abstract algebra
calculus
integration
algebra
function
domain
Ring Theory

Let $ R$ be a commutative ring with identity and $ M$ be a free $ R$-module. As we know the notion of rank of $ M$ is well defined then. Now suppose that $ B$ is a linearly independent subset of $ M$. Is it true that $ |B|\leq \text{rank}(M)$?
PS If $ R$ is an integral domain, then an easy arguement verifies it.

yes, it's true. use the proof here to reduce from the infinite to the finite case http://www.mathreference.com/la,rank.html#infc, it works in your situation as well. the finite case is here: http://www.mathreference.com/mod-pit,basec.html

Follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=124137 , but I wonder whether I'll ever find a write-up-able proof...

[b]EDIT:[/b] I was told that it's not clear how the problem follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=124137 , so here is a brief explanation: If $ B$ is a linearly independent subset of $ M$, then we have an injective homomorphism $ R^{\left|B\right|}\to M$. But $ M\equiv R^{\text{rank }M}$ because $ M$ is free. Thus, we have an injective homomorphism $ R^{\left|B\right|}\to R^{\text{rank }M}$. Thus, $ \left|B\right|\leq \text{rank }M$, because by http://www.mathlinks.ro/Forum/viewtopic.php?t=124137 part [b]b)[/b] a homomorphism $ R^{\left|B\right|}\to R^{\text{rank }M}$ cannot be injective if $ \left|B\right|>\text{rank }M$.

Uhm, of course I have assumed that $ \text{rank }M$ was supposed to be finite here.

  darij


Describe the sets u(n) having the following property:There exist a permutation s of N such that u(s(n)) is monotonic after a certain point.

Since you used the word "set", I'm asuming the terms of the sequence $u_n$ are distinct.

Assume the set $A=\{u_n|n\in\mathbb N^*\}$ has at least two accumulation points (I mean accumulation pts in $\bar{\mathbb R}$), and let $a_1<a_2$ be two such points. Take $\varepsilon>0$ small enough that $(I_1=)(a_1-\varepsilon,a_1+\varepsilon)\cup(a_2-\varepsilon,a_2+\varepsilon)(=I_2)=\emptyset$. There are infinitely many $u_n$ in both $I_1,I_2$, so the set doesn't meet the requirements.

This means that $A$ has $1$ or $0$ accumulation points. If it has $0$ accumulation pts, it's a discrete bounded set which has no limit point, meaning that it must be finite, which can't be true. This leaves the possibility of $A$ having one accumulation point $a$. If elements tend to $a$ from below and above, we get a contradiction, since for every $\varepsilon>0$, we have infinitely many $u_n$ in both $(a-\varepsilon,a)$ and $(a,a+\varepsilon)$. This means that $A$ must have a unique accumulation point s.t. points from $A$ tend to $a$ only from below or only from above (i.e. there are only finitely many $u_n$ in at exactly one of $(-\infty,a),(a,\infty)$).

It's easy to see that all such sets meet our requirements.

Your sequence is not a contradiction to what grobber wrote. You have infinitely many a(n) $<0$ and $>0$ as well.

Misha

Than I want to see the permutation!!!!
Please tell me !!!

grobber has shown that there is no such permutation for sequences with infinitely many numbers $<a$ and $>a$. Your sequence is an example for such a case so there is no required permutation. Please read what grobber wrote with a bit more attention  :? .

Misha


MATHCOUNTS has [url=http://www.mathcounts.org/webarticles/anmviewer.asp?a=510&z=54]posted the 2007 State contests[/url] on their website, so it is now permitted to discuss them.

Where are the Nat's qualifiers and their grades???(I'm asking because I saw them last year)

[url]http://www.mathcounts.org/webarticles/articlefiles/815-State_Winners_07.pdf[/url].

They also have the returning students.

[url]http://www.mathcounts.org/webarticles/articlefiles/815-Returning_Students07.pdf[/url]


geometry
3D geometry
algebra
factorization
difference of cubes
special factorizations

Find the largest prime factor of 2^18+1

How should I approach this problem?  Is there some clever way to factor this?  Can it be rewritten in a better way?  (I don't think (1+1)^18+1 helps, for example). 2^18+1=262145, which is divisible by 5, and 262145/5=52429.  I'm not sure what to do from there.  Maybe setting 2^18+1=x and...I'm not sure

[hide="Hint"]So we essentially have $ (2^6)^3 \plus{} 1^3.$ Use the difference of cubes to factor that further.[/hide]

EDIT: Yikes...I accidentally thought it was difference of squares.

Hello,

Use this:

$ 2^{18} \plus{} 1 \equal{} (2^6 \plus{} 1)(2^{12} \minus{} 2^6 \plus{} 1) \equal{} 65 * 4033 \equal{} 13*5*37*109$ $ \minus{} \minus{} >$ From there we get that the biggest prime is $ 109$.

Edit: Beaten  :P

How do you know $ 4033 \equal{} 37*109$?

hello, since $ \sqrt{4033}\approx 63$ you must test all prime numbers until $ 61$.
Sonnhard.

Dr. Sonnhard

i understand that but it takes a while to get up to 37...

I don't think that there is any special arithmetic trick if that's what you're asking.


algebra
polynomial
ratio
real analysis
real analysis unsolved

I encountered the following problem in a book which prepares high-school
students for the final exam at the end of the 12th form (BAC in Romanian). I
found a solution, but it is quite complicated (for an average student).
Maybe someone has a better idea. Here is the problem:

Let $a_{n}=1+\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{n!}$ and $b_{n}=a_{n}+%
\dfrac{1}{n\cdot n!}.$ It is known that $a_{n}\rightarrow e.$

a) Prove that $a_{n},$ $b_{n}$ are monotonous

b) Prove that $a_{n+1}<e<b_{n}.$

c) Prove that $\dfrac{1}{(n+1)!}<e-a_{n}<\dfrac{1}{n!\cdot n},$ for any $%
n\in N,$ $n\geq 1$

d) Prove that $e\notin Q$

e) Prove that one can not find two polynomials $f,g\in R[X]$ such that $%
a_{n}=\dfrac{f(n)}{g(n)}$ for any $n\in N,$ $n\geq 1$

I am interested in a solution for e), but I wrote the whole problem for
completeness.

So, the first four parts are standard and can be found in a variety of texts.  The last part is a statement about rates of convergence.  Say, to arrive at a contradiction, that we could find such polynomials.

Then $a_{n+1}-a_n=1/(n+1)!=[f(n+1)g(n)-g(n+1)f(n)]/[f(n)g(n)]$.  The latter is a quotient of polynomials, whereas we know that $n!\sim O(\sqrt{n}(n/e)^n)$ and hence for large $n$ it tends to infinity (and accordingly the reciprocal tends to zero) faster than any quotient of polynomials, so we are done.

Well. I want a simple solution. No way to use Stirling! My solution resembles yours, but I avoided to use Stirling. However it is still complicated

But the last step is not necessary at all (the use of Stirling). Al you need to prove is that $n!$ tends to $\infty$ faster than any power of $n$. Say you want $\frac {n!}{n^k}\to\infty$. The ratio is larger than $\frac{n(n-1)\ldots(n-k)}{n^k}$, which goes to $\infty$ as $n\to\infty$.

[quote="costica"]Yes guys, I know. The question is: are there any other solutions? For grobber: Do you think that an average
high school student in Romania can deal with such a solution? The problem was taken from Culegerea pentru BAC de la Sigma. I don't have the book, so I do not know an official solution[/quote]An average student doesn't reach the point where he needs to solve this subpoint (usually it's worth 0.2 points out of 10 and it's the last one in the paper).


Let  a,b,c >0  with  abc=1Show that
Σ[(1+ab 2 )/c^3] \geq 18/(a^3+b^3+c^3)

Of course we have \sum 1/c3 \geq 9/(\sum c3).
So it suffices to prove that \sum ab2/c3 \geq 9/(\sum c3).
By Cauchy, (\sum ab2/c3)(\sum c3) \geq (\sum b\sqrta)2 \geq 32 = 9.
[The last inequality follows from AM-GM and abc = 1.]


(3rd Iberoamerican 1988)

Points A1, A2, ... , An are equally spaced on the side BC of the triangle ABC (so that BA1 = A1A2 = ... = An-1An = AnC). Similarly, points B1, B2, ... , Bn are equally spaced on the side CA, and points C1, C2, ... , Cn are equally spaced on the side AB. Show that (AA1^2 + AA2^2 + ... + AAn^2 + BB1^2 + BB2^2 + ... + BBn^2 + C1^2 + ... + CCn^2) is a rational multiple of (AB^2 + BC^2 + CA^2).

See  attached MS-Word File with solution.

  :cool:

Sorry,in my MsWord file i've done a liitle mistake.I should write
(n+1)AAi instead of AAi.


symmetry
topology
combinatorial geometry
real analysis
real analysis unsolved

Suppose $A \subset \mathbb{R}^n$ satisfies the following:
For each $x \in \mathbb{R}^n$, there is $\epsilon >0$ s.t. the set $B(x,\epsilon)\cap A$  is convex.  (Emty set is considered convex)
( $B(x,\epsilon)$ is the open ball centered at $x$, of radius $\epsilon$ )

Then does it follow that the set $A$ is convex?

hmm. I may have missed something, but can't you just take two points and epsilon half the distance between them? Even if you put in the condition that A must be connected or something, consider a horseshoe-shaped thing. Heck, any open set satisfies this local convexity thing, doesn't it?

[quote="redkimchi"]Then does it follow that the set $A$ is convex?[/quote]

No, otherwise any open set would be convex.

Without assuming connectivity it's false. But I don't think that the horseshoe thing gives a counterexample in the connected case (the horseshoe does not seem to satisfy the hypothesis, pick the internal point right below the symmetry axis of the horseshoe).

Well, actually the horseshoe [b]is[/b] a counterexample (the open horseshoe ... lol). This shows the statement is even true for simply-connected sets.

And here's a counterexample.
Let $A$ be the union of the open lower half-plane (in $\mathbb{R}^2$) with the open segments from $(n-\frac14,0)$ to $(n+\frac14,0)$. For each $x$, the ball centered at $x$ with radius $\frac14$ intersects $A$ in a convex set. The convex hull of $A$ is the closed lower half-plane.

The open horeshoe does not satisfy the hypothesis because it is not required that $x \in A$. It's ok to pick $x$ from the boundary of the open horeshoe.

jmerry's counterexample satisfies the hypothesis, is simply-connected, and is not convex.

What if we add the condition that A is open and connected.

If $A \subset \mathbb{R}^n$ satisfies the following three conditions:

1. For each $x$ from the boundary of $A$, there is $\epsilon >0$ s.t. the set $B(x,\epsilon)\cap A$ is convex. (equivalent to the original condition)
2. $A$ is open.
3. $A$ is connected.

, is $A$ convex?

Oh that's right, $x$ does not necessarily belong to $A$. By the way, it seems that you got the interesting formulation of the problem redkimchi ...


[quote="St. Elena"]Show that:
\[ (1 \plus{} kx_1)(1 \plus{} kx_2)...(1 \plus{} kx_n)\geq (1 \plus{} k)^ n
\]
where $ x_1x_2...x_n \equal{} 1$ and $ x_1,x_2,...,x_n > 0.$[/quote]
It's wrong. Try $ k\equal{}\minus{}1$ and $ n\equal{}2.$ :wink:


We have:
[b](*)[/b] $ a_{n + 1} > a_n$
$ \lim_{n\to \infty}{\frac {a_n}{\sqrt [2007]{a_n}}} = \lim_{n\to \infty}{\sqrt [2007]{{a_{n}}^{2006}} = \sqrt [2007]{2007}}$
This means that: $ \lim_{n\to\infty}{a_n} = \sqrt [2006]{2007}$
But if we put $ a_1 = 10^4$, for examaple, and using [b](*)[/b] we will have for all $ a_n > 10^4$, so check your problem please...  :)

Consider the sequence $ (x_{n})_{n\ge0}$ which verifies $ (1+n!x_n^2)x_{n+1}=e^nx_n$, $ x_{0} > 0$.
Find $ \lim_{n\to\infty}{x_{n}}$.

[quote="Sebi Crisan"]Consider the sequence $ (x_{n})_{n\ge0}$ which verifies $ (1 + n!x_n^2)x_{n + 1} = e^nx_n$, $ x_{0} > 0$.
Find $ \lim_{n\to\infty}{x_{n}}$.[/quote]

If $ x_0 > 0$ then $ x_n > 0$. We have here:
$ e^{n}x_n = x_{n + 1}(1 + n!x_{n}^2)\geq x_{n + 1}\cdot 2\sqrt {x_{n}^{2}n!} = x_{n + 1}x_{n}\cdot 2\sqrt {n!}$
And:
$ x_{n + 1}\leq\frac {e^n}{2\sqrt {n!}}$
We alsow know that $ n! = \sqrt {2\pi n}\left(\frac {n}{e}\right)^{n}\left(1 + O\left(\frac {1}{n}\right)\right)$
So $ \sqrt {n!} = \sqrt [4]{2\pi n}\left(\frac {n}{e}\right)^{\frac {n}{2}}\left(1 + O\left(\frac {1}{n}\right)\right)$

And finally we have:

$ x_{n + 1}\leq\frac {e^{\frac {3n}{2}}}{2\sqrt [4]{2\pi n}\cdot {n}^{\frac {n}{2}}\cdot\left(1 + O\left(\frac {1}{n}\right)\right)} = \frac {1}{2\sqrt [4]{2\pi n}\left(1 + O\left(\frac {1}{n}\right)\right)}\cdot\left(\frac {e^3}{n}\right)^{\frac {n}{2}}$

If we take $ n\to \infty$ then we will have: $ \lim_{n\to\infty}{x_n} = 0$     :w00tb:

I`m drunk a little but still can solve...   :D


pigeonhole principle
number theory
relatively prime

Prove that every selection of $1325$ integers from $M=\{1, 2, \cdots, 1987 \}$ must contain some three numbers $\{a, b, c\}$ which are pairwise relatively prime, but that it can be avoided if only $1324$ integers are selected.

$ 1324$ can be achieved by choosing the mutiples of either $ 2$ or $ 3$
but for the part of $ 1325$,I can't come up with a construction of pigeonholes just like official solution for IMO 1991 problem 3.I think the proofs should be similar,but $ 1987$ is too big to check :(

Looking at the solution at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366445&sid=9556e86316f70b641e898387570238ac#p366445[/url], we see the answer to the general question

[color=#0000FF]Let $M=\{1, 2, \ldots, N\}$, let $2=p_1<p_2 < \cdots < p_n \leq N < p_{n+1}$ be the start of the sequence of primes, and let $1\leq k < n$.
Determine the largest size a subset of $M$ may have, so it does not contain $k+1$ pairwise relatively prime elements[/color]

is given by the size of the subset containing all multiples of $p_1,p_2,\ldots,p_k$, which can be computed by PIE. So the difficulty is bigger when $k$ is large, not when $N$ is large.


function
calculus
derivative
calculus computations

[quote="Azat123"]Differentiate the following
(a)SinH(x^2-3)   (b)Ln(cosh x)  (c) x Sinh x-cosh x
 ;)[/quote]

$\sinh x = \frac{{e^x - e^{ - x} }}{2},\,co{\mathop{\rm sh}\nolimits} x = \frac{{e^x + e^{ - x} }}{2}$

So, derivative of sinhx is coshx and that of coshx is sinhx.

you can use chain rule to solve the problem


If $ a,b,c$ be nonnegative real numbers then

$ \sum\sqrt[3]{\frac{a(a^2\plus{}2bc)}{(b\plus{}c)^3\plus{}16abc}}\le\sum\frac a{b\plus{}c}\le \frac1{2}\sum\frac{a(2a^2\plus{}7bc)}{(b\plus{}c)^3\plus{}abc}$

I think they're interesting. Please tell me your idea, Vasc, arqady, Can_hang... ?   :)

[quote="dduclam"]If $ a,b,c$ be nonnegative real numbers then

$ \sum\sqrt [3]{\frac {a(a^2 \plus{} 2bc)}{(b \plus{} c)^3 \plus{} 16abc}}\le\sum\frac a{b \plus{} c}\le \frac1{2}\sum\frac {a(2a^2 \plus{} 7bc)}{(b \plus{} c)^3 \plus{} abc}$[/quote]
The first part does not hold, for exampl $ a\equal{}b\equal{}1,c\equal{}0.01$. The second part, see here: http://canhang2007.wordpress.com/2008/11/27/an-inequality-duong-duc-lam/ :)

[quote="can_hang2007"][quote="dduclam"]If $ a,b,c$ be nonnegative real numbers then

$ \sum\sqrt [3]{\frac {a(a^2 \plus{} 2bc)}{(b \plus{} c)^3 \plus{} 16abc}}\le\sum\frac a{b \plus{} c}\le \frac1{2}\sum\frac {a(2a^2 \plus{} 7bc)}{(b \plus{} c)^3 \plus{} abc}$[/quote]
The first part does not hold, for exampl $ a \equal{} b \equal{} 1,c \equal{} 0.01$. The second part, see here: http://canhang2007.wordpress.com/2008/11/27/an-inequality-duong-duc-lam/ :)[/quote]

Thank you, Can. I'm sorry for my misstake, the first part must be $ \sum\sqrt [3]{\frac {a(a^2 \plus{}bc)}{(b \plus{} c)^3 \plus{} 8abc}}\le\sum\frac a{b \plus{} c}$  :(

And do you think that, the following inequality is also true

$ \sum\frac a{b \plus{} c}\ge\sum\sqrt {\frac {a(a^2 \plus{}bc)}{b^3 \plus{} c^3 \plus{} 6abc}}$ for all $ a,b,c$ be side lengths of a triangle ?  :)

[quote="dduclam"][quote="can_hang2007"][quote="dduclam"]If $ a,b,c$ be nonnegative real numbers then

$ \sum\sqrt [3]{\frac {a(a^2 \plus{} 2bc)}{(b \plus{} c)^3 \plus{} 16abc}}\le\sum\frac a{b \plus{} c}\le \frac1{2}\sum\frac {a(2a^2 \plus{} 7bc)}{(b \plus{} c)^3 \plus{} abc}$[/quote]
The first part does not hold, for exampl $ a \equal{} b \equal{} 1,c \equal{} 0.01$. The second part, see here: http://canhang2007.wordpress.com/2008/11/27/an-inequality-duong-duc-lam/ :)[/quote]

Thank you, Can. I'm sorry for my misstake, the first part must be $ \sum\sqrt [3]{\frac {a(a^2 \plus{} bc)}{(b \plus{} c)^3 \plus{} 8abc}}\le\sum\frac a{b \plus{} c}$  :(

And do you think that, the following inequality is also true

$ \sum\frac a{b \plus{} c}\ge\sum\sqrt {\frac {a(a^2 \plus{} bc)}{b^3 \plus{} c^3 \plus{} 6abc}}$ for all $ a,b,c$ be side lengths of a triangle ?  :)[/quote]
Sorry, my friend. But these two inequalities do not hold, the same counter example as the previous one. ;)

[quote="can_hang2007"]
Sorry, my friend. But these two inequalities do not hold, the same counter example as the previous one. [/quote]

Oh, I see  :oops:   Thank you, my friend.  :)

[quote="dduclam"][quote="can_hang2007"][quote="dduclam"]If $ a,b,c$ be nonnegative real numbers then

$ \sum\sqrt [3]{\frac {a(a^2 \plus{} 2bc)}{(b \plus{} c)^3 \plus{} 16abc}}\le\sum\frac a{b \plus{} c}\le \frac1{2}\sum\frac {a(2a^2 \plus{} 7bc)}{(b \plus{} c)^3 \plus{} abc}$[/quote]
The first part does not hold, for exampl $ a \equal{} b \equal{} 1,c \equal{} 0.01$. The second part, see here: http://canhang2007.wordpress.com/2008/11/27/an-inequality-duong-duc-lam/ :)[/quote]

Thank you, Can. I'm sorry for my misstake, the first part must be $ \sum\sqrt [3]{\frac {a(a^2 \plus{} bc)}{(b \plus{} c)^3 \plus{} 8abc}}\le\sum\frac a{b \plus{} c}$  :(

And do you think that, the following inequality is also true

$ \sum\frac a{b \plus{} c}\ge\sum\sqrt {\frac {a(a^2 \plus{} bc)}{b^3 \plus{} c^3 \plus{} 6abc}}$ for all $ a,b,c$ be side lengths of a triangle ?  :)[/quote]

Hmm, thay're not true for all $ a,b,c$ be nonnegative. But we hope they correctly with the restrictions conditions of the veriables ?

1/ $ \sum\sqrt [3]{\frac {a(a^2 \plus{} bc)}{(b \plus{} c)^3 \plus{} 8abc}}\le\sum\frac a{b \plus{} c} , \forall a,b,c\in\left[\frac1{4}, 4\right]$
2/ $ \sum\sqrt [3]{\frac {a(a^2 \plus{} 2bc)}{(b \plus{} c)^3 \plus{} 16abc}}\le\sum\frac a{b \plus{} c} , \forall a,b,c\in\left[\frac1{6}, 6\right]$

My conjecture
\[ \sum\sqrt [3]{\frac {a(a^2 \plus{} pbc)}{(b \plus{} c)^3 \plus{} 8pabc}}\le\sum\frac a{b \plus{} c}
\]
for all $ a,b,c\in\left[\frac1{p\plus{}3}, p\plus{}3\right]$. :)

[quote="dduclam"]

Hmm, thay're not true for all $ a,b,c$ be nonnegative. But we hope they correctly with the restrictions conditions of the veriables ?

1/ $ \sum\sqrt [3]{\frac {a(a^2 \plus{} bc)}{(b \plus{} c)^3 \plus{} 8abc}}\le\sum\frac a{b \plus{} c} , \forall a,b,c\in\left[\frac1{4}, 4\right]$
2/ $ \sum\sqrt [3]{\frac {a(a^2 \plus{} 2bc)}{(b \plus{} c)^3 \plus{} 16abc}}\le\sum\frac a{b \plus{} c} , \forall a,b,c\in\left[\frac1{6}, 6\right]$

My conjecture
\[ \sum\sqrt [3]{\frac {a(a^2 \plus{} pbc)}{(b \plus{} c)^3 \plus{} 8pabc}}\le\sum\frac a{b \plus{} c}
\]
for all $ a,b,c\in\left[\frac1{p \plus{} 3}, p \plus{} 3\right]$. :)[/quote]

For the colection  :lol:
1/   $ \sum\sqrt{\frac{a(4a^2\plus{}5bc)}{(b\plus{}c)^3\plus{}abc}} \ge\sum\frac {2a}{b\plus{}c}$
2/   $ \sum\frac a{b\plus{}c}\ge\sum\sqrt[3]{\frac{a(a^2\plus{}2bc)}{2(b\plus{}c)^3\plus{}8abc}}$
3/   $ \sum\frac {2a}{b\plus{}c}\ge\sum\sqrt[3]{\frac{a(7a^2\plus{}2bc)}{(b\plus{}c)^3\plus{}abc}}$
4/   $ \sum\frac {2a}{b\plus{}c}\ge\sum\sqrt[3]{\frac{a(7a^2\plus{}2bc)}{b^3\plus{}c^3\plus{}7abc}}$
for all $ a,b,c\ge0$. :)


inequalities
advanced fields
advanced fields unsolved

Given any three non-empty sets A, B and C, |A| denotes the number of all the elements in A, (A Inter B) denotes the intersection of A and B, (A Union B) denotes the union of A and B, then how to prove:
|A Inter B|/|A Union B|+|B Inter C|/|B Union C|-|A Inter C|/|A Union C|<=1?

In a readable form: $\frac{|A \cap B|}{|A \cup B|}+\frac{|B \cap C|}{|B \cup C|}-\frac{|A \cap C|}{|A \cup C|}\le1$

This is a stupid way to go about it, but it does work: draw a Venn diagram and label each region with its own little variable.  Then, write the expression $\frac{|A \cap B|}{|A \cup B|}+\frac{|B \cap C|}{|B \cup C|}-\frac{|A \cap C|}{|A \cup C|}-1$ in terms of these variables and put it over a common denominator.  When you combine like terms, every single term of the numerator has the same sign, and you're done.

In fact, there is lots of wiggle room, which suggests one can make several simplifying assumptions of some sort.  Also, there should be some smarter combinatorial proof of some sort.

Using $\Delta$ for the symmetric difference of sets, one can rewrite the inequality as follows.
\[\frac{|A \Delta B|}{|A \cup B|}+\frac{|B \Delta C|}{|B \cup C|}\ge \frac{|A \Delta C|}{|A \cup C|}\]
So, $d(A,B): =\frac{|A \Delta B|}{|A \cup B|}$ is a metric. I suppose this formulation should lead to a simpler proof... but I don't see it yet.


If the blanks are replaced with three numbers to create an arithmetic sequence, what is the sum of these three numbers? 18, $ \underline{\ \ \ \ }$, $ \underline{\ \ \ \ }$, $ \underline{\ \ \ \ }$, 54.

$ d$ being the common difference in the sequence
18+4$ d$ = 54
4$ d$ = 36
$ d$ = 9

The three missing numbers are then 27, 36, and 45, which add to $ \boxed{108}$.

Also, the average of the 2 known numbers is 36. This is the middle number. The sum of the middle three then is three times 36, as shown by 36+(36-d)+(36+d). The 'D's cancel, giving 108.


geometry
geometric transformation
rotation
ratio
complex numbers
geometry proposed

Let $ ABCD$ be a cyclic quadrilateral. Let $ P,Q,R,S$ be the orthocenters of the triangles $ BCD,CDA,DAB,ABC$ respectively. Prove that the quadrilaterals $ ABCD$ and $ PQRS$ are congruent.

Lets represent the points $A,B,C,D$ with the complex numbers $a,b,c,d$.  W.L.O.G. we may assume $a,b,c,d$ lie on the complex unit circle.  Using useful formula #40 at

[url]http://mathcircle.berkeley.edu/complexBMC2.pdf[/url]

we have

\begin{eqnarray*}p=b+c+d=(a+b+c+d)-a\\ \ q=c+d+a=(a+b+c+d)-b\\ \ r=d+a+b=(a+b+c+d)-c\\ \ s=a+b+c=(a+b+c+d)-d\\ \ \end{eqnarray*}

Thus $pqrs$ corresponds to rotating $abcd$ by 180 degrees, and then translating it by ammount $a+b+c+d$.  Since rotation and translation are linear transformations, we have $pqrs$ is congruent to $abcd$.

the source of the problem is Balkan Mathematical Olympiad 1984.

Let $G_{1}G_{2}G_{3}G_{4}$ be the quadrilateral formed by the four centroids of the triangles, then by Euler's theorem we can prove that $G_{1}G_{2}G_{3}G_{4}\sim ABCD$ with ratio $\frac{1}{3}$ and $G_{1}G_{2}G_{3}G_{4}\sim PQRS$ with ratio $\frac{1}{3}$. Hence $ABCD \cong PQRS$.

This problem is also used at a Chinese Math League in 1992. The problem asked students to show that $PQRS$ is cyclic and find the location of its circumcentre.


If $ a,b,c>0$, then

$ (a^2\plus{}2)(b^2\plus{}2)(c^2\plus{}2)\geq3\sqrt{3}(a\plus{}b\plus{}c)\sqrt{ab\plus{}bc\plus{}ca}$

[quote="filomen"]If $ a,b,c > 0$, then

$ (a^2 + 2)(b^2 + 2)(c^2 + 2)\geq3\sqrt {3}(a + b + c)\sqrt {ab + bc + ca}$[/quote]

[size=75]Prove (or disprove) that:
 $ (x^2+k)(y^2+k)(z^2+k)\ge{\frac{(k+1)^3}{3\sqrt{3}}(x+y+z){\sqrt{xy+yz+zx}}}$

holds${ \forall{(x,y,z)}\in{R^3_{+}}}$ if and only if $ k=2$.[/size]

##################
read"D:/schur88.txt":

>f:=(x^2+k)^2*(y^2+k)^2*(z^2+k)^2-1/27*(k+1)^6*(x+y+z)^2*(x*y+y*z+z*x):

>symprove(subs(k=2,f));

                         it is symmetric for  {x, y, z}

                            ITS = [Delta]

                            IS = 1, N = 2

                        The inequality holds.

                            ITS = [Delta]

                            IS = 1, N = 1

                        The inequality holds.

                            ITS = [Delta]

                            IS = 1, N = 1

                        The inequality holds.

                               DEGU = 4

                               DEGU = 4

                               DEG = 4

                          ITS = [ee, Delta]

                            IS = 1, N = 2

                        The inequality holds.

                          ITS = [ee, Delta]

                            IS = 1, N = 2

                        The inequality holds.

                          ITS = [ee, Delta]

                            IS = 1, N = 2

                        The inequality holds.

                          ITS = [ee, Delta]

                            IS = 1, N = 2

                        The inequality holds.

                          ITS = [ee, Delta]

                            IS = 1, N = 2

                        The inequality holds.

                  The original form is psd!(SYM3(1))

                           它是平凡非负的!

                             TIME = 5.000

#####################
read"d:/bottema2009":

>xmin(subs(z=y,(x^2+k)^2*(y^2+k)^2*(z^2+k)^2-1/27*(k+1)^6*(x+y+z)^2*(x*y+y*z+z*x))>=0,[],k);

.....................................

`OUTPUT RESULT:`
`The inequality does not hold for any `.k

>xmax(subs(z=y,(x^2+k)^2*(y^2+k)^2*(z^2+k)^2-1/27*(k+1)^6*(x+y+z)^2*(x*y+y*z+z*x))>=0,[],k);

.....................................

`OUTPUT RESULT:`
`The inequality does not hold for any `.k

########################3

[quote="filomen"]If $ a,b,c > 0$, then

$ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2)\geq3\sqrt {3}(a \plus{} b \plus{} c)\sqrt {ab \plus{} bc \plus{} ca}$[/quote]
It's suffices to prove $ (a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1) \ge 3(a \plus{} b \plus{} c)^2$
 :roll:


Euler
geometry
incenter
geometric transformation
reflection
geometry unsolved

Prove that the following two statements are [b]equivalent [/b]
$1.$The angles of the triangle are in A.P.
$2.$ The Euler line is parallel to the common tangent of incircle and nine-point circle.

If $\angle B=60^{o}$,we only need prove that the excenter$O$,incenter$I$ and orthocenter$H$ of $\triangle ABC$and the points A,C are on the same circle the center of which is the reflection of $O$ with regard to $AC$.


Find all non-negative integral solutions $(n_1, n_2, \dots, n_{14})$ to
\[ n_1^4 + n_2^4 + \dots + n_{14}^4 = 1599  \]

Yes, because all perfect 4th powers are equivalent to $0$ or $1 \mod 16$. So since $1599\equiv 15 \mod16$ and we only have $14$ numbers, it is impossible.

[hide]

RHS is $15 \bmod 16$

quartic residues for integers $\mod 16$ are $0$ or $1$, hence the LHS can equal anything but $15 \bmod 16$

so no solutions

[/hide]

hmm what year was this from?

[quote="chess64"][quote="mysmartmouth"]Can someone post a hint?  This one has me stumped.  Could one use mod 16?[/quote]

How do you think of mod 16 and then not see the solution? :P[/quote]

Well, I saw mod 16 after typing my hint request, and then was too lazy to come back and post a solution.  :D

How do you people think of mod 16 though?  Unless you seen those type of problems before......

Almost any sort of equation of this sort on an olympiad has no solutions, except maybe a few trivial cases. There are two major ways to prove that something has no integral solutions: Infinite Descent, or incongruent modulos. In both cases, you use mods. To use mod 16, any power of 2 works great for squares (and fourth powers are squares); why 16 works good here is the number of variables (14) and the fact that 1599 is 15 mod 16.


geometry
3D geometry
number theory unsolved
number theory

Let $ x, y \in Z$ and $ \frac {x^2 \plus{} y^2 \plus{} 6}{xy} \in Z$. Find all $ x, y \in Z$ know that $ \frac {x^2 \plus{} y^2 \plus{} 6}{xy}$ be cube of a natural number.

we can suppose that $ x,y$ are in $ N$.
Let's suppose that $ x,y$ is a solution of $ x^{2}\plus{}y^{2}\plus{}6\minus{}kxy\equal{}0$ where $ k \in N$ and $ x>y>1$.note that if $ x\equal{}y$ then $ \frac{x^{2}\plus{}y^{2}\plus{}6}{xy} \not \in Z$.
let $ x'$ be the other solution of this equation for which $ xx'\equal{}y^2\plus{}6$ and $ x'\plus{}x\equal{}ky$.obviously $ x' \in N$.
then u can easily prove that  $ x'\equal{}\frac{y^2\plus{}6}{x}<y$.so we get a solution $ y,x'$ for which $ y>x'$.
doing this for all solution the process will end with either $ x\equal{}y$ or $ y\equal{}1$.
the first case is easy to check.
for the second one u get $ x\equal{}1,7$. which implies that $ k\equal{}8$ for both cases.
hence all solutions are obtained by these two sequences:(all the solutions are $ (a_n,a_{n\minus{}1})$ and $ (b_n,b_{n\minus{}1})$).
$ a_1\equal{}1,a_2\equal{}1,a_n\equal{}8a_{n\minus{}1}\minus{}a_{n\minus{}2}$, and
$ b_1\equal{}1,b_2\equal{}7,b_n\equal{}8b_{n\minus{}1}\minus{}b_{n\minus{}2}$.

Great solution, and of course there is just one sequence of solutions because $ a_{n\plus{}1} \equal{} b_n$

[quote="shoki"]we can suppose that $ x,y$ are in $ N$.
Let's suppose that $ x,y$ is a solution of $ x^{2} \plus{} y^{2} \plus{} 6 \minus{} kxy \equal{} 0$ where $ k \in N$ and $ x > y > 1$.note that if $ x \equal{} y$ then $ \frac {x^{2} \plus{} y^{2} \plus{} 6}{xy} \not \in Z$.
let $ x'$ be the other solution of this equation for which $ xx' \equal{} y^2 \plus{} 6$ and $ x' \plus{} x \equal{} ky$.obviously $ x' \in N$.
then u can easily prove that  $ x' \equal{} \frac {y^2 \plus{} 6}{x} < y$.so we get a solution $ y,x'$ for which $ y > x'$.
doing this for all solution the process will end with either $ x \equal{} y$ or $ y \equal{} 1$.
the first case is easy to check.
for the second one u get $ x \equal{} 1,7$. which implies that $ k \equal{} 8$ for both cases.
hence all solutions are obtained by these two sequences:(all the solutions are $ (a_n,a_{n \minus{} 1})$ and $ (b_n,b_{n \minus{} 1})$).
$ a_1 \equal{} 1,a_2 \equal{} 1,a_n \equal{} 8a_{n \minus{} 1} \minus{} a_{n \minus{} 2}$, and
$ b_1 \equal{} 1,b_2 \equal{} 7,b_n \equal{} 8b_{n \minus{} 1} \minus{} b_{n \minus{} 2}$.[/quote]

Thank you very much.  :P


Dear Mr. Rusczyk,

I can not find my round 2 score at http://www.usamts.org web site. Can you

check for me?

Thanks,


[quote="Jeremic"]I'm sorry but what do you mean by 5s?[/quote]he's talking about the exponent of $5$ when you prime factorize $50!$, $19!$, and $31!$..
Quevvy.. you have to remember that $25 = 5^{2}$. so whenever you have a multiple of $25$ you want to count that twice. and similarly for $125=5^{3}$ you want to count that three times..

[hide]In 50!, there are 10 multiples of 5 and 2 multiples of 25.
In 19!, there are 3 multiples of 5.
In 31!, there are 6 multiples of 5 and 1 multiples of 25.

Thus, there are 12-3-7=2 zeroes.[/hide]

[quote="Quevvy"][quote="bpms"][quote="ckck"]In how many zeros does the following end when expressed in standard base ten notation:
$\frac{50!}{19!(31!)}$[/quote]
[hide]Looking at the number 5s. There are 12 on top and 10 on bottom. Clearly there are more than 2 2s so our answer is 2[/hide][/quote]
[hide]I think there would be 10 5s on the top and 9 on the bottom and then 25 2s on the top and 24 on the bottom. That makes one zero in the final answer.[/hide]
I think that's right, but I'm not sure... :P

50th post! :D  :lol:[/quote]
Sorry, but you are wrong because
[hide]You didn't account for factors of 4, 8, 16, and 32 when counting the 2s since they give extras and you didn't account for factors of 25 when counting the 5s.[/hide]

Yeah the number is 30405943383200, so there is 2 zeros, to show that 2 is the correct answer.

But without a caluculator, you can do it the way that everyone has been doing it noting the the number of 5s is less than the number of 2s so the answer is the number of 5s.

But sometimes we can't just say the number of 5s, as in this example

$\frac{5!}{4!1!}$.

[quote="bpms"][quote="13375P34K43V312"]Yes, we have to make sure that there are that many number of twos as well.[/quote]
Well in this case it's kinda clear that there are more than two twos...[/quote]

I mean when we actually divide-there are 3 on the top and 3 on the bottom, so when we divide there won't be any left afterward.

[quote="13375P34K43V312"]Yes, we have to make sure that there are that many number of twos as well.[/quote]With factorials, aren't there always enough 2s?

Well, yeah, to some extent, because 1!, there aren't any twos :P

But when we divide, sometimes there will be an excess, but really rarely so we don't need to worry about it that much.

[hide="The solution  :) "]In the numbers fro $1$ to $50$, there are $10$ multiples of $5$, but $25$ counts twice since it is $5^{2}$

So there are actually $11$ $10$'s on the top.

Similarly, there are $3+6=9$ $10$'s on the bottom So I think there are $2$ zeros. My calculator agrees with me. [/hide]


function
inequalities
complex numbers
triangle inequality
AMC

A function $ f$ is defined by $ f(z) \equal{} (4 \plus{} i) z^2 \plus{} \alpha z \plus{} \gamma$ for all complex numbers $ z$, where $ \alpha$ and $ \gamma$ are complex numbers and $ i^2 \equal{} \minus{} 1$. Suppose that $ f(1)$ and $ f(i)$ are both real. What is the smallest possible value of $ | \alpha | \plus{} |\gamma |$?

$ \textbf{(A)} \; 1 \qquad \textbf{(B)} \; \sqrt {2} \qquad \textbf{(C)} \; 2 \qquad \textbf{(D)} \; 2 \sqrt {2} \qquad \textbf{(E)} \; 4 \qquad$

[hide] Let $ \alpha\equal{}a\plus{}bi,\gamma\equal{}c\plus{}di$. We have:

$ f(1)\equal{}4\plus{}i\plus{}a\plus{}bi\plus{}c\plus{}di\equal{}(4\plus{}a\plus{}c)\plus{}(1\plus{}b\plus{}d)i$. Thus, $ b\plus{}d\equal{}\minus{}1$. Additionally,

$ f(i)\equal{}\minus{}4\minus{}i\plus{}ai\minus{}b\plus{}c\plus{}di\equal{}(\minus{}4\minus{}b\plus{}c)\plus{}(\minus{}1\plus{}a\plus{}d)i$. Thus, $ a\plus{}d\equal{}1$. Subtracting the first equation from the second, $ a\minus{}b\equal{}2$.

We wish to minimize $ |\alpha|\plus{}|\gamma|\equal{}\sqrt{a^{2}\plus{}b^{2}}\plus{}\sqrt{c^{2}\plus{}d^{2}}$. Note that there are no restrictions on $ c$, so we can just set $ c\equal{}0$.

We have $ \sqrt{a^{2}\plus{}b^{2}}\plus{}\sqrt{d^{2}}\equal{}\sqrt{4\plus{}2ab}\plus{}|d|\ge\sqrt{4\plus{}2ab}\ge\sqrt{2}$, with equality if and only if $ d\equal{}0,a\equal{}1,b\equal{}\minus{}1$. Luckily enough for us, this is indeed achievable, so our desired minimum is $ \boxed{\sqrt{2}}$.[/hide]

[hide="Alternate Solution Ending"]$ b \equal{} a \minus{} 2$ and $ d \equal{} 1 \minus{} a$.

$ | \alpha | \equal{} \sqrt {a^2 \plus{} (2 \minus{} a)^2} \equal{} \sqrt {2} \cdot \text{QM} [a, 2 \minus{} a] \geq \sqrt {2} \cdot \text{AM} [a, 2 \minus{} a] \equal{} \sqrt {2}$, with equality when $ a \equal{} 1$

$ | \gamma | \equal{} |c \plus{} di| \geq |di| \equal{} |d| \equal{} |1 \minus{} a| \geq 0$, with equality when $ a \equal{} 1$.

So both terms are minimized simultaneously with $ a \equal{} 1$ and thus the minimum is $ \sqrt {2} \plus{} 0 \equal{} \sqrt {2}$ [/hide]

[quote="CatalystOfNostalgia"][hide] Let $ \alpha \equal{} a \plus{} bi,\gamma \equal{} c \plus{} di$. We have:

$ f(1) \equal{} 4 \plus{} i \plus{} a \plus{} bi \plus{} c \plus{} di \equal{} (4 \plus{} a \plus{} c) \plus{} (1 \plus{} b \plus{} d)i$. Thus, $ b \plus{} d \equal{} \minus{} 1$. Additionally,

$ f(i) \equal{} \minus{} 4 \minus{} i \plus{} ai \minus{} b \plus{} c \plus{} di \equal{} ( \minus{} 4 \minus{} b \plus{} c) \plus{} ( \minus{} 1 \plus{} a \plus{} d)i$. Thus, $ a \plus{} d \equal{} 1$. Subtracting the first equation from the second, $ a \minus{} b \equal{} 2$.

We wish to minimize $ |\alpha| \plus{} |\gamma| \equal{} \sqrt {a^{2} \plus{} b^{2}} \plus{} \sqrt {c^{2} \plus{} d^{2}}$. Note that there are no restrictions on $ c$, so we can just set $ c \equal{} 0$.

We have $ \sqrt {a^{2} \plus{} b^{2}} \plus{} \sqrt {d^{2}} \equal{} \sqrt {4 \plus{} 2ab} \plus{} |d|\ge\sqrt {4 \plus{} 2ab}\ge\sqrt {2}$, with equality if and only if $ d \equal{} 0,a \equal{} 1,b \equal{} \minus{} 1$. Luckily enough for us, this is indeed achievable, so our desired minimum is $ \boxed{\sqrt {2}}$.[/hide][/quote]

When and how does the $ z$ from the question come into play?  Thanks!

$ z$ is the input for the function. When you see $ f(1)$, that just means to plug in 1 for $ z$ in the equation $ f(z) \equal{} (4 \plus{} i)z^2 \plus{} \alpha z \plus{} \gamma$, so you get $ f(1) \equal{} (4 \plus{} i) \plus{} \alpha \plus{} \gamma$. Similarly plug in $ i$ for $ z$ to get $ f(i)$. After you have those two expressions, you don't need the function anymore, so $ z$ goes away.

Do we use guess and check to solve     $ \sqrt{a^{2}\plus{}b^{2}}\plus{}\sqrt{d^{2}}\equal{}\sqrt{4\plus{}2ab}\plus{}|d|\ge\sqrt{4\plus{}2ab}\ge\sqrt{2}$?

Thanks!

That's one approach, since this is the AMC and it might be fastest if you guess right. But guess and check, while it might work on this problem because the equality case has such simple numbers, doesn't generally get you the minimum. Here's my explanation of the inequality chain you copied:
[hide="Motivation for finish"]We have $ \sqrt{a^2\plus{}b^2}\equal{}\sqrt{4\plus{}2ab}$ (we make this substitution because it reduces the number of terms with variables from two to one) and $ |d|$ that we're trying to minimize. We notice that the first one is minimized when $ a$ and $ b$ are of opposite sign and have a very negative product, so neither is close to zero, and the minimum comes when $ a\equal{}1$ and $ b\equal{}\minus{}1$. (You should know how to show this rigorously, but since this is the AMC, you just have to convince yourself on the test.) Looking back at our equations, we discover that $ d$ is conveniently zero when this occurs, which means that both parts of $ |\alpha|\plus{}|\gamma|$ are minimized at the same spot. When we write up this solution, we write it in the form you copied, but we don't think of it that way.[/hide]
And if that seems a bit hairy, here's a third finish I thought up:
[hide="Yet another finish"]As in the rest, we note that we can set $ c\equal{}0$, which will minimize $ c^2$, and thus, our expression. Now we have two equations and three variables, so we substitute $ a\equal{}d\plus{}1$ and $ b\equal{}d\minus{}1$ into \[ |\alpha|\plus{}|\gamma|\equal{}\sqrt{(d\plus{}1)^2\plus{}(d\minus{}1)^2}\plus{}\sqrt{d^2}\equal{}\sqrt{2d^2\plus{}2}\plus{}\sqrt{d^2}\] and we see immediately that this is minimized when $ d^2$ is minimized, or when $ d\equal{}0$. Notice that this substitution is a bit cleaner than the one The Zuton Force uses, but all three solutions are equivalent.[/hide]

Sorry for reviving this post, but due to me being so dumb, I fail to see how $ |\alpha|\plus{}|\gamma|\equal{}\sqrt{a^{2}\plus{}b^{2}}\plus{}\sqrt{c^{2}\plus{}d^{2}}$.

Apart from this, I understand the entire solution.

By definition, if $ z\equal{}a\plus{}bi$, then $ |z|\equal{}\sqrt{a^2\plus{}b^2}$.  Remember that $ |z|$ is the distance from $ z$ to the origin in the complex plane.

I don't think that anyone wanted it but here is my solution.

We know that $ 4\plus{}i\plus{}a\plus{}bi\plus{}c\plus{}di$ and $ \minus{}4\minus{}i\plus{}ai\minus{}b\plus{}c\plus{}di$ are real. From here we can infer that $ b\plus{}d\equal{}\minus{}1$ and $ a\plus{}d\equal{}1$. We are trying to minimize $ \sqrt{a^{2}\plus{}b^{2}}\plus{}\sqrt{c^{2}\plus{}d^{2}}$. Because there are no requirements for $ c$, let $ c\equal{}0$. Plugging in that $ b\equal{}\minus{}1\minus{}d$ and $ a\equal{}1\minus{}d$ rearranges the expression to give $ \sqrt{d^{2}\plus{}1} \cdot \sqrt{2}\plus{}|d|$. Both terms are minimized with $ d\equal{}0$, so the answer is $ \boxed{\sqrt{2}}$.

How do you "infer" that b+d=-1 & a+d=-1?  :maybe:

Since the two expressions are real, the imaginary parts of each expression must equal 0.

I understood that, but how did he get b+d=[b]-1[/b] & a+d=[b]1[/b]? That's the only part of the solution that I don't get... Darn I'm dumb...

With $ 4\plus{}i\plus{}a\plus{}bi\plus{}c\plus{}di$, we can rewrite it as $ (4\plus{}a\plus{}c)\plus{}(1\plus{}b\plus{}d)i$. If the imaginary part is $ 0$, then $ 1\plus{}b\plus{}d\equal{}0$ and thus $ b\plus{}d\equal{}\minus{}1$. The same goes for the second equation. Also, we know that since no real number is coefficient to $ i$, that it must be the second part that is $ 0$.

[hide="Another solution"]
Note that $ \text{Im}\{zi\}\equal{}\text{Im}\{ai\minus{}b\}\equal{}\text{Re}\{z\}$

$ \text{Im}\{f(1)\}\equal{}1\plus{}\text{Im}\{\alpha\}\plus{}\text{Im}\{\gamma\} \equal{} 0$

$ \text{Im}\{f(i)\}\equal{}\minus{}1\plus{}\text{Re}\{\alpha\}\plus{}\text{Re}\{\gamma\}\equal{}0$

Multiply the top by $ i$ and add it to the bottom to get:

$ \alpha\plus{}\gamma\equal{}1\minus{}i$ (because $ \text{Re}\{z\}\plus{}i\cdot \text{Im}\{z\}\equal{}z$, clearly).

By the Triangle Inequality, $ |\alpha|\plus{}|\gamma|\geq |\alpha\plus{}\gamma|\equal{}|1\minus{}i|\equal{}\boxed{\textbf{(B)}\ \sqrt 2}$.
[/hide]

This was a fun problem :)


What is the probability that a diagonal randomly selected from a regular octagon connects opposite vertices? Express your answer as a common fraction.

[quote="thdanh90"]There are 20 diagonal totally and there are only 8 "opposite" diagonals. So the answer is $ \boxed 2/5$.[/quote]

You overcounted. There are only 4 opposite diagonals, the answer is 4/20=1/5.


I'm trying to find the remainder. I don't know how to solve this problem, because I was doing a problem, and I had;

9^7 (mod 15)

and then I changed it to

3^14 (mod 15), but I can't do anymore.

Is this how it is done?;

Divide by 3 to get;  3^13 (mod 5)

and then

[3^(2 + 3)] (mod 5)

and then 3^5 (mod 5)

243 (mod 5)

3 (mod 5)

and the remainder would be 3 wouldn't it?

Well, I don't really understand that solution. Isn't it 3^14, not 3^15?

Can you or somebody else please explain it a little bit more clearly?

How did you get from 9(81)^3 to 9(6)(6)(6)?

[hide="Solution"]
\[ 9^7 \equal{} 9^2*9^2*9^2*9 \equal{} 81*81*81*9 \\
81\equiv{6}\pmod{15} \\
6^3*9 \equal{} 36*6*9 \\
36\equiv{6}\pmod{16} \\
6*6*9\equiv{36*9}\equiv{6*9}\pmod{15} \\
54\equiv{\fbox{9}}\pmod{15}
\]
This is something similar to what leoxnlin posted, but I broke it down into steps so that it's as self-explanatory as possible.
[/hide]

You just break it down. There's a bunch of ways to do it.

[quote="rofler"]$ 9^7\equiv3^{14}$ mod 15.

Now $ 3^{14}\equiv 0$ mod 3
And $ 3^{14}\equiv 3^2 \equiv 4$ mod 5 since $ 3^4\equiv1$ mod 5 by fermat's little theorem.

By Chinese remainder theorem, there is a unique number mod 15 that satisfies both equations. Checking, we see that it is 9.

Therefore, $ 3^{15} \equiv 9$ mod 15.[/quote]

How did you get 9 as a solution for both the congruences? I can understand how you used Fermat's Little Theorem and how the Chinese Remainder Theorem shows that there is a number that satisfies both, but where did 9 actually come from?

@grn_trtle
[hide]Let $ 3^{14} \equal{} x$.

Then $ x \equiv 4 \mod 5$ implies that $ x \equal{} 4 \plus{} 5j$ for some integer $ j$.

Now, $ x \equiv 0 \mod 3$, so
$ 4 \plus{} 5j \equiv 0 \mod 3$,\

$ 1 \plus{} 2j \equiv 0 \mod 3$

$ 2j \equiv \minus{} 1 \equiv 2 \mod 3$

$ j \equiv 1 \mod 3$, where we can divide by 2 because 2 is coprime to the mod, 3.

Then $ j \equal{} 1 \plus{} 3k$ for some integer $ k$.  Now plug this back in:

$ x \equal{} 4 \plus{} 5j$

$ x \equal{} 4 \plus{} 5(1 \plus{} 3k)$

$ x \equal{} 9 \plus{} 15k$

$ x \equiv 9 \mod 15$.[/hide]

@archimedes1:
That's pretty cool! Does that sort of substitution method always work when the Chinese Remainder Theorem indicates that there exists a solution to some system of congruences?

powers of 9 give remainders   9,6,9,6.....so on (mod 15)  . so we simply get 9 as answer.

[quote="leoxnlin"]$ 81 \equiv 6 \bmod{15}$.  :wink:[/quote]

Oh, gotcha


Could someone please post/link the solutions for past AIME's to the Contests section?  Some of the AIME's have solutions, but some of them don't.  In particular, I am looking for the 2009 AIME II solutions.

Thanks a lot!

[quote="kunny"][url=http://www.unl.edu/amc/e-exams/e7-aime/archiveaime.shtml]You can look it here.[/url][/quote]
Or goto
http://www.artofproblemsolving.com/Wiki/index.php/AIME_Problems_and_Solutions
AoPS wiki with AIME problem and solution from  1983

If you click on each individual problem, there's usually a solution in the thread. Solution posting is annoying and time-consuming, so many of the solutions just haven't been entered into the database.


Let $ (b_0,b_1,b_2,b_3)$ be a permutation of the set $ \{54,72,36,108\}$. Prove that $ x^5\plus{}b_3x^3\plus{}b_2x^2\plus{}b_1x\plus{}b_0$ is irreducible in $ \mathbb Z[x]$.

Apply the transformation $ x \rightarrow x \plus{} 1$ we have the zero-coefficient is $ 3^2(3.2 \plus{} 2^3 \plus{} 2^2 \plus{} 2^2.3)$ which is divisible by 2 and not divisible by 4.All of the coefficients of $ x,x^2,x^3$ are divisible by 2.Apply the Generalization of Eisenstein's Criterion,then if $ P(x) \equal{} x^5 \plus{} b_3x^3 \plus{} b_2x^2 \plus{} b_1x \plus{} b_0$ is reducible,it is divisible by a polynomial $ R(x)$ of integer coefficients which $ deg(R) \geq 4$.So $ P(x)$ has a rational root,then it has a integer root because $ P(x)$ is monic,call this root $ x_0$.Set $ x_0 \equal{} 2^p.3^q.r$(easy to see that $ r \equal{}\minus{}1$),check all case of permutations we will have a contradiction.
(sorry because of my bad English and explaining)

When you apply the transformation x --->x+1 to $ P(x) \equal{} x^5 \plus{} b_3x^3 \plus{} b_2x^2 \plus{} b_1x \plus{} b_0$

then $ P(x\plus{}1)\equal{}(x\plus{}1)^5\plus{}b_3(x\plus{}1)^3\plus{}b_2(x\plus{}1)^2\plus{}b_1(x\plus{}1)\plus{}b_0$

the coefficient constant is $ 1\plus{}b_3\plus{}b_2\plus{}b_1\plus{}b_0\equal{}1\plus{}54\plus{}72\plus{}36\plus{}108\equal{}271$ and not  $ 3^2(3.2 \plus{} 2^3 \plus{} 2^2 \plus{} 2^2.3)\equal{}270$

It's pretty easy to check that the polynomial cannot have any integral roots (just use Rational Root Theorem and do some case bashing).

Hence, we can let $x^5 + b_3x^3 + b_2x^2 + b_1x + b_0 = (x^2 + ax + b)(x^3 + cx^2 + dx + e).$

Right off the bat, we can let $c = -a$ by looking at the $x^4-$coefficient.

We now have that $\{b + d - a^2, e+ad - ab, ae+bd, be\} = \{36, 54, 72, 108\}.$

If $3 \nmid b$ or $3 \nmid e$, we have a contradiction from a similar argument as Eisenstein. Else, we have $3 | b, e$. This means that $3| ad, d - a^2$ which implies $3| a, d$. This means that $3| a, b, d, e.$

Since $9| e + ad - ab$, we have $9 | e$ and so $27 | be$. This means that $be = 54$ or $108$. If $be = 54$, then we finish from Eisenstein with $p = 2.$ Hence we have $be = 108.$ If $2 \nmid b$ or $2 \nmid e$, we finish from a similar argument as Eisenstein. Else we have $2| b, e$ and so $e = \pm 18, b = \pm 6.$

From here, the problem is a finite case check (it may first help to get $e + ad - ab = 54$ to cut the search space approximately in third).

$\square$


Construct four three sided dice (say dice A,B,C,D) with faces from 1 to 12, each number used once, so that P(A beats B) > 1/2, P (B beats C) > 1/2, P (C beats D) > 1/2, P (D beats A) > 1/2. The dice may be weighted.

[hide]Dice A:
#s: 2, 6, 10
Ps: 1/3, 1/6, 1/2

Dice B:
#s: 1, 5, 9
Ps: 1/6, 1/3, 1/2

Dice C:
#s: 4, 8, 12
Ps: 1/2, 1/3, 1/6

Dice D:
#s: 3, 7, 11
Ps: 1/2, 1/6, 1/3

A beats B: A picks 10 with probability 1/2 and wins whenever that happens, and A wins other ways too.

B beats C: B picks 9 means 5/6 chance of winning, B picks 5 means 1/2 chance of winning, for a chance of $ \frac12 \cdot \frac56 \plus{} \frac12 \cdot \frac13 \equal{} \frac{7}{12}$.

C beats D: D rolls 3 half the time and loses whenever that happens, and C wins other ways too.

D beats A: D always wins on 11, wins half the time on 7, and one third of the time on 3, for $ \frac13 \plus{} \frac16 \cdot \frac12 \plus{} \frac12 \cdot \frac13 \equal{} \frac{7}{12}$ chance.[/hide]


function
trigonometry
limit
absolute value
calculus
calculus computations

Let $ I_n \equal{} (n\pi \minus{} \pi/2,n\pi \plus{} \pi/2)$ and let $ f$ be the function defined by $ f(x) \equal{} \tan x\minus{} x$.
(a) Show that the equation $ f(x) \equal{} 0$ has only one root in each interval $ I_n; n \equal{} 1; 2; 3; : : : .$
(b) If $ c_n$ is the root of $ f(x) \equal{} 0$ in $ I_n$, find $ \lim_{n\to \plus{} \infty}(c_{n \plus{} 1} \minus{} c_n)$.

hint:
for
 :lol: (a) obviusly

for
(b)
show that :
$ c_{n}$~$ \pi(n \plus{} 1/2) \minus{} \frac {1}{\pi n} \plus{} O(n^{ \minus{} 2})$
then $ lim\equal{}\pi$

[quote="N.T.TUAN"]Let $ I_n \equal{} (n\pi \minus{} \pi/2,n\pi \plus{} \pi/2)$ and let $ f$ be the function defined by $ f(x) \equal{} \tan x \minus{} x$.
(a) Show that the equation $ f(x) \equal{} 0$ has only one root in each interval $ I_n; n \equal{} 1; 2; 3; : : : .$
(b) If $ c_n$ is the root of $ f(x) \equal{} 0$ in $ I_n$, find $ \lim_{n\to \plus{} \infty}(c_{n \plus{} 1} \minus{} c_n)$.[/quote]
(a) let us look at $ f(x) \equal{} \frac {\sin\,x}{x}$

$ f'(x) \equal{} \frac {x\,\cos\,x \minus{} \sin\,x}{x^2}$

we know the extreme points (maximums/minimums) of $ f(x)$ are given by equating $ (x\,\cos\,x \minus{} \sin\,x)$ to $ 0$, i.e. solving $ \tan\,x \equal{} x$.

the extreme points of $ \sin\,x$ (of which there is an infinite number of them) occur at odd multiples of $ \frac {\pi}{2}$, i.e. whenever $ x_n \equal{} \frac {(2n \plus{} 1)\,\pi}{2}$.

now, as $ x$ [i]increases [/i]in absolute value, the points where $ \tan\,x \equal{} x$ [i]approach [/i]$ x_n \equal{} \frac {(2n \plus{} 1)\,\pi}{2}$, which as we just saw are the extreme points of $ \sin\,x$. this is obvious because as $ |x|$ increases, $ \frac {1}{|x|}$ decreases, and its effect on the variation of $ \sin\,x$ consequently [i]steadily [/i]diminishes. it therefore follows there can be only $ 1$ point of intersection of $ \tan\,x$ and $ x$ within the interval $ [x_{n \minus{} 1},x_n)$, where $ x_n \equal{} \frac {(2n \plus{} 1)\,\pi}{2},\;\;\;n\in\{\pm\mathbb N\}$

(b) from the first line of the last paragraph above, it follows

$ \lim_{n\to\infty}\;(c_n \minus{} c_{n \minus{} 1}) \equal{} \frac {(2n \plus{} 1)\,\pi}{2}\minus{}\frac {(2(n\minus{}1) \plus{} 1)\,\pi}{2}\equal{}\pi$

[ my earlier answer of $ 0$ is clearly wrong, as is pointed out by extremal below... ]

[quote="misan"]

(b) ... it follows that $ \lim_{n\to\infty}\;(c_n \minus{} c_{n \minus{} 1}) \equal{} 0$[/quote]
it's wrong.


let $ P$ and $ Q$ $ \in C[X]$ and $ PGCD(P,Q) \equal{} 1$ such that there existe $ a \in C$ and he is double root of $ P^2 \plus{} Q^2$

 prove that $ a$ is a root of $ P'^2 \plus{} Q'^2$

I will pose my solution while waiting for another suggestion

$ P^2 \plus{} Q^2$=$ (P \plus{} iQ)(P \minus{} iQ)$ since to $ a$ is a root of $ P^2 \plus{} Q^2$ so is a root of $ (P \plus{} iQ)$ or $ (P \minus{} iQ)$

but it cannot be a root from both in same time because if $ P(a)\plus{}iQ(a)\equal{}0$ and $ P(a) \minus{} iQ(a) \equal{} 0$

so $ P(a) \equal{} Q(a) \equal{} 0$ what is absurde whith $ PGCD(P,Q) \equal{} 1$

so suppose that $ P(a) \plus{} iQ(a) \ne 0$ so a is a double root of $ P \minus{} iQ$ so $ P'(a) \minus{} iQ'(a) \equal{} 0$

 so $ P'^2(a) \plus{} iQ'^2(a)$=$ P'(a) \minus{} iQ'(a)(P'(a) \plus{} iQ'(a)) \equal{} 0$

if  $ P(a) \plus{} iQ(a) \equal{} 0$ so  $ P(a) \minus{} iQ(a) \ne 0$ so a double root of $ P \plus{} iQ$  $ P'(a) \plus{} iQ'(a) \equal{} 0$

so $ P'^2(a) \plus{} iQ'^2(a) \equal{} P'(a) \minus{} iQ'(a)(P'(a) \plus{} iQ'(a)) \equal{} 0$

so in all case $ P'^2(a) \plus{} iQ'^2(a) \equal{} 0$

can  anyone tell me  what is meant by $ PGCD(P,Q)$ ?


let $ (x_1,\dots,x_n)$ be an increasing sequence of integers, $ x_1\ge 3$, $ x_n < 2n$, $ n > 2$.
prove that $ \sum_{k \equal{} 1}^{n \minus{} 1}\frac {\sqrt {x_{k \plus{} 1} \minus{} x_k}}{\sqrt {x_k}}\le\sqrt {n}$


function
logarithms
calculus
derivative
real analysis
real analysis unsolved

Let c > 0 be a positive real number. Prove that the function    f: (-1/c, oo) -> R defined by setting

f(x) = (1 + cx)^(1/x) for x not= 0

or

f(x) = e^c for x = 0

is of class C^oo, and find f ' (0), f '' (0), and f ''' (0).

Use the fact that $f(x)=\exp\left(\frac{\log(1+cx)}{x}\right)$. For $|x|<c^{-1}$ there is an expansion $\log(1+cx)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(cx)^{n}}{n}$, which you can divide by $x$ to get a power series representation for $\log f(x)$. This proves that $\log f$ is analytic in a neighborhood of zero. Insert its expansion into the power series for $\exp$ to find the derivatives of $f$ at zero.


calculus
integration
function
real analysis
real analysis unsolved

Hello Everyone. I'm a metal worker trying to do some layout for a project using some nice curves. To do the layout, I need to find the arc length of a few equations, but I'm finding some of the integrations to be a bit tricky. I was planning on using some nice simple curves for the surfaces, like various logs and simple 2nd and 3rd degree functions, but I'm having a hard time integrating f(x)=x^3 for arc length, and I'm having no luck scouring the net. Any suggestions for integrating (1+(3x^2)^2)^1/2? If that's not simple enough, I'll just cook up some easily integratable fuctions so I can get welding, but I would prefer to work the other way around. Even a point in the right direction would be appreciated.

Thanks in advance,
Stephen.

http://integrals.wolfram.com

Alternately, graphing calculators have a numerical integration function that you can use if all you're after is specific lengths.

That's so easy, it almost seems like cheating. Thanks! I'd still be interested to know the actual method to do it by hand though.

Actually, for that particular integral the answer the Integrator gives is, uh... not very easy to evaluate.  You're better off using numerical integration.


logarithms
real analysis
real analysis unsolved

Prove or disprove :

if $ \sum\limits_{n \equal{} 1}^{\infty} \sum\limits_{m \equal{} 1}^{\infty} a_{nm} < \infty$ , then :

$ \sum\limits_{n \equal{} 1}^{\infty} \sum\limits_{m \equal{} 1}^{\infty} a_{nm} \equal{} \sum\limits_{m \equal{} 1}^{\infty} \sum\limits_{n \equal{} 1}^{\infty} a_{nm}$

I'm sure this is true when $ a_{nm}$ is positive double sequence and we can prove it using Fubini Theorem .

What about if $ a_{nm}$ is not positive ? Is this still true ?.

Opinion Guys.

BTW, we are on measure space $ (\mathbb {Z\times Z} , \gamma, \gamma)$ , where $ \gamma$ is a counting measure .

A counterexample: $ a_{mn}\equal{}\begin{cases}1&m\equal{}n\\\minus{}1&n\equal{}m\plus{}1\\0&\text{otherwise}\end{cases}$. $ \sum_m\sum_n a_{mn}\equal{}0$ and $ \sum_n\sum_m a_{mn}\equal{}1$.

let $ a_{n1}\equal{}\frac{1}{n}\equal{}\minus{}a_{n2}\equal{}\minus{}a_{n3}$ for n odd and
$ a_{n1}\equal{}\frac{1}{n}, a_{n2}\equal{}a_{n3}\equal{}0$ for n even
and $ a_{nk}\equal{}0$ for k>3

then $ \sum_m a_{nm}\equal{} \frac{(\minus{}1)^n}{n} \to \sum_n \sum_m a_{nm}\equal{} \ln 2 < \infty$
but $ \sum_n a_{n1} \equal{} \infty$ clear contradiction

[quote="jmerry"]A counterexample: $ a_{mn} \equal{} \begin{cases}1 & m \equal{} n \\
\minus{} 1 & n \equal{} m \plus{} 1 \\
0 & \text{otherwise}\end{cases}$. $ \sum_m\sum_n a_{mn} \equal{} 0$ and $ \sum_n\sum_m a_{mn} \equal{} 1$.[/quote]
Does this not violate Fubini's theorem then?

no - Fubini rquires something like $ \sum_{m,n} |a_{m,n}| < \infty$, and this condition is clearly not satisfied in the above mentioned example.


[b]My cousin got this wrong in class, can you help me find the right answer?  Find the next 3 numbersin this pattern adn tell what the pattern is.  10,1,6,6,11
Can anyone help?[/b]

I'm not sure of my answer but, the first two add up to the 4th term so the next one will be 1+6=7. Then the next one will be 6+6=12. The next one will be 6+11=17.

Or maybe it is that you add 1 to the first number in the pattern.
10+1=11.
1+1=2.
6+1=7.
6+1=7
11+1=12
And so on.

Either there are infinitely many patterns, or he didn't give enough information.

KIM_8514, is this the exact same question that your brother was given, or a variation? Its seems as though this question is incomplete...


Plese help me with this problem.
Consider the permutation A= (123) (145)

What is A^99
please give the steps

Tried googling but was not of much help.

Write it as a product of disjoint cycles. Cycles commute with each other, so you can apply the exponent to each one separately.

Thanks for the reply.
But how Do you write it as a product of disjoint cycles. Sorry but Im new to this topic.
Started today and day after tommorows my exam  :D


Let ABC be a acute angled triangle .Construct the square outside the ABC on the  BC,construct the reqular m _ gonal  (A regular

 polygon with m side) outside theABC on the side AB and construct the reqular n _ gonal (A regular polygon with n side) outside the

 ABC on the side AC.m,n are natural number and m,n>5.the center of the square,regular m_gonal and regular n_gonal are the vertex

 of the equilateral triangle.Prove that m=n=6 and determine the angle of triangle ABC.


Since the total number of people was 200, and 96 were female, that means $ 200 \minus{} 96 \equal{} 104$ males. We know 26 doesn't listen to the station, so $ 104 \minus{} 26 \equal{} 78$ do. Thus, the percentage is $ 75$ percent.

Alternatively, you could have done $ 100 \minus{} 100\left(\frac{26}{104}\right)$, which would have gotten the same answer.


more generally, start with the substitution $a_{n}=b_{n}+ax^{2}+bx+c$, and choosing constants $a$, $b$, and $c$ to eliminate the quadratic in $n$, then $b_{n}$ is easily evaluated with characteristic equations

[quote="ch1n353ch3s54a1l"]
I don't really get this part... :oops:[/quote]

Look carefully at the first rearrangement. You have $\mathrm{some.function.of}(n+1)=3\cdot\mathrm{the.same.function.of}(n)$. Therefore you have a big big geometric sequence, and that fact is stated by the third equation. The same relationship exists between the second and fourth equations.

Then analyze the third and the fourth equations. If you put $b_{n}=a_{n}+n^{2}$, they become

$b_{n+1}-2b_{n}=3^{n}$
$b_{n+1}-3b_{n}=-2^{n}$

From there just express $b_{n}$ just as you would from any system of linear equations.

After that you're done, since $a_{n}=b_{n}-n^{2}$


A rectangular wall is being covered with 4-inch by 4-inch square tiles. How many tiles are needed to completely cover a wall that is 5 feet by 7 feet?

There are 12 inches in a foot, so the wall's dimensions in inches is $ 60 \times 84$. Thus, its area is $ 5040$. The area of each of the tiles is 16, so $ 5040 \div 16 \equal{} \boxed{315}$.

EDIT: Oops, that sounded weird. Fixed.

12 inches* in a foot.
sorry izzy, had to point that out.

You could also note that 3 segments that are 4 inches long make a foot, so the answer is $ (5\cdot 3)(7\cdot 3)\equal{}315$


Show that x4a + x4b+1 + x4c+2 + x4d+3, a, b, c, d positive integers, is divisible by x3 + x2 + x + 1.

If u is a (complex) root of x^3+x+x^2+x+1 then u^4=1 and u!=1.

Then u is also a root of  x^4a + x^4b+1 + x^4c+2 + x^4d+3 since u^(4m+n)=u^b.

Then x^4a + x^4b+1 + x^4c+2 + x^4d+3 is divisible by x^3+x+x^2+x+1  :)


It is well known that if A and B are two symmetric matrices that commute, then the product AB is symmetric as well.

Suppose we have two skew-symmetric matrices A and B that commute (i.e. AB=BA). Is it true that AB is skew-symmetric?

I can't quite seem to prove it... And it would be good if someone was to provide proofs of the second part (and first part as well if possible).

No; $ AB$ is symmetric in this case as well.  One has $ (AB)^{T} \equal{} B^T A^T \equal{} ( \minus{} B)( \minus{} A) \equal{} BA \equal{} AB$.  (One should think of skew-symmetric matrices as being "negative" and symmetric matrices as being "positive.")


analytic geometry
combinatorics unsolved
combinatorics

In a board of $2000\times2001$ squares with integer coordinates $(x,y)$, $0\leq{x}\leq1999$ and $0\leq{y}\leq2000$. A ship in the table moves in the following way: before a move, the ship is in position $(x,y)$ and has a velocity of $(h,v)$ where $x,y,h,v$ are integers. The ship chooses new velocity $(h^\prime,v^\prime)$ such that $h^\prime-h,v^\prime-v\in\{-1,0,1\}$. The new position of the ship will be $(x^\prime,y^\prime)$ where $x^\prime$ is the remainder of the division of $x+h^\prime$ by $2000$ and $y^\prime$ is the remainder of the division of $y+v^\prime$ by $2001$.

There are two ships on the board: The Martian ship and the Human trying to capture it. Initially each ship is in a different square and has velocity $(0,0)$. The Human is the first to move; thereafter they continue moving alternatively.

Is there a strategy for the Human to capture the Martian, independent of the initial positions and the Martian’s moves?

[i]Note[/i]: The Human catches the Martian ship by reaching the same position as the Martian ship after the same move.

The answer is [b]yes[/b], there is a strategy. Let $(x,y)$ and $(z,w)$ be the initial positions of the Human and Martian ships, respectively. The strategy is the following:

$\bullet$ Choose an integer $K$ such that
\begin{align*}
K &\equiv x-z-1 \pmod{2000}\\
K &\equiv y-w-1 \pmod{2001}
\end{align*}
Such a $K$ exists by the Chinese Remainder Theorem.

$\bullet$ The Human ship chooses velocity $(-1,-1)$ in the first turn. Thus, it moves to the position $(x-1\pmod{2000}, y-1 \pmod{2001})$

$\bullet$ From now on, if the Martian ship chooses velocity $(h_{i}, v_{i})$ in its $i$-th turn, the Human ship chooses velocity $(h_{i}-1,v_{i}-1)$ in the next turn. As velocities change by at most one, the Human ship can always do this.

Now, if the Human ship performs that strategy, it will capture the Martian ship in its $(K+1)$-th turn, as the Human ship will be in the square $(x-1+h_{1}+h_{2}+...+h_{K}-K \pmod{2000}, y-1+v_{1}+v_{2}+...+v_{K}-K \pmod{2001})$ and the Martian ship would be at $(z+h_{1}+h_{2}+...+h_{K} \pmod{2000}, w+v_{1}+v_{2}+...+v_{K}\pmod{2001})$, but $$(x-1+h_{1}+h_{2}+...+h_{K}-K \pmod{2000}, y-1+v_{1}+v_{2}+...+v_{K}-K \pmod{2001})= (z+h_{1}+h_{2}+...+h_{K} \pmod{2000}, w+v_{1}+v_{2}+...+v_{K} \pmod{2001})$$We are done. $\square$


function
combinatorics unsolved
combinatorics

Let ${ A_1, . . . ,A_p }$ and ${ B_1, . . . ,B_p }$ be two partitions of a finite set $M$such that, whenever $A_i \cap B_j = \emptyset$, it holds that $|A_i|+|B_j | \ge p$. Show that:$|M| \ge\frac{p^2+1}{2}$.Can equality hold?

Somewhat weaker bound.

Let $a = \min |A_i|$. Without loss of generality we assume that $|A_1|=a$.

It is clear that $|M|\geq pa$. On the other hand, $A_1$ has a non-empty intersection with at most $a$ different $B_j$ (first type) and, thus, has an empty intersection with at least $p-a$ different $B_j$ (second type). The total size of $B_j$ of the first type is at least $a$ while each $|B_j|$ of the second type is at least $p-a$ (since $p\leq |A_1| + |B_j|=a+|B_j|$). Hence, $|M|\geq a + (p-a)^2 > (p-a)^2$.

Combining these bounds we have $|M|\geq \min_a\max\{ pa, (p-a)^2\}$. Note that both functions are monotonic in $a$ and, hence, the minimax is attained at a point close to some root of equation $pa = (p-a)^2$ or $a^2 - 3pa + p^2 = 0$. An interesting root is $a = \frac{3 - \sqrt{5}}{2}p$ giving the bound $|M|> \frac{3 - \sqrt{5}}{2}p^2$.

firstly there seems to be some problem with the bound; i think it has to be $\frac{p^2}{2}$ instead, since if $p$ is even then if each set has size $\frac{p}{2}$ in each partition then the condition given is satisfied irrespective of whether $A_i\cap B_j$ is empty or not.
 start with the smallest set in both the collections, say $k=|A_1|\le|A_j|,|B_j|\ \forall\ j$. Then as maxal observed, all you need to see is that there are atleast $p-k$ sets of size atleast $p-k$ among the $B_j's$; as for the other $k$ sets among the $B_j's$, they have a size of atleast $k$ and so we have $n\ge (p-k)^2+k^2$. minimize this and we get that this is minimized for $k=\frac{p}{2}$ and the corresponding minimum value is $\frac{p^2}{2}$.

[quote="seshadri"]as for the other $k$ sets, they have a size of atleast $k$ and so we have $n\ge (p-k)^2+k^2$[/quote]
Wow! I was close. How could I overlook this simple observation?  :?


find the surface area of a hemisphere with a radius of 8 cm when a sector of 60 degrees has been removed.

my friends and i got 3 different answers.
its either 192pi, 512/3pi, or 144pi.
please help?


For [tex]x,y,z[/tex] be non-negative numbers st [tex]x+y+z=3[/tex].Prove that:
[tex]\sum\frac{x}{yz+x}\geq \sum\frac{x}{xy+1}[/tex]
  Both two expression are larger than [tex]\frac32[/tex].It's my conjecture.I 've verify it....But i've not solve it yet

The inequality is wrong. Try $(x,y,z)=(1,1.5,0.5)$.


calculus
integration
inequalities
function
real analysis
real analysis unsolved

Suppose g is a continuous, strictly increasing function on [0,infinity) such that g' is continuous on (0,infinity) and g(0) = 0. Show that for any a, b > 0 we have

integral(0,a) g(t)dt + integral(0,b) g^(-1)(t)dt >= ab

where g^(-1) is the inverse of g.

Please take a look this.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=21574&search_id=2045615223[/url]


function
limit
real analysis
real analysis unsolved

I know sin x^2 is not uniform continuous, and it seems to me that cos x^2 should be the same.  However, I cannot come up with a prove the show that.

Try comparing the values of the function at $ \sqrt{\frac{(2k\minus{}1)\pi}2}$ and $ \sqrt{\frac{(2k\plus{}1)\pi}2},$ noting as well the difference between those two values of $ x.$

Try proving this:

Suppose $ f(x)$ is continuous, non-constant, and periodic of period $ P.$

Prove that $ f$ is uniformly continuous on $ \mathbb{R}.$

Now define $ g(x)\equal{}f(x^2).$ Prove that $ g$ is not uniformly continuous on $ \mathbb{R}.$

Finally, let $ \psi(x)$ be an increasing continuous function on $ [0,\infty)$ such that $ \lim_{x\to\infty}\frac{\psi(x)}{x}\equal{}\infty.$

Let $ h(x)\equal{}f(\psi(x)).$ Prove that $ g$ is not uniformly continuous on $ [0,\infty).$


trigonometry
calculus
integration
limit
function
calculus computations

[quote="didilica"]Let $k>0$ be a positive real number. Find the limit:

$\lim_{n\rightarrow \infty}\int_{0}^{1}\frac{dx}{2+\cos^{k}(nx)}$.[/quote]

:( this one is hard, i was only able to do it for k=1 and k=2....
even for k=3 and fractional values like $\frac{1}{2}$ and $\frac{1}{3}$, my complex
variables approach became way [u]too [/u]complex for me to handle !  :rotfl:

for k=1, i got:

$\textbf{I}_{1}\;=\;\lim_{n\to\infty}\;\left\{ \;\frac{1}{\sqrt 3}\cdot\frac{\tan^{-1}\left( \sqrt{\frac{1}{3}}\cdot\tan \frac{n}{2}\right) }{\frac{n}{2}}\;\right\}$

for k=2, i got:

$\textbf{I}_{2}\;=\;\lim_{n\to\infty}\;\left\{ \;\frac{1}{\sqrt 6}\cdot\frac{\tan^{-1}\left( \sqrt{\frac{2}{3}}\cdot\tan n \right) }{n}\;\right\}$

and after this, i give up ! LOL ! :|

i am [b]speculating [/b] here that for k=3, [u]maybe[/u]

$\textbf{I}_{3}\;=\;\lim_{n\to\infty}\;\left\{ \;\frac{1}{\sqrt 9}\cdot\frac{\tan^{-1}\left( \sqrt{\frac{3}{3}}\cdot\tan \frac{3n}{2}\right) }{\frac{3n}{2}}\;\right\}\;=\;\boxed{\frac{1}{3}}$

so, want to give us the answer now ?   :maybe:

More important than the answer is the kind of approach that one can use to solve this problem. Didilica, can you provide an hint on how to solve your problem?

There are two parts here. One is the general fact that for a nice periodic function $f$ of period $T$, $\lim_{n\to\infty}\int_{a}^{b}f(nx)\,dx=\frac{b-a}{T}\int_{0}^{T}f(s)\,ds$.

The second is actually computing $\int_{0}^{2\pi}\frac1{2+\cos^{k}x}\,dx$.

I think that the limit is 0. I do not have a solution for the general $k$, however if $k$ is an even integer the limit is 0.

More precisely for $k>0$ the following limit holds:

$\lim_{n\rightarrow \infty}\int_{0}^{1}\frac{dx}{1+|\cos nx|^{k}}=0$.

Of course it is not important that 2 in the denominator of the original problem is replaced be 1.

The case $k=2$ is simple to prove since the integral can be evaluated directly.

Then study the two different cases when $k<2$ and when $k>2$.

When $k<2$ we have that

$\int_{0}^{1}\frac{dx}{1+|\cos(nx)|^{k}}\leq \int_{0}^{1}\frac{dx}{1+\cos^{2}nx}$ and the limit follows.

The case $k>2$ is solved by a similar trick, namely the convexity of $x\rightarrow x^{k}$.

So based on the fact that for all even integers the limit is 0 I think that the limit of the original problem is 0, however I do not have a solution.

That has to be wrong, Didi. Note that $\frac12\le\frac1{1+|\cos nx|^{k}}\le 1.$ The integral can't go to zero. The function isn't getting smaller, it's just staying roughly the same size and wiggling faster.

You are right Kent, I was thinking at something else, sorry about that...


MATHCOUNTS
geometry
3D geometry
rectangle

If I paint four faces of a cube, then break the cube up into 512 smaller cubes of congruent size, what is the largest and smallest number of cubes that could have exactly one painted face?  Two painted faces?

Here's my answer.

[hide]I think the answer to [/hide]MCrawford wrote:number of cubes that could have exactly one painted face[hide] is constant because the number of cubes with one painted side does not get affected as to which side is painted.

3:sqrt: of 512 is 8, one side of the cube will be divided into 64 squares (8 :^2: )

Since the edges (with two painted sides) and the corners (with three painted sides) doesn't count, it leaves the center part.

Thus the number of cubes with one side painted is 36 (6 :^2:).

There are 4 sides painted, so there are 144 cubes with one painted face.

In the case of two painted faces, there are two possibilities.

The first case is when the painted side forms a "ring."

In this case, the answer will be 24 cubes ((8-2)*4).

Otherwise, the answer is 30 cubes ((8-2)*5).

So the answer is 144 cubes with exactly one painted face, 24 cubes being the smallest number of cubes that have two painted faces, and 30 cubes being the largest number of cubes that have two painted cubes.

The only problem with this is that I didn't actually prove this answer...

[/hide]

By the way, are we allowed to use Emoticons like this?

Use the emoticons however will help you best express your ideas.

But check your thinking for 2 faces painted.  Which cubes along the edge of the larger cube will have exactly 2 painted?

6, I think... (8-2)

I don't see where I went wrong. Can you be more specific about my mistake?

If four faces of the large cube are painted, do you subtract out all of the corners from the edges?

[hide]Since some of the cubes on the corner have 2 faces painted even though they're corners, the answer should be

smallest number of cubes with 2 painted faces:32 (8*4)

largest number of cubes with 2 painted faces:36 (8*4+8-2)[/hide]

Am I right?

When I do this I get these two possibilities for 2 faces painted

[hide]If you don't paint opposite faces, then there are 4 edges with 6 cubes on each that got painted twice, and all 8 corners are only painted twice giving me 32 as stated in the last post

But when I look at it with not painted adjacent sides, I get 5 edges with 6 each that are painted twice and only 4 corners painted on two faces.  2 of the corners are only painted once(where the non painted faces meet), and 2 are painted on three faces as usual.  Giving me 34, not 32.  Am I doing something wrong I can't see???[/hide][/hide]

[quote="frost13"]Am I doing something wrong I can't see???[/quote]

Unless we are both misinterpreting the problem, I believe you are correct, frost13.

[hide]Slicing one cube into 512 smaller cubes results in a 8x8x8 cube.

Each face will look like this:

00000000

01111110

01111110

01111110

01111110

01111110

01111110

00000000

On each face that is painted, you will have 36 cubes with one face painted.  Four faces means 144 cubes automatically have one face painted.

Next, the edges must be checked.

If two opposite sides are left unpainted, there are 8 edges where painted and unpainted sides meet (Two non-adjacent squares of 8x8 are left unpainted).  Excepting the corners (where two sides are painted), there are 6*8 or 48 cubes with one face painted.

If two adjacent sides are left unpainted, there are 6 edges where painted an unpainted sides meet (One rectangle of 8*16 is left unpainted).  Excepting the corners, there are 6*6, or 36 cubes with one face painted.   Two of the corners (in the center of the rectangle), only have one face painted, so we add those two back in, for a total of 38 additional cubes.

Therefore, the number of cubes with one face painted is between 182 and 192.[/hide]


symmetry
abstract algebra
group theory
superior algebra
superior algebra unsolved

http://en.wikipedia.org/wiki/Klein_four-group
"The three elements of order 2 in the Klein four-group are interchangeable: the automorphism group is the group of permutations of the three elements. This essential symmetry can also be seen by its permutation representation on 4 points:
$V = \{ identity, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}$ "

[quote="hzbrl"]if G is the Klein's-4 group, show that, Aut(G)=S_3

****plz reply ......its urgent for me.[/quote]

Step 1: Show that $|\mbox{Aut}(G)|=6$

Step 2: Show $\mbox{Aut}(G)$ is [b]not[/b] abelian.

Step 3: The only non-abelian group up to isomorphism is $S_{3}$


integration
trigonometry
calculus
calculus computations

I doubt that that's even true.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87337]this topic.[/url]

I tried it for $x=10.$ In that case, $\frac2{x^2}=.02$ and my numerical estimate of the integral is $.045.$


Show how you could synthesize aniline from cyclopentanone in only four steps.

(Source: created by Carcul)

Carcul

Ha Ha Ha revived this topic  :D   :P
My method.
1. Convert Cyclopentanone to Cyclohexanone (Demjenov Rearrangement )
2. Convert Cyclohexanone into cyclohexane (Clemmensen )
3. Use Pd/Se/C to convert cyclohexane into benzene
4. Nitrate benzene and use Fe/HCl HO- to convert it into aniline

Is it right?  :maybe:

[quote="valeriummaximum"]hey madness..........
are'nt you using 1 step extra?????[/quote]
what madness has done is perfect i think but i think you can directly treat cyclohexanone with NH3 and form cyclohexylidine amine and then use Pd/Se to get directly aniline???is this right???

Finally someone found the first problem!

Madness proposal has seven steps, not four (the ring expansion comprises at least three steps, and nitration/reduction are two separate steps. Also,

[quote="Madness"]Use Pd/Se/C to convert cyclohexane into benzene[/quote]

A reference please? Because that doesn't look like a laboratory reaction.

This problem number 1 is still not solved!

Isnt what srinitrs done right? I mean cant we react with $ \ce{NH3}$ and then reduce to get cyclohexane amine and then aromatise?  :maybe:

No, its not right. Not only it does not envolve 4 steps, but also aromatization is an oxidation reaction, and our product is aniline, which is very easily oxidizable. :whistling:

and what reagents are we allowed to use?

You have to start with cyclopentanone, but you can use any other organic or inorganic reagent.

1.Add Diazomethane =>cyclohexanone formed
2.Dehydrogenate the cyclohexanone ring(formation of two = conjugated bonds)
3.Add ammonia=>imine formed and it tautomerises to give aniline(enamine)

[quote="Ram"]1.Add Diazomethane =>cyclohexanone formed[/quote]

Unfortunately, there is considerable formation of an epoxide.

I Checked the yield in a book and about 60% of ring expanded product and about 20% of epoxide product are formed.
Well in any case,Can I Know what your solution is?

I will give only the first step:

1) Cyclopentanone + pyrrolidine + acid -----> ?

That's right - enamine formation. What about the next step?

Following a hint from Carcul: [url=http://pubs.acs.org/doi/abs/10.1021/ol0604365]an article[/url] about addition of methylene to an enamine double bond in the presence of TiCl4/Mg in THF. This gives the cyclopropane, but I'm stuck there...


How would you do this problem?

Wroite 0.232 (with the 32 part repeating) as the quotient of two integers.

is there any method to do these types of problems?

[hide="hint"]Note that $\frac1{99}=0.\overline{01}$[/hide]

[hide="solution"]$0.\overline{23}=23\times\frac1{99}=\frac{23}{99}$[/hide]

[hide="more general"]If you have a number, $0.\overline{d_{1}d_{2}d_{3}d_{4}\cdots d_{n}}$, where $d_{k}$ is a digit, then as a fraction, it is $\frac{d_{1}d_{2}d_{3}d_{4}\cdots d_{n}}{10^{n}-1}$.  Of course, this may need to be simplified  :D [/hide]

Might be a little late but here's an algebra way to do this.
[hide]
Let $x=0.2323\ldots$
Then you know $100x = 23.2323\ldots$.
Now just subtract these two equations:
$100x-x=99x=23.2323\ldots-0.2323\ldots = 23$
So now you know $x=23/99$.
The reasoning behind multiplying x by 100 was that this sequence repeats every two decimal places, so multiplying by 100 would shift everything to the right by two places. Then you can just cancel out all of the repeating stuff by subtracting.
[/hide]

yeah that's the way i learned from algebra

*mental high 5*

[hide="A theoretical approach"]$0.\overline{23}= \frac{23}{100^{1}}+\frac{23}{100^{2}}+\frac{23}{100^{3}}+... = 23 \left( \frac{1}{100}+\frac{1}{100^{2}}+\frac{1}{200^{3}}+... \right)$

Consider the right factor as the sum of a geometric series; then its sum is simply

$\frac{ \frac{1}{100}}{1-\frac{1}{100}}= \frac{1}{99}$

So our expression is equal to $\frac{23}{99}$. [/hide]


calculus
integration
calculus computations

Below I did out an integral (from 0 to infinity)

∫x * e^(-px) dx

u=x, dv=e^(-px), du=1, v=(-1/p)e^(-px)

x*(-1/p)e^(px) - ∫ (-1/p)e^(-px)

I integrated the second integral using parts:
u=-1/p, dv = e^(-px), du = 0, v=(-1/p)e^(-px)

This resulted in a value of (1/p^2)e^(-px)

So the value of ∫x * e^(-px) dx turns out to be:

x*(-1/p)e^(-px) minus (1/p^2)e^(-px)

This factors to (-1/p)e^(-px) (x - (1/p)).

According to the back of my book, applying the limits is supposed to give 1/p.  Yet I'm getting 1/p^2 here.  Is there an error in my calculations?

Sign error at the last step: you should have $ ( \minus{} 1/p)e^{ \minus{} px}(x \plus{} (1/p))$. But your answer is correct: $ \int_0^{\infty}xe^{ \minus{} px}dx\equal{}1/p^2$.

(However, $ \int_0^{\infty}e^{ \minus{} px}dx \equal{} \frac {1}{p}$)


Find the radius of convergence of the series

$ \sum_{n = 1}^{\infty}{\frac {x^{n}}{2^{n}} . \frac {n^n}{n!}}$

I find that $ |{x}| < \frac{2}{e}$, but it says the radius of convergence is actually $ \frac{x}{e}$

Why is this? Shouldn't the radius of convergence be $ \frac{2}{e}$?

The radius of convergence is $ \frac 2e$ (with divergence for $ x\equal{}\frac2e$ and conditional convergence for $ x\equal{}\minus{}\frac2e$ ), so you are correct. A radius of convergence cannot contain the variable that this is a power series in, so $ \frac xe$ makes no sense and has to be an error of some sort, likely just someone's bad typing.


What is the number of square centimeters in the area of one
$ 30^\circ$ sector of a circle of radius 12 centimeters? Express your
answer in terms of $ \pi$.

The area of the entire circle is $ 12^2\pi\equal{}144\pi$ and we multiply by $ \frac{30}{360}\equal{}\frac{1}{12}$ to get $ \boxed{12\pi}$.

How to approach this type of problem:

30/360 = 1/12

We are to find 1/12 of the area of the circle with radius 12.

1/12 of 144 pi is 12 pi.

So 12 pi is the answer.
_____________________
I love pi!!  :juggle:  :love:


Prove that no triple $(x,y,z)$ exists such that:
(i) $x$, $y$ and $z$ are positive integers;
(ii) $x^2=y^2+z$; and
(iii) $z$ is a divisor of $y^2$.

Sorry if this problem is too easy, but I think it requires an insightful idea. ;)

Merry Christmas! Happy New Year! :D

Let $\left(x,y \right)=d$
so we have $d^2|z$
let $x=dx_1,y=dy_1,z=d^2z_1$
we have $x^2_1-y^2_1=z_1$, $(1)$
$z_1|y^2_1$,and $\left(x_1,y_1 \right)=1$
so we have $\left(y_1,z_1 \right)=1$(if not,it will contradiction with $\left(x_1,y_1 \right)=1$)
so $\left(y^2_1,z_1 \right)=1$,too.
and with
$z_1|y^2_1$
then $z_1=1$
so with $(1)$, $x^2_1-y^2_1=1$,it is impossible.
nice problem :)   Merry Christmas!!!

Nice solution. :)
Any other solutions, before I post mine?

i haven't really read zhaobin's solution, so I don't know if ours are the same.
Case 1: $p^a || z$, where p is prime and a is odd.
Then obviously $p|\frac{x^2}{z} - \frac{y^2}{z} = 1$, contradiction
Case 2: z is a perfect square.
Then put $z = k^2$, so $k|x$ and $k|y$, so $\left(\frac{x}{k}\right)^2 = \left(\frac{y}{k}\right)^2 +1$ which obviously has no solution.

Another one:

Since z is a divisor of $y^2$, we can write $y^2=sz$ with some positive integer s. Now, $x^2=y^2+z$ yields $z=x^2-y^2$, so that $y^2=sz=s\left(x^2-y^2\right)=sx^2-sy^2$. This transforms into $\left(s+1\right)y^2=sx^2$, and this, upon multiplication with s, becomes $s\left(s+1\right)y^2=s^2x^2$, so that $s\left(s+1\right)=\frac{s^2x^2}{y^2}=\left(\frac{sx}{y}\right)^2$. Thus, the number s (s + 1) must be the square of a rational number. Since it is an integer, it must therefore be the square of an integer. But s (s + 1) cannot be the square of an integer, since it lies between two adjacent squares: $s^2<s\left(s+1\right)<\left(s+1\right)^2$. Contradiction. Thus, three positive integers x, y, z as required by the problem don't exist.

  Darij


limit geometry inequalities integration logarithms calculus derivative

These are from my real analysis homework.

1) Let p be a positive integer. For $ n \in \mathbb{N}$ let $ b_n \equal{} \sum^{pn}_{i \equal{} n \plus{} 1}\frac{1}{i}$. Determine whether or not the sequence converges.

I have already shown that the sequence converges for p = 2 by demonstrating that it was monotone increasing and bounded, and using the monotone convergence theorem. I'm almost positive that it converges for any p; why would it converge for p = 2 and not p = 3 or p any natural number? Try as I might though I can't figure out how to show that it's a Cauchy sequence, I wind up showing that $ |b_n \minus{} b_{n \plus{} k}| < 1$ rather than $ |b_n \minus{} b_{n \plus{} k}| < \epsilon$.

2) Show that for $ x > 1, \lim_{n \to \infty}(\frac{1\plus{}\sqrt[n]{x}}{2})^{2n} \equal{} x$. You may use the fact that $ \lim_{n \to \infty}(1\plus{}\frac{a}{n})^n \equal{} e^a.$

In a previous question I showed that, for x > 1, $ \sqrt[n]{x} \plus{} \frac{1}{\sqrt[n]{x}} < 2 \plus{} (\frac{x\minus{}1}{n})^2$, and I'm pretty certain we have to use that fact.

Here's a hint for one approach to the first one: $ \sum_{i \equal{} n \plus{} 1}^{4n} \equal{} \sum_{i \equal{} n \plus{} 1}^{2n} \plus{} \sum_{i \equal{} 2n \plus{} 1}^{4n}$.

For the second one, you can try using $ \frac {1 \plus{} z^2}{2} \equal{} \frac {1/z \plus{} z}{2z}$ and your other problem.

An alternative approach to the first one:

Sketch a graph of the $ b_n$ as a sum of boxes of width $ 1,$ and compare it to areas under the curve $ y=x.$ We arrive at the following inequalities:

$ \int_n^{pn}\frac1x\,dx>b_n>\int_{n+1}^{pn+1}\frac1x\,dx$

${ \ln p>b_n>\ln(\frac{pn+1}{n+1})>\ln(p\cdot\frac{n}{n+1}})
>\ln p-\ln(1+\frac1n).$

[quote="Kent Merryfield"]An alternative approach to the first one:

Sketch a graph of the $ b_n$ as a sum of boxes of width $ 1,$ and compare it to areas under the curve $ y = x.$ We arrive at the following inequalities:

$ \int_n^{pn}\frac1x\,dx > b_n > \int_{n + 1}^{pn + 1}\frac1x\,dx$

${ \ln p > b_n > \ln(\frac {pn + 1}{n + 1}) > \ln(p\cdot\frac {n}{n + 1}}) > \ln p - \ln(1 + \frac1n).$[/quote]

What's that squiggly line thing? And what does "dx" mean?  :wink: We haven't even defined derivatives yet, let alone Riemann integrals. I got the solution to the second problem if anyone wants to see it.

Well, if you're going to be that way ... !

$ b_{n \plus{} 1} \minus{} b_n \equal{} \left(\frac1{pn \plus{} 1} \plus{} \frac1{pn \plus{} 2} \plus{} \cdots \plus{} \frac1{pn \plus{} p}\right) \minus{} \frac1{n \plus{} 1}$

$ \equal{} \frac1p\left[\left(\frac1{n \plus{} \frac1p} \plus{} \frac1{n \plus{} \frac2p} \plus{} \cdots \plus{} \frac1{n \plus{} 1}\right) \minus{} \left(\frac1{n \plus{} 1} \plus{} \cdots \plus{} \frac1{n \plus{} 1}\right)\right].$

(We're making sure that the two parentheses have the same number of terms, namely $ p$.)

Now, we match up terms.

$ \frac1{n \plus{} \frac kp} \minus{} \frac1{n \plus{} 1} \equal{} \frac {1 \minus{} \frac {k}{p}}{(n \plus{} 1)\left(n \plus{} \frac {k}{p}\right)} < \frac1{n(n \plus{} 1)}.$

Take $ p$ copies, and divide by $ p$ and we conclude that

$ 0 < b_{n \plus{} 1} \minus{} b_n < \frac {1}{n(n \plus{} 1)}.$

Since we know that $ \sum_{n \equal{} 1}^{\infty}\frac1{n(n \plus{} 1)} \equal{} 1,$ we conclude that $ (b_n)$ is increasing and bounded above, hence convergent.

[No integrals in sight, anywhere - not even in the specific claim of convergence of a series, which can be settled by a partial fractions/telescoping series argument.]


geometry
symmetry
integration
calculus
function
calculus computations

hey i have attached an image to the post.
I have to find the area of the shaded region
kindly guide me about how to find it ?

By symmetry the area is
$2 \int_{0}^{\frac{\pi}{2}}sin \ x$

which works out to $2$

$EDIT$ : Wait a minute the original one can be done by integration too can't it?

yea kindly guide me how to handle such questions ?
i am familiar with integration
but how to choose the interval ? n etc
kidnly GUIDE ME

for all normal functions , the integral over the interval will give you the answer. what i mean by [i]normal functions[/i] is
 functions that are nice, continuous, finite and either
 positive definite or negtive definite. then the procedure

 $\int_{a}^{b}f(x) \ dx$ is absolutely OK.

 Problems arise if the function is non-finite at some points in
 the interval. this you had better look up in standard textbooks
 as it is too varied and complex to put up here. Now the point
 is what happens if a nice function crosses the x-axis at a point?

 Then either by symmetry or by breaking into pieces, resolve
 the integral into two or more separate integrals, and
 [b] ADD THEIR ABSOLUTE VALUES[/b].

thanks alot ..............   :oops:
i got it now
hehehehe


I found this problem in one of the Russian olympiad levels. :) ( Most probably it is from the Moscow Olympiad )

Here it is :
You have n - natural number. Two players are placing dots among a circumference, the first is using green color and the second one red. When n dots are marked from each color, the game is over. Every player determines the longest arc with ends in his color, which does not contain any other marked points. The player with bigger/longer arc wins (if the arcs of both players have equal length, or both players do not have such arc, the game is tie). Which one of those has winning strategy?

:)

This problem is form the tournament of towns, it's very nice
When I first saw it, it made me think of
http://www.mathlinks.ro/Forum/viewtopic.php?p=925868#925868
This problem was from the iberoamerican olympiad 2007, and If I recall correctly it wasn't fully solved by any participant during the test
Some friends of mine knew the solution of the iberoamerican problem, and they found that the problem you have posted has a very similar strategy, so you may check post numer 4 for some inspiration.

Daniel


assume that $F$ is family of real sequences.we know that for every $(a_{n}),(b_{n})\in F$ : $\sum a_{n}b_{n}=0$ .does it follow that $F$ is countable.

If the sequences in question were all square summable (that is, members of $\ell^{2}$), then the answer would be yes, countable.

As it is, I'm skeptical, although I don't have an example. Note that there may also be an issue of whether $\sum a_{n}b_{n}$ is well-defined.


four circular pipers with 2inch radii are bound together with a wire so that they are tangent to each other. Find the length of the shortest wire that will go around them.

[hide]Just draw a square-like wire that goes around four circles but with curved corners just like that of the circles. The curved edges will have 1/4 the circumference of one of the circles. The rest of the wire will be made up of lengths with two of the radius of two the circle. The length of the curved parts will be $ 4(1/4)(2^2)\pi$ which will take the value of $ 4\pi$. The straight segments will have lengths $ 2r$ will be equal to 4 and since there are four of them 4X4=16. So the length of the wire is $ 16 \plus{} 4\pi$.[/hide]


1. The harmonic mean of two non-zero numbers $a$ and $b$ is number $h$ such that
\[ \frac{1}{h}=\frac{1}{2}\left (\frac{1}{a}+\frac{1}{b}\right ).  \]
Let $ABCD$ be a trapezoid with $AB||CD$. Let $s$ be the line segment parallel to $AB$ and $CD$ going through the intersection point of the diagonals $AC$ and $BD$. Prove that the length of $s$ is the harmonic mean of $AB$ and $CD$.

Let $s$ meets $AD$ and $BC$ at $E$ and $F$. Let $O= AC \cap BD$.

Since $AB || s || CD$, so $\frac{EO}{CD}=\frac{AO}{AC}$, $\frac{FO}{CD}=\frac{BO}{BD}$. Add them togather we get $\frac{EF}{DC}=\frac{AO}{AC}+\frac{BO}{BD}$.

Similarly we have $\frac{EF}{AB}=\frac{CO}{AC}+\frac{DO}{BD}$. Once again, add them togather we get:
\[ \frac{2}{EF}=\frac{1}{AB}+\frac{1}{CD} \]


Gauss
combinatorics proposed
combinatorics

Given positive integers$ m,n$ such that $ m < n$. Integers $ 1,2,...,n^2$ are arranged in $ n \times n$ board. In each row, $ m$ largest number colored red. In each column $ m$ largest number colored blue. Find the minimum number of cells such that colored both red and blue.

???

I think the answer is $ m^{2}$, which can occur in the case if you simply write down $ 1\sim n^{2}$.
Pick up and mark the largest number in the board. Do this until there exist a row or column which has $ m$ marked number.
Then erase other numbers in that row/column. We call this row or column is full. Now mark numbers same way. When a row or column become having $ m$ marked numbers, erase others in the row or column.
Do this until we have $ m$ full row or $ m$ full column. Then all marked number would be points colored both red and blue. If the process was ended, there are $ m$ full row/column, always we have least $ m^{2}$ marked points.

Little Gauss,  apply your method with m=2 to this board:

5, 4, 3
7, 8, 9
6, 1, 2

it becomes easy when we notice that the number of cells colored red and blue doesn't change, when we exchange the numbers k and k+1


$\frac{x+2y}{xy}=\frac{11}{12}$
$\frac{2x-3y}{xy}=\frac{2}{3}$

Solve for the variables.

[quote="bob123"][quote="dwx314"][hide]$ \frac {x+2y}{xy} = \frac {11}{12}, \frac {2x-3y}{xy}=\frac {2}{3}$
$x+2y=\frac {11xy}{12}, 2x-3y=\frac {8xy}{12}$
$[2x+4y=\frac {22xy}{12}]-[2x-3y=\frac {8xy}{12}] \rightarrow 7y=\frac {14xy}{12}$
$84y=14xy \rightarrow 6y=xy \rightarrow 6=x$
$\frac {12-3y}{6y}=\frac {2}{3} \rightarrow 36-9y=12y$
$36=21y \rightarrow y=\frac {12}{7}$
$>>x=6, y=\frac {12}{7}<<$[/hide]
I think...Check my work for me someone[/quote]

[hide]Make the second one:
$\frac{2x-3y}{xy}=\frac{8}{12}$
So then you have:
$2x-3y=8$
and
$x+2y=11$
Solve...
$2x+4y=22$
$2x-3y=8$
$7y=14$
$y=2$ $\longleftarrow$
$2x+8=22$
$2x=14$
$x=7$ $\longleftarrow$
[/hide]

I messed up... oh well, too lazy to edit it again[/quote]
You can't do that because xy is not 12


trigonometry calculus integration function

[quote="bos1234"]$ \tan4\theta=1$ has soln. $ 4\theta=\frac{\pi}{4}+{n\pi}$ where n is an integral

How do you prove the second part of the equation that $ 4\theta$ will have solutions in that form[/quote]

The equation,
$ \tan u = 1$
Has infinitely many solutions, which can be obtained by addind $ \pi$.
So since, $ \tan \frac{\pi}{4}= 1$ it means,
$ u = \frac{\pi}{4}+\pi n$

Since here $ u=4x$ it means,
$ 4x = \frac{\pi}{4}+\pi n$
In other words,
$ x = \frac{\pi}{16}+\frac{\pi n}{4}$
-------
And this [b]are[/b] solutions. (Meaning its sufficient as well as necessary).
Because,
$ \tan \left( \frac{\pi}{4}+\pi n \right) = \frac{\tan \frac{\pi}{4}+\tan \pi n}{1-\tan \frac{\pi}{4}\tan \pi n}= 1$

Because the tan function is periodic with period $ \pi$, all solutions can be obtained by adding anymupltiple of $ \pi$.


algebra
polynomial
linear algebra
matrix
group theory
abstract algebra
calculus

Here is a collection of very similar problems given at oral examination for agregation in France:
  1) A is a complex invertible matrix. Prove that there exists a polynomial with complex coefficients $P$ such that $e^{P(A)}=A$.
 2) Let $A$ be a complex invertible matrix. Prove that for all $n$ there is a polynomial $P$ such that $P^n(A)=A$.
 3) Let $A$ be a real invertible matrix which admits a real square root. Prove that for all $k\geq 2$ the matrix admits a $k$ root which is real. (this one is really tough)

1/ $e^A = lim_{n-> \infty} E_n$ where $E_n=I+A+..+\frac{A^n}{n!} \in \mathbb{C}[A]$. But  $\mathbb{C}[A]$ is closed.

2/ it suffices to prove it for matrices that have a small norm. And with the power series of $(1+x)^{1/n}$ it should work.

I don't think you understood the first question, Alekk. The problem is not to prove that $ e^A$ is a polynomial in $A$. ;)    And would you like to give more details about the second problem?

:blush:  soory, I read it too quickly. There is a proof of 1/ with dunford decomposition
$A=D+N=D(I+D^{-1}N)$. But $D^{-1}N$ is nilpotent so it is easy to find $P \in C[X]$ such that
$exp^{P(A)}=I+D^{-1}N$. (remember that $D$ and $N$ are in $\mathbb{C}[A]$). Conclusion follows.
Nevertheless I've read a few days ago another way to prove it:

1/$exp: C[A] -> C[A]$ is a continuous morphism of $(C[A], +)$ to $(U,.)$ where $U \subset C[A]$ is a closed subgroup of the invertible matrices of $C[A]$.

2/ local inversion theorem shows that there is a neighboorhood of $I$ included in $U$. So $U$ is also open.

3/ it is not hard to see that $C[A]$ is connex.

Hence $U=C[A]$.

In the second problem, we want the minimal polynomial of $A$ to divide $P^n(x)-x$. This follows from the fact that given $k$ and some distinct complex numbers $\lambda_1,\ldots,\lambda_t$, we can construct a polynomial $P$ with prescribed values for $P^{(i)}(\lambda_j),\ i\in\overline{0,k-1},\ j\in\overline{1,t}$, where $P^{(i)}$ is the $i$'th derivative with $P^{(0)}$ simply being $P$.

In our case we will need the $\lambda$'s (which are the different eigenvalues of $A$) to be non-zero because, for instance, if you want $P(\lambda)=\lambda^{\frac 1n}$ (i.e. $x-\lambda|P^n(x)-x$; we choose one of the $n$'th roots of $\lambda$) and also $nP'(\lambda)P^{n-1}(\lambda)=1\ (*)$ (i.e. $x-\lambda|(P^n(x)-x)'$), we will want to divide $(*)$ by some power of $\lambda$ to get the value of $P'(\lambda)$.

now we've prove 1/, try to find $exp(M_n(R))$. It is the same conclusion that $exp( \mathbb{R} )$  :D

I know, Alekk but I still don't know a solution which takes less than 4 pages for this result!!! And it is clear the third problem is an easy consequence of that. DO you have simple solution?

If $A=exp(R)$ then $A=(exp(\frac{R}{2}))^2$
If $A=M^2$ where $M \in M_n(R)$. Then there exists a complex polnomial $P \in C[X]$ such that
$M=exp(P(M))$. But $M=exp(Q(M))$ where $Q$ is the conjugated polynomial of $P$.
So $A=M^2=exp(P(M)+Q(M)) \in exp(M_n(R))$ because $P(M)$ and $Q(M)$ commute and $P+Q \in R[X]$.

conclusion: $exp(M_n(R)=\{ M^2, M\in GL_n(R)\}$


I created few AMC Fake Questions (because they look like one but they're not from AMC :D )

I would say this is like #4 on AMC-10.

1. Given that ABCDEF is a hexagon, the quadrilateral PBCO with P being midpoint of AB, has area of 10 then area of ABCDEF is what? (O is the center of the hexagon)

2. Every member of the committees of UPC (United Pluto Committee) has to pick from 5 different category:

A) Origin
B) Skin Color
C) Last name
D) Favorite Food
E) God They Serve

In how many ways, can you have each member different?

EDITED: The quadrilateral was falsely labeled. It should've been PBCO, not PBCD.

I don't think there is enough info to solve #1.  If it was regular or something then it would be solvable....

I haven't actually solve # 1 yet but #2 was a bit  easy..

#1 actually does look like an AMC question
because I thinkI saw a problem almost like that before.

though I couldn't solve it :D

[b]Regular[/b] hexagon.

That should be a right word.

Hmm.. I never thought this question, or similar to this one, appeared on AMC. :)

oh there is...

I think it was like AMC10 or something..

I tried to solve the question.., but couldn.t :D

well, i think they are basically the same questions but only the numbers are different...

hm..

Perhaps I'm thinking about this too simply, but isn't the answer to number 1 40? It should be because connecting PBCO creates a midpoint on the top and essentially fills up the upper right corner. (Draw a diagram) Am I right on this?


This inequality is a particular case of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=192767]Janous' conjecture[/url]. It is reminiscent of http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1832 . It has weaker conditions than http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1832 and a weaker assertion.

[quote="arqady"]If some reals $ a > 0$, $ b > 0$, $ c > 0$, $ d > 0$, $ e > 0$, $ f > 0$ satisfy $ a \plus{} b \plus{} c \plus{} d \plus{} e \plus{} f \equal{} 1$, then prove that
$ abc \plus{} bcd \plus{} cde \plus{} def \plus{} efa \plus{} fab\leq\frac {1}{27}$.[/quote]

This follows from

$ abc \plus{} bcd \plus{} cde \plus{} def \plus{} efa \plus{} fab\leq abc \plus{} bcd \plus{} cde \plus{} def \plus{} efa \plus{} fab \plus{} ace \plus{} bdf$
$ \equal{} \left(a \plus{} d\right)\left(b \plus{} e\right)\left(c \plus{} f\right)$
$ \leq\frac {1}{27}\left(\left(a \plus{} d\right) \plus{} \left(b \plus{} e\right) \plus{} \left(c \plus{} f\right)\right)^3$           (by AM-GM)
$ \equal{} \frac {1}{27}\left(a \plus{} b \plus{} c \plus{} d \plus{} e \plus{} f\right)^3 \equal{} \frac {1}{27}$.

  Darij


function algebra domain limit trigonometry calculus derivative

How could you determine whether the following functions are uniformly continuous on the domain, D. What would the epsilon−delta proof look like?

i) f(x) = 1/(x−1) , D = (1, 2).
ii) f(x) = x^2 sin (1/x), D = (0, 1).

An uniformly continuous function on a bounded set is bounded on that set. But $\ \lim_{x\to1}f(x)=\infty$, so the first function is not uniformly continuous on $\ (1,2)$.

For the second function: if the derivate of a function on an interval is bounded, then the function is uniformly continuous. $\ |f'(x)|=|2x\sin\left(\frac1x\right)-\cos\left(\frac1x\right)|\leq3$ so the second function is uniformly continuous on $\ (0,1)$.

Nevermind, I see the derivative is right

That derivative is correct- remember the chain rule.
It's also overkill- can you show that $x\sin\frac1x$ is uniformly continuous on $(0,1]$?

uhm, was that a rethorical question.  If not : yes  :) .

But I will even do it .The function is continuous in zero with value zero.  If it has a continous extension on the compactum [0,1] it is even uniformly continous there.  Works too for the one with $x^2^$

This is a nice criterion for non uniform continuity : it exists if and only if there are two sequences $x_n$ and $y_n$ such that $x_n-y_n$ converges to zero, while $f(x_n)-f(y_n)$ does not.

$x_n=1+\frac{1}{n}$ and $y_n=1+\frac{1}{2 n}$ does the trick


geometry
rectangle
combinatorics unsolved
combinatorics

1.  The numbers from 1 to n > 1 are written down without a break. Can the resulting number be a palindrome (the same read left to right and right to left)? For example, if n was 4, the result would be 1234, which is not a palindrome.

2.  A polygon can be divided into 100 rectangles, but not into 99 rectangles. Prove that it cannot be divided into 100 triangles.

Thanks.  Any help is appreciated.

The first has appeared several times on the forum.  Try doing a search.

And the second has appeared in iran TST 1997.


The roots of the equation $ ax^2 \plus{} bx \plus{} c \equal{} 0$ will be reciprocal if:

$ \textbf{(A)}\ a \equal{} b \qquad\textbf{(B)}\ a \equal{} bc \qquad\textbf{(C)}\ c \equal{} a \qquad\textbf{(D)}\ c \equal{} b \qquad\textbf{(E)}\ c \equal{} ab$

hello
if roots are reciprocal product =$ 1$
$ c\equal{}a$
thank u