$ P$ is the intersection point of diagonals of cyclic $ ABCD$. The circumcenters of $ \triangle APB$ and $ \triangle CPD$ lie on circumcircle of $ ABCD$. If $ AC \plus{} BD \equal{} 18$, then area of $ ABCD$ is ?

$\textbf{(A)}\  36 \qquad\textbf{(B)}\  \frac {81}{2} \qquad\textbf{(C)}\  \frac {36\sqrt 3}{2} \qquad\textbf{(D)}\  \frac {81\sqrt 3}{4} \qquad\textbf{(E)}\ \text{None}$

To start, note that the circumcenter of APB must be the midpoint of arc AB since it is the intersection of the perp bisector of AB with the circle, similar for PCD.


Let $ a > b > c > \minus{} b$ are real numbers such that $ \left(\frac {a \minus{} b}{a \plus{} b}\right)^3 \plus{} \left(\frac {b \minus{} c}{b \plus{} c}\right)^3 \plus{} \left(\frac {c \minus{} a}{c \plus{} a}\right)^3 \geq 0$.

Prove that $ \left(\frac {a \minus{} b}{a \plus{} b}\right)^2 \plus{} \left(\frac {b \minus{} c}{b \plus{} c}\right)^2 \plus{} \left(\frac {c \minus{} a}{c \plus{} a}\right)^2$ $ \geq$ $ 2 \plus{} \left(\left(\frac {a \minus{} b}{a \plus{} b}\right)^3 \plus{} \left(\frac {b \minus{} c}{b \plus{} c}\right)^3 \plus{} \left(\frac {c \minus{} a}{c \plus{} a}\right)^3\right)^\frac {2}{3}$.


A plane parallel to the base of a square pyramid divides the volume into two equal amounts. If the dimensions of the pyramid are six inches for an edge of the base and six inches from the height, how many inches are in the distance from the plane to the base of the pyramid? Express your answer in simplest radical form.

[hide]Let the largest pyramid be P1 and the smaller be P2. P2 has volume 1/2 of P1, so it has linear lengths 1/cubert(2) of P2. The larger has height 6, so the smaller must have height 6/cubert(2). The distance from the base to the plane is thus 6 - 6/cubert(2) = 6 - 6cubert(4)/2 = 6 - 3cubert(4).[/hide]


search geometry AMC USA(J)MO USAMO Putnam probability

Notice that or between Finance & Science in the listing.  I wouldn't say this means finance is important.

It really depends on what you want you role to be and what sort of firm you want to work for.

Regardless, you should study something that interests you in college.  If you're interested in physics/CS/math but also want a job in the financial world, study physics/CS/math.  If you're not, then study finance.  In the latter case, you will find it a little harder to get into a hedge fund, but then, maybe that's not the place for you, anyway.  There are financial firms besides hedge funds.

As well as hedge funds that aren't quantitative. There are hedge funds based on value investment strategies, growth, GARP, macro. Quantitative hedge funds are just one sector in the broad field (a fast growing field albeit).

hi all,

i was doing a search for Jane Street and found AoPS by accident. i think AoPS is great. the only regret i have is that i didn't find it earlier.

about me, i'm one to two years away from a PhD in applied math. my research area is in computational fluid dynamics.

i'm very interested in job opportunities in quantitative finance. i would like to try to get into the field before i finish my PhD. i realize the quant field is very very selective. i dont have a stellar record of accomplishments like USAMO or putnam so i wouldnt even try DE Shaw(although the founder is from my school). their job qualifications are like you have to be a god. but i think there's got to be a quant job somewhere out there for me.

please feel free to offer any advice or comments.

thanks, mike.

Yeah, almost all these "quant" firms are hedge funds.  Ask Richard to tell you the "cake" analogy...it's pretty interesting...

[quote="texas137"]What's the difference btwn Jane Street and DE Shaw? Are they both hedge funds?[/quote]

Yes, they are both hedge funds.  I'd suggest strongly that people who are interested in hedge funds look at both of them to contrast the different characters of the two places.

I am gonna go off topic here but I think that all this stock market stuff is interesting, I am reading a book that describes stocks, bonds, mutual funds (doesnt explain hedge funds, maybe because they are too hard) so I think finance is very fun.

What do you have to do to be a mutual fund manager? Do you also have to have a math degree or finance or does it not matter? Just curious, I am an accounting and economics major, wonder if I can ever become one.

"I am reading a book that describes stocks, bonds, mutual funds"

Michael, what book are you reading?  Any good?  Anyone have any good book recommendations for an intro to the world of finance (for a math student who knows nothing about finance)?

I am reading several books. "Common Sense on Mutual Funds" by John Bole, the founder of Vanguard. This book is about how bad managed mutual funds are in general and explains that index funds that are passive and low-cost a superior investment.

Another book I am reading is "The Intelligent Investor" by Ben Grahame. According to Warren Buffet, the 2nd richest man in the world, this is the best book on investing ever written. I think it is more complicated then the first book which just talks pretty much about the same thing over and over again, how index funds are superior to mutual funds whereas, the intelligent investor delves into more topics and at least for me, is quite a challenge, but it is the classic of investing. I think that a person who understands the book well can do well in investing for the long term.

Do you know any good books? Really, I am a beginner myself.

Thanks.  Nah I don't know any good Finance books myself, but I did just notice another thread has a good list of Finance books.

So many good things to learn...not enough time.

I was referring to the thread called "Quantitative Finance Books."

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49469

This link was especially interesting (although it doesn't tell you where to start):

http://finmath.com/

For the sake of motivation read this: My Life as a Quant, http://finmath.com/DermanMLQ.html

That is what you might know/have about math/cs: analysis, algebra, statistics, probability theory, functional analysis, optimization in continuous spaces, machine learning, statistical pattern recognition, probabilistic graphical models, computer vision, programming experience (C, C++, Java, Matlab, Ocaml)

And here are some suggestions from finance courses if you are coming from a math/cs background like me:

Financial Computing: http://www.cs.ucl.ac.uk/teaching/syllabus/ug/6006.htm
Financial Institutions and Markets: http://www.cs.ucl.ac.uk/teaching/syllabus/ug/4026.htm
Financial Information Systems: http://www.cs.ucl.ac.uk/teaching/syllabus/ug/4027.htm

[hide="Here are the references in case the sites may expire in the future - listed in order of the courses above"]
- S R Pliska, Introduction to Mathematical Finance-Discrete Time Models, Blackwell, 1997, ISBN 1-55786945-6; J C Hull, Options, Futures and other Derivatives, Prentice Hall, 1989, ISBN 013-264367-7.
- Cornelis A. Los, Computational Finance: A Scientific Perspective.
- Jessica Keyes, J. Keyes (Editor), Handbook of Technology in Financial Services.
- Understanding the Markets (Securities Institute Global Capital Markets S.)
- Yannis Avgerinos, Regulating and Supervising Investment Services in the European Union.
- Tony Plummer, Forecasting Financial Markets

- Saunders, ‘Financial Institutions Management’, 2003.
- Valdez, ‘Introduction to Global Financial Markets’, 2002.
- Roberts, ‘Wall Street’, 2002.
- Steiner, ‘Mastering Financial Calculations’, 1998

- Mayer, 'The Bankers: New Worlds of Money, Credit, and Banking in an Electronic Age', 1997.
- Weber, 'Hedge Construction with Option Simulator'.
- Tibeldeo and Buben, 'Cashing in on Technology: A Primer on Electronic Payments Systems', IS Audit and Control, 1996.
- Weber, 'IT in the Major International Financial Markets'.
- Clemons and Weber, 'Turmoil, Transparency & Tea', 1992.
- 'The OptiMark Experience: The First Eighteen Months'.
- Uretsky and Weber, 'Next-Generation Trading in Futures Markets: The Cantor Financial Futures Exchange (CFFE)', 1998.
- Weber, 'Trade Order Management: A FIX Protocol Primer', 1997
- Clemons and Weber, 'The Changing Nature of the Financial Services Industry'.
- Grody and Weber, 'Global Electronics Market Study', Global Custodian, 1994.
[/hide]

go to: QuantFinanceJob.com  to get more detailed guides.

[quote="ffdbzathf"]From what I've heard about becoming a trader, I thought a degree in finance wasn't important and that math, CS, and physics were the way to go.  But on the Jane Street Capital website, I saw this:

Job Listings
Trader: Quantitative Trader (NY, NY) is needed to conduct securities trading with quantitative methods. Must have a Masters degree in Finance or Science, plus skills in quantitative analysis, C and VB .

Finance seems to be important...any comments?[/quote]

How about a serious finance book and a serious finance course?

I confess I don't know anything about finance.  It was not part of my mathematics education.  So I am taking a first finance course.  We are using this textbook:

Principles of Corporate Finance, Eighth Edition, by Brealey, Myers and Allen.

I don't know who is competing for these quant jobs, but I am guessing a math degree with a finance minor would be a pretty attractive candidate.

[quote="joeorez"]

I don't know who is competing for these quant jobs, but I am guessing a math degree with a finance minor would be a pretty attractive candidate.[/quote]

Only if it came with programming skills.


floor function
logarithms
number theory
number theory unsolved

This is really number theory, not analysis. In fact, both of the problems you just posted here are number theory. I'll move them.

We can use Bertrand's Postulate, that there's a prime $p$ between $\lfloor\frac n2\rfloor$ and $n$. $\frac 1p$ is the only fraction in our sum which has denominator divisible by $p$, meaning that the numerator of the sum is not divisible by $p$, while the denominator is.

There's the special case $n=2p$ to take care of (it doesn't work out exactly as above), but that's easy.

How about doing it without any difficult theorems on the distribution of primes?

Let $k=\lfloor \log_2n\rfloor$ and let $M$ be the largest odd factor of $n!$. Then $2^{k-1}MS=\sum_{j=1}^n\frac{2^{k-1}M}j$ and all but one term in that last sum is an integer. The remaining term is equal to $\frac M2$, which is not an integer, so $2^{k-1}MS$ is not an integer and $S$ is not an integer.


limit
integration
calculus
inequalities
real analysis
real analysis unsolved

I proved that $ \sigma(\gamma) : \equal{}\lim_{n\to\plus{}\infty}\frac{1}{n^{\gamma\plus{}1}}\sum_{k \equal{} 1}^{n}k^{\gamma}\equal{}\frac{1}{\gamma\plus{}1}$, for $ \gamma\in\mathbb{N}$. Is this true also for $ \gamma\in\mathbb R$, \gamma >0? Is this true, I mean $ \sigma(\gamma) \equal{}\lim_{n\to\plus{}\infty}\frac{1}{n}\sum_{k \equal{} 1}^{n}(\frac{k}{n})^{\gamma}\equal{}\int_{0}^{1}x^{\gamma}dx \equal{}\frac{1}{\gamma\plus{}1}$ ?

Sorry, I wanted to post this in Unsolved, not here.

the next fact is true :
let $ a > 0$ then
$ \sum^{n}k^{a\minus{}1}$~$ \frac{n^{a}}{a}$ this mean that your limit will be equal to $ \frac{1}{a}$
but when $ a < 0$ then this sum will be equal tu some constant  because
$ \sum\frac{1}{k^{1\plus{}a}}$ converge if $ a > 0$

In fact, $ \sigma(\gamma) \equal{}\frac{1}{1\plus{}\gamma}$ for all $ \gamma >\minus{}1$.

When $ \gamma\geq 0$, $ \sigma(\gamma)$ is precisely the integral $ \int_{0}^{1}x^{\gamma}\, dx$ because $ x^{\gamma}$ is continuous on $ [0, 1]$ for $ \gamma\geq 0$. (Here we regarded $ 0^{0}$ as $ 1$)

For $ \minus{}1 <\gamma < 0$, however, this argument does not work anymore, since $ f(x) \equal{} x^{\gamma}$ is neither continuous nor bounded on $ [0, 1]$.

But you can avoid such singularity, using the inequality

$ \int_{1/n}^{1}x^{\gamma}\, dx\ \leq\ \frac{1}{n}\sum_{k\equal{}1}^{n}\left(\frac{k}{n}\right)^{\gamma}\ \leq\ \frac{1}{n^{1\plus{}\gamma}}\plus{}\int_{1/n}^{1}x^{\gamma}\, dx$.

To understand how this inequality holds, it will be helpful to interprete this geometrically with the graph of $ f(x) \equal{} x^{\gamma}$.


counting distinguishability geometry algorithm search

Machines (computers, robots, etc.) are made by people. Some people think that machines cannot think, they can just run the command but create something… Do you think the machines with A.I. in future can think like human?

I think it is possible that machines can think like a human, but definitely not in this century. :o

Given the rate at which technology is progressing nowadays (and the fact that that rate seems to be constantly increasing), I think that people will be able to create an intelligence that is virtually indistinguishable from a human's.  Debugging in such an effort will be a large task though, that will certainly take time.

I totally thought this discussion was going to be about American Idol when I clicked on it and whether things like VFTW are right from a moral/ethical with some miscellaneous spam of people rooting or booing people thrown in and I had a rant on AI Producers and KiKi and Chris Richardson prepared :(

I actually remember seeing something on a TV show about a robotic secretary that looks and acts a lot like a person.

[quote="K81o7"]In order to design such AI, we still have to understand exactly how the brain works, which we don't.[/quote]

But it may be possible for us to inderstand exactly how the barin works in the future.

psh. Machines take previous examples and use a database to develop a new strategy. That's why Deep Blue, etc won against Kasparov and Kramnik. Definately not this century but it will probably happen later...

Given the rate that humanity is degenerating, I'm pretty sure that in the near future, it won't be too difficult.

Take a look at the beginning of the 20th century.  People thought that everything that could be invented was invented.  Now look where we are!  About 100 years ago, the Wright Bros. built a working airplane.  Now, we can fly into outer space!

I think that we will be able to build advanced A.I. sometime in the future.  As for when, I'm not sure.

[quote="my mother"]A. I. does not need to work exactly like our brain does.  Therefore, we do not need to completely understand the brain to design A. I. that is "virtually" indistinguishable from human intelligence.  This is called the Turing Test:  where it is impossible to tell whether the intelligence is artificial or human by observing its responses.  I believe within the next century, we will reach this point.  And they say girls can't do math :P .[/quote]

[aside]
The forum doesn't like messages that contain only quotes.  Ironically, this particular quote is not restating something already posted, but giving due credit to a third party.
[/aside]

You can put my brain in a computer!

Once I'm done with it.
[quote]"Dude, we're out of funding. The brain in computer thing has gotten old."

"NOOOOOOOOO! I WANNA LIVE!"[/quote]

[quote="zeb"]You can put my brain in a computer!

Once I'm done with it.[/quote]

You can also put my brain in a computer once I'm done with it.
But my brain will not useful at all, since I'm so stupid... :oops:

[quote="K81o7"][quote="Valentin Vornicu"][quote="K81o7"]In order to design such AI, we still have to understand exactly how the brain works, which we don't.[/quote]No, not really :)[/quote]

Well, do you have a different definition of "thinks like a human" than I do?  I'm saying literally, thinks in the same way a human does, so its thought processes are indistinguishable from those of a human.  Unless you propose somehow making a human brain without understanding how it works, or taking a human brain from a human being and putting it in a machine?[/quote]This is a flaw in logic you have. As someone's mother (!! :) ) said above, the AI must not have the same pattern of thinking as the human brain but end up with the same result. For each problem you might have there are different paths one can take to the solution. If I don't take the same path as you, but still reach the same result, does that not mean that I have solved the problem just as you have?

I agree with Valentin: there is no need to duplicate the exact processes in the human brain to devise a powerful thinking machine; moreover, to do so is a path leading nowhere: what we should be after is the machines that transcend our thinking capabilities, not just reproduce them. The way an average human thinks is very far from ideal from both logical and ethical viewpoints :(.

I do not know whether the progress will come from designing purely mechanical structures or whether some simbiosis  will be achieved between the "living" and the "silicon" things (I would prefer the latter because, despite I feel no more horror about humans being outsmarted and ultimately replaced by machines as dominant species than about dinosaurs having been eventually replaced by humans, it would be nice if humanity in some form or other could continue to play some role in the evolution) but I believe that we are slowly getting to the point where such machines could be devised. Whether our civilization shall reach this point or terminate itself in some stupid act depends only on us, but I see no reason why either of those events cannot happen as early as this century or, at most, this millenium.

Another thing I strongly believe is that strict adherence to our biological nature will inevitably start pulling us back at some point: it seems to me that, in order to colonize other planets, for instance, it would be much more logical to adjust the biological nature of the would-be colonists than the conditions of the world to colonize.

What I really hate is the rather widespread "human supremacy complex" that makes us believe that we, as we are, are the ultimate achievement and everything should serve our needs and should stand below us or, at best, be equal to ourselves. I'm afraid that the nature (or the God, if you believe in one) may have a different opinion on the subject.

Anyway, the point is that if the random unthinking evolution has succeeded in designing an electrochemical machine with not too bad thinking capabilities (the one that calls itself a human being), I see no reason why a directed intelligent design cannot achieve the same given sufficient time and resources. It would be interesting to see what the religious beliefs of such an AI would be (if any), especially if it is confined to some virtual reality, but it is another question.

Humans are imaginative and creative. Where does imagination come from and how would a machine gain it?

This is hard to answer because there is no clear definition of "imaginative". Can you give an example of some particular act of human imagination that you think would be hard to achieve in machine thinking? This would clarify your statement/question quite a lot.

All you need to build an AI is to use program to resemble each neurone in brain, sounds not difficult at all.

Imagine you build a world with AIs in your computer. Then you appear as god in the world.

"I, God of Math, order you lowly people to prove Goldbach conjecture in 10 years, or else it will be the end of the world!"

Then you can just press the fast-forward button and drink a cup of tea, the proof will be ready in a minute. Cool!!! :D  :D  :D

Humans can write intricate novels. They can build tall buildings, sculpt sculptures, and discover new areas of mathematics. These require physical action, but first and foremost imagination and creativity.

If we were to create a computer to copy a human, it would need to have this imagination. We cannot understand our imagination, now at least, by some sort of complex algorithm, or by some ridiculously intricate method. If we did understand it that way, we would feel deceived, as it is something we can pursue into different areas. Imagination seems to go beyond what is initially defined in the system.

Does that help at all? I think we are often too boastful about what we know about humans and machines. There are many things we don't come close to understanding.

Hi,

in the field of AI there is a research line named computational creativity. Maybe you would be surprised about all the programs that were written in order to model creativity or simply to get novel and "valuable" results. In musci you have the work of David Cope (EMI - Experiments in Musical intelligence), in design there are tens of programs, in drawing you have the work of H. Cohen (Aaron, pretty old work actually), you have the work of Hofstadterin creating font styles, in literature you have the work of MINSTREL, BRUTUS, MEXICA and so on.
In mathematical discovery you have the old ones (and critized) AM, EURISKO. In mathematical discovery a relatively recent program is GRAFFITI, a conjecture-making program that was able to propose new boundaries for some important quantities (conjectures were proven by mathematician, I read there are several papers on proving conjectures proposed by the program). You also have the program of Simon Colton (sorry, forgot the name of it).
Ok, so...is this creativity. I mean, these programs are creative as humans do? Before answering you have to keep in mind that there is no agreement about what it is creativity: a trait or process by which we get novel and interesting results. Obviously not all the mentioned programs claim that they model the human way of creating, but they definitively are able to produce new things that are even perceived as valuable by some.

Regards.

So far you didn't mention any example of "imagination" that wouldn't just come as a result of observation combined with some rearrangement and/or distortion of observed blocks. What I mean is that I am yet to see a novel that raises above the description of external things the author could see/feel/think in his everyday life. Making sculptures and tall buildings is just about changing scales and proportions of what we see in reality. Mathematical theories are further away from directly observed reality, but still, when you try to solve a problem or come up with a new object, all you can do is to run through the things you learned before and slightly change them or combine them in some new way. I'm inclined to believe that our so called imagination is just the ability to observe the surroundings, to find an external thing that is close to fitting our purpose and to modify it a little bit.  So, our being imaginative may come not as much from our internal properties as from the richness of the universe we live in and are capable of observing. It is driven by the tasks we are given from outside or impose upon ourselves for which no ready solution is present but for which a solution can be achieved by some slight modification or rearrangement of the existing objects. Can we write a program that will look around, see what is there, and try to make a small modification in the surroundings that will achieve (or come closer to) some given purpose? I believe it has been done already in many instances of very restricted "reality". If I don't mistake, this is how the airplane wings, car body shapes, bridges, etc. are nowdays projected. Also, if I don't mistake, the computers have already a few times come up with solutions that surprised their programmers as completely unexpected. So, it may well be that all that is needed is to improve "imagination" of a machine is to give it some richer reality to operate on. I may be completely wrong, of course.

As to the claim that there are many things we haven't come close to understanding, that is 100% true and will remain true even 1000 years from now. What is interesting is that we can do quite a lot with insufficient knowledge. I believe that the first machine capable of thinking at the human level will be constructed long before we fully understand human physiology os psychology. It'll just run on completely different principles but will be able to achieve the same or better results, and it is all that matters.

Of course, my beliefs here are those of a layman. I am very far from being able to evaluate the difficulties of this project with any reasonable accuracy. All I can say is that the way I think of mathematics, at least as I perceive it, has no features beyond those of a blindfolded search in a huge storage room where you take things one by one, evaluate them, and decide in which direction to make the next step based on what you picked up. The only thing that makes it complicated is the set of rules that are used in making that decision and in the process of evaluation of what is in my hands but even those are not my own invention: they were imposed on me by my previous training. That's about as creative as I can get in mathematics and I do not see why a machine cannot eventually do the same. Perhaps, I am just not imaginative enough :(.


set theory
advanced fields
advanced fields unsolved

Hey guys;

How to show this :

if $ X \subseteq Y$ and $ |X| \leq |Y|$ then :

$ |X^c|\equal{}|Y|$ , where $ |X|\equal{}cardinality \ of \ X$ and $ X^c$= Complement of X relative to Y .i.e $ Y\minus{}X$.

Need details please

How do you mean? I must be understanding something wrong here because $ |X^c|\equal{}|Y|\minus{}|X|$. Obviously, if $ |X|\geq 1$, $ |X^c|<|Y|$.  :?:

[quote="mornik"]How do you mean? I must be understanding something wrong here because $ |X^c| \equal{} |Y| \minus{} |X|$. Obviously, if $ |X|\geq 1$, $ |X^c| < |Y|$.  :?:[/quote]

you got misunderstanding of context .

I meant to prove $ |Y\minus{}X|\equal{}|Y|$.

Only $ Y\minus{}X\equal{}X^c$ without cardinality notation

$ X \equal{} Y$ satisfies all your conditions but not the conclusion.  Please try to think about what you write so that you don't post nonsense.

What about $ X \equal{} \mathbb Q$ and $ Y \equal{} \mathbb R$

The reason of this post JBL since I saw some proof of  proving that $ |\mathbb Q^c| \equal{} |\mathbb R|$.

JUst ,I doubt of it man. :maybe:

I think you mean $ X \subset Y, |X| < |Y|$, and in that case we still have a contradiction if $ X, Y$ are (say) finite.

Surely ;it's false when $ X,Y$ are finite .

What about if they are infinite like our  case.

Then I cannot help but suspect that the answer is equivalent to the [url=http://en.wikipedia.org/wiki/Continuum_hypothesis]continuum hypothesis[/url].

Edit:  Or not.  I believe the statement is equivalent to showing that $ |S| \equal{} |S \times S|$.

Even if $ X$, $ Y$ are infinite, what you claim is not true.

Let $ X\equal{}\mathbb Q^c$ and $ Y\equal{}\mathbb R$. Then $ |X^c|\ne|Y|$.

[quote]
The reason of this post JBL since I saw some proof of proving that $ |\mathbb Q^c|\equal{}|\mathbb R|$.[/quote]

You need to understand the definition of [url=http://en.wikipedia.org/wiki/Cardinality]cardinality[/url]. And also understand the context of the post in which the claim was asserted.

Okay, finally from the corrected version of post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1185598#1185598]#5[/url] we can discover what the sensible statement behind the nonsense is:

If $ X \subset Y$, $ |X| < |Y|$ and $ Y$ is infinite then $ |X^c| \equal{} |Y|$.

Here's one not-wonderful way to prove it, using the fact that for an infinite set $ S$, $ |S| \equal{} |S \times \{a, b\}|$: We have that $ |\{a\} \times X| \equal{} |X|$ and $ |\{a, b\} \times Y| \equal{} |Y|$ and so $ |(\{a, b\} \times Y) \setminus (\{a\} \times X)| \equal{} |Y \setminus X|$.  But $ (\{a, b\} \times Y) \setminus (\{a\} \times X) \supset \{b\} \times Y$, so $ |Y| \equal{} |\{a, b\} \times Y| \geq |(\{a, b\} \times Y) \setminus (\{a\} \times X)| \geq |\{b\} \times Y| \equal{} |Y|$ and we're done.

There must be something more appealing than this proof, though.

Erm... If $ X\subset Y$ and $ X_1\subset Y_1$, then $ |X|\equal{}|X_1|$ and $ |Y|\equal{}|Y_1|$ [b]do not[/b] imply $ |Y\setminus X|\equal{}|Y_1\setminus X_1|$ in general ;).

are you sure this goes in advanced mathematics because this kind of set theory doesn't seem college level maybe intermidiate topics?

If you know an intermediate level solution, just post it instead of making rather useless remarks. If I (or other mod) decide that the topic does not belong here, we'll move it, don't worry ;)

[quote="fedja"]Erm... If $ X\subset Y$ and $ X_1\subset Y_1$, then $ |X| \equal{} |X_1|$ and $ |Y| \equal{} |Y_1|$ [b]do not[/b] imply $ |Y\setminus X| \equal{} |Y_1\setminus X_1|$ in general ;).[/quote]

In my first draft I had written "not necessarily rigorous" for "not wonderful" because I wasn't sure how to justify that step.  It [b]is[/b] true when $ |X| < |Y|$, but that statement is a consequence of the result we are trying to show, so my argument is circular.

JBL, you've already pointed out the key fact: $ |2\times A| \equal{} |A|$ if $ A$ is infinite.
now just suppose $ |Y\setminus X| < |Y|$. now $ |Y|\equal{}|Y\setminus X|\plus{}|X| \equal{} \max\{|Y\setminus X|,|X|\}<|Y|$, contradiction.


[quote="pankajsinha"]Let f(x)=g(x)/(1+h(x))

where (i)g(x)=exp(-x) and h(x)=[x]

          (ii)g(x)=exp(x) and h(x)=[x]

where [x] denotes the greatest integer less than or equal to x.[/quote]
I not understand. What is question?  :maybe:

Sorry, for not making the question clear.
Actually we are required to find range of f(x).(exp(x)=e^x)


[quote="cosinator"]a. $71^{43}$ is an 80 digit number in its base ten representation.  Prove that at least one of the ten digits appears at least nine times.

b. If $m$ and $n$ are positive integers and each of the base ten digits appears an equal number of times in the base ten representation of $m^{n}$, then, besides $m$ and $n$ being the right size so that the number of digits in $m^{n}$ is divisible by 10, what other properties must $m$ and $n$ have?[/quote]

[hide="for a"]Use contradiction[/hide]

[hide="a"]Well assume the opposite, then there's 8 of everything. But then it has to be divisible by 9, but 71^43 isn't divisible by 9 so BAM the assumption is false[/hide]


$x=1$, by inspection  :P

note: I think you should be able to isolate the square root, square both sides, isolate the cube root, cube both sides, isolate teh fourth root, and raise both sides to the fourth power. Then you'll have a normal polynomial. But it'll probably be of disgustingly high degree. Knowing you can factor out (x-1) might help a tiny bit  :D

Can you  write it perfect ? I am waitting for true answer from xevarion.I think problem very hard because the equation has stantled degree(say then easy,do then hard)

Geometer's Sketchpad shows that there are two solutions. One is $x=1$ the other one is around $1.58$. I don't know how to find the second one.

I need one true answer from you ,i thinks the problem need be estimated right by everyone

WHY?nobody have true detailed schemed solution????????The problem no easy


Hi every1,

I was reading the proof of the fact that the graph [b]K(n[/b]) has a total of [b]n^(n-2) [/b]spanning trees... Well the proof used the idea of [b]Prufer sequences[/b] and I can reproduce it in case u want me to.

What i want to know is that the authors maintained that [b]many proofs[/b] do exist for [b]counting the number of spanning trees[/b] of any graph (unlike Prufer's proof which counts spanning trees only for K(n) )....

So all I m asking is just a favor. Please tell me about the other proofs which exist for counting the number of spanning trees of a graph.

Thanx and have a nice day and in case u can plzz tell me whether any  ebook on [b]dynamic programming [/b]by Bellman is available on the net..

You could be interested in Kirchhoffs theorem: http://en.wikipedia.org/wiki/Kirchhoff%27s_theorem

I think you can find in Knuth how to count number of spanning tree.. basically it reduces to finding the determinant of a matrix. One Google search would have been enought to find the necesarry information.. for example , read here http://mathworld.wolfram.com/MatrixTreeTheorem.html and remember, Google is your friend  :P

Hi every1,

friends, i have seen the proof given in Knuth, but my knowledge of determinants is horrible. So i was unable to comrehend it. But the proof using Prufer sequences is more than elegant and in case u need I can post it.

The only thing is that I saw the authors maintain that the problem has got many solutions. But i cant find any of them on google (sure enough I didnt use a good search title)... So this inability to find the proof manifested itself in the form of my asking u of the different proofs u know....

Thanx and have a nice day.....

You can also do this, for a fixed degree sequence $(d_1,d_2,...,d_n)$ number of spanning tree is $\frac{(n-2)!}{\prod (d_i-1)!}$. So as $\sum d_i=2n-2$ total number of spanning tree is
\[\sum_{\sum d_i=2n-2}\frac{(n-2)!}{\prod (d_i-1)!}=n^{n-2}\]


[quote="bgbgbgbg"]show that if p is prime and 3p+1 is a complete square then p must be 5[/quote]

$ 3p\plus{}1\equal{}a^2$ $ \iff$ $ 3p\equal{}(a^2\minus{}1)\equal{}(a\minus{}1)(a\plus{}1)$ and so, wlog saying $ a>0$, and since $ 3$ and $ p$ are prime :

Either $ a\minus{}1\equal{}1$ and $ a\plus{}1\equal{}3p$ and sp $ a\equal{}2$ and $ p\equal{}1$, not prime
Either $ a\minus{}1\equal{}3$ and $ a\plus{}1\equal{}p$ and $ a\equal{}4$ and $ p\equal{}5$
Either $ a\minus{}1\equal{}3p$ and $ a\plus{}1\equal{}1$ and $ a<0$

Hence the result

But you should not post this in an olympiad forum, IMHO


limit
quadratics
linear algebra
matrix
geometric sequence
real analysis
real analysis unsolved

Let $ (x_n)_{n \ge 0}$ be defined by $ x_0 \equal{} \alpha > 0$ and $ x_{n \plus{} 1} \equal{} \displaystyle\frac{2x_n \plus{} 1}{2x_n \plus{} 3}$, for any nonnegative integer $ n$. Find, with proof, the limit of $ x_n$.

Let $ y_n \equal{} \frac {x_n \minus{} \frac {1}{2}}{x_n \plus{} 1},$ we have $ y_{n \plus{} 1} \equal{} \frac {x_{n \plus{} 1} \minus{} \frac {1}{2}}{x_{n \plus{} 1} \plus{} 1} \equal{} \frac {1}{4}\cdot \frac {x_n \minus{} \frac {1}{2}}{x_n \plus{} 1}\Longleftrightarrow y_{n \plus{} 1} \equal{} \frac {1}{4}y_n\ \therefore y_n \equal{} \left(\frac {1}{4}\right)^ny_0,$ yielding $ \lim_{n\to\infty} y_n \equal{} 0\ \therefore \lim_{n\to\infty}x_n \equal{} \frac {1}{2}.$

Or just note that $ f(x)\equal{}\frac{2x\plus{}1}{2x\plus{}3}$ maps $ [0,\plus{}\infty)$ to itself, $ |f'(x)|\equal{}\frac 4{(2x\plus{}3)^2}$ is $ \le\frac 49<1$ and use the Banach fixed point theorem. :sleeping:

The sequence is bounded.

You can thus verify that the inferior and superior limits must converge to the same number.

QED. ;)

You are right. I solved the problem for Japanese high school students, because your solution gives no points in Japanese university entrance exam, while that is trivial.

[quote="kunny"]Let $ y_n \equal{} \frac {x_n \minus{} \frac {1}{2}}{x_n \plus{} 1}$[/quote]

Can you please explain how you came up with this fraction for $ y_{n}$?

What kunny did is a classic trick. You look at it and say: Well, if it had a limit, what could it be? Suppose it has a limit l. Since both $ x_{n \plus{} 1} and x_{n}$ tend to it, l must satisfy the equation$ l \equal{} (2l \plus{} 1)/(2l \plus{} 3)$. Solving that gives the solutions 1/2 and -1. Now the trick is that the fraction of this quadratic equation's solution, like kunny used, will always be a geometric progression which allows you to compute directly the limit of $ x_{n}$.

   Also you can take a matrix of the coefficients of the original fraction and use that to compute all the terms of the sequence.

   If you don't know the trick you can also try monotonicity for the sequence and you will have 2 cases, $ 0< \alpha < 1/2$ and $ 1/2< \alpha$. Taking $ f(x)  \equal{} (2x\plus{}1)/(2x\plus{}3)$ with fixed point 1/2 you will find that it is conveniently strictly increasing on a convenient interval and thus once you establish something like $ x_0 < x_1$ in the first case or the reverse in the second case you get complete monotonicity on the whole sequence by applying f to itself. Boundedness is proved easily and then you can go on to find your limit using the same equation as above.


Doc thought of a 200-digit number, added its digits, and whispered

the sum into Happy's ears.

Happy caluculated the sum of the digits of the number he was just told,

and whispered the sum into Grumpy's ears.

Grumpy figured out the sum of the digits of this number, which happened

to be a 2-digit number, and he whispered this number into Sneezy's ears.

Sneezy also calculated the sum of the digits of the number he just heard

and whispered it to Depey.

Which number did Happy whisper to Grumpy, and which number did

Sneezy whisper to Dopey?

The sum of the digits of the first number must be between 200 and 1800 (and any such value is possible).

Call the number Grumpy tells Harry A, the number Harry tells Grumpy B, the number Grumpy tells Sneezy C, and the number Sneezey tells Dopey D.

We know $200 \leq A \leq 1800$.

Therefore $B \leq 27$.

We know that $C \geq 10$, and the only number $\leq27$ with sum of digits $\geq10$ is 19. Hence B=19, C=10, D=1.

Possible minor nit-pick - (Not that it changes the solution in any significant way ..)  [quote="tim1234133"]The sum of the digits of the first number must be between 200 and 1800 (and any such value is possible).
[/quote]
why can't it be less than 200? (Origninal number can have zeros at any place except the left most)

:wallbash: yeah, oops. I don't know what made me think it couldn't have zeros.

I can't edit my previous post anymore, but I don't think it changes much in the end.


In planul euclidian E2, raportat la un reper ortonormat, se considera familia de conice de ecuatie:

ax2 - 2xy+ ay2 - 2x+2y+3=0, a parametru real  (*)
( x2 respectiv y2 a se citi x respectiv y la puterea a doua)

a) sa se discute natura si genul conicelor din familie dupa valorile parametruui  real a
b) Sa se determine locul geometric al centrelor conicelor din familia (*)[/img]


The point marking the bottom of a unit circle is positioned at 0 on a number line.  The circle is rolled along the line until that point has rotated through an angle of $ \frac{5\pi}{6}$.  Find the number line location of the point that marks the new location of the bottom of the circle.

That would be it.  On the unit circle, 1 unit corresponds to 1 radian so if it moves that number of radians, then it moves that many units.


quadratics calculus vector function integration derivative Euler

This is a physics problem? But I *think* it's a quadratic curve (and then you'd have to use some relatively messy calculus, unless we're trying to be all clever or something).

physics yeah partcialy but vectors involve math too. Try it out it is an interesting challange. When u get the right answer all that messy work will be worth it. :flex:  trust me

Physics is noting but math applyed to everyday things.

[quote="MysticTerminator"][quote="dimak11214"]I have never met a person that could solve this problem. I hope one person here can.

There are two poles standing 10 feet heigh each. The distance bettween them is 12 feet. A string with a lenght of 15 feet is tied to the very topes of both poles, and leaves a curve shape between them. What is the height of the lowest point of the string.

       Good luck  ;)[/quote]

If I remember right, stuff like this follows the graph of the cosh function - you then do some calculus for the arc length to figure out the right parameters.[/quote]
Yes - assuming the mass of the string is uniformly distributed, it will form a catenary curve - I think the proof has to do with the force of gravity on each part of the string and the pull of the other parts of the string on it, but I don't remember it very well.

Yay, I get to apply a variational principle! I hope it's right.
Let's say the rope is described by the function $f(x)$ as $x$ goes from $0$ to $L$. The rope will shape itself so that its total gravitational potential energy is a minimum. At this minimum, if we slightly vary $f(x)$ by adding a small function $g(x)$ while keeping the length of the curve constant, the gravitational potential energy will not vary to first order in $g(x)$. If it did, then we could choose a $g(x)$ that would give us a lower energy.

So the minimality condition requires that
$\int_0^L g(x)dx = 0$
But there is a constraint on $g(x)$, the length of the curve can't change, so
$\int_0^L \sqrt{(f'(x)+g'(x))^2+1}dx-\int_0^L \sqrt{f'(x)^2+1}dx=0$
$\int_0^L (\sqrt{(f'(x)+g'(x))^2+1}- \sqrt{f'(x)^2+1})dx=0$
Taylor expand the first square root to first order in $g(x)$, things cancel and the integral becomes
$\int_0^L \frac{f'{x}}{\sqrt{1+f'(x)^2}}g'(x)dx=0$
Do integration by parts--move the derivative from $g(x)$ to the other term. It becomes
$blah-\int_0^L [\frac{f'(x)}{\sqrt{1+f'(x)^2}}]'g(x)dx=0$
$blah$ is a boundary term that equals zero because $g(0)=g(L)=0$
Somehow the constraint $\int_0^L [\frac{f'(x)}{\sqrt{1+f'(x)^2}}]'g(x)dx=0$ has to imply that $\int_0^L g(x)dx = 0$. I'm pretty sure this is only possible if the factor other than $g(x)$ under the integral--the derivative of the bracketed term--is a nonzero constant. Then the bracketed term is of the form $cx+d$.
So
$\frac{f'(x)}{\sqrt{1+f'(x)^2}}=cx+d$
Something is wrong...

It is widely established that the solution is of the form $f(x) = \frac{ e^{Ax} + e^{-Ax} }{2A}$ and the only thing that remains is to calculate the value of $A$ from the givens.

Using variational ideas is definitely the right way to proceed here.  In particular, it's easier if you work directly with the Euler-Lagrange equation.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=53021]here[/url] is a post of mine, regarding the same thing.. i guess it's just another solution :)

Very beautiful problem.  Read [url=http://www.du.edu/~jcalvert/math/parabola.htm]this[/url]

Masoud Zargar

[quote="blahblahblah"]I still think variational methods are a more natural way to tackle this problem.[/quote]

Could you post a solution using such methods, then? Thanks.

I'm also curious where my solution went wrong, certainly somewhere before
$\frac{f'(x)}{\sqrt{1+f'(x)^2}}=cx+d$.


[quote="gauss1181"]Have you ever played other variants of the Rubik's Cube, like 2x2x2 Pocket Cube, 4x4x4 Rubik's Revenge, or 5x5x5 Professor's Cube? If so, how easy are each?[/quote]

2x2x2 is easier than 3x3. I've tried 4x4 (pretty easy) and 5x5 (similar just takes longer).

If you can solve $ 4 \times 4 \times 4$, you can solve any bigger one, too (you can construct a strategy).
Try the ones in $ 4$ or $ 5$ dimensions ;)

I've tried the pocket cube, rubiks cube, and the revenge and can solve them all.

[quote="tjhance"]5x5x5 isn't very hard. But I haven't been able to solve the 4x4x4.[/quote]

what method do you use? if you use the centers, dedges, 3x3 method it's the same

4x4x4 is fairly easy, and 2x2x2 is a joke. Haven't tried the 5x5x5 though.

I can do both 3x3 and 4x4 fairly easily.

Although I don't do 2x2 I'm guessing it's a no brainer if you could do a 4x4.

Someday I wish I can do a 5x5.

[quote="moogra"]way to bring back an old topic.

I did a 7x7 this year at Berkeley Spring 2008. I pretty much failed the pairing the dedges because theres some weird cases.

2x2-5x5 are all easy.[/quote]
*gasp* you can do the 6x6 and 7x7? how long does it take u for each?

Nonono I said I failed. That was back in February though and I barely could do a 5x5.
I got the centers though. Everything above the 4x4 is solved the same way. Also I couldn't take the 7x7 apart because I can't put it back together.

what's your best time for each size of cube?

Well best times don't matter.
I've gotten sub-10 on 3x3 because I've been lucky.
Last layer skips on 2x2 are quite common too.
On 4x4, my last solve was in June and that was 2 minutes something. After that mine broke.
On 5x5, I've average about 3-4 minutes but my cube is horrible and breaks apart too much.
Rubik's magic best time 1.11, best average 1.19, regular average ~1.40 (enough to win a competition where I'm at since there aren't that many people that do it).

About the v-cubes. The 6x6 I used was HORRIBLE. It got stuck more than my 4x4 in the beginning.
The 7x7 was awesome though.

i can solve 1x1 (gasp! OMG SO HARD!) to 5x5

OMG I CANNOT SOLVE A 1x1. TELL ME HOW.
What are your times?


[quote="mateivld"]Let $a,b,c>0,a+b+c=1$.Prove that:\[ \displaystyle \sum\frac{1}{1+ab}\geq\sum\frac{1}{1+a^2}  \].[/quote]
The inequality is equivalent to $\sum_{cyc}\displaystyle\left(\frac{2}{1+ab}- \frac{1}{1+a^2}-\frac{1}{1+b^2}\right)=\sum_{cyc}\displaystyle\left(\frac{(1-ab)(a-b)^2}{(1+a^2)(1+b^2)(1+ab)}\right)\ge 0$

[quote=mateivld]Let $a,b,c>0,a+b+c=1$.Prove that:\[ \displaystyle \sum\frac{1}{1+ab}\geq\sum\frac{1}{1+a^2}  \].[/quote]
Let $a,b,c>0,a+b+c=abc$.Prove that$$\displaystyle \sum\frac{1}{1+ab}\leq\sum\frac{1}{1+a^2}  $$

[b]Remark.[/b] This simplifies to $$\frac{3+2(ab+bc+ac)+(a+b+c)^2}{1+ab+bc+ac+2(a+b+c)^2}\le \sum \frac1{1+a^2}.$$


[quote="kchande"]I kind of did this problem with algebra, but I want to look at if anyone else has more elegant and simple solution to this problem :

Prob : In the first two-mile race, Kendi beats Jerri by 1/3 mile, and in the second two-mile race, Jerry beats Micah by 1/2 mile. If Kendi and Micah run a third two-mile race, and all three runners run at the same speed in every race, by how many miles will Kendi beat Micah ? Express your answer as a common fraction.

Thanks.[/quote]

$K : J = 2 : 1 \frac{2}{3}= 6 : 5$

$J : M = 2 : 1 \frac{1}{2}= 4 : 3$

$K : J : M = 24 : 20 : 15$

$k : M = 24 : 15 = 8 : 5$

$\frac{2}{8} \cdot 5 = \frac{5}{4} = 1\frac{1}{4}$

  So when Kendi finishes the race, Micah will only have completed $1\frac{1}{4}$ mile, hence he wll have beaten her by $2-1\frac{1}{4}=\boxed{\frac{3}{4} \text{mile}}$

You have four unknowns, but need to find just one.

[hide]

$x$ = Kendi's rate
$y$ = Jerry's rate
$z$ = Micah's rate
$w$ = the distance by which Kendi beats Micah

we will use the formula $t = \frac{d}{r}$ to set up our proportions

From the first race between Kendi and Jerry:

  $\frac{2}{x} = \frac{2 - \frac{1}{3}}{y} = \frac{\frac{5}{3}}{y} \ ; \ 2y = \frac{5}{3} x \ ; \ y = \frac{5}{6}x$

From the second race between Jerry and Micah:

   $\frac{2}{y} = \frac{2 - \frac{1}{2}}{z} = \frac{\frac{3}{2}}{z} \ ; \ 2z = \frac{3}{2} y \ ; \ 2z = \frac{3}{2}(\frac{5}{6}x) = \frac{5}{4}x$

From the third race between Kendi and Micah:

    $\frac{2}{x} = \frac{2 - w}{z} \ ; \ 2z = x(2 - w)$

Therefore, $\frac{5}{4}x = x(2 - w)$

so $2 - w = \frac{5}{4} \ ; \ w = \frac{3}{4}$

Kendi beats Micah by $\frac{3}{4}$ mile

[/hide]

eeek - i messed up my arithemtic and switched kendra with jerri, ill go edit..


counting
derangement
function
combinatorics theorems
combinatorics

Can someone help me explain about derangements n Bernoulli-Euler's famous formula of misadressed letters ?? And their function in China 2000; here in the attached file ????

Also in the second attached file...
I don't know exactly about the second part ( m and M )

Thx for helping...  :)

Can anyone help me explain the solution ??
Or maybe anyone have shorter solution than I have....  :maybe:
Please, I really need that, and any help will be greatly appreciated.


i had given the exam laast year. dint bother to write it this year, coz it sux.

Well, yeah I have written the exam this year.
The results come on March 1st.
I am expecting in top 10 ranks in 8th std.

[quote="Cyber-Shreyas"][quote="rituraj007"]Who all gave that exam???
i know my reult
but u will perhaps have to wait for 1-2 days
follow the link for result
[url]http://www.nettlinx.com/unified[/url][/url][/quote]

How do you know your result?? It says Unified Council 2007????[/quote]

Well the top rankers get phonecall at home in advance

[quote="VARUNARASU"][quote="sachin a S"]the exam's as poor as its rewards
i had given d exam last yr too and had got a dvd wit some books for toppin d city. dey announce great prizes but i doubt its validity[/quote]
lotsa posts lately. what happened? finished your tenth revision and bored of doing your eleventh i suppose?javascript:emoticon(':P')
Razz[/quote]

no re started for boards and other competitives
and u know our textbooks r as good as our lecturers at inducin sleep
so evrytime i'm in d mood of sleepin i come over here 4 a change :D

[quote="sicilian c5"][quote="VARUNARASU"][quote="sachin a S"]the exam's as poor as its rewards
i had given d exam last yr too and had got a dvd wit some books for toppin d city. dey announce great prizes but i doubt its validity[/quote]
lotsa posts lately. what happened? finished your tenth revision and bored of doing your eleventh i suppose?javascript:emoticon(':P')
Razz[/quote]

hi never worry henceforth u wont find much posts from him  :D[/quote]
what makes u say that? do you know him? :huh:

yeah of course i know him and i know u too

just check your pm :D

dat's y i asked u to check your pm  :rotfl:  :rotfl:

[quote="Cyber-Shreyas"]Hmm..THe Bangalore guys are trying to take over AOPS. No, we chennaites wont allow it.  :D  :D  :D[/quote]go on...
i m with u
 :mad:

(( i kno i m not gonna find anyone of my close community))

where are u from then??

PS i have a feeling pardesi's gonna move this topic to the meta thread :D

i m from north(india) i can say

 :roll:
to kno exact loc. go to introduction


Ωραίες ασκήσεις βάλατε στο φόρουμ! Συνεχίστε έτσι!
Συμφωνω με την ιδέα να ασχοληθούμε με Ανάλυση.

Να και μια διαφορετική άσκηση για το σαββατοκύριακο!
Καλα κούλουμα!

ΝΔΟ
$[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+...+[\sqrt{n^2-1}]+1^2+2^2+...+n^2=n^3$
$n \in \mathbb{N}$

$[ \ \ ]$   : ακέραιο μέρος

 :)

[quote="silouan"]Είμαι σε καλή φόρμα αλλά δεν ήμουνα όταν έπρεπε. :o .

Τεσπα καταπληκτική άσκηση, thanks Θανάση[/quote]

Να σαι καλα ρε!

Ριξε και εσυ καμια! :)

Κοίτα έχω σκοπό να βάλλω πολύ ωραία πράγματα εδώ http://www.mathlinks.ro/Forum/viewtopic.php?t=78445 , αλλά δεν φάνηκε να

ενδιαφέρθηκαν και πολλοί. Καταρχήν δεν είπε κανένας με τι θέλει να ασχοληθεί. :huh:


u have a black box and outside it 4 terminals two on each side and u don't know how the resistors are connected across these terminals and there is no opening.well what is given is a battery . two ends were connected on one side and other two ends connetced to the battery  of a known emf.then the heat developed eas $ H_{1}$.now the connected ends were disconnected and battery was connected to the same two ends and heat developed was $ H_{2}$.now similar process was repeated by connecting battery across other two ends and the heat measured was $ H_{3}$ and $ H_{4}$.
prove that
               $ H_{1}H_{2}=H_{3}H_{4}$

I am working on it but I am not sure I have got it right.
1, 2, 3, 4 are terminals,
H1: Battery 'A' of emf E btwn 1&2, wire btwn 3&4
H2: A btwn 1&2, nothing btwn 3&4
H3: A btwn 3&4, wire btwn 1&2
H4: A btwn 3&4, nothing btwn 1&2

I solved the problem, but it is very long and messy, and doesn't really involve any physics in it... It is simply math calculation... :(

i don't know about your method but actually the problem has a short and elegant solution

Is it true that for any 2 pts on any circuit of resistors, you can get equivalent circuit with emf's on one side and resistors on other? I have heard that there is a theorem in electronics to this effect. If this is wrong, is there any similar general statement about circuits?

can u be more clear...yes the problem involves certain theorems..


Determine if it is possible to choose nine points in the plane such that there are $n=10$ lines in the plane each of which passes through exactly three of the chosen points. What if $n=11$?


quadratics
number theory
number theory unsolved

Prove that the equation $n$ = $x^{2}$ + $y^{2}$ + $z^{2}$ has solution in $\mathbb N^*$ if and only if $n$ $\neq$ $4^{m}(8k+7)$ for all $m,k$ in $\mathbb N$

This is not Algebra, but number theory! (what do you think number theory is about if not about such things¿)
And the title is also bad, user more meaningful titles, this is not the only "nice and hard result".

Mods.: please move it.

To the problem:
Posted before, see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5269 .
"Just" use Chevalley-Warning and Hensel to get solutions modulo any prime power, then use Hasse-Minkowski to get rational $x,y,z$, then Davenport-Cassels to get integers  :P
(names mean the corresponding theorems ;) )

PS: Assumed that $\mathbb N^*$ are the nonnegative integers, in fact it gets wrong if we don't allow $0$. But the notation $\mathbb N^*$ is really uncommon for that set.

Or otherwise, use reduction of ternary forms, Dirichlet and then quadratic reciprocity. Nice ways of stating big theorems, isn't it?


I am working through aops volume 1. I am at chapter 3. I am having difficulty with the word problems. Where can i find more practice?

Maybe you should look at the solutions to a couple of the problems, learn what you can from them, then try to do the rest again. It works for me. Also you can find a lot of practice problems in the middle school and getting started (and possibly intermediate) areas.


Compare these two number $ \binom{100}{20}0.2^{20}\*0.8^{80}$ and $ 0.2$,which one is larger?

If $ k > > 1,n \minus{} k > > 1$, then $ \binom{n}{k}(\frac kn)^k(\frac {n \minus{} k}{n})^{n \minus{} k} \equal{} \sqrt {\frac{n}{2\pi k(n \minus{} k)}}(1 \plus{} O(\frac 1k \plus{} \frac {1}{n \minus{} k}))$.

[quote="Rust"]If $ k > > 1,n \minus{} k > > 1$, then $ \binom{n}{k}(\frac kn)^k(\frac {n \minus{} k}{n})^{n \minus{} k} \equal{} \sqrt {n}{2\pi k(n \minus{} k)}(1 \plus{} O(\frac 1k \plus{} \frac {1}{n \minus{} k}))$.[/quote]

Sorry, can you explain how you get the equation?
I can't understand it.
And what is "$ O(\frac 1k \plus{} \frac {1}{n \minus{} k})$"?
Thank you.


function
Euler
calculus
integration
logarithms
real analysis
real analysis unsolved

A lot of questions have been asked about its irrationality, but is there an exact formula for this?

$ \sum_{k \equal{} 1}^{n} \sqrt {k}$

I am expecting something like...

$ {2 \over 3}\left(n \plus{} {1 \over 2} \minus{} f(n)\right)^{3 \over 2}$

or

$ {2 \over 3}\left(n \plus{} {1 \over 2}\right)^{3 \over 2} \minus{} {5 \over 24} \minus{} f(n)$

The error term f(n) in the first formula is approximately the reciprocal of (4.81...)sqrt(n) + c.
I can't figure out what this pesky 4.81... coefficient is supposed to be. Then again, it could just be 24/5, and my floating point errors may have caused it to seem greater.

Mathworld recasts the general answer for non-integer exponent $ p$ in terms of the [url=http://mathworld.wolfram.com/HurwitzZetaFunction.html]Hurwitz zeta function[/url].  For a solution more along the lines of your reasoning, see the [url=http://mathworld.wolfram.com/Euler-MaclaurinIntegrationFormulas.html]Euler-Maclaurin integration formula[/url], which gives a series expansion of the error term.

The reason I find the Euler-Maclaurin summation difficult is that I am trying to get the result to end up as a three halves power, $ \sqrt {x}$ doesn't behave nicely at 0, etc.

The only other thing I could think of was writing it as the log series...

$ \sqrt {x} \equal{} 1 \plus{} {1 \over {1!2^{1}}}\ln(x) \plus{} {1 \over {2!2^{2}}}\ln^{2}(x) \plus{} \, ...$

And for an update, at least for the 4.8101, it seems $ {24 \over 5}$ is not the correct constant after all. I am having better luck with $ {25\gamma \over 3}$. Go figure.

I feel a little silly. This number that is about -5/24 is none other than $ \zeta\left( \minus{} {1 \over 2}\right)$

I can only wonder why I didn't see that before...

$ \sum_{x \equal{} 1}^{n} x^k \approx {1 \over {k \plus{} 1}}\left(n \plus{} {1 \over 2}\right)^{k \plus{} 1} \plus{} \zeta( \minus{} k)$

So we basically have an integral term and a Dirichlet series term.

The error term for $ k \equal{} {1 \over 2}$ seems to be something like

$ E \approx \sqrt {n \plus{} {1 \over 2}} \times {1 \over {48(n \plus{} {1 \over 2}) \plus{} {20 \over {21(n \plus{} {1 \over 2}) \plus{} ...}}}}$

[url=http://en.wikipedia.org/wiki/Faulhaber%27s_formula]Faulhaber's formula[/url] says

$ \sum_{k\equal{}1}^n k^p\equal{}\frac{1}{p\plus{}1}\sum_{k\equal{}0}^p (\minus{}1)^k\, {p\plus{}1\choose k}\, B_k\,\, n^{p\plus{}1\minus{}k}$ for $ p\in\Bbb{Z}^{\ge 0}$ .

Due to $ \zeta(\minus{}p)\plus{}\frac{1}{p\plus{}1}\, (\minus{}1)^{p\plus{}1}\, B_{p\plus{}1}\equal{}0$ we can also write

$ \sum_{k\equal{}1}^n k^p\equal{}\zeta(\minus{}p)\plus{}\frac{1}{p\plus{}1}\sum_{k\equal{}0}^{p\plus{}1} (\minus{}1)^k\, {p\plus{}1\choose k}\, B_k\,\, n^{p\plus{}1\minus{}k}$ for $ p\in\Bbb{Z}^{\ge 0}$ .

Thus

$ \sum_{k\equal{}1}^n k^p\equal{}\zeta(\minus{}p)\plus{}\frac{1}{p\plus{}1}\sum_{k\equal{}0}^m (\minus{}1)^k\, {p\plus{}1\choose k}\, B_k\,\, n^{p\plus{}1\minus{}k}\plus{}o(n^{p\plus{}1\minus{}m})$ for $ m,p\in\Bbb{Z}^{\ge 0}$

Remarkably the last formula is still true when $ p$ is any complex number $ \neq \minus{}1$ .

Example: For $ m\equal{}8$ and $ p\equal{}\frac12$ we obtain:

$ \sum_{k\equal{}1}^n k^\frac12\equal{}\zeta\left(\minus{}\frac12\right)$$ \plus{}\frac23 n^\frac32\plus{}\frac12 n^\frac12\plus{}\frac{1}{24} n^{\minus{}\frac12}\minus{}\frac{1}{1920} n^{\minus{}\frac52}\plus{}\frac{1}{9216}n^{\minus{}\frac92}\minus{}\frac{11}{163840}n^{\minus{}\frac{13}{2}}\plus{}o\left(n^{\minus{}\frac{13}{2}}\right)$


limit
induction
quadratics
vector
linear algebra
matrix
function

Let $ x_n$ and $ y_n$ be sequences of positive integers such that $ x_0=y_0=7$ and

\[ x_n=4x_{n-1}+3y_{n-1}\]

\[ y_n=2x_{n-1}+3y_{n-1}\]

Find $ \lim_{n%Error. "arrow" is a bad command.
\infty} \dfrac{x_n}{y_n}$.

[hide="answers"]I say $ \infty$, since $ x_n$ increases faster than $ y_n$, but the real answer is $ \dfrac{3}{2}$. Who's right?[/hide]

[quote="1=2"]Let $ x_n$ and $ y_n$ be sequences of positive integers such that $ x_0 = y_0 = 7$ and
\[ x_n = 4x_{n - 1} + 3y_{n - 1}
\]

\[ y_n = 2x_{n - 1} + 3y_{n - 1}
\]
Find $ \lim_{n%Error. "arrow" is a bad command.
\infty} \dfrac{x_n}{y_n}$.

[hide="answers"]I say $ \infty$, since $ x_n$ increases faster than $ y_n$, but the real answer is $ \dfrac{3}{2}$. Who's right?[/hide][/quote]

I cant give a great answer, but this is how my logic goes

[hide]
The limit will occur when $ \frac {m}{n} = \frac {4m + 3n}{2m + 3n}$
Hence $ 4mn + 3n^2 = 2m^2 + 3mn$
So $ 3n^2 + mn - 2m^2 = 0$
Therefore $ (3n - 2m)(n + m) = 0$
Since both start $ \geq 1$ and increase, $ 3n - 2m = 0$ therefore $ 3n = 2m \Rightarrow \frac {3}{2} = \frac {m}{n}$
[/hide]

[u]Solution.[/u] Solving the equations for $ x_n$ and $ y_n$, we obtain $ x_n\minus{}7x_{n\minus{}1}\plus{}6x_{n\minus{}2}\equal{}0$ and $ y_n\minus{}7y_{n\minus{}1}\plus{}6y_{n\minus{}2}\equal{}0$.  Using characteristic equations, we obtain $ x_n\equal{}\frac{42\cdot 6^n\minus{}7}{5}$ and $ y_n\equal{}\frac{28\cdot 6^n\plus{}7}{5}$.  Hence, $ \lim_{n\rightarrow\infty}\frac{x_n}{y_n}\equal{}\frac{3}{2}$. $ \Box$

$ \frac{x_{n\plus{}1}}{y_{n\plus{}1}}\equal{}\frac{4x_n\plus{}3y_n}{2x_n\plus{}y_n} \equal{} \frac{4\frac{x_n}{y_n}\plus{}3}{2\frac{x_n}{y_n}\plus{}3}$
If we set $ a_n\equal{}\frac{x_n}{y_n}$ we get the recursion $ a_0\equal{}1, a_{n\plus{}1}\equal{}\frac{4a_n\plus{}3}{2a_n\plus{}3}\equal{}2\minus{}\frac{3}{2a_n\plus{}3}$.
By induction we obtain that all $ a_n\le\frac{3}{2}$. Using this we can prove $ a_{n\plus{}1}\ge a_n$ by evaluating the quadratic equation $ 2\minus{}\frac{3}{2a_n\plus{}3}\equal{}a_n$.
As $ a_n$ is bounded from above and monotonically increasing it has a limit which you can easily compute to $ a\equal{}\frac{3}{2}$.

[quote="1=2"]I say $ \infty$, since $ x_n$ increases faster than $ y_n$, but the real answer is $ \dfrac{3}{2}$. Who's right?[/quote]
Just because $ f(x) \equal{} 2x$ increases faster than $ g(x) \equal{} x$ does not mean that the limit of $ f(x)/g(x), x \rightarrow \infty$ is infinity.

Not really.  From the recursion $ a_{n \plus{} 1} \equal{} 2 \minus{} \frac {3}{2a_n \plus{} 3}$, the "correct value" of $ \lim_{n \to \infty} a_n$ (if it exists) is $ \frac {3}{2}$.  By induction we prove that it exists.

How I solved it:

Consider the vector $ \begin{pmatrix}x_n \\
y_n\end{pmatrix}$. Then consider $ \begin{pmatrix}4 & 3 \\
2 & 3\end{pmatrix}\begin{pmatrix}x_n \\
y_n\end{pmatrix}$ This equals $ \begin{pmatrix}4x_n \plus{} 3y_n \\
2x_n \plus{} 3y_n\end{pmatrix} \equal{} \begin{pmatrix}x_{n \plus{} 1} \\
y_{n \plus{} 1}\end{pmatrix}$.

Now the next step has a bit of background behind it, but here's what you would do.

We want to find the eigenvectors of the matrix $ \begin{pmatrix}4 & 3 \\
2 & 3\end{pmatrix}$. That is, we want to find $ x$ and $ y$ such that $ \begin{pmatrix}4 & 3 \\
2 & 3\end{pmatrix}\begin{pmatrix}x \\
y\end{pmatrix} \equal{} \lambda\begin{pmatrix}x \\
y\end{pmatrix}$, or $ 4x \plus{} 3y \equal{} \lambda x$ and $ 2x \plus{} 3y \equal{} \lambda y$.

We can solve these equations by various methods and obtain the solutions
$ \lambda \equal{} 1, x \equal{} \minus{} y$ and $ \lambda \equal{} 6, 3y \equal{} 2x$. Any vector can be written as $ c_1\begin{pmatrix}1 \\
\minus{} 1\end{pmatrix} \plus{} c_2\begin{pmatrix}3 \\
2\end{pmatrix}$. In particular, $ \begin{pmatrix}7 \\
7\end{pmatrix}$ can and $ c_2\neq 0$. Thus, since the $ \lambda$ (called the eigenvalue) associated with $ \begin{pmatrix}3 \\
2\end{pmatrix}$ is $ 6$ while the other is $ 1$, the vector $ \begin{pmatrix}x_n \\
y_n\end{pmatrix}$ will tend towards $ a_n\begin{pmatrix}3 \\
2\end{pmatrix}$, giving the value of $ \lim_{n\to\infty}\frac {x_n}{y_n}$ as $ \frac {3}{2}$ as desired.

[quote="Fraenkel"][quote="bertram"]
By induction we obtain that all $ a_n\le\frac {3}{2}$.[/quote]

Wouldn't you have to know that beforehand? I mean you can guess that but its a long shot.

[quote="boxedexe"][u]Solution.[/u] Solving the equations for $ x_n$ and $ y_n$, we obtain $ x_n \minus{} 7x_{n \minus{} 1} \plus{} 6x_{n \minus{} 2} \equal{} 0$ and $ y_n \minus{} 7y_{n \minus{} 1} \plus{} 6y_{n \minus{} 2} \equal{} 0$.  Using characteristic equations, we obtain $ x_n \equal{} \frac {42\cdot 6^n \minus{} 7}{5}$ and $ y_n \equal{} \frac {28\cdot 6^n \plus{} 7}{5}$.  Hence, $ \lim_{n\rightarrow\infty}\frac {x_n}{y_n} \equal{} \frac {3}{2}$. $ \Box$[/quote]

I don't understand how you solved for $ x_n$.
What are the characteristic equations?
Also please explain how you used the characteristic equations?[/quote]

We have $ a_0\equal{}1\le\frac 32$ and $ a_n\le \frac 32\Rightarrow 2a_n\plus{}3\le 6\Rightarrow \frac 3{2a_n\plus{}3}\ge \frac 12\Rightarrow 2\minus{}\frac 3{2a_n\plus{}3}\le \frac 32 \Rightarrow a_{n\plus{}1} \le \frac 32.$
So by induction all $ a_i$ are $ \le \frac 32$.

@hamster: you only find the limit, but you didn't prove it exists... for instance, consider the function $ f(x) \equal{} 1 (|x|\neq 2), f(2) \equal{} \minus{}2, f(\minus{}2) \equal{} 2$. The only soultion of $ f(x)\equal{}x$ is $ x\equal{}1$, but if we start with $ a_0\equal{}2$ and set $ a_{n\plus{}1}\equal{}f(a_n)$ then $ \lim_{n\to\infty} a_n$ doesn't exist.

I actually do prove that it exists. The key is that the function is a linear transformation of the current vector, which can be expressed as a sum of the eigenvectors of the matrix. Your "counterexample" is not a linear transformation.

Given that $ \begin{pmatrix}7\\7\end{pmatrix} \equal{} c_0\begin{pmatrix}1\\\minus{}1\end{pmatrix} \plus{} c_1\begin{pmatrix}3\\2\end{pmatrix}$, the general form of $ \begin{pmatrix}x_n\\y_n\end{pmatrix}$ is $ c_0\begin{pmatrix}1\\\minus{}1\end{pmatrix} \plus{} c_1\cdot 6^n\begin{pmatrix}3\\2\end{pmatrix}$, which makes it quite obvious that the limit does indeed exist.

[quote="Hamster1800"]I actually do prove that it exists. The key is that the function is a linear transformation of the current vector, which can be expressed as a sum of the eigenvectors of the matrix. Your "counterexample" is not a linear transformation.

Given that $ \begin{pmatrix}7 \\
7\end{pmatrix} \equal{} c_0\begin{pmatrix}1 \\
\minus{} 1\end{pmatrix} \plus{} c_1\begin{pmatrix}3 \\
2\end{pmatrix}$, the general form of $ \begin{pmatrix}x_n \\
y_n\end{pmatrix}$ is $ c_0\begin{pmatrix}1 \\
\minus{} 1\end{pmatrix} \plus{} c_1\cdot 6^n\begin{pmatrix}3 \\
2\end{pmatrix}$, which makes it quite obvious that the limit does indeed exist.[/quote]

Yeah, I kind of figured out that it would work if we have a linear transformation :) But I don't like algebra, so I'll stick to my solution :P

All the solutions here except mine/hamsters are not exact though as noone has proven that the limit actually exists, which is usually the hardest part of finding the limit of a recurive sequence (at least if it's only in one variable).

[quote="Altheman"]Uhh...what? boxedexe's solution is complete (or at least it is trivial to show that the limit exists).[/quote]

oops :)

[hide]
If we first assume the limit exists and is equal to $ L$, then from
\[ \frac{x_{n\plus{}1}}{y_{n\plus{}1}}\equal{}\frac{4x_n\plus{}3y_n}{2x_n\plus{}3y_n}\equal{}\frac{4\cdot \frac{x_n}{y_n}\plus{}3}{2\cdot \frac{x_n}{y_n}\plus{}3}\]
we get
\[ L\equal{}\frac{4L\plus{}3}{2L\plus{}3}\Rightarrow 0\equal{}2L^2\minus{}L\minus{}3\equal{}(L\plus{}1)(2L\minus{}3)\]
and since the sequences are always positive, $ L$ must be nonnegative, so $ L$ must be $ \frac{3}{2}$.

This suggests looking at $ 3y_n\minus{}2x_n$. We have
\[ 3y_{n\plus{}1}\minus{}2x_{n\plus{}1}\equal{}3(2x_n\plus{}3y_n)\minus{}2(4x_n\plus{}3y_n)\equal{}3y_n\minus{}2x_n\]
so $ 3y_n\minus{}2x_n$ is invariant and always equal to $ 3y_0\minus{}2x_0\equal{}7$. This leads to
\[ \frac{3}{2}\minus{}\frac{x_n}{y_n}\equal{}\frac{7}{2y_n}\]
But it is easy to see that $ y_n\rightarrow \infty$ as $ n\rightarrow \infty$ (for example, $ y_{n\plus{}1}\geq 3y_n$ implies $ y_n\geq 7\cdot 3^n$), so $ \frac{x_n}{y_n}$ does converge to $ \frac{3}{2}$.
[/hide]


How many centimeters are in the length of the longest side of a rectangle whose area is 108 square centimeters and whose perimeter is 42 centimeters?

Please tell me a method BESIDES guess and check.

akhil0422

Let the sides be $ a$ and $ b$.

$ a \plus{} b \equal{} 21$ (Since the perimeter is 42.)

and

$ ab \equal{} 108$

From our first equation, we get $ a \equal{} 21 \minus{} b$. Substitute this into our second equation and solve the resulting quadratic.

When low on time, this is not the best strategy. Instead, you could note that the perimeter is too small for the numbers to be far apart(like 2 and 54.) Instead, the numbers will be close together. So it makes sense that 9 and 12 would work. 9 times 12 is 108, and 9+12=21. Thus 12 is the answer.


Can we have a emoticon limit for the shouts? They can get really annoying when someone overdoes it.

Thanks!

Can it be an option for each individual blog?

Ok, eggylv999, why would the admins go through all that extra trouble just to allow people to post random spam in blogs?

Well, they allowed us to customize our scrollbars...

Well, there's a reason someone would want to change their scrollbars; it could sometimes make the blog look nicer. But there is [i]NO[/i] reason that someone would need to post a shout with more than one or two emoticons.

I [i]hate[/i] these spam shouts. They're pointless. If you take pleasure in creating shouts like these, then there's something wrong with you. If you don't, then stop doing it.


function
trigonometry
limit
integration
real analysis
real analysis unsolved

hi,
let $ f: \mathbb{R} \rightarrow \mathbb{R}$ with
\[ f(x) \equal{} \sum_{k\equal{}1}^{n}{\sin{(a_{k}x)}},\]
$ a_{k} \in \mathbb{R}, k \equal{} 1,...,n$. If $ |a_{i}| \not\equal{} |a_{j}|$ for $ i \not\equal{} j$ prove that there exists $ x_{0} \in \mathbb{R}$ s.t. $ f(x_{0}) \not\equal{} 0$.

Compute $ \lim_{T\to\plus{}\infty}\frac 1T\int_0^T f(x)^2\,dx$ ;)

[b]Proposition:[/b]  Let $ \alpha_1, ... \alpha_n$ be distinct complex numbers.  Then the functions $ e^{\alpha_1 z}, ... e^{\alpha_n z}$ are linearly independent over $ \mathbb{C}$.

[i]Proof.[/i]  Their [url=http://en.wikipedia.org/wiki/Wronskian]Wronskian[/url] is a Vandermonde determinant.


hey friends,
I invite you to participate in our college's techfest event- online math.
the link is
[url]http://www.pragyan.org/08/home/events/brainwork/onlinemath/[/url]

there are 5 problems in every problem set. And as the college students are disappointing me right now :D i thought i'd request you guys to participate and earn some easy money (am sure that's not the incentive either way)
the first problems deadline has already finished. the 2nd problem set is up and the third will be put up later. Am sure the problems are easier for you guys

best regards

Wrong timing. We have our boards!! I'd advise you to post this in some of the other forums, we have a lot of math crackers here. :D

thanks for the advice.
i'm sure there are a lot of math crackers here. especially since i have been on this forum since it started.
anyway good luck for your boards

regards


linear algebra
matrix
superior algebra
superior algebra solved

A in M_2(R), with det(A)=d=/=0 and det(A^2+I_2)=0.
 Prove that det(A-(d^n)*A^(-1))=4,for any n in N\{0}
     Cristian Grecu[/quote][/list]

Hello!

This is problem 24969 in GM 10/2003 ... and as I remember problems in journals, gazetts etc must be at least one year old before posting and solving a problem from it/them, on this forum!

Bye!

Hi now you can post your solutions for this problem since october is long gone

Obviously from $\det(A^{2}+I_{2}) = 0$ it follows that $\det A = 1.$ Also from the equality $A^{2}+I_{2}=O_{2}$ it follows that $-A = A^{-1}.$ Now $\det(A-d^{n}A^{-1}) = det(A-1^{n}(-A)) = 4.$

cheers! :D :D


Hello,

I want to discuss performers on string instruments and who they are. I don't particularly know a lot of famous violinists, other than:

Ithak Perlman

Isaac Stern

and

Jascha Heifetz.

Any more?

Plus, I only know one famous orchestra:

Moscow Symphony Orchestra.

Any others?

Feel free to post here.

hwenterprise

Boston Symphony Orchestra and Detroit Symphony Orchestra are lovely orchestras. So is Chicago Symphony Orchestra.  :)

[quote="hwenterprise"]Hello,

I want to discuss performers on string instruments and who they are. I don't particularly know a lot of famous violinists, other than:

Ithak Perlman

Isaac Stern

and

Jascha Heifetz.

...
Feel free to post here.

hwenterprise[/quote]

What about Paganini, though he's not a living violinist. :P

How about cellists...

David Popper, Casals, Yo-yo ma... :?

Yes, Yo-Yo Ma is a celloist...

Have any of you heard of Kyung Wha Chung?

Do any of you know the names of the two violinists in the movie "Together"??

It's in chinese, though...

[url=http://www.joshuabell.com/]Joshua Bell[/url] is a phenomenal violinist. I like his discs a lot.

WOO Gil Shaham!  He actually went to my current school... :D his tone is good
But yeah Paganini was probably the best violinist who ever lived.  Then there come those in the 20th century, like Jascha Heifetz, Nathan Milstein, etc etc...if you want to know a lot of violinists, watch Art of the Violin  :)

Arthur Grumiaux is also pretty well known. A lot of people think his recording of Bach's solo sonatas and partitas is the best ever. Some less alive famous violinists also would be Pablo Sarasate and Fritz Kreisler, although they're more known their compositions than anything else.

A lot of composers also played their instruments (Wieniawski as well, for example)...
I recently saw some videos of David Oistrakh and Henryk Szeryng (hope i spelled that correctly)...pretty good..
Early 20th century people

Yehudi Menuin (Menuhin?), Hillary Hahn, Midori.

Berlin Philharmonic is very good.  So are New York Phil, Cleveland Orchestra, several London orchestras.

Apparently, there recently was, at my school, a thing honoring Gil shaham as a distinguished Alumnus...he went to my school if you didn't read the previous post...
Anyways, I just missed it...and I saw the picture in the newspaper, and right behind him is one guy I know...i soooo wish i had been there... :mad:

[quote="dasherm"]I have a recording of him, he's a great player.  I used the think he was "Gilsha Ham".  :huh:    It sounded like a weird name.[/quote]

Yeah...i have many recordings of him...
I just really wanted to meet him in person.  My violin teacher says he's a really nice guy...

Jacqueline du Pre

Vanessa Mae is famous but I'm not sure how much of that is hype and how much skill...

Gil Shaham is awesome.
Michael Rabin, although he killed himself because he was lovesick :P
Several Chinese violinists are pretty good.

Btw, the orchestras?
Vienna Philharmonic!

Joshua Bell is scary if you actually watch him play. :P

Famous violinists (add on):
-Auer!!
-Francescati
-Kreisler
-Sarasate

Orchestras:
-Berlin Phil = the best
-Vienna phil = in my opinion 2nd.
-The Israelic one right now consists of many of the world's leading instrumentalists, but...they aren't doing so well  :?


inequalities
calculus
real analysis
real analysis unsolved

A stat book of mine asserts the following:

\[ 1\minus{}(1\minus{}p^n)^{\frac {{2}^{n}}{n}}  > 1\minus{}e^{\minus{}\frac {{(2p)}^{n}}{n}}\]

For $ n>0$,$ 0\leq p\leq 1$.  However, it's been a while since calculus, and am not seeing immediately why this inequality holds.  Any pointers would be appreciated.

you actually need $ 0<p\leq 1$.  taking the 1's away, multiplying by $ \minus{}1$ (and reversing the inequality) and then raising both sides to the power $ \frac{n}{2^n}$, what you have to prove is

\[ 1\minus{}p^n<e^{\minus{}p^n}\]

and this is a case of the more general

\[ e^x \geq 1\plus{}x \mbox{ for } x\in \mathbb{R}\]

to prove this, write $ f(x)\equal{}e^x\minus{}x\minus{}1$. we need to show that $ f(x)\geq 0$ for all $ x$. first, $ f(0)\equal{}0$. now, $ f'(x)\equal{}e^x\minus{}1$, which is negative for $ x<0$ and positive for $ x>0$. thus $ f$ is decreasing on $ (\minus{}\infty,0)$ and increasing on $ (0,\infty)$. so $ f(0)\equal{}0$ is the minimum value of $ f(x)$, which is what we needed.


integration
trigonometry
calculus
calculus computations

[quote="Carcul"]Find $ \int (x \minus{} 1)^{3/2}(2 \minus{} x)^{5/2}\,dx$.[/quote]
do substitution $ x \minus{} 1 \equal{} y$
and after some trigonometryc substitution say $ y \equal{} \sin^{2}x$.
 It's standart integral. I am sure, you will take something good  :wink:


If a satellite is given energy E ( say velocity V in direction perpendicular to its motion.), then how do we decide what trajectory it follows. The question says total  energy : positive , negative, zero ; trajectory: hyperbola, ellipse, parabola, circular . We have to match which with which.

[quote="Conjurer"]The question says total  energy : positive , negative, zero ; trajectory: hyperbola, ellipse, parabola, circular . We have to match which with which.[/quote]

Today's FIITJEE question??

[quote="Conjurer"]I am not giving fiitjee man. I found this somewhere else. Did this come in the test?[/quote]

Yes, a similar question! Very similar!


geometry
parallelogram
ratio
geometry proposed

In a parallelogram $ ABCD$, points $ E$ and $ F$ are the midpoints of $ AB$ and $ BC$, respectively, and $ P$ is the intersection of $ EC$ and $ FD$. Prove that the segments $ AP,BP,CP$ and $ DP$ divide the parallelogram into four triangles whose areas are in the ratio $ 1: 2: 3: 4.$

Denote J the intersection of CE and AD.
We have $ \frac{DP}{FP}\equal{}\frac{DJ}{CF}\equal{}4$
$ \Rightarrow S_{APD}\equal{}4S_{BPC} (1)$
Similarly, we can calculate $ \frac{EP}{PC}\equal{}\frac{3}{2}$ then $ S_{APB}\equal{}\frac{3}{2}S_{DPC} (2)$
Let I be the intersection of BP and AC.
Arcording to Menelaus's theorem we obtain $ \frac{AI}{IC}.\frac{PC}{PE}.\frac{BE}{BA}\equal{}1$
Then $ \frac{AI}{IC}\equal{}3$ which follow that $ S_{APB}\equal{}3S_{BPC} (3)$
From $ (1)(2)(3)$ we completed the proof.


Find all functions $ f: \mathbb{R}\to\mathbb{R}$ such that \[ f((x \plus{} z)(y \plus{} z))\equal{}(f(x) \plus{} f(z))(f(y) \plus{} f(z))\,\forall x,y,z\in\mathbb{R}.\]

$ f((x\plus{}z)(y\plus{}z))\equal{}(f(x)\plus{}f(z))(f(y)\plus{}f(z))$

$ \star\;: x \equal{} y \equal{} z \equal{} 0\rightarrow\left[\begin{array}{ll}f(0) \equal{} 0 \\
f(0) \equal{} \frac {1}{4}\end{array}\right.$

-[b]Case 1[/b] $ f(0) \equal{} 0$

$ \star\;: z \equal{} 0\rightarrow f(xy) \equal{} f(x)f(y)\rightarrow f(x)\ge 0;\forall x\ge 0,$

-$ f(1) \equal{} 0$ put $ x \equal{} 0,z \equal{} 1\rightarrow f(y \plus{} 1) \equal{} 0;\forall\;y\in\mathbb{R}\rightarrow f(x) \equal{} 0$

-$ f(1) \equal{} 1\rightarrow f(x) \equal{} \minus{} f( \minus{} x)$

$ \star\;: x \equal{} y\rightarrow f((x \plus{} z)^2) \equal{} (f(x) \plus{} f(z))^2\rightarrow f^2(x \plus{} z) \equal{} (f(x) \plus{} f(z))^2$

For all $ \{(x,z)|x,z\ge 0\}$ then $ f(x) \plus{} f(z) \equal{} f(x \plus{} z)\rightarrow f(x) \equal{} x,\forall\;x\in\mathbb{R},x\ge 0$

But $ f(x) \equal{} \minus{} f( \minus{} x)$ then $ f(x) \equal{} x;\forall x\in\mathbb{R}$

-[b]Case 2[/b] $ f(0)\equal{}\frac{1}{4}$

$ \star\;z\equal{}0\longrightarrow f(xy)\equal{}(f(x)\plus{}\frac{1}{4})(f(y)\plus{}\frac{1}{4})\longrightarrow f(x)>0\;,\forall\;x\ge 0$

$ \star\;y\equal{}z\equal{}0\longrightarrow \frac{1}{4}\equal{}(f(x)\plus{}\frac{1}{4})\frac{1}{2}\longrightarrow f(x)\equal{}\frac{1}{4}\;,\forall\;x\in\mathbb{R}$

Hello !

I don't understand this :
[quote="quangpbc"]-$ f(x) \equal{} 1\rightarrow f(x) \equal{} \minus{} f( \minus{} x)$[/quote]

He mean that $ f(\minus{}1)\equal{}\minus{}1$. I think so.


limit
topology
advanced fields
advanced fields unsolved

Let $Y$ be a subset of a metric space $(X,d)$ such that there is a metric $\rho$ on $Y$, equivalent to $d$, which makes $(Y,\rho)$ a complete metric space.

Prove that $Y$ is a $G_\delta$ set of $X$ (i.e. a countable intersection of open subsets of $X$).

What's $X$? Should it be $Y$ a subset of metric space $(X, d)$?

Of course. I'm terribly sorry about that. It should look fine now.  :blush:

Define $d(x,Y)=\inf\{d(x,y): y\in Y\}.$

Let $Y_{n}=\left\{x: d(x,Y)<\frac1n\right\}.$

We can show that $Y_{n}$ is open.

Let $Z=\bigcap_{n=1}^{\infty}Y_{n}=\{x: d(x,Y)=0\}.$

Clearly $Z$ is a $G_{\delta}$ set. We need to show that $Y=Z.$ $Y\subset Z$ easily, so we need to show the opposite inclusion.

If $x\in Z,$ then $\exists\,x_{k}\in Y$ such that $\lim x_{k}=x.$

Since $(x_{k})$ converges in $X,$ then it is a Cauchy sequence. in $X.$ Since the metric on $Y$ is equivalent to the metric on $X,$ it is a Cauchy sequence in $Y.$ Since it is a Cauchy sequence in $Y$ and $Y$ is complete, then it converges to a point in $Y.$ But limits in a metric space are unique, and we already know that the limit of the sequence is $x.$ We conclude that $x\in Y$ and thus $Z\subset Y$

So $Y=Z,$ which is $G_{\delta}.$

By equivalent metric I mean that it induces the same topology. A sequence in $Y$ which is Cauchy with respect to $d$ need not be Cauchy with respect to $\rho$.


In the three main sections of these forums (getting started, intermediate, advanced), its been stated that when someone asks for help, everyone is able (and actually encouraged) to give help, because the main point of asking is to work out how to solve the problem, and thus you don't just want people who don't know how to solve the problem to try it.

Are the rules different in this forum? I thought they'd be the same, but in a thread I was asked not to answer by someone (not the person asking for help though.)

When I post questions I'm stuck on, I appreciate a quick response, and an accurate one, so I'd say, answer in a spoiler.

I certainly expect ANYONE to be welcome to answer an informational question posted in any forum. I see that there doesn't seem to be the same kind of description of this forum posted as an announcement here on the Mathcounts forum as on the main problem-solving forums. But I would expect that if someone answers in spoiler text, they shouldn't be berated for helping someone with an informational question.

Part of the problem, of course, is that some questioners, whether because of inexperience with the forum or unfamiliarity with the English language, post informational questions in forms that could be mistaken for challenge problems. (Sometimes the reverse happens too.) The usual protocol is that challenge problems are only for people currently working at the level of the forum to answer, while informational questions are for anyone to answer. Some of the confusion about who may answer in a particular thread doubtless arises from confusion about whether the first post in the thread posed a challenge problem or asked an informational question. That's why I have recently been urging participants (of whatever level) on the Getting Started Forum to title their posts with either "Question: . . . " or "Challenge Problem: . . . " as fits each case.

Of course, I would defer to the moderators of this forum regarding the rules of this forum, and if they decide differently, I will do my best to respect their wishes. As a moderator of other forums, I, in turn, defer to the site owner for all questions regarding how the forums I moderate should be moderated.

It's fine to set any reasonable rules, but the problem is that loads of people who post here never bother to read the forum rules. It makes sense to me that anyone should be able to reply to a plea for help on a problem. Having said that, I don't usually reply to posts for help on Getting Started and MATHCOUNTS unless I think I have something to say that no one else will. (Usually that means giving another related problem.)

Yes, it is fine for question problems to post a solution.  Even when it's a challenge question and people are having trouble and or not replying, you can urge on the discussion by hints and stuffs.

I'll try to get people to specify whether they are posting a challenge or actual question.


function
logarithms
real analysis
real analysis unsolved

Let $(f_n; n\in N)$ be a sequence of functions and  $f_n \to f$ by measure $\mu$. Let $g$ be a continous function. Proof that $(g \circ f_n; n\in N)\to g \circ f$ by measure $\mu$.

what about $f_n(x)=x+\frac{1}{n}$ and $f(x) =x.$, $g(x)=e^x.$
Then $\forall \delta \; \mju\{x| |f(x)-f_n(x)|>\delta\} = \mju\{x| \frac{1}{n}>\delta\} \to 0$, when $n\to \infty$.
Then $f_n\to f$ by measure.
Form the other side $\mju\{x| |g(f(x))-g(f_n(x))|>\delta\} = \mju\{x| |e^x-e^xe^{1/n}|>\delta\} =$
$= \mju\{x|\; e^x|1-e^{1/n}|>\delta\} = \mju\{x|\; e^x>\frac{\delta}{|1-e^{1/n}|} \} =$
$= \mju \{ x> \ln \frac{\delta}{|1-e^{1/n}|}\} = \infty$ $\forall n$, so $\nrightarrow 0$... :?:

Sorry  :(
Functions from the sequence must be jointly limited. What is the proof now?

[quote="tom123"]must be jointly limited.[/quote]
$\exists C$ such that $\forall x,n \; |f_n(x)| <C$?

$\exists {A} \forall_{x,n} |f_n(x)|<A$
$g\in C([-A,A])$ then $g$ is uniform continuous by Cantor theorem. It mean \[ \forall \delta \exists \omega \; |x-y|<\omega \Rightarrow |g(x)-g(y)|<\delta \]
Note that  $\{|g(x)-g(y)>\delta|\}\subset \{|x-y|>\omega\}$
Then
$\forall \delta>0 \; \mu\{|g(f_n)-g(f)|>\delta\} \le \mu\{|f_n-f|>\omega\} \to 0$ ,when $n\to \infty.$
Thus $\forall \delta>0 \; \mu\{|g(f_n)-g(f)|>\delta\} \to 0.$


Which exam is usually more difficult: the IMO or the USAMO?

It looks to me like IMO's easy problems are easier and the hard problems are harder compared to USAMO.

[quote="batteredbutnotdefeated"]IMO is harder by a long shot. You have to qualify for IMO by doing well on USAMO.[/quote]

Actually, I'd say USAMO > IMO in terms of difficulty. And China TST >> USA TST >> IMO.

@FightTheTide: The hardest problems vary a lot....

Both exams' difficulty vary from year to year, so it's impossible to give one answer. The difficulty is similar, although the IMO has probably been a tad bit harder in recent years.

I'm not so sure about the IMO hard problems necessarily being harder: USAMO 09/3b, 08/6, and 07/5 were all probably on the level of the last few IMO 6s. As for the easy problems, it depends on what you think of the G1 that pops up on the IMO every year.

The USAMO's since 2002 have most likely been harder than the IMO. And you know, the IMO wasn't always as hard as it is now. Just take a look at the 1959 exam - it asks to prove the relative primality of 21n+4 and 14n+3.

Even the ones from the 1990's are quite easy.

And it's not surprising to find some IMO #1's and #4's easier than some AIME problems.


Find all solutions to $ \sqrt{x}^{log_{10}x}\equal{}100$


We define $d(u)$ as the digit number before the decimal and after the decimal of $x$   i.e $d(123.456)=3.4$ . Compute $d\left(\left(\sqrt{2}+\sqrt{5}\right)^{2000}\right)$

 :)

I think this was on the Irish MO this year - if not, it was something very similar.

The trick is to note that $\left(\sqrt{2}+\sqrt{5}\right)^{2000}+\left(\sqrt{5}-\sqrt{2}\right)^{2000}$ is an integer, and is congruent to $2\left(2^{1000}+5^{1000}\right)\equiv2\mod{10}$, while $\left(\sqrt{5}-\sqrt{2}\right)^{2000}$ is very small (in particular, less than $0.1$).

I think you get $d\left(\left(\sqrt{2}+\sqrt{5}\right)^{2000}\right)=1.9$, but I could be wrong with the details.


prove that the polynomial $P(x)=\frac{x^n}{n!}+\frac{x^{n-1)}}{(n-1)!}+...\frac{x}{1!}+1$ is ireeducible over $\mathbb{Q}[X]$.

Alekk, do you have an easy solution for this problem? I think Sam-n posted it also and if I'm not wrong the proof is quite very difficult.


conics ellipse calculus derivative quadratics

$P(x,y)$ is a point on the ellipse $4x^2+9y^2=36$.
Find the maximum possible value of $x^2+\frac{2}{3}xy+\frac{3}{2}y^2$.

Is this answer right?
[hide]$\frac{75}{8}$[/hide]

All I did was solve the elliptic equation for $y$, plug that in for $y$ in the second equation, and then find the maximum of that graph.

Hmmm...I did it with a new method. Is this right?
[hide]$216$[/hide]

Noooooo. This is the right answer.
[hide]$225$[/hide]

[hide]
hmm? solving for $x$ gives $x = \sqrt{9 - 9y^2/4}$

substituting this into our thing you should get

$9 - 3y^2/4 + 2y\sqrt{1 - y^2/4}$

then i took derivative of this, found the critical value of $y$, put it back in and you get $10$ as the max....

there must be a better way tho using noncalculus methods hmm

[/hide]

gotztahbeazn's answer is correct :)

And yes, there are some "nicer" non-calculus solutions to this problem.

[hide]Let $k = x^2 + 2xy/3 + 3y^2/2$.  Then
\[ x^2 + \frac{2}{3} xy + \frac{3}{2} y^2 = \frac{k(4x^2 + 9y^2)}{36}. \]
Re-arranging, this becomes
\[ (4k - 36)x^2 - 24xy + (9k - 54)y^2 = 0. \]

As a quadratic in $x$, this has a root if and only if its discriminant is non-negative:
\begin{eqnarray*} 24^2 y^2 - 4(4k - 36)(9k - 54)y^2 &\ge& 0 \\ \Leftrightarrow \ -144(k - 5)(k - 10) y^2 &\ge&0. \end{eqnarray*}

If $y = 0$, then $x^2 + 2xy/3 + 3y^2/2 = x^2 = 9$.  Otherwise, we can say $(k - 5)(k - 10) \le 0$, which means that $5 \le k \le 10$.

If $k = 10$, then the quadratic above becomes $4x^2 - 24xy + 36y^2 = 4(x - 3y)^2 = 0$, so $x = 3y$, so
\[ x^2 + \frac{2}{3} xy + \frac{3}{2} y^2 = \frac{25}{2} y^2 = 10, \]
and $y = \pm 2 \sqrt{5}/5$, and $x = \pm 6 \sqrt{5}/5$.  For these values of $x$ and $y$, $x^2 + 2xy/3 + 3y^2/2 = 10$ and $4x^2 + 9y^2 = 36$, so the maximum value is 10.[/hide]


[hide]i try this with $10^{3}-25=975$
23*9+12=something i am too tired to do right now[/hide]

[quote="i_like_pie"]What is the sum of the digits of $100^{25}-25$?[/quote]

hint
[hide]
You are left with a bunch of 9s and 75. [/hide]

Please:  If you aren't going to post a solution, don't post at all.  An answer alone doesn't help.  I won't delete since you're new to this forum, but just remember that please ;).

[quote="mathgeniuse^ln(x)"][quote="funcia"][quote="mathgeniuse^ln(x)"]October 16, 2006, that isn't really new.

[hide]And yeah, we get that there a 50 zeroes after a 1, so when we subtract 25, we have 50 digits total

We see that the last two digits are 75, and the remaining are 9, so the answer is 9*48+7+5=444

$444$[/hide][/quote]
could you explain the first block of words after the hide?[/quote]

[hide]100^25 has 50 zeros after a 1, then we subtract a 25, then the conclusion follows from there.[/hide][/quote]
oh i see  :wallbash: not 10^25

[quote="i_like_pie"]What is the sum of the digits of $100^{25}-25$?[/quote]
[hide]
there are 48 9's followed by a 7 and a 5
the sum is $48*9+12=444$

[/hide]

THe answer is 444. THats what I got. (I'm  a 6th grader!) I calculated that there was 51 digits in 100^25. There are 48 nines in 100^25-25. Then, you still have a 7 and 5 (from 75).
48*9=432+7+5=444.

[hide="or if it was a sprint round mathcounts problem to do in your head"]
Steps: $10^{50}-25 = 1-25 = 3 (9)$, the sum is just below the sum of $50$ 9's clearly, $50*9=450$ use $3 (9)$ so the sum must be $444$.
Of course, buzz in at $3 (9)$ step, answer should be found by the time name is called... if more time, once can check by looking at the units digit of $50(9)-2-4$[/hide]

[quote="ProtestanT"]what is $3(9)$?[/quote]
I think thats short for mod.

The correct answer is $\boxed{444}$, and mathgenius's explanation is exactly the same as mine.

San Jose Tokutomi Junior Division 2003 #15.


How many ounces of 20% acid solution must be mixed with a 30% acid solution to obtain 45 ounces of 24% acid solution?

[hide]457[/hide]

[hide]27

You want 3 parts of 20% for 2 parts of 30%, so 3/5 * 45 is 27.[/hide]

goode job to all of you.  next time put a solution, scrambled.


Here are a few problems in Sequence and Series.
Please do try them.

1) Find the sum to n terms of the series:
$\frac{3}{1^22^2} + \frac{5}{2^23^2} + \frac{7}{3^24^2} + \cdot\cdot\cdot$

2) A multiple choice question (more than one option may be correct):
If a,b,c are in H.P., then :
(a) $a^n +c^n > b^n$             (b) $a^n +c^n > 2b^n$
(c) $a^n +c^n > (2b)^n$         (d) none
where n is a positive integer.

3) Another multiple choice question (more than one option may be correct):
If the system of the equations :
$ax + by + (a\lambda + b)z =0$
$bx + cy + (b\lambda + c)z =0$
$(a\lambda + b)x + (b\lambda + c)y =0$  has a non-trivial solution, then:
(a) $\lambda$ is a root of $ax^2 + 2bx + c =0$
(b) a,b,c are in AP
(c) a,b,c are in GP
(d) a,b,c are in HP

[hide="2"](b) and (c) don't work if $a=b=c=1$.

(a) doesn't work if $a=1/3,\ b=1,\ c=-1$.

Then the answer must be (d).[/hide]


What is the correspondence between molecular geometry and hybridization?  What #s of regions of electron density correspond to what types of hybridizations?

Thanks in advance!

sp       - linear
sp2     - trigonal planar
sp3     - tetrahedral
sp3d    - trigonal bipyramidal                  : lone pairs occupy equatorial positions first
sp3d2  - square bipyramidal (octahedral) : lone pairs occupy axial positions first
sp3d3  - pentagonal bipyramidal              : lone pairs occupy equatorial positions ( i think.....)

......if that's wat u r askin' 4........

Here's a tip I use when generalizing shapes and hybridizations.

Add the "exponents!"

For example, $ sp^3$ would be "$ 4$", meaning that there are [b]four[/b] pairs of electrons around the central atom. (While double and triple bonds also count as one effective "pair") $ \ce{CH4}$, for example.

$ \ce{CO2}$, has two effective "pairs", making the hybridization $ sp$.

From there, the number of [b]lone[/b] pairs around the central atom may be used to determine more general shapes (for example, trigonal pyrimidal instead of tetrahedral for $ \ce{NH3}$). For an organized chart, the Princeton Review Book for AP Chemistry has the best [I've seen] summary of the hybridizations, lone pairs, and shapes (including examples and exceptions to the octet rule).


AMC
USA(J)MO
USAMO
algorithm
AIME
videos
AMC 10

I pray for God’s mercy and blessing.....

I pray that I will stay calm and remain focused in next two days. I pray that no matter what the outcome will be, I will have no regret as I have tried my best and with God’s love I have qualified for USAMO and this is ultimately my goal for this year and I am proud what I have already accomplished. I pray that God will give all of us wisdom so everyone will be at his and her best ability next Tuesday and Wednesday. I wish everyone good luck!

May God bless us all! May God bless the ultimate 12 winners and the USA IMO team! Amen!

Um, I didn't make USAMO, but I'm just hoping that everyone who deserves to make MOP makes it...basically:

[quote]I pray that God will give all of us wisdom so everyone will be at his and her best ability next Tuesday and Wednesday. I wish everyone good luck! [/quote]

rrusczyk recomments this:

[quote]
*******
I know all of us have been studying all year for this test. That being said, if you were given 3 hours to review, how would you spend those 3 hours?
rrusczyk
By playing basketball. [/quote]

I second rrusczyk's recommendation. Everyone should rest tomorrow.

USACO Open tomorrow for me :). Get into shape on those algorithm problems that the USAMO loves so much :).

Homework for me tomorrow...and more after the usamo  :|

Darn I will be doing a 4 page history paper, big physics lab, chemistry homework, spanish project, and english project that includes artwork and an essay...starting now.

[quote="cyberspace"]I'll probably be playing guitar hero for most of tomorrow :lol:[/quote]

That's what I'm talking about.  :)

Good Luck to everyone taking it this week! :lol:  Unless you are one of super genuises, you are probably just going in there proud of your accomplishments and ready to enjoy some very hard-core problems! Get a good night's sleep, and be ready! Congrats to all you guys who made it! I hope I will make it, in ummm, 3 years? It is just my first year making AIME. (got a 3 there....LOL) :rotfl:

I will keep doing math in the vain hope that it will help. And I will sleep more than usual D: and pray. A lot.

[quote="Zhero"]I pray that my getting sick with a high fever, a strong headache, a cough, and a twisted muscle by my lungs will be insignificant during the USAMOs. And that I don't mess up.[/quote]
Eep swine flu?

Anyway, I will attempt to do less starcraft and more sleep...

God really is awesome.

For me, I'm thinking "Sweet, I tried my best. It's been great so far. So now I get to be locked in a tiny room for 4.5 hours bored" (I really couldn't save time for USAMO prep).

But, I'll be reminded of my totally unexpectedly high AMC 10 and AIME scores and praise God just for the experience.

Then I'll spend the other 4.45 hours doodling.

:P

Wait... maybe I'll bring huge amounts of pencils/erasers and make a construction. Which won't even take up half the time. But I totally want to spend 4 hours praying :D.

LOLOL

Good luck all! [I'm seriously gonna try :D]

[quote="Textangle"]God really is awesome.[/quote]
Darn at this rate (notice post time :(), if I somehow do well on USAMO, I really think it'll be an instance of divine intervention, and I'll have to stop being atheist. Only 2 essays to go, and 3 hours till normal wake up time...I hope I'll make it.

I always find that divine intervention comes when I'm high on adrenaline. Maybe you guys should try that too.

[quote="Brut3Forc3"]
Eep swine flu?[/quote]
I think so. =( I went to NYSML, and I read in the newspaper that it was present in Queens, and someone from Queens probably went to NYSML.

And I have so much homework to do. Sigh.

At least I'll probably be able to get a good sleep. And the medicine's curing my symptoms for now, so medicine ftw.

This post is getting off topic, because of my post, but yeh.

lol.

Try your best. Or, if you can't do the problems, take a good nap :D.

For those of you who qualified for USAMO, GOOD LUCK to you all!

Have a good night sleep, be prepared for a hard core battle, enjoy the mathematical exploration process and do your best. You have made this far, and the entire AMC community (parents, teachers, fellow students and friends etc.) is very proud of you. We will cheer you up on Tuesday and Wednesday.

Thanks Dr. Dunbar and his AMC organization for continuing to work hard every year and promoting the mathematical learning and competition in America. Hats off!

GOOD LUCK to Dr. Zuming Feng and his 2009 USA IMO team!!!

What am I going to do... hmm... well I got like almost 5 hours of sleep last night. Hope I get 9 today lmao. yea right T.T...

apparently eating dark chocolate helps so ya =D. lots of whole wheat and other long lasting foods.

I'm just hoping I beat last years score.

Good luck all

WOOOOOOOOOOOOOOOO IM EXCITED!!!!

good luck all :D

I will pose as one of my friends who qualified but isn't taking the USAMO.

I'll be playing video games and writing fanfiction.

Oh, screw that, the contest's tomorrow, and I haven't spent any time preparing for it.

Peanut butter and honey sandwich with 100% orange juice for breakfast apparently works wonders, because you get proteins and carbs with a lower risk of a sugar crash.

so like

I should sleep

now

instead of doing homework

I'm reading a novel (4 chapters left). Loads of fun.

And I'm gonna finish my lunch tomorrow during car ride to another school. Maybe plus a bit caffeine?

[quote="Ubemaya"]Well I can't find a compass or protractor, but at least I'm adequately supplied with candy :) .[/quote]
Compass problem can be fixed by filling up on candy, attaching a pencil to the chest, and spinning on your belly on top of your diagram.  Not sure how to go about measuring angles, though.

Good luck to everyone, and hope you are able to sleep well tonight so you can own tomorrow.

[quote="soltixnj"][quote="Zhero"]I pray that my getting sick with a high fever, a strong headache, a cough, and a twisted muscle by my lungs will be insignificant during the USAMOs. And that I don't mess up.[/quote]

Don't bring the Swine Flu to BCA, Alex.

And good luck all the USAMOers. Tomorrow and Wednesday will be INSANE.[/quote]
It's actually not swine.

2 HOURS LEFT. And I'm doing physics homework. <_<

EDIT: oops. I forgot to change summertime to "yes".  :oops: