induction
number theory proposed
number theory

Prove that there is a multiple of 2^67 that can be written only with digits 6 and 7.

We can show by induction that for each $n\ge 1$, there is a multiple of $2^n$ with $n$ digits, formed only by the non-zero digits $a$ and $b$ of different parity (assume $a$ is even). We do this by induction. Having constructed $k_n$, we go to $n+1$ by putting $a$ to the left of $k_n$ if $2^{n+1}|k_n$ and $b$ otherwise.

Ok, now a little further: show that there is exactly 1 k_n for every k (mod 2^n)

:)

That follows directly from the construction. All we need to change in the proof is keep the uniqueness in mind when we take the induction step.


Find distinct positive integers a, b, c, such that

[b]              a + b + c,     ab + bc + ca,        abc[/b]

forms an arithmetic progression.

Does your problem ask how many pairs of (a,b,c) satisfy the equation? or just find a triplet that works?

alright  0,0,0 is a constant arithmetic progression, so hah (0,0,0)

:spam:

Also, read the question.

Demon: do as much as you can, of course.  (Also, there is never a good reason to quote the entire post directly preceding yours.  Especially when it's the first post on a thread.)

why not find them all?
[hide]
So we have the equation $a+b+c+abc=2(ab+bc+ca)$. If $a,b,c\geq 6$ then there's a problem because
\[3abc=abc+abc+abc\geq 6(ab+bc+ca)\Rightarrow a+b+c+abc>abc\geq 2(ab+bc+ca)\]
WLOG let $a\geq b\geq c$ so $c\leq 5$. If $c=2$ or $c=5$, then equation fails mod 3. If $c=1$, equation becomes $b+c+bc=1$ which is bad.

If $c=4$, equation becomes $2ab+4=7(a+b)$ so $(2a-7)(2b-7)=37$. Since $a,b>4$ and 37 is prime, we have $2a-7=37,\ 2b-7=1$. But this means $b=c$, so $a,b,c$ not distinct!

So $c=3$, equation becomes $(a-5)(b-5)=22$. Then both $a-5,b-5$ positive (if both negative, then $(a-5)(b-5)\leq 16$) so $a-5=22,b-5=1$ or $a-5=11,b-5=2$. Therefore, solutions are $(27,6,3),\ (16,7,3)$ and permutations.
[/hide]

[quote="scorpius119"]why not find them all?
[hide]
So we have the equation $a+b+c+abc=2(ab+bc+ca)$. If $a,b,c\geq 6$ then there's a problem because
\[3abc=abc+abc+abc\geq 6(ab+bc+ca)\Rightarrow a+b+c+abc>abc\geq 2(ab+bc+ca) \]
WLOG let $a\geq b\geq c$ so $c\leq 5$. If $c=2$ or $c=5$, then equation fails mod 3. If $c=1$, equation becomes $b+c+bc=1$ which is bad.

If $c=4$, equation becomes $2ab+4=7(a+b)$ so $(2a-7)(2b-7)=37$. Since $a,b>4$ and 37 is prime, we have $2a-7=37,\ 2b-7=1$. But this means $b=c$, so $a,b,c$ not distinct!

So $c=3$, equation becomes $(a-5)(b-5)=22$. Then both $a-5,b-5$ positive (if both negative, then $(a-5)(b-5)\leq 16$) so $a-5=22,b-5=1$ or $a-5=11,b-5=2$. Therefore, solutions are $(27,6,3),\ (16,7,3)$ and permutations.
[/hide][/quote]
Nice solution, but I was wondering is there anything special that these can be roots of a polynomial in the form of $x^{3}-kx^{2}+(k+d)x-(k+2d)$

wehre k=a+b+c and d is the common difference.

Solving for the integer roots of a polynomial with unknown integer coefficients... not a good idea.  Scorpius' approach is much more sensible.


In the picture one of the 4 equations is the right one. Which one is it and how do I do it.

God I've been stuck doing this for the past 3 and a half hours!!

[quote="t0rajir0u"]Calculate the length of the horizontal parts and the vertical parts separately.

[hide="More hints"] - The length of the horizontal segments on top add up to the bottom (that is, $ 5x^2 \minus{} 3x \plus{} 8$).
- There is only one vertical segment you don't know the length of.  Call its length $ y$.  Can you figure out what relationship $ y$ has to the other vertical segments? [/hide][/quote]

Yeah I did that actually, I forgot to mention it. I'm just stuck on putting them all together.

The vertical segment that I dont know is actually 2x + 3 is it not?

Awww man thats truee!

Can't believe I suck so much at this stuff. 2x + 3 instead of 3x + 3. :roll:

Thanks a lot for the help man.

[quote="t0rajir0u"]Almost.  Note that the segment of length $ x$ goes "down" (if we view the top of the shape as a graph), which changes the length of the segment that we don't know.

[hide="Its actually..."] $ 3x \plus{} 3$. [/hide]
[hide="Which makes the answer..."] $ 2(5x^2 \minus{} 3x \plus{} 8) \plus{} (2x \minus{} 1) \plus{} (x) \plus{} (3x \plus{} 3) \plus{} (4x \plus{} 2)$ [/hide]
[hide="Which is..."] $ 10x^2 \plus{} 4x \plus{} 12 \to \boxed{A}$. [/hide][/quote]
Wait a second... that doesnt really equal that...

2 x 8= 16 + -1= 15 + 3= 18 + 2= 20...

so 10x^2 + 4x + 20... no?

Hmm.  Looks like I'm not getting any of the answers, then.  Are you sure you copied everything down correctly?


calculus integration function geometry rectangle algebra domain

[quote="JRav"]This looks very similar to a Laplace transform.[/quote]

Thanks. But I haven't seen the solution. Would you make it clearer?

As written, it's obviously false- you've got two copies of $ g$ in the second. If you remove one, it's simply a matter of switching variables, which works as long as $ f$ and $ g$ aren't too bad.
Keyword: Fubini's theorem.

I'm sorry for mistyping. I must be:

Let $ f$, $ g: [0,t]\to \mathbb{R}$ such that
    $ I_1 \equal{} \int_0^t\left(\int_0^sf(s \minus{} \tau)g(\tau)d\tau\right)ds$ and
    $ I_2 \equal{} \int_0^tg(\tau)\left(\int_{\tau}^tf(s \minus{} \tau)ds\right)d\tau$
    hold. Is the equality $ I_1 \equal{} I_2$ true?

I appreciate your keyword, but how can we deal with $ [\tau,t]$? I wonder if $ \iint_{[0,t]\times[0,s]} \equal{} \iint_{[0,t]\times[\tau,t]}, 0 \leq \tau,s\leq t$? Hence it seems unnatural to apply Fubini's Theorem.

[quote="JRav"]If I can remember correctly from my Diff Eq course, it has to do with where/how the function is defined.[/quote]

Please suppose $ f,g$ are integrable on $ [0,t].$

Actually the replacement $ [0,s]$ by $ [s,t]$ makes me confused. By eyes, it may be correct but I haven't got the proof.

One way to look at it: extend $ h(s,\tau)\equal{}f(s\minus{}\tau)g(tau)$ to the whole rectangle by defining it to be zero on the triangle where $ \tau>s$. Then you're integrating on a rectangle, and you can simplify by moving the bounds to cut off empty areas.

In general, when you integrate over some non-rectangular shape, it helps to draw a picture of the region.

[quote="jmerry"]One way to look at it: extend $ h(s,\tau) \equal{} f(s \minus{} \tau)g(tau)$ to the whole rectangle by defining it to be zero on the triangle where $ \tau > s$. Then you're integrating on a rectangle, and you can simplify by moving the bounds to cut off empty areas.

In general, when you integrate over some non-rectangular shape, it helps to draw a picture of the region.[/quote]

Thanks for your helpful comments. Indeed, we only need to switch the variables, not applying Fubini's Theorem.

As your suggestion, we have the following equalities:
\[ \int_0^t\left(\int_0^sh(s,\tau)d\tau \right) ds  (s\leq t) \equal{} \iint_{\{(s,\tau): 0\leq \tau\leq s\leq t\}}h(s,\tau)d\tau ds \equal{} \int_0^t\left(\int_{\tau}^sh(s,\tau)ds \right)d\tau.\]
This guarantees "$ I_1 \equal{} I_2$".

[quote="jmerry"]Well, no. For one thing, you haven't correctly written the limits. For another, there is a "nice enough" condition on the function, and you haven't said anything about that.[/quote]

I suppose all integrals hold.

[quote="jmerry"]What do you mean by "all integrals"?[/quote]

I meant that you shouldn't care of the conditions for the existence of the above integrals since I supposed all of them hold. The problem is that what argument makes "$ I_1\equal{}I_2$" hold or not? I am sorry that I haven't been clear until now. :(

The standard necessary condition for the switch to work is that the integrals converge [b]absolutely[/b]. That is exactly Fubini's theorem, in the usual real analysis context. I don't see that assumption in what you started with, and it might be possible to construct a counterexample under those rules, which would necessarily be unbounded and change sign.

Also: [b]never[/b] quote the entire post preceding yours. You are assumed to be replying to it anyway. I'll occasionally quote a sentence or two to focus attention, but any more is just a waste of space.

Thanks for your reminding.

Since the above integrals are originally iterated integrals ($ 0\leq \tau \leq s\leq t$), I think it is confusing to apply Fubini's Theorem!

I agree that the above integrals are iterated, then we cannot apply Fubini's theorem. We may only be able to switch the variables.

And I think leuyenhoc was wrong to write $ \iint_{\{(s,\tau): 0\leq \tau\leq s\leq t\}}$.

You don't understand- Fubini's theorem is the justification to switch the variables. We don't have the hypothesis, so we can't do anything.

I only know that we use Fubini's theorem to change the multi-integrals into single integrals and change the order of integrating. Maybe, I must read and understand more.

In this case, the domain $ \{(s,\tau): 0\leq \tau\leq s\leq t\}$ makes us embarrassed. I guess we cannot have any multi-integral here. Thus, how can we apply Fubini's theorem? Please, suppose all integrals hold as Sir leuyen!

To change multi-integrals into single integrals?  No, Fubini's Theorem takes a double or triple integral and reverses the order of integration so that the computation might actually be possible.  Listen to what jmerry tells you.

Oh, thanks for your opinion towards my confusing statement "To change multi-integrals into single integrals".

However you agreed that Fubini's theorem helps us change "the order of integration", didn't you? So, for this problem, how can we apply Fubini's theorem since we can't change the order of integration?

[quote="LeUyenHoc"]Let $ f$, $ g: [0,t]\to \mathbb{R}$ such that
    $ I_1 \equal{} \int_0^t\left(\int_0^sf(s \minus{} \tau)g(\tau)d\tau\right)ds$ and
    $ I_2 \equal{} \int_0^tg(\tau)\left(\int_s^tf(s \minus{} \tau)g(\tau)ds\right)d\tau$
    hold. Is the equality $ I_1 \equal{} I_2$ true?[/quote]

I guess it should be

 $ I_1 \equal{} \int_0^t\left(\int_0^sf(s \minus{} \tau)g(\tau)d\tau\right)ds$ and
    $ I_2 \equal{} \int_0^tg(\tau)\left(\int_0^{\tau}f(s \minus{} \tau)g(\tau)ds\right)d\tau$

The previous is too hard, but I have not been sure if it is right or wrong.

Has there been the final result for this problem?


hi,
i have some problems with a plot of a function in matlab and i need some help. The function is $f(x,y)=e^{xy}+x^{2}+2y^{2}$. Can somebody give me a good advice or tell me how it works ?

Just try this command : ezmesh('exp(1)^(x*y)+x^2+2*(y^2)')

I don't really know anything about matlab but wouldn't exp(x*y) do the same thing as exp(1)^(x*y) ?


geometry
geometric transformation
geometry unsolved
Tangents
Locus problems
China
TST

Given a fixed acute angle $\theta$ and a pair of internally tangent circles, let the line $l$ which passes through the point of tangency, $A$, cut the larger circle again at $B$ ($l$ does not pass through the centers of the circles).  Let $M$ be a point on the major arc $AB$ of the larger circle, $N$ the point where $AM$ intersects the smaller circle, and $P$ the point on ray $MB$ such that $\angle MPN = \theta$.  Find the locus of $P$ as $M$ moves on major arc $AB$ of the larger circle.

All the triangles $MPN$ are directly similar, since $\angle MPN$ and $\angle PMN$ are constant. $\frac{AM}{AN}$ is constant, so all the triangles $MPA$ are directly similar. This means that $P$ moves on a circle which is the image of the large circle through a spiral similarity $\mathcal S$ centered at $A$, and the actual locus is the image of the large arc $AB$ through $\mathcal S$.


[hide="my method"]
1. Heat with $ \ce{NaNO_2}$ + $ \ce{H_2SO_4}$

2. Then add $ \ce{SO_2 \minus{} Cu}$

3. $ \ce{Cl_2}$ in presence of $ \ce{NaOH}$

4. Then $ \ce{NH_3}$ and then $ \Delta$
[/hide]

summa eruthen nan ella entha NITISH ala :P  :P


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

Let D be an integral domain with positive characteristic. Prove all nonzero elements of D have the same additive order.
Help! Thanks!

Show that the order of a given element is prime. After that: If two elements would have different orders, what does this imply¿


arithmetic sequence
Ramsey Theory
combinatorics solved
combinatorics

Ok, my first proposal :)  :

Van der Waerden's theorem states that for each positive integer k, if the set of positive numbers is finitely colored then we may find a monochromatic arithmetic progression of length k.

So :
a)  Prove that there exists finite coloring of |N such that there is no infinite monochromatic arithmetic progression.
b) Show that the same holds if 'arithmetic progression' is replaced by any sequence of positive integers of the fom (P(n)) where P is a non-constant poynomial.
c) In each case, what is the minimum number required to have such a coloring?

Have fun...

Pierre.

For the first part the minimal number is 2. Why? Put 1 in A, then put 2 and 3 in B, then put 4,5,6 in A, then 7,8,9,10 in B and so on. Neither A nor B contains an infinite arithmetic progression.


Hello, I am taking Algebra II and there is a problem from my book I cannot understand.  Don't worry, it's okay to discuss since its not homework, and the answer is in the back of the book anyways.
The question goes like this:
The sequence $\frac{1}{2},\frac{1}{3},\frac{1}{4},...$ is called the harmonic sequence.  Clearly, $t_{n}$, the term number, gets closer and closer to zero as $n$ becomes every large.  Yet the harmonic series $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$ does not converge! It just keeps getting larger and larger as $n$ increases.  By associating terms, show that:
$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...>\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}...$

The book's solution is that the terms can be rewritten like this:
$(\frac{1}{2})+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})...>(\frac{1}{2})+(\frac{1}{4}+\frac{1}{4})+(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})...$, in which the first is clearly bigger.

My problem is that since each sequence diverges, isn't it impossible to compare their sizes?  And if the comparitive sizes of each partial sum were to be added up, the second would always be bigger anyways.  Can anyone help me explain this?

[quote]The book's solution is that the terms can be rewritten like this:
(\frac{1}{2})+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})...>(\frac{1}{2})+(\frac{1}{4}+\frac{1}{4})+(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})...[/quote]
looking at the second sum:
1/3+1/4 and compare it with the second sum on RHS..
we subtract 1/4 from both sides to get: 1/3 on LHS and 1/4 on RHS... clearly 1/3>1/4...
doing the same with the third sum:
1/5+1/6+1/7+1/8 and 1/8+1/8+1/8 + 1/8... subtracting the 1/8... we get 1/5+16+1/7 on LHS... sionce all of these frations are greater than 1/8, their sum is greater than 3/8.
continuing this idea, we'll get LHS is always greater than RHS.

[quote="eryaman"]My problem is that since each sequence diverges, isn't it impossible to compare their sizes?  And if the comparitive sizes of each partial sum were to be added up, the second would always be bigger anyways.  Can anyone help me explain this?[/quote]Well basically you're showing that the harmonic series can be written as the sum of an infinite number of terms (1/2, 1/3 + 1/4, 1/5 + 1/6 + 1/7 + 1/8, ...) that are each greater than 1/2. This mean thats the series clearly converges.


linear algebra
matrix
limit
linear algebra unsolved

For which $n \times n$ matrices $B$ there exists an $n \times n$ matrix $A$ and a sequence of real numbers $u_k$,such that $u_kA^k \to B$ ($k \to \infty$) ?

one can suppose that $A$ is in Jordan form and work in complex vectorial spaces.
1/ If $A$ is nilpotent then $B=0$
2/ otherwise let us call $a_1,a_2,..,a_n$ the eigenvalues of $A$, with $0 < |a_1|=max |a_i|$ for example.
Then $u_ka_1^k$ tends towards $L$.

*if $l \neq 0$ then for any $|a_i| \neq |a_1|$, $\lim u_ka_i^k=0$. And because the $u_k$ are reals, $a_1$ must be real, and thus $l \in \mathbb{R}$. Moreover it is straightforwards to see that if $|a_1|=|a_i|$ then $a_1=a_i$ (because $l \neq 0$), and that if there is no non diagonal Jordan block with $a_1$ on the diagonal. Easy calculations show that a jordan block $M=aI+J$, with $|a|<|a_1|$ satisfies $\lim u_kM^k=0$.
Hence $\lim u_kA^k=B=diag(l,l,..,l,0,0,..,0)$. Hence $B=lP$ whre $P$ is a projection matrix.

*if $l=0$. The same kind of arguments shows that if a Jordan block $M=aI+J$ satisfies $\lim u_kM^k=T$, then $T$ is a matrix with $0$ everywhere except on the up right corner (if one suppose the jordan block upper triangular). From here the conclusion follows (i think because I've not carefully made the computaions ..)


Given$ a,b,c>0$ and $ abc\equal{}1$. Prove that :
$ \frac{a^2c\plus{}ab\plus{}c}{a^3b(a\plus{}b)^2}\plus{}\frac{b^2a\plus{}bc\plus{}a}{b^3c(b\plus{}c)^2}\plus{}\frac{c^2b\plus{}ca\plus{}b}{c^3a(a\plus{}c)^2} \ge\ \frac{(a\plus{}b\plus{}c)^4}{4(ab\plus{}bc\plus{}ca)^2}$  :)

I think this ineq is not true.Because after use cauchy-schwarz once time,We need to prove
$ \frac{1}{a(a\plus{}b)}\plus{} \frac{1}{b(b\plus{}c)}\plus{} \frac{1}{c(c\plus{}a)} \ge \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca}$
But this ineq is not true.LHS is very weak,and RHS is very strong.

[quote="babyloverain"]I think this ineq is not true.Because after use cauchy-schwarz once time,We need to prove
$ \frac {1}{a(a \plus{} b)} \plus{} \frac {1}{b(b \plus{} c)} \plus{} \frac {1}{c(c \plus{} a)} \ge \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca}$
But this ineq is not true.LHS is very weak,and RHS is very strong.[/quote]
no,this inequality is true :mad:
I think,your proof can't solve this inequality :)

[quote="Potla"][quote="babyloverain"]I think this ineq is not true.Because after use cauchy-schwarz once time,We need to prove
$ \frac {1}{a(a \plus{} b)} \plus{} \frac {1}{b(b \plus{} c)} \plus{} \frac {1}{c(c \plus{} a)} \ge \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca}$
But this ineq is not true.LHS is very weak,and RHS is very strong.[/quote]

Sorry but I did not get your point. How are you using Cauchy?[/quote]
 babyloverain used:
$ \sum{\frac{a^2b^2}{a^2b^2\plus{}a^2\plus{}1}} \equal{}1$ with $ abc\equal{}1$  :)


analytic geometry
vector
trigonometry
calculus
calculus computations

It's easy to see that, in cartesian coordinates, if we take the field $ \mathbf{E} \equal{} (x, \; y, \; z)$, that is, the identity, we get $ \nabla \cdot \mathbf{E} \equal{} 3$.

However, when I write the same thing for spherical coordinates (the field at a vector equals the vector itself), $ \mathbf{E} \equal{} (r , \; \theta , \; \phi )$, I suddenly get something else:

$ 3  \plus{} \frac{1}{r \sin \theta} \plus{} \frac{\sin \theta \plus{} \theta \cos \theta}{r \sin \theta}$

What's going on? Shouldn't the two fields, which are really the same field expressed in different coordinates, have the same divergence?

It probably means you're using the wrong form of the $ \nabla$ operator. If we use $ \mathbf{e}_r,\mathbf{e}_{\theta},$ and $ \mathbf{e}_{\phi}$ be the unit vectors in the direction in which those variables increase, can you write $ \nabla$ in terms of those?

(I can assure you that it's [b]not[/b] $ \nabla \equal{} \frac {\partial}{\partial r}\mathbf{e}_r \plus{} \frac {\partial}{\partial\theta}\mathbf{e}_{\theta} \plus{} \frac {\partial}{\partial\phi}\mathbf{e}_{\phi}.$)

A notational question before we proceed any further: in your notation, which of $ \theta$ or $ \phi$ is the longitude and which is the co-latitude. The book I'm teaching from right now uses $ \theta$ for longitude and $ \phi$ for co-latitude, along with $ \rho$ for the radial distance. That is, $ x\equal{}\rho\sin\phi\cos\theta,y\equal{}\rho\sin\phi\sin\theta,z\equal{}\rho\cos\phi.$ But I know that conventions vary. We should make sure we're using the same convention before getting deeper into the computation.

The other thing to note is that that's not the same vector field. A point is not the same thing as a vector, and this is especially important in nonlinear coordinate systems. The identity vector field in spherical coordinates is $ \mathbf{E}\equal{}r \mathbf{e}_r$.

I'm using standard spherical coordinates.

$ \nabla \equal{} \boldsymbol{\hat r}\frac {\partial}{\partial r} \plus{} \boldsymbol{\hat \theta}\frac {1}{r}\frac {\partial}{\partial \theta} \plus{} \boldsymbol{\hat \phi}\frac {1}{r \sin\theta}\frac {\partial}{\partial \phi}$

So is it like, in spherical, the unit vectors are defined at a particular point in space? Is it like, at the point $ (0,0,1)$, the vector $ \mathbf{e}_r$ would equal $ (0,0,1)$, and then the other two vectors would be perpendicular?

If that's the case, then thanks, I get it now :)

One small comment: what TZF just posted makes clear that he is using $ \theta$ for co-latitude and $ \phi$ for longitude. This is exactly the opposite of the use of $ \phi$ and $ \theta$ in the textbook I'm using this semester.

I wouldn't call either usage right or either usage wrong; it's just a place where commonly used conventions differ.


algebra
polynomial
number theory proposed
number theory

Let $b$, $c$ be integer numbers, and define $f(x)=(x+b)^2-c$.

i) If $p$ is a prime number such that $c$ is divisible by $p$ but not by $p^{2}$, show that for every integer $n$, $f(n)$ is not divisible by $p^{2}$.

ii) Let $q \neq 2$ be a  prime divisor of $c$. If $q$ divides $f(n)$ for some integer $n$, show that for every integer $r$ there exists an integer $n'$ such that $f(n')$ is divisible by $qr$.

Assume that $ p$ divides $ c$ but $ p^{2}$ does not, and such that $ p^{2}$ divides $ f(n)$ for some $ n$.  Then, $ p^{2}$ divides $ (n\plus{}b)^{2}\minus{}c$, or $ p$ divides $ (n\plus{}b)^{2}$.  Thus, $ p$ divides $ n\plus{}b$, or $ p^{2}$ divides $ c\equal{}(n\plus{}b)^{2}\minus{}f(n)$, contradiction.

Second part should read: "Let $ q\neq2$ such that $ q$ DOES NOT divide $ c$.  If $ q$ divides $ f(n)$ for some integer $ n$, then show that for any positive integer $ r$, integer $ n'$ exists such that $ q^{r}$ divides $ f(n')$".

It is a fundamental of quadric reside
$ x^{2}\equiv c(\mod p)$
If $ p$ odd then
There exist
$ x^{2}\equiv c(\mod p^{n})$
More general
$ f(x_{1})\equiv 0(\mod p)$
$ \gcd(f'(x_{1}),p) \equal{} 1$
Then for each n there exist $ x_{n}$ such that
$ f(x_{n})\equiv 0(\mod p^{n})$
(Use Helsen Lemma)
An apply
Prove that for all m then exist $ (a,b,k)\in Z$ such that
$ 2m \equal{} a^{19}\plus{}b^{99}\plus{}k2^{1999}$
Where $ (a,b)$ odd.


analytic geometry
advanced fields
advanced fields unsolved

Suppose we have a topological space $ B$ and its cover $ B'$ with a projection
map $ p : B' \to B$. Suppose we also have a continuous map $ f : I^n \to B$, where
where $ I \equal{} [0,1]$.

I want to prove that the map $ f$ can be lifted to a map $ f: I^n \to B'$. Can this be deduced from the usual lifting theorem when $ n\equal{}1$ ?

[quote="jmerry"]Lift along curves, and prove uniqueness with the monodromy theorem.[/quote]

What curves? Do you mean fixing n-1 coordinates first?

Actually, you can use any curves, and you want to use many for each point. If you don't, continuity is harder.

[quote="jmerry"]Lift along curves, and prove uniqueness with the monodromy theorem.[/quote]

Thanks. The idea in general is as Jmerry said, take a point $ x$ in $ I^n$, fixed base-point $ x_0$, and consider paths $ s$ connecting $ x_0$ to $ x$, take a path $ f \circ s$ in $ B$ and lift it to a path $ \bar{s}: I^n \to B'$, knowing the case of $ n \equal{} 1$, as a value of the lifting map $ \bar{f}$ at $ x$ consider the endpoint $ \bar{s}(1)$. The contractibility of $ I^n$ allows us to say this is well-defined, monodromy theorem does the uniqueness part.


Belgium did good at the IMO, we got 86 points: Here are the results:

BEL1      7 1 1 2 0 0   11   H.M. (Julien Coppe)
BEL2      7 1 1 7 0 0   16   Bronze (David Frenay)
BEL3      4 7 0 0 7 0   18   Bronze (Timothe Marquis, HM at IMO2003)
BEL4      6 7 0 7 6 0   26   Silver (Arne Smeets!!!)
BEL5      0 1 1 0 3 0   05   (Nicolas Van De Rijt)
BEL6      1 5 0 3 1 0   10   ME!

Arne did not yet posted this beacuse he is still in Greece till 30 july...


Let $ABC$ be an acute-angled triangle and consider an inscribed rectangle $DEFG$ such that $D \in AB, E \in AC$, $F \in BC, G \in BC$. As $D$, $E$, $F$, $G$ vary, what is the locus of the intersection of the diagonals of the rectangle? :)

I will change a bit your notations:

Let the altitude from $C$ lie on the $y$ axis and the foot of altitude from $C$ to $BA$ be the origin. Suppose that $B(b,0)$, $A(a,O)$ and two of the vertices of the given rectangle lie on $AB$. Let $K(k,l)$ be the vertex that lies on $BC$. Then the vertex that lies on $CA$ will be: \[ T(\frac{a(c-l)}{c},l).  \]
However, the foot of the perpendicular from $K$ to the $x$ axis will be $U(k,0)$. We conclude that the midpoint of $TU$ will be:
\[ O(\frac{\frac{a(c-l)}{c}+k}{2},\frac{l}{2}) .  \]
Since $k=\frac{b(c-l)}{c}$ we obtain that $O$ verifies the equation of the line that passes through the midpoint of the altitude from $C$ and the midpoint of the side $AB$, which is:
\[ y=\frac{-c}{a+b}x+\frac{c}{2}.  \]

Therefore the locus is the segment $H'M$ (we may also include its endpoints if the rectangle degenerates) , where $H'$ is the midpoint of the altitude from $C$ and $M$ is the midpoint of the side $[AB]$.

A synthetic solution uses a simple lemma: if $X,Y$ are variable points on the sides $AB,AC$ of the triangle $ABC$ such that $XY$ is parallel to $BC$, then the locus of the midpoint of $XY$ is the median $AM$ of the triangle.
Now, let $N$ be the midpoint of $DE$. The locus of $N$ is the median $AM$ of the triangle $ABC$. Let $P$ be the midpoint of $FG$. The center of the rectangle is the midpoint of $NP$, hence its locus is the median $MQ$ of the triangle $AMR$, where $R$ is the foot of the altitude from $A$ in the triangle $ABC$.

The Enescu's proof is nicely and classically. [hide="Here I shall present an another proof."] Supppose $AB<AC$ and define the points: $L\in DF\cap EG$, $P\in BC$, $S\in DE$, $\overline {PLS}\perp BC$;
the middlepoints $M,N$ of the side $[BC]$ and the altitude from the vertex $A$ respectively.
But $PM=\frac 12\cdot (FC-BG)$ and $LP=\frac 12\cdot SP$ $\Longrightarrow$
$\cot\widehat {BML}=$ $\frac{PM}{PL}=$ $\frac{\frac 12\cdot (FC-BG)}{\frac 12\cdot SP}=$ $\frac{FC}{FE}-\frac{BG}{BD}=$ $\cot C-\cot B$.
Therefore, the value of the angle$\widehat {BML}$ is constant and the point $M$ is fixed $\Longrightarrow$
[u]the geometrical locus of the point[/u] $L$ [u]is the segment[/u] $(MN)$.[/hide]


floor function
combinatorics unsolved
combinatorics

A person picks $ k$ distinct numbers from the first $ n$ natural numbers. A second person picks another set of $ k$ distinct numbers from the remaining numbers. What is the minimum possible sum of the numbers the first person picked such that, whatever the numbers the second person picks, the sum of his set of numbers will always be less than that of the first persons.

First, we need $ n\geq 2k$. Clearly the second person will pick the largest $ k$ numbers available to him. Shift to the right the numbers picked by the first person, so that together with those picked by the second person they will make up the last $ 2k$ numbers in the interval, of sum $ \frac {n(n \plus{} 1)} {2} \minus{} \frac {(n \minus{} 2k)(n \minus{} 2k \plus{} 1)} {2} \equal{} k(2n \minus{} 2k \plus{} 1)$. This will only increase the sum $ \sigma$ of the numbers picked by the first person, so we need $ \sigma > \frac {k(2n \minus{} 2k \plus{} 1)} {2}$.
The minimum possible sum is therefore $ k(n \minus{} k) \plus{} \lfloor k/2 \rfloor \plus{} 1$.

EDIT. True, I was hasty. First player needs plug in just the gaps in the second player's choice, not come contiguous from below. I will come back with the calculations ...

Hi mavropnevma
I have problems understanding your solution. But i've an example that contradicts with the answer
Consider $ n\equal{}10, k\equal{}3$ . Suppose the first person picks $ 10, 9, 3$ which sumto $ 22$. The second person can have a maximum of $ 6\plus{}7\plus{}8\equal{}21$
Another exmple
$ n\equal{}20, k\equal{}5$. First person $ 20\plus{}19\plus{}18\plus{}17\plus{}1\equal{}75$. Second person maximium $ \equal{} 16\plus{}15\plus{}14\plus{}13\plus{}12 \equal{} 70$

Ok, here are my results.

Clearly, once the first person (called $ A$) has picked his $ k$ numbers, the second person (called $ B$) will pick the largest $ k$ numbers available to him in the interval $ \{1,2,\ldots,n\}$. Therefore, from $ n$ downwards, there will be picked $ k \plus{} \ell$ consecutive numbers (in some order, $ k$ by $ B$ and $ 0 \leq \ell \leq k$ by $ A$), followed by some other $ k \minus{} \ell$ numbers (all of them picked by $ A$).
Let us denote the sum of the $ k$ numbers picked by $ A$ by $ \sigma_A$, and the sum of the $ k$ numbers picked by $ B$ by $ \sigma_B$. We look for $ \delta \equal{} \sigma_A \minus{} \sigma_B \geq 1$, so we can tentatively start with $ \ell \equal{} k$, the largest $ k$ numbers being picked by $ A$ and the next $ k$ by $ B$. Moreover, $ \sigma_A \equal{} \frac {n(n \plus{} 1)} {2} \minus{} \frac {(n \minus{} k)(n \minus{} k \plus{} 1)} {2} \equal{} \frac {k(2n \minus{} k \plus{} 1)} {2}$ and $ \delta \equal{} k^2$, while $ \sigma_B$ is lowest possible.

The way to decrease $ \sigma_A$ while $ \sigma_B$ increases as slowly as possible, while keeping $ \delta \geq 1$ is by swapping adjacent elements $ a$ picked by $ A$ with $ b \equal{} a \minus{} 1$ picked by $ B$, with the effect that $ \sigma_B$ increases by $ 1$, $ \sigma_A$ decreases by $ 1$, while $ \delta$ decreases by $ 2$; or, if $ a \minus{} 1 \geq 1$ is not picked by either $ A$ or $ B$, decrease it by $ 1$, with the effect that both $ \sigma_A$ and $ \delta$ decrease by $ 1$. Since if we continue this process till the end, $ A$ will end up having picked the least $ k$ elements, with clearly $ \delta < 0$, it means somewhere in the process there was a moment when $ \delta \equal{} 1$ or $ 2$, and that is the optimal position that we need to compute.

The stages and steps by which $ \delta$ decreases can be counted like this:

$ \bullet$ as first stage, the least element picked by $ A$ swaps with the $ k$ elements picked by $ B$, then shifts left until becomes equal to $ 1$ in $ n \minus{} 2k$ steps, so $ \sigma_A$ decreases by $ k \plus{} (n \minus{} 2k) \equal{} n \minus{} k$, while $ \delta$ decreases by $ 2k \plus{} (n \minus{} 2k) \equal{} n$;
$ \bullet$ as next stage, the second least element picked by $ A$ swaps with the $ k$ elements picked by $ B$, then shifts left until becomes equal to $ 2$ in $ n \minus{} 2k$ steps, so again $ \sigma_A$ decreases by $ k \plus{} (n \minus{} 2k) \equal{} n \minus{} k$, while $ \delta$ decreases by $ n$; for $ k^2 \minus{} 1 \equal{} qn \plus{} r$, with $ 0 \leq r \leq n \minus{} 1$, we can do this for $ q$ stages;
$ \bullet$ as last stage, we have to be careful. The $ (q \plus{} 1)^{\textrm{th}}$ least element picked by $ A$ swaps with as many of the $ k$ elements picked by $ B$ as possible: if $ r < 2k$ with $ \lfloor r/2 \rfloor$, so $ \sigma_A$ decreases by $ \lfloor r/2 \rfloor$, while $ \delta$ decreases by $ 2\lfloor r/2 \rfloor$; otherwise with all $ k$, then shifts left for more $ r \minus{} 2k$ steps, so $ \sigma_A$ decreases by $ k \plus{} (r \minus{} 2k) \equal{} r \minus{} k$, while $ \delta$ decreases by $ 2k \plus{} (r \minus{} 2k) \equal{} r$.

To summarize, since $ q \equal{} \left \lfloor \frac {k^2 \minus{} 1} {n} \right \rfloor$, while $ r \equal{} k^2 \minus{} 1 \minus{} n\left \lfloor \frac {k^2 \minus{} 1} {n} \right \rfloor$.

$ \bullet$ if $ r < 2k$, then the minimum for $ \sigma_A$ is $ \frac {k(2n \minus{} k \plus{} 1)} {2} \minus{} q(n \minus{} k) \minus{} \lfloor r/2 \rfloor$;
$ \bullet$ if $ r \geq 2k$, then the minimum for $ \sigma_A$ is $ \frac {k(2n \minus{} k \plus{} 1)} {2} \minus{} q(n \minus{} k) \minus{} (r \minus{} k)$.

Let us try a few examples.

1) $ k \equal{} 2m$, $ n \equal{} 2k$. Then $ q \equal{} \left \lfloor \frac {4m^2 \minus{} 1} {4m} \right \rfloor \equal{} m \minus{} 1$, $ r \equal{} 4m \minus{} 1$, so the minimum value for $ \sigma_A$ is $ m(6m \plus{} 1) \minus{} 2(m \minus{} 1)m \minus{} (2m \minus{} 1) \equal{} 4m^2 \plus{} m \plus{} 1$.
2) $ k \equal{} 2m \plus{} 1$, $ n \equal{} 2k$. Then $ q \equal{} \left \lfloor \frac {4m^2 \plus{} 4m} {4m \plus{} 2} \right \rfloor \equal{} m$, $ r \equal{} 2m$, so the minimum value for $ \sigma_A$ is $ (3m \plus{} 2)(2m \plus{} 1) \minus{} m(2m \plus{} 1) \minus{} m \equal{} 4m^2 \plus{} 5m \plus{} 2$.

As for your examples, these are the results.

3) $ k \equal{} 3$, $ n \equal{} 10$. Then $ q \equal{} 0$, $ r \equal{} 8$, so the minimum value for $ \sigma_A$ is $ 22$, with the example $ A$ having picked $ 3, 9, 10$.
4) $ k \equal{} 5$, $ n \equal{} 20$. Then $ q \equal{} 1$, $ r \equal{} 4$, so the minimum value for $ \sigma_A$ is $ 73$, with the example $ A$ having picked $ 1, 15, 18, 19, 20$.

Thankyou very much for your kind concern. This is a great job!
The procedure is very nice, thankyou. Sorry if this is a dumb question, we say that $ \sigma_{A}$ attains its possible minimum when $ \delta$ is 1 or 2. But if we take the sets {10,7,6} and {9,8,5} their sums differ by one but the first set isn't the one with minimum sum. Could you clear me out please?

That was exactly the [color=red]oversight[/color] in my first post. Not all configuration with $ \delta \equal{} 1$ (or $ 2$) provide minimal sum for $ \sigma_A$. In my solution I reach the right formula because I start with the least possible $ \sigma_B$ and let it increase as slowly as possible, until $ \delta$ becomes $ 1$ (or $ 2$).

One thing is certain: any configuration with $ \delta > 2$ cannot be optimal, since we can reduce $ \sigma_A$ by the very method I exposed.


Starting with the odd number 1, what is the sum total of every other odd number included in the first one million counting numbers ?

That is, 1 + 5 + 9 + … (1,000,000) = ?

CAn you please help me someone? I really need the answer and how to do it.

Do you want to calculate $1+3+5+7+9+...$ or $1+5+9+13+17+...$¿
And shall the summation end before $1000000$ or at the $1000000$-th number of the sequence¿

I think i want to calculate 1+3+5 and end at the highest odd number before hitting the number 1 million.

Also, if you don't mind, how would i figure it out if it was 1+5+9?

thanks a bunch.

Let's do it for the sum of the first $n+1$ terms of the sequence $a, a + d , a+2d , ... , a+nd , ...$ for constant $a,d$.
We have
$\sum_{k=0}^{n} (a+dk) = \sum_{k=0}^{n} (a) + \sum_{k=0}^{n} (dk) =$
$=(n+1)a + d \sum_{k=0}^{n} (k) = (n+1)a + d \frac{n(n+1)}{2}$.
(the last step by $1+2+3+...+s = \frac{s(s+1)}{2}$)

We have $a=1$ in both cases and $d=2$ or $d=4$, depending on what we want to calculate.
Also, for $d=2$ our formula gets $(n+1) + n(n+1) = (n+1)^2$.
So the question left is: what is our $n$ here¿
We want $1+dn$ to be maximal $\leq 1000000$, thus $n \leq \frac{1000000-1}{d}$. Now calculate it to get:
$d=2$: $n \leq \frac{999999}{2} \implies n \leq 499999 \implies n= 499999$.
$d=2$: $n \leq \frac{999999}{4} \implies n \leq 249999 \implies n= 249999$.

Now setting these into the formula above gives:
$d=2$:  $250000000000=500000^2$
$d=4$: $124999750000$

*blinks*, wow, you're brilliant!  thanks a ton!


topology
advanced fields
advanced fields unsolved

[quote="grobber"]Let $ \mathcal M$ be a von Neumann algebra which is a  type $ II_1$ factor, and let $ \mathcal U$ be its unitary group, with the usual, uniform topology (i.e the topology induced by the norm). What is the fundamental group $ \pi_1(\mathcal U)$?[/quote]
It is isomorphic to the additive group of real numbers. The reference is: Araki, Smith and Smith, "On the homotopical significance of the type of von Neumann algebras", Commun. Math. Phys. 22 (1971), 71–88.

It's also the reference I had, but I just wanted to let people enjoy it before they jump to look it up :). I found it as an exercise in Takesaki's "Theory of Operator Algebras" and the proof (the one I had in mind; I didn't read the paper precisely because I didn't want it to spoil my fun :)) isn't trivial, but is still simple enough to post here, I think.


algebra
partial fractions
calculus
calculus computations

Find three different laurent series expansions for \[ f(z)\equal{}\frac{1}{(z\plus{}1)(z\plus{}3)}\]

specify the annulus for each series.

that's easy :D first break into partial fractions to get

 $ f(z)\equal{}\frac{1}{2}\left( \frac{1}{z\plus{}1}\minus{}\frac{1}{z\plus{}3}\right)$

then do

$ (1) \quad|z|<1$

 $ f(z)\equal{}\frac{1}{2}\left( \frac{1}{1\plus{}z}\minus{}\frac{1}{3(1\plus{}\frac{z}{3})}\right)$

$ \equal{} \frac{1}{2}\minus{}\frac{1}{2}z \plus{} \ldots \quad \minus{}\frac{1}{6} \plus{} \frac{1}{18}z \minus{}\ldots$

$ (2) \quad1<|z|<3$

 $ f(z)\equal{}\frac{1}{2}\left( \frac{1}{z(1\plus{}\frac{1}{z})}\minus{}\frac{1}{3(1\plus{}\frac{z}{3})}\right)$

$ \equal{} \frac{1}{2z}\minus{}\frac{1}{2z^2} \plus{} \ldots \quad \minus{}\frac{1}{6} \plus{} \frac{1}{18}z \minus{}\ldots$

$ (3) \quad|z|>3$

 $ f(z)\equal{}\frac{1}{2}\left( \frac{1}{z(1\plus{}\frac{1}{z})}\minus{}\frac{1}{z(1\plus{}\frac{3}{z})}\right)$

$ \equal{} \frac{1}{2z}\minus{}\frac{1}{2z^2} \plus{} \ldots \quad \minus{}\frac{1}{2z} \plus{} \frac{3}{2z^2} \minus{}\ldots$


Ring Theory
algebra
polynomial
superior algebra
superior algebra unsolved

determine all prime ideals in the ring $ \mathbb{R}[X]$ and $ \mathbb{C}[X]$

note that these are UFDS...
also remember that in a UFD: prime <=> irreducible

And much more important: They are PIDs and hence every prime ideal is in fact of the form (p) for some irreducible element p.

In $ \mathbb{C}[x]$ all non-zero proper ideals must be generated by irreducible polynomials $ p(x)$. Since every $ p(x)$ is reducible if $ \deg p(x) > 1$ it means $ p(x)$ must be a linear polynomial. So the ideals are $ \{ 0 \}, \mathbb{C}, \left < x \plus{} a \right >$ where $ a\in \mathbb{C}$. In $ \mathbb{R}[x]$ those intermediate ideals must be monic quadradic polynomials with discrimant negative and monic linear polynomials by the similar idea.

@sylow: All [i]prime[/i] ideals must be generated by an irreducible polynomial. Of course there are other ideals too.
And the polynomials don't need to be monic, since the nonzeor constants are units and hence multiplication with a non-zero element doesn't change the ideal.


:D and the second question:
 Constant force  F  affacts to the soap standing on the ground. Coeficiant of friction between the soap and ground is f. As we know soap isn't a solid matter. and that is why soap is loosing  k=1% of its mass for every 5m of the way. Find the velocity of the soap while the mass is halved(m/2)?

here is a picture for make question understable


inequalities
logarithms
inequalities proposed

[hide="HINT"] Factorize : $ (a_n \plus{} b_n)x_n \minus{} a_nb_n \equal{} (b_n \minus{} x_n)(x_n \minus{} a_n) \plus{} x_n^2$ [/hide]


Solving the inequality $ (x^2\minus{}7x\plus{}6)(x^2\minus{}5x\plus{}6) \leq 0$,  then:

a) $ 2 \leq x \leq 3$.
b) $ 1 \leq x \leq 6$.
c) $ 2 \leq x \leq 6$
d) $ 1 \leq x \leq 2$ ou $ 3 \leq x \leq 6$.
e) $ 1 \leq x \leq 3$.

$ (x\minus{}6)(x\minus{}1)(x\minus{}2)(x\minus{}3)\le 0$

And for it to be negative, either $ 1$ or $ 3$ of the brackets must be negative. This is only possible if $ d) 1\le x \le 2$ or $ 3\le x \le 6$

[quote="EastyMoryan"]$ (x \minus{} 6)(x \minus{} 1)(x \minus{} 2)(x \minus{} 3)\le 0$

And for it to be negative, either $ 1$ or $ 3$ of the brackets must be negative. This is only possible if $ d) 1\le x \le 2$ or $ 3\le x \le 6$[/quote]

Thanks man.


[quote="chien than"]Let  $xyzt=1989^{1991}$ and $x;y;z;t\in \mathbb{N}$
 Prove that:$(x+y+z-t)\vdots 1992$ ?.[/quote]
Are you sure? Example $x=1989^{1991}\;\; ,y=z=t=1$ :D

http://www.mathlinks.ro/Forum/viewtopic.php?t=135914 !
That's the last time I will not lock something just because an answer was given.


integration
function
real analysis
real analysis solved

I'll post some more prbs from the Romanian MO. Lagrangia, I know you have the solns, but could you pleeease not post them so quickly? :D I think these are nice and it's enjoying to think abt them:

P7

f:[0, 1]->R, derivable. There is an a in (0, 1] s.t.  \int {0, a} f(x)dx=0. Prove that | \int {0, 1}f(x)dx| <= (1-a)/2*sup|f'(x)| where the sup is taken for all x in (0, 1).

"Official Solution":

from the fact that \int {0,a) f(x)dx=0 we get that \int {0,1}f(ax)dx=0. Putting M=sup|f'(x)| we get that f(x)-f(ax)<=(1-a)xM, and integrating  we get that:
| \int {0,1}f(x)dx|=| \int {0,1} (f(x)-f(ax))dx|<=(1-a)M \int {0,1}xdx=1/2 (1-a)M.
we have equality for the following functions: f:[0,1]-->R, f(x)=+-M (x-a/2).

cheers! :D :D :D

Sometimes these problems look scary, but they hace short and nice solutions, don't they? I'll post some more tonight.


Welcome to Lucid Dreams.

This game appears to have begun as a rather frivolous mafia mutation, but nobody ever bothers scum hunting anyway (and scum always win), so I decided to eliminate the whole informed minority uninformed majority thingy. Instead you will be divided into some number of roughy equal teams, and you will be given the identities of some number of people on your team (but most definitely not all).

There will be two alternating phases, night and day. I intend to have fast deadlines.
At night, you may submit any action you like to me. ANY action, and it will be carried out. Be creative.
During day, all the players together may vote on an action. Again, ANY action. You will have 2-3 RL days to suggest actions (at most one suggestion per player) and vote. Each player gets one vote, and the (legal) action that has the most votes at the end of the day will be carried out (plurality voting).

When you suggest an action, your vote gets automatically placed on your suggestion. You may not retract your suggestion after it has received any other votes, though are allowed to unvote on your own suggestion.

I'll do my best to handle priority for night actions. Be warned that order of submission can potentially be used to decide. As you may see though, it's a subjective game. Some guidelines:
Thou mayst not give thyself extra night actions.
Thine action may not be a "compound action" (kill X and Y), nor may it target universally (kill everyone but me, kill everyone on X's team).
Standards of acceptability may not necessarily be the same for the day action and night actions.
Thou mayst not give thyself extra votes (important for this game, cuz a lot of things can happen during day).

Basically, anything that's not "broken" or "unfun" is fair game.

Are you ready?

join.
day sounds like nomic
night sounds like...i dunno, i guess i'll find out

\join.

I'll try not to drop out this time.   :P

WARNING:  I will not have Internet access until September 6th.  So if you need to replace me from inactivity, go ahead.

Should be interesting, though I have limited internet for now. JOIN (ok I won't commit suicide as I did last game)

Okay, game starting now.

[b]Alive (7)[/b]
1. zapi2007
2. westiepaw
3. spaceguy524
4. EggyLv.999
5. cf249
6. james4l
7. pythag011

There are 3 teams. But wait, you say, there's 7 players! Oh well.

Role PMs will be going out.

[quote="Generic role PM for everyone"]
You are on Team {1,2,3}.

[i](version A: )[/i] Don't you wish you knew which other person(s) are on your team? Oh well.
[i](version B: )[/i] You know that [person] is on your team.

You have 2 days to get your first night action in. Remember, any action. If you are worried about your action being rejected, you can send two actions, with the second as a backup. No more than one backup though. I don't really care if you talk in thread during this first night. However, no talking outside of thread. Ever.[/quote]

Once again, it is now [b]Night 0[/b]. Deadline is whenever I declare night to be over, which should be in 2 days.

You can keep talking during this night. I don't care.

got the pm

ready for the game!  :)

will night actions be announced?

No they won't be announced. However, if someone dies or gets turned into a ham sandwich or whatever, then yes.

Oh yeah, you can't revive people. Cuz I just don't wanna go there.

[quote="echinodermata"]if someone dies or gets turned into a ham sandwich[/quote]

haha. lol  :rotfl:

[quote="Scamper"]Can I join? To make these even?[/quote]

3 teams, so 8 => not even.

9 would be, I guess.

I assume team of 3 = less info.

As the sun rises, you all assemble at your local starbucks (who needs a town square?). But wait! One two three four five... oh no! Wait, you forgot to count yourself. Six! You all rush to spaceguy's house (what's his cell phone number again?), bust down the door, run up the stairs, to find spaceguy sprawled on the floor. Dead? Nah, just asleep. Hmm. No reason to wake him up. You decide to just hang out in his living room instead.

It is now [b]day 1[/b]. (Thanks for getting your actions in fast.) [u]Spaceguy has no vote, cannot officially suggest a day action, and cannot be targeted today.[/u]

Alive: all 7.

Reminder:
During the day, you suggest actions (at most one suggestion per player) and vote. Each player gets one vote, and the (legal) action that has the most votes at the end of the day will be carried out (plurality voting).

When you suggest an action, your vote gets automatically placed on your suggestion. You may not retract your suggestion after it has received any other votes, though are allowed to unvote on your own suggestion.

Deadline is 3 real life days. Go.

1. you can't propose twice
2. dead players are permanently dead. As in they have no more involvement in this game.

With just the two of you alive, you decide to go out to a fancy restaurant for dinner. It just may be your last, after all. You can't really come to terms on what to do for the day... so you quietly eat, in near silence.

You're not taking home leftovers today.

But what about that wine in front of you?

[b]Day 3 is over[/b]. You have 2 RL days to send in your actions. Good night.

ok, is this day skippable?

I made a pretty stupid move.

Day 5 ends.

Except... seems like no one woke up today.

echinodermata, ace investigator, goes to zapi's house to investigate. He was tied to a chair and bolted down. seems like zapi is still alive.

next, ace investigator goes to eggy's house.

Looks he's asleep... permanently.

Eggy has died for not posting this gameday (as required by the day 2 action).
zapi however, has not died because he had a post restriction preventing him from posting

[b]FINAL[/b]
[b]Alive: (1)[/b]
[color=blue]zapi2007, team 3[/color]

[b]Dead
[color=green]james4l, team 2, smited by the e-police day 1 [/color]
[color=red]cf249, team 1, eaten by a grue day 1 [/color]
[color=blue]westiepaw=vahalla, team 3, tranquilized permanently day 2 [/color]
[color=red]pythag011, [/color][size=150][u][i]correction[/i][/u][/size][color=red] team [u]1[/u] (hope that didn't mess anybody up... probably didn't), eaten by a grue day 2 [/color]
[color=green]spaceguy524, team 2, died of brain freeze day 2[/color]
[color=green]EggyLv.999, team 2, found asleep permanently day 5[/color][/b]

Thanks for playing. I'll be posting all the actions soons.

[quote="echinodermata"]pythag011, correction team 1 (hope that didn't mess anybody up... probably didn't), eaten by a grue day 2 [/quote]

Well, I spend the last 3 game days thinking I actually had a chance to win, but that's no biggie.   :wink:

Congratulations to zapi and Team 3.

Sweet!

did team actually have anything to do with it?

1st night action:  Can't get nked

2nd night action:  First proposer of the day dies.  :)

The purpose of the team mechanic was to give some sort of incentive for people to not just going around killing everybody else. You could very well (and very easily) ignore it altogether.

Comments:
Well played by zapi. Because no one else played agressively.
Hey eggy. I think you'll notice now that there were 0 kill attempts at night. The whole game. On night 2 or 3 or 4, why didn't you just kill zapi? WIFOM?

[b]ACTIONS[/b]
[i]The bracketed numbers (e.g. [5th] represent the send order).
pythag knew cf249's alignment and zapi2007 knew westiepaw's.[/i]

=============================
NIGHT 0 ACTIONS
=============================
Team 1.
cf249* [5th] james4l may not use the letter "e" in any of his posts during the next game day.
pythag011. [3rd] Make james4l be told that he is partners with a random person every other Real Life days. (The random person changes). BACKUP: Make zapi2007 eat james4l.

Team 2.
james4l.[7th] Kill Da Hamma: The last voter dies. This effect is permanent.
EggyLv.999. [4th] I protect self...
spaceguy524. [6th] I sleep

Team 3.
westiepaw* [2nd] Investigate spaceguy524 to find out if he's on my team. BACKUP: Release a virus into zapi2007's computer.
zapi2007. [1st] I choose to make a protective shield around me, preventing me from getting killed at night for the rest of the game. This shield is unretractable and immune to any attempts to remove it. The last statement is also cannot be taken down by a vote or an action and so on... However, this shield is one-way, so I can still execute an action at night. BACKUP: I don't want to die this round, so I put a force field around myself to prevent death at night. [i]Mod Note: I only permitted this action because: 1) He very well could have died during day, and 2) If you turned him into a ham sandwich, then you could have circumvented this action, and he would have been out of the game without being "killed"[/i]

Day 1. Post restrictions only have to be followed every other post.

=============================
NIGHT 1 ACTIONS
=============================
Team 1.
pythag011. [4th] Anyone who uses an action on zapi gets the action redirected to himself or herself

Team 2.
EggyLv.999. [1st] protect self
spaceguy524. [5th] I will go stock up on caffine pills. Mod note: You take a large dose of caffeine, preventing you from sleeping tomorrow night. As this is "lucid dreams" you have lost your next night action. However you do get to be a double voter for day 2.

Team 3.
westiepaw=vahalla. [3rd] Examine Eggy.
zapi2007. [2nd] The first person to propose an action dies, effect permanent, occurs everyday.

Day 2. Posting is required every day from now on unless you have a post restriction that stops you from doing so.

=============================
NIGHT 2 ACTIONS
=============================
EggyLv.999. [1st] protect self
zapi2007. [2nd] nullify james' action that the last voter is killed. [i]Mod Note: There was a little controversy over how zapi knew that it was james that made that action[/i]

Day 3. No action

=============================
NIGHT 3 ACTIONS
=============================
EggyLv.999. [2nd] protect self
zapi2007. [1st] Force silence on Eggylv.999 for one game day

Day 4. No action

=============================
NIGHT 4 ACTIONS
=============================
EggyLv.999. [2nd] protect self
zapi2007. [1st] my next night action is to silence myself/give myself a post restriction that I can't post for the rest of the game during day but I can still make night actions

Will there be a part 2?

preferably with more players?

Yeah, I'll probably run it again sometime. Depends on what's going on in the mafia forum. Cuz I faintly understand that I'm competing for their players. Expect a game very soon if consensus is slow in mafia discussion.

More players, very likely.

well, a lot of people were in both games, and it sorta worked anyway...

If anyone wants to run lucid dreams 2, you're perfectly welcome. (I'd appreciate it if you'd me a heads up about it though.)


calculus
derivative
analytic geometry
graphing lines
slope
calculus computations

For both questions, find an equation of the tangent line to the curve at the given point.

1) y = 2x/(x+1), (1,1)

2) y = x + sqrt{x}, (1,2)

In general for the tangent line of any differentiable curve at any defined point, calculate the derivative and evaluate it at the x-coordinate of the point.  This value is the slope of the line.  Then use the point slope formula to get an equation for the line.

Example 1:  $y=\dfrac{2x}{x+1} \Rightarrow y' = \dfrac{2}{(x+1)^2} \Rightarrow y'(1) = 1$

Then we have $y-1 = 1(x-1) \Rightarrow y=x$


geometry
angle bisector
geometry unsolved

Let $ ABC$ is a triangle. Let $ M$ is the midpoint of $ BC$. Let $ N$ is a point on $ CA$ such that $ \angle ABN\equal{}\angle CBN$
Let $ P$ is the intersection of $ AM$ and $ BN$ and $ T$ is the intersection of $ CP$ and $ AB$
Prove that $ TB\equal{}TN$

Applying [b]Ceva's theorem[/b] to triangle $ ABC$ with the concurrency $ AM$, $ BN$ and $ CT$, we get $ \frac {MB}{MC}\cdot\frac {NC}{NA}\cdot\frac {TA}{TB} \equal{} 1$, i.e. $ \frac {NA}{NC} \equal{} \frac {TA}{TB}$. Which follows that $ TN\parallel BC$, and hence $ \angle TNB \equal{} \angle NBC \equal{} \angle TBN$, i.e. $ TB \equal{} TN$. Done!

[img]http://i213.photobucket.com/albums/cc110/April_ML/t171853.png[/img]

At first:As far as i understand $ BN$ is an angle bisector,so it would be much easier if you will write only $ BN$ is an angle bisector,isn't it?
Solution:
Since $ BN$ is an angle bisector then $ TB\equal{}TN\Longleftrightarrow TN\|BC$.Or which is equivalent to $ \frac{AN}{NC}\equal{}\frac{AT}{TB}$.
Apply Menelaus theorem to $ \triangle CTB$ and transversal $ \overline{APM}$,we obtain that
$ \frac{CP}{PT}\equal{}\frac{AB}{AT}$.But $ \frac{CP}{PT}\equal{}\frac{BC}{BT}$.So
$ \frac{AT}{TB}\equal{}\frac{AB}{BC}\equal{}\frac{AN}{NC}$.Thus we are done. :)


algebra
polynomial
inequalities open
inequalities

Given $ x_i > 0$ ,  $\sum_{i=1}^{n} x_i = c$ and $\prod_{i=1}^{n} x_i = 1$

Find the maximal and minimal value of:

$\sum_{i /=j} x_i x_j$

Please only use Lagrange multiplier in the last attempt . I think that considering a polynomial is a good direction.

Is there anyone who are interested in my problem?


The $ \textit{Twin Prime Conjecture}$, which has not been proven, argues that there are infinitely many primes of the form $ n$ and $ n\plus{}2$. Such a pair is called a $ \textit{prime pair}$. Some examples of prime pairs are 5 and 7, 17 and 19, and 29 and 31. How many prime pairs have both members less than 100?

There is clearly not enough time in FTW to count all of those. Could this problem be solved in a none guess and check way?

[quote=Andrewlu2012][b]/bump[/b] help?[/quote]
Um....you just grind it out by going over each individual prime.

[quote=Lord_Baltimore]Andrewlu2012 means is there a slick trick for this. (At least that's what I think he/she means...)[/quote]

I think there is, but it's very, very complicated so I would not recommend it.[quote=Andrewlu2012]Yeah.

Andrew is a he name :P

Altho I see your logic, males have a knack of playing female.[/quote]

?????

Like, in some video games, males choose female characters for idk what reason.

Maybe smaller hitboxes?

[quote=willmathxu]this is going off topic  :D[/quote]

True, true, I was just finishing my most recent off topic talk.

This question is very outdated. The twin prime conjecture has been proven many times with different number theory proofs, First by Yitang Zhang, who wrote an explanation of how there where twin primes less that 247, but it has been proved again, and again, so I think is a dead thheory that has been proved. Support this comment if you want a mini-history on the twin prime conjecture

It is still an unsolved problem. Yitang Zhang made progress, and other people have made progress as well. However, there is no proof for or against the statement currently.


quadratics
search
algebra
quadratic formula

[quote="LinearBoy"]Two students are playing a game. The first player chooses three
non-zero >> numbers << . The second player substitutes the >> numbers << for the
stars in the equation [img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/0/f/a/0fa20323c1f66f334fb8cd2fcbbf7a006281b560.gif[/img]
in any order he likes. [/quote]

LinearBoy,

*state* the types of numbers to replace the asterisks.  What category of "numbers"
do they belong to?
All you have done toward that so far is to state that they are non-zero.

I honestly don't know, I couldn't solve it from that competition, so i came to you guys for help. I re-typed the question exactly ( I have a copy of the questions)

Seriously I even doubt player 1 always wins.

Is this a faulty question? I keep coming to this conclusion after hours of trying to figure out this problem.

I too am confused by the wording of the problem. I am particularly confused about the phrase
[quote]The first player [b]can [/b]always win. [/quote]
Do we need to prove that with any series of three numbers, there is the possibility that the first player will win?
If so, we can easily show that if the first player picks the numbers $ 1,1,1$, then he will always lose (disproof by counterexample).

Do we need to prove that there are three numbers $ a,b,c$ with the property that $ \sqrt {a^{2} \minus{} 4bc} \equal{} x$ and $ \sqrt {b^{2} \minus{} 4ac} \equal{} y$ and $ \sqrt {c^{2} \minus{} 4ab} \equal{} z$ (with $ x,y,z \in Q$)?  (If this is true, then the first player can pick these numbers and always win.)

Do we need to prove simply that there exist three numbers that, when substituted into the equation in a particular order, yield rational solutions? (This is a ridiculous question).

[quote="mismatchtea"]
Do we need to prove that there are three numbers $ a,b,c$ with the property that $ \sqrt {a^{2} \minus{} 4bc} \equal{} x$ and $ \sqrt {b^{2} \minus{} 4ac} \equal{} y$ and $ \sqrt {c^{2} \minus{} 4ab} \equal{} z$ (with $ x,y,z \in Q$)?  (If this is true, then the first player can pick these numbers and always win.)[/quote]

This is most likely the correct interpretation of the question (adding the provision that $ a,b,$ and $ c$ are not equal to zero).

[hide="Almost brute force"]
$ 3, \minus{}2, \minus{}1$
$ 3x^2 \minus{} 2x \minus{} 1: x \equal{} 1, \minus{}\frac{1}{3}$
$ 3x^2 \minus{} x \minus{} 2: x \equal{} 1, \minus{}\frac{2}{3}$
I'm too lazy to check the rest.
[/hide]

If making the determinant a rational is sufficient (no evil pitfalls?), then I believe the above is a solution.

[hide="@math_explorer:"]
$ \minus{}x^2\plus{}3x\minus{}2\equal{}0 : x\equal{}1, 2 \\ \\
\minus{}x^2\minus{}2x\plus{}3\equal{}0: x\equal{}1,\minus{}3 \\ \\
\minus{}2x^2\minus{}x\plus{}3\equal{}0: x\equal{}1,\minus{}\frac{3}{2} \\ \\
\minus{}2x^2\plus{}3x\minus{}1\equal{}0: x\equal{}1,\frac12$

It was clever how $ 3\minus{}2\minus{}1\equal{}0$, so every case has $ 1$ as a solution.[/hide]

geez, how did u all get that combination?
dont tell me u used trial and error?

[quote="LinearBoy"]Two students are playing a game. The first player chooses three non-zero numbers. The second player substitutes the numbers for the stars in the equation
[img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/0/f/a/0fa20323c1f66f334fb8cd2fcbbf7a006281b560.gif[/img]
in any order he likes. The first player wins if the resulting quadratic equation has two distinct rational roots. Otherwise, the second player wins. Prove that the first player can always win.

Not sure whether this question was asked before, im a newbie here. Great forum![/quote]

I hope this gets us somewhere.

The discriminant of the quadratic formula is $ \sqrt {b^{2} \minus{} 4ac}$ . So, if the first player chooses three non-zero integers, and we plug these values into the discriminant. We can get two rational roots if the discriminant is rational.

EDIT: So therefore, the three non-zero numbers are always going to be rational, making the discriminant rational, and the first player can therefore always win.

Actually, we only get rational roots when the discriminant is rational. This was mentioned by [b]mismatchtea[/b] in a previous post.

[hide="Hint"]Player 2 can only win if the roots of the equation is a double root or complex conjugates.  In the first case, what must the quadratic be?[/hide]

isnt this question already solved?

If player 1 chooses 3, -2, and -1, he will always win, no matter what order player 2 puts the number in.

But how would you come to those three number without guess and check?

[hide="Proof?"]Assume the first player can never win if the second player plays optimally. Let $ a > 0$, $ b > 0$, and $ c < 0$. By our assumption, at least one of $ a^2 \minus{} 4bc \le 0$, $ b^2 \minus{} 4ac \le 0$, and $ c^2 \minus{} 4ab \le 0$. Note that $ 4bc$ and $ 4ac$ are negative, so $ a^2 \minus{} 4bc > 0$ and $ b^2 \minus{} 4ac > 0$. Therefore, $ c^2 \minus{} 4ab \le 0$, so $ c^2 \le 4ab$. Let $ c \equal{} \minus{}(2\sqrt{ab} \plus{} 1)$. Therefore, $ c^2 > 4ab$, so $ c^2 \minus{} 4ab > 0$, yet we already know that $ a^2 \minus{} 4bc > 0$ and $ b^2 \minus{} 4ac > 0$. Contradiction, therefore, the first player can always win.

[/hide]

that proves that we can always pick 3 numbers where the discriminant is greater than 0, but it does not prove that the numbers will yield a perfect square in the discriminant, as we need rational solutions.

Idk how to do this. Would simply saying that if player 1 picks 3,-2, and -1, then he will always win, or does that not count as a real proof.

And how did you arrive at 3,-2, and -1, mathexplorer?

I guess this problem is solved, but not proven.

math_explorer said "brute force", so I think that would be guess and check.

[hide="Continuation"]Well, since the competition says "numbers" arbitrarily, we can select $ c$ in my proof such that $ c < 0$ and $ c^2 \minus{} 4ab \equal{} 1$.

If we restrict our search to integers, then we must prove that $ c^2 \minus{} 4ab \equal{} m^2$ for some integer $ m$. Equivalently, $ c^2 \minus{} m^2 \equal{} 4ab$. We know that any number equivalent to $ 0$ modulo $ 4$ can be expressed as the difference of two integer squares ($ (n\plus{}2)^2 \minus{} n^2 \equal{} 4n\plus{}4$, and as $ n$ varies, $ 4n\plus{}4$ ranges over all numbers divisible by $ 4$.) , so in any case, we are done.[/hide]


induction
inequalities proposed
inequalities

Let $n>2$ be an integer. Suppose that $a_{1},a_{2},...,a_{n}$ are real numbers such that $k_{i}=\frac{a_{i-1}+a_{i+1}}{a_{i}}$ is a positive integer for all $i$(Here $a_{0}=a_{n},a_{n+1}=a_{1}$). Prove that $2n\leq a_{1}+a_{2}+...+a_{n}\leq 3n$.

[quote="N.T.TUAN"]Let $ n > 2$ be an integer. Suppose that $ a_{1},a_{2},...,a_{n}$ are real numbers such that $ k_{i} \equal{} \frac {a_{i \minus{} 1} \plus{} a_{i \plus{} 1}}{a_{i}}$ is a positive integer for all $ i$(Here $ a_{0} \equal{} a_{n},a_{n \plus{} 1} \equal{} a_{1}$). Prove that $ 2n\leq a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n}\leq 3n$.[/quote]
$ k_1\plus{}...\plus{}k_n \equal{}\sum^{n}_{i\equal{}1} \frac{a_i}{a_{i\plus{}1}}\plus{}\frac{a_{i\plus{}1}}{a_i}$
On the other side,we will actually show that $ k_1\plus{}...\plus{}k_n \leq 3n\minus{}2$ for $ n \geq 2$,by induction on $ n$
 :P

@N.T.TUAN, i solved it before and also saw it in book "Andreescu - Contests Around the World 1997-1998.pdf"  :wink:

[url=http://artofproblemsolving.com/community/c6h598545p3551877]Taiwan 2014 TST 3 Quiz 3, P1[/url]


trigonometry
Inequality
Trigonometric inequality
minimum
IMO Shortlist
IMO Longlist

Let $n$ and $k$ be positive integers such that $1 \leq n \leq N+1$, $1 \leq k \leq N+1$. Show that: \[ \min_{n \neq k} |\sin n - \sin k| < \frac{2}{N}. \]

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[quote="orl"]Let $n$ and $k$ be positive integers such that $1 \leq n \leq N+1$, $1 \leq k \leq N+1$. Show that:
\[ \min\limits_{n \neq k} |\sin n - \sin k| < \dfrac{2}{N}. \][/quote]

Assume the contrary: Assume that $\min\limits_{n\neq k}\left|\sin n-\sin k\right|\geq\dfrac{2}{N}$. Then, for any two n and k with $n\neq k$, we have $\left|\sin n-\sin k\right|\geq\dfrac{2}{N}$. Now let p be a permutation of the set {1; 2; ...; N + 1} such that $\sin p\left(1\right)\leq\sin p\left(2\right)\leq ...\leq\sin p\left(N+1\right)$ (in other words, p is the permutation which arranges the numbers sin 1, sin 2, ..., sin (N + 1) in increasing order). Then, it follows that for any integer i with $1\leq i\leq N$, we have $\left|\sin p\left(i+1\right)-\sin p\left(i\right)\right|\geq\dfrac{2}{N}$. Since $\sin p\left(i\right)\leq\sin p\left(i+1\right)$, we have $\left|\sin p\left(i+1\right)-\sin p\left(i\right)\right|=\sin p\left(i+1\right)-\sin p\left(i\right)$, and thus $\sin p\left(i+1\right)-\sin p\left(i\right)\geq\dfrac{2}{N}$. Hence,

$\sin p\left(N+1\right)-\sin p\left(1\right)=\sum\limits_{i=1}^N\left(\sin p\left(i+1\right)-\sin p\left(i\right)\right)\geq\sum\limits_{i=1}^N\dfrac{2}{N}=N\cdot\dfrac{2}{N}=2$.

But this is impossible, since $\sin p\left(N+1\right)<1$ (in fact, $\sin p\left(N+1\right)\leq 1$, and equality can occur only if $p\left(N+1\right)$ is a multiple of $\dfrac{\pi}{2}$, what is impossible since $p\left(N+1\right)$ is a natural number and $\pi$ is irrational) and similarly $\sin p\left(1\right)>-1$. Thus, we get a contradiction, i. e. our assumption that $\min\limits_{n\neq k}\left|\sin n-\sin k\right|\geq\dfrac{2}{N}$ was invalid, and thus we have $\min\limits_{n\neq k}\left|\sin n-\sin k\right|<\dfrac{2}{N}$.

  darij


I am a freshman in high school, and I am most probably going to pursue something in the science fields, likely Bio -med/ -tech related.  My problem is that I cannot take as many different science classes as I want in high school, and I don't know how much of an impact that will make on college apps either.  I can only take one science class per year, so only 4 classes.  I'm taking Bio this year, and Chemistry next year.  I can then take Bio AP in junior year and Chem AP senior year.  That means that I will not be taking any Physics or other various science offerings.  Any suggestions?

I suggest you take bio, chem, physics and an AP.  If you can't take more science classes, you can't take more science classes (although it might be worth your while at some later stage to find out whether you [i]really[/i] can't take more than one science class a year or whether that's something they tell over-ambitious freshmen who are already worrying about college instead of doing something fun with their time) and so there's not much point in worrying about whether it's going to hurt your college applications.  What you can do, however, is find extracurricular opportunities to expand your studies of science -- Science Bowl, the various science olympiads, research projects of some sort, etc., whatever is most available and interesting to you.

I agree with JBL - physics is more important than having both AP classes. Different schools do things differently, but for us, the Chemistry AP gains more time for a Biology major than the Biology AP would. It's a fairly common pattern to have General Chemistry (what AP Chem covers) as a prerequisite for the cell and molecular biology portion of the Biology sequence required for the many different possible majors in the biological sciences. And such majors (and pre-med course patterns) do require college-level Physics courses as well.

How self motivated are you? Some people self study or take online AP classes. Here are some free AP classes:
http://www.ucopenaccess.org/


inequalities
function
algebra
domain
inequalities unsolved

[color=blue][size=150]Find k max satisfying $x^{4}+y^{4}+2kx^{2}y^{2}\geq (k+1)(x^{3}y+yx^{3})$ with $x\geq 2y>0$ [/size][/color]

Maybe you made a typo, anyway the answer is  $\frac{7}{2}.$

Forgive me when i don't know it.Can you say evident ? :blush:

No problem.Don't mind it! imathsvn.

Problem: Find the maximum value of $k$ for which the following inequality holds

\[ x^4+y^4+2kx^2y^2\geq (k+1)(x^3y+xy^3),\ x\geq 2y>0. \]

I will give you a hint.

Divide by $y^4>0$ both sides of the inequalities and set $t=\frac{x}{y}.$

The hint is too strong  :P  The function is homogeneous in all variables so such a simplification is evident, anyway.

$t^4 + 2kt^2 + 1 \ge (k+1)(t^3 + t), t \ge 2$
$f(t) = t^4 - (k+1)t^3 + 2kt^2 - (k+1)t + 1 \ge 0, t \ge 2$

The inequality obviously holds true as $t \to \infty$, so first we examine it at $t = 2$.

$16 - 8k - 8 + 8k - 2k - 2 + 1 = 7 - 2k \ge 0 \implies k \le \frac{7}{2}$.

Now it remains only to prove that $f(t)$ is increasing for $t \ge 2$.

$f'(t) = 4t^3 - 3(k+1)t^2 + 4kt - (k+1)$
$f'(2) = 32 - 12k - 12 + 8k - k - 1 = 19 - 5k \ge 0 \implies k \le \frac{19}{5}$, so the function is increasing for $k \le \frac{7}{2}$ and our solution is complete.

[quote="kunny"]
Problem: Find the maximum value of $k$ for which the following inequality holds

\[ x^4+y^4+2kx^2y^2\geq (k+1)(x^3y+xy^3),\ x\geq 2y>0.  \]

I will give you a hint.

Divide by $y^4>0$ both sides of the inequalities and set $t=\frac{x}{y}.$[/quote]
Then set $u=t+\frac{1}{t}.$
Note that the domain of $u.$


calculus
integration
logarithms
calculus computations

Evaluate $ \int_1^{\sqrt{2}} \frac{x^2\plus{}1}{x\sqrt{x^4\plus{}1}}\ dx$.

The given integral
$ I\equal{}\int_1^{\sqrt{2}}\dfrac{x^2\plus{}1}{x\sqrt{x^4\plus{}1}}\mathrm dx\equal{}\int_1^{\sqrt{2}}\dfrac{1\plus{}\frac{1}{x^2}}{\sqrt{x^2\plus{}\frac{1}{x^2}}}\mathrm dx$
$ \equal{}\int_1^{\sqrt{2}}\dfrac{1\plus{}\frac{1}{x^2}}{\sqrt{\left(x\minus{}\frac{1}{x}\right)^2\plus{}2}}\mathrm dx$
Substitute $ x\minus{}\dfrac{1}{x}\equal{}t$ which gives $ \left(1\plus{}\dfrac{1}{x^2}\right)\mathrm dx\equal{}\mathrm dt$. Also when $ x\equal{}1$, then $ t\equal{}0$ and when $ x\equal{}\sqrt{2}$ then $ t\equal{}\dfrac{1}{\sqrt{2}}$. So we get
$ I\equal{}\int_0^{\frac{1}{\sqrt{2}}}\dfrac{\mathrm dt}{\sqrt{t^2\plus{}2}}\equal{}\left|\ln(t\plus{}\sqrt{t^2\plus{}2})\right|_0^\frac{1}{\sqrt{2}}\equal{}\ln\left(\dfrac{\sqrt{5}\plus{}1}{2}\right)\equal{}\ln\phi$
where $ \phi$ is the golden ratio.


invariant
search
geometry
geometric transformation
homothety
cyclic quadrilateral
angle bisector

Prove that, if a secant to two orthogonal circles passes by the center of one of them, the four points of intersection form a harmonic division.

[hide="Ugly Solution"]

Notice that $ M$ and $ N$ divide segment $ AB$ harmonically if and only if

$ \dfrac{\overline{MA}}{\overline{MB}} = \dfrac{\overline{NA}}{\overline{NB}}$.

Hence

${ \dfrac{2r_1-\overline{MB}}{\overline{MB}} = \dfrac{2r_1+2r_2-\overline{MB}}{2r_2-\overline{MB}}\Longrightarrow \overline{MB}^2-2\cdot(r_1+r_2)\cdot\overline{MB}+2\cdot r_1\cdot r_2=0\Longrightarrow \overline{MB}=r_1+r_2\pm\sqrt{r_1^2+r_2^2}}$

Notice that $ \overline{MB}=\overline{MP}+\overline{PB}$. $ (*)$

From the right triangle $ O_1CP$, we get

$ \overline{O_1P}=\dfrac{r_1^2}{\sqrt{r_1^2+r_2^2}}$

Note that $ \overline{CP}=\dfrac{\overline{CD}}{2}=\dfrac{r_1\cdot r_2}{\sqrt{r_1^2+r_2^2}}$.

Thus,

$ \overline{PB}=r_1-\overline{O_1P}=r_1-\dfrac{r_1^2}{\sqrt{r_1^2+r_2^2}}$.

From the right triangle $ O_2CP$, we get

$ \overline{O_2P}=\dfrac{r_2^2}{\sqrt{r_1^2+r_2^2}}$.

Thus

$ \overline{MP}=r_2-\overline{O_2P}=r_2-\dfrac{r_2^2}{\sqrt{r_1^2+r_2^2}}$.

Therefore,

${ \overline{MB}=r_1+r_2-\sqrt{r_1^2+r_2^2}}$.

And we are donne (really?).

[/hide]

Can we use the fact that each circle is invariant under inversion wrt the other?  If so, this is pretty simple...

Let $ O_1$ be the circle on the left and $ O_2$ be the circle on the right. Suppose that a line through $ O_1$ intersects, from left to right, at $ A$, $ B$, $ C$, and $ D$.  Let the circles intersect at $ P$ and $ Q$. Suppose that $ O_3$ is a circle tangent to $ AD$, at $ C'$, and tangent to $ O_2$ at $ P$. Then $ O_3$ is orthogonal to $ O_1$.

Note that $ \angle O_1C'O_3\equal{}\angle O_1PO_3\equal{}\frac{\pi}{2}$, $ O_1O_3\equal{}O_1O_3$ and $ O_3P\equal{}O_3C'$, so $ \triangle O_1PO_3\cong\triangle O_1C'O_3$. But then $ OC'\equal{}r\equal{}OC$. Hence $ C\equiv C'$.

Since $ BD$ is a chord of $ O_2$ tangent to $ O_3$ at $ P$, $ \angle BPC\equal{}\angle DPC$ by a common lemma.

Notice that $ \angle APC\equal{}\frac{\pi}{2}$. These two facts imply that $ A$, $ B$, $ C$, and $ D$ form a harmonic range by yet another common lemma.

You can use the lemma from http://www.mathlinks.ro/viewtopic.php?search_id=572576844&t=46146 (the equivalence $ A2$ ).

[quote="LoboSolitario"]  [color=darkred]Prove that if a secant to two orthogonal circles passes by the center of one of them,
then the four points of intersection form a harmonic division.[/color]  [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Let $ w_1 \equal{} \mathcal (O_1,r_1)$ , $ w_2 \equal{} \mathcal (O_2,r_2)$ be two circles for which $ \{A,B\} \equal{} w_1\cap w_2$ .

For a line $ d$ , $ O_1\in d$ denote $ \{X,Y\} \equal{} d\cap w_1$ , $ \{Z,T\} \equal{} d\cap w_2$ so that $ Z\in (XY)$ and

$ Y\in (ZT)$ . Denote the projection $ M$ of the center $ O_2$ to the line $ d$, i.e.the point $ M$ is the middle

of the segment $ [ZT]$ . Observe that $ 4\cdot\left(r^2_1 \plus{} r^2_2 \minus{} O_1O_2^2\right) \equal{}$ $ XY^2 \plus{} 4\cdot ZO_2^2 \minus{} 4\cdot O_1O_2^2 \equal{}$

$ XY^2 \plus{} 4\cdot\left(ZM^2 \plus{} MO_2^2\right) \minus{} 4\cdot\left(MO_1^2 \plus{} MO_2^2\right) \equal{}$ $ XY^2 \plus{} ZT^2 \minus{} 4\cdot MO_1^2$ , i.e.

$ \boxed {4\cdot\left(r^2_1 \plus{} r^2_2 \minus{} O_1O_2^2\right) \equal{} XY^2 \plus{} ZT^2 \minus{} 4\cdot MO_1^2}\ \ (*)\ .$

In conclusion, the circles $ w_1$ , $ w_2$ are orthogonally $ \Longleftrightarrow$ $ r_1^2 \plus{} r_2^2 \equal{} O_1O_2^2$ $ \Longleftrightarrow$

$ XY^2 \plus{} ZT^2 \equal{} 4\cdot MO_1^2$ $ \Longleftrightarrow$ the points $ Z$ , $ T$ are harmonical conjugate

w.r.t. the points $ X$ , $ Y$ on the line $ d$ (see the equivalence $ A2$ of the my lemma from

 http://www.mathlinks.ro/viewtopic.php?search_id=572576844&t=46146 ).

[b][u]Remark.[/u][/b] Suppose that the circles $ w_1$ , $ w_2$ are orthogonally, i.e. the division

$ (X,Z,Y,T)$ is harmonically. From a well-known property obtain

$ \left\|\begin{array}{ccccc}
AX\perp AY & \implies & \widehat {YAZ}\equiv\widehat {YAT} & \implies & \frac {YZ}{YT}\equal{}\frac {AZ}{AT}\\\\
BX\perp BY & \implies & \widehat {YBZ}\equiv\widehat {YBT} & \implies & \frac {YZ}{YT}\equal{}\frac {BZ}{BT}\end{array}\right\|$ $ \implies$

$ \frac {AZ}{AT} \equal{} \frac {BZ}{BT}$ , i.e. the cyclic quadrilateral $ AZBT$ is harmonically !  [/color]

[quote="LoboSolitario"][quote="Altheman"]
Since $ BD$ is a chord of $ O_2$ tangent to $ O_3$ at $ P$, $ \angle BPC \equal{} \angle DPC$ by a common lemma.
[/quote]

Here is the figure and $ BD$ is a chord of $ O_2$ tangent to $ O_3$ at $ C$. Could you please explain the lemma? :blush:[/quote]

The lemma is: Suppose that two circles are internally tangent at $ P$ and $ AB$ is a chord of the larger circle that touches the smaller circle at $ C$. Then $ PC$ is the bisector of $ \angle APB$.

I'm searching for a proof for this lemma and i just can't find it :(

Let $ PC$ intersect the larger circle at $ C'$. Since a homothety center $ P$ maps $ C$ to $ C'$ and $ O_3$ to $ O_2$, $ O_3 C// O_2 C'$. But then $ O_2C'$ is perpendicular to $ BD$. It pretty easily follows that $ C'$ is the midpoint of arc $ BD$. Then $ \angle BPC$ and $ \angle DPC$ subtend congruent arcs, so they are congruent.

Let $ \angle PAO_1\equal{}\alpha ,\; \angle PDO_1\equal{} \delta$ . Then $ \angle O_1PB \equal{} \delta$ (perifferial - inscribed). As $ \angle O_1PC\equal{} \angle O_1CP \equal{} 90^\circ \minus{} \alpha \; \Rightarrow \angle BPC \equal{} 90^\circ \minus{} \alpha \minus{} \delta$.
As $ \angle BCP$ (external) $ \equal{}\angle CPD \plus{} \angle CDP$ we have $ \angle CPD \equal{}90^\circ \minus{} \alpha \minus{} \delta \equal{} \angle BPC$.
So $ AP \perp PC$ and $ PC$ is an angle bisector it follows that AP is the external angle bisector hence (A,B,C,D)=1

Thanks to delta, Altheman and Virgil Nicula.  :)


[quote="� luffy�"]let a,b,c be positive numbers such that a^2 + b^2 +c^2 =1
prove that 1/(1-ab) + 1/(1-bc) + 1/(1-ca) =< 9/2[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66936
$ \minus{}9a^2b^2c^2\geq\minus{}abc(a\plus{}b\plus{}c)$ it's the Vasc's hint.


algebra unsolved
algebra
number theory
pen

Let a sequence $\{a_n\}$, $n \in \mathbb{N}^{*}$ given, satisfying the condition

\[0 < a_{n+1} - a_n \leq 2001\]

for all $n \in \mathbb{N}^{*}$

Show that there are infinitely many pairs of positive integers $(p, q)$ such that $p < q$ and $a_p$ is divisor of $a_q$.

[i]How do you get ideas like this one?[/i]

well, compliments, i suppose this is the fardest sequence problem to formalize i've ever seen

Ok, passing on the fact that $O(n^{-s})$ is nonsense (I guess you meant $O(n^{1-s})$), I must disagree with you about the difficulty of the problem, but since everything is relative, I would like to see your proof, because it would be really nice to see that the example of Erdos and Besicovith of a set of positive upper density with no two elements multiple of one another is wrong. Even if it's long, could you please post that proof? Only three lines in which one says the problem is simple, I'm afraid that won't suffice...

Ok! I am sorry for my English.
Let $A=\{a_1<a_2<....\}$ set with terms $a_i$ We chose $b_1=a_1$ and finitly subset $A_1$,suth that $a_i \in A_1$ if and only if $b_1|a_i$. Then we chose minimal element $b_2\in A/A_1$ and finitly subset $A_2$, suth that $a_i \in A_2$ if and only if $b_2|a_i$, e.t.c construct $b_i,A_i$. If we have K pairs $a_i|a_j$, then subset $A_1UA_2U...A_k$ had only K terms (or less). Density all terms $a_i$, suth that $b_j\not|a_i$ is $(\prod_{j=1}^N (1-\frac{1}{b_j}) )^s$. It is long to prove about exist s (For it consider degree prime divisors $b_j$).  Because $b_j$ is are $a_i$ exlution K terms, density $b_j$ is not less 1/2001. Therefore we have density $a_i$ less than $(\prod_{j=1}^N (1- \frac{1}{b_j}) )^s<N^{-s/2001}$. It give contradition.

thanks Rust .I see it is very interesting

[quote="Rust"]It is easy problem (I see solution for 5 sec), but it long to write. Idea is - exist N, suth that if finitely pairs $a_i|a_j, i\le N$, then density of sequense $a_i$ less than $O(N^{-s},s>0$. It give contradition with density not less 1/2001.[/quote]
What's $O(N^{-s},s>0$ ?

[quote="nmtruong1986"]Hence $b_j$ divisible by $b_i$[/quote]

How do you conclude $a_i|a_j$ or sth like that.

For the solution of Rust, which basic concepts did you introduce, why can you conclude such things?

Lemma:
There exist $x,y \in \mathbb {N}, x<y$ such that:
$x|y , x+1|y+1 , x+2|y+2 , ... , x+2000|y+2000$
Proof is easy (Use this: $x+i$ must divide $y-x$)
We name the pairs like (x,y) friends.
Let $x_1 \in \mathbb {N}$ such that $x_1 \ge a_1$
Now we define:
$X_1=\{x_1,x_1+1,x_1+2,...,x_1+2000\}$
$X_2=\{x_2,x_2+1,x_2+2,...,x_2+2000\}$
$X_3=\{x_3,x_3+1,x_3+2,...,x_3+2000\}$
.
.
.
Such that $x_1,x_2,x_3,...$ are friends.
We know that for all $i \in \mathbb {N}$ one of the members of $X_i$ is in $\{a_n\}$
So there exist infinitely $i,j,k \in \mathbb {N}, 0 \le k \le 2000 , i<j $ such that $x_i+k,x_j+k \in \{a_n\}$ and as $x_i+k|x_j+k$ so we're done :)

Soroush's solution is correct and the same as mine: to clarify, there exists an infinite sequence $x_1, x_2, \cdots$ such that $x_i+j|x_k+j$ for all $i<k$.

We focus on constructing $x_{i+1}$ given $x_i$ by considering $x_{i+1}$ mod powers of $p$ for a fixed prime $p$ such that $p\mid \prod\limits_{j=0}^{2000} (x_i+j)$. Let $k$ such that $\nu_p(x_i+k)$ is maximal among $k=0,\cdots,2000$, call $t$. Then set $x_{i+1}\equiv -k(\bmod\; p^t)$ suffices mod $p$. This works because $\nu_p(x_i+m) = u (u\le t)$, then $x_{i+1}+m \equiv m-k(\bmod\; p^t)$. If $u<t$ then $\nu_p(m-k)=u$, and if $u=t$ then this also works because $p^t\mid m-k$.

After finding $x_{i+1}$ mod each prime power, we can see that $x_{i+1}$ exists by Chinese Remainder Theorem.

Now, let $f(n)$ be any number in $\{0,\cdots,2000\}$ such that $x_n+f(n)$ is in the sequence. By pigeonhole principle, there exists infinitely $m,n$ such that $f(m)=f(n)$. It follows that $x_m+f(m)\mid x_n+f(n)$ and both terms are in the sequence, so we are done.

Wait here is a much simpler way to construct $x_{i+1}$ based on $x_i$:

Just do $x_{i+1}=x_i+ \prod\limits_{j=0}^{2000}(x_i+j)$


What's the method for solving this problem using algebra?

A merchant has coffee worth $20 a pound that they wish to mix with 40lbs of coffee worth $80 a pound to get a mixture that can be sold for $60 a pound.
How many pounds of the $20 dollar coffee should be used?


topology
vector
advanced fields
advanced fields unsolved

Ex. 1.
$ l_0$ bounded sequence and onli finite elements $ \not\equal{}0$
$ l_0^*\equal{}?$ Where $ *$ means dual space.
Ex. 2.
On $ C(0,1)$ define $ (A_n x)(t)\equal{}x(t^{1\plus{}\frac{1}{n}})$
Check covergence:
- strong
- weak
- uniform

$ l_0^*$ = all sequences, in a natural way

Actually the first question is meaningless. The notion of the dual space is based on the notion of a [b]continuous[/b] linear functional and we are not given any topology in $ \ell_0$, so we cannot tell which functionals are continuous and which aren't.

Well, there is the notion of "linear dual" or whatever you want to call it. Consider the space in question purely as a vector space, and ask for the space of all linear functionals, with no reference to any topology. Since this space (whatever we call it - I've never found a notation I particularly liked, or that seems standard) has a concretely given countable Hamel basis, we have a nice concrete description of that set of all linear functionals. (It's the "type specimen" for what a vector space with a countably infinite Hamel basis looks like - of course, up to vector space isomorphism, there is only one such vector space, and this is a realization of it.)

This does play an important role in some proofs. For example:

Suppose we want to know the (Banach) dual of $ \ell^p$ for $ 1\le p<\infty.$ We start bt saying that any bounded linear functional on $ \ell^p$ is a linear functional (topology not mentioned) on the finitely supported sequences, which means the the functional $ L(x)$ can be written as $ L(x)\equal{}\sum_{k\equal{}1}^{\infty}a_kx_k$ for some sequence $ a_k,$ at least for finitely supported $ x.$

Once we have that, we can then use the property that $ L$ is bounded on $ \ell^p$ to work out the required norm properties of $ a.$ And then we finish with an argument that uses that the finitely supported sequences are dense in $ \ell^p$ to extend what we have.

Of course there is standard name for it, the "algebraic dual" in contrast to the usual "topological dual" in functional analysis.


limit
trigonometry
conics
parabola
vector
linear algebra
matrix

Here is the list of problems of Samarkand State University 1-year students Olympiad which held on 25-th of March
                                                                                Duration:2 hours

1.Let the polynom P(x) has a distinct n=deg(P(x)) roots         $a_1,a_2,...,a_n$ and let $b_1,b_2,...,b_{n-1}$ be  the roots of the polinom P'(x).
Find the value of the sum:
$\displaystyle\sum_{i,k} \frac{1}{a_i-b_k}$
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=81093[/url]

2.Find the limit :
$\lim_{n\to\infty} \sin^2(2\pi \sqrt[3]{n^3+n^2})$
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=81097[/url]

3. Find the number of all non-negative ordered integer triples (x,y,z) such that x+y+z=2006
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=81094[/url]

4.If $\forall \epsilon>0 \text{ exist } \delta(\epsilon): \text{ } \forall x,y \in (0,+\infty) : |x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon$ ( i dont know how it calls, equally continuous???) then what can we state about f(+0), f(+oo)?
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=81095[/url]

5.Let $f(x) \in C^{(1)}(1,+\infty) \text{ and } \forall x\in (1.+\infty ) \Rightarrow |f(x)| \le \sqrt{x}$
Prove that $\exists x_n\to +\infty \text{ such that } f'(x_n) \to 0$
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=81096[/url]

[b]National University of Uzbekistan (old Tash SU)
                                                         Students Olimpiad 2006/30/03[/b]                                                                                Duration 3 hours
1. Find the $\lim_{n\to\infty}n\sin(2 \pi n!e)$.
smth like that is here [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=69467[/url]
2. Let the polynom $P(x)=a_nx^n+...+a_1x+a_0$ be such that $P(k), P(k+1), ... ,P(k+n)\in\mathbb{Z}$.
Prove that   $\forall x\in\mathbb {Z}$ we have $P(x)\in\mathbb{Z}$.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=81832[/url]
3. Let two parabolas with perpendicular axis intersect each other in 4 points. Prove that this 4 points are cyclic.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=81830[/url]
4. Let $\mathbb{R}^n$ be the Euclidean space. Prove that for any vectors $a_1, a_2, ... ,a_{n+2}$ there exist $i\ne j \text{ such that } (a_i,a_j)\geq 0$.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=81829[/url]
5. Lets call matrix A a chess-matrix if
$A=\begin{pmatrix} *&0&*&0&... \\ 0&*&0&*&... \\ *&0&*&0&...\\ \vdots&\vdots& \vdots&\vdots&\ddots\end{pmatrix}$

or
0*0*...
*0*0...
0*0*...
.....
.....
..... where * may be zero too
Prove that if $detA\ne 0$ then the $A^{-1}$ is also a chess matrix.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=81828[/url]
This problems were given for all 1,2,3,4-th(between all the courses) year students and only one of them will participate in Republic Students Olympiad .
It's incredible! I've won the firts prize!
Thanx for Mathlinks again!!! :winner_first: (I've  wanted to post this emotion for a long time...

[b]The Samarqand State University Students Olympiad [/b](between the winners of The first round of all the courses, Orif Ibragimov won the 1st prize in the 1st Round and the 2nd in this one, Congratulations!)
Held on 3rd of April
Duration: 2 hours.

1. Let the ellipsis with the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Find the point on this ellipsis such that triangle has  the minimal square which is formed with the tangent in this point
and coordinate axis
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=83110[/url]

2.Show that there is real $\alpha$  s.t
                         $\frac13\leq {\alpha^n}\leq \frac{2}{3}$
holds for all positive integers n. (where  {x} is fractional part of x)
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=82564[/url]

3. Let function f(x) is integrareble on the segment [0;1] and $\int_{0}^{1} f(x)>0$
Prove that there exist a,b s.t. 0<a<b<1 and f(x) is positive on this segment.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=83109[/url]

4.On the plane given n points $A_1,A_2,...,A_n$ s.t:
1)all of them are not collinear
2)the line passing through $A_i$ and $A_j$ passes some $A_k$ $\forall i,j \text{ and } i\ne j$
Show that such points are infinitely many.($n\to\infty$)
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=82810[/url]

[b]First competition[/b]:
Held on 16th of May in Urgenj State University
Duration 2.5 hours

1. If continuous function f(x) on [a;b] satisfies the following conditions:
$\frac{1}{b-a}\int_{a}^{b}f(x)dx=1$
$\frac{1}{b-a}\int_{a}^{b}f(x)^2dx=1$

then prove that f(x)=1
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=88040[/url]

2. Find $\lim_{n\to\infty}\int_{0}^{1}(f(\frac{{x}^2}{n})+1)dx$ if $f(x)=(x^2+1){}^e{}^x$
3.Prove that if  $I+AB$ is invertible then $I+BA$ is invertible
[url]http://www.mathlinks.ro/Forum/topic-86370.html[/url] :P

4. Prove the inequality for numbers $x_1,x_2,...,x_n \in [a;b]$ where a is positive:

$(x_1+x_2+...+x_n)(\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n})\leq n^2\cdot \frac{(a+b)^2}{4ab}$
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=88042[/url]

5.Prove the inequality for any point P and triangle ABC:

   $AB^2+BC^2+AC^2\leq 3(PA^2+PB^2+PC^2)$
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=88045[/url].

[hide="Second Competition:"]
Held on 16th of May in Urgenj State University
Duration 1 hour

There was tests with 20 questions whch were basicly easy but i did 3 wrong.[/hide] [hide="Third Competition:"]
Held on 16th of May in Urgenj State University
Duration 1 hour
There was only one question: write all what u know aout the "kreviye vtorogo poryadka i ih klassifikatsiya"
(would be gratefull if sombody translate this from russian to english).[/hide] [hide="Fourth Competition:"]
Held on 17th of May in Urgenj State University
Duration 15 minutes
Write the Mean Value Theorem for the Integrals(i mean frot eh integrals in the from $\int_{a}^{b}$ u call them defined?) and tell it on the blackboard during 10 minute
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=389291#p389291[/url]
 :D [/hide] [hide="About results:"]
1- Khudayar from the Qarshi State University, 3rd year student
2- Shohruh from  the Samarkand State University, 2nd year student (IMO 2003,2004)
3- Frunza from the Urgenj State University, 2nd year student
4- me  from National University of Uzbekistan(old Tashkent State University), 1st year student (IMO 2005)
.........I didn't think that Students National Olympiad this year  would be like that...[/hide]

[quote="rustam"]i mean frot eh integrals in the from $\int_{a}^{b}$ u call them defined?[/quote]

In English, definite integrals, the other kind (anti-derivatives) being indefinite integrals.

Calculate: $\displaystyle\lim_{n\to\infty}\int_{0}^{1}\Bigl(f\bigl(\frac{{x}^2}{n}\bigr)+1\Bigr)dx$, where $f(x)=({x}^2+1) e^{x}$.

I think it should be 2. Just look at what happens to $x^4/n^2$ and $e^{x/n}$ when $x\in[0,1]$ and $n$ gets large. (Maybe I'm mistaken though...)

$\lim_{n\to\infty}\int_0^1 (f(\frac{x^2}{n}) + 1)dx$
$f(x) = (x^2 + 1)e^x$

Which is 2, as I said. Unless I made a mistake, which is possible.

Problems of National University of Uzbekistan Students Olympiad 2007
Though seems like high-school competition
Held on 12th of April
For 1,2,3,4 year students
Duration 90 minutes

1. Prove for all $a,b,c \in \mathbb{N}$
\[\gcd (\text{lcm}(a,b), lcm(b,c), lcm(c,a) = lcm (\gcd(a,b), \gcd (b,c), \gcd (c,a)) \]
2. Prove the inequalities for $0<x\leq \frac{\pi}{4}$
\[( \cos x)^{\cos^{2}x}\geq ( \sin x)^{\sin^{2}x}\]

\[( \cos x)^{\cos^{4}x}\leq ( \cos x)^{\cos^{2}x}\]
3. Prove the inequality for all positive integers n:
\[\frac{1}{n+1}< \ln \left ( \frac{n+1}{n}\right ) < \frac{1}{n}\]
4. Find the determinant of a matrix $A=(min(i,j))$

5. Solve the differential equation
\[y=xy'+x^{2}y'' \]

Problems of National University of Uzbekistan Students Olympiad 2007
Held on 12th of April
For 1,2,3,4 year students
Duration 3 hours.

1. Calculate the $ \displaystyle \lim_{n\to\infty} \frac{1}{n} \sqrt[n]{(n\plus{}1)\cdot (n\plus{}2) \cdot \ldots \cdot (n\plus{}n)}$

2. Let $ A$ be a $ n \times n$ matrix.  Show that $ \det\left[\begin{array}{cc} A & A^2\\ A^2 & A^3
\end{array} \right]$ is zero.

3. Let $ S_n$ be the square of the figure on the plane, bounded with the curve $ x^{2n}\plus{}y^{2n}\equal{}1$. Show that $ \lim_{n \to \infty}S_n\equal{}4.$

4. Let $ P$ be a polynom with integer coefficients, such that $ P(a)\equal{}b, \, P(b)\equal{}c, \, P(c)\equal{}a.$
Prove that there exists an integer solution of $ P(x)\equal{}x.$

5. Let $ G$ be an additive subgroup of $ (\mathbb{R},\plus{}).$ Show that either $ G$ is dence in $ \mathbb{R}$ or it is discrite (дискретная).

[quote="rustam"]15. Solve the differential equation
\[ y \equal{} xy' \plus{} x^{2}y''
\]
[/quote]

 change variable
x=e^t

g(t)=y(e^t) find a differential equation verify by g

[quote="rustam"]Problems of National University of Uzbekistan Students Olympiad 2007
Though seems like high-school competition
Held on 12th of April
For 1,2,3,4 year students
Duration 90 minutes

$ ( \cos x)^{\cos^{4}x}\leq ( \cos x)^{\cos^{2}x}$

[/quote]
The inequality is false when $ x\equal{}30^0$
$ {(\frac {\sqrt {3}}{2})}^{\frac {9}{16}} \leq {(\frac {\sqrt {3}}{2})}^{\frac {3}{4}}$ then $ 1 \leq {(\frac {\sqrt {3}}{2})}^{\frac {3}{16}}<1$ which is a contadiction (I suppose that the sign of the inequality must be changed so it will be true in all value of $ x$ in the interval given)

[quote="rustam"]Problems of National University of Uzbekistan Students Olympiad 2007
Though seems like high-school competition
Held on 12th of April
For 1,2,3,4 year students
Duration 90 minutes

1. Prove for all $ a,b,c \in \mathbb{N}$
\[ \gcd (\text{lcm}(a,b), lcm(b,c), lcm(c,a) = lcm (\gcd(a,b), \gcd (b,c), \gcd (c,a))
\]
2. Prove the inequalities for $ 0 < x\leq \frac {\pi}{4}$
\[ ( \cos x)^{\cos^{2}x}\geq ( \sin x)^{\sin^{2}x}
\]

\[ ( \cos x)^{\cos^{4}x}\leq ( \cos x)^{\cos^{2}x}
\]
3. Prove the inequality for all positive integers n:
\[ \frac {1}{n + 1} < \ln \left ( \frac {n + 1}{n}\right ) < \frac {1}{n}
\]
4. Find the determinant of a matrix $ A = (min(i,j))$

5. Solve the differential equation
\[ y = xy' + x^{2}y''
\]
[/quote]
Solution for problem number 4 can be obtained by substracting n th column to n-1 th column etc so we can get a matrices with the entries of $ a_ij$ is $ 1$ when $ i \leg j$ and otherwise. So we claim the determinant is $ 1.1.1.1...1=1$

[hide="The last ineq for number 3 the same competition problem set I answered above"]
$ ln(\frac {n\plus{}1}{n}) \leq \frac {1}{n}$
$ n.ln(\frac {n\plus{}1}{n}) \leq 1$
$ ln (\frac {n\plus{}1}{n})^n <1$
True
[/hide]


algebra
polynomial
superior algebra
superior algebra unsolved

Given a polynomial $f(X) \in \mathbb{Q}[X]$ irreducible over $\mathbb{Q}$ and two of its complex roots $\alpha$ and $\beta$, and given a Galois extension $E$ of $\mathbb{Q}$ (with $\mathbb{Q} \subseteq E \subseteq \mathbb{C}$) such that $E=\mathbb{Q}[\epsilon]$ where $\epsilon$ is a root of unity, show that $\mathbb{Q}[\alpha] \cap E = \mathbb{Q}[\beta] \cap E$.

Let $\Gamma$ be the isomorphism from $\mathbb{Q}[\alpha]$ to $\mathbb{Q}[\beta]$ taking $\alpha$ to $\beta$.As E is galois(normal) we can see from this map that
$E \cap \mathbb{Q}[\alpha] \simeq E \cap \mathbb{Q}[\beta]$.
Now both of them are subfields of $E$ containing $\mathbb{Q}$ having same degree over
$\mathbb{Q}$.But $E$ is a cyclic Galois  extension of
$\mathbb{Q}$,so it has a unique subfield$F$ containing $\mathbb{Q}$ for each number $n$ which divides $[E: \mathbb{Q}]$ s.t $[F: \mathbb{Q}] = n$.
So $E \cap \mathbb{Q}[\alpha] = E \cap \mathbb{Q}[\beta]$.


I was just wondering when exactly one should use the 'implies' sign (the rightarrow) and the 'implies one another' sign (sign with both left and right arrow), particularly when using algebraic manupulation. For example, would it be appropriate to use one of the signs in the following situation: x/3 = 4/5 => x = 12/5 ?

Fierytycoon

I may be wrong, but I believe that the "implies" and "implies another" are other ways of writing "if...then..." (or "only if") and "if and only if", respectively.  The former is called the conditional and the latter is called the biconditional.

For that reason, I think you should use the "implies another" sign, since x = 12/5 does imply x/3 = 4/5, in addition to the other way around.

The biconditional is used to describe equivalent statements, since it says that one set contains another, and that other set contains the first set, so both sets must have the exact same things in them or else both conditions won't be satisfied.

The conditional is used when one set is a member of another set, but not vice versa.  For instance "if I'm in Massachusetts, then I'm in the USA" is true (Massachusetts --> USA) but "I'm in Massachusetts if I'm in the USA" isn't necessarily true (Massachusetts not <-- USA)

[quote="ComplexZeta"]The sign <==> means that the two statements are equivalent.[/quote]

Yes...biconditional, right?

So you're saying that things like x = 12/5 does imply x/3 = 4/5 require a <=>, correct? And could you give an example or two of the use of => ? thx

Fierytycoon

Yes, you may write x=12/5 <==> x/3=4/5. For the ==> sign, we can say that x=2 ==> x^2=4, but x^2=4 ==> x=2 is false since x=-2 is also possible. I hope this helps.

Here's the => (the conditional):

"If I'm in this room, then I'm in Nevada."  This is true (at least in my case).

However, I can't say "If I'm in Nevada, then I'm in this room" because I could be in a different room and still be in Nevada.

So this is an example of the conditional, since it only works one way.

You could also say any of these equivalent statements:

"I'm in this room only if I'm in Nevada."
"I'm in Nevada if I'm in this room"
"I'm not in this room if I'm not in Nevada"

The last one is called a contrapositive.  It has the same truth value as any of the other forms (If P then Q, Q if P, P only if Q).

___

Another example:

If jojobird likes all vegetables, then he likes lettuce.

This is true, right?

But this isn't:
If jojobird likes lettuce, then he likes all vegetables.

Since lettuce is a member of the set vegetables, if something is true for the set vegetables, then it's true for all of the things like lettuce that are in the set.

Equivalent forms:
"Jojobird likes vegetables only if he likes lettuce."
> that is to say, if he didn't like lettuce, we can't say that he likes all vegetables
"Jojobird likes lettuce if he likes all vegetables"
> again, this makes sense; since lettuce is a vegetable, and he likes all vegetables, then he must like lettuce
"Jojobird doesn't like all vegetables if he doesn't like lettuce"
> Contrapositive.  This is true - see the comment for the first equiv. form.

Whereas for the biconditional, both a statement and the "vice-versa" of the statement must be true.  For instance:

If x = 2, then 2x = 4
If 2x = 4, then x = 2

Thus for this case, we can write x = 2 <==> 2x = 4


Show that the arithmetic progression 1, 730, 1459, 2188, ... contains infinitely many powers of 10.

[hide="Solution"] We will prove an equivalent statement:  that infinitely many powers of $10$ satisfy

$10^{k}\equiv 1 \bmod 729$

But this is obvious from Totient Theorem or similar. [/hide]


Let real sequence {un} n=0,1,2... $\infty$ and such that:
          u(n+1)=2un -n^2
Calculate un follow n and uo
u0=? such that un>0 (n>=0)


Show that for every closed set $A$ there exists a $C^\infty$ function on the line such that $f(x)=0$ iff $x$ is in $A$.

[url=http://en.wikipedia.org/wiki/Mollifier]This[/url] is how you do it when $\mathbb R\setminus A$ is an interval. In general, $\mathbb R\setminus A$ is a countable union of disjoint open interval. Define the function on each interval separately.

Two remarks:
1) The question also makes sense in $\mathbb R^{n}$.
2) Just doing it on each interval separately without much thinking may result in a discontinuous function ;).
Nevertheless, mlok gave you a really good advice; just use it with caution and imagination :).