MATHCOUNTS
email
geometry
AMC
AIME
number theory
AMC 10
\/closed

You should (obviously) take the Do I need this? quizzes for the classes--that's your best bet.
I am currently taking Intro to Geometry and have learned a lot from it. I highly recommend that class.

I am currently taking AMC 10 and Introduction to Geometry, and I think you will learn a lot from both of them.

In intro to geometry, you will find that MOST of the material taught in class is the same as the material taught in the book. However, in each class we delve a little deeper and we do some extra proofs. I would say my skill in geometry has increased ten fold from doing the challenge sets and attending the classes. I also highly reccomend intro to geometry.

As far as AMC 10 is concerned, I think that doing a large amount of AMC10's, 12's, and AIMEs is good enough.

If you can get the Introduction to Number Theory book (as in, you wouldn't mind wasting the money/can borrow it from someone else or something), I would suggest reviewing the last few chapters - that helped me a bit. In Spring, there's an [url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=31]Intermediate Number Theory Seminar[/url] that you will probably be interested in.

You are almost definitely ready for introduction to geometry, because you don't require any geometry background in it, just basic algebra stuff. In fact, I really recommend the class. I'm currently taking it (it's going to end in a week or so), and I started in March, alongside with my school geometry book. By May or so we were already ahead of my school class, plus I obviously learned a lot more. :) Honestly, though, out of all the AoPS classes I've taken (admittedly not too many :P ) this one helped me the most. Geometry used to be my weakest point and now I'd say it's my strongest. I can't emphasize enough how much I strongly recommend you take it.

If you want my opinion, I don't think the problem series classes are all that necessary. You learn best from introduction/intermediate classes and especially by just doing problems. I've been taking Introduction to Geometry and Intermediate Algebra over the summer and my scores went from about 100s (on the AMC 10) to the 130s - 140s.

If I were you, I would look at something like this, in no particular order -

AoPS Classes:
Introduction to Geometry
Intermediate Number Theory Seminar
Intermediate Algebra (or Intermediate C&P)

Books:
Once you finish Volume 1, go through the chapters that gave you most trouble/you're weakest in again
Volume 2
(obviously the books that accompany the AoPS classes)
Introduction to Number Theory??
Intermediate C&P (or Intermediate Algebra)

Other:
Do every single AMC 10 test, AMC 12, AHSME, MATHCOUNTS state and nationals, etc., you can get your hands on, in addition to other materials. But pace yourself with the tests. They run out fast.  :)

Anyway, that's my \$0.02.

How often do most of the intermediate classes run?

From what I observe, is it about every other season?  Spring, not summer, Fall, not spring, Summer, not fall...

I would suggest not taking the Intermediate Number Theory Seminar if you don't have a strong base in the subject. I just took it and found it became very hard at certain points.

How is the Intermediate Number Theory [i]Seminar[/b] different from the other classes? Is it just that there isn't an accompanying book?

Thanks for the help. I will look into the Intro to Geometry class.

Well, it's different in the sense that it is only 8 weeks.

[quote="PowerOfPi"]My background in a paragraph:

I got 5th at ND MathCounts State, and a 20 on the AMC 8. I think I improved a lot and I can get 24's and occasionally 25's on the AMC, and feel confident I can get 1st at state this year. I only have about a half year of experience doing math outside of school, but school math was always easy for me.

I am reading through AoPS Volume 1 and I think I will finish by the end of this month (hopefully). Then I will go through Intro to C&P pretty quickly and Intro to Geo. After that I will probably focus on some competition stuff and may start Volume 2 or review.

I sent an email to classes@artofproblemsolving.com and rruzscyk said I probably wouldn't need any of the intro subject classes except for Intro to Geometry.

My questions:

I feel I am a little weak in Number Theory, but not too weak to get the book or the class. What do you recommend for improving NT?

Am I ready/Do I need Intro to Geo and how much will I learn from it?

Is the AMC 10 problem series good for me?[/quote]

It's you're decision if you think you should just learn from the book or class. I think you should do Volume 1 and the Intermediate series before Vol. 2 because it can get tough. If you finish Introduction Series in a while by yourself, I don't think AMc 10 is necessary

PowerofPi,
I had some  problems in Mathcounts because of my geometry+number theory concepts need to be better.
Seriously, the Intro.to Geometry class+ the textbook helped me a ton.
As for number theory, I felt that AOPs vol.1 did not have that much stuff devoted to it, so I'm reading through Intro.to Number Theory,its a great book and takes you step to step on figuring out harder number theory concepts.


analytic geometry
factorial
AMC
AIME
USAMTS

Sorry to keep you waiting..

I was going to post it on the midnight but I just went bed and woke up at 12:00. :D

Here they are:

The same rules apply so I won't go "blahblahblahbl" on rules.. Just make sure send me 5 correct ANSWERS and to qualify, you have to get them all right.

I also made poll on this thread whether it's harder than A or not.

Can you specify if for question #5, if you do want complex answers or only real ones.

you mean you're gonna change the whole problem?

hope not...a good one would be to ask for the coordinates of where the two lines that are solutions to the problem meet. If thats it, it would be easy :)

where are the problems???? i dont see them.

there were here as PDF, but silverfalcon, took them away to change the #5.

oh...sorry. I thought it was posted on thread just like the COP A.

Problems will be posted within one hour.

I have to do some work so it would take around one hour.

#5 was hard. I'm glad he's changing it  :) but I hope it's not harder now  :(

[quote="YSJeong"]can we take it for fun, even if we got to 2nd round by taking A?[/quote]

Yes. But you won't get double prize or anything like that. :D

#5 was indeed hard but I don't think new #5 will be easy neither. But who knows.. That's what I think. :)

In #5, $p_1$, $p_2$, etc. are distinct primes, right?

This seems quite a bit more difficult than A.  I voted "Yes".

i didn't participate in A, but this was very easy,
the whole thing took me about 25 mins :D

I sent in answers, but I've never PMed before so I dont know if they got in.

Guys, take this seriously.

You have to make sure you check your work. I do not let any small mistakes to happen and if it does, you're OUT. ;)

what does the notation in question 5 mean?

[quote="amirhtlusa"]the notation means Product
by the way, guys, just a piece of advise:
TAKE YOUR TIME AND READ THE PROBLEMS CAREFULLY...
cause i did the whole thing like in 20 mins. and i made a STUPID mistake on #5...just because i didnt read the problem carefully.. ;)
dont hurry...you have one week.
TAKE YOUR TIME...
PROBLEMS ARE EASY...but don't rush... :)[/quote]

by this do u mean its like n factorial blah over something where n=1 times nfactorial blah over something where n=2 etc... til n=8?

$\prod$ is just a way to express the products instead of $\sum$ which gives you the sums.

For example:

$\prod^{2}_{i=1} i^2$ equals $1^2 \cdot 2^2 = 4$.

I take it that, if we answer these problems and send them in, and we get one or more wrong, but we've already qualified on 1A, we're still fine?

so far the only qualifiers for the B test are from Utah.  I'm surprised(I even got in at all).

thats sad. The stupidest people in the AoPS level of ppl in the nation qualafied....

What about
[hide]Mathgeek2006[/hide]
Just kidding.

[quote="lorddude"]Where are the questions, or am I just stupid? :([/quote]

Go to the first page of this thread.

If you still can't, just click [url=http://www.artofproblemsolving.com/Forum/files/round_1b_cop4th_331.pdf]Direct URL of PDF Problems[/url]

that's o.k. I can see them now I'm at home
school pc's are screwed up

nothing to do about school. I did that too, to show my teacher the #5 :) and it only shows up when your loged in.

#5 is actually simpler than it looks.

Hah! At least one non-Utah person has qualified now! :lol: :P

[quote]Hah! At least one non-Utah person has qualified now! Very Happy Razz[/quote]

That's good.  We need more non-utards to qualify!

I said B was harder until #5 was changed.

#5 is definitely not easy but it's not really hard problem neither. It's easy if you see the "key" in the problem although you will eventually have to do some work.

I'm thinking of making Round 2 extremely hard because just from COP, there are 19 people who qualified, and personally, I think all of them are very smart. :)

So, Round 2 has to be very hard so that we can see who can be the "best" :lol:

As long as you are keeping track of states I guess I'm representing Iowa.

If you're from Iowa, where else would you be representing? :lol:

Really hard Round 2, eh? Why can't you make the round after that really hard, and Round 2 just pretty hard? :P

Sounds like fun, I'm up for a challenge, would have passed on the 1A form but AP testing had me distracted :blush:

even harder???  I barely figured out these questions as it is!!  Just kidding.  Yeah you should make extremely hard.  I actually thought this first round was pretty easy.  for some reason I've been craving a really hard question...of course I could just look at an AIME question or a USAMTS question or something....

I hope it's hard enough..

I don't know.. I originally had 7 problems done but that was before I thought this COP was going to be competitive.

Now, I guess it has to be hard. :D

Personally, I think A was way easier than this one :P But it was a challenge. I've got through all of them but one.. I've gotten to $\sqrt{x}$ and the x is some weird number with 5 decimals and it's not a perfect square (without the decimals) so I'm just trying to figure out other ways to approach this, especially without a calc -_-.

EDIT: Well I figured out that one, but I made a really careless mistake on the 4th one, and waiting til SF posts the answer to the 5th one to figure out why I got that one wrong.


How many grams of Barium Phosphate will form when 127.9 grams of Potassium Phoshate reacts with 75.6 grams of Barium Chloride?

Start by writing the chemical equation,

$2K_{3}PO_{4}+3BaCl_{2}\longrightarrow Ba_{3}(PO_{4})_{2}+6KCl$.

Now convert the mass of each reagent in number of moles by using the molar mass of each. Then you have to decide what reagent will be the limiting one.

That's right, barium chloride is the limiting reagent.


eh the carolinas are pretty much the only state forums that are somewhat active in terms of non-spam. Except with Arnav's posts, they have a good deal of spam anyway.

[quote="alanchou"]Why doesn't anything happen in this forum?[/quote]

Because we're too busy doing math.

"We?" :wink:

Actually my impression is that the NC forum has quite a bit of spam. If you look, you'll see lots of random posts in capital letters including words like "dude" and "wtf" and they aren't all from Arnav.

Maybe nothing happens here b/c u haven't taken the initiative to do anything .

Does anybody realize that everything in this topic except for the first post is rated "spam"? (Wasn't me)
There aren't that many people from Mass. on AoPS, are there?
I think MA is more sports-oriented, but that's just my own experience.
My school announced every little local game our hockey team wins, but they didn't even say anything about MC state.
But then, we sucked so badly that it wasn't worth mentioning.

if not for me, my school wouldn't have a math team except a sucky local one, which i pretty much singlehandedly raised about 4 divisions. That's not final yet though-I dont know where there'll be next year. but they'll be screwed. my school isn't sports oriented either. i really wish i'd grown up in lexington.

[quote="MysticTerminator"]you guys are just jealous because spam is awesome and you don't have any[/quote]
So, is that how you got your 3000+ posts?

It would be more if valentin and co. hadn't deleted a lot of them :wink:

admins/mods

apparently he's banned from like half the forums

I think the post above is classified as "spam".  :o

[quote="ZhangPeijin"]Arnav?(I don't even know him)
[/quote]

DUDE THAT IS NOT GOOD
I KNOW YOU'RE LYING
BECUASE IF YOU WEREN'T
YOU WOULD HAVE TO BE KILLED

[quote="The Herald of Doom"][quote="ZhangPeijin"]Jealous of what?
Your 17 posts?
Arnav?(I don't even know him)
Spam?[/quote]

I LIKE BIG BUTTS AND I CANNOT LIE
YOU OTHER BROTHERS CAN'T DENY
THAT WHEN A GIRL WALKS IN WITH AN ITTY BITTY WAIST
AND A ROUND THING IN YOUR FACE
YOU GET SPRUNG, WANNA PULL OUT YOUR TOUGH
'CAUSE YOU NOTICE THAT BUTT WAS STUFFED[/quote]

wtf?

Someone seriously needs to get Herald of Doom out of this place.

Anyways, yeah, seriously, what the [please fill in the blank]?


I know the atomic mass of an element is something like the weight of a sample of an element. It is said that to find the most common isotope of an element, round the atomic mass to the nearest whole number.

But let's say there exists an element with only isotopes with mass numbers of, let's say, 249 and 251. There is about an equal amount of each isotope, so its mass number is like 250.001. Using the rule above, its most common isotope has mass number 250, but that doesn't exist! Is there such a situation?

The random atomic mass is another one of those human inventions to make matters simple. Instead of taking everytime all the different isotopes, we calculate a fictitious Mean Atomic Mass as an average and use it. Such a "element" will never exist. Mean Atomic mass is just a mathematical definiton. period. The independent existence is only of the isotopes.

Hope this is clear.


A wooden cube has edges of length $3$ meters. Square holes, of side one meter, centered in each face are cut through to the opposite face. The edges of the holes are parallel to the edges of the cube. The entire surface area including the inside, in square meters, is

$\textbf {(A) } 54 \qquad \textbf {(B) } 72 \qquad \textbf {(C) } 76 \qquad \textbf {(D) } 84\qquad \textbf {(E) } 86$

Wasn't something like this on a Mathcounts contest sometime?

[hide]
The surface area of the outside of a face is $9-1=8$.  For each face, the hole contributes $4$, so each face contributes $12$ to the surface area.  There are 6 faces, so the total surface area is $72$ square meters.
[/hide]

[hide]This figure kind of scares me once I thought about it.  Like an endless abyss.

$9-1=8$  on the outside surface area. There are $6$ little "cubes" on the inside with the very center poked through.  So the 6 little "cubes" have surface area of $4$.

$8*6+4*6=72$

$\boxed{72} B!$[/hide]

Almost assuredly something similar on mathcounts... I hate these types of problems lol... too much spatial thinking.


Write $ \left(\frac{1 \plus{} i \sqrt{3}}{1 \minus{} i \sqrt{3}}\right)^{20}$ in algebraic form.

[quote="Carcul"]Write $ \left(\frac {1 \plus{} i \sqrt {3}}{1 \minus{} i \sqrt {3}}\right)^{20}$ in algebraic form.[/quote]

[hide]
uhh notice that $ \left(\frac {1 \plus{} i \sqrt {3}}{1 \minus{} i \sqrt {3}}\right)^2 \equal{} \left(\frac {1 \minus{} i \sqrt {3}}{1 \plus{} i \sqrt {3}}\right)$

and that $ \left(\frac {1 \plus{} i \sqrt {3}}{1 \minus{} i \sqrt {3}}\right)^4 \equal{} \left(\frac {1 \minus{} i \sqrt {3}}{1 \plus{} i \sqrt {3}}\right)^2 \equal{} \left(\frac {1 \plus{} i \sqrt {3}}{1 \minus{} i \sqrt {3}}\right)$

that's really cool-looking and it also implies that $ \left(\frac {1 \plus{} i \sqrt {3}}{1 \minus{} i \sqrt {3}}\right)^{20}\equal{}\left(\frac {1 \plus{} i \sqrt {3}}{1 \minus{} i \sqrt {3}}\right)^{2} \equal{}\left(\frac {1 \minus{} i \sqrt {3}}{1 \plus{} i \sqrt {3}}\right)$
[/hide]

[hide="Solution"]

Let $ z \equal{} \frac {1 \plus{} \sqrt {3}i}{1 \minus{} \sqrt {3}i} \equal{} \frac {(1 \plus{} \sqrt {3}i)^2}{4} \equal{} \frac { \minus{} 1 \plus{} \sqrt {3}i}{2}$, we have $ 2z \plus{} 1 \equal{} \sqrt {3}i$. Squaring both sides and arranging the equation gives $ z^2 \plus{} z \plus{} 1 \equal{} 0\ \therefore z^3 \minus{} 1 \equal{} (z \minus{} 1)(z^2 \plus{} z \plus{} 1) \equal{} 0$, yielding $ z^3 \equal{} 1$. Thus $ \left(\frac {1 \plus{} \sqrt {3}i}{1 \minus{} \sqrt {3}i}\right)^{20} \equal{} z^{20} \equal{} \frac {z^{21}}{z} \equal{} \frac {(z^3)^7}{z} \equal{} \frac {1}{z} \equal{} \boxed{\frac { \minus{} 1 \minus{} \sqrt {3}i}{2}}$. [/hide]


If $\sqrt{2 + \sqrt{x}} = 3$, then $x =$

$\textbf{(A)}\ 1 \qquad
\textbf{(B)}\ \sqrt{7} \qquad
\textbf{(C)}\ 7 \qquad
\textbf{(D)}\ 49 \qquad
\textbf{(E)}\ 121$

Hey, can a moderator of this forum fix the answer choices here to the following?  Thanks!

[quote=Silverfalcon]
$\textbf{(A)} 1 \qquad
\textbf{(B)}\ \sqrt{7} \qquad
\textbf{(C)}\ 7 \qquad
\textbf{(D)}\ 49 \qquad
\textbf{(E)}\ 121$[/quote]

faster solution:
[hide=solution(fastway)]We know it cannot be $\sqrt{7}$ because the square root of a square root of a square root of a non-perfect square (that was confusing) is not an integer, so that leaves us with either 1,7,49,121. We also know it cannot be 7, since it is not a perfect square. Testing values of 1,49,121 gets us $49$ or $\boxed{(C)}$.[/hide]

Breakzi has the best solution here.

aaosdfoajdofajo


calculus
integration
rotation
limit
geometry
geometric transformation
reflection

Ok, so I have to do this problem with using integration with volume and with y=(e^x)+1 and x=2 and I have to find the volume of object rotated around y-axis using washers. Well everytime I do it I get a really high number so I know I am doing something wrong. Can anyone help me?

There must be something missing: $y=(e^{x})+1$ and $x=2$ do not enclose any region in the $xy$-plane.

[quote="DanK"]Unless it wants the improper integral $\pi\int^{2}_{-\infty}(e^{x}+1)^{2}dx$, which doesn't seem very likely, especially because it goes to infinity...[/quote]
this integral calculus easy by substitution: $e^{x}=y$. and then easy seen :it's dont goes infinyti

Here's a question that does make sense; why don't you all solve it instead?

A solid is generated by rotating the region bounded by the $x$- and $y$-axes, the line $x = 2$ and the curve $y = e^{x}+1$ about the $y$-axis.  What is the volume of this solid?

Anyway, why don't you calculate the volume again?

[quote="mlok"][quote="JBL"]Here's a question that does make sense; why don't you all solve it instead?[/quote]
Because solving calculus problems that make sense is not nearly as much fun as discussing those that don't.
$\pi\int_{0}^{2}(e^{x}+1)^{2}dx=\pi(\frac{1}{2}e^{2x}+2e^{x}+x)|_{0}^{2}=\pi(e^{4}/2+2e^{2}+2)$[/quote]

I said this result is wrong.

[quote="DanK"]mlok got what I got (which is also how I got that improper integral to be $\infty$ (using a variable rather than 0 and then taking the limit)).  What's wrong with it?

Btw, JBL, could you explain what your post using big O means?  I haven't done anything with that yet...[/quote]

Kunny is complaining because you rotated around the $x$-axis, not the $y$-axis.

There's a sticky in this forum describing what big-O, as well as little-o, and possibly some other similar notations, mean.  $O(1)$ just means "something bounded by a constant."  The actual expression for the integral is somewhat ugly, as you're probably aware, so I was using big-O notation to save some typing.

[quote="JBL"]Um, what?  $\int_{t}^{2}(e^{x}+1)^{2}dx =-t+O(1)$ as $t \to-\infty$.[/quote]
I know all of them
I mean it's don't goes $-oo$. :lol: sorry have mistake

[quote="JBL"]Kunny is complaining because you rotated around the $x$-axis, not the $y$-axis.[/quote]

Oh, that would be a problem in both of my posts in this thread so far.  Never mind... :blush:

[quote="Extremal"][quote="JBL"]Um, what?  $\int_{t}^{2}(e^{x}+1)^{2}dx =-t+O(1)$ as $t \to-\infty$.[/quote]
I know all of them
I mean it's don't goes $+oo$.[/quote]

$\lim_{t \to-\infty}-t =+\infty$.

Edited to reflect a missing - sign.  The thing I was [i]trying[/i] to write was correct ;)

[quote="JBL"]$\lim_{t \to-\infty}t =+\infty$.[/quote]

Umm, how do you get that?


Show that

$ \sum_{k\equal{}1}^\infty \left(\frac{k^k}{e^k\, k!}\minus{}\frac{1}{\sqrt{2\pi k}}\right)\equal{}\minus{}\frac23\minus{}\frac{1}{\sqrt{2\pi}}\, \zeta\left(\frac12\right)$  .

[b]REFERENCES[/b]
Bailey, D. H.; Borwein, J. M.; Calkin, N. J.; Girgensohn, R.; Luke, D. R.; and Moll, V. H. [url=http://www.amazon.com/exec/obidos/ASIN/156881271X/]Experimental Mathematics in Action[/url]. Wellesley, MA: A K Peters, pp. 16 and 221, 2007.

Borwein, J.; Bailey, D.; and Girgensohn, R. "Knuth's Series Problem." §1.5 in [url=http://www.amazon.com/exec/obidos/ASIN/1568811365/]Experimentation in Mathematics: Computational Paths to Discovery[/url].  Wellesley, MA: A K Peters, pp. 15-17, 2004.

Knuth, D. E. "Problem 10832." Amer. Math. Monthly 107, 863, 2000.

By [url=http://mathworld.wolfram.com/KnuthsSeries.html]Knuth's Series[/url] - MathWorld


function
calculus
derivative
calculus computations

find all derivablle  functions $ f$  such that $ f : \mathbb R \rightarrow \mathbb R$ , and ,  $ f'(x)\equal{}f(\minus{}x)$

How about this:

$ f'(x)\equal{}f(\minus{}x)$

$ f'(\minus{}x)\equal{}f(x)$

$ \minus{}f''(\minus{}x)\equal{}f'(x)$

$ \minus{}f''(\minus{}x)\equal{}f'(\minus{}x)$

$ f'(\minus{}x)\plus{}f''(x)\equal{}0$

$ f(x) \equal{} C_1\cos(x)\plus{}C_2\sin(x)$

$ f'(x) \equal{} C_2\cos(x)\minus{}C_1\sin(x)$

$ f(\minus{}x) \equal{} C_1\cos(x)\minus{}C_2\sin(x)$

Therefor, we must have $ C_1 \equal{} C_2$ so

$ f(x) \equal{} C\left(\cos(x)\plus{}\sin(x)\right)$

thank's milin
but $ c$ is $ \in \mathbb R$ or what set

I think it can be any constant! It get's removed on both sides when plugging in the function!

Milin, I think your proof is a little garbled (some extra $ '$s floating around?).  I would have done
$ f'(x) \equal{} f(\minus{}x)$ implies $ f'(\minus{}x) \equal{} f(x)$ (by substituting $ \minus{}x$ for $ x$) and $ f''(x) \equal{} \minus{}f'(\minus{}x)$ (by taking a derivative, since the RHS is differentiable) and so $ f''(x) \plus{} f(x) \equal{} 0$.  Then the end is the same.

You did the same thing.. But mine is messier, I agree.. :)


inequalities
calculus
derivative
function
algebra
polynomial
inequalities proposed

$ n\ge2,a_k,b_k>0, S\equal{}\sum_{k\equal{}1}^n a_k, T\equal{}b_1b_2\cdots b_n$
Prove that :
$ \frac{1}{n\minus{}1}\sum_{i\equal{}1}^n (1\minus{}\frac{a_i}{S})b_i \ge (\frac{T}{S}\sum_{j\equal{}1}^n \frac{a_j}{b_j})^{\frac{1}{n\minus{}1}}$

Assume $ S \equal{} 1$ and donate $ y_i \equal{} \frac {b_i}{\sqrt [n]{T}}$ $ \Rightarrow \prod_{i \equal{} 1}^n y_i \equal{} 1$ .
Therefore your inequality becomes:

$ \frac {\sqrt [n]{T}}{n \minus{} 1}\sum_{i \equal{} 1}^n (1 \minus{} {x_i})y_i \ge \left(\sqrt [n]{T^{n \minus{} 1}}\sum_{j \equal{} 1}^n \frac {x_j}{y_j}\right)^{\frac {1}{n \minus{} 1}}$. $ \Rightarrow$ you can assume $ T \equal{} 1$, too.

For $ n \equal{} 2 :$
  $ x_1 \equal{} (1 \minus{} x_2)$; $ y_1 \equal{} \frac {1}{y_2}$ and so from this, the inequality is an equality: $ [(1 \minus{} x_1)y_1 \plus{} (1 \minus{} x_2)y_2] \equal{} \frac {x_1}{y_1} \plus{} \frac {x_2}{y_2}$, actually.

$ n\ge 3 :$

$ \frac {1}{n \minus{} 1}\sum_{i \equal{} 1}^n (1 \minus{} {x_i})y_i \ge \left(\sum_{j \equal{} 1}^n \frac {x_j}{y_j}\right)^{\frac {1}{n \minus{} 1}}$
$ \Leftrightarrow \frac {\sum_{i \equal{} 1}^n (1 \minus{} {x_i})y_i}{\sum_{i \equal{} 1}^n (1 \minus{} {x_i})} \ge \left(\sum_{j \equal{} 1}^n \frac {x_j}{y_j}\right)^{\frac {1}{n \minus{} 1}}$ $ \equal{} \left(\sum_{i \equal{} 1}^n \frac {1}{y_i} \minus{} \sum_{i \equal{} 1}^n \frac {(1 \minus{} x_i)}{y_i}\right)^{\frac {1}{n \minus{} 1}}$

Donate $ (1 \minus{} {x_i}) \equal{} t_i$.
$ \Rightarrow \frac {\sum_{i \equal{} 1}^n t_iy_i}{\sum_{i \equal{} 1}^n t_i} \ge \left(\sum_{i \equal{} 1}^n \frac {1}{y_i} \minus{} \sum_{i \equal{} 1}^n \frac {t_i}{y_i}\right)$.

From Caushy $ \left(\sum_{i \equal{} 1}^n t_iy_i\right)\left(\sum_{i \equal{} 1}^n \frac {t_i}{y_i}\right)\ge \left(\sum_{i \equal{} 1}^n t_i\right)^2 \equal{} (n \minus{} 1)^2$.
Then $ \left( \minus{} \sum_{i \equal{} 1}^n \frac {t_i}{y_i}\right)\le \minus{} \frac {(n \minus{} 1)^2}{\sum_{i \equal{} 1}^n t_iy_i}$. Let's substitute $ \sum_{i \equal{} 1}^n t_iy_i \equal{} A\ge \frac{n \minus{} 1}{n}\left(\sum_{i \equal{} 1}^n y_i\right)$.

In that case the inequality will be reduced to $ \frac {A}{n \minus{} 1}\ge \left(\sum_{i \equal{} 1}^n \frac {1}{y_i} \minus{} \frac {(n \minus{} 1)^2}{A}\right)^{\frac {1}{n \minus{} 1}}$ or $ \sum_{i \equal{} 1}^n \frac {1}{y_i}\le \frac {A^{n \minus{} 1}}{(n \minus{} 1)^{n \minus{} 1}} \plus{} \frac {(n \minus{} 1)^2}{A}$.

$ F(A) \equal{} \frac {A^{n \minus{} 1}}{(n \minus{} 1)^{n \minus{} 1}} \plus{} \frac {(n \minus{} 1)^2}{A}$ has first derivative $ F'(A) \equal{} \frac {\left(\frac {A}{n \minus{} 1}\right)^n \minus{} 1}{\left(\frac {A}{n \minus{} 1}\right)^2}\ge 0$, so the function is increasing and $ \frac {A^{n \minus{} 1}}{(n \minus{} 1)^{n \minus{} 1}} \plus{} \frac {(n \minus{} 1)^2}{A}\ge \left(\sum_{i \equal{} 1}^n \frac {y_i}{n}\right)^{n \minus{} 1} \plus{} \frac {n \minus{} 1}{\sum_{i \equal{} 1}^n \frac {y_i}{n}}$.

$ \left(\sum_{i \equal{} 1}^n \frac {y_i}{n}\right)^{n \minus{} 1} \plus{} \frac {n \minus{} 1}{\sum_{i \equal{} 1}^n \frac {y_i}{n}} \minus{} \sum_{i \equal{} 1}^n \frac {1}{y_i}$ $ \equal{} \frac {\left(\sum_{i \equal{} 1}^n \frac {y_i}{n}\right)^n \minus{} \left(\sum_{i \equal{} 1}^n \frac {y_i}{n}\right)\left(\sum_{i \equal{} 1}^n \frac {1}{y_i}\right) \plus{} (n \minus{} 1)}{\left(\sum_{i \equal{} 1}^n \frac {y_i}{n}\right)}\ge 0$ .

PS : I'm sorry in advance if there is a mistake again. :ninja:

By the way, very good problem, [b]NextPeace[/b]. If it is your own - congratulations. :thumbup:

[quote="BG Yoda"]The given conditions aren't enough. Assume $ S \equal{} 1$ and donate $ y_i \equal{} \frac {b_i}{T}$ $ \Rightarrow \prod_{i \equal{} 1}^n y_i \equal{} 1$ .
Therefore your inequality becomes:

$ \frac {T}{n \minus{} 1}\sum_{i \equal{} 1}^n (1 \minus{} {x_i})y_i \ge \left(\sum_{j \equal{} 1}^n \frac {x_j}{y_j}\right)^{\frac {1}{n \minus{} 1}}$.

For $ n \equal{} 2 :$
  $ x_1 \equal{} (1 \minus{} x_2)$; $ y_1 \equal{} \frac {1}{y_2}$ and so from this, the inequality could be true only for $ T\ge1 :$ $ T[(1 \minus{} x_1)y_1 \plus{} (1 \minus{} x_2)y_2] \equal{} T\left[\frac {x_2}{y_2} \plus{} \frac {x_1}{y_1}\right]\ge \frac {x_1}{y_1} \plus{} \frac {x_2}{y_2}$. But for $ T\le 1$ it wouldn't hold.[/quote]

Oh, you have a mistake, I think.
If we donate $ y_i \equal{} \frac {b_i}{T}$, we have $ \prod_{i\equal{}1}^n y_i\equal{}\frac{1}{T^{n\minus{}1}}$  :wink:

I' sorry but i'm very poor at math..

i can't understand what you say that

$ \sum_{i \equal{} 1}^n t_iy_i \equal{} A\ge \frac{n \minus{} 1}{n}\left(\sum_{i \equal{} 1}^n y_i\right)$

can you make me understand this?

쨃쨻쨻님 ㅎㅇㅎㅇ

근데 님이 질문하신식은 그 식의 바로 윗줄하고 옆에 식을 보면 자명해요 ^^

그리고 블레임님..  본 증명에선 'yi=bi/루트T'라고 나왔는데 오타가 아닌것같에요   :maybe:

[quote="BG Yoda"]
$ \sum_{i \equal{} 1}^n t_iy_i \equal{} A\ge \frac {n \minus{} 1}{n}\left(\sum_{i \equal{} 1}^n y_i\right)$.
[/quote]
Sorry, but I don't think It is true. try $ n\equal{}3,x_1\equal{}0.5,x_2\equal{}x_3\equal{}0.25,y_1\equal{}4,y_2\equal{}y_3\equal{}0.5$
$ \Rightarrow A\equal{}\sum t_i y_i \equal{} 2.75<\frac{2}{3}(\sum y_i)\equal{}\frac{10}{3}$ :wink:

[quote="BG Yoda"]
By the way, very good problem, [b]NextPeace[/b]. If it is your own - congratulations. :thumbup:[/quote]

Thanks, but it isn't mine. It is (a few) generalize of fkmo 2002.
In this problem, I don't have prefect solution, sorry.  :blush:

To. swanny1080
별로 자명하지 않아요;; 그리고 저 분께서 포스팅을 수정하신거임..

Oh yeah.
I didn't think over that  :blush: I used this method in another similar inequality with other conditions and there, this was obvious :roll:
Well ... I'll see how to correct my solution, but at the weekend. Now, I have to study for some tests at school and I don't have so much time. Just because of that I've made so many silly mistakes :oops:

To 블레임
아 포스팅 수정하신건가요..
그리고 쨃쨃쨻님이 질문하신거 자명한데 -_-;;(제눈에만 그런가요;;)
왜 자명하지 않다고 말씀하신건가요.. 전 레이텍 쓸주몰라서 지금 설명 못드리고있는데요... 쓸수만 있다면 지금 당장 설명드리고 싶어요ㅡ0ㅡ;

p/s:xmo에서 잘지내봐요  :lol:

To swanny1080
작성자가 그게 틀렸다고 인정한건데 자명해요?그럴리가 없는데..

그럼 오타를 어떻게 고쳐야하져?
저님은 그냥 학교시험 준비한다고 간건가요 :?:

[quote="BG Yoda"]Assume $ S \equal{} 1$ and donate $ y_i \equal{} \frac {b_i}{\sqrt [n]{T}}$ $ \Rightarrow \prod_{i \equal{} 1}^n y_i \equal{} 1$ .
Therefore your inequality becomes:

$ \frac {\sqrt [n]{T}}{n \minus{} 1}\sum_{i \equal{} 1}^n (1 \minus{} {x_i})y_i \ge \left(\sqrt [n]{T^{n \minus{} 1}}\sum_{j \equal{} 1}^n \frac {x_j}{y_j}\right)^{\frac {1}{n \minus{} 1}}$. $ \Rightarrow$ you can assume $ T \equal{} 1$, too.

$ \frac {\sum_{i \equal{} 1}^n (1 \minus{} {x_i})y_i }{n \minus{} 1}\ge \left(\sum_{j \equal{} 1}^n \frac {x_j}{y_j}\right)^{\frac {1}{n \minus{} 1}}$
[/quote]

Well - let us look the polynomial $ F(x) \equal{} (x \minus{} y_1)\dots(x \minus{} y_n)\sum_{i \equal{} 1}^n \frac {x_i}{x \minus{} y_i} \equal{} \sum_{i \equal{} 1}^n x_i\prod_{j\not \equal{} i}^{n}(x \minus{} y_j)$. It has $ n \minus{} 1$ roots but some of them can be equal. Actually, the number of the distinct roots is $ s \minus{} 1$, where $ s$ is the number of the pairwise distinct $ y_i$. Here we should conform that if $ y_i \equal{} y_j$, then $ F(x)$ will have multiple roots (I'm not sure if this is the term. I want say roots such that $ (x\minus{}y_i)^t|F(x)$ for $ t\ge 2$).

Let these roots be $ p_1, p_2, \dots, p_{n \minus{} 1}$. $ (y_1\ge p_1\ge y_2\ge \dots \ge p_{n \minus{} 1}\ge y_n\ge 0$ $ \Rightarrow$ the roots are positive $ ).$

From Viete $ \sum_{j \equal{} 1}^n \frac {x_j}{y_j} \equal{} |F(0)| \equal{} \left|\prod_{i \equal{} 1}^{n \minus{} 1} p_i\right|$ and $ \left|\sum_{i \equal{} 1}^{n \minus{} 1} p_i\right| \equal{} \sum_{i \equal{} 1}^{n}x_i\left(\sum_{j\not \equal{} i}y_j\right)$ $ \equal{} \sum_{i \equal{} 1}^{n}y_i\left(\sum_{j\not \equal{} i}x_j\right) \equal{} \sum_{i \equal{} 1}^n (1 \minus{} {x_i})y_i$

But $ \frac {\sum_{i \equal{} 1}^n (1 \minus{} {x_i})y_i }{n \minus{} 1} \equal{} \frac {\left|\sum_{i \equal{} 1}^{n \minus{} 1} p_i\right|}{n \minus{} 1}\ge \sqrt [n \minus{} 1]{\left|\prod_{i \equal{} 1}^{n \minus{} 1} p_i\right|} \equal{} \sqrt [n \minus{} 1]{\sum_{j \equal{} 1}^n \frac {x_j}{y_j}}$  :huh:

Again I'm not sure, but I think this works surely for distinct $ y_i$.

And to swanny1080
$ \sum_{i \equal{} 1}^n t_iy_i \equal{} A\ge \frac{n \minus{} 1}{n}\left(\sum_{i \equal{} 1}^n y_i\right)$
can be transformed equivalent to $ \frac{\sum_{i\equal{}1}^ny_i}{n}\ge \sum_{i\equal{}1}^nx_iy_i$
and it is Chebyshevs' ine form. (just because $ \sum_{i\equal{}1}^nx_i\equal{}1$)
But we can't decide order of x_i and y_i
so there exist many cases that make above inequality be not true

is it wrong?

[quote="NextPeace"]$ n\ge2,a_k,b_k > 0, S \equal{} \sum_{k \equal{} 1}^n a_k, T \equal{} b_1b_2\cdots b_n$
Prove that :
$ \frac {1}{n \minus{} 1}\sum_{i \equal{} 1}^n (1 \minus{} \frac {a_i}{S})b_i \ge (\frac {T}{S}\sum_{j \equal{} 1}^n \frac {a_j}{b_j})^{\frac {1}{n \minus{} 1}}$[/quote]
See here: http://www.mathlinks.ro/viewtopic.php?t=258592, my post #7. :)

[quote="BG Yoda"][quote="BG Yoda"]Assume $ S \equal{} 1$ and donate $ y_i \equal{} \frac {b_i}{\sqrt [n]{T}}$ $ \Rightarrow \prod_{i \equal{} 1}^n y_i \equal{} 1$ .
Therefore your inequality becomes:

$ \frac {\sqrt [n]{T}}{n \minus{} 1}\sum_{i \equal{} 1}^n (1 \minus{} {x_i})y_i \ge \left(\sqrt [n]{T^{n \minus{} 1}}\sum_{j \equal{} 1}^n \frac {x_j}{y_j}\right)^{\frac {1}{n \minus{} 1}}$. $ \Rightarrow$ you can assume $ T \equal{} 1$, too.

$ \frac {\sum_{i \equal{} 1}^n (1 \minus{} {x_i})y_i }{n \minus{} 1}\ge \left(\sum_{j \equal{} 1}^n \frac {x_j}{y_j}\right)^{\frac {1}{n \minus{} 1}}$
[/quote]

Well - let us look the polynomial $ F(x) \equal{} (x \minus{} y_1)\dots(x \minus{} y_n)\sum_{i \equal{} 1}^n \frac {x_i}{x \minus{} y_i} \equal{} \sum_{i \equal{} 1}^n x_i\prod_{j\not \equal{} i}^{n}(x \minus{} y_j)$. It has $ n \minus{} 1$ roots but some of them can be equal. Actually, the number of the distinct roots is $ s \minus{} 1$, where $ s$ is the number of the pairwise distinct $ y_i$. Here we should conform that if $ y_i \equal{} y_j$, then $ F(x)$ will have multiple roots (I'm not sure if this is the term. I want say roots such that $ (x \minus{} y_i)^t|F(x)$ for $ t\ge 2$).

Let these roots be $ p_1, p_2, \dots, p_{n \minus{} 1}$. $ (y_1\ge p_1\ge y_2\ge \dots \ge p_{n \minus{} 1}\ge y_n\ge 0$ $ \Rightarrow$ the roots are positive $ ).$

From Viete $ \sum_{j \equal{} 1}^n \frac {x_j}{y_j} \equal{} |F(0)| \equal{} \left|\prod_{i \equal{} 1}^{n \minus{} 1} p_i\right|$ and $ \left|\sum_{i \equal{} 1}^{n \minus{} 1} p_i\right| \equal{} \sum_{i \equal{} 1}^{n}x_i\left(\sum_{j\not \equal{} i}y_j\right)$ $ \equal{} \sum_{i \equal{} 1}^{n}y_i\left(\sum_{j\not \equal{} i}x_j\right) \equal{} \sum_{i \equal{} 1}^n (1 \minus{} {x_i})y_i$

But $ \frac {\sum_{i \equal{} 1}^n (1 \minus{} {x_i})y_i }{n \minus{} 1} \equal{} \frac {\left|\sum_{i \equal{} 1}^{n \minus{} 1} p_i\right|}{n \minus{} 1}\ge \sqrt [n \minus{} 1]{\left|\prod_{i \equal{} 1}^{n \minus{} 1} p_i\right|} \equal{} \sqrt [n \minus{} 1]{\sum_{j \equal{} 1}^n \frac {x_j}{y_j}}$  :huh:

Again I'm not sure, but I think this works surely for distinct $ y_i$.[/quote]

Yes, this solution is very simillar with mine.(In original problem, condition is $ \prod_{i \not\equal{} j}(b_i\minus{}b_j)\not\equal{}0$, and give the funtion $ \prod_{k\equal{}1}^n (x\minus{}b_k)\sum\frac{a_i}{x\minus{}b_i}$.
But I don't know how can we say the funtion has $ n\minus{}1$roots when some $ b_i$are equal(And I don't know this too,now.). Thanks your post, BG Yoda. :)

And Thanks for your beautiful solution, can_hang!  :lol:


geometry
3D geometry
octahedron
probability
pyramid

Inspired by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213328]this topic[/url]:

A point is chosen at random in the interior of a regular octahedron. What is the probability that it lies closer to the center of the octahedron than to any of the vertices?

[hide]In order to be closer to a vertex, the point must be in one of the pyramids formed by a vertex and the midpoints of the four segments extending from it. These pyramids are of volume 1/16 that of the octahedron. There are 6 of these, so the probability of the point being closer to a vertex than the center is 3/8. The probability that it is not closer to a vertex, or closer to the center, is the complement of this, or 5/8.

edit: Thanks for the correction! :oops: [/hide]

That is correct, math154.

As I did on that other topic, I could also ask for a characterization of the polyhedron around the center. Is it also an Archimedean solid? How many of what kinds of faces does it have?

[hide]A regular octahedron has 8 faces and 12 edges. For each pyramid we "cut off", we add one face and 4 edges. That's 14 faces and 36 edges. However, because the pyramids are formed by the midpoints of the original edges, the original edges will "disappear" when we take away these pyramids, giving us 24 edges instead. By Euler's, we also know that it has 12 vertices. This is a [url=http://en.wikipedia.org/wiki/Cuboctahedron]cuboctahedron[/url], with eight triangular faces and six square faces.[/hide]

edit: Solution corrected. Forgot that the original edges "disappear".

It does have 14 faces. But that is not the correct number of edges and that is not the correct identification of the polyhedron.

Edit: Yes, now that is correct.


limit
inequalities
function
real analysis
real analysis solved

Just take this for granted (we can prove it by L'Hospital):

n(e-(1+1/n)^n)->e/2 as n->infinity. This means that the limit is 0 for  \lambda =1, which means that the limit is zero for all values  \lambda <=1. I'm still thinking abt the other cases.

[quote="pcalin"]Compute the following limit  n \lim \infty n  \lambda   (e-(1+1/n) n-e/2n). Discussion about the real values of \lambda .[/quote]

Use an asymptotic expansion when n--> +oo  of

(1 + 1/n)^n = exp(n.ln(1+1/n)) = e-e/(2n) +10e/(24n^2) + o(1/n^2)

we have

Un = (-10e/(24))1/n^(2- \lambda ) + o(1/n^(2- \lambda ))

If 2- \lambda > 0 the limit is 0
If  \lambda =2 the limit is -10e/(24)
If 2- \lambda <0 the limit is - \infty

Using the inequalities e/(2n+2)<e-(1+1/n) n <e/(2n+1) I got the same results as moubi for  \lambda >2 and  \lambda <2. The problem was the case  \lambda =2, for which I am still looking for a more elementary solution.

[quote="grobber"]I think we can also use L'Hospital (even if it becomes somewhat messy).[/quote]

hi grobber,

How do you use Marquis de L'Hpital theorem, i only know Guillaume L'Hpital for functions not for sequence.

Theorem de L'Hpital : f,g functions continous, derivable on I\{a} ,

f(a)=g(a)=0
for any t \in I\{a} g'(t)=/=0

If  \lim _{t->a, t=/=a} f'(t)/g'(t) = L then  \lim _{t->a,t=/=a} f(t)/g(t) =L

http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/De_L'Hopital.html

yes but if one has continuous functions we know that if f(x) - > l as x -> \infty then we can take n -> \infty and the limit yould be still l.


calculus
integration
floor function
inequalities

$ s\equal{}\frac{1}{1980}\plus{}\frac{1}{1981}\plus{}\cdots\plus{}\frac{1}{2000}$

find the value $ \lfloor \frac{1}{s}\rfloor$

i think we can use tellescopic cancellation to estimate the sum

my method is much less legitimate, but it worked for this problem at least.

$ \frac{21}{2000}<S<\frac{21}{1980}$ so we know $ 94.???<\frac{1}{s}<95.???$
At this point I narrowed my options down to a 50/50 guess between 94 and 95, so I tried one more thing.
I tested out that $ \frac{1}{2}\plus{}\frac{1}{3}\plus{}\frac{1}{4}\plus{}\frac{1}{5}\plus{}\frac{1}{6}>\frac{5}{4}$ so I made a generalization that $ S>\frac{21}{1990}$ so $ \frac{1}{S}<94.???$ Now since the upper and lower bounds were between 94 and 95 I knew it was 94.

[b]Elefont[/b], have you heard of the "arithmetic mean" and the "harmonic mean"? There's an important inequality that will help you formalize your upper bound:

[i]"For any list of positive numbers, the harmonic mean of the list does not exceed the arithmetic mean of the list."[/i]

This is known colloquially as AM-HM (or as part of a larger chain of inequalities, AM-GM-HM, where GM is the geometric mean).

The two means relevant to this problem are defined as below:

$ \text{AM}[a_1, a_2, \ldots a_n ] \equal{} \frac {a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n}{n}$

$ \text{HM}[a_1, a_2, \ldots a_n ] \equal{} \frac {n}{\frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} \ldots \plus{} \frac {1}{a_n}}$

Can you finish your solution from here?


Can someone give me some problems using it? I need practice..

[quote="funtwo"](2e^pi/3)^100
2^100 e^100pi/3
2^100 e^4pi/3
2^100 (-1/2 +\sqrt{3}/2 i)[/quote]

Careful... $ e^{\pi/3}\ne e^{i\pi/3}$.

Anyway, here it is with De Moivre's theorem:

[hide="Solution"]
Plot $ 1+\sqrt3i$ in the complex plane and form a right triangle with sides $ 1$, $ \sqrt{3}$, and $ 2$. This is obviously a 30-60-90 triangle, so the angle opposite from $ \sqrt{3}$ is $ 60^\circ=\pi/3$.

Therefore, $ \arg(1+\sqrt3i)=\frac{\pi}{3}$, so $ 1+\sqrt3i=|1+\sqrt3i|\;\text{cis}\;\frac{\pi}{3}=2\;\text{cis}\;\frac{\pi}{3}$.

Thus, $ \left(2\;\text{cis}\;\frac{\pi}{3}\right)^{100}=2^{100}\;\text{cis}\;\frac{100\pi}{3}$ from De Moivre's theorem.

$ \frac{100\pi}{3}\equiv \frac{4\pi}{3}$, so our answer is:

$ 2^{100}\left(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}\right)=\boxed{-2^{100}\left(\frac{1}{2}+\frac{\sqrt3}{2}i\right)}$.
[/hide]

Here's another good problem (a little too easy if you know what you're doing):
Given that $ z\in\mathbb{C}$ such that $ z+\frac{1}{z}=2\cos3^\circ$, find the least integer that is greater than $ z^{2000}+\frac{1}{z^{2000}}$ (2000 - AIME II #9).


search
real analysis
real analysis unsolved

Let $ a$ , $ b$ be two positive constants. Ascertain $ \min_{\begin{array}{c}
x>0\ ,\ y>0\\\
ax\plus{}by\equal{}1\end{array}}\ \frac {xy}{\sqrt {(1\plus{}x^2)(1\plus{}y^2)}}$ .

Your problem is ill conditioned; the $ min$ doesn't exist because $ f(0,1/b)\equal{}0$ isn't reached.
However if you ask for the local $ extrema$ then we solve the system $ df\plus{}\lambda{d\phi}\equal{}0$ where $ \phi(x,y)\equal{}ax\plus{}by\minus{}1$. We obtain $ b(y^3\plus{}y)\equal{}a(x^3\plus{}x),ax\plus{}by\equal{}1$. This system admits one solution (it's always the case); for example, if $ a\equal{}1,b\equal{}2$; then $ x\approx0.466,y\approx0.264$. It remains to study this point.....
Remark: perhaps you want the $ max$ ???
The $ max$ exists over the compact $ ax\plus{}by\minus{}1\equal{}0,x\geq0,y\geq0$ and the obtained solution reached necessarily this value.

[quote="loup blanc"]Perhaps you want the $ max$ ??? The $ max$ exists over the compact $ ax \plus{} by \minus{} 1 \equal{} 0,x\geq0,y\geq0$ and the obtained solution reached necessarily this value.[/quote]
Sorry, you are right ! I search the maximum ! Thank you !


search
algorithm
calculus
analytic geometry
vector
advanced fields
advanced fields theorems

hi maths people!

i have a simple solution to a problem of mine, but would like to know if there is a more 'elegant' solution.

i have a list of 30 or so places, all identified by their latitude/longitude coordinates.  given a lat/long pair, calculate the nearest place to it (out of the 30 or so in the list).  currently my solution is to start at the first place in the list and calculate the distance between them (i already have a method to  calculate the distance between two lat/long pairs).  then i move to the next place and calculate the distance again - if its any shorter than the previous, then this becomes the closest place.  i repeat until the end of the list is found.  of  course, this works, but something tells me there is a better method.

This is really a question for computer science people. I suggest you to [url=http://www.google.com/search?hl=en&lr=&q=Nearest+neighbor+search&btnG=Search]do this[/url].

But I would not bother looking for a better algorithm with a small dataset as yours.

As an aside, I'd like to see your formula for computing distance given latitude and longitude. (I have assigned this as a homework exercise in a Calculus III course, in the spherical coordinates section. The assignment worksheet contains the coordinates of a number of actual places and asks such "you'd better compute it before you answer" questions such as which of Moscow, Taipei, or Beunos Aires is closest and which furthest from Los Angeles.)

Your question boils down to this computer science/algorithms question: given a list of $n$ numbers (or, in your case, $n$ potential numbers, which can be computed on demand), what is the most efficient way to choose the minimum? In particular, is there any way to improve upon the obvious $O(n)$ algorithm you describe? (As mlok said, for $n\sim 30,$ there's not much motivation to look for anything better.)

I think that to be dead sure that you found the minimum, you would, in fact, have to make all of the comparisons. There are some interesting statisitical questions about predicting the magnitude of that minimum from an incomplete sample.

hi guys (or girls  :wink: )

thanks for your replys.  you are right - this is a computer science question.  i have posted on a c++ forum and they told me to come here!  mlok, thanks for the link, i will look into this.

kent merryfield - i have a couple of methods for working out the distance bewteeen 2 lat/long points.  the first is by using the haversine formula, which i am sure you already know about.  the second is a 'poor  mans' method:

- for LATITUDE, simply calculate the distance in minutes between both lat coordinates.  as 1 minute = 1 nautical mile, so we have the number of nautical miles.  call this y.

- for LONGITUDE, work out the distance of one minute of longitude at the point of latitude by doing cos(latitude) (because the distance of one minute of longitude changes according to where you are latitude-ly (i think i just made this word up)).  this gives you the answer in nautical miles.  simply multiply this number by the actual number of longitudinal minutes (longitude * 60) to get the number of nautical miles between the 2 points.  call this x.

- now we have x and y, do a bit of pythagorus: result = sqrt(x*x + y*y).

- multiply this (which is in nautical miles) by 1.85 to get the result in km.

if you are teaching this though, i would probably not use this method!  i am not on a maths course so i do  not have to be that exact with the actual  formula.

thanks again,
-Paul

One thought: if you had a truly large number of distances to test, you could use some version of your estimate (the second method) on a first pass, then apply the exact formula to the few distances closest to the minimum. This would rely on the simplicity of the second method to speed up calculations; I would use the square of the distance to compare, since computing square roots is relatively slow.

Paul - your "poor man's formula" looks like it would work reasonably well for small distances but be quite inaccurate for large distances. The primary issue with large distances is that the shortest distance between two points on the globe lies along a great circle route, and not a line of constant bearing.

Here's the text of my Calculus III homework exercise; the two pieces of knowledge presumed are how to get an angle between vectors from a dot product and the conversion formulas between rectangular coordinates and spherical coordinates in $\mathbb{R}^{3}.$ If you're wondering what Long Beach is doing in my examples, well, that's where I'm from and where my class is taught. Most of my student's don't really know enough history to catch why Paris to St. Helena should sound familiar. I included Casablanca to Johannesburg just to make the point that Africa is a BIG continent.

[quote="Kent Merryfield's Calculus III Homework"]The problem:

Derive and simplify a formula (that you could use in a calculator or spreadsheet) which accepts as input the latitude and longitude of any two points on the surface of the earth and returns as output the shortest (great circle) distance between those two points. In doing this, assume that the earth is a perfect sphere (it’s not, quite) and that its radius, $R,$ is about $3960$ miles (if you want to express the answer in miles) or about $6373$ kilometers (if you want to express the answer in kilometers).

Demonstrate that your formula works by computing all of the following distances:

Long Beach (33.8° N, 118.1° W)	to San Francisco (37.7° N, 122.4° W)
Long Beach 	(33.8° N, 118.1° W)	to	New York City 	(40.8° N, 74.0° W)
Long Beach 	(33.8° N, 118.1° W)	to	Taipei			(25.1° N, 121.6° E)
Long Beach 	(33.8° N, 118.1° W)	to	Paris			(48.8° N, 2.3° E)
Paris		(48.8° N, 2.3° E)	to	St. Helena		(15.0° S, 5.7° W)
Long Beach 	(33.8° N, 118.1° W)	to	Buenos Aires		(34.6° S, 58.3° W)
Buenos Aires	(34.6° S, 58.3° W)	to	Auckland		(36.9° S, 174.8° E)
Casablanca	(33.2° N, 7.7° W)	to	Johannesburg		(26.2° S, 28.0° E)

Hints: Consider the two vectors from the center of the earth to the two points in question. How is the angle between those vectors related to the distance we seek? How can we find that angle? There are uses in this problem for both rectangular and spherical coordinates. Be careful of degrees and radians - the data is given in degrees but the relationship between angle and arclength involves radians. Be careful also of the conversion between latitude (which is 0° at the equator) and $\varphi,$ or co-latitude, which is 0 at the north pole and $\pi/2$ at the equator. Use at least one trig identity to simplify things somewhat, and test your formula for reasonableness by considering several special cases: What if both longitudes are the same (north-south travel)? What if you shift both longitudes by the same amount? What if both points are on the equator? What if one point is the north or south pole? [/quote]

I would like to know if there is a formula that can be applied to solve the  kind of problems listed  below?

Thanks

Sharma

Problem 01
	An a/c departs from position 2700.0N 17000.0W and travel 3000km on a track of 180º (T), then 3000 km on a track of 090º(T), then 3000 km on a track of 000º and, finally, 3000 km on a track of 270º(T). The a/c final position is?

 Problem 02
	What is the great circle distance btween A and B if A:65ºN 130ºW and B:72ºN 050ºE?

I only can solve problems such as the the two listed above by basic method, like calculating the difference between the coordinates and apply the value of each degree of LAT or LON. ie 1 degree of LAT = 111 km and 1 degree of LON = 49 nm. The value of 1 degree of LON varies as you go north or south of the Equator.

Any one knows of a formula that may be applied to solve those problems?


let $ p_n$ - $ n$th prime.

is this series convergent or not?
\[ \sum\frac{1}{p_{n\plus{}1}\minus{}p_n}\]

It is not converge.
Interested question about $ \sum_n \frac{1}{p_n(p_{n\plus{}1}\minus{}p_n)}.$

It diverges because $ p_{n \plus{} 1} \minus{} p_{n} < p_{n}$ by Bertrand's Postulate and $ \sum \frac {1}{p_i}$ diverges (it is also trivial if the twin prime conjecture is true...).

[quote="bambaman"] [...] (it is also trivial if the twin prime conjecture is true...).[/quote]
it was the reason to ask about it :D
it interested me what can be said about this series while twin-prime-problem is still open

[quote="Rust"]Interested question about $ \sum_n \frac {1}{p_n(p_{n \plus{} 1} \minus{} p_n)}.$[/quote]
Is it known ?

It converged. Consider number of primes $ x<P_n<2x, p_{n\plus{}1}\minus{}p_n<a$.

[quote="Rust"]It converged. Consider number of primes $ x < P_n < 2x, p_{n \plus{} 1} \minus{} p_n < a$.[/quote]
May I ask for a sketch of proof ?


Well we are opening up the Forum for the IMO 2009 discussions. Have fun :D

hey guys,
it is a great honour for me to write the first "real" post in this IMO 2009 forum.
as some of you may know i am one of the organizers for next year IMO in germany. if you have any wishes or suggestions which can help us to "improve the quality" of the event, don't hesitate to ask me (either here or in an extra-thread)^^

see you all next year,

Naphthalin

The content on the web page is still quite sparse. When are you going to do anything about it?  :D

"when are you" -> it will come soon ;) but it isn't actually my work to write the textes...

Naphthalin

Your reply is correct but this little specific answer does not help anybody.  :D

I hope that Vietnamese team will do better in this year than last year.And my dream is take part in IMO 2010 (which is just no more than dream)  :roll:

indian team FTC!

[hide]for the championship[/hide]

really looking forward to the 50th IMO in Germany ~~~~

Will there be any special arrangement for this special IMO that differ it from the past?

[quote="Naphthalin"]i think so ;) e.g. there will be a "jubilary day" - look at the website, there should be more information now...

Naphthalin[/quote]

What is jubilary day? I didn't see it on the official website.

I am sure you are all looking forward to an update of the booklet "[i]The German Teams at the International Mathematical Olympiads 1959-1998[/i]" that discusses some general statistics of the IMO which can partially be found in this attachment at the end of my post [url=http://www.mathlinks.ro/viewtopic.php?p=110248#110248]here[/url] and the results of the German teams.  :D

[quote="rituraj007"]i think jubilary day means the Silver Jubilee celebrations :)[/quote]

Ah... It didn't really help.

No. It should be [url=http://en.wikipedia.org/wiki/Golden_Jubilee]Golden Jubilee[/url]. (Well not necessarily applicable for monarch's reign only, as stated in that wiki article :P  :wink: )

I wish I could enjoy the journey to Germany and make friends there


Heres the question...
Theres a cylindrical vessel whose height and radius is known. It is filled initially with water whose level is also known. A thirsty crow with hope of quenching his thirst approaches the cylindrical vessel. He throws small pebbles, which can be approximated as small spheres, into the water. However he observes that no matter how many pebbles are thrown the water level never reaches till the mouth of the vessel. Assume that there is no leakage anywhere. The radius of the pebbles is also known(i.e can be included in the calculations) Find the condition for this to happen?

This was all that was given.
Its really a good problem. Please state if you are making any assumptions in your solution.
regards

[quote="Adwaitvedant Mathkar"]...water level never reaches till the mouth of the vessel. There is no leakage anywhere....Find the condition for this to happen?[/quote]

Possible if the pebbles float, which is highly unlikely.

Hmm, I had conceived almost exactly the same notion that Rajiv had replied. But to contradict him, why can't pebbles float? ( Is it mandatory that all pebbles of spherical shape must sink?) I assert that I am not an expert in Physics, but I am interested in learning.

[b] Assumption [/b] (It might be incorrect, do correct me if it's incorrect): A possible reason for such a condition is because the pebbles float.

I ask the person who put up this problem to tell whether my mode of thinking about the problem was correct?

Pebbles don't float, true. But there one good reason why the water may not reach the crow. When the crow throws in the pebbles, let us assume, the pebbles get arranged in such a way that they are closest to each other much similar to the [url=http://en.wikipedia.org/wiki/Hexagonal_crystal_system]hexagonal type crystal structure[/url]. However, there remains a void in between the four spheres in contact.
Volume of each small sphere is $ \frac {4}{3}\pi r^3$. if the crow throws in $ n$ pebbles, the volume of the void is $ V\equal{}\pi R^2h \minus{} \frac {4n}{3}\pi r^3$. Let initial level of water be $ l$. So the water never comes to the top if $ \pi R^2l < V$

Well I dont know the answer...but even i solved it on the lines of what rajiv has done....and yes the hexagonal close packing part is essentially correct..thats what I believe....I dont have access to the answer as the question had come in an competitive exam....anymore ideas??

I can't think of any other way this could be possible. Atleast nothing within my scope of knowledge.

When you fill the vessel with pebbles, there is some space between the pebbles.
If the volume of this empty space is larger than the initial volume of water, then the water never reaches the top.
This is how it can happen. The details will depend on how do you pack the spheres. More likely it will be a random packing and not a hexagonal one.

[quote="nasu"]
The details will depend on how you pack the spheres. More likely it will be a random packing and not a hexagonal one.

[/quote]

I know, but I take for granted that the crow has good command on Chemistry. You see, he does appear intelligent enough to be so with that idea of pebbles.

Wow, rajiv, thanks for the comprehensive explanation in your post. I would have never thought of a hexagonal arrangement of the pebbles. Although I did math classes in an [url=http://www.onlinecolleges.net/online-schools-with-degrees-offered/]online college program[/url] this was way too difficult for me. But if you think twice about it, it's logical.

and I'm sure that this crow has good chemistry and maths knowledge ...  :D

Mathematicians have failed to prove the nonexistence of some irregular sphere packing that has higher density.

But a random sphere packing has 64% density. Both close packings will have around 74%. Radius should be largely irrelevant as we assume pebbles have extremely small radii compared with the vessel.

(Wait a minute, this implies that a vessel full of infinitesimally small spheres will be 26% space...)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be an infinite group such that every finitely generated subgroup of $G$ is cyclic. Must $G$ be countably generated?

An equivalent statement: Let $G$ be a group such that any two nonidentity elements of $G$ are commensurable. Then $G$ is countable.

First, we show that $G$ is abelian. If $a$ and $b$ do not commute, they generate a subgroup which is nonabelian and hence not cyclic. We now write the group additively.
Now, we show that there are at most $n$ elements of order dividing $n$ for each positive integer $n$: If not, a set of $n+1$ such elements generates a subgroup with more than $n$ elements such that every element has order dividing $n$. Such a group is not cyclic, and we are done.
Now, we show that for some arbitrary $a\neq0$, only countably many elements may be commensurable with $a$.
Every such element has the form $mb$ for some $b$ such that $nb=a$ for some pair of integers $m,n$. If $nb=nb'=a$, $n(b'-b)=0.$ By the above, there are at most $n$ possible values of $b'-b,$ so there are at most $n$ such elements for each pair $m,n$ and countably many overall.
Since every element of $G$ is commensurable with $a$, $G$ is countable.

If $G$ has nontrivial elements of finite order, it is isomorphic to a subgroup of $\mathbb{Q}/\mathbb{Z}.$
If $G$ has elements of infinite order, it is isomorphic to a subgroup of $\mathbb{Q}.$

Thank you for your nice solution! How about the following more general thing? (to this I don't know a solution):

Let $G$ be a group such that there are only countably many isomorphism classes of subgroups of $G$. Must $G$ be countable?

[quote="vess"]Let $G$ be a group such that there are only countably many isomorphism classes of subgroups of $G$. Must $G$ be countable?[/quote]

The answer is "no". Think about a free group on $\aleph_1$ generators. Every subgroup is a free group, and its isomorphism class depends only on the cardinality of a set of generators, which can be either finite, $\aleph_0$, or $\aleph_1$. Our group is thus an example of an uncountable group having only countably many isomorphism classes for its subgroups.


function
trigonometry
algebra proposed
algebra

A real-valued function $ f$ is defined on the reals and satisfies $ f(xy) = xf(y) + yf(x)$ and $ f(2x) = f\Big( \sin{\frac{\pi}{2}(x + y)}\Big) + f\Big(sin{\frac{\pi}{2}(x - y)\Big)}$ for all $ x, y$. Find $ f(1990 + \sqrt{1990}+ \sqrt[3]{1990})$.

My solution:
-)$ x \equal{} y \equal{} 1: f(1) \equal{} 2f(1)\Rightarrow f(1) \equal{} 0$
-)$ x \equal{} y \equal{} \minus{} 1: f(1) \equal{} \minus{} 2f( \minus{} 1)\Rightarrow f( \minus{} 1) \equal{} 0$
-)$ y \equal{} x \minus{} 1: f(2x) \equal{} f\Big(\sin \frac {\pi}{2}(2x \minus{} 1)\Big)$
-)$ y \equal{} x \plus{} 1: f(2x) \equal{} f\Big(\sin \frac {\pi}{2}(2x \plus{} 1)\Big)$
$ \Rightarrow f\Big(\sin \frac {\pi}{2}(2x \minus{} 1)\Big) \equal{} f\Big(\sin \frac {\pi}{2}(2x \plus{} 1)\Big) \equal{} f\Big(\sin \frac {\pi}{2}(1 \minus{} 2x)\Big)$
$ \Rightarrow f(\sin t) \equal{} f(\sin( \minus{} t))$
Then $ f(2x) \equal{} f(2y)$ for all $ x,y$. It mean $ f(x) \equal{} 0 \forall x$


What is the sum of all integers $x$ that satisfy $-5\le \frac{x}{\pi} \le 10$?

[hide]
Multiply the inequality by $\pi$
$-5\pi\leq x\leq 10\pi$

Approximate $-5\pi$ by multiplying by 3.14 and do the same for $10\pI$.

$-15.7\leq x\leq 31.4$

So the integers from -15 to 31 inclusive work.

The 1-15 and -1 to -15 cancel out. Now we have 16+17+...+30+31.
Compute any way you want to get 376[/hide]


On the front cover of the recent edition I think you should remove the cursor. lol

:rotfl:

Wow!  You are observant!  And I've had the 7th edition for a while, too...

I have the 7th edition, and I've never noticed it!

about an inch below the top of the sphere on the front cover


Yes, it would, because every point on the line will have the same x-coordinate (5).

The equation is x=5. No matter how far you move up or down, the x-coordinate stays the same.

[quote="Inspired By Nature"]It would be undefined.
[hide="reason"]$m=\frac{-15}{0}$[/hide][/quote]
The slope is undefined, but we don't have to write it in slope-intercept form.

[quote="stevenbao"]Problem -
Find the equation of  line that passes through the points (5,7) and (5,-8).
Wouldn't this be x=5?[/quote]
I miss read your question. Yes, the equation would $x=5$. :)


One student in a class of boys and girls is chosen to represent the class.  Each student is equally likely to be chosen and the probability that a boy is chosen is $ \frac23$ of the probability that a girl is chosen.  The ratio of the number of boys to the total number of boys and girls is

$ \textbf{(A)}\ \frac13 \qquad \textbf{(B)}\ \frac25 \qquad \textbf{(C)}\ \frac12 \qquad \textbf{(D)}\ \frac35 \qquad \textbf{(E)}\ \frac23$

$ \frac {2}{3}P(g) + P(g) = 1\implies{P(g) = \frac {3}{5}}$ where $ P(g)$ is the probability that a girl is chosen. $ P(b)=1-P(g)=\frac{2}{5}\implies{(B)}$

[hide="Solution"]Obviously, the ratio of boys to girls is $ \frac23$.  Thus, the ratio of boys to the whole class is $ \frac{2}{2\plus{}3}$ or $ \boxed{\text{(B)}}$.[/hide]


calculus
derivative
real analysis
real analysis unsolved

Let $ y \equal{} arcsin(x)$  Prove the following

$ y^{(2n)} \equal{} 0$ and $ y^{(2n \plus{} 1)} \equal{} 1^{2} 3^{2}...(2n \minus{} 1)^{2}$

$ n\in\mathbb{N}$

[hide="Hint"]You can get the power series centered at zero for the arcsine by integrating the series (binomial series) for $ (1\minus{}x^2)^{\minus{}1/2}.$[/hide]


Έστω $ ABCD$ κυρτό τετράπλευρο και $ A_1, A_2 \in AB$, $ B_1, B_2 \in BC$, $ C_1, C_2 \in CD$ και $ D_1, D_2 \in DA$ και $ A_0, B_0, C_0, D_0$ σημεία εσωτερικά του τετραπλεύρου. Να δείξετε ότι αν τα $ A_0B_0C_0D_0, AA_1A_0D_2, BB_1B_0A_2, CC_1C_0B_2, DD_1D_0C_2$ είναι περιγράψιμα, τότε και το $ ABCD$ είναι περιγράψιμο.

Το μόνο που χρειάζεται είναι θεωρία περιγράψιμου τετραπλεύρου και βγαίνει . Μόλις βρώ χρόνο να κάνω το σχήμα στο Euclidraw θα βάλω τη λύση .

Αρχικά ξέχασες να αναφέρεις 'οτι τα σημεία $ D_2 , A_0 , B_0 , B_1$ είναι σε ευθεία ομοίως και τα $ D_1 , D_0 , C_0 , B_2$ . Λύση :

Εμείς θέλουμε να αποδείξουμε οτι ΑΒ+DC=ΑD+ΒC αλλά εύκολα μετατρέπεται σε $ A_3B_3\plus{}D_4C_4\equal{}A_5D_5\plus{}B_5C_5$ . Λόγω συμμετρίας ισχύει $ A_3B_3\equal{}A_4B_4  , D_4C_4\equal{}D_3C_3 , A_5D_5\equal{}LM , B_5C_5\equal{}KN$. Γνωριζουμε όμως ότι $ A_4A_0\equal{}A_0L , KB_0\equal{}B_0B_4 , C_0C_3\equal{}C_0N , D_0D_3\equal{}D_0M$ αφού ειναι τμήματα εφαπτόμενα από το ίδιο σημείο . Έτσι η σχέση μετατρέπεται σε $ A_0B_0\plus{}D_0C_0\equal{}A_0C_0\plus{}B_0D_0$  που ισχύει αφού γνωρίζουμε από την υπόθεση ότι το τετράπλευρο $ A_0B_0C_0D_0$ είναι περιγράψιμο . Άρα το τετράπλευρο ABCD είναι περιγράψιμο . Συγγνώμη για τη διαφορά γλώσσας ανάμεσα σε σχήμα και τύπους.

Δίκιο έχεις. Είναι συνευθειακά και το ξέχασα.


source: Book: Inegalitati; Authors:L.Panaitopol,V. Bandila,M.Lascu

If $a>b>c>0$, then: $ \frac{a^2}{b^2}+2 \frac{c^2}{a^2} \geq  \frac{b^2}{a^2} + 2 \frac{c^2}{b^2}$.

Multiply by [tex]a^2b^2[/tex] to get [tex]a^4+2b^2c^2 \geq b^4+2c^2a^2[/tex] which simplifies to [tex]a^4-b^4-2(a^2c^2 - b^2c^2 \geq 0[/tex] or [tex](a^2-b^2)(a^2+b^2) - 2c^2(a^2-b^2) \geq 0[/tex] or [tex](a^2-b^2)(a^2+b^2-2c^2) \geq 0[/tex], which is clearly true from [tex]a>b>c>0[/tex]


Let $ G$ be a group. Let $ H_1,H_2 \subset G$ be subgroups with $ |H_1| \equal{} n_1, |H_2| \equal{} n_2$ and $ n_1 \in U(n_2)$. Find $ |H_1 \cap H_2|.$

This is one of my homework questions. I just wanted to clarify my solution. I got that the order of the intersection is 1, by Lagrange's.
If someone could possibly just post if this is right or wrong? If its correct I'll post my solution, if it's wrong... well I will post it anyways and ask one of you to help =).

By the way $ U(x) \equal{} \{ g : \gcd(x,g) \equal{}1 \}$  Group of positive integers coprime to $ x$.

Your solution is [b]right[/b].  :wink:

$ |H_{1} \cap H_{2}|$ is  a commond divisor of $ |H_{1}|$ and $ |H_{2}|$.

Thank you for the very fast response =)
Is there any way I may clarify this.

Since $ |H_1 \cap H_2|$ is a subgroup of $ G$ (we proved this in class long ago, and I am unsure if I should bother for this assignment) its order must divide $ |G|$, by Lagrange's Theorem. Since $ n_1 \in U(n_2)$, $ \gcd(n_1,n_2)\equal{}1$, hence coprime. Since $ H_1 \cap H_2$ must divide both $ n_1, n_2$ again by Lagrange's, we get that the order must equal 1.

Since $ H_1 \cap H_2$ must divide both $ n_1, n_2$
Should I justify this? it's pretty self-explanatory

[quote="Fermat's Little Turtle"] Since $ H_1 \cap H_2$ must divide both $ n_1, n_2$
Should I justify this? it's pretty self-explanatory...[/quote]

That's an immediate consequence of Lagrange's theorem: $ H_1 \cap H_2$ is a subgroup of both $ H_{1}$ and $ H_{2}$.

Regards,

[quote="Fermat's Little Turtle"]

Since $ |H_1 \cap H_2|$ is a subgroup of $ G$ (we proved this in class long ago, and I am unsure if I should bother for this assignment) its order must divide $ |G|$, by Lagrange's Theorem.

[/quote]

Also, this line is not essential to the proof. :)


calculus
integration
number theory unsolved
number theory

Find all integral solutions to the following integral equation :

$\frac{x^{7}-1}{x-1}$ $=$ $y^{5}-1$

[hide="Hint"]There are no integral solutions  :P  :D [/hide]

what the heck... :lol:

where is india going?
copying problems....

maybe it was from the ISL 2006.
the ST was on [b]May 14, 2007.[/b]


calculus
integration
logarithms
function
absolute value
calculus computations

Hi,
I know this is a very basic question but
what is the integral of
$1/(x*ln(x)) dx$

Let $u=ln(x)$, then $du=\frac{dx}{x}$ and we now must integrate u du.  This gives $ln(u)+C$ Substituting for u, we have $ln(ln(x))+C$ :D

but dont u need $ln(abs(ln(x)))$
this is what i was confused about.....

because integral of 1/u du is ln(abs(u)) + c

If $u>0$ on the interval we care about, $\int\frac1u\,du=\ln(u)+C.$

If $u<0$ on the interval we care about, $\int\frac1u\,du=\ln(-u)+C.$

These are two separate formulas. The function $\frac1u$ does not have an antiderivative valid on any interval that contains zero.

Writing $\int\frac1u\,du=\ln|u|+C$ is a handy way of writing down both formulas at the same time but does not change the fact that they are two separate formulas.

If you know which side of the origin you're on, you don't need to worry about this. For example,

$\int\frac{x\,dx}{1+x^2}=\frac12\ln(1+x^2)+C.$ Don't need an absolute value there.

So, if $x>1,\,\int\frac{dx}{x\ln x}=\ln\ln x+C$,

and if $0<x<1,\,\int\frac{dx}{x\ln x}=\ln(-\ln x)+C$,

and if you don't know which side of 1 $x$ lies on, you can write

$\int\frac{dx}{x\ln x}=\ln|\ln x|+C.$


Segundo nivel, problema 5:

En un triángulo $ABC$, $\angle A=2\angle C$ y $2\angle B=\angle A+\angle C$. $E$ es el pie de la bisectriz de $C$, $F$ es el punto medio de $AE$, y $D$ es el pie de la altura por $A$. Si la mediatriz de $DF$ corta a $AC$ en $M$, mostrar que $AM=MC$.

De los datos tenemos: $\measuredangle A= 80^\circ$, $\measuredangle B= 60^\circ$, $\measuredangle C= 40^\circ$

Observemos el triángulo isósceles $ACE$($\measuredangle AEC = \measuredangle EAC = 80^\circ$), entonces:

$\measuredangle ACF = \measuredangle FCE = \frac{\measuredangle ACE}{2}= 10^\circ$

Los puntos $A,F,D,C$ son concíclicos ($\measuredangle AFC = \measuredangle ADC= 90^\circ$), entonces: $\measuredangle ACF = \measuredangle ADF = 10^\circ$

Ahora desde $A$ trazamos una perpendicular a $FD$ en el punto $G$, de la misma forma desde $C$ trazamos una perpendicular a $FD$ en el punto $H$.

Los triangulos $AGD$ y $DHC$ son semejantes por $AAA$, entonces: $\frac{GA}{HD}= \frac{AD}{DC}$
Tambien los triangulos $AGF$ y $ADC$ son semejantes por $AAA$, entonces: $\frac{GA}{GF}= \frac{AD}{DC}$

De lo cual concluimos que: $GF=DH$

Llamemos $N$ al punto en que la mediatriz corta a $F$, tenemos $FN=NC$  :arrow:  $GN=NH$

$AG//MN//CH$ y por el teorema de Thales: $\frac{AM}{MC}=\frac{GN}{NH}$  :arrow:  $\boxed{ \mathbb{AM=MC}}$

Después de verificar que $AFDC$ es cíclico supongo que es suficiente observar que, si $M'$ es el punto medio de $AC$, entonces $M'$ es el circuncentro del cuadrilátero, por lo que $FM'=M'D$. Luego $M'$ está sobre la mediatriz de $DF$, así que $M=M'$. :)

Supongo que se me paso la mano  :rotfl:  :rotfl:

Es un problema muy fácil para ser el 5 del segundo nivel (comparado con otros años) , de hecho, estuvo mucho más fácil que el 5 del nivel 1.

$Tipe$

bueno eso si es cierto, y ahora q recuerdo me falta postear los problemas del nivel 1!! solo q creo q seria mas conveniente colocarlos en el foro de preparatoria no??


OK, this one is very easy ...
(It was my very first problem, it's about 2 years old)

For an integer k the number f(k) is determined as follows :

f(k) = :sqrt:((k + 1)(k + 1) + k)

Determine all integers k such that f(k) is an odd integer.

(k + 1)(k + 1) + k = (1 + k(k+1))

Since k(k+1) is always even, f(k) is always an odd integer  :mrgreen:

[color=#88FF00]belenos, do you know how to use spoiler?[/color]

It's part of the BBCode. It hides the text so that you have to select it (drag your mouse over it) to see the text. For example, [hide]like this![/hide]

All you have to do is click on Spoiler button at the top of the message before and after you write the text you want hidden. You could also select the text you want to hide and click on Spoiler.

I doubt belenos know. He's still new here, remember?

JBL wrote:belenos, do you know how to use spoiler?

[hide]now, yes !  [/hide]:wink:

Btw do you really think Spoiler is usefull (if you don't want to see the solution, just don't scroll down ?) ?

I don't think so but I will try to "spoil" from now anyway.

Yeah, I sorta agree with Belenos. Its a lot easier to simply not read the text if you start seeing a solution.

well, people like me might want a hint, but not a solution.

An untrained pair of eyes can focus on only a very small area... usually about a 1cm diameter... This does not mean that the eyes arent seeing what is not being focused. The brain sees with the eyes subconsciously, and your range of vision is enormous.

I know about this because I studied a lot of speed reading theory when I used to speed read. Most of it is learning how to use your eyes to consciously see more, the human brain is already able to handle it.

So, if you viewing a post directly below one you don't want to see, if the post you don't want to see is on the screen, you are subconsciously taking in everything in that post. This may lead you to think of things you wouldnt have normally thought of... and other things...

Sorry for pontificating... but the moral is, there are good reasons for the spoiler, so just use it.

k, bye

[color=#A000FF]Because you don't know in advance what is a solution and what isn't . . . so for your particular answer, by the time I had time to figure out whether that was a solution or not, I already would have seen the answer and not had a chance to do the problem myself.  Spoiler lets me know that osmething is a hint or answer, so if I want to see it I still can and if I don't want to see it, I won't accidentally.[/color]


Wow, that is quite a piece of pro-Bush propaganda! I don't think I'm going to argue a lot this time but I want to point you out some things. Don't know if you care but anyway

1) [quote="atm"]In my opinion, Bush is a good president because he deals with the terrorist stuff pretty well, which is extremely important because it makes us safe.[/quote]

First of all, remember this is an international forum. Maybe it makes [i]you[/i] safe, I certainly don't see how it makes me safe. Maybe you wanted to talk with US people alone but in such a case, I think it is most appropriate to post it in the local community forums. In my opinion Bush is dealing with the terrorist stuff pretty bad but my opinion is irrelevant since I don't vote in your country. I also don't think he is making you any safer, actually think totally the opposite but uff... I think it is for me a discussion not going anywhere since this is a subject much debated and your opinions are still so diametrically opposite than mine.
Ah don't know if you've heard this but war on Iraq it is probably not to protect you but to control oil. It also serves the purpose of getting rid of some weapons so making some more is profitable.

2)[quote="atm"] Would Bill Clinton, John Kerry, or Hilary Clinton do a better job than Bush if they face such situations like the 911 events,[/quote]

Maybe not. One thing I agree with your friends in this case is that Hillary Clinton or whoever is not going to be very different. I wouldn't know if they would go sneaking right after 911 though.

3)[quote="atm"] Iraq stuff, or nuclear weapons stuff[/quote].

I'm dying of curiosity here, what do you mean by Iraq stuff? or nuclear weapons stuff? Exactly, what are you referring to? I don't think it is correct to say "Bush faces the situation on Iraq" since there was no situation on Iraq, at least not a US related one, until Bush arrived. What nuclear weapons stuff are you talking about?

[quote="atm"] I mean if we dont do so, we are in risk of having another "911" again you know.[/quote]

No, as I said, I dont.

4)[quote="atm"]Would Hilary be able to deal with those *going on* events as well as Bush. We thought Clinton was a good president, but then he hold power in a peaceful time, and thus it is not fair to think that Bush is a bad president just because he causes many wars.[/quote]

If you think Bush dealt with it soo well, then surely Hillary won't be able to equal him. Because he "causes many wars"? You're telling me it is actually OK for a president to go provoking wars. That is so sad.

Ursula

[quote="ursula"]First of all, remember this is an international forum. Maybe it makes you safe, I certainly don't [/quote]
Americans are voting for a president of USA, not United Nations Secretary-General, I don't think they should take your safety into consideration.
The problem that really matters is, did Bush make americans safer? It's not persuasive to say yes.

[quote="Buffalo"]Americans are voting for a president of USA, not United Nations Secretary-General, I don't think they should take your safety into consideration.[/quote]
Just because many USAers [i]don't[/i] have a cosmopolitan view about how our relationship with the rest of the world ought to be doesn't mean we [i]shouldn't[/i]. Furthermore, "many" isn't "all".

[quote="Buffalo"]
Americans are voting for a president of USA, not United Nations Secretary-General, I don't think they should take your safety into consideration.[/quote]

Also, if you think that way, don't speak about "us" in an international forum. The sentence "Bush makes us safer" in this forum I assume it is to be applied to all people using the forum, therefore my words. Go to the US community forum, it is there so you can feel 'us' as a synonym for USAers.

Ursula

Aah I still remember those days when foreign and "us" didn't mean anything. :( ....

well maybe they think the personal pronoun "us" really stands for "united states" :D

What is "stuff"?  What is "terrorist stuff"?   What do you mean, he deals with it well?   We hear about attacks by the Taleban all the time.
Hundreds of thousands of people have lost their lives in Iraq.
You may think it makes you safe, but some estimate that for each US victim on 9/11, 20 Iraqis have died by now.

  A country under a firm dictatorial secular regime that was almost attacked during the Gulf War by Bin Laden and his fundamentalist warriors, now provides an endless supply of terrorists.  We are in a phase in which muslims from for instance Western European countries go to Iraq.  What do you think will happen when the next phase begins and they start coming back?? :huh:

[quote="atm"]Hey guys,
I just debated over with my peers that Hilary wouldnt do a better job than Bush, in case she is elected.

In my opinion, Bush is a good president because he deals with the terrorist stuff pretty well, which is extremely important because it makes us safe. Would Bill Clinton, John Kerry, or Hilary Clinton do a better job than Bush if they face such situations like the 911 events, Iraq stuff, or nuclear weapons stuff. I mean if we dont do so, we are in risk of having another "911" again you know. Would Hilary be able to deal with those *going on* events as well as Bush. We thought Clinton was a good president, but then he hold power in a peaceful time, and thus it is not fair to think that Bush is a bad president just because he causes many wars.

My friends debated that either Bush or Hilary is controlled by the parties (Republic or Democracy) and the great super-beneficial corporations in America. Then no one is better than the other (if Hilary is elected) because they are *just* the representatives of the country, of the parties and corporations hidden behind them. All of their decisions are affected by those things.

So what do you guys think about this about this?[/quote]

In my opinion Bush is a horrible president, the worst ever. If he is doing well with 9-11 why didn't he go after Afghanistan where Bin Laden was and why doesn't he go after him? He might put some troops there but that is only a fraction of the amount in Iraq.
This is something that the average American doesn't know. Bill Clinton had heard that terrorists might attack United States. He had plans including extra security and finding and capturing the terrorists. When Bush came in and saw those plans he ignored them because he thought that terrorists couldn't do anything to our country. He was wrong.

If you think he is doing good in Iraq he isn't.
1.Not saying Saddam is good, but he was doing better then the guys in there now. Now there is murder, killing and noone there can feel safe. Before Saddam might have tortured but at least the country was under peace.
2. There is no such thing as WMD in Iraq he just lied about it. I think that he just went in to get the oil...

Bush is the worst president ever in my opinion.

So I think all of those people could do better because its practically impossible to do worse

[quote="WindSlicer"]Focus, ya'll. This topic is about Hillary, not Bush. I know ya'll love to bash him, but at least find the correct thread to do it on. And who really cares if the OP uses 'us' to refer to USA people? It's just semantics. Move on.

/rant[/quote]

The topic is whether Hillary would do a better job than Bush or not if I read correctly. I believe an evaluation of Bush work falls completely into the topic. And don't know about everyone else, I don't "love" to bash anyone. It is not some kind of sport. The fact that Bush has to be bashed so much is possibly the most unfortunate thing of the recent years history.

Ursula

[quote="Ihatepie"][quote="atm"]

If you think he is doing good in Iraq he isn't.
1.Not saying Saddam is good, but he was doing better then the guys in there now. Now there is murder, killing and noone there can feel safe. Before Saddam might have tortured but at least the country was under peace.
2. There is no such thing as WMD in Iraq he just lied about it. I think that he just went in to get the oil...

Bush is the worst president ever in my opinion.[/quote][/quote]

I just read something interesting in the newspapers :USA has now been active in Iraq for more days than during the Second World War...

I must disagree, the country wasn't in peace at all.  There was a lengthy war against Iran, conflicts with the Kurds and Shias in the south (he drained the waters from the Shias in the south).....
But if you remove a man like Hussein, you should definitely remove Kim Jung Il, the military regime in Myanmar, ....


algebra
polynomial
vector
Ring Theory
superior algebra
superior algebra unsolved

Let $ \mathbb{C}[x]$ be the polynomial ring over the field $ \mathbb{C}$ of the complex numbers. Define a subset $ I$ of $ \mathbb{C}[x]$ by $ I\equal{}\{f(x)\in{\mathbb{C}[x]} | f(1)\equal{}f'(1)\equal{}0\}$. Show that $ I$ is an ideal of $ \mathbb{C}[x]$ and find the dimension of the quotient ring $ \mathbb{C}[x]/I$ as a vector space over $ \mathbb{C}$.

$ f(1) \equal{} 0 \Leftrightarrow f(x) \equal{} (x \minus{} 1)g(x)$, which gives $ f'(x) \equal{} (x \minus{} 1) g'(x) \plus{} g(x)$.  Now $ f'(1) \equal{} 0 \Leftrightarrow g(x) \equal{} (x \minus{} 1) h(x)$.  It follows that $ I \equal{} ((x \minus{} 1)^2)$ and $ \dim \mathbb{C}[x]/((x \minus{} 1)^2) \equal{} 2$.


integration
calculus
function
derivative
real analysis
real analysis theorems

[quote]Here is an alternative way to solve linear first-order nonhomogeneous differential equations like:
dy/dx + P(x)y = Q(x)
without using integrating factors. It is based on the fact that you can solve the associated homogeneous equation
dyH/dx + P(x)yH = 0
easily because it is a separable equation (dyH/yH = −P(x) dx). Notice
that the general solution of this associated homogeneous equation always
contains an arbitrary multiplicative constant A:
yH(x) = Af(x) , where f(x) is any particular solution.
The alternative method of solving the nonhomogeneous equation is just to
try the same formula that worked for the associated homogeneous problem
but to generalize the constant A into an unknown function of x. Therefore
we could try to substitute into the nonhomogeneous problem
y(x) = A(x)f(x)
where f(x) is a known nonzero solution of the associated homogeneous
equation. This amounts to a change of the dependent variable from y(x)
to A(x). The point is, the equation for A(x) that follows from this substitution
turns out to be one that you can solve in your sleep! Once you
have the general solution for A(x), you know the general solution for y(x)
because y(x) = A(x)f(x). Use this method to find the general solution to
the following nonhomogeneous equations:[/quote]

can some1 explain this a little better pls.. maybe with an example.. i kinda get what they are talking about.. but im getting lost in the 2nd part of it where you have to plug it back into the origonal equation or something.... thx for any help and sorry if this is the wrong place to ask this.....

For example, let's try to solve the equation

$dy/dx+xy=x$

Part of what makes this equation more difficult than most is it's the fact the the right hand side is nonzero.  If instead we had the equation:

$dy/dx+xy=0$

we could solve it using separation of variables.

$dy/dx=-xy$

$\int dy/y=- \int x dx$

$ln |y| = \frac{-x^2}{2}+C_1$

$y= C_2 e^{-x^2/2}$

The constant of integration turned into a multiplicative constant when we took exponentials of both sides of the second to last equation.

Although this is a lovely class of functions, unfortunately they do not solve the original equation because we changed the right hand side from $e^x$ to 0 before solving.  However, the above strongly suggests that $e^{-x^2/2}$ should be involved somehow in our final solution.

So let's define a new function $z$, where $z$ is chosen such that $y=e^{-x^2/2} z$.  Note that although we don't know what $z$ is, we can write the derivative of $y$ in terms of that of $z$ using the product rule and the chain rule.  Doing so gives:

\[dy/dx= (dz/dx-zx)e^{-x^2/2}\]

But we also know from the differential equation that $dy/dx=x-xy=x-xze^{-x^2/2}$

Setting these two equal, $dz/dx * e^{-x^2/2}=x$, so $dz/dx=xe^{x^2/2}$

Integrating, $z=e^{x^2/2}+C_3$, so $y=1+C_3 e^{-x^2/2}$.

thank-you very much... last few lines cleared everything up  :)


[quote="236factorial"]What is the sum of the measures of the angles in a 32 sided figure?[/quote]
[hide]
im assuming you mean interior angles sum.

forumla is (n-2)180
so its 30 x 180= 5400
[/hide]

[quote="risinglegend"]32-2=30 30*180=5400  is dat rite?[/quote]

No. Copying and pasting gets a 0.  :roll:

:rotfl: Yeah, that's the second time you do that

[quote="risinglegend"]32-2=30 30*180=5400  is dat rite?[/quote]stop copying peopls's answers!!! its not cool :gleam:

and adding 'is dat rite' doesn't help!! :(

risinglegend, if you want to get better at math then don't copy peoples answers.  If you don't want to get better at math then why did you join this forum :?:

[hide="Original Solution"]
32-2=30 30*180=5400

u get da 32 from magic n u get teh other numbers from magic 2 lol
[/hide]

[quote="236factorial"]What is the sum of the measures of the interior angles in a 32 sided figure?[/quote]

formula: (n-2)*180. 30*180=5400degreezz

[hide]
I got 5400 degrees.
By using the formula: $(n - 2) * 180 =$ degrees.
So, $(32 - 2) * 180 = \framebox{5400}$

[/hide]


factorial
IMO Shortlist
number theory
number theory unsolved

For what  values of n does n! have 6 for its last nonzero digit?

[quote="harazi"]The problem is trivial, because if we denote by vp(n!) the exponent of the prime number p in n!, then we must have v2(n!)<=v5(n!)+1. This means that [n/2]+[n/4]+...<=1+[n/5]+[n/25]+.... Since [n/4]>=[n/25],..., we must have [n/2]<=1+[n/5]. Now everything is clear: n<=6 and the answer is n=3.[/quote]

=========
What about 12! = 479001600? I think is quite serious problem.

Namdung

How could I be so stupid? It isn't necessary that the exponent of 2 is that of 5 plus one.I had a very bad day today: when writing the so called solution (in fact, a stupidity), I was thinking that 4|10. Yes, now I realize that the problem is very difficult.

This is going to be a pretty useless post  :D I tried to find some pattern by calculating factorials for a lot of numbers with Mathematica, but found nothing. They really look to be randomly distributed. Where did yopu find this, Manlio?

Here's another thought: If some1 finds the sequence of n s.t. n! has 6 as its last non-zero digit, maybe it will be such a sequence that it doesn't become periodic, and this would solve that IMO shortlist prb: prove that the sequence d_n doesn't become periodic after some n0, where d_n is the last non-zero digit of n!.

the problem "last nonzero digit of n!" has been discussed many times in Olympics in Informatics not Math Olympiad...

One of the solutions is like this:
Let f(n) be the last nonzero digit of n.
From some calculations, we find out that f((5^4 n)!)=f(n!)
and if we list f(n) of n=(5^i * k)!, where i=0,1,2,3 and k=1,2,...,n....
we will find there are some certain patterns...

i forgot the details but the abstract is shown above.


We have a $3$-variable function
$f(a,b,c)=\frac{(a+b+c)^3}{abc}\forall a,b,c>0$
It is easy to check that
     $1.f(a,b,c)\ge f(a+b,b+c,c+a)$
     $2.f(a,b,c)\le f(a^2,b^2,c^2)\forall a,b,c>0$
$a)$ Can we prove that $\forall a,b,c>0$
$2f(a,b,c)\ge f((a+b)^2,(b+c)^2,(c+a)^2)$ ?
$b)$ With $a,b,c>0$ are the sides of a triangle ,can we prove that
$f(a,b,c)\ge f((a+b)^2,(b+c)^2,(c+a)^2)$  ?
 :)  :blush:

:) [b]here is a likely strategy that might work:

-use ravi substition a = x + y, b = y + z, c = z + x with x,y,z > 0 and regroup the expression into symmetric sums.

-proceed to use muirhead inequality or s.o.s.[/b]

[quote="nthd"]
$b)$ With $a,b,c>0$ are the sides of a triangle ,can we prove that
$f(a,b,c)\ge f((a+b)^2,(b+c)^2,(c+a)^2)$  ?
 :)  :blush:[/quote]
The answer is no,with the example :
$a=2;b=c=1+\epsilon$ let $\epsilon \rightarrow +\infty$
With part a) did you have any idea? :play_ball:


induction
combinatorics unsolved
combinatorics

$x_1+...+x{}_3{}_2=120$
$x_i<60,x_i \in \mathbb{N}$
prove that
$\exists j: \{i_1,...,i_j\} \subset \{1,...,32\},\sum_{k=1}^j x_i__k=60$
in other words prove that there exists some of the $x_i$s such that they add $60$ :?:

I think this problem is one case of the problem that replaced 30 by n.
By using induction, I done successfully with some cases:
1) exist i that xi =2
2) exist i,j that xi=1, xj=3
3) exist i,j that xi=xj=4

But, I haven't solved completely yet.
I hope someone will concern this probem.


Find all solutions of the equation $ sin 3x \plus{} cos 3x \equal{} \frac {\sqrt 5}{2}$ in the interval $ [ \minus{} \pi,0)$.

I know how to do the first part (i suppose). here's my solution...

[hide]
$ sin 3x \plus{} cos 3x \equal{} \sqrt 2 (sin(3x \plus{} \frac {\pi}{4})$

which means, $ sin(3x \plus{} \frac {\pi}{4}) \equal{} \frac {\sqrt {10}}{4}$ how could i solve it? it seems to be a non-special angle, isn't it?
[/hide]

You could just use the inverse sine to get that $ 3x \plus{} \frac {\pi}{4} \equal{} \sin^{ \minus{} 1}\frac {\sqrt {10}}{4}$, so $ x \equal{} \minus{} \frac {\pi}{12} \plus{} \frac {\sin^{ \minus{} 1}\frac {\sqrt {10}}{4}}{3}$.

As far as I know, the solution is not a special angle.

It would be easier to square the equation to get

$ {\sin^{2} {3x}} \plus{} 2\sin {3x}\cos {3x} \plus{} {\cos^{2} {3x}} \equal{} \frac {5}{4}$

So with pythagorean and double angle formulas, we have

$ \sin{6x} \equal{} \frac{1}{4}$.

There are 6 solutions in the interval but all of them are not good angles...


All the $7$ digit numbers containing each of the digits $1,2,3,4,5,6,7$ exactly once , and not divisible by $5$ are arranged in increasing order. Find the $200th$ number in the list.

There are $n! - (n-1)!$ permutations of $1 .. n$ not ending with $5$

$6! - 5! > 200$ so number begins with $1$
$2*(5!-4!) < 200 < 3*(5!-4!)$ so number begins with $14$

$200 - 2*(5!-4!) = 8$ so we just have to count $8$ after $1376542$

If I did it right, it is $1425673$

[quote="Kab^2"]Sorry to post this but I think you might be wrong. I got 1425763. Using a similar method, I got 10 left to count. Remember that 5 is sort of like an exception. I apologise for keeping the thread up.[/quote]Your answer is wrong. At 192nd, the number starts with 142****. Count 8 more and you will get tµtµ's answer.

[quote="Rushil"]All the $ 7$ digit numbers containing each of the digits $ 1,2,3,4,5,6,7$ exactly once , and not divisible by $ 5$ are arranged in increasing order. Find the $ 200th$ number in the list.[/quote]
actually the rmo 2000 q 4 asked to find out the 2000 th no in the list

its easy to find the 200th no. I think......

Here is the question from the question paper(scanned).
ATTACHED.

There are totally $ 6!$ numbers with first digit $ 1$ but $ 5!$ of them are divisible by $ 5$ making totally $ 6! \minus{} 5! \equal{} 600$ numbers. Same goes for 1st digit $ 2$ and $ 3$. That's first $ 1800$ numbers.

Now consider the numbers starting with $ 41$. Analogously with previous case we can count that there are $ 5! \minus{} 4! \equal{} 96$ numbers. Same goes for those starting with $ 42$. Now we have first $ 1992$ (we are getting close   :) ).

There are $ 3! \minus{} 2! \equal{} 4$ numbers starting with $ 4312$ (only $ 4$ more  :lol: ) so $ 2000th$ number must be starting with $ 4315$ and now it's not hard to find that it is number $ \boxed{4315672}$.

The number of 7-digit numbers with 1 in the left most place and containing
each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once is 6! = 720. But 120 of these end in 5 and
hence are divisible by 5. Thus the number of 7-digit numbers with 1 in the left most place
and containing each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once but not divisible by 5 is 600.
Similarly the number of 7-digit numbers with 2 and 3 in the left most place and containing
each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once but not divisible by 5 is also 600 each. These
account for 1800 numbers. Hence 2000-th number must have 4 in the left most place.
Again the number of such 7-digit numbers beginning with 41,42 and not divisible by 5 is
120 − 24 = 96 each and these account for 192 numbers. This shows that 2000-th number
in the list must begin with 43.
The next 8 numbers in the list are: 4312567, 4312576, 4312657, 4312756, 4315267, 4315276,
4315627 and 4315672. Thus 2000-th number in the list is 4315672.


integration
ratio
limit
calculus
calculus computations

Let $ f(x) \equal{} \sum_{n\equal{}1}^{\infty} \frac{ (\minus{}1)^n(x\minus{}2)^n }{n}$.  Find the interval of convergence of $ \int f(x) dx$.

[hide="Solution"]$ \int f(x)\ dx\equal{}\sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^n}{n}\cdot\frac{(x\minus{}2)^{n\plus{}1}}{n\plus{}1}$

Consider the ratio of consecutive terms in the limit as $ n$ grows infinite:

$ \lim_{n\to\infty}\left|\frac{(x\minus{}2)^{n\plus{}2}}{(n\plus{}1)(n\plus{}2)}\cdot\frac{n(n\plus{}1)}{(x\minus{}2)^n}\right|\equal{}|x\minus{}2|\cdot\lim_{n\to\infty}\frac{n}{n\plus{}2}\equal{}|x\minus{}2|$

The sum will only converge if this is less than (or possibly equal to) $ 1$:

$ \minus{}1<x\minus{}2<1\implies 1<x<3$

If $ x\equal{}1$, we have $ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^{2n\plus{}1}}{n(n\plus{}1)}<\sum_{n\equal{}1}^{\infty}\frac{1}{n^2}\equal{}\frac{\pi^2}{6}$, so it converges.

If $ x\equal{}3$, we have $ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^n}{n(n\plus{}1)}$, which again converges for the same reason.

$ \boxed{1\leq x\leq 3}$[/hide]


inequalities
real analysis
real analysis solved

i'll have to think a little bit about that exponent of yours, but about the lagrange multiplicators, as the experience shows, it is often (often means 99% here) much more easier to actually solve the problem elementary than to try to solve it using lagrange multiplicators. Much more, some of my friends, Andrei Jorza for example, missed a gold medal at the IMO twice (in 2000 and 2001) because he tried to solve the inequalities that were proposed with that tehnique. He is more elevate to give you advices, but I strongly recommend that you should not use such superior theoretical results (lagrange multiplicators, sturm's way - you could see me fooled in thinking that i have solved siutz's 2nd inequality, etc. ) in contests. They waste time, and if you do not complete them, you get a big round score.

The power serie of f(z)= exp(exp(z)) is

sum_{p>=0}(A(p).z^p)/p! where

A(p) = sum_{n>=0}n^p/n!

you have also a recurrence formulae from the differential equation

f'(z) = exp(z)f(z)

A(0)= e

A(p+1) = sum_{k=0  n}(p;k)A(k) where (p;k) = p!/(k!(p-k)!)

[quote=vineet]hii
here's a question which is not ur typical imo stuff.but i really got to know this.can we expand e^(e^x) as a power series in x and what will be the coefficient of x^n in the expansion.and another qusetion ,can we apply lagrange multipliers in any inequality problem.i wud like to see an example of proving an inequality using lagrange multipliers.
that's it .
thanx
vineet[/quote]

I don't know, but I think it has to do with calculus.

This might be related but $e^{e^x-1}$ is the EGF of the bell numbers.

[url= https://en.wikipedia.org/wiki/Bell_number#Generating_function] Bell number [/url]
https://en.wikipedia.org/wiki/Bell_number#Generating_function

[url= https://artofproblemsolving.com/community/c7h210519p1159897] see here [/url]
https://artofproblemsolving.com/community/c7h210519p1159897


-$ f$ $ :$ $ R$->$ R$
-$ f$ is a continous function
-$ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$

Prove that f is [b]Linear[/b]
I only know how to prove this in case of:
f : $ Q$->$ R$  :(

generaly if $ f$ is continue in $ \mathbb{R}$ and $ \forall x\in C: \ f(x)\equal{}g(x)$ where ($ C$ dense dans $ \mathbb{R}$, $ \mathbb{R}\subset\overline{C}$) then $ \forall x\in C: \ f(x)\equal{}g(x)$
here we have $ C\equal{}\mathbb{Q}$

[quote="aviateurpilot"]generaly if $ f$ is continue in $ \mathbb{R}$ and $ \forall x\in C: \ f(x) \equal{} g(x)$ where ($ C$ dense dans $ \mathbb{R}$, $ \mathbb{R}\subset\overline{C}$) then $ \forall x\in C: \ f(x) \equal{} g(x)$
here we have $ C \equal{} \mathbb{Q}$[/quote]

Ok I see,
But where did you get this theorem? whats the name of it? Where can I found its proof

And what is $ \overline{C}$
Thanks   :wink:

for $ C\subset \mathbb{R}$
$ \overline{C} \equal{} \{x\in\mathbb{R}|\exists(x_n)_{n\in\mathbb{N}}\in C^{\mathbb{N}},\ \lim_{n\to \plus{} \infty}x_n \equal{} x\}$ it's easy to prove that $ \overline{\mathbb{Q}} \equal{} \mathbb{R}$ because if $ x\in\mathbb{R}$ we take $ x_n \equal{} \frac {E(10^nx)}{10^n} \equal{} x\ with\ n\ digits\ before\ the$ "$ ,$" so we have $ (x_n)_{n\in\mathbb{N}}\in \mathbb{Q}^{\mathbb{N}}\ and\ \lim_{n\to \plus{} \infty}x_n \equal{} x$. we suppose that $ \forall x\in \mathbb{Q}: \ f(x) \equal{} ax$

$ \forall x\in \mathbb{R},\exists(x_n)_{n\in\mathbb{N}}\in \mathbb{Q}^{\mathbb{N}},\ \lim_{n\to \plus{} \infty}x_n\equal{}x$
$ f$ continue gives $ f(x) \equal{} \lim_{n\to \plus{} \infty}f(x_n)$then$ f(x) \equal{} \lim_{n\to \plus{} \infty}f(x_n) \equal{} \lim_{n\to \plus{} \infty}ax_n \equal{} ax$

[quote="aviateurpilot"]for $ C\subset \mathbb{R}$
$ \overline{C} = \{x\in\mathbb{R}|\exists(x_n)_{n\in\mathbb{N}}\in C^{\mathbb{N}},\ \lim_{n\to + \infty}x_n = x\}$ it's easy to prove that $ \overline{\mathbb{Q}} = \mathbb{R}$ because if $ x\in\mathbb{R}$ we take $ x_n = \frac {E(10^nx)}{10^n} = x\ with\ n\ digits\ before\ the$ "$ ,$" so we have $ (x_n)_{n\in\mathbb{N}}\in \mathbb{Q}^{\mathbb{N}}\ and\ \lim_{n\to + \infty}x_n = x$. we suppose that $ \forall x\in \mathbb{Q}: \ f(x) = ax$

$ \forall x\in \mathbb{R},\exists(x_n)_{n\in\mathbb{N}}\in \mathbb{Q}^{\mathbb{N}},\ \lim_{n\to + \infty}x_n = x$
$ f$ continue gives $ f(x) = \lim_{n\to + \infty}f(x_n)$then$ f(x) = \lim_{n\to + \infty}f(x_n) = \lim_{n\to + \infty}ax_n = ax$[/quote]

Nice explanation!!

Thanks a lot  :wink:  :wink:  :D

[quote="aviateurpilot"]
$ with\ n\ digits\ before\ the$ "$ ,$"
[/quote]

You dont have to choose $ n$ like this, you just take $ n\in \mathbb$$ N$