Hi everyone!

I've known that Mathlinks is a good forum for mathematicians. They share excellent problems, that were posted at various contests\Olympiads, or that they simply find really worth solving and they help each other.

I hoped that this Physics forum would act in the same way. I hoped to meet many people who share a passion for Physics and who are willing to post/try to solve good (read as difficult) problems. I was partially right, but only partially.
I've seen topics like "Homework problems" which obviously speak that these problems are not quite... Olympiad problems, or Very difficult problems.
I am not the owner or moderator of the forum, and that is why I am not to decide what should be permitted and what should not, but to me such  topics are disappointing, and I am against them.  Why should this forum turn into one where people post homework problems, and others solve them. Not only is this unfair, but also irrational. It's irrational to solve lots of low difficulty problems, because they worsen your skills and waste your time.

That is why, I ask you all to post only trully challenging problems.

Thank you.

PS - Please express your opinions on the matter.

I quite agree with this complaint, and although I haven't been a member for long, I can see that a large portion of these problems are of easy or (barely) medium difficulty. Now, I appreciate that there are people who need help with their homework -- I don't know whether this forum is intended for them. If it is, then there's nothing to be done about it; everything should remain untouched. If it isn't, then this forum clearly needs more rigorous rules and moderators.

Well, it seems I've been the first one to vote -- I voted that this should be a forum for posting challenging problems. I'd really like to hear the opinions of the other frequent visitors of this forum. I'd also like to see the response of the authorities.

I hope that people will take this matter seriously and decide what should be done with the Physics forum before it becomes a disappointment to those who want challenging instead of school problems (and I hope that the number of these people isn't small!)

We need a subforum "Homework Problems." The Calculus and Real Analysis forum did something similar by creating a "Calculus Computation and Tutorials" forum. AoPS/mathlinks isn't an ideal place for basic-level homework help in physics--physicsforums.com is much better for this. However, we still need to accomodate those who come here for homework help.
We also need a new moderator. xxreddevilzxx isn't active on this forum, so we still have outdated announcements, and the forum needs FAQ to answer the recurring questions about the US screening exams.

2 schulmannerism: Good ideas!  How about writing an e-mail to the owner of forum, or the main moderator, in order to draw his/her attention to this matter.  The only problem would probably be the realtively small number of people who post on this forum. Though improving the forum will help it become more popular.

I've been away for a long time from this forum coz I was (and am still veryyyy busy in IIT-JEE prep - I hope you know about it!!)...

While I was here, I found that people like kubus, yetti , me , daveed etc. (sorry if I missed someone out!) discussed great theoretical problems and some interesting and challenging problems requiring great though as well ( search the forum to find these gems!!) - I even started a weekly contest to encourage and further discuss the great problems of Physics --- but the contest went null as almost no one participated --- now I see that after a break of a month when I return , the forum is littered(no offence please ~!!) with simple "formula based problem"" I do not wish to offend anyone --- it is true that when people who are also studying physics visit mathlinks , they will obviously post all problems -easy or challenging discussions - here !! I strongly second the motion of creation of an appropriately named subforum for simple forums (the title should be such that first time posters should post easy problems in that subforum!)..... We should also suggest when the Physics forum page opens that greater detailed and more specific discussions 'might' be better posted at physicsforums.com (of course along with that  - you should post it here to share it here!!)

Regarding the moderator --- the present one is really helpful when you approach him and always there - he's probably more engaged in his Maths forums and probably gives a little lower priority to Physics(no offence please)... I was very active in this forum and will be 2 months hence once my all important IIT exam(Indian Institute of Technology - the toughest Physics, MAth, Chem combined exam in the WORLD!) is over..... I have always wanted to be a part of moderating this forum.... Valentin told me to wait --- I feel the time has come--- I would like to nominate myself!!  :D (Please note that I will be active only after 2 months -- however, since the issue hs come up - I post my candidacy right now!!  :lol: )

Rushil, I will personally support you in case there will be a voting!

I would prefer it to be a hard problems forum, but then I feel it really doesn't matter if some homework problems are posted once in a while.......Homework problems can, at times, be a key to understanding concepts pretty well.........The problem with problems like those from the physics olympiads is that the minute someone from somewhere near the eigth or ninth grade sees them, they might really feel that the stuff is going above their head and leave....On the contrary, when you have a mixture of both simple and difficult problems, it should lead to a higher number of active participants in the forum.... And well, if some member posts homework problems for say, a couple of months, he's bound to look at the other problems as well....So basically, it's like letting him/her slowly mature and thereby understand things better.......When you have only the tough Olympiad problems, it's somewhat like discouraging young talent......All the same, the more experienced members do need to go on with the olympiad style of problems. I quite look forward to seeing a mixture of both, though I'll prefer a majority of the olympiad problems while at the same time not ignoring the homework problems.

Well, it's fairly clear that the best solution would simply be to have a subforum for "computations and tutorials", as previously suggested, and hopefully the small number of people won't be a problem, as the forum seems to be growing in number (as opposed to the state forums, which have definitely been declining since near the start). Seeing as I don't think anybody has written a PM to Mr. Rusczyk or another administrator yet, as implied from the posts above, I'll do so now... this should probably solve all problems.

Hey... not a seperate forum --- it should be a SUBFORUM !!! Otherwise everyone will be confused!!!

Also, we should start moving topics to the easy forum if we consider it to be so ---- the moderator can do it only!!  :D

[quote="Rushil"]Please do so... Valentin i s usually more active on the forum --- maybe you should PM him too!!![/quote]No need to, my eyes see everything :police: :yoda: :P

Regarding the issues presented in this topic, the following measures will be taken:

- a second moderator will be assigned (by the way, are there any other nominations besides Rushil?)
- a homeworks help subforum will be created and all the homework stuff should be dealt with there (remember still that some people might just be at the "getting started" level in physics and not really want to do homework - don't push people just because they are asking easy questions ;) )

You can further discuss nominations for moderators (though keep in mind that the final decision will [b]not[/b] be based (entirely) on the "democratic" vote :P) and naming of the homework help subforum (for now I'll just name it "Homework Problems & Tutorials").

That's great Valentin!!

Eagerly waiting your decision and the reformating of the Physics forum!!!

Although the creation of the new forum (Introductory Physics) was much needed, the entire Physics section still needs moderating. Seeing that nobody except [b]Rushil[/b] has publicly nominated himself for the position of the new moderator -- together with the fact that even [b]Rushil[/b] will be unable to devote his time to the forum for a few month -- has made me decide to offer myself as the moderator before [b]Rushil[/b] is back.

However, since I am rather new on this forum, I will gladly step back in case that a senior of mine (the number of such people on this forum is easily very large) wishes to take these chores onto himself -- I have no desire to hinder someone's years-old plans to become a moderator :) -- in other words, I am willing even to withdraw when [b]Rushil[/b] (or almost anybody else, for that matter) feels eager to take on moderating. I just want to have some authority  active here (and this authority doesn't need to be me) who'd sort topics into the right forums and so on, just so the chaos in this section does not increase further. A couple of months is not a very long time, but I think that the Physics section has no need to wait that long for attention.

Seeing the lack of response to the Administrator's call, I can only hope that other people who want to moderate the two Physics forums will announce themselves here. In the meantime, I hope that my offer will be considered in due time. Again, I have no pretensions to gain a position on this forum :) -- it's just that there doesn't seem to be anybody else with time and will to do this and I'd like to see these forums in good shape.

Thanks!

That's great , Djole!!

Since youve started the idea of reform of the forum, i'm sure you'll make a great moderator! I support you(as you do me!!  :D )... Anyways, I hope Valentin can make both of us moderators ---- there's no reason for you to step down later --- it'll be great!!

There's no 'year-old' plan to be a moderaot - so dont think you are obstructing anyone!
It'll be great if more people can partcipate in the organisations!!

Still waiting Valentin....

[size=150][color=red]Valentin, Where are you??!!!!!![/color][/size]


algebra
polynomial
inequalities
function
algebra unsolved

Let real coefficient polynomial $f(x) = x^n + a_1 \cdot x^{n-1} + \ldots + a_n$ has real roots $b_1, b_2, \ldots, b_n$, $n \geq 2,$ prove that $\forall x \geq max\{b_1, b_2, \ldots, b_n\}$, we have
\[f(x+1) \geq \frac{2 \cdot n^2}{\frac{1}{x-b_1} + \frac{1}{x-b_2} + \ldots + \frac{1}{x-b_n}}.\]

Put $x-b_i=a_i$. We must prove that for positive numbers $a_i$ we have $ (1+a_1)...(1+a_n)(1/a_1+...+1/a_n)\geq 2n^2$. USe Huygens inequality to find that the LHS is greater than or equal to $n(1+G)^n/G$ where $G$ is the geometric mean of $a_i$. A study of the function $\frac{(1+x)^n}{x}$ shows that it has minimal value in $\frac{1}{n-1}$ and this one is $ (n-1)(n/(n-1))^n$ obviously greater than $2n$.

Well , you can prove the last ineq : $ \frac{(1\plus{}x)^n}{x} \ge 2n$ like this
$ \frac{(1\plus{}x)^n}{x} \ge  \frac{1}{x}\plus{}n\plus{}\frac{ n(n\minus{}1)}{2x} \ge n\plus{}\sqrt{ 2n(n\minus{}1)} \ge 2n$

[quote="math10"][quote="harazi"]Put $ x \minus{} b_i \equal{} a_i$. We must prove that for positive numbers $ a_i$ we have $ (1 \plus{} a_1)...(1 \plus{} a_n)(1/a_1 \plus{} ... \plus{} 1/a_n)\geq 2n^2$. USe Huygens inequality to find that the LHS is greater than or equal to $ n(1 \plus{} G)^n/G$ where $ G$ is the geometric mean of $ a_i$. A study of the function $ \frac {(1 \plus{} x)^n}{x}$ shows that it has minimal value in $ \frac {1}{n \minus{} 1}$ and this one is $ (n \minus{} 1)(n/(n \minus{} 1))^n$ obviously greater than $ 2n$.[/quote]
Can anyone post other solution?(This problem is not difficult but I can not solve this problem by other)[/quote]
$ x\minus{}b_i \equal{} c_i \Rightarrow (c_1\plus{}1)...(c_n\plus{}1)\left ( \frac{1}{c_1} \plus{} ... \plus{} \frac{1}{c_n} \right ) \ge 2n^2$
$ \frac{1}{c_1} \plus{} ... \plus{} \frac{1}{c_n} \ge n\sqrt[n]{\frac{1}{c_1c_2...c_n}}$
$ c_i\plus{}1 \equal{} c_i\plus{}(n\minus{}1)\frac{1}{n\minus{}1} \ge n\sqrt[n]{c_i\frac{1}{(n\minus{}1)^{n\minus{}1}}}$
So $ LHS \ge n^{n\plus{}1}\frac{1}{(n\minus{}1)^{n\minus{}1}} \ge 2n^2 \iff$
$ \left ( \frac{n}{n\minus{}1} \right )^{n\minus{}1} \ge 2 \iff$
$ \left ( 1 \plus{} \frac{1}{m} \right )^m \ge 2, m \equal{} n\minus{}1 \ge 1$
But $ \left ( 1 \plus{} \frac{1}{m} \right )^m \equal{} \sum_{i\equal{}0}^m \binom{m}{i} m^{\minus{}i} \ge \binom{m}{0}\plus{}\binom{m}{1}m^{\minus{}1} \equal{} 2$, and we are done.


Are these worth buying... Are they helpful and what do they cover?

There have been a few discussions about this already:

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=arbelos&t=32200

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=arbelos&t=27684

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=arbelos&t=12785

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=arbelos&t=8937


trigonometry algebra polynomial quadratics

Find the exact value of $\sin1^{\circ}$ (or at least outline a process)

$\sin 15^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}$, then solve a polynomial of degree 15 that you obtain through Euler.  :)

Alternately, if you feel like evaluating the Taylor Series of $\sin \frac{\pi}{180}$...

This must be impossible.  If you could find the exact value of the sin of one degree, you could use addition formulae to find the exact value of sin of every other degree angle, and I know that this is not possible.  Do you mean a process for approximating the number?  In that case, you could either simply use a McLaurin/Taylor series, or figure out a high-degree polynomial, one of whose (non-analytic) roots would be the proper number.

[hide="an outline"]$\sin 45^\circ=\cos 45^\circ=\frac{\sqrt{2}}{2}$.

$\cos 36^\circ=\frac{\sqrt{5}+1}{4}$ (and from this $\sin 36^\circ$ can be found). This can be shown by using $\cos 36^\circ \cos 72^\circ=\frac 14$ (since multiplying by $4\sin 36^\circ$ gives $\sin 36^\circ$), expanding in terms of $\cos 36^\circ$, and getting rid of the extraneous root -1/2.

$\cos 20^\circ$ (and therefore $\sin 20^\circ$) can be found since $4\cos^3 20^\circ -3\cos 20^\circ-\frac 12=0$, and then the cubic can be solved.

Since $1=9(45-36)-4(20)$, the addition formula can be used enough times to find $\sin 1^\circ$.[/hide]

[quote="scorpius119"]
$\cos 20^\circ$ (and therefore $\sin 20^\circ$) can be found since $4\cos^3 20^\circ -3\cos 20^\circ-\frac 12=0$, and then the cubic can be solved.
[/quote]

Not by the cubic formula - that method will force you to calculate a root that evaluates to an expression in terms of $\cos 20^{\circ}$, which takes you right back where you started.  I also don't think a simple linear term can be factored out...

In any case, the polynomial of degree 15 was kind of a joke  :)  If you wanted an approximation, the Taylor series converges rather quickly - the first two terms gives $\sin \frac{\pi}{180} \approx 0.017452406424$ while calculator gives $\sin \frac{\pi}{180} \approx 0.017452406437$ (but of course, the calculator doesn't do more than evaluate the Taylor series to sufficient digits...)

The question, then, is what you mean by "exact value" - a closed radical expression?  This is clearly impossible - at best, finding $\sin \frac{\pi}{180}$ in terms of $\sin \frac{\pi}{5}$ requires dividing by $36 = 2^2 \times 3^2$, a process equivalent to solving two quadratics and two cubics - and both cubics cannot be solved except in terms of the $\sin$ values they are trying to find!

(Edit:  However, by this logic it would be entirely feasible to find a closed form for $\sin 9^{\circ}$.  It would be very ugly  :( )

Edit: My mistake $90 - 1 = 99$ between the hours of 11 p.m. and 1 a.m. Don't mock me, it makes me sad.

$\sin 1^{\circ} = \cos 89^{\circ} = \cos \frac{89\pi}{180}$
$2\cos a = \sqrt{2+2\cos 2a}$ when $\cos a$ is positive and
$2\cos a = -\sqrt{2+2\cos 2a}$ when $\cos a$ is negative
$\Rightarrow 2\cos\frac{89\pi}{180} = \sqrt{2+2\cos\frac{89\pi}{90}} = \sqrt{2-\sqrt{2+2\cos\frac{89\pi}{45}}} = \sqrt{2-\sqrt{2+2\cos\frac{\pi}{45}}}$
It can be proven that $2\cos\frac{\pi}{45}$ is ($\frac{\pi}{45} = \frac{91\pi}{2^{12} - 1}$ is how we find our string of signs) equal to the the infinite radical of the form $\sqrt{2 + \sqrt{2 + \sqrt{2 \cdots}}}$, but it has a string of 12 signs which is too long to actually put under a radical, so we will just say, the repeating string $(\overline {++++---+--+-})$. [hide="(Extra)"]See if you can prove it, it is actually proven generally by me in other [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59094&start=20]post,[/url] sort of.[/hide]
Subbing back in, we get:
$2\sin 1^{\circ} = 2\cos\frac{89\pi}{180} = \sqrt{2-\sqrt{2+2\cos\frac{\pi}{45}}}$
$= \sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots}}}}}}$
$= [-+]\Vee (++++---+--+-) = (-+ \overline {++++---+--+-})$

[quote="Pork_Chop8"]$\sin 1^{\circ} = \cos 99^{\circ}$...[/quote]

Am I missing something here? Some recent groundbreaking research in trigonometry? Last time I checked, $\sin 1^{\circ} = \cos 89^{\circ}=-\cos 91^{\circ}$

$\sin x^{\circ} = \cos (90^{\circ} - x^{\circ})$

so yes

$\sin 1^{\circ} = \cos 89^ {\circ}$

Ah, I realized that trying to find $\cos 20^\circ$ in terms of radicals is useless - the result is $\frac{\sqrt[3]{1+i\sqrt{3}}+\sqrt[3]{1-i\sqrt{3}}}{2\sqrt[3]{2}}$, which is just DeMoivre's Theorem (this would be like saying that $\cos 72^\circ$ is half the sum of two fifth roots of 1, not very useful).

At least it would be possible to find (and, in fact, construct with compass/straightedge) the value of $\sin 3^\circ$.

[quote="Altheman"]well, i was thinking that we have sin 36 and sin 30, so use subtraction to get sin 6, then half to sin 3, then triple angle general cubic (solve with cardano) to get sin 1[/quote]

Cardano would give your answer in terms of $\sin 1^{\circ}$, I have already tried trisecting a similar angle with it  :|

we have sin 18 which is phi/2 and sin 15,
find sin 3 by sin(18-15). Then use triple angle formula for sin (sin 3x = 3sinx -  4sin^3(t))and solve for sin 1 using the cubic formula

But unfortunately, it doesn't work since at some point you have to take a cubic root from a complex number, and using the trigonometric form (or the exponential, which is basically the same) takes you right back at the start. (You end up with $\sqrt[3]{\cos 3^\circ+i\sin 3^\circ}=\cos 1^\circ+i\sin 1^\circ$, and there's no other way to take the root).


The positive integers a, b, c, d are such that
a^2 + b/(a+c) = d
Prove that d≤b+(c-1)^2

Let  a = 5, b = 16, c = 3 then d = 27.
Our inequality 27 =< 16 + 4 = 20 so counterexample.

Ooops typo!!
In fact it's (a 2 +b)/(a+c)=d


topology
advanced fields
advanced fields unsolved

I was following a topic at the following url:  http://mathforum.org/kb/thread.jspa?threadID=1545458&tstart=60

The question was the following:  Define a subspace of the real line that can be represented as the union of two locally compact subspaces, one of which is closed and the other open, and that this is not a locally compact space.

The solution was given as:  Take A = (0,1/2) U (3/4,5/6) U (7/8,9/10) U ... and B = {1}. Then A is open, B is closed, A and B are locally compact, but their union isn't.

I still don't understand why A U B would NOT be locally compact.

I was given the answer that "Because no neighborhood of 1 in A U B is compact."  I don't understand why it wouldn't.

Forgive my ignorance here but could anyone actually provide a proof of the above problem?

R.D.

A neighborhood of $\{1\}$ is a set $N$ such that $1$ is contained in the interior of $N$. Since we are dealing with the topology induced from $\mathbb R$, this means: $\exists\ r>0$ such that $(A\cup B)\cap (1-r,1+r)\subseteq N$. It follows that $N$ contains one (in fact infinitely many) of the open intervals that comprise $A$. Consequently, $N$ is not closed in $\mathbb R$. Finally, recall that a subset of $\mathbb R$ is compact iff it is closed and bounded.


How do you do problems in the form:

What is the smallest $ n$ so that $ 71$ divides $ 7n \plus{} 1$?

First note that $ 7n\plus{}1\ge 71$  obvsiouly for $ n\equal{}10$  we get     $ \frac{71}{71}\equal{}1$  which is the smallest $ n$ for which $ 71|7n\plus{}1$.


Drexel University’s Goodwin College of Professional Studies offers tuition remission for prospective middle- and high-school math and science teachers who enroll in a federally funded program to teach in high-need urban areas.  Successful candidates will earn state teacher certification and can choose to teach in the School District of Philadelphia (SDP), Chester/Upland, or Mobile County, Alabama.  All course work is online and begins this summer.  Candidates will work as classroom teachers while they complete course work towards certification.  Salary bonuses and other financial incentives are available for SDP teachers.

A bachelor’s degree with a minimum GPA of 3.0 from an accredited college or university (with major course work in math or science) and successful completion of the Praxis I (PPST Reading, Writing, Math) exam are required for admission.  Official college transcripts and regular access to the Internet are required.  A three-year commitment to teaching within designated school district is obligatory.

For more information and other program requirements, please contact Tom Yates, Drexel University:  tty22@drexel.edu

I am also in Drexel University!! They are the ones who host the mathforum.

Well, is there anybody who would like to do what Tom Yates proposed?


If you deposit 150 per month on an account that has an annual compound interest of 3percent  how much you will have after 10 years? When will kyou exceed $100,000? How much do you need to put per month if you want to have 100,000 in ten years?

Doesn't it depend on how often it is compounded, and is the interest calculated after the money is deposited or before?

is it compounded monthly, quaterly, or annually??? whatever the compounding period, the best way to do this is by drawing a time line. we did this in math class last year...it's a piece of cake that way...u don't need any formulas or any special equations.

e.g. if u have $X at the beginning and the interest rate is x% per annum and is compounded monthly, first divide x by 12. after the first month, you'll have (1+x/12)X, after second month, you'll have (1+x/12)^2*X, etc. ends up being a geometric sequence...sum it and u'll get the desired number.

cheers!


Let $\triangle ABC$ and the points $M$, $N$, $P$ which belong to the lines $BC$, $CA$, $AB$ respectively. Prove that:

$\blacksquare\ 1^{\circ}.\ \boxed {AM^2+BN^2+CP^2\ge MN^2+NP^2+PM^2+2\left(MB\cdot MC+NC\cdot NA+PA\cdot PB\right)}.$

$\blacksquare\ 2^{\circ}.$ The triangles $ABC$ and $MNP$ have same centroid if and only if exists the following relation: \[ \boxed {\ AM^2+BN^2+CP^2=MN^2+NP^2+PM^2+2\left( MB\cdot MC+NC\cdot NA+PA\cdot PB\right)\ }\ .  \] [hide="Here are some remarks."] See Lemma 3 from the my last reply
http://www.mathlinks.ro/Forum/viewtopic.php?t=67453
and the my proposed inequality from
http://www.mathlinks.ro/Forum/viewtopic.php?t=68437 [/hide]

Let G be the centroid of triangle ABC.From Leibniz we have $\displaystyle MG^2=\frac{MA^2+MB^2+MC^2}{3}-\frac{AB^2+BC^2+CA^2}{9}$,and the analogs for N and P we summ them up and we get $\displaystyle \sum MG^2=\frac{\sum MA^2-2\sum MA\cdot MB}{3}\geq {\frac\sum MN^2}{3}$ so we got the desired ineq.

[quote="mateivld"]Let G be the centroid of triangle ABC.From Leibniz we have $\displaystyle MG^2=\frac{MA^2+MB^2+MC^2}{3}-\frac{AB^2+BC^2+CA^2}{9}$,and the analogs for N and P we summ them up and we get $\displaystyle \sum MG^2=\frac{\sum MA^2-2\sum MA\cdot MB}{3}\geq \frac{\sum MN^2}{3}$,from Leibniz ineq, so we got the desired ineq.


I found this image online;
Read the Poll
[img]http://img002.picture2life.net/5663304/kj_web-large_medium.png[/img]

The image does not move , as u said is an optical effect . :wink:

See the figure closely and then you'll realize that it does not move , It seems with movement when u see it with a certain distance to the scream

I think it's because of your tiny eye movements or something. :huh:

whoa.... that's cool... and kinda creepy

lol... i know but i chose it because i thought that wuz super cool!!!

http://www.michaelbach.de/ot/

There were so many cool ones.

Check it out!!

YAY! Illusions.... I grew up with them but my eyes made the dang images look like magic.  :rotfl: Yeah, I was a strange child.

hmm, should we post more optical illusions?
 :D  :D

COUNT THE NUMBER OF BLACK DOTS!!!!!!! :wink:

i'm up for the task.

Are these rows straight? Or not?

[img]http://ccl.northwestern.edu/netlogo/models/models/Sample%20Models/Art/Optical%20Illusions.png[/img]

haha, nice one! :wink:

Look at this one!!

SAY THE COLOR NAME OF THE WORD OUT LOUD!!!! (NOT WHAT THE WORD SAYS) GL!

Wow, you're good at those kind of stuff.

whoa...how'd you get nine? i can only see 4, including the baby

That's what I once said.
I can faintly see one in the lower left hand corner in the sky...

I don't think lots of people can see what I can see...

[hide="I see 8 people"]
the obvious huge one
the woman in the big man's hair
her baby
the man where his face is the big man's eye
way to the right, at the bottom of the sky, there is a sideways face facing right
same thing for the left side of the arch, between the bird and the arch, facing left
to the right of the black bird is a faint other bird, which is the eye of a person
to the left of the black bird is a faint face facing forward

nine if the thing at the bottom left counts[/hide]

On the right track :).
Anyways,
here's one.

[img]http://www.guy-sports.com/fun_pictures/jf_dots1a.jpg[/img]
Can you control the spinning by relaxing your eyes?

If you look straight at them and don't allow your eyes to twitch or move at all, they stop moving.  :D

[code][img]url[/img]
[/code]
For example:
[img]http://whywedoit.files.wordpress.com/2009/04/smile.jpg[/img]

Also, you can use the File, Browse thing at the bottom as well.

Like this:

oh!!
I will post one!
Alright, stare at the center!!What happens to the circle?
[img]http://www.freebiehome.co.uk/images/stare-optical-illusion.gif[/img]


In how many ways can 6 faces of a cube be painted with 6 different colours????

Please explain the answer also..

Are alowed to paint two different 'faces' with the same colour ?
If yes :there are $ 6^{6}$ ways of painting them
If no : there are $ 6! \equal{} 6\cdot 5\cdot 4\cdot 3 \cdot 2\cdot 1$ ways of painting .

The solution is more complex than that if two "cubes" are said to be equivalent if one can be achieved by rotating the other in three-dimensional space. Then, for instance, there are only $ 5\cdot 3\cdot 2\equal{}30$ possibilities for the "non-repetition" case, and other cases can be found using Polya enumeration (or Burnside's lemma).

It would help hugely if you would write the problem exactly as it is given. They gave the correct answers for the problem you posted, but apparently we were wrong and the answer is 12. I can't see how you could possibly get 12, but maybe I'm mistaken.

I got the answer 15, the number of ways to group 6 colors into groups of 2.


I need some help solving problems that say things like:

If in a game you can score 3 points, 4 points, or 7 points, what is the highest score that can't be obtained?

I don't need to solve the problem, but I need some help with the reasoning. I don't want to do guess and check. I think I have seen a formula for these types of problems, but I can't remember.

I'm not sure if there's a formula, but I think you can generate functions.

Thanks. I did a search for it and found [url=http://www.artofproblemsolving.com/Wiki/index.php/Chicken_McNugget_Theorem]this[/url] form good old AoPSWiki, and it has a proof. Just wondering, what about three numbers?

eh, for two numbers it's
$ \ x*y\minus{}(x\plus{}y)$
well i think :|

That is the expanded version of athunder's.

Assume that $ n$ is the largest number that cannot be achieved only using 3, 4, or 7.

Then, $ n$ must be smaller or equal to $ 3\cdot4 \minus{} 3 \minus{} 4 \equal{} 5$ since anything that can be achieved using 3 and 4 can be achieved using 3, 4, and 7.

Repeating this process, we have $ n\le 5$, $ n\le 11$, and $ n\le 17$.

Hence, we conclude that 5 is the largest number that cannot be achieved using 3, 4, and 7.

[quote="athunder"]Well...its a theorem if you want a proof look it up.

I do not think there is one for any over two.[/quote]
I tried to figure out one for three numbers (and I epic failed).
Is there one for 4 numbers?

There is no known formula for 3 or more - search Frobenius Coin Problem on Wikipedia.

Its also called the dartboard formula or dartboard trick.

To make this not all pointless, i'll make a problem and solve it. In a dart, you can either attain 5 or 3 points. Whats the largest number that cannot be attained. The answer would be $ (5*3)\minus{}5\minus{}3$ or $ 7$.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=43173]this[/url] for a bash of mine.

Basically, you attempt to create every number from 0 to n-1 (mod n) where n is the smallest of the three numbers. Then take the largest number you've created and subtract n from it.

In this case:

0 (mod 3) is 3
1 (mod 3) is 4
2 (mod 3) is 8

The largest of those three numbers is 8. Subtracting 3 from it, our answer is 5.

It's not a formula, more of an algorithm. If you try to create a proof, I bet it's pretty similar to the proof for two variables in Volume 2. Creating the n numbers should be pretty intuitive. Do you see why we subtract n at the end?

EDIT: Yeah, uh, (n-1)+1 is not n-1. I can see why I didn't win Mathcounts.

[quote="xpmath"]
Basically, you attempt to create every number from 0 to n-1 (mod n) where n is the smallest of the three numbers. Then take the largest number you've created and subtract n from it.

In this case:

0 (mod 3) is 3
1 (mod 3) is 4
2 (mod 3) is 8

The largest of those three numbers is 8. Subtracting 3 from it, our answer is 5.
[/quote]

Hold on. For the 2 mod 3 part, how did you get 8 specifically? You could have gotten 5, or 8, or 11 etc. Same with the 0 mod 3. you could have gotten, say, 12, and since that is the largest number you created, subtracted 3 to get 9 (which is wrong)

And would fantasylovers way work for more than 3 numbers?

Sorry, I forgot to mention; you can only use the numbers mentioned in the problem. So, you can only use 3, 4, and 7 to construct the numbers (mod 3). You're trying to make the smallest number possible for each residue. And it'll be pretty much a waste of time if you use anymore 3s after the 0 (mod 3) as it doesn't change the residue.

Fantasy's method works for this since there are small numbers. Try 4, 9, 11 from my link. You'd have to check numbers less than or equal to 23.

I don't think I'm doing a good job of explaining at 10:30, so ask any questions you have.

[quote="xpmath"]For example, consider 4, 9, and 11. We have to find the smallest sum of 4s, 9s, and 11s so we end up with numbers congruent to $ 0, 1, 2, 3 \pmod{4}$.

Smallest 0 (mod 4): 4
Smallest 1 (mod 4): 9
Smallest 2 (mod 4):18
Smallest 3 (mod 4): 11

Our answer is 18-4=14. [/quote]

Ok, Im extremely dumb and have no idea what residue is. Can you try to explain to me how you got the numbers 4,9,18,11, instead of, say, 4,9,14,11 or 4,9,10,7

Err...I'm not sure.

Here, I'll try explaining again.

So, in this case, we have the find the largest number that cannot be created by adding any number of 3s, 4s, or 7s.

Here's an algorithm, that (I think) will work:

[hide="algorithm, on this example"]First, take the smallest of the three numbers you're allowed to use, which is 3.

Next, list the possible remainders a number can have when divided by 3, which are 0, 1, and 2.
Now, find the smallest number that can be created by adding any number of 3s, 4s, and 7s, for each remainder:

The smallest number that we can make with 3s, 4s, and 7s, which leaves a remainder of 0 when divided by 3, is just 3 itself.

The smallest number that we can make with 3s, 4s, and 7s, which leaves a remainder of 1 when divided by 3, is 4.

The smallest number that we can make with 3s, 4s, and 7s, which leaves a remainder of 2 when divided by 3, is 4+4=8.

Now take the largest of the numbers you made; in this case, it's 8. From this, subtract 3 (the smallest of the three numbers you could use), to get your answer, 5.[/hide]

[hide="algorithm with 4, 9, 11"] The smallest of the three numbers is 4.

We now have to make the smallest numbers, $ a, b, c, d$, that leave remainders of 0, 1, 2, and 3 when divided by 4, respectively. Note that we can [b]only[/b] use the numbers 4, 9, and 11 to make these numbers.

We can see that the smallest number that leaves a remainder of 0 when divided by 4 that we can make is 4 itself.

The smallest that leaves a remainder of 1 when divided by 4 is 9.

The smallest that leaves a remainder of 2 when divided by 4 is 9+9=18.

The smallest that leaves a remainder of 3 when divided by 4 is 11.

The greatest of those four numbers that we just got is 18. From here, we subtract 4 to get the answer, 14.[/hide]

Eh you edited. Basically, we can only use the numbers in the problem. You can't make 14 using 4s, 9s, or 11s.

Also a residue (mod m) is just any remainder that's possible when dividing a number by m. [url=http://www.artofproblemsolving.com/Wiki/index.php/Residue_class]Residue class[/url], as said on the Wiki.

Sry one more question. Wouldn't 0 mod n be sorta trivial since it  would obviously be n, and n-n is 0? That would only be the right answer If 1 were one of the numbers,right?

Yeah because if 1 is one of the numbers, you can pretty much make every natural number...

Also, yes, for 0 (mod n), you obviously use n  :P


For a>b>0 and n positive integer > 1 show that for k > = 0
$ (a^n \plus{} k^n)^{1/n} \minus{} (b^n \plus{} k^n)^{1/n} < \equal{} a \minus{} b$
 :P

this has LMV written all over it
define $ f(x)\equal{}(x^{n}\plus{}y)^{\frac{1}{n}}$ where $ x,y \in \mathbb{R}^{\plus{}}\plus{}0$ 0 only for y  :wink:
now $ f'(x)\equal{}\frac{x^{n\minus{}1}}{(x^{n}\plus{}y)^{\frac{n\minus{}1}{n}}} \leq1$
so
$ \frac{f(a)\minus{}f(b)}{a\minus{}b} \equal{}f'(c) \leq 1$
equality only holding when $ k\equal{}0$
result follows

[quote="pardesi"]this has LMV written all over it
define $ f(x) \equal{} (x^{n} \plus{} y)^{\frac {1}{n}}$ where $ x,y \in \mathbb{R}^{ \plus{} } \plus{} 0$ 0 only for y  :wink:
now $ f'(x) \equal{} \frac {x^{n \minus{} 1}}{(x^{n} \plus{} y)^{\frac {n \minus{} 1}{n}}} \leq1$
so
result follows[/quote]

How did ya get f'(x) as that value isn't it 1/$ n^2$ times that value?  :maybe:

No sorry i figured it out  :blush:

More solns in a different approach would be welcome  :)  :wink:
What the hell is the 10 mins rule for???  :mad:

Actually this sum is a sum from an IIT book for inequalities you know so try using inequalities  :P

@ Shreyas : Please see the private msg.


Say we have a number like: 541685414. Can it be written as a sum of three perfect cubes? Can someone show me the steps required to tackle a problem like this, because I'm not really interested in the final answer.

Thanks.

You can try modulo 7 and 9 in general case. But for your particular example it doesn't work :(

Cubes gives remainders -1, 0, +1 modulo 7 or 9. Threfore, sum of two cubes can't be equal to 7k+3, or sum of three cubes can't be equal to 9k+4 and 9k+5.


Just wondering for people who have used the book: First Steps for Math Olympians:

Do you find it useful in preparing for the AMC 10?
If I did AoPS V1 and did all of the problems, would I need it or would it be overkill?

Also, will doing the AMC 10 online class boost my score from a current 110-111 average to 120+?

Thanks guys

Definitely FSMO will help, however i recommend AoPs v1 b/c it boosted my score alot. the class is good to.

You can also find a lot on the AMC website:
http://www.unl.edu/amc.


Find all ordered pairs $ a,\ b,\ c$ of positive integers such that $ \left\{
\begin{array}{ll}
ab\plus{}c\equal{}13 &\quad  \\
 a\plus{}bc\equal{}23 &\quad
\end{array}
\right.$

[hide=" this is too easy"]
$ (a\plus{}bc)\minus{}(ab\plus{}c)\equal{}10$
$ a\minus{}ab\minus{}c\plus{}bc\equal{}10$
$ a(1\minus{}b)\minus{}c(1\minus{}b)\equal{}10$
$ (c\minus{}a)(b\minus{}1)\equal{}10 \implies b\minus{}1\in\{1,2,5,10\}$

checking all 4 cases we get 2 solutions: $ (1,2,11)$ and $ (1,11,2)$[/hide]

Not perfect in a way not entirely correct or in a way this could have been solved in a nicer way?

You have almost solved the problem, but you forgot one more pair.

The only thing that could work is $ (23, 0, 13)$ but $ 0$ is not a positive integer.

[hide="The third"]You neglected to record $ (2,3,7)$, which results from $ b\minus{}1 \equal{} 2 \implies b\equal{}3$[/hide]

Yeah, you are right. I had a typo in a notebook and was too lazy to check XD


A five-legged Martian has a drawer full of socks, each of which is red,
white or blue, and there are at least five socks of each color. The Martian
pulls out one sock at a time without looking. How many socks must the
Martian remove from the drawer to be certain there will be 5 socks of the
same color?

The worst case scenario is if we get $ 4$ of each of the $ 3$ colors.

This gives $ 12$ socks. The next sock must be a matching sock

to one of the colors, giving $ \boxed{13}$ as the answer.


If $ y \minus{} x \equal{} 7$, what is the value of $ x \minus{} y$ ?

There is a very useful identity $ (a\minus{}b)\equal{}\minus{}(b\minus{}a)$.

This can be proven by multiplying $ (a\minus{}b)$ by $ \minus{}1$ to get $ (\minus{}a\plus{}b)$, and then rearranging terms to get $ (b\minus{}a)$.

So, we substitute $ y$ for $ a$ and $ x$ for $ b$; if $ y\minus{}x\equal{}7$, then $ x\minus{}y\equal{}\boxed{\minus{}7}$.

Can you edit it so that the solution is hidden?
[hide="Psst!"]
To hide your message, write it in this format: [quote] [hi de="Title of the message"] "Message" [/hi de][/quote]
ignoring the spaces in between hi and de. [/hide]

[code][hide="Why??????????"]Why?[/hide][/code]
^That's the code for it

[hide="Why??????????"]Why?[/hide]
I did [hide = " Why??????????"]Why?[/hide] or something


find all possible value of $ \LARGE F_{i}(t) (0 \leq t \leq 1,1 \leq i \leq n)$
define:$ \LARGE F \equal{} max_{(0 \leq x_{i} \leq 1)} \left| F_{1}(x_{1}) \plus{} F_{2}(x_{2}) \plus{} ...... \plus{} F_{n}(x_{n}) \minus{} x_{1}x_{2}......x_{n} \right|$
determine $ F_{min}$

not typical problem.Our Question for thinking
hard to me :|

[b] anybody have thoughts on this problem?[/b] :maybe:


Let $ A\equal{}span\{(1,2,3,4), (4,3,2,1), (2,3,4,5), (1,1,1,1)\}\subset R^4$ and $ B\equal{}span\{(\minus{}1,\minus{}2, 1, 0), (1,\minus{}3,0,1)\}$
Do there exist a subspace $ C\subset B$ such that $ A\oplus C\equal{}R^4$?

$ A\oplus C\equal{}R^4$ means that: $ R^4\equal{}A\plus{}C$ and $ A\cap C\equal{}{0}$

Such a subspace $ C$ exists if and only if $ A\plus{}B\equal{}\mathbb R^4$. The necessity should be clear. For sufficiency, suppose that $ A\plus{}B\equal{}\mathbb R^4$. Choose a basis of $ A$ and a basis of $ B$. Add the elements of the basis of $ B$ to the basis of $ A$ one by one, while making sure that the union is linearly independent. When you can't do this any more, you have a basis of $ \mathbb R^4$. The added basis elements from $ B$ will be the basis of $ C$.

Because of the above, the question can be answered by forming a 6x4 matrix of all given vectors and computing its rank.

I forgot the second point of this problem. Here it is:
Find or prove that it's impossible to find such subspaces of $ R^4$- $ U, V$ ,that:
\[ R^4\equal{}A\oplus U\equal{}U\oplus V\equal{}V\oplus A\]
(A is the same as above)

That is possible if and only if $ \mathrm{dim}(A)\equal{}2$. Informally, $ U$ and $ V$ can be two "random" two-dimensional subspaces. Formally, you choose $ U$ to be some complement of $ A$ and then think of what $ V$ should be.


Which of these three numbers is between the other two on the number line? Answer with the letter only.

A) $ 1/\sqrt[3]5$

B) $ 2\minus{}\sqrt2$

C) $ 8\minus{}\sqrt{55}$

[quote="GameBot"]Which of these three numbers is between the other two on the number line? Answer with the letter only.

A) $ 1/\sqrt [3]5$

B) $ 2 - \sqrt2$

C) $ 8 - \sqrt {55}$[/quote]

Lets start by comparing B and C.

$ 2-\sqrt2 \? 8-\sqrt{55}$
$ \sqrt{55}-\sqrt{2} \? 6$
$ 57-2\sqrt{110} \? 36$
$ 21 \? 2\sqrt{110}$
$ \frac{441}{4} \? 110$
So B is larger then C.

$ \frac{1}{\sqrt{3}} \? 2-\sqrt{2}$
$ 1 \? 2\sqrt{3}-\sqrt{6}$
$ 7+2\sqrt{6} \? 12$
$ 2 \sqrt{6} \? 5$
$ 24 \? 25$

So B is larger then A.

$ \frac{1}{\sqrt{3}} \? 8-\sqrt{55}$
$ 1 \? 8\sqrt{3}-\sqrt{165}$
$ 166+2\sqrt{165} \? 192$
$ 2\sqrt{165} \? 26$
$ 660 \? 676$
So C is larger then A

So B>C>A, and our answer is C.

You can just put the numbers into your calculator and then compare them if you wanted the easy way to the answer or if you were lazy.... :wink:


[hide="Answer"]Valves $A$ and $C$ will fill $\frac{2}{3}$ of the tank in 1 hour, and valves $B$ and $C$ will fill $\frac{1}{2}$ of the tank in one hour. $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=1$ and $\frac{1}{A}+\frac{1}{C}=\frac{2}{3}$, so $1-\frac{2}{3}=\frac{1}{3}=\frac{1}{B}$. $\frac{1}{B}+\frac{1}{C}=\frac{1}{2}$, so $\frac{1}{C}=\frac{1}{6} \Rightarrow \frac{1}{A}+\frac{1}{6}=\frac{2}{3}$, so $\frac{1}{A}=\frac{1}{2}$, and $\frac{1}{A}+\frac{1}{B}=\frac{5}{6}$ of the tank filled in one hour, so the entire tank is filled in $72$ minutes.[/hide]

[b]EDIT:[/b] Modified solution to make a little more sense.

[quote="4everwise"]Each valve A, B, and C, when open, releases water into a pool at its own constant rate. With all three valves open, the tank fills in 1 hour, with only valves A and C open it takes 1.5 hours, and with only valves B and C open it takes 2 hours. How long will it take to fill the tank with only valves A and B open?[/quote]

[hide]
To solve this problem, it is essential to know that the time needed to finish filling pool divided by all the members equals 1:
I do realize that this solution is worded well, but you will understand by looking at the solution  :o
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$
$\frac{3}{2a}+\frac{3}{2b}=1$
$\frac{2}{b}+\frac{2}{c}=1$

Solve for a and b in terms of c, and then substitute c back in to equation 2 and 3

Ans. $\frac{3}{2}$ hours  :D [/hide]


search
real analysis
real analysis unsolved

Here is an interesting summation that I came up with.  I am hoping to check to make sure that I evaluated it correctly.  How would you evaluate it?  What did you come up with?

$\sum_{n=2}^{\infty}\frac{(-1)^{n+1}4^{n}(6n^{2}+9n+3){\pi}^{2n+1}}{(2n)!}$

[hide="Here is my answer."]$-30{\pi}^{3}$[/hide]

This is a Ramanujan-like series, similar sums have been obtained by Guillera and others by using integers relation method, you might want to search the Ramanujan journal for papers on such topics.

$6n^{2}+9n+3 = \frac{3}{2}(2n)(2n-1)+6\cdot 2n+3$, so we can re-write the sum as
\[\sum_{n = 2}^{\inf}-\frac{3\pi}{2}\left(\frac{(2\pi i)^{2n}}{(2n-2)!}\right)-6\pi \frac{(2\pi i)^{2n}}{(2n-1)!}-3\pi \frac{(2\pi i)^{2n}}{(2n)!}\]
and then you can calculate it from the power series for $\cosh$ and $\sinh$.


Does anyone here play on chess live?  If you do, please list your handle here. Mine is johnnyfox :) if you need with chess live contact me there im an unofficial SR(too young to be a real one)but i know the site well

smallpawn but hopefully will change it to "whatev" someday

Nah, you'll be too lazy lazy lazy.  Plus, you should change it to ORZNORZ!!!

By the way, ORZNORZ is the name of our iTest team.  Do you or Jeremy want to be on it?  PM me.

chess live is a cool place, its a newer server though, so it has less people than ICC and other places like that :(

Anyone know how we can get a computer account on USCL?

hope this helps... some info on computer applying and computer usage :) hope to play you there!

computer_app
This topic is generated from a server help file and is directed at the command line interface that underlies Chess Live. While it requires more knowledge to use, this text interface allows great flexibility in using the server. See the help on Command Line Interface for more details.

If you want to have a computer account, please send a message to Computers with ALL of the following info (please use this as a form):

Full name: Handle: Requested Handle: (must be available) Engine name: Programmer: Hardware: Software: Variant suported: Manual or Automatic operation: Login times: Estimated rating: Main time control and variant: Reason for having a computer account:

Be sure to read the computers help file for the guidelines of a computer account. Be advised that the decision taken by the computer commitee is final and without appeal.

Here is the basis upon which computer accounts are granted:

- Priority is given to those who can play variants. - Priority is given to automatic computers (not need of an operator.) - Priority is given to those who have had an account here for some time. - Priority is given to those which are built from home grown code and operated by the programmer.

Of course, the main criteria in deciding on which computer accounts are allowed is the current demand.

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

General rules regarding computer accounts

Computers may register and play rated chess matches. However, computers playing rated games are required to be registered so that users can know whether they are playing against humans or computers. Using an unregistered computer to play rated chess matches on this server is considered to be abusive behavior and will result in sanctions; see etiquette.

To request a computer account, visit the Computers section of the Chess Live web-site. At this time, computer accounts will only be granted for computers of U2000 strength! You will find an application form and other information there. It takes some technical skill to get a Computer account running, as you must use WinBoard rather than the Chess Live client to control your chess engine, and you must spend some effort configuring it. There is a How to Get Crafty working on USCL page which can help you out with this. Note that Chess Live staff cannot for the most part help you configure WinBoard and the engine in question.

Special Notes:

Server admins reserve the right to limit the number of computer accounts registered on the server and may seek to ensure a wide range of playing strength of these accounts.
Computer accounts that want to be open for challenges for suicide chess games need to do the following: have the 'f1' formula variable set to "suicide" (with no extra spaces of any kind); you need not use 'f1' as part of the 'formula' setting. Computer accounts without 'f1' set to "suicide" will have challenges for suicide chess declined automatically. In the same fashion as for suicide, set 'f2' to "crazyhouse" if you wish for your computer account to be open for crazyhouse challenges.
Computer accounts may only play in tournaments that are explicitly announced to be open to computers when the tournament is opened. Joining a tourney with a computer account when it was not authorized may result in an immediate ban from tourneys.
If computer account is not used for more than 3 months it may be removed WITHOUT any warning. We want the computers on the server to be active.
Computers should always keep information about hardware/software in finger notes. This must include (at a minimum): the owner of the computer account, the type of machine it is being run on, and the engine that is being used by the account. Providing this information is mandatory, and not having it in your computer's finger notes may result in a loss of the account.
Rated games between a computer and account and its operator are NOT allowed. Play your computer unrated only if you want to test it.
Abuse of any kind by either the owner or the computer account itself may result in the removal of the account in addition to other sanctions.
How to start a game against a computer
Computer accounts on Chess Live consist of a variety of chess programs. There are computers for players of all skill levels. Some computers are intentionally configured to play below their potential strength in order to assist beginning and intermediate players. The majority of computers, however, are very strong.

How to see what computer accounts are currently on-line
If you decided to play a game or few against a computer and want to see which computers are currently available for a game, To do this, check for any players in the Players Tab of the Info Window for any players whose name is followed by a (C) badge.

How to find computer's strength and other qualifications
After you found a computer you might like to play, you would want to check its qualifications to make sure that it suits your playing interests. To do that, right click over their name in the players window and select Profile,
After you did that, on your screen you will see this computer's Profile which will contain it's rating table and any notes that the operator has set for that particular account.

How to start a game against a computer
There are two ways:

Check in the Seek Tab of the Info Window for any square dots, these represent computer seeks. If you hold the mouse over any of these, you will see the name of the computer, its rating, and the type of game being offered. Double clicking on the seek will begin the game.
Challenge the computer directly. However, before challenging the selected computer to a game, make sure it plays the variant that you want. Some computers are created specifically to play wild, suicide, crazyhouse, etc. variants. When you are ready to play, the easiest way to issue a challenge is to go to "Actions" menu, and selecting "Challenge Opponent" option. Follow the same procedure as you would against a regular opponent. Remember that computer accounts have the same privileges as regular user accounts (expect for some technical commands).

If at any time during the game against a computer you find that something goes wrong, please message its operator! The name of the operator can be found in computer's finger notes. It is very important that you let the operator know about anything that goes wrong with the computer. Most of the time, automatic computer accounts are unattended and therefore the operator has no idea if something goes wrong. If the computer seriously misbehaves, contact an on-line administrator.
-----------------------------------------------------------------------------------

I'm going to assume that was copy+pasted.....

I play in FICS, USCL was too empty and I could never find human ppl to play

Yes, please CITE everything you quote off another source. It's the legal and polite thing to do.

Otherwise, I was kinda wondering if anyone has one. I play on USCL sometimes and I would kinda like to have a computer account as well.

i haven't been on in such a long time but i hear my account champchess still has its TM abilities, maybe something to do a little this summer.

How do you get to it? Do you have to download anything to use it? Is it welcome to people under the age of 18? Please give me your answ3ers!

My username is mathprodigy. (Not exactly true, but.... :D)

If you challenge me and tell me you're from AoPS, I always accept.

I'm knightmair. I am an ex-SR and ex-TM (both removed due to inactivity) but I have a good 300 tournaments run and 30 online days logged.

I unfortunately don't play on US chess live.

Does anyone play on FICS? If so, add me! My user name is JavaManIssa

Just downloaded Chess Live. Handle=fireemblem

does anybody know a free bughouse site (preferably no downloads) where 2 people can practice?

Yeah, that will be awesome. Also, do anybody know a download for four player chess? Heard it's really fun.

[quote="JavaMan"]I play on FICS, I added you to my GNotify.. it says you've never connected or played any rated games  :ninja:[/quote]

Lol, that's because I just made my FICS account right before I posted. :lol: I'll try to get on sometime soon. What's your handle?

mine is brick. i couldn't thnk of anything else.  i'm not on much though.  i might be on more in the future, when i get a faster computer.

My handle's JavaManIssa

Jeesh Brick, you last disconnected 1998  :lol:


A rectangular box is 4 cm thick, and its square bases measure 16 cm by 16 cm. What is the distance, in centimeters, from the center point $ P$ of one square base to corner $ Q$ of the opposite base? Express your answer in simplest terms.

[asy]import three;
draw((0,0,1/4)--(1,0,1/4)--(1,1,1/4)--(0,1,1/4)--(0,0,1/4)--cycle3,linewidth(2));

draw((0,1,0)--(1,1,0),linewidth(2));
draw((1,1,0)--(1,0,0),linewidth(2));
draw((0,1,0)--(0,1,1/4),linewidth(2));
draw((1,1,0)--(1,1,1/4),linewidth(2));
draw((1,0,0)--(1,0,1/4),linewidth(2));

dot((1/2,1/2,1/4));
dot((0,1,0));

label("P",(1/2,1/2,1/4),W);
label("Q",(0,1,0),E);[/asy]

[quote="thdanh90"][hide]12 if I am correct.[/hide][/quote]

you know, people would appreciate it if you included HOW you got your answer


trigonometry
function
calculus
calculus computations

Investigate the stability of the constant solutions $ (u_{n\plus{}1} \equal{} u_n)$ of the discrete equation
\[ u_{n\plus{}1} \equal{} 4u_n (1 \minus{} u_n)\]In the case $ 0 \le u_0 \le 1$, use the substitution $ u_0 \equal{} \sin^2 \theta$ to find the general solution and verify your stability results. Can you find an explicit form of the general solution in the case $ u_0 > 1$?

I want to type for you but I don't know how...(I'm a computer illiteracy, sorry.) It's about the logistic map. See the book [Mathematical Methods for Physicists] by Arfken.

 Thinking graph of the function $ y \equal{} 4x(1 \minus{} x)$and $ y\equal{}x$ will help.

The interation $ u_{n+1}=f(u_n)$ can only possibly converge to a fixed point of the function, namely an $ x$ such that $ x=f(x).$

Such a fixed point $ x$ is stable (all sequences starting sufficiently close to $ x$ converge to $ x$) if an only if $ |f'(x)|<1.$ If $ |f'(x)|>1,$ then the fixed point $ x$ is unstable, and sequences that start very close to (but not exactly at) $ x$ do not remain very close to $ x.$

(Point for which $ |f'(x)|=1$ require more delicate analysis; I will not attempt such analysis below.)

The family of logistic mappings $ f(x)=rx(1-x)$ maps $ [0,1]$ into $ [0,1]$ as long as $ 0\le r\le 4.$

Solving $ x=rx(1-x)$ gives us two roots: $ x=0$ and $ x=1-\frac1r.$

$ f'(x)=r$ for $ x=0$ and $ f'(x)=2-r$ for $ x=1-\frac1r.$

For $ 0\le x<1,$ the point $ x=0$ is a stable fixed point and the point $ x=1-\frac1r$ does not lie in $ [0,1]$ and cannot be accessed by a sequence starting inside $ [0,1]$. Hence, $ x=0$ is the unique stable solution, and all sequences starting in $ [0,1]$ tend to zero.

For $ 1<x<3,$ the fixed point $ x=0$ is unstable but the fixed point $ x=1-\frac1r$ is stable. All sequences starting in $ [0,1]$ tend to $ 1-\frac1r.$

But for $ 3<x\le 4$ (which includes the $ r=4$ case that aidan is asking about), both the $ x=0$ and $ x=1-\frac1r$ fixed points are unstable. Unless the sequence starts precisely at one of those fixed points, it will not converge.

The output of the following program (written in pseudo-code) is a famous picture.

Set up a $ [0,4]\times[0,1]$ window for plotting. (Or perhaps $ [3,4]\times[0,1].$)

Set up a loop, incrementing $ r$ by many very small steps from $ 0$ to $ 4$ (or perhaps from $ 3$ to $ 4$).

Let $ x: =\frac12.$

Loop 250 times:

Let $ x: =rx(1-x).$

For the first 50 iterations, plot nothing; for the next 200 iterations, plot a dot at $ (r,x).$

End of loop.

End of loop on $ r.$

Ah this is interesting - I've been reading a wikipedia article about the logistic map too. Thanks!


16777215 posts!?!?  100% of total?

Who is that person...

Use search instead of using more question marks than necessary.

Well, when I click his profile and see "view all posts", he only has about 25...

http://www.mathlinks.ro/viewtopic.php?t=225601

It's a forum bug probably.

Thanks bugi.  Probably some hacker dude.

If you now look on Socc's posts- you'll see that he's not in the first place and he has 25 posts.


group theory
abstract algebra
superior algebra
superior algebra unsolved

suppose $ \ G$ is a group with the oder $ \ p^3$ ($ \ p$ is prime).show:$ \ G^{\prime} \ \equal{}Z(G)$.

The [url=http://en.wikipedia.org/wiki/Commutator_subgroup]commutator subgroup[/url], I presume.
The statement is clearly false if the group is abelian.

ok.this propety is true for the groups that are not abelian.

If G is not abelian, Z(G) is not the whole group and G' is not trivial, so we have $ |Z(G)|<\equal{}p^2$ and $ p<\equal{}|G'|$.

There is a normal subgroup N of order p. Hence G/N is of order $ p^2$ and therefore abelian. Because of that $ G'\subseteq N\Rightarrow G'\equal{}N$.

Z(G) is not trivial since G is a p-group. $ |Z(G)|$ is not $ p^2$ because if it would, $ G/Z(G)$ would be cyclic and hence G would be abelian contrary to the assumption. So we get $ |Z(G)|\equal{}p$ and as we have already seen it follows that $ G'\equal{}Z(G)$.


By Jensen's inequality applied to the convex function $ f(t)\equal{}\frac {1}{\sqrt t}$,we get
$ \frac{x}{\sqrt {x\plus{}y}}\plus{}\frac{y}{\sqrt {y\plus{}z}}\plus{}\frac{z}{\sqrt {z\plus{}x}} \ge (x\plus{}y\plus{}z) \sqrt{ \frac{x\plus{}y\plus{}z}{x\plus{}y\plus{}y\plus{}z\plus{}z\plus{}x}}$
It 's right this?

[quote="soruz"]By Jensen's inequality applied to the convex function $ f(t) \equal{} \frac {1}{\sqrt t}$,we get
$ \frac {x}{\sqrt {x \plus{} y}} \plus{} \frac {y}{\sqrt {y \plus{} z}} \plus{} \frac {z}{\sqrt {z \plus{} x}} \ge (x \plus{} y \plus{} z) \sqrt {\frac {x \plus{} y \plus{} z}{x \plus{} y \plus{} y \plus{} z \plus{} z \plus{} x}}$
It 's right this?[/quote]
I think by Jensen inequality, we have:
$ \frac {x}{\sqrt {x \plus{} y}} \plus{} \frac {y}{\sqrt {y \plus{} z}} \plus{} \frac {z}{\sqrt {z \plus{} x}} \ge (x \plus{} y \plus{} z) \sqrt {\frac {x\plus{}y\plus{}z}{x(x\plus{}y)\plus{} y(y\plus{}z) \plus{} z(z\plus{}x)}}$

[quote="soruz"]By Jensen's inequality applied to the convex function $ f(t) \equal{} \frac {1}{\sqrt t}$,we get
$ \frac {x}{\sqrt {x \plus{} y}} \plus{} \frac {y}{\sqrt {y \plus{} z}} \plus{} \frac {z}{\sqrt {z \plus{} x}} \ge (x \plus{} y \plus{} z) \sqrt {\frac {x \plus{} y \plus{} z}{x \plus{} y \plus{} y \plus{} z \plus{} z \plus{} x}}$
It 's right this?[/quote]
No, it can be:
$ \sum_{cyc}\frac {x}{\sqrt {x \plus{} y}} \equal{} \frac {x}{x \plus{} y \plus{} z}\cdot\sum_{cyc}\frac {1}{\sqrt {\frac {x \plus{} y}{(x \plus{} y \plus{} z)^2}}}\geq\frac {1}{\sqrt {\frac {x(x \plus{} y) \plus{} y(y \plus{} z) \plus{} z(z \plus{} x)}{(x \plus{} y \plus{} z)^3}}} \equal{}$
$ \equal{} \sqrt {\frac {(x \plus{} y \plus{} z)^3}{x^2 \plus{} y^2 \plus{} z^2 \plus{} xy \plus{} xz \plus{} yz}}.$ :wink:


When Jorge took his place in the marching band, he was one person in a rectangular array of musicians. He noticed that he was 10th from the front, 7th from the back, 3rd from the left, and 8th from the right. How many musicians were in the band?

From the given information we can conclude there are $ 16$ rows and $ 10$ columns.
$ 16 \times 10 \equal{} \boxed{160}$


calculus
integration
trigonometry
function
binomial coefficients
absolute value
real analysis

Is the integral \[\int_0^\infty \left((1-\frac{\sin x}{x})^{-\frac{1}{3}}-1\right)dx\] convergent? Is it absolutely convergent? Give proofs to your answers.

The integral converges, but not absolutely.

First, split this into the integral on $(0, 2]$ plus the integral on [$2, \infty$).

As $x\to0^+$, the function is $O(x^{-2/3})$, and the function is otherwise continuous on $(0, 2]$, so that portion of the integral converges absolutely.

For $x\ge2$, expand the integrand out using the binomial series, obtaining:

$\displaystyle \left((1-\frac{\sin x}{x})^{-\frac{1}{3}}-1\right) = \frac13 \frac{\sin x}x + \frac29 \frac{\sin^2x}{x^2} + \frac{14}{81}\frac{\sin^3x}{x^3} + \dots$

For $\displaystyle k\ge2, \int_2^{\infty} \left| \frac{\sin^kx}{x^k}\right| dx \le \int_2^{\infty} \frac{dx}{x^k} = \frac2{(k-1)2^k}$.

Multiply these estimates by the appropriate binomial coefficients and sum as $k$ goes from $2$ to $\infty$. The binomial series has a radius of convergence of 1, and mutliplying by $\frac2{k-1}$ doesn't change the radius of convergence. Since $\frac12$ lies well inside the radius of convergence, this series converges absolutely. (When I made the unusual choice to split the integral at 2 rather than at 1, I was looking forward to this argument about the radius of convergence.)

We have now accounted for everthing except for this: $\displaystyle \frac13 \int_2^{\infty} \frac{\sin x}x dx$. It is well known that this integral converges conditionally. It fails to converge absolutely because its absolute value is essentially comparable to $\frac1x$. The convergence can be proved by integrating by parts, integrating $\sin x$ and differentiating $\frac1x$.

Yes, your answers are correct, Merryfield.

[quote="Kent Merryfield"]We have now accounted for everthing except for this: $\displaystyle \frac13 \int_2^{\infty} \frac{\sin x}x dx$. It is well known that this integral converges conditionally. It fails to converge absolutely because its absolute value is essentially comparable to $\frac1x$. The convergence can be proved by integrating by parts, integrating $\sin x$ and differentiating $\frac1x$.[/quote]

the convergence of $\int_2^{\infty} \frac{\sin x}x dx$ can be determined by Dirichlet's Test.  :maybe:
and its it not absolutely convergent because
\[\frac{|\sin x|}x > \frac{\sin^2 x}x = \frac{1}{2}(\frac{1}x - \frac{\cos 2x}x) \]


To side AC of right triangle (<C=90) parallel line intersects AB and BC at M and N with CN/BN=AC/BC=2. AN and CM intersects at O and K lies on the segment ON with MO+OK=KN. Perpendicular line to AN at K(K is feet of this perpendicular) and bisector of angle <B of triangle ABC intersects at point T. Find angle <MTB.


Find all continuous $g: R \to R$ such that:$g^2(x)-g^2(y)=g(x+y).g(x-y) \forall x,y\in R$

[quote] $(g(x))^2$?[/quote]

This is it,i think  :)

first of  all  we  have    for  x=y=0  that  g^2(0) - g^2(o) = g(0)*g(0) then  g(0) = 0

we  also obtain that the  function  f(x) = ax  with  a is  a real  number  is a  solution  of  this  problem
because  we  have a^2*x^2 - a^2* y^2 = a(x+y)a(x-y)
so 0=0  that  is  true for  all  x,yER

Certainly,g(x)=ax and g(x)=0 is the solution of this problem.But I need a solution of the problem.I think it is very hard

Who can prove that$g(2x)=2g(x)$.I can do but my solution very long


One:

Simplify:
a-5ab+2a+ab+2ab

Two:

Find the greatest common factor:
8c^3dw^5-10c^6d^2w^10

3/4:
Distributive property:

8a(3b-c)

3x62(2xy+4y3)

One: [hide]$3a-2ab$[/hide]

Two:[hide]$2c^3dw^5$[/hide]

Three:[hide]$24ab-8ac$[/hide]

And the other one, I'm not sure if it's 3 times 62 or 3x62, so I won't do it. Put it in LaTeX to claify it.[/hide]

Please read this: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23215]Classroom Forum Posting Guidelines[/url]


It is really annoying to open a problem which doesn't fill the browser window and see the solution accidentally below. To hide a solution, use the following code:
[code]
[hide=Solution]
This is trivial. QED.
[/hide]
[/code]Obviously, replace the interior of the hide tag with your solution.

Credit for above post goes to Worthawholebean, as he posted it first in pre-olympiad.

This is a similar question but not exactly: When one quotes another person, who do you make it say "1=2 said" or something like that, instead of "Quote"?

well when you quote someone, you see {quote="name here"}, and you can change that name to whatever you want, like [quote="rrusczyk"]I'm leaving AoPS[/quote].

(the above is not a real quote, it's just a sample)

hope that answers your question.

[quote="JRav"]
Thanks, that helped.
[/quote]


geometry
3D geometry
sphere
category theory
abstract algebra
advanced fields
advanced fields unsolved

1. For $ n\geq 2$, prove there is no such smooth map $ f: S^n\rightarrow E$($ E$ is the equator of $ S^n$) which has the property that the restriction of $ f$ to $ E$ is the identity map on $ E$.

2. Prove that $ S^n$ is not a retract of $ \mathbb{R}^{n \plus{} 1}$

In both cases, you want to prove something is not a retract.

If $ Y$ is a retract of $ X$, then for any functor $ F$ from spaces to abelian groups, the map $ F(Y)$ to $ F(X)$ must be injective.

So, $ S^{n \minus{} 1}$ (the equator of $ S^n$) cannot be a retract of $ S^n$ since the map $ \mathbb{Z} \equal{} H_{n \minus{} 1}(S^{n \minus{} 1}) \to H_{n \minus{} 1}(S^n) \equal{} 0$ is clearly not injective.

Likewise, $ S^n$ is not a retract of $ \mathbb{R}^{n \plus{} 1}$ since $ \mathbb{Z} \equal{} H_n(S^n) \to H_n(\mathbb{R}^{n \plus{} 1}) \equal{} 0$ is not injective.

[color=green]Redundant quote removed[/color]
Thanks a lot!

Can anyone prove these using cohomology?

It's the same trick. If $ Y$ is a retract of $ X$, for a contravariant functor $ F$ (like cohomology) you still get an injection $ F(Y) \to F(X)$, except this time its induced from the retraction map instead of the inclusion map. There still aren't injections in the cases you've described, since the homology and cohomology groups of these spaces coincide.

[color=green]Redundant quote removed[/color]

Thank you!

Moderator's note: there's no good reason to quote the whole post immediately preceding yours. It's assumed you're responding to it anyway. Excerpts are fine, but the whole thing is a waste of space.


geometry
geometric transformation
rotation

Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius $ 20$ feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point $ 10$ vertical feet above the bottom?

$ \textbf{(A)}\ 5 \qquad
\textbf{(B)}\ 6 \qquad
\textbf{(C)}\ 7.5 \qquad
\textbf{(D)}\ 10 \qquad
\textbf{(E)}\ 15$

All you have to do is draw a picture of the situation.  Mine looks like a 5 year old drew it, but it works.

It's obvious that the 10-x-20 triangle is congruent to the 10-x-? triangle by SAS.  So the triangle inside the circle is equilateral, meaning every angle is 60 degrees.  60/360 = 1/6, so the rider goes 1/6 of the way around.  Since there's one rotation a minute, it takes 60/6 = 10 seconds to go 1/6 of the way around. The answer is D.

[quote=Zanderman]Hello,

I understand @bove's solution (with constructing a rectangle), but not the official aops wiki one. How did they get that $B$ is the midpoint of $OA$, because $OA$ isn't isosceles that way (well they didn't prove it, it turns out it is though).[/quote]

$B$ is defined as $10$ feet above the ground; therefore $AB = 10$. Since $OA$ is a radius it has length $20$, and therefore $OB = 10$. I don't think it is originally defined as the midpoint.


if ac-ad+bd-bc= 36 and c-d=8 what is a-b?


Let triangle $ ABC$ with incircle $ I$ tangent $ BC,CA,AB$ at $ D,E,F$ respectively. Denote $ ID\cap EF\equal{}K,AK\cap BC\equal{}M$. Prove that: $ M$ be midpoint of $ BC$

Already posted. See [url]http://www.mathlinks.ro/viewtopic.php?t=152959[/url].


algebra
polynomial
invariant
Galois Theory
group theory
superior algebra
superior algebra unsolved

For $ K$ a field of characteristic $ p>0$ and $ a\in K$, if $ f(x)\equal{}x^p\minus{}x\minus{}a$ has no root in $ K$, then $ f(x)$ is irreducible in $ K[x]$.

in fact, $ f$ splits into linear factors or it is irreducible. there are two well-known proofs for this:

1) if $ c$ is a root in a splitting field (of $ f$), then obviously $ c \plus{} i$ are the other ones, where $ i$ runs through $ \mathbb{F}_p$. in particular, $ K(c)$ is a splitting field. with $ f' \equal{} \minus{} 1$ we see that $ f$ is separable. hence $ K(c)$ is a galois extension of $ K$. our aim is to prove that the galois group is isomorphic to $ \mathbb{Z}/p$, for then $ p \equal{} |K(c) : K|$ is the degree of an irreducible factor of $ f$, thus $ f$.  observe that $ Gal(K(c) : K) \to \mathbb{Z}/p , \sigma \mapsto \sigma(c) \minus{} c$ is a well-defined injective group homomorphism, with nontrivial image (else $ c \in K$), thus an isomorphism since $ p$ is prime.

2) $ f$ is invariant under the automorphism $ x \mapsto x \plus{} 1$ of $ K[x]$. thus $ \mathbb{Z}/p$ acts on the monic irreducible factors $ g$ of $ f$. if some $ g$ is a fixed point, $ g(x) \minus{} g(0)$ has every $ x \in \mathbb{F}_p$ as a root, thus has degree $ \geq p$ and $ f \equal{} g$ is irreducible. otherwise all orbits have $ p$ elements, which yield $ p$ distinct monic irreducible factors and thus a splitting of $ f$.

Suppose $ f\equal{}gh$ is a nontrivial factorization of $ f$ in $ K[x]$ with $ g,h$ monic. Let $ u$ be a root of $ f$ in some splitting field. The remaining roots (in that splitting field) are $ u\plus{}1,u\plus{}2,\ldots,u\plus{}p\minus{}1$. Write $ g\equal{}x^k\plus{}c_{k\minus{}1}x^{k\minus{}1}\plus{}\cdots\plus{}c_1x\plus{}c_0$ where $ c_i\in K$ and $ 0<k<p$. Since $ \minus{}c_{k\minus{}1}$ is sum of $ k$ of the above roots, we have $ \minus{}c_{k\minus{}1}\equal{}ku\plus{}m$ for some $ m\in Z$. This implies $ u\in K$.


Let $ I$ be the incentre of $ \Delta ABC$.With $ I$ as centre and any arbitrary radius a circle is drawn.Drop perpendiculars from $ I$ on on to $ BC,CA,AB$ which cut the circle at $ L,M,N$respectively.Prove that $ AL,BM,CN$ are concurrent


Given a triangle $ ABC$. $ D$ is arbitrary point on circumcircle of $ ABC$. Let $ H$ is orthocenter of $ ABC$. Consider points $ E,F$ on $ AC$, $ BC$ respectively,  such that $ DE$ and $ DF$ perpendicular with $ AC$ and $ BC$ respectively. If $ M$ is midpoint of $ DH$, then prove that $ E,H,F$ are collinear.


Hello!

Im interested in the general form of the extreme value theorem from topology:

Let $ f: X\to Y$ be continuous, where $ Y$ is an ordered set in the order topology. In $ X$ is compact, then there exist points $ c$ and $ d$ in $ X$ such that $ f(c)\leq f(x)\leq f(d)$ for every $ x\in X$.

The extreme value theorem of calculus is a special case of this theorem that occurs when we take $ X$ to be a closed interval in $ \mathbb{R}$ and $ Y$ to be $ \mathbb{R}$.

I am interested where this generalized form comes to use-there are probally many applications where we would care to show there is an extreme value--maybe in PDE with Hilbert spaces...?

Thanks in advance!

Los

In general this is false. If it was true, this would be a strong assertion about comparability of elements in a compact subset. In particular it would imply that every two elements are comparable: Just take $ A \equal{} \{y_1,y_2\}\subseteq\ Y$ which is finite and hence compact. The inclusion $ A\hookrightarrow Y$ is continuous.
If the claim was true, we would get $ y_1\leq y_2$ or $ y_2\leq y_1$, i.e. $ Y$ would be a total order.

On the other hand: If we assume that $ Y$ is totally ordered, then the proposition becomes true. $ f(X)$ is a compact subset of $ Y$.
Assume that $ f(X)$ is not bounded from below by a $ f(c)$. Then for every $ c\in X$ there is a $ x\in X$ with $ f(x) < f(c)$. Hence $ f(X)\subseteq\bigcup_{x\in X} (f(x), \plus{} \infty)$ is an open covering.
It cannot have a finite subcovering, because $ \min_{i \equal{} 1\ldots n} f(x_i) \notin \bigcup_{i \equal{} 1}^n (f(x_i), \plus{} \infty)$. This is a contradiction to the compactness of $ f(X)$, hence there is a lower bound of the form $ f(c)$.
Analogue there is an upper bound of the form $ f(d)$.

I'm not quite sure where you used total ordering here... I'm pretty sure Munkres proves this result with Y only having the order topology, but I don't have my book close by to check.

I used it everywhere! :-)

That "not bounded from below by a $ f(c)$" is the same as "$ \forall c\in X\exists x\in X: f(x)<f(c)$" uses the fact that the negation of $ a\leq b$ is $ a>b$, in other words it uses that $ Y$ is totally ordered.
I also used that every finite set has a minimum. This is an equivalent formulation of being totally ordered.

We must be using different definitions because in Munkres an ordered set is comparable which is what we are assuming.


I think you made a typo:  If $ p,q\equal{}0$, then $ p^2\equal{}0\equal{}2pq$ and $ q^2\equal{}2pq$.  I believe it should read greater than or equal to, not just greater than.

[quote="7h3.D3m0n.117"]Prove or disprove that for all numbers $ 0 \le p \le 1$ and $ 0 \le q \le 1$, $ 2pq > p^2$ and $ 2pq > q^2$.[/quote]
For p=0.3 and q=0.1
$ p^2\equal{}0.09$ and $ q^2\equal{}0.01$
$ 2pq\equal{}0.06 < p^2$

[quote="7h3.D3m0n.117"]Prove or disprove that for all numbers $ 0 \le p \le 1$ and $ 0 \le q \le 1$, $ 2pq > p^2$ and $ 2pq > q^2$.[/quote]

Since this is Hardy-Weinberg I think 7h3.D3m0n.117 meant that $ p \plus{} q \equal{} 1,\ 0 \le p,q \le 1$. But the inequalities don't make sense no matter how you try to think about them .. (unless its supposed to be really easy to disprove)

[quote="azjps"]Since this is Hardy-Weinberg I think 7h3.D3m0n.117 meant that $ p \plus{} q \equal{} 1,\ 0 \le p,q \le 1$. But the inequalities don't make sense no matter how you try to think about them .. (unless its supposed to be really easy to disprove)[/quote]That is exactly a missing condition ($ p\plus{}q\equal{}1$) in Hardy-Weinberg. For more info, if anyone is interested in Biology, here is a link to it

[url]http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/H/Hardy_Weinberg.html[/url]

[quote="7h3.D3m0n.117"]Prove or disprove that for all numbers $ 0 \le p \le 1$ and $ 0 \le q \le 1$, $ 2pq > p^2$ and $ 2pq > q^2$.[/quote]

[quote="7h3.D3m0n.117"]Since I originally placed this in HSB, I meant to prove that $ 2pq$ (representing the frequency of heterozygous genotypes) would always be greater than $ p^2$ or $ q^2$, since by viewing a monohybrid Punnet Square, the ratio of homozygous dominant:homozygous reccessive:heterozygous would be 1:1:2.[/quote]

Counterexample for quote #1: let $ p\equal{}0.1$. It's an 'and' statement, so one of the inequalities is false implies the claim is false.

Quote #2 is too weak and $ 2pq$ is at least greater than $ p^2$ or $ q^2$

Quote #1 can be true for certain values of $ p$. Find that interval of values for $ p$ for which quote #1 is true, that is, for which $ 2pq > p^2$ [u]and[/u] $ 2pq > q^2$.


Does anyone know when the results for FGH will be released?

i am usually the first to know about these things (my teacher is in the loop and he tells it to us as soon as he hear something)... I'll let you guys know if i hear these things

Ok thx.

For once, I actually think I did pretty good on Fryer that's why I'm getting more and more anxious.

[quote="kool_dudy"]I just got my result from Fryer today.

I scored a 32...in my opinion, pretty bad.

Second in my school.[/quote]

Average for Fryer was nearly 27.

I'm really disappointed now.  I just went to CEMC and found out that gold standard is at 36 and I'm only eligible for bronze...:(


geometry
search
combinatorics proposed
combinatorics

Given 18 points in the plane, no three collinear, so that they form 816 triangles. The sum of the area of these triangles is A. Six are colored red, six green and six blue. Show that the sum of the areas of the triangles whose vertices are the same color does not exceed A/4.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1657994991&t=3300

Pierre.


http://www.zeropages.co.uk/norpton

hope you like it and good luck.

[quote="math92"]i can't find the game.[/quote]

[hide="It is there."]The way to the first level isn't as simple as clicking a link. Unless you're not doing it on your own.[/hide]

[hide="Hint"]Try using the things under "The following things will help you a lot" on the hints page.[/hide]

i am stuck on level 4 cant find the password or username :blush: can anyone help?

This is what I figured out for level 3:
[hide]The stuff in the corner is assembly code, and according to the assembly code tutorial on the main site, it does nothing. Also, there's the title, "dont leave me!" and the picture, in which the word LEAVE can be seen in red. I wonder if this refers to going "up" a page in order to find the tutorial? Or does it refer to some assembly command? I also don't get "the slash goes inbetween" or "the power to create everything."[/hide]

I made it to level 5 of notpron, and am currently at level 2 of norpton. Keep adding more stuff!

Hint for level 3: your on the right track but you need to know what it does

[quote="math92"]i still can't get there :([/quote]

Me too. :( I've tryied to change the URL ... but nothing(level.htm, start.htm, game.htm).  They don't work.

u need to decode the numbers/letters. its one word look at the other hints

Right now my browser is having trouble opening your website...


superior algebra
superior algebra unsolved

If into a finite unitar ring A, 2 is nilpotent, then card(A)=2^n ?

If $ 2$ is nilpotent, the additive order of any element is a power of $ 2$.  Apply Cauchy's theorem.

Why the additive order of any element is a power of 2 ?

Because if $ 2^n\equal{}0$, $ 2^n x \equal{} 0$.


algebra
polynomial
superior algebra
superior algebra unsolved

By Fermats little theorem, $X^{6}\equiv1(\mod7)$ so that
$X^{6}+5\equiv6(\mod7)$. Hence the polynomial is irreducible.

I wonder why there are so many people thinking that irreducible=no_roots ...

Here is the long way to do it.

abdul show it has no linear factors (but it can still be reduccible).

Then there are only two possible cases:

$x^{6}+5 = (x^{2}+ax+b)(x^{4}+cx^{3}+dx^{2}+ex+f)$

Or,

$x^{6}+5=(x^{3}+ax^{2}+bx+c)(x^{3}+dx^{2}+ex+f)$

And compare coffeicients after expanding.

If a is a root of $x^{6}+5$, then the other roots are $2a,3a,4a,5a,6a$.
$\phi_{i}: \mathbb{F}_{p}(a) \Rightarrow \mathbb{F}_{p}(a)$ generated by fixing $\mathbb{F}_{p}$ and $a \Rightarrow ia$ defines a $F_{p}$ automorphism. Therefore all these 6 roots belong to the same irreducible polynomial in F_p[x]. So it is irreducible.

nici one Soarer and thanks abdul, that's what i thought too . the whole problem was something like : find all a in $Z_{7}$ so as $X^{6}+aX+5$ is irreducible . and i said that the only solution was for $a=0$ , because $X^{6}=1$ so the polynom is actually $aX+6$ when x is not 0 . whatever $a$ different from 0 we take , this has a root. so the initial polynom is reducible. am i right ?

[quote="Soarer"]If a is a root of $x^{6}+5$, then the other roots are $2a,3a,4a,5a,6a$.
$\phi_{i}: \mathbb{F}_{p}(a) \Rightarrow \mathbb{F}_{p}(a)$ generated by fixing $\mathbb{F}_{p}$ and $a \Rightarrow ia$ defines a $F_{p}$ automorphism.[/quote]
Can somebody show me that "$a \Rightarrow ia$ defines a $F_{p}$ automorphism" explicitly?

you're just permuting roots of an irreducible polynomial: in this case, it suffice adjoining one single root to have the splitting field, so fixing the image of a root fixes the automorphism of the field..
is this what you wanted to ask, or should i go further into details?

Thank you,mago.
In fact,I want to ask how can I see such automorphism exist?

We have for any field $K$,$u,v$ is algebraic over $K$,there exist a isomorphism $f: K(u) \to K(v)$,such that fixed $K$ and $u \to v$ if and only if $u,v$ are the roots of the same  irreducible polynomial.

So I am sorry to say that $Soarer$ 's proof is nonsense.
If his proof works,then consider the problem:
does $x^{12}+4$ reducible over $Z_{13}$

We can use his way to get $x^{12}+4$ is irreducible over $Z_{13}$.But this is obviously wrong,because:$x^{12}+4=(x^{6})^{2}+4x^{6}+4-4x^{6}=(x^{6}+2)^{2}-4x^{6}=(x^{6}+2x^{3}+2)(x^{6}-2x^{3}+2)$

I am sorry if what I wrote is wrong.

[Edited]


A triangle whose all sides have length not smaller than $1$ is inscribed in a square of side length $1$. Prove that the center of the square lies inside the triangle or on its boundary.

If a triangle XYZ doesn't contain the center of the square, there is a line through the center of the square that doesn't intersect XYZ.
Proof: Suppose that every line through the center intersects XYZ. If the four vertices of a convex quadrilateral are part of a convex figure, then all points that line inside that quadrilateral are part of the convex figure. By taking two lines through the origin and finding their intersections with the triangle, we can find a convex quadrilateral containing the origin, inside of XYZ.

Draw a line perpendicular to the line through the center of the square, as shown in the diagram.

All sides of the triangle must lie in the union of the blue and yellow regions.

The quadrilaterals surrounding the blue and yellow regions are concyclic, inscribable in circles of diameter $\sqrt{2}/2$. Any line segment of length 1 that lies in the union of the colored regions must lie in both, since $1>\sqrt{2}/2$

If X lies in the blue region, then Y lies in the yellow region, then Z lies in the blue region, then X lies in the yellow region, a contradiction.

[quote="schulmannerism"]
 By taking two lines through the origin and finding their intersections with the triangle, we can find a convex quadrilateral containing the origin, inside of XYZ.[/quote]

Is it always that there must be a quadrilateral containing the origin?


For $x;y;z$ are positive real numbers satisfying $3x^{3}+4xy+3y^{2}=14$
Find the least value of the expression:$A=x^{2}+y^{2}$

chien than, Are you sure that $3x^{3}+4xy+3y^{2}=14?$
I think that $3x^{2}+4xy+3y^{2}=14.$ :maybe:


logarithms
trigonometry
calculus
calculus computations

can you show this ? :D [color=white]............ [/color](i know i can't) ! LOL !   :rotfl:

$ \sum_{k=1}^\infty\;\;\frac{(-1)^{k+1}\cdot k^{2}}{1+k^{3}}\;\;=\;\;\boxed{\frac{1}{3}\left( \ln \left( \frac{e}{2}\right) \;+\;\frac{\pi}{\cosh \left( \frac{\pi\;\sqrt 3}{2}\right) }\right) }$


geometry
3D geometry
calculus
calculus computations

The base of a cone-shaped tank is a circle of radius 5 feet, and the vertex of the cone is 12 below the base. The tank is filled to a depth of 7 feet, and water is flowing out of the tank at a rate of 3 cubic feet per minute.

Find the rate of change of the depth of the water in the tank.

i really need someone to show this one to me

Is water flowing out of the tank from the bottom (i.e., from the vertex)?

Have you tried to solve this by your own?


algebra
polynomial
number theory unsolved
number theory

We have two distinct positive integers $a,b$ with $a|b$. Each of $a,b$ consists of $2n$ decimal digits. The first $n$ digits of $a$ are identical to the last $n$ digits of $b$, and vice versa. Determine $a,b$.

[hide="solution"]
Let $ a\equal{}x10^n \plus{}y$ and $ b\equal{}y10^n \plus{}x$. We start by using polynomial long division (with the "variable" being $ 10^n$), which gives $ \frac{y}{x}$ as the quotient and $ x\minus{}\frac{y}{x}$ as the remainder. In order for $ a$ to divide $ b$, the remainder ($ x\minus{}\frac{y}{x}$) must equal zero, so $ x^2\equal{}y$. Remember, however, the requirement that $ x$ and $ y$ have the same number of digits. Since squaring any positive integer greater than 3 results in the number of digits increasing, the only possible values for $ a$ and $ b$ are $ (24,42)$ and $ (39,93)$.
[/hide]

[quote="Valentin Vornicu"][quote="foxjwill"][hide="solution"]
Let $ a \equal{} x10^n \plus{} y$ and $ b \equal{} y10^n \plus{} x$. We start by using polynomial long division (with the "variable" being $ 10^n$), which gives $ \frac {y}{x}$ as the quotient and $ x \minus{} \frac {y}{x}$ as the remainder. In order for $ a$ to divide $ b$, the remainder ($ x \minus{} \frac {y}{x}$) must equal zero, so $ x^2 \equal{} y$. Remember, however, the requirement that $ x$ and $ y$ have the same number of digits. Since squaring any positive integer greater than 3 results in the number of digits increasing, the only possible values for $ a$ and $ b$ are $ (24,42)$ and $ (39,93)$.
[/hide][/quote]Your answer is not correct, 24 does not divide 42, nor does 39 divide 93.[/quote]

Oy! I was so excited that I had found a solution that I forgot to check to see if it was correct! >_<

Using the well-known properties of the decimal expansion of $ \frac{1}{7}$, it's easy to find:

$ 142857 \times 6 \equal{} 857142$.

I'm quite sure there are no other solutions, but I need some more time to find a clean proof.


let $ \\a_n\equal{}\frac{(2\plus{}\sqrt{3})^{2^n}\plus{}(2\minus{}\sqrt{3})^{2^n}}{2}, n\geq1.$

$ \\p|a_n$ and $ 12|p\minus{}1,p\minus{}prime$ then prove that $ 2^{n\plus{}2}|p\minus{}1$

It is an good problem .
If $ p\equiv 1 (\mod 12 )$ then we have :
$ (\frac {3}{p}) \equal{} 1$ so exist an integer m such that $ m^2\equiv 3 (\mod p)$
So $ a_n\equiv (m \plus{} 2)^{2^m} \plus{} (2 \minus{} m)^{2^n} (\mod p)$
$ \Rightarrow p|(2 \plus{} m)^{2^n} \plus{} (2 \minus{} m)^{2^n}$
$ \Rightarrow p|a^{2^n} \plus{} 1$
Where $ a \equal{} (\frac {m \plus{} 2}{2 \minus{} m}) (\mod p )$
Because $ (m \plus{} 2)(2 \minus{} m)\equiv 1 (\mod p)$ so $ a$ is a quadric reside mod p .
It mean that there exist an integer $ x$ such that $ x^2\equiv a (\mod p )$
So $ p|x^{2^{n \plus{} 1}} \plus{} 1$
Because $ ord_x(\mod p) \equal{} 2^{n \plus{} 2}$ so $ 2^{n \plus{} 2}|p \minus{} 1$
 Problem claim .

[quote="TTsphn"]It is an good problem .[/quote]

Maybe, but not sufficientely to be posted three times in a row :wink:
Two times deleted...

Pierre.


Given $ f: (a,b) \to \mathbb{R}$ is both continuous and differentiable, find the following in terms of $ f(x)$ and $ f'(x)$:

[list]$ \lim_{h \to 0} \frac {f(x \plus{} 2h) \minus{} f(x)}{h}$[/list]
Obviously, $ f'(x) \equal{} \lim_{h \to 0} \frac {f(x \plus{} h) \minus{} f(x)}{h}$, but I don't know how to deal with the $ 2h$. Thanks.

What can you do with $ \lim_{h\to0}2\cdot\frac{f(x\plus{}2h)\minus{}f(x)}{2h}\ ?$

[quote="Kent Merryfield"]What can you do with $ \lim_{h\to0}2\cdot\frac {f(x \plus{} 2h) \minus{} f(x)}{2h}\ ?$[/quote]
Got it. Doh! Thanks.


Two six-sided dice are rolled. Given that at least one die shows a one, what is the probability that both of them are ones? Express your answer as a common fraction.

There are eleven ways we get atleast a single one, all of the same probability.
1-1
1-2
1-3
1-4
1-5
1-6
2-1
3-1
4-1
5-1
6-1
Out of these, only the first one, or one choice out of 11, has both dice one's.
Ans: $ \boxed{1/11}$


Hello,

Are all of the seats in MathPath taken up already? I want to know if I can apply if possible. I know this is late but I didn't find out about these camps until this summer, and now I'm regretting it. :(

Thank you
rts2007

Hello,

You can go to http://www.mathpath.org and click on the link to email Prof Maurer. Ask him if seats are still open and application procedures. Since the MathPath website says seats are still open you can probably still apply.

It starts in exactly 14 days (June 28). The deadline was the end of April, but it was extended to the middle of May. You should contact them. I found out in the middle of May, and I only had a few days to decide. I didn't apply, since I wasn't really ready.

My advice is to plan for next year. Find the dates, as other camps my have conflicting dates. Plan so that there is no overlap. Find all the information you will need (recommendations, materials, $ \$\$\$$). Check regularly for the admission test. Apply early, since some camps have an early bird fee.

If I am wrong, please correct me. I have never been to a summer camp  :oops: .

Yes, best of luck, but I would think, with camp just 2 weeks away, they would be set.

Also, levans, are you going to be there?

Dear rts2007,

  levans is right as to what you should do.

  We can still make room for you at camp, but you have to move fast.  You have to do the same application as everyone else, and that usually takes a week to put together.  Even if you see how to solve all the Qualifying Test problems right away, it still takes time to write them up clearly, which counts a lot with us.  And you have to line up your recommenders and get them to submit their recommendations (getting [i]them[/i] to hurry up is often difficult).  And if we do admit you, then unless you live within driving distance of Colorado Springs, you have to book airfare [i]after[/i] we admit you.  Maybe you'll get a great last minute fare, but more likely it will be expensive.

  We are game to consider your application if you and your parents are willing to do it.  But I would start right now, this weekend, working on the Qualifying Test.  And you still have to email me through the MathPath website so I can give you further speed-up suggestions.

  Oh, pinkmuskrat (who will be attending MathPath) asks if levans will be there.  Yes, for the first two weeks.

  Prof Maurer (= Mr M at camp)


abstract algebra
superior algebra
superior algebra unsolved

[quote="artin"]Let $ R$ be an injective $ R$ module ,prove that:

$ Ann(I \cap J ) \equal{} Ann(I) \plus{} Ann(J)$.    $ I$ and $ J$ are ideals of $ R$.[/quote]

$ Ann(I) \plus{} Ann(J) \subseteq Ann(I \cap J)$ is trivial. Now let $ x \in Ann(I \cap J).$ Then the map $ f: I \plus{} J  \longrightarrow R$ defined by $ f(a\plus{}b)\equal{}ax, \ \ a \in I, b \in J,$ is a

well-defined $ R$ homomorphism and thus, since $ R$ is self-injective, there exists $ r \in R$ such that $ ax\equal{}f(a\plus{}b)\equal{}(a\plus{}b)r,$ for all $ a \in I, b \in J.$ Particularly

$ ax\equal{}ar$ and $ br\equal{}0$ for all $ a \in I, b \in J.$ hence $ x\minus{}r \in Ann(I)$ and $ r \in Ann(J).$ Thus $ x \in Ann(I)  \plus{} Ann(J)$ and the proof is complete.


algebra
polynomial
abstract algebra
superior algebra
superior algebra unsolved

Let $\displaystyle \mathcal K$ be a field with $\displaystyle 27$ elements. Prove that there is $\displaystyle a \in \mathcal K$ such that $\displaystyle a^3 = a + 2$.

[i]Mircea Becheanu[/i]

hi,
consider $\mathbb{F}_3$,the polynomial $x^3 -x -2$ is irreducible over it,so the field obtained by adjoining a root of it to $\mathbb{F}_3$ is a field of order 27,now since all fields of order 27 arre isomorphic,every such  field has an element $\alpha$ s.t $\alpha ^3 =\alpha +2$.

I think everybody who knows a little bit about finite fields has thought of rohith's solution ;-)

Is there another "elementary" proof? :-)

Hi.

There is something like an "elementary" proof:

Consider the map $\phi: \mathbb{F}_{27}\rightarrow\mathbb{F}_{27}: x\mapsto x^3-x$ which is $\mathbb{F}_3$-linear.
Since every element a of $\mathbb{F}_{27}$ is a solution of $x^{27}-x=0$, the third power of this map is zero: it is nilpotent.
The kernel of $\phi$ is the subfield $\mathbb{F}_3$. Since $\phi$ ist nilpotent, $ker(\phi)$ is a subset of $im(\phi)$. So there must be an element $y\in\mathbb{F}_{27}$ with $\phi(y)=2$, this is the element we've been looking for.

Is this elementary enough?

mfg Gockel.


vector
algebra
polynomial
linear algebra
matrix
inequalities
induction

Let $ V$ be a finite dimensional vector space and $ T: V\to V$ be a linear transformation such that $ \exists n\in \mathbb{Z}^ \plus{}$ with $ T^n \equal{} 0$. Let $ r(k) \equal{} \text{rk }(T^k)$. Show that $ r(k)$ is strictly decreasing when $ r(k) > 0$.

I can show that $ r(k)$ is non-increasing by showing that $ \text{im }(T^{k \plus{} 1})$ is a subspace of $ \text{im }(T^k)$, but I'm not sure where to go from here. I also considered trying to show that $ \text{dim(ker}(T^k)) < \text{dim(ker}(T^{k \plus{} 1}))$, but I can't find a way to show that $ \exists \vec{v}\in V$ such that $ T^{k \plus{} 1}\vec{v} \equal{} \vec{0}$ and $ T^{k}\vec{v}\neq \vec{0}$.

As a matter of notation and thought pattern, I always prefer working with matrices to working with linear transformations and subspaces.

By Schur's lemma, if all the characteristic polynomial can be completely factored, than any matrix is similar to an upper triangular matrix with the eigenvalues on the main diagonal.

As a corollary of that, any nilpotent matrix is similar to a strictly upper triangular matrix.

Try proving this as a theorem about strictly upper triangular matrices.

Of course there is a minimal n for which $ T^n\equal{}0$. Then we can say there is a $ v$ for which $ T^{n\minus{}1}(v)\ne 0$.

If V has dimension $ N$, in the case $ N\le n$, we have that

$ v$, $ Tv$, ... $ T^Nv$ is a basis for $ V$ from which your claim follows.

Now it suffices to consider the case $ N>n$...

Perhaps in this case we have to be more careful with our choice of v... Say $ \{v_1,...,v_n\}$ is a basis for $ V$. Then we can choose a $ v_i$ such that $ T^{n\minus{}1}(v)\ne 0$. Then I think that the same proof still holds but we note that the dimension can go down by more than 1.

(extent v, Tv,..., T^{n-1}v to a basis of V...)

[url=http://planetmath.org/encyclopedia/FrobeniousInequality.html]The Frobenius inequality[/url] yields $ \mathrm{rank}A^{a\plus{}b}\plus{}\mathrm{rank}A^{b\plus{}c}\leq\mathrm{rank}A^b\plus{}\mathrm{rank}A^{a\plus{}b\plus{}c}$. The rest is combinatorics ;)

  darij

I think it's a rare case of endemic blindness and beating around the bush. The result is trivial.  :)

We have $ r(k) : \equal{} \textrm{rk}(T^k) \equal{} \dim \textrm{Im}(T^k)$, and obviously $ \textrm{Im}(T^{k\plus{}1}) \subseteq \textrm{Im}(T^k)$. Assume $ r(k\plus{}1) \equal{} r(k) > 0$; this means $ \textrm{Im}(T^{k\plus{}1}) \equal{} \textrm{Im}(T^k)$, i.e. $ T^{k\plus{}1}(V) \equal{} T^k(V)$. But then, by trivial induction, $ T^{m}(V) \equal{} T^k(V)$ for all $ m>k$, absurd, since $ T^n(V) \equal{} \{0\}$.

[color=red]In one line, once the sequence of nested subspaces $ T^k(V)$ contains two consecutive equal terms, it becomes stationary, which according to the givens can only occur for $ \{0\}$.[/color]


modular arithmetic
algebra
binomial theorem

Find the remainder when $ 3^{2003}$ is divided by $ 28$.

$ 3^3\equiv \minus{}1\pmod{28}$

$ 3^6\equiv (\minus{}1)^2\equiv 1\pmod{28}$

$ 3^{2003}\equal{}3^{1998}\cdot3^5\equiv 1^{333}\cdot3^5\equiv 3^5\equiv 19\pmod{28}$

[hide=" "]
I bet someone will try to explain this better, except if electron proton posts that he didn't understand the solution right away [/hide]

hello, note that $ 3^3\equiv \minus{}1 \mod 28$ so we get $ 3^{2001}\equal{}3^{3\cdot667}\equiv\minus{}1\mod 28$ and from here we get $ 3^{2003}\equiv\minus{}9\mod 28$
Sonnhard.

[hide="Using binomial theorem"]
Write it as $ 3.(28\minus{}1)^{667}$ and apply binomial theorem.
19 will be the answer.

[/hide]


Given that $ 2^{2004}$ is a 604-digit number with leafing digit 1. Determine the number of elements in the set $ \{2^0,2^1,\cdots,2^{2003}\}$ with leafing digit 4. (The leafing digit is the first digit on the left hand)

Wasn't this problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1368329#1368329]just posted[/url]?


:arrow: Δύο κύκλοι σε ένα επίπεδο τέμνονται. Έστω $ A$ ένα από τα σημεία τομής. Ξεκινώντας ταυτόχρονα από το $ A$ δύο σημεία κινούνται με σταθερές ταχύτητες, και το καθένα ταξιδεύει πάνω στο δικό του κύκλο στην ίδια κατεύθυνση. Μετά από μια περιστροφή τα δύο σημεία επιστρέφουν ταυτόχρονα στο $ A$ . Δείξτε ότι υπάρχει σταθερό σημείο $ P$ στο επίπεδο τέτοιο ώστε κάθε στιγμή η απόσταση των σημείων από αυτό να είναι σταθερή.

Μήπως εννοείς το αθροισμα των αποστάσεων???


This should probably go in the classes forum.

I am not sure if any parents take the classes.  There aren't any that I know of.  I don't really see any problem with teachers/parents/etc taking the classes.  I can definitely see the benefit of teachers and parents taking it.  Then they can teach their children what they have learned.  All in all the classes are a ton of fun and you learn so much.

Moved it.

I think that there should be no problem b/c:
(a) the internet is anonymous, so nobody knows your age
(b) the classes are not geared towards a specific age group, I don't think (just specific tests that HS kids take...)

There are a lot of adults on this forum, but they are mostly the owners of the website or math professors or something like that.

[quote="Thomas"]Hello,

I am new to this forum and would like to know if teachers and/or parents are also taking the AoPS courses.  Although I have been looking through this site for my son, I feel that I could benefit from the courses as well.  Have other adults tried the courses, and if so, what were their thoughts?

Thanks[/quote]

We do have adults take our classes.  We have had at least one teacher take several of our courses.  We have no age limitations.  If you feel our courses would be beneficial to you then we would be happy to have you in class.


Let S(n,p) denote the number of surjections from a set with a cardinal equal to n to a set with a cardinal equal to p.

1/ Calculate S(n,n) and S(p+1,p)
2/ Show that :
$S(n,p) = \sum_{k=1}^{n} \binom{n}{k}S(n-k,p-1)$
3/ Show that :
$\sum_{k=0}^{p} \binom{p}{k}S(n,k) = p^{n}$

Enjoy :D

It's all about Stirling numbers of the second kind.

Hmm... could you explain what you mean by this?

Nota : I know what those numbers are, but I can't see how they help with this problem..  :blush:

S(n,p) = p!*S2(n,p), where S2(n,p) is Striling number of the second kind.

In particular, formula 3) is exacly formula (10) at [url]http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html[/url] for x=m=p.