Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:

To make it short: [b]ALWAYS USE YOUR COMMON SENSE IF POSTING![/b]
If you don't have common sense, don't post.

More specifically:

[u]For new threads:[/u]

a) [b]Good, meaningful title[/b]:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
[u]Bad titles:[/u]
 - "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
 - "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
 - "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
 [u]Good titles:[/u]
 - "On divisors of a³+2b³+4c³-6abc"
 - "Number of solutions to x²+y²=6z²"
 - "Fibonacci numbers are never squares"

b) [b]Use search function[/b]:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

[b]Update (by Amir Hossein). [/b] The best way to search for [i]two keywords[/i] in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".

c) [b]Good problem statement[/b]:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.

[u]For answers to already existing threads:[/u]

d) [b]Of any interest and with content[/b]:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) [b]Well written and checked answers[/b]:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.

To repeat it: [b]ALWAYS USE YOUR COMMON SENSE IF POSTING![/b]

Everything definitely out of range of common sense will be [u]locked or deleted[/u] (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.

Huhhh... :roll: , [b]ZetaX[/b], I see you realy got angree :ninja:. Anyway, I can understand you; It is realy difficult to moderate a forum where there are,hmm...,[b]27567 [/b]users :blush: .

 I am in 100% agreement with you and I share your points. I think you should have done this long before(but better later than never :) ) and it is adviced to other moderators to undertake this kind of actions in other
sections too.

Thank you!

Kudos to ZetaX. These guidelines, and many more, should have been enforced a long time ago and not just here in NT forum, but ALL other forums [u]ought to[/u] follow as well (except maybe G&FF and Test forums). Common sense can be difficult to be defined for certain 'individuals' as they have 'common sense' defined differently.

Some BAD habits seen in some posts that resembles to the ones below:

No self-control: I must post something!!! I must reply! I'm very talkative!

Gibberish: "Yes I solved it hahaha!" "This problem is so easy that it is only for donkeys  :roll: :P "

Expressing own feelings that no one really cares: "I got it right and my friends didn't" "What tha heck I didn't get perfect score in AMC!"

Useless answer replies: "is it 420?"

Faulty/no LaTeX used: $cosx=\frac1$

Abundant use of chat/1337 speak: lol omg wtf |-|4><0r5

Repetitive solutions: someone solved the problem and I am posting my solution too, I don't care if it's the same!

Lack of use of hide tags: *scratches head* I dunno how to use 'hide' (dunno=don't know)

And much much more annoying and stupid things I have seen in AoPS/ML.

I added that contest names aren't good either as titles.

I always wanted to ask this, and I don't know why I have waited so long... How to use search function????
You can't type latex code if you want to search for some expresion, and if you are looking for same problem in 90% cases, text, beside expresion codes, are useles. Is there some way for using this for which I don't know, because, I get very upset when I see locked post, and message use search function, but I can't find problem like that.

For example this: http://www.mathlinks.ro/Forum/viewtopic.php?p=741742#741742

Can you show how to find this problem.
Tnx

Bye

P.S. If this is not the place for this kind of message, pls delete it, or I will, but pls, answer it...

EDIT: I am sorry I didn't mention this at first, I also support this kind of mesurements for making people to use their common sense, this should be used in other forums too, I can't put up with titles like "hard", and all that in sake of mora functional search function :)

The keywords "phi positive integers" allow to find it quite quickly, restricting the search to the Number Theory subforum.

Also, I just wanted to say that I strongly agree with [b]ZetaX[/b]'s decision.

I agree with you ,ZetaX.But sometimes I don't have many words to show my YOUR COMMON SENSE ,because my English is not good. :blush:
I'll try to do better.
I also find that The search function seldom works.........

[quote="Altheman"]@10000th User:

I disagree with the idea of no repeat solutions. I post solutions for problems because it is instructive for me to write out the proof and to get commentary from other users.[/quote]

I think he was referring to the lower fores, where solutions often happen to be merely computations etc. Here in the advanced section, it is always ok to post [b]solutions[/b], even if they happen to be equivalent to solutions already posted (it can always happen that the new post is easier to understand than the older one, and there are other reasons as well). It is also ok to post [b]hints[/b] if no solutions were posted. But if someone posted a complete proof of an inequality using Cauchy-Schwarz, then it is not ok anymore to post the one-liner "Cauchy-Schwarz pwns it".

  darij

[quote="hinhhoc273"]But no ones explain 'what is common sense' ?
Thank you for your ideas. I'll try correct my mistakes.[/quote]

Your common sense is your natural ability to make good judgements and to behave in a pratical and sensible way.

Zetax, a fully agree with you. I'll be good in this site! :lol:

Is this allowed to post a solution which is faulty, or the one who posts is not able to recognize whether his solution is complete or not???
Like those who are new to this IMO LEVEL MATHS?

Of course it is allowed. Sometimes we can actually learn from others' mistakes. But you should post the solution in a way that people can understand it and see whether it is right or wrong. This is what this whole topic is about. Unfortunately, many people here tend to answer topics with one-sentence posts like "the problem is easy using modulo arithmetic", and whenever somebody points out to them that such a posting is completely useless, they flame him for not having understood their ingenious one-sentence solution. As ZetaX writes, use your common sense. Nobody requires you to double-check every solution you write, but you should make sure it is readable.

[quote=jlordhe][quote=LoveMaths26102003]Moderators are the ones with green usernames??[/quote]

Blue usernames are site admins, red usernames are forum admins.[/quote]
Blue usernames with the [aops]X[/aops] at the beginning, that is.

[quote=RopuToran]Question: Is that OK when posting a geometry problem without any figure ?[/quote]

Yes that's perfectly fine. In fact the P6 of IMO 2018 had no figure in the official statement of the problem.

[quote=RopuToran]In IMO the problem was checked for several times :) But here, posting a figure will help other people know if the statement are true quickly.[/quote]

Although I agree that a picture can be worth more than a thousand words, but if someone can give a good description of what's going on in the problem then there is no need to add a picture. But if someone thinks that the problem will be easier to understand from the figure then adding a figure can be golden.
In other words, it is not necessary to add a figure, but if you want you can add one.

I want to disagree with (a), sometimes titles like "Chinese TST 2003, Problem 1" are useful, as it allows other users who took the test to find discussion threads easier.

I personally dislike 'creative titles' because they sometimes spoil the problem. Calling 2018 AIME II #14 "Projective Fun" would be a prime example, as it describes a method people would use to solve the problem.

[quote=hydrohelium]Dear Maththinkers,

I too agree with ZetaX but I believe contest name are good titles.[/quote]

Yeah, I think the same.

@above You can already specify that in "source".

I do agree with ZetaX, but I also agree with @above2 & @above3.

But, there are a lot of questions of contests that are not from the moderators of admins. Then, how can we post the problem in contest?(like amcs)

cause they want to get their HSO posts up. to look like they're smart.

[quote=skywalker321]why would you post amc questions in hso when it says olympiads in it hmm[/quote]

no common sense


a,b,c>0 and abc=1

prove that:

2048(a²+b²+c²)>=(a+b+c)*8

thanks

I think your inequality is not true!!
because we can choose a very great!!!

excuse me
think your kihe_freety5  inequality is

a,b,c>0 and abc=1

prove that:

2048(a²+b²+c²)=<(a+b+c)*8

thanks


calculus
derivative
analytic geometry
graphing lines
slope

please give me some problems of the derivative!

you can make em up easily like find the slope of the tangent of
f(x) = x/(1 + x^2) at point (3,3/10).

Why don't you go look through the posts in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=296]calculus[/url] section of this site?

thanks i am a new member so i don't know enough!

well not all the problems there are as simple as finding derivatives, like most are very complicated, but if u can handle those than yeah..


Within a month or so, the Wii and the Playstation 3 will be coming out into the market.  How do you think they will do, in comparison to each other and the X-Box 360?

I saw in a paper that Nintendo is putting its focus on gameplay more than power.  This will make the Wii substantially cheaper and more affordable than the PS3, which will be at least twice as much.

To me, the quality of the games is the essential part, not how fancy you can get with them and how much power they are backed by.  I mean, games such as Pac-Man and Super Mario Bros. were popular back in the 80s, when technology wasn't as good as it is now.  Even today, they're still good games, even though there are better ones.

What am I saying?  I think the Wii will do pretty well.  What do you think?

I think the Wii will pwn all others, then PS3 will pwn Xbox360.
The ps3, bah it's just a ps2 with better graphics. But wii is like a brand new style of gaming, and it does have goo graphics, too. And some of its games look very nice too. the new super smash brothers, zelda, metroid,blablabla.
About the 360, it's not really that great at all. compared to the PS3, it has worse graphics, and it seems that the PS3 games are going to OWN 360 games.

[quote="PTynan89"]I LOVE HOW SONY IS SO CREATIVE IN NAMING THEIR GAME SYSTEMS!![/quote]
Heh heh...

Anyway, I like the controller idea of the Wii.  That's something that the other two systems don't offer.  I can't wait until it comes out.

What are you guys planning on getting?

Wiiiiiiiiiiiiiiiiiiiii

Mainly because I'm stingy, partly because it's got a cool controller, but mostly because it has Super Smash brothers coming out on it

If I _were_ to get one, which I'm not, because I neither game nor feel like spending that much money on a addon computer, I'd get a PS3. Because Yellow Dog Linux has partnered with Sony to allow people to run Linux on the PS3. Pretty lame reason, but I don't have any other reason to prefer one over the others.

The thing is, while Sony's hardware may be high tech but still breaks on a regular basis, they've had smarter marketing techniques than Nintendo and secured basically most gaming companies and that's the reason the PS2 outperformed the GameCube is because there were no good games for the GC. Also, the Wii's idea may seem innoative, but with all innoative things, there's a risk that your average player may not like the change and then you have a failure system which in turn, will give Sony and Microsoft an even bigger market share for many future generations of consoles as well.

Well, everything has a risk.  I think people will go for the Wii just because it's different.

[quote="goldendomer"]I think the Wii is going to be the best. It's just too expensive though and not really worth it.[/quote]

I don't think the wii is to expensive!!  $250$ extremely cheap compared to the other new gaming systems!!!!!


geometry
rectangle
geometric transformation
rotation
circumcircle
geometry unsolved

PQRS is a rectangle and inside it we draw another rectangle ABCD which is similar to PQRS (The sides of these two rectangle are not necessarily parallel). Let E,F,G,H be the midpoint of segment AP,BQ,CR,DS.

Show that EFGH is also a rectangle similar to PQRS.
[img]http://i5.photobucket.com/albums/y194/jelly119/001.jpg[/img]

First of all, complex number works fine for it.
Here is a hint for a geometric proof: try to work on the special case where the corresponding sides are parallel first.

I am not sure if this was your intention, but the diagonal intersections of the similar rectangles $PQRS \sim ABCD$ do not have to coincide and the rectangle $EFGH$ is still similar to the rectangles $PQRS, ABCD$. The rectangle $ABCD$ is obtained from the rectangle $PQRS$ by a transformation called dilative rotation or spiral similarity. Let $M, N$ be the intersections of the lines $PQ \cap AB$ and $PA \cap QB$. The rotation center $O$ is the intersection of the circumcircles of the triangles $\triangle MPA, \triangle MQB$ other than the point $M$, which is identical with the intersection of the circumcircles of the triangles $\triangle NPQ, \triangle NAB$ other than the point $N$. It follows that the triangles $\triangle OPA \sim \triangle OQB \sim \triangle ORC \sim \triangle OSD$ are all similar and they are obtained from each other by a dilative rotation around the rotation center $O$. The segments $OE, OF, OG, OH$ are medians of these similar triangles. From this, we conclude that the triangle triples $\triangle OPQ \sim \triangle OAB \sim \triangle OEF$, $\triangle OQR \sim \triangle OBC \sim \triangle OFG$, $\triangle ORS \sim \triangle OCD \sim \triangle OGH$, $\triangle OSP \sim \triangle ODA \sim \triangle OHE$ are similar. Since the sides $OP, OA, OE$ are common to the triangle triples $\triangle OSP \sim \triangle ODA \sim \triangle OHE$ and $\triangle OPQ \sim \triangle OAB \sim \triangle OEF$, etc. by cyclic exchange of the vertices $P, Q, R, S;\ A, B, C, D;\ E, F, G, H$, we deduce that

$\frac{OE}{OP} = \frac{OF}{OQ} = \frac{OG}{OR} = \frac{OH}{OS}$ [color=white].[/color] and [color=white].[/color] $\frac{OE}{OA} = \frac{OF}{OB} = \frac{OG}{OC} = \frac{OH}{OD}$

In addition, the following angular sums are equal (in this case all to $90^o$):

$\angle OPS + \angle OPQ = \angle OEH + \angle OEF = 90^o$

$\angle OQP + \angle OQR = \angle OFE + \angle OFG = 90^o$

$\angle ORQ + \angle ORS = \angle OGF + \angle OGH = 90^o$

$\angle OSR + \angle OSP = \angle OHG + \angle OHE = 90^o$

It follows that the quadrilateral $EFGH$ is a rectangle obtained from the rectangles $PQRS$ or $ABCD$ by a dilative rotation with the rotation center $O$. Note that the similar quadrilaterals $PQRS, ABCD$ (with the corresponding sides not parallel and the diagonal intersections not coincident) do not have to be rectangles. The midpoints $E, F, G, H$ of the segments $PA, QB, RC, SD$ form a quadrilateral $EFGH$ similar to the quadrilaterals $PQRS, ABCD$.

OK, here is my proof.
Rotate ABCD about A to obtain a new rectange AB'C'D' s.t. AB'//PQ. E',F',G',H' are defined similarly. A simple angle chasing and segment computation shows that $EFGH\cong E'F'G'H'$.
I think the conclusion of this problem can be generalised to n-gon, but I don't have time to finish now(I'm having computer lesson at my school now :D )


calculus
integration
trigonometry
complex analysis

Compute by residue calculus the integral
\[ \int_{-\pi}^{\pi}\frac{\cos{\theta}}{\sqrt{2}+\cos{\theta}}d\theta \]

universal substitution tg(q/2)=u.
 :wink:

[quote="HilbertThm90"]The standard trick is to let $ z=e^{i\theta}\Rightarrow d\theta=\frac{dz}{iz}$ and now you have $ \cos\theta = \frac{1}{2}\left(z+\frac{1}{z}\right)$ to put this together you get

$ \int_{-\pi}^\pi \frac{\cos\theta}{\sqrt{2}+\cos\theta}d\theta= \int_{|z|=1}\frac{1/2(z+1/z)}{iz (\sqrt{2}+1/2(z+1/z))}dz$ Do you want to see further? I haven't worked this out to know if it works, but if this is a standard by the book problem it should.[/quote]
Yes please explain further

$ I = \int^{\pi}_{-\pi}\frac{\cos \theta}{\sqrt{2}+\cos \theta}\,d \theta$.

Change variables according to $ \theta = \pi-t$ and so $ I = \int^{2 \pi}_{0}\frac{\cos t}{\cos t-\sqrt{2}}\,dt$.
Now, as Hilbert suggested, make $ z = e^{it}$, $ dt = \frac{dz}{iz}$, $ \cos t = \frac{1}{2}(z+1/z)$, and so

$ I = \oint_{C}\frac{z+1/z}{z+1/z-\sqrt{2}}\cdot \frac{dz}{iz}= \frac{1}{i}\oint_{C}\frac{z^{2}+1}{z(z^{2}-\sqrt{2}z+1)}\,dz$,

where C is the unit circle centered in the origin. Now you just have to calculate the poles of the integrand, see if they are included in C, compute the residue on each pole, and then use the residue theorem.


I'm sure most of you guys have seen problems like this, but anyway:

Two real numbers are chosen randomly between 0 and 10, inclusive. What is the probability that they differ by no more than 5.

To people like ComplexZeta who probably know the solution, remember to post in spoiler!!!

I have this feeling that this needs calculus in order to be solved and I don't know calculus...:(

Actaully, there is ABSOLUTELY no calculus needed. you can sovle this one with 7th grade math.... :D

I think I got it.

[hide]Start with 0 as one of the numbers picked. the probability that the second number differ by more than 5 is 1/2.

Then move on to the next number, x, whereas 0:le:x:le:5. If a number, y, which is less than x, is picked, then that combination have already been examined when x=y. Thus you only have to think of numbers greater than x. Since x is less than or equal to 5, the possibility that the second number (greater than x) differ by more than 5 is 1/2. Until here, I'm sure that it's correct, but I get wobbly from here on...

When x is greater than 5, the possibility should be 1/2 because if you graph the possibility, it should come up to be an isosceles right triangle with a leg of 1...(I'm not sure about this part)

Thus the probability that the second number differ by more than 5 is 1/2.[/hide]

A lot of guess work involved with the above process...

HMm, well I did the problem and got a different answer:

[hide]I used a geometrical approach: Let us draw a square composed of 100 unit squares, and let one side represent the first number and let one of its adjacent sides be the second nubmer. If the first number is 0, then the second number cannot be in the range (5,10]. And if the first nubmer is 10, then the second nubmer cannot be in the range [0,5). Given this information, we can mark off an isosceles right triangle of legs 5 in the lower right corner of the square. Similarly, we can also mark off an isosceles right triangle in the upper left corner of the square. Thus, the polygon left in the middle comprises of the total area of all the ordered pairs (1st #, 2nd#) which satisfy our conditions. Therefore, its area is 100-25, or 75. And thus the final probability is 75/100 or 3/4.

[/hide] [/hide]

you click the little "Spoiler" button before you type your explanation... once you are finished, click the "Spoiler*" button.

Dark, your solution seems like it would work, but count your possibilities carefully.. 0 - 10 is 11 numbers, which gives 121 square units .

My answer:[hide] 21/52 [/hide]

O shoot. You got a point

HEy, i just said 3/4 is the answer. I didn't say it is the [i]correct[/i] answer  :D

[hide]The only points in the square that don't work are the points that satisfy x>y+5 and y>x+5 in the square 0<=x<=10, 0<=y<=10. The area of each of those pieces is 12.5 since each is a right isosceles triangle with side 5. Thus the area that doesn't work is 25. The area that does work is 75. The total area is 100. Thus the probability is 75/100=3/4.[/hide]

A related problem that makes routine appearances on competitions is something like this:

You have a piece of string of length 10. You randomly cut the piece of string in two pieces. What's the probability that the three pieces will form a triangle?

[quote="darkquantum"]hmm, apparently, i don't kno how to use the spoilers.

LoL. SOmeone plz enlighten me?[/quote]
You could also highlight the part where you want to use the spoilers and click on "Spoiler"

LoL
ComplexZeta, it appears you made the same mistake I did. The length of each side of the square is 11, not 10. Read the previous posts.

o wait, no

Ur rite.

And I was rite, before. Hmm, Spoon, there are 11 numbers between 0 and 10, but the length of the side of the square is nonetheless 10.

SO, i guess i was rite  :D , and then wrong  :( , and then rite again. :D


Find all the points on earth that when you go to South 10 miles, East 10 miles, and North 10 miles, you are back to the starting point .
Hint: more than one answer  :lol:
This problem is hard and requires some thoughts

[hide="a couple of answers"](1) The north pole.
(2) The set of all points 10 miles north of a latitudinal circle having a circumference of $ \frac {10}{k}$ miles for some $ k\in \mathbb{N}$ (a countably infinite number of latitudinal circles all slightly more than 10 miles north of the south pole).

Depending on how you define 'go east 10 miles' at the poles, and how you define 'go north/south 10 miles' at points closer than 10 miles to the poles, there may be more answers.[/hide]


Find the integer that is closest to $ 1000 \sum_{n=3}^{10000}\frac{1}{n^{2}-4}.$

[quote="4everwise"]Find the integer that is closest to $1000 \sum_{n=3}^{10000}\frac{1}{n^{2}-4}.$[/quote]
[hide]$\frac{1}{(n+2)(n-2)}$
Applying the substitution $n=x+2$
$\frac{1}{x(x+4)}$
Doing a decomposition gives us that
$\frac{1}{x(x+4)}=\frac{1}{4x}-\frac{1}{4x+16}$
So what we want is
$1000((\frac{1}{4}-\frac{1}{20})+(\frac{1}{8}-\frac{1}{24})+...+(\frac{1}{39992}-\frac{1}{40008}))$
Telescoping, we get:
$1000(\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{16}-\frac{1}{39996}-\frac{1}{40000}-\frac{1}{40004}-\frac{1}{40008})$
We can ignore the last four since they don't change the value much. This value is closest to $\boxed{521}$.
[/hide]
EDIT: chinesechess beat me to it barely...

You could also remember the fact that $\frac{1}{n(n+k)} = \frac{1}{k}(\frac{1}{n}-\frac{1}{n+k})$ if you don't want to use partial fraction decomposition.

Here's my full solution which I'll also put in the wiki:
Using the fact that $\frac{1}{n(n+k)} = \frac{1}{k} ( \frac{1}{n}-\frac{1}{n+k} )$ or by partial fraction decomposition, we both obtained $\frac{1}{x^2-4} = \frac{1}{4}(\frac{1}{x-2}-\frac{1}{x+2})$. The denominators of the positive terms are $1,2,..,9998$, while the negative ones are $5,6,...,10002$. Hence we are left with $1000 \cdot \frac{1}{4} (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{9999} - \frac{1}{10000} - \frac{1}{10001} - \frac{1}{10002})$. We can simply ignore the last $4$ terms, and we get it is approximately $1000*\frac{25}{48}$. Computing $\frac{25}{48}$ which is about $0.5208..$ and moving the decimal point three times, we get that the answer is $521$

When I did this problem, I just deemed the last four "terms" useless since they were ridiculously small, which is generally safe but not $100\%$ certain that this will change any potential rounding. Therefore it suffices to find the closest integer to $250\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)$.

[b]Solution:[/b] Use PFD to get that $\frac{1}{n^2 - 4} = \frac{1}{4} \cdot (\frac{1}{n-2} - \frac{1}{n+2})$. Then you can cancel out specific fractions and get that the answer is just $250 \cdot (1 + 1/2 + 1/3 + 1/4 - 1/9999 - 1/10000 - 1/10001 - 1/10002) \approx 250 \cdot (1 + 1/2 + 1/3 + 1/4) \approx \boxed{521}$.


geometry
analytic geometry
circumcircle
LaTeX
complex numbers

I brought all the math problems I had in Edmonton over to my residence in Waterloo this winter. It was enormous amount of problems, 8kgs of problem sheets and about 15kgs of problem books.

I decided to type and categorize the problem sheets, currently working on geometry problems. It is going very slow, but I'm certainly making progress.

I'll post some problems I could't solve once in a while. Please help me by solving these problems.

Note that I already solved UNG0001-03, UNK0002-06, UNK0002-02, UKG0001-11. I just didn't bother to put it out.

Perhaps IMO Shortlist.  One of them is at least from IMO, if I recall correctly.

[quote="lightrhee"][quote="kunny"][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=132953]Double Posting[/url] :mad:[/quote]
Sorry, I didn't mean to. I don't know how it happened.[/quote]
 :)

[quote="rem"]Where are these problems from? :maybe:[/quote]
It's totally random. Some are from math camps, some are from internet, some are from friends, and some are from Dr. Liu.

[b]Solution to #1[/b] Let the lowercase letters represent the complex coordinates of the uppercase letters. Let the circumcenter $O$ of $\triangle ABC$ be the origin of the complex plane.
\[\\d=\frac{b+c}{2}\\e=\frac{1}{2}\left(\frac{b+c}{2}-\frac{ac(\overline{b}+\overline{c})}{2R^{2}}+a+c\right)\\f=\frac{d+e}{2}=\frac{\frac{b+c}{2}+\frac{1}{2}\left(\frac{b+c}{2}-\frac{ac(\overline{b}+\overline{c})}{2R^{2}}+a+c\right)}{2}\]
After some (:P) computations (left for the reader since I am lazy to type it), you obtain
\[(f-a)\cdot(e-b)=0,\]
where $a\cdot b=\frac{1}{2}\left(\overline{a}b+a\overline{b}\right)$ is the real product of the complex numbers $a,b$.$\blacksquare$

[quote="mathgeniuse^ln(x)"]Anyone know how to do #1[/quote]
Non-bashing solution for #1

Let X be the midpoint of BD. Since BE is parallel to FX, it is sufficient to prove that $\angle XFA = 90^{\circ}$.

Let line AF meet BC at Y, XF meet AD at P, BE meet AD at Q, then
DY:YC
= $\triangle$ ADF : $\triangle$ AFC
= $\triangle$ AEF : $\triangle$ AFC (DF=DE)
= AE:AC
= AE:AB (AC=AB)
= EQ:BQ (AD is bisector of A)
= FP:XP

Therefore, $\triangle XFA \sim \triangle CDA$ and we are done.

i don't know if you have done unk0008-04 :)  anyways here is how it goes...
$PM^{2}=AM^{2}=MR*MB$
THEN $\triangle MRP\sim \triangle MPB$,
and $\angle MPR=\angle RBC=\angle CSR$
then CS||AP

edit: lol, in latex now

Here's a complex number solution to UKG0001-07.  I apoligize in advance for my horrible latex skills.  Wolog,  $ABCDEF$ is positively oriented and $r=1$.  Let the lowercase letter denote the complex coordinate of the uppercase letter.  Wolog (in degrees):
$a=1, b=cis(60) , c= cis(60+y), d= cis(120+y), e= cis(120+y+z), f= cis(180+y+z).$

Then $p= (cis(60)+cis(60+y))/2, q= (cis(120+y)+cis(120+y+z))/2$, and $r= (cis(180+y+z)+1)/2$.

Now the problem is really easy if know (or can show) that $PQR$ being equilateral is equivalent to $p+xq+x^{2}r = 0$ where $x=cis(120)$.  I am not sure if this is well-known or not.  However, it is not difficult to prove.  So the proof will end here.

There are two more sets like this on [url=http://etotheipi1.googlepages.com]my website[/url] now. Help is appreciated on these problems.

Hi, David, which ones have u already solved?

Are these problems actually hard??? :roll:


Incircle of$\triangle{ABC}$ tangents side $AB$ at $D$. $E$ is arbitrary point on side $AC$ Prove that incircles of $\triangle{ADE}$ , $\triangle{BDE}$, $\triangle{BEC}$ have a common tangent

See the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=81808]Tangent[/url].


calculus
integration
calculus computations

I have trouble solving these types of integration problems. Please help.

[img]http://img301.imageshack.us/img301/1229/calccgh2.jpg[/img]

I'm pretty sure it deals with ln

[quote="jmerry"]Yes, the antiderivatives involve the natural logarithm.

Write $ \frac1{8x}$ as $ \frac18\cdot\frac1x$; that might help you recognize what's going on.[/quote]

thanks for the hint. I think I got it.


AMC
USA(J)MO
USAMO
AIME
FTW
AMC 12
AMC 8

[quote="SnowEverywhere"]I was just curious as to how many people in this forum has made the listed competitions. Please let me know if you guys feel as though this poll is too invasive or if this is the wrong place to ask this question. Thank you. :)[/quote]

Invasive? I don't think people mind about saying that they made USAMO or IMO or something. :wink:

[quote="SarahAhmed"]Uhhh......Well I haven't! lol

I bet a lot of people on this site probably have. There's a bunch of geniuses on AoPS.[/quote]

Less than you think...

There are 61795 users on AoPS and 40753 of them have 0 posts...

Okay, people join AoPS to improve at math. By the way, that would mean I am way below average, I am capable of scoring 9 on AIME at most.

[quote="CatalystOfNostalgia"][quote="SnowEverywhere"]Wow! 3 IMO team people already (I did not know that that many IMO participants check this forum).[/quote]

Clearly, all three of them [i]actually[/i] made the IMO team.[/quote]

Well duh, I don't think that even needs to be stated.

[quote="not_trig"]lol this is a slightly unfortunate poll[/quote]

QFT

[quote="AIME15USAMO"]I sense sarcasm...but it is highly likely that all IMOers (all over the world) have an AoPS. They aren't necessarily on the USA IMO team. Then again, it could just be random people lying to try to be funny.[/quote]

...then, what would have been the whole point of making this poll anyways? I guess this should be locked.

BTW, on topic of the poll, I only made AIME. Hopefully I should make USAMO once I steer clear of those stupid mistakes.

[quote="AIME15USAMO"]I sense sarcasm...but it is highly likely that all IMOers (all over the world) have an AoPS.[/quote]

wait, were you being serious here?

no i wasnt being serious, were u being serious?

[quote="AIME15USAMO"]A 9 on the AIME isn't something to be ashamed of. That is a score only about the top .1% (probably even less) of US high school students could achieve. This would make you a genius in about 99.99% of American high school students' eyes.[/quote]

I know...

A 9 on AIME...
I meant if all of AIME was IMO qualifiers, THEN I would be terrible...

Also.. from getting a 21 on AMC 8 and 93 on AMC 10 and going to state MC in 2009 (4th chapter) then everyone sort of thinks I am a genius

[quote="CatalystOfNostalgia"]My prediction: more than half of voters will choose "made IMO team"[/quote]
Self-fulfilling prophecies FTW!

Also, to the OP, you could just look at the [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:AoPS_Community_Awards#IMO_Participants_and_Medalists]Wiki.[/url]

Thanks. But how many of them still check this forum and post actively?

Darn, I fail. Currently, my official scores do not entitle me to even vote in this poll.

Although, I am probably at ~6 AIME level and ~116 AMC12 level (maybe more) [well, I did take AMC 12B 2002 and scored a 132 on that, but it was definitely too easy for today's level]

[quote="AdithyaGanesh"]Wait a second AwesomeToad, if you can get a 9 on AIME,  how come you only got 93 on AMC 10?

 :huh:[/quote]

Last year I got a 93. Now I get like 144s or 150. (MY lowest this year was 136.5)

I took an AIME and got an 8.

[quote="An0maly"]Good to know we're up to 16 IMO people xD[/quote]  :huh:
Yes, but most of those people were definately lying.

apparently it is pretty difficult to detect sarcasm here on teh interwebz

I wish I had found out about this whole thing at least a year earlier...

it would sweet to make MOSP... but it's not the type of thing you do on only a few months preparation.  oh well


algebra
polynomial
combinatorics proposed
combinatorics

Let $ a_1,a_2 ,..., a_{100} $ and $ b_1,b_2,..,b_{100} $ be 200 distinct real numbers  . Construct a table $ 100 * 100 $ with $ a_i + b_j $ written in the i-th row and the j-column . Suppose that the product of the entries of each column is $ 1 $ . Prove that : the product of the entries of each row is $ -1 $

Cute :).

From the hypothesis we find the polynomial identity $(x+a_1)\ldots(x+a_{100})=(x-b_1)\ldots(x-b_{100})+1$, and the conclusion follows immediately.

that is because if $(a_1+b_i)(a_2+b_i)\cdots(a_n+b_i)=1$ for each $i$, then the polynomial $(x+a_1)(x+a_2)\cdots(x+a_n)-1$ has $b_1,b_2,\ldots,b_n$ as $n$ distinct roots.


vector
linear algebra
matrix
superior algebra
superior algebra unsolved

Let $V$ be a vectorspace over $F_{p}$. Find conditions on the dimension of $V$ such that the Frobenius automorphism $x \mapsto x^{p}$ is diagonalizable over the algebraic closure of $F_{p}$.

[quote="Soarer"]I'm sorry if this is a stupid question, but what is the Frobenius map? Isn't it defined on a field only? or is V is a field over $\mathbb{F}_{p}$?[/quote]

It is defined for rings with prime characheristics as well.

In any case, $V$ needs multiplication, not just a vector space structure. The OP should clarify.

I implicitly meant: let $V$ be a finite field and consider this as a vectorspace over $F_{p}$. You can pick a basis for this vectorspace, and give sense to the matrix associated with the frobenius map, on this manner


Prove that \[\sqrt[n]{n+1}>1+\frac{1}{\sqrt[n]{n!}}\] for $n\in\mathbb{N}.$


[quote="Ihatepie"]well the answer to that is 67/99

What you do is take the number that repeats (in that case 67) and put it over an equal number of 9s. so 67/99.

.527527527527... would be 527/999[/quote]

THANK YOU!!!!!!!!!!!!!!!!!!!!!!!!!!!!! THIS WUZ GREAT HELP!!!!!!!!!!!!!!!!!!!!!!!!!!

One small note... if there are zero(s) before the repeating portion, multiply the repeating 9 number by the power of ten w/ the same # of zeros...
Example 0.0033333333333333333333333333333..... = 3/900 = 1/300

If you wish, take this to the general case, $ .\overline{a_{1}a_{2}a_{3}...a_{n}}$, where the $ a_{i}$ are digits.


calculus
derivative
limit
calculus computations

Show that $f(x)$ is not differentiable at $x=0$

${f(x) = \begin{cases} |x|^x, &x \ne 0 ;}\\ 1, &x =0. \end{cases}$

this is my solution:

$f(x)$ is differentiable at $0$ if $\lim_{x\to0}f'(x)$exists.
for $x < 0$, $f(x) = (-x)^x$, $f'(x) = (-x)^x (ln(-x) +1)$

for $x >0$, $f(x) = (x)^x$, $f'(x) = (x)^x (lnx +1)$
using L'hopital rule, $\lim_{x\to0}x^x = 1$ and $\lim_{x\to 0}(-x)^x = 1$
so, for both $x >0, x <0$, $\lim_{x\to0}f'(x) = -\infty$
hence the limit does not exist => not differentiable


inequalities
topology
inequalities unsolved

Denote:   f(u,v) =  ((abs(u-v)) / ((sqrt(1+u^2)(1+v^2));
Prove that for all x,y,z are pairwise we have:
         f(x,y)+f(y,z) > f(x,z)

I rewrite the problem:
Denote $f(u,v) = \frac{|u-v|}{\sqrt{(1+u^2)(1+v^2)}}$
Prove that for all $,y,z$ are pairwise we have:
$f(x,y)+f(y,z) > f(x,z)$
I remember that $R$ with mapping f is metric space

We have to prove that

$\frac{\left|v-w\right|}{\sqrt{\left(1+v^2\right)\left(1+w^2\right)}}+\frac{\left|w-u\right|}{\sqrt{\left(1+w^2\right)\left(1+u^2\right)}}\geq\frac{\left|u-v\right|}{\sqrt{\left(1+u^2\right)\left(1+v^2\right)}}$

holds for any three reals u, v, w.

We work in the Euclidean plane. Let g be a line, let O be a point which has the distance 1 to the line g, and let P be the orthogonal projection of the point O on the line g. Then, $\left|OP\right|=1$. Let A, B, C be points on the line g satisfying PA = u, PB = v and PC = w, where we are using directed segments. Then, $\left|BC\right|=\left|PC-PB\right|=\left|w-v\right|=\left|v-w\right|$, and similarly $\left|CA\right|=\left|w-u\right|$ and $\left|AB\right|=\left|u-v\right|$, while in the right-angled triangle OPA, the Pythagoras theorem yields $\left|OA\right|=\sqrt{\left|OP\right|^2+\left|PA\right|^2}=\sqrt{1^2+PA^2}=\sqrt{1+u^2}$, and similarly $\left|OB\right|=\sqrt{1+v^2}$ and $\left|OC\right|=\sqrt{1+w^2}$. Now, applying the Ptolemy inequality to the quadrilateral OACB, we get $\left|BC\right|\cdot\left|OA\right|+\left|CA\right|\cdot\left|OB\right|\geq\left|AB\right|\cdot\left|OC\right|$. Using the formulas above, this becomes

$\left|v-w\right|\cdot\sqrt{1+u^2}+\left|w-u\right|\cdot\sqrt{1+v^2}\geq\left|u-v\right|\cdot\sqrt{1+w^2}$,

or, equivalently, $\frac{\left|v-w\right|}{\sqrt{\left(1+v^2\right)\left(1+w^2\right)}}+\frac{\left|w-u\right|}{\sqrt{\left(1+w^2\right)\left(1+u^2\right)}}\geq\frac{\left|u-v\right|}{\sqrt{\left(1+u^2\right)\left(1+v^2\right)}}$, and we are done.

  Darij


Of the 400 eighth-graders at Pascal Middle School, 117 take algebra, 109 take advanced computer, 114 take industrial technology. Furthermore, 70 take both algebra and advanced computer, 34 take both algebra and industrial technology, and 29 take both advanced computer and industrial technology. Finally 164 students take none of these courses. How many students take all three courses?

Using the Principle of Inclusion-Exclusion (go [url=http://www.artofproblemsolving.com/Wiki/index.php/Principle_of_Inclusion-Exclusion]here[/url] for more information about the principle)
\[ 400 \minus{} 164 \equal{} 117 \plus{} 109 \plus{} 114 \minus{} 70 \minus{} 34 \minus{} 29 \plus{} n
\]

\[ 236 \equal{} 207 \plus{} n
\]
$ n \equal{} \boxed{29}$.


parameterization
induction
real analysis
real analysis unsolved

Hi :)

I have some questions ...

Is given sequence bounded ? convergent ? Discuss.

1) $ x_1 \equal{} x,\quad x_{n \plus{} 1} \equal{} 4x_n(1 \minus{} x_n)$

2) $ a_1 \equal{} a,\quad a_{n \plus{} 1} \equal{} a_n(a_n \minus{} 1)$

The answers should depend on $ a,x$ (for which $ x,a$ is bounded, for which convergent)

The first one is the logistic map; [url]http://en.wikipedia.org/wiki/Logistic_map[/url]. You can consider the more general case where $ x_{n \plus{} 1} \equal{} cx_n(1 \minus{} x_n)$, which exhibits some very interesting and chaotic behaviour depending on the value of $ c$, especially as $ c$ approaches 4. If $ c \equal{} 4$, then for $ x > 1$ the sequence $ x_n$ diverges. If $ x < 1$ then all sorts of weird things can happen, limit cycles and whatnot, or it can converge to a finite limit. I don't know if you can say anything conclusive about it regarding dependence on the initial parameter.

I must correct you hadjimurad. The sequence converges for all $ x \in [0,1]$ and diverges in other cases. The chaotic behavior starts to kick in when approaching the Feigenbaum constant - [url]http://mathworld.wolfram.com/FeigenbaumConstant.html[/url] that is when the number of period doubling bifurcations reaches $ \plus{} \infty$  - this is known as the Fiegenbaum attractor and has a structure like Cantor set. On the note of it converging, for a parameter larger then $ 3$ and a non trivial $ x_0$, it can never converge, but it always converge when the value of the parameter is smaller than $ 3$ (and here, the negative values are excluded). Also, both of them are logistic maps, but the second one has parameter $ \minus{} 1$ - usually the parameter of the logistic map is in the interval $ [1,4]$ - this is where the fun stuff happens.

[quote="milin"]The sequence converges for all $ x \in [0,1]$ and diverges in other cases.[/quote]

Could you explain why $ x_{n \plus{} 1} \equal{} 4x_n(1 \minus{} x_n)$ converges for any $ x_0 \in [0, 1]$? I tried to convince myself why or why not but I couldn't come up with anything right away. For some values of $ x_0$ it settles to a fixed point immediately, but for others it seems to have multiple subsequential limits, although admittedly that's based on some numerical fiddling around.

On another note, Mitchell Feigenbaum is a friend of my family's  :blush:

let me try :)

if $ x\in[0,1]$ then $ 0\le x_n\le 1$ for all $ n$ because we can use Cauchy's ineq. for $ x_n$ and $ 1\minus{}x_n$ and estimate $ x_{n \plus{} 1}$.
and next, if $ x\in[0,1]$ then iterating $ f(x) \equal{} 4x(1 \minus{} x)$ will lead us to zero (I think so, .... thinking on a formal proof now)

I think the proof that it diverges for other $ x$ is much more difficult (I have no idea how to prove it) ... could you show the proof, or any example (special case) ?

Would you care to share what are these points beside $ 0$, $ \frac{1}{2}$ and $ 1$? I must warn you.. When the parameter of a logistic map is higher than Fiegenbaum's constant, there are no fixed points but for some special cases (in windows) because the infinitive number of period doubling bifurcations broke them into infinitely many points. You have a period of size $ \infty$. May be I'm missing something..  :huh:

It does not approach $ 0$ for all $ x\in[0,1]$, and in general does not approach any limit except for countably many initial values.  For one thing, $ x \equal{} \frac {2\pm 1}{4}$  are immediate counterexamples, as are $ \frac {5\pm\sqrt {5}}{8}$, $ \frac {2\pm \sqrt3}{4}$, and infinitely many other algebraic seed values for which the iteration settles into a finite period.  It is also not true, however, that the iteration approaches $ 0$ only for $ 0$, $ \frac {1}{2}$, and $ 1$.  It can be shown by induction that the iteration also approaches $ 0$ for numbers of the form
\[ \left\lbrace\frac {2\pm\sqrt {2}}{4},\frac {2\pm\sqrt {2\pm\sqrt {2}}}{4},\frac {2\pm\sqrt {2\pm\sqrt {2\pm\sqrt {2}}}}{4},...\right\rbrace
\]
Furthermore, it approaches $ 0$ in finitely many iterations only for initial values of this form and $ \left\lbrace0,\frac {1}{2},1\right\rbrace$.  More generally, I suspect it shouldn't be hard to prove that there are only countably many initial values for which the period is finite, and for the rest the orbit is aperiodic.


S  =  Sqrt(1 +2* Sqrt(1 + 3*Sqrt(1+ 4* Sqrt(1+5*sqrt(1+6*sqrt(1+..................... upto infinity

find the sum  S .
please help

[hide="Solution Sketch"]This is Ramanujan's nested radical. The actual proof of the generalization of this problem is a bit long, so here is a solution sketch, which may or may not be satisfying:

We claim that $ \sqrt {1 \plus{} 2\sqrt {1 \plus{} 3\sqrt {1 \plus{} 4\sqrt {1 \plus{} \cdots}}}} \equal{} 3$. Notice that the LHS is bounded from above:

$ \sqrt {1 \plus{} 2\sqrt {1 \plus{} 3\sqrt {1 \plus{} 4\sqrt {1 \plus{} \cdots}}}} < \sqrt {3\sqrt {4\sqrt {5\sqrt {6\cdots}}}} < \sqrt {3\sqrt {9\sqrt {27\sqrt {81\cdots}}}}$ where $ \sqrt {3\sqrt {9\sqrt {27\sqrt {81\cdots}}}} \equal{} 3^{1/2 \plus{} 2/4 \plus{} 3/8 \plus{} 4/16 \plus{} \cdots} \equal{} 3^2 \equal{} 9$, with the penultimate equality following from an evaluation of the sum of the arithmetic-geometric series.

This means that the increasing sequence of nested radicals $ \sqrt {1 \plus{} 2}, \sqrt {1 \plus{} 2\sqrt {1 \plus{} 3}}, \sqrt {1 \plus{} 2\sqrt {1 \plus{} 3\sqrt {1 \plus{} 4}}}, \dots$ is bounded and thus has a limit $ \leq 9$.

Now, just use the identity $ n^2 \equal{} 1 \plus{} (n \minus{} 1)\sqrt {(n \plus{} 1)^2}$:

$ \sqrt {3^2} \equal{} \sqrt {1 \plus{} 2\sqrt {4^2}} \equal{} \sqrt {1 \plus{} 2\sqrt {1 \plus{} 3\sqrt {5^2}}} \equal{} \cdots$, as desired. (Qualitatively speaking, we are getting ever the closer to the desired limit.

If you desire, I (or someone else) may post a more complete proof later (which might consider the solution to $ f(x) \equal{} \sqrt {1 \plus{} x\sqrt {1 \plus{} (x \plus{} 1)\sqrt {1 \plus{} \cdots}}} \Leftrightarrow f(x)^2 \minus{} 1 \equal{} xf(x \plus{} 1)$ by a similar though rigorized approach of iteratively bounding to obtain $ f(x) \equal{} x \plus{} 1$).[/hide]

Thanx for the solution .but i have found a more interesting one . please advice whether the following is correct.

x = sqrt{1+(x-1)(x+1)}
x+1 = sqrt{1+(x)(x+2)}
x+2 = sqrt{1+(x+1)(x+3)}

So
x = sqrt{1+(x-1)*sqrt{1+(x)*sqrt{1+(x+1)...}

Hence x = 3.


When the length of an object is given to the nearest quarter-inch, the measurement could be off by a maximum of what part of an inch? Express your answer as a common fraction.

$ \frac{1}{8}$, which is pretty trivial.


topology
real analysis
real analysis unsolved

I'd like to know if the Borel sets of $ \mathbb{R}^n \minus{} \{0\}$ with the usual topology are exactly the sets $ B \minus{} \{0\}$ where $ B$ is a Borel set in $ \mathbb{R}^n$. I've managed to show that $ B \minus{} \{0\} \in \beta_{\mathbb{R}^n \minus{} \{0\}}$ if $ B$ is a Borel set in $ \mathbb{R}^n$ but couldn't show the other inclusion. Is it true? It seems reasonable to me...

show that the set $ B \setminus \{0\}$ is a sigma-algebra.  it contains all the open subsets of $ \mathbb{R}^n \setminus \{0\}$, so it contains the sigma-algebra generated by them.


inequalities
advanced fields
advanced fields unsolved

Let $M$ be a closed convex set of real normed linear space $X$. Show that
\[\inf_{z\in M}\|x-z\| = \sup_{f\in X^\ast\atop \|f\|=1}\left\{f(x)-\sup_{z\in M}f(z)\right\}, \quad \forall x\in X. \]
$X^\ast$ is the space of all bounded linear functionals on $X$.

===

I have proved this:
For all $x\in X\setminus M$ there exists $f_{1}\in X^\ast$ satisfying $\|f_{1}\|=1$ and $\sup_{y\in M}f_{1}(y) \leq f_{1}(x)-d(x)$, where $d(x) = \inf_{z\in M}\|x-z\|$.

For the other direction, use the inequality $f(x)-f(z)\le \|x-z\|$, in which you should take inf on both sides.

what about $x\in M$?
I don't see why the right-hand side is 0.

In general it isn't (example: $M$ is the closed unit ball, $x=0$). Two ways to fix this: (a) assume $x\notin M$; (b) change $\|f\|=1$ to $\|f\|\le 1$. I prefer (b).


Let triangle $ ABC$ such that
$ \sqrt{2}(a\plus{}b\plus{}c)\equal{}(\sqrt{a\plus{}b}\plus{}\sqrt{a\minus{}b})(\sqrt{a\plus{}c}\plus{}\sqrt{a\minus{}c})$
Find triangle ABC

[quote="Algadin"]Let triangle $ ABC$ such that
$ \sqrt {2}(a \plus{} b \plus{} c) \equal{} (\sqrt {a \plus{} b} \plus{} \sqrt {a \minus{} b})(\sqrt {a \plus{} c} \plus{} \sqrt {a \minus{} c})$
Find triangle ABC[/quote]
 $ \sqrt {2}(a \plus{} b \plus{} c) \equal{} (\sqrt {a \plus{} b} \plus{} \sqrt {a \minus{} b})(\sqrt {a \plus{} c} \plus{} \sqrt {a \minus{} c})\Leftrightarrow b^2 \plus{} c^2 \equal{} a^2.$ Nice problem! :lol:

[quote="great math"]Could you show more steps aquady? It seems that we cannot derive that equality for the first line  :wink:[/quote]

$ \sqrt {2}(a \plus{} b \plus{} c) \equal{} (\sqrt {a \plus{} b} \plus{} \sqrt {a \minus{} b})(\sqrt {a \plus{} c} \plus{} \sqrt {a \minus{} c})\Leftrightarrow$
$ \Leftrightarrow(a\plus{}b\plus{}c)^2\equal{}2\left(a\plus{}\sqrt{a^2\minus{}b^2}\right)\left(a\plus{}\sqrt{a^2\minus{}c^2}\right).$
But $ (a\plus{}b\plus{}c)^2\minus{}2\left(a\plus{}\sqrt{a^2\minus{}b^2}\right)\left(a\plus{}\sqrt{a^2\minus{}c^2}\right)\equal{}$ $ \equal{}b^2\plus{}c^2\minus{}a^2\plus{}2a\left(b\minus{}\sqrt{a^2\minus{}c^2}\right)\plus{}2a\left(c\minus{}\sqrt{a^2\minus{}b^2}\right)\plus{}$
$ \plus{}2\left(bc\minus{}\sqrt{(a^2\minus{}b^2)(a^2\minus{}c^2)}\right)\equal{}$
$ \equal{}(b^2\plus{}c^2\minus{}a^2)\left(1\plus{}\frac{2a}{b\plus{}\sqrt{a^2\minus{}c^2}}\plus{}\frac{2a}{c\plus{}\sqrt{a^2\minus{}b^2}}\plus{}\frac{2a^2}{bc\plus{}\sqrt{(a^2\minus{}b^2)(a^2\minus{}c^2)}}\right).$


The deadline is over. So I think that it is ok to post the problem here:

Fing all positive intergers $ (m,n)$ such that $ \frac {m^4\plus{}n^2}{7^m\minus{}3^n}$ is interger.

[quote="mathVNpro"]The deadline is over. So I think that it is ok to post the problem here:

Fing all positive intergers $ (m,n)$ such that $ \frac {m^4 \plus{} n^2}{7^m \minus{} 3^n}$ is interger.[/quote]
$ 4 \mid 7^m\minus{}3^n$
Assume that $ n$ is odd. Then $ m$ is odd as well and $ m^4\plus{}n^2 \equiv 2 \bmod 4$, contradicition. So $ 2 \mid m,n$.
Let $ m \equal{} 2x,n\equal{}2y$. Then $ 7^m\minus{}3^n \equal{} (7^x\plus{}3^y)(7^x\minus{}3^y)$
Now $ m^4\plus{}n^2 \equal{} 4(4x^4\plus{}y^2)$. Hence $ 7^x\plus{}3^y < 4x^4\plus{}y^2$ - from here it is easy.

[quote="mathVNpro"]The deadline is over. So I think that it is ok to post the problem here:
Fing all positive intergers $ (m,n)$ such that $ \frac {m^4 \plus{} n^2}{7^m \minus{} 3^n}$ is interger.[/quote]
It's IMO 2007 Shortlist N1. See here http://www.mathlinks.ro/viewtopic.php?p=1106459#1106459


Find the exact perimeter of a rectangle whose area is 40 and diagonal is 12.

[hide="Solution"]
We have that \[x=\frac{40}{y}\text{ and }x^{2}+y^{2}=144 \implies (x,y) = \left( \sqrt{ 72+16 \sqrt{14}}, \sqrt{ 72-16\sqrt{14}}\right) \]
\[\implies 2(x+y) = \left( 4 \sqrt{2}\right) \left( \sqrt{ \left( \sqrt{7}+\sqrt{2}\right)^{2}}+\sqrt{ \left( \sqrt{7}-\sqrt{2}\right)^{2}}\right) = \boxed{ 8 \sqrt{14}}\] [/hide]

Different idea
[hide]well, if xy=40 and x^2+y^2=144, then x^2+y^2+2xy=(x+y)^2=224, so x+y=4sqr14 so 2(x+y)=perimeter=8sqr14[/hide]

[quote="rnwang2"][hide="Maybe"]
$ab=40$.

$a^{2}+b^{2}=144$.

$(a+b)^{2}=224\implies a+b=4\sqrt{14}$.

Thus, the perimeter is $8\sqrt{14}$.[/hide][/quote]

Oh, that's a lot easier than mine, though this problem doesn't take too long to do either way.  :)

Hmm, nvm.. I guess rnwang has alreayd covered my solution.. as everyone posted at same time:8:47..


limit
real analysis
real analysis theorems

If $\lim_{x \rightarrow c}f(x)+g(x)$ does exist , it is posible for $\lim_{x \rightarrow c}f(x)$ doesn't exists and $\lim_{x \rightarrow c}g(x)$ exists.

My teacher state(without proof) that there is no such example exists. But i still don't believe her..

EDit: its Alright now i got it (after 45 second posting this  :blush:  :blush:  :blush: )

$\lim_{x \rightarrow c}-g(x)$ exists, and by linearity rules $\lim_{x \rightarrow c}(f(x)+g(x))-g(x) = \lim_{x \rightarrow c}f(x)$ exists


Show that if a and b are positive integers, then

(a+1/2)^n + (b+1/2)^n

is an integer for finitely many positive integers n

note that n can not be even(by considerations modulo 4) :)
so suppose n is odd
x^n+y^n=(x+y)[x^(n-1)-x^(n-2)y+.......+y^(n-1)]
we have to find integers  a,b,n  such that (2a+1)^n+(2b+1)^n is divisble by 2^n
so plugging x=2a+1 and y=2b+1 in the above relation and noting the fact the number of terms in the huge factor on the rhs is odd (sum of odd number of odd terms) we get that 2^n must divide 2a+2b+2 or 2^(n-1) must divide a+b+1.
this relation only gives finitely many n
qed
nice problem by donald newmann :D
cheers


Question. [b]When three numbers $880$, $1238$, and $2133$ are divided by $d$ ($d>0$), they all have remainder of $r$. Find all possible values of $d-r$.[/b]

I believe it is MATHCOUNTS level.

[hide]1238-880=358 which is 2x179, 2133-1238=895 which is 5x179. The GCF of 358 and 895 is 179.179=1x179 So the possible number n is 1 and 179. The remainder for all the numbers divided by 1 is 0 so n-r=1. The remainders of all the numbers divided by 179=164. So 179-164=15. The possible answers are 1 and 15.[/hide]

I find this problem a possible state/national target round question


What is the predominant religion in Africa?  I'm interested to know what most people believe in.  I have a few African friends at school, and they are all Muslim.  Is that the general population?

[url="http://library.stanford.edu/africa/religion.html"]Stanford Research[/url]

From Wikipedia (based on Stanford research):
[quote]Africans profess a wide variety of religious beliefs, with Christianity and Islam being the most widespread. Approximately 46.3% of all Africans are Christians and another 40.5% are Muslims. Roughly 11.8 percent of Africans primarily follow indigenous African religions. A small number of Africans are Hindu, or have beliefs from the Judaic tradition, such as the Beta Israel and Lemba tribes. [/quote]

The religion "make-up" greatly changed after the "so-called" in the late nineteenth century. Scramble for Africa.

In North Afrika are all muslim ( Tunizia, Egypt, Maroko,Algir etc.), in medium of Afrika have muslim, christians, hindu, pagans etc, in South Afrika have very christians!


limit
logarithms
calculus
integration
real analysis

1/1000000+1/1000001+...+1/2000000= (approximately)?

How many numbers are from 1 to 1000000 (one million) such that x^2+x+1 is a a multiple of 12?

[hide="1"] It's easy to show that $\lim_{n \to \infty}\sum_{k=n}^{mn}\frac{1}{k}= \ln m$.  The desired quantity is approximately $\ln 2$. [/hide]
[hide="2"] None.  $x^{2}+x+1$ is always odd. [/hide]

2: Oops. Yeah. I didn't actually solve it.

x^2+x?

[quote="t0rajir0u"][hide="1"]It's easy to show that $\lim_{n \to \infty}\sum_{k=n}^{mn}\frac{1}{k}= \ln m$.  The desired quantity is approximately $\ln 2$.[/hide][/quote]

Just wondering, is it possible to show that without the integral definition of natural logs?

[quote=DanK]t0rajir0u wrote:
Just wondering, is it possible to show that without the integral definition of natural logs?[/quote]
Not that I know of. The best approximation you can get is a minimum of 1/2, which is enough to prove that 1+1/2+1/3+... is infinite, but not close enough for this problem.

You can do a series of approximations bounding the turns a variety of ways, but an integral is the simplest, fastest, and most accurate way.

It's a Riemann sum in disguise, so I doubt it.

[hide="Explanation"] $\begin{eqnarray*}\lim_{n \to \infty}\sum_{k=n}^{mn}\frac{1}{k}&=& \lim_{n \to \infty}\sum_{k=n}^{mn}\frac{1}{ \frac{k}{n}}\cdot \frac{1}{n}\\ &=& \int_{1}^{m}\frac{1}{x}\, dx \\ &=& \ln m-\ln 1 \\ &=& \ln m$ [/hide]


function
real analysis
real analysis solved

Let $a_n$ be a sequence (with $n$ starting at $0$) such that
$a_0 = 1$ and
For $n\ge 1, \sum_{k=0}^n a_{n-k}a_k = \frac{(-1)^n}{n!}$

Find $a_n$.

I found  $a_n= 1/n!.(-1/2)^n$ Am I correct ?

That was quick. Care to show us your explanation?

Nice problem for generating functions.

\[A(x)=\sum_{n=0}^\infty a_n x^n.\]
\[A(x)^2=\sum_{n=0}^\infty \frac{(-x)^n}{n!}=e^{-x}.\]

Therefore $A(x)=e^{-x/2}$, and so $a_n=\frac{(-1/2)^n}{n!}$.

$(\sum a_nx^n)(\sum a_px^p) = \sum_{n\geq 0}(\sum_{k=0}^{n}a_ka_{n-k})x^n=\sum (-1)^n.1/n!x^n=e^{-x}$

According $a_0=1$

$\sum a_nx^n = e^{-x/2}=\sum (-1/2)^n.1/n!x^n$ unicity of the coefficient of

power series => $a_n=(-1/2)^n.1/n!$

How do you go from $\sum a_n = \sum (-1/2)^n \cdot 1/n!$ to $a_n = (-1/2)^n \cdot 1/n!$?

In proposing it, I was thinking of it as a generating function problem, so ComplexZeta's post is exactly what I would have said about it.

Ravi B has a point: there are many different series that add up to the same sum. I think that to nail down the uniqueness, you do have to introduce a variable, making it a generating function problem.

You could change the value of  the convolution, $\sum_{k=0}^na_{n-k}a_k$, to several other sequences, as long as those sequences are the coefficients of functions for which you could find the power series of the square root. The original I copied this from had $\sum_{k=0}^na_{n-k}a_k = 1$.


The [url=http://www.nyas.org/snc/podcasts.asp]NYAS podcast[/url], from which I have posted several links before, has a great segment called "My Life as A Quant" ([url=http://www.nyas.org/podcasts/snc/derman.mp3]direct link[/url]).

This is a great preview of Emanuel Derman's upcoming lecture at NYAS on October 19.  [url=http://www.nyas.org/events/eventDetail.asp?eventID=7442&date=10/19/2006%206:00:00%20PM]Check it out[/url] if you're in the area.

[quote="NYAS"]
In the late 1970s, Emanuel Derman left academia to put his background in theoretical physics to work on Wall Street. Derman, now director of Columbia University's program in financial engineering, spoke with Science & the City in a preview to his upcoming lecture at the New York Academy of Sciences on October 19.
[/quote]

I have read parts, and my son has read all, of Derman's book. It's a great read.


$ \log_ab \equal{} \frac{1}{\log_ba}$

I'm having trouble with the proof, any hints or step-by-step guides would be greatly appreciated.

Also, this problem:
If the value of $ 2^x\plus{}2^{\minus{}x}$ is $ \sqrt[3]{3}$ then find the exact value of $ 8^x\plus{}8^{\minus{}x}$

Give the final result in its simplest form.

$ \log_a{b} \equal{} \frac {\log{b}}{\log{a}} \\
\log_b{a} \equal{} \frac {\log{a}}{\log{b}}\\\frac{\log{b}}{\log{a}}\equal{}\frac{1}{\frac{\log{a}}{\log{b}}}$

[hide="Solution"]Squaring $ 2^x \plus{} 2^{ \minus{} x}$, we have:
$ 4^x \plus{} 4^{ \minus{} x} \plus{} 2 \equal{} 3^{\frac {2}{3}} \Rightarrow 4^x \plus{} 4^{ \minus{} x} \equal{} 3^{\frac {2}{3}} \minus{} 2$

Multiplying the LHS by $ 2^x \plus{} 2^{ \minus{} x}$, and the RHS by $ \sqrt [3]{3}$, we have:
$ 8^x \plus{} 8^{ \minus{} x} \plus{} 2^x\plus{}2^{\minus{}x} \equal{} 3 \minus{} 2\sqrt [3]{3} \Rightarrow \boxed{8^x \plus{} 8^{ \minus{} x} \equal{} 3 \minus{} 3\sqrt [3]{3}}$[/hide]


complex analysis
function
limit
complex analysis unsolved

Suppose $f$ is a holomorphic function from the open unit disc $D$ to $\mathbb{C}$, and $\lim_{r \to 1}f(re^{i\theta}) = 0$ for all $\theta \in [0, 2\pi]$, i.e., the radial limit of the function is everywhere 0. Prove that $f$ is identically 0.

(I saw a sketch of the proof of this today. I think it was first proved by Lusin and Privalov. Anyone interested in trying it? I think it's a cool problem, altho I haven't had a course in complex and hardly understand it. ;) )

Hello,

if I have not missunderstood the conceptual formulation - my proposal:
You can interpret the radial limit for all \theta as an curve; so it's possible to use the identity theorem. That's all.

Regards
?

It's not that simple. Your argument would apply if we had that $f$ were continuous on the closed disc $\overline{B(0;1)},$ but that's not what the hypotheses said. Note that nothing was said about uniform convergence - just that the radial limit exists. That's not a strong enough hypothesis to imply continuity at the boundary - unless you happen to have already proved the theorem at hand.


I've finished my undergraduate program in mathematics, and I'm planning to take a graduate program in a challenging college. However, first, I'm going to Africa for one year, to do some volunteer work, along with my parents. During this year, I'm not in the circumstances to study at all (there are just not the right living circumstances).

Therefore I wanted to know how this will affect my math skills. During the 3 years of college I noticed a hugh progress of problem solving skills (and, secondary, an increased knowledge). Will my problem solving skills fade very quickly, and do I need to start all over again, and will I be way below the requirments of graduate school? Or is learning mathematics, like say riding a bike, it's a learned skill and you never forget how to do it. But you need to freshen it up after a while.

Has someone ever taken a long time off from mathematics and returned?

I'm probably not in the position to give real advice, since I am relatively poor at math.
I have not done any math out of school for a year, and the math in school doesn't really take any thinking, so I basically have not thought about math for a year. Upon coming back to doing math again, I found I actually got better by a little. I can't explain that, but I do think that if you truly did gain a very good understanding of something in math, you probably won't forget it.

Touching up skills is probably important though.

I took off a year before grad school.  I studied for quals before classes started, and by the first day of school, it was like I never left.  I don't recall if I had to shake off much rust when I first started studying for quals.  In short, I wouldn't sweat it.  Bring a couple math books to Africa for light bedtime reading.  Even if you don't *really* learn it, it never hurts to see a little extra math.

A lot of problem solving skills is crystalized; meaning, the ability to recognize patterns and opportunities to apply known facts in interesting ways. You won't lose this, though if you take a long break, you might need a few days to "get back in the mode". But that's not much at all.

Then there's general mental dexterity, agility, and awareness. You won't lose this by not doing [i]math[/i], as long as you're exercising your brain in [i]some[/i] kind of problem solving. Anything intellectual helps. If you find yourself slightly less clever or fast when you get back, this is what you'll need to build back up.

Finally, there's knowledge of the subject matter. You'll probably need a quick refresher when you get back -- leafing through a few books should do it.

I think TZF has the right idea.  I'd also like to add that if you're young enough, your brain might still be developing, which could be a reason some people get better at math after a long break: they mental ability simply increases due to brain development.

A person that is a grad student in my year (and hence in several of my classes) took several years off before coming. The first day of classes things were assumed, like the definition of a group etc. He actually couldn't remember them. Everything came back so quickly for him that by the end of the week (maybe a little longer, but definitely not longer than two weeks) it was like he had taken no time off. Now, he is probably the best person in our year.


number theory
greatest common divisor
analytic geometry
geometry

Can someone explain to me why the number of lattice points in a grid on the line connecting points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is equal to $\mbox{GCD}(x_{2}-x_{1}, y_{2}-y-1)-1$?

This fact was used to solve this problem here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=363529#p363529

[quote="vishalarul"]Isn't a lattice point a point with integer coordinates? :maybe:[/quote]
Yes, it is. :D

Can someone please explain why the above expression holds?

Wlog, let $(x_{1},y_{1}) = (0,0)$ since you can slide the line along lattice points without changing the properties. And for clarity, $(x_{2},y_{2}) = (a,b)$.

This line has the equation $y = \frac{b}{a}x$, which is simplified as $y = \frac{\frac{b}{gcf(a,b)}}{\frac{a}{gcf(a,b)}}$ and will be an integer when x is a multiple of the denominator, $\frac{a}{gcf(a,b)}$. Since x ranges from 0 to $a$, this will occur a number of times equal to gcf(a,b).

For example... x would be a multiple of a/3 at a/3, 2a/3, and 3a/3. $a$ itself is overcounted, so subtract 1.

Thank you, that is a very good explanation.

btw, I have noticed a lot of kids from Vestavia Hill area on AoPS and they are all good at math.

[quote="breez"]
btw, I have noticed a lot of kids from Vestavia Hill area on AoPS and they are all good at math.[/quote]

Alot?

name 4.

90% of the VHHS kids aren't on the math team and probably are terrible at math...the creator was just from Vestavia so he group invited everyone. I'd be surprised if there were more than three people on AOPS from Vestavia.

Well if he is very astute he may notice me, elem, reinhardt, john, wei, etc. But only elem and maybe reinhardt post regularly. Maybe I'll start again...


search
analytic geometry
inequalities
Harvard
college
articles

There are the Taco Bell tomato farmers in the US being treated very harshly.  Have any of you heard of Bond Labor Liberation Front (BLLF)?  This is a group against harsh labor on anyone

Could you perhaps provide a link to what you are talking about?

In the 70's, there was a boycott on lettuce and grapes, because the farmers for grapes and lettuce were treated like slaves and were often sprayed with pesticide from planes with the pilots not realizing the farmers were working there.  Plus, they were supervised by a soldier-like attendant.  If they tried to run away, they would get shot by him.  When American citizens realized this, they refused to eat lettuce or grape for a whole year.  Then, after the bosses of the grape and lettuce changed their policy, people started to eat grapes and lettuce again.

I am not for removing Taco Bell, although I do not go there very often, I am for having their policy changed.

I've never eaten at Taco Bell... not cause i have something against it - i'd actually like to, but I just never have for some reason.

Um, I heard about the 70s thingy but not this one...

personally it really doesn't affect me, there isn't Taco Bell for miles here. (In fact I haven't seen one in over 4 years) so doesn't affect me.

I'd vote for the boycott if I must choose because I have this bad feeling about Taco Bell and McDonalds and Burger King and such...

BOYCOTT TACO BELL!! YEAH!! BETTER WORKING CONDITIONS!

i've never eaten at taco bell, so no real loss to me

[quote="frodo"]For another story, check out this site:

http://www.thirdworldtraveler.com/Heroes/Iqbal_Masih.html[/quote]
That has ABSOLUTELY nothing to do with Taco Bell. However, after some Googling, I found [url=http://www.ciw-online.org/]this link[/url] which appears to be the official site for the workers.

In layout, it's reminiscent of the propaganda-dissemination sites run by the government of the DPRK, anti-war protestors, environmentalists, and others - it's chaotic, and few of the links actually "tell" you anything. Brightly colored images and accusing headlines seek to grab attention and sway people to the cause - but there's a problem.

It took some clicking, but I found a statement on "why you should boycott Taco Bell." There are tomato farmers in Florida who are payed between 40 and 50 cents [i]by their employer[/i] per bucket of tomatoes picked. The site claims that farmers must pick approximately a ton of tomatoes a day to earn $48, making it sound like they are picked by hand, and makes the socialist argument that it isn't fair that big executives make that much before they sit down in the morning. Taco Bell purchases tomatoes from the employer of these farmers. Look carefully. Taco Bell is not the corporation paying these farmers. Also, we live in a country in which those farmers are free to strike or go get a job elsewhere if they don't like the conditions.

As for being payed 40 to 50 cents per bucket of tomatoes picked - that actually sounds [i]pretty darn good[/i] when you consider that [i]tomatoes are not picked by hand commercially[/i]. That would be ridiculous. They are harvested with mechanical pickers (developed in 1959) using harvesters that can pick hundreds of tons of tomatoes in a normal working day.

Taco Bell is not at fault here. Personally, I think it sounds like the farmers are whining - they're not picking those tomatoes by hand, so a ton a day is hardly a big deal. But if it was, they're free to go get another job. With links such as [url=http://news.google.com/news?hl=en&ie=UTF-8&oe=utf-8&q=taco+bell+boycott&btnG=Search+News]this one from Google[/url] (4 links at the time of posting), it's easy to tell that the news media doesn't consider this to be a significant thing either. I'm not going to boycott Taco Bell over this. In fact, I think I might go eat there tomorrow, because of the people that are boycotting it.

[quote]In the 70's, there was a boycott on lettuce and grapes, because the farmers for grapes and lettuce were treated like slaves and were often sprayed with pesticide from planes with the pilots not realizing the farmers were working there. Plus, they were supervised by a soldier-like attendant. If they tried to run away, they would get shot by him. When American citizens realized this, they refused to eat lettuce or grape for a whole year. Then, after the bosses of the grape and lettuce changed their policy, people started to eat grapes and lettuce again. [/quote]

I have some trouble believing that is true, at least if it happened in the U. S. If the soldier - like attendent shot anyone, whether or not they were running away, he would be commiting homocide and would be charged with this crime by the police. So this kind of enforcement was and is impossible (at least in the US.) Also, as for the planes, was it really that difficult to coordinate spraying and other work done on lettuce/grape fields? This seems like an issue of disorganization and a lack of coordination, not cruelty. So maybe you could tell us how you got this information?

[quote="tetrahedr0n"]I have some trouble believing that is true, at least if it happened in the U. S. If the soldier - like attendent shot anyone, whether or not they were running away, he would be commiting homocide and would be charged with this crime by the police. So this kind of enforcement was and is impossible (at least in the US.) Also, as for the planes, was it really that difficult to coordinate spraying and other work done on lettuce/grape fields? This seems like an issue of disorganization and a lack of coordination, not cruelty. So maybe you could tell us how you got this information?[/quote]

Actually, this is quite well known.  Many of the workers were Mexican illegal immigrants or other migrant workers.  [url=http://www.incwell.com/Biographies/Chavez.html]Cesar Chazez[/url] was instrumental in winning some rights for the workers, and was helped by a major boycott.  I don't know that much about this, actually, but it would be worth going and looking information about it up.  Just another lesson in "Bad things happen, even in the United States."

So, as I said in my previous post, I don't know too much about American agriculture/labor history in general, and certainly not about this issue.  Nevertheless, I can give you some non-content related criticism.  For example:

[quote="confuted"][quote="frodo"][b]For another story[/b], check out this site:

http://www.thirdworldtraveler.com/Heroes/Iqbal_Masih.html[/quote]
That has ABSOLUTELY nothing to do with Taco Bell.[/quote]
(emphasis mine.)

Now, granted, he posted an unrelated website, but he did in fact say that it was unrelated in his posting of it.  So your act of pointing out that it was unrelated was silly.

[quote="confuted"]Taco Bell purchases tomatoes from the employer of these farmers. Look carefully. Taco Bell is not the corporation paying these farmers. Also, we live in a country in which those farmers are free to strike or go get a job elsewhere if they don't like the conditions.[/quote]

Now, there are two arguments there, and I'll take the first one on first.  That's just patently absurd.  Most agriculturalists don't sell retail, they only sell wholesale.  It isn't possible to boycott specific farmers, unless they happen to label their goods with their farm name.  The point here is the long-established politics of pressure -- try to get Taco Bell to drop the supplier or force them to set regulations on the treatment of workers in companies supplying their goods.  This is [b]exactly[/b] the tactics that have been used, repeatedly and successfully, to force sweatshops to treat their workers well.

Now, as for the second point, ("we live etc."), that's obviously false.  Imagine you were living so that you were earning just enough money to keep yourself clothed, fed and roomed.  Striking would mean almost certainly being fired, and if not, it would involve going without pay for possibly months.  Workers in this country are decidedly [i]not[/i] free to just form unions when they want, and it's particularly difficult in communities of migrant workers.  (That's what made what Cesar Chavez did so amazing.  But it's not something easily repeated.)  Furthermore, without capital, you can't just get up and move.  You need some method of transport and somewhere to go, where you'll be able to find a place to stay and a job.  I recommend you to Barbara Ehrenreich's book [url=http://www.amazon.com/exec/obidos/ASIN/0805063897/qid=1077294301/sr=2-1/ref=sr_2_1/002-2641935-4512046]Nickled and Dimed[/url], which is an (admittedly very unscientific) look at actually living on a minimum-wage job.

[quote="confuted"]I think I might go eat there tomorrow, because of the people that are boycotting it.[/quote]
Now this is just silly and childish.  If there is a boycott, your money won't help.  And why should Taco Bell need your help anyhow?  If there isn't a serious boycott, you'll be wasting your money.

And while I'm at it,

[quote="confuted"]There are tomato farmers in Florida who are payed between 40 and 50 cents [i]by their employer[/i] per bucket of tomatoes picked. The site claims that farmers must pick approximately a ton of tomatoes a day to earn $48, making it sound like they are picked by hand, and makes the socialist argument that it isn't fair that big executives make that much before they sit down in the morning.
As for being payed 40 to 50 cents per bucket of tomatoes picked - that actually sounds [i]pretty darn good[/i] when you consider that [i]tomatoes are not picked by hand commercially[/i]. That would be ridiculous. They are harvested with mechanical pickers (developed in 1959) using harvesters that can pick hundreds of tons of tomatoes in a normal working day.[/quote]
Now, lets be reasonable for a minute -- do you think, reasonably speaking, that if the site said that farmers had to pick a [i]ton[/i] (or 2000 pounds) of tomatoes every day to make [i]$48[/i], they're trying to make it sound like they're doing it by hand?  Give me a break.  That's (again) just absurd.  So your further insults of the farmers, because they have machines, are just you attacking a position that isn't actually held by anyone, and then claiming you've explained why the farmers are just whining.

Now, backing up a minute and approaching your (implicit) attack on the "socialist argument" that inequality is bad:

economists -- capitalist economists -- think that inequality is bad.  The capitalist ideal, after all, is that through capitalism, [i]everyone[/i] will end up with everything they need.  The real argument among economists here is what the solution is.  Marty Feldstein, a conservative economist here at Harvard, for example, is of the opinion that the "solution" to inequality is to ignore inequality and focus on elliminating poverty instead.  Fine.  But no one thinks that inequality should be ignored -- there isn't a single economist on the face of this earth who thinks that if Bill Gates owned [i]everything[/i] (and I mean here literally every property, corporation and good currently in existence) that, even if the average wealth were much higher than today, that would be a good thing.  So somewhere in the middle is a point where inequality itself becomes a problem -- where exactly is a matter of debate.  I personally think we're at that point now.  I'm sure you disagree.  But don't pooh-pooh inequality as if it doesn't matter for anything.

You'll note that the workers, earning $48 a day, will earn just more than $17,000 -- [i]if they work 365 days a year[/i].  That's not very much, considering they're doing serious manual labor.  And don't kid yourself about tomato-picking machines -- it's not as if these guys are sitting inside drinking and watching the machines pick.  They're doing real work, all day, in the sun, for poverty wages.

Finally, having now gone and taken a good look at their webpage, I am willing to agree with you that they might benefit from some better webdesigners, but I would also like to point out that they have numerous links to a variety of news source's coverage of their actions, such as their [url=http://www.ciw-online.org/tz_site-revision/home/home.html#ASBTBNews]anti-slavery actions[/url] which have helped result in convictions of several business owners, among other things.  They have a [url=http://www.ciw-online.org/tz_site-revision/press/press-main.html#highlights]link[/url] to a collection of news articles about the boycott, for example.  Thus falls another baseless accusation.

Being a moderate - I agree with about half of what confuted said, and half of what JLB said. So I'll just take the position of arguing with both of them. :)

First off, on the utility and morality of Boycotting: In general the decision to boycott something is a much more complicated game-theoretic decision than you might realize. It's effectiveness really depends on how effectively organized it is, and many other people [i]really[/i] care about your message. One of the dangers of boycotting ineffectively is that the institution or business that you're trying to target could decide that it's better for them to suck it up, and end up passing it's burden on to the very people you are trying to help - either by lowering their wages even more or by completely laying them off to avoid the controversy.

A good example of an ineffective boycott is the NAACP's boycott of South Carolina for flying the confederate flag on its Statehouse grounds. First of all, hardly anyone even knows or cares about the boycott; and the little effect that it does have mainly robs small black businesses of the income that the boycotters would otherwise have brought to them.

Confuted makes a good point that in a free society the workers aren't [i]forced[/i] to work there. To a large degree, they work there by choice. For American citizens, there are labor protection laws as well as unemployment and welfare benefits to help people in job transition should they decide to quit. So for American citizens to work in conditions like that, it's either a matter of choice or a matter of unfamiliarity with their labor rights.

The problem, of course, comes with undocumented workers who aren't eligible for such benefits and protections. The argument could be made that they take that risk by choosing to work here as undocumented workers. Many of them are willing to take that risk because, making $48 dollars a day is still more than they would make in their native country. For instance, a friend of mine who is from Mexico, and has a job as what would be considered a professional there, only makes slightly more than that when you figure it out. And that is considered a pretty good wage there.

And if the primary motivation of employers hiring undocumented workder is to get around fair labor practices, then enforcing those practices would only diminish the reason for hiring them in the first place and might just eliminate their jobs altogether. It all depends on how much competition there is for those jobs. But this is a good reason to give undocumented workers legal worker status so that they can at least be eligible for labor law protections, and not subject to human rights abuses here in the United States.

But regardless of untilitarian considerations, boycotting also has a moral element. Just like people who don't eat meat based on moral reasons (it's not like they are going to realisticaly stop the beef industry), but it makes them feel better morally to stand up for that cause. And maybe it will raise at least some awareness of animal rights issues and of what [i]really[/i] goes on behind the scenes of meat processing.

As for the question of whether wealth inequality is bad, or how bad it is: that is a very hotly debated issue among economists and sociologists. It has been for many centuries. By nature, pure capitalism will always allow resourceful or skilled people to accumulate large amounts of wealth. That's because wealth is a positive feedback system; with wealthy, you can easily accumulate more wealth at an exponential rate! But the accumulation of wealth doesn't necessarily come at the expense of other people's livelihood, as is commonly believed.

A good example of this is sports. Most people think that big name professional athletes are paid too much for what they do - as opposed to say, the average everyday guy who sells hotdogs at the games. But the big name athletes are the ones who bring the people into the stands, and sit them down in front of their televisions to watch games! If there weren't any hotdog vendors - or if there were different people doing it, people would still watch the games. But if there weren't any big name athletes, or if you tried to replace them with anyone else, no one would watch. So the athletes unique skills are what create the wealth for the whole system that employs everyone - from the ticket collectors, to the hotdog salesmen, to the advertisers.

So while they're getting paid an extraordinary amount compared to everyone else - they're also increasing the wealth and value of everyone associated with the team or the sport. Basically, someone getting rich does not necessarily come at the expense of anyone else. If you asked most people, however, they would honestly tell you that they believe people are poor because someone else is hogging up all the money and resources! Just listen to politicians rhetoric! Economics isn't a zero sum game, where wealth is a fixed amount that's gets divied up amongst a set number of people. Just ask John Nash!

There are arguments to be made against pure monopolies though. I believe that some monopolies can actually be good in some particular circumstances. But since we can't trust anyone to have the wisdom, or the honesty, to decide which ones are good and which ones are bad - it's good that we have laws restricting them. They can be very dangerous. Just as important as the principle of market freedom, are the principles of checks and balances and fair competition. These are what keep democracy and capitalism healthy.

In the end, would I boycott Taco Bell for these alleged abuses? Probably not (although I don't like TB to begin with), because I don't think what they are doing arises to the level of human rights abuse. Also because I'm not convinced of their level of culpability, or even of how much good a boycott would do in this circumstance. What I [i]would[/i] do, is to lobby my congressman for stricter enforcement of labor laws, better welfare benefits for those in job transition, and for at least some labor protections for undocumented workers.


function
calculus
integration
trigonometry
college
abstract algebra
real analysis

This is false. The functions $ \sin(nx),n\in\mathbb{N}$ do not span a dense subset of $ C[0,2\pi];$ they span a proper subspace that is "about half" of that space.

Let $ f\in\text{span}\{1,\cos x,\cos(2x),\cos(3x),\dots\}.$ Then those integrals are zero but there are surely many nonzero $ f$ in this span.

The problem is from the Pre-lim section of Pugh's Real Mathematical Analysis book.  I believe they are supposed to be from first year Berkeley graduate exams.  Does anyone know the correct version of this problem.  Also, it seems that not sleeping overnight really does make most, if not all, brains not function well.

[quote="boxedexe"]Does anyone know the correct version of this problem..[/quote]

I don't know what the correct version is.  But it should be true if you require $ f$ to be odd as well (then all of the fourier coefficients of $ f$ are $ 0$, so ...)

e: okay here's a silly question.  i don't have any fourier analysis texts handy, so i was asking myself why this was true (that $ 0$ fourier coefficients imply that the function is identically $ 0$) just now.  what i came up with was that fejer's theorem implies that the cesaro means $ s_nf$ converge uniformly to $ f$.  On the other hand, the cesaro means are all $ 0$, so $ f\equal{}0$.  is there any easier proof?

That's the easiest proof I know. You could use Abel means (Poisson kernel instead of Fejer kernel) or you could use some other summability method but this is really the way to go.

The result is similar with it:
If $ f\in C[a,b]$, and if $ \int_a^bx^nf(x)dx\equal{}0$ for each $ n\equal{}0,1,2,...$ then $ f\equal{}0$

Sure.  Here the classical version of Stone-Weierstrass is used in place of the version for trigonometric polynomials.

[quote="boxedexe"]Let $ f\in C^0([0,2\pi],\mathbb{R})$ be such that $ \int_0^{2\pi}f(x)\sin (nx)\ dx \equal{} 0\ \forall n\in\mathbb{N}$.  Prove that $ f\equiv 0$.

The Stone-Weierstrass Theorem is certainly helpful here.[/quote]

The true version is, i think that:
Let $ f\in C^0([0,2\pi],\mathbb{R})$ be such that $ \int_0^{2\pi}f(x)\sin (nx)\ dx \equal{} 0\ \forall n\in\mathbb{N}$ and $ \int_0^{2\pi}f(x)\cos (nx)\ dx \equal{} 0\ \forall n\in\mathbb{N}$.  Prove that $ f\equiv 0$
(this rezult is due to Lebesgue)

again, the hypotheses imply that all fourier coefficients of $ f$ are $ 0$, so...

The proof that I know (that if $ f\in C^0(\mathbb{T})$ and $ \hat{f}(n) \equal{} 0\forall n$, $ f\equal{}0$) is the following:
Define $ F(t) \equal{} \int_0^tf(t)dt$.  Then $ F\in C^1(\mathbb{T})$ and $ F' \equal{} f$.  Also, $ F$ is $ 2\pi$-periodic.  So $ 0 \equal{} \hat{f}(n) \equal{} \hat{F'}(n) \equal{} in\hat{F}(n)\implies F(x) \equal{} \hat{F}(0)\implies h \equal{} 0$.
This may be the same proof that blahblahblah mentioned.


Let $A \in C^{n\times n}$ be an invertible matrix and suppose that, for some $m \geq 1$, $A^{m}$ is diagonalizable.
Show that $A$ is diagonalizable.

This is one of the few problems that really needs $\mathbb{C}$; we must make use of both the fact that the field is algebraically closed and that it has characteristic zero.

Look at the Jordan form for a matrix which isn't diagonalizable. What happens when we raise it to a positive power?

Hi,
I still don't understand what are you trying to say. Could you say more about it?


limit
trigonometry
LaTeX
calculus
calculus computations

Hi
I need help for this problem.
what values of $r$ so that this limit \[ \lim_{x \rightarrow 0} \frac{x^r \sin{\frac{1}{x}}}{\sin^2{x}} = 0  \]

I believe that it is $r > 2$, but i still don't know what happen if $r=2$, it seem that we need to find $r$ so that $x^r$ force the $\sin{\frac{1}{x}}$ tend to zero more nominate  than the $\sin^2{x}$ force it to $\infty$ or undefined, i'm sorry if i can't explain this better. :ninja: .
I don't how to get $r > 2$ with a nice way. it seem reasonable for me.

Is it
$\lim_{x \rightarrow 0} \frac{x^r \sin{\frac{1}{x}}}{\sin^2{x}} = 0?$

yes it is. I frequently miss-typo latex code  :blush:

For $r>2$,
$\lim_{x \rightarrow 0} \frac{x^r \sin{\frac{1}{x}}}{\sin^2{x}} = \lim_{x \rightarrow 0} x^{r-2} \frac{x^2}{}{\sin^2{x}} \sin{\frac{1}{x}}$

$x/\sin x \to 1$ as $x\to 0$
and $\sin \frac 1x$ is bounded, $x^{r-2}\to 0$, hence the limit is 0.

for $r=2$ look for $f(1/(\pi + 2k\pi))$ and $f(1/2k\pi)$ : there is no limit

[quote="liyi"]For $r>2$,
.....
and $\sin \frac 1x$ is bounded, $x^{r-2}\to 0$, hence the limit is 0.[/quote]
how do you get this?
what theorem did you use?

Since $\lim_{x\to0}\frac{\sin^2x}{x^2}=1,$ the original limit might as well be

$\lim_{x\to0}\frac{x^r\sin\frac1x}{x^2}=\lim_{x\to0}x^{r-2}\sin\frac1x.$

Presumably, liyi and tµtµ did this step quickly and automatically.

yes that should help me much :)
an according to liyi's last statement. I have this (please check if this theorem is not true)

if $f(x)$ is bounded above and bounded below, that is $|f(x)| \leq C$ then $\lim_{x \rightarrow 0} x^p f(x) =0$ for $p \geq 1$.
proof: (can someone check this)

since $|f(x)| \leq C$ then $-C \leq f(x) \leq C$ multiply this by  $x^p$ by asuming it is positif. we have $-C x^p \leq x^p f(x) \leq C x^p$ (if x^p is negative we use similar argument) then by pinching theorem we have,
$-C \lim_{x \rightarrow 0} x^p \leq \lim_{x \rightarrow 0} x^p f(x) \leq C \lim_{x \rightarrow 0} x^p$ then $\lim_{x \rightarrow 0} x^p f(x) =0$

it is true or not?


Let ABC be an acute-angled triangle, and let O be its circumcentre. The circle through A, O and B is called S. The lines CA and CB meet the circle S again at P and Q respectively. Prove that the lines CO and PQ are perpendicular.

Since A, B, P, Q are concyclic, the segment PQ is antiparallel to AB and hence perpendicular to CO.

  Darij

[quote="Arne"]This problem was intended for the "beginners" on this forum Darij :D

I don't think you are a beginner actually.[/quote] well at least not in plane geometry :D:D

My apologies - again I was not aware of the intention of this subforum. Although, beginners are encouraged to solve the problem without using antiparallels...

  Darij


Find all pairs of positive integers $ x,y$ satisfying the equation

\[ y^x \equal{} x^{50}\]

[hide="Solution"]

Case I: $ x \equal{} y$ then $ (x,y) \equal{} (50,50), (1,1)$.

Case II: $ x < y$ then $ (x,y) \equal{} (2,2^{25}), (5,5^{10}), (10,10^{5}), (25,25^2)$. These pairs are found by using equation $ y \equal{} \sqrt [x] {x^{50}} \equal{} x^{\frac {50}{x}}$. Clearly, $ x$ must be a divisor of 50.

Case III: $ x > y$ then this means $ x^{50} > y^{50}$ so $ y^x \equal{} x^{50}$ if $ x > 50$. But from the equation in Case II, if $ x > 50$ then $ x^{\frac {50}{x}}$ is not an integer so there is no solutions in this condition.

$ \therefore$ The all pairs of solutions are $ (1,1), (2,2^{25}), (5,5^{10}), (10,10^5), (25,25^2), (50,50)$. [/hide]

[quote="Hamster1800"]Your case 3 doesn't quite work. (100, 10) is a solution since $ 10^{100} \equal{} 100^{50}$
Unfortunately I don't see a way to fix it.[/quote]

The mistake was assuming that if (general variables being used here) $ k$ is not an integer, and x is, then $ x^k$ is not an integer. (100, 10) was missed because $ 100^\frac{1}{2}$ is clearly an integer. It then becomes a problem of finding all integers $ x$ such that $ x$ is a perfect power, with exponent  $ \frac{x}{50}$. Clearly x must be a multiple of 50. We can then proceed by casework

Exponent = 2

Then x = 100, yielding the point (100, 10). Adding in any other prime factors will make the exponent greater than 2.

Exponent > 2

Notice that the prime factorization of 50 = 2 * 5 * 5. For the exponent to be 3 or greater, we will need to add prime factors of 2 * 2 * 5 at least. However, when we divide out 50, we get an exponent of at least 20. It is impossible to reach any exponent greater than 2, because in order to add in the necessary prime factors to make x a perfect power greater than 2, the exponent in the equation will always be greater than this power, if that makes sense.

Another kind of solutions..

The solution $ x \equal{} y \equal{} 1$ is obvious. So we can write
$ x \equal{} \prod p_i^{a_i}$, $ y \equal{} \prod p_i^{b_i}$. (Note that $ p \mid x \iff p \mid y$)
So $ y^x \equal{} x^{50} \iff b_ix \equal{} 50a_i$
Thus $ b_i \equal{} \frac{50a_i}{x} \in \mathbb{N}$ But then $ x \mid 50a_i$
With $ a_i \le 1,2$ we get the solutions already stated by the others.
Assume $ a_i > 2$ we now that there exists $ k$ so $ k^{a_i} \mid 50a_i$, but this is clearly impossible. Contradiction.


analytic geometry
algebra
function
domain
parameterization
Asymptote

[asy]size(200);
  Label f;
  f.p=fontsize(6);
  xaxis(-30,30,Ticks(f, 5, 0.5));
  yaxis(-30,30,Ticks(f, 5, 0.5));
  real f(real x)
  {
  return x^2/(1-x);
  }
  draw(graph(f,-3,3));[/asy]

Shouldn't the one pointing upwards continue to infinity, or at least the end of the coordinate plane? Also shouldn't the bottom one cut off at the bottom coordinate?

ehh.. I'm surprised that your code didn't give an error. You need to manually set the domain so the curve stays reasonably within the desired range, which is in this case [-30,30].

for example,

draw(graph(f,-3,.97)); draw(graph(f,1.03,3));

which will give

[asy]size(200);
Label f;
f.p=fontsize(6);
xaxis(-30,30,Ticks(f, 5, 0.5));
yaxis(-30,30,Ticks(f, 5, 0.5));
real f(real x)
{
return x^2/(1-x);
}
draw(graph(f,-3,.97)); draw(graph(f,1.03,3));[/asy]

Why don't we zoom in, though?

[asy]size(200);
Label f;
f.p=fontsize(6);
xaxis(-10,10,Ticks(f, 5, 0.5));
yaxis(-10,10,Ticks(f, 5, 0.5));
real f(real x)
{
return x^2/(1-x);
}
draw(graph(f,-10,.9)); draw(graph(f,1.1,10));[/asy]

[b]graph(f,a,b)[/b] just returns a broken line passing through many points (the default is something like 100, though I'm not sure and too lazy to look it up now; anyway, you can reset it by adding one extra integer parameter to the command:
[asy]size(150);
xaxis(-0.2,1.2);
yaxis(-0.2,1.2);
real f(real x)
{
return x^3;
}
draw(graph(f,0,1,3),red);
draw(graph(f,0,1,400),blue);[/asy]
The code didn't give an error because the dangerous point was missed in the mesh (though the result was that an almost vertical line was added to the graph).

Ohhh that is nice :D

For images like the above, I was defining each point and stuff.. like this, which is tedious:

[asy]import graph;
size(200);

xaxis(0,6);
yaxis(0,4);
real a(real x) {return 4sin(2x/5);}
D(graph(a,0,6));
D(D((1,a(1)))--D((5,a(5))),orange);
D((1,a(1))--(3,a(3))--(5,a(5)),lightred);
D((1,a(1))--(2,a(2))--(3,a(3))--(4,a(4))--(5,a(5)),red+linewidth(1));
D((1,a(1))--MP('a',(1,0)),dashed); D((5,a(5))--MP('b',(5,0)),dashed);[/asy]

Thanks for the tip! :)


$a,b,c$ positives réels and $\frac{1}{a}+ \frac{1}{b}+\frac{1}{c} =3$
Prove that $a^4+b^4+c^4 \geq\frac{a+b+c}{3}+2abc$

Since
$\frac{1}{a} + \frac{1}{b} +\frac{1}{c} = 3$

from am-hm we have

$a+b+c\geq3$

from gm-hm we have

$abc \geq 1$

now we already have all neccessary ineqs to attack this problem

from power mean we have

$\frac{a^4+b^4+c^4}{3} \geq \frac{(a+b+c)^4}{81} \geq \frac{a+b+c}{3}$

from am-gm we have

$\frac{2(a^4+b^4+c^4)}{3} \geq 2 (abc)^(4/3) \geq 2abc$

just add up the two ineqs above and we got the desired result.

it's very easy. i got it in less than 3 minutes.

[quote="Fermat -Euler"][quote="jmerry"]Couldn't you think of a more distinctive name? A problem in the Inequalities section shouldn't be called "Inequality".[/quote]
sorry i forget and i write it in french[/quote]Inégalité is not more distinctive than "Inequality". It's basically the same thing actually. What jmerry tried to say was that you need to give a more meaningfull name to the topic, such that users know a little bit about what the problem is when they are browsing the forum ...


limit
topology
real analysis
real analysis unsolved

Is it true that any countable topological space has a countable basis?

It's been edited.

I think the answer is yes. Suppose $X=\{x_1,x_2,\ldots\}$ is our topological space and let $B_i=\{U_{i1},U_{i2},\cdots\}$ be a basis at point $x_i$. Claim that $B:=\bigcup_{i=1}^\infty B_i$ is a basis of $X$. Clearly $B$ is countable. Let $U$ is open in $X$ and $x_j\in U$. We have that $U$ is a neighborhood of $x_j$ and therefore by the first countability, there is $U_{jk}\in B_j\subset B$ such that $x_j\in U_{jk}\subset U$.

Of course, that's obvious :). Sorry for that. I wanted to ask something else:

Is it true that any countable topological space has a countable basis? (so we drop the first-countability)

Again, I apologize  :blush: Maybe now it's not that trivial.

[quote="grobber"]Is it true that any countable topological space has a countable basis? (so we drop the first-countability)
[/quote]

Did someone made any progress on this one? :)

Oh my, a 2 year old topic! :roll: I guess that all mathlinkers belonged to one of the 2 categories: for the first one, the question was too obvious to bother to answer and for the second one, it was too difficult. Anyway, take $ \mathbb N$ and declare a set $ E$ open if either $ E \equal{} \varnothing$, or its complement has density $ 0$, i.e., $ \lim_{n\to\infty}\frac 1n\#(E\cap\{1,2,\dots,n\}) \equal{} 1$. Then, clearly, it gives you a topology on $ \mathbb N$ (finite intersections and arbitrary unions are OK). On the other hand, given any countable number of open sets, we can construct an open set that contains neither of them (by making the complement thicker at $ \infty$ than any of the complements of our sets or just by choosing one point in each set so that the resulting sequence has density $ 0$ and considering the set of all not chosen points). That's all.


calculus
integration
trigonometry
logarithms
calculus computations

Calculate
\[ \iiint\limits_{\begin{subarray}{c} 0 < x,y < 1 \\
0 < z \end{subarray} } {\frac {{dx dy dz}} {{\left( {1 \plus{} x^2 z^2 } \right)\left( {1 \plus{} y^2 z^2 } \right)}}}.
\]

$ \frac{1}{(y z)^2\plus{}1} \int_0^1 \frac{1}{(x z)^2\plus{}1} \, dx\equal{}\frac{1}{(y z)^2\plus{}1} \frac{\tan
   ^{\minus{}1}(z)}{z}$

$ \frac{\tan ^{\minus{}1}(z)}{z} \int_0^1 \frac{1}{(y z)^2\plus{}1} \, dy\equal{}\left(\frac{\tan
   ^{\minus{}1}(z)}{z}\right)^2$

$ \int_0^{\infty } \left(\frac{\tan ^{\minus{}1}(z)}{z}\right)^2 \, dz\equal{}\int_0^{\frac{\pi }{2}}
   \left(\frac{\theta }{\sin (\theta )}\right)^2 \, d\theta$

i dont have time now, but after i will calculate this last integral

nicely done , eulerintegral.  :) in case you were wondering what the integral evaluates to, it is $ \boxed{\pi\ln\,2}$

  a good problem, bookworm_vn! i liked it.  :thumbup: