abstract algebra
Ring Theory
superior algebra
superior algebra solved

$A$ a commutative ring.
1/
$I$ an ideal of $A$ and $a \in A$ an element of $A$.
Suppose that the ideals $I+aA$ and $(I: a)=\{x \in A | ax \in I\}$ are finitely generated. Show that $I$ is finitely generated.
2/
Suppose that any prime ideal of $A$ is finitely generated, then $A$ is noetherian.

1) Let $f_1+ax_1,\ldots,f_n+ax_n$ generate $I+aA$, where $f_i\in I$ and $x_i\in A$.
Let $u\in I$. Then $u\in I+aA$, so there exist $\alpha_1,\ldots,\alpha_n\in A$ such that $u=\alpha_1(f_1+ax_1)+\ldots+\alpha_n(f_n+ax_n)=(\sum_{i=1}^n\alpha_if_i)+a(\sum_{i=1}^n\alpha_ix_i)$.
This implies $\sum_{i=1}^n\alpha_ix_i\in (I: a)$, so there exist $\beta_1,\ldots,\beta_m\in A$ such that $\sum_{i=1}^n\alpha_ix_i=\sum_{j=1}^m\beta_jy_j$, where $y_1,\ldots,y_m$ generate $(I: a)$.
It is now easy to see that $u$ is a linear combination of $f_i$s and $y_j$s.

2) This is a theorem due to Cohen.

Suppose there are ideals of $A$ that are not finitely generated.
Let $\mathcal I$ be the set of such ideals. $\mathcal I$ is ordered with respect to inclusion. Let $T$ be a totally ordered subset of $\mathcal I$.
Let $I$ be the union of all the elements of $T$. It is easy to see that $I$ is an ideal of $A$.
Suppose $I$ is finitely generated. Let $x_1,\ldots,x_n$ be a set of generators of $I$. It is easy to prove that there exists an ideal from $T$ that contains all $x_1,\ldots,x_n$. This easily implies that $I$ is in $T$ which is impossible because all the ideals in $T$ are not finitely generated. Therefore $I$ is not finitely generated.
This proves that $\mathcal I$ is inductively ordered. By Zorn, there exists a maximal element $I$ of $\mathcal I$.
I will prove that $I$ is a prime ideal of $A$. Suppose there exist $x,y\in A$ such that $xy\in I$ and $x,y\not\in I$. Then $I\subsetneq I+xA$ and $I\subsetneq I+yA$. By the maximality condition on $I$, $I+xA$ and $I+yA$ are finitely generated.
Obviously, $I\subset (I: x)$. $y\in (I: x)$ and $y\not\in I$, so $I\subsetneq(I: x)$. Just like for $I+xA$, $(I: x)$ is finitely generated. By part 1 of the problem, $I$ is finitely generated, which contradicts $I\in\mathcal I$.
Therefore $I$ is prime.
By hypothesis, $I$ is then finitely generated, thus contradicting $I\in\mathcal I$.

In conclusion, all ideals of $A$ are finitely generated, so $A$ is noetherian.


geometry
modular arithmetic
number theory
arithmetic sequence

How many integers can be expressed as a sum of three distinct integers chosen from the set $ \{ 4, 7, 10, 13, \dots , 46 \}$?

[[PMWC]] problems are great practice for number theory, geometry, and counting problems in MATHCOUNTS.  Please consider adding your solutions to the AoPSWiki.

[hide]
Each integer of the set can be expressed as $ 1\plus{}3n$.  n is a member of the set of integers from 1 to 15.  Let's call it N.
So the sum will be $ 3\plus{}3(a\plus{}b\plus{}c)$, where a, b, and c are members of N.
We can fixify the set so that it consists of only the members of N, and we will get the same number of sums.  With the members of N, we can make the distinct numbers from 6 to 135.  That makes $ 130$ sums.

[/hide]

Yes it does because I'm counting from the minimum to the maximum [b]sum[/b].

Oops, yes, abac is right.  Will edit if possible.  :blush:
edit: the post is uneditable, [hide="the correct answer is "]37 like everyone said.  I need to work on my division.[/hide].

@donkeykong: If this is what you meant, I'm sorry, I thought you meant there was something wrong with my methods, not my division.

[quote="CircleSquared"][hide]
Each integer of the set can be expressed as $ 1 \plus{} 3n$.  n is a member of the set of integers from 1 to 15.  Let's call it N.
So the sum will be $ 3 \plus{} 3(a \plus{} b \plus{} c)$, where a, b, and c are members of N.
We can fixify the set so that it consists of only the members of N, and we will get the same number of sums.  With the members of N, we can make the distinct numbers from 6 to 135.  That makes $ 130$ sums.

[/hide][/quote]

Somehow, both the 6 and the 135 can't be correct unless I'm seriously misunderstanding the solution.  Although we already agreed that there was a mistake...

[quote="CircleSquared"][hide]
Each integer of the set can be expressed as $ 1 \plus{} 3n$.  n is a member of the set of integers from 1 to 15.  Let's call it N.
So the sum will be $ 3 \plus{} 3(a \plus{} b \plus{} c)$, where a, b, and c are members of N.
We can fixify the set so that it consists of only the members of N, and we will get the same number of sums.  With the members of N, we can make the distinct numbers from 6 to 135.  That makes $ 130$ sums.

[/hide][/quote]

Perhaps you multiplied 45 * 3 or added (44 + 45 + 46) instead of 14 *3 or (13+14+15)... let's remember that we fixated the set!  :D

[hide]we get sums from 6 to 14*3 = 42, so we have 42 - 6 + 1 = [b]37 sums[/b]...[/hide]

[quote="DonkeyKong"][quote="CircleSquared"][hide]
Each integer of the set can be expressed as $ 1 \plus{} 3n$.  n is a member of the set of integers from 1 to 15.  Let's call it N.
So the sum will be $ 3 \plus{} 3(a \plus{} b \plus{} c)$, where a, b, and c are members of N.
We can fixify the set so that it consists of only the members of N, and we will get the same number of sums.  With the members of N, we can make the distinct numbers from 6 to 135.  That makes $ 130$ sums.

[/hide][/quote]

So what you're saying is find the least and biggest sum possible out of a set and subtract them inclusively to get the answer?? :huh:

Well, does your solution also consider that some of sum of two numbers in the set can be same??(ex.(46 and 4)&(43 and 7))[/quote]

But since the set has numbers that are in a arithmetic sequence you can use this method. If they aren't, I am quite sure you can't!

[quote="abacadaea"]no, were basicaly counting. the least sum possible is 21, using 4,7, 10. then we can also make 24 by changing the 10 to a 13. then 27 by changing the 10 to a 16. and so on. once we get to 46, we have 57 as the sum. then we change the 7 to a 10 to get 60. ect. then, when 7 becomes 43, the sum is 93. the keep changing te 4. we go all the way up to 129. then we have (129-21)/3  which equals 36 i think. then add one because were including the first one so 37. i think circlesquared made some mistake somewhere or forgot to divide or something[/quote]
I think this is in clearer terms,
[hide]Each number is $ 1\pmod{3}$, so the sum of 3 elements will be divisible by 3.
The smallest is 21, highest is 129. We can find the # of terms because this is an arithmetic progression, and it's $ \frac{129\minus{}21}{3}\plus{}1\equal{}37$.[/hide]

yeah this is actually quite an easy problem due to the sequence the numbers are in (rather, if the numbers were random, the problem might be a bit - or a lot harder)...and so we can just make it a bit simple by just showing the case for 1-15, which i think you can easily figure out, also has 37 possible ways...hold on...why am i getting 37 if everyone else has 36

ok since no one wants to look at this i'll just hide it

[hide]
21 1
24 2
27 3
30 4
33 5
36 6
39 7
42 8
45 9
48 10
51 11
54 12
57 13
60 14
63 15
66 16
69 17
72 18
75 19
78 20
81 21
84 22
87 23
90 24
93 25
96 26
99 27
102 28
105 29
108 30
111 31
114 32
117 33
120 34
123 35
126 36
129 37[/hide]

yeah so i have to go but that should be right


As figure shown below, a hemicircle has diameter $ AB$ and center $ C$. Point $ D$ is on $ AB$; points $ E$ and $ F$ are on the arc of the hemicircle. If $ \angle CED\equal{}\angle CFD\equal{}10^{\circ}, \angle ACE\equal{}40^{\circ},$ find $ \angle BCF.$

$ A,C,D,F$ are on the same circle.

$ \angle EFC \equal{} \angle EDC \equal{} 30^{\circ}$

$ \angle ECF \equal{} 180^{\circ}\minus{} 60^{\circ} \equal{} 120^{\circ}$

$ \angle FCB \equal{} 180^{\circ}\minus{} 40^{\circ} \minus{} 120^{\circ} \equal{} 20^{\circ}$


why did u start a new topic? :o
neways
1). there are too many and i dont remember them
2). it is an antiplatelet
3).definnitely not mine :P
4).bcoz of Cr
5).now dats obvious innit?

3. mrs. carrie everson

and btw, where da hell did u get this on from?? i dont think the yellow outlined boxes are evaluative!!

[quote="valeriummaximum"]Another question guys...

1. Name 2 antihistamines
2.Give reason: Aspirin prevents a heart attack
3.WHose idea was the froth floating method.
4. Whay does ruby have red coluor
5.Why is Ka2 << Ka1 for sulphuric acid?[/quote]

1) BromPheniramine & Terfenadine

3) Innovative Washerwoman not cary evans

5) Text Book says the ans is
H2SO4 is easily dissociated comp to HSO4- so I will write that but the ans can be explained by comparing str of conjugate base.

All the best to all students writing the exam tomorrow.

@Varun: Are your exams over?

[quote]3) Innovative Washerwoman not cary evans[/quote]

[size=200][color=darkred][u][i][b]NV[/b][/i][/u][/color][/size]

All the Best of Luck to ALLLL AOPS'ians.
I guess u will all find the paper jujubi  :D

no, shreyas, my exams havent even started as yet. i lied to pardesi dat my exams were gonna start soon :P . they start on the 17th.

Wow, has pardesi transmorfigured into supernerd???? And is studying so seriously that he doesn't find time to come online. Wow!!!!!

@Varun: when does it end da?

Waiting for pardesi to merge this...

ur rite abt pardesi, earlier he was online all the time, i guess boards bring out the anitsahu in him. :D
i dont know when my exams end, probably 28th i suppose.

[quote="VARUNARASU"]ur rite abt pardesi, earlier he was online all the time, i guess boards bring out the anitsahu in him.
[/quote]

 :rotfl:  :rotfl:  :rotfl:

[quote="VARUNARASU"]ur rite abt pardesi, earlier he was online all the time, i guess boards bring out the anitsahu in him. :D [/quote]

Not sachin a.s ?  :huh:  Im disappointed da varun  :rotfl:

I think somebody should start another thread for the discussion of the exam. :P


geometry
circumcircle
trigonometry
geometry proposed

A triangle $ ABC$ is given and let $ O$ be center of its circumcirle. One intersection of a line through $ O$ parallel to $ BC$ witch a circumcircle is $ M$. If $ BCMO$ has all equal sides find an $ \angle BAC$??
(JBMO=children <15,5 years they dont know trigonometry :D)

we have $ \overline{CB} \equal{} \overline{BO} \equal{} \overline{CO} \equal{} b$, then $ \triangle BCO$ is equilateral and $ \angle BAC \equal{} 30^\circ$ if the triangle is acutangle or $ \angle BAC \equal{} 150^\circ$ if it's obtusangle.

[quote="vedran6"][quote="¬[ƒ(Gabriel)³²¹º]¼"]we have $ \overline{CB} \equal{} \overline{BO} \equal{} \overline{CO} \equal{} b$, then $ \triangle BCO$ is equilateral and $ \angle BAC \equal{} 30^\circ$[/quote]
  Are you sure you completed your solution???  :D[/quote]I'm sure he (or she) has. I'll try to explain more.

Since $ BCMO$ has all equal sides, then $ |BO|\equal{}|BC|$. But $ BO$ is the radius of the circle. So $ BC$ has the same length as the radius. Therefore $ |BO|\equal{}|BC|\equal{}|CO|$, and $ \triangle BCO$ is equilateral. Thus $ \angle BOC\equal{}60^{\circ}$. But, $ \angle BOC$ is the central angle subtending the arc $ BC$ and $ \angle BAC$ is the circumferential angle. So $ \angle BAC\equal{}\frac12\angle BOC\equal{}30^{\circ}$.

[quote="Johan Gunardi"][quote="vedran6"][quote="¬[ƒ(Gabriel)³²¹º]¼"]we have $ \overline{CB} \equal{} \overline{BO} \equal{} \overline{CO} \equal{} b$, then $ \triangle BCO$ is equilateral and $ \angle BAC \equal{} 30^\circ$[/quote]
  Are you sure you completed your solution???  :D[/quote]I'm sure he (or she) has. I'll try to explain more.

Since $ BCMO$ has all equal sides, then $ |BO| \equal{} |BC|$. But $ BO$ is the radius of the circle. So $ BC$ has the same length as the radius. Therefore $ |BO| \equal{} |BC| \equal{} |CO|$, and $ \triangle BCO$ is equilateral. Thus $ \angle BOC \equal{} 60^{\circ}$. But, $ \angle BOC$ is the central angle subtending the arc $ BC$ and $ \angle BAC$ is the circumferential angle. So $ \angle BAC \equal{} \frac12\angle BOC \equal{} 30^{\circ}$.[/quote]
Or, $ \angle BAC \equal{} \frac12\angle BOC \equal{} 150^{o}.$ :wink:

Kostas vittas.

[quote="Johan Gunardi"]Of course not, we are talking about the interior angle. :wink:[/quote]
But, the problem states, ''Finding an $ \angle BAC$.''

Kostas Vittas.

[quote="vittasko"][/quote][quote="Johan Gunardi"][quote="vedran6"][quote="¬[ƒ(Gabriel)³²¹º]¼"]we have $ \overline{CB} = \overline{BO} = \overline{CO} = b$, then $ \triangle BCO$ is equilateral and $ \angle BAC = 30^\circ$[/quote]
  Are you sure you completed your solution???  :D[/quote]I'm sure he (or she) has. I'll try to explain more.

Since $ BCMO$ has all equal sides, then $ |BO| = |BC|$. But $ BO$ is the radius of the circle. So $ BC$ has the same length as the radius. Therefore $ |BO| = |BC| = |CO|$, and $ \triangle BCO$ is equilateral. Thus $ \angle BOC = 60^{\circ}$. But, $ \angle BOC$ is the central angle subtending the arc $ BC$ and $ \angle BAC$ is the circumferential angle. So $ \angle BAC = \frac12\angle BOC = 30^{\circ}$.[/quote][quote="vittasko"]
Or, $ \angle BAC = \frac12\angle BOC = 150^{o}.$ :wink:

Kostas vittas.[/quote]

 Your + Gunardis parts should complete a solution but for your part ofcourse more explanation is needed  :P


Using only the digits 1 through 5, how many even three-digit positive integers less than 500 can be written if digits can be used more than once?

If the number is less than 500 and all of the digits must be one through five, then their are 4 ways for their to be a hundreds digits, 5 ways for a tens digit and since it is even 2 ways for a ones digit. Multiplying the answer would be [b]40[/b].

[quote="ernie"]But wouldn't the answer be 50, 5x5x2?

[color=blue]* ernie is confuzzled[/color][/quote]

some of the numbers you count in those 50 are above 500

ex. 512


It seems to me that the games forum has more and more spam these days. What do you think?

I say two of the threads actually belong in the Fun Factory.


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ K/F$ be a finitely generated extension which is algebraic.
Show this extension is finite.

Let $ K\equal{}F(a)$, and let $ m$ be the minimal polynomial of $ a$ over $ F$. Then $ K \cong \frac{F[X]}{(m(X))}$, then it is of finite degree.

How do we know that such an element $ a$ exists?

What we can do, however, is take a generating set $ a_1,\cdots,a_n$. We can adjoin each one in a succession, and by algebraicity, each adjunction gives a finite extension. By the Tower Law, the total extension is finite.


I was just wondering : do you guys know it?  And do you watch it?  What do you think of it?

My opinion is that now it is all politics : Eastern European countries have been dominating over the past years, and the quality of the winning songs really has declined

"Poupée de cire...", "Waterloo" , "Love shine a light" those were really songs deserving to win in my opinion

So what do you think?

I'll move this to Games and Fun Factory for a better match with the topic of the forum.

Okay that is no problem

However, the fact that it has been moved from' Music and art' to 'Games and fun' does not sound too good :D


I was wndering ,about this  :D
What do you usually do when you are thinking on a problem?

I by myslef:tearing the edges of the papers and chewing them or chewing my pen or pencil

if its a math problem, i just concentrate on it. if i have to study for a test, i play music. it relaxes me.

[quote="G-UNIT"]If Im working on something frustratingly difficult, I talk to myslef as I walk around aimlessly, giving the appearance of someone mentally ill.[/quote]

 :rotfl:  :rotfl:  :rotfl:

On something excrutiatingly difficult, I lie on my back and close my eyes and "say" my calculations out loud.

[quote]I take my right heel off the ground and I sort of vibrate my right leg relatively fast, on my toes. I don't know why. And it SHOULD tire me out, but it can do it forever or something. It's just a habit I have during tests.
[/quote]

i also do that. it helps take your mind off of the testing situation. i also play music in my head. :D

[quote="Drunken_Math"]Whenever I do any type of test... I do this weird thing...

I take my right heel off the ground and I sort of vibrate my right leg relatively fast, on my toes. I don't know why. And it SHOULD tire me out, but it can do it forever or something. It's just a habit I have during tests.

A lot of people think I'm really nervous and that's why I do it, but that's not why.[/quote]

  ha, i just realized I do this too, its one of those obsessive habits like bitting you pen.....which I shamefully find myself doing when theres a pen around...........

I repeat the following quietly:
"I don't talk to myself. I don't talk to myself. I don't talk....."

when im taking a test, i do what drunken math does

then when i get nervous, i close my eyes and i quietly recite different types of rice

"fried rice, steamed rice, hot rice, cold rice," and so on

i dont know why i do this

i put my knees up close to my chest and bring the paper forward and twirl the pencil

I take a deep breath and try to continue, If I can't and I'mat home I leave the problem alone for a little while and then come back to it. My mom does that knee-bobbing hting all the time it drives me crazy!

[quote="Hexaditidom"][quote="Drunken_Math"]Whenever I do any type of test... I do this weird thing...

I take my right heel off the ground and I sort of vibrate my right leg relatively fast, on my toes. I don't know why. And it SHOULD tire me out, but it can do it forever or something. It's just a habit I have during tests.

A lot of people think I'm really nervous and that's why I do it, but that's not why.[/quote]

Knee-bobbing? That's pretty common, I do it too.[/quote]

Wow... I didn't know that it was quite common...

Nobody in my school does it...  :P

I didn't know it had a name!  :huh:

Usually I'm just staring at the paper in concentration.  If I'm not thinking as much (ie., doing calculations as opposed to thinking how to approach a problem), I twirl my pencil around my thumb and do the "knee bobbing thing"  :D

i'll fold the edge of the paper 500 times...dunno why

I wonder why I wonder why I wonder why I wonder why I wonder why I wonder why I wonder....

  zeek.....zeeek..zeek....

[quote="Drunken_Math"]Whenever I do any type of test... I do this weird thing...

I take my right heel off the ground and I sort of vibrate my right leg relatively fast, on my toes. I don't know why. And it SHOULD tire me out, but it can do it forever or something. It's just a habit I have during tests.

A lot of people think I'm really nervous and that's why I do it, but that's not why.[/quote]

Me too. sometimes i jiggle my right foot. sometimes my left. sometimes both. I also play with my hair. If I get really nervous I take off my shoes and tuck my feet under my body.

if im at school, i sit and stare. ive also been told that i have a really blank expression when im concentrating, probably equivalent to whenever i zone out.

if im at home, i stare, slouch and stare, sit on the floor and stare, lie down and stare, get back on the chair and stare, etc...

[quote="L_Li"]

if im at home, i stare, slouch and stare, sit on the floor and stare, lie down and stare, get back on the chair and stare, etc...[/quote]

Isn't that tiring??

[quote="Arne"]Well, I don't know really, but I remember that I ate a lot of sweets at IMO, and that I went to the toilet very often.[/quote]
Was it because you were more of less nervous?

I am a bit different. I ride bike to school everyday. The riding will take me 90 minutes per day. So I think during the ride.

i stare blankly... at whatever was in front of my face when i started thinking  :D

same as bluewinkle and i tap my fingers on the table

[quote="shobber"]
Was it because you were more of less nervous?
[/quote]

No, I wasn't actually that nervous during the contest itself (obviously I was before the contest started). I just consumed a lot of sugar to keep "strong" (I mean, concentrating hard for 4.5 hours isn't easy) and I went to the toilet often to take a 2 - minute break. That actually helps, since it refreshes your mind, sort of. (And you keep thinking, subconsciously, about the problems.) I always had my "bright" ideas after toilet breaks... :D

[quote="Arne"] That actually helps, since it refreshes your mind, sort of. (And you keep thinking, subconsciously, about the problems.) I always had my "bright" ideas after toilet breaks... :D[/quote]
That is true. On IMC, I went to the toilet about every hour, and even in school exams I go at least once or twice. :)

I also look around for people I know, looking how they're doing. You could also consider it sort of a break :)


[quote="Nbach"]Let $ p$ be a prime number which $ p>3$. Prove that there exist $ n$ integers
$ \frac{-p}{2}<a_{1}<a_{2}<...<a_{n}<\frac{p}{2}$ such that
  $ \frac{(p-a_{1})(p-a_{2})....(p-a_{n})}{a_{1}a_{2}...a_{n}}$ is equal to $ 3^{k}$ or $ -3^{k}$ where $ k$ is a natural number.[/quote]

Suppose that $p=6r-1$ for a natural $r$. Choose $a_j=3(j-r)-1$ for $j=1,2,...,2r$. Obviously $a_i<a_j$ when $i<j$. It's not difficult to see that all the values of $|a_j|$ are distinct. Indeed, if $i,j$ are such that $a_i+a_j=0$, then $3(i+j-2r)=2$, but this is imposible. Let $S=\{|a_1|, |a_2|,...,|a_{2r}|\}$. Clearly $|S|=2r$, and there is no $x\in S$ such that $3|x$. Because every $a_j$ belongs to the closed interval $[1,3r-1]$, we have that $S$ is the set that contains $\{1,2,4,...,3r-1\}$ (in other words, $S$ contains every natural in $[1,3r-1]$ not divisible by $3$). For the other hand, note that $p-a_j=3(3r-j)$, and then:

\[\frac{(p-a_1)(p-a_2)\cdots (p-a_n)}{|a_1a_2\cdots a_n|}=\frac{3^{2r}r(r+1)...(3r-1)}{1 \cdot 2\cdot 4\cdot 5....\cdot (3r-1)}=3^{3r-1}\]

If $p=6r+1$ for $r$ natural, it's enough choose $a_j=3(j-r)-2$ and $n=2r$. We use the same argument showed in the previous part and obtain that:

\[\frac{(p-a_1)(p-a_2)\cdots (p-a_n)}{|a_1a_2\cdots a_n|}=\frac{3^{2r}(r+1)...(3r-1)3r}{1 \cdot 2\cdot 4\cdot 5....\cdot (3r-1)}=3^{3r}\]

And we are done :D

P.S: This problem is from MOSP 2007. Is there another link with the same question?

[quote="BlAcK_CaT"]
P.S: This problem is from MOSP 2007. Is there another link with the same question?[/quote]
In fact, it's a CGMO(China Girl's Mathematics Olympiad) problem. I think it's earlier than 2007.


linear algebra
matrix
vector
induction
linear algebra unsolved

Let $A$ be a $n\times m$ matrix with entries from a ring $R$ (here, "ring" means "commutative ring with $1$"). Let $u$ be a positive integer such that $u\leq n$ and $u\leq m$. Assume that all $u\times u$ minors of the matrix $A$ (not only the principial ones) have determinant $0$. Prove that for any $u$ column-vectors $v_{1}$, $v_{2}$, ..., $v_{u}$ of the matrix $A$, we can find $u$ elements $a_{1}$, $a_{2}$, ..., $a_{u}$ of the ring $R$ such that $a_{1}v_{1}+a_{2}v_{2}+...+a_{u}v_{u}=0$, but not all of the $u$ elements $a_{1}$, $a_{2}$, ..., $a_{u}$ equal $0$.

Please take into consideration that $R$ is just a ring and may have zero divisors.

  darij

A generalization and extension (you can easily see that it is a generalization) by Peter Scholze:

Let $A$ be a $n\times u$ matrix (this means, a matrix with $n$ rows and $u$ columns) with entries from a ring $R$. Let $v_{1}$, $v_{2}$, ..., $v_{u}$ be the column-vectors of the matrix $A$. Prove that there exist $u$ elements $a_{1}$, $a_{2}$, ..., $a_{u}$ of the ring $R$ which are not all equal to $0$ and satisfy $a_{1}v_{1}+a_{2}v_{2}+...+a_{u}v_{u}=0$ if and only if there exists a nonzero element $m$ of the ring $R$ such that $m\cdot\det X=0$ holds for every $u\times u$ minor $X$ of the matrix $A$.

  Darij

For Peter's problem:

Let $v$ be the column vector with entries $a_{i}$, and for every $u$-element subset $I$ of $\overline{1,n}$, let $A_{I}$ be the $u\times u$ matrix consisting of the rows of $A$ with indices in $I$ (since you mention $u\times u$ minors, I’m assuming $n\ge u$). For an arbitrary square matrix $X$, let $X^{*}$ be its classical adjoint (a square matrix with the property that $XX^{*}=X^{*}X=$ the diagonal matrix having all diagonal entries equal to $\det X$).

$\Rightarrow$

For every $I\subset\overline{1,n},\ |I|=u$, we have $A_{I}v=0$, so $A_{I}^{*}A_{I}v=0$, or, in other words, $a_{i}\det X=0$ for all $i$ and all $u\times u$ submatrices of $A$. This means that we can take $m$ to be one of the non-zero $a_{i}$.

$\Leftarrow$

We can assume WLOG that the hypothesis no longer holds for the $(u-1)\times(u-1)$ minors, since otherwise we can use induction on $u$. This means that we can take $v$ to be the first column of an adjoint $A_{I}^{*}$ of some convenient $A_{I}$, multiplied by $m$.


They didn't show it in my daughter's school and in the school my friend's children go to. My friend was quite unhappy about it, so he and a few other concerned parents talked to the superintendent and they finally arranged for a compromise: the students wishing to hear the speech were given a school bus that brought them to another school where it was shown and back (but the more obvious solution to put those students in a separate closed room with a TV was out of question for some reason). I was lazier and just played the record at home a few days later.

In my opinion, the speech itself won't make a history in the noble art of
public presentation but it makes perfect sense and there is absolutely no reason
why the children should not hear it.

I just wonder how it went in your school or school district and what you think of it (I mean, both the speech itself and the decision your school made about showing or not showing it).

Our school played it on the smart boards in the classrooms, and the students whose parents wouldn't let them watch it could go to the office during the speech

As usual Obama's speech was fascinating and inspiring...I have no clue why people dont like it.. It tells us to work harder, and excel at school...but some people thinks thats weird...sigh..
I found nothing bad in that speech! Oh yeah...btw..my school never showed it?!?!

[quote="Jongy"]My school did not show it, but it did say that social studies teachers can show it at a later date if they want to (aka they're probably not going to show it).  I did watch it over YouTube cause I was interested to see what Obama said.  I don't think it's "communist, fascist, socialist, Nazi propaganda!" as worthawholebean said, cause Obama in the speech told students to work hard.  If you were the President, are you really going to tell students to keep being lazy?[/quote]

My school didn't show it live, but our social studies teacher let us watch it on Friday if we wanted to.  :)  BTW, I agree with what Jongy said...

thought it was kinda lame...

was expecting something more creative that that...

i thought it was pretty convincing, but apparently my half-asleep World Geo class disagrees

[quote="fedja"][quote]
But it's communist, fascist, socialist, Nazi propaganda!
[/quote]
Erm... This is pretty much the same as to say "muslim, catholic, buddist, orthodox church". I mean, it can, probably, be classified as propaganda, all right (what else do you expect a politician's speech to be? :?), but you'd better choose one of the four to start with if you want us to take your post seriously :P (that is if I am to take your post literally, if it just was an exercise in sarcasm, then I was too dumb to get it :( ). By the way, have you listened to it yourself?[/quote] Yeah, it was sarcasm.

My school didn't let anyone watch it either. I don't really see why people were making such a big fuss about it, especially since the first Bush did it back during his years.

[quote="worthawholebean"]Obama is clearly trying to indoctrinate the youth. Just like Hitler, Mussolini, Stalin, and Mao - clear evidence of Nazism, socialism, fascism, and communism. [/quote]
Why don't you just say: [i]"Obama makes speeches directed at the youth. Just like Hitler, Mussolini, Stalin, and Mao did - clear evidence of Nazism, socialism, fascism, and communism"[/i]. By the way, this is a deadly incriminating evidence against almost everybody here, starting with Mr. Rusczyk. :rotfl:

OK, I've got the idea. :) Now, worthawholebean, what do you really think (if anything)?

worthawholebean, the sarcasm isn't doing it for me today, sorry.  Others, I've deleted the post Fedja was responding to and a couple of others -- there was actually an interesting question here.  (It would be nice if those who did see it could give a more detailed comment, though -- your opinion without an explanation of why you have that opinion (preferably one that refers to actual aspects of the speech) is not very interesting at all.)

We didn't see it at our school. It was never really mentioned at all, except when the calc teacher asked if anyone saw it. I get the impression it wasn't something people took too seriously.

The point I was trying to make is that there isn't a controversy. There never had been. A very small (yet very vocal) fraction of the electorate has decided that Obama is a fascist, socialist, Nazi, communist, and so they have decided to make a big fuss. 70,000 marching in the capital don't mean much out of 300 million. It was inflated by the Glenn Beck demagogue-types into this apparent huge firestorm.

I was trying to demonstrate that.

Language is tricky.  Part of the problem is that the word "socialism," like many political words, has lost any real meaning because of the many ways it is used.

According to dictionary.com, socialism is a theory or system of social organization that advocates the vesting of the ownership and control of the means of production and distribution, of capital, land, etc., in the community as a whole.

On the one hand, I think it's clear that neither modern day Western Europe, nor the Obama administration, is socialist according to that definition.  On the other hand, if one thinks of socialism as a spectrum of viewpoints, it is fair to say that Western Europe is "more socialist" than the US, and that Obama's health plan is "more socialist" than our current system of healthcare.

The brainlessness comes from using the word "socialist" as an epithet, with the underlying idea being that "socialism is evil," and therefore any move in the socialist direction is automatically wrong.  If the health plan is really a bad idea, then one ought to be able to argue this point without calling it socialist.  Our society has many features that were once considered socialist and are now accepted by most Americans:  the legality of unions, Social Security, Medicare, Medicaid, food stamps, minimum wage, the 5-day work week, etc.  Presumably, many of the protesters who derisively cry, "socialism" have no beef with some or all of these things.  For example, the now-famous cry, "Keep your government hands off my Medicare!"

As for Nazi Germany, the fact that it was the called "National Socialist" party doesn't quite answer the question of whether it was "socialist."  (For example, the "Democratic" party believes in a "republican" form of government just as much as the "Republican" party.)  Claiming the mantle of socialism is not the same thing as being a socialist.  The man himself, Hitler, once said, “Our adopted term 'Socialist' has nothing to do with Marxian Socialism. Marxism is anti-property; true Socialism is not.”  On the other hand, he also said, "basically National Socialism and Marxism are the same."  (Quotes from Wikipedia.)  Although one could argue about the ways that Nazi Germany was or was not socialist, on balance it is clear that it wasn't very socialist insofar as the term was generally understood at the time.

The reason why so many educated people bristle over the idea that something can be both "fascist" and "socialist" simultaneously is that, while they are not quite logically contradictory (depending on definition), the word "fascist" is essentially defined by the historical examples of Mussolini, Hitler, and Franco, and these three were avowed enemies of the international forces of Marxist socialism/communism at the time.  Fascism is also thought of as being right-wing while socialism is left-wing.

In any case, you can forgive me if I'm skeptical that some people have any idea what socialism means when they are simultaneously displaying their ignorance by complaining about "death panels."  Calling Obama and his policies fascist and socialist typically just means that the speaker just wants to use every nasty word available.

Here's an actual oxymoron--Van Jones, the communist czar!

In a way I almost think it is positive that there are fundamentalists who think all "socialism" is bad, even if as yenlee point out, they may be incredibly inconsistent and display that they don't really know what they are talking about. It is refreshing because now it seems that almost everybody is some kind of postmodern utilitarian who thinks that all morality is completely relative. Someone has to take a stand  :D

[quote="Kalle"]It is refreshing because now it seems that almost everybody is some kind of postmodern utilitarian who thinks that all morality is completely relative. Someone has to take a stand  :D[/quote]
Hmmm... How does "socialism" relate to "relativeness of morality"? I have always thought that "capitalism" allows for at least as much "moral flexibility". If you want to refer to the bloodshed during the Russian revolution, I can easily refer to the one during the first world war. As to the Stalin and Mao times, that wasn't socialism at all, just a personal dictatorship masked as socialism (if you want a "capitalism" analogue, recall the French revolution and the Reign of Terror). Granted, $ 36$ is $ 9$ times greater than $ 4$ but Russia is $ 9$ times larger than France too and China went through the disastrous times in $ 10$ years.

Capitalism and "democracy" do not promote moral per se at all. It is just a commodity that one tries to use when other means fail to give the advantage over the competitors. When the power is balanced and everything is stable, everybody tries to show how friendly, trusting, and trustworthy he is to attract clients but give it a very minor push away from the equilibrium and banks refuse to give clients their money back, the gas prices skyrocket, checks and credit cards are no longer accepted at many gas stations, credit card interest rates go up, etc. What keeps things at bay in the capitalist society is that if you are too greedy, you'll be undercut by somebody who is slightly less greedy. I have nothing against this "greedy equilibrium" per se but I hardly can call it "moral".

Socialism often promotes moral by means of brute force. Needless to say, this is too blunt a weapon to achieve the desired effect and usually, those who hold the weapon lose all the moral completely. Ironically, exactly this way of promoting "morals" seems to be the dream of most anti-socialists you are referring to (starting with McCarthy).

So, yes, somebody has to "take the stand" but I always prefer people taking stands [b]for[/b] something to the people taking stand [b]against[/b] something. Sometimes one implies the other, of course, but being "against socialism" is very far from being "for morals".

[quote="fedja"]
So, yes, somebody has to "take the stand" but I always prefer people taking stands for something to the people taking stand against something. Sometimes one implies the other, of course, but being "against socialism" is very far from being "for morals".[/quote]
It's always easier to do the latter.  People fear what they don't understand and it's blatantly obvious that most people don't understand socialism.  Call Obama a socialist and it's easy to get people against him (and I do consider myself somewhat well-versed in socialism as I have read the works of Marx, Lenin, Luxemburg, Proudhon,...)

I did not say that the idea of (the economic system) socialism exhibits relative morality, but that [b]people today[/b] make things very easy for themselves by arguing about so many things in terms of utility! Basic philosophical questions are disregarded entirely, because utility-arguments don't need them. In fact, trying to argue something else seems to be very much looked down upon as immature. Unless you have a PhD in philosophy, you are not allowed to try to reason about morality. Even if my own political views are likely sophomoric, why should we dissuade debate by ridiculing our opponents and calling them crazy (as in this thread)?

But I'll still reply to the rest of your post:

The discussion is not only about if agents in capitalism act morally, but also whether or not it can be moral to deny someone the right to exist in a capitalistic system. Banks refusing to give money back are doing something illegal by breaking a contract, but I strongly think we should give a gas station owner the right to set whatever prices he wants to. It can be rational and moral to help someone of "lesser fortune", but that does not imply that it is right to force someone to do it.

I believe fully in the non-aggression principle, and thus many socialistic ideas are suddenly wrong. However, I certainly don't agree that you should use weapons to promote "morals" - rights are something given to you, not taken with brute force. Since socialism is just an economic system, a socialist could say the same thing. The fact is however, that every social liberal I've ever met does agree that we should have taxes and government control even against people's will. Forcing people to do against their will requires violence or threat of such, and so anyone who wants to implement any such policy necessarily has "trigger happy" written into their political ideology! The workaround might be to deny some other premise, denying full property rights being a common example.

[I am implicitly only talking about cases where no one has done anything wrong, i.e. initiated the use of force against someone. I certainly think we can force criminals to do things against their will.]

[quote="Kalle"][W]hy should we dissuade debate by ridiculing our opponents and calling them crazy (as in this thread)?[/quote]  Because, in fact, many of the participants in the "debate" in the U.S. [i]are[/i] crazy or, perhaps worse, willingly make false and inflammatory statements in order to mislead the public about the actual content of policy proposals.  A necessary condition for debate to be an effective means of getting to the truth is that both sides be roughly constrained by the truth, but this is simply not the case in current circumstances.

In parliamentary systems, this is not too much of a problem, because when one party holds 40% of the seats in the legislature and another party holds 60%, the majority party gets to implement its agenda and then people can decide for themselves whether they like it or not.  In the American system, it turns out that a party with 40% of the seats can be remarkably effective at blocking the policies of the majority party.  The Republican party appears to have decided that it will use the heckler's veto at every opportunity, and this leads to a circumstance in which open debate is completely useless for anything.

As an aside, JRav@#35, I think you've confused TARP (a program established under Bush) with the stimulus bill.

[quote="JBL"] In the American system, it turns out that a party with 40% of the seats can be remarkably effective at blocking the policies of the majority party.[/quote] I disagree. The Republicans somehow get everything they want done when they have even a smaller majority. for some reason The Democrats have trouble doing this.

That's your distorted perspective talking. A ton of stuff that the Republicans wanted to do in the early Bush years got blocked.

[quote="JBL"]In parliamentary systems, this is not too much of a problem, because when one party holds 40% of the seats in the legislature and another party holds 60%, the majority party gets to implement its agenda and then people can decide for themselves whether they like it or not. [/quote]
Of course that is complicated by the fact that many parliamentary systems in the world are multiparty systems. There are many cases where the largest party, which is the party of the prime minister, has, say, 38% of the seats. It forms a working majority by entering into a coalition with some smaller parties. Such a party is then somewhat free to rule, but with the limits that it might not be able to pass something antithetical to the values of its minor coalition partners.

Of course, there are things that can happen in the U.S. system that would be unthinkable in a parliamentary system. In the mid 1990's (in the aftermath of the 1994 election), President Clinton was unable to get his budget proposal through the House of Representatives. The resulting showdown over the lack of a budget resulted in the partial shutdown of the federal government. In a parliamentary system, a prime minister who couldn't get a budget passed --- wouldn't be the prime minister any longer. (A prime minister can always be removed for purely political causes - all it takes is a majority of the parliament to vote for his removal.) Of course, the situation in the California state government is even stranger, as it takes a supermajority to pass a budget or to raise any taxes. That's another level of unthinkable in a parliamentary system.

Looking at it that way... the "Blue Dogs" are effectively a minor coalition partner, who have objected. There has been serious debate between them and the bulk of the Democratic Party in addition to the Republican shouting.

[quote="jmerry"]Looking at it that way... the "Blue Dogs" are effectively a minor coalition partner, who have objected.[/quote]  I actually think this is a poor analogy: if the US were a parliamentary system, we would (approximately) have the House but not the Senate.  The House (which also has about 60% Democratic seats at the moment) has already passed a decent climate bill and a health reform bill vastly better than anything Senate Finance has been considering.  This is despite the presence of a large Blue Dog caucus there.  The major obstacle is that in the Senate, very small groups of people have traditionally been able to exert a great deal of power to block legislation.  I don't think there's any doubt that if a decent health-reform bill were put up for a vote in the Senate, it would get 50 votes, regardless of the handful or two of conservative Democrats making tepid noises about various aspects of it.  It's just awfully difficult to get to the point of having the Senate vote on it when a minority (and you may include, say, Senators Conrad et al. from our hypothetical centrist coallition member in the count of the minority here) is really bent on objecting.

(By the way, Fedja, if you'd like to drag this discussion back on topic, just let me know.)

Back to fedja's original point, do you think there would be the same amount of objection if a Republican president made the same speech?  Let's say McCain was president and he wanted to speak to school children.  Will there be widespread anger?

[quote=JBL"]
As an aside, JRav@#35, I think you've confused TARP (a program established under Bush) with the stimulus bill.[/quote]
Correct, I believe I just had my numbers wrong. :)

[quote="JRav"]
Let's say McCain was president and he wanted to speak to school children. Will there be widespread anger? [/quote]
True, most/some of the oppostion may have to do with the fact the President Obama is the one making the speech.

According to my dad though, there was some(but not as much) opposition to George W. Bush's back to school speech.

Reconciliation would allow the bill to pass with 50 votes.  However, there is a bizarre idea that somehow it's better if health care reform passes with "bipartisan" support.  Part of the conventional wisdom for why bipartisan support is so crucial is that it provides "political cover" for moderate Democrats in swing districts.  While it's true that health care reform could be a tough sell for some of these Democrats, I can't imagine why some voter in a battleground district of North Carolina would give a rat's ___ what Olympia Snowe thinks.

Robert Reich has argued that, ironically, it's better if the moderate Republicans walk away from health care reform, because then we'll end up with a better law.

@Ihatepie, worthawholebean:  For example, Bush pushed through the tax cuts using reconciliation.  On the other hand, his proposed Social Security reform went nowhere fast.


Does Muirhead apply also for cyclic sums ,or only for symetric ? :huh:

[quote="Svejk"]only for symetric ? :huh:[/quote]
Yes, that's right :)

[quote="Achu"]Can you show me the different of Muirhead and Karamata[/quote]
The inequalities you are talking about are obviously different:

[b]Muirhead inequality[/b]

If $\vec{a}\succ \vec{b}$ and $x_{1},\ldots,x_{n}>0$, then $\sum_{\sigma\in S_{n}}x_{\sigma(1)}^{a_{1}}\cdot\ldots\cdot x_{\sigma(n)}^{a_{n}}\geqslant \sum_{\sigma\in S_{n}}x_{\sigma(1)}^{b_{1}}\cdot\ldots\cdot x_{\sigma(n)}^{b_{n}}$.

[b]Karamata inequality[/b]

If $\vec{a}\succ\vec{b}$ and $f$ is a convex function, then
$\sum_{1\leqslant i\leqslant n}f\left(a_{i}\right)\geqslant \sum_{1\leqslant i\leqslant n}f\left(b_{i}\right)$.

[quote="Achu"]Karamata is stronger than Muirhead ????[/quote]
Yes, Muirhead can be easily proved by Karamata.

how ?
 :maybe:
which function with ?
please someone to help  :oops:


14. A wooden block (mass $ M$) is hung from a peg by a massless rope. A speeding bullet (with mass m and initial speed $ v_0$) collides with the block at time $ t \equal{} 0$ and embeds in it. Let $ S$ be the system consisting of the block and bullet. Which quantities are conserved between $ t \equal{} \minus{}10 s$ and $ t \equal{} \plus{} 10 s$?

A)The total linear momentum of $ S$.
B)The horizontal component of the linear momentum of $ S$.
C)The mechanical energy of $ S$.
D)The angular momentum of $ S$ as measured about a perpendicular axis through the peg.
E) None of the above are conserved.

I'm guess it's not A because of force gravity, and not B because of the tension of the rope. It can't be C because the collision is inelastic. So I think it's D, but I'm not sure.

[quote="pascal12"] Which quantities are conserved between $ t \equal{}\minus{}10 s$ and $ t \equal{} \plus{} 10 s$?
[/quote]
10 s ?? Are you sure?

Oh wait, yeah, angular momentum isn't conserved either. So it's E, right?

Yes it is E.. but the question doesn't make sense.. In [b]10s[/b] they can be anywhere..

I think it might've been $ t\equal{}\minus{}10ms$ to $ t\equal{}\plus{}10ms$ (ms=milliseconds)
Because I feel they want to ask what would be the case [i]during[/i] the collision (from just before to just after)

Is the angular momentum not conserved because of gravity acting at a time other than $ t\equal{}0$?

Yes, the tension from the rope doesn't affect the angular momentum, only the weight.
There is no information for negative or zero t, but the answer ought to be E.


superior algebra
superior algebra unsolved

Prove that there exist an unitary ring with $ p^3$ elements(p prime) that isn't commutative.

You've got me, [b]LR[/b]!  :blush:

Regards,

Are you just being silly, coquitao?  In any case, there is a correct example that looks [i]extremely[/i] similar to this one...

If I'm not mistaken, LydianRain means the upper-triangular matrices in $ \mathcal{M}_2(\mathbb{F}_p)$.


geometry
parallelogram
geometry unsolved

Given a triangle $ ABC$ with $ \angle A \equal{} 120^{\circ}$. The points $ K$ and $ L$ lie on the sides $ AB$ and $ AC$, respectively. Let $ BKP$ and $ CLQ$ be equilateral triangles constructed outside the triangle $ ABC$. Prove that $ PQ \ge\frac{\sqrt 3}{2}\left(AB \plus{} AC\right)$.

let $ \overrightarrow{QL}\cap \overrightarrow{PK}\equal{}R$ and $ AK\equal{}k, AL\equal{}l$...

we have that $ AKRL$ is a parallelogram and that $ PR\equal{}c\minus{}k\plus{}l$ and $ QR\equal{}b\plus{}k\minus{}l$.. then, we have to prove that

$ PQ^2\equal{}(c\minus{}k\plus{}l)^2\plus{}(b\plus{}k\minus{}l)^2\plus{}(c\minus{}k\plus{}l)(b\plus{}k\minus{}l)\geq \dfrac{3(b\plus{}c)^2}{4}$, or equivalently,

$ (b\plus{}c)^2\minus{}(c\minus{}k\plus{}l)(b\plus{}k\minus{}l)\geq \dfrac{3(b\plus{}c)^2}{4}$...

this one can be rewritten as $ ((b\plus{}k\minus{}l)\minus{}(c\minus{}k\minus{}l))^2\geq 0$ which is true, and we're done  :D


[quote="GoKorea"]$ x(b\plus{}c)/(y\plus{}z)\plus{}y(c\plus{}a)/(z\plus{}x)\plus{}z(a\plus{}b)(x\plus{}y) >\equal{}  \sqrt{3(ab\plus{}bc\plus{}ca)}$
$ a,b,c,x,y,z >0$

help,,, :maybe: [/img][/quote]
See here:
http://reflections.awesomemath.org/2008_1/unexpected_ineq.pdf


LaTeX
trigonometry
vector
geometry
trig identities
Law of Sines

[quote="ckck"]Find the radius of the circle circumscribed about a triangle whose vertices are (5,3),(10,7),(9,-2).[/quote]
[hide="method 1"]Find the length of the circum radius using the formula $R=\frac{abc}{4[ABC]}$[/hide]
[hide="method 2"]Use distance formula on $(x,y)$ and equate your equations.[/hide]
EDIT: $\LaTeX$ fixed

[quote="bpms"][quote="ckck"]Find the radius of the circle circumscribed about a triangle whose vertices are (5,3),(10,7),(9,-2).[/quote]
[hide="method 1"]Find the length of the circum radius using the formula $R=\frac{abc}{4[ABC]}$[/hide]
[hide="method 2"]Use distance formula on $(x,y)$ and equate your equations.[/hide][/quote]

That should fix his LaTeX. And yes, (extended) law of sines might be the easiest way.

[quote="kunny"]Let $A(5,\ 3),\ B(10,\ 7),\ C(9,\-2),$ we have that $\angle{BAC}=90^\circ.$ Thus desired radius is $\frac{1}{2}BC=\frac{\sqrt{82}}{2}.$[/quote]
But why is the radius half of BC?

[quote="max_tm"]Because $\angle BAC$ is subtending half the circumference of the circle (ie. it's a right angle), BC is the diameter of the circle.[/quote]
Oh!!!! :wallbash_red:  :wallbash_red:  :wallbash_red:
I didn't notice it was a right triangle.

But even if it weren't a right triangle, we could still find an angle of the triangle (through vectors) and use the ELoS.

hmm well if it wasn't a right triangle, a decent way is just finding the intersection of two perpendicular bisectors and then finding the distance from that point to a vertex of the triangle

[quote="max_tm"][quote="jli"][quote="max_tm"]I'm not sure if this was bpms' first idea that for some reason didn't show up, but
[hide]
law of sines would work.
[/hide][/quote]

i dont get how law of sines applies here[/quote]

Kunny's solution is in fact quicker than my suggestion, and what I meant was the extended law of sines (as Hamster 1800 pointed out):

$\frac{\sin A}{a}= \frac{\sin B}{b}= \frac{\sin C}{c}= 2R$[/quote]\

I'm pretty sure you meant $\frac{a}{\sin A}= \frac{b}{\sin B}= \frac{c}{\sin C}= 2R$ or else we'd have that $R$ decreases when we scale the triangle up.

Yeah, that is what I meant  :P. I should probably proofread my posts.


group theory
abstract algebra
superior algebra
superior algebra unsolved

[b]Prove [/b]that a finite cyclic group of order n has exactly one subgroup of index m for each positive divisor m of n.

PLEASE HELP!!!!!!!!!!!!!

This is the theorem of Lagrange for groups...
You will have to have knowledge of cosets to do this...

http://web.usna.navy.mil/~wdj/tonybook/gpthry/node22.html

i dont see how it pertains to my question... if naythign coudl you please reword my question maybe that could help me cause i am not sure exactly what it is asking

the solution follows through contradiction.
ok
so we have $G=<x> =\left\{e,x,x^2,...,x^k,...,x^n\right\}$
also $|G|=n$.
Proove that if $m|n;|H|=m$ and $H\leq G$ then $H$ is unique with those
properties.

so $H=\left\{x : ord(x)|m \right\}=\left\{x^k : k|m \right\}$

if there would be $H'$ with the same properties it couldnt be
different from $H$ because it would mean that one of its
elements orders wouldnt divide m so it wouldnt be in $H'$.

Finite cyclic group of order n is isomorphic to $Z_n$
To obtain a group of index m, you only need to pick the multiples of $\frac{n}{m}$. It's easy to show it's a subgroup and it's of order m.


Is there any way to solve this problem with mass points? I don't know how to begin...

In $ \triangle ABC$, let $ M$ be the midpoint of $ AB$ and let $ Q$ and $ P$ be the trisection points of $ CM$ (with $ Q$ closer to $ M$). If the extensions of $ BP$ and $ BQ$ intersect $ AC$ at $ X$ and $ Y$, respectively, then evaluate
\[ \frac {CX \plus{} AY}{XY}.
\]

Here's an approach that's basically mass points.

[hide="Solution"]Because $ M$ is a midpoint, we have
\[ M \equal{} \frac{A \plus{} B}{2}.
\]
Because $ P$ is a trisection point, we have
\[ P \equal{} \frac{2C \plus{} M}{3} \equal{} \frac{A \plus{} B \plus{} 4C}{6}.
\]
Because $ Q$ is a trisection point, we have
\[ Q \equal{} \frac{C \plus{} 2M}{3} \equal{} \frac{A \plus{} B \plus{} C}{3}.
\]
Because $ X$ is the extension of $ P$, we have
\[ X \equal{} \frac{A \plus{} 4C}{5}.
\]
Because $ Y$ is the extension of $ Q$, we have
\[ Y \equal{} \frac{A \plus{} C}{2}.
\]
In other words, $ X$ and $ Y$ split $ \overline{CA}$ in a $ 2: 3: 5$ ratio.  So the desired fraction is $ \frac{7}{3}$.[/hide]

Sorry, I'm not very familiar with mass points. How did you figure out that
\[ X\equal{}\frac{A\plus{}4C}{5}?\]
Also, is there any way for you to label all the points with masses to make it easier for me to understand?

Thanks.

Well, we can rewrite our equation
\[ P \equal{} \frac {A \plus{} B \plus{} 4C}{6}
\]
as
\[ P \equal{} \frac {5}{6} \cdot \frac {A \plus{} 4C}{5} \plus{} \frac {1}{6} \cdot B.
\]
In other words, $ P$ is a convex combination of points $ \frac {A \plus{} 4C}{5}$ and $ B$.  Because $ X$ is the unique point on $ \overline{AC}$ and $ \overrightarrow{BP}$, it must be the point $ \frac {A \plus{} 4C}{5}$.

I know that some folks like to talk about mass points as placing masses at various points.  (Hence the name.)  I think of it instead as algebra.  But if you insist, imagine placing a mass of 1 at $ A$, a mass of 1 at $ B$, and a mass of 4 at $ C$.  The two masses at $ A$ and $ B$ are equivalent to a mass of $ 2$ at $ M$.  So our center of mass is $ P$.  The two masses at $ B$ and $ C$ are equivalent to a mass of $ 5$ at $ \frac {A \plus{} 4C}{5}$.  So $ X$ must be $ \frac {A \plus{} 4C}{5}$.


geometry
circumcircle
cyclic quadrilateral
geometry unsolved

A cyclic quadrilateral $ABCD$ is given. Let's denote with $Y$ the intersection of the diagonals $AC$ and $BD$. The line through $Y$, perpendicular to $OY$ (where $O$ is the circumcenter of $ABCD$), intersects $AD$ and $BC$ at $X$ and $Z$.
Show that $|XY|=|YZ|$.

It's a well known theorem; see

[url]http://www.mathlinks.ro/Forum/topic-18781[/url]

have u ever seen the butterfly theorem?
this theorem is exactly the same as ur question.

here is another solution for ur question, by using pole and polar and harmonic points:

theorem:
(ABCD)=-1 and O is not on line AB. we draw aline from B wich is parallel to OA and intersects OC and OD at points P and Q respectivly.
then we have: PB=BQ
i think it doesnt need proof.

by using it ,the problem wiill easily be solved:

let CB and AD meet each other at point K.
again, BA and CD at point M. we know that KM is parallel to XZ .because KM is the polar of point Y.
and again we know K(MAYB)=-1 (again its a theorem)
and by using the theorem that i told u it will be proven. :D

Solution:
From point  O draw perpendiculars to BC and AD.
Feet of the perpendiculars are F,G respectively.
We know AG=GD,BF=FC,∆AYD≈∆BYC so ∆AYG≈∆BYF→∠AGY=∠BFY
Also YOFZ and YOGX are cyclic quadrilaterals→∠ AGY=∠XOY , ∠BFY=∠ZOY   so ∆XOZ is an Isosceles triangle and YZ=XY


function logarithms calculus derivative algebra polynomial limit

Is there an elementary solution to the differential equation $\displaystyle \frac{d^2}{dx^2} f(x) = f(\sqrt{x})$, $x > 0$? And are there any general methods for solving such a problem?

From what you wrote, I assume that you require your solution to be a $C^2$ and, therefore, $C^\infty$ function on the entire positive semiaxis. Then you have a two-dimensional family of solutions. These solutions are nice real-analytic functions (if you make change of variable $y=\ln x$, they become entire functions of $y$) but I don't think they are elementary functions (except the "trivial" solution $f=0$), though I do not know how to formally prove it. (BTW, I would really appreciate it if someone shows me how to prove that, for example,  $F(z)=\sum_{n\geq 1}\frac {z^n}{n^n}$ is not an elementary function).

As to the methods of solving, you may consider the initial value problem  with given $f(1)$ and $f'(1)$ and use Picard iterations to show that it has a unique solution on sufficiently short interval around $1$. Then this existence and uniqueness will automatically extend to the entire $(0,+\infty)$. You can then use estimates of derivatives to show that the solution is real-analytic near $1$. Again, real analyticity will then automatically extend to $(0,+\infty)$. Actually, with some extra care, you'll be able to show that the solution is an entire function of $\ln x$. (or you can make the change of variable $y=\ln x$ and use the power series method to show it). You can even use the Laplace transform to get the solution. So, whatever you learned about the second order linear ODE can be used here with some modifications. Just don't be afraid to try it. :)

I finally found a proof that the 2-dimensional space of $C^2(0,+\infty)$ solutions to the equation $f''=f(\sqrt x)$ contains no elementary functions :lol:. So the answer to [b]gauss202[/b]'s first question is, indeed, "No". The main idea is that the derivatives of any elementary function are algebraicly dependent, while the derivatives of the solutions of our equation are algebraicly independent over $\mathbb C$. If I have time, I'll post a more detailed proof tomorrow or on Wednesday  (unless somebody (say, [b]grobber[/b], [b]harazi[/b] or [b]Leva1980[/b] or...) beats me to it :P )

[quote="fedja"] If I have time, I'll post a more detailed proof tomorrow or on Wednesday [/quote]
Here goes.

First of all, note that any elementary function belongs to some differential field that is an extension of $\mathbb C$ of finite transcendence degree. So, if $f$ is an elementary function, then, for sufficiently large $m$, the functions $f$, $f'$, $f''$, ... , $f^{(m)}$ are algebraicly dependent over $\mathbb C$, i.e., there exists a non-zero polynomial $P(x_0,\dots, x_m)$ of $m+1$ variables with complex coefficients such that $P(f,\dots, f^{(m)})=0$. So, to prove that the solutions to our equation are not elementary, it suffices to show that their derivatives are algebraicly independent over $\mathbb C$.

The next step is to find the general formula for the solutions. It is not hard because the most naiive version of the Picard iteration scheme works here. (you should run the classical uniqueness proof to show that there are no other solutions though). To make the long story short, just notice that the series $\sum_{j\geqslant 0} a_j x^{p_j}$ is a solution if $p_0=0\text{ or }1$, $a_0=1$, $p_{j}=\frac{p_{j-1}}2+2$  and $a_{j}=\frac{1}{p_j(p_j-1)}a_{j-1}$. In this case the second derivative of each next term is the previous term evaluated at $\sqrt x$ and the second derivative of the term corresponding to $j=0$ is $0$. The powers $p_j$ are easy to find: $p_j=4-2^{-j}(4-p_0)\to 4$. The coefficients $a_j$ are about $12^{-j}$ then, so the series converges uniformly together with all its derivative series on compact subsets of $(0,+\infty)$. Clearly, the functions corresponding to $p_0=0$ and $p_0=1$ are linearly independent, so all possible solutions are just linear combinations of these two.

The third step is to observe that, if the function $f(x)$ is elementary, then so is the function $e^{4z}f(e^{-z})$. Thus, it suffices to show that a series of the kind $F(z)=\sum_{j\geqslant 0}a_j e^{q_j z}$, where all $a_j\neq 0$ and the spectrum $q_j$ considts either of the numbers of the kind $4\cdot 2^{-\ell}$ or the kind $3\cdot 2^{-\ell}$ or both arranged in the decreasing order, has algebraicly independent derivatives (I switched from powers to exponents because the exponential functions are a bit easier to differentiate). We need the following calssical lemma:

If $\Lambda\subset\mathbb R$ is a countable bounded set, $\sum_{\lambda\in\Lambda} |a_\lambda|<+\infty$, $\sum_{\lambda\in\Lambda} |b_\lambda|<+\infty$ and    $\sum_{\lambda\in\Lambda} a_\lambda e^{\lambda z}=\sum_{\lambda\in\Lambda} b_\lambda e^{\lambda z}$ for all $z\in\mathbb C$, then $a_\lambda=b_\lambda$ for all $\lambda\in\Lambda$.

To prove the lemma, it suffices to note that $a_\gamma=\lim_{T\to+\infty}\frac1{2T}\int_{-T}^T e^{-i\gamma t}\sum_{\lambda\in\Lambda} a_\lambda e^{i \lambda t}\,dt$.

Now note that $F{(\ell)}(z)=\sum_j a_j p_j^\ell e^{p_j z}$. Assume that the first several derivatives of $F$ are algebraicly dependent, i.e., a sum of several terms of the kind $c_\ell f^{(\ell_1)}\dots f^{(\ell_m)}$, where $\ell=(\ell_1,\dots,\ell_m)$ is a multiindex with $0\leqslant \ell_1\leqslant\ldots\leqslant \ell_m$ ans $c_\ell\in\mathbb C$, is identically $0$. Our task is to show that then all $c_\ell=0$. Note that each such term is again an absolutely convergent sum of exponents with spectrum consisting of all possible sums of $m$ numbers $q_j$. Let $M$ be the greatest length of the multiindex $\ell$ for which $c_\ell\neq 0$. Using the lemma, we can then make the following observation:

If $Q\in\mathbb R$ can be represented as a sum of not more than $M$ elements $q_j$ in the unique (up to the order of terms) way and this way includes exactly $M$ terms with pairwise different indices $j_1,\dots,j_M$, then the polynomial $S(y_1,\dots,y_M)$ that is equal to the symmetrized sum of $c_\ell y_1^{\ell_1}\dots y_M^{\ell_M}$ taken over the multiindices $\ell$ of length $M$ vanishes on the $M$-tuple $q_{j_1},\dots,q_{j_M}$ (symmetrization here means that we take the sum over all $M!$ rearrangements of $y_1,\dots, y_M$).

It is relatively clear that if not all $c_\ell$ are $0$, then $S$ is not identically $0$. On the other hand, it is not hard to show (I leave it as an exercise to the reader) that, if $J$ is large enough (how large depends on $M$ only), then $Q=q_{j_1}+\dots+q_{j_M}$ satisfies the unique representation property above, provided that the difference between every two $j$'s is at least $J$. But then we can choose an arbitrarily long interval from which to take $j_1$, skip $J$ indices after this interval, choose an arbitrarily long interval from where to take $j_2$, and so on, and for each such choice we will have $S(q_{j_1},\dots,q_{j_M})=0$, which means that $S$ vanishes on a direct product of finite sets of arbitrarily large cardinality. Then $S$ must vanish identically and we obtain a contradiction.

As you may see, I left some details out. Hope that the readers will be able to recover them. As usual, any questions are welcome  :)

P.S. Sorry for posting three times in a row in one thread...:blush:


[quote="7h3.D3m0n.117"]Well I have a HUGE chess tournament on the 31st of March and I need an offensive black opening (I play Sicilian but I find that it is far too defensive and eventually I'll be forced to make an odd pawn move that will cost me the game).

My friend recommends a neat Petroff trick but he says higher ranked players (which I will undoubtedly play) will not fall for it. Petroff is an offensive black opening but it does rely a little to heavily on Nf3 after e4 for white.

I do not know any further lines to the Petroff (does anybody have a link or good site?) so that is of no use (making it up as a I go down a pawn and maybe doubled pawns will be ineffective).

I found the Barnes Defense- Fried Fox variation to be...a little unusual but I played my TD once and he creamed me with it with sound defense and a solid attack.

Any ideas? I will try all ideas online, and if anyone has a Yahoo account, maybe we can do an instructional game (where we can take back and most importantly, chat). I have a FICS account but chat is impossible (unless by AIM) and therefore not good for instructional games.

Thanks.[/quote]

If you have a US Chess Live account, we could play an instructive chatting game. It's free. PM me if you can get on. What's your official rating?

[quote="math92"]I find that I'm using Caro-Kann alot.  I really like it a lot and its a good opening you should learn.  A little on the less-offensive side in my opinion is Kings Indian Defense.  Learn those two and you should do fine.  :)[/quote]

To me, the Caro-Kann is the [i]most[/i] flexible opening for black . It can be played against e4 or d4, or even c4 with a little planning. But other than 1. e4 c6 2. d4 d5 the rest is pretty much exchange and develop. I'm not entirely crazy about the fact that the b8 knight has no decent place to move and will be a disadvantage. I love however, the number of possible pawn breaks the Caro-Kann produces  :lol: . I will ask my instructor about this the next time I meet him (possibly tommorow).

Kings Indian Defense has a bunch of holes in it. For one, the fianchettoed bishop can be attacked by a bishop diagonal with the queen for white, and the center is promptly in White's control. If black does not act immediately after castling to take back the center, the game is lost. It is, however, a usuful counter (other variation: The Grunfeld) against d4.

@goldendomer: The US Chess Live registration is messed up for me. I am a member of the USCF but my registration won't go through  :(

OK well the French defense is a quite solid opening for black, but if you really want to still have attacking or offensive possibilities, just stick to Sicilian?  :maybe:  That's pretty much the best you can do as black, and basically the trick is to not make that odd pawn move that loses ;). It's still pretty hard to play against white's best lines, but if white ever lets up you usually have good chances.

[quote="7h3.D3m0n.117"]French is solid but it depends on white making e4 and d4 on both his first moves and there are many variations for just one of white's wierd moves (like 2 nf3)[/quote]

Umm.. ok wtf any opening depends on white making certain moves, so I mean there's really no getting around that. If you know an opening well enough, then you should do OK against whatever white does (at your level... I'm guessing you're around 1300?).

sicilian is NOT too defensive.

It maintains a good balance of attack and defense. Black CANNOT play offense right off the bat. Lines that are black offense are pretty much unsound.

my favorite is sicillian by far, (english opening if you're white is also strong).

But then again, if you're black, you should probably be looking into counter attacking (especially with sicillian).
French defense is also solid, but is not one of my favorites becuase it will give up some central space and moreover it will block the developement of your bishop; but if you are a patient player; it is a very effective way of playing and it give you many opportunities to put traps and gain control of the center.

[quote="Captain Sugar"]
white       black
1. e4    doesn't matter
2. Bc4  doesn't matter
3.Bxf7
[/quote]
ive seen players do this in like 5 minute games, but its not that effective especially with queens on the board. :wink:
(not to mention you're down a piece).

[quote="Captain Sugar"]i'm another fan of the caro-kann defense.
however, if you're white, you can play the risky
white       black
1. e4    doesn't matter
2. Bc4  doesn't matter
3.Bxf7

now the king can't castle.
It's unexpected, but if it doesn't work, then you're in development trouble.[/quote]
This will cause you to lose against any player rated more than 1300.

The Sicilian I'm working with is the Sveshikov (might have spelled it wrong but it is also called the Pelican). Its semi offensive with an f5 pawn push (supposedly after black castles) to break up white in the center. However, certain white counter-plays can leave double f pawns a missing g pawn on black's kingside, not the best defensive position I would like. Actually my rating is 1100 but I never get out much (the highest I've beaten was a 1400 but that was pure luck...I only made 2 good moves the entire game-he just did stupid stuff).

EDIT:

[b]Concerns:[/b]
After white and black develop our pieces, and exchange a little, leaving few pawn breaks available, I will have no useful move in the Sicilian, either I move a pawn (which causes me to lose because of the holes implemented) or I move a minor piece (which I will probably regret later) and I lose tempo. What should I do?

Also: my club members have played the following counter against my Sicilian, and even though I'm playing at my best, they still manage to beat me...it's an unorthodox and useless counter and theoretically I should be able to win but..

1. e4 c5
2. d4!

obviously I take the pawn, leaving his queen in the middle, so it can be chased away by my knight. Therefore, White loses tempo, is down development, and I should have a chance to win...

[quote="7h3.D3m0n.117"]

obviously I take the pawn, leaving his queen in the middle, so it can be chased away by my knight. Therefore, White loses tempo, is down development, and I should have a chance to win...[/quote]
exactly...thats the entire point of the Sicilian, to trade the c pawn for the central d pawn.
if you end up losing the game it wont be because of this move, but rather because of later mistakes.

if you're playing against Sicilian i highly discourage you to play 2.d4
instead stick to the classic variation:
1.e4 c5
2.Nf3 d6
3.d4 cxd4
4.Nxd4 Nf6
5.Nc3 g6
now, white is doing perfectly fine; and the knight and the pawn in the center are quite strong.

[quote="7h3.D3m0n.117"]1. e4 c5
2. d4!

obviously I take the pawn, leaving his queen in the middle, so it can be chased away by my knight. Therefore, White loses tempo, is down development, and I should have a chance to win...[/quote]

Actually this is a legitimate variation (Smith-Morra), and one of the more annoying ones that white can play. However, I think it has been (relatively) recently refuted, but it would take quite a bit of studying to learn how. Why don't you just play something more normal (Schevenigen, perhaps?) that uses more fundamentals and requires less subtlety? If you want to be more aggressive, maybe try accelerated dragon.

For the problem of having no "useful moves," the answer is to come up with a plan. Think through a sequence of 4-5 potential moves to accomplish a specific goal that will give you an advantage. In the middle game, often single stand-alone moves are never "useful" per se. You have to think of something to do instead of just moving randomly.

[quote="probability1.01"]Actually this is a legitimate variation (Smith-Morra), and one of the more annoying ones that white can play. However, I think it has been (relatively) recently refuted, but it would take quite a bit of studying to learn how. Why don't you just play something more normal (Schevenigen, perhaps?) that uses more fundamentals and requires less subtlety? If you want to be more aggressive, maybe try accelerated dragon.

For the problem of having no "useful moves," the answer is to come up with a plan. Think through a sequence of 4-5 potential moves to accomplish a specific goal that will give you an advantage. In the middle game, often single stand-alone moves are never "useful" per se. You have to think of something to do instead of just moving randomly.[/quote]

I do know the Schevenigen and it is as solid defense (a little [i]too[/i] defensive, you if ask me). It is hard to resolve to an offensive plan with the Schevenigen because of the two pawns that block the bishops.

I think my main mistake with the Schevenigen (and sometimes Sveshikov or Najdorf) I would push a6, then b5, for that Bb7 fianchetto. Then I would maneuver my knight to d7 and then swing the rook to b8. Of course this takes a while and all the Sicilian counters I've played against White always castles kingside...making my fianchettoed bishop even stronger. Of course with the a6 and b5 pawns hanging around there are a lot of holes in my queen side, and with proper planning, white's queen can end up seizing free pawns before I can have a chance to respond. Also, I always end up castling too late (I hate castling too early because of loss of tempo and there is no theat to my king at that moment) and White can check me in ways that would make me lose material (a couple pawns or so, but that still is a lot).

try the Scicilian modern variation
1.c5
2.Nf6
3.g6
4.Bg7
5.e6
6.Ne7
7.o-o or d5

[quote="7h3.D3m0n.117"]The Sicilian I'm working with is the Sveshikov (might have spelled it wrong but it is also called the Pelican). Its semi offensive with an f5 pawn push (supposedly after black castles) to break up white in the center. However, certain white counter-plays can leave double f pawns a missing g pawn on black's kingside, not the best defensive position I would like. Actually my rating is 1100 but I never get out much (the highest I've beaten was a 1400 but that was pure luck...I only made 2 good moves the entire game-he just did stupid stuff).
[/quote]

In my opinion, the Sveshnikov is too deep for someone your rating (this is by no means an insult). A few years ago, the Sveshnikov was my absolute favorite after Leko started pushing it; however, there are many variations where black ends up in the drain or where white can easily come out with a better position. Simply put, a simpler more solid opening line would do better for you, in my opinion, because you would be able to focus more on middle game rather than openings which is where you should be moving (I could be completely wrong, but it's the norm).
I myself found myself in cramped, passive, and destroyed positions when I was playing the Sveshnikov. The two reasons:
1. My opening theory in the line wasn't deep enough (after b5 there are about 10 different variations eaching branching off exponentially) that you would need to study to gurantee that you don't come out of the opening losing.
2. My middlegame knowledge was weak which in turn led to the collapse of my game after my opening knowledge ran out.

If you really like the Sveshnikov and have spent quite some time with it, it can be quite a powerful tool (WIM Vicary completely crushed me recently with it), but from what you said about finding yourself on the defending side it seems like there are still a few lines you don't know.

I'm a Sicilian player as well (I've tried just about every variation  :P, but I think I finally found one that I like). I can't say the exact line (I have the K-9 Nationals too), but it's a Najdorf.

In regards to the Morra Gambit, taking the pawn on d4 followed by c3 is absolutely fine and yields a nice advantage for black if you're familiar with the lines.

[quote="GrantCBHS"]In regards to the Morra Gambit, taking the pawn on d4 followed by c3 is absolutely fine and yields a nice advantage for black if you're familiar with the lines.[/quote]

But instead of c3 white takes with the queen...(in my experiance against other people, anyway)

[quote="GrantCBHS"]In my opinion, the Sveshnikov is too deep for someone your rating (this is by no means an insult).[/quote]

Yeah my 1900 friend recommended I learn this. But I wanted an [i]offensive[/i] black opening, preferably a Sicilian, so the Sveshikov seemed to fit (with the f5 pawn push-supported by the pawn on d5).

Well, the tournament's over. I'd like to thank everyone for their kind help. I managed to get 3rd place :winner_third:  with 4 out of 5 wins (I should have gotten at least 2nd because I played a 1500 with a unorthodox variation of Nimzo-Indian :wacko: against his 1. d4). My friend who got 2nd actually drew the same 1500 I played so that's why he got 2nd...grr. It's okay actually. I knew I was going to lose against the 1500.


1)Though I have assigned a signature it does not show up in my posts. Why is this?

2)When I try to check all the posts by me (_MM_) I get several results(140) with some other authors ( like hmmm ) posts combined with mine. Is this because we have similar usernames?

Please help......

for the signature part, before you post a mesage there are 4 checkboxes above the submit button. one of these says attach signature. click on that and your signature will be shown

It appears that the _ characters in your username are interpreted as wildcards, so searching for your username brings up anything with mm somewhere in the middle.

So how can I change my username apart from registering again, in order to avoid this confusion ?
Thanks for the help.

Searching on your username should now work properly.

Thanks a lot.

I actually have 93 posts ( you can search on my username to check)
But my post count had got stuck on 78 and only resumed recently and thus shows only 82 posts. ( maybe due to the above problem)

Could this be fixed.. :)

Some of your posts are in forums that don't count, like FF. Not a bug.

Thanks didn't know that.
That's quite fair cause one can get loads of posts at FF which don't really count and are meant only for fun


[quote="can_hang2007"]Let $ a,$ $ b,$ $ c,$ $ d$ be nonnegative real numbers such that $ a\plus{}b\plus{}c\plus{}d\equal{}3.$ Prove that

$ \frac{ab}{4\minus{}a}\plus{}\frac{bc}{4\minus{}b}\plus{}\frac{cd}{4\minus{}c}\plus{}\frac{da}{4\minus{}d} \le 1.$[/quote]
$A=\sum \frac{ab}{4-a}=\sum \frac{3ab}{a+4b+4c+4d} \le \sum \frac{3ab}{a+4b+2c+2d}$
Now
$\sum \frac1{a+2c+2(2b+d)} \le \sum \frac19\left(\frac1{a+2c}+\frac2{2b+d}\right)$
So,
\[A \le \frac13\left(\sum \frac{ab}{a+2c}+\sum \frac{2bc}{2c+a}\right)=\frac{a+b+c+d}3=1\]
and equality holds when $a=2,b=1, c=d=0$ and its cyclic permutations. $\blacksquare$


calculus
integration
trigonometry
calculus computations

Couplling with $ \int_0^{\frac {\pi}{4}} x\sin ^ 2 x\ dx$

First hint: use a trig identity to get $ \cos 2x$ before you integrate by parts.

Items for plausibility checking: the integrand is positive, so the integral has to be positive. How big is that integrand? It's $ 0$ at one end of the interval and $ \frac {\pi}{8}$ at the other end; but that's not the max. So the average value - maybe $ \frac {\pi}{10}?$ maybe a little bigger? An average a little around $ .3,$ a length of interval a little under $ .8$ ... so maybe an answer around $ .2$ or $ .3.$

[hide="Yes, that estimate is right"]$ \approxeq 0.22556211$[/hide]

Thanks. I also integrated by parts to get

$ [\frac{x^{2}}{4} \plus{} \frac{x \sin 2x}{4}\plus{} \frac{\cos 2x}{8}]_{0}^{\frac{\pi}{4}}$,

that is,

$ \frac{\pi^{2}}{64} \plus{} \frac{\pi}{16} \minus{} \frac{1}{8}$ which is approximately $ 0.225562$.

Any ideas on using kunny's hint?

Hello !!

By using kunny hint ,

Let  $ I \equal{} \int_{0}^{\frac{\pi}{4}} x \cos^{2} x\ dx$   and $ P \equal{} \int_0^{\frac {\pi}{4}} x\sin ^ 2 x\ dx$

$ I\plus{}P\equal{}  \int_{0}^{\frac{\pi}{4}} x \equal{} \frac{\pi^2}{32}$

And $ I\minus{}P \equal{}  \int_{0}^{\frac{\pi}{4}} xcos(2x) \equal{} \frac{\pi}{8} \minus{} \frac{1}{4}$

Therefor , $ I\equal{} \frac{(I\plus{}P)\plus{}(I\minus{}P)}{2}\equal{}\frac{\pi^2}{64}\plus{}\frac{\pi}{16} \minus{} \frac{1}{8}$


Find all $ n\in N$ such that there exist $ n^2\plus{}n\plus{}1$ subsets consisting of numbers $ 1,2,3,...,n$ so that

(a) Each of the numbers appears in $ n\plus{}1$ subsets
(b) Each subset consists of $ n\plus{}1$ numbers
(c) Each pair of numbers belong to exactly one subset.

This is asking for a finite [url=http://mathworld.wolfram.com/ProjectivePlane.html]projective plane[/url] of order $ n$.  It's known to exist for $ n$ a power of a prime, and conjectured not to exist for any other $ n$, but this is open for $ n \geq 12$.

Wow, thanks a lot.

Can you show the proof for $ n$ power of a prime, or at least $ n$ a prime?  :)

I think one of these should be helpful:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16699
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14949
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=36885


Question 1:
A system consists of three particles, each of mass 6.00 g, located at the corners of an equilateral triangle with sides of 31.0 cm.
(a) Calculate the potential energy of the system. (Assume that U = 0 as r goes to infinity.)

Question 2:

A comet of mass $ 1.20*10^10$ kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between 0.600 AU and 57.0 AU. (Note: 1 AU = one astronomical unit = the average distance from the Sun to Earth = $ 1.496*10^11$ m.)

(a) What is the eccentricity of its orbit?

(b) What is its period?

(c) At aphelion what is the potential energy of the comet-Sun system?

1)The potential energy will have magnitude

$ F=\frac {3Gm^2}{a}$  where a is the side of the triangle.Put the values and you will get the answer.

2)
a)From the general properties of an ellipse,

$ e=\sqrt{1-\frac{b^2}{a^2}}$

Put  $ a=57*1.496*10^{11} m$ and $ b=.6*.496*10^{11} m$

b)Use

$ T=\sqrt{\frac{4{\pi}^2b^3}{GM}}$

c)At aphelion,$ a=57*1.496*10^{11} m$

$ PE=\frac{GMm}{a}$  where M is the mass of the sun and m is mass of comet


geometry
algebra
polynomial
probability
logarithms
function
complex numbers

I need to study for the AMC this weekend, and I know most of the bacis stuff. What would be the 4 most useful chapters in AOPS to prepare me for the AMC? :?

Here are my opinions:

From Volume 1 I think that:

Using the Integers (5)
Triangles a.k.a. Geometry (11)
A Potpourri of Geometry (20)
Learning to Count (25)

are important.  Focus on doing the problems.

From Volume 2:

Not Just for Right Triangles (2)
More Triangles (3)
Polynomials (6)
Equations and Expressions (13)
Combinatorics (15)

Ok, so that's 5 but chapter's 2 and 3 kinda go together.  Since you are studying for AMC don't try any of the olympiad problems.  If you had more time I would recommend them but with only a few days I think you could spend your time a little bit more wisely.

Here are other chapters that I would also consider looking over:

Probability (19)
Logarithms (1)
Complex Numbers (9)
Functions (7)
Sequences and Series (16)

(those aren't in any order expect for the order that I wrote them) ;)


without using a computer or tables or things like that show that tan(50)>1,18.

thank you!!


analytic geometry
modular arithmetic
geometry unsolved
geometry

Show that among any five points $ P_1,...,P_5$ with integer coordinates in the plane, there exists at least one pair $ (P_i,P_j)$, with $ i \not\equal{} j$ such that the segment $ P_i P_j$ contains a point $ Q$ with integer coordinates other than $ P_i, P_j$.

[quote="moldovan"]Show that among any five points $ P_1,...,P_5$ with integer coordinates in the plane, there exists at least one pair $ (P_i,P_j)$, with $ i \not \equal{} j$ such that the segment $ P_i P_j$ contains a point $ Q$ with integer coordinates other than $ P_i, P_j$.[/quote]

Let $ P_i\equal{}(x_i,y_i)$

By Pigeon-hole principle, there exist at least a pair $ (P_i, P_j)$ so that $ x_i \equiv x_j \pmod{2}$, $ y_i \equiv y_j \pmod{2}$.

So $ Q\equal{}(\frac{x_i\plus{}x_j}{2}, \frac{y_i\plus{}y_j}{2})$ is on the segment $ P_iP_j$ and is of integer coordinates.


How do you express $ a^3 \plus{} b^3 \plus{} c^3$ in symmetric sums?

Another way to ask this question is
If you know the values of $ a\plus{}b\plus{}c$, $ ab \plus{} ac \plus{} bc$, and $ abc$, how can you find the value of $ a^3 \plus{}b^3\plus{}c^3$?

I found my answer to be
$ a^3\plus{}b^3\plus{}c^3\equal{}(a\plus{}b\plus{}c)^3\minus{}(a\plus{}b\plus{}c)(ab\plus{}ac\plus{}bc)\plus{}3abc$
but when I implemented this into a problem I came up with the wrong answer. :maybe:

[quote="l Bob l"]
$ a^3 \plus{} b^3 \plus{} c^3 \equal{} (a \plus{} b \plus{} c)^3 \minus{} \boxed{3}(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc) \plus{} 3abc$
 [/quote]

Ok I've got it. Thanks for the help.
This thread can be closed.


What is with the problems in which a ball will bounce? I never seemed to get those very well.  :huh:

umm i don't get it what do you mean?could you give example? :oops: i bet this is going to be way easier for a bunch of people.

Are you talking about those where you bounce a ball, it bounces to some height, then bounces a fraction of the previous height? That's what I think of when I hear "bouncing ball" problems.

You mean like this?

Question. A ball is dropped from a height of $100$ feet. At each bounce it rises to two-thirds of its previous height. If a complete bounce includes the maximum height the ball reaches after its last bounce, how far will it have traveled if it bounces
a) infinite times?
b) three times?
c) $n$ times?

=geometric series sum :D

oh okay then i guess you might not know how to sum these series:
 well for an infinite series
a/1-r if |r|<1
where a is first term and r is the common ratio
finite series,
S_n=a(1-r^n)/(1-r)
where a and r are same as above and n is the number of "bounces" in this problem

[quote="funcia"]so do those formulae help? or did you already get this but you just have adding problems..[/quote]

Just think of it as a geometric series.

i think I just need some practice, I'm kind of slow at doing these problems


One-fourth length of a uniform rod of mass m and length l is placed on a rough horizontal surface and it is held stationary in horizontal position by means of a light thread. The thread is then burnt and the rod starts rotating about the edge. Find the angle between the rod and the horizontal when it is about to slide on the edge. The coefficient of friction between the rod and the surface is $ \mu$.

For diagram, refer http://www.flickr.com/photos/30014811@N06/3502744339/

Suggestion:

Right click on your image, select image location and use the img tags to insert your image.

Like this:

[img]http://farm4.static.flickr.com/3618/3502744339_203e79a4f9.jpg[/img]

Thanks! Any ideas about how to go about the problem?

I would do the following.

The moment of inertia of the rod about the pivot point is $ I_P \equal{} 7ml^2/48$, while its moment of inertia about the center of mass is $ I_{CM} \equal{} ml^2/12$.  Imagine the rod is tilted at an angle $ \theta$ w.r.t. the horizontal.  If you draw an F.B.D., there are three forces on the rod: $ mg$ (downward and acting on the center of mass), $ F_\perp$ (perpendicular to the rod and acting on the pivot point outward), and $ F_{fric}$ (parallel to the rod and pointing to the left, but also upward at an angle of $ \theta$).  Find the torque due to gravity about the pivot point, and hence the angular acceleration as a function of $ \theta$:
\[ \frac{1}{4}mgl\cos\theta \equal{} I_P \alpha(\theta) \implies \alpha(\theta) \equal{} \frac{12g}{7l}\cos\theta.
\]
Now find the torque due to $ F_\perp$ about the CM:
\[ F_\perp \frac{l}{4} \equal{} I_{CM} \alpha(\theta) \implies F_\perp \equal{} \frac{4}{7}mg\cos\theta.
\]
Hence $ F_{fric} \equal{} \frac{4}{7}mg\mu\cos\theta$.  At this point, I believe there are two ways of proceeding.  You could assume that $ \omega(\theta)$ is negligible, so you don't have to worry about the centripetal force that $ F_{fric}$ provides.  In that case, simply impose the condition
\[ F_{fric} \geq mg\sin\theta \implies \frac{4}{7}\mu \geq \tan\theta,
\]
which gives you $ \theta_{max}$.  However, you'd ideally find $ \omega(\theta)$ (and I don't see how to solve the required D.E.) and impose the condition
\[ F_{fric} \minus{} mg\sin\theta \geq m\omega(\theta)^2 \frac{l}{4}
\]
to take into account the centripetal force.

The answer is:
[hide]$ \theta \equal{} \arctan \left(\frac {4\mu}{13}\right)$[/hide]
That doesn't come by neglecting $ \omega(\theta)$, does it?So what to do?

[quote="genericme"]
The moment of inertia of the rod about the pivot point is $ I_P \equal{} 7ml^2/48$,

[/quote]

I got $ \frac{19ml^2}{48}$

[quote="genericme"]
...there are three forces on the rod: $ mg$ (downward and acting on the center of mass), $ F_\perp$ (perpendicular to the rod and acting on the pivot point outward), and $ F_{fric}$ (parallel to the rod and pointing to the left, but also upward at an angle of $ \theta$).

[/quote]

That is what I did at first but I was confused about that. The normal reaction (your $ F_\perp$) would be opposite in direction and equal in magnitude of the resutant of $ mg$ and $ m\omega^2r$ directed along the rod, outwards when the rod is inclined at $ \minus{}\theta$ w.r.t horizontal while $ F_{friction} \perp N$.

By conservation of energy,
$ mg\frac{l}{4}\sin\theta \equal{} \frac{1}{2}I\omega^2$

Ha, I hadn't even thought of using conservation of energy.  I guess my physics is getting rusty.

But anyway, $ I_P \equal{} ml^2/12 \plus{} m(l/4)^2 \equal{} 7ml^2/48$.  And using what you suggested, Rajiv, it turns out that if we consider the centripetal force, we get MNrule's answer:
\[ mg\frac{l}{4}\sin\theta \equal{} \frac{1}{2} I_P \omega^2 \implies \omega^2 \equal{} \frac{24g}{7l}\sin\theta,
\]
and
\[ F_{fric} \minus{} mg\sin\theta \geq m\omega^2\frac{l}{4} \implies \frac{4}{7}mg\mu\cos\theta \minus{} mg\sin\theta \geq \frac{6}{7}mg\sin\theta,
\]
or $ \arctan\left(\frac{4\mu}{13}\right) \geq \theta$.

No, genericme, its good that you didnt think of conservation of energy..., but started examining the real dynamics...

As i have emphasized in many posts, its not good to think of conservation of energy as a "fundamental" thing, but it is much better to understand the real dynamics of the system.., and then if the system has some symmetry, you can conclude to some conserved quantity..

[quote="genericme"]
But anyway, $ I_P \equal{} ml^2/12 \plus{} m(l/4)^2 \equal{} 7ml^2/48$

[/quote]

Sorry! I mesed up in a hurry! I guess my mathematics is getting rusty. :wink:

[quote="yeahnigga"]
No, genericme, its good that you didnt think of conservation of energy..., but started examining the real dynamics...

[/quote]

Do you have some personal enmity with Conservation of Energy principle? I agree that now, I'm not as educated as you are but you are confusing me. Could you please support your arument regarding the failure of energy conservation with an example?


When does the AMC release the cutoff level for qualifying fo the  AIME from performance on the  USAMTS?  Also, approximately what percentage of students who participate in the USAMTS qualify for the AIME?

I'm not sure of the answer to the latter, but I'll check with the AMC for the former.  I think the cutoff in the past has been in the high 60s after 3 rounds.

i'm quite sure the cutoff was about 70, i had a 68 and didn't qualify through USAMTS, though i did qualify through AMC  :)

[quote="ZennyK"]The cutoff score for the 2003-2004 school year was 69 after 3 rounds.[/quote]

...you sure about that?  I'm pretty sure it was 68.  Mariag, you wouldn't know if you qualified for AIME through USAMTS unless you didn't qualify via AMC, so you may have actually qualified through USAMTS.

it was 68 your "score" value translating into the AMC was (USAMTS score/75) * 100. Thus you would be given something between 90 and 100 on the AMC for your performance.

[quote="themonster"]That's kinda odd that your translated score from USAMTS to AMC could be lower than the AMC's cutoff score for qualification for the AIME, even though you qualify for the AIME through USAMTS.  It seems that your at a strong disadvantage to qualify for the USAMO if you qualify for the AIME through USAMTS.[/quote]

Well, yes, there was a distinct disadvantage.  (Although I think the cutoff for at least one AMC12 was below 100...)  The USAMTS --> AIME qualification is essentially a joke to put a feather in the USAMTS's cap, or perhaps to allow students whose schools didn't want to pay for AMC still make AIME.  I don't know.  Anyone serious about USAMO should take AMC, and several times, too.


calculus
integration
geometry
trapezoid
rectangle
real analysis
real analysis unsolved

A certain computer takes two seconds to compute a certain definite integral accurate to 4 digits to the right of the decimal point, using the left rectangle rule. How long (in years) will it take to get 8 digits correct using the left rectangle rule?? How about 12 digits? 20 digits?

The left rectangle rule has pretty lousy precision: the error does not exceed $C\frac{(b-a)^{2}}{n}\max |f'|$, where $C$ is some constant I don't care about. The main point here is that to decrease the error by the factor of $10^{4}$ you need to increase $n$ by the same factor. So, if $4$ digits take $2$ seconds, then $8$ digits will take $2\cdot 10^{4}$ seconds and so on...

If you used the trapezoidal rule instead, you'd have $n^{2}$ instead of $n$ in the denominator. This way, you decrease the error by $10^{4}$ when increasing $n$ just by $10^{2}$. And with Simpson's it's even better.

you mentioned the rule (b-a)^2/n, what formula is this? is this to calculate the error?? where did you get 10^4 from?

so if it's 12 digits then it's 3 x 10^4??

If you want to get $4$ more decimal digits, you must decrease the maximum error by $10^{4}$.

The error bounds with $(b-a)^{2}/n$, etc, can be found in a calculus textbook.

The computer will need $2\cdot 10^{8}$ seconds to find $12$ digits.


Wording has changed again.

Suppose there is 10 distinct points lie in the circle. If you connect some or all points to form a convex polygon, how many distinct polygons are possible?

[hide="Answer"]
968[/hide]

[hide="My_Sol"]
I liked this question since this is easy counting. To form polygon, the number of vertices we need is $\geq 3$ because lesser than 3 makes line or a point. And the maximum we can get is using all 10 points.

So, this is same as evaulating:

$\binom {10}{3} + \binom {10}{4} + \binom {10}{5} + \binom {10}{6} + \binom {10}{7} + \binom {10}{8} + \binom {10}{9} + \binom {10}{10}$

But $\binom {10}{3} = \binom {10}{7}$ and so forth, there aren't actually a lot of calculating we have to do.

And they come out quite nicely.

Evaluating them would give 120+210+252+210+120+45+10+1 = 968 as final answer.[/hide]

Just wondering, is there anyway I can do this without expanding and finding the each value of $\binom {10}{3}, \binom {10}{4}$, etc?

[quote="Silverfalcon"]Just wondering, is there anyway I can do this without expanding and finding the each value of $\binom {10}{3}, \binom {10}{4}$, etc?[/quote]

[hide]well, you could subtract $\binom {10}{2}, \binom {10}{1}, \binom {10}{0}$ from $2^{10}$ :)[/hide]

Yeah. Since we know that $\binom{n}{0} + \binom{n}{1} + \ldots + \binom{n}{n-1} + \binom{n}{n} = 2^n$, we can easily make the conclusion that alan made. The polygon can't have 0, 1, or 2 sides, so subtract $\binom{n}{0} + \binom{n}{1} + \binom{n}{2}$, and that will give you your answer.


induction
combinatorics unsolved
combinatorics

Given 98 points in a circle. Mary and Joseph play alternatively in the next way:

- Each one draw a segment joining two points that have not been  joined before.

The game ends when the 98 points have been used as end points of a segments at least once.  The winner is the person that draw the last segment. If Joseph starts the game, who can assure that is going to win the game.

I think Jose can always win the game.

Jose must only care to notice when there are only 3 unused points, he can almost do wahtever he wants with the other 95 and he will still be the winner.
From here we have two possibilities:

a)If it is Jose's turn when this situation gives, his next turn will join to used points.

b)If it is Maria's turn and she uses one of the 3 points, Jose wins. If she join to used points Jose returns to situation a).

This can continue untill all 95 points have been joined with each other, and as this need $ 95*47$ segments to be drawn, it will be Maria's turn when there are no more points left to joined within this 95 so we are in b) and Jose wins the game.

 :)

Unless both of us are mistaken, ElChapin gets this one right!

I think Joseph can always win. Here's my solution:
Let S be the number of points which are not used in a segment, at the beginning of the game S=n(where n is the number of given points, in this case 98). since none of the points have been "used" at the beggining Jose always decreases S by 2 (joins 2 points that have not been used) so, Mary has n-2 points and can decrease S by -1 or by -2 (-1 if she joins a point that has been used with another un-used and -2 if she joins 2 un-used points).
So we can turn this game into a simple Nim, with 1 heap of n-2 counters and , you can take 1 or 2 counters, where the second player begins. We know that 0 is a losing position to this game , so 1 is winning (just do -1 and lead to a losing position) 2 is also winning (do -2), 3 is losing because it leads to 2 or 1 which are winning positions. and so on... it is easy to prove by induction that 3k for some k is a losing position for the second player, so if n-2=3k (note that we started this game with n-2 counters) it is a losing position for the second player, so n=3k+2 , in this case n=98=3k+2 so it is losing for the second player, winning for the first one. A winning strategy would be: - if the second  player joins a segment with 2 unusued points , the first one joins a used point with one unused and viceversa.