logarithms
Gauss
calculus
integration
limit
real analysis
real analysis unsolved

Determine whether $ \sum_{n\equal{}2}^\infty n^{\minus{}1 \minus{} (\ln(\ln \, n))^{\minus{}2}}$ is convergent.

there is many way to prove this:
also here work Gauss pricipe.
I represent by integral:
$ \sum^{N} ...$~$ \int^{N}\frac {1}{n^{1 \plus{} \frac {1}{(\ln\ln n)^{2} }}}dn$~$ \frac {1}{N^{1/(\ln \ln N )^{2}}}\int^{N}\frac {1}{n}$~
$ \frac {\ln N}{N^{1/(\ln \ln N )^{2}}} \minus{} > 0$ when $ n \minus{} > \plus{} oo$
last step is standart limit.
but this step :
$ \int^{N}\frac {1}{n^{1 \plus{} \frac {1}{(\ln\ln n)^{2} }}}dn$~$ \frac {1}{N^{1/(\ln \ln N )^{2}}}\int^{N}\frac {1}{n}$
is not so obviusly for some one therefore it's need to chek.
if some one can't prove this step and know about Cezaro Stolt's  theorem they can do by next way:
let's prove that this sum
$ \sum^{N} ...$ is aproximately equal to $ \frac {\ln N}{N^{1/(\ln \ln N )^{2}}}$ by using Cezaro theorem you only need to chek that :

$ \lim_{n \minus{} > \plus{} oo} \frac {\frac {1}{n^{1 \plus{} \frac {1}{(\ln \ln n)^{2}} }}} {\frac {\ln n}{n^{1/(\ln \ln n )^{2}}}\minus{}\frac {\ln (n\minus{}1)}{(n\minus{}1)^{1/(\ln \ln (n\minus{}1) )^{2}}}} \equal{} 1$
try show it yourself

That looks like a complete mess, Extremal. You haven't even answered anything.

This is well-suited to the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=157993]condensation test[/url].
If $ f(n)$ is the $ n$th term of this series,
$ 2^nf(2^n)\equal{}\exp(n\ln 2)\cdot\exp\left(n\ln 2\cdot\left(\minus{}1\minus{}(\ln(n\ln 2))^{\minus{}2}\right)\right)$
$ \equal{}\exp(\minus{}n\ln 2\cdot (\ln n\plus{}\ln \ln 2)^{\minus{}2})<\exp\left(\minus{}\frac{n\ln 2}{\ln^2 n}\right)$
Since powers of $ n$ dominate logarithms, we have $ n>\ln^4 n$ and $ 2^nf(2^n)<n^{\minus{}2\ln 2}$ for large enough $ n$ ($ n>10000$ works). The modified series converges (faster than any $ n^{\minus{}k}$), and thus the original series converges as well.

[quote="jmerry"]That looks like a complete mess, Extremal. You haven't even answered anything.
[/quote]
I don't agree with you. My aproximation work.
this is true :
$ \sum^{N} ...$~$ \int^{N}\frac {1}{n^{1 \plus{} \frac {1}{(\ln\ln n)^{2} }}}dn$~$ \frac {1}{N^{1/(\ln \ln N )^{2}}}\int^{N}\frac {1}{n}$~
$ \frac {\ln N}{N^{1/(\ln \ln N )^{2}}} \minus{} > 0$ when $ n \minus{} > \plus{} oo$

another proof:
everyone know that this sum convergent:
$ \sum \frac {1}{n (\ln n )^{2}}$
because
$ \sum^{N} \frac {1}{n (\ln n )^{2}}$~$ \int _{1}^{N}\frac {1}{n(\ln n)^{2}}$~$ \frac {1}{\ln n} \minus{} > 0$
now let's prove :
$ \sum n^{ \minus{} 1 \minus{} (\ln \ln n )^{ \minus{} 2}} < \sum^{N} \frac {1}{n (\ln n )^{2}}$
because for big $ N$ we have

 $ n^{ \minus{} 1 \minus{} (\ln n )^{2}} < \frac {1}{n (\ln n )^{2}}$
or
$ n^{(\ln \ln n)^{ \minus{} 2}} > (\ln n )^{2}$
or
$ \ln n > 2 (\ln \ln n )^{4}$
last is true for big $ N$
maybe it's more simple than my first solution.   :blush:
if you want I can post more solutions

[quote="jmerry"]... and that approximation is wrong, because it assumes the series diverges.[/quote]
OK, Let's chek :
we consider :
$ \frac {\ln N}{N^{1/(\ln \ln N )^{2}}}$ when $ N \minus{} > \plus{} oo$
let $ \ln N \equal{} a$ and now $ a \minus{} > \plus{} oo$ then we have
$ \frac {a}{e^{\frac {a}{(\ln a)^{2}}}}$ now let
$ \ln a \equal{} b$ where $ b \minus{} > \plus{} oo$ then we have :
$ \frac {e^{b}}{e^{\frac {e^{b}}{b^{2}}}} \equal{} \exp (b \minus{} \frac {e{^b}}{b^{2}})$
lets consoder $ b \minus{} \frac {e^{b}}{b^{2}}$when $ b \minus{} > \plus{} oo$
we have
$ \frac {b^{3} \minus{} e^{b}}{b^{2}} \minus{} > \minus{} oo$ because for bi $ b$ $ e^{b} > > > > b^{M}$ for all $ M$ when $ b$ is big therefore
 $ \exp (b \minus{} \frac {e{b}}{b^{2}}) \minus{} > 0$ trust me it's true

You said the sum goes to zero. It doesn't.

Think before you post. Don't just churn out symbols.

Yes, here I agree with you I really forget constant. answer looks like this
$ C \plus{} \frac {\ln N}{N^{1/(\ln \ln N )^{2}}} \plus{} o(\frac {\ln N}{N^{1/(\ln \ln N )^{2}}})$ where $ C > 0$ is  constant.
i must be more correct in my proof, but it's no so important in this case.
Constants don't play main role when we consider convergent of series .  :P

[quote="Extremal"].
I represent by integral:
$ \sum^{N} ...$~$ \int^{N}\frac {1}{n^{1 \plus{} \frac {1}{(\ln\ln n)^{2} }}}dn$[/quote]

$ \int^{N}\frac {1}{x^{1 \plus{} \frac {1}{(\ln\ln x)^{2} }}}dx$

with change variable $ u\equal{}\ln x$ we get

$ \int^{N}\frac {1}{e^{\frac{u}{(\ln u)^2}}}du$

and $ \frac {1}{e^{\frac{u}{(\ln u)^2}}}\equal{}o(\frac{1}{u^2}$

with comparaison Riemann integral we have the convergence of the integral

also the serie


A set of $ 25$ square blocks is arranged into a $ 5\times 5$ square. How many different combinations of $ 3$ blocks can be selected from that set so that no two are in the same row or column?

$ \textbf{(A)}\ 100\qquad
\textbf{(B)}\ 125\qquad
\textbf{(C)}\ 600\qquad
\textbf{(D)}\ 2300\qquad
\textbf{(E)}\ 3600$

[hide]There are $ 25$ possibilities to pick the first block. Since we can't use any of the blocks in the same column or row, we can't pick $ 5 \plus{} 5 \minus{} 1 \equal{} 9$ blocks anymore which leaves $ 16$ possibilities. Picking any of them makes another $ 5 \plus{} 5 \minus{} 2 \minus{} 1 \equal{} 7$ unusable, so we have $ 9$ possibilities to pick the third block, hence we can pick 3 blocks in $ \textbf{(E)}\ 3600$ ways.[/hide]

[hide="Mines different"]
After one of the $ 25$ blocks is chosen there are $ 16$ left that don't share its row or column.  After that, there are $ 9$ left that don't share any of those rows or columns.  These can be chosen in any order, so we have

\[ \frac{25 \times 16 \times 9}{3!}\equal{}25 \times 8 \times 3\equal{}600
\]

or $ \boxed{\textbf{(C)}}$.
[/hide]


MATHCOUNTS
probability
AoPS Books
number theory

I am prepring for the MATHCOUNTS State Competition which AoPS books have you found most helpful for someone in my situation?  Thanks

AOPS Volume 1 is probably the best book, since it covers a wide range of topics.

Try getting the intro books too (counting and probability, number theory, algebra)

Thanks, what exactly are the AoPS Volume 1+2 books,  my friends say they are great but idk what they are.  I just cant get a good understanding by the exerpts or the descriptions in the bookstore.

AoPS vol 2 is considerbly harder than vol 1. If you are buying books only for MC, I'd suggest vol 1 and/or some intro books. Vol 2 topics aren't likely going to be covered in MC questions.

As for how great they are, AoPS books provide challenging math that doesn't require experiance or knowledge. All it requires is interest.

[quote="Bomboyage"]Sorry to interupt the conversation, but which one is the next best to buy?[/quote]

Next best?

Well in my opinion here are the books in order in which I believe you should get

[hide]
Vol 1
Vol 2
(Any order of intro books)
(Any order of intermediate books)
[/hide]

Volumes 1 and 2 are a hodgepodge of things you'll see in the intro books so it's good to get a general idea before going into a subject. Volume 1 is great for MathCounts as it covers pretty much all the material the competition might ask. Volume 2 is a bit more complex and is used by students taking the AMC/AIME. The intro books are great for tackling your weak spot and making it stronger. Intermediate books are just extensions of the intro concepts.

I think the best books to buy are the AoPS Classic Volumes 1 and 2, the Introduction series, Intermediate Counting and Probability, Intermediate Algebra (coming soon), the MATHCOUNTS School Handbooks, and the All-Time Greatest MATHCOUNTS Problems.

regarding the all-time best mathcounts problems, some of their solutions are a bit controversial and you may get a question wrong for believing their solutions

Yeah, the solutions in the all time greatest aren't the best, but I guess those are the ones which come to most peoples' mind firsts?

are there books that you can buy from the AoPS bookstore that contain national MATHCOUNTS tests.  I cant find any of the national test books anywhere

there is "the all time greatest mathcounts problems"
otherwise, you can look on the internet


inequalities
calculus
derivative
geometric inequality
inequalities proposed

$ \triangle\ ABC$ $ \implies$  $ \boxed {\frac {(ab\plus{}bc\plus{}ca)^2}{abc(a\plus{}b\plus{}c)}\ge 3\plus{}\frac {R\minus{}2r}{2(4R\plus{}r)}}\ge 3$ .

It is not hard to prove... We need only Gerretsen's inequality:
\[ r(16R \minus{} 5r)\le p^2\le 4R^2 \plus{} 4Rr \plus{} 3r^2
\]
The inequality is equivalent to:
\[ \frac {(p^2 \plus{} r^2 \plus{} 4Rr)^2}{8p^2Rr}\ge 3 \plus{} \frac {R \minus{} 2r}{2(4R \plus{} r)}
\]

\[ \iff \frac {p^2}{8Rr} \plus{} 1 \plus{} \frac {r}{4R} \plus{} \frac {1}{p^2}\left(\frac {r^3}{8R} \plus{} 2Rr \plus{} r^2\right)\ge 3 \plus{} \frac {R \minus{} 2r}{2(4R \plus{} r)}
\]

\[ \iff \frac {r(16R \minus{} 5r)}{8Rr}\plus{} \frac {r}{4R} \plus{} \frac {1}{4R^2 \plus{} 4Rr \plus{} 3r^2}\left(\frac {r^3}{8R} \plus{} 2Rr \plus{}r^2\right)\ge 2\plus{}\frac {R \minus{} 2r}{2(4R \plus{} r)}
\]

\[ \iff \minus{} \frac {3}{8x} \plus{} \frac {1}{4x^2 \plus{} 4x \plus{} 3}\left(\frac {1}{8x} \plus{} 1 \plus{} 2x\right) \ge \frac {x}{8x \plus{} 2} \minus{} \frac {1}{4x \plus{} 1}
\]

\[ \iff (x\minus{}2)(2x\plus{}1\minus{}4x^3)\ge 0, \forall x \equal{} \frac {R}{r}\ge 2
\]
I have edited something... but it is not alway true... I will post complete solution but not now... hint: using derivative ;)

Moreover, we have:
\[ x^4 \plus{} 52x^3 \plus{} 31x^2 \minus{} 6x \minus{} 4\ge 972
\]

\[ \iff (x \minus{} 2)(28x^3 \plus{} 108x^2 \plus{} 247x \plus{} 488)\ge 0
\]
It is always true.
We have a stronger one:
\[ \frac {(ab \plus{} bc \plus{} ca)^2}{abc(a \plus{} b \plus{} c)}\ge \frac{31}{8} \plus{} \frac {R \minus{} 2r}{2(4R \plus{} r)}\]
Because:
\[ \frac {x^4 \plus{} 52x^3 \plus{} 31x^2 \minus{} 6x \minus{} 4}{2x(4x^2 \plus{} 4x \plus{} 3)(4x \plus{} 1)}\ge \frac {7}{8}
\]
when $ x\in [2,\infty),x\to \infty$

[quote]We have a stronger one: $ \frac {(ab \plus{} bc \plus{} ca)^2}{abc(a \plus{} b \plus{} c)}\ge \frac {31}{8} \plus{} \frac {R \minus{} 2r}{2(4R \plus{} r)}$[/quote]
Observe that $ a \equal{} b \equal{} c\implies R \equal{} 2r\implies 3\ge \frac {31}{8}\ !$

The equality holds on $ x\equal{}\frac{R}{r}\to \infty$  :D I don't see the equality in my solution  :ninja:

I think there are two cases:
+ My solution is correct. Therefore we have:
\[ \frac {(ab \plus{} bc \plus{} ca)^2}{abc(a \plus{} b \plus{} c)}\ge 3 \plus{} \frac {R \minus{} 2r}{2(4R \plus{} r)}\plus{}\frac{486r^4}{(4R^2\plus{}Rr)(4R^2\plus{}4Rr\plus{}3r^2)}
\]
The equality holds on $ x\equal{}\frac{R}{r}\equal{}2\to R\equal{}2r$
+ My solution is wrong.  :D

Sorry, I think my solution have made mistake in computing... but using this way, I am sure that we can prove above inequality very easily... I will comeback as soon as positive with a correct solution (Tomorrow, I will have an exam of Chemistry in school so I should study Chemistry) :D

Virgil and zaizai maybe you are very good at geometric inequality  :) .Can you help me [url]http://www.mathlinks.ro/viewtopic.php?t=183881[/url]
    Thank !  :D


A ball is made of a thick rubber shell and has an empty interior.  The density of the rubber shell alone is $ 1.4$ $ g/cm^{3}$, and the density of air is estimated to be zero.  The radius of the "sphere of air" inside the ball is $ \frac{4}{5}$ the radius of the entire ball.  What is the average density of the entire ball (shell and air together) in $ g/cm^{3}$?  Express your answer as a decimal to the nearest hundredth. [hide="ANS"]$ 0.68$[/hide]

The increasing sequence $ 12, 15, 18, 21, 51, 81, ...$ consists of all the positive multiples of $ 3$ that contain at least one digit that is a $ 1$.  What is the $ 50^{th}$ term of the sequence? [hide="ANS"]318[/hide]

[hide="#1"]We know that the density can be expressed as $ \frac {m} {V}$, where $ m$ denotes the mass and $ V$ denotes the volume.

Let the radius be denoted by $ r$.

We are given that the density of the rubber shell is $ 1.4 g/cm^{3}$.
The volume of the rubber shell is $ \frac {4} {3}\pi(r^{3}\minus{}(\frac {4} {5}r)^{3})\equal{}\frac {4} {3}\pi(\frac {61} {125}r^{3})$

The average density is merely the fractional part of the total volume that the rubber shell contains multiplied by the density of the rubber shell because the air does not contain any mass.

This is $ \frac {61} {125}*1.4\approx 0.68$[/hide]

[hide="#2"]Umm... smart counting... I guess.
So we first note that all multiples of $ 3$ between $ 10$ and $ 20$ work and those between $ 100$ and $ 200$.
Then, the first one not in the first interval is $ 21$. To find the next multiple of $ 3$ with the same units digit, we add $ 10*3$.  So we have $ 12,15,18,21,51,81...$
For those multiples of three between $ 100$ and $ 200$, there are $ \frac {198\minus{}102} {3}\plus{}1\equal{}33$.
So far, we have $ 39$ multiples of $ 3$.  Between $ 200$ and $ 300$, the numbers that work are $ 201,210,213,216,219,231,261,$ and $ 291$.  There are $ 8$ multiples of $ 3$.
So we need $ 3$ more multiples of $ 3$ to get to the $ 50$th.
And this is $ 318$.[/hide]

[quote] The average density is merely the fractional part of the total volume that the rubber shell contains multiplied by the density of the rubber shell because the air does not contain any mass.

This is $ \frac {61} {125}*1.4\approx 0.68$[/quote]

Thanks, mathcrazed!

However, I have a question.  For the first problem, the formula is $ Density \equal{} \frac {m}{V}$.  Thus, why did you multiply in the last step?

The average density is the density of the rubber shell multiplied by the fraction of the volume it contains because the air has no mass, and thus does not affect our final result.
$ d_A \equal{} d_s*\frac {V_s} {V}$

$ d_A$ is the average density, $ d_s$ is the density of the shell, and $ V_s$ is the volume of the shell.

Do you not have school today? 'Cause I don't :D:D.  Hope that makes sense.

EDIT: If you want to use the formula $ d \equal{} \frac {m} {V}$, then the mass is going to be $ d_s*V_s$, correct?
So to find $ d$, we take $ d \equal{} \frac {d_s*V_s} {V}$

Technically, these two solutions are the same.

Thanks a lot, mathcrazed!

No, I don't have school.


summer program
Mathcamp
AMC
AIME
USA(J)MO
USAMO

[quote="MysticTerminator"]MOP is pretty cool. thoughts?[/quote]

Must be. There's a [url=http://moppers.kaseorg.com/]whole Web site about MOP[/url].

Not that I know personally one way or the other.

[quote="Yasha"]Yeah, MOP is cool. Fun people, fun games, etc. But the classes are way too long (2.5hrs.). By the end I'm like "Who cares about those d--n medians and angle bisectors? What do they have to do with anything?" And they're kind of repetitive too. Problem after problem after problem after problem after...

Other than that it's a great camp. In all the ones I have been to it's second only to [url=http://www.mathcamp.org]Mathcamp[/url]. You learn lots of stuff that's helpful on AIME and USAMO and you meet cool people.[/quote]

The best way to get better, once you know the necessary theorems, is to practice their applications. For hours. And hours. And hours. (and hours!!!)

Things go faster when someone is there to explain HOW to arrive at the solution, but this only helps after you've spent a good amount of time on the problem already...

Like in Blue, Titu pretty much just gave us problem sets and has us work on them for about three hours this week already... thing is it's not that bad... you can get a lot out of it if you put a lot into it.

[quote="texas137"]when are they going to pick the IMO team?[/quote]

The Selection Exam is Friday afternoon and Saturday morning, and the team will probably be announced Saturday night.


linear algebra
matrix
complex numbers
linear algebra unsolved

let's consider the square matrices over the complex numbers

it can easily be shown with jordanforms that all invertible matrices have square roots, and in fact, there also is one
if the eigenvalue zero has the same algebraic multiplicity and geometric multiplicity

however, i figure this isn't a necessary condition

is there a way to tell from the jordanblocks for the eigenvalue zero it the matrix has a square root?

thx,

fred

Seeing someoen asking about it, I wondered about this question, and then realised I had already asked.  I am not of course certain that the condition still isn't necessary.

But then what IS the condition?


There are 12 eggs in a carton, 8 cartons in a crate, and 9 crates in a container. How many eggs are in a container?

there are $ 12$ eggs in a carton

there are $ 8$ cartons in a crate, so there are $ 12\times8\equal{}96$ eggs in a crate

there are $ 9$ crates in a container, so there are $ 96\times9\equal{}864$eggs in a container


Here's the problem. I have no clue how to do it.

Which of the following is considered to be an "unlucky" and "extravagant" number?
(A) 2   (B) 7   (C) 8   (D) 9   (E) 14

[hide]
The answer was 8.[/hide]

Explanation would be appreciated.

Hmm, thinking non mathematically I would choose 8 also. The 8 ball in pool is the ball the ball you want to avoid pocketing. And 8 follows 7 (luckiest single digit integer) {cnn.com}

http://en.wikipedia.org/wiki/Extravagant_number

http://en.wikipedia.org/wiki/Lucky_number

I guess an "unlucky" number is a number that doesn't survive that sieve.

2 is not extravagant

7 is a not an unlucky number

9 is not an unlucky number

and 14 is not an extravagant number so that leaves 8. Which fits both.

Sounds like a problem from a competition written by students.  Where is this from?   :P

Ok, thanks.
So basically you need a lot of background knowledge.

The problme is from a Texas math competition called UIL.

I'm a bit curious how someone from another state came across a UIL Mathematics problem. Are you, perhaps, moving to Texas and looking to join UIL?
Most of the problems on the UIL Mathematics test have a mathematical basis in algebra, geometry, trigonometry, or calculus.  However, there are a few problems (like your example) that require a knowledge of number classifications.

No, I just came across it and thought I'd try it for practice.
I haven't learned enough calculus to do a few of the problems though.  :(


I have reproduced the "bug" below:
Also, see the picture
Edit: The picture will make your screen look like it's exploded

The buttons at the top of the forum which deal with save states and expanding and collapsing the forum don't work for me.

What happens & what browser are you using?  Have you tried deleting your cookies?

The buttons at the top of the forum work now, but

is still a problem

[quote="ealex292"]I have a bookmark of the login page. When I go to that, enter my username and password, and press log-in, I get taken to the forum. It tells me when my last visit was, and shows the forums for the classes that I am taking . When I then go to the My Classes page, I am told that I am not logged in. I then return to the forum, and can no longer see my classes listed in the topics. However, when I click on the log-in link this time, I can succesfully log in. Any idea how to fix it, or do I need to log in twice each time?[/quote]

Very weird.  Try deleting your cookies (Tools -> Internet Options -> Delete Cookies on IE).

RC - I'll take a look at that bug (I edited your post to remove it so ealex could read my answer easily).

[quote="rrusczyk"]
Very weird.  Try deleting your cookies (Tools -> Internet Options -> Delete Cookies on IE).[/quote]

I just tried that. It didn't help any. I'm also not using IE, but I figured out [url=http://getfirefox.com]Firefox[/url] quickly enough.


My math team coach gave me this problem:

Tickets to a concert are being sold. A line forms before sales begin, and the same number of people come to buy their ticket every minute. If $ 3$ ticket booths are opened, it will take $ 40$ minutes for the tickets to be sold out. If $ 4$ ticket booths are open, it will take $ 25$ minutes for the tickets to be sold out. How many minutes will it take for tickets to sell out if $ 8$ ticket booths are opened?

Hey brightzhu,

I'm not sure how to do this problem. I first tried doing proportions, but that's obviously not going to work. So then I tried using a linear equation. y = mx + b. I plugged in x = 3, 4 and y = 40, 25 respectively. But that didn't work. Afterwards, I tried switching it but that still didn't work. Is there a possibility that the problem might be impossible because I'm getting negative integers for the answer. Maybe ask your coach how to do this problem? Keep me posted.

Thanks,

rts2007

I just figured it out:
[hide]
Let $ y \equal{}$ number of people already waiting there and $ x \equal{}$ how many people get there every minute. Then, from the given info:$ \frac {40x\plus{}y}{3(40)}\equal{}\frac {25x\plus{}y}{4(25)}$. Simplifying, we get $ y\equal{}50x$. For the eight ticket booths, let $ t \equal{}$ minutes it takes for tickets to sell out, so $ \frac {tx\plus{}y}{8t}$. We can set this expression equal to one of the others because the number of tickets is always the same. To make the problem easier, let $ x \equal{} 1$.We now have $ \frac {t(1)\plus{}(50)}{8t}\equal{}\frac {40(1)\plus{}(50)}{3(40)}$. Simplify, and we get $ t\equal{}10$. So the answer is it takes $ 10$ for the tickets to be sold out if eight booths are opened.
[/hide]

Here's a more complicated version of this problem, that appeared as $ \#4$ in the AIME in $ 2007$, I think.
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=554212177&t=141165


induction
geometry
geometric transformation
reflection
email
AMC
USA(J)MO

Does anybody else think that the AMY induction packet is really hard? I barely got two in the easy questions!

Also, are there any girls 12-14 going to Awesome Math?

Yea...the packet is difficult...I guess you can solve more with experience. I think that the packets reflect the higher difficulties at the camp.

As for making progress, I think one problem might be that you are having trouble starting, and then all the time working on the problems isn't really going anywhere. If there are certain problems that you want hints on or something, you should ask on this forum. There is a lot of good stuff to learn in the induction packet.

And...err...my username should be a sufficient answer to your second question  :P .

yeah I definitely found the packet hard too, even the easy ones >< plus I was trying to do them during finals week and it... didn't work out too well bc I got super overstressed and I pretty much ditched the packet and haven't looked at it in a few weeks.... hahahah.

But hey, I'm a girl and I'm going!  I'm 16 though.  Whatev.

[quote="Altheman"]And...err...my username should be a sufficient answer to your second question  :P .[/quote]
Really?
'altheman' tells us whether there are or aren't any girls 12-14 going?

pretty much

like, just think REALLY REALLY hard about it

and you should figure it out

just keep reciting "altheman"

over and over again

and not only do you miraculously find the answer to "how many girls 12-14 will there be at awesome math this year" but many other mysterious of the universe as well

I hear there's some sort of girls' math olympiad team that's supposed to be there.

some team that returns no results on google... :huh:

anyway...just incase i was not obvious enough al_the_man

so w/e i am not a female [waits for some smart --- to be like OMG UR A GIRL]

anyway yea

oh and try what arnav said...even though he is not even going to amsp

Bit slow Jay, this has been discussed numerous times

mkay, i didn't choose my words carefully...but luckily we are going to a math camp, rather than say, an english camp. Its okay that I suck at english.

[quote="sapphyre571"]According to Dr. A's email:

"For example, all USA Girls Math Olympiad team members will be in track 3."

That means that a) there will be girls there and b) they will all be in track three.[/quote]
not all girls there are on the USA GMO team.

[quote="Fanatic"]Altheman, your username tells us that YOU are a guy but that doesn't tell us whether or not there are any 12-14 year old girls going to amsp.[/quote]
Huh, that was my and Arnav's implication

leave it to me to completely overlook the "USA Girls Math Olympiad" thread (I was gone for the last week or so and I don't normally notice threads without the yellow folder next to them). I did not mean to imply that all girls would be in 3, I meant to say that all GMO girls would be in 3.

[quote="Fanatic"]This USA girls math olympiad... there is a such thing? Or is it simply "girls in the USA math olympiad"... how many?? like 3?[/quote]

dude you are so sexist wtf yes 3 girls in the USA do math
i think it's an international thing and the USA sends a team although i may be wrong

Two teams of four girls each were chosen to represent the USA in a competition in China after the camp. The highest-scoring girls in the USAMO are on this team.

[quote="Fanatic"]I thought it meant girls that made MOP, which is why I said "like 3?" [sarcastic under-exaggeration][/quote]
[correct number]

wheee i'm a girl 12-14! ^.^

and yeah, i thought the packet was pretty hard... >.<

so who here's gonna be 13-14 (guys or girls)?

mkay it seems that i am going to be a bit old here... anybody 17?  :D

I don't think so.
(I'm 13 too, but I'm probably older than you =P)

[quote="lena_siz"]Umm, i'm 13, so yeah.  :)

Do we have to bring solutions of the qualification quiz(A/B/C) to discuss them at camp?[/quote]
I don't think so.

hmmm
rap sucks
one of the main reasons why the music industry is fubar
anyway
i am 15.5
hmmm i should be going to driver school thingy soon

[quote="13375P34K43V312"]i'm 13
i turn 14 like a week after camp

DUDE TWIZ I HOPE THAT'S YOU IN YOUR AVATAR
it's probably some random rapper than i don't recognize, but eh[/quote]

That's Young Jeezy. My friend made me make that my avatar for some reason, and I don't have anything better to change it to at the moment, so that's why it's there.

[quote="13375P34K43V312"]
i turn 14 like a week after camp
[/quote]

i win...i turn 16 a day after camp :) :) :)

I'm Like 15.64 or something... December born.... lol for exact Age

[quote="Aryth"]I turn 14 three weeks after camp :P[/quote]
I turn 14 one week into the camp :P

I turn 14 on the first Saturday of camp (July 21).

I turn 14   162 days after AMSP ends. :)


(1)What happens when cyclopentyl methyl ketone is treated with $Ca(OH)_{2}$and $I_{2}$ and then heated?

(2) Does bronze also contain zinc?What is the composition of type metal?

I don't have much time, but quickly, for 1) look up iodoform reaction, and for 2) just wikipedia bronze.  I always forget the composition of bronze, brass, etc.

(1) The reaction of a methyl ketone with an halogen in the presence of a base is called the haloform reaction, and the products are a carboxilate anion and an haloform. In this case, the products would be calcium cyclopentylcarboxilate (which shlould not be very soluble) and iodoform (a nice yellow solid).

(2) Bronze is an alloy of copper and tin.

Carcul


number theory
least common multiple
binomial coefficients
number theory proposed

Prove that $ lcm(1, 2, ..., n)\equal{}lcm(nC0, nC1, ..., nCn)$ if and only if $ n\plus{}1$ is prime.

[quote="shoki"]the problem is directly related to this problem :
if $ n \in N$ then we have :

$ \frac {1}{n \plus{} 1}(lcm(1,2,...,n \plus{} 1)) \equal{} lcm(\binom{n}{0},\binom{n}{1},...,\binom{n}{n})$
(i saw this problem in this forum but don't remember the source unfotunately)

now the problem is very trivial! :)[/quote]

Yes, you are right. I had invented this problem by that. :lol:

Now let's talk about the problem you had post. :wink:

[quote=shoki]this lemma can easily be proven :
none of the binomial coefficients $ \binom{n}{k}$ where $ 0 \le k \le n$ are divisible by $ p$ if and only if we have $ n\equal{}p^rm\minus{}1$ where $ r,m$ are non-negative integers and we have $ m<p$.
we can easily prove by using the lemma and the fact that
$ v_p( \binom{n}{k})\equal{}\sum_{i\equal{}1}^{\infty} \left([\frac{n}{p^i}]\minus{}([\frac{n\minus{}r}{p^i}]\plus{}[\frac{r}{p^i}]) \right)$
this divisibility :
$ lcm(\binom{n}{0},\binom{n}{1},...,\binom{n}{n}) |  \frac{1}{n\plus{}1}(lcm(1,2,...,n\plus{}1))$
now let $ p^r \le n < p^{r\plus{}1}$ then put $ k\equal{}p^r\minus{}1$ and then it is easy to prove that $ p^r \parallel{} \binom{n}{k}$
and since $ v_p(\frac{1}{n\plus{}1}(lcm(1,2,...,n\plus{}1)) )\equal{}r$ we r done![/quote]

Everything can easily be proven...


Euler
geometry
circumcircle
geometric transformation
reflection

Let ABC be a triangle with euler triangle DEF.
if the vertices D, E, F of euler triangle  are
reflected wrt. euler line of ABC
prove that the images lie on the nine point circle of ABC.

We know that the Euler triangle circumcircle is the nine point circle for the original  triangle and reflection of the euler trinagle vertices will lie on the nine point circle thus.


calculus
integration
logarithms
calculus computations

I was trying to find $ \int \frac {\ln{x}}{(1 \plus{} \ln{x})^2}dx$

but,, I was stuck here
$ \int \frac {\ln{x}}{(1 \plus{} \ln{x})^2}dx \equal{} \int \frac {1}{1 \plus{} \ln x} \minus{} \frac {1}{(1 \plus{} \ln x)^2}dx$

anyone can help me to find this integral?
thanks

$ \int \frac {\ln{x}}{(1 \plus{} \ln{x})^2}dx\equal{}\int \frac {x\ln{x}}{(1 \plus{} \ln{x})^2}\; \frac{dx}{x}$

$ \equal{}\int \frac {x\ln{x}}{(1 \plus{} \ln{x})^2}d\left(\ln{x}\right)\equal{}\int x\;\frac {(1\plus{}\ln{x}\minus{}1)}{(1 \plus{} \ln{x})^2}d\left(\ln{x}\right)$

$ \equal{}\int \frac {x}{1 \plus{} \ln{x}}d\left(\ln{x}\right)\minus{} \int \frac {x}{(1 \plus{} \ln{x})^2}d\left(\ln{x}\right)$

leave the 1st one alone , apply integration by parts to the 2nd one

$ \equal{}\int \frac {x}{1 \plus{} \ln{x}}d\left(\ln{x}\right) \plus{}\frac {x}{1 \plus{} \ln{x}} \minus{}\int \frac {dx}{1 \plus{} \ln{x}}$

now recall $ dx \equiv x\;d\left(\ln{x}\right)$

hence , the 2 integrals cancel each other out and we are left with $ \frac {x}{1 \plus{} \ln{x}}$, which is your answer.

\[ \int \frac{\ln{x}}{(1\plus{}\ln{x})^2} dx \equal{} \int \frac{\ln{x} \plus{} 1 \minus{} x\frac{1}{x}}{(1\plus{}\ln{x})^2} dx \equal{} \int \frac{v(x)u'(x) \minus{} u(x)v'(x)}{[v(x)]^2} dx\]
with $ u \equal{} x$ and $ v \equal{} 1\plus{}\ln{x}$ so
\[ \int \frac{\ln{x}}{(1\plus{}\ln{x})^2} dx \equal{} \frac{x}{1 \plus{} \ln{x}} \plus{} C\]

[quote="misan"]$ \int \frac {\ln{x}}{(1 \plus{} \ln{x})^2}dx \equal{} \int \frac {x\ln{x}}{(1 \plus{} \ln{x})^2}\; \frac {dx}{x}$

$ \equal{} \int \frac {x\ln{x}}{(1 \plus{} \ln{x})^2}d\left(\ln{x}\right) \equal{} \int x\;\frac {(1 \plus{} \ln{x} \minus{} 1)}{(1 \plus{} \ln{x})^2}d\left(\ln{x}\right)$

$ \equal{} \int \frac {x}{1 \plus{} \ln{x}}d\left(\ln{x}\right) \minus{} \int \frac {x}{(1 \plus{} \ln{x})^2}d\left(\ln{x}\right)$

[/quote]

I've done till this step,,
but,,..can you tell me how did you find that::
$ \minus{} \int \frac {x}{(1 \plus{} \ln{x})^2}d\left(\ln{x}\right)\equal{}\plus{} \frac {x}{1 \plus{} \ln{x}} \minus{} \int \frac {dx}{1 \plus{} \ln{x}}$

[quote="beezees222"]
\[ \int \frac {\ln{x}}{(1 \plus{} \ln{x})^2} dx \equal{} \int \frac {\ln{x} \plus{} 1 \minus{} x\frac {1}{x}}{(1 \plus{} \ln{x})^2} dx \equal{} \int \frac {v(x)u'(x) \minus{} u(x)v'(x)}{[v(x)]^2} dx
\]
with $ u \equal{} x$ and $ v \equal{} 1 \plus{} \ln{x}$ so
\[ \int \frac {\ln{x}}{(1 \plus{} \ln{x})^2} dx \equal{} \frac {x}{1 \plus{} \ln{x}} \plus{} C
\]
[/quote]

wow,,it's a very nice idea..
thanks a lot.. :)

$ \int \frac {\ln{x}}{(1 \plus{} \ln{x})^2}dx$ Put $ lnx \equal{} t$

$ \equal{} > x \equal{} e^t \frac {1}{x} dx \equal{} dt$

$ \equal{} > dx \equal{} e^tdt$

So now integral reduces to  $ \int \frac {t}{(1 \plus{} t)^2}e^tdt \equal{} \int [ \frac {1}{1 \plus{} t} \minus{} \frac {1}{1 \plus{} t^2}]e^tdt$

This is of the form $ \int e^t(f(x) \plus{} f'(x))dt \equal{} e^tf(x) \plus{} c$

Oh really?

$ \frac {d}{dx}\bigg[\frac {1}{1 \plus{} x}\bigg] \equal{} \frac { \minus{} 1}{1 \plus{} x^2}$ eh?

it was a typo bro...

$ \int e^t(f(x) \plus{} f'(x))dt \equal{} e^tf(x) \plus{} c$
is still confusing for me..

but,thanks for the idea..

EDIT::ah,,I just realized...sorry for asking to many question

It's the product rule

$ \frac {d}{dx}\bigg[e^xf(x)\bigg] \equal{} e^xf'(x) \plus{} e^xf(x) \equal{} e^x(f(x) \plus{} f'(x))\Rightarrow\int\bigg[e^x(f(x) \plus{} f'(x))\bigg]dx \equal{} e^xf(x) \plus{} C$

There is never a good reason to quote the post immediately preceding yours in full -- it just makes the thread harder to read.  Please don't do it.

I removed the unnecessary quotes. Please, don't do it again.

Sorry, I didn't mean to impose. On other forums not quoting is considered an iniquity. I won't do it again.


((x)/(x-2))-((x-6)/(x-4)) solve for x.

I know I have to get a common denominator, but how do I do that?

[quote="10000th User"]Awfully suspicious about this topic and the one in [url=http://www.mathlinks.ro/viewtopic.php?t=174909]HERE[/url] :mad:[/quote]
someone probably has the same class as me


At McMiddle School, there are 360 seventh-grade and eighth-grade students. Three-fourths of the students get an allowance, and one-third of the students are eighth-graders. What is the smallest possible number of seventh-graders who get an allowance?

There are $ 360\cdot\frac {3}{4} \equal{} 270$ students that get an allowance. There are $ 120$ students are $ 8^{th}$ graders, so to minimize the number of $ 7^{th}$ graders that get an allowance, all of the $ 8^{th}$ graders should get an allowance. So the smallest possible number of seventh-graders who get an allowance is $ \boxed{150}$.


Suppose $ x,y \in \mathbb{Z}^\plus{}$ ,find the numbers of positive integers $ n$ such that $ n$ [color=red]cannot[/color] be represented by $ 5x\plus{}7y$ ?

Is it infinite or finite?

The set is countably infinite.  For example, that equation CAN generate numbers such as 12, 17, 19, 24, 29,... but it CAN'T generate numbers such as 10, 13, and 20.  You should notice that all generated numbers follow the pattern of $ 12\plus{}5n$, so infinitely many numbers will not be generated.  But the integers are countable so this set is countable.

[quote="vinskman"]Suppose $ x,y \in \mathbb{Z}^ \plus{}$ ,find the numbers of positive integers $ n$ such that $ n$ [color=red]cannot[/color] be represented by $ 5x \plus{} 7y$ ?

Is it infinite or finite?[/quote]

the answer is finite

[hide="the answer"]

Note that:
$ 24 \equal{} 2\cdot 7 \plus{} 2\cdot 5$
$ 25 \equal{} 5\cdot 5$
$ 26 \equal{} 3\cdot 7 \plus{} 5$
$ 27 \equal{} 7 \plus{} 4\cdot 5$
$ 28 \equal{} 4 \cdot 7$

So any number $ k \ge 29$ can be written as $ 24\plus{}5a$, $ 25\plus{}5a$, $ 26\plus{}5a$, $ 27\plus{}5a$ or $ 28\plus{}5a$.
And the largest number is $ 23$.

Right ?

[/hide]

[quote="mathson"][quote="vinskman"]Suppose $ x,y \in \mathbb{Z}^ \plus{}$ ,find the numbers of positive integers $ n$ such that $ n$ [color=red]cannot[/color] be represented by $ 5x \plus{} 7y$ ?

Is it infinite or finite?[/quote]

the answer is finite

[hide="the answer"]

Note that:
$ 24 \equal{} 2\cdot 7 \plus{} 2\cdot 5$
$ 25 \equal{} 5\cdot 5$
$ 26 \equal{} 3\cdot 7 \plus{} 5$
$ 27 \equal{} 7 \plus{} 4\cdot 5$
$ 28 \equal{} 4 \cdot 7$

So any number $ k \ge 29$ can be written as $ 24 \plus{} 5a$, $ 25 \plus{} 5a$, $ 26 \plus{} 5a$, $ 27 \plus{} 5a$ or $ 28 \plus{} 5a$.
And the largest number is $ 23$.

Right ?

[/hide][/quote]

A little bit difference from our soltuions , because $ x\ge1 ,y\ge1$ are requested in the question.

This is called the "Chicken McNugget Theorem."

[quote="JRav"]The set is countably infinite.  For example, that equation CAN generate numbers such as 12, 17, 19, 24, 29,... but it CAN'T generate numbers such as 10, 13, and 20.  You should notice that all generated numbers follow the pattern of $ 12 \plus{} 5n$, so infinitely many numbers will not be generated.  But the integers are countable so this set is countable.[/quote]

It generates, for example, all numbers which are larger than $ 5\cdot 7 \minus{} 7 \minus{} 5$, so the set of numbers which cannot be generated is finite.  Perhaps you should read the problem more carefully.

Maybe I'm just missing something but how would it generate, say 30 if $ x,y\geq 1$ are both integers?
[quote="blahblahblah"]Perhaps you should read the problem more carefully.[/quote]
I'm aware of what the question is asking.

every integer over $ 35$ can be represented as $ 5a\plus{}7b$ where $ a,b$ are positive integers, and every integer over $ 23$ can be represented as $ 5a\plus{}7b$ where $ a,b$ are nonnegative integers. i guess it depends what you take as the definition of $ \mathbb{Z}^\plus{}$.


what exactly is a conservative force?

one which when doing work to an object in a system transfers energy so that total mechanical energy of the system stays constant?

one which does the same work on an object when it moves from some position to another regardless of what path that object takes to get from that some position to that other position?

Both definitions are OK, although I prefer the latter one.

[quote="Mango 070"]
one which when doing work to an object in a system transfers energy so that [i]total mechanical energy of the system stays constant?[/i]

one which does the same work on an object when it moves from some position to another regardless of what path that object takes to get from that some position to that other position?[/quote]
The  second one is the most appropriate one that describes a conservative force.

The first one isn't correct because there is no necessity for the total mechanical energy of the [b]system[/b] to be constant


real analysis
calculus
integration
limit
function
trigonometry
real analysis solved

Assume that f is countinous on [0,1], f(0) = 0 , f ' (0) exist .
Prove that the Lebesgue Integral  from 0 to 1  (f(x) * x^(-3/2) ) dx exist

Since $f'(0)$ exists and $f(0)=0$ we conclude that $\lim_{x\to 0}\frac{f(x)}{x}$ exists, it follows that $f(x)x^{-3/2}$ behaives like $x^{-1/2}$ near to 0 and it is continuous on $(0,1)$. It follows that $f(x)$ is integrable.

[quote="sparky_tx"]Thank u again , I have a question , In order to prove Lesbegue integral whaqt I should prove in general ?, thank u again[/quote]
There are two issues:

1. Measurability
2. Size

In many, many cases, you don't worry about measurability; it's very easy for a function to be measurable. Functions given by formulas tend to be measurable - they tend to be continuous except perhaps at finitely many points, and that's stronger than measurable.

That leaves size as your concern. Assuming you're taking measurability for granted, the question of integrability is exactly the same as one of the questions you would ask about a series: does the series converge absoluteley.

The series $\sum_{n=1}^{\infty}\frac{\sin 3n}{n^2}$ does converge absolutely.

The series $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}n$ does not converge absolutely. That we call it conditionally convergent is tied to adding the terms together in a particular order, and is subject to being upset by rearrangements of that order.

The definition of the Lebesgue integral has rearrangement built into its very core; we can't have the Lebesgue integral being affected by rearrangement, so we must have absolute convergence.

You might try using "absolutely integrable" as an approximate synonym for "Lebesgue integrable": it focuses your thoughts in the right direction.

What about $\int_0^{\infty}\frac{\sin x}x\,dx$? The integrand is bounded, but it's behavior as $x\to\infty$ appears to be no better than $\frac1x$, and that means that it is not absolutely (Lebesgue) integrable.

Perhaps the integral exists as an improper integral. We can explore that possiblity with an integral by parts, leaving us with a limit which goes to zero and the integral

$\int_0^{\infty}\frac{1-\cos x}{x^2}\,dx.$

This is a nonnegative integrand, so the Lebesgue integral is unambiguously defined. But to call it integrable, we must rule out $\infty$. The integrand is bounded in a neighborhood of zero (meaning bounded, period), and its behavior as $x\to\infty$ is $O(x^{-2}),$ which is sufficient to declare it (Lebesgue) integrable. But the original integral ($\sin x/x$) still does not exist in the Lebesgue sense.


Can someone help me with the USAMTS? I registered last year with my email mictan@bergen.org but forgot the username and password. I clicked the "forgot your password?" button and it said that it emailed my password and username to me yesterday, but I waited until now and I still don't have an email. Can someone fix this?

Write to usamts@usamts.org with the details.

Check your spam.  I've had password reminder e-mails go there a lot!


Support geometry calculus integration MIT college ratio

I noticed the 'university' thread, has basicaly just become a place for people to ask questions about uwaterloo for next year.

--edit-- were now doing a roll call style thing, on eylot's request, so I will be sorting you all by however I see fit :D

[b]Post if your going to waterloo Next year or currently, and what program.[/b]

the list ( so far ):

Starting September 2006:

    Math:

        1)rzsolt
        2)Sisyphus
        3)Singlular
        7)cheeZZcake
        8)jpark
       10)neil477
       13)lightrhee

    Engineering:

        4)YY (Chemical)
       11)Thahn (Software)

    Bioinformatics:

        5)Sunny

    CS:

        9) Richard Peng

Currently at Waterloo:

    CS:

        6)Elyot
      12)Cherriz

[b][size=150]SPECIAL BONUS PRIZE CATAGORY CATAGORY::::::::::::

    Math:

      13)nf_guy2001 ( HE WILL BE GOING IN SEPTEMBER, PUREMATH, MASTERS DEGREE)!!!!!11111111 [\b][/size] :D
           ( much better?  :D )[/b]

Good idea.

You'll see me around, yay!  So excited!   :D

Sunny, are you doing math or something else?

Bioinformatics.  So more computer science than math.

'K...I'm going to UWaterloo Chemical Engineering.
No idea what to expect. Oh well.

[quote="rzsolt"]The math stream seems nice, becuase you get so much freedom to choose your cources. I would think that bioinormatics would have alot of your classes chosen for you?
I picture your first semester like: calc 1, classical algebra, cs, bio 101 , and one elective?[/quote]

Hmm, I'm not too sure really.  But that sounds right.  I don't mind much though, I'll be taking classes I like anyways, lol bio and math.

Hey Yiyi, whatcha doing for rez?

Good topic for a thread.

rzsolt, what I recommend you do is edit the first post to show a proper "roll call"-style listing of all those at Waterloo... you should have a few numbered columns with names (and possibly programs)... with titles such as people coming to waterloo this september, people currently at waterloo, etc.

I would do this myself, but since you own the first post in this thread, the responsibility to do this now falls upon you!!!

An example of how to write a roll call thread can be found [url=http://www.tranceaddict.com/forums/showthread.php?s=&threadid=334039]HERE[/url]

If you feel like doing this, add me to the "currently at Waterloo" column, in CS (+ Undecided Math). :P  And don't forget to update it often when people keep posting!

[quote="Sisyphus"]I'm actually surprised at how much of the 'Canadian' subforum on here is headed for/already at Waterloo. You'd think that there'd be more people from UofT, UBC, etc. Not that I'm complaining of course. The more, the merrier.   :lol:[/quote]

Heheheh... It's all my doing!  Secretly, I've been plotting to get more smart math people to come to Waterloo, so we can take over the world.  Promoting Waterloo on AOPS/Mathlinks is just one step in my giant, evil plan!

 :P  :P  :P

Well... not really.  But it's nice to see lots of people coming to Waterloo.

~ Elyot

[quote="Sisyphus"]I'm actually surprised at how much of the 'Canadian' subforum on here is headed for/already at Waterloo. You'd think that there'd be more people from UofT, UBC, etc. Not that I'm complaining of course. The more, the merrier.   :lol:[/quote]

Well, more people go to UBC for life sciences and stuff. Honestly, most of my friends going to UBC are just choosing it becuase its the farthest away from there parents.

I'm like the only one of my friends that has chosen there university programs by quality of the program.....!

well, that and half my friends will be doing victory laps next year!

[quote="Sunny"]
Hey Yiyi, whatcha doing for rez?[/quote]

Not a clue. I've been waiting for that package for 3 weeks now, and I'm still waiting. :rotfl:

[quote="rzsolt"]
Well, more people go to UBC for life sciences and stuff. Honestly, most of my friends going to UBC are just choosing it becuase its the farthest away from there parents.
[/quote]

lol...that's actually part of the reason why I chose Waterloo.

[quote="YY"]

[quote="rzsolt"]
Well, more people go to UBC for life sciences and stuff. Honestly, most of my friends going to UBC are just choosing it becuase its the farthest away from there parents.
[/quote]

lol...that's actually part of the reason why I chose Waterloo.[/quote]

Wow, I seem to be that unique person who isn't eager to escape parental grasp. Maybe because I won't get any farther away from my father, since he's in Ukraine anyway. Maybe also because my mother isn't too constraining or protective. I would really miss her and my sisters if I left. Though living completely independently from adults would definitely be interesting. Are you guys finding this to be a dilemma as well, or are you decidedly resolute to get away from home?

are there any dedlines comming up that I don't want to miss? such as when I pay or choose my rez? I have no idea when any of that is, but I think it would be bad for me to remain ignorant.  :oops:

You have to get in your $\$$500 residence deposit and ranking form in by June 12. You can pick your M-section starting on the 12th too. June 15 is the deadline to apply for financial assistance...and that's it for June (I think)

For all you guys currently attending Waterloo, does anyone know what the Physics-based Math 137 course is like? I'm interested in the Mathematical Physics major as well as the Pure Math major..and it looks like the physics based calculus class is something that I should think about.

[quote="lightrhee"][quote="Elyot"][quote="liudaveliu"]Is the president scholarship renewable if you maintain a good academic standing?

Btw. Are there any national scholarship winners here? (Waterloo seems to be very quiet with their scholarship winners)[/quote]

Me!!!  Got a 25k SAS national scholarship from Waterloo last year!

3 of this year's 25k winners are members of this site actually (no idea who won the one for ACTSCI).  I'm not sure if I'm actually supposed to reveal who they are yet until the formal press releases make it out, but nobody told me not to... so yeah... big congrats to SARAH, JENN, and RICHARD!!!  WOOT!!!

(I sincerely hope I didn't get any of that wrong... or angered anyone by giving that away.  You guys deserve all the recognition you can get!)

Several others won 18k and 14k scholarships as well.[/quote]
I'm getting 25K too. :)[/quote]

I'm getting 25K too.

Which one are you getting David?

[quote="TheDreamer"][quote="Elyot"][quote="TheDreamer"]Ooh i haven't been here in ages..but i'm gonna be at AciSci double major pure PURE math next year!! look forward to seeing you guys! :)[/quote]

AciSci?  You traitor!!!  I suppose taking pure math as well redeems you... but... are you masochistic?

Congrats to David as well.[/quote]

Heh I had to look up masochistic, but still don't understand.  You know more words than I can contemplate.  :(

I look forward to seeing you again Elyot!! BTW, the MCTIC@London thing that we went to last year, we went again this year (you should've mentored us)...we came first!! We each won a copy of maplesoft 10, the software developed by yourself!  So sweet to have a piece of Elyot on my harddrive.  We also had a second team that was reasonably competitive :)..bah, we're trying.

Make sure you come to seminar![/quote]

By masochistic, I mean... do you take pleasure in subjecting yourself to the amount of work that ActSci + Math is???  Have fun!

I did do some work on Maple 10, but most of the stuff I did last term will be going into a later version.  In Maple 10, about half of the ArrayTools package was done by me.  I did far more crazy stuff for Maple 11, mostly related to Algebra.

Yes, I will visit you all at Seminar!

Let today denote the [i]n[/i]th of June.
Let [i]x[/i] denote the level of excitement and anticipation.
By the WMCS Theorem, Seminar starts in (11-[i]n[/i]) days.

Attempt to prove that as [i]n[/i] :arrow: 11, [i]x[/i]  :arrow: infinity for all seminar attendees by mathematical induction.

Seminar attendees are well ordered, such as there is a rank in which all attendees are invited.  Let [i]r[/i]1 denote the lowest rank, and let [i]r[/i]60 denote the highest.

As Ethanol scored a 138 on Fermat, Ethanol = [i]r[/i]1.
The postulate is certainly true for [i]r[/i]1.
Assume true for all seminar attendees, such that as [i]n[/i] :arrow: 11, [i]x[/i]  :arrow: infinity for all r.
Now, [i]r[/i]61 will be an unusual invitation because only 60 people normally go to seminar.  Thus, [i]r[/i]61 will be excited, and by the second corollary to the WMCS Theorem, x = [i]r[/i]61/n.  Since this result is of the correct form, the postulate holds true for all [i]r[/i].

And even if the induction is terrible, the result is still true :P

[quote="Sisyphus"]For all you guys currently attending Waterloo, does anyone know what the Physics-based Math 137 course is like? I'm interested in the Mathematical Physics major as well as the Pure Math major..and it looks like the physics based calculus class is something that I should think about.[/quote]
I've heard that it's a bit more interesting than the regular Math 137. However, since you're thinking about Pure Math, I think you should absolutely take Math 147. It's on the same level of mathematical rigor as the PMath major courses, which is much higher than the Math 137 level. Same goes for Math 135/145.

someone should add something to the uw math page at wikipedia

http://en.wikipedia.org/wiki/University_of_Waterloo_Faculty_of_Mathematics

cs folks can add to

http://en.wikipedia.org/wiki/David_R._Cheriton_School_of_Computer_Science

anyone can add to

http://en.wikipedia.org/wiki/Centre_for_Education_in_Mathematics_and_Computing

i already did my share in engineering.   :)

My personal fav!!

Stereotypes
Becoming an engineer means facing a multitude of stereotypes, from other engineers and from non-engineers. While many engineers are "nerds" or "geeks" (or Dilbert-esque), not all appreciate this label, and not all fit this label. Engineers are seen, stereotypically, as antisocial, devoid of spelling abilities and personal hygiene, scared of women and obscenely talented at math and science with godly computer skills. While some of these traits may be true for some engineers, it is hard to find one that fits that whole definition.

Another big stereotype is that engineers work hard and play hard, and can demolish forty beers. The validity of this statement cannot be verified.

Congratulations everyone with the scholarships! You guys rock!

[quote="Elyot"][quote="J-I-M-B-O"]
Another big stereotype is that engineers work hard and play hard, and can demolish forty beers. The validity of this statement cannot be verified.[/quote]

I vote true.

ALL the drunkest people I know at Waterloo are engineers.[/quote]

hahaha i don't drink though...

[quote="YawningLion"][quote="Ethanol"]Let today denote the [i]n[/i]th of June.
Let [i]x[/i] denote the level of excitement and anticipation.
By the WMCS Theorem, Seminar starts in (11-[i]n[/i]) days.

Attempt to prove that as [i]n[/i] :arrow: 11, [i]x[/i]  :arrow: infinity for all seminar attendees by mathematical induction.
[/quote]

 :rotfl: this induction is a new word in mathematics - you could present it at one of the seminar lectures. By the way, according to your theory, what happens to the level of excitement for n>11? Does it stay infinite during the seminar, and then drop to zero after the seminar ends?[/quote]

Hmmmmmm.

Let [i]x[/i] be be a function of are [i]e[/i] (excitement) and [i]a[/i] (anticipation), such that [i]x[/i] = [i]ea[/i] + [i]e[/i] + [i]a[/i].  As [i]n[/i] :arrow: infinity, [i]a[/i] :arrow: infinity, as has been already established by the previous butchering of mathematical induction.  (Even though [i]n[/i] is a discrete variable, the power of the WMCS is enough to give it an asymptote). [i]e[/i]=[i]f[/i]([i]n[/i]), a fifth degree polynomial, since the roots of our excitement are not fathomable by any simple formula.  When 11≤[i]n[/i]≤17, [i]a[/i]=0 (because WMCS has begun).  However, [i]e[/i] becomes a number so high it cannot be computed in polynomial time (thus also proving that [i]p[/i]≠[i]np[/i]).  When [i]n[/i]≥18, [i]a[/i], [i]e[/i] are set to  0.  When [i]n[/i]=1 again, the function of [i]n[/i] repeats.


number theory
least common multiple
number theory unsolved

[quote="steveshank"]Find two positive integers with sum 798 and least common multiple 10,780.[/quote]

[hide]Let the two number be $ax$ and $ay$

$axy=10780$
$a(x+y)=798$

we find that $a=14$,$x=22$,$y=35$ making the two numbers $308$ and $490$[/hide]

Awesome... the book I got this from suggested that you use the fact that $(a,b)=(a+b,[a,b])$, a previous exercise. But this is more efficient. Nicely done.


calculus
integration
limit
trigonometry
function
real analysis
real analysis unsolved

[quote="kunny"]Evaluate

\[\lim_{n\to\infty} \int_0^{\pi} x^2 |\sin nx| dx\][/quote]

That was one of kunny's daily integral problems in the "Calculus Computations and Tutorials" forum. Now try this:

Suppose $f$ satisfies (some reasonable hypothesis which you get to name). Prove that

$\lim_{n\to\infty}\int_a^bf(x)|\sin nx|\,dx=\frac2{\pi}\int_a^bf(x)\,dx.$

This has already been discussed in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21933[/url]. The idea is very useful (that's, probably, why Kent decided to bring it up again). It's a pity nobody answered.  :(

At a bacalaureat test in Romania we have a problem:

$f:R \to R$ $f(x) = \sin x$

compute $\displaystyle \lim_{x \to \infty} \frac{ \int_{0}^{x} |f(t)| dt}{x}$.

I think the formula posted before works also in this case.

How about if we had:

$\displaystyle \lim_{n \to \infty} \int_{a}^{b} f(x) |\cos nx| dx = ?? $

can we find a similar formula? or.. not?

thank you very much!

The answer to Mitzah's second question is yes. That is essentially the same problem and has essentially the same answer. In fact, $|\sin(nx)|$ or $|\cos(nx)|$ can be replaced by $g(nx)$ for any (nicely enough behaved) periodic function $g$; the only thing that matters over the long run will be the average value of $g$ over one period.

Mitzah's first question is a slightly different - and slightly easier - question because the division by $x$ lies outside the integral. Again, all that really matters about $|\sin x|$ is that it is periodic and we know its average value over a period.


ratio algebra polynomial probability function logarithms blogs

[hide]

1. Nope

2. $ \frac{2\sqrt{3}}{\sqrt{3}}$

3. It exists as a value that is infinitely accurate.

4. So bored people have something to memorize and people have ways to calculate how big a nuke is
[/hide]

However, irrational numbers are much more common than rational numbers...

[quote="aufha"]I was just wondering...
These aren't problems I want anyone to try to solve. I'm just curious.

1. Can an irrational number divided by a rational number be rational?

2. Can a irrational number divided by a different irrational number be rational?

3. Do irrational numbers exist? In what sense? How can pi exist? Do the diameter of a circle and circumference both not have rational values? What if the diameter is rational- does that mean the circumference is ... irrational?

4. Why do we define irrational numbers?

When I tried to prove $ \sqrt {2}$ was irrational, I started to feel that many imaginary numbers make much more sense than irrational numbers.[/quote]

Try proving $ \sqrt {2}$ is rational, then reach a contradiction. Start by putting it a lowest term fraction $ a/b$ (since all rational numbers can be represented as a fraction) and go on from there.

[hide="1"] No.  By definition, an irrational number $ j$ is not rational; but if the quotient of $ j$ and a rational number $ p$ were another rational number $ q$, then

$ j \equal{} p \times q$

where $ p \times q$ is rational; contradiction. [/hide]
[hide="2"] Yes.  Any irrational number $ j$ divided by a rational multiple of it $ pj$ gives the rational quotient $ \frac{1}{p}$. [/hide]
[hide="3"] This is actually a fairly deep question.  The Greeks said "no" because, to them, the concept of a number that was not the ratio of two integers was just too strange.  Some people, even today, are of the opinion that real numbers have no physical meaning, being numbers that are infinitely precise.

Let me answer your question with a deeper question:  [b]Do numbers exist?[/b]

As far as your other questions, yes, the diameter and circumference of a circle cannot be simultaneously rational (otherwise $ \pi$ would be rational).  They are known as [b]incommeasurable[/b] quantities because their ratio, $ \pi$, is irrational. [/hide]
[hide="4"] We defined irrational numbers for the same reason that we defined any other mathematical tool; because they are [b]useful tools[/b] to us.  Irrational numbers allow us to investigate geometric quantities that were previously lost to us, the roots of polynomials (this is actually related to the previous concern), probability distributions, differential equations (also part of the motivation for using imaginary numbers)... the list goes on and on.  The uses of the irrational numbers $ \pi$ and $ e$ alone would fill textbooks. [/hide]

[quote="aufha"]When I tried to prove $ \sqrt {2}$ was irrational, I started to feel that many imaginary numbers make much more sense than irrational numbers.[/quote]

What do you mean by this?  In any case, [hide="here is a simple proof"] Suppose by way of contradiction that $ \sqrt{2} \equal{} \frac{p}{q}$, where $ p, q$ are positive integers such that $ \gcd(p, q) \equal{} 1$.  Then

$ 2q^2 \equal{} p^2$

Now, define a function $ e_2(k)$ to be the greatest exponent $ e$ such that $ 2^e | k$.  Then

$ e_2(2q^2) \equal{} 1 \plus{} 2 e_2(q)$
$ e_2(p^2) \equal{} 2 e_2 (p)$

Since $ e_2(x)$ is odd for the LHS but even for the RHS, the two can never be equal, but they must be in order for the LHS and RHS to be equal; contradiction.  We conclude that $ \sqrt{2}$ is irrational. [/hide]

DO numbers exist?

 :huh:

Numbers were invented by people to help them better understand the world, like plenty of other math...

This is also true for irrationals, but for some reason they seem less tangible than numbers.
I understand that my feeling doesn't make any sense. Numbers are no more real than imaginary numbers. They are fuzzy, but useful, concepts that people invented like you said.

Also, how did you get from here:

Now, define a function $ e_2(k)$ to be the greatest exponent e such that $ 2^e | k$. Then

to here:

$ e_2(2q^2) \equal{} 1 \plus{} 2 e_2(q)$

What is LHS? What is RHS?

[quote="aufha"]Also, how did you get from here:

Now, define a function $ e_2(k)$ to be the greatest exponent e such that $ 2^e | k$. Then

to here:

$ e_2(2q^2) \equal{} 1 \plus{} 2 e_2(q)$[/quote]

$ e_2(k)$ gives the power of $ 2$ in the prime factorization of $ k$.  Generally,

$ k \equal{} 2^{e_2(k)} 3^{e_3(k)} 5^{e_5(k)} ...$

Now it should be clear that

$ 2q^2 \equal{} 2^{2e_2(q) \plus{} 1} 3^{2e_3(q)} 5^{2e_5(q)} ...$

Hence that $ e_2(2q^2) \equal{} 2e_2(q) \plus{} 1$.  Generally, $ e_p(k)$ has the property that $ e_p(ab) \equal{} e_p(a) e_p(b)$.

See [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=141316]an old blog post[/url] for further discussion.

Admittedly, this proof is usually phrased in a simpler way, but I prefer this direct method; in any case, they are equivalent.

[hide="The simpler proof"] Assume that $ \sqrt {2} \equal{} \frac {p}{q}$ where $ \gcd(p, q) \equal{} 1$.  Then

$ 2q^2 \equal{} p^2$

Now, $ 2 | p^2$, so $ 2 | p$.  Let $ p \equal{} 2r$.  Then

$ 2q^2 \equal{} 4r^2 \Leftrightarrow q^2 \equal{} 2r^2$

But then $ 2 | q^2$, so $ 2 | q$ and $ \gcd(p, q) \ge 2$; contradiction. [/hide]

[quote="aufha"]What is LHS? What is RHS?[/quote]

The Left Hand Side and Right Hand Side of the equation, respectively.

[quote="CircleSquared"]Notice that your function $ e_2(k)$ is equivalent to $ \log_2{k}$[/quote]

Only for powers of two.  (Please verify that what you are saying is correct before posting it.  You have made a lot of incorrect assertions in recent threads with unwarranted confidence.)  It satisfies similar properties, though.  In fact, the function $ e_p(k)$ gives rise to a rather interesting mathematical tool:  the [url=http://en.wikipedia.org/wiki/P-adic_number]p-adic numbers[/url].


A small mass, attached to the extremity of a spring, oscillates according to equation
\[ f(t)=8 \sin (3 \pi t),  \]
which represents the position of the mass in instant t seconds, measure in centimeters from the equilibrium position.
Counting from $t=0$, in which instant the mass will pass for the seventh time to a distance $|f(t)|$ of 4cm of the equilibrium position?

[quote="joml88"]This is Physics and belongs in the Physics forum rather than Intermediate. [color=red] Moved.[/color][/quote]

Excuse me, but this is about trigonometric functions. In Brazil we study that in the last year before entering the University. Any student in this year has the basic notion of simple harmonic movement. And only this notion is what you need to solve the problem. All the rest is trig.

[quote="daveed"]4 meters is half the amplitude...  and we know it's the first time it reaches it after 1.5 periods...
so... it's pretty much plug 'n chug[/quote]

I'm not sure if i got what you said, but period is $\displaystyle \frac{2}{3}$ and $1.5\cdot \frac{2}{3} = 1$.
And that is not the right [hide="answer."]$\frac{19}{18}$ seconds[/hide]

[quote="daveed"]I meant the first timeafter it completes 1.5 oscillations, meaning 1.5 periods plus whatever time it took to get from the origin to a distance of 4 meters.  That's when $sin(3 \pi t)=1/2$, so $3 \pi t=\pi/6$ so the extra $t$ is $\pi/18$ -- add that to the time spent going through the one and a half periods and you get the same answer..[/quote]

I think you meant $\sin(3 \pi t) = \frac{1}{2} \Rightarrow \sin(3 \pi t) = \sin \frac{\pi}{6} \Rightarrow t = \frac{1}{18} + \frac{2k}{3}$.

Hence, after $k = 1.5$ periods, $t = \frac{19}{18}$ seconds.  :first:

I was making it very complicated because i always thought about $k$ ($t = \frac{1}{18} + \frac{2k}{3}$) as being an integral.

yeah, that's what i meant.  When i was doing the problem I just waved my hand back and forth and counted how many times it'd pass the point, heh.

[quote="joml88"]This problem clearly belongs in the Physics forum.  Just because it involves math doesn't change the fact that the main concept of the problem is Physics related.  As further reasoning, consider how many times you've seen such a problem on a math competition.  Compare that to the number of times you've seen it on a physics competition.[/quote]

That's your point. Nevermind.

As further info, this question is one of the questions of the test of Mathematics for admission in a Brazilian University.


I want two proofs in L  (logic)

B->(BVC)  (-> is implies)

and

B->(CVB)

 :)

I'm willing to accept votes on where exactly this post should be sent.

Is there a cryptography forum on this site?

You should move this to High School Basics.
[hide="Solution"]
Implication isn't true if and only if LHS is true and RHS is true.
That means:
1) B-true and BvC-false - that's not possible
2) Similar
[/hide]
Now move this!

The author may want to say what kind of logic is allowed to be used...
Well, depending on that, either Basics or Advanced ;)

There are mor forum than the HighSchool ones ;) I meant http://www.mathlinks.ro/index.php?f=72 .
(For the reason that university logic is not suitable for any other forum; not for the reason that it's too complex)

By the way, if it _is_ some logical calculus: what axioms are you using¿


Let $ P$ be any point inside a regular polygon.
if $ d_{i}$ is the distance of $ P$ from the $ i^{th}$ side of the polygon, prove that
$ d_{1}\plus{}d_{2}\plus{}....\plus{}d_{n}$ is constant
where $ n$ is the number of sides of the polygon

Let $\Delta$ denote the area of the n-gon and $L$ the lentgh of its side. Then

$ \Delta \equal{} \frac{L ( d_1+d_2+d_3+ ... + d_n)}{2} \Longrightarrow \ d_1 \plus{} d_2 \plus{} d_3 \plus{} ... +d_n= \frac{2 \cdot \Delta}{L}=\text{const}$

Hey i think youre wrong
its not necessary that all the areas together will form the area of the ngon as the altitude can be on the produced side too!

somebody help :(  :o  :maybe:

now i understand
i dont know what happened to me that time...

thanx.


http://www.cartiaz.ro - A nice website where u can find a lot of math pdf books

There are many others such as gutenberg.org and esnips.com

[color=darkblue]Thanks jianuovidiu very much.[/color]

Please don't revive old threads just to say thank you. Thank you.

Hi can anyone get me a copy of Hardy et al, 's Inequalities?

http://www.math.upenn.edu/~wilf/

Herbert Wilf's page links to several of his books available free for download as well as to a Georgia Tech collection of free texts.

There's also numerous texts (mostly classic) on the Internet Archive. For example a search for G H Hardy gives [url]http://www.archive.org/search.php?query=creator%3A%28G%20H%20Hardy%29[/url].

The texts on inequalities and number theory by Hardy et al. are still very much in popular print and do not appear on that list.


a) Let $\{a_n\}_{n=1}^\infty$ is sequence of integers bigger than 1. Proove that if $x>0$ is irrational, then $\ds x_n>\frac{1}{a_{n+1}}$ for infinitely many $n$, where $x_n$ is fractional part of $a_na_{n-1}\dots a_1x$.

b)Find all sequences $\{a_n\}_{n=1}^\infty$ of positive integers, for which exist infinitely many $x\in(0,1)$ such that $\ds x_n>\frac{1}{a_{n+1}}$ for all $n$.

[i]Nikolai Nikolov, Emil Kolev[/i]

It's easy to see that : $a_1x=U_1+x_1 , a_2x_1 = U_2+x_2, ... , a_{n+1}x_n=U_{n+1}+x_{n+1},...$, where $0\leq x_n < 1$ and $U_n$ is an integer such that ${0 \leq U_n\leq a_n-1}$
(1) All $x_n$ are irrational because $x$ is irrational , so  $x_n>\frac{1}{a_{n+1}}$ or  $0<x_n<\frac{1}{a_{n+1}}$
$\exists m\forall n \geq m , \ x_n<\frac{1}{a_{n+1}}\Longrightarrow \forall n \geq m,$$\ U_n =0 \Longrightarrow \forall n \geq m ,\ x_m =\frac{x_{n+1} }{a_{m+1}...a_{n+1}}\leq \frac 1{2^{n-m+1}} \Longrightarrow x_m=0$, manifestly false !!
So $x_n>\frac{1}{a_{n+1}}$ for infinitely many $n$
(2)  Let $x\in(0,1)$ be such that that  $\forall n,\ x_n>\frac{1}{a_{n+1}}$, so $\forall n>1,\ 1 \leq U_n \leq a_n-1$
$x = \frac{U_1}{a_1}+\frac{x_1}{a_1} = \frac{U_1}{a_1}+\frac{U_2}{a_1a_2}+\frac{x_2}{a_1a_2}=...=\sum_{n=1}^{\infty}\frac{U_n}{a_1..a_n}$
So the asked sequences have a infinite sub-sequence $a_{p(n)}$ such that ${a_{p(n)} \geq 3}$
 :cool:

[quote="Diogene"]
So the asked sequences have a infinite sub-sequence $a_{p(n)}$ such that ${a_{p(n)}\geq 3}$
 :cool:[/quote]

What does this mean?

Yeah, what does it mean? Can someone tell?


Mr. Pillot always rides his bicycle to work, and he begins his ride at the same time every day. If he averages 10 miles per hour, he arrives at school 2 minutes late, but, if he averages 15 miles per hour, he arrives 1 minute early. How many miles does Mr. Pillot ride to work? Express your answer as a decimal to the nearest tenth.

$ 10(x\plus{}2)\equal{}15(x\minus{}1) \implies 10x\plus{}20\equal{}15x\minus{}15 \implies x\equal{}7$

so $ 10*\frac{9}{60}$ or $ 15*\frac{6}{60}\equal{}\frac{3}{2}$

answer: 1.5 miles


Prove that,  $(a+2b,b+2a)$ is $1$ or $3$

It is obvioslly $(a+2b,b+2a)=(a+2b,3b)=k(a,b)$, were k=1 or 3.


A wall clock loses 2 minutes in an hour. A table clock gets 2 minutes ahead of the wall clock in an hour. An alarm clock falls 2 minutes behind the table clock in an hour. A wristwatch gets 2 minutes ahead of the alarm clock in an hour.
At noon all four timepieces were set correctly. To the nearest minute, what time will the wristwatch show when the correct time is 7:00 p.m. ?

[size=75]
from book "The Moscow puzzles" by Boris A. Kordemsky

Please hide Your solution . Write inside Hide ... /Hide
[/size]

Please no more watch problems.....

[hide="Solutions"]In one hour, the wall clock will tick away 58 minutes. While the wall clock is ticking down, the table clock ticks away 62 minutes. In 58 minutes, by the wall clock, the table clock ticks away $ 58\cdot\frac{62}{60}$ minutes. And in this time, the alarm clock ticks away $ 58\cdot\frac{62}{60}\cdot\frac{58}{60}$, and in this time, the wristwatch ticks away $ 58\cdot\frac{62}{60}\cdot\frac{58}{60}\cdot\frac{62}{60}$ which is approximately 60 minutes. The wristwatch loses about 1 minute per 7 hours, so at 7, the wristwatch will show 6:59.[/hide]


[quote="sen"]For $a, b, c > 0$, prove that:
\[ a^ab^bc^c \geq (abc)^{\frac{a+b+c}{3}} \][/quote]
It's shown easily by taking log and using Chebyshev's inequality.

http://www.mathlinks.ro/Forum/viewtopic.php?t=82706 (the first posts) :)

[quote="Lovasz"]Well, mathmanman, I cannot understand why many members of this forum, like you, darij, and etc,  always remember the links of problems which were posted, I knew you use "search function", but I'm impossible...[/quote]
You will be able to do that as well if you have stayed long enough in this forum.

[quote="sen"]For $a, b, c > 0$
Prove that

$a^ab^bc^c \geq (abc)^{\frac{a+b+c}{3}}$[/quote]

instead of using natural log ln or AM-GM, here's a simpler way:

reaarange the equation to get $a^{2a}b^{2b}c^{2c} \geq a^ba^cb^ab^cc^ac^b$

assume $a \geq b \geq c$

then ${(\frac{a}{b})}^{a-b} \geq 1$

rearranging that, you'll get $a^ab^b \geq a^bb^a$
the same goes for ${(\frac{b}{c})}^{b-c} \geq 1$
and ${(\frac{a}{c})}^{a-c} \geq 1$

multiply the three results together, and you'll get $a^{2a}b^{2b}c^{2c} \geq a^ba^cb^ab^cc^ac^b$

 :)


trigonometry
inequalities
trig identities
Law of Cosines

[quote="frt"][quote="shobber"][quote="frt"]If $x^2+x+1$, $-2x-1$, and $x^2+2x$ are 3 sides of a triangle, find the measure, in degrees, of the largest angle.

(Don't just plug in numbers! :) )[/quote]
[hide]
Law of cosine: $\cos{\theta}=-\frac{(x^4+4x^3+4x^2)+(4x^2+4x+1)-(x^4+2x^3+3x^2+2x+1)}{2x(x+2)(2x+1)}=-\frac{2x^2+5x+2}{2(2x^2+5x+2)}=-\frac{1}{2}$
So $\theta = 120^o$[/hide][/quote]
Well, your answer is correct, but how do you know that $x^2+x+1$ is the longest side?
(You assumed that the angle that subtends to the side $x^2+x+1$ is the largest angle.)[/quote]

You don't have to assume that $x^2+x+1$ is the longest side in his solution.  You just use the Law of Cosines and it happened that one of the angles turned out to be $120^\circ$ and thus must have been the largest.

[quote="joml88"]You don't have to assume that $x^2+x+1$ is the longest side in his solution.  You just use the Law of Cosines and it happened that one of the angles turned out to be $120^\circ$ and thus must have been the largest.[/quote]
Ok, well, here's the reason why $x^2+x+1$ has to be the longest side :)
[hide]Since $x^2+x+1$, $-2x-1$, and $x^2+2x$ are the three sides of the triangle, we must have these inequalities:

$(x^2+x+1)+(-2x-1)>x^2+2x \Longleftrightarrow -3x>0 \Longleftrightarrow x<0$
$(-2x-1)+(x^2+2x)>x^2+x+1 \Longleftrightarrow -x-2>0 \Longleftrightarrow x<-2$
$(x^2+x+1)+(x^2+2x)>-2x-1 \Longleftrightarrow 2x^2+5x+2>0$
$\Longleftrightarrow x<-2$, $-\frac{1}{2}<x$
From these, we obtain $x<-2$.

Then, we see
$(x^2+x+1)-(-2x-1)=(x+1)(x+2)>0$ and
$(x^2+x+1)-(x^2+2x)=-x+1>0$

So $x^2+x+1$ must be the longest side.
[/hide]

[quote="frt"]Ok, well, here's the reason why  has to be the longest side  [/quote]

joml88's point is that it doesn't matter - you can just try the three sides in certain orders in the Law of Cosines. He got 120, which ends up being the largest angle.


I know that they say that you should have gotten at least a 5 on the AIME, but I have never taken the AIME (or the AMC for that matter) and I think I may want to take WOOT however I don't know if I would be ready.  Are there any other ways to tell if I would be ready for this?

Try taking a previous year's AIME in your spare time.  They're up at the 'Contests' section at the top.

Obviously, taking it by yourself with no pressure, even if you set the right time limit, inflates your score somewhat.  Take from a few different years to get a better idea of how ready you are.


or all positive integers n and any distinct real numbers x and y, x-y divides x^n-y^n

[hide]My approach:
Using binomial theorem we have: x^n-y^n=(x-y)(x^(n-1)+x^(n-2).y+....+y^(n-1)).
Since x#y, so we easily see that x^n-y^n is divisible by (x-y) for all positive integers n(Q.E.D)[/hide]

That's not the binomial theorem.  You're just either dividing polynomials or using the sum of a geometric series.

One way to do the inductive proof is to compute $ (x \minus{} y)(x^n \plus{} y^n)$ and assume that $ x \minus{} y | x^{n\minus{}1} \minus{} y^{n\minus{}1}$.  What can you conclude?


$ x,y$ are integers.
Solve $ y(x^2\plus{}36)\plus{}x(y^2\minus{}36)\plus{}y^2(y\minus{}12)\equal{}0$.

easy but nice..just solve in x as a quadratic equation and find y for which the discriminant is a perfect square!! :wink:


advanced fields
advanced fields unsolved

Let $ C_{p,q}$ be a bicomplex ($ 0 \leq p,q$). Suppose that spectral sequence of this bicomplex (there are actually two, but it probably doesn't matter which one you take?) has a property that $ E^2 _{s,t} \equal{} 0$ for $ 0 < t < n$. Let $ T_{*}$ be a total complex for bicomplex $ C_{p,q}$. Show that there is an exact sequence:

$ H_{n \plus{} 1} (T_{*}) \rightarrow E^2 _{n \plus{} 1,0} \rightarrow E^2 _{0,n} \rightarrow H_{n} (T_{*}) \rightarrow E^2 _{n,0} \rightarrow 0$

Are you sure the directions of the arrows aren't reversed¿ Then it would be the standard five term exact sequence.

Yes, I'm sure it's okay. I have seen a solution today.

How do you even get a map $ E_{n \plus{} 1,0}^2 \to E_{0,n}^2$¿

Ok, also by the way you write the indices, you seem to use some kind of homology (not cohomology) of the complex. How did you define it¿

Ah, that's why the confusion arose: you have spectral sequences in both homological and cohomological situation. The one considered above is a 'homological' s.s  (differentials on level n are of degree (-n,n-1); hence a map from (n+1,0) to (0,n) ). And if you treat this s.s. 'cohomologically' you should get the same diagram but with arrows reversed as you said at the beginning.


geometry
AMC
AIME
ARML
analytic geometry
geometric transformation
rotation

What are everyone's favorite obscure theorems that somehow prove to be extremely helpful on many math competitions like the AMC/AIME? Post theorems that tend to reduce otherwise very difficult questions to trivial answers.

Two of my favorites are Ceva's/Menelaus' Theorem. These are very useful on many of those triangle geometry problems.

http://jwilson.coe.uga.edu/Texts.Folder/ratio/Ceva.html

http://www.ies.co.jp/math/java/vector/menela/menela.html

...Except Ceva's and Menelaus' aren't that obscure.  :mrgreen: It would be a good idea to learn them for the AIME and above.

This isn't a theorem, but the technique that I have found most useful on the AIME is simply assigning variables appropriately.  Sure, there is a lot of knowledge, understanding, and experience associated with assigning variables usefully, but much of the knowledge and understanding need can be summed up in a few simple principles.

If it's big and ugly, give it a nice small name, like x.  Or Bob.  But x will do nicely.

If it's repeated over and over in slightly different forms, see if it can be summed up as xi with a definition of xi that covers all the different forms.

If it can be expressed in terms of something else, name it or the something else and see how they relate.  Many unkowns are solved for by expressing a quantity in terms of a variable in two different ways and then equation both expressions.

If you're after the value of some expression involving multiple variables, see if replacing that entire expression with one variable makes it easier (this can really be summed up with the first suggestion, but it gives a different hint as to the possible applications).

Well Fiery, I'm sure plenty of people on the board know of Ceva's/Menelaus'. And obviously, it's not THAT obscure, or else I wouldn't know what it :).

However, ask a random high school student out of geometry what Ceva's Theorem or Menelaus' Theorem is, and I will give you 50-1 odds that the student will have no idea what you are talking about. These are the type of useful competition-style theorems that I'm thinking about.

Oh yeah, and techniques like MCrawford's variable assigning are good too :).

[quote]However, ask a random high school student out of geometry what Ceva's Theorem or Menelaus' Theorem is, and I will give you 50-1 odds that the student will have no idea what you are talking about. These are the type of useful competition-style theorems that I'm thinking about. [/quote]

Oh. Then the Power of a Point Theorem if you don't know that yet.

Power of a Point is a really great theorem, although it's taught in the standard geometry curriculum, is it not?

I'll have to think about a theorem I particularly like.

I particularly enjoy Pick's Theorem. While it isnt as useful as Ceva's or [b]STEWART'S THEOREM[/b] (notice the bold letters), it is a very surprising one that really makes analytical geom a lot more appealing on some problems.

Also, just for sheer entertainment (no AIME value), the pancake theorem is awesome. I mean, really, the pancake theorem?! that is just so great.

[quote]Power of a Point is a really great theorem, although it's taught in the standard geometry curriculum, is it not?[/quote]

Not in most I think (certainly not in the one I took). Neither are Heron's and Brahmagupta's. But Stuyvesant's math classes include a bit more content than the average class I think.

[quote="JBL"]Power of a Point is a really great theorem, although it's taught in the standard geometry curriculum, is it not?[/quote]

Amazingly enough to all you problem solvers, NO, the power of a point theorem is NOT taught as a unified theorem in many treatments of high school mathematics. Instead, you get several special cases labeled as separate theorems, such as the treatment in the Jurgensen-Brown-Jurgensen textbook section 9-7, with theorem 9-11 (two chords intersecting inside a circle), theorem 9-12 (two secant segments extended to an external point), and theorem 9-13 (secant segment and tangent segment to external point). There is nary an attempt to unify the "separate" theorems in this text.

Thus I would definitely label power of a point an obscure theorem for purposes of this discussion.

[quote="Fierytycoon"]Not in most I think (certainly not in the one I took). Neither are Heron's and Brahmagupta's. But Stuyvesant's math classes include a bit more content than the average class I think.[/quote]

I actually learned geometry in my middle school, where I'm fairly certain we learned power of a point.  It was definitely the first thing that we taught in Freshman math team, though.

[quote="mathfanatic"]There's also the [url=http://mathworld.wolfram.com/HamSandwichTheorem.html]Ham Sandwich Theorem[/url]...[/quote]

Yeah, I love that theorem!

One of my favourites is Lucas' theorem, which says to find the remainder of the binomial coefficient (aCb) when divided by p, write a in base p (amam-1...a0), write b in base p (same thing), then the remainder is amCbm * ... * a0Cb0

In other words, 135C30 mod 7: 135 is 252 in base 7, and 30 is 042 in base 7,  so its 2C0 * 5C4 * 2C2 = 5 mod 7. (if any of the top numbers were smaller, it would be 0, thus divisible by that prime)..

I don't know if this is a theorem and it's probably too easy for AIME but it's for questions involving two relatively prime integers and you want to find the greatest number not attainable w/ the 2 integers. It's ab-a-b

Webhappy, I don't understand what you're trying to say.

And TripleM, could you elaborate? I'm a bit lost, and the Mathworld definition is anything but simple.

Here's an example of a problem that the theorem webhappy mentioned could be used for: (from the first Mock AMC)

12. Bob is throwing turtles at a dartboard with two regions.  If he hits the smaller region, he gets 15 points; if he hits the larger region, he gets 7 points.  What is the largest score that Bob cannot get?
(A) 67
(B) 79
(C) 83
(D) 89
(E) 97

Isn't Apollonius' theorem a special case of Stewart's Theorem?

Erm, mathworld says that they are different names for the same thing.  If you told me there was something more often known as Apollonius' theorem that was a special case of Stewart's, I would believe you.

i read somewhere that apollonius' theorem only gives the length of the median... Stuarts gives the length of any cevian

Woah, so it does. Yep, from other sources, Apollonius is when you take the median.

On a completely offtopic note - does anyone notice something a tiny bit unusual about this page?
http://mathworld.wolfram.com/search/?as_lq=http://mathworld.wolfram.com/ApolloniusTheorem.html

ya, it says results 1 through 1 of 2... there's only 1 result, not 2...

you're right, that was off topic

Okay, I take it back.  If the odd thing you mean is that the search for Apollonius' theorem gives you Stewart's theorem, that's because the Stewart entry mentions Apollonius, and all the Apollonius entry does is link to Stewart.  You might e-mail the head of the website and tell him what it should say for Apollonius.

Oh god I love this theorem.

If you have a triangle ABC, and choose a point O inside the triangle such that angles ABO, CBO, and ACO are congruent with measure x, and a,b,c, are the angles or the triangle, then cot a + cot b + cot c = cot x.

The proof I derived it from was long and complicated, so I'm going to omit it here.

Are you sure you've labelled your angles correctly?  I would expect it to be something more like angles ABO, BCO, and CAO, not the three angles you stated.  (I just want to make sure because I'm lazy and don't care to do proving things that aren't true :))

Hmm. Going back to the ab-a-b thm.

Does anyone know a generalization? i.e. whats the biggest value that cannot be obtained by aX + bY + cZ using positive a,b,c?

I think that's been discussed on these forums several times before, and I believe that general consensus is that there isn't a nice generalization, although I might vaguely remember Ravi B. posting something about there being a not-so-nice generalization.  Try a search of the website.

Yes, the bottom line is that there is an efficient algorithm for 3 variables, but not a formula.  See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9993[/url].

Yep. But remember that any *particular* case is reasonably easy to solve using the method I showed before, with a bit of guess and check. Its the general case that its hard for.

Oh yes; don't we also like:

If A, B, C are three angles of a triangle, then

tan A + tan B + tan C = (tanA)(tanB)(tanC)

...priceless.

yeah those trig identities.  could save your life sometimes if you know them.  ;)


Find the number of those subsets of {1, 2, 3, ..., 2n} in which the equation

x + y = 2n + 1

has no solutions.

[i]Poland Mathematical Olympiad 1995[/i]

Consider the n pairs (x,2n+1-x), for x=1,2,..,n.. For each pair, we have 3 possibilities : both aren't in the set,only x is in the set,only 2n+1-x is in the set. Thus 3^n possibilities.

I didn't know you are a beginner, Andrei :D

I pretend not to be one :D

Will some moderator delete these posts please? I didn't see that it's the beginner section ;)

ya well for some reason I also get cross-linked from times to times to the beginners section here :?

well, at least then I feel smart until I discover it :D


geometry
function
algebra proposed
algebra

Find all functions $f: \mathbb Z \to \mathbb Z$ for which
\[ f(m+f(f(n))=-f(f(m+1))-n\]
holds for all integers $m$ and $n$.

[quote="MJ GEO"]$ f$ from $ Z$ to $ Z$ and $ f(m \plus{} f(f(n)) \equal{} \minus{} f(f(m \plus{} 1)) \minus{} n$[/quote]

Let $ P(x,y)$ be the assertion $ f(x\plus{}f(f(y))\equal{}\minus{}f(f(x\plus{}1))\minus{}y$

$ P(0,x)$ $ \implies$ $ f(f(f(x)))\equal{}\minus{}x\minus{}f(f(1))$ and so $ f(x)$ is a bijection and so is $ f(f(x))$ and $ \exists a$ such that $ f(f(a))\equal{}1$

Let $ x\in\mathbb Z$. Since $ f(x)$ is a bijection, $ \exists y$ such that $ f(y)\equal{}x$. Then : $ P(y\minus{}1,a)$ $ \implies$ $ x\equal{}\minus{}f(x)\minus{}a$ and so $ f(x)\equal{}\minus{}x\minus{}a$

Plugging this back in the original equation, we get $ a\equal{}1$ and the unique solution : $ \boxed{f(n)\equal{}\minus{}(n\plus{}1)}$ $ \forall n\in\mathbb Z$

why there is and a that $ f(f(a))\equal{}1$

[quote="MJ GEO"]why there is and a that $ f(f(a)) \equal{} 1$[/quote]

Because $ f(x)$ is a bijection, and so $ f(f(x))$ is also a bijection, and so is a surjection.

[quote="MJ GEO"]now i confused becasue of geometry problem and cant understand you[/quote]
Let $ P(x,y)$ be the assertion $ f(x \plus{} f(f(y)) \equal{} \minus{} f(f(x \plus{} 1)) \minus{} y$

In a more detailed way :

$ P(0,x)$ $ \implies$ $ f(f(f(x))) \equal{} \minus{} x \minus{} f(f(1))$  So :
1) $ f(x)$ is injective:
$ f(a)\equal{}f(b)$ $ \implies$ $ f(f(f(a)))\equal{}f(f(f(b)))$ $ \implies$ $ \minus{} a \minus{} f(f(1))\equal{}\minus{} b \minus{} f(f(1))$ $ \implies$ $ a\equal{}b$ and so $ f(x)$ is injective.

2) $ f(x)$ is surjective :
Using $ x\equal{}\minus{}u\minus{}f(f(1))$ in $ f(f(f(x))) \equal{} \minus{} x \minus{} f(f(1))$  , we get $ f(f(f(\minus{}u\minus{}f(f(f1))))) \equal{} u$ and so $ \forall u$ $ \exists v\equal{}f(f(\minus{}u\minus{}f(f(f1))))$ such that $ f(v)\equal{}u$ and so $ f(x)$ is a surjection

and so $ f(x)$ is a bijection
Then $ f(x)$ bijection $ \implies$ $ f(f(x))$ is too a bijection.

Since $ f(f(x))$ is a bijection, it's a surjective function and so $ \exists a$ such that $ f(f(a)) \equal{} 1$

Is it OK ?

And the end now :
Let $ x\in\mathbb Z$. Since $ f(x)$ is a bijection, $ \exists y$ such that $ f(y) \equal{} x$. Then : $ P(y \minus{} 1,a)$ $ \implies$ $ x \equal{} \minus{} f(x) \minus{} a$ and so $ f(x) \equal{} \minus{} x \minus{} a$

Plugging this back in the original equation, we get $ a \equal{} 1$ and the unique solution : $ \boxed{f(n) \equal{} \minus{} (n \plus{} 1)}$ $ \forall n\in\mathbb Z$

What was your own solution ?


Hello,

Let $ a,b \in \mathbb{N}\:$ and  $ \: P(x)\equal{} x^2 \plus{} ax \plus{} b$

Prove that : $ \;\forall n\in\mathbb{N},\exists m\in\mathbb{N} \;/ \;p(n)p(n\plus{}1) \equal{} p(m)$

[hide]$ p(n)p(n\plus{}1) \equal{} p(b \plus{} n \plus{} a n \plus{} n^2)$[/hide]

thanks you ,SimonM

but who are you find this ? you can give me the proof


AMC
USA(J)MO
USAMO
calculus
Ross Mathematics Program

Hi - on the AMC site, they have some pamphlets (booklets? books?) by Samuel Greitzer which says they contain problems for gifted students but doesn't state what age group or level of problems -- I mean, there are problems for gifted first graders and problems for gifted USAMO members, but clearly, one volume is not appropriate for another. Has anyone ever seen or used any of these? Are they worthwhile problems? How challenging? For what age group and/or level of math - calculus? Algebra? etc. - unless I was looking at the wrong place, I couldn't find much info on the site.  Just that there were several volumes ...  If anyone purchased, can you please post here?

This summer (2009) we sold the last of the Arbelos volumes.  We do not have any more, and have no plans to reprint them.  When we get some other important projects finished, we will remove the references from our web page.

The level of the material in the Arbelos was at about the upper AIME/beginning USAMO level.  A lot of the content was a mixture of mathematical history, opinions from Sam Greitzer, and mathematical curiosities.  The content and style of the USAMO has changed from the 1980s when Greitzer was the USAMO chair and USA IMO Leader.  In my opinion, the Arbelos had become interesting reading, but not satisfactory for "training", but possibly useful for "gifted enrichment"

If you are looking for comparable gifted enrichment material, I recommend any of the books by Ross Honsberger available from the MAA Bookstore or large online booksellers.  If you are looking for a periodical that is similar, I recommend Math Horizons, from the MAA.  See http://www.maa.org for both.

-- Steve Dunbar
MAA Director, American Mathematics Competitions.


Ring Theory
superior algebra
superior algebra unsolved

How to prove that in a Dedekind ring $ A$, for all ideal $ K$ of $ A$ there exists an integer $ n$ such that $ K^n$ is principal ? thanks for advance.

I think this is false. The class group is not necessarily torsion.

I have probably made a false generalization of this problem : Prove the principal ideal theorem  : for any algebraic number field $ K$ and any ideal $ I$ of the ring of integers of $ K$, if $ L$ is the Hilbert class field of $ K$, then $ IO_L$  is a principal ideal $ \alpha O_L$, for $ O_L$ the ring of integers of $ L$ and some element $ \alpha$ in it.

I don't get how you reduced/generalised this to that problem. The principal ideal theorem is quite hard as it normally needs Furtwaengler's theorem. The rest is a bit of work (with Verlagerung / transfer maps), but quite standard.


summer program
MathPath
articles
search
geometry
similar triangles

Words of the Week this week are [[proportion]] and [[MathPath]].

Please make additions to these pages!  We're trying a bit of an experiment this week -- both of these are nearly blank slates to work with.

[quote="Temperal"]Proportion was completely empty.  :maybe:

Anyways, may I edit the template?[/quote]

Template for the Problem of the Week?  Sure.

[quote="Temperal"]A question: do the sample problems for a math term have to be AMC->IMO chain questions from the wiki, or can we make our own up?[/quote]

I tend to like to pick questions from the AMCs because I can just > WhatLinksHere > copy+paste > save (not even necessarily on the AMC -> IMO chain); as long as the problems are appropriate for the article in question I can't imagine there being any problems.

[quote="Temperal"]Hmm... are there any good AMC-IMO problems involving proportions? I search brings up nothing:
http://www.artofproblemsolving.com/Wiki/index.php/Special:Search?search=proportion&fulltext=Search[/quote]

:lol: I'm tempted to answer all of them, since proportions are such a "basic" concept in the sense of foundation (just like trying to find example problems for addition that are suitable for AoPSers).

Since I sincerely hope that all AoPSers know what proportions are, I don't believe we need to overkill the section with too many problems.  However, we can probably just pull a couple of examples from various fields (similar triangles etc.) to fill the section.

[quote="Temperal"]Hmm... are there any good AMC-IMO problems involving proportions? I search brings up nothing:
http://www.artofproblemsolving.com/Wiki/index.php/Special:Search?search=proportion&fulltext=Search[/quote]Dunno about AMC, but I definitely think there are none at the IMO :)


geometry
rectangle
ratio
analytic geometry
number theory
relatively prime
similar triangles

I have a few problems my team and I don't know how to do... :blush:

1. Rectangle ABCD has points E and F on sides AB and CD, respectively. If AE = $ \frac{1}{3} AB$ and CF = $ \frac{1}{4}$ CD and segments DE and BF intersect diagonal AC at G and H, respectively, what is the ratio AG:GH:HC? Express your answer in the form $ a: b: c$, where a, b and c are relatively prime positive integers.

2. A circle with radius 1 unit lies in the first quadrant and is tangent to both the x- and y-axes. A second larger circle lies in the first quadrant, is tangent to both axes and is externally tangent to the first circle. What is the radius of the second circle? Express your answer in simplest radical form.

3. On an 8 by 8 grid of unit squares, a red marker starts in the unit square called (3,1), which is the unit square in the third row and first column. Each separate move of the red marker is to a neighboring unit square horizontally, vertically or diagonally. How many paths of exactly four moves are there from (3,1) to (4,5)?

ummm... Bash?  :huh:

I don't get it...  :(

Just brute force this problem with the many similar triangles that are present. :wink:

I think you're right, mind explaining your method?  :lol:

[quote="ThinkFlow"]I got 5:11:4. :|[/quote]

Same, $ 5: 11: 4$ by coordinate bashing it.

Put in coordinates for A, B, C, D, E, and F.  Using equations, figure out G and H.  The use the distance formula to find the distances, which were $ \sqrt{18}: \sqrt{87.12}: \sqrt{11.52}$.  Square is and divide by $ 72$ to get $ 25: 121: 16$, and now square root it for the answer of $ \boxed{5: 11: 4}$

You can similar triangle bash it by using AEG similar to GDC, etc.

ohhh... okay! I get it now! thanx everyone!  :)

I think we all forgot about the other questions...

I already asked number 2 but how did you guys do number 3?

[quote="TheCoffeeMonster"]

3. On an 8 by 8 grid of unit squares, a red marker starts in the unit square called (3,1), which is the unit square in the third row and first column. Each separate move of the red marker is to a neighboring unit square horizontally, vertically or diagonally. How many paths of exactly four moves are there from (3,1) to (4,5)?[/quote]

Staying inside that area, we have $ \binom{4}{3} \equal{} 4$ ways.  Now the number of ways to go out is:

DDDU
DUDD
UDDD
DUDD
DDDU
UDDD

Note: The D can mean up-right or up-left.  The U means up.

There are then $ 4 \plus{} 6 \equal{} \boxed{10}$ ways.

[b]Edit:[/b] Forgot the zig-zags inside the rectangle.

Thanks Luminescence.

For 3, I just drew a 5x2 diagram. I realized that I couldn't go down, only diagonally and horizontally, in order to fit into 4 moves. I had to go once diagonally, and I had four choices for where to do that.
But then I realized that I could go in zigzags, which gave me two more options. Also, I could go diagonally out of my diagram, then back in. That gave me 4 more choices (two going to row 5, two to row 2), for a total of ten.

Sorry, this is really hard to explain without a visual aid, and I haven't figured out how to insert those yet. So if someone could do that, that would be great. Meanwhile, did anyone else get ten?

EDIT: Forget this - until there's a visual aide up, 007's explanation is much more helpful.


geometry
3D geometry
real analysis
real analysis solved

suppose that we have countable set A of balls and unit cube in$R^3$.for every B  whichis finit subset of A,it is possible to put the balls of Binside the cube in such a way all their interior are disjoint, prove that it's possible for A too.

You mean to say that it's possible to put them [b]inside the cube[/b] s.t. their interiors are disjoint?

Does a compacity argument (in the 'logical' sense) does not work?

Pierre.

I would not state the compactness principle in all its generality because I do not know clearly about it (the general case use the axiom of choice. Here since the set of balls is countable, we may prove it without using AC). In another hand, I can state its graph theoretic approach :

Let $r$ be given.
Let $H=(V,E)$ be an hypergraph where all hyperedges are finite (but $V$ need not to be).
Suppose that for all finite $W \subseteq V$ the restriction of $H$ to the set of vertices $W$ has chromatic number not greater than $r$, then $H$ itself has chromatic number not greater than $r$.

Let's consider the graph whose set of vertices is $A$, two connected by an edge if and only if they have nondisjoint interior when we put all the balls into the unit cube (it is unclear if we are allowed to deplace the balls or if the set $A$ is given with no possible 'deformations', but it is worthless).
From the hypothesis, for all finite subset $B$ the restriction of the graph to $B$ has chromatic number 1.
From the compactness principle the whole graph has chromatic number 1 too (it cannot be less than 1  :D ). It means that there is no edge and that we may put all the balls into the unit cube.

Pierre.

I was looking through some old Jarnik problems, I saw this, and I remembered it had already been posted. I see Pierre has already posted a proof, but this is what I did:

Let $\mathcal A_n,\ n\ge 1$ be an arrangement of the first $n$ balls, $B_1(O_1^n,r_1),\ldots,B_n(O_n^n,r_n)$ inside the cube, where the exponents of the $O$'s indicates the index of the arrangement.

From the sequence $(O_1^n)_n$ we can extract a subsequence converging to $O_1$. From the subsequence of $(O_2^n)_n$ indexed by the elements of the previous subsequence, we can extract a subsequence converging to $O_2$, and so on. We can now simply place the ball with index $n$ with its center in the point $O_n$, and I think this does it.


let $ P\equal{}1$x$ 2...$ x$ n$ and  $ S\equal{}1\plus{}2\plus{}...\plus{}n$
prove that : $ S$ divide $ P$  <=>  $ n\plus{}1$ isn't an odd prime number.

[quote="ghostchungho"]let $ P \equal{} 1$x$ 2...$ x$ n$ and  $ S \equal{} 1 \plus{} 2 \plus{} ... \plus{} n$
prove that : $ S$ divide $ P$  <=>  $ n \plus{} 1$ isn't an odd prime number.[/quote]
This is equivalent to $ n(n\plus{}1) \mid 2 n! \iff n\plus{}1 \mid 2(n\minus{}1)!$
If $ n\plus{}1$ is an odd prime => $ S$ doesn't divide $ P$.
For an abritary prime let $ p^a \mid n\plus{}1$ and $ p^{a\plus{}1} \nmid n\plus{}1$. Then it suffices to prove that $ p^a \mid 2(n\minus{}1)!$
If $ a > 1$ we just prove that $ a \le e_p(2(n\minus{}1)!)$ ($ e_p(n)$ means the number of times $ p$ divides $ n$)
For $ p \equal{} 2$ we have $ a \le 1 \plus{} [(2^a\minus{}1)/2] \plus{} [(2^a\minus{}1)/4] \plus{} ... \le [2 \plus{} [n\minus{}1/2] \plus{} [n\minus{}1/4] \plus{} ...$ (Let $ [x]$ be the integer part of $ x$)
for $ p \ge 3$ it is obvious that $ a \le [(n\minus{}1)/2] \plus{} [(n\minus{}1)/4] \plus{} ...$

For $ a \equal{} 1$ we know that we can write $ n\plus{}1 \equal{} kj$ with $ (k,j) \equal{} 1$ and thus $ k \mid 2(n\minus{}1)!, j \mid 2(n\minus{}1)! \Rightarrow n \mid 2(n\minus{}1)!$

Done :D