linear algebra
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1/ Prove that $ SL(2,\Bbb{Z})$ is generated by $ \left({\begin{array}{*{20}c}0 & 1\\ 1 & 0\\ \end{array}}\right)$ and $ \left({\begin{array}{*{20}c}0 & 1\\ 1 & 0\\ \end{array}}\right)$

2/ Prove that $ SL(n,\Bbb{Z})$ is finitely generated group

3/ Prove that
$ GL(n,\Bbb{R})$ isn't connect topological groups
$ GL(n,\Bbb{C})$ is connect topological groups
$ U(n)$ is connect topological groups (groups of orthogonal unitary matrix)
$ SL(n,\Bbb{R})$ is connect topological groups.

4/ $ K$ is finte fields find the number of elements of the groups  $ GL(,n,K), SL(n,K),PGL(n,K),PSL(n,K)$ in that  $ PGL(n,K) \equal{} GL(n,K)/Z(GL(n,K))$ is general projective group and $ SGL(n,K) \equal{} GL(n,K)/Z(SL(n,K))$( I don' know its name?)

1. You mean $ \left(\begin{array}{cc}0 & 1\\ 1 & 0\end{array}\right)$ and $ \left(\begin{array}{cc}1 & 1\\ 0 & 1\end{array}\right)$, don't you?

2. Prove that $ SL(n,\mathbb{Z})$ is generated by $ I\plus{}aE_{ij}$ where $ a\in\mathbb{Z}, i\neq j$ (Hint: Gauss elimination)

3. Show that $ SL(n,K)$ is generated by elementary row operations (analogue to 2.) and deduce that $ SL(n,K)$ is pathconnected. Then show that every $ A\in GL(n,\mathbb{C})$ is path-connected to a $ A'\in SL(n,\mathbb{C})$. $ GL(n,\mathbb{R})$ connected, because the matrices with positive resp. negative determinants form open, but disjoint sets.

Something stronger on 2: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=853765928&t=36105

Oh thank you very much Gockel and perfect_radio,
$ SL(2,\Bbb{Z})$ is generated by $ \left({\begin{array}{*{20}c}0 & 1\\ 1 & 0\\ \end{array}}\right)$ and $ \left({\begin{array}{*{20}c}1 & 1\\ 0 & 1\\ \end{array}}\right)$ as you said Gockel.
Actually in $ 4/$ I haven't understand about general projective group much? Is this definition related to the groups of projective tranformations of projective spaces, if there can any body show me the relation  :maybe: ? thank you!

Could anybody please help me? I really don't know :maybe:

Yes, the projective groups have a connection to projective spaces: Consider the Operation of $ GL(n,K)$ (and all its subgroups) on $ PK^{n}$ given gy $ A\cdot\langle v\rangle: \equal{}\langle Av\rangle$. The kernel of this operation (that is the set $ \{A\in GL(n,K) |\forall v\in\ K^{n}: A\cdot\langle v\rangle: \equal{}\langle v\rangle\}$) is exactly the center $ Z\equal{}\{\lambda I |\lambda\in K\}$, so there is a natural Operation of $ PGL(n,K)\equal{}GL(n,K)/Z$ (and the image of all subgroups) on the projective space which is faithful. So $ PGL(n,K)$ is a subgroup of the projective transformations group. In fact this group is $ P\Gamma L(n,K)$: The group of all transformations incuded by linear transformations together with the transformations induced by field-automorphisms of K.


superior algebra
superior algebra solved

a) If $p$ is a prime then no matter how we choose a $p$-cycle and a transposition, they will generate $S_p$.
 b) For any $n$ no matter how we choose a $n$-cycle, a $n-1$-cycle and a transposition they will generate $S_n$.

Note that conjugation with a fixed element of $ S_n $ induces an automorphism of $ S_n $, and the correct conjugation permutes the numbers $ 1, \ldots, n $ in the cycle structure of any element of $ S_n $ in whatever fashion we want, allowing us to conveniently 'relabel' a cycle.

a) Using the above, WLOG, let the cycle of $ p $ elements be $ c = (1, 2, \ldots, p) $, and the transposition be $ t = (i, i + d) $. Then $ c^{-k}tc^k = (i + k, i + k + d) $, so we can produce the set of all transpositions of 'difference' $ d $. Under another conjugation, induced by the permutation $ f((i + jd)_p) = j $ (this is a permutation because $ p $ is prime), this set then becomes $ (1,2), (2,3), \ldots, (p, 1) $, which is a standard set of generators for $ S_p $.

b) Again, using the above, WLOG, let the cycle of $ n $ elements be $ c = (1, 2, \ldots, n) $, and let $ n $ be the only number fixed by the cycle $ d $ of $ n - 1 $ numbers (i.e. each of $ 1, \ldots, n - 1 $ appears in the cycle of length $ n - 1 $). Let the transposition be $ t = (i, j) $. Since $ c $ is a cycle of length $ n $, we must have $ i = t^k(n) $, for some $ k $. Now, the conjugate $ p = t^{-k}dt^k $ of $ d $ must have the same cycle structure as $ d $, i.e. a cycle of length $ n - 1 $. $ i $ is fixed under $ p $. So $ p^{-l}tp^l = (i, p^l(j)) $, and $ \{ p^l(j) : 1 \leq l \leq p-1 \} = \{ 1, 2, \ldots, n-1 \} \backslash \{ i \} $, since the possible values of $ p^l(j) $ must run through the numbers in the cycle of $ p $. Now, the set of all such $ (i, p^l(j)) $ is easily seen to generate $ S_n $.


[quote="tranthanhnam"]prove that :where $a,b,c > 0$ and$a+b+c=1$
$\sum{\frac{a \sqrt{b}}{\sqrt{b+c}}}\leq \frac{3\sqrt{3}}{4}{(a+b)(b+c)(c+a)}$[/quote]

This looks too much like problem 1 of the ongoing 1st Mathematics and Friends contest to me (it was slightly different, but it looks like you got the inspiration from that problem). Please don't discuss this until the deadline: November 15, 2006.

  darij


[b]1.[/b] Demonstrati ca $ 0<\equal{}(z_1\plus{}z_2\plus{}...\plus{}z_n)(\frac{1}{z_1}\plus{}\frac{1}{z_2}\plus{}...\plus{}\frac{1}{z_n})<\equal{}n^2$ unde $ z_1,...,z_n$ sunt numere complexe de modul 1.

[b]2. [/b] $ |z_0\minus{}z_1|^2\plus{}|z_0\minus{}z_2|^2\plus{}...\plus{}|z_0\minus{}z_n|^2 <\equal{} |z\minus{}z_1|^2\plus{}|z\minus{}z_2|^2\plus{}...\plus{}|z\minus{}z_n|^2$ unde $ z_0\equal{}(z_1\plus{}z_2\plus{}...\plus{}z_n)/2$ si $ z_i$ sunt numere complexe.

[b]1. [/b] [hide]\[ (\sum_{j\equal{}1}^n z_j)(\sum_{j\equal{}1}^n \frac{1}{z_j})\equal{}(\sum_{j\equal{}1}^n z_j)(\sum_{j\equal{}1}^n \bar{z_j})\equal{}\sum_{j\equal{}1}^n z_j \cdot \bar{\sum_{j\equal{}1}^n z_j}\equal{} | \sum_{j\equal{}1}^n z_j|^2\] de unde \[ 0<\equal{} | \sum_{j\equal{}1}^n z_j|^2 <\equal{} n^2\equal{}(\sum_{j\equal{}1}^n |z_j|)^2\][/hide]

[b]2.[/b] [hide]\[ \sum_{j\equal{}1}^n | z\minus{}z_j|^2\minus{}\sum_{j\equal{}1}^n |z_0\minus{}z_j|^2 \equal{} \sum_{j\equal{}1}^n (z\minus{}z_j)(\bar{z}\minus{}\bar{z_j}) \minus{}\sum_{j\equal{}1}^n (z_0\minus{}z_j)(\bar{z_0}\minus{}\bar{z_j})\equal{}nz\bar{z} \minus{} z \sum_{j\equal{}1}^n \bar{z_j} \minus{} \bar{z}\sum_{j\equal{}1}^n z_j\minus{} nz_o \bar{z_o} \plus{} z_0 \sum_{j\equal{}1}^n \bar{z_j} \plus{} \bar{z_0} \sum_{j\equal{}1}^n z_j\equal{}n(z\minus{}z_0)(\bar{z}\minus{}\bar{z_o})\equal{}n|z\minus{}z_o|^2 \geq 0\][/hide]


Texas Instruments is annoucing this spring TI-84 which is fully compatible with TI-83 Plus There will be more memory and a built in USB port. They are anno9ucing this summer TI-89 Titanium Edition which also has similar features. It seems like TI is going up to challenge the HP 49G+ which has a flash memory card, and a usb port.

but they are going to stop the production of ti-89's and silver edition's when the newones come to the store. the only one product they aren't goiing to stop is the voyager 200 although many companies stock up on ti-83s i

Most likely you would own one and win one. (in that order)

Interestingly enough, my family owns four (different) TI calculators, only one of which we bought.

In 7th grade I got a (standard black) TI-83+ because I needed one for geometry class.  Then I went to mathcounts nationals and got a Ti-83+ SE.  In 8th grade I got a TI-89 from mathcounts nationals.  In 9th grade I won a Ti-83+ SE from the Ole Miss POW site, but they sent me the wrong calculator.  I figured it would be rude to complain, seeing that they were giving me a free calculator anyway, so I gave the Ti-73 Explorer to my little brother.

Now that I think about it, what justifies the existence of a Ti-73 Explorer?  They're basically Ti-83+s, but dumbed down and made incompatible with the most common TI graphing calculator.

So no, I don't think I'll be getting one of those new calculators.  Besides, a TI-89 Titanium sounds...oh, I don't know, kind of dorkily retro, and I'm just totally worried about my fashion image.  </jk>

[quote="confuted"]You failed to mention that they will be more expensive than the current models. I'll stick with my TI-89 [i]old school edition[/i] unless somebody gives me something else that I like better.[/quote]

Well, the 84+ is supposed to retail at $110, which is about what the 83+ cost for a while.  (The price later dropped to $87.00 at Wal-Mart or somat.)  The TI-89 Titanium is supposed to cost $150, which is about what the current Ti-89 costs.  So the price increase isn't that much.  If I were in the market for a new calculator, I'd probably buy it.  But I'm not.

The HP 49G+ is far superior to any TI calculator...

TI 89 is a close second... (TI-89 Titanium is nothing more than some more memory anyhow)

everything else is far behind....

(I have used every single calculator from the 83, to 83+, to 85, 86, 89, 92, Casios, HP 48G, 49G, and various others insignificant ones)

yeah the hp has a prcocessor that is like six times faster. any way according to the tests at http://alpage.ath.cx/hptute/compare.htm the hp 49+ won 2/3 of them. but ti-89 has more memory and more programs,  the screen is bigger, price
hp has faster processor, memory cards (upt to 512mb)

lol but i don't really care about wich calculator i have virtual ti (82,83,83+,83se,86,89,92) and some virtual hps and it seems that ti-89 is much cooler. but i don't use them maple and mathmeatica wastes any calculator

[quote="churchilljrhigh"]you guys are lucky. i never was rich enough to buy a nice graphing calculator. i had a $40 casio grapher since 6th grade. i then won a ti-83 (state math contest) plus and a ti-89 (i wish i did mathcounts in 7th grade ti-83 se are awesome).
 but one thing is for usre if you don't have a ti-89 get one it'll raise your AP score alot!!!!!!!! the ti-89 gives the most bang for the buck![/quote]

Hmm...that's interesting.  You got a calculator for winning state and a calculator for going to nationals?  That's like getting paid twice.

yeah i should get two another for 8th grade. however the university changed. it seems like private BYU (U.S's most sober and religious univeristy) gives better prizes. they give a math textbook, a fun math adventure book and a calculuator and a $50 check, and have steak at the award's assembly. weber on the other hand gives a $100 barnes and noble card and a free planetarium ticket a trophy and chicken. (they give top schools $300 checks though)
i think that is unecessary lol.

You should buy HP.  I've had an HP 49G for the past 2 years and it  has served me well.  One of my favorite features is the equation writer - it lets you input the equation almost EXACTLY like it looks on the paper.  Another one is the RPN.  This is kinda hard to get used to, but still, its better than normal calculator modes once you get used to it.  Last but not least, there's lots of cool apps and games. Yes, I know Ti has more, but HP's are better.  http://www.hpcalc.org


inequalities
function
inequalities unsolved

Let $ a,b,c > 0$ Prove that \[ \sum_{cyclic}a\sqrt{8a^4\plus{}b^2c^2} \leq 3(a^3\plus{}b^3\plus{}c^3)\]

[quote="can_hang2007"][quote="anonymous111"]Let $ a,b,c > 0$ Prove that
\[ \sum_{cyclic}a\sqrt {8a^4 \plus{} b^2c^2} \leq 3(a^3 \plus{} b^3 \plus{} c^3)
\]
[/quote]
The hint is to use the Jensen Inequality. :)[/quote]
Can you explain?
which function ?

[quote="can_hang2007"][quote="anonymous111"]Let $ a,b,c > 0$ Prove that
\[ \sum_{cyclic}a\sqrt {8a^4 \plus{} b^2c^2} \leq 3(a^3 \plus{} b^3 \plus{} c^3)
\]
[/quote]
The hint is to use the Jensen Inequality. :)[/quote]

Can you post your solution ?  :blush:

Thanks


Is geometric mean defined for nonpositive reals?
If so, either the definition would be different or it would violate AM-GM...

AM-GM is defined as only positive integers  :wink: .

[hide="Logical Progression"]
Let's say $ n\equal{}4$, so we have, on the GM side, 4 numbers that will become positive after simplified. The fourth root will yield a positive GM side. However, on the AM side, it will generically be negative, therefore the inequality is proved false for all negative integers.
[/hide]

Ahh... right.

So geometric mean is still defined as it is for positives but AM-GM only holds for positives.

Integers?  The geometric mean is defined for any odd-numbered collection of [i]reals[/i] and for an appropriate even-numbered collection of [i]reals[/i] (an even number of which must be negative), but the geometric mean loses its meaning except with respect to non-negative reals.

In particular, the "geometric" aspect of the geometric mean refers to lengths of certain line segments, which are clearly non-negative.

[quote="t0rajir0u"]The geometric mean is defined for any odd-numbered collection of [i]reals[/i]....[/quote]

Why can't an even-numbered collection work?

If an odd number of them are negative, we'll need to take the square root of a negative number.

This is somewhat analogous to defining the domain of the function $ f(x) \equal{} x^x$.  Technically, it can be given sensible values for negative rational $ x$ with odd denominator in lowest terms, but for nearly all intents and purposes it makes more sense just to give its domain as $ [0, \infty)$ (the singularity at $ 0$ is removable).


The sides of a triangle ABC, (right triangle) measured 5, 12 and 13. Calculate  $ \vec{AB}.\vec{AC} \plus{} \vec{BA}.\vec{BC} \plus{} \vec{CA}.\vec{CB}.$

Since $ 5^2\plus{}12^2\equal{}13^2$, the triangle is a right angled triangle. Which angle is $ 90^\circ ?$

kunny A is a right angle, $ A \equal{} 90^\circ$

OK. You need not to give the condition right triangle. :lol:
You give that which side is hypotenuse.

Suppose that $ AB \equal{} 5,\ AC \equal{} 12$ and $ BC \equal{} 13$, we have $ AB^2\plus{}AC^2\equal{}BC^2$, so $ A\equal{}90^\circ$.

$ \overrightarrow{AB}\cdot \overrightarrow{AC} \equal{} 0,\ \overrightarrow{BA}\cdot \overrightarrow{BC} \equal{} |\overrightarrow{BA}|^2 \equal{} 5^2,\ \overrightarrow{CA}\cdot \overrightarrow{CB} \equal{} |\overrightarrow{CA}|^2 \equal{} 12^2$, yielding the answer is $ \boxed{169}.$


In a pet store, $ \frac{1}{2}$ of the animals are fish, $ \frac{1}{10}$ are dogs, $ \frac{1}{5}$ are birds and the remaining six are cats. How many animals are in the
store?

Let $ x$ be the total number of animals in the store.

$ x\minus{}\frac{x}{2}\minus{}\frac{x}{10}\minus{}\frac{x}{5}\equal{}\frac{x}{5}\equal{}6$
$ x\equal{}30$

So there are $ \boxed{30}$ animals in the store.


consider a function connected by n variables f(x,y,z,...........) subject to m different constraints.find a maxima or a minima of such function Hence and otherwise maximise the foll. function

x+y+z+a+b=3025

x+y<600
y-z<250
x+y+z<305
a+b<1500
a+b+z<or =520
a,b,x,y,z>0

[quote="JRav"]Looks like you can use Lagrange multipliers.[/quote]
That's a useless comment. Did you notice that all of the functions and constraints were linear? This is a linear programming problem, and I'm moving it out of the calculus section.

Yes, right Ifigured it out myself recently.

Lagranges multipliers help only when no of constraints are less than the the no. of variables.

Thaks anyways. :|


Let $ f$ be a polynomial such that $ f(i) \equal{} a_i$ for all $ i$ such that $ 0 \leq i \leq n$. If $ a_i \in \{0,1\} \forall i$ and $ n\plus{}1$ is a prime, prove that $ \text{deg}(f) \geq n$
[hide="hint"] Interpolation and working in $ \mathbb{Z}/p\mathbb{Z}$[/hide]

You are wrong. Let $ n\equal{}1,a_0\equal{}a_1\equal{}1,f(x)\equiv 1$, then $ p\equal{}n\plus{}1\equal{}2$ is prime.
I think right question is n is prime all $ a_i$ are integer and $ n\not a_n\minus{}a_0.$ then prove, that $ deg(f(x))\ge n$.
Solution.
Let $ D: f(x)\to f(x\plus{}1)\minus{}f(x)$, Obviosly if f(x) is polinom, then $ deg(Df(x)\equal{}deg(f(x))\minus{}1$, therefore $ D^{n}f(x)\equiv 0$ if $ deg(f(x)<n$. Let $ n$ prime, then we had $ D^{n}f|_{x\equal{}0}\equal{}\sum_{k\equal{}0}^{n}C_{n}^k(\minus{}1)^kf(n\minus{}k)\equal{}\sum_{k\equal{}0}^n C_{n}^k(\minus{}1)^ka_k\not \equal{}0$.

Make it odd prime. Sorry for the mistake.

Let $ a_i=1$ for all i, then $ f\equiv 1$, therefore $ deg(f)=0$.
Secjnd condition is:
Let $ a_i\in\{0,1\}$, and $ 0<\sum_{i=0}^n a_i <n+1=p$, p is prime, then $ deg(f)\ge 1$.
Prove. $ D^nf|_{x=0}=\sum_{k=0}^nC_{p-1}^k(-1)^ka_k=\sum_{k=0}^na_k%Error. "modp" is a bad command.
\not =0$.


geometry
analytic geometry
conics
ellipse
geometry proposed

Let $M$ be a point in the plane of $\triangle ABC$ not lying on its sides. If $S_{1}$ , $S_{2}$, $S_{3}$ are the areas of $\triangle AMB$, $\triangle BMC$, $\triangle AMC$, respectively, find the locus of point $M$ satisfying the relation:
 $(MA^2+MB^2+MC^2)^2=16(S_{1}^2+S_{2}^2+S_{3}^2)$.

Answer : two points.
It is possible to prove that there exist two points P, Q such as, for any point M, we have
(MA^2+MB^2+MC^2)^2-16(S1^2+S2^2+S3^2) = 9 MP^2 MQ^2 but the problem is to localize these points.
The following seems very tedious; there is probably a better answer.

I use cartesian coordinates with origin the centroid G, the x-axis and the y-axis being respectively the major axis and the minor axis of the Steiner circumellipse. Let a = semimajor axis, b = semiminor axis.
As the mapping [x,y] -> [x, ay/b] maps ABC to an equilateral triangle inscribed in the circle (G,a), there exists t such as
A = [a cos(t), b sin(t)]; B = [a cos(t+2Pi/3), b sin(t+2Pi/3)]; C = [a cos(t-2Pi/3), b sin(t-2Pi/3)].
If M = [x,y], a little computation gives
MA^2 + MB^2 + MC^2 = 3x^2+3y^2+3/2(a^2+b^2) and
16(S1^2+S2^2+S3^2) = 18(b^2x^2 + a^2y^2) + 9a^2b^2
Putting c = root(a^2-b^2); P = [0, c/root(2)] and Q = [0, -c/root(2)], it follows that
(MA^2+MB^2+MC^2)^2-16(S1^2+S2^2+S3^2) = 9 MP^2 MQ^2
Thus P and Q lie on the minor axis of the ellipse such as GP = GQ = GF/root(2) where F is a focus.
Kind regards


number theory
prime numbers
number theory unsolved

Solve the system

$ x^2\plus{}yu\equal{}(x\plus{}u)^n, x^2\plus{}yz\equal{}u^4$,

where $ x,y,z$ are prime numbers, and $ u$ a positive integer.

Hmm... I think $ n$ is any positive integer

I don't know if it can be solved or not.

Its simple.....search only for $ n\equal{}2,3$, and $ n\equal{}4$. :wink:

[quote="Αρχιμήδης 6"]Its simple.....search only for $ n \equal{} 2,3$, and $ n \equal{} 4$. :wink:[/quote]

Umm... please explain why that is so.

$ x \plus{} yu \equal{} (x \plus{} u)^n$ and $ x^2 \plus{} yz \equal{} u^4$
From the second we can see that $ y < u^4$. So from the first we have:$ (x \plus{} u)^n \equal{} x \plus{} yu < x \plus{} u^5 < (x \plus{} u)^5 \equal{} > 1 < n < 5$....

[quote="Αρχιμήδης 6"]Its simple.....[/quote]

OK, I still don't think it's that simple...... please continue


BTW; The standard equation of a circle is (x-h)²+(y-k)²=r².

r = radius
(x,y) = point on circle
(h,k) = center

Ignore this, meant to hit reply; my mistake.


group theory
abstract algebra
superior algebra
superior algebra theorems

Right now, I'm trying to prove that for a subgroup $H$ of a group $G$, there are an equal number of left and right cosets. In the case where $G$ is an infinite group, and $G$ has an infinite number of left and right cosets, I believe that I have found a proof that there are an equal number of left and right cosets.

However, once I prove that there are an equal number of left and right cosets in the case that there are an infinite number of left and right cosets, does the finite case (that is, there are a finite number of left and right cosets) follow as a special case?

Given two (finite or infinite) sets $A$ and $B$, we may informally say that they "have the same number of elements" if there is a bijection $f\colon A\to B$. A formal statement would be that $A$ and $B$ [i]have the same cardinality[/i].

I don't see why the finite case would follow from the infinite case.  But if you found a bijection between the set of all left cosets and the set of all right cosets, then it's very likely that it will work for both finite and infinite case alike.


trigonometry
linear algebra
matrix
induction
strong induction

Consider the $ n\times n$ matrix $ [a_{ij}]$ where $ a_{11}=\cos \theta,a_{ii}=2\cos \theta (i=2,3,...,n),a_{i,i+1}=a_{i+1,i}=1(i=1,2,...,n-1)$ , and all other elements are zero. Prove that $ |a_{ij}|=\cos n\theta$.

Call the $ n \times n$ matrix $ [a_{ij}]$, $ A_{n}$.  Then since it is tridiagonal, if we expand along the bottom row we have:

$ |A_{n}|=2 \cos \theta |A_{n-1}|-|A_{n-2}|$.

Now we use strong induction and the above relation.

For $ n=2$ we have:  $ |A_{2}|=2 \cos^{2}\theta-1 = \cos 2 \theta$ which is correct.  Assume $ |A_{n}|=\cos n \theta$ for all $ 1 \leq i \leq n$.  We will show $ |A_{n+1}| = \cos (n+1) \theta$.

$ |A_{n+1}|=2 \cos \theta |A_{n}|-|A_{n-1}| = 2 \cos \theta \cos n \theta-\cos (n-1) \theta$

Now we want to conclude the righthand side is equal to $ \cos (n+1) \theta$.   But this follows easily if we write:

$ \cos (n+1) \theta = \Re e^{i(n+1)\theta}$ and expand similarly on the righthand side and compare.


Is there a way of solving this without using guess and check?

A non-square rectangle has integer dimensions. The number of square units in its area is numerically equal to the number of units in its perimeter. What is the number of units in the perimeter of this rectangle?

I don't think so, but it's quite obvious that 6x3 rectangle works, so 18 units.

Let the length be $ l$ and let the width be $ w$.

We have:

$ 2l \plus{} 2w \equal{} wl\implies{wl \minus{} 2l \minus{} 2w \equal{} 0}\implies{(l \minus{} 2)(w \minus{} 2) \minus{} 4 \equal{} 0}\implies{(l \minus{} 2)(w \minus{} 2) \equal{} 4}$

We are given that the rectangle is not a square so $ l \minus{} 2$ cannot equal $ w \minus{} 2$. The only other possiblity is for $ l \minus{} 2$ to equal 4 and $ w \minus{} 2$ to equal 1. Thus $ l \equal{} 6$ and $ w \equal{} 3$.


[quote="fdump"]So...A bear falls into a hole in the ground. The hole is 10  metres deep. After one second the bear reaches the bottom of the hole. What colour is the bear?[/quote]Green?, violet?, fuchsia?, something similar. I am not too good on phisics, but I would say that a body that is "inert" (without initial velocity, or with initicial velocity 0) would fall $\frac{gt^2}{2}$ meters after $t$ seconds, where $g\approx9.79m/s^2$ (with a not that big variance along the globe), hence, the bear would fall $\approx4.9$ meters during the first second (say that you are in a place of the earth that is soo near to the poles, then may be it could reach 6 meters, but I don't believe it). Then what you say is not possible on Earth, then may be the bear lives on Jupiter of some other more masive planet.

haha, this is a famous one, i have heard this before,
i think this problem had like a \$1000 reward or something :yup:
it was the bear falling into a well of water, ...(same thing)

I don't know about the reward, should be nice. :). I haven't won any money myself on solving it :).Still, the problem is real and correct, and it's all happening on Earth.

The bear must tumble a bit in the mud, so it is brown?

is this physics related, or is it just a brain teaser?

it's not a brain teasser; it's the real thing. as for physics the  relation of it to the problem is the one mentioned by djimenez. :)

probability, what do you think????
of course not!!!

Well it made sense... the bear had to tumble a bit instead of falling straight down --> the bear got dirty.

[hide="my guess is"]white[/hide]

[hide="because"]I feel like it has to do something with the north pole(thats the only reason I can come up with why he asked the color of the bear)[/hide]

not white. nothing to do with the north pole. and try to think not guess, it's not that hard as it looks like when you read it.   :)

I was thinking about this.

A bear walked a mile south, then turn 90 degrees and walked a mile east, and then turn another 90 degrees and finally walked a mile north, but came back to where it was before it walked three miles.
What color is the bear?

[hide="of course"]white[/hide]

[quote="Geehoon"]http://www.everwonder.com/david/panda/
Scientists have debated for more than a century whether giant pandas belong to the bear family, the raccoon family, or a separate family of their own.

http://www.worldofboxes.com/animals/pandas.htm
Zoologists have long debated how to classify pandas in the animal kingdom - as part of the bear family, racoon family, or a family all their own.

Yes, now I know that "giant panda" is closer to bear than it is to racoon, but it is not quite a bear.[/quote]

There is no point of arguing about this. We got the answer and that is it.

Since when did pandas climb stuff?  I didn't know that...

[quote="mathnerd314"]I thought other types of bears could climb trees too... :?[/quote]
You are right. Some bears can climb a tree. But very few species can do that(as far as I know). :D

Okay, just for theheck of it, i'll post another bear problem (this one is pretty easy).
There is a house with four south-facing walls which are arranged in a rectangle.  There is a window in one of the walls.  A bear walks by the window.  What color is the bear's fur?

[hide]
Presumably the walls intersect each other. The only place on Earth where every direction is south is in the Arctic. Therefore the bear is white.
[/hide]

There is already a thread discussing this.

unfortunately the north pole is underwater so therefore the bear would be swimming, not walking. However, if we weren't going to take that into account it would be white due to it being at the north pole.

It hasn't completely melted has it?  The very middle, the North Pole where every direction is south, still is ice right?  Besides, the bear I'm talking about swims anyways.
And the fur color isn't white.  :D It's another zoology question

Suppose the spot we are talking about is not ice. Then it is on water, and consequently the house is underwater, and the bear is not walking. This contradicts your statement. Therefore we are talking about a place that is ice. Q.E.D.

Oh, please. Let's stop with this silly argument. The point of the problem was to get the fact that the house is at the north pole. It is not meant to cause arguments on swimming and fur colour. :D

eh? don't polar bears have white fur and black skin? Maybe I'm mistaken but that's what I remember....

Is it peach? or watever his HUMAN skin color is......

Dictionary.com's definition number 2 under nouns:
2.  A large, clumsy, or ill-mannered person.

A clumsy person would fall into a hole.....


vector
linear algebra
linear algebra unsolved

Find all $f$:$M(n,C)\rightarrow M(n,C)$ such that for any $A,B \in M(n,C)$
$f(A+B)+f(A-B)=2f(A)+2f(B)$

The set of such maps is the vector space generated by maps of the form $f(X)=P(X)Q(X)$, where $P$ and $Q$ are both linear maps. Unfortunately, as written, that's $\mathbb{Q}$-linear, which gives everything uncountable dimension.
The answer could be expressed relatively nicely in tensor language, especially if you add a hypothesis to make those maps linear over a nicer field.


the numerator is $ \sqrt(360\plus{}x)\equal{}x  x^2\minus{}x\minus{}360 \equal{}0  x\equal{}\minus{}1\plus{}\sqrt(1441)$
Denominator = $ \frac1{(k\plus{}1)}\minus{}\frac1{(k\plus{}2)}$
after summation, we get $ (\frac12 \minus{} \frac1{102})*(\minus{}1\plus{}\sqrt{1441})*100$
simplified 2500(-1+sqrt(1441)/51


Let f from C^2[a,b], f'(a) = f'(b) = 0, show that there exists c from (a, b) such that   f"(c) >= 4|f(a)-f(b)|/(b-a)^2.


Hello,

This forum seems very helpful. Could somebody spare sometime and help me with this question.

What are different type of sampling?

I know some of them random sampling,stratfield sampling etcc but I don't whats the definition. could you please give me definition for these along with some others.

Thank You.

Charikaar

Hi there.  The forum you have posted this in is for beginning problem solvers in middle school.  Your question would probably belong much better in the "Other Problem Solving Topics" forum of the High School section of this site.

One definition of sampling (from mathworld.wolfram.com , an exceptionally useful site) is "In statistics, sampling is the selection and implementation of statistical observations in order to estimate properties of an underlying population."  However, you really need someone who knows more statistics than I to answer this question for you.

Simple random sampling is choosing names out of a hat. An equivalent is assigning each experimental unit a number and use a random numbr generator to choose the units. Each unit has the same chance of being chosen and each n-sized group has the same chance of being chosen.
Stratified random sampling is dividing your experimental units into groups (stratas) and then applying the simple random sample to each group. You can have a stratified random where each unit has the same chance of being chosen, but each n-sized group can never have the same chance of being chosen.
An example:
Say there are 100 boys and 150 girls. If you divide them into groups according to their gender and pick 10 boys and 15 girls, each person has a 1/10 chance of being chosen. But it is impossible for you to pick anything other than 10 boys and 15 girls, like all boys. This is a stratified random sample.
However, if you put all the names of the boys and girls in a hat, and randomly pick 25 names out of the hat, this is a simple random sample, since each person has the same chance of being chosen and each 25 person group is equally likely to occur.
The other two I know of are systematic random sampling and multistage random sampling.
In systematic random sampling, you say in advance what system you use, like picking every tenth person, and use a simple random sample to pick the first. From the example above, say you list the names alphabetically, and you pick one person with a simple random sample. Then to get a sample of 25 people, you would take that person and every tenth person on the list after him/her. If you picked the 4th person, then your sample would be the 4th, 14th, 24th,...,244th.
Multistage random sampling involves using a simple random sample many times to choose a small sample from a large population.

I'm moving this, you can probably get more answers at Other Problem Solving Topics.

Hello,

Thanks for your replies. I am pretty new to this forum and didn't know where to post.

I am sorry to have cause you any inconvenience.

Charikaar


Let $ ABCD$ be a quadrilateral, $ AC\cap BD\equal{}\{O\}$. $ d,d'$ are two lines contain $ A,C$. $ P\in d,Q\in d'$ such that $ P,O,Q$ are collinear. $ DQ\cap AP\equal{}\{M\},BP\cap CQ\equal{}\{N\},BM\cap CQ\equal{}\{E\},DN\cap AP\equal{}\{F\}$. Prove that $ E,O,F$ are collinear.

Note that, compare with the problem in the post [url=http://www.mathlinks.ro/viewtopic.php?t=319948]Collinear points[/url]

Let $ O' \equiv EF \cap PQ$. Apply Pappus' theorem to $ \{P,M,F\} \subset d$ and $ \{Q,N,E\} \subset d'$ we have $ B \equiv PN \cap EM$, $ O' \equiv PQ \cap EF$ and $ D \equiv MQ \cap NF$ are collinear. Thus $ O' \equiv PQ \cap BD \equiv O$ and we are done.

[img]http://i524.photobucket.com/albums/cc322/khanhto-photo/NewBitmapImage2.jpg[/img]

From[b] Pappus' theorem[/b] ,applied to $ (MDQ)$ and $ (NBP)$ ,we conclude that $ O,E,F$ are collinear .
:)


Two numbers whose sum is $ 6$ and the absolute value of whose difference is $ 8$ are roots of the equation:

$ \textbf{(A)}\ x^2\minus{}6x\plus{}7\equal{}0 \qquad
\textbf{(B)}\ x^2\minus{}6x\minus{}7\equal{}0 \qquad
\textbf{(C)}\ x^2\plus{}6x\minus{}8\equal{}0 \\
\textbf{(D)}\ x^2\minus{}6x\plus{}8\equal{}0 \qquad
\textbf{(E)}\ x^2\plus{}6x\minus{}7\equal{}0$

[hide]
You can easily see that the two numbers are $ 7 \text { and } \minus{}1$.
So the equation is $ (x \minus{} 7)(x\plus{}1) \equal{} x^2 \minus{}6x\minus{}7 \implies \boxed {B}$.[/hide]

$ x_1 \plus{} x_2 \equal{} 6$ and $ |x_1 \minus{} x_2| \equal{} 8 \Rightarrow x_1 \equal{} 6 \minus{} x_2 \Rightarrow |6 \minus{} 2x_2| \equal{} 8 \Longrightarrow 6 \minus{} 2x_2 \equal{} 8$ so $ x_2 \equal{} \minus{} 1$ and $ x_1 \equal{} 7$
So the equation is $ (x \minus{} 7)(x \plus{} 1) \equal{} x^{2} \minus{} 6x \minus{} 7$ or $ \boxed B$

[hide="Alternate Solution"]
Let $ x$ and $ y$ be the numbers mentioned. We have $ x \plus{} y \equal{} 6$ and $ |x \minus{} y| \equal{} 8$. Because we are looking for a polynomial with roots $ x$ and $ y$, we might as well try to find the value of $ xy$ to find such an equation. We can start by squaring the second equation.

$ (|x \minus{} y|)^2 \equal{} x^2 \plus{} y^2 \minus{} 2xy \equal{} (x \plus{} y)^2 \minus{} 4xy \equal{} 6^2 \minus{} 4xy \equal{} 8^2\implies \minus{}4xy \equal{} 28\implies xy \equal{} \minus{}7$.

We have $ x \plus{} y \equal{} 6$ and $ xy \equal{} \minus{}7$ so the polynomial that has roots $ x$ and $ y$ is $ \boxed{x^2 \minus{} 6x \minus{} 7\text{ (B)}}$.
[/hide]


OMGOMGOMGOMGOMGOMGOMGOMG

what's a rapper?

a singer of a type of genre of music called rap.

[quote="1=2"]what's a rapper?[/quote]

um, wtf?


calculus
integration
linear algebra
matrix
limit
function
complex numbers

$A$ matrix such that all its eigenvalues are $<0$

Compute $\int_{0}^{+\infty}e^{At}dt$

It should be sufficient to say that all of the eigenvalues of $A$ have negative real parts. That is enough to conclude that $\lim_{t\to\infty}e^{tA}=0.$

Then, a primitive for $e^{tA}$ is $A^{-1}e^{tA}$ and

$\int_0^{\infty}e^{tA}\,dt= A^{-1}e^{tA}|_0^{\infty}=-A^{-1}.$

Closely related and of considerable practical use: the Laplace transform of the function $e^{tA}$ is $(sI-A)^{-1}.$ This formula is valid as long as the real part of $s$ is greater than the maximum of the real parts of the eigenvalues of $A.$


calculus
integration
trigonometry
absolute value
calculus computations

\[ \int\limits_{ \minus{} 1}^{\frac{3}
{2}} {\left| {x\sin \pi x} \right|}
\]

break the integral into intervals where the argument in the absolute value is only positive or only negative, then eliminate absolute values and integrate by parts.

$ \int_{\minus{}1}^{3/2} |x \sin \pi x| dx \equal{} \int_{\minus{}1}^0 x \sin \pi x dx \plus{} \int_0^1 x \sin pi x dx \plus{} \int_1^{3/2} \minus{}x\sin \pi x dx$

Now
$ F(x) \equal{} \int x \sin \pi x dx \equal{} \minus{}\frac{1}{\pi}x\cos \pi x  \plus{} \int \frac{1}{pi} \cos \pi x dx$
$ \equal{} \minus{}\frac{1}{\pi}x\cos \pi x \plus{} \frac{1}{\pi^2}\sin \pi x$

$ F(\minus{}1) \equal{} \minus{}\frac{1}{\pi}$
$ F(0) \equal{} 0$
$ F(1) \equal{} \frac{1}{\pi}$
$ F(3/2) \equal{} \minus{}\frac{1}{\pi^2}$

The value of the integral is then
$ [F(0)\minus{}F(\minus{}1)]\plus{}F(1)\minus{}F(0)\minus{}[F(3/2)\minus{}F(1)] \equal{} \minus{}F(\minus{}1)\plus{}2F(1)\minus{}F(3/2) \equal{} \frac{3}{\pi}\plus{}\frac{1}{\pi^2}$


I just have noticed this...
since when does "<=" not work in $ \text{\LaTeX}$
I mean untill yesterday (not sure about the exact time) by putting "<=" between the dollar signs we would get what we want,but now it gives us $ <=$...
why is this happening? :maybe:

That's an issue with the forum setup so I've moved this thread to support.

I've completely changed the latexrender class methods, for expansion on the Wiki and other subsystems we have on the site. A few bugs might arrise to the different internal encodings of some of our software. I will see what I can do (though it's highly recommended you use \leq directly).

[quote="i_like_pie"]Should "\le" be used, or "\leq"? The first is obviously easier to type (one less character :D), and they produce the same result.[/quote]They don't produce the same result: $ a\leq b$ and $ a\le b$. Okay it seems that weirdly enough here they produce the same result but they shouldn't.

[quote="Valentin Vornicu"]Okay it seems that weirdly enough here they produce the same result but they shouldn't.[/quote]Are you thinking about the difference between $ \le$ and $ \leqslant$?

Another instance I just encountered: || becomes "parallel". I wanted the vertical bars, of course.

Well, you could have typed "modulus of something or other" as $ |\cdot |.$ Or even $ |\,|$ or $ |\ |$.

The correct syntax is \parallel and \leq. The other ones are parsed just because some people are too lazy or don't know the proper syntax.

If I wanted $ \parallel$, I would have used it; I was trying to do something else and encountered this bug by accident. The bug is that it gets parsed into something nobody wants- the word "parallel" rather than either the symbol or the two vertical bars.
I also tried putting a space in to break up the expression; that gave me $ | |$, which I decided I didn't like.

Well most people use || as \parallel so that's how it's parsed. Visually they are almost the same, but it's better for me to parse it as parallel.

I always thought it was \|.

\|: $ \|$

\parallel: $ \parallel$

There is a considerable difference between \| and \parallel, having to do with spacing:

a \| b is $ a\|b$
a \parallel b is $ a\parallel b$

The symbol \| is used for norms: $ \|x\|_2^2.$ Or $ \left\|\frac{\langle 1,x\rangle}{\sqrt{1\plus{}x^2}}\right\|.$

Basically you shouldn't use || at all :)


calculus
derivative
function
calculus computations

consider f (x) = x^3−x+1

Show that f (x) has exactly one root c, and that c is in (-2,-1).

I did the latter part but can't prove the first part using Rolle's theorem.

Can someone offer any assistance? Thanks.

Since the derivative has two zeros, we don't have a simplistic use of Rolle's Theorem here. Let's move past Rolle, through the Mean Value Theorem, and into using the derivative to determine where a function is increasing and decreasing.

$f(x)$ has a local maximum at $x=-\frac1{\sqrt{3}}$ and a local minimum at $x=\frac1{\sqrt{3}}.$

$f\left(\frac{-1}{\sqrt{3}}\right)=\frac{3\sqrt{3}+2}{3\sqrt{3}}>0$

$f\left(\frac{1}{\sqrt{3}}\right)=\frac{3\sqrt{3}-2}{3\sqrt{3}}>0$

$f$ is increasing on the interval $\left(-\infty,\frac{-1}{\sqrt{3}}\right].$ As an increasing function it cannot have more than one zero on that interval. You have already shown (presumably by the Intermediate Value Theorem) that it does have one zero.

$f$ is decreasing on the interval $\left[\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right].$ Thus in this interval, $f(x)\ge f\left(\frac1{\sqrt{3}}\right)>0$ and is never zero.

We use a similar argument to show there are no zeros in $\left[\frac{1}{\sqrt{3}},\infty\right).$


geometry
3D geometry
algebra
factorization
difference of cubes
special factorizations

How many digits does 1000000....000001 with n zeroes have when raised to the third power? [u]Make sure your solution is rigorous.[/u]

[hide="Click to reveal hidden solution"]
We can start by noting that the number is only 1 away from 100000...0000 with $n+1$ zeros, so 10000....00000 is $10^{n+1}$.  The cube of this is $10^{3n+3}$. which has $3n+4$ digits. Now we can use a little bit of difference of cubes: $a^{3}-b^{3}=(a-b)(a^{2}-ab+b^{2})$.  Someone go on from there.[/hide]

[quote="mathgeniuse^ln(x)"][hide="hint"]Notice that the number is just $(10^{n+1}+1)^{3}=10^{3n+3}+3*10^{2n+2}+3*10^{n+1}+1$[/hide][/quote]

That was the rigorous method I was implying.

[hide="spoiler"]note that the terms other than $10^{3n+3}$ have no effect on othe number of the digits of the sum because they are too small[/hide]

[quote="13375P34K43V312"] [u]Make sure your solution is rigorous.[/u][/quote]

What do you mean by 'rigorous'?

[quote="Math Geek"]I would be terribly tempted to say that the answer is simply 3n+4, because the 1 is quite insignificant, but I could be wrong. :roll:

egghead91, he means that the method you use should work in all cases.[/quote]

No, I mean that it is fully justified without making any assertions that are not justified. But an intuitive explanation should suffice.


MATHCOUNTS
inequalities
algebra
function
domain
complex numbers
imaginary numbers

For those of you who don't get where this is coming from, read [url=http://www.artofproblemsolving.com/wforum/viewtopic.php?t=1351&postdays=0&postorder=asc&start=0]here[/url] (start with MysticTerminator's post, then skip over to the solution at the very beginning of page 3)

Prove, disprove, or partially prove the following: x>x; x>-x; lxl>x; :sqrt:(x:^2:)>x; and x:^2:<0 :twisted:

Tare wrote:Prove, disprove, or partially prove the following: x>x; x>-x; lxl>x; :sqrt:(x:^2:)>x; and x:^2:<0 :twisted:

[hide]

x>x is impossible

x>-x only if x is positive

|x|>x only if x is negative

:sqrt:(x:^2:)>x only if x is negative

x:^2:<0 is impossible (unless ur talking about imaginaries)[/hide]

Um, [quote="I"]Prove, disprove, or partially prove[/quote]

Tare, come on. Some of those are ridiculously trivial.

I know I know, there all just warmups for the big guy at the end there :twisted: , you could just do the last one if you want.

dude...the last one is true if and only if x has imaginaries in it...

Well I never specified x's domain now did I? :twisted:  :wink:

uh...tare normally in MC when a domain isnt specified, we assume u mean the set of all reals...not including i's since most MC kids dont use those...

carrots2 is not a real number, so it cannot be compared with 0 using <, >, or =.

for all we know, carrots :^2: might mean c*a*r :^2: *o*t*s :^2:

true...unless Mystic would give us values for the variables...

Mmm, well you have three options you know, and there might be multiple answers depending on the different domains...

no...unfortunately we have infinite options for the domain. we have the domain of the set of vegetables. we have to domain of the set of complex numbers. we have the domain of the set of pure imaginaries. we have the domain of rotten tomatoes. we have the domain of the set {1}. etc...

OK, fine, since some of you tend to be creative, I'll list all the domains possible:
-real number
-imaginary number (complex number)
-all number

*sigh* sorry I wasn't specific enough...*sigh*

[quote="MysticTerminator"]um...you should. otherwise, i could assume the domain of x is the set of all vegetables. I have no idea whether (carrots):^2:<0 is true or not, so please specify a domain.[/quote]

actually, (Carrots):^2: has been defined as log _5 (tomato).

...what-ever! :roll:

Can we please be mathematical? :)

you still need to be slightly more specific in defining your domain. For example, if you take one domain to be the set of complex numbers, one possible x is 2+i. However, that makes x:^2:=3+4i. It is undefined as to whether this is greater than or less than zero. Sorry, try again.

A number can only be > or < 0 if it is real.

Well guys, the question didnt have too much validity to begin with...

*sigh* Fine, I'm bad at making up problems off the top of my head, but at least it kept the forum occupied for a while along with my other problems...at least give me [b]some [/b]credit for trying...

We need more people to post problems, then I wouldn't have to make them up...it's so hard to come up with original math problems...*sigh* I apologize for this stupid question and wasting your (plural) time.


quadratics
modular arithmetic
number theory unsolved
number theory

Can anyone help me with this problem, please?

Suppose an odd number $ a>1$ is not a perfect square. Show that there exist two distinct odd primes $ p$ and $ q$ such that $ a$ is a quadratic nonresidue modulo $ p$ and also modulo $ q$.

I know I have to use the reciprocity law for the Jacobi symbol somehow, but I still can't figure it out. Any help is appreciated, thanks.

if $ a\equal{}2$ then choose $ p,q$ such that $ p,q \equiv \pm 3 \pmod {8}$.
suppose that $ a\equal{}q_1^{b_1}...q_k^{b_k}$
then we have

$ \left( \frac{a}{p}\right)\equal{}\prod_{i\equal{}1}^{k} \left(\frac{q_i}{p} \right)^{b_i}$

since $ a$ isn't a perfect square between the prime divisors of $ a$,we choose those who have an odd exponent(such prime exists because of the assumption),let these primes be $ q_{i_1},...,q_{i_t}$
we have 2 cases:
1-if one of $ q_{i_1},...,q_{i_t}$ is $ 2$ (WLOG supposing $ q_{i_1}\equal{}2$) then we choose $ p$ such that $ p\equiv \minus{}3 \pmod {8}$ and after this we have.

$ \left(\frac{a}{p}\right)\equal{}\prod_{j\equal{}1}^{t} \left(\frac{q_{i_j}}{p} \right)^{b_{i_j}}\equal{}\minus{}1.\prod_{j\equal{}2}^{t} \left(\frac{q_{i_j}}{p} \right)\equal{}\minus{}1.\prod_{j\equal{}2}^{t} \left(\frac{p}{q_{i_j}} \right)$

now we choose $ p$ such that $ \left(\frac{p}{q_{i_j}} \right)\equal{}1$ for all $ 2 \le j \le t$ and then we get $ \left(\frac{a}{p}\right)\equal{}\minus{}1$.

for this just choose $ p$ such that $ p \equiv 1 \pmod {q_{i_2}...q_{i_t}}$ and we are done!
similarly u can choose $ q$.
so finally for this case we must consider a prime $ p$ such that $ p \equiv 1 \pmod {q_{i_2}...q_{i_t}},p\equiv \minus{}3 \pmod {8}$.
for finding a "prime" answer to this system of congruency at first we choose $ X$ such that :
$ X \equiv \minus{}3 \pmod {8}$
$ X \equiv 1 \pmod {q_{i_2}...q_{i_t}}$
then,due to Dirichlet's theorem , in the sequence $ 8q_{i_2}...q_{i_t}k\plus{}X$ there exists infinitely many primes since $ \gcd(8q_{i_2}...q_{i_t},X)\equal{}1$.

2-if all of $ q_{i_1},...,q_{i_t}$  are odd then choose $ q_{i_1}$ and let $ p$ be a prime such that $ p\equiv 1 \pmod {4}$ and also $ p \equiv r \pmod {q_{i_1}}$  where $ r$ is a quadratic non-residue modulo $ q_{i_1}$,then obviously we have

$ \left(\frac{p}{q_{i_1}} \right)\equal{}\minus{}1$

now choose $ p$ such that $ p \equiv 1 \pmod{q_{i_2}...q_{i_t}}$ and then we'll have $ \left(\frac{p}{q_{i_j}} \right)\equal{}1$ for all $ 2 \le j \le t$ which implies that $ \left(\frac{a}{p}\right)\equal{}\minus{}1$
the only thing that we need to prove is that there exists at least two primes such that $ p\equiv 1 \pmod {4},p \equiv r \pmod {q_{i_1}},p \equiv 1 \pmod{q_{i_2}...q_{i_t}}$ which is again obvious by considering Dirichlet's theorem and crt:

for finding a "prime" answer to this system at first u choose $ X$ such that:
$ X \equiv r \pmod {q_{i_1}}$
$ X \equiv 1 \pmod {4q_{i_2}...q_{i_t}}$

then ,due to Dirichlet's theorem,the sequence $ 4q_{i_2}...q_{i_t}.q_{i_1}k\plus{}X$ contains infinitely many primes since $ \gcd(X,4q_{i_2}...q_{i_t}.q_{i_1})\equal{}1$ .
so we are done.

This is very helpful. I haven't learnt Dirichlet's theorem yet, but I guess i can learn it now!
THank you!!!  :lol:


For $ x \equal{} \sqrt {3 \minus{} \sqrt {8}}$, find the value of $ x^4 \plus{} 2x^3 \minus{} 2x \minus{} 1$.

It helps to know that

$ x^4 \plus{} 2x^3 \minus{} 2x \minus{} 1 \equal{} (x\minus{}1)(x\plus{}1)^3$

[quote="kunny"]Good, but we have a general solution. for such like this.[/quote]

For example,
$ x^2 \equal{} 3 \minus{} \sqrt{8}$ and $ x^4 \plus{} 2x^3 \minus{} 2x \minus{} 1 \equal{} x^2 (x^2 \plus{} 2x \minus{} 2/x \minus{} 1/x^2)$ and plug it in = $ (3 \minus{} \sqrt{8})(3 \minus{} \sqrt{8} \plus{} 2\sqrt{3 \minus{} \sqrt{8}} \minus{} 2/ \sqrt{3 \minus{} \sqrt{8}} \minus{} 1/(3 \minus{} \sqrt{8})$.

Since $ x\equal{}\sqrt{2}\minus{}1\Longrightarrow (x\plus{}1)^2\equal{}(\sqrt{2})^2$ $ \therefore x^2\plus{}2x\minus{}1\equal{}0$

Doing Long Division gives $ x^4\plus{}2x^3\minus{}2x\minus{}1\equal{}(x^2\plus{}2x\minus{}1)(x^2\plus{}1)\minus{}4x\equal{}\minus{}4x\equal{}\minus{}4(\sqrt{2}\minus{}1)\equal{}\boxed{4\minus{}4\sqrt{2}}$.


Is it supposed to be $ sin80$?

If it is then we have $ 1\minus{}sin20$

By making a common denominator, then using the double angle formula for sin.

You can still solve it the way it is.
Just change sin70 to sin(10 + 60) = (sin10 x cos 60) + (cos 10 x sin 60) and continue.

[hide="Slight Variation to the Problem"]$ \frac {1}{2 \sin 10} \minus{} 2\sin 70 \equal{} \frac {1 \minus{} 4\sin 10 \sin 70}{2\sin 10} \equal{} \frac {1 \minus{} ( 2\cos 60 \minus{} 2\cos 80)}{2\sin 10} \equal{} \frac {2\sin 10}{2\sin 10} \equal{} \boxed{1}$.

[/hide]

The actual simplification goes to [hide="Something Not As Clean"]$ \frac{1}{2}(2 \plus{} \csc 10^\circ)$[/hide].


What is the maximum number of square inches in the area of a rectangle with a perimeter of 12 inches?

$ a\plus{}b\equal{}6$
AM-GM, $ ab \leq 9$. 9.


$\boxed{4}$

i assume you're adding the factorials..your notation is a little awkward ;)

assuming that !n is equivalent to n!, then here is the solution to arkmastermind's answer
[hide="solution"]
its pretty clear that for all $n \in \mathbb{Z}, n \ge 5$, the ones digit of $n!$ is $0$ b/c we have a factor of $5$ and a factor of $2$, so therefore we have a factor of $10$, and hence, a trailing 0
this means that we only care about $0! +1! + 2! + 3! + 4!$. this comes out to $1 + 1 + 2 + 6 + 24 = 34$ and a ones digit of $\boxed{4}$
[/hide]

[quote="Elemennop"]Maybe they are subfactorials?[/quote]

subfactorials?

[quote="Elemennop"]Maybe they are subfactorials?[/quote]

And what do you mean with subfactorial???

lol, i remember sometime back finding a recursion for the number of derangements with n objects, but at that time i didnt know that it was a well known fact of mathematics, or even what derangements were..

Yeah, I remember doing the same thing back in the day! :)  It's really not hard to derive $!n=(n-1)(!(n-1)+!(n-2))$ (and from there not difficult to find $!n=n(!(n-1))+(-1)^n$), though you have to use exponential generating functions to derive its closed form.


analytic geometry
combinatorics unsolved
combinatorics

Assume that a grid is made from identical equilateral triangles. A tile is the system of two equilateral triangle with common side. One can put tiles to the plane in three different positions. Prove that if one covers the regular hexagon of $ 6n^2$ triangles with tiles, the number of tiles in different positions are equal.

Hmm..difficult to explain with no drawing... :wink:

Consider a tiling of the hexagon. Assume that the sides of the hexagon are parallel to the lines $ y\equal{}0$, $ y\equal{}x \sqrt 3$ and $ y\equal{}\minus{}x \sqrt 3$.
Tiles with sides parallel to the first two lines are of type $ 1$, those with sides parallel to the last two lines are of type $ 2$ and the others of type $ 3$.

Now, consider the upper side of the hexagon, which is horizontal. This side is divided into $ n$ unit-segments, each of which being a side of a tile of type $ 1$ or $ 3$ (but not $ 2$).

Draw a line from the middle of this side to the middle of the opposite side of the tile. This opposite side is a common side with another tile of type $ 1$ or $ 3$. Then, continue to draw to the middle of the opposite of this second tile, and so on...

Starting from a tile of type $ 1$ or $ 3$ which covers a unit-length segment of the upper side of the hexagon, you will draw a path which goes through only tiles of type $ 1$ or $ 3$, and when you go through one tile, you always decrease your height of one unit. Moreover, you only connect midpoints of parallel segments. This ensures that the path has length $ 2n$ and connect a segment of the upper side of the hexagon to a segment to the lower side of the hexagon.

Clearly, no two paths can intersect, and each tile of type $ 1$ or $ 3$ belongs to such a path (choosing any tile of one of these type, its upper side is one of the hexagon, or is a common side with another tile of one of these two type, then go on until you reach the side of the hexagon, which is the start of the desired path...).
Thus, the paths go through each of these tiles exactly once.

Now, from the left to the right, number midpoints the unit segments of the upper side of the hexagon $ s_1, \cdots, s_n$ and those of the lower side $ s'_1, \cdots , s'_n$, it follows that for each $ i$, the path starting at $ s_i$ ends at $ s'_i$.
Since $ s_i$ and $ s'_i$ have the same $ x$-coordinates, this means that along the path there are the same number of tiles of type $ 1$ (which decrease the $ x$-coordinate by one) than of type $ 3$ (which increase the $ x$-coordinate by one).
Since the paths are pairwise disjoints and that each tile of type $ 1$ or $ 3$ belongs to exactly one path, it follows that the total number of tile of type $ 1$ is equal to the total number of tiles of type $ 3$.

We prove similarly that the total number of tiles of type $ 2$ is the same of the total number of tiles of type $ 1$, and we are done.

Pierre.


Observe $1233 = 12^{2}+33^{2}$.

There is a $14$-digit number. If we halve it into two $7$-digit numbers and compute the sum of squares of these two numbers, we get the original number. Find this number.

Can you show the method too?

[hide="The relevant equation"] $a \cdot 10^{7}+b = a^{2}+b^{2}$ [/hide] [hide="Rewritten"] $a(10^{7}-a) = b(b-1)$ [/hide]

[quote="t0rajir0u"][hide="The relevant equation"] $a \cdot 10^{7}+b = a^{2}+b^{2}$ [/hide] [hide="Hint"] Simon's Favorite Factoring Trick. [/hide][/quote]
How do you use SFFT on that thing? :maybe:

Hang on, that doesn't work at all.  :blush:

BTW, $88^{2}+33^{2}=8833$ too :)

The question is to find factors of $10^{14}+1$ ((and knwoing   101 and 29 (yes, FLT and knowing 29 is prime helps to get this)  are factors of $10^{14}+1$ helps -  , you get the two factors right away :) )  ... and since there are four factors (281, and a biggie)   there  could be 16 solutions in integers (not counting non-positive  values for b)  for $a^{2}+b^{2}=10^{7}a+b$

Some of these   may  not be greater than $10^{6}$ so you may not like to call them 7 digit numbers ...

Anyway  I am listing  the 8 solutions, for each $a$ you can substitute $(10^{7}-a)$ to get another result ..

(Eg If I list,  $9900990^{2}+990100^{2}$ You could also list
$99010^{2}+990100^{2}$

Anyway I happen to have a calculator with me so I can  write down all the solutions:
$9995140^{2}+220401^{2}$
$9110320^{2}+2846976^{2}$
$9231820^{2}+2663025^{2}$
$9900990^{2}+990100^{2}$
$9939870^{2}+773101^{2}$
$8465170^{2}+3604525^{2}$
$8520690^{2}+3448276^{2}$

(And of course, the trivial ones: $(10000000)^{2}+1^{2}$ or $10000000^{2}+0^{2}$)
 You may like to see  why here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=5882353&t=81888

(You may have to follow the link mentioned in the above problem...
As the question has been asked before so wont spoil the fun by giving all the steps._


You know what? That's a good idea; I'm going to purposely fail my classes now.

From [url=https://mathcounts.org/Page.aspx?pid=287]here[/url].

Quote:
Students enrolled in the 6th, 7th or 8th grade are eligible to participate in MATHCOUNTS® competitions. Students taking middle school mathematics classes who are not full‑time 6th, 7th or 8th graders are not eligible. Participation in MATHCOUNTS competitions is limited to [u]three years[/u] for each student though there is no limit to the number of years a student may participate in the school‑based coaching phase.

Good, three years eh? I only participated two years. pwnt.

So if you do MC in 7th and 8th grade, and repeat 8th grade..that is 3 years, correct?

I seriously hope you guys arent considering this...

I am though :lol:

I wonder if you're allowed to do the same thing if you make a stupid mistake and end up getting a 24 on AMC 8?

I am definitely repeating 8th grade.

What does that have to do with anything?

[quote="AIME15USAMO"]If I were to fail all my classes and have to repeat 8th grade, could I still participate in MATHCOUNTS?
Not that I am failing but i was just wondering  :lol:[/quote]

If you are failing your classes in 8th grade, but are still participating in MATHCOUNTS, you need to set your priorities straight.

hmm..set my priorities straight?
MATHCOUNTS > school, AMC series > school
Yeah..that looks right to me

MathCounts focuses on quick calculation and not making errors, rather on deep thinking and creative problem solving. MathCounts is a good start to math, but it isn't everything in math. Math can be much more fun and interesting.

MathCounts helps you begin math, but it won't help you later on. For example, there is no spot for MathCounts score on a college application, but there are spots for your GPA and sometimes more advanced math, like the AMC series, all the way up to IMO.

[quote="pythag011"]Don't do this.

I guess mathcounts seems sort of cool, but seriously, school is definitely much more important than MATHCOUNTS.

There's a lot of cool math beyond mathcounts. After like maybe 1/2 years of doing mathcounts, you should focus more on other math...[/quote]

Im pretty sure this is a complete joke topic :P

Wait, so MathCounts doesn't look good on a college application(aka it won't help me get into college)? :(

I didn't mean it won't help you, I meant that there were more important things (like the priority thing AIME15 was talking about).


trigonometry
calculus
derivative
function
geometric series
real analysis
real analysis solved

developing in power series f(x)=$sin^2(x)$  how do i develop

Use trigonometry  $\sin^{2} x=1/2 - 1/2\cos(2x)$  and power serie of $\cos(2x)$

:D  ok ! but how? So f(x)=$(1-cos(2x))/2$ ;
so :
 $f'(x)= 2*sin(2x)$
$ f''(x)= - (2^2)*cos(2x)$
$ f'''(x= - (2^3)*sin(2x) $
 OK. Ans so on...
 I can't find the exact definition of f(x); This is my big PROBLEM. Pls HEPL me. thank u!

There are a small number of power series that you should have memorized. The list I give to my calculus students has seven: the geometric series, $\ln(1+x),$ the arctangent, the exponential, the sine, the cosine, and the binomial series. We can do a great deal by substitution and other manipulations from that list.

$\cos(u)=1-\frac{u^2}2+\frac{u^4}{4!}+\cdots= \sum_{k=0}^{\infty}\frac{(-1)^ku^{2k}}{(2k)!}$

Now replace $u$ by $2x$ and change the sign:

$-\cos(2x)=\sum_{k=0}^{\infty}\frac{(-1)^{k+1}2^{2k}x^{2k}}{(2k)!}= -1+\frac{4x^2}2-\frac{16x^4}{24}+\frac{2^6x^6}{6!}+\cdots$

Now add 1, which causes the constant term to disappear, and divide by 2:

$\sin^2x=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}2^{2k-1}x^{2k}}{(2k)!}= x^2-\frac{8x^4}{24}+\frac{2^5x^6}{6!}+\cdots$

This, of course, is exactly what Moubinool intended for you to do with his comment.

Now, to solidify your understanding of this:

Let $f(x)=\sin^2x.$ Find $f^{[100]}(0).$ (That is, the hundredth derivative at zero.)

$\left(sin^2(x)\right)'=2sinxcosx=sin(2x)$
but $\left(sin(x)\right)^{(n)}=sin(x+n\frac{\pi}{2})$
so the following should be true if i remember well

$\left(sin(2x)\right)^{(n)}=(-1)^n 2^n sin(2x+n\frac{\pi}{2})$

You're missing the point, spx2. Remember that I didn't ask for the derivative in general, I asked for the derivative [i]at zero.[/i]

Don't look at the formula for the function; look at the series.

Remember Taylor series is $\displaystyle \sum_{k=0}^{\infty}\frac{f^{(k)}(a)(x-a)^k}{k!}$

So, since you're using a form centered around 0, the formula would be $\displaystyle \sum_{k=0}^{\infty}\frac{f^{(k)}(0)x^k}{k!}$

($\displaystyle f^{(k)}(0)$ means the $k^{th}$ derivative of f at zero)


superior algebra
superior algebra unsolved

For a commutaive ring R  find the two sided ideals in $ M_n(R),$ i.e. the ring of $ n\times n\minus{}$matrices.

Let $ J$ be such an ideal.
Set $ E_{ij}\in \mathcal{M}_n(R)$ with $ (i,j)$-entry $ 1$ and $ 0$ otherwise.
Then $ I: \equal{}\{x\in R: xE_{11}\in J\}$ is obviously an ideal of $ R$.
For $ M\equal{}(m_{ij})\in J$ we have $ m_{ij}E_{11}\equal{}E_{1i}ME_{j1}\in J$ and $ m_{ij}\in I$ for all $ i,j$.
So $ J\subseteq \mathcal{M}_n(I)$ and conversely for $ M\equal{}(m_{ij})\in\mathcal{M}_n(I)$ we have
$ M\equal{}\sum_{i,j}m_{ij}E_{ij}\equal{}\sum_{i,j}E_{i1}\cdot m_{ij}E_{11}\cdot E_{1j}\in J$.
So $ J\equal{} \mathcal{M}_n(I)$.

Also $ \mathcal{M}_n(I)$ is obviously an ideal of $ \mathcal{M}_n(R)$ for any ideal $ I$ of $ R$.

How do you get that $ m_{ij}E_{11} \equal{} E_{ii}ME_{j1}$ and that I is an ideal?

We have the relations $ E_{ij}E_{jk}\equal{}E_{ik}$ for all $ i,j,k$
and $ E_{ij}E_{kl}\equal{}0$ for all $ i,j,k,l$ with $ j\not\equal{}k$.
So for $ M\equal{}(m_{ij})\equal{}\sum_{i,j}m_{ij}E_{ij}$ we have
$ E_{1k}ME_{l1}\equal{}\sum_{i,j}m_{ij}E_{1k}E_{ij}E_{l1}\equal{}m_{kl}E_{11}$.

For $ I$ defined as above we have
$ x,y\in I\Rightarrow xE_{11},yE_{11}\in J\Rightarrow (x\plus{}y)E_{11}\equal{}xE_{11}\plus{}yE_{11}\in J\Rightarrow x\plus{}y\in I$
and
$ r\in R,x\in I\Rightarrow xE_{11}\in J\Rightarrow rxE_{11}\equal{}(rE_{11})(xE_{11})\in J\Rightarrow rx\in I$.
So $ I$ is ideal of $ R$.

I thought this was obvious? :maybe:


geometry
3D geometry
pyramid
tetrahedron

What is the name of a twenty-sided polygon?

[quote="SplashD"]I think that only applies to icosahedron. I never of an icosagon, but what do I know?[/quote]
No, icosa=20. You just add the suffix of whichever you are talking about, the 2-d; -gon or the 3-d; -hedron.

Thanks :)

Does that same word "icosa" work to say if you multiply some thing by 20 that you icosatupled it :?:

[quote="kyyuanmathcount"]I think that would be more for geometric things.

If you multiply something by 4, you don't necessarily square it.

Or better, usually, when you multiply something by 6, you don't say I hexed it.

For those of you confused, either refer to AoPS book 1 or just Google it.[/quote]

Multiplying by 4 is quadrupling (the prefix for 4 isn't square...), and multiplying by 6 is sextuple... (which is basically hextuple)

You basically have to use a prefix that sounds correct and means whatever number you want.

For example, a 4 sided polygon is called a quadrilateral, so you would have quadruple.  If you had a 5 sided polygon, it would be pentuple (I think)...

You just have to figure out which prefix is correct (not always easy).

For 6, you have hex and sex to choose from.  Hextuple doesn't sound as good as sextuple, so choose sextuple.

Unless you want a bunch of pointless memorization, just go by what it sounds like.

*2 = double
*3 = triple
*4 = quadruple
*5 = quintuple
*6 = sextuple
*7 = septuple
*8 = octuple
That's as high as I know off the top of my head

you really can't switch between the polygons and the multiplication - like the rest of english its makes no logical sense whatsoever.

http://en.wikipedia.org/wiki/Polygon
Wikipedia says it's icosagon. :D

EDIT: So does MathWorld.
http://mathworld.wolfram.com/Polygon.html

[quote="akalra1"]so what would tri-hedron is the same as a pyramid?[/quote]

Polyhedra must have a minimum of four sides.

Edit:  I should say, non-degenerate polyhedra.

A tetrahedron is a 4-sided poly[b]hedron[/b] or triangular pyramid. The term is usually used to denote a regular tetrahedron.

Wow, my writing style sounds really dictionary-like...

Edit: Aaaargh, I made a glaring error... :wallbash_red: Hope the correction clears stuff up a bit.

[quote="b-flat"][quote="mathnerd314"]A tetrahedron is a 4-sided polygon, or triangular pyramid. The term is usually used to denote a regular tetrahedron.

Wow, my writing style sounds really dictionary-like...[/quote]

Don't you mean a tetrahedron is a 4-faced polyhedron with equilateral triangles as its faces?[/quote]

we did that in our math class

certain 3d shapes can be put together with equilateral triangles as their faces

there are several types of tetrahedrons, and the 4-faced pyramid with the triangle for the base is one of them

They don't have to be [b]equilateral[/b] triangles unless it's a [b]regular[/b] tetrahedron.

tetrahydrons were the real name i think

dunno though

[quote="mathnerd314"]A tetrahedron is a 4-sided polygon[/quote]
I believe the term for 4-sided polygons is quadrilateral...

Oh I get it.

Check AoPS volume 1, page 221.

4 faced [i][color=blue]polygon[/color][/i] is a quadralateral.
4 faced [i][color=red]polyhedron[/color][/i] is a tetrahendron.


Let $ f(x)\equal{}ax\plus{}b$, where $ a$ and $ b$ are positive numbers, and assume that the follow equation holds for all values of $ x$: $ f(f(x))\equal{}bx\plus{}a$. Show that $ a\plus{}b\equal{}1$.

$ f(f(x)) \equal{} f(ax \plus{} b) \equal{} a^2x \plus{} ab \plus{} b \equal{} bx \plus{} a\implies a^2 \equal{} b,ab \plus{} b \equal{} a$

$ \implies ab \plus{} a^2 \equal{} a,a > 0\implies a \plus{} b \equal{} 1$, as desired.


SMO 1999 Open #10

What is the largest power of 3 that divides $10^{3^{1999}}$?

Hint: Very big number.

Sorry, mistake  :blush:

What is the largest power of $3$ that divides $10^{3^{1999}}-1$?

[quote="t0rajir0u"][hide="Solution"] Let $T_{n}= 10^{3^{n}}-1$.  We claim by induction that $3^{n+2}$ is the greatest power of $3$ that divides $T_{n}$.  The base case gives $T_{0}= 2$, true.

Suppose $3^{k+2}| T_{k}$.  Then

$T_{k+1}= (T_{k}+1)^{3}-1 = T_{k}(T_{k}^{2}+3T_{k}+3)$

The right factor is congruent to $3 \bmod 3^{k+2}$ and therefore contains only one factor of $3$.  Hence $3^{k+3}$ is the greatest power of $3$ that divides $T_{k+1}$.  QED.

For this problem our answer is $\boxed{3^{2001}}$.  [/hide][/quote]
can you plz tell me where did you get the number 3^{n+2} or it's your forecast before doing induction   :roll:

[quote="HTA"]can you plz tell me where did you get the number 3^{n+2} or it's your forecast before doing induction   :roll:[/quote]
[quote="t0rajir0u"]We claim by induction that $3^{n+2}$ is the greatest power of $3$ that divides $T_{n}$.[/quote]

Er I think s/he meant how t0rajir0u thought of using $3^{k+2}$ for induction er something...

[quote="t0rajir0u"][hide="Solution"]
Suppose $3^{k+2}| T_{k}$.  Then

$T_{k+1}= (T_{k}+1)^{3}-1 = T_{k}(T_{k}^{2}+3T_{k}+3)$

The right factor is congruent to $3 \bmod 3^{k+2}$ and therefore contains only one factor of $3$. [/hide][/quote]

When you say the right factor is congruent to $3\bmod 3^{k+2}$, I don't get how $3^{k+2}$ is more special than any other power of $3$...wouldn't it be $3\bmod$ any power of 3?  :maybe:

Any power of $3$ less than $3^{k+2}$.  The point is that it's not divisible by $9$.

Nice way to solve this one, the original solution was by binomial theorem  :P

[quote="HTA"]
can you plz tell me where did you get the number $3^{n+2}$ or it's your forecast before doing induction   :roll:[/quote]
You can do it by inspection, trying n=0, n=1, etc. Personally i just knew that a number with $3^{n}$ equal digits is divisible by $3^{n}$, making $3^{n+2}$ a natural guess, since in this case the digits are all 9 :)


A cylindrical tank with radius 4 feet and height 9 feet is lying on its side. The tank is filled with water to a depth of 2 feet. What is the volume of the water in cubic feet?

[hide="hint"]Find the area of the base, using $ r\equal{}4$, and take that things multi by $ 2$(the depth of the water in the tank), I don't know but I don't use the height of the tank[/hide]

@doabli - The tank is lying on its side. ;)

[hide="Hint"]Find the area that the water makes on a base of the tank (the area that you would see if you looked at one of the bases), and multiply that by the height of the tank.[/hide]

[hide="Solution"]Draw the circular side of the tank. Call the center point O. Draw radii to the bottom of the tank and the top of the depth of the water calling the points where the edges of the tank meet the water A and B and where the radii to the bottom of the tank meets the water point C. $ OC \equal{} 2$ and we have 2 30-60-90 triangles $ AOC$ and $ BOC$. Thus we see that AB is $ 4\sqrt3$ and that $ \angle AOC \equal{} 120$. $ \triangle AOB$ has an area of $ 4\sqrt3$ and sector $ AB$ has an area of $ \frac {16\pi}{3}$. So we have the area of the water on the circle is $ \frac {16\pi}{3} \minus{} 4\sqrt3$. We multiply by 9 to get $ 48\pi \minus{} 36\sqrt3$

[img]http://i201.photobucket.com/albums/aa289/crazypianist1116/untitled.jpg[/img][/hide]

Wow, I feel like such a moron. It seems so obvious now. Thanks, though.,


function
real analysis
real analysis solved

[quote="kiemkhach"]Consider a continuous $f:R->R$ satisfies:

$f(x+2002)(f(x)+\sqrt{2003})=-2004$  $\forall x \in R$[/quote]
Is there anything missing? As you wrote "Consider...", there must a second part for the problem lolz

[quote="kiemkhach"]Consider a continuous $f:R->R$ satisfies:

$f(x+2002)(f(x)+\sqrt{2003})=-2004$  $\forall x \in R$[/quote]
and Find $f(x)$?
Thank vuhung :D

[quote="kiemkhach"]Consider a continuous $f:R->R$ satisfies:

$f(x+2002)(f(x)+\sqrt{2003})=-2004$  $\forall x \in R$
find $f(x)$[/quote]
and about this problem ;who did you slove it?[b]vuhung[/b] do you slove one?i haven't sloved

[quote="fedja"]There are no such functions... ;)[/quote]

It is not hard to say that there are not such functions,but how can you prove it?(if you are right?)

OK, here goes.
[hide="click here for the proof"]
First of all, $f$ cannot vanish at any point and therefore, being continuous, cannot change sign. Similarly, $f(x)+\sqrt{2003}$ cannot change sign. The only possible combination of signs that gives a negative product is $f<0$, $f+\sqrt{2003}>0$. But then $-\sqrt{2003}<f<0$ everywhere and the product cannot be as large as $2004$ in absolute value.
[/hide]


inequalities
Cauchy Inequality
inequalities proposed

Let $x,y,z>0$ and $xyz=1$ prove that
$2\sqrt{3(x^{y}+y^{z}+z^{x})}+x+y+z\geq 9$

Could anyone solve it ?  :wink:
Mr Vasc there exist a way to apply EV-theorem to thsi one ?
Thank you

An easier one is the following
$2\sqrt{3(x^{y}+y^{z}+z^{x})}+\frac{y}{x}+\frac{z}{y}+\frac{x}{y}\geq 9$

But I think neither SOS nor EV nor mixing variables could kill the first  :wink:

Could anyone solve the first one ? :wink:

I create the first but I have no solution  :(
Mr Vasc or Arqady could you help me to prove the first inequality ?
For the second I posted I have a very nice solution using Cauchy inequality .If you want I will post my proof .

I believe that a slightly harder problem is true:

[b][color=indigo]Problem 1.[/color][/b] [i]Assume that $a,b,c$ are positive real numbers which satisfy the condition $abc=1.$ Then the following inequality is true:[/i]
\[2(a^{b}+b^{c}+c^{a})+a+b+c\geq 9. \]
[b][color=indigo]Proof.[/color][/b] I don't have a proof of this. My original idea was to find other restraints for $a,b,c$ if the inequality isn't to hold by strengthening the upper and lower bounds of $x\in\{a,b,c\}$ iteratively: If c is the maximum, then $c<9$ and $a,b>1/c^{2}.$ But then we have $c+2/c^{2}<9\ldots$ By continuing this we will obtain
\[\left(c+\frac{2}{c^{2}}\right)+\frac{2}{\left(c+\frac{2}{c^{2}}\right)^{2}}+\cdots<9 \]
This probably converges, however we can also iteratively strenghten the lower bound of $a^{b}+b^{c}+c^{a}$ (for example $a^{b}+b^{c}+c^{a}\geq\sqrt{3}$ and its lower bound gets larger as the gap between $a,b$ and $c$ gets smaller) Can we prove that we will finally obtain $a=b=c?$

I'll try to build a complete solution and post it as soon as possible.


I can't understand my book's explanation of the relationship. Do you know any good websites on this topic that includes examples or summaries? Thanks!

Here are some problems that I was unable to do:

Which function dominates as $ x\rightarrow \infty$?

1. $ x^5$ and $ 0.1x^7$
2. $ 0.01x^3$ and $ 50x^2$
3. $ \ln(x \plus{} 3)$ and $ x^{0.2}$
4. $ x^{10}$ and $ e^{0.1x}$

[url=http://www.math.tamu.edu/~mbelk/LimitsAtInfinity.pdf]Limits at infinity[/url] - this is a pdf file with supplemental notes written by one of my colleagues.


Find all real numbers $x$ for which
\[ 10^x + 11^x + 12^x = 13^x + 14^x .  \]

The graphs of the functions form the two sides will be two exponential curves which will intersect $(-infinity,0)$ when $x=-infinity$(I do not know if this solution counts) and $(2,365)$ when $x=2$. Those are the only solutions. :D

No, $x = -\infty$ shouldn't count (btw, the latex code is \infty).

[hide]The way I would do it is to show that $10^x+11^x+12^x$ does not grow as fast as $13^x+14^x$ because

$10^y+11^y+12^y < 12^{y-x}(10^x+11^x+12^x)$

while

$13^y+14^y > 13^{y-x}(13^x+14^x)$

and clearly $13^{y-x} > 12^{y-x}$ for $y > x$ (both sides are also strictly positive). So once we have equality (easily seen to be $x = 2$), the RHS will continue to grow faster than the LHS and we cannot have any more solutions. The same argument shows that if there existed a solution with $x < 2$ then the solution $x = 2$ wouldn't exist. [/hide]


function
calculus
integration
complex numbers
complex analysis
complex analysis unsolved

hello!
can you help me to evaluate this integral
\[{ \int_0^{+\infty} e^{itx-\frac{1}{2x}}}\frac{1}{\sqrt{x^3}}dx \]

Let us start with the integral $I(t)=\int_0^\infty e^{-tx-\frac{1}{2x}}x^{-\frac 12}\frac{dx}{x}$ ($t>0$).  Making change of variable $x=\frac{y}{\sqrt{2t}}$, we get \[ \begin{aligned} I(t)&=(2t)^{1/4}\int_0^\infty e^{-\sqrt{\frac t2}(y+\frac1y)}y^{-\frac12}\frac{dy}y \\ &=(2t)^{1/4}\int_1^\infty e^{-\sqrt{\frac t2}(y+\frac1y)}(y^{-\frac12}+y^{\frac12}\frac{dy}y=(z=y+\frac1y) \\ &=(2t)^{1/4}\int_0^\infty e^{-\sqrt{\frac t2}z} \frac{y^{-\frac12}+y^{\frac12}}{y-\frac1y}\,dz \\ &=(2t)^{1/4}\int_0^\infty e^{-\sqrt{\frac t2}z}\frac{1}{\sqrt{z-2}}\,dz=(z=2+u^2) \\ &=2\cdot (2t)^{1/4}\int_0^\infty e^{-\sqrt{\frac t2}(2+u^2)}\,du=\sqrt{2\pi}e^{-\sqrt{2t}} \end{aligned} \]
Now note that $I(t)$ is an analytic function in the right half-plane $\Re t>0$ continuous up to the boundary, so the formula we get should work for $-it$ instead of $t$ as well (if we take the appropriate branch of the square root). This gives the answer $\sqrt{2\pi}e^{-\sqrt t}e^{i \sqrt t}$ for $t>0$ (the answer for $t<0$ is just complex conjugate to that for $-t$).
Hope, I didn't make a stupid mistake somewhere in the computation... :)


[b]1.[/b] The inequality is obvious. For proof you can take the square of LHS and RHS to get
$ n < (n!)^2\quad\Leftrightarrow\quad 1 < n!\cdot (n - 1)!$ which holds true for positive integers $ n\ge 2$.
________________________

[b]2.[/b] You can suppose (WLOG) that
$ \frac {a_1}{b_1}\le\frac {a_2}{b_2}\le\cdots\le\frac {a_{j - 1}}{b_{j - 1}}\le\frac {a_j}{b_j}\le\cdots\le\frac {a_n}{b_n}$. Next denote $ m: = \min_{1\le j\le n}\frac {a_j}{b_j}$ and $ M: = \max_{1\le j\le n}\frac {a_j}{b_j}$.

By easy manipulation you obtain the known fact that
$ (a,c,b,d)\in\mathbb{R}^2\times\mathbb{R}_{ + }^2,\quad \frac {a}{b}\le\frac {c}{d}\quad\Rightarrow\quad \frac {a}{b}\le\frac {a + c}{b + d}\le\frac {c}{d}$.

Hence, in our case we have
$ m: = \frac {a_1}{b_1}\le\frac {a_1 + a_2}{b_1 + b_2}\le\frac {a_2}{b_2}\le\frac {a_3}{b_3} \\
m: = \frac {a_1}{b_1}\le\frac {a_1 + a_2}{b_1 + b_2}\le\frac {a_3}{b_3}\quad\Rightarrow\quad m\le\frac {a_1 + a_2 + a_3}{b_1 + b_2 + b_3}\le\frac {a_3}{b_3} \\
\qquad\cdots\qquad \\
m\le\frac {\sum_{j = 1}^{n}a_j}{\sum_{j = 1}^{n}b_j}\le M,\qquad\qed$


Find the radius of convergence of:

$ \sum_{n\equal{}1}^{\infty} \frac{2 \cdot 5 \cdot 8 \cdot ... \cdot(3n\minus{}1)}{3 \cdot 7 \cdot 11 \cdot ... \cdot (4n\minus{}1)}x^n$


vector
limit
function
topology
real analysis
real analysis unsolved

A book I'm working from asks the following question:

Is every point of every open set $E \subset \mathbb{R}^{2}$ a limit point of $E$?

I came up with the following proof:

Let $x$ be a point of an open set $E$.  Then there exists $\epsilon > 0$ such that $N_\epsilon(x) \subset E$.  Hence, any neighborhood smaller than $\epsilon$ is contained in $E$ as well.  Thus, for all $\delta > 0$, $N_\delta(x) \cap E \neq \O$ since $N_\delta(x)$ contains some part of $N_\epsilon(x)$.  So $x$ is a limit point of $E$.

But my proof doesn't use the fact that $E$ is in $\mathbb{R}^{2}$.  So is this statement true for all open sets, or is my proof invalid?

I believe that you were implicitly using that $\mathbb{R}^{2}$ is a metric space.

In which step of the proof?  As far as I know, each step applies to all sets which are subsets of a metric space, not just those which are subsets of $\mathbb{R}^{2}$.

Your proof is invalid.

The problem is that you tacitly assumed that $N_{\delta}(x)\cap E$ contains a point other than $x.$

To illustrate the problem consider the space $\mathbb{Z}^{2}$ of lattice points in the plane, with the usual Euclidean metric. Take any nonempty subset $E$ of $\mathbb{Z}^{2}$ - it doesn't matter, all such subsets are open in $\mathbb{Z}^{2}.$ But in this metric space, $N_{\delta}(x)$ consists only of the single point $x$ whenever $\delta<1.$ No point in this space is a limit point.

You need to find a proof that finds an infinte sequcence of points in $E$ other than $x$ itself that tend to $x.$

Here's a valid proof.

Suppose $x\in E.$

There exists a point $y\in\mathbb{R}^{n}$ such that $y\ne x.$ (We can just take some fixed nonzero vector and add it to $x$ to get $y.$) Note that we don't know whether or not $y\in E.$

Now create a sequence as follows: $y_{n}=\frac1ny+\left(1-\frac1n\right)x.$

Then $d(y_{n},x)=\frac1nd(y,x).$

Now suppose we find $\epsilon$ such that $N_{\epsilon}(x)\subset E.$ There exists an $M$ such that for $n>M,\ \frac1nd(y,x)<\epsilon.$ Hence, for $n>M,\,y_{n}\in E\setminus{x}$ and $\lim_{n\to\infty}y_{n}=x.$ That makes $x$ a limit point of $E.$

This proof would work in any normed vector space of the $\mathbb{R}$ or $\mathbb{C}.$

Ah, I see.  I made the assumption that the image of the distance function for the metric is dense in $\mathbb{R}$, which need not be.

But I don't understand:

[quote]You need to find a proof that finds an infinte sequcence of points in $E$ other than $x$ itself that tend to $x$.[/quote]

You seem to be suggesting that such a proof exists.  But as per your counter example, there are open sets in which every point is not a limit point.

Did my second post make things clear? That the theorem is true in $\mathbb{R}^{n}$ and in some wider settings, such as any normed vector space over $\mathbb{R},$ but that it's not true for every metric space?

Yes, I believe so.  But my proof is also valid once it is assumed that the metric space has the property that for any $\epsilon > 0$ and any $p$, there exists $q$ such that $d(p, q) = \epsilon$.

I believe there is probably a much weaker property for which my proof is valid as well, but I can't quite seem to put my finger on it.

You can't make that much weaker. You can make it slightly weaker, by assuming that for any $p,\epsilon>0$ there exists $p$ with $0<d(p,q)<\epsilon$. This is equivalent to saying that no single-point set is open, or that there are no isolated points.


USAMTS
AMC
USA(J)MO
USAMO
number theory
divisibility tests

So I've gotten a bunch of 5s on problems but no commendeds so far. This is my first year doing USAMTS, so that's not a huge surprise, but I'm trying to identify why I don't. Sometimes my solutions aren't optimal, but even when they are I don't get commended. I want to figure out any problems in my proof-writing before USAMO this year so I don't get points off for that. I attached all my solutions, if anyone wants to read through some and tell me if there's anything obvious, that would be amazing.

[quote="Ignite168"]Straight from the FAQ: "A commended score means that a grader reading your solution thought that besides meeting the critera for a score of "five" (a concise well-written solution), the grader thought that your solution was unusually well-done, perhaps involving a novel argument or an interesting generalization"

It needs to be more than optimal.[/quote]
I've seen commendeds for optimal solutions that use the argument everybody else uses.

I think it just means you wrote your proof very clearly and concisely. Not too much, but not too little. But then, this is my first year too, so I'm not sure.

We graders are asked to commend between about 5% of the correct solutions. This makes commendations a special case. For earning points, the problem-solvers compete against the problem, trying to solve it and prove the solution. We would give everyone five points if all their solutions and proofs were correct. For earning commendations, the problem-solvers compete against each other. Someone could write a wonderful and elegant proof, but if several other people's proofs are even more wonderful and elegant, that first person does not get a commendation.

Because commendations depend on what the other participants submit for the USAMTS, it is almost impossible to deliberately write for a commendation. Luck is a major factor.

[quote="worthawholebean"]I've seen commendeds for optimal solutions that use the argument everybody else uses.[/quote]
Yes, we prefer to give commendations for extra insight into the mathematics. But some problems have no special insights to offer. No-one uses a novel argument. In that case, we commend on the quality of the writing rather than the quality of the mathematics.

I graded Problem 4 in Round 3, so let me measure your solution to that problem against our criteria for commendations. Your writing in the problem is strongly above average. You have an occasional sloppy sentence such as, "We can define it alternatively as $ a_1 \equal{} 2007$ and $ a_n \equal{} 10000a_{n\minus{}1} \plus{} 2007$, because each $ a_n$ is $ a_{n\minus{}1}$ shifted four digits to the left + 2007," but the structure of your explanation is clear and solid.

However, Problem 4 did offer good variety in solutions and insights, so we did not have to commend Problem 4 on writing alone. If your writing had been of sufficiently high quality that we considered your solution as an official solution to be posted on the USAMTS web site, we would have commended it. However, on this problem we did not commend writing that was merely good. Instead, we commended solutions that used elegant divisibility tests, or solutions that did a better job in analyzing the structure of the series modulo 101.

Problem 3 in Round 3 was one of those problems that offered few unique insights, so let me check over your solution to that problem. Uh oh. That is not good writing. You structure your answer as theorem and proof, but your theorem is nothing but a statement that the solution to the problem is f(0)=607. The theorem structure is wasted. And your theorem would benefit from adding the word "then" to separate hypothesis from conclusion. And you spent two lines expanding out f(x) into standard form as $ c_nx^n \plus{} c_{n\minus{}1}x^{n\minus{}1} \plus{} \ldots \plus{} c_1x \plus{} c_0$, yet never make meaningful use of standard form.

You did find the elegant insight that $ f(x) \equal{} (x\minus{}200)(x\minus{}7)h(x) \plus{} 2007$. Unfortunately, so did almost everyone else who solved that problem correctly, which means that that insight is not worth a commendation.

You had good writing on Problem 4, which was commended on mathematical analysis, and weak writing on Problem 3, which was commended on good writing. Luck was against you.

Erin Schram
USAMTS grader

[quote="Erin J. Schram"]
Erin Schram
USAMTS grader[/quote]
Thank you so much!

[quote="Erin J. Schram"]
If your writing had been of sufficiently high quality that we considered your solution as an official solution to be posted on the USAMTS web site, we would have commended it.
[/quote]

I have an official solution, yet the solution was not commended.

[quote="not_trig"]I have an official solution, yet the solution was not commended.[/quote]
That is unusual but not unprecedented.

I had picked the official solutions back in Year 14. One problem had a routine solution that was too boring to be commended. All the commendations went to the more interesting methods of solving the problem. Nevertheless, the dull solution was the most common solution, so I picked a well-written example of it as one of the official solutions.

Erin Schram
USAMTS grader


[quote="new_member"]Probably there is a stricter inequality than the following:
Prove that for non-negative real numbers $ a,b,c$ the following is true:

    $ (a^3 \plus{} b^3 \plus{} c^3)^3 \geq 3(a^8b \plus{} b^8c \plus{} c^8a)$.  $ (*)$

Vasc's inequality is : $ (a^2 \plus{} b^2 \plus{} c^2)^2 \geq 3(a^3b \plus{} b^3c \plus{} c^3a)$.
My question is if we could generalize his inequality?I came up with $ (*)$ but there is probably a stricter one.Forgive me it's too late here :([/quote]

please try $ a\equal{}4,b\equal{}2,c\equal{}1$.

 :|

[quote="new_member"]Hey kuing,your counterexample is valid.However it would be good i think to find a generalization of that inequality :)

~Peace[/quote]

$ (a^{3} \plus{} b^{3} \plus{} c^{3})^{3}\geq 2(a^{8}b \plus{} b^{8}c \plus{} c^{8}a)$

holds

and

$ (a^{3} \plus{} b^{3} \plus{} c^{3})^{3}\geq 4(a^{7}b^2 \plus{} b^{7}c^2 \plus{} c^{7}a^2)$

$ (a^{3} \plus{} b^{3} \plus{} c^{3})^{3}\geq \frac {27}{4}(a^{6}b^3 \plus{} b^{6}c^3 \plus{} c^{6}a^3)$ ($ \frac {27}{4}$ is the best)

$ (a^{3} \plus{} b^{3} \plus{} c^{3})^{3}\geq 7(a^{5}b^4 \plus{} b^{5}c^4 \plus{} c^{5}a^4)$

are true too

Please, tell me how you prove these... Cyclic inequalities of such high degree, i am so curious  :roll:

Seriously people  :), does anyone know what kind of sorcery proves these inequalities?

Even the corresponding symmetric inequalities look tough, as they are stronger than Muirhead...


algebra polynomial quadratics conics parabola

Umm... $ \sqrt {ax^2 \plus{} bx \plus{} c}$?

What exactly do you want us to do with this?

For practical purposes, the form you have it in, and the equivalent format yongyi used, are in simplest form.

an example $ \frac{36p^5q^4}{9p^3q}$ simplied is $ 4p^2q^3$

can we make this $ (ax^2\plus{}bx\plus{}c)^\frac{1}{2}$ simplified as well?

it cant be made any simpler? can you prove that it cant get simpler?

Can you simplify $ \frac {p}{q}$?. It depends. If $ p \equal{} 2q$ it's obviously possible for every q. But generally this isn't possible (p and q may be relatively prime).
You can write that in other forms, such as $ \frac {p}{q} \equal{} 1 \minus{} \frac {q \minus{} p}{q}$ but that's all.
In your formula you can make a simplification if $ ax^2 \plus{} bx \plus{} c$ is a perfect square, otherwise it's generally not possible.

Bye
GM

There's no such thing as a proof that something can't be simplified, only a proof that something can be simplified. For example, prove that $ x^2\plus{}2$ can't be simplified. See, you can't do it.

[quote="binomial_4eva"]rigoruos  mathematical proof please.[/quote]

Rigorous definition of simplification, please. I know it sounds kind of funny, but I just can's see what kind of simplification you seek. If you seek for a transform which will take your expression to a polynomial, then the answer is: you can't if the expression under the root isn't a perfect square. And that is trivial to show.

isnt it possible to make a series with taylors,mclaurent,fourier series and from that show that it cant be transformed to a polynomial? i am looking for that kind of proof.

[u]Four things:[/u]
[b]
1. [/b]The general quadratic is of degree two, so its squareroot must be O(x). If the squareroot is then to be a polynomial, then it must be of the form dx+f. However, the space of quadratics is of dimension 3 whereas the space of monomials is of dimension 2, so there cannot be a continuous bijection between the two.

[b]2.[/b] The square of a monomial will be zero iff. the monomial is zero, meaning that the graph of the parabola must have one double-root, which happens if and only if $ b^2 \minus{} 4ac \equal{} 0$.

[b]3.[/b] Courtesy Mathematica. Let $ y$ be the monomial in $ x$ and $ p$ be the constant such that $ y^2 \plus{} p \equal{} ax^2 \plus{} bx \plus{} c$. Then, $ \sqrt {ax^2 \plus{} bx \plus{} c} \equal{} \sqrt {y^2 \plus{} p} \equal{} \sum_{n \equal{} 0}^{\infty} \frac {\left( \minus{} \frac {1}{p} \right)^n \sqrt {p} \; \; \Gamma \left( n \minus{} \frac {1}{2} \right)}{2\sqrt {\pi} \; \; \Gamma(n \plus{} 1)} y^n$, which does not exactly look simple to me.

[b]4.[/b] Fourier Series are for periodic functions.


calculus geometry rectangle derivative inequalities function quadratics

In a given circle of diameter d inscribe a rectangle with maximal $a b^2$, where a and b are the sides of the rectangle.

This is, of course, an almost trivial calculus exercise (it was, indeed, our last homework), but it is an interesting (and not trivial, although not difficult) problem to solve it elementarily, just by AM-GM. (The proof is by no means easier than the one using derivatives, but imagine an inequality with more than one variable and you'll appreciate the experience with applying AM-GM!)

By the way, there may be a proof using other well-known inequalities, too, but I have not found any.

  Darij

well... ab = 2a(b/ :sqrt: 2) <= 2[(a+2(b/ :sqrt: 2))/3] = 2r/27.
isn't it?
i have no pen nor paper... :D

Very nice solution! (Typos corrected:
[tex]ab^{2}=2a\left( \frac{b}{\sqrt{2}}\right) ^{2}\leq 2\cdot \left( \sqrt{\frac{a^{2}+2\left( \frac{b}{\sqrt{2}}\right) ^{2}}{3}}\right) ^{3}=\frac{2}{9}\sqrt{3}d^{3}[/tex].)

I didn't use AM-QM, hence my solution was much much messier:

I need to find the maximum of $a b^2$, where $a^2 + b^2 = d^2$ is fixed. Hence, I can write $b^2 = d^2 - a^2$, so that $a b^2 = a\left( d^2-a^2\right) =a\left( d+a\right) \left( d-a\right)$. This is a function of the positive variable a, and we must find the maximum of this function. We try applying AM-GM by writing

[tex]a\left( d+a\right) \left( d-a\right) =\frac{1}{\alpha \beta \gamma }\cdot \alpha a\cdot \beta \left( d+a\right) \cdot \gamma \left( d-a\right) \leq \frac{1}{\alpha \beta \gamma }\cdot \left( \frac{\alpha a+\beta \left( d+a\right) +\gamma \left( d-a\right) }{3}\right) ^{3}[/tex],

where $\alpha $, $\beta $, $\gamma $ are some constants. But this only gives the required maximum if [b](a)[/b] the right hand side of the equation is constant, i. e. all terms containing a cancel each other, and [b](b)[/b] there exists a value of a for which equality occurs, i. e. there exists an a fulfilling the equations $\alpha a=\beta \left( d+a\right) =\gamma \left( d-a\right) $. The condition [b](a)[/b] entails $\alpha +\beta -\gamma =0$, while the condition [b](b)[/b] yields the existence of an a such that

[tex]\frac{\alpha }{\beta }=\frac{d+a}{a}[/tex]

and

[tex]\frac{\beta }{\gamma }=\frac{d-a}{d+a}[/tex].

Using $\alpha +\beta -\gamma =0$, we obtain a quadratic equation for $\frac{\alpha }{\beta }$, and thus we can determine the required coefficients $\alpha$ and $\beta$ and subsequently $\gamma$ up to proportionality. Actually:

$\alpha :\beta :\gamma =\left( \sqrt{3}+1\right) :1:\left( \sqrt{3}+2\right) $.

This finally gives

[tex]a\left( d+a\right) \left( d-a\right) =\frac{1}{\left( \sqrt{3}+1\right) \cdot 1\cdot \left( \sqrt{3}+2\right) }\cdot \left( \sqrt{3}+1\right) a\cdot \left( d+a\right) \cdot \left( \sqrt{3}+2\right) \left( d-a\right)[/tex]
[tex]\leq \frac{1}{\left( \sqrt{3}+1\right) \cdot 1\cdot \left( \sqrt{3}+2\right) }\cdot \left( \frac{\left( \sqrt{3}+1\right) a+\left( d+a\right) +\left( \sqrt{3}+2\right) \left( d-a\right) }{3}\right) ^{3}[/tex]
[tex]=\frac{1}{\left( \sqrt{3}+1\right) \cdot 1\cdot \left( \sqrt{3}+2\right) }\cdot \left( \frac{\left( 3+\sqrt{3}\right) d}{3}\right) ^{3}[/tex]
[tex]\overset{\text{after some algebra}}{=}\frac{2}{9}\sqrt{3}d^{3}[/tex].

This is the maximum, and it is obtained for $\alpha a=\beta \left( d+a\right) =\gamma \left( d-a\right) $, i. e. for $a=\frac{\sqrt{3}}{3}3$.

  Darij

:P I did half of my (*bad memories return*) differentiation homework this way, and there wasn't much my teacher could do because technically, it's correct ;) .  I even remember, there was one question where it was faster to prove AM-HM and use it on the question than to differentiate and find the answer :lol:

Well, among the extremal problems we do, I'd say ca. 60% are [i]easier[/i] with AM-GM than with differentiation (i. e., they are just trivial with AM-GM). But the rest 40% are rather tricky, like the one I posted (although I made it much more difficult than it really is).

  Darij


Prove that there do not exist eleven primes, all less than $20000$, which form an arithmetic progression.

Suppose such an arithmetic progression exists, with common difference $d$. We assume wlog it is increasing.

We have $2|d$, since otherwise every other term would be even, and the only even prime is $2$.
We also have $3|d$, since otherwise every third term would be divisible by 3, and the only such prime is $3$.
We also have $5|d$, since otherwise at least two terms would be divisible by 5, but the only such prime is $5$.
If 7 does not divide $d$, then 7 is a member of the arithmetic progression, as at least one of the 11 members must be divisible by 7. But 7 being the fourth prime, it is at most the fourth member of the progression. Hence the eleventh member is $7+7d$, which is divisible by 7 and not 7. Hence $7|d$.

If $11|d$ then $2310|d$, and then clearly the last term is greater than 20000.

Therefore 11 does not divide d, and exactly one member of the progression is divisible by 11. This must be 11.

Clearly 11 is the first term. Now $d<2000$ and $210|d$.
Also, $11+d$ is prime. By calculation, this narrows $d$ down to
420,630,1050,1470,1890.

After a lot more calculation, the fact that $11+2d$ is prime means that $d=1050$. But then $11+3d=3161$ is not prime, and so no such $d$ can exist.

Contradiction.

More generally (cited e.g. by W. Sierpinski in [250 Problems of Elementary Number Theory])

[b]Theorem[/b] ([i]Thébault[/i]). [color=#0000FF]If $n$ primes form an arithmetic progression, then its ratio is divisible by any of the primes $p< n$ (and if $n$ is itself a prime, then either it is the first term, or else the ratio is also divisible by $n$).[/color]

Also, been done here: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=464&t=150563&p=2387904#p2387904]E 40 - PEN[/url]


[quote="mdk"]Let f : (0;1) to R be a decreasing function such that g(x) = f(x)/x is an increasing function. Prove that f is necessarily continuous. Give an example of such a function.[/quote]

We have $ f(x+h)-f(x) \le 0$ for all non-negative h.....($ 1$)
Further $ f(x+h)-f(x) = (x+h)g(x+h)-xg(x) \ge (x+h-x)g(x) = hg(x)$
So, $ \mathop{Lim}_{h \to 0}\left[{f(x+h)-f(x)}\right] \ge \mathop{Lim}_{h \to 0}\left[{hg(x)}\right] = 0$....($ 2$)
From ($ 1$) and ($ 2$), we have
$ \mathop{Lim}_{h \to 0}\left[{f(x+h)}\right] = f(x)$

Hence f is continuous.


What is the smallest possible value of $ x$ if $ 12\div(x\plus{}1)$ is an integer?

smallest factor of 12 is
$ \minus{} 12$
$ x \plus{} 1 \equal{} \minus{} 12$
$ x\equal{}\minus{}13$
-13


group theory
abstract algebra
topology
graph theory
superior algebra
superior algebra unsolved

Is there an elementary proof that subgroups of free groups are free?

i think a nice one could involve graphs: suppose you have a non-free group $ G$ immersed in a free group $ F$.
then you can find a finite-rank free subgroup $ F'<F$ having a non-free subgroup $ G'<G$.
now $ F'$ is the fundamental group of a bunch of $ S^1$ which i'll call $ X$, and every subgroup of $ F'$ is the fundamental group of a covering $ \tilde{X}$ of $ X$. since every covering of a graph is itself a graph, and the fundamental group of a graph is free, you finish.

My current textbook refers to Serre's book [i] Trees [/i]

which is too advanced at the moment (I don't know any algebraic topology or graph theory)

the approach in trees to this theorem does not afford any knowledge about topology. a little bit of graph theory is developped there, so that almost no prerequesites are neccessary.


group theory
abstract algebra
vector
number theory
superior algebra
superior algebra theorems

[color=blue][b]Burnside theorem.[/b] Let $ G$ be a finite group, and let $ X$ and $ Y$ be finite $ G$-sets. Then, the $ G$-sets $ X$ and $ Y$ are isomorphic if and only if every subgroup $ H$ of $ G$ satisfies $ \left|X^{H}\right| \equal{} \left|Y^{H}\right|$.

Here, if $ U$ is a $ G$-set, and $ H$ is a subgroup of $ G$, then $ U^H$ denotes the subset of $ U$ defined by $ U^H \equal{} \left\{u\in U\mid hu \equal{} u\text{ for all }h\in H\right\}$.[/color]

This is an easy exercise. (I am quoting it from Michiel Hazewinkel's wonderful [url=http://arxiv.org/abs/0804.3888]survey on Witt vectors[/url]. Even if you neither care for algebra nor for number theory, this work is way too beautiful to skip!)

[hide="Some absolutely stupid remarks that turned out to be wrong, not unexpectedly"]What I am wondering is: Will this theorem still hold if "every subgroup $ H$ of $ G$" is replaced by "every cyclic subgroup $ H$ of $ G$" (or, what yields the same, by "every one-element subset $ H$ of $ G$")?

[EDIT: This is wrong even for $ G\equal{}\mathbb{Z}\slash\left(2\right)\times\mathbb{Z}\slash\left(2\right)$. Sorry for posting first and thinking second.]

Here is why I care: Recall that $ G$-sets are usually seen as representations of $ G$ over $ \mathbb{F}_1$. The map $ G\to \mathbb{Z}$ which takes every element $ g\in G$ to $ \left|X^{\left < g\right > }\right|$ can be seen as a kind of character of the $ G$-set $ X$ (in fact, it is the character of the corresponding permutation representation of $ G$ on $ k^X$, where $ k$ is some field of characteristic $ 0$). Over fields of characteristic $ 0$, a representation is uniquely determined by its character. Over fields of positive characteristic, it isn't. What is the characteristic of $ \mathbb{F}_1$? (I am perfectly aware that this sounds like "what is the sound of one hand clapping?".)[/hide]

  darij

You may want to look at permutation characters.  Taking H to be a cyclic subgroup generated by h, the permutation character of X at h is |X^H|, the number of fixed points of h.  A permutation representation is not determined up to Sym(n)-equivalence by its permutation character, only up to GL(n,C)-equivalence.

The difference between the Burnside ring and the Green ring is well studied but not simple.

Let me know if you want references or examples, but I don't keep them at my fingertips.  Curtis&Reiner has some reasonably standard material on this, and somewhere I've posted some journal articles detailing some of the differences.

Your edit mentions 2 x 2, and you can work out the characters here fairly easily:  The permutation characters are 4000, 2200, 2020, 2002, and 1111. Then 4000 + 2*1111 = 2200+2020+2002 are equivalent as representations over fields of characteristic not 2, but are not equivalent as permutation representations.  In characteristic 2, the two representations are not equivalent, and (somewhat coincidentally) the decomposition into directly indecomposable modules is exactly the decomposition into transitive constituents (the perm rep decomposition).

It can even turn out that non-isomorphic transitive permutation representations can have equal permutation characters.

Thanks a lot. Which of the three(?) Curtis&Reiner's? (I know of almost no good references for representation theory, except of online textbooks and stuff I randomly found on rapidshare. It's not like our university has anybody interested in representations right now.)

  darij

Chapter 11, "Methods" vol 2, of the 1980s vintage is devoted entirely to the Burnside ring.

I suggested Curtis-Reiner since local students expressing similar interests as you have also expressed an affinity for these books.


[hide]

The last in which n appears is it's triangular # or (n+1)n/3

88(88+1)/2 = 3916

So 89 is the 3917 - 4005th terms...

so the 4000 th term is 89[/hide]

yup. um...neal...what planet do you live on?

He lives in Oklahoma, and the last time I checked that was on Earth :)

[hide]89, the sum of the first 89 numbers is the first greater than 4000 (89*91)/2[/hide]

um. there are 365 days in a year on Earth. I wanna know where he's livin.

1's end at pos 1.  2's end at pos 3... n's end at pos n*(n+1)/2.

a little experimentation reveals that 4000 is in the interval for which n =89, b/c the 89's end at 4005 which is right above 4000.

Nah, he doesn't live in Oklahoma, he was joking  ;)