limit
function
calculus
calculus computations

Caculate:
$ \mathop {\lim }\limits_{N \to \infty } \sqrt N \left( {1 \minus{} \mathop {m{\rm{ax}}}\limits_{1 \le n \le N} \left\{ {\sqrt n } \right\}} \right)$

it's looks more interesting then any other limits'  :) .
ghe question is to study behavior of the function $ f(N) = \max\limits_{1\leq n \leq N} \{\sqrt {n}\}$
it's alo interesting to  study existence of limit $ N^{a}(1 - \max\limits_{1\leq n \leq N} \{\sqrt {n}\})$
I try to post hint:
consider for big $ N$ numbers $ N^{2} - 1$ and try to estimate $ \{ {\sqrt {N^{2} - 1} \}\approx 1-\frac{1}{N}}$
 it's good to use Taylor's approximation

[color=green] Moderator edit: the unnecessary quote was removed; never quote the post preceding yours in full! Also, use normal English, not "web talk", please :) [/color]

 Can you give a detailed solution?
 I guess  the answer to the problem with your hint is
 $ \lim_{N \to \infty } \sqrt N \left( {1 - \max_{1 \le n \le N} \left\{ {\sqrt n } \right\} \right) = 1}$

[quote="Extremal"] $ \{ {\sqrt {N^{2} - 1} \}\approx 1 - \frac {1}{N}}$
[/quote]
Extremal, don't make me feel ashamed for the quality of education in the university I graduated from.  :mad:. Or, in short, [color=red][size=134]Please, spend some time thinking before you post! [/size][/color]

limit don't exist. since if we choose sequance $ x_{n} \equal{} n^{2}$ then limit for this sequance is equal to (as it was noted) $ \frac{1}{2}$
on the other hend if we choose $ x_{n} \equal{} n^{2}\minus{}2$  then limit will equal to $ 1$.

maybe im dumb, but why wouldnt the max of a square root be at the largest value of n?

$ \max_{1\le n \le N} \sqrt{n} \equal{} \sqrt{N}$ ?

[quote="kenn4000"]maybe im dumb, but why wouldnt the max of a square root be at the largest value of n?

$ \max_{1\le n \le N} \sqrt {n} \equal{} \sqrt {N}$ ?[/quote]
because we need to find not max of $ \sqrt{...}$, we need to find maximum of fractional part of $ \sqrt{...}$
in generally $ \{x\} \equal{} x\minus{}[x]$ where $ [x]$ means integer part of $ x$
for example:
$ [2.344] \equal{} 2$ and $ \{2.344\} \equal{} 0.344$

ah, notation i wasnt familiar with. that makes more sense. ty


integration
real analysis
calculus
topology
real analysis unsolved

1) If $ f_{k}\to f$ in $ L^{p}$, $ 1\leq p<\infty$, $ g_{k}\to g$ pointwise, and $ \left\|g_{k}\right\|_{\infty}\leq M$ for all $ k$, prove that $ \int f_{k}g_{k}\to\int fg$ in $ L^{p}$.

Here is what I have:
\begin{align*}
\left\|f_{k}g_{k}-fg\right\|_{p}&\leq\left\|g_{k}(f_{k}-f)\right\|_{p}+\left\|f(g_{k}-g)\right\|_{p}\\
&\leq\left\|g_{k}\right\|_{\infty}\left\|f_{k}-f\right\|_{p}+\left\|f\right\|_{p}\left\|g_{k}-g\right\|_{\infty}
\end{align*}.
The term on the left should go to zero as $ k\to\infty$, but I don't see necessarily why the term on the right should, since I didn't assume that $ f\in L^{p}$ or that $ g_{k}$ converged in $ L^{\infty}$.  That's assuming I even set it up correctly.

2) Find a closed set consisting entirely of irrationals with positive measure.

3) Let $ E\subseteq [a,b]$ in $ \mathbb{R}$.  Show that $ b-a+m^*(E\setminus [a,b])\leq m^*(E)$.  Do not assume $ E$ is measurable, and use only the most basic facts about Lebesgue outer measure (such as the definition, monotonicity).

2) Let $ \mathbb{Q}\cap [0,1] \equal{} \{q_1,q_2,...\}$. Let $ \varepsilon > 0$ such that $ \sum \frac {\varepsilon}{2^{k \minus{} 1}} < 1.$
Consider $ A \equal{} [0,1]\setminus \bigcup_{k \equal{} 1}^{\infty} \left(q_k \minus{} \frac {\varepsilon}{2^k},q_k \plus{} \frac {\varepsilon}{2^k}\right)$. A is closed, because [0,1] closed and union of intervals is open.
Clearly that $ A\subseteq \mathbb{R\setminus Q}.$

1) $ L^p\equal{}L^p(\mathbb{R})$ with standart measure??

In problem 1 to prove that $ \|f(g_k\minus{}g)\|_p \to 0$
one just needs to apply the Dominated convergence theorem.

Yes, but what is the dominating term?  I never assumed that $ f$ was in $ L^{p}$.

And any ideas for the third problem?

[quote="JRav"]Yes, but what is the dominating term?  I never assumed that $ f$ was in $ L^{p}$.[/quote]

What do you think $ f_k\rightarrow f$ in $ L^p$ means?

[quote="JRav"]

And any ideas for the third problem?[/quote]

The problem appears to be misstated.

The problem states: Show that $ m_*(E)\leq m^*(E)$ for any $ E\subseteq [a,b]$, where $ m_*(E)$ is defined as $ m_*(E)\equal{}b\minus{}a\minus{}m^*([a,b]\setminus E)$.  I just realized I wrote $ E$ and $ [a,b]$ in the wrong order when I typed it the first time. :oops:

[quote="JRav"]The problem states: Show that $ m_*(E)\leq m^*(E)$ for any $ E\subseteq [a,b]$, where $ m_*(E)$ is defined as $ m_*(E) \equal{} b \minus{} a \minus{} m^*([a,b]\setminus E)$.  I just realized I wrote $ E$ and $ [a,b]$ in the wrong order when I typed it the first time. :oops:[/quote]

Just use the following elementary property of the outer measure:

$ m^*(E \cup F) \le m^*(E) \plus{} m^*(F)$

That does indeed make the problem trivial... the only thing I have reserves about is that the problem I presented is #7 while subadditivity is #13.  I do know all of these facts-I am going back into my old textbook and trying to solve all of the problems without "cheating" by using techniques before they are introduced.

The solution to the first problem is correct now, yes?  That is, everything is set up correctly and everything on the right goes to zero.  In particular, $ \int f_{k}g_{k}\to\int fg$ in $ L^{p}$ means $ \left\|f_{k}g_{k}\minus{}fg\right\|_{p}$, right?

[quote="JRav"]That does indeed make the problem trivial... the only thing I have reserves about is that the problem I presented is #7 while subadditivity is #13.  I do know all of these facts-I am going back into my old textbook and trying to solve all of the problems without "cheating" by using techniques before they are introduced.[/quote]

The subudditivity follows directly (and easily) form the outer measure definition, so there is no cheating if you first solve #13, and then aplly it to #7: sometimes textbook problems are written in wrong order.

[quote="JRav"]The solution to the first problem is correct now, yes?  That is, everything is set up correctly and everything on the right goes to zero.  In particular, $ \int f_{k}g_{k}\to\int fg$ in $ L^{p}$ means $ \left\|f_{k}g_{k} \minus{} fg\right\|_{p} \to 0$, right?[/quote]

Well, I suppose there are some misprints. For example, $ \int f_{k}g_{k}$ and $ \int fg$ are just numbers, so there is no sence to consider them as functions. I suppose the correct statement of the problem is:

Let $ f_k, f \in L^{p}\equal{}L^{p}(\Omega, \Sigma, \mu)$, $ f_{k}\to f$ in $ L^{p}$, $ 1\leq p<\infty$, $ g_{k}\to g$ pointwise, and $ \left\|g_{k}\right\|_{\infty}\leq M$ for all $ k$, prove that $ f_{k}g_{k} \to  fg$ in $ L^{p}$.

In particular, if the $ \mu(\Omega) < \infty$, this implies that $ \int f_{k}g_{k} \to \int fg$.

Yes, you are correct.  It seems that what happened was when I was reading the problem, since everything was written on one line, the $ f$ looked like an integral, so I saw an integral there when there really was none. :)


Never make a bet like this, no matter how sure it may seem.

http://sports.espn.go.com/nfl/news/story?id=2584148&campaign=rss&source=ESPNHeadlines

yeah, it's like playing russian roulette with a mentally retarded kid...

[quote="Ignite168"]How was it a bet if there wasn't any reward for them not shutting out the Packers?[/quote]

I guess the reward was personal satisfaction...?

AAAAAAUUUUUGGGGGGGGHHHHHHHHH!!!!!!!!!!!!! :wallbash_red:  :wallbash:  :wallbash_red:

PACKERS STILL ROCK!!!!!!

You should be telling us to make bets like this, so one day you may be able to take advantage of it... :D

everyobdy who lives near me, make bets like this.


number theory
combinatorics proposed
combinatorics

An $8\times8$ chessboard is completely tiled by $2\times1$ dominoes. Prove that we can place positive integers
in all cells of the table in such a way that the sums of numbers in every domino are equal and the numbers placed
in two adjacent cells are coprime if and only if they belong to the same domino. (Two cells are called adjacent if
they have a common side.)

Well this can belong to number theory as well...

Place 64 natural numbers on the chessboard in such a way that any two number in the same cell are coprime and any two numbers in different cells are not coprime (it can be done in many ways).
Now choose a giant prime number $p$. For each two numbers $a$ and $b$ in the same domino there is decomposition $p=ax+by$, where $x,y\in\mathbb N$. Place numbers $ax$ and $by$ in that domino.

That's all.

Take 32 numbers $a_1,a_2\dots a_{32}$  with\begin{align*}
a_i=Q\times P_{i,1}P_{i,2}\dots P_{i,32}
\end{align*}
Where $P_{i,i}$ term is missing.For example,\begin{align*}
a_3=Q\times P_{3,1}P_{3,2}P_{3,4}\dots P_{3,32}
\end{align*}.
Here $Q$ and all $P$ are distinct odd primes.
Now,By CRT,take $x$ such that,\begin{align*}
x&\equiv a_k\pmod{P_{i,k}}\ \ \ \forall i\ne k,\forall k\\
x&\not\equiv a_k\pmod{Q}\ \ \ \ \forall k\\
x&\not\equiv a_k\pmod{P_{k,i}}\ \forall i\ne k,\forall k\\
x&\equiv 1\pmod{2}
\end{align*}

Now write $a_i,x-a_i$ in each of the 32 dominoes.

We need to show:[list]
[*]$gcd(a_i,x-a_i)=1\ \ \ \forall i$
[*]$gcd(a_i,a_j)\ne 1\ \ \ \forall i\ne j$
[*]$gcd(x-a_i,x-a_j)\ne 1\ \ \ \forall i\ne j$
[*]$gcd(a_i,x-a_j)\ne1\ \ \ \forall i\ne j$
[/list]

$\bullet$ As all prime divisors of $a_i$ namely,$Q,P_{i,1},P_{i,2},\dots P_{i,32}$ does not deide $x-a_i$ does not devide $x-a_i$ ,hence $gcd(a_i,x-a_i)=1\ \ \ \forall i$.

$\bullet$ As $Q|a_i,Q|a_j$ so $gcd(a_i,a_j)\ne 1\ \ \ \forall i\ne j$.

$\bullet$ As $x$ is odd by construction and $a_i$ are also odd by construction,So $2|gcd(x-a_i,x-a_j)$.

$\bullet$ Finally,by construction $P_{i,j}| a_i$ and $P_{i,j}|x-a_j$.
Hence,$gcd(a_i,x-a_j)\ne1\ \ \ \forall i\ne j$ $\blacksquare$


induction
combinatorics open
combinatorics

Let $m,n\in\mathbb{Z},m+n>2$. Consider the $m\times n$-grid. First we have only the outer borders of grid colored. Players must choose the adjacent or overlapping side of two squares and color that side. If side is already colored this side can't be chosen. If player can color the last non-colored side of a square in his turn he will get one point. Game will end when all sides are colored and the winner is the one who has more points as all the sides of the grid are colored.

Determine when the first player has a winning strategy, when the second player has a winning strategy as

(a) $m=1$

(b) $m$ and $n$ are odd

(c) arbitrary $m$ and $n$.

There is useful graph representation of this game.  From the grid we can build graph as follows: put vertices in centers of grid squares, and connect two vertices with an edge iff corresponding squares share the side (the edges are perpendicular to corresponding sides). Now the game can be described using graph: players take turns removing edges from graph; if after some turn there are isolated vertices, they are also removed and current player gets one point for each removed vertex; the game ends when the whole graph has been removed, and the winner is the player who has most points. Note that this game can be played using any graph (without isolated points). We can define the cost of any graph as difference of final scores of first and second players if both use the best strategy.

Now consider case (a), where m=1. The graph is just n vertices in the line, with n-1 edges between consecutive vertices (call them "n-lines"). We prove that for even n, first player wins, and for odd n, second player wins. To prove this statement, we need to consider more general graphs, consisting of arbitrary number of such "lines" (they are not connected to each other). If such graph contains  only odd lines, we call it "type 1" graph; if there is single even line, it is "type 2" graph (graphs with more than 1 even lines aren't needed). Now we can prove the following statement:
1. Cost of type 1 graph is at most -1
2. Cost of type 2 graph is at least 2
Use induction by k - number of edges.
Base: if k=1, the only possible graph is 2-line, and its cost is clearly 2 (the first player removes the only edge and both vertices). As 2-line is type 2 graph, the statement is proved.
Step: suppose we have proved the statement for all graphs with  k-1 edges. If we have type 1 graph with k edges, then any possible move of first player turns it into type 2 graph with k-1 edges. Futhermore, first player can score at most 1 point with such move. Now, according to our statement, second player can archieve score difference of at least 2 given this graph, so first player can achieve difference of at most 1-2=-1 with initial graph.
If we have type 2 graph with k edges, then first player can cut single vertex from even line, either making it an odd line or deleting it. In any case, first player scores at least 1 point and produces type 1 graph with k-1 edges. Now second player can get difference at most -1, so the cost of initial graph is at least  1-(-1)=2. This finishes the proof.

Now, as even lines are type 2 graphs and odd lines are type 1 graphs, first player wins for even n and second player wins for odd n.

What about the other two cases?

[quote="puuhikki"]
(b) $m$ and $n$ are odd

(c) arbitrary $m$ and $n$.
[/quote]

I may be talking nonsense, but isn't (b) a second-player win because he can guarantee a win by making the symmetric move as player A and thus getting an extra square in the center?

If this is correct I think (c) is also a second-player win, but I can't prove it.


Let $x,y,z$ be non-negative reals such that $x+y+z=3$. Prove that

$\frac{1}{x^p+y+z}+\frac{1}{x+y^p+z}+\frac{1}{x+y+z^p} \le 1$

(a) for $p=2$;
(b) for $p=3$;
(c) for $p \rightarrow \infty$;
(d)* for any integer number $p$, $p \ge 4$;
(e)* for any positive number $p$, $p>1$.

for (a), $\frac{1}{x^2+y+z} \le \frac{1}{9} (-x+4)$ by standard method.

What is the 'standard method'?  :?:

For (b):

If $x,y,z \le 2$,
\[\displaystyle\sum_{cyc} \frac{1}{x^3+y+z}=\displaystyle\sum_{cyc} \frac{1}{x^3-x+3} \le \displaystyle\sum_{cyc} (\frac{-2}{9}x+\frac{5}{9})=1.\]

If one of the $x,y,z$ is greater than 2, let's say it's $x$,
\[\frac{1}{x^3-x+3}+\frac{1}{y^3-y+3}+\frac{1}{z^3-z+3}<\frac{1}{9}+\frac{4}{9}+\frac{4}{9}=1.\]

[quote="yptsoi"]
If one of the $x,y,z$ is greater than 2, let's say it's $x$,
\[\frac{1}{x^3-x+3}+\frac{1}{y^3-y+3}+\frac{1}{z^3-z+3}<\frac{1}{9}+\frac{4}{9}+\frac{4}{9}=1.\][/quote]

why?

Yes, Siuhochung and Yptsoi. Your solutions for $p=2$ and $p=3$ respectively are the same to my solutions.
For $p\rightarrow\infty$ the solution is easy. :lol:


In a right square pyramid all five faces have the same area. If the height of the pyramid is 9in, what is the total surface area in square inches? Explain how you got your answer. Thanks you all who help.

[hide]Let the side length of the square be s. So, we have $ s^2\equal{}sh/2$, where h is the slant height. Thus, $ h\equal{}2s$. The slant height, height, and the s/2 form a right triangle. $ s^2/4\plus{}81\equal{}4s^2$.

$ s^2\plus{}324\equal{}16s^2$
$ s^2\equal{}324/15\equal{}108/5$

So, the total surface area is $ 5s^2\equal{}108$

[/hide]

when you do s^2/4 + 81= 4s^2
why did you do s^2/4 and where did you get 81.

Two regular polygons of the same number of sides have sides 48m and 55m in length respectively.
What is the length of the side of the another regular polygon of the same number of sides, if its area
is equal to the sum of the two ? Please explain how you got the answer. Thanks in advance.


Does anyone happen to know the formal name of the generalized chicken mcnugget theorem: if one has two integers $ a,b$ with $ (a,b)\equal{}k$ for $ k\ge 1$, then all sufficiently large integers divisible by $ k$ can be written in the form $ ar\plus{}bs$ for some nonnegative integers $ r,s$?

This statement is equivalent to the $ k \equal{} 1$ case.  The proof that the [url=http://mathworld.wolfram.com/FrobeniusNumber.html]Frobenius number[/url] exists is very straightforward and I don't know if the statement of its existence has a name, but the formula that exists (in this case) could be called the formula for the Frobenius number in the case $ n \equal{} 2$.

(One can generalize the statement of the existence of the Frobenius number to any number of integers that have no common factor.)


advanced fields
advanced fields unsolved

Let $ p$ be a prime, $ n$ a natural number, and $ S$ a set of cardinality $ p^n$ . Let $ \textbf{P}$ be a family of partitions of $ S$ into nonempty parts of sizes divisible by $ p$ such that the intersection of any two parts that occur in any of the partitions has at most one element. How large can $ |\textbf{P}|$ be?

The answer is $\boxed{\frac{p^n-1}{p-1}}.$

To prove the bound fix an element $e$ and look at the parts containing $e$ over all partitions.

To construct, we use induction. $n=1$ trivial. If $n=2,$ number the elements $0$ through $p^2-1$ and let the first partition $P_1$ be

$$\{0,1, \dots , p-1 \}, \{p,p+1, \dots , 2p-1 \}, \dots , \{p^2-p,p^2-p+1, \dots , p^2-1 \}.$$

Then for each residue $i \pmod{p}$ we define a partition where for any integer $0 \le k \le p-1$ we take the integer $\equiv k + (m-1)i \pmod{p}$ from the $m$th part of $P_1,$ take these over all $m,$ and put these integers all in one part.

For $n>2,$ first number the elements $0$ through $p^n-1$ and let the first partition $P_1$ be
$$\{0,1, \dots , p-1 \}, \{p,p+1, \dots , 2p-1 \}, \dots , \{p^n-p,p^n-p+1, \dots , p^n-1 \}.$$
The key idea is to consider these $p^{n-1}$ parts as their own set of cardinality $p^{n-1}$ which we apply the inductive hypothesis on. There is a family of partitions of these $p^{n-1}$ parts into "groups" of $p$ parts, with size $\frac{p^{n-1}-1}{p-1},$ where any two groups that occur in any of the partitions can only share at most one part in common. And then for each of these partitions of the $p^{n-1}$ parts, we can generate $p$ partitions of the original $S$ using the same argument as the $n=2$ case on each group of $p$ parts. So

$$1 + \frac{p(p^{n-1}-1)}{p-1} =\frac{p^n-1}{p-1} $$ groups, done. $\blacksquare$


Prove that $ x\plus{}\frac{1}{x^3}\geq \frac{4}{\sqrt[4]{27}}$ for any $ x>0$

[hide]Write $ x \equal{} {x\over 3} \plus{} {x\over 3} \plus{} {x\over 3}$ and apply AM-GM. Equality for $ {x\over 3}\equal{}{1\over x^3}\iff x\equal{}\sqrt[4]{3}$[/hide]

[hide="Would this also work?"]
By AM-GM:

\begin{eqnarray*}
\dfrac{x+\dfrac{1}{x^3}}{2} &\ge& \sqrt{x\cdot \dfrac{1}{x^3}}\\
\dfrac{x+\dfrac{1}{x^3}}{2} &\ge& \sqrt{\dfrac{1}{x^2}}\\
\dfrac{x+\dfrac{1}{x^3}}{2} &\ge& \dfrac{1}{x}\\
x+\dfrac{1}{x^3} &\ge& 2\cdot \dfrac{1}{x}\\
x+\dfrac{1}{x^3} &\ge& \dfrac{2}{x}
\end{eqnarray*}

With equality when $ x=\dfrac{1}{x^3}$ or $ x=1$ ($ x\not= -1$) which means that the lowest value for $ x+\dfrac{1}{x^3}$ is $ 2$, and since $ \sqrt[4]{27}>2$ then $ \dfrac{4}{\sqrt[4]{27}} < 2$ which means it's always true for $ x>0$.

[/hide]

[quote="tonypr"][hide="Would this also work?"]
By AM-GM:

\begin{eqnarray*} \dfrac{x + \dfrac{1}{x^3}}{2} & \ge & \sqrt {x\cdot \dfrac{1}{x^3}} \\
\dfrac{x + \dfrac{1}{x^3}}{2} & \ge & \sqrt {\dfrac{1}{x^2}} \\
\dfrac{x + \dfrac{1}{x^3}}{2} & \ge & \dfrac{1}{x} \\
x + \dfrac{1}{x^3} & \ge & 2\cdot \dfrac{1}{x} \\
x + \dfrac{1}{x^3} & \ge & \dfrac{2}{x} \end{eqnarray*}

With equality when $ x = \dfrac{1}{x^3}$ or $ x = 1$ ($ x\not = - 1$) which means that the lowest value for $ x + \dfrac{1}{x^3}$ is $ 2$, and since $ \sqrt [4]{27} > 2$ then $ \dfrac{4}{\sqrt [4]{27}} < 2$ which means it's always true for $ x > 0$.

[/hide][/quote]

Nope, you simply showed that $ x + {1\over x^3} = {2\over x}$ holds ONLY for $ x = 1$, but for all other values you have $ x + {1\over x^3} > {2\over x}$, and from there you can't conclude the minimum is obtained for $ x = 1$ (and it isn't).

Yes, now I see the huge error in what I did.

How did you learn to solve inequalities like that? I would have never guessed to turn $ x\equal{}\dfrac{x}{3}\plus{}\dfrac{x}{3}\plus{}\dfrac{x}{3}$ and then take

$ \dfrac{\dfrac{x}{3}\plus{}\dfrac{x}{3}\plus{}\dfrac{x}{3}\plus{}\dfrac{1}{x^3}}{4}\ge \sqrt[4]{\dfrac{1}{27}}$

Experience, I guess. What you're looking for is a way for all the variables to get cancelled out in the product figuring in the GM. Once you grasp that idea, it's not that difficult to see you need "three x's" to cancel the denominator of the second term.

As an exercise, try to find the minimum of $ 5x^2\plus{}{4\over x^3}, x>0$ and state the value of $ x$ for which it is obtained.

This is what I have:

$ 5x^{2}+\dfrac{4}{x^{3}} = x^2+2x^2+2x^2 +\dfrac{2}{x^3}+\dfrac{2}{x^3}$

And with AM-GM:
$ x^2+2x^2+2x^2 +\dfrac{2}{x^3}+\dfrac{2}{x^3}\ge \sqrt[5]{2^4}$

I haven't checked it yet but for now I think that the value of $ x$ that makes this true is when $ 5x^{2}=\dfrac{4}{x^{3}}$ which is $ \sqrt[5]{\dfrac{4}{5}}$.

Hmm.. well another possibility is:

$ \dfrac{\dfrac{5x^2}{3} \plus{} \dfrac{5x^2}{3} \plus{} \dfrac{5x^2}{3} \plus{} \dfrac{2}{x^3} \plus{} \dfrac{2}{x^3}}{5}\ge \sqrt [5]{\dfrac{5^3\cdot 2^2}{3^3}}$

With equality in $ \dfrac{5x^2}{3} \equal{} \dfrac{2}{x^3}$ which is $ x \equal{} \sqrt [5]{\dfrac{6}{5}}$?

I think this one works.

Edit: hehe, Fixed

[quote="Farenhajt"]Not exactly, but you're getting there :)

What you also must bear in mind is the condition of equality in AM-GM (which is, by definition, the minimal value). ALL the terms must be equal - but CAN you guarantee that with your proposed splitting: $ x^2, 2x^2, {2\over x^3}$?[/quote]
Hi ,
 As I understood by your words,We need to splitt in that way such that all the terms must be equal.But,tonypr didn't splitt in all equal terms.What I am missing ? :blush:

What Farenhajt means is that there must exist a value of $ x$ such that all the terms are equal.  Otherwise, the equality case of AM-GM can't be satisfied.

tonypr, one way to state the general result Farenhajt is using is as [i]weighted[/i] AM-GM:  for non-negative reals $ w_1, ... w_n, x_1, ... x_n$ such that $ w_1 \plus{} ... \plus{} w_n \equal{} 1$, we have

$ w_1 x_1 \plus{} w_2 x_2 \plus{} ... \plus{} w_n x_n \ge x_1^{w_1} x_2^{w_2} ... x_n^{w_n}.$

This can be proven first by proving it for rational $ w_i$ (where it reduces to ordinary AM-GM) and then applying a density argument, or it can be proven directly by weighted Jensen's.  But in practice you will only ever need to use the theorem for $ w_i$ rational and so it's better to think of it as just a fancy application of AM-GM.


Define s(n) to be the number of ways that the positive integer n can be written as an ordered sum of at least one positive integer. For example $4=1+3=3+1=2+2=1+1+2=1+2+1=2+1+1=1+1+1+1$, so s(4)=8. Find and a prove a general formula.

We have $C_{n-1}^{m-1}$ ways to write n as m ordered natural numbers, so $s\left(n\right)=\sum_{m=1}^n C_{n-1}^{m-1}=2^{n-1}$
Throw off the word "ordered", then this problem doesn't have a close-form solution, the best result is a recursive formula.


calculus
derivative
function
calculus computations

Consider the map $ F: (u,v)\rightarrow (x,y,z): \equal{}(u\plus{}v^2,u^2\minus{}v^3,2uv)$. Let $ (u_0,v_0)\equal{}(2,1)$ and $ (x_0,y_0,z_0)\equal{}F(u_0,v_0)$. Prove that the image of an open neighborhood of $ (u_0,v_0)$ is the graph of a mapping $ z\equal{}g(x,y)$. Calculate $ \frac{\partial^2 g}{\partial x\partial y}(x_0,y_0)$.

the proof could be derived from Cauchy's theorem, which is just to claim that the existence of implicit function at the neibough of $ F(u_0, v_0)$, and the caculation need to take the use of chain theorem.


geometry
circumcircle
inequalities
Euler
inequalities proposed

Show that in any triangle we have

$E\leq \frac{3 \cdot \sqrt{3}}{4} \cdot R^2$

where $E$ and $R$ denote its area and the radius of its circumcircle respectively!  :)

Or  you  can  multiply  these  well known  ineqs

1) [tex]r\leq\frac{R}{2}  [/tex]  (Euler)

2) [tex] 4s^2=(a+b+c)^2\leq3(a^2+b^2+c^2)\leq27R^2 [/tex]


function
induction
algebra unsolved
algebra

Find all functions $ f:\mathbb{Q}\to\mathbb{R}$ that satisfy $ f(x\plus{}y)\equal{}f(x)f(y)\minus{}f(xy)\plus{}1$ for every $x,y\in\mathbb{Q}$.

Let $ x\equal{}y\equal{}0$ then $ f^2(x)\minus{}2f(x)\plus{}1\equal{}0$ or $ f(0)\equal{}1$.
Let $ y\equal{}1$ then $ f(x\plus{}1)\equal{}f(x)f(1)\minus{}f(x)\plus{}1\equal{}f(x)(f(1)\minus{}1)\plus{}1$. Now let $ x\equal{}\minus{}1$ then $ f(\minus{}1)(f(1)\minus{}1)\equal{}0$ so $ f(\minus{}1)\equal{}0$ or $ f(1)\equal{}1$. But if $ f(1)\equal{}1$ then let $ y\equal{}1$ we have $ f(x\plus{}1)\equal{}f(x)f(1)\minus{}f(x)\plus{}1\equal{}1$ or $ f(x)\equal{}1,\forall x\in\mathbb{Q}$. Now consider $ f(\minus{}1)\equal{}0$. Put $ x\equal{}y\equal{}\minus{}1$ and $ x\equal{}1;y\equal{}\minus{}2$ then we have $ f(\minus{}2)\plus{}f(1)\equal{}1$ and $ f(1)f(\minus{}2)\minus{}f(\minus{}2)\plus{}1\equal{}0$, which lead to $ f(\minus{}2)\plus{}f(1)\equal{}1$ and $ f(1)f(\minus{}2)\plus{}f(1)\equal{}0$. There are 2 cases here:
1. $ f(1)\equal{}2,f(\minus{}2)\equal{}1$ then $ f(x\plus{}1)\equal{}f(x)\plus{}1$ and by induction we got $ f(x\plus{}n)\equal{}f(x)\plus{}n,\forall x\in\mathbb{Q},n\in\mathbb{N}$ so $ f(n)\equal{}n\plus{}1,\forall n\in\mathbb{N}$.
Now let $ x\equal{}n,y\equal{}\frac{1}{n}$ we have $ f(n\plus{}\frac{1}{n})\equal{}f(n)f(\frac{1}{n})\minus{}f(1)\plus{}1$ and by what said above we conclude $ f(\frac{1}{n})\plus{}n\equal{}(n\plus{}1)f(\frac{1}{n})\minus{}1$ or $ f(\frac{1}{n})\equal{}1\plus{}\frac{1}{n}$. Next, let $ x\equal{}m,y\equal{}\frac{1}{n}$ we have $ f(m\plus{}\frac{1}{n})\equal{}f(m)f(\frac{1}{n})\minus{}f(\frac{m}{n})\plus{}1$ or $ m\plus{}f(\frac{1}{n})\equal{}(m\plus{}1)(\frac{1}{n}\plus{}1)\minus{}f(\frac{m}{n})\plus{}1$ or $ f(\frac{m}{n})\equal{}\frac{m}{n}\plus{}1$, which leads to the function $ f(x)\equal{}x\plus{}1,\forall x\in\mathbb{Q}$.
2. $ f(1)\equal{}0,f(\minus{}2)\equal{}1$, then we have $ f(x\plus{}1)\equal{}\minus{}f(x)\plus{}1,\forall x\in\mathbb{Q}$, replace $ x$ with $ x\plus{}1$ we obtain $ f(x\plus{}2)\equal{}\minus{}f(x\plus{}1)\plus{}1\equal{}f(x)\minus{}1\plus{}1\equal{}f(x),\forall x\in\mathbb{Q}$. Since, we just need to express $ f(x)$ in $ [0;2]$.

At here, I did not find anyway to come through. Could anybody please help me???

Dear Mashimaru,

" 2. , then we have f(x+1)=-f(x)+1,\forall x\in\mathbb{Q}, replace  with  we obtain f(x+2)=-f(x+1)+1=f(x)-1+1=f(x),\forall x\in\mathbb{Q}. Since, we just need to express  in .

At here, I did not find anyway to come through. Could anybody please help me???  "

You almost had the solution.

From $ f(x\plus{}2) \equal{} f(x)$ and $ f(1) \equal{} 0, f(\minus{}2) \equal{} 1$, we have $ f(n) \equal{} 0$ for $ n$ odd and $ f(n) \equal{} 1$ for $ n$ even.
Substituting $ x \equal{} 2k, y \equal{} \frac{1}{2k}$ in the main equation, we get

$ f(2k \plus{} \frac{1}{2k}) \equal{} f(2k)f(\frac{1}{2k}) \minus{}f(1) \plus{} 1 \Rightarrow f(\frac{1}{2k}) \equal{} f(\frac{1}{2k}) \minus{}f(1) \plus{} 1$ because $ f(2k) \equal{} 1, f(2k \plus{} x) \equal{} f(x)$

This gives us $ f(1) \equal{} 1$ which is a contradiction. Hence, case 2 is not possible.


Let $ a$, $ b$, $ c$ and $ n$ be positive integers such that $ a^n$ is divisible by $ b$, such that $ b^n$ is divisible by $ c$, and such that $ c^n$ is divisible by $ a$.
Prove that $ \left(a \plus{} b \plus{} c\right)^{n^2 \plus{} n \plus{} 1}$ is divisible by $ abc$.

An even broader [i]generalization[/i], though not part of the QEDMO problem and not quite number theory either:
If $ u$ and $ n$ are positive integers, and $ a_1$, $ a_2$, ..., $ a_u$ are integers such that $ a_i^n$ is divisible by $ a_{i \plus{} 1}$ for every $ i$ such that $ 1\leq i\leq u$ (we set $ a_{u \plus{} 1} \equal{} a_1$ here), then show that $ \left(a_1 \plus{} a_2 \plus{} ... \plus{} a_u\right)^{n^{u \minus{} 1} \plus{} n^{u \minus{} 2} \plus{} ... \plus{} n \plus{} 1}$ is divisible by $ a_1a_2...a_u$.

Let's write $x\to y$ for "$x$ is divisible by $y$".
We have $a^{n^2}\to b^n\to c$, so $a^{n^2+n+1}\to cba$. (and cyclical)
Now the expression $\left(a+b+c\right)^{n^2+n+1}$ if expanded will be a sum of terms $a^ib^jc^k$ with $i+j+k=n^2+n+1$. We'll prove that each one of these terms is $\to abc$. We only need to consider the cases that one or two of $i,j,k$ are 0. If it's two of them, we have the monomial $a^{n^2+n+1}\to abc$ (and cyclical) by the above.
If it's only one, say wlog a binomial $b^ia^{n^2+n+1-i}$, we have either $i\le n$ (in that case $a^{n^2+n+1-i}\to a^{n^2+1}\to ac$) or $i>n$, in that case $b^i\to b^{n+1}\to bc$ and we are done again.

It is clear that the generalisation goes likewise but would be much harder to write up. :blush:

Nice problem  :P

Some hint for a solution (for the general one, too) that is easier to write up but doesn't show that much:
[hide]It's one thing to show that all such terms are already divisible by $abc$ (lets state it as additional problem ;) ), but there's another solution using $m|n \iff v_p(m) \leq v_p(n)$ for all primes $p$ ($v_p$ the $p$-adic valuation). But this one doesn't show that the summands have that property, too.[/hide]

I had tried that with the prime factors before but didn't succeed. Of course all the $a_i$ must have the same prime factors $p$ for which $\nu_p(a_i)>0$. But how much can we know about the factors of $\sum a_i$? Is it sufficient to start with $\min_i\nu_p(a_i)$? - Oh yeah, (grübel) actually it seems it is  :)  :) Nice :trampoline:

[quote="amirhtlusa"][quote="spanferkel"] Is it sufficient to start with $\min_i\nu_p(a_i)$? -  [/quote]
how do you show that?? :?[/quote]well, obviously if a certain prime $p$ divides one of the $a_i$, it must divide all the others too. So $\min_i\nu_p(a_i)$ divides all of them, so it divides their sum too. For each $p$ that divides the product $a_1\cdots a_u$, there is an $j=j(p)$ such that $\nu_p(a_j)=\min_i\nu_p(a_i)$
Now taking the $({1+n+...+n^{u-1}})$th power of that sum guarantees that the product of the $a_i$ divides it, as for each $p$, starting with $a_{j(p)}$ and then considering its $n$th, $n^2$th... powers, the sum (over $i$) of all the multiplicities $\nu_p(a_i)$ is contained in it.  :D


inequalities trigonometry search logarithms AMC USA(J)MO USAMO

Prove
1.in an acute triangle ,prove that $ {sinA\plus{}sinB\plus{}sinC>2}$.
2.if $ {\sin^2 A\plus{}\sin^2 B\plus{}\sin^2 C<2}$,then prove  $ {\triangle ABC}$ is obtuse.
3.A,B,C are three angles of a triangle ,prove : $ {sinA*sinB*sinC<\equal{}\frac{3\sqrt{3}}{8}}$.
 :)

for 1, use the extended law of sines to rewrite the inequality, then use the triangle inequality to prove it. this problem has appeared before; do a search if you want.

for 3, use Jensen's Inequality on $ \log \sin x$. this has probably appeared before too. it relates to an old USAMO question

It is hard to search,can anyone give a complete solution, dear mathlinkers ?

for the first one you could use jordan's inequality $ \sin \alpha\geq \dfrac{2\alpha}{\pi}$ for $ \alpha \in (0,90)$; another approach is using karamata's inequality applied for sequences $ (A,B,C)$ and $ (\frac{\pi}{2},\frac{\pi}{2},0)$ for the concave function $ \sin x$...

for the second one you could use that $ \sum \sin^2A\equal{}2\plus{}2\cos A\cos B\cos C$...

returning to the first you have that $ \sum \sin A\geq \sum \sin^2A\geq 2$ (for acute-angled triangles)...

for the third one you can use that $ \sum \sin A\leq \frac{3\sqrt3}{2}$ (by jensen's inequality)... and then am-gm...

you can use for example the identity $ \cos^2A\plus{}\cos^2B\plus{}\cos^2C\plus{}2\cos A\cos B\cos C\equal{}1$ to prove that $ \cos A\cos B\cos C\leq \frac{1}{8}$ and from this deriving that $ \sum \sin^2A\leq \frac{9}{4}$... finally from this, concluding again that $ \prod \sin A\leq \frac{3\sqrt3}{8}$...

for 1, see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=317788

for 3, check that $ \log \sin x$ is concave by computing the second derivative and seeing that it's negative. then a direct application of Jensen proves your inequality. very easy.

Thanks a lot  campos,pluristiq. I have learnt a lot !


Αποδείξτε ότι η εξίσωση $ x^2\plus{}y^2\equal{}z^5\plus{}z$ έχει άπειρες ακέραιες λύσεις που είναι πρώτες μεταξύ τους...

Δε βλέπω να απαντά κανείς οπότε παραθέτω 2 hints...

[hide] Υπάρχουν άπειροι πρώτοι αριθμοί της μορφής $ p_i \equal{} 4k_i \plus{} 1$.[/hide]

[hide] Κάθε πρώτος αριθμός της μορφής $ 4k_i \plus{} 1$ μπορεί να γραφεί ως άθροισμα 2 τετραγώνων.[/hide]


algebra
function
domain
ratio
real analysis
real analysis unsolved

if $ I: (l^1,\parallel{}.\parallel{}_2) \to (l^1, \parallel{}.\parallel{}_1)$ is the identity map then
both $ I$ and $ I^{ \minus{} 1}$ are continuous
is it right?

There's one direction in which it's not continuous.

Let's take a specific easy to evaluate example. Let $ f_r(k)\equal{}r^{k\minus{}1}$ for $ 0<r<1.$ (I'm also assuming that the domain for $ \ell^p$ is $ \mathbb{N},$ meaning $ 1\le k<\infty.$)

Then $ \|f_r\|_1\equal{}\frac1{1\minus{}r}$ and $ \|f_r\|_2\equal{}\frac1{\sqrt{1\minus{}r^2}}.$ Hence $ \frac{\|f_r\|_1}{\|f_r\|_2}\equal{}\sqrt{\frac{1\plus{}r}{1\minus{}r}}.$

Note that this ratio is unbounded as $ r\to 1^{\minus{}1}.$

I'll leave to you sorting out whether that means $ I$ is unbounded (disccontinuous) or $ I^{\minus{}1}$ is unbounded.


inequalities
logarithms
calculus
calculus computations

Prove that for $0<x<a$ we have:

$\cosh^{-1}\frac{a}{x}+\sinh^{-1}\frac{x}{a}>\frac{\pi}{2}$

The proper bound is anything smaller than $\log(1+\sqrt{2}) = \sinh^{-1}(1) \approx 0.88137$.


I feel stupid starting a thread for this, but I'm working a log problem (that I have worked before, actually, a while back), and I keep getting the wrong answer. Could someone point out my mistake?....

Problem: logba/logca = 2003/2002 and b/c = c^k...Find k.

My solution:

logbc = 2003/2002

logcb = 2002/2003

b=c^(k+1)

logb=(k+1)(logc)

logcb=k+1

k+1 = 2002/2003, k= -1/2003....the real answer is [hide]1/2002[/hide]...which I got sometime in the past...

[hide]1/2002[/hide] doesn't work. If you sub that into b / c = ck, you get [hide]b = c2003 / 2002, which means logcb = 2003 / 2002[/hide]. That's not what the question gave.

-1/2003 is the right answer.

Ok, thanks...strange, maybe they threw out that question (this was a local contest).

Yeah. I think the original question was meant to be .. = 2002/2003, not the other way around..


trigonometry
limit
logarithms
complex numbers
real analysis
real analysis unsolved

Let $c_{n}=u_{n}e^{i\theta_{n}}$, and you know that $u_{n}\rightarrow u$ and $\theta_{n}\rightarrow \theta$, where $c=ue^{i\theta}$.

The modulus of your complex number is
$\left(1+\frac{u_{n}^{2}}{n^{2}}+2\frac{u_{n}}{n}\cos \theta_{n}\right)^{n/2}\rightarrow e^{u\cos \theta}$.

The argument of the complex number is equal to:

$n\arctan \frac{u_{n}\sin \theta_{n}}{n+u_{n}\cos \theta_{n}}=\frac{\arctan \frac{u_{n}\sin \theta_{n}}{n+u_{n}\cos \theta_{n}}}{\frac{u_{n}\sin \theta_{n}}{n+u_{n}\cos \theta_{n}}}\frac{u_{n}\sin \theta_{n}}{1+u_{n}\cos \theta_{n}/n}\rightarrow u\sin \theta$.
Thus the sequence of complex numbers converges to: $e^{u\cos \theta+iu\sin \theta}=e^{c}$.

[quote="didilica"]
The modulus of your complex number is
$\left(1+\frac{u_{n}^{2}}{n^{2}}+2\frac{u_{n}}{n}\cos \theta_{n}\right)^{n/2}\rightarrow e^{u\cos \theta}$.[/quote]

Could you make this convergence more rigorous please? Which result are you using?

We deal with $1^{\infty}$ case, therefore we have that:

$(1+\frac{u_{n}^{2}}{n^{2}}+2\frac{u_{n}}{n}\cos \theta_{n})^{n/2}\rightarrow e^{\lim \left(\frac{u_{n}}{2n}+u_{n}\cos \theta_{n}\right)}=e^{0+u\cos \theta}$,
since $u_{n}\rightarrow u$ and $\theta_{n}\rightarrow \theta$.

Taylor formula is true also for complex natural log.
therefore we can use Taylor formula.

I mean

because
$\lim_{n\to+oo}c_{n}=c$
therefore
for all $\epsilon >0$
there exist $N$ such as for all $n>N$
$|c|-\epsilon<|c_{n}|<|c|+\epsilon$ is bounded
therefore

$\ln{(1+\frac{c_{n}}{n})}=\frac{c_{n}}{n}+o(\frac{1}{n})$
what does it mean for complex ?
Some definition:

it means that
if $g(x) \in$ "complex"
we wrote
$g(x)=o(x-x_{0})$
if
$\frac{|g(x)|}{x-x_{0}}\to 0$ when $x\to x_{0}$
where $x \in R$
ok
therefore

$(1+\frac{c_{n}}{n})^{n}=e^{n\ln{(1+\frac{c_{n}}{n}})}=e^{n(\frac{c_{n}}{n}+o(\frac{1}{n}))}$
therefore
$\lim_{n\to+oo}e^{n(\frac{c_{n}}{n}+o(\frac{1}{n}))}=e^{c}$

[quote="Arne"]Let $c_{1},c_{2},\,\cdots$ be a sequence of complex numbers converging to a finite limit $c$.

Show that $\left(1+\frac{c_{n}}{n}\right)^{n}$ converges to $e^{c}$.[/quote]

Let me do it for $(s_{n})$ which is a [b]real[/b] sequence. This can then be easily generalized to complex sequences as well.
--------
Since $(s_{n})$ is convergent it is bounded, so, for some $M>0$ we can right $|s_{n}|\leq M$. Choose $N>0$ so that for all $n>N$ we have $\frac{M}{n}< 1$.

We have,
$\left( 1+\frac{s_{n}}{n}\right)^{n}= \exp \left(\ln \left(1+\frac{s_{n}}{n}\right)^{n}\right) = \exp \left[ n \ln \left(1+\frac{s_{n}}{n}\right) \right]$

Consider the exponent. And note that $\left| \frac{s_{n}}{n}\right| \leq \frac{M}{n}<1$ for all $n>N$.
Use infinite series to expand this,
$n \left( \frac{s_{n}}{n}-\frac{s^{2}_{n}}{n^{2}}+\frac{s^{3}_{n}}{n^{3}}-.... \right) = s_{n}-\frac{s^{2}_{n}}{n}+\frac{s^{3}_{n}}{n^{2}}-...$
Since $s_{n}$ is bounded we have $|s^{k}_{n}|\leq M^{k}$ and hence $\frac{s^{k}_{n}}{n^{k-1}}\to 0$ for all $k\geq 2$.
Since this series is uniformly convergent we can pass the limit to each term and we get:
$s_{n}-0+0-0+... = s_{n}$.

Thus, this converges to $\exp (s_{n}) = e^{s_{n}}$ because $e^{x}$ is continous on $\mathbb{R}$.


[hide]x= 49/3
m= -15
y= -2[/hide]
assuming that I didn't make any careless mistakes

Are calculators allowed?

now a ranger, please show a solution

[hide="Hint"]
Both the first and the second equation can be divided by something to make them a bit simpler.[/hide]

After dividing the first two equations like Ignite168 said, use the third equation to find 'm' in terms of x and y. Then subsitute that in for where ever there's a 'm'. Now it's a 2-variable system, and you can solve that.

yeah, i did it on paper, but it kinda got a little messy. But i think i got the right answer

i used the 3 equations , then basically made them into 2 variables, and used system of equations to find the value 1 after another

or you can use Cramer's Rule which makes everything faster and limits the mistakes you will make

[hide]$12x+8y+6m=90;\, 6x+10y+2m=48;\, 3x+4y+m=26$

Multiply the second equation by 2 and sutract it from the first equation to get $-12y+2m=-6,$ or $6y-m=3.$

Multiply the third equation by 4 and subtract it from the first equation to get $-8y+2m=-14,$ or $4y-m=7.$

Now, subtract one of these equations from the other to get $y=-2.$

Then, find $m=-15.$

Plug these two values into the original first equation to find $x=\frac{49}{3}.$

The asnwer is $\left\{\frac{49}{3},\,-2,\,-15\right\}.$[/hide]

[quote="Ihatepie"]What's the answer?

12x+8y+6m=90
6x+10y+2m=48
3x+4y+m=26

Find what x,y,and m equal[/quote]
[hide]\begin{eqnarray}12x+8y+6m &=& 90\\ 6x+10y+2m &=& 48\\ 3x+4y+m &=& 26\end{eqnarray}

Multiply (2) by -2, and add the result to (1):

$-12y+2m =-6$

Multiply (3) by -2, and add the result to (2):

$2y =-4$

$y =-2$

$-12\left(-2\right)+2 m =-6$

Factoring out 2 on both sides above:

$12+m =-3$

$m =-15$

Substituting these results into (3):

$3x+4\left(-2\right)-15 = 26$

$3x-23 = 26$

$3x = 49$

x = $\frac{49}{3}$

$\boxed{\left(x,y,m\right) = \left(49/3,-2,-15\right)}$[/hide]

[quote="Dr. No"][quote="Ihatepie"]What's the answer?

12x+8y+6m=90
6x+10y+2m=48
3x+4y+m=26

Find what x,y,and m equal[/quote]
[hide]
$\boxed{\left(x,y,m\right) = \left(49/3,-2,-15\right)}$[/hide][/quote]

[hide]i think the answer looks better like this:

$\boxed{\left(x,y,m\right) = \left(\frac{49}{3},-2,-5\right)}$[/hide]

[quote="7h3.D3m0n.117"][quote="Dr. No"][quote="now a ranger"]yeah, it looks more professional

not amateurish  :rotfl:[/quote]

Perhaps, except it is incorrect. I'll leave it to you to figure out why.[/quote]

I can't figure out why....why is it wrong?[/quote]
Hint: there is a difference between your answer and mine that has nothing to do with style of presentation.

oh oops  :blush:

sorry im not allowed to edit that post so i'll just put it here

[hide="real answer"]
$\boxed{\left(x,y,m\right) = \left(\frac{49}{3},-2,-15\right)}$
[/hide]


1/   $ \frac{4}{x} \plus{} \sqrt {x \minus{} \frac{1}{x}}  \equal{} x \plus{} \sqrt {2x \minus{} \frac{5}{x}}$

2/   $ x^3  \minus{} \sqrt[3]{{6 \plus{} \sqrt[3]{{x \plus{} 6}}}} \equal{} 6$

3/   $ \left\{ \begin{array}{l}
 x \minus{} \frac{4}{x} \equal{} 2y \minus{} \frac{2}{y} \\
 2x \equal{} y^3  \plus{} 3 \\
 \end{array} \right.$

4/   $ \left\{ \begin{array}{l}
 \sqrt x  \plus{} \sqrt y  \plus{} \sqrt z  \minus{} \frac{1}{{\sqrt x }} \minus{} \frac{1}{{\sqrt y }} \minus{} \frac{1}{{\sqrt z }} \equal{} \frac{8}{3} \\
 x \plus{} y \plus{} z \plus{} \frac{1}{x} \plus{} \frac{1}{y} \plus{} \frac{1}{z} \equal{} \frac{{118}}{9} \\
 x\sqrt x  \plus{} y\sqrt y  \plus{} z\sqrt z  \minus{} \frac{1}{{x\sqrt x }} \minus{} \frac{1}{{y\sqrt y }} \minus{} \frac{1}{{z\sqrt z }} \equal{} \frac{{728}}{{27}} \\
 \end{array} \right.$

Yeah! How do you feel it?

[hide]Perhaps number 4 can be simplified if we substitute $ a \equal{} \sqrt{x}, b \equal{} \sqrt{y}, c \equal{} \sqrt{z}$...[/hide]

hello,let's solve the third system. By substituting $ x\equal{}\frac{y^3\plus{}3}{2}$ in the first equation of our system we get $ \frac{y^3\plus{}3}{2}\minus{}\frac{8}{y^3\plus{}3}\equal{}2y\minus{}\frac{2}{y}$. After a algabraic computation we have
$ \frac{(y\minus{}1)(y^2\plus{}y\minus{}3)(y^4\plus{}3y\plus{}4)}{2y(y^3\plus{}3)}\equal{}0$, From here follows
$ y_1\equal{}1$, $ y_2\equal{}\minus{}\frac{1}{2}\plus{}\frac{1}{2}\sqrt{13}$ and $ y_3\equal{}\minus{}\frac{1}{2}\minus{}\frac{1}{2}\sqrt{13}$. The polynomial $ p(y)\equal{}y^4\plus{}3y\plus{}4$ hase no real solutions. Further we get $ x_1\equal{}2$, $ x_2\equal{}\minus{}1\plus{}\sqrt{13}$ and $ x_3\equal{}\minus{}1\minus{}\sqrt{13}$.
Sonnhard.

[hide="2"]
Re-write this as $ x \equal{} \sqrt [3]{6 \plus{} \sqrt [3]{6 \plus{} \sqrt [3]{6 \plus{} x}}}$, from which we find that $ x \equal{} \sqrt [3]{6 \plus{} x} \Longrightarrow (x \minus{} 2)(x^2 \plus{} 2x \plus{} 3) \equal{} 0$ yields a solution of $ \boxed{x \equal{} 2}$. It is not too difficult to see that this is the only real solution (for example by checking derivatives).
[/hide]

[hide="4"]
Make the substitution $ a \equal{} \sqrt {x} \minus{} \frac {1}{\sqrt {x}}, b \equal{} \sqrt {y} \minus{} \frac {1}{\sqrt {y}}, c \equal{} \sqrt {z} \minus{} \frac {1}{\sqrt {z}},$ and the equations reduce to\[ a \plus{} b \plus{} c \equal{} \frac 83, \quad a^2 \plus{} b^2 \plus{} c^2 \equal{} \frac {172}9, \quad a^3 \plus{} b^3 \plus{} c^3 \plus{} 3(a \plus{} b \plus{} c) \equal{} \frac {728}{27}
\]Let $ e_1 \equal{} a \plus{} b \plus{} c, e_2 \equal{} ab \plus{} bc \plus{} ca, e_3 \equal{} abc$. Then\[ e_1 \equal{} \frac 83, \quad e_1^2 \minus{} 2e_2 \equal{} \frac {64}{9}, \quad e_1^3 \minus{} 3e_1e_2 \plus{} 3e_3 \minus{} 3e_1 \equal{} \frac {728}{27}
\]Solving gives $ e_2 \equal{} 0$ and $ e_3 \equal{} 0$. Thus $ a,b,c$ satisfy the cubic equation\[ (w \minus{} a)(w \minus{} b)(w \minus{} c) \equal{} w^3 \minus{} e_1w^2 \plus{} e_2w \minus{} e_3 \equal{} w^3 \minus{} \frac {8}{3}w^2 \equal{} w^2(w \minus{} 8/3) \equal{} 0
\]Thus $ \{a,b,c\} \equal{} \{0,0,8/3\}$, and we have $ \sqrt {k} \minus{} \frac {1}{\sqrt {k}} \equal{} 0, 8/3 \Longrightarrow \boxed{\{x,y,z\} \equal{} \{1,1,9\}}$.
[/hide]

[hide="1"]
Let $ u \equal{} x\minus{}\frac{4}{x},\ v \equal{} x\minus{}\frac{1}{x}$. Then the equation can be rewritten as $ \minus{}u \plus{} \sqrt{v} \equal{} \sqrt{u\plus{}v}$. Since $ f(x) \equal{} \sqrt{x}$ is strictly increasing, then $ u > 0$ yields $ \sqrt{v} < \sqrt{u\plus{}v} \equal{} \minus{}u \plus{} \sqrt{v} < \sqrt{v}$ and $ u < 0$ yields $ \sqrt{v} > \sqrt{u\plus{}v} \equal{} \minus{}u \plus{} \sqrt{v} > \sqrt{v}$. Thus $ u \equal{} x \minus{} \frac 4x \equal{} 0 \Longrightarrow \boxed{x\equal{}2}$.
[/hide]

[quote="Dr Sonnhard Graubner"]hello,let's solve the third system. By substituting $ x \equal{} \frac {y^3 \plus{} 3}{2}$ in the first equation of our system we get $ \frac {y^3 \plus{} 3}{2} \minus{} \frac {8}{y^3 \plus{} 3} \equal{} 2y \minus{} \frac {2}{y}$. After a algabraic computation we have
$ \frac {(y \minus{} 1)(y^2 \plus{} y \minus{} 3)(y^4 \plus{} 3y \plus{} 4)}{2y(y^3 \plus{} 3)} \equal{} 0$, From here follows
$ y_1 \equal{} 1$, $ y_2 \equal{} \minus{} \frac {1}{2} \plus{} \frac {1}{2}\sqrt {13}$ and $ y_3 \equal{} \minus{} \frac {1}{2} \minus{} \frac {1}{2}\sqrt {13}$. The polynomial $ p(y) \equal{} y^4 \plus{} 3y \plus{} 4$ hase no real solutions. Further we get $ x_1 \equal{} 2$, $ x_2 \equal{} \minus{} 1 \plus{} \sqrt {13}$ and $ x_3 \equal{} \minus{} 1 \minus{} \sqrt {13}$.
Sonnhard.[/quote]

We could also note that the first equation yields $ 0 \equal{} x^2y\minus{}4y\minus{}2xy^2\plus{}2x \equal{} xy(x\minus{}2y)\plus{}2(x\minus{}2y) \equal{} (xy\plus{}2)(x\minus{}2y)$. If $ x\equal{}2y$ we get $ 0 \equal{} y^3\plus{}3\minus{}2x \equal{} y^3\minus{}4y\plus{}3$. If $ xy\plus{}2 \equal{} 0$ we get $ 0 \equal{} y^4\plus{}3y\minus{}2xy \equal{} y^4\plus{}3y\plus{}4$.

[quote="L_Euler"]$ \sqrt [3]{{6 \plus{} \sqrt [3]{{x \plus{} 6}}}} \equal{} y,\sqrt [3]{{x \plus{} 6}} \equal{} z$

so $ \left\{\begin{array}{l} x^3 \minus{} y \equal{} 6 \\
y^3 \minus{} z \equal{} 6 \\
z^3 \minus{} x \equal{} 6 \\
\end{array} \right.$

That equation system have single root, can you prove that? :)  :)[/quote]

Whoa nice.


inequalities
geometry
perimeter
geometry unsolved

Hi,

I found the following geometrical inequality problem at: http://math.stanford.edu/~vakil/putnam06/06mc7.pdf , Problem 5.

5. Let $ ABC$ be a triangle. Points $ D$, $ E$, $ F$ are given on the sides $ BC$, $ CA$, $ AB$, respectively, such that $ DC \plus{} CE \equal{} EA \plus{} AF \equal{} FB \plus{} BD$. Prove that $ DE \plus{} EF \plus{} FD\geq\frac {AB \plus{} BC \plus{} CA}{2}$.

I think i got it when $ DEF$ forms a non-obtuse triangle.

First, it's easy to see that, if $a=BC,b=CA,c=AB$, we have $DC+CE=EA+AF=FB+BD=\frac{a+b+c}{3}$. Now, let $E'$, $F'$ be the ortogonal projections of $E,F$ into $BC$, respectively. From basic trigonometry, we get $BF'=BF.cosB$ and $CE'=CE.cosC$. Therefore:

$$EF\ge E'F' = a-BF.cosB-CE.cosC \Rightarrow EF \ge  a-BF.cosB-CE.cosC...(1)$$

Similarly, we have:

$$FD \ge b-CD.cosC-AF.cosA...(2)$$
$$DE \ge c-AE.cosA-BD.cosB...(3)$$

Summing $(1),(2),(3)$:

$$DE+EF+FD \ge a+b+c-[(AE+AF)cosA+(BF+BD)cosB+(CD+CE)cosC]=$$
$$(a+b+c)(1-\frac{1}{3}(cosA+cosB+cosC)) \ge (a+b+c)(1-\dfrac{1}{2})=\frac{a+b+c}{2}$$

As desired. In the last inequality, we used $cosA+cosB+cosC \le \frac{3}{2}$, a well-known inequality that can be proved using Jensen's Inequality.


limit
real analysis
real analysis solved

I have to solve the following problem,here is it:

$\lim\limits_{x\to\infty} f(x)=0, |f^{``}(x)\leq 1|   $ Prove that $\lim\limits_{x\to\infty} f^`(x)=0$

Let's write, for a certain fixed $h>0$, $f(x+h)-f(x)=hf'(x)+\frac{h^2}2 f''(t_x)$, where $t_x\in (x,x+h)$. Now make $x\to \infty$. We get, for large $x$, $-\varepsilon<hf'(x)+\frac{h^2}2 f''(t_x)<\varepsilon$, and from this we get, for large enough $x$, $-(\frac{\varepsilon}h+\frac h2)<f'(x)<\frac{\varepsilon}h+\frac h2$. Here $h$ is fixed, and the significance of this is that when $x$ is large enough, $f'$ becomes "crammed" between $-\frac h2$ and $\frac h2$.

We can say that there is $x_n(h)$ s.t. $-(h+\frac 1n)<f'(x)<h+\frac 1n,\ \forall x\ge x_n(h)$ for any $n\in\mathbb N^*$ and any $h>0$. If we now simultaneously make $n\to\infty,\ h\to 0$, we get what we want.


Given that Z=cos rº + i sin rº

a) Determine utilizing Newton's binomial theorem the expansion of Z^5
b) Determine utilizing Moivre's theorem, an expression equivalent to Z^5
c) Find equivalent expressions, using complex number equality, to Cos 5rº and Sin 5rº

[u]Discussion.[/u]  Let's work with the generalization of the conclusion you are trying to reach.  Let $ z \equal{} \cos\theta \plus{} i\sin\theta$.  Using the Binomial Theorem,
\[ z^n \equal{} \sum_{k \equal{} 0}^n\binom{n}{k}\cos^n\theta(i\sin\theta)^{n \minus{} k}.
\]
On the other hand, by De Moivre's Theorem, $ z^n \equal{} \cos n\theta \plus{} i\sin n\theta$.  Hence, $ (\cos n\theta,\sin n\theta) \equal{} (\Re z^n,\Im z^n)$, in which you can use the $ z^n$ derived using Newton's Binomial Theorem.  This method allows you to find trigonometric identities for $ \cos n\theta$ and $ \sin n\theta$ in terms of $ \cos\theta$ and $ \sin\theta$.

i already solved it ;) but thanks man i appreciate your help.

Ben


A triangle of sides $\frac{3}{2}, \frac{\sqrt{5}}{2}, \sqrt{2}$ is folded along a variable line perpendicular to the side of $\frac{3}{2}.$ Find the maximum value of the coincident area.

it seems we can use analytic, any better way?

We'll call the vertex A, B and C such that $ |AB|\equal{} {3 \over 2}, |AC| \equal{} {{\sqrt 5}\over 2}$ so $ |BC| \equal{} \sqrt 2$.
Let $ S$ be the area of the overlapping portions of the folded triangle, and $ S_0 \equal{} {1 \over 4}$ the "starting" value of S, when the folding line contains vertex C (or the hight $ h \equal{} 1$, relative to the base AB).

If we now displace the folding line toward the vertex B by an amount of $ x$, we'll add to $ S$ the area of the trapeze with bases $ h, (h \minus{} tan \angle ABC)$ and height $ x$, by one side, but in the other side, subtract the area of the rectangular (thanks to $ \angle ABC \equal{} {\pi \over 4}$) triangle with cathetus $ x. sec \angle ABC$ adjacent to $ \angle BCA$.

So  $ S \equal{} {1 \over 4} \plus{} x \minus{} {{x^2} \over 2} \minus{} 3x^2$, whose maximum of $ {9 \over {28}}$ occurs when $ x \equal{} {1 \over 7}$.


function
absolute value
complex analysis
complex analysis unsolved

True or False? Supply proof or counterexample for each statement

1.  A bounded, infinite subset of Q must have an accumulation point in Q.

2.  let c in R be an accumulation point of a set  .  Assume f: D-->R is a function such that lim x-->c of f(x)=L.  Then lim x-->c absolute value of f(x)= absolute value of L.

1.: no, there are rational sequences converging to irrational numbers (e.g. taking the first $n$ digits of the decimal fraction as the $n$-th term).

2.: yes, the function $|f(x)|$ is a concatenation of two functions that are continous at the important points.

Can you give a proof  for the second parts to show how it is a yes.  Its easier to see it with a proof and easier to understand.  ANd for #1 is there any other counterexample besides the rational and irrational?

Please help me understand a little better how the problem is a yes.  Can someone give a thorough proof of the problem using epsilon?


Prove there exists a one-to-one mapping between $Q$ and $N_+$.

There is a bijection between any two countable sets.

Write $\mathbb{Q}\cap [0,1)=\bigcup^{\infty}_{i=1}\{1/i,2/i,\ldots,(i-1)/i\}$ and

$\mathbb{Q}=\bigcup^{\infty}_{i=-\infty} \mathbb{Q}\cap [i,i+1)$

to show the countability of $\mathbb{Q}$.

[quote="blahblahblah"]There is a bijection between any two countable sets.[/quote]

What if the number of elements of the two sets are different? how can the mapping be a bijection?

It's difficult to speak of "number of elements" of an infinite set. Instead you want to consider cardinality. There is a bijection whenever the cardinality is the same. $N_+$ and $Q$ are both countable. So there is a bijection between them.

Where by "countable," both blahblahblah and Xev mean "countably infinite," if that's what your question meant.


Prove that
$\frac{a^5}{b}+\frac{b^5}{c}+\frac{c^5}{a}\geq\frac{1}{27} \cdot (a+b+c)^4$

for all $a,b,c >0$

Too easy!!! :(  :?  :D .

 Anyway,from Cauchy(or Jensen) we have:

   $\sum\frac{(a^2)^3}{ab}\geq\frac{(a^2+b^2+c^2)^3}{3(ab+bc+ca)}$.But using $3(a^2+b^2+c^2)\geq(a+b+c)^2$ twice and $a^2+b^2+c^2\geq ab+bc+ca$, so we are done. ;)

 Now I'm thinking to find a sharper inequality,because this is too weak. :(

There is another weaker inequality:
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq (a+b+c)^1\]
So can we get the patern from these two?

I'm sorry Shobber but if you say, $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a+b+c$, well,this is not true for any $a,b,c>0$.It is true in some conditions for $a,b,c$ ;)

  Mybe,there is a misunderstanding and you are talking about this one:

                 $\sum\frac{ab}{c}\geq\sum a$

I imagine that he was thinking of
$\sum{\frac{a^2}{b}}\geq{\sum{a}}$


How many integer that have three different digits and divided by 3.

$A = \overline{abc}=100a+10b+c$ such that $a+b+c \equiv 0 \mod 3$ and $a,b,c \in \{0,1,2,...,9 \}$  .

Does this help ??  :)

[hide]
We basically want the number of subsets of $[1,2,3,4,5,6,7,8,9]$ of size $3$ that are divisible by $3$ (call this $N$), and the number of subsets of $[1,2,3,4,5,6,7,8,9]$ of size $2$ that are divisible by $3$ (to account for zeroes, and call this $M$).

Then we want $6N+4M$.

For $M$:
If the number we pick is not divisible by $3$ ($6$ possibilities), there are $3$ other possibilities.
If it is divisible by $3$ ($3$ possibilities), there are $2$ other possibilities. This leads to $6*3+3*2 = 24$ possibilities.

For $N$:
Partition the set into $[1,2,3],[4,5,6],[7,8,9]$. Then take the first non-empty set, and cycle the elements in it. Exactly one part of the cycle will have a sum divisible by $3$, unless the first non-empty set contains all $3$ numbers that are used. Each cycle has length $3$, and there are $\binom{9}{3}$ ways to arrange stuff, and $3$ ways to have full sets, leading to $\frac{\binom{9}{3}-3}{3}+3 = \frac{9*8*7}{18}+2 = 4*7+2 = 30$ possibilities.

Then $6*30+4*24 = 180+96 = 276$ possibilities.
[/hide]

Hi - A nice simple question this I think.

There are certainly 30 elements of N as described and each can take one of 6 permutations giving an initial count of 180, never using zeros.

With a zero, there are only 12 distinct pairs of distinct integers divisible by 3.  They are:  12, 24, 15, 45, 36, 27, 18, 57, 48, 39, 78, 69.
So M = 12.

So the correct answer is 180 + 4 * 12 = 228.

I think where the previous post went wrong was in saying 'pick an element divisible by 3.  Then there are 2 possibilities to choose from for the second element.'  This is true e.g. pick 3.  That leaves 6 or 9.  So you get 3+6 or 3+9.  But then pick 6.  That gives 6+3 or 6+9.  Then multiply M by four.
In particular 3+6 leads to 360, 306, 630, 603.  So does 6+3.  Same for the numbers not divisible by 3.  Hence the total came to 24 instead of 12 because everything was being counted twice.

I don't have my own solution, but when I use the computer to write a program and check it, I see that the value of it is: $228$


geometry
geometric transformation
rotation

In your new job, you are helping to design stunts for a new movie. In one scene the writers want a car to jump across a chasm between two cliffs. The car is driving along a horizontal road when it goes over one cliff. The car is to travel through the air from one cliff to another, lower cliff. The higher cliff is 900 meters above the bottom of the chasm. The movie makers want to know the minimum value of the speed of the car so that it does not fall into the chasm. They have not yet select a location so they want an expression for the difference in height between the upper road and the lower road, h, in terms of the car's mass, m, the width of the chasm, w, and the maximum speed of the car, v. The stunt director will plug in the actual numbersafter the location is selected. Please include a diagram of this scenario.

I get most of it, except for the how the mass is to be put in here. Because of that, I can't solve it at all. Can someone help please? Thanks. BTW, air resistance is neglible.  :)

As far as kinematics is concerned, I don;t think the mass will appear in the equations.
Now, the time for free fall through a height $h$(irrespective of initial horizontal velocity) is $t = \sqrt{\frac{2h}{g}}$
Now, at max velocity, the car should cover a distance of $w$ horizontally in this time.
Thus$v \times t = w$

Thus we get,

$h = \frac{gw^2}{2v^2}$

Yeah, that's what I got, too. But that "m" is worrying me right now.  :ninja:

As I said, you solve all the Newton's laws and do get that most kinematic situations(like yours) do not depend on the mass of the object involved(approx). Of course, if you take into consideration the rotation of the car as it jumps or the effects of turning etc, the problem becomes VERY Complicated. You need even more data than just the mass. But in a very simple analysis, the mass will not appear. Don't worry!!!! :lol:

KK. I should've been confident with your answer in the first place, since you're interest is ASTRONOMY PHYSICS!!!  :P Much thanks!

Thanks!!!  :blush:

Are you participating in the Physics Weekly contest!!! See the Sticky!!!

Be confident in yourself!!

Rushil is correct.

Don't feel bad, this is what was shown in the fameous experiement at the leaning tower.

BTW Does anyone know why he was spending so much time at the top of the tower (it doesn't involve Physics experiements)?  (I'll give you a hint, he had his telescope with him and it was during the day.)

[quote]Be confident[/quote]
Just be sure, someone [b]else[/b] is doing the stunt :)


there are 3001 candies. 2 people alternates. every person every time can take p^n candies(n=0,1,2,....), p is a certain prime number withing 3001. the person to pick up the last candy wins. is there a strategy to make sure you can win?

I think there is no strategy because when there are 5 candies left, how can you pick the last one?($p \ne 1$)

[u]Question[/u]: Does $p$ remain constant until the end of the game?
[hide="If yes, then it is simple"]
At any time, the player can take $1,p, p^{2},p^{3},...$ candies
At the last step, the winner takes $p^{k}$ candies from a total of $p^{k}$, for some $k=0,1,...$

For any $p>2:$

[u]If you take first, then you will win, and you'll not need a strategy.[/u]

Indeed, after you have played, there is an even number of candies, since [b]odd-odd=even[/b]

Then, your opponent will remove an odd number of candies, so the result is [b]even-odd=odd[/b]
And so on...

Your opponent plays always from an even number, so he can't win

Only if $p=2$ there is a strategy. It is: "Avoid multiples of 3"
That is:
If there are [b]3k[/b] candies left, then you'll loose
If there are [b]3k+1[/b], then substract [b]1[/b]
If there are [b]3k+2[/b], then substract [b]2[/b]

Your opponent will never have the ability to remove $2^{n}$ candies, because he will alway have a multiple of [b]3[/b][/hide]


algebra
polynomial
function
calculus
integration
algebra unsolved

Find a polynomial $\large P(x) \in Z[x]$ has minimum
$\large \sqrt{2}$

This polynomial doesn't exist.Is it right?

Hi SuperMath,I supposed $\deg P = 4$ and the polynomial doesn't exist.Your initial statement didn't contain $\ deg P=4$  ;) .Vietnamesegauss89,how did you solve this problem without this condition?

vietnamesegauss89, your solution???
thanks!

[quote="vietnamesegauss89"]Sorry,I only can solve this problem if degP=4 :blush:[/quote]
Let f(x) is polinomial function with whole coefficients and f(x)>0, then Integral of (m+3)!f(x)(x^2-2),m=degrf(x) can be P(x) for any f(x).


[quote="despair"]I'm just wondering.

How do you become a moderator?[/quote]

This is not an official answer (which could only come from the site owner) but it does come from a moderator. The way I have seen most people become moderators of most online forums I have been on is for the moderator candidates to be plain old everyday participants who post USEFUL posts and who over time are noticed by the forum management. Sometimes moderators are asked to be moderators, sometimes they volunteer and are approved by the forum management. This AoPS Web site has had some turnover of moderators since I joined it in September 2003. I expect that there will be some more turnover in the next few years.

Post some interesting, useful posts, and you will attract notice. Welcome aboard.

Oh...are you also need to be fairly active and willing to read a huge variety of posts... :-)

Posting good stuff like tokenadult also helps alot.


What is the value of $ 101^{3} \minus{} 3 \cdot 101^{2} \plus{} 3 \cdot 101 \minus{}1$?


quadratics
algebra
complex numbers
absolute value

Let $ \alpha ,\ \beta$ be the roots of the quadratic equation : $ x^2\minus{}ax\plus{}1\equal{}0\ (a\in\mathbb{R})$. Consider the complex number expressed by $ x\equal{}\alpha ,\ \beta$.

(1) If $ a$ moves in the range of $ |a|\geq 2$, then prove that $ z$ moves in the all positive real numbers.

(2) If $ a$ moves in the range of $ |a|\leq 2$, then prove that $ z$ moves in the all complex numbers whose absolute value is 1.

Use the discriminant $ \sqrt{a^{2}\minus{}4}$ to prove it.

yes, if a>2,  both roots will be real, and if a<2, the roots will be of the form $ \dfrac{1}{2}a\pm \dfrac{1}{2}i\sqrt{4\minus{}a^2}$. The absolute value of this is $ \dfrac{1}{4}a^2\plus{}\dfrac{1}{4}(4\minus{}a^2)\equal{}1$


[quote="DangChienbn"]we use this inequality:
\[ \sum \frac {ab(a \plus{} b)}{c^3 \plus{} abc} \minus{} \sum \frac {a \plus{} b}{2c} \equal{} \sum \frac {(a \plus{} b)(ab \minus{} c^2)}{2(c^3 \plus{} abc)}\]

\[ \equal{} \sum \frac {b(a \minus{} c)(a \plus{} c) \plus{} a(b \minus{} c)(b \plus{} c)}{2(c^3 \plus{} abc)}\]

\[ \equal{} \sum \frac {a(b \plus{} c)(b \minus{} c)}{2}(\frac {1}{c^3 \plus{} abc} \minus{} \frac {1}{b^3 \plus{} abc}) \geq 0\]
[/quote]
Hi DangChienbn,you have nice sulotion! :)
And We have:
$ \sum_{cyc}{\sqrt{\frac{ab(a\plus{}b)}{c^3\plus{}abc}}} \geq 3$
It's easy by AM-GM! :P

[quote="2424"]
Hi DangChienbn,you have nice sulotion! :)
And We have:
$ \sum_{cyc}{\sqrt{\frac{ab(a\plus{}b)}{c^3\plus{}abc}}} \geq 3$
It's easy by AM-GM! :P[/quote]
Nice !  :P
$\sum \sqrt{\frac{ab(a+b)}{c^3+abc}}=\sum \frac{\sqrt{ab}(a+b)}{\sqrt{(c^2+ab)(ac+bc)}}$
$\ge \sum \frac{2\sqrt{ab}(a+b)}{(a+c)(b+c)}$
$\ge \sum \frac{4ab(a+b)}{(a+b)(b+c)(c+a)}$
$\ge 3$
we complete the proof equality occurs when $a=b=c$

[quote="arqady"][quote="2424"]With $ a,b,c>0$.Prove that
  $ \frac{ab(a\plus{}b}{c^3\plus{}abc}\plus{}\frac{bc(b\plus{}c)}{a^3\plus{}abc}\plus{}\frac{ca(c\plus{}a)}{b^3\plus{}abc} \geq 3$[/quote]
The following inequality is stronger.
Let $a$, $b$ and $c$ are positives. Prove that:
\[\frac{a^2(a+b)(a+c)}{(b^2+ac)(c^2+ab)}+\frac{b^2(a+b)(b+c)}{(a^2+bc)(c^2+ab)}+\frac{c^2(a+c)(b+c)}{(a^2+bc)(b^2+ac)}\geq 3\][/quote]
It is equivalent to

$\sum_{cyc} (a-b)^2\left[(a^2+ab+b^2)^2+ab\cdot \frac{7(a+b)^2+(a+b-2c)^2}{4}\right] \geq 0$


So who here is on the ARML team(s) for Ohio?  I'll be wearing my AoPS t-shirt to practice this Saturday; if you have one, wear it!

Yea, go us!

But I don't have an ARML shirt to wear.  Or any other math shirt.

[quote="h_s_potter2002"]Haha, I saw you today.[/quote]

Wow, so did I!  What a coincidence..... :)

[quote="h_s_potter2002"]I think we need a list of Ohio ARML AOPSers, just to keep track.

nr1337
h_s_potter2002[/quote]

Um, that's probably all of us, unfortunately :(


limit
integration
function
calculus
symmetry
real analysis
real analysis solved

$f(x)$ is monotone in $(0,1)$ and [tex]\lim_{n\to\infty} \frac{1}{n}\sum_{i=1}^{n-1} f(\frac{i}{n})[/tex] exists, but $\int_0^1 f(x)dx$ is not convergent.

Isnt $ f(x)=\frac{1}{x} $ an example? I dont understand what this problem want.

hey

$f(x) = 1/x$ doesn't work because $\sum \frac{1}{n}f(\frac {i}{n}) = \sum \frac{1}{n}\cdot\frac{n}{i} = \sum \frac{1}{i}$ doesn't converge.

Now I understand what you want. But I knew that a sequence can have limit although it doesn't converge!

[quote="harazi"]But I knew that a sequence can have limit although it doesn't converge![/quote]

 :shock: what does it look like?

Dear lyi, the sequence $ x_n=n $ has obviously limit infinite, but it does not converge. But it's ok, I understood the question and I will think about.

Well...In most books in China, limit "infinte" is considered not be convergent.

Different books use different definitions.  :D

To make this work, we need a function $f$ whose graph is symmetric about the point $\left( \frac12,0 \right)$ and which goes to $\infty$ at 0 (and at 1) fast enough for the integral to diverge. A good example would be $f(x)=\cot(\pi x)$. Another good example would be $\displaystyle f(x)=\frac{1-2x}{x(1-x)}$. Because of the symmetry $f(1-x)=-f(x)$, the sum $\displaystyle \sum_{i=1}^{n-1}f\left(\frac in\right) = 0$ for all $n$. However, since $\displaystyle \int_0^{\frac12}f(x)dx = +\infty$ and $\displaystyle \int_{\frac12}^1f(x)dx = -\infty$, the integral diverges.

We are relying quite a bit on some exact definitions and semantics. I take "converges" as "tends to a limit which is a number", the key point being that $\infty$ is not a number. If the limit is $\infty$ the correct language is that the limit (or sum, or integral) "diverges to $\infty$".

We also used the convention that in order for an improper integral to be called convergent, you must be able to cut it into pieces each involving just one singular point and have each of those pieces converge. It's worth noting in our case that the Cauchy principal value $\displaystyle \lim_{\epsilon\to 0} \int_{\epsilon}^{1-\epsilon} f(x)dx = 0$ does converge - just not the improper integral itself.

Nice, Merryfield.
Mine is [tex]\frac{1}{x}-\frac{1}{1-x}[/tex] too.  :lol:  :lol:  :lol:


geometry
geometric transformation
rotation
calculus
integration
absolute value
calculus computations

What is the surface area of the solid generated when the curve $3y=\sqrt{x}(3-x)$, from (0,0) to (3,0), is rotated about the y-axis?

Rewriting,
y = (1/3)x^(1/2)(3-x)

dy/dx = (1-x)/(2x^(1/2))

 2*pi*{Integral from x = 0 to x = 3 of x*sqrt(1 + (dy/dx)^2)dx}
will be used to determine the area of revolution.

Integrand = x*sqrt(1 + (1 - 2x + x^2)/(4x))

................ = x*sqrt((1 + 2x + x^2)/(4x))

................ = x*sqrt((1 + x)^2/(4x))

................ = x(1 + x)/(2*sqrt(x)), because x is nonnegative and an absolute value symbol is not needed.

Full form is 2*pi*{Integral from x = 0 to x = 3 of
x(1 + x)/(2*sqrt(x))dx}

= pi*{Integral from x = 0 to x = 3 of sqrt(x)(1 + x)dx}

= pi*{Integral from x = 0 to x = 3 of (x^(1/2) + x^(3/2))dx}

= pi*[(2/3)x^(3/2) + (2/5)x^(5/2)] from x = 0 to x = 3

= 28*pi*sqrt(3)/5

Answer:  28*pi*sqrt(3)/5 square units


Find the last digit of $ 1^{2007}+2^{2007}+3^{2007}....100^{2007}$
How would you do this? My teacher says use logs but i think there's a way by mods...i don't know how..can someone help me please? Thanks

[hide="solution"]Since we are looking for the last digit we take this mod 10, so $ 1^{2007}+2^{2007}+\dots+99^{2007}+100^{2007}\equiv 10(1^{2007}+2^{2007}+\dots+9^{2007}) \equiv 0 (mod 10)$. So the answer is $ \boxed{0}$

The second simplification comes from that $ b^{2007}\equiv (1b)^{2007}\equiv \dots \equiv (9b)^{2007}(mod 10)$ where ab denotes a 2-digit number (with tens of a, units of b)[/hide]


Billy Bob has a pet snail called Larry. Larry wants to climb a wall in Billy Bob's mansion. The wall is 37 feet tall. Larry can climb 3 feet in one day, but at night he slips down two feet. Larry starts the great climb on Sunday, June 2, 2006. On what day (day of the week and date) will Larry finish this great climb?

P.S. Root for Larry along with Billy Bob so that Larry will make it!!! :wink:

[quote="mtms5467"][hide]So basically Larry climbs 1ft/day. The day/date 37 days from June 2. (Oh wait...June 2, 2006 is a Friday...)

[hide="Assuming June 2, 2006 is a Sunday"]Saturday July 8[/hide]

[hide="Assuming he means July 2 (actually a Sun.!)"] Monday Aug 7[/hide][/hide][/quote]
It's a classic trick question. After 34 days, he's at 34 ft. And on the 35th day, he reaches the top, because he won't slide the 2 feet back down.


integration
trigonometry
calculus
calculus computations

For natural number $ n,$ prove that $ 0<\int_{0}^{\plus{}\infty}(1\minus{}e^{\minus{}\frac{t}{n}})e^{ \minus{} t}\sin^{3}t \ d t < \frac {9}{25n}.$

[hide="brute force"]$ B_n\equal{}\int_{0}^{\plus{}\infty}(1\minus{}e^{\minus{}\frac{t}{n}})e^{ \minus{} t}\sin^{3}t \ dt$

$ I_a\equal{}\int e^{\minus{}at}\sin^3 t dt\equal{} \minus{}ae^{\minus{}at}\sin^3 t \plus{}3a \int \sin^2t \cos t e^{\minus{}at} dt$
$ \equal{}\minus{}ae^{\minus{}at}\sin^3 t \plus{}3a[\minus{}ae^{\minus{}at}\sin^2t\cos t \plus{}a\int e^{\minus{}at}(2\sin t \minus{}3\sin^3 t)dt]$
$ \equal{}\minus{}ae^{\minus{}at}\sin^3 t \minus{}3a^2e^{\minus{}at}\sin^2t\cos t\minus{}\frac{6a^3e^{\minus{}at}}{1\plus{}a^2}(\sin t \plus{}a \cos t)\minus{}9a^2I_a$

$ I_a\equal{} \frac{\minus{}ae^{\minus{}at}}{1\plus{}9a^2}[\sin^3t\plus{}3a\sin^2t \cos t \plus{} \frac{6a^2}{1\plus{}a^2}(\sin t \plus{} a\cos t) ]_0^\infty\equal{}\frac{6a^4}{(1\plus{}a^2)(1\plus{}9a^2)}$

$ B_n\equal{}I_1\minus{}I_{\frac{n\plus{}1}{n}}\equal{}\frac{3}{10}\minus{}\frac{6(n\plus{}1)^4}{(n^2\plus{}(n\plus{}1)^2)(n^2\plus{}10(n\plus{}1)^2)}$
$ \equal{}\frac{6n^4\plus{}\frac{81}{5}n^3\plus{}\frac{177}{10}n^2\plus{}\frac{24}{25}n\plus{}\frac{6}{25}}{22n^4\plus{}62n^3\plus{}71n^2\plus{}40n\plus{}10}$[/hide]


Let $\displaystyle ABCD$ be a a convex quadrilateral and $\displaystyle O$ be a point in its interior. Let $\displaystyle a,b,c,d,e,f$ be the areas of the triangles $\displaystyle OAB,OBC,OCD,ODA,OAC,OBD$.

Prove that \[ \displaystyle \left| ac - bd \right| = ef .  \]

Wrinig formula of surface, we only need to prove
\[ |\sin(AOB)\sin(COD)-\sin(BOC)\sin(DOA)|=\sin(AOC)\sin(BOD) \]
But just take a circle with center $O$ and let $A',B',C',D'$ be the intersections of this circle with $OA,OB,OC,OD$.
Now the above equality (in sines) becomes the ptolemy's theorem for the cyclic $A'B'C'D'$.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=83639 .

  darij


how do you solve these sorts of questions?

find the angle formed by the hour and second hand of a clock showing the time 1:46:23 (hour, minute, second).

[quote="mathnerd314"]There's a big long nasty complicated formula you can use. I like to find the angle between 12:00 ("north") and each of the hands, and then subtracting the two as if I were taking a bearing on a compass.[/quote]

I would also use that method.

To find that angle, however, note that the hour hand moves $30^\circ$ every hour, so it moves half a degree each minute. The minute hand moves $6^\circ$ every minute, and the second hand moves $360^\circ$ every minute (or $6^\circ$ every second).


AMC
AIME
probability
modular arithmetic
function
number theory
relatively prime

This is a #15 problem from a Mock AIME; [b]hints only[/b] are greatly appreciated.

Thanks in advance!

Let $ S$ be the set of integers $ 0,1,2,...,10^{11}\minus{}1$. An element $ x\in S$ (in) is chosen at random. Let $ \star (x)$ denote the sum of the digits of x. The probability that $ \star(x)$ is divisible by $ 11$ is $ \frac{m}{n}$ where $ m$ and $ n$ are relatively prime positive integers. Compute the last 3 digits of $ m\plus{}n$.

This seems like a good idea to try.

[hide]

You can represent each number as 11 slots that can be filled with 0's through 9's.  If you filled all but the last slot with numbers, then you know there is exactly one digit from 0 through 9 that could be placed in the last slot to make the final number divisible by 11, unless the sum of the 10 slots you filled in is congruent to 1 mod 10.  (this would require a digit to be congruent to 10 mod 11).

[/hide]

[hide="Trivial hint"]We seek the number of solutions to $ s_1 \plus{} s_2 \plus{} \ldots \plus{} s_{10} \equiv 0 \pmod 11$, where $ s_1, \ldots, s_{10}$ lie between 0 and 9 inclusive. [/hide]

[hide=" Big hint"]As long as $ s_1 \plus{} s_2 \plus{} \ldots \plus{} s_9 \not \equiv 1 \pmod{11}$, there exists a unique value of $ s_{10}$ within the range that makes $ s_1 \plus{} s_2 \plus{} \ldots \plus{} s_9 \plus{} s_{10} \equiv 0 \pmod{11}$. [/hide]

My first idea was to use the generating function $ (1 \plus{} x \plus{} ... \plus{} x^9)^{11}$ and trying to calculate the coefficients of $ x^{11k}$. However, it seemed too much computations. If we use the previous two poster's observations, the only way I see to solve it would be calculating the number of solutions to $ s_{1} \plus{} s_{2} \plus{} \ldots \plus{} s_{9}\equiv 1\pmod{11}$ subtracted from $ 10^{10}$, which seems no easier than the original problem itself... So is there a better approach I'm not realizing?
BTW I could not make sense out of this wik solution at [url]http://www.artofproblemsolving.com/Wiki/index.php/Mock_AIME_1_2006-2007/Problem_15[/url]
but is the answer even correct?
(Oh wait, if we assign the first 8 digits the 9th digit would be unique unless the sum of the previous 8 digits is $ 2\mod11$ and so we could build a recursion... Am I right?)

[quote="mathwizarddude"] If we use the previous two poster's observations, the only way I see to solve it would be calculating the number of solutions to $ s_{1} \plus{} s_{2} \plus{} \ldots \plus{} s_{9}\equiv 1\pmod{11}$ subtracted from $ 10^{10}$
[/quote]
Yeah but we can keep repeating this idea. Thus number of solutions to $ s_{1} \plus{} s_{2} \plus{} \ldots \plus{} s_{9}\equiv 1\pmod{11}$ is the number of solutions to $ s_{1} \plus{} s_{2} \plus{} \ldots \plus{} s_{8}\equiv 2\pmod{11}$ subtracted from $ 10^9$ and so on. (Too lazy to work out the rest).


function
algebra unsolved
algebra
functional equation

Note that if $f(a)=f(b)$, then $2f(a)=2f(b), f(2f(a))=f(2f(b)), a+1998=b+1998, a=b$, so $f$ is injective. Also note that

\[f(2f(x))=x+1998 \\
2f(2f(x))=2x+3996 \\
f(2f(2f(x)))=f(2x+3996) \\
2f(x)+1998=f(2x+3996).\]
Also it is clear from the given condition that for all $x$ at least $1999$, there's some $y$ such that $f(y)=x$. Now note that $f(2x+3996)-2f(x)=1998$ for all $x$, so it is constant. Also, for any $k \geq 1999$, we can write $2k$ as $2(k-1998)+3996$, so that $f(2k)$ is determined by $f(k)$. So we have, $f(2k)=f(2(k-1998)+3996))=2f(k-1998)+1998$, so for all $k \geq 1999$, $f(2k)$ is even.

Now we work with odd $x$. We have $f(2f(x))=x+1998$, so $f(2f(x))$ is odd. However, since $2f(x)$ is even, this implies $f(x) \leq 1998$; but there are at most finitely many values of $x$ for which this can happen, because $f$ is injective and there are finitely many numbers in the range $1, 2, ..., 1998$. so take some odd $x$ such that $f(x)>1998$, and we get a contradiction. So there are no such functions.

[quote=Severius]We have $f(2f(x))=x+1998$, so $f(2f(x))$ is odd. .[/quote]

What?!?!??!

Edit: I see they meant the VALUE is odd, not the function is odd.

[quote=Megus]Find all functions $f : \mathbb N \to \mathbb N$ satisfying, for all $x \in \mathbb N$, \[ f(2f(x)) = x + 1998 . \][/quote]

Let's start by plugging in some small values of $x$ to see if we can find any patterns.

For $x = 1$, we have $$f(2f(1)) = 1 + 1998.$$Solving for $f(1)$, we get $$f(1) = \frac{1999}{2}.$$Since $f$ is a function from the natural numbers to the natural numbers, we must have $f(1) \in \mathbb N$. Therefore, we can conclude that $1999$ must be an even number, so $f(1) = 999$.

For $x = 2$, we have $$f(2f(2)) = 2 + 1998.$$Solving for $f(2)$, we get $$f(2) = \frac{2000}{2} = 1000.$$So we have found that $f(1) = 999$ and $f(2) = 1000$.

Now let's try plugging in $x = 3$. We have $$f(2f(3)) = 3 + 1998$$Solving for $f(3)$, we get $$f(3) = \frac{2001}{2}$$Since $f$ is a function from the natural numbers to the natural numbers, we must have $f(3) \in \mathbb N$. However, $2001$ is an odd number, so there is no natural number that satisfies this equation.

Therefore, there are no functions $f : \mathbb N \to \mathbb N$ that satisfy the given condition.

[quote=Quantum_fluctuations][quote=Megus]Find all functions $f : \mathbb N \to \mathbb N$ satisfying, for all $x \in \mathbb N$, \[ f(2f(x)) = x + 1998 . \][/quote]

Let's start by plugging in some small values of $x$ to see if we can find any patterns.

For $x = 1$, we have $$f(2f(1)) = 1 + 1998.$$Solving for $f(1)$, we get $$f(1) = \frac{1999}{2}.$$Since $f$ is a function from the natural numbers to the natural numbers, we must have $f(1) \in \mathbb N$. Therefore, we can conclude that $1999$ must be an even number, so $f(1) = 999$.[/quote]

how do u solve for $f(1)$???? what 1999 even number?????


Hi, im studying for my quiz and in my review WS there is a question. What are periods?
p.s. We're studying now the periodic table.

they're the horizontal rows on the periodic table


Sami decided to park her car at Park$ \&$Go for her four-day trip.  The cost is usually $ \$8$ per day or partial day.  However, Sami has a $ 20\%$ discount coupon for her entire bill.  How much did Sami's parking cost?

The total amount owed before the discount would be $ 4 \times 8\equal{}32$.  We know it is $ 20\%$ off, so we do $ \frac {100\minus{}20}{100} \times 32\equal{}\frac {4}{5} \times 32\equal{}\boxed {25.6}$

[quote="JSGandora"]A word of caution:
FTW does not accept $25.6$ for their answer, according to FTW, $25.60$ is correct and $25.6$ is incorrect.[/quote]
I believe $25.6$ is incorrect because it is not the formal notation for currency in the US.


$ 3(1 \plus{} 2 \plus{} 3 \plus{} \cdots \plus{} (n \minus{} 1) \plus{} n) \equal{} \frac32n(n \plus{} 1) \equal{} 1584$

$ \implies n(n \plus{} 1) \equal{} 1056,32 < \sqrt {1056} < 33\implies n \equal{} \boxed{32}$.


Ok, I think as unofficial squad training exams this month we should simply take the 2004 UK FSTs. If you want to do them, post in this thread and I'll forward the PDFs to you.

I also thought it might be good if we marked them (although I was unable to get hold of proper mark schemes - they quite possibly don't exist), to encourage good mathematical presentation and highlight possible technical errors (maybe if you missed out a case in your own argument you wouldn't notice).

Anyone disagree with this? If not, any volunteers to mark them this month? (Jack? - clearly it's best if a marker is better than everyone else) If not, I'll happily mark them, purely from a "logic" point of view (I obviously won't be familiar with the possible solutions).

Yeah, that sounds good- though I probably won't be able to take either of them until half term (2 weeks time).

Marking is a sensible idea; would we scan, LaTeX or post them?

Sure, I'll mark them if you like.  If you scan your scripts in and email them to me, that'll do fine (although feel free to LaTeX them if you have copious free time).

Jack

Sure - I am up for it! However, time is always something of an issue for me, so Latexing them might be a bit tricky...
Incidentally, since this is now becoming quite serious, do you think we should email the squad mailing list with details of this bit of the forum. Obviously it would be crazy for Jack to end up having to read 20 scripts, but I am sure we could find alternative arrangements if there are lots of other volunteers.

That would seem a reasonable idea. Do you want to do the honours Dom? :)

Or, as a slightly more plausible alternative, we could find out who was interested, put their names in a cycle, and then send your paper to the person who is next in the cycle (or whatever...). Just like marking the final exam in Hungary. Only actually marking it, rather than just skimming it and giving it 7...

That's true, but my thinking was that the person marking would be better informed about possible approaches having marked several scripts, and be able to compare different approaches. The danger with the cyclical system is that people will see a script and go "this looks nothing like my solution, and he has made a small grammatical error at the start, therefore I doubt this will be correct." (as I did, albeit correctly, with a question I marked at Hungary - although it [i]actually [/i]made no sense).

My "that sounds like a reasonable idea" was referring to informing the squad however. :lol:

Surely they deserved to get 0/7 for a solution which was different to yours and contained a small grammatical error, no?  Obviously this was a foreshadowing of a larger hole in their solution which would have lain unrecognised in their script, flukily having been given 7.
Do you think they do this for BMO2?
Sam

For marking, could we each 'specialize' in a question, so we each get four people's question 1 (for example) to mark?

(If we end up with a lot more people, that is)

[quote]do you think we should email the squad mailing list with details of this bit of the forum[/quote]

Erm, shall I just do it then?

I have just finished FST1 (what better way to cure a cold than do intensive IMO training!). I found it ok. It also occurred to me during one of my write ups that a LaTeXed solution would be much easier to follow, and therefore against the spirit of IMO marking :lol:. In seriousness, it's probably better to learn how to scrawl it in a meaningful way the first time round.

Can you send me the PDF('s) please, since its now the Half term.

Jonathan

Has anyone else finished any of the FSTs yet?

I'm gonna do the 2005s, although I guess maybe some of you already have done them.

I did the first one yesterday, and I'll do the other in a few days.

The deadline for these is relatively soon. Is everyone happy for discussion to commence in March?
Incidentally, I think we'll let the March exams be the 2005 FSTs, with deadline on March 23rd so we can discuss and fill knowledge gaps before Trinity.

I'm happy to discuss them whenever.

I have already done the 2005 FSTs (or at least most of the questions from them), but I suppose thats not a huge problem.


Let $ m$ and $ n$ be positive integers satisfying $ m>n$ and $ m\plus{}n$ even. Prove that the roots of equation $ x^2\minus{}(m^2\minus{}m\plus{}1)(x\minus{}n^2\minus{}1)\minus{}(n^2\plus{}1)^2\equal{}0$ are positive integers, but not squares.

it's clear that the roots if this equation are $ n^2\plus{}1$ and $ m^2\minus{}m\minus{}n^2$ .
$ n^2\plus{}1$ is not a perfect square .
suppose that there exist an integer $ x$ such that $ m^2\minus{}m\minus{}n^2\equal{}x^2$ then $ 4(m^2\minus{}m)\minus{}4n^2\equal{}4x^2$ so $ 4m^2\minus{}4m\minus{}4n^2\plus{}1\equal{}4x^2\plus{}1$.
therfore since $ 4m\equal{}2(m\plus{}n\plus{}(m\minus{}n))$ we get :
$ 4m^2\minus{}4m\minus{}4n^2\plus{}1\equal{}4m^2\minus{}4n^2\minus{}2(m\plus{}n)\minus{}2(m\minus{}n)\plus{}1$
$ \equal{}(2(m\plus{}n)\minus{}1)(2(m\minus{}n)\minus{}1)$.
now let $ m\plus{}n\equal{}2a$ the number $ 4a\minus{}1$ has a prime on the form $ 4k\plus{}3$ but $ 4x^2\plus{}1$ doesn't have such prime which it is a contradiction.


group theory
abstract algebra
linear algebra
matrix
superior algebra
superior algebra theorems

Does any one know how to show the commutator subgroup of the real 2x2 matrices is the special linear group of 2x2 over the reals. One inclusion is trivial, one other much more difficult.

I am a little surprised nobody got this problem. I try to take Sundays off from working on problems, but usually I am not very successful. However,this Sunday, being Father's day here in the US, I had to take the day off. I thought someone would have solved it by now

Nevertheless I figured it out:

G= group of real 2x2 matrices under matrix multiplication

N= subgroup of G containing matrices with determinant 1.

G'=commutator subgroup of is contained in N( EASY PART )

To show N is contained in G' one must show for every n can be expressed as a commutator from G.

To do this show that N is normal in G by mapping elements in G to the multiplicative group of non-zero reals via their determinant. The kernel of this mapping is N itself.So N is normal in G.

So, gNg^1=N for all g in G. With a little more work this will establish the other inclusion. Note for a particular n in N there are many commutators in G equal to n.

BTW this is Herstein p.66 #19, a STARRED PROBLEM

I wrote up the "solution" and ran into a problem with the last step. I spoke too soon. Let me see if I can repair it.....

In general, the commutator subgroup of $G=Gl_n(K)$ is $Sl_n(K)$, for any $n\ge 1$ and any field $K$ different from $\mathbb F_2$ (the field with two elements):

For one inclusion (I suppose it's the one you considered easy?), notice that $G/Sl_n(K)$ is isomorphic to $K^*$ (the multiplicative group of non-zero elements of $K$), which is abelian, so $Sl_n(K)$ contains the commutator subgroup $G'$ of $G$ (because the latter is the smalest normal subgroup $H$ of $G$ such that $G/H$ is abelian).

For the other inclusion, try to prove that every invertible matrix of determinant $d\ne 0$ can be written as a product of transvections times the diagonal matrix having entries $1,1,\ldots,1,d$ (the [i]transvection[/i] $T_{ij}(\lambda),\ i\ne j$ is the matrix having $1$ on the main diagonal and $\lambda$ in position $(i,j)$). Since all $T_{ij}(\lambda),\ \lambda\ne 0$ are similar to one another (they lie in the same conjugacy class of $G$), the result we want will follow if we manage to prove that $G'$ contains at least one transvection other than $1$ itself.

If $K\ne\mathbb F_2$, then we can find $\lambda\ne 0$ with $1+\lambda\ne 0$. Since we can find $S\in G$ such that $ST_{12}(\lambda)S^{-1}=T_{12}(1+\lambda)=T_{12}(1)T_{12}(\lambda)$, we have $T_{12}(1)= ST_{12}(\lambda)S^{-1} T_{12}(\lambda)^{-1}\in G'$, and we’re done.

Grobber, this post goes back to the summer also. I didn't understand your post, but if you have a copy of Artin's Algebra(1991), it is a blue book. The solution to this problem from Herstein's text( section 2.7 ) is found on pages 296-298 in Artin's book. It is a bunch of tedious exercises toward the solution and deserves a star from Herstein.

Mozart


geometry
circumcircle
trigonometry
cyclic quadrilateral
geometry unsolved

[quote="Jumbler"]Let $ABCD$ be a convex quadrilateral. $BC=CD$ and $\angle C+2\angle A = 180$. Let $M$ be the midpoint of $BD$. Show that $\angle MAD = \angle BAC$[/quote]
Just prove that $CB$ and $CD$ are tangent lines to the circumcircle of the triangle $ABD$,and this problem comes to the well-known problem
about the harmonic quadrilaterals. :wink:

[quote="Tiks"]
Just prove that $CB$ and $CD$ are tangent lines to the circumcircle of the triangle $ABD$,and this problem comes to the well-known problem
about the harmonic quadrilaterals. :wink:[/quote]

Yes, I can prove the first, but how to continue?

Let $O$ be the circumcenter of the triangle $\bigtriangleup ABD.$ It is easy to prove that the circumcircle $(O),$ of $\bigtriangleup ABD,$ tangents to $BC,$ $CD,$ at points $B,$ $D.$ We draw a line through the point $C$ and perpendicular to the segment line $AO,$ which intersects the segment lines $AD,$ $AB,$ at points $E,$ $F,$ respectively. So, we can prove the follows results:

[b][size=117] a) [/size][/b] – The point $C,$ is the midpoint of the segment $EF.$ This result is the conclusion of the below well known (at least to me) problem.

[b][size=117]PROBLEM.[/size][/b] - A point $O,$ is moved on a fixed line $(\ell),$ passes through the vertex $A,$ of a given angle $\angle xAy.$ We draw the circle $(O),$ centered at point $O$ with radius $R = OA,$ which intersects the segment lines $Ax,$ $Ay,$ at the points $B,$ $D,$ respectively. Find the locus of the point $C,$ as the intersection point of the tangent lines to the circle $(O),$ at points $B,$ $D.$

We draw an arbitrary line perpendicular to $(\ell),$ which intersects the segment lines $AD,$ $AB,$ at points $E',$ $F',$ respectively. Then the locus of $C,$ is the line connecting the vertex $A,$ of the given angle $\angle xAy,$ with the midpoint of the segment $E'F'.$

If this problem has not already been discussed in the Forum, I will present with pleasure the solution I have in mind.

[b][size=117] b) [/size][/b] – The segment lines $BD,$ $EF,$ are antiparallel each other with respect to the segment lines $AB,$ $AD.$ It is easy to prove that the quadrilateral $BDEF,$ is cyclic $($ We denote as $K,$ $L,$ the intersection points of $(O),$ $EF$ respectively, from the line $AO.$ So, $\angle AEF = 90^{o}-\angle DAK = \angle AKD = \angle ABD$ $).$

Hence, because of the points $M,$ $C,$ are the midpoints of the segments $BD,$ $EF$ respectively, formed by two antiparallel lines with respect to the sidelines $Ax,$ $Ay,$ of the given angle $\angle xAy,$ we conclude that the lines $AC,$ $AM,$ are isogonal each other with respect to the angle $\angle xAy.$

That is we have that $\angle DAC = \angle BAM$ $\Longrightarrow$ $\angle DAM = \angle BAC$ and the proof is completed.

Kostas Vittas.

Or we can use metric argument.

I take for proven that $CB, CD$ are tangents to circumcircle of $\triangle ABD$ (elementary by the theorem on inscribed and tangent angles). Let the circle in question be $\omega$ and let $CM\cap\omega=\{X,Y\}$. Then $XY$ is a diameter of $\omega$. Let $x=CX, y=CY$. Since $CM$ is harmonic mean of $CX$ and $CY$ (by the known theorem), we have $CM={2xy\over x+y}$, giving $XM={x(x-y)\over x+y}, MY={y(x-y)\over x+y}, XY=x-y$ - see the attached diagram. Hence ${YC\over YM}={x+y\over x-y}$

If $\alpha=\angle AXY$, then

\begin{eqnarray*}AM^{2}&=& AX^{2}+XM^{2}-2\cdot AX\cdot XM\cdot\cos\alpha\\ &=& AX^{2}+XM^{2}-2\cdot AX\cdot XM\cdot{AX\over XY}\\ &=& AX^{2}\left(1-{2XM\over XY\right)+{x^{2}(x-y)^{2}\over (x+y)^{2}}\\ &=& AX^{2}\left(1-{2x\over x+y}\right)+{x^{2}(x-y)^{2}\over (x+y)^{2}}\\ &=&-{x-y\over x+y}AX^{2}+{x^{2}(x-y)^{2}\over (x+y)^{2}}}\end{eqnarray*}

\begin{eqnarray*}AC^{2}&=& AX^{2}+XC^{2}-2\cdot AX\cdot XC\cdot\cos\alpha\\ &=& AX^{2}+XC^{2}-2\cdot AX\cdot XC\cdot{AX\over XY}\\ &=& AX^{2}\left(1-{2XC\over XY\right)+x^{2}\\ &=& AX^{2}\left(1-{2x\over x-y}\right)+x^{2}\\ &=&-{x+y\over x-y}AX^{2}+x^{2}\\ &=& \left({x+y\over x-y}\right)^{2}\left(-{x-y\over x+y}AX^{2}+{x^{2}(x-y)^{2}\over (x+y)^{2}}\right)\\ &=& \left({x+y\over x-y}\right)^{2}AM^{2}}\end{eqnarray*}

Therefore ${AC\over AM}={x+y\over x-y}={YC\over YM}$, hence $AY$ is the bisector of $\angle MAC$. By the initial conditions and the construction, $AY$ is also the bisector of $\angle DAB$. Those two facts imply $\angle BAC=\angle DAM$.

Let $P$ be the point in the prolongation of $CM$ such that $<DPM=<MAD$.
Thus $A$, $M$, $D$, $P$ are concyclic and $<APM=<ADM=180^{0}-<ABC$.
Then, $A$, $B$, $C$, $P$ are concyclic and $<BAC=<BPC=<DPM=<MAD$.

Congratulations dear Jonfv, for your clever solution.

Some typo corrections: $($ In my drawing $AB = 7.5,$ $BD =6.4,$ $AD = 6.6$ $).$

From the cyclic quadrilateral $ADMP$ $\Longrightarrow$ $\angle APM = 180^{o}-\angle ADM$ $,(1)$

$180^{o}-\angle ADM = \angle DAB+\angle ABD = \angle DBC+\angle ABD = \angle ABC$ $,(2)$

From $(1),$ $(2)$ $\Longrightarrow$ $A,$ $P,$ $B,$ $C,$ are concyclic and then the conclusion.

Thank you and best regards, Kostas Vittas.

Some typo corrections again: Equation(1) should be $\angle APM = \angle ADM$.  :rotfl:

Thank you OHO for your remark and sory for my inappropriate note. I had made a different drawing $($ where $AB > AD$ $)$ and so, there is no typo corrections in Jonfv’s post $($ where $AB < AD$ $).$

Kostas Vittas.

[quote="silouan"]Nobody said about Tiks solution .It's beautiful without considering a point  :wink:[/quote]
Thanks Silouan :) ,I know that you like the facts about harmonic quadrilateral :lol: .


inequalities
combinatorics proposed
combinatorics

This might be well known, it is not hard, but it appeared on my research.

Given sets $ S\equal{}{(1,...,k)}$ and $ T\equal{}({1,...,m})$ define the following game $ G_n(S,T)$ between two players Alice and Bob:

Alice plays first and she chooses a set $ S$ or $ T$. WLOG, let's say she chose $ S$. Then she chooses an element $ s_1 \in S$ and removes it from the set $ S$. Now, Bob must respond by choosing an element $ t_1\in T$ on the opposite set and removes it from $ T$. They keep doing this until each player has chosen exactly $ n$ elements.
(Note: At each step Alice can choose any set. For example, Alice first play can be $ s_1\in S$ where Bob would have to respond with $ t_1\in T$, and in the next play Alice can choose an element $ t_2\in T$ in which Bob would have to respond with an element $ s_2 \in S$)

Bob wins if for all $ i\neq j$ we have $ s_i<s_j$ if and only if $ t_i<t_j$.
Otherwise, Alice wins.

Prove that if $ k,m\geq 2^n\minus{}1$ then Bob has a winning strategy in $ G_n(S,T)$.

Am I misunderstanding something here?  If we take $ S$ to be a chain and $ T$ to consist of $ m$ incomparable elements, then at the end we'll either have $ s_1<s_2$ or $ s_2<s_1$, but no corresponding inequality for the $ t_i$.

Every two elements of $ T$ are comparable. That's why I wrote $ T$ as a set of  the first $ m$ integers!
$ S$ and $ T$ are finite linear orders. That means, they are isomorphic to a finite set of consecutive integers with the usual $ <$ relation.


Search for some solution to the homogeneous eq.

$ \frac{d^{2}y}{dx^{2}}\plus{}\frac{2}{x}\frac{dy}{dx}\equal{} 0$

of the form $ y\equal{}x^{n}$  (you'll find $ n\equal{}\minus{}1$)

Then search for the solution to the complete one of de form $ y\equal{}\frac{u}{x}$ and ...

[quote="Gibbenergy"]Can anyone help me solve this differential equation:

$ \frac{d^{2}y}{dx^{2}}\plus{}\frac{2}{x}\frac{dy}{dx}\equal{} Ay$
  A is a constant.

Many thanks ! :)[/quote]

$ y(x)\equal{}(C_{1}e^{\minus{}\sqrt{A}x}\plus{}C_{2}e^{\sqrt{A}x})/x$.


$a \equiv b \ mod(m) \longrightarrow m | b-a$

[quote="AstroPhys"]Shouldn't it be $a \equiv b \mod m \Rightarrow m | a-b$?[/quote]
doesn't matter though the first one is by "definition" the right...bot are equivalent

If you would like it in uncomplicated terms, "mod" is simply the remainder when a number is divided by another.

For instance, 3 (mod 2) would be 1 because when you divide 2 into 3, we get 1 as a remainder.

[b]@bos1234[/b]
the best way toi know the power of this expression is to use it solving problems get ypurself a good elementary number theory book and start enjoying.... :)


show that product of 3 consecutive even no's is divisible by 48

pretty stupid , right?

Let the 3 consecutive even numbers be 2n, 2n+2 and 2n+4.
We have to prove that
48 l 2n(2n+2)(2n+4)
48 l 2n(4n^2 + 12n + 8)
48 l 8(n^3 + 3n^2 + 2n)

We can devide by 8 and reduce the problem to proving that
6 l n^3 + 3n^2 + 2n
6 l n(n+1)(n+2)

This means we have to prove that the product of any 3 consecutive integers is divisable by 6. In any 3 consecutive integers there will at least be a 2 and a 3, so it will be divisable by 2x3 = 6.

well, at least one of the numbers is divisible by 4 and the other 2 are divisible by 2

sice there are 3 consecutive numbers, one of them is divisible by 3

therefore 4*2*2*3=48

[quote="shadysaysurspammed"]Nice method using variables,mellissa...

HOw bout just mugging logic?THe smallest possible conecutive even integers are 2,4,6 and 2*4*6=48.,which is divisible by 4....in the rest of the even consecutive nos , there will be 2*4*6(obviously).[/quote]
Im not sure that is allowed in major competitions and $2*4*6=48$ are not the lowest because $0*2*4=0$ are.... unless $0$ is not an even. :?  :(

A simple intuitive solution combining the methods mentioned above:

  Lets say the product of our 3 consecutive even integers is:

$2(n) \cdot 2(n+1) \cdot 2(n+2)$

  This equals:

$8(n)(n+1)(n+2)$

  Setting this up we get:

$8(n)(n+1)(n+2)$ |  48

$(n)(n+1)(n+2)$   |  6

  Among any three consecutive positive integers, there will be at least one 2 and one 3.  Consequently, $2\cdot3=6$ so our original equation is in fact divisible by 48.

[quote="Melissa"]If you look at my post you'll see that this is exactely what I did.[/quote]

  I suppose, but you didnt factor out in the beginning so it gave the impression it was taking a lot longer than it should have been...

yup melissa and G unit , u 2 are quite correct :D

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even numbers:
2k, 2k+2, 2k+4
$(2k)(2k+2)(2k+4)=(4k^2+4k)(2k+4)=(8k^3+16k^2+8k^2+8k)=8(k^3+3k^2+k)$
Now, we need to show that $k^3+3k^2+k$ is a multiple of 3.

$k^3+3k^2+k=k(k^2+3k+1)=k(k+1)(k+2)$

any 3 consecutive integers must contain a multiple of 3, thus $k(k+1)(k+2)$ is a multiple of 3. It is also a multiple of 2 because it must have at least one even term.

The original product, $(2k)(2k+2)(2k+4)$ is divisible by $8*2*3$ or  $48$.

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If you care to read the previous posts in this thread, you might realize that this particular solutions has been given quite a few times  ;)