[quote=can_hang2007]Do you all still remember the following old and hard inequality
\[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}(3\sqrt[3]{4}\minus{}2)\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2} \ge 3\sqrt[3]{4}\plus{}1
\]
We haven't got a complete solution for it yet, and now, here is a hint for us to solve this problem
Lemma. For any $ a,b,c>0,a\plus{}b\plus{}c\equal{}1,ab\plus{}bc\plus{}ca\equal{}q$ $ (1 \ge 3q >0)$. Prove that
\[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge \frac{2(27q^2\minus{}9q\plus{}1)}{9q^2\minus{}2q\plus{}(1\minus{}3q)\sqrt{q(1\minus{}3q)}}\plus{}\frac{1}{q}\minus{}6
\]
Hope you all will interest with it. :)[/quote]

as the equality condition is really ugly... I don't get directly on applying proper formula

[quote=xzlbq][quote=can_hang2007]Do you all still remember the following old and hard inequality
\[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}(3\sqrt[3]{4}\minus{}2)\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2} \ge 3\sqrt[3]{4}\plus{}1
\]
We haven't got a complete solution for it yet, and now, here is a hint for us to solve this problem
Lemma. For any $ a,b,c>0,a\plus{}b\plus{}c\equal{}1,ab\plus{}bc\plus{}ca\equal{}q$ $ (1 \ge 3q >0)$. Prove that
\[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge \frac{2(27q^2\minus{}9q\plus{}1)}{9q^2\minus{}2q\plus{}(1\minus{}3q)\sqrt{q(1\minus{}3q)}}\plus{}\frac{1}{q}\minus{}6
\]
Hope you all will interest with it. :)[/quote]

\[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}(3\sqrt[3]{4}\minus{}2)\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2} \ge 3\sqrt[3]{4}\plus{}1
\]
<=>
$ \left(  \left( 15\,\sqrt [3]{4}+13+9\,{4}^{2/3} \right)  \left( x+y+z
 \right) ^{4}+ \left( -42-63\,\sqrt [3]{4}-54\,{4}^{2/3} \right)
 \left( xy+zx+yz \right)  \left( x+y+z \right) ^{2}+ \left( 36+54\,
\sqrt [3]{4}+81\,{4}^{2/3} \right)  \left( xy+zx+yz \right) ^{2}
 \right) {x}^{2}{y}^{2}{z}^{2}+ \left(  \left( x+y+z \right) ^{7}+
 \left( -11-3\,\sqrt [3]{4} \right)  \left( xy+zx+yz \right)  \left( x
+y+z \right) ^{5}+ \left( -18\,\sqrt [3]{4}-24 \right)  \left( xy+zx+y
z \right) ^{3} \left( x+y+z \right) + \left( 30+15\,\sqrt [3]{4}
 \right)  \left( xy+zx+yz \right) ^{2} \left( x+y+z \right) ^{3}
 \right) xyz+4\, \left( xy+zx+yz \right) ^{5}-4\, \left( x+y+z
 \right) ^{2} \left( xy+zx+yz \right) ^{4}+ \left( x+y+z \right) ^{4}
 \left( xy+zx+yz \right) ^{3}\geq 0---(1)
$

equality hold when $x=1,y=.9330590808,z=1 $ or $x=1,y=1.099101237,z=1 $

.9330590808 Exact representation?

(1)<=>

In triangle,

$g={s}^{8}+{\frac {1}{1439}}\, \left( 3\,\sqrt [3]{4}+11 \right)  \left(
1710\,r\sqrt [3]{4}-514\,r+711\,r{4}^{2/3}-5756\,R \right) r{s}^{6}-
 \left( -16\,{R}^{2}-60\,Rr\sqrt [3]{4}-128\,Rr+54\,{r}^{2}{4}^{2/3}+
11\,{r}^{2}+48\,{r}^{2}\sqrt [3]{4} \right)  \left( 4\,R+r \right) r{s
}^{4}$
$+ \left( -64\,{R}^{2}-128\,Rr-72\,Rr\sqrt [3]{4}+8\,{r}^{2}+36\,{
r}^{2}\sqrt [3]{4}+81\,{r}^{2}{4}^{2/3} \right)  \left( 4\,R+r
 \right) ^{2}{r}^{2}{s}^{2}+4\,{r}^{3} \left( 4\,R+r \right) ^{5}\geq 0
$[/quote]
the equality occur when $a=b=c$ or $a,b,c$ are three roots of following equation
$$t^3-t^2+\frac{2+\sqrt[3]{2}}{10}t-\frac{3-\sqrt[3]{2}}{10}=0$$

[quote=can_hang2007]Do you all still remember the following old and hard inequality
\[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}(3\sqrt[3]{4}\minus{}2)\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2} \ge 3\sqrt[3]{4}\plus{}1
\]
[/quote]

When $a, b, c >0, \quad (a-b)^2+(b-c)^2+(c-a)^2\ne 0,$ write the inequality as
$$\frac{a^4c+a^3b^2-a^3bc-a^2b^2c-a^2bc^2+a^2c^3+ab^4-ab^3c-ab^2c^2-abc^3+b^3c^2+bc^4}{3abc(a^2+b^2+c^2-ab-bc-ca)} \ge \sqrt[3]{4}.$$

$\sqrt[3]{4}$ is the best constant in the sense that
$$\inf_{a, b, c > 0;\atop (a-b)^2+(b-c)^2+(c-a)^2\ne 0} \frac{a^4c+a^3b^2-a^3bc-a^2b^2c-a^2bc^2+a^2c^3+ab^4-ab^3c-ab^2c^2-abc^3+b^3c^2+bc^4}{3abc(a^2+b^2+c^2-ab-bc-ca)} = \sqrt[3]{4}.$$
WLOG, assume $a, b\ge 1, \quad c = 1$.
When $a \approx 1.59487120323811,\quad b \approx 1.10674318178336, \quad c = 1,$ the minimum  $\sqrt[3]{4}$ can be attained.
The exact $a$ is a real root of $a^9-3a^8+21a^7-36a^6-6a^5+27a^4-a^3-3a^2-1 = 0,$ and the exact $b$ is a real root of
$b^9+3b^7+b^6-27b^5+6b^4+36b^3-21b^2+3b-1 = 0.$

[quote=Skravin]…

as the equality conditions are really ugly... I don't get directly on applying proper formula.[/quote]

[url=https://artofproblemsolving.com/community/c6h217169]This one[/url] may be [hide=better][img]https://imgsa.baidu.com/forum/pic/item/730f354e251f95cabd5ec320c9177f3e66095286.jpg[/img][/hide].

[quote=Rhapsodies_pro]
[url=https://artofproblemsolving.com/community/c6h217169]This one[/url] may be [hide=better][img]https://imgsa.baidu.com/forum/pic/item/730f354e251f95cabd5ec320c9177f3e66095286.jpg[/img][/hide].[/quote]
I see it's in can_hang2007 book “[i]Inequalities with Beautiful Solutions[/i]” :rotfl:


what happen with http://www.kalva.demon.co.uk/ ??  :huh:  :huh:

I think its dead. o.O

Click [url=http://web.archive.org/web/20040405065644/http://www.kalva.demon.co.uk/index.html]here[/url] for an older index of the site.


logarithms
trigonometry
calculus
calculus computations

solve for x and y
x+iy=(1+i [(2)^(1/2)])^i

I think you said

$ x + iy = (1 + i\sqrt {2})^i.$

Some general principles:

$ z^w = e^{w\log z}$

In turn, $ \log z = \ln|z| + i\,\text{arg}\,z$ where $ \text{arg}\,z$ refers to the value of $ \theta$ needed to represent the point $ (x,y)$ (from $ z = x + iy$) in polar coordinates.

One note here: $ \text{arg}(1 + i\sqrt {2}) = \tan^{ - 1}\sqrt {2}$ is not a "nice" or special value. (Or it's that plus an integer multiple of $ 2\pi.$ Is argument, hence logarithm, ambiguous? Well, yes. That's an important issue.)

Try to finish this yourself before clicking on the following:

[hide="Short answer"]${ x + iy = e^{ - \tan^{ - 1}\sqrt {2}}(\cos(\frac {\ln 3}2}) + \sin(\frac {\ln 3}2))$

In other words, ${ x = e^{ - \tan^{ - 1}\sqrt {2}}\cos(\frac {\ln 3}2})$ and ${ y = e^{ - \tan^{ - 1}\sqrt {2}}\sin(\frac {\ln 3}2}).$

Except that we could multiply these answers by $ e^{2\pi k}$ for any integer $ k$ and still potentially be correct.
[/hide]


conics
function
trigonometry
parabola
analytic geometry
graphing lines
slope

(yea i am suprised school books have good problems) The book gave away the important idea, but i think it is an interesting problem if you are not given the hint.

A ball is shot with an initial angle of $45^\circ$ with the horizontal and follows a parabolic trajectory. If the max height is $40$ ft, how far did the ball travel?

[hide="Physics"]Vertically we have uniformly deaccelerated motion with the initial velocity $v_{y}=v_{0}\sin\theta$. The height of the rise is given by $h=\frac{v_{y}^{2}}{2g}=\frac{v_{0}^{2}\sin^{2}\theta}{2g}$, and the duration of the rise by $t_{0}=\frac{v_{y}}{g}=\frac{v_{0}\sin\theta}{g}$

Horizontally we have uniform motion with the velocity $v_{x}=v_{0}\cos\theta$. Since the ball first rises and then falls, the horizontal distance of the shot is given by $d=2t_{0}\cdot v_{x}=\frac{2v_{0}^{2}\sin\theta\cos\theta}{g}$.

From $h=\frac{v_{0}^{2}\sin^{2}\theta}{2g}$ we get $v_{0}^{2}=\frac{2gh}{\sin^{2}\theta}$. Plugging that into the expression for $d$, we get $d=\frac{2\sin\theta\cos\theta}{g}\cdot\frac{2gh}{\sin^{2}\theta}=4h\cot\theta$.

With the given data for $h$ and $\theta$, we get $d=160\,\text{ft}$[/hide]

both of those solutions are good (ones which i thought of), but there is another interesting result that the book states to let you solve the problem,

[hide]the focus of the parabola lies on the horizontal[/hide]

which i suppose is another method to solve it

[quote="midorya"][hide="my solution"]
we know what trajectory is graph of the function $f(x)=ax^{2}+bx+c$.
$f(0)=0 \Rightarrow c=0$.
$f'(0)=1 \Rightarrow b=1$.
$f(-\frac{1}{2a})=40 \Rightarrow a=-\frac{1}{160}$.
So our function is $f(x)=-\frac{1}{160}x^{2}+x$. The second root is $\boxed{160}$.
[/hide][/quote]

Nice solution, i especially liked the $f'(0) = 1$.

[quote="SplashD"]What kind of function is $f'$?[/quote]
I believe that's the derivative. Since it's thrown at a 45 degree angle, its slope at $x=0$ is 1. Since the derivative of $ax^{2}+bx+c$ is $2ax+b$, when $x=0$, the derivative is just $b$.

just look at the axis of symmetry, $\frac{-b}{2a}$ for $ax^{2}+bx+c$.  This fact is easily proven by calculus, setting the derivative equal to zero.  From there we can mirror the parabola.

[quote="236factorial"][quote="Altheman"]the focus of the parabola lies on the horizontal
[/quote]

What does the book mean by the "horizontal"?[/quote]

I think it means the ground level.

[quote="demji"]but how does that necessarily help us.  Parabolas only have 1 focus.  Thus it will only be on the horozontal for one particular value of height.  So I don't see how it helps us in the long run on a large variety of these problems.  It must be a specific case where it is so.[/quote]

that is true that it is a specific case, however, using this hint there is a fully geometric solution!

i understand that this might be a specific case but the way it is approached it makes it appear that it is true for all cases.  Just want to clarify that statement.

[quote="SplashD"]Is this like physics? because there's a question I don't know how to solve that's sort of like this.

A projectile is launched upward from ground level with an initial speed of 98 m/s. How high will it go? When will it return to the ground?[/quote]

Yes, that's pure physics, though elementary. You need the equations for uniformly (de)accelerated motion.

No, I got that as a question from my Algebra II teacher though.


any one got a good science fair project idea?

This has been posted at least 3 times recently. Please search before posting.


geometry
area of a triangle
combinatorics unsolved
combinatorics

Assign to each side $ b$ of a convex polygon $ P$ the maximum area of a triangle that has $ b$ as a side and is contained in $ P$. Show that the sum of the areas assigned to the sides of $ P$ is at most three times the area of $ P$.


$ABC$ is a triangle. $D$ is a point on line $AB.$ $(C)$ is the in circle of triangle $BDC$. Draw a line which is parallel to the bisector of angle $ADC,$ And goes through $I$, the incenter of $ABC$ and this line is tangent to circle $(C)$. Prove that $AD=BD.$

[quote="ashegh"]ABC is a triangle.D is a point on line AB.(C) is the in circle of triangle BDC.draw aline wich is paralell to the bisector of angle ADC.and goes through I(the in center of ABC).
this line is tangent to circle (C).prove that:  AD=BD[/quote]

I don't think that there is such a triangle! :huh:

[quote="ashegh"]ABC is a triangle.D is a point on line AB.(C) is the in circle of triangle BDC.draw aline wich is paralell to the bisector of angle ADC.and goes through I(the in center of ABC).
this line is tangent to circle (C).prove that:  AD=BD[/quote]

Please care about correctly copying the conditions of the problem! This one is wrong and I can only guess that it is the problem from http://www.mathlinks.ro/Forum/viewtopic.php?t=27798 with some typos and forgotten conditions.

By the way, this belongs not into Open Questions but rather into Proposed&Own Problems. Here is a quick explanation of the different sections:

Assume that you have a mathematical problem.

- Post it into [b]Unsolved Problems[/b] if you don't have a solution, but you know that there are people in this world who have a solution (e. g., it was a past olympiad question or you found it in a book without solution).
- Post it into [b]Proposed&Own Problems[/b] if you have a solution (by yourself or by somebody else) and want to share the problem with others.
- Post it into [b]Open Questions[/b] if you don't have a solution and you don't know whether anybody in this world has a solution. So, for instance, the Riemann Hypothesis is an open question; but also if you come up with some conjecture and can't prove it, then it belongs into Open Questions (unless you have very good reasons to suspect that it will be easily solved by other people; in this case, it belongs into Unsolved Problems).

As for the other sections:

- [b]Solved Problems[/b] is a kind of archive for the problems that were posted in "Unsolved Problems" and subsequently solved. Don't post new problems in this section! However, if you have some comments on a problem/solution in this section (for instance, an additional solution, or you would like to point out a mistake in a posted solution), feel free to post it.
- [b]Theorems and Formulas[/b] is a place for theorems (and proofs) which are quite important or useful, as well as for general questions (like: what is inversion and how to use it; although I must say such questions are mostly not answered too well on MathLinks, since this would require a lot of time). Don't post just any arbitrary problem here, post only results you find particularly important!

  Darij


geometry
trigonometry
trig identities
Law of Cosines

Point P is inside a square ABCD such that its distance from A is 1, from B is 4, and from C is 5. what is the area of the square?

i know that the distance from D is sqroot(10).  how do you find the area?

Well I can think of a trig way, involving assigning lots and lots of variables for angles and applying the law of cosines, which leads to guess what, ugly algebra.

[quote="Walk Around The River"]this is on a test that is taken by many middle schoolers such as myself.  might there be an easier way?[/quote]

Well, think about the way of deriving the British Flag Theorem.


Is it possible to find uncountably many pairwisely disjoint subsets of $[0,1]$, all with external Lebesgue measure $1$?

Or, more generally, given an arbitrary family $\{F_\alpha\}$ of cardinality continuum consisting of subsets $F_\alpha$ of a set $X$ of cardinality continuum such that each $F_\alpha$ is of cardinality continuum itself, is it possible to find a collection $\{G_\beta\}$ of cardinality continuum of disjoint subsets of $X$ such that $F_\alpha\cap G_\beta\ne \varnothing$ for all $\alpha,\beta$? As usual, it seems a good idea to solve the problem with "of cardinality continuum" replaced by "countably infinite" first. ;)


calculus
integration
number theory proposed
number theory

Let $N$ be the number of intergal solutions of the equation $x^{2}-y^{2}=z^{3}-t^{3}$
satisfying the condition $0\leq x, y, z, t\leq 10^{6}$
And let $M$ be the number of intergal solutions of the equation $x^{2}-y^{2}=z^{3}-t^{3}+1$
satisfying the condition $0\leq x, y, z, t\leq 10^{6}$
Prove that $N>M$


linear algebra
matrix
algebra
polynomial

We know that if A is diagonalizable then A^2 is diagonalizable,too.Is the converse also true? That is, if A^2 is diagonalizable,can we say that A is diagonalizable?

Not over $\bf R$, certainly: try to find a real matrix with pure imaginary eigenvalues.

Assume the eigenvalues exist in the field in question, thus taking away JBL's example. It's still false.

Cheap example: $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}.$

we've a good result , let's put $P$ the characteristic polynomial of the matrix $A$ if we suppoose that $P$ is split  and let  $B$ be an other matrix st : $B^{k}=A$( $k\ge2$) if $A$ is diagonalizable then $B$ is diagonilazable


Divide by 4, take $\tan$ of both sides, and just simplify the $4\tan{\frac{1}{5}}$ using the double-angle formula twice. Then just reduce and simplify, carefully.

an interesting proof would be to do it geometrically...any ideas? first, it would be best to divide by 4, and show that those 2 make a 45 degree angle

Ooh, I have the best solution! :D

[hide]
Notice that \[ (5-i)^4(i+1)=4(239-i). \] $1+i$ has argument $\pi/4$. Also, $\arg (5-i) = -\arctan 1/5$. Recall that $\arg (xy) = \arg x + \arg y$. Hence, \[ \arg [(5-i)^4(i+1)] = -4\arctan (1/5)+\pi/4, \] and \[ \arg [4(239-i)] = -\arg (239-i) = -\arctan 1/239. \] Since the arguments must be the same, we have \[ 4 \arctan 1/5 - \arctan 1/239 = \pi/4, \] as desired.
[/hide]

BTW, where did you see this problem?

chess64: In a physics forum or something.

Wow, the solution was straightforward! Except that complex number solution :D  That was crazy. Never would've thought of that....

I rather enjoy that complex number solution, as it provides the most straightforward way of deriving an endless number of other Machin-like formulas  :D

1234567890:  That expression is very popular because $4 \arctan \frac{1}{239}$ is very small, so evaluating it with Taylor series algorithmically gives very fast approximations very quickly - it is a popular method of calculating $\pi$.  I'm not sure why you would've seen it in a physics forum, it is more likely to crop up in a computer science discussion.

[quote="agolsme"][quote="1234567890"]Prove:
$\pi = 16\arctan(\frac{1}{5}) - 4\arctan(\frac{1}{239})$

I'm stumped. :([/quote]

[hide]

First we manipulate the equations a bit so we can take tan of both sides:
\[ \pi/4 + 4\arctan{\frac{1}{239}} = 4\arctan{\frac{1}{5}}  \]

Now we take $\tan()$ of both sides:
\[ \frac{1 + 1/239}{1 - (1)(1/239)} = \tan(\arctan{5/12} + \arctan{5/12}) = \frac{5/6}{119/144}  \]
which are equal to $\frac{120}{119}$ on both sides. So we have proven the identity.
[/hide][/quote]
It is possible for $\tan$ of two angles to be equal without the angles being necessarily equal. In addition to your solution, you must also control that  $\pi/4 + \arctan{\frac{1}{239}}$ and $4\arctan{\frac{1}{5}}$ lie in the same quadrant (or I suppose the same half of the unit circle is actually enough) by estimations, although I suppose this is quite trivial.

[quote="Elemennop"]Did you actually start with that original expression, or did you get that and are showing the solution backwards?

Because if you got that original expression off your head, that's freakin' amazing.[/quote]

Well it was on the interm trig/complex numbers final exam and the hint was to multiply those complex numbers. :blush:


Simplify
\[\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}\]

$\left(A\right) \sqrt{x}$
$\left(B\right) \sqrt[3]{x^{2}}$
$\left(C\right) \sqrt[27]{x^{2}}$
$\left(D\right) \sqrt[54]{x}$
$\left(E\right) \sqrt[81]{x^{80}}$


function
LaTeX
superior algebra
superior algebra unsolved

By using mobius inversion's formula how can I prove that

P(x) = sum_k 1/kp(x^{1/k})
where p(x) = sum_k u(k)/kP(x^{1/k}), here u is the mobius function.

here both sums are infinte.

[quote="peteryellow"]By using mobius inversion's formula how can I prove that

$ P(x) \equal{} \sum_k \frac{1}{k} p(x^{1/k})$
where $ p(x) \equal{} \sum_k \frac{\mu(k)}{k} P(x^{1/k})$, here $ \mu$ is the mobius function.

here both sums are infinte.[/quote]
Isn't it nicer with LaTeX tags? :)

You'll need some words about convergence to justify these formal manipulations, but here's the idea:

$ \sum_{k\geq 1} \frac{1}{k} p(x^{1/k}) \equal{} \sum_{k\geq 1} \frac{1}{k} \sum_{l\geq 1} \frac{\mu(l)}{l} P(x^{1/{kl}})\equal{}$ $ \sum_{k\geq 1} \sum_{l\geq 1} \frac{\mu(l)}{kl} P(x^{1/kl}) \equal{} \sum_{n\geq 1} \sum_{d | n} \frac{\mu(d)}{n} P(x^{1/n}) \equal{}$ $ \sum_{n\geq 1} \frac{1}{n} P(x^{1/n})\sum_{d | n} \mu(d) \equal{} \sum_{n\geq 1} \frac{1}{n} P(x^{1/n}) \delta_{1 n} \equal{} P(x)$.

Sorry actually it is opposite by using P I want to show that p(x) is given by this formula.

It should be doubly clear from context... $ \delta_{a b} \equal{} 1$ if $ a \equal{} b$ and $ 0$ otherwise.

What you need to complete the proof is a lemma like: if $ \sum_{k\geq 1} \frac {1}{k} p(x^{1/k}) \equal{} 0$ then $ p \equal{} 0$.  You'd better be more specific about what our assumptions on $ p$ are if we're going to prove something like that...

EDIT: Typo.

Well I have shown that the sum is finite, so there are no conmvergence problem, thanks for your idea.

how do you get that

\[ \sum_{k\geq 1} \sum_{l\geq 1} \frac{\mu(l)}{kl} P(x^{1/kl}) \equal{} \sum_{n\geq 1} \sum_{d | n} \frac{\mu(d)}{n} P(x^{1/n})\]

That's just a change of notation (assuming absolute convergence); we let $ kl \equal{} n$ and note that $ l$ runs over the divisors of $ n$ for fixed $ n$.


geometry
probability
summer program
MathPath
Support
3D geometry
pyramid

[quote="SirErnest"]What does everyone think about aliens? Are they out there?  If so, what do they look like?  Are the friendly?[/quote]

Define alien.  If you mean `foreigner', then yes, certainly.  If you mean the little green men from mars, perhaps.

ya, after writing that i was debating whether or not i should change that to extraterrestrials but i wasnt sure how to spell it.

I think he means the little green men from mars but Im not for sure lol. Yes we do plutonian math which is totally different than regular math. We use it to plot crimes against people and with our inginuitive idea's we cant be defeated (me, joe bob, jewls, booky, donut man, and duck foot man that is) We are undefeatable!!! We shall rule the world!!!!!!!!!!!!

If Plutonians are smart enough to get to the earth, then surely they use hyperbolic geometry.

[quote="SirErnest"]What does everyone think about aliens? Are they out there?  If so, what do they look like?  Are the friendly?[/quote]
If y=f(x) where y is the probability of aliens existing and x is the amount of matter in the space we exist in, then lim(x :arrow: :inf:)>0 or something like that. Basically what it's saying is if you assume the space is infinitely large then there must be something alive out there. It even tells us that there must be our counterparts somewhere out there, and if we keep looking we'd find infinite amount of us (called the "Agent Smith Syndrome" :lol:), isn't infinity great? (and if any of you steal this then I'd have to do something really bad to you :wink:)

As for the friendly question it depends on the anatomy and the society of the aliens, so we never know. Heck we might just all kill each other before we could understand what they're saying if they are not developed enough, but then again if they come to us then they must be pretty developed :wink:

Course all this is under the assumption that space is infinite, which contradicts the Big Bang theory, so meh :wink:

P.S. I'm winking a lot because I got something in my eye...*scrub scrub* aaaah an alien! lol

we looked at hyperbolic geometry a little at mathpath, but i forgot what it was (i didnt really understand it, so i didnt pay alot of attention).  whats hyperbolic geometry? is it more complicated?

Given a line and a point not on the line, there are two lines through the point parallel to the line.

Non-euclidean geometry is geometry as usual (euclidean geometry) except that instead of a postulate that given a line and a point not on the line, there exists exactly one line through the point parallel to the line, we have either two parallels (hyperbolic or Lobachevskian geometry) or zero parallels (elliptic, spherical, or Riemannian geometry).

Don't mind aalindsey theres a thing with our science teacher and space...it involes space and the library(don't ask).

I think that it is entirely possible that there are aliens and by that I mean forien people from different planets, or different universes.

I think that it is rather arrogant to believe that there is no life anywhere else in the universe. Why would this one planet be so special that it is the only one anywhere that could support life? Surely there are other ``solar systems'' that are quite similar to this one, and almost certainly a planet that can support life will actually support life. Furthermore, I think it has been discovered from examining metiorites that have hit the earth and also Mars that at one point there were some very simple life forms there, but they seem to have died out.

Let me propose a thought. What if there is an advanced life form out there that is even beyond us. What if we were created by another life form rather than the big bang theory or by God. What if rather we are the technology of a life form even beyond belief. I mean there is that possibility right? Also another proposal, what if space is merely just a portion of whats out there I mean we could be just in a world within a tiny portion of someone elses world. It's totally possible if the universe is really equaled to infinity. So I mean how can someone not believe with all the knowledge there is out there to support a theory such as this. Knowledge that has yet to be discovered.

[quote](and if any of you steal this then I'd have to do something really bad to you ) [/quote]
Ooooh what if I stole it what would happen? What bad thing would you do to me :wink:?

[quote="aalindsey100"]Let me propose a thought. What if there is an advanced life form out there that is even beyond us. What if we were created by another life form rather than the big bang theory or by God. What if rather we are the technology of a life form even beyond belief. I mean there is that possibility right? [/quote]
we've talked about that kind of thing in science class.  who would hav built the pyramids and what about stonehenge?  how did ancient civilizations become so advanced? thats kinda what i was thinking about when i posted this topic.

[quote="aalindsey100"]I am an alien. I come from Pluto. My school contains 16 different universes and many different planets. The halls are space. Im an alien. Im everyones favorite extraterrestrial at school probably because Im the only one. One day I will rule the world!!!![/quote]

You like Pluto, WOW I love Pluto! It's the coolest planet; who cares about life on Mars when we don't even know what color the surface of Pluto is! Because of budget cuts NASA might cancel the Kiuper Belt express! Show your support for Pluto lets complain to NASA :twisted: ! And now you are going to think I'm insane--everybody does :cry: .

And now on topic: I don't think there could be any intelligent life any place--the odds of having the right habitat are tiny. And by the way I said any place because humans don't count any more :) .

Of course they do. Haven't you heard of Yuggoth and the Old Ones?  :wink:

[quote]Course all this is under the assumption that space is infinite, which contradicts the Big Bang theory, so meh  [/quote]
Actually, it doesn't.  The big bang theory's primordial fireball is infinitely dense and infinitely hot.  Space was infinite around it.

[quote="SirErnest"]another thought:
what if these highly intelligent aliens are out there,  watching us?  what if they are controlling us, like puppets?  what if were all part of this really big play and our lives and actions planned out?  what if it is some kind of dream or illusion?  just a thought[/quote]

All the world's a stage, and the men and women merely players...

Or, what if the white mice are financing the Earth as a huge computer, which is calculating the Ultimate Question to Life, the Universe, and Everything, and we're all part of the computer?

Oh wait, that's just what the Hitchhiker's Trilogy is about....:P

How would anyone know if aliens exist or not now???

[quote="JBL"][quote="Combinatoricsman"]Actually, it doesn't.  The big bang theory's primordial fireball is infinitely dense and infinitely hot.  Space was infinite around it.[/quote]

That can't be true, because we have a finite universe with a finite amount of matter and a finite amount of energy.[/quote]

How do you know?

"According to current cosmological theory."

we don't know if aliens exsit.  but when we were first sending probs into Jupiter we sent out the pioneer probs to make sure they wouldn't get destroyed.  On the probs were plates that welcomed aliens and asked them to visit and comunicate with us.  those probs are outside this galexy now and people still speculate that they could turn up something.

This is a deceivingly complicated question - because there are a lot of factors thatpeople never seem to consider. First of all, what does one mean by life? Would you count anything, even things like viruses, or do you only mean things like plants and animals, or do you only mean intelligent life? How do you define intelligent? Or do you mean life in the form of
complex "civilizations"?

These are all very different things. If you look at the entire history of our solar system, we are possibly the only planet with ANY life at all. And then there has only been plant and animal life for a fraction of the Earth's existence; "intelligent" life for an even smaller fraction; and "civilization" for an infinitessimally small fraction. If the time line of the Earth were the length of a football field (100 yards), the entire history of human civilization would be on the last millimeter of it. And of our billions and billions of species only one has achieved civilzation.

So this leads to the second question: what do you mean by [i]is[/i] there life? Do you mean at this exact instant in time, or was or ever will be?

Relativity makes this question very complicated - because our "is" is other parts of the Universe's past!! Suppose you look up through a telescope and see a planet with some form of life on it. If they're x-light years away you're seeing them as they were x years ago. So even if you tried to send them an SOS message, by the time it gets to them, their whole species may have gone extinct (counting the time it took the light to get to you and then your message to get to them).

The fact that the Universe is so large may work to increasing the chances that there ever was, or will be, intelligent life somewhere. But the sheer magnitude of the distances (both time and space wise) between them may prohibit them from having any contact with each other. For another civilization to have even detected the existence of civilization on our planet, they would have to be within say 8,000 light years of us (which cosmologically speaking is very small). And likewise, the chances of us looking up through a telescope in our brief existence on this Earth and seeing something alive at that exact moment of time is infinitessimally small.

Okay, we're getting way~~ off the subject here
I'll bet my entire fortune (40$) that the aliens exist
Everyone knows about the theory of life being created on the Earth?
Chances are, there could have been similar tings happening in other places.[/b]


Prove that-
$(a+b+c)^(333)-a^(333)-b^(333)-c^(333)$
is divisible by-
$(a+b+c)^3-a^3-b^3-c^3$

Heheh,this is my 333rd post,what a coincidence.... :rotfl:

To make the whole $333$ be in exponent, you have to write x^{333}

[hide="Sol"]
Factoring,
$(a+b+c)^3-a^3-b^3-c^3$
$3(a+b)(b+c)(c+a)$
Now it is trivial to prove that these individualy are factors of the 1st expression
[/hide]

Shady where do u get all those questions?tell me.... c'mon :lol:


[quote="spider_boy"]Tarlan Nazarov(me) 10 grade
           Jafar Jafarov 9 grade
           Farhad Aghayev 11 grade(will participate 3rd time)
           Asgar Cavadov 10 grade
           Tural Amirov 11 grade[/quote]

Does your team only have five members?


function
limit
induction
algebra unsolved
algebra

Find all functions $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that

$ (f(x)\plus{}f(y))(f(z)\plus{}1)\equal{}f(xz\minus{}y)\plus{}f(x\plus{}yz)$ for all reals $ x,y,z$

mszew gives a good discussion. Here is the complete solution.
From mszew, $ f(0)\equal{}0$. We know if $ f(x)\equal{}0$ holds for all $ x$. It will be one solution.
Now assume $ f(x)\neq0$ for some $ n\neq0$. By $ f(xz)\equal{}f(x)f(z)$, $ f(x)\neq0$ for any $ x\neq0$.
$ P(x,y,\frac yx): (f(x)\plus{}f(y))(f(\frac yx)\plus{}1)\equal{}f(x\plus{}\frac {y^2}x)$.
Multiply $ f(x)$ on both sides, $ (f(x)\plus{}f(y))^2\equal{}f(x^2\plus{}y^2)$. We know $ f(x)>0$ for $ x>0$. mszew shows $ f(y)\equal{}f(\minus{}y)$. So $ f(x)>0$ for $ x\neq0$. Then $ f(x)\plus{}f(y)\equal{}f(\sqrt{x^2\plus{}y^2})$. This implies $ f(x)$ is strictly increasing when $ x>0$. mszew shows $ f(x)\equal{}x^2$ for all rationals. Then for any irrational $ x>0$, we can find series $ y_n$ increasingly converging to $ x$, $ z_n$ decreasingly converging to $ x$. $ \lim f(y_n)\leq f(x)\leq \lim f(z_n)$ implies $ f(x)\equal{}x^2$. Then we can use $ f(x)\equal{}f(\minus{}x)$ to prove $ f(x)\equal{}x^2$ for all $ x$.

The answer is $ f(x)\equal{}0$ or $ f(x)\equal{}x^2$.

[quote="micf"][quote="mszew"]
$ P(x,1,1): f(x \plus{} 1) \equal{} 2f(x) \minus{} f(x \minus{} 1) \plus{} 2$
$ f(x) \equal{} x^2$ for integers
[/quote]

Can anybody tell me where is it from?[/quote]

Use induction on non-negative integers:

$ f(0)\equal{}0^2\equal{}0$, $ f(1)\equal{}1^2\equal{}1$ asume that it is true for $ x\minus{}1$ and $ x$ then
$ f(x\plus{}1)\equal{}2 x^2 \minus{} (x\minus{}1)^2\plus{}2\equal{}(x\plus{}1)^2$
and remember that $ f(x)\equal{}f(\minus{}x)$


How many digits are there in $2^{100}$ ?

[hide="Click to reveal hidden text"]$2^{10}$ is roughly equal to 1000 (it's 1024) and there's 3 digits following the initial one. So for something like $2^{20}$, logically there would be six digits following the initial one (similiar to a million). Following this trend, you would get $\boxed {31}$ digits (the initial one and the 30 following)[/hide]

[quote="robinhe"]Except it's not.  Sometime, it will gain a digit or two, so it's a couple more than 30.  However, I don't know the exact answer without a calculator.  :?[/quote]

Well it's 31. I forgot the initial digit.

In reality, you're questioning wether $1.024^{10}$ is gonna be greater than 10 or not. Using a calculator, $1.024^{10}=1.26765$

[quote="robinhe"]Really?

That's weird.  I thought it might be 32-ish...  :oops:[/quote]

I did too, but you still gotta figure problems like these are for people to realize the pattern. Then again, it could be a trick to make me say 31  :roll:

I still think I'm right  :|


Can anyone give me correct answer and explanations on the following two questions?

1. In quadrilateral ABCD, m :ang: A = x, m  :ang: B = 2x, m  :ang: C = 3x, and m  :ang: D = 4x.  Find the value of x and then state which pair of sides of ABCD must be parallel.

Here's how far I got:

[hide]x+2x+3x+4x=360[/hide]

[hide]10x=360[/hide]

[hide]x=36[/hide]

But I can't do the second part, stating the parallel thingie...

2. ABCDEFGHIJ is a regular decagon.  If sides AB and CD are extended to meet at K, find the measure of  :ang: K.

I got nice picture of decagon at:

http://mathforum.org/dr.math/gifs/decagon.gif

But I still can't figure out this question and everytime I do, I get wrong answer...

Big help here

Thanks!!!

On #1, I'd suggest sketching a picture - freehand, but with some vague attempt to get the angles right and with the angles labeled. You should be able to see what's going on. On #2, here's an exercise: Take every other side - AB, CD, EF, GH, IJ - and extend each such side in both directions until it meets the nearest other exteded sides. What would that figure be?

2)[hide]After extending AB and CD, you get an isosceles triangle KBC, with angles B and C being congruent. These angles are the exterior angles of the decagon and the equation for each exterior angle of an n-gon, is 360/n, so Angle b and c are 360/10 = 36 degrees. There are 180 degrees in a triangle thus

angle K + 36 + 36 = 180

K = 108[/hide]

theone853 wrote:Hint for 1:[hide]Supplementary angles are what makes parallel lines in a quadrilateral[/hide]

Cool...how do you prove that?

Using Kent Merryfield's hint for no. 2:

[hide]Extending every other side in both directions makes a regular pentagon, which makes sense because if you draw it, every two angles are replaced by one, and 10/2=5 angles and 5 sides. K is one of the angles in the pentagon, whose interior angles add up to 540 degrees. (I could prove that if I wanted...should I?) K therefore equals 540/5=108 degrees.[/hide]

Thanks for number one but I still confuse on second one:

Can anyone do somethingl like picture of that quadrilateral by Geometry Press?

[quote="Silverfalcon"]Thanks for number one but I still confuse on [b]second one[/b]:

Can anyone do somethingl like picture of that [b]quadrilateral[/b] by Geometry Press?[/quote]

The second one didn't contain a quadrilateral?

#1: Two pairs of angles add up to 180. ABCD is a trapezoid, just draw it out and the supplementary angles form two parallel lines with a line cutting through them.

#2: Just draw three sides and extend two of them.  The extended sides and the side between them form an isoceles triangle.  A regular decagon has angles of 144, the congruent angles (36 degrees) of the isoceles are supplementary to them, so angle K is 180-36-36=108.


limit
superior algebra
superior algebra unsolved

Let $ R$ be a ring complete with respect to an ideal $ \mathfrak{m}$. Let $ M$ be separated over $ R$, meaning $ \cap_k\mathfrak{m}^kM\equal{}0$. Show that if $ m_1,...m_n$ generate $ M/\mathfrak{m}M$, then it generates $ M$.

My idea was to first show that if $ \mathfrak{m}M\equal{}M$, then $ M\equal{}0$. This I know how to do, but I cannot conclude the Nakayama's Lemma. (My argument essentially reduces to showing that if $ N\equal{}M/\Sigma Rm_i$, $ N$ is reduced which I do not know how to do).

Input is appreciated.

Let $ f: R^n\to M$ be the homomorphism induced by the $ m_i$'s. Then $ f_0: (R/\mathfrak{m})^n\to M/\mathfrak{m}M$ is surjective, by assumption, which implies that each of the maps \[ f_k: \mathfrak{m}^kR^n/\mathfrak{m}^{k+1}R^n\to \mathfrak{m}^kM/\mathfrak{m}^{k+1}M\]with $ k>0$, are surjective, too.

To show that $ f$ is surjective, take an arbitrary element $ m\in M$. Step by step we will lift $ m$ modulo powers of $ \mathfrak{m}$ to $ R^n$ hoping that the approximating "error" in $ M$ will ultimately go to zero and the approximating preimages will converge in $ R^n$: Define $ m_0=m$ and construct inductively elements $ m_k\in \mathfrak{m}^{k}M$ and elements $ x_k\in\mathfrak{m}^{k-1}R^n$ with \[ m_{k+1}=m_k-f(x_{k+1})\] for $ k\ge 0$ using that each of the maps $ f_k$ is surjective. The partial sums of sum $ \sum x_k$ are a Cauchy sequence in $ R^n$, hence convergent, by completeness of $ R$. Call the limit $ x^*\in R^n$. Because $ M$ is $ \mathfrak{m}$-adically separate, we have $ \lim_{k\to\infty}m_k=0$. Now $ f$ is continuous w.r.t. $ \mathfrak{m}$-adic topologies such that we have \[ f(x^*)= f(\sum_{k\ge 1} x_k)=\sum_{k\ge 1} f(x_k)=\sum_{k\ge 1} (m_{k-1}-m_{k})=m_0-\lim_{k\to\infty} m_k=m\,.\]


The Fibonacci sequence $ 0, 1, 1, 2, 3, 5, 8, 13,...$ is defined as a sequence whose
two first terms are $ F_{0}= 0, F_{1}= 1$, and each subsequent term is the sum of the two previous
ones: $ F_{n}= F_{n-1}+F_{n-2}(n\geq 2)$. Prove that $ F_{10^{k}}$ is odd for every $ k\geq 0$

[hide="Umm,"]
the Fibonacci sequence is $ 1,1,0,1,1,0, \ldots$ in $ \bmod 2$. So, all we need is $ 3 \not | 10^{k}\ \forall k \in \mathbb{N}^{0}$, which is true.[/hide]


Which of the following three numbers is largest? $ 20^{30}, 40^{20}, 50^{10}$


It's now permitted to discuss the 2008 AIME II contest.

A reminder: the [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]AIME II Math Jam[/url] will be on Friday (April 4) at 7 PM Eastern / 4 PM Pacific.  We will discuss solutions to all 15 problems on the contest.

There's one other item that I'd like to remind everyone about.

From the USAMO Teachers' Manual, available on the [url=http://www.unl.edu/amc]AMC website[/url]:
[quote]The selection of the USAMO participants will be made on or before April 11th. All selected students and their School Manager will receive a letter of invitation, to be mailed out no later than April 11th. Please do not inquire about the USAMO invitations with anyone except the AMC Director, and do not phone prior to Friday, April 11th.[/quote]
In particular, this means [b]do not post on the Forum asking about the USAMO indexes, floor scores, or invitations[/b], and please do not ask or complain about when it will be available.  We do not have this information and we do not know exactly when it will come out, except for what is shown in the quote above.

I also strongly discourage speculation about what the USAMO qualifying index or floor will be.  What useful purpose does it serve to speculate about it?  You'll know for sure in a little over a week.  Just be patient.

[quote="Silverfalcon"][quote="Valentin Vornicu"][quote="DPatrick"]What useful purpose does it serve to speculate about it?  You'll know for sure in a little over a week.  Just be patient.[/quote]Actually when you are involved in a competition, discussing the possible cut-offs can be very exciting, and will make the time pass faster.

As long as they don't ask [b]us[/b] about the cut-offs, I guess all is good :)[/quote]

True but if you monitored AMC forum in this time of year for past couple years, posts like that appear every single time and they are very annoying spam-like threads.

So wait, does this mean I lock the topic asking for them?[/quote]
This year it seems like all the speculation is contained in one thread.  I think this is perfectly fine, as it prevents spamming in other random topics, and people who don't care to read the discussion don't have to.  In addition, people are discouraged from making their own pleas for floor score predictions in separate threads.


Hi there,

As we all know that the lottery is a 6 digit game.

My question is what is the odds of winning in a 2-digit game? You will just pick a number ranging from 1 to 31.

Is there  a formula on this one?
What is the probability on this one?

Thanks guys.

It should be that the number of possible lottery "draws" is $ 31^{2}$, and you are picking one of those choices, so the probability of winning is $ \frac{1}{31^{2}}=\frac{1}{961}\approx .104\%$. This was assuming that the same number could be picked for both choices example: 12 and 12.

Thanks for the reply.
Sir how can I increase my chances of winning with that solution?
How many tickets should I purchase?
What formula did you used there?

Thanks..

I was assuming it was like NY lotto in that you have balls numbered $ 1-31$ in a jar and then a second jar with balls number $ 1-31$ so that you can get two of the same numbers (instead of both draws coming from the same jar).

Let $ A$ be the event that the first number is the same as your pick.
Let $ B$ be the event that the second number is the same as your pick.

You want $ P(A\cap B)=P(A)\cdot P(B)$, but by definition it should be clear that the chances of $ A$ happening $ P(A)=\frac{1}{31}$ and likewise $ P(B)=\frac{1}{31}$ so $ P(A\cap B)=\frac{1}{31}\cdot\frac{1}{31}=\frac{1}{31^{2}}$.

Hmm... how many tickets should I purchase. I'm assuming this means how many do I need to buy in order to have a probability of greater than $ .5$. Well, since each ticket has a $ \frac{1}{31^{2}}$ chance of winning you need $ \frac{x}{31^{2}}=.5$ which means you need to buy $ (.5)(31^{2})=480.5$ so $ 481$ tickets.

But then again, you need to consider the [b]expected value[/b] of the lottery: if you buy the $ 481$ tickets, as [b]HilbertThm90[/b] suggests, is it really worth it? Will the payout be high enough to warrant buying all of those tickets? In the example, if say, the payout is $ 800$ dollars, and ticket costs $ 1$ dollar, is it worth it, given that we have about an even chance of winning or losing?

The answer here would be no: We have a probability of $ \frac{1}{2}$ of gaining the payout, but we have a probability of $ 1$ of losing the $ 481$ dollars in the tickets' price (you've already paid, and can't get that money back). Now, to calculate the expected value, you sum up each "payout" (either positive, for gain, or negative, for loss) multiplied by the probability of each. In the example, we would have:

 \[ E = \left( \frac{1}{2}\right) (800)+(1)(-481) =-81 \]
So, on average, you would lose $ 81$ dollars each time you bought these $ 481$ tickets. That is to say, were you to continue playing this game indefinitely. Since the probability of winning at each attempt is random (it most likely is, if the set-up for the lottery is fair), you may in fact win a couple times before beginning to lose your money again; or you may begin by losing a lot of money, and gaining it back. If you begin with gains, then you can stop early, though this is unlikely to happen; but if you begin with losses, it may be harder to stop. If you lose all of your money, then you have no way of getting it back.

You should also consider the [b]expected value[/b] of buying each [u]individual[/u] ticket; is it really worth it at all to play? (The answer, almost always, is [b]no[/b], as the organization running the lottery has of course already gone through all of this, and made it so that the expected value is negative for you.)

You can also read up on this sort of scenario at this [url=http://en.wikipedia.org/wiki/Gambler%27s_ruin]Wikipedia article[/url].

I did think about that, but it wasn't really the question. Plus, I didn't know if we were assuming $ \$1$ per ticket. In fact, from the question we really don't know very much. We've had to make numerous assumptions. Oh well. I know I wouldn't play no matter what.

@Lazarus: I agree with the picture comment.

[quote="Lazarus"]Wait, wait, all math is irrelevant until this very important question is answered:

Is that you in your picture?

@cincodemayo5590: This is why the lottery is called a tax on people who are bad at math.[/quote]

No that's not me dude.
That's Misa Campo a filipina/dutch.
Import model (import cars stuff like that)
Just like her face and body though.

Anyway back to the topic.
Btw the reason is I am doing a research on this one.
Also can we consider the screening of the machines that will be used on the draw?
How about the other stuff on screening before the draw?

Is there a formula or theories on this one?

Thanks


Let x_1, x_2, ..., x_8 be integers. A new sequence is constructed: lx_2-x_1l, lx_3-x_2l, ..., lx_1-x_8l. Find the sequences that a certain point after repeating constucting new sequences, all the number in the sequence are equal.

Solution outline - the largest number keeps decreasing, unless there are zeros, but the zero effect is finite. so it must reach 0,0,0,0,0,0,0,0 and so the case before is all numbers being equal.

[quote="grobber"]I think a more general version has been posted a very long time ago by orl. It asked to find all $n$ for which this conclusion holds when we work with $n$ numbers (here, $n=8$). I think I showed in that thread that those $n$ are precisely the powers of $2$. I didn't get any feedback, so the solution could be wrong :).[/quote]

Could you post a link to the relevant thread?

Sorry, I can't think of any good keywords right now.


Can someone (especially from English speaking countries) help me paraphase the following, clarify it?

Symmetries mean transformations which, when applied to dependent and independent
variables, produce another solution to the same differential system.

I need this for an English course.

I would just change which to that and singular to plural, resulting in

By symmetries we mean transformations that, when applied to dependent and independent variables, produce other solutions to the same differential system

but I'm not a native speaker either :P

Thank you fedja. I have another

Transformations that generate other solutions to the same differential system when applied to dependent and independent variables are refered to as symmetries.

Is it mathematically sound?

Well, sounds good to me (I mean I do not see any grammatical mistakes and the meaning is clear) but let's see what Kent or JBL will say; those two are real experts :).


Here's two factoring problems...
$ a^{2}\minus{}b^{2}\minus{}c^{2}\plus{}2bc\plus{}a\plus{}b\minus{}c$
$ 1\plus{}a\plus{}b\plus{}c\plus{}ab\plus{}bc\plus{}ac\plus{}abc$
Yeah I honestly don't know how you would factor those.

The second one factors into
[hide="2"]
$ (a\plus{}1)(b\plus{}1)(c\plus{}1)$
[/hide]

As for the first, I got it down to $ a(a\plus{}1)\plus{}(b\minus{}c)(b\minus{}c\minus{}1)$, but that's as far as it gets, I think, and my calculator won't factor it either.

second is variation of simon's fave
first i'm pretty sure is unfactorable (into two terms anyways)

Simon's Fave? Never heard of that... *Google Time*

you won't get anywhere using google, its an aops thing
simon's favorite factoring trick
on a sidenote, this shouldn't be under mathcounts, it should be under classroom math?

It's basically a factoring trick that turns an expression into $ (a \minus{} b)(c \minus{} d)$
Besides, I can't factor for my life.

[quote="uldivad9"]1. I honestly have no idea how I got this, but I got it in about 40 seconds.
[hide](a+b-c)(a-b+c+1)[/hide][/quote]

huh, it looks like I forgot to distribute a negative again

but anyways...

$ a^2\minus{}(b^2\minus{}2bc\plus{}c^2)\plus{}a\plus{}b\minus{}c$
$ a^2\minus{}(b\minus{}c)^2\plus{}a\plus{}b\minus{}c$
$ (a\minus{}b\plus{}c)(a\plus{}b\minus{}c)\plus{}a\plus{}b\minus{}c$
$ (a\minus{}b\plus{}c\plus{}1)(a\plus{}b\minus{}c)$

dang, I forgot about difference of squares.

ahahahaha
stupid me
yeah that works, but seriously, why is this under mathcounts?

Because I want you to find me a 8th grade or below Math Class in the US that teaches this to their students, then you can tell us to move it to Classroom Math.

Well, it's a concept from Algebra I, which is Classroom Math... but w/e, it doesn't matter.

Edit: Ra5249 beat me to it.

I think we did this problem a while ago in math for trivial reasons...I saw this problem recently.


function
trigonometry
real analysis
real analysis unsolved

I was studuing the absolutic continuous functions and got some difficulties

1. Why every absolutely continuous function has a bounded variation?

2. Why every Lipschitz function is absolutely continuous?

3. If $f: [a,b]\to [c,d]$ is absolutely continuous and strictly increasing and $g: [c,d]\to\mathbb{R}$ absolutely continuous, why $g\circ f$ is absolutely continuous?

4. What is an example of absolutely continuous functions $f: [a,b]\to [c,d]$ and $g: [c,d]\to \mathbb{R}$ such that $g\circ f$ is not absolutely continuous?

Solving #1 and #2 is only a matter of reading both definitions and putting them together. #3 is not much harder. A suggestion for #4: $f(x)=x^{2}\sin^{2}(1/x)$, $g(x)=\sqrt{x}$.

If $[a_{i},b_{i}]$ is a system of disjoint intervals of small total length, then so is $[f(a_{i}),f(b_{i})]$. That (and the definition of absolute continuity) is all you need for #3.


prove the sum of the first n odd integers is $ n^2$

[quote="MrRubix"][color=blue]look at the proof on the aops wall [/color][/quote]

wheres that? i usually use mathlinks.com except for class

MrRubix is talking about the animated proofs under the navigation bar in the AoPS skin.

The "proofs" shown on the side of the AoPS skin are not rigorous proofs; they only serve to provide a visual explanation as to why it should be true.

Anyway, if you actually want to see the correct one, find it on [url=http://www.usamts.org/Gallery/G_Gallery.php]this[/url] page.

[hide="another proof"]The first number is 1, and the last number is $ 2n\minus{}1$, so the average of every pair of numbers is $ n$.  There are $ n$ numbers, so their sum is $ n\times n\equal{}n^2$.[/hide]

[hide="another way  i found somewhere"]well work backwards
we have $ n^2$
subtract n, $ n^2\minus{}n$
subtract n-1, $ n^2\minus{}n\minus{}(n\minus{}1)$
factor $ n(n\minus{}1)\minus{}(n\minus{}1)\equal{}(n\minus{}1)^2$ now we have $ n^2\minus{}(2n\minus{}1)\equal{}(n\minus{}1)^2$
note how we subtracted $ 2n\minus{}1$ total, which was the nth odd number. then we can repeat the process for $ (n\minus{}1)^2$ to $ (n\minus{}2)^2$, and so on, until you get to $ 0^2$ notice youve subtracted the first n odd numbers. now we have $ n^2$-(sum of 1st n odd numbers)$ \equal{}0$
therefore $ n^2$=(sum of 1st n odd numbers)[/hide]

[hide="Does This Hold Valid As A Proof?"]
Notice that the sum of the $ n$ natural odd numbers, written as $ 1 \plus{} 3 \plus{} 5 \plus{} ... \plus{} (2n \minus{} 1)$, can be rewritten as
$ \sum_{k \equal{} 1}^{n}(2k \minus{} 1)$ where $ n$ is the $ n$th odd natural number.

$ \sum_{k \equal{} 1}^{n}(2k \minus{} 1) \equal{} 2\sum_{k \equal{} 1}^{n}k \minus{} \sum_{k \equal{} 1}^{n}1$

$ \equal{} n(n \plus{} 1) \minus{} n$
$ \equal{} n^2 \plus{} n \minus{} n \equal{} \boxed{n^2}$

Therefore
$ 1\plus{}3\plus{}5\plus{}...\plus{}(2n\minus{}1)\equal{}n^2$[/hide]

[quote="Patterns_34"][hide="Does This Hold Valid As A Proof?"]
Notice that the sum of the $ n$ natural odd numbers, written as $ 1 \plus{} 3 \plus{} 5 \plus{} ... \plus{} (2n \minus{} 1)$, can be rewritten as
$ \sum_{k \equal{} 1}^{n}(2k \minus{} 1)$ where $ n$ is the $ n$th odd natural number.

$ \sum_{k \equal{} 1}^{n}(2k \minus{} 1) \equal{} 2\sum_{k \equal{} 1}^{n}k \minus{} \sum_{k \equal{} 1}^{n}1$

$ \equal{} n(n \plus{} 1) \minus{} n$
$ \equal{} n^2 \plus{} n \minus{} n \equal{} \boxed{n^2}$

Therefore
$ 1 \plus{} 3 \plus{} 5 \plus{} ... \plus{} (2n \minus{} 1) \equal{} n^2$[/hide][/quote]

This is valid, however, you would still have to show that $ \sum_{k\equal{}1}^{n}k\equal{}\frac{k(k\plus{}1)}2$, which is probably just as well known as this, so you should prove that as well.


Find positive integers $ a$ and $ b$ such that
\[ (\sqrt[3]{a} \plus{} \sqrt[3]{b} \minus{} 1)^2 \equal{} 49 \plus{} 20\sqrt[3]{6}
\]

Problems in Elementary Number Theory

[url]http://www.mathlinks.ro/index.php?f=456[/url]

Here is this problem:

[url]http://www.mathlinks.ro/viewtopic.php?t=150693[/url]


Show that polynomial $ P(a,b,c)\equal{}(a\plus{}b\plus{}c)^5\minus{}(a^5\plus{}b^5\plus{}c^5)$ divide with polynomial $ Q(a,b,c)\equal{}(a\plus{}b\plus{}c)^3\minus{}(a^3\plus{}b^3\plus{}c^3)$.Determine the ratio.

we'll factor $ P(a.b.c)$. Note that $ P(a,\minus{}a,c)\equal{}0$ hence $ P(a,b,c)$ is divisible by $ (a\plus{}b)(b\plus{}c)(c\plus{}a)$, by symmetry.
Hence,we can factor easily $ P(a,b,c)\equal{}5(a\plus{}b)(b\plus{}c)(c\plus{}a)(a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca)$.
also, $ Q(a,b,c)$ analogously can be factorized as $ Q(a,b,c)\equal{}3(a\plus{}b)(b\plus{}c)(c\plus{}a)$.

Show that polynomial $ P(a,b,c)\equal{}(a\plus{}b\plus{}c)^{2n\plus{}1}\minus{}(a^{2n\plus{}1}\plus{}b^{2n\plus{}1}\plus{}c^{2n\plus{}1})$ is divisable by $ (a\plus{}b)(b\plus{}c)(c\plus{}a)$. Determine the ratio.

[quote="soruz"]Show that polynomial $ P(a,b,c) \equal{} (a \plus{} b \plus{} c)^{2n \plus{} 1} \minus{} (a^{2n \plus{} 1} \plus{} b^{2n \plus{} 1} \plus{} c^{2n \plus{} 1})$ is divisable by $ (a \plus{} b)(b \plus{} c)(c \plus{} a)$. Determine the ratio.[/quote]

We know that the $ (a \plus{} b \plus{} c)^3 \minus{} a^3 \minus{} b^3 \minus{} c^3$ differs from the product $ (a \plus{} b)(b \plus{} c)(c \plus{} a)$ only by a constant factor. Thus we only need to prove that $ (a \plus{} b \plus{} c)^{2n \plus{} 1} \minus{} (a^{2n \plus{} 1} \plus{} b^{2n \plus{} 1} \plus{} c^{2n \plus{} 1})$ is divisible by $ a \plus{} b,b \plus{} c,c \plus{} a$.

It's easy to see that $ \left((a \plus{} b \plus{} c)^{2n \plus{} 1} \minus{} a^{2n \plus{} 1}\right) \minus{} (b^{2n \plus{} 1} \plus{} c^{2n \plus{} 1})$ and cyclic is divisible by $ b\plus{}c,a \plus{} c,a\plus{}b$


inequalities
Cauchy Inequality
inequalities proposed

Let $ a,b,c$ be the real number satisfying $ a,b,c \geq 1$ and $ (ab\plus{}bc\plus{}ca\minus{}2abc)^2\equal{}2abc$.Prove that $ (a\minus{}1)(b\minus{}1)(c\minus{}1) \leq 1$

No one likes my problem?The equality occurs when $ a\equal{}b\equal{}c\equal{}2$.

It can be proven by using contradiction and following transformation

$ a\minus{}1\equal{}x$
$ b\minus{}1\equal{}y$
$ c\minus{}1\equal{}z$

Therefore, we assume that $ xyz\equal{}1$ and we are done!

Sorry,I don't understand,how can we assume $ xyz\equal{}1$?What can we prove and what do we need to prove after we assume that?

[quote="great math"]It can be proven by using contradiction and following transformation

$ a \minus{} 1 \equal{} x$
$ b \minus{} 1 \equal{} y$
$ c \minus{} 1 \equal{} z$

Therefore, we assume that $ xyz \equal{} 1$ and we are done![/quote]
I have tried like you but this seems to be complicated.I don't think putting $ a\equal{}x\plus{}1,b\equal{}y\plus{}1,c\equal{}z\plus{}1$ in the given condition and expanding is a good idea.

[quote="great math"]The fundamental way to attack this kind of problem is to use the contradiction. After plugging all the transformations to the given problem. In stead of expanding, why don't you give it a go further by another transformation such as

$ x \equal{} \frac {m}{n}$

$ y \equal{} \frac {n}{p}$

$ z \equal{} \frac {p}{m}$

These are hints for you,  KDS.[/quote] This would mean $ xyz\equal{}$1 but we can't assume it since we have the condition $ xyz\le 1$

[quote="great math"]The fundamental way to attack this kind of problem is to use the contradiction. After plugging all the transformations to the given problem. In stead of expanding, why don't you give it a go further by another transformation such as

$ x \equal{} \frac {m}{n}$

$ y \equal{} \frac {n}{p}$

$ z \equal{} \frac {p}{m}$

These are hints for you,  KDS.[/quote]
Thanks but:
If we use the contradiction,we have to prove the problem which is equivalent to the problem we are given.
So If you assume $ (a\minus{}1)(b\minus{}1)(c\minus{}1)\equal{}1$,so what do we need to prove?
$ (ab\plus{}bc\plus{}ca\minus{}2abc)^2 \leq 2abc$ or this one $ (ab\plus{}bc\plus{}ca\minus{}2abc)^2 \geq 2abc$?

[quote="great math"]you are there, KDS
The second inequality of your suggestions is absolutely right since we assume that $ abc \equal{} 1$ and there exists a solution only by cauchy inequality as quoted in my signature :)[/quote]
Why $ abc\equal{}1$?You first assume that $ (a\minus{}1)(b\minus{}1)(c\minus{}1)\equal{}1$.Even if the second ineq is true when $ (a\minus{}1)(b\minus{}1)(c\minus{}1)\equal{}1$,you cannot use the contradicting method in this case,so post your full solution,and then we will continue.

that is a mistake in my typo. It should be $ xyz \equal{} 1$

By the way, after the above transformations, we need to prove that

$ (m^2n \plus{} n^2p \plus{} p^2m \plus{} mpn)^2 \ge 2mnp(m \plus{} n)(n \plus{} p)(p \plus{} m)$

I think you understand a little bit more about what I want to demonstrate here.

By rearranging the inequality and suppose that $ n \equal{} max\{m,n,p\}$ then

$ (mn(m \plus{} p) \plus{} p(n^2 \plus{} mp))^2\ge 4mnp(m \plus{} p)(n^2 \plus{} mp) \equal{} 4mpn(m \plus{} p)(n \plus{} p)(n \plus{} m)$

We complete our proof!

We can assume like that because  $ xyz \equal{} 1$ is the least value for which the above inequality occurs that leads to the contradiction to the given condition and if we assume that $ xyz >1$ then the above inequality still happens so in brief, we get the desired result.

[quote="SUPERMAN2"]Let $ a,b,c$ be the real number satisfying $ a,b,c \geq 1$ and $ (ab \plus{} bc \plus{} ca \minus{} 2abc)^2 \equal{} 2abc$.Prove that $ (a \minus{} 1)(b \minus{} 1)(c \minus{} 1) \leq 1$[/quote]
We can also solve this by trigonometric method.


u know in christmas the point in it is getting and giving and getting and worshiping God and getting and eating good christmas food and getting and getting and i guess u guys know the point.  is there any other point of Chrismas than just that?

santa clause? or just a good reason to buy all ur frends expensive presents and puting lites up and.......well it just looks good. besides religion.....some ppl think of it as a moment of peace.

you should already know. THERE'S NO SANTA!!!!!!!!

yea just post it here on this forum where sum little kids can have access too. YEA GO AHEAD AND DESTROY A UNIVERSAL CHILDHOOD BELIEF. but anyone ever talked to those mall santas before? i did once. and the easter bunny.

[quote="gameworld7"]celebrating God, celebrating Jesus, that's the same thing, right? and Easter is Jesus' death right?[/quote]
no.
good friday is jesus's death (i think but im no christian)
easter is some roman holiday that has been changed to some weird one in america
WHERE THE HECK DID SANTA CLAUS COME FROM!?!?!?!?
i dont see any point in christmas either, seeing a good deal of non-christians celebrate it

random comment: why are there only miller ppl posting on a good deal of the recent threads here?

because, we have very imaginative minds while other people don't

Christmas supposedly celebrates Jesus's birth, but Jesus was most likely born sometime during the summer. Christmas ended up on December 25 because that coincides with some ancient Pagan holiday, and priests long ago thought having a holiday then would appeal to potential converts. Santa Claus is from Saint Nicholas. All I remember about him is that he was a Christian saint, but Google will probably turn up some better info. And Christmas does have meaning, but what that meaning is depends on whether you celebrate it as a religious or secular holiday (me=secular). For me it's about having an excuse to put up lots of pretty decorations and to get together with my cousin's family practically every day over winter break. For Christians, well, ask a Christian.

are you sure he was born in the summer? i thought that he was only born on Christmas day.  get it? [b]Christ[/b] mas day.

no one knows, really

all I "know" is that he was born in a manger

I love getting presents for Christmas. (Santa would be bankrupt if all the stuff about Santa's presents were real, so I dont believe it)

One more thing about Christmas- no school! (Winter Break)

Why not everyday be a holiday? Then I dont have to go to school!

Christmas has been so commerialized that people just use it as an excuse to get presents.

easter is the day Jesus was resurrected ( i dont think its the EXACT date either)

its 3 days after His death, lyk in the bible.

also, no one knows the EXACT date of his birth... the day ...
magi came to see him, wise men came to see him, the day many prophesies concerning this "baby boy" were fullfilled(with more fulfilled later-all of them, fullfilled).

i dont kno why u cant say "Christmas" at school, but "Winter Holiday,"

BUT

u can say "Halloween" even tho it offends some ppl (lyk me)

What...since when? Christmas isn't exactly a swear word or anything...why would you not be able to say it?

its supposed to offend jews... and buddhists... etc :maybe:

my teachers would haf to correct themselves

well i think wut she means is some people say Christmas break instead of Winter Break and that isnt "politically correct" and maybe her school enphasizes you dont say that or something

Oh...well that's just your school being idiotic then. Nothing weird like that happens here. The official name is Winter Break but nobody cares...all holidays are discussed freely and if someone's offended (which never happens anyway), too bad. I cannot understand why the name of a holiday would offend anyone...and probably it's not the kids that are offended anyway, but their parents...am I right?

yes you are
i remember my social studies teacher got in trouble because he said christmas break and some guy's parents found out he said that and made a big deal about it

[quote="bob123"]Saint Nicolas... according to a Cartoon Network "Winter holiday" special I saw 4-5 years ago, he was a guy (name Nicolas) who gave toys to poor kids, after a while, he started making all kinds of toys and he would distribute them on December 24 evening... and now he's called Santa Clause.[/quote]

when he was (supposedly) grown-up, he put some bags of money down the chimney and it "miraculously" landed in some stockings drying by the fireplace.

my mom says that it used to be a holy day, until people started using it for bad reasons(since people thought that the dead saints would lyk be reaising from the dead, kids thought it was fun to dress up, make a lot of noise, and paints someone's house unless they were given free food)

and Catholic is different from Christianity

[quote="sunnyray"]and Catholic is different from Christianity[/quote]

How so? I was under the impression that Catholicism is a branch of Christianity.

[quote="gameworld7"]um only 3 people that are 11 and under go to this forum (check on my topic "when were you born?") so what's the chance they'll look at my post?[/quote]

Not everyone voted in that poll you know.

Almost anything you do/say can be interpreted as something religious.  I made a list of holidays and showed how each one can be religious (even mother's day or flag day).

[quote="RobinHood3000"][quote="sunnyray"]u can say "Halloween" even tho it offends some ppl (lyk me)[/quote]

Why does it offend you?  If I'm not mistaken, Hallowe'en originated as a feast on All Hallows Eve, the day before All Hallows Day (duh, right?), or All Saints Day, a Catholic holiday.[/quote]

wasn't it from before that from a celtic holiday called Samhain, when the dead walk as if they were alive again

[url]http://en.wikipedia.org/wiki/Salem_witch_trials[/url]

Proof you don't have to be a blasphemous heathen to believe in the occult.

[quote]
[quote="sunnyray"]
and Catholic is different from Christianity[/quote]
How so? I was under the impression that Catholicism is a branch of Christianity.[/quote]

The short, somewhat cynical story of Catholicism vs. Christianity (according to the Christian faith and history textbooks)

Once upon a time about 2000 years ago, a [Jewish] guy named Jesus was born.  He preached about his dad, and performed a lot of miracles.  After he was executed, some of his friends started talking to other people about what he did.  These freinds were [u]Christ[/u]ians because they believed that Jesus was the Christ.  They converted a whole bunch of other people, and Christianity was born.

Jump ahead about 1600 years to a guy named Martin Luther (Not to be confused with Martin Luther King, Jr).  He didn't like that the Catholic Church wasn't following what was written in the bible.  He posted a list of all the wrongdoings on the church door and started his own branch of Christianity, Lutheranism.  At this point, there were two branches of Christianity: Catholicism and Lutheranism.

In the centuries following, Lutherans got slightly different ideas, and started their own branch of Lutheranism (Calvanism, Mormonism, etc.).  Lutheranism and these other branches came to be known as PROTESTantism.  Nowadays, there is Christianity, and under that there are the divisions of Protestantism and Catholicism.  Therein lies the difference between Christianity and Catholicism.

[quote="gameworld7"]wow i'm surprised.  3 people don't celebrate christmas. :([/quote]Back in the communist era in Romania, celebrating Christmas was kinna forbidden. But kids wanted the "capitalist" way for holiday celebrating: gifts :P So they invented a "Santa Cold" (or something like that, I can't really translate it) which came and brought presents to the good kids :)

Well Easter is on the real date, it happens to come after Passover which has always been celebrated based on the Jewish calendar, and its very hard to really remember how it goes, but anyway it has been carefully dated since a long time before Jesus' death (on Good Friday) and his resurrection on Easter Sunday.
Christmas day really happened some time between June and August. This time frame has been determined based on some old records (kinda slept through this lesson).

Now Christianity is a very broad term that includes al people who believe the Jesus was our saviour. Even Catholicism is not completely specific as their are various differences between different denominations. The largest Christian denomination is Roman Catholicism. The differences range from some calendar ideas (Orthodoz are 13 days ahead) to how much we glorify Mary to beliefs on divorce and female priests (there are a lot of differences). This is much like Judaism is not just one belief but there are Orthodox, Hasidic, Reform or the different forms of Buddhism (can't remember them, remind me to study them before our test).

I think thats enough for now

[quote="tetrahedr0n"][quote]Santa Cold[/quote]

Mos Gherila?[/quote]Yes, that would be the Romanian name for it :)

[quote="Jhyon8149"]The reason why you are not suppsosed to say "Christmas" is because it offends the people from Hebrew origins as some people said...... That's why in my middle school days, we had to put up a menorah (sorry if I splled it wrong) and the thing for Kwanzaa. It is like school regulations or something.....[/quote]

if they can do something for jews, why not christians?

[quote="yif man 12"]How so? I was under the impression that Catholicism is a branch of Christianity.[/quote]

their different in small ways, but still different...

its lyk if u had a sibling, u'd be somewhat the same, but still different.

They shouldn't be able to do anything for any religion/political group/whatever. As long as nobody's being hurt, it's rather un-Constitution-y. Just don't blame the non-Christian kids for it...most likely it's not their fault...

And wouldn't it be Catholicism and Protestantism instead of Catholicism and Christianity?

I agree, I was under the impression that Catholicism is a sub-sect of the many, MANY religions classified as Christianity.

I agree. I'd like to see those Catholics who don't consider themselves Christians.


Let an odd integer $ a>17$ satisfying $ 3a\minus{}2$ is a square. Prove that there exists positive integers $ b,c$ $ (b\ne c)$ such that $ a\plus{}b, b\plus{}c, c\plus{}a, a\plus{}b\plus{}c$ are squares.


algebra
polynomial
function
induction
limit
search
number theory unsolved

Given $ \alpha$ irrational and $ k \geq 2$ integer, prove/disprove that sequence $ \{ n^k \alpha \}$ is dense in $ [0,1]$

[b]Generalisation:[/b]
Let $ f(x) \equal{} a_n x^n \plus{} ... \plus{} a_0$ be a real polynomial such that at least one of $ a_1, a_2, ..., a_n$ is irrational (we call these polynomials [i]nice[/i]). Then $ f(x) \mod 1$ is equally distributed in $ [0,1]$ (especially, dense).

Proof:

Lemma: For every nice polynomial $ f(x)$ and every integer $ m > 0$, there is an $ x < 1$ such that $ F(n) : \equal{} \left| \sum_{s \equal{} 0}^n e^{2 \pi i m f(s)} \right| \in O(n^x)$ (functions bounded by $ C \cdot n^x$ for some $ C$).

Proof by induction:
If $ f(x) \equal{} a_1x \plus{} a_0$ has degree $ 1$, then $ a_1$ is irrational and we easily get
\[ F(n) \equal{} \\
\equal{} \left| \sum_{s \equal{} 0}^n e^{2 \pi i m (a_1 s \plus{} a_0)} \right| \equal{} \\
\equal{} \left| e^{2 \pi i m a_0} \sum_{s \equal{} 0}^n e^{2 \pi i m a_1 s} \right| \equal{} \\
\equal{} \left| \frac {e^{2\pi i m a_1 (n \plus{} 1) } \minus{} 1}{e^{2\pi i m a_1} \minus{} 1} \right| \leq \\
\leq \left| \frac {2}{e^{2\pi i m a_1} \minus{} 1} \right| \in O(x^0)
\]
(we used that $ e^{2\pi i m a_1 x} \neq 0$, meaning that $ a_1$ is irrational).

Induction step:
We easily see that if $ f(x)$ nice, then $ g_d(x) \equal{} f(x \plus{} d) \minus{} f(x)$ is nice for every $ d \in \mathbb Z \backslash \{0\}$, too. As it has lower degree, we can use the induction hypothesis.
Now:
\[ F(n)^2 \equal{} \\
\equal{} \left| \sum_{s \equal{} 0}^n e^{2 \pi i m f(s)} \right|^2 \equal{} \\
\equal{} \left( \sum_{s \equal{} 0}^n e^{2 \pi i m f(s)} \right) \cdot \left( \sum_{s \equal{} 0}^n e^{ \minus{} 2 \pi i m f(s)} \right) \equal{} \\
\equal{} \left( \sum_{s_1 \equal{} 0}^n \sum_{s_2 \equal{} 0}^n e^{2 \pi i m f(s_1) } \cdot e^{ \minus{} 2 \pi i m f(s_2) } \right) \equal{} \\
\equal{} \left( \sum_{s_1 \equal{} 0}^n \sum_{s_2 \equal{} 0}^n e^{2 \pi i m \left( f(s_1) \minus{} f(s_2) \right) } \right) \equal{} \\
\equal{} \sum_{d \equal{} \minus{} n}^{ \minus{} 1} \sum_{s \equal{} \minus{} d}^n e^{2 \pi i m \left( f(s \plus{} d) \minus{} f(s) \right) } \plus{} \sum_{s \equal{} 0}^n e^{2 \pi i m \left( f(s) \minus{} f(s) \right) } \plus{} \sum_{d \equal{} 1}^{n} \sum_{s \equal{} 0}^{n \minus{} d} e^{2 \pi i m \left( f(s \plus{} d) \minus{} f(s) \right) } \equal{} \\
\equal{} (n \plus{} 1) \plus{} \sum_{d \equal{} 1}^{n} \sum_{s \equal{} 0}^{n \minus{} d} e^{ \minus{} 2 \pi i m g_d(s) } \plus{} \sum_{d \equal{} 1}^{n} \sum_{s \equal{} 0}^{n \minus{} d} e^{2 \pi i m g_d(s) } \leq \\
\leq (n \plus{} 1) \plus{} \sum_{d \equal{} 1}^{n} \left| \sum_{s \equal{} 0}^{n \minus{} d} e^{ \minus{} 2 \pi i m g_d(s) } \right| \plus{} \sum_{d \equal{} 1}^{n} \left| \sum_{s \equal{} 0}^{n \minus{} d} e^{2 \pi i m g_d(s) } \right| \in \\
\in O(n) \plus{} \sum_{d \equal{} 1}^{n} O((n \minus{} d)^{x_d^\prime}) \plus{} \sum_{d \equal{} 1}^{n} O((n \minus{} d)^{x_d}) \subset \\
\subset O(n) \plus{} n O(n^{x_D}) \plus{} n O(n^{x_D}) \subset \\
\subset O \left( n^{x_D \plus{} 1}\right)
\]
for some $ x_d, x_d^\prime < 1$ and $ x_D$ being the maximum of the $ x_d$ and $ x_d^\prime$.
Taking square roots on both sides, we get $ F(n) \in O \left( n^{\frac {x_D \plus{} 1}{2}}\right)$ with $ \frac {x_D \plus{} 1}{2} < 1$ and are done.

Now using Weyl's criterion (see below), the original statement is easy:
$ \left| \frac {1}{n} \sum_{s \equal{} 0}^n e^{2 \pi i m f(s)} \right| \in O(n^{x \minus{} 1}) \subset o(1)$, giving that
\[ \lim_{n \to \infty} \frac {1}{n} \sum_{s \equal{} 0}^n e^{2 \pi i m f(s)} \equal{} 0.
\]
By the mentioned criterion of Weyl, we are done.

[b]Weyl's criterion:[/b]
Let $ x_n$ be a sequence of reals. Then $ x_n \mod 1$ is equally distributed in $ [0,1]$ iff for all $ m \in \mathbb{Z} \backslash \{ 0 \}$ we have that
\[ \lim_{n \to \infty} \frac {1}{n} \sum_{s \equal{} 0}^n e^{2 \pi i m x_s} \equal{} 0.
\]

I can post a proof if needed.

Thank you very much ZetaX for a detailed answer. Just to give some more insight:

The proof of the special case of Weyl's criterion you use is already on the forum: http://www.mathlinks.ro/viewtopic.php?search_id=72051046&t=152605

Also, just a few minutes ago I discovered that topic of equidistribution (with a proof of a theorem you prove) is covered in depth in a book of H. Iwaniec "Analytic number theory".

[quote="ZetaX"]

[b]Weyl's criterion:[/b]
Let $ x_n$ be a sequence of reals. Then $ x_n \mod 1$ is equally distributed in $ [0,1]$ iff for all $ m \in \mathbb{Z} \backslash \{ 0 \}$ we have that
\[ \lim_{n \to \infty} \frac {1}{n} \sum_{s \equal{} 0}^n e^{2 \pi i m x_s} \equal{} 0.
\]
I can post a proof if needed.[/quote]

Can you post it？


how do you factor something like $x^3 + 3x^2 + 4x - 28 = 0$?
I know quadratic factoring, but not equations with $x^3$ or above

You need to find a root of this equation. We can find $x=2$, so $x^3+3x^2+4x-28=(x-2)(x^2+ax+14)$.

wait, there's different factorings of this equation? The two replies I've gotten have different answers on them.

236factoral I guess the other one just missed the 5 inplace he wrote a

[quote="Devendra Patel"]Using the Factor Theorem W e can say that for x=2 the equation is equal to zero
Therefore x-2 is a factor and then using long divnsion we get the other factor to be x^2+5x+14
therefore x³+3x² +4x -28=(x-2)(x² +5x+14)[/quote]
can u explain what the factor theorem is??

I just use instinct to find which terems might work and then use long division. I think u can find long division in an algebra textbook or on the internet. I can't type equations on the forum.


A drum in the shape of a right circular cylinder is required to have a volume of 500 cubic centimeters.  The top and bottom are made of material that costs 6 cents per square centimeters, while the sides are made of material that costs 4 cents per square centimeters.

(a) Express the total cost C of the material as a function of the radius r of the cylinder.

(b) What is the cost if the radius is 4cm?

NOTE: What exactly is a cubic centimeter anyway?

[quote="b-flat"]what is the height?  A big part of this problem resolves around height.

If the radius is 4, then the height will be $\frac{4}{\pi}$, so the vertical surface area will be $\frac{4}{\pi}\cdot8\pi=32$, costing $4\cdot32=\$1.28$.  The top and bottom will have surface area $32\pi$, so it will cost $192\pi$ cents.

A cubic centimeter is a unit cube with edge length 1 centimeter.[/quote]

you can find the height in terms of the readius like in the last post "steel drum"


is our performance bad this year.
5 bronze no silver.

I'm not from India but while it may not be a top top score, consider the fact that it's still far better than most countries do.

yes but comparing with many small countries like vietnam etc,we are nowhere

hats off to syllabus in maths at +2 level :mad:

The performance is quite good this year... @skand you are mistaken. the reason for our overall not so great show in maths olympiads is not because of our maths +2 syllabus!!! Consider the fact tht USA does not have such a challenging syllabus yet their students do very well...!! The blame is on the students and not on the school system. And moreover in India people are so taken in by this IIT craze. And preparing for IIT, leaves your maths olympiad skills handicapped.

smaller countries train their people better.

do u think more olympiads should be held for lower classes and more good books must be written by indian authors?

@thealchemist:i say better syllabus for math in scools improve motivation to think better!!!!!

thats a good point but u should remember that they are already good books present for olympiads but ur point on mathematicl tactical skills being inoculated from the lower class is quite i[mpressive!

@thealchemist thats what i was trying to say

IIT syllabus is quite vast and leaves no way for other things

for example geometry(2 questions are compulsorily from it)is taught only till class 10

very few students are interested in geometry even after 10th
i've heard that in many countries geometry is a part of syllabus for +2 stage

basically its IIT craze that is responsible for our bad performance
normally india performs better in physics olympiad(syllabus is similar to JEE syllabus)

and i don't consider students responsible for that
career matters for everybody

we need to train students for math oly from 6 or 7. we should first train them in basics and how to improve thinking.

books are not plenty.
considering the fact that BJV CRP KNR and Krishnamurty have written a wonderful book called challenges and thrills but this is their only major book.(there r other books like problem primer etc. but for hardcore training only few materials are available.

I was in IMOTC this year .I could see the level of coaching and motivation given to guys from bhaskaracharya Pratishthana(pune)

sadly guys from other places stand to miss out on training or even motivation.

CORRECT!
EVEN I COSIDER THAT AS AN INSPIRATIONAL TOPIC!


modular arithmetic
Pascal\u0027s Triangle
representation
number theory
IMO Shortlist

Let $p$ be a prime number. Prove that there is no number divisible by $p$ in the $n-th$ row of Pascal's triangle if and only if $n$ can be represented in the form $n = p^sq - 1$, where $s$ and $q$ are integers with $s \geq 0, 0 < q < p$.

The condition is $0 < q <p$.

This result is a direct consequence of a well-known theorem of Lucas, which states that :
If $p$ is a prime, $n,k$ are nonegative integers with base $p$ representations (some digits may be zero, even the leading ones for $k$)
$n = n_0 + n_1p + ... + n_tp^t$ and $k = k_0 + k_1p + ... + k_tp^t$ then
$C_n^k \equiv \prod_{i = 0}^{t} C_{n_i}^{k_i}$ mod[$p$].

From that theorem, we deduce that $C_n^k$ is divisible by $p$ if and only if $C_{n_i}^{k_i}$ is equal to zero for some $i$, that is $n_i < k_i$ for some $i$.
Note that for $0 \leq k \leq n$, we have $k_t \leq n_t$, but for $k = (p-1) + (p-1)p + ... (p-1)p^{t-1}$, if $C_n^k$ is not divisible by $p$ that leads to $n_i = p-1$ for  $i = 0,1,...,t-1$ so that, if none of the $C_n^k$ is divisible by $p$, we must have $n = q \times p^t - 1$ for some $q = n_t \in \{0,1,...,p-1\}$.
The converse is easy.

Pierre.

I just want to point out a solution without using Lucas's theorem or anything tricky.

We have \[\binom nk \frac{n-k}{k+1}=\binom n{k+1}\]
Suppose $p^s<n\leq p^{s+1}$. Then, no binomial coefficent is a multiple of $p$ iff whenever $p$ divides $n-k$, it also divides $k+1$, and also whenever $p^2$ divides $n-k$ it divides $k+1$ and so on up to whenever $p^s$ divides $n-k$, it divides $k+1$. The last part is true iff $n-k\equiv -(k+1)\mod p^s$, or $n\equiv -1 \pmod p^s$. This is equivalent to the problem statement.

[quote=pbornsztein]but for $k = (p-1) + (p-1)p + ... (p-1)p^{t-1}$, if $C_n^k$ is not divisible by $p$ that leads to $n_i = p-1$ for  $i = 0,1,...,t-1$[/quote]
What if $k\neq (p-1) + (p-1)p + ... (p-1)p^{t-1}$?

[quote=caffeineboy]whenever $p$ divides $n-k$, it also divides $k+1$, and also whenever $p^2$ divides $n-k$ it divides $k+1$ and so on up to whenever $p^s$ divides $n-k$, it divides $k+1$. The last part is true iff $n-k\equiv -(k+1)\mod p^s$[/quote]
As far as I'm concerned $\forall_{l\in\lbrace 1,2,...,s\rbrace} \ [p^l|n-k\implies p^l|k+1]$ isn't equivalent to $p^s|(n-k)+(k+1)$


The Island's travelling sales monkey is selling gold watches that washed up on the beach. The islanders use green stones, pink clamshells and ox horns for money. Sixty-two green stones have the same value as 17 ox horns and 258 pink clamshells have the same value as 89 green stones. The monkey sells each watch for what is equivalent to 92 dollars and 95 cents. One ox horn is equal to 6 cents. How many green stones does the monkey get for each gold watch?

[hide]
62g=17h
258p=89g
w=92.95
h=.06

Since each ox horn is 6 cents, each green stone would equal 51/31 cents after plugged into equation (1).

Then divide 92.95 by 51/31 to get approximately 5649.9 or [b]5650[/b] green stones.

[/hide]


Let m be a positive integer having a prime divisor greater than  \sqrt(2m) + 1.
Find the least positive integer M for which there is a set T,
consisting of a finite number of distinct positive integers, such that
(a) m and M are the minimum and maximum of T, respectively;
(b) the product of all numbers in T is a perfect square.

[i]Vietnam IMO team selection test, 2002 -  Problem 3[/i]

M is the value we have to find. The question is right, although not very clear. M can depend not only on m but also on prime divisor of m (the one that had been mentioned in condition).

Namdung

Let $m=p\cdot s$, where $p$ is prime divisor of $m$ and $p>\sqrt{2m}+1$. It is clear that $m$ is not divisible by $p^2$, hence $T$ contains at least one more number which is divisible by $p$. It follows that $M\geq p\cdot(s+1)$. We will show that for $M=p\cdot(s+1)$ there exist desired set $T$.

Consider two cases.\\
1) $p=4s+1$ or $p=4s+3$. In this case $T=\{p\cdot s,\,p\cdot(s+1),\,4s\cdot(s+1)\}$;\\
2) $p\neq 4s+1$ and $p\neq 4s+3$. In this case
\[T=\{p\cdot s,\,p\cdot(s+1),\,(p+1)\cdot s,\,(p-1)\cdot (s+1),\,\dfrac{(p-1)\cdot(2s+1)}{2},\,\dfrac{(p+1)\cdot(2s+1)}{2}\}.\]
It can be shown that $T$ consists of distinct numbers and satisfies (a) and (b).


function
real analysis
real analysis unsolved

Let $\ { \ f_{\alpha} \ }$ be a family of analytic functions such that for each $z$ the set of values $f_{\alpha}(z)$ is countable .does it follow that the family is countable ?,what if we replace analytic with infinitly diffrentiable?

The infinitely differentiable version is trivial: just take a $C^\infty$ function $f$ supported by some compact set lying deep inside the unit square and, for every collection $\alpha=\{\alpha_k\}_{k\in \mathbb Z^2}$ of $\pm 1$, put $f_\alpha=\sum_{k\in\mathbb Z^2}\alpha_k f(z-k)$. Then the functions $f_\alpha$ take at most $2$ different values at each point but the entire family is of cardinality continuum. The question about analytic functions seems really interesting. One can also ask a meaningful question about continuous function: if $\{f_\alpha\}$ is an uncountable family of continuous functions on $\mathbb C$ (or on $\mathbb R$, it, probably, does not matter) such that for each $z$ the family $\{f_\alpha(z)\}$ is countable, is it true that some two functions in our family coincide on an uncountable set?

Amazingly, Erdos proved that if the continuum hypothesis is true, then there exists such a family of cardinal $c$ and otherwise, the family must be countable.


Jose has 6 favorite DVDs. If each DVD is 20 minutes long, and Jose wants to watch all 6 DVDs over and over in every conceivable order, how long (in hours) will it take him to accomplish this?

[hide]You need to find the number of permutations (number of arrangements) of the 6 DVDs.  There are two ways to find this:
1:  You have 6 choices for the first, 5 for the second, etc.  So, $ 6*5*4*3*2*1 \equal{} 720$.
2:  The formula for a permutation is $ nPr \equal{} \frac {n!}{(n \minus{} r)!}$, where $ n$ is the total number of items, and $ r$ is the number of items you are choosing.  So, in this case, $ n \equal{} r \equal{} 6$, so it's $ \frac {6!}{(6 \minus{} 6)!} \equal{} 720$.

Therefore, there are 720 ways you can watch the DVDs.  Since each movie takes 20 minutes to watch, 6 movies take 120 minutes (or 2 hours) to watch.

$ 720*2 hours \equal{} \boxed{1440 hours}$[/hide]


Can you find the smallest positive integer with factors 1-100?
Hint: It's smaller than $100!$ by a lot.

Yeah, I think it is just the products of all the primes with their powers as well.


Solve for $ x$: $ 3^{2x\plus{}1} \minus{} 10\cdot 3^x \plus{} 3 \equal{} 0$.

Hello
$ 3^{2x}.3\minus{}103^{x}\plus{}3\equal{}0$
$ 3^{x}\equal{}3 or \frac{1}{3}$
thus answer is $ 1 or \minus{}1$.
Thank u

hello, rewriting your equation in the following form $ (3^x)^2\minus{}\frac{10}{3}3^x\plus{}1\equal{}0$ and solving the quadratic equation we get $ x_1\equal{}1$ or $ x_2\equal{}\minus{}1$.
Sonnhard.


Is there a general expression (and proof) for the sum of n-digit subsets of a set?  Does it only work with a set with consecutive numbers?

[edit] - An example for this is:
(2003 Aime/3) S = {1, 2, 3, 5, 8, 13, 21, 34}. Find ∑ max(A) where the sum is taken over all 28 two-element subsets A of S.

For sum of subsets I think they (and I) mean that the sum of subset (1,2) and (2,3) is 1+2+2+3=8.  Is that right?

You should clarify that problem statement.  What does it mean to sum subsets?  What do you mean by n-digit subsets?  (You don't need to actually answer these questions, just rephrase your problem so that it is clear what you're asking about.)

I am not sure, but I believe that this question is asking for the sum of the larger elements in each of the 28 2-element subsets, and that should probably solve your problem (pun intended).
(If I remember correctly from taking this AIME last year)

ya, they are asking for the max of the subset, namely the larger number... so for (1,2) and (2,3) it would be 2+3 = 5.

That problem seems extremely easy, actually.

[hide]There are going to be 28 subsets, and each number is going to show up seven times (makes sense..7 * 8 = 56, 56/2 = 28). If you want the maximum in each subset, then 34 is going to show up the most, 21 second-most, etc. You get 34 * 7 + 21 * 6 + 13 * 5 + ... + 2*1 as the answer.[/hide]


articles conics ellipse ratio algebra function domain

How to play:

1: Click on the link to a wikipedia article last posted.
2: Click on any one link the [b]body of the article[/b] that [b]links to another page on Wikipedia[/b].
(that is, don't click on "Log In", "Upload File", an outside reference in the citation, or anything like that). Also [b]avoid stub articles[/b].
3: Copy the URL of that page
4: Go back to this topic and paste the URL here.

You are supposed to be able to get from any article by clicking on links from other articles. Let the fun begin.

Oh yeah, and feel free to post quotes from the article that you found interesting/amusing/inaccurate.

ARTICLE #1
http://en.wikipedia.org/wiki/Mathematics

#3:
http://en.wikipedia.org/wiki/Windows_XP

[quote]In the mass market, however, the Starter Edition has not had much success. Many markets where it is available have seen the uptake of illegally cracked or pirated versions of the software instead.[/quote]

#4:
http://en.wikipedia.org/wiki/Internet_Explorer

Go Firefox!
[quote]As of November 7, 2006, Secunia reports 3 unpatched vulnerabilities in Internet Explorer, of which the most severe vulnerability is rated "moderately critical".[22]. In contrast, Mozilla Firefox, the main competitor to Internet Explorer, is reported to have no unpatched security vulnerabilities.[23] Opera, another competitor to Internet Explorer, is also reported to have no unpatched security vulnerabilities.[24]

[b]In May 2006, PC World rated Internet Explorer 6 the eighth worst tech product of all time.[/b][/quote] (My bold)

Hehe!  Yes.  VERY inferior.
[quote]Internet Explorer's ubiquity, in spite of its [b]inferiority [/b]in this area, frustrates developers who want to write standards-compliant, cross-browser code and the advanced functionality it provides, because they are often stuck coding pages around Internet Explorer's bugs, proprietary featureset, and missing standards support instead.[/quote]

Wow...Microsoft must have very poor coding...
[quote]The idea of removing Internet Explorer from a Windows system was first proposed during the United States v. Microsoft case. Critics felt that users should have the right to uninstall Internet Explorer freely just like any other application software. [b]One of Microsoft's arguments during the trial was that removing Internet Explorer from Windows may result in system instability.[/b][/quote]

#6
I hope you know what rRNA is...
http://en.wikipedia.org/wiki/Ribosomal_RNA

#8
http://en.wikipedia.org/wiki/Trimethoprim-sulfamethoxazole

By the way, I forgot to mention, also avoid linking to articles that are stubs.

http://en.wikipedia.org/wiki/Category:Antibiotics

I found this interesting

[img]http://upload.wikimedia.org/wikipedia/commons/3/3b/Cutaneous_abscess_MRSA_staphylococcus_aureus_7826_lores.jpg[/img]

http://en.wikipedia.org/wiki/Krispy_Kreme

I found this insightful

[img]http://upload.wikimedia.org/wikipedia/en/9/95/Krispykremesbeingmade.JPG[/img]

[url]http://en.wikipedia.org/wiki/Wikipedia:Citation_needed[/url]
mwahahaha...now you can only click on 1 linki there

[quote="ashpotter"][url=http://en.wikipedia.org/wiki/Lust]lust[/url][/quote]
There are some links in there I don't exactly want to click on  :huh:

Back on topic:

http://en.wikipedia.org/wiki/Wikipedia:Verifiability

No more style pages!

http://en.wikipedia.org/wiki/Jimmy_Wales

http://en.wikipedia.org/wiki/Mysticism

my religion. :P

http://en.wikipedia.org/wiki/Portal:Spirituality

you should get plenty of topics


Carl is lost in the coordinate plane. Every minute, he moves one unit up, to the left, to the right, or down. How many ways can he be back where he started after 8 minutes?

[hide="hint"]If we make any given move, we have to reverse that move at some point, whether it be before or after the move.[/hide]

It should be more than [hide] 4^4, right[/hide]

[hide="I think"]So we have x+y=4 (because N and S are equal and W and E are equal)

SO we can have (0,4) (1,3) (2,2) (3,1) (4,0)

So if its (0,4) thats like 0 N 0 S 4 W 4 E.  Thats 8C4=70
If its (1,3) thats like 1 N 1 S 3 W 3 E.  THats 8!/(3!3!)=1120
If its (2,2) tahts like 2 N 2 S 2 W 2 E.  Thats 8!/(2!2!2!2!)=2520

the others are repeats so 70*2+1120*2+2520=4900[/hide]

But you didn't count paths like N S S N N S S N. Did you?

Its just 4 W and 4 E (same so i doulbed it)

[quote="13375P34K43V312"][quote="diophantient"][hide]
$\sum^{4}_{k=0}\frac{8!}{(2k!)((8-2k)!)}$?[/hide][/quote]

Yeah. Post your solution.[/quote]
Sweet pwnage right there.
That evaluates to [hide]128[/hide] by the way.
[hide="solution"]Okay, let's say that he makes k moves south.
Then he also makes k moves north.
He then makes 8-2k moves in other ways.
$\frac{8!}{2k!(8-2k)!}$ where k=0, 1, 2, 3, 4.
[/hide]

That can't be right. It must be at least 256, for sure.

Hmm yeah, I think you're missing something, but as far as I know, you're on the right track.

So whats the right answer? Because I get 4900

I don't know, but I think it should be a little lesser...

um what solution? can you not be useless lumps and actually post solutions?

I think this is what Madskillz meant.
[hide]

If there are are 4-N moves and 4-S moves, there are 8C4 = 70
If there are are 4-w moves and 4-E moves, there are 8C4 = 70
If there are 1-W,1-E,3-S,3-N moves, then there are 8!/(3!*3!) = 1120
If there are 3-W,3-E,1-S,1-N moves, then there are 8!/(3!*3!) = 1120
If there are 2-W,2-E,2-S,2-N moves, then there are 8!/(2!*2!*2!*2!) = 2520

Total=4900[/hide]

ya..I already posted a solution earlier like that...but oh well..yea thats what I meant.


No matter how hard I try I cannot solve this:

[i]Let ABC be any triangle. The two medians BN and CM intersect at the point P. Given that the area of the triangle BPC is 22 square units, how many square units are in the area of the triangle ABC?[/i]

Thanks

http://en.wikipedia.org/wiki/Centroid

Read carrefully this page.

With that we can easily show that the area of the triangle over the area of your triangle is equal to 3


$ \\a,b,c>0$ then prove that $ \\frac{2(a^3\plus{}b^3\plus{}c^3)}{abc}\plus{}frac{9(a\plus{}b\plus{}c)^2}{a^2\plus{}b^2\plus{}c^2}\geq33$

[quote="boroo12"]$ \\
a,b,c > 0$ then prove that $ \\
frac {2(a^3 \plus{} b^3 \plus{} c^3)}{abc} \plus{} frac{9(a \plus{} b \plus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2}\geq33$[/quote]
The inequality is equivalents with
\[ \Leftrightarrow \left[ {\frac{{2\left( {\sum {a^3 } } \right)}}{{abc}} \minus{} 6} \right] \minus{} \left[ {27 \minus{} \frac{{9\left( {\sum a } \right)^2 }}{{\sum {a^2 } }}} \right] \ge 0
\]
\[ \Leftrightarrow \frac{{\left( {\sum a } \right)\left( {\sum {\left( {b \minus{} c} \right)^2 } } \right)}}{{abc}} \minus{} \frac{{9\left( {\sum {\left( {b \minus{} c} \right)^2 } } \right)}}{{\sum {a^2 } }} \ge 0
\]
\[ \Leftrightarrow \left( {\frac{{\sum a }}{{abc}} \minus{} \frac{9}{{\sum {a^2 } }}} \right)\left( {\sum {\left( {b \minus{} c} \right)^2 } } \right) \ge 0
\]
which is true since $ \frac{{\sum a }}{{abc}} \minus{} \frac{9}{{\sum {a^2 } }} \ge 0$.  :)


prove that

$ \sum_{k\equal{}1}^n {2n \choose n\plus{}k} k^2\equal{}n\cdot 2^{2n\minus{}2}$

$ \sum_{k\equal{}1}^{n}{2n\choose n\plus{}k}k^{2}\equal{}\sum_{i\equal{}0}^{n\minus{}1}{2n\choose i}(n\minus{}i)^2\equal{}\sum_{i\equal{}0}^{n\minus{}1} \left({2n\choose i}n^2\plus{}{2n\choose i}i^2\minus{}2in{2n\choose i} \right)$.
for the rest use these equalities:

1-$ \binom{2n}{i}i\equal{}\binom{2n\minus{}1}{i\minus{}1}2n$.

2-$ \binom{2n\minus{}1}{i\minus{}1}i\equal{}\binom{2n\minus{}1}{i\minus{}1}(i\minus{}1)\plus{}\binom{2n\minus{}1}{i\minus{}1}$.

3-$ \sum_{i\equal{}0}^{n\minus{}1}{2n\choose i}\equal{}2^{2n\minus{}1}\minus{}\binom{2n}{n}$.


Let $ a$, $ b$, $ c$ be positive real numbers. Prove that

$ \left[\left(a^2b \plus{} b^2c \plus{} c^2a\right)\left(ab^2 \plus{} bc^2 \plus{} ca^2\right)\right]^{1/2}$
$ \geq abc \plus{} \left[\left(a^3 \plus{} abc\right)\left(b^3 \plus{} abc\right)\left(c^3 \plus{} abc\right)\right]^{1/3}$.

There's an ugly solution of just expanding and applying bunching

Here is a not so ugly solution: divide the inequality by abc and denote x=a/b, y=b/c, z=c/a. The inequality becomes: [(x+y+z)(1/x+1/y+1/z)]^1/2>=1+[(x/z+1)(y/x+1)(z/y+1)]^1/3=1+[(x+y)(y+z)(z+x)]^1/3. But (x+y+z)(1/x+1/y+1/z)=(x+y)(y+z)(z+x)+1, and the inequality becomes [(x+y)(y+z)(z+x)+1]^1/2>=1+[(x+y)(y+z)(z+x)]^1/3, which can be easily proved.

Thank - This is my solution
Multiplying 1/abc, then inequality is equivalent:
 (a/c + b/a + c/b)1/2(b/c + c/a + a/b)1/2 >= 1 + [(a2/bc + 1)(b2/ac + 1)(c2/ba + 1)]1/3
 Put x=a/c, y= b/a, z=c/b, then x,y,z>0 and xyz=1, we get: (x + y + z)1/2(1/x + 1/y + 1/z)1/2 >= 1 + [(x/y +1)(y/z +1)(z/x +1)]1/3
 Put t=[(x/y +1)(y/z +1)(z/x +1)]1/3, then (x + y + z)( 1/x + 1/y + 1/z)= t3 + 1, and applying Cauchy's inequality, we get t>=2
 We need prove: (t3+1)1/2 >= 1 + t.
 Squaring, we get: t3>= t2 + 2t (1)
  On the other hand, we have t(t+1)(t-2)>=0 (since t>=2), then (1) is true.


I was beyond bored in math today and was trying to figuring how to explain base numbers to my friends and I figured out something.
x= any number
y=base
z= any number

x + y == x*z mod 10 and y. Is this equation true are not?

what if $ x\equal{}z\equal{}10\equal{}1010_2$ and $ y\equal{}2$?


geometry
3D geometry
trigonometry
calculus
complex numbers

I've never met anyone who can raise i to the i, and most people have no clue of where to start.  It took me over a day to figure out the solution to this, but here you go ... if anyone has a better idea, then please tell me.

Want i ^ i.
e ^ ix = cos(x) + i sin(x)
Let x = (pi / 2)

e ^ i(pi / 2) = cos(pi / 2) + i sin(pi / 2)
e ^ i(pi / 2) = 0 + i(1)
e ^ i(pi / 2) = i
e ^ i(pi / 2) * i = i ^ i
e ^ (i^2)(pi/2) = i ^ i

SOLUTION:  [ e ^ - (pi / 2) = i ^ i ]

this is too complicated for classroom math.  mod, please move this.

Not appropriate belongs in one of the highschool forums!  I don't get over half of it!

That way is the best way and, yes, you are correct.  But you haven't quite found all values of $i^i$.  To do so:

\[i=e^{\left(\frac{\pi}2+2k\pi\right)i}\Rightarrow i^i=e^{-\left(\frac{\pi}2+2k\pi\right)} \ \ \text{for} \ k\in \mathbb{Z}.\]

I don't understand the e^itheta notation. What we're taught in Australia is cisx, which is an abbreviation for cosx + isinx

eg. If z^3 = 8, then:  z^3 = 8cis(0 + 2kpi)   in polar form.

Basically any complex number in polar form is expressed as |r|cisx, where |r| is the magnitude of the distance of the point from the origin, and x is the angle which the line connecting the origin to the point makes with the positive real axis.

(because angle 0 lies on the positive real axis of the argand plane)

Hence taking cube root of boths sides:

z = 2cis (0 + [2kpi]/3)
and you let k = -1, 0 and 1 to obtain 3 roots of 8.

So can some one please explain to me why other places use e^iptheta notation, they don't teach that theory in Australia (or atleast in the state I live in).

$e^x, \sin x,\cos x$ are defined by their Taylor series.  These Taylor series converge for all real numbers, and, more importantly, for all complex numbers.  It's an easy exercise in calculus to show that $e^{i\theta}=\cos \theta+i\sin \theta$ (everywhere).  Advantages of the exponential form are clarity of notation, and also that it explains the 'geometry' of multiplying complex numbers in a much more satisfactory fashion.

Very interesting :)

Just curious, I got $i^{i^i}=-i$. Is it correct?


quadratics
LaTeX
algebra
quadratic formula

Here's a problem for you to solve (alright, copied it from math club, but changed the numbers)

Find the value of

${ \frac{4}{4+\frac{4}{4+\frac{4}{4+\frac{4}{\frac{4+4}{...}}}}}}$

Ok if we set that value equal to zero and then notice that since it repeats we have $ x = \frac {4}{4 + x}$.  Multiplying each side by $ 4 + x$ yeilds $ x^2 + 4x - 4 = 0$ quadratic formula gives us $ \frac { - 4 plus or minus \sqrt {16 - 4(1)( - 4)}}{2}$ which equals $ \frac { - 4 \pm 4\sqrt {2}}{2}$ which equals $ - 2 \pm 2\sqrt {2}$.

EDIT: changed latex for plus or minus

I think there's a typo at the end of the question.  It should be ${ \frac{4}{4+\frac{4}{4+\frac{4}{4+\frac{4}{4+\frac{4}{...}}}}}}$.
I used that in my solution.
[hide="Solution"]
Let ${ x=\frac{4}{4+\frac{4}{4+\frac{4}{4+\frac{4}{4+\frac{4}{...}}}}}}$
Therefore, $ x=\frac{4}{4+x}$
$ (4+x)x=\frac{4}{4+x}(4+x)$
$ 4x+x^2=4$
$ x^2+4x-4=0$
Use the quadratic formula to get $ x=-2\pm2\sqrt{2}$.
Since the answer must be positive, we only take the positive solution:  $ \boxed{-2+2\sqrt{2}}$.[/hide]