function
calculus
derivative
algebra
domain
polynomial
calculus computations

Prove that there exists a differentiable function f such that:

 [f(x)]^5 + [f(x)] + x = 0

for all x

 Any ideas? I'm stuck on this..

thanks

 it would probably help if I mentioned that this question has something to do with inverse functions

[quote="mche"]it would probably help if I mentioned that this question has something to do with inverse functions[/quote]
Ya think? Yes, it was obvious that that's what this is about, whether you'd said that or not.

Define $ g(y) \equal{} \minus{} y^5 \minus{} y.$

Some questions:

What do you know about $ g'(y)?$ Is that derivative ever zero? Is it either always positive or always negative? What does that tell you about whether $ g$ is monotone or whether it has an inverse? Also: is the range of $ g$ all of $ \mathbb{R}?$

Note that what you have is $ g(f(x)) \equal{} x.$ If $ g^{ \minus{} 1}$ makes any sense, then we should have $ f(x) \equal{} g^{ \minus{} 1}(x).$

What theorems do you have that say anything about the derivatives of inverse functions, and what hypotheses or conditions must be satisfied in order to quote these theorems?

The one thing you [i]don't[/i] want to do is waste your time trying to compute exact algebraic expressions.

thanks for the reply, I  thought about f^-1(y) = -y^5 - y when I was doing this problem and I think I have a pretty good argument:

  let 'x' = f^-1(y) ,  f^-1(y) = -y^5 - y for all 'x'. -y^5-y is one to one, continuous and has R as its domain (since (f^-1(y))' is never zero. Therefore, f is also one to one, continuous and takes values on R .. and so it is a differentiable function. (one to one and continuous makes it differentiable, since it will be monotonic on any interval)

  Then we have a function f such that  (y)^5 + y + f^-1(y) = 0 that satisfies our condition. It's also important that f^-1(y) takes values on R, since the question asks for all x, if we set x = f^-1(y), we will have no problem, since f^-1(y) is defined on R

 how's that?

thanks a lot

Almost: you missed one thing. You haven't shown that the [i]range[/i] of the function which I called $ g$ is all of $ \mathbb{R}.$ The range of $ g$ will be the domain of $ g^{\minus{}1}.$

You said several times that the domain of $ g$ is all of $ \mathbb{R},$ which is true, but that's not the range.

[quote="Kent Merryfield"]Almost: you missed one thing. You haven't shown that the [i]range[/i] of the function which I called $ g$ is all of $ \mathbb{R}.$ The range of $ g$ will be the domain of $ g^{ \minus{} 1}.$

You said several times that the domain of $ g$ is all of $ \mathbb{R},$ which is true, but that's not the range.[/quote]

 ah yes, I thought I had said that explicitly - but after reading my post, maybe not. It's true though, since 'g' has to be able to take on any value since we're saying x = f^-1(y) = g(y) for any x, f^-1(y) has to be able to take on any value, right?

thanks a lot!

You still haven't given a reason why $ g$ has range that is all of $ \mathbb{R}.$

Or, let's put it this way: $ h(y)\equal{}e^y$ is a one-to-one function whose range is not all of $ \mathbb{R}.$ What have you said that wouldn't also apply to this function?

[quote="Kent Merryfield"]You still haven't given a reason why $ g$ has range that is all of $ \mathbb{R}.$

Or, let's put it this way: $ h(y) \equal{} e^y$ is a one-to-one function whose range is not all of $ \mathbb{R}.$ What have you said that wouldn't also apply to this function?[/quote]

   well, d e^x / dx  = e^x approaches 0 as lim x -> -infinity so the rate of growth of this function becomes stagnant, or even at larger values, it might as well behave like 0 itself. on the other hand,
-5x^4 - 1 is always  < 0, so it is defined on all R. It is never zero or around zero, so the function will just keep growing. Is that it?

thanks

[quote="Kent Merryfield"]No. That's not it.[/quote]

 ah, it is one to one and invertible , therefore if it has a domain of R it must have a range of R

[quote="Kent Merryfield"]Still wrong.[/quote]

 okay, then it's just because it's an odd degree polynomial with domain of R haha..
 It goes to - infinity as x goes to -infinity and infinity as x goes to infinity. So, it is onto

thanks


AMC
AMC 10
geometry
rectangle
3D geometry
perimeter
quadratics

i am tutoring this middle school kid who has desire to learn... (not just school work, but also competition-type)

what would be good formulas for a student whose level is just below AMC10?

Ah, i forgot to mention, formulas about Algebra... i dont mind if you have some Geometry formulas...

P.S. I am in OLYMPIAD LEVEL, but it has been a long time since ive studied at his level. I am not sure what to tell him, because if i tell him hard formulas, he wont get it, and as you might have noticed, i am too lazy to dig up my old works.

Go to my site http://yashar.biz/mathcounts
and check out the bible.

I have also posted some really good formulas on my blog.

Reccomend the AoPS book to the kid too.  That is the best guide ever.

[quote="Farazy"]
Reccomend the AoPS book to the kid too. That is the best guide ever.
[/quote]

i did.and he ordered it. but shippig is taking too long and he is becoming inescapably impatient.

Well:
Area of a Square: $A=s^{2}$
Area of Rectangle: $A=lw$
Area of Circle: $A=\pi r^{2}$
Volume of Cube: $V=S^{3}$
Perimeter of Rectangle: $2(l+2)$
Pythagorean Theorem: $a^{2}+b^{2}=c^{2}$
45-45-90 Triangle: $x,x,x\sqrt{2}$
30-60-90 Triangle: $x, x\sqrt{3},2x$
Plus volume/S.A. of cones, cylinders, etc.
Sum of first $n$ natural numbers: $\frac{n(n+1)}{2}$
Quadratic Formula: $\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
Geometric/Arithmetic sequence formulas
The choosing formula (not sure what its called):$\frac{n!}{m!(n-m)!}$
How to factor, ie. :$x^{2}+5x+6=(x+3)(x+2)$
Common factorizations:
$(a+b)^{2}=a^{2}+2ab+b^{2}$
$a^{2}-b^{2}=(a+b)(a-b)$
$(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}$
$a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$
These are some I just thought of off the top of my head.
Is he doing Mathcounts?
Heron's formula might help as I've seen it show up before: $\sqrt{s(s-a)(s-b)(s-c)}$

I would definitely recomment AOPS Volume 1 to highly motivated middle school students, it is not too advanced yet requires a lot of thinking and working, and not just reading through the book.

[quote="eryaman"]
The choosing formula (not sure what its called):
[/quote]

it is called combinatoric.

btw, all the formulas you gave are elementary type things. Everyone on this site should know that, I think. He is getting the AoPSbook, but he is just too impatient to wait, so he asked me. Thanks anyways.

[quote="Treething"]My belief (the same as many other people) is that memorizing formulas isn't nearly as useful as knowing how to [i][b]derive[/b][/i] them.[/quote]

deriving is good...but could take too long in competitions, I think you should know how to derive the formula and memorized the formula.

If all that stuff is too basic for him yet he is still below AMC10 level, I think he probably just needs more practice attacking problems and learning how to use the knowledge he has.  Maybe you should tell him to work through theMathcounts Handbook and you can give hints on the questions he does not know.


Express $F = \frac{{\sqrt[3]{2}}}{{1 + 5\sqrt[3]{2} + \sqrt[3]{4}}}$
with a rational denominator

I managed to factorise the denominator but then got demoralised when I forgot what I was trying to achieve by doing it...

still,

[hide]let the cube root of 2 equal x and complete the square to find roots to the quadratic. Im not entirely convinced that this helps though...[/hide]

Use the proof in the link below (details in portuguese)

http://cienciasexatas.blog.uol.com.br/images/racion.jpg

(if direct link doesnt works then copy and paste link on address bar)


[quote="superwizard188"]If x + 1/x = 2, what is x^5 + (1/x)^5?[/quote]

If x=1, then the second equation is equal to 2.

[quote="alanchou"][hide]It's kinda funny, because this one has x=1, so
$x^{5}+\frac{1}{x^{5}}=2$ as well.[/hide][/quote]

But the expression is $x^{5}+(\frac{1}{x})^{5}=2$

(Which is the same thing, but oh well.)

[quote="xpmath"]If x+1/x=2 then
square that, subtract 2, square that, subtract 2 again, then multiply by whatever will get it to the desired format.
There was a problem kinda' like this on the AMC 10A[/quote]
 Look at the first post. THis is a trick, the obvious answer is one.

You made a fatal mistake! You put 0, when the answer is ±0!  :rotfl:

[quote="xpmath"]Which is the same thing...[/quote]

It was a joke...  :roll:


let $a,b,c$ are positive numbers such that $ab+bc+ca=1$ prove that  this inequality
$\sqrt [3]{\frac{1}{a}\ + 6b} + \sqrt [3]{\frac{1}{b}\ + 6c} + \sqrt [3]{\frac{1}{c}\ + 6a} \leq\ \frac{1}{abc}\$

It has been posted before at least 2 time. Try to search.


Sometimes I want to include some formulae in article section titles. They show perfectly well on the page, but they disappear in the "contents" box. Is it something that is going to be fixed later or does it mean that using formulae in section titles is strictly prohibited?

I'm not sure I understand.  Can you link me to a page so that I can see what you mean contextually?

[url]http://www.artofproblemsolving.com/Wiki/index.php/Pi_is_not_Liouvillian[/url] is a good example

LaTeX isn't really necessary (you could just write R(x,y)) but it would certainly be nice to get it fixed.


Evo i zadaci sa prvog dana Drzavnog takmicenja BiH.

Evo zadaci i sa drugog dana Drzavnog takmicenja BiH. Good luck :lol:

Rezultati Drzavnog takmicenja BiH  :D Za ovo 6. ce biti dodatni test u srijedu.

Zadaci + rjesenja (na cirilici) Drzavnog takmicenja BiH

[url]http://rapidshare.com/files/115927250/Bosnian_TST_2008..pdf.html[/url]

Dodatno takmicenje za izbor 6. clana ekipe:

[url]http://www.mathlinks.ro/viewtopic.php?t=205988[/url]
[url]http://www.mathlinks.ro/viewtopic.php?t=205992[/url]
[url]http://www.mathlinks.ro/viewtopic.php?t=205989[/url]
[url]http://www.mathlinks.ro/viewtopic.php?t=205993[/url]

Evo i zadaci sa prvog dana priprema nase ekipe Prvi je iranski a ova dva slovenski IMO shortlist Taman nas je Krivosija (autor testa) sa ova dva "poznata " pravo zezno a test mi se pravo dopao.

Zadaci sa drugog pripremnog testa BiH IMO'08 tima Test je sastavio prof.dr.Sefket Arslanagic.

Zadaci sa treceg testa BiH IMO'08 tima. Test je sastavio Irfan Glogic, student prve godine prirodno matematickog fakulteta (bronzana medalja na IMO'06)

Zadaci sa 4. dana pripremnog testiranja BiH IMO '08. tima. Test sastavio Damir Hasic, deputy leader.

Zadaci sa 5. dana priprema BiH IMO '08 tima. Test sastavio doc.dr.Hasan Jamak, proslogodisnji deputy leader BH tima.


integration
calculus
calculus computations

[color=darkblue]Ascertain $\{\begin{array}{c}I=\int\frac{x}{(x^{4}+bx^{2}+1)\sqrt{x^{4}+ax^{2}+1}}\mathrm{\ dx}\\\\ J=\int\frac{x^{3}}{(x^{4}+bx^{2}+1)\sqrt{x^{4}+ax^{2}+1}}\mathrm{\ dx}\end{array}\ \ ,$ where $\{a,b\}\subset \mathbb R\ .$

[hide="Indication."]Calculate $I+J$ and $I-J\ .$[/hide][/color]

The substitution $x=\frac{1}{t}$ shows that

$I+J=0$.

On the other hand

$I-J=\int\frac{\frac{1}{x^{2}}-1}{(x^{2}+b+\frac{1}{x^{2}})\sqrt{x^{2}+a+\frac{1}{x^{2}}}}dx=$
The substitution $x+\frac{1}{x}=y$ shows that

$I-J=\int\frac{-dy}{(y^{2}+b-2)\sqrt{y^{2}+a-2}}$, and this integral can be evaluated by elementary calculations.


ratio
arithmetic sequence
complex numbers
algebra solved
algebra

[b]1.[/b]
a) Let a, be a real number such that a>1 then:
1+1/a+1/a+...+1/a^n<a/(a-1) for any natural n.
b) if p_n is the nth prime number and (a_n)n is a sequence, such that a_n=[p_1/(p_1-1)]...[p_n/(p_n-1)] demonstrate that for any M>0 there exists a natural n, such that a_n>M.
***
An object is projected on the plane alfa in the square P and on the plane beta in the square Q. Proove that if the square P has a side parallel with a side of the square Q then the squares have the same area.
				[i]Ion Savu[/i]
[b]3.[/b]
Let f from C[x] be a polinom, f=a_0+X*a_1+...+X^n*a_n with the property that |f(z)|<=1 for any z from C, |z|=1. Proove that |a_0|+|a_1|+...+|a_n|<=1.
				[i]Dan Marinescu and Ioan Serdean[/i]
[b]4.[/b]
We consider the sequence (a_n)n limited, not constant of real numbers which has the property that in a vicinity of 0 it has a finite number of therms.
a) Prove that this sequence doesn't contain an infintite arithmetic progression.
b) Prove that this sequence doesn't contain an infintite geometric progression.
				[i]Radu Gologan[/i]

contest "Alexandru Papiu Ilarian"- Targu Mures- 8th November-11th grade
soon as i have time i'll post the subjects for 12th grade.

_____________________________________________________
in the honour of the one which invented the beeeer!!!! :D

For 1:

1+1/a+1/a^2+.. = 1/(1-1/a) = a/(a-1) (sum of the infinite geometric progression). It's pretty clear now that the property at a) is true. For b):
You can write a_n as (1+1/p1+1/p1^2+..)*...*(1+1/pn+1/pn^2+..) where each sum is infinite. To make it short, let me make some writing abuse: a_00 = 1+1/2+1/3+1/4+... = 00, so a_n->00.                           a_00 is lim {n->00} a_n

this is the easiest way to do it
but i wanted to do it like this
b_n = 1/a_n
and i wanted to proov that b_n->0, n->oo but i couldn't
can u help me?

Hmmm... 4 shouldn't be too hard either. It's well known that an infinite arithmetic progression is non-bounded, but our sequence is bounded (I assume that's what 'limited' means). For (b) it should be easy to prove that a geom progression with ratio >1 is non-bounded and a geom progression with ratio <1 has infinitely many terms 'near' 0 (in the vicinity of 0) and infinitely 'close' to 0. It's all obvious, there's no point in writing it all in formal language.

I see that noone speaks about the most interesting and beautiful problem: nr 3. I will present a solution to this elegant problem. Let x1, x2,...,x(n+1) the roots of order n+1 of the unity. Then one can easily check that (n+1)a0=sum p(xi), (n+1)a1=sum xi^n *p(xi), (n+2)a2=sum xi^n-1*p(xi),...etc. So it is enough to prove that if a1,...,an are complex numbers which verify |ai|<=1 ( put ai=p(xi) for simplicity), then
 |a1+a2+...+an+1|^2+|a1*x1+...+an+1*xn+1|^2+...+|a1*x1^n+...+an+1*xn+1^n|^2<=(n+1)^2. But the LHS is exactly (n+1)(|a1|^2+...+|an+1|^2) ( this can be proved using the fact that |z|^2=z*conj(z).

indeed the number 3 is the most beautifull pb
a solution:
Let Un={z|zn=1}(the roots of order n of the unity.
one can check that sum{z \in Un}zk=0 if n|k and n else(k is not divisible by n)  (*)
|f(z)| \leq 1=>f(z)f(z.conj) \leq 1<=>
[sum{k=0;k<=n}akzk][sum{j=0;j<=n}aj.conj zj.conj]<=>sum{k=0,j=0}akaj.conj zk-j \leq 1
now we have that sum{z\in Un+1}sum{k=0,j=0}akaj.conj zk-j \leq n+1 and (*)=>sum{k=0;k \leq n}|ak| \leq1

i like very much the problem 2 to  :D i'm waiting for different solutions of this pb :D
Kuba


probability
inequalities
topology
advanced fields
advanced fields unsolved

Here is a very usefull lemma to compute Hausdorf dimension:
$A \subset \mathbb{R}^{n}$ a closed set such that $H^{\alpha}(A)>0$, where $H^{\alpha}$ is the $\alpha$-hausdorf measure. Prove that there exists a probability measure $\mu$ supported on $A$ and a constant $C>0$ such that for all Borel set $B$ the following inequality holds: \[\mu(B) \leq C diam(B)^{\alpha}\]

How do you use this lemma to compute Hausdorff dimension?

And, more importantly, where do you use the constant $C$? :D

Yes, this does not allow to compute the dimension  :lol:  But, it is clear that if $\mu(B) \leq C diam(B)^{\alpha}$ for borel $B$ subsets of $A$, then the Hausdorff dimension of $A$ is at least $\alpha$. Frostman's gives a kind of converse to this result, which shows that a good approach to compute an Hausdorff dimension is to try to find a probability measure on $A$ that is "well" distributed on $A$. (this gives lower bounds) And upper bounds are often easy to obtain.


Ring Theory
algebra
function
domain
number theory
greatest common divisor
superior algebra

Prove that Dedekind domain is a PID iff it is a UFD.

(Tomorrow is the algebra final exam, and I got stuck in this problem...)

It suffices to prove that an ideal generated by two elements is principal.

Let $\{p_i\}_{i\in I}$ be the set of primes, and let $a,b\ne 0$ be elements of our domain $A$. Clearly, $(a)+(b)=d((a')+(b'))$, where $d$ is the GCD of $a,b$, and $a=da',b=db'$, so that $a',b'$ are coprime. If we prove that $(a')+(b')=A$, we're done: it will then follow that $(a)+(b)=(d)$. We thus want to show that if $0\ne a',b'\in A$ and $a',b'$ are coprime, then $(a')+(b')=A$.

First we determine the prime (or maximal, sicne it's the same thing) ideals of $A$: if $p_{i_1}\cdot\ldots\cdot p_{i_k}$ is an element of a maximal ideal, then one of the $p_{i_j}'s$ must also be an element, and then our maximal ideal contains the maximal ideal $(p_{i_j})$, meaning that they are equal, so we have our answer: the maximal ideals are precisely those generated by a prime element.

Now, if $(a')+(b')\ne A$, then it's contained in a maximal ideal, which must be generated by some prime $p$ according to the paragraph above, but this means that $p|a',b'$, contradicting the coprimality of $a',b'$.


induction
logarithms
ceiling function
geometric sequence
geometric series

Is there a better way to solve this instead of guessing and checking?

When I purchassd my first catapult for $2000, it needed repairs. The first repair costs$200. Each susequent repair costs 10% more than the repair just before. After how many repairs could I first correctly cry, "I already spent more for catapult repairs than I spent for the catapult!"

(from Math Team team round)

Yes, you need to develop an infinite series with an $ a_{1}$, an $ r$-value, and a way to determine $ a_{n}$.  Once you do that, you can find the value of $ n$ such that the sum exceeds 2000.

[hide="Highlights"]Start out with a clear notation: Let $ a_n$ be the $ n$-th successive repair cost. Map out your costs:

$ a_1\equal{}200$

$ a_2\equal{}a_1\plus{}10\% (a_1)\equal{}a_1(1.1)$

$ a_3\equal{}a_2\plus{}10\% (a_2)\equal{}a_2(1.1)\equal{}a_1(1.1)^2$

At this point, induction should help you get the general [u]geometric sequence[/u].

Let $ S_n$ be the sum of all the repair costs, ie $ S_n\equal{}\sum_{k\equal{}1}^na_n$. To finish up the part "I already spent more for catapult rparis than I spent for the catapult!", set $ S_n>2000$. Use the geometric series formula here and solve for $ n$.[/hide]

JRav, I think you mean [b]finite[/b] geometric series?

[hide="Solution"]The question simplifies to:
$ 200 \plus{} 200(1.1) \plus{} 200(1.1)^2 \plus{} 200(1.1)^3... > 2000$

Using the formula for a finite geometric sequence, we solve:
$ \frac {200(1 \minus{} (1.1)^{n \plus{} 1})}{1 \minus{} 1.1} > 2000$

Solving, we get:
$ \frac {200(1 \minus{} (1.1)^{n \plus{} 1})}{1 \minus{} 1.1} > 2000$
$ \Rightarrow 200(1 \minus{} (1.1)^{n \plus{} 1}) < \minus{} 200$
$ \Rightarrow 1 \minus{} (1.1)^{n \plus{} 1} < \minus{} 1$
$ \Rightarrow (1.1)^{n \plus{} 1} > 2$
$ \Rightarrow n \plus{} 1 > \log _{1.1} \left( 2 \right)$
$ \Rightarrow n > \log _{1.1} \left( 2 \right) \minus{} 1\approx 6.272540897$

So the amount of repairs needed is $ 6.272540897\plus{}1$, so:

The answer is $ \lceil 7.272540897 \rceil \equal{} \boxed{8}$.[/hide]

[quote="10000th User"]Unfortunately 123456789, your answer is off by 1. Pay attention at the exponent of your formula and how you are using it. To make sure if you got it right, add the individual costs to find out.

EDIT: there is a typo in that quote, it's not 'rparis', it's 'repairs'[/quote]

ya, the correct answer is [hide="this"] 8[/hide]

also, thanks for noticing the typo, i edited it


algebra
polynomial
vector
superior algebra
superior algebra unsolved

Suppose $ N\left(n,q\right)$ is the number of all monic irreducible polynomials of degree $ n$ over the field $ \mathbb F_q$ (this is the field with $ q$ elements). Prove that
\[ N(n,q) \equal{} \frac1n\sum_{d|n}\mu\left(\frac nd\right)q^d,
\]
where $ \mu: \mathbb{N}\to\mathbb{Z}$ is the Mobius function.

The most essential in this proof is the proof of the equality
$ x^{p^n} \minus{} 1 \equal{} \prod_{d|n}F_d(x)$.  (*), where $ F_d(x)$ is the producty of all the monic irreducible polynomials of degree $ d$.
As soon as we've managed to prove (*) it is only left write the identity related with the degree in it's left and right sides:
$ p^n \equal{} \sum_{d|n} dN(d,q)$ and applying Moebius formula $ N(n,q) \equal{} \frac {1}{n}\sum_{d|n}\mu(\frac {n}{d}) p^d$.

It is intetersing exercise to prove (*), if you want i'll write the proof i know

First I think you meant $\prod_{d|n}F_d(x)=x^{p^n}-x$

I think I've proved that $\prod_{d|n}F_d(x)|x^{p^n}-x$. For this if every 2 irreducible polynomials have distinct roots, and if  $\alpha$ is a root of polynomial of degree $d|n$ then $F(\alpha)$ has $p^d$ elements. So $\alpha^{p^d-1}=1$. So $\alpha^{p^n-1}=1$ and $\alpha$ is also a root of $x^{p^n}-x$

Also if $\alpha$ is a root of $x^{p^n}-x$ then $(\alpha)$ is a field with $p^d$ elements, I think we can prove that there exist a $\beta$ that $(\alpha)=F(\beta)$

The easiest way to prove that identity with $x^{p^n}-x$ is just to see that when taking the splitting field of that polynomial, you get the field with $p^n$ elements.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=30473]here[/url] for a different approach, which only needs to know the size of the field.

I completed my proof and this is my final proof. Suppose $p(x)=x^{p^n}-x$

Suppose $\alpha$ is a root a polynomial with degree $d$ that $d|n$. So $F(\alpha)$ is a field that is a vector space of dimension $d$ on the field $F$. So $|F(\alpha)|=p^d$. By the lagrange theorem in the group $F(\alpha)^{\times}=F(\alpha)\backslash\{0\}$,it's obvious that $\alpha^{p^d-1}=1$. So $\alpha$ is a root of $p(x)$. Now suppose $\alpha$ is a root of $p(x)$, you see there is a $d$ that $d|n$ and $d$ is the least number that $\alpha^{p^d}=\alpha$. You see there are $p^d$ numbers like $t$ that $t^{p^d}=t$.  Easily set of these numbers is a field containing $F$(Call it $E$). ($(x+y)^{p^d}=x^{p^d}+y^{p^d}=x+y$ and $(xy)^{p^d}=x^{p^d}.y^{p^d}$,$\forall u\in F u^{p^d}=u^p=u$. So it is a vector space of dimension $d$ on $F$.You see that $E=F(\alpha)$. So $\alpha,\alpha^2,\dots\,\alpha^{d-1}$ are independent on $F$ (Because $[E: F]=d$),and $\alpha,\alpha^2,\dots,\alpha^{d}$ are dependent on $F$. So there exist a monic polynomial $g(x)\in F[x]$ with degree $d$ that $g(\alpha)=0$. So $g$ is the minimal polynomial of $\alpha$, it is irreducible,monic and of degree $d$ that $d|n$. Everything has been proved and proof is complete, So the two sides of the equation are equal.(Every root of each side is the root of the other side, and there is no multiple root.)


calculus
integration
logarithms
absolute value
calculus computations

[quote="kunny"]Find the minimum value of $\int_{x}^{x+l}\left(t+\frac{1}{t}\right)dt \ (x>0,\ l>0).$[/quote]
Hope I am not wrong on this one.

I assume, in the problem, $l>0$ is some fixed constant, and, $x$ is the independent variable.

$I(x) = \int_{x}^{x+l}\left(t+\frac{1}{t}\right)\,dt = \frac{t^{2}}{2}\Big|_{x}^{x+l}+\ln |t|\Big|_{x}^{x+l}$
$\Rightarrow I(x) = \frac{l^{2}}{2}+lx+\ln (x+l)-\ln x$ ...(Note that we can drop the absolute value signs.)
$\Rightarrow I'(x) = l+\frac{1}{x+l}-\frac{1}{x}$ and $I''(x) = \frac{1}{x^{2}}-\frac{1}{(x+l)^{2}}$

Solving $I'(x) = 0$ for positive $x$, we get $x = \frac{-l+\sqrt{l^{2}+4}}{2}$ (which is a point of local minimum because $I''(x) > 0\ \forall x > 0$.)

Now, note that as $x\rightarrow 0^{+},\ I(x) = \frac{l^{2}}{2}+lx+\ln (1+\frac{l}{x}) \rightarrow \infty$, and
as $x\rightarrow \infty,\ I(x) = \frac{l^{2}}{2}+lx+\ln (1+\frac{l}{x}) \rightarrow \infty.$

Thus $x = \frac{-l+\sqrt{l^{2}+4}}{2}$ is also a point of absolute minimum.

So, $I_{min}= \frac{l^{2}}{2}+l\left( \frac{-l+\sqrt{l^{2}+4}}{2}\right)+\ln \left( 1+\frac{2l}{\sqrt{l^{2}+4}-l}\right)$
$\Rightarrow I_{min}= \frac{l\sqrt{l^{2}+4}}{2}+\ln \left( 1+\frac{2l}{\sqrt{l^{2}+4}-l}\right)$


function
group theory
abstract algebra
algebra proposed
algebra
ENS

i see that there is a lot  of probleme s about periodic  and continious functions. me to i have a nice problem :
propve that doesnt exist continous function and periodec of period  $1;and \sqrt{2}$  :P

You must be kidding, they never give at oral examinations such trivial problems! This is immediate, since the additive subgroup generated by $1$ and $\sqrt{2}$ is dense.  :P

i am not kidding .  j ai trouve cette exercice dans  un livre d oral 1987(crus). je vais revoir le livre .

[hide=Triviality is subjective]
I think this is an even more trivial solution.
PS : In many places group theory is not a part of High School syllabus, so only a handful of students who know what to read about group theory can read it.
[/hide]
A non constant function continuous periodic function $f: \mathbb{R} \to \mathbb{R}$ will have a smallest positive period. Say this be $p$.
Then, we claim that every period of $f$ will have to be integer multiple of $p$.
Let $f$ have another period $p' >0$.
Then,
$f(x+p')=f(x)=f(x+p)$ for all $x$.
Now, let $kp$ be the greatest integer multiple of $p$ strictly less or equal to $p'$. If $p'=kp$ then done.
So assume, $p'-kp = d >0$
Then $0<d <p$
$f((x+kp)+d)=f(x)=f(x+kp)$ for all $x$
And hence we get, $f(x+d)=f(x)$ for all $x$, which contradicts the minimality of $p$.  $\blacksquare$

Thus we conclude that, if $p_1,p_2$ are positive periods of such a function $f$ then $\frac{p_1}{p_2}$ is rational.

Let $f: \mathbb{R} \to \mathbb{R}$ be a non constant Darboux function (that is it maps each connected set to a connected set). Can $f$ have periods $1, \sqrt{2}$ simultaneously ?
[hide=.......]And of course in general, can $f$ have periods $p_1,p_2$ where $\frac{p_1}{p_2}$ is irrational ?
[/hide]


Fie $\left( G, \cdot \right)$ un grup. Stim ca exista $a \in G$ astfel incat $x^{3}= a x a, \, \forall x \in G$.
Demonstrati ca $\left( G, \cdot \right)$ e abelian.

Din relatia data avem, pentru $x=1$ ca
\[a^{2}=1\,. \]
Atunci
\[(xy)^{3}=axya=axa\cdot aya=x^{3}y^{3}\,,\forall x,y\in G\,. \]
(Aplicatia $x\mapsto x^{3}: G\longrightarrow G$ coincide cu automorfismul interior asociat elementului $a$).
Rezulta atunci ca
\[x^{2}y^{2}=(yx)^{2}\,,\forall x,y\in G\,. \]
In particular,
\[x^{2}=x^{2}a^{2}=(ax)^{2}=axa\cdot x=x^{4}\,,\forall x\in G\,, \]
de unde $x^{2}=1\,,\forall x\in G$, deci $x=x^{-1}\,,\forall x\in G$.
Dar atunci
\[xy=(xy)^{-1}=y^{-1}x^{-1}=yx\,,\forall x,y\in G\,. \]

da... astea-s probleme d-alea la care solutia pare mult prea ascunsa. incerc ceva mai scurt, poate merge :

$a^{2}=e$ , unde $e$ este elementul neutru. asta e evident
mai departe, inlocuim pe x cu ax, si avem asa
$(ax)^{3}=axaxax=aaxa=xa$, deci $axaxax=xa$. inmultim treaba asta la stanga cu a, si avem $xaxax=axa$. conform relatiilor de mai sus, inlocuim pe $axa$ cu $x^{3}$ si obtinem $x^{5}=x^{3}$, de unde prin simplificare $x^{2}=e$. si de aici iese ca e abelian. evident , avem mereu oricare ar fi $x$ and stuff. unde am gresit ?


geometry
geometric transformation
dilation
symmetry
function

[b]Edit:  Nevermind, I figured this out[/b]

Let's say you travel away from Earth at velocity such that the lorentz factor is 2 for 1 year, then you stop, turn around and return to Earth at the same speed.  Then, by time dilation 2 years would have passed for you, but 4 for the Earth.  By symmetry, if the Earth is travelling away from you at the same velocity 2 years, then the Earth turns around and travels back to you at the same speed, 4 years would have passed for the Earth but 8 for you (by time dilation).  In Earth's reference frame, both scenarios are identical, yet in one scenario 2 years pass for you whereas 8 years pass in the other scenario, how is this possible?

Einstein's equations are not intitutive to the casual observer, as we don't observe relativistic effects.

The way I think about it is that the time dilation, length shortening and mass increasing are a function of the acceleration, which won't be the same in the two cases you suggest, but the accelerating object will have the relativistic effects.

I hope this helps you sort out this effect.

Keep in mind that Albert said that his math wasn't strong enought to allow him to ask the hard questions, so he asked the easy ones that others had neglected.  (This means if you find yourself lost and things seem complicated, you are probably on the wrong track, go back to basic principles and don't go too far away in your thinking.)  Then you have to do the math following your "simple" thinking.

You're not traveling in an inertial frame of reference, so predictions made from you (the traveling guy) is not accurate. In fact, the scenarios aren't identical

looks similar to mthe famous twin paradox....never mind.


The diameter of a circle is 26 cm. A chord is drawn such that its midpoint is 12 cm from the center of the circle. What is the number of centimeters in the length of the chord?

OK, so the triangle as a radius of $ 13$.  Let's label the circle $ A$ and the chord endpoints $ B$ and $ C$.  The distance from $ BC$ and the center is $ 12$.  Now, draw the radius $ AB$.  It forms a right triangle!  So the answer is simply $ 5 \times 2$, or $ \boxed{10}$.


Dear all,

Many of you have noticed difficulty with the application site in (at least) the past 24 hours.  This appears to be due to [i]very[/i] high traffic on the site: for example, the site had 48 users logged in at one point this morning.    As a result, the application program has run out of memory a few times.

We have tweaked the program, and we hope that this will reduce the memory usage.  The site has now been stable since the early afternoon, and I'm keeping my fingers crossed that that won't change.  We will be monitoring it throughout the evening and tomorrow.

If you do experience further problems, please email us (camp@mathcamp.org) quickly at the first sign of trouble.

thanks,
Dave

Yeah, the site still doesn't work.  I sent an email a couple hours ago, but still haven't had any success.  Suggestions?

Well, if you have a fax machine you could fax in your application. (At least, I think you can.)

Yet for the faxing, there is no way to know whether the solutions were received, because on the website there were only 2 options: Postal Mail and Web.

I faxed my solutions in, although I really should have $\LaTeX$'d them, yet I did not really know how to do diagrams. Hopefully, they will be received. If not, I have a verification transcript that I sent them. :D

[quote="mcalderbank"]Yet for the faxing, there is no way to know whether the solutions were received, because on the website there were only 2 options: Postal Mail and Web.[/quote]

Mark as postal mail in your app, then we'll mark it as received manually (just as if we had gotten it by mail).
In any event -- we got your fax.

Nice. I apologize in advance for my bad handwriting. I need to learn to $\LaTeX$ diagrams....

Lol, I did almost my entire mathcamp quiz, including subscripts, superscripts, math signs, etc. except diagrams on Microsoft Word.  I'm still learning Latex.

Last year I decided to learn TeX specifically for my mathcamp app. Of course, I had no idea how to use it effectively, so I'm sure the poor guy who read my solutions had a lot of problems - the margins were huge, there was no spacing, no size alterations. For the sequence problem from last year, I'd feel bad if I had to read my solution.

[quote="tetrahedr0n"]Last year I decided to learn TeX specifically for my mathcamp app. [/quote]
That's what I did this year... But any sort of TeX would have been better than my handwriting. lol.

last week end the registration was down, and today when i first logged in it was fine until I tried to modify my phone number and it crashed when I reloaded when I go from the welcome to the personal info there is an error.


geometry
3D geometry
tetrahedron
AMC
USA(J)MO
USAMO
inequalities

A given tetrahedron ABCD is isosceles, that is, AB=CD, AC=BD, AD=BC. Show that the faces of the tetrahedron are acute-angled triangles. [hide]USAMO 1972[/hide]

What is an acute angled right triangle? That seems somewhat contradictory to me. If it's a right triangle, the other angles are automatically acute. If all the angles are acute, then it isn't right.

erm. ya sorry about that. i'm not exactly sure why i put the word "right" in the problem statement.

I like this one, but I've seen it before.  So I restrict myself to one hint: [hide]what can you say about the three angles at any vertex of a tetrahedron?  One thing is that they must add up to less than 360 degrees, but are there any other relationships between them?[/hide]

bringing it back up top for another round of eager young minds who wish to delve into the mysterious depths of this problem

To be specific, this is USAMO 1972 #2.

[hide="Solution"] All of the faces are clearly congruent.  In particular, they are all triangles with side lengths $ a, b, c$, where (say) $ a \equal{} AB, b \equal{} AC,  c \equal{} AD$.  Taking JBL's hint, the angles formed at a given vertex are the three angles in any face.  However, again taking JBL's hint, the angles must satisfy the triangle inequality, so none of them can be greater than $ 90^{\circ}$.  [/hide]

A friend of mine suggested that I do all of the past USAMOs in order, and I looked up this question to see if there were any threads about it on AoPS.  :)

[quote="K81o7"][quote="ZzZzZzZzZzZz"][quote="t0rajir0u"]the angles must satisfy the triangle inequality, so none of them can be greater than $ 90^{\circ}$.[/quote]

But this doesn't prove that it's acute because one angle could still be $ 90^\circ$.[/quote]

Well then it would be degenerate...
If we assume it's non-degenerate (which is a pretty reasonable assumption I think) then the sum of any two of the angles exceeds the third, and so none of the angles are greater than 90 degrees.

EDIT: By the way, t0r, how did you find such an old thread?[/quote]

Yeah, I agree with the proof that none of the angles are greater than 90 degrees, but there could be an angle that is equal to 90 degrees, making the triangle right.

As Krishanu already pointed out, that would make the tetrahedron degenerate.


Put n lines on a plane to cut it. ($ n \ge 3$) for any three of them $ a,b,c$, $ a \cap b \cap c\equal{}\emptyset$ and for any two, $ a \cap b \ne \emptyset$,

Try to prove, there're at least $ n\minus{}2$ triangles (in which without line segments)

Er...for every set of $ 3$ lines, there is exactly one triangle, and three are no other triangles otherwise, so there are exactly $ \binom{n}{3}$ triangles...or are we not allowed to count overlapping triangles?

A triangle here shouldn't be "cut"... it means that only "small" triangles (no line segment inside) count. for examples, there're exactly 2 triangles when n=4.


linear algebra
matrix
linear algebra unsolved

Give $ A \in M_n$ is a matrix such that $ rank(A \plus{} E_n) \plus{} rank(A \minus{} E_n) \equal{} n$
Prove that $ A^2 \equal{} E_n$

I prefer to write $ I$ for the identity matrix rather than $ E,$ so that's the notation I'll use.

We're working over a field $ K$ which isn't quite arbitrary: we need the condition that $ \text{char}(K)\ne 2.$

Let $ S \equal{} \text{nul}(A \minus{} I)$ and $ T \equal{} \text{nul}(A \plus{} I).$

If $ s\in S,$ then $ As \equal{} s.$ If $ t\in T,$ then $ At \equal{} \minus{} t.$ If $ x\in S\cap T,$ then $ Ax \equal{} x \equal{} \minus{} x,$ and (as long as the characteristic of the field is not 2), that implies that $ x \equal{} 0.$

Hence $ S\cap T \equal{} \{0\}.$ But then

$ \dim(S\cup T) \equal{} \dim(S) \plus{} \dim(T) \equal{}$

$ \equal{}(n \minus{} \text{rank}(A \minus{} I)) \plus{} (n \minus{} \text{rank}(A \plus{} I)) \equal{} 2n \minus{} n \equal{} n.$

This means that $ S\cup T \equal{} K^n.$

Let $ x$ be an arbitrary element of $ K^n.$ We can write $ x$ uniquely as $ x \equal{} s \plus{} t$ for $ s\in S,t\in T.$

$ Ax \equal{} As \plus{} At \equal{} s \minus{} t.$

$ A^2x \equal{} As \minus{} At \equal{} s \plus{} t \equal{} x.$

Since $ A^2x \equal{} x$ for all $ x\in K^n,$ we have that $ A^2 \equal{} I.$

The failure of this result in characteristic $ 2$ should be obvious.

This condition is possible, as shown by the example of $ \begin{bmatrix}1 & 0 \\
0 & \minus{} 1\end{bmatrix}.$


Do you think it is a good or bad idea to include in the forum a space dedicated to the most interesting formulae or theorems of geometry, number theory, combinatorics,....?
It could be a list in which everybody can include its favourite formulae with or without proof.

and where should the list be positioned? It is a good idea in my opinion ..

I think it could be positioned in problem directory main page.

i think this is a good idea
maybe we should post them in a separately directory in the forum and sort them like: Analisys, Combinatorics, Theory Numbers,...


I'm using Firefox 2.0. All of the images have disappeared. These include all of the small images next to My Blog, Calendar, etc. The MathLinks logo has also disappeared. I cannot see the emoticons.

Never mind. I fixed it.  :) It was a local problem on my end. Somehow, it had blocked all images from AoPS.


If $n$ divides one Fibonacci number (the sequence $1,1,2,3,5,8,13,21,...$), show that it will divide infinitely many of them

I think this works...

[hide]Let us assume that $n|a$, where $a$ is some number in the Fibonacci sequence. Let $b$ be the number directly after.

Thus, the sequence will go as follows:

$a, b, a + b, a + 2b, 2a + 3b, \cdots$

The coefficients of $a$ and $b$ will always be two consecutive Fibonacci numbers (and every Fibonacci number will be used at some point in time). Eventually, there will be two coefficients, $x$ and $y$, such that $y = a$. Thus, this number will be

$xa + yb = xa + ab = a(x + b)$

Since $n|a$, $n|a(x + b)$.

Now, we just redefine $a$ to be this new multiple of $n$ and $b$ to be the next Fibonacci number. We can then follow this method over and over to find an infinite number of multiples of $n$.[/hide]

We can prove that $F_k|F_{lk}$ where $F_x$ denote $x$th Fibonacci number.
As we know \[F_k=\frac1{\sqrt5}((\frac{1+\sqrt5}2)^k-(\frac{1-\sqrt5}2)^k)\]
and \[F_{lk}=\frac1{\sqrt5}((\frac{1+\sqrt5}2)^{lk}-(\frac{1-\sqrt5}2)^{lk})\] in the futur we will use $a=\frac{1+\sqrt5}2$ and $b=\frac{1-\sqrt5}2$ so as we know number like $a^x+b^x$ and $a^xb^x$ are integers ....($*$).
Now let's divide $F_{lk}$ and $F_{k}$ that is equal to \[\frac{a^{lk}-b^{lk}}{a^k-b^k}\] wich gives \[a^{kl-1}+a^{kl-2}b^k+...+a^kb^{kl-2}+b^{kl-1}\] wich is integer because of ($*$). So there you go...
This is much stronger stuff :D


Vasily Astakhov wrote about results of MSU team:
A. Efimov - 11 problems,
V. Asthakhov - 10,
A. Perepechko - 9,
D. Baranov - 8.5,
S. Smirnov - 6.5(unfortunately again the last in our team  :( )

I'll try to remember Princeton's results. Official results will come tomorrow afternoon I think (but I won't have a chance to post until Thursday evening, so probably someone else will post them).

Andrei Negut -- about 10.5
Simion Filip -- about 8.5
Kenneth Tay -- about 5
Marius (I forgot his last name but he's Romanian and probably went to IMO 2005) -- about 5
Aleksander Lishkov -- about 4
Nathan Savir -- about 3.5? :-(

By the way, I think the top 3 people were Mr. Efimov and Andrei Negut and Nima Ahmadi Pouri something from Sharif University of Technology (in that order). But that is before the challenges. Official results tomorrow.

Probably Ungureanu (but we call him Andrei, not Marius).

[quote="Xevarion"]No, not him. His employers wouldn't let him leave, or something.

Marius Pachitariu.[/quote]Ah, right :) I forgot he was at Princeton too (ikap is his username on AoPS - MathLinks).

Now, who can solve problem 6 on day 2?
Can you help me?

Can anybody post the official results of IMC?????????

[quote="Xevarion"]By the way, I think the top 3 people were Mr. Efimov and Andrei Negut and Nima Ahmadi Pouri something from Sharif University of Technology (in that order)[/quote]
its Nima Ahmadi Pour (you should loose the "i" at the end) with same username in ML...

[quote="vess"][quote="Xevarion"]By the way, no contestant solved problem 6 on day 2. The highest score given was 3 (out of 20). I heard that the problem was inspired by a recent journal article....[/quote]

Interesting... It's the only problem that interested me, but I couldn't solve it either. Do you know a solution?[/quote]
We were given an official solution of two pages...

ZetaX how did you do on the competition?

I got information about results of [b]Indonesian[/b] universities at IMC 2007

Fajar Yuliawan: [b]Bandung Institute of Technology[/b](Informatics Engineering): 86--- third prize,
M. Hafiz Khusairi: [b]Bandung Institute of Technology [/b](Informatics Engineering): 61,--HM
Prastudy Mungkas Fauzi:[b]University of Indonesia[/b](Computer Science): 52--HM
Albert Gunawan: [b]Gajah Mada University [/b](Mathematics Science):45---HM
Ajat Adriansyah:[b]University of Indonesia[/b] (Mathematics Science):25

Is anybody here meet any of them at the Competition? How do you feel? What is your impression? :wink:

[quote]By the way, I think the top 3 people were Mr. Efimov and Andrei Negut and Nima Ahmadi Pour something from Sharif University of Technology (in that order).[/quote]

first of all ,I say congratulate to him for his great success  :lol: , of course Nima Ahmadi Pour is one of the genuis men that I have ever seen, and it goes without saying he would be successful in future. :)

We're expecting him in this thread now what is he doing? :P

He has just returned from Bulgaria, and as I talked with him 2 hours ago, he is in Tehran. He got 4th (not 3rd)
I informed him.

Well, thank you guys. As Omid told before, I became 4th. Although because I didn't see the final results (after the contestations) I don't know who became 3rd (In the closing ceremony it was hard to hear the names!), but I think he must have been from Etovos University.
IMC 2007 was very nice, the organizers, the competitors and all the people involved. After all its more informal and more friendly than IMO is, and the ranking and all these stuff are the last matters of importance.

Sharif University result in IMC
Nima Ahmadipur 4th 1st Prize
Aliakbar Daemi  6th 1st Prize
Erfan Salavati 8th 1st Prize
Nase Talebizadeh 1st Prize
Mohammad Gharakhani 1st Prize
Roozbeh Farhoodi 1st Prize
Behzad Mehrdad 2nd Prize
Sam Nariman 3rd Prize

Princeton University results

Andrei Negut (21?) 2nd place, 1st Prize
Simion Filip (170) 10th place, 1st Prize
Kenneth Tay (104) 2nd Prize
Marius Pachitariu (102) 2nd Prize
Aleksander Lishkov (8?) 3rd Prize
Nathan Savir (64) Honorable Mention (missed 3rd prize by 1 point!)

team:  4th place

Congratulations Nima Ahmadi Pour and all the team from Sharif !

 :clap2:

Where can I find all the results? Or the results of the romanians. Thank you.

IMC 2007 results is in attached file.
Disclaimer: table was scanned and OCRed so file may contain little mistakes. :wink:

If you look at result of Cyprus University you will found this funny or strange. Everyone  solved 3 problem in first day. It seems to me that no one team do this and
Angela	Evagorou	University of Cyprus	49	20	3	20	3	0	0	46	12		20	20	2			42	88	Third Prize

Ermioni	Ioannou	University of Cyprus	100	20	3	20	3	0	0	46	192	20	2	20	0	0	0	42	88	Third Prize

Andri	Eftychiou	University of Cyprus	44	20	3	20	3	0	0	46	1	20	1	20	0			41	87	Third Prize

Cheating is dark side of teaching. :wink:

[quote="BanishedTraitor"]What's the disqualified about? :maybe:[/quote]I wonder about that too... does anyone know?

Where is Alexander Rybak? Can`t find him.  :(

Alexander Rybak is too old:) He is 5 years student. So he can't participate, as I know ha was team laeder of KPI.

[quote="BanishedTraitor"]What's the disqualified about? :maybe:[/quote]
[quote="Peter"]I wonder about that too... does anyone know?[/quote]
Cheating. :mad:

What kind of cheating in fact?  :maybe:  I cannot imagine how they can find it out in such a competition (or they'd have to be really stupid)...

[quote="xavi"]They were talking all time during the exam... in fact, first day I was near to them and that was what I could see, I suppose that someone saw that too and they were disqualified[/quote]

That is unbelievable (unbelievably stupid I mean).


Consider a triangle ABC and a arbitary point P.$ PA,PB,PC$ meet $ BC,CA,AB$ at $ A_1,B_1,C_1$.The circumcircle $ (W)$ of triangle $ A_1B_1C_1$ meet $ BC,CA,AB$ at $ A_2,B_2,C_2$. $ A_2P,B_2P,C_2P$ meet $ (W)$ at second points $ A_3,B_3,C_3$.Prove that $ AA_3,BB_3,CC_3$ are concurrent.

Dear Mathlinkers,
see for example
Nguyen K. L., Salazar J. C., On Mixtilinear Incircles and Excircles, Forum Geometricorum vol. 6 (2006) 1-16.
Sincerely
Jean-Louis

Sorry I can't see where's my problem on that article? :maybe:

see theorem 15 p. 12
Sincerely
jean-Louis


[hide]Hi !

You have $3x+1=10$ and then $3x=9$ then $x=3$. :)[/hide]

[hide]or you can distribute, getting 6x+2=20
subrract 2, getting 6x=18
divide by 6, getting x=3
[/hide]

Hey, are you in Algebra?  What grade 2

[quote="beto"]please help to solve 2(3x+1)=20[/quote]

[hide="are u in pre-alg or something"]

just expand and get $6x+2=20$. $6x=18$. So, $x=3$. :roll: [/hide]

Expanding...Read a pre-algebra textbook or something.

distributing...something to that extent

[hide][quote="beto"]please help to solve 2(3x+1)=20[/quote]

6x+2=20
6x=18
x=3[/hide]

[hide]Divide by 2 on each side. The rest is easy.[/hide]

[hide]Do you know about the distributive property?

a(b+c)= ab+ac

So....

2(3x+1)=20
3x+1= 10
3x= 9
x=3[/hide]

[quote="beto"]please help to solve 2(3x+1)=20[/quote]

[hide]
6x+2=20
6x= 18
x = 18/6
x= 3
[/hide]


I'm wondering about what people think of chess. Many people say chess can't be a science because it is too much fun.(I think science is fun)Others say it can't be called just a game. I want to know what the masses have to say!

[quote="math92"]In my opinion, its just a strategy game. Science is defined as a branch of knowledge or study dealing with a body of facts or truths systematically arranged and showing the operation of general laws. Of course the pieces obey laws of gravity and friction, but other than that, its just a fun strategy game, and if one practice's hard enough, can become very good at.[/quote]

Chess... the ultimate turn-based strategy game! :D

Yeah... you can use all kinds of mathematical language and group theory and game theory and other stuff to describe chess and possibly come up with a winning strategy, but in the end it's still a game.

Since every chess game must end in a finite number of moves (32 x 50 = 1600 is an upper bound) game theory can provide an exact winning strategy with a complete rollback analysis, but perhaps we shall need a quantum computer for that.

Chess is a deterministic discrete strategic game (so no fuzziness).
The study of chess is a science of course (maths is much easier - I have been through both).

That's all there is to it - people's opinion doesn't matter. (It would matter for example on the question of chess being an art and so on, but there can be no aruments at all.)

Why must every game end in a finite number of moves? Can't one just use the knights in multiple positions to avoid the repetition rule and have an infinite game?

Finitely many pieces on a finite board implies finitely many positions, so the repetition rule plus pigeonhole gives a finite upper bound on game length.

It's technically possible for a chess game to go on forever, I suppose, since even though there's a finite number of positions, these positions can follow in such a pattern as no rule would come into play.

Still, defnitely a game. There's at least 2 famous people quoted as saying "Chess in not a science!".

[quote="Temperal"]even though there's a finite number of positions, these positions can follow in such a pattern as no rule would come into play.[/quote]

If any board state is repeated 3 times, either player may claim a draw.  For purposes of this discussion, if any board state is repeated 3 times, the game ends.  There are only finitely many states, so some board state must eventually be repeated 3 times.

I thought the rule was if a position is repeated three times regularly. (e.g. same span of moves between each turn)

Hmm, I really do suck at chess. I wikipediaed it and JBL is right.

[url=http://en.wikipedia.org/wiki/Threefold_repetition]Link.[/url]

Okay, then, I guess there are a finite number of positions.

I'm quite sure that bubka's limit of 1600 is far too low, though, seeing as pawns can promote.

I believe there's also a 200 move rule or something to that extent: after 200 moves the game is automatically a draw.

Don't you mean:

After 50 moves with no captures, pawn moves, or promotions, the game is a draw?

There's been games longer that 400 moves, I think...

50 consecutive moves with no pawn moves, no promotions, and no captures -->

50 * 16 * 6 (pawn moves/promotions) + 50 * 15 * 2 (captures) = 6300. I am undercounting a little bit I think but I am too lazy to be careful. So, that rule gives a max of like 6350 moves before a draw is possible. (Note that this maximum can't actually be obtained, because the pawns can't all get past each other except by capturing other pieces. But that means a pawn move and a capture occur simultaneously, wasting 50 moves.)

Board state repeated 3 times --> very large number. There are TONS of possible board states. I am not even going to try to estimate it but the maximum game length according to this rule is going to be unbelievably huge. Even with just two kings on the board, there are still more than 64 * 55 board positions, which means it could take over 10,000 moves before a draw could be claimed for that reason. And in fact I think it has to be the same player's move, which doubles the number, right? I have ignored the fact that such a sequence may not be possible to construct (an interesting problem in its own right -- anyone want to try it for two kings on a smaller board?)

Although the board state repeated 3 times rule has been used in chess tournaments, that is not a rule instated to limit game length. Instead, that's an extra little thing that a skilled player can use to pull off a draw when he should lose (like if the opponent is chasing his king and not being careful enough, or something like that). The 50 move rule is what really limits game length. (Also, sometimes one player can trap another player but not completely, so that a certain cycle of moves repeats and only one player ever has a choice. If the player with the choice is in a significantly worse position if he gives up this control, then he can just force the moves indefinitely, so they might as well declare a draw.)

PS I think it's not a draw automatically in either of these cases but a player must claim the draw. Otherwise the play will continue. However in most cases the player probably wants the draw, so the game will end there anyway.

[quote="Temperal"]

After 50 moves with no captures, pawn moves, or promotions, the game is a draw?

[/quote]

technically, a promotion is a pawn move, so it should be "50 moves with no captures or pawn moves" :P

btw, though i agree that chess is not a science, it has some scientific aspects to it.
I remember once reading a book written by Yasser Seirawan (Winning Chess Strategies, cmiiw) where he said that sometimes chess can be treated as a science, an art (combinations can be really beautiful), and also a sport.

In my opinion, chess is a game. It could be studied and one might find the winning strategy or something similar to it, but that takes the fun out of the game. The nice thing about chess is that nearly every game is different, and new combinations may come up, involving some beautiful moves. Chess is played for fun, and with two people -- it is not one person sitting in front of the board and studying it (unless it's a grandmaster preparing for a tournament or something).

I don't see why chess is a science. It is just a game and should be enjoyed like it. The only time one should study it like rem said is if you are a grandmaster and heading to a tournament.

[quote="987"]JBL, I do not think you are correct about chess ending after finitely many moves. In fact I think this is a  consequence of Thue's Theorem stating that there exist two letter (0 and 1 for the sake of this post) alphabets in which no block occurs three times in succession.[/quote]
It doesn't have to be three times in succession. It just has to be three times, period. There can be intervening moves.

[quote="Temperal"]There's been many a tournament where that rule is not observed, though.[/quote]
What do you mean? If you mean the game doesn't end after that happens, it's because of what I mentioned before: someone has to actually claim the draw; it's not forced. Also, the same board position may occur 3 times but with it being different players' moves. The board position has to be the same INCLUDING which player's move it is at the time of that position. So it might not always be legit to claim the draw then. (Also in lower level play it is not too surprising if the players don't notice.)

[quote="Temperal"]I mean that many TDs don't observe that rule, due to either blitz nature or something.[/quote]
Oh, well blitz chess isn't really the same anyway. You can really only have it if the board positions are recorded; and also if the time is very limited then the other draw-by-repetition type rules become less significant anyway. (If you can force an unending game then you can instead try to win on the clock by making your moves very quickly.)

One bullet tournament (>=2 minutes, >=1 increment) tried to observe that rule, surprisingly. But there are standard time tournaments that don't observe the rule either.

I see. Well, I am not an expert chess player so I can't really give a good explanation of that. You would probably have to ask the tournament director who decided not to apply that rule.

My only guess would be that they thought that the situation comes up so rarely that the effect of the rule is way less than the additional complication of having another rule. Generally, it's nicer when games have fewer rules...

i call 3 fold repetition! it occured at...umm....move...4587328? and....512340982174?   must be some guy with awesome memory :)

it's a the max is VERY high since no previous 3-fold repetition can occur after any capture or promotion, so......yeah.......

[quote="Temperal"]Really? Where'd you get that?[/quote]

A board position can only be repeated if all the same pieces are still on the board. Furthermore, a pawn move and a castle are not reversible moves, so if any of those occurs then you're "starting again" also. So, the upper bound for the number of moves before a repetition-draw could be claimed is the sum of four times the number of board positions for each given set of pawn positions, castle statuses, and remaining other pieces. (Four times because each position should occur twice for each player to achieve maximum.) As I pointed out earlier, even in the final case where there are only the kings remaining, there are still thousands of board positions. There are also many thousands of pawn arrangements etc and many of those are going to be far more complicated than the case where the pawns are all gone.

I invite anyone to attempt to make an estimate of the maximum number of moves. If you provide an upper bound and demonstrate that your bound is within a factor of 10 of being right, I'll be hugely impressed. ;)

Well, I'd estimate 1,638,400, though I have no way to prove it, of course.

Chess is an art, science, and a game wrapped into a set of pieces laying on a board with 64 little squares on it.
Tis amazing!

chess isn't exactly a science.  it's an expertly crafted strategy game that has been played for thousands of years and still has not yet lost its flavor. I don't know how you would call chess a science.... but no matter what everyone must agree it is a game.

Why is the premise of the the thread that "game" and "science" are mutually exclusive?  Game theory is a branch of mathematics, hence in some sense a branch of science.

Well, game theory isn't necessarily relating directly to games. It's just a convenient label applied to something that could be called any number of things.

for clarification:
  sci·ence (sī'əns) pronunciation
n.

   1.
         1. The observation, identification, description, experimental investigation, and theoretical explanation of phenomena.
         2. Such activities restricted to a class of natural phenomena.
         3. Such activities applied to an object of inquiry or study.
   2. Methodological activity, discipline, or study: I've got packing a suitcase down to a science.
   3. An activity that appears to require study and method: the science of purchasing.
   4. Knowledge, especially that gained through experience.
   5. Science Christian Science.

Chess is a game but It will become a science!! :lol:


$ a_1, a_2,..., a_n$ be positive real numbers with $ a_1 \le a_2 \le .... a_n$ such that the arithmetic mean of $ a_1^2,a_2^2,...,a_n^2$ is 1. If the arithmetic mean of $ a_1,a_2,...,a_n$ is $ m$.. Prove that $ a_i \le m$ for some$ 1\le i \le n$, then $ n \minus{} i \ge n(m \minus{} a_i)^2$

http://www.mathlinks.ro/Forum/viewtopic.php?t=34202

  darij


The base of a rectangle is $ 19 \plus{} 2x$ cm and the height is $ 57$ cm.
How many centimeters are in the perimeter when the rectangle
is a square?

since its square, the perimeter of it is one side *4.

one side=57

57*4=228

answer : 228


hi,
can anyone solve this problem ? :blush:

[u]problem:[/u]

a,b,c $\geq0$ and $(a+)(b+c)(c+a)=1$
prove that : $ab+bc+ca\geq\frac{3}{4}$

thanks in advance

The sign is exactly opposite, and see http://www.mathlinks.ro/Forum/viewtopic.php?t=37173 .

  darij


The crystalline hydrate cadmium nitrate tetrahydrate loses water when placed in a large, closed, dry vessel. This process occurs even though it is endothermic; that is the change in enthalpy is positive. Is this process an exception to Bertholet's generalization? Explain.
Cd(NO3)2.4H20(s)=Cd(NO3)2(s)+4H20(g)

And what exactly is the "Bertholet's generalization"?


probability
absolute value
probability and stats

There are 1001 red marbles and 1001 black marbles. Let P denote the probability that two marbles drawn from the box are the same color, and Q denote the probability that the two marbles are not the same color. Find the absolute value of (P-Q).

A) 0
B) 1/2002
C) 1/2001
D) 2/2001
E) 1/1000

[hide]For P, we have to find the probability of drawing either two red marbles or two black marbles. The probability of picking two red marbles is $ \frac {\binom{1001}{2}}{\binom{2002}{2}}$, which is equal to $ \frac {500}{2001}$. Since we can draw either two red or two black marbles, then we multiply this value by 2, which is $ \frac {1000}{2001}$.

For Q, we know the probability of drawing two marbles of the same color  is $ \frac {1000}{2001}$, so the probability of them NOT being the same color is $ 1 \minus{} \frac {1000}{2001} \equal{} \frac {1001}{2001}$

$ |P \minus{} Q| \equal{} |\frac {1000}{2001} \minus{} \frac {1001}{2001}| \equal{} | \minus{} \frac {1}{2001}| \equal{} \frac {1}{2001} \Rightarrow C$
[/hide]
Of course, I could have made a mistake, so if anyone gets a different answer or believes I'm wrong, let me know.


Hello,

I have some homework that i need to do and am having trouble doing the following problems. all of which have to do with simplifing and the answer with postitive exponents only. also if it is not to much trouble can you go through the exact steps to solve them so i can learn how to do them.
thanks

1. (-1/2)^-2 + 4^-1
2. 2^-4 * 5^2 * 2^2 * 5^-1
3. (-4x^-3y)^-2 * (xy^3)^0 / 2x^-5
4. 4r^3s(r+5s)(r-5s)
5. (5v+7)^2 -(1-6v)^2
6. (x+6)(3z+4y)-(z-6y)(6+x)
7. -116 X 10^-5
8.  8^-2 R^-3 / 8^-12 R^4
Solve
9. y^2(3-y)+5y(3-y)=14(3-y)
10. r^4 + r^2 - 20 = 0

Actually this is way too easy for this forum, some mod please move it. Also, you are not supposed to post homework problems. But anyway, here goes, hope this helps:

[hide="1"]$\left(\frac{-1}{2}\right)^{-2}+4^{-1}$
$=\frac{1}{\left(\frac{-1}{2}\right)^2}+\frac{1}{4}$
$=1\div\frac{1}{4}+\frac{1}{4}$
$=4\frac{1}{4}$[/hide]

[hide="2"]$2^{-4}\times{5}^2\times{2}^2\times{5}^{-1}$
$=\frac{1}{16}\times{25}\times{4}\times\frac{1}{5}$
$=\frac{1}{4}\times{5}\times{1}\times{1}$
$=\frac{5}{4}$[/hide]

[hide="6"]$(x+6)(3z+4y)-(z-6y)(6+x)$
$=(x+6)(3z+4y-z+6y)$ (sometimes it helps to factorise first before simplifying)
$=(x+6)(2z+10y)$
$=2xz+10xy+12z+60y$[/hide]

Maybe with these done you will be able to figure out some of the others yourself. I'm not going to do all your homework for you ;)


Let f(x) = x^2, g(x) = 3x and h(x) = sqrt{x} + 1.
Express each function as a composite of f, g, and/or h.

1) q(x) = x + 2$\sqrt{x}$ + 1

2) p(x) = x^4

Chess64,

I want to thank you for your quick reply.  However, Iam a bit confused about why you decided to square the value for h(x) for the first question.  See it?

How can I solve these questions on my own?
What are the steps or formulas needed?

[quote="blueshark"]I am a bit confused about why you decided to square the value for h(x) for the first question.[/quote]

Notice that $(\sqrt{x}+1)^2=x+2\sqrt{x}+1$, which is $h(x)$. :)


Talk about anything you wish, but I will talk about Brawl.

Who is your favorite character?

Kirby.
That's a no-brainer. He's my favorite... not because he's cute and all... it's like because he can absorb other player's powers... which is freakishly awesome!  :lol:

Paper Mario thousand year door is pretty awesome too!

I'm a samus player myself; too bad getting a final smash makes you suck.

Go Mr.Game&Watch!!!

I love Octopuses and cheap Side-B attacks

Ice Climbers!

Ice Climbers are about the hardest to master, but since i'm so pro and stuff i rock with ice climbers. I'm awesome with using nana and stuff, and Ice Climbers+fan+fan+wall=999% in less than 10 seconds

falco is epic. start with an easy chain grab to get damage to 30-50%, then use down+b spike. the speed of falcos spike makes it much easier to hit with than marth's, plus you don't need a tipper.

lucas, spamming with up+b and pk fire is awesome. Also, pk freeze and using your up+b to burst into other players is awesome.

that technique with lucas only works on noobs :)

FALCO... PUNCH!

Yes, Captain Falcon is like the only character I know how to play. Not that I play that much, so...

I prefer Pit because his blocks can prevent other Super Smash's against him.

I like Ike he is my favorite. I llike his up-b and his counter.

I'd sa PIKACHU or toon link.
i like toon like, cuz well overall, he's just plain good.
PIKACHU, oh the other hand, is a counter person. i like his up+b very affective agains flying ppls/ jumping ppls.

this is a discussion of games. technically this is the only type of thread that should be in this forum ,and you want it to be moved to FF?

a lot more people look in the fun factory than in this section, i think.

This discussion shall stay in Games thread. Tourney signups will be moved to FF.

ok....
signup locked...
signups ended...

[quote="Berzerk"]meta knight but the guy dies before reaching 100% for me or even less than 90%.[/quote]

you mad?
ok...against snake...that's about it
and lucario is good

Meta Knight's the best guy in the game bye far if you're good with him. Unluckily, I'm not  :(

GO MARTH
even though marth doesn't have FIRE POWERS ohhhh!
he still pwns ike and the rest.
Most people haven't figured out his side b comos, all 12 of them
and they just trade him for ike

also

mewto does destroy lucario

Fox PWNS if you know how to use him. which i do by the way. omg. i beat my 18 year old brother in 1 minute on stock 5.

Please! don't revive.

[size=92][color=red]Hi, please PM instead of post on the thread next time you witness a revival. Thanks. [/color][/size]

yeah, reviving things that are over a month old mess up the forum i think.
I accidentally did it once :P

[size=92][color=red]Hi, please PM instead of post on the thread next time you witness a revival. Thanks. Also please don't say something that doesn't contribute to anything in the discussion. [/color][/size]


The Sagebrush student council has 6 boys and 6 girls as class representatives. Two different subcommittees, each consisting of 2 boys and 2 girls, are to be created. If no student can serve on both subcommittees, how many different combinations of subcommittees are possible?

[hide]
Let's name the committees A and B.
For committee A, there are ${6\choose 2}$ ways to choose 2 boys, and ${6\choose 2}$ ways to choose 2 girls, making $15\cdot 15=225$ ways total.
For committee B, there are ${4\choose 2}$ ways to choose 2 boys from the remaining 4, and ${4\choose 2}$ ways to choose 2 girls from the remaining 4, making $6\cdot 6=36$ ways total.
Thus, the number of ways is 1350.

I'm not sure if this is right.
[/hide]

[quote="ch1n353ch3s54a1l"][quote="mathnerd314"][hide]
Let's name the committees A and B.
For committee A, there are $\choose 62$ ways to choose 2 boys, and $\choose 62$ ways to choose 2 girls, making $15\cdot 15=225$ ways total.
For committee B, there are $\choose 42$ ways to choose 2 boys from the remaining 4, and $\choose 42$ ways to choose 2 girls from the remaining 4, making $6\cdot 6=36$ ways total.
Thus, the number of ways is 1350.

I'm not sure if this is right.
[/hide][/quote]

Everything was right until you said "Thus, the number of ways is 1350."  :) How did you get that?[/quote]

Shouldn't it be [hide] $36\times225=\boxed{8100}$ :D [/hide]

[quote="mathnerd314"][quote="lotrgreengrapes7926"]Shouldn't it be [hide] $36\times225=\boxed{8100}$ :D [/hide][/quote]

Yeah, I accidentally typed in 225*6 instead of 225*[b]3[/b]6. (Annoyed Grunt)! :oops_sign:[/quote]

No calcs on sprints ;)

Yeah, I multiplied $225*36$ afterwards realized $15^2*6^2=15*6*15*6=90^2=8100$  I wasted so many seconds of my life!


Triangle $ ABC$ with area $ S$ satisfy :$ \angle A<90,\angle B <90 ,\angle C < 90$  prove that:
$ 4S(\cot A \plus{} \cot B \plus{} \cot C) \equal{} |AB|^2 \plus{} |BC|^2 \plus{} |AC|^2$

[hide="solution"]

For a triangle $ abc$ with angles $ ABC$ and an area $ S$

Cosine law: $ c^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab\cos{C}$
Area: $ S\equal{}\dfrac{1}{2}ab\sin{C}$

Rearrangement yeilds
$ \cot{C} \equal{} \dfrac{a^2\plus{}b^2 \minus{} c^2}{4S}$

$ \Rightarrow \cot{A}\plus{}\cot{B}\plus{}\cot{C} \equal{} \dfrac{(a^2\plus{}b^2 \minus{} c^2)\plus{}(a^2\minus{}b^2 \plus{} c^2)\plus{}(\minus{}a^2\plus{}b^2 \plus{} c^2)}{4S}$

$ \Rightarrow 4S(\cot{A}\plus{}\cot{B}\plus{}\cot{C})\equal{}a^2 \plus{} b^2 \plus{} c^2$
Q.E.D.[/hide]

[quote="ocha"][hide="solution"]

For a triangle $ abc$ with angles $ ABC$ and an area $ S$

Cosine law: $ c^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab\cos{C}$
Area: $ S \equal{} \dfrac{1}{2}ab\sin{C}$

Rearrangement yeilds
$ \cot{C} \equal{} \dfrac{a^2 \plus{} b^2 \minus{} c^2}{4S}$

$ \Rightarrow \cot{A} \plus{} \cot{B} \plus{} \cot{C} \equal{} \dfrac{(a^2 \plus{} b^2 \minus{} c^2) \plus{} (a^2 \minus{} b^2 \plus{} c^2) \plus{} ( \minus{} a^2 \plus{} b^2 \plus{} c^2)}{4S}$

$ \Rightarrow 4S(\cot{A} \plus{} \cot{B} \plus{} \cot{C}) \equal{} a^2 \plus{} b^2 \plus{} c^2$
Q.E.D.[/hide][/quote]

Thanx ... I have along solution  :blush:


algebra
polynomial
search
algebra proposed

[color=blue]

   Prove that the polynomial $f=X^{n}+5X^{n-1}+3$ is irreductible over $\mathbb{Z}[X]$.

  [/color]

If I'm not mistaken, you can substitute $X=t-5$ and apply Eisenstein (all the coefficients except the leading one are divisible by $5$ and the constant one is not divisible by $5^2$)

That's not my solution. But, please feel free to post full solutions. :)

On the second view, I'm almost surely mistaken :)

[quote="silouan"][quote="harazi"]What about Perron's criterion?  ;)[/quote]

What is Perron's criterion ?[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=150 :)

Harazi, why shoul we apply Peron's criterion when it can be easily solved. :)

Cezar, please, I really don't need your remarks. I just indicated a solution in a single word. And don't believe that all your life you will continue to play around with 2 relations thinking this is maths.

[quote="enescu"]This is problem 1, IMO 1993 :)[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=62187&start=0

Harazi can you post your solution, please? :)

Oki, sorry. From now on I won't do this. :)

There is no solution to post. Just look in the theorems and formulas section for the statement of Perron's criterion. I won't check here that $5>3+1$.

lemma : for all $ P$ with integer coefficients , if $ a_0,a_1,P(1)$ are odd ,then $ P$ is irreducitible in $ Q[X]$
prove:use Einstein criteria

So $ x^8\plus{}x^7\plus{}x^6\plus{}x^5\plus{}x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1 \equal{} (x^2\plus{}x\plus{}1)(x^6\plus{}x^3\plus{}1)$ is irreducible¿


geometry
perimeter
inequalities
inradius
geometry unsolved

Given are $\triangle ABC,X,Y,Z$ are arbitrary point lied in segment $BC,CA,AB$ prove $p(XYZ)\ge\min\{p(AYZ),p(BZX),p(CXY)\}$
here $p(XYZ)$ show perimeter of $\triangle XYZ$

Very nice and what's the solution in this AMM

I see it is nice problem too, but really I don't know its solution in AMM, I only know it from the credits of my teacher and  he said with me its solution is bad ,because I hope every body are interested in and give a nice solution!
Thanks  :)

The inequality remains true even if we replace the perimeter with area or with inradius :wink:
BTW the proof of the inequality with perimeter is indeed ugly :(

Can you plesase post solutions of inradius?  :) I known a solution but I lose it!


two questions
1. What is the geometrical representation of x^2 + y^2 = z^2 (spherical coord here)?

I knew that it was a cone, but when i plugged in the spherical values for x, y, and z, i got that sin (phi) = cos (phi), which only happens at 45  and 225 deg. So wouldnt that mean that this is two cones? One with phi = 45 and one with phi = 225? The book just said that it is the representation of a cone.

2. Heh... While typign question 2 up, I realized that I had read the explanation incorrectly and saw OX instead of OY.

Well, I guess I would say it's two cones, of identical shape, joined at the vertex.  This is also sort of like one cone, but extended in both directions.  The two cones comes from the bit where it's squared -- you get both positive and negative values of z, which correspond respectively to 45 and 225 degrees.   Capish?


How do u insert numbers into the chat box? i seen people insert like 4th power and stuff. How do u do that?

PS: can anyone find the radius of circle "a" that John has a circle "b" where "b" is 4/3 the area of a. b overlaps a where the inner side of b to the inner side of a is 3pi. (pi=3.14=180, not the food). Don't bame me, I need to insert pictures.

click the LaTeX tab on the left; it'll give you a tutorial on how to make nice looking math symbols

Also, based on the posts you've made, I think you might want to consider starting in a lower-leveled forum such as [url=http://www.artofproblemsolving.com/Forum/index.php?f=149]High School Basics[/url].


function
floor function
LaTeX
limit
algebra
calculus
calculus computations

If [[x]] is the greatest integer function,prove that: lim(x)/[[x]]=1 when x approaches to infinity.
Edit: lol I knew how to do it. Here's my solution(pleased check it for me!):
Since we have:x={x}+[[x]],{x} is a fractional part of x if x is not integer(we just care about the case when x is not integer though)
Therefore: [[x]]=x-{x}
Now: x/[[x]]=x/(x-{x})=1/(1-{x}/x).Since 0=<{x}<1,we have: lim{x}/x=0 when x approaches to infinity(both sides). So,lim(x)/[[x]](x approaches to infinity)=1/1-0=1(Q.E.D)

Here's my problem in LATEX(I try using it!):
If $ \lfloor x \rfloor$ is the greatest integer function, prove that: lim x/($ \lfloor x \rfloor$ =1 when x approaches to infinity

[b]Solution:[/b]
Since we have:x={x}+$ \lfloor x \rfloor$,{x} is a fractional part of x if x is not integer(we just care about the case when x is not integer though)
Therefore: $ \lfloor x \rfloor$=x-{x}
Now:x/$ \lfloor x \rfloor$=1/(1-x/{x}).Since 0=<{x}<1,we have: lim{x}/x=0 when x approaches to infinity(both sides). So,lim(x)/$ \lfloor x\rfloor$ (x approaches to infinity)=1/1-0=1(Q.E.D)
Does anybody check my solution for me? :)

When I said "(dollar sign)" in my message to you, I meant the actual symbol $ \$$. I didn't write it that way in my message because that would have activated the $ \text{\LaTeX}$ math mode, and I didn't want to do that. I edited the symbol into what you wrote above. Click edit (or if it's too late for that, click quote) to see what it looks like.

See items 1 and 2 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]here[/url] for more.

Here's a rewrite of what you did that makes more use of $ \text{\LaTeX}.$ (Again, click quote to see what the code looks like):

[b]Problem:[/b] If $ \lfloor x\rfloor$ is the greatest integer function, prove that: $ \lim_{x\to\infty}\frac{x}{\lfloor x\rfloor}=1.$

[b]Solution:[/b] We have $ x=\{x\}+\lfloor x\rfloor,$ with $ \{x\}$ as the fractional part of $ x.$ Therefore $ \lfloor x\rfloor=x-\{x\}.$

Now: $ \frac{x}{\lfloor x\rfloor}=\frac{1}{1-\frac{\{x\}}{x}}.$ Since $ 0\leq \{x\}<1,$ we have $ \lim_{x\to\infty}\frac{\{x\}}{x}=0.$ So, $ \lim_{x\to\infty}\frac{x}{\lfloor x\rfloor}=\frac{1}{1-0}=1.$

(I did fix one mistake you had - you had $ \frac{\{x\}}{x}$ upside down at one point.)


limit
trigonometry
LaTeX
calculus
calculus computations

Evaluate $ \lim_{x\to0}\frac{\sqrt{1\plus{}\tan x}\minus{}\sqrt{1\plus{}\sin x}}{x^3}$

cant solve it with l'hospital...

Please help me!

CountDown : YES! NAVIER-STOKES EQUATION!!

$ \lim_{x\to 0}\frac{\sqrt{1\plus{}\tan x}\minus{}\sqrt{1\plus{}\sin x}}{x^3} \equal{} \lim_{x\to0}\frac{1\plus{}\tan x\minus{}(1\plus{}\sin x)}{x^3(\sqrt{1\plus{}\tan x}\plus{}\sqrt{1\plus{}\sin x})}\equal{}$

$ \lim_{x\to0}\frac{\sin x}{x} \frac{ \sec x\minus{} 1}{x^2(\sqrt{1\plus{}\tan x}\plus{}\sqrt{1\plus{}\sin x})}\equal{}$

$ \lim_{x\to0}\frac{\sin x}{x} \frac{ \sec^2 x\minus{} 1}{x^2(\sqrt{1\plus{}\tan x}\plus{}\sqrt{1\plus{}\sin x})(\sec x \plus{}1 )}\equal{}$

$ \lim_{x\to0}\frac{\sin x}{x} \frac{ \tan^2 x}{x^2(\sqrt{1\plus{}\tan x}\plus{}\sqrt{1\plus{}\sin x})(\sec x \plus{}1 )}\equal{}$

$ \lim_{x\to0}\left(\frac{\sin x}{x}\right)^3 \frac{ \sec^2 x}{(\sqrt{1\plus{}\tan x}\plus{}\sqrt{1\plus{}\sin x})(\sec x \plus{}1 )}\equal{}\frac{1}{4}$

[quote="FantasyLover"]Evaluate $ \lim_{x\to0}\frac {\sqrt {1 \plus{} \tan x} \minus{} \sqrt {1 \plus{} \sin x}}{x^3}$

cant solve it with l'hospital...

Please help me!

CountDown : YES! NAVIER-STOKES EQUATION!![/quote]
Why ,I think it can,we use l'hospital three time (although it is very ugly),I cant post it,write it in $ LaTex$ is too hard :blush:

$ \sqrt {1 \plus{} \tan x} \minus{} \sqrt {1 \plus{} \sin x} \equal{} (\tan(x) \minus{} \sin(x))\frac {1}{2\sqrt {1 \plus{} \xi_x}}$ with $ \xi_x$ between $ \sin(x)$ and $ \tan(x)$ (Mean value theorem).
$ \tan(x) \minus{} \sin(x) \equal{} \frac{\sin(x)}{\cos(x)}(1 \minus{} \cos(x)) \equal{} \frac {\sin^3(x)}{\cos(x)(1 \plus{} \cos(x))}$
$ \lim_{x\to0}\frac {\sqrt {1 \plus{} \tan x} \minus{} \sqrt {1 \plus{} \sin x}}{x^3} \equal{} \lim_{x\to0}\left(\frac {\sin(x)}{x}\right)^3\frac {1}{2\cos(x)(1 \plus{} \cos(x))\sqrt {1 \plus{} \xi_x}} \equal{} \frac14$

Prove [u]without l'Hospital[/u] that $ \alpha \ne 0\ \implies\ \lim_{x\to 0}\frac {\tan^{\alpha}x\minus{}\sin^{\alpha}x}{x^{\alpha \plus{}2}}\equal{}\frac {\alpha}{2}$ .

By proof $ \lim_{x\rightarrow 0}\frac {1 \minus{} (cos(x))^\alpha}{x^2}\equal{}\frac{\alpha}{2}$.it's easier.


geometry
analytic geometry
vector
calculus
integration
absolute value
AMC

Consider the triangle with vertices $ (0,0), (x_1, 2009\minus{}41x_1), (x_2,2009\minus{}41x_2)$. The area is given by the absolute value of $ 1/2[x_1(2009\minus{}41x_2) \minus{} x_2(2009\minus{}41x_1)] \equal{} 1/2[2009(x_1\minus{}x_2)]$. Now it is clear that $ x_1$ and $ x_2$ have to be of the same parity for the area to be integral. $ x_i$ ranges from $ 0$ to $ 49$, so there are $ 25$ odd and $ 25$ even numbers. So the ans is $ 2(25C2) \equal{} 600$.

[quote="aznlord1337"]Consider the triangle with vertices $ (0,0), (x_1, 2009 \minus{} 41x_1), (x_2,2009 \minus{} 41x_2)$. The area is given by the absolute value of $ 1/2[x_1(2009 \minus{} 41x_2) \minus{} x_2(2009 \minus{} 41x_1)] \equal{} 1/2[2009(x_1 \minus{} x_2)]$. Now it is clear that $ x_1$ and $ x_2$ have to be of the same parity for the area to be integral. $ x_i$ ranges from $ 0$ to $ 49$, so there are $ 25$ odd and $ 25$ even numbers. So the ans is $ 2(25C2) \equal{} 600$.[/quote]

wait i don't get why you multiply 2 with $ (25C2)$??

[quote="E^(pi*i)=-1"]Because we can choose two numbers from either of two different sets of $ 25$: the even numbers, or the odd numbers.[/quote]

ahhh... thanks

shoot i missed that :(

[quote="aznlord1337"]Consider the triangle with vertices $ (0,0), (x_1, 2009 \minus{} 41x_1), (x_2,2009 \minus{} 41x_2)$. The area is given by the absolute value of $ 1/2[x_1(2009 \minus{} 41x_2) \minus{} x_2(2009 \minus{} 41x_1)] \equal{} 1/2[2009(x_1 \minus{} x_2)]$. Now it is clear that $ x_1$ and $ x_2$ have to be of the same parity for the area to be integral. $ x_i$ ranges from $ 0$ to $ 49$, so there are $ 25$ odd and $ 25$ even numbers. So the ans is $ 2(25C2) \equal{} 600$.[/quote]

I did $ \binom{50}{2}\minus{}\frac{50*25}{2}$, which is also correct  :)

Is there any possible way to this if you didn't know shoelace? (Other than doing ugly algebra with distance to find height and base which I ended up doing and failed :\.) I read about shoelace but never used it in a problem so I ended up forgetting shoelace when it came to the actual test.

[hide="no shoelace"]Let the x- and y- intercept point be $ X\equal{}(49, 0)$ and $ Y\equal{}(0, 2009)$ respecitvely. WLOG $ P$ lies above $ Q$. We have $ [OPQ]\equal{}[OQY]\minus{}[OPY]\equal{}\frac{x_{Q}\cdot y_{Y}}{2} \minus{} \frac{x_{P}\cdot y_{Y}}{2} \equal{} \frac{2009(x_{Q}\minus{}x_{P})}{2}$. Then just proceed as the shoe-lace solution does. (* $ x_A$ means the x-coordinate of A, etc)
[/hide]

Stupid: I said there were 24 even numbers (didn't count x=0)

[quote="feuxfollets"].......didn't this problem say noninteger coordinates?.....[/quote]
No...
[quote="abacadaea"]Consider the set of all triangles $ OPQ$ where $ O$ is the origin and $ P$ and $ Q$ are distinct points in the plane with [b]nonnegative integer coordinates[/b] $ (x,y)$ such that $ 41x \plus{} y \equal{} 2009$. Find the number of such distinct triangles whose area is a positive integer.[/quote]
"Nonnegative" is there, but not "noninteger".

If you want, you can use vectors and use the cross product to find the area in terms of a point $ a$ and $ b$ where the $ y$ coordinates are given by $ y\equal{}2009\minus{}41x$.

Very similar to post #2 in this thread, but I think it's a bit simpler to understand.

If you consider the triangle with coordinates A(0,0), B(0,2009), C(0,41), it starts with a non-integer area, something and a half. When you pick your 2 points P and Q, say P is the one with the greater Y coordinate, then your area of triangle APQ = triangle ABC - triangle BPA - triangle QCA. Triangle BPA has area 2009*(x-coord of P)/2 and QCA has area 2009*(y-coord of Q)/2.

All your choices of P and Q are either (odd x, even y) or (even x, odd y), and there's 25 of each. You realize you need to pick two that are either both (odd x, even y) or (even x, odd y), and so you get 25*24/2*2 = 600

Triangle OPQ has area of 49*2009/2.
Segment PQ is divided into 49 smaller equal segments with lattice points.
Therefore each of the 49 smaller triangles have an area of 2009/2, and it is obvious that in order to have an integral area you need an even number of triangles, and answer follows to be 600.

Lol. I tried this and failed... I did 25^2 not even realizing that the two points have to be distinct, and when i tried to divide that by 2, bad things happened. This was a nice problem too :).

(So, yes, I'm ashamed and stuff now. Not really.) xD

I should have got this one. Only 2 times 25 choose 2 is 625, not 600...O__O

Terrible Problem in my opinion.
Let $P(\alpha,2009-41\alpha), Q(\beta,2009-41\beta)$ be the two points. Observe $OQ=\sqrt{\beta^2+(2009-41\beta)^2}$ and by the point-line formula, the altitude ito $OQ$ is $\frac{|(2009-41\beta)\alpha-\beta(2009-41\alpha)|}{\sqrt{\beta^2+(2009-41\beta)^2}}$. Thus, after simplifying $$[OPQ]=\frac{2009}{2}|\alpha-\beta|.$$ Now, $2009-41\alpha,\beta \ge 0 \implies \alpha,\beta \in [0,49]$. There are $25$ odds, and $25$ evens, thus we just have $\dbinom{25}{2}*2=\boxed{600}$.

[quote=NuclearFusion]wait so why'd you bump it after almost 10 years then xd[/quote]

Because I like to post solutions to some problems that I solved

Kinda odd.

[hide=Solution]Let $P=(a, 2009-41a)$ and let $Q=(b, 2009-41b)$. Shoelace gives an area of $\frac{2009}{2}(a-b)$, meaning that either $a,b$ are both even or both odd. But $a,b$ can only go from $0$ to $\frac{2009}{41}=49$, making $25$ even numbers and $25$ even numbers. Order doesn't matter since swapping $a,b$ still gives the same coordinates, meaning the answer is $2\binom{25}{2}=25 \cdot 24 = \boxed{600}$.[/hide]

[quote=abacadaea]Consider the set of all triangles $ OPQ$ where $ O$ is the origin and $ P$ and $ Q$ are distinct points in the plane with nonnegative integer coordinates $ (x,y)$ such that $ 41x\plus{}y \equal{} 2009$. Find the number of such distinct triangles whose area is a positive integer.[/quote]

Honestly, this was just a coordinate bash soup problem - the coordinates are so messy. Took me a while to finish it.

[hide = Solution]

The distance from $(0,0)$ to the line is

$$D = \frac{|41(0)+1(0)-2009|}{\sqrt{41^2+1}} = \frac{2009}{29\sqrt{2}}$$

Also, $P$ and $Q$ are distinct points on the line. So, we may say that $P=(x_1, 2009-41x_1)$ and $Q=(x_2, 2009-41x_2)$ for some distinct $x_1, x_2$.

Notice that this height is for the same no matter what points we pick on the line. However, it does matter which points we pick on the line because it determines the base of the triangle. The distance between $P$ and $Q$ is therefore

$$L=\sqrt{(x_2-x_1)^2 + (41x_1-41x_2)^2} = \sqrt{(x_2-x_1)^2+41^2(x_2-x_1)^2} = 29(x_2-x_1)\sqrt{2}$$

Thus we may compute the area of the triangle:

$$[\triangle OPQ] = \frac{1}{2}DL = \frac{1}{2} \cdot  \frac{2009}{29\sqrt{2}} \cdot  (x_2-x_1)29\sqrt{2} = \frac{2009}{2} (x_2-x_1)$$

WLOG $x_2>x_1$. Obviously, $x_2-x_1$ must be even, which means that $x_1$ and $x_2$ must have the same parity. Hence $x_1,x_2$ are distinct integers from the set

$$S_1 = \{1,3,5,7,\cdots,49\}$$

$$S_2 = \{0,2,4,6,\cdots, 48\}$$

Each selection of two integers from one of these sets gives us a unique pair $(x_1,x_2)$, which also determines a unique triangle. Hence, there are $2 \cdot \binom{25}{2} = \boxed{600}$ distinct triangles with a positive integer area.

[/hide]

And yes, I'm definitely not failing AMCs next year - going to actually start prepping for AIME + olympiad. The perfect combination.

I suppose that spamming problems in the Contest Collections webpage is helpful too. Keep the grind going!

Haha I just did this again but I misremembered Shoelace as something else so I thought it would be bashy. So this time my solution involved finding the distance from $O$ to the line $41x+y = 2009$, which is $\frac{2009}{29\sqrt2}$, and taking two points $(a,2009-41a), (b,2009-41b)$ + finding their distance, giving $29\sqrt2 |a-b|$ so $|a-b|$ must be even, giving $2\binom{25}{2}$ ways.


Zdravo!  :)

A zna kdo povedat, koliko je takih celoštevilskih parov (x, y), ki rešijo
enačbo m/x^2+n/y^2=1/p^2 za dane m, n, p?  :read:

Zdej mam že zgruntan...  :play_ball:
V bitvu je zadeva bol enostavna, kot se zdi...


conics
hyperbola
ellipse
geometry unsolved
geometry

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=13948[/img]

 :(

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=13953[/img]

is the same constrution of the ellipse

 :thumbup:


A particular cubic box has edge length two inches. If the box
has no top, what is the external surface area of the box?

$ 5\cdot 2^2\equal{}\boxed{20}$, because it has no top.


limit
function
quadratics
calculus
derivative
trigonometry
calculus computations

$ \lim_{x \rightarrow 0} \frac{\sqrt{1\minus{}e^{\minus{}x^2}}}{x}$

I don't find any application of L'Hopital's rule for this problem convincing. Try to prove the two one-sided limits with other methods. That numerator is about $ \sqrt{x^2}$...

[quote="Icosahedroid"][hide]There is no limit. The function is not continuous at x=0. Using L'Hospitals rule, you can show that the limit x=>0 of the function is equal to the limit x=>0 of the reciprocal of the function. This is true only for 1 and -1. The limit as x approaches 0 from the right is 1 and the limit as x approaches 0 from the left is -1.[/hide][/quote]

What L'Hospital's rule told you is that if $ \lim \frac{f'}{g'}$ EXISTS, then you can say something about $ \lim \frac{f}{g}$. Else it tells you nothing. And of course, jmerry's way is better.

[quote="jmerry"]Try a series expansion.[/quote]

the series expansion of it seems quite complicated?

other easier way to solve it?

i think the answer is 1 .

$ \lim_{x\rightarrow 0}\frac {\sqrt {1 \minus{} e^{ \minus{} x^{2}}}}{x}$ = $ \lim_{x\rightarrow 0}\frac {\sqrt {e^{x^{2}} \minus{} 1}}{x\sqrt {e^{x^2}}}$

as jmerry suggested using series expansion i get it as 1 .

$ \sqrt{x^2}$ is $ |x|$, not $ x$. It matters that this is a two-sided limit.

No, the series expansion isn't that complicated, not if you only derive the information you really need.

$ e^{\minus{}x^2}\equal{}1\minus{}x^2\plus{}O(x^4),$ so

$ 1\minus{}e^{\minus{}x^2}\equal{}x^2\plus{}O(x^4)\equal{}x^2(1\plus{}O(x^2)).$

Hence $ \sqrt{1\minus{}e^{\minus{}x^2}}\equal{}\sqrt{x^2(1\plus{}O(x^2))}\equal{}|x|\sqrt{1\plus{}O(x^2)}$

So $ \frac{\sqrt{1\minus{}e^{\minus{}x^2}}}x\equal{}\frac{|x|}{x}\,\sqrt{1\plus{}O(x^2)}.$

Since $ \sqrt{1\plus{}O(x^2)}\to 1$ as $ x\to 0,$ this is controlled by the factor $ \frac{|x|}x.$ Hence, it tends to $ 1$ as $ x\to0^\plus{}$ and $ \minus{}1$ as $ x\to0^\minus{}.$

In my mind (and in jmerry's mind) this is, by far, the easiest way to solve it. It's easy enough that I saw the answer for  myself without writing anything down. In particular, we get quite directly to why this should be different for $ x>0$ and $ x<0,$ once we see that we essentially have $ \frac{\sqrt{x^2}}x\equal{}\frac{|x|}x.$

The function is (even)/(odd) = (odd), so the limit is either 0 or nonexistent.

The numerator of the square-limit is quadratic, so the square-limit is non-zero, and the original limit does not exist.

[quote="kelvin911"]$ \lim_{x \rightarrow 0} \frac {\sqrt {1 \minus{} e^{ \minus{} x^2}}}{x}$[/quote]

Letting $ f(x)$ denote this function, we have  $ f(x)^2 \equal{} \frac {1 \minus{} e^{\minus{}x^2}}{x^2}$. Because the derivative of $ e^h$ at $ 0$ is $ 1$ , we have $ \frac {1 \minus{} e^{\minus{}h}}{h} \equal{} \frac {e^{\minus{}h}\minus{}e^0}{\minus{}h} \rightarrow 1$ as $ h \rightarrow 0$. So $ \lim_{x\rightarrow0}f(x)^2 \equal{} 1$. Now $ f$ is positive to the right of $ 0$, negative to the left. Therefore the right and left limits of $ f$ at $ 0$ are, respectively, $ 1$ and $ \minus{}1$.

[quote="TZF"]

The numerator of the square-limit is quadratic, so the square-limit is non-zero, and the original limit does not exist.[/quote]

TZF, what is the square limit you are referring to? Thanks.

The limit of the square of the given expression, i.e., $ \lim_{x\to 0}\frac{1\minus{}e^{\minus{}x^2}}{x^2}$.

[quote="kunny"]Agree with you.

At first sight of the problem, I solved in 3 seconds in my head.

Why do they need to use l'Hopital or Taylor expansion for the easy problem?

Anyway, Some people may forget $ \sqrt {x^2} \equal{} |x|.$ :wink:[/quote]

nani?  3 seconds desu ka?

anyway I have another 2 limits.  I manage to L'Hopital the 1st one 4 times to get $ \frac{1}{6}$.  I wanna see if anyone can have a better solution

(1)  $ \lim_{x \rightarrow 0} \frac{\cos(\sin{x})\minus{}\cos{x}}{\sin^4 x}$

(2)  $ \lim_{x\rightarrow 0} \frac{\sin(\tan{x})\minus{}\tan(\sin{x})}{\arcsin(\arctan{x})\minus{}\arctan(\arcsin{x})}$

You don't need to be surprised about getting it in 3 seconds if you are more used to using power series.  :D

That said, it's also better to use power series expansion for the two limits you mentioned instead of using L'Hospital.

e.g., a sketch of your first limit.

$ \sin x \sim  x \minus{} \frac{x^3}{6}$, $ \cos x \sim  1 \minus{} \frac{x^2}{2}$, so
$ \cos (\sin x) \sim  \cos(x \minus{} \frac{x^3}{6}) \sim  1 \minus{} \frac{1}{2} \left(x \minus{} \frac{x^3}{6}\right)^2 \sim  1 \minus{} \frac{1}{2} \left(x^2 \minus{} \frac{1}{3}x^4\right)$
$ \cos(\sin x) \minus{} \cos x \sim  1 \minus{} \frac{1}{2} \left(x^2 \minus{} \frac{1}{3}x^4\right) \minus{} 1 \plus{} \frac{x^2}{2} \equal{} \frac{1}{6}x^4$
$ \frac{\cos(\sin x) \minus{} \cos x}{\sin^4 x} \sim  \frac{\frac{1}{6}x^4}{x^4} \equal{} \frac{1}{6}$

i guess i aint good at power series.  its been like 5 years havent done math

i will try pick it up again

[quote="Soarer"]You don't need to be surprised about getting it in 3 seconds if you are more used to using power series.  :D

That said, it's also better to use power series expansion for the two limits you mentioned instead of using L'Hospital.

e.g., a sketch of your first limit.

$ \sin x \sim x \minus{} \frac {x^3}{6}$, $ \cos x \sim 1 \minus{} \frac {x^2}{2}$, so
$ \cos (\sin x) \sim \cos(x \minus{} \frac {x^3}{6}) \sim 1 \minus{} \frac {1}{2} \left(x \minus{} \frac {x^3}{6}\right)^2 \sim 1 \minus{} \frac {1}{2} \left(x^2 \minus{} \frac {1}{3}x^4\right)$
$ \cos(\sin x) \minus{} \cos x \sim 1 \minus{} \frac {1}{2} \left(x^2 \minus{} \frac {1}{3}x^4\right) \minus{} 1 \plus{} \frac {x^2}{2} \equal{} \frac {1}{6}x^4$
$ \frac {\cos(\sin x) \minus{} \cos x}{\sin^4 x} \sim \frac {\frac {1}{6}x^4}{x^4} \equal{} \frac {1}{6}$[/quote]

so you only consider the first two terms of the power series?

[quote="kelvin911"]
so you only consider the first two terms of the power series?[/quote]

In this problem, yes, since that suffices. (You can continue to write down the remaining terms, but they are of order $ o(x^4)$ as $ x \rightarrow 0$ so they do not contribute to the limit.)

Of course, what I wrote was not too rigorous. (Rigorously I guess one shouldn't leave out the remainder terms, meaning that I should have put all those small o-terms back there) But you get the idea.

[quote="kunny"]I have just found another solution.  :lol:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=321685]Hint[/url][/quote]

Since the problem of [color=blue]Hint[/color] has been almost solved, I show the clue to solve to the limit.
\[ \boxed{\boxed{\frac {1}{1 \plus{} x^2}\leq \frac {\sqrt {1 \minus{} e^{ \minus{} x^2}}}{|x|}\leq 1}}\]


calculus
derivative
function
quadratics
number theory open
number theory

Define the [i]number derivative[/i] $ ' : \mathbb{Z}^{ \plus{} } \cup \{ 0 \} \to \mathbb{Z}^{ \plus{} } \cup \{ 0 \}$ to be the unique function satisfying

$ 0' \equal{} 0$
$ p' \equal{} 1 \forall p \in \mathbb{P}$
$ (mn)' \equal{} m'n \plus{} mn' \forall m, n \in \mathbb{Z}^{ \plus{} } \cup \{ 0 \}$

1)  Find a nontrivial family of infinitely differentiable functions $ f_n(x)$ such that $ f'_n(0) \equal{} f_{n'}(0) \forall n$ and $ f_{mn}(x) \equal{} f_m(x) f_n(x)$, or prove that none exist.

2)  If the above family of functions does not exist, can you find a family of infinitely differentiable functions $ f_n(x)$ such that $ f'_n(0) \equal{} f_{n'}(0)$ and $ f_{mn}(x)$ is expressible in terms of $ f_m(x), f_n(x)$?

[hide="Remark"] I believe I have solved 1), but am currently working on 2) myself.  [/hide]

Nobody?  My work on 1):

Given differentiability, the first condition is equivalent to $ f_n'(x) \equal{} f_{n'}(x)$.  Hence we have $ f'_0(x) \equal{} f_{0'}(x) \equal{} f_0(x) \Leftrightarrow f_0(x) \equal{} C_0 e^x$ for some constant $ C_0$.

Additionally, since $ (m \cdot 1)' \equal{} m' \cdot 1 \plus{} m \cdot 1' \equal{} m'$, we conclude that $ 1' \equal{} 0$, so $ f'_1(x) \equal{} f_{1'}(x) \equal{} f_0(x) \equal{} C_0 e^x \Leftrightarrow f_1(x) \equal{} C_0 e^x \plus{} C_1$ for some other constant $ C_1$.

For a prime $ p$, $ f'_p(x) \equal{} f_{p'}(x) \equal{} f_1(x) \Leftrightarrow f_p(x) \equal{} C_0 e^x \plus{} C_1 x \plus{} C_p$ for some constant $ C_p$.

So much for the first condition.  The second condition allows us to derive a contradiction.  Since

$ f_{n \cdot 0}(x) \equal{} f_n(x) f_0(x) \equal{} f_0(x)$

we conclude that in fact $ f_0(x) \equal{} C_0 \equal{} 0$.  Similarly, since

$ f_{n \cdot 1}(x) \equal{} f_n(x) f_1(x) \equal{} f_n(x)$

we conclude that in fact $ f_1(x) \equal{} C_1 \equal{} 1$.  It follows that $ f_p(x) \equal{} x \plus{} C_p$ for some constant $ C_p$.

However:  consider the following two calculations.

$ f_{9}(x) \equal{} f_3(x)^2 \equal{} (x \plus{} C_3)^2 \equal{} x^2 \plus{} 2C_3 x \plus{} C_3^2$
$ f_9''(x) \equal{} f_6'(x) \equal{} f_5(x) \equal{} x \plus{} C_5 \Leftrightarrow$
$ f_9(x) \equal{} \frac{x^3}{6} \plus{} C_5 \frac{x^2}{2} \plus{} C_{9,1} x \plus{} C_{9,0}$

In one calculation $ f_9(x)$ is quadratic and in the other it is cubic!  Contradiction.

This is my motivation for 2).  Can we replace the condition $ f_{mn}(x) \equal{} f_n f_m(x)$ by a more sensible condition?

A thing that I never understand is why you post the question and why you yourself also replied. If you knew the solution then it means you try to measure the others all, and if not so it means you try to the solution... but why on the online, not on the paper? So you wanna make us measure you? Then I proved in any cases, either you measure us or we measure you; so I proved you and we are in mutual relationship.
More thanks.

He obviously told us that problem because he considers it interesting. And as after one day no solution was given, he showed his work on part 1, e.g. to be checked by us or to give an idea how one could tackle it.

And nobody "measures" anyone...

And don't quote such a bi post for such a short comment, especially if it corresponds to the post above yours.


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You could earn a bit of my respect bye proving that the absolute value of secant(n)/n² for n an integer larger than 1, IS ALWAYS SMALLER THAN 1 !

its seams so simple .... but is it ? be carefull for proving with series expansions since taylor series dont represent the entire function ( radius of convergenge ! )

im sorry if this doesnt fit well in this type of forum because its too hard or not in the books or belongs in an other section of AoPS.

im new here  :oops:

greetings

Wait so you want us to prove that for any integer n>1, the absolute value of (n/n^2)<1? That's fairly obvious isn't it?

[quote="Palytoxin"][quote="AntonioMainenti"][quote="tommy1729"]the absolute value of secant(n)/n²[/quote]\[\left|\frac{\sec{n}}{n^2}\right|\][/quote]

yea but doesn't have to specify the $n$ in secant is [b]radian[/b] or [b]degree[/b]? im very  :?
i've come back to edit and realize this problem is too hard for me[/quote]Oh yeah. He probably meant degrees, so \[\left|\frac{\sec{n^\text{o}}}{n^2}\right|.\] That reminds me, what's the latex for a degree symbol?  :D

edit: yeah, there it is \/

[quote="AntonioMainenti"]Oh yeah. He probably meant degrees, so \[\left|\frac{\sec{n^\text{o}}}{n^2}\right|.\] That reminds me, what's the latex for a degree symbol?  :D[/quote]

i have no idea, i do this 90^\circ which produces $90^\circ$, which is the closest i have gotten.

[quote="NoSoupForYou"]Whenever the measure of an angle does not say if it is in radians or in degrees, it is assumed that it is in radians.[/quote]That's true, but for this problem it seems like that would be strange, but I really have no idea how to solve it so I could be wrong.

I've just started trig so this seems interesting.

Clearly we can see that $n^2 | \cos n | > 1$.  It looks true, but I don't know how to prove it...

yea but $n$ is a positive integer and so $n^2|\cos n|\neq 0$

[quote="NoSoupForYou"]Don't forget that $n^2 | \cos n | =0$ when $n=\frac{\pi}{2}+\pi k$, for all integer $k$.[/quote]But of course $n$ is integral, so it will never quite be equal, but if $\frac{n}{k}$ approximates $2\pi+\frac{\pi}{2}$ for some odd integer $k$, then $\cos{n}$ will be near $0$. If this is always greater than $\frac{1}{n^2}$ then we have what we want.


inequalities
geometry
incenter
inradius
function
trigonometry
geometry proposed

Let $I$ be incenter of triangle $ABC$ with inradius $r$ prove that
$IB.IC+IC.IA+IA.IB\ge 12r^{2}$

First off, let's find $IA$ in function of $a$, $b$, $c$. We know that the bisector $AA'$ satisfies \[AA'^{2}=\frac{bc(b+c+a)(b+c-a)}{(b+c)^{2}}\] In addition, \[\frac{AI}{AA'}=\frac{b+c}{a+b+c}\] Hence \[IA=\sqrt{\frac{bc(b+c-a)}{a+b+c}}\] So \[\sum_{cyc}IA \cdot IB=\frac{1}{a+b+c}\sum_{cyc}c\sqrt{ab(b+c-a)(c+a-b)}\geq 12r^{2}\] Now we use Ravi's substitution $a=y+z$, $b=z+x$, $c=x+y$. Taking into account that \[r^{2}=\frac{(s-a)(s-b)(s-c)}{s}\] where $s$ is the semiperimeter of the triangle, we get the following equivalent algebraic inequality: \[\sum_{cyc}(y+z)\sqrt{yz(x+y)(x+z)}\geq 12xyz\] Using Jensen's inequality with the function $f(x)=1/\sqrt{x}$, which is convex in $(0,\, \infty)$ we obtain \[\mbox{lhs}=\sum_{cyc}(y+z)f\left(\frac{1}{yz(x+y)(x+z)}\right) \geq 2(x+y+z)f\left(\frac{S}{2(x+y+z)}\right)\] where \[S=\sum_{cyc}\frac{y+z}{yz(x+y)(x+z)}=\frac{x(y+z)^{2}+y(z+x)^{2}+z(x+y)^{2}}{xyz(x+y)(y+z)(z+x)}\] Hence \[f\left(\frac{S}{2(x+y+z)}\right)=\sqrt{\frac{2xyz(x+y)(y+z)(z+x)(x+y+z)}{x(y+z)^{2}+y(z+x)^{2}+z(x+y)^{2}}}\] So it suffices to show that \[2(x+y+z)\sqrt{\frac{2xyz(x+y)(y+z)(z+x)(x+y+z)}{x(y+z)^{2}+y(z+x)^{2}+z(x+y)^{2}}}\geq 12xyz\] Squaring and rearranging we get \[(x+y)(y+z)(z+x)(x+y+z)^{3}\geq 18xyz[x(y+z)^{2}+y(z+x)^{2}+z(x+y)^{2}]\] I'm not that good at inequalities :roll: So the only idea I came up with was multiplying out everything! :D After some minutes of brute-force algebra we see that this is equivalent to \[\sum_{sym}(4x^{4}yz+x^{3}y^{2}z+x^{5}y+4x^{4}y^{2}+3x^{3}y^{3}-13x^{2}y^{2}z^{2}) \geq 0\] And finally, noting that $4+1+1+4+3=13$, clearly the result follows by Muirhead's theorem :P

Although I guess there is a better solution to the algebraic inequality, I had a lot of fun solving this nice problem :wink:

thank Jutaro for your algebraic solution,it can be solved by trigonometric method:
note that $\frac{r}{IA}=\sin\frac{A}{2}$ thus we need prove that
$\sum\frac{1}{\sin\frac{B}{2}\sin\frac{C}{2}}\ge 12$
but $\frac{9}{4}\ge (\sum\sin\frac{A}{2})^{2}\ge 3(\sum\sin\frac{B}{2}\sin\frac{C}{2})\Rightarrow \sum\sin\frac{B}{2}\sin\frac{C}{2}\le \frac{3}{4}$
and from
$(\sum\sin\frac{B}{2}\sin\frac{C}{2})(\sum\frac{1}{\sin\frac{B}{2}\sin\frac{C}{2}})\ge 9$ we have the solution.