Asymptote
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Hi, I am on a mac computer right now and I am trying to install asymptote. I have followed all the steps outlined, except the terminal throws an error message when I enter in the configure command.

r236059174:asymptote-1.85 myName$ ./configure
checking for gcc... no
checking for cc... no
checking for cl.exe... no
configure: error: in `/Users/myName/Desktop/asymptote-1.85':
configure: error: no acceptable C compiler found in $PATH
See `config.log' for more details.
r236059174:asymptote-1.85 myName$

Thanks!

Hi,
I don't know Mac but I understand that you must install a C compiler; gcc for example.
You may take a look [url=http://www.tech-recipes.com/rx/726/mac-os-x-install-gcc-compiler/]here[/url], google is your friend.

I think it installed properly, but when I tried using a segment of asymptote code, I get an error.

[5])
Runaway argument?
! File ended while scanning use of \next.
<inserted text>
                \par
<*> Section01.tex

?

[quote="Calculus"]but when I tried using a segment of asymptote code, I get an error.[/quote]
Can you post the [b]entire code[/b] you try to commpile, please ?

I tried the smilie face sample code on latex. Here is the error

This is pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6)
 \write18 enabled.
 %&-line parsing enabled.
entering extended mode
(./smilelatex.tex
LaTeX2e <2005/12/01>
Babel <v3.8h> and hyphenation patterns for english, usenglishmax, dumylang, noh
yphenation, arabic, basque, bulgarian, coptic, welsh, czech, slovak, german, ng
erman, danish, esperanto, spanish, catalan, galician, estonian, farsi, finnish,
 french, greek, monogreek, ancientgreek, croatian, hungarian, interlingua, ibyc
us, indonesian, icelandic, italian, latin, mongolian, dutch, norsk, polish, por
tuguese, pinyin, romanian, russian, slovenian, uppersorbian, serbian, swedish,
turkish, ukenglish, ukrainian, loaded.
(/usr/local/texlive/2007/texmf-dist/tex/latex/base/article.cls
Document Class: article 2005/09/16 v1.4f Standard LaTeX document class
(/usr/local/texlive/2007/texmf-dist/tex/latex/base/size10.clo))
(/usr/local/texlive/2007/texmf-dist/tex/latex/graphics/graphicx.sty
(/usr/local/texlive/2007/texmf-dist/tex/latex/graphics/keyval.sty)
(/usr/local/texlive/2007/texmf-dist/tex/latex/graphics/graphics.sty
(/usr/local/texlive/2007/texmf-dist/tex/latex/graphics/trig.sty)
(/usr/local/texlive/2007/texmf/tex/latex/config/graphics.cfg)
(/usr/local/texlive/2007/texmf-dist/tex/latex/pdftex-def/pdftex.def)))
(/usr/local/texlive/2007/../texmf-local/tex/latex/asymptote/asymptote.sty
(/usr/local/texlive/2007/texmf-dist/tex/latex/base/ifthen.sty)
(/usr/local/texlive/2007/texmf-dist/tex/latex/graphics/color.sty
(/usr/local/texlive/2007/texmf/tex/latex/config/color.cfg))) (./smilelatex.aux)
(/usr/local/texlive/2007/texmf-dist/tex/context/base/supp-pdf.tex
[Loading MPS to PDF converter (version 2006.09.02).]
)

Package asymptote Warning: file smilelatex-1.pdf does not exist on input line 1
5.

[1{/usr/local/texlive/2007/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]
(./smilelatex.aux) )</usr/local/texlive/2007/texmf-dist/fonts/type1/bluesky/cm/
cmex10.pfb></usr/local/texlive/2007/texmf-dist/fonts/type1/bluesky/cm/cmmi10.pf
b></usr/local/texlive/2007/texmf-dist/fonts/type1/bluesky/cm/cmmi7.pfb></usr/lo
cal/texlive/2007/texmf-dist/fonts/type1/bluesky/cm/cmr10.pfb></usr/local/texliv
e/2007/texmf-dist/fonts/type1/bluesky/cm/cmr7.pfb>
Output written on smilelatex.pdf (1 page, 19082 bytes).
Transcript written on smilelatex.log.

Here is the code

\documentclass{article}
\usepackage[pdftex]{graphicx}
\usepackage{asymptote}
\begin{document}
Hello.
I like to make pics with Asymptote like this one:
\begin{figure}[h]
 \begin{asy}
  include graph;
  size(1inch);
  filldraw(circle((0,0),1),yellow,black);
  fill(circle((-.3,.4),.1),black);
  fill(circle((.3,.4),.1),black);
  draw(arc((0,0),.5,-140,-40));
\end{asy}
\end{figure}
\par It makes me happy,
since I can still type my normal LaTeX stuff around it:
\(\int_0^{\pi}{\sin{x}}\,dx=2\)
\end{document}

I don't see any "Runaway argument?".
All seems right.

Oh, that was with a different code. When I tried the sample code above, the above error was what I got and the image would not display.

Be sure that there are no space before \begin{asy} and \end{asy}.

@Calculus, what error are you referring to?  In the log you posted, I see a warning, but no error.

There are no spaces in front of those commands.

The error I was referring to was that the image does not appear. The text does, but no asymptote stuff.

Thanks

If the name of your file is foo.tex you must do:
latex foo.tex
asy foo (or asy foo.asy)
latex foo.tex

See if [url=http://www.math.ualberta.ca/~mss/misc/TeXShopAndAsymptote.pdf]this pdf[/url] helps. I used it to get Asymptote to work with TeXShop on a Mac.

So I followed the directions in the PDF above, and I still get an error. The code I used was the code in the PDF. Here is the error

This is pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6)
 %&-line parsing enabled.
entering extended mode
(./Simlie.tex
LaTeX2e <2005/12/01>
Babel <v3.8h> and hyphenation patterns for english, usenglishmax, dumylang, noh
yphenation, arabic, basque, bulgarian, coptic, welsh, czech, slovak, german, ng
erman, danish, esperanto, spanish, catalan, galician, estonian, farsi, finnish,
 french, greek, monogreek, ancientgreek, croatian, hungarian, interlingua, ibyc
us, indonesian, icelandic, italian, latin, mongolian, dutch, norsk, polish, por
tuguese, pinyin, romanian, russian, slovenian, uppersorbian, serbian, swedish,
turkish, ukenglish, ukrainian, loaded.
(/usr/local/texlive/2007/texmf-dist/tex/latex/base/article.cls
Document Class: article 2005/09/16 v1.4f Standard LaTeX document class
(/usr/local/texlive/2007/texmf-dist/tex/latex/base/size12.clo))
(/usr/local/texlive/2007/texmf-dist/tex/latex/amsmath/amsmath.sty
For additional information on amsmath, use the `?' option.
(/usr/local/texlive/2007/texmf-dist/tex/latex/amsmath/amstext.sty
(/usr/local/texlive/2007/texmf-dist/tex/latex/amsmath/amsgen.sty))
(/usr/local/texlive/2007/texmf-dist/tex/latex/amsmath/amsbsy.sty)
(/usr/local/texlive/2007/texmf-dist/tex/latex/amsmath/amsopn.sty))
(/usr/local/texlive/2007/texmf-dist/tex/latex/amsfonts/amssymb.sty
(/usr/local/texlive/2007/texmf-dist/tex/latex/amsfonts/amsfonts.sty))
(/usr/local/texlive/2007/../texmf-local/tex/latex/asymptote/asymptote.sty
(/usr/local/texlive/2007/texmf-dist/tex/latex/graphics/keyval.sty)
(/usr/local/texlive/2007/texmf-dist/tex/latex/base/ifthen.sty)
(/usr/local/texlive/2007/texmf-dist/tex/latex/graphics/color.sty
(/usr/local/texlive/2007/texmf/tex/latex/config/color.cfg)
(/usr/local/texlive/2007/texmf-dist/tex/latex/pdftex-def/pdftex.def))
(/usr/local/texlive/2007/texmf-dist/tex/latex/graphics/graphicx.sty
(/usr/local/texlive/2007/texmf-dist/tex/latex/graphics/graphics.sty
(/usr/local/texlive/2007/texmf-dist/tex/latex/graphics/trig.sty)
(/usr/local/texlive/2007/texmf/tex/latex/config/graphics.cfg)))) (./Simlie.aux)
(/usr/local/texlive/2007/texmf-dist/tex/context/base/supp-pdf.tex
[Loading MPS to PDF converter (version 2006.09.02).]
)

Package asymptote Warning: file Simlie-1.pdf does not exist on input line 9.

(./Simlie.aux) )
No pages of output.
Transcript written on Simlie.log.
error: could not load module './‘basename Simlie.tex .tex‘.asy'
This is pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6)
 %&-line parsing enabled.
entering extended mode
(./Simlie.tex
LaTeX2e <2005/12/01>
Babel <v3.8h> and hyphenation patterns for english, usenglishmax, dumylang, noh
yphenation, arabic, basque, bulgarian, coptic, welsh, czech, slovak, german, ng
erman, danish, esperanto, spanish, catalan, galician, estonian, farsi, finnish,
 french, greek, monogreek, ancientgreek, croatian, hungarian, interlingua, ibyc
us, indonesian, icelandic, italian, latin, mongolian, dutch, norsk, polish, por
tuguese, pinyin, romanian, russian, slovenian, uppersorbian, serbian, swedish,
turkish, ukenglish, ukrainian, loaded.
(/usr/local/texlive/2007/texmf-dist/tex/latex/base/article.cls
Document Class: article 2005/09/16 v1.4f Standard LaTeX document class
(/usr/local/texlive/2007/texmf-dist/tex/latex/base/size12.clo))
(/usr/local/texlive/2007/texmf-dist/tex/latex/amsmath/amsmath.sty
For additional information on amsmath, use the `?' option.
(/usr/local/texlive/2007/texmf-dist/tex/latex/amsmath/amstext.sty
(/usr/local/texlive/2007/texmf-dist/tex/latex/amsmath/amsgen.sty))
(/usr/local/texlive/2007/texmf-dist/tex/latex/amsmath/amsbsy.sty)
(/usr/local/texlive/2007/texmf-dist/tex/latex/amsmath/amsopn.sty))
(/usr/local/texlive/2007/texmf-dist/tex/latex/amsfonts/amssymb.sty
(/usr/local/texlive/2007/texmf-dist/tex/latex/amsfonts/amsfonts.sty))
(/usr/local/texlive/2007/../texmf-local/tex/latex/asymptote/asymptote.sty
(/usr/local/texlive/2007/texmf-dist/tex/latex/graphics/keyval.sty)
(/usr/local/texlive/2007/texmf-dist/tex/latex/base/ifthen.sty)
(/usr/local/texlive/2007/texmf-dist/tex/latex/graphics/color.sty
(/usr/local/texlive/2007/texmf/tex/latex/config/color.cfg)
(/usr/local/texlive/2007/texmf-dist/tex/latex/pdftex-def/pdftex.def))
(/usr/local/texlive/2007/texmf-dist/tex/latex/graphics/graphicx.sty
(/usr/local/texlive/2007/texmf-dist/tex/latex/graphics/graphics.sty
(/usr/local/texlive/2007/texmf-dist/tex/latex/graphics/trig.sty)
(/usr/local/texlive/2007/texmf/tex/latex/config/graphics.cfg)))) (./Simlie.aux)
(/usr/local/texlive/2007/texmf-dist/tex/context/base/supp-pdf.tex
[Loading MPS to PDF converter (version 2006.09.02).]
)

Package asymptote Warning: file Simlie-1.pdf does not exist on input line 9.

(./Simlie.aux) )
No pages of output.
Transcript written on Simlie.log.

Your problem is here:
error: could not load module './‘basename Simlie.tex .tex‘.asy'

Try to replace in the file [i]asyEngine.engine[/i] the line
[code]basefname="${location}/`basename "$1" .tex`"[/code]
by this one (cleaner):
[code]basefname="${1%.tex}"[/code]
(Make a Copy and Paste the code please)

Great! It works now!

Thanks for all of the help.

The [url=http://www.math.ualberta.ca/~mss/misc/TeXShopAndAsymptote.pdf]pdf[/url] is now updated.


Let be $ a,b,c>0$ such that $ a^2\plus{}b^2\plus{}c^2\equal{}x$ and $ ab\plus{}bc\plus{}ca\equal{}y$ . Prove that :

$ \frac{1}{x\plus{}y\plus{}2ab}\plus{}\frac{1}{x\plus{}y\plus{}2bc}\plus{}\frac{1}{x\plus{}y\plus{}2ca}\le \frac{1}{2}\left(\frac{1}{a^2\plus{}y}\plus{}\frac{1}{b^2\plus{}y}\plus{}\frac{1}{c^2\plus{}y}\right)$

the right-hand side is equal to $ \frac{a\plus{}b\plus{}c}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$

denote (a+b)(b+c)(c+a)=P

the first fraction on the left-hand side can be multyplied and divided by (2c+a+b) so to have

$ \frac{(2c\plus{}a\plus{}b)}{(x\plus{}y\plus{}2ab)(2c\plus{}a\plus{}b)}\equal{}\frac{2c\plus{}a\plus{}b}{4P\plus{}2c^3\plus{}a^3\plus{}b^3\minus{}bc^2\minus{}b^2c\minus{}ac^2\minus{}a^2c}\le\frac{2c\plus{}a\plus{}b}{4P}$.

Doing similar transformations with the other two fractions and then performing the summation prooves the desired inequality.


Prove that

$ \frac{3^{n\plus{}1} \minus{} (\minus{}1)^{n\plus{}1}}{4} \equal{} \sum_{k\equal{}0}^{n} \sum_{x_1 \plus{} ... \equal{} n\minus{}k} \left( F_{x_1\plus{}1}^2 \plus{} ... \right)$

where the inner sum runs over all partitions of $ n\minus{}k$ and $ F_n$ is the usual Fibonacci sequence.

[hide="Hint 1"] $ F_{n\plus{}1}$ is the number of ways to tile a $ 1 \times n$ rectangle with $ 1 \times 1$ and $ 1 \times 2$ tiles. [/hide]
[hide="Hint 2 (large)"] $ \frac{3^{n\plus{}1} \minus{} (\minus{}1)^{n\plus{}1}}{4}$ is the number of ways to tile a $ 2 \times n$ rectangle with $ 1 \times 1$ and $ 1 \times 2$ tiles. [/hide]

My apologies:  the inner summation should actually look like

$ \sum_{x_1 \plus{} ... \plus{} x_k \equal{} n} F_{x_1}^2 \plus{} ... \plus{} F_{x_k}^2$;

in other words, the entire sum just runs through the partitions of $ n$.


Is it trues with $ a, b, c \geq\ 0$ ?
I have a solution by SOS but it's only trues with a, b, c are three sides of triangle.
@arqady: Can U post your solution, please ?

[quote="nguoivn"]
@arqady: Can U post your solution, please ?[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=224112

[quote="can_hang2007"][quote="manlio"][quote="ely_en"]Let $ a, b, c > 0$ Show that:
$ \frac {a^2 \plus{} bc}{b \plus{} c} \plus{} \frac {b^2 \plus{} ca}{c \plus{} a} \plus{} \frac {c^2 \plus{} ab}{a \plus{} b}\ge \frac {a(b^2 \plus{} c^2)}{a^2 \plus{} bc} \plus{} \frac {b(c^2 \plus{} a^2)}{b^2 \plus{} ca} \plus{} \frac {c(a^2 \plus{} b^2)}{c^2 \plus{} ab}$[/quote]

It is so nice that I think it is better to post it in this forum so everybody can enjoy himself with it

P.S. Arqady found a full expansion solution which posted in referred forum[/quote]
Hint: Think of Chebyshev's Inequality. :)[/quote]

How to use Chebyshev's ineq?

Thank you very much.

Can you please post your solution, can_hang2007?

Thank you very much. :)


MATHCOUNTS quadratics calculus integration algebra polynomial function

Let x be a number such that the additive inverse of x equals the multiplicative inverse of x. f(x) is a quadratic such that all coefficients are integral and x is a zero. If the sum of the coefficients of f(x) is x^y, what is y?

I made it myself so please don't overcriticize...*braces*

Nice problem.  One suggestion -- don't say "express the coefficients in terms of x."  Ask for the polynomial as a whole or the coefficients as numbers.

Thank you :)

Mmm thought it would be interesting to start with a number x and end with it too...

My only objection is that because one can solve explicitly for x and the coefficients, there are many right answers.

For example, suppose that in a question you found that y = 1 and z = 3 and you are asked to express z in terms of y.  One obvious answer would be z = 3y.  Another would be z= y2 + y + 1.  And so on.

Oh I see what you're saying, it's fixed now, thanks.

Nice rephrasing.  Now, we just need someone to solve it :).

I'm thinking that this may actually be a bit advanced for mathcounts ... but we'll see.

Um one more suggestion to reword the problem.. you are using the variable x twice :S. Which makes f(x) not a quadratic at all since you already decided that x was a number. but yeah. that needs changing.

Yeah, but that's one of the thing to make it complicated...

Explanation: [hide]One's a variable and one's a number but the solver need to think about which is which...[/hide]

[hide]so from the first piece of info i get that -x = 1/x which means x :^2: = -1 and x = i.

So, a quadratic for which x is a 0 and all coefficants integers would be (call the function f(t) to clear things a bit) f(t) = t :^2:+1. add the coefficients and you get 2. But i^y can never equal 2...[/hide]

Mmm, I suggest you look up the definition of coefficients, I think they only apply when there's a variable (for example they are the a in the form ax^b)

I'm sorry Tare, but you can only have it standing for one at once. At the moment the problem is unsolvable. And the constant coefficient is a coefficient nonetheless.

well the the x term is simply x^0 which equals one.

so a*x^0 = a, which becomes the constant term.

You still just can't let x stand for two different things in one problem.. at the moment, we know x is a number, so f(x) isn't a quadratic at all, but it equals 0, as we were given, so the sum of the coefficients is 0, so now we want to solve the equation i^y = 0. hmm..

the problem still needs rewording.

Oh, quit worrying about little things :)

I agree with Stephen that it's pretty important. I won't try to evaluate integrals with the variable of integration as a bound, for example. It's polite to write well-formulated problems.

[quote="TripleM"]JBL, I wouldn't call it a little thing, I honestly have no idea what the whole question is talking about. Not a clue, nor what you are being asked to find.[/quote]
Then I guess the problem served its purpose. After all the whole point of this problem is to confuse people and I see I've successfully confusd two, so congrats to myself! *pat pat* :)

I can do that too :P. Prove that 1=2. Basically uses the same sorta ideas as the last puzzle :P

I am confused whether or not to snip this thread for it serving no purpose.

If you do that I'll be so mad at you I'll **** *** your ****. This is like the first successful problem that I made up myself and if anyone so much as lay a finger on it I'll get very emotional.

why has this been moved from mathcounts to f&g?

OMG...this is a blatant attempt to p*** me off for those of you who don't understand what this is...I've sent it back, and I expect better of you, come on.

You all need to relax --
Stephen and Simon, I think you're being hyper-literal.  Agreeing that the problem was, strictly speaking, not rigorously written: isn't it obvious what he was asking?  I mean, come on.  He used the same letter twice for two different things -- it was still quite evident what he meant.

No. I'm being 100% serious. I have no idea what the problem is talking about. Tare, I wouldn't call this problem successful.

while not taking sides, i hope this clears it up:

Let t be a number such that the additive inverse of t equals the multiplicative inverse of t.  The polynomial f(x) is a quadratic such that all coefficients are integral and t is a zero. If the sum of the coefficients of f(x) is t^y, what is y?

That's a better reformulation of the problem, though the answer to it is much more complicated than anyone is suspecting.

As the problem is stated by zscool, t = :pm: i, and f(x) = x2 + 1. The sum of the coefficients of f(x) is then 2 (not 1!!). So you're trying to solve (:pm: i)y = 2, for y. This does have solutions, but they are imaginary!

The original formulation of the problem was incorrect. You absolutely cannot use the same variable in two different ways like that. But zscool's reformulation of the problem is good - and it leads to a very interesting, but very difficult problem. I certainly wouldn't expect for many mathcounts students to be able to solve it. Look below if you'd like to see the solution in spoiler.

[hide]We're trying to solve (:pm: i)y = 2, for y. Taking the ln of both sides, we get y*ln (:pm: i) = ln 2. So y = ln 2/ln (:pm: i). The question then becomes, what in the heck is ln (:pm: i)??

Well any complex number, z, can be written as z = r*ei*v, where v is the arguement of z. So ln z = ln r + i*v (assuming the usual branch of the ln function). Using this on z = :pm: i, we get ln (:pm: i) = :pm: (2n+1)*:pi:*i/2, where n is any integer. So y = :pm: 2*ln 2*i/((2n+1)*:pi:). Whew![/hide]

So in the end it turned out to be a nice problem. It was just poorly worded at first, and turned out to be way more difficult than Tare might have suspected.

Well, gauss, I got to say even you are wrong :)

f(x) = k(x^2 + 1). Its sum of coefficients is 2k. So you are trying to solve the equation ( :pm: i ) = 2k.

Ah! In deed you're right. It doesn't say that the quadratic has to be monic. I also didn't account for the multiple complex branches of ln 2; and the :pm: signs weren't necessary because I said that n could be any integer. Sooo the total set of possible solutions should be:

[hide]y = 2*ln (2k)*i/((2n :pm: 1)*:pi:) + 4m/(2n :pm: 1), where k is an integer not equal to 0, and m and n are any integers.[/hide]


EDIT: The "Introduce Yourself!" Topic replaces this thread.

December 26, 2006: This day will be remembered as the birth of the Iowa AoPS Local Community! (Put it on your calendar)

Welcome, fellow comrades!

I'll reiterate a few things from the other forums:

(1) Regular forum rules apply (we all know that that means...).
(2) Please make this forum active! I know it won't be as exciting as other parts, but it's worth something.
(3) Tell all your friends to come here; this is place is going to get about four posts/year if we don't get larger. Post some regional competition times (i.e. ARML, GPML), problem sets, etc. that will make this subforum actually interesting and worthwhile.

There's not much else to say...anyone?

[quote="hillarryous"][quote="kevinkore3"]I'll bet larry will never reply to this post :P[/quote]

Congrats—you just got me to log in and write my first post in almost two years.[/quote]
Darn, I thought you had ditched AoPS, but apparently the Iowa forum is still on your watch list :O

This forum has gotten well over 4 posts per year :-D


An hollow sphere weights 47 grams. We put 203 mL of water inside that sphere and next we put it floating in water: one observe that half the sphere is submerged.

(a) Compute the volume of the sphere.
(b) From that sphere another one, with equal volume, is suspended. What should be the weight of this new sphere in order for the two spheres to be floating but completely submerged?

\[\rho \frac{V}{2}g = m_{0}g+\rho V_{1}g,\] where $m_{0}= 47\text g$ and $V_{1}= 203\text{ml}$; $\rho = 1000 \text{kg/m}^{3}$ is the density of the water. So \[V = 2 \frac{m_{0}+\rho V_{1}}{\rho}= 0.5 \text l.\]

As for (b), the weight of the new sphere, $Q = mg-\rho V g$, must equal the additional buoyant force $\rho\frac{V}{2}g$, so \[m = \frac{3\rho V}{2}= 0.75\text{kg}.\]

Do check my calculations, they're often wrong ;)


Can someone explain me how to take this equation :$ n^{5}\minus{}5n^{3}\plus{}60n^{2}\minus{}56n$ in module 3 and 5? Please explain it in very simple words because i have problems with notation. Thanks! :D

$ n^5 \minus{} 5n^3 \plus{} 60n^2 \minus{}56n \equiv n^5 \plus{} n^3 \plus{}n \mod 3$

$ n^5 \minus{} 5n^3 \plus{} 60n^2 \minus{} 56n \equiv n^5 \minus{}n \mod 5$

Simply reduce the coefficients in each mod. So why do you want to do this? Are you trying to see if the polynomial is always divisible by 3 and 5, or are you learning about irreducible polynomials?

jeje, i was proving that it is always divisible by 15  :P  It is very clear. Thank you!
Do you know of some resources for learning ALL about modular algebra from the begining??

Here are a couple of links:

http://en.wikipedia.org/wiki/Modular_arithmetic

http://mathworld.wolfram.com/ModularArithmetic.html

http://www.math.rutgers.edu/~erowland/modulararithmetic.html

http://www.cut-the-knot.org/blue/Modulo.shtml


Find the min of 1/x + 4/y + 9/z, if x + y + z = 1.


well i am a new user and i couldnt find posts on how to crack kvpy so aif anyone can help i will appreciate

I don't think you'll find 'cracks' to anything on this site  :rotfl:
But yeah, just search KVPY here- http://www.artofproblemsolving.com/Forum/search.php and check out the former discussions in case you need to give yourself satisfaction of being on the right tracks.

well i meant what basic knowledge should one have
what the prerequisites to get selected

[quote="eldiablo"]well i meant what basic knowledge should one have[/quote]
Class 10th science. Mostly NTSE stuff, a little 11-12.
[quote="eldiablo"]what the prerequisites to get selected[/quote]
Being reasonable smart and passionate about science

my teacher said that they might ask 9th std cbse portions as well

[quote]arrey! 10th means 1-9 as well![/quote]how is this?

I mean, it doesn't mean....but pretty obvious yaar! Forgive my lil fingers for not explicitly saying it earlier.


The sequence $ (a_n)$ is defined by $ a_1\equal{}a_2\equal{}a_3\equal{}1$ and $ a_{n\plus{}1}a_{n\minus{}2}\minus{}a_n a_{n\minus{}1}\equal{}2$ for all $ n \ge 3.$ Prove that $ a_n$ is a positive integer for all $ n \ge 1$.

PMI on $ a_{n\plus{}2} \equal{} (3 \plus{} ( \minus{} 1)^n)a_{n \plus{}1} \plus{} a_n$.


Find all the integers x,y,z,t such that

$x^y+y^z+z^t=x^{2005}$

[quote="Rust"]Infinetely series: $(x,2005,-2005,-2005),(1,-1,1,t),(1,1,-1,t \is \ odd)$.[/quote]

How do you get that? :blush:

I think you guess the answer , don't you ?  :D
I love math.

I think this one should be solutions too.
$(x,2005,-2005^{2005},-2005^{2004})$
And like this so many.


One night, 18 people go out to dinner.  When it comes time to pay their tip, all of them reach into their pockets and take out 3 coins: two half-dollars and one quarter.  To their surprise, they realize that all of Alice's coins were minted in 1991, all of Bob's coins were minted in 1992, all of Cedric's coins were minted in 1993, ... and all of Rodrigo's coins were minted in 2008.  In their excitement, they each color one of their half-dollars red and the other one blue.  Remembering that they have to pay their tip, they each throw one coin into the middle of the table.  How many possible ways are there for them to pay a $ \$6.00$ tip?

This is probably completely and utterly wrong, but:

[hide]
Using generating functions, we get:

$ (2x^{50} \plus{} x^{25})^{18}$

The term with $ x^{600}$ will have a coefficient of:

$ C(18,6)(2^6)$
$ \equal{} 1188096$?[/hide]

[hide]The possible methods of obtaining 6.00 are:

12 HD 0Q
11 HD 2Q
...
3  HD 18Q

However, each person throws 1 coin, thus # of HD + # of Q = 18, so only
6 HD 12 Q works.

For each of the 6 Half Dollars, there can be 0-6 that are blue.
Thus there are $ 7$ possibilities[/hide]

[hide="But..."]All the people are distinct, so you must choose 6 out of 18 of them to throw a half dollar. Then among these 6, each can throw a red or blue half dollar. So there are $ \binom{18}{6} \cdot 2^6$ ways, which agrees with Carbon57's answer.[/hide]

oh yea  :D
my solution only works for clones then I guess


Three siblings have a gift of $ \$169$ to split in the ratio of $ \frac12 : \frac13 : \frac14$. What is the greatest number of dollars that any of the siblings will receive?

Since 6+4+3 = 13,

the first should actually take 6/13 * $ \$$ 169 = $ \$ 78$.


I'm looking for an estimate for V(2, n). Is there any known, simple one?

The fact that there is none listed in the [url=http://www.research.att.com/~njas/sequences/A005346]OEIS[/url] suggests that none is known.  According to [url=http://mathworld.wolfram.com/SzemeredisTheorem.html]Mathworld[/url], the best bound on van der Waerden numbers are derived from Szemeredi's Theorem.


Let $x^2-a(a-2)x-(a-1)^2=0$ with $a\in R$

$a)$ Show that the ecuation har real roots for any real a and solve it.
$b)$ Find all the valuse of a which verify this ineq: $2\cdot\sqrt{x_{1}+x_{2}-2\cdot(a-2)}-3\cdot\sqrt{-x_{1}x_{2}}\geq 1$, whwre $x_{1,2}$ are the roots of the given ecuation.

$a$:
$(-a(a-2))^2+4(a-1)^2>0$ so you will always have real roots.
i don't have time to do the second one know

$ a)\ x^2-a(a-2)x-(a-1)^2=0 \Longleftrightarrow (x+1)(x-(a-1)^2)=0\Longleftrightarrow x=-1,(a-1)^2$ Q.E.D.

$ b)$ Using the relation of root and coefficient for quadratic equation, $x_1+x_2=a(a-2),\ x_1x_2=-(a-1)^2$.

Therefore we have $2\sqrt{x_1+x_2-2(a-2)}-3\sqrt{-x_1x_2}\geqq 1\Longleftrightarrow 2|a-2|\geqq 3|a-1|+1$

Solving this with sketching two graphs $b=2|a-2|$ and $b=3|a-1|+1$ on the $a-b$ plane, we obtain $0\leqq a \leqq \frac{6}{5}$.

[quote="shobber"]Kunny, your solution to the first part has a small mistake.[/quote]

Yes. You are right. I will edit. Thank you for pointing out my mistake.

kunny

yeah, just noticed i forgot my ||  :(  :blush:
*is going to redo it*

or were you talking about the 3? that should be a 9 of course  ;)


Only CTC Members may sign up. You will all be broken up into teams of two. Here's how it will work:

[u]Team AB vs. Team CD[/u]

A vs. C
A vs. D
B vs. C
B vs. D

Best of 5 Countdowns.

If the total amount of matches one is tied after 2-2, there will be a Team Battle.

Two teams of two members, one from each team, find a time to meet up on FTW and play one game (15 questions and 20 seconds) so two total normal games are played. A and B's scores are then summed and compared the to the sum of C and D's scores.

[hide="People Eligible To Participate"]
Skaldjfhrulez
Violin321
Kl2836
Xoangieexo
Yoh2241
Jongao
Dojo
Pacman2812
Myyellowducky82
Vamathletes
Guineapiggies
Pinkpiglet
Jain_harshi
Splatyango
Cherry_pi
RomanianGenius
Weird888
Dragonlaird7
Dragon96
Dannyhamtx
Kookamunga
Nikeballa96
Pi guy
27r
Yclarinet
Chenhsi
Textangle
Jjx1
Aime_is_hard
Pi-dimension
Mind Storm
Bisbmard9
Emilylambkin
Johnnybeard
Discrete_math
[/hide]

[hide="People Who Have Signed Up So Far"]
Isabella2296
Mewto55555
superllamel
AIME15
Lotsofmath
Vallon22
Ernie
Mathblitz
FantasyLover
Mathnerd711
[/hide]

izzy didn't mention this, but i think it's vital to keep spam at a controllable level

pm your "i'll join's" to her

[quote="vallon22"]izzy didn't mention this, but i think it's vital to keep spam at a controllable level

[b][size=200]pm your "i'll join's" to her[/size][/b][/quote]

anyway, i think it'll be unfair if super and fantasy are together
however, we'll discuss the teams after the deadline for joining
BTW izzy, when is the deadline?

I posted here because I didn't just post "I'll join." I also made a request for what team I'm on.

I'll join.  But how will the teams be decided?

[quote="superllamel"]I would like to join and I want to be on Fantasy's team (if it's ok with him).[/quote]

ill join.

why would it be not okay with me  :juggle:

I don't know Fantasy, but I think we should get some say in the teams. I would have way more fun playing if I got to choose my team, and others might feel the same way. :juggle:

Oh, all right. If it's fine with Mewto and Ernie, I guess you guys could choose teams. I never really did understand why you want it fair. Is Bobby Shen's team not allowed to participate in MathCounts together because it's "unfair"?

I call superllamel, then rg, then fantasy (in that order,based on availability)

Whoa.  What about ernie?

I remember I pwned you twice last time we cd-ed.

And I can pwn Fantasy.

I call super, mewto, fantasy, rg, izzy (in that order)

I call Fantasy (in that order).  :lol:

And I suppose then it goes mewto, izzy.

Wow.

Okay, so here are teams based on who called whom first. (To be constantly updated, until I have to make a new post.)

Team Fantasyllamel

Team mewtoGenius

Team IzzyErnie

Team vallonofmath

Team AIMEblitz

So far, Mathnerd711 doesn't have a team.

You must first join CTC.  Please look at the Introducing CTC thread.  IDC if u don't like reading online.  :wink:  :P  :rotfl:

Where are we in this tournament? Has this been mod-abandoned?

Yes, we are. Guys, [b]play your games[/b].

I haven't seen irsmart or mathblitz in weeks.

...and i dont think CA Math was on for a while.

Private message your opponents. If you make sure to let me know that you pmed your opponents and they didn't reply before the deadline, it will count as a win for you. The deadline, by the way, is Sunday, October 5th, midnight. This should be plenty of time, assuming your opponent is cooperative.

I forfeit against mewto55555, which is to say I also forfeit against romainiangenius.

So people don't have to go to the last page for matchups...

A. romanianmewto vs. jamesangle
1. mewto55555 defeats james4l by forfeit.
2. RomanianGenius defeats james4l by forfeit.
B. AIMEnerd vs. dragonx1
C. ernzzy vs. pinkpiggies
1. ernie defeats guineapiggies by forfeit.
2. isabella2296 defeats guineapiggies by forfeit.
D. lotsofvallon vs. irblitz
E. CA Math vs. Fantasyllamel

[quote="ernie"]So people don't have to go to the last page for matchups...

A. romanianmewto vs. jamesangle
1. mewto55555 defeats james4l by forfeit.
2. RomanianGenius defeats james4l by forfeit.
B. AIMEnerd vs. dragonx1
1. AIME15 defeats dragon96 by forfeit.
2. AIME15 defeats jjx1 by forfeit.
3. mathnerd711 defeats dragon96 by forfeit.
4. mathnerd711 defeats jjx1 by forfeit.
TEAM AIMEnerd DEFEATS TEAM dragonx1.
C. ernzzy vs. pinkpiggies
1. ernie defeats guineapiggies by forfeit.
2. isabella2296 defeats guineapiggies by forfeit.
D. lotsofvallon vs. irblitz
E. CA Math vs. Fantasyllamel[/quote]

Sorry izzy, but rg and i are dropping out. this tournament is just log string of forfeits, no actual playing.

[quote="isabella2296"]This Tourney is officially canceled. No one is playing games, and the school year is a terrible time. Perhaps to be continued in the summer.[/quote]

Recycle time! [b]RECYCLE! RECYCLE! RECYCLE! RECYCLE![/b] RECYCLE THS THREAD IZZY!!!!

hmmm. Wasn't CTC suppose to make tournies that actually ended. I mean one of teh reasons you made CTC was because there were lots of tournies that did not finish. Actually, after I think about it not one tourney from CTC has finished

Andy, if you have complaints about CTC, talk to Mewto. Besides, those were [b]summer tournies[/b]. It's not summer anymore.

Locked.


By Pythagorean Theorem, $ \overline{AD}$ and $ \overline{DC}\equal{}9$ and $ 8$ respectively.

Find the area of the triangles and add \[ \frac{9\cdot12}{2}\plus{}\frac{8\cdot15}{2}\equal{}\boxed{114}\]


a,b,c>0
=>
(a+b)(b+c)(c+a) >= 8^1/2 * [abc(a+b+c)]^4/3
ha ha very nice and hard


Based on an article about WorldCup team elimination:
"The second round, or knockout stage, is a single-elimination round in which teams play each other in one-off matches, with extra time and penalty shootouts used to decide the winner. It begins with the round of 16 (or last 16) in which the winner of each group (from the group stage) plays against the runner-up from another group. This is followed by quarter-finals, semi-finals and a final. The losing semi-finalists contest a third place match. Up until 2002, there were nearly 2000 goals that had been scored in the finals tournament", (assume that the position in the tournament ladder is randomly selected).

Question:
1) If 16 teams in the second round are evenly matched, and they
include Brazil and England. What is the probability that both team
meet in a match during the tournament.

2) Replace 16 by 2^n in the above question. What is the probability
that they meet?

I have no idea how to solve the questions above. Please help! Thanks in advance!

[size=59]
Grammar Ninja
Stealthily edits the post
and loves haiku
[/size]

Since ther are 16 teams and two one desired matchup, consisting of a group of two. I would use 16 P 2 to find the total # of ways in which these teams can match up, this goes in the denominator. now I also want to know one possibility so 2 divided by total # of outcomes, gives you the probability. which is .008. I think this mitght me wrong . you could alos try 1/16 simply, which is a more realistic answer.

If you want to know the answer you've got to assume that every team has a same chance to win a match despite the opponent, but this assumption is absolutly not true in reality.

@kaca123

It wouldn't be 16_P_2 becuase you can only use each team once.   :ninja: For example if you have Ukraine vs. Australia then that eliminates two teams you can use.
By saying 16_P_2 you would be saying that you can use each team more than once. I think.....<<;;

Can't we find the probability that they meet in the Round of 16, the Quarterfinals, Semifinals, and Finals separately, then add them?

[quote="jhollenbeck"]Probability they meet in the first round=$\frac{1}{15}$ Probability they meet in the second round=$\frac{1}{4}$ (probability they both make it) time $\frac{1}{7}=\frac{1}{28}$  Probability they both make the third round=$\frac{1}{16}$  Probability they meet in the third round=$\frac{1}{3}*\frac{1}{16}=\frac{1}{48}$  Probability they meet in the final round is $\frac{1}{64}$  The probability that they will make it there because they will face off there if they both make it.  Now add up the fractions, $\frac{1}{15}+\frac{1}{28}+\frac{1}{48}+\frac{1}{64}=\frac{14923}{107200}$  Someone check with a calculator, I probably messed up doing it by hand.[/quote]
[hide] I got $\frac{1}{15}+\frac{1}{4}\cdot\frac{2}{15}+\frac{1}{16}\cdot\frac{4}{15}+\frac{1}{64}\cdot\frac{8}{15}=\frac{1}{15}+\frac{1}{30}+\frac{1}{60}+\frac{1}{120}=\frac{1}{8}$, which is nicer, but I don't know if it is right. [/hide]


Let a,b,c,k,q>0;and ab+bc+ca=1;and 2k>q.
Find the max of A:
A=$ k*\frac{1\minus{}a^2}{1\plus{}a^2}\plus{}q*(\frac{1\minus{}b^2}{1\plus{}b^2}\plus{}\frac{1\minus{}c^2}{1\plus{}c^2})$

If you want to find the maximum, then there should be constant value for $ k\plus{}q$ or something like that, because otherwise you could take $ k,q \to \infty$ and then $ A \to \infty$?  :roll:

No one in this world can solve my problem? :D

hint:
A max=$ \frac{2k^2\plus{}q^2}{2k}$


In quadrilateral $ ABCD$, angle $ DAC$ is congruent to angle $ DBC$, and $ \frac{[ADB]}{[ABC]}\equal{}\frac{1}{2}$. Let $ O$ be the intersection of $ BD$ and $ AC$. If $ AD\equal{}4$, $ BC\equal{}6$, $ BO\equal{}1$, and the area of quadrilateral is $ \frac{a\sqrt{b}}{c}$ in simplest form.  Find $ a\plus{}b\plus{}c$

[quote="unimpossible"]In quadrilateral $ ABCD$, angle $ DAC$ is congruent to angle $ DBC$, and $ \frac {[ADB]}{[ABC]} \equal{} \frac {1}{2}$. Let $ O$ be the intersection of $ BD$ and $ AC$. If $ AD \equal{} 4$, $ BC \equal{} 6$, $ BO \equal{} 1$, and the area of quadrilateral is $ \frac {a\sqrt {b}}{c}$ in simplest form.  Find $ a \plus{} b \plus{} c$[/quote]
There might be a nicer solution...
[hide="Solution"]
\[ \frac{[ADB]}{[ACB]}\equal{}\frac{1}{2}\iff \frac{AD\cdot BD\sin \angle ADB}{AC\cdot BC\cdot \angle ACB}\equal{}\frac{1}{2}\] \[ 2\cdot 4\cdot BD\equal{}6\cdot AC\implies BD\equal{}\frac{3AC}{4}\] Let $ AC\equal{}x$, so $ BD\equal{}\frac{3x}{4}$.  Now, $ \triangle AOD\sim \triangle BOC$ since $ \angle ADO\equal{}\angle OCB$ and $ \angle BOC\equal{}\angle AOD$ since $ ABCD$ is cyclic.  Thus, $ \frac{BC}{BO}\equal{}\frac{AD}{AO}\implies 6\equal{}\frac{4}{AO}\implies AO\equal{}\frac{2}{3}$.  Hence, $ OC\equal{}x\minus{}\frac{2}{3}$ and $ OD\equal{}\frac{3x}{4}\minus{}1$.  By Power of a Point, $ AO\cdot OC\equal{}BO\cdot OD\iff \frac{3x}{4}\minus{}1\equal{}\frac{2}{3}(x\minus{}\frac{2}{3})\implies x\equal{}\frac{20}{3}\implies OC\equal{}6, OD\equal{}4$.  There are several ways to find the area of $ \triangle BOC$ and $ \triangle AOD$, but either way, we get that $ [BOC]\equal{}\frac{\sqrt{143}}{4}$ and $ [AOD]\equal{}\frac{\sqrt{143}}{9}$.  Then, $ [AOB]\equal{}\frac{AO}{OC}\cdot [BOC]\equal{}\frac{1}{9}\cdot \frac{\sqrt{143}}{4}$.  In the same fashion, we find that $ [DOC]\equal{}\frac{OC}{OA}\cdot [DOA]\equal{}9\cdot \frac{\sqrt{143}}{9}$.  Hence, the total area is \[ \frac{\sqrt{143}}{9}[9\plus{}1\plus{}\frac{9}{4}\plus{}\frac{1}{4}]\equal{}\frac{25\sqrt{143}}{18}\] Therefore, the answer is $ 25\plus{}143\plus{}18\equal{}161\plus{}25\equal{}\boxed{186}$. [/hide]

[hide="a different way"]
Quadrilateral $ ABCD$ is cyclic.
$ \frac{[ADB]}{[ABC]}\equal{}\frac{1}{2}$
$ \Rightarrow 4BD\equal{}3AC$

Since $ \angle DAO \equal{} \angle CBO$, $ \triangle OAD \sim \triangle OBC$.
Therefore, $ AO\equal{}\frac{2}{3}$.
We can easily find $ DO\equal{}4$ and $ CO\equal{}6$ if we use $ 4BD\equal{}3AC$ in conjunction with the triangle similarity.

By similar reasoning as before, $ \triangle OAB \sim \triangle ODC$
Let $ [OBC]\equal{}z$  and $ [OAB]\equal{}x$. Then, $ [OAD]\equal{}\frac{4z}{9}$ and $ [ODC]\equal{}36x$

We can use $ \frac{[ADB]}{[ABC]}\equal{}\frac{1}{2}$ to solve for $ z$ in terms of $ x$.
We find $ z\equal{}4x$. Now we have the area of the entire quadrilateral in terms of $ z$, specifically
$ [ABCD] \equal{} \frac{50z}{9}$.

$ z$ is easily calculated by Heron's (because $ \triangle OBC$ has integer side lengths).
$ z\equal{}\frac{\sqrt{143}}{4}$.

Therefore,
$ [ABCD]\equal{}\frac{25\sqrt{143}}{18}$
and our answer is $ \boxed{186}$

[/hide]


calculus
integration
inequalities
function
real analysis
real analysis unsolved

Prove that there exists a constant $ C<\infty$ such that for any $ \gamma>1$ which is not an integer and for any $ \lambda, b>1$ we have

\[ \left|\int_0^be^{i\lambda t^{\gamma}}dt\right|\leq\frac{C}{t^{1/\gamma}}.\]

[quote="Cezar Lupu"]
\[ \left|\int_0^be^{i\lambda t^{\gamma}}dt\right|\leq\frac {C}{t^{1/\gamma}}.
\]
[/quote]

 Cezar could you correct RHS ?

Probably the correct inequality is $ \displaystyle \Big|\int_0^b \exp (i \lambda t^{\gamma}) \, dt \Big| \le \frac{C}{\lambda^{1/\gamma}}$, in which case by redefining variables, we see it suffices to take $ \lambda \equal{} 1$. The conditions that $ \lambda, b > 1$ or that $ \gamma \not \in \mathbb{Z}$ should not matter. So, we wish to show that we may uniformly bound $ \int_0^b \exp (i t^{\gamma})\,dt$ or, again by redefining variables, $ \frac{1}{\gamma} \int_0^b \exp(is) s^{1/\gamma\minus{}1} ds$. If we let $ \alpha \equal{} \frac{1}{\gamma}$, we wish to uniformly bound $ \alpha \int_0^b s^{\alpha\minus{}1} \exp(is) \, ds$ where $ 0 < \alpha < 1$. Well, we may consider this function as a continuous function on $ \alpha \in [0, 1], b \in [0, 2\pi]$ by defining its value for $ \alpha \equal{} 0$ as the limit, i.e. identically one, and then a continuous function on this compact space has some uniform bound. Now in fact I think that for all values of $ \alpha$, the value of $ b$ which maximizes the norm does in fact occur in $ [0, 2\pi]$ but I do not see how to prove this claim elegantly. Fortunately this step does not seem to be necessary as I may bound the rest using standard integration-by-parts.


abstract algebra
probability
superior algebra
superior algebra unsolved

In any non-abelian group G with n elements, the number of pairs $ (a,b)$ such that $ab=ba$ is at most $\frac{5n^2}{8}$.

amazing isn't it. Moreover the bound is achieved ..

It seems it was in

RMS Revue Mathmatiques Spciales
Anne 2001-2002
volume 2
page 275-279

This problem is known in probability.

$G$  finite group, the probability of $G$ to be commutative is $p(G)=\frac{card[(a,b) \in G^2;ab=ba]}{card(G^2)}$

If $G$ is not commutative then $p(G)\leq 5/8$

$C(G)$ the center of $G$; $C_{G}(a)=[b\in G; ab=ba]$ centraliser of $a$

If the group is not commutative

1/Compare $card\;(G/C(G))$ with $4$
2/ Use classes equation for group to get a relation beetween  $p(G)$ and
$\frac{card\;C(G)}{card\;(G)} + \sum_{a\in G-C(G)}\frac{card\;C_G(a)}{(card\;G)^2}$

And get the upper bound.

Yes, moubi, I know this solution, because it is the one I also found. But where can I find this journal Revue de...?

[quote="harazi"]Is there any place where I can find this magazine, moubi?[/quote]

I will get a photocopy for you, when it is ready will let you know  :cool:

Thanks a lot, Moubi, you're most kind. It's nice to be here in France, with so nice people arround. Thank you very much.

hi Harazi, you can also find this journal at the library of your school ..

Yes probably in library of LLG they have it.

RMS has web page, some sample of math-question is available

http://www.rms-math.com/

[quote="Moubinool"]This problem is known in probability.

$ G$  finite group, the probability of $ G$ to be commutative is $ p(G) \equal{} \frac {card[(a,b) \in G^2;ab \equal{} ba]}{card(G^2)}$

If $ G$ is not commutative then $ p(G)\leq 5/8$

$ C(G)$ the center of $ G$; $ C_{G}(a) \equal{} [b\in G; ab \equal{} ba]$ centraliser of $ a$

If the group is not commutative

1/Compare $ card\;(G/C(G))$ with $ 4$
2/ Use classes equation for group to get a relation beetween  $ p(G)$ and
$ \frac {card\;C(G)}{card\;(G)} \plus{} \sum_{a\in G \minus{} C(G)}\frac {card\;C_G(a)}{(card\;G)^2}$

And get the upper bound.[/quote]

I'm trying to find the solution of this problem .

Could you please write the solution of $ 1$ and $ 2$ (Need details ).

 :roll:

Let $ Z(G)$ be the center of $ G$.  Note that for any $ b\not\in Z(G)$, $ C_G(b)\neq G$.  But $ b\in C_G(b)$, so $ C_G(b)\neq Z(G)$ and we have two proper inclusions: $ Z(G) < C_G(b) < G$.  It follows that if $ G$ is nonabelian, $ Z(G)$ cannot have prime index.  In particular, the index is at least $ 4$.

Now, it's easy to write $ |\{(a,b)\in G^2; ab \equal{} ba\}| \equal{} \sum_{a\in G} |C_G(a)| \equal{} |G|\cdot |Z(G)| \plus{} \sum_{a\in G \minus{} Z(G)} |C_G(a)|$.

Since all the centralizers in the sum are proper, they all have index at least $ 2$ and we can write: $ |G|\cdot |Z(G)| \plus{} \sum_{a\in G \minus{} Z(G)} |C_G(a)| \leq |G|\cdot |Z(G)| \plus{} (|G| \minus{} |Z(G)|) \cdot |G| \cdot \frac {1}{2} \equal{}$ $ |G|^2 (\frac {1}{2} \cdot \frac {|Z(G)|}{|G|} \plus{} \frac {1}{2})\leq \frac {5}{8} |G|^2$

We would have equality with any nonabelian group whose center has index $ 4$, by reversing the above logic.  Both nonabelian groups of order $ 8$ have this property.  Is it easy to construct infinitely many?


In the expansion (a+b)to the Nth power, there are n+1 dissimilar terms. The number of dissimilar terms in the expansion of (a+b+c) to the Nth power is:
a) 13
b) 66
c) 55
d)91
e) 144.

[hide]the answer is D. but can someone explain why to me?[/hide]

Well it'd be helpful to know what $N$ was... but here's a general method of solving it. It's the same problem as finding the number of ways to distribute $N$ things (powers in this case) among three places (variables in this case).

Consider the following diagram: OO...OOXOO...OOXOO..OO where there are $N$ O's and $2$ X's. Everything before the first X is a power on $a$, between the $2$ X's is a power on $b$, and after the second X is a power on $c$. This way they always add up to $N$, which must be the case.

We have $N+2$ total X's and O's of which we want to choose two spots to be X's. So the desired number is $\binom{N+2}{2}$. I assume the $N$ in question is $N = 12$, which would give $\binom{14}{2} = 91$.

oh ooops, i forgot to type in 12. lol, thanks for explaining it to me


calculus
integration
parameterization
geometry
3D geometry
sphere
real analysis

I have to evaluate the integral of [b]F[/b] = (x+y,x+y,x+y+z) along the curve that is defined as the intersection of surfaces x^2 + y^2 + z^2 = 1 and 2x + y + z = 0 in two ways: a) directly and b) with Kelvin-Stokes theorem.

What is the parametrization of the curve in question?

This is the intersection of a plane and a sphere- the curve is a circle. There are many different parametrizations.

The easiest parametrization to find probably uses one of $ x,y,z$ as the parameter, and has two parts. It's not the easiest to calculate with.


[u][b]I want to write a thesis about applied mathematics,but i don't know what to write about.Could you give me some ideas ? Thanks![[/b][/u][/b]

You should simply try Google. There are some free available ebooks about applied mathematics.

You'll have to be more specific in what you're asking. What aspect of applied mathematics are you interested in writing about?


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

I believe she means this:

If $(0)=\bigcap_{i=1}^n\frak q_i$ is a minimal primary decomposition of $(0)$ (in a Noetherian ring) where $\frak q_i$ is $\frak p_i$-primary, then for every $i$ there is an $r$ such that $\frak p_i^r\subseteq\frak q_i$.

Of course, most of the hypotheses are superfluous :). The relevant fact here is that in a Noetherian ring, every ideal $\frak a$ contains a power of its radical $\sqrt{\frak a}$. Indeed, if $\sqrt{\frak a}$ is generated by $x_1,\ldots,x_t$ such that $x_j^{n_j}\in\frak a$, then $\left(\sqrt{\frak a}\right)^{n_1\cdot\ldots\cdot n_t}\subseteq\frak a$.

I hope I did not misunderstand the problem.


Is math jams class a class you don't have to sign up for and it is free? So everybody who can go?

That is correct.  We mainly use the Math Jams for information about our classes, but we also have one with an MIT Admissions officer and MIT Professor about applying to MIT, and we have Math Jams for the AMC and AIME tests.


calculus
integration
number theory proposed
number theory

For any given integer A , prove that the equation

(A^2 -1)(B^2 -1)  = (C^2 - 1) has at least two  integral solutions.

Actually there is a two-sided infinite sequence of poisitive integers which satify the equation.

Bomb


Putnam
modular arithmetic
inequalities
college contests
Putnam number theory

As you all know, there's a long tradition of "year" problems in mathematical contests. It's strong enough that you can often know when some otherwise unlabeled test was given just from the appearance of such a problem. More often than not, the year is simply an awkwardly large number. Sometimes it matters that the year is in some particular range - between certain powers of 2, between certain Fibonacci numbers. Sometimes it matters whether the year is even or odd, or what it is mod 3. There's a special place for problems in which the exact value of the year really, really matters. I'm posting this just because such a fine year problem should be in this forum somewhere.

[b]A5.[/b] Prove that there are unique positive integers $a,n$ such that $a^{n+1}-(a+1)^{n}=2001.$

First of all looking at the expression modulo a, gives us: $ a^{n+1}-(a+1)^{n}\equiv 2001$ (mod a) $ \Leftrightarrow 2001 \equiv-1$ (mod a) so: $ a|2002=2*7*11*13$
Now looking at it modulo 3: $ a^{n+1}\equiv (a+1)^{n}$ (mod 3) $ \rightarrow a \equiv 1$ (mod 3) and $ 2|n$ so we can do: $ n=2k$ and $ b=a+1$ and we get: $ (b-1)^{2k+1}-b^{2k}=2001$
Looking at it modulo b: $ 2001 \equiv-1$ (mod b) so $ (a+1)|2002$ after some inspection we see that the only possible pair of consecutive positive integers that both divide 2002 are (1,2) and (13,14), (1,2) obviously doesn't work, so $ a=13$ now it's easy to see that $ n=2$ is the only solution and $ 13^{3}-14^{2}=2001$

[quote="Afonso Bandeira"]... now it's easy to see that $ n=2$ is the only solution and $ 13^{3}-14^{2}=2001$[/quote]
You've made a good argument, and your solution is correct as far as it goes, but it also incomplete. I doubt that it would earn more than 2 points in the strict Putnam grading system. The issue: you have not yet ruled out the possibility of a solution with $ a=13$ and $ n>2.$ ("It's easy to see" isn't a phrase likely to impress Putnam graders.)

From my 2001-2002 writeup for my class (I don't remember how much jmerry had to do with this):

[hide="The rest of the solution."]The only thing we have left to prove is that this is the only answer.
Since $ 2001$ factors as $ 3\cdot23\cdot29$ and we had some success mod $ 3,$ we consider the equation mod $ 23$ and mod $ 29.$ In both cases, the equation becomes  $ 13^{n+1}\equiv 14^{n}\pmod{23 \text{ or }29},$ which we rewrite as  $ 13 \equiv (14/13)^{n}.$ Whatever $ 14/13$ is (it turns out to be $ 17 \pmod{23}$ and $ 10 \pmod{29}$ but we don’t really nead to know that), by Fermat’s little theorem, its multiplicative order mod $ 23$ must divide $ 22$ and its multiplicative order mod $ 29$ must divide $ 28.$ Consider it mod $ 23.$ The order of $ 14/13$ must be $ 2, 11,$ or $ 22,$but since $ (14/13)^{2}= 13,$ the order is not $ 2.$ Thus the order of $ 14/13$ must be $ 11$ or $ 22,$ and $ 13 = (14/13)^{n}$ can only happen if $ n \equiv 2 \pmod{11}.$ Consider it mod $ 29.$ The order of $ 14/13$ must be $ 2, 4, 7, 14,$ or $ 28.$ But $ (14/13)^{2}\equiv 13$ and $ (14/13)^{4}\equiv 13^{2}\equiv 24 \pmod{29},$ so the order of $ 14/13$ must be $ 7, 14,$ or $ 28$ and $ 13 = (14/13)^{n}$ can only happen if $ n \equiv 2 \pmod{7}.$ We have a Chinese remainder problem whose solution is $ n \equiv 2 \pmod{77}.$ But the additional consideration that $ n$ be even forces $ n \equiv 2 \pmod{154}.$ That means that the next possible solution after $ n = 2$ is $ n = 156.$ But $ 13^{157}-14^{156}= 13^{156}(13-(14/13)^{156}).$ Now use Bernoulli's Inequality: $ (1+x)^{n}\ge 1+nx$ for $ n$ a positive integer, $ x \ge-1:$ $ (14/13)^{156}\ge 1+\frac{156}{13}> 13.$ Hence,
$ 13^{157}-14^{156}< 0$ and cannot be $ 2001.$ Even stronger versions of this estimate hold for larger values of $ n,$ so $ n = 2$ represents the only solution. 					[i]q.e.d.[/i]
[/hide]

I think that the specific trick of working mod 23 and 29 was my idea, but I'm not sure either.

In the actual contest, I didn't have time for that, and left an obviously incomplete solution, which was probably worth one or two points.

[hide="A better argument for the last part"][quote="JAM"]Now is not hard too see that only $ n \equal{} 2$ can work: If $ n\ge 3$, then $ 8|14^{n}$, so looking mod $ 8$,
$ 13^{n\plus{}1}\minus{}14^{n}\equiv 2001\pmod{8}\Rightarrow 5^{n\plus{}1}\equiv 1\pmod{8}$
But $ n$ is even and $ 5^{2}\equiv 1\pmod{8}$, so
$ 5^{n\plus{}1}\equiv 5\cdot 5^{n}\equiv 5\pmod{8}$

This is a contradiction, so $ n\le 2$. And now obviously only $ n \equal{} 2$ works. So $ a \equal{} 13,n \equal{} 2$ is indeed the only solution.[/quote][/hide]


Prove that for every distinct and rational $ x,y,z$ the expression

$ \frac{1}{(y\minus{}z)^2}\plus{}\frac{1}{(x\minus{}y)^2}\plus{}\frac{1}{(z\minus{}x)^2}$

is a perfect square.

Let $ a\equal{}y\minus{}z$, $ b\equal{}x\minus{}y$, $ c\equal{}z\minus{}x$ such that $ a\plus{}b\plus{}c\equal{}0$.

Then we have

$ \frac1{(y\minus{}z)^2}\plus{}\frac1{(x\minus{}y)^2}\plus{}\frac1{(z\minus{}x)^2}\equal{}\frac1{a^2}\plus{}\frac1{b^2}\plus{}\frac1{c^2}\equal{}\frac{(ab)^2\plus{}(bc)^2\plus{}(ca)^2}{(abc)^2}$

$ \equal{}\frac{(ab\plus{}bc\plus{}ca)^2\minus{}2(ab\cdot bc\plus{}bc\cdot ca\plus{}ca\cdot ab)}{(abc)^2}$

$ \equal{}\frac{(ab\plus{}bc\plus{}ca)^2\minus{}2abc(a\plus{}b\plus{}c)}{(abc)^2}\equal{}\left(\frac{ab\plus{}bc\plus{}ca}{abc}\right)^2$,

as desired.


I have a minor in Applied Mathematics received in conjunction with my Bachelor's of Engineering a few years ago from an accredited University.  I know every school has a slightly different credit system but, generally speaking, how many credits would I have to take to receive a Bachelor's in Mathematics?  Would I have to take classes I've already taken inorder to receive the full degree or could I simply continue where my minor left off and build upon that?

The is greatly dependent on the institution you attend for your BS in math. Typically transfer credit is determined on a case by case basis. I would assume they'd ask for course titles, and textbooks and the grades you received and then decide whether or not the department feels you sufficiently satisfied their requirements and standards.

Yes, Spoon's answer tells the basic story. Some colleges accept transfer credits freely from some other colleges, and some accept transfer credits very sparingly. It may also matter that you have already applied your previous credits to the minor subject on your first degree--usually there is a tendency not to let courses count "twice," although you sometimes don't have to start from the absolute beginning to get a second undergraduate degree. Because what each college will do differs according to its policies, the best advice we can give you here is to inquire at more than one college and see what answers you get that are specific to your situation.

Thanks, I sort of figured the answer would be something along those lines but I wanted a second opinion from those more in the know.

I'll give you my own institution's answer to that. For several years, we were not accepting any applicants for second bachelors degrees, but now we are accepting applications in a few fields, math among them.

We specify a minumum of 30 additional semester units at our institution. Of those, at least 24 must be upper division and at least 12 must be in the major. (For all practical purposes, "12 units in the major" is so low that it never matters - see below.)

If your degree is from an accredited college or university in the U.S., we would not hold you to our general education requirements; if your degree is from outside the U.S., then you would need to satisfy our general education requirements, which would add quite a bit.

Whether transferred courses suffice to satisfy major requirements is ultimately up to the department's undergraduate advisor. (In this case, me.) It would be typical for a student with an engineering degree to still need somewhere between 27 and 36 additional units of mathematics courses. A previous minor would take case of some of that - as for how much, I'd have to see the details. Whether transferred units were part of a previous minor or major is not an issue.

wow thats great information, thank you Dr Merryfield


If 3$ \%$ of 400 is equal to $ 5n \plus{} 2$, then what is the value of $ n$ ?

hello, $ 3%
$ of $ 400$ is $ 12$ so we get $ 12=5n+2$ and from here we have $ n=2$.
Sonnhard.


inequalities
geometry
3D geometry
prism
geometry unsolved

Let $ ABCA'B'C'$ be a triangular prism. We choose two points $ M$ and $ N$ on $ AA^{\prime}$ and $ CB$, respectively, such that  $ \frac {AM}{AA'} \equal{} \frac {CN}{CB}$. Prove that the length of the segment $ MN$ is no less than the distance from the point $ A^{\prime}$ to the line $ C^{\prime}B$.

again,this problem is a problem issuse in M&Y

The deadline being over, this problem can be discussed again.

  darij


Hi
show that the map f : IN* --> IN*
                                n    --> E ( 1 + 1/2 + ... + 1/n )
is surjective. (non inj)

$ f(1)\equal{}f(2)$ gives that $ f$ isn't injective.
we take $ u_{n}\equal{}1\plus{}\frac{1}{2}\plus{}...\plus{}\frac{1}{n}$, we have $ 0<u_{i\plus{}1}\minus{}u_i\equal{}\frac{1}{1\plus{}i}<1$
then $ \forall k\in \mathbb{N}^*,\exists j: \ u_j\in[k,k\plus{}1[$ so $ f(j)\equal{}[u_j]\equal{}k$
then $ f$ is surjective

thanks for your fast reply .
Indeed and actually , every integer has many antecedents , how many are they ? :|


inequalities
linear algebra
matrix
linear algebra unsolved

Is that true that   $rank\begin{pmatrix}A & 0\\ 0 & B \end{pmatrix}$ $\leq$ $rank\begin{pmatrix}A & 0\\ C&B \end{pmatrix}$  ?

Yes, any set of linearly independent rows in the left matrix is also linearly independent in the right matrix.

:blush:  yes, it was obvious... :blush:
thank you mlok


[quote="hoasua"]A combinatorical problem![/quote]
To Mods: Please lock this topic, this is a problem from M&Y, issue 9, and we can't discuss it now


Daca adunam sau scadem doua inegalitati adevarate, vom obtine mereu un adevar?
Daca scadem urmatoarele inegalitati:
$ 9<10$
$ 4<7$
ajungem la $ 5<3$ care este fals.
Ati putea sa-mi explicati in ce conditii putem aduna sau scadea doua inegalitati, si daca ajungem mereu la un adevar? multumesc :wink:

Nu ai voie sa scazi doua inegalitati.De ce ? Ti-ai gasit singur un exemplu.Insa ai voie sa aduni doua inegalitati. De exemplu:

$ 2ab\leq a^2\plus{}b^2$
$ 2x\leq x^2\plus{}1$

Ai voie sa le aduni si sa afirmi ca :

$ 2ab\plus{}2x\leq a^2\plus{}b^2\plus{}x^2\plus{}1$

Ai grija cand le aduni, sensul ambelor inegalitati sa fie acelasi.Aici in ambele inegalitati sensul era $ "\leq "$.


linear algebra
matrix
absolute value
linear algebra unsolved

If a third order determinant has elements $1, 2, . . . , 9$, what is the maximum value it may have?

Hi mayhem,
the computers made progress since 1955.
We may assume that $1$ is the element $A_{1,1}$ of the matrix $A$.
1) With Maple one finds (time:20") that the maximum of all the $det(A)$ is $412$ and is obtained with 4 matrices one of which is:\[\begin{pmatrix}1&7&5\\4&2&9\\8&6&3\end{pmatrix}\]
We can see that the filling of matrix $A$ in the order $1,2,3,\cdots$ is done in an order similar to the filling by the successive powers (that is to say $I,J,J^{2}$) of the permutation matrix: \[J=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}\]
2) For the dimension $n=4$ the computer is impotent (with the brute force) but one can seek matrices $A$ in the form obtained in the case $n=3$. One chooses the diagonal equal to $1,2,3,4$ and we test all the permutations of order 4 for the 4 elements $5,6,7,8$ associated to $J$,  all the permutations of order 4 for the 4 elements $9,10,11,12$ associated to $J^{2}$,  all the permutations of order 4 for the 4 elements $13,14,15,16$ associated to $J^{3}$. The best result (IN ABSOLUTE VALUE) for these $24^{3}$ matrices is  $37918$ and is obtained (time:1'16") only with the matrix \[\begin{pmatrix}1&13&11&7\\8&2&15&9\\10&6&3&16\\14&12&5&4\end{pmatrix}\]
The last line $[14,12,5,4]$ isn't clear in the preview.
Who can do better ?


AMC
USA(J)MO
USAMO
MATHCOUNTS
AIME
articles
AMC 12

Are you sure you were accidently qualified for the USAMO?  As a ninth grader, it would be exceptional to qualify on your own.  If you haven't gone way beyond classroom mathematics, I'm afraid that its too late to prepare you now, but go ahead and check out this great tutorial [url=http://www.matholymp.com]here[/url].

Rui...

you got 15th at Mathcounts Nationals last year.  You're not accidentally qualified!

:-o. :?

[quote="MithsApprentice"]You dont just accidentally qualify for USAMO.  :no:[/quote]

I wish I did last year :(

[quote="MithsApprentice"]Well...if you guys don't want to make it, donate a part of ur index to me. :-)[/quote]

I'm taking donations for MOP index... anyone? ;)

[quote="paladin8"]I'm taking donations for MOP index... anyone? ;)[/quote]

What grade are you in?

[quote="henryyangrui"]How do i prepare for the USAMO?
BACKGROUND INFORMATION: I am in ninth grade and was ACCIDENTALLY qaulified for the USAMO. I don't think i have much chance but i will try to do my best.[/quote]

Since the USAMO isn't too far from now, you can't learn many new things between now and the USAMO.  Just relax and maybe practice somee old USAMOs (http://www.kalva.demon.co.uk).

What'd you get on the AIME?  How can you 'accidentally' qualify?

lol, oops...i qualified, haha, yeah, if anyone doesn't want some of their USAMO index when they take it (all you upper thirties and lower forties people) you're welcome to donate it to me.

[quote="DPopov"](all you upper thirties and lower forties people)[/quote]

I'm not sure "all you" is the correct word there...Assuming you mean 35 - 42, there were only 4 last year. The cutoff for the winners was at 24...just over 3 problems correct.

WOW. thx for all these GREAT replies. I am currrently preparing for the Mu Alpha Theta FLorida State competition, too. I said i was accidentally qualified because i didn't prepare for the AIME and i didn't think I even had a chance to be qualified before i took the test.

Sadly I'm not going to the Florida Mu Alpha Theta State Convention...:-(.

I might have accidentally qualified for USAMO too, depends on if the AIME floor will be 10 or not.  I'm in 9th grade too.

Next year i will be tenth grade. Is it better to take AMC 10 or AMC12? and can i take both?

Better to take both. If you can only choose one, take 12 if you believe you can qualify for AIME with 12. If not, choose 10.

They choose people for the AIME differently on the 10 and 12.  The top 1% from the AMC 10 advance.  Anybody with a score above 100 on the AMC 12 advances.

[quote="henryyangrui"]Where did u get the AIME II problems?[/quote]

http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=51

[quote="MCrawford"][quote="henryyangrui"]How do i prepare for the USAMO?
BACKGROUND INFORMATION: I am in ninth grade and was ACCIDENTALLY qaulified for the USAMO. I don't think i have much chance but i will try to do my best.[/quote]

You might want to hunt through our articles area.  Last year Richard and I wrote several articles to help students prepare for the USAMO (How to Write a Solution).

You might also consider building up with some of the "soft" olympiad books such as the USSR Olympiad Problem Book and 500 Mathematical Challenges (ranges from AIME to USAMO level).

You are probably aware of this by now, but we teach several Olympiad level classes here at AoPS that many USAMOers and MOPpers take.[/quote]

There are full solutions to those problems in those books right?

I feel that I will be needing solutions to about 90% of them.

That's a good start! I need solutions to about 98% of them.

AHH Don't even tell me it was an accident! You WERE studying for it! I saw you studying!

Hey umm I'm REALLY new unlike rei... Can you guys tell me where I can find things to help me starting with AMC and up from there? I do have the volume of the AoPS that Jason let me borrow but what from there?

Yes, I am also going to Florida State competitions.... But only for the free food... And to leech off Rei's math skills.

hey are you yang rui of florida, 15th place in national mathcounts 2004?

EWW that's him? wow ive been having the wrong friend for like 6 months! He can't have been 15th! He musta been like 1st


conics
ellipse
geometry
geometric transformation
reflection
analytic geometry

We have an ellipse with the foci $F,F^\prime$.

Can you prove without analytic geometry that if the tangents through a point $M$ touch the ellipse at $T$ and $T^\prime$, then $\angle TMF = \angle T^\prime M F^\prime$?

Personally, I don't think it is possible. (I'm reffering now to the "every beam that goes through a foci goes through the other as well" statement") But I'll be highly impressed if someone can prove me wrong  :D

suppost that we have 2 foci and a tangent, and we are trying to find the ellipse, we know that for the ellipse to be tangent, the point at which it intersects the line is the minimum sum of the distances from that point to the line and to the other focus, this is true since we have to shrink the ellipse and that corresponds to a smaller sum

now that we have that point of tangency is the minimum sum of distances, reflect one of the foci over the line, and since it is the min sum, the first foci, the point of tangency and the reflection are collinear since a line is the min sum of distances

looking at the vertical angles, and the reflection gives you your answer  ;)

For proving analytic problems with geometry methods, I suggest the book:"Geometric Properties of Conics" by a British preffessor.

A short synthetic proof to the problem in post 1 can be found at this site:

http://www.mathematik.uni-bielefeld.de/~rehmann/DML/dml_links_title_A.html

This is the site of the Digital Mathematical Library DML

Go to 137 "An Elementary Treatise on the Geometry of Conics" by Asutosh Mukhopadhyay

Then go to page 79,Prop XX and you will find a proof

Thanks, Altheman. It's a really nice idea. Now I can prove the problem in the first post using your reflection and congruent triangles.

@Michael Niland: I just checked now your link, and it looks like pure gold. Thanks !

[quote="Michael Niland"]A short synthetic proof to the problem in post 1 can be found at this site:

http://www.mathematik.uni-bielefeld.de/~rehmann/DML/dml_links_title_A.html

This is the site of the [u]Digital Mathematical Library DML[/u]

Go to [b]137[/b] "[u]An Elementary Treatise on the Geometry of Conics[/u]" by [b]Asutosh Mukhopadhyay[/b]

Then go to [b]page 79[/b],[b]Prop XX[/b] and you will find a proof[/quote]Thanks, Michael Niland ! Your address is a great and precise information and I respect you for it. Again, thank you !

doesnt projection help?

because we can change each ellipse to a circle with a parallel projection.

(im not sure if it works) ;)


Let $a_{0}=1$ and $a_{n}=1+\frac{1}{a_{n-1}}$. Find a formula for $a_{n}$.
This is actually the continuous fraction: $1+\frac{1}{1+\frac{1}{1+\cdots}}$.

$a_{n}= \frac{F_{n}}{F_{n-1}}$ where $F_{n}$ is the Fibonacci sequence.

Pierre.

[quote="Rust"]Exactly $a_{n}=\frac{F_{n+2}}{F_{n+1}}.$[/quote]

aargh..you are right.
I always forget the initial $0$'s... :wink:

Pierre.


Camp for Freshmen...

 N boys studying physics and N girls studying liberal arts are sitting by the flickering
campfire. Just by them are N, strictly coed tents (for two people each). Egon Quark has to arrange the sleeping
order.
It is known that on the average each person is willing to sleep in one tent with m members of the other sex - but
only Egon knows who with who. How should m(N) be chosen so that Egon would have a 50% chance to arrange
a sleeping order that is satisfying for everyone? Give the asymptotic behaviour of m(N) as N → ∞.
For N fixed, define the width of the transition “Egon flunks − Egon does not flunk” (f(N)). How does f(N)
behave as N → ∞?

[quote="preetham"]How should m(N) be chosen so that Egon would have a 50% chance to arrange
a sleeping order that is satisfying for everyone? Give the asymptotic behaviour of m(N) as N → ∞.
For N fixed, define the width of the transition “Egon flunks − Egon does not flunk” (f(N)). How does f(N)
behave as N → ∞?[/quote]
50% chance of getting ALL right or 50% of the partners are right?
The last part is really OTH for me. I'm gonna fail JEE big time :(

its 50% chance of gettin all  right

actually no official solution exists

i got  ans for part 1 ( m(N) thing)

but i donno wats the meaning of  [b]the width of the transition[/b]


[quote="tranvanluan"]Có ai có đáp án 3 cuộc thi olimpic sinh viên gần đây nhất không cho mình xin với ![/quote]
Cos 1 cuốn viết về mấy cái này còn những năm gần đây anh thử lên http://diendantoanhoc.net/forum/index.php?,xem sao,có lần em thấy thì phải


integration
calculus
calculus computations

how to solve this problem$\int\frac{x^2}{(1+x^2)^2} dx$use integration by parts

i use trigonometric subsitution get the answer is $\frac{1}2(tan^{-1}x-\frac{x}{1+x^2})+C$
but how about using integration by parts??

[quote="shyong"]notice that

$\int \frac{x}{(1+x^2)^2}dx=-\frac{1}{2(1+x^2)}$

so you do integration by part using $\int x\cdot\frac{x}{(1+x^2)^2}dx$[/quote]
thanks your help


Any ideas on how to make an egg safely accelerate from the top of an 8m height without breaking on impact with the ground?  I was thinking of making a propellor but I do not want to use anything that cushions the egg or else that destroys the purpose.  How does a propellor slow the egg down?

[quote="math92"]Last year, when I did this, the person who won had a huge parachute with a basket like thing made out of paper to hold the egg (sort of like a hot air baloon).  You can also surrond the egg with paper cushioning to decrease impact force.[/quote]

Any other ideas, since cushioning and parachutes are very common ways?


An ant is walking up a soda can 6 inches tall at a 45 degree angle to the can.  (Like a helix)  How far will the ant have walked when he reaches the top of the soda can? Is the answer [hide]6*sqrt 2, using a 45-45-90 triangle[/hide]? Or something else?

Wow... I got the same thing, but somehow it seems too easy. Yes.... I'm inclined to agree with you.

I think so, because if you unwound the path, you would get that answer.

If it seems too easy, you can always do it with a piece of paper.

[hide]why do you post the answer in the question? if u guess, the chances you getting the question right is 50%[/hide]


It's less than a week until the AIME and my school has yet to receive ANYTHING regarding my qualification via the USAMTS.

Does that mean I didn't qualify? If not, when should my school receive the notification?

hmm...I would recommend contacting USAMTS office and/or U of Nebraska...

very (and i mean [i]very[/i]) few people get into AIME thru USAMTS...I dont have the numbers but I know of like no one that does it that way b/c most ppl that do USAMTS do AMCs anyways...

Also, you might want to talk to your school math coach :-)

Just checking, but is your score above 68?

Yes, I got above a 68 on the USAMTS. I did not take the AMC 12

How is my school supposed to be notified in the first place? When does the USAMTS send out the lists of qualifying people? Is it supposed to be included in the email summary of results each school receives? Or are additional forms for the USAMTS-qualifying students included in the package of AIME tests? Or both?

My math coach knows nothing about the matter.

I've emailed Mr. Kelly, no response yet...

Seriously, no one here has qualified for the AIME through the USAMTS?

[quote="tAsDoOm"]Seriously, no one here has qualified for the AIME through the USAMTS?[/quote]

I've qualified through the USAMTS, but I also took the AMC10 and qualified through that, so that doesn't apply to me.

What score did you need and by what round to quali?

the maximum score each round is 25, so the maximum thru 3 rounds is 75. i think many people have this score. as for average, this is unknown. i dont think they do statistics.

Hi,
I am in exactly the same situation. I really wonder whats going on. I have contacted the people from AMC and have yet to receive a reply. If I get anything ill tell you.

I'm glad that I'm not the only one in this situation. If you don't receive a response (which will most likely be the case), what are you going to do?

The following is a clip from a USAMTS newsletter from three years ago.

[i]After three rounds of the USAMTS, 127 participants have a cumulative score of 68 or better. This is the cutoff determined for this year by the American Mathematics Competitions (AMC) for USAMTS participation in the American Invitational Mathematics Examination (AIME). These students will be invited by the AMC to take the AIME, the second step in the process of selecting the USA Mathematical Olympiad Team. Other students, who have done well on the multiple-choice AMC-10 and AMC-12 exams (given 13 February), will also be invited.

The AIME will be given on 27 March. Qualifying USAMTS students will be contacted by the AMC. We wish all these students good luck in taking the difficult three-hour AIME exam![/i]

I haven't been contacted by the AMC. I am really beginning to wonder what form of contact this is.

Well, my school got my AIME, today. So, I guess ill be taking it on tuesday. I hope it came for you too.


We need to prove that $(ax+by)^2+(cx+dy)^2\leq\frac{1}{16}$, where $a^2+b^2+c^2+d^2\leq\frac{1}{16}$ and $x^2+y^2=1$.
Hence, we need to prove that
$(ax+by)^2+(cx+dy)^2\leq(a^2+b^2+c^2+d^2)(x^2+y^2)$, which is equivalent to
$(b^2+d^2)x^2-2(ab+cd)xy+(a^2+c^2)y^2\geq0$, which is true because
$(ab+cd)^2-(b^2+d^2)(a^2+c^2)\leq0$ is true by C-S.

Hello.
What's ik mean? I don't get the Sum lines.


let the system of equations
$ ax\plus{}(a\minus{}1)y\equal{}2\plus{}4a$
$ 3|x|\plus{}2y\equal{}a\minus{}5$
What values of a, the system of equations has only root?? :oops:

:P sorry
what values of a, the system of equations has one root? :roll:

[quote="hell_ever"]You're not being very clear....what is the meaning of "has only root"?[/quote]
I believe that he means "one solution."
[hide="Solution"]
Mulptiply the first equation by $ a \minus{} 1$ and the second by $ 2$ to get that $ 2ax \plus{} 2(a \minus{} 1)y \equal{} 4 \plus{} 8a$ and $ 3(a \minus{} 1)|x| \plus{} 2(a \minus{} 1)y \equal{} (a \minus{} 5)(a \minus{} 1)$.  Subtract them to get that $ 3a|x| \minus{} 3|x| \minus{} 2ax \equal{} a^2 \minus{} 14a \plus{} 1$.

[b]Case 1: $ x\ge 0$[/b]
Thus, we obtain that $ x(3a \minus{} 3 \minus{} 2a) \equal{} a^2 \minus{} 14a \plus{} 1\implies x \equal{} \frac {a^2 \minus{} 14a \plus{} 1}{a \minus{} 3}$.

[b]Case 2: $ x\le 0$[/b]
Now, we obtain that $ x( \minus{} 3a \plus{} 3 \minus{} 2a) \equal{} a^2 \minus{} 14a \plus{} 1\implies x \equal{} \frac {a^2 \minus{} 14a \plus{} 1}{3 \minus{} 5a}$.

Now, we obtain that there is only one value of $ x$ if $ x$ is negative in Case 1 and Case 2 or if $ x$ is positive in both.  Thus, $ a \minus{} 3$ and $ 3 \minus{} 5a$ are both above or both below $ 0$.  Hence, $ a > 3$ and $ a < \frac {3}{5}$ or $ a < 3$ and $ a > \frac {3}{5}$.  The first possibility yields two disjoint sets, giving no solution.  The second possibility gives that $ \boxed{\frac {3}{5} < a < 3}$ [/hide]

[quote="MellowMelon"][hide="Possible problem with above"]What if $ a^2 \minus{} 14a \plus{} 1 \equal{} 0$? This gives a unique value for $ x$ (0). I haven't checked this fully, so I don't think it gives more answers, but it is a hole to plug.[/hide][/quote]
Whoops! :ewpu:

Sorry about that!  That does indeed yield another pair of solutions, which are $ a\equal{}7\pm 4\sqrt{3}$ after plugging $ a^2\minus{}14a\plus{}1\equal{}0$ into the quadratic formula.


geometry
circumcircle
trigonometry
cyclic quadrilateral

I am trying the problem for a long time but I cannot solve it. Please help me ( :blush: ).
[b]Problem[/b]
The radius of the circumcircle of $ \Delta ABC$ with area $ \Delta$ is $ R$; and $ D$ is the midpoint of $ BC$, and $ E$ abd $ F$ are the feet of the perpendiculars from $ D$ to $ AB$ and $ AC$. If $ S$ is the area of $ \Delta DEF$, prove that :
$ R \equal{} \frac {a}{4}\sqrt {\frac {\Delta}{S}}$.
[hide]
Rudiments of mathematics, Part I, Pg-528, Prob 15[/hide]

We have $ DE\equal{}{h_c\over 2}, DF\equal{}{h_b\over 2}$ and $ \angle EDF\equal{}\pi\minus{}\alpha$ (either by noting that $ AEDF$ is a cyclic quadrilateral, having two opposite angles right, or by arguing that $ \angle EDF\equal{}\pi\minus{}\angle EDB\minus{}\angle CDF\equal{}\pi\minus{}\left({\pi\over 2}\minus{}\beta\right)\minus{}\left({\pi\over 2}\minus{}\gamma\right)\equal{}\beta\plus{}\gamma\equal{}\pi\minus{}\alpha$)

Therefore $ S\equal{}{h_b\over 2}\cdot{h_c\over 2}\cdot{\sin\alpha\over 2}\equal{}{\Delta\over b}\cdot{\Delta\over c}\cdot{\Delta\over bc}\equal{}{\Delta^3\over b^2c^2}$

Since $ R\equal{}{abc\over 4\Delta}$, we get $ bc\equal{}{4R\Delta\over a}$. Plugging that in, we get

$ S\equal{}\frac{\Delta^3}{{16 R^2\Delta^2\over a^2}}\equal{}{a^2\Delta\over 16R^2}$

From there $ R^2\equal{}{a^2\Delta\over 16S}\iff R\equal{}{a\over 4}\sqrt{\Delta\over S}$

By $ \alpha, \beta ,\gamma$ do you represent the respective angles of $ \Delta DEF$?

This is a particular case of Euler's theorem.

$P$ is a point on the plane of $\triangle ABC$ with circumcircle $(O,R)$ and $X,Y,Z$ are the orhogonal projections of $P$ on the sidelines of $\triangle ABC.$ Then we have
\[{\frac {[\triangle XYZ]}{[\triangle ABC]} = \frac {p(P,(O))}{4R^2}}, \]
where $p(P,(O))=|PO^2-R^2|$ stands for the power of $P$ with respect to $(O,R).$


function
real analysis
real analysis unsolved

Find the differentiable functions $f(x),\ g(x)$ which satisfy the following conditions.

(a) $f(1)=2006,\ g(1)=-2005$
(b) $f(x+y)=f(x)f(y),$
     $\ \ g(x+y)=2006g(x)g(y)+{2006}^2\{g(x)+g(y)\}+2005\cdot 2006\cdot 2007,\ \forall x,y\in \mathbb{R}.$

Can anyone solve this problem? :)

kunny

Various versions of (a) have been discussed on this forum many times. The answer is $2006^x$. As to (b), just consider the function $u(x)=2006g(x)+2006^2$. It satisfies $u(x+y)=u(x)u(y)$, so we get the same old story again. :roll:


polynomial
algebra
function
domain
number theory
Ring Theory
superior algebra

Definitions:

Call an element $p$ of a ring $R$ "prime over $R$" if, whenever $p$ evenly divides the product $x \times y, x, y \in R$, $p$ must divide either $x$ or $y$.  For the ring $\mathbb{Z}$ this corresponds to the usual definition of primality.

Define the ring $\mathbb{Z}[k]$, for $k$ such that $k^n \in \mathbb{Z}$ for some $n \in \mathbb{N}$, as the set of all numbers representible in the form $a_0 + a_1 k + a_2 k^2 + ... + a_{n-1} k^{n-1}$, where $a_i \in \mathbb{Z}$.  Multiplication and addition are done normally, as expected.

Problems:

1.  Find the set of numbers prime over $\mathbb{Z}[i]$, the Gaussian integers, in terms of the numbers prime over $\mathbb{N}$.

2.  Find the set of numbers prime over $\mathbb{Z}[ \omega ]$, where $\omega = e^{ \frac{2 \pi i}{3} }$, the Eisenstein integers.

(The first one is interesting, and corresponds to a more mundane number theory problem in $\mathbb{N}$.  The second one... I have no idea :P )

This is in the wrong forum. It should probably be in the College Playground, Superior Algebra Forum.

[hide]First note that all elements of $\mathbb{Z}[\omega]$ can be expressed as $a + {\omega}b$, because ${\omega}^2 = -{\omega} - 1$.

Then note that $\mathbb{Z}[\omega]$ is an euclidean domain, because the same euclidean valuation (or norm, $|\cdot|$) that exists on $\mathbb{Z}[i]$ also works on $\mathbb{Z}[\omega]$. But on $\mathbb{Z}[\omega]$ it takes the form $|a + {\omega}b| = a^2 - ab + b^2$.

Also note that $a^2 - ab + b^2$ factors as $(a + {\omega}b)(a + {\omega}^2b)$. [Similar to the way that $a^2 + b^2$ factors as $(a + ib)(a - ib)$].

It then becomes a similar problem to that of finding the primes in the Gaussian Integers. Namely, it becomes a number theoretic question of which primes in $\mathbb{Z}$ can or cannot be written as $a^2 - ab + b^2$ for integers $a$ and $b$.

Basically, $a + {\omega}b$, $a,b \neq 0$, is prime in $\mathbb{Z}[\omega]$ iff $a^2 - ab + b^2$ is a prime in $\mathbb{Z}$. And a real number $z$ is a prime in $\mathbb{Z}[\omega]$ iff $z$ is a prime in $\mathbb{Z}$ which cannot be expressed as $a^2 - ab + b^2$ for integers $a$ and $b$.

These by the way are called Eisenstein Primes.[/hide]


i was thinking maybe someone on here would know what to do...

when i plug my ipod vid in and turn on itunes, it keeps saying "iTunes has detected an iPod in recovery mode.  You must restore this iPod before it can be used with iTunes"

i restored it a bunch of times but it keeps popping up.

does anyone know that to do?

thanks
Jli

Try removing the porn - it might be carrying spyware.

[quote="math92"]Try holding the center button and the menu button for like 10 seconds on your ipod. Something funky will happen and it will restart. That usually works for me when my ipod doesnt connect or my computer doesnt recognize it.[/quote]

i restored it a bunch of times but it keeps popping up.

Learn to read.

Why are you asking this here rather than the apple forums?

Because in the Apple forums, everybody would be like "APPLE PRODUCTS NEVER BREAK, YOU MUST'VE STEPPED ON IT"?

[quote="Ignite168"][quote="math92"]Try holding the center button and the menu button for like 10 seconds on your ipod. Something funky will happen and it will restart. That usually works for me when my ipod doesnt connect or my computer doesnt recognize it.[/quote]

i restored it a bunch of times but it keeps popping up.

Learn to read.
[/quote]

What I suggested isnt restoring.  Restoring is when all your iPod settings get returned to normal and all your songs get deleted. My method keeps your songs and settings, but it just reboots the iPod in a funky way.

Yeah, exactly.

But my ipod just takes 5 seconds...???
It usually works for me. If it doesn't, just post it on the Apple forums, and capitalize :

I DID NOT BREAK IT!!!!

it works now...you just have to restart the computer with the ipod plugged in :D


Can we get a forum just for Math Jams? I think it would be good for people to put their thoughts after the Jam and for people to ask for possibilities of new Math Jams.

Isn't there a suggestion forum for that?

Well, then a math jam forum can be just for discussing Math Jams right after.


Let $ a,b,c$ be positive real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that:
\[ 8\prod {(2 \minus{} a)}\ge \prod {(a \plus{} bc)}\]

$ 2\minus{}a \equal{} \frac{b^2\plus{}c^2\plus{}1}{2\plus{}a}$, so we can rewrite the ineq to be
$ 8 \prod(b^2\plus{}c^2\plus{}1) \ge \prod (2\plus{}a)(a\plus{}bc)$
By Cauchy Schwarz, we know that $ (a^2\plus{}b^2\plus{}1)(a^2\plus{}c^2\plus{}1) \ge (a^2\plus{}bc\plus{}1)^2$, so
$ LHS \ge 8 \prod(a^2\plus{}bc\plus{}1) \equal{} \prod 2(a^2\plus{}bc\plus{}1)$
However,
$ 2(a^2\plus{}bc\plus{}1) \ge (2\plus{}a)(a\plus{}bc)$
$ \Leftrightarrow 2a^2\plus{}2bc\plus{}2 \ge a^2\plus{}2a\plus{}2bc\plus{}abc$
$ \Leftrightarrow (a\minus{}1)^2 \plus{} (1\minus{}abc) \ge 0$.
which is true since $ 1 \ge abc$ by AMGM.


limit
integration
geometry
3D geometry
calculus
analytic geometry
calculus computations

Find $\lim_{n\to+\infty}\int_{Q_{n}}e^{-\sqrt{x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2}}}dx_{1}dx_{2}...dx_{n}$,
where $Q_{n}$ is a n-dimensional cube in $R^{n}$ with centre $(n,n,...,n)$, side $n$, with faces parallels to the axes.

So, that's the integral of $e^{-r}$ over one "quadrant"; it's $\frac1{2^{n}}$ of the integral over all of $\mathbb{R}^{n}$.
Let $V_{n}$ be the volume of the $n$-dimensional unit ball; the volume of the ball of radius $r$ is $V_{n}r^{n}$. The surface of a ball of radius $r$ thus has "area" $nV_{n}r^{n-1}$, so in "spherical" coordinates, $\int_{\mathbb{R}^{n}}e^{-\sqrt{x_{1}^{2}+\dots+x_{n}^{2}}}\,dx_{1}\,\dots\,dx_{n}= \int_{0}^\infty \int_{|x|=r}e^{-r}dA(x)dr=\int_{0}^\infty nV_{n}r^{n-1}e^{-r}\,dr$.
This last integral is $nV_{n}\Gamma(n)=\Gamma(n+1)V_{n}$.

The sequence $V_{n}$ has been calculated on this forum before, but I can't find the threads right now.

And here's one of the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=682876]threads[/url].


[quote="namelessguy"]Let m be in the set of natural numbers, let ~ be the equivalence mod m relation on Z. Let a, b in Z and prove that if a~b then a+c~b+c and ac~bc.

What does it mean to have a~b? I know that ~ means m divides a-b, so a-b = km. I don't know how to interpret a+c~b+c and ac~bc. Anyone can help please.[/quote]
[hide]
We have that $ x\equiv y\bmod m$ if $ x \minus{} y$ is divisible by $ m$, as you said.  So, we have that $ (x \plus{} c)\equiv (y \plus{} c)\bmod m$ if $ (x \plus{} c) \minus{} (y \plus{} c)$ is divisible by $ m$.  Notice that $ (x \plus{} c) \minus{} (y \plus{} c) \equal{} x \minus{} y$ and $ x \minus{} y$ is divisible by $ m$.  So, $ (x \plus{} c) \minus{} (y \plus{} c)$ is divisble by m and thus, $ (x \plus{} c)\equiv (y \plus{} c)\bmod m$.

Now, $ xc\equiv yc\bmod m$ if $ xc \minus{} yc$ is divisible by m.  Now, notice that $ xc \minus{} yc \equal{} c(x \minus{} y)$.  Denote $ x \minus{} y \equal{} km$ (as you said), so $ xc \minus{} yc \equal{} (x \minus{} y)c \equal{} (km)c \equal{} m(kc)$.  So, $ xc \minus{} yc$ is a multiple of m and therefore, $ xc\equiv yc\bmod m$.  [/hide]

Thanks a lot for your quick reply QuattoMaster. I have never learned about the congruence notation before, so I just looked it up. So basically, in my problem a+c~b+c means that m divides (a+c) - (b+c) and ac~bc means that m divides ac-bc. If I had known that, I would have been able to approach this problem. Greatly appreciate your help.


linear algebra
matrix
vector
inequalities
linear algebra unsolved

An $ nxn$ real matrix $ T$ is positive definete $ T$ is symmetric and $ < Tx,x > > 0$ for all nonzero vectors $ x \in R^n$,Where $ < u,v >$ is the standard inner product .Suppose that $ A$ and $ B$ are two positive definite real matrices .
1)Show that there is a basis {$ v_1,v_2,...,v_n$} of $ R^n$ and real numbers $ \lambda_1,...,\lambda_n$ ,such that , for $ 1 \leq i,j \leq n$ :
          $ < Av_i,v_j >$ =$ \begin{cases} 1 & \text {if } i\equal{} j \\
0 & \text{if } i\neq j \end{cases}$
and      $ < Bv_i,v_j >$=$ \begin{cases} \lambda_i & \text{if } i \equal{} j \\
0 & \text{if } i \neq j \end{cases}$
2) Deduce from part 1) that there is an invertible real matrix $ U$ such that $ U^tAU$ is the identity matrix and $ U^tBU$ is diagonal ,
Thanks ! :roll:

This is a result very interesting (I will add it with  matrix theory and it is diferent proof for this theory). It is correct with Hermitian matrix . And show need $ A$ is positive definite , $ B$ real symmetric matrix . $ \lambda_i$ is real .(Here I am proof with real symmetric matrix , With Hermitian matrix are same but note $ < , >$)
  You can check my proof .Thanks !
Put $ e_1 \equal{} \begin{bmatrix}1 \\
0 \\
0 \\
... \\
0\end{bmatrix}$,$ e_2 \equal{} \begin{bmatrix}0 \\
1 \\
0 \\
... \\
0\end{bmatrix}$,$ e_3 \equal{} \begin{bmatrix}0 \\
0 \\
1 \\
... \\
0\end{bmatrix}$,$ e_n \equal{} \begin{bmatrix}0 \\
0 \\
0 \\
... \\
1\end{bmatrix}$ are basis of $ \mathbb{R}^n$
 Then we can  perform $ v_i,v_j \in \mathbb{R}^n$ for all $ 1 \leq i, j \leq n$ as :
  $ v_i \equal{} \sum_{k \equal{} 1}^{n}{u_{ki}e_k}$, $ v_j \equal{} \sum_{l \equal{} 1}^{n}{u_{lj}e_l}$
eassy $ < Ae_k,e_l > \equal{} a_{lk} \equal{} a_{kl}, < Be_k,e_l > \equal{} b_{lk} \equal{} b_{kl},$ Hence
 $ < Av_i,v_j > \equal{} \sum_{l \equal{} 1}^{n}\sum_{k \equal{} 1}^{n}{u_{ki} < Ae_k,e_l > u_{lj}} \equal{} \sum_{l \equal{} 1}^{n}\sum_{k \equal{} 1}^{n}{u_{ki}a_{kl}u_{lj}} \equal{} \sum_{l \equal{} 1}^{n}\sum_{k \equal{} 1}^{n}{u^{t}_{ik}a_{kl}u_{lj}}$
 Let $ U \equal{} (u_{ij})_{ij},U^{T} \equal{} (u^{t}_{ij})_{ij},u^{t}_{ij} \equal{} u_{ji}$,
then  $ U^{T}AU \equal{} (a^{"}_{ij})_{ij}$ have $ a^{"}_{ij} \equal{} \sum_{l \equal{} 1}^{n}\sum_{k \equal{} 1}^{n}{u^{t}_{ik}a_{kl}u_{lj}} \equal{} < Av_i,v_j >$
 From hyppothesis we have :$ U^{T}AU \equal{} I_n$  Then $ \det(U) \neq O,\det(U) \in \mathbb{R}$ and from here $ \{v_1,v_2,...,v_n\}$  is basis of $ \mathbb{R}^n$
The same : we have $ U^{T}BU \equal{} diag(\lambda_1,...,\lambda_n)$
 [color=blue]If $ A$ is Hermitian matrix and positive definite , $ B$ is Hermitian matrix then the exist $ U \in GL_n{(\mathbb{C})}$ such that :
 $ U^{*}AU \equal{} I_n$ and $ U^{*}BU \equal{} diag(\lambda_1,..,\lambda_n)$ ,$ \lambda_k \in \mathbb{R} , 1 \leq k \leq n$[/color]
1) This Theory for a some inequality If $ A,B$ are real  symmetric  matrices and semipositive definite (or with Hermitian matrix and semipositive definite) then
a) $ \sqrt [n]{\det(A \plus{} B)} \geq \sqrt [n]{\det(A)} \plus{} \sqrt [n]{det(B)}$
b) If $ A \geq B$ then $ \det(A) \geq \det(B)$
...................


algebra
polynomial
real analysis
real analysis unsolved

Consider the endomorphism $f$ of $\mathbb{R}_n[X]$-the space of polynomials of degree $\leq n$, $f: \mathbb{R}_n[X] \to \mathbb{R}_n[X]$. Define norm in $\mathbb{R}_n[X]$ as $||P(x)||=\sum_{k=0}^n |P(k)|$. Let $||| \cdot |||$ be an operator norm induced by $||\cdot||$, $|||f|||=sup_{x \neq 0 \in \mathbb{R}_n[X]} \frac{||fx||}{||x||}$.
Endomorphism $f: P(x) \to P(x+2)$. Find norm $|||f|||$

Let $P_k(x)=\prod_{j\neq k} \frac{x-j}{k-j}$; $P_k(j)=0$ if $j\neq k$, and $P_k(k)=1$. These $P_k$ are a basis for the space, and the norm is the 1-norm with respect to this basis. By the nature of a 1-norm, $\|f\|=\max_k \|f(P_k)\|$.

Calculating, $\|f(P_k)\|=\sum_{j=0}^n |P_k(j+2)|=|P_k(-2)|+|P_k(-1)|+(1\text{ or }0)$, where the last term is 1 if $k\le n-2$ and zero otherwise.
$P_k(-2)=\prod_{j\neq k}\frac{-2-j}{k-j}=(-1)^r\frac{(n+2)!}{2(k+2)}\cdot \frac1{k!\cdot (n-k)!}$
$=(-1)^r\frac{(n+1)(n+2)}{2(k+2)}\cdot\binom{n}{k}$, where $r$ is some irrelevant integer.
$P_k(-1)=\prod_{j\neq k}\frac{-2-j}{k-j}=(-1)^r\frac{(n+1)!}{k+1}\cdot \frac1{k!\cdot (n-k)!}=(-1)^r\frac{(n+1)}{k+1}\cdot\binom{n}{k}$
$=(-1)^r\binom{n+1}{k+1}$.

That's an ugly mess, but it's pretty obvious that it's maximized for $k$ about half of $n$. I don't have time for the details right now.


abstract algebra
integration
geometry
vector
calculus
complex numbers
combinatorics solved

Let z1,z2,..., zn, n>=4 be complex numbers (not necesarely different), for which we have: |z1|+|z2|+..+|zn|=1. Prove that we can choose some of these numbers so that the module of their sum is strictly bigger than 1/4.

cheers! :D :D

Lemma: If z_1,...,z_n complex numbers then there is a subset A of
{1,...,n} such that | \sum_{k\in A} z_k |  \geq 1/ \pi \sum_{k=1 to n}|z_k|

proof: z_k =|z_k|e <sup>iy_k </sup>

for - \pi  \leq  \emptyset  \leq  \pi  let B( \emptyset )= { k such that cos(y_k - \emptyset )>0}

|\sum_{B (\emptyset)}z_k |= |\sum_{B( \emptyset)}e<sup>-i\emptyset </sup>z_k| \geq Real (\sum_{B(\emptyset)}e<sup>-i\emptyset </sup>z_k)
=\sum_{k = 1 to n}|z_k|max(cos(y_k- \emptyset);0)

take  (\emptyset)_0 to maximise the last sum and consider
A=B((\emptyset)_0)  \geq  the average of the sum on [- \pi ; \pi ]

the average = 1/(2\pi)\int_[- \pi ; \pi ]max(cos(y-s);0)ds = 1/ \pi
for any y

So in our problem

| \sum_{k\in A} z_k |  \geq  1/ \pi ~0,318...> 1/4=0,25

This lemma is used to proove the
theorem: If m is a complex measure on E then |m|(E) <\infty

actually my idea was a little bit different (a lot different !!! :D:D) ... it's more of a combinatorial problem. but the solution moubi posed it quite nice.

hmm.. nice solution moubi, but what is your solution Valentin?

cheers! :D :D

This problem published in russian greate geometry book "Planimetry Problems". Solution is the same as moubi's solution.

I know similar problem.
Let z1,z2,...,zn are complex numbers, |zi| \leq 1. Prove we can choose signs in sum z= \pm z1 \pm z2 ... \pm zn such way |z| \leq \sqrt 2.

But for the initial problem , a simpler idea, like pigeon-hole, is sufficient.
Of course don't forget : IF (x(k),y(k))  are m couples such that the x(k) numbers are zero or have the same sign,
the y(k) numbers are zero or have the same sign, and between x(1),..,x(m) or y(1),..,y(m) there are at least 2 non zero numbers, THEN  Sum{x(k)^2 + y(k)^2} < (Sum{x(k)})^2 + (Sum{y(k)})^2

This problem has a generalisation in dimension n

Let x_1,...,x_m vectors of R^n, then there exist a subset A of {1,2,...,m}
such that

||\sum_{k\in A}x_k||  \geq 1/(2n.K_n)\sum_{k\in A}||x_k||

with K_n = \int_[0; \pi /2] (cos(a))^n and the constant 1/(2n.K_n)

is the best possible

Reference:
Bourbaki book
Topologie gnrale, chap 5-8, page.85

I thank Pierre.B for the reference

It's 8 pm i am going to eat some chinese food
Have a good diner everyone
Moubinool Omarjee

there's a similar problem in Chinese Team Selective Contest of a year..well, i cannot remember what exactly the year is...the problem used in the contest asks us to proof |z1|+|z2|+...+|zn|>=1/6, which is weaker.
A contestant proved it is not smaller than 1/4, and he got a "honorable prize".

For the problem Myth posted: it was given at the final round of the national Romanian olympiad (9'th grade) in 2001 (I think). I remember it because I was in the 9'th grade then. It uses induction. I don't know if I can remember the soln exactly, and that's why I won't attempt to post it here (I did solve it during the competition, however  :) )

it was in 1986 i think. anyway the solution with 1/6 was based on a more trigonometric reasoning rather than combinatorial. a combinatorial approach leads to 1/4 (my prob, at least at some point - I found out later that the much more powerfull problem appeared with 1/ \pi :) ) and the calculus approach leads to the best 1/ \pi.

well, it is so. 1986 was the first year CMO held.

[quote="Myth"]I know similar problem.
Let z1,z2,...,zn are complex numbers, |zi| \leq 1. Prove we can choose signs in sum z= \pm z1 \pm z2 ... \pm zn such way |z| \leq \sqrt 2.[/quote]
where did you find this one?
because the same problem was given at the Romanian NMO in 2001, proposed by Mihai Baluna.

It is an old well known problem in Russia. I saw it in many sources (for example, geometry book "Planimetry Problems" which I mentioned early).


Ring Theory
superior algebra
superior algebra unsolved

Let $ K$ be a field, $ \nu : K^* \rightarrow %Error. "texbb" is a bad command.
{Z}$ a discrete valuation on $ K$, and $ R=\{x \in K^* : \nu(x) \geq 0 \} \cup \{0\}$ the valuation ring of $ \nu$. For each integer $ k \geq 0$, define $ A_k=\{r \in R : \nu(r) \geq k \} \cup \{0\}$.

(a) Prove that for any $ k$, $ A_k$ is a principal ideal, and that $ A_0 \supseteq A_1 \supseteq A_2 \supseteq\ldots$
(b) Prove that if $ I$ is any nonzero ideal of $ R$, then $ I=A_k$ for some $ k \geq 0$.

(a) Take $ p\in R$ with $ v(p)\equal{}1$, then $ A_k\equal{}p^kR$.
(b) Consider the minimum of $ v(I\setminus\{0\})$.


calculus
integration
function
real analysis
real analysis unsolved

let f(x) be a positive monotone decreasing function on {1,infinity). Show that ((f(n)) is summable if and only if

integral of f(x) from 1 to infinity is less than infinity

Hint:

show that

summation of (f(n)) from n=2 to k+1 <integral (f(x)) from x=1 to k+1 < summation of (f(n)) from n=1 to k

Well, by monotonicity, we have

$f(n+1)<\int^{n+1}_n f(t)dt<f(n)$.

Sum this.

You should be able to bound $f(n)$ by some function of $\int^{\infty}_1 f(t)dt$

Similarly, you can bound $\int^{\infty}_1 f(t)dt$ by some function of $\sum^{\infty}_{i=1} f(i)$

Monotone bounded sequences converge.  Do you see how to complete it?

Thanx , how do u type the mathematical symbols?

I am still figuring how to complete the solution u gave


Let n be a natural number. Define t(n)= the number of positive divisors of n. Prove that  $ t(n) < <$ln(n)$ ^h$
is false for every positive integer  h.

[/hide] For h=1 consider the set of numbers of the form (2^(2m+x))(3^(2m+x))  where m is the bound

 t(n)/ln(n) approaces to.[hide]