RMO result of Chhattisgarh are out. I am selected. Thanks to mathlinks for making me prepare well.

from where did you get to know of it....??? any site???

And of course.... Congrats...

Even the Delhi Results are out !! I have also been selected.. Thanks to all of you for helping me, and clearing all my doubts  !!

Well they Called me up and informed that i have been selected and have to attend a camp next weekends. I don't think that the results have been put up on an y site....BY THE WAY ARE YOU FROM DELHI REGION???

*** INMO IS ON  JAN 18 ,2009

oh yes....... but never mind... I didnt make it......

[quote="fbaggins"]@amangupta,I telephoned MP RMO co=ordinator and he said that due to counting of votes, results will take another 8-10 days. SO, can u tell me what is ur expected score seeing the official solutions!![/quote]
Hi fbaggins. Just in case you were wondering the actual RMO scores you get does not depend on whether your solution is matching to the official solution. All that matters is whether your proofs are rigorous and you get the same final answer (matching with the official answer). Since  you got all atleast 3 question fully right don't be so anxious. (I understand the tension you must be  feeling though to know if you cleared RMO but there is a 99% chance that you will clear RMO, becuase you are in ninth grade.)

@befuddlers;

Thanks very much for motivating me but i feel more depressed now because I got only 2 fully right!!!!!! Q5 and Q6!!!!!!

Now plz tell if i have any chance!

BTW , it is my personal opinion that being in class IX doesn't help very much in clearing RMO!!

fbaggins, how does the counting of votes have relationship with the RMO results?
See befuddlers, it is not like 9th stds have an edge. It is the realtive performance. If the other 9th stds have given a mediocre performance and u have solved 2 with excellent presentation(I suppose-fbaggins) , U will surely get through.It is like tht. If the other 9thers have done horribly and 1 medium kinda performance, he has a better chance. Its like that. :D   :rotfl:

well! @ aravind,
well i was also in doubt as to what is the connection of counting of votes with RMO?But my dear frnd,
votes are counted by the teachers of the sise institute where RMO papers are checked!So u see.............

[quote="fbaggins"]@amangupta,I telephoned MP RMO co=ordinator and he said that due to counting of votes, results will take another 8-10 days. SO, can u tell me what is ur expected score seeing the official solutions!![/quote]

I was able to get 3.5 questions correct .however i attempted 4.5 questions.I am not sure how much scores i would have scored .
By the way, being able to solve 2/6 question while still being in ninth grade surely separates you from the crowd.

Aravind has rightly said, it is all about relative performance.There's always hope and chances are that you too might get selected  .moreover i am sure that you will do much much better the next time.
 :wink:  :)

Aravind , can you please suggest me some books / sites etc. form where i could prepare for INMO . Coz' if i am not mistaken , you are an INMO Awardee and must be aware of it . SO , can you please  help me ???

[quote="Akashnil"]There's no pressure if you don't care about marks, they won't give assignments etc. too much.You have to study yourself. If you don't your marks will be bad that's all.[/quote]

:Wow: thats really supercool man!!! :blush:

[quote="anurags92"][quote="amangupta2"] Yes I do. I have joined Vidyamandir Classes ,Delhi (You [/quote]

Even i am in vidyamandir... Aim iit Group A....[/quote]
Wow !! That's Great , I am in Regular JEE 2010 Batch 2 !! Great to Hear this .

Bye the way do you all know when will be the Results of ZIO 2009 declared !! They said that they will be out by 15th dec.???

I am not worried about zio at all coz i know i wopnt qualify at all... :|    :|

Well , there's some good news , I have cleared The ZIO 2009 from Delhi.....The results were out today.....I dIdn't Expect things to be this way...Bu it's Good

Hello! @amangupta,

first RMO then ZIO, keep rocking!!!!!!!!!!!! :lol:  :D

well ! I am interested in ZIO but it is not held anywhere near! :(

SO.......................

Anyway,  congrats again!!!!

BTW, anyone else Qualified in this forum?

@fbaggins Thanks !!!

 Sorry !! I Do not Have any idea , The Result have been declared today itself  and not much people have posted since then.....

Bye  the way , which region do you live in ??
 :wink:

Gaurav Patil has got through.
I haven't got through.

ny1 who knows whether or not rajasthan RMO results have been declared?

@amangupta,
well i live in a very very remote place called vindhyanagar in dist.-singrauli in madhya pradesh

@fbaggins , Oh Yes !! I have studies about Singrauli in 10th , If I am not wrong , it is famous for rich mineral deposits  , right???

@rituraj , This  time In delhi INMO TC will be held for 4 days but In Two Distinct Phases.

Does anyone here know , whether any camp is organised before the INOI (like INMO TC for Maths) , If Yes, Then What all is taught there , and when is this camp organised ?????

well,correct amangupta,  Asia's LARGEST THERMAL POWER PLANT is located here,NTPC(National Thermal Power Corporation)!!!!

hey what is this zio u all are talking about?

ZIO stands for Zonal Informatics olympiad. for more info, go to http://www.iarcs.org.in


We're not sure.  It won't be soon, but hopefully within a year.


algorithm
combinatorics unsolved
combinatorics

There are $ 63$ points arbitrarily on the circle $ \mathcal{C}$ with its diameter being $ 20$. Let $ S$ denote the number of triangles whose vertices are three of the $ 63$ points and the length of its sides is no less than $ 9$. Fine the maximum of $ S$.

Lemma: consider a graph with $ 63$ vertices, each two connected by (one) blue or red edge. Suppose there are no [i]blue[/i] $ 7$-cliques in the graph. Define

$ S_1 \equal{} \{V_1, V_2, ... V_{11}\}$
$ S_2 \equal{} \{V_{12}, V_{13}, ..., V_{22}\}$
$ S_3 \equal{} \{V_{23}, V_{24}, ..., V_{33}\}$
$ S_4 \equal{} \{V_{34}, V_{35}, ..., V_{43}\}$
$ S_5 \equal{} \{V_{44}, V_{45}, ..., V_{53}\}$
and
$ S_6 \equal{} \{V_{54}, V_{55}, ..., V_{63}\}$

Then the maximimum number of [i]blue[/i] triangles is attained when two vertices are connected by a blue edge if and only if they lie in different $ S_i$'s (this distribuition is the same in Turan's theorem).

Proof: take a graph $ G$ and an arbitrary red clique $ C$. Take in $ C$ the vertex that lies in more blue triangles and call it $ V$. We can change our graph by making each vertex in $ C$ be linked with the vertices already linked with $ V$ (and only with this vertices). We are going to prove that this does not make a blue $ 7$-clique appear in the graph and does not decrease the number of triangles. First, notice that a blue triangle can obviously have at most one vertex in $ C$. Thus, since a blue $ 7$-clique with vertices in $ G \minus{} C$ cannot have appeared (the edges in this graph are the same), any blue $ 7$-clique has exactly one vertex in $ C$. However, if we change this vertex by $ V$, the colors of the edges will be the same (by construction), therefore we had a $ 7$-clique before. Contradiction! On the other hand, the number of blue triangles with no vertices in $ C$ didn't change, and the number of blue triangles having a given vertex of $ C$ became the number of blue triangles having the vertex $ V$. By the maximality of $ V$, we get the result (there is no decrease).

Now we are going to show an algorithm which makes any graph become the described graph (or a graph with less blue triangles), based on the operation $ O$ above. Take a red clique $ R$ (with at least $ 2$ vertices) and make the operation $ O$. If some vertex in $ R$ is connected with some vertex $ A \in G \minus{} R$ by a red edge, then all the vertices in $ R$ are. Thus, $ R \bigcup \{A\}$ is a red clique. Make the operation $ O$ and repeat this procedure, until we get a red clique $ R_1$ such that any edge between $ R_1$ and $ G \minus{} R_1$ is blue. Now we make the same thing with $ G \minus{} R_1$, getting a red clique $ R_2$ and another graph $ G \minus{} R_1 \minus{} R_2$. Now there are no red edges between $ R_1$ and $ R_2$ or $ R_1$ and $ G \minus{} R_1 \minus{} R_2$ or $ R_2$ and $ G \minus{} R_1 \minus{} R_2$. We can make this until we get $ R_6$, at most (if we had $ 7$ red cliques as we said, taking a vertex in each of them, we would get a blue $ 7$-clique). Our algorithm only stops when all the remaining edges are blue (in particular, if $ |G \minus{} R_1 \minus{} ...| \le 1$, i.e., if there are no edges in the remaining graph). In any case, in the end we will have a partition $ G \equal{} R_1 \bigcup ... \bigcup R_k \bigcup B$, where $ B$ is a (maybe empty) blue clique and each $ R_i$ is a red clique (with at least two vertices). Case $ |B| \plus{} k \ge 7$, it is obvious that there is a blue $ 7$-clique, which we are assuming false. Else, we can think the vertices in $ B$ as unitary red cliques, therefore having a partition $ G \equal{} R_1 \bigcup R_2 \bigcup ... R_n$, with $ n \le 6$. Moreover, if now we accept empty cliques, we may assume $ n \equal{} 6$.

Now the result is trivial! We just need to prove that, if $ a \equal{} |R_1| > |R_2| \plus{} 1 \equal{} b \plus{} 1$, moving a vertex from $ R_1$ to $ R_2$ is going to increase the number of blue triangles. This would imply that, in the maximum, the best distribuition for the points is the most balanced one. Moving a vertex $ X$ from $ R_1$ to $ R_2$ makes the blue triangles with vertices in $ X$ and some point in $ R_2$ disappear and the triangles with vertices in $ X$, some point in $ R_1$ and some point in $ R_3$, $ R_4$, $ R_5$ or $ R_6$ become blue. Since $ a \minus{} 1 > b$, we have

$ (a \minus{} 1)(\sum_{i \equal{} 3, 4, 5, 6} |R_i|) > b(\sum_{i \equal{} 3, 4, 5, 6} |R_i|)$,

id est, the number of triangles becomes greater.

Back to the problem. Two vertices of the $ 63$-agon are going to be connected with a blue edge if and only if the distance between them is $ \le 9$. The other pairs of vertices are going to be connected by a red edge. Suppose now we have a blue $ 7$-clique, say $ P_1P_2...P_7$, with $ P_i$ between $ P_{i \minus{} 1}$ and $ P_{i \plus{} 1}$. We know that an arc has always a length greater than the corresponding chord. Then the circle has length $ \widehat{P_1P_2} \plus{} \widehat{P_2P_3} \plus{} ... \plus{} \widehat{P_7P_1} > 9 \plus{} 9 \plus{} ... \plus{} 9 \equal{} 63 > 20\pi$, absurd! It follows that the maximum number of triangles is obtained when $ P_1 \approx ... P_{11} \approx A$, $ P_{12} \approx ... P_{22} \approx B$, $ P_{23} \approx... P_{33} \approx C$, $ P_{34} \approx ... P_{43} \approx D$, $ P_{44} \approx ... P_{53} \approx E$ and $ P_{54} \approx ... P_{63} \approx F$ and $ ABCDEF$ is a regular hexagon (its side must be the radius of $ \mathcal{C}$, that is $ 10$, that is $ > 9$, which means the maximum is really attained). It is easy to see that this number is $ 11^3 \plus{} 9.11^2.10 \plus{} 9.11.10^2 \plus{} 10^3 \equal{} 23121$.

Did somebody read my post?

I want to know if it is right...

Hi Renan, I think that your proof is not right. I have the number 25515 to be the answer. The example would be when the 63 points are divided into 7 groups of 9 points each(the points in a group are closely together), 6 form a regular hexagon and the other is the center of the hexagon. It is easy to see that the example verify.

You made a great mistake when saying that the arc is greater to the corresponding chord leads to an absurd.

My proof is actually quite similar to that of the Turan Theorem. I will post it later (it is 0:00 and I am sleepy).

No,Cuenca,it says 63 points ON the circle.

But i think Feliz should connect a blue edge if and only if the distance >=9


It is given $ A_1 A_2 A_3...A_n$ a polygon and it's diagonals such as it is imposible to find 3 diagonals that concur.Find the number of triangles having only one vertice of the polygon in their construction.(explain it)

Do you mean, "Find the number of triangles whose edges belong to the diagonals of the $ n$-gon and exactly one of whose vertices is a vertex of the $ n$-gon"?

I thought there were $ \dbinom{n}{2} \minus{} n \equal{} \frac {n(n \minus{} 3)}{2}$ diagonals, and the number of the intersection points is $ \dbinom{n}{4}$, so I do not think your result is correct.

It's absolutely incorrect, since I didn't put almost any thought to it. Deleted.

and now you can give a final response and as an extension: find in how many triangles a $ n\minus{}$gon can be divided with help of its diagonals

[hide="Hopefully correct now"]From each vertex (call it $ A$) we choose $ 2$ diagonals as two sides of our triangle (call those $ AB$ and $ AC$). Now we must "cut" those two diagonals by the third one, and that one may not contain either $ B$ or $ C$. Obviously, all triangles obtained in such a way are unique (no overcounting).

There are $ n$ vertices. There are $ \binom{n\minus{}3}2$ ways to choose first $ 2$ diagonals. The third must connect any two vertices but $ A, B, C$, hence there are $ \binom{n\minus{}3}2$ of those. So our final count is $ N\equal{}n{\binom{n\minus{}3}2}^2$.[/hide]


If $ a,b$ and $ c$ be any real numbers where $ a\plus{}b\plus{}c \neq 0$ then prove that

$ \sum \frac{(b\plus{}c\minus{}a)^2}{a^2\plus{}(b\plus{}c)^2}\geq \frac{3}{5}$

[quote="pankajsinha"]If $ a,b$ and $ c$ be any real numbers where $ a \plus{} b \plus{} c \neq 0$ then prove that

$ \sum \frac {(b \plus{} c \minus{} a)^2}{a^2 \plus{} (b \plus{} c)^2}\geq \frac {3}{5}$[/quote]

This problem is similar with Japan 1997, and (I think) solution is similar, too.
See here, http://www.mathlinks.ro/viewtopic.php?p=537 .


calculus
integration
logarithms
function
calculus computations

Somehow my calculus textbook doesn't mention a bit on $\displaystyle\int_{-x}^{x}ln x$.
If anyone knows the answer, plz respond.
Also, I'm only a 4th grader and trying to study calculus, so I don't know much about it.

It is possible to calculate $\displaystyle\int\ln{x} dx$, but you must use a technique called integration by parts.

See [url]http://en.wikipedia.org/wiki/Integration_by_parts[/url].

Integration by parts: $\displaystyle\int u dv = uv - \displaystyle\int v du$
where $u$ is a differential function and $dv$ is an integrable function.

In this example:
Let $u = \ln{x}$, $du = \frac{1}{x} dx$
 Let $v = x$, $dv = 1 dx$.

It simplifies out to $x \ln{x} - x + c$.

Also, since $\ln{x}$ is not defined over negative values, $\displaystyle\int_{-x}^{x}\ln{x} dx$ doesn't exist.

It's also better not to have the same variable in your bounds as your variable of integration. Is it the same $x$? If so, the expression is hard to interpret -- how can you evaluate between bounds that aren't fixed? If it's not the same $x$, why call them the same thing?

I agree with you, Xevarion. I add to say that we can't define the definition for which the logarithm is defined, that is to say the condition of anti-logarithm.

[quote]Integration by parts:  $\int u dv = uv - \int v du$
where $u$  is a differential function and $dv$ is an integrable function.

In this example:
Let $u = ln x$ , $du = \frac{1}{x} dx$.
Let $v=x$ , $dv= 1 dx$ .[/quote]

The standard presentation of integration by parts bothers me.  I think it's unrigorous.  What is "$dx$"?  Is it a number?  If it's a number, how large is it?  (Some people try to say $dx$ is an "infinitely small real number", which is nonsense.)  If $dx$ is not a number, then what is it?

Back in the old unrigorous days, people used to loosely speak of "infinitesimal" quantities, like $dx$ and $dy$.  When mathematicians got more careful, "infinitesimals" were banished from Calculus because the idea doesn't make sense, and everything was explained by taking limits of finite quantities and never referring to an "infinitesimal."  And yet, after all this time, infinitesimals still linger here in the standard presentation of Integration by Parts.

(Sure, the idea of an "infinitesimal" may have been brought back and made precise in some parts of advanced math, but that's advanced math and not basic Calculus.  And in any case, if $dx$ is going to be referred to and manipulated as its own entity, then it needs to be defined precisely, which is never done in a basic Calculus course.)

It's easy to modify the standard presentation of integration by parts to make it rigorous:

$\int u(x)v'(x) dx = u(x)v(x) - \int u'(x)v(x) dx$.

In this example:
Let $u(x) = ln(x)$, $u'(x)= \frac{1}{x}$.
Let $v(x)=x$, $v'(x)=1$.

Then $\int ln(x) dx = \int u(x)v'(x)dx = u(x)v(x) - \int u'(x)v(x) dx$

= $x ln(x) - \int \left(\frac{1}{x} \right)x dx = x ln(x) - \int 1 dx = x ln(x) - x + C$.

(By the way, integration by parts is just the product rule in reverse.)


Find all functions $f : R \longmapsto R$ such that
f(x+y) = f(x)f(y)f(xy) for all real x and y.

The solutions are $f(x) \equiv -1$ ; $f(x) \equiv 0$ or $f(x) \equiv 1$
But the proof seems to be very long  :D

f(x+y)=f(x)f(y)f(xy)
let f(0)=a then f(x)=f(x) a^2 for all x. so either f(x)=0 or a^2=1
Let y=x/(x-1)
Then f(x^2/x-1) =f(x)f(x/x-1) f(x^2/x-1)
So again either f(x)=0 or f(x) f(x/(x-1)) =1. (*)
Also Let f(1)=b
Then f(x)= b [f(x-1)]^2 (**)
So f(x) has the same sign as b, and therefore f(x) has one definite sign.
Case1. a=1, and therefore f(x)>0.
From equation (*) we get that f(2)=1. from (**) then we get b^3 = 1 so b=1.
hence f(0)=f(1)=f(2)=1. This together with (**) implies that f(z)=1 for all integers z.
Now f(1/(x-1))^2=f(1)f(1/x-1)^2 = f(1+1/x-1) = f(x/x-1) =1/f(x) (***)
Also f(x/(x-1) -1) =f(x/x-1)^2 = 1/f(x)^2
So f(1/x-1) = 1/f(x)^2 = f(1/x-1)^4 [ from equation (***)]
Thus f(x) =1 for all x.
case 2. a=-1 can be handeled in the same manner, resulting in f(x)=-1 as the solution.

-Ali

$f(x+y) = f(x) f(y) f(xy)$

If there exists $y \in \mathbb{R}$ such that $f(y) = 0$, then $f(x+y) = 0, (\forall x)$, so f is identically $0$. Now, let us assume that f takes only nonzero values.

$f(x+y+z) = f(x + (y+z)) = f(x)f(y+z)f(xy+xz) = $
$ f(x)f(y)f(z)f(yz)f(xy+xz) = f(x)f(y)f(z)f(yz)f(xy)f(xz)f(x^2 yz)$

In a similar manner, we can prove that, for example:

$f(x+y+z) = f((x+y) + z) = \cdots = f(x)f(y)f(z)f(xy)f(xz)f(yz)f(xyz^2)$

Comparing the last two relations we get:

$ f(x) f(y) f(z) f(xy) f(xz) f(yz) f(xyz^2) = f(x) f(y) f(z) f(xy) f(xz) f(yz) f(x^2 yz)$, so dividing by $f(x)f(y)f(z)f(xy)f(xz)f(yz) \neq 0$ we have:

$f(xyz^2) = f(x^2yz)$

Now let's take $y = 1$ and two real non-zero numbers $u, v$. The system $x^2 z = u$ and $x z^2 = v$ has the solution $x = \frac{u}{\sqrt[3]{uv}}, z = \frac{v}{\sqrt[3]{uv}}$. So plugging in these values for $x$ and $z$ we get $f(u) = f(v)$. So, $f$ is constant over all non-zero reals. Let $f(x) = c$ for $x\neq 0$. Then $c^3 = c$. Plug in the initial equation $x=-y$ for $x\neq 0$ and we get $f(0) = f(x)f(-x)f(-x^2) = c^3 = c$. Thus, $f$ is constant. It is easy to see that $c\in \{-1,0,1\}$, but we have already ruled out the case $c=0$, so $f(x) = -1, (\forall)x$ or $f(x) = 1, (\forall) x$.  It is easy to see that all three solutions verify the initial equation, so we are done.


When I log in, I enter my username and password, and check the "Log me in automatically for future visits" box. But for some reason, that box doesn't work. Every single time I access the site, I have to log in all over again. Any ideas why?

Check your cookies.

A cookie was probably not created to hold your username and password.


Is it true that there exist 100 lines in the plane, no three concurrent, such that they intersect in exactly 2002 points?

this sort of problem has been posted earlier and solved as well(possibly similar to what i am going to say).
the basic idea is to note that if we have $r$ parallel classes of lines with the $i$th parallel class having $k_i$ lines, then the number of intersection points(since there are no three lines concurrent) is simply $\sum_{i<j} k_ik_j$. So all we need to do is to see if we can solve for $\sum_{i=1}^r k_i=100,\sum_{i<j} k_ik_j=2002$. we can change this second equation to the more simple looking $\sum_{i=1}^r k_i^2=(100)^2-2\cdot 2002=5996$. Now by fiddling around one cn find that $5996=77^2+7^2+2^2+14,100=77+7+2+14$. so the answer is, yes.


abstract algebra
induction
vector
analytic geometry
modular arithmetic
linear algebra
group theory

Let $(G, \cdot)$ be a group so that $x^{2}=e$ for any $x \in G$.  (It's easy to see that this group is comutative from this property) Prove that if $G$ is finite then:

a) $|G| = 2^{n}$.

b) $G \simeq \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \cdots \times \mathbb{Z}_{2}$ (n terms)

thank you very much!

[quote="julien_santini"]Use Cauchy theorem and the representation of a commutative group as the direct product of its p-Sylows (each of them written as the product of cyclic $p$-groups).[/quote]
A total overkill... :roll:

p-Sylows  .... why not using directly decomposition of finite abelian groups  :P

You can also consider $G$ as a $\mathbb{Z}/2\mathbb{Z}$ linear space

or :

$G = < x_1, .., x_n >$  with $n$ minimal and $f : \mathbb{Z}/2\mathbb{Z}^n \rightarrow G$, $(a_1, ..., a_n) \rightarrow (x_1^{a_1}, ..,x_n^{a_n})$. Then $f$ is well defined (must be proved), surjective by definition and injective. So $f$ is an isomorphism

or :

It's also possible to proceed by induction assuming that $G$ has an element of order $2$ (by Cauchy or better by hypothesis ;) )

[quote="tµtµ"]It's also to proceed by induction assuming that $G$ has an element of order $2$ (by Cauchy)[/quote]
Why you all try to use Cauchy¿ It's [u]given[/u] that all nontrivial elements have order $2$!

OK, here's my solution (actually, it's the induction suggested by some of you):

First, let's prove that $G$ is abelian: $x^2y^2 = (xy)^2 = e\iff xxyy = xyxy \iff xy = yx$.

Suppose $G \neq \{e\}$. Pick $x_1 \neq e$ and consider $\langle x_1\rangle$. Since $x_1^2 = e$, $\langle x_1\rangle$ has two elements, namely $e$ and $x_1$. If there are no more elements on $G$ then $|G| = 2$.

Else, pick a element $x_2 \in G\setminus \langle x_1\rangle$ and consider $\langle x_1, x_2\rangle$. There are two types of elements in this set: the ones that have $x_2$ as factor and the ones that don't have. There are, of course, two elements of the latter type, namely the elements of $\langle x_1\rangle$. But, since $G$ is abelian and $x_2^2 = e$, either you have one factor $x_2$ or you don't have it at all. So there is a bijection between numbers of the two types: just multiply the elements of $\langle x_1\rangle$ by $x_2$. Thus $\langle x_1, x_2\rangle$ has $2\cdot 2 = 4$ elements.

Now it's not hard to see that we can proceed by induction: suppose we already found $x_1,x_2,\ldots,x_k$ and $X_k = \langle x_1,x_2,\ldots,x_k\rangle$, $|X_k| = 2^k$. If $G\setminus X_k =\emptyset$, then $|G| = 2^k$. Else, consider $x_{k+1} \in G \setminus X_k$ and do pretty much the same we did in the last paragraph: classify each element of $X_{k+1} = \langle x_1,x_2,\ldots,x_k,x_{k+1}\rangle$ in $X_k$ and $Y_k = X_{k+1} \setminus X_k$. We have again a bijection between $X_k$ and $Y_k$ (guess what bijection?) and then $|X_{k+1}| = 2|X_k| = 2^{k+1}$.

Since $G$ is finite, this procedure will stop at some moment, so $|G| = 2^n$. Furthermore, we even have found the generators of $G$: $G = \langle x_1,x_2,\ldots,x_n\rangle$. And now the isomorphism between $G$ and $Z_2^n$ is clear: $\sigma\colon G \to Z_2^n$, $\sigma(x) = (a_1,a_2,\ldots,a_n)$, where $x = x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}$.

If I'm not mistaken (and I may be), this same reasoning proves that if $G$ is abelian and finite, them $G$ has a finite number of generators and $|G|$ is equal to the product of the orders of these generators and, thus, $G$ is isomorphic to a product of $Z_m$'s.

[quote="Mitzah"][quote="cyshine"] (guess what bijection?)[/quote]

Can you please post which is the bijection?  :blush:

Thank you very much![/quote]

Hi, and sorry from replying so late. Believe it or not, I've been busy these days (I will give classes to the national olympiad winners, so I'm preparing them, plus some material).

The bijection is the analogous to case $k=1$: multiply each element of $\langle x_1,x_2,\ldots,x_k\rangle$ by $x_{k+1}$. The proof that this is indeed a bijection is also analogous to the case $k=1$: since $G$ is abelian and $x_2^2 = e$, either you have one factor $x_{k+1}$ or you don't have it at all. And since the operation "multiply by $x_{k+1}$" is inversible, we have a bijection.

[quote="fredbel6"]cyshine ,from what i see you give the longest solution, but the most straightforward elementary one (it's either that or overkill :blush: )
is my assumption true that with some modifications you have proven that any group where ever element has order p with p prime, and abelian, is such a direct product of cyclic groups of order p?[/quote]

Your assumption is true.

I wanted to give an "straightforward elementary" solution (as you wrote) because a lot of people don't know Cauchy's theorem or the theorem on representation of finite abelian groups. And these theorems have, as far as I know, long proofs (that is, if you want to start from first principles of group theory). And, most incisively, one doesn't need to know these theorems to solve the problem (I didn't remember them myself). And I actually just expanded tµtµ's induction idea.

Then again, the most straightforward solution is probably linear algebra: since $G$ is abelian, it is naturally a module over $\mathbb{Z}$. Multiplication by 2 gives zero, so, $G$ is a $\mathbb{Z}/2\mathbb{Z}$-module. That quotient is a field, and $G$ is a vector space over the field with 2 elements. Since $G$ is finite, it is spanned by finitely many elements and is finite-dimensional.
Going back to the theorems everyone learned in linear algebra, every finite-dimensional vector space over a field $F$ has a basis and is isomorphic to $F^n$, with coordinates chosen with respect to this basis. The additive structure of $F^n$ is the product of $n$ copies of the additive group of $F$, and we have the desired result.

In that first linear algebra course, the field was implicitly assumed to be $\mathbb{R}$- but none of the proofs used anything other than field operations, so they all apply to (finite-dimensional) vector spaces over an arbitrary field. My first paragraph uses some relatively advanced language, but we could skip right to defining the vector space operations once we know that $G$ is abelian:
$x+y=x*y$, where $*$ is the group operation. $0=e$. $0\cdot x=0$, $1\cdot x=x$. That's it for scalar multiplication, since there aren't any other scalars. The axioms to check:
The space is an abelian group under $+$: Already done.
$ab\cdot x=a\cdot(b\cdot x)$: All four cases are obvious.
$0\cdot x=0,1\cdot x=x$: True by definition.
$a\cdot x+b\cdot x=(a+b)\cdot x$: Three cases are trivial because 0 is an identity. If $a=b=1$, we have $a+b=0$ and $x+x=0$, so this checks.
$a\cdot x+a\cdot y=a\cdot(x+y)$: Both cases are obvious.

There's nothing wrong with invoking a few standard theorems from an earlier course.

Hi,

This solution is correct, but exactly the construction of a basis of a general finite-dimensional vector space $V$ is essentially inductive.

This proof is from Herstein, Topics in Algebra. The definition of 'finite-dimensional' is that there exists a finite subset $S$ of $V$ that span $V$ ('span' is a technical term: this means that every element of $V$ is a linear combination of a finite number of elements from $S$). Let $S = \{v_1,v_2,\ldots, v_n\}$. If these $n$ vectors are linear independent, $S$ is a basis of $V$. If not, weed out the first $v_j$, which is a linear combination of its predecessors. This set still span $V$: just replace $v_j$ by the corresponding linear combination. And continue until there are no more $v_k$'s that are linear combination of its predecessors. Then this set is a basis of $V$.

Let me stress that jmerry's solution is OK but I still think one would need to write some additional steps, since the proof above (which is instructive itself) is, in my humble opinion, not significantly easier than my solution. I just feel that this solution is like proving that $2^6 \equiv 1\pmod 7$ using Fermat's theorem. It's correct, but I don't feel quite right proving a result using another that is not easier to prove.

I agree that there's nothing wrong in using facts from earlier courses. But, strangely enough, I studied (I mean, with proofs) group theory before linear algebra (I studied using Herstein's book), so group theory is, to me, and maybe, only to me, a earlier course.


AMC
AIME
geometry
geometric transformation
rotation
linear algebra
matrix

When do we often see to use them in math contests?

Some AMC and AIME problems contain rotation matrices in disguise and ask you to identify some large iteration of the rotation.  Matrices can also be useful for doing computations with sequences such as the Fibonacci sequence, as well as for solving certain types of combinatorics problems.

I see. The system of equations problem in this forum would be an example, I guess.
Have a sample problem on you? (I have absolutely no practice with them except how to find the determinant).

I don't think I've ever seen a problem that explicitly required the use of matrices.  As t0rajir0u pointed out, [url=http://en.wikipedia.org/wiki/Rotation_matrix]rotation matrices[/url] are useful in determining what happens when a point is reflected through a line (which is really just a change of basis).


Hello ppl,
I have been a devoted zune user for several months but just recently my dad got a free 8gig touch when he bought a macbook. I want to be able to move my files from the zune to my computer but I'm not sure how. I think the sync might work but I have no idea how to copy and paste the songs from the zune onto the hard drive. Any help would be appreciated. Even if you jsut have a suggestion. O yah, and just fyi I am planning to jailbreak my touch so any advice on that area would help too.

Thank you very much
~zhang~ :D

1. Go to my computer
2. Click on the Zune device, assuming its already connected.
3. Copy the files to your desktop or somewhere.
4. Your done :)


Um... it seems to me that at move 4, you can't move a bishop to g5... because none of the black bishops have moved yet...

Also:
[quote] 5.d3 Bh5 6.d3 h6 [/quote]
How does a pawn move from d3 to d3??

[quote="mathelete"]Check this out: 1.e4 e5 2.Nf3 d6 3.Bc4 Nc6 4.Nc3 Bg5 5.d3 Bh5 6.d3 h6 7.b3 g5. What is White's best move?[/quote]
This is what I meant:

Check this out: 1.e4 e5 2.Nf3 d6 3.Bc4 Nc6 4.Nc3 Bg4 5.d3 Bh5 6.h3 h6 7.b3 g5. What is White's best move?


[quote="mathemagician1729"]Find the angle $ \theta$ between the two lines 2x and 3x.

I made this into a really messy deal with tangents and what not...

Is their a more efficient way without a calculator?

Thanks in Advance[/quote]
[hide]The line $ y\equal{}2x$ makes a $ \tan^{\minus{}1} 2$ with the x-axis and the line $ y\equal{}3x$ makes a $ \tan^{\minus{}1} 3$ with the x-axis. Thus the angle between these two lines is $ \tan^{\minus{}1} 3\minus{}\tan^{\minus{}1} 2$, setting this to $ \theta$ and taking the tangent of both sides results in $ \tan\theta\equal{}\frac{1}{7}$.
I don't know what else you could do without a calculator other then putting it in the form $ \theta\equal{}\tan^{\minus{}1} \frac{1}{7}$. On the other hand, using a calculator gives $ \theta$ to be approximately $ 8.13^{\circ}$[/hide]

The reason I asked was because they wanted you to be able to find the range of theta as between 0 and 10 degrees without a calculator.


circle $c_1$ and $c_2$ are externally tangent at $T$. A line passing through point $T$ cuts $c_1$ at $A$ and   $c_2$ at $B$. the line tangent to $c_1$ at $A$ cuts $c_2$ at $D$ and $E$.

If $D$ $\in$ $|AE|$, $|TA|$=a $|TB|$=b find $|BE|$

Let centres of $c1$ and $c2$ be $P$ and $Q$. Let angle $TAD$ be $r$, and angle $ATD$ be $s$. Because $AE$ is tangent to $c1$ then angle $APT$ is $2r$. $T$ is pt at which two circles are tangent so angle $PTD$ = angle $BTQ$. Hence angle $BQT$ is also $2r$. Then angle $TDB$ is $r$.Then also angle $BET$ is $r$.
If angle $ATD$ is $s$, then angle $BED$ is $s$.
Angle $TDE$ is equal to $r+s$. But angle $BDT$ is $r$ so angle $BDE$ is $s$. Hence $DB=BE$.
In triangle $TDB$ have $DB/sin(180-s)=TB/sin(r)$ or $DB/sin(s)=TB/sin(r)$.
In triangle $ABE$ have $BE/sin(r)=AB/sin(s)$.
Multiplying the 2 equations we get above, and cancelling out $sin(s)$ and $sin(r)$ we get:
$DB*BE=AB*TB$. Subst $DB=AE$, $AB=a+b$, $TB$=b get:
$BE^2=(a+b)b$.
So $BE = \sqrt{(a+b)b}$ and we are done  :)


Is the following true?: for each positive integer $ k$ there exists a positive integer $ m$ such that all the numbers $ m,2m,3m,...,m^2$ have exactly $ k$ nonzero digits.

What makes you think it might be true?  Can you give one example of such an $ m$ for $ k\geq 2$?

Not only is it not true for every $ k$, unless I've misunderstood your problem at first glance I don't think it's true for [i]any[/i] $ k\geq 2$.


inequalities
function
integration
advanced fields
advanced fields theorems

We all know Jensen's inequality.

We have got a measure space $ (X,A,\mu)$ with $ \mu(X)\equal{}1$, and a real-valued function $ f$ that is $ \mu$-integrable.
So if $ \phi$ is a measurable convex function (with $ \phi : \mathbb{R} \rightarrow \mathbb{R}$), then

$ \displaystyle \phi\left(\int_X f d\mu\right) \leq \int_X (\phi \circ f) d\mu$

But when does equality occur?

A typical proof involves an affine function $ \psi\colon\mathbb R\to\mathbb R$ such that $ \psi\le \phi$ and $ \phi(a)\equal{}\psi(a)$ where $ a\equal{}\int_X f\,d\mu$.
From this we find: equality occurs if and only if there exists $ \psi$ as above such that $ \phi(f(x))\equal{}\psi(f(x))$ for a.e. $ x$.
If $ \phi$ is strictly convex, the above simplifies to $ f(x)\equal{}a$ for a.e. $ x$.


En liten presentasjon av Latex  :)
Det skal ikke mye til for å bruke Latex på Mathlinks (ML).
Når du skal skrive en matteformel skal du sette to dollartegn rundt uttrykket.
For eksempel:   $\$$ $a + b = c$ $\$$. (dollar er ctrl+alt+4)
Fra nå av skal jeg bare skrive det som står mellom disse dollartegnene.
De mest brukte ting:

   $x^n$ --- du skriver  x^n  og $x^{123}$  ---  x^{123}
(skal du ha mer enn 1 siffer eller bokstav når du bruker funksjoner må du ha uttrykket i slike paranteser: {}  )

   $a_1$ og ${AC}_{bc}$ ---  a_1          og             {AC}_{bc}

   $a \geq b$, $b \leq c$ og $m \equiv n$ ----   a \geq b   ,          b \leq c          og        m\equiv n

   $a \not= b$ -----           a \not= b

   $\frac{a^2+3}{b^2+c}$ -------      \frac {a^2+3}{b^2+c}

   $\sum_{i=0}^{n}x^i$   -------   \sum_{i=0}^{n}x^i

   $\prod _{j=1}^{n} \frac{j+1}{j}$ ------    \prod _{j=1}^{n} \frac{j+1}{j}

   $f''(x+y)$ ------     f''(x+y)

Denne er av og til nyttig og: $3\cdot5^3$-------3\cdot5^3  .

Her finn ein ei større oversikt over latex-språket. Men det er lett å gå seg vill, mesteparten treng ein ikkje til skriving på Mathlinks
[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]


geometry
complex numbers
absolute value
algebra unsolved
algebra

A,B,C are complex numbers such that absolute value of them are equal 1
and A+B+C =0 Find the area of triangle ABC in the complex graph

Hint No. 1: What does the absolute value of those three points imply (got something to do with the unit circle, doesn't it)?
Hint No. 2: Can $ A\plus{}B\plus{}C\equal{}0$ tell you something about the shape of the triangle?

[hide]BTW, I'm not really sure where this topic belongs... Maybe Intermediate topics?[/hide]


let $ s\equal{}sin{x}$

How many solutions does the equation

$ 2\equal{} s\plus{}{s}^2 \plus{}{s}^3 \plus{}{s}^4\plus{}...$

have in the range 0 (less than or equal to) x (less than 2pi)?

Would this be 0 solutions because the series must be less than one?

[hide="hint"]$ |\sin x| \le 1$, think of a geometric series.[/hide]

[hide="solution"]
Since $ |\sin x| \le 1$ the series converges.

In this case $ a \equal{} 1$, $ r \equal{} \sin x$
So $ 1 \plus{} s \plus{} s^{2} \plus{} s^{3} \plus{} ... \equal{} \frac {1}{1 \minus{} \sin x} \equal{} 3$
$ 1 \equal{} 3 \minus{} 3\sin x$
$ \sin x \equal{} \frac {2}{3}$
There are two solutions, one in quadrant 1, and one in quadrant II.[/hide]

[quote="JasonSullivan"]Where have you gotten the 3 from and why does the series suddenly start with 1 as opposed to sinx?[/quote]
he has added one to both sides :)

[quote="JasonSullivan"]Where have you gotten the 3 from and why does the series suddenly start with 1 as opposed to sinx?[/quote]

You can start with $ \sin x$ if you wanted to:

$ \sin x \plus{} \sin^2 x \plus{} \sin^3 x \plus{} \dots \equal{} 2$
$ \implies \frac {\sin x}{1 \minus{} \sin x} \equal{} 2$
$ \implies 3\sin x \equal{} 2$
$ \implies \sin x \equal{} \frac {2}{3}$

[quote="JasonSullivan"]
Would this be 0 solutions because the series must be less than one?[/quote]

Just because the terms are all $ \leq 1$ doesn't mean the series must be -- consider:
$ 1\plus{}\frac{1}{2}\plus{}\frac{1}{4}\plus{}\frac{1}{8}\plus{}\dots \equal{} 2$


inequalities
binomial coefficients
rearrangement inequality

Help me please :

Let $n$ be a positive integer. Determine the smallest possible value of the sum \[a_1b_1+a_2b_2+...a_{2n+2}b_{2n+2}\] , where $(a_1,a_2,...a_{2n+2})$ and $(b_1,b_2,...b_{2n+2})$ are rearrangements of the binomial coefficients \[\left(\begin{array}{cc}{2n+1}\\0\end{array}\right) , \left(\begin{array}{cc}{2n+1}\\1\end{array}\right) , ... , \left(\begin{array}{cc}{2n+1}\\2n+1\end{array}\right)\]

Well, by the rearrangement inequality (what a coincidence), you just pair up the biggest with the smallest, 2nd biggest with 2nd smallest, etc.

So one possible arrangement is $a_k = \binom{2n+1}{k-1}$ and $b_k = \binom{2n+1}{n-k+1}$ if $k \le n$ or $b_k = \binom{2n+1}{3n+2-k}$ if $k \ge n+1$ (I think that works).


If the equality $ x^k\plus{}a_1 x^{k\minus{}1} \plus{}...\plus{}a_k\equal{}0$ has $ k$ distinct real roots then prove that
$ a_1^2 >\frac{2ka_2}{k\minus{}1}$

[quote="pradmath"]the equality $ x^k \plus{} a_1 x^{k \minus{} 1} \plus{} ... \plus{} a_k \equal{} 0$ has $ k$ distinct real roots then prove that
$ a_1^2 > \frac {2ka_2}{k \minus{} 1}$[/quote]

Here is my solution:

Assume that the equality $ x^k \plus{} a_1 x^{k \minus{} 1} \plus{} ... \plus{} a_k \equal{} 0$ has $ k$ distinct real roots $ x_1,x_2,...,x_k$.

By Viete theorem, we have:
$ a_1 \equal{} (\minus{}1)^{k \minus{} 1}.(\sum_{i \equal{} 1}^k x_i)$ (1)
$ a_2 \equal{} \sum_{i \neq j}^k x_ix_j$

By squaring, (1) becomes $ a_1^2 \equal{} (\sum_{i \equal{} 1}^k x_i)^2$

Now we must prove

$ a_1^2 > \frac {2ka_2}{k \minus{} 1}$

$ \leftrightarrow (\sum_{i \equal{} 1}^k x_i)^2 > \frac{2k}{k \minus{} 1}\sum_{i \neq j}^k x_ix_j$

$ \leftrightarrow (k \minus{} 1)\sum_{i \equal{} 1}^k x_i^2 > 2\sum_{ i \neq j}^k x_ix_j$

$ \leftrightarrow \sum_{i \neq j}^k (x_i \minus{} x_j)^2 > 0$

which is clearly true because $ x_1,x_2,...,x_k$ are distinct real numbers.


[quote="nthd"][quote="ductrung"][quote="nthd"]Prove that $\forall a,b,c>0$and $k\ge 0$ we have:
a)$\sum\frac{a^2-bc}{b^2+c^2+ka^2}.\frac{b^2-ca}{c^2+a^2+kb^2}\le 0$ :)  :) [/quote]

Let
$f(a,b,c)=\sum\frac{a^2-bc}{b^2+c^2+ka^2}.\frac{b^2-ca}{c^2+a^2+kb^2}$
Then
$f(1,1,3)=\frac{4(k-78)}{(k+10)^2(9k+2)} > 0$ for $k>78$.
So where am I wrong?[/quote]I think you are wrong because with $k=0$ it's true.[/quote]
But doesn't it say "for all $a,b,c>0$ and $k\ge 0$"? What if we fix $k > 78$, then put $a=b=1$ and $c=3$?

Or do you mean for all $a,b,c>0$, there exists $k\ge 0$ such that
$\sum\frac{a^2-bc}{b^2+c^2+ka^2}.\frac{b^2-ca}{c^2+a^2+kb^2}\le 0$ ?

[quote="ductrung"][/quote][quote="nthd"][quote="ductrung"][quote="nthd"]Prove that $\forall a,b,c>0$and $k\ge 0$ we have:
a)$\sum\frac{a^2-bc}{b^2+c^2+ka^2}.\frac{b^2-ca}{c^2+a^2+kb^2}\le 0$ :)  :) [/quote]

Let
$f(a,b,c)=\sum\frac{a^2-bc}{b^2+c^2+ka^2}.\frac{b^2-ca}{c^2+a^2+kb^2}$
Then
$f(1,1,3)=\frac{4(k-78)}{(k+10)^2(9k+2)} > 0$ for $k>78$.
So where am I wrong?[/quote]I think you are wrong because with $k=0$ it's true.[/quote][quote="ductrung"]
But doesn't it say "for all $a,b,c>0$ and $k\ge 0$"? What if we fix $k > 78$, then put $a=b=1$ and $c=3$?

Or do you mean for all $a,b,c>0$, there exists $k\ge 0$ such that
$\sum\frac{a^2-bc}{b^2+c^2+ka^2}.\frac{b^2-ca}{c^2+a^2+kb^2}\le 0$ ?[/quote]
I think your transformation is wrong! :)

I used a computer algebra system to check before I posted - but as you insist: ok, the software may be buggy though! I will check my calculation an extra time and be more careful.

Oh, BTW, I didn't do any transformation. I just put $a=b=1$, $c=3$ and, say $k=80$.

[quote="ductrung"][quote="nthd"]Oh, ductrung, I think you want to say $k \le 78$ but you wrote $k>78$ :lol:[/quote]

What a lame excuse! No matter which I wrote: $k\le 78$ or $k>78$, the expression
$f(1,1,3)=\frac{4(k-78)}{(k+10)^2(9k+2)}$
should have rung a bell.[/quote]
Because you type wrong $k>78$ so I didn't understand you. :mad: .I don't sorry. :)

[quote="ductrung"][quote="nthd"]Because you type wrong $k>78$ so I didn't understand you.
[/quote]
Where did I type wrong?

[quote="nthd"]
 :mad: .I don't sorry. :)[/quote]
I don't expect so :-)[/quote]
I think I don't understnad you. :)

[quote="nthd"]I think I don't understnad you. :)[/quote]

That's no crime :-)

So are you going to rephrase your problem? Something like "find all $k$ such that ..."?


The directions are to find all real solutions of the equation.

$\dfrac{x}{2x+7} - \dfrac{x+1}{x+3} = 1$

I found the LCD and got

-x + 12x + 6 = 0.

Then I found that the discriminate is 168.

x = 6 - sqare root of 42

x - 6 + square root of 42

Are these right? I checked them, and they didn't work, so did i do something wrong? I know it's supposed to have at least two solutions, right? Please help me.

Thanks, I forgot to muliply the 1 by the least common denominator :blush:


real analysis
integration
calculus
function
real analysis unsolved

How to prove that the $\int_{0}^{1}\frac{1}{\sqrt{x}}$ is Lebesgue integrable? And how about $\int_{0}^{1}\frac{cos(1/x)}{\sqrt{x}}$?

If someone could kindly point me to some reference for this kind of problems, it will be greatly appreciated.

Thanks a lot!

It's Lebesgue integrable because it's absolutely integrable. Evaluate the improper integral; since the function is positive, that also works Lebesgue.

In the second case, use comparison.

Hi, could you please tell me why the function $1/\sqrt{x}$ is absolute integrable? Somehow our prof and textbook never mention this concept.

Thanks!

Chapter 10 of Rudin's "Principles of Mathematical Analysis" covers Lebesgue integration. If $f\ge 0$, then $\int_{0}^{1}f(x)dx$ (in the Lebesgue sense) is the supremum of $\int_{0}^{1}g(x)dx$, where $g$ is a simple function and $0\le g\le f$. In general, $f$ is integrable iff both $f_{+}$ and $f_{-}$ are integrable.

Since $f_{+}\le |f|$ and $f_{-}\le |f|$, it follows that $f$ is integrable whenever $|f|$ is. This can be used to reduce your second question to the first one, because $\left|\frac{\cos(1/x)}{\sqrt{x}}\right|\le \frac{1}{\sqrt{x}}$.

To answer the first question, you need a theorem: if $f\ge 0$ and $f$ is Riemann integrable, then $f$ is Lebesgue integrable, with the same value of the integral.

Hi mlok, thank you very much for the reply. Now I get that "if $f$ is  measurable, then $|f|$ integrable => $f$ integrable".

However, for the case that $f = \frac{1}{\sqrt{x}}$ is integrable, since it is a improper Riemann integral on [0,1], I am not sure whether I can apply the Riemann integration result directly. Anyhow, now I am able to construct an integrable function $f_{n}$ that monotonically converges to $f$, I think that can prove that $f$ is Lebesgue integrable, where $f_{n}= \frac{1}{\sqrt{x}}$ if $1/n^{2}< x \leq 1$, or $f_{n}= n$ when $0 \leq x \leq 1/n^{2}$.


Hello all.. I'm 15-years old Korean student.

I have some questions for you...

Please solve these problems...

Thank you..  Problems are here...

(1) If ax+by=3,  a x^2 + b y^2  = 7,  ax^3 + by^3 = 16,  ax^4 + by^4 = 42, then find the value of ax^5 + by^5.

(2) Find all x, y, z that satisfied (x+y)^3 = z,  (y+z)^3 = x, (z+x)^3 = y.

(3) Find all x, y, z that satisfied x+[y]+{z}=1.1,  [x]+{y}+z=2.2,  {x}+y+[z]=3.3.
      ( [n] is a integer part of n, and {n}=n-[n] )

(4) Find all x, y, z that satisfied x^3 -9(y^2 -3y+3)=0, y^3 -9(z^2 -3z+3)=0,  z^3 -9(x^2 -3x+3)=0.

(5) What is the nearest value of  \sqrt 65 -  \sqrt 63?
     1: 0.12    2: 0.13    3: 0.14    4: 0.14    5: 0.15

(6) Find (x,y) that satisfied 1!+2!+3!+...+x! = y^2    (x, y \in Z)

??? why until 3pm tomorrow? Is it homework?

Piere.

not only that, aren't these problems the same ones you posted to another forum (i think i have seen these somewhere else) the same ones, that you mentioned were homework...and that you needed to have done for you? I am pretty sure that is what you did :(

as you will notice, and other ppl might say this too, but this is not a homework solving site but mainly an olympiad problem solving site!

-cheers
:D :D

No, It' s not a homework.

I mean it ask you to post until 3 p.m. is just I want to know the solution early!!

I think it for over 2 hours.

So, I found some solutions about that but I can't certain that is correct.

Please Please.... just post its solution...

Then I'll check it and solve again myself..

Why are you thinking that ways??. :(

first of all no-one says like solve these problems untill 3pm unless there is a deadline of some sort and you need them by then.

splintercell is right, you cannot ask people to do your job instead. asking for help or advice is one, but what you did is something else.

next time, please post one problem at a time, in it's apropiate subforum, without giving any deadlines.

people will gladly help you if you do not force them to.


LaTeX
calculus
derivative
function
Support
advanced fields
advanced fields unsolved

I was looking at the exercises for the first chapter of this book, and I saw:

1.6. Let $u \in D'(R^{n})$ [$u$ is a distribution] have the property that $\langle u, \phi \rangle \ge 0$ for all real valued nonnegative $\phi \in C_{c}^\infty(R^{n})$. Show that $u$ is of order 0.

I read the chapter, but somehow I didn't notice and now can't find what is the order of a distribution! Can someone please explain (use this example for an explanation or show me how to do it, I'll probably learn it either way).

Also, how do you make that curly D in LaTeX, the one they use for the space of distributions?

I'm not sure, but I think that, intuitively, the order is the number of derivatives that the distribution "sees."  For example, a 5th order distribution can be evaluated on any C^5 function, but it cannot be evaluated on a function that is only C^4.  Delta function is zero order, while $\delta'$ would be first order.  I don't remember the precise definition, but there are probably several equivalent ways to say it.

As for the curly D, perhaps you mean $\mathcal{D}.$

No, \mathcal{D} ($\mathcal{D}$) isn't what Xevarion is looking for. I don't know how to get the font he wants.

One could find and quote the topological technicalities behind yenlee's explanation, but in this case I'd start with the intuition, just as yenlee expressed it.

Well now I understand what is the [i]definition[/i], but I have no idea how to approach the problem from the text. I mean, intuitively, it seems like if we allow $u$ to "see" the derivative, then we can pick a test function that has arbitrarily large negative derivative on some interval (tho still satisfying the original condition) and contradict the condition on $\langle u, \phi \rangle$. But I can't tell what information I get from assuming $u$ is of order $\ge 1$ that allows me to pick an appropriate test function for the contradiction...

Any suggestions? Thanks again for your help so far! :)

If you have a test-function $\varphi$ supported by a compact $K$, then $|\langle u,\varphi\rangle|\le C(K)\max|\varphi|$. Indeed, just take some test-function $\psi$ that is non-negative everywhere and identically $1$ on $K$ and consider the functions $M\psi\pm\varphi$ where $M=\max|\varphi|$. They are both non-negative whence we can take $C(K)=\langle u,\psi\rangle$. How to finish depends on the definition of order you use.

fedja, I'm sorry, but you seem to have mistaken me for somebody who understands math. It's probably not your fault, but I really have absolutely no idea what you said (except at least I understand what are $\varphi, \psi, K$, little help though that may be). Don't worry too much about explaining it, I'm getting tempted to give up and try to practice basic analysis more before I try to go adventuring into this Advanced Topics forum again.

OK, let me try to explain it in more detail.
[b]Claim:[/b] For every compact $K$ there exists a finite positive constant $C(K)$ depending on $K$ such that, for every test-function $\varphi$ supported by $K$, we have $|\langle u,\varphi\rangle|\le C(K)\max|\varphi|$.
I don't know what definition of order you have in your book, but one of them says that this property is exactly what it means that $u$ has order $0$ and all other reasonable definitions should be equivalent to it.
[b]Proof:[/b]. Fix any $C_{0}^\infty$ function $\psi$ that is non-negative everywhere and that is identically $1$ on $K$. Let $M=\max|\varphi|$. The function $M\psi-\varphi$ is non-negative. Hence $\langle u,M\psi-\varphi\rangle=M\langle u,\psi\rangle-\langle u,\varphi\rangle\ge 0$, i.e., $\langle u,\varphi\rangle\le \langle u,\psi\rangle\max|\varphi|$. Similarly, the function $M\psi+\varphi$ is non-negative. Hence $\langle u,M\psi+\varphi\rangle=M\langle u,\psi\rangle+\langle u,\varphi\rangle\ge 0$, i.e., $\langle u,\varphi\rangle\ge-\langle u,\psi\rangle\max|\varphi|$. Thus, $|\langle u,\varphi\rangle|\le \langle u,\psi\rangle\max|\varphi|$ and we can put $C(K)=\langle u,\psi\rangle$.

Is it clearer now?

OH! Well, I guess it helps to understand when I know what you're actually trying to prove. So yes, I totally understand what you did. Thanks. However! I still have absolutely no idea how that could possibly be equivalent to the thing about derivatives...  :( The reason I couldn't understand before is that I thought you were trying to suggest that there was one trivial step left to get to THAT definition of order.  :maybe:

Also, using your way of defining order, what do the definitions of orders greater than 1 look like? I guess $|\langle u, \phi \rangle| \le C(K_{0})\max|\varphi|+C(K_{1})\max|\varphi'|+...$ where $C(K_{i})$ is a constant depending on the support of the $i^{th}$ derivative (can the derivative of a $C^\infty$ function of compact support have different support from that of the original function? I shouldn't have to think about that. This is why I should go back and actually learn analysis before I try this stuff...)? Can we always separate an order $n$ distribution into $n$ parts each of which depends only on the $i^{th}$ derivative?

Sorry, I'm just trying to see what's going on. You don't need to answer all the questions, the only essential one is how yenlee/professor Merryfield's definition is equivalent to yours.

[quote="Xevarion"] How is yenlee/professor Merryfield's definition  equivalent to yours?[/quote]
It is not completely trivial but not very hard either. Indeed, if we have the property that $|\langle u,\varphi\rangle|\le C(K)\max|\varphi|$, then we can extend the functional $u$ to a continuous linear functional on the space $C_{0}$ of continuous functions with compact support. To do it, just take any $f\in C_{0}$ and approximate it in the uniform norm by infinitely smooth functions $\varphi_{k}$ such that all the supports of $\varphi_{k}$ are contained in some compact $K$ (possibly a bit larger than $\mathrm{supp\,}\varphi$). Then the sequence $\langle u,\varphi_{k}\rangle$ is a Cauchy sequence and, therefore, it has a limit. We define $\langle u,f\rangle$ to be that limit. One needs to check that it doesn't depend on the approximating sequence and that it is a continuous linear functional on $C_{0}$ but this is a fairly straightforward exercise, so try to do it yourself. Similarly, if $u$ has order $k$, it can be extended to a continuous linear functional on $C^{k}_{0}$.
As to the last question, the answer is again "Yes" in the following sense: if $u$ is of order $k$, then there are distributions $u_{0},\dots,u_{k}$ of order $0$ such that $\langle u,\varphi\rangle=\sum_{j=0}^{k}\langle u_{j},\varphi^{(j)}\rangle$ for every test function $\varphi$, though this representation is by no means unique.


function
absolute value
arithmetic sequence
algebra unsolved
algebra

Let f be a function from the reals to reals s.t. absolute value of f(x)-f(y) is smaller than or equal to the basoltue value of x-y. Prove that if the sequence x,f(x),f(f(x)),... is an arithmetic progression for any real x , then f(x)=x+a for some real a.

Bomb

This one already appeared on ML and was solved by Myth.

Pierre.


in $ \Delta$ $ ABC$, we have $ AC\equal{}2BC$. the tangents at $ A$ and $ C$ to the circumcircle of $ \Delta$ $ ABC$ meets at $ P$.
prove that the line $ BP$ bissects the arc $ BAC$ of the circumcircle.

[quote="saif"]in $ \Delta$ $ ABC$, we have $ AC \equal{} 2BC$. the tangents at $ A$ and $ C$ to the circumcircle of $ \Delta$ $ ABC$ meets at $ P$.
prove that the line $ BP$ bissects the arc $ BAC$ of the circumcircle.[/quote]

do u mean the arc ABC or AIC  or BAC  :read: ???

:o  you see that i write $ BAC$ so it's $ BAC$ and it means the biggest arc $ BC$.

[quote="saif"]in $ \Delta$ $ ABC$, we have $ AC \equal{} 2BC$. the tangents at $ A$ and $ C$ to the circumcircle of $ \Delta$ $ ABC$ meets at $ P$.
prove that the line $ BP$ bissects the arc $ BAC$ of the circumcircle.[/quote]

ok  mr saif
this is  a geometrical construction to the question and realy i want to know how will the line PB bisects the arc  BAC  :huh:
thanks again for reply
almonkez :
 [url=http://www.servimg.com/image_preview.php?i=313&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/geomet10.gif[/img][/url]

you see now that the arc $ CBA$ is divided into to equal arcs so the result still hold  :wink:

sorry for the mistake i mean $ AC\equal{}2AB$ .

[quote="saif"]sorry for the mistake i mean $ AC \equal{} 2AB$ .[/quote]
oh , now my point of view is right mr saif :mad:
ok we had now to redraw and solve
thanks for all
mr : mohamed
almonkez

$ PB$ is the symmedian .thus , if $ M$ is the midpoint of $ AC$ and $ K\equal{}(PB,BC)$,  then  $ \angle MBC\equal{} \angle LBA\equal{} \angle LCA$.it remains to prove that $ \angle LBM\equal{}\angle MCB \implies \angle MBA\minus{}\angle LBA\equal{}\angle MCB,\angle LBA\equal{}\angle MBC \implies  \angle MBA\minus{} \angle MBC\equal{}\angle MCB \implies \angle MBA \equal{}\angle MCB\plus{} \angle MBC,\angle MCB\plus{}\angle MBC\equal{}\angle BMA \implies \angle BMA\equal{}\angle ABM$.
which is directly deduced from the fact that $ AB\equal{}AM$.

[quote="saif"][color=darkred]Let $ w$ be the circumcircle of $ \Delta ABC$ for which $ AC \equal{} 2\cdot AB$. Denote

$ P\in AA\cap CC$ and $ \{B,D\} \equal{} PB\cap w$ . Prove that $ DB \equal{} DC$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ S\in PB\cap AC$ and the midpoint $ M$ of $ [AC]$ . Is well-known that the ray $ [BS$ is the $ B$-symmedian in $ \Delta ABC$ . Therefore,

$ AM \equal{} AB$ and $ \widehat {AMB}\equiv\widehat {ABM}\equiv\widehat {CBS}$ , i.e. $ \widehat {AMB}\equiv\widehat {DBC}$ . Since $ \widehat {BAM}\equiv\widehat {CDB}$ obtain $ \Delta AMB\stackrel{(a.a.)}{\sim}\Delta DBC$ , i.e. $ DB \equal{} DC$ .[/color]

[quote="Virgil Nicula"][color=darkred][b][u]An easy extension[/u].[/b] Let $ ABC$ be a triangle inscribed in a circle $ w$ . Denote

$ P\in AA\cap CC$ and $ \{B,D\} \equal{} PB\cap w$ . Prove that $ \frac {DB}{DC} \equal{} \frac {2\cdot AB}{AC}$ . [/color][/quote]

[color=darkblue][b][u]Proof[/u].[/b]. The quadrilateral $ ABCD$ is harmonically, i.e. $ AB\cdot DC \stackrel{(*)}{ \equal{} } BC\cdot AD$ . From the [u][b]Ptolemeu[/b]'s theorem[/u]

obtain $ AC\cdot DB \equal{} AB\cdot DC \plus{} BC\cdot AD$ , i.e. $ AC\cdot DB \equal{} 2\cdot AB\cdot DC\ \implies\ \frac {DB}{DC} \equal{} \frac {2\cdot AB}{AC}$ .

[b][u]Remark[/u].[/b] Observe that $ 2\cdot AB \equal{} AC\ \implies\ DB \equal{} DC$ (the proposed problem).

We can obtain easily the relation $ (*)$ . Indeed, $ \left\|\begin{array}{ccc} PAB\sim PDA & \implies & \frac {PA}{PD} \equal{} \frac {AB}{DA} \\
 \\
PCB\sim PDC & \implies & \frac {PC}{PD} \equal{} \frac {CB}{DC}\end{array}\right\|\ \stackrel{(PA \equal{} PC)}{\implies}\ (*)$ .[/color]


inequalities
trigonometry
triangle inequality
trig identities
Law of Cosines
IMO Shortlist
inequalities proposed

For  $a,b,c$ positive reals such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} =1$ prove that

$\sum \sqrt{a^4-a^2b^2+b^4} \geq \frac{\sqrt{3}}{2}abc $

i have a question.. why do you use $\sum$ in this case?
do you mean

$\sqrt{a^4-a^2b^2+b^4} \geq \frac{\sqrt{3}}{2}abc $

Hmmm...looks strange....where did I miss something?

$a^4 - a^2b^2 + b^4 \geq a^2b^2$ so that the lhs is at least $ab+bc+ca$, and this last term is $abc$.
Since $\frac {\sqrt {3}} 2 < 1$, we have a stronger result, and best possible since equality occurs if and only if $a=b=c = 3$.

Pierre.

I am very,very,very stupid to post this inequality :( .

This was my bad reasoning.

At first I proved that

$ \sqrt{x^2+xy+y^2} +\sqrt{x^2+xz+z^2 } \geq \sqrt{(x+y)^2 +(x+z)^2 +(x+y)(x+z)} $

then I made the substitution $x+y=c,x+z=b,x=a$ to obtain

$\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2} \geq\frac{1}{2}(\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2})$

Now I used

$(\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}$$ \geq \sqrt{3ab} +\sqrt{3bc} +\sqrt{3ca}  =\sqrt{3}\sqrt{abc}(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}})$

and finally I have substituted $a,b,c$ with$a^2,b^2,c^2$ and imposed the condition $ \frac{1}{a} +\frac{1}{b}+\frac{1}{c} =1 $

but in my reasoning there is a big mistake :blush: !!!!!!!.
It was better to post my initial inequality

[quote]$\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2} \geq\frac{1}{2}(\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2})$[/quote]

which I hope it is not banal.

Thank you very much Pierre to have pointed it.

Sorry very much again :blush: .

[quote="manlio"]
$\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2} \geq\frac{1}{2}(\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2})$
[/quote]

This follows from the triangle inequality:

$\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2} \geq \sqrt{c^2+ca+a^2}$ ---- (*)

You could see (*) by Minkowski, Law of Cosines or complex numbers.

Manlio, what is the original IMO SL problem?

It is problem g.2 of IMO shortlist 2002 you can find at

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15588&highlight=shortlist


calculus
integration
trigonometry
logarithms
calculus computations

I'm not sure about answer.It looks very strange.
$ \int_{1}^{e}\frac {dx}{x\sqrt {1 \plus{} ln^2x}}$

for u=lnx-->u'=1/x

$ \int \frac {du}{\sqrt {1 \plus{} u^2}}$

substituting $ u \equal{} tan\theta$

$ \equal{} \int \frac {d\theta}{cos\theta} \equal{} ln|sec\theta \plus{} tan\theta|$

$ \int_{1}^{e}\frac {dx}{x\sqrt {1 \plus{} ln^2x}} \equal{} ln|\sqrt { \minus{} 1}|$

Let $ w \equal{} lnx$, then $ xdw \equal{} dx$
We get:

$ \int_{1}^{e}\frac {dx}{x\sqrt {1 \plus{} ln^2x}} \equal{} \int_{0}^{1} \frac {1}{\sqrt {1 \plus{} w^{2}}}dw$
With $ w \equal{} \tan(\theta)$ and $ dw \equal{} \sec^{2}(\theta)d\theta$

We get:
$ \int_{0}^{\frac{\pi}{4}} \frac {\sec^{2}(\theta)d\theta}{\sqrt {1 \plus{} \tan^{2}(\theta)}} \equal{} \int_{0}^{\frac {\pi}{4}} \sec\theta d\theta$

$ ln|sec\frac{\pi}{4}\minus{}sec\frac{\pi}{4} tan\frac{\pi}{4}|\minus{}ln|sec0\minus{}sec0tan0|$
$ ln|\sqrt{2}\minus{}\sqrt{2}|\minus{} ln|1\minus{}0|\equal{}ln0$:oops_sign:

$ \int_0^{\pi/4}\sec \theta\,d \theta \equal{} [\ln|\sec \theta \plus{} \tan \theta|]_0^{\pi/4} \equal{} \ln(\sqrt{2} \plus{} 1) \minus{} \ln(1 \plus{} 0) \equal{} \ln(\sqrt{2} \plus{} 1)$.


Hello, This is a stupid question but I don't really know the answer:
How is 12,988,816 pronounced? :maybe:
Thanks!

I would pronounce it as:

"Twelve million, nine hundred eighty-eight thousand, eight hundred sixteen"

where commas indicate pauses.

At least in American English, there are also optional "ands" that isabella omitted.  Including them gives, "Twelve million nine hundred and eighty-eight thousand, eight hundred and sixteen."

My teachers always told me to never say "and" in a number. But I guess that was their opinion.


geometry
circumcircle
vector
geometry solved

Obviously, the lines through $B,D,F$ perpendicular to $AC,CE,EA$ respectively, are concurrent (in the circumcenter of $ACE$). This means that the triangles $ACE,BDF$ are [url=http://mathworld.wolfram.com/OrthologicTriangles.html]orthologic[/url], i.e. the desired conclusion holds: the lines through $A,C,E$ to $FB,BD,DF$ are concurrent.

Thanks Grobber for quick response: actually I've come up with my solution just after posting - one simple lemma solves all:

In convex quadriteral $ABCD$ we have: diagonals are perpendicular if and only if the following holds:

$|AB|^2+|CD|^2=|BC|^2+|AD|^2$

sorry for bothering

Suppose we have proven lemma  :D  (actually proof of it is not hard but I dont like it - it's just some calculations on vectors)

Let $X$ be the point of concurrency of perpendiculars from $A$ and $C$. So we have to prove that $EX$ is perpendicular to $FD$. Now our lemma come handy. Consider quadriterals $FXAB$ and $BXDC$ - use lemma to them and you obtain as a result an equality (after using equality of some sides) which induce that quadriteral $EFXD$ has perpendicular diagonals. Sorry for my messy language above - I tried to elaborate a bit and avoid calculations simultanously  :lol:

[quote="Megus"]Let $ABCDEF$ be a convex hexagon with $AB = BC, CD = DE$ and $EF = FA$. Prove that the lines through $C,E,A$ perpendicular to $BD,DF, FB$ are concurrent.[/quote]

I know this is hyper-useless now, but let me note that the same problem has been discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=1240 .

  darij


Beautiful inequality, thanks Lagrangia!

The inequality can be generalized to n positive sequences (and since I am a masochist, I write it up using the $\sum$ notation):

If $a_{1,1},$ $a_{1,2},$ $...,$ $a_{1,n},$ $a_{2,1},$ $a_{2,2},$ $...,$ $a_{2,n},$ $...,$ $a_{u,1},$ $a_{u,2},$ $...,$ $a_{u,n}$ are positive numbers, then

$\Huge \sum_{i=1}^{u}\frac{\sum_{k=1}^{n}a_{i,k}^{2}}{\sum_{k=1}^{n}a_{i,k}}\geq \frac{\sum_{k=1}^{n}\left( \left( \sum_{i=1}^{u}a_{i,k}\right) ^{2}\right) }{\sum_{k=1}^{n}\sum_{i=1}^{u}a_{i,k}}$.

My [i]proof[/i] uses two lemmata:

[b]Lemma 1 (Cauchy-Schwartz inequality in the Arthur Engel form).[/b] If $b_{1},$ $b_{2},$ $...,$ $b_{u}$ are arbitrary reals, and $c_{1},$ $c_{2},$ $...,$ $c_{u}$ are positive reals, then

$\Huge \sum_{i=1}^{u}\frac{b_{i}^{2}}{c_{i}}\geq \frac{\left(\sum_{i=1}^{u}b_{i}\right) ^{2}}{\sum_{i=1}^{u}c_{i}}$.

[i]Proof.[/i] This is quite an unspectacular application of the Cauchy-Schwartz inequality:

$\Huge \sum_{i=1}^{u}\frac{b_{i}^{2}}{c_{i}}\sum_{i=1}^{u}c_{i}=\sum_{i=1}^{u}\left( \left( \frac{b_{i}}{\sqrt{c_{i}}}\right) ^{2}\right) \sum_{i=1}^{u}\left( \left( \sqrt{c_{i}}\right) ^{2}\right) \geq \left( \sum_{i=1}^{u}\left( \frac{b_{i}}{\sqrt{c_{i}}}\sqrt{c_{i}}\right) \right)^{2}=\left( \sum_{i=1}^{u}b_{i}\right) ^{2}$.

[b]Lemma 2 (Minkowski inequality).[/b] If $a_{1,1},$ $a_{1,2},$ $...,$ $a_{1,n},$ $a_{2,1},$ $a_{2,2},$ $...,$ $a_{2,n},$ $...,$ $a_{u,1},$ $a_{u,2},$ $...,$ $a_{u,n}$ are arbitrary real numbers, then

$\Huge \sum_{i=1}^{u}\sqrt{\sum_{k=1}^{n}a_{i,k}^{2}}\geq \sqrt{\sum_{k=1}^{n}\left( \left( \sum_{i=1}^{u}a_{i,k}\right) ^{2}\right) }$.

[i]Proof[/i] (for the sake of completeness)[i].[/i] By squaring, our inequality becomes

$\Huge \left( \sum_{i=1}^{u}\sqrt{\sum_{k=1}^{n}a_{i,k}^{2}}\right) ^{2}\geq \sum_{k=1}^{n}\left( \left( \sum_{i=1}^{u}a_{i,k}\right) ^{2}\right)$,

Using the formula

$\Huge \left( \sum_{i=1}^{u}x_{i}\right)^{2}=\sum_{i=1}^{u}x_{i}^{2}+2\cdot\sum_{1\leq i<j\leq u}\left( x_{i}x_{j}\right)$,

this is rewritten as

$\Huge \sum_{i=1}^{u}\sum_{k=1}^{n}a_{i,k}^{2}+2\cdot \sum_{1\leq i<j\leq u}\sqrt{\sum_{k=1}^{n}a_{i,k}^{2}\cdot \sum_{k=1}^{n}a_{j,k}^{2}}\geq \sum_{k=1}^{n}\left( \sum_{i=1}^{u}a_{i,k}^{2}+2\cdot \sum_{1\leq i<j\leq u}\left( a_{i,k}a_{j,k}\right) \right)$,

or

$\Huge \sum_{i=1}^{u}\sum_{k=1}^{n}a_{i,k}^{2}+2\cdot \sum_{1\leq i<j\leq u}\sqrt{\sum_{k=1}^{n}a_{i,k}^{2}\cdot \sum_{k=1}^{n}a_{j,k}^{2}}\geq \sum_{k=1}^{n}\sum_{i=1}^{u}a_{i,k}^{2}+2\cdot \sum_{k=1}^{n}\sum_{1\leq i<j\leq u}\left( a_{i,k}a_{j,k}\right)$.

However, obviously

$\Huge \sum_{i=1}^{u}\sum_{k=1}^{n}a_{i,k}^{2}=\sum_{k=1}^{n}\sum_{i=1}^{u}a_{i,k}^{2}$,

so that it remains to prove

$\Huge \sum_{1\leq i<j\leq u}\sqrt{\sum_{k=1}^{n}a_{i,k}^{2}\cdot \sum_{k=1}^{n}a_{j,k}^{2}}\geq \sum_{k=1}^{n}\sum_{1\leq i<j\leq u}\left( a_{i,k}a_{j,k}\right)$,

or, equivalently,

$\Huge \sum_{1\leq i<j\leq u}\sqrt{\sum_{k=1}^{n}a_{i,k}^{2}\cdot \sum_{k=1}^{n}a_{j,k}^{2}}\geq \sum_{1\leq i<j\leq u}\sum_{k=1}^{n}\left( a_{i,k}a_{j,k}\right)$.

But actually, for any i and j, we have

$\Huge \sqrt{\sum_{k=1}^{n}a_{i,k}^{2}\cdot \sum_{k=1}^{n}a_{j,k}^{2}}\geq \sum_{k=1}^{n}\left( a_{i,k}a_{j,k}\right)$

by the Cauchy-Schwartz inequality. Hence Lemma 2 is proven.

Now to the proof of the main inequality: By Lemma 1, we have

$\Huge \sum_{i=1}^{u}\frac{\sum_{k=1}^{n}a_{i,k}^{2}}{\sum_{k=1}^{n}a_{i,k}}=\sum_{i=1}^{u}\frac{\left( \sqrt{\sum_{k=1}^{n}a_{i,k}^{2}}\right) ^{2}}{\sum_{k=1}^{n}a_{i,k}}\geq \frac{\left( \sum_{i=1}^{u}\sqrt{\sum_{k=1}^{n}a_{i,k}^{2}}\right) ^{2}}{\sum_{i=1}^{u}\sum_{k=1}^{n}a_{i,k}}$.

But Lemma 2 yields

$\Huge \left( \sum_{i=1}^{u}\sqrt{\sum_{k=1}^{n}a_{i,k}^{2}}\right) ^{2}\geq \sum_{k=1}^{n}\left( \left( \sum_{i=1}^{u}a_{i,k}\right) ^{2}\right)$,

so that we get

$\Huge \sum_{i=1}^{u}\frac{\sum_{k=1}^{n}a_{i,k}^{2}}{\sum_{k=1}^{n}a_{i,k}}\geq \frac{\left( \sum_{i=1}^{u}\sqrt{\sum_{k=1}^{n}a_{i,k}^{2}}\right) ^{2}}{\sum_{i=1}^{u}\sum_{k=1}^{n}a_{i,k}}\geq \frac{\sum_{k=1}^{n}\left( \left( \sum_{i=1}^{u}a_{i,k}\right) ^{2}\right) }{\sum_{i=1}^{u}\sum_{k=1}^{n}a_{i,k}}=\frac{\sum_{k=1}^{n}\left( \left( \sum_{i=1}^{u}a_{i,k}\right) ^{2}\right) }{\sum_{k=1}^{n}\sum_{i=1}^{u}a_{i,k}}$,

and the inequality is proven.

Now I hope to have convinced everybody of the superiority of the $\sum$ notation :-)

  Darij


Can someone please explain the concept of why the dot product of the normals of two planes is 0 if they are perpendicular? Our teacher told us that it is so, but he can't explain why.

[quote="t0rajir0u"]The intersection of two planes that aren't parallel is a line, and it can be parameterized.  You can find information on this in any introductory textbook on linear algebra; we can, for example, set $ z \equal{} t$ which gives $ x \equal{} 5 \plus{} 3t, y \equal{} 5 \minus{} 4t$.[/quote]

oops, sorry, i already knew the answer, so i changed the question. I didn't know someome would post immediately.

[hide="Hint"]The dot-product of two vectors is zero iff. they are perpendicular.[/hide]
[hide="Off Topic"] :wink: Of course, I won't draw any premature conclusions, I'm just pickin'
[quote="guile in another thread"][quote="JBL"]1) Do your own homework.[/quote]

This is not a homework. Im studying linear algebra on my own in preparation for the next semester. Im just answering exercises on the lecture notes i have borrowed from my friends who have already taken linear algeba.  :D[/quote]
[quote="guile in this thread"]Our teacher told us that it is so, but he can't explain why.[/quote][/hide]


Given a natural sequence $ \{a_n\}$ such that
$ (a_{2n}\plus{}a_{2n\plus{}1}\plus{}1)(a_{2n\plus{}1}\minus{}a_{2n}\minus{}1)\equal{}3(1\plus{}2a_n)$
$ a_{2n} \geq a_n$ with all $ n \in N$
Find all $ n \in N$ such that $ a_n \leq 2009$

[quote="hcccccccccc"]Given a natural sequence $ \{a_n\}$ such that
$ (a_{2n} \plus{} a_{2n \plus{} 1} \plus{} 1)(a_{2n \plus{} 1} \minus{} a_{2n} \minus{} 1) \equal{} 3(1 \plus{} 2a_n)$
$ a_{2n} \geq a_n$ with all $ n \in N$
Find all $ n \in N$ such that $ a_n \leq 2009$[/quote]

It seems this is a problem from the still running  MYM 45th anniversary contest.
see http://www.mathlinks.ro/Forum/viewtopic.php?p=1568836#1568836

If so, posting the problem to find solution from others is a rather unfair behaviour, IMHO.


function
integration
real analysis
real analysis solved

Given a sequence of continuous functions $f_n:[0,1],\ n\ge 1$ s.t. $\int^1_0f_i(x)f_j(x)dx=1$ if $i=j$ and $0$ otherwise and $\sup\{f_n(x)|x\in[0,1],n\ge 1\}<\infty$, show that it has no pointwise convergent subsequence.

What is $ \sup\{f_n(x)|x\in[0,1],n\ge 1\}<\infty$ in your opinion?

I have just seen official source. I can't understand why such obvious problem was used for IMC as B6 for 20 points?

Yeah, it does seem a bit obvious. Might as well give an answer: the condition of uniform boundedness is more then enough to invoke the Dominated Convergence Theorem, so a pointwise convergent subsequence would also converge in $L^2.$ But every function in an orthonormal set must be at a distance of $\sqrt{2}$ from every other function in that set, and hence an orthonormal set can have no convergent subsequences.


This is my first game, and these are the rules.
The first person will post a word in BOLD in a certain topic, like board games (I will post the topic), and the second player will post a word with the 1st letter being the same as the last letter of the previous post.

Example:
Player 1: Monopoly
Player 2: Yahtzee
Player 3: Ummm.....
So since Player 2 got 3 stumped, he/she gains 2 points.
If Player 3 has any points (which is not the case), he/she loses 3 points because he/she couldn't think of a word fitting the topic. There is no penalty if the player has 0 points. If you post with a word, then you get 1 point.

Scoreboard:
Player 1: 1
Player 2: 2

Please post the scoreboard.  Are the rules clear enough?

I don't get any points for starting us off.

The topic is..... food. (Every once in a while I'll change the topic.)

Ok.....
[b]Hamburger[/b]

[u]Scoreboard[/u]
Nobody yet...

*SIGH*..... Somebody please delete this.

remember: DO NOT GIVE UP HOPE,

especially within 30 minutes of posting your game.

After all, these games require time, LOTSA time.

-Raspberries-

Scoreboard:

louiethechewy123-0

tcjy-1

Uhhh, the game is in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=283787]this[/url] topic.

Yeah, randomly caught the FAMOUS Virus called copied posting. It breaks the "Cannot post consecutively for 10 min" law, too.


I'm just curious about the heights of everyone.

I'm only 5'4"-ish...i'm pretty short for an 8th grader.... :(

sorry I have no idea what everyone's height is
in fact, I doubt anybody has
it's not something you commonly take pains to ask and note down, you see

4'6

[color=white]The message is too small. Please make the message longer before submitting.[/color]

[quote="1=2"]7'2. I'm tall for my age.

ok, ok. I'm 5'2.[/quote]
yippeee i think now i am not abnormally short....wait, how old were you again?
 :(

im 5 feet 3 inches...
if it makes any of you feel better my grandmother is 4 feet 3 inches

[quote="funcia"]
yippeee i think now i am not abnormally short....wait, how old were you again?
 :([/quote]He's 11.

*curses under breath* and i'm almost 14(in june hahahaha got june 93)

About 5'8''. i'm tall and thin compared to those in my school.

lol. everyone thinks that BOB's actually short even though he's not. BOB YOU WERE SHORT LAST YEAR. I told him that at a nats practice last year and he got really p^$\$\$$ed.

Oh ya and I'm 5'11"

you're still safe from stalkers BOB

5'3"-5'4" ish

Funcia, don't be so sad. Aren't you a girl? (terribally sorry if I am wrong)
Girls are naturally shorter.

[quote="pianoforte"]5'7" about

man, eating cheese will make u tall or what? or is it that only white ppl live in cheese country cause we all know azns are short[/quote]
well i guess maybe if i ate cheese i could be taller but cheese is umm really icky
all dairy products are errr like cheese :what?:

and yeah i'm a girl but i still am really short

There was a thread about this a long time ago. I remember because Ashegh said he was taller than me! :o

I'm 6'5" $\approx$ 197 cm.

[quote="spix"]1.81 m , but I can jump high enough.[/quote]Can you dunk a regular basketball? :D Cheese doesn't help you: I ate lots of cheese while I was young, but I'm still at a measly 5'10'' (1.78m) :(

[quote="asmallgoat"]wow you tall! I'm around Funcias height.[/quote]
you forget to add the 1 & 1/2 years younger part  :(

[quote="mustafa"][quote="pianoforte"]5'7" about

man, eating cheese will make u tall or what? or is it that only white ppl live in cheese country cause we all know azns are short[/quote]

Milk is good for your bones :P . There are some asians in Madison, but I don't think that there are many farther north.[/quote]

??I'm azn and i live farther north!

around 5 feet last time i measured i was 4 11

never mind, I just measured myself. I'm 6'7'' counting my hair. Without my hair, I'm a little shorter.

like 6'4" if i stick my hair straight up?

sadly i'm only 5'11" normally

[quote="footballrocks41237"]
Milk is good for your bones :P . There are some asians in Madison, but I don't think that there are [b][size=150]many[/size][/b] farther north.[/quote]

??I'm azn and i live farther north![/quote]

Ahem.


For a ellipse

$\frac{x^{2}}{4}+y^{2}$ = 1

how do you know that the auxillary circle has eqn $x^{2}+y^{2}= 4$

--------------------

The radius of the auxiliary circle would be the same as the semimajor axis of the ellipse. The ellipse has axes  length 4 (x-direction) and length 2 (y-direction), So the major axis is 4 and the semimajor axis is $\frac{4}{2}= 2$. Thus the circle is $x^{2}+y^{2}= 2^{2}$.

In general, for an ellipse with equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1$, we get an auxiliary circle of $x^{2}+y^{2}= \max\{a^{2},b^{2}\}$


search
Ring Theory
superior algebra
superior algebra unsolved

Show that the ring $\mathbb Z \left[ \tfrac{1+\sqrt{-19}}{2}\right]$ is not euclidian to any norm.

[hide="hint"]prove first it's not euclidian to complex norm. what do we know about the ring of units?[/hide]

is it easy? :o
I think it was an ariticle. :ninja:
do you know an easy proof?
I am eagerly waiting for it.

It was a class exercice, so it's definitely not unsolvable or a research article. But I admit I haven't found its solution yet.

As my friend has mentioned, this is due to [i]J.C Wilson[/i], A principal ideal ring that is not a Euclidean ring, Math. Mag $46, 34-8 (1973)$. Do a search on the web and you'll certainly find lots of related links.

check out this "instruction" :-)

http://math.berkeley.edu/~gbergman/grad.hndts/nonEucPID.ps


Sequence
algebra
algebraic identities
IMO
IMO 1967

Let $a_1,\ldots,a_8$ be reals, not all equal to zero. Let
\[ c_n = \sum^8_{k=1} a^n_k\]
for $n=1,2,3,\ldots$. Given that among the numbers of the sequence $(c_n)$, there are infinitely many equal to zero, determine all the values of $n$ for which $c_n = 0.$

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]  :)

$c_n$ must be zero for all odd $n$.

Proof:
WLOG suppose that $a_1 \geq a_2 \geq ... \geq a_8$. If $a_1+a_8 > 0$ then for sufficiently high odd $n$, $c_n$ will be dominated by $a_1$ alone i.e. it will always be positive. Similarly if $a_1+a_8 < 0$; hence $a_1=-a_8$. Now for odd $n$ these terms cancel, so we can repeat for the remaining values. Now all the terms cancel for all odd $n$. Since some $a_i$ is nonzero $c_n > 0$ for even $n$.

QED
Many early IMO problems are simply jokes in comparing with present problems.

$n$ can't be even since that would mean that all $a_1,...,a_8=0$ which goes against the conditions.
If n is even.
Now we must prove that for all odd n the statement is true.
Assume the opposite that:
$a_1>=a_2>=.....>=a_8$
Case 1: $a_1+a_8>0$
Then we would have that $c_n$ is infinite for huge n and then there wouldn't be infinitely many $c_n$ which are equal to 0.
Case 2: $a_1+a_8<0$
Then we would have that $c_n$ is infinite for huge n and then there wouldn't be infinitely many $c_n$ which are equal to 0.
Therefore, $a_1+a_8=0$.
Now we can repeat this process for the pairs.


[quote=Gibbenergy]RHS- LHS  $= (x+y+z) \left(\sum x^{2}(y+z)(y-z)^{2}\right) \geq 0$
Not hard ! :lol:[/quote]

Very long

[quote=cefer]Let $x,y,z \geq 0$.Prove that
$(2x+y+z)(2y+x+z)(2z+x+y)xyz \leq (x+y)^{2}(y+z)^{2}(z+x)^{2}$
P.S.I have a solution with opening brackets,does anyone have an elegant solution? :maybe:[/quote]
Put $p=a+b+c,\,q=ab+bc+ca$ and $r=abc$ inequality become
\[(pq-r)^2 \geqslant r(2p^3+pq+r),\]
or
\[pq^2 \geqslant (2p^2+3q)r.\]
Which is true because $r \leqslant \frac{q^2}{3p},$ and
\[pq^2-(2p^2+3q) \cdot \frac{q^2}{3p} = \frac{q^2(p^2-3q)}{3p} \geqslant 0.\]
The proof is complete.

[quote=cefer]Let $x,y,z \geq 0$.Prove that
$(2x+y+z)(2y+x+z)(2z+x+y)xyz \leq (x+y)^{2}(y+z)^{2}(z+x)^{2}$
P.S.I have a solution with opening brackets,does anyone have an elegant solution? :maybe:[/quote]
Suppose $x \geqslant y \geqslant z,$ apply AM-GM Inequality, we have
\[xyz(2x+y+z)(2y+x+z)(2z+x+y) \leqslant \frac{1}{16}[4zx(2y+z+x)+y(2z+x+y)(2x+y+z)]^2.\]
We need to prove
\[4(x+y)(y+z)(z+x) \geqslant 4zx(2y+z+x)+y(2z+x+y)(2x+y+z),\]
equivalent to
\[y(2x-y-z)(x+y-2z) \geqslant 0.\]
Which is true.

[quote=tarzanjunior][quote=henderson]Since the inequality is homogenous we can assume that $x+y+z=3.$ Then the inequality turns to $(3+x)(3+y)(3+z)xyz\leq (3x+yz)(3y+zx)(3z+xy)$ $\iff$ $(3+x)(3+y)(3+z)\leq (3+\frac{yz}{x})(3+\frac{zx}{y})(3+\frac{xy}{z}).$ By $Cauchy-Schwarz \ inequality$ we have $(3+\frac{yz}{x})(3+\frac{zx}{y})\geq (3+z)^2.$
Multiplying similar inequalities we get the result.

[color=#0f0]Note:[/color] $(x+y)^2(y+z)^2(z+x)^2=(x+y)(y+z)\cdot (y+z)(z+x)\cdot (z+x)(x+y)=((y(x+y+z)+zx)((z(x+y+z)+xy)(x(x+y+z)+yz)=(3x+yz)(3y+zx)(3z+xy).$[/quote]

I cannot understand your method. How can we assume the sum of x,y,z ?No such statement is given in the question?[/quote]

Let $x=kx_1, y=ky_1$ and $z=kz_1$ such that $x_1+y_1+z_1=3.$ Then substituting $x,y,z$ in the given inequality we have to prove:

$(2x_1+y_1+z_1)(2y_1+x_1+z_1)(2z_1+x_1+y_1)x_1y_1z_1 \leq (x_1+y_1)^{2}(y_1+z_1)^{2}(z_1+x_1)^{2}$ where $x_1+y_1+z_1=3.$ (you can observe that this is same with the given inequality with additional condition $x+y+z=3.$) So, we can assume that $x+y+z=3.$
[color=#f00]Note:[/color] If the inequality is homogenous, we can also give any certain value to $x+y+z,$ $xyz,$ $xy+yz+zx,$ ...

[quote=cefer]Let $x,y,z \geq 0$.Prove that
$(2x+y+z)(2y+x+z)(2z+x+y)xyz \leq (x+y)^{2}(y+z)^{2}(z+x)^{2}$

P.S.I have a solution with opening brackets,does anyone have an elegant solution? :maybe:[/quote]

[img]http://s1.sinaimg.cn/middle/006ptkjAzy77zpNEEUw40&690[/img]

[quote=cefer]Let $x,y,z \geq 0$.Prove that
$(2x+y+z)(2y+x+z)(2z+x+y)xyz \leq (x+y)^{2}(y+z)^{2}(z+x)^{2}$

P.S.I have a solution with opening brackets,does anyone have an elegant solution? :maybe:[/quote]

oneline down

It is well-known that in triangle holds sin<A+sin<B+sin<C<=3sqrt(3)/2. Use that for a=x+y... then calcul all sines in function of a,b,c.

[quote=cefer]Let $x,y,z \geq 0$.Prove that
$(2x+y+z)(2y+x+z)(2z+x+y)xyz \leq (x+y)^{2}(y+z)^{2}(z+x)^{2}$

P.S.I have a solution with opening brackets,does anyone have an elegant solution? :maybe:[/quote]

stronger:

\[{\frac { \left( y+z \right) ^{2} \left( z+x \right) ^{2} \left( x+y
 \right) ^{2}}{xyz \left( 2\,y+z+x \right)  \left( 2\,z+x+y \right)
 \left( z+2\,x+y \right) }}\geq {\frac {27}{64}}\,{\frac { \left( y+z
 \right) ^{2} \left( z+x \right) ^{2} \left( x+y \right) ^{2}}{
 \left( x+y+z \right) ^{3}xyz}}\geq 1\]

Left is am-gm.
please nice do right.

[quote=xzlbq]

stronger:

\[{\frac { \left( y+z \right) ^{2} \left( z+x \right) ^{2} \left( x+y
 \right) ^{2}}{xyz \left( 2\,y+z+x \right)  \left( 2\,z+x+y \right)
 \left( z+2\,x+y \right) }}\geq {\frac {27}{64}}\,{\frac { \left( y+z
 \right) ^{2} \left( z+x \right) ^{2} \left( x+y \right) ^{2}}{
 \left( x+y+z \right) ^{3}xyz}}\geq 1\]

Left is am-gm.
please nice do right.[/quote]

$ \left( x+y \right) ^{2} \left( y+z \right) ^{2} \left( z+x \right) ^{
2}-{\frac {64}{27}}\, \left( x+y+z \right) ^{3}xyz$
$={\it ssgm} \left(
 \left( y-z \right) ^{2} \left( {x}^{4}+5/7\,{x}^{3}y+5/7\,{x}^{3}z
 \right)  \right) +{\it ssgm} \left( {\frac {83}{189}}\, \left( {x}^{2
}-yz \right) ^{2}yz \right) +{\it ssgm} \left( {\frac {25}{189}}\,
 \left( 3\,{x}^{2}-xy-zx-yz \right) ^{2}yz \right)\geq 0 $

ssgm=sum.

it equals with
$$\prod \left(\frac{1}{x+y}+\frac{1}{y+z}\right) \leq \frac{1}{xyz}$$

and just AG


Show that for all positive reals $ x_i$ we have $ \frac {1}{x_1} \plus{} \frac {2}{x_1 \plus{} x_2} \plus{} ... \plus{} \frac {n}{x_1 \plus{} x_2 \plus{} ... \plus{} x_n} < \frac {4}{x_1} \plus{} \frac {4}{x_2} \plus{} ... \plus{} \frac {4}{x_n}$.

[color=red]Edited by N.T.TUAN[/color]

[quote="Myszon11"]Here is a stronger one  :wink: http://www.mathlinks.ro/viewtopic.php?t=173813[/quote]

Thank you very much.  :D


real analysis
function
topology
advanced fields
advanced fields unsolved

Today I heard that it is possible to find two sets $A$ and $B$ so that
$A \cup B = \mathbb{R}$
$A \cap B$ is empty
$\mu(A) = 0$ (lebesgue measure)
$B$ is of Baire category 1

How to construct such sets? :D

Let $(q_{n})_{n\ge 1}$ be an enumeration of all the rationals, $A_{n}=\bigcup_{k\ge 1}I_{k}^{n}$, where $I_{k}^{n}$ is an open interval of length $\frac 1{2^{n+k}}$ around $q_{n}$, and $B_{n}=\mathbb R\setminus A_{n}$. Now take $A=\bigcap_{n}A_{n},\ B=\bigcup_{n}B_{n}$.

$A_{n}$ is an open set with measure going to $0$ as $n\to\infty$, so the intersection $A$ of all $A_{n}$ is a $G_\delta$ set of measure zero. $B_{n}$ is a nowhere dense closed set, so the union $B$ of all $B_{n}$ is an $F_\sigma$ set of category $1$, and it's clear that they form a partition of $\mathbb R$.

This is a rather indirect argument from real one-variable analysis. Try to fill in the details if you like.

Fact #1:  every increasing function is differentiable everywhere except for a set of measure zero.

Fact #2: if a function is differentiable on a set of second category, then it is Lipschitz on some interval. (Being Lipschitz means that $|f(x)-f(y)|\le L|x-y|$ for some $L$ independent of $x$ and $y$.)

Fact #3: there exists an increasing function that is not Lipschitz on any (nontrivial) interval.

[quote="cyqwhu"]There is a book named "Measure and Category" in Graduate Texts in Mathematics,[/quote]

The complete reference is
J. C. Oxtoby, Measure and Category, Springer-Verlag, 1980.

Note that the set $A$ in grobber's example has Hausdorff dimension $0$. Here is an obvious generalization: Suppose that $\{q_{n}\colon n\ge 1\}$ is dense in $\mathbb R$, and $\sum_{n\ge 1}c_{n}$ is a convergent series with positive terms. Then the set $A=\bigcap_{m\ge 1}\bigcup_{n\ge 1}\left(q_{n}-\frac{c_{n}}{m},q_{n}+\frac{c_{n}}{m}\right)$ has measure zero and $\mathbb R\setminus A$ is of first category.

Is it possible to choose $q_{n}$ and $c_{n}$ so that $\mathrm{dim}\,A>0$? Or even $\mathrm{dim}\,A=1$?


i participate in mathcounts and was wondering if there are any good sites that show shortcuts and tricks for mathcounts problems.

Ah yes... I remember a few sites when I was doing Mathcounts:

[url]http://www.mathcounts.org/Problems/strategies01.html[/url]

[url]http://www.unidata.ucar.edu/staff/russ/mathcounts/diaz.html[/url]

Hope these help!  :D

-interesting_move

Yep, I remember reading Nick Diaz' mathcounts bible when I was little...well not that little, maybe 2 years ago.  But it's kind of common sense, you know...memorize squares thru 30, cubes thru 15, square roots of 2, 3, 5, 6, 10.  Memorize fractions (especially 1/7 - very helpful).  I don't know if those are [u]shortcuts[/u] but they're things to memorize.  I'll post more soon - after I find them again.

Google is king!

MATHCOUNTS TRICKS AND TIPS at http://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=5939 Yeah that should help


Veamos cómo les va con este  :)

[img]http://img49.imageshack.us/img49/2136/trianglezy7.png[/img]

En la figura, sea $\triangle{ABC}$ acutángulo, donde $O$ es el centro de su circunferencia circunscrita. Sean $x,y,z$ segmentos perpendiculares a los lados, como muestra la figura. Sean $r$ y $R$ los radios de las circunferencias inscrita y circunscrita, respectivamente. Demuestre que $x+y+z=r+R$

Saludos

Aki mi solucion:

Como O es circuncentro entonces sabemos que x, y, z cortan a BC(a), AC(b) y AB(c) en sus puntos medios.
Ahora, aplicamos el Teorema de Ptolomeo en cada uno de los tres cuadrilateros ciclicos formados por x,y, z.

$(xc)/2+(ya)/2 = (Rb)/2$   ... $(*)$
$(yb)/2+(zc)/2 = (Ra)/2$  ...  $(**)$
$(za)/2+(xb)/2 = (Rc)/2$   ...  $(***)$

Por suma de areas tenemos:

$(xa)/2+(yb)/2+(zc)/2 = (ABC)$ ...  $(****)$

Sumando $(*), (**), (***) y (****)$ tenemos:

$(x+y+z)(a+b+c)(1/2) = R(a+b+c)(1/2)+(ABC)$
$(x+y+z) = R+(ABC)(2/(a+b+c)) = R+r$

Lo ultimo lo deduzco de que $(ABC)= sr$ donde $s= semiperimetro$ y $r=inradio$

Muy buena solución, bastante mejor que la mía, que es por trigonometría

Hoy en el colegio obtuve esta solución, por casualidad aprendí Ptolomeo  :wink:

Saludos  :P


Find all real numbers $a$ such that the equations below have a common root.

$x(x^4 - a)^2 - x^3(1-a)^2 + x^2 + 4 = 0$

and

$x^3 + x^2 + x + a = 0$

How do I find the roots of the second equation? I've gotten as far as $a = -rst$ and $-r-s-t = rs + st + tr = 1$ where $r$, $s$, and $t$ are roots of the second equation, but I can't see how to substitute or work that into the first equation. What do you do with the first equation to solve it?

It looks like a manipulation of the second equation will do the trick, \[ -a=x^3+x^2+x.  \]

[hide]Solving for $a$ in terms of $x$ in the second equation, $x^3+x^2+x=-a$. Then substitute. Since $x^3(x^3+x^2+x+1)^2=x(x^4+x^3+x^2+x)^2$, the first equation is just $x^2+4=0$.

Either $2i$ or $-2i$ must be a root of the second equation. By complex conjugates, if one of them works, then both will, so just plug in $x=2i$. Then $-8i-4+2i+a=0$. But this makes $a$ imaginary. Therefore $\boxed{\text{no solution}}$.[/hide]


The numbers 3, 5 and 7 are important in many aspects of Japanese life. Most positive integers can be expressed as a sum of only 3's, 5's and 7's. For example, 15 can be expressed as 5 + 5 + 5 or 3 + 5 + 7. What is the greatest even integer that cannot be expressed as a sum of 3's, 5's and 7's?

6=3+3
8=3+5
10=3+7

Now you can add 3+3=6 to one of those numbers repeatedly to get any even number greater or equal to 6. Thus, the answer is 4.


let A and B is a point of circle w. and C is the point of the circle diameter. and the radius of this circle is 4√7. D is the piont of the line AB. angle ADC is 90 degree, and angle CAB=60. then find the value of CD.

Could you, please, post a sketch?
Thanks!

Best regards,
sunken rock

sorry for this sketch, angle CAB=60, and angle ADC=90 find the value of CD


Euler function logarithms inequalities limit

Prove that $ \sum_{i\equal{}1}^{\infty} \frac1{p_i}$ diverges, where $ p_i$ is the $ i$th prime.

hello, here you will find the answer
http://everything2.com/title/Erdos%2527%2520proof%2520that%2520the%2520sum%2520of%2520the%2520reciprocals%2520of%2520the%2520primes%2520diverges
Sonnhard.

For reference, the standard proof is by the Euler product formula

$ \zeta(s) \equal{} \sum_{n \ge 1} \frac{1}{n^s} \equal{} \prod_{p \in \mathbb{P}} \frac{1}{1 \minus{} p^{\minus{}s}}$.

The divergence of the product is equivalent to the divergence of the sum $ \sum_{p \in \mathbb{P}} \frac{1}{p^s}$ as $ s \to 1^{\plus{}}$ by a standard lemma.  One says that the zeta function has a simple pole at $ s \equal{} 1$.

In Euler's proof we have $ \prod_{i \equal{} 1}^{\infty}1 \plus{} \frac {1}{p_i \minus{} 1}$ diverges

but $ 1 \plus{} x\le e^x$ so the product is bounded by $ \exp\left(\sum \frac1{p_i \minus{} 1}\right)$, but can this be modified to

$ \sum \frac1{p_i}$?

[b]Lemma:[/b]  Let $ (a_n)$ be a sequence of real numbers such that $ \sum_n a_n^2$ converges.  Then $ \prod_n (1 \plus{} a_n)$ converges if and only if $ \sum_n a_n$ converges.

The proof is by the approximation $ \ln (1 \plus{} a_n) \equal{} a_n \plus{} O(a_n^2)$.

So from that lemma we have $ \sum_{i\equal{}1}^{\infty}\frac1{p_i\minus{}1}$ diverges, but how would this result prove the divergence of $ \sum_{i\equal{}1}^{\infty}\frac1{p_i}$?

(hmm if we use that lemma how would we deal with the convergence of $ \sum a_n^2$ first?)

Use the inequality $ \frac{1}{p} \ge \frac{1}{2(p\minus{}1)}$.

[quote="mathwizarddude"](hmm if we use that lemma how would we deal with the convergence of $ \sum a_n^2$ first?)[/quote]
Compare it to $ \sum_n \frac{1}{n^2}$.

The version of that lemma I prefer:
Let $ a_n$ be a sequence of (real or complex) numbers. The series $ \sum |a_n|$ converges if and only if the product $ \prod(1\plus{}|a_n|)$ converges.
If that series $ \sum |a_n|$ converges, the series $ \sum a_n$ and the product $ \prod (1\plus{}a_n)$ also converge*.

* If $ 1\plus{}a_k\equal{}0$ for some $ k$, we say that $ \prod(1\plus{}a_n)$ converges to zero. If none of those terms is zero but $ \lim_{N\to\infty}\prod_{n\le N}(1\plus{}a_n)\equal{}0$, we say that the product diverges to zero.
That technicality in the definition also matters for t0rajir0u's version, which needs to exclude the cases of $ a_k\equal{}\minus{}1$. It doesn't matter for application to this problem, which has $ a_n>0$ everywhere.

I saw a beautiful solution.I am posting it.According to PNT ,$ln n +ln ln n -1\leq \frac{p_n}{n}\leq ln n+ln ln n$
for $n\geq 6$ .
Now $\sum_{n=1}^{\infty}\frac {1}{p_n} > \sum_{n=6}^{\infty} \frac {1}{p_n} \geq  \sum_{n=6}^{\infty} \frac {1}{n ln n+ n ln ln n} \geq  \sum_{n=6}^{\infty} \frac {1}{2n ln n}=\infty$

from the well known fact that  $\sum_{n=2}^{\infty} \frac {1}{n (ln n)^p}$ diverges for $p\leq 1$ and converges for $p >1$. So done!.


The set of ordered pairs below define a function  { (3,1), (-1,1), (2,4), (-2,4) }

What is the range of the function?

a. { 3,4 } b. { -2,3 } c. { -2,4 } d. { 1,4 } e. NG

[hide] The range is just the set of $ y$ values in a function. Hence, in this case, the range is $ {(1,4)}$, so the answer is D.[/hide]


If $ a$ and $ b$ are the catetes in a right triangle and $ c$ is the hypothenuses.
Prove that $ c^{3}>a^{3}\plus{}b^{3}$

The following is my solution. Is it right?Give you your own solution.

$ c^{3}\equal{}(a^{2}\plus{}b^{2})\sqrt{a^{2}\plus{}b^{3}}$
$ c^{6}\equal{}(a^{2}\plus{}b^{2})^{2}(a^{2}\plus{}b^{2})$

Since a,b,c are length is true that $ (c^{3})^2>(a^{3}\plus{}b^{3})^{2}$
so:
$ a^{6}\plus{}3a^{4}b^{2}\plus{}3a^{2}b^{4}\plus{}b^{6}>a^{6}\plus{}2a^{2}b^{2}\plus{}b^{6}$

$ 3a^{2}b^{2}(a^{2}\plus{}b^{2})>2a^{3}b^{3}$
$ 2(a^{2}\plus{}b^{2})\plus{}a^{2}\plus{}b^{2}\minus{}2ab>0$
$ 2c^{2}\plus{}(a\minus{}b)^{2}>0$

The last inequality is always true :$ 2c^{2}\plus{}(a\minus{}b)^{2}>0$  :)

[hide="Another solution."]
First, start off with the statement that $ c > b$.  Then it follows that $ c > \frac {b^3}{b^2}$ and that $ c > \frac {b^3 \minus{} 2ab^2 \minus{} a^3}{a^2 \plus{} b^2}$.

Factoring out a negative sign and rewriting the expression in another way, you obtain:

$ c > \minus{} \frac {a^2b \plus{} ab^2 \plus{} a^3 \plus{} ab^2 \minus{} a^2b \minus{} b^3}{a^2 \plus{} b^2}$

$ c > \minus{} \frac {ab(a \plus{} b) \plus{} (a \minus{} b)(a^2 \plus{} b^2)}{a^2 \plus{} b^2}$

$ c > \minus{} \frac {ab(a \plus{} b)}{a^2 \plus{} b^2} \plus{} a \minus{} b$

$ c \minus{} a \plus{} b > \minus{} \frac {ab(a \plus{} b)}{a^2 \plus{} b^2}$.

Since $ c^2 \equal{} a^2 \plus{} b^2$, you can multiply both sides by $ c^2$:

$ c^2(c \minus{} a \plus{} b ) > \minus{} ab(a \plus{} b)$
$ c^3 \minus{} c^2(a \plus{} b) > \minus{} ab(a \plus{} b)$
$ c^3 > c^2(a \plus{} b) \minus{} ab(a \plus{} b)$.

Substituting $ a^2 \plus{} b^2$ back for $ c^2$:

$ c^3 > (a^2 \minus{} ab \plus{} b^2)(a \plus{} b)$
$ c^3 > [ (a \plus{} b)^2 \minus{} 3ab ]( a \plus{} b )$
$ c^3 > (a \plus{} b)^3 \minus{} 3ab(a \plus{} b)$
$ c^3 > (a \plus{} b)^3 \minus{} 3a^2b \minus{} 3b^2a$
$ c^3 > a ^ 3 \plus{} b^3$.
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If I understand your solution correctly, when you expanded $ (a^3 \plus{} b^3)^2$, the middle term should have been $ 2a^3b^3$, I believe.

Where did the square term in $ (x^3 \plus{} y^3)^2$ come from?