Prove that for all positive a, b and c such that abc = 1 we have

1/(a(b + c)) + 1/(b(c + a)) + 1/(c(a + b)) :ge: 3/2.

This is a bit more difficult ... :wink:

dunno how hard this is .. but.. dig here in solutions...
one even given by me! :D :D :D

http://www.mathlinks.ro/phpBB/viewtopic.php?t=932

cheers! :D

Here's a solution:

Let x = 1/a, y = 1/b, and z = 1/c.  The inequality then becomes:

(x^2) / (y+z) + (y^2) / (x+z) + (z^2) / (x+y)  :ge: 3/2

Cauchy's inequality tells us that:

(a^2 + b^2 + c^2)*(u^2 + v^2 + w^2)  :ge: (au + bv + cw)^2

We plug in the following values:

a = x / sqrt(y+z)
b = y / sqrt(x+z)
c = z / sqrt(x+y)

u = sqrt(y+z)
v = sqrt(x+z)
w = sqrt(x+y)

to get

[(x^2) / (y+z) + (y^2) / (x+z) + (z^2) / (x+y)]

:ge: [(x+y+z)^2 / (2(x+y+z))]

= (x+y+z) / 2  :ge: 3/2, AWD.

Let X be the given expression. Also by Cauchy we have

X(a(b +c) + b(c +a) + c(a + b)) :ge: (1/a + 1/b + 1/c) = (bc + ca + ab).
Hence X :ge: (ab + bc + ca)/2 :ge: 3/2 (abc)^(2/3) = 3/2.

IMO 1995.

Yes I posted some solutions myself on mathlinks

I came into the world of proofs only about 2 months ago.... and I've read about 40+ of these abc=1 inequality proofs, and I have yet been able to do one...

what on earth is Cauchy? and how do you guys suggest I gain more knowledge on these types of problems.

CAUCHY !

The best inequality ever :

(a(1) + a(2) + ... + a(n))(b(1) + b(2) + ... + b(n))
:ge: (a(1)b(1) + a(2)b(2) + ... + a(n)b(n)).

You wanna gain more knowledge for inequalities ?
Well, some hints :
(1) EXERCISE ! For example, here can you find 30 inequalities.
(2) A good link : on the amc website you can find a good document concerning inequalities in the topic resources. It's written by Kiran Kedlaya and it was used for the MOP.

Prove or disprove that for all positive $ a, b$ and $ c$ such that $ abc \equal{} 1$ we have

$ \frac {1}{a^b(b \plus{} c)} \plus{} \frac {1}{b^c(c \plus{} a)} \plus{} \frac {1}{c^a(a \plus{} b)}\ge\frac {3}{2}.$

This is a bit more difficult …:wink:

I agree, but it would have been nicer if you had made a new thread and linked to this thread, rather than revive a six year old thread...

A proof with using Andreescu for the first one (though I'm not that good at ineqs yet, so I'd like anybody to point out any mistakes (?)):

$ \frac{1}{a^3(b\plus{}c)}\plus{}\frac{1}{b^3(a\plus{}c)}\plus{}\frac{1}{c^3(a\plus{}b)} \equal{} \frac{b^2c^2}{a(b\plus{}c)} \plus{} \frac{c^2a^2}{b(a\plus{}c)} \plus{} \frac{a^2b^2}{c(a\plus{}b)} \geq \frac{{(ab\plus{}bc\plus{}ca)}^2}{2(ab\plus{}bc\plus{}ca)} \equal{} \frac{ab\plus{}bc\plus{}ca}{2} \geq \frac{3 \sqrt[3]{a^2b^2c^2}}{2} \equal{} \frac{3}{2}$


invariant
algebra
polynomial
linear algebra
matrix
vector
geometry

Given $ a_0 \equal{} 1$, $ a_1 \equal{} 3$, and the general relation $ a_n^2 \minus{} a_{n \minus{} 1}a_{n \plus{} 1} \equal{} (\minus{}1)^n$ for $ n \ge 1$. Then $ a_3$ equals:

$ \textbf{(A)}\ \frac{13}{27}\qquad
\textbf{(B)}\ 33\qquad
\textbf{(C)}\ 21\qquad
\textbf{(D)}\ 10\qquad
\textbf{(E)}\ \minus{}17$

hello, for $ n \equal{} 1$ we get $ a_1^2 \minus{} a_0\cdot a_2 \equal{} ( \minus{} 1)^1$, hence $ a_2 \equal{} 10$, for $ n \equal{} 2$ we get
$ a_2^2 \minus{} a_1\cdot a_3 \equal{} ( \minus{} 1)^2$ hence $ a_3 \equal{} 33$.
So the right answer is $ \boxed{B}$.
Sonnhard.

Interesting -- this sequence appears to be the same as [url=http://www.research.att.com/~njas/sequences/A006190]A006190[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].  However, the definition there is that $ a_n \equal{} 3a_{n \minus{} 1} \plus{} a_{n \minus{} 2}$.  Can anyone prove that these are equivalent?

[quote="mathwizarddude"]Is there a standard technique for solving recurrence of this sort (like characteristics eq for linear recurrence)?[/quote]

well, the technique for this one was fairly straightforward. Just plug in the numbers, and see if you can boil down the whole question to one single equation with one variable that can be solved.

[quote="mathwizarddude"]Is there a standard technique for solving recurrence of this sort (like characteristics eq for linear recurrence)?[/quote]
No -- as with differential equations, general recursive equations are very hard to solve and/or understand.  (I mean, it's totally non-obvious from the definition of this sequence that it even consists of integers, right?  And this property is highly dependent on the initial values.)

By the way, here's a (totally unsatisfying, to me) answer to my question:  let $ a_n$ be defined as in the question and let $ b_0 = 1$, $ b_1 = 3$ and $ b_{n + 1} = 3b_n + b_{n - 1}$ for $ n \geq 1$.  We claim that $ b_n = a_n$ for all $ n$.  We have that it's true for $ n = 0, 1$ by definition and $ a_2 = \frac {3^2 - ( - 1)}{1} = 10 = 3 \cdot 3 + 1 = b_2$, so it's true for $ n = 2$, as well.  Suppose that $ b_n = a_n$ for all $ n \leq k$.  Then
\begin{align*} a_{k + 1} & = \frac {a_k^2 - ( - 1)^k}{a_{k - 1}} \\
& = \frac {b_k^2 - ( - 1)^k}{b_{k - 1}} \\
& = \frac {9b_{k - 1}^2 + 6b_{k - 1}b_{k - 2} + b_{k - 2}^2 - ( - 1)^k}{b_{k - 1}} \\
& = 9 b_{k - 1} + 6b_{k - 2} + \frac {b_{k - 2}^2 - ( - 1)^k}{b_{k - 1}} \\
& = 9 b_{k - 1} + 6b_{k - 2} + \frac {a_{k - 2}^2 - ( - 1)^{k - 2}}{a_{k - 3}} \cdot \frac {b_{k - 3}}{b_{k - 1}} \\
& = 9 b_{k - 1} + 6b_{k - 2} + a_{k - 1} \cdot \frac {b_{k - 3}}{b_{k - 1}} \\
& = 9 b_{k - 1} + 6b_{k - 2} + b_{k - 3} \\
& = b_{k + 1}, \end{align*}
as needed.  On the other hand, these computations seem like they would carry over for many similar kinds of recurrences, so it's possible that there's a known general approach for this specific type of recursive equation.

Hmm so in this case it's possible to transform the original recurrence into an equivalent linear recurrence... do you know of any books that are devoted to this (just wondering do strange recurrences appear very often in olympiad problems?)? Also how did you realize those two are actually equivalent?
(also feel free to check out my infinite nested radicals problem in pre-olympiad section on strange recurrences) :!:

Here is an explanation of the recurrences being equivalent that is hopefully more satisfying.

[hide]The equations for the recurrence for $ n$ and $ n \plus{} 1$ are
\[ a_n^2 \minus{} a_{n \minus{} 1}a_{n \plus{} 1} \equal{} ( \minus{} 1)^n,
\]

\[ a_{n \plus{} 1}^2 \minus{} a_n a_{n \plus{} 2} \equal{} \minus{} ( \minus{} 1)^n.
\]
Add and simplify.
\[ a_n(a_n \minus{} a_{n \plus{} 2}) \equal{} a_{n \plus{} 1}(a_{n \minus{} 1} \minus{} a_{n \plus{} 1}),
\]

\[ \frac {a_n \minus{} a_{n \plus{} 2}}{a_{n \plus{} 1}} \equal{} \frac {a_{n \minus{} 1} \minus{} a_{n \plus{} 1}}{a_n}.
\]
So $ \frac {a_{n \minus{} 1} \minus{} a_{n \plus{} 1}}{a_n}$ is an invariant; it is constant for all $ n$. Let it be $ k$. Then $ a_{n \minus{} 1} \minus{} a_{n \plus{} 1} \equal{} ka_n$, so $ a_{n \plus{} 1} \equal{} \minus{} ka_n \plus{} a_{n \minus{} 1}$. It remains to show $ k \equal{} \minus{} 3$. By computing that $ a_2 \equal{} 10$, we get $ \frac {a_0 \minus{} a_2}{a_1} \equal{} \frac {1 \minus{} 10}{3} \equal{} \minus{} 3$, as desired.

A lot of sequences of this form, with $ a_n^2 \plus{} c a_{n \minus{} 1} a_{n \plus{} 1}$ on one side ($ c$ fixed), have this invariant.[/hide]

Yes, that's much more satisfying, and provides a good explanation of the observation that "these computations seem like they would carry over for many similar kinds of recurrences."  Thanks.

[hide="Another explanation"] Let a sequence $ a_n$ satisfy a recurrence with characteristic polynomial $ P(x)$ and let $ \mathbf{A}$ denote its [url=http://en.wikipedia.org/wiki/Companion_matrix]companion matrix[/url].

[b]Theorem:[/b]  There exist vectors $ \mathbf{u}, \mathbf{v}$ such that $ a_n \equal{} \mathbf{u}^T \mathbf{A}^n \mathbf{v}$.

Now one changes to an appropriate basis and computes $ \det \mathbf{A}^n$.  For $ n \equal{} 2$, the answer is the most satisfying.  In particular, for the Fibonacci sequence we obtain

$ \mathbf{A} \equal{} \left[ \begin{array}{cc} 1 & 1 \\
1 & 0 \end{array} \right], \mathbf{A}^n \equal{} \left[ \begin{array}{cc} F_{n \plus{} 1} & F_n \\
F_n & F_{n \minus{} 1} \end{array} \right]$

and the determinant computation gives $ F_{n \plus{} 1} F_{n \minus{} 1} \minus{} F_n^2 \equal{} ( \minus{} 1)^n$.  Here the relevant matrix is

$ \mathbf{A} \equal{} \left[ \begin{array}{cc} 3 & 1 \\
1 & 0 \end{array} \right], \mathbf{A}^n \equal{} \left[ \begin{array}{cc} a_n & a_{n \minus{} 1} \\
a_{n \minus{} 1} & a_{n \minus{} 2} \end{array} \right]$

and the determinant computation gives $ a_n a_{n \minus{} 2} \minus{} a_{n \minus{} 1}^2 \equal{} ( \minus{} 1)^n$, which is equivalent to the desired result.  One obtains a similar result by starting with any matrix in $ SL_2(\mathbb{Z})$.  Changing the $ 3$ to a variable gives the [url=http://mathworld.wolfram.com/FibonacciPolynomial.html]Fibonacci polynomials[/url]. [/hide]

Nonlinear recurrences are [b]quite[/b] subtle; [url=http://faculty.uml.edu/jpropp/somos.html]Somos sequences[/url] and some related nonlinear recurrences coming from the theory of cluster algebras are an area of active research.

[b]Edit:[/b]

[hide="Third explanation"] A different direction in which this result generalizes is via the [url=http://en.wikipedia.org/wiki/Fundamental_recurrence_formulas]fundamental recurrence formulas[/url] of continued fraction theory.  This case corresponds to the computation of the convergents of the continued fraction

$ 3 \plus{} \frac{1}{3 \plus{} \frac{1}{3 \plus{} ...}}$

which is equal to one of the roots of the characteristic polynomial $ x^2 \equal{} 3x \plus{} 1$.  The connection to the second explanation is given by a well-known relationship between continued fractions and $ SL_2(\mathbb{Z})$. [/hide]


logarithms
function
limit
symmetry
algebra unsolved
algebra

Find all the pairs of integers positives, sucht that

a^b = b^a

1) a=b any integer,
2) a=2,b=4 or a=4,b=2

Ok,your solution is correct, but...
Do you can explain your solution??

Put $b=qa$ where $q$ is rational different from $1$. Then

$a^{b}=b^{a}\iff a^{qa}=(qa)^{a}\iff a^{q}=qa\iff a^{q-1}=q\iff a=q^{1\over q-1}$

If $q$ is a proper fraction, $q^{1\over q-1}$ can't be integer, hence $q$ is integer, but then $q^{1\over q-1}$ must always be irrational except if $q-1=1\iff q=2$. Then $a=2, b=qa=4$

Here's the solution I was taught. Similar to medicu's but less elegant. The technique might be helpful for other problems (like the famous one: show that $e^\pi > \pi^{e}$), so I'll post it anyway.

Taking logarithms, $b\ln a=a\ln b$, $\frac{\ln a}{a}=\frac{\ln b}{b}$.

Let's define the function $f(x)=\frac{\ln x}{x}$.

It can be proved that $f$ is continuous for $x>0$ and has a maximum at $x=e$:
$f'(x)=\frac{\frac{1}{x}x-\ln x}{x^{2}}=0$ if $1=\ln x$, that's $x=e$,
and additionally
$f(x)<0$ if $0<x<1$,
$f(1)=0$,
$\lim_{x\to\infty}f(x)=0$,

Following this facts, if we want $f(a)=f(b)$ with a fixed $a>1$ (except for $a=e$), there are exactly two solutions in $\mathbb{R}$:
- one with $a=b$ (trivial),
- and the other with $1<a<e<b$ or $1<b<e<a$.

If we restrict ourselves to positive integers,
- If $a=1$, then $1^{b}=b^{1}$ has only one solution $b=1$.
- If $1<a<e$, the only integer is $a=2$ then we've showed that there are exactly two solutions, and we can check that $b=2$ and $b=4$ are solutions.
- If $a\geq 3$, then $a=b$ or $1<b<e$, the only integer is $b=2$, and by symmetry the only solutions are with $a=2$ and $a=4$.

Therefore, the only nontrivial solutions ($a\neq b$) are $a=2,b=4$ and $b=2,a=4$.


probability
combinatorics theorems
combinatorics

There are 6 positive and 8 negative integers. 4 numbers are drawn without replacement. Find the probability that their product is positive.

why is this in "theorems and formulas"? a positive product means an even number of negative numbers. so the probability is

$ \frac{\binom{8}{0}\binom{6}{4}\plus{}\binom{8}{2}\binom{6}{2}\plus{}\binom{8}{4}\binom{6}{0}}{\binom{14}{4}}$


Find A,B contained in a Set of 2x2 matrices of real numbers, such that (e^A)(e^B) does not equal e^(A+B)

Sorry for the bad notation, hopefully you guys can get it!

Thanks for your help!

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

$ exp(A)exp(b) \equal{} exp(A\plus{}B)$ if $ AB \equal{} BA$ so to satisfy your problem you can easily find a set of 2x2 matrices such that $ AB$ $ \neq$ $ BA$

$ AB\not\equal{}BA$ doesn't imply that $ e^{A\plus{}B}\not\equal{}e^Ae^B$; cf for example:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=191160


[quote="shobber"]Find all positive integers $n$ such that $\dfrac{n^3+3}{n^2+7}$ is a positive integer![/quote]

$\dfrac{n^3+3}{n^2+7} =n+\dfrac{3-7n}{n^2+7}$

Since it is an integer  we have $\mid 3-7n \mid\geq n^2+7$.....

 :)

As socrates said, $\dfrac{n^3+3}{n^2+7} = n + \dfrac{3-7n}{n^2+7}$

We examine $f: N\rightarrow N : x \rightarrow n^2+7$ and $g: N\rightarrow N : x\rightarrow 3-7n$

When $x=0$, $f(0) > g(0)$

We search the intersection of the two graphs: $n^2+7n+4=0$
There are no solutions of n in this equation, so the two graphs never intersect.
\[ \Rightarrow n^2+7 > 3-7n \]

So $\dfrac{3-7n}{n^2+7} \notin N$

There exist no positive integers $n$ such that $\dfrac{n^3+3}{n^2+7}$ is an integer.

Jan
I'm afraid you are not right.
Just take n=2....

What I said is to solve the inequation $\mid 3-7n \mid\geq n^2+7$
Do you get it??
If not tell me again about it  :)  :)

Oh, I'm sorry...  It's wrong because I said that the equation $n^2+7n+4=0$ has no integer solutions for $n$. (I had to be quick because my mom already called me twice to come and eat.  :) )
So factoring $n^2+7n+4$ gives $(x-x_1)(x-x_2)$
Between these two points, $3-7n \geq n^2 + 7$,
We notice that $-1<x_1<x_2<0$ so there are no integer values of n between them.

If $3-7n < 0$ , we get the equation $3-7n \leq -n^2-7$
Working out gives: $n \in [2,5]$
Bruteforcing tells us that $2$ and $5$ are the only possibilities:
$\dfrac{2^3+3}{4+7}=1$ and $\dfrac{5^3+3}{32}=4$

Socrates, that's your solution, only in a long and less elegant way. I'm just posting this to show that my solution wasn't completely useless (if I only calculated the quadratic correctly...  :( )

[hide="solution"]
We have \[ \frac{n^3 + 3}{n^2 + 7} = n + \frac{3 - 7n}{n^2 + 7} .\]
Thus \[ 7n - 3 \ge n^2 + 7 \] and trying all possibilities in the solution range gives $ x = 2, 5 $ as the only integers that work.
[/hide]

[quote=shobber]Find all positive integers $n$ such that:
\[ \dfrac{n^3+3}{n^2+7} \]
is a positive integer.[/quote]

Note that:
$$\frac{n^3+3}{n^2+7}=\frac{n^3+7n-7n+3}{n^2+7}=n+\frac{3-7n}{n^2+7}$$
Now since $\frac{n^3+3}{n^2+7}$ is an integer, hence $\frac{3-7n}{n^2+7}$ is an integer:
That means that $3-7n \ge n^2+7$ or $7n-3 \ge n^2+7$.
Now if $3-7n \ge n^2+7$.
$$n^2+7n+4 \le 0 \implies \; \text{no such} \; n$$
This is becuase $n \in \mathbb N$ implies $n \ge 1$ and then at least $12 \le 0$, contradiction!.
Hence $7n-3 \ge n^2+7$ implies:
$$n^2-7n+10 \le 0 \implies (n-2)(n-5) \le 0 \implies n \in (2, 5)$$
By some calculations you get that $n=2$ or $n=5$.
Thus we are done  :blush:


If a stick is broken in two at random, what is the average ratio of the length of the shorter piece to the length of the larger?

I don't see how you can do this without integration...

[hide="Answer"]$ \frac{1}{1/2 \minus{} 0} \int_{0}^{\frac{1}{2}} \frac{x}{1\minus{}x} dx \equal{} \ln 4 \minus{} 1$[/hide]


linear algebra
matrix
induction
invariant
inequalities
linear algebra unsolved

Let $ A_1, \cdots, A_m$ be $ n\times n$ complex matrices. For positive integer $ k\ge 1$, show $ M \equal{} [a_{ij}]_{m\times m}$, where $ a_{ij} \equal{} tr\{(A_i^*A_j)^k\}$ is positive semidefinite.

Remark. When $ k \equal{} 1$, $ M$ is a Gram matrix.

I like matrix this !!!
Case $ k = 1$ Let $ A_i = [a_{km}^{(i)}]_{1 \leq k,m \leq n}$ for all $ i = 1,2,..,n$
We have : $ a_{ij} = Tr(A_i^{*}A_j) = \sum_{m = 1}^{n}\sum_{k = 1}^{n}{\overline{a_{km}^{(i)}}a_{km}^{(j)}}$
  Hence $ \forall x \in M_{n1}(\mathbb{C})$ We have :
    $ x^{*}Ax = \sum_{i = 1}^{n}\sum_{j = 1}^{n}\sum_{m = 1}^{n}\sum_{k = 1}^{n}{\overline{a_{km}^{(i)}}a_{km}^{(j)}\overline{x_i}x_j = \sum_{m = 1}^{n}\sum_{k = 1}^{n}(\sum_{i = 1}^{n}{|x_i a_{km}^{(i)}|})^2 \geq 0}$
 QED $ A \geq O$
 case $ k = 2$   $ a_{ij} = Tr((A_i^{*}A_j)^{2}) = \sum_{m = 1}^{n}\sum_{k = 1}^{n}{\overline{b_{km}^{(j,i)}}b_{km}^{(i,j)}}= \sum{\overline{f_i}f_j}$ , $ b_{km}^{(i,j)}$ from $ A_i^{*}A_j$ you can think simple $ f_i$ OK !
 The same QED $ A \geq O$
 Whith $ k \geq 3$ use induction method .....
 Some case .Example : $ A_i = A^{i}$ Here $ A \in M_n{\mathbb{(R)}}$ for all $ i \in \{1,2,..,n\}$ then
 $ A = [Tr(A^{i + j})]_{1\leq i,j \leq n}$ is positive semidefinite ....

Gram matrices are positive semidefinite + [url=http://www.mathlinks.ro/viewtopic.php?t=23591]the Hadamard product of two positive semidefinite matrices is positive semidefinite[/url].

[EDIT: Wrong. Thanks for noticing the mistake, miwalin.]

[quote="darij grinberg"]Gram matrices are positive semidefinite + [url=http://www.mathlinks.ro/viewtopic.php?t=23591]the Hadamard product of two positive semidefinite matrices is positive semidefinite[/url].[/quote]
You are saying $ k \equal{} 1$ and the matrix with $ a_{ij} \equal{} \{tr(A_{i}^{*}A_{j})\}^{k}$ but not what problem states.

QuyBac, can you please explain what you do in the $ k\equal{}2$ case?

  darij

[quote="darij grinberg"]QuyBac, can you please explain what you do in the $ k \equal{} 2$ case?

  darij[/quote]
 Oh ... I can  talk that :
 $ a_{ij} \equal{} Tr((A_{i}^{*}A_{j})^{k}) \equal{} \sum\sum_{i \equal{} 1}^{n}\sum_{j \equal{} 1}^{n}{\overline{f_{i}}f_{j}}$

With "explain" I mean "explain", not "write down a formula full of undefined symbols".

What are your $ b_{km}^{(i,j)}$ ? What are your $ f_i$?

  darij

Remark. When $ m\equal{}2$, it suffices to show $ |Tr\{(A_1^*A_2)^k\}|^2\le Tr\{(A_1^*A_1)^k\}\cdot Tr\{(A_2^*A_2)^k\}$, which is a consequence of a unitarily invariant norm inequality appeared in p.81 of X.Zhan, Matrix inequalities, Springer, 2002.

[quote="darij grinberg"]Nicely disproved by Pavel Etingof [url=http://mathoverflow.net/questions/16983/on-a-positivity-of-a-matrix-with-trace-entries]on MathOverflow[/url].

However, the $ m \equal{} 2$ case is nice. Actually, Zhan proves the more general inequality:

$ \left|\mathrm{Tr}\left(\left(AXB^{\ast}\right)^k\right)\right|^2 \leq \left|\mathrm{Tr}\left(\left(A^{\ast}AX\right)^k\right)\right| \cdot \left|\mathrm{Tr}\left(\left(XB^{\ast}B\right)^k\right)\right|$.

EDIT: Wait, I am reading Zhan wrong. $ \left|U\right|$ means $ \sqrt {U^{\ast}U}$ in his text. Miwalin, how do you conclude your $ m \equal{} 2$ case from his results?[/quote]
Set $ X \equal{} I$, $ A \equal{} A_1$ and $ B \equal{} A_2$, note that $ |Tr\{(A_{1}^{*}A_{2})^{k}\}|^{2} \equal{} |Tr\{(A_{1}A_{2}^*)^{k}\}|^{2}$. This is because for any $ M$, $ |TrM|\equal{}|TrM^*|$

[quote="darij grinberg"]Nicely disproved by Pavel Etingof [url=http://mathoverflow.net/questions/16983/on-a-positivity-of-a-matrix-with-trace-entries]on MathOverflow[/url].

However, the $ m \equal{} 2$ case is nice. Actually, Zhan proves the more general inequality:

$ \left|\mathrm{Tr}\left(\left(AXB^{\ast}\right)^k\right)\right|^2 \leq \left|\mathrm{Tr}\left(\left(A^{\ast}AX\right)^k\right)\right| \cdot \left|\mathrm{Tr}\left(\left(XB^{\ast}B\right)^k\right)\right|$.

EDIT: Wait, I am reading Zhan wrong. $ \left|U\right|$ means $ \sqrt{U^{\ast}U}$ in his text. Miwalin, how do you conclude your $ m\equal{}2$ case from his results?[/quote]

Can one find a explict counterexample? That is , to find some matrix explicitly.


modular arithmetic
number theory unsolved
number theory

[hide]Assume the contrary that $ a^2\minus{}3a\minus{}19$ is divisible by 289. Note that $ a^2\minus{}3a\minus{}19\equal{}(a\minus{}10)(a\plus{}7)\plus{}51$. If this integer is divisible by 289, it is divisible by 17. So $ (a\minus{}10)(a\plus{}7)$ is divisible by 17. Since $ a\minus{}10\equiv a\plus{}7\pmod{17}$, then $ a\minus{}10$ and $ a\plus{}7$ are both divisible by 17, which implies that $ (a\minus{}10)(a\plus{}7)$ is divisible by 289. But then $ (a\minus{}10)(a\plus{}7)\plus{}51$ is not divisible by 289. A contradiction.[/hide]

also u can just analyze the dicriminant which gives $ 17^2k.4\equal{}z^2\minus{}85$ and we have $ 17 \parallel{} 85$.

assume $ 17|4(a^2 \minus{} 3a \minus{} 19)\implies 17|(2a \minus{} 3)^2 \minus{} 85\implies 17|(2a \minus{} 3)^2$
$ \implies 17^2|(2a \minus{} 3)^2\implies 17^2\not |(2a \minus{} 3)^2 \minus{} 85\implies$ contradiction. :lol:

[quote="keyree10"]Show that there is no integer $ a$ such that $ a^2 \minus{} 3a \minus{} 19$ is divisible by $ 289$.[/quote]
$ 289\equal{}17^2$
$ a^2\minus{}3a\minus{}19 \equiv a^2\plus{}14a\minus{}2 \equiv (a\plus{}7)^2\minus{}51 \equiv (a\minus{}10)^2 \bmod 17$ so $ a \equiv 10 \bmod 17$.
$ a \equal{} 10\plus{}17b$:
$ a^2\minus{}3a\minus{}19 \equal{} (10\plus{}17b)^2\minus{}3(10\plus{}17b)\minus{}19 \equiv$ $ 100\plus{}20*17b\minus{}30\minus{}51b\minus{}19 \equal{} 51\plus{}17^2b \equiv 51 \bmod 289$.
So $ 289 \mid a^2\minus{}3a\minus{}19 \Rightarrow 17 \mid a^2\minus{}3a\minus{}19 \iff$ $ a \equiv 10 \bmod 17 \Rightarrow a^2\minus{}3a\minus{}19 \equiv 51 \bmod 289$. And we obtain a contradiction. Hence $ 289 \nmid a^3\minus{}3a\minus{}19$

ah this is the only question i m confident of........ can easily be solved by number theories

We have $ a^{2}\minus{}3a\minus{}19\minus{}289k\equal{}0$
discriminant has to be a perfect square,
$ r^{2}\equal{}85\plus{}289\cdot4k$
$ r^{2}\equal{}17(5\plus{}17\cdot4k)$
$ 17|5\plus{}17\cdot4k$ contradiction

One more solution:

$ 289|(a^2 \minus{} 3a \minus{} 19) \Longrightarrow 17|(a^2 \minus{} 3a \minus{} 19)$

$ \Longrightarrow 17|((a^2 \minus{} 3a \minus{} 19) \minus{} 17(a \minus{} 7)) \Longrightarrow 17|(a^2 \minus{} 20a \plus{} 100)$

$ \Longrightarrow 17|(a \minus{} 10)^2 \Longrightarrow 17|(a \minus{} 10) \Longrightarrow 289|(a \minus{} 10)^2$

$ \Longrightarrow 289|((a^2 \minus{} 3a \minus{} 19) \minus{} 17(a \minus{} 7))$

but $ 289|(a^2 \minus{} 3a \minus{} 19) \Longrightarrow 289|17(a \minus{} 7) \Longrightarrow 17|(a \minus{} 7)$

but $ 17|(a \minus{} 10) \Longrightarrow 17|(a \minus{} 7) \minus{} (a \minus{} 10) \Longrightarrow 17|3$

Contradiction!!! :P

Let 289 divide the given expression
a^2 -3a -19=289k for some k in integers
a^2-3a-2 =17(17k+1)
Rhs can resolved into linear factors whereas lhs cannot be resolved into linear factors hence contradiction

n^2 - 3n - 19 is congruent to (n + 7)^2 mod 17 therefore n = 17k - 7. Put in the original equation.
We are done.

[quote=John-Wick]$n^2 - 3n - 19\equiv (n + 7)^2 \pmod{17}$ therefore $n = 17k - 7$. Put in the original equation.
We are done.[/quote]
@above latex version

Note that $a^2 - 3a - 19 = (a - 10)(a + 7) +51$.

Say that for the sake of contradiction, there is a solution to this equation for$a$.
Clearly $289\equiv0\pmod{17}$.
$\implies (a - 10)(a + 7)\equiv0\pmod{17}$

We know that $(a + 7) - (a - 10) = 17$, which implies that both of them divide $17$. This implies that:
$\implies (a + 7)(a - 10)\equiv0\pmod{289}$
$\implies 51\equiv0\pmod{289}$

We arrive at a contradiction.

um, we assumed for the sake of contradiction that there are solutions to $a$

Assuming that $\exists$ $a$ such that $289\vert a^2-3a-19 \Rightarrow a^2-3a-19=289k$.
As $a$ is an integer then the discriminant of the quadratic must be a perfect square.

$\Rightarrow 9+76+1156k= m^2$
$\Rightarrow 85+1156k=m^2$
$\Rightarrow 17(68k+5)=m^2$
$\Rightarrow 17\vert m^2 \Rightarrow 17^2\vert m^2 \Rightarrow 17\vert 68k+5$.
But, as $17\vert 68k \Rightarrow 17\vert 5$, which is not possible. Thus our assumption was wrong. $\boxed{\text{Contradiction!}}$.


Ben bought a fish tank of dimension 100*30*50,and poured the water upto the brim,At the end of each month ,  Ben takes out one-quarter of the tank water and replaces it with fresh water, He does this for 10 years. Approximateley what volume of original water is left in the tank?

ans( 1.5 * 10^(-10) )cubic cms.

After one month, you have $ 75%
$, or $ .75$, of the original water. So at the end of 10 years, you'll have $ .75^{120}$ of the original water. Assuming your given dimensions are in cm, then you'll have $ 150000 * .75^{120} cm^{3} = 1.526*16^{-10} cm^{3}$

Ok, the original volume is $ 150000cm^{3}$ from multiplying the dimensions $ 100*30*50$.
At the end of one month, because you're taking out $ \frac{1}{4}$ of the original water, you'll have $ \frac{3}{4}$ of it left. In volume, that'll be $ 150000*\frac{3}{4}$. At the end of the second month, you take out another $ \frac{1}{4}$ of the original water, so in volume, that'll be $ 150000*\frac{3}{4}*\frac{3}{4}$, or $ 150000*(\frac{3}{4})^{2}$.
Since there are 120 months in 10 years, you're left with $ 150000*(\frac{3}{4})^{120}$ of the original water.

i have got upto this much,but in the olympiad which was multiple choice type had several options and one was 1.5 *10^(-10)which was the correct answer ,since we were not allowed to use the calculator thats why I asked how to calculate  150000*(3/4)^120

Oh, how to calculate it without a calculator? I have absolutely no clue.


geometry
3D geometry
octahedron
ratio
analytic geometry

The centers of a cube are connected to form an octahedron. The centers of the ocanedron are then connected to form a smaller cube. What is the ratio of the side of the smaller cube to he side of the original cube?

If by "centers" you mean the centers of the faces of the polyhedron...

[hide]The ratio between the side length of the octahedron and the side length of the outer cube is $ \frac{\sqrt{2}}{2}$.  The horizontal cross-section of the octahedron with a vertex held to the top is square.  When such a cross-section is made so that it includes the centers of the top four faces of the octahedron, it is two-thirds the way down each face, so it has a side length of two-thirds the side length of the octahedron.  The cube's top face is inscribed in the cross-section, with vertices on the mid-points of the square cross-section, so the ratio between the side length of the smaller cube and the side length of the octahedron is $ \frac{2}{3}/\sqrt{2}\equal{}\frac{\sqrt{2}}{3}$.  Multiplying these ratios gives the ratio between the two cubes:

$ \frac{\sqrt{2}}{2}*\frac{\sqrt{2}}{3}\equal{}\frac{1}{3}$[/hide]

[hide]Let the vertices of the cube be $ (\pm 1, \pm 1, \pm 1)$. The centers of the faces are clearly $ (\pm 1, 0, 0)$ and permutations.

Now take the octahedron vertices $ (1,0,0)$ and permutations. The centroid of this is $ (1/3, 1/3, 1/3)$. Doing similarly, we find that the 8 centroids of the faces of the octahedron are $ (\pm 1/3, \pm 1/3, \pm 1/3)$. But these coordinates are $ \boxed{1/3}$ of those of the original cube's vertices.[/hide]


It seems that the AOPS textbooks are on a sort of different level of Latex than your garden variety. I am looking to do something similar and find some professional latex layout templates. If anyone could direct me to some I could modify or that are freely available I would appreciate it.

Thanks,
Jack

I don't have a template, but it shouldn't be too hard to reproduce, with some experience.  You can get headers and footers with fancyhdr.  You can put things in the margins with \marginpar{}.  You can adjust paper size as an optional argument in the \documentclass command, e.g., \documentclass[12pt,a4]{book} .  One good resource for these kinds of things is lshort.pdf (first google hit).  For specialized things, look in the documentation.  If you're going to get into LaTeX, it wouldn't be a bad idea to read lshort.pdf cover to cover (well, except maybe the section on LaTeX graphics, since it's usually a better idea to create an image somewhere else and then import it).

Also, in general, almost everything to do with LaTeX is available for free at the TeX Users Group, http://tug.ctan.org .  If there's something to do that seems tricky, somebody's probably already written a package that does it.


Welcome to my riddle tourney. due to my dumbness  :rotfl:  i will have "helpers" to make riddles
to become a helper pm me a riddle and an answer the 5 riddles i like the best will be my "helpers" and all riddles will be posted in first round!!!

pm me your answers for first round of elimination and after that i will choose a helper to post a riddle. the first to post the correct ans wins a point (added to the points of the first round) at the end of 10 rounds whoever has the most points wins!!!!!!! :first:

i no this is confusing if u have any qs just post and ill answer them. FOR THOSE OF U WHO WANT TO POST RIDDLES YOUR PM IS EXCPECTED THIS FRIDAY AT MIDNIGHT FORUM TIME!!!!!! JOIN NOW!!!!!!!! :D

Mods, please move this to Puzzles and Brainteasers.


Got this cryptology problem here:

I have three (3) vaults which is locked by four (4) numbers.
Intricately engraved in each vault are these numbers:

[b]Vault #1[/b]
0260059300820968  0264035402640593  0992W0968  18690593067805930264

[b]Vault #2[/b]
1024072311230822  0822082205221123  1024072311230822  032111230822

[b]Vault #3[/b]
28597377  058664  28597377  91860059

Each set of number (there are four sets) represents one number (hence there are 4 numbers per vault).
What are those four (4) numbers which un-locks the vaults?

[b]By the way, the letter "W" in Vault #1 is not a typo.[/b]

Could you please help me out with the solution to the cryptology problem above (the answer and the process to get those numbers)?

Thanks a lot guys.

hey guys... anyone?... please! help!....

[quote="mbucko"]@CrystalClear, Please contact me via PM. A few questions though:

1. What was the time the code was created? (timestamp)
2. Is this software-related problem, ie. is it very possible that crypto is confined to standard (shuffle, rotate, etc.)?

--
mishabucko.wordpress.com[/quote]

@mbucko: Please, don't revive old and ancient topics. If you see the user CrystalClear registered on 2005 and his/her last visit is on 2005, so...

You can make new threads in your blog, or on forums. People will answer you.

Thanks.


induction
calculus
derivative
function
inequalities proposed
inequalities

Find min $ M$ with $ M\equal{}\sqrt{a_1\plus{}1}\plus{}\sqrt{a_2\plus{}1}\plus{}...\plus{}\sqrt{a_n\plus{}1}$ where $ a_1,a_2,...,a_n\ge 0$ such that $ a_1\plus{}a_2\plus{}...\plus{}a_n\equal{}n$

More general .     :lol:  If $ b < 1 , \ c > 0,\ a_1,...,a_n \geq 0$  then $ (a_1 \plus{} c)^b \plus{} ... \plus{} (a_n \plus{} c)^b \geq (n \minus{} 1)c^b \plus{} (a_1 \plus{} ... \plus{} a_n \plus{} c)^b$
[hide]By induction . The case $ n \equal{} 1$ is trivial .
Case $ n \equal{} 2$. Let's denote $ x \plus{} y \equal{} S$, with $ x,y \geq 0$. By using derivative,  we prove that  the  function
$ (x \plus{} c)^b \plus{} (S \minus{} x \plus{} c)^b$ is increasing  on $ [0,\frac S2]$ and decreasing on $ [\frac S2,S]$, so $ (x \plus{} c)^b \plus{} (y \plus{} c)^b \geq c^b \plus{} (x \plus{} y \plus{} c)^b$
Let's suppose that $ (a_1 \plus{} c)^b \plus{} ... \plus{} (a_m \plus{} c)^b \geq (m \minus{} 1)c^b \plus{} (a_1 \plus{} ... \plus{} a_m \plus{} c)^b$ for all $ m \leq n$. We can write :
$ (a_1 \plus{} c)^b \plus{} .. \plus{} (a_n \plus{} c)^b \plus{} (a_{n \plus{} 1} \plus{} c)^b\geq (n \minus{} 1)c^b \plus{} (a_1 \plus{} .. \plus{} a_n \plus{} c)^b \plus{} (a_{n \plus{} 1} \plus{} c)^b$$ \geq nc^b \plus{} (a_1 \plus{} .. \plus{} a_n \plus{} a_{n \plus{} 1} \plus{} c)^b$ . (q.e.d)[/hide]
 :cool:

how about with $ M\equal{}\frac{\sqrt{a\plus{}1}\plus{}\sqrt{b\plus{}1}\plus{}\sqrt{c\plus{}1}}{\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}}$ and $ a\plus{}b\plus{}c\equal{}3$

This problem is in M&Y,a contest on math magazine in Vietnam,please don't help him!
Mod should lock this topic!

I think to some extent, he came up with the question because he found out that the denominater is easily solved but actually, he could not find the minimum value of the numerator. What a trick! Some problems posted on this forum may be generated by other approaches to the actual questions.


calculus
AMC
AIME
real analysis
real analysis theorems

I'm going to be taking AB Calculus starting in about two weeks, and I would really like to get a solid foundation in calculus.  However, my school only focuses on preparing for the AP exam, so its not rigorous at all.   I'm currently going through AoPS Vol I trying to learn some of what I didn't get in school; I really don't want to repeat the experience so I'm looking for some supplemental materials that I could do alongside my class.  Note I'm not that great at math (in fact I would be hard pressed to solve a single AIME problem), but I would like to truly understanding everything about Calculus.

Get the book [u]Calculus[/u] by Michael Spivak

Should I read through all the lessons, or just do the exercises?  (Ideally I would love to read the whole book, but time is an issue).

To get the most out of it you should read the book in addition to doing the exercises. The quality of the writing and explanations is one of the strengths of the book. I don't see why time would be an issue. You'll be studying calculus in your class for the next year right? You could read the book throughout the year as you're taking your class (assuming you have the time).


[quote="gr8sk8r"]5 ROMS=2 HIJS and 3 HIJS=4 NIMS and 2 NIMS=6 SOLS
if there were 26 SOLS, how many ROMS would there be?[/quote]

$N=3S$, $5R=2H$, $3H=4N$. So $3H=12S$, and $H=4S$. $5R=8S$.

$\frac{5}{8}=\frac{x}{26}$
$26 \cdot 5 = 8x \Rightarrow x=\frac{13 \cdot 2 \cdot 5}{2^3}=\frac{65}{4}=16.25$.


geometry
inequalities proposed
inequalities

If $a , b , c$ are the sides of triangle $ABC$, prove that :

   $\sum \frac{(a-b+c)4^{}}{a(a+b-c)}\geq ab+bc+ca$,

   where the sum is cyclic .

 [u]Babis[/u]

[quote]$\sum \frac{(a-b+c)^{4}}{a(a+b-c)}\geq ab+bc+ca$
[/quote]
$\sum \frac{(a-b+c)^{4}}{a(a+b-c)}\geq \frac{(\sum(a-b+c)^{2})^{2}}{\sum(a^{2}+ab-ac)}=\frac{(\sum(a-b+c)^{2})^{2}}{a^{2}+b^{2}+c^{2}}\geq a^{2}+b^{2}+c^{2}\geq ab+bc+ca$
So we have to prove that $\frac{(\sum(a-b+c)^{2})^{2}}{a^{2}+b^{2}+c^{2}}\geq a^{2}+b^{2}+c^{2}$ $\leftrightarrow$ $\sum(a-b+c)^{2}\geq a^{2}+b^{2}+c^{2},(*)$
But because $a,b,c$are triangle sides we can put $a=x+y$ ,$b=y+z$ and $c=z+x$ and using those notations the (*) relations is equivalent to
$x^{2}+y^{2}+z^\geq xy+yz+xz$ which is trivial

Interesting, but if you substitute $x,y,z$ first the solution becomes nearly impossible to spot nicely in the step where you tighten to $\sum a^{2}\geq \sum ab$..
I ended up proving the weaker one but it took messy work

Why should $ a,b,c$ be the sides of a triangle?  silouan's solution doesn't use this fact, does it?
@Svejk: how do we prove that $ a,b,c$ can be written in the form $ x \plus{} y,y \plus{} z,z \plus{} x$?

[quote="mathwizarddude"]Why should $ a,b,c$ be the sides of a triangle?  silouan's solution doesn't use this fact, does it?
[/quote]
If a,b,c weren't sides of a triangle then $ a\plus{}b\minus{}c$ could be negative  :wink:

Oh yeah, I didn't notice that.  :wink:
So how do we prove that $ a,b,c$ can be written as $ x\plus{}y,y\plus{}z,z\plus{}x$?

Let $ s \minus{} a \equal{} z,s \minus{} b \equal{} x,s \minus{} c \equal{} y$ then you'll get $ a \equal{} x \plus{} y,b \equal{} y \plus{} z,c \equal{} z \plus{} x$ (here $ s$ denotes the semi-perimeter). Geometrically, if $ X\in (AB), Y\in (BC), Z\in (CA)$ are the points of tangency of the incircle with the sides of triangle $ ABC$, then $ x \equal{} BX \equal{} BY,$ $ y \equal{} CY \equal{} CZ$, $ z \equal{} AZ \equal{} AX$ implies $ a \equal{} x \plus{} y,b \equal{} y \plus{} z,c \equal{} z \plus{} x$.


inequalities proposed
inequalities
geometry

[quote=nguoivn]Given $ A, B, C$ are three angles of a triangle satisfy: $ 5cosA \plus{} 6cosB \plus{} 7cosC \equal{} 9$
Prove that: $ sin^2\frac {A}{2} \plus{} sin^3\frac {B}{2} \plus{} sin^4\frac {C}{2} \geq\ \frac {7}{16}$

Althought the coefficients is diffirent but equality occurs when $ A \equal{} B \equal{} C \equal{} 60^0$.
 :)[/quote]
$ 5cosA \plus{} 6cosB \plus{} 7cosC \equal{} 9\iff  5sin^2\frac {A}{2} \plus{} 6sin^2\frac {B}{2} \plus{} 7sin^2\frac {C}{2}=\frac {9}{2},$
$sin^3\frac {B}{2} \plus{}\frac {1}{16} \geq \frac {3}{4}sin^2\frac {B}{2},$   $ sin^4\frac {C}{2}  \plus{}\frac {1}{16} \geq \frac {1}{2}sin^2\frac {C}{2},$
$ sin^2\frac {A}{2} \plus{} sin^3\frac {B}{2} \plus{} sin^4\frac {C}{2} \geq sin^2\frac {A}{2} \plus{} \frac {3}{4}sin^2\frac {B}{2} \plus{} \frac {1}{2}sin^2\frac {C}{2} -\frac {2}{16}$
$=\frac {9}{4}\left( sin^2\frac {A}{2} \plus{} sin^2\frac {B}{2} \plus{} sin^2\frac {C}{2}\right) -\frac {2}{16} -\frac {9}{8}\geq \frac {9}{4}\cdot\frac {3}{4}-\frac {20}{16}=\frac {7}{16}.$


calculus integration LaTeX derivative function logarithms

I am trying to find an analytical solution for this infinite sum, I know it converges by integral test to a value under 1. however I am not able to analytically (not approximately) solve it. Any help ??
$ \sum^{\infty}_{n\equal{}1} \frac{ln(n)}{n^2}$

PS. please dont laugh at me, I suck at these kind of stuff....

Btw, just add cash signs (on US keyboard, shift+4) signs around the expression and it'll look nice :)

$ \Sigma^{\infty}_{n\equal{}1} \frac{ln(n)}{n^2}$

Also you can type \sum instead of \sigma and it works just as well.
It'll look like $ \sum^{\infty}_{n\equal{}1} \frac{ln(n)}{n^2}$

Cheers :)

Does this really converge to a sum less than 1?

Hi, sorry new to the forum and latex skills are sloppy.

yes it does $ \int^{\infty}_{1} \frac {ln(x)}{x^2} dx \equal{} \minus{} \frac {1}{x}(1 \plus{} ln(x))|^{\infty}_{1} \equal{} 1 < \infty$, so according to the integral test (if I remember correctly from many years ago) it converges, I think to a number less than one based on just intuition.

This is the negative of the derivative of the Riemann zeta function at $ 2$, and I see no reason why that should have a closed form.

hello, Maple 12 says $ \sum_{n\equal{}1}^{\infty}\frac{\ln(n)}{n^2}\equal{}\minus{}\zeta(1,2)$.
Sonnhard.

Hi!

Thanks, it is the derivative of zeta @ 2 indeed! And it appears that it actually does have a closed form [url]http://mathworld.wolfram.com/RiemannZetaFunction.html[/url] formula (45).

$ \zeta'(2)\equal{}\frac{\pi^2}{6}[\gamma\plus{}\ln(2\pi)\minus{}12\ln A]$ in which $ \gamma$ and $ A$ are two constants I never heard about before. Its value is about -0.93
Does anyone have a link to a paper or publication that has its derivation in closed form ?

Well, if you look at the article about $ A$, it says
[quote]
It has closed-form representations
$ A	\equal{}	e^{1/12\minus{}\zeta'(\minus{}1)}
	\equal{}	(2\pi)^{1/12}[e^{\gamma\pi^2/6\minus{}\zeta'(2)}]^{1/(2\pi^2)}
	\equal{}	1.28242712...$
[/quote]
With such an approach, everything has a closed form :P. What really happens is that one inexpressible in normal terms constant is related to another. There are great results of this sort like Euler's identity $ \sum_{n\ge 1}\frac1{n^2}\equal{}\frac{\pi^2}{6}$ (not to mention various supernatural identities by Ramanujan) and there are just trivial reformulations like that formula (45). All references are right there in the MathWorld articles, so I do not see any reason to copy them here.


[quote="juicybooty911"]UNC rules
yaaaaaaaaaaa[/quote]

So what? did you really have to post a tipic on [b]The Round Table[/b] just for that comment? It seems kind of out of place, don't you think? It does not have like the feeling to be a serius topic for discussion. It could may be be acceptable on Fun and Games.

Any way, it is just an opinion. Live and let die as Guns n' Roses said!

There is a topic on the NCAA Tournament here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28757


Do we automaticlaly get all the videos or do we have to "unlock" them?  Also, when is the second generation of testers losing ALcumus?

EDIT: first post in the forum!

Videos become available as you are able to receive problems from a topic. Some topics require you to have passed a particular topic before seeing them.

How does one "unlock"topics? By doing well on previous ones?

is this just counting and probability or does it go into geometry as well?

AIME15, what do you mean by "second generation of testers losing Alcumus"?

Note that this is a thread from prior to public release of Alcumus.  AIME15 was referrring to the set of testers we added to Alcumus at the end of the summer.

oh ok i was confused i thought for a second that once we finish alcumus we get new videos

@below I'm sorry, I didn't notice the date. I will be more careful next time. Thanks for the answer!

[quote=yeskay]oh ok i was confused i thought for a second that once we finish alcumus we get new videos[/quote]
No, the completion of Alcumus does not provide new videos (8 year bump).

[quote=green_dog_7983]10 year... Unless of course this is 2016 which I believe it isn't...[/quote]
Can't math :blush:.


linear algebra
matrix
linear algebra unsolved

Assume $ A,B,C\in M_{n}(\mathbb{R})$ and $ A^T\equal{}A$, $ B^T\equal{}B$, $ C^T\equal{}C$ and $ A, B, C$ are all positively definite matrixs

can we have all the eigenvalues of $ ABC$ are positive?

[quote="flydata"]Assume $ A,B,C\in M_{n}(\mathbb{R})$ and $ A^T \equal{} A$, $ B^T \equal{} B$, $ C^T \equal{} C$ and $ A, B, C$ are all positively definite matrixs

can we have all the eigenvalues of $ ABC$ are positive?[/quote]
All the eigenvalues of $ ABC$ could be positive, take for instance $ A\equal{}B\equal{}C\equal{}I_n$.
But I guess the question you really want to ask is:
Are the eigenvalues of $ ABC$ necessarily positive?

An example where the eigenvalues of $ ABC$ are $ <0$ !
$ a: \equal{}\begin{pmatrix}4293&5742\\5742&7720\end{pmatrix}$
$ b: \equal{}\begin{pmatrix}17117&\minus{}8180\\\minus{}8180&4573\end{pmatrix}$
$ c: \equal{}\begin{pmatrix}2080&900\\900&14305\end{pmatrix}$

With $ 3$ matrices we don't obtain all the matrices with positive determinant. But we have:
If $ A\in\mathcal{M}_n(\mathbb{R})$ is s.t. $ det(A)>0$  and $ A$ is not a scalar matrix $ \alpha{I}$ with $ \alpha<0$, then $ A$ is the product of $ 4$ positive definite matrices.


Let $ P$ be the area of triangle $ ABC$ ($ AB \equal{} a, \ BC \equal{} b, \ CA \equal{} c$) and let $ Q$ be the area of triangle $ XYZ$ ($ XY \equal{} x, \ YZ \equal{} y, \ ZX \equal{} z$). Prove that:

$ a^2( \minus{} x^2 \plus{} y^2 \plus{} z^2) \plus{} b^2(x^2 \minus{} y^2 \plus{} z^2) \plus{} c^2(x^2 \plus{} y^2 \minus{} z^2) \ge 16 \cdot P \cdot Q$

http://www.mathlinks.ro/Forum/viewtopic.php?t=2958

  darij


For any integer k >= 2, find the smallest n such that there exists a graph with exactly n nodes, each of degree 3, containing no cycle of length <= k.

Well, it's easy to prove that:
if $k$ is even, then $n \geq 3 \cdot 2^{\frac k 2}-2$;
if $k$ is odd, then $n \geq 2^{\frac {k+3} 2}-2$.
For $k<6$ equality holds, while for $k=6$ it doesn't.
Any ideas to complete the solution?

A $(d,g)$-cage is a regular graph of degree $d$ (that is : each vertex has degree $d$) and girth $g$ (the length of the shortest cycle), with the least number, say $p$, of vertices.

You are asking for the size of a $(3,k)$-cage.

It is more generally known that :
$p \geq \frac {d(d-1)^r-2} {d-2}$ if $g=2r+1$
$p \geq \frac {2(d-1)^r - 2} {d-2}$ if $g=2r$.

If $g=5$ then it is further known that the above becomes $p \geq d^2 + 1$ and equality holds only if $d=2,3,7,57$. (Behzad and Chertrand, 1971)

Moreover, if $g \geq 3$ :
$p \leq \frac {d-1} {d-2} [d(d-1)^{g-2} + d - 4]$ (Tutte, 1961).

It has been shown that for $d-1$ a prime power (which is the case here), $(d,g)$-cages can be obtained from finite projective geometries (Bondy and Murty, 1976).

I do not have the papers. Perhaps a search on the web with the above key-words will give you more.

Pierre.

Have a look here :
http://mathworld.wolfram.com/CageGraph.html

Pierre.


How to insert preface for a report? The preface page numbers should be in Roman Numeral.

I presume that your problem is getting roman numerals for a few pages then starting from 1 with arabic numerals.
In the preface put [code]\pagenumbering{roman}[/code]and then in the main section put [code]\pagenumbering{arabic}
\setcounter{page}{1}[/code]

but what is the command for preface? Is there something like \begin{preface}?

Why should there be a command? A preface is just a chapter which happens to be called Preface. However if you use
[code]\frontmatter
\chapter{Preface}
this is the preface
\mainmatter[/code]then you will find the change of page numbering is automatic.

Thanks. But when I type that into texnic center, it gave me two errors(\frontmatter and \mainmatter). Should there be usepackage something?

There's no package required. You must have typed something wrongly as \frontmatter and \mainmatter are already defned in the book class.
[code]\documentclass{book}
\begin{document}
\frontmatter
\chapter{Preface}
this is the preface
\mainmatter
\end{document}[/code]should work fine.

Ok, now it works. My original documentclass was report.

Does each chapter of a book must have an even number of pages? It gives me a blank page at the end of the chapter if the chapter has an odd number of pages.

As the preface is a chapter it won't work in the report class. The book class uses the usual practice of starting the chapter on the right-hand page but can be overridden with [code]\documentclass[openany]{book}[/code]


What is the tens digit in the sum $ 11^1 \plus{} 11^2 \plus{} 11^3 \plus{} \ldots \plus{} 11^9$?

For 11^n (n is from 1 to 9), the tens digit is always n. Why? Because we can start with 121 (11^2). Then using long multiplication, you keep adding 1 to the tens digit every time you multiply. Since the ones digit is 1, there is no carrying. Thus, the tens digit is the units digit of $ (1\plus{}2\plus{}3\plus{}4\plus{}5\plus{}6\plus{}7\plus{}8\plus{}9)\equal{}55$, which is 5.

$1+2+3+4+5+6+7+8+9$ is 45, not 55. But as usual you can ignore the tens digit and get the answer of 5.


quadratics
number theory theorems
number theory

I want some advice. What topic I should read in this book for prepare IMO.
I have finished reading indices in Quadratic Residues topic.

I read it to prepare my exam in school.
You have to read the section 1, 2 ,3, some of 4(about Pell's equation), and 5 to prepare IMO.
Reading all of section 4 may help you.

[quote="tkhoa2711"]could you tell me where can I find that book?[/quote]
Is this what you meant: [url]http://www.amazon.com/Higher-Arithmetic-Introduction-Theory-Numbers/dp/0521634466[/url] (I just googled it)?


function
integration
inequalities
logarithms
real analysis
real analysis unsolved

Real function $f$ is defined on an interval $I \in R$,
If $\int_{I}|f|(log^{+}|f|)^{1+\delta}dx<\infty$,
where $\delta>0$ and
$log^{+}(y)=log(y)$ if $log(y)>0$, $log^{+}(y)=0$ if $log(y)\leq 0$,

how to show:$f \in L^{2}(I)$?

[quote="fedja"]There is no way to do it because it is false: consider $f(x)=1/\sqrt x$ on $I=[0,1]$.[/quote]

sorry, I meant: How to show all $L^{2}(I)$ function satisfy this inequality?

The fact that $x(\log^{+}x)^{1+\delta}<cx^{2}$ for some constant $c$ which depends on $\delta$ and all $x>0$ is all you need.


Hey Guys!
I think one of the posters of a similiar question is something I know, but I cannot find a proper answer

Can some explain to me the formula for deriving a Hardy-Ramanujan number?
I know that 4104 is one of them  ( i THINK).. but what is the biggest number before 1 million?

Thanks
May Hannah Bless you all

psst... find a similiar topic and the poster is my ex-girlfriend!!


mr rrusczyk previously posted the useful chapters for aime preparation.

but what about AMC preparation. which chapters are useful?

[quote="pkothari13"]Would you mind posting the link to that thread about AIME preparation?[/quote]

Please use search button. I know this is a tedious job and people do not prefer it but search button is really nice if you use it.  :)

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=70754#p70754

[quote="pkothari13"]Meh, I tried searching before I asked, but I guess my searching skills aren't up to par.   :(  What did you search?[/quote]

AIME Preparation.

It didn't come as "ONLY" one. That rarely happens so after you searched for a broad thing, you should go through each smaller threads.

I think all of the same are just fine for AMC as well as AIME.

Which reminds me, I should do at least a few more chapters of AoPS Vol. II before the AMC rolls around, and try to finish the book by AIME...


function
group theory
abstract algebra
algebra
functional equation
algebra unsolved

Find all functions $ f : \mathbb{R} \to \mathbb{R}$ such that
\[f(x - f(y)) = f(x) + f(f(y)), \quad \forall x,y\in\mathbb R.\]

[quote="thachpbc"]Find all functions f : R-> R and:
f(x -f(y)) = f(x) + f(f(y)) all x,y in R.[/quote]

So $ f(x \minus{} f(y)) \equal{} f(x) \plus{} f(f(y))$ $ \forall x,y\in\mathbb R$

Not very hard, but interesting ...

[hide="My solution"]
1) [b][u]basics elements [/u][/b]:
$ P(x,y)$ : $ f(x \minus{} f(y)) \equal{} f(x) \plus{} f(f(y))$
Let $ f(0) \equal{} a$

$ P(f(x),x)$ $ \implies$ $ f(f(x)) \equal{} \frac a2$ and so $ f(x) \equal{} \frac a2$ $ \forall x\in I \equal{} f(\mathbb R)$  (** edited **)
$ P(2f(x),x)$ $ \implies$ $ f(2f(x)) \equal{} 0$ and so $ 0\in I \equal{} f(\mathbb R)$ and so $ f(0) \equal{} \frac a2$ but $ f(0) \equal{} a$ $ \implies$ $ a \equal{} 0$

So we have :
$ f(f(x)) \equal{} 0$ $ \forall x$
$ f(x \minus{} f(y)) \equal{} f(x)$ $ \forall x,y$

From the second equation, it's easy to conclude that $ f(x \plus{} a) \equal{} f(x)$ $ \forall x\in\mathbb R$ and $ \forall a\in A$ where $ A$ is the smallest additive subgroup generated by elements of $ I \equal{} f(\mathbb R)$ and this is enough to establish the general solution :

2) [b][u]general solution [/u][/b]:
Let $ A$ any additive subgroup of $ \mathbb R$
Let $ \sim$ the relation $ x\sim y$ $ \iff$ $ x \minus{} y\in A$
Since $ A$ is an additive group, $ \sim$ is an equivalence relation. Let then $ c(x)$ any choice function which associates to a real $ x$ a representative (unique per class) of it's equivalence class.
Let then $ h(x)$ a function from $ c(\mathbb R)\to A$ such that $ h(c(0)) \equal{} 0$

Then $ f(x) \equal{} h(c(x))$ $ \forall x$

3) [b][u]proof that any function defined as in 2) is a solution [/u][/b]:
$ f(\mathbb R)\subseteq A$ $ \implies$ $ f(y)\in A$ $ \forall y$ $ \implies$ $ c(f(y)) \equal{} c(0)$ $ \implies$ $ f(f(y)) \equal{} h(c(f(y))) \equal{} h(c(0)) \equal{} 0$
$ f(y)\in A$ $ \implies$ $ \minus{} f(y)\in A$ $ \implies$ $ x \minus{} f(y)\sim x$ $ \implies$ $ c(x \minus{} f(y)) \equal{} c(x)$ $ \implies$ $ f(x \minus{} f(y)) \equal{} f(x)$

And so $ f(x \minus{} f(y)) \equal{} f(x) \plus{} f(f(y))$
Q.E.D

4) [b][u]proof that any solution is in the form defined in 2[/u]).[/b]
Just choose as $ A$ the smallest additive subgroup generated by elements of $ I \equal{} f(\mathbb R)$ and the conclusion is immediate.

5) [b][u]some examples[/u].[/b]
5.1) $ A \equal{} \{0\}$, then $ c(x) \equal{} x$ and $ h(x) \equal{} 0$ $ \implies$ $ f(x) \equal{} 0$ $ \forall x$

5.2) $ A \equal{} \mathbb Z$, with $ c(x) \equal{} x \minus{} [x]$ and $ h(x) \equal{} [10\sin(x)]$ $ \implies$ $ f(x) \equal{} [10\sin(x \minus{} [x])]$

5.3) $ A \equal{} \mathbb Q$ with $ h(c(\pi)) \equal{} 1$, $ h(c(\sqrt 2)) \equal{} 2$ and $ h(x) \equal{} 0$ anywhere else $ \implies$
$ f(\pi \plus{} a) \equal{} 1$ $ \forall a\in\mathbb Q$
$ f(\sqrt 2 \plus{} a) \equal{} 2$ $ \forall a\in\mathbb Q$
$ f(x) \equal{} 0$ anywhere else.

...

[/hide]

[quote="pco"] [b][u]basics elements [/u][/b]:
[...]

$ P(f(x),f(x))$ $ \implies$ $ f(f(x)) \equal{} \frac a2$ [/quote]

Is P(f(x),x)?

However too beautiful, i think you are the best with functional equation  :o

[quote="bboypa"] Is P(f(x),x)?[/quote]
Sure, you're right. Thanks for the remark. I've edited my post.

[quote="bboypa"] However too beautiful, i think you are the best with functional equation  :o[/quote]
Thanks  :blush:


calculus
integration
function
derivative
real analysis
real analysis solved

Here's a nice one. It's not actually an 'unsolved problem', because I have a solution, but I didn't know where else to post it:

f:[0,1]->R is a derivable function, and let F be a primitive of f. There exist a, b in (0,1) such that F(a)-F(0)= F(1)-F(b)=0. Let M be sup |f'(x)|, where f' is the derivative of f. show that |F(1)-F(0)|<= (1+b-a)*M/4.

okay so here's my solution:

we have \int10 f(x)dx = \intba f(x)dx = F(1)-F(0)=F(1)-F(b)+F(b)-F(a)+F(a)-F(0) from the Leibnitz-Newton relation and the first two relations in the hypothesis.

therefore it is sufficient to prove that

4|\intba f(x)dx | \leq (1+b-a)M

We know from the mean-value Theorem that  \int1b f(x)dx = (1-b)c1, where c1 \in [b,1]. Same goes for c2 \in [0,a].

It is easy to check that for every x in [a,b] at least one of the numbers |x-c1| and |x-c2| is smaller than 1/2.

Let us use again this Theorem, this time to compute the integrals above:

|  \int10 f(x)dx|=(1-0)|f(c)|, where c \in [0,1] and |\intba f(x)dx | = (b-a)|f(c')|, where c' \in [a,b].

Now using Lagrange Theorem (is kinna the same with the Mean Value, the only difference is when f' is not integrable) we get for any x \in [a,b]
f(x)-f(ci)=(x-ci)f'(di), where di \in [x,ci] (or possibly [ci,x]), for each i \in {1,2}.

As f(ci) is 0, and |f'(di)|\leq M we have that

|f(x)| \leq 1/2 M for each x \in [a,b] which leads to the fact that

| \intbaf(x)dx |\leq (b-a)/2 M.

All we are left to prove is that 2(b-a) \leq 1+b-a which is an obvious fact.

Nice problem Grobber, where did you find it?

Tahnks! My solution is exactly the same. It was proposed by Marcel Chirita for the final round of the olympiad in Romania in 2002.

do u mean it was an assumption or you can deduce it from the given assumptions?

I think it wasn't specified in the problem, but I think you need a<b. If a>b can the solution be saved?

I've cheked up the problem's statement and it is indeed specified there that a<b. The problem remains for a>b.

You're absolutely right, Valentin. I looked it up and a<b is indeed mentioned. I guess I just missed it when I posted the problem. Sorry! :D


[quote="MCrawford"]Let $m$ be the number of positive integer factors of 876, 876, and let $n$ be the number of positive integers factors of 678,678.  What is $m-n$?[/quote]

[hide="solution"]$876876=2^2*3*7*11*13*73$

there are 3*2*2*2*2*2=96 divisors

$678678=2*3*7*11*13*113$

there are 2*2*2*2*2*2=64 divisors

96-64=32[/hide]


[quote="nayel"]If $ a,b,c$ are positive real numbers prove that
\[ \left(\frac ab \plus{} \frac bc \plus{} \frac ca\right)\left(\frac {a}{c \plus{} a} \plus{} \frac {b}{a \plus{} b} \plus{} \frac {c}{b \plus{} c}\right)\ge 3\left(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)
\]
[hide="After expansion"]this becomes
\[ 2\sum a^2b^4 \plus{} abc\sum a^2b \plus{} \sum a^3b^3\ge 2abc\sum a^3 \plus{} 6(abc)^2
\]
which seems to be true but I can't prove it. :( [/hide][/quote]
I think it's wrong. Try $ a\rightarrow\plus{}\infty$ and $ b>c.$ :wink:

[quote="nayel"]Can you please give a counterexample? I can't find one. :blush:[/quote]
you are right ,arqudy
Please check with $ a\equal{}100,b\equal{}1,c\equal{}0,1$
I think it will true if $ a,b,c$ are 3 sides of a triangle


Which program are you doing or did you do in High School? IB or AP?

I'm in a gifted program for middle school and we are considering extending the program into High School. There are two schools of choice. The first, has only AP programs while the second offers both AP and IB. I would like to know which program is more math and science oriented than the other ones. Thank you.

My school does AP.  Don't quote me on this one cuz I don't know much about IB but I thought IB was like enriched everything, where AP you can pick and choose what you want to be enriched or not.

Yeah I think there's something like that.  IB is only done by a few schools, AP's are much more common.

yup. IB tries to create well rounded students while you pick what AP classes you want to take. my school does AP.

So, am I the only one to take IB in MathLinks? :D
My school offers partial IB, so I only take math IB. I regret that I only chose math IB, since most of other classes are too easy. I had to make this choice shortly after I immigrated to Canada, so I felt that taking full IB was a crazy thing to do. However, it turned out that it isn't. :(

Hmm, another question.  Someone once told me that AP was just a test, while IB is a course.  Anyone wanna refute/support this?

IB is awesome, I'm loving it.  The people in it make the classes great, and the material is a lot more interesting than the regular stream.

Well AP is a test, so you "pass" as long as you can pass the test.  AP Courses are designed by the teacher, school, whatever and are designed to allow you to pass the test.  So you can self-study for the AP Test and still get college credit without taking the course.


Consider the polynomials P(x) = x^6 - x^5 - x^3 - x^2 - x and Q(x) = x^4 - x^3 - x^2 - 1.  Given that z_1, z_2, z_3, and z_4 are the roots of Q(x) = 0, find P(z_1) + P(z_2) + P(z_3) + P(z_4).

I'll rewrite it in $ Latex$:
Consider the polynomials $ P(x) \equal{} x^6 \minus{} x^5 \minus{} x^3 \minus{} x^2 \minus{} x$ and $ Q(x) \equal{} x^4 \minus{} x^3 \minus{} x^2 \minus{} 1$. Given that $ z_1, z_2, z_3,$and $ z_4$are the roots of $ Q(x) \equal{} 0$, find $ P(z_1) \plus{} P(z_2) \plus{} P(z_3) \plus{} P(z_4).$

Original thread is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=713214#p713214]here[/url]. In the future, use the "Resources"/"Contests" section to find the original thread for the problem before posting it a second time. This keeps the discussion on that problem from becoming fragmented across the forum.


What follows is an easy problem, but I was charmed by it because I'd never heard about it before.

Show that a connected graph with $n$ vertices has a connected subgraph with $k$ vertices for all $1\le k\le n$.

Isn't it cute ? :)

Isn't it trivial????
Take a sub-tree of this connected graph. Now eliminate the leaves one by one. You will get a sub-tree of any dimension you would ever wish....:)

Well, there is a step between easy and trivial ;)

Pierre.


calculus
integration
LaTeX
conics
ellipse
geometry
calculus computations

What is the volume between the elliptical paraboloid $x=\frac{y^{2}}{2p}+\frac{z^{2}}{q}$ and the plane $x=a$?

[hide="Well, the setup at least..."]We will use a triple integral to determine this volume.

First note that it is easy to see that the solid in question, which we shall call G, is a simple yz-solid.

$\int \int_{G}\int \,dV$ (sorry, I'm not too sure how to use $LaTeX$ to represent multiple integrals very well) gives the volume of the solid. We wish to find the iterated integral equal to this triple integral.

Assuming both $p$ and $q$ are positive, (in order to get a reasonable result) projecting the intersection of the paraboloid and the plane onto the yz-plane yields:

$a = \frac{y^{2}}{2p}+\frac{z^{2}}{q}$

To find the limits of integration, we can solve for either variable. Let's just solve for $y$:

$y = \pm \sqrt{2p(a-\frac{z^{2}}{q})}$. From this we can easily determine the limits of integration on $y$.

To determine the limits of integration on $z$, we simply look at the graph of our projection and see that $z$ ranges from $-\sqrt{q}$ to $\sqrt{q}$.

Obviously the limits of integration on $x$ are $0$ and $a$.

Thus, our desired iterated integral is:

$\int^{a}_{0}\int^{\sqrt{q}}_{-\sqrt{q}}\int^{\sqrt{2p(a-\frac{z^{2}}{q})}}_{-\sqrt{2p(a-\frac{z^{2}}{q})}}\,dy\,dz\,dx$.

However, I'm not seeing how to evaluate this integral after integrating with respect to $z$... I may be spacing out, but I see no elementary antiderivative to $2\sqrt{2p(a-\frac{z^{2}}{q})}$...[/hide]

try z = Cos[m]*Sqrt[a*p]

this helps a lot.

but whitout knows q is dificult to solve...

because the simplyfication of this integral need to sin be positive...

i did not verificate if this equation is rigth.

but i think it is...

You're making this much more complicated than it needs to be. The slice at $x=c$ is an ellipse, with area proportional to $(\sqrt{c})^{2}$. Just find the scaling constant- the rest is easy.

Hiiiii Sir Jmerry

I think we are doing the same think

because the integral of the Sir I Am Me we just need to integrate one time, because the anothers 2 are just multiply by a constant

and if we do like you saied we need to integrate too. because we want the volume, not the area.

in the end you will have the same integral, just one, not 3.
hum....

sorry if i'm wrong. but i think that it will end in the same way


ratio algebra polynomial quadratics induction

solve for $x$ if

$1+\frac{1}{(1+\frac{1}{1+\frac{1}{1+...\frac{1}{1+\frac{1}{x}}}})}=x$

where there are exactly $n$ total fractions on LHS

[hide="um"]
that strongly resembles the golden ratio ! :wink: [/hide]

[hide="I think"]$x=\frac{-1+\sqrt{5}}{2}$
Because  $\frac{1}{1+x}=x$  so that     $x=\frac{-1\pm\sqrt{5}}{2}$    but  $x$  should be positive.   So   $x=\frac{-1+\sqrt{5}}{2}$[/hide]

[quote="digger"][hide]I prefer to write this fraction as $[0,1,1, ... , 1, x]$ where $0$ and $x$ ocuppy places numbered by $1$ and $n+2$ respectively.

We have ($F_{0}=0, F_{1}=1, F_{n+1}=F_{n}+F_{n-1}$ - Fibonacci numbers):
$[0,1,1, ... , 1]=\frac{F_{n}}{F_{n+1}}$
and
$\frac{xF_{n}+F_{n-1}}{xF_{n+1}+F_{n}}=[0,1,1, ... , 1,x]=x.$
This implies
$x^{2}=\frac{F_{n-1}}{F_{n+1}}$ and $x=\pm \sqrt{\frac{F_{n-1}}{F_{n+1}}}$
P.S. Alternative approuch (with same result) is to write the equation in the form $[1,1,...1,x]=\frac{1}{x}$[/hide][/quote]
sorry it should have been $1+\frac{1}{(1+\frac{1}{1+\frac{1}{1+...\frac{1}{1+\frac{1}{x}}}})}=x$

[1,1,1,...,x] i guess?

[quote="digger"]$\frac{xF_{n}+F_{n-1}}{xF_{n+1}+F_{n}}=[0,1,1, ... , 1,x]=x.$
[/quote]

could you show the details for this please?

i wuld have thought there would have been a lot more solutins, given you have a polynomial of high degree if you multiply out everything.

[quote="math227"][1,1,1,...,x] i guess?[/quote]
Yes, with n units.

I guess that [b]jdai_ming[/b] have "seen" 1 (which was absentfor that moment)  but made mistake considering that $x$ must be positive - there is no reason for such assumption.

As we know that $[1,1,1,...,1,x]_{n+1}=x$ is quadratic equation and evidently golden ratio $x=[1,1,1,...] = \frac{-1+\sqrt{5}}{2}$ is one of it's roots then $x'= \frac{-1-\sqrt{5}}{2}$ must be root too. Here [b]jdai_ming[/b] made mistake - there is no reason for assumption that $x>0$.

The same can be calculated as above:

$[1,1,1,...,1]_{n}= \frac{F_{n}}{F_{n-1}}$,
where subindex n shows how many places there are in the fraction

$[1,1,1,...,1,x]_{n+1}= \frac{xF_{n}+F_{n-1}}{xF_{n-1}+F_{n-2}}$

and we have $\frac{xF_{n}+F_{n-1}}{xF_{n-1}+F_{n-2}}=x$ yielding
$x^{2}-x-1=0$ for $n>1$ because we must cancel $F_{n-1}$. But for $n=1$ there is nothing to cancel - we have $x^{2}-x-1=0$ immediately.

Edit: improved typo there was $\frac{-1\pm{5}}{2}$  instead of $\frac{-1\pm\sqrt{5}}{2}$

[quote="seamusoboyle"]could you show the details for this please?[/quote]
This is simply how to calculate $P_{n}$ and $Q_{n}$ if we denote $\frac{P_{n}}{Q_{n}}: =[q_{1}, q_{2}, ... , q_{n}]$
They are (easily proved by induction):
$P_{0}=1, P_{1}=q_{1}, P_{k+1}=q_{k+1}P_{k}+P_{k-1}$
$Q_{0}=0, Q_{1}=1, Q_{k+1}=q_{k+1}Q_{k}+Q_{k-1}$

Well $x=\pm \sqrt{2}$ and $x=\pm \sqrt{3}$ are solutions to:

$1+\frac{1}{1+x}=x$ and $1+\frac{1}{1+\frac{1}{1+x}}=x$

respectively

define $f_{i}(x)$ to be the contiued fraction above with i fraction signs.

notice: $f_{1}(x)=1+\frac{1}{x}$
the equation $x=f_{1}(x)$ has solution $x=\phi$

$f_{2}(x)=1+\frac{1}{1+\frac{1}{x}}=f_{1}(f_{1}(x))$
obviously $x=f_{2}(x)$ also has solution $x=\phi$

the value of x is $\phi$ for all n.

alternatively from $x=\frac{xF_{n}+F_{n-1}}{xF_{n-1}+F_{n-2}}$ we have $F_{n-1}(x^{2}-x-1)=0$

edit: which is what digger has 3 posts above :)

[quote="seamusoboyle"]given you have a polynomial of high degree if you multiply out everything.[/quote]

Actually not true.  You'll find by induction that the polynomial is always quadratic.  :)

My God! I should stop and think before posting! You'd think the fact i actually did 2 examples (wrong examples, to be sure) and got quadratics in both would have said something!


What is the sum of all the positive factors of 360?

I donno why nobody answered this one...... :huh:

[hide="With my fingers crossed....."] $ 636$  :maybe:

 I'll post the solutions if it's correct..
I don't want to embarass myself... :wink: [/hide]

[quote="danb80"]Sorry.

See my latest addition under trick and tips for help[/quote]

That's exactly what I did :o .......

It was a miscalculation, probably.....

[hide="ALL RIGHTY"]Since $ 360$ is equal to $ {2}^{3}$+$ {3}^{2}$+$ {5}^{1}$ ,

I took the base number and added the powers of each prime number up to its exponent (start with 0) and multiplied the resulting sums together, which gave me

 $ ({2}^{0}\plus{}{2}^{1}\plus{}{2}^{2}\plus{}{2}^{3})$ $ ({3}^{0}\plus{}{3}^{1}\plus{}{3}^{2})$ $ ({5}^{0}\plus{}{5}^{1})$. = $ \boxed {1170}$  :maybe:[/hide]

[quote]I took the base number and added the powers of each prime number up to its exponent (start with 0) and multiplied the resulting sums together [/quote]

so thats the formula for finding all positive factors of a number?

[quote="alanchou"][quote="jogaman323"][quote="danb80"]Sorry.

See my latest addition under trick and tips for help[/quote]

That's exactly what I did :o .......

It was a miscalculation, probably.....

[hide="ALL RIGHTY"]Since $ 360$ is equal to $ {2}^{3}$+$ {3}^{2}$+$ {5}^{1}$ ,

I took the base number and added the powers of each prime number up to its exponent (start with 0) and multiplied the resulting sums together, which gave me

 $ ({2}^{0} \plus{} {2}^{1} \plus{} {2}^{2} \plus{} {2}^{3})$ $ ({3}^{0} \plus{} {3}^{1} \plus{} {3}^{2})$ $ ({5}^{0} \plus{} {5}^{1})$. = $ \boxed {1170}$  :maybe:[/hide][/quote]

I don't remember $ 360$ being equal to $ 22$...[/quote]

Oh, you know that he meant $ {2}^{3}\times{3}^{2}\times{5}^{1}$

[quote="alanchou"][quote="jogaman323"][quote="danb80"]Sorry.

See my latest addition under trick and tips for help[/quote]

That's exactly what I did :o .......

It was a miscalculation, probably.....

[hide="ALL RIGHTY"]Since $ 360$ is equal to $ {2}^{3}$+$ {3}^{2}$+$ {5}^{1}$ ,

I took the base number and added the powers of each prime number up to its exponent (start with 0) and multiplied the resulting sums together, which gave me

 $ ({2}^{0} \plus{} {2}^{1} \plus{} {2}^{2} \plus{} {2}^{3})$ $ ({3}^{0} \plus{} {3}^{1} \plus{} {3}^{2})$ $ ({5}^{0} \plus{} {5}^{1})$. = $ \boxed {1170}$  :maybe:[/hide][/quote]

I don't remember $ 360$ being equal to $ 22$...[/quote]
Try base $ 179$ :P


function
number theory unsolved
number theory

do there exist a positive integer $m$ , such that for any $n\in{N}$ :
$n|1!+2!+...+(n-1)! - m$.

Related problem: is the function $f(n) = (1!+2!+\dots+(n-1)!)\bmod n$ bounded as $n\to\infty$?
Negative answer to this one implies negative answer to the original problem.

See also sequence [url=http://oeis.org/A049782]A049782[/url] and [url=http://mathpages.com/home/kmath064.htm]Can n Divide !n ?[/url]


function
induction
algebra
floor function
combinatorics unsolved
combinatorics

n people stand in a straightline .With a given positive integer $k \leq [\frac{n}{2}]$ , How many ways there are to pick k people out of n people so that no two adijacent people are chosen

Let this number be $f(n,k)$. Then for $k > [\frac{n}{2}]$ f(n,k) = 0.
and we have obvious identity
$f(n,k)=f(n-1,k)+f(n-2,k-1)$

... and from the recursion it is easy enough to deduce that $f(n, k) = \binom{n-k+1}{k}$.  Which is also easy enough do discover directly, by gluing all but the rightmost of our selected individuals to the person to his right.  (Actually, Baronov, if by $[\cdot]$ you mean the floor function, then you are not quite correct in your first line.)

To JBL : I don n't think it is easy to prove the formula of $f(n;k)$ :D

You're welcome to think that, but you are wrong.  Induction is essentially trivial: the base cases are $f(n, 0) = 1 = \binom{n+1}{0}$, $f(1, 0) = 1 = \binom{2}{0}$, $f(1, 1) = 1 = \binom{1}{1}$, and for $k > 1$ $f(1, k) = 0 = \binom{2-k}{k}$.  Then $\binom{n-k}{k-1}+\binom{n-k}{k}= \binom{n-k+1}{k}$ is the inductive step.

Non-inductively is simple as well: there is an easy bijection between choices of the given form and choices of $k$ people from $n-k+1$ with no requirement on adjacencies.

Yes i understand thank you verry much :D
But Can you show that :$f(n,k)=f(n-1,k)+f(n-2,k-1)$ ?

There are $f(n-1, k)$ ways if we do include the last person, and $f(n-2, k-1)$ ways if we don't.


Call n be the amount of the permutations of  {1,2,3,...,2005} such that : the algebraic difference of two consecutive numbers is at most 2 . Find the remainder when n divides into 3 ?

I think we can generalize this problem , but I still coludn't solve it


Let [x] be floor(x).

Given the following recurrence:

T(1) = 1
T(n) = T([n/2]) + n , n > 1

What is the general function for this recurrence, that is, T(n) = f(n) for some function f(n).

I'll just rewrite the problem, I have to go to my class now. haha, I'll answer it later.

Let $ [x]$ be a Floor function(greatest integer function) of $ x$.

Solve the recurrence relation

$ T(1) \equal{} 1$
$ T(n) \equal{} T\left(\left[\dfrac{n}{2}\right]\right) \plus{} n ; n>1$

What is the general function for this recurrence, that is, $ T(n) \equal{} f(n)$ for some function $ f(n)$.

if i haven't made any mistakes it should be like this...
t(n)= n + t(floor(n/2)) = n + floor(n/2) + t(floor(n/4)) = n + floor(n/2) + floor(n/4) + ... + floor(n/(2^k))

where k + 1 happens to be the number of terms, which equals floor(log_2(n)) + 1

so t(n) = sum( floor( n/ (2^i)), i , 0 , k -1)

where did you get this problem - it was nice!

To make the above a bit cleaner:

[hide="Answer"]Observe that we can redefine the recurrence using simply the second equation, since $ 1 \plus{} \left \lfloor \frac{1}{2} \right \rfloor \equal{} 1$.

Therefore, the answer is $ \sum_{k\equal{}0}^{\infty} \left \lfloor \frac{n}{2^k} \right \rfloor$, which should look familiar to anyone who's done those problems about trailing zeros at the end of a factorial[/hide]

Tsk.  There is a simpler formula available.

[hide="Hint"] What does the recurrence do to the binary digits of $ n$? [/hide]

Hello t0rajir0u!

I didn't understand your hint. Could you explain it? What function f(n) would represent T(n)?

Thanks!

[hide="Second hint"] The function $ n \mapsto \left\lfloor \frac {n}{2} \right\rfloor$ chops off the last binary digit of $ n$.  So how can you represent the summation $ \sum_{k \equal{} 0} \left\lfloor \frac {n}{2^k} \right\rfloor$ in terms of the binary digits of $ n$? [/hide]
[hide="Answer"] Let $ s_2(n)$ denote the sum of the binary digits of $ n$.  For example, $ s_2(25) \equal{} s_2(11001_2) \equal{} 3$.  Then

$ T(n) \equal{} 2n \minus{} s_2(n)$. [/hide]
[hide="Generalization"] Similarly, verify that

$ \sum_{k \equal{} 0} \left\lfloor \frac {n}{p^k} \right\rfloor \equal{} \frac {p}{p \minus{} 1} n \minus{} \frac {s_p(n)}{p \minus{} 1}$

for any positive integer $ p$.  It follows that $ \sum_{k \equal{} 0} \left\lfloor \frac {n}{p^k} \right\rfloor \sim \frac {p}{p \minus{} 1} n$.  (We say that $ f(n) \sim g(n)$ if $ \lim_{n \to \infty} \frac {f(n)}{g(n)} \equal{} 1$.) [/hide]
[hide="Application"] Find a concise expression for the largest power of $ p$ dividing $ {n \choose k}$ for prime $ p$. [/hide]


let $f$ be polynomial of degree $n$ wit complex coefficients prove that the polynomial $F=f.f'\dots \cdot f^{n-1}$ has at least $n+1$ disticnt roots.

What about the polynomial $f(x)=x^n$?
We have $F(x)=cx^k$ for some $c,k$ and it has only one root (no two distinct roots)

i think it should be mentioned that f has more than one root although it was not specifically mentioned in original source.


induction
number theory unsolved
number theory

Find all $m,n$ in positive integer such that:

a)$mn\mid m^n+n$ and $mn\mid n^m+m$

b)$mn\mid m^n+n^m$

a)$mn\mid m^n+n$ and $mn\mid n^m+m$

[hide]$mn\mid m^n+n => $
$m\mid m^n+n => $
$m\mid n  $

$mn\mid n^m+m => $
$n\mid n^m+m => $
$n\mid m  $

Hence $ n=m $
[/hide]

$b)$: Very easy or I missed something? :?

$lemma$
Let $a>0$, $b>0$ and $p>1$ integers. Then $a+b \leq bp^a$
$Proof$
by induction it's easy to show that $a \leq 2^a-1$ so $a \leq p^a-1 \leq b(p^a-1) \Rightarrow a+b \leq bp^a$.

$Theorem$
Let $(m,n) \in (\mathbb{N}_*)^2$:
$mn \mid m^n+n^m$ iff $m$ and $n$ have the same prime factors.

$proof$
- If $mn \mid m^n+n^m$ then $m \mid n^m$ and $n \mid m^n$ so each prime factor of $m$ is a prime factor of $n$ and reciprocally.
- Assume that $m$ and $n$ have the same prime factors.
Let $p$ a prime factor of $m$. We have $p \mid n$.
Let $a=ord_p(m)>0$ and $b=ord_p(n)>0$.
$ord_p(m^n)=na \geq p^ba \geq a+b$ (by lemma) and by the same way: $ord_p(n^m) \geq a+b$.
So $ord_p(m^n+n^m) \geq ord_p(mn)$ and we are done.


f(f(x-y)) =f(x)-f(y)-f(x)f(y)-xy for all x,y belong to R.
and x,y#0,and x#y.  Then find f(x).

You can not post this problem here, wrong place!  :lol: But this is a hint

[hide]See at $ \text{Im}\, f$.[/hide]


geometry
geometric transformation
rotation
trigonometry
cyclic quadrilateral
geometry proposed

I have in front of me a roumanian book with olympiad problems for classes 9 and 10. The problem below is referd as INDIA 2003. In the book there is a trig. solution. The greek IMO constestand and medalist Silouan  found a nice solution with harmonics. If someone has the official solution or some more elementery , please  send it.

Dera Stergiu and Mathlinkers,
I present my synthetic proof.
Let C' be the midpoint of AB, N the midpoint of BH, P the foot of the A-altitude and 1 the Euler's circle.
In the triangle ABH, NC' // AHP.
TR is the Pascal's line of DMPEC'ND and TR // AP
We have AP orthogonal to BC
then TR orthogonal to BC.
Sincerely
Jean-Louis

[b][size=100][color=DarkBlue]GENERAL PROBLEM. – A triangle $ \bigtriangleup ABC$ is given and let $ \bigtriangleup DEF$ be, its Cevian triangle with respect to an arbitrary point $ P,$ inwardly to it. We denote as $ L$ the midpoint of the segment $ PC$ and let be the point $ M\equiv AC\cap DL.$ Prove that $ TM\parallel PC,$ where $ T\equiv AP\cap EF.$[/color][/size][/b]

[b][size=100]PROOF.[/size][/b] – We consider $ M,$ as the intersection point of $ AC,$ from the line through the point $ T$ and parallel to $ PC$ and it is enough to prove that the points $ D,\ L,\ M$ are collinear, or equivalently $ MT \equal{} MQ$ $ ,(1)$ where $ Q\equiv BC\cap TM.$

In the triangle $ \bigtriangleup DTQ$ with transversal $ AMC,$ applying the [b][size=100]Menelaus theorem[/size][/b],

we have that $ \frac {MT}{MQ}\cdot \frac {CQ}{CD}\cdot \frac {AD}{AT} \equal{} 1$ $ \Longrightarrow$ $ \frac {MT}{MQ} \equal{} \frac {CD}{CQ}\cdot \frac {AT}{AD}$ $ ,(2)$

From $ (1),$ $ (2),$ it is enough to prove that $ \frac {CD}{CQ} \equal{} \frac {AD}{AT}$ $ ,(3)$

But from $ TQ\parallel PC$ $ \Longrightarrow$ $ \frac {CD}{CQ} \equal{} \frac {PD}{PT}$ $ ,(4)$

From $ (3),$ $ (4)$ it is enough to prove that $ \frac {AD}{AT} \equal{} \frac {PD}{PT}$ $ ,(5)$

But the relation $ (5)$ is true, because of the points $ A,\ T,\ P,\ D$ are in harmonic conjugation,
from the complete quadrilateral $ AEPFBC.$

So, the relation $ (1)$ is also true and then, we conclude that the points $ D,\ L,\ M$ are collinear and the proof is completed.

Kostas Vittas.

[quote="Virgil Nicula"][size=117][color=darkblue][b]Nice problem, Stergiu ! Thank you.[/b][/color][/size]

[quote="Stergiu"][color=darkred]In $ \triangle ABC$ with the orthocenter $ H$ denote the projections $ D\in BC$ , $ E\in CA$ , $ F\in AB$ of $ H$ on

its sidelines and the midpoint $ X$ of $ [HA]$ . Define $ \left\|\begin{array}{c} T\in EX\cap AB \\
\ R\in BE\cap DF\end{array}\right\|$ . Prove that $ TR\perp BC$ .[/color][/quote]

[color=darkblue][b][u]My proof.[/u][/b] Denote the midpoint $ Y$ of $ [HB]$ . Prove easily that  $ AXF$ , $ DYF$ are isosceles ( $ XA \equal{} XF$ , $ YD \equal{} YF$ )

and $ \triangle AXF\stackrel {(*)}{\ \sim\ } \triangle DYF$ . Observe that $ \widehat {FXT}\equiv\widehat {FYR}\equiv\widehat {FDE}$ because $ DEXFY$ is cyclically (in the [b]Euler's circle[/b]) .

Thus $ T$ , $ R$ are omologously in the similarity $ (*)$ , i.e. $ \frac {TF}{TA} \equal{} \frac {RF}{RD}$ what means $ TR\parallel AD$ $ \implies$ $ TR\perp BC$ .[/color][/quote]

[color=red][i]Dear Virgil , your solution is very simple , elementary and very nice  :) ! [/i][/color]

 [u]Babis[/u]

AH meets BC at F.
We see that $ \triangle ADE$ and $ \triangle FBE$  are similar (both are similar with $ \triangle ABC$)
And $ \angle ADT$ = $ \angle MAD$ = $ \angle FBR$
Hence by rotation at $ E$, $ EF$ becomes $ EA$, $ R$ becomes $ T$
(Or by similarity of $ \triangle ADT$ and $ \triangle FBR$)
Thus $ RE/RF\equal{}TE/TA$  then $ TR//AD$
So $ TR$ and $ BC$ are perpendicular $ q.e.d$

[quote="Stergiu"][color=darkred]In $ \triangle ABC$ with the orthocenter $ H$ denote the projections $ D\in BC$ , $ E\in CA$ , $ F\in AB$ of $ H$ on

its sidelines and the midpoint $ X$ of $ [HA]$ . Define $ \left\|\begin{array}{c} T\in EX\cap AB \\
\ R\in BE\cap DF\end{array}\right\|$ . Prove that $ TR\perp BC$ .[/color][/quote]

[color=darkblue][b][u]My proof.[/u][/b] Denote the midpoint $ Y$ of $ [HB]$ . Prove easily that  $ AXF$ , $ DYF$ are isosceles ( $ XA = XF$ , $ YD = YF$ )

and $ \triangle AXF\stackrel {(*)}{\ \sim\ } \triangle DYF$ . Observe that $ \widehat {FXT}\equiv\widehat {FYR}\equiv\widehat {FDE}$ because $ DEXFY$ is cyclically (in the [b]Euler's circle[/b]) .

Thus $ T$ , $ R$ are omologously in the similarity $ (*)$ , i.e. $ \frac {TF}{TA} = \frac {RF}{RD}$ what means $ TR\parallel AD$ $ \implies$ $ TR\perp BC$ .

[b][u]Shortest metrical proof.[/u][/b]

$ \left\|\begin{array}{ccccc} \frac {TF}{TA} = \frac {EF}{EA}\cdot\frac {\sin\widehat {TEF}}{\sin\widehat {TEA}} & = & \frac {a\cos A}{c\cos A}\cdot\frac {\sin [B - (90^{\circ} - C)]}{\sin (90^{\circ} - C)} & \implies & \boxed {\ \frac {TF}{TA} = \frac {a\cos A}{c\cos C}\ } \\
 \\
\frac {RF}{RD} = \frac {BF}{BD}\cdot\frac {\sin\widehat {RBF}}{\sin\widehat {RBD}} & = & \frac {a\cos B}{c\cos B}\cdot\frac {\sin (90^{\circ} - A)}{\sin (90^{\circ} - C)} & \implies & \boxed {\ \frac {RF}{RD} = \frac {a\cos A}{c\cos C}\ } \end{array}\right\|$ $ \implies$ $ \frac {TF}{TA} = \frac {RF}{RD}$ $ \implies$ $ TR\parallel AD$ .

[b][u]Remark.[/u][/b] Prove analogously $ \frac {RB}{RH} = \frac {TB}{TA} = \tan A\tan C$ , i.e. $ TR\parallel AD$ .[/color]

[quote="vittasko"][color=Darkred][b][u]GENERAL PROBLEM.[/u][/b] A triangle $ \bigtriangleup ABC$ is given and let $ \bigtriangleup DEF$ be its Cevian triangle

w.r.t. a point $ P$ inwardly to it. We denote $ Y\in BE\cap DF$ , $ X$ the midpoint $ X$

of the segment $ [PA]$ and let be the point $ Z\equiv AB\cap EX$ . Prove that $ YZ\parallel AD$ .[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Suppose w.l.o.g. $ EC > EA$ . Denote $ \left\|\begin{array}{c} \frac {DB}{m} = \frac {DC}{1} = \frac {a}{m + 1} \\
 \\
\frac {EC}{n} = \frac {EA}{1} = \frac {b}{n + 1} \\
 \\
\frac {FA}{p} = \frac {FB}{1} = \frac {c}{p + 1}\end{array}\right\|\ (\ mnp = 1\ )$ . Observe that ([b]Van Aubel's relation[/b])

 $ \frac {PC}{PF} = \frac {mn + 1}{m}$ , i.e. $ \frac {PC}{mn + 1} = \frac {PF}{m} = \frac {CF}{mn + m + 1}$ . Denote $ T\in EX\cap CF$ . Apply the [b]Menelaus' theorem[/b]

to $ \overline {TXE}/\triangle APC$ and obtain $ \frac {TP}{TC} = \frac {EA}{EC}$ , i.e. $ \frac {TP}{1} = \frac {TC}{n} = \frac {PC}{n - 1}$ $ \implies$ $ TP = \frac {1}{n - 1}\cdot PC$ and $ TF = TP - FP =$

$ \frac {1}{n - 1}\cdot PC - \frac {m}{mn + 1}\cdot PC =$ $ \frac {m + 1}{(n - 1)(m + 1)}\cdot PC$ . Therefore, $ \frac {TF}{TP} = \frac {m + 1}{mn + 1}$ . Apply the [b]Menelaus' theorem[/b]

to $ \overline {TZX}/\triangle AFP$ and obtain $ \frac {ZF}{ZA} = \frac {TF}{TP}$ . Thus, $ \boxed {\ \frac {ZF}{ZA} = \frac {m + 1}{mn + 1}\ }\ (1)$ . Apply an well-known relation

$ \frac {YF}{YD} = \frac {EA}{EC}\cdot\frac {BF}{BD}\cdot\frac {BC}{BA} =$ $ \frac 1n\cdot\frac {\frac {c}{p + 1}}{\frac {ma}{m + 1}}\frac ac =$ $ \frac {m + 1}{mn(p + 1)}$ , i.e. $ \boxed {\ \frac {YF}{YD} = \frac {m + 1}{mn + 1}\ }\ (2)$ .

In conclusion, from the relations $ (1)$ , $ (2)$ obtain $ \frac {ZF}{ZA} = \frac {YF}{YD}$ . i.e. $ YZ\parallel AD$ .[/color]

Sorry to revive, but a proof by angle chasing only..

If $ AH $ meets $ BC $ at $ F $ then in right $ \triangle ADH, DM $ is a median hence $ MD=MH \implies \angle MDH=\angle MHD=\angle BHF=\angle BEF \implies TERD $ is a cyclic quadrilateral thus, $ \angle TRH=\angle TED=\angle AHD \implies TR \parallel AH \implies TR \perp BC $.


If you could be one math operation/symbol, what would you be?
I would be $\pi$ because i never end and everyone would try to calculate me.

[quote="xxreddevilzxx"]$\phi$ looks the coolest.[/quote]

Agreed! $\mathbb{GO} \ \mathbb{PHI} \phi$

[quote="math92"]i also like i because its imaginary[/quote]

narcissist... :P  :D

-1 because it combines $e,\ \pi,\ i$  :D

[quote="xdl615x"]Who cares anyway?[/quote]
it's just for fun

wow talk about spam, do we really need to know whether you haven't made up your mind or not? sheesh, and "who cares" well i think the 5 or so people who have posted their preferance on mathematical operation can answer that..........................

myn: "$!$"  In my opinion its the most exciting mathematical operation!!!!!!!!!!!!!!!  :rotfl:  :rotfl:

[quote="TheVelho"]Yeah, it does look cool too.

This is not maths but is nice: हुई[/quote]
What's this?

[quote="math92"]i looks like a letter in some language other than english :?[/quote]
I agree with you. I don't think it belongs to the European language. Maybe some Asian one...

I just think Capital Sigma looks the coolest and looks the hardest to work with...

$\sum_{i=1}^{\infty}coolness$

[quote="math92"][quote="236factorial"]my class used to think that the variable x was a high school topic  :D[/quote]
lol,  :rotfl:
that's really funny...when did they think this?[/quote]

Beginning of 6th grade, before my class learned that it was called a variable. They still don't know what y is, i don't think.  :roll:

[quote="236factorial"][quote="math92"][quote="236factorial"]my class used to think that the variable x was a high school topic  :D[/quote]
lol,  :rotfl:
that's really funny...when did they think this?[/quote]

They still don't know what y is, i don't think.[/quote]
lol, your class is hilarious

[quote=dragonlouis]$\sum_{}{}$[/quote]

[quote=UnbeatableJJ]$pi$ :)[/quote]


induction
function
logarithms
integration
limit
real analysis
real analysis unsolved

Let $ f(x)\equal{}\sum_{n\equal{}1}^{\infty} a_n/(x\plus{}n^2), \;(x \geq 0)\ ,$ where $ \sum_{n\equal{}1}^{\infty} |a_n|n^{\minus{} \alpha} < \infty$ for some $ \alpha >  2$. Let us assume that for some $ \beta > 1/{\alpha}$, we have $ f(x)\equal{}O(e^{\minus{}x^{\beta}})$ as $ x \rightarrow \infty$. Prove that $ a_n$ is identically $ 0$.

[i]G. Halasz[/i]

If this is true for some $\alpha$, then this is true for $\forall \alpha'>\alpha$, so for $\forall\beta>0$, so conditions $\beta>1/\alpha, \alpha>2$ are not necessary.

Also condition $\sum_{n=1}^{\infty}\frac{|a_n|}{n^{\alpha}}<\infty$ is not necessary, because if exists $\sum_{n=1}^{\infty}\frac{a_n}{n^2}$, then $\exists N>0: \forall n, \frac{|a_n|}{n^2} \leq N$, so $\sum_{n=1}^{\infty}\frac{|a_n|}{n^4}\leq \sum_{n=1}^{\infty}\frac{N}{n^2}<\infty$


calculus
limit
real analysis
real analysis unsolved

[img]http://i173.photobucket.com/albums/w47/ishti10/yo.jpg[/img]

The above problem could not be solved by anyone in my math department. I am admittedly stumped.

We have solved 19 a) but 19 b) is proving to be a hard nut to crack. I hope you guys can offer some help.[/img]

Split the series into blocks $b_{j}=a_{n_{j}}+\dots+a_{n_{j+1}-1}$ such that $|b_{j}|\ge\epsilon>0$ for all $j$. Choose $(q_{j})$ so that $|q_{j}b_{j}|>2\sum_{i<j}|q_{i}b_{i}|$. Observe that $\left|\frac{1}{q_{j}}\sum_{i=1}^{j}q_{j}b_{j}\right|>\frac12 |b_{j}|$. Finally, let $p_{n}$ be equal to $q_{j}$ within the $j$th block.

While I have written my solution ,mlok posted his solution and it is the same as my one.

I am starting to understand this solution.

A couple of points though:
1) How do you know that you may have  all   $|b_{j}| \geq  \epsilon$ ?
2) Also don't I have to ensure that $(p_{n})$ is increasing and unbounded.

Sorry to bother, but-

1) How do you know that you may have all $|b_{j}| \geq  \epsilon$  ?

A Calculus II lecture might go like this: "$a_{n}\not\to 0$ implies  that $\sum a_{n}$ diverges, but the converse is false. False, I'm telling you. I'm really not kidding."

And students would be like this: :what?: :sleeping: :gathering: :whistling:

However, the converse [b]is[/b] true in the following sense: a series is divergent if and only if its terms can be grouped into blocks $b_{k}$ such that $b_{k}\not\to 0$. Proof: let $s_{n}=a_{1}+\dots+a_{n}$. Since $\lim s_{n}$ does not exist, there exists $\epsilon>0$ such that for any $n$ there is $m>n$ such that $|s_{m}-s_{n}|\ge\epsilon$. From this you can build the blocks inductively: let $n_{0}=0$, then apply the above with $n=n_{k}$ and let $n_{k+1}=m$.

As for your second question: there is nothing to stop you from choosing really big values for $q_{j}$ in the above proof.

lol hillarious.

Thanks for the help.
You really must be a fun professor.

Sorry to bother again!

1) How do I prove the existence of such $b_{k}$s?
2) How do I construct the blocks inductively?

I have never dealt with blocks so this is unfamiliar to me physically and conceptually. I tried to solve the following questions but I do believe I have made a mistake along the way. If one of you guys can show me the proper way I could use it to compare it to what I have and learn something new!

1) How do I prove the existence of such $b_{k}$s?
2) How do I construct the blocks inductively?

Thanks in advance!

Sorry to revive this, but I just had to know this.

How do you claim that for ANY n there exists m>n?

Did you mean to use the negation of cauchy convergence?

Do the blocks you constructed cover the entire sequence of a_n?
Also what is $s_{0}$ ?


Let x1, ..., xn be arbitrary real numbers.
Prove that |{(i, j)|1 < |xi - xj| < 2}| \leq n2/4.

Direct application of Turan's theorem. Let us consider a graph G with the vertices in x_1,...,x_n (we can suppose they are distinct) , puting an edge between x_i and x_j if and only if |x_i-x_j| is in (1,2). One can easily check that G does not contain triangles, so it has at most n^2/4 edges and the conclusion follows.


geometry
circumcircle
3D geometry
tetrahedron
sphere
national olympiad

Day I

1. In acute triangle $ABC$ point $O$ is circumcenter, segment $CD$ is a height, point $E$ lies on side $AB$ and point $M$ is a midpoint of $CE$. Line through $M$ perpendicular to $OM$ cuts lines $AC$ and $BC$ respectively in $K$, $L$. Prove that $\frac{LM}{MK}=\frac{AD}{DB}$

2. Positive integer will be called white, if it is equal to $1$ or is a product of even number of primes (not necessarily distinct). Rest of the positive integers will be called black. Determine whether there exists a positive integer which sum of white divisors is equal to sum of black divisors

3. Plane is divided with horizontal and vertical lines into unit squares. Into each square we write a positive integer so that each positive integer appears exactly once. Determine whether it is possible to write numbers in such a way, that each written number is a divisor of a sum of its four neighbours.

Day II

4. Given is an integer $n\geq 1$. Find out the number of possible values of products $k \cdot m$, where $k,m$ are integers satisfying $n^{2}\leq k \leq m \leq (n+1)^{2}$.

5. In tetrahedron $ABCD$ following equalities hold:
$\angle BAC+\angle BDC=\angle ABD+\angle ACD$
$\angle BAD+\angle BCD=\angle ABC+\angle ADC$
Prove that center of sphere circumscribed about ABCD lies on a line through midpoints of $AB$ and $CD$.

6. Sequence $a_{0}, a_{1}, a_{2},...$ is determined by $a_{0}=-1$ and
$a_{n}+\frac{a_{n-1}}{2}+\frac{a_{n-2}}{3}+...+\frac{a_{1}}{n}+\frac{a_{0}}{n+1}=0$ for $n\geq 1$
Prove that $a_{n}>0$ for $n\geq 1$

My comments: I messed up terribly  :blush: . I've done 3 problems first day and 2 problems second day, but instead of checking my solutions I spent 3 hours in vein on problem 5 and hence I omitted a small bug in 4. They gave my 0 points for that problem in result and I ended sixth ex equo with some other guys, but due to my bad results at second round they didn't choose me for IMO  :(  :blush: Still I'll try to get at least 2 points for problem 4 as a bug is minor and easy to repair without changing anything in whole reasoning (this would qualify me for IMO). The final results, after checking all recalls, will be up by 20th May. All in all problems are easy (maybe beside 5, but I'm terrible at stereometry)

links:
1 - http://www.mathlinks.ro/Forum/viewtopic.php?p=821644#821644
2- http://www.mathlinks.ro/Forum/viewtopic.php?p=821648#821648
3- http://www.mathlinks.ro/Forum/viewtopic.php?p=821650#821650
4- http://www.mathlinks.ro/Forum/viewtopic.php?p=821652#821652
5 - http://www.mathlinks.ro/Forum/viewtopic.php?p=821647#821647
6- http://www.mathlinks.ro/Forum/viewtopic.php?p=821653#821653

[quote="Megus"]
 Still I'll try to get at least 2 points for problem 4 as a bug is minor and easy to repair without changing anything in whole reasoning (this would qualify me for IMO). The final results, after checking all recalls, will be up by 20th May.
[/quote]

It is possible to "protest" in Poland?

Yes, but they answer positively really seldom.

Hey good luck with that Megus. I hope you make it to the IMO! :)

[quote="_el_doopa"]well well well, but I think they check your solution many times,because it decide if you go on imo or not. and they couldn't give you 2 points now, because they decided that you have 0.
I'm six ex aequo and i have the best result on second stage so I'm polish member for imo, and if they give you the 2 points I will protest too :P.[/quote]LOL :D Yeah I know what you guys feel. I've been on 7 position at 0.5 points of the Balkan MO team and then I moved up to 6th position for the IMO team, 2 TSTs later ... :)

[quote="_el_doopa"]well well well, but I think they check your solution many times,because it decide if you go on imo or not. and they couldn't give you 2 points now, because they decided that you have 0.
I'm six ex aequo and i have the best result on second stage so I'm polish member for imo, and if they give you the 2 points I will protest too :P.[/quote]

This was posted on 23 Apr 2007.
I am very curious: What happened since then?  Did Megus get his 2 points?


Translate this: κατεβάζω καντήλια

google tells me download sacred lamps... :rotfl:

yea it says download sacred lamps...but im indian i used google translate

Yahoo Babel Fish says I lower vigil candle.

κατεβάζω καντήλια is pronounced "Katibeazo Kanthlia," I think, but I have no idea what it means.

[quote="bbgun34"]how'd the phrase come up?[/quote]

someone visited my blog and his/her computer died of dnkywin's spam

[quote="maybach"][quote="EggyLv.999"]:rotfl:  :rotfl:  :rotfl: this probably has something to do with an insult or the word spam.[/quote]
Moderators, please lock.[/quote]

don't lock it, lock it when i get a working answer  :mad:

hmm.. i got download secret lamps too..
wierd..


During my active years of  tournament chess playing(1979-1982.),i read in some chess books about '[b] chess[/b] correspondence[b] tournaments[/b].[b]'[/b] Just how such
correspondence chess really works? Can anyone participate in the said chess tournament? Please inform me.

                                             Respectfully,
                                            A.Eddington

i might do it. do you have a uscf rating


Substitute each of the capital letters by a different base 6 digit from 0 to 5 to satisfy this cryptarithmetic equation, where $ F$ is nonzero.

$ FIXED \equal{} FEE^2$

[hide]Perfect squares $ \not\equiv2,5\pmod6$ so $ D\ne2,5.$ As $ D\ne E,$ $ D\ne0,3$ either. Hence either $ D \equal{} 1\ \mbox{and}\ E \equal{} 5$ or $ D \equal{} 4\ \mbox{and}\ E \equal{} 2.$ However, $ (5\times6 \plus{} 5)^2\not\equiv5\times6 \plus{} 1\pmod{6^2}.$ Thus $ D \equal{} 4\ \mbox{and}\ E \equal{} 2.$ And testing with $ F \equal{} 1$ immediately gives
\[ 15324\ \equal{} \ 122^2\quad\mbox{(base 6)}\]
:-D[/hide]


Let $A, B\in M_{n}(\mathbb{C})$. Show that if $AB-BA$ is invertible and \[A^{2}+B^{2}=\sqrt{3}(AB-BA)\] then $n$ is a multiple by 6.

Do you mean A,B are matrices with complex elements of order n?

I can prove it for real matrices (it's similar to an IMC problem :D):

We have that $(A+iB)(A-iB) = (AB-BA) \left( \sqrt 3-i \right)$. The $\det$ of this must be real (since $A+iB = \overline{A-iB}$). But $\det (AB-BA)$ is a non-zero real, so $\left( \sqrt 3-i \right)^{n}$ must be real. Since $\sqrt 3-i$ has an argument of $-\frac{\pi}6$, the conclusion follows immediately.

For complex matrices, I have no idea  :maybe:

(A+iB)(A-iB)=(sqrt(3)-i)(AB-BA),
(A-iB)(A+iB)=(sqrt(3)+i)(AB-BA).
We may deduce that (sqrt(3)-i)^n=(sqrt(3)+i)^n
and exp(5*i*Pi*n/3)=1; therefore n is a multiple by 6.
   Can somebody find a such couple A,B for n=6?

[quote="me@home"]$(A+iB)(A-iB) = (AB-BA) \left( \sqrt 3-i \right)$
How do you show this...[/quote]

\begin{eqnarray*}(A+iB)(A-iB) &=& A^{2}-i AB+i BA-i^{2}B^{2}\\ \ &=& (A^{2}+B^{2})-i (AB-BA) \\ \ &=& \left( \sqrt 3-i \right) (AB-BA) \end{eqnarray*}
(keep in mind that we have that $A^{2}+B^{2}= \sqrt 3 \left( AB-BA \right)$)

But my proof works only for real matrices $A,B$.

I've written a stupidity here. See below.

wow completely forgot about that handy given... anyway does distributive law and all that really work with matrices? are they only funky because they are non commutative?

Hi perfect_radio.
My proof works for complex matrices. You can't find a such counterexample in dimension 2.


Hello. Please post here names of youngest perfect scorers or gold medallist  at IMO. Comments would be welcomed . THANKS

Terence Tao got gold medal when he was 13 i guess. When he was 12 he got 40 but only perfect scorers got gold medals. So he has bronze (11),silver(12),gold(13) and he received Fields Medal in 2006.

22 contestants got $ 42$ !! An if one look at the statistics this was the most common result :)
Must have been easy since many hit the ceiling.

Anyway perhaps there are some young people in this [url=http://www.imo-official.org/year_individual_r.asp?year=1987&column=total&order=desc]list[/url]?


i need help solving this proof using vectors:

Given: triangle ABC
D and E are midpoints of AB and AC respectively
F is the midpoint of EC
Segment DF is extended to meet BC extended at P
Prove: CP is half as long as BC

(As a side challenge for our beginning Intermediate students - try solving it without vectors).

[b]
Do we have to use vectors?
[hide]
$CP=DE=\frac{1}{2}BC$[/hide][/b]


algebra
polynomial
complex numbers
superior algebra
superior algebra unsolved

[quote="Xevarion"]I know if $p(x)$ is a polynomial with complex coefficients we can always factor it to $c(x-r_{1})(x-r_{2})\cdots(x-r_{n})$ where the $r_{j}$ are the complex roots.

In the same vein: What is the most general form into which we can factor a polynomial of 2 variables, $p(x, y) = \sum_{j, k}c_{jk}x^{j}y^{k}$?[/quote]

There is no such form for more than one variable.  For one variable over an algebraically closed field (e.g. complex numbers) irreducible polynomials are linear.  For two variables almost every polynomial is irreducible.


Let $S$ be such that $S=\sum_{k=1}^n a_k$ . Manifestly the $a_k$ numbers have to be positive&real so from $AM/GM$ I have $\displaystyle  n \leq S \Longleftrightarrow \frac{n}{S} \leq \frac{n}{n-1 + \frac{S}{n}}$
Because of $\frac{1}{n-1 + x}$ is convexe I write : $\frac{n}{n-1 + S/n} \leq  \sum_{k=1}^n \frac{1}{n-1 +a_k}$
Because of  $\frac{1}{A - e^ x}$ is concave for $e^x < A$,  I write :$ \sum_{k=1}^n \frac{1}{S+1 - a_k} \leq  \frac {n}{S+1 -( \prod_{k=1}^na_k)^{1/n}} = \frac{n}{S}$ , and here you are !

 :cool: I hope I am right! .  Do you follow me ? :cool:

[quote="Diogene"]Because of  $\frac{1}{A - e^ x}$ is concave for $e^x < A$,[/quote]

Alas, bad try. $\frac{1}{A-e^x}$ is convex for $e^x<A$.

  darij

I too made that mistake  :blush:

I think it is very hard to find a simple solution :(

[quote="harazi"]For a very long time I am trying to prove that if $a_1\cdo a_2\cdot...\cdot a_n =1$ then
  $\sum{\frac{1}{1+a_2+...+a_n}}\leq \sum{\frac{1}{n-1+a_1}}$. Does anyone have any solution for this extraordinary problem?[/quote]

It is ambiguous what is inside the $\sum$.   What is the precise statement?
(Maybe one needs to know the Bulgarian olympiad problem; I did not see it.)

[quote="fleeting_guest"][quote="harazi"]For a very long time I am trying to prove that if $a_1\cdo a_2\cdot...\cdot a_n =1$ then
  $\sum{\frac{1}{1+a_2+...+a_n}}\leq \sum{\frac{1}{n-1+a_1}}$. Does anyone have any solution for this extraordinary problem?[/quote]

It is ambiguous what is inside the $\sum$.   What is the precise statement?[/quote]

Well, that's just shorthand notation; the inequality is meant like this:

$\sum_{i=1}^{n}\frac{1}{1+S-a_i}\leq\sum_{i=1}^{n}\frac{1}{n-1+a_i}$,

where $S=\sum_{i=1}^n a_i$.

  darij

A similar ineq:With$a_1,a_2,...,a_n>0: a_1a_2...a_n=1$prove that
  $\sum \frac{1}{n-2+a_i+a_{i+1}} \le \sum \frac{1}{n-1+a_i}$
But I cann't solve it,too.  :(  :blush:

[quote="Mildorf"]I seem to remember proving the result that $a_1a_2\cdots a_n = 1$ implies

\[ \sum_{i=1}^{n} \frac{1}{n-1+a_i} \le 1  \]

So I assume that somewhere on the forum it has been shown that

\[ \sum_{i=1}^n \frac{1}{1+S-a_i} \le 1  \]

Does anyone know where that is?[/quote]
but I don't think it is useful. :)


Here is an interesting problem that I came up with... let me know if it can't be done for some reason...

Let $a$, $b$, and $c$ be nozero real numbers such that their sum is zero.  If the sum of their squares equals their product, find the value of the sum of their reciprocals.