function algebra polynomial vector induction logarithms factorial

[i]The three problems that follow are closely related, but have the property that solving the first one makes the other two much easier and also that seeing either of the second two provides part of the answer to the first.  Therefore, if you would like to maximize difficulty, solve either of the second two before seeing the first, but if you would like to maximize the unity of this thread, solve them in order.  Try also to give bijective proofs if possible; pair them with a generating functions proof if you'd like. [/i]

[hide="Problem 1."] The family of polynomials $ {x \choose k}$ forms a basis for the vector space of polynomials over some field.  Since this vector space comes equipped with a ring structure, it follows that there exist coefficients $ c_{a, b, j}$ such that

$ {x \choose a} {x \choose b} \equal{} \sum_{j \equal{} 0}^{a \plus{} b} c_{a,b,j} {x \choose j}$.

Compute $ c_{a,b,j}$.  (Remember that proving that a polynomial identity holds for non-negative integers is the same thing as proving that it holds identically.) [/hide]

[hide="Problem 2."] Let $ a, b, n$ be non-negative integers.  Show that

$ {n \plus{} a \choose a} {n \plus{} b \choose b} \equal{} \sum_{j \equal{} 0}^{a \plus{} b} ( \minus{} 1)^j {a \plus{} b \minus{} j \choose a \minus{} j, b \minus{} j, j} {n \plus{} a \plus{} b \minus{} j \choose a \plus{} b \minus{} j}$. [/hide]

[hide="Problem 3."] Let $ a, b$ be non-negative integers.  Show that

$ {2a \choose a} {2b \choose b} \equal{} \sum_{j \equal{} 0}^{a \plus{} b} ( \minus{} 4)^j {a \plus{} b \minus{} j \choose a \minus{} j, b \minus{} j, j} {2(a \plus{} b \minus{} j) \choose (a \plus{} b \minus{} j)}$. [/hide]

[hide="The answer (but not the solution) to Problem 1."] In the indexing of the other two problems (and not this one),

$ {x \choose a} {x \choose b} \equal{} \sum_{j\equal{}0}^{a\plus{}b} {a \plus{} b \minus{} j \choose a \minus{} j, b \minus{} j, j } {x \choose a \plus{} b \minus{} j }$. [/hide]

[hide="A solution (but not an answer) to Problems 2, 3."] Problem 2 follows from Problem 1 upon the substitution $ x \equal{} \minus{}n\minus{}1$.  Problem 3 follows from Problem 1 upon the substitution $ x \equal{} \minus{} \frac{1}{2}$.  I don't know how to turn either of these manipulations into combinatorial proofs, which is my real motivation for posting this. [/hide]

For 2:
[hide]
Suppose you want to tile a strip of length $ n \plus{} a \plus{} b$ with some mixture of red, yellow, and green $ 1 \times 1$ squares and blue $ 1 \times 2$ dominoes, subject to the following two constraints:

1. You must use exactly $ n$ green squares.
2. The difference between the number of red squares and the number of yellow squares must be exactly $ a \minus{} b$.

If we use $ j$ dominoes, then we use $ a \minus{} j$ red squares and $ b \minus{} j$ yellow squares, so the number of ways of doing this is exactly the summand (up to the $ ( \minus{} 1)^j$ sign term).

We now define a pairing between the colorings with $ j$ even and $ j$ odd as follows:  Scan your coloring from left to right until you find the first occurrence of either a domino or a yellow square immediately to the right of a red one.  Replace the domino by a RY pair and vice-versa.  The only place this operation is undefined is if you have a coloring with no dominoes and no instances of $ RY$.

In other words, the right hand side is equal to the number of arrangements of $ n$ green tiles, $ a$ red tiles, and $ b$ yellow tiles so no yellow tile appears immediately to the right of a red one.  We can think of this as taking one of the $ \binom{n \plus{} a}{a}$ arrangements of green and red tiles and adding the yellow tiles in.  Since no yellow tile can follow a red tile, we can effectively think of the $ n \plus{} a$ blocks so far placed as consisting of $ n$ larger blocks of the form $ RRR\dots RG$, along with a few rogue red tiles at the end (the number of red tiles in a block could be $ 0$).  This implies that there are $ \binom{n \plus{} b}{b}$ choices for how to arrange the $ b$ yellow tiles and the $ n$ blocks, which gives the left hand side.[/hide]

For 3 (not a full answer, just some thoughts):

[hide]Suppose you want to tile a strip of length $ 2(a\plus{}b)$ with a mixture of red, yellow, and green $ 1 \times 1$ squares and blue, pink, orange, and black $ 1 \times 3$ triominoes, subject to the following two constraints:

1. Exactly half of the tiles used must be green squares.
2. The difference between the number of red squares and the number of yellow squares must be exactly $ a\minus{}b$.

Then you can work out that again (up to sign) the number of tilings with $ j$ triominoes is equal to the right hand side.  If we wanted to imitate our argument for (2), we would scan from left to right until we found either a triomino or a corresponding set of three tiles, replacing one by the other, e.g.

blue triomino $ \leftrightarrow$ RYG
pink triomino $ \leftrightarrow$ GRY
orange triomino $ \leftrightarrow$ YRG
black triomino $ \leftrightarrow$ GYR
(YGR and RGY would also be possibilities for the right hand side here)

The fixed points of this mapping are the tilings with $ a\plus{}b$ green tiles, $ a$ red tiles, and $ b$ blue tiles such that none of the four above patterns occur.  However, I see no obvious reason why this should equal $ \binom{2a}{a}\binom{2b}{b}$.[/hide]

For the sake of completeness (though this is the easiest part):

[hide="To problem 1"]
[color=blue][b]Theorem.[/b] For any nonnegative integers $ n$, $ a$ and $ b$, we have

$ \binom{n}{a}\binom{n}{b}\equal{}\sum_k\binom{n}{k}\cdot\binom{n\minus{}k}{a\minus{}k}\cdot\binom{n\minus{}a}{b\minus{}k}$
$ \equal{}\sum_k\binom{n}{a\plus{}b\minus{}k}\cdot\binom{a\plus{}b\minus{}k}{a\minus{}k,b\minus{}k,k}$.[/color]

Here, we use Knuth's sum convention: Whenever $ v$ is a variable, we use $ \sum_v$ as an abbreviation for $ \sum_{v\in\mathbb{Z}}$. Of course, a sum $ \sum_{v\in\mathbb{Z}}A\left(v\right)$ (where $ A: \mathbb{Z}\to\mathbb{Q}$ is a map) only makes sense if $ A\left(v\right)\equal{}0$ for all but finitely many $ v$. Hence, we can speak of $ \sum_{v}A\left(v\right)$ only if $ A\left(v\right)\equal{}0$ for all but finitely many $ v$. For instance, we may write $ \sum_k\binom{n}{k}\cdot\binom{n\minus{}k}{a\minus{}k}\cdot\binom{n\minus{}a}{b\minus{}k}$ because $ \binom{n}{k}\cdot\binom{n\minus{}k}{a\minus{}k}\cdot\binom{n\minus{}a}{b\minus{}k}\equal{}0$ for all but finitely many $ k$ (namely, for all $ k<0$ and for all $ k>n$, at least (since $ \binom{n}{k}\equal{}0$ for all $ k<0$ and for all $ k>n$)).

[i]Proof of the Theorem.[/i] Let $ N\equal{}\left\{1,2,...,n\right\}$. Then, $ \left|N\right|\equal{}n$, so that by the definition of binomial coefficients,

$ \binom{n}{a}\equal{}\left|\left\{A\in\mathcal{P}\left(N\right)\mid \left|A\right|\equal{}a\right\}\right|$ and
$ \binom{n}{b}\equal{}\left|\left\{B\in\mathcal{P}\left(N\right)\mid \left|B\right|\equal{}b\right\}\right|$.

Hence,

[color=green][b](1)[/b][/color] $ \binom{n}{a}\binom{n}{b}\equal{}\left|\left\{A\in\mathcal{P}\left(N\right)\mid \left|A\right|\equal{}a\right\}\right|\cdot\left|\left\{B\in\mathcal{P}\left(N\right)\mid \left|B\right|\equal{}b\right\}\right|$
$ \equal{}\left|\left\{\left(A,B\right)\in\left(\mathcal{P}\left(N\right)\right)^2\mid \left|A\right|\equal{}a\text{ and }\left|B\right|\equal{}b\right\}\right|$
$ \equal{}\sum_{C\in\mathcal{P}\left(N\right)}\left|\left\{\left(A,B\right)\in\left(\mathcal{P}\left(N\right)\right)^2\mid \left|A\right|\equal{}a\text{ and }\left|B\right|\equal{}b\text{ and }A\cap B\equal{}C\right\}\right|$
$ \equal{}\sum_k\sum_{C\in\mathcal{P}\left(N\right),\ \left|C\right|\equal{}k}\left|\left\{\left(A,B\right)\in\left(\mathcal{P}\left(N\right)\right)^2\mid \left|A\right|\equal{}a\text{ and }\left|B\right|\equal{}b\text{ and }A\cap B\equal{}C\right\}\right|$.

Now, for any integer $ k$ and for any set $ C\in\mathcal{P}\left(N\right)$ with $ \left|C\right|\equal{}k$, there exists a bijection

$ \phi: \left\{\left(A,B\right)\in\left(\mathcal{P}\left(N\right)\right)^2\mid \left|A\right|\equal{}a\text{ and }\left|B\right|\equal{}b\text{ and }A\cap B\equal{}C\right\}$
$ \to\left\{\left(X,Y\right)\in\left(\mathcal{P}\left(N\setminus C\right)\right)^2\mid \left|X\right|\equal{}a\minus{}k\text{ and }\left|Y\right|\equal{}b\minus{}k\text{ and }X\cap Y\equal{}\emptyset\right\}$,

given by $ \phi\left(A,B\right)\equal{}\left(A\setminus C,B\setminus C\right)$ for all $ \left(A,B\right)$ (its inverse $ \phi^{\minus{}1}$ is given by $ \phi^{\minus{}1}\left(X,Y\right)\equal{}\left(X\cup C,Y\cup C\right)$ for all $ \left(X,Y\right)$). Thus,

$ \left|\left\{\left(A,B\right)\in\left(\mathcal{P}\left(N\right)\right)^2\mid \left|A\right|\equal{}a\text{ and }\left|B\right|\equal{}b\text{ and }A\cap B\equal{}C\right\}\right|$
$ \equal{}\left|\left\{\left(X,Y\right)\in\left(\mathcal{P}\left(N\setminus C\right)\right)^2\mid \left|X\right|\equal{}a\minus{}k\text{ and }\left|Y\right|\equal{}b\minus{}k\text{ and }X\cap Y\equal{}\emptyset\right\}\right|$
$ \equal{}\sum_{X\in\mathcal{P}\left(N\setminus C\right),\ \left|X\right|\equal{}a\minus{}k}\left|\left\{Y\in\mathcal{P}\left(N\setminus C\right)\mid \left|Y\right|\equal{}b\minus{}k\text{ and }X\cap Y\equal{}\emptyset\right\}\right|$
$ \equal{}\sum_{X\in\mathcal{P}\left(N\setminus C\right),\ \left|X\right|\equal{}a\minus{}k}\left|\left\{Y\in\mathcal{P}\left(N\setminus\left(X\cup C\right)\right)\mid \left|Y\right|\equal{}b\minus{}k\right\}\right|$
(since $ Y\in\mathcal{P}\left(N\setminus C\right)\text{ and }X\cap Y\equal{}\emptyset$ is equivalent to $ Y\in\mathcal{P}\left(N\setminus\left(X\cup C\right)\right)$)
$ \equal{}\sum_{X\in\mathcal{P}\left(N\setminus C\right),\ \left|X\right|\equal{}a\minus{}k}\binom{\left|N\setminus\left(X\cup C\right)\right|}{b\minus{}k}$
(since $ \left|\left\{Y\in\mathcal{P}\left(N\setminus\left(X\cup C\right)\right)\mid \left|Y\right|\equal{}b\minus{}k\right\}\right|\equal{}\binom{\left|N\setminus\left(X\cup C\right)\right|}{b\minus{}k}$ by the definition of binomial coefficients)
$ \equal{}\sum_{X\in\mathcal{P}\left(N\setminus C\right),\ \left|X\right|\equal{}a\minus{}k}\binom{n\minus{}a}{b\minus{}k}$
(since $ X\in\mathcal{P}\left(N\setminus C\right)$ yields $ X\cap C\equal{}\emptyset$, so that $ \left|X\cup C\right|\equal{}\left|X\right|\plus{}\left|C\right|\equal{}\left(a\minus{}k\right)\plus{}k\equal{}a$ and thus $ \left|N\setminus\left(X\cup C\right)\right|\equal{}\left|N\right|\minus{}\left|X\cup C\right|\equal{}n\minus{}a$)
$ \equal{}\left|\left\{X\in\mathcal{P}\left(N\setminus C\right)\mid \left|X\right|\equal{}a\minus{}k\right\}\right|\cdot\binom{n\minus{}a}{b\minus{}k}$
$ \equal{}\binom{\left|N\setminus C\right|}{a\minus{}k}\cdot\binom{n\minus{}a}{b\minus{}k}$
(since $ \left|\left\{X\in\mathcal{P}\left(N\setminus C\right)\mid \left|X\right|\equal{}a\minus{}k\right\}\right|\equal{}\binom{\left|N\setminus C\right|}{a\minus{}k}$ by the definition of binomial coefficients)
$ \equal{}\binom{n\minus{}k}{a\minus{}k}\cdot\binom{n\minus{}a}{b\minus{}k}$
(since $ \left|N\setminus C\right|\equal{}\left|N\right|\minus{}\left|C\right|\equal{}n\minus{}k$).

Thus, [color=green][b](1)[/b][/color] becomes

$ \binom{n}{a}\binom{n}{b}$
$ \equal{}\sum_k\sum_{C\in\mathcal{P}\left(N\right),\ \left|C\right|\equal{}k}\left|\left\{\left(A,B\right)\in\left(\mathcal{P}\left(N\right)\right)^2\mid \left|A\right|\equal{}a\text{ and }\left|B\right|\equal{}b\text{ and }A\cap B\equal{}C\right\}\right|$
$ \equal{}\sum_k\sum_{C\in\mathcal{P}\left(N\right),\ \left|C\right|\equal{}k}\binom{n\minus{}k}{a\minus{}k}\cdot\binom{n\minus{}a}{b\minus{}k}$
$ \equal{}\sum_k\left|\left\{C\in\mathcal{P}\left(N\right)\mid \left|C\right|\equal{}k\right\}\right|\cdot \binom{n\minus{}k}{a\minus{}k}\cdot\binom{n\minus{}a}{b\minus{}k}$
$ \equal{}\sum_k\binom{\left|N\right|}{k}\cdot \binom{n\minus{}k}{a\minus{}k}\cdot\binom{n\minus{}a}{b\minus{}k}$
(since $ \left|\left\{C\in\mathcal{P}\left(N\right)\mid \left|C\right|\equal{}k\right\}\right|\equal{}\binom{\left|N\right|}{k}$ by the definition of binomial coefficients)
$ \equal{}\sum_k\binom{n}{k}\cdot\binom{n\minus{}k}{a\minus{}k}\cdot\binom{n\minus{}a}{b\minus{}k}$
$ \equal{}\sum_k\frac{n!}{k!\left(n\minus{}k\right)!}\cdot\frac{\left(n\minus{}k\right)!}{\left(a\minus{}k\right)!\left(n\minus{}a\right)!}\cdot\frac{\left(n\minus{}a\right)!}{\left(b\minus{}k\right)!\left(n\minus{}a\minus{}b\plus{}k\right)!}$
$ \equal{}\sum_k\frac{n!}{\left(a\plus{}b\minus{}k\right)!\left(n\minus{}a\minus{}b\plus{}k\right)!}\cdot\frac{\left(a\plus{}b\minus{}k\right)!}{\left(a\minus{}k\right)!\left(b\minus{}k\right)!k!}$
$ \equal{}\sum_k\binom{n}{a\plus{}b\minus{}k}\cdot\binom{a\plus{}b\minus{}k}{a\minus{}k,b\minus{}k,k}$.

This proves the Theorem.[/hide]

Now a question: Is there any proof of

$ \binom{n}{a}\binom{n}{b}\equal{}\sum_k\binom{n}{a\plus{}b\minus{}k}\cdot\binom{a\plus{}b\minus{}k}{a\minus{}k,b\minus{}k,k}$

for all $ n\in\mathbb{R}$ that does not require the "polynomial identity" argument, except for induction over $ a$ and $ b$? Not only cannot I find any combinatorial proof of this kind, I don't see any with Vandermonde and alike. Any ideas?

  darij

I'll unpack your argument a little, since the notation is somewhat dense.

[hide="Briefly..."] Picking two subsets of size $ a, b$ from $ [n]$ is the same thing as picking their union, which has size $ a \plus{} b \minus{} k$ where $ k$ is the size of their intersection, and then partitioning this set into the $ k$ elements in the intersection, the $ a \minus{} k$ elements in the first set only, and the $ b \minus{} k$ elements in the second set only.  This generalizes to an arbitrary product $ {n \choose a_1} {n \choose a_2} ...$. [/hide]

[hide="Now for all n..."] Apply the general binomial and multinomial theorem to $ (1 \plus{} x)^n (1 \plus{} y)^n \equal{} (1 \plus{} x \plus{} y \plus{} xy)^n \equal{} \sum_k {n \choose a \plus{} b \minus{} k} (x \plus{} y \plus{} xy)^{a \plus{} b \minus{} k}$. [/hide]

A proof done by comparing coefficients would be interesting; it implies something nontrivial about [url=http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind]Stirling numbers of the first kind[/url].

[quote="t0rajir0u"][hide="Now for all n..."] Apply the general binomial and multinomial theorem to $ (1 \plus{} x)^n (1 \plus{} y)^n \equal{} (1 \plus{} x \plus{} y \plus{} xy)^n \equal{} \sum_k {n \choose a \plus{} b \minus{} k} (x \plus{} y \plus{} xy)^{a \plus{} b \minus{} k}$. [/hide][/quote]

This is formally correct, but is this a proof? I mean, you [i]define[/i] $ \left(1\plus{}x\right)^n$ as $ \sum_k \binom{n}{k}x^k$, and other than for $ n\in\mathbb{N}$, this does not directly correspond to anything like "multiply $ 1\plus{}x$ with itself $ n$ times". So I don't see a direct reason why $ \left(1\plus{}x\right)^n\left(1\plus{}y\right)^n$ should be equal to $ \left(\left(1\plus{}x\right)\left(1\plus{}y\right)\right)^n$ except of proving it by comparison of coefficients. For $ n\in\mathbb{C}$, you could try some complex analysis, though not too trivial (note that we are working in several variables), and most importantly, this is not better than the polynomial identity argument.

  darij

Well, we can also define $ (1 \plus{} x)^n \equal{} e^{n \ln (1 \plus{} x)}$, but I agree that this is no better than the polynomial argument.  If we write $ {x \choose a} \equal{} \frac {(x)_a}{a!}$, then the coefficient of $ x^m$ on both sides is

$ \sum_{i \equal{} 0}^{m} \frac {s(a, i) s(b, m \minus{} i)}{a! b!} \equal{} \sum_k \frac {s(a \plus{} b \minus{} k, m)}{(a \minus{} k)! (b \minus{} k)! k!}$

or

$ \sum_{i \equal{} 0}^{m} s(a, i) s(b, m \minus{} i) \equal{} \sum_k s(a \plus{} b \minus{} k, m) {a \choose k} {b \choose k} k!$

which is interesting:  $ |s(a, i) s(b, m \minus{} i)|$ counts the number of pairs of permutations of $ [a]$ and permutations of $ [b]$ with a total of $ m$ cycles and $ |s(a \plus{} b \minus{} k, m)| {a \choose k} {b \choose k} k!$ counts the number of pairs of permutations of $ [a \plus{} b \minus{} k]$ with $ m$ cycles and permutations of $ [k]$, although I'm not clear on what the binomial coefficients are for and the coefficients alternate.  It might be easier to apply this approach to Problem 2, since the falling factorials are replaced with rising factorials.


induction
geometry
combinatorics unsolved
combinatorics

Prove that the regions formed by [i]n[/i] circles in the plane, each with one chord, can be colored with three colors such that any neighboring region are colored differently.

[quote="a_vakilian"]each with one chord[/quote]

What does it mean?

Pierre.

I mean that each circle has also one chord(exactly one)

Let's number the colors 0 to 2 and do it by induction. For 1 circle it's obvious.
Let there be k circles in the plane, each with one chord, such that the coloroing of them satisfies the task. Now it's enough to prove that if we add arbitrary (k+1 st) circle, we are able to change colors of some areas to fulfil the given condition. The chord in the k+1 st circle divides its interior into LEFT and RIGHT part. For each A from 0 to 2 in the LEFT part we change color A with A+1 (mod 3), in the RIGHT part we change A with A-1 (mod 3).
Now we should prove that no two neigboring regions have the same color. Because we changed colors of all regions inside the interior of the circle, no two regions which have border on the circle can be the same color. Because we changed it in the LEFT and RIGHT parts in different ways, no two regions which have border on the chord can be the same color.
QED

Oh it is wrong :?
If the chord in $k+1$ circle is on another chore($i$th) and consider that right of $i$th chord is 2 and left part is 0.When you consider $k+1$th chord they become both 2!


Θεωρούμε ένα τυχαίο μετρικό χώρο (X,d) κα έστω το Α να είναι υποσύνολο του X. Η κλειστή θήκη του Α είναι κλειστό σύνολο. Και μάλιστα το ελάχιστο κλειστό υποσύνολο του Α.

[quote="τρικλινο"]Εαν Α={1/n : nεN} ,τοτε η κλειστοτητα του Α ΕΙΝΑΙ ανοικτο η κλειστο συνολο στο R??[/quote]

"Κλειστότητα" το λέμε και όχι "κλειστή θήκη"? Huh...

Anyway, από τη στιγμή που το closure ενός συνόλου $ A$ είναι το μικρότερο [i]κλειστό[/i] σύνολο που περιέχει το $ A$... ;)

Cheerio,

Durandal 1707

Κλειστή Θήκη=Κλειστότητα (closure)


Regular polygon $ ABCDEFGHIJKLMNOPQR$ has side length $ 1$. The area of polygon $ AMC$ can be expressed as $ \frac {\sqrt {3}(1 \minus{} x)(x \plus{} 1)}{2x}$. What is the value of $ 16x^3 \minus{} 12x \plus{} 15$?

What do you mean by "I just put this in"?

Please don't post random problems in the AMC forum unless they are directly relevant to the AMC contests.

Ok, so it's a regular 18-gon.
I THINK it's well known that the area of a regular N-gon with side length S is
$ \frac{N \cdot S^2}{4\tan\left(\frac{\pi}{N}\right)}$

In this case, $ S \equal{} 1, N \equal{} 18$ and the formula works out to
$ \frac{9}{2} \cdot \cot{\frac{\pi}{18}}$

Actually, I'm not sure if there's an easy way to work out $ \cot{\frac{\pi}{18}}$...

[quote="DiscreetFourierTransform"]Ok, so it's a regular 18-gon.
I THINK it's well known that the area of a regular N-gon with side length S is
$ \frac {N \cdot S^2}{4\tan\left(\frac {\pi}{N}\right)}$

In this case, $ S \equal{} 1, N \equal{} 18$ and the formula works out to
$ \frac {9}{2} \cdot \cot{\frac {\pi}{18}}$

Actually, I'm not sure if there's an easy way to work out $ \cot{\frac {\pi}{18}}$...[/quote]But he's asking about the area of triangle $ AMC$... and $ \cot\left(\frac {\pi}{18}\right)$ is the root of a cubic with rational coefficients...

Oh, I missed AMC. I thought he was looking for the whole 18-gon. :)


function
integration
limit
algebra
floor function
real analysis
real analysis unsolved

Consider the function $\displaystyle f:R\to R;f(x)=3+\{x\}(1-\{x\})$
and $F:R\to R ;F(x)=\int_{0}^{x} f(t)dt$
(by {} we understand fractionary part
 by []  we understand floor function
).

a)verify that $f(x+1)=f(x) (\forall )x\in R$
b)proove that f is continous in $x=1$
c)proove that $F(x)=3x+\frac{x^2}{2}-\frac{x^3}{3} (\forall)x\in [0,1)$
d)proove that $3\leq f(x) \leq 4 (\forall)x \in R$
e)proove that any primitive of $f$ is strictly increasing on $R$
f)find $L=\lim_{x\to \infty} F(x)$
g)proove that $(\exists) a\in R$ so that function $G:R\to R,G(x)=F(x)-ax$
   to be periodical and having a period of $1$.

Every sentence behind the other ones can be used in prooving
the current sentence even if it was not prooven.
(ex:i can use points a),b),c) and/or d) in solving point e) ).

So this is a subject i recieved at an exam.I will try to explain what i did for point
e and f and i would like to know if it is or not correct.

Firstly i used point a) and i said that :
I have $x \in R$ and the property $f(x)=f(x-1)$ wich is wrote a little different
from point a.
Now we can use it succesively like this:
$f(x)=f(x-1)=f(x-2)=...=f(x-k)$ until $x-k\in [0,1)$ .
So this prooves that the values of function $f$ for $x \in R$
can be reduced to the values of $f$ for $x\in[0,1)$.
This property we can write $f((-\infty,+\infty))=[0,1)$

I wrote on paper that we can break the interval $[0,x];x\in R$ in a multitude of
intervals of the form $[k,k+1],k\in Z$ so that we can use the above property on it.
so we can say that

$F(x)=[x]\cdot \int_0^{1}\left(3t+\frac{t^2}{2}-\frac{t^3}{3}\right)dt + \int_{0}^{\{x\}} \left( 3t+\frac{t^2}{2}-\frac{t^3}{3} \right)dt =
[x]\cdot \frac{19}{6} + \int_{0}^{\{x\}} \left( 3t+\frac{t^2}{2}-\frac{t^3}{3} \right)dt $

So now we can say that $F$ is strictly incresing on $R$ wich is exactly what
point e) asks.
And also we can easily calculate L from point f) wich is $\infty$.

What i want to ask is if you can help me find if these here things i said are true or
not and help me solve point g).

by your solution.can't we get if $a=\frac{19}{6}$ and g) is true :)

are you sure i'm right in what i'm sayin there ?
and if a=19/6 the term -ax+[x]19\6 =0 so that G is periodic...
but i still may be wrong

so do you think my solution is correct or not ?

$F(x)=[x]\cdot \int_0^{1}\left(3t+\frac{t^2}{2}-\frac{t^3}{3}\right)dt + \int_{0}^{\{x\}} \left( 3t+\frac{t^2}{2}-\frac{t^3}{3} \right)dt =
[x]\cdot \frac{19}{6} + \int_{0}^{\{x\}} \left( 3t+\frac{t^2}{2}-\frac{t^3}{3} \right)dt$
I think this one is true.and all can be sloved.what do you think :)

the curious stuff is that i prooved that $a=\frac{19}{6}$ and the answers to this exercice state that $a=0$
so wich is right and wich is wrong ?

also the second question,is $F(x)$ periodic of period $1$ on $R$


Let $ a,b,c,x,y,z \in R$ satisfy $ (x \plus{} y)c \equal{} (a \plus{} b)z \plus{} \sqrt {6}$.Find the minimum value of $ F \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} x^2 \plus{} y^2 \plus{} z^2 \plus{} ax \plus{} by \plus{} cz$

The problem has been posted before : http://www.mathlinks.ro/viewtopic.php?t=308076
No one has written a solution though yet.

Lagrange multipliers yields some simple equations for the partial derivatives.


geometry
analytic geometry
geometry proposed

in Hyacinthos@yahoogroups.com

Eric Danneels wrote

"the perpendiculars to the angular bisectors through the
corresponding excenters form a triangle A*B*C* perspective with ABC

The perspector has barycentric coordinates

(a/(3a^2 -(b-c)^2 - 2.b.(c+a)) : ....)"

I found the following result (of which the above claim is a corollary):

If A'B'C' is perspective with ABC through P and A' lies on BC, B' lies on CA and C' lies on AB and if A"B"C" is perspective with A'B'C' through Q and A" lies on B'C', B" lies on C'A' and C" lies on A'B', then  A"B"C" is perspective with ABC through R.

[quote="sprmnt21"]If A'B'C' is perspective with ABC through P and A' lies on BC, B' lies on CA and C' lies on AB and if A"B"C" is perspective with A'B'C' through Q and A" lies on B'C', B" lies on C'A' and C" lies on A'B', then  A"B"C" is perspective with ABC through R.[/quote]

Unfortunately, this result is old, more than a century old, I believe. It's the so-called [i]cevian nests theorem[/i], and it was proved in http://www.mathlinks.ro/Forum/viewtopic.php?t=3536 posts #2-#3. Many of us have rediscovered this theorem in their life...

[b]PS.[/b] As for the Hyacinthos message you referred to ([url=http://groups.yahoo.com/group/Hyacinthos/message/11080]Hyacinthos message #11080[/url]), perhaps the most important thing to say is that the triangle A*B*C* is the antimedial triangle of the excentral triangle of triangle ABC. Using this observation, all properties of this triangle mentioned in that message can be proved synthetically.

  darij


Let $ x,y,z \geq 0$ such that $ x \plus{} y \plus{} z \equal{} 2$
Prove that $ (x^2 \plus{} y^2)(y^2 \plus{} z^2)(z^2 \plus{} x^2) \geq 2(x^2y^2 \plus{} y^2z^2\plus{}z^2x^2)^2$

[quote="2424"][quote="caubetoanhoc94"][quote="2424"]Let $ x,y,z \geq 0$ such that $ x \plus{} y \plus{} z \equal{} 2$
Prove that $ (x^2 \plus{} y^2)(y^2 \plus{} z^2)(z^2 \plus{} x^2) \geq 2(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2)^2$[/quote]

We have:
$ (x \plus{} y \plus{} z)^2(x^2 \plus{} y^2)(y^2 \plus{} z^2)(z^2 \plus{} x^2) \ge 8(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2)^2$  :P[/quote]
yes,but I think you can't pove this problem! :P
$ x,y,z \geq 0$ such that $ x \plus{} y \plus{} z \equal{} 2$
prove that $ x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3 \plus{} 2xyz \leq 1$
You can't prove this inequality? :P[/quote]
Why?
After expansion we obtain:
$ (x \plus{} y \plus{} z)^2(x^2 \plus{} y^2)(y^2 \plus{} z^2)(z^2 \plus{} x^2) \ge 8(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2)^2\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(x^6y^2\plus{}2x^5y^3\minus{}3x^4y^4\plus{}2x^5y^2z\plus{}2x^4y^3z\plus{}2x^3y^3z^2\minus{}6x^4y^2z^2)\geq0,$
 which is obviously true. :wink:

[quote="2424"]Let $ x,y,z \geq 0$ such that $ x \plus{} y \plus{} z \equal{} 2$
Prove that   $ x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3 \plus{} 2xyz \leq 1$[/quote]
It's an old problem of hungkhtn. How about the stronger, 2424:
$ x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3 \plus{} \frac{9}{4}xyz \leq 1$
We can solve it by Am-Gm  :)

[quote="nguoivn"][quote="2424"]Let $ x,y,z \geq 0$ such that $ x \plus{} y \plus{} z \equal{} 2$
Prove that   $ x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3 \plus{} 2xyz \leq 1$[/quote]
It's an old problem of hungkhtn. How about the stronger, 2424:
$ x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3 \plus{} \frac {9}{4}xyz \leq 1$
We can solve it by Am-Gm  :)[/quote]

How?

[quote="nguoivn"][quote="2424"]Let $ x,y,z \geq 0$ such that $ x \plus{} y \plus{} z \equal{} 2$
Prove that   $ x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3 \plus{} 2xyz \leq 1$[/quote]
It's an old problem of hungkhtn. How about the stronger, 2424:
$ x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3 \plus{} \frac {9}{4}xyz \leq 1$
We can solve it by Am-Gm  :)[/quote]
If so, $ x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3 \plus{} \frac {179}{72}xyz \leq 1$ also true by AM-GM.


function
search
number theory
modular arithmetic
relatively prime

[hide]All you should have to do is find how many numbers in S are < 10.
[hide]$4: 1, 3, 7, 9,$[/hide]
Then multiply that number by 10
[hide]$4 \times 10 = 40$[b] D[/b] :?: [/hide][/hide]

[hide="Edit: Solution"]
$\underbrace {11\ldots 11} _{n \text { ones}} = \frac{10^n - 1}{9}$ for some integer $n$, which is equal to the number of ones.

$10^n - 1 \equiv 0 \bmod d$ for the divisor $d$ we wish to find.  This means
$10^n \equiv 1 \bmod d$ which can occur for $n = \phi (d)$ by Euler's Theorem for any $(d, 10) = 1$, so we want $d < 100$ such that $(d, 10) = 1$ - in other words, $d$ is not divisible by 2 or by 5.

This gives us $99 - 49 - 19 + 9 = 40$, so our answer is $\boxed{D}$.

(fireforce, even though your method gave the right answer, I don't think your reasoning was right.)
[/hide]

I'm interested in finding out a non-number-theoretic solution to this. One that doesn't involve Euler's function and modulus :P.

-BLt

[quote="thebrokenlight"][quote="t0rajir0u"]But number theory is clearly the correct way to approach the problem.  Why not?[/quote]

For an unfortunately selfish reason - there are a handful of people in Getting Started who haven't an elementary understanding of modulus or $\phi (n)$ yet... including me :P. Out of curiosity, though: could there exist another way to approach this problem without the use of number theory? If not, I'll be quiet... but should number theory-specific problems belong in the Getting Started forums anyhow?

And about the previous solution. There exists only one number belonging to S that is $<10$, that number being 1. I am curious, though, as to how you arrived at your (correct) answer.

-BLt[/quote]

1.  The only ways to approach this problem short of actually checking every number less than 100 all involve modular arithmetic at the very least, and as far as I can tell they are all equivalent to a proof of Fermat's Little Theorem.  So yes, I don't think this belongs in Getting Started.

2.  Not so - $3 | 111, 7 | 111111, 9 | 111111111$.  Try it on your calculator.

The reason his method arrived at the correct answer is that between the numbers $10k + 1$ and $10k + 10$ there are exactly four numbers that are not divisible by 2 or 5, so he simply counted the ones between $1$ and $10$ and then multiplied, which gives the total number of numbers in the range not divisible by 2 or 5.

[quote="t0rajir0u"]2.  Not so - $3 | 111, 7 | 111111, 9 | 111111111$.  Try it on your calculator.[/quote]

Ahhh! Duh!! I originally misinterpreted what he meant. My bad. Thanks.

-BLt

[quote="t0rajir0u"]...so we want $d < 100$ such that $(d, 10) = 1$ - in other words, $d$ is not divisible by 2 or by 5....[/quote]

$\phi(100)$ is the number of integers less than or equal to 100 that are relatively prime to it..... $\phi(100) = 100\left(1-\frac{1}{2}\right)\left( 1-\frac{1}{5}\right)=40$

[quote="boxedexe"][quote="236factorial"]Let S be the set of positive integers that divide at least one of the numbers 1, 11, 111, 1111, ... for example, 3 is in S since 3 divides 111. The number of elements in S that are less than 100 is

a) 22
b) 28
c) 34
d) 40
e) 48[/quote]

Where can I get more problems of this type?

Masoud Zargar[/quote]

Do a google search on the source provided.


Find all positive integers $ n$ such that $ n^2\mid 3^n\minus{}1$

There are infinetely many solutions.
If $ n_0^2|3^{n_0}\minus{}1$ and $ p|3^{n_0}\minus{}1, p\not |n_0$, then for $ n_1\equal{}n_0p$ we get $ n_1^2|3^{n_1}\minus{}1$.
$ n_0\equal{}1\to n_1\equal{}2\to n_2\equal{}4\to n_3\equal{}20\to ...$


I hope I'm allowed to post one of the probles we did in the trig/complex number class.

Find all the roots of $z^{9}+z^{6}+z^{3}+1=0$.


algebra polynomial Vieta function induction

I'm stuck on these algebra problems... Could someone point in me in the right direction?

[b]1. Show that if $ P(x) \equal{} a_0x^n \plus{} a_1x^{n \minus{} 1} \plus{} ... \plus{} a_n$ and $ Q(x) \equal{} a_nx^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} ... \plus{} a_0$, and furthermore if $ a_0, a_n$ don't equal 0, then P(x)=0 if and only if $ P(\frac {1}{x}) \equal{} 0$. What is the relation between P(x) and Q(x)?[/b]

When I look at the problem, I see that the coefficients are flipped and I think of Vieta's. How do you spell Vieta by the way? I've seen it spelled like this before: Viete

[b]2. We say that P(x)|Q(x) if there exists some R(x) such that P(x)R(x)=Q(x). Show that if P(x)|Q(x) and P(x)|R(x), then P(x)|aQ(x)+bR(x), where a and b are integers.[/b]

What I don't understand about this problem is that that statement holds true for numbers in general. Am I supposed to do something differently just because they are functions?

[b]3. Suppose that P is a 10th degree polynomial such that P(0)=P(1)=P(2)=P(3)...=P(9)=1 and P(10) equals 1. Find P(11). Can you generalize this?[/b]

Exactly what am I supposed to generalize? Will it become clear if I can find P(11)? I know that if I plug in any integer from 0 to 10 in P, I get 1. Should I just set up an 11 line system of linear equations using what I know and try to solve for the coefficients in the polynomial?

[b]4. For what x is P(x) equal to the sum of the roots of the polynomial?[/b]

One idea I had was to have every term except the coefficient of the term raised to the second highest power somehow be eliminated so I'm left using Vieta's. I had no luck with that idea.

[b]5. Show that x-1|$ x^n$-1 and find $ \frac {x^n \minus{} 1}{x \minus{} 1}$. [/b]

When we use long division (or synthetic I think) to start to divide this out, we *see* that x-1 will divide into it (try it). Is there a more rigorous way to prove this?

[b]6. When does x+1|$ x^n$-1?[/b]

I think that if n is a positive even integer, I can always factor the difference of two squares until I see that two factors are (x+1) and (x-1). Can I prove that this will always be the case? Is this even the only case when x+1|$ x^n$-1?

[b]7. Can x-r|$ x^n$-1?[/b]

[b]8. When does $ x^a$-1|$ x^b$-1?[/b]

[b]9. If a monic polynomial has only real roots, and the $ x^{n \minus{} 1}$ term is zero, what is true about the sign of the $ x^{n \minus{} 2}$ coefficient?[/b]

I know that the sum of the roots is zero. What do I know about the plus or minus sum of the roots taken two at a time?

[hide="1"]
In th fist one isn't it $ P(x) \equal{} 0$ if and only if $ Q(\frac {1}{x}) \equal{} 0$?? I think it is.

... from $ Q(\frac {1}{x}) \equal{} 0$ to $ P(x) \equal{} 0$ it is easy, only evaluate $ Q(\frac {1}{x})$ and work with it

... from $ P(x) \equal{} 0$ to $ Q(\frac {1}{x}) \equal{} 0$, introduce $ x \equal{} \frac {1}{y}$ and work with it

 :)
[/hide]
[hide="2"]
The idea is the same as for integers, express $ Q(x)$ and $ R(x)$ in therms of $ P(x)$ and work with it

 :)
[/hide]
[hide="3"]
How many roots has 10th degree polynomial?
Can you somehow "transform" the "one" into a "zero"??

I think you have to generalize for a polinomial of degree $ n > 1$ such that $ P(i) \equal{} k$ for $ n \plus{} 1$ different values of $ i$, and find $ P(n \plus{} 1)$

 :)
[/hide]
[hide="5"]
I suppose you've just start learning about the theme, so in order to understand and get familiar with the behavor of expressions like this, I will suggest you the explorative solution: try small values of $ n$ and look up for a relation, then you can use the definition in $ 2$ to prove it, or induction if you know it. (then try to explain yourself the relation)
This may not be the most elegant solution but it helps to profoundly understand what you are doing.

 :)
[/hide]
[hide="7"]
If you know Factor 's Theorem, use it.
If not use the definition in $ 2$ and pay attention to the independent therms.

 :)
[/hide]
[hide="8"]
Similar to $ 7$, but now the independent therm does not deserve the main attention.
Also try to use, or derive, a generalization of $ 5$.

 :)
[/hide]
[hide="9"]
Try to use the fact that you already know, the sum of the roots is zero.

 :)
[/hide]

I hope this helps you.
I know this doesn't have to due with me but as a friendly suggestion: try to exhaust your ideas, as crazy as they may be, before posting a problem you must solve (like your homework, ...)

Good luck!!

 :)


Evaluate $(x^x)^{(x^x)}$ at $x = 2$.

$\textbf{(A)} \ 16 \qquad \textbf{(B)} \ 64 \qquad \textbf{(C)} \ 256 \qquad \textbf{(D)} \ 1024 \qquad \textbf{(E)} \ 65,536$

Or, if you know your power of 2's by heart  :P :
$(2^2)^{(2^2)}=2^{2 \cdot 4}=256 \leadsto \boxed{C}$


Suppose that I have 6 different books, 2 of which are math books.  In how many ways can I stack my 6 books on a shelf if I do not want the math books to be next to each other?

since it says 6 different books, they are all distinguishable.

Call 1st math book A
2nd math book B
other 4 books C

you can have
ACB
BCA
and thats all.

But for C there are 24 combos to arrange the 4 books that make up C

so 2*24=48


LaTeX
number theory unsolved
number theory

Prove that there are infinitely many composite positive integers $n$ with the property : $n\mid 3^{n-1}-2^{n-1}$. ;)

:idea: well you know  p|a^{p-1}-1(fermat) if (a,p)=1 and p prime,so if n is prime and n>3  then we have  n|(3^{n-1}-1-(2^{n-1}-1))  and then it's easy  :D  (and by the way can you teach me how to write math formulas like fractions and powers couse i'm new here  and i couldn't figure it out  :oops:  pls write me a message when you read this  :?

The problem asks for composite $n$ not $n$ prime which is an easy solution...Now about the other think, check out some latex manual...

I remember posting this a very long time ago (and not being able to solve it :blush: ).

Anyway, just take all $n$ of the form $3^{2^k}-2^{2^k}$. It's easy to prove that it works for all $k$.


topology
analytic geometry
function
real analysis
advanced fields
advanced fields unsolved

Let $ X$ be a smooth manifold of dimension at least 1. Let $ w_{p}$ be a nonzero element of $ T^{*}_{p}X$ - cotangent space at $ x$ for some $ p$ in $ X$.

a). Construct a smooth 1-form $ w$ on $ X$ whose value at $ p$ is $ w_{p}$.

b). Can we always construct $ w$ in part a) to be closed. Proof or counterexexample.

For a) consider a coordinate patch around $ p$ and use a smooth function supported on the patch.

Not sure about b). Something to try: let $ M$ be the Mobius band, i.e. $ [-1,1]^{2}$ with vertical edges glued with opposite orientation. Let $ p=(0,0)$ and $ w_{p}=dy$. Is this a counterexample?   :maybe:

mlok, I do not understand why this should be a counterexample. However, I think you can do this for arbitrary $ X,p,w_{p}$: choose coordinates $ (x_{i})$ around $ p$, write $ w_{p}=\sum a_{i}\, dx_{i}$ and choose a smooth global function $ f$ which coincides locally with $ \sum a_{i}x_{i}$. Then, $ (df)_{p}=w_{p}$. owk


$a ,b ,c$ positives réels prove that
$\sqrt\frac{2a}{a+b}+\sqrt\frac{2b}{b+c}+\sqrt\frac{2c}{c+a}\leq3$

$\sum \sqrt\frac{2a}{a+b} \leq \sum \sqrt\frac{2a}{b+c}$ by rearrangement

Fix  a+b+c = 1, then required to show $\sum \sqrt{\frac{a}{1-a}} \le_? \sqrt{\frac{9}{2}}$

By jensen, $\sum \sqrt\frac{a}{1-a} \le 3\sqrt{\frac{1/3}{1-(1/3)}} = \sqrt{\frac{9}{2}}$ as desired

Folosind media aritmetică - media geometrică inégalité,

$ \sum_{cyc}{\sqrt {\frac {2a}{a \plus{} b}}}\leq\sum_{cyc}{\left[\frac {3(c \plus{} a)}{4(a \plus{} b \plus{} c)} \plus{} \frac {2a(a \plus{} b \plus{} c)}{3(c \plus{} a)(a \plus{} b)}\right]}$

$ \equal{} \frac {17}{6} \plus{} \frac {4abc}{3(b \plus{} c)(c \plus{} a)(a \plus{} b)}\leq 3.$

A se vedea : http://www.mathlinks.ro/viewtopic.php?&t=125084


x different y

Be $x+y=x.y$
$Find$  $the$ $minor$ $natural$ $value$    to        $x.y$

[quote="chess64"]Do you mean: If $x \ne y$, and $x+y=xy$, then find the minimum value of $xy$ for $x,y \in \mathbb{N}$? If so, then what is your definition of $\mathbb{N}$? Should we include $0$ or not?[/quote]

$xy$ non consider 0, x,y E $R$

[quote="cezar lupu"][quote="felipesa"]x different y

Be $x+y=x.y$
$Find$  $the$ $minor$ $natural$ $value$           $x.y$[/quote]

Felippesa,let's consider that $x,y>0$ and from $x+y=x\cdot y $ we have $\frac{1}{x}+\frac{1}{y}=1$ so from
AM-GM inequality we have $1=\frac{1}{x}+\frac{1}{y}\geq\frac{2}{\sqrt{x\cdot y}}$ so the minimal value of $xy$ is $4$ ;)[/quote]

Nice work, but note : $x$is different from $y$,  and $x+y=xy=4$ for $x=y=2$

[quote="prototypex24"]I think there is no solution.

$y=\frac{x}{x-1}$ so $xy=\frac{x^2}{x-1}$ Since $x^2$ is not divisible by $x-1$ there are no natural values[/quote]

there are no natural value for x and y( x,y E R), but xy is a natural  value

[quote]Nice work, but note : $x$is different from $y$,  and $x+y=xy=4$ for $x=y=2$[/quote]
ok then
[quote]AM-GM inequality we have $1=\frac{1}{x}+\frac{1}{y}\geq\frac{2}{\sqrt{x\cdot y}}$ so the minimal value of xy is 4[/quote]
the min xy=5

gunes is right, solving the system

x+y=5
xy=5

x=$\frac{5+\sqrt5}{2}$
y=$\frac{5-\sqrt5}{2}$

Can be switched.

The system of x+y=3 and xy=3 has imaginary answers, and if xy is any natural number below 3, the answer is also imaginary.

Just make use of quadratic equation

Setting $x+y$=$xy$=$p$ where $x$ and $y$ are the roots for the quadratic equation $t^2-pt+p=0$ .Using discriminant ,

$b^2-4ac$=$p^2-4p$=$p(p-4)>0$ .From this we can get $p>4$ and hence the min natural value for $xy$=$p$=$5$  :D


[quote="duc 95"]Try to solve it!!This is very fun[/quote]

what conditions of $ n$ ?

[quote="duc 95"]Try to solve it!!This is very fun[/quote]

if $ n\geq 1$

Notes  $ \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\le 1$

we have  $ \sqrt [n]{\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}}\ge\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}$

we only prove

$ \frac {a^3 \plus{} b^3 \plus{} c^3}{3abc} \plus{} 2\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\geq 3$

The following inequality is equivalent to

$ (b^2 \minus{} ab \plus{} a^2 \minus{} bc \minus{} ca \plus{} c^2)(b^3 \plus{} b^2a \plus{} b^2c \minus{} 6abc \plus{} ba^2 \plus{} bc^2 \plus{} ca^2 \plus{} c^2a \plus{} a^3 \plus{} c^3)\geq 0$

from AM-GM the inequality holds.

since $ \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\le 1$ then  $ n \ge1$ we have $ \sqrt [n]{\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}} \ge \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}$

Thus it sufficent to show the inequality when $ n \equal{} 1$

Or ,

$ \frac {a^3 \plus{} b^3 \plus{} c^3}{3abc} \minus{} 1 \plus{} \frac {2(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2} \minus{} 2 \ge 0$
$ \sum_{cyc}\frac {(a \plus{} b \plus{} c)(a \minus{} b)^2}{6abc} \minus{} \sum_{cyc}\frac {(a \minus{} b)^2}{a^2 \plus{} b^2 \plus{} c^2} \ge 0$
$ \sum_{cyc}\frac {(a \minus{} b)^2((a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) \minus{} 6abc)}{(a^2 \plus{} b^2 \plus{} c^2)6abc}\ge 0$

which is true since $ (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) \ge 9abc > 6abc$

P/s : sorry hedeng i don't seen your post  :blush:

\[ a,b,c > 0 \\
\frac {{a^3 \plus{} b^3 \plus{} c^3 }}{{3abc}} \plus{} 2\sqrt [n]{{\frac {{ab \plus{} bc \plus{} ca}}{{a^2 \plus{} b^2 \plus{} c^2 }}}} \ge 3.\]
if $ n \ge 1$, because $ \frac {{ab \plus{} bc \plus{} ca}}{{a^2 \plus{} b^2 \plus{} c^2 }} \le 1$, we have

$ \sqrt [n]{{\frac {{ab \plus{} bc \plus{} ca}}{{a^2 \plus{} b^2 \plus{} c^2 }}}} \ge \frac {{ab \plus{} bc \plus{} ca}}{{a^2 \plus{} b^2 \plus{} c^2 }},$

so we just need to prove
\[ \frac {{a^3 \plus{} b^3 \plus{} c^3 }}{{3abc}} \plus{} 2 \cdot \frac {{ab \plus{} bc \plus{} ca}}{{a^2 \plus{} b^2 \plus{} c^2 }} \ge 3\]
that's very easy by SOS...

you so fast...  :lol:

[quote="duc 95"]Can you solve it only by AM-GM[/quote]

3#'s used AM-GM at last...

I like using simple inequalities and I always try to use ithem


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a group, and $ N, K$ normal subgroups of $ G$ such that $ N \subseteq K$. Prove that $ \overline{K}\equal{}K/N$ is normal in $ \overline{G}\equal{}G/N$ and that $ G/K \cong \overline{G}/\overline{K}$.
Attempt
Take any $ k\plus{}N \in K/N, g_1 \plus{}N, g_2\plus{}N$, and show that $ (g_1 \plus{}N)(k\plus{}N)(g_2\plus{}N)^{\minus{}1} \in K/N$. This seems that it is trivially true. Is that correct? For the second part, use the third isomorphism theorem, and it follows from this?

Thanks in advance.

[quote="mingcai6172"]Let $ G$ be a group, and $ N, K$ normal subgroups of $ G$ such that $ N \subseteq K$. Prove that $ \overline{K} \equal{} K/N$ is normal in $ \overline{G} \equal{} G/N$ and that $ G/K \cong \overline{G}/\overline{K}$.
Attempt
Take any $ k \plus{} N \in K/N, g_1 \plus{} N, g_2 \plus{} N$, and show that $ (g_1 \plus{} N)(k \plus{} N)(g_2 \plus{} N)^{ \minus{} 1} \in K/N$. This seems that it is trivially true. Is that correct? For the second part, use the third isomorphism theorem, and it follows from this?

Thanks in advance.[/quote]

The first part is obviously trivial. Anyway you cannot use the third isomorphism theorem to prove the third isomorphism theorem... Anyway it's not difficult. Find an homomorphism between $ \overline{G}$ and $ G/K$ with kernel $ \overline{K}$.


which reacts faster $HC(CH_{3})(Br)CH((CH_{3})_{2})$ or $C(Br)((CH_{3})_{2})CH_{3}$ via $E2$ in Alcholoic KOH and why?

The second alkyl halide should react faster because 1) the hydrogen to be eliminated is from a primary carbon, and 2) the base, because it is solvated by etanol molecules (and so it is a very large species), can attack more easily a primary carbon than a secondary one.

but the first one forms a trisubstituted more stable alkene wjile the second gives less stable disubstituted alkne.....

That's right, but the second one reacts faster.

but how  do you know which factor outweighs which  :?:


Define the hyperfactorial $n^{!}$ for integers $n\geq1$ as follows:

$n^{!}=\displaystyle\prod_{k=1}^n k!$

For how many positive integers n is it true that $(n!)!=n^{!}$ ?

[hide="Solution"]
There are only two solutions:

$1^! = 1!!$

$2^! = 2!1! = 2!!$

Proof:

If $n = 3$ then the statement $n^! < (n!)!$ is true: (3^! = 3!*2!*1! = 24, (3!)! = 720)

Now if we assume that for all $k$, the statement $k^! < (k!)!$ is true, it follows that:

$k^! < (k!)!$

$k^!*(k+1)! < (k!)!*(k+1)!$

$(k+1)^! < (k!)!*(k+1)! < ((k+1)!)!$

Which means, by mathematical induction that for $n>3$, $(n+1)^! > ((n+1)!!$, so the only solutions are the ones above.[/hide]

[hide="addition to the proof"]
In my last post I somewhat excluded the justification for the line $(k!)!*(k+1)! < ((k+1)!)!$, for $k>3$ because I found it somewhat intuivitve at first, but I will prove it here:

From the statement that $((n+1)!-1)((n+1)!-2)\dotsm((n!)!+2)((n!)!+1) > 1$, it follows that

$\frac{(n+1)!((n+1)!-1)((n+1)!-2)\dotsm((n!)!+2)((n!)!+1)}{(n+1)!} > 1$

$\frac{(n+1)!((n+1)!-1)((n+1)!-2)\dotsm4*3*2}{(n+1)!(n!)!} > 1$

$(n+1)!((n+1)!-1)((n+1)!-2)\dotsm4*3*2 > (n+1)!(n!)!$

$((n+1)!)! > (n+1)!(n!)!$

Which completes the proof.[/hide]

Ok, that's a valid solution (from what I see). That's good you provided the second proof, because I noticed that it wasn't entirely intuitive in the first proof. The second proof was a bit more difficult to follow, but it seems correct.
Do you like the way I defined the hyperfactorial? That is my original idea.


inequalities
calculus
calculus computations

Given the inequality

$\sum_{m=-\infty}^{\infty}\vert v_{m}^{n+1}\vert^{2}+\vert v_{m}^{n}\vert^{2}+a\lambda (v_{m}^{n+1}v_{m+1}^{n}-v_{m+1}^{n+1}v_{m}^{n})=\sum_{m=-\infty}^{\infty}\vert v_{m}^{n}\vert^{2}+\vert v_{m}^{n-1}\vert^{2}+a\lambda (v_{m}^{n}v_{m+1}^{n-1}-v_{m+1}^{n}v_{m}^{n-1})$

Show it's stable for $\vert a\lambda \vert < 1$

I was told to use the Cauchy-Schwarz Inequality, but I'm stuck. Please help. Thanks in advanced.

First, I am not sure what those $v's$ mean. Also, could you explain what "stable" means in this context? Finally, I don't see any inequality. Am I missing something?


geometry
analytic geometry
3D geometry
tetrahedron
octahedron
prism
pyramid

Do you know any tips for 3d geometry problems?

Some things I have difficulty with are drawing diagrams, finding ways to change the problems into 2d geometry problems, and working with 3d coordinate geometry and drawing out a diagram with the three axes.

[hide="Specific Questions"]
To draw a cube, you can draw a square, then draw one overlapping it to the right and up a bit. Then you connect the vertices. How would you go about drawing a tetrahedron or an octahedron? How do you go about drawing and working with any of these on a 3d coordinate system?

When is it a good idea to try drawing top views or side views of a 3d figure?
[/hide]

I actually never draw anything when doing 3d geometry; it always gets messed up and I can visualize it better in my mind.

I draw 3d geo...
Only basic stuff like prisms and tetrahedrons and the such.
For other stuff I try to do it in my head, because in those problems they basically only ask you something like how many edges are there or something.

On the problems where it asks you about lines connecting faces of tetrahedrons or the sort, I just draw one face and try making lines that are only specific to the problem.

[quote="aufha"]How do you draw a tetrahedron?[/quote]

You can draw a quick sketch (assuming you're a good artist); otherwise you'll just have to know the properties (most of the time they don't even help) and solve it from there.

[quote="aufha"]How do you draw a tetrahedron?[/quote]

A tetrahedron is a pyramid with a triangular base.

[quote="happyme"]A tetrahedron is a pyramid with a triangular base.[/quote]

I think he might have meant a polyhedron in general. Those are often complicated to draw if not to understand.

[quote="aufha"]How do you draw a tetrahedron?[/quote]

Honestly, I know how to draw a tetrahedron (not a very good one) but it didn't help much as for other polyhedron.

[hide="Drawing tetrahedron"]
Start with an equilateral triangle. On the right side of the triangle, pick a point and make two lines by connecting the top vertex of an initial equilateral triangle to that point and the bottom right vertex of an initial equilateral triangle to that point. Then, make a dotted line from bottom left vertex of an initial equilateral triangle to that point. There, you have four faces of equilateral triangle, or tetrahedron. [/hide]

I don't disagree with too much with Temperal but how sometimes, you can visualize better with mind. I wouldn't go far as to "never" though. There are instances when drawing polyhedrons help to see where to start. That brings up to the hidden question. There really isn't any good tip to see when you should look at side view or from the top view.  That depends a lot on the problems and on the contestants. I say just do many problems.

And with 3D, just remember that 90% of them are actually 2D geometry problems so yeah, it is just matter of getting your work started.


vector
factorial
number theory solved
number theory

It smells like vectors or at least complex numbers, but i dont know, maybe ill try it later

let x,y,z be positive integers.and we know that xy= z 2 +1

prove that there exists nonnegative integers a, b, c,d such that
x=a 2 +b 2 and y=c 2 + d 2 and z=ac+ bd.

(a^2+b^2)(c^2+d^2)=(ac+bd)^2+1
(ad-bc)^2=1
ad-bc=1 or -1
For example, if you take ad-bc=1
You may take, b=md-1, c=nd+1, then
ad-(md-1)(nd+1)=1
ad-(mnd-nd+md-1)=1
a=mn-n+m
Then for m,n E N, we have
a=mn-n+m
b=md-1
c=nd+1
d=d
as a particular solution, so solution exists. Can I prove it like this?

I don't have time for all the details right now, but it can b shown by using the fact that $Z[i]$ is factorial (i.e. every element can be uniquely written as the product of finitely many prime elements in $Z[i]$, without considering the invertible elements in $Z[i]$). We just write $z^2+1=(z+i)(z-i)=xy$, then we decompose $x,y$ in their prime factors and when we write $x=a^2+b^2=(a+bi)(a-bi),\ y=c^2+d^2=(c+di)(c-di)$ in various ways there must be a way of choosing the factors of $a+bi,\ c+di$ in such a way that $(a-bi)(c+di)=z+i$, and we're done$.

Sorry if it's not clear and/or wrong :D. I'm not very sure about it, and I know it's not really asolution, but I think it's the right track.

that's sort of dodgy ...
 but if we can prove this: all numbers of the form k^2+1 do not have any prime factors of the form 4m+3 (which i dont know how to) we are done bcoz: xy is the sum of 2 squares and have no factors of the form 4m+3, so each x and y dont have any such factors and can thus be expressed as the sum of 2 squares and the conclusion follows.
the lemma might be proven using gaussian integers but i dont really c how: we know (z+i)(z-i)/4k+3 is element of Z[i], where z is some integer. so 4k+3 divides z+i or 4k+3 divides z-i. i dont c how that gives us a contradiction ...

oh actually i have a way of proving the lemma now:
if 4k+3 | z+i, let z+i=(4k+3)(a+bi)=a(4k+3)+i*b(4k+3), so b(4k+3)=1, and since b and a are integers we need b=-1 or +1, 4k+3 = +1 or -1, contradiction.
we have a similar contradiction for 4k+3 | z-i and we are done i fink  :P

That's very well-known. I assumed it known when I thought about what I've written in my previous post.

Let's assume $p$ is a prime of the form $4k+3$ s.t. $p|a^2+b^2$ Let's assume now that $p$ doesn't divide $a$. In that case $b^2\equiv -a^2\ (mod\ p)\Rightarrow (\frac{-a^2}p)=1\Rightarrow (\frac{-1}p)=1\Rightarrow (-1)^{\frac{p-1}2}=1\Rightarrow p$ has form $4k+1$, which is false, so we have a contradiction. This shows that if a prime of the form $4k+3$ divides $a^2+b^2$ then it divides $a,\ b$ and by applying this to $a=z,\ b=1$ we reach the conclusion that $z^2+1$ can have no prime factor of the form $4k+3$, because it would have to divide $1$.

Sorry! I was typing my mssage while you were typing yours, and yours was shorter, soyou beat me to it :).

Sorry...
I dont think it is complete.We have only shown that x and y can be expressed as sums of two squares(e.g. x=a 2 +b 2 ,
y=c 2 +d 2 ) but I cant understand why z=ac+bd.
 :?  :?

sorry i'm such an idiot.
proving z=ac+bd is obviously equivalant to proving ad-bc = 1 or -1. i have no idea how to do this ...
maybe we could use the construction for expressing a number as the sum of  2 squares but that wouldnt be very nice and prolly wouldnt work  :?

I still think that what vinoth described as "dodgy" (:D) works, but, again, I have no time to write the details. We have to use the fact that primes in $Z[i]$ are numbers of the form $a+bi$, where $a^2+b^2$ is a prime in $Z$ (of the form $4k+1$). We use the fact that $z^2+1$ only has prime factors (in $Z$) of the form $4k+1$, so the decomposition of $x,y$ (in $Z[i]$) only contains primes of the form stated above.

I'm sorry if what I'm trying to do isn't clear, but I'll be back on the topic (I'm during my computer science class right now :)).

it is quite easy to show that one can choose a,b, such that a+bi and a-bi have no common divisors(suppose they have one and then take that divisor which is 1 mod 4, factor it and twist the factors). Since z+i and z-i have no common divisor, one of them must be divisible by a+bi, the quotient of them has no common divisor with a-bi, thus must divide y. Let this quotient be c-di, then (c-di)(c+di)=y, so we're done, because in (a+bi)(c-di)=z+i.

Damn! I forgot to get back on this topic :D. Anyway, this is pretty much what I was referring to, but I would've said it less efficiently, I'm sure :).


Hi, I just wanted to know if anyone else on this forum speaks tamil?

Hello..
I live in Tamil Nadu and can understand some amount of Tamil..
Many of the others on this forum who are shifting to IIT Madras soon will be speaking tamil in another few months as well  :D

It's his abbreviated user name. :D
How many tamils are there in this forum?

I'm a tamil person who lives in VA and found out about AoPS recently! Most of my relatives live in tamil nadu too. :D

last yr tamils use to come on aops in vast nos.
m not sure abt this yr  but i think the size of tamil family on aops is growing as fast as indian population :D
though conversations in tamil seems drastically reduced :)

I'm Tamil too.
There a lot of Tamilians here.
Keyree10, Phinphinity, Agr 94 math, beffudlers, srivatsan, ram, thealchemist, sanjith, chapplika, chemrock, chronoz, adharshraghavan, unearthly, saikrishna, harekrishna, louis123,madness,valeriummaximum, ashwath.rabindranath .
Our mod, pardesi is also a Tamilian.
And lots more. :) Most are from Chennai.
Most of the seniors are not active now.If you see previous posts, it will be mostly by them. :)

A wonderful comprehensive list,good for surveys :)

Me tooo!!
But I think that list is more than enough!!

I am tamil but not very much, literally :p

I am Tamil. I also know that k00lperson is Tamil.

The list forgot geniusbliss and srinath.r

[quote="arjunarul"]Keyree10, Phinphinity, Agr 94 math, beffudlers, srivatsan, ram, thealchemist, sanjith, chapplika, chemrock, chronoz, adharshraghavan, unearthly, saikrishna, harekrishna, louis123,madness,valeriummaximum, ashwath.rabindranath .
Our mod, pardesi is also a Tamilian.
[b]And lots more.[/b][/quote]
The list was never claimed to be comprehensive.

I speak [i]lil[/i] Tamil ... having stayed in chennai for 4 years

I am an American Indian who understands but can't speak tamil...

i am tamil too... and active too...
the number of active ppl has certainly gone down...
the following batches arent as aware...

i am from andhra pradesh and ive been to tamil nadu before. i am also aiming for IIT Chennai and understand a bit of tamil, as its almost like telugu, my native language.

Iam tamil 2....
luv the fact dat lots of dudes in chennai r comin 2 dese forums !

BTW ny1 preparin for JEE here?
(guess most of u are :P)

About 75% of indians who wants to join Olympiads try for IIT as well. It is that they won't get good ranks in IIT but they would get good ranks in CMI or Bits and go to that.

Im telegu.
I was born in Andhra Pradesh and brought to America.


calculus integration function derivative trigonometry geometry rectangle

$ \int^{b}_{a} f(x) dx$ is the notation for an integral. Now, my teacher explained that the dx is supposed to merely show that f(x) is a function dependent on x, otherwise it's a constant. I went along with this until u substitution was introduced, as this requires usage of differentials. So then, what on Earth is the purpose of dx there? I have a little theory about it, but it's probably absurd.

So, $ dy \equal{} f'(x)\triangle x$ and $ dx \equal{} \triangle x$, such that $ \frac{dy}{dx}$ gives you the derivative of f(x), which is consistent with Leibniz notation. We also know that:

$ \int{ \frac{d f(x)}{dx} } \equal{} f(x)$ which suggests that when you take the antiderivative of the derivative, the dx dissapears, which would explain why the dx is present in the integral notation, since the "dx"es cancel out. This is the best explanation I can muster up that doesn't make the use of u substituion differentials completely fluky.

if f is integrable on [a,b] 、F(x)=antiderivative and f is continuous at c in [a,b], then  F is differentiable at c and  F'(c)=f(c)

That's not at all what I was asking, so perhaps I should give an example to clarify my point.

We want to integrate $ 2x\cos{(x^{2})} \; dx$. We let $ u \equal{} x^{2} \iff \frac{du}{dx} \equal{} 2x \iff \underline{du \equal{} 2x \; dx}$ so our original statement can be re-written as:

$ \int_{a}^{b} \underline{ \cos{(u)} \; du } \equal{} \sin(u) \equal{} \sin{(x^{2})}$

Why are differentials even involved here? It's the fact that they're being used that makes the dx in $ \int_{a}^{b} f(x) dx$ more significant than simply symbolic value, such as showing $ f(x)$ is not a constant. This is why I'm wondering how the dx ever got there, if it has any relevance when algebra is involved... not sure how else to explain this.

f(x)dx is the differential form (in current parlance). Best to think of it either that way (look it up) or not at all. ;)

Some authors don't even use that notation. They just put f beside the integral sign. It becomes a little important when dealing with something like Fubini's theorem as a way of book keeping but if you are learning about integration for the first time I wouldn't worry about it since there's quite a bit to differential forms.

The dx tells you which variable you're integrating with respect to.

$ \int_a^b f(x) dx$ really means $ \int_{x \equal{} a}^{x \equal{} b} f(x) dx$

The dx tells you that that x is the variable that you're varying between the bounds.

Most of the time, it doesn't affect your calculations, but if you had to do something like
$ \int_a^b x^3t^4dx$, suddenly the dx becomes a big deal. If you were to change the dx to a dt, you'd get a different answer. It IS important to always have it there.

About u substitutions:
When you do a u substitution, you're changing the variable of integration to a u. You can't integrate with respect to u unless you also change the dx to a du. In order to do that, you must take the derivative of your substitution to get the proper conversion between dx and du.

Like for example: if $ u \equal{} x^2$, then $ du \equal{} 2x dx$.

If you want a more nitty gritty explanation, I will do my best to explain:

An integral is really a sum of infinitesimal slices. Its like a big sigma (Riemann Sum) except there are an infinite number of infinitesimal terms. When you are taking your slices, the $ f(x)$ represents the size of the slice. The dx represents the infinitesimal thickness of each slice.

In your standard integral problems like $ \int_0^2 x^2 dx$, you are taking the area and slicing it vertically into slices. The height of each slice is $ x^2$ while the thickness of each slice is $ dx$, an infinitesimal width. The integral sums all of these slices from 0 to 2 along x, the variable of integration.

Whenever I'm unsure how to apply an integral to a word problem, I always think of what shape the slice would be. This lets me figure out what to integrate.

I hope I didn't confuse you more

Ok, I might as well try something. We look for $ \int 2x\cos(x^{2})$, with $ u \equal{} x^2$ as before, and this time, I'll write $ u' \equal{} 2x$. Thus;

$ \int \cos(u)u'$ then, reversing the chain rule, we have: $ \sin(u) \equal{} \sin(x^{2})$ as desired. Is this acceptable?

You should use the differential notation.  In particular, you should write $ \int 2x \cos (x^2) \, dx$, substitute $ u \equal{} x^2 \implies du \equal{} 2x \, dx$ and conclude that the integral is $ \int \cos u \, du$.

The notation $ \int \cos u \, u'$ is troublesome, since what you really mean is $ \int \cos u \, \frac {du}{dx} \, dx \equal{} \int \cos u \, du$.

Personally, I like to think of the $ dx$ in terms of Riemann sums.  It's the "width" of one of the rectangles into which we partition the area under a curve, and $ f(x)$ is the "height."  This interpretation is particularly instructive when you start doing volume integrals.

There are several more ways to look at it.

1) $ dx$ is just a symbol as well as $ du$. Both are just there for historical reasons and we can completely dispose of them. $ du\equal{}2x\,dx$ is just a way to write $ (x^2)'\equal{}2x$. The change of variable theorem is $ \int_a^b (f\circ g)g'\equal{}\int_{g(a)}^{g(b)}f$.

2) We talk about Lebesgue-Stiltjes integral. In general it is $ \int f\,dg$ which means the integral of $ f$ with respect to the measure generated by the function $ g$. $ du\equal{}2x dx$ is just a way to write the fact that the density of the measure generated by $ x^2$ with respect to the measure generated by $ x$ (which is the usual Lebesgue measure on the line) is $ 2x$.  The change of variable formula is $ \int_a^b (f\circ g)\,dg\equal{}\int_{g(a)}^{g(b)}f\,dx$ where $ x$ is the identity function.

3) We are talking about an integral of a differential form $ f(x)dx$ where $ dx$ is the identity linear mapping on $ \mathbb R$. $ du\equal{}2x\,dx$ is the expression of the differential of the $ 0$-form $ u$ as some function times the form $ dx$. The chande of variable formula is $ \int_{a}^b (f\circ g)\,dg\equal{}\int_{g(a)}^{g(b)}f\,dx$

There are a few more viewpoints but they are close to one of these three or the one mentioned by t0rajir0u. As I said in another thread, don't pay too much attention to the notation if it is consistent and you understand what is written. I usually think of Lebesgue integration, so I prefer to write $ \int f(x)\,dm(x)$ or just $ \int f\,dm$ even for line integrals ($ m$ here stands for the Lebesgue measure, of course and I use the form with $ x$ mainly if my function depends on several arguments and I want to integrate with respect to one of them).

In teaching this in a calculus class, I say that the $ dx$ is part of the integral symbol; it closes the bracket and tells us what variable we're integrating with respect to. In other contexts, I might do things differently. I'm fine with $ \int_a^b f(x)\,dx$ or $ \int_a^b f$, but not $ \int_a^b f(x)$; if we mention the variable at all, we need to mark it as a dummy variable.

Yes, but when I was a student, my professor said that $ x$ stood just for the identity mapping, after which he happily wrote $ \int_0^1 x^2$ on the blackboard without any $ dx$. :)

Ok, what if we say $ \int_{a}^{b} f(x)$ where $ f(x) \equal{} x^{2} \plus{} 3x \plus{} 5$, in other words, we know $ f(x)$ is a function...?

The same: if $ x$ is the identity mapping, you just have a composition :lol: Of course, you can then just write "consider $ \int_a^b f$ where $ f\equal{}x^2\plus{}3x\plus{}5$". As I said, any notation is fine as long as it is consistent and understandable. So choose whichever you like. If somebody insists that it is easier for him to understand a different one, just shrug your shoulders and rewrite the formula the way he wants it.


rotation
combinatorics proposed
combinatorics

Given a 6x6 square. Can we cover this square by some of four kinds of block as bellow picture?
[img]http://files.myopera.com/The%20soul%20of%20rock/albums/147179/IMGUJBYYDN3PM.jpg[/img]

Of course! And, I think we don't need C and D-kind. :D  Sorry for my mistake!

Color the tiles of the $ 6\times 6$ block alternately black and white by columns- not a chessboard pattern. There are 18 black and 18 white tiles. Each L-shaped block covers either 3 white and 1 black tile, or 3 black and 1 white. This makes the covering impossible; we can't balance black and white with exactly 9 tiles.


geometry
angle bisector
geometry proposed

Construct the bisector of a given angle using a ruler and a compass, but without marking any auxiliary points inside the angle.

Let the angle be $ \angle AOB$. We assume that $ \angle AOB<180^{\circ}$, otherwise the word "inside" is meaningless. Let $ C$ be on the extension of $ BO$. Now we construct the angle bisector of $ \angle BOC$. Put the center of the compass on $ O$, construct the circle intersecting $ OB$ and $ OC$ at $ X$ and $ Y$ respectively. Construct an arbitrary line through $ X$. Take a point $ M$ on the line, and take a point $ N$ on the extension of $ XM$ such that $ XM\equal{}MN$. Join $ NY$, construct a line through $ M$ parallel to $ NY$. This line passing through the midpoint of $ XY$, call this point $ Z$. Therefore $ OZ$ is the angle bisector of $ \angle BOC$. Now it remains to construct a line passing through $ O$ perpendicular to $ OZ$, which is trivial. To avoid marking points inside the angle, we construct the line 'outwardly'.


Στην προσπάθειά μου να αποδείξω ένα πρόβλημα έχω την εξής απορία :

[b] ΑΣΚΗΣΗ[/b]

 Έστω κύκλος με κέντρο[b] Ο [/b], ένα σημείο [b]Γ [/b]εκτός αυτού , οι εφαπτομένες [b]ΓΒ,ΓΔ [/b], η διάμετρος [b]ΒΑ[/b] και μια τέμνουσα [b]ΓΕΖ[/b] . Να αποδειχθεί ότι οι δέσμη [b]Α. ΑΔ,ΑΕ,ΑΒ,ΑΖ [/b] είναι αρμονική .

 Αν αυτό είναι σωστό , τότε δίνω λύση στην άσκησή μου.

[b] Μπάμπης[/b]

Διορθώστε με αν κάνω κάποιο λάθος αλλά νομίζω ότι αυτό ισχύει γαι κάθε σημείο $ A$ επί του κύκλου (κι όχι μόνο για το αντιδαιμετρικό του $ B$), γιατί το $ BEZC$ είναι [i]αρμονικό τετράπλευρο[/i].

Το παραπάνω κατ' αρχάς προκύπτει (είναι βασική ιδιότητα ενός αρμονικού τετραπλέυρου) απ' το ότι οι εφαπτομένες στα $ B,D$ τέμνονται επί της διαγωνίου $ EZ$ (στο Γ). Για να αποδείξω ότι η παραπάνω ιδιότητα καθιστά ένα τετράπλευρο αρμονικό απλά θέτω $ M$ την τομή των $ BD,EZ$ και έχουμε $ DZ,DM,DE,DD$ είναι αρμονιική δέσμη κι άρα η τομή τους με τον κύλκο είναι αρμονική, δηλαδή $ BEZC$ αρμονικό τετράπλευρο.
Τώρα αφού το τετράπλευρο είναι αρμονικό απευθείας έχουμε ότι γαι οποιοδήποτε $ A$, η δέσμη $ AZ,AB,AE,AD$ είναι αρμονική.

[quote="bilot4"]Διορθώστε με αν κάνω κάποιο λάθος αλλά νομίζω ότι αυτό ισχύει γαι κάθε σημείο $ A$ επί του κύκλου (κι όχι μόνο για το αντιδαιμετρικό του $ B$), γιατί το $ BEZC$ είναι [i]αρμονικό τετράπλευρο[/i].
([color=red][b] Εννοείς το $ BEDZ$ ?[/b][/color] )

Το παραπάνω κατ' αρχάς προκύπτει (είναι βασική ιδιότητα ενός αρμονικού τετραπλέυρου) απ' το ότι οι εφαπτομένες στα $ B,D$ τέμνονται επί της διαγωνίου $ EZ$ (στο Γ). Για να αποδείξω ότι η παραπάνω ιδιότητα καθιστά ένα τετράπλευρο αρμονικό απλά θέτω $ M$ την τομή των $ BD,EZ$ και έχουμε $ DZ,DM,DE,DD$ είναι αρμονιική δέσμη κι άρα η τομή τους με τον κύλκο είναι αρμονική, δηλαδή $ BEZC$ αρμονικό τετράπλευρο.([color=red][b] Εννοείς το $ BEDZ$ ?[/b][/color] )
Τώρα αφού το τετράπλευρο είναι αρμονικό απευθείας έχουμε ότι γαι οποιοδήποτε $ A$, η δέσμη $ AZ,AB,AE,AD$ είναι αρμονική.[/quote]

Δημήτρη σε ευχαριστώ  :) . Την ιδέα για το αρμονικό τετράπλευρο την είχα κατά νου , αλλά προσπαθούσα να την αποφύγω και να βρω πιο κάτι άμεσο ή  κάτι πιο στοιχειώδες(λόγους ή βασικές σχέσεις στην αρμονικότητα). ΑΝ βρεις και κάτι άλλο , θα χαρώ να μας το στείλλεις. Θα σας στείλλω κάποια στιγμή και το πρόβλημα που λύνει αυτή η διαπίστωση.
 [u]Μπάμπης [/u]

Θα χρησιμοποιήσω το αγγλικό αλφάβητο προς χάριν των αλλοδαπών φίλων του φόρουμ.

Ας δούμε τα όσα γράφει ο Δημήτρης, με μια στοιχειώδη προσέγγιση της θεωρίας περί πολικών.

Η ευθεία $ CO,$ όπου $ O$ είναι το κέντρο του κύκλου $ (O),$ τέμνει τον $ (O)$ στα σημεία έστω $ K,$ $ N$ και τις ευθείες $ AE,$ $ AZ,$ στα σημεία $ E',$ $ Z'$ αντιστοίχως.

Εάν $ L\equiv BD\cap CO,$ αποδεικνύεται εύκολα ότι το σημείο $ L,$ είναι το αρμονικό συζυγές του $ C,$ ως προς τα σημεία $ K$ και $ N.$

Πράγματι $ \frac {CK}{CN} \equal{} \frac {KK'}{NN'} \equal{} \frac {DK'}{DN'} \equal{} \frac {LK}{LN}$ $ \Longrightarrow$ $ \frac {CK}{CN} \equal{} \frac {LK}{LN}$ $ ,(1)$ όπου $ K'$ και $ N',$ είναι τα σημεία τομής της ευθείας $ CD,$

από τις εφαπτόμενες του κυκλου $ (O),$ στα σημεία $ K$ και $ N$ αντιστοίχως.

Ορίζουμε το σημείο $ P\equiv KZ\cap EN$ και εύκολα συμπεραίνουμε ότι το σημείο $ M'\equiv KE\cap NZ,$ είναι το ορθόκεντρο του τριγώνου $ \bigtriangleup PKN,$ γιατί $ KE\perp EN$ και $ NZ\perp ZK,$ λόγω της διαμέτρου $ KN$ του κύκλου $ (O).$

Επομένως έχουμε ότι $ PM'\perp KN\equiv CO.$

Από το πλήρες τετράπλευρο τώρα, $ PEM'ZKN,$ συμπεραίνουμε ότι η διαγώνιά του $ PM',$ διέρχεται δια του σημείου $ L,$ ως το αρμονικό συζυγές του σημείου $ C\equiv EZ\cap KN,$ ως προς τα σημεία $ K,$ και $ N.$

Ακόμα, από το πλήρες τετράπλευρο $ PEM'ZKN,$  έχουμε ότι το σημείο $ M\equiv EZ\cap PL,$ είναι το αρμονικό συζυγές του σημείου $ C,$ ως προς τα σημεία $ E$ και $ Z.$

$ \bullet$ Αποδεικνύεται εύκολα ότι η ευθεία $ PL,$ ταυτίζεται με την ευθεία $ BD,$ αν σκεφτούμε ότι οι ευθείες $ CB,$ $ CD,$ είναι οριακές θέσεις της τυχούσας τέμνουσας $ CEZ,$ κατά την περιστροφή της περί το σημείο $ C$ και όπως έχει αποδειχθεί, το σημείο $ M'\equiv KE\cap NZ,$ ανήκει πάντοτε στην ευθεία $ PL\perp KN\equiv CO.$

Δηλαδή το σημείο $ B,$ είναι το οριακό σημείο $ B\equiv E\equiv Z\equiv M'.$

Άρα η ευθεία $ PL,$ διέρχεται δια του σημείου $ B$ και ομοίως δια του σημείου $ D.$

[b][size=100]Σημείωση.[/size][/b] – Αποδείχτηκε έτσι με στοιχειώδη μέσα, η βασική ιδιότητα από τη θεωρία περί πολικών, όπου η ευθεία $ BD,$ ως πολική του σημείου $ C$ ως προς τον κύκλο $ (O),$  τέμνει την τυχούσα τέμνουσα $ CEZ,$ στο σημείο $ M,$ ως το αρμονικό συζυγές του $ C,$ ως προς τα σημεία $ E$  και $ Z.$

$ \bullet$ Επειδή τα συνευθειακά σημεία $ Z,$ $ E,$ $ M,$ $ C,$ συνιστούν αρμονική σημειοσειρά, συμπεραίνουμε ότι η δέσμη $ D.ZEMC,$ είναι επίσης αρμονική.

Επειδή οι δέσμες  $ A.ZEMC,$ $ D.ZEMC,$ έχουν ίσες τις γωνίες των ομόλογων ακτινών τους, ως εγγεγραμμένες σε κοινά τόξα στον κύκλο $ (O),$ συμπεραίνουμε επίσης ότι η δέσμη $ A.ZEMC,$ είναι αρμονική και αυτό πολύ σωστά όπως λέει ο Δημήτρης, ισχύει για κάθε σημείο $ A,$ του κύκλου $ (O)$ και όχι μόνο για το αντιδιαμετρικό του $ B.$

Τέλος, αφού αποδείχτηκε ότι η δέσμη $ A.ZEMC$  είναι αρμονική και έχουμε $ AD\parallel CO,$ ως κάθετες επί την ευθεία $ BD,$ συμπεραίνουμε ότι $ OE' \equal{} OZ',$ όπου $ E'\equiv AE\cap CO$ και $ Z'\equiv AZ\cap CO,$ αν μάντεψα σωστά, την άσκηση στην οποία αναφέρθηκε ο Μπάμπης.  :wink:

Κώστας Βήττας.

Kατ' αρχάς κ. Στεργίου πράγματι εννοώ το τετράπλευρο $ BZDE$ .

Κατά δεύτερον κ. Βηττα (συγχωρέστε με που δεν έχω χρόνο να δαιβάσω όλο το post) αλλά περιληπτκά που το είδα απλά ήθελα να σημειώσω κάτι. Νομίζω ότι το $ O$ είναι το μέσο του $ E'Z'$ κι όχι το $ L$ κι αυτό προκύπτει λόγω του συμπεράσματος ότι $ AZ,AB,AE,AD$ είναι αρμονική δέσμη και λόγω του ότι η διατέμνουσα $ CO//AD$ (Κάνω κάτι λάθος...?)

Δημήτρη σ' ευχαριστώ για την επισήμανση του λάθους και το διόρθωσα ήδη.

Κώστας Βήττας.

Κώστα , όντως αυτή είναι η άσκηση που ήθελα να αποδείξω !!!

 Φαίνεται όμως ότι η άσκηση βασίζεται αποκλειστικά  στη θεωρία των αρμονικών και των πολικών και δεν επιδέχεται πιο απλή προσέγγιση.
 Ωραία λοιπόν , θα την καταχωρίσω σε αυτήν ομάδα του συμπληρώματος της γεωμετρίας.

 Σας ευχαριστώ και τους δύο  :) !

 Μπάμπης


integration
logarithms
trigonometry
calculus
calculus computations

First, make the substitution $x=\sin^{-1}\left(\cos u\right)$, then our integral becomes:

$-\int_{\frac{\pi}{2}}^{0}\ln\cos u \ du = \int_{0}^{\frac{\pi}{2}}\ln\cos u \ du$.  Now denote the desired integral as $I$, then

$2I = \int_{0}^{\frac{\pi}{2}}\left(\ln \sin x+\ln \cos x\right) \ dx = \int_{0}^{\frac{\pi}{2}}\ln\left(\frac{1}{2}\sin 2x\right) \ dx$

$=\int_{0}^{\frac{\pi}{2}}\ln\frac{1}{2}\ dx+\int_{0}^{\frac{\pi}{2}}\ln \sin 2x \ dx$

$=-\frac{\pi \ln 2}{2}+\int_{0}^{\frac{\pi}{2}}\ln \sin 2x \ dx$

Now, note that $\ln \sin 2x$ is symmetric on the interval $\left(0,\frac{\pi}{2}\right)$, thus

$\int_{0}^{\frac{\pi}{2}}\ln \sin 2x \ dx = 2\int_{0}^{\frac{\pi}{4}}\ln \sin 2x \ dx = \int_{0}^{\frac{\pi}{2}}\ln \sin u \ du = I$.

So, finally we have $\boxed{I =-\frac{\pi\ln 2}{2}}$.


Who should admit his/her fault and leave his/her office?
And who would actually do this?

The director of FEMA; have you heard about his resume-padding?

There is already a topic about this. Use that.


Isn't it easy. Just put
$ xy^2 + 1 = xy^2 + xyz\geq 2xy\sqrt{yz}$
and then you need to prove
$ \sum \frac {\sqrt{z}}{\sqrt{y}z^3}\geq \frac {9}{(\sum x^3y^3)} $

Applying Cauchy and then AM-GM we directly get the rezult
or I misunderstood something.

Yes . Indeed , this is my younger brother 's problem (8th grade) . I have another solution :
  Let  x= 1/a , y = 1/b , z = 1/c . After that we use Cauchy with 6  numbers , 3 numbers ===> result  :lol:


[quote="jelyman"]And salut to all. This is, as you know, the private delegate forum for the student body representatives of FAMAT. Say what you like, but be aware that anything overly offensive or rude will be dealt with swiftly (and with consequences). Considering this is a private forum where only the delegates can get in, I dont expect that to happen.

So yes, go and speak your mind. Make your opinions known and send forth your ideas, for we are a democracy of fellows, and only in a marketplace of ideas can a fellowship survive![/quote]
Finally! Something about muAlphaTheta on this website!!


Let $n$ be a 5-digit number, and let q and r be the quotient and remainder, respectively, when $n$ is divided by 100.  For how many values of $n$ is $q+r$ divisible by 11?

\[ \text {(A) } 8180 \qquad \text {(B) } 8181 \qquad \text {(C) } 8182 \qquad \text {(D) } 9000 \qquad \text {(E) } 9090  \]

[hide]From the given information, we can write $n=100q+r$.  Then $n\equiv q+r\pmod{11}$, so $q+r$ is divisible by $11$ iff $n$ is divisible by $11$.  The least 5-digit multiple of $11$ is $10010$, and the greatest 5-digit multiple of $11$ is $99990$.  The answer is $\frac{99990-10010}{11}+1=\boxed{8181}$ , (B).[/hide]

[hide]$n=100q+r$, so $q+r$ is divisible by 11 if n is. There are $8181$ 5-digit multiples of 11. So $\boxed{B}$.[/hide]

[quote="OVB"]How did you determine that statement "q + r is divisible by 11 if n is?"[/quote]
Since $n=100q+r=99q+(q+r)$, if $11|n$ then $11|(q+r)$.


floor function
search
function
logarithms
real analysis
real analysis unsolved

If $ (x_{n})_{n}$ is a sequence of real numbers such that for any $ c>1$ the sequence $ x_{[c^n]}$ converges, prove that $ x_{n}$ converges.

I don't think this is true. I'm assuming that you mean $ c \in \mathbb{N}$. Take the sequence $ x_n \equal{} (\minus{}1)^n$. If c is even, then $ x_{c^n} \equal{} 1$ identically, and if c is odd, $ x_{c^n} \equal{} \minus{}1$ identically. Both of these subsequences converge, but $ x_n$ does not.

[quote="t0rajir0u"]$ c$ is real.  $ [c^n]$ means $ \lfloor c^n \rfloor$.[/quote]

Ah, nevermind then, didn't understand the notation.

Correct me if I am wrong but, if we take $ c \equal{} 1 \plus{} \epsilon$ for $ \epsilon > 0$ small enough than the series $ x_{[c^n]}$ will contain $ x_n$ as a sub-sequence, hence $ x_n$ must converge to the same limit.

You are wrong: every exponential grows (eventually) faster than any polynomial.  In particular, for fixed $ \varepsilon$, we have for some $ n$ that $ (1 \plus{} \varepsilon)^n > \frac {2}{\varepsilon}$, so $ (1 \plus{} \varepsilon)^{n \plus{} 1} \equal{} (1 \plus{} \varepsilon)^n \plus{} \varepsilon (1 \plus{} \varepsilon)^n > (1 \plus{} \varepsilon)^n \plus{} 2$, i.e. we eventually skip some value.  A similar argument shows that the gaps eventually grow arbitrarily large.

do you think we can find $ c, d > 1$ such that $ \{\lfloor c^k\rfloor\}\cap \{\lfloor d^k \rfloor\}$ has infinitely many elements?

Of course, let try $ \displaystyle{d\equal{}c^{n}}$  :D

one of my good friends points out... if this problem is true, then for every infinite subset $ S$ of the natural numbers, there must exist some $ c > 1$ such that the sequence $ (\lfloor c^n \rfloor)$ meets $ S$ infinitely times. for otherwise, for every $ c > 1$, $ (\lfloor c^n \rfloor)$ meets $ S$ in finitely many places. we could then define our sequence $ (x_n)$ as $ x_i \equal{} 0$ if $ i\in S$, $ x_i \equal{} 1$ otherwise, and then $ (x_{\lfloor c^n \rfloor})$ converges to 1 for every $ c$, but the sequence is obviously not convergent.

Thor: are you sure that this problem is correct?

Sorry, I'm not sure if we know that all such sequences have the same limit or not. You can assume that.

well they do in the "counter-example" above. the issue is something else...as a corollary of this problem, *if* it is true, we get the existance of a $ c$ for every subset of the natural numbers, as explained in my last post. so i'm wondering, what's the source of this problem? and do you know if it's definitely true??

I know it's true but I haven't solved it. My analysis professor gave it to me.

Thor, we want to see a solution to this problem. the suspense is KILLING us. intelegi frate? :)

[quote="pleurestique"]the suspense is KILLING us[/quote]
Come on, it appeared on the forum more than once! :) The main idea is that if you know that $ n_1,\dots,n_k$ can arise as integer parts of powers of some non-integer $ c$, then [b]every sufficiently large[/b] number $ n$ can arise as an integer part of a power of $ c$ in addition to $ n_1,\dots,n_k$. Then you have to use a few standard things like the nested interval lemma, etc. (Or just search the forum for old threads :P).

ha, are you serious?? i've never seen it before on the forum and all this time no one's said anything

Hi,fedja,
Can you explain more?
I can't understand your argument with the words "in addition to".

Yes, pleurestique, I'm serious. See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70016[/url], for instance. I know that the problem there appears somewhat different but if you just think a bit... ;) And even back then it was considered something "discussed many times already" (see alekk's remark near the end).

Hi,fedja.
If I understand properly,I think you mean consider the function $ f(t)\equal{}x_{\lfloor t \rfloor}$ and change $ c^n$ to $ n \ln c$,using right continuous instead of continuous.
But How to show for every $ c,x_{\lfloor c^n \rfloor}$ converges to the same limit,I have thought for a long time,but no result.Or have I missed something?

You have to mimic the proof rather than just use the result ;) But you are on the right track. Just assume that there are two subsequences with different (possibly infinite) limits and alternate them in the construction.

But I am sorry,I still don't get your mean.How to "alternate them in the construction"?
I also wonder if the following theorem is true?
Let $ f$ be a continuous function such that $ f(nx)$ converges when $ n \to \infty$ for any $ x > 0$, Can we prove that $ f(x)$ converges as $ x$ tends to infinity?
Should it use the hamel basis of $ \mathbb{R}$ over $ \mathbb{Q}$?

Just look at anik_andrew's argument in the topic I mentioned above. What he does is to take a sequence $ \xi_n$ where $ |f(\xi_n)|>\varepsilon$. Instead, you need to take two sequences $ \xi_n$ and $ \zeta_n$ such that $ f(\xi_n)$ and $ f(\zeta_n)$ tend to different limits and choose some $ \xi$, then some $ \zeta$, then some $ \xi$, then some $ \zeta$ and so on in anik_andrew's proof instead of going $ \xi,\xi,\xi,\xi,\dots$.

I hope that the following hint can solve the proposed problem. Let $ \left\{ {a_{p_n } } \right\},\left\{ {a_{q_n } } \right\},...,\left\{ {a_{s_n } } \right\}$ be subsequences of $ \left\{ {a_n } \right\}$ such that $ \left\{ {p_n } \right\},\left\{ {q_n } \right\},...,\left\{ {s_n } \right\}$ are pairwise disjoint and form the sequen $ \{a_n\}$. If $ S,S_p,S_q,...,S_s$ are the sets of all the limit points of the sequences $ \left\{ {a_n } \right\},\left\{ {a_{p_n } } \right\},\left\{ {a_{q_n } } \right\},...,\left\{ {a_{s_n } } \right\}$ then
\[ S \equal{} S_p  \cup S_q  \cup ... \cup S_s .
\]

Oh,Thank you,fedja.I think I get it.Which your proof to contradict with is that for some point $ x$ the limit $ f(nx)$ will not exist.Thanks again.


In a recent wedding, the groom's age was fifteen years more than half the bride's age. If their ages added to 51 years, how old was the groom?

equations

brides age=x

grooms age=$ \frac{1}{2}x\plus{}15$
add together and set equal to 51

$ \frac{3}{2}x\plus{}15\equal{}51$
x=24
so grooms age=51-24=27

Thanks mz. The reason I reported this is because ... (see it yourself)
[center][img]http://i96.photobucket.com/albums/l172/thdanh90/2-3.jpg[/img][/center]


function
trigonometry
integration
trig identities
calculus
calculus computations

Please help me calculate fourierseries of the following functions

a) $\sin{(x-1)}\cos{2x}$

b)  $\sin^{3}{(x)}$

c)$\cos^{4}{x/2}$

d)$|\sin{(x)}|$

especially d and a. thanks.

Hint: Use trigonometric identities.

d) Note that $sin(x) = \pm \sqrt{1-cos^{2}(x)}$
And $cos^{2}(x) = \frac{1}{2}+\frac{cos(2x)}{2}$

Therefore, $|sin(x)| = \sqrt{1-cos^{2}(x)}= \sqrt{1-(\frac{1}{2}+\frac{cos(2x)}{2})}= \frac{\sqrt{2}}{2}\sqrt{1-cos(2x)}.$

[quote="jmerry"]Forget the trig identities for $|\sin|$; calculate directly from the definition, recalling that $|\sin|$ is an even function with period $\pi$.[/quote]

Could anyone show this more explicit, please?

Since $f(x) = |sinx|$ is even, we get $f(x) = a_{0}+\sum_{n =1}^{\infty}a_{n}cos(\frac{n\pi x}{a})$, $-a < x < a$
where $a_{0}= \frac{1}{a}\int_{0}^{a}f(x)dx, a_{n}= \frac{2}{a}\int_{0}^{a}f(x)cos(\frac{n\pi x}{a})$
$f(x)$ is periodic with period $\pi$, so $a = \frac{p}{2}= \frac{\pi}{2}$.

You only need to compute $a_{0}$ and $a_{n}$

$a_{0}= \frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}sin(x)dx$
$a_{n}= \frac{4}{\pi}\int_{0}^{\frac{\pi}{2}}sin(x)cos(2nx)dx$


I didn't know where to post this, so I'm posting it here.  I attempted problem #420 in AoPS Volume 2:  "Are there integers m and n such that 5m :^2: - 6mn + 7n :^2: = 1985? (IMO 1985)"  Here's the solution I came up with and want to know if it is right (my solution is different from the answer book's solution):  (in spoiler)

[hide]

Subtracting 5m :^2:  from both sides of the equation, I got -6mn + 7n :^2: = 1985 - 5m :^2: .  Applying mod 5 to it, I got

-6mn + 7n :^2:  :equiv: 1985 - 5m :^2:  :equiv: 0 (mod 5).  Thus, 7n :^2: - 6mn must be divisible by 5.  I did the division and got n :^2:  with a remainder of 2n :^2: - 6mn.  In order for that to be zero, n must be zero.  Putting that into the original equation, we get 5m :^2:  = 1985.  Dividing be five gets m :^2: = 397, but since 397 is not a perfect square, m is not an integer, so there is no ordered pair (m, n) where m and n are integers that is a solution to the eqution.

[/hide]

nr1337 wrote:[hide]...I did the division and got n :^2:  with a remainder of 2n :^2: - 6mn.  In order for that to be zero, n must be zero.[/hide]

This part isn't true. You're right that 2n2 - 6mn must be 0 mod 5; but that doesn't necessarily mean that n has to be zero. First, just because something is 0 mod 5 doesn't mean it is zero - it just has to be a multiple of 5; and second, n in this case doesn't even necessarily have to be 0 mod 5. For example, take n = 1, m = 2.

Oh... that's a big mistake.  I'm going to try to fix it.  Thank you.


It's really bad and gloomy that such a massive, devstating earthquake has his Northern India and Pakistan! Nearly [size=200][b]30,000[/b][/size] have died in Pakistan with numbers rising!!!! and nearly 700 in Indian Kashmir!!!!! It's so Bad!!!!! Let's bring our thoughts together for all those affected !!! India and Pakistan should come closer together!!!!!

Please post your thoughts and views!

Is anyone's schools doing something or anyones neighborhood doing anything?  We have a fund at school, but after the tsunami, Hurricane Katrina, Hurricane Rita, the earthquake, I wish we would actually have a car wash or bake sale or something instead of just giving a couple of dollars...

Hello.
I am an Indian citizen living in Luxembourg, Europe.
The earthquake situation is worsening everyday and as you might have heard in the news that more than 3 million people are displaced. The greater problem, of course, is that those who have not died yet, but are injured, cannot find quality treatment for recovery. And now the hostile winter is approaching.
I am heading a huge clothes drive (blankets and sleeping bags) here at the International School of Luxembourg and have established contact with the Pakistani consulate. There is a very small South Asian community here compared to countries like UK and USA but we are trying our best.

Hope help reaches our people more quickly than possible.


vector
superior algebra
superior algebra unsolved

In the category of $ \mathbb{F}_p$-vector spaces, give an example where $ \oplus \mathrm{Hom}(A, B_i) \hookrightarrow \mathrm{Hom}(A, \oplus B_i)$ isn't surjective.

this exists over any field $ K$. putting $ B_i \equal{} K$ and choosing a basis $ J$ for $ A$, the map is essentially

$ \bigoplus_{i \in I} \prod_{j \in J} K \to \prod_{j \in J} \bigoplus_{i \in I} K , ((a_{ij})_j)_i \mapsto ((a_{ij})_i)_j$

on the left hand side you have matrices with a finite number of (nonzero) columns and arbitrary entries in those columns, and on the right hand side you have matrices with an arbitrary number of columns, having each only a finite number of entries. the map is then usual transposing. clearly it is not surjective if $ I$ and $ J$ are infinite, because in the images the number of entries per column is bounded by a single number for all columns.


Let be $ \Delta ABC$ . Prove that $ A\equal{}\frac{\pi}{2}\vee B\equal{}\frac{\pi}{2}\vee C\equal{}\frac{\pi}{2}$ if at least one of the following inequalities is true :

1)$ \sin \ B\equal{}\frac{\sin \ B\plus{}\sin \ C}{\cos \ B\plus{}\cos \ C}$

2)$ \mathrm{tg} \ B\equal{}\frac{\cos (C\minus{}B)}{\sin A\plus{}\sin (C\minus{}B)}$

hello, where is the angle $ \alpha$ in your first formula
$ \sin(\beta)\equal{}\frac{\sin(\beta)\plus{}\sin(\gamma)}{\cos(\beta)\plus{}\cos(\gamma)}$?
Sonnhard.

Maybe ... $ \sin \ A \equal{} \frac {\sin \ B \plus{} \sin \ C}{\cos \ B \plus{} \cos \ C}$ and in this case is very easily !

hello, and what is with the other case?
Sonnhard.

hello, let's have a look at your first formula, with $ \sin(\gamma)\equal{}\frac{c}{b}\sin(\beta)$ we get $ \cos(\beta)\plus{}\cos(\gamma)\equal{}1\plus{}\frac{c}{b}$, with the law of cosine we get $ \frac{a^2\plus{}c^2\minus{}b^2}{2ac}\plus{}\frac{b^2\plus{}a^2\minus{}c^2}{2ab}\equal{}1\plus{}\frac{c}{b}$ or $ (b\plus{}c)(\minus{}b^2\plus{}2bc\minus{}2ac\plus{}a^2\minus{}c^2)\equal{}0$ from here follows
$ a^2\minus{}(b\minus{}c)^2\equal{}2ac$ which is equivalent with $ \sqrt{\frac{(s\minus{}b)(s\minus{}c)}{ac}}\equal{}1$ i.e. $ \sin(\frac{\alpha}{2})\equal{}1$ and $ \alpha\equal{}\pi$ ist the solution of our equation.
Sonnhard.


The sequence $(p_n)$ is defined as follows: $p_1=2$ and for all $n$ greater than or equal to $2$, $p_n$ is the largest prime divisor of the expression $p_1p_2p_3\ldots p_{n-1}+1$.
Prove that every $p_n$ is different from $5$.

$p_{1}= 2 \equiv 2 (mod 5)$
$p_{2}= p_{1}+1 = 3 \equiv 3 (mod 5)$

Note that $(p_{n}-1)p_{n}+1 = p_{n+1}$

If $p_{n}\equiv 2 (mod 5)$ and $p_{n+1}\equiv 3 (mod 5)$, then $p_{n+2}\equiv (p_{n+1}-1)p_{n+1}+1 \equiv (3-1)*3+1 \equiv 2 (mod 5)$, and $p_{n+3}= (p_{n+2}-1)p_{n+2}+1 \equiv (2-1)*2+1 \equiv 3 (mod 5)$.

So if any two consecutive terms are $2$ and $3$ in $mod 5$, respectively, the next two must also be. So in $mod 5$, the sequence goes $2, 3, 2, 3, 2, 3, 2, 3$ ... and no term is divisible by $5$.

[quote="neelnal"]Starting...
[hide]
The multiplication of p1p2p3...pn is always even, so we know the in order to be divisible by 5, the last digit of this multiplication must be 4. Does this happen?[/hide][/quote]
no because you have $p_{1}p_{2}...p_{n-1}+1$ and now $p_{1}p_{2}...p_{n-1}$

@ZutonForce: the problem is, you take the largest prime divisor of  $p_{1}p_{2}...p_{n-1}+1$  and not $p_{1}p_{2}...p_{n-1}+1$  itself. So the mod will be different.

I will post the sol-n in a few days if nobody solves the problem.

[quote="rem"][quote="neelnal"]Starting...
[hide]
The multiplication of p1p2p3...pn is always even, so we know the in order to be divisible by 5, the last digit of this multiplication must be 4. Does this happen?[/hide][/quote]
no because you have $p_{1}p_{2}...p_{n-1}+1$ and now $p_{1}p_{2}...p_{n-1}$
[/quote]

I was not trying to say that your problem was wrong. I was just eliminating the possibility the p1p2...p(n-1) + 1 has a 0 as its last digit since p1=2, making p1p2...p(n-1) + 1 odd.

Oh sorry I did not quite understand what you wrote. :blush:
Yes that's a start: the last digit cannot be 0.

Some more thoughts:

[hide]
2a+1 can be divided by either 3, 5, or 7; where a is any positive integer.

If 2a+1 can be divided by 5, then we have to prove that 5 will never be the largest prime divisor.

We know that p1p2..pn +1 will never equal 5 itself because
p1=2 and p2=3, so p3=7>5

Maybe there is a pattern that can be followed.

p1=2, p2=3, p3=7, p4=43, no too hard to find pattern

We can just stick to examining the last digits of
p1p2p3...pn+1.

p1 - 2; p2-3; p3 - 7, p4-3, p5 -7
Still very hard

p1p2...pn cannot have a last digit of 4.

2(2) and 2(7) provide a last digit of 4.

So, p2p3..pn cannot have a last digit of 2 or 7.
p2p3...pn will never have last digit of 2.

Now we have to examine last digits of 7.
Case 1:
If we were to break p2p3...pn into two factors, those two factors would have to have onesdigits of 1 or 7.
Then, the factor with the ones digit of 7, when broken into two factors, will have to have a factor with digit of 1  and 7, and so on, we continue.

Case 2:
The two factors can have ones digit of 9 and 3.
A similar, but more complex method can be used for this.

I don't know. It seems a lot of work and too brute force. There is probably a more elegant solution.[/hide]

Please do not post ideas that lead nowhere and are irrelevant to the problem.
The problem does have a very nice and simple solution.
And by the way, the problem is number theory and it is not proposed, it appeared on Iberoamerican olympiad 1987, question 5.

but it's very easy and nice!
2,3 are the first two terms in the sequence and a prime can appear only once in $p_{1},p_{2},...$ so for 5 to appear we need $p_{1}p_{2}\cdots p_{n}=5^{n}-1$ at some stage but 4 doesn't divide LHS :P

Yes that's the nice solution I was talking about. Good job!
I really like this beautiful rpoblem. :lol:

[quote="Albanian Eagle"]but it's very easy and nice!
2,3 are the first two terms in the sequence and a prime can appear only once in $p_{1},p_{2},...$ so for 5 to appear we need $p_{1}p_{2}\cdots p_{n}=5^n-1$ at some stage but 4 doesn't divide LHS :P[/quote]

Why is $5^n$?
__________________________________________
I thinks the solution is:

$a_i=2 \Leftrightarrow  i=1$ because $a_1 a_2 ... a_n +1$ is odd with every $n > 1$.

So $a_1 a_2 ... a_n$ isn't divisible by $4$. On the other hand, $a_1 a_2 ... a_n \neq 4 \Rightarrow$ Q.E.D

$p_1=2$, therefore $p_n,n>1$ are odd and $a_n=p_1....p_n+1=3\mod 4$. Because $p_2=3$ we get $a_n=7\mod 12\to  p_n\ge 7$  ($a_n$ had prime divisor form $3\mod 4$ greater than 3. It is at least 7).

$ a_{2}=3$ , then $2$ and $3$ won't appear on the sequence again. For $5$ to appear we need $\prod_{i=1}^{n}a_i\ = 5^{w} - 1$ so $ \prod_{i=1}^{n}a_i\ \equiv 24 (mod\ 100) $ which means $\prod_{i=3}^{n}a_i\ \equiv 4 (mod\ 50)$, but $\prod_{i=3}^{n}a_i\ \equiv 1 (mod\ 2)$ so is odd, so contradiction.

$a_2=3$. Now suppose that for some $n, p_n=5$ then the number $x=p_1p_2...p_{n-1}+1$ could be divisible by 2, 3 or 5. Notice that since $gcd(p_1p_2...p_{n-1},x)=1$  it is impossible for $x$ to have a 2 or 3 in its prime factorization thus;
$x= p_1p_2...p_{n-1}+1=5^a$ and therefore $p_1p_2...p_{n-1}=5^a-1=4(5^{a-1}+...+5+1)$. Since all $p_n$ are odd for $n>1$ this equality cannot happen and thus  by contradiction, no $p_n$ is equal to $5$


[hide="Hint"]
$7$ is a multiple of $7!$ and on.
[/hide]

Oh, so you only need the first six terms. So the answer is B?


inequalities
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inequalities unsolved

If $a,b,c$ are sides of a triangle of circumradius $R$, prove the inequality
$\frac{a^2}{b+c-a}\ + \frac{b^2}{c+a-b}\ + \frac{c^2}{a+b-c}\geq\ 3 R \sqrt{3} .$

Come on, not this again. I posted two solutions some time ago. Try to search it. ;)