Ring Theory
superior algebra
superior algebra unsolved

$\mathbb{K}$ is an infinite field of caracteristic 0.
Show that $\mbox{SL}_{n}(\mathbb{K})$ is not finitely generated.

Hint?[hide]I think I can show it for $\mathbb{K}=\mathbb{Q}$, but not sure it helps![/hide]

I think it's just the infinitude of $K$ that matters, not the characteristic.

If a linear group is generated by finitely many matrices $A_{1},\ldots,A_{t}$, then the entries of all the matrices in the group are elements of the $\mathbb Z$-algebra generated by the elements of $A_{i}$. In our case, we want this algebra to be the entire $K$. Consequently, $K$ is a finitely generated algebra over its prime field $F$.

There is a theorem (I know I read it in Atiyah and Macdonald once; there’s also a discussion on the forum, started by alekk, if I remember correctly, but no complete proof was posted) stating that if the field $K$ is a finitely generated algebra over a subfield $F$, then it is a finite algebraic extension of $F$. Applying this to our $K$ and $F$, we have two cases: either $F$ is finite, and this means that $K$ is a finite algebraic extension of a finite field and is thus finite, contradicting the hypothesis, or $F=\mathbb Q$, and $K$ is an algebraic number field. In this latter case, if $K$ is generated as a $\mathbb Z$-algebra by, say, $k_{1},\ldots,k_{u}$, then these elements are quotients of elements in the ring $A$ of integers of $K$. Let $\frak p$ be a prime ideal of $A$ which contains none of the denominators of $k_{i}$. If $a\in\frak p$, then $\frac{1}{a}$ cannot belong to $\mathbb Z[k_{1},\ldots,k_{u}]$, and this contradicts the fact that $K=\mathbb Z[k_{1},\ldots,k_{u}]$.


inequalities
combinatorics unsolved
combinatorics

Problem1 If graph$G_n$does not contain$K_{2,3}$,prove that the number of its edges is not greater than$\frac{n\sqrt{n}}{2}+n$

Have a look http://www.mathlinks.ro/Forum/viewtopic.php?highlight=graph&t=22391.

Ok.

Since there is no $K_{2,3}$, we know that there are at most 2 paths of length 2 from any vertex to any other vertex.

The number $p$ of such paths is $\sum_v d(v)(d(v) -1),$ as is seen by counting paths through each vertex

Then $p \le 2 {n \choose 2} = n(n-1) = n^2 - n.$

Therefore,
\[ \sum_v d(v)^2 - 2e = \sum_v (d(v)^2 - d(v)) \le n^2 - n, \]

where $e$ is the number of edges in the graph.

Now, by Cauchy-Schwarz (or your choice of inequality)

\[ \sum_v d(v)^2 \ge \frac{ (\sum_v d(v))^2 }{n} = \frac{4e^2}{n}. \]

So, we have

\[ 4e^2 - 2ne \le n (\sum_v d(v)^2 - 2e) \le n^3 - n^2, \]

which implies that $4e^2 - 2ne + \frac{n^2}4 \le n^3 - \frac{3n^2}{4} \le n^3.$

This is equivalent to

$(2e - \frac{n}{2})^2 \le n^3 \Rightarrow 2e \le n\sqrt{n} + \frac{n}{2} \Rightarrow e \le \frac{n\sqrt{n}}2 + \frac{n}4,$

which is a slightly stronger result than we wanted.

[quote="Nukular"]Ok.

Since there is no $K_{2,3}$, we know that there are at most 2 paths of length 2 from any vertex to any other vertex.

The number $p$ of such paths is $\sum_v d(v)(d(v) -1),$ as is seen by counting paths through each vertex

[/quote]

10x for the fast reply but I just wanna ask isn't the number of such paths $\frac{1}{2}\sum_v d(v)(d(v) -1),$ as in your way of counting you count each path twice.
Of course I might be wrong :)
Anyway, thank you for your solution :)

Ah, yes.

That explains why I the inequality was so strong. I'm pretty sure it follows through still, just pushing the appropriate constatns through correctly.

Let me make sure though.

Yes, I think that your solution is right, my way is almost the same as yours but I wanted to check any other ideas because I couldn't get the final answer :)
Anyway, 10x for the reply :)


trigonometry
calculus
probability
analytic geometry
graphing lines
slope
geometry

I will be starting a new science fair project next year and would like to try something realted to mathematics. Please post some ideas that you think would make an excellent project.

how to prove 1 = 0.99.....

good topic...
(if you got see some of the past topics about 0.9999...=1 as a reference, it would be very useful)

[quote="PenguinIntegral"]Fractals[/quote]

like what about them? i want to do one too  :)

Mathematicaly proving is complex
Disapproving is more complex

[quote="4everwise"]I need somethin more complex, like algebra 2 and trig complex[/quote]Based on this comment, I've moved this topic out of the Middle School Basics forum, over to Other Problem Solving topics.

If I can safely assume you are in middle school, I would suggest doing something about calculus.  You don't have to master all of it; just learn some applications.  It would be really cool to, say, model the cooling of something by differential equations if you're in middle school.  One specific aspect of calculus is not hard to learn (well, the same really can be said of all of calculus AB ;) ).

[quote="probability1.01"]It would be really cool to, say, model the cooling of something by differential equations if you're in middle school.[/quote]

Yeah, because most people learn differential equations in a couple weeks for a science project, not in a semester-long college-level course or anything  :?

Well, 4everwise does have all summer and about half of the school year to at least get somewhat familiar with the material... i guess it all depends on how much effort he is willing to invest in such a sci project, i spent 6 months prior learning all the prerequisite math in order to do mine... but that's because i am not as bright as you guys are naturally and have to try harder

[quote="4everwise"]I need somethin more complex, like algebra 2 and trig complex[/quote]

I'm only half kidding, but if you're serious, you can always look here:

http://www.springeronline.com/sgw/cda/frontpage/0,11855,5-10042-72-29396256-0,00.html
or
http://www.springeronline.com/sgw/cda/frontpage/0,11855,5-10042-72-34511272-0,00.html

Write a full solution and you're GUARANTEED an A.  :D

[quote="bubala"]My guess would be that probability was referring to the diffeq that you learn in calculus in about two weeks--just solving first-order homogeneous eqns, slope fields, etc.  I don't think he meant the college course, otherwise it would be impossible for 4everwise to do his project on it.[/quote]

Yeah, it turns out (I think) that temperature can be modeled by a simple differential equation something like y' = ky or something.

try something with number theory, and represent it using geometry
or you can try game theory, because thats always interesting


I need to verify this Inequality :

$ (\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}) \ge \frac {1}{3}(a \plus{} b \plus{} c)(\frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{a})$

Please someone prove / disprove this inequality.

And give me explanation about Chebyshev

Thanks

I think that you should start off by multiplying everything on the right side of the equation, but I am not sure.

The inequality

             $ (\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}) \ge \frac {1}{3}(a \plus{} b \plus{} c)(\frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{a})$

 does not hold for  $ a \equal{} 8,b \equal{} 3,c \equal{} 1.$

            If a, b, and c are the three sides of a triangle, then the inequality  holds.

i think the lhs should be $ \frac{a}c\plus{}\frac{b}b\plus{}\frac{c}a$.

Yes, it should be $ \frac{a}{c}\plus{}\frac{b}{b}\plus{}\frac{c}{a}$ because $ (a,b,c)$ and $ (\frac{1}{c},\frac{1}{b},\frac{1}{a})$ are similarly arranged.

Thank you very much..
I have many problems with LHS=$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$

Like this : $ abc \equal{} 1$
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\ge a \plus{} b \plus{} c$

Someone in Indonesia (maybe he's here) give a proof like this :  :D
$ LHS \ge \frac {1}{3}(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})\ge a \plus{} b \plus{} c$
Because $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\ge 3$

That's why I ask you this inequality..

[quote="Ronald Widjojo"]Thank you very much..
I have many problems with LHS=$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$

Like this : $ abc \equal{} 1$
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\ge a \plus{} b \plus{} c$[/quote]
You can solve it by using weighted AM-GM :)
Multiplying by $ (abc)$ on the left side and by $ (abc)^{2/3}$ right side we get:
$ \sum_{cyc} a^2c \ge \sum_{cyc} a^{5/3}b^{2/3}c^{2/3}$

But this is equivalent to:
$ \sum_{cyc} x \cdot a^2c \plus{} y \cdot b^2a \plus{} z \cdot c^2b \ge \sum_{cyc} a^{5/3}b^{2/3}c^{2/3}$
When $ x \plus{} y \plus{} z \equal{} 1$.

Using weighted AM-GM we get:
$ \sum_{cyc} x \cdot a^2c \plus{} y \cdot b^2a \plus{} z \cdot c^2b \ge \sum_{cyc} a^{2x\plus{}y}b^{2y\plus{}z}c^{2z\plus{}x}$
Solving the equation system:
$ x \plus{} y \plus{} z \equal{} 1$

$ 2x \plus{} y \equal{} \frac{5}{3}$

$ 2y \plus{} z \equal{} \frac{2}{3}$

$ 2z \plus{} x \equal{} \frac{2}{3}$

Which has the solutions:
$ x \equal{} \frac{2}{3}$, $ y \equal{} \frac{1}{3}$, $ z \equal{} 0$

And thus
$ \sum_{cyc} a^2c \equal{} \sum_{cyc} \frac{2}{3} \cdot a^2c \plus{} \frac{1}{3} \cdot b^2a \ge \sum_{cyc} a^{5/3}b^{2/3}c^{2/3}$

You can practice the method on this inequality:
$ \frac{a^{101}}{b^{100}} \plus{} \frac{b^{101}}{c^{100}} \plus{} \frac{c^{101}}{a^{100}} \ge a \plus{} b \plus{}c$

Case$ \sum a\geq\sum ab:$
$ \sum\frac{a}{b}\geq\frac{(\sum a)^2}{\sum ab}\geq\sum a$
Case$ \sum a\leq\sum ab:$
$ \sum(a^2c\plus{}c)\geq 2\sum ac\geq 2\sum a\Rightarrow\sum\frac{a}{b}\geq\sum a$


AMC
AIME
USA(J)MO
USAMO
AMC 12
MATHCOUNTS
probability

I got a 132.5 (20 / 20 answered.)  I wanted to know how this compares to the other sophomores that took the AMC 12A.  If poissible, can you post your score if you are a soph, or scores of other sophs you know?

Thanks a lot.  Good job to all who participated, and best of luck on the AIME.
JB

118.
The other sophomore at my school that took it got 101

[quote="jackb1115"][quote="math_u_matics"]I got a 98.5 on the AMC 12A...congratulations anyway, you're probably among the top sophomores in the nation. ;)[/quote]

Thanks
98.5 is a burn.  One of my friends ended up with it.  He thought he had made it at first with a 104.5, but then re figured out he got a prob wrong that he should have gotten right.

JB[/quote]

i know someone who got a 99.5 on his first AMC-12 and 100.5 on his second ;)  ;)

99.5 ~ 12 right, 11 skipped, 2 wrong
100.5 ~ 13 right, 9 skipped, 3 wrong

I know a sophomore who got a 99.5 and another sophomore who got a 114.5 on the AMC 10 after missing #2.  Ouch  :blush:

i barely got in with a 104 (14 correct, 3 wrong, 8 omit)

108.5

Way too many errors. Carelessness, really.

123.  19 answered, 1 wrong, 6 blank.  I missed the one about numbers that look prime.  :D

126.5.....i thought some of them were too easy to answer so I didn't take a risk...ex: 17, odometer question, and I made a super careless error on the one circle question...oh well, at least it doesn't really matter.

lol, although I did answer #20-23...haha...i wasn't afraid to go for those but yet I didn't answer #17 where they drew such a perfect diagram....haha and btw they draw all of the pictures to scale....after i solved the big circle tangent to the two small circles I checked by measuring and it was drawn to scale. :P  :P  :P

It's impossible for 16 not to be drawn to scale and still have those congruent circles and points of tangency. I have a friend who got the correct answer just by marking the length of the radius of the large circle and cutting it into small radii. But then, he got different answers when he tried again after the test...

104.5, one of only two sophomores from the first testing date who qualified for the AIME (the other went to MOSP).

If you take the total number of sophomores who took the AIME last year and divided it by 50, that's an average of 28 qualifiers by state.

99.5 :( on 12A. I was very disappointed... so I skipped volleyball practice to take the 12B, and got a 109. Incidentally, my coach is a math teacher, so he understood.

120.5 on the 12B.  Eh, up from a 90-something on the AMC-10 last year, anyways :)

I got a 129 on the AMC 12B. Didn't think it was terribly bad.

131 on the 12B. Screwed myself over by reading questions incorrectly. Should've been 137 at least. Oh well.

Kaimor --

You got 150...AS A SOPHOMORE??

I must congratulate you on a job well done...

By the way, I heard that anyone 10th grade or under who qualifies for AIME from the AMC12 automatically gets a "150" used as their AMC score for their USAMO index.  Is that true?

JB

I kind of doubt it; wouldn't that make it a bit unfair ?

Nice job Adam. But I know you, so its no surprise =P.

I've never heard of the automatic 150 thing, and I agree that it would be quite unfair. I still believe they use the floor value system.

[quote="Kamior"]150. That should make qualifying for the USAMO a bit easier...

-Adam Hesterberg[/quote]

Ah, no surprises there.. I know who you are, too. Congratulations!

No surprise you got 150...i think I remember u from mathcounts, you got 1st right?

Anyways i got 127.5 ................... (kicks myself for my incompetency at arithmetic)..............should have been 139.5.......

About the floor value thing, people in 10th grade and under get an automatic 150 on the AMC, if they qualify for the AIME. That way, they only need to get the floor value or higher to be able to qualify for USAMO.  I guess it's kind of unfair, but I think they do it to get more younger kids interested in math.

But getting an automatic 150 is not the same as getting the floor value. For example, if all 11/12th graders got 130 AMC12 and 8 AIME then the cutoff would be 210 index but the floor value would be 8. However, if 9/10th graders got an automatic 150, they would only need to get 6. So is it the floor value or an automatic 150? Because it may be different depending on which system they are using.

I got a 98.5  on AMC 12 and I'm a junior. I answered 12 (I know I'm dumb) but missed that money probability problem on 12B.

I answered 11 questions and found out one of them was wrong so I ended up with 95 =/

i got a 111.5,  on the 12A, but im a complete idiot and i didnt mark one of the problems, that i solved right, and then thought i did wrong, so i didnt take the chance, but of course i should have marked it- always go with your first choice. i guess i qualified- but i really want that perfect by the time im a senior... think i have a chance in a million?


1. 10
2.10
3.5th
4.yes
5.10
6. how old are you?
7.10
8. how do you do
9. good
10. good-bye

1. 14 (will be a soph next year)
2. 13 I think...
3. will be a soph next year, 10th grade
4. yes
5. same as number 2...13
6. what is my username?
7. eryaman
8. what is 1+1? what is 2x2?
9. 2, 4
10. yay

[quote="Spencer"]Answer these questions!

1. How old are you right now?
2. When you first logged on, how old were you?
3. What grade are you in?
4. Do you like this Forum?
5. When you first posted somethng in this Forum, how old were you?
6. Ask your own Question.
7. Answer your Question.
8. Make some more Questions.
9. Answer more Questions.
10. You can leave now![/quote]

um ok...

1. 13
2. 13
3. 7th.. well for like 3 days.. then im officially 8th.
4. yes
5. 13
6. That isnt a question. But, if you want me to.. "What's up?"
7. Not much.
8. <insert "some more questions"/>
9. too lazy.
10. Thanks for the permission...

[quote="eryaman"][quote="math92"]1.  12
2.  12
3.  7th
4.  yes
5.  12
6.  um... Do you like chocolate?
7.  yes
8.  ...You told me to answer these questions and #8 isn't a question...
9.  see #8
10.  see #8[/quote]numbers 6 and 7 aren't questions either... ;)[/quote]
yeah i know, i just love chocolate and i wanted to write that ;)

[quote="Spencer"]Answer these questions!

1. How old are you right now?
2. When you first logged on, how old were you?
3. What grade are you in?
4. Do you like this Forum?
5. When you first posted somethng in this Forum, how old were you?
6. Ask your own Question.
7. Answer your Question.
8. Make some more Questions.
9. Answer more Questions.
10. You can leave now![/quote]

1. 12
2. 11
3. going to 7
4. yes
5. 11
6. How old am I?
7. 12
8. How old are you one year from now? Why are you answering these questions?
9. 13 , I don't know
10.  yes!!!

[quote="Spencer"]Answer these questions!

1. How old are you right now?
2. When you first logged on, how old were you?
3. What grade are you in?
4. Do you like this Forum?
5. When you first posted somethng in this Forum, how old were you?
6. Ask your own Question.
7. Answer your Question.
8. Make some more Questions.
9. Answer more Questions.
10. You can leave now![/quote]

1) 15
2) 15
3) Going to be in 11th after summer is over.
4) NO..jk :rotfl:
5) 15...obviously
6) $2+2=x$, what is the value of $x$?
7) $5$
8) Do you like math?
    What's your favorite NBA team?
    What's your favorite NFL team?
    Which sport do you play?
9) No
    Philadelphia 76ers
    Philadelphia Eagles
    Tennis
10) Bye

1. 15
2. 14
3. Going to be in 10th
4. Yes
5. 14
6. Who will win Wimbledon?
7. Federer!!!
8. Too lazy
9. Way too lazy
10. thanks a lot


Let be given real number A. Find all functions f:R+ -->R+             ( R+ means all real x such as x>0)
 satisfying for all x>0

 A x^2 f(1/x) + f(x) = x/(1+x)

technique: use  a transformation x|-> 1/x.
note: when A is 1 or -1, there is no f.

Could You explain it more clearly because your idea seems to be wrong.


let $ a;b;c \geq 0$; prove that:
$ a^5b \plus{} ab^5 \plus{} b^5c\plus{}bc^5\plus{}ca^5 \plus{} c^5a\plus{}6a^2b^2c^2 \geq a^4b^2\plus{}a^2b^4 \plus{} b^4c^2 \plus{} b^2c^4 \plus{} c^4a^2 \plus{} c^2a^4 \plus{} 2abc(a^3 \plus{} b^3 \plus{} c^3)$


geometry
perimeter
inequalities
triangle inequality
number theory
prime numbers

The length of each side of an isosceles triangle is a prime number. The perimeter of the triangle is greater than 21 units and is not a prime number. What is the least possible perimeter of the triangle?

[hide="Solution"]Let the perimeter of the triangle be $ 2a\plus{}b$. By the triangle inequality and the restriction on the perimeter, $ a>5$.

We note that $ a\equal{}11$ and $ b\equal{}2$ is a valid pairing, yielding a perimeter of $ 24$. Since the perimeter must be composite, the only other possible perimeter is $ 22$. Three prime numbers summing to an even number implies that one of the primes is $ 2$. However, a triangle of side lengths $ 2$, $ 2$, and $ 18$ violates the rule, as does a triangle of side lengths $ 2$, $ 10$, and $ 10$. Therefore, the minimum perimeter is $ \boxed{24}$.
[/hide]


Find the minimum value of $x^{3}+y^{3}$ $\textrm{if}$ $\frac{x}{y^{2}}+\frac{y}{x^{2}}=2$

Looking at the code, it seems that vishalarul meant
$\frac{x}{y^{2}}+\frac{y^{2}}{x}=2$.
[hide="Solution"]We know that for $z>0$, $z+\frac{1}{z}\geq 2$ and the equality only holds for $z=1$, so $x=y^{2}$.
Now we must find the minimum value of $y^{6}+y^{3}$, which is equivalent to the minimum value of $t^{2}+t$ because $t=y^{3}$ is an increasing bijective function.
The minimum of the parabola is at $t=\frac{-1}{2}$ and its value is $\boxed{\frac{-1}{4}}$.[/hide]


If  $f(n)=\frac{2n+1+\sqrt{n(n+1)}}{\sqrt{n+1}+\sqrt{n}}$  for all  positive  integers $n$, evaluate

(a) $f(1)$,

(b)  the  sum  $A=f(1)+f(2)+...+f(400)$.

By multiplying nominator and denominator by $\sqrt{n+1}-\sqrt{n}$ we get:
$f(n)=(n+1)\sqrt{n+1}-n\sqrt{n}$
$f(1)+f(2)+\cdots+f(400)=2\sqrt{2}-1\sqrt{1}+3\sqrt{3}-2\sqrt{2}+4\sqrt{4}-3\sqrt{3}+\cdots+401\sqrt{401}-400\sqrt{400}$
So the sum $A=401\sqrt{401}-1$

(a) Sub n=1,we get f(1)=2sqrt2-1.

(b) Rationalising f(n),we get f(n)=(n+1)sqrt(n+1)-(n)sqrt(n).
f(2)=3sqrt3-2sqrt2
f(3)=4sqrt4-3sqrt3
...........
f(400)=401sqrt401-400sqrt400.
Summing up the telescoping series,we obtain A=401sqrt401-1.  :)


What is your favorite antivirus? What antivirus are you using? Why?

http://xkcd.com/391/

THE best antivirus!

I use avast.It's quite good.But a russian one (don't remember the name)is the best.

A russian anti-virus, starts with C .Sorry I don't remember it's full name.


find the value of $cos2x(1+tan^2x)+\frac{1}{cos^2x}$.

excellent. :D
here is mine:
$cos2x(1+tan^2x)+\frac{1}{cos^2x}=(2cos^2x-1)( \frac{1}{cos^2x})+\frac{1}{cos^2x}$ $=2-\frac{1}{cos^2x}+\frac{1}{cos^2x}=2$


induction
number theory solved
number theory

Prove that for each odd  $n \in \mathbb{N}: 256 |n^6+3n^4+7n^2-11$
I tried to do it by induction, by it started to become more and more complicated.

If everything else doesn't work, you can kill such problems with the brute of fork by verifying it for all different odd residue classes $\mod 256$.
But easier is to consider
$x^6+3x^4+7x^2-11=(x-1)(x+1)(x^4+4*x^2+11)$
and to see that $8 | (x-1)(x+1)$.
Now its enough to handle the third factor $\mod 32$ where we get
$x^4+4x^2+11-32=(x^2-3)(x^2+7)$
the second factor of this is clearly divisible by $8$ and the first by $2$, so there is only one factor $2$ missing, which we get by looking $\mod 16$ on $(x^2+7)$ and $(x-1)(x+1)$.


[quote]
AMCDirector can post a new thread if and when he responds.[/quote]

[quote]To anyone who knows: How can I submit a formal protest?[/quote]
[quote]There is no official procedure for submitting protests, as far as I know.[/quote]

On the back of each and every contest pamphlet, in italics, at the top, is the statement:
"Correspondence about the problems and solutions for this [Contest] ... should be addressed to:
and then, our postal address, our phone number, fax number, and email address.

Steve Dunbar
MAA Director, American Mathematics Competitions

Thank you for the information.  I can only assume this is, at least partially, directed toward the thread I made about #11 on the AMC 12 B.  I emailed my concern to the appropriate address on Friday morning, and received notification that it would be forwarded to the appropriate people.

Are there any other steps I need to take?


Not sure if which forum this would go into, or even if its a valid question, but :

What are some good science fair topics (preferably dealing with math) that I could go into  :?: I'm completely out of ideas, so if any of you have some topic you considered but threw away that would be great.

How super cooling wire effects resistance
 or try using trig with something


What movies did you watch lately? I want something different: neither comedies nor horrors. Maybe history, S.F., action movies, or something else.


For a sale, a store owner reduces the price of a $ \$10$ scarf by $ 30\%$.
Later the price is lowered again, this time by one-half the reduced
price. What is the current price?

[quote="pubado"]You mean $ \boxed{3.50} $ (it's money we are dealing with)[/quote]
I believe FTW will sometimes accept $3.5$

An answer of 3.5 for money would have been accepted in FTW1, but it is no longer accepted in FTW2. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1856728#p1856728]this post[/url].

Note that nikeballa96's post was written before FTW2 came along.


Give $ a,b,c \ge 0$ and $ a^2\plus{}b^2\plus{}c^2\equal{}3$.Prove that:
$ \frac{3\plus{}4ab}{\sqrt{c^2\plus{}3ab}}\plus{} \frac{3\plus{}4ac}{\sqrt{b^2\plus{}3ac}}\plus{} \frac{3\plus{}4bc}{\sqrt{a^2\plus{}3bc}} \ge (2\plus{}\sqrt{3})(a\plus{}b)$

[quote="secrets"]Give $ a,b,c \ge 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.Prove that:
$ \frac {3 \plus{} 4ab}{\sqrt {c^2 \plus{} 3ab}} \plus{} \frac {3 \plus{} 4ac}{\sqrt {b^2 \plus{} 3ac}} \plus{} \frac {3 \plus{} 4bc}{\sqrt {a^2 \plus{} 3bc}} \ge (2 \plus{} \sqrt {3})(a \plus{} b)$[/quote]
Is the RHS $ (2\plus{}\sqrt{3})(a\plus{}b\plus{}c)?$ :oops:

$ (2\plus{} \sqrt{3})(a\plus{}b)$,my friend.


Let A, B, C be the vertices of an equalateral triangle and let Q be the circle through these vertices. Let R be a point on the arc AB. Then RC=RA+RB. HELPP!! It works when R is the midpoint of arc AB, but does it still work if R is randomly chosen? Can you guys help me!!!??

Yes, it still works when $ R$ is chosen randomly. Here's an outline of a proof using trigonometry and analytic geometry. WLOG let

$ A \equal{} \left(\cos \left(\frac{2 \pi}{3} \right), \sin \left(\frac{2 \pi}{3} \right) \right)$

$ B \equal{} \left(\cos \left(\frac{4 \pi}{3} \right), \sin \left(\frac{4 \pi}{3} \right) \right)$

$ C \equal{} \left(1, 0 \right)$

Then by definition $ R \equal{} \left( \cos(\theta), \sin(\theta)\right)$ for some $ \theta \in \left(\frac{2 \pi}{2},\frac{4 \pi}{3}\right)$. Using the coordinate representations of these points, you can explicitly calculate $ RC$, $ AR$, and $ RB$ and show using trigonometry that $ RC \equal{} AR\plus{}RB$.

Let me know if that makes sense to you.

Then by definition $ R \equal{} \left( \cos(\theta), \sin(\theta)\right)$ for some $ \theta \in \left(\frac {2 \pi}{2},\frac {4 \pi}{3}\right)$. Using the coordinate representations of these points, you can explicitly calculate $ RC$, $ AR$, and $ RB$ and show using trigonometry that $ RC \equal{} AR \plus{} RB$.

I don't quite understand this, mainly the little E looking letter thing.  Could you break it down more for me?

Well， it seems a little too complicated to me   :( , can you explain it without trigonometry? thank you very much

By ptolemy's theorem, letting $ s$ be the length of a side of the equilateral triangle,

$ s*RA\plus{}s*RB\equal{}s*RC$

$ RA\plus{}RB\equal{}RC$

OH!! Thank you very much dude. Ｙour are smart! :D


geometry
geometric transformation
calculus
function

Prove that a statement in propositional logic is not a law of logic.

Prove which arguments are valid or not.

1)if 2+2=5 then 2+3=6.But $ 2\plus{}2\neq 5$ hence $ 2\plus{}3\neq 6$

2)if 2+2=5 then 2+3=6 .But 2+2=5 ,hence 2+3=6.

3) if 2+3 =5 then 2+4 = 7.But $ 2\plus{}4\neq 7$,hence $ 2\plus{}3\neq 5$

Arguments would need to be broken down somewhat; removing statements such as 2+2=5, and what not, do you mean to suggest the following arguments?

Letting F be false, T be true.

1)
Given: $ p \implies q$, $ p \equal{} F$
Then: $ q \equal{} F$

2)
Given: $ p \implies q$, $ p \equal{} T$
Then: $ q \equal{} T$.

3)
Given: $ p \implies q$, $ q \equal{} F$
Then: $ p \equal{} F$.

Is this a correct translation of what you mean? If so, it is easily checked by truth table.

let us see what other people in the forum have to say.

In the 1st part of the thread it is asked to [b]prove [/b] that a statement is not a law of logic.

Showing that the thread " is that proof correct" was locked in a hurry ascertaining the ignorance w.r.t the propositional calculus of those who did it

1),2) are true implications, 3) is false

Why? Logic says that $ T\equal{}>T,F\equal{}>T,F\equal{}>F$ are true implications, and $ T\equal{}>F$ is a false implication.

1) $ F\equal{}>F$, so the implication is true

2)=1)

3) $ T\equal{}>F$, the implication isn't true

Do not mix implication with the arguments on LHS and RHS!

[quote="jmerry"][quote="τρικλινο"]Showing that the thread " is that proof correct" was locked in a hurry[/quote]
I assure you, that was no hurry. If I were in a hurry, I would have locked it at post #23. Instead, I gave you a chance to acknowledge your errors and bring sensible discussion back into the picture. You didn't.[/quote]

Do you know how one [b] proves [/b] that a statement in propositional calculus is not a law of logic.

Because that was the central issue of the thread " is that proof correct" and not unfounded examples that you claim as an answer to the said thread.

What is sensible discussion in mathematics??

In mathematics we have valid or non valid arguments nothing more nothing less.

My errors ,how about your errors ??

[quote="τρικλινο"]
Do you know how one [b] proves [/b] that a statement in propositional calculus is not a law of logic?
[/quote]
Frankly speaking, I do not even know what you mean by this and have a strong suspicion that nobody here does :roll:.
[quote]
What is sensible discussion in mathematics??
[/quote]
I would say that it is a discussion in which people have at least some idea of what the others are talking about and try to explain their points so that everybody would understand them. I don't know if you understand what [i]everybody else[/i] is saying but I can assure you that nobody currently has the slightest idea of what [i]you[/i] are looking for or driving at (not in this thread, at least).
[quote]
In mathematics we have valid or non valid arguments nothing more nothing less.
[/quote]
Not really. If it were true, we would just sit and fill out the addition table starting from $ 1\plus{}1$ and going up, and up, and up, and up, (well, you get the idea) or writing down some set of axioms and applying modus ponens to them in some random order to derive various statements until we either get a contradiction or just get tired of this game.

That's not quite how it works. I am reluctant to post a long explanation of my viewpoint on what mathematics really is (or, if you prefer, may be) and what we have and do not have in it, but the formal validation of the arguments by writing them in the symbolic form and checking that every line is either an axiom or follows from the previous ones by applying one of the laws of logic is very, very low on the list of priorities for most mathematicians.

If you want an example of something that is crucial to the development of mathematics but cannot be obtained by pure logical derivation from axioms, it is the choice of objects to deal with. Yes, we have a formal definition of, say, continuity, but the reason we bother to prove theorems about continuous functions rather than about the functions such that for every $ \varepsilon>0$ there exists $ \delta>0$ such that there there exists a pair of points $ x,y$ with $ |x\minus{}y|\equal{}\delta$ and $ |f(x)\minus{}f(y)|\equal{}\varepsilon$ or something else like that is not within the realm of formal logic.

[quote]
My errors ,how about your errors ??
[/quote]
The situation here is somewhat similar to the one when everybody around you is speaking English and you are speaking Greek. From your perspective, all people around you just talk gibberish and only your own words make any sense at all. From every other person's perspective, everybody talks normally and you can neither understand anything, nor say anything meaningful yourself.

Generally speaking, I'm quite willing to accept the idea that you are talking to a bunch of idiots (including myself) who cannot even understand what a proof is but, as you know, there are 2 ways for a single clever person to deal with a bunch of idiots:

1) To learn their language and to explain things in terms they can understand
2) Not to waste time on them at all and (2a) just make them shut up if it is within his powers or (2b) to move as far away from them as possible if it isn't.

Unless #1 or #2 is executed, he risks
3)To be suppressed by an irritated mob.

If you can do #1, we'll have a meaningful discussion. It is certainly not within your current abilities to execute option 2a. But it is certainly within our abilities to execute #3. Being rather good-natured idiots, we are trying to avoid this for as long as possible, but, you know, nobody's patience is infinite. Jmerry locked some of your threads and I moved some other ones from College Playground to Fun Factory. These are just the first signals.

As I said, it all may be just due to the lack of mutual understanding, but it seems that, either because of being such poor idiots as we are, or for some other reason, we are just unable to learn your language, so we humbly ask you to have some mercy on us and to try to learn ours ;).

[quote="fedja"][quote="τρικλινο"]
Do you know how one [b] proves [/b] that a statement in propositional calculus is not a law of logic?
[/quote][/quote]
Frankly speaking, I do not even know what you mean by this and have a strong suspicion that nobody here does :roll:.
[quote]

O.K ,let us take things step by step.

Would you be so kind and take ANGELO'S  MARGARIS book ,FIRST ORDER MATHEMATICAL LOGIC , and read page 69 and then come back and tell me ,whether you understand or not ,when i ask how can one prove that a statement is NOT a law of logic??

[quote="τρικλινο"]
Would you be so kind and take ANGELO'S  MARGARIS book ,FIRST ORDER MATHEMATICAL LOGIC , and read page 69 and then come back and tell me ,whether you understand or not ,when i ask how can one prove that a statement is NOT a law of logic??[/quote]

That's disgustingly authoritarian, and conjures up images of religious zealots quoting verse from the Bible.

[quote="Bugi"]1),2) are true implications, 3) is false

![/quote]

According to propositional calculus the above is wrong.

Arguments are named in propositional calculus as valid or non valid and not false or true.

But even if you mean that arguments (1) and (2) are valid and (3) is non valid ,this is still wrong


function
inequalities unsolved
inequalities

$ a^4 \plus{} b^4 \plus{} c^4 \equal{} 3,a,b,c > 0$  prove that
$ \frac {1}{4 \minus{} ab} \plus{} \frac {1}{4 \minus{} bc} \plus{} \frac {1}{4 \minus{} ca}\le1$

[quote="alex2008"]I already posted a solution for this one  :D , see [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1456436&search_id=611476114#1456436]here[/url] post #77[/quote]

Can you tell me,$ \frac{1}{4\minus{}a^{2}}\le\frac{a^{4}\plus{}5}{18}$ how to obtain it?
thanks.

[quote="sxgfly"][quote="alex2008"]I already posted a solution for this one  :D , see [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1456436&search_id=611476114#1456436]here[/url] post #77[/quote]

Can you tell me,$ \frac {1}{4 \minus{} a^{2}}\le\frac {a^{4} \plus{} 5}{18}$ how to obtain it?
thanks.[/quote]

It is equivalent with $ (a^2\minus{}1)^2(2\minus{}a^2)\ge 0$ .

@sxgfly: you wanted to know how to obtain it: the function $ g(x)\equal{}\frac{x\plus{}5}{18}$ is the tangent to the function $ f(x)\equal{} \frac{1}{4\minus{} \sqrt{x}}$ in the point $ x\equal{}1$ (where the equality is attained).

If he wanted to know that , see also [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1606793#1606793]here[/url] dgreenb801's post .


1/Given m natural number dont exceed n at all and n natural numbers dont exceed m at all.Show that we can find two subsequences of 2 this sequences sums the same!
2/Show that this conclusion is still true for the condition:a1+a2+..+am=b_1+b_2+...+b_n<mn.
   Theyare both ASU problems!

for problem2 let the same sum is V and consider $\{\sum_{i=1}^{j}a_{i}-\sum_{i=1}^{k}b_{i}(mod V)\}$.
  Furthermore:$-V+1\leq\sum_{i=1}^{j}a_{i}-\sum_{i=1}^{k}b_{i}\leq V-1$ and there are $mn>V$ such values!
And pegionhole principle!


This is simple, but I don't know how to approach it with the square root.

-----The limit of ((3+4x)^(1/2))/(10+7x) as x approaches infinity.

I tried doing the conjugate but that didn't get me anywhere.

Thanks for helping.

It's about size. The numerator grows like $ \sqrt{x}$. The denominator grows like $ x$. As $ x\to\infty$, $ x$ is bigger than $ \sqrt{x}$, and the limit is zero.

If those were the same exponent, I'd have to pay attention to the constant multipliers.

I thought it was zero as well, but it is an online problem and it won't accept zero.  I have tried 10 different times with anything and everything.  Could it be something else?

Nope, it's pretty much just zero.  A bound:

$ \frac{\sqrt{3\plus{}4x}}{ 10\plus{}7x}<\frac{\sqrt{4\plus{}4x}}{ 7\plus{}7x}\equal{}\frac{2}{7\sqrt{x\plus{}1}}$

Then the numerator would look like $ 2x$ for large $ x$ and the denominator would look like $ 7x,$ and the limit would be $ \frac{2}7.$

Advice: de-clutter. Look for the important parts of your expressions. (Which is what jmerry was saying above.)


algebra
polynomial
algorithm
calculus
calculus computations

Numerically.  Quartic equations are ugly and there's no reason to solve them algebraically.

... but do the substitution anyway. Polynomials are much nicer to work with numerically than exponentials. There is exactly one valid (positive) $ x$-value.

An iteration that works pretty fast in this case:
$ x_{n\plus{}1}\equal{}\sqrt[4]{\frac{98.39\minus{}3x_n\minus{}3x_n^2\minus{}3x_n^3}{103}}$
It's first-order, but with a very good coefficient of about $ \frac1{21.6}$.

The root is approximately $ x\equal{}0.96676573$, or $ y\equal{}\minus{}0.0675982$

Also, I'm moving this to calculus.

Also, I'm pretty sure that $ 98.39$ means "something between $ 98.385$ and $ 98.395$", so the digits beyond the fifth decimal place in the answer can hardly be justified.

At last, since, looking at the equation, one would immediately suspect that $ y\approx 0$, you can just linearize at $ 0$ and get $ 112\plus{}215y\approx 98.39$, i.e., $ y\approx \minus{}\frac{13.61}{215}\approx \minus{}0.063$, which is only $ 6\%$ off the exact solution jmerry provided. Of course, this is not the technique you should use for high precision computations but it allows you to get the idea of what the correct order of magnitude must be almost without any computations at all so if your computer implementation of the iteration algorithm is faulty, you'll be able to see that something is fishy immediately.

[quote="Keno20"]I wasn't sure which forum to post this in so please forgive me if this is misplaced.

Solve 3e^.5y+3e^y+3e^1.5y+103e^2y=98.39

I've tried substituting e^.5y for x. But I'm also not sure how to solve that either. I also know that log(a+b) is not log(a) +log(b) so that's my problem trying to use logs thus far.

Thanks for the help.[/quote]

Why did Riemann calculate square roots to 20 decimal places?  It's a quartic right?  Let $ x \equal{} e^{y/2}$ then we have $ x^4 \plus{} \frac {3}{103}x^3 \plus{} \frac {3}{103}x^2 \plus{} \frac {3}{103}x \minus{} \frac {a}{103} \equal{} 0$.  The first step is to convert this to a reduced quartic by letting $ x \equal{} z \minus{} 103/4$.  This changes it to $ z^4 \plus{} qz^2 \plus{} rz \plus{} s \equal{} 0$.  Already makin' progress.  Next would be to convert it to a resolvent cubic and then go from there.  If you want to . . . , the book I have is old and dusty:  "College Algebra" by Rosenbach et.al. but I'm sure you can find the steps elsewhere.


This was a question in one of the threads a while ago, but I forgot where :P

An equilateral triangle has the same perimeter as a square. If the area of the triangle is 1, what is the area of the square?

Everytime I do this problem I end up with [3*sqrt(3)]/4
but the people who solved it ended up with [4*sqrt(3)]/3

What am I doing wrong?
[x^2*sqrt(3)]/4 = 1 (area of triangle where x = side)
the side of the squares are (3/4)*x so the area of the square should be (9/16)*x^2

Solving for x^2 in the triangle I get x^2 = [4*sqrt(3)]/3
Then multiply both sides by (9/16) to get the area of the square  I get (9/16)*x^2 = [36*sqrt(3)]/48 = [3*sqrt(3)]/4

What am I doing wrong, or do I have the right answer and they did it wrong
Also is there any better way to do this problem?

[hide]$\frac{s^{2}\sqrt{3}}{4}=1$

$s^{2}\sqrt{3}=4$

$s^{2}=\frac{4}{\sqrt{3}}$

$s^{2}=\frac{4\sqrt{3}}{3}$

$s=\frac{4\sqrt[4]{3}}{3}$

$3s=4\sqrt[4]{3}$

The perimeter of the triangle is $4\sqrt[4]{3}$.

And then each side of the square is $\frac{4\sqrt[4]{3}}{4}=\sqrt[4]{3}$

So, the area of the square is $\left(\sqrt[4]{3}\right)^{2}=\sqrt{3}$.[/hide]

[hide]The area of the triangle is 1, and $A=\frac{x^{2}\sqrt{3}}4,$ so the side length of the triangle is $\frac{2\sqrt[4]{3}}3.$

The perimeter is $2\sqrt[4]{3},$ so the side length of the square is $\frac{\sqrt[4]{3}}2.$

$A=\left(\frac{\sqrt[4]{3}}2\right)^{2}=\boxed{\frac{\sqrt{3}}4}$[/hide]

@ anirudh - Check your work and hide it. :wink:

Oops. :blush: I messed up. there's not supposed to be a 4 there, so it'll be multiplied by 1/3 at the end, so the answer is $\boxed{\frac{2\sqrt[4]{3}}3}$

[quote="anirudh"]the answer is $\boxed{\frac{2\sqrt[4]{3}}3}$[/quote]
That's the side length of the triangle, we're finding the area of the square. :roll:

[quote="Vorfield"]I think you guys messed up in your solving
When you take the square root of s you have to take the square root of the whole $\frac{4\sqrt{3}}{3}$ Not just a part of it

anirudh you only took the sq root of $sqrt{3}$
and i_hate_pie you forgot to take the square root of the denomination 3

it should be

$s^{2}=\frac{4\sqrt{3}}{3}$

$s=\frac{2\sqrt[4]{3^{3}}}\sqrt{3}$

right?

I need to learn how to use Latex :P lol[/quote]
I'm not i_hate_pie, I'm i_like_pie. :roll:

[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]

[hide="Steps"]$\frac{x^{2}\sqrt{3}}4=1$

$x^{2}\sqrt{3}=4$

$x^{2}=\frac{4}{\sqrt{3}}$

$x=\frac{2}{\sqrt[4]{3}}$

$x=\frac{2\sqrt[4]{27}}3$

$p=2\sqrt[4]{27}$

Area of square $=\left(\frac{2\sqrt[4]{27}}{4}\right)^{2}=\frac{4\sqrt{27}}{16}=\boxed{\frac{3\sqrt{3}}4}$[/hide]


calculus
integration
trigonometry
real analysis
real analysis unsolved

Ok, so Spivak did it in 2 steps ... unfortunately, I did not.  Is there something I did not see?  My answer seems so messy.

$ \int\frac {x\tan^{ \minus{} 1}x}{(1 \plus{} x^2)^3}dx$

$ \int\frac {\tan^{ \minus{} 1}x}{(1 \plus{} x^2)}\frac {x}{(1 \plus{} x^2)^2}dx\left\{\begin{array}{cc}x \equal{} \tan{t} \leftrightarrow t \equal{} \tan^{ \minus{} 1}x \\
dt \equal{} \frac {dx}{1 \plus{} x^2}\end{array}\right\} \\
 \\
 \\
\int t\cos^{3}t\sin{t}dt\left\{\begin{array}{cc}u \equal{} t & dV \equal{} \cos^{3}t\sin{t}dt \\
du \equal{} dt & V \equal{} \frac {1}{4}\cos^{4}t\end{array}\right\} \\
 \\
 \\
\int t\cos^{3}t\sin{t}dt \equal{} \frac {1}{4}t\cos^{4}t \minus{} \frac {1}{4}\int\cos^{4}tdt \\
 \\
 \\
\int t\cos^{3}t\sin{t}dt \equal{} \frac {1}{4}t\cos^{4}t \minus{} \frac {1}{32}\int(3 \plus{} 4\cos{t} \plus{} \cos{4t})dt \\
 \\
 \\
\int t\cos^{3}t\sin{t}dt \equal{} \frac {1}{4}t\cos^{4}t \minus{} \frac {1}{32}\left(3t \plus{} 2\sin{2t} \plus{} \frac {1}{4}\sin{4t}\right)$

Let $ u \equal{} tan ^{ \minus{} 1}x\Longrightarrow x \equal{} \tan u$.

$ \therefore \int \frac {x\tan ^ { \minus{} 1}}{(1 \plus{} x^2)^3}dx \equal{} \int \frac {u\tan u}{(1 \plus{} \tan ^ 2 u)^3}\cdot \frac {1}{\cos ^ 2 u}du$

$ \equal{} \int u\cos ^ 3 u\cdot \sin u\ du$

$ \equal{} \minus{} \frac {1}{4}u\cos ^ 4 u \plus{} \frac {1}{4}\int \cos ^ 4 u du$

$ \equal{} \minus{} \frac {1}{4}u\cos ^ 4 u \plus{} \frac {1}{4}\int \left(\frac {1 \plus{} \cos 2u}{2}\right)^2 du$

$ \equal{} \minus{} \frac {1}{4}u\cos ^ 4 u \plus{} \frac {1}{16}\int \left(1 \plus{} 2\cos 2u \plus{} \frac {1 \plus{} \cos 4u}{2}\right)du$

$ \equal{} \minus{} \frac {1}{4}u\cos ^ 4 u \plus{} \frac {1}{16}\sin 2u \plus{} \frac {1}{128}\sin 4u \plus{} \frac {3}{32}u \plus{} Const.$


What does the graph $y^2 > x^2+1$ look like?

Tell me if I'm wrong. To graph this, you solve for $y$. This gives $y>\sqrt{x^2+1}$ and $y>-\sqrt{x^2+1}$. It seems to me like the shading for $y>-\sqrt{x^2+1}$ would be above not below?

You're solving is wrong. First, when you take a square root from the square of a not-necessarily-positive number, you must ALWAYS take absolute values into account. Hence, the right solving is $|y|>\sqrt{x^2+1}$, and that comes down to two cases: $y>\sqrt{x^2+1}$ or $y<-\sqrt{x^2+1}$, which means that the shading is above the upper branch and below the lower branch, as shown.


[quote="10000th User"]Here is an induction problem for beginners to think and prove rigorously apart from the overly used statement $1+2+3+\dots+n=\frac{n(n+1)}2$

If you are beginner and want to learn induction with me, let's start with different ways of using induction in proofs. To keep it at the Getting Started level, the difficulty is kept at minimum.

Start to plug in a few integers to convince yourself. Please write out the steps so you can be corrected if it is needed.

1) Prove that $1\cdot 1!+2\cdot 2!+\cdots+(n-1)\cdot(n-1)!=n!-1$ for all positive integers $n\ge 1$[/quote]

[hide="1"]Well start with $n=1$
$(0)(1)=1!-1=0$.

So then ... I added $n(n)!$ to both sides to get $1\cdot 1!+2\cdot 2!+\cdots+(n-1)\cdot(n-1)!+n\cdot (n)!=n(n)!+n!-1$

So then the RHS is $n!(n+1)-1$ which is what you get by replacing n with $n+1$: $(n+1)!-1=(n)!(n+1)-1$
Does this prove it?[/hide]

[quote="10000th User"]That's good, you checked the base case and then you used the hypothesis that it is true for $n$ and proved that it is also true for $n+1$. So by induction hypothesis, you proved it for all positive integers $n\ge 1$.

Let's try another easier one and then we go and try some more subtler ones:

2) What is the sum of the first $n$ odd positive integers?

Again, try to come up with a formula (it's easy :) ) and then show that your formula is true for all positive integers $n$? :)[/quote]

[hide="2"]
Adding the first few $2n-1$, we get: 1, 4, 9, 16, ...
This looks like $n^2$

We wish to prove that $\sum^n_{i=1}2n-1=n^2$.  Adding $2(n+1)-1$ to each side, we have $\left(\sum^n_{i=1}2n-1\right)+2n+1=n^2+2n+1$ so $\sum^{n+1}_{i=1}2n-1=(n+1)^2$.  We have that this is true for $n=1$, so we've proved it for all positive integers.
[/hide]

No because this problem involves a sum of $n$ positive terms and it inherently uses induction.

You are correct, ccy.

Here is a subtle one that can put off some people (I hope you guys do well):

3) For what positive integers $n$ is the following statement true: $n^3\ge 4n^2+1$ ?

The answer is of the form like $n\ge 2006$, something like that; and just like before, use induction to prove your answer :)

[quote="srulikbd"][quote="10000th User"]Here is an induction problem for beginners to think and prove rigorously apart from the overly used statement $1+2+3+\dots+n=\frac{n(n+1)}2$

If you are beginner and want to learn induction with me, let's start with different ways of using induction in proofs. To keep it at the Getting Started level, the difficulty is kept at minimum.

Start to plug in a few integers to convince yourself. Please write out the steps so you can be corrected if it is needed.

1) Prove that $1\cdot 1!+2\cdot 2!+\cdots+(n-1)\cdot(n-1)!=n!-1$ for all positive integers $n\ge 1$[/quote]
can someone prove it in other way (not induction...)[/quote]

the 1) can be proved wothout induction...

also without using induction one can try to relate
the sequence $n+...+2+1$ to $C_{n}^{2}$ this way:
it can be seen that from the set $\{ 1,2,...,n \}$
1).we can extract $n-1$ sets of 2 components
 of the form $(n,k)$ with $1\leq k \leq n-1$
2).we can extract $n-2$ sets of 2 components
of the form $(n-1,k)$ with $1\leq k \leq n-2$
.
.
.
n-2). we can extract $2$ sets of 2 components
of the form $(3,k)$ with $1\leq k \leq 2$
n-1). we can extract only (2,1) with the conditions as above.

now it is obvious that the reunion of the sets
from 1),2),...,n-1) are actually the sets that
$C_{n}^{2}$ refers to but $C_{n}^{2}=\frac{n(n-1)}{2}$
so the equality is prooven

[quote="10000th User"]Here is a subtle one that can put off some people (I hope you guys do well):

3) For what positive integers $n$ is the following statement true: $n^3\ge 4n^2+1$ ?[/quote]
[hide="hint"]
It might be a bit harder to prove the inequality in this form.
Move everything to the LHS ($n^3-4n^2-1\ge 0$) and try to prove that the LHS is positive (for all $n$ satisfying your answer).
[/hide]
[quote="shadysaysurspammed"][quote="srulikbd"][quote="10000th User"]1) Prove that $1\cdot 1!+2\cdot 2!+\cdots+(n-1)\cdot(n-1)!=n!-1$ for all positive integers $n\ge 1$[/quote]
can someone prove it in other way (not induction...)[/quote]

the 1) can be proved wothout induction...[/quote]
I understand that this thread is for practicing/learning induction, so I'll post an easy non-induction proof just for reference.
[hide]
Note that $k\cdot k!=((k+1)-1)k!=(k+1)k!-k!=(k+1)!-k!$.

So $1\cdot 1!+2\cdot 2!+...+(n-1)\cdot(n-1)! = (2!-1!)+(3!-2!)+...+(n!-(n-1)!) =n!-1$
[/hide]

Nice topic! Induction is extremely nice! But be carefull and do not start a marathon.


linear algebra
matrix
invariant
algebra
polynomial
linear algebra unsolved

Show that if $A$ and $B$ are square matrices where
\[ A = \left( {\begin{array}{*{20}c} B & C \\ 0 & D \\ \end{array}} \right)  \]
, then the Jordan form of $B$ is a [b]principal submatrix[/b] of (EDIT) [b]the jordan form of[/b] $A$.

'principal submatrix' is defined as follows:
If a submatrix of n by n matrix $J$ is obtained by selecting a subset $S$ of {1,2,3,...n} and then discarding the rows and columns that correspond to (the same) $S$, then it is called a principal submatrix of $J$.

The above problem is equivalent to this problem.
"Show that if $V$ is an $A$-invariant subspace of $\mathbb{R}^n$, then the Jordan form of $A$ restricted to $V$ is a principal submatrix of the Jordan form of $A$."

I wonder if there is another solution.

Without loss of generality, assume that the characteristic polynomial of $A$ is $x^n$ and that the sizes of the jordan cells of $A$ are $a_1,a_2, \cdots ,a_s$ in nonincreasing order. The characteristic polynomial of $B$ is $x^m$ and the sizes of the jordan cells of $B$ are $b_1,b_2,\cdots,b_t$ in nonincreasing order.
Let $V = R^{m} \times 0 \subset R^n$. $B$ is a restriction of $A$ to $V$.
For each $k$, Let $U_k = nullspace(A^k)$.
Let $c_k$ ($d_k$) be the number of items in $a_1,\cdots,a_s$ (in $b_1,\cdots,b_s$ ,respectively) that are greater than or equal to $k$.
\[ c_k \\ = nullity(A^k) - nullity(A^{k-1}) \\ = dim(U_k) - dim(U_{k-1}) \\ \geq dim(U_k \cap V) - dim(U_{k-1} \cap V) \\ = nullity(B^k) - nullity(B^{k-1}) \\ = d_k  \]
But that $c_k \geq d_k$ for all $k$ is equivalent to that $a_k \geq b_k$ for all $k$ and $s \geq t$. This is stronger than that the jordan form of $B$ is a principal submatrix of the jordan form of $A$.


hi.

i have two questions.

1.) what is the latest version of latex? where can i get them? when you say latest version, does that mean the compiler like Bibtex?

2.) has anyone use beamer? i tried to follow many tutorials on the net but no success yet? what's the easiest thing to do. i want my presentations to be on beamer.

thanks.

[quote="stevem"]1)You don't say what operating system you use. [url=http://miktex.org/]MikTeX 2.8[/url] (Windows) and [url=http://www.tug.org/texlive/]TexLive2008[/url] (Windows, Mac, Linux, Unix) are the latest versions. The complete versions (rather than the basic ones) will include BibTeX. The downloads will be very large - more than 1GB.

2) [url=http://www.math.umbc.edu/~rouben/beamer/]A Beamer Quickstart[/url] is a good introduction to Beamer with an easy to follow [url=http://www.math.umbc.edu/~rouben/beamer/quickstart-Z-H-3.html#node_sec_3]A bare-bones sample[/url].

I would strongly suggest that you are familiar with LaTeX first before you use Beamer or BibTeX and have read the main parts of [url=http://www.amazon.com/Guide-LaTeX-4th-Helmut-Kopka/dp/0321173856/sr=8-1/qid=1161947605/ref=pd_bbs_sr_1/102-6787070-3364907?ie=UTF8&s=books]Guide to LaTeX by Kopka & Daly[/url].[/quote]

thank you very much.

[quote="stevem"]1)You don't say what operating system you use. [url=http://miktex.org/]MikTeX 2.8[/url] (Windows) and [url=http://www.tug.org/texlive/]TexLive2008[/url] (Windows, Mac, Linux, Unix) are the latest versions. The complete versions (rather than the basic ones) will include BibTeX. The downloads will be very large - more than 1GB.

2) [url=http://www.math.umbc.edu/~rouben/beamer/]A Beamer Quickstart[/url] is a good introduction to Beamer with an easy to follow [url=http://www.math.umbc.edu/~rouben/beamer/quickstart-Z-H-3.html#node_sec_3]A bare-bones sample[/url].

I would strongly suggest that you are familiar with LaTeX first before you use Beamer or BibTeX and have read the main parts of [url=http://www.amazon.com/Guide-LaTeX-4th-Helmut-Kopka/dp/0321173856/sr=8-1/qid=1161947605/ref=pd_bbs_sr_1/102-6787070-3364907?ie=UTF8&s=books]Guide to LaTeX by Kopka & Daly[/url].[/quote]

i tried to copy the example in barebones sample but here is what i got.

That's because you haven't got the pgf package installed probably because you don't have a [i]complete[/i] version of MiKTeX. MiKTeX is offering to install pgf for you so go ahead.


In triangle $ABC$, let $D$ be the midpoint of $BC$.

Find $\frac{\cot \angle DAC-\cot \angle DAB}{\cot \angle ADB}$

$\measuredangle DAC= \alpha$
$\measuredangle DAB= \beta$
$\measuredangle ADB= \theta$

By sine law: $\frac{sin(\alpha+\beta)}{sin(\beta+\theta)}=2\frac{sin\alpha}{sin\theta}$

Then: $(sin\alpha cos\beta+sin\beta cos\alpha)sin\theta=2sin\alpha(sin\beta cos\theta+sin\theta cos\beta)$

$cot\beta+cot\alpha = 2(cot\theta+cot\beta)$

$\frac{cot\alpha-cot\beta}{cot\theta}= 2$

Draw perpendicular lines from $B,C$ to $AD$ at $P,Q$ respectively.

$\frac{cot\angle DAC-cot\angle DAC}{cot\angle ADB}=\frac{CQcot\angle DAC-BPcot\angle DAC}{BPcot\angle ADB}=$$\frac{AP-AQ}{DP}=\frac{PQ}{DP}=2$

Note: If $\angle ADB>90^\circ, \frac{CQcot\angle DAC-BPcot\angle DAC}{BPcot\angle ADB}=\frac{AP-AQ}{-DP}=\frac{PQ}{DP}$.

Denote $m(\widehat{DAC})=x$ , $m(\widehat{DAB})=y$ , $m(\widehat{ADB})=z\ .$ Thus, $1=\frac{DB}{DC}=\frac{AB}{AC}\cdot\frac{\sin y}{\sin x}=\frac{\sin (z-x)}{\sin (z+y)}\cdot\frac{\sin y}{\sin x}$

$\Longrightarrow$ $\sin y\sin (z-x)=\sin x\sin (z+y)$ $\Longrightarrow$ $\cot x-\cot z=\cot y+\cot z$ $\Longrightarrow$ $\boxed{\ \frac{\cot x-\cot y}{\cot z}=2\ }\ .$


The tangent TP of length 35 cm is drawn on a circle, with the point P being 37 cm from the centre of the circle.

a) Find the length OT (radius of circle)
b) Find the angle theta (angle made at the centre of circle)
c) Find the area of the sector OAT.
d) Find the shaded area.

[hide=answer]
a)$\text{radius of circle}=37^2-35^2=\sqrt{144}=12$
b)$\sin{\theta}=\frac{35}{37}$                                 $\theta=71.07535558$
c)Where did the $A$ come from?
d)What is the shaded area?

I believe a diagram is needed. I'm assuming $A$ is the point on the circumference of the circle between $O$ and $P$.  The area of sector OAT = $(\frac {71.07535558}{360} \times \pi \times 12^2) cm^2$ = $89.31592598 cm^2$.

Shaded area= $(210 - 89.31592598) cm^2 = 120.6840704 cm^2$.


calculus
integration
LaTeX
calculus computations

take $ tanx\equal{}t$ you will get an expression of the form $ (1\plus{}(bt/a)^{2})/(1\plus{}t^{2})^{2}$ now split the numerator as $ 1\minus{}(b/a)^{2}\plus{}(b/a)^{2}(1\plus{}t^{2})$ then u r done with with the resulting second expression which requires the integration of $ 1/(1\plus{}t^{2})$
and the first expression requires the integration of square of it which can be done by further splitting its numertor.
P.S: sry for incomplete non latex solution

[quote="v.dinesh"]take $ tanx \equal{} t$ you will get an expression of the form $ (1 \plus{} (bt/a)^{2})/(1 \plus{} t^{2})^{2}$ now split the numerator as $ 1 \minus{} (b/a)^{2} \plus{} (b/a)^{2}(1 \plus{} t^{2})$ then u r done with with the resulting second expression which requires the integration of $ 1/(1 \plus{} t^{2})$
and the first expression requires the integration of square of it which can be done by further splitting its numertor.
P.S: sry for incomplete non latex solution[/quote]

Wonderful, thank you


[hide="blah"]We have $a>(\frac13)(\frac35)(\frac57)\cdots(\frac{1997}{1998})=\frac1{1998}$

Also, $\frac1{1998}>\frac1{1999}$, so we have that $a>\frac1{1999}$

$a<(\frac23)(\frac45)\cdots(\frac{1998}{1999})$

Multiplying this by $a$, we have:

$a^{2}<(\frac23)(\frac34)\cdots(\frac{1998}{1999})$

$a^{2}<\frac1{1999}$

$a^{2}<\frac1{1999}<\frac1{1936}$

$a<\frac1{44}$[/hide]

Most proofs, unless absolutely basic do not belong in HSB.  Moved to Intermediate.

Hmm.  Those bounds are disappointing.  Can anyone come up with clever ways to strengthen it?

[hide="Say..."] By induction on the product $a_{2k}= \frac{1}{2}\cdot \frac{3}{4}\cdot ... \cdot \frac{2k-1}{2k}$.  Hmm?[/hide]


$ \spadesuit$. Let $ a,b,c>0$ such that $ 9\plus{}3abc\equal{}4(ab\plus{}bc\plus{}ca)$. Prove that $ a\plus{}b\plus{}c\ge 3$

Let $ p\equal{}a\plus{}b\plus{}c,q\equal{}ab\plus{}bc\plus{}ca,r\equal{}abc$.From the conditon we have
\[ (i)\quad 9\plus{}3abc\equal{}4(ab\plus{}bc\plus{}ca)\leq \frac{4}{3}(a\plus{}b\plus{}c)^2\Longrightarrow p^2\geq \frac{27\plus{}9r}{4}\]
and from AM-GM
\[ (ii) 9\plus{}3r\equal{}4q\geq 4\cdot 3\sqrt[3]{r^2}\Longrightarrow r\leq 1\]
Also we get the following one from Schur
\[ (iii)\quad 9\plus{}3abc\equal{}4(ab\plus{}bc\plus{}ca)\leq\frac{p^3\plus{}9r}{p}\Longrightarrow (p\minus{}3)(p^2\plus{}3p\minus{}3r)\geq 0\]
So we just need to show that $ p^2\plus{}3p\minus{}3r\geq 0$ which is obvious from $ (i)$ and $ (ii)$.

[quote="Tourish"]
and from AM-GM
\[ (ii) 9 \plus{} 3r \equal{} 4q\geq 4\cdot 3\sqrt [3]{r^2}\Longrightarrow r\leq 1\]
[/quote]
Explain this step, please. Thank you.


It's just a minor question because they didn't have such sequence that worked as an example.

Would a sequence with at least one A by itself, e.g. ABAA, work or would it not?  I think it would be easier to count the necessary sequences if it doesn't work.  Then again, it says "[i]consecutive[/i] A's."

$ ABAA$ has a run of $ 1$ consecutive $ A$. $ 1$ is an odd number and not an even one. So it breaks the restriction on the length of consecutive $ A$s and thus is not valid.


Solve the system:
$\sqrt{x^2-2x+6}.log_3(6-y)=x\newline \sqrt{y^2-2y+6}.log_3(6-z)=y\newline \sqrt{z^2-2z+6}.log_3(6-x)=z$
              [url=http://www.mathlinks.ro/Forum/topic-76271.html]View all problem from Vietnam NMO 2006[/url]

hello, i think $ x\equal{}y\equal{}z\equal{}3$ is the only real solution.
Sonnhard.

[hide]Let $ f(u) \equal{} \dfrac{u}{\sqrt {u^2 \minus{} 2u \plus{} 6}}$.

It is easy to prove that $ f(u) < 1$ if $ u < 3$, $ f(u) > 1$ if $ u > 3$, and $ f(3) \equal{} 1$.

Rewrite the system as:
$ y \equal{} 6 \minus{} 3^{f(x)}$
$ z \equal{} 6 \minus{} 3^{f(y)}$
$ x \equal{} 6 \minus{} 3^{f(z)}$

If $ x < 3$, then $ f(x) < 1$, and $ y > 3$.
If $ y > 3$, then $ f(y) > 1$, and $ z < 3$.
If $ z < 3$, then $ f(z) < 1$, and $ x > 3$.
This is a contradiction.

If $ x > 3$, then $ f(x) > 1$, and $ y < 3$.
If $ y < 3$, then $ f(y) < 1$, and $ z > 3$.
If $ z > 3$, then $ f(z) > 1$, and $ x < 3$.
This is a contradiction.

If $ x \equal{} 3$, then $ y \equal{} 3$, and $ z \equal{} 3$.
Therefore, $ (x,y,z) \equal{} (3,3,3)$ is the only solution. [/hide]


Αν $ a,b,c$ 3 μη αρνητικοί πραγματικοί αριθμοί να δειχτεί η ανισότητα:

$ a^2\plus{}b^2\plus{}c^2\plus{}2abc\plus{}1\geq 2(ab\plus{}bc\plus{}ca)$

Βγαίνει με ΑΒC...δεν είχα αυτό στο μυαλό μου... Η λύση που έχω στο μυαλό μου πάντως χρησιμοποποιεί πιο απλές ανισότητες :wink:

Nick με το ΑΒC είναι σαν να λύνεις τη γεωμετρία του αρχιμήδη με μυγαδικούς  :P

Λοιπόν να η λύση μου (mim την αφιερώνω σε σένα)
Από την ανισότητα Schur kai met;a AM-GM έχουμε

$ 2(ab \plus{} bc \plus{} ca) \minus{} a^2 \minus{} b^2 \minus{} c^2\leq\frac {9abc}{a \plus{} b \plus{} c}\leq\frac {9abc}{3\sqrt [3]{abc}} \equal{} 3\sqrt [3]{(abc)^2}$

Και $ 2abc \plus{} 1 \equal{} abc \plus{} abc \plus{} 1\geq 3\sqrt [3]{(abc)^2}$   and we are done  :)

Thnx Σιλουανέ για την αφιέρωση και την όμορφη λύση :)  Αυτή είχα κι εγώ στο μυαλό μου.


is anyone here attending fiitjee classes

i am

so now those going must have given FTRE or FTSE

especially rajiv

plz tell me what all they r going to ask in the second level i.e. interview stage before providing scholarships

result of ftre is round the corner so i must start preparing for interview

so i want ur suggestions(rajiv)

i had given FTRE for class 10th

so should i be preparing some specific topics or wat

plz reply :help:  :help:  :help:

I am a Fiitjee student and took the FTRE for Class 10. Last time I went for the FTRE interview(in 8th).... I was asked the basics in Physics/Chemistry/ Math. The interview lasted for about 2 hours.
How much are you expecting in the writtten round out of 405 this time? You can PM me. Good Luck.

First of all, thanks to address me individually!  :D
I dont attend classes at the FIITJEE, but I've only enrolled for the All India Test Series for IIT-JEE. So I'm sorry I cant help you much about the FTRE :( .

yeh rajiv

i am expecting a score of about 360-370

what r my chances of getting selected for interview

how much did u score

plz reply :)

Expectations very rarely turn into the same results.
But it depends on how others have performed.
What is so special about the interview?
I have written the FTRE for Class 9 and I am expecting the same range of marks as Jaswinder though dont know whether it will be true.

@aravind

so u wanna say i am a lier

the answers have been discussed in fiitjee

and after that i have given my expected marks

i am already attending fiitjee classes

u r so rude

better be polite if u want to continue on this forum

 :harhar:  :furious:

aravind we r not here to fight

in first post u sounded rude in the next me

so both of us should apologize

i am apologizing though u r junior to me

i never meant that ur continuation of this forum is in my hands

i said if u want to be friendly with me

but nonetheless we r stopping this matter here

lets be friends again....................

afterall we r here to discuss math and nothing else so let's do that

plz give me the links to some real nice problems and some files if u have any............... :)  :thumbup:

[quote="Aravind Srinivas L :bruce: "]Excuse me Jaswinder but I never said that you are a liar.
I said that what one expects are not fully what one gets in omst cases. Why do you have to take that in thw wrong manner?
If you ask me, I would say that you are the one being rude though I dont want to stress on that fact which is out of discussions now.
And who are you to decide my continuation on this forum?Let the moderator decide ok. THs might sound rude but I just have to say it. And stop this immediately after this post without a reply. Maybe PM me if u want to continue this but dont continue it in this topic as we dont want out of topic discussions.[/quote]

[quote="Jaswinder  :weightlift: "]
@aravind

so u wanna say i am a lier

the answers have been discussed in fiitjee

and after that i have given my expected marks

i am already attending fiitjee classes

u r so rude

better be polite if u want to continue on this forum
[/quote]

Jaswinder v/s Aravind
 :starwars:

 :clap2:  :thumbup:  :clap:

More updates available shortly!!!

 :rotfl:  :rotfl:  :heli:

[quote]in first post u sounded rude in the next me

so both of us should apologize

i am apologizing though u r junior to me

i never meant that ur continuation of this forum is in my hands

i said if u want to be friendly with me

but nonetheless we r stopping this matter here

lets be friends again.................... [/quote]
[quote][b]lets be friends again[/b].................... [/quote]  I never thought I was your friend before.

[quote]in first post u sounded rude in the next me

so both of us should apologize [/quote] I nevr thought I sounded rude so I am not going to apologze neither did I ask you to as I do not feel serois for these childish things like you are a so called senior to me.

[quote]Jaswinder v/s Aravind

More updates available shortly!!! [/quote]

Sorry Rajiv for this late update but I forgot about this topic altogether.  Ony now got reminded of it.
Anyway, I did this for [b]chumma jujubi[/b] only  :rotfl:   :rotfl:   :rotfl: .
Jaswinder my SUPER SENIOR, could you please enlighten us with news of your FTRE rank. I think I remember the results to have been announced on 20th the January itself.AIR 1 or AIR 2? But sadly there is no interview this year Jaswinder Pehredar who claims himself to be ther head of maths city and wants to be called if there is a calamity. WIll come in a jetplane  :pilot:   :rotfl: .

[/quote]

And yeah finally Par could u move this topic to the general discussion as our SO CALLED SUPER SENIOR didnt know where to post a discussion topic too sad.
And Jaswinder, please take it in the right spirit ok. Dont get angry.

sorry if i am interupting but i was just wondering what is a GOER??   :huh:
and cheers for the fight :thumbup:    :clap2:

may those who go to FIITJEE :rotfl:  :oops:

Anyway, if the post of Jaswinder the seniro read for those who go to FIITJEE, I wouldnt have posted a single reply as I dont go to tht stpid institute{Stupid atleast in TN}.
But goers?Who are goers? If I see the very reliable dictionary, I see that goers means those who leave or depart.
That means FIITJEE goers means those leave or derpart from FIITJEE or something easier those who dont go to FIITJEE. Then why have D1994 and Jaswinder posted here?
 :rotfl:  :rotfl:  :rotfl:
goer - someone who leaves
departer, leaver
migrant, migrator - traveler who moves from one region or country to another. This is posted as a result of my looking up the dictionary as well as the thesaurus.

 :rotfl:

[quote="Aravind Srinivas L"]
goer - someone who leaves
departer, leaver
migrant, migrator - traveler who moves from one region or country to another. This is posted as a result of my looking up the dictionary as well as the thesaurus.

 :rotfl:[/quote]

That should go to the English Literature forum, if there's any!

[quote="Rajiv"]
That should go to the English Literature forum, if there's any!

[/quote]
Now that somebody has mentioned English Literature :wink: , I'd like to know who has taken the IELTS? I'm planning to take the test soon.

Why did the fighting stop? Keep going! Remember, I'm your ref. :rotfl:

I think someone has taken it to his heart :noo:

Aravind pls dont leave ,we really want future geniuses like you in this forum.

hey,u hav almost won the fight ,,no need to leave so soon :P  :lol:

[quote="Aravind Srinivas L"]Anyway, I must stop my frequent visits to this forum with so many INMO Awardees and I must become one to have eligibilty to talk with them.[/quote]
Bro, here I am....shamelessly admitting I couldn't even manage a 90 in my ICSE Boards! By that measure, I have no right to be anywhere. But we are all proud aopsians! There are no educational qualifications for this great community. A burning passion for science and maths is al that needs to glue us together. And that you have in heaps. C'mon! How many of us even thought of giving RMO in 10/9? You are a champion! Don't feel disappointed so soon. You have a long way to go :)

Well said man ... I second gaurav in this regard ...

Now i have done it i am really really sorry,i dont mean a thing of it please forgive Aravind and get back here  :(
and talking of your eligiblity you know it better than everyone
once again i am really sorry

just to avoid confusion ... "gaurav" in my last post was Gaurav Shukla ...

[quote="gauravpatil"]
Sad Rajiv the other team is making you look like a fool

[/quote]
I dont think so...i :wink:  :P If you read post before it, I think the insult had hit the target it was aimed at. Therefore such a reaction is natural.

[quote="gauravpatil"]
this whole post has a largely sarcastic touch to it (see the enough) ,you cannot accept such an appreciation can u?I find it quite surprising that you with the intelligence of yours accepted such an insult as a praise.

[/quote]

Well, I agree I'm not intelligent, but I'm not that foolish either (I can understand this post of yours too has a "sarcastic touch" which proves my latest claim :rotfl: ) not to understand such taunts :rotfl:. So you can see there were no points awarded for that!

[quote="Rituraj"]
LOL! I'd give it a $ 6$ star!

[/quote]

I too wanted to do the same [hide="but..."]I RESERVE SUCH GENEROUSITY ONLY FOR THE FAIRER SEX... :rotfl:  :D [/hide]

[quote="Rituraj"]
Please post our points

[/quote]

Then.....

[quote="Aravind"]
Bye.

[/quote]

Can I take it as a :surrender:?

That way I'd be saved from the pains of counting your points. :rotfl:

In case others want to continue on his behalf, let me announce the revised marking scheme:

[b]Each Post:[/b] 10pts-boys, 20-girls (dnt ask why, I already explained it b4).
[b]I like your post:[/b] 50pts-boys. 100-girls(I usually like all of their posts, good or bad. :D )

I m fiitjee integrated batch .I m in class 9th. If u guyz hve any wat's app group add me . .  If not can we make one and ask doubts

I used to go to Fiitjee, Now I think aops is better. Who is with me!

[quote=nbala2k2]add me too 9445821795
balaji[/quote]

Just look at the year of posts.
Everything is settled down :)


What exactly is nuclear binding energy?  How does one calculate it?

My book mentions it, but I'm confused where it is getting its numbers from... is it positive or negative (can it be either or must it always be + or - )?

Thanks!

By the way, this is in my chemistry book... but is this more of a Physics topic?  Should this threat be moved to a more appropriate board?

Nuclear binding energy is, well, the energy that holds a nucleus together.  Not very informative, is it?  Here's a practical way of thinking about it, that also explains how one calculates it:

Start with a pair of neutrons completely separated from one another.  Now, as we bring them together, closer and closer, what happens?  Well, there is a force acting to attract them: the strong interaction (really, it acts on their quarks, but that's ok).  So this situation is analogous to taking two planets and bringing them together (analog of strong force in this case is gravity)--and we know what happens there, we release a lot of stored potential energy.  Same thing happens here.

So we release the potential energy... but recall, energy is mass!  So what happens is, the neutrons lose some mass, which is converted into energy, which is released... this is the mysterious binding energy.  A similar thing happens when we throw more nucleons into the mix, just protons also have a small repulsion due to electrostatics, but it is weak compared to the strong interaction.

So to determine it, all one has to do is check the mass of a nucleus (which can be done in a number of ways) and compare to the total mass of the nucleons... the difference is the binding energy.  It is as simple as that!

[quote="gregx007"]So to determine it, all one has to do is check the mass of a nucleus (which can be done in a number of ways) and compare to the total mass of the nucleons... the difference is the binding energy.  It is as simple as that![/quote]

Hmmm.... yeah... I guess I was getting confused because my book just put together some random numbers together that I had no idea where they were coming from... like, it has no references for where to find the masses of nuclei and masses of nucleons...

Binding energy is a positive value, right?

I don't know the sign convention... it doesn't matter much, all binding energy is released when nuclei form.  Otherwise, they would just immediately fall apart.  Note even with things that do break apart, such as U, there is still a binding energy released when they form: just less than the elements that they form when breaking apart.


limit
real analysis
real analysis unsolved

Assume that $\sum_{j=1}^n a_j=\sum_{j=1}^n b_j$, for $a_j \in N,b_j \in N\ (j=1,2\cdots)$.

Evaluate

\[\lim_{n\to\infty}\prod_{j=1}^n\frac{\sum_{k=1}^n k^{a_j}}{\sum_{k=1}^n k^{b_j}}\]

I think that it goes using Stolz-Cesaro:
denoting $\lim_{n\to\infty}\frac{\sum_{k=1}^n k^{a_j}}{\sum_{k=1}^n k^{b_j}}=\lim_{n\to\infty}(n+1)^{a_j-b_j}$ and multiplying such result  for j=1,2,...,n we obtain
that our limit is equal to $(n+1)^{\sum_{k=1}^n a_k - \sum_{k=1}^n b_k}=1$


$ a,b \in N$ so that $ a^n \plus{} n|b^n \plus{} n$ for all $ n \in N$.Prove that $ a \equal{} b$.

I'm waiting nice solutions...


Let $m,n$ be positive integers such that \[ \sqrt 7-\frac mn>0.\] Prove that \[\sqrt 7-\frac mn>\frac{1}{mn}.\]
(grade 9)

If $0< \frac m n < \frac {\sqrt 7} 2$ then $\sqrt 7 - \frac m n \geq \frac {\sqrt 7 } 2 > 1 \geq \frac 1 {mn}$ and we are done in this case.

Now assume that $\rac {\sqrt 7 } 2  < \frac m n < \sqrt 7$ then $m^2-7n^2 < 0$ and $n \sqrt 7 + m < 3m.$
But, for all integer $x$, we have $x^2 = 0,1,2,4 \mod[7]$. Thus for all $m,n$ we have $m^2-7n^2 \neq 5,6 \mod [7]$. It follows that $m^2-7n^2 \geq -3$.
Therefore  :
$n\sqrt 7 - m \geq \frac 3 {n \sqrt 7 + m} > \frac 1 m $ which leads to $\sqrt 7 - \frac m n > \frac 1 {mn}$ as desired.

Pierre.


[quote="ashwinrk_jain"]answer is $1$.If the answer is true then i will submit the soluion.[/quote]
You should be more confident with yourself! Go ahead to post your solution!  :D

Denote the function $f: \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\rightarrow R$, where $f(x)=x^{3}+\sin x\ .$ For $\{x,y\}\subset \left[-\frac{\pi}{4},\frac{\pi}{4}\right]$ we observe that $|2y|\le\frac{\pi}{2}$, $f(-x)=-f(x)$ and $f'(x)=3x^{2}+\cos x>0$, i.e. the function $f$ is $\nearrow$ (in particularly, $f(x)=f(y)\Longleftrightarrow x=y$). From the system results $f(x)+f(2y)=0$ $\Longrightarrow$ $f(2y)=f(-x)$ $\Longrightarrow$ $2y=-x$ $\Longrightarrow$ $x+2y=0$ $\Longrightarrow$ $\cos (x+2y)=1\ .$


i really need help
i installed miktex 2.7 according to AoPS recommendations
when i used fancyhdr, texnic center prompted me to install the package
what directory should i use

i have fancyhdr.sty and extramarks.sty in c:\\program files\miktex 2.7\tex\latex\fancyhdr

That message is from MiKTeX. It will know where to install fancyhdr.
Did you install fancyhdr yourself? If so, did you remember to refresh the database (MiKTeX, Settings, Refresh FNDB)?


abstract algebra
limit
superior algebra
superior algebra unsolved

If we take $ \widehat{M}$ the completion of $ R \minus{}$ module $ M$ with respect to $ \mathfrak{a} \minus{}$adic topology. Then how can we have that $ \widehat{M}\equal{}0\Leftrightarrow M \equal{} \mathfrak{a}M$?
Thanks for helping me.

if $ M \equal{} \mathfrak{a} M$, then $ M \equal{} \mathfrak{a}^n M$ for all $ n \geq 1$ and thus $ \widehat{M} \equal{} \lim M / \mathfrak{a}^n M \equal{} 0$. to show the converse, assume $ M \neq \mathfrak{a} M$, i.e. $ \exists m \in M \backslash \mathfrak{a} M$. then $ (m,m,...)$ defines a nontrivial element in $ \widehat{M}$.


Select numbers a and b between 0 and 1 independently and at random. Let c be their sum. Let A, B, and C be the results when a, b, c, respectively, are rounded to the nearest integer. What is the probablity that A+B=C?

[hide] well (a,b) corresponds to a point in a unit square,

if a < 1/2 and b > 1/2 (or vise versa)  then a gets rounded to 0 and b goes to 1,  and since 1 < a+b < 3/2   c is rounded to 1.  so 0+1=1. and these sections make up for 1/2 of the total area

now consider if a+b< 1/2  (or similarly a+b> 3/2),  clearly a and b are less thanh 1/2 so they get rounded to zero, and since the sum is less than on half, so does c, giving us 0+0=0.  These two sections make up another 1/4 of the area giving us a total of 3/4[/hide]


A sequence $ (a_n)$ of integers is defined by $ a_1\equal{}1,a_2\equal{}2$ and $ a_n\equal{}3a_{n\minus{}1}\plus{}5a_{n\minus{}2}$ for $ n\equal{}3,4,5,...$

Does there exist an integer $ k\geq 2$ such that $ a_k$ divides $ a_{k\plus{}1}a_{k\plus{}2}$?

Hmm, you're posting almost all the problems that have appeared on Polish Mathematical Olympiad this year.
It was easy, just transform the sufficient condition to
$ a_{k}|75(a_{k\minus{}1})^{2}$
And we can even prove that $ GCD(a_{k},a_{k\minus{}1})\equal{}1$ which finishes the problem :wink:


AMC
USA(J)MO
USAMO
blogs
geometry
geometric transformation
reflection

See for yourself.

http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/06usamo-results.html

Very disappointed.... 10 :( I was hoping for at least 14...

[quote="Phelpedo"]Er, how exactly did you find this site? The URL isn't linked from anywhere else...[/quote]

Dnas gave me the link on AIM.

[quote="nr1337"][quote="chess64"]Who's the perfect? :)[/quote]

Why, number 188 of course![/quote]

And who's #188? :P

[hide="DO NOT CLICK"]WAIT HUH I GOT A 3???

WTF??? NO THAT'S IMPOSSIBLE
LIKE ABSOLUTELY IMPOSSIBLE FOR WHAT I DID

AND ANYWAY I GOT A 3 LAST YEAR... I WAS HOPING FOR LIKE 21 WHICH WAS TOTALLY REACHABLE GIVEN HOW I DID ON PRACTICES... BUT AT LEAST RED MOP

NO NOW I CAN'T MAKE RED MOP

NO THIS IS IMPOSSIBLE
WAIT HUH HOW DID THIS HAPPEN

WTF WTF WTF
OMG OMG OMG
WAIT A 3?????

HOW IS THAT POSSIBLE??

LIKE A 3???
CAN YOU EVEN GET A 3???
WAIT HUH HOW DID THIS HAPPEN???
NO THIS ISN'T POSSIBLE

IT JUST ISN'T POSSIBLE
ARE YOU SURE THERE ISN'T SOME MISTAKE
OF COURSE THERE ISN'T, BUT OF COURSE THERE IS
A 3??????
THAT'S NOT POSSIBLE!!!!!!!!!!!!!!!

THIS IS IMPOSSIBLE!!!!!!!!!!!!!!
NO
NO
NO
NO
NO
HOW DID I NOT MAKE RED MOP
HOW HOW HOW
I WAS HOPING FOR BLUE BUT AT LEAST RED
I'VE NEVER NOT COME IN THE TOP 5 FOR ANY STUPID GRADE-WISE RANKING OVER THE PAST 4 YEARS
AND NOW I'M NOT IN THE TOP 30???

OMG OMG OMG
WAIT HUH THIS IS IMPOSSIBLE
HOW DID THIS HAPPEN

NO THIS JUST CAN'T BE TRUE
NO IT JUST CAN'T
NO HOW DID I GET A 3???
HOW DID I NOT MAKE RED MOP
NO I WAS UBER EXCITED FOR MOP
NO
HOW DID THIS HAPPEN
NO NO NO

NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO!!!!!!
*SIGHS WITH ANGUISH*
NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO!!!!!!

--collapses-- [/hide]

YEAH BUT IT WAS MY SECOND TRY  ;)

MY FIRST TRY I WAS LIKE WHOA I GOT A 43.... ERRR THEN I ENDED UP WITH A 3... BUT I DIDN'T CARE 'CAUSE I WAS LIKE HUH 8TH GRADE USAMO DOESN'T MATTER

NOW I SORT OF LIKE... CARE
;)

HMMM THAT RANT TURNED OUT PRETTY LONG... I SHOULD PROBABLY HIDE IT OR SOMETHING

OMG :furious:  :wallbash_red:
I seriously thought I did way better.
Is it possible that they didn't receive some of my papers or messed sth up?Because I solved 4 problems (with 1 minor error)and did partial stuff for number 3. BUT MY SCORE WAS 17.
I overestimated my score cause it's y 1st time but i never even thought of getting sth this low.

I might not make Can IMO team this year. feel very depressed.

[quote="chess64"]Who's the perfect? :)[/quote]

 Brain Lawrence ;)

winners and HM:
http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/06usamowinners.html

Just so you know... capital letters are very annoying.  And I'm still confused about the point phelpedo made, in regard to why these would be up so early and why there's no link on the AMC site to reach it  :?  Also, there is a huge difference in the number of significant digits in both the percentile groups and average scores of this year's in comparison to past years.  I was going to say that they listed 450 numbers when less than that number took the test... but past years also seemed to round up to the nearest multiple of 25.

Sorry if I seem skeptical but after the fake USAMO list it's hard to believe anything but a phone call  ;) Just because Dnas said it doesn't make it true... and that's assuming that he DID say it.

[quote="darktreb"][quote="h_s_potter2002"]So I'm thinking MOP cutoffs for non-freshman MOP is a 22 or 23.[/quote]

Honorable Mention cutoff is already 23. Don't forget a lot of those scores are put up by decent Seniors. The MOP cutoff is probably going to lower than 22....

As for bubala ... welcome to the club of people who dramatically overestimate their USAMO scores on their first try. Writing a rigorous proof is harder than most people think. I've yet to meet someone who underestimated their score on their first USAMO try.[/quote]

I seriously thought I would get a 12 and I got an 18.

This might be a dumb question, but: do they take off marks if you used separate sheets of paper to write up solutions, and for one of problems you never used the original sheet(the one in teacher's manual)?????? :?

[quote="chess64"][quote="jrshoch"]Sorry if I seem skeptical but after the fake USAMO list it's hard to believe anything but a phone call  ;) Just because Dnas said it doesn't make it true... and that's assuming that he DID say it.[/quote]

Huh what fake USAMO list

anyway it's on the AMC website so yeah.

And bubala, I'm very surprised that you got a 3 too. maybe some papers didn't get sent in? :?[/quote]
I think not all of mine were marked. :( Oh well. Maybe I just can't write up solutions.whatever

[quote="jrshoch"]Sorry if I seem skeptical but after the fake USAMO list it's hard to believe anything but a phone call  ;) Just because Dnas said it doesn't make it true... and that's assuming that he DID say it.[/quote]

I certainly didn't make it... or else I wouldn't have given myself a 10! :D  But seriously, DPatrick deleted this thread for a while, then brought it back up, so maybe the AMC people said it was ok?  I don't know.  I just posted what I saw on impulse.  I just hope I didn't do something wrong.

[quote="amirhtlusa"]no its all on the AMC website right now:
http://www.unl.edu/amc/e-exams/e8-usamo/usamo.html  ;)  ;)[/quote]

Erm.... I don't see it.

[quote="ThAzN1"][quote="darktreb"]I've yet to meet someone who underestimated their score on their first USAMO try.[/quote]

Hah, you sure?  ;)

Hmm, I guess the Blue MOP cutoff will probably be around what it was last year... 20-21 seems reasonable...[/quote]

Disclaimer is that this doesn't include people who don't even manage to realize their "invalid" solution was actually valid :P

Why are the student numbers starting from 11 and not 1???????

response to post # 24:

ummm...
its says on the left side  ;)  ;)

also rruszyk posted it his blog (sorry if i spelled it wrong)

The results have been officially released.  Go to the [url=http://www.unl.edu/amc]AMC home page[/url], click on the USAMO tab near the top, then look in the left column.

If you don't see it at first, click the refresh button and it will appear. I don't really know why, but...

In response to why do student numbers start at 11...

Looking at past years, the student numbers always start at 11 and go to a number a little bit higher than the number of students who took it.... no idea why.

And sorry about the whole skepticism, I've been thoroughly disproved.  Lol if anyone can make a fake bunch of pictures of USAMO graders I think they should be invited to MOSP for creative talent  :lol: .  And I guess this does make everything Dnas says true (sorry if I offended anyone by questioning that before).

to all wondering if their papers didn't get marked, when I was a freshman, I know that I solved #1 completely and got partial credit on some other problems, yet I got a 5. Huh? Yea, so in general, the USAMO graders know what they're doing, and I think that in the manual, they specifically say to fax in all papers (even if you haven't written anything) to be SURE that they've gotten all the papers. Thus, you probably just overestimated your score.

[hide="jrshoch only:"] no matter what your aesthetic views on capitalization, that remark was slightly insensitive. You probably didn't mean to be so harsh, but yeah. [/hide]

[quote="jrshoch"]Ok in response to mysticterminators truthful response and a bit of self-reflection:

Sorry to anyone any of my comments have made feel bad.  I am very tense (as I'm sure everyone else is)... I hate first experiences, waiting, and cutoffs; so as you can imagine this is a bad situation for me[/quote]
Ha we have very similar personalities ;)

Btw, somebody please answer this: do they take marks off if you use extra sheets of paper and sometimes don't use the original sheets(the ones provided in teacher's manual)?

[quote="rem"]Btw, somebody please answer this: do they take marks off if you use extra sheets of paper and sometimes don't use the original sheets(the ones provided in teacher's manual)?[/quote]

No. Why would they do that?

hah, I got a 4...that's four points higher than I thought it would be.

hehe bubala dont worry
i thought i got at least 15. i did 1,2,5 perfectly. but i forgot to mention a point for Q5.
also i was hoping to get partial mark for Q4.
but the score came out was 7 hehe.
jeez how could someone get perfect on it?

I was predicting a 3 or so...1 point for each of 1, 2, and 4. I guess I got higher...dont' want to say how much higher...but higher nonetheless.


I was thinking about another fun problem on this board that involved adding up the squares of the cosines of the degree measures between 1 and 90. I figured out how to do it but then I thought of some alternate methods, and I came up with this.

[img]http://i33.tinypic.com/2z9ffaf.jpg[/img]

Obviously that's not too good, but I'm not sure where I made the mistake. Could anyone shed any light on this?

Also, the limits of the summation are measured in degrees.

EDIT: Nevermind, the summation on the right on the top line should enclose the 1 as well as the sin^2(n). My bad. ;_;

I think $ \sum\limits_{n\equal{}1}^{90}$cos$ n^2\equal{}90\minus{} \sum\limits_{n\equal{}1}^{90}$sin$ n^2$

I think
you can't take $ sin  {n}^2\plus{}cos  {n}^2\equal{}1$

Instead it should be $ sin^2{n}\plus{}cos^2{n}\equal{}1$

Sorry for bad latexing

:blush: Oh sorry about my mistake :roll:

except for that, i am pretty sure the rest of steps have the correct concept...


modular arithmetic
number theory proposed
number theory

Prove that $2^{147}-1$ is divisible by $343.$

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

This is just $7^3|2^{3\cdot 7^2}-1=7\cdot (8^{7^2-1}+8^{7^2-2}+\ldots+8+1)$. This is equivalent to $7^2|8^{7^2-1}+8^{7^2-2}+\ldots+8+1$. The RHS is a sum consisting of $7^2$ terms, all of them being $1\pmod 7$, so it's divisible by $7^2$, Q.E.D.


Hello,

can someone give some hints to solve the following situation

* P vector represents the length between two points (Z and Y) and it's a position
(P = X + Y)
* v vector is the velocity at a certain point
then it's said that X (0) and V (0) are so that a rule is followed. From that rule its possible to reach
v<sqrt(2*G*m/2)
However if
V = VX + VY
and
howto get VX and VY?
Using V^2 = VX^2 + VY^2 ?
The problem is we dont know the angle and the trajectory is eliptic.

Thanks.

I'd love to help you out, but I simply don't understand your question! I think that you've made several typos and now I'm not sure what is what. Could you please edit your message so that the question is understandable? (What is Z, what is Y and what is X? And if $\vec p$ represents the distance between two points, $X$ and $Y$ (which are given by their vectors $\vec x$ and $\vec y$), then $\vec p = \vec x - \vec y$ instead of what you said...)

Hello,

P is a vector and represents the distance from planet P1 to planet P2. P2 is gravitating around P1. P has two coordinates, X and Y.

From a given expression (related to energy) it's possible to reach the following expression

v<sqrt(2*G*m/d)

G = Gravitational constant
m = planet mass (P1)
d = distance, d = sqrt ( Px^2 + Py^2 )

It's necessary to find the Initial P and V (vectors) using that velocity expression.

P = Px + Py (vector)
V = Vx + Vy (vector)

The problem is getting those components (for P and V vectors) for the initial moment wich has v < sqrt ( 2*g*m/d). From that expression is possible to get the "distance" and the "velocity" scalar values. The problem is getting the components.

Thanks.

Hmmm...

The problem is clear now, but it's a bit awkward. Let's see what's wrong with this reasoning:

The gravitational force acting on planet $P_2$ is $\vec F = G\frac{mm_0}{p^3}\vec p$, where $p$ is the distance between these planets and $m_0$ is the mass of planet $P_2$. As far as I can see, we're working in the approximation $m\gg m_0$ meaning that $P_1$ doesn't move significantly, whereas $P_2$ rotates around $P_1$, following an elliptical trajectory. Sweet; this is what the Earth does as it revolves around the Sun. Now, the "steady state" of this planetary motion is given by $\frac{m_0v^2}{p^2}\vec p = \vec F$ -- the centrifugal force acting on $P_2$ must be equal to the gravitational force. Using the above expression for $\vec F$ we get that $v = G\frac{m}{p}$. This means that the speed of $P_2$ will [i]always[/i] be equal to $G\frac{m}{p}$, so the condition $v < \sqrt{2G\frac{m}{p}}$ will be satisfied for any distance $p$ between the planets.

I don't think that there are any mistakes in my reasoning. What do you think? If you can, post the entire problem (I mean the "given expression related to energy") -- maybe there's something you missed and we can help you find what.

Hello,

the energy expression is simple.

It says e < 0 and that e = 1/2 v^2 - G M / d

From that is possible to reach the v < sqrt (2*G*M/d) expression.

Thanks for helping.

Right! You don't need any equations, actually. The condition $E<0$ always means that the planet is "trapped" in a gravitational field; the only way for the planet to get out of the field is to obtain sufficient kinetic energy so that its total energy becomes positive...


If $f(x)=\frac{(x^{3}+8)(x^{2}-5x-6)}{(x^{2}-2x+4)(x^{2}-4)}$

Find $f(2001)$ without a calculator

DO NOT JUST SUBSTITUTE IN AND MULTIPLY OUT

[quote="Treething"]It's not turning out nice, and it looks like a contest math problem. Did I make a mistake?[/quote]
No, and you can't simplify that (3993990 isn't divisible by 1999, and 1999 is prime).

[quote="i_like_pie"][quote="Farenhajt"]If there were $x^{2}-5x+6$ in the numerator, the result would be nice. Perhaps a typo sneaked in?[/quote]
It would be $(x-6)(x-1)(x-3)=(1995)(2000)(1998)=7972020000.$[/quote]

[hide]No, you wouldn't keep the typoed part in there, so it'd be $2001-3=1998$[/hide]

[quote="i_like_pie"][quote="Farenhajt"]If there were $x^{2}-5x+6$ in the numerator, the result would be nice. Perhaps a typo sneaked in?[/quote]
It would be $(x-6)(x-1)(x-3)=(1995)(2000)(1998)=7972020000.$[/quote]

Well... I might be blind, but I think it would be

$\frac{(x^{3}+8)(x^{2}-5x+6)}{(x^{2}-2x+4)(x^{2}-4)}=\frac{(x+2)(x^{2}-2x+4)(x-2)(x-3)}{(x^{2}-2x+4)(x-2)(x+2)}=x-3$

so $f(2001)=1998$

EDIT: robinhe already got it.

[quote="i_like_pie"]Oh, you meant change the minus to a plus. I thought you wanted to add another trinomial to the numerator. :ninja:[/quote]

Oh, I see. Sorry for being unclear  :(