calculus
vector
analytic geometry
geometry
graphing lines
slope
trigonometry

the algebra gets really messy, but the idea is that you have a perpendicular line to the given line that goes through the given point, and you find the intersection and do the distance formula

I think you can prove that using calculus (that's one way) or using vectors.

Unfortunately, I never mastered either of them so show the proof.

yes in the AOPs they show u the proof using vectors which is kinda neat  :P

Yes, vectors is one choice.  I don't know a calculus solution and I don't think there is one that reduces the work.  You could also use an algebra appoach, such as:

Draw in the segments parallel to the coordinate axes that go from the point to the line.  Then calculate the area in two ways.  You then have a lot of algebra.  But I thinks its a pretty decent problem to go through, if you have the patience.

I don't think that Algebra is messy.

The line passing through the point $(x_0, y_0)$ has the slope $\frac{B}{A}$ since it is perpendicular to $y=-\frac{A}{B}-\frac{C}{B}$.
Plug $(x_0, y_0)$ in we get the equation of the perpendicular line: $y=\frac{B}{A}+y_0-\frac{B}{A}x_0$.

Now we have two equation. Solve them and the solutions will be the point of their intersection:

\[(\frac{B^2x_0 - AC - ABy_0}{A^2+B^2}, \frac{A^2y_0-BC-ABx_0}{A^2+B^2})\]
Now use the distance formula:
\[x_0-x=\frac{A^2x_0 + B^2x_0 - Bx_0 + AC + ABy_0}{A^2+B^2}= \frac{A}{A^2+B^2} \cdot (Ax_0+By_0+C)\]

\[y_0-y=\frac{A^2y_0+B^2y_0-A^2y_0+BC+ABx_0}{A^2+B^2}=\frac{B}{A^2+B^2} \cdot (Ax_0+By_0+C)\]

Hence:
\[d=\sqrt{(x-x_0)^2+(y-y_0)^2}=\sqrt{\frac{(Ax_0+By_0+C)^2(A^2+B^2)}{(A^2+B^2)^2}}=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}\]

[quote="gotztahbeazn"]Prove the point to line distance formula:

$\displaystyle\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}} = d$[/quote]

Let the line Ax+By+C=0 makes an angle $\alpha$ with X-axis.  So, the line perpendicular to it through origin makes an angle $\alpha  - 90^0$ ( I am taking an ideal case with the line meeting positive X and Y axes).

If the distance from origin to the line is 'd', the the point $(dcos(\alpha  - 90^0), dsin(\alpha  - 90^0))$  i.e., $(dsin(\alpha), -dcos(\alpha))$ lies on the line.

Hence we have ${A(d\sin \alpha ) - B(d\cos \alpha ) + C = 0}$

$ \Rightarrow d = \left| {\frac{C}{{B\cos \alpha  - A\sin \alpha }}} \right|$

As we have slope of the line = $\tan \alpha  =  - \frac{A}{B}$

we take $sin\alpha  = \frac{A}{{\sqrt {A^2  + B^2 } }},\,\,\cos \alpha  =  - \frac{B}{{\sqrt {A^2  + B^2 } }}$ (in this ideal case $\alpha$ is obtuse and A, B are assumed positive)

This gives $d = \left| {\frac{C}{{\sqrt {A^2  + B^2 } }}} \right|$

Now, shift the origin to the point $(x_0 ,\,\,y_0 )$.

The equation of line now is of the form $A(x + x_0 ) + B(y + \,\,y_0 ) + C = 0$

i.e., $Ax + By + (Ax_0  + By_0  + C) = 0$

Hence the distance from the origin(new) is given by $\displaystyle\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$

if I remember correctly, in my 10th grade textbook there was a proof of this formula using scalar product, but i can't remember it right now  :blush:

maybe someone else can find it

There is a short proof using the Cauchy-Schwarz inequality.  I'll leave it as a challenge problem.

[quote="gotztahbeazn"]Prove the point to line distance formula:

$\displaystyle\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}} = d$[/quote]

Hold on.... I've never even heard of this. :blush:  What does everything represent?

We are given a line $Ax + By + C = 0$.  We are given a point $(x_0, y_0)$.  Let $d$ be the distance from the point to the line.  The problem is to prove that the formula above gives us the distance $d$.

It's a cool formula.

[quote="chess64"]:o NO RADICALS IN THE DENOMENATOR!! :o jk, jk :D :D

I guess it's because it would be extremely messy and this is much easier to remember too. :)[/quote]

I dont think there is any great difficulty in remembering the formula too.

Numerator is the value you get when the point is substituted in the line equation.
You may need a  little effort in remebering the denominator only.

[quote="gotztahbeazn"]proof using cauchy? im not sure of how to get rid of the inequality sign at the end though

you have point $(x_0,y_0)$ that is not on line $Ax+By+C=0$, and you have the normal line that goes through $(x_0,y_0)$, intersecting line $Ax+By+C=0$ at $(x_1,y_1)$

thus, you want to find the distance between $(x_0,y_0)$ and $(x_1,y_1)$ and you can setup the inequality like so:

$((x_0 - x_1)^2 + (y_0 - y_1)^2)(A^2 + B^2) \ge (A(x_0-x_1) + B(y_0 - y_1))^2$

$(d^2)(A^2+B^2) \ge (Ax_0 + By_0 - (Ax_1 + By_1))^2$

Noting that $Ax_1 + By_1 = -C$, you get:

$(d^2)(A^2+B^2) \ge (Ax_0 + By_0 + C)^2$

$(d)\sqrt{(A^2+B^2)} \ge |Ax_0 + By_0 + C|$

$d \ge \displaystyle\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2+B^2}}$

and.. yea... im gonna assume somewhere in there ur supposed to make the $\ge$ sign into an $=$ sign??

im not sure how to do that tho.... :blush:[/quote]

The inequality is true for any point  $(x_1,y_1)$ on  $Ax+By+C=0$.  The equality happens when
$(x_0 - x_1)^2 + (y_0 - y_1)^2=d^2$ becuase d is the minimum value for  the distance between  $(x_0,y_0)$ and any $(x_1,y_1)$ on $Ax+By+C=0$.

oh... well then that... completes my proof... haha XP

gotztahbeazn, your proof is fine.  You proved that every point on the line is distance at least $\frac{\left| Ax_0 + By_0 + C \right|}{\sqrt{A^2 + B^2}}$ from the point $(x_0, y_0)$.  To show that equality is attained, we recall the equality conditions for the Cauchy-Schwarz inequality.  Equality in your use of Cauchy will occur when
\[
  \frac{x - x_0}{A} = \frac{y - y_0}{B} \, .
\]
That equation describes a line with slope $\frac{B}{A}$.  The original line has slope ${}-\frac{A}{B}$, so the two lines are perpendicular.  In particular, the two lines will intersect.  The point where the two lines intersect is an equality case, so its distance to $(x_0, y_0)$ is indeed $\frac{\left| Ax_0 + By_0 + C \right|}{\sqrt{A^2 + B^2}}$.

(This argument assumed that $A$ and $B$ are nonzero.  We can handle the zero cases separately.)

the proof using cauchy is absolutely superb...

 :starwars:  :surf:  :w00tb:  :thumbup:


calculus
integration
limit
function
logarithms
real analysis
real analysis unsolved

I found this integral on some of websites in the past;

I tried a lot of time to solve it , but I couldn't  :

Here the integral I want the solution for it :

$ T\equal{}\int\limits_{0}^{t} \frac{\ln(1\plus{}\frac{x}{t})^x}{t(1\plus{}x)} dx$ :?:

Note that :

$ \lim\limits_{t\rightarrow \infty } T \equal{} \ln(4)\minus{}1$

how we could solve the integral only  :roll:  :?:

Are you sure that the integral is not
\[ \int_0^t\frac {\ln(1 \plus{} \frac {x}{t})^x}{t(1 \plus{} \frac {x}{t})}dx
\]
or
\[ \int_0^t\frac {\ln(1 \plus{} \frac {x}{t})^x}{t(t \plus{} x)}dx
\]or
\[ \int_0^t\frac{\ln(1\plus{}\frac{x}{t})^x}{t\plus{}x}dx\]
?

I'm pretty sure about  :wink: , and we use Dominated Convergence theorem to solve the limit of expression.

But; could you tell me the solution of any of the integrals you wrote may it will help me by my problem if you don't get the solution of $ T$.??

I appreciate your concern  :wink:

Basically, my idea in the proposed variations, or other similar integrals, is to write $ y \equal{} \ln(1 \plus{} \frac {x}{t})$, such that the integral limits become 0 and $ \ln2$, while $ \frac {x}{t} \equal{} e^y \minus{} 1$ and $ \frac {dx}{t \plus{} x} \equal{} dy$.  From here on, use standard techniques (possibly including integration by parts or use of some especial function such as gamma, zeta or so on depending on the exact form of the integrand) since you have an integral of $ y$ and $ e^y$...

Edited note: in proposed variations as the ones I mentioned in my post above, you end up having an integral
\[ \int_0^{\ln2}y(e^y \minus{} 1)dy \equal{} \left(\int_0^{\ln2}ye^ydy\right) \minus{} \frac {\ln^22}{2},
\]
and the remaining integral is obviously trivial by parts...

The integral becomes (after your substitutions) :

$ \int\limits_{0}^{\ln(2)} \frac{y.t.\exp{y}.(\exp{y}\minus{}1)}{1\plus{}t(\exp{y}\minus{}1)} dy$

I think is gonna be more hard to distinguish ....but it's good try.

We still work on it....

Thank you

[quote="Hidden Scofield"]The integral becomes (after your substitutions) :

$ \int\limits_{0}^{\ln(2)} \frac {y.t.\exp{y}.(\exp{y} \minus{} 1)}{1 \plus{} t(\exp{y} \minus{} 1)} dy$

[/quote]

That is precisely why I thought there may be a typo in your statement.  Now, if there was no typo, perhaps the whole point of the exercise WAS to use dominated convergence.  Otherwise, the integral is indeed not trivial...

You may want to try then $ y \equal{} \frac {x}{t}$, yielding
\[ T \equal{} \int_0^1\frac {ty\ln(1 \plus{} y)}{1 \plus{} ty}dy \equal{} \int_0^1\ln(1 \plus{} y)dy \minus{} \int_0^1\frac {\ln(1 \plus{} y)}{1 \plus{} ty}dy...
\]
Your first integral is going to produce $ (y\plus{}1)\ln(y\plus{}1)\minus{}y$ evaluated in 1 and 0 (yielding your $ 2\ln(2)\minus{}1$) and your second integral tends to 0 when $ t\to\infty$, so your dominated convergence result seems to be correct  :)

[quote="daniel73"][quote="Hidden Scofield"]The integral becomes (after your substitutions) :

$ \int\limits_{0}^{\ln(2)} \frac {y.t.\exp{y}.(\exp{y} \minus{} 1)}{1 \plus{} t(\exp{y} \minus{} 1)} dy$

[/quote]

That is precisely why I thought there may be a typo in your statement.  Now, if there was no typo, perhaps the whole point of the exercise WAS to use dominated convergence.  Otherwise, the integral is indeed not trivial...

You may want to try then $ y \equal{} \frac {x}{t}$, yielding
\[ T \equal{} \int_0^1\frac {ty\ln(1 \plus{} y)}{1 \plus{} ty}dy \equal{} \int_0^1\ln(1 \plus{} y)dy \minus{} \int_0^1\frac {\ln(1 \plus{} y)}{1 \plus{} ty}dy...
\]
Your first integral is going to produce $ (y \plus{} 1)\ln(y \plus{} 1) \minus{} y$ evaluated in 1 and 0 (yielding your $ 2\ln(2) \minus{} 1$) and your second integral tends to 0 when $ t\to\infty$, so your dominated convergence result seems to be correct  :)[/quote]

I checked it out again , and there is no typo ., The question was to solve the problem using DCT , by interchanging between the integral and limit After that,to try solve the integral and  calculating  the limit of the result.

The integral exist ...I'm pretty sure about it.

Your second try is nice ....

But I still need the exact solution of it.

I appreciate your help dude.. :wink:

I still waiting ..

Thank you

Mathematica claims \[ T\equal{}\int\limits_{0}^{t} \frac{\ln\left((1\plus{}\frac{x}{t})^x\right)}{t(1\plus{}x)} dx \equal{} \minus{}1\plus{}\log 4 \plus{}\frac{\text{Li}_2\left(\frac{t}{t\minus{}1}\right)\minus{}\text{Li}_2\left(\frac{2t}{t\minus{}1}\right)\minus{}\log 2 \log \frac{1\plus{}t}{1\minus{}t}}{t}\]
where $ \text{Li}_2$ is the dilogarithm function.  You can show that the numerator in the fraction approaches a constant as $ t \to \infty$, which proves the desired result.

[quote="JoeBlow"]Mathematica claims
\[ T \equal{} \int\limits_{0}^{t} \frac {\ln\left((1 \plus{} \frac {x}{t})^x\right)}{t(1 \plus{} x)} dx \equal{} \minus{} 1 \plus{} \log 4 \plus{} \frac {\text{Li}_2\left(\frac {t}{t \minus{} 1}\right) \minus{} \text{Li}_2\left(\frac {2t}{t \minus{} 1}\right) \minus{} \log 2 \log \frac {1 \plus{} t}{1 \minus{} t}}{t}
\]
where $ \text{Li}_2$ is the dilogarithm function.  You can show that the numerator in the fraction approaches a constant as $ t \to \infty$, which proves the desired result.[/quote]

Thank you , but the problem not in result . The result is definitely true when we take the limit of $ T$.

I think we need to make some substitution and using zeta function ..that what I feel..may it's wrong. :P

 :D

Well you already have a closed form expression for your integral, albeit in terms of non-elementary functions.  If not that, what exactly are you looking for?

I'm looking for the solution of this integral .

I believe there is some way to solve it.

 :huh:

O.K

What about if we part the expression inside the integral , to put in inequality. :maybe:

Substitute  $ ty\equal{}x$  .


The squares of a chessboard are labeled arbitrarily from 1 to 64, and 63 knights are placed on the board. The square labeled 64 is the only square without a knight on it. Each knight is then labeled with the number of the square on which it stands. Some "legal" chess moves are made, and after some time the 64th square turns out to be empty again. Let $n(k)$ denote the number of the square where the knight labeled $k$ now stands. Maximize the sum \[\sum_{i = 1}^{63}|n(k) - k|.\]

$ \sum_{i = 1}^{63}|n(k) - k| $
  where does the i=1 come in. I mean there is no 'i' in the given expression is there?
 I'm gonna assume that k=1.Ok.

I'm not sure if this is entirely correct but here goes.
    [b]In the end the 64th square is empty. So the sum of all the n(k) terms will be the sum of the numbers 1 to 63. similarly as k varies from 1 to 63 the sum of all the k terms is also gonna be the sum of the numbers between 1 and 63. Hence the maximum value of the given expression is zero.
                Ashwath[/b] :)

0 is the minimum, but the problem asks about the maximum.

The point is to understand, first, which permutations are possible (I think all of them are, because the graph of knight's moves is sufficiently connected and contains cycles of even length, which are odd permutations).   Second, to find the permutation that maximizes the sum.


$n$ persons sit around a round table. The number of persons having the same gender than the person at the right of them is the same as the number of those it isn't true for.
Show that $4|n$.

[hide]Call a pair of people "interesting" if

They sit next to each other.
They are of opposite genders.

The number of interesting pairs encountered in one trip around the table is clearly even, because we must return to the same person who has the same gender as when we started.  Thus, the nubmer of people with the person to the right having the same gender is even, and so $4|n$ (because the two groups of people given are clearly disjoint and makes up everyone at the table)[/hide]


calculus
floor function
logarithms
function
algebra

Ok so my high school calculus teacher just showed the class the movie PI which i trust most of you have seen but in case you haven't, it is about a man going crazy because of math.

What i want is a very hard math question that would make him start pulling his hair in frusteration. But of course, it has to be something that he could solve in maybe a half an hour tops

Sounds evil. Why do you want your math teacher to pull his hair out?  :P

1) I have not seen the movie Pi. I generally hate math movies.
2) This is a very silly question.
3) Look in the resources for an AMC/AIME problem. [size=67]Most high school teachers cannot solve olympiad problems. [/size]

Heres a hard one!!

 Define f(1) = 1, f(n) = n/10 if n is a multiple of 10 and f(n) = n+1 otherwise. For each positive integer m define the sequence a1, a2, a3, ... by a1 = m, an+1 = f(an). Let g(m) be the smallest n such that an = 1. For example, g(100) = 3, g(87) = 7. Let N be the number of positive integers m such that g(m) = 20. How many distinct prime factors does N have?

[hide="The Answer"]Most integers have two possible predecessors. Let M be the operation of moving back from n to n-1 and D the operation of moving back from n to 10n. There are two related difficulties. We cannot take M10 on a multiple of 10 (because M does not apply to multiples of 10 plus 1); we cannot even take M9 on 10 (because having reached 1 we must stop). The predecessor of 1 must be 10. Then we need a further 18 moves back. So, leaving aside the difficulties above, there are 218 possibilities. We must exclude the 29 M9X...X, where X denotes M or D. Then we must exclude the 27 DM10X...X. Similarly, we must exclude the 27 XDM10X...X, the 27 XXDM10X...X, ... , and the 27 X...XDM10. Thus we are left with 218 - 29 - 8·27 = 218 - 3·29 = 29509.
[/hide]
Ps. This is not a olympiad question!!

[quote="mdk"]What is the answer to the problem?[/quote]

[hide="(Long-winded) Explanation"]

Let's look at $100!$ first, as it's easier to illustrate with a smaller number. The number of terminal zeros depends on the number of times $10$ will divide the number, which is the same as the number of pairs of $2$s and $5$s in that number's prime factorisation. We will have more $2$s than $5$s, so we just need to find the number of $5$s in this number. If we expand out $100!$ into$(1)(2)(3)(4)(5)(6)...(100)$ we see that 5 will divide every fifth number. So there are at least $100/5 = 20$ zeros at the end.

But what about numbers that $5$ will divide twice, such as $25$, $50$, $75$ and $100$? So there are $100/{5^{2}}= 4$ more zeros. So our answer is $24$. We can do much the same for $123456789!$, but we'll need to take into account greater powers of $5$.

So we get:
$\lfloor{123456789 \over 5}\rfloor+\lfloor{123456789 \over 5^{2}}\rfloor+\lfloor{123456789 \over 5^{3}}\rfloor+...$
[/hide]

dude....

why don't you just ask your teacher to prove fermat's last theorem?

..Or Goldbach's conjecture, I mean, it seems easy enough  :lol:

[quote="DiscreetFourierTransform"]Ask him to find the amount of zeroes on 123456789!
It's easier than it looks, but it might stump him for a while.[/quote]

Actually, you could possibly have zeros that are not in the end of the number, so it is more than whatever you guys got.

This is a really pointless thread. There are hard problems all over the place, including right in this forum.  :(

Sorry, I meant the number of terminal zeroes:
The solution is always:
$\sum_{i=1}^{\lfloor{\log_{5}N}\rfloor}\left\lfloor \frac{N}{5^{i}}\right\rfloor$

Where the floor function rounds the number down to the greatest integer less than or equal to the argument.

Thus, for $123456789!$,
we have:
$\sum_{i=1}^{\lfloor{\log_{5}123456789}\rfloor}\left\lfloor \frac{123456789}{5^{i}}\right\rfloor$
$\sum_{i=1}^{11}\left\lfloor \frac{123456789}{5^{i}}\right\rfloor$

Which works out as a previous poster put it.

EDIT: Or better yet: find a closed form of the inverse gamma function (to my knowledge, no such closed form function exists)
Ask him to solve this: (approximate x)
$\Gamma(x) = 115$
That requires the result of the original problem.


A number $p$ is $perfect$ if the sum of its divisors, except $p$ is $p$. Let $f$ be a function such that:

$f(n)=0$, if n is perfect
$f(n)=0$, if the last digit of n is 4
$f(a.b)=f(a)+f(b)$

Find $f(1998)$

Huh... nevermind.

Edit:  [hide="Solution"]Let us prove a stronger statement:  $5 \not | n \implies f(n) = 0$.

$f(6) = f(2) + f(3) = 0$
$f(4) = 2 f(2) = 0$
$f(2) = f(3) = 0$

[b]Lemma:[/b]  Given an odd integer $k, 5 \not | k$, then some $n$ exists such that $2^n k \equiv 4 \bmod 10$.

Proof:  Clearly $2^n k$ is even, so we only require $\equiv 2 \bmod 5$.  If $k \not \equiv 0 \bmod 5$ then such $n$ obviously exists.

From our lemma, for such $n, k$ we have

$f(2^n k) = n f(2) + f(k) = f(k) = 0$

From which our claim follows.

Therefore, $f(1998) = \boxed{0}$.
[/hide]

See also : http://www.mathlinks.ro/Forum/viewtopic.php?t=81054 . (Spanish section)

i think that the exercice is obviously not hard because the fact that we don't know any    f(x) # 0 $\Rightarrow$ the result must be 0  :rotfl: .

[quote="M4RI0"]Find $ f(1998)$[/quote]
It's not $ f(1998)$ but $ f(1988)$.
Anyway, it doesn't change the problem. The result is 0.
But it's a bit strange that we can easily solve the problem without making reference to perfect numbers.


Bill buys a 32 oz bottle of soda for 3 cents per ounce. Jill buys a 40 oz bottle of soda for 2.5 cents per ounce. How many more cents does Jill's bottle of soda cost?

Let me just $ \text{\LaTeX}$ - ify that for you.

$ (40 \times 2.5) - (32 \times 3) = 100 - 96 = \boxed{4}$


geometric sequence
algebra proposed
algebra

1)show the existance of (at least) 4 strictly increasing infinite aritmetic sequence such that:
   i)they have no terms in common with each other
   ii)they have no terms in common with Fibonacci sequence
   iii)each sequenze is "primitive" in the sense that is not a subset of another

2)show the existance of (at least) 4 strictly increasing infinite geometric sequence such that:
    i)they have no terms in common with each other
    ii)they have no terms in common with Fibonacci sequence
    iii)each sequenze is "primitive" in the sense that is not a subset of another

1) $ 4\plus{}11\mathbb{N}$, $ 6\plus{}11\mathbb{N}$, $ 7\plus{}11\mathbb{N}$ and $ 9\plus{}11\mathbb{N}$.
2) $ \{7^n \}_{n\in\mathbb{N}}$, $ \{2\cdot 7^n\}_{n\in\mathbb{N}}$, $ \{4\cdot 7^n\}_{n\in\mathbb{N}}$ and $ \{5\cdot 7^n\}_{n\in\mathbb{N}}$.

Are those "primitive"? Not sure what you mean by that.

yes,perfect  olorin :wink:
(but in 2) we must let $ n>0$)


Who all is going to CCU tomorrow?  :)  I'll be there for Wilson High School, though unfortunately I have to be in the lower division since I haven't officially taken Precalc yet...

Anyways... see ya guys there! I'll be the Indian kid with longish hair on my team...  :)

I'll be there with AC Flora... the Indian kid with the really short hair (cut it yesterday for the first time this year haha).

And jb05 will be there with Hoggard...

Wait, I thought your name was Sean Soni? That sounds Italian...

But anyway, see ya guys! :)

[quote="gauss202"]I wasn't able to attend today. What were the results of the contest?[/quote]

In Large School Level 1, it was:

1) me (Wilson)
2) kid from AC Flora who's name I can't remember
3) Aman (Wilson)

and we won for school in that division. Of course AC Flora ended up winning it all. I remember Ben Lippen won small schools, I think both Level 1 and 2... I forgot level 2 results otherwise...

Large Schools Level 2

1) Harry
2) Ian
3) John Berman

What was the quality/difficulty of the tests like this year?

They were actually pretty good. It could have been much harder, though. Both my mistakes were pretty silly (externally instead of internally tangent and drew a line between the wrong two edges of a triangle). Judging by the test difficulty and morning lectures (sudoku), the test writers really underestimate us.

I was really happy though because Hoggard got 3rd for both varsity and jv. For us, that's exciting.


calculus
integration
trigonometry
calculus computations

Using the generalized mean value theorem for integrals, prove that $\int_{\sqrt{n\pi}}^{\sqrt{(n+1)\pi}}\sin (x^{2})dx=\frac{(-1)^{n}}{c}$ where ${\sqrt{n\pi}}\leq{c}\leq{\sqrt{(n+1)\pi}}$.

[i]Hint [/i]:[hide]Evaluate the integral $\int_{\sqrt{n\pi}}^{\sqrt{(n+1)\pi}}x\sin (x^{2})dx$ with the substitution $u=x^{2}$.[/hide]


geometry
inradius
trigonometry
circumcircle
symmetry
congruent triangles
perpendicular bisector

Let ABCDE a cyclic pentagon, if rabc = raed and rabd = race, prove that AB=AE, and BC=DE. [u]NOTE[/u]: rxyz is the lenght of the inradius of the triangle XYZ  :rotfl:

[b]Lemma[/b] Given two points $X,Y$ on a circle $\Gamma$ (with radius $R$) and a real positive number $r$, there exist at most two points $Z$ on $\Gamma$ such that the triangle $XYZ$ has inradius $r$.

[b]Proof[/b] : The inradius of a triangle is given by the formula \[r=4R\sin(x/2)\sin(y/2)\sin(z/2)\] (calling $x$ the angle in $X$ and so on); we have fixed $R,r$ and the angle at $Z$, then also the sum $x+y=\pi-z=\alpha$.
So \[\sin(y/2)\sin(\alpha/2-y/2)=\frac{r}{4R\sin(z/2)}=\textrm{const.}\]  \[\sin(y/2)(sin(\alpha/2)\cos(y/2)-\cos(\alpha/2)\sin(y/2))=\textrm{const.}\]  \[\frac{1}{2}\sin(\alpha/2)\sin(y)+\cos(\alpha/2)\cos(y)-\frac{1}{2}\cos(\alpha/2)=\textrm{const.}\]  \[\cos(y-\alpha/2)=\textrm{const.}\] but $|y-\alpha/2|=|y/2-x/2|\leq \pi/2$ so there are at most 2 solutions at the above equation, in the form \[y=\frac{\pi}{2}-\frac{z}{2}\pm\beta\quad(\star)\] QED

Now, from the so called Japanese Theorem, it follows that \[r_{ADE}+r_{ABD}+r_{BCD}=r_{ABC}+r_{ACE}+r_{CDE}\] then $r_{BCD}=r_{CDE}$; the points $E,B$ are uniquely determined from the lemma and are such that $BCD$ and $ECD$ are congruent triangles (from $(\star)$). So $DE=BC$.
By hypotesis $r_{ABC}=r_{AED}$; there are at most two points $P$ on the circumcircle of $ABCDE$ such that $PED$ has fixed inradius and, from $(\star)$, they're symmetric wrt the perpendicular bisector of $ED$; a symmetry in the perpendicular bisector of $CD$ exchanges $BC$ and $DE$ and takes $A$ to a point $A'$ on the circle such that $r_{A'ED}=r_{ABC}=r_{AED}$, so $A'$ is either $A$ or its symmetric wrt the perpendicular bisector of $DE$, but this latter case is impossible (unless $CD$ and $DE$ coincide, so $A'=A$ and $A$ is on the axis of $CD$ and of $BE$, that is $AB=BE$.

PS : Now I've written it all, I realize it's not the clearest solution I ever wrote ... if it's too messy to be readable, tell me and I'll rewrite it, maybe not past midnight. ;)


What are the rules for making avatars/signatures?

[quote="Avatar Control Panel"]Only one image can be displayed at a time, its width can be no greater than 80 pixels, the height no greater than 100 pixels, and the file size no more than 21 KB.[/quote]
And I think its obvious that nothing offensive or obscene should be posted...

[quote="Signature Control Panel"]There is a 255 character limit on 3 line(s), font size 10 to 12. In your text, no more than 75 characters without space too. No images. No more than 1 image(s). No more than 2 url(s). Only internal links allowed. Only 3 smilies are allowed. No code tags. No hide tags. No quote tags. No list tags. No colors. No centered LaTeX tags. No youtube tags.[/quote]
Again, pretty obvious that nothing offensive or obscene should be in there...

And don't talk LiKe tHiS, kay?

Please do not make such a title, it's distracting. As sated in the User CP:

[quote]Displays a small graphic image below your details in posts. Only one image can be displayed at a time, its width can be no greater than 80 pixels, the height no greater than 100 pixels, and the file size no more than 21 KB.[/quote]
[quote]This is a block of text that can be added to posts you make. There is a 255 character limit on 3 line(s), font size 10 to 12. In your text, no more than 75 characters without space too. No images. No more than 2 url(s). Only internal links allowed. Only 3 smilies are allowed. No code tags. No hide tags. No quote tags. No list tags. No colors. No centered LaTeX tags. No youtube tags.[/quote]

EDIT: Too late.


Read the "Advice" thread by me in this forum.

[quote="saturnlife"]I am a freshman in high school, and I'm hoping to do well enough on the AIME test in late March, so I can qualify as one of the 30 freshmen to be chosen to be in the summer program. I currently have access to almost all the past AIME tests and I also have both Art of Problem Solving books. Can anyone please give me some advice on how to study for the AIME during this next crucial month?[/quote]

As far as I understood, it's not just AMC and AIME scores that will determine which freshmen go to MOP, particularly if more than 30 make USAMO.  Is this correct?

Almost certainly USAMO score will count. I'm not sure if anything else will in fact.

It would make logical sense that they select the top 30 performing freshman on the USAMO, without regarding their AMC and AIME scores, seeing as how MOP is geared toward IMO that type of math.

That would make sense to me at least...

That's not how they determine who goes to MOP though. The normal way of determining who goes to MOP is by the index AMC12(10?)+10*AIME+9*USAMO.


function
number theory proposed
number theory

f(n) = gdc(1,n) + gcd(2,n) + ... + gdc(n,n).
prove that f(n^k) = (n-1)kn^(k-1) + n^k where n is prime and k naturel number

Define $ f(n) = gcd(1,n) + gcd(2,n) + \ldots + gcd(n,n) $.
Fix a prime $ p $ and natural number $ k $. Then for $ 0 \leqslant l \leqslant k $
there are exactly $ p^{l} $ natural numbers between $ 1 $ and $ p^{k} $, which
are divisible by $ p^{k-l} $. Therefore for $ 1 \leqslant l \leqslant k $
there are exactly $ p^{l} - p^{l-1} $ natural numbers $ a $ between $ 1 $ and $ p^{k} $,
such that $ gcd(a,p^{k}) = p^{k-l} $, and the only number $ a = p^{k} $ for which
$ gcd(a,p^{k}) = p^{k} $. Therefore $ f(p^{k}) = p^{k} + \sum_{l=1}^{k}
(p^{l} - p^{l-1})p^{k-l} = p^{k} + \sum_{l=1}^{k} (p^{k} - p^{k-1}) =
p^{k} + k(p-1)p^{k-1} $.

I think it is from here: http://www.mathlinks.ro/Forum/viewtopic.php?t=30906
(I think you should ask there next time)


A disk is divided into $2n$ equal sectors. Among these sectors, $n$ are colored in blak and $n$ in white. Let us number the white sectors from $1$ to $n$ clockwise, and the blak ones from $1$ to $n$ counterclockwise. show that there is a half-disk consisting of $n$ sectors which contains all numbers from $1$ to $n$.


A triangle $ABC$ is given. A circle $\Gamma$, passing through $A$, is tangent to side $BC$ at point $P$ and intersects sides $AB$ and $AC$ at $M$ and $N$ respectively. Prove that the smaller arcs $MP$ and $NP$ of $\Gamma$ are equal iff $\Gamma$ is tangent to the circumcircle of $\Delta ABC$ at $A$.

Let $\angle MAP=\angle PAC=\frac{\angle A}{2}$. Because $\Gamma$ is tangent to $BC$ then $\angle BPM=\angle CPN=\frac{\angle A}{2}$. So $\angle AMP=\angle B+\frac{\angle A}{2}$ and $\angle ANP=\angle C+\frac{\angle A}{2}$. SO $\angle MPA=\angle C$ and $\angle APN=\angle B$. Let's $O'$ be the center of $\Gamma$. Then $\angle O'AP=\frac{\pi}2-\frac{\angle A}{2}-\angle C$ (asume that $AB<AC$), and $\angle O'AB=\frac{\pi}2-\angle C=\angle OAB$ (where $O$ is circumcentre). Hence $\Gamma$ is tangent to circumcircle.

Let $\Gamma$ be tangent to the circumcircle ($\Omega$) at $A$. Let $Q$ be the second intersection of line $AP$ with $\Omega$. Then $\angle PAO'=\angle APO'=\angle AQO$ which implies that $OQ\parallel O'P$ or $Q$ is midpoint of arc $BC$. Hence $\angle BAP=\angle BAQ=\angle CAQ=\angle CAP$.

The problem becomes very easy if you invert through $A$.


LaTeX
combinatorics proposed
combinatorics

The plane is partitioned into regions by a finite number of lines no three of which are concurrent. Two regions are called "neighbors" if the intersection of their boundaries is a segment, or half-line or a line (a point is not a segment). An integer is to be assigned to each region in such a way that:

i) the product of the integers assigned to any two neighbors is less than their sum;
ii) for each of the given lines, and each of the half-planes determined by it, the sum of the integers, assigned to all of the regions lying on this half-plane equal to zero.

Prove that this is possible if and only if not all of the lines are parallel.

Valentin, sorry but what does  "a less than b" mean : a is STRICTLY SMALLER THAN b or a is SMALLER OR EQUAL to b? :)

less than is not less than or equal (remember \leq in LaTeX ? :D) ...

Hy Mith!
Considering you are from Russia, I am sure you already met the problem.
I am sure I did, but I made a mistake during the contest :(

My name is Myth, not Mith!
I never met this problem before.

Sorry Myth!
BTW, somebody was right, the second part of the problem is Tournament of the towns 1989, the first is Yaglom. Not very fair for a BMO.

[quote="kueh"]I've got a different solution: the abs value of every segment = number of sides (-1 if it is open on one side). We can verify that this works.[/quote]

Can you explain a bit more?

I think the solution of Myth and DusT are both incorrect.

I thinl the right solution is:

We give each region a integer number:
Firststep:number each region the number of its sides(segment or half-line).

Secondstep:give"+" or "-"sigh to the numbers,
                  S.T.the sign of two neighbored region is opposite.

then we got the idea.

Oh no!Sorry.

My statement is correct if and only if the number of the lines is even.

if the number of lines is odd

we can do like this:

if the region is closed, number it the number of its sides
if the region is open,   number it the number of its sides and minus 1.

then give "+"or "-"sign to those regions
S.T.the sign of two neighbored  region is opposite.

[quote="georgew_bush"]I think the solution of Myth and DusT are both incorrect.

[/quote]

No, they are correct.  :)

Myth's solution do not satisfied " the product of the integers assigned to any two neighbors is less than their sum"
kueh's solution do not satisfied "for each of the given lines, and each of the half-planes determined by it, the sum of the integers, assigned to all of the regions lying on this half-plane equal to zero. "

In the post above I give my solution in Chinese.

[quote="georgew_bush"]
In the post above I give my solution in Chinese.[/quote]
Do you think that anyone can understand it? ;)

I give the graph to show the example of numbering.
I have post the english solution above. :P

[quote="Myth"]Let L be arbitrary line and P1,...,Pk be points of intersection L with other lines. These points divide L onto k+1 "segments". We assign positive integer number to each segment such way that sum of even-placed segments equals sum of odd-placed segments.

Now we color all regions with black and white such way that any two neighbors are colored with different colors. It is well known problem.

For each region we calculate sum of numbers which assigned to his sides. If region is white then we simply place this sum in this region, otherwise we invert sum.
It is easy to show i) and ii) are satisfied.[/quote]

Maybe my word is wrong.
To read your solution,what's yout definition of "even-placed segments" and "odd-placed segments"?

[quote="Myth"][quote="georgew_bush"]Myth's solution do not satisfied " the product of the integers assigned to any two neighbors is less than their sum"[/quote]
So many posts and so few meaning!  :mad: My solution is correct![/quote]
Myth!!! Finally you are back!! So long time not seeing your posts!

Another solution(similar to Myth's but it has one difference).
We color the regions white and black.
Let we call a good point, point which is intersection of two lines.
If region is white then we calculate number of good points on boundary of that region, and we assign that number to region, if region is black, we multiply that number by $ \minus{}1$,and then we assign product to that region. If all lines are parallel,we see that each region must be 0, and therefore first condition isn't ok.


Let $ a,b,c,x,y,z>0$ s.t. $ a\plus{}b\plus{}c\equal{}x\plus{}y\plus{}z$.Prove that:
$ ax^2\plus{}by^2\plus{}cz^2\plus{}xyz\geq 4abc$

[quote="Inequalities Master"]Let $ a,b,c,x,y,z > 0$ s.t. $ a \plus{} b \plus{} c \equal{} x \plus{} y \plus{} z$.Prove that:
$ ax^2 \plus{} by^2 \plus{} cz^2 \plus{} xyz\geq 4abc$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31344


AMC
USA(J)MO
USAMO
Putnam
AMC 10
ARML
AIME
\/closed

AoPS now offers [url=http://www.artofproblemsolving.com/Books/AoPS_B_CP_AMC.php]AMC books[/url] (past AMC/AHSME and USAMO problems), [url=http://www.artofproblemsolving.com/Books/AoPS_B_CP_Putnam.php]Putnam books[/url] (undergraduate proof competition), and the two [url=http://www.artofproblemsolving.com/Books/AoPS_B_CP_Mand.php]Mandelbrot Competition books[/url] (a high school contest we recommend to anyone who enjoys math).

WOHOOOO!!!

More great AoPS, we must all rejoice.

Yay!

I need those AMC problem sets.
 :lol:

Yay, AMC practice :o

maybe it'll help me break a 150 on the AMC10 :\

I JUST ORDERED AN AMC TITLE FROM AMAZON YESTERDAY.  DANG.

I have been wondering what things like AMC, ARML, AIMS, and the rest stand for ever since I came to AoPS. Could someone help me?

I have read about Putnam in Mathematical Gems though.

AMC = American Math Competition. You don't have to be smart to take it.
AIME = American Invitational Math Examination (?). You have to be pretty smart to qualify.

ARML=American Regional Mathematics Leagues
A competition where each state sends several teams to compete at one of 3 locations.  It is both individual and team round stuff.  The states choose whatever teams they want.  For more information, check out: [url=http://www.artofproblemsolving.com/Forum/forum-348.html]www.artofproblemsolving.com/Forum/forum-348.html[/url]

AMC=American Mathematics competion
link:[url=http://www.artofproblemsolving.com/Forum/forum-133.html]www.artofproblemsolving.com/Forum/forum-133.html[/url]
There are 3 competitions: AMC 8/10/12
The grade limit on each is 8/10/12, respectively
They are each 25 question exams of varrying difficulty, and if you do well enough, you can qualify for AIME (American invitational mathematics exam), which is a 3 hour 15 question test.  All answers are integers from 000 to 999.  Doing well here qualifies you for the USAMO, and from there, MOP.

Look around the forums for more information.

Here's a link to a topic in the AMC forum:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11693]FAQ: Description of the AMC Examinations[/url]

[quote="mathninja"]Can we pre-order the intro number theory book in any way?[/quote]

No, but they'll be available to order in just a few weeks.

[quote="Phelpedo"]Would it be worthwhile to read the geometry and number theory books in addition to AoPS volumes 1 & 2, considering that both of the former are covered in the latter?[/quote]

There's not all that much number theory in the AoPS books and the Intro Geometry book is a far more complete treatment of the subject that in the AoPS books.  The new books are intended to be a full curriculum whereas the original AoPS books simply didn't have enough pages to cover all the material or even contain enough problems to cover all related problem solving techniques.

[quote="d.lam.86"]Are there any confirmed dates of when the AoPS Number Theory and Algebra textbooks will be released? Could someone from AoPS PM me if there's any tentative dates? Thanks.[/quote]

Number Theory: end June, this year.

Algebra: Early 2007.

[quote="maokid7"]would it be worth getting the intro. to nuber theory book if i can pretty much already do most of the problems in the number theory anddiophantine equations chapters in AoPS books 1&2?[/quote]

Maybe and maybe not.  The topics covered in the Intro NT book will be broader than in the AoPS books, but it may be that you can already solve many or most of those problems.  What might be a good idea for a student in your situation is to encourage your math team to buy a copy and just study the few sections you can learn the most from before passing it on to other math teamers.

It's not likely there's much new material for you in the Intro NT book, but there may be a few concepts and problem types that you haven't encountered.

But as for whether or not that makes it worth purchasing for you, only you can decide.  Perhaps that decision will be easier when we publish a few pages of the book in our bookstore area online (as we do for the other intro books).


Buenas tardes,

encontre este link:

http://www.divulgamat.net/weborriak/RecursosInternet/Recaula/Geometria/viviani.htm

aparece el enunciado y su demostracion ...

conocia el teorema pero no el nombre  :wink:

carlos

[quote="MathTeX"]Podían colocar el enunciado de este teorema y su demostración por favor?

Gracias.[/quote] No entiendo el motivo de este post, si es fácil buscar en google: Teorema de Viviani, y ver el primero de la lista...¿¿??. Nada más.

$Tipe$

Tambien observemos que este teorema se sigue cumpliendo para cualquier poligono regular.


geometry
incenter
circumcircle
trigonometry
geometry solved

let I be the incenter of triangle ABC let AI meet the circumcircle of ABC at D .Denote the feet of the perpendicular from I to BD,CD by E,F respectively. if EF+IF=1/2AD, calculate <BAC.

[color=red][Moderator edit: Also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=34374 .][/color]

90 if you ment IE+IF, not EF+IF.
All you have to do is compute AD using Ptolemeu, BD using the sine law, use BD=DC=DI and write the area of BICD in two ways.


It shames me to admit that I am still unsure of the exact positions of these two political parties, and I am thus unsure of which party I stand in. So I was wondering if you guys could help me. I will list some beliefs of mine, and see if anyone can categorize me. (I am not advocating any of these beliefs).

I AM PRO-LIFE (against abortion)

I AM AGAINST GAY MARRIAGE

I BELIEVE THAT THE GOVERNMENT SHOULD BE ACTIVE IN HELPING THE CITIZENS

I BELIEVE THAT WE SOULD NOT GET INTO FOREIGN AFFAIRS WITHOUT THERE BEING A DIRECT THREAT TO THE HEALTH OF THIS COUNTRY

I BELIEVE THAT RAISING THE LOWER CLASS IS MORE IMPORTANT THAT ENRICHING THE UPPER CLASS

If there are any more topics that I should put my position on in this thread please let me know but that's all I can think of for now!

I fail to see any good reason for your use of capital letters.  Please don't do that.

Being a Democrat or Republican is a matter of officially registering as a member of a political party.  So, in all likelyhood, you are neither.  If you're interested in which party shares the majority of your views, you should probably go look at what they claim their beliefs are (e.g. by visiting their respective party websites, either for the national party or for your state or local branch).  Then, of course, you'll discover that both parties try to appeal to everyone, but that (in general) one party is much closer on some of your beliefs and the other party on others.  So, then you should consider what your beliefs actually mean.  For example, does "opposing abortion" mean that you think the medical procedure of abortion should be outlawed?  That's a position shared by one party but not the other.  On the other hand, one party supports comprehensive sex education, including contraceptive education, that results in a better-educated populace having safer sexual relations and fewer unwanted pregnancies, so possibly fewer abortions.  Then, of course, once you've decided [i]that[/i], you have to decide how that compares in importance to you with the question of whether government should help transfer wealth down the income scale or not.  Etc.  And, of course, you can carry this analysis out not just for the parties but for each individual politician running for office.

If you wanted my personal opinion on which party you should vote for, I would tell you to select the party that [i]I[/i] vote for :P

In my opinion, you're somewhere in the middle.  If 0 represents a democrat and 100 represents a republican, and 40-60 is in the middle, you're about a 40.

Take this quiz:

http://www.politicalcompass.org/

It will tell you where you are.

As for your positions, I'm also interested in your stance on free trade (NAFTA e.g.)


[quote="brettr"]I see LaTex images in many of the post and the image link shows they are hosted on mathlinks.ro.  The FAQ says mathlinks.ro doesn't host images.  What is the easiest way to use LaTex in posts?

Thanks,
Brett[/quote]
You just need to type it directly into your post, using dollar signs ("$\$$") to surround the LaTeX. For example,
[code]$\$$1+2^{2}=5$\$$[/code]
becomes
$1+2^{2}=5$.
That's all there is to it!  :lol:

Also, here is a helpful topic:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5261

Thanks.  What is required for a website to render that way?

Brett

[quote="brettr"]What is required for a website to render that way?[/quote][url=http://www.mayer.dial.pipex.com/tex.htm]LatexRender[/url] which requires a LaTeX installation on the server plus ImageMagick & GhostScript.

If that's not possible then [url=http://www.forkosh.com/mimetex.html]MimeTeX[/url] uses a CGI program, at the cost of flexiblity and image quality; if CGI programs aren't possible then [url=http://en.dahnielson.com/2006/09/mimetex-plugin.html]Dahnielson[/url] has a script for WordPress that caches mimeTeX images on a public server; there may be speed issues but I haven't tried it. This could no doubt be rewritten for any PHP page.

I have typed according to the user guide as following but unable to generate graphics. Could anyone tell me why I failed?

[quote]\documentclass{article}
\usepackage[pdftex]{graphicx}
\begin{document}

This is my first image.

\includegraphics{myimage.png}

That's a cool picture up above.
\end{document}[/quote]

[quote="bchui"]Could anyone tell me why I failed?[/quote]How? You haven't given any information on what failed. Fortunately, you give the error in another thread http://www.artofproblemsolving.com/Forum/viewtopic.php?p=652681#652681 where the problem is answered.


All diagonals of a convex $n$-gon $F$, where $n \ge 4$, have equal length.  Which of the following statements is true?
(A) $F$ must be a quadrilateral.
(B) $F$ must be a pentagon.
(C) $F$ is either a quadrilateral or a pentagon.
(D) $F$ is either an equilateral or equiangular polygon.

[quote="pilot"]The diagonals of a square or a regular pentagon have equal lengths and if n>=6 this isn't true, so it should be C.[/quote]

Same thing with any rectangle.


Carl Marx's Capital and other works
Vladimir Ilich Lenin's works
Mao, Zedong's works

If the answer is yes, what will happen if you read them?

I can feel the hostility against communism around, or in other words, the idea that "communism is evil, maybe even worse than terrorism" is well established in the deep mind of most American. I often hear such words: China? Oh, it's changing, no longer communism, things are much better now, blah blah.
I write the above beacause of this [url=http://www.southcoasttoday.com/daily/12-05/12-17-05/a09lo650.htm]news[/url]

[quote="Jan Tokk"]Marx was a German, so it should be pretty easy for me to get one of his books...
If you want one too, just tell me :D[/quote]
Thank you for your kindness, unfortunately, I can't read German. These books, in different languages, can be found on web easily. What I mean is the traditional paper-base form.

[quote="bleumoose"]librarians across the country have fought to keep books from being banned. i'm sure they're available, but let me double check.

EDIT: the first two are available in my public library, however, Mao Zedong's stuff isn't there. Although this isn't surprising. Our library system is pretty good, but not stupendous.[/quote]
What a precisian! Good job.
What level is the library that you checked in? I mean, is it serving a university, or a city or the capital of a province/state/country?

my relativity professor told us he once went to poland during communism, and that he could be buy for four euros or something , a decent physics book.  However, included in the price, where works of Lenin.

 ghent public libraryhas books by marx : das kapital, communist manifest...
By Mao zedong : red book-thoughts on cultural revolution, on guerilla warfare

they also have books of stalin  and the infamous 'mein kampf' by hitler

the university also has libraries scattered throughout ghent, a bit harder accessible, but they have many books like that

Bottomline : I could easily get my hands on a copy
However, which ones are worth it? I heard Hitler's writing style is a bit weird in mein kampf...

In Greece you can  - easy and everywhere  -  find these books (in greek) as well.Many people between 1970 - 1990 have read them , me too. Now ,  nobody reads! All watch tv, top models and other  fully series(Sorry). But I think that everybody must read K. Marx, Lenin and MAO - other books , too.If these books dissapear , then a greater terror will rise!
[u]Babis[/u]

I think there is a difference between Karl Marx and the rest
most people here in belgium would consider the first to be 'an author', while they would consider mao , stalin,... dangerous men, whose books they would see more as a warning

Is communism popular in Greece (that is the image I get from november 17,etc...)

Marx gave the motivation and the principle, the others practised his idea in different circumstances, thus the latters are more dangerous, because people not only know what to do, but also how to do.

By the way, many ideas of Socialism/Communism have been partly realized in most well-developed capitalistic countries, like planned economy, labour union, welfarism.....

fredbel6, what happened on november 17?

I've read Hitler's "My Struggle" before, but didn't find the legendary Hitler's instigat-ivity(what should the word be?), maybe that's the problem of translation, translation can make you understand what he said, not how he spoke. Anyway, he did exactly on the way to what he claimed, quite honest a bastard.

Belgium has had its share of communism too, socialists have a much bigger influence

Right now extreme right, occassionally racist, Vlaams Belang, (Flemish Interest) is being blocked from power by all other parties.  They take an extremely firm stance against immigrants, have been linked to  neo nazism, and they feel that Belgium should split.  I wrote occassionally because a couple of years ago they were still the Flemish Block, but because a coupled of linked associations were convicted for racism they just changed their name  :huh:

Back on topic: many people  feel the block from power is undemocratic, but many people point out that communists, while successful in the past, have also been blocked from power

Also, in 1984/1985 we had a Belgian terrorist Carette, bombing NATO installations and banks and stuff, he wounded several, and killed two firemen, surprisingly he was released a couple of years ago

Sorry for writing so much, communism and radical regimes , I have an interest for them for some reason.

November 17 : Once in Greece, on november 17, 1973, students protested against a military coup (greece seemed to be struggling?)
in 1975, november 17, a group that called itself marxist, began acts of terror
as of now they killed at least 24 people, and did more than hundred bombings.
Also, until 2002, none of them had been apprehended

They also killed a British diplomat Stephen Saunders, for some reason they thought he had to do with Kosovo bombings I believe.
I discovered them because I am a huge 24 fan , and Stephen Saunders is the name of a character there
I wonder what his family thinks now, 24creators didn't do this by accident?

"Claimed to be communist" and "Be communist" are different things.

I didn't even try to purchase these books in the main library of our country, although they have entries in the catalogue. I'm almost sure they will be "written off" and the fact of your purchase can even have bad consequences for you.

I think so, because once I purchased a Soviet astronomy journal and it was "written off". And about 2 days ago I purchased a Soviet mathematical journal, which was "written off" too. I don't even know, which Soviet periodical isn't "written off" yet. In contrast, some books are still available [b]so far[/b].

This is a sort of inquisition to Soviet editions here and I feel like living in the Middle Ages.

Aren't Russian recalling Maxism in some sence?

Well, my friends, this a great story. I can not speak  english well, so I must not write. Of  course there are some men in history which have influensed or inspirited the people to do real and human history. Every attempt needed blood.Unfortunatly the death is a friend of the changes and revolutions.In France(1789) only a revolution brought human rights, in USA , too. All our countries have fighted against some "enemy".Simple people have not  enemies!But then why there are war and terror? You can find easily an answer , the true answer, as well.
Karl Marx, a great philosopher, tried to explain the lows of societies and to give an orientation , which could make the life better.The result is still unknown.But these are political questions , really usefull.
  Terror is something else. In Greece the most people believe always(and now) in Democracy.The left side has 10%. In the past , as in whole Europe , there where some people who believed that society will change , if they kill some important persons.17 November was such a group. But Marx never said that we'll change our life or our society by killing but only by changing the way of producing the social product.Very difficult to understand or to explain. But without real and deep democracy nothing can be done.This is proven by the history in the so called << East Europe>> between 1945 and 1990.
   To young MLinkers: It is really good to read not only the books we agree with but books with the oposite opinion  too.Perhaps this will add to our civilization something good. In Athens of Pericles(5 centuries BC) they had democracy , because there where many opinions and people knew how to conversate.

 [u]Babis[/u]
 ( Never stop wondering: Why do there is terrorism ? Are terrorists communists or are capitalists -and some strong counties - the greater terrorists? And how?)

[quote="stergiu"]  It is really good to read not only the books we agree with but books with the oposite opinion  too.[/quote]
Good point.

[quote="Buffalo"]What do you mean by "Our Country", nonIMO?[/quote]
It's an ex-Soviet republic. I don't want to speak more concretely. I just wanted to note, that countries with such a situation exist.

[quote="fredbel6"]Why don't you wanna say which ex soviet state you live in?  Is it a Baltic state?[/quote]

Why should I want this?  :)  For me is important, that I was born in the USSR, which is therefore my Motherland, regardless of whether my birthplace was near the Baltic Sea, near the Black Sea or in the south. It's as if a foreigner would ask, in which part of Belgium were you born.

can't some of those be found on amazon...
for example i've definitely seen the little red book and on guerilla warfare by mao

[quote="fredbel6"]nonIMO, actually you (accidentally?) struck a sensitive chord
Belgium consists of Flanders and Wallonia, first part borders with Netherlands and speaks Dutch, south part speaks French and borders mainly with France.  There is not much tension (really, Belgium is a safe country, nobody is taking arms or fists up for it, the battle is fought out on paper) but there is not much unity, I'm not close with anyone in the south and I don't have any peers who are either.  Both parts selected their greatest Belgian recently (you know that television show, many countries do it) and guess what?  we selected Father Damien, they chose Jacques Brel, as expected we both tok them from our own group

So if someone might ask me where I'm from, I might even add Flanders myself :oops:

I didn't know this forum still provided a flag for USSR, strange I thought that most non Russians would have prefered strongly to mention their own country.  But hey, that is all fine by me, certainly didn't mean to offend you.  By the way, when people live in USA, they often do mention too which state they are from, it is just nice to have a better idea of that person.

I understand now why you oppose (you do right?) the witch hunt for Soviet era books[/quote]

Interesting, I understand now, that I do really need to fill a gap in my knowledge of the Belgian history  :)  By the way, I've always liked united countries (you can find this in my earlier posts in other threads of the ML).

I didn't think, that you wanted to offend me. A calm discussion is going on for such a controversial topic.

[quote="Buffalo"]Carl Marx's Capital and other works
Vladimir Ilich Lenin's works
Mao, Zedong's works

If the answer is yes, what will happen if you read them?[/quote]

Books by all of those authors are available on the open shelves of my county's public library system. (I just checked the computerized catalog to be sure.) Nothing would happen to anyone who reads them here--other than the reader wasting some time on silly communist ideas.

Have you seen the new book [url=http://www.amazon.com/gp/product/0679422714/]Mao: The Unknown Story[/url] yet? I'm checking, because books by the authors you mention are widely available in both English and in Chinese in Taiwan, where I lived just more than four years ago, but I get the impression that [url=http://www.cpj.org/regions_06/asia_06/asia_06.html#china]press freedom is still quite limited in China[/url], as it was when I last traveled to China.

Just one question, what percentage of what Chang Jung written in her book do you think could possibly be true?
Could you please explain to me what is demonization?

[quote="shobber"][quote="Buffalo"]Chang Jung[/quote]
What's this?[/quote]
ZHang Rong

Well, I live in Romania (see flag) and I can't think of a reason anyone here would want to read such books, let alone buy them.
Romania is an ex soviet Republic and all the people who might have wanted to read these "works" have had ample opportunity during the former regime. In fact, many of those who did not want to read them have also done so. Of contrition if nothing else.
Nowadays nobody really reads much of anything anyway, let alone these. It's hard to make ends meat here, and it takes up most of one's time.

I can only imagine westerners, few of those anyway, possibly wanting to read Marx, Engels or Lenin and also having the time and patience to do so. This is because your peoples (most of them anyway) have not had the "glorious opportunity" to live under communism for half a century. You would not be able to believe your luck if you knew anything about the human, practical nature of the implementation of these thinker's "utopical" ideology. You have not experienced it, so you seek this knowledge in books. This is what it means to be curious. But my personal opinion is that these works are not only unsavoury but quite dangerous. They have often had strong and malignant influence on the reasoning of impresionable youths.

So, in conclusion, why don't you read something useful, meaningful, able to teach you something, instead? There are other, much more productive things to do with your time...

[quote="tokenadult"]but I get the impression that [url=http://www.cpj.org/regions_06/asia_06/asia_06.html#china]press freedom is still quite limited in China[/url], as it was when I last traveled to China.[/quote]

I was speaking to a friend who spent her Christmas break in China, and apparently wikipedia was blocked by her ISP there.  She was quite traumatized as a result.

maersklandro, maybe i will read some stuff by mao
but one important note, i don't think many people here who might read them also agree with them
and neither do I
Ceaucescu is quite famous in Belgian, because when he was removed from power, a belgian camera man was killed while filming (yep that is how news works here, the number of belgians involved pulls it up, distance pulls it down)


In the picture, AC = 8 cm and CD = 4 cm. Find the measure of BD:

[img]http://img230.imageshack.us/img230/9803/cals.jpg[/img]

a) 9
b) 10
c) 12
d) 15
e) 16

By AA, $ \triangle ABC \sim \triangle DAC$. So,

$ \frac {AC}{DC} \equal{} \frac {BC}{AC} \rightarrow \frac {8}{4} \equal{} \frac {4 \plus{} BD}{8}$

$ 2 \equal{} \frac{4 \plus{} BD}{8}$

$ 16 \equal{} 4 \plus{} BD$

$ BD \equal{} 12$

So the answer is $ \boxed{C) \ 12}$.


Suppose we have 2 numbers a,b s.t. (a,b)=1. A set S is called good if (1) 0 is in S
(2) if an integer x is in S, then x+a and x+b are in S. Find the number of good sets.

Is there omething missing? I don't see whay there can't be infinitely many such sets.

Because a good set contains at least all numbers of the form $xa+yb$ where $x,y$ are non-negative integers.
Since each nonnegative integer not less than $ab-a-b+1$ can be written as above, it follows that there are only finite number of good sets.
Moreover, it is well-known (Sylvester, 1884) that the number of nonnegative integers which cannot be written in the above form is $N = \frac {(a-1)(b-1)}{2}$, which gives an upper bound for the number of good sets as $2^N$.

Pierre.

This is last combinatorics problem from ISL 2000. It was also proposed for Romanian TST 2001( Valentin knows..)
This is just a sketch of the proof:
To solve it, write the natural numbers in the lattice plane: in (m,n) wih  write number mb+na. color red the numbers from the set.
The if (x,y) is red the (x+1,y) and (x,y+1) is red
We can see that a good set is uniquely detemined by the path of lowest red points joining (0,0) and (a,b).(such a path has a+b points)
Moreover this path must pass over the line joining them,
Now define a path ULULLUU( up-left-up etc.). We can prove that from the set consiting of a path and its a,b-1 shifts exactly one passes above desired line.
hence our number is (a+b a)/(a+b)

you can find the sol fro the the third round prob3 here:
http://ajorza.tripod.com/selection2001sol.html


calculus
derivative
probability and stats

let p and  b an integers such that b<p , let x1,x2,...,xp  p integers in {0,1,...,b} , let X=Sum xi, i=1...p , what is the value of  IE(X) ? ( I tried to fund the number of p uplet (x1,...xp) such that X=n with n<= p.b but the problem is that the xi are laying in {0,1,...,b} , i can solve it when the xi are <=n !! ) any help ?

I think that I solve the problem , using combinatory, the number of solution of X=n is the the coefficient of X^n in the polynome Q^p where Q=1+...+X^b , so the expectation we le be the derivative of Q^n in x=1 divided by Q^n(1) so the result is pQ'(1)/Q(1) =pb/2  , sorry  I'm so stupide, using that they are independant gives the result immediatly !!


For those of you that attend NYC Pubic High Schools, I would imagine that you took the NYC Interscholastic Mathematics League contest (1)
The questions are:

1.) The average of five numbers is 1/3 and the average of three of those numbers is 1/5. Find the average of the other two numbers.
2.) One leg of a right triangle has a length of 20. If the hypotenuse and leg are consecutive odd integers, find the hypotenuse.

3.) Compute the two digit number that is 9 times the sum of it's digits.
4.) A boat can travel 8 miles per hour in still water. If it can travel 15 miles with the current in the same time it travels 9 miles AGAINST the current, compute the rate of the current in mph.

5.) There are 4 different science books, 3 different math, and 2 different history. How many ways can you arrange these books on a shelf such that the books with the same subject are next to each other?
6.) ABCD is a rectangle and P is a point inside the rectangle. If PA= 3. PB=4.
PC=5. Find PD.

[hide="Solution to 1"]The five numbers sum to $ 5/3$. The three numbers sum to $ 3/5$. The two other numbers sum to $ 5/3 \minus{} 3/5 \equal{} 16/15$, so their average is $ \boxed{8/15}$.[/hide]

[hide="Solution to 2"]$ x^2 \minus{} (x\minus{}2)^2 \equal{} 4x\minus{}4 \equal{} 400$, so $ x \equal{} \boxed{101}$.[/hide]

[hide="Solution to 3"]The number must be divisible by $ 9$. $ \boxed{81}$.[/hide]

[hide="Solution to 5"]$ 3! \times 4! \times 3! \times 2! \equal{} \boxed{1728}$ ways.[/hide]

4)
$ v_b\equal{}8$

$ \frac{15}{v_b\plus{}v_c}\equal{}\frac{v_b\minus{}v_c}{9}$

$ 9(v_b\plus{}v_c)\equal{}15(v_b\minus{}v_c)$

$ 9v_b\plus{}9v_c\equal{}15v_b\minus{}15v_c$

$ 24v_c\equal{}6v_b$

$ 4v_c\equal{}v_b$

$ 4v_c\equal{}8$

$ v_c\equal{}2$

For #6, use the "British Flag Theorem," if that's what is called :P

[hide="6"]
Pretty direct application of the British Flag Theorem

$ 3^2\plus{}5^2\equal{}4^2\plus{}PD^2\implies PD \equal{} \boxed{3\sqrt{2}}$.
[/hide]

Hey grnturtle, can you explain what the British flag theorem is?
Also, if possible can someone give an answer with a diagram if thats possible?

All the answers are right though.

[url=http://en.wikipedia.org/wiki/British_flag_theorem]This[/url] can probably explain the British Flag Theorem better than I can. It's really just a convenient result from the Pythagorean theorem.


vector
linear algebra
matrix
abstract algebra
search
linear algebra unsolved

Let $ \mathcal{A}$ be a subalgebra of $ \mathcal{M}_3(\mathbb{C})$. What are the possible dimensions of $ \mathcal{A}$ (as a vector subspace) ? Give an example for each dimension.

OK Olorin.
$ n\equal{}8$ is impossible. Why ?

Because $ \mathbb{C}^3$ would then have to be a simple $ \mathcal{A}$-module,
and by  [url=http://en.wikipedia.org/wiki/Artin%E2%80%93Wedderburn_theorem]Artin–Wedderburn[/url] we would then have $ \mathcal{M}_3(\mathbb{C})$ as factor of $ \mathcal{A}/J(\mathcal{A})$.
So $ \dim_\mathbb{C}\mathcal{A}\ge 9$.

Hi Olorin;
I don't understand your first sentence:
"Because  $ \mathbb{C}^3$ would then have to be a simple $ \mathcal{A}$-module".
Is it because $ \mathcal{A}$ cannot be written in a form described in your first post ?
Now about your second sentence: does $ J(A)$ denote the radical of $ \mathcal{A}$ ?
Do you apply the Artin-Wedderburn theorem to $ \mathcal{A}/J(\mathcal{A})$ (which has a zero-radical) ?

[quote="loup blanc"]Because  $ \mathbb{C}^3$ would then have to be a simple $ \mathcal{A}$-module[/quote]

If $ \mathbb{C}u$ is a 1-dim. $ \mathcal{A}$-submodule of $ \mathbb{C}^3$, then by taking some basis $ \{u,v,w\}$ of $ \mathbb{C}^3$
we get $ \mathcal{A}$ conjugate to a subalgebra of $ \left(\begin{matrix}*&*&*\\
&*&*\\&*&*\end{matrix}\right)$ of dim. $ 7$.

If $ \mathbb{C}u+\mathbb{C}v$ is a 2-dim. $ \mathcal{A}$-submodule of $ \mathbb{C}^3$, then by taking some basis $ \{u,v,w\}$ of $ \mathbb{C}^3$
we have $ \mathcal{A}$ conjugate to a subalgebra of $ \left(\begin{matrix}*&*&*\\
*&*&*\\&&*\end{matrix}\right)$ of dim. $ 7$.

So $ \mathbb{C}^3$ doesn't have non-trivial $ \mathcal{A}$-submodules, i.e. it is simple.

[quote="loup blanc"]Now about your second sentence: does $ J(A)$ denote the radical of $ \mathcal{A}$ ?
Do you apply the Artin-Wedderburn theorem to $ \mathcal{A}/J(\mathcal{A})$ (which has a zero-radical) ?[/quote]
Yes and yes.

Pretty proof olorin (as usual).

I have an other proof.
Using Burnside theorem we can prove the following (cf also a Harazi's post few weeks ago):
$ max\{k;\exists\,\mathcal{A} \,proper\,subalgebra\, of \,\mathcal{M}_n(\mathbb{C})\,of\,dimension\,k\}\equal{}n^2\minus{}n\plus{}1$.
Proof: Let $ \mathcal{A}$ be a proper subalgebra of $ \mathcal{M}_n(\mathbb{C})$; $ \mathbb{C}$ is algebraically closed, then by Burnside, there exists a proper vector subspace $ V$ which is $ \mathcal{A}$-invariant. If $ V$ is of dimension $ k$ then (cf olorin last post) $ \mathcal{A}$ is a subalgebra of an algebra of dimension $ n^2\minus{}k(n\minus{}k)$. The max of this dimension is obtained for $ k\equal{}1$. Moreover let $ u$ be a non zero vector; the matrices  which admits $ u$ as an eigenvector is an algebra of dimension $ n^2\minus{}n\plus{}1$.     QED
If $ n\equal{}3$ then the maximum value is $ 7$.

Now I have a question (I don't know the answer):
If $ n\equal{}4$ then the maximum value is $ 13$; if $ k\leq13$ then there exists a subalgebra of dimension $ k$ (as for $ n\equal{}2,3$).
I don't think that this result can be generalized for $ n>4$. Does somebody know a counterexample ?

I believe this is actually the same proof, as Burnside probably uses Artin–Wedderburn the same way I did.

Now for $ n=4$ we have
$ 0,\mathbb{C}I_4,\mathbb{C}I_4+\left(\begin{matrix}*&&&\\&&&\\&&&\\&&&\end{matrix}\right),\mathbb{C}I_4+\left(\begin{matrix}*&&&\\&*&&\\&&&\\&&&\end{matrix}\right),\left(\begin{matrix}*&&&\\&*&&\\&&*&\\&&&*\end{matrix}\right),\left(\begin{matrix}*&*&&\\&*&&\\&&*&\\&&&*\end{matrix}\right)
,\left(\begin{matrix}*&*&&\\&*&&\\&&*&*\\&&&*\end{matrix}\right),$
$ \left(\begin{matrix}*&*&*&\\&*&*&\\&&*&\\&&&*\end{matrix}\right),\left(\begin{matrix}*&*&*&\\ *&*&*&\\&&*&\\&&&*\end{matrix}\right),\left(\begin{matrix}*&*&&*\\ *&*&&*\\&&*&*\\&&&*\end{matrix}\right),\left(\begin{matrix}*&*&*&*\\ &*&*&*\\&&*&*\\&&&*\end{matrix}\right),\left(\begin{matrix}*&*&*&*\\ *&*&*&*\\&&*&*\\&&&*\end{matrix}\right),\left(\begin{matrix}*&*&*&*\\ *&*&*&*\\&&*&*\\&&*&*\end{matrix}\right)$
and $ \left(\begin{matrix}*&*&*&*\\ *&*&*&*\\ *&*&*&*\\&&&*\end{matrix}\right)$ all dimensions $ \le 13$.

And for $ n\ge 5$ there exists no subalgebra $ \mathcal{A}$ of $ \mathcal{M}_n(\mathbb{C})$ with $ n^2-2n+4<\dim_\mathbb{C}\mathcal{A}<n^2-n+1$.
By Burnside we find some non-trivial $ \mathcal{A}$-submodule $ U$ of $ \mathbb{C}^n$, and if $ 2\le k=\dim_\mathbb{C}U\le n-2$,
then by taking some basis $ v_1,\ldots,v_n$ of $ \mathbb{C}^n$ with $ v_1,\ldots,v_k\in U$,
we get $ \mathcal{A}$ conjugate to a subalgebra of $ \left(\begin{matrix}\mathcal{M}_k(\mathbb{C})&*\\
&\mathcal{M}_{n-k}(\mathbb{C})\end{matrix}\right)$
of dimension $ n^2-k(n-k)\le n^2-2(n-2)=n^2-2n+4$. Contradiction!

Now take a simple $ \mathcal{A}$-submodule $ U$ of $ \mathbb{C}^n$, so $ \dim_\mathbb{C}U=1$ or $ n-1$.
If $ W: =\mathbb{C}^n/U$ is not a simple $ \mathcal{A}$-module with non-trivial submodule $ V/U$,
then the tower $ 0\subsetneq U\subsetneq V\subsetneq\mathbb{C}^n$ gives $ \dim_\mathbb{C}U=1,\dim_\mathbb{C}V=n-1$.
Then by taking some basis $ v_1,\ldots,v_n$ of $ \mathbb{C}^n$ with $ v_1\in U,v_2\ldots,v_{n-1}\in V$,
we get $ \mathcal{A}$ conjugate to a subalgebra of $ \left(\begin{matrix}\mathbb{C}&*&*\\
&\mathcal{M}_{n-2}(\mathbb{C})&*\\
&&\mathbb{C}\end{matrix}\right)$
of dimension $ n^2-(n-1)-(n-2)= n^2-2n+3$. Contradiction!

So $ U,W$ are simple $ \mathcal{A}$-modules of both dimensions $ 1,n-1$.
Again by Artin–Wedderburn we get different factors $ \mathbb{C}$ and $ \mathcal{M}_{n-1}(\mathbb{C})$ of $ \mathcal{A}/J(\mathcal{A})$ corresponding to $ U,W$.
So we get some ideal $ J$ of $ \mathcal{A}$, which annihilates both $ U,W$, s.t. $ \mathcal{A}/J\cong \mathbb{C}\times\mathcal{M}_{n-1}(\mathbb{C})$
of dimension $ 1+(n-1)^2=n^2-2n+2<\dim_\mathbb{C}\mathcal{A}$.

So $ J\not=0$, and $ JU=0$ and $ JW=0$ gives $ J\mathbb{C}^n\subseteq U$ and $ J^2\mathbb{C}^n=0$, so $ J^2=0$.
We also get the $ \mathbb{C}$-linear map $ \Phi: J\to \text{Hom}_\mathbb{C}(W,U),A\mapsto {\mathbb{C}^n/U\to U\atop v+U\mapsto Av}$
and for $ A\in\ker \Phi$ we have $ Av=0$ for all $ v\in \mathbb{C}^n$, so $ A=0$.
This gives $ \Phi$ injective.

If $ \Phi$ is also surjective, then $ \dim_\mathbb{C}J=\dim_\mathbb{C}\text{Hom}_\mathbb{C}(W,U)=\dim_\mathbb{C}W\cdot\dim_\mathbb{C}U=n-1$,
which gives the contradiction $ \dim_\mathbb{C}\mathcal{A}=\dim_\mathbb{C}\mathcal{A}/J+\dim_\mathbb{C}J=n^2-n+1$.

Assume therefore we have $ f\in \text{Hom}_\mathbb{C}(W,U)\setminus \Phi(J)$.
Also take $ 0\not= B\in J$ and set $ g=\Phi(B)\not=0$.

Assume first $ \dim_\mathbb{C}W=1,\dim_\mathbb{C}U=n-1$.
Then $ W=\mathbb{C}w$ for some $ 0\not=w\in W$ and set $ u_1: =f(w),v_1: =g(w)\in U\setminus\{0\}$.
Then take two bases $ \{u_1,\ldots,u_{n-1}\}, \{v_1,\ldots,v_{n-1}\}$ of $ U$
and $ h\in \text{End}_\mathbb{C}(U)$ with $ h(v_i)=u_i$ for $ i=1,\ldots,n-1$.
But by Artin–Wedderburn $ \text{End}_\mathbb{C}(U)\cong \mathcal{M}_{n-1}(\mathbb{C})$ is the factor of $ \mathcal{A}/J(\mathcal{A})$ corresponding to $ U$,
so we find some $ A\in \mathcal{A}$ with $ h(u)=Au$ for all $ u\in U$.
$ J$ is ideal of $ \mathcal{A}$, so we get $ AB\in J$ with $ \Phi(AB)=f$. Contradiction!

Finally assume $ \dim_\mathbb{C}W=n-1,\dim_\mathbb{C}U=1$.
Then $ U=\mathbb{C}u$ for some $ 0\not=u\in U$ and we find two bases $ \{v_1,\ldots,v_{n-1}\}, \{w_1,\ldots,w_{n-1}\}$
of $ W$ with $ f(v_1)=g(w_1)=u, f(v_i)=g(w_i)=0$ for $ i=2,\ldots,n-1$.
Then take $ h\in \text{End}_\mathbb{C}(W)$ with $ h(v_i)=w_i$ for $ i=1,\ldots,n-1$ and again
by Artin–Wedderburn we find $ A\in \mathcal{A}$ with $ h(w)=Aw$ for all $ w\in W$.
But this gives $ BA\in J$ with $ \Phi(BA)=f$. Contradiction!

Now that's an ugly proof.  :blush:

Hi Olorin,
bravissimo for this hard proof.
Remark: there exists a simple proof (less than 1 page) of the Burnside's theorem (without Artin–Wedderburn's theorem)
You can obtain this proof by this way: Search in google:
"burnside theorem halperin rosenthal"


real analysis
MIT
college
trigonometry
function
real analysis unsolved

Im learning some more calculus/analysis on my own using rudin and I just covered some stuff on bounded variation. i have been attempting some homework sets i find on websites (mit ocw for example). this one stumped me: using only the definition of bounded variation, show that $x^2\sin 1/x^2$ is not of bounded variation, but $x^2 \sin 1/x$ is. this is easy using the lipschitz condition, and is how I approached at first.

That had better be local at the origin: $x^2\sin\frac1x$ doesn't have bounded variation on the whole line.

define each to be zero at x=0, and consider only on [-1,1].

i believe I have proven that $x^2 \sin 1/x^2$ is not of bounded variation by showing a sequence of partitions which produces larger and larger variations. but showing that the other is of bounded variation seems more difficult.

$f(x)=x^2\sin\left(\frac1x\right)$ (and $f(0)=0.$)

$f$ is differentiable on $[-1,1]$. Hence by the Mean Value Theorem, for any $a$ and $b$ in that interval, $f(b)-f(a)=f'(c)(b-a)$ for some $c$ between $a$ and $b$.

But (for $x\ne0$), $f'(x)=2x\sin\left(\frac1x\right)-\cos\left(\frac1x\right).$ For $|x|\le1,$ we have $|f'(x)|\le 3.$ (Cheap estimate, not trying for anything optimal.)

Hence $|f(b)-f(a)|\le3|b-a|.$ $f$ satisfies a Lipschitz condition. You can show that over a bouned interval, any function satisfying a Lipschitz condition must be of bounded variation.


algebra
polynomial
function
superior algebra
superior algebra unsolved

This is an exercise 13.6.7 of Dummit & Foote.

Prove that

$ \Phi_m(x)\equal{}\prod_{d|n}(x^d \minus{} 1 )^{\mu(\frac{m}{d})}$, where $ \Phi_m(x)$ is m-th cyclotomic polynomial and $ \mu$ is a Mobius $ \mu$-function.

I've learnt the definition of cyclotomic polynomial and Mobius function, but I don't have any clue how to start this problem.

Any hint?

Thanks.

Well, you know that $ x^m \minus{} 1 \equal{} \prod_{d|m} \Phi_d (x)$.
Then try Mobius inversion.

Edit: oops I didn't see you include mobius inversion formula in your title. Maybe then you want to try taking log on both sides. Everything should then be clear.

Just follow [b]Soarer[/b].In fact,I think it is just Mobius Inversion Formula changes to the mutiply case.What I do following is to write it explicitly.

Bacause:
$ x^{d}\minus{}1 \equal{}\prod_{k|d}\Phi_{k}(x)$

Thus
\[ \prod_{d|m}(x^d \minus{} 1 )^{\mu(\frac {m}{d})}\equal{}\prod_{d|m}\prod_{k|d}\Phi_{k}^{\mu(\frac {m}{d})}(x)\equal{}\prod_{k|m}\prod_{k|d|m}\Phi_{k}^{\mu(\frac {m}{d})}(x)\equal{}\prod_{k|m}\prod_{i|\frac{m}{k}}\Phi_{k}^{\mu(i)}(x)\equal{}\prod_{k|m}\Phi_{k}^{\sum_{i|\frac{m}{k}}\mu(i)}(x)\equal{}\Phi_m(x)\]


Outline a method for make each of the following conversions, using the starting molecule as [i]the only organic material allowed[/i]:

1) 1-Bromopropane to 2-bromopropane.

2) Ethanol to propan-1-ol.

3) Propan-1-ol to n-hexane.

1.a.$ \text{EtOK \minus{} Ethanol}$
    b.$ \text{HBr}$

2.a.$ \text{TsCl}$
   b.$ \text{HBr}$
   c.$ \text{KCN}$
   d.$ \text{HCl,SnCl}_2$,$ \text{H}_2 \text{O}$
   e.$ \text{LiALH}_4$

3. a.$ \text{TsCl}$
   b.$ \text{HBr}$
   c. $ \text{Na}$

[quote="pardesi"]1.a.$ \text{EtOK \minus{} Ethanol}$[/quote]

What is it?

$ \text{Et}$ is the short form for ethyl group

What is the use (learning as I go along).

saves space time and more imp is used frequently
u have
$ \begin{array}{c} \text{Me \equal{} Methyl} \\
\text{Et \equal{} Ethyl} \\
\text{Ph \equal{} Phenyl} \\
\text{Ts \equal{} Tosyl} \end{array}$

Some other short forms:

Substituents: n-Pr, i-Pr, n-Bu, i-Bu, s-Bu, t-Bu, Ac, Bz, Bn, Ms, Mes, BOC, Tf, TMS, TBDMS, Tol, Tr.

Reagents: AIBN, 9-BBN, DBN, DBU, DEAD, DIBALH, LDA, LTA, m-CPBA, NBS, PCC.

Solvents: DME, HMPA, HMPT, TFA, THF, DMF, DMSO.

What I actually meant was how does it help in the synthesis.

It is a strong base, but also a strong nucleophile. That's why the proposal of [b]Pardesi[/b] for the first conversion is actually incorrect.

Even so your proposal is not reasonable. The problem is not in the temperature.

Because your purpose is to trigger an elimination on a primary alkyl halide. Ethoxide ion is a strong base but also a strong nucleophile, and because its nucleophilic/basic site is not crowded, substitution is a much more easy path than elimination. To force elimination reaction on a primary substrate, one should use a strong and crowded base, such as t-BuOK.


inequalities
trigonometry
geometry proposed
geometry

[quote="Virgil Nicula"]Let $ \triangle ABC$ . Denote the length $ m_a$ of $ A$-median and the length $ l_a$ of $ A$-bisector of $ \widehat {BAC}$ a.s.o.

Prove that there is the inequality $ \boxed {\ 4\cdot\left(\frac {m_a}{l_a} \plus{} \frac {m_b}{l_b} \plus{} \frac {m_c}{l_c}\right)\ \ge\ 3 \plus{} (a \plus{} b \plus{} c)\cdot\left(\frac 1a \plus{} \frac 1b \plus{} \frac 1c\right)\ }$ .[/quote]

$ m_a \ge \frac{1}{2}(b\plus{}c)\cos(\frac{\alpha}{2})$ and $ l_a \equal{} \frac{2bc\cos(\frac{\alpha}{2})}{b\plus{}c} \Rightarrow \frac{m_a}{l_a} \ge \frac{(b\plus{}c)^2}{4bc}$

$ LHS \ge 6 \plus{} \frac{a}{b} \plus{} \frac{b}{a}\plus{}\frac{b}{c}\plus{}\frac{c}{b}\plus{}\frac{c}{a}\plus{}\frac{a}{c}\equal{}RHS$


geometry
3D geometry
number theory unsolved
number theory

find all natural n such that:
$ n^2\plus{}n\plus{}1$ and $ n^2\minus{}n\plus{}1$ are cube.

Let $ n^2\minus{}n\plus{}1\equal{}x^3$, then $ (n^2\plus{}n\plus{}1)\minus{}(n^2\minus{}n\plus{}1)\equal{}2n \ge (x\plus{}2)^3\minus{}x^3>6x^2$(because both $ x^3$ and $ n^2\plus{}n\plus{}1$ are odd) $ \Rightarrow n^3>27(x^3)^2$
$ \Rightarrow n^3\plus{}1>27(n^2\minus{}n\plus{}1)^2 \Rightarrow n\plus{}1>27(n^2\minus{}n\plus{}1)$, which is impossible for all $ n$. So no solution for it.


How do we find the sum of all positive divisors of an integer.

[hide]Positive divisors of an integer are of the form $ 2^i3^j5^k\cdots$, depending on the prime factors of the integer. All possible combinations of these can be written as $ \left(2^0 \plus{} 2^1 \plus{} \ldots \plus{} 2^i\right)\left(3^0 \plus{} 3^1 \plus{} \ldots \plus{} 3^j\right)\cdots$, again depending on the integer's prime factors.

1. Determine the prime factorization, for example: $ 468 \equal{} 2^2\cdot3^2\cdot13$.

2. Write out the formula: $ \left(2^0 \plus{} 2^1 \plus{} 2^2\right)\left(3^0 \plus{} 3^1 \plus{} 3^2\right)\left(13^0 \plus{} 13^1\right)$.

3. Expand. (I'll leave this part to you.) :D[/hide]

[quote="i_like_pie"]
3. Expand. (I'll leave this part to you.) :D[/quote]
Just in case you don't know how to expand, he is telling you to use distributive property.
ex. (1,2)(1,3)=1,2,3,6.


function
real analysis
real analysis unsolved

[color=darkblue]Ascertain the continue function $f: R\rightarrow R$ for which $(\forall)x\in R\ ,\ f(f(x))-3f(x)+2x=0\ .$
[hide="Here is some solutions of the given functional equation."]$f_{m,n}(x)=\{\begin{array}{ccc}2x-m & , & x<m\\\ x & , & m\le x\le n\\ 2x-n & , & n<x\end{array}$, where $\{m,n\}\subset \overline R$ and $m\le n\ .$ These are only solutions ?[/color][/hide]

Obviously $ f$ is injective. Then continuity implies strict monotonicity.We see that $ f(x)\equal{}x$ is one solution. Now we assume there exists $ f_0\equal{}f(x_0)\neq x_0$.
$ f(f_0)\equal{}3f_0\minus{}2x_0$ and $ (f(f_0)\minus{}f_0))*(f_0\minus{}x_0)>0$ implies $ f$ has to be strictly increasing.

Note that if there exists root to $ f(x)\equal{}b$, $ f(f(x))\minus{}3f(x)\plus{}2x\equal{}0$ implies $ x\equal{}\frac{3b\minus{}f(b)}2$.

To show $ f$ is surjective, it suffices to disprove $ \inf_xf(x)>\minus{}\infty$ and $ \sup_xf(x)<\infty$.
Assume $ M\equal{}\inf_xf(x)>\minus{}\infty$. Then there exists series $ y_n$ in the range of $ f$ which decreasingly converges to $ M$.
Then we can find $ x_n\equal{}\frac{3y_n\minus{}f(y_n)}2$ such that $ f(x_n)\equal{}y_n$. Hence, the continuity implies $ x\equal{}\frac{3M\minus{}f(M)}2$ solves $ f(x)\equal{}M$. In other words, $ M$ is in the range of $ f$. Then we can find $ x\neq M$ such that $ f(x)<M$. Contradiction! Similarly we can disprove $ \sup_xf(x)<\infty$.

Now we fix pair $ (x_0,f_0)$ with $ f_0\equal{}f(x_0)\neq x_0$. We can define a two-direction series $ x_n$ by
$ x_{n\plus{}1}\equal{}f(x_n)\equal{}3x_n\minus{}2x_{n\minus{}1}$ and $ x_{\minus{}n}\equal{}f^{\minus{}1}(x_{\minus{}n\plus{}1})\equal{}\frac{3x_{\minus{}n\plus{}1}\minus{}x_{\minus{}n\plus{}2}}2$.
It is easy to see that for any integer $ n$,
$ x_n\equal{}(f_0\minus{}x_0)2^n\plus{}2x_0\minus{}f_0$.
Since $ f$ is continuous, we can find $ 0<\epsilon<\frac{|f_0\minus{}x_0|}2$ such that, for any $ x_1\in(x_0\minus{}\epsilon,x_0\plus{}\epsilon)$,
$ (f(x_1)\minus{}x_1)(f_0\minus{}x_0)>0$.
Now we pick another pair $ (x_1,f_1)$ with $ f_1\equal{}f(x_1)$ and $ x_1\in(x_0\minus{}\epsilon,x_0\plus{}\epsilon)$.
Similarly, we can define another series $ y_n\equal{}(f_1\minus{}x_1)2^n\plus{}2x_1\minus{}f_1$. It is easy to prove the two series have no intersection.
Now monotonicity implies, for integer $ n$ and $ m\equal{}n\plus{}k$,
$ (x_n\minus{}y_m)(x_{n\plus{}1}\minus{}y_{m\plus{}1})>0$, or
(1) $ ((a\minus{}a'2^k)2^n\plus{}b\minus{}b')((a\minus{}a'2^k)2^{n\plus{}1}\plus{}b\minus{}b')>0$,
where $ a\equal{}f_0\minus{}x_0$, $ a'\equal{}f_1\minus{}x_1$, $ b\equal{}2x_0\minus{}f_0$ and $ b'\equal{}2x_1\minus{}f_1$.
Suppose $ b\neq b'$ and let $ z\equal{}\frac{a\minus{}a'2^k}{b\minus{}b'}$, (1) implies
$ (z\plus{}2^{\minus{}n})(z\plus{}2^{\minus{}n\minus{}1})>0$.
If $ z<0$, we can always find $ n$ such that $ \minus{}2^{\minus{}n}<z<\minus{}2^{\minus{}n\minus{}1}$ and (1) will not hold then.
To sum up, we have, for any integer $ k$
$ (a\minus{}a'2^k)(b\minus{}b')\geq0$, or
(2) $ (f_0\minus{}x_0\minus{}(f_1\minus{}x_1)2^k)(2x_0\minus{}f_0\minus{}2x_1\plus{}f_1)\geq0$.
Since $ k$ can be arbitrarily large or small, which implies
(3) $ (f_0\minus{}x_0)(2x_0\minus{}f_0\minus{}2x_1\plus{}f_1)\geq0$.
(4) $ (f_1\minus{}x_1)(2x_0\minus{}f_0\minus{}2x_1\plus{}f_1)\leq0$.
Since $ (f_0\minus{}x_0)(f_1\minus{}x_1)>0$, we have
(5) $ 2x_0\minus{}f_0\minus{}2x_1\plus{}f_1\equal{}0$.
Since $ x_1$ is arbitrary in $ (x_0\minus{}\epsilon,x_0\plus{}\epsilon)$, we have $ f'(x_0)\equal{}2$.

Now we conclude either $ f(x)\equal{}x$ or $ f'(x)\equal{}2$. It is easy to see the only possible solution is $ f_{m,n}(x)$.

No matter if this solution is correct, I will be more than happy to get some comments.


Solve inequality

$ (x\plus{}a)^{\log_a (x\plus{}a)} >a^3  ,a\in \mathbb{R}$

Take logs of both sides to the base a, to get

$ (\log_a (x\plus{}a))^{2} \ge 3$

$ x\plus{}a \le a^{\minus{}\sqrt{3}}$ or $ x\plus{}a \ge a^{\sqrt{3}}$

$ x\le a^{\minus{}\sqrt{3}} \minus{} a$ or $ x \ge a^{\sqrt{3}} \minus{} a$

[quote="AndrewTom"]Take logs of both sides to the base a, to get

$ (\log_a (x \plus{} a))^{2} \ge 3$

$ x \plus{} a \le a^{ \minus{} \sqrt {3}}$ or $ x \plus{} a \ge a^{\sqrt {3}}$

$ x\le a^{ \minus{} \sqrt {3}} \minus{} a$ or $ x \ge a^{\sqrt {3}} \minus{} a$[/quote]

Incomplete. You must discuss $ 0<a<1$ and $ a>1$

Thanks, Farenhajt. Was rushing, as I am now.


calculus
integration
function
trigonometry
LaTeX
calculus computations

Easy function: compute $\int \cos { \sqrt x } \; dx$

edit: sorry for posting it in the wrong section

There is a [url=http://www.mathlinks.ro/Forum/index.php?f=296]Calculus Computations and Tuturials[/url] forum on this board. Use it to ask (easy) questions about calculus.

Moved there.

Try a "backwards" substitution. You want to use $w=\sqrt{x}$ but you don't see the right ingredients for $dw.$ So write it backwards:

$x=w^2$
$dx=2w\,dw$

The primitive becomes $\int2w\cos w\,dw.$

Now what? (And I used $w$ rather than $u$ to leave $u$ available for another use.)

thanks for the hint, i solved it :)

Primitive of:
2w.cos(w)

becomes:
2w.sin(w)+2.cos(w)

Replace w with sqrt(x) and theres my answer

Sorry but im not very handy with latex yet

This is the solution in LATEX[hide]
$\int 2\,w \cos{w}\,dw=2w\sin{w}- 2\,\int \sin{w}\, dw$
$= 2\,w\,\sin{w}+2\,\cos{w} + \, C$ so
$\int \cos{\sqrt{x}}\;dx= 2(\sqrt{x}\;\sin{\sqrt{x}}+ \cos{\sqrt{x}}) + \;C$
[/hide]

[quote="nemesis"]${2(\sqrt{x}-1)} e^{\sqrt{x}}$[/quote]

 :)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ acts transitively on a finite set $S$ and let $H$ be a normal subgroup of $G$. Prove that if we consider $H$ acts on $S$ then all orbits of $H$ on $S$ are all of the same cardinality.

[quote="grobber"]So, in particular, all the orbits under the action of $G$ have the same number of elements (since $G$ is normal in itself)? :? This can't be right.. Maybe by "orbit" you mean something other than the orbit of an element of $S$ with respect to the group acting on $S$?

Take, for instance, $G=\mathbb Z_2,\ S=\{1,2,3\}$, and let the non-trivial element of $G$ send $1$ to $2,\ 2$ to $1$, and $3$ to $3$. This action has an orbit of size $2$ and an orbit of size $1$. :?[/quote]

Sorry, $G$ acts on $S$ transitively. I will edit the previous post.

Given any action of $G$ on a set $S$ and a normal subgroup $H$ of $G$, we can define an action of $G/H$ on the set of orbits generated by the action of $H$ on $S$ in an obvious way: for $\bar x\in G/H$ ($\bar x$ is the image of $x$ through the canonical projection $G\to G/H$) and $s\in S$, the orbit of $s$ (under the action of $H$) is taken by $\bar x$ to the orbit (again, under the action of $H$) where $s$ is taken by $x$. It's easy to check that this is indeed an action. It's also easy to see that if $S$ is finite, as in our case, every two orbits under $H$ which belong to the same orbit under $G/H$ have the same cardinal, since an element of $G$ which maps them onto one another acts as a bijection.

In our case, the action defined as above is also transitive, so every two orbits under $H$ must have the same cardinal, according to the observations above.

Thanks a lot grobber, your method much more simpler than mine. Do you know any problem which is an application of the principle you have mentioned above??

Not really. I didn't even explicitly know that "principle". It just sort of "fell into place" when I saw the problem :).


[quote="nhat"]$a,b,c >0$ satisfy $ab+bc+ca=3$.Prove that:
                    $a^2+b^2+c^2+abc \geq a+b+c+1$  :)  :)[/quote]
http://www.mathlinks.ro/Forum/topic-19666.html. See Darij Grinberg's mention about the inequality:
$a^2+b^2+c^2+2abc+3 \geq (1+a)(1+b)(1+c)$ . Nhat, are you attempting to prove this inequality? :)

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \minus{} (a \plus{} b \plus{} c \plus{} 1) \minus{} (bc \plus{} ca \plus{} ab \minus{} 3)$

$ \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \plus{}3\minus{}(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)$

$ \equal{}\sum{\frac{4(a\minus{}1)^2(a\plus{}2bc)\plus{}3a(b\plus{}c\minus{}2)^2\plus{}a(b\plus{}c\minus{}2a)^2}{8(a\plus{}b\plus{}c)}}\geq 0.$

See also : http://www.mathlinks.ro/viewtopic.php?t=40979


Post what you know about them, or state some questions!
Any information.   :D

Question #1:
What is a toroidal chessboard?

I believe it's...and don't laugh at me for this...

But I think an example is like this

board #1

board #2

board #3

Pretend there in mid-air above each other.  If a queen at the left hand corner on board 1, and a piece on the right hand corner on board 2, the queen moves to the right hand corner on board 1 to knock out the piece on board 2 (if theres nothing in the way)

So don't laugh at me if i'm wrong...

You only need one board. The squares just "wrap around" for example c1 is considered adjacent to c8 and a2 adjacent to h2. I don't know what the initial setup would be, though, as the kings could obviously not start adjacent to each other.

heh, perhaps i should've explained the basics  :lol:

like jmadsen said, you only need one board (la mafioso, i think you're thinking of 3-D chess) and the squares "wrap around." In this way, the four corners are adjacent.
I'm not too concerned with the initial setup, just the general idea of this board format, but the setup would include an extra row of pawns below the regular pieces.

[quote="Hellstar"][quote="la mafioso"]

Pretend there in mid-air above each other.  If a queen at the left hand corner on board 1, and a piece on the right hand corner on board 2, the queen moves to the right hand corner on board 1 to knock out the piece on board 2 (if theres nothing in the way)
[/quote]
That's a neat idea actually 3-D chess.  Maybe they should make a new game similiar to chess where you have three normal chess boards stacked on top of each other, and bishops can move in space diagonals and stuff. That would be fun.  :D[/quote]

I actually thought there was 3-D chess, an 8x8x8 thing...
What about chess on a spherical surface?  That would be more realistic in this day and age in simulating warfare.

You can also go to
http://www.bonus.com/bonus/nav/scooter/scooterize.htmp?loc=http%3A%2F%2Fwww%2Ebonus%2Ecom%2Findex%2Ehome%2Ehtml%3Fscooterize%3D1&useragent=Mozilla%2F4%2E0+%28compatible%3B+MSIE+6%2E0%3B+Windows+NT+5%2E1%3B+SV1%3B+%2ENET+CLR+1%2E1%2E4322%29&refv=1&referrer=&entry=bonus&SERVER=www%2Ebonus%2Ecom

kind of a long link...
essentially, go to bonus.com, and in the top right, you can search. search for chess, and the first hit should be chess variant. there are so many variants here it's crazy, and the computer is decent.


induction
modular arithmetic
algebra proposed
algebra

Prove that for each $n\geq 2$, there is a set $S$ of $n$ integers such that $(a-b)^2$ divides $ab$ for every distinct $a,b\in S$.

[hide="solution"]
We will prove by induction on $n$, that we can find such a set $S_n$ with non-negative elements. For $n=2$ take $S_n=\{0,1\}$
Now suppose that for some $n\geq 2$, the desired set $S_n$ of $n$ non-negative integers exists. Let $L$ be the least sommon multiple of those numbers $(a-b)^2$ and $ab$ that are non-zero, with $(a,b)$ ranging over pairs of distince elements from $S_n$.  Now define:
$S_{n+1}=\{L+a: a\in S_n\}\cup \{0\}$
Then $S_{n+1}$ consists of $n+1$ non-negative integers, since $L>0$.  If $i,j\in S_{n+1}$ and either $i$ or $j$ is zero, then $(i-j)^2$ divids $ij$.  If $L+a, L+b\in S_{n+1}$ with $a,b$, distinct elements of $S_n$, then
$(L+a)(L+b)\equiv ab\equiv 0 \pmod{(a-b)^2}$
so $[(L+a)-(L+b)]^2$ divides $(L+a)(L+b)$ thus completeing the induction step
$\mathbb{QED}$[/hide]

[quote="MithsApprentice"]Prove that for each $ n\geq 2$, there is a set $ S$ of $ n$ integers such that $ (a \minus{} b)^2$ divides $ ab$ for every distinct $ a,b\in S$.[/quote]
I am not sure if the following is right:
[hide="Solution"]
We define $ a!^n$ to be $ (a!)!^{n \minus{} 1}$.  For example, $ 3!^3 \equal{} (3!)!^2 \equal{} 6!^2 \equal{} 720!$.  Now, let $ g \equal{} \gcd (a,b)$ where $ a$ and $ b$ are part of $ S$, so $ a \equal{} a_1g$ and $ b \equal{} b_1g$.  We then have that $ \frac {ab}{(a \minus{} b)^2} \equal{} \frac {a_1b_1}{(a_1 \minus{} b_1)^2}$.  Yet, $ \gcd (a_1 \minus{} b_1, a_1) \equal{} 1$ and $ \gcd (a_1 \minus{} b_1, b_1) \equal{} 1$, so $ a_1 \minus{} b_1$ can only divide $ a_1b_1$ if $ a_1 \minus{} b_1 \equal{} 1$.  This means that we must have $ (a \minus{} b)|a, b$.  We now consider the numbers:\[ n!^{2n}, n!^{2n} \plus{} n!^{2n \minus{} 2}, n!^{2n} \plus{} n!^{2n \minus{} 2} \plus{} n!^{2n \minus{} 4}, \cdots , n!^{2n} \plus{} n!^{2n \minus{} 2} \plus{} n!^{2n \minus{} 4} \plus{} \cdots \plus{} n!^2\]We will show that these numbers work.  Define $ a_i$ to be $ n!^{2n} \plus{} n!^{2n \minus{} 2} \plus{} \cdots \plus{} n!^{2n \minus{} 2i \plus{} 2}$.  We notice that $ a_i \minus{} a_j$ for $ i > j$ is equal to \[ n!^{2n \minus{} 2i \plus{} 2} \plus{} n!^{2n \minus{} 2i \plus{} 4} \plus{} \cdots \plus{} n!^{2n \minus{} 2j} < n\cdot n!^{2n \minus{} 2j} < n!^{2n \minus{} 2j \plus{} 1}\]  Thus, $ (a_i \minus{} a_j)|(n!^{2n \minus{} 2j \plus{} 2})|a_j$ as desired.  This set works, so we are done.  [/hide]

$ S_2 \equal{} \{1,2\}$
suppose we have constructed set $ S_n \equal{} \{a_1,a_2,...,a_n\}$
let $ x \equal{} \prod_{i \equal{} 1}^n a_i$
construct $ S_{n \plus{} 1}$ as follows: $ S_{n \plus{} 1} \equal{} \{x, x \plus{} a_1,x \plus{} a_2,...,x \plus{} a_n\}$

[hide="Does this work?"]We use induction. The base case is trivial. Let the elements of our set be $ a_1, a_2, \ldots, a_n$. Note that if we scale all the elements by some amount, the condition is still satisfied. Hence, we may scale all the variables by $ m \equal{} \prod_{1 \leq i \leq n} a_i \prod_{1 \leq j < k \leq n} (a_i \minus{} a_j)^2$. It follows that $ m$ and all numbers of the form $ a_i$ and $ a_j \minus{} a_k$ for all $ i$, $ j$, $ k$, $ j \neq k$, between 1 and $ n$ inclusive have the same set of prime factors.

Let $ c \equal{} m \plus{} 1$. We claim that $ m$ and all numbers of the form $ c \minus{} a_i$, $ 1 \leq i \leq n$, are relatively prime. If $ p | m$ and $ p |c \minus{} a_i$ for some $ i$, then $ p | m \plus{} 1 \minus{} a_i$. Since $ m$ and $ a_i$ have the same set of prime factors, $ p | 1$, a contradiction.

If $ p | c \minus{} a_i, c \minus{} a_j$ for some distinct $ i$ and $ j$, $ p | a_i \minus{} a_j$ as well. But $ a_i \minus{} a_j$ and $ a_i$ have the same prime factors, so $ p | a_i$. Therefore, $ p | c \equal{} m \plus{} 1$. But $ p | m$ as $ m$ and $ a_i$ have the same set of prime factors, so $ p | 1$, again a contradiction.

Hence, by the Chinese remainder theorem, we can find some $ k$ such that $ k \equiv 0 \pmod m$, and $ k \equiv \minus{}a_i \pmod{(c \minus{} a_i)^2}$ for all integers $ i$ with $ 1 \leq i \leq n$.

Let $ b_i \equal{} a_i \plus{} k$, and let $ b_{n\plus{}1} \equal{} c \plus{} k$. For all $ i, j$ with $ 1 \leq i  < j \leq n$, $ (b_i \minus{} b_j)^2 \equal{} (a_i \minus{} a_j)^2$, and $ b_i b_j \equal{} (a_i \plus{} k)(a_j \plus{} k) \equal{} a_i a_j \plus{} k(a_i \plus{} a_j \plus{} k)$. Since $ (a_i \minus{} a_j)^2 | m | k$, and our inductive hypothesis tells us that $ (a_i \minus{} a_j)^2 | a_i a_j$, we have that $ (b_i \minus{} b_j)^2 | b_i b_j$.

It remains to show that $ (b_{n\plus{}1} \minus{} b_i)^2 | b_{n\plus{}1} b_i$. But $ k$ was constructed so that $ (b_{n\plus{}1} \minus{} b_i)^2 \equal{} (c \minus{} a_i)^2 | a_i \plus{} k \equal{} b_i | b_i b_{n\plus{}1}$, so our proof is complete. [/hide]

[hide=" Solution"]

We induct on $n$. For the base case $n=2,$ just pick two numbers which differ by 1.

Let $a_1, a_2, \cdots , a_j$ be $j$ numbers which satisfy $(a_m-a_n)^2 \mid a_ma_n \quad \forall \ m \neq n.$ Let $M = \displaystyle \prod_{m \neq n} (a_m - a_n)^2 \cdot a_1^2 a_2^2 \cdots a_j^2.$ Notice that the set $\{M+a_1, M+a_2, \ldots, M+a_j\}$ satisfies the property too since for any $m \neq n,$
\[(a_m-a_n)^2= ((M+a_m) - (M+a_n))^2 \mid a_ma_n \\ \implies ((M+a_m) - (M+a_n))^2 \mid (M+a_m)(M+a_n).\]
We need to append another number to this set. Appending $M$ to this set works, since $((M+a_i)-M)^2 = a_i^2 \mid M(M+a_i) \quad \forall \ i.$ Hence, we have found a $j+1$ element set of integers with the required property. By induction, we are done. $\blacksquare$  [/hide]

EDIT: Oops, sorry to revive. Didn't notice. :oops:

Another approach:
Let $a_1=x$ and set $a_t=\frac{k^{t-1}}{(k-1)^{t-1}}x$ for all $t \ge 2$.
Choose $x$ divisible by $(k-1)^{n-1}$ and we get an $n$-term sequence.


I would like to be a moderator for the Graphical Design and Puzzles and Brainteasers forums. I believe that I will make a fairly capable mod, and I greatly contributed to both forums' popularity, with my Flash/Art Contest and Puzzle/Brainteaser Marathon. Also, both forums only have one moderator, and I feel that only one mod per forum is insufficient, even for small forums.

Thanks,
NoSoupForYou

[quote="Arne"]You should rather sent a PM...[/quote]
Oops. I didn't didn't know you have to send a PM. Sorry.

Yes, that would probably be the best way. Directing it to Valentin probably be the correct direction as he seems to be most visible in the organizational workings of the forum.


geometry
3D geometry
sphere
analytic geometry
function
geometric transformation
rotation

We move a solid sphere of radius $R$ with velocity $\vec{u}$ in a liquid of viscosity $\eta$. Find the retro-force of the liquid on the sphere.

stokes esperimented and found that F=c*n*d*v ,c-a constant depending on body's form ,n-viscosity and d-transversal dimension,v-velocity

for a sphere we d=2*R and c=3*pi

F=6*pi*n*R*u
is this the force you are talking about?

I think that you have to prove this :) I have no idea how to, though...

well the ball interacts with fluid molecules and changes their position but i can't see how viscosity is related to intermolecular forces.i think it can only be found the constant which depends on body's form,other variables are just experimental found

Stokes' Law can be derived from purely theoretical considerations.

A sample derivation can be found at:
http://aemes.mae.ufl.edu/~uhk/STOKESDRAG.pdf

That being said, this derivation is not accessible to most people.  Far more people know the quoted formula than the derivation, and I wouldn't bother trying to understand it until you are well into college.

Here is a nice solution:
In the coordinate system moving with the sphere the flow is stationary, and $div\vec{v}=0$ so we can $\vec{v}$ as $rot\vec{A}$. $\vec{A}$ is an axialvector which is the function of only $\vec{u}$ and $\vec{r}$, and because its linear in $\vec{u}$ its form should be as $\vec{A}=({\vec{u}}\times{\vec{r}})f(r)$.
Now making the rotation of the equation of motion we can eliminate the pressure:
$rot\bigtriangleup \vec{v}=0$
and:
$\bigtriangleup rot rot [({\vec{u}}\times{\vec{r}})f(r)]=0$
this is a fourth order differential equation for $f(r)$, the solution:
$\vec{v}=a\vec{u}+b\frac{r^{2}\vec{u}+(\vec{r}\vec{u})\vec{r}}{r^{3}}+c\frac{3(\vec{r}\vec{u})\vec{r}-r^{2}\vec{u}}{r^{5}}$
We can consider the liquid to have a big expansion so, we get the following conditions:
$a=-1$
the velocity disapeares on the sphere so:
$b=\frac{3R}{4}$ and $c=-\frac{R^{3}}{4}$
so the solution:
$\vec{v}=-\vec{u}+\frac{3R}{4}\frac{\vec{u}+(\vec{n}\vec{u})\vec{n}}{r}-\frac{R^{3}}{4}\frac{3(\vec{n}\vec{u})\vec{n}-\vec{u}}{r^{3}}$
where $\vec{n}=\frac{\vec{r}}{r}$
So we get:
$\bigtriangleup \vec{v}=\frac{3R}{2}\frac{\vec{u}r^{2}-3(\vec{u}\vec{r})\vec{r}}{r^{5}}\equiv \frac{3R}{2}grad \frac{\vec{u}\vec{r}}{r^{3}}$
integrating the $grad p =\eta \bigtriangleup \vec{v}$ equation we get the pressure:
$p=p_{0}+\frac{3\eta R}{2}\frac{\vec{u}\vec{r}}{r^{3}}$
The pressure tension component of the total retro force:
$\vec{F_{1}}=\oint \oint pd\vec{F}=p_{0}\oint \oint d\vec{F}+\frac{3\eta R}{2}\oint \oint \frac{\vec{u}\vec{r}}{r^{3}}d\vec{F}$
The first part of the integral is 0, and the second one using Gauss-Osztrogradszkij theorem ($d\vec{F}$ is a vector pointing to the center of the sphere):
$\vec{F_{1}}=\frac{3 \eta}{2R^{2}}\oint \oint (\vec{u}\vec{r})d\vec{F}=-\frac{3 \eta}{2R^{2}}\int \int \int grad(\vec{u}\vec{r}) dV=-2\pi R\eta \vec{u}$
The component of retro force from the shearing tension can be calculated similarly:
$\vec{F_{2}}=\eta \oint \oint (\vec{D}+\vec{\tilde{D}})d\vec{F}=-4\pi R\eta \vec{u}$
Where $\vec{D}\equiv \frac{d\vec{v}(\vec{r})}{d\vec{r}}$ is the differentiate tensor.
So the total retro force is:
$\vec{F}=\vec{F_{1}}+\vec{F_{2}}=-6\pi R\eta \vec{u}$
I hope everyone understands it cause my english is not the best.


Determine all possible values of $n$ such that $n!-1$ is a prime number.

Is it a joke?
http://primes.utm.edu/top20/page.php?id=30

Not intended to be one but turned out to be one....by the way if I give a bound on n like say n< 5000 is there a systemic procedure to determine the primes

Systematic procedure is simple:

for n=1 to 5000 do
  test if n!-1 is prime
endfor

If one do sieving in advance (to eliminate n's for which n!-1 has a small prime factor), it may speed up things a bit.