Solve $ \left\{ \begin{array}{ll} \frac {dx}{dt}\plus{}\frac{2}{t}(x\minus{}y) \equal{}1 & \quad \\
\frac {dy}{dt}\plus{}\frac{1}{t}(x\plus{}5y) \equal{}t & \quad \end{array} \right.$

hello, it must be
$ x(t)\equal{}\frac{1}{30}\left(\frac{30C_2\plus{}30C_1\plus{}2t^6\plus{}9t^5}{t^4}\right)$
$ y(t)\equal{}\minus{}\frac{1}{60}\left(\frac{30C_2\plus{}60C_1\minus{}8t^6\plus{}3t^5}{t^4}\right)$
Sonnhard.

[quote="kunny"]Your answer is incorrect.[/quote]

Are your sure?


Does $ \frac{a\minus{}bc}{a^2}\plus{}\frac{b\minus{}ac}{b^2}\plus{}\frac{c\minus{}ab}{c^2}\leq6$ holds?
with a+b+c=1 and a,b,c are positive reals.

[quote="bearkiller"]Does $ \frac {a \minus{} bc}{a^2} \plus{} \frac {b \minus{} ac}{b^2} \plus{} \frac {c \minus{} ab}{c^2}\leq6$ holds?
with a+b+c=1 and a,b,c are positive reals.[/quote]
Yes, of cause! It's Schur. :wink:

can you explain it a bit more? I don't understand it.

Let $ ab\equal{}z,$ $ ac\equal{}y$ and $ bc\equal{}x.$
$ \sum_{cyc}\frac {a \minus{} bc}{a^2}\leq6\Leftrightarrow\sum_{cyc}\frac {a^2\plus{}ab\plus{}ac \minus{} bc}{a^2}\leq6\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}b^2c^2(ab\plus{}ac\minus{}bc)\leq3a^2b^2c^2\Leftrightarrow\sum_{cyc}(x^3\minus{}x^2y\minus{}x^2z\plus{}xyz)\geq0,$ which is Schur.


Conside  a sequence as follow
$ a_{0}\equal{} 1$
$ a_{n\plus{}1}\equal{} sin{a_{n}}$
Find$ \lim n.a_{n}^{2}$

This solution use
1)Stolz lemma
2)L'Hospital law


AwesomeMath
summer program
AMC
USA(J)MO
USAMO
blogs
Mathcamp

Anyone hear about it? Sounds like good stuff, something to help bridge USAMO and MOSP.

One of my math teachers forwarded me an e-mail about it - it seems interesting, though I'm not sure I see how it's much different from, say, Math Camp.

Mr. Rusczyk wrote a short blog about it [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=3633]here[/url].  You can find AwesomeMath's website [url=http://www.awesomemath.org/]here[/url].

It sounds like it will be a great program.  Titu Andreescu and Zuming Feng are very experienced when it comes to problem solving.  I'm not sure how it will compare to Mathcamp though (never been there).

It will be TOTALLY different from mathcamp.  From what I've heard, Awesomewhatev is a camp focused solely on problem-solving mathematics.  Mathcamp is focused on a much broader range of math, from problem-solving to complex analysis to game theory to bayesian statistics, etc.

Apart from that, WhatevMath has a very rigorous program that students must follow.  Mathcamp grants students complete freedom--if you feel like no doing math at all, you don't have to (why you would choose this, I don't know--just mentioning this as a possibility).

Hmm how do you apply? Is there an application test?

I think there is an application test, just hasn't been released, though. I am going to apply and if I get in, I'll probably go if I can fundraise to pay at least 3/4 of the amount.

Should be awesome! Problem solving and problem solving! What more could you ask for?


algebra
polynomial
number theory
prime factorization

1. Prove that the equation [tex]y^2=x^3+23[/tex] has no integer solutions.

2. What is the remainder when dividing [tex]x^{2001}-1[/tex] by  [tex](x^2+1)(x^2+x+1)[/tex]?

3. For what integer a does [tex]x^2-x+a[/tex] divide [tex]x^{13}+x+90[/tex]?

[hide]1. put this on the pythagorean theorum. the hypotonuse is y^2. one side is x^3 and the other is 23.  Plug this into the formula .y^2 is a^2+b^2.  the side that is 2ab must be x^3.  The 23 side is a^2-b^2. subtract 2ab from a^2+b^2. now that side is a^2-2ab+b^2. That is (a-b)^2 = 23. so a-b=root 23.  but can (a+b)(a-b) be whole if a and b arent? the second equation is a+b*(root 23).  It must be a multiple of root 23.  but then but it must be cuberootable.  so a+b must be root 23 ^n if n is 5 mod 6. and now im stuck[/hide] im probably going in the wrong direction.

Scrambled -- just because three integers satisfy a + b = c does not imply that their square roots are integers, which is what you assumed when you plugged them into the formula for primitive pythagorean triples.

sorry, u didnt' go wrong, I did... the solution I though I had is faulty. OOPS!

Answer to #2:

[tex](x+1)(x^2+x+1)[/tex]

Hint: [tex]x^2+x+1[/tex] divides [tex]x^{2001}-1[/tex]. Let f(x) be the polynomial
[tex]\display{f(x)=\frac{x^{2001}-1}{x^2+x+1}=\frac{(x-1)(x^{2001}-1)}{x^3-1}}[/tex]

Let [tex]r(x)=ax+b[/tex] be the remainder when f(x) is divided by [tex]x^2+1[/tex] Then

[tex]ai+b=r(i) = f(i) = i+1[/tex] and [tex]a(-i)+b=r(-i) = f(-i) = -i+1[/tex]

So [tex]r(x)=x+1[/tex]

For #1, maybe we could express x as n and y as n+d and investigate the roots of the resulting polynomial...

Also, can you clarify #3?  What exactly is a?

edit: oops, misread the question.

hint for number 1: [hide]if a prime congruent to 3 mod 4 divides the sum of two squares, its exponent in the prime factorization must be even.[/hide]

and a is stated as being an integer.

#3. The only possible value for a is 2. But I am too lazy to verify [tex]x^2-x+2[/tex] divides [tex]x^{13}+x+90[/tex]. Could somebody verify this using Mathematica? Here is why a=2 is the only possible value. Plug in x=0, x=1 we find that a divides both 90 and 92. So a can only be +1, -1, +2, -2.  Plug in x=-1, we get a+2 divides 88. This excludes the cases a=1 and -2. Plug in x=-2 would exclude the case a=-1.

Rep:  when mods tell you something is impossible (such as you would need 1 + 2 :equiv: 0 (mod 4) or something), that means that it really is impossible.  However, there are lots of times when mods won't tell you something is impossible.  That doesn't mean that it is possible, necessarily -- it just means that mods aren't a powerful enough tool when employed so straighforwardly.


algebra
polynomial
complex numbers
number theory unsolved
number theory

Prove that if the number 1^n+2^n+4^n is prime n has to be a power of 3

Nice solution maxal!! very elegant...  :)

I like your solution too. Mine is more "brut force" but I have to admit yours is nicer  :)  :)  :)  :)

[quote="M@re"]I think i solve this problem but is very complicated.
This are the steps that you should folow...
First, prove that $3|n$ that is easy part, if 3 doesn't divide $n$ than $1^n+2^n+4^n$ is divisible with 7. Then let $m$ be such that $3^m|n$ but $3^{m+1}$ doesn't divide n.
Now write $n$ like $n=3^mn_1$, $3^m=M$ so $n=Mn_1$ and now show that $1^M+2^M+4^M|1^n+2^n+4^n$...(*) and that yeilds a contradiction.

(*) $1^n+2^n+4^n=1+2^{Mn_1}+2^{2Mn_1}$ now you shoul divide polynoms
$1+2^{Mn_1}+2^{2Mn_1}$ and $1+2^M+4^M=1+2^M+2^{2M}$.
when you are dividing them you should see period.... because $3$ doesn't divide $n_1$ you obtain $1^M+2^M+4^M|1^n+2^n+4^n$.
Dividing part is tricky one but i can't write it because... well you know how you divide polynoms i really don't have idea how to write it here so that part you should do yourself.. good luck

m@re :lol:  :lol:[/quote]
Thanks for the solution, my idea was similar but I probably made some silly mistakes in deviding polinomials.

Well,
as I told you, that part is very ugly...
:D

I know you probably want something elementary, but with a bit of algebra (Grobber, where are you? :) ) the problem can be solved in a much nicer way.

Consider the polynomial $\Phi_3\left(x\right)=x^2+x+1$. We will first show:

[color=blue][b]Lemma 1.[/b] If k is a positive integer not divisible by 3, then $\Phi_3\left(x^k\right)$, written as a polynomial of x, is divisible by the polynomial $\Phi_3\left(x\right)$ over $\mathbb{Z}\left[x\right]$; in other words, there exists a polynomial u with integer coefficients such that $\Phi_3\left(x^k\right)=u\left(x\right)\cdot \Phi_3\left(x\right)$.[/color]

[i]Proof of Lemma 1.[/i] The complex zeroes of the polynomial $\Phi_3\left(x\right)=x^2+x+1$ are the two primitive 3rd roots of unity, i. e. the two complex numbers z such that $z^3=1$ but $z\neq 1$ (each being a zero with multiplicity 1). Let u be one of these two zeros. Then, $u^3=1$ but $u\neq 1$. Now, since k is an integer, $\left(u^k\right)^3=u^{3k}=\left(u^3\right)^k=1^k=1$. On the other hand, since k is not divisible by 3, it is coprime with 3, so we can find integers $\alpha$ and $\beta$ such that $\alpha\cdot k+\beta\cdot 3=1$. Hence, if we had $u^k=1$, then we would have $u=u^1=u^{\alpha\cdot k+\beta\cdot 3}=u^{\alpha\cdot k}\cdot u^{\beta\cdot 3}=\left(u^k\right)^\alpha\cdot\left(u^3\right)^\beta=1^{\alpha}\cdot 1^{\beta}=1$, contradicting $u\neq 1$. Thus, we must have $u^k\neq 1$. [Well, I know this argumentation was overkill, but it's generalizable.] Thus, the number $u^k$ satisfies $\left(u^k\right)^3=1$ but $u^k\neq 1$. Hence, $u^k$ is also a primitive 3rd root of unity, i. e. a root of the polynomial $\Phi_3\left(x\right)$. In other words, $\Phi_3\left(u^k\right)=0$. But this can also be interpreted as follows: The number u is a root of the polynomial $\Phi_3\left(x^k\right)$. Hence, we have showed that, if u is a root of the polynomial $\Phi_3\left(x\right)$, then u is also a root of the polynomial $\Phi_3\left(x^k\right)$. Thus, the polynomial $\Phi_3\left(x^k\right)$ must be divisible by the polynomial $\Phi_3\left(x\right)$. Since both polynomials have integer coefficients and the leading coefficients are both 1, it follows that the quotient is a polynomial with integer coefficients, i. e. there exists a polynomial u with integer coefficients such that $\Phi_3\left(x^k\right)=u\left(x\right)\cdot \Phi_3\left(x\right)$. Lemma 1 is proven.

As a consequence of Lemma 1, if x and k are positive integers and k is not divisible by 3 and satisfies k > 1, then the number $\Phi_3\left(x\right)$ is a proper divisor of $\Phi_3\left(x^k\right)$. In fact, since x is an integer and $\Phi_3$ and u are polynomials with integer coefficients, the equation $\Phi_3\left(x^k\right)=u\left(x\right)\cdot \Phi_3\left(x\right)$ yields that the number $\Phi_3\left(x\right)$ divides the number $\Phi_3\left(x^k\right)$, and in fact it is a proper divisor since it is not equal to 1 (in fact, $\Phi_3\left(x\right)=x^2+x+1>1$) and not equal to $\Phi_3\left(x^k\right)$ (since $\Phi_3\left(x\right)=x^2+x+1<\left(x^k\right)^2+x^k+1=\Phi_3\left(x^k\right)$).

Now, if the number n is not a power of 3, then it has a prime divisor $\neq 3$. Let q be this prime divisor. Then, q is a positive integer, q > 1 (since q is a prime), and q is not divisible by 3 (since q is a prime $\neq 3$), and also, the number n can be written in the form n = qr, where r is some positive integer. Hence, by the above, the number $\Phi_3\left(2^r\right)$ is a proper divisor of $\Phi_3\left(\left(2^r\right)^q\right)$. Thus, $\Phi_3\left(\left(2^r\right)^q\right)$ cannot be a prime. But $\Phi_3\left(\left(2^r\right)^q\right)=\Phi_3\left(2^{qr}\right)=\Phi_3\left(2^n\right)=\left(2^n\right)^2+2^n+1=1^n+2^n+4^n$ must be a prime by the condition of the problem. Hence, n must be a power of 3.

That's all.

  darij

[b]PS:[/b] The same problem appeared in http://www.artofproblemsolving.com/Forum/viewtopic.php?f=59&t=291 .

When there is solution like maxal's one, why do you even care for another one, especialy like this :D :D

Because this one shows how to generalize it!

Hmmm... I think [i]darig[/i]'s solution uses nice techniques that we can learn. We shouldn't just find a nice elegant solution and stop there but seek to find generalisations to the problem and see if the techniques used in the original solution can be generalised too. Perhaps the motivation behind the problem can then be understood even better. :)

As you like it, but darij's solution is far too comlicated for me :)


Whoz going 2 Rockdale? I am...ya. :spider:  :starwars:  :ddr:

Um, gotta do All-State finals.  Too bad though... :(

I'm going as the only person on my team.

We actually will have a full team for the cherokee county tournament (and we may even have one extra)! Rockdale was kind of tough because we needed six people and that's almost an impossibility for my team.

I'm gang to Rockdale, I believe with a team [it's relatively easy for me to get a team for there, it's only a hop-skip-jump-run away from my home stomping grounds. Although with Val being with SDG it's not near everyone's stomping grounds.]. I'm going to be studying programming beforehand, not math, as an important compsci competition for me happens between individual round and awards :|.

Hey, can anyone tell me what it is like? i.e., what is the format of Rockdale? I heard that it's ARML style and stuff in terms of individual round, but I also heard there was a proof round. Also, if it was ARML style, how would they break ties (becuase unless there is about 50+ ARML style problems, there's bound to be some sort of tie)? Thanks.

Tiebreakers are tiebreakers, of course; Rockdale follows ARML very closely except in team size.

Also, given the weather forecast for Saturday, are you _still_ coming?

Umm, it's more like im coming BECAUSE of the weather forecast.

Snow =  :thumbup: snowball fight!!!

So yeah, I'm coming.

No more rockdale for northview, i'm not coming anymore.... :mad:  :wallbash_red:   :noo:

Fulton County schools has canceled all activities for the weekend. So Northview, Alpharetta, Centennial, and Milton are all down to parent transportation if they are coming. From Arvind, it sounds like Northview is staying home.

We have a bus for Walton, Dodgen, and Dickerson. It Cobb closes down, we'll stay home also. It's too much to try to coordinate alternate transportation at this point. Even if we come, we may leave right after the tournament. It's supposed to get worse as the day goes on.

Tom

Hi Mr. Fulton!  This may seem to others as spam, but this is urgent!  Could you check your private messages and/or email?  They're is something important I must ask you.  Thanks!

Sry for spamming, everyone.  Man, I wonder if Westminster is closing, because if that's true, then All-State finals will be postponed, which I do not know is bad or good.:maybe:

It already is--not like you don't know that.  :lol:

Man, for the first time in my life, I chose the right choice!!! We still had prelims for my swim meet(though finals were cancelled.)  :rotfl:

My school has not been cancelled in two years due to weather, but when there's something I actually want to go to, the snow comes pouring down. I wonder if Rockdale will be rescheduled. I would hate to plan a whole tournament and have it be cancelled.

[quote="gtf"]What meet? (Rays Invite?) Where do you swim? Walton was at the meet in Athens and I heard that they canceled the meet after 6 events.[/quote]

Yes, Rays Invite(well, that's what I'd call it, but the coach's[coaches'?, w/e]), and now I swim for SWAT-John's Creek. It's sort of sad that state is coming up(well, age group.) If I had two more meets I might have been glad to get to do something other than finals. :D Problem is, I'm so close to a few state cuts...and this is my last age group year. :|

I'm guessing not many freshman went to the Athens meet?

Oh well, this is a math forum, not a swim forum.  :rotfl:

And if Rockdale is scheduled, chances are I'd be able to go.

I was in an especially iteresting predicament as my math team sponsor convinced our superintendent to let us share a bus to Rockdale. Thus we woke up at around 5:00, left at 6:30, and arrived at 9:05 to find out the tournament had been cancelled at 7 am the day of.

The upside is I got to see it snow (Columbus may have had some snow at night, but it all melted in that case). I miss snow...

[quote=Dr Garner]Please accept my apologies for the abrupt cancellation of the Rockdale Mathematics Competition.  We received no indication that cancellation was imminent on Friday, and it surprised us to get the news Saturday morning.

We have looked into the possibility of rescheduling the competition for later in the semester, and discovered that, in order to avoid conflicts with other Math Team events, our facility is not available on any Saturdays before the various state math tournaments begin.

Our plan now is to reschedule the competition for the following school year in November 2008.

Therefore, your registration is rolled over to November 2008.  Additionally, you may bring additional teams in November at no extra cost.  (If your school was one of the few to show up Saturday morning, only to be turned away, then please contact me so that we may discuss other possibilities.)

I hope this in some small way makes up for any inconvenience this cancellation caused.
[/quote]


Write it as the sum of a geometric series.

[quote="DaReaper"]Find 0.462462... as a simplified fraction.

how do you find this???[/quote]
[hide]$0.\overline{462}=\frac{462}{999}=\frac{154}{333}$[/hide]

[quote="DaReaper"]do you always put repeating decimals over 999?[/quote]
Nope. You see how many numbers are under the repeating bar. That's how many 9's there is going to be in the bottom of the fraction. Since there were three numbers that were under tha repeating bar there was 3 9's.

[quote="DaReaper"]do you always put repeating decimals over 999?[/quote]
If there are $n$ digits repeating, the denominator will be $10^{n}-1$. The numerator will be the repeating digits.


Of the numbers $ \frac{7}{10}, \frac{4}{5}$ and $ \frac{3}{4}$ which number is the arithmetic mean of the other two?

Convert to common denominator.

7/10=14/20
4/5=16/20
3/4=15/20

So the answer is 3/4.


function
limit
calculus
calculus computations

Let f(x) and g(x) be two piecewise continuous functions with exponential orders $ \alpha$ and $ \beta$, respectively, on $ [0,\infty)$. What can then be said about the continuity and exponential order of the functions

(a) $ c_1f(x) \plus{} c_2 g(x)$, with c1 and c2 being constants,

(b) $ f(x) \cdot g(x)$?

$ \lim_{t \to \infty} |f(t)e^{ \minus{} \alpha t}| \equal{} 0$

$ \lim_{t \to \infty} |g(t)e^{ \minus{} \beta t}| \equal{} 0$

ill assume $ c_1 \ne 0 \ne c_2$

$ \lim_{t \to \infty} |[c_1f(t) \plus{} c_2g(t)]e^{ \minus{} max(\alpha, \beta ) t}| \le \lim_{t \to \infty} \left(c_1|f(t)e^{ \minus{} \alpha t}| \plus{} c_2|g(t)e^{ \minus{} \beta t}| \right) \equal{} 0$

and $ \lim_{t \to \infty} |f(t)g(t)e^{ \minus{} (\alpha \plus{} \beta )t}| \equal{} \lim_{t \to \infty} |f(t)e^{ \minus{} \alpha t}| |g(t)e^{ \minus{} \beta t}| \equal{} 0$

as for continuity look at sets

$ D_f \equal{} \{ t\in [0, \infty ) : \lim_{t' \to t} f(t') \ne f(t) \}$
$ D_g \equal{} \{ t\in [0, \infty ) : \lim_{t' \to t} g(t') \ne g(t) \}$

both of which are at most countable

then $ D_{c_1f \plus{} c_2g} \equal{} \{ t\in [0, \infty ) : \lim_{t' \to t} c_1f(t') \plus{} c_2g(t') \ne c_1f(t) \plus{} c_2 g(t) \} \subset D_f \cup D_g$
 and $ D_{fg} \equal{} \{ t\in [0, \infty ) : \lim_{t' \to t} f(t')g(t') \ne f(t)g(t) \} \subset D_f \cup D_g$

if you want a little more info than subset for the second one.. consider the zero sets of f and g:

$ Z_f \equal{} \{ t\in [0, \infty ) : f(t) \equal{} 0 \}$, $ Z_g \equal{} \{ t\in [0, \infty ) : g(t) \equal{} 0 \}$

and the define $ Z'_f \equal{} Z_f \setminus D_f$, $ Z'_g \equal{} Z_g \setminus D_g$

then $ D_{fg} \subset [D_f \setminus Z'_g] \cup [D_g \setminus Z'_f]$

in retrospect... i dont know if i believe myself for the multiplication part of the exp order


Let $ y\equal{}f(x)$ is differentiable on $ [0,1]$ and $ f(0)\equal{}0$ , $ f(1)\equal{}1$. Prove that there are real numbers $ a,b \in [0,1]$ such that
$ \frac{1}{f'(a)}\plus{}\frac{1}{f'(b)}\equal{}2$

This is one of frequently asked questions here, so you don't need to wait for an answer. -o- Just refer to [url=http://www.mathlinks.ro/viewtopic.php?p=1436448#1436448]here[/url].


Find the roots of $ x^3\plus{}a[rx\plus{}(1\minus{}r)]^2\equal{}0$, $ a\neq0$, $ r\neq0$ in terms of $ a$ ,$ r$.
If you have got a solution, please let me know
 miwalin@163.com

hello, do you mean the equation $ x^3\plus{}a\left(rx\plus{}(1\minus{}r)\right)^2\equal{}0$?
Sonnhard.

Is it not the same thing except with parentheses instead of brackets?

Right, square brackets are often used to represent things like the floor function, or the nearest-integer function, so it is important to clarify whether they represent a function or are just being used for grouping.

hello, you will need these formulas
http://mathworld.wolfram.com/CubicFormula.html
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, do you mean the equation $ x^3 \plus{} a\left(rx \plus{} (1 \minus{} r)\right)^2 \equal{} 0$?
Sonnhard.[/quote]
Yes, I mean this.


Sad... Mccain won by 0.2% (5886 votes). At least we have the honor of being the state with the closest election in 2008.... :(

They aren't doing a recount because they don't think MO is an important enough state and it won't affect whos the president.... even thought it's required by LAW...

So lets write an angry letter to someone.

Let's write an angry letter to our Secretary of State... who I believe is Robin Carnahan.... or did she lose re-election?


let $ ABC$  be a triangle and $ M$ an interior point.prove that$ min(MA,MB,MC) \plus{} MA \plus{} MB \plus{} MC < AB \plus{} BC \plus{} CA$

Please do not post known IMO problems, just visit the [url=http://www.mathlinks.ro/Forum/resources.php]Resources Section[/url] and find the link. Also, the search engine is useful in case of known problems.
By the way, here is the link to this problem :
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=131828#131828

[quote]Please do not post known IMO problems, just visit the Resources Section and find the link.[/quote]
fine. i was in a hurry yesterday and didnt have time to search


modular arithmetic
number theory proposed
number theory

Determine the positive integers $ n$ which satisfy: $ n|1^n\plus{}2^n\plus{}...\plus{}n^n.$

i have obtained this
\[ \sum_{k\equal{}1}^{n}k^p\equal{}\frac{1}{p\plus{}1}.\left ((n\plus{}1)^{p\plus{}1}\minus{}1\minus{}\sum_{k\equal{}1}^{n}\sum_{i\equal{}2}^{p\plus{}1}{{p\plus{}1}\choose {i}}k^{p\plus{}1\minus{}i}\right).\]

in the question, $ p\equal{}n$ as of above. so
\[ \sum_{k\equal{}1}^{n}k^n\equal{}\frac{1}{n\plus{}1}.\left ((n\plus{}1)^{n\plus{}1}\minus{}1\minus{}\sum_{k\equal{}1}^{n}\sum_{i\equal{}2}^{n\plus{}1}{{n\plus{}1}\choose {i}}k^{n\plus{}1\minus{}i}\right).\]
we could proceed further with above.

also, by fermat's theorem, \[ \sum_{k\equal{}1}^{p\minus{}1}k^p\equiv \sum_{k\equal{}1}^{p\minus{}1}k \equiv \frac{p(p\minus{}1)}{2} \equiv 0 \mod p.\]

it is true for odd $ n$.
now let's suppose that $ n$ is even and $ 2 |n$.
then define $ a \equal{} v_2(n) \ge 1$ and $ 2^ak \equal{} n$ where $ k$ is odd.
obviously we have
$ 1^n \plus{} .... \plus{} n^n \equiv k(1^n \plus{} 2^n \plus{} ... \plus{} (2^{a})^n) \pmod {2^a}$
but we must have
$ 2^a | 1^n \plus{} .... \plus{} n^n$
or
$ 2^a | k(1^n \plus{} 2^n \plus{} ... \plus{} (2^{a})^n)$
but we know that $ k$ is odd so we must have
$ 2^a | 1^n \plus{} 2^n \plus{} ... \plus{} (2^{a})^n$
if $ l$ is an even number then obviously $ 2^a | l^n \equal{} (2l')^{2^ak}$
and if $ l$ is an odd number then $ l^n \equal{} l^{2^ak} \equiv 1 \pmod {2^a}$
so from the last divisibility we get
$ 2^a | 1 \plus{} 0 \plus{} 1 \plus{} 0 \plus{} ... \plus{} 1 \plus{} 0$
where the number of $ 1's$ is $ 2^{a \minus{} 1}$ which implies
$ 2^a | 2^{a \minus{} 1}$
which is a contradiction.
so we are done!
another question is to find the needed and sufficient condition than an even number like $ n$ must have to satisfy this :
$ \frac {n}{2} | 1^n \plus{} ... \plus{} n^n$

[quote="phymax"]sorry, i do not understand $ v_2(n)$..what is it?[/quote]
it is the non-negative number $ a$ satifying:
$ 2^a |n$ and $ 2^{a\plus{}1} \nmid n$


\[x_1^{x_1} \cdot x_2^{x_2} \ldots x_n^{x_n} \geq (x_1 \cdot x_2 \ldots x_n) ^ {\frac{x_1+x_2+\ldots+x_n}{n}}\]

I know proofs this inequality, but I can't prove it with Jensen's inequality:) If anyone has any idea, post it please:)

Take the log of both sides, and then a few Jensen approaches should appear.

But in both sides we have sums of logs.... :)


algebra
polynomial
abstract algebra
geometry
advanced fields
advanced fields unsolved

I'm have read the definition of an elliptic curve. I have seen two definitions:
1. (Silverman) An elliptic curve is a smooth projective variety of genus 1.
2. (Liu) Let $ k$ be a field. We define an elliptic curve over $ k$ to be a smooth projective curve $ E$ over $ k$, isomorphic to a closed subvariety of $ \mathbb{P}_k^2$ defined by a homogeneous polynomial $ F(u,v,w)$ of the form
\[ F(u,v,w) \equal{} v^2w \plus{} (a_1u \plus{} a_3w)vw \minus{} (u^3 \plus{} a_2u^2w \plus{} a_4uw^2 \plus{} a_6w^3)
\]
with the priviledged rational point $ o \equal{} (0,1,0)$.

Why are these two definitions equivalent?

For 1 (an elliptic curve [i]over $ k$[/i]; more generally over a scheme: a section), you also need a $ k$-rational point $ O$.

Then, 1 => 2 follows from Riemann-Roch ($ g \equal{} 1 \implies \ell(D) \equal{} \mathrm{deg}(D)$) applied to $ \mathcal{L}(2O),\mathcal{L}(3O), \mathcal{L}(6O)$ (I think you can find it in Silverman, The Arithmetic of Elliptic Curves)

2 => 1: you can show that the genus of a smooth plane projective curve of degree $ d$ is $ (d \minus{} 1)(d \minus{} 2)/2$, e.g. by Riemann-Hurwitz or by writing down a canonical divisor. (exercise in Hartshorne)

exercise:
3. Abelian variety of dimension $ 1$.

There is also a book of Silverman and ??? on diophantine geometry (including Bombieri's simplification of Vojta's proof of the Mordell conjecture) which covers among other things (curves of arbitrary genus, abelian varieties) also this AFAIR.

[quote="myan"]
There is also a book of Silverman and ??? on diophantine geometry (including Bombieri's simplification of Vojta's proof of the Mordell conjecture) which covers among other things (curves of arbitrary genus, abelian varieties) also this AFAIR.[/quote]

Hindry and Silverman. Great book, but not quite easy reading.


$A,B,C,D$ are cyclic, respectively. $AD$ meets $BC$ at $E$. $AB$meets $CD$ at $F$. $EM$ is perpendicular to $AB$ ($M$ is on $AB$ ).$FN$is perpendicular to $AD$ ($N$ is on $AD$ )Prove that $M$,$N$ and the midpoint $K$ of $BD$ are collinear

Triangles $AME$ and $ANF$ are similar.So we have:
$\frac{NA}{MA}=\frac{NF}{ME}$.Also, triangles $NFC$ and $MEB$ are simlar, so we have: $\frac{NC}{MB}=\frac{NF}{ME}$, that yields to $\frac{NA}{NC}\cdot\frac{BM}{MA}=1$.

Now we apply Menelaus Theorem for the triangle $ACB$ and  for the points $M,N,K$ and we are Q.E.D.

[i]Nick Rapanos [/i]

Good solution
this problem is similar to the probem above but it more difficult
$A,B,C,D$ are cyclic, respectively. $AD$ meets $BC$ at $E$. $AB$meets $CD$ at $F$. $EM$ is perpendicular to $AB$ ($M$ is on $AB$ ).$FN$is perpendicular to $AD$ ($N$ is on $AD$ ). $I, J$ are the midpoint of $AC,EF$. Prove that $IJ, MN$ and $BD$ are concurrent

Another version of the original problem, with an application of Gauss' line.

Hi treegoner!Can u post your solution here, because i tried Gauss's Line without success and i would really like to see your solution!

It is true in general that the midpoints of $EF, BD, AC$ are collinear. Probably someone can help me verify the name of theorem (I think it is Gauss' theorem but probably a theorem of Newton)

treegoner i am not stupid...i know Gauss's line!but the question is how do u apply the theorem here????

I think treegoner used Gauss line to prove the second problem posted by  Huyền Vũ  :)

i hadnt even read the second problem,  because treegoner said [u]<<the [/u][u]original>>[/u] problem...But if it is for the second problem-then it's ok!

I don't understand the second Huyen Vu's problem.

It has been proved that the segment line $MN,$ passes through the midpoint $K,$ of $BD$ $($ first problem $).$

But the segment line connecting the midpoints $I,$ $J,$ of $AC,$ $EF$ respectively $($ Gauss's line $),$ also passes through the point $K.$

kostas Vittas.

I used Gauss' line for the second problem, not the original one.


[b]If h (x) is the product of the integers from 1 to x, the least prime factor of
h(100)+1 is in which of the following ranges?

A)   2 to 10
B)  10 to 20
C)  20 to 30
D)  30 to 40
E)  above 40[/b]

blahblahblah wrote:Hint:  [hide]Say that a number is a prime under 40.  What can you say about h(100) and h(100)+1?[/hide]

That's a difficult hint!

I would say primes 2 and 5 are out. But then what?

Done wrote:If h (x) is the product of the integers from 1 to x, the least prime factor of
h(100)+1 is in which of the following ranges?

A)   2 to 10
B)  10 to 20
C)  20 to 30
D)  30 to 40
E)  above 40

[hide]

A)

 2*3*4*5*6*7*8*9*10=2*3*2*3*5*2*3*7*2*2*4*3*3*2*5

=(2e6)*(3e5)*(5e2)*7

2*3*5*7=210   min prime for A) is 211

a is the answer since it has the fewest primes.  Furthermore, these are the smallest primes[/hide]

Done wrote:If h (x) is the product of the integers from 1 to x, the least prime factor of
h(100)+1 is in which of the following ranges?

A)   2 to 10
B)  10 to 20
C)  20 to 30
D)  30 to 40
E)  above 40

[hide]E

h(x) is divisible by all the integers from 1 to x, inclusive.From that we can see that h(x)+1 is not divisible by any integer from 3 to x. Because h(100) is even, h(x)+1 is not divisible by two.  1 is not a prime factor, and it doesn't fit into the choices anyways, so all integers from 1 to 100 don't work. [/hide]

i think i know what you mean now.  100!+1=x  then the smallest prime divisor of x is in which range.  Is that what you mean?

your wording was kinda funny.  I found it hard to understand.  Hope it clears it up for others

Hmmm...I would do it like this:
[hide]E.
Since every number less than 100 wouldn't be divisible by something with that factor (for example 100*99!+1 wouldn't be divisible by 99 because of the addition of the 1)
[/hide]
I got the same answer you guys did, but I'm not sure if my logic is correct or not.

[hide]h(100) is divisible by all of the numbers from 1 to 100.  So if 1<x<101, h(100) = 0 mod x and h(100) + 1 = 1 mod x.  Therefore, x has to be greater than 100, or (e).[/hide]


I hope you'll understand what I mean.
I'm not good at English.  :P

Let A(1;1), B(a;3), C(b;0)
Find a,b that make ABC is a ... triangle (I don't know what the word is, but means the ABC triangle has AB=BC=CA)

Could you be a little more clear?
what is a:1...all that junk.

[quote="i_like_pie"]$\triangle ABC$ is a triangle with $A\,(1,1),\, B\,(a,3),\, C\,(b,0)$

Find $a$ and $b$ that make $\triangle ABC$ an equilateral triangle.[/quote]

Yes that's exactly what I want to say.
Ah, ... and who can teach me how to use the Latex, and solve my problem?

A(1;1), B(a;3), C(b;0)

first plot A in the coordinates plane.

extend your y-axis to 3 and draw a line where x=3

and now observe where could B be

if C was somewhere in the x-axis line and B is somewhere in the x=3 line.

This is how you would solve it in middle school level.

don't know if they teach the distance formula in middle school... if they do use it

Answer[hide]
Anyways if you use it you get a=1
and there for b is 2

i don't know LaTeX and I'm still learning and this would probably be my last non-LaTeX

so u know the distance formula is
sqrt[[ (x1-x2) + (y1-y2) ]]
so if you do it for all three you get

0 =2a-b
1-b+1=2a-b
2=2a
and a=1
and therefore b=2

but as I said this method isn't neccesary, just use the drawing method in the top. If the problem gets complicated the distance formula will be needed.[/hide]

oh I'm sorry
but I don't understand anything.
But I think It cannot be that easy.
I've try is solve it.
This is my think.

[/url]

This problem doesn't look like a classroom math level but the use of the distance formula will make it a lot easier.
In analytic geometry, the distance from $A(x,y)$ to $B(p,q)$ is given by
$AB=\sqrt{(x-p)^{2}+(y-q)^{2}}$
for this problem, we can use the fact that $AB=BC=AC$.
I'm a little busy right now, so if you require a proof, just ask.

[quote="bibobeo"]Yes that's exactly what I want to say.
Ah, ... and who can teach me how to use the Latex, and solve my problem?[/quote]
[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]

[hide]To solve for $a$ and $b,$ use the distance formula (we know the three sides are equal).[/hide]
[quote="bibobeo"]I know how to prove that.
However, can you show me the way to the equation system when we have AB=BC=CA[/quote]
[hide]$\sqrt{(3-1)^{2}+(a-1)^{2}}=\sqrt{(0-3)^{2}+(b-a)^{2}}=\sqrt{(0-1)^{2}+(b-1)^{2}}$[/hide]

i_like_pie do you participate in any math competitions? i think you'll do well. You should enter in mathcounts. it's a really great math competition, and i think you'll do very well

-soccerstar123

I participate in as many as I have time for, but this isn't the place to discuss them. :)


geometry
circumcircle
symmetry
parallelogram
geometric transformation
reflection
radical axis

Let A' be the symmetric of A wrt M and O the circumcenter of ABC. Let AO meet the circumcircle (O) on F. As know the circumcircle (O') of BHC (with center O') is the symemtric of (O) wrt BC. As know with a composition 2 symmetry with axes BC and MO, H go on F, then M is the midpoint of FH. Then AFA'H i sa parallelogram and $ \overline{AF} \equal{} \overline{A'H}$, then $ O' \in HA$, so $ K \in (O')$.

The second part is a particolar case of this know problem:

Let ABCD a cyclic quad and E, F in it such that ABEF and EFCD are cyclic quads. Then AB, CD, EF are concurrent.

proof: Call $ P: AB \cap CD$, call $ \Gamma$ the circumcircle of ABC, $ \Gamma_1$ of ABE and $ \Gamma_2$ of CDE. Then P stay on the radical axes of $ \Gamma_1$ and $ \Gamma_2$ (EF) iff $ pow_{\Gamma_1} (P) \equal{} pow_{\Gamma_2} (P)$, but $ pow_{\Gamma_1} (P) \equal{} PA \cdot PB \equal{} pow_{\Gamma} (P) \equal{} PC \cdot PD \equal{} pow_{\Gamma_2} (P)$

[quote="stergiu"]If $ K$ is the projection of the orthocenter $ H$ of an acute trianlge $ ABC$ on the median $ AM$ , prove that points

 $ B,C,K,H$ are concyclic. Moreever, if $ BD,CE$ are altitudes in triangle $ ABC$ , then lines $ DE,BC,KH$

are concurent .

[color=blue][i]Which basic olympiad problem reminds you the second question ?[/i][/color]

Babis[/quote]
[hide="Solution"]
Let the intersection between the circumcircles of $ \triangle ADH$ and $ \triangle BCH$ be $ L$.  Let $ AL$ meet $ BC$ at $ N$.  Then, $ \angle LCB \equal{} \angle DHL$ as $ BHLC$ is cyclic.  Since $ \angle AEH \equal{} 90 \equal{} \angle ADH$, we see that $ AEHLD$ is cyclic, so $ \angle DHL \equal{} \angle LAD$.  Thus, $ \angle NAC \equal{} \angle LCN$.  Then, $ \triangle LNC\sim \triangle CNA$ since $ \angle LNC \equal{} \angle NCA$.  This means that $ \frac {NC}{NA} \equal{} \frac {LN}{NC}$, so $ NC^2 \equal{} LN\cdot NA$.  Similarly, $ NB^2 \equal{} LN\cdot NA$, so $ NB \equal{} LN$, so $ N \equal{} M$.  This means that $ L$ is the intersection between the circumcircle of $ ADLHE$ and $ AM$.  Yet, since $ \angle ADH \equal{} 90$, we have that $ AL\perp AM$, so $ L \equal{} K$.  Then, $ BHKC$ is cyclic.  Furthermore, $ KH$, $ BC$, and $ DE$ are radical axes of the circumcircles of $ BEDC$ (notice that $ \angle BEC \equal{} 90 \equal{} \angle BDC$ so $ BEDC$ is cyclic), $ AEHKD$, and $ BHKC$, so $ KH$, $ BC$, and $ DE$ concur. [/hide]

Both are new solutions and I add them to my archiev. :) .
 Thank you very much.Of course , as you mentioned , the second part is based on the radical center of three circles.
The olympiad problem that I have in mind is this :

[b] Problem[/b]

 If $ BD,CE$ are altitudes in triangle $ ABC$  and line $ DE$ meets line $ BC$ at $ R$, then $ RH$ is perpendicular to the median $ AM$ of $ \triangle ABC$ .

[b] HINT[/b]

 [color=blue]We draw $ HK \perp AM$ . Then , as in the problem above , $ KH$ passes through $ R$.
 We have discussed this problem in this forum again and we have given many solutions.It has  also apeared in CRUX.[/color].

[quote="stergiu"]

[b] Problem[/b]

 If $ BD,CE$ are altitudes in triangle $ ABC$  and line $ DE$ meets line $ BC$ at $ R$, then $ RH$ is perpendicular to the median $ AM$ of $ \triangle ABC$ .
[/quote]
Ah, yes, thank you for pointing this hint to another solution out, dear [b]stergiu[/b]!
[hide="Solution"]
The following lemma is quite well-known (if you want a proof, then ask):

For a cyclic quadrilateral $ ABCD$ with $ AB$ and $ CD$ meeting at $ P$, $ AD$ and $ BC$ meeting at $ Q$, the center of it being $ O$, and $ AC$ and $ AD$ meeting at $ R$, then we have that $ PO\perp RQ$.

Applying this to the cyclic quadrilateral $ BEDC$ with center $ M$, we have that $ AM\perp RH$.  Then, the intersection between $ AM$ and $ RH$ is the foot of the perpendicular from $ H$ to $ AM$, which is $ K$.  Thus, $ \angle AKH \equal{} 90 \equal{} \angle AEH$, so $ AEKH$ is cyclic.  Similarly is $ AKHD$, so $ AEKHD$ is cyclic.  Then, $ RK\cdot RH \equal{} RE\cdot RD \equal{} RB\cdot RC$, so $ KBCH$ is cyclic, as desired. [/hide]

Thank you for the second solution!
Ah! For the proof of the lemma I suppose you use polars .Do you have any more elementary approach for it?
 Babis

[quote="stergiu"]Do you have any more elementary approach for it?
[/quote]
I don't know polars, but here is my proof:
[hide="Proof"]
Let the circumcircles of $ \triangle APB$ and $ \triangle CPB$ meet at $ Z$.  Also, let the midpoints of $ AC$ and $ BD$ be $ M$ and $ N$ respectively and let the feet of the perpendiculars from $ P$ to $ AC$ and $ BD$ be $ X$ and $ Y$ respectively.  First, notice that $ RZ$, $ AB$, and $ DC$ are radical axes of the circumcircles of $ ABCD$, $ \triangle ABP$, and $ \triangle CPD$.  This means that they concur at $ Q$.   Now, we have that $ \angle ZAB \equal{} \angle BRZ \equal{} \angle ZCD$, and similarly, $ \angle ZBA \equal{} \angle ZDC$, so $ \triangle ZDC\sim \triangle ZBA$, so a spiral similarity maps $ \triangle ZAC\sim \triangle ZBD$.  As $ M$ and $ N$ are the respective midpoints of $ AC$ and $ BD$, we have that $ \triangle ZAM\sim \triangle ZBN$, so $ \angle RNZ \equal{} \angle RMZ$, so $ RZMN$ is cyclic.  Yet, $ O$ is the circumcenter of $ ABCD$, so $ \angle ONR \equal{} 90 \equal{} \angle OMR$, so $ ONRM$ is cyclic, so $ ONRZM$ is cyclic, so $ \angle OZR \equal{} \angle OMR \equal{} 90$.  Furthermore, $ \angle PCA \equal{} \angle PDB$, $ \angle PYD \equal{} 90 \equal{} \angle PXC$, and $ \angle BPD \equal{} \angle APC$, so $ \triangle PAC\sim \triangle PBD$ and $ \triangle PXC\sim \triangle PYD$.  These yield that $ \frac {XC}{YD} \equal{} \frac {PC}{PD} \equal{} \frac {AC}{BD}$, so $ \triangle ZXC\sim \triangle ZYD$.  This gives that $ \angle ZYR \equal{} \angle ZXR$, so $ XYZR$ is cyclic.  Yet, $ \angle PYR \equal{} 90 \equal{} \angle PXR$, so $ PXRY$ is cyclic, so $ PXRZY$ is cyclic, so $ \angle PZR \equal{} \angle PYR \equal{} 90$.  This means that $ \angle PZO \equal{} 90 \plus{} 90 \equal{} 180$, so $ P$, $ Z$, and $ O$ are collinear, and $ \angle QZP \equal{} 90$, so $ QR\perp OP$, as desired. [/hide]

[b] To Quatto MAster 6000 only[/b]

 [hide]It looks strange  :) : You know radical axes but not polars.Any way, I think that you will become soon an expert in plane geometry, especially if you study polars , harmonic division and perspectivity(pappus and Desargues).
 Of course , I prefare your nice  proofs with standard theory and basic theorems as above , because in Greece we do not teach polars, harmonics ect , but only the very basic geometry.
 I wish you to be soon an IMO gold medalist!(if you are still a high school student !)[/hide] Babis

I think its not hard to use complex nubers for first part :wink:

Here is my solution to the first part:
Let D be the foot of altitude from A to BC. Draw the circumcircle of $ \triangle ABC$, extend AD to hit the circle at H', we know H' is the reflection of H over BC. So we need to show the reflection of K over BC also lies on the circumcircle, to show BHKC is cyclic. But since M is the midpoint of BC, by symmetry, this is equivalent to showing $ KM \equal{} MP$, where P is the intersection of AM with the circle. By Power of a point, $ MP \cdot AM \equal{} (\frac {a}{2})^2$, so $ MP \equal{} \frac {a^2}{4AM}$.
Now, $ \triangle HAK$ and $ \triangle DAM$ are similar, so $ \frac {AK}{AD} \equal{} \frac {AH}{AM}, or AK \equal{} \frac {AD \cdot AH}{AM}$. But $ KM \equal{} AM \minus{} AK$, so $ KM \equal{} \frac {AM^2 \minus{} AD \cdot AH}{AM}$
So we have to show $ a^2 \equal{} 4(AM^2 \minus{} AD\cdot AH)$. But
$ AM^2 \minus{} AD \cdot AH \equal{} AM^2 \minus{} AD(AD \minus{} HD) \equal{} AM^2 \minus{} AD^2 \plus{} AD \cdot H'D \equal{} AM^2 \minus{} AD^2 \plus{} BD \cdot DC \equal{} DM^2 \plus{} (\frac {a}{2} \minus{} DM)(\frac {a}{2} \plus{} DM) \equal{} \frac {a^2}{4}$.
We can use the radical axis theorem to finish the problem as Quatto did.

Here is a solution to the second problem (not second part of first problem) stergiu wrote:
Let K be the perpendicular from H to AM, and N the foot of the altitude from A to BC (passes through H). Then HKMN is cyclic ($ \angle HKM\equal{}\angle HNM\equal{}90$), AEHKD is cyclic ($ \angle AEH\equal{}\angle AKH\equal{}\angle ADH\equal{}90$), and EDMN is cyclic (they lie on the nine-point circle). Therefore, their radical axis, DE, HK, and MN concur, so R,H, and K are collinear, so $ RH \perp AM$

Let $ Q \equiv DE \cap BC.$ Line $ AQ$ is the polar of $ H$ WRT the semicircle $ (M)$ with diameter $ BC.$ Thus, $ QH$ is the polar of $ A$ WRT $ (M)$ $ \Longrightarrow$ $ AM \perp QH.$  Let $ K \equiv AM \cap QH.$ Note that $ H$ is common orthocenter of $ \triangle AQM$ and $ \triangle ABC,$  thus $ KH \cdot HQ=DH \cdot HB= CH \cdot HE.$ In other words, the inversion with center $ H$ and power $ KH \cdot HQ$ takes $ K,B,C$ into the collinear points $ D,E,Q$ $\Longrightarrow$ $ B,C,H,K$ are concyclic, as desired.


function
analytic geometry
inequalities
calculus

[quote="fjwxcsl"]Let  x1,x2,...,xn>0,x1^2+x2^2+...+xn^2=n, prove (or disprove) that:

        1/x1+1/x2+...+1/xn+2*(x1+x2+...+xn)>=3*n.[/quote]

OBSERVATION: Consider the function f(x) = 1/x + 1/(s-x) for some positive s.
It is easy to see that on the interval (0,s), function f(x) takes its unique minimum at x=s/2.

Now consider the set P of all points x=(x_1,...,x_n) with non-negative coordinates
and   x1^2+x2^2+...+xn^2=n   for which
g(x):=  1/x1+1/x2+...+1/xn+2*(x1+x2+...+xn)  <=  3*n.

Since g(x) is continuous, P is a closed set.
Furthermore, all coordinates of points in P are strictly positive.
Hence, there exists a point p in S that minimizes g over P.
If p has two non-equal coordinates, we could get a smaller function value
by applying the above OBSERVATION; contradiction.

Hence all coordinates in p are equal.
Hence all coordinates in p are +1.
Hence g(p)=3n.


For $ a_1,\cdots,a_n>0$, prove that
$ \frac{1\cdot 2}{a_1\plus{}a_2}\plus{}\cdots\plus{}\frac{(n\minus{}1)\cdot n} {a_1\plus{}\cdots\plus{}a_n}<3\left( \frac{3}{a_1}\plus{}\cdots\plus{}\frac{n\plus{}2}{a_n}\right).$

[quote="ringos"]Stronger inequality
\[ \frac {1\cdot 2}{a_1 \plus{} a_2} \plus{} \cdots \plus{} \frac {(n \minus{} 1)\cdot n} {a_1 \plus{} \cdots \plus{} a_n} \plus{} \frac {n^3} {a_1 \plus{} \cdots \plus{} a_n} < 3\left( \frac {3}{a_1} \plus{} \cdots \plus{} \frac {n \plus{} 2}{a_n}\right)
\]
is easily proved by induction.[/quote]
I'm not pretty sure about your inequality because the inductive step seems to be ... :huh: Your idea is nice though.  :)

Anyway, I've verified that the following one is okay.
$ \frac {1\cdot 2}{a_1 \plus{} a_2} \plus{} \cdots \plus{} \frac {(n \minus{} 1)\cdot n} {a_1 \plus{} \cdots \plus{} a_n} \plus{} \frac {n^3} {3(a_1 \plus{} \cdots \plus{} a_n)} < 3\left( \frac {3}{a_1} \plus{} \cdots \plus{} \frac {n \plus{} 2}{a_n}\right)$


algebra
polynomial
function
calculus
derivative
real analysis
real analysis unsolved

Prove that no polynomial function of degree greater than 1 is uniformly continuous on R.

   May somebody help me?

if the degree is greater than 1 then $ f'$ is not constant and thus unbounded. Given $ \epsilon > 0 , \delta > 0$
Then $ |f(x \plus{} \delta) \minus{} f(x)| < \epsilon$ can not hold for all $ x$ otherwise by $ |f'(c_x)| \equal{} |\frac {f(x \plus{} \delta) \minus{} f(x)}{\delta}| < \frac {\epsilon}{\delta}$ for some $ c_x\in [x,x \plus{} \delta]$ which contradicts the fact that $ f'$ is unbounded.

An unbounded derivative is not enough- it's possible to be uniformly continuous anyway with an unbounded derivative.
Work with differences over a fixed length instead.

The proof for x^2 can be modified for this. If you know how to show it for x^j, j>1, you are essentially done

Sorry, yes. I meant We can fint an $ x_0$ so that $ x>x_0\rightarrow f'(x)>\frac{\epsilon}{\delta}$

in fact,f:UC(R)[i.e.uniformly continuous on R]
existA,B>0.for-all(x:R).|f(x)|<A|x|+B;
so,if f:R[x]andUC(R),deg(f)<=1.

[quote="lambda"]in fact,f:UC(R)[i.e.uniformly continuous on R]
existA,B>0.for-all(x:R).|f(x)|<A|x|+B  [/quote]

Can someone prove this?
Thanks.


5. If $A$ and $B$ are sets such that $A=B$, prove that $A=B$ implies $B=A$.

6. Prove: ${\{a,b\}=\{a,b,a}\}$.

7. Prove: If $n$ is a nonnegative integer, then $\omega_{n+1}=\omega_{n}\cup\{n\}$.

If $n\in\mathbb{Z}_{0}$, we say that $\omega _{n}$ is a set such that
\[ \omega_{n}=\{k\in\mathbb{Z}_{0}|k<n\} \]
Each $\omega_{n}$ is called an initial segment of $\mathbb{Z}_{0}$.
Examples would be, $\omega_{0}=\emptyset ,\omega_{1}=\{0\}, \omega_{2}=\[\{0,1\}$.

Masoud Zargar

I really don't get question 1 and 2. I know you said in an earlier post "using only mathematics", but you REALLY need to state what can be assumed to have been shown already, so that we know what we can use. The first one should be obvious from the reflexive property of the equal sign; is that not enough?

For the third we see that $LHS = \omega_{n+1}=\{0,1,2,...,n\}$ and $RHS = \omega_{n}\cup\{n\} = \{0,1,2,...n-1\}\cup\{n\} = \{0,1,2,...,n\} = LHS$. What is there to show really? You need to clarify.

6 isn't true. The first set has two elements, the second has three.
Unless you're working under some different definitions. Then you would have to tell us what those are, if you expect to get some answers.

What does EMT stand for anyway?

6 is not really correct because when you write a set, you don't repeat elements
and for 5, equality is commutative

5. According to $A=B$ and/or $B=A$, we deduce that $A\subseteq B$ and $B\subseteq A$.  This implies that for all $x\in A$, $x\in B$.  This means that $A$ and $B$ contain all $x\in A,B$.  Since sets $A$ and $B$ have exactly the same elements, $A=B$ and/or $B=A$.

6. Prove: $\{a,b\}=\{a,b,a\}$

Let $A=\{a,b\}$ and $B=\{a,b,a\}$.

We can see that there is such an $x\in A$ such that $x\in B$.  As a result, we say that $A\subseteq B$.

When we say that $x\in A$ such that $x\in B$, we are saying that every element of $A$ is contained in $B$.  This obviously implies that it does not matter if "a" is mentioned  in set $B$ more than once.  According to the above, we can say that $A=\{a,b\}$ and $B=\{a,b,a\}=\{a,b\}$.  As a result, $A=B$ or $\{a,b\}=\{a,b,a\}$.  By the way, $\{a,b,a\}$ is a valid set because you are allowed to repeat anything as much as you want in a set; it will still imply one of each variable.

7. $\omega_{n+1}=\omega_{n}\cup\{n\}\rightarrow\omega_{n+1}=\{0,1,2,...,n\}$ and $\omega_{n}\cup\{n\}=\{0,1,2,...,n-1\}\cup\{n\}=\{0,1,2,...,n\}$.  As a result, $\omega_{n+1}=\omega_{n}\cup\{n\}\rightarrow\{0,1,2,...,n\}=\{0,1,2,...,n-1\}\cup\{n\}=\{0,1,2,...,n\}$.  $\therefore\omega_{n+1}=\omega_{n}\cup\{n\}$.

Masoud Zargar

In a sense, $\{a,b\}=\{a,b,a\}$ because they contain the same elements, but your last notation is incorrect, because you do not write the repeats in a set.

Similarly, equality essentially has the property that $A=A$, $A=B\implies B=A$, and $A=B$ and $B=C\implies A=C$, as long as $A,B,C$ are well-defined.  So that question, and all of the other questions you are posting like prove $0=-0$, $1=1$, are just nonsense, and you should probably move onto posting reasonable questions, not quasi-rigorous set-theory trivialities.


Find the maximum value of:

$ tg3\theta \cdot ctg^3\theta$

If: $ 0<\theta <\displaystyle\frac{\pi}{2}$

Greetings  :D

hello, are you sure that there exists a maximum?
Sonnhard.

[quote="Immanuel Bonfils"]Should be ~ .029437[/quote]

plot tan(3x)cot(x)^3, x=0..pi/2
http://www.wolframalpha.com/input/?i=plot+tan%283x%29cot%28x%29^3%2C+x%3D0..pi%2F2

minimize tan(3x)cot(x)^3 over[0,0.523]
http://www.wolframalpha.com/input/?i=minimize+tan%283x%29cot%28x%29^3+over[0%2C0.523]
maximize tan(3x)cot(x)^3 over[0.524,1.57]
http://www.wolframalpha.com/input/?i=maximize+tan%283x%29cot%28x%29^3+over[0.524%2C1.57]

$ 0 < 0.523 < \frac {\pi}{6} < 0.524 < 1.57 < \frac {\pi}{2}$

There are a local minimum at $ \theta \equal{} {\pi \over 8}$ of $ y_1 \equal{} 33.97$ and a local maximum at $ \theta \equal{} {{3\pi} \over 8}$ of $ y_2 \equal{} {1 \over {y_1}} \equal{} 0.0294$, but no absolute maximum.


Let $a_{i}= b_{i}c_{i}$. Then $c_{i}$ is the least divisor of $a_{i}$ which is greater than $1$. It follows that $c_{i}$ is prime.
Moreover, from the given inequalities, we have $c_{1}<c_{2}<...<c_{10}$.
Thus $c_{i}\geq p_{i}$ where $p_{i}$ is the $i^{th}$-prime in increasing order.
In particular, $p_{9}\geq 23$.

In another hand, from the minimality of $c_{i}$ we have $b_{i}=1$ or $b_{i}\geq c_{i}$. Since $b_{9}> b_{10}\geq 1$, we then have $b_{9}\geq c_{9}\geq 23$, hence $a_{9}\geq 23^{2}= 529$.
Therefore $a_{10}\geq 530$ and we are done.

Pierre.

[quote="pbornsztein"]
Thus $c_{i}\geq p_{i}$ [/quote]

Some one can explain to me this ?

[quote="pbornsztein"]

In another hand, from the minimality of $c_{i}$ we have $b_{i}=1$ or $b_{i}\geq c_{i}$. Since $b_{9}> b_{10}\geq 1$, we then have $b_{9}\geq c_{9}\geq 23$

Pierre.[/quote]

And this ?

Solution is correct $c_{i}>1$ and prime. Therefore $c_{1}\ge 2 , c_{2}>c_{1}\Longrightarrow c_{2}\ge 3$ et.c.

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=86768 .

  darij


There have been some wierd names as i recall.

there was a city called IMAFREEZIN and another called nikpath.

names of people: Boon Tee and Ilian.

sorry if i am offending anyone but i highly doubt anyones names are Boon Tee or Ilian

exactly the point. i dont think they want to offend anyone or anyplace by using their name in a problem, especially if it involves something bad

Errr.. there's another thread that's almost exactly the same in G&FF.....

I heard that MC wants to "diversify", so they do names from different ethnicities and stuff...o.0

There was also Howgarts. During the actual test, I was like not doing well in general, and thought it said Hogwarts...

Recent problem on a test I wrote:

Ahmed, Bhat, Carl, and Delfonso are interrogated about a murder. They each give a statement, but exactly two of them is lying.

Ahmed: I did it.
Bhat: Ahmed did not do it.
Carl: Either Bhat or Delfonso did it.
Delfonso: Carl and Ahmed are lying.

Who committed the murder?

Actually, I just realized this problem is messed up. If Bhat and Carl are telling the truth, no contradictions arise and you can't tell who committed the murder. I have to change it.

From a SoCal local competition: Yvonne and Zahra.

In this case, they're named after competitors from a few years back.

The standard is of course to use initials that fit the letters you want. If you can get diversity along the way or throw in a reference, so much the better. I remember Walter showing up on late 90's AMC contests.

Well I don't think anyone has a name in all caps such as IMAFREEZIN


Show that there exists a positive integer $ k$ such that $ k \cdot 2^{n} \plus{} 1$ is composite for all $ n \in \mathbb{N}_{0}$.

If $p$ is a prime dividing $2^{2^k}+1$, then $2$ has order $2^{k+1}$ $\mod p$. This helps us finding the following list:

We have primes $p_1=3, p_2=5,p_3=17,p_4=257,p_5=65537,p_6=641,p_7=6700417$ such that the order of $2$ modulo those primes is $2^1 , 2^2, 2^3, 2^4 , 2^5 , 2^6 , 2^6$ respectively.

We now can write $\mathbb{N}_0$ as the union of arithmetic sequences $A_1, A_2, ... , A_6 , A_7$ with differences $2^1 , 2^2, 2^3, 2^4 , 2^5 , 2^6 , 2^6$ (e.g.: the sequences of numbers $\equiv 2^{i-1} \mod 2^i$ for $i=1,2,..,6$ together with those $\equiv 0 \mod 2^6$) and call the starting terms $a_1, a_2, ... , a_6 , a_7$ and the differences $b_1, b_2, ... , b_7$ respectively (so we could take $a_i=2^{i-1}, b_i=2^i$ for $i=1,2,...,6$ and $a_7=0, b_7=2^6$).

We then choose an integer $k$ by the chinese remainder theorem such that $k \cdot 2^{a_i} \equiv -1 \mod p_i$ . Then for each $n \in \mathbb{N}_0$, $n$ is contained in one of the sequences $A_i$, and additionally $2^n \mod p_i$ only depends on $n \mod b_i$, yielding $k\cdot 2^n+1 \equiv k \cdot 2^{a_i}+1\equiv 0 \mod p_i$.
Thus if we've chosen $k$ big enough, too (which we can), none of the numbers$ k\cdot 2^n+1$ is prime, each is divisible by one (and exactly one) of the primes listed above.

if $ n\equiv 0 \pmod2$ then we can choose $ k \equal{} 4$, otherwise $ k \equal{} 2$, in fact we have the form $ a^4 \plus{} 4b^4$ that is composite.

t0rajir0u yes, i've read too fast the text, you're rigth, scuse me

The problem asks for a single integer $ k$ that works for all values of $ n$.

Let $F_n = 2^{2^n} + 1$ for each $n \in \mathbb{N}$. We claim that if $F_n$ is composite for some $n \in \mathbb{N}$, then the assertion in the problem is true. Assume that $m$ is the smallest positive integer satisfying that $F_m$ is composite and let $F_m = a \cdot b$. Now note that $F_0, F_1, \dots, F_{m-1}$ are all distinct prime numbers. Furthermore, note that
\[F_0 \cdot F_1 \cdots F_{m-1} = (2-1)(2+1)(2^2 + 1)\cdots (2^{2^{m-1}}+1)=2^{2^m}-1=F_m - 2\]
which implies that $F_0, F_1, \dots, F_{m-1}$ are coprime to $F_m$ and therefore to $a$ and $b$. Now consider a $k \in \mathbb{N}$ that satisfies
\[k \cdot 2^{(2^{i})-1} \equiv -1 \pmod{F_i}\]
for all $0 \le i \le m-1$, and which satisfies
\[k \cdot 2^{(2^{m})-1} \equiv -1 \pmod{a},\]
\[k \equiv -2 \pmod{b}.\]
Note that such a $k$ exists by the chinese remainder theorem. Now we will prove that $k\cdot 2^n +1$ is divisible by one of $F_0, F_1, \dots, F_{m-1}, a, b$ for all $n \in \mathbb{N}_0$. Suppose that the largest power of $2$ dividing $n+1$ be $i$ and let $n=2^{i} (2q+1)-1$ where $q \in \mathbb{N}$. Now if $0 \le i \le m-1$, then since $F_i | 2^{2^{i+1}}-1$,
\[k\cdot 2^n +1 \equiv k\cdot (2^{2^{i+1}})^q \cdot 2^{2^{i}-1} +1 \equiv k \cdot 2^{2^{i}-1} +1 \equiv 0 \pmod{F_i}.\]
If $i=m$, then
\[k\cdot 2^n +1 \equiv k\cdot (2^{2^{m+1}})^q \cdot 2^{2^{m}-1} +1 \equiv k \cdot 2^{2^{m}-1} +1 \equiv 0 \pmod{a}.\]
If $i \ge m+1$, then let $n=2^{m+1}\cdot r-1$ where $r \in \mathbb{N}$. Now, since $b| F_m | 2^{2^{m+1}}-1$,
\[k\cdot 2^n +1 \equiv (-2) \cdot 2^{2^{m+1} \cdot r -1} +1 \equiv -(2^{2^{m+1}})^r +1 \equiv 0 \pmod{b}.\]
Hence $k\cdot 2^n +1$ is divisible by one of $F_0, F_1, \dots, F_{m-1}, a, b$ for all $n \in \mathbb{N}_0$. Let $2^l$ be such that $l \in \mathbb{N}$ and $2^l > F_0, F_1, \dots, F_{m-1}, a, b$. As proven previously, $(k\cdot 2^l) \cdot 2^{n} +1 = k \cdot 2^{l+n}+1$ is divisible by one of $F_0, F_1, \dots, F_{m-1}, a, b$ for all $n \in \mathbb{N}_0$. Since $(k\cdot 2^l) \cdot 2^{n} +1> 2^l > F_0, F_1, \dots, F_{m-1}, a, b$, it follows that $(k\cdot 2^l)\cdot 2^n +1$ is composite for all $n\in \mathbb{N}_0$, thus proving the claim.

Since Euler proved that $F_5$ is composite, the claim yields the desired result.

The topic is also covered by R.K. Guy in his Unsolved Problems in Number Theory, [b]B21[/b].

[quote="SnowEverywhere"]Since Fermat proved that $F_5$ is composite[/quote]
Euler did, Fermat's work was mainly making conjectures and completely unbased claims of having a proof.

[url="https://www.artofproblemsolving.com/community/c6h424755p2403712"] 1982 USAMO #4  [/url]

Here is a sketch of a bit easier proof, (still not really easy)
Looking k in mods 3,5,7,13,19,37 and 73 ,  So for [b]n[/b] ,
You can eliminate 1 mod from 2 (because we choose k in mod 3) , 1 mod from 4 (k in mod 5)
 ,1 mod from 3( k in mod 7 realise 2 is quadratic) ...... going on like this you can eliminate all mods from 36 ,
( You need to see 2 is quadratic for 2,19 and 73.) For example we look in 36 instead of 72 for prime 73. So for all n's , The expression divides one of 3,5,7,13,19,37 or 73. Since we choose appropriate mod for k in 3.5.7.....73 there are infinetly many of k's ,so we choose k>73. And now n is divisible by one of them and it can't be equal to them so n is not prime for all integers greater than 0.


Let $ a;b;c\geq 0$ satisfying $ \sum a\equal{}1$.Prove that:
        $ \sum\frac{1}{a\plus{}b}\plus{}2abc\geq\frac{247}{54}$

Oh, nobody like it
I think it have a nice proof


calculus
derivative
function
induction
calculus computations

This is for people who have just begun calculus.  Since an equation for the n-th derivative of a function is not usually found in textbooks, I have discovered this very simple equation for the n-th derivative of a function.

[b][i][u]Lemma.[/u][/i][/b] The n-th derivative of $f(x)=ax^p$ is $f^n(x)=a\frac{p!}{(p-n)!}x^{p-n}$.

[b][i][u]Proof.[/u][/i][/b] If $f(x)=ax^p$, $f'(x)=apx^{p-1}$, $f''(x)=ap(p-1)x^{p-2}$, ...
$f^n(x)=a\frac{p!}{(p-n)!}x^{p-n}$.  This is true due to the fact that after n times of differentiation, the coefficeint of the term becomes $a\frac{p!}{(p-n)!}$.  The power of the term after n times of differentiation is $p-n$.  If $n>p$, $f^n(x)=0$. And the conclusion follows.

$\mathbb{QED}$

Masoud Zargar

Any sort of 'proof' here should really use induction.

Isn't correct reasoning sufficient proof?

You just provide a couple of examples and deduce the general case from that.  I hardly call that 'correct reasoning'.

The next two lines are the actual reasons, but I agree that a proof via induction is more acceptable.

Not really, you are just putting the formula in words.  I mean, on one hand, you may look at this formula as self evident (it pretty much is) but your proof contains no rigor whatsoever.


If you want to graph the level curves of f(x,y) x/xy, could you rearrange the equation of k=x/xy to y= (-kx+x)k?

Also, how would you find the limit of that function as f(x,y) of x/xy approaches (0,0)?  What would be the limit for approaching along the path of the x axis (y is 0) and approaching along the y-axis (x is 0)?

Yes, x/(xy).

For the graphing,  how is the correct algebra manipulation of putting k= x/(xy)  in terms of y?

I have no idea what your question means.  Isn't x/(xy) the same as 1/y (for x <> 0)?

[quote="Mathhelp"]And you claim to be an expert on the Riemann hypothesis.[/quote]

I'm as confused as Ravi.. are you saying that x/(xy) is not equal to 1/y ?!

It is a function of two variables. f(x,y) = x/(xy).

Could you tell me if I rearranged the equation with correct algebra in terms of y?

[quote="JBL"]What does x/xy mean?  x/(xy) ?  Something else?  I don't understand it.

The general idea you stated for drawing the level curves is correct.[/quote]

If you have something like f(x,y) = xy/x, then that could be equal to y/1.  When there's two variables in the numerator but only one in the denominator, you could cancel out.  When you have one variable in the numerator and two variables in the denominator, how could you eliminate the x like that?

I'm as confused as Ravi.. are you saying that x/(xy) is not equal to 1/y ?![/quote]

if you have k= x/(xy),  then  xkyk=x,  yk= x(xk),  y= x/(xk)/(k) = x/ (xk* 1/k) ?

[quote="Ravi B"]Sure, you can eliminate x from top and bottom, as long as x is not 0.  So that gives k = 1/y.  Solving that gives y = 1/k.  I don't understand where you got y = (-kx+x)k.[/quote]

whoa ...

Are there supposed to be plus signs in there?  It looks like you wrote it correctly and took out all of the plus signs or something.  Are you sure you're typing this in correctly?

I thought that when you have two variables in a denominator that are multiplied together, to get one variable to another side, you must first multiply both variables to the other side.

[quote="zscool"]I think you are making it needlessly complicated, if x/(x*y) = k, then multiply y to both sides, so that k*y = x/x and divide by k, then

y = x/(x*k) which should be your level curves.[/quote]

Well, you could if you wanted.  But if you have an x in the numerator and an x in the denominator, they cancel out.  Also, when you have k*(x*y), this is [i]not[/i] equal to kxky.  You're trying to distribute multiplication over multiplication, which you're not allowed to do.  If it were addition, you could do k*(x + y) = kx + ky (which is why I thought you were missing + signs), but it doesn't work in that way for multiplication.

No, what he wrote is correct.  Try specific values for a, b and c if you don't believe it.

I apologize.  This theorem works.  With these variables a, b, c, I don't see how it applies to the function I had asked about.

[quote="Mathhelp"]Shouldn't the c be multiplied to the other side?

[quote="zscool"]you can, (a/(b*c))*c = a/b, because: a/(b*c) is a*1/b*1/c

so (a/(b*c))*c = a*1/b*1/c*c = a*1/b*c/c = a*1/b*1 = a/b[/quote][/quote]

Multiply the expression through on both sides by y, as zscool suggested.


Just wondering, should I get both AOPS's? My math experience includes up till ap calc (though that doesn't mean much in this type of competition). I'll also be studying kendlaya's geometry packet. I would prefer to just get one AOPS....if possible. I'm looking to do decently on the AIME, plus get a high score on the AMC 12

Have you looked through Kedlaya's packet?  It's more aimed toward the Olympiad level, iirc.  That's pretty random to study AoPS and Kedlaya's packet at the same time.  I'm not really sure what to say since it depends a lot on your current skill level.  If you can give a better idea of how much you know (e.g. how many problems you get on average on an AMC or AIME) that would be helpful.

Since you've studied up till AP Calc, you might not need vol. 1.

Definetly get volume 2. It is very helpful.

Some chapters of volume 1 might help you, so get that only if you want to.

If you can understand Kedlaya's Geometry Packet, consider getting  Engel's Problem Solving Strategies or Zeitz's Art and Craft of Problem Solving. But for AIME preparation, AoPS 1 and 2 are ideal.


abstract algebra
group theory
calculus
integration
invariant
number theory proposed
number theory

$L$ is a fullrank lattice in $\mathbb R^{2}$ and $K$ is a sub-lattice of $L$, that $\frac{A(K)}{A(L)}=m$. If $m$ is the least number that for each $x\in L$, $mx$ is in $K$. Prove that there exists a basis $\{x_{1},x_{2}\}$ for $L$ that $\{x_{1},mx_{2}\}$ is a basis for $K$.

A generalization:

Let $G$ be a free abelian group of rank $n$, and let $H$ be a subgroup of $G$ such that $G/H$ is a cyclic group of finite order $m$. Then there is a basis $x_{1},\ldots,x_{n}$ of $G$ such that $x_{1},\ldots,x_{n-1},mx_{n}$ is a basis for $H$.

Since the index of $H$ in $G$ is $m,\ H$ is also free of rank $n$, and a basis $f_{1},\ldots,f_{n}$ of $H$ is obtained from some basis $e_{1},\ldots,e_{n}$ of $G$ through a linear transformation $A$ with integral coefficients and determinant $1$. By the Invariant Factor Theorem, there are invertible integral matrices $P,Q$ such that $PAQ$ is diagonal, and its diagonal entries are $d_{1}|d_{2}|\ldots|d_{n}$ with $\prod d_{i}=m$ (we can adjust $P,Q$ such that $d_{i}\ge 1$, of course). The exponent of $G/H$ is thus $d_{n}$, and since we know that it is $m$, we must have $d_{1}=\ldots=d_{n-1}=1$, and $d_{n}=m$. This finishes the proof.

P.S.

I hope I understood the problem correctly. I don't know what $A(K), A(L)$ are, but I interpreted $\frac{A(K)}{A(L)}=m$ as $[G: H]=m$, i.e. the same thing as saying that $H$ has index $m$ in $G$, and the next statement about $m$ as saying that $G/H$ has exponent $m$. This information is enough to conclude that $G/H$ is cyclic of order $m$.


find f(x) such that: Derivation of f(x) is  $e{}^x{}^2$

You probably mean $e^{x^2}$? I don't think a "nice" antiderivative exists :)

Of course, you can express the antiderivative in terms of a power series.


function algebra domain limit quadratics polynomial calculus

find all continuous function $f: R\rightarrow R,\forall x,y\in R$IF :
$f(x+y)=f(x)+f(y)+xy(x+y)$

[hide="Approach 1"]Putting $y=0$ we get $f(0)=0$

Restricting the domain to the integers and putting $y=1$ we get the recursive equation

$f(n+1)=f(n)+f(1)+n^{2}+n$

which can be rewritten as

$f(n+1)-\frac{(n+1)^{3}-(n+1)}{3}-(n+1)f(1)=f(n)-\frac{n^{3}-n}{3}-nf(1)$

giving

$f(n)-\frac{n^{3}-n}{3}-nf(1)=\mathrm{const.}=0$, which is obtained putting $n=0$

Therefore $f(n)=\frac{n^{3}-n}{3}+nf(1)=\frac{n^{3}}{3}+Cn$ where $C=f(1)-{1\over 3}$.

Hence $f(x)={x^{3}\over 3}+Cx, C\in\mathbb{R}$[/hide]

[hide="Approach 2"]Rearrange the equation thus:

$\frac{f(x+y)-f(x)}{y}=\frac{f(y)}{y}+x(x+y)$

Assuming the function is differentiable at $x=0$ and taking $\lim_{y\to 0}$, we get

$f'(x)=f'(0)+x^{2}\iff f(x)={x^{3}\over 3}+Cx$

where $C=f'(0)$[/hide]

Unfortunately, your approach is not entirely correct (although I am sure you can easily adapt it to find a complete proof): continuous functions are not determined only by their values on the integers. However, you can note that $f(x)-x^{3}/3$ is an additive continuous function and now you are done.

Thanks for the input, harazi. I've found another approach which also seems a bit shaky, but hey... :) The bottom line is that both approaches lead to what you said about studying $g(x)=f(x)-x^{3}/3$.

[hide]Let y = 1. This rearranges to $f(x+1)-f(x) = f(1)+x^{2}+x$. This is the finite difference function of $f(x)$ and is quadratic, so $f(x)$ is a cubic polynomial. Let $f(x)=ax^{3}+bx^{2}+cx+d$.

When $x=0$, the finite difference function becomes $f(1)-f(0)=f(1)$,so $f(0)=0$, $d=0$.

The finite difference function becomes $3ax^{2}+(3a+2b)x+(a+b+c) = f(1)+x^{2}+x$. Setting the coefficients equal, $a=1/3$ and $b = (1-3a)/2 = 0$ and $c=f(1)-a-b=f(1)-1/3$.

Substitution gives $f(x)=x^{3}/3+(f(1)-1/3)x$.

Since this function satisfies the initial requirement no matter what the value of $f(1)$ is, it cannot be simplified further.
[/hide]

$f(x)=\frac{1}{3}\cdot x^{3}+g(x)$, then

$g(x+y)=g(x)+g(y)$

and there is no general solution for this, i forget the name of this property, it might be a distributive function,kx, e^x, etc are solutions for g(x)

$f(x)=\frac{1]}{3}\cdot g(x)$

[quote="Altheman"]$g(x+y)=g(x)+g(y)$

and there is no general solution for this[/quote]

That's just what harazi wrote about in his post. Such function is called additive and the functions galore you offered as solutions simply don't work. The only continuous solution is $g(x)=Cx$ and the proof is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=115188]here[/url]

oh woops...i got confused g(a+b)=g(a)g(b)...
a more simple solution to:
$f(a+b)=f(a)+f(b)$,

$f(0)=0$, $\lim_{h\rightarrow 0}f(h)=0$ by continuity

$\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=\frac{f(h)}{h}$

this is a 0/0 limit so it is indeterminant, but we know that the limit is independent of x, so f'(x) is a constant, that implies f(x)=kx+c

1. $g(a+b)=g(a)g(b)$ also has a unique solution $g(x)=c^{x}$

2. There ARE functions which are continuous but nowhere differentiable, hence the derivative argument is generally a bit flawed.

[quote="Altheman"][quote="Farenhajt"]1. $g(a+b)=g(a)g(b)$ also has a unique solution $g(x)=C^{x}$

2. There ARE functions which are continuous but nowhere differentiable, hence the derivative argument is generally a bit flawed.[/quote]

i don't believe that i assumed that the function is differentiable, but rather i showed that it exists (the derivative is the same throughout the entire function) it makes no sense that the tangent is infinity everywhere, then it is not continuous, otherwise it is a constant[/quote]

By assuming that the limit on the LHS exists, you assumed that $f$ is differentiable - that's the very definition of the differentiability. So your logic is circular: in order to prove that the limit exists, you assumed it exists, in spite of ${0\over 0}$ form.

The fact the $\lim_{h\to 0}{f(h)\over h}$ doesn't depend on $x$ says nothing about its very existence. By the way, that limit, IF EXISTS, is equal to $f'(0)$ - but as a counterexample take $f(x)=|x|$, which is continuous but non-differentiable at $x=0$.

The functional equation that Altheman obtained $g(x+y) = g(x)+g(y)$, is one of the many Cauchy's Functional Equations, and the only solution is :$g(x)=cx$.

This topic is really popular for repeating each other's posts, isn't it? :lol:

[quote="me@home"]How is your first statement proven? Are you assuming that it is a polynomial? (Because that makes the problem trivial). Please explain![/quote]

Finite differences can be interpreted as slopes of periodically taken secant lines (differences in y is the same as differences in y over an difference of 1 in x). For this reason, the behaviour of finite differences is very similar to the behaviour of derivatives.

So, if the differences in y values of a function taken at periodic x intervals are characterized by a polynomial of degree n, then the function itself must be a polynomial of degree n+1.


function
real analysis
probability and stats

Hi, can someone help me here? Let X[size=75]1[/size], X[size=75]2[/size], ... , X[size=75]n[/size] be random variables with moment generating functions M[size=75]X[/size](t). Let N be independent of the X[size=75]i[/size]'s with non-negative integer values.
S[size=75]N[/size] =
0 if n = 0
X[size=75]1[/size]+X[size=75]2[/size]+...+X[size=75]n[/size] if n>= 1

[size=117]I need to prove that [/size]M[size=92]S[/size][size=75]N[/size](t) = P[size=84]N[/size](M[size=84]X[/size](t))

Hi JimmyT

I suggest  you try with N=2 and make a conditionning over the sum $ X_1\plus{}X_2$ (try to write things in terms of convolution product).
From there you should be able to conclude.

After it's only a recurence argument for the general case.


Hi.

I have 2 questions:

1: if $ a^2\equiv \minus{} 1 (mod 401)$ and $ b^2\equiv \minus{} 1 (mod 5)$ then according to the chinesere remainder theorem:

$ X\equiv a (mod 401)$ and $ X\equiv b(mod 5)$

Can someone explain how it applies here?

2: Let $ m \equal{} \displaystyle\prod_{n \equal{} 1}^{t_i}p_i^{r_i}$.

   We know that if $ (\frac {n}{p_i}) \equal{} 1$ for each i (Legendre symbol), then $ (\frac {n}{m}) \equal{} 1$ (Jacobi).

Is the coverse also true? meaning that if the jacobi symbol is -1, then every legendre symbol that "splits" is also -1 ?

Thanks.

RealTime:
Can you please be more specific, e.g. explain what your exact question is.
1.
You didn't say what $ X$ is supposed to be (probably $ X^2 \equiv \minus{}1 \mod 2005$; then you especially get $ X^2 \equiv \minus{}1 \mod 401$, thus $ X \equiv \pm a \mod 401$).
2.
WHat do you mean by "split"¿ But take a look at $ \left( \frac{10}{21} \right)$.

aadil: Use Google or Wikipedia  :wink:


linear algebra
matrix
vector
linear algebra unsolved

For 1). If we denote by $f$ the linear operator on $M_n(K)$ $f: B \to AB$ then $dimImf+ dimKerf=n$ , but $dimImf=dimImA \leq n-1$ since $A$ is not invertible, which follows that $KerB$ is non-empty

I have sol for 1) im n't sure it's right!?
We always have 2 matrixs P,Q are invertible such that : $A=P.\left(\begin{array}{cc}I_r&0\\0&0\end{array}\right).Q$ where r=rank(A);
let $B=Q^-1.\left(\begin{array}{cccc}0&a12&...&a1n\\a21&0&...&a2n\\&......&\\an1&a_n2&...&0\end{array}\right).P^-1.$
We can choose $aij$,with i#j such that B #0.
Easily to see that : BA=AB=0;

for 2/, prove that if $T$ is upper triangular inversible, so is $T^{-1}$.

[quote="viking"]1/ Let A in M_n(K). Prove that if A is not invertible then exist B in M_n(K),B #0 such that : AB=BA=0.
2/Let A,B in M_n(K).
 a)Suppose A is a triangle matrix and A.A^T=I_n.Prove that A=diag(a1,..,an) where a_i=1 or -1
 b) Let A=A^T,B=-B^T, (A - B) is invertible,AB=BA. Put C=(A+B)(A-B)^-1 then C.C^T=I_n[/quote]
For 1/, find a non-zero vector $u$ such that $Au=0$ and another vector $v$ such that $vA = 0$. Then consider the matrix $uv$.

For 2/ a), see the hint by alekk.

2/b)  is a calculation.


Let $ x,y$ are real numbers such that :$ x^3\plus{}3x^2\plus{}4x\plus{}5\equal{}0$ and $ y^3\minus{}3y^2\plus{}4y\minus{}5\equal{}0$ determine $ (x\plus{}y)^{2008}$

Summing up the two given expressions gives
$ \frac{1}{2}(x\plus{}y)((x\minus{}y\plus{}2)^2\plus{}(x\plus{}1)^2\plus{}(y\minus{}1)^2\plus{}2) \equal{} 0$
Thus $ x\plus{}y \equal{} 0$

This was on the most recent Olymon (in fact, I emailed them the correction for this problem).

I just differentiated the first one, to show that it had only one real soln.

f'(x)=3x^2+6x+4

Now, f'(x)=0 means that 3(x+1)^2+1=0, so it is strictly increasing.

Therefore, x axis is crossed only once (it is crossed since as x aporaches negative infinity, we have negative infinity, and as x approaches positive infinity, we have positive infinity, and f(x) is continuous).

If the first eq. is f(x), then the second is f(-x), and also crosses x-axis only once, but this time at -x.

Therefore, y=-x, and x+y=0.

I remember this from the Olymon... here is my solution:
Let the two equations be f(x) and g(x), thus,
Assume that the real root is a, so that $ f(a) \equal{} a^3 \plus{} 3a^2 \plus{} 4a \plus{}5 \equal{} 0$ then,
$ \minus{}f(a) \equal{} \minus{}a^3 \minus{}3a^2 \minus{} 4a\minus{}5 \equal{} 0 \equal{} g(\minus{}a)$
Therefore, the $ (x\plus{}y)^{2008} \equal{} 0$

QED.
 :D

But you don't know if it has more than one real root, so the proof isn't valid (you still have to show that (x+y)^2008 is a unique number).

Sry  :(


Let $P^{*}$ be the set of primes less than $10000$. Find all possible primes $p\in P^{*}$ such that for each subset $S=\{p_{1},p_{2},...,p_{k}\}$ of $P^{*}$ with $k\geq 2$ and each $p\not\in S$, there is a $q\in P^{*}-S$ such that $q+1$ divides $(p_{1}+1)(p_{2}+1)...(p_{k}+1)$.

[quote="HUY VAN"]Who can post the solution of this problem?[/quote]
This is set of all number stisfying
[hide]Set of Mersenne primes less than $10000$ is $\{M_{2},M_{3},M_{5},M_{7},M_{13}\}$[/hide]

We claim that, all such primes $p$ are Mersenne primes, namely, $p=2^k-1$ for a suitable $k$ (which itself is a prime). For convenience, recall that, the set $M_p$ of all Mersenne primes, less than $10000$ are $\{3,7,31,127,8191\}$.

First, let $S=M_p\setminus \{p^*\}$. Note that, $q+1\mid \prod_{k}(p_k+1)=2^u$ for some $u$. In particular, $q+1$ is a Mersenne prime. If $p^*\notin S$, it holds that $S\cap M_p = \varnothing$, which yields that, it is impossible for such a $q$ to exist. Hence, $p^*\in S$, and therefore, $q=p^*$.

Now, conversely, we shall prove that, all $p^*\in M_p$ satisfies the conditions. We begin with $p^*=3$. Note that, since $k\geq 2$, we immediately have $(p_1+1)(p_2+1)\cdots (p_k+1)$ is divisible by $3$, thus it suffices to take $q=3$. If $p^*=7$, take a subset $S$, not containing $7$. If $3\notin S$, we can simply take $q=3$. If $3\in S$, then note that $(p_1+1)(p_2+1)=4(p_2+1)$, and is divisible by $8$. Hence, we can take $q=7$. Now, suppose $p^*=31$, and take a subset. Again, if $3\notin S$, take $q=3$. If $3\in S$, and $7\notin S$, then take $q=7$. If $3,7\in S$, then $(p_1+1)\cdots (p_k+1)$ is divisible by $(3+1)(7+1)=32$. Thus, we may take $q=31$. Now, let $p^*=127=2^7-1$. Again, if $3\notin S$, take $q=3$, if $3\in S$, and $7\notin S$, take $q=7$. Now, suppose $3,7\in S$. If $31\notin S$, take $q=31$. Otherwise, if $31\in S$, then $(3+1)(7+1)(31+1)=2^{10}$ divides $(p_1+1)\cdots (p_k+1)$. Thus, we can take $q=127$. Finally, if $q=8191=2^{13}-1$, then do the same analysis, if $3,7,31,127\in S$, then $2^{17}\mid (p_1+1)\cdots (p_k+1)$, hence we can take $q=8191$.


[quote="khanhsy"]$ \forall a,b,c > 0$ prove that $ \sum_{cyclic} \frac {a^2}{\sqrt {2a^2 \plus{} ab \plus{} b^2}}\ge \frac {a \plus{} b \plus{} c}{2}$

[hide="note"] marry christmas :D  [/hide][/quote]

 :lol:


Yale
college
geometry
geometric transformation
reflection
calculus
function

[url=http://www.theamericanscholar.org/su08/elite-deresiewicz.html]The Disadvantages of an Elite Education[/url] by William Deresiewicz who taught English at Yale University from 1998 to 2008. "Our best universities have forgotten that the reason they exist is to make minds, not careers."

Maybe you'll find someone who will care to read the entire article. Too pompous for my taste. He could have learned how to talk to Joe Sixpack much earlier. All he had to do is ask the chauffeur, or the butler, or the maid, then listen. After listening, he could even have talked to them without talking down to them. I loose my patience with people who look down upon others based on age, gender, nationality, socio-economic status, or academic attainment. The authors still writes as if he believes his opinion matters more than that of Joe Sixpack, and we should read it. I read three fourths of it and decided that it was not idea-dense enough to be worth my time. The ideas presented were not so novel and earth-shattering either. I'll go ask a high school drop-out about this issue. I do not automatically expect a less enlightened opinion.
I do not believe that it takes a stellar SAT to go into those schools. What did George W. Bush get for an SAT? I do not believe that they teach how to think or that they increase "analytic" ability, either.
I do agree that the "education" machine results in a form of social engineering, separating an elitist privileged class from the rest of society. One problem in the US is that the sorting of people into winners and losers starts in elementary school, when the teachers pick the gifted students for the gifted programs. Influential parents can prevail and get their kids in. Others need to be well liked by the teachers. Once students have been sorted, their standing and "achievement" becomes a self-fulfilling prophecy. "A Brave New World" by Aldous Huxley comes to mind.
Grading is a scam everywhere, to various degrees. A grade should reflect knowledge attained, but that is rarely the case. Teachers like to dole out grades based on class participation, or perceived effort. Pleasing and/or flattering the teacher may work too. Does the final grade show what the student knew as of the last day of the course? Not in the US. The final exam may be worth at most 20% or 30% of the grade. Does the grade show what the student will remember a year later? Not usually. Are GPA's based on comprehensive cumulative exams? Not in the US. Is an A from one teacher equivalent to an A from another teacher? How about A's from different schools? Does a GPA of 4.0 mean that the student took Calculus? Chemistry? Physics? Is one 3.5 GPA of one student equivalent to the 3.5 GPA of another student? Of course not; that's why we look at the SAT and the ACT scores. What is the value of a GPA, if graduating from high school does not guarantee that the student can read well enough to function in everyday life?

a comment on the facebook event page for this

"The advantages are myriad, the disadvantages are trivial. This much is obvious, and the details are too annoying to waste valuable math time on."

Every now and then, someone tries to get attention by writing an article about the shortcomings of a good education. While it is true that a lot of education takes place on the streets, it is really bogus to call that a shortcoming of a good academic education.

Here's a look at some of the points that this guy makes.

[quote]The first disadvantage of an elite education, as I learned in my kitchen that day, is that it makes you incapable of talking to people who aren’t like you.[/quote]
[b]Verdict:[/b] False. Just because you can't hold a conversation with a normal person, doesn't mean other Ivy grads can't.

[quote]these schools tend to cultivate liberal attitudes[/quote]
[b]Verdict:[/b] True. Even though I'm in many ways a social liberal, it bothers me that the environment in many elite schools is so vehemently liberal. A little more balance would be nice! Well, maybe it's just Cornell.

[quote]The second disadvantage, implicit in what I’ve been saying, is that an elite education inculcates a false sense of self-worth.[/quote]
[b]Verdict:[/b] False. This depends mainly on a person's maturity and predisposition to these feelings than their school.

[quote]An elite education gives you the chance to be rich—which is, after all, what we’re talking about—but it takes away the chance not to be.[/quote]
[b]Verdict:[/b] False. This is such a bogus argument.

[quote]But if you’re afraid to fail, you’re afraid to take risks, which begins to explain the final and most damning disadvantage of an elite education: that it is profoundly anti-intellectual ... students who get into elite schools are precisely the ones who have best learned to work within the system, so it’s almost impossible for them to see outside it, to see that it’s even there.

Long before they got to college, they turned themselves into world-class hoop-jumpers and teacher-pleasers, getting A’s in every class no matter how boring they found the teacher or how pointless the subject, racking up eight or 10 extracurricular activities no matter what else they wanted to do with their time. [/quote]
[b]Verdict:[/b] Misleading. Just because that's how some people get into elite schools, doesn't mean that that the environment fosters this kind of thinking, or that the environment is dominated by these kinds of students.


trigonometry
limit
algebra unsolved
algebra

Determine the values of $ \alpha$ with $ \alpha \in \mathbb{R}$ such as $ \alpha \neq \frac {\pi}{2} \plus{} k \pi$ for wich we can verify the equalty $ \sum_{i \equal{} 1}^{n}  \binom{n}{i} (sec \alpha \minus{} tan \alpha)^{n\minus{}i} \frac {1}{(sec \alpha \plus{} tan \alpha)^i} \equal{} \frac {255}{(sec \alpha \plus{} tan \alpha)^n}$. Consider that $ k \in \mathbb{Z}$ and $ n \in \mathbb{N}$.

[quote="VSelkoR"]Determine the values of $ \alpha$ with $ \alpha \in \mathbb{R}$ such as $ \alpha \neq \frac {\pi}{2} \plus{} k \pi$ for wich we can verify the equalty $ \sum_{i \equal{} 1}^{n} \binom{n}{i} \frac {1}{(sec \alpha \plus{} tan \alpha)^i} \equal{} \frac {255}{(sec \alpha \plus{} tg \alpha)^n}$. Consider that $ k \in \mathbb{Z}$ and $ n \in \mathbb{N}$.[/quote]

Let $ \frac {1}{\sec \alpha \plus{} \tan \alpha}$ $ \equal{} \tan(\frac {\pi}4 \minus{} \frac {\alpha}2)$ $ \equal{} u\neq 0$

The equation is $ (1 \plus{} u)^n \minus{} 1 \equal{} 255u^n$ $ \iff$ $ f(u) \equal{} 255u^n \minus{} (u \plus{} 1)^n \plus{} 1 \equal{} 0$

If $ n \equal{} 1$, this equation is $ 254u \equal{} 0$ and so $ u \equal{} 0$, which is impossible, so no solution.

If $ n > 1$, we have $ f(0) \equal{} 0$ and $ f'(0) \equal{} \minus{} n < 0$ and $ \lim_{x\to \plus{} \infty}f(x) \equal{} \plus{} \infty$ and it is easy to show that we have exactly one positive root for any value of $ n>1$ plus exactly one negative root for any odd $ n>1$.

So I dont know what you are waiting for. We need to solve for any $ n$ and it seems to me that there is no general solution :

$ n \equal{} 1$ $ \implies$ no solution

$ n \equal{} 2$ $ \implies$ $ 254u^2 \minus{} 2u \equal{} 0$ and so $ u \equal{} \frac 1{127}$ and $ \alpha \equal{} \frac {\pi}2 \minus{} 2\text{atan}(\frac 1{127}) \plus{} 2k\pi$

$ n \equal{} 3$ $ \implies$ $ 254u^3 \minus{} 3u^2 \minus{} 3u \equal{} 0$ and so $ u \equal{} \frac {3\pm\sqrt {3057}}{508}$ and $ \alpha \equal{} \frac {\pi}2 \minus{} 2\text{atan}(\frac {3\pm\sqrt {3057}}{508}) \plus{} 2k\pi$

$ n \equal{} 4$ $ \implies$ $ 254u^4 \minus{} 4u^3 \minus{} 6u^2 \minus{} 4u \equal{} 0$ and so $ u \equal{} 0,287858873...$ and $ \alpha \equal{} \frac {\pi}2 \minus{} 2\text{atan}(0,287858873...) \plus{} 2k\pi$

...

There is a funny trivial solution for $ n \equal{} 8$ : $ u \equal{} 1$ and so $ \alpha \equal{} 2k\pi$

I'm surprised that such a problem was submitted in an olympiad contest or training. Where is it coming from ?

I need to apologize. I was studying that late at night with my mother commanding me to turn the computer off and I didn't type the problem correctly. I has been fixed already.

Sorry for the mistake and the waste of your time and work. I will watch better from now on. I appreciate your math abbilities along your posts, Patrick!

New version of the problem :

[quote="VSelkoR"]Determine the values of $ \alpha$ with $ \alpha \in \mathbb{R}$ such as $ \alpha \neq \frac {\pi}{2} \plus{} k \pi$ for wich we can verify the equalty $ \sum_{i \equal{} 1}^{n} \binom{n}{i} (sec \alpha \minus{} tan \alpha)^{n \minus{} i} \frac {1}{(sec \alpha \plus{} tan \alpha)^i} \equal{} \frac {255}{(sec \alpha \plus{} tan \alpha)^n}$. Consider that $ k \in \mathbb{Z}$ and $ n \in \mathbb{N}$.[/quote]

$ sec \alpha \minus{} tan \alpha\equal{}\frac {1}{sec \alpha \plus{} tan \alpha}$ and so $ LHS\equal{}\frac {2^n\minus{}1}{(sec \alpha \plus{} tan \alpha)^n}$ and so :

If $ n\equal{}8$, the equation is true $ \forall \alpha\neq \frac{(2k\plus{}1)\pi}2$

If $ n\neq 8$, the equation is false $ \forall \alpha$


quadratics
algebra
quadratic formula
continued fraction

Let 3 + 1/(3 + 1/(3 + 1/...))=x.
Since the entire expression is infinite, we can replace the 3 + 1/(3 + 1/(3 + 1/...)) in the second part of the original expression by x, thus leaving us with the equation:

x=3+1/x. This leaves us with the quadratic
x^2-3x-1=0.

x=(3 :pm: sqrt[13])/2. Since we know x>3, we take the postive value,

so x=(3+sqrt[13])/2

That's a good question, but not for the reason you think. This depends on whether i is $i = \sqrt{-1}$ or not. I'll assume it is.

$i+\frac1{i+\frac1{i+\frac1{i+\frac1{i+\ldots}}}}=x$

$i+\frac1{x}=x$

$ix + 1 = x^2$

$0 = x^2 - ix -1$

using the quadratic formula:

$x=\frac{-b+\sqrt{b^2-4ac}}{2a}$

$x=\frac{i+\sqrt{-1-4(1)(-1)}}{2}$

$x= \frac{i+\sqrt{3}}2$

you can also apply this method to things like:

$x^{x^{x^{x^{x^{x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}}}}}}=2$

challenge: find the maximum value that $x^{x^{x^{x^{x^{x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}}}}}}$ can converge to.

a well known continued fraction is:

$\phi = 1+\frac1{1+\frac1{1+\frac1{1+\frac1{1+\ldots}}}}$

[quote="Hamster1800"]That's a good question, but not for the reason you think. This depends on whether i is $i = \sqrt{-1}$ or not. I'll assume it is.

$i+\frac1{i+\frac1{i+\frac1{i+\frac1{i+\ldots}}}}=x$[/quote]

You have to be careful. By setting an expression equal to x you're assuming that it equals some number, and then (with the assumption that it equals [i]something[/i]) you're solving to find what that number is. But what if your original assumption is wrong? What if it doesn't equal anything?

But just because I write a repeated expression doesn't mean that it has to be equal to something. For example, if I said 1 - 1 + 1 - 1 + 1 - 1 + ... = x, then -x = x - 1. Therefore x = 1/2, which is absurd.

Technically with these questions, you have to deal with something called convergence. But in most contest situations the problem writers just overlook this issue. In this problem though, its unavoidable, because how do you know that it doesn't oscillate like the 1 - 1 + 1 - 1 + ... problem? And even if it doesn't oscillate, how do you know which sign to use in the quadratic equation (+ or -)?

Actually, that one runs into divison by zero. It has a three-step cycle with values $i,0,\infty$.

your right! so let's assume i is a variable! Then we again have the quadratic $x^2-ix-1 = 0$ so applying the quadratic formula again we get:

$\frac{i+\sqrt{i^2+4}}{2} \text{ or }\frac{i-\sqrt{i^2+4}}{2}$

[quote="Hamster1800"]find the maximum value that $x^{x^{x^{x^{x^{x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}}}}}}$ can converge to.[/quote]

[hide="hint"]$n=x^{x^{x^{x^{x^{x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}}}}}}$
Notice that $n=x^{n}$.[/hide]

well, if you look at it from the inside, smallerlooking numbers and work it out you get:1+i,.5+.5i,1,1+i...  If you assume that the last denominator in the infinite chain is i+1 and not i, it gets to a finite loop

the max value that x^x^x^x^x... can finitely converge to is euler's number.  Let x be e^1/e.


trigonometry
algebra
polynomial
geometry unsolved
geometry

Let $ABC$ be a isosceles triangle $(AB=AC)$ with the bisector $BD=AC=AB$. Denote $AB=AC=BD=b, BC=a$. Prove that:
$(1+\frac{a}{b})(\frac{a}{b}-\frac{b}{a})=1$

[quote="tunganh"]Let $ABC$ be a isosceles triangle $(AB=AC)$. Denote $AB=AC=b, BC=a$. Prove that:
$(1+\frac{a}{b})(\frac{a}{b}-\frac{b}{a})=1$[/quote]

you not forgott anything, my friend?  :wink:

Oh, sorry. I forgot that the bisector BD is equal to the side AB and AC :D

[quote="derezin"]@yetti:
Polynomial $z^{7}-1$ has only one root, $1$!
What's more, I don't understand what do u mean by '$i\sin(x)$'...[/quote]
$i = \sqrt{-1}$. I think he means "$7$ roots" by "$7$ complex roots"  :)
Another solution is use the formula: $BD^{2}= \frac{(a+b+c)(a+c-b)ca}{(c+a)^{2}}$, with $b=c$ and calculate to have $(1+\frac{a}{b})(\frac{a}{b}-\frac{b}{a})=1$  :D

[quote="PDatK40SP"]
$i = \sqrt{-1}$[/quote]
As far as i know we define only square roots of positive numbers, while -1 isn't...

no but he is not finding the roots, he is finding the x+1/x, which is real

But he uses square roots of negative numbers! It's like dividing by 0 :wink:

[quote="tunganh"][color=darkred]Let $ABC$ be an isosceles triangle $(AB=AC)$ with the bisector $BD=AC=AB.$
Denote $AB=AC=BD=b,\, BC=a\,(b<a).$ Prove that: $(1+\frac{a}{b})(\frac{a}{b}-\frac{b}{a})=1$[/color][/quote]
[color=indigo][b]Proof.[/b] Take the point $E\in BC$ so that: $BE=BD=BA=b$
Thus, easy to show that: $\triangle CDE\sim \triangle CBD$
So: $CE\cdot CB=CD^{2}\Longrightarrow (a-b)a=\left(\frac{ab}{a+b}\right)^{2}\Longrightarrow\frac{a+b}{b}\cdot\frac{a^{2}-b^{2}}{ab}=1$ $\Longrightarrow\left(1+\frac{a}{b}\right)\left(\frac{a}{b}+\frac{b}{a}\right)=1\quad\mathbb{QED}$[/color]

[color=darkred][u][b]Similar problem.[/b][/u] Let $ABC$ be an isosceles triangle $(AB=AC)$ with the bisector $BD+DA=BC.$
 Prove that: $\angle BAC=100^\circ$[/color]


geometry
rectangle
ratio
analytic geometry
conics
parabola
similar triangles

A rectangle is inscribed in a isoceles triangle. Prove that when the area of the rectangle is maximised, its area is half of the isoceles triangle

The rectangle with maximum area is a square.

That's just not true.

For purposes of this problem, "inscribed in an isosceles triangle" means that one side of the rectangle is contained in the base of the triangle.  (In case this was making anyone uncomfortable.)  In fact, it is possible to achieve a ratio of $ \frac{1}{2}$ for any triangle.

i think my proof may be circular... but here it goes

[hide]

we first label any generic isosceles triangle on the coordinate plane, letting it have vertices $ (a,0) , (-a, 0), (0,b)$. Then we inscribe a rectangle into the isosceles triangle and let it have coordinates $ (x,0), (-x,0), (x,-bx/a+b), (-x,-bx/a+b)$ (we obtain the last two points by finding the equation of the line connecting $ (0,b)$ and $ (a,0)$ and then by plugging in $ x$). Hence we know that the area of the triangle is $ ab/2$ and the area of the rectangle is $ x(-bx/a+b)$. We wish to prove that:

$ 2x(-bx/a+b) \le ab/2 \implies 0 \le ab/2-2x(b-bx/a)$

simplifying this we obtain:

$ 0 \le \frac{x^{2}}{a^{2}}-\frac{x}{a}+\frac{1}{4}= \Big( \frac{x}{a}-\frac{1}{2}\Big)^{2}$

therefore we know the area of the rectangle is maximized when the equality holds and $ x/a = 1/2$

hence we know $ x = \frac{a}{2}$ which means that $ -bx/a+b = b/2$

and so the area of the rectangle is $ \frac{a}{2}\cdot \frac{b}{2}= \frac{ab}{4}$ which is half that of the triangle

[/hide]

That's not circular, it's correct  :wink:

I came up with a proof for any triangle. An attachment has been added.

Proof

We have two cases for a rectangle being inscribed within a rectangle.

Case 1 (see Figure 1)

The triangle is right angled, and the inscribed rectangle has a vertex at this right angle.

Case 2 (see Figure 2)

One side of the rectangle lies upon one side of the triangle (think of this as sharing four of the rectangle's vertices between three of the triangles sides). The altitude which is perpendicular to this side always passes through the rectangle. If this were not the case then the four rectangle's vertices would have to be shared between two of the triangle's sides. This is only possible when these two sides are parallel - an obvious contradiction.

Lets begin with Case 1

By similar triangles:

$ \frac{y}{b-x}= \frac{a}{b}\Rightarrow y = \frac{a}{b}(b-x)$

$ A_{rectangle}= xy = \frac{a}{b}x(b-x)$

If we were to sketch a graph of $ A_{rectangle}$ against $ x$ a concave down parabola would be obtained which intersects the $ x$-axis at $ x = 0$ and $ x = b$. The maximum value of $ A_{rectangle}$ is found midway between these roots so corresponds to $ x = \frac{b}{2}$

Hence $ A_{rectangle\, max.}= \frac{a}{b}\cdot \frac{b}{2}\left(b-\frac{b}{2}\right) = \frac{a}{2}\cdot \frac{b}{2}= \frac{1}{2}\left(\frac{1}{2}ab\right) = \frac{1}{2}A_{triangle}$

Now lets turn to Case 2:

By similar triangles:

$ \frac{c}{y}= \frac{e}{a-y}\Rightarrow c = \frac{ye}{a-y}$

$ \frac{d}{y}= \frac{f}{a-y}\Rightarrow d = \frac{yf}{a-y}$

$ \Rightarrow c+d = \frac{y(e+f)}{a-y}$

$ \Rightarrow b-x = \frac{yx}{a-y}$

$ \Rightarrow (b-x)(a-y) = xy$

$ \Rightarrow ab-ax-by = 0$

$ \Rightarrow y = \frac{a}{b}(b-x)$

This is the same expression for $ y$ as in Case 1, and by applying the same logic, the maximum rectangle which can be inscribed is half the area of the triangle. QED.

EDIT: Attachment won't "Attach". Anyone got any suggestions?


calculus
integration
inequalities
geometry
trapezoid
analytic geometry
graphing lines

Prove the following inequality.

\[ \frac{2}{2\plus{}e^{\frac 12}}<\int_0^1 \frac{dx}{1\plus{}xe^{x}}<\frac{2\plus{}e}{2(1\plus{}e)}\]

Hi kunny.

For the right-hand inequality, we see that the curve lies below the top of the trapezium; the line segment joining $ (0, 1)$ to $ (1, \frac {1}{1 \plus{} e})$. The arrea of this trapezium is $ \frac {2 \plus{} e}{2(1 \plus{} e)}$.

For the left-hand inequality, we see that $ (\frac {1}{2}, \frac {2}{2 \plus{} e^{\frac {1}{2}}})$.

Differentiating, we find the slope of the tangent at this point and thus its equation. Substituting $ x \equal{} 0$ and $ x \equal{} 1$, we get a trapezium whose area is less than the under the curve and this works out nicely to be $ \frac {2}{2 \plus{} e^{\frac {1}{2}}}$.

I've just noticed that the tangent gives two congruent triangles, so the area of the trapezium is the same as the area of the rectangle of width $ 1$ and height $ \frac{2}{2\plus{}e^{\frac{1}{2}}}$ and we have result without differentiating.

I didn't apply any theorems. For the right-hand inequality, I just drew a rough sketch and joined the end-points. For the left-hand one, I noticed that the value for $ x\equal{}0.5$ is what is needed, took the long way round and then saw an easier way.

We can apply hermittes hadamards inequality for convex funtions. Put a=0, b=1 and you are ok :wink:
http://en.wikipedia.org/wiki/Hermite%E2%80%93Hadamard_inequality


algebra
polynomial
trigonometry
inequalities
algebra open

What can be said about a polynomial $ P(x)$ that satisfies: $ p(\cos x)\equal{}p(\cos (x\minus{}\theta))$ for some $ \theta$, for every $ x \in [0, \pi]$?

Let $ f(x) \equal{} P(\cos x)$

Then $ P(\cos x) \equal{} P(\cos(x \minus{} \theta)) \iff f(x) \equal{} f(x \minus{} \theta)$

So $ f$ is a polynomial of $ \cos x$ and its period is $ \theta$ (accurately its period divides $ \theta$)

One example is $ P(x) \equal{} 16 x^5 \minus{} 20 x^3 \plus{} 5 x$

Then $ f(x) \equal{} 16 cos^5x \minus{} 20 cos^3x \plus{} 5 cosx \equal{} cos(5x)$

$ f$ has period $ T \equal{} \frac {2\pi}{5}: \ f(x) \equal{} f(x \minus{} T) \ (1)$

For any $ k\in \{1,2,3,4\}$ define $ \theta \equal{} kT$. Then, according to $ (1)$ we have $ f(x) \equal{} f(x \minus{} \theta) \iff P(\cos x) \equal{} P(\cos(x \minus{} \theta))$

* Instead of $ f$ we can use any $ F(x) \equal{} af(x) \plus{} b$

* We can use $ f(x) \equal{} \cos(mx)$ for any $ m$, because it is a polynomial of $ \cos x$

* Here, $ \frac {\theta}{\pi}$ is a rational. I don't know if there are other solutions with $ \frac {\theta}{\pi}$ being irrational  :huh:

[quote="pontios"]* Here, $ \frac {\theta}{\pi}$ is a rational. I don't know if there are other solutions with $ \frac {\theta}{\pi}$ being irrational  :huh:[/quote]
There are not!

If $ f(x) \equal{} f(x \minus{} \theta)$ for any $ x$, then $ \theta \equal{} aT$ for some integer $ a$

But $ f$ is a polynomial of $ \cos x$, so $ f(x) \equal{} f(x \plus{} 2\pi), \ \forall x$

So, its period is $ T \equal{} \frac {2\pi}{m}$ for some $ m \in \mathbb{N}$

Consequently $ \theta  \equal{} \frac {2a\pi}{m}$

Man, I liked very much your problem! It also gave me the opportunity to learn more about Fourier transforms

Here we go!

I'll find the polynomials $ P$ which satisfy $ P(\cos x) \equal{} P(\cos(x \minus{} \theta)), \ \forall x\in\mathbb{R}$

Let $ T_k(x)$ be all the [url=http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html]Chebyshev Polynomial of the First Kind[/url]. Then

$ T_0(x) \equal{} 1$
$ T_1(x) \equal{} x$
$ T_2(x) \equal{} 2x^2 \minus{} 1$
$ T_3(x) \equal{} 4x^3 \minus{} 3x$
$ T_4(x) \equal{} 8x^4 \minus{} 8x^2 \plus{} 1$
$ T_5(x) \equal{} 16x^5 \minus{} 20x^3 \plus{} 5x$
$ \cdots$

Then $ T_k(\cos x) \equal{} \cos(kx)$ for any $ k$

Let $ P(x) \equal{} \sum^\infty_{k \equal{} 0}{a_k T_k(x)}$ where $ a_k$ are arbitrary chosen reals

Then we define $ f(x) \equal{} P(\cos x) \equal{} \sum^\infty_{k \equal{} 0}{a_k \cos(kx)}$

$ f$ is periodic. Its period is $ T \equal{} \frac {2\pi}{m}$ for some $ m\in \mathbb{N^\star}$

[hide="ok"]If:
$ a_0 \equal{}$ arbitrary
$ a_m\neq 0$
For any $ s$ such that $ m\not | s$ we have $ a_{s} \equal{} 0$
then the period is $ T \equal{} \frac {2\pi}{m}$[/hide]

Now for any $ w\in\{1,2,...,m \minus{} 1\}$ the number $ \theta_w \equal{} wT$ has the attribute $ f(x) \equal{} f(x \minus{} \theta_w) \Rightarrow P(\cos x) \equal{} P(\cos(x \minus{} \theta_w)),\ \forall x$

[size=150]Conversely[/size]

Suppose that there is a polynomial $ P$ and a real number $ \theta \in(0,2\pi)$ such that $ P(\cos x) \equal{} P(\cos(x \minus{} \theta)),\ \forall x$

Let $ f(x) \equal{} P(\cos x)$. Then $ f(x) \equal{} f(x \minus{} \theta),\ \forall x$

Then (according to my last post) the period of $ f$ is $ T \equal{} \frac {2\pi}{m}$ where $ m\in N^\star$ and $ \theta \equal{} \frac {w}{m}2\pi$, with $ w\in\{1,2,m \minus{} 1\}$

$ \star$ Notice that $ f( \minus{} x) \equal{} P(\cos( \minus{} x)) \equal{} P(\cos x) \equal{} f(x)\Rightarrow f$ even (very important fact)

Fourier's theorem says that:
[quote]Any periodic function f(x) can be expressed as a sum of sine and cosine functions[/quote]
And if $ f$ is [u]even[/u] then we have only [u]cosines[/u]

So $ f(x) \equal{} \sum^\infty_{k \equal{} 0}{a_k\cos\left(\frac {2\pi k}{T}x\right)} \equal{} \sum^\infty_{k \equal{} 0}{a_k\cos(mkx)}$

Or simply $ f(x) \equal{} \sum^\infty_{s \equal{} 0}{c_{s}\cos(sx)}$, where $ c_s \equal{} 0$ if $ m\not|s$

Then we define the polynomial $ P(x) \equal{} \sum^\infty_{s \equal{} 0}{c_{s}T_s(x)}$, which has the property $ P(\cos x) \equal{} f(x), \forall x$

[size=150]Conclusion[/size]

A polynomial $ P(x)$ has the property $ P(\cos x) \equal{} P(\cos(x \minus{} \theta))\ \forall x\in\mathbb{R}$ iff $ P(x) \equal{} \sum^\infty_{k \equal{} 0}{a_k T_k(x)}$