ratio
floor function
logarithms
calculus
calculus computations

discuss the convergence of series
$ \sum_{n \equal{} 1}^{\infty}( \minus{} 1)^{[n\sqrt {2}]}\frac {1}{n}$
where []-floor

Let $ a_n\equal{}(\minus{}1)^{[n\sqrt{2}]}\frac{1}{n}$.

$ \frac{\left|a_{n\plus{}1}\right|}{\left|a_n\right|}\equal{}\frac{n}{n\plus{}1}<1$ for all natural number $ n$, since $ (\minus{}1)^{[n\sqrt{2}]}$ is $ 1$ or $ \minus{}1$. So by the ratio test, the given infinite series converges.

[quote="aktyw19"]something is wrong[/quote]

Not 'something', my approach was entirely wrong.
The least upper bound of $ \frac{\left\|a_{n\plus{}1}\right\|}{\left\|a_n\right\|}\equal{}\frac{n}{n\plus{}1}$ is $ 1$, we can't apply the ratio test in this problem.  Sorry for my big mistake.

There was a recent topic (someone else will have to find it) about estimating the size of $ \sum_{k\equal{}1}^n(\minus{}1)^{\lfloor k\sqrt{2}\rfloor}.$

If we can show that that sum is $ O(n^a)$ for some $ a<1,$ we will be able to prove that the series in this problem converges conditionally, using a summation by parts argument.

this problem, apparently, is quite difficult
I do not have solutions

The topic Kent Merryfield is probably alluding to is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314595]this one[/url], and another thread (some four and a half years old) discussing your exact problem is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39398]here[/url].  A non-elementary (i.e. involving Fourier series) solution is cited in the first, and the second thread gives an elementary (but tricky) argument that $ \sum_{k \equal{} 1}^n( \minus{} 1)^{\lfloor k\sqrt {2}\rfloor} \equal{} \mathcal{O}\left(\log n\right)$, which together with Kent's suggestion yields a solution to your problem.


My son and I sometime uses the same computer to access AoPS site.  The trouble begin when he clicked automatically log on box when he logged on my computer.  No matter what I do, I cannot log him off until I found this option of deleting all cokies.

As AoPS wants the site as a resources not just for students, I imagine this problem will happen in a number of situations: family share the same computer, school lab computers, etc.

SOLUTION: Add an new option: Login As Another User will solve the problem.

The logout link should log you out usually, but we just made some changes to the cookie system so it might be buggy for a while.

Quicker (and already available) solution: in such houses, do not use log on automatically, and if you accidentally do, the link in each forum (including main page) in the lower right corner to delete all cookies set by this site (as you have already seen it) deletes all cookies (only for this board) and the automatic login is set to none (again).

Another solution (if there are 2 users of the site in one household) would be each to have the log on automatically set, but one to navigate using mathlinks url and the other one through aops url. Both can have the AoPS skin if they wish, but only the url will be different: instead of artofproblemsolving.com you'll have mathlinks.ro

[quote="Valentin"]Quicker (and already available) solution: in such houses, do not use log on automatically,...[/quote]
That's what we do. We just have to keep remembering our own passwords. We've both settled on Firefox, Mac version, as the browser and haven't encounted the problem mentioned here, with both of us using AoPS skin. I think there have been occasions (probably not since you made those recent changes) in which one of us clicked the logout button, leaving the computer connected and the browser on AoPS, after which the other logged in, and I think that worked OK.

The best idea however (don't know how Mac is) is to have different Windows (or Linux for that matter) accounts. This solves all problems. Use quick switch to go from an account to the other, and this way you can have your own browsers' settings and all ...

[quote="Valentin Vornicu"]I forgot to ask you a very important thing: what browser do you use? Normally the logout button should work as such. It has known problems with Firefox unfortunately (alternatively a solution of this problem would be that the father/brother users another browser, or better yet - another windows account!).[/quote]

IE 6.0

On my IE 6.0 (sp2 fully-patched) using the logout link/button logs me out (even I set it to log me in automatically) ... I understand that this doesn't work for you?!

[quote="Stun"]Dear Valentine
I have to ask you a question : Why I cannot log out the user name Stun? My computer is being uses by 2 another people and they all access the Mathlinks Forum and they don't want to post under the name Stun[/quote]
Please post your OS and Browser Version first.

PS You can always [url=http://www.mathlinks.ro/Forum/remove_cookies.php]remove the cookies[/url] from this site (only!) :)

Same problem(can't log out).

Windows XP, Firefox 1.0(too bad Firesomething extension doesn't work anymore, or I would have Thunderiguana or something!). I know that there is a newer version of FF but it is too buggy with my essential extesnions.

There is something wrong with your cookies. Delete them and you will have no problem. I just tested on my Firefox 1.0.1, IE 6 Sp2, and Opera 7.54. Works fine ...  :?

I see what the Opera problem is. It doesn't show me in the list of active users, but it doesn't have a login link either.

The logout feature worked once in FF just now!

Download the new version of Firefox and it should work fine.

Thanks.

[quote="Valentin Vornicu"]This seems to be a Firefox issue. I can logout perfectly on my IE 6.0. But if you can't just click on the link above (remove cookies for this site - which you also find in the lower right corner of each index.php page on the navigation bar) and you will be logged out :P[/quote]

I always use Firefox, and it works fine. What I recommend if you use firefox, you go to the mozilla.org and downlaod firefox 1.6. it fixes up all the probelms,a nd the forum runs faster than ie 6.0.

________________


geometry
circumcircle
power of a point
radical axis
geometry proposed

[quote="Allnames"]Let $ ABC$ is a triangle and its circumcircle $ (O,R)$,$ (O_1)$ is a circle which tangents $ AB,AC$ and inner tangents with $ (O)$ at $ X$,we have $ Y,Z$ points similarly with vertex $ B,C$Prove that $ AX,BY,CZ$ concur .
P/s I have a proof by Menelaus theorem (show that they concur at $ K$ where $ K$ such that $ \frac {KI}{KO} \equal{} \frac {r}{R}$,with $ (I,r)$ is incircle of $ ABC$)but I think inversion will work well  :lol:[/quote]
No idea???

[quote="The QuattoMaster 6000"]Look here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=243889

Using inversion, I prove that if $ D$ is the point of tangency between the $ A$-excircle and $ BC$, then $ AD$ and $ AX$ are isogonal conjugates, which would imply that $ AX$, $ BY$, and $ CZ$ concur at the isogonal conjugate of the Nagel Point of $ \triangle ABC$.[/quote]
Oh,can you give more detail,I want to improve my skill in inversion ,thank you very much
so how about this general one :
[hide="General one for it"]Let $ ABC$ is a triangle ,a circle  $ (O,R)$ is given,$ (O_1)$ is a circle which tangents $ AB,AC$ and inner tangents with $ (O)$ at $ X$,we have $ Y,Z$ points similarly with vertex $ B,C$Prove that $ AX,BY,CZ$ concur [/hide].
Now $ (O)$ is not circrumcircle of $ ABC$,can inversion work now?

[quote="Mr Green"][quote="Allnames"]Let $ ABC$ is a triangle and its circumcircle $ (O,R)$,$ (O_1)$ is a circle which tangents $ AB,AC$ and inner tangents with $ (O)$ at $ X$,we have $ Y,Z$ points similarly with vertex $ B,C$Prove that $ AX,BY,CZ$ concur .
P/s I have a proof by Menelaus theorem (show that they concur at $ K$ where $ K$ such that $ \frac {KI}{KO} \equal{} \frac {r}{R}$,with $ (I,r)$ is incircle of $ ABC$)but I think inversion will work well  :lol:[/quote]
This problem equivalent with problem

Let triangle $ ABC$ and $ I$ is its incenter.The perpendicular with $ AI$ at $ I$ intersect $ BC$ at $ D$.$ E,F$ determine similarly.Prove that $ D,E,F$ are collinear

This problem is easy :D[/quote]
Yes,I see,thank you,with inversion it is really easy
How about the general one above

Let triangle $ ABC$ and $ I$ is its incenter.The perpendicular with $ AI$ at $ I$ intersect $ BC$ at $ D$.$ E,F$ determine similarly.Prove that $ D,E,F$ are collinear

$ D,E,F$ lie in radical axis of $ (O,R)$ and $ (I,0)$

Dear Mathlinkers,
for a complete proof of the initial problem, sse
http://perso.orange.fr/jl.ayme   vol. 4  A new mixtilinear incircle adventure p. 20-27.
Sincerely
Jean-Louis


I'm from Connecticut and 3 of us got pins from all 57 places :D Not neccesarily each pin that was given out from each place, but 3 of us got at least one pin from each place. The last member of our team thought he forgot his pins at home, but then Saturday, he found them.

Also, CT got 100 pins each of the most worthless one, yet we managed to get full collections :P I have about 50 pins to spare, one team member has about 5, another 10, and the one that thought he forgot his pins has 60 left.

[quote="lotrgreengrapes7926"]I got around 40.  :roll: No offense, but NO ONE wanted Connecticut pins after around Thursday night. :D Don't trust Virgin Islands people. I got a "Roller Coaster Designer" pin from them.[/quote]
Haha, really? I got the real virgin islands pin.

Anyway, I got 56, as did another guy on the team. I doubt you got 57 unless you're counting the little league DoD pin as the true DoD one... There were only 19 of the true DoD pins at nats...

Uh... yeah... they were pretty much trading freely on Saturday...

[quote="lotrgreengrapes7926"]So California really did have pins... :ninja:[/quote]

Yeah, they said I was the one of the first to get one XD

The fact that no one wanted CT pins makes it more amazing. The reason I have 50 left and my teammates have 5 and 10 was because I only traded 2:1 once total. I have 3 VI pins, all 3 of the one with 2 flags though.

[quote="Treething"]Lol, after that Georgia kid stole my pins, I started giving away free NJ pins.[/quote]

You mean Florida? And I accidentally stepped and broke one of your NJ pins by the elevator.

I got around 48 of the pins, and ended up with 14 extra KS and 8 Georgias. One of my teammates has almost all of them, and 14 Georgia pins. heh, KS was one of the most worthless pins  :P .

I got a black virgin island pin with a skull... Darn, I traded away my Idaho potato... And yes, California did have pins...

[quote="Treething"][quote="Phelpedo"][quote="lotrgreengrapes7926"]I got around 40.  :roll: No offense, but NO ONE wanted Connecticut pins after around Thursday night. :D Don't trust Virgin Islands people. I got a "Roller Coaster Designer" pin from them.[/quote]

No one has wanted Connecticut pins since like 1990. We've been "full of surprises" for almost the entire history of mathcounts. Add that to the fact that they give out 400 pins every year, and you have about 7000 connecticut pins in circulation.[/quote]

how surprising, I was really wondering what they were going to do this year...[/quote]

Ironic, eh? I wonder if the Connecticut Society of Professional Engineers had it printed to say "We're full of surprises" on purpose just to laugh at us.

One kid was like, unstable and started yelling at one of my teammates after he realized CT had the same pin last year. But my teammate is an idiot (same one that cost us a question on team round), 90% of the people there were willing to trade 1:1, he starts trading 3:1 right when we got in the hotel even though there was no reason to. We were going to trade 1:1 and our circulation was even diluted that year because one of us thought he forgot his pins but he screws us up. In the end though, he was basically out of pins and I have enough to go to nats again next year and get 57 along with my extras from other states :P

And pianoforte, did Georgia give you all of those for free? CT has way more worthless pins than KS. Ours are plastic, we have 400 every year, and the people that give the pins have been "full of surprises" like Phelpedo says for the last 20 years or so.

I think the Connecticut team should make their own pins...out of cardboard or something :lol: as they'd be worth more than the other ones :P at least for returning national competitors.

Uh...most national competitors aren't returners  :(

Our Massachusetts pins were the worst!!
They didn't say Massachusetts on them!!!
They were gotten for free by are state coordinator from Raytheon!!!
Our state coordinator has promised us good pins next year.
I was still able to make a bunch of 1:1 deals though  :) .

I thought the AoPS pin looked the coolest...

Also the Missouri one :rotfl: Of course, I may be a little biased...

I only had like 38 pins to begin with...

The South Carolina wristbands were pretty cool...

I think I'm using ellipses (...) way too much...

The michigan one I think had the most colors, so I give it :winner_first:

Virgin islands had a :coool: pin.

I found a nebraska on stuck into the bottom of my shoe, luckily, it was the thick part, and I couldn't feel it.

SC wristbands were...creative.  I wear mine every other day while im practicing.

I put my pins on a random handkerchief, and used my hat as temporary storage.

The Idaho pin (POTATO!!) entertains me, as does the Alaskan walrus (Teeth).

I gave all my pins to kyyuanmathcount, because I didn't care and he did, so this is why he has like practically all the pins.  Plus from hunting down people himself, he was able to get pins.

I also had a pin stuck into the bottom of my shoe...

We kept our pins on these maps with our names and a pouch for extra pins. It was very hard to tell what pins I had and didn't have for the Eastern states, since there was like no room, while other states we're completely surrounded with space, like Texas.

[quote="Fanatic"]Did anyone see a kid with a big map full of pins?

Moi.

The sad thing is that I still don't have WA or MN.[/quote]

hehehe...i am on the mn team, u should have come and found me, i had lyk no pins at all...they are all gone now, tho.

washingtons was just the state and it said washington.  mn's was cool..of course...lol

 :rotfl:

yea, i think i am pretty much insane....

[quote="kyyuanmathcount"]I met Minnesota at the airport, and gave 50 cents for one because the state coordinator hadn't arrived yet.

I got around 50.[/quote]

oh yea, that was u...ur pins wer pretty cool...our team was the first to get them

[quote="Sly Si"][quote="k8reindeer"]Minnesota's pin is shaped like minnesota, and it says minnesota mathcounts on it with a picture of a duck and a flower.[/quote]

Sounds awesome!  My first year, it was pretty cool because we had four different ones and they were all nice.  My second year, they were all the same and not all that interesting; I think we got them for free from the state tourism department or something.

Also, the "duck" of which you speak is almost certainly the common loon, the state bird.[/quote]

o yea!!!..lol  And the flower was the ladyslipper...go us!!
btw, did u catch the random red band along the top right corner? anyone tell me what that is and i'll owe them forever :p

I traded pins when I was going to get a new roomkey (mine wasn't working and my roommate had left hers in the room.)  :)
I also got 2 Mississippi's right after they got there, because their rooms where just down the hallway, and I happened to be in the hall when they came.  :)

[quote="mathnerd314"][quote="Ignite168"]Instead of making 5 posts in a row, try using the edit button ;)[/quote]

That's [b]3[/b] posts... ;)

I just kept a list of all the pins I had on the back of my hotel registration sheet... the one that had my room key in it.[/quote]

There were 5, I'm sure of it.

I had a list of all 57 places but I lost it with like 20 to go and that's probably why I realized I'm missing 1 or 2 after nats. :(

I only got about 45 different pins though *sighs*

I did get over 20 georgias, 30 connecticuts and 10 delawares. Pretty good since georgia had only 20 pins total *cough*. BUT I EVENTUALLY GOT THE MICHIGAN PIN. HAHAHAHA

[quote="kyyuanmathcount"][quote="mathnerd314"][quote="Ignite168"]Instead of making 5 posts in a row, try using the edit button ;)[/quote]

That's [b]3[/b] posts... ;)

I just kept a list of all the pins I had on the back of my hotel registration sheet... the one that had my room key in it.[/quote]

Give the poor thing a break, MATHCOUNTS is over for the year  ;) .[/quote]

No, actually, there were 5 posts... Apparently 2 of them were deleted.

Did you hear what the moderator said, he said shut it.

Now for the pins.  A lot of people kept asking us do you have any pins every minute and we say no, and one average the same person comes to us 5 times and we say no.  But then we finally got pins and then we started trading them.


There are few things you can do:
\cdot yields $ \cdot$
\cdots yields $ \cdots$
\ldots yields $ \ldots$
\ddots yields $ \ddots$
\vdots yields $ \vdots$

[quote="ZhangPeijin"]Yeah, whenever I use the dot command in my TeXnicCenter, it gives me a big splotchy mess that doesn't really resemble a dot at all.
Is there any way for it to look better like this one?
[asy]dot((0,0));[/asy][/quote]
$ \bullet$ [b]\bullet[/b]
$ \cdot$ [b]\cdot[/b]

To easily remember these commands, think about this

\cdot  (central dot)
\cdots (central dots)
\vdots (vertical dots)
\ldots (... :D)

OK, what do you want from your dots in Asymptote?
[asy]size(120);
for(int k=1;k<20;++k) dot((k,k),red+linewidth(k/4));[/asy]

In Asy, my dots look like a big splotch, unlike the dots that I make on AoPS.
I was asking how do I make the dots look better.

But AoPS uses the same Asymptote! Something may be wrong with either your version of Asymptote or your GSviewer (if you do .eps files) or Adobe Acrobat (if you do PDF). Also, remember that if your picture contains just one dot or nothing but a few dots on the horizontal or vertical line, Asymptote may go nuts and scale your picture in a strange way. For "normal" pictures you should be fine nevertheless, so try to reinstall your viewers if you get starange results.

Well, I don't have GSviewer installed, but I do have GS...
Eh, I guess I could live with it.

Just to make this wild shot. Can you post an example of code that produces those ugly blotches instead of neat dots? I'll try it on my PC and see if I get the effect. GS you mentioned is probably the same as GSview (ghostscript is the full name for it). It is free, so it'll cost you only five minutes of your time to reinstall it and see if it helps. The same with Asymptote and Acrobat Reader. Unless you did something strange in your code, it must be one of them that creates the problem.

dot((0,0));
Anyways, here, it looks like:
[asy]dot((0,0));[/asy]
But on my computer it looks different each time I use it, sometimes it's a square rotated 45 degrees, sometimes its a big sloppy ink blotch, etc.

Ah, I warned you about this. As long as your picture is 0 or 1-dimensional instead of 2-dimensional, Asymptote goes nuts trying to scale to size. Do you have any problem with
dot((0,0)); dot((1,1)); ?
This normally should give you two nice looking dots.

Yeah I do actually.
One looks like a diamond, while the other is a splotch.

Then the only thing I can say is "spend 15 minutes reinstalling everything". I cannot guarantee it'll help, of course, because it is completely unclear why it happens, but it definitely won't hurt :).

Are you looking at your document with a zoom other than 100%? This sometimes messes things up.

Hmm, I've noticed that the splotches look different at different magnifications...
Hmm...

It would help to post
a) a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=198207]minimal example[/url] of a complete document;
b) attach the output file showing the blotch;
c) any error messages that you get.

[code]
\documentclass{report}
\usepackage{amsmath}
\usepackage[pdftex]{color,graphicx}
\usepackage{makeidx}
\usepackage{asymptote}

\pdfpagewidth 8.5in
\pdfpageheight 11in

\setlength\topmargin{0in}
\setlength\headheight{0in}
\setlength\headsep{0in}
\setlength\textheight{9.0in}
\setlength\textwidth{6.5in}
\setlength\oddsidemargin{0in}
\setlength\evensidemargin{0in}
\setlength\parindent{0.25in}
\setlength\parskip{0.25in}

\begin{document}
	\begin{asy}
		unitsize(1cm);
		draw(circle((0,0),5));
		dot((0,0));
		dot((5,0));
		dot((0,5));
		dot((-5,0));
		dot((0,-5));
	\end{asy}
\end{document}
[/code]
EDIT: Hmm, that's weird.
When I look at it after I attach it, it looks fine, but in my preview thing on TeXniC Center it has the blotches...
Maybe my Adobe Reader is just bad. Since I used Foxit Reader online.

Looks fine on my computer (a bit irregular dots in normal size but nice and round when I look at them at high magnification). So, it is a viewer problem for sure.

I can confirm that it looks fine in Adobe Acrobat 7 here. Try going into Adobe's Edit, Preferences, Page Display and tick Smooth Text, Smooth Line Art, Smooth Images. If it's a laptop try CoolType on that page as well.

It looks fine on Preview.
Crystal-Clear Dots.
 :lol:

maybe you should post what your problem looks like...

[quote="ZhangPeijin"]Eh, is there any way to change my default previewer?[/quote]Yes. In TexnicCenter, Build, Define Output Profiles, Viewer tab, Path of executable


Hi guys,

Didn't know where exactly to post this message and decided to do it here. Anyways ... Does anyone of you know of a good book about functional equations solving techniques, which is also available for a free download?

What I am looking for is a detailed treatment of the subject, perhaps inculding some advanced techniques in algebra/analysis, etc. I know that there is such a book in bulgarian from Liubomir Davidov, but it's impossible for me to buy/steal/borrow it right now, since I live abroad for awhile (I don't know of any bulgarian books listed @ amazon). Does anyone have that book scanned?

Many thanks in advance!
Zhivko

There is courses on functionals  equations

http://www.animath.fr/tutorat.html.fr#Les%20cours%20de%20l'Olympiade%20française%20de%20mathématiques

It is in Cours Olympiades française de mathématiques

- Les équations fonctionnelles in pdf or ps

Downloaded Moubi! Many thanks once again for the link!

Any basic courses in French which come together with this course? Just joking :)  ... hopefully the maths will be understandable.

I see that you are one of the authors!! Very good!  ;)

Best regards!
Zhivko


Find all ordered triples $(a,b,c)$ satisfying $a^2+b^2+c^2=2005$ and $a\leq b\leq c$,
 where a, b and c are natural numbers.

$26\le c \le 44$. It is easy found all solutions $a^2+b^2=N=2005-c^2$.

Rust, can you solve the generalization?  :P

What generalisation¿ With general $2005$¿ Then good luck with class numbers already for their number ;)

Yes.

Yes, I know.  :D

(I changed the smiley to a ":P" in my previous post)

$2005=5(mod \ 8)$. Therefore equation have solutions and one of them odd, one divisible to 4, one =2(mod 4).

This has been discussed some time ago.
See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=44856]link[/url].


I try to make a signature with a quote (13 lines total), and I get this message:

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

Signature: the value is too long.

Click Here to return to your profile.

Click Here to return to the Index

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

I've seen longer signatures. :huh:

max sig length=255 characters.
that includes ANYTHING that goes in the sig(like the hide tag counts too)

what sigs are longer than 13 lines? :huh:

[quote="junggi"]what sigs are longer than 13 lines? :huh:[/quote]
Oh. Never mind. :ninja:

Well, Vally has special admin powers if that's what you are talking about.  Also, I believe having links and pictures takes up LOTS of space.

premium members (I guess that's what they are called, seen that used in various forums) and mods can have special features

Also we aren't allowed to have sigs over 3 lines anymore

[quote="nutz_for2.718281828"]They took away our hide text in the signature!!! :noo:

BUT you can use white text. :thumbup:[/quote]

lol thats really annoying ive seen someone use it and almost went blind...

plus i think that counts as spam


How many positive integers less than 50 have an odd number of positive integer divisors?
\[ \text{(A)} 3\ \ \ \text{(B)} 5\ \ \ \text{(C)} 7\ \ \ \text{(D)} 9\ \ \ \text{(E)} 11 \]

[hide]The answer is [b]7.[/b]  A number will always have an even number of divisors unless it is a perfect square, because there will always be a "pair" with a divisor.  For example 3*4=12, 4 is the other number.  However with a perfect square one of the "pairs" will be the number itself, such as 7*7=49.  So for perfect squares there wll be some pairs and one more number, which is the square root.  Because there are seven perfect squares under 50 (1-7) the answer is 7.[/hide]

[hide]Those are all the square numbers; there are 7 of them[/hide]


superior algebra
superior algebra solved

Let G be a group. Show that there exists a monoid which isn't a group, M, s.t. G is the group of the invertible elements of M.

Take M=GxZ2.
Define the operations
(g,a)*(g'*a')=(gg',aa'), where g and g' are multiplied with respect to the operation in G and a and a' are multiplied (not summed) in Z2.

The operation "*" is associative because so are the multiplications in G and Z2.
The neutral element is (e,1).
(g,a) is invertible iff a=1, so Gx{1} is the group of invertible elements in M and it is isomorphic to G.

Thanks. There are, of course, lots of solutions. I took M=GxN (N is the set of naturals, on which the operation is "+"). If we solve it for G={e} then we're done.


Give $ a.b.c \geq 0$ such that $ a\plus{}b\plus{}c\equal{}1$. Prove that
          $ \frac2{a\minus{}a^2}\plus{}\frac2{b\minus{}b^2}\plus{}\frac2{c\minus{}c^2} \geq 5(\sqrt{\frac{a}{bc}}\plus{}\sqrt{\frac{b}{ca}}\plus{}\sqrt{\frac{c}{ab}})$

[quote="dduclam"]Give $ a.b.c \geq 0$ such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that
          $ \frac2{a \minus{} a^2} \plus{} \frac2{b \minus{} b^2} \plus{} \frac2{c \minus{} c^2} \geq 5(\sqrt {\frac {a}{bc}} \plus{} \sqrt {\frac {b}{ca}} \plus{} \sqrt {\frac {c}{ab}})$[/quote]
$ \frac2{a \minus{} a^2} \plus{} \frac2{b \minus{} b^2} \plus{} \frac2{c \minus{} c^2} \geq 5\left(\sqrt {\frac {a}{bc}} \plus{} \sqrt {\frac {b}{ca}} \plus{} \sqrt {\frac {c}{ab}}\right)\Leftrightarrow$
$ \Leftrightarrow\frac{1}{a(b\plus{}c)}\plus{}\frac{1}{b(a\plus{}c)}\plus{}\frac{1}{c(a\plus{}b)}\geq\frac{5}{2\sqrt{(a\plus{}b\plus{}c)abc}}.$
Let $ a\equal{}\frac{1}{x},$ $ b\equal{}\frac{1}{y}$ and $ c\equal{}\frac{1}{z}$ and see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101476


Lately, every time I type something in LaTeX I get an error message saying something like: TexnicCenter has found an error, would you like to report it? When I click "Don't Send", TexnicCenter closes and I lose my work if I haven't saved it? Could this have anything to do with other programs that re running on my computer (firewall, antivirus, etc.)? Do I need to update my version of TexnicCenter? Thx.

normally, you don't even need latex or technixcenter to view this site. Something that certainly helps - if you can miss it - is exitting (or uninstalling if necessary) the program you're talking about.

I believe he is talking about a program completely seperate from these boards.  I don't know how to fix the problem but try looking up the problem in the help file if there is one...if not try to find the webpage where you got the program and see if they have a troubleshooter or a FAQ.  Good luck with fixing your problem.

All the stuff about an error and don't send etc is just some weird microsoft thing, its like that for all programs. All that is happening is that the program is crashing, you can ignore the rest. Probably the best fix would be to update/reinstall the program.


http://lunarwars.net/

Join Lunar Wars Round 2 today!

And join the Manticore alliance!

xscapezaer is in the alliance as well.

Please join!

Awesome guys...

Some general advice:

You should buy your AI in chunks of 10 (since that is the most you can buy at a time at full discount), and save up your AI discount, and but just before you get to 20% (any more is wasted)

Focus on getting AI for now. Once AI gets to expensive, start buying AI. You should also get improvements, except for Life Support.

I'm waiting for our members to accept your alliance request.

Don't worry about getting attacked yet, you are too weak (and in Peace Mode) to get attacked. There is a requirement that you have to be in the same range of strength to attack someone.

Here's our alliance forum.

http://z10.invisionfree.com/Manticore/

EBWOP:

You need to make an account in Lunar Wars, apply for our alliance, and after you do so, please apply on our forums as well.

http://z10.invisionfree.com/Manticore/index.php?showtopic=2

I applied (as aleph0) but I still have not been put in yet.

...

Thank you for applying, but our alliance leader has stated that you MUST apply in both the game AND the forum for him to accept your request.

So make an account in our forum, and POST in the registration thread or whatever.

That's for all of you. (1=2, you still need to post in the application thread, not just make an account)

dangit i sold all my minerals! DX

Now I have to wait three hours!

Moral if this story: DO NOT SELL ALL OF YOUR MINERALS UNLESS YOU DO NOT WANT TO BUILD!

how come i can't register?
the verification code part is completely blank

Applied on the forum...hopefully in soon

Well, interesting chenhsi you actually did advertise like you said you would xD

aleph0 and eggylv999, I have masked you. 1=2, I don't see any registration from you... :(

Could everyone post their lunar wars account name and colony name?
I'm aleph0 and my colony is also aleph0.  :D

I'm jjx1 and my colony is jjx1

okay I've done everything. Applied on lunarwars.net and that forum.

I GOTZ TOO MUCH COPPERZ AND TOO LITTLE MONEYZ AND ILMENITZ!!!

I wanna get more AI!

[quote="aleph0"]I got a message that said look at the manticorean creed but I can't access it.[/quote]

me too.

Then you're probably not logged into the forums. Also, the point in the creed is that you shouldn't raid other colonies  :o

It is hard to maintain your money. My mining saved me from getting $ 68 dollars per day until I got out. I built houses for$800 income. Switching to war mode does a lot.

Edit: WOW LATEX

Um hi guys, a few things...

First, we're pretty much done recruiting, we didn't expect this much response and I didn't really expect chenhsi to even try recruiting on AoPS.

Second, you're not obligated to join Manticore if you join the game, but most of the AoPSers who have joined Lunar Wars have joined us. Feel free to join some other alliance if you wish :P

Third, if you are a Manticore member, please DO NOT USE THIS THREAD FOR IN-GAME DISCUSSION! That is why we have our own forums, please talk there or register if you haven't already. I may sound paranoid, but you could give away sensitive information. Teh internetz have eyez :o  :maybe:

umm i signed up as mathkid601, but if your done recruiting I will withdraw my request

No, it's good you can still sign up if you already were going to. I'm just saying we don't really need more people anymore.

I just transfered my account from Round 1 to Round 2. I'm XRCatD. I'm applying for Manticore.

Applying under name: "jamesbondtriple07"

I've kinda abandoned this game. Havn't went on it in like 20 days or longer. so yeah.


Let $ a,b,c$ be [b]real[/b] numbers such that $ (a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}8$. Prove that
$ ab\plus{}bc\plus{}ca \le 3.$

[quote="can_hang2007"]Let $ a,b,c$ be [b]real[/b] numbers such that $ (a + b)(b + c)(c + a) = 8$. Prove that
$ ab + bc + ca \le 3.$[/quote]
Assume:$ a \ge b \ge c$,
-If $ b+c < 0$ then $ (a+b)(a+c) < 0$ ,so $ ab+bc+ca < -a^2 <0$  (Q.E.D)
-If $ b+c> 0$ then $ a+b \ge a+c \ge b+c >0$
Put $ a+b=2x ,b+c=2z, c+a=2y$,then $ xyz=1 ,a=y+z-x ,...$
This inequality become:
$ x^2+y^2+z^2+3 \ge 2(xy+yz+zx)$
$ \leftrightarrow (x-y)^2+(z-1)^2+2z(x-1)(y-1) \ge 0$
Assume$ (x-1)(y-1) \ge 0$, we have Q.E.D  :)

Let $ a\plus{}b\equal{}x$ $ b\plus{}c\equal{}y$ $ c\plus{}a\equal{}z$. The condition turns into $ xyz\equal{}8$

We need to show that $ 2(xy\plus{}xz\plus{}yz)\minus{}(x^2\plus{}y^2\plus{}z^2)\le 12$

Let $ x\equal{}2p^3$, $ y\equal{}2q^3$, $ z\equal{}2r^3$ The condition turns into $ pqr\equal{}1$

$ 2(p^3q^3\plus{}q^3r^3\plus{}r^3p^3)\le 3\plus{}p^6\plus{}q^6\plus{}r^6$

$ 2(p^3q^3\plus{}q^3r^3\plus{}r^3p^3)\le 3p^2q^2r^2\plus{}p^6\plus{}q^6\plus{}r^6$

By schurs, we know that $ p^6\plus{}q^6\plus{}r^6\plus{}3p^2q^2r^2\ge p^4q^2\plus{}q^4p^2\plus{}p^4r^2\plus{}r^4p^2\plus{}q^4r^2\plus{}r^4q^2$

Then the ineq. turns into

$ p^4q^2\plus{}q^4p^2\plus{}p^4r^2\plus{}r^4p^2\plus{}q^4r^2\plus{}r^4q^2 \ge 2(p^3q^3\plus{}q^3r^3\plus{}r^3p^3)$

Or $ p^2q^2(p\minus{}q)^2\plus{}p^2r^2(p\minus{}r)^2\plus{}q^2r^2(q\minus{}r)^2\ge 0$. Q.E.D.

[quote="rofler"]Let $ a \plus{} b \equal{} x$ $ b \plus{} c \equal{} y$ $ c \plus{} a \equal{} z$. The condition turns into $ xyz \equal{} 8$

We need to show that $ 2(xy \plus{} xz \plus{} yz) \minus{} (x^2 \plus{} y^2 \plus{} z^2)\le 12$

Let $ x \equal{} 2p^3$, $ y \equal{} 2q^3$, $ z \equal{} 2r^3$ The condition turns into $ pqr \equal{} 1$

$ 2(p^3q^3 \plus{} q^3r^3 \plus{} r^3p^3)\le 3 \plus{} p^6 \plus{} q^6 \plus{} r^6$

$ 2(p^3q^3 \plus{} q^3r^3 \plus{} r^3p^3)\le 3p^2q^2r^2 \plus{} p^6 \plus{} q^6 \plus{} r^6$

By schurs, we know that $ p^6 \plus{} q^6 \plus{} r^6 \plus{} 3p^2q^2r^2\ge p^4q^2 \plus{} q^4p^2 \plus{} p^4r^2 \plus{} r^4p^2 \plus{} q^4r^2 \plus{} r^4q^2$

Then the ineq. turns into

$ p^4q^2 \plus{} q^4p^2 \plus{} p^4r^2 \plus{} r^4p^2 \plus{} q^4r^2 \plus{} r^4q^2 \ge 2(p^3q^3 \plus{} q^3r^3 \plus{} r^3p^3)$

Or $ p^2q^2(p \minus{} q)^2 \plus{} p^2r^2(p \minus{} r)^2 \plus{} q^2r^2(q \minus{} r)^2\ge 0$. Q.E.D.[/quote]
This is exactly the same as my solution. :)


logarithms
function
limit
calculus
calculus computations

1)Find  all a such that
$ \text{ln}(x\plus{}1)\geq x\minus{}ax^{2},\forall x\in [0,\plus{}\infty)$
2)Find all $ c\in R$such that
$ e^{x}\plus{}e^{\minus{}x}\leq 2e^{cx^{2}},\forall x\in R$

$ \log (x\plus{}1)\geq x\minus{}x^{2}\alpha$

after some modification $ \frac{x\minus{}\log (x\plus{}1)}{x^{2}}\leq\alpha$

So, the problem is analogous to find the maximum of that function.

That function is decreasing, so the Maximum is on $ \lim_{x\to 0}\,\frac{x\minus{}\log (x\plus{}1)}{x^{2}}\equal{}\frac{1}{2}$

The second problem

$ \cosh (x)\leq e^{x^{2}\alpha }$ so $ \frac{\log (\cosh (x))}{x^{2}}\leq\alpha$

This function has a maximum on zero and the limit is half unit too.

I will omit the proof that this is a maximum


Let $a$, $b$, $c$ be positive real numbers such that $a^2+b^2+c^2=1$. Prove the inequality \[a+b+c+\frac{1}{abc} \geq 4\sqrt{3}.\]

Thanks, now we know the source. ;)

http://www.mathlinks.ro/Forum/viewtopic.php?t=26913 .

  darij


inequalities
calculus
function
algebra
polynomial
inequalities proposed

Let $ x,y,z$ be positive integers such that $ x\plus{}y\plus{}z\equal{}3$. Prove that

$ \sum_{cyc} \frac{x}{3x^2\plus{}xyz\plus{}27}\leq \frac{3}{31}$.

Come on. Let's look for a nice solution!  :)

[quote="easternlatincup"]Let $ a,b,c$ be positive integers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ \sum_{cyc} \frac {a}{3a^2 \plus{} abc \plus{} 27}\leq \frac {3}{31}$.[/quote]
I got 2 solutions for this one, the first is by CYH technique. The second is SOS. Let me post the solution by SOS :)
By Schur Inequality, we get $ 3abc \ge 4(ab \plus{} bc \plus{} ca) \minus{} 9$. It suffices to prove that
\[ \sum \frac {3a}{9a^2 \plus{} 4(ab \plus{} bc \plus{} ca) \plus{} 72} \le \frac {3}{31}
\]

\[ \sum\left( 1 \minus{} \frac {31a(a \plus{} b \plus{} c)}{9a^2 \plus{} 4(ab \plus{} bc \plus{} ca) \plus{} 72}\right) \ge 0
\]

\[ \sum \frac {(7a \plus{} 8c \plus{} 10b)(c \minus{} a) \minus{} (7a \plus{} 8b \plus{} 10c)(a \minus{} b)}{a^2 \plus{} s} \ge 0
\]
where $ s\equal{} \frac {4(ab \plus{} bc \plus{} ca) \plus{} 72}{9}$.
\[ \sum (a \minus{} b)^2\frac {8a^2 \plus{} 8b^2 \plus{} 15ab \plus{} 10c(a \plus{} b) \plus{} s}{(a^2 \plus{} s)(b^2 \plus{} s)} \ge 0
\]
which is true. :)

[quote="can_hang2007"][quote="easternlatincup"]Let $ a,b,c$ be positive integers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ \sum_{cyc} \frac {a}{3a^2 \plus{} abc \plus{} 27}\leq \frac {3}{31}$.[/quote]
I got 2 solutions for this one, the first is by CYH technique. The second is SOS. Let me post the solution by SOS :)
By Schur Inequality, we get $ 3abc \ge 4(ab \plus{} bc \plus{} ca) \minus{} 9$. It suffices to prove that
\[ \sum \frac {3a}{9a^2 \plus{} 4(ab \plus{} bc \plus{} ca) \plus{} 72} \le \frac {3}{31}
\]

\[ \sum\left( 1 \minus{} \frac {31a(a \plus{} b \plus{} c)}{9a^2 \plus{} 4(ab \plus{} bc \plus{} ca) \plus{} 72}\right) \ge 0
\]

\[ \sum \frac {(7a \plus{} 8c \plus{} 10b)(c \minus{} a) \minus{} (7a \plus{} 8b \plus{} 10c)(a \minus{} b)}{a^2 \plus{} s} \ge 0
\]
where $ s \equal{} \frac {4(ab \plus{} bc \plus{} ca) \plus{} 72}{9}$.
\[ \sum (a \minus{} b)^2\frac {8a^2 \plus{} 8b^2 \plus{} 15ab \plus{} 10c(a \plus{} b) \plus{} s}{(a^2 \plus{} s)(b^2 \plus{} s)} \ge 0
\]
which is true. :)[/quote]
can you talk abuot CYH tecnique and post your sencond solution,Can?

[quote="tronghieu"]
can you talk abuot CYH tecnique and post your sencond solution,Can?[/quote]

The one he posted is his second solution, and it seems nice. Anyway, I'm interested in more "elementary and nice" solution  :)  :D

I saw this line in the question:
[size=150]Let  x ,y ,z be positive integers.[/size]
So is it a trick?

[quote="fjwxcsl"]It`s  equivalent  to
$ 23G91 \plus{} 145G92 \plus{} 1624G93 \plus{} 2568G94 \plus{} 9784G95 \plus{} 5544G96 \plus{} 6844G97 \plus{} 880G98 \plus{} 405G99 \minus{} 123G910 \plus{} 3045G911\geq0,$
in which
$ G91 \equal{} \sum{x^7(x \minus{} y)(x \minus{} z)}\geq0$,
$ G92 \equal{} \sum{x^6(y \plus{} z)(x \minus{} y)(x \minus{} z)}\geq0$,
$ G93 \equal{} \sum{x^5yz(x \minus{} y)(x \minus{} z)}\geq0$,
$ G94 \equal{} \sum{x^4yz(y \plus{} z)(x \minus{} y)(x \minus{} z)}\geq0$,
$ G95 \equal{} \sum{x^3(yz)^2(x \minus{} y)(x \minus{} z)}\geq0$,
$ G96 \equal{} \sum{x^2(y \plus{} z)(yz)^2(x \minus{} y)(x \minus{} z)}\geq0$,
$ G97 \equal{} \sum{x(yz)^3(x \minus{} y)(x \minus{} z)}\geq0$
$ G98 \equal{} \sum{(y \plus{} z)(yz)^3(x \minus{} y)(x \minus{} z)}\geq0$,
$ G99 \equal{} \sum{(x \plus{} y \plus{} z)^3x^2(y \minus{} z)^2(x \minus{} y)(x \minus{} z)}\geq0$,
$ G910 \equal{} \sum{(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx)x^2(y \minus{} z)^2(x \minus{} y)(x \minus{} z)}\geq0$,
$ G911 \equal{} \sum{yzx^3(y \minus{} z)^2(x \minus{} y)(x \minus{} z)}\geq0$,
$ G99\geq3G910$.[/quote]
How'll you do with the following stronger:
$ \sum_{cyc} (\frac {a}{3a^2 \plus{} abc \plus{} 27})^{\frac{10}{9}}\leq \frac{3}{31^{\frac{10}{9}}}$
You can also see here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=199850&start=20
It can be solved easily by Am-Gm  :)

[quote="easternlatincup"]Let $ x,y,z$ be positive integers such that $ x \plus{} y \plus{} z \equal{} 3$. Prove that

$ \sum_{cyc} \frac {x}{3x^2 \plus{} xyz \plus{} 27}\leq \frac {3}{31}$.[/quote]
If $ x, y, z$ are positive INTEGERS and $ x \plus{} y \plus{} z \equal{} 3$, the only values for $ x, y, z$ are :  $ (x, y, z) \equal{} (1, 1, 1)$ :D

So I think your question should be : "For POSITIVE QUANTITIES x, y, z......."

Don't you think so? :)
Edit your question. However, I think Can_hang 's solution is nice! :lol:

[quote="fjwxcsl"]It`s  equivalent  to
$ 23G91 \plus{} 145G92 \plus{} 1624G93 \plus{} 2568G94 \plus{} 9784G95 \plus{} 5544G96 \plus{} 6844G97 \plus{} 880G98 \plus{} 405G99 \minus{} 123G910 \plus{} 3045G911\geq0,$
in which
$ G91 \equal{} \sum{x^7(x \minus{} y)(x \minus{} z)}\geq0$,
$ G92 \equal{} \sum{x^6(y \plus{} z)(x \minus{} y)(x \minus{} z)}\geq0$,
$ G93 \equal{} \sum{x^5yz(x \minus{} y)(x \minus{} z)}\geq0$,
$ G94 \equal{} \sum{x^4yz(y \plus{} z)(x \minus{} y)(x \minus{} z)}\geq0$,
$ G95 \equal{} \sum{x^3(yz)^2(x \minus{} y)(x \minus{} z)}\geq0$,
$ G96 \equal{} \sum{x^2(y \plus{} z)(yz)^2(x \minus{} y)(x \minus{} z)}\geq0$,
$ G97 \equal{} \sum{x(yz)^3(x \minus{} y)(x \minus{} z)}\geq0$
$ G98 \equal{} \sum{(y \plus{} z)(yz)^3(x \minus{} y)(x \minus{} z)}\geq0$,
$ G99 \equal{} \sum{(x \plus{} y \plus{} z)^3x^2(y \minus{} z)^2(x \minus{} y)(x \minus{} z)}\geq0$,
$ G910 \equal{} \sum{(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx)x^2(y \minus{} z)^2(x \minus{} y)(x \minus{} z)}\geq0$,
$ G911 \equal{} \sum{yzx^3(y \minus{} z)^2(x \minus{} y)(x \minus{} z)}\geq0$,
$ G99\geq3G910$.[/quote]

I did not understand ANYTHING?! :huh:
Could you please elaborate a little?

anyone please help me understand this. :(

Dear my friend, it's only calculus in computer, I think you shouldn't try to understand it.  :)
[quote="nguoivn"]
How'll you do with the following stronger:
$ \sum_{cyc} (\frac {a}{3a^2 \plus{} abc \plus{} 27})^{\frac {10}{9}}\leq \frac {3}{31^{\frac {10}{9}}}$
You can also see here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=199850&start=20
It can be solved easily by Am-Gm  :)[/quote]

[quote="Gibbenergy"][quote="fjwxcsl"]It`s  equivalent  to
$ 23G91 \plus{} 145G92 \plus{} 1624G93 \plus{} 2568G94 \plus{} 9784G95 \plus{} 5544G96 \plus{} 6844G97 \plus{} 880G98 \plus{} 405G99 \minus{} 123G910 \plus{} 3045G911\geq0,$
in which
$ G91 \equal{} \sum{x^7(x \minus{} y)(x \minus{} z)}\geq0$,
$ G92 \equal{} \sum{x^6(y \plus{} z)(x \minus{} y)(x \minus{} z)}\geq0$,
$ G93 \equal{} \sum{x^5yz(x \minus{} y)(x \minus{} z)}\geq0$,
$ G94 \equal{} \sum{x^4yz(y \plus{} z)(x \minus{} y)(x \minus{} z)}\geq0$,
$ G95 \equal{} \sum{x^3(yz)^2(x \minus{} y)(x \minus{} z)}\geq0$,
$ G96 \equal{} \sum{x^2(y \plus{} z)(yz)^2(x \minus{} y)(x \minus{} z)}\geq0$,
$ G97 \equal{} \sum{x(yz)^3(x \minus{} y)(x \minus{} z)}\geq0$
$ G98 \equal{} \sum{(y \plus{} z)(yz)^3(x \minus{} y)(x \minus{} z)}\geq0$,
$ G99 \equal{} \sum{(x \plus{} y \plus{} z)^3x^2(y \minus{} z)^2(x \minus{} y)(x \minus{} z)}\geq0$,
$ G910 \equal{} \sum{(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx)x^2(y \minus{} z)^2(x \minus{} y)(x \minus{} z)}\geq0$,
$ G911 \equal{} \sum{yzx^3(y \minus{} z)^2(x \minus{} y)(x \minus{} z)}\geq0$,
$ G99\geq3G910$.[/quote]

I think it is a nice solution after all.  :)
To Potla, basically, he expanses everything and use Schur inequality to prove that the numerator term should be positive.
I like this solution because I saw his technique. As you can see, he formed a set of independent function of degree 9 to represents 9 th degree polynomial of the numerator.
I guess you can also do it by hand if you have enough patience to solve a system of 9 equations.
Of course, from Schur form, one can easily transform it to SOS form, which is more natural to think of. [As reader sees Can's proof]

It is just my opinion.  :)[/quote]
Thanks for your elaboration. Actually I could not understand the notation of G93, G911 , ....... :roll:

I think both are nice! :rotfl:

[b]can_hang2007[/b] yours proof is magnificent, in my opinion. I like SOS although it's difficult, especially for me... Could you tell me something about CYH technique, please ?
Maybe you have a nice example for this method?  :roll:

And an interesting question: Find the best constant k satisfy: (with the same condition)
$ \sum_{cyc} (\frac {a}{3a^2 \plus{} abc \plus{} 27})^{k} \leq \frac {3}{31^k}$
With $ k \leq\ \frac {10}{9}$, as I have said, we can prove it easily by Am-Gm  :)


what is the equivalience point pH for HCOOH + NaOH titration

When you have added as many moles of NaOH as there are moles of HCOOH.

By definition equivalence point is when there is stoichiometric amount of titrant added. SO in this case it will be solution of pure sodium formate.

The pH of the resulting mixture with equal moles of both HCOOH and NaOH is the same as if you started with a solution of Na+ HCOO-.

The weak base HCOO- will take some H+ from the water, forming some HCOOH.  The exact pH is dependent on the concentration of the HCOO-, but the solution will be weakly basic.

Weak base hydrolysis. Check [url=http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base]weak base pH calculation lecture[/url] for theory and [url=http://www.chembuddy.com/?left=BATE&right=pH-cheat-sheet]pH cheat sheet[/url] for practical solution.

Hm... the replies grow more and more technical as the thread continues.  I'll have to balance it.

The point on the titration curve midway along the really steep part.

Just the following:
HCOOH+NaOH---> HCOONa +H2O
 HCOONa----> HCOO- +Na+
HCOO- +H2O ---> HCOOH + OH-
then Kb=[HCOOH][OH-]/[HCOO-]<=> [OH-]=...
pH=...


integration
calculus
calculus computations

Compute $\int \sqrt{\frac{x}{1-ax}}\, dx$, where $a \in \mathbb R$.

Hi,
We consider the case when $a\neq 0$.
Use the substitution $y^{2}=\frac{x}{1-ax}$. You get that

$I=\int\frac{2y^{2}}{(1+ay^{2})^{2}}dy=\frac{2}{a}\left(\int\frac{dy}{1+ay^{2}}-\int\frac{dy}{(1+ay^{2})^{2}}\right)$.
The first integral is an arctangent while for calculating the second one
you can use this trick

$\int\frac{dy}{(1+ay^{2})^{2}}=\frac{y}{2(1+ay^{2})}+\frac{1}{2}\int\frac{dy}{1+ay^{2}}$.
I hope that there are no mistakes.


Find the sum of all 3 digit positive integers divisible without remainder by exactly 8 out of 9 single digit positive integers.

[hide="Solution"]Firstly, everything is divisible by 1, so one of our 8 digits is 1.
We claim that one of our digits is 2. Suppose this were not the case; then the rest of the digits would have to be 3,4,5,...,9. However, 4,6, and 8 are multiples of 2, and therefore, any integer divisible by those would be divisible by 2: contradiction. Therefore, 2 is another of the 8 digits.
By a similar argument, 3 and 4 must be one of our digits as well.
If 6 were not one of our digits, that would imply that the 3 digit number were not divisible by 2 and/or 3, which is not the case, so  6 is one of our digits. By a similar argument, 8 and 9 are one of our digits as well.
We are left with 5 and 7, both primes whose multiples are greater than 9.

Let the three digit number be $ N$.

[b]Case 1:[/b] [i]The last digit is 5[/i].
In this case, $ N|\text{lcm}(1,2,3,4,5,6,8,9) \equal{} 360$. The only multiples of 360 that are 3 digits long are 360 and 720.

[b]Case 2:[/b] [i]The last digit is 7[/i].
In this case, $ N|\text{lcm}(1,2,3,4,6,7,8,9) \equal{} 504$. The only multiple of 504 that is 3 digits long is 504.

Therefore, our answer is $ 360 \plus{} 720 \plus{} 504 \equal{} \boxed{1584}$.[/hide]

Edit: Oh, yes, ffao is right. Sorry.  :blush:

[hide="Little correction"]
[quote="grn_trtle"]
If 6 were not one of our digits, that would imply that the 3 digit number were not divisible by 2 and/or 3, which is not the case, so  6 is one of our digits. By a similar argument, 8 and 9 are one of our digits as well.
[/quote]

The "similar argument" does not hold. For a simple example, 15 is a multiple of 3 but not of 9. So we have two more possibilities:

Case 3. (the last digit is 8)
In this case, $ N|lcm(1, 2, 3, 4, 5, 6, 7, 9) \equal{} 1260$. As 1260 is 4 digits long, there are no solutions in this case.

Case 4. (the last digit is 9)
In this case, $ N|lcm(1, 2, 3, 4, 5, 6, 7, 8) \equal{} 840$. The only multiple of 840 that is 3 digits long is 840, and that isn't divisible by 9.

Therefore, the answer is $ 360 \plus{} 720 \plus{} 504 \plus{} 840 \equal{} \boxed{2424}$
[/hide]


I cannot do it even using Pythagoras theorem and similarity of triangles...
In a quadrilateral ABCD, it is given that AB is parallel to CD and the diagonals AC and BD are perpendicular to each other. Show that:
(a) $ AD.BC\ge AB.CD$
(b) $ AD \plus{} BC\ge AB \plus{} CD$
[hide="My progression"]Can you give me hints?
I saw that (a) implies (b) and vice versa.
MY PROGRESSION:
Let AC and BD intersect at O.
$ OA^2 \plus{} OB^2 \equal{} AB^2$
And similar for the other sides.
Hence $ AD^2 \plus{} BC^2 \equal{} AB^2 \plus{} DC^2$
Also $ \Delta OCD$ is similar to $ \Delta AOB$.
Implying that $ \frac {OA}{OB} \equal{} \frac {OC}{OD} \Longrightarrow OA\cdot OB \equal{} OB\cdot OC$
Hence $ AB^2 \plus{} DC^2\ge 4 OB\cdot OC$,
         $ AD^2 \plus{} BC^2\ge 4 OA\cdot OD$

I have even thought about the extension of ptolemy:
$ BC.AD \plus{} AB.CD > AC.BD$

But I cannot get any further!!

Please help me :( [/hide]
[hide="P/s:"]
Can anyone please teach me how to draw images or diagrams using geogabra or whatever??[/hide]

see here ==> [url]http://www.mathlinks.ro/viewtopic.php?p=570007#570007[/url]

[hide="number one"]
Let $ AC$ and $ BD$ intersect at $ O$. then we have two similar triangles, $ \bigtriangleup ACO \sim \bigtriangleup BDO$. Let $ \frac {AB}{DC} \equal{} k$, the sclaing factor of the two triangles.
Let $ AO \equal{} a, OD \equal{} ka, BO \equal{} b, OC \equal{} kb$
Then we have a bit of a bash with pythag

$ AB^2 \equal{} a^2 \plus{} b^2$
$ DC^2 \equal{} (ka)^2 \plus{} (kb)^2$
$ AB^2 \equal{} (ak)^2 \plus{} b^2$
$ BC^2 \equal{} (bk)^2 \plus{} a^2$

After multiplying the relevant terms, it suffices to prove that
$ k^4(ab)^2 \plus{} (ab)^2 \ge 2k^2(ab)^2$
$ k^4 \minus{} 2k^2 \minus{} 1 \equal{} (k^2 \minus{} 1)^2 \ge 0$

$ \Rightarrow AB^2\cdot DC^2 \le AD^2\cdot BC^2$
$ \Rightarrow AB\cdot DC \le AD\cdot BC \qquad QED$[/hide]

[hide="number two"]
Using the same notation as above
$ AD^2 \plus{} BC^2 \equal{} (k^2a^2\plus{}b^2) \plus{} (b^2k^2\plus{}b^2) \equal{} k^2(a^2\plus{}b^2) \plus{} (a^2\plus{}b^2) \equal{} AB^2\plus{}DC^2$
So using our last result
$ AD^2 \plus{} BD^2 \plus{} 2AD\cdot BC \ge AB^2 \plus{} DC^2 \plus{} 2AB\cdot DC$
$ (AB\plus{}BD)^2 \ge (AB\plus{}DC)^2$
$ (AB\plus{}BD) \ge (AB\plus{}DC) \qquad QED$

[/hide]

Thanks to you both for helping me out!!!!!!!!!! :lol:


algebra
polynomial
real analysis
real analysis theorems

let $T_{n}g(x;a)$ be the Taylor polinomial of $g$ at $a$

if $g(x)=f(cx)$ then $T_{n}g(x;a)=T_{n}f(cx;ca)$

what can we say if $g(x) = f(cx+b)$?


integration
real analysis
real analysis solved

Let f:[0,00)->R be a continuous function.

(a) If f is increasing show that \int{a->b}f(x)dx<=bf(b)-af(a) for all b>=a>=0.
(b) Study the converse sentence (is it true?).

a) f is increasing  \int _[a,b]f(x)dx \leq \int _[a,b]f(b)dx
 \int _[a,b]f(x)dx \leq f(b)(b-a)=bf(b) - af(a) + a(f(a) - f(b))

b \geq a  \geq 0 then a(f(a)-f(b))  \leq 0

\int _[a,b]f(x)dx \leq bf(b) - af(a) + 0

Let b \geq a \geq 0

bf(b)-af(a) \geq  \int _[a,b]f(x)dx  \geq \int _[a,b]f(a)dx
bf(b) -af(a)  \geq bf(a) - af(a)
b(f(b) - f(a))  \geq 0  but b is positive => f(b) - f(a)  \geq 0

f is increasing

I thought it was untrue
10x grobber


The inradius of triangle $ ABC$ is $ 1$ and the side lengths of $ ABC$ are all integers. Prove that triangle $ ABC$ is right angled.

we obtain $ 4(a \plus{} b \plus{} c) \equal{} (a \plus{} b \minus{} c)(a \plus{} c \minus{} b)(b \plus{} c \minus{} a)$
the cases $ a \equal{} b \equal{} c$ and $ a \equal{} b$ are easy to check.
now suppose that $ a > b > c$ then we'll get $ a \plus{} b \minus{} c < a \plus{} b \plus{} c < 3(a \plus{} b \minus{} c)$ and since $ a \plus{} b \minus{} c |4( a \plus{} b \plus{} c )$ we have these case  which are again easy to check:
$ i(a \plus{} b \minus{} c) \equal{} 4(a \plus{} b \plus{} c)$ where $ 5 \le i \le 11$ by using this we get $ (a \plus{} c \minus{} b)(b \plus{} c \minus{} a) \equal{} i$ the rest is trivial.


Find all the positive integers x,y,z satisfiing : x^{2}+y^{2}+z^{2}=2xyz

[color=red]Edit by Megus: I've changed your topic's title - please name it next time appropriately[/color]

By looking at parity, and using infinate descent, the only solution is $(0,0,0)$

Sorry if i am brief, i must run for a train!

That is called Markov`s equation in case n=2
$x^2+y^2+z^2=nxyz$

It only has solution for $n\in{1,3}$

I think anyway... it only has an infinate number of solutions in 1 and 3, but im not sure if any other solutions exist...


AMC
AIME
USA(J)MO
USAMO
AMC 12
blogs
geometry

What are you guys (and girls) doing to study for the AIME?
Post your daily schedule.

I'm not sure if I'm going to pass AIME this year, but I'll pass next year for sure... I heard that the AIME Problem Series really helps. Also, just do a lot of the past years' problems. That helps as well. Go through AoPS Volumes 1 and 2 and REALLY REALLY REALLY study them (especially Volume 2). And pray a lot. I find that to help on the AMC especially  ;)   :D

No specific daily schedule...just try to do some aops v. 2 and practice aime problems.

every weekend i do either some amc or aime practice tests, or i work on olympiad stuff, but recently, i have been doing many tests to improve my speed, and also i do some easier tests like there is this north carolina test that i go through every once in a while

[quote="pkerichang"]how do you really study for USAMO, tho??? (AoPS is too easy I heard...)[/quote]

Past USAMO/IMO/Other Olympiad Problems. Exhaust them, and you should be good ;).

[quote="JesusFreak197"]Will any USAMO problems be posted here, usually?[/quote]I'm not sure what you mean. They're all posted in the archive. :)

Actually, I take that back. Several years have not been posted yet. Maybe I should post some, as I have all the original problems from 1972-1986. (A book my teacher gave me) :)

Do you guys think I should try a few USAMO problems once in a while? Let me re-word that. Do you think it will help me improve my Aime scores?

I usually get Aime scores from, 6-8, so they'll be challenging. However, I enjoy challenges.  :)

Well, USAMO is not really the same kind of thing as AIME, but it's possible that they'd still help some.

I was asking if people would post USAMO problems in this forum or Intermediate so that I can practice writing up solutions. I'd rather not handwrite them until the actual test, because it's a lot more work, and I don't want to bump old threads. ;)

That's what I was talking about. :)

Check [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=27]this link.[/url]

[quote="JesusFreak197"]Well, USAMO is not really the same kind of thing as USAMO...[/quote]

Oh. I didn't know that.

He means USAMO is different from AIME.  ;)

You know what he means...  :P

I am going to attempt to exhaust all of the aimes on kalva by next year.  There is no way I will make it this year.. lol I might get like a 3 or a 4 at best.

[quote="Karth"]Would you (anyone here) say that studying ONLY the previous AIMEs from past years (and perhaps mock AIMEs from the forums) would help you get a decent score on the AIME (passing the AIME, as in 12+)?[/quote]

Nothing is ever guarenteed.

Just do problems.

[quote](passing the AIME, as in 12+)?[/quote]

That's not passing the AIME. That's [i]crushing[/i] it.

[quote="Karth"]Would you (anyone here) say that studying ONLY the previous AIMEs from past years (and perhaps mock AIMEs from the forums) would help you get a decent score on the AIME (passing the AIME, as in 12+)?[/quote]

Of course not (it certainly wont hurt though).  Your going to improve your score if the math you are doing is forcing you to bang your head against the wall and try new things regardless of whether they are AIME problems or not.

[quote="lowest_reynolds"]
Just do problems.
[/quote]

Yea,
Just do math.

Sounds like it should be on a shirt....lol

[quote="SnowStorm"][quote="Karth"]Would you (anyone here) say that studying ONLY the previous AIMEs from past years (and perhaps mock AIMEs from the forums) would help you get a decent score on the AIME (passing the AIME, as in 12+)?[/quote]

Of course not (it certainly wont hurt though).  Your going to improve your score if the math you are doing is forcing you to bang your head against the wall and try new things regardless of whether they are AIME problems or not.

Sounds like it should be on a shirt....lol[/quote]

So, we have to learn how to approach the problems in an new way?

@ tetrahedr0n: What would someone consider a passing score on the AIME to be?  :D

I'd say passing would be the floor. Were you considering passing to be what would be a C or higher, percentage-wise?

Well, I guess passing would be to make the USAMO... what's the range usually?

Scoring the floor or higher is making USAMO if you're a sophomore or younger. This varies between years depending on the number of qualifiers for USAMO (which I believe is the same or approximately the same each year, except that this year it has been increased by ~50% due to AoPS) and depending on the difficulty of the test. The floor has only been in existence since 2002(?), and so far they have been 6, 8, 7, and 9, respectively. Last year's was unusually easy; expect it to be around 7 this year, possibly 6 due to the expansion, unless the test is unusually hard or easy.

However, it's entirely different for juniors and seniors, who score by index, which is $\text{AMC}+10\cdot \text{AIME}$. So, it'll normally be approximately the same, but if you scored higher or lower on the AMC, you'll need to score lower or higher, respectively, on the AIME to qualify for USAMO.

[quote="h_s_potter2002"][quote]However, it's entirely different for juniors and seniors, who score by index, which is \text{AMC}+10\cdot \text{AIME}. So, it'll normally be approximately the same, but if you scored higher or lower on the AMC, you'll need to score lower or higher, respectively, on the AIME to qualify for USAMO.[/quote]

Wouldn't you always have to score the same, or higher as the floor on the AIME as a junior or senior, because the floor is the lowest AIME score out of all the qualifiers.[/quote]

Yes.  You have to score as high as the [i]lowest[/i] junior/senior who makes the index which is generally considerably easier than making the index.  For example, the index was a 210 in 2002.  Somehow someone managed to get a 150 on the AMC 12 and then get a 6 on the AIME (really good day on the AMC or really bad day on the AIME, apparently).

[quote]@ tetrahedr0n: What would someone consider a passing score on the AIME to be?  [/quote]

Haha, I don't know. I'd like to think I passed it at least once though. I like the floor suggestion.

i started doing past AIMES again....
just finished the 2001B test...

Ahh... I'm only a sophmore this year... do you guys think I still have much time? (I don't think I will make it past AIME this year :( )

[quote="lfm"]Not quite. Anyone who took the AMC12 are selected first based on index. Then the index is applied to the sophmores and younger who did not make the first draft.[/quote]

Oh, okay. I had heard that sophomores and younger could qualify by index, too; I guess that's how it works.

[quote="lfm"][quote="JesusFreak197"]That's what I meant. Juniors and seniors are accepted first based on index; after that, sophomores and younger are accepted based on floor, which is determined by the juniors and seniors who already qualified.[/quote]

Not quite. Anyone who took the AMC12 are selected first based on index. Then the index is applied to the sophmores and younger who did not make the first draft.[/quote]

Haha, it'd be really funny if it was all freshman and sophomore who qualified for USAMO.

[quote="h_s_potter2002"]Haha, it'd be really funny if it was all freshman and sophomore who qualified for USAMO.[/quote]

From time to time, an 8th grader gets in "the hard way," and once in a while even a 7th grader. This year should be a reasonably good year for 8th graders to qualify for USAMO's expanded eligibility by attaining the AIME floor score.

How am I preparing?
I average maybe 1 AIME a week.  That's about it.  I don't want to go overboard with practice, having a lot of other commitments...it might just be a waste of time, better spent doing usamo stuff anyways.  But I have been too lazy to do usamo stuff...

[quote="chess64"][quote="matt276eagles"]I don't know about you guys, but I find that just doing practice tests doesn't really improve my score.  I generally use practice tests just to see if I'm improving.[/quote]

Obviously you aren't doing it right. Make sure to read solutions to every single problem you didn't get. And really understand it. Like you have to explain it to your friends the next day. You should also consider dedicating a blog for the purpose of posting solutions of AIME problems. *cough* my blog *cough* My score has improved more than 4 points by doing past AIME problems. (5 to 9+, \/\/00+)[/quote]

Well I don't think it's OBVIOUS that he's not doing it right, but I would agree that doing practice tests does serve as a good way to improve one's AIME score. It's also helpful to get the foundations - Algebra, Counting, Geometry, Number Theory are the big four to know. Make sure you know the general theorems/ideas of each and how to apply them effectively (DO PRACTICE PROBLEMS). That should get you a long ways.

[quote="paladin8"]Well I don't think it's OBVIOUS that he's not doing it right...[/quote]

lol, I was just kidding. :)

Eat, sleep and do math  :D

It was on the back of our school mathcounts shirts in middle school...

[quote="randomdragoon"]Eat, sleep[/quote]
What? That's weak.
[quote]and do math  :D[/quote]
OK that's allowed... oh yeah I think one thing that really helped me get better is to not give up on AIME problems you can't do for at least half an hour (often an hour or more, even).


[quote="Klebian"]I was going over AOPS bk 1 and on ch 4, there's a problem:

68) x is directly proportional to y and inversely proportional to z. If x = 1/2 when y = 3/4, find x when y = 7/8 and z = 7/9.

The solution tells to take xz/y as constant, but what I was wondering was why can't we ignore y and just take xz as a constant and solve like that? It shouldn't be incorrect, as, after all, x [i]is[/i] inversely proportional to z.[/quote]
x is inversely proportional to z while being directly proportional to y at the same time. You'd have to change the wording of the problem to get your reasoning.

[quote="lotrgreengrapes7926"][quote="Klebian"]I was going over AOPS bk 1 and on ch 4, there's a problem:

68) x is directly proportional to y and inversely proportional to z. If x = 1/2 when y = 3/4, find x when y = 7/8 and z = 7/9.

The solution tells to take xz/y as constant, but what I was wondering was why can't we ignore y and just take xz as a constant and solve like that? It shouldn't be incorrect, as, after all, x [i]is[/i] inversely proportional to z.[/quote]
Well, if you ignore $z$ and take $\frac{x}{y}$ to be constant, you'll get a different answer.[/quote]

I know that, but it's still answering the queston...


There are eight circles in the picture below. The shaded circles
are congruent. If each circle in the picture is tangent to each of
its neighbor circles, what is the ratio of
the total area of the seven shaded
regions to the area of the largest circle?
Express your answer as a common
fraction.

[asy]draw(Circle((0,0),30));
fill(Circle((0,0),10),gray(0.7));
fill(Circle((17.3,10),10),gray(0.7));
fill(Circle((17.3,-10),10),gray(0.7));
fill(Circle((-17.3,10),10),gray(0.7));
fill(Circle((-17.3,-10),10),gray(0.7));
fill(Circle((0,20),10),gray(0.7));
fill(Circle((0,-20),10),gray(0.7));
draw(Circle((0,0),10));
draw(Circle((17.3,10),10));
draw(Circle((17.3,-10),10));
draw(Circle((-17.3,10),10));
draw(Circle((-17.3,-10),10));
draw(Circle((0,20),10));
draw(Circle((0,-20),10));[/asy]

If a small circle has radius $ r$, then the large circle has radius $ 3r$, and the desired ratio is \[ \frac{7r^2}{(3r)^2}\equal{}\boxed{\frac{7}{9}}\]


limit
number theory unsolved
number theory

Suppose $ \pi(x)$ is number primes less than $ x$
Prove that:
$ \lim_{x\to \infty}\frac{\pi(n)\log(n)}{n}\equal{}1$


Ok, so I went to my blog options, and for replies permission, I selected "Registered Users".  But for some reason, nobody can post a comment...

help!

Mine is also set as registered users. Maybe try setting as guests first. Then see what happens

Ok, I deleted my blog and tried to make a new one, but I couldn't...

ya, i've read about some other complaints about pplz not being able to get their blog back

ask one of the admins to undelete your blog

Ok, now someone help me with the inability to make comments?????????

Hi, sorry no one responded sooner. I don't know much about the blogging system but I think I figured out the problem.

You had the Post Reply Comment field left blank in your profile, so no link was showing. I put in something there and users can now post replies.

Problem fixed. If field is left blank, the default text is now put in place.


Is there any easy way to find certain terms of the sequence that are too hard to find just by writing the sequence out?

can you give an example of exactly what you are looking for? :?:

[quote="indianamath"]where n is the nth term:

lol i'm too lazy to put it in Latex.[/quote]

Is that remotely practical if you're finding any term beyond the 3rd term or so?

The specific problem that inspired me to ask this question is:

What is the remainder when the 100th term of the Fibonacci sequence is divided by 4?

Even if the problem ends up not needing to find the 100th term, I still wondered if there was any way to find such a term.

for there is a pattern in any modulo. here it is modulo 4:

[b]1,1,2,3,1,0,[/b]1,1

the bold part is what repeats.

i remember doing this problem last year in 6th grade, i had no clue either. :lol:  :)

hehe bad grammar on last post. i was starting to write something and went onto something esle


geometry
geometric transformation
reflection
Support
inequalities
search
articles

What do you think of the culture clash Europe vs. Middle East? Or in other words how to value freedom of press vs. freedom of religion? The origin are insulting caricatures of Prophet Muhammad. Does personal religious belief legitimate riots against Northern European embassies in the Middle East? What do you think about the recent statements of Iran's president vs. those mentioned cartoons?

- http://www.nytimes.com/2006/02/05/international/middleeast/05cnd-beirut.html?hp&ex=1139202000&en=3ae1e60f699adbaf&ei=5094&partner=homepage

- http://news.ft.com/cms/s/b90d681a-9685-11da-a5ba-0000779e2340.html

I am quite interested in your opinions as I am thinking to go to Haifa, Israel for some time.

Well, also here in Belgium there is fierce debate about that.  It really is a major issue right now.

Let's get one thing straight : where are the original cartoons?  I would like to see them.( I saw the one with the bomb on top of the head)

I personally think that this is a slap in the face of Europe.  The one with the bomb isn't that bad, it would not pick it out right away as over the top.  However, publishing them like in france for the kick out of it is asking for problems, and as at that time was already clear, very risky for people of the same country.

When one says islam is to be associated with violence, muslims are offended by the generalization.  But when one Danish makes a cartoon they are very quickly to generalise themselves.

It also irks me that is substantial proof that Danish imams went on a 'hate rally' in middle eastern countries with the cartoons...and with a whole bunch of neonazi work that had nothing to do with the cartoons in the paper at all, thus misinforming the people.

How did the western world feel when we saw trains exploding, towers collapsing,etc...  Americans for instance love their flag very much, it is a sport to burn it in muslim countries.  How would they react if the Spanish made banners :"Slay all those muslims'.  If the US government would demand the arrest of an Iranian or Palestinian flag burner they saw on tv, they'd you have nothing to say here.

Hasn't Europe done anything valuable in the middle east?  Palestine basically lives of aid from the European Union (which might now diminish as fatah has lost).

Maybe those muslims acting violently don't understand our culture very well.  Papers are not controlled by the government.  Here in Belgium, we make a national sport of mocking politicians, our monarchs,etc...

They should really just let it go.  Don't they have other worries?  In Palestine and Israel, the old leaders are being replaced.  I don't understand how they can keep going about it.

One tip to muslims reacting violently, like burning the consulate : try thinking why the cartoon was made, and realise that this violent reaction is the ideal proof for the idea behind the cartoon

Sorry for what may seem a far left view.  I do not consider myself racist.  However, I think this whole incident shows a dangerous culture clash indeed.  It shows Muslims in the middle east have completely different priorities.  Also, after the murder of theo van gogh, nobody here is willing to give more freedom of speech away.

[quote="orl"]Or in other words how to value freedom of press vs. freedom of religion? The origin are insulting caricatures of Prophet Muhammad.[/quote]There should be no debate here. Freedom of speech/press is about voicing opinions and facts that are to benefit people, not about saying whatever you feel like saying. For example, if a government is corrupt and no one is allowed to speak of the government's corruption -  this is where freedom of speech is important. It is not the purpose of freedom of speech and freedom of press to make jokes about a religious figure.

When are people allowed to make racist comments and claim 'freedom of speech'?

On the other hand, the people, the extremists, who are committing these acts of violence do not reflect the views and beliefs of the Muslim community, though they claim their actions to be in the name of Islam.

I stick with my opinion.  I just read that in Turkey, a christian priest was killed.  I don't care anymore about the original cartoons.

I still call this a slap in the face of Europe.  Many countries, like Belgium and Denmark, still try hard to be a tolerant country.  There are mosques here, the children can take islamic courses in public schools, etc...

How hateful and intolerant can you be?  Don't they realise all this will do is pull a huge amount of voters to far left all over Europe.

Simply, I cannot find words to express how twisted and intolerant i find this.[url=http://edition.cnn.com/2006/WORLD/europe/02/06/london.cartoon.protests/index.html]picture[/url]
take a look at this picture...

Please don't desecrate the word "tolerance" before you understand the following proverb by Confucius, "What you do not want to done to yourself, do not do to the others."
The generalized version is, "Press nothing on others, even your most favourite."

[quote="orl"]Or in other words how to value freedom of press vs. freedom of religion? [/quote]

-Just want to point out that the publication of those cartoons does not inhibit freedom of religion-- nobody is any less free to practice his or her religion now than they were before the publication.

-In case anyone is interested, here is a compendium of depictions of mohammed throughout the ages (including the aforementioned cartoons): [url=http://www.zombietime.com/mohammed_image_archive/]www.zombietime.com/mohammed_image_archive/[/url]

-The whole point of having freedom of speech is to be able to say something regardless of whether it might offend someone. Otherwise, it is not free speech! Recall the following aphorism:
[quote]I do not agree with what you say, but will defend to the death your right to say it[/quote]

Personally, I believe that there is nothing wrong with criticism of Islam, and it is in fact necessary to preserve Western values (like freedom, tolerance, rationality, etc.). However, those who may differ on this point should nonetheless support free speech over censorship-- it is all too easy to take our freedom for granted. Moreover, saying what speech is 'good' and what is not requires a value judgement, and needless to say, one's values might not alway agree with one's censors' values!

[quote]When are people allowed to make racist comments and claim 'freedom of speech'? [/quote]

Any time, at least here in the USA. For example, there are several organizations with the explicit purpose of promoting racism (KKK, Neo-nazi.) Their existence is entirely legal and protected by law. They can also gather and demonstrate.

[quote] It is not the purpose of freedom of speech and freedom of press to make jokes about a religious figure. [/quote]

There is no "purpose" to freedom of speech. It's a freedom. Making jokes about any religious figure is entirely covered by it.

[quote="rookworm"]The whole point of having freedom of speech is to be able to say something regardless of whether it might offend someone. Otherwise, it is not free speech![/quote]Where are you getting this from? You talk about Western values. Don't you know that in Western society, certain rights are considered more important than others? You are from Canada? Let me quote something for you: [quote]The Canadian Charter of Rights and Freedoms guarantees the rights and freedoms set out in it subject only to such [b]reasonable limits[/b] prescribed by law as can be demonstrably justified in a free and democratic society.[/quote]I've bolded the keywords to help you out.

[quote="rookworm"]Personally, I believe that there is nothing wrong with criticism of Islam, and it is in fact necessary to preserve Western values (like freedom, tolerance, rationality, etc.).[/quote]You are so sure that these Western values that you speak of need to be preserved. But, don't we all remember a time when Western values included racial discrimination? People tried to preserve those Western values too. Yet, from the beginning of Islam, Prophet Muhammed put an emphasis on equality. I think you are mistaken on which values need to be preserved.

The freedom of speech does not preclude freedom from discrimination.

In Christianism, can you find equality between men and women, God's select people and those reject, Christianism and the other religions? At least Muhammad said that Jesus is a prophet, who was sent to this world by Yahveh, which is called Allâh in Islam. If I'm not mistaken, Islam accept the old testment too, so, as a fact, Judaism, Catholicism and Islam are the three branches of the same thing.

Anyway, Why are you so sure about your "Western values"'s correctness? Is "Western values" so sweet that it's worth pressing it on the others at the cost of tear, blood(of course, mainly the others')? Where do you get the authorization? If you think the others are uncivilized, ignorant, could you please just let them be?

Please notice the distinction between Islam and the extremism.
Fundamentalism is a word from Christianism, I think.

[quote="Buffalo"]

Anyway, Why are you so sure about your "Western values"'s correctness? Is "Western values" so sweet that it's worth pressing it on the others at the cost of tear, blood(of course, mainly the others')? Where do you get the authorization? If you think the others are uncivilized, ignorant, could you please just let them be?

.[/quote]

So what are you saying.   Our values may be wrong, so may theirs.  So either we follow our values and they do too, which in our case is , oh no!!! , a few cartoons, and in some muslim countries, means amputation of limbs of minors for petty theft, honor killings,...

Or do you mean, nobody can do anything anymore?

[quote="Valentin Vornicu"][quote="The Master"]Yet, from the beginning of Islam, Prophet Muhammed put an emphasis on [b]equality[/b]. I think you are mistaken on which values need to be preserved.[/quote]Are you sure about equality? It seems to me that there is no equality between men and women in Islam  :?[/quote]You are mistaken, my friend. The Qur'an says in Chapter 3, Verse 195:
"Their Lord responded to them: "I never fail to reward any worker among you for any work you do, be you male or female, you are equal to one another..."
Many places in the Qur'an, you will see that when a reference is made to people in general, it often states, "man or woman" to a degree that would seem unnecessarily repetitive.

Indeed the Prophet Muhammed did teach equality, in all senses of the word - between races, sexes, national/ethnic orgins, even religions. This does not mean that every religion is the same, just as all races are not the same. It means equal treatment to all people, independant of those things.

Well as far as I know, women in muslim countries aren't allowed to walk outside without covering their faces, and also a man can marry up to 4 women. How can you describe this as equality? Can a women marry 4 men? ...

http://www.cnn.com/2006/WORLD/asiapcf/02/07/cartoon.protests/index.html

Very interesting, a retaliation so far.  An Iranian newspaper is holding a contest for the best cartoons depicting the Holocaust...

I think that's awful.  However, it is a better answer to the caricatures than the rioting in my opinion, it brings the point across.

What I also don't understand, is my teacher tells me there have been cartoons depicting Jesus constantly.  Look no farther than some popular tv shows.  However...you don't exactly see Christians rioting in the streets over it.  If Islam is supposed to be a peaceful religion....why all the violence?  The caricatures depicted them in a violent light...they are only confirming the cartoons now.

[quote]What I also don't understand, is my teacher tells me there have been cartoons depicting Jesus constantly.[/quote]

Yup. Check out if you're interested iidb.org (an atheism and freethought forum.) They have [i]photoshop contests[/i] of Jesus. Offensive? Highly. But also terribly amusing...

[quote]If you think the others are uncivilized, ignorant, could you please just let them be?[/quote]

Um... I'm lost. Is anyone forcing muslims to read the paper? To look at the cartoons. Your statement applies backwards terribly well. If we are indeed all blasphemers/heretics/ignorant, why not just let us be?

Could be the fact that the Qur'an promotes violence against non-believers.

[quote]Many places in the Qur'an, you will see that when a reference is made to people in general, it often states, "man or woman" to a degree that would seem unnecessarily repetitive. [/quote]

Right... then how about...

[quote="Qur'an"]
The male shall have the equal of the portion of two females;
[/quote]

[quote="Qur'an"]
and call in to witness from among your men two witnesses; but if there are not two men, then one man and two women from among those whom you choose to be witnesses
[/quote]

[quote="Qur'an"]
Men are the maintainers of women because Allah has made some of them to excel others and because they spend out of their property; the good women are therefore obedient, guarding the unseen as Allah has guarded; and (as to) those on whose part you fear desertion, admonish them, and leave them alone in the sleeping-places and beat them; then if they obey you, do not seek a way against them; surely Allah is High, Great.

[/quote]

These are just a few examples; there are many more. "Equality" in Islam is sheer fantasy.

[quote="smbelcas"][quote="rookworm"]
It is also laughable to think that `the media' or `beauty myths', or the scapegoat of the day causes anorexia, since this condition is present in equal proportions in societies where larger women are considered to be beautiful, moreover, there is evidence that the basis for eating disorders is mainly physiological.

To compare the media's influence on fashion trends (!) to the abominable conditions that women endure in other countries is, frankly, demeaning to us all.
[/quote]

(1) Do you have any references (even pop media refs) to the claim that anorexia (and bulimia?) occur uniformly across cultures or that they are physiological?  This is the first I've heard of either claim.

(2) I do not believe that anyone was making the comparison you charge.  The comparison was between covering and not-covering.  Covering in and of itself is definitely not an abominable condition.  In fact, lots of women think that covering is a good thing.[/quote]

1-for references, you can try  [url=http://google.com]here[/url] or [url=http://scholar.google.com]here[/url] or [url=http://jstor.org]here[/url]-- specifically look for stuff published in the last ten years (perhaps this is why the mass hysteria over this issue has subsided). I heard about the anorexia in other societies study some time ago on a tv documentary. Clearly, you will not see similar rates of bulimia in societies where food is scarce, so that is irrelevant.

Anyway, eating disorders are not typified by cases like Princess Diana or Calista Flockhardt  (sp?)-- they are serious medical condtions that should be treated as such. Cynically using the victims of these conditions towards political ends only hurts them and the public at large.

2-Western women have the freedom to choose to cover or not (some months ago, there was a silly story on the BBC about a non muslim woman who liked to wear a burqua around London). The point is that women in the muslim world do not have this freedom.

[quote]
Every culture sees social problems and tries to solve them; the attempted solutions beget other problems. In the Arab world, there are currently some countries in which women are required to cover their hair and others in which they're required to cover their bodies. Most of us in the West think of this as a restriction of freedom. However, our culture allows women to dress "as they please"---put in quotation marks because what pleases us is influenced by the media, etc. There is a significant problem in the U.S. (which does not appear to exist in the Arab world) of eating disorders related to expectations to be thin. [/quote]

I take this to mean that you do not consider women to really be free to dress as they please. But there is a big difference! If you chose to only wear mumus and doc martens, then the worst consequence would probably be that you wouldn't get in with the popular crowd. On the other hand, if a woman in a muslim country decided to wear a miniskirt, then she would probably be sentenced to death by stoning. [i]That[/i] is what is at issue.

Just to pile on a bit, that is obnoxious -- if you make a factual claim and someone requests a reference, you should either have one, find one, or be grown up enough to note that you don't have one.

[quote="Dog of Justice"][quote="rookworm"]1-for references, you can try  [url=http://google.com]here[/url] or [url=http://scholar.google.com]here[/url] or [url=http://jstor.org]here[/url]-- specifically look for stuff published in the last ten years (perhaps this is why the mass hysteria over this issue has subsided).[/quote]

When you made your claim, presumably you already had some references in mind.  It is rude to ask that we search for them on our own when you have them at hand, and your claim goes that much against conventional wisdom.  You could have at least filled in appropriate search terms in the URLs you provided.[/quote]

Some snippits from the wikipedia entry on anorexia nervosa:

[quote]Neurochemistry abnormalities

There is increasing speculation that the onset of anorexia has a genetic component. It has been shown that a certain gene linked to abnormalities with the neurotransmitter chemical serotonin is more common amongst sufferers of anorexia than in the general population. Such genetic characteristics might potentially equate to an easier path towards overly high serotonin levels, thus instilling heightened levels of anxiety and the like. Biologically, when a person is in a state of starvation, their levels of serotonin decrease, and thence increase again upon the consumption of food because of the tryptophan amino acids contained therein (tryptophan is used by the body to synthesise serotonin). This raises the spectre that the anorexic is conditioned into avoiding food to reduce his or her anxiety, and that there may be yet another layer of complexity with respect to the cause/effect relationship between physiological factors and the mental beliefs of the anorexic.[/quote]

Also, links have been made between eating disorders and depression and OCD. I believe that the former can be successfully treated with SSRIs (a type of antidepressant medication)

[quote]
In addition, it has recently come to light that there appear to be girls exhibiting anorexic behaviours in remote parts of Africa that have not been exposed to modern forms of advertising. These girls link their self-starvation to religious causes.[/quote]

this can be a good starting point for further research

Anyway, my rationale for posting links to search engines rather than actual articles is that I am not an authority on the subject of eating disorders, and searching through volumes of medical literature is really not my cup of tea. On the other hand, I do have worthwhile information to contribute to this debate (that is, a discussion among non-experts), and researching eating disorders might be someone else's cup of tea. I am not a Psychologist, merely a clear thinker.

However, to appease my critics, I say: ignore the `laughable' comment I made, and instead of having denied the identification of "the media" with the cause of eating disorders, merely consider me to have cast doubt upon it.

[quote="coolmathematician"]you the one who opened the topic:
you say "About that last post, let's not forget about the prophet's unbelievably young wife (nine years?)"
when will you learn to speak scientificly? here is some sort of science forum. Do you have any kind of document or proof in the history of this event?
you are blaming our prophet without any proofs but with only prejudices...
but I never do the same thing because my holly book means "love all the prophets because all was sent by the same god (Allah)"
So I love the Jesus, so I love the Moses, so I love the David and the others...
If cartoons like these were made about the JESUS, I would have the same anger.
what you mean tolerance I am full of tolerance to all religions...
I don't need to find lies about prophets of the other religions because this is a shame for me
and last thing  SHAME OF YOURSELF.[/quote]

If you believe what he said was false and stupid, why did you just bump up a four month old topic to allow more people to see it?

[quote="coolmathematician"]I saw the topic yesterday...
what should I do? wait?
someone is lying about my holly values and will I wait because someone will see these lies?
not pal... this is not right... wherever you see the wrong you should say NO
because if we don't say after a while lies are believed to be right
if nobody rejects they were thought to be right
this way is better I guess[/quote]

Hello coolmathematician,

you were not wrong in bumping up, and I am doing it again now, you probably want a reaction from me.

First of all, what you say is completely correct 'after a while lies are believed to be right'.  But... aren't you the very same person who disagreed with my comments on the Pamuk case in Turkey in my "Freedom of speech : in general" topic?  Ever thought of the possibility that your government is just doing that?

About the comment about Mohammed's too young wife : I don't know how old she was. I want to apologize if it sounded like I was sure of it.  I only read accusations.  I am not a muslim, haven't read much of the Quran, so I am no expert.
How old was Aisha according to you when she married Mohammed?

coolmathematician, I can tell you are a person who dislikes injustice.  You believe the Armenian genocide is wrong for instance, and therefore think it is injust to claim it's true.
You should seriously consider the possibility of something being uncertain.  Even though you believe in it, doesn't mean all other parties should.

i will not participate in the larger part of this disussion, but i would like to imphasize on some important points.
Firstly, this is a sensitive topic, and i would advise everyone in general to speak only of what they know for sure; Do not make stuff up. And to rookworm, Muslim mathematicians have given a lot to mathematics. for example, the law of sines was discovered by a muslim mathemtician.
http://www.answers.com/topic/abul-w-fa-1
Abul made the law and oher trig identites.
http://www.answers.com/topic/list-of-muslim-mathematicians
List of muslim mathematicians with their contributions. I respect your ideas and option rookworm, but please do not say things you haven't researched enough about. Do not take me in the wrong sense, you are, ofcourse, entitled to your own point of view and way of looking at things. Kent Merryfield has defined quite a few things that Muslims have contributed, and I thank him for setting them out. :)
A quick note on another issue, is that everything in the quran has very deep meaning. It can't be just taken literaly. There is so much to say, so many misunderstandings, yet all i can say for now is... i'm not sure what to say. :(
This is a very vast topic, with many misunderstandings, and many views of islam and muslims that aren't true.
I can not address them all; just please be respectful and serious when you are talking in this topic...

[quote]and i would advise everyone in general to speak only of what they know for sure; Do not make stuff up.[/quote]

Try following your own advice. The sine law, which you mentioned, for instance, appears in Ptolemy’s Almagest (which predates mohammed by centuries). The sources you cite (answers.com) are taken from Wikipedia, which is frequently edited by people with an agenda, and may be edited by anyone, regardless of whether he/she is "making stuff up". In general, there has been a highly politicized trend in recent decades to try to trump up the acheivements of Islamic culture, as though the existence of some kind of past "golden age" lessens its current pitifulness. There was a recent exhibit, called "1001 Islamic inventions", or some such, that made claims as outlandish as that article of yours, such as "muslims invented cybernetics, chemistry, medicine", and so on.

Serious historians of science usually do not bother to refute such claims, because to do so would be politically incorrect, and generally, there are more important issues to deal with.

[quote]And to rookworm, Muslim mathematicians have given a lot to mathematics. for example, the law of sines was discovered by a muslim mathemtician.
http://www.answers.com/topic/abul-w-fa-1
Abul made the law and oher trig identites.
http://www.answers.com/topic/list-of-muslim-mathematicians
List of muslim mathematicians with their contributions. I respect your ideas and option rookworm, but please do not say things you haven't researched enough about. Do not take me in the wrong sense, you are, ofcourse, entitled to your own point of view and way of looking at things. Kent Merryfield has defined quite a few things that Muslims have contributed, and I thank him for setting them out.[/quote]

well said kent. i would sugggest ending this topic now. It hasn't really solved many problems, and it hasn't changed many people's minds. rookworm, i am sure you have some clever way of coming to anything i or someone else says, but so do we. this wont solve anything, only create more problems. I will form a conclusion now; if you wish to continue this topic then do so for as long as you please:
In conclusion, there are many differences between the european and the widdle eastern cultures. there, simple.
I repect your beliefs rookworm, and i hope in time you will respect mine. if you already do, then i am happy. :)

[quote="Kent Merryfield"][quote]The sine law, which you mentioned, for instance, appears in Ptolemy’s Almagest[/quote]
Hmmm ... Ptolemy had only one trigonometric function, and that function was "chord." The half-chord, or sine (the word "sine" is a mistranslation), was an Indian invention, and appears in the works of Aryabhata, about 300 years later than Ptolemy. (Did Aryabhata know of Ptolemy's work? That's not known; any statement would be sheer speculation. Personally, I think it's likely.) Note that you gave the title of Ptolemy's principal work as [i]Almagest[/i]. That's not Ptolemy's own title. In fact, we do not know what Ptolemy's own title was. [i]Almagest[/i] is an Arabic title, meaning approximately "the greatest." Ptolemy's works were preserved only in the Moslem world, primarily by the scholars of Baghdad in the 800's or 900's. Those scholars synthesized Ptolemy's system with the astronomy and trigonometry inherited from Indian sources, and improved on Ptolemy's calculations based on better observations and records.

Yes, I do regard Ptolemy as the earliest significant developer of trigonometry, and Ptolemy's Theorem about the sides and diagonals of a cyclic quadrilateral as essentially a trigonometric identity. And trigonometry was always developed by astronomers, for astronomers. But in tracing the development of trigonometry, it is impossible to ignore the further contributions of both Indians and Arabs.[/quote]

Perhaps they selectively synthesized, improved, preserved, and generally made minor "contributions" to some existing bodies of knowledge in a small space of a few hundred years (by the way, most of the people doing this were non-muslims living as dhimmis, and some converts). But this only reinforces my point that Islamic culture was (and is) intellectually repressive. There is no need for free inquiry when you are in possesion of a Total Explanation of the Universe (i.e. Islam). If you can't deal with a couple of cartoons, how are you going to deal with philosophy or science? In fact, we can see some of the same trends happening under other totalitarian systems. Where are the [i]original[/i] contributions?

[quote="Inspired By Nature"]I repect your beliefs rookworm, and i hope in time you will respect mine. if you already do, then i am happy. :)[/quote]
I guess this isn't going to be happening. :(
You do not understand islam rookworm; i would say more, how its been misinterpreted for example, but whats the point :?:


There are 25 students in a class and they all have to turn their reports in.  The first five turn theirs in on Monday, the next five on Tuesday, etc.  What is the probability that the shortest kid will turn his in on Thursday?

[quote="lotrgreengrapes7926"][quote="ragnarok23"][hide]There is a 1/5 probability that the shortest student (wtf?) will turn his report in on Monday. He has a 4/5 chance of not turning it in.
On Tuesday, there are 20 students left. He has a 3/4 probability of not turning it in.
Wednesday: 15 students left. 2/3 probability of not turning it in.
Thursday: 10 students left. 1/2 probability of turning it in.
$\frac{4}{5}*\frac{3}{4}*\frac{2}{3}*\frac{1}{2}=\boxed{\frac{1}{5}}$[/hide][/quote]
[hide] Isn't it easier to say $\frac{1}{5}$ of the students are turning it in on Thursday, so he has a $\frac{1}{5}$ probability.  :?  :D [/hide][/quote]
I know, I just love the hard way.  :whistling:  :D


Find all positive integers $a$, $b$ such that:

$a^3+b^3+1=(ab)^2$

Clearly $ a \neq b$. WLOG, $ a > b$, then it's easy to see that $ a^2|b^3\plus{}1$

So $ a^2 \le b^3\plus{}1$, also $ a^2b^2\equal{}a^3\plus{}b^3\plus{}1 < 2a^3$  $ \Rightarrow$  $ b^2<2a$

$ \Rightarrow b^4<4a^2 \le 4b^3\plus{}4 \Rightarrow b\equal{}1,2,3,4$

When $ b\equal{}1$, $ a^2|(1)^3\plus{}1$, no solution.
When $ b\equal{}2$, $ a^2|(2)^3\plus{}1 \Rightarrow a\equal{}3$, so $ (a,b)\equal{}(2,3),(3,2)$ are solutions.
When $ b\equal{}3$, $ a^2|(3)^3\plus{}1\equal{}28$, no solution.
When $ b\equal{}4$, $ a^2|(4)^3\plus{}1\equal{}65$, no solution.

So $ (a,b)\equal{}(2,3),(3,2)$ are the only solutions.


x1*x2*x3*x4*x5=1050
find how many + integers solution are there in this product .where x1 x2, x3 ,x4 ,x5 ,, are +ve integers.
_________________
Q2 in how many ways a cube can be painted with six colors.

when i solve this question i do like that
for any face we have five option(ie five pair). corresponding to each of these options there are there options . at last there there is only one option for each of these options . so answer should we 15. but answer given is 30


limit
logarithms
integration
real analysis
real analysis unsolved

Suppose $A>0 ,\; A\ne 1 .$ Using Riemann sums prove the well-known result $\lim\limits_{n\to \infty}n \left(\sqrt[n]{A} - 1\right)=\ln{A}\; .$

I've done this myself- it's a very nice calculation, and you can get at properties of the logarithm without needing any form of the fundamental theorem.

[hide="hint"]Integrate $\frac1x$ using a geometric partition. Try the same idea for $\int_A^B x^k\,dx$ ($0<A<B$) in general- unlike the traditional uniform sums, this works with no fuss.[/hide]

My solution is following: [b]Suppose $A>1$, $f\in C[1,A]\; ,\; f(x)=\frac{1}{x}$ and  consider  the division $(\delta)$ of $[1,A]$ where
\[ (\delta): \; \; \; 1=x_0 <x_1<\ldots < x_k<\ldots <x_n =A\; \; ,\; \; x_k: =\sqrt[n]{A^k}\; ,  \]
having the norm $\nu(\delta): =\max\limits_{1\le i \le n-1 }(x_{i+1}-x_i ) \; .$ Because \[ \begin{array}{l} \nu(\delta)=A(\sqrt[n]{A}-1)= A(\sqrt[n]{A\cdot 1\cdots 1}-1)<A\left(\frac{A+(n-1)}{n}-1\right)=\frac{A(A-1)}{n} \\ n\left(\sqrt[n]{A}-1\right)=\sum\limits_{k=1}^{n}(\sqrt[n]{A}-1)=\sum\limits_{k=1}^{n}f(x_{k-1})(x_k -x_{k-1})=: \sigma_{n,\delta}(f)\\ \end{array}  \]
we conclude that $\lim\limits_{n\to \infty}\sigma_{n,\delta}(f)= \lim\limits_{\nu(\delta)\to 0}\sigma_{n,\delta}(f)=\int\limits_{1}^{A}f(x)\; dx =\ln{A} \; .$ [size=75]A  similar proof may be performed  in case $0<A<1 .$  [/b] [/size]