geometry
geometric transformation
rotation
trigonometry
calculus

What level of math and/or physics do Nasa people need to know in order to work with planet rotation and orbit questions?

Is there a special planetary geometry course they take?

How much physics do they really need to know?

Planetary geometry?  You mean the kind of things covered in high school physics courses - Kepler's laws, and so forth?  You just take a physics class for those.

But that is besides the point.  You seem to have a very narrow view of what NASA does.  I suggest you take a look at

http://www.nasajobs.nasa.gov/jobs/civil_service_jobs/career/page10.htm

I will look into the links later tonight.

Why did you say narrow view?

I would think Nasa scientists are not "normal" people with simple intelligence, right?

I read somewhere that complicated geometry, trigonometry and calculus is needed to understand the orbit of all planets around the Sun.

[quote="Interval"]Why did you say narrow view?  [/quote]

Because you seem to assume that studying orbits is the only thing people do at NASA.  You're missing entire fields - for example, planetary geology.  Studying the geology of Mars, or Mercury.  It's an entirely different discipline.

[quote="Interval"]I read somewhere that complicated geometry, trigonometry and calculus is needed to understand the orbit of all planets around the Sun.[/quote]

Fully, yes.  That's because the planets exert gravitational pulls on each other.  Studying the orbits of just two planets around the Sun is known as the "three-body problem," which was studied in the 1700s but not solved until 1912.

And still only in terms of convergent infinite series, which i feel is cheating a little.  :lol:


[quote="alan"]an apple vendor sold large apples at 3 for a dollar and small apples at 5 for a dollar. sometime later, he had 30 apples of each size left, and went to lunch, telling his son to take over for him.  when he came back, all of the apples were sold and his son handed him 15 dollars.  the guy said to his son, "you should have received 10 dollars for the large apples and 6 dollars for the 30 small apples, making 16 dollars in all." The son replied, "There were an equal number of each sized apple left, I sold them all at the average price of 4 for a dollar. 4 goes into 60 fifteen times so I am sure 15 dollars is correct."  where did the $1 go?[/quote]

[hide="Well..."]I would say that the average price was 80/3 cents by apple, but if you sell four for a dollar, then you have 25 cents as the average, but 80/3>25!![/hide]

The "average" that the son speaks of is the harmonic mean, not the arithmetic mean.

The average of 40 and 60 is always 48, remember.


algebra
polynomial
combinatorial geometry
IMO Shortlist
Ring Theory
combinatorics proposed
combinatorics

We have $n$ points in the plane, no three on a line.
We call $k$ of them good if they form a convex polygon and there is no other point in the convex polygon.
Suppose that for a fixed $k$ the number of $k$ good points is $c_k$.
Show that the following sum is independent of the structure of points and only depends on $n$ :
\[ \sum_{i=3}^n (-1)^i c_i \]

We call a set of $k$ points [i]bad[/i] if this set is not good (guessable definition :)). Given $k$, the number of bad $k$-subsets of our initial set $S$ will be denoted by $b_k$. When the $n$ points are vertices of a convex polygon, all subsets are good, so in order to reach the conclusion we need to show that for all possible arrangements of the $n$ points, we have $\sum_{k=3}^n(-1)^kb_k=0$. We can prove this by finding a correspondence among bad sets, pairing each bad set to another bad set whose size has the opposite parity.

Suppose our $n$ vertices do not form a convex $n$-gon, and let $A_1,\ldots,A_t$ be the points lying inside the convex hull of $S$. Any bad set which does not contain $A_1$ but whose strict convex hull contains $A_1$ will be paired to the set we get from it by adjoining $A_1$. This way, we have found several pairs of bad sets, and several bad sets have been left unpaired. Among these, pair the ones which do not contain $A_2$ but whose strict convex hull contains $A_2$ to the set we get by adjoining $A_2$, and so on.

I hope it's Ok.

Grobber's pairing seem not to be correct.. There can be a concave quadrilateral which does not contain any other points in it's convex hull except for its vertices. Maybe I'm wrong.. Can you explain why your solution works? What is the one-line solution?

[quote="Nima Ahmadi Pour"]Yes, its correct.
But try to find a one line proof!
One of my friends solutions is just one line.
I will post it if no one finds it.[/quote]

Could you post your friend's solution :?:

grobber there must be a mistake in your solution
let the n points be vertices of a convex quadrilateral
then the answer is $ \frac{\minus{}(n\minus{}1)(n\minus{}2)}{2}$

grobber's solution seems rather correct to me. xirti didn't correctly understand the problem statement, and gauravpatil didn't correctly understand the solution.

In other news: The problem is a consequence of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=874985]IMO Shortlist 2006 problem C3[/url]. In fact, let $S$ be the set of our $n$ points. Using the notations of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=874985]IMO Shortlist 2006 problem C3[/url], this problem C3 says that every real $x$ satisfies

$\sum_P x^{a\left(P\right)} \left(1-x\right)^{b\left(P\right)} = 1$, where the sum is taken over all convex polygons $P$ with vertices in $S$.

Since two polynomials which take the same value on every real must be identic, this yields that

$\sum_P X^{a\left(P\right)} \left(1-X\right)^{b\left(P\right)} = 1$ in the polynomial ring $\mathbb Z\left[X\right]$, where the sum is taken over all convex polygons $P$ with vertices in $S$.

Now consider the polynomial $\sum_P X^{a\left(P\right)} \left(1-X\right)^{b\left(P\right)}$. This polynomial clearly has degree $\leq n$ (since $a\left(P\right)+b\left(P\right)\leq n$ for every $P$), and its coefficient before $X^n$ is

$\sum_{\substack{P;\\ a\left(P\right)+b\left(P\right)=n}} \left(-1\right)^{b\left(P\right)} = \sum_{\substack{P;\\ a\left(P\right)+b\left(P\right)=n}} \left(-1\right)^{n-a\left(P\right)} = \left(-1\right)^n  \sum_{\substack{P;\\ a\left(P\right)+b\left(P\right)=n}} \left(-1\right)^{a\left(P\right)}$.

Since the polygons $P$ satisfying $a\left(P\right)+b\left(P\right)=n$ are exactly the good polygons, this means that its coefficient before $X^n$ is $\left(-1\right)^n\sum_{P\text{ is a good polygon}} \left(-1\right)^{a\left(P\right)}$. But since this polynomial equals $1$, its coefficient before $X^n$ is $0$ if $n\geq 1$ and $1$ if $n=0$. Comparing these results, we see that

$\left(-1\right)^n\sum_{P\text{ is a good polygon}} \left(-1\right)^{a\left(P\right)} = 0$ if $n\geq 1$,
and
$\left(-1\right)^n\sum_{P\text{ is a good polygon}} \left(-1\right)^{a\left(P\right)} = 1$ if $n = 0$.

In other words,

$\sum_{P\text{ is a good polygon}} \left(-1\right)^{a\left(P\right)} = 0$ if $n \geq 1$,
and
$\sum_{P\text{ is a good polygon}} \left(-1\right)^{a\left(P\right)} = 1$ if $n = 0$.

Before you blindly apply this to our problem, notice that IMO Shortlist 2006 problem C3 allows polygons to have $0$ vertices (empty set), $1$ vertices (point) or $2$ vertices (line segment). All of these polygons with $\leq 2$ vertices are convex and good, so it is clear how to remove them from the equations above.

Alternatively :

We can show that $\sum_{P \subseteq  S, |P| \ge 3}^{{}}(-1)^{|P|} = \sum_{i=3}^{n}(-1)^{i}c_{i} = - \binom {n-1}{2}$, where $P$ is any subset of $S$.

Let $P_a$ be those subsets which form vertices of a convex polygon and there is no point inside it. (Those are the good polygons)

Let $P_b$ be those subsets which don't form a convex polygon

Let $P_c$ be those  subsets which form a convex polygon, but there is some points inside it, that is they are convex hull of some sets in $P_b$

Take $\mathcal{P} \in P_c$, and make a set $M$ with sets in $P_b$ which has convex hull as $\mathcal{P}$.

Now it is easy to see that $\sum_{P \in M \cup \mathcal{P}, |P| \ge 3}^{{}}(-1)^{|P|} = 0 $

So $ \sum_{P \subseteq  S, |P| \ge 3}^{{}}(-1)^{|P|} = \sum_{P \in P_a |P| \ge 3}^{{}}(-1)^{|P|} =  \sum_{i=3}^{n}(-1)^{i}c_{i}$

And it is immediate that  $\sum_{P \subseteq  S, |P| \ge 3}^{{}}(-1)^{|P|} = - \binom {n-1}{2}$

And this confirms [b]darij[/b]'s computation, since that offers $0 = 1 - \binom {n} {1} + \binom {n} {2} + \sum_{i=3}^{n}(-1)^{i}c_{i}$, whence $\sum_{i=3}^{n}(-1)^{i}c_{i} = -1 + \binom {n} {1} - \binom {n} {2} = -\binom {n-1} {2}$.

A direct counting approach involving (two!) inductions works.

We will consider the empty set, individual points, and segments as good polygons. For a set of points $A$, write $f_A$ to denote the sum $\sum_{X \subseteq A} (-1)^{|X|}$ across all good polygons $X$ whose vertices are in $A$. Also, for points $P_1, \dots, P_m \in A$, we write $f_A(P_1, \dots, P_m)$ to restrict $X$ to those polygons containing $P_1, \dots, P_m$.

So, if $S$ is our given set of $n$ points, we are interested in $f_S$. We claim that $f_S=0$, which will clearly solve the problem.

The proof is by induction, with small cases (e.g. $n=1, 2, 3$) trivial. Consider the case for a general $n$ and pick a point $O$ on its convex hull. Pivoting at $O$, sweep clockwise through the remaining points and label them $A_1, \dots, A_{n-1}$ in sequence, where $\angle A_1OA_{n-1} < 180^\circ$. No polygon whose vertices belong to $A=\{A_1, \dots, A_{n-1}\}$ can contain $O$, so by hypothesis, $f_A=0$. All remaining good polygons have $O$ as a vertex, so we wish to show that $f_S(O)=0$.

Let $S_i$ denote the set of points $\{O, A_1, \dots, A_i\}$. We claim that
\[
  f_{S_i}(O, A_i)=
  \begin{cases}
        1 & i=1, \\
        0 & 2\le i \le n-1.
  \end{cases}
\]
This is clear when $i=1$, so suppose the claim holds for $i=1, 2, \dots, j-1$. We'll show that it holds for $i=j$, i.e. $f_{S_j}(O, A_j) = 0$.

Take the ray pointing in the direction of $\overrightarrow{OA_j}$ and pivot it counterclockwise about $A_j$; we'll sweep through the points $A_1, \dots, A_{j-1}$ in some order and hit $A_L$ last. Each time we hit $A_k$, we'll add to our count all the good polygons with $O, A_j, A_k$ as consecutive vertices. There are a few cases:
[list]
[*]Suppose $\triangle OA_jA_k$ contains a point in its interior. Then there are no such good polygons.
[*]Suppose $\triangle OA_jA_k$ has no interior points and $k \neq L$. Let $S_k'$ be the set of $A_i$'s with $i \le k$ that lie between or on rays $\overrightarrow{A_jO}$ and $\overrightarrow{A_jA_k}$, excluding $A_j$. Note that every good polygon with $O, A_j, A_k$ as consecutive vertices can be built by appending $A_j$ to a unique good polygon counted by $f_{S_k'}(O, A_k)$. The key point is that $|S_k'| \ge 3$ since it contains $O, A_k, A_L$! So, by assumption that our claim holds for smaller values of $i$ (smaller configurations, really), we must have $f_{S_k'}(O, A_k) = 0$. In particular, when we append $A_j$ to those polygons, we just flip the parity of the number of sides of each one, meaning we still add $0$ to our count in this case.
[*]If $k=L$, then all we have is the triangle $OA_kA_L$, so we add $-1$ to our count.
[/list]
Note that the above three cases cover all good polygons counted by $f_{S_j}(O, A_j)$ except for the segment $\overline{OA_j}$ itself. Thus, we compute $f_{S_j}(O, A_j)=-1 + 1=0$, and the claim is proved.

Finally, we may compute $f_S(O)$ by summing over all $f_{S_i}(O, A_i)$, and then subtracting $1$ for the good polygon consisting of just $O$. By the claim, this is just $0$, which concludes our induction.


1)Let a,b,c>0.abc=1.Prove that:
$ \frac{1}{{a^4  \plus{} b^4  \plus{} 2abc^2 }} \plus{} \frac{1}{{a^4  \plus{} c^4  \plus{} 2acb^2 }} \plus{} \frac{1}{{c^4  \plus{} b^4  \plus{} 2cba^2 }} \le \frac{9}{{4(a \plus{} b \plus{} c)}}$
2)Let a,b,c>0.abc=1.Find the best constant for k such that:
$ \frac{1}{{a^4  \plus{} b^4  \plus{} kabc^2 }} \plus{} \frac{1}{{a^4  \plus{} c^4  \plus{} kacb^2 }} \plus{} \frac{1}{{c^4  \plus{} b^4  \plus{} kcba^2 }} \le \frac{9}{{(2 \plus{} k)(a \plus{} b \plus{} c)}}$ :)


Let $p,q\geq 2$ be coprime integers. A list of integers $(r,a_{1},a_{2},...,a_{n})$ with $|a_{i}|\geq 2$ for all $i$ is said to be an expansion of $p/q$ if
$\frac{p}{q}=r+\frac{1}{a_{1}+\frac{1}{a_{2}+\frac{1}{...+\frac{1}{a_{n}}}}}$.
Now define the weight of an expansion $(r,a_{1},a_{2},...,a_{n})$ to be the product
$(|a_{1}|-1)(|a_{2}|-1)...(|a_{n}|-1)$.
Show that the sum of the weights of all expansions of $p/q$ is $q$.

Can someone find solutions for it?
I haven't got good ideas.What is the topic of the problem(....... fractions).Can someone suggest articles on it

Article is basically called continued fractions. I'll perhaps post solution by tomorrow(right now I am working on it).

I have a question: can $r$ be negative, because if that is the case, isn't there an infinite number of expansions for $\frac{p}{q}$?

@above no, since $\frac{1}{a_1 + \frac{1}{a_2 + \cdots}}$ is bounded above due to $|a_i| \ge 2.$ In fact, it's pretty easy to see that $r = \left \lfloor \frac{p}{q} \right \rfloor$ or $\left \lfloor \frac{p}{q} \right \rfloor + 1.$

Define the [b] height [/b] of an expansion $(r, a_1, a_2, \cdots, a_n)$ to be $(r-1)$ times the weight.

[b] Lemma. [/b] The sum of the heights of all expansions of $\frac{p}{q}$ is $p-q.$

[b] Proof. [/b] The proof goes by induction on $p + q.$ If $p+q = 5$, it's easy to check this directly. Suppose it holds when $p+q < k.$ If $p+q = k,$ then set $t = \left \lfloor \frac{p}{q} \right \rfloor.$ We have in any expansion $(r, a_1, a_2, \cdots, a_n)$ that $r = t$ or $r = t+1.$ Note that:

$$\frac{p}{q} = t + \frac{1}{\frac{q}{p \% q}},$$

and

$$\frac{p}{q} = (t+1) + \frac{1}{- \frac{q}{q - p \% q}},$$

where $a \% b$ denotes the remainder when $a$ is divided by $b.$

The sum of the heights when $r = t$ is $(t-1)$ multiplied by the sum of the heights of the expansions of $\frac{q}{p \% q},$ where we need not worry about $a_1 = 1$ because those have height zero anyways. By induction, this is precisely $(t-1)(q - p \% q).$

The sum of the heights when $r = t+1$ is $t$ multiplied by the sum of the heights of the expansions of $\frac{q}{q - p \% q}$ (just negate all the $a_i$'s). This is equal to $t (p \% q)$ by induction.

Summing these together, we get that the sum of the heights of expansions of $\frac{p}{q}$ is $(t-1) (q - p \% q) + t (p \% q) = tq + p \& q - q = p - q,$ completing the induction.

$\blacksquare$

We will now use the lemma to finish the problem. To compute the sum of the weights of all expansions of $\frac{p}{q}$, we will use casework on whether $r = \left \lfloor \frac{p}{q} \right \rfloor$ or $\left \lfloor \frac{p}{q} \right \rfloor + 1.$ If $r = \left \lfloor \frac{p}{q} \right \rfloor$ then the sum of the weights there is simply the sum of the heights of all expansions of $\frac{q}{p \% q}$ because

$$\frac{p}{q} = \left \lfloor \frac{p}{q} \right \rfloor + \frac{1}{\frac{q}{p \% q}}.$$

This is $q - p \% q$ by the lemma.

If $r = \left \lfloor \frac{p}{q} \right \rfloor + 1$, then the sum of the weights there is simply the sum of the heights of all expansions of $\frac{q}{q - p \% q}$ because

$$\frac{p}{q} = \left \lfloor \frac{p}{q} \right \rfloor + 1 + \frac{1}{-\frac{q}{q - p \% q}},$$

and we can negate all the $a_i$'s in an expansion of $- \frac{q}{q- p \% q}$ to get one of $-\frac{q}{q - p \% q}.$

This is $p \% q$ by the lemma.

Summing the two cases above, it's clear that the sum of the weights of all expansions of $\frac{p}{q}$ is $q,$ as desired.

$\square$

[hide=Directed to Pathological]What was your motivation to define the height of an expansion, and give the following ?

Lemma: The sum of the heights of all expansions of $\frac{p}{q}$ is $p-q.$[/hide]


A triangle has three different integer side lengths and a perimeter of 20 units. What is the maximum length of any one side?

By triangle inequality, the maximum length is $ \boxed{9}$.

Could someone explain? I do not entirely get the triangle inequality.

The triangle inequality states that the sum of any 2 sides in a triangle must be larger than the third side.

So let us take 10 units as one of the side lengths, that must mean the other side lengths must equal to 20-10 = 10 units. This triangle doesn't satisfy the triangle inequality hence it is not a real triangle.

Therefore, the largest side that does satisfy this condition is a side with 9 units. The rest of the sides must equal 20 - 9 = 11. 11>9 so this triangle is a real triangle and satisfies all conditions.


Let the sequence (a(n)) be define d by a(1)=t and (a(n+1))=4(a(n))(1-a(n)) for n>=1. How many possible values of t are there if a(1998)=0 ?
my solution [hide]I have got a solution
say that there are b(n) possible values of t if a(n)=0.
1)if a(n-1)=0 there are b(n-1) possible values of t
2)if a(n-1)=1 then
a(n-2)=1/2
lemma 1
a(j) (j: 1->n-3) has got two values for fixed a(j+1)
proof: the question gives that |1-2a(n-1)|=sqrt(1-a(n))
if a(k) is not equal to 1 for k<n-1 the statement is true
say that for such j, a(j)=1
a(j+1)=0 and a(n-1) must equal to 0 but it is not it gives contradiction
lemma 2
however we take one of the two values of a(j)'s. we have different values of t
proof: this is true because when we have a fixed a(1) we have a fixed sequence a(n)
from lemma 1 and 2 we have 2^n-3 different values of t
finally we have b(n)=b(n-1)+2^n-3 different values of t
from a basic induction we have b(n)=2^n-2+1
for 1998 this is 2^1996+1[/hide]

I dont know it is a question from turkey tst's but in imo.org.yu this is completely the same so I think this is in the complex plane
please check my solution i think this is right but  i cant trust

Yes, my question is incorrect. If $f(x)=4x(1-x)\in [0,1]$, then $x\in [0,1]$.
Therefore we must calculate solutions $f(x)=y$, $f(x)=x$ (give not new solution - $x=0$ or $x=\frac{3}{4}$). If $t=0$ or $t=\frac{3}{4}$ then for $a_{n}=t$ for any n.
1. $f(t)=y$ has exactly 2 solution if $y\not = 1$.
2. Let $f^{(n_{1})}(t_{1})=0=f^{(n_{2})}(t_{2})$, were $n_{1},n_{2}$ minimal natural numbers satisfyed this condition. It give $t_{1},t_{2}$ are distinct, when $n_{1},n_{2}$ are distinct.
3. Let $b(n)$ is number of t, with condition $f^{(n)}(t)=0\not =f^{(n-1)}(t).$ It give $b(0)=1$ (t=0), $b(1)=1$ (t=1), $b(2)=1, (t=\frac{1}{2})$( $b(n)=2^{n-2},n\ge 2$.
Therefore it give $1+1+1+2+...+2^{1996}=2^{1997}+1$ solutions.

1)the sequence doesn't contain a(0) so there is nothing like b(0)
2)b(1)=1 b(2)=1 anda also b(3) must be 1
because if b(3)=0 b(2)=1 b(1)=1/2
so the answer is 1+1+1+2^1+ ......  +2^1995=2^1996+1
rust what is this questions level?

I don't say, that sequence contains a(0). I calculate number of solutions $f^{(n)}(t)=0$, it contains $1+1+1+2+...+2^{n-3}=1+2^{n-2}, n\ge 1$ old solutions (solutions of equations $f^{(n-1)}(t)=0$) and $2^{n-2}=b(n)$ (n>1) new solutions. It give $N(n)=1+2^{n-1},n>1$ solutions of equation $f^{(n)}(t)=$. For your case $N(1998)=1+2^{1997}$.

if b(n-1)=1 b(n-2)=1/2 from n-3 to 1 there are 2 values of b(n) so there are 2^n-3 values of t but you say 2^n-2 possible values

i can't understand why N(2)=3
a(2)=0 =>
1)a(1)=1
2)a(1)=0 N(2)=2
you didn't say anything for the level of the problem

$N(n)$ is number of solutions $f^{(n)}(t)=0$. We have $N(n)=N(n-1)+b(n)$, were b(n) -is number of solutions t, suth that $f^{(n-1)}(t)\not =0$. We have b(2)=1 (solution $t=\frac{1}{2}$), and $b(n)=2^{n-2}$. It give (for example by induction) $N(n)=1+2^{n-1}$.

$b(2)$ is number of solutions $f(f(t))=0,f(t)\not =0$. It is only one solution (t=1/2). Therefore b(2)=1. It give $b(n)=2^{n-2},N(n)=1+2^{n-1}$.

according to your formula
if a(3)=0 the number of the possible values of t must be 5
but if a(2)=0 a(1) have 2 values
if a(2)=1 a(1) has got one value
so possible values of t is 3 instead of 5
the formula must be 2^(n-2)+1

Formula for number of solutions $f^{(n)}(t)=0$ given $N(n)=1+2^{n-1}$ is true.
But I forget, that you needed number of solutions $a(1998)=f^{(1997)}(t)=0 \ (a_{1}=t)$, therefore needed $N(1997)=1+2^{1996}$. I am sorry.

ohh :lol: but of course this is not important. you are so good at algebra
when i ask a problem your sol is ready for max 1 hour


I've just upgraded the Wiki. Because of the huge amount of difference between this and the previous version it is possible some of the edited scripts and hooks that tie the rest of the site with the Wiki (this includes among others the $ \text{\LaTeX}$ and the common login features, and the AoPS/MathLinks skins) might not work perfectly.

Let me know what displaying bugs you find and I'll see what I can do about them.

I tried to access it on IE but couldn't. :(

On Firefox 1.0.7 with Windows XP, the default font size is now very small -- please increase it for readability.

The small font size is there in multiple browsers (Safari and Firefox 3 on a Mac).

I can't access it at all when logged in on the forum- I just get a white screen.

Hmm it doesn't even load for me!

I was on the mathlinks skin

[url=http://img244.imageshack.us/my.php?image=aopswikihs6.png][img]http://img244.imageshack.us/img244/104/aopswikihs6.th.png[/img][/url]

FF 3 beta 5 on Ubuntu Hardy (8.04)

Try it while not logged in- I could at least see the wiki that way.

Unfortunately, there's now an empty toolbox that' says "search" as well as the normal search toolbox, and the "toolbox" and "languages" toolboxes are empty.

EDIT: Additionally, sysop and bureaucrat tools are not working.

EDIT 2: Firefox 3.0.0.1, Windows XP, Mathlinks skin.

Ok, MathLinks skin has not been optimized (I didn't have time to fix that one yet). Let me know what is wrong with the AoPS skin. I found everything to be working fine there but if there is something that's not working let me know. I will let you all know when the MathLinks skin is going to work fine too.

When I'm logged in, I can't see the wiki - just a blank white page. I can get to the wiki while not logged in (annoyingly small font). I can then log in and still be in the wiki (and remain logged in when coming back to the forum). Hence it appears to have something to do with login.

AoPS skin
Mac OS 10.5.4
Firefox 2.0.0.16 for Mac

Edit: having logged in while in the wiki, I can now go back and forth freely in both directions.

The navigation panel seems to shove all the article text down below it for me.  I use IE7 on XP.

I still can't access the Wiki...

What's going on? :(

Sysop/b'crat tools seem to work on neither AoPS nor ML skin, though I'm still listed as both, checking Special:Userrights.

on the MonoBook skin(dunno if this is important, but hey), I go to the recent changes page, I click on the Show Minor Edits button and it doesn't show the minor edits.

[quote="#H34N1"]Archimedes, how'd youg et access?

edit: it works when you login from http://www.artofproblemsolving.com/Forum/login.php?redirect=wiki.php[/quote]

Yeah, now it works if you login from there.

did you logout and then click the link?

Do that, and the aops wiki should work.

it worked but the pages were not positioned right :(

It should work now normally. Log out and re-log back in or clear your cookies (either one is fine) to be on the safe side. I think I've also gotten rid of the bug with the session that made you have to click twice in order to save an edit :)

The internet explorer view though is still pretty bad on the AoPS skin. Any CSS suggestions to fix it? (I'm trying my own stuff but it's not working great so far).

[quote]The internet explorer view though is still pretty bad on the AoPS skin.[/quote]
I assume that's just an IE problem? It looks just fine to me in Firefox (what I usually use) or Safari (I tried that just to see.) The earlier font size problems are gone now, and the access/login problems seem to be gone (I just did a little editing.)

As far as I know neither Firefox nor Safari = IE :)

YAY! It's finally back to normal for me!

Hmm... I just went there and everything is in thick big letters now (compared to Forum posts), so boldface hardly stands out at all. Is it "normal" or I should play with some preferences (Firefox, Windows XP)?

Well you guys wanted the larger font didn't you? :) It's at 140% right now. I can make it at little bit smaller if you want.

Smaller would be nicer.
Well, I checked on Opera also, and it's a bit smaller there, but could still use some shrinking.
(I use Firefox 3.)
I don't know about IE or Safari though.

[quote="fedja"]Is it possible to make the font size an user option?[/quote]Possible, not that easy. I'll make it smaller tomorrow, and we'll see how that goes for now.

Yay, tools are working now.

ML skin looks pretty good to me overall, actually, other than the strange empty boxes.

[quote="Temperal"]ML skin looks pretty good to me overall, actually, other than the strange empty boxes.[/quote]Yeah didn't have time to modify it yet to make it compatible with 1.13 version.


geometry
perimeter
AMC
AIME
ratio
trigonometry
AIME I

14.  Triangle APM has ∠A = 90o and perimeter 152. A circle center O (on AP) has radius 19 and touches AM at A and PM at T. Find OP.

[img]http://www.kalva.demon.co.uk/aime/soln/diag0214.jpg[/img]

My primary reason for posting this problem is that is there an easy way to do this?

[hide]I saw that triangle $PTO$ and $PMA$ were similar, and the sidelengths of triangle

 $PTO$ are $x+19, 19, \sqrt{x^2+38x}$, and the sidelengths of $PMA$ are

 $\sqrt{x^2+38x}+y, y, 38+x$, then by setting up a proportion:

$\frac{\sqrt{x^2+38x}}{19}=\frac{x+38}{MA}\Longrightarrow MA=\frac{19(x+38)}{\sqrt{x^2+38x}}$. From there, I had:

$\frac{38(x+38)}{\sqrt{x^2+38x}}+38+x+\sqrt{x^2+38x}=152$, by letting

$a=\sqrt{x^2+38x}$, I eventually simplified to $(x+38)^2=a(114-x)$, thus:

$\frac{(x+38)^2}{114-x}=\sqrt{x^2+38x} \Longrightarrow \frac{(x+38)^4}{(114-x)^2}=x(x+38) \Longrightarrow (x+38)^3=x(114-x)^2 \Longrightarrow x^3+114x^2+4332x+54872=12996x-228x^2+x^3 \Longrightarrow 342x^2-8664x+54872=0 \Longrightarrow x=\frac{38}{3} \Longrightarrow \boxed{OP=\frac{95}{3}}$

 Now my answer is correct but I really needed a calculator
 to solve this and considering we aren't allowed to use calculators on the AIME, I was wondering if anyone had an easier solution or if anyone good at algabraic manipulation could tell me how to simplify this without a calculator. [/hide]

[quote="boxedexe"]Well, you can do all of those calculating operations by hand :P[/quote]

No, it will take ages, and usually AIME problems won't be that calculator dependent ;) so I'm thinking there must be an easier method to solve tihs

using the angle biscector theorem (draw OM, and AMO=TMO), and the perimeter thing, you can get it down easily to
$(x-133)^2(x-19)=(x+19)^3$ where $x=OP$, and then do a substitution of $n=19+x$, then you can get it down to
$9n^2-16*19*3n+19^2*8^2=0$
$(3n-19*8)^2=0$
$n=\frac{19*8}{3}$
$x=\frac{19*5}{3}=\frac{95}{3}$

This can all be done using similar triangles.

Extend O to T, and notice AM = TM.

The ratio of the sides are equivalent to the ratio of the triangles' perimeters.

First thing I  will notice, is 152 is divisible by 19, so (without loosing anything) I would simplyfy and take AO=1, perimeter=8 (and later multiply my answer by 19).
If I further assume that AIM problem setters are nice (kind or lazy) and like 3-4-5 type rightangles, I would further assume (ratio wise)   OT=3, PT=4, PO=5.. (Making OA=OT=3)  so AP=8,AM=TM=6 sp PT=4+6=10.. right! Check! double check!! (perimeter  here is indeed 8*3=24) ..everything fits. so OA=5..
Adjusting for the ratio -- the answer has to be $5*(19/3)$.. And 30 seconds leter I am on the next problem ...

Of course, I would be lucky to just find the answer looking straight at me .. , but in any case, if I took the ratio (19) out and  tried to solve equation with smaller numbers I would have an easy time. ..:)

taking out the 19 is definately a good step (i.e. dialating the figure by a ratio of 1/19), but noticing the 3-4-5 triangle is not something that you would really do

Ataully, if I recall correctly, it was one of AIME Q in my son's.(Yes, I know, many years ago) and it was noticed right away :)
In any case,  if you notice the ratio  (perimeter and AO) is 8, it is very easy to deduce it is a 3-4-5 triangle.
For example if you take the angle $\angle ANP = \theta$ you get right-away:
$ratio = 8 = (1+\cos \theta + \sin \theta) : (1-\cos \theta)\cdot \cot \theta$
Now easiest to solve this, is the oldtrick, let $\tan (\theta/2) = t$ (so that  sin, cos, tan all are nice functon of t

[hide]If you don't know this, remember $\sin \theta = 2t/(1-t^2)$ ; $\cos \theta= (1-t^2)/(1+t^2)$ and   $\tan \theta = 2t/(1-t^2)$
so you get something like $8 = \frac{(1+t^2 + 2t + 1-t^2) } {1+t^2} \cdot \frac{(1+t^2)(2t)}{(1+t^2-1+t^2)(1-t^2)}$
which simplifies to $8t(1-t)=2$ gives $t=1/2$ and  $tan \theta = 4/3$ [/hide]
  you get $t=(1/2)$  and 3-4-5 triangle etc ..

Hope it is helpful ...

Thank you, I liked Altheman's method a lot, but Gyan, thank you for your advice since it will REALLY help on other AIME problems :)


[hide="My solution"]

(I guess you're looking for the remainder in division of first two polynomials).

The remainder has the degree of at most $ 1$ ($ deg(r(x))<deg(h(x))$).

So, we can write $ r(x)\equal{}Ax\plus{}B$,and determine the coefficients using, for instance, Bezout's theorem:

$ p(x)\equal{}f(x)\minus{}r(x)$ is divisible by $ x^2\minus{}1$, hence, $ p(1)\equal{}0$ and $ p(\minus{}1)\equal{}0$.

Form there we get $ r(1)\equal{}f(1)$ and $ r(\minus{}1)\equal{}f(\minus{}1)$, $ A\plus{}B\equal{}7$, $ \minus{}A\plus{}B\equal{}\minus{}5$, hence $ A\equal{}6$, $ B\equal{}1$.

$ r(x)\equal{}6x\plus{}1$.[/hide]

Here $ x^2 \equiv 1 \bmod h(x)$, so we can reduce the exponents of every term $ \bmod 2$ and conclude immediately.


Find all real pairs $ (x;y)$ in this system:

$ xy(x+y)=30$

$ x^{3}+y^{3}=35$

Expanding xy(x+y) and multiplying by 3 we get:

  3(x^2y)  +  3(xy^2) = 90

Noticing that these are values contained in (x+y)^3 along with x^3 + y^3 we get:

(x + y)^3 = 125

which gives us (x+y) = 5

Setting x = (y-5) and substituting into the first equation we get the quadratic:

y^2 -5y  + 6 =0   which yields as solutions among the reals:

(3,2) , (2,3)

I think there is a mistake in your answer: (0, 5) is not a solution, for example.

Since $ xy(x+y) = 30$, we find $ xy = 6$ (because x+y=5).
This gives us $ x(5-x) = 6$, so $ x^{2}-5x+6=0$ which leads to $ x=2 \vee x=3$.

So the solutions would be $ (2;3)$ and $ (3;2)$.

EDIT: now it's ok


inequalities
trigonometry
geometry
perpendicular bisector
IMO
IMO 1961

Construct a triangle $ABC$ if $AC=b$, $AB=c$ and $\angle AMB=w$, where $M$ is the midpoint of the segment $BC$ and $w<90$. Prove that a solution exists if and only if \[ b \tan{\dfrac{w}{2}} \leq c <b  \] In what case does the equality hold?

Suppose that we have constructed the triangle $ABC$

First of all, it's obvious that $b>c \ \ \ \ (1)$.

Indeed, the triangles $AMC,AMB$ have $AM=AM, MC=MB$ and $\widehat{AMC}>90^0>\widehat{AMB}$

Now, we bring a parallel from $C$ to $AM$, which intersects the line $AB$ at the point $B'$.

$MC=MB \Rightarrow AB=AB'$

The triangle $BB'C$ have median $CA=b$, a side $BB'=2c$ and $\widehat{BCB'}=w$

[u][b][color=blue][size=150]Construction[/size][/color][/b][/u]

We construct the segment $BB'=2c$, and let $A$ its midpoint.

Then, we construct the arc $h$ of the circle, which pass through $B,B'$ and $\forall P \in h,\ \widehat{BPB'}=w$.
(There is also an arc $h'$ on the same circle of $h$, such that $h \cap h' = \{A,B\}$ and $h \cup h'$ is a circle. But $\forall P \in h', \ \widehat{BPB'}=180^0-w$.
Also, since $w<90^0$, $h$ is greater than $h'$. That is, the [u]ray[/u] $AO$ intersects the arc $h$ (at its midpoint), where $O$ is the center of the arc)

For every $P \in h$, the triangle $BPB'$ has $\widehat{BPB'}=w$ and $BB'=2c$. Now we have to find a point $P$, such that $AP=b$.

Let $K$ be the intersection point of $h$ with the perpendicular bisector of $BB'$ ($\equiv$ the line $AO$).
Of course, the max length of $AP$ is when $P \equiv K$ (from the triangle $AOP$ we have $AP\leq OA+OP=OA+R$, so we have a max value for $AP$ when $A,O,P$ are collinear).

There exists a solution, if the given $b$ is not greater than $AK$.
If $b>AK$ then there are not solutions.

Let's find the length of $AK$.

$\triangle AKB : \widehat{AKB} = \frac{w}{2}$

$\tan{\frac{w}{2}}=\frac{AB}{AK} =\frac{c}{AK} \Rightarrow$

$AK \cdot \tan{\frac{w}{2}}=c$

$b\leq AK \Rightarrow$

$b \cdot \tan{\frac{w}{2}} \leq AK \cdot \tan{\frac{w}{2}} \Rightarrow$

$b \cdot \tan{\frac{w}{2}} \leq c \ \ \ \ \ (2)$

If both conditions $(1),(2)$ hold, then we construct a circle $C1=(A,b)$ and let $C$ an intersection point of $C1$ and $h$ (the other is symmetric to $C$ w.r.t. $AK$, so it makes a symmetric triangle)

The triangle $BCB'$ has side $BB'=2c, \ \ \widehat{BCB'}=w$ and median $AC$.

Let $M,M'$ the midpoints of $BC,BC'$ respectively. We have

$\widehat{AMB}=\widehat{AM'B}=\widehat{BCB'}=w \ \ (CMAM'$ is a parellelogram $)$

So there are [b]two[/b] triangles with the specified properties:

The triangle $ABC$ has $AB=c, AC=b$, and $\widehat{AMB}=w$

The triangle $AB'C$ has $AB'=c, AC=b$, and $\widehat{AMB'}=w$

If it is $b \tan{\frac{w}{2}}=c$, then $b=AK$, so we have only one solution
(the triangles $ABC,AB'C$ are equal) and $\triangle ABC$ is a right triangle $(A=90^0)$


Really, don't.

HOW COME MYSMARTMOUTH IS ALLOWED TO HAVE ALL CAPS TITLES?

BECAUSE HE IS A MODERATOR

OMG THE ARNAV TREND IS CATCHING ON

:spam:
DPatrick made a topic about spam this summer.  There was even a lockdown.  Has everyone forgotten about it?

Man is so forgetful at times.

And pulak is too lazy to lock this, so this shameless spam must go on.  :mad:


inequalities proposed
inequalities
algebra

[quote="N.T.TUAN"]Suppose that positive numbers $x$ and $y$ satisfy $x^{3}+y^{4}\leq x^{2}+y^{3}$. Prove that $x^{3}+y^{3}\leq 2.$[/quote]
It's stronger:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111000
and still unsolved. :D

Suppose that positive numbers $x$ and $y$ satisfy $x^2+y^3\leq x+y^2.$ Prove that $$x^3+y^3\leq 2$$
Suppose that positive numbers $x$ and $y$ satisfy $x^2+y^4\leq x+y^3.$ Prove that $$x^4+y^4\leq 2$$
Suppose that positive numbers $x$ and $y$ satisfy $x^3+y^5\leq x+y^3.$ Prove that $$x^5+y^5\leq 2$$
Suppose that positive numbers $x$ and $y$ satisfy $x^3+y^5\leq x^2+y^4.$ Prove that $$x^6+y^6\leq 2$$

[quote name="Quantum_fluctuations" url="/community/p16775767"]
[quote=sqing][quote=N.T.TUAN]Suppose that positive numbers $x$ and $y$ satisfy $x^{3}+y^{4}\leq x^{2}+y^{3}$. Prove that $x^{3}+y^{3}\leq 2.$[/quote]
$x^{2}+2y^{3}-y^{4}\leq \frac{2x^{3}+1}{3}+2y^{3}-\frac{4y^{4}-1}{3}$
[/quote]

How is that true?
[/quote]
[quote name="Brudder" url="/community/p23463372"]
Sorry could you please explain $x^{2}+2y^{3}-y^{4}\leq \frac{2x^{3}+1}{3}+2y^{3}-\frac{4y^{3}-1}{3}?$ I just can't seem to grasp my head around that : (
[/quote]
$x^{2}+2y^{3}-y^{4}\leqslant \frac{2x^{3}+1}{3}+2y^{3}-\frac{4y^{3}-1}{3} \Leftrightarrow 0 \leqslant (x-1)^2(2x+1) +(y-1)^2(3y^2+2y+1).$

[quote=Brudder]Sorry could you please explain $x^{2}+2y^{3}-y^{4}\leq \frac{2x^{3}+1}{3}+2y^{3}-\frac{4y^{3}-1}{3}?$ I just can't seem to grasp my head around that : ([/quote]
By AM-GM.$$x^{3}+y^{3}\leq\frac{2x^{3}+1}{3}=\frac{x^{3}+x^{3}+1}{3}$$
$$ 4y^{3}\leq y^{4}+ y^{4}+ y^{4}+1$$
[quote=sqing]Suppose that positive numbers $x$ and $y$ satisfy $x^2+y^3\leq x+y^2.$ Prove that $$x^3+y^3\leq 2$$[/quote]

Let $a,b,c$ are positive real numsers such that $x^2+y^3+z^4 \geqslant x^3+y^4+z^5.$
(1) Prove that $x^3+y^3+z^3 \leqslant 3.$
(2) Prove that $x^2+y^2+z^2 \leqslant 3.$
[url]https://artofproblemsolving.com/community/c6h3030623p27265297[/url]


The question is:

Let A, B, and C represent three terms of an arithmetic sequence.  If we were to decrease A by 1 OR increase C by 2 the result would be a geometric sequence of numbers.  What is B in this case?

Thank you for taking the time to read this.

The definition of an arithmetic sequence is $c-b=b-a$.

The definition of a geometric sequence is $\frac cb=\frac{b}{a-1}$, or similarly, $\frac{c+2}{b}=\frac{b}{a}$.

Solving these three equations will lead you to your answer.

Thanks so much for the hint, worked like a charm.


In Mathematica, I have successfully programmed a plot of a quadratic function where the coefficients can be manipulated; but I now wish to display the changing vertex quantity while the coefficients are manipulated. Here's what I currently have:
f[a_, b_, c_] := a + b*y + c*y^2
quad[a_, b_, c_] :=
 Plot[f[a, b, c], {y, -10, 10},
  PlotLabel -> Solve[D[f[a, b, c]] == 0, y]]
Manipulate[quad[a, b, c], {a, -10, 10}, {b, -10, 10}, {c, -10, 10}]

Any ideas?
Thanks in advance.

Sorry for the stupidity, I have now provided adequate info. Thanks for pointing that out.


[quote="kirwa01"]suppose that $ \{a_1 , a_2, .. a_n \}$ and $ \{ b_1 , b_2, .. b_n \}$ are two permutations of $ \{1, \frac {1}{2}, .., \frac {1}{n} \}$
such that : $ a_1 \plus{} b_1 \ge a_2 \plus{} b_2 \ge ... \ge a_n \plus{} b_n$

show that for every $ m \le n$  $ a_m \plus{} b_m \le \frac {4}{m}$[/quote]

No one wante to solve this problem ?

[quote="Dumel"]where is n?[/quote]

$ n$ is an integer greater than $ 2$.

[url]http://ilib.mirror1.mccme.ru/djvu/olimp/vsesojuznye.htm[/url]
Problem № 109. Page 145


linear algebra
matrix
probability
probability and stats

A real $ m \times n$ matrix $ A \equal{} (a_{i,j})$ is said to have a saddlepoint at $ (k,l)$ if $ a_{kj}\le a_{kl} \le a_{il}$ for all $ 1 \le i \le m, \, 1 \le j \le m$. Put differently, the element $ a_{kl}$ is maximal in its row and minimal in its column.

Given an $ m \times n$ matrix $ A$ with i.i.d. entries from $ N(0,1)$, find the probability that it has a saddlepoint.

Since only the order of the matrix elements matters and for i.i.d. random variables all $ (mn)!$ orders are equiprobable, the normal law is completely irrelevant to the question. All we need is some continuous distribution, so that the probability that some two elements are equal is $ 0$. The idea is to notice that you may have at most one saddle point in the entire matrix and that the probability that a given point is a saddle point is $ \frac{(m\minus{}1)!(n\minus{}1)!}{(m\plus{}n\minus{}1)!}$. So the answer is just the product of this number and $ mn$.


calculus
integration
real analysis
real analysis unsolved

$ \int_{ \minus{} \infty} ^{\infty} {\frac {b} {a \plus{} x}\cdot e^{ ( \minus{} \frac {x^{2}} {c})} dx} \equal{}$

$ (a, b, c \equal{} constant)$

May I ask you a favor ?
Thank you  :blush:

p.s. Is this easy ? I feel like so hard . Can you give me some formulas (about integrals) like this problem or links ? :(

This diverges for real $ a.$ Note the singularity in the integral around $ x\equal{}\minus{}a.$

Oh , I have missed the point . Thank you !
 :wallbash_red:

Regarded as a singular integral, we can evaluate its value. For the convenience of calculation, we may assume $ b, c = 1$ and let $ I = I(a)$ be the given integral. Then

\begin{eqnarray*} I(a) & = & PV \int_{ - \infty}^{\infty} \frac {e^{ - x^2}}{x + a} \, dx \\
& = & PV \int_{ - \infty}^{\infty} \frac {e^{ - (x - a)^2}}{x} \, dx \\
& = & \int_{0}^{\infty} \frac {e^{ - (x - a)^2} - e^{ - (x + a)^2}}{x} \, dx \\
& = & e^{ - a^2} \int_{0}^{\infty} \frac {2 \sinh (2ax)}{x} e^{ - x^2} \, dx. \end{eqnarray*}

By Libneiz's integral rule, we have

\begin{eqnarray*}\frac {d}{da} (e^{a^2} I(a)) & = & \int_{0}^{\infty} 4 \cosh(2ax) e^{ - x^2} \, dx \\
& = & \int_{0}^{\infty} 2 e^{ - x^2 + 2ax} \, dx + \int_{0}^{\infty} 2 e^{ - x^2 - 2ax} \, dx \\
& = & 2 \int_{ - \infty}^{\infty} e^{ - x^2 + 2ax} \, dx \\
& = & 2 e^{a^2} \int_{ - \infty}^{\infty} e^{ - (x - a)^2} \, dx \\
& = & 2 \sqrt {\pi} e^{a^2}. \end{eqnarray*}

Together with $ I(0) = 0$, this implies $ I(a) = 2 \sqrt {\pi} e^{ - a^2} \int_{0}^{a} e^{t^2} \, dt$.

[color=gray]p.s. 혹시 지식인에 같은 질문 올리시지 않았나요?[/color]


I thought it would be interesting so I put an option for about everyone...

Show me a short story. Quick! I voted Yes.

I'll kill you if you don't let me...

That's what I voted.

Okay but first I have to restore my files!  :roll:

EDIT: I found my works on an old site I published the on and rolling my eyes looks nice not trying to be mean. :roll:

AND anyone who said yes PM me..or post here which is WAY more efficient.

I voted no because I want the whole thing

But you are unfortunate b/c I am still working on it. :wink:

I have voted as same as nutz...

Btw, you are using caps. And dots, too.

Okay NEW question(not for the poll) IS there anyone I can read there works or do you guys just read?

I wrote some stupid thing once. I might be able to find it.

I assume that by "funcia occidere debeo" you mean "I have to kill funcia"?

Well that case, treating funcia as a noun like rosa, I think you have to use the accusative in this case:" funciam occidere debeo"

Yes and then I should have my name as funcius so it should be:
Funcio occidere debeo.

Meaning: I ought to kill funcia.

EDIT: Or if you please I must kill Funcia.

Hmm, i'm pretty sure that if you go for "funcius" and treat is as "avus", it should be "funcium" for the singular accusative.

[quote="funcia"]Oh well maybe I am wrong :
nom us i
gen i orum
dat o is
acc um os
abl o  is

 :blush: oops i was..[/quote]
That's it.  Don't forget the vocative though.  Usually that is just the same as nominative, but for words like "avus" it becomes "ave", so I would have to address you with "funcie".

Yes I haven't gotten there in latin but it was like that in greek..


let a,b,c are positive such that abc=1 .prove that :
\[ \frac{a}{b^2(c\plus{}1)} \plus{}\frac{b}{c^2(a\plus{}1)} \plus{} \frac{c}{a^2(b\plus{}1)}  \geq \frac{3}{2}\]

Substitute $ a \equal{} \frac {x}{z},b \equal{} \frac {y}{x},c \equal{} \frac {z}{y}$
And use $ \sum_{cyc}\frac {a}{b^2(c \plus{} 1)} \equal{} \sum_{cyc}\frac {x^3}{yz(y \plus{} z)}\geq \sum_{cyc}\frac {x^3}{y^3 \plus{} z^3}$
and use Nesbitt's inequality.

we can prove it with caushy and chebychev.
suppose that $ x \geq y \geq z$ then $ x^3 \geq y^3 \geq z^3$ and $ \frac{1}{y^3 \plus{} z^3} \geq \frac {1}{x^3 \plus{} z^3} \geq \frac {1}{y^3 \plus{} x^3}$
with chebychev
$ S\equal{}\sum_{cyc}\frac{x^3}{y^3 \plus{} z^3} \geq \frac{1}{3}(x^3\plus{}y^3\plus{}z^3)\left(\sum_{cyc}\frac {1}{x^3 \plus{} z^3}\right)$
with caushy:
$ \left(\sum_{cyc}\frac {1}{x^3 \plus{} z^3}\right)\geq\frac{9}{2(x^3\plus{}y^3\plus{}z^3)}$
then:$ S\geq \frac{1}{3}(x^3\plus{}y^3\plus{}z^3)\left(\frac{9}{2(x^3\plus{}y^3\plus{}z^3)}\right)\equal{}\frac{1}{3}\times\frac{9}{2}\equal{}\frac{3}{2}$


number theory
prime numbers
number theory theorems

For all prime numbers p there exist a positive integer r such that r is primitive root in modulo [img]http://www.mathlinks.ro/Forum/latexrender/pictures/3d57128c18c793b56feae10c03222544.gif[/img] . I think this is very usefull.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=55473 , Lemma 3 essentially proves this.

PS: $p$ has to be odd.

I know this teorem's proof. You can prove it with two lemma.

  [b]Firs lemma:[/b]If r is primitive root in modulo p, then r or r+p is primitive root in modulo [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/ca77baa174984994b648741752abfe84.gif[/img].
  [b]Second lemma:[/b]If r is primitive root in modulo [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/ca77baa174984994b648741752abfe84.gif[/img], then for all k>1, r is primitive root in modulo [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/7904d09549ad423513d48045a00dab62.gif[/img] .

$p$ must be odd prime because primitive root doen't existe for the numbers $2^{n}$ for $n\ge3$.in this cas we have $a^{1/2\phi(2^{n})}\equiv1(mod2^{n})$

You might find this interesting

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=nonquadratic+primitive&t=107636


Given a,b,c >= 0, ab+bc+ca =< 3abc. Prove that:
 a^b + a^c +b^c + b^a+c^a+c^b >=6
[u][i] From: http://www.diendan3t.net[/i][/u]


algebra
polynomial
number theory proposed
number theory

Find all polynomials $f$ with integer coefficients such that there exists $a,b$ natural numbers such that $ f(n)$ divides $ a\cdot\frac{(bn)!}{n! (bn-n)!}$ for all n. I have a monstruous solution for this one. Anyone has something easier?

Since you have had a solution,could you post it to share with us?

No, because in this case I would spoil the problem. ;)


trigonometry
geometry
parallelogram
geometry unsolved

Suppose we are given a convex quadrilateral $ABCD$ (labeled clockwise), and we are given angles $ABC, BCD, ADC, BAD$, and $BAC$.  The question is to determine the measure of angle ABD.  For example, if $ABCD$ is cyclic, it's easy to see that
\[ABD = ACD = \pi-ADC-BAD+BAC\]
since 1) $ABD$ and $ACD$ both subtend the arc $AD$, and 2) angles of a triangle sum to $\pi$.  The proof for an arbitrary convex quadrilateral, however, escapes me.

Any ideas?

You cannot find < ABD without trigonometry - and I mean ugly trigonometry. It is easy to find quadrilaterals ABCD where all angles < ABC, < BCD, < ADC, < BAD and < BAC are rational multiples of $\pi$, while < ABD is not. (For instance, you can take a triangle ABC whose angles are rational multiples of $\pi$, and construct D in such a way that ABCD is a parallelogram; then, the angle < ABD will be the angle between a median and a side of triangle ABC, and such angles need not be rational multiples of $\pi$.)

  Darij


limit
logarithms
calculus
derivative
calculus computations

$ \lim_{x \to 0} \frac { x \plus{} \ln (\sqrt {1 \plus{} x^2} \minus{} x)}{x^3}$
I tried  to find answer and found $ L \equal{} \frac {1}{2}$.
But  answer in  book $ L \equal{} \frac {1}{6}$

hello, using L'Hospital's rule twice we get
$ \lim_{x \to 0}\frac{\sqrt{1\plus{}x^2}\minus{}1}{3x^2\sqrt{1\plus{}x^2}}\equal{}\lim_{x\to 0}\frac{1}{3(2\plus{}3x^2)}\equal{}\frac{1}{6}$
Sonnhard.

Thanks, who can solve by method of Taylor's expansion? I am interested :lol:

Being bad at differentiation, I did solve it by Taylor expansions.

I think Hong Quy was probably interested in [i]seeing a solution by Taylor methods[/i], not in taking a poll of who knows how to use those methods.  :roll:

We have $ \sqrt {1 + x} = 1 + \frac {x}{2} + O(x^2)$, so $ \sqrt {1 + x^2} = 1 + \frac {x^2}{2} + O(x^4)$ and $ \sqrt {1 + x^2} - x = 1 - x + \frac {x^2}{2} + O(x^4)$.  Also $ \ln(1 + x) = x - \frac {x^2}{2} + \frac {x^3}{3} + O(x^4)$, so
\begin{align*} \ln(\sqrt {1 + x^2} - x) & = - x + \frac {x^2}{2} + O(x^4) - \frac {( - x + \frac {x^2}{2} + O(x^4))^2}{2} + \frac {( - x + \frac {x^2}{2} + O(x^4))^3}{3} + O(x^4) \\
& = - x + \frac {x^2}{2} + O(x^4) - \frac {x^2 - x^3 + O(x^4)}{2} + \frac { - x^3 + O(x^4)}{3} + O(x^4) \\
& = - x + \frac {x^3}{6} + O(x^4) \end{align*}
and probably you can finish from here.

Thank JBL,  you accuratly understood me!!!
When i solved it, i also worked like you


geometry
3D geometry
algorithm
algebra
factorization
sum of cubes

I used the method from Art and Craft to come to this:

$\sum_{k=1}^{n} k^4 = \frac{k(k+1)(6k^3+9k^2+k-1)}{30}$

but i am certain that i am wrong, since it should be all factorable. I will go back and see what i did wrong. (probably arithmetical error :blush: )

EDIT:  FIXED

You can see that something is wrong with the cubic, since at low values of $k$, the whole thing will be negative. I would guess that there should be more adding than subtracting in there.

I haven't read Art and Craft, but is it like this?

\begin{eqnarray*}
\sum_{k=1}^n \left((k+1)^5-k^5\right)&=&\sum_{k=1}^n\left(5k^4+10k^3+10^2+5k+1\right)\\
(n+1)^5-1&=&5\sum_{k=1}^n k^4+10\sum_{k=1}^n k^3+10\sum_{k=1}^n k^2+5\sum_{k=1}^n k+\sum_{k=1}^n 1\\
&=&5\sum_{k=1}^n k^4 +10\frac{n^2(n+1)^2}{4}+10\frac{n(n+1)(2n+1)}{6}+5\frac{n(n+1)}{2}+n\\
5\sum_{k=1}^n k^4&=&n^5+\frac 52 n^4+\frac 53 n^3-\frac 16 n\\
\sum_{k=1}^n k^4&=& \frac{n(n+1)(6n^3+9n^2+n-1)}{30}
\end{eqnarray*}

Basically the first sum is expressed in two ways, one of which telescopes. Then you plug in the formulas for sum of squares, sum of cubes, etc. and work out the algebra.

thats exactly what i was doing. and yes, that is what my corrected sum came out to be  :D

[quote="pkothari13"]xxreddevilzxx, where did you find that method in art and craft? do you have a page number or chapter or something?[/quote]

page 174, the thing about telescoping. It takes the sum $x^2$ for example. I just used its algorithm and applied to $i^4$.

Try evaluating it in this way, which is oh-so cool.

Take the set $ ( 1,2,...,n+1 ) $. Take $x,y,z,w,t$ from the set such that $x,y,z,w$ are all less than $w$. Count how in two ways.

[quote="gotztahbeazn"]this was on the algebra midterm and i think i got it but i thought it was really neat and wanted to know if anyone else got what i got

$\displaystyle\sum_{i=1}^{n}i^4$[/quote]
heres a geometric way

$s=\sum_{k=1}^{n}{k^4}$

$s=\frac{n^3(n+1)(2n+1)}{6}-\sum_{k=1}^n{\frac{(2k-1)(k-1)(k)(2k-1)}{6}}$

$s=\frac{n^3(n+1)(2n+1)}{6}-\frac23{\sum_{k=1}^{n}{k^4}}+\frac43{\sum_{k=1}^n{k^3}-\frac56\sum_{k=1}^n{k^2}+\frac16\sum_{k=1}^n{k}}$

$\frac53s=\frac{n^3(n+1)(2n+1)}{6}+\frac{n^2(n+1)^2}{3}-\frac{5n(n+1)(2n+1)}{36}+\frac{n(n+1)}{12}$

$s=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$

now what about $i^5$?
You can also make a more general statement that
$\sum_{i=1}^{n}{i^x}=n^{x+1}-\sum_{i=1}^{n}{(i-1)(i^x-(i-1)^x)}$
From this you can just use the formulas for $\sum_{i=1}^{n}{i^{\{1,2,3,...,x-1\}}}$ that you have gotten fom using the formula above.  It could be called recursive

[quote="Rushil"]Does anyone here know about Bernoulli numbers. These questions are special cases og general expressions involving Bernoulli numbers!!![/quote]
No, Ive never heard of them what are they? Please explain


yah, got everything right except 25, which I skipped 'cause I read the question wrong :(

@Smart...k

I think you are not suppose to discuss the solutions until tomorrow afternoon..

So there are still schools writing Cayley? Hm. I hope they try to change the contest writing date to just one day. Maybe for emergencies?

as I was saying earlier, anybody have a solution to 25?
THanks

Ok, so here are my answers, I think that al of them are right, except 25... I just drew that one in GeoGebra, and found that the answer was E (6.0), unless I drew it wrong...

If anyone has a nice solution to 25, please post it...

Here are my answers:
1. A
2. C
3. B
4. E
5. B
6. D
7. C
8. D
9. C
10. D
11. A
12. E
13. A
14. A
15. C
16. E
17. C
18. B
19. A
20. D
21. C
22. D
23. C
24. B
25. D
I know I probably did really bad.... but whatever.....

- Nawahd Werdna

I am sure that 23 is B, not C.

it's 203.1........a lot of my peers got the same answer

I think for you, it's just a calculation error

lol...23 is 203.4...and 25 is 6...this year's cayley is so easy...especially the last question...compared to previous years...

[quote="appiejam"]IF 23 is C, and

25 is E,

I ACED the TEST !!!YES!![/quote]

lol ...congrats...haha...im pretty sure thats right...

[quote="appiejam"]i calculated it some way and figured out that 6 works..

since 6 works, any of the other answers cannot be right..

cuz 6 is the biggest number there[/quote]

um...actually its pretty straght forward...the locus of A is a semi circle...so the farthest distance from P just has to pass through the center of the circle...which is the midpoint of segment XW(wutever letters they r...)

i get it!

that is a brilliant way!

i calculated it by creating a right angle triangle APT

(T is a point somwhere on the right segment of the big square)

[quote="BFG"][quote="bryant124"]
the locus of A is a semi circle[/quote]
 I think it is a semi ellipse.[/quote]
no...um cuz A is always a right angle...and the angle always passes through 2 points...

ah i realy messed cayley up.
got question 20 wrong. the total time i spent on it was 20 seconds, i could've read it carefully=(

was the cayley for grade 10 students? some people in my schools math club wrote the cayley instead of the fermat, ive been wondering why heh..

thanks

lol...got my cayley result...darn...should have taken fermat...influenced by friends...

Does anyone know when the official Cayley results will come out?  :huh:

your teacher can check for them online with your school ID and password..

[quote="SM4RT"]y? was cayley too easy for u?[/quote]

yea...only spent 20 minutes on it... :( so sad...

:rotfl:

i sucked pretty much

i don't know which one i got wrong

but i made a mistake on one of the C block questions

142.. ROFL faliure


The equalities   $\frac{n}{2^{nm}} \sum\limits_{j=0}^{m}{nm \choose nj}=   \sum\limits_{k=0}^{n-1}\left(\cos{\frac{k\pi}{n}\right)^{nm} ,\; n=1,2,..., }$  are  verified for all positive integers $ m $ ?
[or only for $ m $ =even ?]

If someone interested in the idea, then write $2\cos(k\pi/n)=e^{ik\pi/n}+e^{-ik\pi/n}$.


Express the length of each pipe in feet using improper fractions  6'4" and 8'1"

I may be completely off topic here, but assuming you are talking about feet, and 12 inches equals 1 feet.

$\displaystyle 6'4\"$

$\displaystyle 6 \frac{4}{12} = 6 \frac 13 = \frac{19}{3}$

$\displaystyle 8'1\"$

$\displaystyle 8 \frac{1}{12} = \frac{97}{12}$

Is this what you wanted?

If it is, why would you post a question like this?

Moved, because this is usually material covered in classroom discussion.

[hide]19/3
97/12[/hide]
Please hide your answers/solutions!


I was wondering if any of you have ever read Ann Coulter's books. Her books are politically oriented and it's mostly from a US perspective only although it may be interesting for people in other countries to read. Some of the books she's written off the top of my head are:

Godless: The Church of Liberalism
Slander: Liberal Lies about the American Right
How to Talk to a Liberal (If You Must)

Since Godless is new, the price is high so I have no purchased it yet, however, I have recently purchased Slander: Liberal Lies about the American Right.

While Ann Coulter does have some radical ideas that many people are uncomfortable with, she does make some good points as well, she just takes them to an extreme. What's everybody else's opinions on her books/views?

I like [url=http://www.youtube.com/watch?v=Fv9xa-VxchM&mode=related&search=]Ann Coulter[/url] in the same manner I like Borat.  She's not a politician, she's a comedian.

But as far as taking her seriously, the world needs people to throw out crazy ideas.  I will go as far as to say that I even agreed with some of her ultra-conserative ideas on taxes.

But you know what?  Ann Coulter is nothing compared to the crazy religious lady from CNN.  If you didn't see that, I won't post a link, but you can google "crazy religious lady" and find a video, if you want to be...amused.

[quote="mdk"]she wasn't crazy. Just because she believed that Bush signing the terrorrist detainee bill was immoral does not make her crazy.[/quote]

Are you talking about Ann Coulter or the crazy religious lady?  They are both crazy, the latter for obvious reasons.

Ann Coulter, I guess, isn't crazy, but her views are pretty far out and offensive to a lot of people.


The igits 1,2,3......up to n can be arranged in the form of n!
Find the sum for 4! and 5!  Then, find a formula that would work for all n!

For Exp:2!=2x1 12+21=33
          :3!=3x2x1 321+312+213+231+123+132=1332
          :4!=?
          :5!=?

Formula=?

[hide]Well, every number is going to appear in each place $\frac{n!}{n}=(n-1)!$ times. And the sum of all of the numbers is $\frac{n(n+1)}{2}$ .So the sum of any place is:

$\frac{n(n+1)}{2}\cdot(n-1)!$

Next, all we need to do is multiply it by  $1+10+100...10^{n}=111...$ ($n$ one's). To generate this sequence of ones, we observe that $n$ nine's is just $10^{n}-1$. We divide this by 9 to get $n$ one's. So $n$ one's is just $\frac{10^{n}-1}{9}$. So our final formula, which I am too lazy to simplify is:

$\frac{n(n+1)}{2}\cdot(n-1)!\cdot\frac{10^{n}-1}{9}$

Plugging in $n=4$ and $n=5$, we get 66660 and 3999960.[/hide]

you have a good answer, except if you try plugging in 2 or 3, u added a "n" in the 2nd last spot. Plugging 3 in yours we get 13332 instead of 1332 if you add em up the slow way. And plugging in 2 in we get 333 instead of 33. So I think you made a little mistake in one of your spots.


algebra
polynomial
AMC
AIME
number theory
relatively prime
geometric series

The polynomial \[P(x)=(1+x+x^2+\cdots+x^{17})^2-x^{17}\] has 34 complex roots of the form $z_k=r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34$, with $0<a_1\le a_2\le a_3\le\cdots\le a_{34}<1$ and $r_k>0$. Given that $a_1+a_2+a_3+a_4+a_5=m/n$, where $m$ and $n$ are relatively prime positive integers, find $m+n$.

[quote="scorpius119"][hide]Using geometric series for the summation, $P(x)=\frac{(x^{18}-1)^2}{(x-1)^2}-x^{17}$.

Under a common denominator this is $P(x)=\frac{(x^{18}-1)^2-x^{17}(x^2-2x+1)}{(x-1)^2}$.

The numerator simplifies to $(x^{36}-2x^{18}+1)-(x^{19}-2x^{18}+x^{17})=x^{36}-x^{19}-x^{17}+1=(x^{19}-1)(x^{17}-1)$.

The roots are given by the 19th and 17th roots of unity (except 1), so the sum is $\frac{1}{19}+\frac{1}{17}+\frac{2}{19}+\frac{2}{17}+\frac{3}{19}=\frac{3}{17}+\frac{6}{19}=\frac{159}{323}\Rightarrow \boxed{482}$.[/hide][/quote]

Nice solution, same as mine. Though when I actually took the 2004 AIME I was like :o *faints*.

Yeah, it is a lot harder taking the test than seeing the questions here... *remembers this year's AIME*

I did this the same way too (is there really any other way) except screwed up on the roots of unity... for some reason I thought it was $e^{i\pi \frac{2k+1}{n}}$.  And after I found out it was wrong, trust ol' AoPS told me to just do deMoivre on $w^{\frac 1n}$...  :blush:

[hide=Solution]

First, note that $1$ clearly is not a root; $18^2 - 1 \neq 0$.

Note that by the finite geometric series formula, $P(x) = \left(\frac{x^{18} - 1}{x-1}\right)^2 - x^{17}$. Because we saw this on a roots of unity handout from AMSP long back we decide to guess random roots of unity.

The $18$ looks really annoying; we can get rid of that by plugging in a non-unity $17$th root of unity $\omega$ and since $\omega^k = \omega^{k\pmod{17}}$ we have $P(\omega) = \left(\frac{\omega^{18}-1}{\omega-1}\right)^2 - \omega^{17} = \left(\frac{\omega-1}{\omega-1}\right)^2 - 1 = 1 - 1 = 0$, so all non-unity $17$th roots of unity are roots, so $e^{\frac{2\pi i k}{17}}$ are roots for $1\leq k\leq 16$.

By the Fundamental Theorem of Algebra, there are $34$ roots of $P$ and we already found $16$ of them so there are $18$ left. If we find all $k$th non-unity roots of unity, we will get $k - 1$ root, so a good candidate is $k = 19$.

Let $\zeta$ be a non-unity $19$th root of unity, then $\zeta^k = \zeta^{k\pmod{19}}$, then $P(\zeta) = \left(\frac{\zeta^{18} - 1}{\zeta-1}\right)^2-\zeta^{17} = \left(\frac{\zeta^{-1}-1}{\zeta-1}\right)^2 - \zeta^{-2} = \left(\frac{\frac{1-\zeta}{\zeta}}{\zeta-1}\right)^2 - \zeta^{-2} = \left(-\frac{1}{\zeta}\right)^2 - \frac{1}{\zeta^2} = \frac{1}{\zeta^2} - \frac{1}{\zeta^2} = 0$, so all non-unity $19$th roots of unity are roots, so $e^{\frac{2\pi i k}{19}}$ are roots for $1\leq k\leq 18$.

Thus, we have found all roots of $P$, and note that since $e^{ip} = \cos(p) + i\sin(p)$,  the $a_k$ in $e^{\frac{2\pi i k}{n}}$ is $\frac{k}{n}$. Thus, it is not hard to see that $a_1 = \frac{1}{19}, a_2 = \frac{1}{17}, a_3 = \frac{2}{19}, a_4 = \frac{2}{17}, a_5 = \frac{3}{19}$. Adding yields $\frac{159}{323}$, so $m + n = \boxed{482}$. [/hide]

@below it's in contest collections, so when I wanted to post my solution to this, I went on contest collections and found the designated thread(it's a past aime problem)

This is off-topic, but how did this thread get reopened after 15 years?

I have no idea, but the problem was pretty interesting lol

[hide=Solution]We have $\frac{(x^{18}-1)^2}{(x-1)^2}  = x^{17}$ after simplification we get $(x^{19}-1)(x^{17}-1)=0$. disregarding $1$ as a root since it would not work in the original equation, we just have the roots of unity. Since $19$ and $17$ are close we can just eyeball the fractions to get the first $5$ as $\left(\frac{1}{19},\frac{1}{17},\frac{2}{19},\frac{2}{17},\frac{3}{19}\right)$ which gives
$$\frac{1}{19}+\frac{1}{17}+\frac{2}{19}+\frac{2}{17}+\frac{3}{19}=\frac{6}{19} + \frac{3}{17} = \frac{159}{323} \Rightarrow 159 + 323 = 482$$[/hide]

Solved from [b]Polynomials in the AIME.[/b]
Factoring and expanding yields $$\frac{x^{36}-2x^{18}+1}{x^2-2x+1}-x^{17}=\frac{x^{36}-x^{19}-x^{17}+1}{(x-1)^2}=\frac{(x^{19}-1)(x^{17}-1)}{(x-1)^2}.$$ Indeed, these are the 19th and 17th roots of unity without 1 (no other overlap since 2kpi/n has denominator 17 vs 19). The smallest such angles are 1/19,2/19,3/19,1/17,2/17. Summing them yields a final answer of $\boxed{482.} \blacksquare$


abstract algebra
group theory
number theory
relatively prime
superior algebra
superior algebra unsolved

Everyone is aware of Herstein's famous problem: If G is abelian and has subgroups of order m and n, prove G has a subgroup of order l.c.m of m and n.

Earlier last summer I proved the previous and somewhat similiar question pertaining to order of elements in the same context. However, we still don't have a solution to this "famous" problem. I have a solution, but I am not content with it because it is not elementary enough for me. I'll outline the proof below.

1) Consider the case when G has subgroups of order m and n and (m,n)=1. Let M and N be these subgroups respectively. Then the set  MN is a subgroup because G is abelian and furthermore one can show their intersection is (e).( Hint: assume x is an element of the interesection and show x=e
by writing out (m,n)=1 as a linear combination of m and n).

2) Write out the l.c.m. of m and n in terms of the max function.

3) Prove Sylow's theorem for abelian groups. This is done in Herstein in section 2.7( before Sylow's Theorem)  as an application using the idea of Quotient groups and normality of subgroups in section 2.6. Note before this Herstein also proves Cauchy's theorem for abelian groups and then our needed result.

4) With 1) and 3) at hand one can piece the appropiate parts to conclude that the subgroup does indeed exist.

Hence, if we can do 3) without relying on "forward" results in Herstein we are done. Has anyone seen a proof of Sylow's theorem for abelian groups more elementary that the one offered in Herstein. This would indeed settle the issue once and for all.

MOZART

In 1959 Helmut Wielandt proved Sylow's theorem with some number theory, combinatorics, equivalence relation theory and familiarity with cosets. This proof finishes it....

Moreorless this problem is asking the reader to prove Sylow's first theorem without understanding what is a normal subgroup. Wielandt did it in 1959, and the proof is in Herstein in the section on Sylow's Theorem.

Anyone have other elementary proofs of Sylow's first Theorem???

The crux here is the converse to #1: If $G$ is abelian and has a subgroup $H$ of order $mn$ with $m$ and $n$ relatively prime, $G$ has subgroups $K$ and $L$ of order $m$ and $n$ respectively. This gives the Sylow theorem for abelian groups immediately.
We have obvious choices for $K$ and $L$: $K=nH=\{x\in H: mx=0\}$ and $L=mH=\{x\in H: nx=0\}$ (writing $G$ additively).
Let $K_{1}$ and $L_{1}$ be the first definitions, and let $K_{2}$ and $L_{2}$ be the second definitions. They are easily verified to be subgroups. We have $K_{1}\subset K_{2}$ and $L_{1}\subset L_{2}$ by Lagrange (the order of any element in $H$ divides $mn$). Since $m$ and $n$ are relatively prime, there are integers $a$ and $b$ with $ma+nb=1$; this gives $K_{2}\cap L_{2}\subset \{x\in H: (ma+nb)x=0\}=\{0\}$. In addition, $K_{1}+L_{1}\supset maH+nbH\supset H$.

Now $K_{1}+L_{1}=K_{2}+L_{2}=H$ and $K_{1}\cap L_{1}=K_{2}\cap L_{2}=\{0\}$, so $|K_{1}|\cdot |L_{1}|=|K_{2}|\cdot |L_{2}|=mn$ and $K_{1}=K_{2},L_{1}=L_{2}$. By Lagrange again, the order of $K_{2}$ divides some power of $m$; since it also divides $mn$, that order divides $m$. Similarly, the order of $L_{2}$ divides $n$. Since the product of these orders is $mn$, $|K_{2}|=m$ and $|L_{2}|=n$.

The only reason this fails for nonabelian groups is that $K_{1},L_{1},K_{2},L_{2}$ are likely not to be subgroups. It doesn't use normal subgroups, so it's at least as elementary as the Herstein proof.

Jmerry, I am not following your reply starting from the second to last paragraph. I am little confused by the subscripts 1 and 2 and changing from products to sums. We just need one subgroup of order m and n. Enlighten me, I am probably not seeing the obvious....I understand the definitions given and the implications, but the last step because of Lagrange's Thm, how do we get the equalities of the set orders....

Jmerry, I can repair your proof,but I will need Cauchy's Thm for abelian groups. Herstein's proof uses normality, but there is a proof by Mckay that is more elementry, like Wielandt's proof it is "group-action" like. I believe Mckay published it in 1958, so we are still in the 1950's....


Suppose Yi ~ Poisson(lambda*Xi)

Show the maximum likelihood estimate of lambda is (sample mean of Y/sample mean of X)

Calculate Variance of this MLE.

Okay, I have proved the MLE part. But what is its variance?

So, I have Var(sample mean of Y) = lambda*Xi,

Should I treat sample mean of X as a constant and just square it out in the variance expression?

Thanks.


Let $A_1A_2B_1B_2C_1C_2$ be an inscribed hexagon on a circle $(O)$. Suppose $M$ is an arbitrary point. Let $A'_1$ and $A'_2$ be the second intersections of $A_1M$ and $A_2M$ with the circle $(O)$. Let $B_a$ and $C_a$ be the intersections of $A'_1A'_2$ with $C_1C_2$ and $B_1B_2$. Define $C_b, A_b$ and $A_c, B_c$ analogously. Show that $A_bA_c, B_cB_a, C_aC_b$ are concurrent at $M$.


function
calculus
derivative
topology
real analysis
real analysis unsolved

Let $ f: X\mapsto Y$. We call $ f$ open if $ f(V)$ is open for every open set $ V\subset X$. Prove that every continuous open $ f: \mathbb{R}\mapsto\mathbb{R}$ is monotonic.

Hint please, no solution.

Can a function be open if it has a local maximum?

Ok, so from that I get the following.

Suppose that a continous open function $ f: \mathbb{R}\mapsto \mathbb{R}$ has a max/min on $ (a,b)\subset\mathbb{R}$.

Let $ f(\xi)$ with $ \xi\in(a,b)$ be max/min point of $ f$ on $ (a,b)$ we would then have that if $ f(\xi)$ is a max that $ (a,b)\mapsto (c,f(\xi)]\text{ or }[c,f(\xi)]$ either way $ f(\xi)$ is

 included violating $ f$'s openess.

Min implies $ (a,b) \mapsto[f(\xi),c)\text{ or }[f(\xi),c]$.

Thus $ f$ cannot have a max or min on any closed interval. This implies that $ f'(x)\ne{0}\text{ } \forall{x}\text{ }\in(a,b)$,

so assuming connectendess of $ (a,b)$ we have that

$ f'(x) > 0\text{ }\forall{x}\in(a,b)$ or $ f'(x) < 0\text{ }\forall x\in(a,b)$ which is the definition of monotonicity.

If this is what you were talking about this is how I would have done it, but my book has not yet covered maximums and minimums so I do not think this is the intended method. Also I had to assume connectedness of $ (a,b)$

Thank you very much by the way

Don't talk about derivatives- we're not assuming differentiability. The definition of monotonicity doesn't say anything about differentiability.

We want to directly show that a function which isn't monotonic has a local maximum or a local minimum somewhere; that should come from the Extreme Value Theorem. If $ a<b<c$, $ f(a)<f(b)$, and $ f(c)<f(b)$, there is a local maximum somewhere in $ (a,b)$.

[quote="jmerry"]Don't talk about derivatives- we're not assuming differentiability. The definition of monotonicity doesn't say anything about differentiability.

[/quote]
Thank you sir. Correct me if I am wrong I did correctly show that a local max/min cannot occur in an open function. My problem is how I used that result, right?

Yes, that part was correct. (You are implicitly using the Intermediate Value Theorem; we don't actually need that the image is an interval).

Ok so we have established that for a continuous open function $ f: \mathbb{R}\mapsto\mathbb{R}$ there cannot be a maximum in any $ (a,b)$. Now let us consider the interval

$ [a,b]$, by the Heine-Borel theorem we know that this interval is compact. Now it is known that if $ f: X\mapsto Y$ and $ X$ is compact that the extreme

value theorem is applicable, therefore there must exist a $ f(c)\equal{}\sup_{c\in[a,b]}f\left([a,b]\right)$ and a $ f(c_1)\equal{}\inf_{c_1\in[a,b]}f\left([a,b]\right)$. Now the lack of relative extrema forces

 either one of two cases, the first is that $ \sup f\left([a,b]\right)\equal{}f(a)$ and $ \inf f\left([a,b]\right)\equal{}f(b)$, or the second being $ \sup f\left([a,b]\right)\equal{}f(b)$

and $ \inf f\left([a,b]\right)\equal{}f(a)$. This translates to either $ f(b)\leqslant f(c) \leqslant f(a)$ or $ f(a)\leqslant f(c) \leqslant f(b)$ with $ c\in(a,b)$. Therefore either $ f(a)\leqslant f(b)$

or $ f(b)\leqslant f(a)$, and since $ a,b$ were arbitrary this completes the proof.

How does that look sir?

[quote="jmerry"]That's one way to put it. I was really thinking of the form of the extreme value theorem you see in a calculus class, but there's nothing wrong with being more technical.[/quote]
Thank you sir.


[quote="icx"]Is there any formula for something like 1*2*3*4*...*n? Where '*' is the sign for multiplication. How can i solve 1*2*3*...*2006= ... ?

By the way, where i can find the mathematical signs you use in your messages?

And not to create another thread for a small thing:

what means n!

eg. 2006!

I have some books but i don't really get the point.[/quote]

You seem to have answered your own question.  In fact,

$1 \cdot 2 \cdot 3 \cdot ... \cdot 2006 = 2006!$

:D

The notation $n!$ denotes the value $1 \cdot 2 \cdot 3 \cdot ... \cdot n$ for positive integers $n$.  Generally $0! = 1$.

I thought so, thank you very much.

But then, how can you solve $n!=2003$ or $1\cdot2\cdot3\cdot...\cdot2003$ ? There must be a special formula. Or if not, how to find out the result?

PS: The latex thing rocks :)

Unfortunately, there's not a simple closed formula (using the [url=http://en.wikipedia.org/wiki/Gamma_function]Gamma function[/url] would be too complicated).
But you have [url=http://230nsc1.phy-astr.gsu.edu/hbase/math/stirling.html]Stirling's approximation[/url].


Hi!
I would like to find macednian guide. If someone has his e-mail please send me it.
Thank you:).

Do you know his/her name? That would really help

[url=http://imo.wolfram.com/contestants/MKD5/]Here[/url] you may have what you need...

tstoleimo2001=yahoo.com

[Moderator edit: I have replaced the @ sign by = for spam protection.]

Is this the person you were looking for?

[quote="José"]
Is this the person you were looking for?[/quote]
Yes, thanks:):):).


Two players play a game with 2 6-sided dice. Before they roll the dice, they have to guess the sum of the 2 dice. They can pick any number between 2 and 12, but they cannot be the same. Then the 2 dice are rolled. The player whose guess is closer to the actual sum of the 2 dice wins the game. What number would you guess to maximize your chance of winning should you go first and what is the probability of winning?

[hide="Hint"]
What is the expected value of the sum of the dice? Can this value be achieved, and if so, what's special about it?
[/hide]

Edit: Sorry, blanked out. :oops:

Well, not all sums have equal probability of occurring:

2: 1,1
3: 2,1; 1,2
4: 2,2; 1,3; 3,1
5: 3,2; 2,3; 4,1; 1,4
6: 3,3; 4,2; 2,4; 1,5; 5,1
7: 3,4; 4,3; 5,2; 2,5; 6,1; 1,6
8: 4,4; 2,6; 6,2; 3,5; 5,3
9: ....

Because the sum of seven can be reached through the maximum number of ways, the player should guess a sum of 7. Also, seven is close to both 6 and 8, the two sums with the second greatest number of possibilities.

If you guess 7, the other player has 11 possibilities to choose from.
Probability of each sum being correct:
2, 12: 1/36
3, 11: 1/18
4, 10: 1/12
5, 9: 1/9
6, 8: 5/36
7: 1/6

When the sum is two, you will win if the other player selects 8-12
When the sum is three, ~
When the sum is four, ~
When the sum is five, you will win when the other player selects 2 or 8-12
When the sum is six, you will win when the other player selects 2, 3, 4, or 8-12
When the sum is seven, you win (other player selects anything).
When the sum is eight, same probability as six...........

So....
2*(1/36)*(5/11) + 2*(1/18)*(5/11) + 2*(1/12)*(5/11) + 2*(1/9)*(6/11) + 2*(5/36)*(8/11) + (1/6)*(11/11)

Note: I am multiplying the probability that it is a certain sum by the probability that you will win if it is that sum (and sometimes I multiply that by 2 b/c the same thing occurs for two cases).

probability of winning = 127/198

Also, because the probability of winning > .5, no other sum can be as great (all must be less than .5), therefore you know correct sum (the sum with the greatest probability of winning) was selected.

I agree that one should guess 7. But I wasnt clear about the probability. First I thought it was 21/36. But I think it is wrong. Thanks for any help.


This kind of helps me, but I don't really get how the word problems at the end of each chapter help me, because I barely know how to do many of them, and they don't really fit with the chapter...

The problems at the end of the chapters give you practice in applying the basics that you learn in the chapter to creative problems such as those that appear on contests.  Don't worry about not knowing where to start on any of the problems; it comes with practice.

Yeah, many of the problems in the back of the chapters come from popular contests, such as the AHSME, ARML, etc. It helps you learn a new way to apply what you learned in the chapters, instead of just endlessly doing the same boring exercises like you do in school math classes. You can't really learn "problem solving", you just get better at it with practice. As archimedes said, just keep practicing those problems, and eventually you'll understand them better. Look at the answers to get an idea of how to solve those types of problems if you can't solve them.


limit
topology
real analysis
real analysis unsolved

Let $X$ be a compact metric space and $f: X\to X$ be a continuous function. Prove that $\exists Y\subset X \text{st}\ f(Y)=Y.$

Choose an arbitrary point $x_{0}\in X.$ Let $x_{1}=f(x_{0})$ and in general $x_{n+1}=f(x_{n}).$ Let $Y$ be the set of all accumulation points of the sequence $\{x_{n}\}.$ By the compactness of $X,$ we know that $Y$ is nonempty.

$y\in Y$ iff there exists a subsequence $x_{n_{k}}$ that converges to $y.$

First, consider the sequence $\{f(x_{n_{k}})\}=\{x_{n_{k}+1}\}.$

$f(y)=\lim f(x_{n_{k}})=\lim x_{n_{k}+1}$ so that subsequence also converges, and to another member of $Y.$ Hence we can say that $f(Y)\subseteq Y.$

Now, consider the subsequence $z_{k}=x_{n_{k}-1}.$ There's no reason to assume that $\{z_{k}\}$ converges but, again by compactness, some subsequence $\{z_{k_{j}}\}$ does converge to a point $z\in Y.$ Making one more use of the continuity of $f,$ we see that $f(z)=y.$ We conclude that $Y\subseteq f(Y).$

It seems that we do not actually need the condition that $X$ is a metric  space. The theorem holds for Hausdorff topological spaces.

Here is another proof. Define the sequence of sets $S_{i}$ by $S_{0}= X$, $S_{i+1}= f(S_{i})$. Since $f$ is continuous, $f$ maps a compact set to a compact set. Therefore, all sets $S_{i}$ are compact (and therefore they are closed sets). Also since $S_{1}=f(X)\subset X = S_{0}$, we have $S_{i+1}\subset S_{i}$. Therefore, $S_{0}\supset S_{1}\supset S_{2}\supset \dots$ is a sequence of nested closed sets - so it has a non-empty intersection. Denote the intersection by $Y$.
It easy to see that $f(Y) = Y$. We found the desired set.

yet another proof using Zorn's lemma.

Let I be the family of non-empty closed subsets Y of X such that f(Y) is a subset of Y. I is not empty because it contains X. If (Y_a) is an indexed chain in I, then the intersection of all Y_a is a closed non-empty set because X is compact. This intersection is again in I.

By Zorn's lemma, there is a minimal element Z of I. If f(Z) were a proper subset of Z, then f(Z) would be a smaller element of I. Therefore f(Z) = Z.


Let $ n$ be an integer.

For any integers $ a$ and $ b$ with $ b\geq 0$, let $ a\bmod b$ be the integer $ k$ satisfying $ b \mid a \minus{} k$ and $ 0 \leq k < b$.
For every integer $ a$ with $ 0 \leq a \leq 2n \plus{} 1$, define an integer $ Z\left[a\right]$ as follows:
$ Z\left[a\right] \equal{} a$ if $ 0 \leq a \leq n$, and
$ Z\left[a\right] \equal{} 2n \plus{} 1 \minus{} a$ if $ n \plus{} 1 \leq a \leq 2n \plus{} 1$.
(Note: from this, it follows that $ 0 \leq Z\left[a\right] \leq n$ for every integer $ a$ with $ 0 \leq a \leq 2n \plus{} 1$.)

Let $ \lambda$ be an arbitrary integer such that $ 1 < \lambda \leq n$, and assume that $ 2n \plus{} 1$ is an odd prime for some integer $ n \geq 2$.

Prove or disprove that
$ \sum_{i \equal{} 1}^{n}\left|i \minus{} Z\left[\left(\lambda i\right)\bmod (2n \plus{} 1)\right]\right| \equal{} \frac {n\left(n \plus{} 1\right)}{3}$.

is $\frac{n(n+1)}{3}$ always an integer? :P

[quote="kathi;)"][quote="RDeepMath91"]Yup, since if $ n \equiv 1(mod 3)$, then $ 2n \plus{} 1 \equiv 0 (mod 3)$ which is impossible since $ 2n \plus{} 1$ is a prime greater than $ 3$..[/quote]

hm? isn't it an integer when $ n\equiv 2 (mod3)$ or $ n\equiv 0 (mod 3)$ ?[/quote]

Sure, that's why I prove $ 3 \not| n \minus{} 1$..

[quote="RDeepMath91"][quote="kathi;)"][quote="RDeepMath91"]Yup, since if $n \equiv 1(mod 3)$, then $2n+1 \equiv 0 (mod 3)$ which is impossible since $2n+1$ is a prime greater than $3$..[/quote]

hm? isn't it an integer when $n\equiv 2 (mod3)$ or $n\equiv 0 (mod 3)$ ?[/quote]

Sure, that's why I prove $3 \not| n-1$..
[/quote]

oh,thats right. sorry

I have moved this to Research Problems, lest this forum completely dies out.

If anyone cares, here is a pretty badly-written java code to verify this conjecture. It will yield exactly those integers which are prime [i]xor[/i] satisfy the conjecture. In quite a large interval of integers tested, it only outputted 4 and 8, so it seems like the conjecture not only holds, it even has a kind of converse (though this wouldn't be particularly interesting - it's surely not a quick way to primality testing).

[hide]
[code]
public class anamazing1 {

	public static void main(String[] args) throws Exception {

		for (int p = 3; p < 100000; p++) {

			boolean everythingok = true;
			boolean isprime = true;

			int n = (p - 1) / 2;
			for (int l = 2; l <= n; l++) {
				int sum = 0;

				for (int i = 1; i <= n; i++) {
					int k = (l * i) % p;
					if (k > n) {
						k = p - k;
					}
					sum += Math.abs(i - k);
				}

				if (sum != n * (n + 1) / 3) {
					everythingok = false;
				}

			}
			for (int u = 2; u <= p - 1; u++) {
				if (p % u == 0) {
					isprime = false;
				}
			}

			if (everythingok ^ isprime) {
				System.out.println(p);
			}

		}
	}
}
[/code]
[/hide]

Does anyone else sense a relation to Gauss' lemma about Lagrange symbols?

  darij

I [b]think[/b] I have a proof for this problem. I must say working it out was a lot of fun. :)

Please see [url]http://btw.blobtrox.net/~btw/modabs.pdf[/url] if you are interested.

I hope it's not too tedious to read, I haven't written too many papers and
I'm not a native English speaker.

If you find any errors in the text, please tell me.

hipsishopsis


[quote="Redriver"]Give $x_1,x_2,.......,x_n> 0$
Prove that : $\large\frac{1}{1+x_1} + \frac{1}{1+x_1 + x_2} + ...+ \frac{1}{1+x_1+...+x_n} \leq \frac{1}{3} (\frac{1}{x_1} + ...+ \frac{1}{x_n})$[/quote]
Not true.
For example, for $n=2$ take $x_1=x_2=1$.
LS=$5/6$, RS=$4/6$. :(
I think the sign is supposed to be the other way around. :?

No, the reversed inequality holds neither :(
(Consider the case $n=1$.)

oh ! I see but I write like this means $n \ge 3$

Well, it still doesn't hold:

For arbitrary $n$ take $x_2 = \ldots = x_n$ to $\infty$  (for fixed $x_1$)
then the given inequality would imply
$\frac 1 {1 + x_1} \le \frac 1 {3x_1}$,
which is violated for big $x_1$.

Similarly one can see that for arbitrary $n$ the reversed inequality
doesn't  hold either.

I actually remember this ineq from somewhere and I think it should be something like $4$ instead of $\frac{1}{3}$. Maybe

Oh I see ! But then I should give the problem folllowing :

Fing the smallest natural number $k$ such as the inenquality :$\large\frac{1}{1+x_1} + \frac{1}{1+x_1 + x_2} + ...+ \frac{1}{1+x_1+...+x_n} \leq k (\frac{1}{x_1} + ...+ \frac{1}{x_n})$ hold for all $n \in N^*$ and $x_1,x_2,.....,x_n>0$

OK, now the problem makes sense. However, it still is not too interesting:

It is obivous that $k=1$ works - just compare the $k$-th term on the left hand
side and the $k$-th term on the right hand side. So you see that the
inequality is far away from being sharp.

On the other hand for ever real $k < 1$ the inequality doesn't hold in general:
There is a counterexmaple for every $n \ge 1$, as can be seen
just as in my last post.


In Texnic center, how do you change page margins, and how do you get rid of page numbers at the bottom of the page?? Thanks.

If you want no page number at all, put this command in your preambule:

\pagestyle{empty}

If you want a single page with no number, put this command where you want no number:

\thispagestyle{empty}

I think the default style is [b]plain[/b] but I'm not sure.  To have more control on numbering, headinds and footers, try this:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35119]www.artofproblemsolving.com/Forum/viewtopic.php?t=35119[/url]

For the margins, there's a lot of thingd to change, so I'll just give you this link:

[url=http://www.ctan.org/tex-archive/info/lshort/english/lshort.pdf]www.ctan.org/tex-archive/info/lshort/english/lshort.pdf[/url]

Look in section "6.4 Page Layout"

ledurt

In automatically generated TOC, LOT, LOF, etc., at the first page,  I got page number in the footer though I used fancystyle with fancy heading. How to get rid of page number in the first page of those objects, or, how to change those first pages into fancyheading (i.e. useer defined heading in the header and nothing in footer)?

Thanks a lot

vnns

did you check the first link in my previous post?  I think that answer your question, if not, give us more detail.

ledurt

Hi Ledurt,

I checked your link already but it did not help. I still cannot find where to change it. Could you have any idea?

Thanks a lot

vnns

try this:

\newpage
\thispagestyle{empty}
\tableofcontents

if that doesn't work or you still want to have some header/footer in your page, load the package "fancyhdr" and then:

\newpage
\pagestyle{fancy}
\lhead{\slshape \rightmark}
\chead{}
\rhead{}
\lfoot{}
\cfoot{}
\rfoot{}
\tableofcontents

but, i'm not sure what you want, do you want to have to page number trough the whole document or just in the TOC, LOF and LOT? Or do you want to have a different numbering (like roman) in the beginning?

ledurt

Your solution didnt work, mate. Sorry for not stating the problem clearly.

Some of lines in my preamble are as follows:

%--------------------------------------------------------------------
\pagenumbering{roman}
\addcontentsline{toc}{chapter}{TABLE OF CONTENTS}
\tableofcontents
\addcontentsline{toc}{chapter}{LIST OF TABLES}
\listoftables
\addcontentsline{toc}{chapter}{LIST OF FIGURES}
\listoffigures

% restart page numbering from 1 in arabic
\pagenumbering{arabic}

% include chapters of the thesis
\include{intro}
\include{chapt2}
%--------------------------------------------------------------------

The problem is: while in all pages of the document, there are header and page number (they are all in the header position as I used fancyheading), in the first page of TOC, LOF, LOT, there is also page number in the footer position (the subsequent pages of TOC, ... return to have fancyheading). What I want is to get rid of page number in that first page.

Your suggestions didnt work because TOC, LOT, LOF are automatically generated.

Thanks a lot

vnns

Ok, now I get it.  Personnally, I use the package fancyhdr so i'm not really familiar with fancyheading.  In my documents, I have a structure very similar to yours but I have no trouble at all with TOC, LOF and LOT.

May be you can post what you do with fancyheading, I could try to help you with that or you can look on google for something like  fancyheading+latex.  Usually, there's a pdf file about packages and maybe you get your answer there.

ledurt


How many words can you find inside of other words?
For example CATTLE

some words include
ATE
EAT
CAT
TEAL
TALE
LATE
ALE
LACE

This thread will be run in the following way:
1. One person finds words inside a previous word.
2. If he/she chooses, they may post a new word

The first word is BRONTOSAURUS

[quote="kstan013"]The first word is BRONTOSAURUS[/quote]
Run, ran, rot, rat, rats, not, son, tab, bot, bat, sat, ton, star, stars, post, posts, oat, oats, tar, bar, boat, tub, stub, rub, soar, root, roots, sons, bun, buns, tubs, boats.

That's 32, I'm too lazy too think anymore.

The next word is ELECTRICITY. I got 120+ once.

ELECTRICITY:

let, lit, cry, try, celt, leet, eel, reel, rely, rite, tire, lee, city, electric, elite, tile, tele, yet, lice,

I would find many more, but must run for computer theory class now. I leave it open yet. I have not even started on big words.

REST
ATE
RATS
RAT
TARE
TEAR
STARE
STAR
ARE
SEAT
SAT
RATE
RATES
TASER
SEAR
EAR
EARS
that was a good word w/good letters! :D

so or to soy pot toy rot port sort sport

That's it???

dyslexia

bwahahaha!

[quote="modeler"]day dye dale yes six sled slide sly seal sale lead led lad lax is aid

i intentionally leave out one word :D[/quote]
:roll:

Next word: ASTRONOMY.

ok. the s word and as. That means that he left out two words.

It's probably sex, the other word for gender. Though he left out as as well, this was the one he [i]intentionally[/i] left out.

A S T R O N O M Y

rot, not, nor, torn, ton, Troy, Ron, Roy, Tom, my, may, nay, toy, say, soot, toon, art, tar, root, moot, moon, mart, mat, tan, tray, stay, too, moo, star, storm, story, stormy

and many plurals.

how about:
FLOCCINAUCINIHILIFILIFICATION

[quote="Koda"]how about:
FLOCCINAUCINIHILIFILIFICATION[/quote]

It's Floccinaucinihili[b]p[/b]ilification. You put an F. :harhar:

float
talc
cat
fat
rat
my head hurts...

STETHOSCOPE


[hide] Since Triangle ABC is congruent to Triangle A'B'C', we can say that Angle C equals Angle C'.  Since Angle C'=3x-40 and Angle C=2x-10, we have that 3x-40=2x-10.  Solving for x, we get that x=30.  Now, plug 30 into x for the equation 2x-10, so we get (2)(30)-10=50, which is Angle C.  [/hide]

[quote="Interval"]Triangle ABC is congruent to triangle A'B'C'.   If angle C is represented by 2x - 10
and angle C' is represented by 3x - 40,
find the measure of angle C.[/quote]

[hide="solution"]$2x-10=3x-40$

$x=30$

$2\cdot 30-10=\boxed{50^\circ}$[/hide]


6!+7!+8!= (n)(6!) what is n?  :blush:

im new here so hi everyone!

thanks forte I really get it now! thank you too fantasy!


algebra
function
domain
logarithms
complex analysis
complex analysis unsolved

I know how to show that -Log(z) = Log(1/z) (i.e. the principal value). How do I find a non-principal branch of log(z) such that -log(z) = log(1/z) is not satisfied somewhere in the domain of analyticity of log(z)? I can't really find any information on non-prinicpal branches on the internet. For arg(z) + 2pi*k for some k as an integer, -log(z) will equal log(1/z). Do I take k as a non-integer? Bascially, how do I define a non-principal branch in this case?

You add a multiple of $ 2\pi i$, and/or move the branch cut. On any (open, connected) region in which $ \log z$ and $ \log \frac1z$ are both defined, $ \log z\plus{}\log\frac1z$ is a constant. That constant is zero with the principal branch, and often nonzero with any other choice. If we change things only by putting the branch cut on the positive imaginary axis, we get $ \log z\plus{}\log\frac1z\equal{}\minus{}2\pi i$ on the left half-plane.


function geometry blogs calculus integration logarithms

[color=darkblue][size=150]Consider the serial binary operations: +, . , and power. What is the next operation? Are you interested to know and research about this interesting topic?  If yes go to  http://en.wikipedia.org/wiki/Ultra_exponential_function. [/size][/color]

See [url=http://en.wikipedia.org/wiki/Hyper_operator]hyper operator[/url], [url=http://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation]Knuth's up-arrow notation[/url], [url=http://en.wikipedia.org/wiki/Conway_chained_arrow_notation]Conway chained arrow notation[/url] and the [url=http://en.wikipedia.org/wiki/Ackermann_function]Ackermann function[/url].  How familiar are you with other work in this area?

Based on his blog I get the impression that he's writing his graduate thesis on something to do with extending these functions to the reals instead of just integers.

Wikipedia discourages original research.  That article appears to have been written just by you.  Please remove it.

[quote="JSteinhardt"]Hm is it still considered original research if he has published papers in the area?  His work appears in a journal (although I don't have access to more than the abstract).[/quote]

Dear friends,

A new paper, related to ultra power and ultra exponential function (unique extension of tetration),  has been published in the journal
Integral Transforms and Special Functions. In it, another new function "[b]Ultra power part function[/b]"  and the dual of Uxp  namely "[b]Infra logarithm function (Iog)[/b]" ,that  for a>1 is its inverse,  have been introduced. You can see its abstract in

http://www.informaworld.com/smpp/content~content=a901619070~db=all?alerttype=ifirst_author_alert,email&jumptype=alert

[color=brown]M.H. Hooshmand, August 2008, "Infra logarithm and ultra power part functions", Integral Transforms and Special Functions, Vol. 19, No. 7, 497-507.[/color]


Given $ a, b, c > 0$. Prove that: $ \frac{a^4}{b}\plus{}\frac{b^4}{c}\plus{}\frac{c^4}{a} \ge 3. \sqrt{\frac{a^6\plus{}b^6\plus{}c^6}{3}}$

A similar and well-known result: $ \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a} \ge 3. \sqrt[6]{\frac{a^6\plus{}b^6\plus{}c^6}{3}}$  :)

I think we can multiply by $ (a\plus{}b\plus{}c)$ use am-gm on the left side , and holder on the right one , I'm lazy now to do computations , perhaps later :wink:  ( sorry if this is a spam )

[quote="anas"]I think we can multiply by $ (a \plus{} b \plus{} c)$ use am-gm on the left side , and holder on the right one , I'm lazy now to do computations , perhaps later :wink:  ( sorry if this is a spam )[/quote]
Your hint is for the first or the second, anas?
The second is very stronger ineq and still unsolved  :maybe:

Another result: With $ a, b, c > 0$ we also have:
$ \frac{  a^5}{b}\plus{} \frac{  b^5}{c}\plus{} \frac{  c^5}{a} \ge 3. \left(\frac{a^6\plus{}b^6\plus{}c^6}{3}\right)^{\frac{2}{3}}$  :)

[quote="nguoivn"]
The second is very stronger ineq and still unsolved  :maybe:[/quote]

http://www.mathlinks.ro/viewtopic.php?t=184602

[quote="shaaam"]Hi. Maybe, we should use the following inequality of Arqady?

[url]http://www.mathlinks.ro/viewtopic.php?t=296853[/url]

which is still unsolved.[/quote]
Dear Shaam, indeed, it's strong but also doesn't help us to prove the second ineq  :maybe:
You should try with the easier  :) :
[quote="nguoivn"]Another result: With $ a, b, c > 0$ we also have:
$ \frac { a^5}{b} \plus{} \frac { b^5}{c} \plus{} \frac { c^5}{a} \ge 3. \left(\frac {a^6 \plus{} b^6 \plus{} c^6}{3}\right)^{\frac {2}{3}}$  :)[/quote]

dear brother , I discovered a fatal error in my proof  , I'm really ashamed from u :blush:  , how ever  I will try later :wink:

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \frac {a^4}{b} \plus{} \frac {b^4}{c} \plus{} \frac {c^4}{a} \ge 3. \sqrt {\frac {a^6 \plus{} b^6 \plus{} c^6}{3}}$

[/quote]

It is equivalent to : $ a^2b^{10} \plus{} 2ab^6c^5 \plus{} 2a^6b^5c \plus{} b^2c^{10} \plus{} 2bc^6a^5 \plus{} c^2a^{10} \minus{} 3a^8b^2c^2 \minus{} 3a^2b^8c^2 \minus{} 3a^2b^2c^8\geq 0$

It is true :)

Done !

Quách Thị Hồng Dung

Any another idea for these ineqs ?
[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \frac {a^4}{b} \plus{} \frac {b^4}{c} \plus{} \frac {c^4}{a} \ge 3. \sqrt {\frac {a^6 \plus{} b^6 \plus{} c^6}{3}}$

A similar and well-known result: $ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \ge 3. \sqrt [6]{\frac {a^6 \plus{} b^6 \plus{} c^6}{3}}$  [/quote]
I just have proof for this one  :maybe: :
[quote="nguoivn"]: With $ a, b, c > 0$ we also have:
$ \frac { a^5}{b} \plus{} \frac { b^5}{c} \plus{} \frac { c^5}{a} \ge 3. \left(\frac {a^6 \plus{} b^6 \plus{} c^6}{3}\right)^{\frac {2}{3}}$  :)[/quote]


inequalities
vector
quadratics
function
inequalities unsolved

Can anyone prove these inequalities.
1. for any real number a,b,c : |a + b| + |b + c| + |c + a| <= |a| + |b| + |c| + |a + b + c|
2. for counting number n : (n + 1) (n - 1)/2 <= n!
3. for all real number a1,a2, ... ,  an > 0 and S = a1 + a2 + ... + an : (1 + a1)(1 + a2)...(1 + an) <= 1 + S + S 2 /2! + ... + S n/n!

Yes. It's very hard for me. It's may be too simple for you. :D

I " new for somethat IMO style problem.Can anyone help me.?  :lol:

1) it was posted some time ago for vectors in the plane. it is easy to reduce it to your problem by the following theorem:
let $A_1, ..., A_n$, $B_1, ..., B_n$ be vectors. assume that for all lines through the origin, the projections of $A_1, ..., A_n$, $B_1, ..., B_n$(call them $a_1, ..., a_n$, $b_1, ..., b_n$) satisfy $|a_1|+...+|a_n|\geq |b_1|+...+|b_n|$. then $|A_1|+...+|A_n|\geq |B_1|+...+|B_n|$.

it follows from parametrizing the lines by an angle $\alpha$ and then integrating the inequality from 0 to $2\pi$.

your inequality just follows from considering thousands of cases...rather unexciting.

2) it follows from $(n-k)(2+k)=2n-2k+kn-k^2=(n+1)+(-k^2+k(n-2)+(n-1))\geq n+1$ for $0\leq k\leq n-1$ since the second term is a quadratic function in $k$, thus is concave(the leading coefficient is negative), thus the minimum is assumed at $0$ or $n-1$, where the inequality is true.

3) multiply out and group terms of equal degree. seperating the inequality into inequalities for terms of each degree, we see that each inequality is true, as you may trivially see if you know muirhead(it's not even sharp).

Peter

[quote="phongthong"]1. for any real number a,b,c : |a + b| + |b + c| + |c + a| <= |a| + |b| + |c| + |a + b + c|[/quote]

You can also prove 1. by applying Popoviciu's inequality to the convex function f(x) = |x|. What you get is

$\left\vert x\right\vert +\left\vert y\right\vert +\left\vert z\right\vert
+3\left\vert \frac{x+y+z}{3}\right\vert \geq 2\left\vert \frac{x+y}{2}%
\right\vert +2\left\vert \frac{y+z}{2}\right\vert +2\left\vert \frac{z+x}{2}%
\right\vert $,

i. e.,

$\left\vert x\right\vert +\left\vert y\right\vert +\left\vert z\right\vert
+\left\vert x+y+z\right\vert \geq \left\vert x+y\right\vert +\left\vert
y+z\right\vert +\left\vert z+x\right\vert $.

This is exactly your inequality.

You can find a discussion of Popoviciu's inequality in http://www.mathlinks.ro/Forum/viewtopic.php?t=430 .

And yes, the similar assertion about vectors was discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=16600 .

  Darij

I was too lazy to write solutions. :blush:

Yo. ! many thanks for your solution and good link. !!!!  :cool:

Peter Scholze and darij grinberg. I hope 2 of you can prove Riemann Hypothesis and Navier - Stokes Equations soon  :D . (Is it possible .????)

But for problem 2,3. Can anyone help me. ??

[quote="phongthong"]3. for all real number a1,a2, ... ,  an > 0 and S = a1 + a2 + ... + an : (1 + a1)(1 + a2)...(1 + an) <= 1 + S + S 2 /2! + ... + S n/n![/quote]

The problem 3) is trivial:

If we have a set of numbers, the [i]product[/i] of the set will simply mean the product of all numbers in the set.

Now, in order to prove the inequality

  $\left(1+a_1\right)\left(1+a_2\right)...\left(1+a_n\right) \leq 1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+...+\frac{S^n}{n!}$

for nonnegative reals $a_1$, $a_2$, ..., $a_n$, expand the left hand side, $\left(1+a_1\right)\left(1+a_2\right)...\left(1+a_n\right)$. What you get is the sum of the products of all possible subsets of the set {$a_1$; $a_2$; ...; $a_n$}. Now, for every j with $0\leq j\leq n$, consider only those subsets which have j elements. The sum of the products of these j-element subsets will be called $T_j$ (hereby, we set $T_0=1$). Then, clearly, the left hand side equals

  $\left(1+a_1\right)\left(1+a_2\right)...\left(1+a_n\right) = T_0+T_1+T_2+...+T_n$.

On the other hand, the right hand side of the inequality is

  $1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+...+\frac{S^n}{n!} = \frac{S^0}{0!}+\frac{S^1}{1!}+\frac{S^2}{2!}+...+\frac{S^n}{n!}$.

Hence, in order to prove the inequality, it is enough to show that $T_j \leq \frac{S^j}{j!}$ for every j with $0\leq j\leq n$. But this can be shown as follows: For j = 0, we have $T_0=1$ and $\frac{S^0}{0!}=1$, so this is clear. For $j\geq 1$, we proceed as follows: If one expands $S^j=\left(a_1+a_2+...+a_n\right)^j$, one gets a sum which contains (among other things) the products of all j-element subsets of the set {$a_1$; $a_2$; ...; $a_n$}, and each of these products j! times, and hence, $S^j$ is greater or equal to j! times the sum of these products, so that $\frac{S^j}{j!}$ is greater or equal to the sum of these products, i. e. to the sum $T_j$. And we are done.

This was the most unsharp inequality I have ever seen, by the way. For $n\geq 2$, equality holds only if the numbers $a_1$, $a_2$, ..., $a_n$ are all 0. However, I understand what it is useful for: In the case $a_1=a_2=...=a_n=\frac{1}{n}$, it yields $\left(1+\frac{1}{n}\right)^n \leq 1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$, what can be probably used to show some properties of e.

  Darij


Prove that
$\frac{3\sqrt{3}}{4}\leq \frac{yz}{x(1+yz)}+\frac{xz}{y(1+xz)}+\frac{xy}{z(1+xy)}\leq \frac{x+y+z}{4}$
where x;y;z are positive real numbers satisfying the condition x+y+z=xyz


Hey, they [i]FINALLY[/i] got the MathCounts website redone! I really think it's cool. Now, you have to have a username and password to use the site. The problem is, I can't find anywhere to register! Does anyone kno how you can get to the registration part of it?

http://www.mathcounts.org

Fiend, your link is wrong, its mathcounts.org not .com :P

I don't see anything.  Am I just looking at a cached copy?

Can someone post a link to the 'students' area?  Maybe the main page is just funny.

It doesn't work for me either...maybe they temporarily took the new site down? When I go there I just get the old site.

It's up again.  And it looks [i]nice[/i].  (I still think they should have just surrendered to the AoPS Forum though...)

Finally!  After about 6 months of waiting...

Hmm, pretty cool, although I could've sworn I've seen that format somewhere before. Also, pretty important.

[color=green][b]stickificated![/b][/color]

Hm, the site doesn't work for me.  It's probably down again.

yeah. it's not working for me either. they must still be working out glitches and stuff, cuz it worked fine this morning.

the stretches are [i]very[/i] boring...that's what we're doin now