function algorithm calculus derivative Euler inequalities algebra

Determine the exact value of

$.5^{.5^{\ldots}}$

[hide="partial solution"]
Let this power be $x$. Then $.5^x=x\Rightarrow x^\frac 1x =.5$.

Let $t=\frac 1x$ and take reciprocal of both sides to get $t^t=2$.

This solution is uniquely determined because if $t$ were less than 1, then $t^t<t^0=1<2$. When $t\geq 1$, $t^t$ is increasing because if $u>v\geq 1$, then $u^u>u^v>v^v$, which makes $t^t$ one-to-one for $t>1$. I have no idea how to get an exact value for $t$, though.[/hide]

That's good, that is the exact result I got. However, I don't know how to analytic solve $t^{t}=2$. Either the solution is really elegant, or requires methods we havn't learned yet.

What about utilizing limits? I tried using a limit of the $ln$ of the the function $x^x$ and then tried to use L'Hospital's rule, but I couldn't because the conditions required to use it were not satisfied.

Well you can find that it is approximatly 1.5596105... by trying numbers to see which are the closest
I am not sure how you would find an exact solution or if that is even possible.

Let $x = \frac{1}{2}^{\frac{1}{2}^{\frac{1}{2}^{...}}}$
Since the repeating exponent is between $0$ and *** $\sqrt{2}$ ***, it converges.
By replacement, $x = \frac{1}{2}^x$ and $\frac{1}{x}=2^x$

This is approximately $\boxed{0.641185744505}$ after several recursions of Newton's recursion.

*** [b][u][i]EDIT[/i][/u][/b]***: The correct value is actually $e^{1/e}$

So you used Newton's method to approximate the zero of $2^{x}-x$. Let me try to remember that algorithm.

You start by picking a value close to the root, then using a tangent line approximation for the function. Finding the root of this tangent line gives you a new value of x, which should be closer to the root. Continued indefinitely, this algorithm gives you the answer to a desired accuracy.

I know there is a formula for the accuracy right? Also, there is a relation between the derivatives that tells you whether picking a certain x value will cause the algorithm to fail. Can someone post them and does anyone know how to prove them?

Also, didn't Euler develop a similar procedure named Euler's Method. I havn't studied before, so if anyone knows about it, post what you know.

we want to solve the equation $k=x^{x^{x^{...}}}$, the large part of this is that you have to show the sequence is convergent, stated earlier, $(0,\sqrt{2}]$ is not the correct range, can you find and prove the correct one?

I think this is right. Correct me if you see any mistakes.
[hide]The range is $(1/2,1)$

Proof: First we will show that $.5^{.5^{...}}<1$. First note that this is equivalent to $2^{.5^{.5^{...}}}>1$ which is true since $.5^{.5^{...}}>0$ and $2^{k}>1$ if $k>0$.
Now to show that $.5^{.5^{...}}>.5$. Let $x=.5^{.5^{...}}$. We know from previous results that $0<x<1$ and it is easy to see that $x=.5^{x}$. To continue, we first show that $f(x)=.5^{x}$ is a decreasing function. We know that $f'(x)=.5^{x}\ln.5$. $ln.5<0$ since $.5<1$ and $.5^{x}>0$ for all x, so $f'(x)<0$ for all x.
Thus we know that, on the interval $(0,1)$, $f$ attains a maximum value at $x=0$ and a minimum at $x=1$. Then $.5^{1}=.5 < .5^{x}=x < .5^{0}=1$[/hide]

I understand that the proof is a bit tedious at times, but I'm trying to get accustomed to rigour.

You can show that $.5^{.5\cdots}<1$ much easier. Simply, if $k<1$, $k^n<1 \ \ \forall n>0$

Continuously raising the $n^{1/n}$ to it's own power converges by induction.

$(n^{1/n})^{n^{1/n}}<(n^{1/n})^n=n$, so the result is less than $n$.
Let the result be $k1$

$(n^{1/n})^{k1}<(n^{1/n})^n$, so the result is less than $n$.
Call this $k2$

$(n^{1/n})^{k2}<(n^{1/n})^n$, so the result is again less than $n$.
etc.

For all $n>1$:

$(n^{1/n})^{n^{1/n}}< (n^{1/n})^{kn} < (n^{1/n})^{k(n+1)} < (n^{1/n})^n=n$

So everything up to and including the $n^{1/n}$ definately converges.

Every number that can be expressed as $n^{1/n}$ converges.

Maximizing $n^{1/n}$, we get $n = e$, so the maximum value of $n$ for which $n^{n^{n^{n^{n^{n^{n...}}}}}}$ converges is $e^{1/e}$, which converges to $e$.

The range of convergence is $(0, e^{1/e}]$

Your interval of convergence is *very* close to the correct one Zuton, but still not entirely correct. The bottom endpoint is not right.

The error is in the inequalities leading up to the statement: "Every number that can be expressed as $n^{1/n}$ converges."

For values of n that are too small this is not true. Try some numerical examples to see why.

Finding the bottom endpoint of convergence is pretty difficult.

By range of congvergence, I meant values of $n$, not what the tower actually converges too. I managed to figure out the upper bound of the tower to be $e$ where each term is $e^{1/e}$ but am stumped as to how to prove the lower bound. Fooling around with some numbers is suggesting to me that the range of the power-tower is $(1/e,e]$ but I have no idea how to prove this

I don't think it's possible to find the exact value of the number, unfortunately...  Here's more digits tho: http://www.research.att.com/~njas/sequences/A104748 .

I know what you meant Zuton. We are talking about the same thing. We should just call it the domain the the tower function.

In that regard, the range that you found is basically correct. (Actually it's $[1/e, e]$, the lower bound is attainable). And now that you've found the correct range, and you understand where the values come from, you can probably guess what the correct domain is!

[hide]$[1/e^e, e^{1/e}]$[/hide]

Oh! That makes sense now  :) Thanks gauss

[quote="gauss202"]and you understand where the values come from, you can probably guess what the correct domain is!
[/quote]

But how the values come from , I don't understand  , could you say it clearly

The Zuton Force's post a few posts up explains it.  But if it is necessary:

[hide="Another discussion"] Suppose $y = x^{x^{x^{...}}}$.  Then $y = x^{y}\implies x = y^{ \frac{1}{y}}$.  So the domain is all possible values of $y^{ \frac{1}{y}}$.

Using some calculus, it turns out that we have a maximum at $y = e$, giving $x = e^{ \frac{1}{e}}$ (also the corresponding maximum) and a minimum at $y = \frac{1}{e}$, giving $x = \frac{1}{e^{e}}$.

It also turns out that for $x > e^{ \frac{1}{e}}$ the power tower diverges to infinity, and for $x < \frac{1}{e^{e}}$ the power tower oscillates.  [/hide]
Edit:  And as for the original problem...
[hide="An expression"] $t^{t}= 2 \implies$
$t \ln t = \ln 2 \implies$
$\ln t = W(\ln 2) \implies$
$t = \boxed{ e^{W (\ln 2)}}$

Where $W(x)$ is the [url=http://mathworld.wolfram.com/LambertW-Function.html]Lambert W-Function[/url].  [/hide]
Edit:  Oh, and one more thing...
[hide="It turns out that..."] $x^{x^{x^{...}}}=-\frac{W(-\ln x)}{\ln x}$. [/hide]

But I think the domain of $x$ is $(0, e^{1/e}]$ , because the domain of $y^{ \frac{1}{y}}$ is $(0, e^{1/e}]$

How it could be the $[1/e^{e}, e^{1/e}]$ ? :huh:

That point has already been discussed.  The power tower does not converge for $0 < x < \frac{1}{e^{e}}$.

That is to say, the sequence

$x_{1}= x$
$x_{k+1}= x^{x_{k}}$

Does not converge when $x < \frac{1}{e^{e}}$.  Rather, it oscillates between two values.  (Verify this experimentally using some small $x$, say, $x = 0.001$.)

Has that point already been discussed ? but I can't find it above all of these posts
they haven't said how the lower bound of $x$ come from , Help me ! Please ! I desire to know .

And one more thing , It seem that when $x = \frac{1}{e^{e}}$,the power tower oscillates
[img]http://img011.photo.21cn.com/photos/user/24454028/20070406201320/o/8424081.jpg[/img]

Rounding errors will cause the calculation to oscillate.

t0rajir0u , Would you please tell me what I really want to know , I just a boy of High School , Of course what I have learnt is few than you

You have said the domain of $x$ is all possible values of $y^{ \frac{1}{y}}$.but I figured out the domain of $y^{ \frac{1}{y}}$ is $(0, e^{1/e}]$
Of course I know I'm wrong , but I don't know why , I just want you to give me the process

The domain of $x^{1/x}$ contains the numbers such that the power tower will not increase to infinity.

For example, if we pick something higher than this, such as $5$, and do $5^{5^{5^{5^{5^{5^{5^{5^{\ldots}}}}}}}}$, this infinite tower is infinitely large.

However, for positive numbers that are very very close to zero, the tower will not go to infinity, but will not settle on a single number - it will oscillate between two numbers. The numbers that oscillate have to be included from our domain also.

Here's the thing with that though:

You had to assume that the power tower converges to a single number to make that substitution. Once you made that, you came out with a single number, but only because you had to assume that to make the substitution in the first place. You end up with circular reasoning.

Here's a list of the first few iterations of the power tower (left column is iteration number, right column is value).

It seem that the power tower is converges from the two sides , are you sure it will not converges to $0.3502248527432$ ?

No, it settles out between oscillating between $0.13735939573795616 \ldots$ and $0.6626608389005485 \ldots$

OK I know for $x > e^{ \frac{1}{e}}$ the power tower diverges to infinity, and for $x < \frac{1}{e^{e}}$ the power tower oscillates
But I don't understand how to figure out the lower bound of $x=\frac{1}{e^{e}}$ , I have no idea how to prove it ?
And you have not prove it in this topic yet

I want discuss another thing but related this topic.  What can we say about $ i^{i^{i^...}}$ . $ [i\equal{} \sqrt{\minus{}1}]$  I know it's a komplex number. I wonder can we found this complex number by calculus or another elemantary  methods?

The answer is $ \frac {2 i}{\pi} \textbf{W} \left( \minus{} \frac {\pi i}{2} \right) \approx 0.438283 \plus{} 0.360592 i$, where $ \mathbf{W} (x)$ is the Lambert function.

This answer can be derived fairly easily using the definition of the Lambert function, given that the sequence converges.

The real problem is showing that the sequence $ I_n \equal{} \underbrace{i^{i^{i ^{i \ldots}}}}_{\text{n times}}$ converges.

Defining the full complex exponential is a somewhat thorny issue.  We take the stance that $ a^b \equal{} e^{a \ln b}$ and therefore write

$ i^i \equal{} e^{i \ln i}$;

in other words, defining the complex exponential for arbitrary bases requires a consistent definition of the complex logarithm (of $ i$), so the answer depends on what branch you pick.

Why should that affect convergence if the logarithm is anyway undone?  :maybe: If you write the infinite tetration as the limit of a sequence of finite tetrations, in no term do you end by taking the logarithm... perhaps I'm missing something obvious.

I mean that on complex numbers the exponential is unambiguous; it is defined by a power series that converges everywhere.  One therefore computes $ e^{a \ln b}$ by first computing $ a \ln b$ and then taking its exponential; the logarithm is not automatically "undone" in a complex setting.  In other words,

$ i^i \equal{} \sum_{k \equal{} 0}^{\infty} \frac {(i \ln i)^k}{k!}$,

a sum that cannot be evaluated without evaluating $ i \ln i$.

Edit:  I should be more specific.  We know that $ e^{i \pi (1 \plus{} 2k)} \equal{} i$; depending on which branch of the logarithm we pick, we can therefore write $ \ln i \equal{} i \pi (1 \plus{} 2k)$ for any integer $ k$ (in some loose sense).  It then follows that $ i \ln i \equal{} \minus{} \pi (1 \plus{} 2k)$ and that

$ i^i \equal{} e^{i \ln i} \equal{} e^{\minus{} \pi (1 \plus{} 2k) }$

for any integer $ k$.  (Interestingly enough, this quantity is purely real!)  Obviously the choice of $ k$ is very important in determining what value the eventual exponential tower will have.  One usually chooses $ k \equal{} 0$, but of course even once the choice has been made it is not at all obvious (at least to me) that this sequence converges; for example, we now have

$ i^{i^i} \equal{} e^{e^{\minus{}\pi} \ln i} \equal{} e^{e^{\minus{}\pi} (i \pi)} \equal{} \cos (e^{\minus{}\pi} \pi) \plus{} i \sin (e^{\minus{}\pi} \pi)$.

How to prove $ x<e^{\minus{}e}$ then we get $ x^{x^{x^{.^{.^{.}}}}}$ oscillate 2 values ???

I think this can be done in this way
Let,
$ .5^{.5}...$ be x
So,
x=$ .5^x$
Solving using a scientific calculator,
we get,
x=.6411857445(approx.)

when I solved the original equation I got $ x \equal{} \frac {W(\ln 2)}{\ln 2}$, which is as far as I can tell, the reciprocal of your $ t \equal{} e^{W(\ln 2)}$.  But why does that happen?  That is, how can we show that $ \frac {W(\ln 2)}{\ln 2} \equal{} e^{ \minus{} W(\ln 2)}$ without reverting to the original problem?

EDIT: nevermind, I see it now... rearranging we have the definition of $ W(\ln 2)$...

Hi t0rajir0u.

You say that "It also turns out that for $ x>e^{\frac{1}{e}}$ the power tower diverges to infinity, and for $ x<\frac{1}{e^{e}}$ the power tower oscillates."

How can we prove this?

There's something wrong with my computer: now I can see more of this post. I was referring to your post of Wednesday April 04, 2007. I get the maximum at $ y\equal{}e$ but can't see how you get the minimum at  $ y\equal{}1/e$.

Should this be in a different forum?

It seems at first sight that the power tower diverges to infinity for $ x>1$; for $ x\equal{}1$ it is the constant sequence $ 1$ and that for $ 0<x<1$ it is an increasing sequence oof numbers converging to $ 1$. What's wrong?

(Would this get some response if it were in a different forum?)

I don't really know what to say because I don't really know why you're making the assumptions that you are.

If an infinite tetration converges, say $ y \equal{} x^{\ldots}$, then by substitution, you learn that $ (x,y)$ must satisfy $ y \equal{} x^y$, or equivalently $ x \equal{} y^{1/y}$. Note that the infinite tetration is the [i]definition[/i] of $ y$, and that the equations derived from it are only necessary conditions on $ (x,y)$ [i]given that the tetration converges in the first place[/i].

With calculus, we can find that the maximum value of $ x \equal{} y^{1/y}$ is $ e^{1/e}$, achieved when $ y \equal{} e$. From our [i]definition[/i] of $ y$, we require that for $ x > 1$, $ y$ increases with $ x$. We know that the maximum value of $ y$ is $ e$, and thus the maximum value of $ x$ is $ e^{1/e}$ (if x were greater, y would have to be greater, which is not possible).

The deal with the lower bound is entirely different. We can solve $ y \equal{} x^y$ on this interval just fine -- for example, $ y \equal{} 1/10$ and $ x \equal{} 1/10^{10}$ solves the equation. Unfortunately, the sequence is moving [i]away from[/i] the value of $ y$ obtained from solving this equation. And you're right, this is a point that hasn't been dealt with rigorously at all in this topic. I still don't have a convincing explanation for this, apart from just looking at the sequences numerically.

Thanks, TZF - and for teaching me a new word: I'd never heard of tetration before.

Maybe this article by Louis Talman will answer some of the questions people are having about the lower bound on the interval of convergence.


Let $ a,b,c\ge 0$ such that $ a\plus{}b\plus{}c\equal{}3$. Find the maximum value of:
\[ a^2b\plus{}b^2c\plus{}c^2a\]

I don't think so because it was appeared in hungkhtn's book before (in hungkhtn's book, I don't see any proof for $ k \equal{} 2$ but having a proof by Lagrange's multiplier for general case:
\[ a^kb \plus{} b^kc \plus{} c^ka
\]
Anyway, I don't care general case because I haven't found any solution without using Lagrange's multiplier as hungkhtn did. For k=2 my proof is strange (I think so  :P )
Okie, I will post my solution and hoping someone will give some new proof
$ \bullet$ We have:
\[ a^2b \plus{} b^2c \plus{} c^2a\le 4 \Leftrightarrow 4(a \plus{} b \plus{} c)^3\ge 27(a^2b \plus{} b^2c \plus{} c^2a)\]

$ \bullet$ Put  $ \frac {b \plus{} c}{a} \equal{} x,\frac {b \minus{} c}{a} \equal{} y$ then:

\[ a^2b \plus{} b^2c \plus{} c^2a\le 4\]
\[ \Leftrightarrow 4(a \plus{} b \plus{} c)^3\ge 27(a^2b \plus{} b^2c \plus{} c^2a)\]
\[ \Leftrightarrow 27x \plus{} 9\left(x \minus{} 1\right)^2 \plus{} \left(y \plus{} 1\right)^2\left(\frac {5x \plus{} 3y}{2} \minus{} 5\right)\ge 0\]
which is alway true if we assume that $ a \equal{} min(a,b,c)$.  We are done  :)

[quote="zaizai-hoang"]Let $ a,b,c\ge 0$ such that $ a \plus{} b \plus{} c \equal{} 3$. Find the maximum value of:
\[ a^2b \plus{} b^2c \plus{} c^2a
\]
[/quote]
The stronger Inequality is
\[ a^2b \plus{} b^2c \plus{} c^2a \plus{} abc \plus{} \frac {1}{2}abc(3 \minus{} ab \minus{} bc \minus{} ca) \le 4
\]
:)

purpose   that :$ a\leq  b \leq c$.   then   we have :
 $ a(b\minus{}a)(b\minus{}c)  \leq 0$  hence :
$ \sum ab^2  \leq b(a\plus{}c)^2 \leq 4.$
equality hold when  and only when :$ a\equal{}0,b\equal{}1,c\equal{}2$

[quote="nkht-tk14"][i] Bach Ngọc Thanh Cong is the author of this problem but his proof
       is very hard
  zaizai Hoàng can you post your solution[/i] :lol:[/quote]
I don't think so,this problem is very old and is Canada Olympiad problem (1998?)

[quote="Hong Quy"]purpose   that :$ a\leq b \leq c$.   then   we have :
 $ a(b \minus{} a)(b \minus{} c) \leq 0$  hence :
$ \sum ab^2 \leq b(a \plus{} c)^2 \leq 4.$
equality hold when  and only when :$ a \equal{} 0,b \equal{} 1,c \equal{} 2$[/quote]

I think you can't purpose   that :$ a\leq b \leq c$ because this isn't symmetric inequality ;)
We only can purpose that b is between a and c then
$ c(a \minus{} b)(b \minus{} c)\ge0$  => $ a^2b \plus{} b^2c \plus{} c^2a \le a^2b \plus{} bc^2 \plus{} abc \equal{} b(a \plus{} c)^2 \minus{} abc \le 4$ (AM-GM)

(We proved stronger problem: $ a^2b \plus{} b^2c \plus{} c^2a\plus{}abc\le 4$ )

*We have got a nother proof for that problem,very simple:
Purpose $ a\equal{}max(a,b,c)$ we have

$ P\equal{}a^2b\plus{}b^2c\plus{}\frac{c^2a}2\plus{}\frac{c^2a}2\le a^2b\plus{}abc\plus{}\frac{c^2a}2\plus{}\frac{ca^2}2\equal{}a(a\plus{}c)(b\plus{}\frac c{2})$
$ \equal{}4.\frac a{2}.\frac{a\plus{}c}2.(b\plus{}\frac c{2})\le4$ (AM-GM)

so we are done! ;)

[quote="dduclam"][quote="nkht-tk14"][i] Bach Ngọc Thanh Cong is the author of this problem but his proof
       is very hard
  zaizai Hoàng can you post your solution[/i] :lol:[/quote]
I don't think so,this problem is very old and is Canada Olympiad problem (1998?)

[quote="Hong Quy"]purpose   that :$ a\leq b \leq c$.   then   we have :
 $ a(b \minus{} a)(b \minus{} c) \leq 0$  hence :
$ \sum ab^2 \leq b(a \plus{} c)^2 \leq 4.$
equality hold when  and only when :$ a \equal{} 0,b \equal{} 1,c \equal{} 2$[/quote]

I think you can't purpose   that :$ a\leq b \leq c$ because this isn't symmetric inequality ;)
We only can purpose that b is between a and c then
$ c(a \minus{} b)(b \minus{} c)\ge0$  => $ a^2b \plus{} b^2c \plus{} c^2a \le a^2b \plus{} bc^2 \plus{} abc \equal{} b(a \plus{} c)^2 \minus{} abc \le 4$ (AM-GM)

(We proved stronger problem: $ a^2b \plus{} b^2c \plus{} c^2a \plus{} abc\le 4$ )

*We have got a nother proof for that problem,very simple:
Purpose $ a \equal{} max(a,b,c)$ we have

$ P \equal{} a^2b \plus{} b^2c \plus{} \frac {c^2a}2 \plus{} \frac {c^2a}2\le a^2b \plus{} abc \plus{} \frac {c^2a}2 \plus{} \frac {ca^2}2 \equal{} a(a \plus{} c)(b \plus{} \frac c{2})$
$ \equal{} 4.\frac a{2}.\frac {a \plus{} c}2.(b \plus{} \frac c{2})\le4$ (AM-GM)

so we are done! ;)[/quote] sory   i write english badly  hence i forget  let b is between a and c .
  and   my solution has a little not exactly

All above solutions are extracly  :D Sorry for above inequality because it is very easy to prove. Let try for another one which is harder than above.
$ \bigstar$ Let $ a,b,c\ge 0$ such that $ a\plus{}b\plus{}c\equal{}3$. Find the maximum value of:
\[ V\equal{}a^3b\plus{}b^3c\plus{}c^3a\]

[quote="can_hang2007"]
The stronger Inequality is
\[ a^2b \plus{} b^2c \plus{} c^2a \plus{} abc \plus{} \frac {1}{2}abc(3 \minus{} ab \minus{} bc \minus{} ca) \le 4
\]
:)[/quote]

Hard for me  :blush:

[quote="zaizai-hoang"]All above solutions are extracly  :D Sorry for above inequality because it is very easy to prove. Let try for another one which is harder than above.
$ \bigstar$ Let $ a,b,c\ge 0$ such that $ a \plus{} b \plus{} c \equal{} 3$. Find the maximum value of:
\[ V \equal{} a^3b \plus{} b^3c \plus{} c^3a
\]
[/quote]
Hoang,this problem is very old and easy,too.

Purpose $ a \equal{} max(a,b,c)$ we have

$ P \equal{} a^3b \plus{} b^3c \plus{} \frac {c^3a}2 \plus{} \frac {c^3a}2\le a^3b \plus{} a^2bc \plus{} \frac {c^2a^2}2 \plus{} \frac {ca^3}2 \equal{} a^2(a \plus{} c)(b \plus{} \frac c{2})$
$ \le a^2(a \plus{} c)(b \plus{} \frac {2c}3)\equal{} 3^3.\frac a{3}.\frac a{3}.\frac {a \plus{} c}3.(b \plus{} \frac {2c}{3})\le 3^3.(\frac{3}{4})^4$ (AM-GM)

The general problem: $ a^nb\plus{}b^nc\plus{}c^na \le 3^{n\plus{}1}\frac {n^n} {(n\plus{}1)^{n\plus{}1}}$ (also old and easy,and i also post in this forum long times ago ) be proved same too.  :)


[color=blue][size=150]Rumania[/size][/color]

Encontrar todas las funciones continuas que:

$f: \mathbb{R} \longrightarrow \lceil0; \infty \rangle$, tal que para todo $x,y \in \mathbb{R}$,
se cumple que:
$f(x^2 +y^2) = f(x^2 - y^2) + f(2xy).$

(Sugerencia)

[hide="Hint"]Mostremos que $f(t) = f(-t)$ para todo $t \in \mathbb{R}.$
Definamos $g(x) = f(\sqrt{x}),$ y mostremos que $g(a+b) = g(a) + g(b),$ para todo $a,b \geq 0.$[/hide]

[hide="English Version"]
Find all continuous functions, definited as :
$f: \mathbb{R} \longrightarrow \lceil0; \infty \rangle$, for all $x,y \in \mathbb{R}$,
such that:
$f(x^2 +y^2) = f(x^2 - y^2) + f(2xy).$[/hide]

Carlos Bravo :)
Lima -PERU

[color=blue][size=150]Solucion:[/size][/color]

1) En primer lugar, calculemos $f(0)$, para ello, tomemos: $x=0, y=0$
entonces:               $f(0)=0$
luego, tomemos: $x=0$, de aqui:
                 $f(y^2) = f(-y^2)$, para $y^2 \in \mathbb{R}^+$

Tomemos $a,b>0$
tal que: $x^2 - y^2 = a$ y $2xy=b$, tendriamos en la expresion inicial:
  $f(a) + f(b) = f(\sqrt{a^2 + b^2}) \cdots (\alpha)$
 Definimos una funcion $g$ tal que $g(x)=f(\sqrt{x}),$
de cuerdo a esto, tendriamos en $(\alpha)$

$g(a + b) = g(a) + g(b),$ para todo $a,b \geq 0.$

Notemos q $g$ es continua, lo que generaria una funcion multiplo $g(a) = ka,$
por tanto:

                                                       $f(x)=kx^2.$

Carlos Bravo :)
Lima- PERU


geometry
geometric transformation
vector
geometry unsolved

In an equilateral triangle ABC,suppose P is a point on AB,R is a point on BC and Q,S are points on AC such that PA+AQ=RC+CS=AC. Prove that PS ,QR intersect at an angle of 60 degree

See that AP = CQ and AS = CR, therefore the triangles APS and CQR are equal. As a translation of vector AC followed by a 60 degs clockwise will map the triangle APS over CQR, hence the angle between PS and QR is 60 degs.

Best regards,
sunken rock

Here s a slightly different way of putting things.
As per the problem,
$ PA \plus{} AQ \equal{} AC \equal{} AQ \plus{} QC \Rightarrow PA \equal{} CQ$
Similarly $ RC \equal{} AS$.
Hence by $ SAS$ congruence criterion, $ \triangle APS \cong \triangle CQR$ forcing $ \angle APS \equal{} \angle CQR$ and $ \angle ASP \equal{} \angle CRQ$.
Hence, comparing the angles in the little triangle $ QTS$  and $ CQR$ ; we are forced to conclude that $ \angle QTS \equal{} 60$ where I have assumed that $ T$ is the point of intersection of $ QR$ and $ PS$.
Best.


function
integration
real analysis
real analysis unsolved

Is there any solution to the following differential equation : $f''.f=\lambda$ ??
where $\lambda$ is a real constant number

What do you mean? Of course, by general existence theorems there is a local solution to any initial value problem $f(x_0)=a\ne 0$, $f'(x_0)=b$. If, say, $\lambda>0$ and $b=0$, the solution is even defined on the entire real line. On the other hand, if you ask whether there is a nice simple formula for $f(x)$, then, if I don't mistake, the answer is No.

$f$ function of variable $x$

If  the contant C  is zero the equation is $f''.f=0$
$f$ is identically zero function or $f(x)=ax+b$

Suppose now $C \neq 0$

(1) $y''.y-C =0$ is a incomplete equation

change variable  $z=y'$ gives $y''=z\frac{dz}{dy}$

(1)  becomes $z\frac{dz}{dy}y=C$ or

this equation is separable variable

$z.dz=\frac{C}{y}dy$

we have to work with

(2) $z^{2}=2C\ln(|y|)+k$ where k is real constant

(2bis) $y'^{2}=2C\ln(|y|)+k$ again an incomplete equation of order one.

We work on the appropriate interval where every thing is defined(should see where Cauchy-Lischitz theorem is valid here)

You should study constant solution of (2bis)

let $y=a(s)=s$ and $y'=b(s)=\epsilon\sqrt{2C\ln(|s|)+k}$

$\frac{dx}{ds}=\frac{dx}{dy}\frac{dy}{ds}=\frac{1}{\epsilon\sqrt{2C\ln(|s|)+k}}$

this gives a parametric representation of the solution

 $x(s)=\int_{a}^{s}\frac{1}{\epsilon\sqrt{2C\ln(|w|)+k}}dw+x(0)$
$y(s)=s$

If you want to see the curve solution  it is an exercice of parametric curve
I let someone to study and post the curve  on the forum.


algebra
polynomial
Vieta
LaTeX
Online Math Open
inequalities

[quote="mostel"]Σας παραθέτω τα θέματα του 67ου Πανελλήνιου Μαθηματικού Διαγωνισμού στα Μαθηματικά "ΕΥΚΛΕΙΔΗΣ" για τη Β' Λύκειου.

1) Δίνεται ότι το πολυώνυμο $P(x)=x^{3}+\kappa x+\lambda$ με $\kappa, \lambda\in\mathbf{R}$ έχει τις πραγματικές ρίζες $x_{1}, x_{2}, x_{3}$ που ανά δύο είναι διαφορετικές μεταξύ τους. Να εκφράσετε την παράσταση:

$A=(1-x_{1}^{2})(1-x_{2}^{2})(1-x_{3}^{2})$ συναρτήσει των $\kappa, \lambda$.

2) Θεωρούμε τόξο $AB=90^\circ$ και προεκτείνουμε τη χορδή $AB$ κατά ευθύγραμμο τμήμα $BC=AB$. Ονομάζουμε $D$ το σημείο επαφής της εφαπτομένης του τόξου $%Error. "curvearrow" is a bad command.
{AB}$ από το $C$ και $K$ το ίχνος της κάθετης από το $A$ προς τη $BD$. Να αποδείξετε ότι $KB=2KA$.

3) Αν $a,b$ θετικοί ακέραιοι, να αποδείξετε ότι:

$\sqrt[3]{\left(\frac{a^{4}+b^{4}}{a+b}\right)^{a+b}}\geq a^{a}b^{b}$.

4) Δίνεται τρίγωνο $ABC$. Από σημείο $M$ της πλευράς $BC$ φέρουμε παράλληλες προς τις $AC$ και $AB$ που τέμνουν τις $AB$ Και $AC$ στα $K$ και $\Lambda$, αντίστοιχα. Αν είναι $MK=x$, $M\Lambda=y$, να βρείτε το ελάχιστο της παράστασης:

$S=x^{2}+y^{2}$

και τη θέση του $M$ για την οποία λαμβάνεται αυτό.

[u]Διευκρίνιση[/u]: Το 1ο θέμα ήταν ίδιο με αυτό του Ευκλείδη της 3ης Λυκείου.[/quote]

[b] Αφού εδώ υπάρχουν τα θέματα , ας συνεχίσουμε σε αυτό το topic  τα σχόλια και τις λύσεις.[/b]
 Μπάμπης

Ναι Αλέξανδρε, όντως. Ήταν μία από αυτές που έλυσα  :wink:

Gia tin 3h askisi:

Apo tin Anisotita Arithmitikou-Gewmetrikou mesou exoume : (Oi a,b einai thetikoi akeraioi)

$\frac{\overbrace{a^{3}+a^{3}+\cdots+a^{3}}^{ a \-\ fores}+\overbrace{b^{3}+b^{3}+\cdots b^{3}}^{b \-\ fores}}{a+b}\geq \sqrt[a+b]{\underbrace{a^{3}\cdots a^{3}}_{ a \-\ fores}\underbrace{b^{3}\cdots b^{3}}_{b \-\ fores}}=\sqrt[a+b]{a^{3a}b^{3b}}$

Ara $\left(\frac{a^{4}+b^{4}}{a+b}\right)^{a+b}\geq a^{3a}b^{3b}$ kai kataligoume sto zitoumeno.

[color=red]EDIT: [/color]

ASKHSH 1

Ap'tous tipous Vieta exoume
$S_{1}=p_{1}+p_{2}+p_{3}=0$
$S_{2}=p_{1}p_{2}+p_{2}p_{3}+p_{1}p_{3}=k$
$S_{3}=p_{1}p_{2}p_{3}=-\lambda$

Ara i parastasi (meta apo tis prwtes prakseis)
$A=1-(p_{1}^{2}+p_{2}^{2}+p_{3}^{2})+(p_{1}p_{2})^{2}+(p_{2}p_{3})^{2}+(p_{1}p_{3})^{2}-(p_{1}p_{2}p_{3})^{2}$
logw twn tautotitwn (me poli eukoles prakseis)

$p_{1}^{2}+p_{2}^{2}+p_{3}^{2}=S_{1}^{2}-2S_{2}=-2k$
$(p_{1}p_{2})^{2}+(p_{2}p_{3})^{2}+(p_{1}p_{3})^{2}=S_{2}^{2}-2S_{3}S_{1}=k^{2}$

ginetai

$A=1+2k+k^{2}-\lambda^{2}=(k+1-\lambda)(k+1+\lambda)$

Alexandros

Y.G. Mporei kapoios na kanei post kai ta ipoloipa themata tou Likeiou? Distixws exw tin epomeni evdomada eksetaseis kai (logw diavasmatos) den mporesa na ta parw...

[quote="cretanman"][quote="mostel"]Σας παραθέτω τα θέματα του 67ου Πανελλήνιου Μαθηματικού Διαγωνισμού στα Μαθηματικά "ΕΥΚΛΕΙΔΗΣ" για τη Β' Λύκειου.

1) Δίνεται ότι το πολυώνυμο $P(x)=x^{3}+\kappa x+\lambda$ με $\kappa, \lambda\in\mathbf{R}$ έχει τις πραγματικές ρίζες $x_{1}, x_{2}, x_{3}$ που ανά δύο είναι διαφορετικές μεταξύ τους. Να εκφράσετε την παράσταση:

$A=(1-x_{1}^{2})(1-x_{2}^{2})(1-x_{3}^{2})$ συναρτήσει των $\kappa, \lambda$.
[/quote]

Xairomai idiaitera pou eixe sizitithei palaiotera tetoio thema kai aforouse ta simmetrika poliwnima. Mostel nomizw se eixa parapempsei kiolas na deis kati antistoixo pou eixe graftei palaiotera... Nomizw den einai katholou eukoli askisi an den ksereis tous genikeumenous tipous tou Vieta kai apli thewria simmetrikwn poliwnimwn.

Alexandros[/quote]

Βασικά εγώ την έλυσα χρησιμοποιώντας σε κάποιο σημείο το γεγονός ότι αν x+ y+z= 0 τοτε x^3 + y^3 + z^3 = 0

Μπορείς αν γίνεται να γράψεις αυτούς τους γενικευμένους τύπους;

foti mporeis na koitakseis edw

http://mathworld.wolfram.com/VietasFormulas.html gia na deis tous genikeumenous KAI POLY xrisimous tipous tou Vieta (tin genikeusi ekeinwn pou emathes stin A' Likeiou gia to athroisma kai to ginomeno twn rizwn) kai mporeis episis na koitakseis sto parakatw post pou egine prin apo merikous mines kai afora ta simmetrika poliwnima. Tha vreis arketa endiaferontes askiseis kai genikeuseis gia provlimata pou aforoun kai anwtero epipedo.

http://www.mathlinks.ro/Forum/viewtopic.php?t=110880  kai

http://www.mathlinks.ro/Forum/viewtopic.php?p=705280#705280

Alexandros

Σε ευχαριστώ!!! :D  :D

ξερει κανεις πως στο %$@&) λυνεται το 3ο θεμα της Α λυκειου...
τα ελυσα ολα σε μια ωρα εκτος απο αυτο...

x,y>0 & x^3+y^2<=64

να αποδειχθει οτι x^4+y^3<512

εκανα χιλιαδες τροπους αλλα κανενας δεν απεδωσε τα επιθυμιτα...το μονο ππου καταφεραα ηταν να αποδειξω οτι ισχυει για χ>=1...μετα σαν περετερω για να μου πιασει και κατι σωστο εγραψα οτι αφαιροντας το χ^3 και απο τα 2 μελη της ανισωσης βρισκουμε οτι το χ<4 και μετα αφαιροντας το y^2 οτι το y<8...μετα ομως...

α και ακουσα οτι ενα απο αυτα της 3ης τους αφησε  :stretcher:  :alien: ...το εχει κανεις ευκαιρο?

Σιλουανέ; Δοκίμασες καθόλου με ΑΜ-ΓΜ;

Και [b]ΜΗΝ[/b] αποκαλείς τον εαυτό σου loser!

Να θυμάσαι:

Ο μαθηματικός κάνει τον Ευκλείδη, όχι ο Ευκλείδης τον μαθηματικό.

Βασικά ή κάπου έχω κάνει λάθος, ή η λύση της άσκησης είναι πολύ απλή.

Έχουμε από υπόθεση:

$x^{3}+y^{2}\leq64$ (1)

και ζητείται να αποδειχθεί ότι:

$x^{4}+y^{3}<512$

H (1) μπορεί να γραφτεί και:

$8x^{3}+8y^{2}\leq 512$

Οπότε αρκεί να δειχθεί ότι:

$x^{4}+y^{3}<8x^{3}+8y^{2}\implies x^{3}(8-x)+y^{2}(8-y)>0$ που ισχύει..

Οκ.. ομολογώ τότε πως δεν ήταν επιπέδου Ευκλείδη! Η παράσταση που βάλαν στον Θαλή ως 4ο θέμα στη Γ' Γυμνασίου ήταν πιο δύσκολη !!

thano symfono sto oti itan eukola alla diafono gia auto pou les gia to teleutaio thema dioti pisteuo oti itan mia klassiki geometriki anisotita (pithanon na yparxei kai se kapoio vivlio) pou vasizetai se dyo gnosta limatakia pou sou erxontai kateutheian sto myalo molis vlepeis tin askisi.
auta einai:
1)MB=MC=MI
2)B,I,O,C OMOKYKLIKA GIA <A=60

emena pantws de mou erxontai lolol

ennow pws linontai eukola kai me gnwseis lykeiou

[quote="silouan"]Βάζω μια ανισότητα που μοιάζει με αυτή που μας βάλλαν σήμερα .
Κατασκευή :Nick Rapanos-Silouan

Νδο για κάθε $a_{i}\in R^{+}$ και $n\in N^{*}$  ισχύει :
\[(\frac{\sum_{i=1}^{k}a_{i}^{n}}{\sum_{i=1}^{k}a_{i}})^{\frac{\sum_{i=1}^{k}a_{i}}{n-1}}\geq\prod_{i=1}^{k}a_{i}^{a_{i}}\]
[/quote]

Loipon ksexasa tote na valw tin lisi mou gia tin anisotita pou proteine o Silouan kai Nikos.

(katarxin ean ta $a_{i}$ einai fisikoi tote i lisi einai auti pou evala pio panw kai einai akrivws i genikeusi tis anisotitas pou evalan stin B' Likeiou ston Eukleidi. Kathws omws genikeuetai gia $a_{i}$ thetikous pragmatikous, den exei noima na milame gia tin proigoumeni apodeiksi...)

Loipon, arxika as ipothesoume oti $a_{1}+a_{2}+\cdots+a_{n}=r$

Tote $\frac{a_{1}}{r}+\frac{a_{2}}{r}+\cdots+\frac{a_{n}}{r}=1$
opote apo tin anisotita twn varwn exoume oti:

\begin{eqnarray}&&\frac{a_{1}}{r}a_{1}^{n-1}+\frac{a_{2}}{r}a_{2}^{n-1}+\cdots+\frac{a_{n}}{r}a_{n}^{n-1}\geq a_{1}^{(n-1)\frac{a_{1}}{r}}a_{2}^{(n-1)\frac{a_{2}}{r}}\cdots a_{n}^{(n-1)\frac{a_{n}}{r}}\nonumber \\ &\Leftrightarrow& \frac{a_{1}^{n}+a_{2}^{n}+\cdots+a_{n}^{n}}{r}\geq \left(a_{1}^{(n-1)a_{1}}a_{2}^{(n-1)a_{2}}\cdots a_{n}^{(n-1)a_{n}}\right)^{\frac{1}{r}}\nonumber \\ &\stackrel{r=a_{1}+a_{2}+\cdots+a_{n}}{\Leftrightarrow}& \left(\frac{a_{1}^{n}+a_{2}^{n}+\cdots+a_{n}^{n}}{a_{1}+a_{2}+\cdots+a_{n}}\right)^{a_{1}+a_{2}+\cdots+a_{n}}\geq \left(a_{1}^{a_{1}}a_{2}^{a_{2}}\cdots a_{n}^{a_{n}}\right)^{n-1}\nonumber \\ &\Leftrightarrow& \left(\frac{a_{1}^{n}+a_{2}^{n}+\cdots+a_{n}^{n}}{a_{1}+a_{2}+\cdots+a_{n}}\right)^{\frac{a_{1}+a_{2}+\cdots+a_{n}}{n-1}}\geq a_{1}^{a_{1}}a_{2}^{a_{2}}\cdots a_{n}^{a_{n}}\nonumber \end{eqnarray}

h opoia einai kai i zitoumeni

Gia opoion den gnwrizei tin anisotita twn varwn, ton parapempw sto poli omorfo post tou silouan : http://www.mathlinks.ro/Forum/viewtopic.php?t=131000

Alexandros

auto einai Aleksandre!!!an de to skeftomoun etsi den ksero an tha tin ellina....alla meta kathomoun arketa ora kai prospathousa na katalavo giati lene <<akeraioi>> stin ekfonisi....

[quote="silouan"]Ωραία Αλέξανδρε.... :)   .Αυτή ήταν και η λυση μας .Η ανισότητα των βαρών φαίνεται ως ένα πολύ ισχυρό εργαλείο για την επίλυση εκθετικών ανισοτήτων .[/quote]

 Στο βιβλίο του [color=red]DRIΜBE (Ανισότητες : Μέθοδοι και ιδέες , Ρουμάνικο )[/color] ,στο καλύτερο δηλαδή , από διδαδκτική άποψη ,  βιβλίο ανισοτήτων που έχω δει ποτέ, ο συγγραφέας εφαρμόζει την ανισότητα των βαρών με ένα καταπληκτικό τρόπο για την εύρεση βοηθητικής ανισότητας  και με κυκλική εφαρμογή  αυτής αποδεικνύει την αρχική. Νομίζω ότι σε κάποια άσκηση στο βιβλίο [color=blue]<< Αλγεβρικές Ανισότητες >> [/color]έχω μια εφαρμογή , αλλά δε θυμάμαι που.Αν τη βρει κάποιος , ας τη βάλει για να τη μελετήσουν και οι άλλοι.

 Αυτό το γράφω ως ενισχυτικό της άποψης που εκφράσατε ότι η ανισότητα των βαρών είναι πολύ γερό εργαλείο !
              ΜΠΑΜΠΗΣ

ε παιδες ενα φιλο μου τον ειδοποιησανε οτι περασε μεσω τηλεφωνου πριν καμια ωρα...εσας?νιμιζω πως δεν περασα για να μην μου εχουν πεια ακομα :(

Παλιότερα σε είχαν πάρει ποτέ τηλέφωνο;Εμένα προσωπικά ποτέ....Στο internet πάντα το έβλεπα...

Τα αποτελέσματα πάντως έχουν βγει .Εγώ πήρα τηλ και μου είπαν ότι πέρασα.... :D

paidia ta apotelesmata vgikan . k egw perasa!!! :D sygharitiria se olous ! kali epityhia se olous mas k ston arhimidi :first:

μπραβο σε ολους ! καλη επιτυχια στον αρχιμηδη  :)

Καλή επιτυχία σε όλους σας παιδιά!

@silouan
[hide]Έδωσες μία υπόσχεση, ελπίζω να την κρατήσεις!  :lol: [/hide]

Παιδιά θα χαρώ να σας δω όλους στον Αρχιμήδη  :D   και εκεί να δώσουμε έναν αγώνα άμιλλας με σκοπό το καλύτερο και την επανασυνάντηση στην Αίθουσα Τελετών του Πανεπiστημίου να είμαστε όλοι . :)

Καλή επιτυχία....

@mostel
[hide]Δεν μου αρέσουν οι υποσχέσεις γιατί είναι δύσκολο να τις τηρήσεις . Αλλά θα κάνω ότι καλύτερο μπορώ.[/hide]

Λοιπόν παιδιά παρακολουθώ καιρό τις συζητήσεις σας στο forum, όμως μόνο πριν λίγο καιρό έγινα μέλος. Πέρασα και εγώ στον Αρχιμήδη και ελπίζω να σας δω εκει  :wink: (Αν και εγώ θα πάω κυρίως για την εμπειρία, και ό,τι βγει.)

Πιστευω πως το 3 του ευκλειδη ηταν μετριο αρκει κανεισ να ηξερε οτι
          αν α,β>0 και α+β<1 τοτε α<1 και β<1


induction geometry rectangle algebra polynomial

Hw many squares are there in the picture:

----------------
-    -     -      -
----------------
-    -     -      -
----------------
-    -     -      -
----------------
-    -     -      -
----------------

this is actually a big square which is divided into 16 smaller squares ( i hope the picture I made is good)

cheers lads!

the number of squares should be between:

29 and 33 (29 or 30 or... or 33)

well it's not such a good picture!  :oops:

I would be nice to find a closed form of the general case. Does anyone know (or better find !) it ?

I wrote a small computer program to find the 20 first values (no obvious pattern emerges, except that it gently increases) :

1 : 1
2 : 5
3 : 14
4 : 30
5 : 55
6 : 91
7 : 140
8 : 204
9 : 285
10 : 385
11 : 506
12 : 650
13 : 819
14 : 1015
15 : 1240
16 : 1496
17 : 1785
18 : 2109
19 : 2470
20 : 2870

By the way, if my program is correct the answer for 4 is 30.

Actually a closed form is not too difficult to find :

let's call u(n) the number of squares one can count in a square cut in n*n equal smaller squares.

Starting the induction :
u(1) = 1 (!)
u(3) = 3
..

Now we add one row and one column of small squares (small square = 1x1 square) to a n*n square :

u(n+1) = u(n) + (number of squares "using" one small square in the upper-most row or right-most column)

(number of squares "using" one small square in the upper-most row) = C(n+2,2) because with just have to choose upper-left and upper-right corners

There are (n+1) squares sharing a small square in both upper-most row or right-most column.

So (number of squares "using" one small square in the upper-most row or right-most column) = 2*C(n+2) - (n+1) = ... = (n+1)^2 !

So u(n) is the sum of the squares and u(n) = n(n+1)(2n+1)/6

And u(4) = 30.

Well, if your intention is to have 16 small squares, then I'll make a slightly different picture:

--------------------
|....|.....|.....|.....|
--------------------
|....|.....|.....|.....|
--------------------
|....|.....|.....|.....|
--------------------
|....|.....|.....|.....|
--------------------

In this case, there are 16 1x1s, 9 2x2s, 4 3x3s, and 1 4x4.

16+9+4+1 = 30

So the formula ought to be something like:

# squares in an nxn array = n:^2: + (n-1):^2: + (n-2):^2: + ... + 1

For which you can use the sum of squares formula, x=n(n+1)(2n+1)/6

So the formula is [b]x=n(n+1)(2n+1)/6[/b]

Of course, this only works for squares.  But it's probably possible to derive a formula that would work for rectangles.

To find the number of rectangles based on a rectangular grid of m*n squares, what we can do is first consider all the vertices. There are (m+1)(n+1) of these, and we pick one. Now we remove the row and column from which the original point was chosen. Then we are left with an m*n set of points. We select another. The number of ways to do this is m(m+1)*n(n+1), but remember that it is possible to select them in the other order, or it is possible to select the opposite vertices of the rectangle, so we must divide by 4, giving m(m+1)/2 * n(n+1)/2 ways to pick rectangles from an m*n grid of squares.

You disappoint me.

First we need to find what degree the polynomial is.

1 : 1
2 : 5
3 : 14
4 : 30
5 : 55
6 : 91
7 : 140
8 : 204
9 : 285
10 : 385
11 : 506
12 : 650
13 : 819
14 : 1015
15 : 1240
16 : 1496
17 : 1785
18 : 2109
19 : 2470
20 : 2870

The 1st degree differences are:
4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400
The 2nd degree differences are:
5,7,9,11,13,15,17,...
And the 3rd degree differences are 2 throughout.

Therefore we can conclude that it is a cubic formula. The general form is ax^3 + bx + cx + d = 0. 4 points define a cubic equation, so select the first four.

1 = 1a + 1b + 1c + d
5 = 8a + 4b + 2c + d
14 = 27a + 9b + 3c + d
30 = 64a + 16b + 4c + d

Solving, we have a = 1/3, b = 1/2, c = 1/6, and this the equation is

a-sub-n = (n^3)/3 + n/2 + n/6.

You don't need to go down that far. The first-order differences should be numbers we recognize, since they are perfect squares. We can then rewrite it as the sum of various falling powers (a ``quadratic'' and a ``linear'') and sum each of those by the power rule of finite (or umbral) calculus. We don't need to go through the painful processes of finding the coefficients by solving a system of equations.

Otherwise, after seeing that the first-order differences are perfect square, we may use the Bernoulli numbers by calculating the first few terms of the power series of x/(e^x-1).

[quote="Lord Venom"]Solving, we have a = 1/3, b = 1/2, c = 1/6, and this the equation is a-sub-n = (n^3)/3 + n/2 + n/6.[/quote]

Which can be also written as: x = n(n+1)(2n+1)/6

Using ComplexZeta's formula "m(m+1)/2 * n(n+1)/2"

In the case of a 2x2 array, we have 2*3/2 * 2*3/2 = 3*3 = 9, but there are only 5 squares!  Did I miss something?

Oh... I feel stupid answering my own question, but there are 9 if you also count the rectangles.  However, I believe the question was "how many squares are there?"  Still, the rectangles are useful...

So a formula could read:

For an array of n x n squares, there are n(n+1)(2n+1)/6 squares.

For an array of m x n rectangles, there are m(m+1)/2 * n(n+1)/2 rectangles.

n(n+1)(2n+1)/6 is for squares on an n*n square grid. m(m+1)/2 * n(n+1)/2 is for the number of rectangles on an m*n rectangular grid. So you didn't miss anything.

Oop, sorry Simon. I replied to my own message while you were typing an answer out, as you'll see...

8/3 + 4/2 + 2/6 = 5, so I don't see where is the confusion.

The confusion was whether we were referring to squares or rectangles. All has been cleared up now.


[b]Consider $ a$,$ b$, and $ c$ such that $ a^2\plus{}b^2\equal{}c^2$ Prove that $ a$ and $ b$ can not both be odd.[/b]

[hide]Then $ a^2$ and $ b^2$ are both 1 (mod 4), and therefore $ c^2$ is 2 (mod 4), which is not a quadratic residue mod 4.[/hide]

EDIT-Wrote c instead of $ c^2$.

[hide="Another one"]All Triple Pythagoras $ (a,b,c)$ can be written as $ (2mn,m^2\minus{}n^2,m^2\plus{}n^2)$, so one of a or b must be 2mn, which is even. Hence proved.[/hide]

The correct expression is $ (2mnk, km^2 \minus{} kn^2, km^2 \plus{} kn^2)$ for $ k$ a product of distinct primes congruent to $ 3 \bmod 4$.  The expression you gave only produces all primitive triples and some non-primitive ones.

If both a and b are odd, then they can be written in the form $ a \equal{} 2m \plus{} 1$ and $ b \equal{} 2n \plus{} 1$, for some naturals m,n. Then

$ a^2 \plus{} b^2 \equal{} (2m \plus{} 1)^2 \plus{} (2n \plus{} 1)^2 \equal{} 4m^2 \plus{} 4m \plus{} 1 \plus{} 4n^2 \plus{} 4n \plus{} 1 \equal{} 4(m^2 \plus{} m \plus{} n \plus{} n^2) \plus{} 2$ *

which means that $ a^2 \plus{} b^2 \equal{} c^2$ is an even number. But that implies that c is also an even number, because the square of an odd number x is still an odd number ($ x \equal{} 2k \plus{} 1 \equal{} > x^2 \equal{} 4k^2 \plus{} 4k \plus{} 1$ which is odd). But if c is even then it can be written in the form c = 2y, and so $ c^2 \equal{} 4y^2$ which means that $ c^2$ is divisible by four. But looking at the expression * above we see that $ a^2 \plus{} b^2$ is not divisible by 4. Since we arrived at a contradiction, then our hypothesis must be false: that means a and b cannot both be odd. It can also be shown that a and b cannot both be even. For three pythagorean numbers a, b, and c such that $ a^2 \plus{} b^2 \equal{} c^2$ it can also be shown that:

- one of a and b must be a multiple of 3;
- one of a and b must be a multiple of 4;
- one of a, b, and c must be a multiple of 5.


calculus
number theory unsolved
number theory

[quote="aktyw19"]find $ a,b,c \in \mathbb{R}$

$ a^{3(b \plus{} c)} \equal{} 36$

$ b^{3(c \plus{} a)} \equal{} 81$

$ c^{3(a \plus{} b)} \equal{} 169$[/quote]

A quick check shows that we have no integer solutions. So I consider we are looking for solutions in $ \mathbb R^ \plus{}$, in order to have power defined.

It's rather quick to show that values are all in $ (1,2.5)$ but from there, I found no clever method to find the solution.

Basic calculus with standard spreadsheet gives the solution, likely unique : $ (a,b,c) \equal{} (1.428940700651...,$ $ 1.581987578530...,$ $ 1.764592790526...)$

Since this is an olympiad training exercise, it is surely not intended to be solved with basic brute calculus. So, aktyw19, could you please give us some information about the origin (a book, a teacher, a national contest). Thanks in advance ...

[quote="aktyw19"]sory , of corse  a,b,c-intiger[/quote]

Are you joking ????? You wrote $ a,b,c\in\mathbb R$ !!!! and I spent some hours on this problem!

If $ a,b,c\in\mathbb Z$, we obviously have no solution : look at the third equation : $ c^{3(a \plus{} b)} \equal{} 13^2$ so :

either $ c \equal{} \pm 13$ and $ 3(a \plus{} b) \equal{} 2$, impossible if $ a,b\in\mathbb Z$

either $ c \equal{} 169$ and $ 3(a \plus{} b) \equal{} 1$, impossible if $ a,b\in\mathbb Z$

Where did you find this problem ?

[quote="aktyw19"]I always write down badly, but I had a, b, c real numbers
Because you always have reasons :lol:[/quote]

???? I dont understand

What is exactly the problem, please ?

And where is it coming from ?


how do you find the centroid of a certain traingle in a corrdinate gragh given the cordinates of the three vertices

If the coordinates are $(a,b)$, $(c,d)$, and $(e,f)$, then the centroid (of center of gravity as I am used to it) is $(\frac{a+c+e}{3},\frac{b+d+f}{3})$.

That formula would have made my last math test so much easier... :( at least now I don't have to spend all day trying to find the centroid in a triangle!

[quote="tiredepartment"]But remember to check if your points actually make a triangle and are NOT collinear.  For example, if someone gave you (0,0), (0,1), (0,2).  By the formula, you would get (0,1), but there is no triangle.[/quote]

Yep
That's definitely something you need to look out for
since you can't have a centroid of a degenerate triangle

Well, its degenerate, so its still a triangle ;)


linear algebra
matrix
abstract algebra
induction
linear algebra unsolved

Let A be a symmetric matrix such that the sum of elements of any row is zero.
Show that all elements in the main diagonal of cofator matrix of A are equal.

I check on several matrices with reals coefficients wich satisfy the statement I found all entries
of the comatrix are equal.

 $A$  a matrix wich satisfy the statement, then all $det(A_{ij})$ are equal
where   $det(A_{ij})$ is the determinant obtained by delating the i-row and
j-column from $A$

This is not surprising. Let $n$ be the size of $A$. Two cases are possible: $\text{rank}A<n-1$ and $\text{rank}A=n-1$. In the first case all entries of the comatrix $\widetilde A$ are $0$. In the second case, the kernel of $A$ is one-dimensional and we know that $(1,1,\dots,1)^T$ is in the kernel. Now, using the identity $A\widetilde A=(\det A) I=0$, we see that all columns of $\widetilde A$ are proportional to $(1,1,\dots,1)^T$, so all elements in one column are equal. Since $\widetilde A$ is symmetric, it follows that all its elements are equal.

Some observations :
1/ $0$ is eigenvalue of $A$ for the eigenvector $^{t}(1,...,1)$
2/ $A$ is diagonalisable if the entries are reals
3/ $Com(A)=bJ$ where $J$ is nxn matrix with all coefficients are 1

Actually this problem is a generalization of the following fact:

[b]Theorem.[/b] Let $G=(V,E)$ a multigraph (that is, we can have more than one edge connecting two vertices) with $|V| = n$ vertices. Its [i]adjacency matrix[/i] is the matrix $A_{n\times n}$ such that $a_{ij} = -k$ if $i\neq j$, where $k$ is the number of edges connecting vertices $i$ and $j$ and $a_{ii} = d(i)$, where $d(i)$ is the degree of vertex $i$, that is, the number of edges going through $i$.

Thus the number of spanning trees with labeled vertices is $\det \tilde A$, where $\tilde A$ is a $(n-1)\times (n-1)$ matrix obtained by deleting the $i$-th row and the $i$-th column for some $i$ (or the cofator $A_{ii}$).

The proof is by induction on the number of edges of $G$ (roughly speaking, remove one edge and count the number of trees going through this edge and the number of trees not going through this edge, applying the induction hypothesis).

The generalization which implies the desired result is obtained assigning weights on each egde and defining "degree" of a vertex as the sum of the numbers of the edges going through it. I believe the same induction works for this problem.

Maybe it's a more lenghty way to solve it, but I really think it's nice!

I know the author of this problem and he told that's how he came up with this problem.


it is in my experience that Geometry is so much easier then algebra[/list]

Well I like them both, and I thought Geometry was easy and all (just repetitive), but then I got into MathCounts and Trig, and now I would say Algebra is a lot better.  It seems more like solid data, instead of abstract thinking and nonverbal skills.

Geometry in American schools is pathetically easy.  I do much better on algebraic probelms in contests than actual competition geometry,which is much more difficult than the classroom version.

[quote="chicconspiracy"]it is in my experience that Geometry is so much easier then algebra[/quote]

From your posts/signature/location flag I'll assume you are or have been partially raised in China. Please excuse me if this is incorrect. However, if my assumption is accurate it should not be surprising that geometry is relatively easy for you, especially geometry in America.

I assume you mean not school.  So, algebra is much easier for me than geometry.

both subject can be really easy or really hard (obviously).

But relatively, everyone has strengths and weaknesses. Well not everyone. But most people.

Actually, I still don't know what the question is exactly. It just says Algebra or Geometry. Are we voting on which is harder or which is easier?

Analysis $\subset$ Algebra $\cup$ Geometry

Hi Everyone,

Both algebra and geometry are beautiful subjects.  One might not appreciate these subjects much until one is exposed to university level geometry and algebra.  When you study university level mathematics, you will have a better understanding of how beautiful both subjects are.   High school level mathematics does these subjects no justice.

Yours truly,

M.E.B.

Don't forget:

Algebraic Geometry = Algebra $\cap$ Geometry

[quote="JBL"]Analysis $\subset$ (Algebra $\cup$ Geometry)^c[/quote]

Fixed.

For me, algebra is easier. I imagine algebra in the later years will get tougher but right now, algebra is easier. Whenever I see a geometry question, I just freeze because I don't know where to start. *sigh* I'm feeling nostalgic because I want to go back to 6th grade where the geometry questions were like "Find the area of the circle with radius of 2"  :P But I think it's safe to say that geometry in America is weaker compared to other places.

I've always found geometry easier.  But then again, I'm in the easy stuff right now.  I'm one of the few people in the world who love and are fascinated by proofs.  Go figure.   :lol:


inequalities
trigonometry
inequalities unsolved

For $ 0<b<a<\frac{\pi}{4}$ prove this inequality:
$ \tan^2{a} > \tan{(a\plus{}b)} \cdot \tan{(a\minus{}b)}$
Can somebody help me...  :?:

Have you tried expanding with

$ \tan(A\plus{}B) \equal{} \dfrac{\tan(A)\plus{}\tan(B)}{1\minus{}\tan(A)\cdot\tan(B)}$

hello, applying the addition formulas your assertion is equivalent with
$ \minus{}2\sec^2(a)\sec(a\minus{}b)\sec(a\plus{}b)\sin^2(b)\sin(a\minus{}\frac{\pi}{4})\sin(a\plus{}\frac{\pi}{4})>0$ and this is true because of $ 0<b<a<\frac{\pi}{4}$.
Sonnhard.

Or even simpler, it is equivalent to

$ \tan^4(a) < 1$

Which is true because $ a < \frac {\pi}{4}$


induction
real analysis
real analysis unsolved

Let $ (X,\mu)$ be a measure space.
If the ststement" $ a\leq \mu (E)$ for some $ a>0$ for all measurable set $ E$ with $ \mu(E)>0$" is false.
Show that there exists a sequence of measurable sets $ \{E_n\}$,which is ptwise disjoint and $ 0<\mu(E_n)<3^{\minus{}n}$ for all $ n\in \mathbb{N}$.

For every $ \epsilon > 0$ there exists $ A$ with $ \mu(A) < \epsilon$. (All sets mentioned have postive measure.) So by induction choose $ A_n$ with $ \mu(A_1) < 1/3$, $ \mu(A_2) < \mu(A_1)/3$, $ \mu(A_3) < \mu(A_2)/3$, ... From the $ A_n$ you can find the desired pairwise disjoint sets.


Dell, 1.86 GHz
EDIT: Forgot to mention, it's a laptop, since the processor speeds for laptops are generally lower than those for desktops.

sponge008 brand
uuh, A64 3200+ is...ping! 2.01 gHz.

Dell, 1.6 GHz, laptop, the cd drive doesn't work...

mines a dell inspiron 2200 ( laptop) how do you know the  Ghz?

ditto i have a dell inspiron 2200 laptop

haha its one of those Great Quality cheap kinds

Athlon 1500+ I think...or maybe its 2200+

what computer do u think is the best brand

[quote="solafidefarms"]*My* last branded computer was a Gateway Solo, Pentium 3 processor. *The family's* last branded computer is a Dell Dimension 8200. The newest computer is mine, built by me, with a AMD Athlon XP processor. Does that answer your question?[/quote]

Sure does, but if you know the processor speeds it would be helpful.

If anybody has other computers in their house please post the same info.

Thanks to everyone who has responded so far. :)

two dell laptops 2650 (my dad from his office) mine 2200 (from nats) home computer no name brand built with different parts but not by our family or anything

[quote="tcs09"]What brand is your computer? (e.g Dell, HP,...)

What is the speed of your processer? (e.g. 2GHz)

Please answer because I need 40 answers for my statistics class.[/quote]

Why dont you do a poll?

my computer is a averatec.
my son's is a dell (he shares it with his siblings)

2005, and 2GHz processors. How far we've come. I am writing this on a Notebook 10. An Intel CORE i7. But back then, Intel i3. Remember.

What the heck I put the tag revivethepast on and someone else did...
Well just look.


Well he knows that 2 cities have not been raided. That means of the 18 remaining cities, 4 are his. Thus our answer is $ \frac {4}{18} \equal{} \boxed{ \frac29}$

Copied from the Solution page:

Problem:
The Vikings raid one of 20 coastal cities in Scotland at random every month. A lord has recently gained control of six of these cities. One month, he receives word from two of these six cities that they have not been raided. What is the probability that none of his six cities were raided that month?

Solution:
Besides the two cities he knows about, the other 18 cities were each raided with equal probability. Fourteen of these are not under this lord's control. The probability that one of these was raided, and hence the probability that none of his six cities were raided, is $14/18=\fbox{7/9}.$

[quote="Conjurer"]Why dont we use Bayes theorem in this problem?[/quote]

You can:

I think it helps to think of the cities being raided as being chosen by a spinner with 20 choices.

Let A be the event the 4 other cities were not raided and B be the event that the two the lord knew of were not invaded.

Then:

P(A) = 16/20
P(B) = 18/20
P(B|A) = 14/16

So from Bayes Theorem:

P(A|B)=P(B|A)*P(A) / P(B) = (14/16)*(16/20) / (18/20) = 14/18 = 7/9

Sorry for my formatting, I haven't learned how to do it so well yet.


Let be $ a,b,c$ the sidelenghts of a triangle . Prove that :
$ \frac{a\plus{}b}{c^3\plus{}abc}\plus{}\frac{b\plus{}c}{a^3\plus{}abc}\plus{}\frac{c\plus{}a}{b^3\plus{}abc}\ge \frac{1}{2Rr}$

hello, we have $ 2Rr\equal{}2\frac{abc}{4A}\frac{A}{s}\equal{}\frac{abc}{a\plus{}b\plus{}c}$ then your inequality is equivalent to
$ \frac{a\plus{}b}{c^2\plus{}abc}\plus{}\frac{b\plus{}c}{a^3\plus{}abc}\plus{}\frac{c\plus{}a}{b^3\plus{}abc}\minus{}\frac{a\plus{}b\plus{}c}{abc}\geq0$  and this is equivalent to
$ \frac{(a^2\minus{}b^2)^2\plus{}(a^2\minus{}c^2)^2\plus{}(b^2\minus{}c^2)^2}{2(b^2\plus{}ac)(a^2\plus{}bc)(c^2\plus{}ab)}\geq0$
which is obviously true.
Sonnhard.


limit
integration
calculus
calculus computations

Let $a_{n}= \sum_{i=1}^{n}\exp \left( \frac{i}{n}\right)$. Study the existence (and find its value, if it's the case) of the following limit: $\lim_{n \to \infty}\left( a_{n+1}-a_{n}\right)$.

I think the limit, if there are no computational mistakes in my calculations, is $e-1$.

I think it's better to post some hints or a solution rather than just say what the answer is  :D

Here are my calculations:

$a_{n+1}-a_{n}=e+\sum_{k=1}^{n}e^{\frac{k}{n+1}}\left(1-e^{\frac{k}{n(n+1)}}\right)$
Let $S$ be the preceding sum.
Then

$S=\sum_{k=1}^{n}e^{\frac{k}{n+1}}\frac{1-e^{k/n(n+1)}}{k/n(n+1)}\frac{k}{n(n+1)}$
Then we have that

$\min_{k}\frac{1-e^{k/n(n+1)}}{k/n(n+1)}\sum_{k=1}^{n}e^{\frac{k}{n+1}}\frac{k}{n(n+1)}\leq S \leq \max_{k}\frac{1-e^{k/n(n+1)}}{k/n(n+1)}\sum_{k=1}^{n}e^{\frac{k}{n+1}}\frac{k}{n(n+1)}$
and since$\frac{1-e^{x}}{x}\rightarrow-1$, we get that $S\rightarrow \int_{0}^{1}xe^{x}dx=-1$
therefore the limits is $e-1$.


A positive integer is called ascending if, in its decimal representation, there are at least two digits and each digit is less than any digit to its right. How many ascending positive integers are there?

[hide="Solution"]The largest ascending number is $123456789$. Each of the digits in this number may be present or not be present, giving us $2^9=512$ ascending numbers. However, the problem asks us not to include the numbers 0 to 9, so our final answer is $512-10=\boxed{502}$.[/hide]

Yeah, that's nice 4everwise.

[hide="Not as slick but works fine"]
$\binom{9}{2}+\binom{9}{3}+\cdots +\binom{9}{9}=2^9-10=\boxed {502}$.
[/hide]

Just as a note, you can find the sum chess's way or 4everwise's way, and that's why $\binom{9}{0} + \binom{9}{1} + \binom{9}{2} +\cdots +\binom{9}{9} = 2^{9}$.

Sorry to revive such an old topic, but could someone explain what you are choosing when you write ${9\choose 2}$? Is it just because you are choosing some two digits and then you know that in some way you can arrange them in ascending order?

Yeah, you're basically choosing which digits 1 through 9 you are including in your number. You know that no matter which ones you choose there is only one way to arrange them. So you just have to find all the ways of choosing the numbers.

Basically when you have couple of digits there is only 1 way to arrange acsending. So you don't need to worry about cases. Now this problem is: How man ways to choose subset of (1,2,3,4,5,6,7,8,9). Thats 9 choose 2 + 9 choose 3, etc. all the way to 9 choose 9. The last case (9 choose 9) would be 123456789. Don't add (9 choose 0) because that would be empty or (9 choose 1) because there would only be 1 digit and problem asks for at least two. A faster way to add all combinations of choosing subset of (1 thru 9) would be $2^9$ (aka 10th row Pascals) but you have to remember to subtract 1+9.

Combo AIME grind: Let’s say our number has $n$ digits. These digits can only go in one order and they must all be different, so there are $\binom{9}{n}$ ways. Summing from $0$ to $9$ gives $2^9=512$, but we must subtract $\binom{9}{0}+\binom{9}{1}=10$, which gives $\boxed{502}$.


Determine all functions $ f,g$ $ \mathbb R \to \mathbb R$ such that
\[ f(x \plus{} yg(x)) \equal{} g(x) \plus{} xf(y)
\]
for all
a)real numbers $ x,y$
b) positive numbers $ x,y$
@Add:
Yes Yuriy Solovyov,It is also my solution.How about b)

Solution to case (a):

1) Put $ y\equal{}0$: $ f(x)\equal{}g(x)\plus{}xf(0)$ and now $ x\equal{}0$, get $ f(0)\equal{}g(0)$.

2) Put $ x\equal{}0$: $ f(yg(0))\equal{}g(0)$. Here are two cases: $ g(0)\ne 0$ or $ g(0)\equal{}0$.

If $ g(0)\ne 0$ then $ f(x)\equal{}g(0)$ and $ g(x)\equal{}\minus{}g(0)x\plus{}g(0)$.

If $ g(0)\equal{}0$ then $ f(0)\equal{}0$ and $ f(x)\equal{}g(x)$.

Then $ y\equal{}\minus{}\frac{x}{f(x)}$ and get $ f\left(\minus{}\frac{x}{f(x)}\right)\equal{}\minus{}\frac{f(x)}{x}$. And we get $ f(x)\equal{}\frac{1}{x}$, but it does not satisfy equation, this means that $ y\equal{}\minus{}\frac{x}{f(x)}\equal{}\text{const}$ and $ f(x)\equal{}x$, which is a solution :)

How can you get $ f(x)\equal{}\frac1x$ from $ f(\minus{}\frac x{f(x)})\equal{}\minus{}\frac{f(x)}x$?

It seems you need conditions like $ \minus{}\frac x{f(x)}$ is surjective.


Helo  :)
   I am Vardan Verdiyan (vardanch)

         This is Armenia team for IMO 2006
1.  Sloyan Tigran   (tiks)
2.  Aramyan Nerses
3.  Verdiyan Vardan   (vardanch)
4.  Karagulyan David
5.  Tovmasyan Narek
6.  Tigran Hakobyan

   :) Hope to see you at IMO 2006  :)

In this year Armenia team selected by
      National olympiad (max 42 points) + TST ( max 42 points)

National olympiad results

David Karagulyan---42 points  (10 gr.)
Sevada Hovsepyan---42 points  (10 gr.)
Nerses Aramyan---42 points  (Limp Biscuit)  (10 gr.)
Tigran Sloyan---41 points (Tiks) (10 gr.)
Vardan Verdiyan---42 points (vardanch) (9 gr.)
Tigran Hakobyan---42 points (9 gr.)
Karapet Kostandyan---42 points (9 gr.)
Narek Tovmasyan---38 points (9 gr.)
Martun Karapetyan---34 points (8 gr.)
Artur Darbinyan---28 points (8 gr.)

This year National Olympiad was too easy and the points were very high.

TST results

Tigran Sloyan---42 points (10 gr.)
Nerses Aramyan---41 points (10 gr.)
Vardan Verdiyan---37 points (9 gr.)
David Karagulyan---28 points (10 gr.)
Narek Tovmasyan---27 points (9 gr.)
Artur Darbinyan---26 points (8 gr.)
Tigran Hakobyan---21 points (9 gr.)
Karapet kostandyan---0 point (9 gr.)

=>
   SUM
Tigran Sloyan---83 points (10 gr.)
Nerses Aramyan---83 points (10 gr.)
Vardan Verdiyan---79 points (9 gr.)
David Karagulyan---70 points (10 gr.)
Narek Tovmasyan---65 points (9 gr.)
Artur Darbinyan--- 54 points (8 gr.)
Tigran Hakobyan---63 points (9 gr.)

For the 6th place there was TST 2 were
   Hakobyan Tigran wins.

[quote="vardanch"]Barev joghovurd   ;)

   Inch ka chka ?
      Ekeq menq mer hamar aistex mez <<huzox>> harcer@ tanq.
      Es lini mer forumi tex@.[/quote]

English please.

[quote="sevakm"]GOOD LUCK to the Armenian team from an armenian IMO veteran.[/quote]
 :o How I can guess you are famous [b]Sevak Mkrtchyan [/b],from Armenian teams for IMOs 1998(Bronze),1999(Bronze),2000(Silver with  29 point :!: ).
I am very glad that you are on this forum too,how I know you are student in USA now,and I would like to know how are you doing?
Thank you !!!!!!!!!!!!

Lol, it is so nice that this year the IMO site offers pictures of everyone, not only contestants, but guides and leaders too.

In 2002 there weren't a lot of pictures, and actually not a lot of live coverage at all.

MInor question  :wink: : that list of participants makes we wonder : is there anyone in Armenia with a family name not ending in -yan? :)

[quote="fredbel6"]

MInor question  :wink: : that list of participants makes we wonder : is there anyone in Armenia with a family name not ending in -yan? :)[/quote]
Actually there were some but they have died out :rotfl: :D .

It seems to me that Vardanch forgot to add the final results here(results in IMO),so let me to do that! :wink:
1. Sloyan Tigran (Tiks)-24(silver) :P
2. Aramyan Nerses-15 (bronze)
3. Verdiyan Vardan (vardanch)-14(honor)
4. Karagulyan David-9(honor)
5. Tovmasyan Narek-14(honor)
6. Tigran Hakobyan-14(honor)


AMC
AIME
counting
distinguishability
USA(J)MO
USAMO
AIME II

From the AIME II 2008, #12
There are two distinguishable flagpoles, and there are $ 19$ flags, of which $ 10$ are identical blue flags, and $ 9$ are identical green flags.  Let $ N$ be the number of distinguishable arrangements using all of the flags in which each flagpole has at least one flag and no two green flags on either pole are adjacent.  Find the remainder when $ N$ is divided by $ 1000$.

I thought my idea wasn't too bad...

I'd like to know the following:
1) Where I went wrong
2) If any of you guys tried this approach (so I can see if my intuition in finding effective ways to solve problems is getting better)

My incorrect answer for $ N$ was $ 6,248$, making the remainder $ 248$.  The correct answer is $ 310$.

Designate the two flagpoles Pole $ A$ and $ B$.  Since the green flags have the most restrictions on how they must be placed, I looked at them first.  If we let the number of green flags placed on $ A \equal{} n$, then the number of green flags placed on $ B \equal{} 9 \minus{} n$.  Because no two green flags can be touching, we must [b]at least[/b] insert one blue flag between each green flag (in both poles).  It ends that if we place $ n$ green flags on $ A$, then we must insert $ n \minus{} 1$ blue flags onto $ A$ and $ 8 \minus{} n$ flags onto $ B$ for a total of $ 7$ blue flags.  That is, no matter what $ n$ equals, we know that at least $ 7$ blue flags must be added to satisfy the condition that there are no two green flags touching.  Therefore, we have $ 10 \minus{} 7 \equal{} 3$ blue flags left.

$ n$ is a case (the number of green flags on pole $ A$.  Let the number of blue flags placed on $ A$ be a subcases.  There are four subcases, each corresponding to placing $ 0,1,2,3$ out of the $ 3$ remaining blue flags onto $ A$.  The number of possible places where we can insert a blue flag is always one greater than the number of green flags for both flagpoles (and this number doesn't change if we add multiple blue flags to the flagpole) so the following cases are like the following

Subcase 1: 0/3 blue flags are placed on $ A$. $ (10 \minus{} n)^3$ ways.
Subcase 2: 1/3 blue flags are place on $ A$. $ (10 \minus{} n)^2(n \plus{} 1)$ ways.
Subcase 3: 2/3                                       $ (10 \minus{} n)(n \plus{} 1)^2$ ways.
Subcase 4: 3/3                                        $ (n \plus{} 1)^3$ ways.

Each subcase is mutually exclusive, so if we add up the algebraic expressions, then we get the total number of subcases for each case (the case = the value of $ n$).  The sum of the expressions is $ S(n) \equal{} 22n^2 \minus{} 198n \plus{} 1111$.

The possible values for $ n$ range from $ 1$ to $ 8$.

$ S(1) \plus{} S(2) \plus{} ... \plus{} S(8) \equal{} 6,248$, and the remainder when divided by $ 1,000$ is $ 248$...doh

[quote="t0rajir0u"][quote="supery00n"]The possible values for $ n$ range from $ 1$ to $ 8$.[/quote]

There's nothing wrong with the cases $ n \equal{} 0, 9$, and in those cases you need $ 8$ blue flags.  I think this is the only mistake, though.

In any case, a word of advice:  AIME problems tend to have relatively nice solutions.  If you're putting this much effort and casework into counting you're probably missing a nicer solution (recursion, for example).  I can't tell you how many times I've worked through a casework solution just to have someone else give an exact solution in terms of binomial coefficients...  :P[/quote]

Wow, thanks so much!!!
I wish I knew how to use recursion better, hopefully that will be covered in the AIME class I'm taking. =)

Wow I'm actually getting good progress on problems after #10 =O

Hmm, would this be an "easy" problem for you?  How much harder are #1 and #3 olympiad problems compared to  the last 10 AIME problems?

Can we not just count the (difference between the two poles) and the blue flags as separators, and use stars and bars and pre-add a separator between each green flag? Or what am I missing?

[quote="supery00n"]How much harder are #1 and #3 olympiad problems compared to  the last 10 AIME problems?[/quote]

The main difference, to me, is the proof aspect.  Getting the answer and showing that it must be correct are two different things; for example, USAMO 2007 #4 was a question many people got the correct answer to, but with a bogus proof.  It was easy to see intuitively that the answer is correct (which can be done in AIME problems, although admittedly not often) but very hard to show that it had to be the case.


geometry
circumcircle
analytic geometry
geometry unsolved

Let $ ABC$ is a triangle. Let $ D$ is a point on $ BC$ and $ AD \perp BC$, let $ E$ is a point on $ CA$ and $ BE \perp CA$, let $ F$ is a point on $ AB$ and $ CF \perp AB$. Let $ H$ is orthocenter of $ ABC$ triangle. Let $ M$ is midpoint of $ BC$. Let $ EF$ meet $ BC$ at $ P$. Prove that $ MH \perp AP$.

Let $ Q$ be the intersection of $ AP$ and the circumcircle of $ \triangle{ABC}$.

$ PQ \cdot PA \equal{} PB \cdot PC \equal{} PF \cdot PE$, so $ A$, $ Q$, $ F$ and $ E$ are concyclic. Since $ H$ is on the circumcircle of $ \triangle{AEF}$, so $ A$, $ Q$, $ H$ and $ E$ are concyclic, which gives $ AQ \perp HQ$.

By the Nine Point Circle theorem, $ M$ is on the circumcircle of $ \triangle{EFD}$. So $ PD \cdot PM \equal{} PF \cdot PE \equal{} PQ \cdot PA$. Hence $ A$, $ Q$, $ D$ and $ M$ are concyclic. So $ MQ \perp AQ$. Hence $ Q$, $ H$ and $ M$ are collinear, which means $ HM \perp AP$.

Thank you very much. Your solution very nice, I think we can be done this problem by ues harmonic. So it's a nice proof.

Since the pencils $A(B,C,H,P)$ and $P(C,E,H,A)$ are obviously harmonic, it follows that line $AP$ is the polar of the intersection $H$ of the diagonals of the quadrilateral $BCEF.$ Now, it's well known that the line connecting the center of the circle and the pole is perpendicular to the polar $\Longrightarrow$ $MH \perp AP,$ as desired.

P.S. An alternate way is using barycentric coordinates.

Yes. I proved this problem by barycentric coordinates because it's easy. But it's for only hight school of VietNam. So I want find other way to prove this problem.


The landlord of an apartment building needs to purchase enough digits to label all of
the apartments from 100 through 125 on the first floor and 200 through 225 on the
second floor. The digits can only be purchased in a package which contains one of
each digit 0 through 9. How many packages must the landlord purchase?

[quote="X_wing1redleader"][hide]it is $3*26+3*26$, or $52*3=156$[/hide][/quote]

Can you please explain that? I don't understand it.

[quote="X_wing1redleader"][hide]well, there are 26 digits from 100 to 125, inclusive. Same goes for 200 to 225. There are 3 digits in each of these 52 numbers, yielding your answer $\boxed{156}$[/hide][/quote]

First of all there are way more than 26 digits.  You probably meant numbers (which you stated later).

Secondly, there aren't three digits of each number.  For instance, the number 1 appears 38 times from 100-125 (I think).

The question asks for packages, so I would to what leonisdabomb did.  ;)

[hide]
The digit 1 appears the most,
For the 100s
26 times as the hundreds
10 as the tens
3 as a one
For the 200s
0 times as a hundred
2 as a ten
3 as a 1
44 total
44 packages
$\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{44}}}}}}}}}}}}}}}}$[/hide]

[quote="leonisdabomb"][hide]Using logic, there will be the most 1s and 2s, and the same number.
26 1s appear in the hundreds place, 20 in the tens, and 6 in the units.
26+20+6=52.[/hide][/quote]
[b]52[/b]
I got the same answer as you! :lol:

[hide]I got 52 boxes.

From 100-125
26 from the hundreds place (one's only- 100-125)
10 from tens place (one's only 10-19)
3 from ones place (101, 111, 121)

From 200-225
0 1's in hundreds place
10 in ten's place (210-219)
3 in units digits (201, 211, 221)

By adding those up, you get 52.

by only counting the 1's. 2's are the same as one's so you don't have to count those. There should be the same amount of each. How did so many people get 156?[/hide]

(This is the same as all the digits between 100-125 inclusive. That would be 25*3=75 TOTAL digits in 100-125.)+(The number of digits in 200-225, which would be 75 again)=150 Okay, divide by ten to get [hide][b]15 boxes.[/b][/hide]

they're asking for number of packages, not digits

[quote="biffanddoc"][hide]
The digit 1 appears the most,
For the 100s
26 times as the hundreds
10 as the tens
3 as a one
For the 200s
0 times as a hundred
 :arrow: 2 as a ten
3 as a 1
44 total
44 packages
$\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{44}}}}}}}}}}}}}}}}$[/hide][/quote]

You said 2 as a ten for the 200s.
There are 210, 211, 212, etc... 10 times, not 2 times.
Add 8 more, and you get 52 like me.  :lol:

uh huh yeah. how many boxes does he need?


I noticed on WOOT there's a discount for MOP qualifiers.  Is there any discount for USAMO qualifiers?

No.  Maybe rruscyzk should make one though :P

i KNOW moppers are invited for free.. :)

MOPpers are invited for free; there are no other discounts.


inequalities
Putnam
number theory open
number theory

Basicly i'm not gonna post a related problem. I'm just so excited with that i have just read and i would like to put it in the table.

In the book of Martin Aigner- Gunter M.Ziegler it is written that Jeff Lagarias has very resently proved  that Riemann's Hypothesis is equivalent to the following inequality :
$\sum_{d|n}d \leq H_{n}+e^{H_{n}}\cdot lnH_{n}$, for all $n\geq 1$ and equality only if $n=1$ .
where $H_{n}= \sum_{i=1}^{n}\frac{1}{i}$.

Isn't it unbelievable ?
Is it so hard to attact this inequality ? (obvious...  :P )

[quote="Anto"]In the book of Martin Aigner- Gunter M.Ziegler it is written that Jeff Lagarias has very resently proved  that Riemann's Hypothesis is equivalent to the following inequality :
$ \sum_{d|n}d \leq H_n + e^{H_n}\dot lnH_n$, for all $n\geq 1$ and equality only if $n=1$ .
where $H_n = \sum_{i=1}^n \frac{1}{i} $.(obvious...  :P )[/quote]

Jeffrey Lagarias (a Putnam fellow years ago) prove this in 2002, but mainly this strong result came from the Robin's criterion (1983) which claims that Riemann's Hypothesis is equivalent to the following inequality:
$ \sum_{d|n}d \leq e^{\gamma} n ln(ln(n)) $
for all integers excepting a finite set.
This seems to hold for all $n \geq 5041$ or something like that.
If there is an infinte set o integers such that this inequality does not hold, the Riemann's Hypothesis will fail.
It worths only 1,000,000 $.

So, the proof in question is valid + correct?

BOmb


Define [i]trio[/i] numbers as numbers that have at least 1 pair of threes in it's digits such that the digits are consecutive. So 12334 is a trio number, 133312 is also a trio number, while 1313131 is not. Find the number of four digit trio numbers.

I'll be working on it too.

edit:

I've already found a solution! I thought this one was tough...

[hide="my solution"]So there are 8 types of trio numbers of that number of digits:

33ab
a33b
ab33
333a
33a3
3a33
a333
3333

where a and b are digits, while they both aren't three.

Therefore, there are 81 numbers for the first three types, 9 for the next 4, and then 1 for the last.

[b]280[/b][/hide]

P.S. I've made a better problem. Find the number of trio numbers in a n digit number there can possibly be. Also, prove it.

You are overcounting quite a bit. For example, for numbers of the form 'a333', you said that there are 9 trio numbers of that type. But if a=3, then you have '3333' which you already covered in the last case, and if a=0 then its '333' which is not a four-digit number.

This is what makes your extension such a difficult problem :) You have to account for overlaps like that.


modular arithmetic
algebra
binomial theorem

1) If f(x) = ax^7 + bx^5 + cx - 5, and f(-5) = 5, what is f(5)?

2) What is the remainder when 2^84 is divided by 5.

Answers:
[hide]
1. -15
2. 1
[/hide]

Hints:
[hide="1"]
$ ax^7\plus{}bx^5\plus{}cx\equal{}\minus{}(a(\minus{}x)^7\plus{}b(\minus{}x)^5\plus{}c(\minus{}x))$
[/hide]
[hide="2"]
$ 2^4\equiv1\pmod5$
[/hide]

[hide="Hint for 1"] f(x)+5 is an odd function.[/hide]
[hide="Solution for 1"]f(-5)+5=10, so f(5)+5=-10, so f(5)=-15[/hide]
[hide="Solution for 2"]2^84=16^21 which is congruent to 1^21=1[/hide]

An non-mod solution for number 2:
[hide]The units digit of powers of 2 cycles, 2,4,8,6, because 84 is divisible by 4, its units digit is the fourth in the sequence, 6.  A number with a units digit of 6 will have a remainder of 1 when divided by 5.  [/hide]

i need help!!

If x + 1/x= 4, then what is the value of xcubed + i/xcubed??

By the binomial theorem, $ (x \plus{} \frac{1}{x})^3 \equal{} x^3 \plus{} 3x \plus{} \frac{3}{x} \plus{} \frac{1}{x^3} \equal{} 4^3 \equal{} 64$.  But we have to rid of the $ 3x \plus{} \frac{3}{x}$.  If you notice, $ 3x \plus{} \frac{3}{x} \equal{} 3(1 \plus{} \frac{1}{x})$, but we already know that's $ 3(4)$!  So we subtract $ 12$ from $ 64$ to get $ \boxed{52}$.


Let $a\in \mathbb{N}$. Consider the sequence $a_n$ defined as follows: $a_0=0$, $a_1=1$, $a_{n+1}=2a_n+(a-1)a_{n-1}$.
For given prime $q$ define the minimal $a$ s.t. for all primes $p\leq q$ we have $p\mid a_p$ and for all primes $p>q$ we have $p\nmid a_p$.

Let $s=1+(a-1)^2$. So, we have that $a_n=\frac 1{2\sqrt s}\left((1+\sqrt s)^n-(1-\sqrt s)^n \right)$, and this becomes $a_p= {p \choose 1}+ {p \choose 3}s+ \dots +{p \choose p}s^\frac{p-1}2$, and this is congruent modulo p to $s^\frac{p-1}2$. So for $p<q$ we have $p \mid 1+(a-1)^2$ but this could be true only if $\left( \frac {-1}p \right) =1$ and this doesn't hold for $p=4k+3$. So, if I didn't make any mistake, there is no $a$ such $p\mid a_p$ for all primes $p<q$...

Then the answer should be the product of all the primes $\le q$, right? :?


Prove that :
$ sin{\frac {A}{2}} \plus{} sin{\frac {B}{2}} \plus{} sin{\frac {C}{2}}\geq\frac {4}{3}(1 \plus{} sin{\frac {A}{2}}sin{\frac {B}{2}}sin{\frac {C}{2}})$

[quote="mehdi cherif"]$ 0\leq\,A,B,C\leq\,\frac {\pi}{2}$   ?????[/quote]
Yes! This inequality is wrong for all triangle.

[quote="mehdi cherif"]$ 0\leq\,A,B,C\leq\,\frac {\pi}{2}$   ?????[/quote]
Sorry .
$ 0\leq\,A,B,C\leq\,\frac {\pi}{2}$ is right


If $ x\equal{}2$ and $ y\equal{}3$ then compute:

$ a)$ $ x\plus{}y$
$ b)$ $ x\minus{}y$
$ c)$ $ 2 \times x \minus{}3 \times y$

hello,
a) $ x\plus{}y\equal{}2\plus{}3\equal{}5$
b) $ x\minus{}y\equal{}2\minus{}3\equal{}\minus{}1$
c) $ 2x\minus{}3y\equal{}2\cdot 2\minus{}3\cdot3\equal{}4\minus{}9\equal{}\minus{}5$
Sonnhard.

Wait, is that parentheses, exponents, multiplication, division, addition, subtraction?(Order of operations.)

Of course, cgyao15.  It's just that thing every fourth grader in America learns.

Hehe, i was just checking(btw, i didn't learn it in 4th grade.)

There is no reason for so much discussion on this topic.


Using vectors, prove that the medians of a triangle are concurrent.
Show all your work.

[hide="Solution"]Let the vertices of the triangle be $\vec{a},\vec{b},\vec{c}$ and $\vec{m}= \frac{\vec{a}+\vec{b}+\vec{c}}{3}$. The medians are fragments of lines connecting the endpoints of vectors $\vec{a},\vec{b},\vec{c}$ with the endpoints of vectors $\frac{\vec{b}+\vec{c}}{2}, \frac{\vec{c}+\vec{a}}{2}, \frac{\vec{a}+\vec{b}}{2}$ respectively. If we prove that the endpoint of $\vec{m}$ lies on the first of mentioned medians, by the symmetry it will lie also on the other medians, so they will be concurrent. Thus it sufficies to prove that the endpoints of vectors $\vec{a},\vec{m},\frac{\vec{b}+\vec{c}}{2}$ are collinear, i.e. the vectors $\vec{a}-\vec{m}$ and $\vec{m}-\frac{\vec{b}+\vec{c}}{2}$ are parallel. But  $\vec{a}-\vec{m}= \frac{2\vec{a}-\vec{b}-\vec{c}}{3}= \frac{2\vec{a}+2\vec{b}+2\vec{c}}{3}-\vec{b}-\vec{c}=$ $2\cdot\left(\frac{\vec{a}+\vec{b}+\vec{c}}{3}-\frac{\vec{b}+\vec{c}}{2}\right)= 2\cdot \left(\vec{m}-\frac{\vec{b}+\vec{c}}{2}\right)$ which ends the proof. [/hide]

[quote="mathgeniuse^ln(x)"]Say, we didn't know that m=a+b+c all over 3.

Then what would we do.

I like your proof though.[/quote]
I defined a mark $\vec{m}= \frac{\vec{a}+\vec{b}+\vec{c}}{3}$, then claimed that all medians pass through the endpoint of this vector and proved it. I did not use any facts about the properties of $\vec{m}$, so the proof is complete.

[quote="JavaMan"]Nice proof.

Do you know where I can get other vector proof problems?[/quote]
Just search for [i]vector[/i] in the Geometry forum; it will give lots of results.
Besides of it, here is a beautiful problem, which is very hard by elementar geometry, but can can be easily done by vector dot product:

Given four lines $k,l,m,n$ such that every two of them are concurrent, but no three of them are concurrent. We consider four triangles created by triples of lines $k,l,m$; $l,m,,n$; $m,n,k$; $n,k,l$. Prove that orthocentres of these triangles are collinear.

[quote="mathgeniuse^ln(x)"]Random, but say I have three vectors that when you add them up together, they equal 0.  Does that mean that the three vectors are concurrent?

It seems like that is true for medians.[/quote]
How would you define concurrent vectors?

[quote="mathgeniuse^ln(x)"]Three vectors that meet at a single point.

Because in your problem, we get the three medians as

b-(a+c)/2, c-(a+b)/2, a-(b+c)/2, add them up, 0.[/quote]
Any $n>0$ vectors are then concurrent, because their startpoints can be set as $0$. To describe a fragment of a line using a vector, you need to specify its startpoint. If you only know that some vectors sum up to $0$, you do not have enough information to say anything about the concurrence of fragments of lines they are represented by.

Well, yeah, I know.

But say that in your problem, my vectors started at the midpoints and ended at the vertices.  Those were my vectors.  they didn't start at the origin, I believe.

But yeah, I know what you are sort of saying.

[quote="mathgeniuse^ln(x)"]Okay.

I just want to also know if there is a way of doing the same for angle bisectors and altitudes.[/quote]

For altitudes: let $H$ be the intersection of the altitudes through $A$ and $B$.  Then $(A-H)\cdot (B-C) = (B-H)\cdot (A-C) = 0$.

$(C-H) \cdot (A-B)= C\cdot A-C \cdot B-H\cdot A+H\cdot B = (A \cdot B-A \cdot H-B \cdot C+C \cdot H)-(A\cdot B-A \cdot C-B \cdot H+C \cdot H)$.

But the right-hand side is just $(B-H)\cdot (A-C)-(A-H)\cdot (B-C) = 0$, so $(C-H)$ and $(A-B)$ are orthogonal, so the line $CH$ is perpendicular to the line $AB$, so it's the altitude, and the altitudes are concurrent.

[quote="mathgeniuse^ln(x)"]But say that in your problem, my vectors started at the midpoints and ended at the vertices.  Those were my vectors.  they didn't start at the origin, I believe.

But yeah, I know what you are sort of saying.[/quote]

More likely, you sort-of know what he's saying.  The sum of three vectors is independent of their locations, so it isn't going to provide you any useful information about how they lie in space relative to each other.  Your question only makes sense in a situation where you have locations for the vectors, and it should be clear that given three vectors (summing to 0 or otherwise), there are some locations I can place them where the lines they define are concurrent and many others where the lines they define won't be concurrent.


[quote="epatjn"]For 20 holes, I have to fill them with different items.

I have 12 different sizes of marbles, and want only 8 in the 20
I have 10 different sizes of  bouncy balls; want only 10
I have 6 different types of yo-yos, want only 5

How do I figure out how many distinct combinations I have[/quote]
p=
i fear the wording is confusing me, by different sizes you mean they all are different right? and you have one of each.
and may more than one ball go into a hole? if not:(you see i think i know what you really want to know)

[hide="if p this might help"]
since they are all distinct we
find combos of the marbles separately and the bouncy balls separately and the yo-yos separately as well.
then we multiply these numbers
EXAMPLE OF WHAT I SAID:
say we had two holes with
two marbles: B and Y there are two combos definitely.
then we have one jack in the box: Bob
he has two combos
so we have 4 in all! I'll list them to show it brute force
when B Y
Bob -
- Bob
when Y B
   Bob -
   - Bob
[/hide]

All 20 holes have with no duplicates.   [u][b]NOTE ERROR on the 10...[/b][/u]I have 12 different sizes of marbles, and want only 8 (no duplicate marbles) in 8 of the 20  holes
I have 10 different sizes of bouncy balls; want only [u][b]7[/b][/u]I have 6 different types of yo-yos, want only 5

How do i figure out the distinct combinations.

I think the wording of this problem is somewhat ambiguous.  However, this is *very similar* to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=139964]a problem[/url] you posted in the MATHCOUNTS Forum.  If you intend these to be the same problem, please avoid double-posting.  If you intend this to be a distinct problem, I think it needs some clarification.

I think I understand the problem.

There are 12 different marbles, of which 8 are randomly selected.
There are 10 different bouncy balls, of which 7 are randomly selected.
There are 6 different yo-yos, of which 5 are randomly selected.

Then, these 20 objects are arranged randomly. How many such arrangements are there?

[hide="Answer"]$\binom{12}{8}\cdot\binom{10}{7}\cdot\binom{6}{5}\cdot20!=867086275714154496000000$[/hide]


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ n$ be a nonzero integer. An abelian group $ A$ is called $ n$-divisible if for every $ x \in A$, there exists $ y \in A$ such that $ x\equal{}ny$. An abelian group $ A$ is called n-torsionfree if $ nx\equal{}0$ for some $ x \in A$ implies $ x\equal{}0$. An abelian group $ A$ is called uniquely $ n$-divisible if for any $ x \in A$, there exists exactly one $ y \in A$ such that $ x\equal{}ny$.
Let $ \mu_n : A \rightarrow A$ be the map $ \mu_n(a)\equal{}na$
(a) Prove that $ A$ is $ n$-torsionfree iff $ \mu_n$ is injective and that $ A$ is $ n$-divisible iff $ \mu_n$ is surjective.
(b) Now suppose $ 0 \rightarrow A \overset{f}{\rightarrow} B \overset{g}{\rightarrow} C \rightarrow 0$ is an exact sequence of abelian groups. It is easy to check that the following diagram commutes: [img]http://lh3.ggpht.com/_miqRA1rlzTU/ST31heIElKI/AAAAAAAAACY/jEM6DbRBqmE/1.PNG[/img]
Suppose that $ B$ is uniquely $ n$-divisible. Prove that $ C$ is $ n$-torsionfree if and only if $ A$ is $ n$-divisible.

(a) trivial
(b) snake lemma

ps: Is this homework?

how do i use the snake lemma for (b)?

Here is the picture, it was not showing.

[img]http://lh3.ggpht.com/_miqRA1rlzTU/ST31heIElKI/AAAAAAAAACY/jEM6DbRBqmE/1.PNG[/img]

Nothing attached. Try to apply the [url]http://en.wikipedia.org/wiki/Snake_lemma[/url] where the vertical arrows are multiplication with $ n$.

So here is what I have for the first part of a).

$ A$ is $ n$-torsionfree
$ \Leftrightarrow \forall x\in A\ (nx\equal{}0 \Rightarrow x\equal{}0)$
$ \Leftrightarrow \forall x\in A\ (\mu_n(x)\equal{}0 \Rightarrow x\equal{}0)$
$ \Leftrightarrow \mu_n$ is injective

I am confused now on how to do the second part of a).

How do I do the second part with the snake lemma? I do not know how to do this part.

Here is the picture: [img]http://i719.photobucket.com/albums/ww191/xianghu21/2.png?t=1229062259[/img]
[url]http://i719.photobucket.com/albums/ww191/xianghu21/2.png?t=1229062259[/url]

[I do not know why my image is not showing]

Take the kernels and cokernels of the vertical mappings and apply the snake lemma.

Here is how to do the second part of part a):

$ \mu_n \text{ is surjective}$
$ \Leftrightarrow \forall y\in A,\, \exists x\in A \, (\mu (x) \equal{} y)$
$ \Leftrightarrow \forall y\in A,\, \exists x\in A\, (nx \equal{} y)$
$ \Leftrightarrow \text{ A is n \minus{} divisible}$

For part b):
[b]I am still not sure how to do part b).[/b]

But, you do need to take the kernels and cokernels of the vertical mappings and apply the snake lemma.  Also, write down the exact sequence to use the snake lemma.  I am not sure how to actually do that though.

[quote="xianghu21"]Let $ n$ be a nonzero integer. An abelian group $ A$ is called $ n$-divisible if for every $ x \in A$, there exists $ y \in A$ such that $ x \equal{} ny$. An abelian group $ A$ is called n-torsionfree if $ nx \equal{} 0$ for some $ x \in A$ implies $ x \equal{} 0$. An abelian group $ A$ is called uniquely $ n$-divisible if for any $ x \in A$, there exists exactly one $ y \in A$ such that $ x \equal{} ny$.
Let $ \mu_n : A \rightarrow A$ be the map $ \mu_n(a) \equal{} na$
(a) Prove that $ A$ is $ n$-torsionfree iff $ \mu_n$ is injective and that $ A$ is $ n$-divisible iff $ \mu_n$ is surjective.
(b) Now suppose $ 0 \rightarrow A \overset{f}{\rightarrow} B \overset{g}{\rightarrow} C \rightarrow 0$ is an exact sequence of abelian groups. It is easy to check that the following diagram commutes: [img]http://lh3.ggpht.com/_miqRA1rlzTU/ST31heIElKI/AAAAAAAAACY/jEM6DbRBqmE/1.PNG[/img]
Suppose that $ B$ is uniquely $ n$-divisible. Prove that $ C$ is $ n$-torsionfree if and only if $ A$ is $ n$-divisible.[/quote]

I still do not know how to solve (b) using the snake lemma or any other ways. help!

just solved it. :-)


Hohoho.hey guys this is my first post on this site.let me give an easy one-

If a>0<2,show that a+(1/a)=>2

chao ;)

[hide]
so if a is greater than 0 and less than 2 it has to be one
plugging 1 into the equation
$a+\frac{1}{a}=2\\\\1+\frac{1}{1}=2\\\\1+1=2\\\\2=2$
correct?[/hide]

[quote="smellyfart"]ofcourse.........................not!!!!
arent there any other values betwwen 0 and 2?[/quote]He was just kidding.

[hide="solution"]$(a-1)^2\ge0\Longrightarrow{a}^2-2a+1\ge0\Longrightarrow{a}^2+1\ge2a\Longrightarrow{a}+\frac{1}{a}\ge2$.

This is true for all positive $a$.[/hide]


How many quintiple of integer $(a;b;c;d;e)$ sastifying $a.b.c.d.e=750$
 :D  :blush:

Why not $\dbinom{5}{1}\dbinom{5}{1}\dbinom{7}{3}$?

If they are integers, we may need to multiply with $\dbinom{5}{0}+\dbinom{5}{2}+\dbinom{5}{4}$.


calculus
integration
trigonometry
limit
real analysis
real analysis unsolved

[quote="misan"]can you solve :D

$ \int_{\minus{}\pi}^{\pi}\;\;\frac{\cos\,(n\cdot x)}{5\plus{}4\,\cos\,(x)}\;\;\textbf dx$

$ \forall\,n\in\Bbb{N}^{\plus{}}$[/quote]

Instead of finding $ \int_{\minus{}\pi}^{\pi}\frac{\cos nx}{5\plus{}4\cos x}dx$ rather find $ \int_{\minus{}\pi}^{\pi}\frac{e^{inx}}{5\plus{}4\cos x}dx$.

If $ |z|\equal{}1$ is a unit circle from $ \minus{}\pi$ to $ \pi$ we have that,
$ \int_{\minus{}\pi}^{\pi}\frac{e^{inx}}{5\plus{}4\cos x}dx \equal{}\oint_{|z|\equal{}1}\frac{z^{n}}{5\plus{}2(z\plus{}z^{\minus{}1})}\cdot\frac{1}{iz}dz$
Which works out to be,
$ \frac{1}{i}\oint_{|z|\equal{}1}\frac{z^{n}}{(2z\plus{}1)(z\plus{}2)}dz$

This contour integral has only a single pole within $ |z|\equal{}1$ which is simple $ z\equal{}\minus{}1/2$.

So, $ \mbox{res}(\minus{}1/2) \equal{}\lim_{z\to\minus{}1/2}(z\plus{}1/2)\cdot\frac{z^{n}}{(2z\plus{}1)(z\plus{}2)}\equal{}\frac{\left(\minus{}1/2\right)^{n}}{3}\equal{}\frac{(\minus{}1)^{n}}{3\cdot 2^{n}}$

By residue theorem together with the $ 1/i$ factor gives us,
$ \int_{\minus{}\pi}^{\pi}\frac{\cos nx}{5\plus{}4\cos x}dx \equal{} (\minus{}1)^{n}\frac{\pi}{3\cdot 2^{n\minus{}1}}$

I can only reconfirm the result of sylow_theory.

For $ |b| < 1\quad\sum_{k \equal{} 0}^{\infty}b^{k}\cos kx \equal{}\frac{1\minus{}b\cos x}{1\minus{}2b\cos x\plus{}b^{2}}\;\Rightarrow\; 2\sum_{k \equal{} 0}^{\infty}b^{k}\cos kx\minus{}1 \equal{}\frac{1\minus{}b^{2}}{1\minus{}2b\cos x\plus{}b^{2}}$

$ \int_{\minus{}\pi}^{\pi}\frac{\cos nx}{1\minus{}2b\cos x\plus{}b^{2}}\equal{}\frac{1}{1\minus{}b^{2}}\int_{\minus{}\pi}^{\pi}\cos nx\left(2\sum_{k \equal{} 0}^{\infty}b^{k}\cos kx\minus{}1\right) dx$

$ \equal{}\frac{1}{1\minus{}b^{2}}\left[2\sum_{k \equal{} 0}^{\infty}b^{k}\underbrace{\int_{\minus{}\pi}^{\pi}\cos nx\,\cos kx\, dx}_{\pi\,\delta_{nk}}\minus{}\underbrace{\int_{\minus{}\pi}^{\pi}\cos nx dx}_{ \equal{} 0}\right] \equal{}\frac{2\pi\, b^{n}}{1\minus{}b^{2}}$

In particular for $ b \equal{}\minus{}\frac{1}{2}\quad\int_{\minus{}\pi}^{\pi}\frac{\cos nx}{1\plus{}\cos x\plus{}\frac{1}{4}}\, dx \equal{}\frac{2\pi}{1\minus{}\frac{1}{4}}\left(\minus{}\frac{1}{2}\right)^{n}$

Here's another solution, which I learned from kunny.

Let $ I_{n}\equal{}\int_{\minus{}\pi}^{\pi}\frac{\cos nx}{5\plus{}4\cos x}\, dx$. Using some trigonometry, $ \cos (n\plus{}1)x\plus{}\cos (n\minus{}1)x \equal{} 2\cos nx\cos x$ and

$ I_{n\plus{}1}\plus{}I_{n\minus{}1}\; \equal{}\;\int_{\minus{}\pi}^{\pi}\frac{2\cos nx\cos x}{5\plus{}4\cos x}\, dx\; \equal{}\;\minus{}\frac{5}{2}I_{n}$

Two roots of $ 2x^{2}\plus{}5x\plus{}2 \equal{} 0$ is $ x \equal{}\minus{}2\text{ or }\minus{}\frac{1}{2}$. Then we can write $ I_{n}\equal{} A (\minus{}2)^{n}\plus{}B\left(\minus{}\frac{1}{2}\right)^{n}$.

But since $ I_{n}$ is bounded, $ A \equal{} 0$ and $ I_{n}\; \equal{}\; I_{0}\left(\minus{}\frac{1}{2}\right)^{n}\; \equal{}\;\frac{2\pi}{3}\left(\minus{}\frac{1}{2}\right)^{n}$.


trigonometry
Trigonometric Equations
algebra
factorization
IMO Shortlist
IMO Longlist

Decompose the expression into real factors:
\[E = 1 - \sin^5(x) - \cos^5(x).\]

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

This is most likely going to be my last post before the Bayreuth seminar with the 2nd QED matboj. Expect some nice stolen problems when I am back. ;)

$1-\sin ^{5}x-\cos ^{5}x=\left( \sin ^{2}x+\cos ^{2}x\right) -\sin ^{5}x-\cos ^{5}x$
$=\left( \sin ^{2}x-\sin ^{5}x\right) +\left( \cos ^{2}x-\cos ^{5}x\right)$
$=\sin ^{2}x\left( 1-\sin x\right) \left( 1+\sin x+\sin ^{2}x\right) +\cos ^{2}x\left( 1-\cos x\right) \left( 1+\cos x+\cos ^{2}x\right)$
$=\left( 1-\cos ^{2}x\right) \left( 1-\sin x\right) \left( 1+\sin x+\sin ^{2}x\right) +\left( 1-\sin ^{2}x\right) \left( 1-\cos x\right) \left( 1+\cos x+\cos ^{2}x\right)$
$=\left( 1-\cos x\right) \left( 1+\cos x\right) \left( 1-\sin x\right) \left( 1+\sin x+\sin ^{2}x\right)$
$+\left( 1-\sin x\right) \left( 1+\sin x\right) \left( 1-\cos x\right) \left( 1+\cos x+\cos ^{2}x\right)$
$=\left( 1-\cos x\right) \left( 1-\sin x\right) \left( \left( 1+\cos x\right) \left( 1+\sin x+\sin ^{2}x\right) +\left( 1+\sin x\right) \left( 1+\cos x+\cos ^{2}x\right) \right)$
$=\left( 1-\cos x\right) \left( 1-\sin x\right) \left( 2+\left( \sin x+\cos x\right) \left( 2+\sin x+\cos x+\sin x\cos x\right) \right)$.

The same factorization is given in the official solution, so I won't bother proving that it can't be improved (anyway, this is not part of the problem).

  Darij


calculus
integration
limit
logarithms
real analysis
real analysis unsolved

If $x \ge 1$, let $\pi (x)$ denote the number of primes ${ \le x}$, that is, $\pi (x) = \sum_{p \le x}1$, where the sum is extended over all primes ${p \le x}$

The prime number theorem states: $\mathop{\lim }_{x \to \infty }\pi (x)\frac{{\log x}}{x}= 1$

Take $\vartheta (x) = \sum_{p \le x}{\log p}$, where again the sum is extended over all primes ${p \le x}$. Apostol shows that $\pi (x)$ and $\vartheta (x)$ may be related by Stieltjes integrals and then he lets the issue go with a "these equations* may be used to  prove that the prime number theorem is equivalent to the relation $\mathop{\lim }_{x \to \infty }\frac{{\vartheta (x)}}{x}= 1$"

Of course, I am not satisfied with that.  :P Can anyone prove the equivalence that he suggests?

*He's referring to:

$\vartheta (x){\rm = }\pi (x)\log{\rm }x-\int_{\rm 2}^{\rm x}{\frac{{\pi (t)}}{t}{\rm dt}}$

$\pi (x) = \frac{{\vartheta (x)}}{{\log x}}+\int_{\rm 2}^{\rm x}{\frac{{\vartheta (t)}}{{t\log^{2}t}}{\rm dt}}$

x>=2

I am confused.  :oops: The first equation says \[\frac{\pi (x)\log x}{x}=\frac{\vartheta (x)}{x}+\frac{\log x}{x}\int_{2}^{x}\frac{\pi (t)}{t}dt\] Assuming the PNT and letting $x\to\infty$, we get $1$ on the left. But \[\frac{\log x}{x}\int_{2}^{x}\frac{\pi (t)}{t}dt\approx \frac{\log x}{x}\int_{2}^{x}\frac{dt}{\log t}\approx 1\]
hence $\frac{\vartheta (x)}{x}\to 0$  :?:

:blush: You are right. I had better fix my typo.

Now from the first equation we get \[\frac{\pi (x)\log x}{x}=\frac{\vartheta (x)}{x}+\frac{\log x}{x}\int_{2}^{x}\frac{\vartheta(t)}{t}dt\] Assuming $\vartheta(x)/x\to 1$, the integral term contributes $\sim \frac{\log x}{x}\int_{2}^{x}1 dt\sim \log x$. Still confused.  :?:

Let's try a different approach. Assume the PNT holds. By definition of $\pi(x)$ and $\vartheta(x)$ we have $\vartheta(x)\le \pi(x)\log x$, hence $\limsup \frac{\vartheta(x)}{x}\le 1$. Also, for any $\epsilon>0$ we have $\vartheta(x)\ge (\pi(x)-\pi(x^{1-\epsilon}))\log (x^{1-\epsilon})$, which implies $\liminf \frac{\vartheta(x)}{x}\ge 1-\epsilon$. Look, no integrals!

:rotfl:

Excuse me. It is morning now and I swear I will get it right. I had a 50-50 chance when I switched the variables in the early hours and I got it wrong. This time I swear it is right. I usually use a program to make up the Text before putting it on here but I tried to manually change the variables in here and failed...

I think you could have proved the equivalence yourself in less time that it took to post the correct formulas. Let's look at the second one:
\[\pi (x)\frac{\log x}{x}= \frac{\vartheta (x)}{x}+\frac{\log x}{x}\int_{2}^{x}\frac{\vartheta (t)}{t\log^{2}t}\, dt \]
Suppose we know that $\frac{\vartheta (x)}{x}\to 1$. Then the last term tends to $0$, which can be proved either by L'Hospital's rule, or (better) by noticing that
\[\int_{2}^{x}\frac{\vartheta (t)}{t\log^{2}t}\, dt\sim \int_{2}^{x}\frac{dt}{\log^{2}t}\sim \frac{x}{\log^{2}x}\]
I write $f(x)\sim g(x)$ to indicate that $\frac{f(x)}{g(x)}\to 1$ as $x\to\infty$.