function
trigonometry
algebra
polynomial
linear algebra
algebra unsolved

if two functions share the same values over all the integers are they the same function?

In terms of linear algebra, any class of functions that forms a subspace of countable dimension will do. (The polynomials are one such subspace, with countable basis $ 1, x, x^2, \ldots$)

The trigonometric polynomials of the form $ \sin 2 \pi nx, \cos 2 \pi nx$ also span a subspace of countable dimension and they don't have the desired property.  This is a result that depends on the property of the basis itself:  each element is determined by its values at the integers.

[quote="t0rajir0u"]$ f(x) \equal{} \sin \pi x$ is zero at the integers.

This is true of polynomials and certain other types of nice functions.[/quote]
so if the function is $ f(x) \equal{} 0$ the other could be $ f(x) \equal{} sin\pi x$ are there any other nice functions that this hold for? so are you saying that unless one function is f(x) = 0 my statement is true? would the other functions be cheap things like $ k^{0\cdot u}$?

[quote="t0rajir0u"]$ f(x) \equal{} \sin \pi x$ is zero at the integers.

This is true of polynomials and certain other types of nice functions.[/quote]
sorry for the double post but by this is true of polynomials .... you mean that if two polynomials share values over all the integers etc. they are the same

Sorry, I meant to stipulate that with respect to their values on integers the functions in the space were linearly independent.

[quote="t0rajir0u"][quote="Dr:D"]so are you saying that unless one function is f(x) = 0 my statement is true? would the other functions be cheap things like $ k^{0\cdot u}$?[/quote]
No.  The only important case is that $ f(x) \equal{} 0$ and $ g(x)$ is a function that is zero on the integers, since every other case reduces to this one after you subtract the two functions, and I gave you an example of such a $ g$.  There are other examples, but the sines and cosines are the easiest to write down.  A continuous function zero on the integers can still take on any values between the integers.

[quote="Dr:D"]if two polynomials share values over all the integers etc. they are the same[/quote]
Yes.  This is because a polynomial of degree $ n$ is defined by $ n \plus{} 1$ of its values.[/quote]
Oh I get it I think, for every g(x) that is zero on the integers we may construct two polynomials that share all the same values on the integers! smart.
So the answer to Dr:D's Q was no, cept for polynomials

[quote="mathemonster"] for every g(x) that is zero on the integers we may construct two polynomials that share all the same values on the integers! [/quote]
Not quite.  What I meant was the following:

$ f(n) \equal{} g(n) \forall n \in \mathbb{Z} \Leftrightarrow f(n) \minus{} g(n) \equal{} 0 \forall n \in \mathbb{Z}$

so everything depends on the existence of a function $ h(n) \equal{} f(n) \minus{} g(n)$ that is zero on the integers but that is not identically zero.  So for example, the functions

$ f(x) \equal{} x^2, g(x) \equal{} x^2 \plus{} \sin \pi x$

work because $ f \minus{} g$ is the same example as before.  Note that $ g$ is not a polynomial.

For what it's worth, there are other classes of functions besides the polynomials for which it's true that they're determined by their values at the integers, such as the algebra generated by the polynomials and the (real) exponentials.

Ok so the functions f(x) = x^2 + e^{0*k} and g(x) = x^2 are considered the same

Well, $ e^{0} \equal{} 1$ everywhere, not $ 0$ and not just at the integers, so if you meant $ g(x) \equal{} x^2 \plus{} 1$ then yes; they're the same function identically.

are there any other functions that work like this except for $ sin(\pi\cdot{u})$ etc?

Do you mean, other continuous functions that are 0 on the integers? Of course.
Some examples: $ f(x)\equal{}0.5\minus{}|0.5\minus{}x|$ for $ 0 \le x \le 1$, $ f$ is periodic with period 1 - $ f(x)\equal{}0.5\minus{}|0.5\minus{}\{ x \}|$.
In general, you can "glue" a sequence $ f_n$ of continuous functions, each defined on $ [n,n\plus{}1]$ and get 0 on $ n$ and $ n\plus{}1$.


geometry
trigonometry
trapezoid
projective geometry
parallelogram
cyclic quadrilateral
power of a point

$ ABCD$ is a convex quadrilateral with $ AB$ not parallel to $ CD$. A circle passing through $ A$ and $ B$ is tangent to $ CD$ at $ P$, and a circle passing through $ C$ and $ D$ is tangent to $ AB$ at $ Q$. Prove that the common chord of these two circles bisects $ PQ$ if and only if $ AD\parallel{}BC$.

It seems nice,but i don't know where to start.
Could anyone give me a hint please？

[quote="plane geometry"]I have found a solution,I will post it later![/quote]

Please do so  :)

I have also in mind a sulution of this nice problem and I will post it here later, if it is different than the one by [b][size=100]plane geometry[/size][/b].

Kostas Vittas.

Denote M , K is the intersection of CD,BA and PQ,EF respectively. PQ meet circle O1,O2 at S,R (Circle O1 passing through A,B and  tangent to CD at P,Circle O2 passing through C,D and  tangent to AB at Q)
MA=a MB=b,MC=c,MD=d,MP=p,MQ=q
KP*KS=KE*KF=KQ*KR
PR/KP=KR/KP-1=KS/KQ-1=SQ/KQ
KP=KQ  〈=〉 pr=qs 〈=〉 pr*pq=qs*pq 〈=〉(p-d)(c-p)=(q-a)(b-p) 〈=〉p2(c+d)2=q2(a+b)2 〈=〉ab(c+d)2=cd(a+b)2
〈=〉ac=bd 〈=〉bc∥ad
done!

This problem can also be solved in this way:
Let $ AB$ and $ CD$ intersect in point $ E$ and denote the circles tangent to $ CD$ and $ AB$ by $ C_{1}(O_{1},R_{1})$ and $ C_{2}(O_{2},R_{2})$, respectively. Let also $ PQ$ intersect $ C_{1}$, $ C_{2}$ and the common chord in points $ F$, $ G$ and $ K$ respectively.

$ PK\equal{}QK  \Rightarrow  FP\equal{}GQ  \Rightarrow \frac{EP}{EQ}\equal{}\frac{R_{1}}{R_{2}}  \Rightarrow  \stackrel{\triangle}{EPO_{1}}\sim  \stackrel{\triangle}{EQO_{2}} \Rightarrow  \stackrel{\triangle}{EAO_{1}}\sim  \stackrel{\triangle}{EDO_{2}} \Rightarrow$ $ ADPQ$ is a cyclic quadrilateral $ \Rightarrow BCPQ$ is a cyclic quadrilateral $ \Rightarrow AD\parallel BC$

[quote="Zagros"]

$ FP \equal{} GQ \Rightarrow \frac {EP}{EQ} \equal{} \frac {R_{1}}{R_{2}}$[/quote]

Why is this?  :maybe:

Let $ AB$ and $ CD$ intersect in point $ E$ and denote the circles tangent to $ CD$ and $ AB$ by $ C_{1}(O_{1})$ and $ C_{2}(O_{2})$, respectively. Let also $ PQ$ intersect $ C_{1}$ and $ C_{2}$ and the common chord in points $ F$, $ G$ and $ K$ respectively.

$ K$ is on the radical axis of $ C_{1}$ and $ C_{2}$ $ \Rightarrow KP.KF\equal{}KQ.KG$.
$ KP\equal{}KQ \Rightarrow KF\equal{}KG \Rightarrow PF\equal{}QG$.
$ EQ$ is tangent to $ C_{2}$. Thus, $ \sin\widehat{EQP}\equal{}\frac{QG}{2R_{2}}$. Similarly, $ \sin\widehat{EPQ}\equal{}\frac{PF}{2R_{1}}\Rightarrow \frac{\sin\widehat
{EQP}}{\sin\widehat{EPQ}}\equal{}\left(\frac{R_{1}}{R_{2}}\right)\left(\frac{QG}{PF}\right)\Rightarrow \frac{\sin\widehat{EQP}}{\sin\widehat{EPQ}}\equal{}\frac{R_{1}}{R_{2}}$. By using law of sines in $ \triangle{EPQ}$, we get $ \frac{EP}{EQ}\equal{}\frac{R_{1}}{R_{2}}\Rightarrow \frac{EP}{PO_{1}}\equal{}\frac{EQ}{QO_{2}}\Rightarrow \triangle{EPO_{1}}\sim\triangle{EQO_{2}}$  $ (1)$.

$ (1)\Rightarrow \widehat{PEO_{1}}\equal{}\widehat{QEO_{2}}\Rightarrow \widehat{AEO_{1}}\equal{}\widehat{DEO_{2}}$  $ (2)$.
$ (1)\Rightarrow \frac{EO_1}{PO_{1}}\equal{}\frac{EO_{2}}{QO_{2}}$.
We have $ PO_{1}\equal{}AO_{1}$ , $ QO_{2}\equal{}DO_{2}$ and $ \frac{EO_1}{PO_{1}}\equal{}\frac{EO_{2}}{QO_{2}}\Rightarrow \frac{EO_1}{AO_{1}}\equal{}\frac{EO_{2}}{DO_{2}}$  $ (3)$.

Note that by $ (2)$, the bisector of $ \widehat{O_{1}EO_{2}}$ is the bisector of $ \widehat{AED}$. Note also that $ \widehat{AO_{1}E}$ and $ \widehat{DO_{2}E}$ are both acute or both obtuse (this depends on whether $ A$ is closer to $ E$ than $ B$ or not). Assume that by reflecting $ \triangle{EDO_{2}}$ against the bisector of $ \widehat{AED}$, we obtain $ \triangle{ED^{'}O^{'}_{2}}$. It is concluded that ($ E$,$ O_{1}$,$ O^{'}_{2}$) and($ E$,$ A$,$ D^{'}$) are collinear. By $ (3)$, and by noting that $ \widehat{AO_{1}E}$ and $ \widehat{D^{'}O^{'}_{2}E}$ are both acute or both obtuse, it is concluded that $ AO_{1}$ is parallel to $ D^{'}O^{'}_{2} \Rightarrow \triangle{EAO_{1}}\sim\triangle{ED^{'}O^{'}_{2}}\Rightarrow \triangle{EAO_{1}}\sim\triangle{EDO_{2}}$ $ (4)$.

$ (3),(4) \Rightarrow \frac{EA}{ED}\equal{}\frac{EO_{1}}{EO_{2}}\equal{}\frac{EP}{EQ}\Rightarrow \frac{EA^{2}}{ED^{2}}\equal{}\frac{EP^{2}}{EQ^{2}}\Rightarrow \frac{EA^{2}}{ED^{2}}\equal{}\frac{EA.EB}{ED.EC}\Rightarrow
 \frac {EA}{ED}\equal{}\frac{EB}{EC}\Rightarrow AD\parallel BC$

$ X \equiv AB \cap CD.$ $ (O_1), (O_2)$ are tangent to $ CD, AB$ at $ P, Q$ $ \Longrightarrow$

$ \frac {\overline{XA} \cdot \overline {XB}}{\overline{XD} \cdot \overline {XD}} \equal{} \frac {\overline {XP}^2}{\overline {XQ}^2}\ \ \ (1)$

$ PQ$ cuts $ (O_1), (O_2)$ at $ E, F.$ $ M$ is the midpoint of $ PQ$ on the radical axis od $ (O_1), (O_2)$ $ \Longrightarrow$ $ \overline{MP} \cdot \overline{ME} \equal{} \overline{MQ} \cdot \overline{MF}\ \Longrightarrow$

$ \overline{QA} \cdot \overline{QB} \equal{} \overline{QE} \cdot \overline{QP} \equal{} \overline{PF} \cdot \overline{PQ} \equal{} \overline{PD} \cdot \overline{PC}\ \Longrightarrow$

$ (\overline{XA} \minus{} \overline{XQ}) \cdot (\overline{XQ} \minus{} \overline{XB}) \equal{} (\overline{XD} \minus{} \overline{XP}) \cdot (\overline{XP} \minus{} \overline{XC})\ \Longrightarrow$

$ (\overline{XA} \plus{} \overline{XB}) \cdot \overline{XQ} \minus{} \overline{XQ}^2 \minus{} \overline{XA} \cdot \overline{XB} \equal{} (\overline{XD} \plus{} \overline{XC}) \cdot \overline{XP} \minus{} \overline{XP}^2 \minus{} \overline{XD} \cdot \overline{XC}\ \Longrightarrow$

$ \frac {\overline{XA} \plus{} \overline{XB}}{\overline{XD} \plus{} \overline{XC}} \equal{} \frac {\overline{XP}}{\overline{XQ}}\ \ \ (2)$

Substituting (2) into (1),

$ \left(\frac {\overline{XA} \plus{} \overline{XB}}{\overline{XD} \plus{} \overline{XC}}\right)^2 \equal{} \frac {\overline{XA} \cdot \overline{XB}}{\overline{XD} \cdot \overline{XC}}\ \Longrightarrow$

$ \left(\frac {\overline{XA} \plus{} \overline{XB}}{\overline{XD} \plus{} \overline{XC}}\right)^2 \equal{} \left(\frac {\overline{XA} \minus{} \overline{XB}}{\overline{XD} \minus{} \overline{XC}}\right)^2\ \Longrightarrow$

$ (\overline{XA} \plus{} \overline{XB}) \cdot (\overline{XD} \minus{} \overline{XC}) \equal{} (\overline{XA} \minus{} \overline{XB}) \cdot (\overline{XD} \plus{} \overline{XC})\ \Longrightarrow$

$ \overline{XB} \cdot \overline{XD} \equal{} \overline{XA} \cdot \overline{XC},\ \ \frac {\overline{XD}}{\overline{XC}} \equal{} \frac {\overline{XA}}{\overline{XB}}\ \Longrightarrow\ BC \parallel DA$

[quote="Zagros"] by noting that $ \widehat{AEO_{1}} \equal{} \widehat{DEO_{2}} \equal{} \alpha$ and $ \frac {AO_{1}}{EO_{1}} \equal{} \frac {DO_{2}}{EO_{2}} \equal{} \sin(\widehat{O_{1}EO_{2}} \minus{} \alpha)$, [/quote]

I'm sorry, but I can't see why...  :maybe:

Let $ S$ be, the point of intersection of the sidelines $ AB,\ CD,$ of the given trapezium $ ABCD,$ with $ AD\parallel BC.$

Let also be the points $ X\equiv AP\cap DQ,\ Y\equiv AC\cap BD,\ Z\equiv BP\cap CQ$ and based on the [b][size=100]Pappus theoerem[/size][/b],

it is easy to show that these points are collinear, applying the triads of the points $ A,\ Q,\ B$ and $ D,\ P,\ C,$
on $ SB,\ SC,$ respectively.

From the circle $ (O_{1},)$ taken as cord the segment $ AB$ and tangents to $ CD$ at point $ P,$

we have that $ (SA)\cdot (SB) \equal{} (SP)^{2}$ $ ,(1)$

From the circle $ (O_{2},)$ taken as cord the segment $ CD$ and tangents to $ AB$ at point $ Q,$

we have that $ (SD)\cdot (SC) \equal{} (SQ)^{2}$ $ ,(2)$

From $ AD\parallel BC$ $ \Longrightarrow$ $ \frac {SA}{SB} \equal{} \frac {SD}{SC}$ $ \Longrightarrow$ $ \frac {(SA)\cdot (SB)}{(SB)^{2}} \equal{} \frac {(SD)\cdot (SC)}{(SC)^{2}}$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac {SP}{SB} \equal{} \frac {SQ}{SC}$ $ \Longrightarrow$ $ \frac {SP}{SQ} \equal{} \frac {SB}{SC}$ $ ,(4)$

From $ (4)$ and because of $ \angle BSP\equiv \angle CSQ,$ we conclude that the triangles $ \bigtriangleup SPB$ and $ \bigtriangleup SQC,$

are similar and so, we have $ \angle SBP \equal{} \angle SCQ$ $ ,(5)$

$ \bullet$ It is easy to show now, that $ AP\parallel CQ$ and $ BP\parallel DQ,$ from $ (5)$ and because of $ \angle SBP \equal{} \angle SPA$ $ (6)$ and $ \angle SCQ \equal{} \angle SQD$ $ (7)$

So, $ XPZQ$ is a parallelogram and then, the line segment $ XY,$ bisects the segment $ PQ.$

Hence, it is enough to prove that the points $ X,\ Z,$ lie on the line segment $ EF,$ as the radical axis of the circles $ (O)_{1},\ (O)_{2}.$

But, it is easy to show that the quadrilaterals $ ADPQ$ and $ BCPQ,$ are cyclic, from $ (5),$ $ (6),$ $ (7).$

Hence, we have that $ (XA)\cdot (XP) \equal{} (XD)\cdot (XQ)$ and then, we conclude that the point $ X,$ lies on the line segment $ EF.$

Similarly, the point $ Z,$ lies also on $ EF$ and the proof is completed.

Kostas Vittas.


Does there exist a $4$-digit integer (in decimal form) such that no replacement of three of its digits by any other three gives a multiple of $1992$?

[quote="nicetry007"]The multiples of 1992 less than 10000 are as follows:$ \{1992, 3984, 5976, 7968, 9960\}$.

Essentially, we need to find a 4 digit number whose leading digit differs from the leading digit of all the multiples, whose second digit differs from the second digit of all the multiples and so on.

The $ i$-th set in the following collection of sets is the set of feasible digits for the $ i$-th position.

$ \{ \{2,4,6,8\}, \;\{0,1,2,3,4,5,6,7,8\}, \;\{0,1,2,3,4,5\}, \;\{1,3,5,7,9\}\}$.

The smallest number satisfying the property is 2001 and there are $ 4 \cdot 9 \cdot 6 \cdot 5 = 1080$ numbers with the required property.[/quote]

There is no stated requirement that, after the replacement, the resulting number must still be a $4$-digit integer. This being said, if we replace in $2001$ the first and last digits with $0$, we get $0$ (a quite valid number, I'd say), which is a multiple of $1992$.

So either add that requirement to the statement, or remove the $0$'s from those lists. Anyways, $2111$ always works.


limit
number theory proposed
number theory

Let $ a;b$ be two real numbers in the open interval $ (0;4)$.A sequence $ (a_n)(n\equal{}0;1;...)$ is constructed by the following recursive fomula:
$ a_o\equal{}a;a_1\equal{}b;a_{n\plus{}2}\equal{}\frac{2(a_{n\plus{}1}\plus{}a_n)}{\sqrt{a_{n\plus{}1}}\plus{}\sqrt{a_n}}$
Prove that the sequence $ (a_n)(m\equal{}0;1;...)$ convers and find limit.

$ a_{n\plus{}2}\equal{}\frac{2(a_{n\plus{}1}\plus{}a_n)}{\sqrt{a_{n\plus{}1}}\plus{}\sqrt{a_n}}.$ It give $ 2\sqrt{min(a_n,a_{n\plus{}1})}\le a_{n\plus{}2}\le 2\sqrt{max(a_n,a_{n\plus{}1})}$<4. Because $ x<2\sqrt x$ if $ 0<x<4$ we get $ b_n\equal{}min(a_{n},a_{n\plus{}1})$ increase and $ \lim_{n\to \infty}b_n\equal{}4\equal{}\lim_{n\to\infty}a_n.$


[quote="crazyman"]A circle with center O pass through the vertices B and C of the triangle ABC and intersects AB,AC again at distinct points N and M respectively,let R be the points of intersection of the circumcircles of triangles ABC and AMN (apart from A),K be the point of intersection of BM and CN.D be the point of intersection of AK and BC. Prove that RD pass through common point.[/quote]

[img]http://img132.imageshack.us/img132/1203/63800591.png[/img]
[color=blue]Let $ L \equiv MN \cap BC$. Because$ LN.LM\equal{}LB.LC$ and $ (AMN) \cap (O) \equiv A$,$ R$ $ \Longrightarrow$ $ L$,$ R$,$ A$ are conllinear .

Let $ E \equiv RD \cap (O)$. Since $ (L,D,B,C)\equal{}\minus{}1$ then $ (RL,RD,RB,RC)\equal{}\minus{}1$ , but the intersections of $ RL$,$ RD$,$ RB$,$ RC$ and $ (O)$ are $ A$,$ B$,$ E$,$ C$ $ \Longrightarrow$ Quadrilateral $ ABEC$ is harmonic $ \Longrightarrow$ $ E$ is a fixed point.[/color]


integration
trigonometry
function
calculus
calculus computations

for ${2 \pi k-\frac{\pi }{2}<x<2 \pi k+\frac{\pi }{2}}$
${\int |\cos (x)| \, dx=\sin (x)+C_{1}}$

for ${2 \pi k+\frac{\pi }{2}<x<2 \pi k+\frac{3 \pi }{2}}$
${\int |\cos (x)| \, dx=-\sin (x)+C_{2}}$

where $k$ is a integer

or alternatively, using the Sign function and simplifying (note: you can do it easily using the 2 previous formulaes)
${\int |\cos (x)| \, dx=|\cos (x)| \tan (x)+C_{3}}$

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

My answer would be
\[F(x)=(-1)^{k}\sin x+2k+C,\quad \text{for }-\frac{\pi }{2}+\pi k\le x\le \frac{\pi }{2}+\pi k,\ k\in\mathbb Z \]
because an antiderivative of $|\cos x|$ must be continuous and increasing on $\mathbb R$. But don't try telling that to Maple.
[img]http://www.math.tamu.edu/~lkovalev/files/abscos.JPG[/img]

[quote="mlok"]because an antiderivative of $|\cos x|$ must be continuous and increasing on $\mathbb R$. [/quote]

Why is that so?

An antiderivative of $f$ is a function $F$ such that $F'=f$. In particular, this means that $F$ is differentiable. A differentiable function is continuous.
Since $f(x)=|\cos x|$ is a nonnegative function, $F$ is nondecreasing. In fact, it is strictly increasing, because $f$ does not vanish on any interval.

${\int \sqrt{f(x)^{2}}\, dx=\frac{\sqrt{f(x)^{2}}}{f(x)}\int f(x) \, dx}$

we can use this relation

the unique problem is find a good constant

[quote="yagaron"]we can use this relation[/quote]
We surely can. Unless we want to obtain a correct answer.

$\int |\cos x|\ dx=\pm \int cos x\ dx=\pm \sin x+Const$, right?

Depends on how you choose the signs and the constants.

I do understand what you say, but in this case, we don't need to take into cosideration the sign of $\cos x$ as a result, I think.

For example, $\int \sqrt{1+\sin x}\ dx=\int \left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|\ dx=\pm \int \left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\ dx$.

What exactly is the meaning of $\int \sqrt{1+\sin x}\ dx=\pm \int \left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\ dx$? Both sides can be understood as certain classes of functions from $\mathbb R$ to $\mathbb R$. The functions on the left side are unbounded and increasing, but the functions on the right side are bounded and oscillating. How can they be equal?

I agree that $1+\sin x=\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}$. Taking square roots, we obtain $\sqrt{1+\sin x}=\phi(x)\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)$, where $\phi$ is a [b]function[/b] (not a constant) that assumes only values of $1$ and $-1$. When integrating the righthand side, we should keep in mind that antiderivative of a product is not necessarily the same as the product of antiderivatives.


modular arithmetic
induction
number theory
relatively prime
number theory unsolved

[hide="Solution"]
Instead of 1996 let $ m$ be the desired number of integers. Let $ N\equal{}m!$. If $ p>N$ is a prime, then $ \gcd(p,N)\equal{}1$ and so for some $ k>1$ we have $ p^k \equiv 1 \pmod{N}$ (if $ p$ happens to be 1 mod $ N$, we can still take $ k\equal{}2$). Now choose $ m$ distinct primes $ p_1, \ldots, p_m$ and let $ a_i\equal{}p_i^{k_i}$ where $ k_i$ is chosen as above so that $ a_i \equiv 1 \pmod{N}$.

Now:

1. The $ a_i$ are pairwise relatively prime: each $ a_i$ is divisible by a unique prime $ p_i$.

2. The $ a_i$ are composite. Each $ a_i$ is a prime raised to a power greater than 1.

3. For and $ d$, $ 2 \leq d \leq m$, the sum of $ d$ elements from $ \{a_1,\ldots,a_m\}$ is congruent to $ d$ mod $ N$. In particular, that sum is divisible by $ d$, but it's certainly larger than $ d$, so it's composite.
[/hide]

[quote="outback"]Prove that there exists a set $ X$ of $ 1996$ positive integers with the following properties:
(i) the elements of $ X$ are pairwise coprime;
(ii) all elements of $ X$ and all sums of two or more distinct elements of $ X$ are composite numbers.[/quote]
[hide="Solution"]
We prove this with induction for a set of $ n$ numbers.  Clearly, if $ n \equal{} 2$, then we are fine as we can just choose $ 20$ and $ 121$.  Now, say that we can choose a set with the desired properties of $ n$ integers, say $ a_1, a_2, a_3, \cdots, a_n$ (without loss of generality, $ a_1 < a_2 < a_3 < \cdots < a_n$.)  We will consider a number $ a_{n \plus{} 1}$ with the following property: for each sum, $ S$, that includes $ a_{n \plus{} 1}$, we will let $ S\equiv 0\bmod p_i$ for different $ p_i$ and $ a_{n \plus{} 1}\equiv 1\bmod a_{i}$ for $ 1\le i\le n$.  We put the restriction on $ p_i$ so that they are primes and $ p_i > a_n$ for all $ i$.  From this, we see that there is a solution for $ a_{n \plus{} 1}$ by the Chinese Remainder Theorem (notice that all $ a_i$ are pairwise relatively prime and since $ p_i$ are prime and greater than $ a_j$, we find that the $ p_i$ are pairwise prime and pairwise prime with the $ a_i$.)  By choosing a sufficiently large $ a_{n \plus{} 1}$, we can guarantee that $ S\ne p_i$ for any sum $ S$ and any considered prime $ p_i$.  Yet, for each $ S$, there exists a $ p_i$, $ p_i|S$.  Then, all $ S$ are composite and $ a_{n \plus{} 1}$ is relatively prime with $ a_i$ since $ a_{n \plus{} 1}\equiv 1\bmod a_i$.  Then, we have a new set, so induction is compete.  Setting $ n \equal{} 1996$, we have the desired set. [/hide]


Find all positive integers $ n$ such $ 20n\plus{}2$ can divide $ 2003n \plus{} 2002.$

So there is no solution in pozitive integers.

I just made a table on my calculator; there are indeed no positive integer solutions.

Making a table on one's calculator is unnecessary - I also assume it was done to check that $10n + 1 \mid 18017 = 43\cdot 419$ has no solution, since $20n + 2 \mid 2003n + 2002$ cannot be checked on a calculator.

Replay. If $20n + 2$ divides $2003n + 2002$, then it also divides $20(2003n + 2002) - 2003(20n+2) = 36034 = 2\cdot 43\cdot 419$. Thus we need $20n + 2 \mid 2\cdot 43\cdot 419$, or $10n + 1 \mid 43\cdot 419$.

Thus $10n + 1$ must be equal to one of $1, 43, 419$ or $43\cdot 419 = 18017$, and none is possible for a positive integer $n$.

[quote=balaylay]Can you explain how those simplifications work?[/quote]

Through the Euclidean Algorithm, which can be found in the Introductory to Number Theory book.

For the division to occur $n$ must be even say
$n=2k$ ,$k\ge 1$
$20m+1|100(20m+1)+3n+901$
$\implies 20m+1|3n+9$ bcoz if $a|b+c$
and $a|b$,$\implies a|c$
But the fraction must be an integer if it divides
And $\frac{3m+9}{20m+1}$ does  not an integer for $m=1,2,3,4$ .
$\frac{3m+9}{20m+1}\ge 5$
$\implies m\le 10$
And checking cases for $m=1,2,...10$
There is no solution.

$20n+2\textnormal{ }|\textnormal{ }2003n+2002\textnormal{ }\Leftrightarrow\textnormal{ } 20n+2\textnormal{ }|\textnormal{ }2003n+2002 - 100(20n+2)\textnormal{ }\\ \Leftrightarrow20n+2\textnormal{ }|\textnormal{ }3n+1802\\\Rightarrow 20n+2\textnormal{ }|\textnormal{ }20(3n+1802)=60n+36040\\\Leftrightarrow 20n+2\textnormal{ }|\textnormal{ }60n+36040-3(20n+2)=36034=2*43*419$
Checking all factors of $36034$, we see that none can be written in the form $20n+2$ for positive $n$ (For nonegative $n$, we see that $0$ works). Hence, there are no solutions.

it is better to use euclidean algorithm as follows
2003n+2002=(100)(20n+2)+3n+1802
hence on dividing 2003n+2002 by 20n+2 we get remainder as 3n+1802 which on checking is not congruent modulo 20n+2[color=#f00] note that remainder must be lesser than divisor or quotient [/color] we must verify for n<=105


Source: Math Magic [url]http://www.stetson.edu/~efriedma/mathmagic/0699.html[/url]

Are there palindrome numbers in base 10 that are the product of 3 or more consecutive integers? I have thought of this problem for a long time, but I just can't find the solution. The only thing I know is that a base 10 palindrome cannot be the product of 5 or more consecutive integers, but that's trivial.

yes: 77*78*79=474474 a palindrone

I would like to see a nice way to do this problem, I just made a program in my calculator...

[quote="Iltanen"]Only 2 such examples are known: 1x2x3=6, and the one you mentioned.  Patrick De Geest has shown that there are no such products less than 6×10^29. So can anyone prove that there are no larger examples?[/quote]

This is a question for Number Theory: Open Problems rather than Pre-olympiad!


linear algebra
matrix
linear algebra unsolved

Is it true?
For any matrix $ A\in M_{n}(\mathbb C)$,there exist a symmetric invertible matrix $ T$ such that $ TAT^{\minus{}1}$ is the Jordan canional form.
 :o

If $ n\geq3$ itsn't true.
Let $ A\equal{}\frac{1}{27}\begin{pmatrix}73&28&\minus{}34\\\minus{}17&154&\minus{}79\\\minus{}28&152&\minus{}65\end{pmatrix}\equal{}Pdiag(1,2,3)P^{\minus{}1}$ where $ P\equal{}\begin{pmatrix}1&4&\minus{}5\\2&7&1\\3&8&2\end{pmatrix}$. Suppose that $ Q$ is a symmetric matrix such that $ Q^{\minus{}1}AQ\equal{}diag(\sigma(1,2,3))$ where $ \sigma$ is a permutation. Then the $ Q\minus{}columns$ are proportional to the $ \sigma(P\minus{}columns)$.
For example if $ \sigma\equal{}id$ then the best that we can do is:
$ Q\equal{}\begin{pmatrix}1&2&3\\2&7/2&\minus{}3/5\\3&4&\minus{}6/5\end{pmatrix}$ and $ Q$ isn't symmetric. The 5 other cases are similar.

Thank you very much,loup blanc.That is an good example.


[quote="davidyko"]Hmm, could you say "let $a = b^{2}$" and solve for $a$, and then take the square roots of $a$ for the values of $b$?[/quote]

That is to solve for the roots of $b$.

To factor, add and subtract $b^{2}$ from the polynomial, you get: $b^{4}+b^{2}+1+b^{2}-b^{2}=(b^{4}+b^{2}+1)-b^{2}=(b^{2}+1)^{2}-b^{2}=(b^{2}-b+1)(b^{2}+b+1)$.

[quote="1=2"]How do you factor a polynomial like $b^{4}+b^{2}+1$?

I know the factorization, I've proved it, but could someone give me a method of finding it?

Is it just finding a root of it, then another polynomial that has the same root, and then dividing?[/quote]

$a^{4}+a^{2}+1=\frac{a^{6}-1}{a^{2}-1}=\frac{a^{3}-1}{a-1}*\frac{a^{3}+1}{a+1}=(a^{2}+a+1)(a^{2}-a+1)$

(to 1=2) The hard way you were talking about, finding the roots, is: you can plug $a = b^{2}$ in and solve the quadratic, then you have square root and say you have 4 roots $x_{1}, x_{2}, x_{3}, x_{4}$, then you know that two of them are conjugates and you have $(b-x_{1})(b+x_{1})(b-x_{3})(b+x_{3})$, but I don't know why you would want to take that long route. Use one of the two routes posted here.


vector
analytic geometry
induction
algebra proposed
algebra

Here is a marvelous problem:
  Given are $2n+1$ irrational numbers. Prove that we can select $n+1$ of them such that no matter how we choose some of these $n+1$ numbers (but without repetitions) their sum is irrational.

If I am not mistaken, problem A3 from IMC 1994 is stronger (or maybe your initial $2n+1$ numbers are not necessarily distinct? :?; then it would be a different problem):

[url]http://www.kalva.demon.co.uk/imc/imc94.html[/url]

I'll have to rememebr my proof, but I'm afraid I don't have enough time for that.. :) I'm not typing from my house right now, but from an ICafe.

Here's a possible solution:
In the solution, whenever I say or write a set, I mean multiset.
We say that $A \subset \{1,\ldots,2n+1\}$ is good for $x_i$'s iff for all $B\subset A$, $\sum_{i\in B} x_i \not \in \mathbb Q$.
We say that $A \subset \{1,\ldots,2n+1\}$ is nice for $v_i$'s iff for all $B\subset A$, $\sum_{i\in B} v_i \neq 0$.

Let $S = \{x_1,\ldots, x_{2n+1}, 1\}$, our set of irrationals together with 1.
Let $V = \text{span } S$. Choose a basis for $V$ of the form $B = \{b_1,\ldots, b_k, 1\}$.
Clearly, for all $i$, $b_i \not \in \mathbb Q$.
For each $x_i$, assign $v_i = (a_{i1},...,a_{ik})^\bot$ such that $\sum_j a_{ji} b_{i} + q_i = x_i$, where $q_i$ is the coefficient of $1$ of the vector $x_i$.

Then $A$ is good for $x_i$'s iff $A$ is nice for $v_i$'s.

So the question is equivalent to a question in finite dimensional vector spaces: if $\{x_i\}_{i=1}^{2n+1} \subseteq W$, then we can choose $n+1$ of them that any sum of them (as in the question) does not equal to zero. [The map $i\to x_i$ is not necessarily one to one].

Lemma: If $x_i \in W$ $\forall 1\leq i \leq m$, then, then there is a set of coordinates in which the first component of each $x_i$ in this coordinate representation is non-zero.

Proof of main result from lemma: First find a coordinate system in which each vector's first coordinate is non-zero. Then:
Let $A = \{j : x_j^{(1)} > 0\}$, $B = \{j: x_j^{(1)} < 0\}$. Note that since $x_i^{(1)}$ is never zero, $|A \cup B| = 2n+1$.
One of these must have cardinality $\geq n+1$, say $A$. Then any sum of elements from $A$ will be non-zero first coordinate, thus will be non-zero.

Proof of lemma: Choose an orthogonal basis $\{b_i\}$ for $V$, such that no $x_i \in b_1^\perp$.
This clearly can be done since the number of $x_i$'s is finite.

I hope that there are no mistakes. If there are none, I'd love to see a non-vector space solution.

[quote="harazi"] Given are $2n+1$ irrational numbers. Prove that we can select $n+1$ of them such that no matter how we choose some of these $n+1$ numbers (but without repetitions) their sum is irrational.[/quote]

The combinatorialization is maybe the following:
given some subsets of $\{ 1,2, \dots , 2n+1 \}$ that cover (in the sense of reverse inclusion) all subsets of size $n+1$.  Then we can choose some of these subsets whose symmetric difference  $S_1 \Delta S_2 \Delta ... S_k = \{ i \}$ for some $i$ (i.e. the symmetric difference contains exactly one element).

For the extended version I was talking about:

It says that given $2n+1$ irrational numbers, we can find $n+1$ of them s.t. every linear combination with non-negative rational coefficients (not all zero) is irrational.

Suppose for each $n+1$-uple we can find a linear combination as above which is rational (call such a combination good). Write the good combinations for $x_1,\ldots,x_n$, and each of $x_{n+1},x_{n+2},\ldots,x_{2n+1}$. In this way, we can write $x_{n+1},\ldots,x_{2n+1}$ as a rational number plus a linear combination of $x_1,\ldots,x_n$ with non-positive rational coefficients, not all zero. By also writing the good combination for $x_{n+1},\ldots,x_{2n+1}$, we will get one such combination for $x_1,\ldots,x_n$. Now, since $x_1,\ldots,x_n$ was an arbitrary $n$-uple, we can conclude that every $n$-uple has a good combination, and we can use induction (the initial assumption was that every $n+1$-uple has a good combination, and this is the assumption we have to prove wrong).

I have a solution not sure that it's correct.
[i]Problem:[/i] Suppose that  Given are $2n+1$ numbers not all of them rational. Prove that we can select $n+1$ of them such that no matter how we choose $n+1$ numbers (but without repetitions) their sum is irrational.

Suppose every $n+1$ that we choose their sum is rational. Suming all equations and taking average, you see that sum of all numbers is rational. So if we choose 2 sets with cardinal $n+1$ that their intersection is one irrational element then one of them must have the irrational sum.

This problem was given in Bulgarian TST 2004 as problem 3, third day. The solution which is proposed is the following one:
If the given numbers are $\alpha_{1},...\alpha_{2n+1},$ we choose the set $S$ with the following property: $1\in S;$ if $\alpha_{i}\not =\sum_{b_{j}\in S}a_{j}b_{j}$ where $a_{j}\in\mathbb{Q}$ we set $\alpha_{i}\in S.$ So we can consider $S=\{\alpha_{0}=1,\alpha_{1},...\alpha_{k}\}$ where $k\leq 2n+1.$ It's easily seen that the representation $\alpha_{i}=\sum_{i=0}^{k}a_{ji}b_{i}$ is unique. Thus a sum of $a_{i}$-s is a rational number iff the sum of $\alpha_{i}-a_{i0}$ is zero. But $\alpha_{i}$ is irrational so $\alpha_{i}-a_{i0}\not = 0$ and by so we can choose at least $n+1$ of $\alpha_{i}-a_{i0}$ with the same sign.


I would agree (A) seems to make a lot of sense but to me, I'm choosing (D) because I think "prospects" seems a bit better than "horizon."

Yeah, I agree that "horizon" is not a very good word choice. But does "disparaging" fit well? That's the only thing that bothered me about (D).

I don't have a problem with "disparaging."  Also, the definition of "grim" is not quite what you're thinking; grim is more of a stern or harsh aura, and I don't think that accurately characterizes the statement.

Doesn't "disparage" mean the same thing as "belittle"?

Yes, they have similar meanings, but I believe "belittle" could also fit in the blank.  Either word indicates that the comments seem to "put down" or show a lowered expectation of the company's future.

First of all, "belittle" isn't one of the choices.  Second, I don't think "belittle" works because although it means to put down, it has a much more negative and sinister connotation.  It's like insulting or deliberately putting down to in order to induce misery and I don't see that fitting.

If "belittle" doesn't work, then how can it be a synonym of "disparage"? Unless they are pretty distant synonyms...

I guess one could make an argument for "belittle" but it's completely irrelevant because it isn't one of the choices.

Okay, I guess I understand why disparage works now. I was only worried because when I look [url=http://www.google.com/search?rlz=1C1GGLS_en-USUS300US304&aq=f&sourceid=chrome&ie=UTF-8&q=define:+disparage]here[/url], the majority of the definitions have something to do with condescension or insult.

Which is a better opposite for "optimistic":  "grim", or "disparaging"?

I would say either works.  If I had to choose between those two on an SAT question, I would choose "disparaging".

[quote="buzzer11"]Taken out of context, the Chairman's comments on the company's ------- might seem -------, but overall he painted a fairly optimistic picture of the future.

(A) horizon..grim
(B) profits..convivial
(C) success..unremarkable
(D) prospects..disparaging
(E) activities..guarded

The answer is (D). But the second blank should be the opposite of "optimistic", and I think "grim" works better than "disparaging". And "horizon" fits alright, too... So why is (D) better than (A)?[/quote]

First, you seem to be taking the rule that "but" negates the clause occurring immediately before it far too literally. In natural language no rule is entirely precise, as it would be in mathematics.

With that in mind, while "grim" might be more nearly the logical opposite of "optimistic," the sentence, "Taken out of context the Chairman's comments on the company's [b]horizon[/b] might seem [b]grim[/b], but overall he painted a fairly optimistic picture of the future" is verbally awkward. The meaning of "horizon" as it refers to the company is unclear. It is also unclear whether the Chairman made said comments from "the horizon", i.e. while standing on it, or concerning the horizon.

Considering answer (D), your benchmark, that "optimistic" should be the opposite of "disparaging," is valid. Ask yourself if it is possible to be optimistic and disparaging simultaneously? If your answer is no then (D) is a possible correct answer.


[color=red][b]Which are the best online-available free Chemistry software for drawing structures, lab equipment, presentations etc. and analysing reactions, products , viewing molecules etc.??? [/b][/color]

[size=150]Please suggest!!![/size]

Step 1.  Go to chemistry forum
Step 2.  Scroll down
Step 3.  Note topic, "Software."

Proceed as you see fit.

Oops sorry!!! Anyways, if somebody wants to recommend software not suggested in the forum above , please recommend here! Thanks!


Give $ \triangle{ABC}$. Find the area of the triangle if AC=6, BC=4, and $ \angle{C}$=45

a. 6$ \sqrt{2}$
b. 8$ \sqrt{3}$
c. 12$ \sqrt{2}$
d. 12

$ [ABC]\equal{}\frac{1}{2}\cdot4\cdot6\cdot \sin 45\equal{}12\cdot\dfrac{\sqrt2}{2}\equal{}6\sqrt2 \rightarrow A$

$ A_{ABC} \equal{} \displaystyle\frac {6 \cdot 4 \cdot \sqrt {2}}{4} \equal{} 6 \sqrt {2}$


Area AEB=1/3area ABC
Area AEB=1/2*AE*EB*sin(angleAEB)=1/3*3*5^.5  So angle AEB is determined
Applying law of cos to triangle AEB we find side AB
Applying STEWART to triangle AEB you find x.So CF=3x

My answer is x=3.892348.....

Again, there is a solution without trig. Find it :D I'll post a solution by Sunday if there isn't any posted.

Please post it  :(  1000th user. I´m unexperienced in geometry and i would like to improve my knowledges in this area.

[quote="Chalao"]In the following figure with sides unequal, given that AG,BH, and CF are the medians of the triangle, and that AG=3, BH=6, and the area of the triangle is $ 3\sqrt {5}$ find the lenght of the remaining median. Please explain it step-by-step....thanks! :lol:  :lol:

[img]http://i204.photobucket.com/albums/bb126/agerador/medians.jpg[/img][/quote]
[b]
Solution without trig[/b]

Firstly, join $ G$ to $ F$. In triangles $ BFG$ and $ BAC$ note that $ BF \equal{} AF$, hence $ AB \equal{} 2BF$. And also $ BG \equal{} CG$, hence $ BC \equal{} 2BG$. Furthermore $ \angle ABC$ are common, hence the two triangles are similar and have a scale factor of 2 (ie ratio of sides are 2). Furthermore note by the equal intercepts theorem that $ GF//AC$.

Join $ F$ to $ H$ and $ H$ to $ G$. In the previous way it can be shown that a similar relationship holds true.

In triangles $ GEF$ and $ AEC$ as $ GF//AC$ then they are similar (AAA). Furthermore we know from before that $ \frac{AC}{GF} \equal{} 2$. From similarity $ \frac{CE}{FE} \equal{} 2$ ie $ CE \equal{} 2 FE$. This property is true for all medians.

From this we can determine the following:

$ AE \equal{} 2, \,\,GE \equal{} 1,\,\, HE \equal{} 2,\,\, BE \equal{} 4$

Now, join $ A$ to $ D$ and $ B$ to $ D$. Consider triangles $ DFB$ and $ AFE$ we have 2 sides and the included angle equal, hence they are congruent.

$ \therefore AE \equal{} DB \equal{} 2$. Similarly in triangles $ AFD$ and $ BFE$ we have $ AD \equal{} BE \equal{} 4$.

Consider triangle $ AED$ with sides $ 2, 4, 2x$. Firstly may i mentioned the well known theorem that the medians of a triangle divide the triangle into 6 triangles of equal area. From this we see that $ |AEF| \equal{} \frac{1}{6} \times 3\sqrt{5} \equal{} \frac{1}{2} \sqrt{5}$.

As triangles $ AEF$ and $ AFD$ have equal bases and share the same vertex then they have the same area. Furthermore as the area of $ AEF$ is $ \frac{1}{2} \sqrt{5}$ then the area of $ AED$ is $ \sqrt{5}$

Using herons formula,

$ A \equal{} \sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}\,\,\,;\,\, s \equal{} \frac{a\plus{}b\plus{}c}{2}$

we have

$ s \equal{} \frac{2 \plus{} 4 \plus{} 2x}{2} \equal{} x \plus{} 3$

$ \therefore \sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)} \equal{} \sqrt{5}$

$ s(s\minus{}a)(s\minus{}b)(s\minus{}c) \equal{} 5$

$ (x \plus{} 3)(x \plus{} 3 \minus{} 2)(x \plus{} 3 \minus{} 4)(x \plus{} 3 \minus{} 2x) \equal{} 5$

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

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

$ (x^2 \minus{} 9)(x^2 \minus{} 1) \equal{} \minus{}5$

$ x^4 \minus{} 10x^2 \plus{} 9 \equal{} \minus{}5$

$ \therefore x^4 \minus{} 10x^2 \plus{} 14 \equal{} 0$

Letting $ u \equal{} x^2$,

$ u^2 \minus{} 10u \plus{} 14 \equal{} 0$

$ u \equal{} \frac{10 \pm \sqrt{100 \minus{} 46}}{2}$

$ \equal{} \frac{10 \pm 4\sqrt{11}}{2}$

$ \equal{} 5 \pm 2\sqrt{11}$

As $ 5 \minus{} 2\sqrt{11} < 0$, then

$ x^2 \equal{} 5 \plus{} 2 \sqrt{11}$

$ \therefore x \equal{} \sqrt{5 \plus{} 2 \sqrt{11}}$

From before we determined that $ CE \equal{} 2FE \equal{} 2x$. Therefore $ CF \equal{} x \plus{} 2x \equal{} 3x$

$ \therefore CF \equal{} 3 \sqrt{5 \plus{} 2\sqrt{11}} \approx \boxed{10.23}$


real analysis
symmetry
probability
real analysis unsolved

Let I be a nonempty open interval. Let A and B be disjoint subsets of I such that the union of A and B is I, and A and B are of positive Lebesgue measure. Let 0<p<1, (p is not 1/2) be a given real number.
Prove that there exist x in A and y in B such that px+(1-p)y is element of A and (1-p)x+py is element of B .

As written, it is false. Just take $ I\equal{}(0,2)$, $ A\equal{}(0,1]$, $ B\equal{}(1,2)$, and $ p< \frac 12$. This counterexample is rather stupid, of course ;)

Sorry, you are right.
What we should prove is: there exist x in A and y in B such that px+(1-p)y  and (1-p)x+py are not in the same chosen subset (A or B).

Or merely add the condition $ p>1/2$. Anyway, let's call the points in the configuration $ XYZT$ (lying on the line in this order from left to right) and consider 2 cases:

1) There is an open interval whose intersection with one of the sets $ A$ and $ B$ has measure $ 0$. Take the maximal (with respect to inclusion) open interval $ (a,b)$ with this property. It cannot be the entire interval $ (0,1)$, so either $ a>0$ or $ b<1$. By symmetry, it is enough to consider the latter case. Also, let us assume that $ |(a,b)\cap B|\equal{}0$. The key point is that the intersection of an arbitrarily small interval $ (b,b\plus{}\delta)$ with $ B$ has positive measure. In particular, there is a point of $ B$ arbitrarily close to $ b$ from the right. Now, choose $ \delta<\min\{1\minus{}b,(b\minus{}a)(1\minus{}p)\}$. Let $ m\equal{}|(b,b\plus{}\delta)\cap B|>0$. Fix $ Z\in (b,b\plus{}\delta)$ so close to $ b$ that $ X,Y\in(a,b)$ whenever $ T\in (b\plus{}\tfrac m2,b\plus{}\delta)$. Then, when $ T$ is chosen randomly and uniformly on that interval, we have $ T\in B$ with positive probability but $ X\in B$ or $ Y\in B$ with probability $ 0$, so we can ensure that $ X,Y\in A$, $ Z,T\in B$.

2) Every open interval intersects both $ A$ and $ B$ by a set of positive measure. In this case we can actually make any assignment we want. Let's make $ X,Z\in A$, $ Y,T\in B$ for instance. Choose any density point $ x$ of $ A$ and any density point $ y$ of $ B$ somewhere near $ \frac 12$. If $ \delta>0$ is small enough, then the vast majority of points in $ (x\minus{}\delta,x\plus{}\delta)$ are in $ A$ and the vast majority of points in $ (y\minus{}\delta,y\plus{}\delta)$ are in $ B$.

Now, if we put $ X\equal{}x$, $ Y\equal{}y$, then $ Z$ will be in some position $ z$. Choose $ z'$ to be the density point of $ A$ very close to this $ z$ (much closer than $ \delta$) and put $ Z\equal{}z'$. With that $ Z$, we can vary the size of the configuration so that $ X$ and $ Y$ run over intervals of lengths comparable to $ \delta$ containing $ x$ and $ y$ respectively. For the vast majority of such sizes, we'll have $ X\in A$, $ Y\in B$ and, moreover, $ X$ and $ Y$ will be the density points of the corresponding sets. Thus, we can find $ x',y',z'$ that can be a part of our configuration so that $ x',z'\in A$, $ y'\in B$ and each point is a density point of the corresponding set. Now it remains to do exactly the same trick again.

Here is an interesting question I cannot answer yet. Suppose that $ |A|\equal{}m$. Is it true that the measure of "suitable" pairs $ (x,y)\subset[0,1]^2$ is bounded from below by a constant depending on $ m$ only?


function
superior algebra
superior algebra unsolved

Let A be a set. I need to prove that the cardinality of A < the cardinality of (2^A), where 2^A is the set of all subsets of A.  I have to show that the cardinality of A <= the cardinality of (2^A) by constructing an injection from A to 2^A. How do I define this injection?
Then, suppose there is a bijection A to 2^A and define a subset U of A by U = {a : a is not in U[size=75]a[/size]) Show that if U = U[size=75]b[/size], we get a contradiction.
Together, this implies the original statement.

How do I show these?

Thanks in advance.

$ a \to \{a\}$ defines an injection $ A \to 2^A$. if $ f : A \to 2^A$ is a surjection, in particular $ \{a \in A : a \notin f(a)\}$ has a preimage, say $ b$, i.e. $ f(b) \equal{} \{a \in A : a \notin f(a)\}$. then $ b \in f(b) \Leftrightarrow b \notin f(b)$, which is a contradiction.

Sorry if I'm missing something, but what does the {a} from the  function: a to {a} denote?

:wink: I suppose this cannot be proved, right? Cantor's Contiuum Hypothesis is still a hypothesis, and should have been proved not "provable".

@BrainMan1: $ \{a\}$ denotes the set whose only element is $ a$. You're constructing a map from $ A$ to $ 2^A$ which consists of subsets of $ A$, so one simple way to do this is to send each element to the subset consisting of just that element.

@KenHungKK: I think you're a bit confused :) It's well-known and easy that $ \kappa < 2^\kappa$. The GCH is the statement that there's nothing else in between.


No I don't talk to myself - and I find it annoying to be in the company of those who do.

i talk to myself when i'm studying by myself with nobody around... cuz it's scary quiet :rotfl:

I mutter angrily to myself...But I don't entirely talk to myself out loud.  :|

I put no, but that's only true if saying stuff like "f***ing dumba**" or "what an idiot" directed at others but inaudible doesn't count.

no, not aloud, but what goes on in my mind is a different matter... :rotfl:

always. and aloud.

people think i'm crazy. which i am somewhat.

[quote="hunter34"]It gets pretty lonely being homeschooled. And I won't talk to my parents about popular things.[/quote]

SWEET I AM HOMESCHOOLED TOO OMG OMG OMG

I do it when I am alone. I go into fantasy worlds and I find myself talking. I try to stop.

I dont, (i am in a public school system [its quite good, actually]) but there is a girl who sat in front of me in math and english and reading who would talk to herself and i found it odd.....

I do sing to myself though. My dog likes it but my family doesnt  :D !!!!

Yeah, I'm homeschooled also. I talk to myself all the time. While reading something interesting (& hard to understand), I start explaining it to myself out loud. My mom is like ...  :huh:  :| 'Are you talking to yourself?' Mm.. I curse to myself sometimes. Lawl.

[quote="dynamo729"]
I do sing to myself though.[/quote]

lol same here! I sing to myself and my whole familiy gets really annoyed, but I can't really help it.

I don't really talk to myself unless I'm cursing at something, though.

Does like, injuring yourself when nobody is around count as talking to yourself? Otherwise, pretty much no. My little sister does, though.

OK Now I'm freaking myself out. I talk to myself all the time since I read this topic.  :rotfl:

I was talking to myself in the halls today...

today inpre-algebra i started cursing at the teacher under my breath, and he caught part of it... he didnt know who i was cursing about, tho.

I talk to myself when everybody is already tired to listen to me.  so i've no other choise :(


algebra
polynomial
Rational Root Theorem

For how many positive integers does $ 3x^{3}\minus{}4x^{2}\minus{}3x\plus{}4 \equal{} 0$?

[quote="GameBot"]For how many positive integers does $ 3x^{3} \minus{} 4x^{2} \minus{} 3x \plus{} 4 \equal{} 0$?[/quote]

A quick check reveals $ \minus{}1$ works. $ (x\plus{}1)(x^2\minus{}7x\plus{}4)$. Since $ x^2\minus{}7x\plus{}4$ isn't a perfect square, there are 3 roots.

I don't know how to do this problem, but I do know the answer is 1 from ftw.
Can someone tell me how?

First, factor $3x^3-4x^2-3x+4 = 0$ as $(x-1) (x+1) (3 x-4) = 0$ (a quick check (checking $-4$, $-2$, $-1$, $1$, $2$ and $4$ by Rational Root Theorem) allows you to factor easily) Since, $(x-1) (x+1) (3 x-4) = 0$, either $x=1$, $x=-1$ or $x=\frac{4}{3}$. However, since $x$ must be a positive integer, there is only $\boxed{1}$ solution for $x$.

Or..

By Rational Root Theorem, the only rational roots are when $x$ is $-4$, $-2$, $-1$, $1$, $2$ or $4$. Since $x$ must be positive, we can eliminate this down to $1$, $2$ or $4$. Plugging in reveals that $1$ is the only solution that works.

Think about this particular problem because you can group variables:

$3x^3 - 3x - 4x^2 + 4 = 3x(x^2 - 1) - 4(x^2 - 1) = (3x - 4)(x^2 - 1) = (3x - 4)(x + 1)(x - 1)$

The soutions are $\frac{4}{3}, \pm1$. $-1$ is negative so it does not work and $\frac{4}{3}$ is not an integer so it does not work. Since $1$ is the only one that works the answer is $\boxed{1}$

*Make sure you understand it is $1$ because it is the only positive integer*


MATHCOUNTS
summer program
MathPath
AMC
AMC 8

I heard that if you get a perfect on the AMC 8, you'll get invited to MathPath. Does this mean you get a discount?

[quote="MathPath Website"]1. AMC8 Perfect Scorers
All students with perfect scores in the AMC8 competition are eligible for a fee reduction of $ \$$500.[/quote]

i need to have text more than quote. darn.

well, i didn't get a perfect score on AMC 8, and i was invited... (i got the brochure sent to me..) so yeah, idk.... i think you get a discount for perfects on AMC 8 but idk about a full scholarship... -_-

I think you get something like a $ 500 cut from the fees if you get 25 on AMC 8, and$250 if you are something like the top for MathCounts or something. I never bothered with that MathCounts one because I'm not from US. You could get more info at the MathPath official site at http://www.mathpath.org.

I don't know whether or not there is financial aid, so you could look into that. MathPath isn't run for profit or anything, so I don't think they have that much extra money for discounts. Also, what you pay for camp covers field trips, meals, dorms, classes, etc., which is a good bit.

Camp is totally worth the money. Trust me :-).

There have been campers that have done local fund raising to go to camp. You might want to look into that. Discounts are hard to come by because the camp does not make any money off the fee.

[quote="isabella2296"]Only 500 USD discount? Is there any other way to get a discount?  :maybe:[/quote]

lol, its free if you place first in written round mc nats.

Okay, thanks everyone!

I think I'll look into that local fund raising...  :)

One girl this year got a full discount because she was the best 7th grade girl at Mathcounts.

Local Fundraising is quite popular and some kids did that this year.

i heard the same, that if u get a perfect you'll get in 4 free.

Well I got a perfect AMC 8 in 7th grade and didn't get squat.
Oh well.

Hi guys, this is Mr M.  The answer to your question is posted on the MathPath website, which an earlier respondent listed.  The answer sometimes changes from year to year; check the "last updated" date on the bottom of each webpage.  Anyway, the answer for 2009 is Yes, a $ \$$250 discount for a perfect score of 25 on the AMC8 in Dec 2008, if you attend for all 4 weeks.  Of course, you have to [i]report[/i] the perfect score to us on your registration form to get the discount.

There are clearly some rumors circulating, for instance, what CyberGenius says.  So when in doubt, check the website, or ask us, the people who run MathPath.

Best wishes!

[quote="BOGTRO"]I can't find the page, but:

Discounts:

1. $ \$250$ for 25 on AMC 8
2. $ \$500$ for making being on the winning state team
3. FULL scholarship for winning Nats written round.[/quote]

Yeah, I got an invitation and discount in the mail for making mathcounts nats, but I wasn't on the winning state team (wasnt even on the official school team).

The discounts changed somewhat this year, especially in light of the new 2-week option.  Any year they might change.  The website always has the correct information.  Goto http://www.mathpath.org/fees/schol.htm

This will be our 3rd time at Colorado College.  It's a nifty location.


I recently changed my primary ISP from a DSL-provider to Comcast.  I am now unable to remain connected to the classroom for an extended period, because the connection is dropped.

Has anyone else had this experience with Comcast, and is there any known workaround?

For the record, I have a similar issue when I attempt to connect to Schwab's StreetSmart Pro service.  The connection disconnects randomly, with a mean time to disconnect of, typically, 15 to 30 minutes.

[quote="Valentin Vornicu"]I think you have to take that up with your provider. That's the best chance of getting anything solved (or change it again).[/quote]
Thanks, Val.  I have no doubt the problem lies with my Internet provider and/or my equipment, and I'm not trying to suggest there is any problem with AoPS hardware, software or connectivity.  However, I am contemplating changing providers, again.  Before I make that decision, I'm simply trying to determine if this problem is unique to me and my connection, or if AoPS and/or your users have noticed a general problem staying connected to the classroom via this ISP.

Comcast has been filtering various stuff (mostly related to illegal stuff). Perhaps it has AoPS marked down for that?

I'm pretty sure that was mostly BitTorrent files (and only uploading...). Thats over anyhow; Comcast and BitTorrent have officially made up...

They were using Sandwine or something like that. Haven't read a lot about it. Do you have any firewalls/

My comcast was fine during the whole AIME math jam.


this one make me confused
prove the ineq
$ \ln{(1+\sqrt{1+x^2})}<\frac{1}{x}+\ln{x}   (x>0) $

It is equivalent to $e^{\frac 1 x} > \frac {1 + \sqrt {1+x^2} } x.$
Now, it is well-known that $e^X \geq 1 + X + \frac 1 2 X^2$, with equality iff $X=0$, thus it suffices to prove that $1 + \frac 1 {2x^2} \geq \sqrt {\frac {1+x^2} {x^2}}$ which is trivial by squaring...

Pierre.

$e^x \geq 1+x+\frac{1}{2} x^2$, can you please tell me why?

yes,it is true. I have two solutions for this one.First one is base on the expansion in power series of $e^x$ and the other solution is based on Bernoulli's inequality.If you want to post one of them or both,I'm here. ;)

shobber: [tex]e^x = \sum_{j = 0}^{\infty} \frac{x^j}{j!} = 1 + x + \frac{x^2}{2} + \ldots[/tex]  :)

[quote]yes,it is true. I have two solutions for this one.First one is base on the expansion in power series of  and the other solution is based on Bernoulli's inequality.If you want to post one of them or both,I'm here[/quote]
can you show me the solution based on bernouli 's ineq? :lol:


geometry
perimeter
inequalities
AMC
AIME
triangle inequality

Hey, we went over this in math team today and both my Math Team coach and I don't really understand the answer.

I won't post the answer here so you guys can give a whack at it, but can any of you explain the solution given here? ( http://www.artofproblemsolving.com/Wiki/index.php/2002_AMC_12B_Problems/Problem_24 ). We are looking for a rigorous explanation to the first conclusion of the solution (which is the part we don't understand).

The problem:

"A convex quadrilateral ABCD with area 2002 contains a point P in its interior such that PA = 24, PB = 32, PC = 28, PD = 45. Find the perimeter of ABCD.

$ A. 4 \sqrt {2002}$ $ B. 2 \sqrt {8465}$ $ C. 2(48 \plus{} \sqrt {2002})$ $ D. 2(\sqrt {8633})$ $ E. 4(36 \plus{} \sqrt {113})$

"

Heh, I was the one who posted that solution but was too lazy to explain the first inequality.  :oops:

Let $ H_1,H_2$ be the feet of the altitudes from $ B,D$ to $ \overline{AC}$ respectively; then $ [ABCD] \equal{} [ACB] \plus{} [ACD] \equal{} \frac 12 \cdot AC \cdot (BH_1 \plus{} DH_2)$. Furthermore, $ BH_1 \le BO$ and $ DH_2 \le DO$ (altitude is the shortest cevian, or a segment from vertex to base), with equality for both statements occuring only if $ H_1 \equal{} H_2 \equal{} O$ and $ \overline{AC} \perp \overline{BD}$.

[b]Edit[/b]: Sorry, I used poor notation here. I changed it: $ O$ is the intersection of the diagonals $ AC$ and $ BD$ - the location of $ P$ with respect to the vertices of $ ABCD$ has not yet been used.

Thanks! It's much clearer now (though still took a bit to grasp)...

I don`t get why the segments $ AP$ and $ CP$ form one line (the same goes with $ BP$ and $ DP$). Isn't it possible to get a figure like this one? [geogebra]11f67fb8f42689cbe95cd3a5d9bb7ba38eb96a20[/geogebra]

That's what my coach and I couldn't figure out.

Azjps explained it in his post (I'll try to explain in other words, this will help me understand as well): the area of the quadrilateral is equal to the sum of the two triangles if you drew $ \overline{AC}$. To get this, you need to multiply $ \overline{BD}$ by the altitudes to $ \overline{BD}$ from $ A$ and $ C$ (and by $ \frac {1} {2}$ as well), or alternatively, $ \overline{AC}$ by $ \overline{BP}$ and $ \overline{DP}$. By definition, an altitude is the shortest cevian possible, so $ \overline{AP}$ and $ \overline{BP}$ (not collinear) were less than or equal to the altitudes. If they were equal to the altitudes, $ \overline{BP}$ and $ \overline{DP}$ would be collinear, as would the others.

Now applying the triangle inequality, you can see that the largest number $ \overline{AC}$ can be is $ 52$ and $ \overline{BD}$ is $ 77$. In that case, the diagonals would be perpendicular (and the lines would be collinear).

In this case, our area of $ ABCD$ would also be at its greatest possible value. So, let's solve for the maximum:

$ [ABCD] \le \frac {1}{2}\ (52)(77) = 2002$. Since the area is given as $ 2002$, our area must be at its maximum value, and the only way to achieve that is to have $ \overline{AC}$ and $ \overline{BD}$ be perpendicular (if they weren't, the area would be less). Thus, the segments are equal to the altitudes and all the segments are collinear.

The rest of the problem is easy as you just apply Pythagorean theorem.

I must say, explaining it myself made me understand it much better. I have to try this more. :o
This was surprisingly fun.

Let me read it slowly, it is quite difficult for me. However, thanks!

Sorry, I edited, my first try was incorrect.

Chalao, this just took me like 2 hours to fully conceptualize and understand. Don't worry about it.

Ok, I've edited it about 6 times now. Sorry if my mistakes confused you.

I wonder if there is an easier way to solve it. This solution make this problem look more like an AIME problem.   :huh:

And only 4% of AMC 12 takers got the problem right that year. I think it's meant to be AIME level.


geometry
3D geometry
sphere
trigonometry
function
geometry unsolved

[b]Euclidean Sphere[/b]
I have posted before but no reply :maybe:
So,I post here for professionals' help.

I hope your exclamation is more of dismay rather than frustration, since the answer is not that complicated.

From usual knowledge, $ <\vec{OA}, \vec{OB}> \equal{} \cos \angle AOB$ et.al. So we have to find the extremes of $ f(x,y,z) \equal{} \cos x \plus{} \cos y \plus{} \cos z$, with restrictions $ x,y,z \geq 0$ and $ x\plus{}y\plus{}z \leq 2\pi$ (since the sum of angles at vertex $ O$ of (possibly degenerate) tetrahaedron $ OABC$ is at most $ 2\pi$).
Now the maximum is easily seen to be $ 3$, when $ A \equiv B \equiv C$, so the three angles are all equal to $ 0$.
For the minimum, we can guess it is $ \minus{}3/2$, when the three angles are all equal to $ 2\pi/3$. Any combination with two of the angles not larger than $ \pi/2$ yields a value not less than $ \minus{}1$, so we need at least two of the angles at least $ \pi/2$. Using the convexity of the $ \cos$ function on $ [\pi/2, 3\pi/2]$, first for only two angles, then for all three, yields the verification of our guess.


function
algebra
domain
complex analysis
vector
linear algebra
matrix

Let $ f_1,f_2,\dots,f_n$ be regular functions on a domain of the complex plane, linearly independent over the complex field. Prove that the functions $ f_i\overline{f}_k, \;1 \leq i,k \leq n$, are also linearly independent.

[i]L. Lempert[/i]

A generalization:

Let $ f_i,\ i \equal{} \overline{1,n}$ be linearly independent holomorphic functions on a domain and let $ g_j,\ j \equal{} \overline{1,m}$ be linearly independent antiholomorphic functions on the same domain. Then the functions $ f_ig_j$ are linearly independent.

$ \partial$ and $ \bar\partial$ denote, as usual, the differential operators $ \frac 12\left(\frac\partial{\partial x} \minus{} i\frac\partial{\partial x}\right)$ and $ \frac 12\left(\frac\partial{\partial x} \plus{} i\frac\partial{\partial x}\right)$ respectively. Now assume we have a relation of the form $ \sum a_{ij}f_ig_j \equal{} 0$ (where $ a_{ij}$ are complex numbers). This can be written as $ \sum h_if_i \equal{} 0\ (*)$, where the $ h_i$ are linear combinations of $ g_j$'s with the coefficients $ a_{ij}$ (and are hence antiholomorphic). Because $ \partial f_i \equal{} 0$, applying a power $ \bar\partial^k$ of $ \bar\partial$ to $ (*)$ yields the equation $ \sum\bar\partial^k(h_i)f_i \equal{} 0$. If not all $ h_i$ are zero, $ k$ can be chosen such that $ \bar\partial^k h_i$ are constants and not all zero; this contradicts the linear independence of the $ f_i$'s, so all $ h_i$ must be zero. But as noted before, $ h_i \equal{} \sum_j a_{ij}g_j$, so, since $ g_j$ are linearly independent, all $ a_{ij}$ must be zero.

Edit:

Mine is also flawed, apparently :). I kept thinking about the functions as being polynomials. Sorry about that; I'll try to repair it.

Let $ D\subset\mathbb C$ be the domain the functions are defined on and $ V$ the $ n$-dimensional $ \mathbb C$-vector space spanned by $ f_1,\ldots,f_n$. The vector space $ V_z \equal{} \{f\in V\mid f(z) \equal{} 0\}$ (for $ z\in D$) is either $ V$ or 1-codimensional in $ V$. Since $ \cap_{z\in D}V_z \equal{} \{0\}$ there are pairwise different points $ z_1,\ldots,z_n\in D$ with $ \cap_{i \equal{} 1}^nV_{z_i} \equal{} \{0\}$. This implies that $ (f_i(z_j))_{i,j}$ is a regular $ n\times n$-matrix (*).

The preliminary remark reduces the setting to check everything on finitely many points, namely $ z_1,\ldots,z_n\in D$.

To prove the original claim let $ \sum_{i,j}\lambda_{ij}f_i\overline{f}_j \equal{} 0$ with $ \lambda_{ij}\in\mathbb C$. Plugging in $ z_k$ ($ k \equal{} 1,\ldots,n$) gives $ \sum_i\left(\sum_j\lambda_{ij} \overline{f_j(z_k)}\right)f_i(z_k) \equal{} 0$. Applying (*) twice,[color=red] this yields $ \sum_j\overline{\lambda_{ij}} {f_j(z_k)} \equal{} 0$ (for all $ i,k$)[/color] and finally $ \lambda_{ij} \equal{} 0$ (for all $ i,j$).

EDIT: Flaw marked [color=red]red[/color].

[quote="grobber"]If not all $ h_i$ are zero, $ k$ can be chosen such that $ \bar\partial^k h_i$ are constants and not all zero; ...[/quote]This is not clear to me at all, since the divisor of $ \overline{h_i}$ can be unbounded. But I suspect that this can be amended by evaluating the functions at finitely many points.

PS: I posted this just a minute ago, but it did not show up in the forum. Strange ...  :?:

[quote="grobber"]So you're saying that the statement holds for [i]any[/i] $ n$-uple of linearly independent functions $ f_i$, not necessarily holomorphic? Because this is certainly not true. You can find plenty of pairs of linearly independent functions $ f,g$, for example, such that $ |f| \equal{} |g|$ (so $ f\bar f \equal{} g\bar g$). The fact that they're holomorphic (or regular) is crucial.[/quote]My proof was flawed (cf. edited version: The dependency on $ k$ is quadratic, hence the flaw.) Thanks for pointing that out.

[quote="grobber"]
If not all $ h_i$ are zero, $ k$ can be chosen such that $ \bar\partial^k h_i$ are constants and not all zero [/quote]
Why? :? They are not polynomials!

I did aknowledge the mistake in my previous post, but didn't have time to post this reply. I hope this will (finally) take care of things :).

I'll change the notations a bit, but I'm basically proving the same thing, namely the slightly more general statement in my previous message. I'll restate that as follows: given a domain $ D$, the fields of meromorphic and antimeromorphic functions on $ D$ are linearly disjoint over $ \mathbb C$. Another restatement: given linearly independent antiholomorphic functions $ f_i,\ i\equal{}\overline{1,n}$ on a domain, they are also linearly independent over the ring of holomorphic functions on the same domain. It's under this last form that I'm going to try to prove it. Notice that this time $ f_i$ are antiholomorphic, whereas in the previous post they were holomorphic. Of course, this isn't important at all. We can always interchange the two words in the statement/proof, and nothing changes.

The proof is by induction on $ n$, the case $ n\equal{}1$ being obvious. Now assume we have $ \sum g_if_i\equal{}0\ (*)$, where $ g_i$ are holomorphic with $ g_1\ne 0$, say. Making the domain smaller, if necessary, we can assume that $ g_1$ has no zeros. By dividing $ (*)$ by $ g_1$, we can assume that $ g_1\equal{}1$. Now there are two possibilities: if the other $ g_i$ are constants, we're done, because we have just contradicted the fact that $ f_i$ are linearly independent; if the other $ g_i$ are not all constant, apply $ \partial$ to $ (*)$. You get $ \sum_{i\ge 2}g_i'f_i\equal{}0$, with at least one $ g_i'$ non-zero. This is taken care of by the induction hypothesis.

[quote="grobber"]By dividing $ (*)$ by $ g_1$, we can assume that $ g_1 \equal{} 1$.[/quote]Why are $ f_i/g_1$ ($ i \equal{} 2,\ldots,n$) antiholomorphic and linear independent in order to apply the induction hypothesis later on?

Happy new year 2009 to everybody.

I wasn't implying that they were. Leave the $ f$'s as they are, and instead replace the $ g_i$'s with $ \frac{g_i}{g_1}$, which are still holomorphic, of course.

Another approach is to use the power series and to observe that if $ \sum_{m,n\ge 0}c_{mn}z^m\bar z^n\equal{}0$, then all $ c_{mn}\equal{}0$. Once you know that, you can easily conclude that $ \sum_{i,j} a_{ij}f_i(z)g_i(z)\equal{}0$ actually implies $ \sum_{i,j}f_i(z)g_j(w)\equal{}0$ for all $ z,w$, from where the desired conclusion follows exactly like grobber tried to do in his first post.


algebra
polynomial
number theory unsolved
number theory

Find all polynomials $p(x)$ with integer coefficients such that $p(n)$ divides $2^{n}-1$ for $n = 1, 2, 3, ...$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=1607
http://www.mathlinks.ro/Forum/viewtopic.php?t=23419
http://www.mathlinks.ro/Forum/viewtopic.php?t=19079

  darij


calculus
integration
function
real analysis
limit
real analysis unsolved

If f is a real Riemann integrable function that is strictly greater than 0 on [a,b]. Is it true that

Int f(x) dx > 0

where the integral is taken over [a,b]?

Let the minimum value of $ f$ on $ [a, b]$ be $ m$; then

$ \int_{a}^{b} f(x) \, dx \ge m(b \minus{} a) > 0$.

Answering the question using non-Reimann tools:

A Riemann integrable function must also be Lebesgue integrable. So if we prove this result for Lebesgue integrals, we will also have it for Riemann integrals.

Suppose that $ f$ is measurable on a set $ E$ of positive measure and that $ f>0$ everywhere on $ E.$

Let $ E_n\equal{}\left\{x: f(x)>\frac1n\right\}.$ $ (E_n)$ is an increasing sequence of sets whose union is $ E.$

Hence, $ \lim_{n\to\infty}m(E_n)\equal{}m(E)$ (where $ m$ denotes measure.)

That means that for some $ n,$ $ m(E_n)>0.$ Then

$ \int_Ef\ge\int_E\frac1n\,\chi_{E_n}\equal{}\frac1n\,m(E_n)>0.$

So that gives the answer to weic's question: the integral must be positive.

Challenge: can anyone prove this without the use of Lebesgue theory?

Sorry; I assumed that $ f$ was continuous and implicitly evoked the [url=http://en.wikipedia.org/wiki/Extreme_value_theorem]extreme value theorem[/url].  If I recall correctly, $ f$ being Riemann integral means $ f$ is piecewise continuous and has at most a finite number of pieces, so we can pick a minimum among the pieces.

[quote="t0rajir0u"]If I recall correctly, $ f$ being Riemann integral means $ f$ is piecewise continuous and has at most a finite number of pieces, so we can pick a minimum among the pieces.[/quote]
Your recall is not correct. Those are sufficient conditions for Riemann integrability, but not necessary conditions.

For instance, $ f(x)\equal{}\begin{cases}\frac1q&\text{if }x\equal{}\frac{p}{q}\text{ in lowest terms}\\ 0&\text{if }x\text{ is irrational}\end{cases}$

is Riemann integrable over any bounded interval; it is certainly not piecewise continuous.

The true necessary and sufficient condition for a bounded function on a bounded interval to be Riemann integrable is that it be continuous almost everywhere - that is, the set of its discontinuities has Lebesgue measure zero.

[quote="Kent Merryfield"]
Challenge: can anyone prove this without the use of Lebesgue theory?[/quote]
It's enough to show that $ f$ has a point of continuity. Here's a proof inspired by one in a book, but I will avoid mentioning Lebesgue measure. A disclaimer is that it's 4 AM and I am very tired, so there is a large possibility that it does not check out (because it turns out quite different from the original proof, basically just the definition of the $ E_k$ is stolen). EDIT: Pretty sure it's right

Suppose $ f$ is discontinuous everywhere. For each positive integer $ k$ let
\[ E_k \equal{} \left\{x\in [a,b] : \limsup_{\delta \to 0} \sup_{|y \minus{} x| < \delta , |z \minus{} x| < \delta} |f(y) \minus{} f(z)| > \frac {1}{k}\right\}
\]
Since the union of all $ E_k$ is $ [a,b]$, by the Baire category theorem, one $ E_k$ is not nowhere dense. So, we choose an interval $ (c,d) \subset \overline{E_k}$.

Take any partition $ a \equal{} x_0 < x_1 < \ldots < x_n \equal{} b$. We can suppose the points $ c$ and $ d$ are included,  say $ x_o \equal{} c$, $ x_p \equal{} d$.

Let $ M_i \equal{} \sup_{x_{i \minus{} 1} \leq x \leq x_i} f(x)$, $ m_i \equal{} \inf_{x_{i \minus{} 1} \leq x \leq x_i} f(x)$

Now
\[ \sum_{j \equal{} 1}^{n} (M_i \minus{} m_i)(x_i \minus{} x_{i \minus{} 1}) \geq \sum_{j \equal{} o}^{p} (M_i \minus{} m_i)(x_i \minus{} x_{i \minus{} 1}) \geq \frac {d \minus{} c}{k}
\]
and this contradicts Riemann integrability.

You can make everything a bit shorter if you try to prove the contrapositive: if a non-negative function $ f$ is Riemann integrable and the integral over $ I$ is $ 0$, then $ f(x)\equal{}0$ for some $ x$.

Indeed, if the integral is zero, there exists an upper Riemann sum that is less than $ |I|$. Hence, there is a closed subinterval $ I_1\subset I$ such that $ \sup_{I_1}f\le 1$. Since $ \int_{I_1}f\equal{}0$, we can now find the upper Riemann sum for $ I_1$ that is less than $ \frac 12|I_1|$. Thus there exists a subinterval $ I_2\subset I_1$ with $ \sup_{I_2}f\le\frac 12$. And so on. The intersection $ \bigcap_n I_n\ne\varnothing$ and any point $ x$ in that intersection satisfies $ f(x)\equal{}0$. This proof seems absolutely minimalistic in the sense that using completeness of $ \mathbb R$ is inavoidable. You can construct Riemann integral for the functions on $ \mathbb Q$ instead of $ \mathbb R$ and the result will be false.

I didn't notice this yesterday, but Kent Merryfields proof can also be translated into a proof with category instead of lebesgue measure, just like I did for the stronger statement of continuity.


Find all primes that can be represented as $ 8n^2\plus{}10n\plus{}3,n\in\mathbb{Z}$.

[quote="Bugi"]Find all primes that can be represented as $ 8n^2 \plus{} 10n \plus{} 3,n\in\mathbb{Z}$.[/quote]

$ 8n^2 \plus{} 10n \plus{} 3 \equal{} p$ $ \iff$ $ 8p \plus{} 1 \equal{} (8n \plus{} 5)^2 \equal{} m^2$ $ \iff$ $ 8p \equal{} (m \minus{} 1)(m \plus{} 1)$ and, since $ p$ is prime :

Either $ m \minus{} 1 \equal{} 2$ and $ m \plus{} 1 \equal{} 4p$ $ \implies$ $ p \equal{} 1$
Either $ m \minus{} 1 \equal{} 4$ and $ m \plus{} 1 \equal{} 2p$ $ \implies$ $ p \equal{} 3$
Either $ m \minus{} 1 \equal{} 4p$ and $ m \plus{} 1 \equal{} 2$ $ \implies$ $ p \equal{} 0$
Either $ m \minus{} 1 \equal{} 2p$ and $ m \plus{} 1 \equal{} 4$ $ \implies$ $ p \equal{} 1$
Either $ m \minus{} 1 \equal{} \minus{} 2$ and $ m \plus{} 1 \equal{} \minus{} 4p$ $ \implies$ $ p \equal{} 0$
Either $ m \minus{} 1 \equal{} \minus{} 4$ and $ m \plus{} 1 \equal{} \minus{} 2p$ $ \implies$ $ p \equal{} 1$
Either $ m \minus{} 1 \equal{} \minus{} 4p$ and $ m \plus{} 1 \equal{} \minus{} 2$ $ \implies$ $ p \equal{} 1$
Either $ m \minus{} 1 \equal{} \minus{} 2p$ and $ m \plus{} 1 \equal{} \minus{} 4$ $ \implies$ $ p \equal{} 3$

And so the only prime which may be written as $ 8n^2 \plus{} 10n \plus{} 3 \equal{} p$ is  $ \boxed{3 \equal{} 8\cdot 0 \plus{} 10\cdot 0 \plus{} 3}$

[quote="Theorem1"]Nice! but it's little long!
\[ 8{n^2} \plus{} 10n \plus{} 3 \equal{} (2n \plus{} 1)(4n \plus{} 3) \in P\,\,\,\,\,\,\,\,\,
\]
Then n=0 and so
\[ 8{n^2} \plus{} 10n \plus{} 3 \equal{} 3
\]
[/quote]

Ah yes  :lol:
(btw, you forgot $ n\equal{}\minus{}1$, but it does not change anything)

-Yes I know that! I wrote the solution after calculation and cases!


A ball is shot from one corner of a 6 foot by 10 foot pool table at a 45 degree angle. How many times will it bounce off the sides before landing in a corner pocket?

[hide]
An easy way to do this is to imagine a grid of $ 10 \times 6$ boxes, each one representing a pool table. Starting from the bottom left corner draw a line a $ 45^o$ until it meets another corner of the grid.

So we look for the minimum $ (k,n)$ such that $ 6k \equal{} 10n$
Obviously the minimum is $ (5,3)$

Therefore the ball crosses $ 5$ vertical lines and $ 3$ horizontal lines,
But we exclude the final point because the ball does not bounce off it

So the answer is $ (5\minus{}1) \plus{} (3\minus{}1) \equal{} 6$[/hide]


[hide="Hint"]If I'm not mistaken, you can use Euler's formula.

$ e^{ix} \equal{} cos  x \plus{} i sin  x$
[/hide]

[hide="Horrid Attempt"]
From this [url=http://www.mathpropress.com/stan/bibliography/complexSquareRoot.pdf]pdf[/url] we get that the complex number is $ \boxed{1\plus{}2i}$
[/hide]

hello, make the ansatz $ \sqrt{4i\minus{}3}\equal{}a\plus{}bi$.
Sonnhard.

[hide]Let $ \smile$ equal $ \sqrt{4i-3}$.
Thus, $ \smile^{2}=4i-3$.
We guess that $ \sqrt{4i-3}$ will
be a complex number, so
we know then that $ \smile=a+bi$ for
$ a,b \in \R$.
Thus, we have that:
$ a^{2}-b^{2}+2abi=4-3i$.
Equating like parts, we get 2 equations:
$ 2abi=-3 \longmapsto a=\frac{-3}{2bi}$.
We can now plug this into our next equation:
$ a^{2}-b^{2}=4$, to get our
values for $ a$ and $ b$.
Finally, we have to plug these $ a$ and $ b$
into our first equation: $ \smile=a+bi$ as
our answer.[/hide]

@z-coli: $ (\sqrt{4i\minus{}3})^2\equal{}4i\minus{}3\ne 4\minus{}3i$.

[hide="Solution"]Let $ a\plus{}bi \equal{} \sqrt{4i\minus{}3} \implies a^2 \minus{} b^2 \plus{} 2abi \equal{} \minus{}3 \plus{} 4i$.
Equating the real and imaginary parts, we get
$ a^2\minus{}b^2\equal{}\minus{}3$, and $ 2ab\equal{}4\implies ab\equal{}2\implies a\equal{}\frac{b}{2}$.
Therefore, $ \left(\frac{b}{2}\right)^2 \minus{} b^2 \equal{} \minus{}3$. Solve this equation to find that $ b\equal{}2\implies a\equal{}1$. Hence, $ \sqrt{4i\minus{}3}\equal{}\boxed{1\plus{}2i}$.[/hide]


Find with proof all integer solutions $ (x,y,z)$ of the following equation: $ x^{2}+y^{2}+z^{2}=2xyz$

[hide="solution"]$ x=y=z=0$ is a solution. Then we assume that all $ x,y,z$ are positive.

All $ x,y,z$ must be even numbers. If two of them were odd then $ x^{2}+y^{2}+z^{2}$ won't be divisible by $ 4$ and $ 2xyz$ will be.
So: $ x=2x_{1},y=2y_{1},z=2z_{1}$, and we have:
$ x_{1}^{2}+y_{1}^{2}+z_{1}^{2}=4x_{1}y_{1}z_{1}$... analogically we have that $ x_{1},y_{1},z_{1}$ must be even number... and so on... and so on... :) So we have infinity solution in positive integers that $ x>x_{1}>x_{2}>x_{3}>...$, but it's impossible. So only $ (0,0,0)$ is a solution.[/hide]

What do you think about this one?
$ x^{2}+y^{2}+z^{2}= xyz$

[hide]It's more special than the last one, since it has infinite solutions[/hide]

and also another (somehow) similar problem is the following one,which is very famous and also beautiful:
find all natural numbers $ (x,y,z,n)$ s.t $ x^{3}+y^{3}+z^{3}=nx^{2}y^{2}z^{2}$
and the solution can be found at [url]http://mathlinks.ro/Forum/viewtopic.php?t=152272[/url] second post...

Im fairly sure that these equations only have solutions when $ n \in (1,3)$

I'm working on this problem. It's my homework.  :D
I think the equation $ x^{2}+y^{2}+z^{2}=(2k+1)xyz$ with integer $ k\geq 2$ has only one root (0,0,0)
Still working.

[quote="kristysophie"]I've considered this problem
$ x^{2}+y^{2}+z^{2}=3xyz$
It also has infinite roots $ (1,1,1),(1,1,2),(1,2,5)...$

What about $ x^{2}+y^{2}+z^{2}=5xyz$ and $ x^{2}+y^{2}+z^{2}=7xyz$ and more ...  :?:
Anyone has an idea?[/quote]
for $ x^{2}+y^{2}+z^{2}=3xyz$ there is a solution as follow:
note that the equation-as kristysophie said- has at least one root,for example $ (x,y,z)=(1,1,1)$ now we will prove that it has infinitly many solutions...
[hide]we`ll prove that by contradiction,so assume that the number of solutions to the equation are finite,now along all these solutions,choose the one that has the maximum value of $ \max(x,y,z)$...
now WLOG assume that $ \max(x,y,z)=x$,i.e. $ x\geq y,x\geq z$ now we have:
$ x^{2}+y^{2}+z^{2}=3xyz$
$ \Rightarrow \boxed{y^{2}+y(-3xz)+z^{2}+x^{2}=0 (I)}$
now this equation is a second degree equation wrt $ y$...
but in the equation $ ax^{2}+bx+c=0$ let $ x',x''$ be our roots,then we have $ x'+x''=\frac{-b}{a}$
so the sum of the roots of our equation is equal to $ 3xz \in \mathbb{Z}$ and also $ y$ is a root,so the other root is an integer,which is $ 3xz-y$,so $ x,3xz-y,z)$ is also a solution...
but we had $ y\leq x$ i.e. $ xy\leq x^{2}$ so $ xy\leq x^{2}+z^{2}$ which gives us:
$ xy+y^{2}<x^{2}+y^{2}+z^{2}=3xyz$
$ \Rightarrow x+y<3xz \Rightarrow x<3xz-y$
$ \Rightarrow \max(x,y,z)<\max(x,3xz-y,z)$
which is a contradiction...[/hide]

Yes, I've proved that if the equation $ x^{2}+y^{2}+z^{2}= kxyz$ has root $ (a,b,c)$ with $ a\leq b\leq c$ then it also has root $ (kbc-x,b,c)$
This means we only need to find one root $ (a,b,c)$ not $ (0,0,0)$, then the equation has infinite number of solutions.

[hide]
We have $ a$ is one root of the equation $ x^{2}-kxbc+b^{2}+c^{2}=0$
And $ a^{2}+b^{2}+c^{2}= kabc$
That gives $ x^{2}-kxbc+kabc-a^{2}=0$
And we get $ x=a$ or $ x=kbc-a$
[/hide]


This morning when I signed on Cut-the-Knot.org site
   a message came up stating that CTK did try phishing
   and a report was created for Internet to investigate
   this incident.

   Any explanation for such curiousity???

   M.T.


this is supposed to be got in a couple of mins, but i cant get it for ages

In a room there are 100 bulbs , all numbered  1 to 100. Initially they are all off.Outside the room there are 100 people wearing T shirts bearing numbers from 1 to 100 respectiviely . Now the person with number 1 , goes  into the room and switchs on all bulbs  which are $multiples$ of his number. then the person no. 2 goes and switches  off the on bulbs of his number's multiples. So basically every person after going in , all bulbs of his no's multiples he finds off ,he puts them on , and all bulbs of his no's multiples he finds on , puts them off

find ,after the 100 th person has come out , what are the number of 'on' bulbs

Perfect squares have an odd number of divisors.  There are 10 perfect numbers $n \leq 100$, thus $\boxed{10}$ is the answer.  Approximately 2 seconds to get the answer.  :D

Masoud Zargar

wow,  :(  you made it look dumb , damn I couldnt get it for ages

guess i shouldnt have put the "challenge" icon beside it  :blush:

This is a common problem. Such an incredibly elegant solution. First time I encountered it I tried a smaller case and recognized the pattern, at which point Masoud's insight hit me. Very beautiful problem for problem solving.

[quote="eminem"]wow,  :(  you made it look dumb , - I couldnt get it for ages

guess i shouldnt have put the "challenge" icon beside it  :blush:[/quote]

Don't worry, boxedexe has obviously seen that type of problem before[size=59], he's just bragging.[/size]


Find all primes p such that $\frac{2^{p-1}-1}{p}$ is a perfect square.

[quote="hydro"]
from (2): consider $2^{\frac{p-1}{2}}$: it must be $\equiv 1 \mod 3$
[/quote]

Why?  :?

Already posted many times, incl [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=341880#p341880]here[/url]

it says in the other post, by fermat's quotient, what is exactly this?

You can find more information about Fermat Quotient [url=http://primes.utm.edu/glossary/page.php?sort=FermatQuotient]here[/url]


Find all  $f$ of reals into reals such that $f(f(x )+y)=f(x^2-y)+4f(x )y$

Take $y = -f(x)$ and $y = x^2$ and see what you get :)

\[f(0) = f\left(x^2 + f(x))\right) - 4f(x)^2\]

\[f\left(f(x) + x^2)\right) = f(0) + 4x^2f(x)\]

Hence \[4f(x)^2 = 4x^2f(x)\] for all $x$ or \[f(x)(f(x) - x^2) = 0.\]

That means that for each real $x$ we have $f(x) = 0$ or $f(x) = x^2$. Be careful, that doesn't necessarily mean that $f(x) = 0$ and $f(x) = x^2$ are the only solutions...  However it is easy to show that a "mix" of solutions (for example $f(a) = 0$ and $f(b) = b^2$ for nonzero $a$ and $b$) cannot occur (just plug in $a$ and $b$ and so on and that will become obvious). So we are done and the only solution are $f(x) = 0$ and $f(x) = x^2$.


[quote="CT_14"]Let AbCD be a square have center E.M is the midpoint of Ab.G is a point on side BC,H is a point on CD satisfying MG and AH parallel.Find angle GEH :)[/quote]
[hide] Assuming $\angle{GEH}$ is the same for all cases, we can set $H=C$. It is easy to see that $\angle{GEH}=\boxed{45^{\circ}}$. [/hide]

[quote="TRAN THAI HUNG"]How do you know that is the same? :?[/quote]

he did not really solve the problem; he used the fact that the question implies there there is one value...

to actually solve it, you can make side=2, and GC=x, and brute force...but i expect that there is an elegant solution too


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a finite group and let $ X$ be the set of all subgroups of $ G$. Show that $ G$ acts on $ X$ by $ g : H \mapsto gHg^{\minus{}1}$ for $ g \in G$ $ H \in X$, where $ gHg^{\minus{}1} \equal{} \{ ghg^{\minus{}1} : h \in H\}$.

Show that the orbit containing $ H$ in this action of $ G$ has size at most $ \frac{|G|}{|H|}$.

If $ H$ is a proper subgroup of $ G$, show that there exists an element of $ G$ which is contained in no conjugate $ gHg^{\minus{}1}$ of $ H$ in $ G$.

I don't really know how to begin this problem because we're dealing with actions on subgroups - fundamentally I'd like to know what constitutes a group action as well!

for the first problem, note that $ gHg^{ \minus{} 1} \equal{} (gh)H(gh)^{ \minus{} 1}$ for every $ h\in H$, so in listing all $ gHg^{ \minus{} 1}$ for $ g\in G$, every $ gHg^{ \minus{} 1}$ occurs at least $ |H|$ times, so there are at most $ \displaystyle \frac {|G|}{|H|}$ different $ gHg^{ \minus{} 1}$'s.

for the second question... by the first problem, the union of all the $ gHg^{ \minus{} 1}$ has size $ \leq |G|$. Equality can only occur if distinct $ gHg^{ \minus{} 1}$'s are disjoint. but that's impossible because they have at least the identity in common. thus there is an element in $ G$ which is not in the union of the $ gHg^{ \minus{} 1}$'s, and the result follows.

There is another problem before that - how do you show that $ G$ acts on the set $ X$?

It also appears counterintuitive when we talk about the orbit of $ H$ under this action - the set $ X$ contains all subgroups of $ G$ but they may be different sizes. How could $ gHg^{\minus{}1}$ take $ H$ to another group then (unless $ H$ is normal)?

[quote="pleurestique"]for the first problem, note that $ gHg^{ \minus{} 1} \equal{} (gh)H(gh)^{ \minus{} 1}$ for every $ h\in H$, so in listing all $ gHg^{ \minus{} 1}$ for $ g\in G$, every $ gHg^{ \minus{} 1}$ occurs at least $ |H|$ times, so there are at most $ \displaystyle \frac {|G|}{|H|}$ different $ gHg^{ \minus{} 1}$'s.[/quote]

Also for this part, why not at most $ |G| \minus{} |H|$ different $ gHg^{\minus{}1}$'s?

[quote="aidan"]There is another problem before that - how do you show that $ G$ acts on the set $ X$?

It also appears counterintuitive when we talk about the orbit of $ H$ under this action - the set $ X$ contains all subgroups of $ G$ but they may be different sizes. How could $ gHg^{ \minus{} 1}$ take $ H$ to another group then (unless $ H$ is normal)?[/quote]

to show that $ G$ acts on the set $ X$ by $ H\mapsto gHg^{\minus{}1}$, verify that the group action properties are satisfied (easy exercise). part of verifying this is showing that $ gHg^{\minus{}1}$ is a subgroup if $ H$ is a subgroup (this, again, is an easy exercise which i suggest you try yourself).

[quote="aidan"][quote="pleurestique"]for the first problem, note that $ gHg^{ \minus{} 1} \equal{} (gh)H(gh)^{ \minus{} 1}$ for every $ h\in H$, so in listing all $ gHg^{ \minus{} 1}$ for $ g\in G$, every $ gHg^{ \minus{} 1}$ occurs at least $ |H|$ times, so there are at most $ \displaystyle \frac {|G|}{|H|}$ different $ gHg^{ \minus{} 1}$'s.[/quote]

Also for this part, why not at most $ |G| \minus{} |H|$ different $ gHg^{ \minus{} 1}$'s?[/quote]

think about it. i said "every $ gHg^{\minus{}1}$ occurs at least $ |H|$ times," the key word being "times"...

I think the two properties are $ ex \equal{} x \quad \forall x \in X$, $ g(h(x)) \equal{} (gh)(x) \quad \forall h, g \in G$?

Then I have $ eHe^{\minus{}1} \equal{} H$, so check; also, $ (gh)H(gh)^{\minus{}1} \equal{} ghHh^{\minus{}1}g^{\minus{}1} \equal{} g(h(H))g^{\minus{}1} \equal{} (gh)(H)$, so that checks as well. I think the notation can be improved, though.

Suppose $ h_1$, $ h_2$ are 2 elements of $ H$. Then $ gh_1g^{\minus{}1} \cdot gh_2g^{\minus{}1} \equal{} gh_1h_2g^{\minus{}1} \in gHg^{\minus{}1}$, where $ \cdot$ denotes the binary operation wrt the group $ G$. In particular every element $ gh_1g^{\minus{}1}$ has an inverse $ gh_1^{\minus{}1}g^{\minus{}1}$, because their product is $ e$. $ (gh_1g^{\minus{}1})(gh_2g^{\minus{}1}gh_3g^{\minus{}1}) \equal{} (gh_1g^{\minus{}1})(gh_2h_3g^{\minus{}1}) \equal{} (gh_1g^{\minus{}1}gh_2h_3g^{\minus{}1})$ This equals $ (gh_1h_2h_3g^{\minus{}1})  \equal{} (gh_1h_2g^{\minus{}1})(gh_3g^{\minus{}1}) \equal{} (gh_1g^{\minus{}1}gh_2g^{\minus{}1})(gh_3g^{\minus{}1})$, so the product is associative. Closure is inherited, so $ gHg^{\minus{}1}$ is a group.

I think that nails it... and I get the last part about $ \frac{|G|}{|H|}$ now :)


Let $ x_1,...,x_n\ge0$. Define $ p_k: \equal{} \frac1n\sum_{i \equal{} 1}^nx_i^k$.

For two partitions $ a \equal{} (a_j)$ and $ b \equal{} (b_j)$ of $ m\in \mathbb N$, such that $ a\succ b$, prove
.
\[ \prod p_{a_j}\ge \prod p_{b_j}.
\]
[hide]For example $ \sum x^4(\sum x)^2\ge (\sum x^2)^3$ because $ (4,1,1)\succ (2,2,2)$ :)[/hide]

Yes,I think so,too.
We can use Karamata inequality to prove this.

Exactly.
By the way, the converse is true as well.

We can formulate the whole as this: [hide](Note that $ n$ has nothing to do with the parts of $ a$ and $ b$.) [/hide]
For a given $ n\ge 2$ and two partitions $ a \equal{} (a_j)$ and $ b \equal{} (b_j)$ of $ m\in \mathbb N$, we have
\[ \boxed{a\succ b\Longleftrightarrow \prod p_{a_j}\ge \prod p_{b_j}\ \ \forall x_1,...,x_n\ge0.}
\]
If there are more than two partitions, we can extend the "$ \Longrightarrow$" but not the converse, e.g. if $ a,b,c,d\vdash m$ and $ [a] \plus{} [b]\ge[c] \plus{} [d]$, this implies $ \prod p_{a_j} \plus{} \prod p_{b_j} \ge\prod p_{c_j} \plus{} \prod p_{d_j}\ \ \forall x_1,...,x_n\ge0$.

The ineq with the products tends to be weaker than the original one. Writing for the occasion (for a fixed $ n$) $ \{a\}: \equal{} \prod p_{a_j}$ etc., and starting for example with Schur which states $ [3,0,0] \plus{} [1,1,1]\ge2\cdot [2,1,0]$, we'll get the following:

$ \{3,0,0\} \plus{} \{1,1,1\}\ge2\{2,1,0\}$.
But this is weak, as in fact, we have e.g. for $ n \equal{} 3$ the stronger

$ \{3,0,0\} \plus{} \frac32\{1,1,1\}\ge\frac52\{2,1,0\}$.

So this little theorem won't yield interesting (i.e. very sharp) inequalities  :roll:

[quote="gb2124"]Then what is $ [a]\plus{}[ b ]>\equal{}[c]\plus{}[d]$[/quote]
This is the "[b]symmetric sum notation[/b]", e.g. see here for a definition : http://www.mathlinks.ro/Forum/viewtopic.php?t=167253. A statement like the above is supposed to mean "... for all $ x_1,x_2,x_3\ge0$" etc.
see [url=http://www.mathlinks.ro/viewtopic.php?t=56369]here[/url] for more :)


greatest common divisor
number theory unsolved
number theory

a_1 =1 ,a_2 =2 ,a_3=3
For n>3 : let a_n be the smallest positive integer which is different from a_i ,i=1,2,...,n-1
and the greatest common divisor of  a_n ,a_(n-1)  >=3

Prove that : the sequence a_n contains all positive integer

Sloane's database lists this as the "EKG sequence" because of its electrocardiogram-like graph:

http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A064413

http://arxiv.org/abs/math.NT/0204011


[quote="James Bond007"]I don`t think that it`s that hard a problem.I suppose you did not work on it for sometime.However here is your sol.

it is easy to prove that 1<a+b<2....(1);1<b+c<2...(2);1<c+a<2....(3) also observe that

1+1/2<1+c<2...(4);1+1/2<1+a<2...(5);1+1/2<1+b<2.....(6)

divide the first ineq. by the fourth

2/3<a+b/1+c<1 and continue similarly, after which add them you will get

2<.......<3. N.B.`<` means less than or equal to.[/quote]

I don't think you can divide one inequality by another.
If $ x\le y \le z$, then $ \frac{1}{x}\ge\frac{1}{y}\ge\frac{1}{z}$

Inequalities can be multiplied, not divided.

[quote="karis"]$ \sum(\frac {a \plus{} b}{1 \plus{} c} \plus{} 1) \equal{} \sum(1 \plus{} c)\sum\frac {1}{1 \plus{} c} \minus{} 2\sum\frac {1}{1 \plus{} c}\geq 9 \minus{} 4 \equal{} 5$[/quote]

Thanks!  :)
Now for the other inequality.

[quote="thomaskurian89"]If $ a, b, c$ are real numbers such that $ \frac {1}{2} \le a \le 1, \frac {1}{2} \le b \le 1, \frac {1}{2} \le c \le 1$;

prove that $ 2 \le \frac {a \plus{} b}{1 \plus{} c} \plus{} \frac {b \plus{} c}{1 \plus{} a} \plus{} \frac {c \plus{} a}{1 \plus{} b} \le 3$[/quote]
This problem, the solution is available at our new book: [b][i]Old And New Inequalities II[/i][/b], Vo Quoc Ba Can - Cosmin Pohoata, GIL publishing house, 2008. :)

karis has proved that

$ \frac {a \plus{} b}{1 \plus{} c} \plus{} \frac {b \plus{} c}{1 \plus{} a} \plus{} \frac {c \plus{} a}{1 \plus{} b} \ge 2$

Now how do I prove the other half, namely:

$ \frac {a \plus{} b}{1 \plus{} c} \plus{} \frac {b \plus{} c}{1 \plus{} a} \plus{} \frac {c \plus{} a}{1 \plus{} b} \le 3$?

After expanding we get its equivalent to:
$ 3abc\plus{}ab\plus{}bc\plus{}ca\plus{}a\plus{}b\plus{}c\plus{}3 \geq \sum ab(a\plus{}b) \plus{} 2(a^2\plus{}b^2\plus{}c^2)$
using the fact $ 2 \geq a\plus{}b$, $ 2 \geq c\plus{}b$, $ 2 \geq a\plus{}c$ we get $ 2ab\plus{}2bc\plus{}2ac \geq \sum ab(a\plus{}b)$, so we have its enough to prove:
$ 3abc\plus{}a\plus{}b\plus{}c\plus{}3 \geq 2(a^2\plus{}b^2\plus{}c^2)\plus{}ab\plus{}bc\plus{}ca$, which we get after using the fact $ \sum (1\minus{}a)(1\minus{}bc) \geq 0$ and $ a\plus{}b\plus{}c \geq a^2\plus{}b^2\plus{}c^2$.
Equality iff $ a\equal{}b\equal{}c\equal{}1$.

[quote="thomaskurian89"]If $ a, b, c$ are real numbers such that $ \frac {1}{2} \le a \le 1, \frac {1}{2} \le b \le 1, \frac {1}{2} \le c \le 1$;

prove that $ 2 \le \frac {a \plus{} b}{1 \plus{} c} \plus{} \frac {b \plus{} c}{1 \plus{} a} \plus{} \frac {c \plus{} a}{1 \plus{} b} \le 3$[/quote]
We have

$ .\frac {a\plus{}b}{1\plus{}c} \plus{} \frac {b\plus{}c}{1\plus{}a} \plus{} \frac {c\plus{}a}{1\plus{}b} \ge\ \frac {a\plus{}b}{a\plus{}b\plus{}c} \plus{} \frac {b\plus{}c}{b\plus{}c\plus{}a} \plus{} \frac {c\plus{}a}{c\plus{}a\plus{}b} \equal{} 2$

$ .\frac {a\plus{}b}{1\plus{}c}\equal{}\frac {a}{1\plus{}c}\plus{}\frac {b}{1\plus{}c} \le\ \frac {a}{a\plus{}c}\plus{}\frac {b}{b\plus{}c}$

Similar we have

$ \frac {a \plus{} b}{1 \plus{} c} \plus{} \frac {b \plus{} c}{1 \plus{} a} \plus{} \frac {c \plus{} a}{1 \plus{} b} \le\ \frac {a}{a\plus{}c}\plus{}\frac {b}{b\plus{}c}\plus{} \frac {b}{b\plus{}a}\plus{} \frac {c}{c\plus{}a}\plus{} \frac {c}{c\plus{}b}\plus{} \frac {a}{a\plus{}b}\equal{}3$

We have done

[quote="onlylove_math"][quote="thomaskurian89"]If $ a, b, c$ are real numbers such that $ \frac {1}{2} \le a \le 1, \frac {1}{2} \le b \le 1, \frac {1}{2} \le c \le 1$;

prove that $ 2 \le \frac {a \plus{} b}{1 \plus{} c} \plus{} \frac {b \plus{} c}{1 \plus{} a} \plus{} \frac {c \plus{} a}{1 \plus{} b} \le 3$[/quote]
We have

$ .\frac {a \plus{} b}{1 \plus{} c} \plus{} \frac {b \plus{} c}{1 \plus{} a} \plus{} \frac {c \plus{} a}{1 \plus{} b} \ge\ \frac {a \plus{} b}{a \plus{} b \plus{} c} \plus{} \frac {b \plus{} c}{b \plus{} c \plus{} a} \plus{} \frac {c \plus{} a}{c \plus{} a \plus{} b} \equal{} 2$

$ .\frac {a \plus{} b}{1 \plus{} c} \equal{} \frac {a}{1 \plus{} c} \plus{} \frac {b}{1 \plus{} c} \le\ \frac {a}{a \plus{} c} \plus{} \frac {b}{b \plus{} c}$

Similar we have

$ \frac {a \plus{} b}{1 \plus{} c} \plus{} \frac {b \plus{} c}{1 \plus{} a} \plus{} \frac {c \plus{} a}{1 \plus{} b} \le\ \frac {a}{a \plus{} c} \plus{} \frac {b}{b \plus{} c} \plus{} \frac {b}{b \plus{} a} \plus{} \frac {c}{c \plus{} a} \plus{} \frac {c}{c \plus{} b} \plus{} \frac {a}{a \plus{} b} \equal{} 3$

We have done[/quote]

Thanks so much!  :)

That was a truly elegant proof.


Let $V$ and $V'$ be vector spaces over $K$ and let $W$ be a subspace of $V$

Show that $\Gamma = \{T \in L(V,V') \vert T(w)=0$ for all $w \in W \}$ is a subspace of $L(V,V')$ and this subspace is isomorphic to $L(V/W, V')$.
Deduce that if $V$ and $V'$ are finite dimensional, then
dim$\Gamma$ = (dim$V$ - dim$W$)dim$V'$

The isomorphism:
Map $f\in \Gamma$ to $f'\in L(V/W,V')$ by
$f(x)=f'(x+W)$. By the definition of $\Gamma$, this is independent of the choice of $x$ within a coset, and is therefore well-defined.

This is all very abstract; the verification that $\Gamma$ is a subspace and that the map we have defined is a linear isomorphism are routine from the definitions.


Daca f este derivabila pe $[a, b]$ si $f`(a) = f`(b)$ atunci exista c din (a, b) astfel incat $f(c) - f(a) = (c-a) f`(c)$
Si inca o intrebare: Continuitate + Darboux = Darboux ? (presupun ca nu, dar vreau sa ma asigur)

[size=117][color=red][u][b]Teorema lui T. M. Flett (Math. Gaz., 42(1958)), p.38-39:[/b][/u][/color][/size]
Dacă funcţia reală $f$ este diferenţiabilă pe $\left[ {a,b} \right]$  şi $f'\left( a \right) = f'\left( b \right)$, atunci există cel puţin un punct $\xi  \in \left( {a,b} \right]$ pentru care avem:
	\[
f'\left( \xi  \right) = \frac{{f\left( \xi  \right) - f\left( a \right)}}{{\xi  - a}}
\]

[img]http://img.photobucket.com/albums/v218/hudrea/Flett.jpg[/img]

Yeah, it is a very good reference. :lol:  :lol:  :lol:

Frozen, nu am un contraexemplu pentru continuu+Darboux=Darboux, dar o derivata+o functie continua are proprietatea lui Darboux, pentru ca este tot o derivata. Poti sa folosesti asta. Uite inca o problema, care e tot cam asta de aici. Cred ca a fost propusa la o nationala mai veche:

Se da functia derivabila reala $f:[a,b]\to\mathbb R$ astfel incat pe dreapta determinata de punctele $(a,f(a)),(b,f(b))$ mai exista un punct $(c,f(c))$ din graficul functiei. Atunci exista o tangenta la graficul functiei care trece prin $(a,f(a))$.


Tried noobing it out with quadratic formula, no success... looking for clever substitutions.

Mathematica trial can't noob it out either

Could I get a hint?

[hide="Solution"]$ x\neq 0$, so the equation is equivalent to $ 4 \equal{} 2^{1/x}x\plus{}\frac{2^{x}}{x}$. Notice that $ x > 0$ (otherwise, $ RHS < 0 < 4$), so according to AM-GM we have $ 2^{1/x}x\plus{}\frac{2^{x}}{x}\geq 2\sqrt{2^{x\plus{}1/x}}\geq 2\sqrt{2^{2}}\equal{} 4$. Equality occurs, so $ x \equal{} 1$ (to get equality in the second inequality). Therefore, $ x\equal{}1$ is the only solution.[/hide]


Two circles, one of radius 5 inches, the other of radius 2 inches, are tangent at point P. Two bugs start crawling at the same time from point P, one crawling along the larger circle at 3π inches per minute, the other crawling along the smaller circle at 2.5π inches per minute. How many minutes is it before their next meeting at point P?

[hide="Answer"]Ant $A$ makes it to point $P$ every $\frac{10}{3}$ minutes, and ant $B$ makes it to point $P$ every $\frac{8}{5}$ minutes. $LCM(\frac{10}{3} ,\frac{8}{5})=\boxed{40}$ minutes.[/hide]

[hide]

The bug on the larger circle takes [tex] \frac {10}{3}[/tex] minutes to go one revolution.

The bug on the smaller circle takes [tex]\frac {4}{2.5}[/tex] minutes to go one revoulition.

so for every revolution the bug on the larger circle takes, the bug on the smaller circle

travels [tex]\frac{\frac {10}{3}}{\frac {4}{2.5}} = 2 \frac {1}{12}[/tex] revolutions.

for every revolution that the bug on the larger circle takes, the bug on the smaller circle will gain [tex] \frac {1}{12}[/tex] of a revolution on the smaller circle.

So it will take the bug on the larger circle 12 complete revolutions before both bugs meet again at point P.

Therefore the time that elapses before the next meeting is [tex](\frac {10}{3} \frac {min}{rev})(12 \ rev) = 40 \ min[/tex][/hide]

What do you know! I agree with JesusFreak197!!