My sister is a year, a month, and a day older than me. She was born on a day of September and I was born on the 17th of a certain month. If you add all the numerical values of our birthdates, the sum is 4029. When Was I born? Give the exact date, month, and year

[hide="Solution"]Let the narrator's name be $ n$, his sister's name $ s$ and the year the narrator was born be $ y\plus{}1$.

Since $ n$ was born on the $ 17$th, $ s$ is one day older than $ n$, and $ s$ was born in September, $ s$ was born on September $ 18$th. Since $ n$ is one month and one day younger than $ s$, $ n$ was born on August $ 17$th.

Adding these numerical values up, we get $ 8 \plus{} 17 \plus{} 9 \plus{} 18 \equal{} 52$. Hence, the sum of the years of their birth must be $ 4029 \minus{} 52 \equal{} 3977$. So now, we have $ 2y \plus{} 1 \equal{} 3977$ so $ y \equal{} 1988$.

Thus, $ n$ was born on $ \boxed{\text{August 17, 1989}}$.[/hide]

[hide="Solution"]
"Your" sister was born a month before you, in September, which means she were born on September 16, and you were born on October 17. If you were born in year $ y,$ she was born in year $ y-1.$ Taking the numerical values of the dates, \begin{align*}9+6+(y-1)+10+17+y&=4029\\2y+45&=4029\\y&=1992.\end{align*} The final answer is $ \fbox{October 17, 1992}.$[/hide]

wouldnt the 6 be a 16?
wouldnt the answer be 17 October 1989
if not please explain


Hi everybody!!!!

Suppose that the cuadrilateral PQRS is cyclic such that the sides PQ, QR and RS have integer lengts and the diagonals QS and PR also have integer lengts. Moreover, angles QPR and QSR are right angles. My question is: Can we conclude that the fourth side PS also have integer lengt?

we can immediately conclude that it is rational by ptolemy [that is if such a quad exists]...i know that such a quadrilateral exists...12,12,15,13 times a sufficient multiple, but i am not sure that your implication is true...i suppose one should search for counter examples


function
trigonometry
calculus
algebra
domain
topology
calculus computations

Hi
I've got a very basic doubt...i read in my textbook that ' a function is said to be continuous when it is continuous at every point in its domain'....
Now, going by the above $ f(x)\equal{}tanx$ should be continuous at (say) $ pi/2$
However i read elsewhere that it isn't...
Now which of the above is correct?? :maybe:

Function $ y\equal{}tanx$ is not determinable at the points $ x\equal{}\frac{\pi}{2}\plus{}k\pi \equal{}>$ It is not continuous at those points.

You can not consider continuity of function in the points that are not in its domain.

Some books give different definitions for continuity with regard to domain.

For example, under the definition quoted by shyamkumar the function $ f(x) \equal{} \tan x$ [b]is continuous[/b], because it is indeed continuous at every point in its domain. It is not, however, continuous at $ x \equal{} \pi/2$.

But according to the definition given in most introductory calculus classes, f(x) = tan x is not considered a continuous function.

The distinction being that in an introductory calculus course you usually only talk about continuity on an interval. And if a function is not continuous at some point in an interval (regardless of whether the point is in the domain or not), the function is said to be not continuous on that interval.

Whereas in more advanced courses, like topology for instance, the continuity of a function is only determined by its continuity within its domain.

The domain of tanx is -pi/2<x<pi/2
As pi/2 is not in the Domain we need not bother about its continuity

Could someone give me an example of a function that is not continuous over a point in its domain?

EDIT: Ahh nvm, many piecewise functions will do =]


invariant
topology
advanced fields
advanced fields unsolved

If $ G$ is a compact (Hausdorff) topological group, then every neighborhood of $ e$ contains a neighborhood which is invariant under conjugation.

Let $ V$ be an arbitrary neighborhood of the identity. Consider the continuous map $ f: G\times G\to G$ defined by $ f(x,y)=xyx^{-1}$. For every $ x\in G$ we have $ f(x,1)=1$, so by continuity we can find open neighborhoods $ U_{x},V_{x}$ of $ x$ and $ 1$ respectively such that $ f(U_{x},V_{x})\subset V$.

The family $ (U_{x})_{x}$ covers $ G$, so we can find a finite covering $ G=\bigcup_{i=1}^{n}U_{x_{i}}$. It follows from this and from the way we chose the sets $ V_{x}$ that the neighborhood $ V'=\bigcap_{i=1}^{n}V_{x_{i}}$ of the identity is mapped inside $ V$ by every conjugation. Now we can take the open set we're looking for to be the union of all the conjugates of $ V'$.


Can anyone hand me AIME 2004 B solutions????

Thanks in advance,

If you click on the number of the problem where it is posted under "Contests/Resources", that will take you to the discussion of that problem.  If you just want answers, you should go to [url=http://www.unl.edu/amc/]the AMC Website[/url], click on AIME, and go to "Archives."


if:f is increasing on $ \left(  0,\infty\right)  ,S\equal{}\left\{  f\left(  x\right)
{|}\text{ }x>0\right\}$,and$ \underset{x\rightarrow0^{\plus{}}}{\lim}f\left(
x\right)  \equal{}c$,then:$ \left(  \underset{x\rightarrow0^{\plus{}}}{\lim}f\left(
x\right)  \equal{}\inf S\right)$

give a proof plz

Observe that
\[ s: \equal{} \mathop {\lim }\limits_{x \to 0 \plus{} } f\left( x \right) \geq \inf S: \equal{} \inf \left\{ {f\left( x \right): x > 0} \right\}.
\]
If $ s > \inf S$ then there exists $ x_0>0$ such that $ s > f\left( {x_0 } \right) \geqslant \inf S$. Now from the definition, there exists $ \delta >0$ such that for all $ 0<x<\delta$ we have
\[ \left| {f\left( x \right) \minus{} s} \right| \leqslant \frac{1}{2}\left( {s \minus{} f\left( {x_0 } \right)} \right).
\]
Since $ f$ is increasing, then  :D .

Sir, bookworm_vn

                           thankQ ,nice proof

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

can i use below?  is right one?

$ \left\vert f\left( x\right) \minus{} s\right\vert < \left( s \minus{} f\left( x_{0}\right) \right)$,for all $ 0 < x < \delta$

$ \Longrightarrow0 < f\left( x\right) \minus{} f\left( x_{0}\right)$,for all $ 0 < x < \delta$

$ \Longrightarrow Contradiction,$since $ f$ is increasing


Misha's favorite number is even. Her favorite number is also non-negative and less than 1. What is Misha's favorite number?

the only non-negative number less than 1 is 0. cahs


$x,y,z$ are positive such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ prove that
$\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge \sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}$

After squaring both sides, it remains to show $(x+yz)(y+zx) \ge (\sqrt{xy}+z \sqrt{xy})^2$ which is exactly cauchy

It's APMO 2002 question nr 4. A similar solution can be found here:
http://www.kalva.demon.co.uk/apmo/asoln/asol024.html


Find all the roots $ r_{1}$, $ r_{2}$, $ r_{3}$ y $ r_{4}$ of the equation $ 4x^{4}\minus{}ax^{3}\plus{}bx^{2}\minus{}cx\plus{}5 \equal{} 0$, knowing that they are real, positive and that:
\[ \frac{r_{1}}{2}\plus{}\frac{r_{2}}{4}\plus{}\frac{r_{3}}{5}\plus{}\frac{r_{4}}{8}\equal{} 1.\]

The product of the roots is $\frac{5}{4}$. Use AM-GM to get that $1 = \frac{r_1}{2} + \frac{r_2}{4} + \frac{r_3}{5} + \frac{r_4}{8} \geq 4 \cdot \sqrt[4]{\frac{r_1}{2} \cdot \frac{r_2}{4} \cdot \frac{r_3}{5} \cdot \frac{r_4}{8}}=1$.
Thus $\frac{r_1}{2} = \frac{r_2}{4} = \frac{r_3}{5} = \frac{r_4}{8}$ (equality only for equality in AM-GM) and the problem gets trivial.

additional condition that $ a,b,c$ are positive is missing,is it?
[i]Edited:[/i]
yup,my mistake.
I didn't note the inscription below "real".

Given that the roots are positive, Vieta's formula implies the positiveness of $ a,b,c$. Anyway, they are not required, are they?


probability
expected value
combinatorics unsolved
combinatorics

A player is playing the following game. In each turn he ﬂips a coin and guesses the outcome. If his guess is correct, he gains $ 1$ point; otherwise he loses all his points. Initially the player has no points, and plays the game
until he has $ 2$ points.
(a) Find the probability $ p_{n}$ that the game ends after exactly $ n$ ﬂips.
(b) What is the expected number of ﬂips needed to ﬁnish the game?

[hide="hint"]Let us assume the outcome can be 0 or 1, in order to use base 2,
WLOG he always guess 1
therefore in order to win he needs 011 as the last three results.[/hide]

This puts me in mind of this recently posted thread: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157886 .  The "solution method" to the first part is the same: calculate the first 5 values, realize what the answer is, and then prove it with an inductive method (e.g. consider whether the sequence of $ n \geq 3$ flips began with a 1 or with a 0).  The second part of the problem then reduces to a standard exercize in computing sums of series involving linear recurrences.


My dear Mango(my mouse )was on world tour.He took only one day to do that.(he is very much inspired by Usain Bolt)
After his world tour
he asked me,"Kabi,during my world tour there were at least two places which were having same temperature."
How could he say this without any instrument and knowledge about science?
But let me remind u that Mango is a mathematics genius :)

temperature is a continuous function, and the temperature of his starting place and the ending place(which is the starting place) are the same. This implies the result.

but what if the weather is different that day? Not it is becuase temperature on earth is never greater than 120 degrees and if he was talking in whole numbers of deg by the pigeonhole principle there must be 2 that have the same temp.

yes, but if he is talking in whole number of deg, then that the problem is extremely easy because if he only moves a centimeter, the temperature will be the same.

Temperature is a continuous function so it must be strictly increasing or our result is true. I can't see how to prove it mathematically if the temperature is different...

The temperature could also be strictly decreasing as well.

when I have did this ques we have considered the temperature to be same.
thus
simply if Mango is coming to same point then at least there would two place where he could find same temp.
TWO WAYS
by Pigeon hole principle.
also Graphically if u plot temp-dist graph, every graph will give u the same result.
*MANGO'S MYSTERY OVER*

and ya thanks to all u guys :clap:  :clap:


1. Find all triples m, n, l of natural numbers such that

m + n = gcd(m, n)^2 ; m + l = gcd(m, l)^2 ; n + l = gcd(n, l)^2 :

2. The equation ax^3 +bx^2 +cx+d = 0 is known to have three distinct real roots. How many real roots are there of the equation

4(ax^3 + bx^2+ cx + d)(3ax + b) = (3ax^2+ 2bx + c)^2 ?

[quote="msg_1989"]
2. The equation ax^3 +bx^2 +cx+d = 0 is known to have three distinct real roots. How many real roots are there of the equation

4(ax^3 + bx^2+ cx + d)(3ax + b) = (3ax^2+ 2bx + c)^2 ?[/quote]
Set $ f(x)\equal{}ax^3\plus{}bx^2\plus{}cx\plus{}d$ has three distinct real roots $ m;n;p;m<n<p$
$ f'(x)\equal{}3ax^2\plus{}2bx\plus{}c$
$ f''(x)\equal{}6ax\plus{}2b$
$ f'''(x)\equal{}6a$
Equation <=> $ 2f(x)f''(x)\equal{}(f'(x))^2$
Set $ g(x)\equal{}2f(x)f''(x)\minus{}(f'(x))^2$
$ g'(x)\equal{}2f(x)f'''(x)\equal{}12a^2(x\minus{}m)(x\minus{}n)(x\minus{}p)$
Note that $ g(m)\equal{}\minus{}(f'(m))^2<0;g(n)\equal{}\minus{}(f'(n))^2<0$ so this equation has two roots!


Find all positive integers n and d such that both of the following are true:
i) $d$ divides $2n^{2}$.
ii) $n^{2}+d$ is a perfect square.

If $p>2$ is a prime divisor of $d$ and $e$, $\alpha$ positive integers such that  $d=e.p^{\alpha}$ , $(p,e)=1$;
we have $p|n$ and $n=m.p^{\beta}$ for some positive integers $m$ ,$\beta$ with $(m,p)=1$, $\alpha\leq 2\beta$;
Now, because  $n^{2}+d=p^{\alpha}(m.p^{2\beta-\alpha}+e)$ is a perfect square, $\alpha$ is even.
Then $d=2^{r}a^{2}$ with $a$ odd, and $n=2^{s}ab$ with $b$ odd and $r\leq 2s+1$.
We have two situations:
If $r$ is odd, we have $n^{2}+d=2^{r-1}a^{2}(2^{2s-r+1}b^{2}+2)$, and, because there are not two perfect squares with diference 2, there are not such $n$, $d$.
If $r$ is even, we have $n^{2}+d=2^{r}a^{2}(2^{2s-r}b^{2}+1)$, and because there are not two positive perfect squares with diference 1, there are not such $n$, $d$.


advanced fields
advanced fields theorems

are there any sets $A$ such that $A$ is a subset of $A\times A$ ?

$A\times A$ consists of ordered pair of elements of $A$. so how can A be a subset of $A\times A$? if you equip the set with some other additional structure then it may be possible ( by identifying A with some set of elements of AxA or some such thing)

$\emptyset \subset \emptyset = \emptyset\times\emptyset$.

knoppix in what context did you encounter this "misunderstanding" ?

Actually, there may be other examples as well -- it depends how you interpret $A \times A$.  "Ordered pairs" are not a natural set-theoretic notion, but if you define them set-theoretically (for instance, $(a, b) : = \{a, \{b\}\}$ seems like one possibility, although I have some issues with it) then you should be able to derive a list of conditions on a non-empty set $S$ for $S \subset S \times S$, and then either show that they are never satisfied or find examples.

Ordered pairs $(a,b)$ are usually defined as $\left\{ \{ a \}, \left\{ a, b \right\}\right\}$.

so it seems more a matter of convention than a real problem...

If we assume the Axiom of regularity (also known as the Axiom of foundation) then there is no such set except for $A=\emptyset$. This axiom is not particularly important axiom of the Zermelo-Fraenkel axiom system. But it is usually added to the list of other axioms to avoid annoying sets like $A=A\times A$.

I'm sorry for the mistake - there should be :

is there any NONEMPTY set $A$ ......

This is a logical problem (the foundations of math, axioms, etc)  and as I know it's quite hard
I assume Zermelo-Fraenkel System and the Axiom of Regularity

Yury, could you tell something more ?
how AR implies $A\neq A\times A$ for nonempty $A$

The proof is based on the notion of the rank of a set. The rank of a set is an ordinal defined as follows:
i. The rank of the empty set is $0$.
ii. The rank of a non-empty set $a$ is the least ordinal which is greater than the rank of each element of $a$.

(See also
http://en.wikipedia.org/wiki/Rank_%28set_theory%29
http://en.wikipedia.org/wiki/Von_Neumann_universe )

Assume that there exists set $A$ s.t. $A\subset A\times A$. Let $a$ be an element of $A$ of minimal rank. We will prove now that $a\notin A\times A$. The rank of every element $(b,c) \in A\times A$ is at least $\mathrm{rank}(a)+1$:
\[\mathrm{rank}((b ,c)) \geq \mathrm{rank}(b)+1 \geq \mathrm{rank}(a)+1.\]
First we use property (ii) of the definition; then we use that $a$ is the element of minimal rank in $A$. We conclude, that $a\notin A\times A$.


topology
real analysis
real analysis unsolved

Prove that for any nonempty compact metric space $E$, the max{$d(p,q): p,q\in E$ exists.

My idea is to start with a sequence of pairs of points $(p_{n}, q_{n})_{n\in I}$, where $I$ is an index set. I want this sequence to have the property that as n goes to infinity, $d(p_{n}, q_{n})$= least upper bound of $d(p,q)$. We know any compace metric space is closed and bounded, so that this least upper bound must also be included in the metric space.

Can I use contradiction and assume that there is no max?

Thanks.

Well, as for the existence, this boils down to the Least upper bound axiom, doesn't it (knowing that a compact metric space is bounded) ?

If you want to show that the max is attained, you can use the continuity of the distance on the compact product space ...


trigonometry
inequalities unsolved
inequalities

let ABC and DEF be triangles
max{A,B,C} \geq max{D,E,F}
min{A,B,C} \leq min{D,E,F}
prove that sinA+sinB+sinC \leq sinD+sinE+sinF
my teacher have proved that
but it's too long
can you have me?

I have seen a sol of this one in a book of MR. Nguyen Van Mau, you can buy this book.

please use the FULL NAME and AUTHORS of the Book when using the problems source, not just "my book", ok? :)

no idea about this one
i think it is a big research of mt Nguyen Van Mau

i think we should arrange A,B,C and D,E,F

now i have the solution
we can arrange
A>=B>=C
D>=E>=F
so we have (A+B)>=(D+E)
A>=D and C<=F
the first case B>=D
we will prove that
sin A+sinB<=sinD+sin(A+B-D)
then we will prove that
sin C+sin(A+B-D)<=sin E+sin F
the second case B<=D
we do as the same :D
is that right?
thanks a lot for mr NVM


geometry
linear algebra
matrix
real analysis
probability
college
MIT

Bear with me, as this may be extremely unorganized.
So I just finished the college admissions process and am looking forward to being a freshman in college next year. As you can probably tell, from the title, I decided to go to a liberal arts college: Williams.

Before you roll your eyes, let me clarify a couple misconceptions about liberal arts colleges:
1. Liberal arts colleges focus in the humanities: This is simply untrue. A liberal arts college is meant to be a more well-rounded experience. While they may have strengths in the humanities, they also do in the sciences, and even mathematics. In fact, many liberal arts colleges are stronger in those areas than in the humanities (look at harvey mudd, for instance).
2. A small school is bad: The smaller the school, the more attention is placed on each individual student. At a liberal arts college (note: college, not university, implies no grad program) there are no TA's, only profs. They are there not only to do research, which may be the case at large research universities, but are there first and foremost to TEACH YOU!

Let me tell a little bit about my experiences in the college admissions process.
At the end of the day, I was down between Williams and CalTech. Upon visiting both, I felt Williams provided a more well-rounded undergraduate education, and I had plenty of time in grad school to focus in on one thing. Furthermore, the social atmosphere is more diverse. Beyond this, the profs were really great at Williams (I talked to some of them for a long time after sitting in on classes on my visit).

Bottom line, I just want to put it out there that liberal arts colleges are a very welcoming and inviting community that is often forgotten about amongst the talk of big research institutions. They provide just as good (if not better) undergraduate experiences both academically and socially and are great stepping stones to grad or professional schools.
If you want a more well-rounded educational experience and want to experiment in different subjects (who knows? maybe you find out you love anthropology or philosophy or art history?), you may want to consider a liberal arts college.

I'll close by saying this:
Liberal arts colleges will cater to your needs. They will give you attention. You will be able to interact with professors on a very very personal level. At the same time, research institutions, by sheer size, have more opportunities to do things. However, you are much more on your own. If you want to do something, it's doable, but you have to be very active in seeking out what you want to do.
I once heard someone say that it's like comparing a nice dine-in resaurant to a big buffet. The restaurant has waiters to serve you. They have delicious food. They want you to be comfortable. The buffet has more offerings, but you have to get up and get them yourself.

Two different styles. Both great educations.
But I want to make it clear that big name universities aren't the only way.
There are also big name liberal arts colleges that are just as good.

I will say that [url=http://www.williams.edu/Mathematics/]Williams[/url] in particular is a very good place to study mathematics. They have excellent teachers who pride themselves on teaching and student interaction. The abilities of the students in their math department is generally very high, and almost every year the very best students get into top tier graduate schools and get good jobs (see [url=http://www.williams.edu/Mathematics/majorplans.html]here[/url] what this year's math graduates will be doing). The only problem is that students who are extremely advanced and have studied a lot of college level math in high school might have a hard time finding enough math courses to keep them interested (especially since they don't have graduate courses). But the professors there could lead you in independent study if that were to happen.

You say that liberal arts colleges offer a well-rounded experience, and that's true. For this community, it's not what many of us are looking for. We focus intensely on mathematics, and are already specialized to some degree; depth is more important than breadth, and a strong graduate program makes a difference.
What course experience do I remember best from my undergraduate years at Berkeley? Taking the graduate algebra class from Ken Ribet. That sort of thing is impossible at a liberal arts college.

The main difference I see between a Liberal Arts college and a more science/engineering school are the people they attract.  You can still take humanities wherever you go.  For example, here at MIT I can still take a bunch of politics classes if I wish.  Additionally, I doubt the quality of classes will be too incredibly different had I taken them at Harvard, say (especially since I would only be taking a few intro courses).  Now, the main difference I'd imagine would be outside of the classroom.  If I want to talk politics, then there'd be a lot less people willing/knowledgeable around MIT than Harvard.

Anyway, this once again goes along with my views on what is important to picking a college.  I hold that the type of culture there is one of the top three factors to consider.  There are tons of reasons for this.  One that is not often considered is mental health.  College, for most people, is a really rough time mentally.  You need to make sure you're somewhere where you feel comfortable and able to do what you really want to (although you'll spend much of your time trying to figure out what you really want to do...).

It's almost unfair to lump Williams in with liberal arts colleges when talking about studying math.  Williams is noticeably ahead of its liberal arts college peers in math research.  Williams may not be the best choice for a lot of math-minded folks, but Nairbil is certainly right that it's a good fit for *some* and is worth considering.  For one thing, the math teaching quality at Williams is phenomenal.  Without exaggeration, their worst instructor would be a star instructor at any state school.  If the whole liberal arts college thing really appeals to you, and you really like math, it's very likely a good fit for you.

This is sort of tangential, but it seems relevant:  IF you are serious about math grad school, and you go to ANY undergraduate school where the typical math majors don't go to grad school (which is most schools),  you should make sure to give yourself a more thorough education than the typical math major at your school.  Sometimes this means being a go-getter.

[quote="yenlee"]IF you are serious about math grad school, and you go to ANY undergraduate school where the typical math majors don't go to grad school (which is most schools),  you should make sure to give yourself a more thorough education than the typical math major at your school.  Sometimes this means being a go-getter.[/quote]

I agree (and it might be worth knowing, when considering where to apply, which schools don't fall under the "most schools" in your parenthetical -- some schools (e.g. Oberlin, although I don't know specifically about math) send many more of their students to graduate school than other, otherwise similar schools), and in an off-topic plug I'd like to add that in addition to "being a go-getter," for some people it means "going to the Budapest Semesters in Mathematics program" (or Math in Moscow, about which I know far less) in order to demonstrate a high level of mathematical competence.

[quote="blahblahblah"]i tend to think that not being able to take graduate courses as an undergrad is not particularly good for one's mathematical development.  it looks like Williams has a reasonable selection of 400 level courses, but they are not offered every year, nor do they cover the same ground as standard first year graduate courses, and some subjects are not represented at all (PDE, for example).  in comparison, many 400 level courses at research universities are crosslisted graduate courses.  for example, there were something like 12 or 13 of these offered at my school last year, as well as another 10 or so non-crosslisted 400 level courses.  williams, by comparison, offered 400 level courses last year in

geometric group theory
matrix groups
algebraic number theory
'the big questions' (famous unsolved problems in math)

i think there is a large subset of math majors who would find none of those courses particularly compelling.

i didn't have a particularly advanced math background as a high schooler but i think i ended up taking 8 grad courses in my 3 years of undergrad, and my experiences are similar to a lot of motivated math majors.  after a while the natural thing to do is to start taking graduate courses and putting them off is a waste of time.

for example, a lot of people take a course out of Rudin in their first or second year.  the obvious followup course is a course in measure theory in your second or third year, and then you can take a course in functional analysis or harmonic analysis or PDE or (measure-theoretic) probability or whatever.  but williams (for example) doesn't have courses in any of those four.[/quote]

I just want to address this by mentioning that a lot of 400 level courses at williams rotate through the years. So some will be offered, say, this year, and others the next year. Here is a full list of 400 level courses offered in the next two years:

functional analysis with applications to mathematical physics
measure theory and probability
irrationality and transcendence
ergodic theory
analysis and number theory
commutative algebra
intro to p-adic analysis
galois theory
geometric group theory
diophantine analysis
algebraic error correcting codes
algebraic number theory
algebraic geometry
riemannian geometry
hyperbolic 3-manifolds
mathematical modeling and control theory
graph theory with applications
the big questions (about poincare, FLT, riemann hypothesis, P vs. NP, Navier stokes, etc)
there's also the senior project or honors thesis

so the choices aren't as sparse as they may seem (although, granted, much less than a large research institution)

and yes, this is for those folks who are looking for a well rounded experience, but with math as well.
i just dont want people in that category to get swept away by the big names you hear so often.... that's all.
i just want students of that type to realize there are other places out there :) namely, LAC's.

i too can look at the williams course calendar!  it doesn't really change the fact that (apparently) they offered exactly four 400 level courses this year, none of which correspond well to a first-year graduate course.  it is also likely that a significant number of their scheduled 400 level courses get cancelled due to insufficient class sizes (by comparing the number of 400 level courses scheduled for 07/08 with the number scheduled for 08/09)  as i mentioned, my school offered 13 400 level courses corresponding exactly to a first-year graduate course, as well as 10 additional 400 level courses not corresponding to a graduate course.  i'm sure many public research schools have an even greater selection, to say nothing of schools like harvard or princeton or ...

look, i'm sure williams is a good school and the instruction is top-notch (i've personally heard frank morgan speak and read one of his books -- tbh i didn't find his talk particularly compelling but i'm sure many people do), but the truth of the matter is that if you are a bright, motivated math major, the lack of advanced undergraduate courses should be viewed as a significant disadvantage (that, in my opinion, can't really be compensated for).

jbl makes a good point that there are programs which would allow one to supplement one's studies at (say) williams with advanced courses abroad, but i'm sure that there are a lot of people that don't particularly want to be 'forced' to study abroad in eastern europe to obtain an equivalent education.

lol blahblahblah, i think we're speaking to different audiences here.
I've mentioned a couple times that I'm addressing students who want, perhaps, a more well-rounded education with experimentation in various areas. I see that you're talking to the hardcore math types who are determined to go to a grad school for math, etc.
As I've said, I just don't want people in that first category to be pulled away by those in the second. If you need to experiment, LAC's are a great place to do so.
Furthermore, I'd just like to add that when you say classes are cancelled due to insufficient class size, class sizes can be surprisingly small at Williams. Williams offers tutorials, which are classes of two. You, another student, and the professor. That's it.
Now I respect those who are bent on getting a math PhD and so forth, but I'm not talking to you guys per se (as I've mentioned before). If you're a student like me, who likes math, but wants to try other stuff, LACs maybe a good place to go. I know I'm looking forward to taking Chinese and psychology and econ and a writing course on science fiction and so forth.
Not to say I'm bad at math (missed the usamo barely all my years of HS :/), but I'm not as hardcore as many of you. I feel that perhaps there are people viewing this forum who are like me (love math, but perhaps want to work in other areas or simply explore), and I want to help them make the right choice. Sure, I may be a math major undergrad, but I don't plan on getting a PhD in math. Maybe I'd rather go to a professional school or get a grad degree in something else. I don't know.


algebra
difference of squares
special factorizations
square of a binomial

[hide]2000, i did paper and pencil.  Is there a formula for this?[/hide]

[hide]
Please don't do this by arithmetic!  If the problem is in one of these forums, chances are that there's a nice way to do it.  The heart of mathematics lies in providing insights that other people couldn't provide and would thus be unequipped for solving problems or would solve them inefficiently (like using arithmetic on this one).  Say $n = 499$.  Then the difference $D$ in question is:

$D = (n+2)^2 - n^2$
$D = (n^2 + 4n + 4) - n^2$
$D = 4n+4$
$D = 4(n+1)$
$D = 4(499+1) = 4\cdot 500$
$\boxed{D = 2000}$

After you've seen this type of problem enough, you can usually read off the difference just by looking at the numbers.
[/hide]

Thanks Barnacle for explaining that to me. :)

Sure.  Once you get this basic idea of simplifying arithmetic with clever algebra, things really work out nicely.  As a test problem, you can try the following, although its a bit more challenging:

Simplify for $S$.

$S = \frac{2005^2 + 4011}{2006}$.

for the original problem
[hide="original"]$501^2-499^2=(501-499)(501+499)=2\cdot 1000=2000$
difference of squares[/hide]

for Barnacle's problem
[hide="Barnacle"]$S=\frac{2005^2+4011}{2006}=\frac{2005^2+2\cdot 2005+1}{2005+1}=2005+1=2006$
square of a binomial[/hide]

[hide]nice problem.  I'll show the work later because i have to go soon but the answer is 2006[/hide]

[quote="Drunken_Math"]Don't you guys take algebra?[/quote]
this forum is open to all middle schoolers including STARTERS so this problem may not be obvious to some students...

[hide]I used Palytoxin's method..... wow..... i agree. Don't you guys take algebra?  :P  5012 - 4992 = (501-499)(501+499) = 2 * 1000 = 2000[/hide]


algebra
polynomial
calculus
integration
quadratics
quadratic formula

Suppose that t is an integer, $1 \le t \le 100$, and that the equation $x^{2}+tx+(\frac{t^{2}-t}{4})=0$ has an integral root.  Then the number of possible values of t is:

A) 6
B) 7
C) 8
D) 9
E) 10

[hide] The answer is E). In fact, in this equation $\Delta=t$. Clearly, $t$ and $\sqrt{t}$ have the same parity, so all squares $\le 100$ are good. Thus, the answer is 10[/hide]

[quote="mysmartmouth"]Sorry, I don't understand your solution.  What do you mean by: $\Delta = t$?[/quote]

he means the discriminant

[hide]The quadratic formula tells us that

$x=\frac{-t\pm\sqrt{t}}{2}$. $t$ can be any perfect square from 1 to 100, and the division by two doesn't pose a problem since if t is odd, then $\sqrt{t}$ is odd as well, and $-t\pm\sqrt{t}$ will be even, and x will be an integer; the same goes for if t is even. So there are 10 values of t.

[/hide]

[hide="solution"]$x^{2}+tx+(\frac{t^{2}-t}{4})=0$
$x=\frac{-t\pm\sqrt{t}}{2}$
We now know that we want squares from 1 to 100 inclusive. Therefore,there are [b]E)10[/b] solutions.[/hide]


Let

A = number of characters of the Source field that are visible in the list of topics
B = number of characters of the Source field that are visible when one reads the actual
        messages (View Topic)

Then  A < B  by at least some 5-10 characters.  In other words, if the Source field has some
long text, it will be truncated when viewing the messages.  I can find a link to some recent threads where this happens, if it is unclear what I mean.

Let   C = number of characters that can be typed into the Source field when posting a message.

What should be the case is B = C with C being generously large, and A < C.

Yes it is natural for the Source to be truncated (to remain on one line)... we'll see about the rest.

I understand truncating the Source in the list of topics, where space is limited.
But in displaying the actual message, where space is not so limited?

i.e. my intuition is that SourceLength(TopicList) << SourceLength(Displayed Message),
but it is the opposite in the current system.

Actually, I'd like to see more truncation in the topic list. When the source is something like this:
http://www.xthost.info/skydragun/Picture%20170.jpg
(currently on page 7, Calculus Unsolved and Proposed), the source line does not truncate, and the page widens to accomodate the text. The source field can handle multiple lines, but it doesn't break anything without spaces.


There is a vessel withe opening at the top of area $ \text{A}$ and botom closed with the same area $ \text{A}$.The height of the top from bottom is $ \text{H}$ .the volume of the vessel is $ \text{V}$.The density of the liquid in the vessel is $ \rho$.
1.What is the force of the liquid on the vessel's side.
2.What is the force exerted by the bottom of the vessel on liquid.

For 2, find the pressure of the liquid at the bottom; then use the definition of pressure; finally, use Newton's third law.

For 1) consider the forces acting on the water:
$ mg \equal{} F_{\text{bottom}} \plus{} F_{\text{wall,v}}$ in the vertical direction and $ F_{\text{wall,h}} \equal{} 0$ in the horizontal.
Hence, $ F_{\text{wall}} \equal{} \rho Vg \minus{} A\rho gH$

Now use Newton's third law.

thanks. i got the answer after i posted it


geometry
circumcircle
symmetry
geometry solved

In cyclic quadrilateral ABCD, AD = AB + CD.  Prove that the angle bisectors B and C intersect on line AD.

Take $P\in AD$ s.t. $AP=AB,\ DP=DC$. Now take $Q$ to be the second intersection of the circumcircle of $BPC$ with $AD$. I think a pretty basic angle chase shows that $Q$ is the point of intersection of the bisectors:

$\angle CBQ=\pi-(\pi-\angle CPD)=\angle CPD=\frac{\pi-\angle ADC}2=\frac{\angle ABC}2$, so $Q$ lies on the bisector of $\angle B$, so by symmetry it also lies on that of $\angle Q$, Q.E.D.

I knew that I solved it before (7 years ago when I was preparing to IMO).
Source: IMO 1985, 1


Prove that if a,b,c>0 verifiy 3abc=1/a+1/b+1/c, then
   b(1-a^2)/[a(a+b)]+c(1-b^2)/[b(b+c)]+a(1-c^2)/[c(a+c)]>=0.
  God knows I love this one.

Gift to my friends Manlio and Harazi

The main question is: how to use the condition?
My approach: break the inregularity. Let's begin

Put  x= 1/a, y = 1/b, z = 1/c then xyz(x+y+z) = 3 (*). The initial ineq becomes  Sum (x^2-1)/(x+y) >= 0 or Sum x^2/(x+y) >= Sum 1/(x+y).
By Cauchy-Schwarz we have Sum x^2/(x+y) >= x+y+z)/2. Dehomogenusize by (*), is sufficient to show that
        3(x+y+z)/2 >= Sum [sqrt(3xyz(x+y+z))/(x+y)] (**)
(**) is symmetric, homogenus and, the most important, adequate:
Using AM-GM we have sqrt(3xyz(x+y+z)) = sqrt(xy).sqrt(3z(x+y+z)<=(x+y)(4z+x+y)/4 =>  sqrt(3xyz(x+y+z))/(x+y) <= (4z+x+y)/4. Similarly for two other expession. Sum up this ineq's we have (**).

Namdung

Your solution is much more natural and beautiful than mine. Congratulations and thank you for your gift. My solution:
   Obviously, the inequality can be written sum (1/a-(ab+1)/(a+b))>=0, or 3abc>=sum (ab+1)/(a+b). It is enough to prove that (ab+1)/(a+b)<=abc. But (ab+1)/(a+b)<=(ab+1)/(2sqrt(ab))=1/2(sqrt(ab)+sqrt(1/ab))<=1/4( abc+1/a+1/b+1/c)=abc because abc+1/c>=2sqrt(ab) and 1/a+1/b>=2 sqrt(1/ab).

Two wonderful solutions for a wonderful inequality.

Thank you very much.


Diophantine equation
algebra
binomial theorem
continued fraction
superior algebra
superior algebra unsolved

Is it true that all of the units of $ \mathbb{Z}\left[\sqrt{2}\right]$ are contained within the set $ \left\{\pm\left(1\minus{}\sqrt{2}\right)^{n}\ : \ n\in\mathbb{Z}\right\}$?  If so, how do we prove this?

Look for Pell equations. You want $ c\equal{}a\plus{}b\sqrt 2$ such that $ a^2\minus{}2b^2\equal{}\pm 1$. We can restrict to $ a,b>0$ and thus can take the smallest such $ c$. Then show that all other such solutions are powers of $ c$. By considering negative powers and signs, we get all solutions back (and your representation).

I didn't know that this had a name (Pell Equation) but I figured that we would need

\[ \frac{1}{a\plus{}b\sqrt{2}}\equal{}\frac{a\minus{}b\sqrt{2}}{a^{2}\minus{}2b^{2}}\]
so therefore $ a^{2}\minus{}2b^{2}\equal{}\pm 1$ as you suggested.  So given the smallest $ c\equal{}1\plus{}\sqrt{2}$ with $ a,b >0$, the claim is that if $ x\plus{}y\sqrt{2}$ is a unit then $ x\plus{}y\sqrt{2}\equal{}k\left(1\plus{}\sqrt{2}\right)^{n}$ for some $ n\in\mathbb{Z}$ and $ k\equal{}\pm 1$.  By the Binomial Theorem,

\[ \left(1\plus{}\sqrt{2}\right)^{n}\equal{}\sum_{k\equal{}0}^{n}\binom{n}{k}2^{k/2}.\]
How do we continue from here?

Theorem: If $ \left|x\minus{}\frac pq\right|\le\frac1{q^2}$, $ \frac pq$ is a convergent of the continued fraction for $ x$.
As a corollary, if $ d$ is an integer (and not a perfect square), every $ \frac pq$ with $ p^2\minus{}dq^2<\sqrt{d}$ is a convergent of the continued fraction for $ \sqrt{d}$. We can calculate the continued fraction explicitly, and see that our family already contains every possible unit.

Well, you will not need that theorem (nonetheless, it's essential for most elementary proofs of the existence of a solution).

Assume that $ a^2\minus{}2b^2 \equal{} \pm 1$ for $ a,b>0$. Assume that $ a\plus{}b \sqrt 2 \neq ( 1\plus{} \sqrt 2)^n$ for all $ n$. But then you may find an $ n$ such that $ ( 1\plus{} \sqrt 2)^n < a\plus{}b \sqrt 2 <  ( 1\plus{} \sqrt 2)^{n\plus{}1}$. Divide this  by $ ( 1\plus{} \sqrt 2)^n$ (this is the same as multiplying with $ (\minus{}1\plus{}\sqrt 2)^n$!), you get $ 1 < a_0 \plus{} b_0  \sqrt 2 < 1\plus{} \sqrt 2$ for some integers $ a_0,b_0$. They still fulfill $ a_0^2\minus{}2b_0^2\equal{}\pm 1$. Now you have to show that this is not possible (e.g. you can show that $ a_0, b_0>0$).

[quote="jmerry"]Theorem: If $ \left|x \minus{} \frac pq\right|\le\frac1{q^2}$, $ \frac pq$ is a convergent of the continued fraction for $ x$.
As a corollary, if $ d$ is an integer (and not a perfect square), every $ \frac pq$ with $ p^2 \minus{} dq^2 < \sqrt {d}$ is a convergent of the continued fraction for $ \sqrt {d}$. We can calculate the continued fraction explicitly, and see that our family already contains every possible unit.[/quote]
I know how to prove the converse of this theorem, but I'm not quite sure about the approach to this one.  If we assume $ \frac{p}{q}\neq\frac{p_{k}}{q_{k}}\equal{}\left[a_{0},a_{1},\dots ,a_{k}\right]$ for any $ k\in\mathbb{N}$ and that $ \frac{p_{k\minus{}2}}{q_{k\minus{}2}}<\frac{p}{q}<\frac{p_{k}}{q_{k}}$ (the argument would be the same if $ k$ were odd instead of even as assumed here), how do we reach a contradiction?  I was trying to use that fact that

\[ \frac{1}{q_{k}q_{k\minus{}2}}\leqslant\frac{a_{k}}{q_{k}q_{k\minus{}2}}\equal{}\left|\frac{p_{k}}{q_{k}}\minus{}\frac{p_{k\minus{}2}}{q_{k\minus{}2}}\right|,\]
but didn't really get anywhere.  Is this the right approach?  Thank you for your help :).


trigonometry
limit
real analysis
real analysis unsolved

Let x_o be a point in (0,1], and define a sequence {x_n}_n>=0 recursively by setting x_n = sin(x_n-1), for all n>=1.

Prove that lim_n->oo sqrt(n)x_n = sqrt(3).

Say $a_{n}= \frac{\sin x_{n}}{x_{n}}$.
Then, using Cesaro-Stolz,
$\lim_{n \rightarrow \infty}n{x_{n}}^{2}= \lim_{n \rightarrow \infty}\frac{n}{\frac{1}{{x_{n}}^{2}}}= \lim_{n \rightarrow \infty}\frac{1}{\frac{1}{{x_{n+1}}^{2}}-\frac{1}{{x_{n}}^{2}}}= \lim_{n \rightarrow \infty}\frac{{x_{n}}^{2}}{1-a_{n}}\cdot \frac{{a_{n}}^{2}}{1+a_{n}}= 3$

[quote="petruspan"]$\lim_{n \rightarrow \infty}\frac{{x_{n}}^{2}}{1-a_{n}}\cdot \frac{{a_{n}}^{2}}{1+a_{n}}= 3$[/quote]
How did you get that last step?

$\lim_{n\rightarrow\infty}x_{n}= 0$ because $(x_{n})$ is a descending sequence and 0 is the only fixpoint of $x_{n}= \sin x_{n-1}$.
According to Taylor is $\sin x= x-\frac{x^{3}}{3!}+{\cal O}(x^{5})$
So $\frac{{x_{n}}^{2}}{1-a_{n}}= \frac{1}{\frac{1}{6}-{\cal O}(x^{2})}$


If the ten letters in MATHCOUNTS are arranged at random, what is the probability that the work MATCH will be spelled out anywhere in the arrangement? Express your answer as a common fraction.

[hide]
I got $\dfrac{1}{5040}$ :?

total possible, 10x9x8x7x6x5x4x3x2x1= 3628800

if M is the first letter, then 1x1x1x1x1x5x4x3x2x1, since the next four letters have to be A,T,C,H and that leaves 5 letters, and that equals 120

M can be any of the first 6 letters so 120x6=720

$\dfrac{720}{3628800}=\dfrac{1}{5040}?$ :? [/hide]

careful! Is the total amount of ways 10!?

[hide="answer"]The total number is $\dfrac{10!}{2}=1814400$ because there are two of the letter T. There are $6$ places for match to occur (M is first second third fourth fifth sixth). No matter what position MATCH is in, there are $5$ other letters in the arrangement. The ways for this arrangement is $5!=120$ because there is just one T (the other is in MATCH). There are $6$ ways for places for MATCH so $6*120=600$ ways total. $\dfrac{720}{1814400}= \boxed{\dfrac{1}{2520}}$[/hide]

[hide]
The bottom of the fraction is $\dfrac{10!}{2}$, since there are 2 Ts. The top is $5!*6=720$ since there are 5 slots for the words that are not used in the word MATCH, multiplied by 6 because there are 6 ways to arrange MATCH. $\dfrac{6\cdot5!}{\dfrac{10!}{2}}=\boxed{\dfrac{1}{2520}}$[/hide]


Hey I was just wondering if anyone has ever heard of or been to or is going to this National Junior Leaders Confrence thingy.  I got a letter and got accepted for one of them and I'd was curious if anyone from here was going or thinking about it.

Are these the same thing as the forums they have for technology, medicine, law etc.?  If so then I ve been invited to quite a few of them.  I havent signed up for any due to financial resources, time and interest conflicts.  They seem pretty cool though.

[quote="perplex"]yes, i have heard that they're a lot of fun, but in my opinion (being merely an upper middle class public high school student with a pretty tight budget), it's a waste of money. one thing is for sure: it won't give you any sort of boost whatsoever during college admissions because practically all of the prestigious college applicants receive invitations.[/quote]

Okay, that definitely depends on what perspective you're coming from. As a debater, I see more "point" in something like this than a math camp since I definitely won't be majoring in math or sciences. (No offense to anyone) And perhaps not everyone does everything for their college application. Sure colleges want to know what you do during the summer, and if debate/leadership/politics happens to be part of your lifestyle and is expressed through your extra-curricular activities, awards, etc. then I see no reason why you shouldn't go. After all, it would seem a lot less like application padding than studying number theory in Boston.

One other thing...most camps you pay several thousand dollars for will not be prestigious. A math camp like USA-Canada or PROMYS is probably valued at the same level as this is.


function
real analysis
real analysis unsolved

[b]Suppose that  f n   and   f  are measurable complex-valued functions and  ϕ: C →C .
	a . if  ϕ  is continous and fn→ f a.e. ,then  ϕ o  fn → ϕ o f
	b . ………………
	…[/b]

fix an $ x\in X$, $ \epsilon>0$
$ f_n \to f$ on $ X\setminus N$ for measure zero set N
and $ \exists \delta >0$ s.t. if $ |z\minus{}f(x)|<\delta$
then $ |\phi (z)\minus{}\phi(f(x))|<\epsilon$
and since $ f_n(x) \to f(x)$, $ \forall \delta>0, \exists N_\delta$s.t.
$ n>N_\delta \to |f_n(x)\minus{}f(x)|<\delta$

all you wanted was pointwise!

[quote="kenn4000"]fix an $ x\in X$, $ \epsilon > 0$
$ f_n \to f$ on $ X\setminus N$ for measure zero set N
and $ \exists \delta > 0$ s.t. if $ |z \minus{} f(x)| < \delta$
then $ |\phi (z) \minus{} \phi(f(x))| < \epsilon$
and since $ f_n(x) \to f(x)$, $ \forall \delta > 0, \exists N_\delta$s.t.
$ n > N_\delta \to |f_n(x) \minus{} f(x)| < \delta$

all you wanted was pointwise![/quote] :o

very intresting.........

thanks i could solve this problem


Prove that every 4-chromatic graph contains a subdivision of K4

What does "subdivision" mean?  (There are 4-chromatic graphs with arbitrarily large girth, so it certainly doesn't mean "induced subgraph.")


linear algebra
matrix
probability
probability and stats

A Markov chain on state space {1,2,3,4,5} has transition matrix

$ \begin{pmatrix} 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 4/{5} & 1/{5} & 0 \\
0 & 1/{6} & 2/{3} & 0 & 1/{6} \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
\end{pmatrix}$

The process starts in state 1.

a) Which states are absorbing?

States 4 and 5.
b) Calculate the probability that the process is absorbed at state 4 (ends up at state 4).

c) Calculate the expectation of the time of absorption.

d) Calculate the expectation of the number of visits to state 2 before absorption.

e) Suppose that you lose £5 for each visit to state 2 and gain £10 for each visit to state 3.
Use first step analysis to calculate the expectation of the amount you gain.

f) How could you work out the answer to e) from c) and d) without doing the first step
analysis of part e)?

I am missed my lectures due illnes so I don't have much understanding of this topic. I would be grateful for some help with this revision question.

Thanks

Hi;

By direct calculation of the transition matrix to large powers.

b) The P(ending up in state 4) = 1 / 6


In the adjoining figure, $AB$ is a diameter of the circle, $CD$ is a chord parallel to $AB$, and $AC$ intersects $BD$ at $E$ with $\angle AED = \alpha$. The ratio of the area of $\triangle CDE$ to that of $\triangle ABE$ is

$\text{(A)} \ \cos \alpha \qquad \text{(B)} \ \sin \alpha \qquad \text{(C)} \ \cos^2 \alpha \qquad \text{(D)} \ \sin^2 \alpha \qquad \text{(E)} \ 1 - \sin \alpha$

[hide]$AD \perp BD \Rightarrow  \triangle DAE$ is right $\Rightarrow cosa = \frac{DE}{AE}$

$\frac{(\triangle CDE)}{(\triangle ABE)} = \frac{DE \cdot CE}{AE \cdot BE} = \frac{DE^2}{AE^2} = cos^2a$[/hide]

[quote="adidasty"]I understand Iverson's method^^^^  but I'm not getting pontios's last step.  I'd like someone to clarify that for me.[/quote]

What don't you understand about his last step?  :huh: , $DE=EC$ and $AE=EB$, and cosine is adjacent/hypotenuse

By the way, I made a mistake in my solution, I assigned $\alpha$ for the wrong angle, but either way, it comes out the same, since $\cos^2(60)=\frac{1}{4}$.

[quote="beta"][quote="Iversonfan2005"]Another way:
[hide]
Let $\alpha=120$, so therefore $\triangle{ABE}$ and $\triangle{DEC}$ are composed of two $30,60,90$ triangles. By letting $AB=2$ we see that $EB=\frac{2\sqrt{3}}{3}$ and since $\triangle{ABD}$ is a right $30,60,90$ triangle (since it's hypotenuse is the diameter), we know $DB=\sqrt{3}$ and $DE=\frac{\sqrt{3}}{3}$. Since the two triangles are similar and the side ratio is $1: 2$, the area ratio is $1: 4$ and $\cos^2(120)=\frac{1}{4}$ so that is our answer.[/hide][/quote]

lol taking advantage of the multiple choices... Can you generalize though?[/quote]

Yeah, I shouldn't have done it that way, oh well, my chance to redeem myself :P
We know $\triangle{ABE}$ is similar to $\triangle{DEC}$ so the ratio of their areas is the ratio of their sides squared, so let's pick $\frac{DE^2}{AE^2}$, and $\cos{\alpha}=\frac{DE}{AE}$, so the answer is $\cos^2{\alpha}$, which is pretty much what pontios did

adidasty, use the fact that if two figures are similar with ratio $a: b$, their area ratio is $a^2: b^2$.


function
induction
algebra
domain
complex numbers
number theory
number theory proposed

Our function f(x) is defined by two identities:
f(ab) = f(a) + f(b)
f(a+b) = f(a)f(b)

First of all,

f(x) = f(1x) = f(1) + f(x), so f(1) = 0.
0 = f(1) = f(-1 * -1) = f(-1) + f(-1) = 2f(-1), so f(-1) = 0.
f(x-1) = f(x)f(-1), so f(x-1) = 0.
f(x+1) = f(x)f(1), so f(x+1) = 0.

By induction, f(x) = 0 at all integers.
But what happens when we consider the rationals?
Let rational r = a/b with a and b being coprime integers.

0 = f(a) = f(r * b) = f(r) + f(b) = f(r), so f(r) = 0.

Imaginary numbers?

0 = f(-1) = f(i * i) = f(i) + f(i) = 2f(i), so f(i) = 0.
f(ri) = f(r) + f(i) = 0.

Complex numbers? Let a and b be real rationals.

f(a + bi) = f(a)f(bi) = 0.

Algebraic numbers? This is where I need help.

This is not number theory.  You are also thinking too hard.  Letting $ a \equal{} b \equal{} 0$ in the first identity gives $ f(0) \equal{} 2f(0) \implies f(0) \equal{} 0$.  Letting $ b \equal{} 0$ in the second identity gives $ f(a) \equal{} f(a) f(0) \equal{} 0$.

Wow, I guess that is a lot shorter. :oops:

Actually, take note of this; an even shorter proof!

$ f(0a) \equal{} f(0) \plus{} f(a) \rightarrow f(0) \equal{} f(0) \plus{} f(a) \rightarrow f(a) \equal{} 0$

So it seems that any function that:

1. is "additive" in the sense of the first identity
2. includes zero in its domain

is zero everywhere. So the second identity isn't even needed.
If we specifically restrict zero from the domain, that would be when we could use my first post.

The reason I put this in number theory was because I was attempting to construct a number theoretic additive function (like $ \ln(n)$ or $ \Omega(n)$) for which the second identity held; then I realized it would necessarily be zero for every integer, and so on.


hi everyone,

i'm taking the GRE subject test in math in about 8 days. right now the only resources i have are 1) a book by princeton review that has 1 full practice test and 2) the practice test given by the ETS to all those who registered.

can anyone here help me to get some more practice papers( may be give links or something like that)  for the math gre?   thanks for any kind of help

focus focus and ummm.... practice... and focus focus and... FOCUS! and practice many practice tests that are challenging.... not boring easy stupid ones that won't help you learn anything  :)

search online, there are bound to be challenging practice tests.


MATHCOUNTS
geometry
trigonometry
3D geometry
algorithm

I am currently an average MATHCOUNTS-level student. I want to improve. How should I learn and master geometry, trigonometry, summations, and Algebra II?

I've been at this level for quite some time and am frustrated by it.

I need to get better!

1. Get books from the library or buy them.

2. Do lots of problems.

3. Do more problems.

4. Make sure you understand why you got problems wrong.

5. Read all the solutions.

6. Post your questions on the forum and ask for hints, not solutions.

7. Do even more problems.

8. See if you can take a class at school.

9. Do even more than more problems.

Go [url=http://www.artofproblemsolving.com/Classes/AoPS_C_About.php]here[/url] and take a class.  Go [url=http://www.artofproblemsolving.com/Books/AoPS_B_Texts.php]here[/url] and buy some books.

AoPS Vol. 1 and 2 are probably a good choice to take, but, for me, they go to fast and I can't understand them tat well until I read it over and over and over again.  The introductions are probably nice (i don't have any introductions).

What you said here:

[quote="akhil0422"]What stuff should I memorize? I mean, I don't just want the basics; I want a SUPER FIRM standing like multiplication up of 50x50 (or more, Im just guessing), or cubes uptil 40 (im just guessing i dont know how much to memorize)

I don't want to depend on brute calculations unless they are exceptionally huge or something/[/quote]

sounds like a different question than what you originally asked:

[quote="akhil0422"]How should I learn and master geometry, trigonometry, summations, and Algebra II?[/quote]

Which question do you mean?

Um... both. I want knowledge to do stuff, and I also want knowledge to do it faster.

Brute computation is the most time-consuming work of all so... both.

[quote="akhil0422"]Brute computation is the most time-consuming work of all so... both.[/quote]

For multiplying numbers, use those Algebra I factor skills. :)

i.e. Evaluate $ 87 \times 93$.

$ 87 \equal{} 90\minus{}3$

$ 93 \equal{} 90\plus{}3$

$ 87 \times 93 \equal{} (90 \minus{} 3)(90 \plus{} 3) \equal{} 90^2 \minus{} 3^2$.

This is a lot easier to compute!

What about radicals such as $ \sqrt[4]{5}$? I meant many things that are useful, not only squares.

Here are [url=http://en.wikipedia.org/wiki/Methods_of_computing_square_roots]square root approximation methods[/url] and here is an [url=http://en.wikipedia.org/wiki/N-th_root_algorithm]$ n^{th}$ root approximation method[/url].

Fast calculation is less important than the ability to solve the problems. Calculation can be improved by FTWing occasionally; problem-solving is only improvable by problem-solving.


function
complex analysis
complex analysis unsolved

Consider a function analytic in the unit disk satisfying $f(2z)=\frac{f(z)}{1+f(z)^{2}}$ and $f(0)=0$.

1. If such a function exists can it be continued to a meromorphic function on $\mathbb{C}$?

2. Does such a function exist?

Consider the Taylor expansion of $f(z)$ at $0$: $f(z) = A z^{k}+O(z^{k+1})$ (where $A\neq 0$, $k\geq 1$). Then
\[A (2z)^{k}= \frac{A z^{k}}{1+A^{2}(2z)^{2k}}+O(z^{k+1}) = A z^{k}+O(z^{k+1}).\]
Therefore, $A=0$. We conclude that no such function exists.

I don't undestand the second equality (and the first seems to me wrong since 2z is z).


Find all pair of integers $ (x,y)$ satisfying $ x^2\minus{}y^3\equal{}1$

Catalan's conjecture states that for $ x^a\minus{}y^b$ with $ x,y,a,b > 1$, the only solution is $ 3^2\minus{}2^3\equal{}1$. But we are working in integers here, so $ (\minus{}3)^2\minus{}2^3$ also works. since $ x^2$ is greater than or equal to $ 0$, the only solution where $ y$ is negative is $ x\equal{}0, y\equal{}\minus{}1$. If $ y\equal{}0$, we have $ x\equal{}1, y\equal{}0$ is a solution. Otherwise there are no solutions.

Catalan's Conjecture's proof has been completed by Mihailovski just in 2004. (Look here for a more detailed but brief information, [url]http://mathworld.wolfram.com/CatalansConjecture.html[/url].) So I think it's not so approriate to use this strong theorem in olympiad questions. There are lots of questions that can be solved easily, maybe trivially, with the use of 'former' Catalan's Conjecture, but it doesn't fit in with the spirit of mathematical olympiads. Actually, "What is the limit when we are using a theorem in an olympiad?" is a really serious problem for mathematical olympiads, in my opinion.


geometry
MATHCOUNTS
rectangle
linear algebra
matrix

[color=blue]kyyuanmathcount: Look below, meenamathgirl posted the problem.[/color]

Hey I found it now!

[url]http://www.mnspe.org/mathcounts/2005statemeet/510-05SprintS.pdf[/url]

[url]http://www.mnspe.org/mathcounts/2005statemeet/510-05AnswerKeyS.pdf[/url]

Yeah, me neither.

Could you give the url that you used?

[quote="mathgeniuse^ln(x)"][quote="meenamathgirl"]The question goes:
What is the area of the region bounded by the three lines with equations $2x+y=8$, $2x-5y=20$, and $x+y=10$?[/quote]

[hide]Finding the three intersection points we get

$(5,-2)$, $(-2,12)$, and $(10,0)$

Now we draw the smallest rectangle that surronunds the triangle, which has vertices $(-2,12)$, $(10,12)$, $(10,-2)$, and $(-2,-2)$

So the area of the rectangle is 168 and then you subtract the three triangles, which you can figure out.

After that you will get the answer of 42.

$42$[/hide][/quote]
Or you could do shoestring.

[quote="Klebian"][quote="mathnerd314"]I have a copy of all the tests from 1997 to 2005 (haven't downloaded the 2006 ones yet :roll:) on my computer.

Apparently Mathcounts never removed the 2001-2005 tests from the site, they just removed the links to them. Just magical URL-changing (change 2005 to 2003 or whatever)[/quote]

Are you sure? I can't get the other tests by changing URL..[/quote]

Weird, last time (i.e. spring 2006) I just googled "2003 state sprint round" or something and got them--maybe they took them down already. I have the pages bookmarked on another computer.

[hide]Solving them pairwise obtains the intersections (5,-2), (10,0), and (-2,12). The line 2x-5y=20 intercepts the x-axis at (4,0). The area of the triangle is the sum of two triangles with base (4,0) to (10,0), base length 6, and combined heights length 14.

$6\times 14/2=42$.[/hide]

[quote="bpms"][quote="mathgeniuse^ln(x)"]Is that the one which involves a determinant?

I think that the best way of doing a problem is without a complicated formula.[/quote]
What is a determinant?[/quote]

A matrix

It is not that helpful, thoguh.

[quote="bpms"][quote="mathgeniuse^ln(x)"]Is that the one which involves a determinant?

I think that the best way of doing a problem is without a complicated formula.[/quote]
What is a determinant?[/quote]

A determinant is a number that is used with matrices (you learn about them in Algebra II) that is used to find solutions to systems of simultanious equations.

Example of a determinant:
$\text{det}(\left[\begin{array}{cc}a&b\\c&d\end{array}\right])=ad-bc$

On your calculator, it is the command det(.

[hide]
If you draw the lines, the three lines form a triangle. It turns out to not be a right triangle, but if you cut the triangle up in to three pieces, it is easy to add the areas up to get the answer.

Using the x-axis, you can split the triangle into another triangle.
Then, draw a line from where the line $2x+y=8$ intersects $2x-5y=20$ to the x-axis.
That line divides the triangle lower than the x-axis into two triangles with sides 1 and 2, and 5 and 2.
The upper part's area is 36, because the triangle formed with the x-axis and the line down to the x-axis from the intersection
of the lines $x+y=10$ and $2x+y=8$ has an area of 72. and the triangle with the same line for the same leg that isn't the x-axis has and area of 36.
$72-36=36$
So, $36+5+1=42$, the answer to life, the universe, and everything. [/hide]


Let $ \alpha$ be the positive root of the equation $ x^{2} \equal{} 1991x \plus{} 1$. For natural numbers $ m$ and $ n$ define
\[ m*n \equal{} mn \plus{} \lfloor\alpha m \rfloor \lfloor \alpha n\rfloor.
\]
Prove that for all natural numbers $ p$, $ q$, and $ r$,
\[ (p*q)*r \equal{} p*(q*r).
\]

Easily change it into the form $ [(\alpha \minus{} 1991)\{p\alpha\}\{q\alpha\}] \equal{} 0$ which is obviously true

I dont know why we have $ [(\alpha \minus{} 1991)\{p\alpha\}\{q\alpha\}] \equal{} 0$ ? :?:

Okay my friend,here is the sketch of my solution :)

in the fifth line you've been producted the term $[\alpha r]$ to the second statement.... am I right?


Do you guys have an official usatt rating?

Lol I have started 3 months and I simply love the game. Was trying the backhand wrist shot today. A beauty no doubt if properly executed with a neat follow through.


algebra
polynomial
topology
advanced fields
advanced fields unsolved

Let $S$ be the set of monic polynomials of degree $n$ over $\mathbb{R}[X]$.
Is $\{ (P,Q) \in S \times S ~|~ P \text{ and }Q \text{ coprimes}\}$  open? Dense?

In what topology? I don't see an obvious default choice for a topology on $\mathbb{R}[X].$

Kent,

It's not just $\mathbb{R}[X]$, $S$ is the set of monic polynomials of degree $n$ over $\mathbb{R}[X]$, and $n$ is fixed of course :wink:  in other words, $S$ is a subset of $\mathbb{R}_{n}[X]$, which is a of finite dimension over $\mathbb{R}$.

Let $R(P,Q)$ be the resultant of $P$ and $Q$. It is a polynomial in the coefficients of $P$ and of $Q$ that vanishes exactly when $P$ and $Q$ are not coprime.
So your set is open and dense both in the usual and Zariski topology.


Putnam
vector
linear algebra
matrix
modular arithmetic
college contests

Let $1,2,3,\dots,2005,2006,2007,2009,2012,2016,\dots$ be a sequence defined by $x_{k}=k$ for $k=1,2\dots,2006$ and $x_{k+1}=x_{k}+x_{k-2005}$ for $k\ge 2006.$ Show that the sequence has 2005 consecutive terms each divisible by 2006.

Consider the vectors $\vec{v}_{k}=(x_{k},x_{k+1},x_{k+2},\dots,x_{k+2005})^{T}.$
We claim that $\vec{v}_{k+1}=A\vec{v}_{k}$ where
\[A=\begin{bmatrix}0&1&0&0&\cdots&0\\ 0&0&1&0&\cdots&0\\ 0&0&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&0&\cdots&1\\ 1&0&0&0&\cdots&1\end{bmatrix}\]
We can explicitly show that $A^{2006}-A^{2005}=I$ which shows that $A^{-1}=A^{2005}-A^{2004}.$ We can compute this explicitly as
\[A^{-1}=A^{2005}-A^{2004}= \begin{bmatrix}0&0&0&\cdots&-1&1\\ 1&0&0&\cdots&0&0\\ 0&1&0&\cdots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&1&0\end{bmatrix}\]
We're going to be working $\pmod{2006}.$ We have to be somewhat careful becuase $\mathbb{Z}_{2006}$ is not a field, but by direct inspection of the arithmetic, we see that our expression for $A^{-1}$ works $\pmod{2006}$ as well as in $\mathbb{Z}.$

Considered $\pmod{2006},$ the vector $\vec{v}_{k}$ can assume one of a finite nubmer of possible values. This number is $2006^{2006}-1,$ with the $-1$ in that the zero vector could not give rise to a nontrivial sequence. Since this sequence, given by a one-step recurrence, has a finite range, then it must be eventually periodic (with period less than $2006^{2006}$). However, since the matrix $A$ is invertible $\pmod{2006},$ the sequence can be run in reverse as well. There is no initial segment; the sequence $x_{k}$ is periodic over $k\in\mathbb{Z}.$ The reverse recurrence is that $x_{k-1}=-x_{k+2004}+x_{k+2005}.$ From this we find that $x_{0}=x_{-1}=x_{-2}=\cdots=x_{-2004}=1.$ That is, we have $2006$ consecutive $1's.$ Now, reach back before that with the backwards recurrence and we find that $x_{-2005}=x_{-2006}=x_{-2007}=\cdots=x_{-4009}=0,$ which is our requisite $2005$ consecutive zeros. Since the sequence is periodic, we will also encounter this stretch of zeros ($\pmod{2006}$), going forward, Hence there are $2005$ consecutive numbers that are divisible by $2006.$

I did just about the same thing, but with no matrix -- you can go backwards, and any such sequence is cyclic in any mod, and that's all you need.

[quote="1234567890"]Does anyone have a more elementary solution?[/quote]
I'd say the matrices are optional, but you are going to have to do arithmetic mod 2006 and I think the idea of running the recursion backwards is utterly vital - no way is there enough time to go forwards.

While thinking about this problem, I replaced 2006 by a smaller number. I used 4, in fact. (Motivated by the fact that 4 is, like 2006, a composite number.) If you're working mod 4, the period has to be no larger than $63=4^{4}-1.$ I did try running this forwards - which is very hazardous using hand arithmetic. My first two attempts went past 64 without repeating. When I finally got it cleaned up, I found that the period was 30. The period for the mod 2006 case is extremely likely to be an absurdly large number.

I didn't get this problem, but someone mentioned that the sequence has to cycle mod $2006$. Then before it repeats, it must look like:
\[\underbrace{0,0,\ldots,0}_{2005 \text{ zeroes}}\underbrace{,1,1,\ldots,1,}_{2005 \text{ ones}}1,2,3,\ldots,2005,2006,2007,2009,\ldots, \]
so there are $2005$ consecutive terms divisible by $2006$.

That is the proof that JBL and I are talking about. However, you do have to prove that there's no "initial segment" before the periodicity kicks in. Hence it's vital to be able to run the recursion backwards as well as forwards.

However, the more I look at it, the more superfluous the matrices in my proof seem. You can live without them.

when i saw this my first thought was that you would in some way have to use the Chinese Remainder Theorem.
is it possible to solve this problem with that method?

[quote="Xevarion"]I think my solution was effectively the same, but less clever and more brutish. I observed that if $k = 2006n+k'$ with $1 \le k' < 2006$, then
$x_{k}\equiv \sum_{j=1}^{k'}x_{2006(n-1)+k'}\pmod{2006}$
From that recursion you can find $x_{k}$ explicitly as a binomial coefficient (mod 2006 that is). Finally you just show that if $n$ is chosen properly then all but the first one will be 0 mod 2006 ($x_{2006n+1}$ is always 1 mod 2006).

I'm not totally confident of my proof though because my argument about the binomial coefficient in the end was a little sketchy.[/quote]

Yes.

You should get something like

$x_{2006m+\ell}=\binom{m+\ell}{m+1}$

Use the Chinese remainder theorem, or take $m=2006!$ or something.

when I was doing this problem, i had the misfortune to choose 3.  I ran a sequence of 1,2,3, 3+1, 3+1+2, 3+1+2+3, etc.  and I never found three consecutive numbers in my sequence that are divisible by 3.  Does the claim that there exist a n consecutive numbers in such a seq as stated in problem hold only for even numbers?

That's an off-by-one error in the question you're asking. There should be two consecutive numbers divisible by 3, not three.

Consider a sequence {x[size=60]k[/size]}k[u]>[/u]1 made from the residue of the given sequence modulo 2006.

Now observe that this sequence is bounded from above and also periodic there are only 2005 residues modulo 2006 and after that every  term of this sequence  will keep repeating.
Writing the recursion backwards as:
x[size=60]k-2005[/size]=x[size=60]k+1[/size]-x[size=60]k[/size] which is also periodic.

Now, notice that this sequence get extended to a doubly infinite sequence {x[size=60]k[/size]} where k is any integer.

So now it sufficient to find 2005 consecutive terms with residue zero in this doubly infinite sequence.
Now writing the recursion backwards we found that:
x[size=60]1[/size]=x[size=60]0[/size]=x[size=60]-1[/size]=.......x[size=60]-2004[/size]=1
x[size=60]-2005[/size]=x[size=60]-2006[/size]=[size=60]-2007[/size]=.......x[size=60]-4009[/size]=0

Therefore there exist 2005 consecutive numbers in the given sequence each divisible by 2006.


integration
function
real analysis
real analysis unsolved

a).Let $g \in L^{1}([0,1])$.Prove that there exists $f \in L^{\infty}$ such that $\int fg = \parallel g\parallel _{1}\parallel f\parallel _\infty$.

b).Let $g \in L^{\infty}([0,1])$.Prove that for every $\epsilon > 0$ there exists $f \in L^{1}$ such that $\int fg \geq (\parallel g\parallel _{\infty}-\epsilon)\parallel f\parallel _{1}$.

Are your functions real or complex valued? That makes for some differences in the details of the arguments.

(a) Assume that $\|g\|_{1}>0/$ Just let $f=\text{sgn}(g).$ That is, let
$f(x)=\begin{cases}1,&g(x)>0\\-,&g(x)<0\\0,&g(x)=0 \end{cases}.$
Then $f(x)g(x)=|g(x)|$ everywhere and $\int fg=\int |g|=\|g\|_{1}.$ But we also have that $\|f\|_{\infty}=1$ by construction.

(b) Assume that $\|g\|_{\infty}>0.$ Given $\varepsilon>0,$ there exists a set $E$ of positive measure such that $|g(x)|>\|g\|_{\infty}-\varepsilon$ for $x\in E.$ We can also assume that $m(E)<\infty,$ by restricting to some subset of $E$ if necessary. Now let $f(x)=\frac1{m(E)}\text{sgn}(g(x))\chi_{E}.$

$\|f\|_{1}=\int |f|=\frac1{m(E)}\int_{E}1=1$ while

$\int fg=\frac1{m(E)}\int_{E}|g|>\frac1{m(E)}(\|g\|_{\infty}-\varepsilon)m(E)=\|g\|_{\infty}-\varepsilon$

Thanks.

Just a minor typo:

[quote="Kent Merryfield"](a) Assume that $\|g\|_{1}>0/$ Just let $f=\text{sgn}(g).$ That is, let
$f(x)=\begin{cases}g(x),&g(x)>0\\-g(x),&g(x)<0\\0,&g(x)=0 \end{cases}.$
[/quote]

$f(x)=\begin{cases}1,&g(x)>0\\-1,&g(x)<0\\0,&g(x)=0 \end{cases}.$


If $ X$ be metric space, and $ Y \subset X$ , $ K \subset Y$, then $ K$ is compact wrt $ Y$ iff $ K$ is compact wrt $ X$.

A set is open in $ Y$ if and only if it is the intersection of $ Y$ and a set open in $ X$, so there is a one-to-one correspondence (more or less) between open covers of $ K$ with respect to $ X$ and open covers of $ K$ with respect to $ Y$.  (In other words, this is a consequence of the definition of subspace topology.)


There are 2 players in this game. The first player starts, and only positive integers can be written down on a piece of paper. If the numbers a,b have been written on the piece of paper any number written after those two cannot be of the form ax+by where x,y are integers. The first player that writes down 1 loses the game.
Determine if either 1st player or second player has a winning strategy.

I do not understand. Do $x,y$ have to be positive?
Because otherwise, the game will be a draw : Assume that there is a winning strategy for the first player. Then, the second one plays a number which is coprime to the first which has been written, so that from Bezout's theorem, any positive integer has form $ax+by$ and the first player cannot play anymore. And then cannot win. A contradiction.
The same is true if this is the second player who has a winning strategy. Since he is supposed to win, he will not write a number coprime with $a$, so that the first player can playfor his second turn and choose a number which is coprime with $a$ and $b$.

Pierre.

[quote="pbornsztein"]I do not understand. Do $x,y$ have to be positive?[/quote]
It is exactly my question too. ;)

ok, then assume, x,y>0. I think the idea is that the winer must let the pair (2,3) written, because the secound player can only write 1. If a,b are coprime then ab-a-b is the largest number wich is not a linear combination of them.

[quote="pascual2004"]ok, then assume, x,y>0. I think the idea is that the winer must let the pair (2,3) written, because the secound player can only write 1.[/quote]
Or 4, since $x,y>0$!  ;)  :D

yes, x,y>=0 or otherwise the problem is trivial. Sorry for the ambiguity.

No, otherwise I would have posted it in the Proposed Problems section...
[Anyway, i have proven this (probably insignificant result): if P1:4 then P2 can win because ::
P2:6. then, if:
P1: 2 or 3, P2: 3 or 2 and P1 loses
P1: 5 or 7, P2: 7 or 5 and P1 loses
P1: 9, P2: 11 and P1 loses
P1: 2k+1, k>4 --> P2: 9 and P1 loses.]


The number of hours of daylight, d, in Hartsville can be modeled by

d= 35/3 + (7/3)sin((2pi/365)*t)

where t is the number of days after March 21. The day with the greatest number of hours of daylight has how many more daylight hours than May 1? (March and May have 31 days each. April and June have 30 days each.)

I dont know how to: a) find the day with the greatest number of hours of daylight b) calculate all this without a calculator and c) Im not sure if 2pi/365 = 360/365 ?

Please explain, thanks.

for (c) it would be, yes, $ \frac {360}{365}$ but it would be in [b]degrees[/b].

the greatest amount of daylight is given at a maximum of the sine function, $ 1$, so the max hours of daylight is $ d\equal{}35/3\plus{}7/3 \equal{} 42/3\equal{}14$

Now lets count till may 1st
March 31st - $ t\equal{}10$
April 30th - $ t\equal{}40$
May 1st-> $ t\equal{}41$ -> $ d\approx13.18013564$

So our answer is about $ 14\minus{}13.18016564 \equal{} .8198643576$ hours


Twenty-seven small cubes are stacked to make a large cube and then the outside of large cube is painted.
How many of the small cubes have at least two faces painted?
A) 8 B) 18 C) 20 D) 22 E) 26

Since $ 3^3\equal{}27$, an edge of the cube has length $ 3$. We can then see that the only cubes which will not have faces painted are the ones in the middles of the $ 6$ faces of the large cube (which will have $ 1$ face painted), and the $ 1$ in the middle of the large cube (which will have no faces painted). Thus, we have $ 27\minus{}6\minus{}1\equal{}20$. The answer is thus C.


A majority of the people who have high intelligence, that I've met, are shy.

Just wondering if that random sample was significantly biased.

I guess it depends on the situation. I'm outgoing when I'm with people around my intelligence and that talk about things that I like.

I'm shy though (well, maybe not shy-but quiet) when I'm around a lot of the other girls in my class, because even though we're all in this advanced math class, they care more about boys and makeup.... while I don't care about the girly thing. I'd rather spend my time solving math problems.

So really the only reason I'm quiet is because I don't relate to them. I just spend most of my class time in la la land, because the majority only take the la la land kind of brainpower (coughlikespanishcough), or I study for my math class, or I read. Most teachers don't even notice because I still have like 100% anyway in their class.

coughspanishcough

i'm rather quiet when i'm around "normal" girls (i.e. those who don't spend most of their free time doing math like me) but otherwise i'm very outgoing. i used to be quiet all the time but lost my shyness in when i joined math team because it got me to be more involved in school and in competitions outside of school. that way i got to meet people and was forced to talk to them (otherwise i would just be left out, and that's no fun) then i got to like it. i usually keep my mouth shut, however, when i'm with girls talking about how hot the guys standing next to us are or how nice someone's dress looks (i mean, who cares about these things??)

lol, I share sentiments with the girls answering here.

"I don't care if I don't wear makeup.  Yes...nice dress...anyways..."

'Smile and nod' is my policy around them....

but I've met a lot of highly intelligent people that are shy, but a very large number who are very outgoing too.  I guess it differs.

I'm very outgoing.  Sometime a little too much for my good, I guess.

I'm shy around people that I don't know that well, but with my friends I'm outgoing.. Or obnoxious. Or hyper. :trampoline:  But only when I can relate to what they're talking to.. which is sometimes. About math, lord of the rings, star wars, video games.. uhh yeah. or other stuff I do...

I'm with bubula

more than once, my eccentricity has gotten me into some sort of situation I didn't need to be in....

Live and learn....or in my case

just live....

im usually shy, but around friends and family i can be very outgoing and hyper just as  ambierona said  :jump:

[quote="eryaman"]For me, it really depends on who I am with.  I am usually very shy around new people, but if I know someone or am comfortable or trust that person, then I'm very outgoing.[/quote]

This seems to be how a lot of the posters are.  I also am this way.

Ya right sheepwarrior, even I am louder than you  in school... if I am your friend that is:P

i'm really outgoing with my friends, but i'm a little shy with people i don't know.

I have to admit I was extremely outgoing, that is until I became facsinated with mathematics.  Thats not to say, mathematics has completely suppressed my desire for socialization, but it has certainly limited it an considerable amount.  Plus the fact that my fellow peers are more interested in frolicing about, waving their hands across their face, and shouting out the phrase, " You can't See me!!!!".  However, mathematics is not the most popular sbject is most North American societies, compared to countries in Europe where mathematics is discussed in coffee shops, parks, and just about anywhere, so it is easy to see why mathematics would easily pull someout from the ideal "scoial-circles" and make them, in some cases, a social-outcast.

for me it was the opposite of G-unit... being a recent immigrant who refuses to assimilate into amercan culture, i wasn't very well liked by my peers. this changed when i joined math team and was able to have conversations about things i like- as oppsed to sitting next to people i should be friends with trying to figure out what they're talking about when i don't care for it

for me, it really depends on the mood i'm in, who i'm with, and the situation.

I'm not shy....but i dont think im outgoing either.... :huh:

i'm sure that most people are somewhat outgoing with their friends and shy with people they don't know...like me :)

well....most of us seem to be this way...shy with little-known people..outgoing with the others.... when I told my mother I am capable of not speaking for an entire hour when it happened to go out with people that I didn't know that much, she thought of it as impossible....also I'm shy if others are bold for me....if they aren't and I'm in a difficult situation I become bold...I can't afford something bad happening because of my being shy!
as for maths in Europe....it depends...I live in a small town with almost 0 interest for maths....so...people look at me differently....or at least they consider me weird....I'm not that way...or at least I like to think I'm not....just the regular :)

for me, when i'm around new people or just people that don't like math, it's not that i'm shy, but that we can't find common ground for talking. i'm just quiet because i dont' care to know who's wearing what pretty things or what makeup, not that i'm shy or anything like that.

I find that using chatting devices (such as MSN, AIM, Yahoo, Forums) makes us talk a lot more, and a lot differently, than in real life. I believe it's because that in words, we can't express our emotions like with our voice, so we can be more bold. I practically never talk in public, and am also pretty quiet with my best friends, but on chat, I talk a lot, with a dry sense of humour and a sarcastic wit.


for how many positive integers n>1 is it possible to express 2005 as the sum of n distinct positive integers?

[hide="hint"]find the minimum sum of n distinct integers[/hide]

can you explain your solution because I get some strange answers


I have a question?

I see that the first three people that would win the contest would receive surprize prizes! that is pretty nice! but what about the first three most active people on mathlinks ... what about them? shouldn't they receive something?

just a thought.. don't kill me now... :D

[quote="bugzpodder"]what are the suprise prizes  :D

i know, if you tell me then it wouldnt be suprise anymore!!  (but still, i am curious, its not like i would even come close to top 3  :D )[/quote]

virus~ :D just kidding...

i won't kill you mizath (not yet anyway - just wait untill the Olympiad - I'm sure that as I did this year I can manage to propose a problem that will kill most of you :D :D :D)

about the most active users as I have said (or not) for now they have higher ranks than everyone else. I will think about it more in the future. :) thanks for the suggestion though ...

this would just encourage spam to your board, please have careful consideration before taking actions. :D

I hope that I won't get that much spam on the board ! :)  ... but thanks for your warning!

I was wondering if there may be soon a contest for each clases 9th, 10th,11th and 12th grade students on mathlinks?! not only a training contest for the IMO but also for the national contest..  what do you say ?

cheers! :D :D

Hey Lagrangia it seems that you've read my mind. That's exactly what I was going to say.
That would be nice!

unfortunately I don't see that coming soon ... because:

1) the math education system in Romania is different from most of the countries in the world (not too many people make IMC - level problems at school :D )

2) there wouldn't be many people interested in such a separate contest (remember most countries have only one set of questions for all people involved in the National Olympiad)

3) and most important I don't have time :D :D

well.. when I said active I ment those who post the most solutions, comments, things like that , not spam ..

I hope I made myself clear this time!!

g'luck!

yeah I know what you ment and I've already given you the answer :)


calculus
derivative
invariant
real analysis
real analysis solved

a) Given that f(x) and g(x) have continuous third derivatives and satisfy f(x)g(x)=1 for all x Element R, prove that:
[img]http://www.torontosop.com/q1.gif[/img]
holds at all those points where the denominators are nonzero.
b) State and prove a formula for the nth derivative of the product of 2 functions.

For (b) there is Leibniz-formula.

$(f\cdot g)^{(n)} = \sum_{k=0}^n {n\choose k}\cdot f^{(n-k)}\cdot g^{(k)}$

:) im working through that lili thanks.
anyone have any input or hints on part 1 of this question?

[quote="Moubinool"]for a) this is Schwarzian derivative
you can use
$Sf=\frac{6}{f'^2}det\left(\begin {array}{cc}f'&f''/2\\f''/2&f'''/6\end{array}\right)$[/quote]

Moub :( i am not familiar with this sort of formula can u suggest an alternative method?

for a/ yes plug g,g',g'' in Sg and simplify you get Sf, the calculus is ugly

Is there an elegant way ?

How about setting $f(x)=e^{h(x)}$ and $g(x)=e^{-h(x)}$ (Assume WLOG that both are positive, since the expression is invariant in negating the functions).
Then $f'=h'f,\, f''=h''f+(h')^2f,\, f'''=h'''f+3h'h''f+(h')^3f$ and $g'=-h'g,\, g''=-h''g+(h')^2g,\, g'''=-h'''g+3h'h''g-(h')^3g.$

Then $\frac{f'''}{f'}=\frac{h'''+3h'h''+(h')^3}{h'}$ and $\frac{f''}{f'}=\frac{h''+(h')^2}{f'}$, so the Schwarzian derivative is
$\frac{h'''+3h'h''+(h')^3}{h'}-\frac32\frac{(h'')^2+2h''(h')^2+(h')^4}{(h')^2}$.
Negating $h$ to get the Schwarzian derivative of $g$, we find that
$Sf-Sg=2\left(\frac{3h'h''}{h'}-\frac32\frac{2h''(h')^2}{(h')^2}\right)=0$.
Every term which is a product/quotient of an even number of funcions $h^{(k)}$ is unchanged, and only the two odd terms contribute to the difference.

Here another way

Plug $g',g'',g'''$ obtained by jmerry post 6 in $S(g)$

$S(g)=\frac{6f^4}{f'^2}\left|\begin{array}{cc}-f'/f^2&-f''/(2f^2)+f'^2/f^3\\-f''/(2f^2)+f'^2/f^3&-f'''/(6f^2)+f'f''/f^3-f'^3/f^4\end{array}\right|$

Factorise $1/f^2$ in the determinant

$S(g)=\frac{6}{f'^2}\left|\begin{array}{cc}-f'&-f''/2+f'^2/f\\-f''/2+f'^2/f&-f'''/6+f'f''/f-f'^3/f^2\end{array}\right|$

elementary operation on the row change $L_2$ into $f'/fL_1+L_2$

$S(g)=\frac{6}{f'^2}\left|\begin{array}{cc}-f'&-f''/2+f'^2/f\\-f''/2&-f'''/6+f'f''/(2f)\end{array}\right|$

another elementary operation on column change $C_2$ into $f'/fC_1+C_2$

$S(g)=\frac{6}{f'^2}\left|\begin{array}{cc}-f'&-f''/2\\-f''/2&-f'''/6\end{array}\right|=S(f)$


Let $X$ be a Banachspace and $w: X \to \mathbb{C}$ a linear map. Prove that $w$ is continuous if and only if $Ker(w)$ closed

This problem is classical.
One implication is simple so I assume that you are interested in the following one:

$ker w=w^{-1}(0)$ is closed, implies that $w$ is bounded.

it suffices to show that $w$ is continuous, since if it is continuous it is bounded.
Let $e\in X$ such that $w(e)=1$, please observe that such an $e$ can be found, for example take $e=\frac{a}{w(a)}$ for some $a$ with $w(a)\neq 0$.

then assume that $w$ is not bounded. It folows that $\parallel w\parallel =\infty$.

$\parallel w\parallel =\sup_{\parallel x\parallel =1}|w(x)|$, so for  $n\in N$, let $x_{n}\in X$ such that $\parallel x_{n}\parallel =1$ with $|w(x_{n})|\geq n$.
Let $y_{n}=e-\frac{x_{n}}{w(x_{n})}$.
It follows that $w(y_{n})=0$, hence $y_{n}\in ker w$ and
$\parallel y_{n}-e\parallel =\frac{\parallel x_{n}\parallel }{|w(x_{n})|}=\frac{1}{|w(x_{n})|}\leq \frac{1}{n}$.
It follows that $y_{n}\rightarrow e$ hence $e \in ker w$ which implies that $w(e)=0$ which is definitely false.

some variations of the same ideea:

let $f$ be a non zero linear functional.
Then $f$ is discontinuous if and only if $ker f$ is dense.