group theory
abstract algebra
graph theory
superior algebra
superior algebra theorems

Can someone help me to find all non-isomorphic cubic Cayley graphs of the alternating group A_5
(or just a way how to find them)

There is only one according to mathworld.  I think they mean un-colored, un-directed.

If you keep track of color and direction, I think there are 2 or 3.  The generating sets are

{ (12)(34), (235) }, { (12)(34), (12345) }, and { (12)(34), (13245) }, where I'm not sure if the last two are isomorphic at the (colored, directed) graph level.  The three generating sets are in different orbits of the automorphism group of the group at least.

How I find those three generating sets? Where is the logic?
thank you for your help

Consider the set X={ {x,y} : x,y in A5 and <x,y>=A5 and xx=1 }.  This is the set of all generating sets giving cubical Cayley graphs.  An automorphism f of A5 induces a permutation on X taking {x,y} to {f(x),f(y)} and these two generating sets produce isomorphic (directed, colored) Cayley graphs.  The automorphism group of A5 acts on X, and you are only interested in (at most) one generating set {x,y} from each orbit.  There are three orbits, and I gave one generating set from each.

1. you gave one generating set from each orbit. but does there exist an other generating set from the same orbit?

2. how I know that there are just 3 orbits?

3. you said <x,y>=A_5. Is there an easier way to check that x and y are generating A_5 or I have to multiply all elements? (How I see which are those x and y that are generating A_5?)

tnx

BTW I omitted the cubical Cayley graphs whose corresponding generating set is a set of three involutions.  There are six orbits of those, of sizes 120, 60, 60, 60, 60, and 20.  At least five distinct colored (directed or undirected does not matter, as all generators are involutions) cubical Cayley graphs are produced this way.

So that is at least seven and at most nine such colored, directed cubical Cayley graphs for A5.

EDIT:  And apparently all nine of them are distinct, even omitting color and direction.

[quote="tinaa"]1. you gave one generating set from each orbit. but does there exist an other generating set from the same orbit?[/quote]

Yes, of course.  Each orbit has 120 generating sets in it.

[quote="tinaa"]2. how I know that there are just 3 orbits?[/quote]

It is a small finite set.  Just count.

[quote="tinaa"]3. you said <x,y>=A_5. Is there an easier way to check that x and y are generating A_5 or I have to multiply all elements? (How I see which are those x and y that are generating A_5?)[/quote]

There are many ways to check.  A5 is not exactly large.  If a subgroup contains x of order 2 and y of order greater than 2, then y has odd order, and so the subgroup they generate is either S3, D10, A4, or A5.  In particular, if x does not normalize <y>, then you have either A4 or A5, and if {x,y} moves all 5 points, then it cannot be A4.

EDIT (continued): I misinterpreted Foster's Census in my first reply (which was quoted too briefly by mathworld, but I did look at the "original" and did not notice the problem).  Only one of the nine graphs has an automorphism group that is transitive on the set { (h,hx) : h in A5 } ∪ { (h,hy) : h in A5 } ∪ { (hy,h) : h in A5 }, and so only one is listed in Foster's census.  That set is not very natural from my point of view, but is natural from a graph theory point of view.  It is the set of directed edges induced by the undirected, uncolored Cayley graph.

The other eight have automorphism groups that act transitively on the set { {h,hx} : h in A5 } ∪ { {h,hy} : h in A5 } of undirected edges of the undirected, uncolored Cayley graph.

At any rate, orbit representatives of the 3-involution generating sets are:

* { (2,3)(4,5), (2,4)(3,5), (1,2)(4,5) },
* { (2,3)(4,5), (2,4)(3,5), (1,2)(3,4) },
* { (2,3)(4,5), (1,2)(4,5), (1,2)(3,4) },
* { (2,3)(4,5), (1,2)(4,5), (1,3)(2,4) },
* { (2,3)(4,5), (1,2)(4,5), (1,4)(3,5) },
* { (2,3)(4,5), (1,2)(3,4), (1,4)(2,5) }

I used a full computation of the graph automorphism group to distinguish the fourth and fifth, but the others are easily distinguished by invariants.

I do not understand anything any more!

I thinkg that A5 is not such a small group. It has 60 elements. I tried to found which 3 elements are generating A5 but when I started to calculate... there are too much posibilities.
(I take one transposition and an element and its inverse, I choose them in this way that I have all numbers from 1 to 5 and then I start calculating...crazy :lol: )

Do you know if there exist a program which can tell you if some elements are generting a group?

thank you

Take for instance X={(1,2)(3,4),(1,2,3)}.  This cannot generate A5 because no element moves 5.

Take for instance X={(1,4)(2,3),(5,1,2,3,4)}.  This cannot generate A5 because (1,4)(2,3) normalizes <(5,1,2,3,4)>, and so the generated subgroup is contained in a dihedral group on 5 points.

Take for instance X={(1,2)(3,4),(3,4,5)}.  This cannot generate A5 because (1,2)(3,4) normalizes <(3,4,5)>, and so the generated subgroup is contained in a copy of the symmetric group on 3 points.

Take for instance X={(1,2)(3,4),(2,4,5)}.  Then (1,2)(3,4) does not normalize <(2,4,5)> so the subgroup generated must be A4 or A5, yet it moves all 5 points, so it must be A5.

Take for instance X={(1,2)(3,4),(1,2,3,4,5)}.  Then (1,2)(3,4) does not normalize <(1,2,3,4,5)> so the subgroup generated must be A4 or A5, yet it moves all 5 points, so it must be A5.

Roughly speaking, I have now handled every generating set (one has to be careful with the 2 conjugacy classes of 5-elements, and be careful to calculate conjugacy in C_A5(<(1,2)(3,4)>), but these are just details).

The GAP computer algebra system can answer such questions fairly easily:

gap> a5 := AlternatingGroup(5);;
gap> gens := Filtered( Combinations(AsSet(a5),2), gen -> 2 in List(gen,Order) and Group(gen)=a5 );;
gap> orbs := Orbits( SymmetricGroup(5), gens, OnSets );;
gap> List( orbs, orb -> orb[1] );
[ [ (3,4,5), (1,3)(2,4) ], [ (2,3)(4,5), (1,2,3,4,5) ], [ (2,3)(4,5), (1,2,4,3,5) ] ]
gap> gen3 := Filtered( Combinations(AsSet(a5),3), gen -> [2,2,2] = List(gen,Order) and Group(gen)=a5 );;
gap> orb3 := Orbits( SymmetricGroup(5), gen3, OnSets );;
gap> List( orb3, orb -> orb[1] );
[ [ (2,3)(4,5), (2,4)(3,5), (1,2)(4,5) ], [ (2,3)(4,5), (2,4)(3,5), (1,2)(3,4) ],
  [ (2,3)(4,5), (1,2)(4,5), (1,2)(3,4) ], [ (2,3)(4,5), (1,2)(4,5), (1,3)(2,4) ],
  [ (2,3)(4,5), (1,2)(4,5), (1,4)(3,5) ], [ (2,3)(4,5), (1,2)(3,4), (1,4)(2,5) ] ]
gap> LoadPackage("grape");;
gap> graphs:=List( Concatenation(orbs,orb3), orb -> CayleyGraph(a5,orb[1]) );;
gap> ForAny(Combinations(graphs,2), xy->IsIsomorphicGraph(xy[1],xy[2]));
false
gap> List(graphs,g->Size(OrbitsDomain(AutomorphismGroup(g),DirectedEdges(g),OnTuples)));
[ 2, 2, 2, 3, 2, 2, 2, 2, 1 ]

This shows that all 9 Cayley graphs are pairwise non-isomorphic as undirected, uncolored graphs, but that only the last one is "symmetric" as in its automorphism group acts transitively on its arcs (directed versions of its undirected edges) and hence only one appears in Foster's census.  The commands take a few seconds to run, total.

GAP is available from http://www.gap-system.org/

[quote="tinaa"]what that mean: (14)(23) normalizes <(51234)>?
(how I check this)? because I have never heard about normalizing ...

thanks[/quote]

If g is an element of a group G and H is a subgroup of G, then "g normalizes H" if and only if gH = Hg, where gH = { gh : h in H } and Hg = { hg : h in H }.  This means that the subgroup generated by g and H can be described as { g^i * h : h in H, i in Z }, and so if g has order n and H has order k, then <g,H> has order dividing n*k.

Ideas like this are introduced fairly early in a study of group theory.  They are in many abstract algebra textbooks.  I think understanding Cayley graphs will be very difficult without understanding generating sets, so it might be a good idea to find all subgroups of:

* the non-abelian group of order 6
* the dihedral group of order 8
* the quaternion group of order 8
* the alternating group of order 12
* the symmetric group of order 24
* the special linear group SL(2,3) of order 24

These should give you a feel for what subgroups of solvable groups look like.  Then to understand the subgroups of A5 you only need to be a little careful that maximal subgroups are smaller than one might expect (which just means you are done earlier than expected).

Ok. I have found 8 generating sets. But I have no idea how to check if I have just 3 involutions in my set.
I also found on interenet one of these graphs called buckyball of the generating set <(1 2 3 4 5),(2 3)(4 5)> :oops:

thank you once again


There are 5 points on a circle. Through the center of gravities of each 3 points, draw a lines which are perpendicular to the lines which pass through remaining two points. Prove that those 10 lines are concur.

Position the center $ O$ of the circle at the origin, and let the five points be $ A_1,A_2,A_3,A_4,A_5$ (synonymously denote by these their respective vectors from origin); then we claim that each line passes through the point $ X = \frac 13 \sum A_i$. Indeed if $ \sigma$ denotes a permutation $ \{1,2,3,4,5\}$, then \begin{align*}\left(X - \frac{A_{\sigma(1)}+A_{\sigma(2)}+A_{\sigma(3)}}{3}\right) \cdot \left(A_{\sigma(4)} - A_{\sigma(5)}\right) &= -\frac 13 (A_{\sigma(4)} + A_{\sigma(5)}) \cdot (A_{\sigma(4)} - A_{\sigma(5)})\\ &= - \frac 13 \left(\| A_{\sigma(4)} \| - \| A_{\sigma(5)} \|\right)\\ &= 0.\end{align*}


analytic geometry
Pythagorean Theorem
geometry
Putnam
college contests

A particle moves in the Euclidean plane. At time t (taking all real values) its coordinates are $x = t^3 - t$ and $y = t^4 + t$ . Show that its velocity has a maximum at $t = 0$, and that its path has an inflection at $t = 0$.

$v(t)_{x-direction} = \frac{dx}{dt} = 3t^2 -1$, and similarly, $v(t)_{y-direction} = 4t^3 + 1$.  Using the pythagorean theorem, we thus have $v(t) = \sqrt{v(t)_x^2+v(t)_y^2} = \sqrt{9t^4 - 6t^2 + 1 + 16t^6 + 8t^3 + 1} = \sqrt{16t^6 + 9t^4 +8t^3 - 6t^2 + 2}$.

For convenience, let $P(t) = v(t)^2$.  Then $v'(t) = \frac{1}{2}P(t)^{-\frac{1}{2}}(96t^5+36t^3+24t^2-12t)$, which clearly has a zero at $t = 0$, meaning its a critical point of v(t).  Its easy enough to check that its a maximum using a monotonicity plot.

To calculate the inflection point, we have, from the above compuations, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t^3+1}{3t^2-1}$.  Thus, $\frac{d^2y}{dx^2} = \frac{12t^2}{3t^2-1} - \frac{6t(4t^3+1)}{(3t^2-1)^2}$, which is 0 @ $t=0$


For each positive integer $ m > 1$, let $ P(m)$ denote the greatest prime factor of $ m$. For how many positive integers $ n$ is it true that both $ P(n) \equal{} \sqrt{n}$ and $ P(n \plus{} 48) \equal{} \sqrt{n \plus{} 48}$?

$ \textbf{(A)}\ 0\qquad
\textbf{(B)}\ 1\qquad
\textbf{(C)}\ 3\qquad
\textbf{(D)}\ 4\qquad
\textbf{(E)}\ 5$

If the greatest prime factor of $ n$ is $ \sqrt{n}$, $ n$ must be the square of a prime number. In this case we want to find the number of sets of two primes, whose squares differ by 48.

Calling $ x$ and $ y$ the larger and smaller primes respectively, we want to find the prime integer solutions to $ (x\plus{}y)(x\minus{}y) \equal{} 48$. By testing a few cases we see that only $ (13,11)$ works, so there is one integer $ n$, namely $ 121$.

[quote=sudhakarr2287]yes, but so does 1 and 7.[/quote]

1 isn't prime


The number 121 is a palindrome, because it reads the same
backwards as forward. How many integer palindromes are
between 100 and 500?

This is so annoying!
Why don't they count
111
222
333
444
????
I've missed problems like this a bout 5 times now.
 :mad:

[color=red]^Lol you newb[/color]

It's easier to just say that there are 4 choices for the first digit and 10 for the second so 40 palindromes between 100 and 500.

[hide]There are 4 choices for the hundreds place and 10 choices for the tens place. 4*10=40.[/hide]


Suppose we push the floor with a force $ F$. We use energy, but we don't do work (because the floor doesn't move). Where does that energy go?

Increasing of internal energy of both you and the floor.

Well, what you are doing is applying a pressure, both to your hands and to the floor. As you push, the atoms in your hands and on the floor a compressed closer together. The electron clouds on the atoms repel each other, so work is done against these repulsive electrostatic forces when the atoms are squashed together. So the work done by the muscles is converted to electrostatic potential energy.


function
trigonometry
topology
real analysis
real analysis unsolved

let (X,d) be a metric space and let E be subset of X.call E path connected iff.
for any x,y in E there exists continuous function f from [0,1] to E satisfying
f(0)=x and f(1)=y.show every connected set is path connected but converse
isnt true. :)

You got it exactly backwards. :) I personally prefer $\{re^{i/(1-r)}\colon 0\le r<1\}\cup\{z\colon |z|=1\}$ to the topologist's sine curve.


Find $ x, y \in N$ of this equation: $ x^3 \minus{} y^3 \equal{} xy \plus{} 61$

it's a well known problem from romania.

Let $ x,y$ be positive integers for which $ x^3\minus{}y^3\equal{}xy\plus{}61$

$ (3x)^3\plus{}(\minus{}3y)^3 \plus{}(\minus{}1)^3 \minus{} 3(3x)(\minus{}3y)(\minus{}1)\equal{}1646$.

$ (3x\minus{}3y\minus{}1)(9x^2\plus{}9y^2\plus{}1\plus{}9x\plus{}3x\minus{}3y)\equal{}2\cdot823$.

the rest is easy.

Nice solution. Here is another one using bounds:
[hide]
Clearly $ x^3 \minus{} y^3 \equal{} xy \plus{} 61 \ge 61$. So, $ x > y$, and thus, $ y \le x\minus{}1$. Also, $ x \ge 4$.

Thus: $ x^3 \minus{} y^3 \ge x^3 \minus{} (x\minus{}1)^3 \equal{} 3x^2\minus{}3x\plus{}1$.

And: $ xy\plus{}61 \le x(x\minus{}1)\plus{}61 \equal{} x^2 \minus{} x \plus{} 61$.

Therefore, we must have:
$ 3x^2\minus{}3x\plus{}1 \le x^2 \minus{} x \plus{} 61$

$ 2x^2 \minus{} 2x \minus{} 60 \equal{}  2(x\minus{}6)(x\plus{}5) \le 0$

$ \minus{}5 \le x \le 6$

Thus, the only possible values of $ x$ are $ x\equal{}4$, $ x\equal{}5$, and $ x\equal{}6$.

$ x\equal{}4$ yields: $ 64\minus{}y^3 \equal{} 4y\plus{}61$, which has no integer solution for $ y$.

$ x\equal{}5$ yields: $ 125\minus{}y^3 \equal{} 5y\plus{}61$, which has no integer solution for $ y$.

$ x\equal{}6$ yields: $ 216\minus{}y^3 \equal{} 6y\plus{}61$, which has only one integer solution: $ y\equal{}5$.

Therefore, the only solution is $ (x,y) \equal{} (6,5)$
[/hide]


Given that $ ln(cis\theta)\equal{}i\theta$, evaluate $ e\cdot(cis(\minus{}i)\minus{}cis(i))$
A) $ 0$
B) $ 1\minus{}e$
C) $ e\minus{}\frac{1}{e}$
D) $ e^{2}\minus{}1$
E) NOTA=None of the Above

$ \operatorname{cis}(\minus{} i) \equal{} e^{i\cdot \minus{} i} \equal{} e$

$ \operatorname{cis}(i) \equal{} e^{i\cdot i} \equal{} \frac1e$

$ e\left(e \minus{} \frac1e\right) \equal{} e^2 \minus{} 1\implies\boxed D$

What's cis?
Sorry for my ignorance, but...

$ \text{cis} \theta \equal{} \cos\theta \plus{} i\sin\theta$. It should be noted that $ e^{i\theta}\equal{}\text{cis}\theta$, which I often find a lot easier to use.


geometry calculus integration MATHCOUNTS inequalities

[quote="nutz_for2.718281828"]IT'S ANNOYING!!!!!!!!!! you may not use calculators, even though it's tempting.

Equilateral triangle ABC with side lenght 2 has sectors of circles of radius 1 at the corners of the triangle. Find the area of the region inside the triangle but outside the sectors? Express your[hide="answer"]as a decimal to the nearest hundredth.[/hide]

The hide is part of the question.[/quote]

[hide]This is just $\sqrt{3}-\frac{\pi}{2}$.

This is just $1.73-1.57 \approx 0.16$

$0.16$[/hide]

[quote="nutz_for2.718281828"]IT'S ANNOYING!!!!!!!!!! you may not use calculators, even though it's tempting.

Equilateral triangle ABC with side lenght 2 has sectors of circles of radius 1 at the corners of the triangle. Find the area of the region inside the triangle but outside the sectors? Express your[hide="answer"]as a decimal to the nearest hundredth.[/hide]

The hide is part of the question.[/quote]
Every nerd should know pi to at least 5 digits, but if you don't then you can approximate it in the middle of the contest. As for the sqrt3, that isn't that bad either.

[quote="Ubemaya"]how do you use binomial on non integral powers?[/quote]
It's messy and in retrospect I shouldn't have mentioned it because it's beyond the scope of mathcounts. I edited what I mentioned out. Anyways, I was suggested by a coach of mine that for mathcounts one should memorize sqrt 2, 3, 5, 6, 7, 10 to a few decimals just in case.

Or, if you really like multiplication, you can get $\sqrt{3}$ in the following way:
It's between $\sqrt{1}=1$ and $\sqrt{4}=2$, so it's between 1 and 2
$(1.5)^{2}=2.25<3$, so $1.5<\sqrt{3}$

Then do like $1.8^{2}$ and get some more inequalities, etc.


algebra
system of equations
algebra unsolved

Given integer $ n$ larger than $ 5$, solve the system of equations (assuming $x_i \geq 0$, for $ i=1,2, \dots n$):
\[ \begin{cases} \displaystyle x_1+ \phantom{2^2} x_2+ \phantom{3^2} x_3 + \cdots + \phantom{n^2} x_n &= n+2, \\ x_1 + 2\phantom{^2}x_2 + 3\phantom{^2}x_3 + \cdots + n\phantom{^2}x_n &= 2n+2, \\ x_1 + 2^2x_2 + 3^2 x_3 + \cdots + n^2x_n &= n^2 + n +4, \\ x_1+ 2^3x_2 + 3^3x_3+ \cdots + n^3x_n &= n^3 + n + 8. \end{cases} \]

I suppose the system is supposed to have $ n$ equations, and the last one is
$ x_1\plus{}2^{n\minus{}1}x_2\plus{}3^{n\minus{}1}x_3\plus{}\cdots\plus{}n^{n\minus{}1}x_n\equal{}n\plus{}2^{n\minus{}1}\plus{}n^{n\minus{}1}$

There's a simple solution
[hide]$ (x_1,x_2,x_3,\cdots,x_n)\equal{}(n,1,0,\cdots,1)$[/hide]

Now try and find a reason why there's no other solution
[hide]The determinant in Cramer's rule happen's to be a non-zero Vandermonde determinant.[/hide]

[quote="lordWings"]I suppose the system is supposed to have $ n$ equations, and the last one is
$ x_1 \plus{} 2^{n \minus{} 1}x_2 \plus{} 3^{n \minus{} 1}x_3 \plus{} \cdots \plus{} n^{n \minus{} 1}x_n \equal{} n \plus{} 2^{n \minus{} 1} \plus{} n^{n \minus{} 1}$
[/quote]

There are only 4 equations for this problem in the official book. So I don't understand what you mean.

denote $ a_i = \sum_{k = 3, k\neq i}^{n - 1}x_ k$, $ b_i = \sum_{k = 3, k\neq i}^{n - 1}kx_ k$, $ c_i = \sum_{k = 3, k\neq i}^{n - 1}k^2x_ k$, $ d_i\sum_{k = 3, k\neq i}^{n - 1}k^3x_ k$... and consider the systems (for all $ 3\leq i\leq n - 1$)

$  \{\begin{array}{c} \displaystyle x_1 + x_2 + x_i + x_n = n + 2 - a_i \\
 \\
x_1 + 2x_2 + ix_i + nx_n = 2n + 2 - b_i \\
 \\
x_1 + 2^2x_2 + i^2 x_i + n^2x_n = n^2 + n + 4 - c_i \\
 \\
x_1 + 2^3x_2 + i^3x_i + n^3x_n = n^3 + n + 8 - d_i \end{array} $

after solving this system we have that $ x_i = \dfrac{n( - 2a_i + 3b_i - c_i) - ( - 2b_i + 3c_i - d_i)}{(n - i)(i - 1)(i - 2)}$, for all $ 3\leq i\leq n - 1$... so we have that

$ n( - 2a_i + 3b_i - c_i) - ( - 2b_i + 3c_i - d_i)\geq 0$...

this implies that $ \sum_{k = 3, k\neq i}^{n - 1}x_k( - 2n + 3kn - k^2n + 2k - 3k^2 + k^3) = \sum_{k = 3, k\neq i}^{n - 1}(k - 1)(n - k)(2 - k)x_k\geq 0$

clearly this implies that $ x_k = 0$, since $ (k - 1)(n - k)(2 - k) < 0$ for all $ 3\leq k\leq n - 1$... taking two distinct indexes, say $ i = 3, i = 4$, it follows that $ x_3 = x_4 = ... = x_{n - 1} = 0$...

from this it's easy to conclude that $ x_1 = n, x_2 = 1, x_n = 1$, and we're done  :D

We have $(n-2)(n^2+n+4)<n^3+n+8$. Therefore, $x_{n-1}+x_n>0$. But if $x_{n-1}$ or $x_n$ are greater than $1$, we immediately arrive at a contradiction. Thus, there are two cases: If $x_n=1$, it follows that $x_1=n$ and $x_2=1$ (subtract the second equation from the first). If $x_{n-1}=1$, it follows that $x_3=1$, $x_1=n$ or $x_2=2$, $x_1=n-1$. Both cases clearly lead to a contradiction. Thus, $(x_1$, $\dots$, $x_n)=(n$, $1$, $0$, $\dots$, $0$, $1)$. :)

I just used Gaussian elimination method to get the solution.


What is the number of units in the distance from the origin to the point $ (\minus{}8,6)$ in a coordinate system?


ratio
search
limit
geometry
3D geometry
tetrahedron
calculus

OK, the Golden Ratio:

$\frac{\sqrt{5}+1}{2}$

I heard it has a bunch of special properties. The only one I think might be true is that the recipricol of the golden ratio is the golden ratio minus 1. *right?*

Are there any other special properties?

well, if you search "golden ratio" under Google, then you could get some pretty detailed explanations.

Yup, for example : http://en.wikipedia.org/wiki/Golden_ratio

[quote="tetrahedr0n"]You could try evaluating:

$1+\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}$

and see what happens.

Let $F_n$ be the $n$th Fibonacci number. Then

$\displaystyle{\lim_{n \rightarrow \infty} F_n/F_{n+1}} = $?[/quote]

I learned in Algebra II that the ratios of succesive Fibonacci terms come closer and closer towards the Golden Ratio, correct?

not appropriate topic in this forum...

might want to change this to getting Started, or MC

[quote="Drunken_Math"]
I learned in Algebra II that the ratios of succesive Fibonacci terms come closer and closer towards the Golden Ratio, correct?[/quote]
Correct. :)

[quote="tetrahedr0n"]$\displaystyle{\lim_{n \rightarrow \infty} F_n/F_{n+1}} = $?[/quote]

Isn't it $F_n/F_{n-1}$?

[quote="Drunken_Math"]I learned in Algebra II that the ratios of succesive Fibonacci terms come closer and closer towards the Golden Ratio, correct?[/quote]

Is that so? Oh yeah, that's what tetrahedron was talking about.

welcome to pre-calculus, math92  :rotfl:

In simpler terms, what tetrahedron wrote just means that the ratio of consecutive fibonacci numbers gets closer to the golden ratio as you use larger fibonacci numbers.

I thought someone just said something about imperial conversions, where did that go?

[quote="math92"][quote="tetrahedr0n"]
Let $F_n$ be the $n$th Fibonacci number. Then
$\displaystyle{\lim_{n \rightarrow \infty} F_n/F_{n+1}} = $?[/quote]
this is too hard for classroom math and i don't even understand it. :?[/quote]

Not really mostly just the notation. F_n is the Nth fibbonacci. Lim_infinity just means as the number approaches infinity.

Cool thing, the metric-imperial conversion factor is very close to $\phi$ therefor you can convert a imperial number into metric if one is a fibbonacci.

Example:

55 mph approx. 89 kph

There's also the connection to geometry....

If you divide a recctangle 's sides in some special way related to golden ratios(I don't remember) , you keep getting beautiful squares


Hey, simple question...

How do we find square roots of purely complex numbers such as -6i

Thanks  :)

I undestand it, don't worry :D

instead of doing a^2 - b^2 = 0
I was doing a^2 + b^2 = 0

:D

[hide="Simple way that only works for square roots, basically"]
$-6i = (a+bi)^2$
$-6i = a^2 - b^2 + 2abi$
$a^2 = b^2$
$-6 = 2ab$
$ab = -3$
$a = \pm \sqrt{3}, b = \mp \sqrt{3}$
$\sqrt{-6i} = \pm \sqrt{3} \mp i \sqrt{3}$
[/hide]

[hide="General way that works for nth roots"]
Euler's Formula:

$e^{ix} = \cos x + i \sin x$
$e^{inx} = \cos nx + i \sin nx = (\cos x + i \sin x)^n$

Set $-6i = re^{ix}$, and because $|-6i| = 6$ we set $r = 6$.

$\cos x + i \sin x = -i$
$\sin x = -1, \cos x = 0$
$x = \frac{3\pi}{2} + 2\pik$

$\sqrt{-i} = \cos \frac{x}{2} + i \sin \frac{x}{2} = \cos \frac{3\pi + 4\pik}{4} + i \sin \frac{3\pi + 4\pik}{4} = - \frac{ \sqrt{2}}{2} + i\frac{ \sqrt{2}}{2}, \frac{ \sqrt{2}}{2} - i \frac{ \sqrt{2} }{2}$

$\sqrt{-6i} = \sqrt{6} \sqrt{-i} = \pm \sqrt{3} \mp i \sqrt{3}$ as before, but this is a powerfully general method that will work for cubic roots, quartic roots...
[/hide]


Let $a, b, c$ three non-negative real numbers. prove that the following inequality holds:

\[ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq 1+\frac{3(a^{3}+b^{3}+c^{3})}{2(a+b+c)(a^{2}+b^{2}+c^{2})}.  \]

[i] Cezar Lupu & Ductrung [/i]

[quote="Cezar Lupu"]Let $a, b, c$ three non-negative real numbers. prove that the following inequality holds:

\[ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq 1+\frac{3(a^{3}+b^{3}+c^{3})}{2(a+b+c)(a^{2}+b^{2}+c^{2})}.  \][/quote]

L-R=$\frac{(a^3+b^3+c^3-2abc)( \sum (a+b)(a-b)^2) }{2(a+b)(b+c)(c+a)(a+b+c)(a^2+b^2+c^2)} \geq 0$

[quote="urun"][quote="ductrung"]I composed a similar problem sometime ago:
Find the maximal real number $k$ for which
\[ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}+k\left( \frac{3(a^3+b^3+c^3)}{(a+b+c)(ab+bc+ca)}-1\right) \]
for all positive real numbers $a,b$ and $c$.[/quote]

Very insight thought. Let
\[A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \]
\[B= \frac{3(a^3+b^3+c^3)}{(a+b+c)(ab+bc+ca)}-1 \]I prove that
\[\boxed{\frac{3}{2}+\left(\frac72-\sqrt{10}\right) B\geqslant A \geqslant \ \frac{3}{2}+\frac{1}{4} \cdot B }\]
is the best inequalities of your type, although I used a powerful method to prove these results :blush[/quote]
游老师：

Nice！

$E(A)=E(B)$

BQ


trigonometry
parameterization
calculus
calculus computations

SOlve the following differential equation:

$y^{''}+3y^{'}+2y=e^{x}\left(x^{2}+1\right)\sin2x+3e^{-x}\cos x+4e^{x}$

:)

i suppose that one could use the method of unsolved coefficients but that would get really ugly very fast. I suggest using the Laplace transform. For the initial conditions just use $y(0)=y_{0}$ and $y'(0)=y_{1}$ I haven't actually tried to solve this yet but when i do that is the first thing i would try.

Hint: use variation of parameters. It should work because I have done so.

yes that works a lot better. wow i actually forgot about that.

I like variation of parameters better than the method of undetermined coefficients because the later is more like a guess work.


modular arithmetic
number theory
relatively prime
algebra
polynomial
Rational Root Theorem

[quote="shaun ong"]x^3+x+1= 0. prove that x is not a rational number. What does rational number means and whats the solution pls?[/quote]

rational number means it can be expressed as a/b where a and b are relatively prime integers
[hide]
assume $ x\equal{}\frac{a}{b}$
then $ \frac{a^3}{b^3}\plus{}\frac{a}{b}\equal{}\minus{}1$
$ \frac{a^3\plus{}ab^2}{b^3}\equal{}\minus{}1$
$ a^3\plus{}ab^2\plus{}b^3\equal{}0$

someone finish pls.
[/hide]

[u]Note:[/u] It's better if you use $ x \equal{} \minus{} \frac {a}{b} \; : \; a, b \in \mathbb{N}$

[hide="Solution"]$ x^3 \plus{} x \plus{} 1 \equal{} 0 \iff a^3 \plus{} ab^2 \equal{} b^3 \implies b^3 \equal{} ka, \; a^3 \equal{} hb^2$

Cubing the second and using the first gives $ a^9 \equal{} h^3 k^2 a \implies a^8 \equal{} h^3 k^2$

Hence, $ h^3 \equal{} \frac {a^8}{k^2} \equal{} \frac {a^{10}}{b^6}$. However, we know that $ h \equal{} \frac {a^3}{b^2} \iff h^3 \equal{} \frac {a^9}{b^6}$. Dividing the two gives $ a \equal{} 1$.

$ \minus{} \frac {1}{b^3} \minus{} \frac {1}{b} \plus{} 1 \equal{} 0 \iff b^3 \equal{} b^2 \plus{} 1$, which is impossible for a positive integer.[/hide]

[quote="CircleSquared"][quote="shaun ong"]x^3+x+1= 0. prove that x is not a rational number. What does rational number means and whats the solution pls?[/quote]

rational number means it can be expressed as a/b where a and b are relatively prime integers
[hide]
assume $ x \equal{} \frac {a}{b}$
then $ \frac {a^3}{b^3} \plus{} \frac {a}{b} \equal{} \minus{} 1$
$ \frac {a^3 \plus{} ab^2}{b^3} \equal{} \minus{} 1$
$ a^3 \plus{} ab^2 \plus{} b^3 \equal{} 0$

someone finish pls.
[/hide][/quote]
[hide="Completion"]
\[ a^3 \equal{} \minus{} b^3 \minus{} ab^2
\]
\[ a^3 \equal{} b^2( \minus{} b \minus{} a) \equal{} \minus{} b^2(a \plus{} b)
\]
Since $ a$ and $ b$ are relatively prime, we see that $ b^2$ and $ a^3$ are relatively prime, so $ b^2$ cannot be a factor of $ a^3$ unless $ b \equal{} 1$.  In this case, we see that $ a(a^2 \plus{} 1) \equal{} a^3 \plus{} a \equal{} \minus{} 1$.  Thus, $ a \equal{} \pm 1\implies a^2 \plus{} 1 \equal{} 2$, which is a contradiction since that means that $ |(a^2\plus{}1)(a)|\equal{}2$ when $ (a^2\plus{}1)a\equal{}\minus{}1$.  [/hide]

with mods
[hide]
Using the previous notation for relatively prime $ a,b$:
$ a^3 \plus{} ab^2 \plus{} b^3 \equal{} 0$
$ a^3 \plus{} ab^2 \plus{} b^3\equiv 0 \pmod b$
$ a^3 \equiv 0\pmod b$
$ b|a$. Contradiction.
[/hide]

$ a^3 \equiv 0 \bmod b$ does [i]not[/i] imply that $ b | a$.  What it does imply, together with the assumption that $ \gcd(a, b) \equal{} 1$, is that $ b \equal{} 1$, which leads to [b]The Zuton Force's[/b] solution.

[hide="We could always just use..."] The [url=http://en.wikipedia.org/wiki/Rational_roots_theorem]rational root theorem[/url].  ;) [/hide]

I am unsure, as I have not worked out many problems using the Rational Root theorem, but could someone help?

[hide="Proof"]The rational root theorem states that all possible rational roots can be listed using $ \pm\dfrac{p}{q}$, where $ p$ is an integer factor of constant term $ a_0$, and $ q$ is an integer factor of leading coefficient $ a_n$.

This means that the total number of rational roots for the equation is $ \pm\dfrac{1}{1}$, so there are two rational roots that are possible - 1 and -1.

Plugging in 1 does not work, as $ 1^3\plus{}1\plus{}1\neq 0$. Plugging in $ \minus{}1$ does not work either, as $ (\minus{}1)^3\plus{}(\minus{}1)\plus{}1\neq 0$. Therefore, there are no rational roots.
[/hide]

Correct.  The proof of the Rational Root Theorem is essentially a generalization of the specific arguments used here.  :)


function
algebra
polynomial
algebra unsolved

Let $F$ be the set of all polynomials $\Gamma$ such that all the coefficients of $\Gamma (x)$ are integers and $\Gamma (x) = 1$ has integer roots. Given a positive intger $k$, find the smallest integer $m(k) > 1$ such that there exist $\Gamma \in F$ for which $\Gamma (x) = m(k)$ has exactly $k$ distinct integer roots.

Since $\Gamma (x)=m$ has $k$ distinct integer roots,we have $\Gamma (x)= (x-x_1)(x-x_2)....(x-x_k)Q(x)+m$ for distinct integers $x_1,x_2,...,x_k$ and $Q \in Z[X]$
Let $t$ be a solution of $\Gamma (x)=1$.Then
$(t-x_1)(t-x_2)...(t-x_k)Q(x)=1-m$
=> $|1-m|=|t-x_1||t-x_2||t-x_3|....|t-x_k||Q(x)|$
=> $|1-m| \geq 1.2.3...k.1$
=> $m-1 \geq k!$
=>$m \geq k!+1$
Thus,the minimum value of $m$ is $k!+1$.
The equality occurs when $\Gamma (x) = (x-1)(x-2)...(x-k+1)(x+k) + k!+1$ .

the moduli $|t-x_i|$ need not be distinct. for example to prove $m(3)=3$ consider the polynomial $X^3+2X^2-X+1$

Indeed, the answer is $1 + \left( \left \lfloor \frac{k}{2} \right \rfloor \right)! \left( \left \lceil \frac{k}{2} \right \rceil \right)!$. To show that this is optimal do the same proof as [b] KDS [/b] did above, with the minor fix that at most two of the $|t-x_i|$'s can be equal to each other. To construct this, consider the polynomial $\Gamma(x) = 1 + \left( \left \lfloor \frac{k}{2} \right \rfloor \right)! \left( \left \lceil \frac{k}{2} \right \rceil \right)! + (-1)^{\left \lfloor \frac{k+2}{2} \right \rfloor}(x-1)(x-2) \cdots \left (x- \left \lfloor \frac{k}{2} \right \rfloor \right) (x+1) (x+2) \cdots \left( x + \left \lceil \frac{k}{2} \right \rceil \right).$ Then $\Gamma(x) = 1$ has the root $x = 0$, and $\Gamma(x) = 1 + \left( \left \lfloor \frac{k}{2} \right \rfloor \right)! \left( \left \lceil \frac{k}{2} \right \rceil \right)!$ has  exactly $k$ distinct integer roots.

$\square$


function
real analysis
real analysis unsolved

f:R^n to R^m is function under which the image of any path-connected set is path-connected and the image of any closed and bounded set is closed and bounded. Show that f is continuous

PS: I dont think this is that tough to be put in this section but my friend needs it quite urgently and I dont have enough time to do, so people, plz help out. Thanks a lot

Suppose $ f(0)\equal{}0$ and $ f$ is discontinuous at $ 0$. Take a sequence of points $ x_k$ tending to $ 0$ such that $ |f(x_k)|>\delta$. Consider the intervals $ [0,x_j]$. Choose points $ \xi_k$ on those intervals with $ |f(\xi_k)|\equal{}\frac{k}{k\plus{}1}\delta$. Consider the image of the set $ \{0\}\cup\{\xi_j: j\ge 1\}$. You should be able to finish from here.

Finally, if your friend needs it for preparing to the upcoming test or something like that, give him the full solution but if it is a part of his take-home exam due tomorrow, just give him a slight hint. We should do our best to help people study mathematics but we are not here to help people cheat on their exams :).


inequalities
calculus
integration
function
vector
linear algebra
matrix

This is about cyclic fractional inequalities like the one from http://www.mathlinks.ro/Forum/viewtopic.php?t=83217

Let $x,y,z>0$. For $a,b,c\ge0$ define $\boxed{\{a,b,c\}: =\sum_{cyc}\frac1{ax+by+cz}}$, e.g. $\{4,2,1\}=\frac1{4x+2y+z}+\frac1{4y+2z+x}+\frac1{4z+2x+y}$. So the ineq from the above link $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+ 4\left(\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\right)\geq 12\left(\frac{1}{3x+y}+\frac{1}{3y+z}+\frac{1}{3z+x}\right)$ could be written as $\{1,0,0\}+4\{1,1,0\}\ge12\{3,1,0\}$ or as well as $\{12,0,0\}+\{3,3,0\}\ge\{3,1,0\}$. Or in a "homogenous" way as $\{4,0,0\}+2\{2,2,0\}\ge3\{3,1,0\}$.
Now, we don't have a Muirhead-like theorem for [i]cyclic [/i]sums $\sum_{cyc}x^ay^bz^c$, even where it holds for symmetric sums. But the Harazi type integration sort of "levels out" the values where it holds and where it fails, and "integrating" a false inequality may yield a true one. I dare to suggest the following [b]conjectures[/b]:
If $a\ge b\ge c$, then prove (or disprove :) ):[b]
(1) for all $0\le t\le a-b$, we have $\{a,b,c\}\ge\{a-t,b,c+t\}$.
(2) for all $0\le t\le b-c$, we have $\{a,b,c\}\ge\{a,b-t,c+t\}$.[/b]

There seems to be strong numerical evidence, but I have no clue how to tackle that - unless the cases of course where the corresponding cyclic power sums yield a true ineq: for example $\{4,2,0\}\ge\{3,2,1\}$ which follows by Harazi's method from $\sum_{cyc}x^4y^2\ge\sum_{cyc}x^3y^2z$ which in turn can be proven by AM-GM.

This conjecture seems quite deep (well, unless somebody comes up with a simple counterexample :oops: ), and maybe it is possible to find a new method for proving it. Certainly not by expanding :blush: Maybe Karamata can be tweaked to prove it?
[hide]Even if this conjecture is not sure, at least one thing is sure: JESUS CHRIST IS RISEN! Happy Easter everybody :clap: [/hide]

Well, actually it seems to me the following more general conjecture might hold:
Let $x_1,...,x_k>0.$ For a partition $\alpha=(a_1,...,a_k)\vdash n$ define the function $\{\alpha\}=\sum_{cyc}\frac1{a_1x_1+...+a_kx_k}$.
Note that this depends on the order of the $a_i$.
[b]Conjecture: If $\alpha, \beta\vdash n$ and $\alpha\succ\beta$, then $\{\alpha\}\ge\{\beta\}$.[/b] (Note: if the cyclic sums are replaced by symmetrical ones, the corresponding inequality follows from Muirhead by Harazi's method.)
Does anybody fully understand the proof of the original Muirhead theorem using double stochastic matrices? I have the feeling that it might be possible by that same method and some (lots of?) linear algebra to prove the above conjecture. This would yield a wonderful result. :roll:

For 3 variables, we know that $\sum_{cyc}a^{u}b^{v}\geq \sum_{cyc}a^{p}b^{q}c^{r}$ if and only if $pu+qv \geq uv, qu+rv \geq uv, ru+pv \geq uv$. (There was a discussion on this: http://www.mathlinks.ro/Forum/viewtopic.php?t=103761 )

And we know that if an inequality holds, then the integrated form (consists of fractions) of inequality holds too. (For some examples, see http://www.mathlinks.ro/Forum/viewtopic.php?t=105062 )

By these two results, we obtain: if $pu+qv \geq uv, qu+rv \geq uv, ru+pv \geq uv$, then $\sum_{cyc}a^{u}b^{v}\geq \sum_{cyc}a^{p}b^{q}c^{r}$ holds for any positive reals $a,b,c$. Applying $a=t^{x-\frac{1}{u+v}}$, $b=t^{y-\frac{1}{u+v}}$, and $c=t^{z-frac{1}{u+v}}$, we get \[f(t)=t^{ux+vy-1}+t^{uy+vz-1}+t^{uz+vx-1}-t^{px+qy+rz-1}-t^{qx+ry+pz-1}-t^{rx+py+qz-1}\geq 0\] for all $t>0$. This means $\int_{0}^{1}f(t) \geq 0$ or \[\frac{1}{ux+vy}+\frac{1}{uy+vz}+\frac{1}{uz+vx}\geq \frac{1}{px+qy+rz}+\frac{1}{py+qz+rx}+\frac{1}{pz+qx+ry}\] for all $x,y,z>0$.

So the integration can be a way to attack this problem. However, by this method we cannot get necessary condition, as integrating method does not satisfy both direction. (it is also discussed in http://www.mathlinks.ro/Forum/viewtopic.php?t=105062 ) Moreover, we do not know yet whether $w$ is minimum among $u,v,w,p,q,r$ in $\sum \frac{1}{ux+vy+wz}\geq \sum \frac{1}{px+qy+rz}$.

Oh, I was idiot. You already knew this method.  :blush:

I don't know if it can be useful, but I found a method which can be applied to some special cases of the conjecture.

Let's define some notations. We will numerate entries of vectors and matrices modulo $n$, i.e. using numbers $0,\ldots,n-1$ and we will identify $\left[\vec{u}\right]_{i}$ with $\left[\vec{u}\right]_{i+n}$ for any vector $\vec{u}$ and $\left[A\right]_{i,j}$ with $\left[A\right]_{i+n,j}$ and $\left[A\right]_{i,j+n}$. For any $n$-dimensional vector $\vec{u}$ we define $\mathbf{M}\left(\vec{u}\right)$ as a matrix of dimension $n\times n$ satisfying $\left[\mathbf{M}\left(\vec{u}\right)\right]_{i,j}=\left[\vec{u}\right]_{j-i}$. All sums will be evaluated from $0$ to $n-1$.
Assume we have two vectors: $\vec{p}=\left[\begin{array}{c}p_{0}\\ \vdots\\ p_{n-1}\end{array}\right]$ and $\vec{q}=\left[\begin{array}{c}q_{0}\\ \vdots\\ q_{n-1}\end{array}\right]$ such that $p_{i},q_{i}\geqslant 0$ for any $i$ and $\vec{p}\succ\vec{q}$. Then let the variables in the inequality be represented by a vector $\vec{x}=\left[\begin{array}{c}x_{0}\\ \vdots\\ x_{n-1}\end{array}\right]$ such that $x_{i}>0$ for any $i$. The inequality $\left\{\vec{p}\right\}\geqslant \left\{\vec{q}\right\}$ can be written as
\[\sum_{i}\frac{1}{p_{0}x_{i}+p_{1}x_{i+1}+\ldots+p_{n-1}x_{i+n-1}}\geqslant \sum_{i}\frac{1}{q_{0}x_{i}+q_{1}x_{i+1}+\ldots+q_{n-1}x_{i+n-1}}. \]
Let's notice that $p_{0}x_{i}+p_{1}x_{i+1}+\ldots+p_{n-1}x_{i+n-1}= \sum_{j}p_{j}x_{i+j}=$ $\sum_{j}p_{j-i}x_{j}= \sum_{j}\left[\mathbf{M}\left(\vec{p}\right)\right]_{i,j}\left[\vec{x}\right]_{j}= \left[\mathbf{M}\left(\vec{p}\right)\cdot\vec{x}\right]_{i}$. We can transform in such way all constituents of both sides of the ineqaulity:
\[\sum_{i}\frac{1}{\left[\mathbf{M}\left(\vec{p}\right)\cdot\vec{x}\right]_{i}}\geqslant \sum_{i}\frac{1}{\left[\mathbf{M}\left(\vec{q}\right)\cdot\vec{x}\right]_{i}}. \]
Of course all numbers $\left[\mathbf{M}\left(\vec{p}\right)\cdot\vec{x}\right]_{i}$, $\left[\mathbf{M}\left(\vec{q}\right)\cdot\vec{x}\right]_{i}$ are positive and the function $f(t)=\frac{1}{t}$ is convex on $\mathbb{R}^{+}$, so by Karamata inequality it is sufficient to prove that $\mathbf{M}\left(\vec{p}\right)\cdot\vec{x}\succ \mathbf{M}\left(\vec{q}\right)\cdot\vec{x}$. Using the doubly stochastic matrix criterion of majorization, this condition is equivalent to fact that for any vector $\vec{x}$ we can find a doubly stochastic matrix $\mathbf{A}$ such that $\mathbf{M}\left(\vec{q}\right)\cdot\vec{x}= \mathbf{A}\cdot \mathbf{M}\left(\vec{p}\right)\cdot\vec{x}$. We can take a very special case: matrix $\mathbf{A}$ doesn't depend on the vector $\vec{x}$. Then it sufficies to prove that there exists such doubly stochastic matrix $\mathbf{A}$ that $\mathbf{M}\left(\vec{q}\right)= \mathbf{A}\cdot \mathbf{M}\left(\vec{p}\right)$ or simply the matrix $\mathbf{M}\left(\vec{q}\right)\cdot \mathbf{M}\left(\vec{p}\right)^{-1}$ is doubly stochastic. Of course there are many vetors $\vec{p}$, $\vec{q}$ such that $\left\{\vec{p}\right\}\geqslant \left\{\vec{q}\right\}$ holds and $\mathbf{M}\left(\vec{q}\right)\cdot \mathbf{M}\left(\vec{p}\right)^{-1}$ isn't doubly stochastic (even if $\mathbf{M}\left(\vec{p}\right)$ is invertible), but this criterion allows to prove some special cases. For example, taking $\vec{p}=\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]$ and $\vec{q}=\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$ with $a+b+c=2$ we get $\mathbf{M}\left(\vec{q}\right)\cdot \mathbf{M}\left(\vec{p}\right)^{-1}= \mathbf{M}\left(\left[\begin{array}{c}\frac{a+b-c}{2}\\ \frac{a+c-b}{2}\\ \frac{b+c-a}{2}\end{array}\right]\right)$. It is obvious that sum of entries in any column or row of this matrix is $1$ and fact that all entries are nonnegative is equivalent to fact that $a,b,c$ are sides of a triangle. We proved:

[i]If $a+b+c=2$ and $a,b,c$ are sides of a triangle, then $\{1,1,0\}\geqslant \{a,b,c\}$.[/i]

I wonder if we can find some weaker conditions for statement that $\mathbf{M}\left(\vec{p}\right)\cdot \vec{x}\succ \mathbf{M}\left(\vec{q}\right)\cdot \vec{x}$ for any $\vec{x}$. Unfortunately, it isn't true for any $\vec{p}$, $\vec{q}$ such that $\vec{p}\succ\vec{q}$, because if we assume that if it is true, we can prove Muirhead for cyclic sums, which is false (this gives a counterexample for $\mathbf{M}\left(\vec{p}\right)\cdot \vec{x}\succ \mathbf{M}\left(\vec{q}\right)\cdot \vec{x}$: we take $\vec{p}= \left[\begin{array}{c}4\\3\\0\end{array}\right]$, $\vec{q}= \left[\begin{array}{c}4\\1\\2\end{array}\right]$ and $\vec{x}= \left[\begin{array}{c}\ln 6\\0\\\ln 2\end{array}\right]$; then $\mathbf{M}\left(\vec{p}\right)\cdot \vec{x}\not\succ \mathbf{M}\left(\vec{q}\right)\cdot \vec{x}$, but the inequality holds).

Looks very nice.  :yup:  :yup: I would indeed expect it to be useful :)
 [quote="Rzeszut"]For example, taking $\vec{p}=\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]$ and $\vec{q}=\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$ with $a+b+c=2$ we get $\mathbf{M}\left(\vec{q}\right)\cdot \mathbf{M}\left(\vec{p}\right)^{-1}= \mathbf{M}\left(\left[\begin{array}{c}\frac{a+b-c}{2}\\ \frac{a+c-b}{2}\\ \frac{b+c-a}{2}\end{array}\right]\right)$. [/quote]
So it might be worth trying some other special cases. Is it possible that we can always write $\mathbf{M}\left(\vec{q}\right)\cdot \mathbf{M}\left(\vec{p}\right)^{-1}= \mathbf{M}\left(\vec{r}\right)$ for a suitable $\vec r$, and then check if the conditions for $r_{i}\ge0$ are fulfilled if $\vec{p}\succ\vec{q}.$ Just an idea - maybe a dream - . I don't have much time right now.

[quote="Rzeszut"]Using the doubly stochastic matrix criterion of majorization, this condition is equivalent to fact that for any vector $\vec{x}$ we can find a doubly stochastic matrix $\mathbf{A}$ such that $\mathbf{M}\left(\vec{q}\right)\cdot\vec{x}= \mathbf{A}\cdot \mathbf{M}\left(\vec{q}\right)\cdot\vec{x}$. We can take a very special case: matrix $\mathbf{A}$ doesn't depend on the vector $\vec{x}$. Then it sufficies to prove that there exists such doubly stochastic matrix $\mathbf{A}$ that $\mathbf{M}\left(\vec{q}\right)= \mathbf{A}\cdot \mathbf{M}\left(\vec{q}\right)$ ...[/quote]
Little typo here. Shouldn't it be twice be $\mathbf M(\vec{p})$ instead of $\mathbf M(\vec{q})$ on the RHS (?)

the final remark is really intriguing... :huh:

[quote="spanferkel"][quote="Rzeszut"]Using the doubly stochastic matrix criterion of majorization, this condition is equivalent to fact that for any vector $\vec{x}$ we can find a doubly stochastic matrix $\mathbf{A}$ such that $\mathbf{M}\left(\vec{q}\right)\cdot\vec{x}= \mathbf{A}\cdot \mathbf{M}\left(\vec{q}\right)\cdot\vec{x}$. We can take a very special case: matrix $\mathbf{A}$ doesn't depend on the vector $\vec{x}$. Then it sufficies to prove that there exists such doubly stochastic matrix $\mathbf{A}$ that $\mathbf{M}\left(\vec{q}\right)= \mathbf{A}\cdot \mathbf{M}\left(\vec{q}\right)$ ...[/quote]
Little typo here. Shouldn't it be twice be $\mathbf M(\vec{p})$ instead of $\mathbf M(\vec{q})$ on the RHS (?)[/quote]
Thank you for noticing; I corrected the typo.
[quote="spanferkel"][quote="Rzeszut"]For example, taking $\vec{p}=\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]$ and $\vec{q}=\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$ with $a+b+c=2$ we get $\mathbf{M}\left(\vec{q}\right)\cdot \mathbf{M}\left(\vec{p}\right)^{-1}= \mathbf{M}\left(\left[\begin{array}{c}\frac{a+b-c}{2}\\ \frac{a+c-b}{2}\\ \frac{b+c-a}{2}\end{array}\right]\right)$. [/quote]
So it might be worth trying some other special cases. Is it possible that we can always write $\mathbf{M}\left(\vec{q}\right)\cdot \mathbf{M}\left(\vec{p}\right)^{-1}= \mathbf{M}\left(\vec{r}\right)$ for a suitable $\vec r$, and then check if the conditions for $r_{i}\ge0$ are fulfilled if $\vec{p}\succ\vec{q}.$ Just an idea - maybe a dream - . I don't have much time right now.[/quote]
It is possible, however the explicit form of $\vec{r}$ doesn't look very nice (at least for me): $\vec{r}= \mathbf{V}^{-1}\cdot\left[\begin{array}{c}\frac{\left[\mathbf{V}\cdot\vec{q}\right]_{0}}{\left[\mathbf{V}\cdot\vec{p}\right]_{0}}\\ \vdots \\ \frac{\left[\mathbf{V}\cdot\vec{q}\right]_{n-1}}{\left[\mathbf{V}\cdot\vec{p}\right]_{n-1}}\end{array}\right]$, where $\mathbf{V}$ is a $n\times n$ matrix satisfying $\left[\mathbf{V}\right]_{k,l}= \mathrm{e}^{\frac{2\pi i}{n}\cdot k\cdot l}$ for any $k,l$; it is easy to see that $\left[\mathbf{V}^{-1}\right]_{k,l}= \frac{1}{n}\cdot \mathrm{e}^{-\frac{2\pi i}{n}\cdot k\cdot l}$. I have no idea how to check if the conditions for $r_{i}$ are fullfilled.

[quote="Rzeszut"] I have no idea how to check if the conditions for $r_{i}$ are fullfilled.[/quote]
Well, I've calculated it for $n=3$. Suppose $\vec{p}=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ and $\vec{q}=\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$. Let $\omega: =\mathrm{e}^{\frac{2\pi i}{n}}$. Then by using the identity
\[\frac1{x+y+z}+\frac1{x+\omega y+\omega^{2}z}+\frac1{x+\omega^{2}y+\omega z}= 3\frac{x^{2}-yz}{x^{3}+y^{3}+z^{3}-3xyz},\]
we find

\[\vec{r}= \frac13\mathbf{V}\cdot\left[\begin{array}{c}\frac{a+b+c}{x+y+z}\\ \frac{a+\omega b+\omega^{2}c}{x+\omega y+\omega^{2}z}\\ \frac{a+\omega^{2}b+\omega c}{x+\omega^{2}y+\omega z}\end{array}\right]=\frac{1}{x^{3}+y^{3}+z^{3}-3xyz}\cdot\left[\begin{array}{c}a(x^{2}-yz)+b(y^{2}-zx)+c(z^{2}-xy)\\ c(x^{2}-yz)+...\qquad\qquad\qquad\qquad\quad \\ b(x^{2}-yz)+...\qquad\qquad\qquad\qquad\quad\end{array}\right].\] [i]Déjà vu...[/i] the components of $\vec{r}$ are all three $\ge0$ iff the ineq can be proved by AM-GM which is the case iff the [i]cyclical sum[/i] $(x,y,z)$ majorizes $(a,b,c)$ (see http://www.mathlinks.ro/Forum/viewtopic.php?p=586225#586225). :wallbash:
I would guess the situation is similar for $n\ge4$.

So this approach, which was [quote="Rzeszut"] a very special case: matrix $\mathbf{A}$ doesn't depend on the vector $\vec{x}$,[/quote]doesn't bring us any further than we are already :( .

Well guys,

this thread is going to ressemble more and more [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=103761]Xevarion's one[/url]. :yankchain: [i]I just found the conjecture is wrong. [/i]For instance, $\{5,1,0\}\succ \{3,3,0\}$ is wrong for the numbers taking values (5,1,0)... :wallbash:
It remains to [b]conjecture that $\{\alpha\}\ge\{\beta\}$ holds iff the [i]cyclical sums[/i] satisfy $(\alpha)\ge(\beta)$ [/b][size=75](as in the other thread!!)[/size][b] iff the problem can be solved by AM-GM[/b]. Which would mean that after all, the Harazi integration [i]doesn't[/i] increase the "range of validity" as I had thought in the initial post.  :mad:

[quote="spanferkel"]For instance, $\{5,1,0\}\succ \{3,3,0\}$ is wrong for the numbers taking values (5,1,0)... [/quote]
For this counterexample we have $(5,1,0) >(3,3,0)$.
On the other hand, the inequality holds $\{5,5,0\}\succ \{6,2,2\}$, but  $(5,5,0)\geq(6,2,2)$ is not true. :roll:
  The proposed inequality is:

$\sum \frac 1{3a+b+c}\leq \frac 2{5}\sum\frac 1{b+c}$.

[quote="Vasc"]
On the other hand, the inequality holds $\{5,5,0\}\succ \{6,2,2\}$, but  $(5,5,0)\geq(6,2,2)$ is not true. :roll:
  The proposed inequality is:

$\sum \frac 1{3a+b+c}\leq \frac 2{5}\sum\frac 1{b+c}$.[/quote]
Thank you for this example.

So it does happen that the fractional inequality holds where the corresponding one for cyclical sums is false. :)

Looking at it again after a long time, inspired by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=452394]this special case[/url]:
[quote="spanferkel in post#1"] [b]conjecture[/b]:
If $a\ge b\ge c$, then prove (or disprove :) ):[b]
(1) for all $0\le t\le a-b$, we have $\{a,b,c\}\ge\{a-t,b,c+t\}$.[/b]
[/quote]In fact, this one can be proved by Harazi integration because the inequality of the corresponding cyclic sums holds. ;)
To establish the latter, by the method given [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=586225#p586225]here[/url], it suffices to show that the three scalar products of $\vec v :=\left[\begin{array}{c}a^2-bc\\ b^2-ca\\ c^2-ab\end{array}\right] $ with $\vec v_1:=\left[\begin{array}{c}a-t\\ b\\ c+t\end{array}\right]  ,\vec v_2:=\left[\begin{array}{c}c+t\\ a-t\\ b\end{array}\right] ,\vec v_3:=\left[\begin{array}{c}b\\ c+t\\a-t\end{array}\right]$ are non negative.

But we have \[\vec v\vec v_1=(\sum a^3)-3abc-t(a-c)(a+b+c)\\ \\ =(a+b+c)\left(\frac{\sum(a-b)^2}2-t(a-c)\right)\\ \\ \mathop\ge \limits^{\text{remember }t\le a-b }\frac12(a+b+c)\Bigl((b-c)^2+[(a-b)-(a-c)]^2\Bigr)\ge0.\]
Further, \[\vec v\vec v_2=\left[\begin{array}{c}a^2-bc\\ b^2-ca\\ c^2-ab\end{array}\right]
\left[\begin{array}{c} t\\  -t\\ 0\end{array}\right]=t(a-b)(a+b+c)\ge 0\] and \[\vec v\vec v_3= \left[\begin{array}{c}a^2-bc\\ b^2-ca\\ c^2-ab\end{array}\right]
\left[\begin{array}{c} 0\\ t\\  -t \end{array}\right]=t(b-c)(a+b+c)\ge 0.\]

Note that for the next conjecture [quote="spanferkel in post#1"][b]conjecture[/b]:
If $a\ge b\ge c$, then prove (or disprove :) ):[b]
(2) for all $0\le t\le b-c$, we have $\{a,b,c\}\ge\{a,b-t,c+t\}$.[/b]
[/quote]the same method shows that the corresponding one for the cyclical sums is [i]always [/i]false.

And I see that (2) itself is generally false as well. :huh:


inequalities
function
algebra
system of equations

[quote="lephuocdaint1987"]find x, y in R satisfies
   *) 2^x + 2^y >=4;
   *) 3^x + 3^ y <=6;
[/quote]
x=y=1

:D yeah my little toe saw that too

I guess the problem is to [b]prove[/b] that :?

no!
you read my answer for my teacher Namdung ?
the answer for my problem is similar it
sorry for my bad English

[quote="siuhochung"]6/log3 <= x+y <=4/log2
any reals satisfying this inequality are ok?[/quote]

no, this would've been correct if there was written 2^(x+y) >= 4

you are abusing the log function :?

http://www.mathlinks.ro/viewtopic.php?t=3640

The above link may lead some head way to solving this system of inequalities since the system of equations is already solved, so I'm just trying to link this topic to that other topic to see if it helps you out solving this problem. :)


ratio
arithmetic sequence
geometric sequence
algebra proposed
algebra

Let $ (a_k)_{k \ge 1}$ be an arithmetic progresion with the ratio $ r>0$ and the first term $ a_1 \ge \frac{1}{2}$. Find the integer part of the number: $ s_n\equal{}\sqrt{1\plus{}\frac{r}{a_1a_2}}\plus{}\sqrt{1\plus{}\frac{r}{a_2a_3}}\plus{}...\plus{}\sqrt{1\plus{}\frac{r}{a_{n\minus{}1}a_n}}\plus{}\sqrt{1\plus{}\frac{r}{a_n}}$, for all $ n \in \mathbb{N}, n \ge 2$.

I believe you mean geometric progression?

What is the "ratio" then?
An arithmetic progression is of the form

$ a, a\plus{}b, a\plus{}2b, a\plus{}3b, ...$

Whereas a geometric progression has a ratio, and is of the form

$ a, ar, ar^2, ar^3, ...$

where $ r$ is the ratio.

This arithmetic progression has the form:

$ a_1$

$ a_2 \equal{} a_1 \plus{} r$

$ a_3 \equal{} a_1 \plus{} 2r$

$ ..................$

$ a_n \equal{} a_1 \plus{} (n \minus{} 1)r$

where $ r$ is the [color=orange][b]common difference[/b][/color]

Moreover, $ \frac {r}{a_k a_{k \plus{} 1}}$$ \equal{} \frac {a_{k \plus{} 1} \minus{} a_k}{a_k a_{k \plus{} 1}}$

Infact you were somehow right, because "ratio" means something else, but in Romanian, the term "ratie", is used both for "ratio" and "common difference"

So, it was my mistake.


[quote="ryan22"]A wave traveling along a string is described by $y(x,t) = 0.0037 sin(72.1x-2.72t)$ in which the numerical constants are in the SI units $(0.0037 m, 72.1 rad/m,$ and $2.72 rad/s)$

A) $ym = 0.003245 m = 3.45$ B)$ym = 0.000321 m = 3.21$ C) $no solution$[/quote] $\pi$ 3.14159265358979323


How does your score this year compare to your score last year?

Subtract your score last year from your score this year, for example $+1$ would mean you did 1 better this year.

Exactly the same.

I solved 4 more problems than last year and made 4 more stupid mistakes than last year  :lol:

I solved one more problem than last year...but I should have gotten a 7 instead of a 5 :blush: .  I forgot to divide by 4 on number 4 (210 instead of 105) and overlooked problem 8 completely.

5 -> 7, but only because I didn't know much of what I know now last year.

I got a 0 last yr. 11 this time. Thought it was kind of a lame AIME with too many counting/brute force type problems. I missed the first question and got the last 5 right


You have a lot of homework one weekend. If you do one-fourth of the homework on
Friday and one-sixth of the homework on Saturday, how much of your homework is left to do on Sunday?

[hide]${\frac{1}{4} + \frac{1}{6} = \frac{6}{24} + \frac{4}{24} = \frac {10}{24} = \frac {5}{12} \text{When we subtract } \frac{5}{12} \text{ from 1, we have } \frac{7}{12}}$ There's your answre ;)[/hide]

lol
how about you do 1/3 of your homework Thursday, 1/4 of whats left Friday and 1/5 of whats left Saturday, how much is left?

[hide]He did 1/6 friday, so he did 1/2 up to then. Then, he did 1/10 on saturday, so that's a total of 3/5, leaving 2/5 left. [/hide]


induction
inequalities
function
logarithms
inequalities unsolved

[quote="harazi"]Yes, it is true, but what do you say when we have n variables?[/quote]
I think we can use induction + mixing variables (make use of the inequality valiowk mentioned)

[quote="spdf"]I think we can uses mixxing variables,let [tex] \ f(a)=\frac{1}{(1+a)^2} [/tex].We have if [tex] ab\geq\ 1 [/tex] then [tex] f(a)+f(b)\geq\ 2f(\sqrt{ab})[/tex](check it).Then you can easily see that the minimum of LHS is attained in (a,a,a,1/a^3).The rest is easy.[/quote]
are you sure about it?

[quote="Valiowk"]Vasc has a nice generalisation (somewhere on this forum) which says that
\[\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \geq \frac{4}{(1+r)^2}\] where $r = 4\sqrt{abcd} \geq 1$.  A technique rather similar to the one that siuhochung showed can be used.[/quote]
sorry.I can't understand the inequality you posted.
is there something wrong with me? :)
or there should be $r = {abcd}^\frac{1}{4}$

I think the problem follows from the following observation. Notice that the function $f(t)=\frac{1}{(1+e^t)^2}$ is convex. So
\[
\sum_{\text{cyc}}\frac{1}{(1+a)^2}=\sum_{\text{cyc}}\frac{1}{\left(1+e^{\ln a}\right)^2}\geq4\frac{1}{(1+e^{(\sum\ln a)/4})^2}=1.
\]
And it can be generalized to any number of variables.   :)

I don't think it's convex for all $t$
Notice that maybe $t<0$.

Ooops... Yes we need $t\geq0$. Thanks hungkhtn  :blush:

[quote="hungkhtn"][quote="spdf"]I think we can uses mixxing variables,let [tex] \ f(a)=\frac{1}{(1+a)^2} [/tex].We have if [tex] ab\geq\ 1 [/tex] then [tex] f(a)+f(b)\geq\ 2f(\sqrt{ab})[/tex](check it).Then you can easily see that the minimum of LHS is attained in (a,a,a,1/a^3).The rest is easy.[/quote]

It's not complete.I don't sure about this.Please Expand.To have f(ab,c,d) max if a=b=c I must prove quite long.

 :(
[quote="siuhochung"]This one is quite classical,
First,
$\frac{1}{(1+a)^2} + \frac{1}{(1+b)^2} \ge \frac{1}{1+ab}$
it's same as $1+ab(a^2+b^2) \ge a^2b^2+2ab$
which is true as $1+ab(a^2+b^2)-a^2b^2-2ab \ge 1+2a^2b^2-a^2b^2-2ab = (ab-1)^2$
then,
the given expression $\ge 1/(1+ab) + 1/(1+cd) = 1$[/quote]

Why the given expression $\ge 1/(1+ab) + 1/(1+cd) = 1$
You  haven't $ab \ge 1$ and $cd \ge 1$
 :(

I agree to Harazi.Complete the solution with $n$ variable is difficult.
 :)[/quote]

1. why do i need $ab \ge 1$ and $cd \ge 1$ ? I haven't used this or assumed this in my proof.

2. actually, what is harazi's general statement? The r.h.s. of the inequality is $\frac{n}{4}$ or $\frac{n}{(1+(x_1...x_n)^{\frac{1}{n}})^{2}}$ ?

Sorry Siuhochung.I 'm wrong.I think you say:

$f(a)+f(b)\ge f(\sqrt{ab})$

Where $f=\frac{1}{(a+1)^2}$

[quote="sjk"]You can check by differentiating $f$ twice.  :D[/quote]

$f=\frac{1}{(1+e^x)^2}$
$f'(x)=-2\frac{e^x}{(1+e^x)^4}$
$f''(x)=-2\frac{e^x(1+e^x)^4-e^x.4(1+e^x)^3e^x}{(1+e^x)^8}$

Then $f''(x) \ge 0$ if and only if
$3e^x \ge 1$

And $f$ doesn't convex for all $x$

[quote="Valiowk"]Vasc has a nice generalisation (somewhere on this forum) which says that
\[\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \geq \frac{4}{(1+r)^2}\] where $r = 4\sqrt{abcd} \geq 1$.  A technique rather similar to the one that siuhochung showed can be used.[/quote]
can somebody show the solution of this one for me??

[quote="hungkhtn"]It's proved by same way as Siuhochung.
Other hand,I think your Proof is not true,Kunny.[/quote]

can you show it?

[quote="harazi"]Well, I am not so sure. I would like very much to see a complete solution for this problem.[/quote]

Let's try the general case $a_1a_2\cdots a_n = 1$. What RHS do you suggest? 1? Evidently we need consider the near-equality case for 4 variables, $a, b, c \rightarrow +\infty$, $d \rightarrow 0$. I think the $n-1$ equal value principle should suffice. We get

\[
k\left(1+x^k\right)^2 - (1+x)^2(1+2x^k) \ge 0
\]

where $k = n-1$. Is this obvious for $k \ge 3$?

I've found a solution based on an idea from the hardest inequality I've ever seen (it really is impossible, in my opinon!). First, I'll post the original inequality by Vasc, from which I have taken the idea.

Let $a,b,c,d>0$ be real numbers for which $a^2 + b^2 + c^2 + d^2 = 1$. Prove that the following inequality holds: $(1-a)(1-b)(1-c)(1-d) \geq abcd$

I'll leave its proof to the readers. A little historical note on this problem: it was proposed in some "Gazeta Matematica Contest" in the last years and while I was still in high-school and training for mathematical olympiads, I tried to solve it on a very large number of occasions, but failed. So I think I will always remember its difficult and smart solution, which I'll leave to the readers.

Now, let us get back to our original problem:

Let $a,b,c,d>0$ with $abcd=1$. Prove that: $\displaystyle \frac{1}{(1+a)^2} + \frac{1}{(1+b)^2} + \frac{1}{(1+c)^2} + \frac{1}{(1+d)^2} \geq 1$.

Although this inequality also belongs to Vasc (he published it in the Gazeta Matematica), it surprisingly made the China 2005 TSTs, thus confirming (in my opinion) its beauty and difficulty. Now, on to the solution:

Let $\displaystyle x = \frac{1}{1+a}, y = \frac{1}{1+b}, z = \frac{1}{1+c}, t = \frac{1}{1+d}$. Then $\displaystyle abcd = 1 \Rightarrow \frac{1-x}{x} \cdot \frac{1-y}{y} \cdot \frac{1-z}{z} \cdot \frac{1-t}{t} = 1 \Rightarrow (1-x)(1-y)(1-z)(1-t) = xyzt$. We have to prove that $x^2 + y^2 + z^2 + t^2 \geq 1$. We will prove this by contradiction. Assume that $x^2 + y^2 + z^2 + t^2 < 1$.

Keeping in mind that $x^2 + y^2 + z^2 + t^2 < 1$, let us assume that $(1-x)(1-y) \leq zt$ and prove that it is not true (I'm talking about the last inequality here, which we assumed to be true). Upon multiplication with 2 and expanding, this gives: $1 - 2(x+y) + 1 + 2xy \leq 2zt$. This implies that $2zt > x^2 + y^2 + z^2 + t^2 - 2(x+y) + 1 + 2xy$. So, $2zt > (x+y)^2 - 2(x+y) + 1 + z^2 + t^2$, which implies $(z-t)^2 + (x+y-1)^2 < 0$, a contradiction. Therefore, our original assumption implies $(1-x)(1-y) > zt$. In a similar manner, it is easy to prove that $(1-z)(1-t) > xy$. Multiplying the two, we get that $(1-x)(1-y)(1-z)(1-t) > xyzt$, which is a contradiction with the original condition $abcd=1$ rewritten in terms of $x,y,z,t$.

Therefore, our original assumption was false and we indeed have $x^2 + y^2 + z^2 + t^2 \geq 1$, qed. The conclusion is that if an inequality is difficult, there are two possibilities : (1) it belongs to Vasc (2) it is a consequence of one of Vasc's inequalities. In this case, both are true :)

Your solution is correct, iandrei2! :lol:
Indeed, I obtained the problem
[i]Let $a,b,c,d>0$ be real numbers for which $a^2 + b^2 + c^2 + d^2 = 1$. Prove that the following inequality holds: $(1-a)(1-b)(1-c)(1-d) \geq abcd$[/i]
starting to my old problem
[i]Let $a,b,c,d>0$ with $abcd=1$. Prove that: $\displaystyle \frac{1}{(1+a)^2} + \frac{1}{(1+b)^2} + \frac{1}{(1+c)^2} + \frac{1}{(1+d)^2} \geq 1$.[/i]

[quote="iandrei2"] The conclusion is that if an inequality is difficult, there are two possibilities : (1) it belongs to Vasc (2) it is a consequence of one of Vasc's inequalities. In this case, both are true :)[/quote]This is a good joke. But you are not right now. ;)

[quote="Vasc"]
[quote="iandrei2"] The conclusion is that if an inequality is difficult, there are two possibilities : (1) it belongs to Vasc (2) it is a consequence of one of Vasc's inequalities. In this case, both are true :)[/quote]This is a good joke. But you are not right now. ;)[/quote]

I don't think it is a joke.  :P

[quote="iandrei2"]I've found a solution based on an idea from the hardest inequality I've ever seen (it really is impossible, in my opinon!). First, I'll post the original inequality by Vasc, from which I have taken the idea.

Let $a,b,c,d>0$ be real numbers for which $a^{2}+b^{2}+c^{2}+d^{2}= 1$. Prove that the following inequality holds: $(1-a)(1-b)(1-c)(1-d) \geq abcd$

I'll leave its proof to the readers. A little historical note on this problem: it was proposed in some "Gazeta Matematica Contest" in the last years and while I was still in high-school and training for mathematical olympiads, I tried to solve it on a very large number of occasions, but failed. So I think I will always remember its difficult and smart solution, which I'll leave to the readers.

Now, let us get back to our original problem:

Let $a,b,c,d>0$ with $abcd=1$. Prove that: $ \frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}}\geq 1$.

Although this inequality also belongs to Vasc (he published it in the Gazeta Matematica), it surprisingly made the China 2005 TSTs, thus confirming (in my opinion) its beauty and difficulty. Now, on to the solution:

Let $ x = \frac{1}{1+a}, y = \frac{1}{1+b}, z = \frac{1}{1+c}, t = \frac{1}{1+d}$. Then $ abcd = 1 \Rightarrow \frac{1-x}{x}\cdot \frac{1-y}{y}\cdot \frac{1-z}{z}\cdot \frac{1-t}{t}= 1 \Rightarrow (1-x)(1-y)(1-z)(1-t) = xyzt$. We have to prove that $x^{2}+y^{2}+z^{2}+t^{2}\geq 1$. We will prove this by contradiction. Assume that $x^{2}+y^{2}+z^{2}+t^{2}< 1$.

Keeping in mind that $x^{2}+y^{2}+z^{2}+t^{2}< 1$, let us assume that $(1-x)(1-y) \leq zt$ and prove that it is not true (I'm talking about the last inequality here, which we assumed to be true). Upon multiplication with 2 and expanding, this gives: $1-2(x+y)+1+2xy \leq 2zt$. This implies that $2zt > x^{2}+y^{2}+z^{2}+t^{2}-2(x+y)+1+2xy$. So, $2zt > (x+y)^{2}-2(x+y)+1+z^{2}+t^{2}$, which implies $(z-t)^{2}+(x+y-1)^{2}< 0$, a contradiction. Therefore, our original assumption implies $(1-x)(1-y) > zt$. In a similar manner, it is easy to prove that $(1-z)(1-t) > xy$. Multiplying the two, we get that $(1-x)(1-y)(1-z)(1-t) > xyzt$, which is a contradiction with the original condition $abcd=1$ rewritten in terms of $x,y,z,t$.

Therefore, our original assumption was false and we indeed have $x^{2}+y^{2}+z^{2}+t^{2}\geq 1$, qed. The conclusion is that if an inequality is difficult, there are two possibilities : (1) it belongs to Vasc (2) it is a consequence of one of Vasc's inequalities. In this case, both are true :)[/quote]

I am slightly confused... can you always assume that one of $x^{2}+y^{2}+z^{2}+t^{2}\geq 1$ or $x^{2}+y^{2}+z^{2}+t^{2}<1$? What if $f(x,y,z,t)$ is sometimes greater than and sometimes less than one? Or is there some theorem (cuz it seems like most expressions are either greater than or less than)


Find  all  integers  $ a$  , such that $ \frac{a^{8k}\minus{}1}{a\minus{}1}$ is square , where $ k$ is given natural number.

Write it as $ (a^{4k}\plus{}1)\cdot(a^{2k}\plus{}1)\cdot\left(\frac{a^{2k}\minus{}1}{a\minus{}1}\right)$ and consider gcds.


Please solve this problem ! NoOne can solve?


conics
algebra
difference of squares
special factorizations

Given condition should be $ a^2 \boxed{\minus{}} b^2 \equal{} c^2$

Sketch: start from the first equation, move one root to the other side, square, simplify, express the remaining root, square again, simplify and take the given condition into account.

Thank you for the correction  :)

I like to do this when there are multiple square-roots:
[hide]
We start with:
$ (I) \qquad \sqrt {(x \plus{} c)^2 \plus{} y^2} \plus{} \sqrt {(x \minus{} c)^2 \plus{} y^2} \equal{} 2a$

Subtracting twice the second term gives
$ (II) \qquad \sqrt {(x \plus{} c)^2 \plus{} y^2} \minus{} \sqrt {(x \minus{} c)^2 \plus{} y^2} \equal{} 2a \minus{} 2\sqrt {(x \minus{} c)^2 \plus{} y^2}$

Multiplying $ (I)$ with $ (II)$ produces a difference of squares:

$ (\sqrt {(x \plus{} c)^2 \plus{} y^2})^2 \minus{} (\sqrt {(x \minus{} c)^2 \plus{} y^2})^2 \equal{} 2a(2a \minus{} 2\sqrt {(x \minus{} c)^2 \plus{} y^2})$

$ (x \plus{} c)^2 \plus{} y^2 \minus{} (x \minus{} c)^2 \minus{} y^2 \equal{} 4a^2 \minus{} 4a \sqrt {(x \minus{} c)^2 \plus{} y^2}$

$ 4cx \equal{} 4a^2 \minus{} 4a \sqrt {(x \minus{} c)^2 \plus{} y^2}$

Now isolate the square-root and square:

$ \sqrt {(x \minus{} c)^2 \plus{} y^2} \equal{} a \minus{} \frac {c}{a} x$

$ (x \minus{} c)^2 \plus{} y^2 \equal{} \frac {c^2}{a^2} x^2 \minus{} 2\frac {c}{a} a x \plus{} a^2$

$ x^2 \minus{} 2cx \plus{} c^2 \plus{} y^2 \equal{} \frac {c^2}{a^2} x^2 \minus{} 2cx \plus{} a^2$

$ \frac {a^2 \minus{} c^2}{a^2} x^2 \plus{} y^2 \equal{} a^2 \minus{} c^2$

And now, we apply the other equation:

$ \frac {b^2}{a^2} x^2 \plus{} y^2 \equal{} b^2$

$ \boxed{\frac {x^2}{a^2} \plus{} \frac {y^2}{b^2} \equal{} 1}$[/hide]

It is necessary for one to go about this derivation at least once while learning about conic sections.

huh?

you are doing

$ p\plus{}q\equal{}r$

$ p\minus{}q\equal{}r\minus{}2q$
$ p^2\minus{}q^2\equal{}r^2\minus{}2rq$

You might as well have just moved over $ p\equal{}r\minus{}q$ then squared it...

You're right, i may as well have, they're identical. I just tend to do it that way because i see the LHS faster, but i guess it sounds more complicated that way.

yes its a^2 - b^2 = c^2 .  Thank you for correction


geometry
3D geometry
pigeonhole principle

The vertices of a cube are arbitrarily labeled from $ 1$ to $ 8$. Prove that there exists a three-edge path joining two diagonally opposite vertices, along which the assigned numbers(end-points included) add up to at least $ 21$.

[RMO, District Round, 2005, grade 8]

Note: You can use that [hide=" hint"] What happens when $ 7$ and $ 8$ are on the same face? But when tto opposite vertices contain are labeled with $ 7$ and $ 8$ ? [/hide]

very nice problem!  :lol:

It suffices to prove that the number of combinations of four of {1, 2,...., 8} that sum to greater than 21 is greater than 36.

but I can't understand your hint. can you please type it in proper grammar.  :blush:

Of course.

[u][b]Hint[/b][/u]: Analize the cases:

1.  7 and 8 are on the same face of the cube
2.  7 and 8 are on opposite vertices

I`m sorry. My english isn`t very good.  :(

[quote="Attila the Hun"]It suffices to prove that the number of combinations of four of {1, 2,...., 8} that sum to greater than 21 is greater than 36.[/quote]  It's true that this suffices in the sense that any statement can be derived from a false statement ;)

I was wrong earlier. but I got it now. we have to analyse the cases where

(i) 7 & 8 are on an edge
(ii) 7 & 8 are on opposite vertices on the same face
(iii) 7 & 8 are diagonally opposite.

analysing the cases, we see that there are 4 paths connecting diagonally opposite points and passing through 7 & 8 in both case (i) and (ii). the only pairs of vertices that sum to less than 6 are (4, 1), (3, 1), (2, 1), (3, 2). we can then prove that if two the the above pairs can be passed through, the remaining two cannot be passed through. so, the two remaining paths have a sum greater than 21. and we are done.
when 7 & 8 are diagonally opposite, there are 6 paths connecting them and only 4 pairs of other vertices which sum to less than 6. so, by pigeonhole principle, there must be at least two other paths which sum to greater than 21 and we are done.  :clap2:


[quote="Utkirstudios"]let abc are non-negative integers. prove hat:
\[ \frac{{a^{3}}}{{2a^{2}-ab+2b^{2}}}+\frac{{b^{3}}}{{2b^{2}-bc+2c^{2}}}+\frac{{c^{3}}}{{2c^{2}-ac+2a^{2}}}\ge \frac{{a+b+c}}{3}\]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66271


(Sorry if this turns out to be trivial)

Consider the function $f : \textbf{Q}\rightarrow \textbf{R}$ where $\textbf{Q}$ are the rationals and:

$f(x) = \left\{ \begin{array}{ll}x & \mbox{if}\: x < \pi \\ x-\pi & \mbox{if}\: x > \pi \end{array}\right.$

Is this function continuous in $\textbf{Q}$ ? (not $\textbf{R}$)

Certainly no. f is continuous at all $x \neq \pi$, but not at $\pi$

That's why I asked about its continuity on its domain.

$\pi \not \in \textbf{Q}$. I know it can be expressed as the [i]limit[/i] of rationals (as they're a dense set), but I don't know if that says much about f's continuity in $\textbf{Q}$.

It is continuous.

We can even modify this example to construct a continuous function $f$ from $\mathbb{Q}$ to $\mathbb{R}$ such that whenever $g$ is a function from $\mathbb{R}$ to $\mathbb{R}$ that agrees with $f$ on the rationals, $g$ is not continuous on any interval.


From izzybarish.com. Apparently my nickname is also some sort of brand. I should copyright the name "Izzy" and then sue them.

This is the pinnacle of pwn. Admit it. I actually set it as my desktop background.  :P

Thanks Flower. :)

What's a Flower?

An pwetty flowah person.

:P

hmmm wow.

maybe THIS was what you wanted to animate on people's desktop backgrounds for your mock test rule breakers :rotfl:

[quote="AIME_is_hard"]Izzy's an awesome person  :D[/quote]
 :D  :D  :D  :D  :D  :D  :D  :D

[quote="gauss1181"]heh, wow.

"I got it from izzybarish.com, which is obviously a website completely dedicated in worshipping moi. :P " ~Izzy

 :rotfl:  :rotfl:  :rotfl:[/quote]

No, seriously, they have a whole page full of pretty pictures that say stuff like "Izzy's World", "Izzy's Restaurant", "Izzy's Steakhouse". Obviously all about me, of course.

Yup, it is indeed. I set it as my desktop background for fun, and every time I see it I twitch.

:rotfl:  :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl:

That was izzy.

(Ha ha, what a knee slapper that was).

It looks very pleasant and peaceful.

:)

hm, lol I'm scared of dogs, but I feel somewhat pitiful for this one..who is about to be eaten :P
it's so cute!


*=multiply

How many positive integer solution are there to the equation:
x+y+z=20

How many positive integer solutions are there to the equation:
x*y*z=20

How many integer solution are there to the equation:
x*y*z=20

just wonderin how you solve these problems with facotrials/combination/permutations

[quote="Logg"]How many positive integer solution are there to the equation:
$x+y+z=20$
[/quote]
[hide]
Take all the pairs $(x,y)$ such that $x+y\leq 19$

We have $1\leq y\leq 19-x$, total $19-x$ solutions for $y$

If $x=1$, we have $18$ solutions for $y$
If $x=2$, we have $17$ solutions for $y$
...........................
If $x=18$, we have $1$ solution for $y$

Since $x+y\leq 19$ we can find $z=20-(x+y)$, which is positive

$1+2+...+18 = 18\cdot \frac{1+18}{2}=171$

In this sum we count the triplets more than one times

$\blacksquare$ For example,
$1-1-18$
$1-18-1$
$18-1-1$

These triplets are in the form $(a,a,b)$.
Every such triplet counted 3 times

$\blacksquare$ There is another kind of triplets, $(a,b,c)$, where $a,b,c$ are distinct.
We have 6 permutations of this triplet

$\blacksquare$ Of course there is not a triplet in the form $(a,a,a)$

Suppose that we have $P$ triplets in the form $(a,a,b)$ and $Q$ triplets in the form $(a,b,c)$. Then $3P+6Q=171$

We can find that there are only 9 triplets on the form $(a,a,b)$. Indeed:
$a+a+b=20 \Rightarrow  b=20-2a$

For $a=1,2,...,9$ we get positive results for $b$.
But $a\geq 10$ yields $b\leq 0$
So $P=9$

$3P+6Q=171 \Rightarrow $

$3\cdot 9+6Q=171 \Rightarrow $

$Q=24 \Rightarrow $

total $P+Q=9+24=33$[/hide]


Show that for all positive reals a,b,c,d with abcd=1,
1/(1+a)^2+1/(1+b)^2+1/(1+c)^2+1/(1+d)^2≥1

$ \frac{1}{(1\plus{}a)^2}\plus{}\frac{1}{(1\plus{}b)^2}\geq \frac1{1\plus{}ab}$ and $ \frac{1}{(1\plus{}c)^2}\plus{}\frac{1}{(1\plus{}d)^2}\geq \frac1{1\plus{}cd}$

 suming them up and using $ \frac{1}{1\plus{}ab}\plus{}\frac{1}{1\plus{}cd}\equal{}1$ we get your inequality.

[quote="Albanian Eagle"]$ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2}\geq \frac1{1 \plus{} ab}$ and $ \frac {1}{(1 \plus{} c)^2} \plus{} \frac {1}{(1 \plus{} d)^2}\geq \frac1{1 \plus{} cd}$

 suming them up and using $ \frac {1}{1 \plus{} ab} \plus{} \frac {1}{1 \plus{} cd} \equal{} 1$ we get your inequality.[/quote]

Another prove :Let $ x\equal{}\frac{bcd}{a^3},y ...$
Now we must prove that
\[ \sum \frac{a^6}{(a^3\plus{}bcd)^2}\ge 1\]
but by Titu inequality(or cauchy) it's enough to prove that
\[ \frac{(\sum a^3)^2}{\sum (a^3\plus{}bcd)^2}\ge 1\]
by using murhead the problem will be solved :wink:


$ ABC$ is a triangle, right-angled at $ C$. The internal bisectors of angles $ BAC$ and $ ABC$ meet $ BC$ and $ CA$ at $ P$ and $ Q$ respectively. $ M$ and $ N$ are the feet of the perpendiculars from $ P$ and $ Q$ to $ AB$. Find angle $ MCN$.

$ \angle MCN \equal{} 45$ :).I'm lazy to type

$ P$ and $ Q$ being on the angle bisectors, they are equally distant from $ AB$ & $ AC$, and respectively from $ AB$ and $ BC$, hence the triangles $ CNP$ and $ CMQ$ are isosceles, so $ \widehat {MQA} \equal{} 2 \widehat {ACM}$ and $ \widehat {NPB} \equal{} 2 \widehat {BCN}$, but $ \widehat {AQM} \plus{} \widehat {BPN} \equal{} 90^\circ$, consequently $ \widehat{QCM}\plus{}\widehat{PCN}\equal{}45^\circ$ and, therefore, $ \widehat{MCN}\equal{}45^\circ$.

Best regards,
sunken rock


My first is in horse but not in pony?
My second is in coyote but not in jackal?
My third is in shrew and in vole?
My fourth is in panda but not in bear?
My last is in baboon but not in monkey?
What am I?


What is the greatest number of whole 4-inch by 6-inch notecards that can be cut from a 25-inch by 27-inch sheet of poster board if every cut must be made paralled to a side of the poster board? No taping or gluing is allowed. Is there a formula for this problem, not just an orientation of placing cards horizontally and vertically?

$ \left \lfloor \frac{25}{4} \right \rfloor \times \left \lfloor \frac{27}{6} \right \rfloor$ or $ \left \lfloor \frac{25}{6} \right \rfloor \times \left \lfloor \frac{27}{4} \right \rfloor$

In this case, both of them equal $ \boxed{24}$.

lol that's not the right answer though, its 26

Put 3 rows of vertical cards with 6 in each row, then put 2 rows of horizontal cards with 4 in each row. That's $ \boxed{26}$. I don't know of any shortcut though...

Is the answer $ \lfloor (25 \cdot 27) \div (4 \cdot 6) \rfloor \equal{} \boxed{28}$?

That's not possible without taping or gluing.

No, I don't think there is a formula. The logical method is quite fast for MATHCOUNTS purposes.


i was looking at some other forums that use phpBB, and a cool feature i saw was the "quick reply" thing. when you click onto a thread they have a box at the bottom with "quick reply" written on the top. it doesnt have the spoilers and emoticons etc. but most posts don't utilize that, so it would save time. especially for those people with 56k

its just a thought

Can you point me to a forum that uses that?

I dont get it...is it still the same reply screen except with less options?

I think that'd be an excellent thing to put in *nods self*

[quote="confuted"]Naw, there's a little reply box at the bottom of each thread where you can type a message. Basically, it's just one fewer page to load. However, those boxes frequently result in spamming because it makes it so easy to make one line comments.[/quote]
Although you have a point might've well try it out and see if that'll happen.

How do you use this function? You're link didn't work Neal...

The way I've seen it in another forum is just that you click fast reply and it opens a little box on the same page where you can type in, instantly, you don't have to load another "reply" page. Saves a lot of time sometimes..


geometry
circumcircle
trigonometry
trig identities
Law of Cosines

$\Delta ABC, AB = 24, BC = 32, AC = 40$
$\Delta ABC \sim \Delta ACD$
The circumcircle of $\Delta ABC$ intersects $AD$ at $T$.
Line $T$ is drawn tangent to the circle.
$P$ is the intersection of Line $T$ and $CD$.
Compute $PD$.

The circumcenter is the intersection of the three perpendicular bisectors of the triangle.

Solution:

[hide]
The triangle clearly is 8 x (3,4,5), so for now I will reduce the problem to a 3-4-5 right triangle and multiply by 8 at the end, which we can do since the end measurement measures a linear feature of the triangle (as opposed to area or volume).

Since we have $b^{2}=a^{2}+c^{2}$, ABC is a right triangle, so its hypotnuse is a diameter of its circumcircle.

We are given similarity, so clearly points B and T are the same point. To understand this, see that only by making AD an extension of AB can we make triangle ADC similar to triangle CBA.

Forming ACD as shown, we are given similarity, so $CD=\frac{5}{3}4$ and $AD=\frac{5}{3}5$.

Draw tangent line T and let PB=x. Since PB and PC are both tangent lines from point P, PC=PB=x.

Since $AD=\frac{5}{3}5$. and AB=3, $DB=\frac{5}{3}5-3$

Since <DPB=180-<CPD, the cosine of one angle is negative the cosine of the other angle. Letting the cosine of DPB equal K, We can then apply the law of cosines to solve for x:

$4^{2}=2x^{2}-2x^{2}K$
$(\frac{5}{3}5-3)^{2}=x^{2}+(\frac{5}{3}4-x)^{2}-2x(\frac{5}{3}4-x)(-K)$

Solving each for K and then setting them equal and simplifying, we get $x=\frac{10}{3}$ and then $PD=\frac{20}{3}-x=\frac{10}{3}$

Multiplying by 8, the answer is $\boxed{\frac{80}{3}}$
[/hide]

Good solution.
[quote="bthings_2000"]
[hide]
Since <DPB=180-<CPD, the cosine of one angle is negative the cosine of the other angle. Letting the cosine of DPB equal K, We can then apply the law of cosines to solve for x:

$4^{2}=2x^{2}-2x^{2}K$
$(\frac{5}{3}5-3)^{2}=x^{2}+(\frac{5}{3}4-x)^{2}-2x(\frac{5}{3}4-x)(-K)$

Solving each for K and then setting them equal and simplifying, we get $x=\frac{10}{3}$ and then $PD=\frac{20}{3}-x=\frac{10}{3}$

Multiplying by 8, the answer is $\boxed{\frac{80}{3}}$
[/hide][/quote]
[hide="Alternatively"]Note that $\angle CBP\cong\angle BCP$ is complementary to $\angle PBD$ and $\angle PDB$. So $CP=PB=PD=CD/2$.[/hide]


calculus
integration
logarithms
calculus computations

[hide]$ \int e^{\minus{}2x} \ln(1\plus{}e^x)\,dx \equal{} \underbrace{\minus{}\frac{1}{2}e^{\minus{}2x} \ln(1\plus{}e^x)}_{\equal{}\,A} \plus{} \frac{1}{2} \int e^{\minus{}2x} \frac{e^x}{1\plus{}e^x}\,dx$

$ \equal{} A \plus{} \frac{1}{2} \int \frac{e^{\minus{}x}}{1\plus{}e^x}\,dx \equal{} A \plus{} \frac{1}{2}\int \left(e^{\minus{}x} \minus{} 1 \plus{} \frac{e^x}{1\plus{}e^x}\right)\,dx$

$ \equal{} \frac{1}{2} \ln(1\plus{}e^x)(1\minus{}e^{\minus{}2x}) \minus{} \frac{e^{\minus{}x} \plus{} x}{2} \plus{} C$[/hide]

Thank you guys! When I have time and luck I'll find more interesting integrals.

[quote="Xevarion"][quote="kunny"]Evaluate $ \int_{\ln 2}^{\ln 3} \frac {\ln (e^x \plus{} 1)}{e^{2x}}\ dx.$[/quote]
Let $ u \equal{} e^x$. Then $ du \equal{} udx$ and $ e^{2x} \equal{} u^2$, so we want to evaluate
$ I \equal{} \int_2^3 \frac {\ln(u \plus{} 1)du}{u^3}$
Integrate by parts:
$ I \equal{} \left[\ln(u \plus{} 1) \cdot \frac { \minus{} 1}{2u^2}\right]_2^3 \plus{} \int_2^3 \frac {du}{2u^2(u \plus{} 1)}$
Now,
$ \frac {1}{u^2(u \plus{} 1)} \equal{} \frac {1}{u^2} \minus{} \frac {1}{u} \plus{} \frac {1}{u \plus{} 1}$
so
$ \int_2^3 \frac {du}{u^2(u \plus{} 1)} \equal{} 1/6 \plus{} 3\ln(2) \minus{} 2\ln(3)$
and then
$ I \equal{} \frac {25}{18}\ln(2) \minus{} \frac {7}{8}\ln(3) \plus{} \frac {1}{12}.$

I'm not too sure about my arithmetic but it seems like a reasonable answer...  :maybe:[/quote]

That's the correct answer. :)

P.S. I was planned to post the following problem.

$ \int_{\ln 2}^{\ln 3} \frac{\ln (e^x\plus{}1)}{e^{3x}}\ dx.$


Hey guys,

I'm a sophomore who just interviewed with Jane Street Capital. I only did so-so on the interviews, so I did not get an internship offer. Can I bother to apply again next year, or will they have my average interview on file and not let me interview again?

Thanks

I don't speak for Jane Street Capital, but I highly doubt that on your application, they wrote "... had a terrible interview.  Don't accept should he ever apply again."  Rather, I think that if you apply again, it will give you a chance to have a better interview, learn from your mistakes, and strengthen your application.


Unbelievable, but I guess bhorsemath won that election of yours...hehe "God does exist, and she has a wicked sense of humor" (Business 2.0 February 2004) :lol:

I didn't run in that election.

Whats so unbelieveable?

Mrph, it's just that the entire thing is so funny in a screwed up sort of a way...I mean think about it, people complain that were slacking off, they decide to take matters into their own hands, the epic battle takes place, and voila! here we are. It's so...amazing, it's like a movie but it's real life. Amazing, I could write a book about this, for real :lol:

It's good to have another member on board as moderator. Yes, we are kinda overstaffed atm, but we can have stricter enforcement this way.

Rep- thats not a good idea.  No offense but it wouldn't work.

Hmm...I think it's a marvelous idea. It would take quite a bit more scripting to enforce...but overall, it's not bad...

However, I think mods and admin deserve some leniency on that rule...

I don't know if that would work though.  Think about when those really popular threads get going every once in awhile.  People watch them and replie quickley to them.  The count would have to be watched too.   It might work, but I think your right Miths the mods and admins should have a little leniency on it.

Hmmm should one of us pm the admins with that idea?

I'm moving this into suggestions for further attention.

On second thought, I should have split the top part of the thread off...

In any case, I propose post limit 25 per day...

Hmm do we need this rule? I think none of us posts 25 a day in F&G anymore, and I only see two or three orange thingies everytime I look at it. Besides I think F&G is much more desolate now anyway, and I have no idea why some people think it's "crowded" along with other forums...:?:

Besides compared to the "neo-jojobird era" there's almost no activity in Etc. :)

I believe it should be put into law :-) Maybe 25 is too much...15 perhaps?

Do we really need this? I mean think about it,
a.) There's not a lot of posts anyways
b.) There's no spam
c.) It's deleted anyways monthly
and
d.) Think about how much work is necessary to altar phpBB to actually perform this task, do you know how much programming that is?

[quote="Tare"]b.) There's no spam[/quote]
I beg to differ.

I think that if someone were to write a script to create daily post limits in Fun and Games, then it would be implemented. It sounds like a lot of work to me, but I don't know anything about scripting languages.

I agree with confuted that there's still plenty of spam here.

That's kinda why i think it's a good idea...

[quote="JamesMatthews"]Would it? Seems like a simple task to me... I've done much harder things...[/quote]

Hey, Kiwi :) now theres two of us..


[quote="THECHAMP"]Use Matrices to find the area of a figure with vertices (7,0), (-4,-1), (-3,2) and (-6,3).[/quote]
\[ \left|\frac {\left| \begin{array}{cccc} 7 & 0 & 1 \\
\minus{} 4 & \minus{} 1 & 1 \\
\minus{} 3 & 2 & 1 \\
\minus{} 6 & 3 & 1 \end{array} \right|}{2}\right|\]
This is right, right?

EDIT: Just wondering, are these from UIL?

EDIT2: And I believe you are right.

According to the shoelace theorem, Im assuming it should be 1 in place of 0


[u]xref4search [/u]: sin (2 x - Pi/4 ) = sqrt (3) / 2 ; equation; trigonometry[quote="abzalserekov"]please solve this one!!!!!!!!

sin(2x-(Pi/4))=sqrt(3)/2

in form like this: power(-1,k)*(Pi/2)...., k is Z.

"sqrt(x)" means "square root of x".
"power(x,y)" means "x to the power of y".[/quote]
It's a rather strange school exercice for an olympiad preparation  :huh:

$ \sin(2x \minus{} \frac {\pi}4) \equal{} \sin(\frac {\pi}3)$ and so :

either $ 2x \minus{} \frac {\pi}4 \equal{} \frac {\pi}3 \plus{} 2k\pi$ $ \iff$  $ x \equal{} \frac {7\pi}{24} \plus{} k\pi$

either $ 2x \minus{} \frac {\pi}4 \equal{} \frac {2\pi}3 \plus{} 2k\pi$ $ \iff$ $ x \equal{} \frac {11\pi}{24} \plus{} k\pi$

And I dont know what "in form like this: power(-1,k)*(Pi/2)...., k is Z" could mean

You're welcome.

Glad to have helped you.


A positive integer $ k$ greater than 1 is given. Prove that there exist a prime $ p$ and a strictly increasing sequence of positive integers $ a_{1}$, $ a_{2}$,$ \dots$, $ a_{n}$,$ \dots$, such that the terms of the sequence
\[ p\plus{}ka_{1},p\plus{}ka_{2},\dots,p\plus{}ka_{n},\dots\]
are all primes.

Not necessarily. $ a_{2}>a_{1}$. So, the sequence could be like this: 1, 4, 7, 22, 199, 2849, ...

Basically, we want to show that for every $ k$, there exists an infinite number of prime numbers equivalent mod $ k$.


MIT college Columbia Harvard Princeton Yale Duke

[quote="SOUMYASHANT NAYAK"]I guess by now the results for all US univs are out.... :)

Do you have any idea about who all got through...
As far as I know, Rik got to MIT...

GV, do you know abt udi's results??....[/quote]

Only Caltech and MIT resjavascript:emoticon(':maybe:')
maybeults are out.
Rik got into both ,Udi was waitlisted by both and I was rejected by both.

We are now waiting for the Ivy League (Brown, Columbia, Cornell, Dartmouth, Harvard, UPenn, Princeton and Yale) results on the 31st. :maybe:

http://en.wikipedia.org/wiki/Ivy_League#Origin_of_the_name

Vihang: Whats your pref now? US Univ's vs CMI/ISI?

They are a set of 8 colleges united by their varsity sports - A sports league of sorts, that have now become synonymous with the best education available for college studies, advanced technology, state of art facilities, and of course billion dollar endowments, from aristrocrats and legacies - Harvard, Princeton, Yale, Cornell, Pennsylvania, Columbia, Brown, Dartmouth constitute the 8 colleges. Highly selective admission with admit rates of around 8-10% and very very rough workload once through, but in the end, an Ivy League Degree is worth all that..

I hope that answers your question

hey,i never knew harvard is also under IVY!!

[quote="Chapplikka"]It seems Udi has been waitlisted at both MIT and Caltech!![/quote]

Vihang has already posted that..

[quote="shruthibaskaran"]Sadly, didnt get through to MIT.. As of now thats all I know about. Rik got through to both MIT and Caltech!! And Ivies release their decision on the 31st mostly, Duke on 25th. And Stanford seems to be leaning a lot towards the 28th.. Thats that then....[/quote]

Is there a good bio-engineering option in the other universities??
(I might be sounding stupid)

i couldn't resist saying this considering it's soumya:
but what do you mean 'sounding'?  :rotfl:  :D

Yes. They have good bio engineering programs. Ranked within the top 15 in the world mostly.

hey who is that rik i think he must be damn brilliant enough to get into MIT and Caltech thats means he is sure all india rank 1 in this years IIT JEE 2008 ah??? :o

AoPSer [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=22989]Abel[/url](Soumitra Kulkarni) got into Duke  :clap:  :clap:

shruti- good u have realized how much i mess p. i think i will leave very soon, it's not like i'm any good..... :(

Hey, hey! It wasn't mean to be the way it turned out. I am really sorry!  I don't think you mess up, actually. See, we used to mess around a lot too. But that was after our JEE and AIEEE days, so it was justified. Or at least alongside other meaningful stuff. It's becoming slightly confusing these days, and it's not because of anyone in particular. Don't leave. That'd be too early a decision to take. Useful stuff happens around here, which will definitely be helpful to you :)

Sorry if I'd said something offensive!

no it's probably me...i had a loww self esteem buzz for a few hours. now i'm okay. i don't think i'm helpful though.

whether you are helpful or not don't depends on your contribution now(most of people here are a lot senior)

it'll be a matter after some years :)

plz plz plz dont leave this forum

(mods and others plz dont consider it as spam)

its like a similar bloody hell discussion going on everywhere :mad:  :mad:  :mad:

hey ritu wats ur CNN matter!
that was a hytre sarcastic remark...........

RITU!!!!!! that is inapropriate context!!!!! swearing is not at the least bit math-y. that is not well- behaved are any of u goody goody's???

just watch CNN-IBN at 11:30 today and u'll know!!!!!!!! :P

does it is it local??? still work if u don't live in india???

it will b in india only :roll:  :oops:
but i can mail u

he might have thats is y he is saying to mail it??
plz mail it to me rituraj

GOOD JOB u have achieved some thing great ritu, .....good job!!!!! YAY!!!! :thumbup:


probability
pigeonhole principle
function
Putnam
conditional probability
graph theory
AoPS Books
\/closed

[i][b]Intermediate Counting & Probability[/b][/i] by DPatrick is now available!

[url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=302]Click here to order![/url]

[i]Intermediate Counting & Probability[/i] is the follow-up to [i]Introduction to Counting & Probability[/i], also by David Patrick.

Topics covered in this book include inclusion-exclusion, 1-1 correspondences, the Pigeonhole Principle, constructive expectation, Fibonacci and Catalan numbers, recursion, conditional probability, generating functions, graph theory, and much more.  The book is ideal for students in grades 9-12 who have mastered basic counting principles such as combinations, permutations, and basic probability.

David Patrick will also be teaching our Intermediate Counting & Probability class, which has been expanded to include the many additional topics he included in the new text.  [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=3]Click here for course details[/url].  There is a discount on the text & solutions if you enroll in the class when ordering the text.  (To all of you who already enrolled in the class: your texts should be mailed today!)

[i]About the authors[/i]: Dr. David Patrick earned his Ph.D. in mathematics from MIT. He is a past USA Mathematical Olympiad winner, a two time Math Olympiad Program invitee, a perfect scorer on the AHSME (predecessor of the AMC 12), and top 10 scorer on the Putnam Exam.  The solutions are co-authored by Naoki Sato, who won first place in the 1993 Canadian Mathematical Olympiad and is a 2-time medalist at the International Mathematical Olympiad.

yes! i have been waiting for it!

it's sooo hard to find a good counting and probability book.

I(my dad) ordered it at like 7:00 AM, we were so excited!

Nice cover art!

[quote="FOURRIER"]Counting is combinatorics??[/quote]Basically. It's a very good combinatorics book :)

Depends on the method you choose to ship it with. Mediamail (the cheapest) will be the slowest. Between 7-10 days within US I think. Other ways are faster.

[quote="diophantient"]About how long will delivery take :D ? (just wondering)[/quote]
Priority shipping takes 2-5 business days.  Media (economy) shipping takes 2-3 weeks.  (These are within the U.S. -- International shipping takes longer.)

although, i often go with media mail (the slower one)..and it still arrives 2-4 days after i order it..i ordered a book sunday night. it came thursday afternoon!

talk about great service!  :lol:

than again, i do live like 2 hours away from san diego.

[quote="xpmath"]A lot of those topics are covered in Volume 2 also...I think they'll be in more detail, right?[/quote]
Definitely.

Does the book go only in Counting and Probabillity or a bitmore?

On the books page there is a link to the [url=http://www.artofproblemsolving.com/Books/IntermCount/toc.pdf]Table of contents[/url], Index and some excerpts from the book.

It doesn't need to be available on amazon, just order it here  :wink:

[quote="james4l"]I searched through Amazon.com and could not find it.

When will it be available on Amazon?[/quote]We actually kinna ask you not to order it on Amazon, but on our website: http://www.artofproblemsolving.com

Oh, OK.

We actually bought quite some AoPS books here. but we have a \$30 rebate from a recent purchase of the ACoPS book at Amazon.  That's why I asked about that question.

From business point of view, it may make sense to publish the  book to wider audience.

[quote="james4l"]From business point of view, it may make sense to publish the  book to wider audience.[/quote]We do belive that there are more (gifted) math interested people roaming on our website than on Amazon :)

In plus Amazon asks for a huge percentage of the book sell, so it's actually not that profitable to sell books on Amazon (except when you have no other choice obviously).

what is the font used for the P in the power sets


greatest common divisor
number theory proposed
number theory

Determine the greatest real number $ s$ such that there exists an infinite sequence of positive integers $ (a _n)$ such that for each positive integer $ n$  we have $ a_n > 2008^n$ and $ a_n^s \le u_n$, where $ u_n$ is the greatest common divisor of the set of integers $ \{a_i \plus{} a_j \mid i \plus{} j \equal{} n \}$.

This is a problem form Vietnam TST in 1997.

Result: max s=1/2

It 's not hard to prove that s does not exceed 1/2. But it's too hard to show a clear sample when s=1/2