I'm just curious about whether or not anyone has gotten phone calls because my parents said that phone calls went out yesterday and today.  Now I'm scared, because I was hoping to make it.   :(

So, if anyone could clarify when they get out or if they've already gotten calls...

Thanks.

They are grading yesterday and today. As I understand, phone calls go out on Monday (at the very least tomorrow)

[quote="solafidefarms"]Out of curiosity, for you previous MOP invitees, is there a ban on admitting to having gotten a call about MOP for a little while (as in, till the cutoff is publicly announced)?[/quote]
No, but why don't we [b]wait until they've finished the grading[/b] to carry on about all this?  Also, according to the S-I letter sent to all USAMO participants, grading continues tomorrow as well.  And remember, that in the past few years, the AMC director has posted the cutoffs on this forum.  When I went to MOP, I didn't find out until the Tuesday after grading (after the scores had been posted, incidentally).  So be patient!  In all probability, we'll know within four days.  And finally, remember :
[quote="The USAMO manual"]Students and teachers should [u]not[/u] inquire about their results before Friday, May 16, 2008.[/quote]

Oh right it does continue tomorrow. But they say they will send out the information on Tuesday, so they will have to call before that. Phone calls Monday :).

Yeah this is a bit early to be thinking about it.

This has been discussed ad nauseam. Locked.


function
modular arithmetic
ceiling function
inequalities

let $p(n)$ be defined by the function that maps the positive integers $n$ to the product of its digits
(i.e $p(1123)=1*1*2*3=6$, $p(31)=3$, $p(2005)=0$).
 find all postive integers $n$ so that
\[ 11p(n)=n^2-2005 \]
 :lol: ;)

If d(n) is the number of digits of n,

$ n \geq 10^{d(n)-1} $ and $ 11 p(n) - 2005 \leq 99 d(n) - 2005 $

Put together, this proves that $d(n) \leq 2$

Now we just have to solve $(10a+b)^2 = 11ab + 2005$.

Only solution I found was 49.

[quote="tµtµ"] If d(n) is the number of digits of n,

.....and $ 11 p(n) - 2005 \leq 99 d(n) - 2005 $
[/quote]
that's seem magically...
explain your solution better.  ;)
I'm searching for non-bruteforcing solution.
the answer is correct.

yes that's right  :lol:
and that the same as mine. I was using bruteforce with some number between 45 and 60 with have the last digit 6 or 9. and the only solution is 49.

[quote="tµtµ"]$p(n) \leq 9d(n)$ : p(n) is max iff all digits are '9', no ?[/quote]

if all the digits are $9$ then $p(n)=9^n$, so what's $d(n)$?

[quote="tµtµ"]Now we just have to solve $(10a+b)^2 = 11ab + 2005$.[/quote]

$100a^2+20ab+b^2-11ab=2005\\
100a^2+9ab+b^2=2005\\
ab+b^2 \equiv 1 \pmod {4}\\
\text {If } b \equiv 0 \pmod {4}, \text { then it's impossible, so } b \equiv 1\\
\text {If } b \equiv 1 \pmod {4}, \text { then } a \equiv 0 \pmod {4}$

Therefore our choices are $41, 45, 49, 81, 85, 89$. Also $10a+b \ge \lceil \sqrt{2005} \rceil = 45$, and also $(10a+b)^2-2005 \equiv 0 \pmod {11} \implies (10a+b)^2 \equiv 3 \pmod {11} \implies 10a+b \equiv 5 \text { or } 6$, and the only one that fit is $49$, though the last part is unnecessary anyway :D all you need is trial and error :P

chess64, may you could explain to me why $p(n) \leq 9d(n)$... i just don't get it

I hope this help.
First let $d(n)$ be the function that defined by the number of digit n.And $p(n)$ is defined like above.
so    $max(p(n))=9^{d(n)}$  and from the inequality $n>10^{d(n)-1}$ and
$n^2=11p(n)+2005 \leq (11)9^{d(n)}+2005$
$ = (99)9^{d(n)-1}+2005<99(10)^{d(n)-1} +2005<99n+2005$
and we have inequality
$ n^2-99n-2005 < 0 $ and also we have $n^2>2005 \rightarrow n>45$ now as the rest you can  do that trial and error method.(don't brute) :D

thanks. i get it now, the correct inequality was $p(n)<9n$  :blush:


integration
calculus
calculus computations

can you show :|

$ \sum_{m\equal{}1}^{\infty}\sum_{n\equal{}1}^{\infty}\;\frac{1}{\left(m\plus{}n\right)^{s}}\;\equal{}\;\boxed{\zeta{(s\minus{}1)}\minus{}\zeta{(s)}}$

[hide]
For each $ m,n$, let $ k\equal{}m\plus{}n$. Then the $ \frac{1}{k^{s}}$ term appears exactly $ k\minus{}1$ (the number of solutions to $ m\plus{}n\equal{}k$) times for each $ k\geq 2$, so
\[ \sum_{m\equal{}1}^{\infty}\sum_{n\equal{}1}^{\infty}\frac{1}{(m\plus{}n)^{s}}\equal{}\sum_{k\equal{}2}^{\infty}\frac{k\minus{}1}{k^{s}}\]
\[ \equal{}\sum_{k\equal{}1}^{\infty}\frac{k\minus{}1}{k^{s}}\equal{}\sum_{k\equal{}1}^{\infty}\frac{1}{k^{s\minus{}1}}\minus{}\frac{1}{k^{s}}\equal{}\zeta (s\minus{}1)\minus{}\zeta (s)\]
[/hide]


calculus
integration
calculus computations

I don't know also, and so i want to ask you:
In which form do you want to get answer? Maybe the answer contains the integral?
Or this is must be a simple formula without $\sum$ and $\int$?
If the last, then do you sure,  that thats formula exists?
Because it's remind a sum $\sum_{k=1}^{n}\frac{1}{k}$ for which i don't know such formula... :?
I think if one can copmute $\sum_{k=1}^{n}\frac{2^k}{k}$ then one can copmute $\sum_{k=1}^{n}\frac{1^k}{k}$. But as far as i know there is no such simple formula :?

There is a closed form for $\sum_{k=0}^\infty \frac{a^k}{k+2}$ when $|a|<1$. There is almost certainly no closed form for the partial sums of the series for any nonzero value of $a$.

I don't know whether such formula exist. I invented this sum by myself during solving some problems. By the way. When the sum has formula and when it's impossible to find one?

[quote="jmerry"] $\sum_{k=0}^n \frac{a^k}{k+2}$  There is almost certainly no closed form for the partial sums of the series for any nonzero value of $a$.[/quote]
We can copmute only whole $\sum_0^{\infty}$ for $|a|<1$.

That's what I meant- it was only a typo ($k$ in place of $a$).

I only repeat your words one more time :roll:


advanced fields
advanced fields unsolved

i) Prove that $ \pi$ is not an algebraic number.
ii) Prove that $ e$ is not an algebraic number.
Thanks!

Both being corollaries of Lindemann's theorem (or more general the one by Lindemann and Weierstrass).
If I remember correctly, it can be found in Lang's "Algebra".

Thank you very much!
I'll try to find Sergey Lang's book and to check it.


floor function
number theory
algebra proposed
algebra

Prove the following (for integers $n$):

$n - [n/2] + [n/4] - [n/8] \pm ...
= [(n+1)/2] + [(n+4)/8] + [(n+16)/32] + ...$,

where $[x]$ means the largest integer $\leq x$.

[EDIT: corrected now.]

Just use the well-known identity $\lfloor x\rfloor+\lfloor x+\frac 12\rfloor=\lfloor 2x\rfloor$ for $\displaystyle x=\frac n2,\frac n8,\ldots,\frac n{2^{2k+1}},\ldots$.

ok ... i didnt realize it was that simple.
There's also a combinatorics solution by considering the following problem in two different ways:
What is the smallest number of integers you need to cross out from the list {1,2...n} so that no integer is twice another?

Well you posted it into the algebra section the it is not surprising that Grobber solved it with algebric tools  ;)
If you wanted the problem to be with a more combinatorical flavor, you should post it into the combinatorics section  :D

Pierre.

[quote="pbornsztein"]Well you posted it into the algebra section the it is not surprising that Grobber solved it with algebric tools  ;) [/quote]

This is so cool! Next time I will post a geometric problem into the number theory section and it will be solved using number theory...

  Darij

Nttu has already done that but in the other direction :
http://www.mathlinks.ro/Forum/viewtopic.php?t=15051  :D

Pierre.


There are $n$ people, and any pair is either friend or enemy. Each person has more friends than enemies. Show that they can be arranged in a circle in such a way that no two enemies are adjacent.

Well, I have an idea.

Consider the arrangement with the least pair of enemies, and suppose there is still some pairs, and try to find a contradiction.

I can't complete it yet; maybe someone can  ;)

[quote="grobber"]Someone did a very long time ago:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=knights&t=21903[/url][/quote]

And that "someone" is you!  :)

See also : http://www.mathlinks.ro/Forum/viewtopic.php?t=11792 . :)


Phase space is basically just a cartesian system where the conventional y axis is replaced with a dx/dt axis, the x axis is still the x axis.

Thus a harmonic oscillator is elliptical, for example, because the energy is conserved and is in the form for an ellipse.

Anyways can you prove that the area of a flow is constant for a particle being acted on by a conservative force. A flow would be the result of starting two particles off with two different initial conditions and noticing that over some time interval, they cover same area.

For example the period of the harmonic oscillator is the same for particles under the same conservative force, ie. the spring constant is the same, but the initial conditions may vary. Thus the area of the flow is constant.

In the case of friction, it is evident that the particle will spiral towards the origin of the plane as energy is not conserved.

I have been working on this for a while, but haven't made significant progress towards realizing my goal.  It is helpful to know, I think that a conservative force constraint can be implemented a few ways:

Sum of kinetic and potential energy is constant, integral of work over a closed loop is zero, force is negative gradient of potential.  These are all basically the same, but something the step I am missing in my work is implementing this constraint. As this is 1D the gradient is less constraining.

I think the best way to start would be to integrate a flow, but I am a little rusty on that.  I was hoping that I could convert it to some polar form, but that also requires proving that if I have some flow starting at $\theta$, then both points are still at the same $\theta$ after a given time.  In addition, I don't think it is too rigorous to just pick initial conditions as such.

This [url=http://planetmath.org/encyclopedia/LiouvillesTheorem3.html]proof of Liouville's theorem[/url] could help.


trigonometry
geometry
circumcircle
trig identities
Law of Sines

For any triangle ABC inscribed in a circle, with radius r, show that

a/sinA=b/sinB=c/sinC=2r

[hide]
To prove $\frac{a}{\sin{A}}=\frac{b}{\sin{B}}=\frac{c}{\sin{C}}$, consider writing the area of the triangle in several different ways; do one equality at a time and combine them.

To prove that those all equal $2R$, consider constructing a right triangle within the circumcircle with one side $2R$ and another side $a, b,$ or $c$. You can also first show this and the first part immediately follows.[/hide]

Actually, it does work. Take an equilateral triangle inscribed in a circle, and draw the angle bisectors. They meet at the center of the circle. Now, draw the perpendicular bisectors for each of the three new triangles formed, making a total of six 30-60-90 triangles, with side opposite the 60 as $\frac{s}{2}.$ This means that the long side is $\frac{s}{\sqrt3}.$ So, the diameter of the circle is $2r=\frac{2s}{\sqrt3}.$ Amazingly enough, $\frac{s}{sinA} = \frac{s}{\frac{\sqrt3}{2}}=\frac{2s}{\sqrt3}.$

The easiest way to prove it is probably to use a right isosceles triangle. Designate sides $a$ and $b$ as the two shorter sides. First off, $2r=a\sqrt2.$ Then, $\frac{a}{sinA} = \frac{a}{\frac{1}{\sqrt2}} = a\sqrt2.$ Voila.

Of course it's true. First of all, it's the Extended Law of Sines (if it's named, it must be true) and you can pretty much invoke it without proof whenever you want.

[quote="diophantus"]The statement is not true.  Assuming it is for an equilateral triangle would give s=4s (s is the side length of the triangle) which is a contradiction.[/quote]
no, it's true.... ;)

[quote="Philmac"]For any triangle ABC [b]inscribed in a circle, with radius r[/b], show that

a/sinA=b/sinB=c/sinC=2r[/quote]

I think it's clear... oh well. Now everyone gets what it means, so we are happy. :D


geometry function trigonometry conics ellipse hyperbola logarithms

During the summer, I've been working non-stop to create a math paper which contains derivations/proofs of virtually every high school formula I've ever encountered, as well as a number of formulas I found myself.

For those of you who have already seen my work at [url]http://www.artofproblemsolving.com/viewtopic.php?t=160871[/url], here are the changes I've made (in the red paragraph).

[color=red]

-I merged the "distance formulas" chapter and the "geometry" chapter together because I found that the distance formulas were too short as a chapter by itself.
-I removed the "functions" chapter because I thought it was overly long and redundant. However, I replaced it with a much better chapter called "elementary algebra". If you've already seen all my work, this is probably the only section you'd want to look at (since it is new).
-Made some very minor changes, very discreet stuff like capitalizing letters, centering pictures, changing the spacing in a few areas, etc...
[/color]

If you haven't seen my work but are still interested, read the blue part.

[color=blue]

Topics covered are:

-Geometry theorems such as the inscribed angle theorem, perpendicular bisector theorem, Appolonius' Circle, etc...
-Trigonometry formula derivations for sum and difference formulas, half-angle formulas, double angle formulas, inverse trigonometric formulas, etc...
-Conic section, which includes the derivation of the circle, ellipse and hyperbola formula based on function constants.
-Elementary algebra, such as factoring methods, a detailed set of logarithm formulas, and analysis of the floor,ceiling, fractional part and integer part function, among other things.
-Keep in mind that I'm not trying to get my work published. It really just serves as a reminder on where certain formulas come from, in case one ever forgets. However, I do think that the derivations do demonstrate a fair deal of problem solving ability (don't mean to sound arrogant).
-There are no problem questions in my document.
-No calculus since this is high school level.
-I'm open to any constructive criticism, whether it be about my formatting, subject matter, or whatever it may be. In fact, I'm anticipating what you guys have to say!
-If anyone finds an error in my document, please tell me about it! Editing a 120 or so page document is a long and gruelling process, and as such it is possible that I overlook a thing or two.

[/color]

And, without further ado, here's the attachment.

Really nice, takes some common formulas and explains them in a clear perspective way.

prime factorization, i can only give u praise (cause im too much of a noob to give u constructive criticism), this is a truly commendable thing uv done...ur on a good track my friend..

Wow dude - if this was a book I'd pay for it. Though I have seen another proof of AAA... but still man - sweet  :D  :D

On page 7 i think it should be $ 1^{m}$, not $ 0^{m}$ for Exponent Rule for Base 1

And I dont think you should use logarithms to prove exponentiation identities - think it should be the other way around. No offense

That was great!

I read the whole thing, and I think it was great!

It's meant to cover basic theorems, right?
There isn't much stuff in there that's really complicated or unique to certain problems, and I think you did a great job covering the basic high school curriculum.

It's like a Mini AoPS book 1 summary, honestly. I loved it!

Aside from some minor errors, that's a great paper, and all that work you put into it was definitely worth it.

Congratulations Prime factorization and thank you for your hard work  :)  this paper is excellent  :10:

This is quite heartwarming to hear this kind of praise. It really makes me feel like all those hours spent learning latex and asymptote and all the frustration I had when I got stuck was really worth it. Also, for some of the proofs I got some help from you fellow forum members so thank you so much for that. Without that help, I wouldn't have completed the document as quickly as I did.

[quote]

Wow dude - if this was a book I'd pay for it. Though I have seen another proof of AAA... but still man - sweet

On page 7 i think it should be $ 1^{m}$ , not $ 0^{m}$  for Exponent Rule for Base 1

And I dont think you should use logarithms to prove exponentiation identities - think it should be the other way around. No offense
[/quote]

Odd. I checked the document and surely enough, it said $ 0^{m}\equal{} 0$ and  $ 1^{m}\equal{} 1$ in the right section.

As for the logarithms, I only used them once to prove an exponential identity. I also referenced the logarithm formula I used, but I suppose I could've also referenced the definition. Maybe there's another way to prove it, so I'll recheck that.

And again, does anyone know how to define multiplication and exponentiation for all real values and not just integers?

Just on a final note, if anyone has a request for a basic high school theorem or topic that I may have forgot, please ask and I'll probably write a chapter about it sometime soon. Once again, thank you for your wonderful responses!

im about half way through looking through ur book and i just noticed two mistakes. one, in the geometry section, u were proving that particular angles from perpendicular lines cut by a transversal are congruent and/or supplementary. anywayz, u said in one of the pairs that they were called alternate interior angles. just this last year, i had geometry and i believe those are called alternate [b]exterior[/b] angles. maybe im wrong or maybe different books teach different ways.
secondly, i believe u misspelled isosceles. two times i read [b]isoceles[/b] when it should really be [b]isosceles[/b]
other than that, good job man!!!!! i look forward to reading the rest of ur manuscripit(would u call it?)

great work. :) .really inspiring...it would have taken real work to write those in $ \text{\LaTeX}$ and probably every necessary pre-calculus  thing is there  :10:

[quote="Prime factorization"][quote="mihail911"] anywayz, u said in one of the pairs that they were called alternate interior angles. just this last year, i had geometry and i believe those are called alternate [b]exterior[/b] angles. maybe im wrong or maybe different books teach different ways.
secondly, i believe u misspelled isosceles. two times i read [b]isoceles[/b] when it should really be [b]isosceles[/b]
other than that, good job man!!!!! i look forward to reading the rest of ur manuscripit(would u call it?)[/quote]

I just checked the alternate interior angles. They seem correct to me  :maybe:

And I'll change the spelling for isosceles. Thanks  :D

[quote="pardesi"]

probably every necessary pre-calculus thing is there

[/quote]

Well actually, I missed complex numbers and binomial theorem. Of course, that will be covered in the near future as I'm going to college in 3 days  :)[/quote]

really? what college?
btw, i may be wrong about the alt. int. angles..its just i learned that the angles on opposite sides of the transversal on the outside of the two parallel lines are called alternate exterior angles.. :maybe:

[quote="Prime factorization"][quote="13375P34K43V312"]2 things i noticed so far:

-your definitions of multiplication and exponentiation are incorrect; they only work for integers.
-you can't use the sines to prove AA similarity, because the notion of sine requires AA similarity to be consistent.[/quote]

1) how would you define multiplication and exponentiation for all real values?

2) I'm not sure what you mean. I'm pretty sure I said that given 2 congruent angles, you can find the third angle and then using trig ratios, you can prove that the sides of the two triangles will be in proportion.[/quote]

For 2), it's pretty much circular logic.

You need AA similarity to prove that the definition of sine is well defined. For example, consider this:

[asy]import cse5; pair A=MP("A",D((0,0))), B=MP("B",D((3,0))), C=MP("C",D((3,4))), D=MP("D",D((6,0))), E=MP("E",D((6,8))); D(A--D--E--cycle); D(B--C);[/asy]

(they're right triangles)

From the definition of sine, we see that
$ \sin A \equal{}\frac{BC}{AC}$
$ \sin A \equal{}\frac{DE}{AE}$

But you need AA similarity to prove that these two ratios are indeed equal.
AA Similarity is usually given as a postulate to avoid these kinds of things.

As for the first question, I think it's a lot more complicated than it looks. It should be easy to extend your definition for rational numbers, but how would you define for irrational numbers?

How would you define something like $ p\times\sqrt{47}$?

[quote="Prime factorization"]Ok , I see now. I hope this doesn't mean that my proof for SAS and ASA (congruency only) are false as well  :( I'm guessing that's why all the similarity theorems are postulates (can't really be proven). Ah well, I tried  :)

Wait, hold on. If the SSS proof holds, the AA holds because it is the converse of the SSS theorem (I even mentioned this in my paper). So I guess the proof with trig ratios isn't even necessary.

and I suppose for multiplication, I'll just switch the $ m\in\mathbb{R}$ for $ m\in\mathbb{Z}$ and say that multiplication and exponentiation are hard to define for fractional and irrational values.

Finally, can anyone confirm whether mihail's right about the alternate interior angle mixup? Cuz that's starting to bother me a little  :huh:[/quote]

i can quote my geometry textbook...if that helps.. :oops:

I downloaded ur previous work and now ill also download this work. Brilliant as usual with very thorough proofs!!!

Interesting. Overall, great job. As always, there are little things that should be addressed. I guess I will give specific comments:

p6. For m<0 there should be |m| terms or -m terms

p7 0^0 is undefined, you also didn't prove anything for real numbers, you proved it for integers

p. 9 equivalent exponent theorem is false.

p.13 $ \sqrt{4a^{2}}\equal{}2a$?

p.27 some weird brackets there...?

p.36 you assumed that B,O,C are collinear, assuming reflection preserves angles is a bit strange...that implicitly uses triangle congruence...

p.39 you cannot just cite that dilations map lines parallel to lines...this is circular logic. hint: prove that [BDE]=[CDE]

p.40-2 no trig when you do similarities...circular

p. 51 explain why the perpendicular line gives the shortest distance

p 64 state your cases more clearly

p 68 you didn't prove the statement you end with...you have to start with AF=FE

p73 thats not harmonic division...thats the angle bisector theorem

p74 you didn't prove that the ratio is constant?

p. 84 that doesn't prove that it is an asymptote...generally the definition is that as we get really far, the graph gets really close to the line

p. 95 you don't say that the values are infinity, you say that they are undefined, furthermore, the limits are undefined as well

p.99 only works for acute triangles

p.112 $ \sqrt{sin^{2}(x)}\equal{}\sin x$?

p. 116 line 74 is wrong...next few pages have inaccuracies that stem from this

p.117 needs more justification...consider domains or inverse trig functions...

so again, generally, nice job

some circular logic in the geo section...sss, etc theorems need to be redone

you should explicitly state your givens for each problem...like thm: blah, pf: blah

you have nothing about domains of inverse trig functions, this makes many of the things incorrect. this needs to be addressed...

wow! thank you so much for doing this for us.

[quote="Anaxerzia"]Wow dude - if this was a book I'd pay for it. Though I have seen another proof of AAA[/quote]

Since it is pretty obvious my AA proof is false, do you mind posting yours? It would be of great help  :blush:

And again, I am up for requests of topics I missed. I'll just give a few examples of stuff I have in mind.

Bramaghupta's formula
Ptolemy's Theorem
Proving Geometric Constructions (can't wait to do this!)
directrix definition for ellipses and hyperbolas

really useful. thanks alot for sharing it.

WOW!!! THIS IS REALLY USEFUL!! AND NICE JOB!!! :10:  :winner_first:

Wow. Not quite book-level, but.... mayhaps we can see the source code?

Here's some transcripts. If you want the rest, PM me I guess.

LOL what's the book comment for? I'm only 17 years old you know, I'm not trying to make this work a critically acclaimed math textbook or anything...


Let given a sequence $ a_{n}$ whose elements are all positive such that $ \sum_{n \equal{} 1}^\infty\ a_{n}^{1 \plus{} \frac {1}{n}}$ converges. Prove that $ \sum_{n \equal{} 1}^\infty a_{n}$ converges too.

[url=https://artofproblemsolving.com/community/c7h423718p2396702] see here [/url]


conics ellipse trigonometry quadratics function

If $x$ and $y$ are real numbers satisfyin: $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, then try to find the ranges of $x+2y$ and $\frac{y-4}{x+3}$.

Go to the parametric equations of the ellipse: $x=3\cos t, y=2\sin t$. Then

$x+2y=3\cos t+4\sin t=5({3\over 5}\cos t+{4\over 5}\sin t)=5\sin(t+\arctan{3\over 4})$, which obviously has the range $[-5,5]$

For the second expression we'll use the substitution $u=\tan{t\over 2}$ which gives $\sin t=\frac{2u}{1+u^{2}},\cos t=\frac{1-u^{2}}{1+u^{2}}$:

$\frac{y-4}{x+3}=\frac{2(\sin t-2)}{3(1+\cos t)}=\frac{2(\frac{2u}{1+u^{2}}-2)}{3(1+\frac{1-u^{2}}{1+u^{2}})}=-{2\over 3}(u^{2}-u+1)$

If $t$ runs through the whole $\mathbb{R}$, then $u$ obviously does too. Therefore we can argue as follows: The quadratic function $f(u)=u^{2}-u+1$ has the range $[-\frac{(-1)^{2}-4\cdot 1\cdot 1}{4\cdot 1},+\infty)=[{3\over 4},+\infty)$, which means that $-{2\over 3}f(u)$ has the range $(-\infty,-{1\over 2}]$


[quote="Αρχιμήδης 6"]I post this theme in english.
Show without multiplying out that :$ (b \minus{} c)/a \plus{} (c \minus{} a)/b \plus{} (a \minus{} b)/c \equal{} [(a \minus{} b)(b \minus{} c)(a \minus{} c)]/abc$
THANKS!!![/quote]

Ωραίο!

Αν τα κάνουμε όλα ομώνυμα κατα φέρναμε στο πρώτο μέλος θα προέκυπτε πολυώνυμο δευτέρου βαθμού ως προς οποιαδήποτε μεταβλητή.

Στη μορφή που μας δίνεται όμως, θεωρώντας το α μεταβλητή έχει προφανείς ρίζες τις b,c και το (b+c). Άρα είναι ταυτότητα και ισχύει η δοσμένη...

Λαθος και με πολλες ελλειψεις.FALSE!

[quote="Αρχιμήδης 6"]Λαθος και με πολλες ελλειψεις.FALSE![/quote]

Χμμ..
Για κάτσε:

Φέρνοντας στο πρώτο μέλος έχουμε:

$ \frac {b\minus{}c} {a} \plus{} \frac {c\minus{}a} {b} \plus{} \frac {a\minus{}b} {c} \plus{} \frac {(a\minus{}b)(b\minus{}c)(c\minus{}a)} {abc} \equal{}0$

Ο πρώτος όρος μετά την απαλοιφή παρονομαστών δεν θα έχει α. Οι δεύτεροι δύο θα έχουν το α σε 3 δύναμη (εδώ είχα λάθος όντως) και το τρίτο θα είναι σε δεύτερη δύναμη. Άρα αρκεί να βρούμε τέσσερις ρίζες ως προς α...

Μια είναι το b, μία το c, μία το b+c και πάρε ότι συνδυασμό θες για τέταρτη (b-c πχ). Τώρα θα έχει 4 ρίζες άρα είναι ταυτοτικό πολυώνυμο... (Προφανώς θεωρούμε a,b,c διάφορα του 0...)

Εστω οτι $ f(x): \equal{}(b\minus{}c)/x\plus{}(c\minus{}x)/b\plus{}(x\minus{}b)/c$
Παρατηρω οτι $ f(b)\equal{}f(c)\equal{}0$. Απο το θεωρημα των συντελεστων αν γραψω το $ f(x)$ ωσ πηλικο πολυωνυμων τοτε $ f(x)\equal{}P(X)/xbc$ τοτε το $ P(X)$ πρεπει να εχει τους $ x\minus{}b,x\minus{}c$ διαιρετες.Επισης το $ P(X)$ εχει βαθμο $ 3$. Θεωρω οτι $ x\equal{}a$ στο $ f(x)$ και τοτε
$ (b\minus{}c)/a\plus{}(c\minus{}a)/b\plus{}(a\minus{}b)/c\equal{}(a\minus{}b)(a\minus{}c)R(x)/abc$ oπου $ R(x)$ γραμμικο πολυωνυμο.Απο συμμετρια οριζω την συναρτηση
$ g(x): \equal{}(x\minus{}c)/a\plus{}(c\minus{}a)/x\plus{}(a\minus{}x)/c$ και εχουμε οτι $ g(a)\equal{}g(c)\equal{}0$ οποτε ισχυει οτι
$ (b\minus{}c)/a\plus{}(c\minus{}a)/b\plus{}(a\minus{}b)/c\equal{}(b\minus{}a)(b\minus{}c)Q(x)/abc$ οπου $ Q(x)$ ενα αλλο γραμμικο πολυωνυμο.τελος ισχυει οτι:
$ (b\minus{}c)/a\plus{}(c\minus{}a)/b\plus{}(a\minus{}b)/c\equal{}K[(a\minus{}b)(b\minus{}c)(c\minus{}a)/abc]$ για καποια σταθερα $ K$ .θετωντας τιμες πχ $ a\equal{}1,b\equal{}2,c\equal{}3$ προσδιοριζω οτι $ K\equal{}\minus{}1$.
ΕΥΧΑΡΙΣΤΩ!!!

[quote="Demetres"]Ισούνται και τα δύο με $ \frac {1}{abc} \begin{vmatrix} 1 & a & a^2 \\
1 & c & c^2 \\
1 & b & b^2 \end{vmatrix}$.

Το αριστερό μέλος από τον κανόνα του Cramer και το δεξί μέλος γνωρίζοντας ότι ο πιο πάνω είναι πίνακας Vandermonde.  (Ή διαφορετικά, παρατηρώντας ότι τα $ (a \minus{} b),(b \minus{} c),(c \minus{} a)$ διαιρούν την ορίζουσα που άρα πρέπει να ισούται με $ K(a \minus{} b)(b \minus{} c)(c \minus{} a)$ όπου $ K \in \mathbb{Q}$ αφού κάθε όρος της ορίζουσας είναι βαθμού 3 στα $ a,b,c$.)[/quote]

Αααυτό!!! Και αν δεν είναι ξεκάθαρος ο τρόπος με τον οποίο εννοώ αυτό το σχόλιο, τότε ο παρακάτω σύνδεσμος το διευκρινίζει απόλυτα:

http://www.youtube.com/watch?v=Onv0u_W7CFI

Πολύ χαριτωμένη λύση, Δημήτρη. (Νομίζω ότι η επεξήγηση για την ορίζουσα Vandermonde είναι περιττή. Ο τύπος της σου δίνει απευθείας τον αριθμητή του δεξιού μέλους.)

demetres εχω καποιες ενστασεις περι <πληροτητας>της αποδειξης σου.Η γενικευση μου σε αυτη την ασκηση  βασιζεται στο θεωρημα Hawk-lewinski.

[quote="Αρχιμήδης 6"] βασιζεται στο θεωρημα Hawk-lewinski.[/quote]

Μπορείς να μας δώσεις κάποιες πληροφοριες γι' αυτο το θεώρημα μιας και δεν κατάφερα να βρω τίποτα μέσω γούγλη;


function
calculus
derivative
integration
linear algebra
matrix
probability

I have maple, mathematica, matlab, mathCAD, derive, mathtype, mathXpert, MuPad, SPSS, Fortran, plus a small grab bag of no name prodcts, but what have I missed (that's good)? Also, of these, which ones do you use?

Thanks,
    -Ben

my question is how were u able to afford all of these programs?

and why would you WANT all those programs??

what would you do with them most of them have the same functions anyway y do u need all of them?

I assume you also have a spreadsheet program (Excel or one of its competitors.) Don't underestimate how much numerical mathematics you can do, and quite quickly, with a spreadsheet. It's particularly good with recursion. (Binomial coefficients mod $n$, Euclidean algorithm/continued fractions, getting a first glimpse of the likely asymptotic behavior of some recursively defined sequence, and so on.)

[quote="phils"]For the poorer (and smarter  :P) among us: [url=http://maxima.sourceforge.net/]Maxima[/url] is a nice, free "all-purpose" CAS. It is not as powerful as Mathematica and Maple, but it comes pretty close. Plus it is open source. For the algebraists among us, check out the extremely nice (and also free) [url=http://www.singular.uni-kl.de/]Singular[/url] and [url=http://www.math.uiuc.edu/Macaulay2/]Macaulay 2[/url]. Macaulay in particular blew my mind.

For number theory, [url=http://pari.math.u-bordeaux.fr/]PARI-GP[/url] is the way to go.

And finally, if you want to write your own math code, you might be interested in [url=http://www.netlib.org/lapack/]LAPACK[/url].

Personally, I use the (functional) programming language Haskell for my "numeric" code. In my opinion, it is an excellent (the best?) choice for math code. Moreover, functional programming is tangent to a lot of advanced and interesting math (though most of it is algebra/logic rather than analysis).[/quote]

Thank you phils. This was a welcome post.

Here are a few websites that have online calculators (all free):

[url]http://www.quickmath.com[/url] - powered by webMathematica: this site is very handy and will do algebra (factoring, simplify, partial fraction decomposition, expand), solve equations/inequalities (single or a system), plot equations/inequalities (single or a system), derivatives, definite or indefinite integrals, and do matrix arithmetic/inverses/determinants.

[color=red]Suggestion[/color]: for those who post "Will someone please check this [problem/integral/derivative/calculation]..." or the like, cut and paste the above paragraph as a reply.

[url]http://www.integrals.com[/url] - powered by the Mathematica engine: just what it seems, but it will only do indefinite integrals.

[url=http://www.livephysics.com/ptools]www.livephysics.com[/url] - has these tools: Evaluate Function, Function Grapher, 3D Function Grapher. Very nice since the [url=http://www.livephysics.com/ptools/online-3d-function-grapher.php]3D Function Grapher[/url] allows for you to plot a 3D function and then cut and paste customized UBB code from the page into a forum post to embed the plot graphic you made into your post (see [url=http://www.livephysics.com/ptools/online-3d-function-grapher.php#add]add[/url]).

[url=http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/utilsindex.html]Online Mathematics Utilities[/url] - includes these categories of utilities/tools: Graphers, Regression Tools, Matrix and Linear Algebra Tools, Probability and Statistics Tools, Calculus Tools, Math of Finance Tools, and Miscellaneous.

Another useful website [url=http://library.wolfram.com/webMathematica/Education/WalkD.jsp]here[/url]. It does derivatives step-by-step, explaining each step by the derivative rules, very handy.

One more website: [url=http://calc101.com/]calc101.com[/url]. Some of this site's features require you to purchase a membership ($\$$ 25 per year) but much of it is free. Check out the [url=http://calc101.com/webMathematica/partial-fractions.jsp]partial fraction decomposition page[/url], or the [url=http://calc101.com/webMathematica/long-divide.jsp]polynomial long division page[/url]: very nice.

[quote="phils"]For the poorer (and smarter  :P) among us: ...[/quote]

You might be interested in [url=http://modular.math.washington.edu/sage/]SAGE[/url] as well. It's free, open source, and it includes most of what you listed. There are also components that allow you to call the non-free programs if you have them, so you can do everything in one interface.


i want to do HMMT really bad, but my school doesnt offer it and i dont think there is one in my area, what should i do?

HMMT is only held at Harvard and MIT. To participate you need to travel to Boston, so the fact that there are no "local HMMTs" is not an unfortunate circumstance but rather a consequence of the fact that it is only held at one location.

This year, it will be held at Harvard University. See also: http://web.mit.edu/hmmt/www/faq.shtml.

EDIT: Also, I see that you live in Virginia. If you live in Northern Virginia, then you can probably find a team of students to go with from local math clubs, since there are a reasonable number of people who participate in math contests, even outside of TJ. Also there are probably people from Maggie Walker (the Governor's school in Richmond) that would want to go. Or you can just go as an individual and get teamed up with other people on the spot.

If you want to talk about information specific to your situation, I recommend e-mailing hmmt-request@mit.edu so that you are talking directly to the contest organizers.

Thanks. Yea, i needed a team around my area actually, sorry if the question wasnt worded right, i do realize that you do have to go to harvard/MIT for this. I needed to know how to find such teams, would the program coordinators have a list of teams in the VA area?

If you're in Northern VA, Fairfax Math Circle might have team--this is totally a guess on my part, though, I'm not associated with them or anything. But wherever you are in VA, you might be able to go with the Chesapeake math team, which I'm on. For ARML, we officially cover all of MD and VA that isn't covered by another team, but for HMMT/PUMaC we can take any other Marylanders/Virginians (and last year we also had a Pennsylvanian and Georgian.) I don't know how much room there is on the team this year, but there probably is some--if you're interested, PM me and I could get in touch with our coach. I'm from near Richmond, VA, btw.

Doesn't everyone have to go to school in the same state, though?

For 2008-2009 PUMaC, there were actually no restrictions on teams (unless the website is mistaken/outdated), except that the coordinator of PUMaC had the final say on the legitimacy of our team.

HMMT doesn't have a state-by-state rule; as long as everyone was within a specific distance, we were fine. Of course, that ruled out having the person from Georgia compete with us again, so we simply found someone else from Maryland.


inequalities
algebra
polynomial
algebra proposed

Let $P(x)=ax^2+bx+c$ with real coefficients. By assuming that  $\sqrt{x(1-x)}\; \cdot P(x) \le 2005\; ,\; \; \forall x\in [0,1],$ prove that   $| P(1)| \le 12030 .$

The inequality can be written like

(1) $\frac{1}{4010}\sqrt{1-(2x-1)^{2}}P(x)\leq 1$

Let denote $y=2x-1$ and $Q(y)=\frac{1}{4010}P(\frac{y+1}{2})$

(1) becomes $\sqrt{1-y^{2}}Q(y)\leq 1$

Now we apply the
Lemma: $R$ complex polynomial with $deg(R)\leq n-1$,suppose also
$\forall x \in [-1,1] ,\; \sqrt{1-x^2}|R(x)|\leq 1$
Then for any $x\in [-1,1]$ we have $|R(x)|\leq n$

Choose $R(y)=Q(y)$ with degree $\leq 2=3-1$

$|Q(y)|\leq 3$  and finally with $y=1$

$|P(1)|\leq 4010 \times 3=12030$

[quote="Moubinool"]
..Now we apply the
Lemma: $R$ complex polynomial with $deg(R)\leq n-1$,suppose also  $\forall x \in [-1,1] ,\; \sqrt{1-x^2}|R(x)|\leq 1$
Then for any $x\in [-1,1]$ we have $|R(x)|\leq n$ Choose $R(y)=Q(y)$ with degree $\leq 2=3-1$  $|Q(y)|\leq 3$  and finally with $y=1$   $|P(1)|\leq 4010 \times 3=12030$[/quote]
Yes, this is the desired solution !


Show that $y = 3\sin x + 2\cos x$ is a sinusoid in the form:

$y = A\sin(x+\delta)$

Find $A$ and $\delta$

yea i've never been good at these problems... XP should've listened during pre-calc class

i think i know how to get $A$, something like $\sqrt{3^2+2^2}$ or something... but yea help plz

think about the expansion of $y = A\sin(x+\delta) $

oh, i think i get it, is it something like....

$y = A(\sin x\cos \delta + \sin \delta \cos x)$

So,

$A\cos \delta = 3$ and $A\sin \delta = 2$

Dividing 2nd by 1st, you get:

$\tan \delta = \frac{2}{3}$, and i did this part on my calculator even though i think there might be a way i can do this by hand, but yea

doing that $\delta \approx 33.69 ^\circ$

we also know $\sin \delta = \frac{2}{\sqrt{13}}$ and $\cos \delta = \frac{3}{\sqrt{13}}$

So, multiplying our original two equations, we have:

$A^2 (\sin \delta)(\cos \delta) = 6 \implies A = \sqrt{13}$

is that right?


Find all continuous solutions to:

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

 :)

[hide="Hint"]

Set $ f(0) \equal{} a$, $ g(0) \equal{} b$ and $ h(0) \equal{} c$ and work with it

[/hide]

 :)

As nobody post a solution I'm having doubts about mine. So if somebody can point me if there is a mistake, I'll be very thankful.

Lets set $ a$, $ b$ and $ c$ as the post above. So we have that $ E(0) \equal{} a \plus{} b \plus{} c$.
Plugging into the original equation $ (x,0,0)$, $ (0,y,0)$ and $ (0,0,z)$ and solving for $ f(x)$, $ g(y)$ and $ h(z)$ respectively, we get:

$ f(x) \equal{} E(x) \minus{} b \minus{} c$
$ g(y) \equal{} E(y) \minus{} a \minus{} c$
$ h(z) \equal{} E(z) \minus{} a \minus{} b$

Substitute this in the original:
$ E(x \plus{} y \plus{} z) \equal{} E(x) \minus{} b \minus{} c \plus{} E(y) \minus{} a \minus{} c \plus{} E(z) \minus{} a \minus{} b$
We can set $ z \equal{} 0$ and sum $ \minus{} a \minus{} b \minus{} c$ to each side:
$ E(x \plus{} y) \minus{} a \minus{} b \minus{} c \equal{} E(x) \minus{} a \minus{} b \minus{} c \plus{} E(y) \minus{} a \minus{} b \minus{} c$

So lets introduce $ J(x) \equal{} E(x) \minus{} a \minus{} b \minus{} c$:
$ J(x \plus{} y) \equal{} J(x) \plus{} J(y)$ wich is very known to have continuous solutions $ J(x) \equal{} xJ(1) \equal{} dx$

This lead us to:

$ E(x) \equal{} dx \plus{} a \plus{} b \plus{} c$
$ f(x) \equal{} dx \plus{} a$
$ g(x) \equal{} dx \plus{} b$
$ h(x) \equal{} dx \plus{} c$

Wich are the solutions for the original equation.

 :)


Let $ AB,CD$ be diameter of circle $ (O)$. $ AB$ perpendicular to $ CD$. $ M,N$ be two point on $ CD$ such that  $ CM\equal{}ON$. let $ K$ be the intersection of $ AN$ with $ (O)$. Prove that $ \triangle AKM$ is acute triangle


[quote="coffeym"]I found this puzzle very good fun:
Snow White wakes up on Christmas morning to find her 7 dwarves have already opened their presents, however, none of them like what they got, but all of them like the presents of the others.
(i) How many ways can Snow White redistribute the presents so that [b]each[/b] dwarf gets a different present to what they had before?
(ii) How many ways could Snow White do this if there were [i]n[/i] dwarves instead of 7.

Just answers will do; no solutions necessary[/quote]

[hide]for (ii) i get $n^2$
then for (i) i obviously get $49$, even though i doubt theyre right[/hide]

[hide]Wouldn't #1 be 42, since each dwarf has to get a different gift? So it would be $7\times6$? Likewise, #2 would be $n(n-1)$. I'm not sure though...[/hide]

Need to try again  :oops: Try imagining first that there are only 3 dwarves, then right down all the possibilities. Then more to 4, and see if you can spot a pattern.

[hide]For 3 dwarves, I get 4 ways. For 4, I get 18 ways. Maybe $(n-1)!(n-1)$?[/hide]

Remember its not enough for one dwarf to have a different present.

[hide]Okay, so I think that gets rid of half the possibilities, so $\frac{(n-1)!(n-1)}{2}$? Am I getting any closer? :? [/hide]

What I think is right:
[hide](i) 2160
(ii) (n!(n-1))/2n[/hide]

[hide]Yay, so I got it right? :D $\frac{n!(n-1)}{2n}=\frac{(n-1)!(n-1)}{2}$. Your way was better to represent it though. However, I noticed that when $n=2$, we would get 0.5, but there should be 1 way to redistribute the gifts. Maybe this only works for $n\ge2$?[/hide]

[hide]
DWARF #     1         2         3         4          5

GIFT             A        B        C       D          E

This is the arrangement for five dwarves that are not happy with their gifts.

I came up with only 36 arrangements where all the dwarves were happy with their gift.

I’ll begin with present A going to dwarf # 5.

These are the arrangements I came up with:

DWARF #        1         2        3       4          5

                       B        C        D       E          A

                       B        E        D       C          A

                       B        D        E       C          A

                       C        D        E       B          A

                       C        E        D       B          A

                       C        D        B       E          A

                       D        C        B       E          A

                       D        E        B       C          A

                       D        C        E       B          A

By putting A in the fifth position we come up with 9 possible combinations.

Then by putting letters B, C and D in the fifth position they will each yield 9 possible combinations.

We don’t put E in the fifth position, otherwise we will have an unhappy dwarf.

This gives you a total of  36 arrangements for 5 dwarfs to all be happy.

According to your general solution, for 5 dwarves you would come out with 48 possible arrangements.

I haven't been able to come up with a general solution so far.

I would welcome anyone's input and check to see if I am correct.

But for 3 dwarves I get 2 possible arrangements and for 4 dwarves I get 9 possible arrangements

[/hide]

Isn't question one simply asking for how many derangements of the set  there are (a derangement is an arrangement of a set where none of the elements are where they started)?

Which in that case the answer would be [hide]1854[/hide]

I have a partial solution:

For 6 dwarves there will be $180$ possible arrangements.

For 7 dwarves there will be $1080$ possible arrangements.

Still no general solution though.

[hide="Theoretical explanation"]Well, if we have $n$ dwarves, we have to count like this:

Take all the permutations - $n!$. Subtract those in which one dwarf got back his present - ${n \choose 1}(n-1)!$ (the first term is the number of choices for that dwarf, and the second is the number of permutations for the rest of the lot). But now we must add back those in which two dwarves got their presents back - ${n \choose 2}(n-2)!$, et cetera.

Having in mind that ${n \choose k}(n-k)!={{n(n-1)\dots(n-k+1)}\over k!}(n-k)!={n! \over k!}$, we have:

$N=n!-{n! \over 1!}+{n! \over 2!}-\dots+{{{(-1)^n}n!} \over n!}=n!\sum_{k=0}^n{{(-1)^k}\over k!}$. This number is specially denoted as $!n$.

For $n=7$ we can calculate that $!7=4420$.[/hide]

Farenhajt is correct - the subfactorial function $!n$ is, in fact, defined as the number of permutations such that no element is in its original position, exactly the desired quantity.   :)

Woops, I meant to post my answer as the subfactorial of $7$...(aka derangement)...dunno how I came up with what I did.

edit> I dunno where I'm messing up; I calculated it again, and I got $5040-5040+2520-840+210-42+7-1=1854$ ? Anyone care to point out where I'm messing up?

hi Elemennop:

 You didn't mess up. Look at this web page and notice the value for $!7$

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

[quote="Elemennop"]...the subfactorial of $7$...$5040-5040+2520-840+210-42+7-1=1854$ ? Anyone care to point out where I'm messing up?[/quote]

[b]SORRY[/b], you're right. The mistake is [u]mine[/u]. Your number is correct.

[quote="Farenhajt"][quote="Elemennop"]...the subfactorial of $7$...$\underline{5040-5040}+2520-840+210-42+7-1=1854$ ? Anyone care to point out where I'm messing up?[/quote]

(See whats underlined.) You don't subtract the factorial itself - you go straight to the next term (i.e. $5040-2520+840-210+\dots$)[/quote]

Actually, yes you do.  The sum starts $k = 0$ so the first two terms of $\sum_{k=0}^{n} \frac{ (-1)^k }{k!}$ are $\frac{1}{0!}$ and $\frac{-1}{1!}$.  :)

Yes, i've realized that after I pressed "submit", but the post is now edited.

Ah, ok...I was just so confused since finally they came up with the same solution as I did and they had a different number.

I hate to belabor this topic but when I looked at all the possibilities for $n = 5$ to satisfy the requirements of the problem I could only come out with $36$ (from my first post). I checked and double checked to make sure I didn't overlook any "eligible" arrangements and each time I came out with $36$.

According to the laws of subfactorials the quantity should be $44$ for $n = 5$.

If this figure  is correct could anyone tell me what the other eight arrangements are that I missed?


trigonometry
limit
real analysis
real analysis unsolved

Hi, everyone  :)

$ \sin_{n}{\theta} \equal{} \sin{(\sin{(\cdots(\sin{\theta})})\cdots )}$ (n times)

Find $ \lim_{n\to\infty}{\frac {\sin_{n}{(2)}}{\sin_{n}{(1)}}}$

Thanks  :blush:

Numerically, it looks like this limit is $ 1.$ Numerator and denominator both behave like $ \frac{C}{\sqrt{n}}.$ The question to be resolved: why is that the same constant in both cases?

Thanks, Kent Merryfield  :)
Well, there is no special reason why that is constant.
Just my wonder...
May I ask you help for the proof ?
Because I am poor at math  :( , I can seldom deal with many questions by myself.

[quote="Kent Merryfield"]Numerically, it looks like this limit is $ 1.$ Numerator and denominator both behave like $ \frac {C}{\sqrt {n}}.$ The question to be resolved: why is that the same constant in both cases?[/quote]

More precisely, if $ x_{n\plus{}1} \equal{} \sin x_n$, starting in $ (0,1)$, then $ x_n \sim \big(3n\plus{}c\big)^{\minus{}1/2}$ ... both numerator and denominator obey this (with different constants $ c$) so their quotient approaches $ 1$.

That should be $ x_n\sim \sqrt{\frac{3}{n\plus{}c}}.$ You had the 3 in the wrong place.

Could you please explain more in detail about your $ x_{n}$?

Starting with $ x_1 \equal{} 1$, we can show the following:

$ \frac {\sqrt {3}}{\sqrt {n} \plus{} 1} \leq x_{n} \leq \frac {\sqrt {3}}{\sqrt {n}}$ (for all $ n > 0$)

Using the fact that for $ x \in (0,1)$

$ x \minus{} \frac {x^3}{6} < \sin x < x \minus{} \frac {x^3}{6} \plus{} \frac {x^5}{120}$

Since starting with $ x_1 \equal{} 2$ will reach the same bounds only a constant number of terms of the sequence later as starting with $ x_1 \equal{} 1$, that the limit is 1 follows.

(In fact limit of $ \sqrt{n} x_{n}$ is $ \sqrt{3}$ and so multiplying and dividing by $ \sqrt{n}$ yields $ a_{n}/b_{n}$ where both $ a_{n}$ and $ b_{n}$ converge to $ \sqrt{3}$)


Find a solution to this recurrence relation:

\[ a_{n} = 2a_{n-1}+8a_{n-2}+81n^{2} \]


function algebra polynomial trigonometry

The function $f$ has the property that, for each real number $x,$ \[ f(x)+f(x-1) = x^2. \] If $f(19)=94,$ what is the remainder when $f(94)$ is divided by 1000?

The function that satisfies $f(x)+f(x-1)=x^2$ is $f(x)=\frac{x^2}{2}+\frac{x}{2}$, but $f(19)\not=94$. What is the problem?

[quote="joml88"]yassin, you incorrectly assumed it had to be a polynomial function.  If you want one such function that works, \[ f(x)=\frac{x^2}2+\frac x2+96\cos(\pi x).  \]  That's one that I came up with.  Of course, there could be (and probably are) other functions that satisfy those constraints.[/quote]

Yeah, I see. Stupid mistake  :)

[hide="Answer"]We have $f(x)=x^2-f(x-1)$, from which we find alternating positive and negative squares. As we are dealing with an even $x$, we will have $94^2-93^2+92^2-91^2+...+20^2-94$. Given that the sum of the first $n$ squares is $\frac{n(n+1)(2n+1)}{6}$, we have $8\left(\frac{47(48)(95)}{6}\right)-\frac{94(95)(189)}{6}- \left(8\left(\frac{9(10)(19)}{6}\right)-\frac{19(20)(39)}{6}\right)-94=4561$ (two times the sum of the first $47$ even squares minus the sum of the first $94$ squares, minus two times the sum of the first $9$ even squares minus the sum of the first $19$ squares, minus $94$), and from this our answer is $\boxed{561}$.[/hide]

JesusFreak, you really didn't need to use the sum of squares formula, you just needed to rewrite $94^2 - 93^2 + 92^2 - 91^2 +-... = (94+93)(94-93) + (92+91)(92-91) + ...$ and use sum of integers instead  :D

Oh, right... I could have done $4\left(\frac{47(48)}{2}\right)-4\left(\frac{10(11)}{2}\right)-37+20^2-94$. :oops: I knew that it simplified to that, but couldn't think of a good way to get the answer from there without a calculator.

[quote="joml88"]yassin, you incorrectly assumed it had to be a polynomial function.  If you want one such function that works, \[ f(x)=\frac{x^2}2+\frac x2+96\cos(\pi x).  \]  That's one that I came up with.  Of course, there could be (and probably are) other functions that satisfy those constraints.[/quote]

Can you please explain how you were able to arrive at that function which, to my amazement, works perfectly, given just the original conditions of the problem? Or at least a proof that this function always works?

Thanks :)

I was wondering if anyone would ask  :D

At first, I only considered if $f$ is a polynomial.  But after a little bit of thought, I came to the realization that $f$ would have to be of degree 2 and proceeded to verify what yassin had already asserted, i.e. $f(x)=\frac{x^2}2+\frac x2.$  But, of course, if that were $f$ then $f(19)$ would equal 190...not 94.  I didn't want to mess with what I already had for $f$ so I considered adding a non-polynomial function $g$ to what we have for $f$ right now.  Thus we would have $f(x)=\frac{x^2}2+\frac x2+g(x).$  We also need $g$ to satisfy $g(x)+g(x-1)=0.$

I thought immediately of cosine when I got to that point.  But if you don't immediately see that, here's something that might make it a little more clear.  We have $g(x)=-g(x-1)$.  So then $g(x-1)=-g(x-2)$ which implies $g(x)=g(x-2).$  So the function is [b]periodic[/b] with period 2.  Of course, the trig functions are periodic and we can just make the coefficient of $x$ equal to $\pi$ so that the period will be 2.

[hide="Verification that my function works."]
We have $f(x)=\frac{x^2}2+\frac x2+96\cos(\pi x).$  Thus

\begin{eqnarray*} f(x)+f(x-1) &=& \frac{x^2}2+\frac x2+96\cos(\pi x)+\frac{(x-1)^2}2+\frac{x-1}2+96\cos(\pi(x-1))\\ &=& x^2+96(\cos(\pi x)+\cos(\pi x-\pi)))\\ &=& x^2+96(\cos(\pi x)+\cos\pi x\cos\pi+\sin \pi x\sin \pi)\\ &=& x^2+96(\cos(\pi x)-\cos \pi x+(0))\\ &=& x^2 \end{eqnarray*}[/hide]

Finding the function was much harder than the original problem :P but it was a lot of fun.  I wouldn't have done it unless yassin asked.  Thanks yassin! :)

Wonderful, thanks joml88 :) I learned something.

[quote="Iversonfan2005"]We have $ 94^2 \minus{} 93^2 \plus{} 92^2 \minus{} 91^2..... \plus{} 20^2 \minus{} f(19) \equal{} 187 \plus{} 183 \plus{} 179..... \plus{} 400 \minus{} 94 \equal{} \frac {94(95)}{2} \minus{} \frac {20(21)}{2} \plus{} 400 \minus{} 94 \equal{} 4255 \plus{} 400 \minus{} 94 \equal{} 4561$ So the answer is $ \boxed{561}$
[/quote]
Hmm, I summed the alternating series in a different way and obtained the same answer, but I don't see how you've got $ \frac {94(95)}{2} \minus{} \frac {20(21)}{2}$...


Im confused what a regular octahedron looks like.  Is a octahedron inscribed in a cube with its vertices being the midpoints of the cube's faces, is that a regular one?


algebra
polynomial
function
algebra unsolved

$ (1)$.Find all polynomials for which if $ ab\plus{}bc\plus{}ca\equal{}0$,then $ P(a\minus{}b)\plus{}P(b\minus{}c)\plus{}P(c\minus{}a)\equal{}0$
$ (2)$.The polynomial $ P$ of degree $ n$ satisfies the condition $ P(i)\equal{}\dfrac{1}{i}$,for $ i\equal{}1,2,...,n\plus{}1$.Find $ P(0)$.

(2)
$ Q\left( X\right) \equal{}$ $ XP\left( X\right) \minus{}1,$
Then for $ i\in \left\{ 1,\cdots ,n\plus{}1\right\}$, $ Q\left( i\right) \equal{}0$
$ deg(Q)\equal{}n\plus{}1$,  $ Q\left( X\right)\equal{}a\prod\limits_{k\equal{}1}^{n\plus{}1}\left( X\minus{}k\right)$.
 $ Q\left( 0\right) \equal{}\minus{}1$ so
$ Q\left( 0\right) \equal{}a\prod\limits_{k\equal{}1}^{n\plus{}1}\left( \minus{}k\right) \equal{}\left(
\minus{}1\right) ^{n\plus{}1}\left( n\plus{}1\right) !\equal{}\minus{}1\Longrightarrow a\equal{}\dfrac{\left(
\minus{}1\right) ^{n}}{n!}$

$ XP\left( X\right) \equal{}\dfrac{\left( \minus{}1\right) ^{n}}{n!}\prod
\limits_{k\equal{}1}^{n\plus{}1}\left( X\minus{}k\right) \plus{}1$

Polynom on right has 0 for root, so it can be write
 $ \alpha X\plus{}\cdots \plus{}\dfrac{\left( \minus{}1\right) ^{n}}{n!}X^{n\plus{}1}$.
then  $ \alpha \equal{}P\left(0\right)$
then $ P(0)\equal{}\frac{\left( \minus{}1\right) ^{n}}{n!}\sum\limits_{k\equal{}1}^{n\plus{}1}\dfrac{\left(
\minus{}1\right) ^{n\plus{}1}\left( n\plus{}1\right) !}{\minus{}k}\equal{}\left( n\plus{}1\right)
\sum\limits_{k\equal{}1}^{n\plus{}1}\frac{1}{k}$.

Has anyone got a solution for the first one?

$ (1)$.Find all polynomials for which if $ ab \plus{} bc \plus{} ca \equal{} 0$,then $ P(a \minus{} b) \plus{} P(b \minus{} c) \plus{} P(c \minus{} a) \equal{} 0$

$ (a,b,c) \equal{} (0,0,0)$ fullfils $ ab \plus{} bc \plus{} ca \equal{} 0$
then $ P(0) \equal{} 0$ then $ P(x) \equal{} x Q(x)$

$ (a,b,c) \equal{} (\frac {a}{3},\frac { \minus{} 2a}{3},\frac { \minus{} 2a}{3})$ fullfils $ ab \plus{} bc \plus{} ca \equal{} 0$
then $ P(a) \plus{} P( \minus{} a) \equal{} 0$, $ P(x)$ is an odd polynomial.

Suppose that exists a least one $ d > 0$ not zero such as $ P(d) \equal{} 0$ pick the maximum of all $ d$

There is a $ b$ such $ (b \plus{} d,b,b \minus{} d)$ fullfils $ ab \plus{} bc \plus{} ca \equal{} 0$
$ b^2 \plus{} bd \plus{} b^2 \minus{} bd \plus{} b^2 \minus{} d^2 \equal{} 0$
$ 3b^2 \equal{} d^2$ then:
$ P(d) \plus{} P(d) \plus{} P( \minus{} 2d) \equal{} 0$, $ P( \minus{} 2d) \equal{} 0$ then $ P(2d) \equal{} 0$ Contradiction

then $ P(x) \equal{} ax^{2k \plus{} 1}$

$ (a,b,c) \equal{} (6,3, \minus{} 2)$ fullfils $ ab \plus{} bc \plus{} ca \equal{} 0$
then $ P(3) \plus{} P(5) \plus{} P( \minus{} 8) \equal{} 0$
$ 3^{2k \plus{} 1} \plus{} 5^{2k \plus{} 1} \plus{} ( \minus{} 8)^{2k \plus{} 1} \equal{} 0$
$ 3^{2k \plus{} 1} \plus{} 5^{2k \plus{} 1} \equal{} 8^{2k \plus{} 1}$
then $ k \equal{} 1$
$ P(x) \equal{} ax$

similar to IMO 2005\2

choose a=2x, b=2x, c=-x, and you get $ P(0)\plus{}P(3x)\plus{}P(\minus{}3x)\equal{}0$ for all x. x=0 gives $ P(0)\equal{}0$, hence P is an odd function: $ P(x)\equal{}\minus{}P(\minus{}x)$
choose a=6x, b=3x, c=-2x, and you get $ P(3x)\plus{}P(5x)\minus{}P(8x)\equal{}0$
If you compare coefficients, you get $ 3^{i}\plus{}5^{i}\equal{}8^{i}$ or $ a_{i}\equal{}0$. if you compare that leading coefficient, we must have $ 3^{n}\plus{}5^{n}\equal{}8^{n}$ (unless P is a constant - 0). One solution is n=1, so degP=1 and 0 is a root so $ P(x)\equal{}ax$ (easy to verify that it works for all a)
for n>1, there's no solution: $ (\frac{3}{8})^n\plus{}(\frac{5}{8})^n\equal{}1 \implies$for n>1:$ (\frac{3}{8})^2\plus{}(\frac{5}{8})^2> 1$, a contradiction

$ P(x)\equal{}ax, a \in R$ are the only solutions

[quote="mszew"]
then $ P(x) \equal{} ax^{2k \plus{} 1}$[/quote]
I think your reasoning doesnt require it...for example if $ p(x)\equal{}x^3\plus{}x$ then  $ P$ dont have nonzero real root...
but if the condition of the problem be still true for complex number $ a,b,c$ you can correct it...


Όλα τα μαθηματικά προβλήματα από τη στήλη Προκλήσεις στα Μαθηματικά και τη Φυσική της ελληνικής έκδοσης του περιοδικού Quantum (44 τεύχη, 1994-2001), δίχως τις λύσεις τους.

[url]http://rapidshare.com/files/54242841/quantum.djvu.html[/url] (2.5mb)

Νομίζω ότι είναι μια καλή συλλογή προβλημάτων.

Καλή διασκέδαση!

Μας στείλατε και το αρχείο στο e-mail του stfu. Λόγω υποχρεώσεων , δε προλαβαίνω να το καταχωρήσω αυτές τις μέρες  :( . Αν θέλετε , ρίξτε μια ματιά στο site στο section των διαγωνισμών μετά τις 11 Σεπτεμβρίου !

Ευχαριστούμε πάντως για τη συλλογή των ασκήσεων,
Στέλιος


calculus
integration
trigonometry
calculus computations

A solid is formed by rotating the curve y = (x^(3/2))/((1 + x^2)^2) between x = 0 and x = 1 through 360 degrees about the x-axis. Find the volume of this solid.

The volume is given by the integral
\[\pi \int_{0}^{1}\frac{x^{3}}{(1+x^{2})^{4}}\ dx\]
You'll have to use several integrations by parts to deal with the [i]four[/i] factors of $1+x^{2}$ in the denominator...

Actually there is a much easier way to do it...

Yes, just let $t=1+x^{2}$ or $x=\tan \theta$.

Both those substitutions would make the integral more complicated.

Actually, $t=x^{2}+1$ does work very well!  :)

(A factor of $x$ gets removed, leaving $x^{2}$, or $t-1$.)

oops...yeah, t = 1 + x^2 is the way i would do it. I was just being silly.


function
algorithm
number theory unsolved
number theory

Prove that for all integers $ a > 1$ and $ b > 1$ there exists a function $ f$ from the positive integers to the positive integers such that $ f(a\cdot f(n)) \equal{} b\cdot n$ for all $ n$ positive integer.

[hide="This seems sketchy"]Letting $ n \equal{} af(n)$ shows that $ f(abn) \equal{} ab f(n)$.

Set $ f(1) \equal{} 1$ arbitrarily, implying that $ f(a) \equal{} b$ (plus other stuff about multiples of $ ab$ which we ignore). For each integer $ n$ for which $ f$ is not yet defined, set it equal to the least positive integer that is
- not in the range of $ f$,
- not a multiple of $ a$ or $ b$,
- at least $ n$.
Then $ f(af(n))$, which has not yet been defined since $ af(n) > f(n) \geq n$ and $ f(n)$ is not already in the range of $ f$, is $ bn$. Note that $ f$ is defined for all multiples of $ ab$ before this algorithm reaches them by the above identity.

Now we have to show the equation is satisfied for all $ n$. If $ n$ is not a multiple of $ a$, then $ f$ was manually made to work in the algorithm. If it is $ ak$ for some $ k$, then either
1. $ k$ is in the range of $ f$, i.e. $ k \equal{} f(m)$ for some $ m$. Then $ f(ak) \equal{} bm$, and $ f(a f(ak)) \equal{} f(abm) \equal{} ab f(m) \equal{} abk$ as desired.
2. $ k$ is not in the range of $ f$. Then $ ak$ was manually worked out in the algorithm.[/hide]


Anyone know how I could count the words in a LaTeX document without copying the text into word and taking the word count there?

I'm not sure how word count is defined for mathematics eg how many words would you say $ \textstyle \int^{\infty}_{\infty}e^{x^2}dx\equal{}\sqrt{\pi}$ is? But if you really need something then the best advice is [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=wordcount]How many words have you written?[/url].
If you are using Winedt it already has this facility, but it wouldn't be difficult to add [i]latexcount[/i], once installed, to TexnicCenter's menu.

I'm actually using WinEdt 5 right now, how do I use latexcount with WinEdt?

I haven't got a copy of WinEdt so I am quoting elsewhere and hope they are correct:

[i]On the "Document" menu, Word Count. [/i]

[i]WinEdt has word-counting. It can be set to do whatever you want.
The default is to count words sans LaTeX markups, in the usual
manner. If you look under the headinf Document, you will see
the logo of a page with a bent corner, containing an italic "i"
and the phrase "Word count".[/i]


let $ x;y;z$ be positive real numbers such that $ x^2\plus{}y^2\plus{}z^2\equal{}3$,prove that
$ \frac{x}{{\sqrt {{y^2} \plus{} 3}  \plus{} x}} \plus{} \frac{y}{{\sqrt {{z^2} \plus{} 3}  \plus{} y}} \plus{} \frac{z}{{\sqrt {{x^2} \plus{} 3}  \plus{} z}} \le 1$
 :)  :)

[quote="Toan_VN_LC"]let $ x;y;z$ be positive real numbers such that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3$,prove that
$ \frac {x}{{\sqrt {{y^2} \plus{} 3} \plus{} x}} \plus{} \frac {y}{{\sqrt {{z^2} \plus{} 3} \plus{} y}} \plus{} \frac {z}{{\sqrt {{x^2} \plus{} 3} \plus{} z}} \le 1$
 :)  :)[/quote]
Under the given hypothesis, we have $ 3 \ge x \plus{} y \plus{} z.$ Therefore
$ \sum \frac {x}{x \plus{} \sqrt {y^2 \plus{} 3}} \le \sum \frac {x}{x \plus{} \frac {y \plus{} 3}{2}} \le \sum \frac {x}{x \plus{} \frac {y \plus{} (x \plus{} y \plus{} z)}{2}} \equal{} 2\sum \frac {x}{3x \plus{} 2y \plus{} z}.$
So, it suffices to prove that
$ \sum \frac {x}{3x \plus{} 2y \plus{} z} \le \frac {1}{2},$ or $ \sum \left( 1 \minus{} \frac {3x}{3x \plus{} 2y \plus{} z}\right) \ge \frac {3}{2},$ that is $ \sum \frac {2y \plus{} z}{3x \plus{} 2y \plus{} z} \ge \frac {3}{2}.$
Now, the Cauchy Schwarz Inequality implies that
$ LHS \ge \frac {\left(\sum (2y \plus{} z)\right)^2}{\sum (2y \plus{} z)(3x \plus{} 2y \plus{} z)} \equal{} \frac {9(x \plus{} y \plus{} z)^2}{5\sum x^2 \plus{} 13 \sum xy},$
and so, we only need to prove
$ 6(x \plus{} y \plus{} z)^2 \ge 5\sum x^2 \plus{} 13\sum xy,$ which is equivalent to $ x^2 \plus{} y^2 \plus{} z^2 \ge xy \plus{} yz \plus{} zx$ (trivial).
This completes our proof. :)


[quote="DaMasterbomb"]This is my first time posting a new topic on a forum on AoPS, so please feel free to give constructive criticism. (and tell me if I posted on the right forum). Any $LATEX$ tips would be welcome as well.

Given: $r,d,s > 0$ such that $r>s$ and  $r-s<d<r+s$
Prove: $s^{2} - ((d^{2}-r^{2}+s^{2}) / 2d)^{2}>0$[/quote][hide="Click me to reveal YAY"]We are given $d<r+s$.

Next, $r-s<d\Rightarrow r<d+s$.

Lastly, $s<r<r+d$.

Therefore, $r,d,s > 0$ forms a triangle with sides $r,d,s$. Now, since the triangle exists, the angle between sides $d$ and $s$ is not equal to zero; let this angle be $\theta>0$. By definition of cosine, $|\cos\theta|\le1$ or $\cos^2\theta\le1$. But $\theta>0$ ($\theta\ne0$), so $\cos^2\theta\ne1$ and we get new inequality $\cos^2\theta<1$. Using the law of cosine with respect to angle $\theta$ we get

$\cos^2\theta=\left(\frac{d^2+s^2-r^2}{2ds}\right)^2<1$

QED[/hide]

$((d^{2}-r^{2}+s^{2}) / 2d)^{2}$ can be better written as $(\frac{d^{2}-r^{2}+s^{2} }{2d})^{2}$ using \frac{numerator}{denominator}


I was thinking about applying to Sumac this summer and was wondering if there is anybody here that has attended the program in the past. How hard is it to get in to the program? What was the experience like?

Yes I also intend to go there, but just with full scholarship since I can't pay all of that money for 4 weeks(too shorter than any other math camps)! Anyway, how do you guys do the admission test? I think I got all of them :lol:

I sent in an application, as well. I think I answered every question correctly;however, the proof that I provided for the last question was not very elegant.

When will we find out if we get in?

Well, I also got all the answer but I couldn't prove the last question's answer is 7. I just use computer to rum all of combinations and it gives me 7. I'm just on the track to prove that 6 can't be the answer.Did anybody get completely this question?It's about covering space and I don't have any prior knowledge abput it. Also what you guys got on question 6D?I used the number theory theorem to prove that n have the even number of divisors formed 4k+3 and sth more. wELL, THE RESULT IS TOO LONG!

First of all, I'm not sure we should be discussing the problems because I've read that they are reused each year.

On the last question I proved that it was not possible with 6 tickets and then gave an example of winning with 7 tickets, which I believe is adequate. I believe problem 6D was about 'lattice squares', and I'm not sure exactly what trouble you had with it. My proof was fairly short.

Oh that's not what I meant! Well, you got 10d :o . Nice job dude! I don't have any trouble proving 6d. Just concern whether I got sufficient condition or not(since I just prove one lemma and that's all for 6d) :)

I'm fairly sure that there are only 8 questions, but I'll assume that was just a typo.

Can anyone comment on how long it will be before we know if we got in or not?

I believe that at the end of April you will know the result of application!


[quote="Leonhard Euler"]Let $ A,B,C,D$ be four point. line $ AB,CD$ meet at $ P$.Denote midpoint of $ AC,BD$ as $ M,N$ and orthocenter of triangle $ PAD,PBC$ as $ H_{1},H_{2}$. Prove theat $ H_{1}H_{2}$ is perpendicular to $ MN$.[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=109505 (and all links from that topic).

  Darij


Find the number of real solutions to the equation $ 567^x\plus{}678^x\equal{}789^x$.

Divide both sides by $ 789^x$

$ a^x \plus{} b^x \equal{} 1$ Now, since both $ 0 < a,b < 1$ we have $ a^x$ and $ b^x$ are both strictly decreasing functions. Hence, $ a^x \plus{} b^x$ is a strictly decreasing function, which means that $ a^x \plus{} b^x \equal{} 1$ can have at most 1 solution. Since $ x \equal{} 0$ gives $ a^x \plus{} b^x \equal{} 2$ and $ x \equal{} 1000$ gives $ a^2 \plus{} b^x < 1$, then since $ a^x \plus{} b^x$ is continuous, then we must have a value of x for which $ a^x \plus{} b^x \equal{} 1$ by the intermediate value theorem. Hence, there is a solution, and since we already said that it cannot have more than 1 solution, we have 1 solution.


Can you show that given a polynomial $ f(x) \equal{} x^3 \plus{} ax^2 \plus{} bx \plus{} c$, with $ b > 0$ and $ ac > 0$ and 3 real roots, that

$ \frac {ab}{c}\geq{9}$

What about this polynomial? All real roots? Any reals $ a,b,c$ can make a polynomial like the one you described.

Yes I knew I forgot to include something; thank you MellowMelon. it's been edited.


Find the minimum value of:
$ P\equal{}\frac{5\minus{}8\sin^2\frac{x}{2}}{\sin2x}$
for $ x\in (0,\frac{\pi}{2})$

[hide]
Sine and cosine are positive in the interval.
Expand the sin(something other than x), and apply AM-GM:
$ \frac {1 \plus{} 4cosx}{4cosxsinx} \geq \sqrt {csc^2xsecx}$
We want the equality case because it is the minimum.
$ \frac {1 \plus{} 4cosx}{4cosxsinx} \equal{} cscx \sqrt {secx}$
$ \frac {1 \plus{} 4cosx}{4cosx} \equal{} \sqrt {secx}$
$ \frac14secx \plus{} 1 \equal{} \sqrt {secx}$
$ (\frac14secx \plus{} 1)^2 \equal{} secx$
$ \frac {1}{16}sec^2x \plus{} \frac12secx \plus{} 1 \equal{} secx$
$ \frac {1}{16}sec^2x \minus{} \frac12secx \plus{} 1 \equal{} 0$
$ (\frac14sec^2x \minus{} 1)^2 \equal{} 0$
$ \frac14sec^2x \minus{} 1 \equal{} 0$
$ cos^2x \equal{} \frac14$
$ cosx \equal{} \frac12$
$ x \equal{} \frac {\pi}{3}$
When we plug it in, we get:
$ \frac {5 \minus{} 2}{\frac {\sqrt3}{2}} \equal{} \frac {6}{\sqrt3} \equal{} 2\sqrt3$.
[/hide]
Isn't this a bit easy for preolymp?

[quote="serialk11r"][quote="Kyoshiro"][quote="serialk11r"][hide]
$ \frac {1 \plus{} 4cosx}{4cosxsinx} \geq \sqrt {csc^2xsecx}$
We want the equality case because it is the minimum.
$ \frac {1 \plus{} 4cosx}{4cosxsinx} \equal{} cscx \sqrt {secx}$
[/hide][/quote]
I think your solution is wrong from this step.
Both sides of your inequality are variables, so you can't imply that the LHS is minimal when there is equality. :huh:[/quote]
How about if I replace all the equalities with $ \geq$ until the part where cosx is isolated? All "variables" are positive so there is no worry about changing direction of the inequality.
In addition I don't think it matters here because we can describe either side as a function of x, and so it implies that the function is always less than the other. The conditions for AM-GM are only that all the numbers you are adding/multiplying/whatever are homogenous in the equation.[/quote]
If you replace all the equalities in your solution with $ \ge$, we get
$ (\frac{1}{4}\sec^2x\minus{}1)^2\ge 0$
which is obvious true and we can't imply that $ \frac{1}{4}\sec^2x\minus{}1\ge 0$

And about your mistake, let's consider a simple problem:
Find the minimum of $ P\equal{}x^2\plus{}1$ for $ x\in\mathbb{R}$
We have a well-known inequality: $ x^2\plus{}1\ge 2x$
Is P minimal when there is equality? :huh:
Of course not! When there is equality, $ P\equal{}2$ but it's clearly that $ \min_{x\in\mathbb{R}}P\equal{}1$ :lol:


inequalities
geometry
geometric transformation
reflection
geometry proposed

Let $ABC$ be a triangle and $P$ be any point. Call $A', B', C'$ be the projections of $P$ on $BC, CA, AB$. Prove that

$2(AA'+BB'+CC') \leq (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(aPA+bPB+cPC)$

The equality happens if and only if $P$ is the orthocenter of $ABC$.

Here is really nice inequality I found a sequence inequality related it
Let $ABC$ be a triangle and $P$ be any point. Call $A', B', C'$ be the projections of $P$ on $BC, CA, AB$. Prove that
$2(PA+PB+PC+PA'+PB'+PC')\ge (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(aPA+bPB+cPC)\ge 2(AA'+BB'+CC') \ge(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})4S$
The equality happens if and only if $P$ be the orthocenter of $ABC$.
Now with the main problem let $P'$ is reflection of $P$ through $BC$ apply Ptoleme's inequality with triangle $ABC$ and $P'$ we have
$bP'B+cP'C\ge aP'A$ but $P'B=PB,P'C=PC$ thus we will have
$aPA+bPB+cPC\ge a(AP'+AP)\ge 2aAA'$ the last inequality is well know by $A'$ is midpoint of $PP'$ and now we have $\frac{aPA+bPB+cPC}{a}\ge 2AA'$ similarly for $B,C$ and add them we are done.
And remark equality holds iff triangle is [u]acute[/u] and $P$ is orthocenter.
 :)

Yup.  :wink: That is also my solution. Very neautifully done, gemath.

Also, your leftmost inequality isn't true. Take $O$ as an example.


[quote="tdl"]Consider the sequence $ a_1 \equal{} 1$ and for $ n\geq 1$ we have $ a_{n \plus{} 1} \equal{} 2a_n \plus{} \sqrt {3a_n^2 \minus{} 3}$. Prove that $ a_{3n \plus{} 1} \equal{} a_{n \plus{} 1}(a_{n \plus{} 1}^2 \plus{} 1)$.[/quote]

That's wrong :
$ a_1\equal{}1$, $ a_2\equal{}2$, $ a_3\equal{}7$ and $ a_4\equal{}26$ and the property $ a_{3n \plus{} 1} \equal{} a_{n \plus{} 1}(a_{n \plus{} 1}^2 \plus{} 1)$ is not true for $ n\equal{}1$

Sorry! I have repaired it! It my guess! May be it wrong!

[quote="tdl"]Consider the sequence $ a_1 \equal{} 1$ and for $ n\geq 1$ we have $ a_{n \plus{} 1} \equal{} 2a_n \plus{} \sqrt {3a_n^2 \minus{} 3}$. Prove or disprove that $ a_{3n \plus{} 1} \equal{} a_{n \plus{} 1}(2a_{2n \plus{} 1} \minus{} 1)$.[/quote]

Writing $ a_n\equal{}\cosh(x)$, we get $ a_{n\plus{}1}\equal{}2\cosh(x)\plus{}\sqrt{3}\sinh(x)\equal{}\cosh(x\plus{}u)$ With $ \cosh(u)\equal{}2$ and $ \sinh(u)\equal{}\sqrt{3}$

And so $ a_n\equal{}\cosh((n\minus{}1)u)\equal{}\frac{(2\plus{}\sqrt{3})^{n\minus{}1}\plus{}(2\minus{}\sqrt{3})^{n\minus{}1}}{2}$

So $ 2a_{2n\plus{}1}\minus{}1\equal{} (2\plus{}\sqrt{3})^{2n}\plus{}(2\minus{}\sqrt{3})^{2n}\minus{}1$

So $ a_{n \plus{} 1}(2a_{2n \plus{} 1} \minus{} 1)\equal{} \frac{(2\plus{}\sqrt{3})^{n}\plus{}(2\minus{}\sqrt{3})^{n}}{2}$ $ ((2\plus{}\sqrt{3})^{2n}\plus{}(2\minus{}\sqrt{3})^{2n}\minus{}1)$

So $ a_{n \plus{} 1}(2a_{2n \plus{} 1} \minus{} 1)\equal{}$ $ \frac{ ((2\plus{}\sqrt{3})^{3n}\plus{}(2\minus{}\sqrt{3})^{n}\minus{} (2\plus{}\sqrt{3})^{n})\plus{}((2\plus{}\sqrt{3})^{n}\plus{}(2\minus{}\sqrt{3})^{3n}\minus{}(2\minus{}\sqrt{3})^{n})}{2}$

So $ a_{n \plus{} 1}(2a_{2n \plus{} 1} \minus{} 1)\equal{}$ $ \frac{ ((2\plus{}\sqrt{3})^{3n}\plus{}(2\minus{}\sqrt{3})^{3n}}{2}$ $ \equal{}a_{3n\plus{}1}$

Q.E.D.


A case of Xu's theorem
Prove that an $ uniquely$ $ 3 \minus{} colorable$ $ graph$ with n vertices $ (n \geq 3)$ has at least $ 2n \minus{} 3$ edges

The question should say "uniquely 3-colorable," not "uniquely $ k$-colorable."

Yes, i have edited it. So can u post the solution?


Find a satisfy the system equation folow has roots:
$ \left\{ \begin{array}{l}x^3  \minus{} ay^3  \equal{} \frac{1}{2}(a \plus{} 1)^2  \\ x^3  \plus{} {\rm{ax}}^2 y \plus{} xy^2  \equal{} 1 \\ x \plus{} y \equal{} 0 \\ \end{array} \right.$

[hide="just a try"]

From the third equation, we get $ x\equal{}\minus{}y$
Substitute this to the other equations, we get
$ x^3\plus{}ax^3\equal{}\frac{1}{2}(a\plus{}1)^2$
$ x^3\equal{}\frac{1}{2}(a\plus{}1)$ (1)

And

$ x^3\minus{}ax^3\plus{}x^3\equal{}1$
$ x^3\equal{}\frac{1}{2\minus{}a}$ (2)

(1)=(2)

$ (a\plus{}1)(2\minus{}a)\equal{}2$
$ a\minus{}a^2\plus{}2\equal{}2$
$ a(a\minus{}1)\equal{}0$

We get $ a\equal{}0$ or $ a\equal{}1$

[/hide]


inequalities
function
integration
calculus
calculus computations

(a) Prove that $ |A \minus{} \gamma B|^2 > 0 \implies |A \cdot B| \le |A\parallel{}B|$

where $ \gamma \equal{} \frac{A \cdot B}{B \cdot B}$.

(b) Express the inequality using both

$ A \cdot B \equal{} \sum_{n\equal{}1}^{\infty} a_n b_n \equal{} \sum_{n\equal{}1}^{\infty} a_n c_n \frac{b_n}{c_n}.$

(c) Generalize the Schwarz's inequality to functions.

This is a theoretical question, that may be found in many books of undergraduate math.  Anyway, here is the answer:

(a) Simply write that
\[ |A\minus{}\gamma B|^2\equal{}|A|^2\minus{}\gamma A\cdot B\minus{}\gamma^* B\cdot A\plus{}|\gamma|^2|B|^2\ge0,\]
where the asterisk denotes complex conjugation (just you wait, you'll see soon why), where
\[ \gamma\equal{}\frac{B\cdot A}{B\cdot B},\qquad \gamma^*\equal{}\frac{A\cdot B}{B\cdot B},\]
(ditto), and where equality is achieved obviously iff $ A\equal{}\gamma B$.  Substitute $ \gamma$ and $ \gamma^*$ and multiply throughout by $ |B|^2\equal{}B\cdot B$, find
\[ |A|^2|B|^2\ge|B\dot A|^2,\]
take square root and done!

(b) For real numbers $ a_i,b_i,c_i$ the distinction between real and complex $ \gamma$, and the order of $ A\cdot B$ and $ B\cdot A$ does not matter, you just find that
\[ \sqrt{a_1^2\plus{}a_2^2\plus{}\dots\plus{}a_n^2}\sqrt{b_1^2\plus{}b_2^2\plus{}\dots\plus{}b_n^2}\ge|a_1b_1\plus{}a_2b_2\plus{}\dots\plus{}a_nb_n|,\]
with equality iff $ a_i\equal{}\kappa b_i$, with $ \kappa$ the same real constant for all $ i\equal{}1,2,\dots,n$.  Similarly,
\[ \sqrt{a_1^2c_1^2\plus{}a_2^2c_2^2\plus{}\dots\plus{}a_n^2c_n^2}\sqrt{\frac{b_1^2}{c_1^2}\plus{}\frac{b_2^2}{c_2^2}\plus{}\dots\plus{}\frac{b_n^2}{c_n^2}}\ge|a_1b_1\plus{}a_2b_2\plus{}\dots\plus{}a_nb_n|,\]
with equality iff $ a_ic_i^2\equal{}\kappa b_i$, again with $ \kappa$ the same real constant for all $ i\equal{}1,2,\dots,n$.

(c) For COMPLEX functions $ f(x)$ and $ g(x)$, define a "function dot product" as
\[ f\cdot g\equal{}\int_a^bf(x)g^*(x)dx,\qquad,g\cdot f\equal{}\int_a^bg(x)f^*(x)dx\equal{}(f\cdot g)^*,\qquad f\cdot f\equal{}\int_a^b|f(x)|^2dx,\]
where $ (a,b)$ is an interval where $ f,g$ are absolutely integrable.  It is not hard to verify that this operation indeed satisfies all criteria to be a scalar product of complex functions of real variable, yielding trivially as per (a) the conclusion
\[ \left|\int_a^bf(x)g^*(x)dx\right|\equal{}\left|\int_a^bg(x)f^*(x)dx\right|\le\sqrt{\int_a^b|f(x)|^2dx}\sqrt{\int_a^b|g(x)|^2dx},\]
with equality iff a constant $ \lambda$ (in general complex) exists such that $ f(x)\equal{}\lambda g(x)$.


Where can I get the package called graphicx to draw graphs etc? Please help

It is already provided as part of MiKTeX - all you need do is to add \usepackage{graphicx} to the beginning of your document. But graphicx is for adding graphics ie images to your documents, it has nothing to do with drawing graphs.


x ; sıfıra sağdan yaklaşırken (tamdeğer x)üssü(tamdeğer x)=?

bu limit yoktur.çünkü, fonksiyonu parçalı olarak yazmak istediğimizde 0<=x<1 için f tanımlı değildir.limiti bulacağımız nokta fonksiyoun tanım kümesinin bir yığılma noktası olmalıdır. herbir 0<=x<1 değeri için f(x)=[|x|]^[|x|] tanmzısdır ve bu noktalarda limit yoktur.


let a and b be positive integers such that a/b² ,b²/a^3,a^3/b^4,b^4/b^5.....
prove that a=b

Have seen this several times...

[hide="Solution"] Let $r = \frac{a}{b}$.  Then

$a^{k}| b^{k+1}\implies a_{k}\le b^{k+1}\implies r^{k}\le b$
$b_{k}| a^{k+1}\implies b_{k}\le a^{k+1}\implies \frac{1}{r^{k}}\le a$

Either $r^{k}$ or $\frac{1}{r^{k}}$ tends to infinity unless $r = 1$. [/hide]


Can the plane be coloured in 2000 colours so that any nondegenerate circle contains points of all 2000 colors?

Obviously yes, or did I miss something?

Pierre.


Given triangle $ ABC$ and its incircle $ (I). (I)$ touches $ BC, CA, AB$ at $ A_1,B_1,C_1$, respectively. Let $ A_2,B_2,C_2$ be the points on $ IA_1,IB_1,IC_1$ such that $ \frac{\overline{IA_2}}{\overline{IA_1}}\equal{}\frac{\overline{IB_2}}{\overline{IB_1}}\equal{}\frac{\overline{IC_2}}{\overline{IC_1}}$. Prove that 3 circles $ (AIA_2),(BIB_2), (CIC_2)$ concur at the point $ Q\neq I$.

Denote $ \frac {\overline{IA_2}}{\overline{IA_1}} \equal{} \frac {\overline{IB_2}}{\overline{IB_1}} \equal{} \frac {\overline{IC_2}}{\overline{IC_1}} \equal{} k \ \Longrightarrow \ \overline{IA_2} \equal{} \overline{IB_2} \equal{} \overline{IC_2} \equal{} kr.$

Inversion with respect to the incircle $ (I)$ takes the vertices $ A,B,C$ into the midpoints $ M,N,L$ of $ \overline{B_1C_1},$ $\overline{C_1A_1},$ $\overline{A_1B_1}.$ If $ A_3,B_3,C_3$ are the inverse points of $ A_2,B_2,C_2,$ it follows that circles $ \odot(AIA_1),\odot(AIB_1)$ and $ \odot(AIC_1)$ are transformed into the lines  $ MA_3,NB_3,LC_3.$ Moreover, we get  $ \overline{IA_3} \cdot \overline{IA_2} \equal{} \overline{IA_3} \cdot kr \equal{} r^2$ $ \Longrightarrow$ $\overline{IA_3} \equal{} \frac {_r}{^k}.$ By similar reasoning we have that  $ \triangle A_3B_3C_3$ and $ \triangle A_1B_1C_1$ are centrally similar with similarity coefficient $ \frac {_1}{^k},$ which also implies that the medial triangle $ \triangle MNL$ of $ \triangle A_1B_1C_1$ and $ \triangle A_3B_3C_3$ are centrally similar with similarity coefficient $ \minus{} \frac {_2}{^k}$ $ \Longrightarrow$ $MA_3,NB_3,LC_3$ concur at their similarity center. Hence the primitive figures, namely, $ \odot(AIA_1) , \odot(AIB_1)$ and $ \odot(AIC_1)$ concur at another point $ Q$ lying on the diacentral line $IO$ of $\triangle ABC.$


trigonometry
inequalities
inequalities unsolved

[quote="magicgeometry"]Prove that for all acute triangle $ ABC$ we have
$ \left(\prod\sin\frac {A}{2}\right)\left(\sum\tan A\right)\geq\frac {3\sqrt3}{8}$[/quote]

USe $ \left(\prod\sin\frac {A}{2}\right)\equal{}\frac 12 \minus{} \frac 12 * \sum \sin ^2 \frac A2$
and then jensens as tan is concave in 0 to pi/2 and sin^2 is convex in that interval.

[quote="hell_ever"][quote="magicgeometry"]Prove that for all acute triangle $ ABC$ we have
$ \left(\prod\sin\frac {A}{2}\right)\left(\sum\tan A\right)\geq\frac {3\sqrt3}{8}$[/quote]

USe $ \left(\prod\sin\frac {A}{2}\right) \equal{} \frac 12 \minus{} \frac 12 * \sum \sin ^2 \frac A2$
and then jensens as tan is concave in 0 to pi/2 and sin^2 is convex in that interval.[/quote]

how?  :blush:
I got $ \sum\tan A\geq \left(\sum\tan A\right)\cdot\left(\sum\sin^2\frac{A}{2}\right)\plus{}\frac{3\sqrt3}{4}$ and how do we continue?

[quote="magicgeometry"]Prove that for all acute triangle $ ABC$ we have
$ \left(\prod\sin\frac {A}{2}\right)\left(\sum\tan A\right)\geq\frac {3\sqrt3}{8}$[/quote]
$ \sum_{cyc}\tan A = \prod_{cyc}\tan A = \frac {64S^3}{(a^2 + b^2 - c^2)(a^2 + c^2 - b^2)(b^2 + c^2 - a^2)}.$
$ \prod_{cyc}\sin{\frac {A}{2} = \prod_{cyc}\sqrt {\frac {(a + b - c)(a + c - b)}{4bc}} = \frac {(a + b - c)(a + c - b)(b + c - a)}{8abc}.}$
Thus, we need to prove that
$ 64^3S^6\prod_{cyc}(a + b - c)^2\geq27a^2b^2c^2\prod_{cyc}(a^2 + b^2 - c^2)^2.$
But $ \prod_{cyc}(a + b - c)^2\geq\prod_{cyc}(a^2 + b^2 - c^2)$ (a very known inequality ).
Hence, it remains to prove that $ \left(\sum_{cyc}(2a^2b^2 - a^4)\right)^3\geq27a^2b^2c^2\prod_{cyc}(a^2 + b^2 - c^2).$
Let $ a^2 + b^2 - c^2 = z,$ $ a^2 + c^2 - b^2 = y$ and $ b^2 + c^2 - a^2 = x.$
Then we need to prove that $ 8(xy + xz + yz)^3\geq27xyz(x + y)(x + z)(y + z),$ which true by Muirhead.


Μία για προθέρμανση :D

Αν είναι :$ x,y,z$ πραγματικοί, με  $ z\equal{}x\plus{}y$ και  $ \frac{x^2}{4} \plus{}\frac{y^2}{5}\plus{} \frac{z^2}{25} \equal{} 1$ , να βρείτε τη μέγιστη κ την ελάχιστη τιμή της παράστασης $ A\equal{}x^2\plus{}y^2\plus{}z^2$

Έχουμε
$ x^2 \plus{} y^2 \plus{} (x \plus{} y)^2 \equal{} A\Leftrightarrow x^2 \plus{} xy \plus{} y^2 \equal{} \frac {A}{2}$

and $ \frac {x^2}{4} \plus{} \frac {y^2}{5} \plus{} \frac {(x \plus{} y)^2}{25} \equal{} 1\Leftrightarrow29x^2 \plus{} 8xy \plus{} 24y^2 \equal{} 100.$
άρα
 $ (29A \minus{} 200)x^2 \plus{} (8A \minus{} 200)xy \plus{} (24A \minus{} 200)y^2 \equal{} 0.$
άρα
 $ (4A \minus{} 100)^2 \minus{} (29A \minus{} 200)(24A \minus{} 200)\geq0.$
και επομένως
$ \frac {75}{17}\leq A\leq10.$

Έτσι που το είδα το ελάχιστο μου φάνηκε λάθος αλλά προφανώς εννοείς $ \frac {75}{17}$. Μπράβο ρε συ πάντως... Ωραία λύση :lol:

Είχα καιρό να γράψω κλάσματα όπως στο δημοτικό και είπα να γράψω ένα  :P
(Τώρα το fixarisa για να είναι κατανοητό) . Η λύση σου ποια είναι ?

Είναι από αύτες που λέγαμε ότι αντενδείκνυνται :rotfl: (Ανάλυση δλδ.... :lol: )


Gauss
function
floor function
algebra
polynomial
number theory unsolved
number theory

Does there exist irrational numbers $a,b>1$ such that $[a^n]$differs$[b^m]$for all pairs of natural number $(m,n)$

Let $a=e,b=e^q,q=\frac{\sqrt 5 +1}{2}$. If $[a^k]=[b^m]$ then $\frac mk =\frac{P_l}{Q_l}\Longrightarrow |b^m-a^k|>a^k|a^{\frac{1}{k\sqrt 5 }}-1|>1$.

Yes, such irrationals do exist, for instance (note that they are algebraic numbers, when Rust, gave example of transcendental numbers)
\begin{eqnarray*} a = \left( \sqrt{\frac{17 \sqrt{5} + 38}{4 \sqrt{5}}} \cdot \sqrt{15} + \frac{5 \sqrt{5}}{2} + 6 \right)^4 & \mbox{and} & b = \left( \frac{3 + \sqrt{5}}{2} \right)^4 \end{eqnarray*}
Let $\theta_0, \theta_1, \theta_2, \theta_3$ be the four roots, of
\begin{eqnarray*} x^4 - 24 x^3 + 26 x^2 - 9 x + 1 & = & P(x) \end{eqnarray*}
We observe that $\theta_0^4 = a$ and $0 < \theta_1, \theta_2, \theta_3 < 1$. One can prove that for $n > 1$,
\begin{eqnarray} \lfloor \theta_0^n \rfloor & = & \theta_0^n + \theta_1^n + \theta_2^n + \theta_3^n - 1 \end{eqnarray}
Similarly let $\rho_0$ and $\rho_1$ be roots of
\begin{eqnarray*} x^2 - 3 x + 1 & = & Q(x) \end{eqnarray*}
Then $\rho_0^4 = b$ and $0 < \rho_1 < 1.$Similarly we prove that for $n \geq 0$,
\begin{eqnarray} \lfloor \rho_0^n \rfloor & = & \rho_0^n + \rho_1^n - 1 \end{eqnarray}
Then using some algebra (field extensions), by relaying on $( 1 )$ and $(2)$, we prove that
\begin{eqnarray*} \lfloor \rho_0^{4 n} \rfloor & \equiv & 1 ( \mbox{ mod } 3 )\\ \lfloor \theta_0^{4 n} \rfloor & \equiv & 0 ( \mbox{ mod } 3 ) \end{eqnarray*}
This clearly gives the desired numbers, $a = \theta_0^4$ and $b = \rho_0^4$.
Note: If you ever want to check these answers with a computer algebra program, be sure to compute, the exact value of $( 1 )$ and $( 2 )$. Otherwise using approximations like float() will lead to erroneous results.

If the proof are of any interest I can give them ...

very nice solution.but I think the number $a$ can be created more easily.

Thank you ! At first I tried investigating $\varphi$, since it has the proprety that
\begin{eqnarray*} \lfloor \varphi^n \rfloor & = & \ell_n - \frac{( - 1 )^n + 1}{2} \end{eqnarray*}
for all positive integers $n$, where $\ell_n$ is the $n$-th Lucas number.
From this we obtain
\begin{eqnarray} \lfloor \varphi^{3 n} \rfloor & = & \ell_{3 n} - \frac{( - 1 )^n + 1}{2} \nonumber\\ \lfloor \varphi^{3 n} \rfloor & = & \left( \sum_{k = 0}^n \left(\begin{array}{c} n\\ k \end{array}\right) 2^k \ell_k \right) - \frac{( - 1 )^n + 1}{2} \end{eqnarray}
By a known reccurence relation of Lucas numbers (see Mathworld).
Now for all positive integer $k$ we have $2^k \cdot \ell_k \equiv 2 ( \mbox{mod } 10 )$. Therefore $( 1 )$ considered $\mbox{mod } 10$ gives
\begin{eqnarray*} \lfloor \varphi^{3 n} \rfloor & \equiv & 2 \sum^n_{k = 0} \left(\begin{array}{c} n\\ k \end{array}\right) - \frac{( - 1 )^n + 1}{2}\\ \lfloor \varphi^{3 n} \rfloor & \equiv & 2^{n + 1} - \frac{( - 1 )^n + 1}{2} \end{eqnarray*}
By simple enumeration we conclude that $\lfloor \varphi^{3 n} \rfloor \in \left\{ 1, 4, 6, 7 \right\} ( \mbox{mod } 10 )$.
Hence it suffice to find an irrational $b$ such that $\lfloor b^n \rfloor \in \left\{ 2, 3, 5, 8, 9 \right\} ( \mbox{mod } 10 )$ for all $n \in \mathbb{N}$, to solve your problem.

Actually each of $\varphi, \frac{3 + \sqrt[]{5}}{2}$ or the $\sqrt[4]{a}$ considered in the post above is a Pisot number.
A Pisot number is an algebraic number (a number which is a root of a polynomial with integer coefficents) with one additional proprety. Namely, if $P ( x )$ is the minimal polynomial of $\alpha$ (a normed polynomial with integer coefficents, of smallest possible degree, such that $P ( \alpha ) = 0$), then any other root $\rho$ of $P$ different from $\alpha$, has modulus $< 1$.
The second important proprety is that if $\rho_0, \ldots, \rho_n$ are all the roots of $P$ (this actually holds for any polynomial with integer coefficents), then
\begin{eqnarray} \rho_0^m + \rho_1^m + \ldots + \rho_n^m & \in & \tmmathbf{Z} \end{eqnarray}
for all positive integers $m$. And since there is only one $| \rho_i | > 1$ (we assume there is no root with $| \rho | = 1 )$ it follows that for
sufficently large $n$, we have
\begin{eqnarray*} \rho_0^m + \ldots + \rho_n^m - \max \left( \rho_i \right)^n & \longrightarrow & 0 \end{eqnarray*}
This allows us, to obtain some nice functional relations for $\lfloor \max \left( \rho_i \right)^n \rfloor .$
For example we can prove that $\lfloor \varphi^n \rfloor = \ell_n - \frac{( - 1 )^n + 1}{2}$ where $\ell_n$ is the $n$-th lucas number.
First of all $\varphi$ is a Pisot number. It is a root of $P ( x ) = x^2 - x - 1 = \left( x - \varphi \right) \cdot \left( x - \bar{\varphi} ) \right.$
The 'other root', $\bar{\varphi}$ has modulus $< 1$, by virtue of $| \bar{\varphi} | = | \frac{1 - \sqrt[]{5}}{2} | < 1$. Hence $\varphi$ is a Pisot number.
Now $\varphi^n + \bar{\varphi}^n$ is an integer, by the proprety $( 2 )$. This in turn implies, that
\begin{eqnarray*} \lfloor \varphi^n \rfloor & = & \varphi^n - \left\{ \varphi^n \right\}\\ & = & \left. \varphi^n - \left\{ \varphi^n - ( \varphi^n + \bar{\varphi}^n \right) \right\}\\ & = & \varphi^n - \left\{ - \bar{\varphi}^n \right\} \end{eqnarray*}
Since $\bar{\varphi} < 0$, if $n$ is even, then $\bar{\varphi}^n > 0$, so we obtain $\left\{ - \bar{\varphi}^n \right\} = 1 - \left\{ \bar{\varphi}^n \right\} = 1 - \bar{\varphi}^n$.
Conversely if $n$ is odd then $\bar{\varphi}^n < 0$, so $\left\{ - \bar{\varphi}^n \right\} = - \left\{ \bar{\varphi}^n \right\} = - \bar{\varphi}^n$.
(Wrtitting the equality $\left\{ \bar{\varphi}^n \} = \bar{\varphi}^n \right.$ is possible since $| \bar{\varphi}^n | < 1$)
This leads to
\begin{eqnarray*} \lfloor \varphi^n \rfloor & = & \varphi^n + \bar{\varphi}^n - \frac{( - 1 )^n + 1}{2} \end{eqnarray*}
(Just check the odd and even case for $n$ to see it is correct) Now it suffice to have in mind that $\varphi^n + \bar{\varphi}^n$ is in fact the $n$-th lucas number $\ell_n$, so that
\begin{eqnarray*} \lfloor \varphi^n \rfloor & = & \ell_n - \frac{( - 1 )^n + 1}{2} \end{eqnarray*}
Note that generally computations are easier when all the conjugates (the 'other roots') of a Pisot number are positive. Then we don't need to handle the positive or negative case. In that case the 'remainder' is always -1, as in the post above.

Thank you very much.
By the way,I like Poland National football Team,though do not play very well in this world cup


MATHCOUNTS probability Euler calculus ARML inequalities

Is the Level 9-12 grades? As I thought Allen was going into 8th grade next year and he's competiting with high schoolers.  :o

The levels are divided by difficulty level of the pieces you play, not the school grades. The highest level is 12.  Upper age limit for Young Artist division is 17.

Wow!
Many people that are good at math are good at piano (or another instrument)!
Like me, (except not as good, in both)  :blush:
I think the 9-12 means just levels (of the music), not grades.

Congratulations. I just woke up one day and called up for pinao lessons one day last summer, so I haven't been playing very long. Must be that math getting to my head.  :D

So, a few questions:
1)How long have you been playing?
2)What pieces did you play?

(1) 7  + (a little) years.

(2) Schubert Impromptu op.142 no.4
+
     Mozart Sonata K279 I and II.

and for the problem>> Its geometric probabilty.

[quote="mysmartmouth"]Congratulations, this is a great accomplishment.  I used to play piano, but I quit...I was never very good.  It takes a HUGE amount of dedication.[/quote]
Same here :lol:

Wow, not only you are good at Math, but you are good at Piano.

I hate piano and always will, but I still get myself to practice 20 minutes or more a day .-. (i hate revolutionary etude)

[quote="MHamlin"]Hello, MathCounts AoPSers, my name is Kevin Yuan. Since my son Allen has been very enthusiastic about this community, I am too happy for Allen not to let you know something wonderful about him: Allen won Grand Prize for Level 9-12 Piano Solo in the Young Artist Division of World Piano Competition last Wednesday in Cincinnati.  I'm very proud of him for this achievement. As soon as he came back from MathCounts National, he has been putting in 3 hours a day practicing piano. And now his hardworking is handsomely rewarded.

Here's the link for the event if you'd like to know more:
http://www.cincinnatiwpc.org/[/quote]

Congratulations, Allen! This is a great achievement!!

[quote="kyyuanmathcount"]:)

50 dollar tossup (to "un-spammatize" this thread):

Let's say I have three pieces, A, B, and C, and I always practice them in that order.  Assuming I practice 3 hrs, what is the probability that, if I switch songs randomly during my practice, I will practice more of piece C than A and B combined?[/quote]

Congratulations Allen!!! :jump:  :first:  :clap2:  :first:  :clap2: :jump:

So this is not spam (as you wish and so I can put in more smileys), here is the answer to your question.

[hide]The only way that he could practice more of piece C than A and B is if C was chosen in the first $1\frac12$ hours.  The switch to piece B must also be here, so the probability is $\frac12\cdot\frac12=\frac14$.  However, piece B must be switched before piece C can be started, so the probability is $\frac12\cdot\left(\frac12-B\right)$, where B is the position of the switch to the B piece.  Here I stop, because I don't know what to do  :D (other than sum an infinite sum, which I don't feel like doing).[/hide]

[quote="mimar"]Nice job Alllen! Its nice to know that people can be good at more than one thing![/quote]
Actually, it is quite reccomended that a person be good at more than one thing. There are well documented cases of people who played multiple games of chess blindfolded and generally worked their brains to the breaking point(Morphy and Alekhine) dying in their thirties. Same with Euler; he made vast contributions to calculus but went blind early in his life, kept doing math, and died. Now, someone like Lasker, who played chess but was also a PhD in math, lived much longer. There are documented cases of straining one part of your brain again and again over a long period of time causing internal bleeding.

Also, the vast majority of Moppers are very good pianists. The ARML writers were too.

Allen, do you have any tips for the pianists who got going rather late in the game?

[quote="b-flat"][quote="kyyuanmathcount"]:)

50 dollar tossup (to "un-spammatize" this thread):

Let's say I have three pieces, A, B, and C, and I always practice them in that order.  Assuming I practice 3 hrs, what is the probability that, if I switch songs randomly during my practice, I will practice more of piece C than A and B combined?[/quote]

Congradulations Allen!!! :jump:  :first:  :clap2:  :first:  :clap2: :jump:

So this is not spam (as you wish and so I can put in more smileys), here is the answer to your question.

[hide]The only way that he could practice more of piece C than A and B is if C was chosen in the first $1\frac12$ hours.  The switch to piece B must also be here, so the probability is $\frac12\cdot\frac12=\frac14$.  However, piece B must be switched before piece C can be started, so the probability is $\frac12\cdot\left(\frac12-B\right)$, where B is the position of the switch to the B piece.  Here I stop, because I don't know what to do  :D (other than sum an infinite sum, which I don't feel like doing).[/hide][/quote]

[hide="hint"] Try setting variables, a and b for when he changes, then drawing a graph using inequalities you can find.[/hide]

To PenguinIntegral: Enjoy it.

[quote="Ignite168"]Someone move this topic to Getting Started, the posted problem seems to hard for MathCounts and let's get Allen some more recognition.[/quote]
Well, the thread is kind of more related to MathCounts than Getting Started, but the problem in it is more like high school level. If it's necessary, I could split the problem from the topic and move it to a high school forum (either Getting Started or Intermediate, whichever is more of the problem level).

And of course, congratulations to Allen on his amazing accomplishment!  :D  :clap:

Also, congratulations to Ryan Liu (a.k.a. asdf4) who was selected as the 2006 National Suzuki Convention piano soloist! :clap2:   :)  (more info here: http://www.suzukiassociation.org/conference/students/pianoconcerto/)

This just goes to show that the people we have here on AoPS are talented in many different ways!

There is a similar problem in [i]Introduction to Counting and Probability[/i].  It's #16 in chapter 10.


Determine all functions f:N->N such that
f(3x + 2y) = f(x) f(y) for every x,y.

I think I  can prove this problem .f(0)=f ^2(0)
             then f(0)=0 or f(0) =1 :D  :?


What is the units digit of $ 1998^{1998}$?

You could also notice a pattern in the units digit.

$ 8^1 \equiv 8 \pmod {10}$

$ 8^2 \equiv 4 \pmod {10}$

$ 8^3 \equiv 2 \pmod {10}$

$ 8^4 \equiv 6 \pmod {10}$

$ 8^5 \equiv 8 \pmod {10}$

$ ...$

Since $ 1998 \pmod 4 \equal{} 2$, the answer is the second number in the pattern, or $ \boxed{4}$.


Αν $ A,B,C$ τα μετρα των γωνιων τριγωνου $ ABC$ σε ακτινια, τοτε
$ aA \plus{} bB \plus{} cC\geq\frac {1}{2}(aB \plus{} aC \plus{} bC \plus{} bA \plus{} cA \plus{} cB)$

Mε διαταξη των $ a,b,c$ επεται και ομοιοστροφη των  $ A,B,C$ εχουμε
 $ aA\plus{}bB\geq{aB\plus{}bA}$
$ aA\plus{}cC\geq{aC\plus{}cA}$
$ bB\plus{}cC\geq{bC\plus{}cB}$
και προσθεση των προηγουμενων


A box is made from a tin sheet (1 X 1 m) that has a congruent square ( X by X) taken off from EACH corner. The tin sheet is then folded up to create the open topped box. Find the max volume of the open topped box. (The problem should be solved through derivatives).

This exercise does not belong here; it should have been posted in the Calculus and Computations section.  As to the exercise itself, set up a function $ V$ in terms of $ x$ that describes the volume of the box.  Then Find the extreme values of the function.  Then find which extreme values lead to a maximum volume, as required.

This is not really calculus level, why should it belong in Calculus Computations? It's really a highschool problem. I don't think that every problem that could be solved using calculus should be in that forum... after all this problem is of a high school nature. Anyway I obviously know that I have to set up 2 equations (the most taught manner of solving this) my problem is that I couldn't isolate the height for the box to set up a cubic function. So I'm really asking what are the equations that need to be set up (I made obv made a mistake somewhere) and how to isolate height.

If the sheet is $ a\times a$, then the volume of the box is $ V(x) \equal{} (a \minus{} 2x)^2x$. (The base is a square whose side is $ a\minus{}2x$, and the height is $ x$.) Write this as $ {1\over 4}(a \minus{} 2x)^2(4x)$ and apply GM-AM:

$ {1\over 4}(a \minus{} 2x)^2(4x)\leqslant {1\over 4}\left({a \minus{} 2x \plus{} a \minus{} 2x \plus{} 4x\over 3}\right)^3 \equal{} {2a^3\over 27}$, and the maximum is reached for $ a \minus{} 2x \equal{} 4x\iff x \equal{} {a\over 6}$

For the given numericals, $ V_{\max} \equal{} {2\over 27}\,\mathrm{m}^3$ for $ x_0 \equal{} {1\over 6}\,\mathrm{m}$

[quote="stack3d"]This is not really calculus level, why should it belong in Calculus Computations? It's really a highschool problem. I don't think that every problem that could be solved using calculus should be in that forum... after all this problem is of a high school nature.[/quote]

Some of calculus is "of a highschool nature" -- in fact, a great many people take one or two years of calculus while in highschool.  Nevertheless, calculus problems belong in the calculus forum.  (Routine exercizes, such as this one, belong in the Calculus Tutorials & Computations section.)  While this question can be solved without calculus (see, e.g. Farenhajt's solution), its most natural solution by far is by application of a derivative.  If you specifically seek a solution not using calculus, then you could include a phrase in the problem to that effect, e.g. "I know this problem has a routine calculus solution, but I'm interested in methods that do not require calculus."  With such a comment, this is an acceptable forum for such a question.  The fact that you included a comment to the exact opposite effect makes abundantly clear that this is the wrong forum for your problem.


limit
calculus
derivative
LaTeX
calculus computations

How to find limx->0 (e^x + x)^(1/x)? I used ln, so it becomes 1/x ln (e^x+x) but I still couldn't find the answer, hope you can help. thanks.

By L'Hopital's rule. $ \lim_{x\to 0}\frac{\ln(e^x\plus{}x)}{x}\equal{}\lim_{x\to 0}\frac{e^x\plus{}1}{(e^x\plus{}x)}$. Or just notice that $ \lim_{x\to 0}\frac{\ln(e^x\plus{}x)}{x}$ is the derivative of $ f(x)\equal{}\ln(e^x\plus{}x)$ at $ x\equal{}0$.

Start by using Latex: $ L \equal{} \lim_{x \to 0} (e^x \plus{} x)^{1/x}$. As an alternative, let's use the known result $ \lim_{x \to 0} (1 \plus{} x)^{1/x} \equal{} e$:

$ L \equal{} \lim_{x \to 0} (e^x \plus{} x)^{1/x} \equal{} \lim_{x \to 0} \left[[1\plus{} (e^x \plus{} x \minus{} 1)]^{1/(e^x \plus{} x \minus{} 1)}\right]^{(e^x \plus{} x \minus{} 1)/x} \equal{} e^{\lim_{x \to 0}[(e^x \plus{} x \minus{} 1)/x]}$.

When x goes to zero, $ e^x \approx 1 \plus{} x$ and so $ \lim_{x \to 0} \frac{e^x \plus{} x \minus{} 1}{x} \equal{} \lim_{x \to 0} \frac{2x}{x} \equal{} 2$. Hence, $ L \equal{} e^2$.

My correct alternative posted solution, posted before Carcul's was wrongly removed.

I will reenter it as best I can:

Let y = lim x-> 0 (e^x + x)^(1/x)

ln y = ln [lim x-> 0 (e^x + x)^(1/x)]

ln y = lim x-> 0 ln(e^x + x)^(1/x)

ln y = lim x-> 0 (1/x)ln(e^x + x)

ln y = lim x-> 0 (e^x + 1)/(e^x + x) by L'Hopital's Theorem

ln y = 2/1

ln y = 2

e^(ln y) = e^(2)

y = e^2 = the desired limit