modular arithmetic
function
least common multiple
number theory
totient function
relatively prime
abstract algebra

1) What is the smallest positive integer $n>1$ such that $a\equiv a^{n}\pmod{100}$ for all positive integers $a<100$

Is there a generalization?

2) What is the smallest positive integer $n>1$ such that $a\equiv a^{n}\pmod{10^{x}}$ for all positive integers $a<10^{x}$

[quote="t0rajir0u"][hide="Remarks"] $n-1$ is called the "exponent" of $\mathbb{Z}_{10^{x}}$.  It is the least common multiple of the orders of every element in $\mathbb{Z}_{10^{x}}$.

In $\mathbb{Z}_{k}$ for $k = 1, 2, 4, p^{n}, 2p^{n}$ for $p$ an odd prime, we can show that the exponent is $\varphi(k)$; in other words, there exists a primitive root.

I seem to recall ZetaX gave me a method for calculating the exponent in the general case, but I don't remember it.  In any case, the important thing to remember here is that $n-1 | \varphi(10^{x})$.  [/hide][/quote]

What do you mean by the notaton "$\mathbb{Z}_{10^{x}}$" and what is the function $\varphi(k)$?

$\mathbb{Z}_{n}=\left\{0,1,2,\dots ,n-1\right\}\implies\mathbb{Z}_{10^{x}}=\left\{0,1,2,\dots ,10^{x}-1\right\}$

$\varphi\left(n\right)$ is the totient function, which gives the number of numbers relatively prime to $n$.  Moreover, we have \[n=\prod_{k}{p_{k}}^{e_{k}}\implies\varphi\left(n\right)=n\prod_{k}\left(1-\frac{1}{p_{k}}\right).\]

[quote="computer_nuke"]What do you mean by the notaton "$\mathbb{Z}_{10^{x}}$" [/quote]

$\mathbb{Z}_{m}$ denotes the mathematical system of arithmetic $\bmod m$.

[hide="More clarification"] The [b]order[/b] of $a \bmod m$ is the smallest positive integer $k$ such that $a^{k}\equiv 1 \bmod m$.

If you're familiar with Euler's Totient Theorem, it should strike you that the order of every element of $\mathbb{Z}_{m}$ must divide $\varphi(m)$, as I've said before.  However, this cannot be proven with Euler's Totient Theorem, since in fact it is the necessary lemma in the proof of the Totient Theorem itself!

The rigorous way to go about proving this lemma is left as an exercise. [/hide]

Edit:  [hide="And more"] You will notice that the order of $a \bmod m$ is undefined if $gcd(a, m) > 1$.  Integers $a$ such that $gcd(a, m) = 1$ are called [b]units[/b] in $\mathbb{Z}_{m}$.

The precise definition of a unit is this:  A unit is an element that has a multiplicative inverse.

When we speak of inverses $\bmod m$, we say that

$a^{-1}\equiv b \bmod m \Leftrightarrow ab \equiv 1 \bmod m$

It should be clear that if $gcd(a, m) > 1$ then this cannot occur.  Again, the proof that this is possible if $gcd(a, m) = 1$ is left as an exercise.

We said before that we defined $\varphi(m)$ to be the integers less than $m$ relatively prime to it.  Note that $\varphi(m)$ is the number of units in $\mathbb{Z}_{m}$.

The set of units in $\mathbb{Z}_{m}$ is denoted $U_{m}$, and it is a very important set to consider when dealing with problems of this type.   Since every element is a unit, and therefore has an inverse, it follows that $U_{m}$ is closed under addition, subtraction, multiplication, and division. [/hide]

[quote="t0rajir0u"]If you're familiar with Euler's Totient Theorem, it should strike you that the order of every element of $\mathbb{Z}_{m}$ must divide $\varphi(m)$, as I've said before.  However, this cannot be proven with Euler's Totient Theorem, since in fact it is the necessary lemma in the proof of the Totient Theorem itself!

The rigorous way to go about proving this lemma is left as an exercise.[/quote]
Let $\left|\varphi\left(n\right)\right|=\phi\left(n\right)$ be the set of numbers relatively prime to $n$.

$\varphi\left(n\right)\subset\mathbb{Z}_{n}$ so by Lagrange's Theorem, $\left|\mathbb{Z}_{n}\right|\equiv 0\bmod \left|\varphi\left(n\right)\right|$.  This implies that for $a\in\mathbb{Z}_{n}$ we have \[a^{\phi\left(n\right)}\equiv 1\bmod\left|\mathbb{Z}_{n}\right|\implies a^{\phi\left(n\right)}\equiv 1\bmod n,\] which is the Totient Theorem.  Also, noting that for $n=p\in\mathbb{P}$, we have \[\phi\left(p\right)=p\left(1-\frac{1}{p}\right)=p-1\implies a^{p-1}\equiv 1\bmod p,\] which is Fermat's Little Theorem.


Stanford
college
Princeton
MIT
geometry
articles
summer program

Hey guys, I am an international student from India who has been admitted to both Caltech and Stanford(and a few others -- I think my username suggests that!)
I'm in a real dilemma and I know its really tought to decide -- but please suggest based on my situation as below.

Note:--- This may get long!  :D

1) I'm absolutely fascinated by Science. But if you know about the IIT-JEE in India, you might know that my 'interest' in science is very much a result of the studies we do here , which for the record dont exactly foster research but do train you. So, I definitely want to study the tough and interesting things in Math and Science , but I'm not too confident about pursuing hardcore research right now -- I may later. I've heard that Caltech pushes you to being excellent in pure science. So if I decide to go to Stanford, will I loose out on this rigour, exposure and experience in pure and applied science? People say that Stanford is also great in Science, but how does it compare in terms of rigour of couse, facilities, faculty and intelligent students as compared to Caltech? How rigorous is Stanford as compared to Caltech --- can I still experience the tough core curriculum at Caltech also at Stanford? Specifically, would going to Stanford be a 'bad compromise' or is its program as good(if not better) than Caltech in Science?

2) In engineering and Computer Science, is Stanford better than Caltech? I've heard Caltech, like say Princeton, is more into theory while Stanford and MIT focus more on practical appls. ? Also, Ive an 75-25 chance of research vis a vis entrepreneurship --- Stanford exposes you to being an innovater --- is this true of Caltech?

3) How do undergrad research opps at Stanford and Caltech compare in terms of resources of faculty, equipment, etc ?

4) If Caltech is say top 1% of cram of crop, what figure would apply to Stanford? Also, while Caltech's huge advantage for me is that its close knit and collaborative, is Stanford collaborative or cut throat competitive? How are the people like in the 2 places?

5) Are the exams and courses at these places bookish ans just simple, those-who-know-will-score type exams or more though provoking and 'interesting'? Im asking for math Stanford and Caltech?

Thanks a lotttttt!!! This, and probably more later will help me reach my decision!

Thanks!

Hey man, I'm in the same dilemma , so I cant help.... But I have a question similar to (1) above...

6) What attracts me most to Caltech is its rigorous Core curriculum and the small size that gives you the chance to interact with the best! I just wanted to know whether I can experience that rigour at Stanford as well and also have the chance to interact with the best in Science , Engg, and other fields? Will Stanford offer me the things (at least upto 90%) that attract me most to Caltech?

Thanks!

Of course as an undergraduate you can do very well at either Stanford or Caltech. If you want rigorous courses and potential research opportunities, there won't be much of a difference between the two. You'll probably just have to ask.

I visited Caltech over the summer since I was potentially interested in going there for graduate school (in mathematics), and I found myself disappointed with the selection of courses and their likely difficulty. (Most of the courses were using the same textbooks as the courses I am taking or have taken at UC Santa Barbara, which suggests that I might be familiar with most of the course material already.) Another problem with Caltech is its small size. There's a much better chance that you'll find someone whose research interests coincide with your own at a larger university.

But actually where you go as an undergraduate doesn't really matter. The undergraduate curriculum is more or less the same anywhere. And after you finish the basic lower division classes at any reasonable university, the exams and homework assignments will start to become more interesting and less routine. (Well, one of my classes assigns very easy homeworks and exams, but you'll find a few professors who give you easy work to do anywhere.) So you will do fine at either place. It's probably better to decide which one to go to based on non-academic criteria (such as would you rather live in Pasadena or Palo Alto, which in my opinion is a really easy choice).

[quote="Rushil"]Hey guys, no help for us!!! Please offer your opinions on this! I know maybe FarmTecher has asked too much .... but PLEEEAAAASSEE!![/quote]

"us"?  Hmm...you two seem very similar.  If the same person is posting as both rushil and farmtecher I would suggest you stop.  Or at least don't go around acting as two different people.  If you are different people, then forgive me.

Good luck with your college search.

Well I have been visiting a friend in Stanford two weeks ago and I was pretty impressed with the campus. It is really huge for just those few students. Well their sports facilities are amazing in number and state. Sometimes it is even ridiculous how the money of the alumni is used, e.g. they are going to have two very big fitness centers just 50 metres apart now. But on the other hand some of the student flats are small and some are located in the basement and they charge you $\$$600/month for it. Even professors and their families are granted homes on the campus due to high house prices in Palo Alto. Though Carmel is even more expensive.  :lol:
I did not have any problems due to negative weather influence but right now some thunderstorms and heavy rains have been announced in California. Governor A. Schwarzenegger referred to extreme danger of people and building in 9 districts of California. And yes there is always the danger of earthquakes, especially in San Francisco which is not too far from Palo Alto and UC Berkeley.
As Simon said there is not much difference among the universites in the undergraduate degree. Well I liked Santa Barbara. But yet it is a question what kind of reputation the university (future job options etc.) represents and how good your fellow students are. But in the end you have to read your books and solve the problems. That's not much different at UC Santa Barbara unless you need particular devices, fluids etc. as in chemistry or physics which will really give you an edge in elite private colleges as Stanford.
I have been in LA but I did not have the time to go to Pasadena and to see LA's sights. But what I figured from browsing their web site some time ago it is a really good school with interesting course.
Unless you don't bother about going to LA or Palo Alto it is up to the courses in my opinion. When I have been a visiting student I mostly based my decision on the courses offered. Lots of courses are shared by both colleges. But look at the "special/unique" courses and decide which courses seem to be most exciting to you.

First of all, please make sure you understand I and FarmTecher are DIFFERENT people ---- I need not say more -- I hope that's settled!

Your advice is GREAT --- I've never heard more convincing arguments For or against the two places - amazing -- thanks and I hope this discussion follows!!

I have a few doubts and clarifications to seek based on your response :-

I agree that at either place, you can get what you want -- its just that Caltech makes it mandatory whle Stanford has wider options in learning and research ( a big +++) -- I mean if you can still do QM coz u always loved it in high school but can still focus on CS as your major! Nice arguments on courses etc, ComplexZeta! that sort of brings to earth what Techers call their 'rigor'! That probably is the 'truth'! I just didnt understand this : "such as would you rather live in Pasadena or Palo Alto, which in my opinion is a really easy choice" .... I'm from india so dont know the local culture! What exactly are you trying to say here, ComplexZeta? Thanks!

orl, thanks for your advice too! Its really great . I just want to ask to which place are you pointing when you say,: " But yet it is a question what kind of reputation the university (future job options etc.) represents and how good your fellow students are" ??? By the way, where are you right now?

Thanks a lot!!

Please keep the suggestions flowing!

Someone showed me the list of mandatory courses at Caltech, and I was completely horrified by how many science classes are required. (My opinion is that knowing about physics, chemistry, and biology do not help you as a math major, but if you aren't a math major, perhaps it's fine.)

As for locale, I'm from the Bay Area, and I think Palo Alto is a terrific place to live (if you can afford it). There's something special, in my opinion, about having a real city (as opposed to just a college town) so centered around learning and knowledge. If you walk around, you'll hear lots of people talking about mathematics and sciences and literature and art. Pasadena doesn't seem like such a nice area. There's too much smog and nasty stuff in the LA area.

The difference in reputation and job opportunites between Stanford and Caltech is probably less than $\varepsilon$ for every $\varepsilon>0$, so don't let that concern you.

[quote="Rushil"]First of all, please make sure you understand I and FarmTecher are DIFFERENT people ---- I need not say more -- I hope that's settled![/quote]

I don't believe you.

[quote="ComplexZeta"]My opinion is that knowing about physics, chemistry, and biology do not help you as a math major, but if you aren't a math major, perhaps it's fine.[/quote]

I really disagree with this. I think you'll find that a lot the world's best mathematicians have a surprisingly wide range of humanistic and scientific knowledge. In particular I've noticed that some of the best mathematicians tend to have really strong knowledge in foreign languages, history, physics, and computer science. Biology and genetics also seem to be becoming increasingly large sources of inspiration for purely mathematical questions.

Having a broad base of knowledge can give a mathematician a greater source of inspiration and flexibility when it comes to thinking about mathematical questions.

I don't know whether most mathematicians learn these other subjects in formal classroom settings, or whether they learn them somewhere else -- but I do think its very important to combine a study of pure mathematics with *at least* one other body of knowledge (be it computer science, physics, biology, economics, finance, history, philosophy, or whatever).

And of course if you want to go into a scientific field, then a breadth of knowledge in all aspects of science is very much important.

[quote="gauss202"][quote="ComplexZeta"]My opinion is that knowing about physics, chemistry, and biology do not help you as a math major, but if you aren't a math major, perhaps it's fine.[/quote]

I really disagree with this.[/quote]

I really disagree with that too, so I'll pile on here. I saw a very interesting comment once somewhere online contrasting mathematicians who "die" (in Erdos's sense of ceasing to do mathematics) at a young age with mathematicians who stay productive in research into their nineties. The discussion I saw used Grothendieck as the example of early death, and Gelfand (my avatar) as the example of productivity continued over many decades. The bad example was described as a great but ultimately too narrow mathematician, who didn't like science. Gelfand, by contrast, was described as an astute scholar of physics too (which is quite typical for a Russian mathematician of his era) who gained insight into mathematical problems by considering research issues in physics.

You can probably Google key terms in the paragraph above to find the source for the idea that science knowledge is useful for pure mathematicians. A published source that says the same thing is

[url=http://www.amazon.com/gp/product/3540669132/]Mathematics Unlimited: 2001 and Beyond[/url]

a collection of articles published by Springer Verlag, in which a couple of articles say that the only way for mathematical research to go forward is for more mathematicians to know more mathematics, and for more scientists to know more mathematics. I think that is the correct direction for curriculum reform. The best universities already pile more physics or other science into a math major's major sequence than the average universities do for a physics major. Similarly, the physics majors at the best universities get lots and lots and lots of math that even math majors don't get at lesser colleges. That's the way to see farther--to have more shoulders of giants to stand on.

I haven't read the article that you're refering to Tokenadult, but I think Grothendieck may be a bad example for that theory. He had fairly varied intellectual interests, including very strong political opinions. His decision to "die" from mathematics probably had more to do with psychological disorders and personal protests against war than from a lack of mathematical inspiration.

However another reason to get a broadly based education is reinforced by this advice from MIT: [url=http://math.mit.edu/undergraduate/major.html#mathematician]Advice for Undergraduate Majors[/url]

I don't argue that it isn't important to study [i]something[/i] other than mathematics; I just don't think it needs to be science. (I study music as well as mathematics.)

Indeed, everyone esp. mathematicians need to have a wide base so that they have great exposure and who knows when something somewhere may give you that 'spark' --- but should it really be so mandatory and hard and fast as in Caltech?!!! I mean , its better to do what [i]you[/i] want as in Stanford... dont you think?

I certainly agree, as does my college as a whole.

[quote]4) If Caltech is say top 1% of cram of crop, what figure would apply to Stanford? Also, while Caltech's huge advantage for me is that its close knit and collaborative, is Stanford collaborative or cut throat competitive? How are the people like in the 2 places? [/quote]

Caltech and Stanford [i]students[/i] are generally collaborative, but that's where the similarity ends.  Caltech's students have a reputation of being extremely stressed, while Stanford's students are generally perceived as having more time to do other activities.  However, both schools are evenly matched in terms of math.

I would advise you to choose Stanford, because it is a more relaxing environment with more relaxed people.  However, some people thrive high-stress surroundings or are completely dedicated to science and math.  Then Caltech would be the better choice.

Yea I also think diversity is a big issue.  I mean lets be real, how many english majors are you going to find at Caltech as opposed to Stanford.

The Caltech science requirements seem very reasonable for someone going into science, math, or engineering. All they require is
-two years of math
-two years of physics
-one year of chemistry
-one term of biology.
(see the catalog: http://pr.caltech.edu/catalog/ )

Simon, would you agree that a certain amount of science should be required, even for math majors? After all, math isn't done only for it's own sake. Mathematician ought to know what has applications in science and technology.

A certain Mathcamp staff member told me that he regrets that he took only one science course during his undergrad at Harvard. I think Caltech's policy of requiring these courses and making everyone take them at the same time actually ends up benefitting everyone. With no requirements, students would tend to gravitate towards their areas of comfort and expertise; whether this is biology at the expense of math and physics, or pure math at the expense of chemistry and other science. Caltech guarantees that everyone gets a broad scientific education. While some of the core curriculum is likely painful and frustrating, the fact that everyone is going through it at the same time should make it much more bearable (or so I've heard.)

Yeah, I agree with Simon. Well LA is hard to reach by public transport, traffic jam, has lots of slums etc. But on the other hand it is superior wrt Palo Alto by being culturally top-notch. Just look at the [url=http://www.getty.edu/art/]Getty Museum of Modern Art.[/url] NBA clubs. You name it. What I found weird about Palo Alto is that Stanford has one of the most exclusive shopping centers. Expensive like Bloomingdale's.  :lol: But well lots of Sillicon Valley executives are quite wealthy. Just search for LA and Palo Alto. After all it is just a matter of personal preference.
When I talked about reputation I meant Caltech/Stanford vs. UC Santa Barbara. But yet UC Santa Barbara has especially excellent students as Simon.  :)
I agree with Simon wrt. the mandatory science courses. Well he likes music. So what is wrong about combining math with something else than science as gauss202 mentioned? Maybe you should be forced to combine it with a minor subject but not necessarily science.
Lots of students think like Simon, for example former Vietnam IMO team member Le Hung Viet Bao who chose the pure math part in [url=http://www.cam.ac.uk/admissions/undergraduate/courses/maths/index.html]Cambridge.[/url]

[i]There are three options for Part IA in Cambridge: [/i]

"[b]Option (a)[/b] Pure and Applied Mathematics, for students intending to continue with Mathematics
[b]Option (b)[/b] Mathematics with Computer Science (under review)
[b]Option (c)[/b] Mathematics with Physics

Options (b) and (c) are intended for students with a strong interest in mathematics who may want to change to Computer Science or Physics after the first year. You can continue with Mathematics after taking Options (b) or (c), or change to Computer Science or Physics after taking Option (a), but this may involve extra work over the long vacation."


[quote="236factorial"]A five-digit number is called a mountain number if its first three digits increase and the last three digits decrease. For example, 12321 is a mountain number but 13342 isn't. How many mountain numbers greater than 70,000 are there?[/quote]

[hide]Let's look at the individual digits of the possibilities:

The first digit has to be 7. Because if it's 8, the third digit will violate the rule. If it's 9, it's self explanatory.

The second digit can only be 8, since it must be bigger than 7. and if it's 9, the third digit can't be bigger than it.

The third digit must be 9.

So far, we've shown that the number must be 7 8 9 _ _

The number of possibilities for the last two digits is $8+7+6+5+4+3+2+1=36$

So there are 36 different mountain numbers greater than 70000.[/hide]


Let [tex]x_1,...,x_n[/tex] are positive such that [tex]x_1...x_n\leq 1[/tex]. Prove that following inequality holds:
[tex]\frac{1}{x_1}+...+\frac{1}{x_n}\leq \frac{x_1+...+x_n+n}{4}[/tex]

Excuse me, did you mistype?

$n=1$
$x_1=1$
$1 \leq \frac{1}{2}$

I'm sorry, this inequality is wrong again.
For $n=1$ the inequality is equivalent to $x_1\geq 1$
which inequality-symbol did you contort?

i think n>1 or there is something like that,then your inequality will be true!please check it again! ;)

It seems a bit ambitious to be placing this in the solved problems forum when we still haven't agreed on a formulation... As far as I can see, the LHS should be corrected as indicated, and then the direction of the inequality flipped to give a correct statement: $\sum_{i=1}^n\frac{1}{1+x_i}\ge\frac{n+\sum_{i=1}^nx_i}{4}$.

Letting $y_i=1+x_i$, we get $y_i\le2\Rightarrow\left(\frac{\sum_{i=1}^ny_i}{n}\right)^2\le4$, and $\sum_{i=1}^ny_i\sum_{i=1}^n\frac{1}{y_i}\ge n^2$ (from AM-HM).

Combining the two gives $\sum_{i=1}^n\frac{1}{y_i}\ge\frac{n^2}{\sum_{i=1}^ny_i}\ge\frac{\sum_{i=1}^ny_i}{4}$ as required.

Very strange... this problem was posted by me so long ago that I just don't remember. But I had some very nice solutions posted and Namdung gave another one. The statement of my inequality is:
   If $ x_1\cdot x_2\cdot...\cdot x_n=1$ then $\sum{\frac{1}{1+x_i}}\leq\frac{n+x_1+...+x_n}{4}$. Is this what you want?

Oh, right - I thought it was $x_i\le1$ for each $i$, not the product.  I don't think I made a mistake though - I guess they're just two very similar inequalities.  I suppose it's not too uncommon for a small change in the conditions like that to reverse the direction of an inequality.


IN THE NAME OF GOD
HIIIIII

Solve the following Puzzles :

ABCDEF × 3 = BCDEFA

CAT = (C + A + T) × C × A × T

REGARDS ,SALAHCOOL
 :first:

Hi salahcool.

[hide=" Solution to first part"]

For the first one, since $ \frac {1}{7} \times 3 \equal{} \frac {3}{7}$ and the decimal representation of the sevenths is periodic and cyclical, we have $ 142857 \times 3 \equal{} 428571$.

[/hide]

[hide=" Solution to second part"]

$ 135 \equal{} (1 \plus{} 3 \plus{} 5) \times 1 \times 3 \times 5$.

[/hide]

Another solution

for part one:

285714*3=857142

for part two:

144=(1+4+4)*1*4*4

thank you very much
AndrewTom

An interesting solution

For both: [hide]If you make all letters equal to 0, the equation is solved[/hide]

obviously, different letter represent DIFFERENT numbers :)

Hi salahcool.

Your second solution follows from the fact that $ \frac{2}{7} \times 3 \equal{}\frac{6}{7}$, that is, on multiplying the period of $ \frac{2}{7}$ we get the period of $ \frac{6}{7}$, that is $ 285714 \times 3 \equal{} 857142$


Pairs of positive integers $ (n,\ a_1,\ a_2,\ \cdots ,\ a_n)$ satisfy $ a_1\plus{}a_2\plus{}\cdots a_n\equal{}2008.$ Let $ A_k\equal{}a_1a_2\cdots a_k.$ Find the maximum value of $ A_1\plus{}A_2\plus{}\cdots A_n.$

[quote="kunny"]Pairs of positive integers $ (n,\ a_1,\ a_2,\ \cdots ,\ a_n)$ satisfy $ a_1 \plus{} a_2 \plus{} \cdots a_n \equal{} 2008.$ Let $ A_k \equal{} a_1a_2\cdots a_k.$ Find the maximum value of $ A_1 \plus{} A_2 \plus{} \cdots A_n.$[/quote]
[hide="Solution"]
We claim that all of the numbers are $ 2$s and $ 3$s.  Assume that there is another number $ p$ in the product such that the product is biggest (with no other product being as big.)  Now, since $ p\ge 4$, we have that $ p(p\minus{}2)3)\ge 2\implies p(p\minus{}2)\ge p$.  So, we can partition $ p$ into $ p\minus{}2$ and $ 2$ without minimizing the product, which is a contradiction.  Therefore, all of the numbers are $ 2$s and $ 3$s.

Now, we will show that there are no more than $ 2$ twos.  Assume that there are at least $ 3$ twos in the largest product.  We take $ 3$ of the twos and turn them into $ 2$ threes, thus forming a product of $ 9$.  This is larger than the other product, which was $ 8$.  This shows that the new product is larger than the previous, which is a contradiction.  Hence, there are no more than $ 2$ twos.

Now, notice that there are either $ 0$, $ 1$, or $ 2$ twos.  With a quick check, we see that there are $ 2$ twos and $ 668$ threes.  We now claim that the $ A_1\plus{}A_2\cdots A_n$ is largest when $ a_{667}\equal{}a_{668}\equal{}2$ and the rest of the $ a$s are $ 3$.  Now, assume that the product is larger when $ a_i$ and $ a_j$ are $ 2$ where $ i<j<668$.  Therefore, \[ 3\plus{}3^2\plus{}3^3\cdots 3^{668}\plus{}2\cdot 3^{668}\plus{}4\cdot 3^{668}\le 3\plus{}3^2\plus{}3^3\cdots\plus{}3^{i\minus{}1}\plus{}2\cdot 3^{i\minus{}1}\plus{}2\cdot 3^i\cdots \plus{}2\cdot 3^{j\minus{}2}\plus{}4\cdot 3^{j\minus{}2}\plus{}4\cdot 3^{j\minus{}1}\cdots \plus{}4\cdot 3^{668}\] This turns into \[ 3^i\plus{}3^{i\plus{}1}\cdots \plus{}3^{668}\plus{}2\cdot 3^{668}\le 2\cdot 3^{i\minus{}1}\plus{}2\cdot 3^i\cdots 2\cdot 3^{j\minus{}2}\plus{}4\cdot 3^{j\minus{}2}\plus{}4\cdot 3^{j\minus{}1}\cdots 4\cdot 3^{667}\] which is clearly false after looking at each corresponding pair of terms.  Hence, the largest $ A_1\plus{}A_2\cdots A_k$ is \[ 3\plus{}3^2\plus{}3^3...\plus{}3^{668}\plus{}2\cdot 3^{668}\plus{}4\cdot 3^{668}\equal{}\frac{3^{669}\minus{}3}{2}\plus{}6\cdot 3^{668}\equal{}\frac{3^{669}\minus{}3}{2}\plus{}2\cdot 3^{669}\equal{}\boxed{\frac{5\cdot 3^{669}\minus{}3}{2}}\][/hide]

I haven't tried to figure out what you did wrong, but the best case (if we replace $ 2008$ with $ i$) frequently involves more than two $ 2$s, e.g. for $ i \equal{} 7$.


Let $f(x)$ be a polynomial of degree n such that $f(x) = 2^x$ for $x \in{0, 1, ..., n}$.  Find $f(n+1)$.

Look out, 42 pigs!
http://www.mathlinks.ro/Forum/viewtopic.php?t=1885
http://www.mathlinks.ro/Forum/viewtopic.php?t=23007
http://www.mathlinks.ro/Forum/viewtopic.php?t=23679
http://www.mathlinks.ro/Forum/viewtopic.php?t=43127

[quote="M4RI0"]Why is the polynomial unique :?:[/quote]

$n+1$ given points define a unique $n$-degree polynomial because it produces a system of $n+1$ linear equations in $n+1$ variables (the $n+1$ coefficients of the polynomial).  The actual polynomial can be found generally through the method of Lagrange Interpolation, but here there is an easier way.

I noticed that while I was going to school, sorry by the dumb question :D .


Time has come around in my school to start signing up for classes, and I'm faced with a dilema.  Since I live in a college town juniors/seniors who have a college admissions index over 70 can get free credit for college AND high school by taking classes at the U.  My problem is that I'm the only one of my friends besides me is considering this.  On the one hand there's the fact that it's FREE college credits, on the downside most small private colleges won't accept credit or even placement, and there's the fact that if I did this I'd never be in (high) school, so I'd not see much of my friends, any ideas?

I would go to the college program in a heartbeat. You'd make new friends there, just as I did when I went off to my university program.

Friendly advice: it looks like you have developed a finger habit of typing "collage" for "college," because the error was repeated several times in your post. The great majority of the time, online spelling mistakes are really typing mistakes, but you could edit your own post to fix the spelling of the word "college," which would be an opportunity for you to practice.

*shrug* Not to be too pessimistic, but your friends in high school probably aren't going to last past graduation anyway. Based on that, I would rather go to college. Besides, it'll give you a solid idea of what a real college environment is like.

Go for it.  It's free, you learn real stuff (not high school cram-down-your-throat/random junk), you can hang out with your friends after school, etc.

I'm sick of high school already. If I were you, I'd take it without thinking about it twice. Now, I'm recommending this assuming that you can handle the courseload. If you think you might get swamped, I might hold off.

You wouldn't be taking [i]all[/i] your classes at the University, would you?

[quote="JBL"]You wouldn't be taking all your classes at the University, would you?[/quote]
Naw, probably just calculus and some science class.
I think that's what I'll do, college in the afternoon, and high school in the morning.
Thanks for the advice!

I'm taking math at UIC next year, which is the best thing in the world.  Other than the fact that I don't have to put up with b.s. high school math (though I know UIC will be far from the best place to go), I get a ton of free time.  It's three days a week, which means two days a week I get about two and a half hours free.  And I got a gym waiver, which is also great because at my school you have to take 7 semesters of gym, now I only have to take 5.  </bragging>

[quote="Caroline"]I agree with what most people have said - go for the university classes.  Especially since you'll still be at your high school some of the time, and therefore you'll get to see your friends, as well as make new friends.  It sounds like a great opportunity that you definitely shouldn't pass up.[/quote]

Amen.

Grahhh. The "collage" admission is killing me.


[quote="rakesh2721"]If 2(a+b+c)=p^2+q^2+r^2,and the roots of x^2+px-a=0 are q,r  and the roots of  x^2+qx-b=0 are p,r,show that the equation whose roots are p,q is x^2+rx-c=0.[/quote]

[hide="solution"]
We have to prove that $ p\plus{}q\equal{}\minus{}r$ and $ pq\equal{}\minus{}c$
From viettas relations in
$ x^2\plus{}px\minus{}a\equal{}0$,we have that $ q\plus{}r\equal{}\minus{}p \Longleftrightarrow r\equal{}\minus{}p\minus{}q$ $ \boxed{1}$

Also:
$ qr\equal{}\minus{}a$, and
$ pr\equal{}\minus{}b$. So $ r(p\plus{}q)\equal{}\minus{}(a\plus{}b) \Longleftrightarrow (p\plus{}q)^2\equal{}a\plus{}b$ $ \boxed{2}$

From $ 2(a\plus{}b\plus{}c)\equal{}p^2\plus{}q^2\plus{}r^2$,using  $ \boxed{1},\boxed{2}$ we have:

$ 2(p\plus{}q)^2\plus{}2c\equal{}p^2\plus{}q^2\plus{}(p\plus{}q)^2 \Longleftrightarrow c\equal{}\minus{}pq$
So $ c\equal{}\minus{}pq$ and $ r\equal{}\minus{}q\minus{}p$
[/hide]


Can someone tell me why the latex is not showing up on my computer, all I see is a bunch of $ signs.  I'm using internet explorer and a pc computer.

thanks

Not sure what you mean. Is this in your latex documents or the forum? Please explain in more detail and give examples, screen shots etc.

You might want to try to "always allow BBCode" in Profile->Preferences->posting a message ;)


Solve the functional identity for all forms of $ f$: $ f(1\minus{}x)\equal{}f(x)\plus{}2x\minus{}1$

[hide="Complete Solution Set"]
Let $ g(x)$ be an arbitrary function.

Let $ f(x)\equal{}g(x)$ on $ (\frac{1}{2},\inf)$

And let $ f(x)\equal{}g(1\minus{}x)\plus{}1\minus{}2x$ when f is on $ (\minus{}\inf,\frac{1}{2})$

Lastly, define $ f(\frac{1}{2})$ arbitrarily.

This is the set of all functions, since the values of f on $ [\frac{1}{2},\inf)$ define the remaining values of f without contradiction.[/hide]

[hide="Alternately"] Let $ g(x) \equal{} f(x) \plus{} x$.  The given identity says that $ g(1 \minus{} x) \equal{} g(x)$, so $ g(x)$ is any function symmetric about $ \frac{1}{2}$.  In other words, $ g \left( x \plus{} \frac{1}{2} \right)$ is any even function. [/hide]


How would you find the equation of a line in a plane, say the plane $x+3y-5z=7$?

How would you find the equation of the line formed by the intersection of two planes, say $x+3y-5z=7$ and $2x-y+z=4$? (I hope I picked two planes that meet in a line...)

Take two vectors normal to each plane.  In this case, the two vectors are $\left< 1,3,-5\right>$ and $\left<2,-1,1\right>$.  Find their cross product to find a direction vector of the line of intersection.  Take a point on both planes, and the rest follows.

For your first question, there are many lines on a plane.  :wink:

"The equation of a line" is not clear.  A line in 3-space is most simply described parametrically; boxedexe shows you how to do this.  As a relationship among the variables $x, y, z$, the most natural representation is by [i]two[/i] equations, not one.  (It's always possible to artificially collapse many equations (in real variables) into one: if our equations are, say $a = b$ and $c = d$, we can write our new equation $(a-b)^{2}+(c-d)^{2}= 0$.)

If you want to find the equation of [i]any[/i] line in a plane, you should pick two (different) points in the plane and consider the line between them.


[quote="orl"]A family of sets $ A_1, A_2, \ldots ,A_n$ has the following properties:

[b](i)[/b] Each $ A_i$ contains 30 elements.
[b](ii)[/b] $ A_i \cap A_j$ contains exactly one element for all $ i, j, 1 \leq i < j \leq n.$

Determine the largest possible $ n$ if the intersection of all these sets is empty.[/quote]
It seems you should use Inclusion–exclusion principle:
http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle


probability
parameterization
function
integration
calculus
Support
probability and stats

Let $ A$ be a Gamma random variable with parameters $ \lambda$ and $ \alpha$ ($ A$~Gamma($ \lambda$,$ \alpha$)).  If we condition on $ A$, then a random variable $ X$ will have a Poisson distribution with parameter $ A$.  What is the (unconditional) probability mass function of random variable $ X$?

Hint:  [hide]It should be a "recognizable" discrete distribution.[/hide]

for example:
\begin{eqnarray*}
P[X=k] &=& C \cdot \int_{a > 0} P[X=k |A=a] a^{\lambda-1} \exp(-\frac{a}{\alpha}) \,da \\
&=&C \cdot \int_{a > 0} \{e^{-a}\frac{a^k}{k!} \} a^{\lambda-1} \exp(-\frac{a}{\alpha}) \,da \\
&=&\frac{C}{k!} \cdot \int_{a > 0}  a^{k+\lambda-1} \exp(-a(\frac{1}{\alpha}+1)) \,da \\
\end{eqnarray*}
do you see how to conclude from that ?

Im still a little confused how to obtain a discrete mass function for X if the integration is done over a continuous function.  Could you show a few more steps?

Thanks a lot.

The law of total probability for discrete conditional distributions says that for an event $ X$,

$ \Pr[X] \equal{} \Pr[X|Y \equal{} y_1]\Pr[Y \equal{} y_1] \plus{} \Pr[X|Y \equal{} y_2]\Pr[Y \equal{} y_2] \plus{} \cdots$

where $ Y$ is a discrete random variable that takes on values $ y_1, y_2, \ldots, y_n, \ldots$.  In the continuous case, the sum is replaced by an integral and the probability mass function is replaced by the probability density function:

$ \Pr[X] \equal{} \int_{y \in \Omega(Y)} \Pr[X|Y \equal{} y] f_Y(y) \, dy$,

where $ \Omega(Y)$ is the support of $ Y$.  If $ X$ is a discrete random variable, then its probability mass function can be described as

$ \Pr[X \equal{} x] \equal{} \int_{y \in \Omega(Y)} \Pr[X \equal{} x | Y \equal{} y] f_Y(y) \, dy$.

This is the formula that is used to solve this question, with X ~ Poisson(Y) and Y ~ Gamma.


Prove that for all positive real numbers a, b, c we have the inequality:
a^5c^8/(a^3b^2c^4 + 2b^6) + b^7c^5/(a^4b^5c^3 + 2a^13) + a^8/(a^6c^4 + 2bc^13) ≥ 1
 :D

That is weird.I think that the author have just picked out an old one then make it new by change the variable. :) And I think that it will take a lots time to find out the root inequality .  :(

[quote="nguoivn"]Yes, I also think so.
Dear quykhtn-qa1, can you write the original ineq ?
We can creat many ineqs as it but I think it really spend much time for the reader  :wink:[/quote]

BUT THE WAY TO SET THE VARIABLE OF THIS PROBLEM IS VERY NATURAL
THEN THE PROBLEM BECOME:
let a,b,c>0,such that abc=1;prove that:
∑c/(a^2.c+2.a^6)≥1.


[quote="dragon96"]When we have the solution after the problem, sometimes you just need to comment the problem or offer another solution etc. Is it possible to add a comment box (like for the videos)?[/quote]
We do have plans for something like this in the future.


quadratics
modular arithmetic
partition
number theory
IMO Shortlist

Write

$ x^2 \plus{} Ax \plus{} B \equal{} 2y^2 \plus{} 2y \plus{} C \Leftrightarrow$
$ 4x^2 \plus{} 4Ax \plus{} B \equal{} (2x \plus{} A)^2 \plus{} (B \minus{} A^2) \equal{} 2(2y \plus{} 1)^2 \minus{} 2 \plus{} 4C \Leftrightarrow$
$ (2x \plus{} A)^2 \minus{} 2(2y \plus{} 1)^2 \equal{} 2(2C \minus{} 1) \plus{} A^2 \minus{} B$.

[b]Lemma:[/b]  No integer of the form $ a^2 \minus{} 2b^2$ is congruent to $ 3 \bmod 9$.

Proof:  Suppose that $ a^2 \minus{} 2b^2 \equiv 3 \bmod 9$.  Clearly $ a, b$ are not both divisible by $ 3$.  If one of $ a, b$ was divisible by $ 3$, then the other would also have to be, so neither is.  Then $ a^2 \equiv 2b^2 \bmod 3 \Leftrightarrow \left( \frac {a}{b} \right)^2 \equiv 2 \bmod 3$, but $ 2$ is not a quadratic residue $ \bmod 3$; contradiction.

Since $ C$ allows us to increase or decrease the value of the RHS by $ 4$ and $ 4$ is invertible $ \bmod 9$, it follows that for any $ A, B$ we can find $ C$ so that $ 2(2C \minus{} 1) \plus{} A^2 \minus{} B \equiv 3 \bmod 9$.

another solution:
note that $ 2x^2 \plus{} 2x$ is divisible by $ 4$ for all $ x$.
let:   $ N_1\equal{}$  the set $ M_1$ mod 4.
therefore:
$ N_1 \equal{} \{B,A \plus{} B \plus{} 1,2A \plus{} B,3A \plus{} B \plus{} 1\} \mod 4$
if $ A$ is odd,then all elements of $ N_1$ have the same parity,therefore $ N_1$ is not complete residue system mod 4.
and if $ A$ is even then obviously $ B\equiv2A \plus{} B \mod 4$ therefore $ N_1$ is not complete residue system mod 4.
since $ N_1$ is not complete residue system mod 4 and $ 4|2x^2 \plus{} 2x$ obviously there exist $ C$ such that $ M_1$ and $ M_2$ do not intersect.

[quote="orl"]Let $ \mathbb{Z}$ denote the set of all integers. Prove that for any integers $ A$ and $ B,$ one can find an integer $ C$ for which $ M_1 \equal{} \{x^2 \plus{} Ax \plus{} B : x \in \mathbb{Z}\}$ and $ M_2 \equal{} {2x^2 \plus{} 2x \plus{} C : x \in \mathbb{Z}}$ do not intersect.[/quote]
I think this works:
[hide="Solution"]
Assume that $ x^2\plus{}Ax\plus{}B\equal{}2x^2\plus{}2x\plus{}C$ has at least one solution for all $ C$.  Then, $ x^2\minus{}(A\minus{}2)X\plus{}(C\minus{}B)\equal{}0$.  This is true if and only if its discriminant, $ (A\minus{}2)^2\minus{}4(C\minus{}B)$, is a perfect square (notice that the quadratic is monic.)  Yet, if we let $ C\equiv (\frac{(A\minus{}2)^2\minus{}3}{4}\plus{}B)\bmod 7$ (which is possible since $ 4$ is invertible $ \bmod 7$), then we have that $ 4C\minus{}4B\equiv (A\minus{}2)^2\minus{}3\bmod 7$, so $ (A\minus{}C)^2\minus{}4(C\minus{}B)\equiv 3\bmod 7$.  Yet, $ 3$ is not a quadratic residue $ \bmod 7$.  Thus, we have a contradiction, so there exists a desired $ C$.  [/hide]

[quote="The QuattoMaster 6000"][quote="orl"]Let $ \mathbb{Z}$ denote the set of all integers. Prove that for any integers $ A$ and $ B,$ one can find an integer $ C$ for which $ M_1 \equal{} \{x^2 \plus{} Ax \plus{} B : x \in \mathbb{Z}\}$ and $ M_2 \equal{} {2x^2 \plus{} 2x \plus{} C : x \in \mathbb{Z}}$ do not intersect.[/quote]
I think this works:
[hide=Solution]
Assume that $ x^2 \plus{} Ax \plus{} B \equal{} 2x^2 \plus{} 2x \plus{} C$ has at least one solution for all $ C$[/quote]
I think you must consider $ x^2\plus{}Ax\plus{}B\equal{}2y^2\plus{}2y\plus{}C$

Why is every solution keeping b? Can't we just assume b is 0 since we could adjust c to make up for it?

The problem states that for any $ A,B$..., i.e., given fixed $ A,B$...

Right, but even if b is fixed we could choose an appropriate c to make up for it. For example, if when b=0 we would choose c=3, then when b=4 we would choose c=7, right?

dgreenb801 is saying that $ \{ x^2 \plus{} Ax \plus{} B \}$ and $ \{ 2x^2 \plus{} 2x \plus{} C \}$ intersect if and only if $ \{ x^2 \plus{} Ax \}$ and $ \{ 2x^2 \plus{} 2x \plus{} (C \minus{} B) \}$ intersect, which is true but does not significantly shorten the solution (as far as I can see).

Hey, this is rather trivial... all you need to notice is that $x(x+A)+B$ can't form a CRS(complete residue system) modulo $4$ (just check for different values of $x \pmod{4}$ to get $B,1+A+B, 2A+B, 1-A+B$ and now check for different values of $A \pmod{4}$) and $2x(x+1) \equiv 0 \pmod{4} => 2x^2+2x+C \equiv C \pmod{4}$, hence there's always a way to choose the required $C$


If the mininum value of 2x^2-bx-c= 0
then c equals

-b.4
b^2/4
c/4
b/4
none of these

[quote="iamhe"]2x^2-bx-c=[/quote]

Equals what?

[hide][hide][hide]i dont know the answer cause theres no question sorry, ok its your fault if you suicideare u sure u want know the answer?you really should figure it out yourselfheres the answer for real this time[/hide]heres the answer![/hide]the answer[/hide]


limit
probability
calculus
calculus computations

Prove that :
$ \lim_{n \rightarrow \plus{}\infty}\sum_{k\equal{}0}^{n}\frac{n^{k}}{k!}e^{\minus{}n}\equal{}\frac{1}{2}$

... but without using probability and central limit theorem.
I know that there is a proof without these things.
If anyone knows the proof , please post it   :)

Hi Extremal. You wrote in your old solution:
[quote="Extremal"]your sum is equivalent for big n = N to $ \frac {1}{2}\frac {N^{N}}{N!}\sqrt {2\pi N}$[/quote]
Could you explain it?

I don't remember how I get such expansion, i think it follows from my last post there. (see my proof)


Duke
college
number theory
modular arithmetic
AMC

Refer to title! The AMC is the perfect test for my level and ability now so I want some more fun problems!

Let's try this one, from the Duke Math Meet 1999:

tan(11x) = tan(34 degrees) and tan(19x) = tan(21 degrees).  Compute
tan(5x).

very nice problem..

[hide]

We know that 11x-34 is divisible by 180, and so is 19x-21. Then we just need to find a combination of 11 and 19 which makes 5, ie 11a+19b=5, and b=2, a=-3 works fine. In other words, -3(11x-34) + 2(19x-21) is divisible by 180, and thus 5x + 60 is divisible by 180, and tan (5x) = tan(-60) = -rt(3).

[/hide]

Just over 8 years too late! For cases like this, I think AoPS should implement an anti-Thanks button.

Why shouldn't qpqpgolf have revived an 8-year-old thread? The question was a good one, and as a result of his action, people can now see the problem that they most likely would not have seen.

EDIT: People rate my post 2 stars just because they disagree with me. Can we stop that childish behavior?

He could have just posted problems from the Duke Math Meet or something.

I believe the problem is not valid.

[hide]
Note: $ x $ is expressed in degrees.

By modular arithmetic, it is easy to establish that if $ x $ is an integer, then $ x \equiv 134 (\mod 180) $ and $ x \equiv 39 (\mod 180) $.

This cannot occur for the same integer, hence $ x $ is not an integer. However, $ 11x $ and $ 19x $ are both integers.

From the fact that $ 11x $ is an integer, while $ x $ is not an integer, $ x $ is a fraction with $ 11 $ as the denominator. However, this implies that $ 19x $ would not be an integer, since $ 19 $ is not a multiple of $ 11 $.

This leads to a contradiction. Thus, there exists no $ x $ such that the given is true. [/hide]


superior algebra
superior algebra solved

$G$ is a group, but not commutative. $[G:1]>2$, prove that there exist $a,b$ (neither of them is the identity element) such that $ab=ba$.

Case 1: There is an element $a$ of order not dividing 2. Then $a$ and $a^2$ commute with each other, and neither is the identity.

Case 2: All elements of the group have order 1 or 2. Then, for any $a,b$, $(ab)(ba)=ab^2a=a^2=e$ and $(ab)(ab)=e$, so by the cancellation law $ab=ba$. Since the group has more than two elements, we can find two elements other than the identity, and they commute.


AMC
AIME
algebra
polynomial
absolute value

I do not understand the official solution or the kalva solution to this problem.  If anybody could explain it it would be much appreciated:
Find the absolute value of the coefficient of x^2 in the polynomial (1-x)(1+2x)(1-3x)...(1+14x)(1-15x).

Thanks!

You can easily multiply it out in a couple minutes. You don't really need anything fancy(though im sure theres a better way.)

basically, the problem is asking for the coefficient of x^2.  Now, when you multiply it out, you get a term with an x^2 if two terms of with an X are multipied with a number of 1s.
Therefore, the answer would be the sum of all the product of two pairs of coefficients with the x term from the equation.
You can brute force it , or you can use a more elegant solution:

You can find the sum of the possible two terms with this:
((-1+2-3+4-5+6-7...-15)^2-(1^2+2^2+3^2+4^2...15^2))/2

(We then divide by two, because each possible pair is counted twice)

Use: 1^2+2^2+3^2+4^2...n^2=(n)(n+1)(2n+1)/6
to find (1^2+2^2+3^2+4^2...15^2)=15*16*31/6

Then you can figure out the answer, which is -588

whose value is 588

Does this help?

hey, check this post out....

http://artofproblemsolving.com/Forum/viewtopic.php?p=148569&highlight=#148569

if that doesnt help, you could always see the mathjam transcript...

although i don't quite understand how they got (-1+2-3..-15)*(-1+2-3..-15)...


number theory
prime numbers
prime factorization

[quote="isabella2296"]Define a 1337 number to be a number that can be written as a sum of 4 consecutive prime numbers and a product of any amount of consecutive numbers. What is the smallest 1337 number?

Source: Me[/quote]

Umm...I'm not sure, but is it 60?

$ 11 \plus{} 13 \plus{} 17 \plus{} 19 \equal{} 60$
$ 3\times4\times5 \equal{} 60$

I used guess and check to find it.

@matt919: $ 7\plus{}11\plus{}13\plus{}17$  does equal 48, but what numbers multiplied equal 48?  The prime factorization of 48 is $ 2^4\times3$, so we can get $ 2\times3\times(2\times2)..$, but the next number has to be 5, not 2.

I would think 17.

17 = 2+3+5+7

17 = 17 (product of 1 consecutive number)

Although that's like cheating.

Since she made it up herself, I think that's just a loophole/typo... :wink:

I got 60. If you add up 4 consecutive primes you get 17, 26, 36, 48, 60, etc... Then you just try them out and 60 is the first one that can be written as he product of more than one consecutive number.

Well, we want the sum to be as small as possible, so we have; 17, 26, 36, 48, 60

i believe the answer is 60, because 60 is 3 x 4 x 5.

Didn't james4l just say that?

[quote="james4l"]

2 consecutive number product: 2, 6, 12, 20, 30, 42, 56, 72, ...
3 consecutive number product: 6, 24, 60, 120, ...
4 consecutive number product: 24, 120, 360, ...
5 consecutive number product: 120, 720, ...

Therefore 60 is the smallest if we only have more than 1 consecutive number product.[/quote]

Well, sometimes people understand things when it is put in different ways.

And you aren't here to think do people get james4l's solution or not.
He wrote it nicely enough :wink:


AMC
USA(J)MO
USAJMO
USAMO
AMC 10
AIME
AMC 12

Since qualifying for the USAMO will almost exclusively be through AMC 12 now, people will be competing for about 275 spots instead of 500. Wouldn't this make it significantly harder to qualify for USAMO? And if so, how much would this affect the index required for the USAMO? Thanks in advance!

I really doubt it. There might be a small jump in one direction, but it should stay small. I think the number of people switching to AMC 12 would've gotten higher scores anyways on the AMC 10 so that compensates for the loss of people.

I'll estimate that the AMC12/AIME index required to qualify for the USAMO will go up by 10-12 points.

It will probably increase since most want to make USAMO more than USAJMO so more will take AMC 12

So why would that make it increase? People get a lower average score on the AMC 12 than the AMC 10, so it shouldn't make much of a difference.

Well, it pretty much solidifies the fact that you need at least 120 on AMC 12 to even have a shot at qualifiying for USAMO.

You mean 100?

While the AMC 10 one was 120, much higher.

Ihatepie:

In the last couple of years, as it became clear that there was an incentive (a perverse incentive, in my mind) to take the AMC10 to qualify for the USAMO, then there were many, many students who did exactly that. That is, there are many people who qualified for the USAMO last year with scores like AMC10 150, AIME 9, or even AMC10 138, AIME 8.

Those people were not counted among the 330 or so who qualified via the AMC12 - they weren't even so counted if they took one AMC10 and one AMC12, since their AMC10 score was higher than their AMC12 score.

Now, assume that this year's rules had applied. If such a student really wanted to qualify for the USAMO, that student would have taken the AMC12, not the AMC 10. In most cases, that student would have scored about 20 points lower on the AMC12 than the AMC10. (There's a spread in this, of course. Individuals differ.) If the first student cited scored 130/9 that would be an index of 220. And it would have displaced some 11th/12th grader who scored at least 201 but nowhere near 220. And remember that the we're going from 330 AMC12 takers qualifying for the USAMO to 270, and with more high-end people taking the AMC12. The index has to go up. How far? My estimate of 10-12 points is really just a guess.

As for what Caelestor said: I'm pretty sure he wasn't referring to the minimum required to take the AIME but rather what it practically takes to qualify for the USAMO. And I'm going to say that what he said is not literally true. Suppose the minimum index for the USAMO is 213. The 114/10 would suffice. Or 104/11. Now it is true that most people capable of scoring 10 or 11 on the AIME tend to score higher than 114 on the AMC12 - but there are exceptions, and I have known some of those exceptions.

Now for one big warning: the difficulty of the AMC/AIME is not standardized from year to year and has never been standardized. The AMC test writers aren't ETS. They don't - and can't - statistically validate their questions before they use them. They do get surprised, one way or another. There were more 150's than usual on the 2009 AMC10. The 2001 AMC12 was so ridiculously hard that the AIME qualifying level was down in the 80's. The AIME bounces up and down in difficulty. So if I say that I think the qualifying index under 2009 conditions would be about 211 or 213, that doesn't mean that's where the index will be in 2010. It could be 229; it could be 198. We won't know until they score the exams.

Kent Merryfield:

This may be an exaggeration but I think the USAMO index will go up by at least 20. I know plenty of underclassmen (10th below) who will be switching to USAMO because they don't want to do the USAJMO; some have even called the USAJMO pointless. The dropping from 330 to 270 also comes into play. And there seem to be more middle schoolers that are amazing at amc12 math now than ever before.

[quote="soltixnj"]And there seem to be more middle schoolers that are amazing at amc12 math now than ever before.[/quote]
There are, and I know some of them. But the people you know (and the people that I know) aren't representative of the U.S. as a whole. And 20 points is a lot. I'm going to stick with a more conservative guess - but we're both guessing.

I think of it as this: I think out of the 500 USAMO qualifiers, 150 of them are 9th/10th graders. I think that half of them will take the amc12 and make it on to usamo. So, we have slightly worse statistics to make it into the USAMO, since less spaces, but more qualifiers (hypothetically). Therefore, I think it's somewhat comparable to the old USAMO, where 250-270 made it, or maybe slightly harder.

In the old qualifying system (such as year 2005), only about [b]170[/b] students got selected through index; additional 70 underclassmen through floor and around 10 through state representation. The index for that year was 233 if you took the AIME I, or 220.5 if you took the AIME II and the floor was 9. That specific year, AIME was considered not very hard. Since there were no incentives to take AMC10 for that year, many underclassmen took AMC12. A few of them qualified through index and many qualified through floor. When an underclassman qualified through index, he/she would take an upperclassman's spot and not the floor spots. [b]Since there are 270 slots through index now (which is 100 more for index qualifiers[/b] compared to 2005), I don’t think that the index will be more than 15 points higher.

Quote:

"Any tenth grader who takes the AMC 12 will qualify for the USAMO by their AMC 12 score only."

How about that?

This year's AIME I was easier than some of the other AIME's, and AIME II was considerably harder than AIME I. The AMC committee can't make the AIME I/II difficulty the same every year, so the USAMO/USAJMO thing is not going to determine the rise/fall of the index


This is one of my first works with Wings 3d (http://www.wings3d.com), a "fractured fractal".

Anyone else have any 3d art?

[quote=MathFiend]stop reviving.

[b]LOCKED[/b][/quote]

are you sure about that?

[quote=happyhippos]lol great bump[/quote]

ikr

[quote=karthik_k]Hey , that looks good....I have some other 3d works , but I'm on a dialup connection and it's painful to upload the stuff...[/quote]

You realize this was a 1 year bump too?  :wow:


HCSSiM
summer program
PROMYS
Mathcamp
MathPath
search
articles

Has anyone been to the Hampshire College program and if you have, can you tell me how the math component compares to other programs such as PROMYS and MathCamp? I was at MathPath two years ago, so that would be my frame of reference. Thanks!

There are definitely some AoPS folk who have been to HCSSiM, but I haven't seen any of them posting lately, which is why I'm replying...  (Do a search of the forums for HCSSiM and that'll turn up a few usernames.)  I'm not quite sure what you mean by "the math component," as ALL of HCSSiM is a math component!  But maybe this information will help---there have been students who went to MathPath and then to HCSSiM and were very happy.  PROMYS concentrates on Number Theory whereas MathCamp and HCSSiM both deal with lots and lots of different branches of mathematics.  At HCSSiM, everyone does similar mathematics for the first three weeks (it's called workshop) so that everyone will have roughly equivalent background for the second half of the program, and chooses the courses they wish take for the last three weeks (maxi and mini courses).  At MathCamp, there are many different courses offered, at different levels.  Oh, and I might add that the directors of HCSSiM and of MathCamp encourage students to apply to both programs (and to others as well).  So if you haven't applied yet, go for it!

I highly recommend the Hampshire College Summer Studies in Math program.  Check out their website at [url=http://hcssim.org]www.hcssim.org[/url], as well as the article [url=http://hcssim.org/commentary.pdf]Supporting a National Treasure[/url], published in Notices of the AMS.

You'll spend 6 weeks living and breathing math, all in a relaxed, fun environment.  No dry, stuffy lectures here.  Course topics include combinatorics, number theory, complex numbers, probability, four-dimensional geometry, fractals and chaos, graph theory, topology, and cellular automata.  Mornings are spent in class.  Evenings offer problem sessions.  Afternoons are free except for the daily Prime Time Theorem.  Most of the faculty members live in the dorms with the students so you'll get to know your teachers really well.

Awards won by HCSSiM alumni include: Morgan Prize, Guggenheim Fellow, MacArthur Fellow, Putnam Fellow, and Ultimate Player of the Year.  Guest lecturers visit throughout the summer.  Weekend activities usually include a trip to Boston.  Frisbee and soccer are popular afternoon activities.

My high schooler and I both had a fantastic time there (oh so many years apart) so if you're looking for a stimulating, mind-stretching math-filled summer, you should definitely consider Hampshire.

I was at the Hampshire College program in 1978, and had the time of my life.  I know that was a long time ago, but judging from the two previous posts, the format seems to have remained approximately the same.

I went in 2003. We covered a wide breadth of topics. I was in the topology maxi, but I was also introduced to complex geometry, number theory, graph theory and other things rarely found in high school.
It was a very intimate, closely-knit community because the teachers do share meals and dorms with the students.
There's a two-three hour session an hour after dinner to solve some problems that remind me of the ARML Power Round. They are more proof-oriented than computational.
It's not all about the math either. I learned how to solve the Rubik's cube, play SET, and throw a Frisbee underhand. I brought all of these back to school and started a trend. There are plenty of things to do in the daily 5 hours between lunch and Prime Time Theorem.

I am also an alum of HCSSiM (2000).  (And I'm sure Sarah-Marie can atest to my terrible presence in the program).

It's definitely one of the best summers I've ever had.  The Math there will be absolutely intense (yet fun!) and the friends and memories you bring back are second to none.  I am now 5 years detached from the program and am still very much in touch with more than a handful of friends I made at that program.

I can tell you definitively that HCSSiM and something Kelly (program director) said to me in self-evaluation has made a big impact in my life since then....


geometry
rectangle
analytic geometry
inequalities

(a) Color the plane in 2 colors. Show that there exists a rectangle, all of whose vertices are the same color.

(b) Show that it is possible to color the plane in 2 colors in such a way that there cannot exist an equilateral triangle of side 1 with all vertices the same color.

[hide="2"]

Let's say there is NO such equilateral triangle if we color 2 different colors in the plane. If there is an equilateral triangle on the plane, however, then it must have two vertices colored the same. This contradicts the statement, because then we can find another equilateral triangle with THOSE two vertices and the third vertex which will have the same color.

[/hide]

I think you misread the question, it asks to prove that such a coloring CAN be done.

a) Let's consider the set of integer points with coordinates $(a,b)$ where $1 \leq a,b \leq n.$
Wlog, we may assume that there are as many blue points than red points in this set. Thus, the number of blue points is $b \geq \frac {n^2}2.$
Let $b_i$ be the number of blue points in the row number $i$.

If $\sum_{i=1}^n C_{b_i}^2 > C_n^2$, (1)
it would force two distinct rows to have blue points in the same columns and then we would have a blue rectangle.
But, using Jensen's inequality (sinc $f(x) = C_x^2 = \frac {x(x-1)} 2$ is convex, we deduce that
$\sum_{i=1}^n C_{b_i}^2 \geq \frac {b(b-n)} {2n} \geq \frac 1 8 n^2(n-2)$. Thus, for $n$ sufficientely large (1) is satisfied (in fact, according to my computations, $n=6$ should be enough...).

b) Divide the plane into parallel bands bounded by parallel lines $\Delta_k$ with equation $x=n \cdot \frac {\sqrt 3} 2$, where $k$ varies over the set of all integers.
Now for each $k$, color blue all the bands delimited by $\Delta_{2k}$ (included) and $\Delta_{2k+1}$ (excluded), and the rest in red.
To verify that this coloring has the desired property is straightforward.

Pierre.

More generally, for a), Gallai's theorem states that given any finite configuration $S$ and any finite coloring of the plane (this is also true in any finite dimension), we may find a monochromatic set which is homothetic to $S$.

(I just love this result :)  )

Pierre.

Does it have some relation to Van der Warden's theorem? Because I though that the result belongs to Van der Warden. :?

Gallai's theorem can also be proved for $\mathbb {Z}^n$ or $\mathbb {N}^n$.
In the case $n=1$, and if the given configuration is a finite set of consecutive integers, then you will have Van der Waerden' theorem about arithmetic progressions (the homothetic of a string of consecutive integers).

Pierre.

I know that. Moreover, in my book Van der Warden's theorem is initially stated for $\mathbb N^n$.
I asked how it is related to van der Warden theorem, i.e. can Gallali's result be derived from van der Warden's theorem or it has completely independent proof, or it shares the same ideas?

As far as I know, I mean the proof I know, the more general version of Gallai's theorem is 'easily' proved as a corollary of Hales-Jewett's theorem.

Pierre.


integration
calculus
derivative
function
real analysis
real analysis unsolved

If $ \Omega\subseteq{R}^n$ is open and $ f\in L^{1}_{\text{loc}}(\Omega)$ satisfies
\[ \int_{\Omega} f\varphi dx \equal{} 0
\]
for every $ \varphi\in C^{\infty}_c(\Omega)$, then $ f\equal{}0$ a.e.

Suppose not. Then [url=http://en.wikipedia.org/wiki/Lebesgue%27s_density_theorem]Lebesgue's density theorem[/url] tells us how to choose $ \varphi$.

The idea here is to prove that $ \int_{B(x,r)} f \equal{} 0$ for all $ x \in \Omega$ and for all $ 0 < r < d(x, \partial \Omega)$. Then the Lebesgue's differentation theorem tells us that $ f \equal{} 0$ almost everywhere. So we need to approximate the characteristic functions $ \chi_{B(x,r)}$ smoothly and use the given condition.

Here we need a lemma:
If $ 0 < r < R$ and $ x_0 \in \mathbb{R}^n$, we can find $ f \in C_c^{\infty}(\mathbb{R}^n)$ so that $ 0 \le f \le 1$, $ f(x) \equal{} 1$ for all $ x \in B(x_0,r)$ and spt$ \,f \subset B(x_0, R)$.
To prove this, assume $ x_0 \equal{} 0$, choose such an $ \epsilon > 0$ that $ r\plus{}2\epsilon < R$ and set $ f \equal{} \chi_{B(0, r\plus{}\epsilon)}*\varphi$, where $ \varphi$ is some variation of the standard moffilier, spesifically $ \varphi \in C_c^{\infty}(\mathbb{R}^n)$, $ \int \varphi \equal{} 1$ and spt$ \,\varphi \subset \overline{B}(0, \epsilon)$.

Proof of the theorem:
Let $ x_0 \in \Omega$ and $ 0 < r < d(x_0, \partial \Omega)$. For all $ i \equal{} 1,2, \ldots$ large enough (so that $ r \plus{} 1/i < d(x_0, \partial \Omega)$) choose $ u_i \in C_c^{\infty}(\mathbb{R}^n)$ so that $ 0 \le u_i \le 1$, $ u_i(x) \equal{} 1$ for all $ x \in B(x_0, r)$ and spt$ \,u_i \subset B(x_0, r\plus{}1/i)$. Now we have
$ 0 \equal{} \int_{\Omega} fu_i \equal{} \int_{B(x_0,r)} f \plus{} \int_{B(x_0, r\plus{}1/i) \setminus B(x_0, r)} fu_i$. (*)
Here
$ \left| \int_{B(x_0, r\plus{}1/i) \setminus B(x_0, r)} fu_i\right| \le \int_{B(x_0, r\plus{}1/i) \setminus B(x_0, r)} |f| \to \int_{S(x_0,r)} |f| \equal{} 0$ because $ 0 \le u_i \le 1$ and the convergence results of measures. Letting $ i \to \infty$ in (*) then shows that $ \int_{B(x_0,r)} f \equal{} 0$ which is what we wanted.


Let be $ a,b,c,d\in \mathbb{N}^*$ . Solve this equation :

$ x^2\minus{}(a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}1)x\plus{}ab\plus{}bc\plus{}cd\plus{}da\equal{}0$ ,

if it has an integer root .

x^2-(a^2+b^2+c^2+d^2+1)x+(b+d)(a+c) = 0

The roots have to add up to (a^2+b^2+c^2+d^2+1) and multiply to
(b+d)(a+c). If the two roots were (b+d) and (a+c), then:

a^2+b^2+c^2+d^2+1 = a + b + c + d

a^2 - a + b^2 - b +c^2 - c + d^2 - d + 1 = 0

a(a-1) + b(b-1) + c(c-1) + d(d-1) = -1

But ultimately i got notin...


If $ [a]$ represents the greatest integer less than or equal to $ a$, what is the value of $ [\pi] \plus{} [\minus{}\pi]$?

$ [\pi]\equal{}3$ and $ [\minus{}\pi]\equal{}\minus{}4$ so the sum is -1.


inequalities
real analysis
real analysis unsolved

Let $ \{x_n\}_{n \equal{} 1}^{\infty}$  be the sequence of real numbers satisfying  $ 0 < x_1 < 1$  and  $ x_{n \plus{} 1} \equal{} x_n \plus{} \frac {x_n^2}{n^2}$  for  $ n\geq 2$.
Show that  the sequence $ \{x_n\}_{n \equal{} 1}^{\infty}$  is bounded above.

[b]Lemma[/b] Let $ \alpha \equal{} x_1$. Then $ x_n \leq \alpha n^{\alpha}$.

[i]proof.[/i] We prove it by induction. The base step is automatically satisfied, so all we need to prove is the inductive step. Assume $ x_n \leq \alpha n^{\alpha}$. Then by Bernoulli's inequality, we have

$ 1 \plus{} \frac {\alpha n^{\alpha}}{n^2} \leq 1 \plus{} \frac {\alpha}{n} \leq \left( 1 \plus{} \frac {1}{n}\right)^{\alpha}$,

which consequently yields

$ x_{n \plus{} 1} \equal{} x_n \left( 1 \plus{} \frac {x_n}{n^2} \right) \leq \alpha n^{\alpha} \left( 1 \plus{} \frac {\alpha n^{\alpha}}{n^2} \right) \leq \alpha (n \plus{} 1)^{\alpha}$.

Thus the inductive step is satisfied and the proof is completed. ////

Therefore we have $ x_{n \plus{} 1} \equal{} \alpha \prod_{k \equal{} 1}^{n} \left( 1 \plus{} \frac {x_k}{k^2} \right) \leq \alpha \prod_{k \equal{} 1}^{\infty} \left( 1 \plus{} \frac {\alpha}{k^{2 \minus{} \alpha}} \right) < \infty$.


The 8-centimeter radius of a circle is increased by one-fourth of its length to form a new circle. What is the circumference, in terms of $ \pi$, of the new circle?

The radius is now 8+8/4=10 so the circumference is 2x10xpi=20pi


I know that there are problems of Romanian Olympiad of this year on http://www.epsilon.ro. I could understand a little. I downloaded some files with problems but there weren't any new problems. Could anybody say the files that I should download? I need the district, national, JBMO training problems of the last year.

Tnx.

there aren't any real math problems on that site. any Romanian member of this forum will confirm it.

most of the problems are for 5th - 8th grades (ages 10-14 years, that is very very very easy level), and are not olympiad type.

unfortunately it is erronately indicated on some websites as the offcial site of the romanian olympiad.

hopefully that will change.

also I was the one who selected the training problems, and posted them on my ex-site, but I wrote them in Romanian. I doubt they will use you too much. Most of them have already been discussed on the forum.

I could find the Bucharest round in Romanian. Now could someone say, is jedetene = city or district and  ONM=Olympiad National Math? :D
And BTW what's your ex-site? :)


How many teams got the cryptograms?  My team got them all.  The last one was evil though!  I don't think any library within a 20 mile radius of my house has [i]The Man Who Loved Only Numbers[/i]...

I think we did (Vestavia). There was one copy of the book in a downtown Birmingham library.

I managed to get 3 of the cryptograms by myself.  Our team didn't really show much teamwork though, so I really don't know what everyone else got.

There were 13 schools (out of 116) that got all five cryptograms.


geometry
vector
parallelogram
ratio
geometry proposed

Let $ABCD$ be a cyclic quadrilateral. Show that the orthocenters of the triangles $ABC,BCD,CDA,DAB$ are vertices of a quadrilateral congruent to $ABCD$ and show that the centroids of the same triangles are the vertices of a cyclic quadrilateral.

I denote $H_A,H_B,H_C,H_D$ the orthocenters of the triangles $BCD,CDA,DAB,ABC$ respectively, and respectively the centroids $G_A,G_B,G_C,G_D$.

Everything here are vectors

In the triangle $BCD$ we have:

$OH_A = 3OG_A = 3 \frac{OB+OC+OD}{3}\Rightarrow$

$OH_A = OB+OC+OD \Rightarrow$

$OA+OH_A = OA+OB+OC+OD \Rightarrow$

$OA+OH_A = 4OG$

where $G$ is the centroid of $ABCD$

So, the parallelogram formed by the vectors $OA,OH_A$ has center in the point $K$ defined by $OK=2OG$.

Hence the points $A,H_A$ are symmetric w.r.t. $K$.

Similarly, we have
$H_B$ is symmetric to $B$ w.r.t. $K$
$H_C$ is symmetric to $C$ w.r.t. $K$
$H_D$ is symmetric to $D$ w.r.t. $K$

So, the quadrilateral $H_AH_BH_CH_D$ is congruent to $ABCD$, so it is cyclic too

For the second question, using the equations:

$OH_A = 3 \cdot OG_A$
$OH_B = 3 \cdot OG_B$
$OH_C = 3 \cdot OG_C$
$OH_D = 3 \cdot OG_D$

we find that the quadrilateral $M_AM_BM_CM_D$ is similar to $H_AH_BH_CH_D$ with ratio of similarity $\frac{1}{3}$. Hence it is cyclic too.


geometry
incenter
inequalities
geometry solved

I is the incenter of ABC. N, M are the midpoints of sides AB, CA. The lines BI, CI meet MN at K, L respectively. Prove that AI + BI + CI > BC + KL.

If we use the fact that $BNK$ and $CML$ are isosceles triangles, after some computations we find $BC+KL=p$ ($p$ is the semiperimeter). Now it's easy because $IB+IC>BC,\ IC+IA>CA,\ IA+IB>AB$. We add these three inequalities and we divide what we get by $2$.

Thanks grobber! :) I understood it perfectly.thanks for your help.


logarithms
limit
irrational number
complex analysis

I was told that (-1)^pi is undefined, is it really true? :maybe:
How can we (or can we) defined the power of an irrational number? :huh:

Thank you very much! :P

This can be answered in various ways. Since you decided to post this question in the complex analysis section, I'll attempt to give it the answer suitable to this section. (You would get a different answer in a different part of the site.)

The general principle is that $a^{b}=e^{b\log a}.$ The complication in that general principle is that $\log a$ is not well-defined. If $a$ is a positive real number, then there is an unambiguous conventional value for $\log a,$ namely the standard natural logarithm. In all other cases, there is ambiguity.

$e^{\pi i}=-1$ so one possible value for $\log(-1)$ is $\pi i.$ But there are infinitely many other possibilities; we can list them all as $(2k+1)\pi i,\ k\in\mathbb{Z}.$

Then the possible values of $(-1)^{\pi}$ are $e^{(2k+1)\pi^{2}i}=\cos((2k+1)\pi^{2})+i\sin((2k+1)\pi^{2}).$ These are all points on the unit circle; the set of all of them will be dense in the unit circle.

Kent Merryfield, thank you very much! :P
Your explanation is very clarified.
I just wonder what another way can approach this question?

Thank you very much! :P

And, as puuhikki is well aware, that procedure won't work for $x<0.$ In trying to approximate a $y$ by rational numbers, we get some with even denominators, some with odd denominators and odd numerators, some with odd denominators and even numerators, and such different things happen with each of those that we don't get a limit.

[quote="puuhikki"]In real line one can define $x^{y}$ if $x\geq 0$. There are two ways I know. First way is to take a sequence $(y_{i})$ of rationals converging to $y$ and define $x^{y}=\lim_{i\to \infty}x^{y_{i}}$. The another way is to use supremums and infimums. Namely
\[x^{y}=\inf\{x^{q}|q\in \mathbb{Q}\text{ and }q>y\}\text{ as }x>1 \]
and
\[x^{y}=\sup\{x^{q}|q\in \mathbb{Q}\text{ and }q>y\}\text{ as }0<x<1 \]
[/quote]

I think this is quite advanced for me, I have no idea what are supremums and infimums, but thank you all for posting anyway, now at least I understand that (-1)^pi is a set. :)

Thank you very much! :P

[quote="puuhikki"]Let $A$ be a set of real numbers. Then $\sup A$ is a real number $r$ such that for every $a\in A$ $r\geq a$ and for every $\epsilon>0$ there exists $a_{0}\in A$ such that $r-\epsilon<a_{0}$. If such a real number does not exists, then $\sup A=\infty$.

$\inf A$ is a real number $r'$ such that for every $a\in A$ $r'\leq a$ and for every $\epsilon>0$ there exists $a_{0}\in A$ such that $r'+\epsilon>a_{0}$. If such a real number does not exists, then $\inf A=-\infty$.

In particular, if $A$ is a set of real numbers then $\sup A$ and $\inf A$ are not sets. They are either real numbers, $-\infty$, or $\infty$.

Note that if we take a finite set $A$ of reals then $\sup A=\max A$ and $\inf A=\min A$ (this can be happen for infinite sets too). But supremum and infimum are more general than max and min because the set $\{x\in \mathbb{Q}|x<\sqrt{2}\}$ has not a maximal element but its supremum is $\sqrt{2}$.[/quote]

puuhikki, thank you very much! :P This epsilon method looks more familiar to me, however it is still very advanced to me.
Kent Merryfield's first post has explained everything I need, I now have more precise understanding of $(-1)^{\pi}$.

Thank you very much! :P


function
inequalities
inequalities proposed

Let $ I$ be an interval in the real numbers, and let $ f: I\to\mathbb{R}$ be a function such that for all $ a,b\in I$ and $ t\in(0,1)$, $ tf(a)\plus{}(1\minus{}t)f(b)\geq f(ta\plus{}(1\minus{}t)b)$

Let $ n$ be a positive integer, and let $ x_1,x_2,\dots,x_n, w_1,w_2,\dots,w_n$ be real numbers such that $ x_1,x_2,\dots,x_n\in I$, $ w_1\plus{}w_2\plus{}\dots\plus{}w_n\equal{}0$ and for all $ x\in\{x_1,x_2,\dots,x_n\}$, $ w_1|x\minus{}x_1|\plus{}w_2|x\minus{}x_2|\plus{}\dots\plus{}w_n|x\minus{}x_n|\geq0$

Prove that $ w_1f(x_1)\plus{}w_2f(x_2)\plus{}\dots\plus{}w_nf(x_n)\geq0$

Hello Liam

There's a proof in this paper by Darij Grinberg: http://www.cip.ifi.lmu.de/~grinberg/Popoviciu.pdf
But I suspect that that's probably where you found the result anyway.

Actually Dylan, I didn't see this problem anywhere else. I was thinking about majorization and convex functions a while ago and this is one of the problems I came up with.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $A\in\mathcal M_n(R)$, if $\det(A+X)=\det(A)+\det(X)$ , $\forall X \in\mathcal M_n(R)$ then $A=O_n$.

if $r$ is the rank of $A$, $A=PJ_rQ$ where $P,Q$ are invertible matrices. From here, you can suppose that $A=J_r$ and the conclusion will follow.

first of all let $J$ be
\[ \begin{pmatrix} 0 & 1 & 0 & ... & 0\\ 0 & 0 & 1 & ... & 0\\ ...&...&...&... & ...\\ 0&...&...&...& 1\\ 1&0&...&...& 0\\ \end{pmatrix}  \]
we can easily prove that $det(A)=0$ by considering $X=2A$ (of course the case of $n=1$ is omitted) ,then we find that $Sp(A)=\{0\}$ (A is considered a complex matrix) then let $T$ be the triangular matrix similar to $A$ we have
$\forall \lambda \in C$ & $\forall m \in N$ : $det(J^mT+\lambda I_n)=\lambda^n$ (btw the condition related to $A$ might be generalise to all complex matrices $X$ :polynomials ;) ) then we conclude that $T=0_n\to A=0_n$ :).


I had problems creating this diagram for a long time:
[asy]import markers;

unitsize(1pt);
size(600,400);
pair A, B, C, P, PA, PB, PC, X, Y, Z, D, E, F, Q;
A=(0,0); B=100*dir(60); C=100*dir(0); P=35*dir(17);
Y=100*dir(120);
X=100*dir(-60);
Z=B+100*dir(0);
PA=35*dir(-43);
PB=Y+35*dir(-43);
PC=B+35*dir(-43);
D=A+200*dir(180);
E=B+200*dir(180);
F=C+200*dir(180);
Q=P+200*dir(180);
draw(D--E--F--D--Q--E);
draw(Q--F);
draw(A--B--C--A--P--B);
draw(P--C--X--A--Y--B--Z--C);
draw(A--PA--P--PB--A);
draw(B--PB--Y);
draw(B--PC--Z);
draw(X--PA--C--PC--P);
label("$A$", A, W);
label("$B$", B, N);
label("$C$", C, ESE);
label("$D$", D, W);
label("$E$", E, N);
label("$F$", F, ESE);
label("$a$", A--P, NNE);
label("$b$", B--P, ESE);
label("$c$", C--P, N);
label("$d$", D--Q, NNE);
label("$e$", E--Q, ESE);
label("$f$", F--Q, N);
label("$Z$", Z, ESE);
label("$X$", X, S);
label("$Y$", Y, W);
label("$P_A$", PA, 2*ESE);
label("$P_B$", PB, 2*N);
label("$P_C$", PC, 2*ESE);
markangle(radius=10pt, PA, A, P);
markangle(radius=10pt, PB, B, P);
markangle(radius=10pt, PC, C, P);[/asy]

If you look at the code and switch any of the labels with an ESE to an E, this causes the symbol to appear radically elsewhere -- about 1000 pts away. The other cardinal directions work perfectly for positioning, but using East -- even once! -- seems to cause my output to get distorted. Any help would be appreciated :)

I fell into this trap once.  You defined E in your code (it's one of your points), which will overshadow the direction E.  To refer to the direction now, type plain.E.  Alternatively, rename your point.

This trap (and solution) is mentioned in the Asymptote manual.


calculus
integration
function
symmetry
calculus computations

Are you doing an integral over a symmetric interval? That is, $\int_{-a}^{a}f(x)\,dx\,?$

If that's the case, always ask whether $f$ is even, odd, or neither. And if neither, can it be readily split into an even part and an odd part.

Provided the integrals in question converge, we have the following:

If $f$ is odd, $\int_{-a}^{a}f(x)\,dx=0.$

If $f$ is even, $\int_{-a}^{a}f(x)\,dx=2\int_{0}^{a}f(x)\,dx.$

The advantages of the first identity, for odd functions, are obvious. The advantages of the second identity, for even functions, are less obvious. In fact, if you're a machine, it hardly matters at all. But if you're human, I find that it's very much worthwhile. The advantage shows up at the last stage, in evaluating the antiderivative at the limits of integration, especially the negative endpoint. Was that three minus signs or just two? There are so many ways (as a human) to make sign errors; taking advantage of symmetry cuts back on that considerably.

no i want to know the answer to that problem. but when i put it in calculator, it gives me a decimal, which i don't want. So, i want it to be in a fraction from, or ratical form.


Are you asking about the EPGY distance learning classes (in which case, which class?) or about the new EPGY Online High School? I am very familiar with the distance learning classes, but can't quite figure out how your question relates to those. Our family has an applicant to the online high school, and we have sent in the full paper application to EPGY.

[quote="nigma"]When are you supposed to send in your score report for EPGY? In the email that contained my student number, I was told that I can send them to XXX, but they don't mention when and how? Am I just supposed to fax my scores down there?

I'm hoping the answer to this question is no, and I'm hoping someone is going to tell me that once you register for the course, you receive instructions on how to mail your scores...[/quote]

Last year, when I was signing up my son, I know I was given instructions on how to send them in, but then he got pneumonia, and then so did I. Somehow they never got sent in. I kept expecting them to send me something demanding the scores, yet it never happened.  I suspect, if you are not asking for any aid, they dont really care as much as they say.

I meant the distance learning courses, particularly AP Physics A-C. I asked because I had registered and paid for the class, yet had not sent in my scores. Later, I saw a message that lead me to believe that I would not get a refund if my PSAT scores were too low. Naturally, this freaked me out, but in the end, my PSAT scores were more than high enough, and so everything worked out.

Thanks for the responses.

Oh, and tokenadult: have either of your sons taken any of the courses in the AP Physics A-C series? How'd it go?

[quote="nigma"]
Oh, and tokenadult: have either of your sons taken any of the courses in the AP Physics A-C series? How'd it go?[/quote]

That was the one and only class he took. It seemed fine. We havent received his AP scores yet, but we are assuming a 5. He did put a fair amount of work into the reading and watching the online class, and of course, the tests. Other than that, he didnt seem to study much, and found it easy to get an average in the high 90s.

I expect my oldest son to start the EPGY physics sequence this fall, but there is some possibility we will arrange a local class. I'm not completely sure whether he will start with an AP B course or an AP C course in physics. He's already had single-variable calculus, and will study more calculus next year.

I have never heard of an ap physics A. Is there such a class? Also, what are the advantages of taking physics C over physics B? For the more in depth material and for more college credit?

In the AP designations for physics there is only a "B" course (with mathematics level matching high school algebra) and a "C" course (with mathematics level gradually including quite a bit of calculus). The topics are not quite the same between the courses, and the AP B course matches the syllabus for the SAT II test in physics better than the C course.

[url=http://apcentral.collegeboard.com/repository/05824apcoursdescphysi_4325.pdf]AP Physics Course Description[/url]


geometry
rectangle
pigeonhole principle
combinatorics proposed
combinatorics

Color the plane with $ n$ colors. If $ n$ is finite, it is easy to show, by Pigeonhole, that there must exist a monochromatic rectangle. (One exists in $ \mathbb{Z}^2$ as well.)

What if $ n$ is countably infinite? (I believe it is true.)

I am not sure but I think that if n is infinite then you cannot assure even that there exists a monochromatic line!!!

By "monochromatic rectangle" it is meant that the 4 vertices are colored the same.  (Also, even 2 colors is enough to prevent any monochromatic lines.)

It is impossible to prove that there is always a monochromatic rectangle parallel to X and Y coordinates. see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=137999]monochromatic corner[/url].

That is what I was thinking about, but now... is there [i]any[/i] rectangle with all vertices the same color?


algebra
polynomial
function
quadratics
superior algebra
superior algebra unsolved

Let's work in $\mathbb{H}[x]$. We have the following polynomial identity: $X^2 + 1 = (X - i)(X + i)$. Since $j$ is a root of $X^2 + 1 = 0$, we get that $(j - i)(j + i) = 0$. Since $\mathbb{H}$ is a division ring, we must have $j = i$ or $j = -i$.

What's wrong? (The "point" is somewhat subtle. Try to give a detailed explanation :))

Main reason: a polynomial in a division ring can have more solutions than his degree says (in the exact prove, commutativity is needed).
Secondary reason: $x^2+1$ doesn't factor into $(x+i)(x-i)$, since that would give that $ix=xi$ for all $x$.

Isn't very detailed indeed ;)

I presented those reasons to the assistant who was in charge of the "practice session" (where this problem was given) but he wasn't satisfied, he said I should be more specific. So, I'm still wondering what he actually means... :D

Well uhm Arne, that really is a nice question, last year in algebra I have been wondering about that too.  Thinking about why this kind of stuff works  in fields and not in quaternions, makes you realise how many times you use commutativity

It is strange that your assistent does not accept the answer $(x+i)(x-i)$ being different from $x^2+1$ is the reason

However perhaps this will help you : even the most basic facts of polynomials over the fields use the fact that it is a field

In general , a polynomial is not a function, but a set of coefficients that is only a finite times different from zero

multiplication is defined such that $a_{n}=b_{n}c{0}+....+b_{0}c_{n}$
in this case ,the functions f , g and h associated respectively with a, b and c, satisfy $f(x)=g(x) h(x)$ for all x in your field

you use commutativity when proving this trivial fact

I  am interested in more discussions about this actually...

And a polynomial in noncommuting variables is a monster; $ax$ is not the same as $xa$, and the general quadratic term has to be written as $axbxc$.

... and $X^2 + 1$ has infinitely many roots in $\mathbb{H}$.

Any more ideas?

If your assistant does not accept it, just ask him what he wanted to see, then give him that on the exam. For here, you are correct, as that polynomial identity is simply false over $\mathbb{H}$... filling in X=j indeed shows it's false. I'm afraid there are not "more ideas"

how about $x=cos(\theta) i + sin(\theta)j$ aren't those all roots (could be wrong :oops: )

The roots are exactly that quaternions $a+bi+cj+dk$ with $a=0$ and $b^2+c^2+d^2=1$.
(prove by calculating $(a+bi+cj+dk)^2$ and setting it's nonreal components to $0$)

ZetaX  , I'm pretty sure you meant $a=0$ and $b^2+c^2+d^2=1$ right?
true, it is indeed quite easy, my solutions are only a small part of that

Sorry, yes, that was what intentionally was meant.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Find a non-abelian group $ G$ such that $ g.x\equal{}g^{\minus{}1}xg$, for all $ g,x \in G$ does define a group action of $ G$ on itself.

[quote="georgech"]Find a non-abelian group $ G$ such that $ g.x \equal{} g^{ \minus{} 1}xg$, for all $ g,x \in G$ does define a group action of $ G$ on itself.[/quote]

I assume you want that (gh).x = g.(h.x) for all g,h in G.  This happens if and only if the inner automorphisms induced by gh and hg are equal, that is, if and only if the inner automorphism group is abelian.  In other words, G/Z(G) is abelian, so G is nilpotent of class at most 2.  If you want G non-abelian, then this is exactly the requirement that G is nilpotent of class exactly 2.

For instance G nonabelian of order 8 works.

[quote="georgech"]Find a non-abelian group $ G$ such that $ g.x \equal{} g^{ \minus{} 1}xg$, for all $ g,x \in G$ does define a group action of $ G$ on itself.[/quote]
(exactly G that are nilpotent of class 2, for instance nonabelian groups of order 8)
[quote="georgech"]There are only two non abelian groups of order 8, $ D_8$ and $ Q_8$. Could you please exain a bit more what does it mean nilpotent of class exactly 2? i haven't met this definition yet..[/quote]

Yes, both D8 and Q8 are examples.  Nilpotent of class exactly 2 means a nonabelian group such that every element of the form g^-1*h^-1*g*h commutes with every other element.