$ etc.  Then $ U_{ab}$ and $ U_{cb}$ can intersect only if $ r(T,U_{cb})>r(T,U_{ac})$.  This inequality cannot hold for all permutations of $ a,b,c$.

Next, I claim that all triangles not in $ S_a$ intersect.  By Helly's Theorem it suffices to check that any three or fewer of them intersect.  The problem statement implies that any two of them intersect.  If three triangles fail to intersect then the "hole" between the triangles is bounded by a slope $ a$ edge of one triangle, a slope $ b$ edge of another triangle, and a slope $ c$ edge of the third triangle.  Then the first triangle must be in $ S_a$.  So any three triangles not in $ S_a$ must intersect.

So choose one point in the intersection of all of the triangles not in $ S_a$, one point in the intersection of all of the triangles not in $ S_b$, and one point in the intersection of all of the triangles not in $ S_c$.  Each triangle will contain at least one of these points.


A tray with a volume of 36 cubic inches was made from a square of tin by cutting one-inch squares from the corners and turning up the sides.  How many square inches are in the area of the original square of tin?

[hide]64

By drawing a diagram, the sides of a tin are (x+2)2
And the volume is (1)(x)(x)=x2=36
So x=6
and  (6+2)2=82=64.[/hide]


Find all integers $ x,y$ such that the number $ (x^2 \plus{} y)(y^2 \plus{} x)$ is the fifth power to a prime.

[quote="triplebig"]Find all integers $ x,y$ such that the number $ (x^2 \plus{} y)(y^2 \plus{} x)$ is the fifth power OF A prime.[/quote]

Did you instead mean this (in the highlighted area here)?

(x^2 + y)(y^2 + x) = p^5 . . . where p belongs to the prime numbers

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Otherwise, then you are meaning this (if you really meant
"to a prime")...

(x^2 + y)(y^2 + x) = (m^5)^P, for some integer  m and
prime p, OR...

(x^2 + y)(y^2 + x) = m^(5p)

Which is it please?

I had to translate the problem, and in Portuguese there isn't that difference. The problem wants the first instance, OF a prime.
Sorry for the ambiguity.

Because p is a prime $ p^5$ can only be decomposed into powers of p and 1. WLOG assume $ y \qeq x$, and we have a few checkable cases


:oops: Out of [i]n[/i] straight lines of length 1,2,3.....n units, find the no. of ways in which four may be chosen such that it willl form  a quadrilateral in which a circle can be inscribed???

 [hide]Pls give ur ans. with detailed explanation.
Ty[/hide]

for a quad. in which circle can be inscribed, sum of each pair of opp. sides is equal.
(can be easily proved using 'equal tangents' property)
so we just have to find the no. of ways in which each of the nos. from 10(=1+2+3+4) to
4n-6[=n+(n-1)+(n-2)+(n-3)] can be split into 4 distinct nos.(each <= n).
This should just involve some simple combinatorics (i hope).
Actually, I'm just too lazy to work it out, or think about it. :oops:

I forgot that we also have to include the condition that sum of opp. side-pairs is equal. :oops:
that will further complicate the combinatorics.

seems the problem is not as simple as I thought it was.
could someone please work it out? (my mind doesnt seem to be functioning properly today)

[hide]
Is it
$ \sum_{i\equal{}5}^{2n\minus{}3} \binom{[\frac{(i\minus{}1)}{2}]}{2}$,[x] is gif

As keyree10 told,If the sum of opposite sides of a quadrilateral are equal ,then it is a tangential quadrilateral.I was not able to prove it  :blush: ,but saw the result in Wikipedia.Someone please post the proof.
Now,we have to choose 4 numbers,such that sum of two is equal to other two.
Minimum sum  is 5.
5=1+4,2+3    ---2 ways
6=1+5,2+4    ---2 ways
7=1+6,2+5,3+4    ---3 ways
and so on.
for i,it can be reprsented a sum of
$ [\frac{(i\minus{}1)}{2}]$  pairs.
where,[x] is gif.
we can choose any two of the $ [\frac{(i\minus{}1)}{2}]$  pairs and form a tangential quad with those four sides.

So,we just add up all from 5 to (2n-3).
2n-3=(n)+(n-3),(n-1)+(n-2)

I am not sure about the answer.
[/hide]

@ Arjun

Draw the inscribed circle

Then u will get 4 pairs of equal tangents from each of the 4 vertices.

Then u observe that the sum of opposite sides is indeed equal.

@befuddlers
No,I meant the proof for the converse of the theorem
(ie)If the sum of opposite sides of a quadrilateral are equal,then it is a tangential quadrilateral (or a circle can be inscribed in it)


Let $a,b,c$ be the length of the sides of triangle $ABC$.
Prove or disprove the following inequality:
$\displaystyle a^6+b^6+c^6+3a^2b^2c^2\geq a^3b(b^2+ac)+b^3c(c^2+ab)+c^3a(a^2+bc)$.

I like this one. Take $ abc=1 $ and denote $ x=a^3,... $. The inequality reduces to $ x^2+y^2+z^2+3\geq xy+yz+zx+x+y+z $ which was already discussed several times. You don't need any triangle.

My soln is different :D , but your soln is nice :)

Can you please post your solution, Hxtung ?

Thank you very much.

WLOG, we suppose $a\geq b\geq c$;
Put $F(a,b,c)=\displaystyle a^6+b^6+c^6+3a^2b^2c^2 - (a^3b(b^2+ac)+b^3c(c^2+ab)+c^3a(a^2+bc)).$
We prove that: $F(a,b,c)\geq F(a,b,b)\geq 0$ :D

Hey manlio!
Try to solve :D .It's very simple

$F(a,b,b) \geq0 $ is like

$a^6+b^6+3a^2b^4 \geq2ab^5 +2a^3b^3+a^4b^2$

as $a \geq b$ we can choose $ k \geq1$ and wite $a=kb$ so the inequality becomes

$ k^6 -k^4-2k^3 +3k^2-2k +1 = (k-1)^2(k^4+2k^3+2k^2+1) \geq 0 $

I hope this is right ! Have you a better proof !

$F(a,b,c) \geq F(a,b,b) $ is like

$c^6 +3a^2b^2c^2+a^3b^3+a^4b^2 +ab^5 +ba^5 \geq b^3c^3 +3a^2b^4 + c^3a^3 + abc(a^3+b^3+c^3) $

I have tried but I am not so clever to solve this inequality.
Unfortunately it is very hard for me. Can you please post your detailed solution hxtung.

Thank you very much.

I am again waiting for your two different solutions hxtung.

Thank you very much.

I think this inequality is weaker than Schur.
   We have
   $a^6+b^6+c^6 + 3a^2.b^2.c^2\geq \sum(a^4b^2+b^4a^2) \geq RHS$

I don't understand.
  The first inequality is Schur:
   $%Error. "suma" is a bad command.
^6 +3a^2b^2c^2 \geq %Error. "suma" is a bad command.
^4b^2+a^2b^4 $
  Then the Murhead is applied :
   $%Error. "suma" is a bad command.
^4b^2+a^2b^4\geq 2%Error. "suma" is a bad command.
^3b^3$
   $%Error. "suma" is a bad command.
^4b^2+a^2b^4\geq 2%Error. "suma" is a bad command.
^4bc $.
   Am I wrong anything?

I 'm sorry. I 'll post it again
   According to Schur inequality we have
     $ LHS \geq \sum(a^4b^2+a^2b^4)$
    By applying Muirhead inequality we have
     $ \sum(a^4b^2+a^2b^4) \geq 2(\sum a^3b^3)$
     $\sum(a^4b^2+a^2b^4)\geq 2(\sum a^4bc)$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let f be automorphism of $G$ that has not non trivial fixed point then there exists p-sylow subgroup of $G$ such that will be fixed under f.

Oh yeah! :) I got scared at first, and spent a lot of time on it, but it turned out not to be so tough (unless I'm making some stupid mistake, of course) :).

First of all, a lemma: if $f\in Aut(G)$ has no non-trivial fixed points, then the map $g\mapsto f(g^{-1})g$ is surjective (in the case of a finite group $G$). Since $G$ is finite, it suffices to prove tht $g\mapsto f(g^{-1})g$ is injective. Assume $f(g_1^{-1})g_1=f(g_2^{-1})g_2$. This translates to $f(g_2g_1^{-1})=g_2g_1^{-1}\Rightarrow g_1=g_2$, and we're done with the lemma.

Fix a prime $p|\ |G|$, and let $H\le G$ be a Sylow-$p$ subgroup of $G$. Since all the Sylow-$p$ subgroups of $G$ are conjugates of $H$, we have $f(H)=\alpha H\alpha^{-1}$ for some $\alpha\in G$. We want to prove that we can find some $g\in G$ s.t. $f(gHg^{-1})=gHg^{-1}$, which is equivalent to $f(g)\alpha H\alpha^{-1}f(g^{-1})=gHg^{-1}\iff \alpha^{-1}f(g^{-1})g\in N(H)$ (the normalizer of $H$ in $G$), and this, in turn, is equivalent to $f(g^{-1})g\in\alpha N(H)$. By employing the lemma above, we immediately find that we can always find such a $g$, so this finishes it, I guess.

Is it ok? :? If it is, now we can solve [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38968]this one[/url] :).


function
calculus
integration
geometry
geometric transformation
topology
complex analysis

Hello here is an exercice that makes me completely mad :w00t: :

Let $ f$ be an entire function , $ c>0$ such that \[ f(z\plus{}c)\equal{}f'(z)\] for all z.

discuss in function of $ c$  the solutions of the above differential equation.

I've solved when $ f$ is bounded on $ \mathbb{R}$.
fedja??...

I'm not Fedja :wink: , but maybe this can be a helpful [hide="hint"]Try the Ansatz $ f(z)\equal{}e^{\lambda z}$ and derive a characteristic equation for $ \lambda$. A discussion on its solutions can give some insight on the solutions of the delayed differential equation. [url=http://en.wikipedia.org/wiki/Lambert%27s_W_function]Lambert's function[/url] can be useful here.
For the uniqueness of the solution some type of Picard-Lindelöf theorem and a discussion about the spectrum of the related integral operator might be helpful.[/hide]

Nevertheless it seems that a picard lindelof type theorem is here impossible.
The integral operator is

\[ \phi \longrightarrow T(\phi)(x)\equal{}\int_{0}^{x} \phi(t\minus{}c) dt\]
The main problem is that there is no interval stable by translation , so i don't know which topology is good to study $ T$

One can solve the problem if $ f$ is dominated by a polynom $ P$ on $ \mathbb{R}$. In that case we can prove that $ f$ has an exponential type and paley-wiener theorems help a lot to solve. I believe that delayed holomorphic differential equation forces solution to have an exponential type.

To show uniqueness is a bit difficult here. The characteristic equation
\[ \lambda \equal{} e^{c\lambda}
\]
has two real solutions if $ 0 < c < 1/e$ and no [b]real[/b] solution if $ c > 1/e$.

In case $ 0 < c < 1/e$ let $ \lambda_1,\lambda_2$ be the real solutions, then the functions $ f_{\lambda_j}(x) \equal{} e^{\lambda_jz}$ are linearly independent over $ \mathbb C$ ($ \lambda_1\neq\lambda_2$!). This implies that some initial condition like $ f(z_0) \equal{} z_1$ does not pick a unique solution of the differential equation, moreover the solution set $ S_c(z_0,z_1)$ is an affine vector space having at least dimension 1.

Complex solutions of the characteristic equation can always be paired by conjugation which implies that the (affine) dimension of $ S_c(z_0,z_1)$ is positive or infinite. In other words, the existence of a strictly complex solution $ \lambda$ implies that there are infinitely many solutions. ... and the uniqueness is gone. ;)

The existence of complex solutions needs further considerations. Especially when $ c > 1/e$.

EDIT: Generically there is a complex solution of the characteristic equation.

More precisely the characteristic equation is equivalent to $ F(c\lambda)\equal{}c$ with the entire function $ F(z)\equal{}ze^{\minus{}z}$ which leaves out at most one point in the complex plane, by Little Pciard. The exceptional point - if it exists - is invariant under conjugation, hence real. This implies that $ S_c(z_0,z_1)$ is infinite for all real $ c>0$ but possibly one single $ c>1/e$. ... hmm, having written this, maybe the case $ c\equal{}1/e$ needs to be dealt with separately, such there are at most two exceptional points.

Hope that this is enough on the existence.

The general answer in the class of all entire functions is $ \sum_{\lambda : \lambda\equal{}e^{c\lambda}}a_\lambda e^{\lambda z}$ where $ a_\lambda\in \mathbb C$ decay faster than $ e^{\minus{}R|\lambda|}$ for every $ R>0$, which will guarantee absolute and uniform convergence of the series on any bounded domain. After the usual story about closed shift-invariant linear subspaces being closures of exponential polynomials contained in them (with respect to the topology of uniform convergence on compact sets), you need to prove that the roots of $ \lambda\equal{}e^{c\lambda}$ are well-separated and therefore a convergent sequence of exponential polynomials must be really a convergent series.

As a side remark, I should say that I know only two places on Earth where this rather involved theorem can be assigned as an [i]exercise[/i]. But if you are at one of them, (judging from your username, I would surmise that it must be the one in the southern France) you should be able to find the proof. ;)

Dear fedja thank you for your help, your approach of the "exercise" is great and i am now looking for a proof concerning your theorem about closed (for the usual topology) - shift invariant- subspaces of entire functions. Nevertheless it seems that additional properties are needed for that like exponential growth for instance. Could you please advices me a reference or a book you know dealing about such theorems?
It is exagerated indeed to qualify this question of "exercise" even if i am (indeed) from southern france  :D .

OK, here is a sketch of the proof. Sorry for the delay.

[b]Part 1[/b]. [i]The dual space to the space $ \mathcal H$ of entire functions with the topology of uniform convergence on compact sets.[/i]

Let $ \varphi\in\mathcal H^*$ be a continuous functional on $ \mathcal H$. Then there exists $ R > 0$ such that $ |\varphi(f)|\le Ap_R(f) \equal{} A\max_{|z|\le R}|f(z)|$. Thus, by the Hahn-Banach theorem, this functional can be extended to a continuous linear functional on the space of continuous functions on the disk $ D_R$ of radius $ R$, centered at the origin. By the Riesz theorem, the latter can be represented by integration against a finite Borel measure $ \mu$ supported by $ D_R$. Note that the same functional on $ \mathcal H$ can be represented by many measures.

[b]Part 2[/b]. [i]Fourier transform[/i]

Let $ \varphi\in \mathcal H^*$ be represented by a measure $ \mu$ with compact support. Put $ \Phi(w) \equal{} \widehat\varphi(w) \equal{} \widehat\mu(w) \equal{} \varphi(e^{wz}) \equal{} \int_{\mathbb C}e^{wz}\,d\mu(z)$ ($ w\in\mathbb C$). Note that $ \Phi$ is an entire function of finite exponential type. Moreover, $ \Phi$ is uniquely determined by $ \varphi$ and we can write the duality in the way
\[ \varphi(f) \equal{} \sum_{k\ge 0}k!a_kb_k
\]
where $ \Phi(w) \equal{} \sum_k a_k w^k$, $ f(z) \equal{} \sum_k b_kz^k$. Once we know the constant $ C$ in the bound $ |\Phi(w)|\le Ce^{C|w|}$, we can estimate $ R$ and $ A$ such that $ |\varphi(f)|\le Ap_R(f)$ (just use Cauchy inequalities for the coefficients). The main morals of the story are that
1) $ \mathcal H^*$ can be identified with the space of functions of finite exponential type
2) If $ |\Phi_k(w)|\le Ce^{C|w|}$ for all $ k$ and $ \Phi_k\to\Phi$ uniformly on compact subsets of $ \mathbb C$, then $ \varphi_k(f)\to \varphi(f)$ for every $ f\in\mathcal H$.

[b]Part 3[/b]. [i]Shift and convolution[/i]

If you shift a measure by $ u\in \mathbb C$, its Fourier transform is multiplied by $ e^{uz}$. If you convolve two measures, their Fourier transform multiply.

[b]Part 4[/b]. [i]Theory of one convolution equation, extracting exponents[/i]

Now we are ready to handle a convolution equation $ f*\mu \equal{} 0$ where $ \mu$ is a measure (or generalized function) with compact support and $ f\in\mathcal H$. Note that $ w\in\mathbb C$ is a root of $ \Phi \equal{} \widehat\mu$ if and only if $ e^{wz}$ is a solution of this equation. Moreover, assuming that the equation is non-trivial on $ \mathcal H$, we see that the set of roots $ w$ is discrete. Assume that all roots are simple (I'll let you guess what should be done with multiple roots ;)). For  every root $ \lambda$, consider $ \Phi_\lambda \equal{} \frac {\Phi}{w \minus{} \lambda}$. It is again an entire function of exponential type, so it is a Fourier transform of some bounded linear functional that is represented by some measure $ \mu_\lambda$. Now, since $ w \minus{} \lambda$ is the Fourier transform of $ f\mapsto f'(0) \minus{} \lambda f(0)$, we see that $ g \equal{} f*\mu_\lambda$ satisfies the differential equation $ g' \minus{} \lambda g \equal{} f*\mu \equal{} 0$, i.e., $ g$ is a single exponent with some coefficient. Now, if we have a finite set $ \Lambda$ of roots, we can similarly divide by $ \prod_{\lambda\in \Lambda}(w \minus{} \lambda)$ and conclude that the convolution of $ f$ with the resulting measure $ \mu_{\Lambda}$  is a linear combination of exponents $ e^{\lambda z}$, $ \lambda\in\Lambda$.

Now we want to show that every solution is in the closure of the linear span of the exponential solutions. Suppose that $ f$ is a solution and $ \nu$ is a measure representing a functional that vanishes on every $ e^{\lambda z}$ where $ \lambda$ is a root of $ \Phi$. Our aim is to show that $ g \equal{} f*\nu \equal{} 0$. For every finite set $ \Lambda$ of roots, we have $ g*\mu_\Lambda \equal{} (f*\mu\Lambda)*\nu \equal{} 0$ because $ f*\mu\Lambda$ is a linear combination of exponents each of which is killed by $ \nu$. The next ingredient is

[b]Part 5[/b]. [i]Hadamard factorization theorem[/i]

$ \Phi(w) \equal{} ae^{bw}\prod_{\lambda}\left(1 \minus{} \frac w\lambda\right)e^{w/\lambda}$. Moreover, the tails of the canonical product $ \Phi_R(w) \equal{} \prod_{|\lambda| > R}\left(1 \minus{} \frac w\lambda\right)e^{w/\lambda}$ admit a bound $ Ce^{C|w|}$ with $ C > 0$ independent of $ R$ and tend to $ 1$ as $ R\to \plus{} \infty$.

[b]Part 6[/b]. [i]The end of the proof that $ f$ is in the closure of the linear span of exponential solutions[/i]

Note that the functional $ \varphi_R$ whose Fourier transform is $ \Phi_R$ is essentially the same as the one represented by $ \mu_\Lambda$ where $ \Lambda \equal{} \{\lambda: |\lambda|\le R\}$. Indeed, the exponential factors are just shifts and $ 1 \minus{} \frac {w}{\lambda} \equal{} \minus{} \frac 1\lambda(w \minus{} \lambda)$. Thus, $ varphi_R$ applied to any shift of $ g$ is $ 0$. On the other hand, $ \Phi_R\to 1$ as $ R\to\infty$ with uniform control, so $ \varphi_R(h)\to h(0)$ as $ R\to\infty$. Thus $ g$ is identically $ 0$ and we are done.

[b]Part 7[/b]. [i]From closure to series[/i]

Let $ p(z) \equal{} \sum_\lambda c_\lambda e^{\lambda z}$ be an exponential polynomial solution. Convolving with $ \mu_\lambda$, we get $ p*\mu\lambda \equal{} c_\lambda\Phi'(\lambda)e^{\lambda z}$. Note that if $ p$ converge to $ f$, we have uniform control for $ |p|$ on any compact set. Also, we have a uniform control for $ \mu_\lambda$. Thus, we have a uniform bound for the convolution in every disk of radius $ R$. Now, we also have $ \Phi(\lambda) \equal{} \lambda \minus{} e^{c\lambda} \equal{} 0$, whence $ \Phi'(\lambda) \equal{} 1 \minus{} ce^{c\lambda} \equal{} 1 \minus{} c\lambda$ is at least $ 1$ in absolute value when $ \lambda$ is large. Thus the monomials $ c_\lambda e^{\lambda z}$ are bounded by some number in every disk of radius $ R$ independently of $ \lambda$. The rest should be clear.

Sorry for any misprints :) Feel free to ask questions, as usual.


Find all polynomial $P(x)$ with real coefficients such that : $(x-16)P(2x)=16(x-1)P(x)$.


Central force F=-ar/(r^3) & Central potential energy,U=-(a/r)
(not U_eff)
Find the nature of orbits if (i)a>0 and (ii)a<0

2. Relevant equations
3. The attempt at a solution

If we remember the attractive central force E=E(r) diagram,i.e.the one showing the graph of U_eff,we only need to know E_total=K+U.
Where only PE is given.
We see,

U= -integration[F.dr]=integration[dW]=-integration[dK]=-K

Then K=a/r and U=-(a/r)

So,the E=K+U=0

Then in positive and negative both caes we get a parabolic orbit.

Please check if i am correct.

Due you mean:
$F=-\frac{a}{r^{2}}\frac{\vec r}{r}$
as the force?

use LaTeX please.

Yes.
Oh! I got it...

Consider a>0.
Then, -(m(v^2)/r)=-(a/r^2)
or,m(v^2)=(a/r)
Hence total energy E=K+U= -(1/2)(a/r)

=>elliptic orbit.

Consider a<0.

Then,F=+b/r where b=-a>0
and U_p=b/r

K is itrinsically positive.So,total energy positive.

From e=sqrt[1+{(2L^2*E)/(m*b^2)}]
e>1.
hence hyperbolic trajectory.


How can i add geometrical pictures or figures to my latex document?? i want it to look like how it would do in a geometry book... i am using TeX .... any ideas?? thanks in advance..

Use Asymptote. Instructions are [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_(Vector_Graphics_Language)]here[/url] and an AoPS forum to ask questions is [url=http://www.artofproblemsolving.com/Forum/index.php?f=519]here[/url].


geometry
geometric transformation
rotation

1) The rotation circle forms a cycloid. If the length of one of its cycles is 104 units what is the radius of the rotating circle rounded to the nearest integer?

2) Discover the pattern to solve for x:
0, -15, 30, -105, -12, 21, -78, 219, -8, x, -42, 111

well what i get from the question is the total length a point has to cover before it reaches the same phase as  it's previous is $ 104$ also the length is $ 8r\equal{}104$ $ \rightarrow r\equal{} 13$
if this is not then please do correct me. :)


Η ερώτηση που έχω να κάνω είναι η εξής πως βρίσκονται τα sup με αυστηρή μαθηματική θεμελίωση και όχι διαισθητικά
 Παράδειγμα(εύκολο) Δίνεται το σύνολο Α=(1,5) να βρεθεί το supA

Η παρακάτω λύση είναι σωστή;
Επειδή 2εA αυτό είναι μη κενό. Έχουμε 1<χ<5 επομένως το 5 είναι άνω φράγμα, άρα σύμφωνα με το αξιώμα του συνεχούς υπάρχει supA. Θα δείξω ότι supA=5

Έστω β ένα άλλο άνω φράγμα τότε έχουμε υπάρχει μ ε R: x<=5 για κάθε χ ε (1,5) και μ<5

μ<5<=>μ<(μ+5)/2 ανήκει (1,5). Άτοπο γιατί μ άνω φράγμα και (μ+5)/2>5 (Δηλαδή βρίκαμε στοιχείο του A μεγαλύτερο του β)

Είναι σωστό; μπορεί να γίνει με άλλον τρόπο; Επίσης έχετε να μου προτείνετε βιβλία (όλων των επιπέδων) για Διαφορικό Λογισμό;

 Ευχαριστώ προκαταβολικά

Η λὐση είναι σωστή. Αν θες κοίταξε το attachment, θα στο επιβεβαιώσει

Μπορεισ να βρεις για τα sup,imf κ οτιδηποτε αλλο για τον διαφορικο λογισμο στισ παρακατω ιστοσελιδες:
[url]http://eclass.uoa.gr/courses/MATH130/[/url]  εδω θα βρεισ υλικο στα εγγραφα
παρομοιως κ εδω:[url]http://eclass.uoa.gr/courses/MATH141/[/url]
και για ε-books δες εδω:[url]http://math.stuff.gr/gr/index.php[/url]

[quote="apeiros"]Σε ευχαριστώ πολύ η εποπτική/διαισθητική έννοια των sup/inf είναι εύκολη αλλά θα πρέπει να γίνεται με έναν πιο αλγεβρικό τρόπο γι'αυτό και η ερώτηση το doc που επισήναψες είναι ωραίο. Μπορείς να μου πεις που το βρίκες, καθώς και αν έχει οποιοσδήποτε να προτείνει. Βιβλία, ebooks, docs Κλπ που να αφορούν τον Διαφορικό Λογισμό;
 Θα με ενδιέφερε για να λύσω ασκήσεις και να δώ παραδείγματα για θεωρία είμαι καλυμένος από διάφορα βιβλία

 Ευχαριστώ πολύ ξανά για την επιβεβαίωση, για το συνημένο αρχείο και περιμένω απόψεις για βιβλία.
Θα ποστάρω και γώ βιβλία μόλις βρώ κάτι το ενδιαφέρον[/quote]
το attacment είναι από ενα βιβλίο που εχω γράψει παλιά Αν θες κάτι από κει στείλε μου μια διεύθυνση να στο στείλω . Λέγεται οπισθόφυλλο και περιέχει μια διαπραγμάτευση των εννοιών του απειροστικού μιας μεταβλητής

DIMITRIS T ευχαριστώ τα συγκεκριμένα sites τα έχω στα bookmarks θα προτιμούσα όμως βιβλία για αγορά, με αρκετές λυμένες, μια πιο βαθιά ανάλυση και ασκήσεις προς λύση.

Για λυμενες ασκησεις το καλυτερο κατα τη γνωμη μου ειναι το τριτομο Problems in Mathematical Analysis 1,2,3 Kaczor-Nowak της AMS (Μπορεισ να το αγοράσεις απο το αμαζον)
Οι αρκετες ασκησεις ειναι τσιμπιμενες και ο πρωτος τομος εχει καλες ασκησεις για τα sup, imf πυκνοτητα κτλ
Τςρα για θεωρια καλο ειναι το βιβλιο του Spivak.Εχω ακουσει καλα λογια κ για το βιβλιο του Apostol αλλα εμενα προσωπικα δν μου αρεσε...

[quote="bouzoukman"][quote="DIMITRIS T"]Για λυμενες ασκησεις το καλυτερο κατα τη γνωμη μου ειναι το τριτομο Problems in Mathematical Analysis 1,2,3 Kaczor-Nowak της AMS (Μπορεισ να το αγοράσεις απο το αμαζον)[/quote]

Έχει αρκετά παραδείγματα???Και κάτι βασικό,έχει υποδείξεις για όλες τις ασκήσεις που παραθέτει δίοτι πιστεύω ένα καλό βιβλίο πρέπει να έχει καλές υποδείξεις απ' όλες τις ασκήσεις που παρουσιάζει...Ρωτάω γιατί θα ήθελα ένα καλό βιβλίο με ασκήσης στην ανάλυση να το έχω και μήπως προλάβω να το κοιτάξω το καλοκαίρι....

Είναι ένα από τα κομμάτια των μαθηματικών που ένας καλός μαθηματικός οφείλει να γνωρίζει σε βάθος...[/quote]
Ναι εηει αρκετες ασκησεις και καλυπτει ολα τα κεφαλαια του απειροστικου(αναλυση μιας μεταβλητης)
Ολες οι ασκσεις  εηουν λυσεις κ πιο συγκεκριμενα η δομη του βιβλιου ειναι η εξης:ο κα8ε τομος εχει 2 μερη.Στο πρςτο μερος εηει ασκησεις ταξηνομημενες σε κεφαλαια κ στο δευτερο εχει τις λυσεις.
Απο τα καλυτερα βιβλια με ασκησεις αναλυσης.
Για περισοτερες πληροφοριες δες εδω:[url]http://www.amazon.com/Problems-Mathematical-Analysis-Numbers-Sequences/dp/0821820508[/url]

PS:Εδω ειναι ενα βιβλιο με καλες ασκησεις [url]http://www.openmathtext.org/lecture_notes/david_santos_calculus1.pdf[/url]

[quote="DIMITRIS T"]PS:Εδω ειναι ενα βιβλιο με καλες ασκησεις [url]http://www.openmathtext.org/lecture_notes/david_santos_calculus1.pdf[/url][/quote]

Από λίγο που το κοίταξα νομίζω ότι έχει καλές ασκήσεις και γενικά φαίνεται καλό βιβλίο.Όποτε βρω χρόνο θα το κοιτάξω με προσοχή....
Πάντως έχω υπόψιν μου και το τρίτομο που πρότεινες που μου φαίνεται πολύ καλό...
Ευχαριστώ που το έστειλες....

πολύ ωραία τα Links παιδιά, ελπίζω να βρώ κάτι χρήσιμο να βοηθήσω και γώ όσο μπορώ


Sa se arate ca seria $\sum_{n\geq 1} (-1)^n/n$ NU este neconditionat convergenta.

Consecinta imediata a criteriului lui Leibniz. ;) care afirma urmatorul lucru:

 Seria $\sum_{n\geq 1} (-1)^{n}\cdot{a_{n}}$ este convergenta daca $a_{n}$ este un sir descrescator cu limita nula. Deci, concluzia este evidenta. ;)

Apropo, pentru convergenta absoluta, aplica definitia, adica treci la modul si din $|(-1)^{n}|=1$ rezulta ca seria este convergenta, dar nu absolut convergenta. ;)

[quote="catalin_dumitru"]De fapt, problema era: Sa se arate ca seria NU este neconditionat convergenta (adica: schimband ordinea termenilor, obtinem o alta suma a seriei). Asta cum se arata?[/quote]
Avem $\sum_{n\ge 0}{\frac{(-1)^n}{n+1}}=\ln{2}$. Sa schimbam ordinea termenilor. Putem arata, de exemplu ca
\[ 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\ldots=\frac{3}{2}\ln2. \]
Intr-adevar, suma seriei de mai sus este limita sirului
\[ x_n=1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\ldots+\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}. \]
Consideram sirul
\[ c_n=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}-\ln{n}. \]
Se stie ca $(c_n)$ este convergent, si notam $c=\lim_{n\to\infty}c_n.$
Observand ca \[ x_n=c_{4n}-\frac{1}{2}c_{2n}-\frac{1}{2}c_n+\ln{4n}-\frac{1}{2}\ln{2n}-\frac{1}{2}\ln{n}, \] obtinem concluzia dorita.
In general, se poate arata ca pentru orice $A\in \mathbb{R}$, putem ordona termenii seriei initiale astfel incat suma ei sa fie $A$. Demonstratia este naturala. De exemplu, daca $A>0$, luam primii termeni $1,\frac{1}{3},\frac{1}{5},\ldots,\frac{1}{2k+1}$, astfel incat $k$ sa fie minim cu proprietatea
\[ 1+\frac{1}{3}+\frac{1}{5}+\ldots+\frac{1}{2k+1}\ge{A}. \]
Apoi adaugam termeni negativi pana cand suma partiala scade sub $A$, apoi iar termeni pozitivi pana cand suma depaseste iar $A$, etc. Nu e greu de aratat ca acest procedeu conduce la o serie cu suma $A$.

[quote="mchis68"]Teoria este foarte clara: O serie este neconditionat convergenta daca si numai daca este absolut convergenta.(respectiv pentru o serie semiconvergenta, poti permuta termenii in asa fel incat sa obtii orice suma ti-ai fixa dinainte)[/quote]
Asa e, teoria este foarte clara. Totusi, un exemplu concret, precum cel de mai sus, nu strica :)


calculus
integration
number theory proposed
number theory

Find all possitive integers (a,b,c) satisfying :
  $ a^3+b^3+c^3 $ is divisible by $ a^2b , b^2c , c^2a $

Firstly,assume that [tex] (a,b)=(b,c)=(c,a)=1 [/tex] and [tex] a \geq b \geq c [/tex] then
[tex] a^3+b^3+c^3 [/tex] is divided by [tex] a^2b^2c^2 [/tex]
[tex] \Rightarrow b^3+c^3 \geq a^2 [/tex].
In addition,we have [tex] 3a^3 \geq a^2b^2c^2 \Rightarrow 3a \geq b^2c^2 [/tex].
Hence [tex] b^4c^4 \leq 9a^2 \leq 9(b^3+c^3) \leq 18b^3 [/tex]
[tex] \Rightarrow bc^4 \leq 18 \Rightarrow c=1 [/tex] and [tex] b \leq 8 [/tex].
And it is easy from now. :)


What is the smallest positive integer that has factors of 16, 18 and 24?

[hide] The LCM of the three integers is $ 2^4(3^2)\equal{}\boxed{144}$.[/hide]


Suppose $n$ is an integer $\geq 2$. Determine the first digit after the decimal point in the decimal expansion of the number

$\sqrt[3]{n^{3}+2n^{2}+n}$

The most natural thing to do is [hide]to cube out $\left(n+\frac{6}{10}\right)^{3}$ and $\left(n+\frac{7}{10}\right)^{3}$ and compare them to $n^{3}+2n^{2}+n$.  [hide="Why those values?"]Because $\frac{6}{10}< \frac{2}{3}< \frac{7}{10}$[/hide][/hide]


inequalities
calculus
derivative
inequalities proposed

So you had a solution,Albanian Eagle  :wink: . It is nice to know that this one
is wrong for $n=9$  :) .

Albanian Eagle, I think you should post solution of you  :P Did you have a nice solution? :D

Zaizai, is the picture above your photograph ?

Sorry, I haven't been around these last 10 days. Now I'm back.
 :)
About this problem my solution is not "nice" :
I firstly show that for an extremum we need 6 equal variables(inflection point) then we must have one of two cases 7 equal variables or 6 equal variables and the other two also equal (derivatives or langrange multipliers ). both are not difficult one-variable inequalities...


calculus
integration
function
trigonometry
calculus computations

I would like to evaluate this integral:
\[ \int_0^\infty \frac{\sin(t+\frac{c}{t})}{t}dt\]
where $ c\in\mathbb{R}$.

I plug this in Mathematica and it gives me some expression in terms of Bessel function of the first kind:

\[ \int_0^\infty \frac{\sin(t+\frac{c}{t})}{t}dt=\frac{\pi}{2}(1+sign(c)) J_0(2\sqrt{|c|})\]

in particular, it is 0 when c is negative, so we can simplify it as
\[ \int_0^\infty \frac{\sin(t+\frac{c}{t})}{t}dt=\left\{\begin{array}{cc}\frac{\pi}J_0(2\sqrt{c})& c>0\ \frac{\pi}{2}& c=0 \ 0 & c<0\end{array}\right.\]

I have searched and checked most integral representations of the Bessel function, but I couldn't find any relevant formula that gives this result.

Anybody could help?

OK, I think I got it.
Assume $ c>0$. Then we can make a substitution $ t\mapsto t\sqrt{c}$,
and another substitution $ t\mapsto e^u$,
and use the formula $ \frac{1}{\pi}\int_{\minus{}\infty}^\infty \sin(x\cosh u)du \equal{} J_0(x)$

For $ c<0$, the $ \cosh$ term will become $ \sinh$ and the integral gives 0 since the integrand is odd.

For $ c\equal{}0$, the well known formula gives $ \int_0^\infty \frac{\sin t}{t}dt \equal{} \frac{\pi}{2}$


Let's $a,$ $b$ and $c$ are non-negative numbers. Prove that
$2^{a+b}+2^{a+c}+2^{b+c}\leq2^{a+b+c+1}+1$

Put $p=2^{a}, q=2^{b},r=2^{c}$. Then for $p,q,r\geqslant 1$ we have to prove

$pq+pr+qr\leqslant 2pqr+1$

Starting from $(p-1)(q-1)\geqslant 0$ we get

$pq-p-q\geqslant-1\qquad/+pq$

$2pq-p-q\geqslant pq-1$

Both sides are non-negative. Multiplying that with $r\geqslant 1$ we get

$2pqr-pr-qr\geqslant pq-1\iff pq+pr+qr\leqslant 2pqr+1$ QED.

The equality holds iff $[(p-1)(q-1)=0\land r=1]\lor(pq-1=0)$, or when at least two of $p,q,r$ are equal to $1$, that is when at least two of $a,b,c$ are zero.


Prove that $ 2^{n}\plus{}1$ is divisible by $ n$ for infinitely many natural numbers $ n$.

[hide="Solution"]
Let $ n=3^k$ for a nonnegative integer $ k$,
Next we need to prove that,
\[ \dfrac{2^{3^k}+1}{3^k}\in \mathbb{N}\]
We can do this with Induction,
Step 1: $ k=0$,
\[ \dfrac{2^{3^0}+1}{3^0}=\dfrac{3}{3}=1\in \mathbb{N}\]
Step 2: Assume it is true for $ k$
\[ \dfrac{2^{3^k}+1}{3^k}\in \mathbb{N}\]
Step 3: Prove it is true for $ k+1$
\begin{eqnarray*}
\dfrac{2^{3^{k+1}}+1}{3^{k+1}}&=&\dfrac{2^{3\cdot 3^k}+1}{3\cdot 3^k}\\
 &=&\dfrac{(2^{3^k})^3+1}{3\cdot 3^k}\\
 &=&\dfrac{(2^{3^k}+1)((2^{3^k})^2-2^{3^k}+1))}{3\cdot 3^k}
\end{eqnarray*}
Since $ (2^{3^k})^2-2^{3^k}+1\equiv 1+1+1\equiv 0\pmod{3}$, we are done. :ninja:
[/hide]


function
integration
inequalities
real analysis
real analysis unsolved

Let $f: [a,b]\to R$ be a continuous convex function. Show that it exists $c\in [a,b]$ such that $\int_a^b f(x)dx = \frac{b-a}{4}f(\frac{a+b}{2}) + \frac{3}{4}(b-a)f(c)$

My oppinion is that we must show that $min f(x)\leq\frac{4}{3(b-a)}\int_a^b f(x)dx - \frac{1}{3}f(\frac{a+b}{2})\leq max f(x)$ and because f has Darboux, QED. I managed to show that it is $\geq min f$, but I can't do it for$\leq max$

$\le \max$ is easy. Just note that if you fix $m=f(\tfrac {a+b}2)$ and $M=\max f$, then, because of convexity, $f$ is bounded from above by the piecewise linear function that decreases from $M$ to $m$ on the left half of the interval $[a,b]$ and increases from $m$ to $M$ on the right half. To check the inequality for this function is a straightforward computation.

Another aproach:

We have that $f(\frac{a+b}{2})\le \frac 1{b-a}\int_a^b f(x)dx\le \frac{f(a)+f(b)}{2}$.
I will show that we connot have :
\[ \int_a^b f(x)dx < \frac{b-a}{4}f(\frac{a+b}{2}) + \frac{3}{4}(b-a)f(x),\forall x \]

Assume that we have that. Integrating and dividing by $b-a$ we get $f(\frac{a+b}{2})> \frac 1{b-a}\int_a^b f(x)dx$ Contradiction.

We cannot have neither
\[ \int_a^b f(x)dx > \frac{b-a}{4}f(\frac{a+b}{2}) + \frac{3}{4}(b-a)f(x),\forall x\Rightarrow \]
\[ \int_a^b f(x)dx > \frac{b-a}{4}f(\frac{a+b}{2}) + \frac{3}{4}(b-a)f(a)\text { (1) } \]
(Since $f$ is convex it's maximum is $f(a)$ or f(b). I chosed it to be $f(a)$ WLOG).
Assume that (1) is true.
But we have $\int_a^bf(x)dx=\int_a^{\frac{a+b}{2}}f(x)dx+\int_{\frac{a+b}{2}}^bf(x)dx\le \frac{f(a)+f(b)}{4}+\frac{f(\frac{a+b}{2})}{2}$
The last statemant combined with $(1)$ gives us that $2f(a)< f(b)+f(\frac{a+b}{2})$ Contradiction, since $f(a)$ is the maximum. Hence there exists a point $c$ such that the equality is fulfiled. :lol:

because

I find the best constant in the equality is $\frac 1 2$
Which means the equality can be :
\[ \int_a^b f(x)dx = \frac{b-a}{2}f(\frac{a+b}{2}) + \frac{1}{2}(b-a)f(c) \]
Am I right?Anybody find it,too :D

[quote="xirti"]Congratulations!!  :lol:  I foud it too. The example for which the equality holds is
\[ f(x)=\begin{cases} 0 \text{ if } x\in [a,\frac{a+b}{2}]\\ f(x)=c(x-\frac{a+b}{2}) \text { if } x\in [\frac{a+b}{2},b]\end{cases}  \][/quote]
 :roll: I find the example is $f(x)=c|x-\frac{a+b}{2}|$,am I wrong or are you wrong ;)


prove that all numerous faces  accept at least 2 faces have the same numbers of edges  .

take in the center of every face $ F_{i}$ a point $ A_{i}$ ,and we consider the graph formed by the set of this points {$ A_{i}$} ,$ A_{i}$ and$ A_{j}$ are connected if and only if $ F_{i}$ and$ F_{j}$ have one comon edge ,so the degree of the vertex$ A_{i}$ is the number of edges of the face $ F_{i}$,
it's welle know  that in evry simple graph there is at least two vertexes wich have the same degree, we are done  :wink:


can someone like give me a tutorial about this place and how to use it? it would help a bunch

1. you already know how to post..

2.dont say anything bad

3.ask questions according to the level of the forum you are in (example: in the middle schoool forum dont ask: what
\[\int{xdx}\]
equals.. thats calculus)

     you can ask that in the highschool forum

3.look at the top where it says "posting guidlines and how to use this forum"   (its a post) for more details...

***pm me your name and grade***

   (by the way pm me by clicking on "PM" under my post)

cant still gotta wait almost 2 more hrs before doing something like that.

why?

PM means personal message....only i get the info...

are you in 8th grade?

I don't think they let you pm until a certain time after you join...

Oh, for more important tips, checkout LaTeX and AoPSWiki links on the left side of the screen.

[quote="buzzer11"]yes, latex is very helpful.  :)
it took me a while to learn, but... its worth the effort. trying to read a mathematical problem with stuff like:

x^2/3*sqrt3x

is very confusing. Instead you can say,

$\frac{x^{2}}{3}\cdot \sqrt{3x}$[/quote]
If you want to know how someone made the symbol on $\LaTeX$, just hold your mouse over to see the code.

[quote="buzzer11"]Yeah, but I see the weird W looking things instead of the \.  :)[/quote]
Is your font set to Batang? That's what has happened to me before.

(Is it like a W with a horizontal dash?)


What is the best way to send update notifications to a userlist via email? I am complete beginner, so if more info is needed please tell me.

Depends on what your site is written in.  If it's PHP, check out PHP's [url=http://us2.php.net/manual/en/function.mail.php]mail function[/url].  Either BCC everyone or send out one email per user.


inequalities
calculus
integration
logarithms
function
real analysis
real analysis theorems

Hölder's inequality (integral version or sum version) is proved using the lemma $ ab \le a^p/p \plus{} b^q/q$. If someone gave you this lemma and told you to prove Hölder's inequality, you would be able to prove the lemma and the Hölder's inequality. But coming up with this lemma (or similar) on your own is a nontrivial magic part. any idea? how to resolve that magic part?

This lemma is quite natural, it is just about convexity. The condition $ \frac{1}{p}\plus{}\frac{1}{q}\equal{}1$ suggests to apply the definition of convexity to $ \ln$ with $ \lambda \equal{} \frac{1}{p}$. I think this is not magic.
One more thing, there are many other ways to prove the Holder inequality which you may find more direct. For example, using the "convexity inequality"...

We must prove that $ \left|\int fg\right|\le \phi\left(\int f^p,\int g^q\right)$, where $ \phi(u,v) \equal{} u^{1/p}v^{1/q}$. Integrals play well with linear functions, which $ \phi$ is not. But $ \phi$ is concave, which means it can be written as infimum of a family of linear functions (i.e. tangent planes to its graph). The equations of tangent planes are all of the form $ \phi_t(u,v) \equal{} \frac {u}{p}t^p \plus{} \frac {v}{q}t^{ \minus{} q}$, $ t > 0$. Thus, $ \phi(u,v) \equal{} \inf_{t > 0}\phi_t(u,v)$, and it suffices to prove that $ \left|\int fg\right|\le \phi_t\left(\int f^p,\int g^q\right)$ for any $ t > 0$. In fact, $ t \equal{} 1$ is enough, because we can incorporate $ t$ into $ f$ and $ g$. Because of linearity, Young's inequality $ \left|\int fg\right|\le \phi_1\left(\int f^p,\int g^q\right)$, unlike Hölder's inequality, can be proved [b]pointwise[/b].

The authors of the magnificent paper  :) [url=http://www.math.brown.edu/~treil/papers/bellman/abstract/abstract.html]The hunt for a Bellman function[/url] stated this principle as
[quote]get the quantities which seem hard to handle directly as solutions of extremal problems[/quote]