parameterization
probability
geometry
parallelogram
integration
calculus
probability and stats

Let X and Y be independent random variables with an exponential distribution with parameters µ and λ.  Let U = min{X,Y} and V = max{X,Y}.  Also, let W = V - U.

Show that U and W are independent.

I'm aware that in order to show this, I have to show that E(UW)= E(U)*E(W).  But I'm unsure of how to go about it so any help would be great.  Thanks in advance!

[quote="SharkBites77"]I'm aware that in order to show this, I have to show that E(UW)= E(U)*E(W).[/quote]

Actually, no. Showing that $ E(UW)\equal{}E(U)(E(W)$ would show that $ U$ and $ W$ are [i]uncorrelated[/i], but it would not show that they are [i]independent.[/i] Independent does imply uncorrelated, but uncorrelated does not imply independent.

Here's a small example: Let $ (X,Y)$ be random variables distributed in such a way that the points $ (1,0),(0,1),(\minus{}1,0),$ and $ (0,\minus{}1)$ each have probability $ \frac14$. Then $ E(X)\equal{}0,E(Y)\equal{}0,$ and $ E(XY)\equal{}0.$ Hence $ \text{Cov}(X,Y)\equal{}0$ and they are uncorrelated. But they are distinctly not independent. For instance, $ P(X\equal{}0)\equal{}\frac12,P(Y\equal{}0)\equal{}\frac12,$ but $ P(X\equal{}0\text{ and }Y\equal{}0)\equal{}0\ne\frac12\cdot\frac12.$

A suggestion: show that $ X\minus{}U$ and $ Y\minus{}U$ have the same joint distribution as $ X$ and $ Y$.

[quote="mlok"]A suggestion: show that $ X \minus{} U$ and $ Y \minus{} U$ have the same joint distribution as $ X$ and $ Y$.[/quote]

Thanks for the suggestion!  Thank you to Kent as well for correcting my assumption.

I'm guessing that to show that X-U and Y-U have the same joint distribution as X and Y, I have to perform some kind of integration.  I'm unsure of what expression I am integrating, though.

I take my suggestion back: $ (X\minus{}U,Y\minus{}U)$ is not distributed in the same way as $ (X,Y)$. Let's use this definition of [url=http://en.wikipedia.org/wiki/Statistical_independence]independence[/url]: $ U$ and $ W$ are independent if for any real numbers $ a$ and $ b$ we have
$ P[U\le a\, \&\, W\le b]\equal{}P[U\le a]P[W\le b]$.

Now, $ P[U\le a\, \&\, W\le b]\equal{}\mu\lambda \iint_{D}e^{\minus{}\mu x}e^{\minus{}\lambda y}\,dx\,dy$ where $ D\equal{}\{(x,y)\colon \min(x,y)\le a, |x\minus{}y|\le b\}$. Since $ D$ is the union of two parallelograms, we have
$ P[U\le a\, \&\, W\le b]\equal{}\mu\lambda\int_0^{a}e^{\minus{}\lambda y}dy\int_{y}^{y\plus{}b}e^{\minus{}\mu x}dx\plus{}\mu\lambda\int_0^{a}e^{\minus{}\mu x}dx\int_{x}^{x\plus{}b}e^{\minus{}\lambda y}dy$
The first integral is equal to $ \frac{\lambda}{\mu\plus{}\lambda}(1\minus{}e^{\minus{}\mu b})(1\minus{}e^{\minus{}(\mu\plus{}\lambda)a})$, the second is $ \frac{\mu}{\mu\plus{}\lambda}(1\minus{}e^{\minus{}\lambda b})(1\minus{}e^{\minus{}(\mu\plus{}\lambda)a})$.
Letting $ b\to \infty$, we get $ P[U\le a]\equal{}1\minus{}e^{\minus{}(\mu\plus{}\lambda)a}$. Keeping $ b$ fixed and letting $ a\to\infty$, we get a formula for $ P[W\le b]$ which finishes the proof.

Thanks!  But if you let a approach infinity while keeping b fixed, you just get

1-e^(-µb), right?

No, you get $ \frac{\lambda}{\mu\plus{}\lambda}(1\minus{}e^{\minus{}\mu b})\plus{} \frac{\mu}{\mu\plus{}\lambda}(1\minus{}e^{\minus{}\lambda b})$.


hameye tavabe'e $ f :\mathbb{R}\to\mathbb{R}$ ra peyda konid ke baraye har $ x,y$ haghighi :
$ f(yf(x)\plus{}x)\equal{}xf(y)\plus{}f(x)$

soale ghashangie...

aval az hame gharar bedid $ y=0$ pas darim:

$ f(x)=f(x)+xf(0)$ baraye har $ x$e haghighi,pas $ f(0)=0$...

hamchenin age baraye $ t\in\mathbb{R}$ dashte bashim $ f(t)=0$ oon moghe age gharar bedim $ x=t$ natije mishe:

$ f(t)=tf(y)\Rightarrow tf(y)=0$ baraye har $ y$e haghighi,pas ya $ t=0$ ya $ f\equiv 0$ pas tabe $ f\equiv 0$ ye javabe...

hala farz konid $ f\not\equiv0$ oon moghe natije mishe ke $ t=0$ yeni $ f(t)=0\iff t=0$...

khob hala eddea mikonim ke $ f(1)=1$e...

[hide="esbate eddea"]
age $ f(1)\not= 1$ bashe oon moghe gharar bedid $ x=1$ va $ y=\frac{1}{1-f(1)}$ ke chon $ f(1)\not= 1$ pas $ y\in\mathbb{R}$ hala deghat konid ke ba in maghadire $ x,y$ darim:

$ x+yf(x)=y$

pas moadele aslimoon baraye in $ x,y$ be shekle zir dar miad:

$ f(y)=f(1)+f(y)$ pas $ f(1)=0$ amma tebghe oon chizi ke dar avvale esbatemoon goftim natije mishe ke $ 1=0$ ke inam tanaghoze pas farze $ f(1)\not=1$ ghalat boode yeni $ f(1)=1$...
---------------------------------------------------------
[/hide]

khob pas ta inja natije shod ke $ \boxed{f(t)=0\iff t=0}$ va $ \boxed{f(1)=1}$...

khob hala too soorate masale gharar bedid $ x=1$,pas darim:

$ f(y+1)=f(y)+1$ pas $ \boxed{f(y+1)-f(y)=1}$

hamchenin be komake esteghra natije mishe ke

$ \boxed{\forall y\in\mathbb{Z}: f(y)=y}$

khob hala too soorate masale age farz konim $ x\in\mathbb{Z}$ oon moghe khahim dasht:

$ f(x+xy)=x+xf(y) (I)$ baraye har $ x$e sahih va $ y$e haghighi...

amma deghat konid ke ghablan dashtim $ f(y+1)-f(y)=1$ pas azin natije mishe ke $ f(x+xy)-f(xy)=x (II)$ baraye $ x$haye sahih va $ y$haye haghighi...

hala az $ (I),(II)$ natije mishe ke:

$ f(xy)=xf(y) (*)$

khob hala gharar bedid $ x=-1$ pas $ \boxed{f(-y)=-f(y)}$ baraye har $ y$e haghighi...

khob hala eddea mikonim ke tabamoon jamie yani $ f(a+b)=f(a)+f(b)$...

[hide="esbate eddea"]

bebinid darim:

$ a=\frac{a+b}{2}+\frac{a-b}{2}=\frac{a+b}{2}+\frac{\frac{a-b}{2}}{f(\frac{a+b}{2})}.f(\frac{a+b}{2})$

amma deghat konid ke too soorate masale age gharar bedim $ x=\frac{a+b}{2}$ va $ y=\frac{\frac{a-b}{2}}{f(\frac{a+b}{2})}$ natije mishe ke:

$ f(a)=f(x+yf(x))=f(x)+xf(y)$

be hamin tartib darim:

$ b=\frac{a+b}{2}+\frac{\frac{b-a}{2}}{f(\frac{a+b}{2})}.f(\frac{a+b}{2})$

pas ba hamoon estedlale ghabli natije mishe ke:

$ f(b)=f(x-yf(x))=f(x)+xf(-y)=f(x)-xf(y)$

pas khahim dasht:

$ f(a)+f(b)=2f(x)=2f(\frac{a+b}{2})$

amma too rabete $ (*)$ age gharar bedim $ x=2,y=\frac{a+b}{2}$ natije mishe ke:

$ f(a)+f(b)=2f(\frac{a+b}{2})=f(a+b)$ yeni hamoon chizi ke mikhastim...

--------------------------------------------------------
[/hide]

khob pas didim ke $ f(a+b)=f(a)+f(b)$ baraye har $ a,b$e haghighi,pas soorate masalaro age bazesh konim mishe:

$ f(x+yf(x))=f(x)+f(yf(x))=f(x)+xf(y)$

pas darim:

$ f(yf(x))=xf(y)$hala gharar bedid $ y=1$ natije mishe:

$ f(f(x))=x$ khob hala biaid too soorate masale bejaye $ x$ gharr bedid $ f(x)$ mishe:

$ f(f(x)+yf(f(x)))=f(f(x))+f(x)f(y)$

ke age bazesh konim mishe:

$ f(f(x))+f(xy)=f(f(x))+f(x)f(y)$ ke age sadash konim mishe:

$ f(xy)=f(x)f(y)$ baraye har $ x,y$e haghighi,yeni tabamoon zarbi ham hast...

pas sabet shod ke tabamoon zarbie ghablanam sabet shod ke jamie pas tabamoon tabe hamanie yeni:

$ \forall x\in\mathbb{R}: f(x)=x$...
pas in mnasale dota javab dare:
$ f\equiv 0,f(x)=x$...


This is the thread for it. Players and rules for it are on the bottom of the second page of the main thread. Remember, only posts before 5:00 PM EDT by the AoPS Clock count (as in, 4:59 counts, 5:00 doesn't).


This is probably not that hard though... but I need help.

This is the question:

"Let m and n be the roots of the quadratic equation 4x^2-5x+3=0 Find (m+7)(n+7)"

You only need to do the part before "Find (m+7)(n+7)

For these types of questions in general, you don't need to [hide="do this"]solve for the actual roots.[/hide]

[hide="Solution"]By Vieta's Equations, $ m \plus{} n \equal{} 5/4$ and $ mn \equal{} 3/4$. Therefore, $ (m \plus{} 7)(n \plus{} 7) \equal{} mn \plus{} 7(m \plus{} n) \plus{} 49 \equal{} 3/4 \plus{} 35/4 \plus{} 49 \equal{} \frac {117}{2}$.[/hide]

[hide]

By using the quadratic equation you can find that the roots are $ \frac{5\plus{}\sqrt{23}i}{8}$ and $ \frac{5\minus{}\sqrt{23}i}{8}$. adding 7 to each and multiplying yields
$ (\frac{61\plus{}\sqrt{23}i}{8})(\frac{61\minus{}\sqrt{23}i}{8})$

$ \frac {61^2 \plus{} 23}{64}$

$ \frac {3744}{64}$

$ \frac {117}{2}$
[/hide]

Yes in this case you could do both things (darkdieuguerre's and zserf's), but in other problems that give you a polynomial with a greater degree (cubic,quartic,and so on), you won't be able to find the roots at all! In those cases, you want to use Vieta's formulas (and later, Newtons Sums).

[url]http://mathworld.wolfram.com/VietasFormulas.html[/url]

[hide="Yet another solution"]If the roots of $ 4x^2\minus{}5x\plus{}3$ are $ m$ and $ n$, then $ m\plus{}7$ and $ n\plus{}7$ are the roots of $ 4(x\minus{}7)^2\minus{}5(x\minus{}7)\plus{}3\equal{}4x^2\minus{}61x\plus{}234$.
Then by Vieta's, the product of the roots of that thing is $ \frac{234}{4}\equal{}\boxed{\frac{117}{2}}$.[/hide]

darkdieuguerre's way is cleaner though.
And zserf's way is the ugliest possible approach, though it still works ;)

[hide="Even Another Solution"]
We can write $ f(x) \equal{} 4x^2 \minus{} 5x \plus{} 3 \equal{} 4(x \minus{} m)(x \minus{} n) \equal{} 4(m \minus{} x)(n \minus{} x)$. We have $ f(\minus{}7) \equal{} 4(\minus{}7)^2 \minus{} 5(\minus{}7) \plus{} 3 \equal{} 196 \plus{} 35 \plus{} 3 \equal{} 234$. Thus, $ (m \plus{} 7)(n \plus{} 7) \equal{} \frac{4(m \minus{} (\minus{}7))(n \minus{} (\minus{}7))}{4} \equal{} \frac{f(\minus{}7)}{2} \equal{} \frac{234}{4} \equal{} \boxed{\frac{117}{2}}$.
[/hide]


Let $ x_1,x_2,\ldots,x_n$ be real numbers greater than 1. Show that \[ \frac{x_1x_2}{x_3}\plus{}\frac{x_2x_3}{x_4}\plus{}\cdots\plus{}\frac{x_nx_1}{x_2}\ge4n\] and determine when the equality holds.

[quote="Johan Gunardi"]Let $ x_1,x_2,\ldots,x_n$ be real numbers greater than 1. Show that
\[ \frac {x_1x_2}{x_3} \plus{} \frac {x_2x_3}{x_4} \plus{} \cdots \plus{} \frac {x_nx_1}{x_2}\ge4n
\]
and determine when the equality holds.[/quote]

try $ x_1\equal{}x_2\equal{}\cdots \equal{}x_n\equal{}2$, $ LHS\equal{}2n<4n$..

The correct problem is:

Let $ x_1,x_2,\dots,x_n$ be real numbers such that $ x_i>1$ for $ i\equal{}1,2,3,\dots,n$. Prove that
\[ \frac{x_1x_2}{x_3\minus{}1}\plus{}\frac{x_2x_3}{x_4\minus{}1}\plus{}\frac{x_3x_4}{x_5\minus{}1}\plus{}\dots \plus{} \frac{x_nx_1}{x_2\minus{}1} \ge 4n\] and determine when the equality occurs.

I've ever posted it before...

I think n should be greater than 2 :blush:

Let $ y_i\equal{}x_i\minus{}1$ for $ i\equal{}1,2,...,n$. Then the inequality is

\[ \sum \frac{y_iy_{i\plus{}1}\plus{}y_i\plus{}y_{i\plus{}1}\plus{}1}{y_{i\plus{}2}}\equal{}\sum \frac{y_iy_{i\plus{}1}}{y_{i\plus{}2}}\plus{}\frac{1}{y_{i\plus{}2}}\plus{}\sum \frac{y_i}{y_{i\plus{}2}}\plus{}\frac{y_{i\plus{}1}}{y_{i\plus{}2}}\]Using the AM-GM twice, \[ \ge 2\sum 2\sqrt{ \frac{y_iy_{i\plus{}1}}{y_{i\plus{}2}^2} }\ge 4n\sqrt[n]{\prod\frac{y_iy_{i\plus{}1}}{y_{i\plus{}2}^2}}\equal{}4n\] by the AM-GM on each sum.

(an example of equality is when $ y_i\equal{}1$ or $ x_i\equal{}2$  $ i\equal{}1,2,...,n$.

[quote="Altheman"]Let $ y_i \equal{} x_i \minus{} 1$ for $ i \equal{} 1,2,...,n$. Then the inequality is
\[ \sum \frac {y_iy_{i \plus{} 1} \plus{} y_i \plus{} y_{i \plus{} 1} \plus{} 1}{y_{i \plus{} 2}} \equal{} \sum \frac {y_iy_{i \plus{} 1}}{y_{i \plus{} 2}} \plus{} \frac {1}{y_{i \plus{} 2}} \plus{} \sum \frac {y_i}{y_{i \plus{} 2}} \plus{} \frac {y_{i \plus{} 1}}{y_{i \plus{} 2}}
\]
Using the AM-GM twice,
\[ \ge 2\sum 2\sqrt {\frac {y_iy_{i \plus{} 1}}{y_{i \plus{} 2}^2} }\ge 4n\sqrt [n]{\prod\sqrt{\frac {y_iy_{i \plus{} 1}}{y_{i \plus{} 2}^2}} }\equal{} 4n
\]
by the AM-GM on each sum.

(an example of equality is when $ y_i \equal{} 1$ or $ x_i \equal{} 2$  $ i \equal{} 1,2,...,n$.[/quote]

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=240954


modular arithmetic
number theory
divisibility tests

So I have this problem:
Show that n(n^2-1) is divisible by 24 when n is an odd number.
An odd number is always in the form 2k±1.
So (2k±1)(4k^2±4k+1-1) = (2k±1)(4k^2±4k) = (8k^3±12k^2±4k).
I know that's divisible by 8, 2, and 1. But how do I finally prove that it is divisible by 24?

Here is a hint:

[hide]Check $n(n^2-1) \pmod 3$[/hide]

Solution:

[hide]Check it mod 3. We get 0 for n=0, 1(1-1)=0 for n=1, 2(4-1)=6=0 for n=2, so its always divisible by 3.[/hide]

Considering the post by younglee1989, maybe he hasn't learned modulus yet...
[hide=""Here is an old-fashioned way to check divisibility by 3""]
Write $n(n^2-1)=(n-1)n(n+1)$. So we have a product of 3 consecutive integers. But, among any 3 consecutive integers one of them is always bound to be divisible by 3!!![/hide]
Hopefully my observation helps out...

I see... I will try to work this one through using the second post.

here's how i did it...

u can factor this into (n-1)n(n+1). Since n is an odd number, therefore (n-1) and (n+1) must both be even. We know that these 3 numbers are consecutive integers, so amongst them one must be divisible by 3. Also, with 2 consecutive even integers, one must be divisible by 4. Thus, u have one number that's divisible by 2, one by 3 and another by 4...therefore multiplying these together, the number (n-1)n(n+1) must be divisible by 2*3*4=24.

HOpe this helps.
Cheers!

P.S: how do you hide answers, like tetrahedr0n and 3X.lich did above?

[quote="mandy_pal"]P.S: how do you hide answers, like tetrahedr0n and 3X.lich did above?[/quote]
Just highlight your answers and click Hide (right above where you're typing).

Not the best solution, but a solution none the less.

n(n 2 -1) = 0 mod 24 is what we are trying to prove for odd n.

If you plug 2k+1 = n then you can simplify n(n 2 -1) to 4k(k+1)(2k+1).
If you recognize that k(k+1)(2k+1)/6 is the sum of all squares you can see that        k(k+1)(2k+1) is always divisible by 6.

Obviously any multiple of 6 multiplied by 4 will always be evenly divisible by 24.

Since n(n^2-1)=(n-1)n(n+1)
When n is odd, (n-1) and (n+1) are even. One of them is divisible by 4. So it is divisible by 8.
Since n-1 n and n+1 are 3 consective numbers, one of them must be divisible by 3.
So it is divisible by 24.


If $ (x \plus{} y)^2 \equal{} 105$ and $ x^2 \plus{} y^2 \equal{} 65$, what is the value of $ xy$?

Alternatively, note that $ x^2 \plus{} y^2$ factorizes as $ (x \plus{} y)^2 \minus{} 2xy$, so we can substitute this in our second equation to get $ 105 \minus{} 2xy \equal{} 65 \implies \minus{} 2xy \equal{} \minus{}40 \implies xy \equal{} \boxed{20}$.


calculus
integration
inequalities
logarithms
real analysis
real analysis unsolved

Prove the following  inequality.

\[\frac{e-1}{n+1}\leqq\int^e_1(\log x)^n dx\leqq\frac{(n+1)e+1}{(n+1)(n+2)}\  (n=1,2,\cdot\cdot\cdot)  \]

How did you prove upper bound?
I can only obtain something like $\frac{e}{n+1}-\frac{1}{e^n(n+1)}$. And I don't want to make further attemps too complicated.

Let $I_n=\int^e_1( \log x)^n dx$,by integral by parts,we have $I_{n+1}=e-(n+1)I_n$.  As $I_{n+1}\leqq I_{n }\eqq\cdots\leqq I_{2}\leqq I_{1}$,

we obtain $I_{n+1}\leqq 1$,thus $I_{n+2}=e-(n+2)I_{n+1}=e-(n+2)\left[e-(n+1)I_{n}\right]\leqq 1$,we are done.

kunny


[b]ABRORBEK wrote:[/b]
$ 9.R^2\le a^2 \plus{} b^2 \plus{} c^2 \plus{} (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2. \ \ \ (1).$
If denote: $ p\equal{}\frac{a\plus{}b\plus{}c}{2}$ and  $ p \minus{} a \equal{} x > 0, p \minus{} b \equal{} y > 0, p \minus{} c \equal{} z > 0$, then:
$ (1) \Leftrightarrow 9.(x \plus{} y)^2.(x \plus{} z)^2.(y \plus{} z)^2\le 64.xyz.(x \plus{} y \plus{} z).(x^2 \plus{} y^2 \plus{} z^2); (\forall)x,y,z > 0.$    (2)
[b](2)  TRUE???[/b]

[quote="ABRORBEK"]:arrow: every triangle prove that: $ 9R^2$\leq$ a^2 \plus{} b^2 \plus{} c^2 \plus{} (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2$[/quote]
I have found that $ 9R^2\geq a^2 \plus{} b^2 \plus{} c^2 \plus{} (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2$ is wrong too.  :wink:

every triangle prove that: $ a^2 \plus{} b^2 \plus{} c^2 \plus{} (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2$\leq$ 9R^2$


[quote="ashrafmod"]ABC is an isosceles right-angled triangle. K and M are given points on the hypotenuse AB. K separates A and M, and .<kcm =45  .Prove that $AK^{2}+MB^{2}=KM^{2}$[/quote]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=27981 (scroll down to the end of the topic), and see http://www.mathlinks.ro/Forum/viewtopic.php?t=371 for a converse.

  darij


Is there a way to block certain people from blogs? Like to make it so that they cannot view what is in the blog?

Umm...I may be wrong here, but I believe if you scroll to the top-left ot the AoPS page there should be a rectangle with "Profile" written in it. From there I believe you got to blog options or something and then it should be self-explanatory with a block list somewhere on that page...

Hope this helps


Or, more generally, find the determinant of $I_{n}+a^{T}b$ where $a=(a_{1},\dots,a_{n})$ and $b=(b_{1},\dots,b_{n})$ are two rows.


Each cat caught one mice in 10 minutes. It's easy to think that it's one cat catches one mice in 1 minutes, but that means 10 cats would take turns catching mice.  :D

One cat catches 10 mice in 100 minutes. 100 mice is ten-fold, so just get 10 cats.

Sorry if there is math involved, but I solved it this way.  :oops:

i don't think ten cats is the answer becuase 10 cats cathc 10 mice in 10 minutes.

It is right. 10 cats catch 10 mice in 10 minutes. The number of cats will not change b/c there are more mice, so the time it'll take to catch them will be longer.

ONE cat cathes ONE mice in 10 minutes.

one cat will catch TEN in ten times the length (100 minutes).

to catch 100 mice, you'll need 10 cats.

[quote="risinglegend"]If ten cats can catch 10 mice in 10 minutes, how many cats are required to catch 100 mice in 100 minutes?[/quote]

[hide=" simple reasoning"]1 mouse per cat in 10min
10 mice per cat in 100 min
so 100/10 = 10 cats[/hide]

[quote="risinglegend"]If ten cats can catch 10 mice in 10 minutes, how many cats are required to catch 100 mice in 100 minutes?[/quote]
[hide]
This means the cats catch one mouse in ten minutes.

If one cat catches one mouse every ten minutes, then all you need is ten cats:

$10 \text{ cats} \times 10 \text{ minutes...}$

$ \Rightarrow  100 \text{ mice.}$

$\boxed{\boxed{\boxed{\boxed{\text{Answer: 10 cats}}}}}$[/hide]

easy, if it takes a 10 times as long to get 10 times as many stuff we divide both sides of the algebra problem to get 10 cats, same thing! :D

-----------------------------------------------------------------------------------
why don't we have a Mrs Green as a smiley?


$ \text{Suppose }  p(x)\equal{}6x^2\minus{}65x\plus{}117, x\in \mathbb{Z}$

$ \text{Let  }  p(x)\equal{}m(m\plus{}1) ,m\in \mathbb{N}$

$ \text{ Other than }  (m\plus{}1,m)\equal{}(9,8 ) \text{ or }  (13,12)\text{  are there  other possible solutions to m ?}$

$ \text{Has the equation many solutions?}$

There have been many questions like this posted recently.  Rather than give you the full solution, I will tell you that this is a [url=http://en.wikipedia.org/wiki/Pell_equation]Pell's equation[/url] and that it should have infinitely many solutions.


geometry
3D geometry
real analysis
real analysis unsolved

Find the value(s) of $ x,y$ in $ \mathbb Z$ of the following equation :

$ \sum\limits_{t\equal{}0}^{7} (x\plus{}t)^3 \equal{} y^3$

 :D

I will solve it for $ x$ positive. Expanding the RHS we get: $ 784 \plus{} 420 x \plus{} 84 x^2 \plus{} 8 x^3 \equal{} A$ consider $ (2x \plus{} 7)^3 \equal{} 343 \plus{} 294 x \plus{} 84 x^2 \plus{} 8 x^3 \equal{} B$, $ A > B$ because $ A \minus{} B \equal{} 441 \plus{} 126 x$ that is positive if $ x\geq \minus{} 3$ and let $ (2x \plus{} 8)^3 \equal{} 512 \plus{} 384 x \plus{} 96 x^2 \plus{} 8 x^3 \equal{} C$, $ C \minus{} A \equal{} \minus{} 272 \minus{} 36 x \plus{} 12 x^2$. $ C \minus{} A$ has irrational roots and $ C \minus{} A\geq 0$ if $ x\geq 7$, $ \Rightarrow$ $ A$ is between two cubes if $ x\geq 7$

I'll continue later....

Consider $ (2x\plus{}11)^3\equal{}D$ then $ D\minus{}A\equal{}547 \plus{} 306 x \plus{} 48 x^2$, $ D\minus{}A>0$ Because it has imaginary roots. Consider $ (2x\plus{}3)^3\equal{}27 \plus{} 54 x \plus{} 36 x^2 \plus{} 8 x^3\equal{}E$, $ A\minus{}E\equal{}\minus{}776 \minus{} 396 x \minus{} 60 x^2<0$  because it has imaginary roots. Then, if $ A$ is a cube it must be one of: $ (2x\plus{}4)^3,(2x\plus{}5)^3,(2x\plus{}6)^3,(2x\plus{}7)^3,(2x\plus{}8)^3,(2x\plus{}9)^3,(2x\plus{}10)^3$. Making all equations.

We get the only solutions: $ (x,y)\equal{}(\minus{}5,\minus{}6), (\minus{}4, \minus{}4), (\minus{}3,4), (\minus{}2,6)$


[I hope to occasionally post problems from the New York State Mathematics League (NYSML).  Here is my first such post.]

Points $P$ and $Q$ are on circle $O$, and chord $\overline{PQ}$ is drawn.  A second circle is drawn with diameter $\overline{OP}$, crossing the chord at point $S$.  If $OP = 7$ and $PQ = 12$, compute $PS$.

[hide]Draw a perpendicular to PQ from O. This bisects PQ, so let's call PQ's midpoint M. Right triangle OPM is inscribed in the circle whose diameter is OP. Therefore M=S, so PS=PM=6.[/hide]

(Edited by moderator to hide solution.)


linear algebra
matrix
linear algebra unsolved

Prove that any real matrix $A$ of trace 0 can be written as $XY-YX$ where $X$ is real symetric and Y has trace 0.

Let $tr(A) =0$ and we can use the known fact, duscussed many times on the forum that $A$ is orthogonally similar to the matrix which has zeros on the main diagonal(proof can be followed for example by induction). Let $O A O^{t}= B$ with $b_{ii}= 0$. Take any diagonal matrix $D = diag(d_{1},d_{2},...,d_{n})$ with different $d_{i}$'s and put $t_{ii}= 0, t_{ij}=\frac{b_{ij}}{d_{i}-d_{j}}$ for $i \neq j$. Then we have $DT-TD = B = O A O^{t}$, thus $A = O^{t}DTO-O^{t}TDO = XY-YX$, where $X = O^{t}DO$, $Y = O^{t}T O$, $X$ is symmetric and $tr(Y) = 0$.


Find all continuous functions $ f: R \rightarrow R$ such that $ 9f(8x)\minus{}9f(4x)\plus{}2f(2x) \equal{} 100x$ , $ \forall x \in R$

Let $ h(x) \equal{} f(x) \minus{} \frac {5x}{2}$
We has :$ 9h(8x) \minus{} 9h(4x) \plus{} 2h(2x) \equal{} 0$
so $ 3(3h(8x) \minus{} 2h(4x)) \equal{} 3h(4x) \minus{} 2h(2x)$
Let $ m(x) \equal{} 3h(2x) \minus{} 2h(x)$
$ 3m(2x) \equal{} m(x)$
So $ m(x) \equal{} \frac {m(\frac {x}{2})}{3}$ so $ m(x) \equal{} \frac {m(\frac {x}{2^n})}{3^n}$
$ m(x)$ is continious so $ m(x) \equal{} m(0) \equal{} 0$
So $ 3h(2x) \equal{} 2h(x)$ easy to check $ h(x)\equiv 0$
SO $ f(x) \equal{} \frac {5x}{2}$


Plane geometry just doesn't seem like it would fit with orals,  and while inversion would indeed be a very interesting topic, I don't see it working very well with the orals format.  If you really wanted a geometry orals topic, I think that elliptical or hyperbolic geometry would work well as an orals topic.

[url]http://www.viebach.net/ictm/HS/oral06.pdf[/url]

Here are the oral topics for the 2008 season. 1A and 2A are both getting fair division, as was expected.

Now, divisions 3AA and 4AA are both getting Voting Methods. Any idea what this would entail?

Here's the link to the material: [url=http://bcs.whfreeman.com/fapp6/default.asp?s=&n=&i=&v=&o=&ns=0&uid=0&rau=0]Book 3AA and 4AA will be using[/url].

It looks like we might be scratching the surface of game theory, though this also seems like an extension of fair division.[/url]

[quote="Altheman"]There has been no geometry in orals topics. Therefore, we should do inversion.[/quote]

Actually, the very first orals topic (NSML, not ICTM) I had was taxicab geometry. I think that almost qualifies.

Kinda, but it does have the word "geometry" in the title at least.

This year's oral topic is Graph Theory: Networks and Circuits.

Which is actually a good topic. Who knew.


Prove that the set of all points mapped similarly from a topological space to a $ T_{1}$ space under two different continuous maps  form a closed set.

This is not true, as there are $ T_1$ spaces that are not Hausdorff, and this condition would imply the opposite.

For, if $ X$ is a $ T_1$ space, we would have then that $ \{(x,y) \in X \times X : y \equal{} x\} \subset X \times X$ is closed, which would imply that $ X$ is Hausdorff.

The claim is true for Hausdorff ($ T_2$), spaces, however, and is an alternate definition of being $ T_2$.

How does one prove that for a $ T_{2}$ space? Assuming i define a Hausdorff space to be a topological space where distinct points are separated by disjoint open sets

It's not too hard to show that if $ X$ is hausdorff then $ \Delta \equal{} \{(x,x) \in X \times X\} \subset X \times X$ is closed. If $ x \ne y$ then there are disjoint open sets $ U,V$ such that $ (x,y) \in U \times V$. Since $ U,V$ are disjoint we have $ U \times V \cap \Delta \equal{} \emptyset$. So $ \Delta$ is closed.

From there, standard argument. Let $ f,g : Y \to X$. Then $ (f \times g) : Y \to X \times X$ is induced, and $ (f \times g)^{\minus{}1}(\Delta) \equal{} \{y \in Y : f(y) \equal{} g(y)\}$ is closed.


function
algebra
functional equation
algebra unsolved

Find all monotonous functions $ f: \mathbb{R} \to \mathbb{R}$ that satisfy the following functional equation:
\[f(f(x)) \equal{} f( \minus{} f(x)) \equal{} f(x)^2.\]

[quote="FelixD"]Find all monotonous functions $ f: \mathbb{R} \to \mathbb{R}$ that satisfy the following functional equation:
$ f(f(x)) \equal{} f( \minus{} f(x)) \equal{} f(x)^2.$[/quote]

This implies $ f(x)\equal{}f(\minus{}x)$ $ \forall x\in f(\mathbb R)$. So, since $ f(x)$ is monotonous, $ f(x)\equal{}c$ $ \forall x\in f(\mathbb R)$ and, since $ f(x)\equal{}x^2$ $ \forall x\in f(\mathbb R)$, we get $ c\equal{}c^2$

And the only solutions are :
$ f(x)\equal{}0$
$ f(x)\equal{}1$

put x=f(x),we get f(x^2)=f^2(x)=f(-x^2)
thus f(x)=f(-x), $ \forall x\in \mathbb{R}$
but f(x) is monotonous,so we get f(x)=c,c=0 or 1

[quote="keyree10"][quote="pco"]This implies $ f(x) \equal{} f( \minus{} x) \forall x\in f(\mathbb R)$[/quote]

Why is this true?[/quote]

Hello keyree10!

If $ x\in f(\mathbb R)$, $ \exists y$ such that $ x\equal{}f(y)$ and so $ f(f(y))\equal{}f(x)$ and $ f(\minus{}f(y))\equal{}f(\minus{}x)$ and, since $ f(f(y))\equal{}f(\minus{}f(y))$, we get $ f(x)\equal{}\minus{}f(x)$ $ \forall x\in f(\mathbb R)$

[quote="pco"][quote="FelixD"]Find all monotonous functions $ f: \mathbb{R} \to \mathbb{R}$ that satisfy the following functional equation:
$ f(f(x)) \equal{} f( \minus{} f(x)) \equal{} f(x)^2.$[/quote]

This implies $ f(x) \equal{} f( \minus{} x)$ $ \forall x\in f(\mathbb R)$. So, since $ f(x)$ is monotonous, $ f(x) \equal{} c$ $ \forall x\in f(\mathbb R)$ and, since $ f(x) \equal{} x^2$ $ \forall x\in f(\mathbb R)$, we get $ c \equal{} c^2$

And the only solutions are :
$ f(x) \equal{} 0$
$ f(x) \equal{} 1$[/quote]

It seems to me that all you proved was that $ f(x)\in \{0,1\}$ for all $ x\in f(\mathbb{R})$ and also since $ f(\mathbb{R})\subset\{0,1\}$ it doesn't seem like much has been done.

I see that you are very good at these problems but could you give a solution with more detail?

One step:
WLOG $ f$ is non decreasing.

Suppose that $ x,y\in f(\mathbb{R})$ such that $ x\ge y$. Then $ f(x)\ge f(y)$, also $ \minus{}y\ge \minus{}x$ so $ f(\minus{}y)\ge f(\minus{}x)$ so $ f(y)\ge f(x)$. Therefore $ f(x)\equal{}f(y)$ for all $ x,y\in f(\mathbb{R})$.

[quote="Altheman"] [I see that you are very good at these problems but could you give a solution with more detail?
[/quote]

 :blush: Youre're right, I was a little bit too quick. Sorry for this. I forgot $ \minus{}1$ :

I think you agree with $ f(x)\equal{}f(\minus{}x)$ $ \forall x\in f(\mathbb R)$ so we have, since $ f(x)$ is monotonous, $ f(y)\equal{}f(x)$ $ \forall y\in [\minus{}|x|,\plus{}|x|]$ and $ f(0)\equal{}f(x)$

So, if $ a \in f(\mathbb R)$ and $ b \in f(\mathbb R)$ we have $ f(0)\equal{}f(a)$ and $ f(0)\equal{}f(b)$ and so $ f(a)\equal{}f(b)$

So $ f(x)\equal{}c$ $ \forall x\in f(\mathbb R)$
And, since $ c\in f(\mathbb R)$, $ f(c)\equal{}c$ but also $ f(c)\equal{}c^2$ so either $ c\equal{}0$, either $ c\equal{}1$

1) if $ c\equal{}0$ : we know that :
$ \forall x\in f(\mathbb R)$, $ f(x)\equal{}0$
But, $ \forall x\in f(\mathbb R)$, $ f(x)\equal{}x^2$
so $ \forall x\in f(\mathbb R)$, $ x\equal{}0$ and so $ f(\mathbb R)\equal{}\{0\}$ and so $ f(x)\equal{}0$ $ \forall x$

2) If $ c\equal{}1$ : we know that :
$ \forall x\in f(\mathbb R)$, $ f(x)\equal{}1$
But, $ \forall x\in f(\mathbb R)$, $ f(x)\equal{}x^2$
so $ \forall x\in f(\mathbb R)$, $ x^2\equal{}1$ and so $ f(\mathbb R)\subseteq\{\minus{}1,\plus{}1\}$

and , since $ f(\minus{}1)\equal{}f(1)\equal{}1$, we have 5 solutions (remember $ f(x)$ is monotonous):

$ f(x)\equal{}1$ $ \forall x$
Let $ a<\minus{}1$ and then $ f(x)\equal{}\minus{}1$ $ \forall x\leq a$ and $ f(x)\equal{}1$ $ \forall x>a$
Let $ a\leq \minus{}1$ and then $ f(x)\equal{}\minus{}1$ $ \forall x< a$ and $ f(x)\equal{}1$ $ \forall x\geq a$
Let $ a>1$ and then $ f(x)\equal{}\plus{}1$ $ \forall x<a$ and $ f(x)\equal{}\minus{}1$ $ \forall x\geq a$
Let $ a\geq 1$ and then $ f(x)\equal{}\plus{}1$ $ \forall x\leq a$ and $ f(x)\equal{}\minus{}1$ $ \forall x>a$

I think it's ok now  :blush:


This is a simple problem... yet I can't find a simple solution  :D
Find the hundreds digit of 4^309. any suggestions?

[quote="pkothari13"]Find the hundreds digit of 4^309[/quote]
Do you mean to find 3 final digit of this number?It's $ 144$

what are the fermat theorems?

the only way i do this is:
i know 2^32 = 4294967296
we only want the hundreds so just use 296 and take only the last 3 digits for the calculations
2^64 = 296*296 = 616
2^128 = 616*616 = 456
2^256 = 456*456 = 936
2^512 = 936*936 = 96
2^576 = 96 * 616 = 136
2^608 = 136 * 296 = 256
2^618 = 256 * 24 = 144

so answer is 1
not too bad

[quote="joml88"]The way I found the last three digits is:
[edit] My solution was wrong...I think...sry bout that.[/quote]

turns out right though @@

Another way to solve this, with less calculation, is to invoke Euler's theorem: if $a$ and $n$ are relatively prime, $a^{\phi(n)}\equiv 1 \mod n$, where $\phi(n)$ is the number of integers between 0 and $n$ which are relatively prime to $n$.

For this problem, we need to calculate $4^{309}$ mod $1000$. We know $1000=2^35^3$, and we calculate each part separately. Obviously $4^{309}\equiv 0 \mod 8$.
Since $\phi(125)=100$, $4^{309}\equiv 4^9 \mod 125$
$4^0\equiv 1$
$4^1\equiv 4$
$4^2\equiv 16$
$4^3\equiv 64$
$4^4\equiv 256\equiv 6$
$4^5\equiv 24$
$4^6\equiv 96$
$4^7\equiv 384\equiv 9$
$4^8\equiv 36$
$4^9\equiv 144\equiv 19$

Therefore, $4^{309}\equiv 144 \mod 125$ and $4^{309}\equiv 144 \mod 8$, so $4^{309}\equiv 144 \mod 1000$.

Some comments:
(The answer, if you are used to these kind of problems, could be found  very fast,  even without pen  or paper)

 With the help of Little Theorem of Fermat (Or Euler's):
And noticing  10=2*5  and

phi(5) = 4
phi (25) = 20
phi (125) = 100  etc...

and phi(2)= 1, phi(4)=2, phi(8)=4 etc...

This means, if you want to see just the last digit the cycle will repeat after in 4, for last 2 digits it will repeat in 20, for last 3 digit it will repeat in 100 etc. (You have to be a liitle careful here  the cycle  could start but  there could be a few "stray terms" in the begining)

[b]This means,  all you have to find is the last 3 digit of 4^9  [/b]..
(As long as 4^9 >1000, you have entered the cycle)
and last 3 digits of 4^109, 4^209, 4^309 etc ... are same!

If you are confused about 'stray terms'  or when to start the cycle  just find the value  , both modulo  8 and 125
For 4^ 9 , obviously the answer is 0 mod 8
To find 4^9 mod 125, One can notice,  all mod 125:
2^7 = 128 ==3   so 2^9 == 4*3 = 12
and 4^9 = 12^2 = 144.
(144 happens to be 0 mod 8 so it is the correct answer, else you look for numbers which are  19 mod 125  which are divisible by 8)
Hope this helps.
Edited later: Sorry  did not see the post above,  it is essentially the same method, explained above..


Let $ f(x)\equal{} \frac{1}{4a^2(1\minus{}x)}$  find all a for which there exists an integer n such that $ f(f(f(...f(0)...))) \equal{}1$

[quote="kihe_freety5"]Let $ f(x) \equal{} \frac {1}{4a^2(1 \minus{} x)}$  find all a for which there exists an integer n such that $ f(f(f(...f(0)...))) \equal{} 1$[/quote]

Here is a rather ugly method. Now that I've seen the result, I'm sure there is a pretty one. Maybe somebody else will propose it.

It's rather easy to show that : $ f^{[n]}(x)^ \equal{} \frac {a_nx \plus{} b_n}{c_nx \plus{} d_n}$ with :

$ b_0 \equal{} 0$, $ b_1 \equal{} 1$, and $ b_{n \plus{} 2} \equal{} 4a^2b_{n \plus{} 1} \minus{} 4a^2b_{n}$
$ d_0 \equal{} 1$, $ d_1 \equal{} 4a^2$, and $ d_{n \plus{} 2} \equal{} 4a^2d_{n \plus{} 1} \minus{} 4a^2d_{n}$
$ a_0 \equal{} 1$ and $ a_{n \plus{} 1} \equal{} \minus{} 4a^2b_n$
$ c_0 \equal{} 0$ and $ c_{n \plus{} 1} \equal{} \minus{} 4a^2d_n$

The problem is to find $ a,n$ such that $ b_n \equal{} d_n$. Let us now consider $ \Delta_n \equal{} b_n \minus{} d_n$ as :

$ \Delta_0 \equal{} \minus{} 1$, $ \Delta_1 \equal{} 1 \minus{} 4a^2$ and $ \Delta_{n \plus{} 2} \equal{} 4a^2\Delta_{n \plus{} 1} \minus{} 4a^2\Delta_{n}$

If $ a^2 \equal{} 1$, we get $ \Delta_n \equal{} \minus{} (n \plus{} 2)2^{n \minus{} 1}$ and $ \Delta_n\ne 0$ $ \forall n\ge 0$
If $ a^2\ne 1$, $ X^2 \minus{} 4a^2X \plus{} 4a^2 \equal{} 0$ has two distinct roots $ r_1$ and $ r_2$ $ \in\mathbb C\backslash\{0,1\}$ and we get :

$ \Delta_n \equal{} \frac {1 \minus{} 4a^2 \plus{} r_2}{r_1 \minus{} r_2}r_1^n$ $ \plus{} \frac {1 \minus{} 4a^2 \plus{} r_1}{r_2 \minus{} r_1}r_2^n$

Then $ \Delta_n \equal{} 0$ $ \iff$ $ (1 \minus{} 4a^2 \plus{} r_2)r_1^n \equal{} (1 \minus{} 4a^2 \plus{} r_1)r_2^n$

Using $ 4a^2 \equal{} \frac {r_1^2}{r_1 \minus{} 1}$ (got from $ r_1^2 \minus{} 4a^2r_1 \plus{} 4a^2 \equal{} 0$) and $ r_1 \plus{} r_2 \equal{} 4a^2$ and $ r_1r_2 \equal{} 4a^2$, we get

$ \Delta_n \equal{} 0$ $ \iff$ $ (1 \minus{} r_1)r_1^{2n} \equal{} (1 \minus{} \frac {r_1^2}{r_1 \minus{} 1} \plus{} r_1)(\frac {r_1^2}{r_1 \minus{} 1})^n$

$ \Delta_n \equal{} 0$ $ \iff$ $ (r_1 \minus{} 1)^{n \plus{} 2} \equal{} 1$ and so $ r_1 \equal{} 1 \plus{} e^{i\frac {2k\pi}{n \plus{} 2}}$

And so $ 4a^2 \equal{} \frac {r_1^2}{r_1 \minus{} 1}$ $ \equal{} \frac {(1 \plus{} e^{i\frac {2k\pi}{n \plus{} 2}})^2}{e^{i\frac {2k\pi}{n \plus{} 2}}}$ $ \equal{} 2(1 \plus{} \cos\frac {2k\pi}{n \plus{} 2})$ $ \equal{} 4(\cos\frac {k\pi}{n \plus{} 2})^2$

So $ a \equal{} \pm\cos\frac {k\pi}{n \plus{} 2}$ with $ \frac {2k}{n \plus{} 2}$ not an integer (in order to avoid $ a \equal{} 0$ and $ a\equal{}\pm 1$)

And, with such a $ a$, $ f^{[p]}(0) \equal{} 1$ where $ p \equal{} \frac {n \plus{} 2}{\gcd(n \plus{} 2,k)} \minus{} 2$ (and consider that $ f^{[n]}(0)$ is not defined for $ n > p$)

This may be simplified as :

$ \boxed{a \equal{} \cos(\frac pq\pi)}$ where $ \gcd(p,q)\equal{}1$, $ p,q\in\mathbb N$ and $ q>2$ and then $ f^{[q\minus{}2]}(0) \equal{} 1$


How many numbers are in the list \[ 6,7,10,11,14,15,\ldots,94,95,98?\]

[hide]For every even element $ n$, the number $ 2n \plus{} 1$ is on the list (except for $ n \equal{} 98$).

Subtracting the list of even elements by $ 2$ and dividing by $ 4$, we find that there are $ 24$ pairs of elements, which means that there are $ 24\cdot2 \equal{} 48$ elements total.

However, $ 98 \plus{} 1 \equal{} 99$ is not on the list, so our final answer is $ 48 \minus{} 1 \equal{} \boxed{47}$.

Edit: Hahahaha. I was being mental.[/hide]

That's impossible.  How can you have 185 terms in a sequence of integers from 6-98?

[hide]I have another solution:
First, subtract five to get $1, 2, 5, 6, 9, 10 ... 89, 90, 93$. Notice that the first two numbers of every four are in the list, up to $92$. Thus, half of those numbers are in the list, giving $92/2=46$. Throwing in 93 gives $\boxed{47}$.[/hide]

[quote="infinity1"][hide="I have another solution:"]
First, subtract five to get $1, 2, 5, 6, 9, 10 ... 89, 90, 93$. Notice that the first two numbers of every four are in the list, up to $92$. Thus, half of those numbers are in the list, giving $92/2=46$. Throwing in 93 gives $\boxed{47}$.[/hide][/quote]

I like infinity1's solution better than the solution given by Alcumus; his (or her) approach is more direct and is easier to follow.

[hide="Here's how I approached this problem:"]
First, $98-6+1=93$ tells us that there "should" be 93 terms in this sequence, but it's easy to see that a bunch of the terms are missing. Which ones are missing? Every term that is missing is a multiple of 4, and that multiple's immediate successor (i.e. 8 and 9, 12 and 13, 16 and 17, etc.) So how many multiples of 4 are there between 6 and 98? Those multiples would be $8, 12, 16, 20, ... \text{ }, 92, 96$. Divide each term by 4 to get $2, 3, 4, 5, ... \text{ },  23, 24$. We then subtract $24-2+1=23$ to see that there are 23 multiples of 4 missing from the sequence.

But the sequence is not only missing its multiples of 4, it is also missing the next number after each multiple (i.e. 9, 13, 17, etc.) Since there are 23 multiples of 4, there are also 23 of their successors. Therefore, the number of terms in the sequence is $93-23-23=\boxed{47}$.
[/hide]

Tim

[quote="infinity1"]I have another solution:
First, subtract five to get $1, 2, 5, 6, 9, 10 ... 89, 90, 93$. Notice that the first two numbers of every four are in the list, up to $92$. Thus, half of those numbers are in the list, giving $92/2=46$. Throwing in 93 gives $\boxed{47}$.[/quote]

I actually took the time to start counting the pattern (thats how slow I am)...until i realized infinity1's solution...:D

[quote=PizzaPerson]The Newport Crab Company marks the location of its crab pots with colored balloons. They mark every fourth pot with a red balloon, every sixth pot with a blue balloon, and every tenth pot with a yellow balloon. After placing 600 crab pots in the bay, how many pots have three different colored balloons attached?[/quote]

[quote=ibmo0907]Please stop posting this question everywhere. You must solve it yourself. Or you can make a new thread and ask it in there. But dont keep asking it everywhere.[/quote]

Yes, he's right. Also if you're stuck then asking questions on the message board of the course you are taking is helpful sometimes. :)

[quote=OlympusHero]how is this level 14?

pair up the numbers as follows

6 7
10 11
14 15
...
94 95
98

up till 98 there would be $2 \cdot \frac{94-2}{4} = 46$ terms. adding the $98$ term gives $46+1=\boxed{47}$.[/quote]

Well maybe some people make off by one errors.

Here's my solution:
Notice that every two numbers it increases by 4, so the answer is $\dfrac{98-6}{4} \cdot 2 + 1 = 47.$


Let $ f : R^3 \rightarrow R^4$ be as follows:
\[ f((x_1, x_2, x_3))\equal{}(x_1\plus{}2x_2\plus{}x_3, x_2\plus{}x_3, x_1\plus{}2x_2\plus{}3x_3, x_1\plus{}x_2\plus{}2x_3)\].
Do exist bases $ A$ and $ B$ of $ R^3$ and $ R^4$ respectively such that:
\[ {M(f)^{B}_A}\equal{}\begin{pmatrix}1&\minus{}1&0\\
0&1&\minus{}1\\
1&0&\minus{}1\\1&\minus{}2&1\\
\end{pmatrix}\]

$ {M(f)^{B}_A}$ is a matrix representation of linear transformation $ f$ from base $ A$ to $ B$ I hope you understant my notation.
Please explain your reasoning. I'm interested in comprehensing the method rather than the result.

If you get to choose the bases on both sides, the only invariants are the size of the matrices and the rank of the transformation. What is the rank of $ f$, and what is the rank of that matrix $ M$?

[hide="Proof of the claim"]Suppose $ M$ is a $ m\times n$ matrix of rank $ r$. There exist invertible matrices $ A$ and $ B$ ($ m\times m$ and $ n\times n$ respectively) such that $ A^{\minus{}1}MB\equal{}\begin{bmatrix}I_{r\times r}&0_{r\times n\minus{}r}\\0_{m\minus{}r\times r}& 0_{m\minus{}r\times n\minus{}r}\end{bmatrix}$; the standard rank $ r$ matrix given is the matrix of $ M$ with respect to $ A$ and $ B$.

To construct $ A^{\minus{}1}$, row-reduce $ M$ and track the row operations. This can be done computationally by working with $ \begin{bmatrix}M&I_{m\times m}\end{bmatrix}$. Then construct $ B$ by column-reducing the remaining matrix; computationally, we stick a $ n\times n$ identity below it.

Since any $ m\times n$ matrix of rank $ r$ can be transformed into our standard matrix by a change of basis on each side, any such matrix can be transformed into any other by two such steps.[/hide]

Just to put the particular answer up here:

The rank of $ \begin{bmatrix}1&2&1\\ 0&1&1\\ 1&2&3\\ 1&1&2\end{bmatrix}$ is $ 3.$

The rank of $ \begin{bmatrix}1&\minus{}1&0\\ 0&1&\minus{}1\\ 1&0&\minus{}1\\ 1&\minus{}2&1\end{bmatrix}$ is $ 2.$

By jmerry's result, rank is the one thing that is invariant. These cannot be two matrices for the same linear transformation.

I have never taken a course in Linear Algebra (hopefully I understood things properly), but the following is my understanding and results regarding the above problem.

[u]Lemma 1.[/u] If $ f: K^n\rightarrow K^m$, $ K$ a field, is a linear mapping, then it follows that $ f$ can be represented using the matrix
\[ \left(\begin{array}{ccc}| & & | \\
f(v_1) & \hdots & f(v_n) \\
| & & |\end{array}\right),
\]
where $ \{v_1,\hdots,v_n\}$ is a basis of $ K^n$.

[u]Proof.[/u] Left as exercise. $ \Box$

[u]Lemma 2.[/u] If $ \{f(a_i)\}$ is a basis of $ \text{im}f$. where $ f: V\rightarrow W$ is a linear mapping, then $ V \equal{} \text{span}\{a_i\}\oplus\ker f$, that is, $ \dim\ker f \plus{} \dim\text{im}f \equal{} \dim V$.

[u]Proof.[/u] Use the definitions of direct sum. $ \Box$

Since for a matrix $ U$ that represents $ f: K^n\rightarrow K^m$, $ \dim\text{im}f \equal{} \text{rank} U$, we have $ \text{rank} U \equal{} \dim K^n \minus{} \dim\ker f$.  In our case, $ \dim\ker f \equal{} 0$, that is, $ \text{rank} U \equal{} n$ for all matrices $ A$.  Since $ \dim\text{im}f \equal{} 3$ and $ \text{rank} M(f)_A^B \equal{} 2$, $ M(f)_A^B$ cannot be a representation of $ f$.  Perhaps, a more experienced individual could corroborate the validity of my steps (if correct).


Solve the equation $ 2x^3\minus{}7x^2\plus{}10x\minus{}6$ knowing that one of the roots is rational.

hello, one solution is $ x\equal{}\frac{3}{2}$.
Sonnhard.

[quote="triplebig"]Solve the equation $ 2x^3 \minus{} 7x^2 \plus{} 10x \minus{} 6$ knowing that one of the roots is rational.[/quote]
triplebig,

in your headline and in the body of your post, you did not present
an equation (nor an inequality).  It is an expression.  For instance,
there is no equals sign as part of it.  You did not give an equation
to solve, but the polynomial you did give has roots which are also
the solutions to the related equation $ 2x^3 \minus{} 7x^2 \plus{} 10x \minus{} 6 \equal{} 0$.

[quote="b.s.o"]The equation is equivalent to:
[url=http://www.monsterup.com][img]http://www.monsterup.com/upload/1258317411195.gif[/img][/url]
so : [url=http://www.monsterup.com][img]http://www.monsterup.com/upload/125831746485.gif[/img][/url][/quote]

b.s.o.,

you made an error twice in the factor by grouping.  It appears to be a typo.

The *expression* factors into

$ x^2(2x \minus{} 3) \minus{} 2x(2x \minus{} 3) \plus{} 2(2x \minus{} 3)$

and it becomes

$ (2x \minus{} 3)(x^2 \minus{} 2x \plus{} 2)$,

which factors into

$ (2x \minus{} 3)[x \minus{} (1 \minus{} i)][x \minus{} (1 \plus{} i)]$.

The roots are $ \frac32, 1 \minus{} i, 1 \plus{} i$


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Let me talk a bit about the UK academic perspective. Below I appended an article by Stephen Court who discusses the history of UK academic tenure until up the present. Especially interesting is the section on "current trends in employment of academic staff" starting at page 7/9. It discusses the casualization of academic staff. For a large part of the 20th century academics of all levels had job security of tenure except for certain legal cases, but this certainty to be most likely employed until retirement is gone now for most academic staff. Before 1990 86% academics were employed on permanent contracts. In 1994 60% of the academic staff had permanent contracts. And this number is actually assumed to be even lower as Higher Education Statistics Agency (HESA) does not consider staff to be contracted on less than a quarter of a full-time contract. And for the first time there were more new appointments on fixed-term than permanent contracts (49.5 percent vs 48.3 per cent). If this development continues most of staff will be employed on fixed-term contract basis and the shift from job security to insecurity will have taken place. Will these fixed-term contracts make academia less attractive in the long run?

My experience from talking to UK academic staff is that they even can dismiss whole departments if it does not make enough money. And you are most likely not appointed as academic staff if you did not prove as post-doc to be able to secure grants in academia or funded private/government projects in industry. And even at the level of senior lecturer, reader or even professor you will run the risk of getting kicked out if you do not succeed in securing enough incoming grants. A good quantity of papers and placing papers at good conferences or journals is mostly not enough. Unless you are "exceptionally" accomplished at doing that. Another opportunity to be able to stay on is to get involved with politics, administration, teaching etc. You may teach a wide range of courses and many of them a term, become an (under-)graduate tutor, a departmental expert on computing issues as $ \text{\LaTeX}$ or use of grid infrastructure for computationally expensive techniques, tutor and supervise many students etc.

Of course there is also a cost of kicking out staff members for the department. It costs effort, time and money to recruit new staff though there might be plenty of people who are interested in that job. At my university teaching assistants below the lecturer level are allowed to contribute at most 10% to the students' final grade. But at other universities as the London School of Economics & Politics graduate students are asked to teach full courses due to lack of staff, e.g. a friend of mine got asked to teach a course on real analysis. But then preparing such a course requires a significant investment of time, especially if you are not "too fluent" in a certain course anymore you are asked to teach. And it may not coincide with your research area at all. And then again teaching in front of possibly 200+ students for the first time in your life can be quite a daunting experience. To be sufficiently assertive to deal with the students in this novel situation can be really challenging.

But this lack of loyalty may also turn out to be mutual that current staff members have less objections to move on to the next university upon arrival of a better offer. IBM experienced a similar fate. Most staff members were hired straight from university and mostly stayed on for a long time, often even until retirement. People knew they were welcome though at some point there skills may not meet the demands of the economy and they could get training to adopt to this new situation. But this also meant that many staff members were reluctant to easily move on to a different employer if a different company tried to headhunt you.

I am not sure about US but I heard a considerable number of professors are tenured starting at the assistant professor level. Even with the higher-ranked universities that may often start at the associate professor level though they know there often is a long queue of highly accomplished candidates. Maybe somebody from the US can shed more light on the situation there?

A related ML/AoPS thread discusses the [url=http://www.mathlinks.ro/viewtopic.php?t=242401]Economic Downturn in Academia[/url] which makes it even harder to secure an academic position these days.

In the US, the first tenure-track job for most fresh PhDs is at the Assistant Professor level.  You'll get about 6 or 7 years before you must apply for tenure and if you don't make it, you leave the institution.  When tenure is awarded, this Assistant Professor is generally promoted to Associate Professor.  A tenured Assistant Professor would be a rare bird in the US.

Experienced faculty members can be appointed at the Associate Professor or Full Professor levels - either with tenure or without.  Generally, those hired without tenure are expected to apply for the permanent appointment within a few years.

The terms "Instructor" and "Lecturer" at US institutions are typically reserved for non-tenure track appointments - in many cases people without PhDs.  Typically, they spend most of their time teaching and have minimal research responsibilities.  The use of these titles is quite different in the US than in many other countries.

Some disciplines in the US expect tenured faculty to support themselves through grant funding.  Certainly medical / health sciences faculty and some Engineers need to get grants or their jobs are in peril.

Tenured professors in the US can also lose their jobs if the institution has financial difficulties or if the prof works in a "low productivity" program (not enough students to be economically viable.)  Many US colleges have instituted "post tenure review" processes within the past 5-10 years.  A few tenured faculty members have lost their positions as a result of negative evaluations.

In most previous economic downturns, academic jobs in the US have held up a bit better than those in industry - mostly because many displaced workers return to university schooling.  In this downturn, some state-funded institutions are already raising admission standards so that costs may be saved through reduced enrollments (which require fewer faculty.)

Erik Ringmar has had an interesting blog entry on what PhD life is like. [url=http://ringmar.net/forgethefootnotes/?p=97]Here[/url] is the article. He is currently employed as professor by the [url=http://www.srcs.nctu.edu.tw/srcs/member/ab-11.htm]Dept of General Education, National Chiao Tung University[/url] and got well-known for this dispute on the teaching quality of the [url=http://www.lse.ac.uk/]London School of Economics and Political Science (LSE)[/url] which led him to resign from his senior lecturer position at LSE. He describes the case in the [url=http://ringmar.net/forgethefootnotes/?p=124]The Times Higher Education Supplement[/url] but it is also independently discussed in the newspaper the [url=http://www.guardian.co.uk/education/2006/may/03/highereducation.economics]Guardian.[/url]


I got a 41 on r2, which means it was worth more than my 120 on r1, whee. :)

[As in last year: let us not discuss rankings, even if we do know them?]

Hey Billy did ur skul teachers receive ur score as nothing is posted on the website.

I got a 32 on r2, which means it was [i]not[/i] worth more than my 138 on r1, rawr.  :P

By the way, what's "skul"?

Um, I'm pretty sure "skul" means "school."  ;\

Let's not discuss rankings, as we didn't do it last year and it was more fun that way.

[quote="befuddlers"]Sitan cud u post the names of guys who placed from Northview

Befuddlers[/quote]

No, because I'm listening to Billy and because I don't know, actually.

Hi this is Heng, I have the results for Northview.

Krishna got 39 on proof.
Sitan got 35 on proof.
Mohan got 31 on proof.
Ben got 29 on proof.
(Krishna and Mohan are our beastly Indian twins.)

The rest of us...let's not talk about it.

Uh, for some reason, the graders took off that many points on my paper for no apparent reason, methinks.  I saw my score today, and I wasn't too pleased, because I thought that all four of my proofs for #1,2,4, and 5 were completely legit.  My solutions on #2 and #4 were different from the official solutions, but they were complete and correct.  Are there elegance points involved, too?


Right folks, if you're a Harry Potter fan, here's a forum you could/should/need to go to and sign up for.

Dumbledore's Army:

http://www.dumbledores-army.net

Okay, when you sign up (which I know you will), your beloved *cough* moderator MithsApprentice is the referrer. Got that? That's M-i-t-h-s-A-p-p-r-e-n-t-i-c-e. Awesome :-) Hope to see some of you folks there.

Oh, and don't join the Death Eaters. Do yourselves a favor and don't join the Death Eaters. Please dont join the Death Eaters. It's for your own good. Really. *nods to self*

cool.  heres sumthing else i found......ppl wrote this and stuff and its actually well written. oh, and these are separate stories and each of them starts from the first chapter so just click on the "next chapter" button until its done.

http://www.schnoogle.com/authors/veela1287/JLAH01.html

http://www.schnoogle.com/authors/kira/M01.html

[quote="MathFiend"]Ok. I joined cuz I'm a HP maniac. :) I'm MathFiend on there, and gave u credit miths. don't worry. :) Could you explain how the site works? What are the"items" in teh item shop for? I'm just a lil' but confuzzled. lol.[/quote]

Alright. Everytime you post, you get 2 casualties. And you also get other casualties for doing other stuff: contests, designs, etc. Basically you can buy items with those casualties in the itemshop and eventually you choose a side: either DA or DE (as explained) and the point is to help your team win the war.

*bump* I'm double-posting to bump this and add something.

Okay, eventually they're going to add an RPG to the current board and stuff. That's what the items are actually for. When that time comes, the items will help you fight. *nods*

Athena! *glares* Why haven't you signed up yet?...lolol...jkjk :P Go sign up!

okay I joined and have almost no clue what its about can someone clue me in?


integration
trigonometry
real analysis
real analysis unsolved

[color=darkred]Ascertain $ \int\frac {x^2}{(x^2\minus{}4)\sin x\plus{}4x\cos x}\ \mathrm dx$ on the interval $ (0,\pi )\ .$[/color]


induction
quadratics
algebra
number theory unsolved
number theory

IF xy| x^2 + y^2 +1 , prove that 3xy = x^2 + y^2 +1

I have started by saying that let there exists a number g such that

xyg = x^2 + y^2 +1
xyg + 2xy = x^2 + 2xy + y^2 +1
xyg + 2xy -1 = (x+y)^2
xy(g+2) -1 = (x+y)^2

but that has gotten me nowhere.

Any pointers would be appreciated

thanks

A couple tips:

-Instead of working with the equation given, you may want to show that it's equivalent to working with the simultaneous equations

$x|y^2+1$ and $y|x^2+1$.

-You might want to try a proof by induction on whichever of $x$ and $y$ is larger (the only time they'll be equal is when $x=y=1$).  Assume $x>y$, and try to construct another pair $(x', y')$ satisfying the pair of equations, but with $x'$ and $y'$ both less than $x$.

I hope it's right (made quickly ;) ):
Let $t=\frac{x^2+y^2+1}{xy}$.
We have the quadratic equation: $X^2-(ty)X+y^2+1=0$. This equation has two integer roots: $x$ and $\frac{y^2+1}{x}$. If we assume $x>y$, then $x'=\frac{y^2+1}{x}<x$. So we have a descent algorithm. Finally we will come to $y=1$. So $x^2-tx+2=0 \Rightarrow \Delta=t^2-8$ is a square so $t=3$.

I have found out general roots of this equal:
[tex](1,1)[/tex] and [tex](a_{n-1},a_n)[/tex] with [tex]{a_n}[/tex] is a sequence st:[tex]a_n=3a_{n-1}-a_{n-2}[/tex].Or:
[tex](F_1,F_2),...(F_{2n-1},F_{2n+1})[/tex]
[tex]F_n[/tex] is n-th Fibonaxi number..


:w00tb:    This is very beautiful!!!!!!!!!!!!
It seems to be a very difficult problem!

Silouan ,who is the constructor of this inequality?

Yes, thank you for this nice one. It seems to be hard.

it's posted before by harazi
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=cool&t=13862


summer program
Mathcamp
PROMYS
Ross Mathematics Program
calculus
MathPath
number theory

For those who didn't make into MOSP and for those who made into MOSP, what is your plan for this summer? MathCamp in Maine or PROMYS at BU, or others?

If you go and look in the Other Programs forum, you can see a variety of topics related to this question.

[quote="beta"]Maybe Ross, since PROMYS doesn't accept 8th graders.  I felt mathcamp is too late for me to apply[/quote]

If Mathcamp is the one you are most interested in, go ahead and contact them and ask if it is too late. The director posted here about a week or so ago saying they would accept late applications.

I am taking calculus I and II at my state school and getting a job.

My son is too young to apply to MathPath (by a matter of days, because of when his birthday is), so we are saving MathPath for next year (2005). The year after that (2006) he will be old enough to get into MathCamp, and maybe will be able to get a waiver to get into Ross, if he writes a good set of application problem solutions. This year he is going to places he can go on the basis of SAT I scores, namely the Centre for Talented Youth Ireland [url=http://www.dcu.ie/ctyi/summer/academic/c_desc04.htm#Modern%20Mathematics]2004 Summer Programme[/url] and the JHU [url=http://www.jhu.edu/gifted/summer/catalogs/osmath.html]CTY summer program[/url] (second session) in Lancaster, Pennsylvania.

2007 is his ninth-grade summer, so that will be a year for trying to get into MOP. I hope everyone is having an enjoyable spring and will have a summer packed with learning math.

I'm going to PROMYS as well as MOP, for a returning year.

They have good advanced lectures on top of the preliminary number theory course (which is also very good), but you have to be willing to work if you want to go at all. It does pay off, though.

I didn't apply to any any math programs.  I will still try to learn more math on my own.  I am signed up to take AMC/AIME Algebra and the AMC 12 Prob series on this site.  I hope that I learn a lot there.  Most of the summer I will be getting up at about 6 to go to cross country training.  There's nothing like running in the mornings!  It's optional but I love running and I want to get better just like math.  Near the end of the summer I am signed up for a one week running camp in NC with my team.  Back to academic.  I am going to try to get myself to study some stuff over the summer(easier said than done).  In particular I want to look at some science stuff(physics and chem. to be specific).  Maybe some Latin and of course everyone's summer past time - reading.  Mostly just ind. studying.

after MOP, I'm going to a camp in China (which is language-based rather than math-based)... :-D

[quote="MithsApprentice"]after MOP, I'm going to a camp in China (which is language-based rather than math-based).[/quote]

So  you're going to a language program in China? How did you sign up for that? I figured you were going over for a family visit, but I'm curious about language programs for Chinese-Americans who want to learn more Chinese.

Yeah I must admit that I am definitely not good at Chinese... which is embarrassing... but I don't have time either :-(

[quote="tokenadult"]
So  you're going to a language program in China? How did you sign up for that? I figured you were going over for a family visit, but I'm curious about language programs for Chinese-Americans who want to learn more Chinese.[/quote]

Haha...it's also an excuse for my family to go back and visit relatives. But in any case, you qualify for the camp thru the HSK (HSK= [b]H[/b]an4 Yu3 [b]S[/b]hui3 Ping2 [b]K[/b]ao3 Shi4) Profiency Exam (administered annually at most Chinese embassies...we happen to have one here in Houston). The program in relatively new, in fact I believe last year was the first year they invited you to a summer vamp. In any case, they pick a certain number of top-scorers and send u off for a month to one of the universities in Beijing.

(The numbers after the PingYin is the accent number)


logarithms
integration
limit
calculus
calculus computations

Prove that

$ \frac {\minus{}1}{10} < \sum_{k \equal{} 1}^{n}{\frac {k}{k^2 \plus{} 1}} \minus{} \ln{n}\leq\frac {1}{2},$

with equality if and only if $ n \equal{} 1.$

[quote="Ji Chen"]Prove that

$ \frac { \minus{} 1}{10} < \sum_{k \equal{} 1}^{n}{\frac {k}{k^2 \plus{} 1}} \minus{} \ln{n}\leq\frac {1}{2},$

with equality if and only if $ n \equal{} 1.$[/quote]
for right
$ f \equal{} \sum_{k \equal{} 1}^{n}(\frac {k}{k^2 \plus{} 1} \minus{} \ln{\frac {k}{k \minus{} 1}}) \equal{} \frac {1}{2} \plus{} \sum_{k \equal{} 2}^{n}(\frac {k}{k^2 \plus{} 1} \minus{} \ln{\frac {k}{k \minus{} 1}})$
$ g \equal{} \frac {k}{k^2 \plus{} 1} \minus{} \ln{\frac {k}{k \minus{} 1}}$
$ g' > 0$ $ \lim_{k \minus{} > \plus{}\infty}{g} \equal{} 0$
so $ f \leq \frac {1}{2}$

[quote="Ji Chen"]Prove that

$ \frac { \minus{} 1}{10} < \sum_{k \equal{} 1}^{n}{\frac {k}{k^2 \plus{} 1}} \minus{} \ln{n}\leq\frac {1}{2},$

with equality if and only if $ n \equal{} 1.$[/quote]
for left,need to prove
$ \frac {1}{2} \plus{} \frac {2}{5} \minus{} \ln{2} \plus{} \sum_{k \equal{} 2}^{n \minus{} 1}(\frac {k \plus{} 1}{(k \plus{} 1)^2 \plus{} 1} \minus{} \ln{\frac {k \plus{} 1}{k}}) > \frac { \minus{} 1}{10}$
we just need to prove $ \frac {1}{2} \plus{} \frac {2}{5} \minus{} \ln(2) \minus{} \frac {1}{2}\sum_{k \equal{} 2}^{n \minus{} 1}(\frac {1}{k \minus{} 1} \minus{} \frac {1}{k}))$
$ > \frac {1}{2} \plus{} \frac {2}{5} \minus{} \ln(2) \minus{} \frac {1}{2}\frac {1}{2*1} > \frac { \minus{} 1}{10}$ that it is.
[hide="proof"]
$ f \equal{} \frac {n \plus{} 1}{(n \plus{} 1)^2 \plus{} 1} \minus{} \ln(1 \plus{} \frac {1}{n}) \plus{}\frac {1}{2n(n \minus{} 1)}$
$ f' < 0,\lim_{n \to \plus{} \infty}{f} \equal{} 0$
[/hide]


[quote="easyas3.14159..."][hide]Pages 1-9=45
Pages 10-19=55
Pages 20-29=65.
45+55+65=165, so there are 29 pages in the book.[/hide][/quote]

Similar to what I did, except after finding the values of the first nine pages, I set up an equation.

[quote="vamathletes"]The sum of the digits of the page numbers of a book is 165.
How many pages are in the book?[/quote]
[hide="my answer"] 1 thru 9 is 45
10 (1) thru 19 (10) is 55
20 (2) thru 29 (11) is 65
therefore there are 29 pages
gauss method of summation is the only thing you need. or a calc[/hide]

[hide]THE number OF pages ARE 29[/hide]

[color=red][size=75][Answer hidden by moderator][/size][/color]

[quote="solver123"]THE number OF pages ARE ...[/quote]

1. if you give an answer, give a solution, it doesnt really help people to give the answer, they want to know HOW to do it.

2. hide your answers and solutions that way someone who wants to solve it themselves doesnt have an answer thrown in their face.  you can hide stuff like this...

[code][hide]ANSWER answer, bla bla bla, solution solution[/hide][/code]


geometry
circumcircle
perpendicular bisector
geometry solved

1. Let ABC be an acute-angled triangle, and let M be a point on the side AC and N a point on the side BC. The circumcircles of triangles CAN and BCM intersect at the two points C and D. Prove that the line CD passes through the circumcentre of triangle ABC if and only if the right bisector of AB passes through the midpoint of MN.

First of all, let's prove this

Lemma: Given pts M on AC and N on BC, the perpendicular bisector of AB passes through the midpt of MN iff AMcosA=BNcosB. Here ABC is an acute triangle.

Proof:
Let X and Y be the projections of M and N respectively on AB, and let Z be the midpt of AB. The projection of the midpt of MN on AB must be the midpoint of XY, but it also must be Z because the perpend on AB through Z passes through the midpt of MN. This means that ZX=ZY, and because ZA=ZB, we get XA=YB. But XA=AMcosA and YB=BNcosB. We got what we wanted.

Now for the problem itself:
It's easy to see (good old angle-chase in the cyclic quadrilaterals MDBC and ADNC) that triangles DMB and DAN are similar (in that order of vertices). We write Ptolemy for the 2 quadrilaterals I mentioned:
BM*DC=DM*BC+DB*MC (*)
NA*DC=DA*NC+DN*AC; we multiply this last one by k=BM/NA=DM/DA=DB/DN to get
BM*DC=DM*NC+DB*AC (**)

If we subtract (**) from (*) we get DM*BN=DB*AM, iff DM/DB=AM/BN

If the perpend. bisector of AB passes through the midpt of MN we get, from the Lemma, DM/DB=AM/BN=cosB/cosA=sin(Pi/2-B)/sin(Pi/2-A)=sinDBM/sinDMB, but we always have DBM+DMB=C=Pi/2-B+Pi/2-A, and by combining these 2 we get DBM=Pi/2-B=OCA (O is the circumcenter of ABC)=DCM=DCA (because MDBC is cyclic), so we found D, O, C collinear.

On the other hand, if we assume DCA=DCM=DBM=Pi/2-B, we get DMB=Pi/2-A, so we find AM/BN=DM/DB=cosB/cosA, and we use the Lemma again to find that the perpendicular bisector of AB passes through the midpt of MN.

I hope it's Ok.
See ya! :D  :D

[quote="grobber"]sin(Pi/2-B)/sin(Pi/2-A)=sinDBM/sinDMB[/quote]
how do you find this, Grobber? please someone explain it to me.

Ph-An could reply my question to Grobber ?

If $C(O,R)$ is the circumcircle of the triangle $ABC$ then

$D\in [CO\Longleftrightarrow m(\angle DCA)=90-B\Longleftrightarrow m(\angle DCB)=90-A$,

a.s.o.


A tank contains the following quantities of these gases: He = 0.6 g; dihydrogen = 0.6 g; Ar = 0.6 g. The total pressure exerted by the mixture of gases at 0 degrees Celsius is 4.6 atmospheres.

Pick the statements that are true in the following questions.

The partial pressure exerted by:

1) each gas is 4.6 atmospheres;  2) each gas is 1.03 atm;    3) the hydrogen is 0.60 atm.  4) the argon is 10 times that of helium;    5) the hydrogen is 2 times that of helium;

Increasing the temperature of the mixture from 0 degrees Celsius to 273 degrees Celsius without change in volume will cause total pressure to:

1) decrease by 1/2 to 2.3 atm;   2) increase by a factor of 2 to 9.2 atm;     3) remain unchanged;   4) increase by a factor of 273;            5) change by a factor which cannot be predicted without knowing the volume.

I solved these two questions, but don't have the answers. Could anyone check them for me?  :)
[hide]
First question:

Only choice 5 is true. (I just used the ideal gas equation to find the independent pressure of each).

Second question:

Only choice 2 is true. [/hide]


If $ n$ runs through all the positive integers, then $ f(n) \equal{} \left \lfloor n \plus{} \sqrt {3n} \plus{} \frac {1}{2} \right \rfloor$ runs through all positive integers skipping the terms of the sequence $ a_n \equal{} \left \lfloor \frac {n^2 \plus{} 2n}{3} \right \rfloor$.

Denote increasing positive integer sequence $\left\{ {{b_n}} \right\}_{n = 1}^{ + \infty }$ runs through all positive integers skipping the terms of the sequence of $\left\{ {{a_n}} \right\}_{n = 1}^{ + \infty }$.
Let ${b_n} = k$, ${a_m} < k < {a_{m + 1}}$, then there are $m$’s $\left\{ {{a_n}} \right\}$ and $n$’s $\left\{ {{b_n}} \right\}$ in $\left\{ {1,2,...,k} \right\}$. So $k = n + m$, and we have
$\left[ {\frac{{m(m + 2)}}{3}} \right] < n + m < \left[ {\frac{{(m + 1)(m + 3)}}{3}} \right]$, and so
$\frac{{m(m + 2) - 2}}{3} + 1 \leqslant \left[ {\frac{{m(m + 2)}}{3}} \right] + 1 \leqslant n + m \leqslant \left[ {\frac{{(m + 1)(m + 3)}}{3}} \right] - 1 \leqslant \frac{{(m + 1)(m + 3)}}{3} - 1$
From this we have
${\left( {m - \frac{1}{2}} \right)^2} < {m^2} - m + 1 \leqslant 3n \leqslant {m^2} + m < {\left( {m + \frac{1}{2}} \right)^2}$,
Hence $m - \frac{1}{2}<\sqrt {3n} <m + \frac{1}{2}$, so  $m<\sqrt {3n} +\frac{1}{2}<m + 1$, and we have  $\left[ {\sqrt {3n}  + \frac{1}{2}} \right] = m$.
So ${b_n} = n + \left[ {\sqrt {3n}  + \frac{1}{2}} \right]$.

Conjecture: Can we replace 3 in the above result by any arbitrary natural k and 2 by
k-1?


superior algebra
superior algebra solved

Find a homomorphism phi from U(30) to U(30) with Ker = {1,11}

and phi of 7 = 7.

I need help showing how to do this please.
Thanks Beth

First, you want to understand the structure of your group. $\mathbb{Z}_{30}^*=\mathbb{Z}_2^*\times\mathbb{Z}_3^*\times \mathbb{Z}_5^*$. The first group is trivial, so the group is naturally isomorphic to the product $\mathbb{Z}_3^*\times \mathbb{Z}_5^*$.

Now $7=(1,2)$ maps to $7=(1,2)$ and $11=(2,1)$ maps to $1=(1,1)$. Clearly, 7 and 11 together generate the whole group (since 2 is a primitive root both mod 3 and mod 5), so their images determine everything: $(x,y)\mapsto(1,y)$. Now all that's left is interpreting this in the original form.

Also, if you're asking a question, you shouldn't put it in "Solved problems".

U(30) can not be isomorphic to Z3 x Z5 under multiplication because they have a different number of elements

I agree they aren't. That's not what I said.


Hi,

I think it's time for a serious public policy topic here. This topic also has personal relevance to most participants on this forum. I have read that as a result of female infanticide and sex-selective abortion in several of the world's most populous countries that there is a worldwide shortage of women in the generation now growing up. I have seen estimates that there are 40 MILLION young men, already born, in one country who cannot reasonably expect to find a woman to marry in their own country, because there girls are missing to that great a degree in the age cohort now becoming of marriageable age in that country. What [url=http://www.reuters.com/locales/newsArticle.jsp?type=worldNews&locale=en_IN&storyID=4516952]news[/url] do you hear about this issue? What implications does it have for social stability and economic development around the world? What's your plan (if you are a single male who desires to marry a woman) to not be part of the unmatched group of young men in the next generation?

I don't think a slight imbalance between the number of men and women is in itself that serious of a social problem. It happens all the time in waring nations (which leaves more women than men). And it's something that can correct itself naturally within one generation. And an imbalance in certain parts of the world doesn't necessarily mean an imbalance in other parts. For example, I think there are more women in both the United States and in Canada than there are men right now.

Anyway, what [i]is[/i] a serious social problem are the ongoing human rights abuses against women. The issues of female infanticide and sex-selective abortion are not new, but are still ongoing in some parts of the world. That's because women as a whole continue to be victims of human rights abuses of various kinds, and devalued in certain societies. To me, that's the more pressing issue to be addressed.

I've heard that there are more things that induces birth of a female than male such as radiation or chemical in green tea leaves...that may help in the future

For most part I agree this isn't a problem, there aren't a huge difference between the number of males and the number of females and some male might choose to remain single or marry another male so I think it'll all work out in the end.

[quote="churchilljrhigh"]in the u.s. there are more women, so men have the benefits of choosing women. (that's why so many women are trying to make themseleves look good)[/quote]

I'm going to have to disagree here.  There aren't significantly more women in the U.S. for it to matter much.  Besides, your point of view (and the point of view of tokenadult) assumes that every man and woman will be paired off and their relationship is going to be concrete.  In reality, relationships are constantly changing, so that even if there is a shortage of women, that just means that some men or women may be single for a few weeks longer.

Besides, there is such a large pool of single men and women that there will always be available men or women, even if there are more people available from one sex than from the other.

Finally, your assertion that "so many women are trying to make themseleves [sic] look good" does not follow from the fact that there are slightly fewer males.

First of all, if that were the case, how do you explain the fact that American men also try and look their best if there are fewer of them?

Second, even if there were ten times more males than there are, many women would still try and look their best because they want to compete for the best males.

Third, there are numerous other factors that cause women to want to look good.  Perhaps women try to look their best because our culture is largely based on appearances.  Perhaps they try and look their best because they see so many glamorous stars on TV.  Perhaps they try and look their best because they are accustomed to.  I don't know the answer to this question, but I don't believe it is because they slightly outnumber the male population.

And Tare, exposing pregnant women to radiation is not the way to go.  Like gauss202 said, the population will even itself out in one generation.

Here's a news update about the situation in China, based on a report in the state-controlled [i]China Daily[/i] newspaper.

[url=http://abclocal.go.com/kabc/story?section=nation_world&id=4930634]China: Fewer Women a Threat to Stability?[/url]

This type of problem has occurred many times in the past.  Migration of people to new fronteers (West, in general for US and Gold Miners in particular), wars, "loss" of women to multiple marriages, in general or a king like Soloman who took a very large number.

I don't think that this will lead to militiary instability, as it can be addressed by less severe methods, such as bride prices, kidnapping and raids.  I don't see the nation of China doing anything to address concerns of poor farmers not having women.

There may be a larger problem with the urbanization of women in developing countries (India) where the young women are allowed to support themselves far from the area where they can marry their neighbor men.

I'm not aware of reading about any concerns in the post-Civil War time.  We lost 30,000 men in three days in Gettysburg as a part of a long war.  Is there anything about this in any history books you are aware of?

actually, I heard last year, 52% of the world population are women

Yeah, competition gets stiff.
[quote]There is an incredible push for men to smell good, to have smoother skin and coincidentally to pay more attention to how they look[/quote]
[url=http://news.yahoo.com/s/afp/20070114/lf_afp/afplifestyleushealth]More American men seeking nip n' tuck[/url]

At the same time, [url=http://www.iht.com/articles/2007/01/16/news/web.0116census.php]51% of U.S. women are now living without spouse[/url].


Let circle O1 and circle O2 be tangent internally to circle O at A,B, respectively. Line g1 and g2 are internal tangent of circle O1 and O2. g1 intersects g2 at C. Line g3 touches circle O1, circle O2 at D,E, respectively. M is midpoint of the line segment DE. N is foot of perpendiculer from C to g3. D,N,M,E in order. AM intersects g1 at P. BN intersects g2 at Q. prove that PQ is parallel to g3.

I'm sorry! this problem can be false. [/quote]

This seems very similar to a problem you just posted a couple days ago... I seem to remember making a diagram
here it is:
[img]6857[/img]
do you remember seeing this?

Yes, I remember. but this problem is different from the problem that I posted a couple day ago. this problem can be open. How to stop it?

I don't think it's correct.

You didn't define $g_{1}\ ,g_{2}$ clearly. but anyway the problem is wrong .

see the pic Below:


let n be a positive integer >=3.
Show that n^(n^(n^n))-n^(n^n)
is divisible by 1989.

Just a qustion: how do you find the questions which weren't used in the Balkan olympiad? Are there some sort of 'Balkan shortlists' :D ?

What we need to do is prove that  (1) 9/E (where E is that nasty expression  ); (2) 13/E; (3) 17/E because 1989=3^2*13*17.

(1)	Lemma 1 3/A=n^(n^n-n)-1 if n<>0(mod 3).

If 3/n then we get 9/E so it?s enough to prove 9/E for n=1,2 (mod 3). For the proof of the lemma, let?s take n=3k+1 first, and it?s obviously true that 3/A. If n=3k+2: n^n-n is even for any n and since 2^2a=1 (mod 3) is obvious, we get 3/A. Let?s use the notation a~b if a=b(mod 9). Then if n<>0 (mod 3) we have E~n^(n^n*A)-1 (we can divide E by n^(n^n)). It?s clear that 2/n^n*A (just consider the cases n odd, n even), and because 3/n^n*A (from the lemma) we have 6/n^n*A=6a. So E~n^(6a)-1=(n^(2a)-1)(n^(4a)+n^(2a)+1). Since 3 divides each of the 2 factors we have 9/E.

(2)	If 13/n then we are finished. Let?s use another notation: a~b if (this time) a=b (mod 13). Then E~n^(n^n*A)-1. We can easily prove that 12/n^n*A=12a, so E~(n^a)^12-1, which is divisible with 13 by Fermat?s thm.

       I don?t have a proof for (3). If we try the same way (Fermat's thm) we should prove that 16/n^n*A, which I?m not sure about. Maybe we have to consider some cases. I?ll try to solve it.

Yeah, it's true. 16/n^n*A for any n. If you try to prove it you'll see it's not hard. So for (3) we just repeat the proof for (2). And now: the already legendary 'hope it's correct'  :D

hi grobber,

I found this problem in a set of number theory problems available at
http://www.problem-solving.be/pen/
by hojoo lee. There are a lot of interesting problems extract from
various maths olympiads. Download it ;-)

[Moderator edit: Link modified, since the old one was not accessible anymore.]

bye.


Let x,y>0. Prove the inequality is true,

$ \frac{3}{2}x^2\plus{}\frac{3}{2}y^2\plus{}2xy\minus{}x\minus{}y\plus{}1 \ge (x\plus{}y)^2$

$ \frac {3}{2}x^2 \plus{} \frac {3}{2}y^2 \plus{} 2xy \minus{} x \minus{} y \plus{} 1 \ge (x \plus{} y)^2$

$ \Longleftrightarrow 3(x^2\plus{}y^2)\plus{}4xy\minus{}2(x\plus{}y)\plus{}2\geq 2(x\plus{}y)^2$

$ \Longleftrightarrow x^2\plus{}y^2\minus{}2(x\plus{}y)\plus{}2\geq 0$

$ \Longleftrightarrow (x\minus{}1)^2\plus{}(y\minus{}1)^2\geq 0$

one other solution is :

          3/2 ( x^2+y^2) + 2xy - x - y +1>= 3/2 2xy + 2xy-x-y+1=3xy+xy-x-y+1=5xy-x-y+1=5xy-(x+y)+1>= 5xy-(2ριζαxy)+1 LET ριζαxy=u then we have 5u^2-2u+1>=0 which is true because D=4-21=-17<O [/i][/hide][/list]

[quote="petros r"]one other solution is :

$ 3/2 ( x^2 \plus{} y^2) \plus{} 2xy \minus{} x \minus{} y \plus{} 1\geq \frac {3}{2} 2xy \plus{} 2xy \minus{} x \minus{} y \plus{} 1 \equal{} 3xy \plus{} 2xy \minus{} x \minus{} y \plus{} 1 \equal{} 5xy \minus{} x \minus{} y \plus{} 1 \equal{} 5xy \minus{} (x \plus{} y) \plus{} 1\geq 5xy \minus{} 2\sqrt {xy} \plus{} 1$
LET $ \sqrt {xy} \equal{} u$ then we have $ 5u^2 \minus{} 2u \plus{} 1\geq 0$ which is true because $ D \equal{} 4 \minus{} 21 \equal{} \minus{} 17 < 0$ [/i][/hide][/list][/quote]

is that what you wrote :?:

Because if it is then the second line is wrong, since the $ 5xy\minus{}(x\plus{}y)\plus{}1\leq 5xy\minus{}2\sqrt{xy}\plus{}1$ :wink:


[hide]Here's my solution:
Applying Cauchy-Schwartz we have:
LHS>=(a+b+c)^2/3(ab+bc+ac)>=1(since a^2+b^2+c^2>=ab+bc+ac)(Q.E.D)
[/hide]

easy  :)
with caushy:
$ \left(\frac{c}{a\plus{}2b}\plus{}\frac{a}{b\plus{}2c}\plus{}\frac{b}{c\plus{}2a}\right)(3(ab\plus{}bc\plus{}ac))\geq (a\plus{}b\plus{}c)^2$
$ \Leftrightarrow \left(\frac{c}{a\plus{}2b}\plus{}\frac{a}{b\plus{}2c}\plus{}\frac{b}{c\plus{}2a}\right) \geq \frac{(a\plus{}b\plus{}c)^2}{3(ab\plus{}bc\plus{}ac)}\geq1$

[quote="Aravind Srinivas L"]Wasn't the same thing written by ghjk? Well one more method: Using Cauchy Lema
We get the given expression $ \ge \dfrac{(\sum a)^2}{3\sum ab} \ge 1$ from the well known inequalities. Though this is the same, just posted it for those who wanted to see it done by this way.[/quote]
You post was also the same as ghjk ;) I see no difference in Cauchy Schwartz and Cauchy Lemma.

[quote="Johan Gunardi"]For the positive real numbers $ a,b,c$ prove the inequality
\[ \frac {c}{a \plus{} 2b} \plus{} \frac {a}{b \plus{} 2c} \plus{} \frac {b}{c \plus{} 2a}\ge1.
\]
[/quote]
Let $ x\equal{}a \plus{} 2b,y\equal{}b \plus{} 2c,x\equal{}c \plus{} 2a$ then we must prove $ \sum_{cyc} \frac{4y\plus{}z\minus{}2x}{9x}\ge 1$ or
$ \sum_{cyc}\frac{4y\plus{}z}{x} \ge 15$ its obviously true by AM-GM

[quote="kunny"]For $ a,\ b,\ c$, prove that $ \frac {2007c}{2008a \plus{} 2009b} \plus{} \frac {2008a}{2009b \plus{} 2007c} \plus{} \frac {2009b}{2007c \plus{} 2008a}\geq \frac {3}{2}$.[/quote]

Let $ 2007c\equal{}x$, $ 2008a\equal{}y$, and $ 2009b\equal{}z$, then our inequality takes form \[ \frac{x}{y\plus{}z}\plus{}\frac{y}{z\plus{}x}\plus{}\frac{z}{x\plus{}y} \ge 1.5\]

a,b,c are positive real numbers ???
... sorry I'm not good at English

@kfc9512: it should be positive real lha..
assume that it's not positive, then let $ x\equal{}0,y\equal{}\minus{}1,z\equal{}2$ then $ LHS\equal{}0\minus{}\frac{1}{2}\minus{}2<\frac{3}{2}$.

Sorry .... Also.. i'm not good at math..  :blush:  thank you.. :blush:

(and...
\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \ge 1.5by caushy)-just practice :wink:

I've just found a really beautiful proof to this problem (inspired by a proof of Nesbitt's Inequality):

Let $ P\equal{}\sum\frac{c}{a\plus{}2b}$, $ Q\equal{}\sum\frac{a}{a\plus{}2b}$, $ R\equal{}\sum\frac{b}{a\plus{}2b}$. By AM-GM, we have $ P\plus{}2Q\equal{}\sum\frac{c\plus{}2a}{a\plus{}2b}\ge3$ and $ 2P\plus{}R\ge\frac{2c\plus{}b}{a\plus{}2b}\ge3$. So $ 9P\plus{}2Q\plus{}4R\equal{}4(2P\plus{}R)\plus{}(P\plus{}2Q)\ge15$, but $ 2Q\plus{}4R\equal{}2(Q\plus{}2R)\equal{}6$, so $ P\ge1$. Q.E.D. :D

Another approach is by brute force, after computations we get
$ 2\sum_{cyc}{a^3} \geq 3abc + \sum_{cyc}{ a^2 (3c - 2b)} \Leftrightarrow {\underbrace{\sum_{cyc}{a^3} - 3abc}}_{\geq 0 } +  {\underbrace{\sum_{cyc}{a^2(a + 3c -2b)}}_{\geq 0} \geq 0}$
for second sum we must prove somehow that is greater than zero

[quote="Johan Gunardi"]For the positive real numbers $ a,b,c$ prove the inequality
\[ \frac {c}{a \plus{} 2b} \plus{} \frac {a}{b \plus{} 2c} \plus{} \frac {b}{c \plus{} 2a}\ge1.
\]
[/quote]
$ \sum\frac{a}{b\plus{}2c}\equal{}\frac{a^2}{ab\plus{}2ca}\geq \frac{(a\plus{}b\plus{}c)^2}{3\sum ab}\geq 1$ from the trivial ineq
 :D

See #2 popst. You should see the former post before you post. :wink:

am i watching the similar problem too many times? I saw it above which the difference is just the number of variant became 4 instead of 3


Let $ n$ be a positive integer such that $ n^2 \minus{} n \plus{} 11$ is the product of 4 prime numbers, some of which may be the same.
What is the minimum value of n?
[color=white]275 but how do we get it?
Edit : 275 is incorrect answer. Correct answer is 132[/color]

[quote="malcolm"]I don't think $ 275$ is right. $ 132$ gives $ 11^3 * 13$

To find this, we denote $ n \equal{} pk \plus{} r$. Then we have $ n^2 \minus{} n \plus{} 11 \equal{} p^2k^2 \plus{} 2pkr \plus{} r^2 \minus{} pk \minus{} r \plus{} 11$
So $ p$ divides this if and only if $ p$ divides $ r^2 \minus{} r \plus{} 11$, where $ r < p$. Checking this for primes up to $ 13$, we find $ p \equal{} 11$, $ r \equal{} 0$, $ p \equal{} 13$, $ r \equal{} 2$ and $ p \equal{} 13$,$ r \equal{} 12$ are the only times where $ p$ divides $ r^2 \minus{} r \plus{} 11$. So our number is of the form either $ 11k$ and of the form $ 13k \plus{} 2$, or of the form $ 11k$ and of the form $ 13k \plus{} 12$. Checking $ 11k$ and of the form $ 13k \plus{} 2$, we find $ 132$ is of both these forms, and gives $ 11^3 * 13$. Checking  the form $ 11k$ and of the form $ 13k \plus{} 12$, we find $ 77$ is the only number less than $ 132$, but this does not give a product of 4 primes. so $ 132$ is the minimum.[/quote]

Oh 132 is less than 275. I'm sorry ! Thanks too much !

I have my method !
First $ n^2\minus{}n\plus{}11$ can't divisible by $ 2,3,5,7$. We get $ n^2\minus{}n\plus{}11 \ge 11^4$
It is $ n \ge \frac{(\sqrt{58521}\plus{}1)}{2}$ then $ n \ge 122$
- $ n\equal{}122$ We get $ 122^2\minus{}122\plus{}11\equal{}(11)(17)(79)$
Let $ f(n)\equal{}n^2\minus{}n\plus{}11$
If $ f(n)$ isn't divisible by 11. We have $ f(n) \ge 13^4\equal{}28561$
Since $ f(i)$ isn't divisible by 11 for $ i\equal{}123,124,...,131$.
We have $ f(i) \ge 13^4$ for $ i\equal{}123,124,...,131$ which give a contradiction.
Consider $ f(132)$; We have $ f(132)\equal{}(11)^3(13)$
Thus, $ min(n)\equal{}132$  :)

[quote="HelloWorld"]
First $ n^2 \minus{} n \plus{} 11$ can't divisible by $ 2,3,5,7$.[/quote]

how do you know that $ n^2 \minus{} n \plus{} 11$ can't be divisible by $ 2,3,5,7$ :huh:

let $ n\equal{}pk\plus{}r$, where $ p$ is $ 2,3,5,7$ and $ 0$ is less than or equal to $ r$, and $ r$ is strictly less than $ p$. Then expanding gives several terms with $ p$ in them, and $ r^2\plus{}r\plus{}11$ is left over when you remove these terms, so this will only be divisible by $ p$ if $ r^2\plus{}r\plus{}11$ is. checking each possible value of $ r$ gives that this is never divisible by $ p$.


trigonometry
number theory unsolved
number theory

Find all $ (r, q) \in \mathbb{Q}^2$ s.t. $ tan \text{ } \pi r \equal{} q$.

It's a standard result that this is possible only when $ q \equal{} 1, 0, \minus{} 1$.  Write $ \sin 2 \pi r \equal{} \frac {2q}{1 \plus{} q^2}$ and $ \cos 2 \pi r \equal{} \frac {1 \minus{} q^2}{1 \plus{} q^2}$, which must both be rational.  The only Gaussian rationals that are also roots of unity are $ \pm 1, \pm i$, and this is not hard to prove using unique factorization in the Gaussian integers or standard facts about cyclotomic polynomials.


function
limit
real analysis
real analysis unsolved

If $ f: [a, b]\to [c, d]$ is a monotone and bijective function, then can we say that $ f$ and $ f^{ \minus{} 1}$ are both continuous? If so, can someone prove this?
Thanks.

We suppose that $ f$ is increasing.Since $ f$ is surjective we have that $ \forall p,q \in [a,b] , p < q$ and $ \forall \lambda \in (f(p),f(q))$ there is $ c \in [a,b]$, more precisely $ c \in (p,q)$ such that $ f(c) \equal{} \lambda$.Therefore $ f$ has the Darboux property.Because $ f$ is increasing we have that there exist $ \lim_{x\to x_0, x < x_0}f(x) \equal{} f(x_0 \minus{} 0)$ and $ \lim_{x\to x_0,x > x_0}f(x) \equal{} f(x_0 \plus{} 0)$, and in addition $ f(x_0 \minus{} 0) \leq f(x_0) \leq f(x_0 \plus{} 0)$.

Try to prove that if the function has $ f(x_0 \minus{} 0)$ (or $ f(x_0 \plus{} 0)$) and f has the Darboux property then it is necessary that $ f(x_0 \minus{} 0) \equal{} f(x_0)$ (respectivly $ f(x_0 \plus{} 0) \equal{} f(x_0)$)

If the function has the Darboux property and  $ f(x_0 \minus{} 0)$ (or $ f(x_0 \plus{} 0)$) exists,  then is it necessary that $ f(x_0 \minus{} 0) \equal{} f(x_0)$ (respectivly $ f(x_0 \plus{} 0) \equal{} f(x_0))$?

Moreover how to conclude from this that $ f^{\minus{}1}$ is also continuous?

Every discontinuous Darboux function has an essential discontinuity but a monotone function has only nonessential discontinuities, hence the function is continuous.

[quote="limsup"]If the function has the Darboux property and  $ f(x_0 \minus{} 0)$ (or $ f(x_0 \plus{} 0)$) exists,  then is it necessary that $ f(x_0 \minus{} 0) \equal{} f(x_0)$ (respectivly $ f(x_0 \plus{} 0) \equal{} f(x_0))$?

Moreover how to conclude from this that $ f^{ \minus{} 1}$ is also continuous?[/quote]

Since $ f$ is continuous it follows that $ f^{\minus{}1}$ must be continuous.See the following arguments:

$ \lim_{x\to x_0} f^{\minus{}1}(x)\equal{}\lim_{x\to y_0}f^{\minus{}1}(f(x))\equal{}y_0$ , where $ \lim_{x\to y_0}f(x)\equal{}x_0$, because $ f$ is surjective.so we conclude that $ \lim_{x\to x_0}f^{\minus{}1}(x)$ exists and is finite.

Now, suppose that there is $ x_0 \in [a,b]$ such that $ \lim_{x\to x_0} f^{\minus{}1}(x) \neq f^{\minus{}1}(x_0)$.Because $ f$ is injective we deduce that $ f(\lim_{x\to x_0} f^{\minus{}1}(x)) \neq f(f^{\minus{}1}(x_0))$ $ \Leftrightarrow$ $ \lim_{x\to x_0} f(f^{\minus{}1}(x)) \neq f(f^{\minus{}1}(x_0))$ (because f is continuous) $ \Leftrightarrow$ $ \lim_{x \to x_0} x \neq x_0$ , contradiction.Hence $ f^{\minus{}1}$ is also continuous.

Thank you very much Svejk and Nemion for your great help. Now I ve completely understood with your arguments and by Rudin.


Here is a nice problem(actually two of them)
Find $x$ for which the series defined by $f(x)$ converges
a). $f(x) = \sum_{n=1}^\infty \frac{ \{ nx\}}{n}$.
b). $f(x) = \sum_{n=1}^\infty \frac{ \{ nx \}-\frac{1}{2}}{n}$.
Here ${x}= x-[x]$ is the fractional part of $x$.

Actually, the first one is easy using Weyl's theorem, but i think i'm not sure what to do with b).

Well as you say it is well known for (a) if $x$ is irrational then the limit is 1/2 almost surely. And obviously it is 0 for $x$ integer.

I think for (b) it is also easy. Since $\{nx\}$ is uniformly distributed in $[0, 1]$ for $x$ irrational, then $\{nx\}-\frac{1}{2}$ is uniformly distributed in $[-1/2, 1/2]$, and then the series has limit 0 almost surely. Now we don't like $x$ integer though.

For some reason it is not obvious to me what happens when $x$ is rational, not integer (in either case). Are the fractions still uniformly distributed in $\mathbb{Z}_{q}/ q$ when $q$ is denominator? Then we get 1/2 and 0 (for (a) and (b) respectively) again for same reason.

[quote="Xevarion"]Well as you say it is well known for (a) if $x$ is irrational then the limit is 1/2 almost surely. And obviously it is 0 for $x$ integer.
[/quote]

It seems i don't understand you. What do you mean by saying that "the limit is 1/2 almost surely" . Did you mean the sum instead of limit ?
In this case, i think you are not  right : if $x \in \mathbb{R}\setminus \mathbb{Z}$  the the sum is $\infty$, if $x \in \mathbb{Z}$, then the sum is $0$. That's what i've got for a).

[quote="Xevarion"]
For some reason it is not obvious to me what happens when $x$ is rational, not integer (in either case). Are the fractions still uniformly distributed in $\mathbb{Z}_{q}/ q$ when $q$ is denominator? Then we get 1/2 and 0 (for (a) and (b) respectively) again for same reason.[/quote]

In a) as i said above in case $x \in \mathbb{Q}$ $f(x)$ is $\infty$ in a) and i think for b) as well.

I'm sorry, I have no idea what I was thinking. Now that I look back, none of what I said makes sense at all.  :blush:


$ 0<x_0<y_0$
$ x_{n\plus{}1}\equal{}\sqrt{x_ny_n}$
$ y_{n\plus{}1}\equal{}\frac{x_n\plus{}y_n}{2}$
Prove the rows $ (x_n)_n$ and $ (y_n)_n$ have a limit. They have a common limit?

could you explain please why the two rows have the same limit?....I can't understand why :blush:

x is an ascending row while y is a descending row
while x is ascending, y is descending and in addition $ x_n\le\ y_n$
this means that x and y will be nearly equals $ x_n\equal{}y_n$
we apply limit on both and will have

$ \sqrt{L1 \bullet\ L2}\ \equal{} (L1 \plus{} L2)/2$
   $ L1^2 \minus{} 2L1\bullet\L2\plus{}L2^2\equal{}0$
     $ (L1\minus{}L2)^2\equal{}0$
     $ L1\equal{}L2$


$R^{*}=\left\{\frac{x'}{y'}|x',y'\in \left\{\R\backslash \{np|n\in \Z\}\right\}\right\}$

Above, the leftmost vertical bar is too small.  How can I make it the full height of the brackets?

You use \right | for the vertical bar then put \left. (note the . which means nothing is to be printed) to the left of the size you want it to be. If you put it before the fraction it will be the size of the fraction, if you put it before the { (ie before left\{) then it will be the size of the {, though they are the same height in this example.
$R^{*}=\left\{\left.\frac{x'}{y'}\right|x',y'\in \left\{\mathbb{R}\backslash \{np|n\in \mathbb{Z}\}\right\}\right\}$

By the way shouldn't \R and \Z be \mathbb{R} and \mathbb{Z}?

Thanks!  I use
\DeclareSymbolFont{AMSb}{U}{msb}{m}{n}
\DeclareMathSymbol{\R}{\mathbin}{AMSb}{"52}

In the preamble and I forgot to change it.  Actually, in this context it should be $\mathbb{Z}$.