topology
Functional Analysis
advanced fields
advanced fields unsolved

[b]Definition.[/b] Von Neumann algebra or W*-algebra is a *-algebra of bounded operators on a Hilbert space that is closed in the weak operator topology and contains the identity operator.

http://en.wikipedia.org/wiki/Von_Neuman_algebra
http://en.wikipedia.org/wiki/Weak_operator_topology

[b]Definition.[/b] For $ S\subset B(\mathcal{H})$ denote $ S'\equal{}\{T\in B(\mathcal{H}): Ts\equal{}sT\ \mathrm{for\ every}\ s\in S\}$.

[b]Problem.[/b] Let $ \mathfrak{a}_1,\ \mathfrak{a}_2$ be Von Neumann algebras in $ B(\mathcal{H})$. Show that

\[ (\mathfrak{a}_1\cup \mathfrak{a}_2)'\equal{}\mathfrak{a}_1'\cap\mathfrak{a}_2'\]

\[ (\mathfrak{a}_1\cap\mathfrak{a}_2)'\equal{}(\mathfrak{a}_1'\cup\mathfrak{a}_2')''\]

the first equation is trivial, and the second follows by applying the commutant ' to the first equation.

That was really easy :oops: ... Now I can't understand what was my problem :oops: ... Maybe someone can use a solution so here it is:

1)
\[ (\mathfrak{a}_1\cup\mathfrak{a}_2)'\equal{}\{T\in B(\mathcal{H}): Ts\equal{}sT\ \forall s\in\mathfrak{a}_1\cup\mathfrak{a}_2\}\equal{}\]
\[ \equal{}\{T\in B(\mathcal{H}): Ts_1\equal{}s_1T\ \wedge\ Ts_2\equal{}s_2T\ \forall s_1\in\mathfrak{a}_1\ \forall s_2\in\mathfrak{a}_2\}\equal{}\]
\[ \equal{}\{T\in B(\mathcal{H}): Ts\equal{}sT\ \forall s\in\mathfrak{a}_1\}\cap\{T\in B(\mathcal{H}): Ts_2\equal{}s_2T\ \forall s_2\in\mathfrak{a}_2\}\equal{}\]
\[ \equal{}\mathfrak{a}_1'\cap\mathfrak{a}_2'.\]

2) From 1) we have $ (\mathfrak{a}_1\cup\mathfrak{a}_2)'\equal{}\mathfrak{a}_1'\cap\mathfrak{a}_2'$. Let us put commutant to both sides. We obtain
\[ (\mathfrak{a}_1\cup\mathfrak{a}_2)''\equal{}(\mathfrak{a}_1'\cap\mathfrak{a}_2')'.\]
Now lets substitute $ \mathfrak{a}_1$ by $ \mathfrak{a}_1'$ and $ \mathfrak{a}_2$ by $ \mathfrak{a}_2'$ what gives us
\[ (\mathfrak{a}_1'\cup\mathfrak{a}_2')''\equal{}(\mathfrak{a}_1''\cap\mathfrak{a}_2'')'\equal{}(\mathfrak{a}_1\cap\mathfrak{a}_2)'.\]
The last equality holds because $ \mathfrak{a}_1$ and $ \mathfrak{a}_1$ are Von Neumann algebras.

Thanks -oo-


Let $ n$ be a positive integer. Prove that there are $ n.n!$ nonnegative integers $ a_{1,1},a_{1,2},\dots,a_{1,n},a_{2,1},a_{2,2},\dots,a_{2,n},\dots,a_{n!,1},a_{n!,2},\dots,a_{n!,n}$ such that  for all $ 1\leq j \leq n!$, $ \sum_{i\equal{}1}^n a_{j,i}\equal{}n$ and for all $ 1\leq k\leq n$, $ \sum_{i\equal{}1}^{n!} a_{i,k}$.

[quote="bakerbakura"]Let $ n$ be a positive integer. Prove that there are $ n.n!$ nonnegative integers $ a_{1,1},a_{1,2},\dots,a_{1,n},a_{2,1},a_{2,2},\dots,a_{2,n},\dots,a_{n!,1},a_{n!,2},\dots,a_{n!,n}$ such that  for all $ 1\leq j \leq n!$, $ \sum_{i \equal{} 1}^n a_{j,i} \equal{} n$ and for all $ 1\leq k\leq n$, $ \sum_{i \equal{} 1}^{n!} a_{i,k}$.[/quote]I'm curious to see what is missing after the last sum.  :maybe:

Sorry  :blush:

The two sums are supposed to be $ \sum_{i \equal{} 1}^{n} ia_{j,i}\equal{} n$ for all $ 1\leq j\leq n!$ and $ \sum_{i \equal{} 1}^{n!}a_{i,k} \equal{}\frac{n!}{k}$ for all $ 1\leq k\leq n$.

[quote="bakerbakura"]Sorry  :blush:

The two sums are supposed to be $ \sum_{i \equal{} 1}^{n} ia_{j,i} \equal{} n$ for all $ 1\leq j\leq n!$ and $ \sum_{i \equal{} 1}^{n!}a_{i,k} \equal{} \frac {n!}{k}$ for all $ 1\leq k\leq n$.[/quote]
but that's very easy. Take the following lines:

$ 0\ 0 \cdots 0\ 0\ 1\qquad \times\frac {n!}n$
$ 1\ 0 \cdots 0\ 1\ 0\qquad \times\frac {n!}{n \minus{} 1}$
$ 2\ 0 \cdots 1\ 0\ 0\qquad \times\frac {n!}{n \minus{} 2}$
$ \cdots \cdots \cdots \cdots \cdots \cdots$
$ (n \minus{} 2)\ 1\ 0 \cdots 0\quad \times\frac {n!}{2}$
$ n\ 0 \cdots 0\qquad\qquad for$ the remaining lines. Let their number be $ x$ for lazyness  :blush: .

Now, (total of row sums) $ \equal{} n!\left(\frac1n \plus{} \frac2{n \minus{} 1} \plus{} \cdots\frac {n \minus{} 1}2\right) \plus{} nx$

 and

 (total of column sums)
$ \equal{} n!\left(\frac1{n \minus{} 1} \plus{} \frac2{n \minus{} 2} \plus{} \cdots\frac {n \minus{} 2}2\right) \plus{} nx$ (first column)
$ \plus{} n!\left(\frac1n \plus{} \frac1{n \minus{} 1 } \plus{} \cdots\frac {1 }2\right)$ (other columns from right to left)
=(total of row sums)  :)

so the first column sum is also correct :lol:

In your solution there are $ n!(\frac {1}{n} \plus{} \frac {1}{n \minus{} 1} \plus{} \dots \plus{} \frac {1}{2} \plus{} x)\geq n!(\frac {1}{n} \plus{} \frac {1}{n \minus{} 1} \plus{} \dots \plus{} \frac {1}{2}) > n!$ rows in the array whenever $ n > 3$, whereas in my problem statement there are $ n!$ rows.


What is the rightmost nonzero digit of $2006!$?

[hide]Use the recurrence Ravi B (Edit: I just noticed that you posted this question  :D )posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=577096#577096]here[/url] to get that \[\begin{aligned}L(2006) & \equiv 2^{401}L(401)L(1) \equiv 2^{401}2^{80}L(80)L(1)L(1) \\ & \equiv 2^{401}2^{80}2^{16}L(16)L(0)L(1)L(1) \\ & \equiv 2^{401}2^{80}2^{16}2^{3}L(3)L(1)L(0)L(1)L(1) \equiv 2^{500}\cdot 6\cdot 1\cdot 1\cdot 1\cdot 1 \\ & \equiv 2^{500}\cdot 6 \equiv 32^{100}\cdot 6 \equiv 2^{100}\cdot 6\equiv 32^{20}\cdot 6\equiv 2^{20}\cdot 6\\ & \equiv 32^{4}\cdot 6\equiv 2^{4}\cdot 6 \equiv 16\cdot 6\equiv \boxed{6}\mod{10}\end{aligned}\] [/hide]

Yep.  Another way to calculate $2^{500}\bmod{10}$ is to observe that the powers of 2 mod 10 repeat in a cycle of 4.  Thus $2^{500}\equiv 2^{4}\equiv 6 \pmod{10}$.

Note that the numbers in your calculation arise because $2006!$ has 500 terminating zeros, and because the base 5 representation of 2006 is $(31011)_{5}$.


MATHCOUNTS calculus AMC AIME ARML USA(J)MO USAMO

This is primarily for American AoPSers, as I have no familiarity with what the rural/urban/suburban distinction is like in almost any other country.

I've been wondering for a while where you guys are from. I go to an urban public school in Columbus, OH. We have almost no extracurricular mathematics, and I am without a doubt the best and most interested at math in my school, but that's not saying much. Naturally, when I discovered the AoPS forum, I was taken aback at several things: (a) the fact that there are so many of you out there who are as interested in math as me and understand it far beyond the high school level; (b) your ability to solve problems that I would have no idea how to solve; and (c) actually just those two.

My background with "problem solving" goes as follows: in 8th grade, I was in mathcounts. The math teacher there was horrible, and mathcounts ended up being a place for people to hang out and joke with the math teacher. So much for that. In 9th grade, I was asked if I wanted to take the AMC. Sure. I did, and although I was very intrigued by many of the problems, I could only solve a few of them. In 10th grade, I had a really great calculus teacher to whom I will always be indebted. I got better at math and my understanding of math in general grew enormously. I qualified for the AIME this time, had a great time doing it, but only got two right. Between 10th and 11th, I did number theory with a graduate student at Northwestern (I'll be indebted to him too, long story), and I finally understood the notion of rigor in math, which encouraged a lot of reading into the history and philosophy of math. I find it both embarassing and hilarious (mostly embarassing) that I failed to qualify for the AIME in 11th grade (I blame the tiny desk, my being overdressed, the glare from the window, etc.). I naturally became angry and began searching online, when I found the AoPS, and then the forums (or fora, whatever). And this is where my story began.

So basically my question is, where are you guys from, anyway? I doubt that many of you could have had a background like mine and still become the problem solvers you are today, or maybe you could. What do you think?

My school is a large public school in SC.  I actually moved to SC right before my freshman year when I had previously lived in Central Florida.  I actually moved from an awesome middle school environment (Mathcounts wise) to just an "OK" high school (Nimish was at my middle school and WarpedKlown1335 went to our rival middle school).  I did not do any competitive mathematics in middle school however, except for one small contest.  When I came to the high school in SC I was in a totally new enviornment-new school, new kids.  Since I no one knew me, I decided to go along a new, more self-guided path.  Back in Florida, I would have most likely stuck to what my close friends were doing, but maybe I would've done math back there I also...I dunno.

Anyway, 9th grade I was not good at all in math.  10th grade I slowly improved during the year as my love and dedication for math grew.  By the end of the year I was placing in the JV division at a couple of contests.  Then I went to ARML at the end of my 10th grade year after I found AoPS.  During the summer after ARML I starting doing math like crazy.  I started on my journey through the AoPS books, took many classes, and participated on the forum.  Then last year, my junior year, I had an extraordinarily successful year.  I got first at all but one local competitons, made USAMO, and did decent on a bunch of other contests.

Well, that about covers my math life thus far.  Right now, I'm not doing very much math...not as much as I would like.  I hope to find time to fit it in more, however.

I'm from Columbus, Ohio too.  Well, Ii live in the suburbs (Reynoldburg).  I'm only in 9th grade, so I don't really have a math history, but I'm thinking that my school is worse than AntonioMainenti's, math-contest-wise.  I found out about MathCounts in 8th grade through AoPS, and even asked my teacher during the summer if she could do it, but then it turned out that she didn't get the package in the mail, or something.  But yea, our school didn't even have MathCounts.  And that was also the first year our school offered the AMC 8, whcihc just had 3 after school practices before the test, where we went over the problems.   So that's all the math contests that I could have done without me asking anyone if I do anything.  But I asked the teacher who was in charge of the AMC 8 at our school if I could take the AMC 10, and he said yea, and I just prepared for that through AoPS, and then took the AIME too.

So yea, except for the AMC 8, every math contest I'm in, I've probably found out about through AoPS.

i live in the New Jersey suburbs, about 15-20 miles from Manhattan. the high school i'll be going to will have 2000 students. its most accelerated sequence of math has you doing multivariable calculus as a senior. i dont think it sponsors any math competitions. as for now, i'm in an 800 student middle school with no MATHCOUNTS.

I go to a private school (its religious) but.. its not prestigious really, its about 900 students. There isnt much on accelerated math - only an honors-type class (the math is tracked, 4 classes). There is a math team, but we dont do any big competitions.. odd, i knwo.

BTW, h78reg, around which county in NJ do you live? I also live 15-20 miles i think from manhatten..

i live in Middlesex county. if only i lived near Hackensack.............. :roll:

hehe, I live like... 2-3 miles from hackensack  :)  - Bergen County

[quote="jimli"]I live in suburbs next to Stanford.  Still our math schoolwise sucks.[/quote]
Whoa! Let's stop to consider this.

A listing of number of AIME qualifiers, by school, for school ZIP codes 94000 through 94403:

 3 Castilleja School, Palo Alto
56 Palo Alto HS, Palo Alto
21 EPGY-Stanford, Palo Alto
54 Henry M. Gunn HS, Palo Alto
 7 Aragon HS, San Mateo
 1 Hillsdale HS, San Mateo

Maybe that's not the most concentrated patch of AIME qualifiers by geographic area - for instance, there are the neighboring ZIP codes 95000-95199:

  7 Cupertino HS, Cupertino
 21 Homestead HS, Cupertino
 36 Monta Vista HS, Cupertino
 12 Los Gatos HS, Los Gatos
  6 Milpitas HS, Milpitas
  6 Adrian C. Wilcox HS, Santa Clara
  2 Santa Clara HS, Santa Clara
  1 Santa Cruz HS, Santa Clara
  6 Scotts Valley HS, Scotts Valley
  3 Redwood MS, Saratoga
 28 Saratoga HS, Saratoga
 10 Leland HS, San Jose
  2 Leigh HS, San Jose
  2 Bellarmine College Prep School, San Jose
 13 Archbishop Mitty HS, San Jose
 85 Lynbrook HS, San Jose
  5 Miller MS, San Jose
 52 The Harker School, San Jose
  1 Evergreen Valley HS, San Jose

(I don't know all of these schools, but I would assume that Castilleja, Bellarmine, Mitty, and Harker are private, the listing for EPGY reflects students who attend a variety of schools, and the rest are all public.)

These are your neighbors. You live in one of the most active areas of the U.S. for the mathematical problem-solving culture, so take advantage of it. As an instructive contrast: ZIP codes 90000 through 90250, which is to say the portion of the city of Los Angeles south of the Santa Monica Mountains, plus Beverly Hills, Culver City, Santa Monica, and a number of other cities - probably more than 2 million people - totaled  just 35 AIME qualifiers, 14 of them from Beverly Hills HS and most of the rest from private schools.

[quote]hehe, I live like... 2-3 miles from hackensack  - Bergen County
[/quote]

do you go to AAST then?

I live in the suburbs in Austin, Texas.  My middle school was so-so mathwise, basically the only math thing we did was Mathcounts.  I did okay at that, I made state in 7th and 8th grade.  During that time I scored 20's in chapter and mid-high 30s in state. (yeah, weird I know).
The high school I go to now isn't exactly a math school.  I am by far the best student at my school problem-solving wise (I'm in 10th grade), but those on this forum know that I'm really not that great.  My school did not do the AMC competitions in the past, but we will this year through my school's Mu Alpha Theta, of which I am an officer.  So yeah, I'm making my school have more math activities, and maybe I can start a math club or something.

i lived in the suburbs all my life, from china (smally-city equivalent) to arlington, TX, to clarksville, MD. ive always attended publc schools, though my parents took care to move to areas with more competitive and academically challenging schools.

math history:
didnt start til i moved to MD in fifth grade, when i joined a talent pool program and entered a periodic contest (not unlike the Math League)
middle school didnt offer anything but AMC-8, but i joined an accelerated program out of school to crunch 3 years of math into two.
high school - started off with AMC12->AIME, Marlyand Math League, and Howard County Math League-> (ARML toward April) for the first two years.
last yr, having had the chance to distinguish myself as a top scorer in the county and gained some upper-classmen power, i dragged the team into the AHSIMC and a University of Maryland competition.
starting this fall i plan to do as much as i can competition-wise, to take advantage of this one last year i have in high school.

In China(at least my hometown), students in public schools alway do much better than ones from private schools. In my view, students in Chinese private schools are those who have a rich family, and behave very very very badly in public schools, so their parents send them to private schools with a lot of money expecting the teachers would be more responsible(of course in fact not), actually, I can  only see them just paly in private schools and learn nothing, most of them do even worse than they were in public schools

Not all the private schools are like what you say qxy. There is a private school in Beijing named Ke Di Shi Yan School, it is one of the most competitive middle school in Beijing. It sends 7 students to Beijing Number 4 High School and a lot of other students to High School Affiliated to Remin University.

[quote="bleumoose"]are you asking my neighborhood or my school? i go to a public magnet school, but i live in a rural community.[/quote]I guess I mean moreso your school. Sorry for the ambiguity.

Anyway, I'd like to hear from more of you; not just where you go to school but also your math backgrounds.

Holy crud, a lot of private schools and stuff seriously do have "academy" in their name. I never noticed that before. Hrm.

West Windsor, suburbs near Princeton, and I go to a middle school with about... probably closer to 1,500 students.

I go to a medium-sized Catholic high school. Math-wise, it's decent, nothing spectacular. It offers the usual algebra, advance math, pre-calc, and calculus sequence. Unfortunately, if a student isn't in the accelerated track, meaning they took Alg. 2H as a freshmen, they miss out. A lot of my friends feel unchallenged because they took Alg. 1 in eighth grade but had to repeat it as freshmen. Luckily, I'm in the accelerated track.

As far as math history goes, I'm a joke. Hence and for other reasons, my name is joker. I have no experience what so ever is this kind of math. I've never participated in math contests nor do I have the skill to do the type of math in this forum. It's ironic to find myself in the company of such elites. :D  I'm pathetic. I know. But, I want to change all that. Someone in another board I frequent directed me to this board and...here I am. I'm entering 10th grade going into Adv. Math H. Hopefully, in a few years, I won't be in the position I am now.

My school is a medium-sized (1000-1100 students) school in northern Mississippi.
I am currently in 9th grade, but in 7th grade I sucked horribly at math.
I did Mathcounts in 7th grade, but I failed very badly. (I don't think I even knew what binomial coefficients were back then, furthermore, I didn't learn the quadratic formula until the spring of my 7th grade year.)
My 8th grade year can be divided into the following:
First half of 8th grade: Still very bad at math
Second half (more like last third) of 8th grade: Improved at math
I was far from being considered OP in 8th grade, but at least I made National Mathcounts.
During the end of 8th grade/the summer between 8th and 9th grade, I learned some math on my own. (Alcumus, etc.)
Now, compared to myself as a 7th grader, I am much better at math now. However, I still would "suck" compared to most of the AoPS Community, as I can only get 4's on past AIMEs. (I think if I did the 2017 AIME I in 7th grade, in which I got a 4 on, I would've only gotten a 1.)
Concerning my math teachers, I would say most of them up to now are/were OK. However, our school's math curriculum up to now has just been focusing on "school math" and not "competitive math" (there's a (big) difference), so to prepare for my AMCs I would have to study math on my own.

I go to a rural school. that is a great place to do math.

[quote=moab33]I am homeschooled. My house is in the suburbs.[/quote]

same

I live in a certain suburb off a particular city in Iowa. (I'll give you a hint: It has the most uncreative name EVER). I have three schools really close to my neighborhood, and for some reason I go to the farthest. (School admin logic is totally crazy.)

suburban/urban, i dont really know how to explain

live in the 'burbs (Middleton, WI) but go to school in madison. (more urban)

more like 15 cuz my dad drives kinda fast so...
but yeah
 :-D

I live in the 'burbs as well

20 min drive to school

small school tho

Mission San Jose Attendance Area (94539) (which includes Hopkins) has like at least 50 for sure, and like so many JMOs. Just in middle school, there's like 6 AIMEs, 3 JMOs, and another who would have made if he hadn't misread the num base problem on the 10A.

I live in the suburbs too. I have like a 5-10 minute drive to my school.

[quote=Eric2009]same. i like surburbs[/quote]

Please don't bump 1 year old threads. At that point it's usually better to make a new one I think


Simplify and factor completely in (mod 6):

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/e6628fcae63a8c728c3a464436131619.gif[/img]

What does factoring mean $\mod 6$ here¿
The factoring is not near to anything unique, nor is it well defined by such identities like \[(2x+3)(3x+2) \equiv x \mod 6.\]

It says "simplify and factor completely". That would mean getting it down to

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/2a963e409c4848dbfc27b5447e1208fc.gif[/img]

for some a and b (in mod 6)

[quote="ZetaX"]Whatever "simplify" means ( :huh: ), you still gave no definition of factoring! At least we need degree $5$, so your term is inappropriate...[/quote]
Yes but notice that $x^{5}$ is same as $x$ mod $5$. This way you can simplify the expression to form $ax^{2}+bx$ because $x^{4}$ is same as $x^{2}$ mod 6 ( we use the lemma that $x^{3}$ is $x$ mod $6$). The expression reduces to $2x-5x^{2}=x(2-5x)$ and you are done.

So not as polynomial, but as function. Would be nice to state that next time ;)


function
real analysis
real analysis solved

Find $f:R\rightarrow R$ derivable such that for any x,y reals

$f(x+y)=f(x)+yf(x+y)+yf(x+y)f(x)$

There are two possible solutions, both constants. Either $f(x)=0$ or $f(x)=-1.$

We have two contradictory calculations. The first is algebraic. Let $x=0$ and let $f(0)=A.$

$f(y)=A+yf(y)+yf(y)A$ from which $f(y)=\frac{A}{1-(1+A)y}.$

The other calculation is to first divide by $y$:

$\frac{f(x+y)-f(x)}y=f(x+y)(1+f(x))$ and then let $y\to0$:

$f'(x)=f(x)(1+f(x)).$

This separable differential equation has two constant solutions $f(x)=0$ and $f(x)=-1.$ its nonconstant solutions are:

$f(x)=\frac{Ce^x}{1-Ce^x}.$

Note that the two nonconstant forms of the solution disagree. Worse, note that either one, if nonconstant, fails to be defined on $\mathbb{R}.$ So the only solutions are the two constant solutions.

[quote="Kent Merryfield"]$f'(x)=f(x)(1+f(x)).$

[/quote]

$y'=y+y^2$ is Bernouilli differential equation

----------------------------------------------------------------
For solving Bernouilli ifferential equation
(1) $y'=a(t)y+b(t)y^k$ where $a,b$ are continous functions, $k$ real number $\neq 0;1$

Do the change variable $y=\frac{1}{z^{k-1}}$ you have a linear
differential equation of order one.


calculus
integration
limit
function
derivative
calculus computations

(1)  I'll just do the brute force way of pluggin' and chuggin':

$ f_n'(x) \equal{} nx^{n\minus{}1}e^{\minus{}x} \minus{} x^ne^{\minus{}x} \equal{} x^{n\minus{}1}e^{\minus{}x}(n \minus{} x)$

The critical points are at $ x \equal{} 0$ and $ x \equal{} n$.

Evaluate $ f_n(x)$ at these points and check which is larger...

$ f_n(0) \equal{} 0$

$ f_n(n) \equal{} n^ne^{ \minus{} n} > 0$ if $ n$ is positive.  This is thus the maximum value.

(2)

$ \lim_{x\to\infty} f_n(x) \equal{} \lim_{x\to\infty} x^ne^{\minus{}x}$

Because no power of $ x$ can increase faster than the decreasing awesomeness that is the exponential function with a negative power, it follows that the $ e^{\minus{}x}$ in this product dominates the $ x^n$ term, resulting in a limit of $ 0$.  (To see the rate of changes, take derivatives!)

That's rather intuitive, but is there really any "formal"/"rigorous" way of evaluating it?

(3)

$ I_n \equal{} \int_0^x f_n(t)\ dt \equal{} int_0^x t^ne^{\minus{}t}\ dt$

Unfortunately, I'm having a brain fart right now >_> and can't remember how I used integration by parts to do this.

If university students see the problem, they would notice what's the problem is related to, but needless to say, Few Japanese high school students understand the theme of the problem.

[quote="Zakary"]

(2)

That's rather intuitive, but is there really any "formal"/"rigorous" way of evaluating it?[/quote]

From the result of (1), there exists the maximum value of $ f_{n \plus{} 1}(x),$ denote it by $ M$, we have ... , then you can use the squeezing theorem.

[quote="Zakary"]

(3)

$ I_n \equal{} \int_0^x f_n(t)\ dt \equal{} int_0^x t^ne^{ \minus{} t}\ dt$

Unfortunately, I'm having a brain fart right now >_> and can't remember how I used integration by parts to do this.[/quote]

Express $ I_n(x)$ in terms of $ I_{n \minus{} 1}(x).$

[quote="kunny"]Answer the following questions for positive integer $ n.$
(1) Find the maximum value of $ f_n(x) \equal{} x^ne^{ \minus{} x}$ for $ x\geq 0.$
(2) Show that $ \lim_{x\to\infty} f_n(x) \equal{} 0.$
(3) Let $ I_n (x) \equal{} \int_0^x f_n(t)\ dt,$ find $ \lim_{x\to\infty} I_n(x).$[/quote]
[b]$(1)$[/b]$ f'_n(x) =-x^{n-1}e^{-x}(x-n)$

$f$ max is $n^{n}e^{-n}$ at $x=n$

[b]$(2)$[/b] $\lim_{x\to\infty} I_n(x)=\lim_{x\to\infty}\frac{x^n}{e^x}=\lim_{x\to\infty}\frac{n!}{e^x}=0$ from [b]L-hopital[/b]

[b]$(3)$[/b]$ L(n)=\lim_{x\to\infty} I_n(x).=\int_{0}^{\infty}x^{n}e^{-x}dx=\mid{-x^{n}e^{-x}}\mid_{0}^{\infty}+nL(n-1)$

$\implies{L(n)= nL(n-1)}$ $\implies{L(n)= n(n-1)(n-2)...3.2.1=n!}$   as $L(0)=1$


trigonometry
inequalities
inequalities unsolved

Maximise $\cos x+\cos y$ subject to the constraint $y-x=\frac{\pi}{4}$.

$cosx+cosy = cosx+cos(x+\frac{\pi}{4}) = 2 cos(x+\frac{\pi}{8})cos\frac{\pi}{8}\leq 2cos\frac{\pi}{8}= \sqrt{\sqrt2+2}$


geometry
inradius
power of a point
area of a triangle

29. A circle is tangent to two sides of a square and its diagonal. Given that the length of a side of the square is four units, how many units are in the radius of the circle? Express your answer as a common fraction in simplest radical form.

I got pwned by this one...
[hide="Answer"]$ 4\minus{}2\sqrt{2}$[/hide]

[hide="one way"]
Draw in a square from the NW corner of the large square to the center of the circle.  The side length of this square is r, and we can see that the diagonal of this square ($ r\sqrt {2}$) plus the radius of the circle ($ r$) will equal half of the diagonal of the large square ($ 2\sqrt {2}$).  Solve for r.
[/hide]
[hide="another way"]
You are looking for the radius of the inscribed circle of a isosceles right triangle.  Since the circle is inscribed, we know that the lines from the vertices of the triangle to the center of the circle are angle bisectors.  Thus, the distances from the vertices of the triangle to the tangent points on the circle must be the same.  We know the distance from the right angle to the tangent points must be r; let the other distance be x.  The hypotenuse of the triangle is 2x, so x must be $ 2\sqrt {2}$.  Then the side of the square is $ 4 \equal{} x \plus{} r \equal{} 2\sqrt {2} \plus{} r$.
[/hide]

[hide="not to tricky"]
i saw this one on ellis saabs page
we can forget about the bottom triangle because it does matter. when you draw radii perpendicular to the sides of the square, we get another square. since this triangle is iscoseles, the point of tangent bisects the hypoenuese. therefore, from there to the end of the hypotenuese is $ 2\sqrt{2}$ we can use something, power of a point i think, to get the point of tangent on the side of the square, to the vertex of the square is alsio $ 2\sqrt{2}$ so $ r\equal{}4\minus{}2\sqrt{2}$[/hide]

couldnt you have just found the inradius of the triangle formed by 2 sides of the square and the diagonal to find the radius of the inscribed circle.  Maybe this would have been harder to do than the other methods shown above.

I heard that there was equation. I think it was A = rs with s being the semiperimiter of the triangle, a being area of the triangle, and r being the radius. Could some one shed some light on this?

[quote="mathcrazed"]IN CIRCLE RADIUS = r!!

rs=A. SO TRUE. i hate pie is right........ darn.....
lol

soooooo

$ r*(4 \plus{} 2\sqrt {2}) \equal{} 8$
WHY? BECAUSE ITS the SEMI perimeter.

done.[/quote]

i love your spammy posts -_-.

basically because the inradius is perpendicular to all the sides, you can express the area of the triangle in interms of an altitude to all the bases, r.

then divide by 2 for the area of a triangle and you get the semiperimeter

and if you know that the inradius of a right triangle is $ \frac{a \plus{} b \minus{} c}{2}$ that makes it even easier giving you $ \frac{8 \plus{} 4\sqrt{2}}{2}$  Which simplifies to $ \boxed{4 \plus{} 2\sqrt{2}}$.

Oh, I see. I am surprised I couldn't think beyond the picture.. Thanks.


Happy Norooz to all the Persian members on AoPS mathlinks! (Persian new year, beggining of spring) I know its a bit late...but whatever!
http://www.persiancards.com/images/ecard/151.swf
and Happy Purim to all other Jewish members on AoPS and mathlinks!
http://www.yahoo.americangreetings.com/display.pd?bfrom=1&prodnum=3066919&Searchstr=purim&path=42809&st=t

 :lol:

Also a Happy Easter to the majority of you guys!

It makes sense to have the new year start in the spring; it is when everything in nature comes to life again.

I also say Happy norooz to all Persian members of the AOPS mathlinks

Any Jewish members have any humentashen left over? JSRosen3 won't give me any.  :(


how to write a prove in...

Q: prove that if $ 3|(a^2\plus{}b^2)$, then $ 3|a$ and $ 3|b$.

Please use more sensefull titles (this one here is completely useless). Additionally, you can use $ \text{\LaTeX}$ to write formulas.
There is also an Edit-Button, you for example seem to have some additional tags...
And next time, please consider which forum to post in  :wink:

PS: Thread edited and moved.

[hide="Hint"]
1) Evaluate the different cases writing $ a$ and $ b$ in the form $ 3m\plus{}n$ where $ n$ is less than $ 3$.
2) Modular arithmetic

The first is basically the 2nd in a different way[/hide]

In fact 3 can be replaced by any prime $ p \equiv 3 \pmod 4$.
We first prove the following lemma

Lemma: If $ p \equiv 3 \pmod 4$ then the equation $ x^2 \equiv \minus{}1\pmod p$ has no solutions.

Proof of the lemma: Suppose if there is a solution (say $ y$).
$ y^2 \equiv \minus{}1 \pmod p$ implies $ (y^2)^{\tfrac{p\minus{}1}{2}} \equiv \minus{}1 \pmod p$ (Since $ \tfrac {p\minus{}1}{2}$ is odd ). Therefore $ y^{p\minus{}1} \equiv \minus{}1 \pmod p$ a contradiction since by Fermat's little theorem $ y^{p\minus{}1} \equiv 1 \pmod p$. Hence there exists no $ y$ satisfying the given equation.

Theorem: If  $ p \equiv 3 \pmod 4$ and $ p|(a^2\plus{}b^2)$ then $ p|a$ and  $ p|b$.

Proof: Suppose $ p \nmid a$ then  $ p \nmid b$ (Since $ p|(a^2\plus{}b^2)$)Hence $ b$ has an inverse say $ c$ in modulo $ p$.
$ a^2\plus{}b^2 \equiv 0 \pmod p$ implies $ c^2(a^2\plus{}b^2) \equiv 0 \pmod p$. Hence $ (ac)^2 \equiv \minus{}1 \pmod p$. A contradiction to the above lemma. Hence the assumption that  $ p \nmid a$ is false. Hence $ p|a$ and  $ p|b$. $ QED$

Things can be simplified a lot if you know that all squares are either 0 or 1 modulo 3. Considering all the cases, $ a^2 \plus{} b^2$ can only be divisible by 3 if both $ a$ and $ b$ are 0 modulo 3.

Let $ a\equal{}3p\plus{}r,\ b\equal{}3q\plus{}s$ for some integers $ p,\ q,\ r,\ s$ such that $ r,\ s\equal{}0,\pm 1$.
We have $ a^2\plus{}b^2\equal{}3(3p^2\plus{}3q^2\plus{}2pr\plus{}2qs)\plus{}r^2\plus{}s^2$.
$ \therefore 3|a^2\plus{}b^2\Longleftrightarrow r^2\plus{}s^2\equal{}0$, yielding $ r\equal{}s\equal{}0$ which completes the proof.


Exits on a highway are numbered consecutively from 1 to 50. The distance from exit 41 to exit 50 is 100 km. If each exit is at least 6 km from the next exit, what is the longest possible distance, in kilometers, between any two consecutive exits from exit 41 to exit 50?

[quote="RoFlLoLcOpT"]There are a total of $ 10$ exits along the highway. We want to find the maximum possible distance between two consecutive exits. We are told that the minimum distance is $ 6$ km. Suppose we let $ 9$ of the highways have a $ 6$ km distance between them. This way we can maximize the last exit distance. Evaluating, we get $ 100 \minus{} 9(6) \equal{} 100 \minus{} 54 \equal{} \boxed{46}$ km.[/quote]
Remember that there are $10$ exits, but there are $9$ distances between the exits.  So the answer should be $100-8(6)=100-48=\boxed{52}$.


Look in your profile. You can become one too.

[url=http://www.artofproblemsolving.com/wforum/faq.php#3]Look in the FAQs first[/url]

Basically hidden just takes you off of the "people logged on" list

What's the point of being a hidden user???

Privacy I guess, one thing is maybe people don't want to be seen logging on here or maybe they used their real name for their username. They could also not want PMs... I'm frankly not sure why any of these is a strong reason to stay hidden...oh well.

Why's that?

I was hidden way before mainly because I didn't know how to become un-hidden (I started a thread in here about it...)

i did it manily so that others don't know what i'm doing on this site.
so, yeah it's for privicy.


parameterization
conics
ellipse
geometry unsolved
geometry

Consider all line segments of length 4 with one end-point on the line $y=x$ and the other end-point on the line $y=2x$. Find the equation of the locus of the midpoints of these line segments.

Let A = (a, a), B = (b, 2b) be 2 segment endpoints and C = (x, y) the midpoint of AB. From $AB^2 = 16,$

$(a - b)^2 + (a - 2b)^2 = 16$

$5b^2 - 6ab + 2a^2 - 16 = 0$

$b = \frac{3a \pm \sqrt{80 - a^2}}{5}$

$x = \frac{a + b}{2} = \frac{8a \pm \sqrt{80 - a^2}}{10}$

$y = \frac{a + 2b}{2} = \frac{11a \pm 2\sqrt{80 - a^2}}{10}$

Eliminating the arbitrary parameter a,

$2(10x - 8a) = 10y - 11a,\ \ \ a = 2(2x - y)$

Substituting this to either the equation for x or y (say x), we get the locus equation.

$x = \frac{8(2x - y) \pm \sqrt{20 - (2x - y)^2}}{5}$

$(8y - 11x)^2 = 20 - (2x - y)^2$

$25x^2 - 36xy + 13y^2 = 4$

WHICH IS AN ELLIPSE CENTERED AT THE

[quote="ashwinrk_jain"](4x-3y)^2 + (3x-2y)^2  = 4[/quote]

You are right.

Here is my solution.

Let $P(s,s)\in y=x,\ Q(t,2t)\in y=2x,$ we have $\overline{PQ}^2=(s-t)^2+(s-2t)^2=16\ \cdots [*]$

Set the midpoint of $PQ$ be $M(X,Y),$ we have $x=\frac{s+t}{2},\ Y=\frac{s+2t}{2}\Longleftrightarrow s=4X-2Y,\ t=-2X+2Y.$

Plugg these into the equation $[*],$ we obytain $(3X-2Y)^2+(4X-3Y)^2=4,$ which is [b]ashwinrk_jan [/b]obtained.

Expand this we obtain $25X^2-36XY+13Y^2=4,$ which [b]yetti[/b] obtained.


I'm having difficulty with this one -- can anyone help?

Let x be an integer greater than 25.  What is the least value of x for which the number 378(base x) is a perfect square?   I've tried using discriminants and mods, but haven't had any success.   Ideas anyone?

Here's what I did:
[hide]First I wrote it as $3x^2+7x+8=n^2$ then $3x^2+7x+8-n^2=0$.
Quadratic gives you:
$x=\frac{-7\pm\sqrt{49-12(8-n^2)}}{6}$
which simplifies into (and the minus isn't going to work since the base, x, is >25)
$x=\frac{-7+\sqrt{12n^2-47}}{6}$
Then I tried some things with mods before just putting it in my calculator and watch the values for integer x's.  The first one after 25 is x=127 (with n=222), so the base is 127.  I checked it and it worked.[/hide]

I wish I could've done it without a calculator.

Edit: I have a start on a non-calculator solution:
[hide]Since the $\sqrt{12n^2-47}$ part will be 1 mod 6, then
 $\sqrt{12n^2-47}=1+6m$ for an integer m, so  $12n^2-47=1+12m+36m^2$
then that reduces to $n^2-4=m+3m^2$

Which if you can solve that you've solved the original problem.  I note one thing, that is that n is even since $(n-2)(n+2)=m(1+3m)$ and the right side is even so the left must be even, hence n is even.[/hide]

Good beginning Wumbate, but your second part is flawed: [hide]$12n^2 - 47$ will be congruent to 1 mod 6, but that doesn't mean its square root will be -- the square root will be congruent to 1 or 5.[/hide]


There is a square 100cu întocmirea rapoartelor de audit 100x100, and all its fields are coloured in one of three given colours. We can recoloure a square 2x2 in that colour which is prevalent, otherwhise if there is no such color - in the color which is absent. Prove that the entire square can coloured in one color.

English? anyone?

If it means "a square 100cu întocmirea rapoartelor de audit 100x100"=100X100 square grid;"fields"=cells;"colour which is prevalent"=Color appearing most
I think I have a solution.

I interpret the question like this :

We have a $ 100X100$ board with every box filled with one of the three colours. For any $ 2X2$ square in the board, we may colour it either with

(i) a colour present in one of the $ 1X1$ square.
(ii) the colour that is not present in any of the $ 1X1$ squares.

In this case, the colouring is trivial. Choose a colour say 'red' and a $ 2X2$ square. If $ 2X2$ has a red $ 1X1$ block , we can colour it all with red. Else if none of the $ 1X1$ blocks have red, then we can colour all with red. This also means we can colour the $ nXn$ board with any colour.

I dont think your interpretation is correct(though I dont know any russian at all).
Note that its a russian problem. And as far as I know they dont make problems like that.

I interpret:

There is a $ 100X100$ square grid.We can recolor a $ 2X2$ block with the color that appears more than other colors in that $ 2X2$ block.If there is no such color(when there are two colors each having two $ 1X1$ blocks.) then we can recolor it with the color which is absent.

I solve:

divide the board into $ 2500$ $ 2X2$ blocks. Apply recoloring to all of them. They becomes monochromatic. Now, for two adjacent $ 2X2$ blocks of different colors we can swap the colors of the two $ 2X2$ blocks because---
[R=Red,B=Blue,G=Green]
[code]+--+--+--+--+
|R1|R2|B3|B4|
+--+--+--+--+
|R5|R6|B7|B8|
+--+--+--+--+[/code]
Applying recolor in $ 2367$ we get
[code]+--+--+--+--+
|R1|G2|G3|B4|
+--+--+--+--+
|R5|G6|G7|B8|
+--+--+--+--+[/code]
Applying recolor in $ 1256$ and $ 3478$ we get
[code]+--+--+--+--+
|B1|B2|R3|R4|
+--+--+--+--+
|B5|B6|R7|R8|
+--+--+--+--+[/code]
There will at least be three $ 2X2$ blocks of one color(PHP)-- say Red.
Suppose we want to convert one Blue into one red.
Then we can get it like this
[code]+---+---+---+---+
|R1 |R2 |B3 |B4 |
+---+---+---+---+
|R5 |R6 |B7 |B8 |
+---+---+---+---+
|R9 |R10|R11|R12|
+---+---+---+---+
|R13|R14|R15|R16|
+---+---+---+---+[/code]
Applying recoloring in $ [6,7,10,11],[2,3,6,7],[7,8,11,12],[3,4,7,8]$
Thus we can convert all into red in this way.

Thus we can color all red.


what people keep posting when they no they aren't supposed to

Do you mean the type of forum spamming that has plagued the Games and Fun Factory, or spamming in general?

Because forum spamming, I would say, is any post that contributes nothing to the actual discussion (i.e., the recent outbreak of $3$, but also any posts that consist of only "I agree" or  " :rotfl: ".  Actually, regular (e-mail and blog comment) spam is like that too, except that that type of spam is almost always trying to accomplish something (specifically, trying to sell you sketchy products.)

spam:
- useless
- annoying for most people
- no message to tell
- often occurs in masses
- absolutely unnecessary

I think everybody knows what spam is so we probably do not require a formal definition, it is rather common sense. The round table is supposed for serious discussion. But then what is the purpose of the "Games & Fun Factory" section? I assumed for most people fun = spam and it is supposed to keep people from spamming other forums. It seems some people are rather smart but keep spamming as Arnav. But as Daniel pointed out this should not offer him privileges to spam. He should better be concerned to get a good score in the Putnam undergraduate competition than wasting his time posting 3s. Several of his posts, even outside GFF, are pretty annoying. Either he posts stupid remarks, writes in cap lock etc. I think people will find a way to spam anyway, so it is okay for them to do it in GFF. But please could you exclude posts from this GFF forum from the latest post review at:

http://www.mathlinks.ro/Forum/portal.php

As B-tone_Party said, around the forum generally, when people make random(ish) comments along the lines of 'I agree' or 'Nice.' this is as good as spamming, and possibly worse because GFF is, like orl said, almost made for spamming, whilst the other forums aren't. (ok yeah I'm saying I agree, so maybe I'm spamming here?).
Also when people post an identical proof to 5 before it, in the same thread, it adds nothing to the discussion and can be annnoying to have to sift through. However, they are  answering the question in the thread, and doing mathematics.  Is this a kind of spamming?

[quote]I'm sure you've all noticed that posts on this board are no longer a part of the post count. However, spam will still be deleted. Although we will be more lenient on what defines spam, please continue to observe the following rules:

1) No commercial advertising. No exceptions.

2) Posts that are > 50% smiley are not good ideas.

3) Every post should be productive in some way, shape, or form. No one wants to scroll through pages of "HAHAHAHA..."

4) Except in cases where the theme of the thread allows (e.g. Mafia "aye" or "nay"), posts should always be longer than 5 words and 1 sentence.
[/quote]

Thats quoted from MithsApprentice from his topic entitled "New definition of spam"

[quote="bubka"]It's best to start at the roots. What [i]is[/i] spamming?[/quote]

see

http://www.mathlinks.ro/Forum/viewtopic.php?t=91178

conceive

[quote="tim1234133"]As B-tone_Party said, around the forum generally, when people make random(ish) comments along the lines of 'I agree' or 'Nice.' this is as good as spamming, and possibly worse because GFF is, like orl said, almost made for spamming, whilst the other forums aren't.
[/quote]

Of course, it no problem to express you admiration from time to time. But then again lots of people only do that but almost all the time don't contribute anything. But they are rather concerned about telling people to lock threads etc. Probably boredom drives them that far. But why don't you guys go out and enjoy the nature outside, read some books, solve some problems etc.?

[quote="tim1234133"]Also when people post an identical proof to 5 before it, in the same thread, it adds nothing to the discussion and can be annnoying to have to sift through. However, they are  answering the question in the thread, and doing mathematics.  Is this a kind spamming?[/quote]

I do not see any problem with it. First they improve their problem-solving, secondly they feel they can get recognition/feedback by fellow users from time to time and quite important they train their proof-writing skills though this is less important in lower level forums. By this I mean writing up their reasoning and solutions. Not just answers.

[quote="tim1234133"](ok yeah I'm saying I agree, so maybe I'm spamming here?). [/quote]

I don't say we should be such strict that only entropy-maximizing posts are permitted. First it is hard to accomplish and you want to write in a convenient and relaxed way. The same goes for reading posts. This spam discussion should not lead to the state that nobody is posting because he is scared of being accused of spamming. And you, for instance, contribute interesting ideas to this discussion. So why should you feel guilty of spamming? Everybody spams from time to time because you are very excited about a certain issue. But as long as you don't exaggerate it and annoy other people, you are fine.

I think these wide discussions about spam will rank ML/AoPS quite highly at Google wrt this issue.  :wink:

Now, what would be the possible solution to avoid spamming in GFF?

Two ideas I think are:

_Banning users.

_Deleting GFF.

I oppose that idea, there are many interesting topics in Games and Fun... like the "Movies you just saw" and "Girlfriend" topic.

Ironically, both topics don't belong there in my opinion.

The first should be in Music & Art, but movie lovers almost got wiped out by the Music fans.

The second is an interesting example : it was supposed to be a serious topic, there were serious replies, then silly replies came.  This urged moderators to move it to Games and Fun Factory, and the result was obvious, whoever was still showing some self-restraint, stopped doing it then and the topic became a sitting duck for spammers, a situation I deplored in this response :

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=591229#591229[/url]

I am afraid that if you DELETE G&FF you will get a very logic reaction : a lot of spam-refugees will flee to the next best thing : Round Table, only making the situation worse.

dont delete GFF!

instead ban people from gff
itll be better

[quote="akalra1"]i think bubka is talking about the $3$s which were really annoying

his own topic, "NO 3s please," even had $3$s in it[/quote]

If you tell someone specifically [i]not[/i] to do something, it invites other people to do it.  Espeicially when it's a topic in Fun and Games.

true

example:
when i tell my brother not to read my work, he reads it
when i tell my brother nothing about reading my work, he doesn't read it

i have tons of more examples..........

i guess that rule is applied on aops too

[quote="bubka"]G&FF was plagued with posts like this:

$3$
[/quote]

I always wondered why people posted things like

$3$

So I, like bubka, started a topic for the purpose of asking why people posted things like

$3$

But even that topic became rapidly flooded with posts like

$3$

Does anyone know where that phenomenon came from?

That is a very good question, it's just unbelievable spam.

If I can speak freely, I think we shouldn't make garbage topics.  I don't disagree with the decision to throw Games and Fun off the counter, I even support it, but that doesn't mean we should surrender it to banditism like that.  We are allowing spam to become serious in Games and Fun, and if you DO decide to shut it down, they won't just go away, you will have- I have said this before-  "spam-refugees" spreading all over the place, first to Round Table, and once we close that down... :o

What will THAT look like?  I don't know for sure, probably something like this :

[img]http://upload.wikimedia.org/wikipedia/commons/f/f0/Rwandan_refugee_camp_in_east_Zaire.jpg[/img]

Let me guess--that's an image of a Rwandan refugee camp in east Zaire (now Dem. Rep. of Congo) uploaded from the Wikimedia Foundation. :rotfl: Looked at the filename...

Sorry...

I agree--when I joined this site, most of the topics in G&FF actually had at least some significant content.

Just a thought: Perhaps the fact that posts in G&FF no longer count might have given people an excuse for mass $3$ing.

[quote="Inspired By Nature"][quote="paladin8"][quote="akalra1"]yea, it has to stay[/quote]
I think a post like this can be considered borderline spam.
1) It probably didn't take more than 30 seconds to conceive of.
2) It doesn't add much content.
3) It's not written in proper english (capitalization, punctuation, etc.).
I don't frequent G&FF anymore, but there is definitely spam in all the math forums with posts that look like something someone put together on their way past a computer or something. The most annoying thing is posts without CONTENT.[/quote]
Oh, come on, akalra1 was just showing he agreed with MithsApprentice. There is nothing wrong with that. As for number 3, you're right. But sometimes people just don't have time, and they just want to get a word in before the discussion changes direction. :)
2 is correct too, but again, it was just a show of opinion. I don't think a post like this should be considered spam.[/quote]

That's a difference in opinion then. I do think it should be considered *nearly* spam; a good post should contain content that in some way contributes to the thread. If reasons were given, I could understand the point of the comment, but as it is the post doesn't seem necessary...

Posts expressing "I agree" ( or the opposite) (and nothing else) are appropriate in threads that specifically asked people's opinions about something. I'd say in this case it's borderline -- we are having a discussion about spam. But nobody asked for opinions about shutting down GFF, so paladin8 has a point.

Keep in mind, when posting: "Could anyone possibly care about what I'm about to post? Is the information in my post useful in the context of this thread?"
That should answer the question of whether "I agree" is a legitimate post.

[quote="ZetaX"]While I wrote this, fredbel6 answered, so an addendum:

I'm not for banning for small reasons (everyone has some chances). But what else doing with people that don't follow the rules¿ (so everyone can loose his chances)
I would not make these bans permanent at first, more like:
- warning of what will happen
- then a one-week-ban
- then a two-week-ban
- then a one-month-ban
- then permanently [/quote]
Oh I am totally pro whatever you say here. But the point that I must emphasize over and over again is :
- deletion of the entire Games and Fun will lead to that picture(but I am assured for now that that is not on the agenda)
- more important : undoing the damnage the burglar has done, by deleting his spam posts

Warning and eventually punishing a spammer is not enough.  He has gotten away with his "crime" and has downgraded the quality of a topic, inviting others to do so as well.  That Girlfriend topic is such a fine example, people with serious intention started it, moderators moved it to Games and Fun which resulted in a disaster.  Do keep in mind that nonspammers value their posts more than spammers, when you move a topic like that, above of all you punish the serious people.

I think every f & g thread should have a little box just made so that people can spam there instead of on the main thread (sort of like one of those shout box thingies).

That would be pretty awesome, as the people who like spam can just look there, while everyone else could just ignore it.

[quote="José"][quote="fredbel6"]That Girlfriend topic is such a fine example, people with serious intention started it, moderators moved it to Games and Fun which resulted in a disaster.[/quote]
True, but then I made a new Girlfriend topic here in the round table to start the discussion again and it was moved to gff, so they really didn't see the mistake[/quote]
That IS a mistake, your second attempt was a lot more succesful and was not a "Game".

I think there are two fundamentally different approaches :
evaluating a topic on the basis of its posts, and evaluating it on the basis of the original post(er)(s).

I support the second.

Yup, because when the secondary posters are spammers, it puts a bad impression on the topic. Mods should first see the first and second pages to see what the topic was originally about, and then see what the last few pages are about. If there is a big difference, it should be locked. If the whole topic was filled with spam from front to end, it should be locked. If it has transformed into a clean topic, it shouldn't be locked. You can build off this.

if you want to post spam, get on aim.

if you still want to post spam, get on aops.

No no. Spamming is bad you know. You should never spam.

Well, almost never that is.

[i]Bubka asked where one of his posts went.  Moderator responds:
You had already made a nearly identical post earlier in the thread.  I have no particular interest in entertaining meta-3s, any more than I am planning on having 3s.[/i]

Ok ok. I am not interested in $3$s either. They are very bad.


Let $M$ be the set of the functions which satisfy
\[\exists T\neq0,\forall x\in\mathbb{R}: f(x+T)=Tf(x). \]
(1) Is the function $f(x)=x$ an element of $M$? Give reasons to your answer.
(2) Suppose that there exists intersection to the graph of the function $f(x)=a^{x}\,(a>0,a\neq1)$ and the graph of $y=x$. Prove that $f(x)=a^{x}\in M$.
(3) Suppose that $f(x)=\sin kx\in M$. Find the range of the real number $k$.

[hide="1"]
No, as if $x=1$, then $1+T=T$ for some $T$, which clearly isn't true.
[/hide]

[hide="2"]
Let's call the point of intersection of the two graphs $(b,b)$.
So $a^{b}=b$.

$f(x+T)=a^{x+T}=a^{x}a^{T}$
Let $T=b$.
$f(a+b) = a^{x}a^{b}= ba^{x}$
So:
$f(x+b)=bf(x), \forall x \in \mathbb{R}$.
[/hide]

[hide="3"]
$\sin{k(x+T)}= T\sin(kx), \forall x$, for some $T$
$\sin{k(x+T)}=\sin(kx+kT)=\sin(kx)\cos(kT)+\cos(kx)\sin(kT)$

$T\sin(kx)=\sin(kx)\cos(kT)+\cos(kx)\sin(kT)$
$\sin{kx}(T-\cos{kT})=\cos(kx)\sin(kT)$
$\tan{kx}={{\sin{kT}}\over{T-\cos{kT}}}$
But the RHS is a constant.
Thus the LHS is a constant, so $k=0, 2\pi$ (or a multiple of $2\pi$).

[/hide]

[hide="another way to do 3"]
If $f(x+T)=Tf(x)$, then plug in $x=0$, so that $f(T)=0$. So there exists a nonzero integer $n$ such that $T=\frac{n\pi}{k}$. Then plug in $x=\frac{\pi}{2k}$, then $f(x)=\sin \frac{\pi}{2}=1$ but $f(x+T)=\sin \frac{(2n+1)\pi}{2}$ is either 1 or -1. So $T=\pm 1$.

If $T=1$ then $f(x+1)=f(x)$, true if and only if $k$ is divisible by $2\pi$. Otherwise, $f(x-1)=-f(x)$, $f(x-2)=f(x)$, $f$ has period two, and $k$ is divisible by $\pi$ (In particular $k$ is an odd multiple of $\pi$ for this case. This works because if $k=\pi m$ and $m$ is odd, let $g(x)=\cos kx$,
\[f(x-1)=f(x)g(1)-g(x)f(1)=f(x)(-1)-g(x)(0)=-f(x)\]

So the $k$ that work are precisely the multiples of $\pi$.[/hide]

[quote="rjt"]
$\sin{k(x+T)}= T\sin(kx), \forall x$, for some $T$
$\sin{k(x+T)}=\sin(kx+kT)=\sin(kx)\cos(kT)+\cos(kx)\sin(kT)$

$T\sin(kx)=\sin(kx)\cos(kT)+\cos(kx)\sin(kT)$
$\sin{kx}(T-\cos{kT})=\cos(kx)\sin(kT)$
$\tan{kx}={{\sin{kT}}\over{T-\cos{kT}}}$
But the RHS is a constant.
Thus the LHS is a constant, so $k=0, 2\pi$ (or a multiple of $2\pi$).
[/quote]
The function $\tan kx$ is a constant implies $k=0$ but $k=2n\pi\,(n\in\mathbb Z)$.

[quote="ifai"][quote="10000th User"][quote="1=2"]I've found f(x)  !!!!

[hide="in here"]

$f(x+1)=f(x)$

$f(x+2)=f(x+1)$

$f(x+2)=f(x)=2(f(x))$

$f(x)=0$[/hide][/quote]
To prove you wrong, see the following:

$f(x+1)=f(x)$ holds when $T=1$

$f(x+2)=f(x+1)$ is the same as above....

$f(x+2)=f(x)=2(f(x))$ here you are mixing a function $T=1$ with another one whose $T=2$

For you to practice, find a non-constant function $f$ when $T=1$. Very very easy.[b] This problem is only for 1=2[/b][/quote]
But his/her name is 1=2... :rotfl:[/quote]
That was his intent...


[color=blue] Two small balls which have the same mass and charge are on a same vertical line at the height $h_1$ and $h_2$.
They're thrown simultaneously in the horizontal direction with the same velocity $v$.
The $1^{st}$ ball collides the ground with the horizontal displacement $l$.
Find the height $h_2$ of the $2^{nd}$ ball at that moment.
Ignore the resistance of the air and the effect of the faradaic charge of the ground.
The gravitational acceleration magnitude is $g$.
[/color]

since both balls experience the same acceleration due to gravity, wouldnt the height of the second ball just be h2 - h1?

[color=blue]You forgot the static electrical force between the two balls, didn't you?[/color]

ah, i knew this problem seemed to easy. I just thought you were saying the balls have equal charge for the sake of being rigorous. This means its calculus time... and I'm lazy.

alright, after being moronic and trying calculus to no avail for around 10 minutes (how do you solve a differential equation like ma = kq^2/x^2???) anyway, I decided to take a more sensible approach involving center of mass.

In the reference frame in which the balls experience no gravitational acceleration, there are no net forces acting on the balls but the electrostatic force. Thus, center of mass equations hold and for every x meters that one ball is away from the center, the other ball is -x meters away. Now go back to the accelerating reference frame. The center of mass, having no net force but gravity acting on it, will have the position function (H1 + H2)/2 - .5g* t^2 meters. t= L/v (this was staring me in the face and i couldnt see it, argh!). Plugging this value into our equation for the position of the center of mass, we get CM = (H1 + H2)/2 - .5g(L/v)^2. Since the bottom ball is on the ground, the top ball will have a height of 2 times this center of mass, so the final answer is, after much frustration,

H1 + H2 - g(l/v)^2

all in all, as the thread title suggests, a nice little problem, despite my own experience with it.

[color=blue]Congratulation man. After all, you got the correct answer.
Isn't the your solution nice?[/color]

ha ha, yeah, the solution is nice. I just get irritated whenever calculus fails me (I <3 calculus).


calculus
integration
trigonometry
calculus computations

Find $ \int e^{x}\frac{(1\plus{}\sin x)}{(1\plus{}\cos x)}\ dx$ and deduce $ \int e^{x}\frac{(1\minus{}\sin x)}{(1\minus{}\cos x)}\ dx$.

hello, we get
$ \int e^x\left(\frac{1\plus{}\sin(x)}{1\plus{}\cos(x)}\right)\,dx\equal{}e^x\tan\left(\frac{x}{2}\right)\plus{}C$
Sonnhard.

Yes, and so the other must be $ \minus{}e^{x} \cot (\frac{x}{2}) \plus{}c$.


Bob has dropped his fiddlesticks!  He was holding 7 different fiddlesticks. In how many unique wasys can he pick them back up?

Is this just $7!=\boxed{5040}$ ?

Yes.
[hide] He can pick any of the 7 the first time, then any of the remaining 6, then any of the remaining 5, all the way up to the last one.  This is 7*6*5...*1=7!=5040 [/hide]

Yes. Think about it. What does $7!$ mean, exactly? It means: \[7\times 6\times 5\times 4\times 3\times 2\times 1\] He has $7$ different selections for the first stick. Since this stick is gone, there are $6$ sticks left and he has to pick his second stick from that pile. Repeating this process gets the above.

What if he took one stick, dropped it back down, and picked up the second stick from this new pile? Assume that he does this process $7$ times.

[quote="myspacesucks"]Vishalarul's methoding is completely incorrect.  If you were to listen to what he believes to be an explanation.  If you listened to him, it would be 7^7, but the answer is 7!.  I have a challenge, can anyone prove this, it's easy if you know some theorems and formulas and functions.[/quote]
His reasoning explains the solutions of 7! because he talks about 7 and then 6 and so on.

[quote="myspacesucks"]Vishalarul's methoding is completely incorrect.  If you were to listen to what he believes to be an explanation.  If you listened to him, it would be 7^7, but the answer is 7!.  I have a challenge, can anyone prove this, it's easy if you know some theorems and formulas and functions.[/quote]

What?

I told them how to do it and then gave them a question to think about that tested their understanding of the problem:
[quote="vishalarul"]
Yes. Think about it. What does $7!$ mean, exactly? It means:

$7\times 6\times 5\times 4\times 3\times 2\times 1$
He has $7$ different selections for the first stick. Since this stick is gone, there are $6$ sticks left and he has to pick his second stick from that pile. Repeating this process gets the above.

What if he took one stick, dropped it back down, and picked up the second stick from this new pile? Assume that he does this process $7$ times.[/quote]

um...
unless i'm missing something, those are asking a new question  :ninja:

[quote="myspacesucks"]Its just the last two sentences I have problems with.[/quote]
That wasn't an interpretation, that was an extension.

You people have somehow made this a very difficut problem...it's just 7!...ok?


trigonometry
inequalities
inequalities open

let $ \alpha,\beta\in(0,\frac{\pi}{2})$
prove that
\[ \frac{\sin\alpha\plus{}\sin\beta}{2} \le \frac{1}{\sqrt{1\plus{}\cot\alpha\cot\beta}}\]

[quote="rex88"]let $ \alpha,\beta\in(0,\frac {\pi}{2})$
prove that
\[ \frac {\sin\alpha \plus{} \sin\beta}{2} \le \frac {1}{\sqrt {1 \plus{} \cot\alpha\cot\beta}}\]
[/quote]
It's wrong. Try $ \alpha\equal{}1^{\circ}$ and $ \beta\equal{}2^{\circ}.$

hello arqady, how did you computed it, with a calculator?
Sonnhard.


real analysis
topology
real analysis unsolved

Let $X$ and $Y$ be metric spaces with $X$ compact, and let $f: X\rightarrow Y$ be a continuous function.  Use the existence of Lebesgue numbers for open covers of $X$ to prove that $f$ is uniformly continuous.

What is  Lebesgue numbers for open covers of $X$? Sorry for this my question  :oops:

For each $x\in X$ there exists $\mathcal{U}\in\mathcal{F}$ (an open set in the open cover) such that

\[B_{\delta}\left(x\right)\subset\mathcal{U},\]
where $\delta$ is the Lebesgue number.

Hint: $X$ is covered by the open sets $f^{-1}(B_{\epsilon}(y))$, $y\in Y$.

[quote="amcavoy"]For each $x\in X$ there exists $\mathcal{U}\in\mathcal{F}$ (an open set in the open cover) such that
\[B_{\delta}\left(x\right)\subset\mathcal{U}, \]
where $\delta$ is the Lebesgue number.[/quote]
But what is  Lebesgue number?  :maybe:

Let's try it again: [b]a[/b] Lebesgue number for an open cover $\mathcal F$ is a number $\delta>0$ such that for each $x\in X$ there exists $\mathcal{U}\in\mathcal{F}$ such that $B_{\delta}\left(x\right)\subset\mathcal{U}$.


modular arithmetic
number theory
greatest common divisor
induction
AMC
USA(J)MO
USAMO

According to my individual score, my #1 apparently either got a 0 or 1 (since my total score is 1...). I've re-read my solution for several times and thought it was fairly legitimate, and someone on this forum had confirmed the correctness of its main content also. Below is a word-for-word transcription of my solution. Thanks for your time and patience.
[hide]
We need to prove that for any $ n \in \mathbb{Z}^ \plus{}$, we can find integer $ a$ and $ n$ pairwise relatively prime integers $ k_0,k_1,...,k_n$ such that
\[ k_0k_1...k_n \minus{} 1 \equal{} a(a \plus{} 1)
\]
Letting $ f(a) \equal{} a(a \plus{} 1) \plus{} 1 \equal{} a^2 \plus{} a \plus{} 1$, this is equivalent as
\[ f(a) \equal{} k_0k_1...k_n
\]
It suffices to prove that $ f(a)$ can have as many distinct prime factors as needed; in the other words, for any $ n \in \mathbb{Z}^ \plus{}$ we can find an $ a \in \mathbb{Z}$ so that $ f(a)$ has at least $ (n \plus{} 1)$ distinct primes. If so, we can always re-write $ f(a)$ as $ (n \plus{} 1)$ pairwise relatively prime integers.
To prove our claim, we need to ensure that
\[ f(a) \equiv 0 \pmod p
\]
(*)
has solution for infinitely many distinct prime $ p$. If this holds, since each prime is relatively prime to another prime, by Chinese Remainder Theorem, we can find a solution $ a$ for any subset of these congruences as a system. In the other words, if the solution to $ f(a)\equiv 0 \pmod{p_i}$ is $ a_i (i \equal{} 1,...,m)$,
$ \begin{cases} a & \equiv a_i \pmod{p_i} \\
... \\
a & \equiv a_m \pmod{p_m} \end{cases}$
will have a solution for $ p_1,...,p_m$ in the set of those infinitely many primes. Then this $ a$ will make $ f(a)$ have $ p_1,...,p_m$ as its distinct prime factors. We'll prove this by introduction.
Suppose (*) has a solution for $ m$ primes $ p_1,p_2,...,p_m$, then there exists $ a \equal{} a'$ such that $ f(a')$ is divisible by $ p_1,p_2,...,p_m$.
Now, consider $ f(a' \plus{} 1) \equal{} a'^2 \plus{} 3a' \plus{} 3$. Because
$ \gcd(f(a'),f(a' \plus{} 1))$
$ \equal{} (a'^2 \plus{} a' \plus{} 1,a'^2 \plus{} 3a' \plus{} 3)$
$ \equal{} (2a' \plus{} 2,a' \plus{} a' \plus{} 1)$
$ \equal{} (2(a' \plus{} 1),a'(a' \plus{} 1) \plus{} 1) \equal{} 1$
(since $ a'(a' \plus{} 1) \plus{} 1$ is always odd)
$ f(a' \plus{} 1)$ is not divisible by $ p_1,p_2,...,p_m$. But $ f(a' \plus{} 1) > 1$, so it must contain another prime $ p_{m \plus{} 1} \neq p_1,...,p_m$. Hence there exists a solution $ a \equal{} a' \plus{} 1$ for
\[ f(a)\equiv 0 \pmod{p_{m \plus{} 1}}
\]
By CRT, there exists a solution $ a''$ for simultaneous congruences
$ \begin{cases} a & \equiv a' \pmod{p_1} \\
... \\
a & \equiv a' \pmod{p_m} \\
a & \equiv a' \plus{} 1 \pmod{p_{m \plus{} 1}} \end{cases}$
Hence there exists a solution for $ (m \plus{} 1)$ primes of (*).
As for our base, consider when $ m \equal{} 2$, $ p_1 \equal{} 3,p_2 \equal{} 7$,in which case
$ f(4) \equal{} 4^2 \plus{} 4 \plus{} 1 \equal{} 21 \equal{} 3\cdot 7$.
Therefore, (*) has solution for infinitely many distinct primes $ p$,and we are done. $ \boxed{QED}$
[/hide]

BTW, for problems like #4 this year, which asks you "to find all these numbers", if you got the answer correct but failed to prove that no others work, how many points do you usually get?

I don't see any big holes in your proof. I did nearly the same thing and seemed to get perfect 7 for it, although I included a proof of the Chinese Remainder Theorem.

Here's the first significant mistake I found:

[quote="timwu"]$ \equal{} (2(a' \plus{} 1),a'(a' \plus{} 1) \plus{} 1) \equal{} 1$
(since $ a'(a' \plus{} 1) \plus{} 1$ is always odd)[/quote]

This line is incorrect.  $ (2(a'\plus{}1),a'(a'\plus{}1)\plus{}1)$ is indeed equal to 1, but your proof seems to indicate that you think odd numbers are always relatively prime to even numbers.  This is not true, take 3 and 6 for the simplest example.

To prove that $ (2(a'\plus{}1),a'(a'\plus{}1)\plus{}1)$, you can note that $ a'(a'\plus{}1)\plus{}1$ is odd, so any number dividing $ a'(a'\plus{}1)\plus{}1$ is odd.  Therefore if the GCD divides $ 2(a'\plus{}1)$ it divides $ a'\plus{}1$, so you can reduce the GCD to $ (a'\plus{}1,a'(a'\plus{}1)\plus{}1)\equal{}(a'\plus{}1,1)\equal{}1$.

I'm not sure this would be worth losing six points, though.  :(

It's the crux of the proof; without it, you can't ensure the existence of a new prime factor.

A stylistic issue:  you write your induction backwards.  You should introduce the $ \gcd$ argument first and the CRT argument second.

[quote="timwu"]wow, I never think in that way. I assumed people would understand I mean they don't have $ 2$ as a CD, plus an obvious(?) fact they don't have $ a' \plus{} 1$ or any its factor as CD as well, they must be coprime. Yeah, this probably counted against me (but at the cost of 6 points?...).[/quote]
I agree. It seems to me that, from what you reduced it to, it was obvious that they were co-prime.

But nothing was explicitly stated; that step wasn't justified, even if it was "obvious."  By the way, the answer to your other question is 0 or 1.

Edit:  one more note about these CRT solutions.  You don't seem to prove that if $ f(a) \equiv 0 \bmod p$ and $ b \equiv a \bmod p$ then $ f(b) \equiv 0 \bmod p$.

I see. For that proof, I suppose (probably wrong again) it is implied in that solutions are are modulo $ p$ (written in congruences), and, although it's bad to say this, it's kind of obvious...

You need to at least cite $ a \minus{} b | f(a) \minus{} f(b)$.  It's not that these are difficult steps; you simply ignore them.

True. I always think that you need to be clear in the main idea, but that seems not enough; any key step should be proved explicitly, even it's "obvious". Thanks for your advice. (And you're right, I'm horrible at style...time to work on it besides doing problems)

That's the main thing I've learned from the USAMO, because I was doing that too when I took it a few years ago.  Even if something seems intuitive, you always need to at least state it--since you have 4.5 hours, you can always spare 30 seconds or so to write down the additional statement.

I think at some point between last USAMO and this one I realized that as well, and as a result my solutions were really long this time (I think I had 9 pages on day 1 and 12 on day 2).

Hm I am also rather confused as to why you got a $ 0$, although that last $ gcd$ reduction is not really immediately clear (also you have a few typos, but that shouldn't matter). However, on problems like this where the proof is only really one or two ideas, I try to state everything as explicitly as possible (I would have, for example, shown why being divisible by AT LEAST $ n$ distinct primes guarantees that we can pick $ n$ relatively prime factors). This probably isn't as necessary on harder problems.


function
induction
algebra unsolved
algebra

Find all functions $f: \mathbb N \cup \{0\} \to \mathbb N \cup \{0\}$ such that $f(1)=1$ and
\[f(m^2+n^2)=f(m)^2+f(n)^2,\]
for all $m,n \in \mathbb N \cup \{0\}$.

[quote="jbourne"]2. n=0
f(m^2)=f(m)^2 =>

answer:
f(x)=x[/quote]

How can you directly deduce $f(x)=x$ from $f(m^{2})=f(m)^{2}$ ???????

--
Patrick

Hello N.T.TUAN
[quote="N.T.TUAN"]...by this method we have $f(k)=k\forall 0\leq k\leq 26$.[/quote]

I'm sorry but this seems not so simple.
Finding $f(7)$, for example, need to use $7^{2}+24^{2}=25^{2}$ and hence to know $f(24)$. this is possible (not immediate) thru $24^{2}+10^{2}=26^{2}$ and $10=3^{2}+1^{2}$ and $26=5^{2}+1^{2}$

But, for computing $f(11)$, the only equation available is $11^{2}= 61^{2}-60^{2}$ so we need to know $f(61)$ and $f(60)$. For $f(61)$, the only equation is $61^{2}=1861^{2}-1860^{2}$ !

So, in order to compute $f(11)$, we need to compute $f(60)$, $f(1861)$ and $f(1860)$ !

So, how could you show that $f(11)=11$ ?

--
Patrick

Now, you are true, i think so. Congratulate for new solution !  :wink:


floor function
function
algebra
polynomial
summer program
MathPath
Euler

Hi, I dont know how to check if 1733 is prime or not. Can sombody hlep me?

You just need to divide by all prime numbers up to the square root of 1733.

Not divisible by 2, because it's odd.
Not divisible by 3, because digits don't add to a multiple of 3.
Not divisible by 5, because it doesn't end in 5 or 0.
etc.

Checking up through 41, 1733 is not divisible by any prime, so it is prime.

As for checking divisibility of each of those primes, there are somewhat complicated rules you can use, or you can just write it all out. No quicker way. :(

It is prime, according to a calculator program I made.

We only have to test primes $p \leq \lfloor \sqrt{1733} \rfloor = 41.$ We test: $\{ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41 \}.$ We can eliminate some of the obvious ones, leaving $\{ 7, 11, 13, 17, 19, 23, 29, 31, 37, 41 \}.$ To test for 7, just substract twice the units digit  from the remaining number, so 173-6=167, and 16-14=2, which is not divisible by 7. For 11, just add or substract alternatingly, like 1-7+3-3=-6, which isn't divisible by 11. For 13 and greater, you can just test on a calculator, or use the divisibility rules for them if you know them.

EDIT: JF197 beat me to it...

But, why square root? 43 is prime, 59 is prime, 89 is prime, why not test those too? Im sorry Im new to this

If you use any number larger than the square root, it must be multiplied by a number smaller than the square root to equal that number.

[quote="10000th User"]ok so for example 37 is prime and 47 is prime, 37x47=1739, $37<\sqrt{1739}<47$. so if no primes below $\sqrt{1733}$ then no primes above $\sqrt{1733}$ and so 1733 is prime?[/quote]
exactly, if $n$ is not divisible by any prime $<\sqrt n$, then $n$ is not divisible by any prime $>\sqrt n$

[quote="Psycho"]exactly, if $n$ is not divisible by any prime $<\sqrt n$, then $n$ is not divisible by any prime $>\sqrt n$[/quote]
OK Tank You

If you happen to have a computer handy you can always check using an online prime factorization machine:  http://britton.disted.camosun.bc.ca/jbprimefactor.htm

[quote="Psycho"]Or if you have a Texas-Instrument 83, you can make a program (it's not that difficult) :)

[quote]
$0\rightarrow \text{dim}(L_1)$
$\text{Prompt N}$
$1\rightarrow S$
$2\rightarrow F$
$0\rightarrow E$
$\sqrt N \rightarrow M$
$\text{While} F \leq M$
$\text{While fPart}(N/F)=0$
$E+1 \rightarrow E$
$N/F \rightarrow N$
$\text{End}$
$\text{If} \ E>0:\text{Then}$
$F\rightarrow L_1 (S)$
$E\rightarrow L_1 (S+1)$
$S+2\rightarrow S$
$0\rightarrow E$
$\sqrt N \rightarrow M$
$\text{End}$
$\text{If} \ F=2$
$\text{Then}:3\rightarrow F$
$\text{Else}: F+2\rightarrow F$
$\text{End}$
$\text{End}$
$\text{If} \ N\not =1:\text{Then}$
$N\rightarrow L_1 (S)$
$1\rightarrow L_1 (S+1)$
$\text{End}$
$L_1$[/quote]

edit-if you enter a number then, you get a "weird" result...
let's say you enter $825$
then you'ld get $\{3 \ \ 1 \ \ 5 \ \ 2 \ \ 11 \ \ 1 \}$
which means that $825=3^15^211^1$[/quote]

Yay! It works. Thanks :)

89s actually have a function:

isPrime(x)

x= whatever number you want to put in

 :starwars:

[quote="someperson01"]89s actually have a function:

isPrime(x)

x= whatever number you want to put in
[/quote]

i've read that too, but they don't give prime factorization if i'm not mistaken :)

they do, factor() does prime factor for numbers and also factors polynomials, etc.

Just checked my TI-89, and yes, there is a isPrime() function. It will simply say "true" if a number is prime, and "false" if a number is composite. factor() will factor whatever you put inside the parentheses, if it can possibly be factored; this includes both composite numbers and polynomials.

well... for those people who don't have a TI-89, my program is quite easy ;)

[quote="236factorial"]Am I correct that no one has found a way to generate primes?[/quote]

There are ways to generate certain primes (Mersenne primes), 'cept they're very far apart.

But it has been proven that there is no polynomial that always generates primes, except something ridiculous like f(x)=5. :lol: (Quote in Mathpath yearbook!!!)

Treething, $2^{32}+1$ is composite.

[quote="shyong"][quote="236factorial"][quote="mathnerd314"]Treething, $2^{32}+1$ is composite.[/quote]

When did he ever say it was prime?  :huh:[/quote]

It is not a prime ! This is fermat numbers when $n=5$ where euler prove that it can be divisible by $641$  :D[/quote]

When did he ever say it was prime? :huh:

[quote="chess64"][/quote][quote="shyong"][quote="236factorial"][quote="mathnerd314"]Treething, $2^{32}+1$ is composite.[/quote]

When did he ever say it was prime?  :huh:[/quote]

It is not a prime ! This is fermat numbers when $n=5$ where euler prove that it can be divisible by $641$  :D[/quote][quote="chess64"]

When did he ever say it was prime? :huh:[/quote]

He never said that it was prime ....

[quote="shyong"]He never said that it was prime ....[/quote]

You said it wasn't prime. When did 236! ever say it was? ;)

[quote="Treething"][quote="236factorial"]Am I correct that no one has found a way to generate primes?[/quote]

[b]There are ways to generate certain primes (Mersenne primes), 'cept they're very far apart.[/b][/quote]

Emphasis mine.

Unless I'm totally mistaken and Mercenne primes are something totally different... But the point is that there is no polynomial that will generate only primes.

EDIT: Well, of course, you couldn't exactly call $2^{2^n}+1$ a polynomial :D

No one said ALL numbers of the form $2^{2^n}+1$ were primes. Well, Fermat did, but what does he know.

[quote="Psycho"]in case you're still wondering: [url=http://www.win.tue.nl/~jessers/aansluiting/priemgetallen.htm]click here[/url] :P (the first 10.000 primes, don't mind the intro: it's dutch ;) )[/quote]

its up to 100.000

[quote="236factorial"]Am I correct that no one has found a way to generate primes?[/quote]

Im not sure, but isn't 10001 a prime? it doesnt show up in the list, and as for the other quote, try

10^X+1

Dunno if it works, something i made up myself

$10^x+1$ does not generate primes.  In fact, it's pretty easy to show that for every odd $x$ it is composite.  Just consider factoring $x^n+1$ when $x$ is odd or, more simply, just show that it is divisible by 11.

Well, there are ways to generate primes (like the Sieve of Eratosthene's) it's just that no one has found a really quick and easy way to do it (like a formula that spits out only primes [i]that is easy to compute[/i]).

I refer you to this quote from http://www.mathworld.com

[quote="mathworld"]
Prime numbers can be generated by sieving processes (such as the sieve of Eratosthenes), and lucky numbers, which are also generated by sieving, appear to share some interesting asymptotic properties with the primes. Prime numbers satisfy many strange and wonderful properties. Although there exist explicit prime formulas (i.e., formulas which either generate primes for all values or else the nth prime as a function of n), they are contrived to such an extent that they are of little practical value.[/quote]


close, but you didn't square the second term on the left side of the equation.

h_s_potter2002 wrote:WarpedKlown1335 wrote:Solve for x:
[tex]x^3 - 3x = \sqrt{x + 2}[/tex]

[hide][tex]{x(x^2-3)} = \sqrt{x + 2} [/tex]
[tex]{x^2(x^2-3)} =  {x+2}[/tex]
[tex]{x^4-3x^2} = {x+2}[/tex]
[tex]{x^4-3x^2-x+2} = 0[/tex][/hide]

Can someone tell me if this is the right track?  I'm pretty sure you're supossed to factor this, but I don't know how to.

You can't just square x in [tex]{x(x^2-3)}[/tex]

That's just multiplying that expression by x, not squaring it.

[tex]x^3  - 3x - \sqrt {x + 2}  = 0 \Leftrightarrow x^3  - 2x^2  + 2x^2  - 4x + x - \sqrt {x + 2}  = 0 \Leftrightarrow [/tex]

[tex] \Leftrightarrow \left( {x - 2} \right)\left( {x^2  + 2x + \frac{{x + 1}}{{x + \sqrt {x + 2} }}} \right) = 0 \Leftrightarrow [/tex]

[tex] \Leftrightarrow x = 2[/tex]    or    [tex]x^2  + 2x + \frac{{x + 1}}{{x + \sqrt {x + 2} }} = 0[/tex]

[b]It is obvious that this equation  [/b][tex]x^2  + 2x + \frac{{x + 1}}{{x + \sqrt {x + 2} }} = 0[/tex][b]  has not a positive solution.[/b]
[b]Then, we shall seek a solution to following ecuation: [/b][tex]x^2  + 2x + \frac{{x + 1}}{{x + \sqrt {x + 2} }} = 0[/tex] [b]only in [/b][tex]D = \left( { - 2,0} \right)[/tex]

[b]We have:[/b]
[tex]x^2  + 2x + \frac{{x + 1}}{{x + \sqrt {x + 2} }} = 0 \Leftrightarrow [/tex]

[tex] \Leftrightarrow x^2  + 2x + \frac{{\left( {\sqrt {x + 2}  - 1} \right)\left( {\sqrt {x + 2}  + 1} \right)}}{{\left( {\sqrt {x + 2}  + 2} \right)\left( {\sqrt {x + 2}  - 1} \right)}} = 0 \Leftrightarrow x^2  + 2x + \frac{{\sqrt {x + 2}  + 1}}{{\sqrt {x + 2}  + 2}} = 0 \Leftrightarrow [/tex]

[tex] \Leftrightarrow x^2  + 2x + 1 - \frac{1}{{\sqrt {x + 2}  + 2}} = 0[/tex]

[b]We denote by  [/b][tex]t = \sqrt {x + 2} [/tex][b]  end ecuation becomes:[/b]

[tex]t^5  + 2t^4  - 2t^3  - 4t^2  + t + 1 = 0 \Leftrightarrow [/tex]

[tex] \Leftrightarrow \left( {t^2  + t - 1} \right)\left( {t^3  + t^2  - 2t - 1} \right) = 0\mathop  \Leftrightarrow \limits_{\mathop {t > 0}\limits^ \uparrow  } [/tex]

[tex] \Leftrightarrow \left( {t^2  + t - 1 = 0} \right) \vee \left( {t^3  + t^2  - 2t - 1 = 0} \right) \Leftrightarrow [/tex]

[tex] \Leftrightarrow t \in \left\{ {t_0  = \frac{{ - 1 + \sqrt 5 }}{2},t_1 } \right\}[/tex]
[b] where  [/b][tex]1.24 < t_1  < 1.26[/tex]

[b]Conclusion:[/b]
[tex]x^3  - 3x = \sqrt {x + 2}  \Leftrightarrow x \in \left\{ {2,t_0^2  - 2,t_1^2  - 2} \right\} \Leftrightarrow x \in \left\{ {2,\frac{{ - 1 - \sqrt 5 }}{2},t_1^2  - 2} \right\}[/tex]

I was wondering how you got the factorization of the 5th degree polynomial with "t". Are there any tricks that I can use in the future?

Oh try finding the exact value of all three solutions  :D .

*Hint, read billzhao's post*

Hem hem billzhao, your solution was the one I read before, the problem was posted on a website somewhere, Mathematical Excalibur I think.

Interesting note: One of metru's answers is phi+1.
Trig sub?  Doesn't get me anywhere...
Also, metru: I'm also wondering how you factored the t^5 polynomial, that is, if you did it quickly.

We assume that: $\exists a,b,\alpha ,\beta ,\gamma  \in \mathbb{Z},$ such as:
$t^5  + 2t^4  - 2t^3  - 4t^2  + t + 1 = \left( {t^2  + at + b} \right)\left( {t^3  + \alpha t^2  + \beta t + \gamma } \right),\forall t \in \mathbb{R} \Leftrightarrow $
$ \Leftrightarrow \left\{ \begin{array}{l}
 a + \alpha  = 2 \\
 b + a\alpha  + \beta  =  - 2 \\
 \alpha b + a\beta  + \gamma  =  - 4 \\
 b\beta  + a\gamma  = 1 \\
 b\gamma  = 1 \\
 \end{array} \right.$

If $b = 1,\gamma  = 1$ arrive at untruth.
But $b =  - 1,\gamma  =  - 1$ arrive at:
$\left\{ \begin{array}{l}
 a = 1 \\
 b =  - 1 \\
 \alpha  = 1 \\
 \beta  =  - 2 \\
 \gamma  =  - 1 \\
 \end{array} \right.$
then we have:
$t^5  + 2t^4  - 2t^3  - 4t^2  + t + 1 = \left( {t^2  + t - 1} \right)\left( {t^3  + t^2  - 2t - 1} \right),\forall t \in \mathbb{R}$

Solve the equation: $x^3  - 3x = \sqrt {x + 2} $

[u]Solution:[/u](H.O.deCastro and Y. Zhao)
If $x <  - 2$ then the right side is not defined.
If  $x > 2$  then
$x^3  - 3x = \frac{{x^3  + 3x\left( {x^2  - 4} \right)}}{4} > \frac{{x^3 }}{4} > \sqrt {x + 2} $
So the solutions must be in $\left[ { - 2,2} \right]$
Write $x = 2\cos t,0 \le t \le \pi $
The equation becomes:
$8\cos ^3 t - 6\cos t = \sqrt {2\cos t + 2} \mathop  \Leftrightarrow \limits_{\mathop {\cos 3\alpha  = 4\cos ^3 \alpha  - 3\cos \alpha }\limits^ \uparrow  } $
$ \Leftrightarrow 2\cos 3t = 2\cos \frac{t}{2} \Leftrightarrow 3t =  \pm \frac{t}{2} + 2k\pi $
$ \Leftrightarrow t \in \left\{ {\frac{{4k\pi }}{5},\frac{{4k\pi }}{7}} \right\}_{k \in Z}  \cap \left[ {0,\pi } \right] \Leftrightarrow t \in \left\{ {0,\frac{{4\pi }}{5},\frac{{4\pi }}{7}} \right\}$

[u]Conclusion:[/u]
$x^3  - 3x = \sqrt {x + 2}  \Leftrightarrow x \in \left\{ {2\cos 0,2\cos \frac{{4\pi }}{5},2\cos \frac{{4\pi }}{7}} \right\}$

[i]Admin edit: Please do not write using bold font. Use bold only if you really want to stress a word or a phrase out. Thanks. [/i]

How can you state that $\frac{x^3}{4}>\sqrt{x+2}$ when $x>2$?

$x > 2 \Rightarrow x^2  > 4 \Rightarrow \frac{{x^2 }}{4} > 1\mathop  \Rightarrow \limits_{\mathop {x > 0}\limits^ \uparrow  } \frac{{x^3 }}{4} > x$

But
$x > \sqrt {x + 2} \mathop  \Leftrightarrow \limits_{\mathop {x > 2}\limits^ \uparrow  } x^2  > x + 2 \Leftrightarrow x^2  - x - 2 > 0 \Leftrightarrow x \in \left( { - \infty , - 2} \right) \cup \left( {1, + \infty } \right)$

Conclusion
If $x > 2 \Rightarrow \left( {\frac{{x^3 }}{4} > x} \right) \wedge \left( {x > \sqrt {x + 2} } \right) \Rightarrow \frac{{x^3 }}{4} > \sqrt {x + 2} $


For all $ a$ solve the equation
$ x\minus{}\sqrt{x^{2}\minus{}a^{2}}\equal{}\frac{(x\minus{}a)^{2}}{2\cdot (x\plus{}a)}$

Let $ u \equal{} \sqrt{x \plus{} a}, v \equal{} \sqrt{x \minus{} a}$.  Then $ u^2 \minus{} v^2 \equal{} 2a$ and

$ u^2 \plus{} v^2 \minus{} 2uv \equal{} \frac{v^4}{u^2} \Leftrightarrow$
$ u^2(u \minus{} v)^2 \minus{} v^4 \equal{} 0 \Leftrightarrow$
$ (u^2 \minus{} uv \plus{} v^2)(u^2 \minus{} uv \minus{} v^2) \equal{} 0$

Because $ u, v$ are non-negative (or else the original expression is not well-defined), the left factor is never zero unless $ u \equal{} v \equal{} 0$, which occurs only when $ a \equal{} 0$.  Simplifying the right factor, we obtain

$ 2a \equal{} uv \equal{} \sqrt{x^2 \minus{} a^2} \implies$
$ 4a^2 \equal{} x^2 \minus{} a^2 \Leftrightarrow$
$ x^2 \equal{} 5a^2 \Leftrightarrow$
$ \boxed{x \equal{} \pm a \sqrt{5} }$.


inequalities limit function search calculus derivative LaTeX

how to give a proof ?

if : $ f$ is strict increasing and continuous on A, then : $ f^{\prime}\left(x\right)  \ >0\,$,for all $ x\in A$

You can use Lagrange theory .
$ f(x)$ is monotone then $ f'(x)>0,\forall x\in A$

[quote="milin"]For every f that is strict increasing and continuous on A, the following inequality is true:

f(x+h)>f(x)

for every x E A and h such that x+h E A!

Now, we have:

f(x+h)>f(x)
f(x+h)-f(x)>0
(f(x+h)-f(x))/h>0

as h->0 we get:

f'(x)>0[/quote]

as h->0we get:

$ f^{\prime}\left(x\right) \ \geqslant0\,$

==============================

f>g ,for all x ,then it not necessarily follow that

$ \underset{x\rightarrow p}{\lim}$ $ f\left( x\right) > \underset{x\rightarrow p}{\lim}g\left( x\right)$

for example f=$ \left\vert x\right\vert$ g=0

Let $ f(x)$ has deriverty on $ (a,b)$ then for every $ x_1,x_2\in (a,b)$ then exist $ c\in (a,b)$ such that
$ \frac{f(x_1)\minus{}f(x_2)}{x_1\minus{}x_2}\equal{}f'(c)$

[quote="TTsphn"]Let $ f(x)$ has deriverty on $ (a,b)$ then for every $ x_1,x_2\in (a,b)$ then exist $ c\in (a,b)$ such that
$ \frac {f(x_1) \minus{} f(x_2)}{x_1 \minus{} x_2} \equal{} f'(c)$[/quote]

but how prove that no exit $ f^{\prime}\left(x\right) \ \equal{} 0\,$,for all $ x\in A$ ?

First of all, you should ask yourself what f' tells you about f. Whenever f' is positive on a given interval, f is increasing on that interval and vice versa. In your problem you have three cases: f' is positive, negative, or zero. You are asked to rule out the last two cases. You could do this by contradiction. Assuming f'(x) = 0 on a given interval, then f is a constant function over that interval and continuous, but not increasing, which is a contradiction. If f' is negative on a given interval, then f is decreasing there, a contradiction.

[quote="luckman168"]if : $ f$ is strict increasing and continuous on A, then : $ f^{\prime}\left(x\right) \ > 0\,$,for all $ x\in A$[/quote]
It's hard to find a valid proof of that, because it is in fact false.

$ f(x) \equal{} x^3$ is strictly increasing on $ \mathbb{R}$ and yet $ f'(0) \equal{} 0.$

$ f(x) \equal{} 2x \plus{} |x|$ is strictly increasing and continuous on $ \mathbb{R}$ and yet $ f'(0)$ does not exist.

I could construct an [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1037519831&t=118219]example[/url]of a strictly increasing continuous function on $ [0,1]$ such that everywhere the derivative exists, it is zero.

---

This would be a true theorem: if $ f$ is nondecreasing, then everywhere that $ f'(x)$ exists, $ f'(x)\ge0.$

The proof is uninspiring. $ \frac {f(x \plus{} h) \minus{} f(x)}{h}\ge0$ by the definition of nondecreasing, and the limit of nonnegative quantities, if it exists, is nonnegative.

Strengthening the hypothesis from "nondecreasing" to "strictly increasing" does not result in a stronger conclusion - see the examples above.

[quote="luckman168"]how to give a proof ?

if : $ f$ is strict increasing and continuous on A, then : $ f^{\prime}\left(x\right) \ > 0\,$,for all $ x\in A$[/quote]

As written is false.This is correct:
Let f be a continuos function in $ [a,b]$ and differentiable in $ (a,b)$ then
$ f$ is strict increasing on $ [a,b]$ $ \leftrightarrow f'(x)>\equal{}0$   $ \forall x\in(a,b)$ and $ f'$ is not identically null in any intervall contained in $ (a,b)$

[quote="Kent Merryfield"]
It's hard to find a valid proof of that, because it is in fact false.

$ f(x) \equal{} x^3$ is strictly increasing on $ \mathbb{R}$ and yet $ f'(0) \equal{} 0.$

$ f(x) \equal{} 2x \plus{} |x|$ is strictly increasing and continuous on $ \mathbb{R}$ and yet $ f'(0)$ does not exist.

I could construct an [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1037519831&t=118219]example[/url]of a strictly increasing continuous function on $ [0,1]$ such that everywhere the derivative exists, it is zero.
[/quote]

thnak's  all , and specially Mr. Kent Merryfield

you are right.

=============================================================================================
[quote="Vinxenz"]
As written is false.This is correct:
Let f be a continuos function in $ [a,b]$ and differentiable in $ (a,b)$ then
$ f$ is strict increasing on $ [a,b]$ $ \leftrightarrow f'(x) > \equal{} 0$   $ \forall x\in(a,b)$ and $ f'$ is not identically null in any intervall contained in $ (a,b)$[/quote]

$ x^{3}$  is continuos differentiable and strict increasing  on $ [ \minus{} 1,1]$

but  f'=0 if: x=0

[quote]
$ x^{3}$  is continuos differentiable and strict increasing  on $ [ \minus{} 1,1]$

but  f'=0 if: x=0[/quote]
Note I wrote $ f'(x) > \equal{} 0$

$ \text{\LaTeX}$ tip: don't write >= but use \ge (or \geq) instead: f'(x) \ge 0 becomes $ f'(x)\ge 0.$


Ss it possible to device a multidigit number with digit $ 7$ as the last digit, so that if this last digit $ 7$ is brought and placed in front of the rest, the new number is five times the original ?

Yes.

Let the number be expressed in the form $ 10m \plus{} 7$ and assume that $ m$ has $ n$ digits. Then from the condition we have $ 5(10m \plus{} 7) \equal{} 7\cdot 10^n \plus{} m$, which can be simplified to $ 7m \equal{} 10^n \minus{} 5$. But $ 10^0\equiv 1, 10^1\equiv 3,10^2\equiv 2,10^3\equiv 6,10^4\equiv 4,10^5\equiv 5\pmod 7$. Thus we can take $ n \equal{} 5$ so that $ m \equal{} \frac {99995}{7} \equal{} 14285$ and therefore the smallest such number is $ 142857$.

This is very similar to http://usamts.org/Tests/USAMTSProblems_18_1.pdf
problem 1.


From a point P ,two tangents PA and PB are drawn to a circle with centr O.if OP=diameter of the circle show that triangle APB is equilateral.

See my solution in the intermediate topic:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179399


Helo
proof:
$ If\ (a,b)\in\mathbb{N}^2\ such\ as\ a>b: \frac{a^2\plus{}b^2}{a^2\minus{}b^2}\notin\mathbb{N}$

Here is one of the 6 or so times that this problem has been posted.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167204


This is a very nice problem:

Let $ X$ be a Hausdorff topological space such that the cardinality of $ X$ is infinite, i.e. $ |X|\geq \omega$
Suppose $ X$ has the property that for any subspace $ Z$ of $ X$ such that $ |Z| \equal{} |X|$, $ Z$ is homeomorphic to $ X$.
Prove that $ X$ has an infinite set of isolated points.

Moreover, it will follow from this that if $ X$ is countable then $ X$ has the discrete topology.

no one?  :maybe:

its a cool problem, you guys should try it.

I'm new to the forums, but I'll give this a shot. I'm assuming that we are using the subspace topology on $ Z$.

[quote="manuel"]Let $ X$ be a Hausdorff topological space such that the cardinality of $ X$ is infinite, i.e. $ |X|\geq \omega$
Suppose $ X$ has the property that for any subspace $ Z$ of $ X$ such that $ |Z| \equal{} |X|$, $ Z$ is homeomorphic to $ X$.
Prove that $ X$ has an infinite set of isolated points.[/quote]

First, we note a theorem: if $ x\in X$ is isolated and $ \phi : X\rightarrow Y$ is a homeomorphism, then $ \phi (x)\in Y$ is isolated.

Proof by induction: prove that $ n \equal{} 1$ isolated point exists, then prove that if $ n$ limit points exist, then $ n \plus{} 1$ limit points exist.

$ n \equal{} 1:$ Suppose $ X$ contains no isolated points. Let $ u,v\in X$ be arbitrary and let $ U,V$ be open in $ X$ s.t. $ u\in U,v\in V,$ and $ U\cap V \equal{} \O$. By the properties of cardinality, there are two cases: (i)- $ |U| \equal{} |X|$ (ii)- $ |X \minus{} U| \equal{} |X|$. (i): Let $ Z \equal{} U\cup \{ v\}$. Then $ |Z| \equal{} |X|$, so $ \exists \phi : Z\rightarrow X$ s.t. $ \phi$ is a homeomorphism. Then, since $ v\in Z$ is isolated, $ \phi (v)\in X$ is isolated, a contradiction. (ii): Let $ Z \equal{} (X \minus{} U)\cup \{ u\}$. Then $ |Z| \equal{} |X|$, so $ \exists \phi : Z\rightarrow X$ s.t. $ \phi$ is a homeomorphism. Then, since $ u\in Z$ is isolated, $ \phi (u)\in X$ is isolated, a contradiction.

$ n\Rightarrow n \plus{} 1:$ Let

$ I \equal{} \{ x\in X|\, x$ is an isolated point of $ X\, \} \equal{} \{ x_1,x_2,\ldots ,x_n\}$

Let $ \{ U_i\}$ be open sets such that $ U_i\cap X \equal{} x_i$ for all $ 1\le i\le n$ (which exist because the $ x_i$ are all isolated). Then, let $ Z \equal{} X \minus{} I$. Since $ I$ is finite, it is clear that $ |X| \equal{} |Z|$, so that $ Z$ is homeomorphic to $ X$. Let $ \phi : X\rightarrow Z$ be a homeomorphism. Since the homeomorphic image of isolated points are isolated in the image set, we know that $ \phi (x_i)$ is isolated in $ Z$ for $ 1\le i\le n$. Take any of the $ \phi (x_i)$, and let $ V \equal{} \phi (U_i)$. It is clear that since $ U_i\cap X \equal{} x_i$, $ V\cap Z \equal{} \phi (x_i)$, and $ V$ is open in $ Z$. Then, since $ Z\subset X$, $ \phi (x_i)\in X$ and $ \phi (x_i)\neq x_j$ for all $ 1\le j\le n$, since $ Z$ does not contain any of $ X$'s isolated points. Now, since $ V$ is open in $ Z$, and since $ Z \equal{} X \minus{} I\Rightarrow X \equal{} Z\cup I$, there exists $ W$ open in $ X$ s.t. $ W\cap X \equal{} V$. Then $ W\subset V\cup \{ x_1,x_2,\ldots ,x_n\}$. Since $ X$ is a Hausdorff space, all of the $ x_i$ are closed, so that $ \{ x_1,x_2,\ldots ,x_n\}$ is closed in $ X$ and $ V \equal{} W \minus{} \{ x_1,x_2,\ldots ,x_n\}$ is open in $ X$. Finally, $ V\cap X \equal{} \phi (U_i)\cap (\phi (X)\cup \{ x_1,x_2,\ldots ,x_n\} ) \equal{} \phi (U_i)\cap \phi (X) \equal{} \phi (U_i\cap X) \equal{} \phi (x_i)$ so that $ \phi (x_i)$ is isolated. Thus, there are $ n \plus{} 1$ isolated points.

By induction, this implies that there must exist infinitely many isolated points. $ \Box$

It may not be the most elegant proof possible, but it looks to me as though it does the trick. Constructive criticism would be greatly appreciated.

Yes, that was my idea.

There is a related question which is open:

Suppose $ X$ is a Hausdorff space which has the property such that every subspace of equal cardinality is homeomorphic to $ X$.
If $ X$ is uncountable, does $ X$ have the discrete topology?

Note that if we drop the Hausdorff condition we can find a counterexample which does not have the discrete topology, for example, the real numbers with the finite complement topology.

When I was trying to prove the $ n \equal{} 1$ case, I realized that the Hausdorff assumption was necessary when I thought of the finite compliment topology... Just thought that was an interesting coincidence.

For the first corollary that you mentioned (countable $ \Rightarrow$ discrete topology), there is a very simple way of doing it. Let $ I$ be the set of all isolated points in $ X$. Since $ |X| \equal{} \aleph _0$, $ |I|\le \aleph _0$. But since I is infinite, $ |I|\ge \aleph _0$. Then

$ \aleph _0\le |I|\le \aleph _0\Rightarrow |I| \equal{} \aleph _0 \equal{} |X|\Rightarrow |I| \equal{} |X|$

So that $ X$ is homeomorphic to $ I$. Since $ I$ has the discrete topology (isolated points are open), $ X$ has the discrete topology. $ \Box$

As to the second part, it seems like proving that $ |I| \equal{} |X|$ might be the best way of going about it, because from there $ X$ has to have the discrete topology. I will think about where to go from here, but nothing is coming to me at the moment.


I want to ask a clarification question on number --, but I don't know if I can ask a clarification question on it. It would only clarify the basic part of the problem. Should I a) ask it here, b) e-mail USAMTS with the question, or c) just state my assumption at the beginning of the question?

You should probably do none of those things.

Our standard reply to all "clarification" question is:

In fairness to all participants, we cannot clarify the problems while the round is in progress.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=294949]this thread[/url].

oops, I meant to say for choice c, "state my assumption [interpretation] of the question at the beginning of my solution."

[quote="DPatrick, in another thread, "]...All USAMTS problems have only 1 correct interpretation.  It would be a substantial failure of our internal vetting process if we had a problem on the contest with more than one valid interpretation.  The problems are read by several experienced people--both USAMTS staff members and people outside the contest who volunteer to read the problems--before a round is made public to try to eliminate any ambiguities in wording...

...Part of the problem is figuring out what the problem means...

...Read the problems carefully. Think about them carefully. There's no rush to solve them: the Round 1 deadline is not until mid-October. We believe that the problem statements are as unambiguous as possible, and we are definitely not trying to trap you with tricky wording...[/quote]
You basically have to keep reading it over and go with your gut instinct. You're on your own.

lets assume the answer to a usamts problem is 3.1415926535.... however i do not believe that it satisfies the problem's wording so i say.

"the only solution to the equation is 3.1415926535.... however this does not solve the problem and so there are no solutions."

and it turns out that the answer was actually

3.1415926535....

would i still get a 5? cause thats pretty much happeneing right now :P


At one of George Washington's parties, each man shook hands with everyone except his spouse, and no handshakes took place between women. If $13$ married couples attended, how many handshakes were there among these $26$ people?

$\text{(A)} \ 78 \qquad \text{(B)} \ 185 \qquad \text{(C)} \ 234 \qquad \text{(D)} \ 312 \qquad \text{(E)} \ 325$

[hide]Each of the 13 men shake hands with 24 other people (excluding wife and self). Each of the 13 women shake hands with 12 other people (excluding husband)

Therefore, $13(24+12)$ shakes take place. However, we forgot to divide by two, so the answer is $13\cdot 18=234$[/hide]

[hide]
There are 26 people at the party. The first man shakes hands with $24$ people because he doesn't shake hands with his wife or himself. The next $23$ so on until the last man has shaken hands which gives you the arithmatic sequence $12+13+14+...24$. Apply the formula and you get $234$.[/hide]

Alternatively...
We have 26 distinct people.  We can pick two of them two shake hands in $\ \binom{26}{2}$ ways.  Then we must exlude handshakes between two women which can happen in $\ \binom{13}{2}$.  Also we must exlude handshakes between a man and his wife which can happen in $\ 13$ ways.  So our final answer is: $\ \binom{26}{2} - \binom{13}{2} - 13 = 204$

Our answer is simply $13*12$ (representing handshakes between men and women)
+ $\frac{13*12}{2}$ (shakes between men), which is simply $234 \implies \boxed{C}$

[hide]With no restrictions, there would be $\frac{25(26)}{2}=325$ handshakes.
Taking away the spouse handshakes, we get 325-13=312.
Taking away the women handshakes ($\frac{12(13)}{2}=78$), we get 312-78=234 handshakes. [/hide]


Jarblek, Belgar, Polgar, and Garion were each paid the same hourly rate to build a bridge. One day Jablek worked the full day, Belgar worked half a day, and Polgar worked half as long as Belgar and a third as long as Garion. Together the four earned 40 lucs. How many lucs did each recieve?

[i]from "Fantastic book of Math Puzzles" by Margaret C. Edmiston[/i]

Please hide Your solution . Write inside Hide ... /Hide

[hide="What I got"]
let $ x$ = a full work day
Jarblek worked 1 full day, or $ x$
Belgar worked $ \dfrac{1}{2}$, or $ \dfrac{1}{2}x$
Polgar worked $ \dfrac{1}{2}$ Belgar's time, or $ \dfrac{1}{4}x$
Garion worked 3 times as long as Polgar, or $ \dfrac{3}{4}x$
They earned a total of 40 lucs, so
$ x \plus{} \dfrac{1}{2}x \plus{} \dfrac{1}{4}x \plus{} \dfrac{3}{4}x \equal{} 40 \dfrac{5}{2}x \equal{} 40 x \equal{} 16$
From that you can find how long each person earned by multiplying each person by the fraction of a full work day they worked.
Jarblek - 16 lucs
Belgar - 8 lucs
Polgar - 4 lucs
Garion - 12 lucs[/hide]
Is it right?
EDIT: oops. Typo I think, or maybe I divided wrong. anyway, thanks latex

Checking tennis work

[hide="Check"]
If you add everything together, you should get $ 40$. Polgar should have gotten $ 4$ lucs. Other wise you're correct.

BTW, I did it the same way as you did.
[/hide]


algebra
polynomial
inequalities
inequalities proposed

a,b,c positive real number such that $abc=1$prove the inequality

 $a^5+b^5+c^5+\frac{ab}{2}+\frac{ac}{2}+\frac{bc}{2} \geq \frac{9}{2}.$

[quote="Ottem"]a,b,c positive real number such that $abc=1$prove the inequality

 $a^5+b^5+c^5+\frac{ab}{2}+\frac{ac}{2}+\frac{bc}{2} \geq \frac{3}{2}.$[/quote]
$a^5+b^5+c^5+\frac{ab}{2}+\frac{ac}{2}+\frac{bc}{2}\geq\frac{9}{2}>\frac{3}{2}.$

the constant $3/2$ should be replaced with $9/2$ and the inequality would still be very easy
[edit]sorry arqady.i did not see your post

Correct. I should have put the 9/2 there as well.  :)


A palindrome is a number that reads the same forwards and backwards. How many palindromes are between 1000 and 10000?

Ways to choose first digit: $ 9$.  Ways to choose second digit: $ 10$.

The third and fourth digits are uniquely determined by the first and second, so the answer is $ 9 \times 10\equal{}\boxed{90}$.

Another way is to think that there are exactly $ 10$ palindromes per hundred, and in $ 9$ hundreds, therefore, there are $ \fbox{90}$ palindromes.

which is exactly what I did?  your "palindromes per hundred" is my second digit and in "9 hundreds" is the first digit.