calculus MIT college function search topology geometry

I was searching for some books on Calculus with basically the following features :-

1) It should [b]not be too basic and trivial [/b]with questions not just on simple computations and just one ot two good questions per exercise.

2)It should cover all the theory from [b]basic to advanced level[/b] lower than real analysis. i.e. it should not cover topics like metric spaces, or completeness properties of real nos, advancd concepts of series etc(basically it should [b]not be like Water rudin's [/b]book on mathematical analysis)

3)It should have [b]tough theoretical problems [/b]as well as interesting and [b]difficult problems [/b]on limits etc. Basically, it should have more problems in general terms which are tough and considerably difficult.

So basically it should not be a good book. ;)

I knew somebody would remark bluntly to my "guidelines". Please don't be offended. I just wrote in such a way to make things cxlear. Actually, the "simple" kind of books I mentioned above are easily available in India. I was wondering whether "Intermediate" books are there???

Still no suggestions. Again, I plead yo to help me. Please suggest some good books(which basically have the above characteristics) or some websites etc(like MIT OCW) where I can get the pertibebt material.  :help:   :read:  :stretcher:

[quote="Rushil"]Is nobody ready to help me?????   :(  :help:  :juggle:  :jump:[/quote]
Try following
[1] Author ? . Hobson (2 volumes) ,   ( title =,, Real Functions and Fourier series" ?? ) [an old book,  but interesting]
[2] R.Boas ,, A primer of real Functions"
[3] Sz.Nagy ,, Introduction to rela functions" [very nice ! , clear exposition ]
Perhaps, help

I don't know if this also would fall under the "Not available in India and/or too expensive" category, but there's a two volume text by Apostol that might meet your criteria.  It does cover some "analysis" topics like completeness of the real numbers, but only so far as they are necessary to provide proofs of the main theorems of calculus, and as I recalled it had a nice mix of theoretical and computational stuff.

You can "search inside the book" on Amazon for each volume, so that would probably be the best way to get a feel for what the text is like.

Yes, I have heard about apostol and fortunately it is easily available. Do you have any idea aboyt the level of problems in this book???

[quote="Rushil"]I checked that they are either too old books or published by MAA - in both circumstances the books are NOt avaialble in India or tooo expensive!!!!
I'm sorry, but I'll have to ask for easily available web resources!!! I think that'sthe only thing I can get in India. the standard calculus books here are worthless. The net is my best source!!![/quote]
Here are most exact references of some books which I like :
[1][b]Tom M. APOSTOL[/b], [i]Calculus (vol.I-II)[/i], Second Edition, John Wiley & Sons,New York,1967,1968.
[2] [b]Ralph P. BOAS [/b], [i]A Primer of Real Functions[/i], Published by MAA, John Wiley and Sons,The Carus  Mathematical Monographs,1960.
[3] [b]E.W. HOBSON[/b], [i]The Theory of functions of a Real variable & The Theory of Fourier's series, Vol.I-II, [/i], Dover Publications ,New York,Vol. I, Third Edition 1927,  Vol. II, Second edition 1926.
[4] [b]Bela  SZ.-NAGY[/b] , [i]Introduction to Real Functions and Orthogonal Expansions[/i], Akademiai Kiado, Budapest, 1964.
[5][b]Georg POLYA[/b] and [b]Gabor SZEGO[/b], [i]Aufgaben und Lehrsaetze aus der Analysis (Band I-II),[/i]
Springer-Verlag Berlin ,Heidelberger Taschenbuecher, 4. Auflage,1970-1971, (there is also available an English translation).
[6][b]David v. WIDDER[/b], [i] Advanced Calculus [/i], Prentice-Hall,Inc.,New York 1947.
[7] [b] E.T.WHITTAKER [/b] and [b] G.N.WATSON [/b], [i]A Course of Modern Analysis [/i], Cambridge University Press, Fourth Edition (reprinted),1969.
[8] [b] Robert WREDE [/b] and [b] Murray R. SPIEGEL [/b], [i] Theory and Problems of Advanced Calculus[/i], McGraw-Hill, 2002.
Supplementary:
[1][b] Casper GOFFMAN, Togo NISHIURA[/b]and [b]Daniel WATERMAN[/b], [i]Homeomorphisms in Analysis[/i],
 Amer.Math.Soc., Mathematical Surveys and Monographs, vol. 54, 1998. (Higher level, but very nice !)

[b]Please send me, by e-mail,  your postal address ! [/b]  Regards,Alex

[quote="Rushil"]Why do you need my postal address???[/quote]
I intend  to send you a CD with some works which perhaps are of   interest for you.

That's an excellent list. Have listed them in order of best(according to you...)??Which are the best ones?? I'll have to check which are available in India. Anyways, thanks!!

Also, we are right now on the move and shifting as my father has been transferred.(he's a government officer). I'd appreciate if you can put the material on the net in parts. That will be the best possible scenario for I also live in india and the cost factor of posting would also come in. Again, it will be best if you can either e-mail that stuff to me!!!

My two cents, guys, in this discussion (btw, I syudied hindi at school for the first 5 years, then moved to another school though :)).

Apostol is an interesting book, but from my point of view absolutely  unsuitable neither for serious nor for the intial studying of analysis. Well, if's funny and maybe one who knows the subject quite good can benefit from reading it, but it's no more than that. From the list above I think "Watson, Whitteker - A course of modern analysis" is ok. At least it's kinda modern. Well, frankly speaking, I don't know an analysis course which was really good. In russian there is a Zorich's book (2 volumes) that is not bad, but again imho there is no good book on analysis in the world. Try to choose the ones with geometrical attitude and may be read some books on differential geometry and differential equation as well, but only the ones with a lot of pictures.

Thx,

[quote="stan"]Well, frankly speaking, I don't know an analysis course which was really good. In russian there is a Zorich's book (2 volumes) that is not bad, but again imho there is no good book on analysis in the world.
 Try to choose the ones with geometrical attitude...Thx,[/quote]
What about Smirnov or the old volumes by Fichtenholtz ? There are also two ,,good" old books by Edwards :
"Differential Calculus" and/or "Integral Calculus" , 2 volumes. Likewise,  I appreciate very much  the books by Hobson or by Widder (e.g. see D.Widder  "Laplace Transform" =very instructive).

Hey flip2004, what's that stuff you were talking about. Can you send them to me as a mail or post them here plz???

Man, you know Fichtenholtz's course?? Cool, I din't know it was translated. Well, it's a bit too large and it's really boring. What I'd expect from a good analysis course: no real number theory (or let it be in a afterword as something that will never be used), early introduction of the complex analysis and riemann surfaces and their relation to integrability of the functions, series etc., early intorduction of manifolds and intuitive ideas of differential geometry and topology (homology, etc.), good physics reference, i.e. not only the statements like the velocity is derivative, but also an elementary (level of the general physics course) ideas of work, power, oscillations, elect. and magnetic fields. I'd also like to have a good vector analysis and differential forms reference. I don't see any reason to treat sequences separately from other functions and I think that measure theory can be read on theory of probabilities courses in the amounts needed there, and so on and so forth. Well, and the most important thing this book should be relatively small, with a lot of pictures and practical examples and written for mathematicians, i.e. with very high requirements to the intellect of the readers. :)

M.Spivak, Calculus, Cambridge University Press
N.L.Carothers, Real Analysis, Cambridge Univeristy Press

A Course in Pure Mathematics by G.H. Hardy, the famous number theorist is a good book and can be downloaded for free from the internet archive. However, it is a bit old, and it is better if it were supplemented by a some newer book such as Apostol's two volume calculus (who also happens to be a number theorist!)

Hardy's book is Classic. But not for an introduction is too heavy...

I have the books by Apostol, and it sounds like they would be right up your alley.  They're significantly more challenging and deep than the books used at my university, but not nearly so much as Rudin.  They almost blur the line between introductory calculus and analysis.  The books are used in the math course that Caltech students take during their first year.  Skipping the course is discouraged, and that's for a student body the majority of which enters with 800's on the Math SAT and 5's on the Calculus AP test, so it is a challenging book.  I don't think it would be too difficult to be an introduction to calculus for talented, motivated students, however.  The book by Spivak is generally regarded as being of equal quality and to be about equivalent in terms of difficulty, so you might want to check that one out as well, but I've never used it.

I should note that I only began to work through Spivak's book before deciding to go through Rudin instead, so I don't know the book extremely well, and I haven't known anyone else who's gone through the book, so I don't know for sure whether or not it would be too much.  I can say for sure that it won't be too little, however, and that most of the students at my university, even the ones who take the honors calculus courses, would struggle with this text, but it sounds like your course is probably meant to be at a higher level than the ones at my university.

@steaxauce
He posted the question 4 years ago.I think he is now teaching calculus. :rotfl:


Hi

I am new so I wanted to know who all have gone to 12th this time and will be giving JEE in 2009.

I am. Arijit(arijit91) and Sanjith(thealchemist) are as well.

me tooo(gone 2 12th)--   but  i m not an aspirant 4 jee

@rituraj are you not writing JEE?

i'm writing JEE 2009 :)

So howz it going ppl. I guess you ppl must be through with every topic unlike me I have hardly done anything except Mechanics,wave,thermo excluding heat ,Co-geo,some algebra,trig eq.  :D  :blush: and ofc nothing in chem except redox/mole ,gas and bonding. :)

What about u :maybe:

ya iam mukunth from PS. Sr , IIT-2009 aspirant ~

[quote="anandghegde"]@rituraj are you not writing JEE?
[/quote]
no
& i mayyy not b writing JEE 2009 as well :oops:

ofc i have not even started yet

do u have any plans of writing? or are you not interested in engg? :)

hey how do i quote someone? :maybe:

And how do i hide some text if i want to? :maybe:

every post has a "quote" option at end and u can use backspace in the post to hide...... :blush:

i m a medical student :(  :(

u kno all that mugging up and trbls like that  :P
but doesn't matter-in the end i love it :D  :D

JEE 2010 FOR ME ALONG WITH NNNNNNNNNNNNNNN

OMG.......

they are going to give JEE 2009 :o
i thought them to have given JEE 2008


Let $ ABCD$ be a quadrilateral. The lines $ a,b,c$ passing through $ D$ are perpendicular to the lines $ DA,DB,DC$ respectively. The lines $ a,b,c$ intersect $ BC,CA,AB$ at $ A',B',C'$ respectively. Prove that $ A', B'$ and $ C'$ are collinear. :roll:

[quote="forevercontinue"][quote="mihai miculita"]$ \mbox{The points A', B', C' is in the Simson line of the point D respect to triangle ABC.}$
[b]view: [/b] [i]Nathan Altshiller-Court[/i], [b]COLLEGE GEOMETRY[/b], sec.ed. p.140[/quote]
mihai miculita,thanks a lot but can you tell me how can I got this book , "College Geometry" .  :P
I 've heard about it before but I don't know where can I find it ? :([/quote]

It is on Dover aditions  and in a very nice price. Amazon also provides it !

College Geometry: An Introduction to the Modern Geometry of the Triangle and the Circle (Dover Books on Mathematics) (Paperback)
by Nathan Altshiller-Court (Author)

 Serch at  http://www.amazon.com
--------------------------------------------------------------------------------

  Babis


Let $ k\leq n^2\plus{}1$. Find all $ (m,n)$ natural number such that $ m|n$, $ 2007m\minus{}1|2007n\minus{}1$ and $ m\equal{}2006k\plus{}1$

whoa man, you really have a lot of tasks for us mathgoers...every night a new batch. where, pray tell, do they flow from?

Are you sure there isn't a typo?

For any integer $ m$, $ n$ satisfies the conditions if and only if
$ n \equiv 0\mod{m}$
$ n \equiv \frac{1}{2007}\mod{2007m\minus{}1}$
By the Chinese remainder Theorem, this holds true if and only if
$ n \equiv m\mod m(2007m\minus{}1)$...

[quote="K81o7"]Are you sure there isn't a typo?

For any integer $ m$, $ n$ satisfies the conditions if and only if
$ n \equiv 0\mod{m}$
$ n \equiv \frac {1}{2007}\mod{2007m - 1}$
By the Chinese remainder Theorem, this holds true if and only if
$ n \equiv m\mod m(2007m - 1)$...[/quote]
Other represent:
Let $ m=kn$
So $ 2007m-1|2007n-1$
$ \equiv alent 2007m-1|2007km-1$
$ equivalent 2007m-1|2007k-1$
Because $ \gcd(2007m-1,2007)=1$ so for each n exist k such that
$ 2007m-1|2007k-1$


Two factors, each with no digit greater than 5, have a product of 16,848. What is the largest possible 3-digit factor satisfying these
conditions?

Factoring gives $ 2^4\cdot3^4\cdot13$. A little thinking tells us that the units digits have to be $ 4$ and $ 2$, and also our smaller factor has to be at least $ \frac{16848}{555}\approx31$, but also keep in mind that it can't be more than $ 55$. So doing stuff we find the answer is $ \boxed{324}$. Sorry for being vague but it's basically brute force on the smaller number.


probability topology Gauss Princeton college

Which are your favourite math books?
(Suggest me some good statistic and probability books,too)

You have a book?

Edit:

Found it,„Olimpiada de matematica… De la provocare la experienta”
Nice!

Edit 2:

You could give us the book as a gift,for being fellow members  :D

[quote="grobber"]How about some books to get me started in topology? I have the feeling I'd like it, but no knowledge :D.[/quote] just don't book you tickets yet to that :D

Valentin, is your book available in english?

PiDelta, In what languages are you looking for books? A classic book in probability theory is W. Feller's [u]An Introduction to Probability Theory and It's Applications[/u]. I also like the book [u]Probability: A First Course[/u] by Frederick Mosteller.

A good bibliography of books in different subjects can be found here: http://www.ericweisstein.com/encyclopedias/books/

Valentin, what makes you say that? Topology is one of the most fascinating branches of mathematics.

A classic book in topology is [u]Topology[/u] by James Munkres. But a good introduction to the foundations of topology is a very inexpensive paperback book called [u]Elementary Topology[/u] by Michael Gemignani.

Thx gauss for the reference.I need the books in romanian,german or english (even french)  :)

Edit:Damn,Valentin ignored the "give me a book for free NOW"-request,its such shame  :?

[quote="gauss202"]Valentin, is your book available in english?

Valentin, what makes you say that? Topology is one of the most fascinating branches of mathematics. [/quote]

it may be fascinating at Princeton, but here at Univ. Bucharest it is not :? most probably because of the teachers :(

No, my book is not (yet) available on English. However I can give you a link with an online Romanian -> English dictionary :P just kiddin'

[quote="Valentin Vornicu"][quote="PiDeltaPhi"]Damn, Valentin ignored the "give me a book for free NOW"-request,its such shame  :?[/quote] I'll give you one, but I don't plan on a trip to Germany pretty soon, so if you want it, come and get it first hand. :D:D

The closest I'll get to Germany in the next couple of months will probably be Prague, at the Ostrawa contest for University students.[/quote]

I will try to get it first hand,or i will wire you the shipping costs  :D

[quote="PiDeltaPhi"]You have a book?
Found it „Olimpiada de matematica… De la provocare la experienta”
Nice!
[/quote]
what is "De la provocare la experienta"?

Hmmm... I imagine it's not possible to get it outside of Romania? My brother speaks Romanian... though he probably doesn't have the math vocabulary...

of course it is possible. I will make the arrangements for all of you people who want it.

please give me a couple of days to figure it out ok?

Hehe :-)

Anyway, what's the book like? Is it a math book, or is it more of a... I don't know... story book.

I.e., would it be worth it for me to get it? I can understand, for example, a lot of Russian math books even though I don't speak much Russian. But if it's a not just a math book, I'd probably not be able to understand a thing.

[quote="Valiowk"]I think any English book that Valentin writes will become an instant bestseller (that is, 400 orders straight away :P ).  Guess how I got the number? ;)[/quote] hopefully we won't have to wait long to check if this is correct :lol:

anyway the book deals with math preparation for the olympiad.
as I am still participating at the olympiads in one way or another, correcting papers, proposing problems, instructing the team, etc. I cannot yet write one about the olympiads :D but I will do it later. hopefully I won't forget all the fun stuff that happened :D.

so I guess you could read it :D.


can you find :D

$ \sum_{k\equal{}0}^\infty\;\dfrac{1}{k^2\plus{}4k\plus{}8}$

[hide="Possible argument"]$ S \equal{} \sum_{k\equal{}0}^{\plus{} \infty}\frac{1}{k^2 \plus{} 4k \plus{} 8} \equal{} \sum_{k\equal{}0}^{\plus{} \infty}\frac{1}{(k\plus{}2)^2 \plus{} 2^2}$.

Now let k + 2 = n, and so $ S \equal{} \sum_{n\equal{}2}^{\plus{} \infty}\frac{1}{n^2 \plus{} 2^2}$. Now, from the application of residues we have

$ \sum_{n\equal{}1}^{\plus{} \infty}\frac{1}{n^2 \plus{} a^2} \equal{} \frac{\pi a \coth(\pi a) \minus{} 1}{2a^2},\,\,a > 0$. Hence:

$ S \equal{} \frac{2 \pi \coth(2 \pi) \minus{} 1}{2 \cdot 2^2} \minus{} \frac{1}{5} \equal{} \frac{10 \pi \coth(2 \pi) \minus{} 13}{40}$.[/hide]

[quote="Carcul"]

$ \sum_{n \equal{} 1}^{ \plus{} \infty}\frac {1}{n^2 \plus{} a^2} \equal{} \frac {\pi a \coth(\pi a) \minus{} 1}{2a^2},\,\,a > 0$. [/quote]
as I know this is not so obviusly, How many proof is known of this ?

That simply follows from this result:

$ \sum_{\minus{} \infty}^{\plus{} \infty}f(n) \equal{} \minus{}[sum\,\,of\,\,residues\,\,of\,\,\pi \cot(\pi z)f(z)\,\,in\,\,the\,\,poles\,\,of\,\,f(z)]$

as long as f(z) is such that, along a square with vertices at (n + 1/2)(1 + i), (n + 1/2)(-1 + i), (n + 1/2)(-1 - i), and (n + 1/2)(1 - i), it verifies $ |f(z)| \leq \frac{M}{|z|^k}$, with k and M constants.


superior algebra
superior algebra unsolved

Suppose $ F$ is a finite field of characteristic 3 and $ \phi : F \rightarrow F$ is the map defined by $ \phi(a)\equal{}a^3$. Prove that $ \phi$ is surjective.

How do I do this? I know how to prove that it is an injective ring homomorphism. But I don't know how to prove it is surjective. Thanks.

If a map from a finite set to itself is injective, what else can you say? :wink:


Let be given $n$ real numbers $t_1,t_2,...,t_n$ such that: $\sum_{i=1}^nt_i=0$
Prove that:
$\forall$ sets of $n$ real numbers $x_1,x_2,...,x_n$ we have:
$\sum_{i,j=1}^n t_it_j|x_i-x_j|\le 0$

Adding a number to each $x_i$ doesn't change the summation in the problem, so we may assume that $x_i>0$ for all $i=1,\ldots,n$.

The inequality follows from:

(1) $|x_i-x_j| = (x_i+x_j) - 2\min\{x_i,x_j\}$;

(2) $\sum_{i,j=1}^n t_it_j(x_i+x_j) = 0$;

(3) $\sum_{i,j=1}^n t_it_j\min\{x_i,x_j\}\ge0$;

___________________________________________

(1) is trivial.

(2) $\sum_{i,j=1}^n t_it_j(x_i+x_j) = 2 \big(\sum_{i=1}^nt_i\big)\big(\sum_{i=1}^nt_ix_i\big)=0$

(3) You can  find a proof  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=USAMO&t=15292]here[/url] (see the lemma in Ravi's post).


Find the sum of the squares of the roots of the equation:
$ 14x^{5}\minus{}63x^{4}\minus{}147x^{3}\plus{}581x^{2}\plus{}21x\minus{}630\equal{}0$

Need some help, not an answer, but the solution.

[hide="hint"]how would you find the sum of squares of the roots of the polynomial $ ax^2\plus{}bx\plus{}c$?[/hide]

Try [u]Newton's Sums[/u]. These give a direct answer to the problem you're working on, and generalize really well.

[hide="Newton Sums"]Using Newton Sums, we have $ a_nS_1 \plus{} a_{n \minus{} 1} \equal{} 14S_1 \minus{} 63 \equal{} 0$, so $ S_1 \equal{} \frac92$. Now, $ a_nS_2 \plus{} a_{n \minus{} 1}S_1 \plus{} 2a_{n \minus{} 2} \equal{} 14S_2 \minus{} 63(\frac92) \minus{} 2(147) \equal{} 0$, so $ S_2 \equal{} \boxed{\frac {165}{4}}$.[/hide]

Note: $ S_k$ denotes the $ \displaystyle\sum_{i \equal{} 1}^{5}r_i^k$.

Edit: fixed arithmetic.

[hide] I got $ \frac{165}{4}$

That factors to $ 7(x\minus{}5)(x\minus{}2)(x\plus{}1)(x\plus{}3)(2x\minus{}3)$

so $ 25 \plus{} 4 \plus{} 1 \plus{} 9 \plus{} \frac{9}{4} \equal{} \frac{165}{4}$
[/hide]

How did you factor that? Maybe there is something wrong with your factorization.
[hide="I agree with perfect628"]From Newton's Sums, we have $ 14s_1\minus{}63\equal{}0\rightarrow{s_1\equal{}\frac{9}{2}}$. Now, $ 14s_2\minus{}\frac{861}{2}\minus{}147\equal{}0\rightarrow{s_2\equal{}\frac{123}{4}}$.[/hide]

Newton's Sums is a better approach than factoring.
[url=http://www.artofproblemsolving.com/Wiki/index.php/Newton%27s_Sums]Newton's Sums[/url]

No I'm pretty sure that's the factorization, I used a lot of synthetic division to get it and checked it with Maxima.

Edit: I got the same solution with Newton Sums as with factoring check your arithmetic.

[hide]$ 14s_{1} \minus{} 63 \equal{} 0$
$ s_{1} \equal{} \frac {9}{2}$
$ 14s_{2} \minus{} 63s_{1} \plus{} 2( \minus{} 147) \equal{} 0$
$ 14s_{2} \equal{} \frac {1155}{2}$
and $ s_{2} \equal{} \frac {165}{4}$

Where are you getting the $ \frac {861}{2}$ from?

Shouldn't it be $ \frac {567}{2}$?

It should also be $ \minus{} 294$ instead of $ \minus{} 147$ because Newton Sums is $ a_nS_2 \plus{} a_{n \minus{} 1}S_1 \plus{} 2a_{n \minus{} 2} \equal{} 0$
[/hide]

Thanks for showing me an easier way to solve it!   :) Took a lot less time then all that synthetic division.


The Seat Shop makes 3-legged stools and 4-legged chairs, and each requires one seat. One day they used 6 seats and 20 legs. How many stools did they make?

If they had all chairs, there would be 24 legs, exceeding the amount in reality by 4. The difference between the number of legs in a chair and a stool is 1;divide the exceeded amount of four by that 1, and you find that there were 4 stools.

If they used all stools they have 2 extra legs. Thus they made 2 chairs so 4 stools.

You could set up a system of equations:

$ a\plus{}b\equal{}6$
$ 3a\plus{}4b\equal{}20$

And solve from there.  You get a=4 and b=2, so the answer is 4.


The numbers -2, 4, 6, 9 and 12 are rearranged according to these rules: \\
1. The largest isn't first, but it is in one of the first three places. \\
2. The smallest isn't last, but it is in one of the last three places. \\
3. The median isn't first or last. \\ \\
What is the average of the first and last numbers?  Express your answer as a decimal to the nearest tenth.

Well, -2, 12, and 6 aren't in either the first or last places.  So the answer is $ \frac{4\plus{}9}{2}$, or $ \boxed{6.5}$.


inequalities
integration
real analysis
real analysis unsolved

Dear everyone.

please read the following

$ f(x)$ is continuous on [0,1] and $ f(0) \equal{} 0$.

$ f'(x)\geq{1}$ on (0,1)

prove that

$ \left(\int_{0}^{1}{f(x)dx}\right)^{2}\leq\int_{0}^{1}{\{f(x)\}^{3}dx }$.

thanks for reading. :)


Is there a perfect square with exactly 900 digits that contains exactly 300 digits 0, 300 digits 1 and 300 digits 2? :wacko:

The answer is no.

A perfect square is always congruent to $ 0,1,4$ modulus $ 8$.

a number ending in $ 001$ is also congurent 1 to modulo 8... so this does not proove anything.

This problem is from the book mathematical circles (Russian experience). This book doesn’t contain complicated problems, and most of them are very easy. but this problem doesn’t seem so easy.


symmetry
LaTeX
number theory open
number theory

Look, this is a waste of everybody's time.  Thinking that a proof this short exists that does nothing nontrivial is an insult to the generations of mathematicians who worked on this problem for 350 years without solving it.  "Proofs" of this nature [b]almost certainly[/b] contain unjustified assumptions or hidden algebraic mistakes, and there is no reason to pay them any attention.

Without reading all of it, it at most shows that $ p|xyz$ (by reading the last line). In fact, I doubt that, too (there are some steps that are pretty hard to follow: it's written very badly). By a one-minute look at it:
Where does [18] come from and where does that symmetry between [22] and [23] come from¿

For my total oppinion, read the post above this one  :wink:

Sorry, guys, but I don't understand your reactions. Don't your math teachers/profs ask you sometimes to work on false proofs in order to sharpen your proof-methods. My prof does that from time to time, and sometimes I spend a lot of time to find the flaws, but I do benefit greatly.

If I went against the rules of this forum, then please let me know, and I won't do it again.

Cheers, and Happy Holidays

It's much easier to do proof-reading if it's written out well (with every step explained). Otherwise it's just a pain in the neck.
And I already gave some objections...

of course the proof is correct. he ended it with QED. clearly he knows what he is talking about  :)

[quote="Jean-Louis"]http://groups.google.com/group/fermats-last-theorem-flt-2009[/quote]
Could you write it in LaTeX, or something else readable?

If I follow his thinking correctly, by the time [17] and [18] are reached, the factor k has been eliminated, so that z = 1 (mod p). I don't think the assumption that y = x implies y = z = -x is correct, but y = x implies y = z = x, so that x + y = 2z = z (mod p).


Eight first graders, 4 girls and 4 boys, arrange themselves at random around a merry-go-around.  What is the probability that boys and girls will be seated alternatively?

This is what I have:
probability of the first person is a girl= $\frac{1}{2}$
probability of the second person is a boy= $\frac{1}{2}$
probability of the third person is a girl= $\frac{3}{8}$
probability of the third person is a boy= $\frac{3}{8}$.... and so on.
I multiply the probabilities together:
$\frac{1}{2}*\frac{1}{2}*\frac{3}{8}*\frac{3}{8}*\frac{2}{8}*\frac{2}{8}*\frac{1}{8}*\frac{1}{8}=\frac{9}{8^{6}}$

The answer is however: [hide]$\frac{1}{35}$[/hide]  i understand how to get the answer, but i don't understand why my way is wrong.  Can anyone tell me what I did wrong?

Thank you so much :)

Eight first graders, 4 girls and 4 boys, arrange themselves at random around a merry-go-around. What is the probability that boys and girls will be seated alternatively?

Ok. I believe you forgot that once one boy/girl is seated on the merry-go-round, there are 7 left, then 6, and so on.

So,

Say we go girl, boy, girl, boy...

1. 1 (girl: It can actually be either. So there is a 100% chance of a kid getting onto the merryground)
2. 4/7 (boy)
3. 3/6 = 1/2 (girl)
4. 3/5  (boy)
5. 2/4 = 1/2 (girl)
6. 2/3 (boy)
7. 1/2 (girl)
8. 1 boy

4/7 x 1/2 x 3/5 x 1/2 x 2/3 x 1/2 = 1/35


Prove that if $a,b,c$ are non-negative real numbers such that $ab+bc+ca=1$, then $18abc\le \sqrt{1+a^2}+\sqrt{1+b^2}+\sqrt{1+c^2}\le 2(a+b+c)$

[quote="Beat"]Why $\frac{a+b}{2}\cdot\frac{b+c}{2}\cdot\frac{c+a}{2}\ge \frac{a+b+c}{3}\cdot\frac{ab+bc+ca}{3}$[/quote]

$9(a+b)(b+c)(c+a) \geq 8 (a+b+c)(ab+bc+ca)$
But $9(a+b)(b+c)(c+a)-8(a+b+c)(ab+bc+ca)= \sum b(a-c)^2 \geq 0$ :)


Να βρείτε όλους τους πρώτους p,q για τους οποίους ισχύει $ a^{3pq}\equiv a(mod 3pq)$ για όλους τους ακεραίους a.

[quote="maniopas"]Βασικά αυτό δεν είναι το μικρό θεώρημα του Fermat στη γενική του έκφραση;[/quote]

Εγώ ξέρω ότι το μικρό θεώρημα του Fermat αποτελεί ειδική περίπτωση του θεωρήματος του Euler σύμφωνα με το οποίο αν $ gcd(a,n)\equal{}1$ τότε $ a^{\phi(n)}\equiv1(modn)$ όπου $ \phi(n)$ παριστάνει την συνάρτηση euler.Θέτοντας τώρα όπου n έναν πρώτο έστω p προκύπτει το  μικρό θεώρημα του Fermat
Εσύ τώρα όταν λές ότι η άσκηση αποτελεί γενίκευση τη εννοείς? Εξάλλου το πρόβλημα  ζητάει να βρεις συγκεκριμένες τιμές για τους πρώτους p,q για τους οποίους ισχύει :wink:


geometry
rectangle
symmetry
integration
vector
3D geometry
sphere

nope sorry the answer is wrong
check it again
i got a pretty boring vector solution

why i don't find a hell flaw with my logic :mad:

i double checked my method...do check urs i think i am right :wink:

p lies on which axes
i got the ans as 3
not 2
and the sorce also tells the answer to be 3 not 2

so the Answer is 2 and [b] not[/b] 3...
Anit confirmed it :D

[quote="pardesi"][hide="solution"]
Let the axis be called $ A$.Consider another axis similarly inclined at $ \frac { \minus{} \pi}{4}$ with the plane called $ A^{'}$ .Hence these two axes are perpendicular .
Consider any point $ P$ on the ring now drop perpendiculars from that point $ P$ to both the axes at $ L$ and $ M$ say of  length $ a$ and $ b$
now the quadrilateral $ POLM$ is a rectangle where $ O$ is the centre of the ring and $ a^{2} \plus{} b^{2} \equal{} R^{2}$
so by symmetry both axis give same MOI so
$ 2I \equal{} I \plus{} I^{'} \equal{} \int dm a^{2} \plus{} \int dm b^{2} \equal{} \int dm R^{2} \equal{} MR^{2}$ so
$ \boxed{I \equal{} \frac {MR^{2}}{2}}$
so the answer is $ 2$ $ Kgm^{2}$
[/hide][/quote]

Nice method da... I was thinking more along the lines of the 3rd axis for the perpendicular axis theorem. :D

BTW you should try deriving M.O.I for a Solid Sphere. Its very interesting, I guarentee.

[quote="pardesi"]thanks
i have already done btw it's for a shell not sphere using symmetry considerations[/quote]

Errr... No I was referring to MI of a solid sphere da.. I believe you have to substitute radius of gyration of hollow at a point, or something like that.. Kinda forgot.

[quote="pardesi"]
now the quadrilateral $ POLM$ is a rectangle where $ O$ is the centre of the ring and $ a^{2} \plus{} b^{2} \equal{} R^{2}$

[/quote]

Correct me if I'm wrong but I'm getting $ a^2 \equal{} \frac{R^2}{2}$ by considering the triangle with the sides OP, OM, PM.

no that isn't sight why shud $ OP\equal{}OM$ :?: ...OPLM is just a rectangle

[quote="pardesi"]no that isn't sight why shud $ OP \equal{} OM$ :?: ...OPLM is just a rectangle[/quote]

Turns out that it is... Dont blame me.. The angle that OP makes with OL is 45 degrees. Because OL being the axis makes 45 with normal to the plane.

And can you exp how else you wrote $ a^2 \plus{} b^2 \equal{} R^2$

[quote="pardesi"]of it turns out then ok...but i wrote it considering the rectangle POLM  :wink:[/quote]

I think it will always be R/Root(2) if you consider the different triangles from a different plane. :D

I still dont understand how you got a2+b2=R2. Explain da.

ok consider the reactangle PLOM .In this we have $ PL\equal{}a$ and $ PM\equal{}b$ clearly both axes are perpendicular to each other because they make $ \frac{\pi}{4}$ with the normal.
so we have three hence all the four angles of $ PLOM$ $ \frac{\pi}{2}$ so it's a rectangle now
$ PM\equal{}LO$ then because it's a rectangle
hence we have $ PM^{2}\plus{}PL^{2}\equal{}PM^{2}\plus{}LO^{2}\equal{}PO^{2}\equal{}R^{2}$ by pythagoras  :wink:

Pardesi, the answer is $ \frac {3} {4} MR^2$
This was got by no new techniques and simple integration in 3d geo.. The mistake with your logic is that everytime you change your point of consideration, your axis changes, so your integral with $ b^2$ will not be abt one axis,but will be sum of $ b^2 's$ abt many axis which has no significance of a moment of inertia..

firstly my answer is right anit pmed me :wink:

secondly
there is no flaw in my logic..ofc one of the axes is fixed and the other traces out a cone of semivertical angle $ \frac{\pi}{4}$ because i have  considered a rectangle in a plane....but that's not a worry because what i am basically doing is just pairing of the points that is
something like the gaussian logic
$ \sum_{i\equal{}1}^{100} i \equal{}S \Rightarrow \sum_{i\equal{}100}^{1} i\equal{}S$ thsu $ 2S\equal{}\sum_{i\equal{}1}^{100}101$
what i have done is just added each element with it's copartner on the other side whose distance from the axis is $ b$ which si such taht $ a^{2} \plus{} b^{2}\equal{}R^{2}$ :wink:

[quote="pardesi"]firstly my answer is right anit pmed me :wink:

secondly
there is no flaw in my logic..ofc one of the axes is fixed and the other traces out a cone of semivertical angle $ \frac {\pi}{4}$ because i have  considered a rectangle in a plane....but that's not a worry because what i am basically doing is just pairing of the points that is
something like the gaussian logic
$ \sum_{i \equal{} 1}^{100} i \equal{} S \Rightarrow \sum_{i \equal{} 100}^{1} i \equal{} S$ thsu $ 2S \equal{} \sum_{i \equal{} 1}^{100}101$
what i have done is just added each element with it's copartner on the other side whose distance from the axis is $ b$ which si such taht $ a^{2} \plus{} b^{2} \equal{} R^{2}$ :wink:[/quote]

Firstly anits source maybe wrong.

Secondly my doubt above was wrong. Not all lines make 45 degrees with radius vector. This holds good for only 90.

Thirdly What chappli means to say is that if you consider the Quad OPLM it may not be a rectangle for all radius vectors when viewed from 3 D space. He says the angle between the 2 axis may not be 90 when viewed in a diff plane.

[quote="Cyber-Shreyas"]

Firstly anits source maybe wrong.

Secondly my doubt above was wrong. Not all lines make 45 degrees with radius vector. This holds good for only 90.

Thirdly What chappli means to say is that if you consider the Quad OPLM it may not be a rectangle for all radius vectors when viewed from 3 D space. He says the angle between the 2 axis may not be 90 when viewed in a diff plane.[/quote]
no comments on anit; source :P
secondly i answwered chappli's doubt already u have to change the other axis so that the quadrilateral lies in single plane and such that its a rectangle
 i wrote it above see the locus is a cone
further what i am basically doing is just pairing off points nothing else :)

$ 2I \equal{} I \plus{} I^{'} \equal{} \int dm a^{2} \plus{} \int dm b^{2} \equal{} \int dm R^{2} \equal{} MR^{2}$

How valid is this step, when one of the axis itself keeps on changing?

well that was a short form with essence the same....
ok if u r confused with those notation...what i have actually shown is to every point $ P$ there exists a unique point $ P^{'}$ on the other side such that $ a^{2}\plus{}b^{2}\equal{}R^{2}$...i used $ I^{'}$ to prove such a point exists because both the aix are same under reflection so $ P^{'}$ is nothing but the point under reflection.of $ P$...

[quote="pardesi"]well that was a short form with essence the same....
ok if u r confused with those notation...what i have actually shown is to every point $ P$ there exists a unique point $ P^{'}$ on the other side such that $ a^{2} \plus{} b^{2} \equal{} R^{2}$...i used $ I^{'}$ to prove such a point exists because both the aix are same under reflection so $ P^{'}$ is nothing but the point under reflection.of $ P$...[/quote]

Agreed... According to you $ dI \plus{} dI' \equal{} dma^2 \plus{} dmb^2$

Lets come from this step. Now, you can equate I and I'. But you cant do that to dI and dI'. Moreoever since the second axis keeps on varying how can you say that $ \int dI' \equal{} I'$???

no no that's not the point...
what i am telling is what point $ P$ is to $ I$ $ P^{'}$ is to $ I$...so since $ P$ is at a distance of $ b$ from $ I^{'}$ ...$ P^{'}$ will be at a distance of $ b$ from $ I$
now let $ I\equal{}\int_{clockwise} dm a^{2}\equal{}\int_{anticlockwise} dm b^{2}$ so $ 2I\equal{}\int (a^{2}\plus{}b^{2})dm\equal{}MR^{2}$
so the result


Is it possible to have an element $ a$ in a ring s.t. there exists $ b,c$, $ c$ not null with $ ab\equal{}1$ and $ ac\equal{}0$?

yes, it's possible.
consider the ring of bounded operators on $ \ell^2(\mathbb{R})$, and the three operators:
$ L(x_1,x_2,\dots) \equal{} (0,x_1,x_2,\dots)$, $ R(x_1,x_2,\dots)\equal{}(x_2,x_3,\dots)$, $ P(x_1,x_2,\dots) \equal{} (x_1,0,0,\dots)$.
you can check that $ RL \equal{} 1$, $ PL\equal{}0$.

@ma_go: $ B(l^2(N))$ oder simply $ End(R^N)$.

sure, no need to take out any measure theory.. of course i meant $ \ell^2(\mathbb{N})$, anyway, and it's shorter than saying "sequences of real numbers :p


geometry
inequalities unsolved
inequalities

What is the largest area of a square that can be covered with two circles with radii 1/2


Find all functions $f:N \rightarrow N$ such that
$f(m+n)f(m-n) = 2f(2m)$
for all natural numbers $m,n$ such that $m \geq n+1$


Fie $f: R\to R$ o functie continua cu proprietatea ca $\lim_{x\to\infty}f(x)$ si $\lim_{x\to - \infty}f(x)$ exista si sunt egale. Aratati ca $\forall d >0$ exista $x_1,x_2$ astfel incat $x_1-x_2=d$ si $f(x_1)=f(x_2)$.

Problema e clasica. Considera functia $g(x)=f(x+d)-f(x)$. Daca $g(x)$ nu se anuleaza, atunci are semn constant. Sa alegem $g(x)>0$ pt. orice $x$. Atunci $g(0)>0,g(-d)>0,$ deci $f(d)>f(0)>f(-d)$. Se obtine usor $f(nd)>f(d)>f(0)>f(-d)>f(-nd)$, pentru orice $n$. Prin trecere la limita se obtine contradictia.


[quote="MJ GEO"]there is a A that F(A)=1 find all function from N TO N that satsfiy   F(N+f(N))=F(N)[/quote]

I suppose that $ \mathbb N$ is the set of positive integers ($ 0\notin\mathbb N$).

Since $ f(a)\equal{}1$, we get $ f(a\plus{}f(a))\equal{}f(a)$ and so $ f(a\plus{}1)\equal{}1$ and so $ f(n)\equal{}1$ $ \forall n\ge a$

Let then $ b\equal{}\inf\{n\in\mathbb N$ such that $ f(k)\equal{}1$ $ \forall k\ge n\}$ ($ b$ exists since $ f(n)\equal{}1$ $ \forall n\ge a$).

If $ b>1$, then $ f(b\minus{}1\plus{}f(b\minus{}1))\equal{}f(b\minus{}1)$ and, since $ b\minus{}1\plus{}f(b\minus{}1)\ge b$, $ f(b\minus{}1)\equal{}1$, which is impossible (definition of $ b$). So $ b\equal{}1$

Hence the answer : $ f(n)\equal{}1$ $ \forall n\in\mathbb N$


geometry
3D geometry
tetrahedron
induction
group theory
combinatorics unsolved
combinatorics

[quote="DusT"](I am not very sure about the exact dates above)
We have a polyhedral. On every face, we have some ant moving, with constant speed. Every ant continuously moves on the edges of it's own face. Prove that after some time, some 2 of the ants will colide.
I am not sure, but I think that the speeds of the ants is the same. At least, try to solve it like this.....[/quote]
Each ant moves in clockwise direction. Ant's speed is not constant, but it can't be below 1cm/s.

I don't want to spoil but you may look at :
http://mathcentral.uregina.ca/MP/previous2001/
Problem from February 2002.

Pierre.

why 1cm/s?

or it is just to say they cannot stop?

[quote="pbornsztein"]I don't want to spoil but you may look at :
http://mathcentral.uregina.ca/MP/previous2001/
Problem from February 2002.[/quote]
They consider very particular case, don't they?

Because of the tetrahedron? I didn't notice that it was a polyhedra in this thread  :?

Pierre.

As I understood, they consider constant speeds. :?

But this hypothesis is in Dust's initial post.

Pierre.

No, he said "I think". Actually, it was his supposition only.

Myth....
[quote]
We have a polyhedral. On every face, we have some ant moving, with constant speed.
[/quote]

That is exactly what is given in the problem from Regina university, but with a tetrahedron.
And :

[quote]
I am not sure, but I think that the speeds of the ants is the same. At least, try to solve it like this....
[/quote]

So... :P

Pierre.

I said statement, which was proposed for LMO, since Dust refer to it.

Yes, I actually meant the LMO problem, but I could remeber the text exactly. Now, what about polyhedrals????
Also, Myth, do you remember exactly if the ants moved in clockwise directions??? Because if it would be so, maybe it wouldn't be too difficult to prove that at some time 2 ants will be on the same edge.....Maybe not too difficult, I am not sure....

I took text from LMO Book. So it is exact statement: polyhedral+clockwise movement+non-constant speeds

[quote="Pascual2005"]ah, so can anyone post a solution, i am tired of this problem, i cant think in 3 dimensions! :([/quote]
It can't be valid argument. You are too smart man for that. ;)

Myth! this one has been unsolved for a long time! I have been thinking on it and I have some ideas...

What about induction on the number of faces? I mean, we can destroy a face and make it a vertex, and reassign speeds in edges adjacento to this vertex and we could find great conclusions...However i dont still get it!, I just have some thoughts about it... ;)

Myth, Do you have any idea?

The proper statement of the problem is:
--------------------------------------------------------------
An ant is crawling clockwise along the contour of each face of a convex polytope (so there are as many ants as there are faces). It is known that their velocities are always at least 1mm/hour. Prove that sooner or later some two ants will run into each other.
--------------------------------------------------------------

There is a rather interesting (although incomplete) story about origin of this problem - an acquaintance of one of the jury members brought it from Moscow where a slightly reformulated version of it was a crucial lemma in a Ph.D. thesis of his friend - topic of the dissertation was combinatorial group theory if I am not mistaken. It would be interesting to find out what exactly this was used for...

One mm/hour here simply gurarantees that the velocities are "separated" from zero - that is each ant is actually making circles around assigned face (and will make infinitely many of them unless it is stopped). Velocities are not required to be constant, of course.

The solution is pretty long so I will skip some of the most obvious pieces.

First, within each face (# i) we select its center - A_i, and then we connect centers of every two adjacent faces with a segment. For any fixed moment of time t in the resulting graph G we will orient some of the edges as follows:

If i-th ant is currently on the common edge of i-th and j-th face then we will orient edge A_iA_j in the direction from A_i to A_j.

Now there is exactly one arrow coming out of each vertex of G. Therefore some of the arrows form a simple cycle which splits surface of the polytope into two regions (there might be more than one such a cycle, of course). Let us measure "area" of any such region by the number of vertices of G inside or on the boundary of the region. Since this quantity can take only finite number of values, there will be moment of time t' such that area within one of such cycles Z is minimal.

There are two cases here:

1) Edges within Z do not repeat. Then there are again two cases - when going along Z we circle the minimal region (a) clockwise or (b) counterclockwise. In case (a) we can find first moment of time after t' when one of the arrows of the cycle Z (coming out from, say, vertex A) "jumps" onto next edge coming out of A (that corresponds to the ant assigned to A moving thru one of vertices on the corresponding face's contour), counting clockwise.... I would gladly insert a picture here but ... and if we follow this new arrow (all the other arrows are still the same) it leads us inside Z and as we follow arrows we will eventually return to a vertex we visited before and the formed cycle will have smaller "area" which is impossible. In case (b) we need to reverse time flow and do the same - that is to find last moment before t' when arrow "jumped" etc.

2) Some edges in Z repeat. But then obviously some edge A_iA_j is being traveled twice in opposite directions which means that this edge has two arrows on it and it follows that ants #i and #j have either just collided or are going to inevitably collide on the common edge of faces #i and #j.

There is one tiny "hole" in the above proof of case 1 but it is easily fixable: basically you need to find region Z with area which is minimal starting after a couple of "full timecycles" when all ants have circled their faces fully at least once - that will save you from using a region which has minimal area in the very beginning of the process when you cannot go back in time to fully prove case (1b).

That's it - hope you liked the proof. I certainly did and still do. Enjoy!

Dmitri Fomin

[quote="dvf"]The proper statement of the problem is:
--------------------------------------------------------------
An ant is crawling clockwise along the contour of each face of a convex polytope (so there are as many ants as there are faces). It is known that their velocities are always at least 1mm/hour. Prove that sooner or later some two ants will run into each other.[/quote]
That is exactly what I wrote above.

Right. I just wanted to precede the solution with the proper copy of the problem's statement.


Here's some old Russian problem.

Let a1, a2, ..., an be positive integers not exceeding 1000, such that lcm(ai, aj) > 1000 for any pair of distinct indices i, j. Prove that \sum 1/ai < 2.

We have a lack of responses from beginners. They are so quiet and shy...

The problem is that this site is so good that beginners grow expert way too fast :D

Surprise surprise... an old unsolved problem! :P


Find  the  values  of  n , such  for any  positive  real numbers  [tex]x_i , i=1,2,...,n [/tex]  the  following  inequality  holds :
[tex] x_1x_2+x_2x_3+...+x_{n-1}x_n\leq\frac{n-1}{n}(x_1^2+x_2^2+...+x_n^2)  [/tex]  Thank  you  :)

Something similar to reaarangement inequality hmmm...

$n \ge 2$. Darij has proven on the forum (sorry I forgot which thread it is), that for all n, $\sum x_i^2 \ge 4(x_1x_2+x_2x_3+...+x_nx_1)$ Using this result you have $n \ge 2$


calculus
trigonometry
Pythagorean Theorem
geometry
calculus computations

Your help will be appreciated :D

a) y= tan x +2cos x find dy/dx at x= pi/4

b) Given that x= tan (y/2), prove that dy/dx = 2(1+x^2)

a=alpha

c)Given that y = e^(-x) sin2x , show that dy/dx can be expressed in the form  Re^(-x) cos(2x+a) .

Thanks

$ y \equal{} tanx \plus{} 2cosx$
Find $ \frac {dy}{dx}$ at $ x \equal{} \frac {\pi}{4}$

[quote]$ y \equal{} tanx \plus{} 2cosx$
$ \frac {dy}{dx} \equal{} sec^2x \minus{} 2sinx$
At $ x \equal{} \frac {\pi}{4}$:
$ m \equal{} sec^2 \left (\frac {\pi}{4}\right ) \minus{} 2sin \left (\frac {\pi}{4} \right )$
$ m \equal{} 2 \minus{} \sqrt {2}$[/quote]

Given that $ x \equal{} tan\left (\frac {y}{2} \right )$, prove that $ \frac {dy}{dx} \equal{} \frac {2}{1 \plus{} x^2}$.

[quote]Either we could rewrite the equation as $ y \equal{} 2arctan(x)$, in which case $ \frac {dy}{dx} \equal{} \frac {2}{1 \plus{} x^2}$.
Alternatively, we can use implicit differentiation.
Differentiating with respect to $ x$:
$ 1 \equal{} sec^2\left (\frac {y}{2} \right )\frac {1}{2}\frac {dy}{dx}$
$ \frac {dy}{dx} \equal{} \frac {2}{sec^2\left (\frac {y}{2} \right )}$
Looking at the attached diagram, $ sec\left (\frac {y}{2} \right ) \equal{} \sqrt {1 \plus{} x^2}$
Substituting this value in:
$ \frac {dy}{dx} \equal{} \frac {2}{1 \plus{} x^2}$[/quote]

Given that $ y \equal{} e^{ \minus{} x}sin(2x)$, show that $ \frac {dy}{dx}$ can be expressed in the form $ Re^{ \minus{} x}cos(2x \plus{} \alpha)$

Stuck on #3.

$ y\equal{}e^{\minus{}x}\sin 2x\vspace{1pt}$

$ \frac{dy}{dx}$

$ \equal{}e^{\minus{}x}\left( 2\cos 2x\minus{}\sin 2x\right)$

$ \equal{}Re^{\minus{}x}\cos (2x\plus{}\alpha )$

$ \equal{}\allowbreak e^{\minus{}x}\left( R\cos 2x\cos \alpha \minus{}R\sin 2x\sin \alpha \right)$

Comparing, we have

$ R\cos \alpha \equal{}2,\,\,R\sin \alpha \equal{}1$

Hence $ R\equal{}\sqrt{5}$ and $ \alpha \equal{}\tan ^{\minus{}1}\frac{1}{2}$

Thus, $ \frac{dy}{dx}\equal{}\sqrt{5}e^{\minus{}x}\cos \left( 2x\minus{}\tan ^{\minus{}1}\frac{1}{2}\right)$

b) $ \boxed{\frac {dy}{dx}\equal{}\frac {1}{\frac {dx}{dy}}}$

[quote="EastyMoryan"][/quote]
I didn't get the whole triangle thing. Could you explain it a little?

I drew a triangle with a base of length $ 1$, and an interior angle of $ \frac{y}{2}$.
Then, since I know that $ tan\left (\frac{y}{2}\right )\equal{}x$, I drew $ x$ as the opposite side of the triangle.
Then, I used the pythagorean theorem to find the hypotenuse.

Using that, I could find the secant of $ \frac{y}{2}$.


Bertha has $ 6$ daughters and no sons. Some of her daughters have $ 6$ daughters and the rest have none. Bertha has a total of $ 30$ daughters and granddaughters, and no great-grand daughters. How many of Bertha's daughters and granddaughters have no daughters?

$ \textbf{(A)}\ 22\qquad
\textbf{(B)}\ 23\qquad
\textbf{(C)}\ 24\qquad
\textbf{(D)}\ 25\qquad
\textbf{(E)}\ 26$

[hide]
6 + 6x = 30, where x is the number of daughters who have granddaughters, so there are 4 daughters w/ grandaughters. Together they have 24 granddaughters who have no children, and there are 6-4=2 daughters with no children. So 24+2 =26 (E) [/hide]

[quote="#H34N1"]Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-grand daughters. How many of Bertha's daughters and granddaughters have no daughters?

A. 22
B. 23
C. 24
D. 25
E. 26[/quote]

[hide="solution"]Bertha has 6 daughters so that means whe has 24 grand-daughters.  4 of her daughters have daughters so 2 of her daughters dont have daughters.  24 of her grand-daughters dont have daughters so 24+2=26[/hide]

Bertha has $30 - 6 = 24$ granddaughters, none of whom have
any daughters. The granddaughters are the children of $24/6 = 4$ of
Bertha's daughters, so the number of women  having no daughters is
$30 - 4 = 26$ so the answer is $\boxed{E}$.


[hide]You have 12 selections separated into three different types of donuts:
_ _ _ _ _ | _ _ | _ _ _ _
To simplify the problem, we can get rid of the leftmost and rightmost spots because we must have at least one of each type. Then, we must find how many ways we can put the two separaters in the remaining spots without them being right next to eachother. That is $\frac {12!}{2!10!} - 11 = 55$. I'm not sure though.[/hide]

Who said he has to have 1 of each type  :D

Oh then
[hide] The answer is $\frac {14!}{2!12!} = 91$.[/hide]

[quote="jb05"][hide]
Tarquin's almost got it right. You have 12 blanks and want to choose places to put the seperators, but you can have 0 donuts of 1 type.

a+b+c=12 a,b,c [u]nonnegative[/u].

----|-----|--- means 4 chocolate, 5 sprinkles, 3 bacon.
|------------| means 12 sprinkles.

If instead you want them positive, add 1 to each to get a problem with a one-to-one correspondence:

a+b+c=15 a,b,c [u]positive[/u].

-----|------|---- means 4 chocolate, 5 sprinkles, 3 bacon.
-|-------------|- means 12 sprinkles.

Given 15 blanks, there are 14 places to put each seperator, and you want to place two seperators (which represent the two addition signs in the above equation), so the answer is 14 choose 2 or [b]91[/b].
[/hide][/quote]

How do you make 12 bacons?

Would it be ||------------?

What about 12 chocolates?

------------||?. Wouldn't you do 15 c 2=105?

no theres only 2 dividers so that makes it 12+2 c 2.

[quote="Press alt f4"]Drew is going to buy a dozen donuts for himself. He will buy chocolate donuts, ones with sprinkles, and ones with bacon. How many different combinations of the different types of donuts can be bought?[/quote]

[hide="Solution"]
Set the donuts up like this.

O O O O O O O O O O O O

There are eleven places to put the first divider and ten to place the second. Divide by two to un-doublecount.

55, I think o_o
[/hide]

[quote="mathgeek2006"][quote="Phelpedo"]why'd you say "tarquin almost got it right"?[/quote]

Probably because he was on the right track...[/quote]

But you got the same answer as him and you put that he "almost got it right". Where's the logic in that?


Find the maximum natural number which is divisible by 30 and has exactly 30 different positive divisors.


group theory
abstract algebra
superior algebra
superior algebra solved

Imbed $S_n$ as a subgroup of $A_{n+2}$ but show,for $n\geq 2$,that $S_n$ cannot be imbedded in $A_{n+1}$.


I was posed this problem by a friend, and i cannot find out why this holds.

Find examples of elements $ x,y \in \mathbb{R}^2$ such that $ \|x\|_1 < \|y\|_1$, while $ \|x\|_2 > \|y\|_2$.
The 1-norm is the sum ofabsolute values, while the 2-normis the euclidean norm.

Can this happen?
Thanks

Consider the blue dot and the red dot.
[img]http://farm3.static.flickr.com/2533/4099910509_8929a38280_o.png[/img]

For instance, $ x \equal{} \left(0, \frac56\right)$ and $ y \equal{} \left(\frac12, \frac35\right)$.


Let $ x,y,z$ be positive reals such that $ xyz\geq x\plus{}y\plus{}z$. Prove that

$ \sqrt{2x^2\plus{}yz}\plus{}\sqrt{2y^2\plus{}zx}\plus{}\sqrt{2z^2\plus{}xy}\geq 9$.

I will have a try,correct me if i am wrong:

By AM-GM we get $ xyz\geq x \plus{} y \plus{} z\geq 3\sqrt [3]{xyz}$

so $ xyz\geq \sqrt {27}$

Now by Minkowsky:

$ \sqrt {2x^2 \plus{} yz} \plus{} \sqrt {2y^2 \plus{} zx} \plus{} \sqrt {2z^2 \plus{} xy} \equal{}$

$ \sqrt {(\sqrt {2}x)^2 \plus{} \sqrt {yz}^2} \plus{} \sqrt {(\sqrt {2}y)^2 \plus{} \sqrt {zx}^2} \plus{} \sqrt {(\sqrt {2}z)^2 \plus{} \sqrt {xy}^2}\geq$

$ \sqrt {2(x \plus{} y \plus{} z)^2 \plus{} (\sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx})^2}\geq$

$ \sqrt {2(3\sqrt [3]{\sqrt {27}})^2 \plus{} (3\sqrt [3]{\sqrt {27}})^2 }\equal{}$

$ 9$

[quote="jbmorgan"]Let $ x,y,z$ be positive reals such that $ xyz\geq x \plus{} y \plus{} z$. Prove that

$ \sqrt {2x^2 \plus{} yz} \plus{} \sqrt {2y^2 \plus{} zx} \plus{} \sqrt {2z^2 \plus{} xy}\geq 9$.[/quote]
AM-GM :)

$ \sum_{cyc} \sqrt {2x^2 \plus{} yz} \ge \sqrt{3} \sum_{cyc} \sqrt {x} \cdot \sqrt[6]{xyz}$ $ \ge 3\sqrt{3} \sqrt[3]{xyz} \equal{} \frac{9 \sqrt{xyz}}{\sqrt{3} \cdot \sqrt[6]{xyz}} \ge$ $ \frac{9\sqrt{xyz}}{\sqrt{x\plus{}y\plus{}z}} \ge 9$


projective geometry
geometry unsolved
geometry

let ABCD be a convex quadrilateral and let AC and BD meet at P. consider an arbitrary line l through P. let it intersect AB and CD at P1 and P2 respectively. let the harmonic conjugate of P1 wrt AB be Q1. and the harmonic conjugate of P2 wrt CD be Q2. prove Q1, Q2, and P are collinear

The division $ (AP_1BQ_1)$ is harmonic, so the pencil $ P(AP_1BQ_1)$ is harmonic, so $ Q_1P$ meet CD on the harmonic conjugate of $ P_2$ wrt CD that is $ Q_2$.


Hi,

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If possible also include a print screen.
[hide="To learn how to take a print screen click here"]In the top right corner of your keyboard (or at least on regular 104 US Keyboards) there will be a key called Prt Scr (short for Print Screen). Press this key any time you want to capture a screen, then go to Start >> All Programs >> Accesories >> MS Paint, open Paint, go to the Edit menu, and click Paste (or alternatively press and hold CTRL-V). Then go to the File menu and click on Save As, and give it a name (you will also want to encode it as a .JPG rather than a BMP) and save.[/hide]

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What about Internet explorer  8? Is it surported ?


Solve in $\mathbb{R}$:
\[ \begin{displaymath} \left\{ \begin{array} x^2+y^2=1 \\ 125x^5-125x^3+16\sqrt{5}=0 \end{array} \right \end{displaymath}  \]

What is Llc, and why do you give equations such as this one which lack nice solutions?  In this case, the unique real root of the second equation has absolute value greater than 1.

Edit: I partially retract my initial complaint: one can in fact solve this non-numerically, by making the substitution $x = \cos t$, converting the second equation to the form $\cos^3t\sin^2t = -\frac{16\sqrt5}{125}$ and then showing via calculus that the minimum of the expression on the left is larger than the expression on the right.  The equation would have roots if 16 were replaced by anything between $6\sqrt3$ and $-6\sqrt3$.


I have been looking for a research topic in combinatorics for a few months now; but every time I think of something, I do the usual check online to see if its been done before, and often someone already wrote a paper on it within the last two years or so...

how frustrating! I guess I'll find something eventually.

Do any of the recently published articles mention extensions of the results published that are still unsolved problems?

Even if you do not find something online about a topic that you want to research, how do you know that it isn't already published in some scientific magazine or textbook?


Let $ a,b,c,d,e >0$, $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2\equal{}1.$ Find min of $ P$:
$ P \equal{} \frac{{a^2 }}{{b \plus{} c \plus{} d}} \plus{} \frac{{b^2 }}{{c \plus{} d \plus{} e}} \plus{} \frac{{c^2 }}{{d \plus{} e \plus{} a}} \plus{} \frac{{d^2 }}{{e \plus{} a \plus{} b}} \plus{} \frac{{e^2 }}{{a \plus{} b \plus{} c}}$
I don''t think it's easy although I found $ inf P$. :(

[quote="tronghieu"]Let $ a,b,c,d,e > 0$, $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2 \equal{} 1.$ Find min of $ P$:
$ P \equal{} \frac {{a^2 }}{{b \plus{} c \plus{} d}} \plus{} \frac {{b^2 }}{{c \plus{} d \plus{} e}} \plus{} \frac {{c^2 }}{{d \plus{} e \plus{} a}} \plus{} \frac {{d^2 }}{{e \plus{} a \plus{} b}} \plus{} \frac {{e^2 }}{{a \plus{} b \plus{} c}}$
I don''t think it's easy although I found $ inf P$. :([/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63850


trigonometry
quadratics
AoPSwiki
trig identities
Law of Cosines
geometry
cyclic quadrilateral

Two pedestrians standing $ 10$ feet and $ 40$ feet away from the base of a billboard structure have equivalent views of the billboard, meaning that the angels marked in the diagram are congruent. If the billboard is $ 9$ feet tall, what is its height $ h$ off the ground?

(I did this using the Law of Cosines, but it got really messy...Is there an easier way to do it?)

[img]http://img140.imageshack.us/img140/9103/mandelbrot7ff8.gif[/img]

[hide]Taking the power of the intersection of the $ h$ and the $ 10$, I got the quadratic $ h^2\plus{}9h\equal{}400$, which I completed the square with to get $ 20.5$. I might have done it wrong, though...[/hide]

[quote="Temperal"][hide]Taking the power of the intersection of the $ h$ and the $ 10$, I got the quadratic $ h^2 \plus{} 9h \equal{} 400$, which I completed the square with to get $ 20.5$. I might have done it wrong, though...[/hide][/quote]

$ h^2 \plus{} 9h \minus{} 400 \equal{} (h\plus{}25)(h\minus{}16)$  :wink:

[quote="azjps"][hide="Idea?"]
There is a cyclic quadrilateral (using the angles equal condition), then Power of a point.

I get $ h \equal{} 16$, but I didn't use the right angle condition, so I might be wrong ..
[/hide][/quote]

i dont think it has to be a right angle becuase it can be a slanted board.

I also just inscribed it in a circle and used power of  the point to get 16.

The answer is $ h \equal{} 16$. Where did you circumscribe the circle? How did you solve it after inscribing it in a circle?

Sorry, but I still can't really imagine where the circle fits?

[quote="azjps"][quote="Temperal"][hide]Taking the power of the intersection of the $ h$ and the $ 10$, I got the quadratic $ h^2 \plus{} 9h \equal{} 400$, which I completed the square with to get $ 20.5$. I might have done it wrong, though...[/hide][/quote]

$ h^2 \plus{} 9h \minus{} 400 \equal{} (h \plus{} 25)(h \minus{} 16)$  :wink:[/quote]

Whoops, I forgot to subtract $ 4.5$ from my final answer when completing the square.  :oops:

Here's how I do it (with similar triangles)

[url=http://img143.imageshack.us/my.php?image=mandelbrot7ff8uw1.gif][img]http://img143.imageshack.us/img143/6167/mandelbrot7ff8uw1.gif[/img][/url]

First I showed that the red outlined triangles were similar through the given angles and the vertical angles in the middle.

Then I showed the blue ones were similar through the info from the red similar triangles

That showed that the two angles labeled '3' were congruent.
Then I used another pair of similar triangles (not shown) to do

$ \frac{h}{10}\equal{}\frac{40}{h\plus{}9}$

[quote="123456789"][quote="Zellex"]at the two people + the two end points on the billboard

They used power of a point.  Sorry, I can't really describe it with words, so maybe AoPSWiki will help.[/quote]

How can you have the circle touch at all four points? I'm confused. Could somebody possibly draw a diagram?[/quote]

Remember that if you have a chord AB on a circle so that minor arc AB has measure 2X, then if we put any point C on major arc AB, the measure of angle ACB is X.

In this case, let the billboard be the chord, and let the two ground points be two possible values of C, since the viewing angle is the same for both. In other words, you want a circle that passes through the two people and the two ends of the board.

How did you know that it was possible to draw a circle like that, that touches all the vertices you want it to? Is it always possible?

No, it is not possible to draw a circle through any four given points  :P

Given two fixed points $ A, B$ in the plane and two points $ C, D$ such that $ C, D$ lie on the same side of line $ AB$ and $ m\angle ACB \equal{} m\angle ADB$, the points $ A, B, C, D$ are concyclic.

This is the converse of the theorem that given a chord $ AB$ in any fixed circle $ O$ and a variable point $ C$ on the circle, the angle $ ACB$ depends only on which arc of the circle (the minor arc $ AB$ or the major arc $ AB$) $ C$ lies on (and in fact the two possible angles are supplementary).

[quote="chickendude"]

First I showed that the red outlined triangles were similar through the given angles and the vertical angles in the middle.

Then I showed the blue ones were similar through the info from the red similar triangles

That showed that the two angles labeled '3' were congruent.
Then I used another pair of similar triangles (not shown) to do

$ \frac {h}{10} \equal{} \frac {40}{h \plus{} 9}$[/quote]

What you did was prove a specific case of power of a point.

So, in all, we know that $ h \equal{} 16$ since $ (h \plus{} 25)(h \minus{} 16) \equal{} 0$, and $ h$ can't be negative.


limit
function
real analysis
real analysis unsolved

#1.
Prove thet if $ f$ is differentiable at c, then
$ f'(c) \equal{} \lim_{h\to 0} \frac {f(c \plus{} h) \minus{} f(c \minus{} h)}{2h}$

#2.
Prove that if the limit value on RHS of the equation #1 is exist, then $ f$ is differentiable at $ c$

[quote="GTK X Hunter"]Prove that if the limit value on RHS of the equation #1 is exist, then $ f$ is differentiable at $ c$[/quote]
Oh really ?

[quote="alekk"]Oh really ?[/quote]
To make alekk's objection more concrete:

Let $ f(x)\equal{}|x|$ and let $ c\equal{}0.$

We have $ \frac{f(h)\minus{}f(\minus{}h)}{2h}\equal{}0$ for all $ h\ne 0,$ so the limit in question exists and is zero. But $ f$ is clearly not differentiable at the origin.

How about

\[ f(x) \equal{} \begin{cases} 1 &\text{if }x\equal{}0 \\ 0 &\text{otherwise} \end{cases}\]

This $ f$ isn't even continuous at $ 0$.

or take the most horrible  function $ f: (0;\infty) \to \mathbb{R}$ you can think off, and then define $ f(x)\equal{}f(\minus{}x)$.


A random positive integer $ \mod {14}$ is chosen. I am told that if it is congruent to 1, 3, 5, 7, 9, 11, or 13 $ \pmod {14}$, then I'll get a piece of shiny stuff. I'm told that I didn't get a piece of shiny stuff. What's the probability it's not congruent to 2 or 10 $ \pmod {7}$?

[quote="Temperal"]
A random positive integer$ \mod {14}$ is chosen. I am told that if it is congruent to $ 1, 3, 5, 7, 9, 11, or 13 \pmod {14}$, then I'll get a piece of shiny stuff. I'm told that I didn't get a piece of shiny stuff. What's the probability it's not congruent to 2 or 10 $ \pmod {7}$?
[/quote]

[hide="ok that makes more sense"]
ok, so you don't get a piece of shiny stuff
that means your number must be $ 2, 4, 6, 8, 10, 12, 14 \pmod{14}$
$ 2, 10 \pmod{7}$ is the same thing as $ 2, 9, 3, 10 \pmod{14}$
so the probablility that it's NOT $ 2, 10\pmod {7}$ is $ \frac {5}{7}$

[/hide]

[quote="gauss202"][quote="CircleSquared"]
$ 2, 10 \pmod{7}$ is the same thing as $ 2, 4, 10, 6 \pmod{14}$
[/quote]

How is this so?[/quote]

uhh hold on
something funny thar.

[quote="CircleSquared"][quote="DanK"]Wait, shouldn't the mods of 14 that it could be only go up to 12?  $ 14\pmod{14} \equiv 0\pmod{14}$.  If that's the case, then it's just [hide]$ \boxed{\frac {2}{3}}$[/hide][/quote]
its 0 to 13
so theres 14 values
right?[/quote]

Yeah, it is 0 to 13.

How are there 14 possible modular residues if they have to be positive integers?
I'm feeling abnormally retarded today (sleep deprivation) so I'm probably doing something stupid.


[quote="nguoivn"]I know the way you created it. It's an old hint  :wink: . See here: http://www.mathlinks.ro/viewtopic.php?t=305439
And how will you solve with the similars:
Let $ a,b,c$ be real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 0$. Prove that:
i) $ a^{2} \plus{} b^{2} \plus{} c^{2}\ge a^{2}b \plus{} b^{2}c \plus{} c^{2}a \plus{} abc(a \plus{} b \plus{} c)$
ii) $ a^2 \plus{} b^2 \plus{} c^2 \ge abc(a \plus{} b \plus{} c) \minus{} (a^2b \plus{} b^2c \plus{} c^2a)$
...
All of them can be solved easily by AM-GM.[/quote]

Thanks nguoivn.Can you post your solution (by AM-GM) for my problem? :wink:

[quote="vuthanhtu_hd"][b]Problem 1[/b](Vu Thanh Tu)
Let $ a,b,c$ be real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 0$.Prove that

$ a^2 \plus{} b^2 \plus{} c^2 \ge a^2b \plus{} b^2c \plus{} c^2a \plus{} abc(a \plus{} b \plus{} c \minus{} 3)$

When does equality occur?

My problem isn't  difficult.But it's very nice  :) .I have 2 solutions for it.[/quote]
The inequality can be written as
$ 2(a^2\plus{}b^2\plus{}c^2) \ge 2(a^2b\plus{}b^2c\plus{}c^2a)\plus{}2abc(a\plus{}b\plus{}c)\minus{}6abc.$
Since $ (ab\plus{}bc\plus{}ca)^2\equal{}0,$ we have
$ 2abc(a\plus{}b\plus{}c)\equal{}\minus{}(a^2b^2\plus{}b^2c^2\plus{}c^2a^2).$
Therefore, the above inequality is equivalent to
$ 2(a^2\plus{}b^2\plus{}c^2)\plus{}(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)\minus{}2(a^2b\plus{}b^2c\plus{}c^2a) \ge \minus{}6abc,$
or
$ (a^2\plus{}b^2\plus{}c^2)\plus{}[a^2(1\minus{}b)^2\plus{}b^2(1\minus{}c)^2\plus{}c^2(1\minus{}a)^2] \ge \minus{}6abc.$
By the AM-GM Inequality, we have
$ a^2\plus{}b^2(1\minus{}c)^2 \ge 2ab(1\minus{}c)\equal{}2ab\minus{}2abc,$
$ b^2\plus{}c^2(1\minus{}a)^2 \ge 2bc(1\minus{}a)\equal{}2bc\minus{}2abc,$
$ c^2\plus{}a^2(1\minus{}b)^2 \ge 2ca(1\minus{}b) \equal{}2ca\minus{}2abc.$
Adding these three inequalities and recalling that $ ab\plus{}bc\plus{}ca\equal{}0,$ the conclusion follows.

[quote="vuthanhtu_hd"]
Thanks nguoivn.Can you post your solution (by AM-GM) for my problem? :wink:[/quote]
Dear my friend, my solution is similar to Can_hang's solution  :) . You can try with the similar results ( I created with the same way):
[quote="nguoivn"]
Let $ a,b,c$ be real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 0$. Prove that:
i) $ a^{2} \plus{} b^{2} \plus{} c^{2}\ge a^{2}b \plus{} b^{2}c \plus{} c^{2}a \plus{} abc(a \plus{} b \plus{} c)$
ii) $ a^2 \plus{} b^2 \plus{} c^2 \ge abc(a \plus{} b \plus{} c) \minus{} (a^2b \plus{} b^2c \plus{} c^2a)$
...
All of them can be solved easily by AM-GM.[/quote]

[quote="can_hang2007"][quote="vuthanhtu_hd"][b]Problem 1[/b](Vu Thanh Tu)
Let $ a,b,c$ be real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 0$.Prove that

$ a^2 \plus{} b^2 \plus{} c^2 \ge a^2b \plus{} b^2c \plus{} c^2a \plus{} abc(a \plus{} b \plus{} c \minus{} 3)$

When does equality occur?

My problem isn't  difficult.But it's very nice  :) .I have 2 solutions for it.[/quote]
The inequality can be written as
$ 2(a^2 \plus{} b^2 \plus{} c^2) \ge 2(a^2b \plus{} b^2c \plus{} c^2a) \plus{} 2abc(a \plus{} b \plus{} c) \minus{} 6abc.$
Since $ (ab \plus{} bc \plus{} ca)^2 \equal{} 0,$ we have
$ 2abc(a \plus{} b \plus{} c) \equal{} \minus{} (a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2).$
Therefore, the above inequality is equivalent to
$ 2(a^2 \plus{} b^2 \plus{} c^2) \plus{} (a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \minus{} 2(a^2b \plus{} b^2c \plus{} c^2a) \ge \minus{} 6abc,$
or
$ (a^2 \plus{} b^2 \plus{} c^2) \plus{} [a^2(1 \minus{} b)^2 \plus{} b^2(1 \minus{} c)^2 \plus{} c^2(1 \minus{} a)^2] \ge \minus{} 6abc.$
By the AM-GM Inequality, we have
$ a^2 \plus{} b^2(1 \minus{} c)^2 \ge 2ab(1 \minus{} c) \equal{} 2ab \minus{} 2abc,$
$ b^2 \plus{} c^2(1 \minus{} a)^2 \ge 2bc(1 \minus{} a) \equal{} 2bc \minus{} 2abc,$
$ c^2 \plus{} a^2(1 \minus{} b)^2 \ge 2ca(1 \minus{} b) \equal{} 2ca \minus{} 2abc.$
Adding these three inequalities and recalling that $ ab \plus{} bc \plus{} ca \equal{} 0,$ the conclusion follows.[/quote]

Very nice.We have equality holds if only if $ (a,b,c) \equal{} (0,0,0)$ or $ (a,b,c)\sim (t,f(t),g(t))$.
with $ t \equal{} 2cos\frac {k \pi}{9} \plus{} 1$  ,$ k \in \{2;4;8\}$

and $ f(t) \equal{} \frac {t^2 \minus{} t}{t^2 \minus{} t \minus{} 1}$ ,$ g(t) \equal{} \frac { \minus{} t}{t^2 \minus{} t \minus{} 1}$

:)

[quote="vuthanhtu_hd"][quote="can_hang2007"]
By the AM-GM Inequality, we have
$ a^2 \plus{} b^2(1 \minus{} c)^2 \ge 2ab(1 \minus{} c) \equal{} 2ab \minus{} 2abc,$
$ b^2 \plus{} c^2(1 \minus{} a)^2 \ge 2bc(1 \minus{} a) \equal{} 2bc \minus{} 2abc,$
$ c^2 \plus{} a^2(1 \minus{} b)^2 \ge 2ca(1 \minus{} b) \equal{} 2ca \minus{} 2abc.$
Adding these three inequalities and recalling that $ ab \plus{} bc \plus{} ca \equal{} 0,$ the conclusion follows.[/quote]

Very nice.We have equality holds if only if $ (a,b,c) \equal{} (0,0,0)$ or $ (a,b,c)\sim (t,f(t),g(t))$.
with $ t \equal{} 2cos\frac {k \pi}{9} \plus{} 1$, $ k \in \{2;4;8\}$

and $ f(t) \equal{} \frac {t^2 \minus{} t}{t^2 \minus{} t \minus{} 1}$ ,$ g(t) \equal{} \frac { \minus{} t}{t^2 \minus{} t \minus{} 1}$

:) [/quote]

More precisely (note that the inequality is not homogenous):
the equality case is

$ a \equal{} b(1 \minus{} c)$
$ b \equal{} c(1 \minus{} a)$
$ c \equal{} a(1 \minus{} b)$

which holds iff
$ (a,b,c) \equal{} \left(2cos\frac {k \pi}{9} \plus{} 1, 2cos\frac {2k\pi}{9} \plus{} 1, 2cos\frac {4k\pi}{9} \plus{} 1\right)$, $ k \equal{} 2,4,6,8$ :D.  The solutions for $ k \equal{} 2,4,8$ are cyclics of each other, and $ k \equal{} 6$ yields the trivial $ (0,0,0$) solution.

Note that $ t \equal{} 2cos\frac {k \pi}{9} \plus{} 1$,  $ k \in \{2;4;8\}$ are the three solutions of $ t^3 \minus{} 3t^2 \plus{} 3 \equal{} 0$.  :)


calculus
derivative
inequalities
inequalities unsolved

I tried to solve it by taking k=1/x, but didn't help me too much. Any ideas?

hello, do you meant $ \frac{x\plus{}1}{x^x}<2$ for $ 0<x<1$?
Sonnhard.

No, he means $ x \plus{} \frac {1}{x^x} < 2$
The derivative is $ 1\minus{}x^{2x\minus{}1}(lnx\plus{}1)$.
1 is clearly a zero, it is easy to show $ x^{2x\minus{}1}(lnx\plus{}1)$ is strictly increasing, hence the answer.

[quote="LaPlanta"]Prove that $ x \plus{} \frac {1}{x^x} < 2$ for $ 0 < x < 1.$[/quote]Using the weighted arithmetic-mean — geometric-mean inequality,

$ \left(\frac {1}{x}\right)^{x}\cdot 1^{1 \minus{} x} < x\cdot \frac {1}{x} \plus{} (1 \minus{} x)\cdot 1 \equal{} 2 \minus{} x.$


A and B are positive integers. xi = (A + 1/2)^i + (B + 1/2)^i. Prove that only finitely many of x1, x2, x3, ... are integers.

Let $x=2A+1$ and $y=2B+1$, which both are odd integers.
The number $x_n$ is an integer if and only if $2^n$ divides $x^n+y^n$.
But, if $n>0$ is even $x^n+y^n = 2 \mod [4]$ thus $x_n$ is not an integer.
If $n$ is odd then $x^n+y^n = (x+y)(x^{n-1} - x^{n-2}y + \cdots + y^{n-1})$ and the second factor is clearly odd. Thus $x_n$ is an integer if and only if $2^n$ divides $x+y$ which can only occur a finite number of times.

Pierre.

I didn't got the third line
and also why the second factor is odd (It is not clear to a dumb)

Oh I got it, since $x$ and $y$ are both odd, their product will be always odd


If a quadrangle is inscribed in one circle and circumscribed about a second circle, its area equals $K = \sqrt{abcd}.$

Are a, b, c and d the sides or what are they?

Yes, a, b, c, d are the sides of the quadrilateral, and the formula is actually $K = \sqrt{abcd}$.

Mod's note -

Solution temporarily removed to give the less experienced a chance.


Princeton
college
geometry
Euler
email
number theory

Well, the full results are posted on the Princeton Math Club website now, right [url=https://cgi.math.princeton.edu/mathclub/index.php/Princeton_Math_Competition_2008_Results]here[/url]!

I'm afraid I don't have time right now to write up the full results and post them here. I'll just give some highlights:

Division A Sweepstakes
$ \begin{tabular}{|c|c|} \text{North Carolina} & 599 \\
\text{TJ A} & 579 \\
\text{Albany Area} & 530 \\
\text{Chesapeake} & 524 \\
\text{Quagga} & 518 \\
\text{GrandmAASTers} & 504 \\
\text{United} & 480 \\
\text{Lehigh Valley} & 479 \\
\text{Deerfield} & 450 \\
\text{Magictones} & 418 \end{tabular}$

Divison A Top Power Test: United
Division A Top Team Test: Where's Waldo?

Division A Individual Finals results
1. Damien Jiang
2. Matt Superdock
3. Brian Hamrick
4. Sterling Chu
5. Andrew Ardito
6. Qinxuan Pan
7. Wenyu Cao
8. Alex Kendell
9. Ben Kraft
10. Mitchell Lee

Division A Individual Test winners
Algebra: Sergei Bernstein with a 23
Geometry: Qinxuan Pan with a 19 (tie broken by finals score)
Number Theory: John Berman with a 32 (tie broken by finals score)
Combinatorics: Matt Superdock with a 23

Division B Sweepstakes
$ \begin{tabular}{|c|c|} \text{DynAASTy} & 550 \\
\text{AAST Alpha} & 526 \\
\text{Lawrenceville D} & 474 \\
\text{Worcester A} & 462 \\
\text{CHE} & 344 \\
\text{Worcester B} & 324 \\
\text{WWPHSS B} & 300 \\
\text{Ridgewood 1} & 263 \\
\text{Ducks} & 248 \\
\text{Eulers} & 243 \end{tabular}$

Division B Top Power Test: Sharper, Better, Fiercer
Division B Top Team Test: Worcester Academy A

Division B Individual Finals results
1. Michael Tan
2. Robert Weckstein
3. Jordan Moldow
4. Utsarga Sikder
5. Jay Yang
6. Chester Caswell
7. Alex Radek
8. Peter Humanik
9. Andrew Lai
10. Jongwhan Park

Division B Individual Test winners
Algebra: Alex Zhu with a 22
Geometry: Stephen Ahn with a 14 (tie broken by finals score)
Number Theory: Alex Zhu with a 30
Combinatorics: Andrew Cai with a 16

Congratulations everyone!

wow worcester A good job. I wish I was div. B i would have beasted it

Sorry. I was in a rush and probably committed more typos than just that one. You were indeed 10th.

[quote="yankeesrule007"]Is there any way to see scores on the individual finals too?[/quote]
I don't think we're publishing those. The finals were graded olympiad-style, 0-7. I seem to recall that the top 10 in A all had at least two full questions right, with the winner close to perfect, and in B the scores were slightly lower.

Okay, thanks.  Do you also know if we'll get to see our graded tests eventually?

Where can I find the full rankings for Power Contest?

[quote="CatalystOfNostalgia"]Is there any way we can get scores checked? I'm pretty sure I got an 18 on Algebra and have my answers with me, unless I copied down my answer to #2 incorrectly (which is entirely possible). Also how were some of these ties broken? They look suspiciously alphabetical...[/quote]
[quote="Xevarion"](tie broken by finals score) [/quote]

(this message only contains quote tags)

[quote="CatalystOfNostalgia"]So apparently power was not graded out of 75. And NC got 96 on Number Theory? That's insane.

Is there any way we can get scores checked? I'm pretty sure I got an 18 on Algebra and have my answers with me, unless I copied down my answer to #2 incorrectly (which is entirely possible). Also how were some of these ties broken? They look suspiciously alphabetical...[/quote]
It's possible that that page doesn't actually list the ties broken in the right order. We broke them with the individual final results, and I definitely noticed at least one that seemed to be wrong.

I suspect that the organizers aren't planning on doing anything else with the tests (rechecking scores, returning individual final or power tests, etc), but I'm not in charge anymore so don't take my word as official.

edit: btw 这个白人看懂了。

You guys misread my name as "Patrick Hulch" at the competition. Fail.

One quick note: my name is spelled wrong on the Individual Finals results (it's Kandell, as on the Combinatorics list, not Kendell).

Also, those Number Theory scores are unbelievable.  I thought that test was hard . . .  :o

Hmm so

[quote=Rules]
1. Each team will send a sub-team of up to 4 members to participate in the Math Bowl, an elimination-style tournament with buzzers. A full list of Math Bowl rules will be sent to teams in a packet after registration.
   2. The winning team in each division will receive 15 points to be added to the team’s total score. The second place team will receive 10 points. Semi-finalists will receive 5 points.
[/quote]

But it seems that Chesapeake/AAST/the semifinalists' score don't have the requisite 10/15/5 points...

[quote="solafidefarms"]But it seems that Chesapeake/AAST/the semifinalists' score don't have the requisite 10/15/5 points...[/quote]
 :o

I don't see how points could be awarded for Math Bowl after about 80 people couldn't participate because of individual finals, though.

That was what I believe the explanation is/was/w/e.

In theory, I definitely agree with Hexahedron - awarding points for an event in which a team's strong performance on the individual rounds is a handicap seems odd.

But I would love to see Chesapeake move into third place . . .

I'm pretty sure I got questions 4,5,6,7 (missed 1,2,3 though,lol) on algebra in division A, which would have given me 16 and put me in that tie.

Also any chance we'll get our graded power tests back? My team solved every question and we are pretty curious how we ended up with 61. What was the highest possible score on power?

When will the solutions be posted? I'd love to learn how to solve algebra 7-9 and geometry 8-10 and team test 7-10...

[quote="Xevarion"]edit: btw 这个白人看懂了。[/quote]

really?! :D

[quote="greentreeroad"]When will the solutions be posted?[/quote]
Hm, I'm surprised they aren't already. I'll ask the guys who wrote them about it.

[quote="greentreeroad"][quote="Xevarion"]edit: btw 这个白人看懂了。[/quote]
really?! :D[/quote]
是真的。可是我的中文不太好。

I'm listed as getting 16 in algebra, but I also thought I got 18.

Solutions are posted along with the tests [url=https://cgi.math.princeton.edu/mathclub/index.php/Princeton_Math_Competition_2008_Tests]here[/url].

If the score you thought you should have gotten doesn't match what's listed, here are some possibilities:

1) We made an error, either in grading or data entry.
2) Your handwriting was sloppy and the graders couldn't tell what you wrote for one of your answers.
3) You solved the problem correctly but wrote down the answer incorrectly or in a more complicated form than expected.

If you're really sure that it was #1 and you should have placed higher, PM me and I'll find out whether there's anything we can do about it.

are the power contest results supposed to be posted?

How many teams/people participated in each division?

Corrected Kendell to Kandell.

At this point, we can't check scores on the short-answer tests. We won't send out the graded proof tests, but if you'd like to know the graders' comments (if any) on your own tests, email ahesterb@princeton.edu sometime before May 1.

Solutions are posted, except Power Test solutions, which will be posted as soon as I can track down the authors.

About 23 teams participated in division B and about 29 in A.


function
trigonometry
complex numbers
real analysis
real analysis unsolved

Can anyone help me on these questions please, I have an exam soon and I would really appreciate your help.

1) There are expansions for tan z and for cot z in terms of Bernoulli numbers. Find a similar expansion for cosec z = 1/ sin z. In particular, write down the coefficient of z^5 in this expansion.

2) Verify that u(x, y) = e^(−x)(x sin y − y cos y) is harmonic on R^2. Find a function v(x, y) so that f(x + iy) = u(x, y) + iv(x, y) is differentiable (in the complex sense) at each point of C, and
express f(x + iy) in terms of z = x + iy.

Thank you in advance.

[quote="Eugene123"]2) Verify that u(x, y) = e^(−x)(x sin y − y cos y) is harmonic on R^2.[/quote]

A function $u(x,y)$ is said to be harmonic if it verifies the Laplace differential equation:

$\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}= 0$.

So in your case we have:

$\frac{\partial u}{\partial x}= e^{-x}\sin y-e^{-x}(x\sin y-y\cos y)$

$\frac{\partial u}{\partial y}= e^{-x}[(x-1)\cos y+y\sin y]$

$\frac{\partial^{2}u}{\partial x^{2}}= e^{-x}(x\sin y-y\cos y)-2e^{-x}\sin y$

$\frac{\partial^{2}u}{\partial y^{2}}= 2e^{-x}\sin y-e^{-x}(x\sin y-y\cos y)$

and so the function $u(x,y)$ is harmonic.

[quote="Eugene123"]Find a similar expansion for cosec z = 1/sin z.[/quote]

Do you mean that we are to find an expansion for cosec z, also in terms of the Bernoulli numbers?

[quote="Eugene123"]Find a function v(x, y) so that f(x + iy) = u(x, y) + iv(x, y) is differentiable (in the complex sense) at each point of C[/quote]

For a complex function $f(x+iy) = u(x, y)+iv(x, y)$ to be differentiable it has to obey the Cauchy-Riemann equations:

$\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$.

[quote="Eugene123"]
1) There are expansions for tan z and for cot z in terms of Bernoulli numbers. Find a similar expansion for cosec z = 1/ sin z. In particular, write down the coefficient of z^5 in this expansion.
[/quote]
Just a little piece of trigonometry: $\frac 1{\sin z}=\frac 1{\tan \frac{z}{2}}-\frac 1{\tan z}$


geometry
geometric transformation
induction
probability

What is the  principle involved in the cooling of a gas by the process of adiabatic magnetization?Some cycles of repeated magnetization and demagnetization is done as far as I know but I haven"t been able to grasp the essence of the process.

Plzz help.

cooling a gas means lowering it's temperature(macro-speaking) and that means decreasing the kinetic energy (of translation) of the molecules so the process can be explained if you can tell how does magnetic induction influence the speed of a gas molecules.i think the induction has the greatest influence upon the nucleus of the atom where are protons and the only atoms that will be influenced are the ones who do not move paralel with induction (B) direction,and due to lorentz force some of the particles will have a rotational trajectory(elicoidal and even circulary) so translation kinetic energy will transform into circulary kinetic energy and that will reduce the probability of a molecule of hiting walls(wich explains how pressure is involved-if you know the micromolecular formula of pressure) so the temperature is lower(PV=yRT).This is how i understand the process.maybe there is something else involved


Let ABC be a triangle with H and O as its orthocenter and circumcenter. Let CM be a altitude in ABC and let MN and OM form an angle of 90 degrees where N lies on AC. Compute angle MHN in terms of angle BAC.

EDIT: I edited the text

Nobody can solve this?
Maybe it's for PO...


Solve this equation if x,y are  integer numbers
\[x^{y^{2}}=y^{x}\]

I introduce a problem to become reference as follows.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=3845]IMO 1997, Problem 5[/url]


A stack of 45 dimes is divided into three piles in the ratio $\frac{1}{6}: \frac{1}{2}: \frac{1}{4}$. How many dimes are in the pile with the least number of dimes?

[hide]

well...you first get rid of the denominators

(multiply by 12)

so you have
$2$:$6$:$3$

wait, i remember how to do this, but now im stuck, because im having a brain shut down :wink:

[/hide]-jorian

There is a typo, the real ratios are $\frac{1}{6}: \frac{1}{3}: \frac{1}{4}$.

lol that makes sense
[hide="from there... :) "]
you multiply by 12 to get
$2$:$4$:$3$

you know u can make that into....
10, 20, and 15 to make 45

so the lowest amount is $10$[/hide]
-jorian


algebra
polynomial
Vieta
number theory unsolved
number theory

(a) Does there exist a polynomial $ P(x)$ with coefficients in integers, such that $ P(d) \equal{} \frac{2008}{d}$ holds for all positive divisors of $ 2008$?
(b) For which positive integers $ n$ does a polynomial $ P(x)$ with coefficients in integers exists, such that $ P(d) \equal{} \frac{n}{d}$ holds for all positive divisors of $ n$?

Let $ Q(d) \equal{} dP(d) \minus{} n$; then $ Q(d) \equal{} 0$ for all $ d|n$, so $ Q(x) \equal{} R(x)\left[\prod_{d_i|n} (x \minus{} d_i)\right] \equal{} dP(d) \minus{} n$ for some integer polynomial $ R(x)$; then checking constant terms, vieta's gives us that $ \left. \prod_{d_i|n} d_i \equal{} n \times \prod_{d_i|n, d_i < n} d_i \right| n$, so $ \prod_{d_i|n, d_i < n} d_i \le 1$; hence $ n$ is either $ 1$ or a prime.

[quote="April"](a) Does there exist a polynomial $ P(x)$ with coefficients in integers, such that $ P(d) \equal{} \frac {2008}{d}$ holds for all positive divisors of $ 2008$?
(b) For which positive integers $ n$ does a polynomial $ P(x)$ with coefficients in integers exists, such that $ P(d) \equal{} \frac {n}{d}$ holds for all positive divisors of $ n$?[/quote]
Use that $ a \minus{} b \mid p(a) \minus{} p(b)$. If $ n \equal{} a \cdot b$ where $ a,b > 1$ then $ ab \minus{} b \mid p(ab) \minus{} p(b) \equal{} 1\minus{}a \iff b(1\minus{}a) \mid 1\minus{}a$ which is not possible. Thus $ n$ must be either $ 1$ or a prime. (It is easy to see that this is sufficient)

Observe that if such a polynomial $P$ exist, then for any  distinct positive divisors  $a$ and $b$ we must have $$-\frac{n}{ab}=\frac{\frac{n}{a}-\frac{n}{b}}{a-b}=\frac{P(a)-P(b)}{a-b}\in\mathbb{Z}$$
In other words,  for any  distinct divisors $a$ and $b$ of $n$,   $ab$ is also a divisor of $n$.  This happen precisely when  $n$ is not composite, since otherwise if $n$ were composite we could take $a=n$ and $1<b<n$ to reach a contradiction.  On the other hand, if $n$ is not composite (i.e. is a prime or 1), then the polynomial $P(x)=-x+n+1$ satisfies the condition of the problem.


On the forum, there are YouTube BBcode tags. Is there a similar thing on Wiki?

(My first guess would be <youtube>).

I was just thinking there should.

but just copy and paste and PM that friend.


Somewhat convoluted:

[hide]
Working out the simultaneous equation of the two lines, we get:
$x={ab - c^2} \over {ac-b^2}$ (I)
$y={a^2-bc} \over {b^2-ac}$ (II)
Where $(x,y)$ is the point of intersection.

We know that $ab-c^2 > 0$ and $a^2-bc>0$
Now there are two cases:

(1) $ac-b^2>0$
In this case we can see that $x>1$ {from (I), and given $b>c$}
Also, y is negative from (II)
So  $l$ is the line joining some point $(x>1, y<0$, and it's easily seen that $l$ will never be 1 away from the origin.

(2)$ac-b^2<0$
Similarly, in this case we get $y>1$ and x is negative, and as in (1), $l$ will never be 1 from the origin.
[/hide]

[quote="kunny"]Let $a,b,c$ be constant numbers such that $a>b>c>0$ and let $l$ be the line passing through the intersection point of two lines $ax+by+c=0$ and $bx+cy+a=0$ and the point $(1,1).$ Can it be possible that the distance between the origin and $l$ is $1.?$ Justify your answer.[/quote]

The line $l$, passing through the intersection of the given lines and (1, 1) is
$(a-b)x+(b-c)y+(c-a)=0$.
Distance from origin to $l$ will be equal to $1$ if,
$\left| {\frac{{c - a}}{{\sqrt{(a - b)^2 + (b - c)^2 } }}} \right| = 1$

$\Rightarrow (c - a)^2 = (a - b)^2 + (b - c)^2$
which will be possible only when either $b=a$ or $b=c$.
But $a>b>c>0$ given.  Hence the distance from origin to the line $l$ will never be equal to $1$

My solution is exactly same as yours, vidyamanohar. :)

kunny


ratio
geometry
3D geometry
sphere
function
quadratics

[quote="SnowStorm"]If the volume of a sphere doubles, what is the ratio of the area of the new sphere to the area of the old sphere?[/quote]Er.. how do you define the area of a sphere?

The volume of a sphere is a cubic function of the radius. The surface area of a sphere is a quadratic function of the radius. Therefore, if the volume doubles, the surface area's scalar is 2^(2/3).

[quote="SnowStorm"]If the volume of a sphere doubles, what is the ratio of the surface area of the new sphere to the surface area of the old sphere?

Edit:
Surface area of the new and old sphere,
also:
[hide]
The formulas are (if you've forgotten):
$\frac{4}{3}\pi r^3$ for volume
$4\pi r^2$ for surface area
[/hide][/quote]

$\frac{dA}{dr} = 8\pi r$
$\frac{dV}{dr} = A$

\[\frac{dA}{dr} = (\frac{dV}{dr})(\frac{dA}{dV})\]
\[\frac{dA}{dV} = \frac{8\pi r}{4\pi r^2}\]
\[\frac{dA}{dV} = \frac{2}{r}\]

so when A doubles, ...
oh wait I've done this the wrong way.
I was trying to practice some new methods I learnt from a textbook :?

There are related rate problems which you need that formula for, but I think there's an easier method =P .