geometry
3D geometry
tetrahedron
ratio
pyramid
function
probability

Just as the coefficients of exponential binomials make up the rows of Pascals Triangle, so the coefficients of exponential trinomials make up the layers of values of a tetrahedron where each value is the sum of the 1, 2, or 3 values that connect with it from above.  This tetrahedron is very like a molecular model where the connecting rods are all of the same length.  Each of the top 3 sides of the tetrahedron is a Pascals Triangle - the rest of the values are inside the tetrahedron.

Unproven properties of this tetrahedron:

If n is a layer of the tetrahedron and m is a row in the Triangle, then the sum of n is 3n.  (The sum of each row m in the Triangle is 2m.)  In the Triangle, m (with second, third, fourth, etc. digits of values carried to values to the left) equals 11m.  (When m=4, the values are 1, 4, 6, 4, 1. 114=14641. When m=5, the values are 1, 5, 10, 10, 5, 1. 115=161051.) In the tetrahedron, patterns are found when such a power of 11 from m on Triangle face 1 is compared to lower rows that are the same distance away from the edge joining faces 2 and 3 as m is from the edge.

Please see top table on the attachment.

The following patterns disregard the first two columns:

1)  Each set (discounting those of coefficients) is the truncated version of the set to its top left in the table.

2)  Each set of coefficients is made up of the (m+1)th digits of all sets of coefficients, including itself.

3)  The first values of the sets of a column compose part of the set that starts the row from which the values in that column start.

4)  The first values in each row make up the power of 11 upon which that row is based.

5)  The first values of the sets found in diagonals pointing to the top left make up part of the beginnings of the sets the furthest up the diagonals.

There are also patterns in the sums of row m in Triangle face 1 and in the sums of the rows
both below m and the same distance away from the edge joining faces 2 and 3 as m is from
the edge (sorry to repeat the description, but it isnt clear what rows Im talking about otherwise).

Please see middle table on the attachment.

The sets in the third column of the middle table are the same as the sets in the top table.

You might also want to see the bottom table on the attachment.

Interesting...I knew there had to be someone out there...where did you find this?

I worked it out - though probably someone else has worked it out ages ago.  I wonder what the model of an exponential quadronomial is.  It might be in four dimensions.  If the fourth dimension really is time, than the model might be Tetrahedra in motion - interacting.  (The Tetrahedron includes Triangles.  The Triangle includes Pascal's Lines, to coin a phrase, lines of 1's where each "1" is the sum of another "1".  [Tetrahedron - sum of 3 values, Triangle - sum of 2 values, Line - sum of 1 value.])  The Tetrahedra could be growing as they are moving - maybe doing both in jerks.  (If it isn't jerks, then that means a constant ratio - and constants are tedious to work out!)  Some of the Tetrahedra might grow normally - at the base, but one non-Pascal tetrahedron might grow in all directions - somehow.  (The Tetrahedron has 3 normally growing Triangles, but another non-Pascal triangle molts, as it were, getting bigger in all directions.  The same thing is true of the Triangle and Lines.)  What do you think?

Hm, I just did a fairly in depth rport over Pascal's traingle, identifying patterns and what not, and then moved on to Pascal's Pyramid or Tetrahedron, whichever you prefer, worked stuff out for that one, but not as in depth. For factoring Pacals Pyramid i used (X + Y + Z)^N to check my work of making the layers of it. I drew out a couple of layers, Designating each corner a variable, and since the main goal was a trinomial expansion through Pascal's Pyramid i designated each corner a letter. To find the value of that spot it was simply the numerical value and each letter tot he power of the row minus the steps away from its desiganted corner. (keep in mind sx is steps from X, same with w y and z)(also C= Current Location value and N=row and V=Numerical value)) C=V(W^(N-sw)*X^(N-sx)*Y^(N-sy)*Z^(N-sz)) Im sure there is a more simple way tofigure it out but eh, im only in 10th grade. After that i wondered what would it be liek instead of using binomials or trinomails to use quad/tetra nomials. So i used (W + X + Y + Z)^N and the only type of model i could really come up with is that each layer is a pyramid or (still working on this one) each layer a square. with both, designate each corner a Letter/variable. the only problem i have with the square is the not all Variables are equal distance from each other (Z would be equally close to X and Y, but be further away from W ( Sqrt(2)N to be exact lol) and im not sure if this matters entirely or not. The nice thing about drawing it as a square though, is its so much easier then trying to draw a 3d pyramid with vlaues in it. (for this i just *unfolded* the pyramid visually, and let the sides that would connect still function as if connected)

Pascal's triangle (drawn as extended out form Y-axis)

1
1 1
1 2 1
1 3 3   1
1 4 6   4   1
1 5 10 10  5    1
1 6 15 20  15   6    1
1 7 21 35  35  21   7    1
1 8 28 56 70   56   28  8   1

and in probablity its just the 2^N as the denominater

Layer of Pascal's Pyramid (X + Y + Z)^N

               6th Row/Layer

                   Z^6
                     1
                   6  6
               15 30 15
           20   60 60   20
        15   60  90  60   [u]15[/u]
      6    30   60 60   30    6
    1   6    15   20   15   6    1
X^6                                      Y^6

So the underlined 15 would be 15(X^(6-6) * Y^(6-2) * Z^(6-4) or 15 Y^4 Z^2
And with probablity the total number of occurences in this layout is just 3^N

And finally for what i came up with for (W + X + Y + Z) ^N

The Pyramid style i Used...

            3rd layer

W^3        Z^3           W^3
   1   3   3   1   3   3   1
     [u]3[/u]   6   3   3   6   3
       3   3   6   3   3
X^3    1   3   3   1    Y^3
            3   6   3
              [u]3 [/u]  3
                1
             W^3

Its a bit harder to see here, it was easier on paper because i had lines drawn to show what connected and what didn't in finding how the pieces are connected so I can use
C=V(W^(R-sw) * X^(R-sx) * Y^(R-sy) * Z^(R-sz))
To get alues of each location.
If lines where drawn to whats connected it would be like a normal Layer of Pascals Pyramid but the 6's would all be connected to each other.
And keep in mind that if you where to add up the total of this..layer?.. you would get a different answer, the total events should just be 4^N if you added this up the wya it was you would get a larger answer due to the reapeated numbers on sides.
For example the Undelrined 3's actually are the same location

Now when i work it out as a square for this layer i come up with...

X^3               Y^3
      1   3   3   1
      3   6   6   3
      3   6   6   3
      1   3   3   1
Z^3               W^3

This is much more simple and easy on the eyes. The Only Problem is I'm not sure how well this would continue to work. Because using this pattern For Layer 1 and 2 i get...

1   1
1   1
 now on this i wouldn't mind some help.
W^2 + X^2 + Y^2 + Z^2 + 2wx + 2wy + 2wz + 2xy + 2xz + 2 yz is what the layers value is
so what i show is what i would come up with from looking at the privious cube. But its 6 2's not 4 2's and 1 4. and i cant help but thikn this is because they are not equal seperated liek they should be so there is no connection between the diagnals. Is this because its drawn 2D in a different shape? The only solution i can think of is split the 4 into 2's and put them on a Z axis. Make one 2 connect to 1 set of diagnals and the other 2 connects to the other set of diagnals.

1   2   1
2   4   2
1   2   1

so would this work better... (if it makes sense)
  ________
 /              2
1---2---1    |
|      // |     |
2    2   2    |
|  //     |     |
1---2---1----

but after seing these three rows and not have working (W  X  Y  Z)^N out further, I' wondeirng if this 2d Square set up only works easily on rows to an odd number power, or Since how far i worked it out wouldn't have any numbers contained inside the Pyrmaid, is any part greater then row 3 going to need to be split like this?

I'm not so sure im going to test this rihgt now until i find a much faster way to do (W + X + Y + Z)^4 I might make a program to work it out, if i do i will reply again with what i found. Any tips on anything ive done would be appreciated. Keep in mind i'm only in 10th grade so please dont be to harsh if i messed up somewhere or if this is simple to you.

Use the multinomial coefficient formula $ \binom{n}{n_1\,;n_2\,;\dots\,;n_k}$ to help you find the coefficient of $ x_1^{n_1}x_2^{n_2}\dots x_k^{n_k}$ and construct 'triangles or pyramids', where $ n \equal{} \sum_{j \equal{} 1}^kn_j$.

EDIT: I didn't check that this is HSB... ok so this is simply $ \binom{n}{n_1\,;n_2\,;\dots\,;n_k}\equal{}\frac{n!}{n_1!n_2!\dots n_k!}$

Thankyou so much for that. I appreciate your help greatly


abstract algebra
superior algebra
superior algebra unsolved

As an example of the tensor product of artinian algebras not being artinian, consider a division ring A with center K which is a field. Let B be a maximal subfield and A infinite dimensional over B. Show that the tensor product of A and B over K is not artinian.

Thank You in advance.

I have found 99% of the solution now. If anyone is interested, I can post it ( I would have to learn how to put math symbols on the page first ).


inequalities
trigonometry
inequalities unsolved

This one may be a problem of Russia 2005

This one looks easy but ,,,.Anyone has the solution can help me ?

[quote="bui_haiha2000"]Prove that:
      $sin$ $\sqrt{x}$ $<$ $\sqrt{sinx}$
                                                   with $0$ $<$ $x$ $<$ $\frac{pi}{2}$[/quote]
For $x\geq0$ $\sin x\leq x-\frac{x^{3}}{6}+\frac{x^{5}}{120}.$
Hence, $\sin\sqrt{x}\leq\sqrt x\left(1-\frac{x}{6}+\frac{x^{2}}{120}\right).$
And remain to prove that $\sqrt x\left(1-\frac{x}{6}+\frac{x^{2}}{120}\right)\leq\sqrt{\sin{x}}\Leftrightarrow$
$\Leftrightarrow\sin{x}\geq x\left(1-\frac{x}{6}+\frac{x^{2}}{120}\right)^{2}$ for $x\in(0,\frac{\pi}{2}).$
But $\sin x\geq x-\frac{x^{3}}{6}\geq x\left(1-\frac{x}{6}+\frac{x^{2}}{120}\right)^{2}$ for $x\in[0,\frac{\pi}{2}].$
Check it.:wink: :)

[quote="arqady"]
For $x\geq0$ $\sin x\leq x-\frac{x^{3}}{6}+\frac{x^{5}}{120}.$[/quote]

Can you explain why do you have this ?


hi,
let f be a function such that : $f(x)=\frac{x-1}{x-|x|+1}$

find : $lim_\frac{-1}{2}{f(x)}$

The limit is undefined.

$\lim_{x\rightarrow-\frac{1}{2}^{+}}\frac{x-1}{x-\left|x\right|+1}=-\infty$

$\lim_{x\rightarrow-\frac{1}{2}^{-}}\frac{x-1}{x-\left|x\right|+1}=+\infty$

[quote="toetoe"]thanks very much for you answer amcavoy.I will be very happy

if you explain more pleaze how you did find the answers .

thanks in advance.[/quote]
Well, the numerator of the fraction stays constant as $x$ approaches $-\frac{1}{2}$, and the denominator approaches $0$ as $x$ approaches $-\frac{1}{2}$. Also, as $x$ approaches from the left of $-\frac{1}{2}$, the denominator is positive, leading to a positive answer, and as $x$ approaches from the right of $-\frac{1}{2}$, the denominator is negative, leading to a negative answer.
Also, since this topic includes limits, it might not be in the right forum.  :wink:

denominator approaches 0 while numerator approaches a constant = gg


Denote that
P(x)=$ \prod_{1 \leq k \leq n \minus{} 1 ; gdc(k,n) \equal{} 1} (x \minus{} \epsilon_n^k)$

with $ \epsilon_n \equal{} cos \frac {2\pi}{n} \plus{} i.sin \frac {2\pi}{n}$

Prove that P(1)= p when n is a power of a prime p
         or P(1)=1 in otherwise

If $ n\equal{}p^m$, then $ P_n(x)\equal{}\frac{x^n\minus{}1}{x^{n/p}\minus{}1}\equal{}x^{(p\minus{}1)p^{m\minus{}1}}\plus{}x^{(p\minus{}2)p^{m\minus{}1}}\plus{}...\plus{}1$ and $ P_n(1)\equal{}p$.
Else $ P_n(x)\equal{}\prod_{d|n}(x^{n/d}\minus{}1)^{\mu (d)}$ and $ P_n(1)\equal{}\prod_{d|n}(\frac nd )^{\mu (d)}\equal{}1$.

[quote="Rust"]Else $ P_n(x) \equal{} \prod_{d|n}(x^{n/d} \minus{} 1)^{\mu (d)}$ and $ P_n(1) \equal{} \prod_{d|n}(\frac nd )^{\mu (d)} \equal{} 1$.[/quote]
i think this line is not true.
Because if $ P_n(x) \equal{} \prod_{d|n}(x^{n/d} \minus{} 1)^{\mu (d)}$ ,$ P_n(x)$
have one root is 1 and it 's impossible.
Can you explain more clearly ?

Let $ f_d(x)\equal{}\frac{x^{n/d}\minus{}1}{x\minus{}1}$, then $ P_n(x)\equal{}\prod_{d|n}f_d(x)^{\mu (d)}$.
Therefore $ P_n(1)\equal{}\prod_{d|n}(n/d)^{\mu (d)}\equal{}\prod_{d|n}d^{\minus{}\mu (d)}\equal{}1$ if n had at least two prime factors.

[quote="Rust"]Therefore $ P_n(1) \equal{} \prod_{d|n}(n/d)^{\mu (d)} \equal{} \prod_{d|n}d^{ \minus{} \mu (d)} \equal{} 1$ if n had at least two prime factors.[/quote]

as above line $ \frac{n}{d} >1$ because d|n then
$ \prod_{d|n}(n/d)^{\mu (d)} >1 and \prod_{d|n}d^{ \minus{} \mu (d)} < 1$
It 's nothing.
Can you post more carefully ?

$ \mu (d)\equal{}(\minus{}1)^k$ if $ d\equal{}P\minus{}1...p_k$ is produrt of k distinkt primes, and $ \mu (d)\equal{}0$ for other cases.
For example, when n had 2 prime factors $ n\equal{}p^mq^l$ we had
$ P_n(1)\equal{}\prod_{d|n}(n/d)^{\mu (d)}\equal{}n*(n/p)^{\minus{}1}(n/q)^{\minus{}1}(n/pq)^1\equal{}n^{1\minus{}1\minus{}1\plus{}1}p^{\minus{}1\plus{}1}q^{\minus{}1\plus{}1}\equiv 1$.

[quote="Rust"]$ \mu (d) \equal{} ( \minus{} 1)^k$ if $ d \equal{} P \minus{} 1...p_k$ is produrt of k distinkt primes, and $ \mu (d) \equal{} 0$ for other cases.
For example, when n had 2 prime factors $ n \equal{} p^mq^l$ we had
$ P_n(1) \equal{} \prod_{d|n}(n/d)^{\mu (d)} \equal{} n*(n/p)^{ \minus{} 1}(n/q)^{ \minus{} 1}(n/pq)^1 \equal{} n^{1 \minus{} 1 \minus{} 1 \plus{} 1}p^{ \minus{} 1 \plus{} 1}q^{ \minus{} 1 \plus{} 1}\equiv 1$.[/quote]

thank you so much.
Sorry about my silly


Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=144403 Locked!


Find all truples of  integer numbres $(x,y,z)$ such that
$x^3-4x^2-16x+60=y$
$y^3-4y^2-16y+60=z$
$z^3-4z^2-16z+60=x$


For the arithmetic sequence $ 1000, \sim 987, \sim 974, \sim 961, \ldots ,$ what is the least positive integer in the sequence?

$ 1000 \minus{} 13 \times \left\lfloor\frac{1000}{13}\right\rfloor \equal{} 1000 \minus{} 13\times76 \equal{} 1000 \minus{} 988 \equal{} \boxed{12}$.

[hide="solution with more explanation..."]Instead of writing out all the terms (gosh I wonder who would do that), we see every term is congruent to 1 modulo 13. For example, $ 1000\equiv12\pmod{13}, 987\equiv12\pmod{13},$ and so on. Therefore, the smallest positive number in the set is the smallest number that is one less than a multiple of $ 13.$ The smallest multiple of $ 13$ is (duh) $ 13,$ and the smallest term is $ 13\minus{}1\equal{}\boxed{12}.$[/hide]


probability
geometry
3D geometry
frustum
LaTeX

Choose a point $ O$ at random insie a give circle with radius $ 3$, and also select a real number $ r$ at random between $ 0$ and $ 2$.  What is the probability that the circle with center $ O$ and radius $ r$ intersects the original circle?

I think this works, tell me if something looks terribly (or even slightly) wrong...
[hide]
The total possibles for r and O: $ (2)(9\pi)$
Thinking about expanding up based on r...
Any point with distance of 1 or less to the center has a height of 2 at that point.
Any point outside that distance gives a height of 3-distance at that point.
These points create a frustum, with lower radius 3, upper radius 1, and height of 2.

Solving for this volume gives:
Volume = $ \frac{2\pi}{3}*(1^2\plus{}1*3\plus{}3^2)$
Volume = $ \frac{26\pi}{3}$
And dividing we get:
$ p\equal{}\frac{26\pi}{3*18\pi}$
$ p\equal{}\frac{13}{27}$

[/hide]

Looks good -- I have just one LaTeX suggestion, which is to use \cdot for the multiplication dot.


linear algebra
matrix
superior algebra
superior algebra solved

I have a problem from my homework, one with matrix:

If X, Y are in M_n(C) with the property that X-Y is invertile and (X+Y)^k=X^k+Y^k for k=2 and k=3. Prove that (X+Y)^n=X^n+Y^n, for any n natural.
Thank you very much!!!

Please help! :D

it's quite simple actually:

we have XY+YX=0, and also from X3+Y3=(X+Y)3=(X+Y)2(X+Y)=X3+Y3+X2Y+Y2X =>  X2Y + Y2X = 0.

But Y2X=-XYX so the latter equality becomes X2Y -XYX = 0 =>  (X-Y)XY=0 => XY=0 => YX=0 => by Newton (X+Y)n = Xn+Yn.

Valentin, why is it that: Y^2X=-XYX?

don't we have that:
Y^2X=-X^2Y=-X(XY)=-X(-YX)=XYX?

cheers! :D

I think this is what Valentin meant:

Y^2X=-YXY, so X^2Y-YXY=0, so (X-Y)XY=0, so XY=0, so YX=0 (from XY+YX=0), and this is it.

ups spelling error :D yes grobber you are right :) (X-Y)XY=0, and this comes from X2Y=YXY :D:D


Maybe we should have like so many accounts for 1 email.  Make sense?
Just an Idea. :D
By the way, they link you to a 404 error!   :huh:

[quote="minicon"]Maybe we should have like so many accounts for 1 email.[/quote]
AoPS does not allow different accounts to have the same email.

[quote="Anaxerzia"]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=166733

These 'Mindzinger' advertising people are at it again. As last, time, they are obviously creating new accounts new advertise their homework helpers company. Topic should be locked and you should give a warning.  :ninja:[/quote]

If they keep doing it and you tell them not to, ask an admin to ban their IPs.

[quote="7h3.D3m0n.117"]..ask an admin to ban their IPs.[/quote]

Dynamic IPs, proxies, etc. render this method ineffective. If someone banned my IP, I could easily bypass it.


I am supposed to be writing an article for the school math journal about an introduction to the theory of partitions (in number theory). Does anyone have any ideas about what I could include (other than the basic definitions and how to define the number of partitions with generating functions)? The stuff should be simple and should be understandable by most high schoolers.

Thanks.

You should discuss some of the history of partition theory. A lot of great mathematicians have contributed to the theory.

A good introduction to partition theory (and a little bit about its history) can be found in Andrew's book [u]Number Theory[/u].

Hmm..history? You mean like "so-and-so discovered this about partitions in year xxxx, and then so-and-so proved this theorem about partitions in the year xxxx, etc..."? Unless there was something really interesting about the history, don't you think that this would be sort of...boring to most high-schoolers?

Don't you think partitions would be boring to most high schoolers? ;)

I think the history could be interesting.  Don't present it like you described it -- work it in to the fabric of the article.  Anything would be boring if you want to present it like that :)

Almost anything can be interesting when it is written by a talented writer. I find that especially true of history.

[quote="JBL"]Don't you think partitions would be boring to most high schoolers? ;)
[/quote]

Hmm...then what would you suggest instead?

I think Joel is just saying that most high school students (at least pretend that they) don't like math.

No, I wasn't serious, I think partitions are interesting -- I was just teasing you.  I agree with Simon -- there are plenty of interesting ways of working history in to something like this, although listing names and dates and achievements in quite the format you suggested isn't one of them.  And if Gauss is correct that there is interesting history to toss in, then it probably would in the end improve your article.  (Especially since you aren't going to get into the nitty-gritty of the math of partitions.)

[quote]No, I wasn't serious, I think partitions are interesting -- I was just teasing you.[/quote]

Aww..you gotta make these things clear Joel. Some folk like me don't get it sometimes.  :mrgreen:

In case you haven't already seen it, you might get some ideas from the first two sections of Hebert Wilf's [i]Lectures on Integer Partitions[/i] which is available here :-

[url]http://br.endernet.org/~drini/books/Wilf%20-%20Lectures%20on%20integer%20partitions.pdf[/url]

(but note that from Section 3 onwards it gets very advanced).

Also, Ferrers diagrams can be used to give intuitive visual proofs of some quite surprising partition identities.


calculus
integration
logarithms
linear algebra
matrix
trigonometry
Gauss

Compute

$ I \equal{} \int_0^{\infty}\sin(x^2) \cdot (\ln x)^2\,dx$

For $ a>0,b\in\Bbb{R}$ and $ \text{Re}(s)>0 \quad \int_0^\infty e^{\minus{}(a\plus{}ib)x}\, x^{s\minus{}1}\, dx\equal{}\frac{\Gamma(s)}{(a\plus{}ib)^s}$

whereat $ a\plus{}ib\equal{}\sqrt{a^2\plus{}b^2}\, e^{i \,\arctan\left(\frac{b}{a}\right)}$ .

So $ \int_0^\infty e^{\minus{}ax}\, \begin{matrix} \sin \\ \cos \end{matrix}(bx)\, x^{s\minus{}1}\, dx\equal{}\frac{\Gamma(s)}{\sqrt{a^2\plus{}b^2}^s}\, \begin{matrix} \sin \\ \cos \end{matrix}\left(s\, \arctan\left(\frac{b}{a}\right)\right)$ .

For $ 0<s<1 \, , \, a\to 0^\plus{}$ and $ b\equal{}1$ we get $ \int_0^\infty \sin(x)\, x^{s\minus{}1}\, dx\equal{}\Gamma(s)\, \sin\left(\frac{s\pi}{2}\right)$ .

Differentiate twice with respect to $ s$ : $ \int_0^\infty \sin(x)\, x^{s\minus{}1}\, \ln^2(x)\, dx$

$ \equal{}\Gamma''(s) \, \sin\left(\frac{s\pi}{2}\right)\plus{}2\,\Gamma'(s)\cos\left(\frac{s\pi}{2}\right) \, \frac{\pi}{2}\minus{}\Gamma(s)\, \sin\left(\frac{s\pi}{2}\right) \, \left(\frac{\pi}{2}\right)^2$ .

Now set $ s\equal{}\frac12$ : $ \int_0^\infty \frac{\sin x}{\sqrt{x}}\, \ln^2(x)\, dx$

$ \equal{}\underbrace{\Gamma''\left(\frac12\right)}_{\left(\frac{\pi^2}{2}\plus{}(2\ln 2\plus{}\gamma)^2\right)\sqrt{\pi}}\, \frac{1}{\sqrt{2}}\plus{}\underbrace{\Gamma'\left(\frac12\right)}_{\minus{}(2\ln 2\plus{}\gamma)\,\sqrt{\pi}}\, \frac{\pi}{\sqrt{2}}\minus{}\underbrace{\Gamma\left(\frac12\right)}_{\sqrt{\pi}}\, \frac{1}{\sqrt{2}}\, \frac{\pi^2}{4}$ .

Finally substitute $ x\to x^2$ to obtain $ \int_0^\infty \sin(x^2)\, \ln^2(x)\, dx\equal{}\frac{\sqrt{2\pi}}{64}\left(4\ln 2\plus{}2\gamma\minus{}\pi\right)^2$ .

everything is good , but it's not obviusly that :
$ \Gamma''(1/2)$ are equal $ \sqrt{\pi}(\pi^{2}/2 \plus{}(2\ln2\plus{}\gamma)^{2})$ or same about $ \Gamma'(1/2)$
there is many way how to show it , for example use Gauss formula for $ \Gamma^{(n)}(p/q)$ or use this presentation
$ \Gamma(x) \equal{} \lim_{n \to \infty} \frac{x^n x!}{\prod_{j}^{n} (x\plus{}j)}$ take both side $ ln$ and differentiate this expression one or two times as you want , after use some simple statement about series likes harmonic and you are done.


The set of problems are attached; answers not included.


calculus
integration
geometry
rotation
vector
analytic geometry
ratio

[b]Suppose we are given a rod with electic charges distributed uniformly from end A to end B. Show that the electric Field at an arbitrary pt C is along the bisector of $ \angle ACB $ outwards.[/b]

[hide]Don't get inti heavy integral calculus although my soln uses similar ideas [/hide]

I dunno if you consider this integral calculus but this is my solution

[hide]
Let's denote the distance of C from line AB as $h$. Let's take a small part of the charged line segment $dx$, which is seen from C by a small angle $d\phi$. Its contribution to total field will be proportional to

$dE \sim \frac{dq}{r^2}$
$dq = \lambda .dx \sim dx$

(lambda is some kind of charge density)

Now $\frac{dx}{r.d\phi} = \frac{r}{h}$, just using similar right triangles (sorry I'm too lazy to provide a picture, but you can figure that out if you make one;). Then $dx = \frac{r^2d\phi}{h}$, plugging back into the first equation

$dE \sim \frac{dx}{r^2} = \frac{d\phi}{h}$

which does NOT depend on the position of the considered $dx$ part within the line AB, only on its size and the distance of the line from C. You can therefore cancel out the perpendicular-to-the-bisector-of-ACB contributions of segments that are equally (angularly) spaced from the bisector. This is not very precisely stated, but I hope it's enough to be clear...
[/hide]

kubus , i just have a feeling that either you have the book [hide]200 puzzling Physics problems[/hide] or you have done the problem before. Most probably, an immediate soln. can be presented if and only if you have seen the problem before. Anyways , your soln. is exactly the one given in the book. :)  ;)  :thumbup:

I'm sorry Rushil to tell you that I do not have that book and I have never seen this problem. Since you gave the answer in the problem statement (that the field points along the angle bisector), it was clear that the contribution of E from different parts seen in the same angle is the same, and the rest was just geometry and simple physics work.

But, mentioning the book, I have heard some of it, I encountered it on amazon.com while looking for a book "123 puzzling physics problems" (or something like that - it is supposed to be really good and written by some hungarian author) ... I am really thinking about buying it. Do you like it, is it good? Could you write a little about it please? Thanks!

The answer given is wrong.
[hide]
Since the point is arbitrary, it need not be along the axis
perpendicular to the rod and passing through its geometric centre.
In this case, the 'approximation' (think of the Centre of Mass method in gravity theory)
is inaccurate, and based upon false assumptions.

If the point C is not on a perpendicular axis through the midpoint,
then the 'near' end of the rod exerts a greater force than the far end.
In this case, the direction of the force experienced by the test-particle (whether mass or charge) will be biased toward the near end.

You can approximate this without any math by imagining a simplified version of the experiment:  Take a barbell made from equal masses (or charges) and a rigid connecting rod of negligible mass or charge.  Rotate it end over end at a fixed test-particle:

[img]http://lentils.imagineis.com/rouge/Image32.gif[/img]

You can see in this diagram that although the ends of the barbell are equal vertical distances away from the centreline, they are different angles from the test-particle on the x-axis, because they are different distances away.

In order for the far end to exert enough force to counter the pull of the near end and keep the direction horizontal (through the origin), it would have to be stronger in pull
than the near end!  But with equal charges this is impossible because the force varies according to the Inverse Square Law.

Think about it.

The line through the geometric centre does not bisect the total end-to-end angle, and the direction of force line doesn't do this either, because even though the charges are equally spaced, they do not form a continuum, but are discretely located quantized points.

The proof is elegant:  Since one pair of charges equidistant from the centre will not give a direction of force that bisects the angle, the sum of all the discretely placed pairs cannot do so either, since they all 'err' in the same direction away from the bisector.

[/hide]

kubus, The book is excellent. Its got problems which require effectively zero math (only algebra - very simple) but great insight . You have to really think out of the box to do them . [hide]Consider for eg. the mountain question I have given in this forum[/hide]
It's not for some course coz the problems do not correspond to any undergrad or other course . They are just interesting , non -routine problems which really test your insight etc. Its a great book . The authors say the questions correspond to some Eotvos competition in hungary.

I'd say, go for it!!

exactly , he's got his math (a bit ) wrong :lol:  :D

[quote="Rushil"]The answer is absolutely correct , Rogue Physicist and one can barely make out what you are talking about. What is your point ? Can you clarify[/quote]

[quote="schulmannerism"]Rogue physicist seems to think that the angle bisector of a triangle bisects the opposite side, i.e. it is the median.  :rotfl:[/quote]

Hello, Rushil and schulmannerism,

While Rogue Physicist was wrong in this instance, the manner of your replies was unnecessary. When a team is trying to solve a problem that has never been tried before, each member makes suggestions with a little chance of success and expects the other team members to shoot them down. That way, the correct answer or the correct way to obtain the answer is found much faster then if one guy struggles with the problem alone. It does not have to be anything as high minded as discovering the Moon or quantum states of neutrons jumping in the gravitational field. Ususally, it is an experimental setup that is really pushing things in some way. On rare occassions, even the most unlikely suggestions are successful. If the incorrect suggestions (a majority of them) were met with derision, the team members would shut up like clams and the whole process would collapse. Please, don't do it any more :).

Yetti

Yes I do understand that our responses were too rash.

Lets see if this helps you confirm that I know the difference between a median and a bisector:

Lets look at the proposal of the question in a diagram:

[img]http://lentils.imagineis.com/rouge/ChargedRod1.jpg[/img]

Now lets look at a simple theorem about circles:

[img]http://lentils.imagineis.com/rouge/CircleTheorem.jpg[/img]

Now lets choose a test-point which is on the rim of the circle of unit radius
around the geometric centre (midpoint) of the rod.

First lets take the most simple case, of two equally spaced like-charges on a rod,
or else a string made of two Hydrogen atoms (a diatomic molecule of H gas):

[img]http://lentils.imagineis.com/rouge/ChargedRod2.jpg[/img]

That is, by simple inspection, since the force from the upper charge is greater than the lower, and the total angle is 90 degrees, the bisector would be 45 degrees, but the resultant force is NOT 45 degrees, since this would only be possible if the forces were equal.
The pull from each end of the rod is covered by the same formula for charge or mass:

[tex] \vec{F} = \frac{Gmm}{d^2}  [/tex]  or,

[tex] \vec{F}  = \frac{Kqq}{d^2} [/tex]

We can set the masses or charges = 1, and choose units of distance to make the constants = 1 as well, simplifying both equations to:

[tex]F = \frac{1}{d^2} [/tex] in the direction of each charge (or mass-particle)

And the total force upon a test particle will be:

[tex]\vec{F} = \vec{F_{top}} + \vec{F_{bot}} [/tex]

by a vector addition of the force from each end of the rod (barbell/molecule).
When we convert this to a single vector, e.g. by conversion from Cartesian form to Polar form, we will find that in general it will *NOT* be a 45 degree ([tex]\frac{\pi}{4}[/tex]) vector (from either component vector).
[img]http://lentils.imagineis.com/rouge/ChargedRod3.jpg[/img]

Now let me quote my original (correct) statement:

[quote]The line through the geometric centre does not bisect the total end-to-end angle, and the direction of force line doesn't do this either, because even though the charges are equally spaced, they do not form a continuum, but are discretely located quantized points. [/quote]
To which I can only add that the bisector APPROXIMATION is only valid when the # of charges or mass-points is large enough to be treated as a continuum.

I'd like to thank [b]yetti[/b] for the following intelligent and important point:

[quote][b]While Rogue Physicist was wrong in this instance[/b], the manner of your replies was unnecessary. When a team is trying to solve a problem that has never been tried before, each member makes suggestions...  If the incorrect suggestions (a majority of them) were met with derision, the team members would shut up like clams and the whole process would collapse...  [/quote]
...which is perhaps even more important when the person who was wrong is not the person you thought.

QED.

I think that you are off the topic. The problem called for a rod with uniform, i.e., continuous charge distrubution. Nobody ever claimed that the electric field at a test point C of 2 point charges at points A, B has the direction of the bisector of the angle $\angle ACB$. The continuous charge distribution is a useful abstraction (like in Maxwell's equations), but no more real than the charge or mass points that you talk about. Moreover, the uniform charge distribution is the appropriate abstraction in this case, while the charge or mass points are not. The reason is that each pair of point charges equally spaced from the rod midpoint, say at points A', B', forms a different triangle with the test point C, with a different bisector of the angle $\angle A'CB'$. It is not clear at all, why the electric fields [b]not[/b] directed along the different bisectors of the different angles $\angle A'CB'$ could not add up to an electric field along the bisector of the angle $\angle ACB$. Just to make sure, I am going to calculate the directions of the angle bisector and of the electric field the hard way:

Put the A-and of the charged rod into the origin and the B-end along the positive x-axis. Let $C = (u, v)$ be the coordinates of the test point C. The side lengths of the triangle $\triangle ABC$ are $L = AB,\ b = AC = \sqrt{u^2 + v^2},\ a = BC = \sqrt{(L - u)^2 + v^2}$. The bisector of the angle $\angle ACB$ cuts the opposite triangle side AB at a point P in the ratio of the adjacent sides:

$\frac{AP}{BP} = \frac{AC}{BC} = \frac b a$

$p = AP = AB \cdot \frac{AC}{AC + BC} = \frac{Lb}{b + a} = \frac{L \sqrt{u^2 + v^2}}{\sqrt{u^2 + v^2} + \sqrt{(L - u)^2 + v^2}}$

and the equation of the bisector of the angle $\angle ACB$ passing through the points P = (p, 0), C = (u, v) is

$y = \frac{v}{u - p}\ (x - p) = \tan \phi\ (x - p)$

$\tan \phi = \frac{v}{u - p} =  \frac{v\left(\sqrt{u^2 + v^2} + \sqrt{(L - u)^2 + v^2}\right)}{(u - L) \sqrt{u^2 + v^2} + u \sqrt{(L - u)^2 + v^2}}$

By Coulomb's law, the electric field $\text d\vec E$ at the test point C = (u, v) due to the charge $\text dQ = \lambda\ \text dx$ at a point $X = (x, 0),\ 0 \le x \le L$ is equal to

$\text d\vec E  = \frac{\text dQ}{4 \pi \varepsilon_0} \cdot \frac{\vec r}{r^3} = \frac{\lambda}{4 \pi \varepsilon_0} \cdot \frac{\vec r\ \text dx}{r^3} \sim \frac{\vec r\ \text dx}{r^3}$

where $\lambda$ is the linear charge density along the line segment AB (a rod of zero radius), $\vec r = (u - x, v)$ is the radius vector from the point X to the point C and $r = \sqrt{(u - x)^2 + v^2}$. Calculating separately the x- and y-components $E_x, E_y$ of the electric field,

$E_x \sim \int_0^L \frac{u - x}{r^3}\ \text dx = \int_0^L \frac{u - x}{\left[(u - x)^2 + v^2\right]^{3/2}}\ \text dx = \int_{u-L}^u \frac{\xi}{(\xi^2 + v^2)^{3/2}}\ \text d\xi =$

$= \left[-\frac{1}{\sqrt{\xi^2 + v^2}}\right]_{u-L}^u = \frac{1}{\sqrt{(u - L)^2 + v^2}} - \frac{1}{\sqrt{u^2 + v^2}}$

$E_y \sim \int_0^L \frac{v}{r^3}\ \text dx = \int_0^L \frac{v}{\left[(u - x)^2 + v^2\right]^{3/2}}\ \text dx = \int_{u-L}^u \frac{v}{(\xi^2 + v^2)^{3/2}}\ \text d\xi =$

$= \left[\frac 1 v \frac{\xi}{\sqrt{\xi^2 + v^2}}\right]_{u-L}^u = \frac{u}{v \sqrt{u^2 + v^2}} - \frac{u - L}{v \sqrt{(u - L)^2 + v^2}}$

The angle formed by the electric field $\vec E = (E_x, E_y)$ with the positive x-axis is given by

$\tan \psi = \frac{E_y}{E_x} = \frac{\frac{u}{v \sqrt{u^2 + v^2}} - \frac{u - L}{v \sqrt{(u - L)^2 + v^2}}}{\frac{1}{\sqrt{(u - L)^2 + v^2}} - \frac{1}{\sqrt{u^2 + v^2}}} = \frac{u \sqrt{(u - L)^2 + v^2} - (u - L) \sqrt{u^2 + v^2}}{v\left(\sqrt{u^2 + v^2} - \sqrt{(u - L)^2 + v^2}\right)}$

If we now expand the last fraction by the expression

$\left(u \sqrt{(u - L)^2 + v^2} + (u - L) \sqrt{u^2 + v^2}\right) \cdot \left(\sqrt{u^2 + v^2} + \sqrt{(u - L)^2 + v^2}\right)$

the numerator and denominator become

$N = \left[u^2 \left((u - L)^2 + v^2\right) - (u - L)^2 (u^2 + v^2)\right] \cdot \left(\sqrt{u^2 + v^2} + \sqrt{(u - L)^2 + v^2}\right) =$

$ = v^2 \left(2uL - L^2\right) \left(\sqrt{u^2 + v^2} + \sqrt{(u - L)^2 + v^2}\right)$

$D = v \left[u^2 + v^2 - \left((u - L)^2 + v^2\right)\right] \cdot \left(u \sqrt{(u - L)^2 + v^2} + (u - L) \sqrt{u^2 + v^2}\right) =$

$= v \left(2uL - L^2\right) \left(u \sqrt{(u - L)^2 + v^2} + (u - L) \sqrt{u^2 + v^2}\right)$

For the direction of the electric field, we then get

$\tan \psi = \frac{E_y}{E_x} = \frac N D = \frac{v \left(\sqrt{u^2 + v^2} + \sqrt{(u - L)^2 + v^2}\right)}{u \sqrt{(u - L)^2 + v^2} + (u - L) \sqrt{u^2 + v^2}}$

which is exactly the same expression as for the slope $\tan \phi$ of the bisector of the angle $\angle ACB$. There is no doubt that the solution of the original problem posted by Kubus is correct. Since a mass line segment is the same kind of abstraction as a mass point, the solution is valid as long as the perpendicular distance of the test point from the axis of a real rod is much larger than the rod diameter. I cannot say that your solution is incorrect, because you posted only objections (not much relevant, in my opinion), but not a solution.

Yetti

I cant quite understand where all of you are heading. I asked for the solution to a simple problem . Also i asked for a soln without calculus as kubus gave, not the one you gave yetti(no offence meant).

Anyways, can you guyz see my other physics posts

First of all, congrads to [b]yetti[/b] for a marvelous and excellent piece of analysis.  Your perceptive insight that there are multiple angles and bisectors involved is the beginning of the true solution to the problem.  Again hats off and well done.

I now only have a few remarks, although I may post my own solution later:

[quote="yetti"]I think that you are off the topic. The problem called for a rod with uniform, i.e., continuous charge distrubution. Nobody ever claimed that the electric field at a test point C of 2 point charges at points A, B has the direction of the bisector of the angle \angle ACB. [/quote] It is a very common and practical matter that rods can be charged with everything from NO charge right up to a few million.  Obviously we wan t a general solution for all levels of charge, and a clear understanding of the mechanics of the problem.  Of course the bisector is a reasonable approximation for large numbers of charges, but *most* static charge work is actually very delicate, and the most interesting applications are with very low charges in lab conditions, where precise measurements and a real understanding are needed.  This is how we set accurate standards for all our electrostatic units!

[quote]The continuous charge distribution is a useful abstraction (like in Maxwell's equations), but no more real than the charge or mass points that you talk about. Moreover, the uniform charge distribution is the appropriate abstraction in this case, while the charge or mass points are not. [/quote] I don't see this.  The equations for charge and mass are the same, and it would be bizarre not to point it out and look at the parallel cases.  Why not solve two birds with one stone, and point out the differences as well?  As it turns out the most common gases are diatomic and isometric.  My approach is general enough for both types of forces and all cases.

[quote]The reason is that each pair of point charges equally spaced from the rod midpoint, say at points A', B', forms a different triangle with the test point C, with a different bisector of the angle \angle A'CB'. [b]It is not clear at all, why the electric fields not directed along the different bisectors of the different angles \angle A'CB' could not add up to an electric field along the bisector of the angle \angle ACB. [/b]Just to make sure, I am going to calculate the directions of the angle bisector and of the electric field the hard way: [/quote]
Actually, it is quite clear from my second post that they [b]don't [/b]add up to the bisector, but approach it as a limit.   [b]You have done excellent work here,[/b] but only for one end of the problem, namely, highly charged rods of  Q >> 2000.

A complete solution will give the correct answer for all cases, and derive the error term as a function of the number of masses or charges and the angle of incidence.

I must make one more serious correction to your statements:

[quote]The problem called for a rod with uniform, i.e., continuous charge distrubution. ... the uniform charge distribution is the appropriate abstraction in this case, while the charge or mass points are not.[/quote]In fact, [b] uniform [tex] \neq [/tex] continuous.[/b]
Two different words, two different meanings, and this will just confuse many other students and lead to similar errors.  It is clear from your usage in the second sentence that you think they are equivalent or synonymous.
At high-school level (skipping relativistic quantum mechanics) the Classical Electrostatic theory is assumed.  And this means point-particles exactly.  [b]Uniform[/b] in this context means uniformly distributed or[b] uniformly spaced [/b]point-particles, not a continuum, which is a highly advanced concept from calculus.
Look at the wording of the original question:[b]"with electic charges distributed uniformly"[/b] .  There is nothing of a continuum here, but rather a discrete arrangement of point-charges.  Discussion of the treatment of point-particles as a continuum is usually introduced at university level, where it is clearly pointed out that the continuum model is an approximation.

[quote]I asked for the solution to a simple problem . Also i asked for a soln without calculus as kubus gave, not the one you gave yetti(no offence meant). [/quote]Actually no.  You posted an answer, and asked for a 'proof'. I posted a good disproof, without calculus.  My disproof gave insight into the problem without heavy math, and led to an unexpected (and apparently unwelcome) result.

Some of this problem could have been avoided if you had worded the question in a way that reflected both the true answer, and your knowledge of it, as well as prepared the student to provide the right 'answer' you were seeking:

[quote]Suppose we are given a rod with electic charges distributed uniformly from end A to end B. Show that the electric Field at an arbitrary pt C is  [b]approximately[/b]  along the bisector of \angle ACB . [/quote]If we gave you the benefit of the doubt, and assumed you posted the question exactly as you found it in a book, then the poor choice of wording is the fault of the author.  But it would still seem to be your responsibility to check that the question (and answer) was right, well worded and not misleading.

[quote]Nobody ever claimed that the electric field at a test point C of 2 point charges at points A, B has the direction of the bisector of the angle \angle ACB.[/quote] Actually, that is precisely what was claimed, explicitly, and exactly.  Read the original question again.
[quote]There is no doubt that the solution of the original problem posted by Kubus is correct. Since a mass line segment is the same kind of abstraction as a mass point, the solution is valid [b]as long as the perpendicular distance of the test point from the axis of a real rod is much larger than the rod diameter.[/b][/quote]When the distance is much larger than the rod length  [b]( d >> l )[/b] any error in the bisector angle vanishes because it has virtually become invisible!  At large distances the rod acts as a point-particle, and just about *any* ridiculous statement about the bisector will be 'true'.

This is just a generalization of the Centre of Mass Approximation:  Distant objects trivially act like point-masses.  How that 'demonstrates' Kubus' solution or any solution is true or false is beyond me.  You have excused yourself from error by invoking 'large distances'.   That's about as disengenious as it gets.

The problem itself however, far from assuming 'large distances' actually presumes the opposite: that we are within ranges where it is significant to talk of a 'rod' and not a 'point-mass' approximation.  That is, close range.

regards, RP


Your answer is correct partly. First remark $|x|\geq -x$ for arbiterary real numbers $x.$

[hide="Not so"]$x={1\over 3}$ or $x=-1$[/hide]

hmm...

I put this originally
[hide]$\pm\frac{1}{3}$[/hide]

[b]EDIT[/b]

but it should be
[hide]$\frac{1}{3}$ and $-1$[/hide]

sorry, i'm kinda new to all this stuff.

First we have:
$|x + |x|| = 1-x$ and $|x + |x|| = -1-x$
then:
$|x| = 1 - 2x$
$|x| = 1 \implies x = -1,1$ (We're done with that one)
$|x| = -1-2x$
$|x| = 1 \implies x = -1,1$ (Again, done with this case)
then:
$x = 1 - 2x \implies 3x = 1 \implies x = \frac{1}{3}$
$x = 2x - 1 \implies x = 1$
$x = -1 - 2x \implies 3x = -1 \implies x = \frac{-1}{3}$
$x = 1 + 2x \implies x = -1$

So, $x = \boxed{1, -1, \frac{1}{3}, \frac{-1}{3}}$

So... how exactly do you do this? I get how $1/3$ is derived, but $-1$?

edit: nvm, got it.

[quote="kunny"]$x=-1,\ \frac{1}{3}$ are not solution.[/quote]

Yes, they are. plug them in.
Upon inspection, though, it seems that the other two don't work. However, the algebra says otherwise.... Hmm....

Absolute value problems often lead to extraneous answers.  You always need to plug in all solutions to make sure they work.

[quote="DanK"][quote="LordoftheMorons"]First we have:
$|x + |x|| = 1-x$ and $|x + |x|| = -1-x$[/quote]

How do we get that?[/quote]
It's pretty obvious that
from the definitio nof absolute value, we may deduce that
$|x+|x + |x|||=1\implies x+|x + |x||=1, -(x+|x + |x||)=1\\\implies |x + |x||=1-x, |x + |x||=-1-x$

[quote="DanK"][quote="LordoftheMorons"]First we have:
$|x + |x|| = 1-x$ and $|x + |x|| = -1-x$[/quote]

How do we get that?[/quote]
To take away absolute value bars you have to make two equations, one in which the other side is negative, and one positive. Then I simply subtracted the $x$.

[hide]
solving this is a pain, so...

Case 1: Assume $x$ is positive
$\\ |x + |x + |x||| = 1 \\ x + x + x = 3x = 1; x = \frac{1}{3}$

Case 2: Assume $x$ is negative
$\\ |x + |x + |x||| = 1 \\ |x + |x + opposite \ of \ negative \ x| = |x + |0|| = |x| = 1 \\$
because $x$ is negative, $x = -1$
[/hide]

Aaah, just didn't think of the moving $x$ to the other side there...


Let $a_1=1-i$, $a_n=r+si$ and $a_{n+1}=(r-s)+(r+s)i$ $(n>1)$ terms of a sequence.

Find, in terms of $n$, $r$ and $s$ which made that sequence an arithmetical progression, where $r$ and $s$ are reals and $i=\sqrt{-1}$.

Hello LoboSolitario,i checked the IME's test today morning and i'm still trying to do the last problem...

[hide]To make the sequence an arithmetical progression we must have that:

$a_n=a_1+(n-1).d$ in which $d=a_{n+1}-a_n=-s+ri$ then:

$r+si=1-i+(n-1)(-s+ri) \implies r+si=1-s(n-1)+[(n-1)r-1]i$

$\implies \{\begin{array}{cc}r=1-s(n-1)\\ s=(n-1)r-1\end{array}$
Solving the sistem:

$s=\frac{n-2}{(n-1)^2+1}$ and $r=\frac{n}{(n-1)^2+1}$ [/hide]

Good Job. :)

I'm going to try last problem today.


function
combinatorics proposed
combinatorics

Let $ \mathcal{A}_n$ denote the set of all mappings $ f: \{1,2,\ldots ,n \} \rightarrow \{1,2,\ldots, n \}$ such that $ f^{-1}(i) :=\{ k \colon f(k)=i\ \} \neq \varnothing$ implies $ f^{-1}(j) \neq \varnothing, j \in \{1,2,\ldots, i \} .$ Prove \[ |\mathcal{A}_n| =
\sum_{k=0}^{\infty} \frac{k^n}{2^{k+1}}.\]

[i]L. Lovasz[/i]

This means that if the pre-image of $i$ is non-empty then the function is surjective. Am I right?
Is the $i$ fixed?

Indeed, the statement is very unclear!  Does anyone have access to the original statement?

There is a typo here. The red  [color=#FF0000][i] i[/i][/color] should be read instead of "$n$" :

Let $ \mathcal{A}_n$ denote the set of all mappings $ f: \{1,2,...,n \} \rightarrow \{1,2,...,n \}$ such that $ f^{-1}(i) : = \{ k : \;f(k)=i\ \} \not= \emptyset$ implies $ f^{-1}(j) \not= \emptyset , j \in \{1,2,\ldots,$ [color=#FF0000][i] i[/i][/color]  $\}$. Prove \[ |\mathcal{A}_n|=  \sum_{k=0}^{\infty} \frac{k^n}{2^{k+1}}.\]

[i]L. Lovasz[/i]

Thank you.  In other words, the question asks for the number of functions $[n] \to [n]$ with image $[i]$ for some $i$.

Denote $s_n= \sum_{k=0}^{\infty}\frac{k^n}{2^{k+1}}$ and let's try to calculate or rather obtain some recurrence relation for $s_n$.
We have:
\[\sum_{k=0}^{\infty}\frac{(k+1)^n}{2^{k+1}}=2 s_n\,\,\,\,\,\,\,\]
It follows:
\[s_n = \sum_{k=0}^{\infty}\frac{(k+1)^n-k^n}{2^{k+1}}=\sum_{k=0}^{\infty}\frac{1}{2^{k+1}} \sum_{j=0}^{n-1}\binom{n}{j}k^j  =\]
\[=\sum_{j=0}^{n-1}\binom{n}{j}\sum_{k=0}^{\infty}\frac{k^j}{2^{k+1}}=\sum_{j=0}^{n-1}\binom{n}{j}s_j\]
Now, let us denote $f_n=|\mathcal{A}_n|$. It remains to establish that $f_n$ comply with the same recurrence relation.
Consider the set $A_1=f^{-1}(1)$ and fix it. Suppose $|A_1|=j$. It's clear that $1\le j\le n$. The restrictions of $\{f\mid f\in\mathcal{A}_n, f^{-1}(1)=A_1\}$ on the set $[n]\setminus A_1$ are exactly $f_{n-j}$ different functions. Hence, we get:
\[f_n= \sum_{j=1}^{n} \binom{n}{j}f_{n-j}= \sum_{j=0}^{n-1} \binom{n}{j} f_j\]
where, for convenience, we have set $f_0=1$.
Since $s_0=f_0=1$, the result follows.


function
Functional Analysis
real analysis
real analysis unsolved

Let $ X$ be a Banach space  and $ N \subset X$. Suppose that any linear function $ f: N \longrightarrow \mathbb{R}$ is continuous then $ f$ is bounded. Prove $ N$ is compact.

if X is finite dimensional normed linear space then N is compact.otherwise we cannot say


function
parameterization
real analysis
real analysis unsolved

Q1 A function y=f(x) is given by x=1/(1+t ^2)  and y=1/t(1+t^ 2)  for all x >0 then show that f is increasing in (0,1).

Q2 Find range of g(x)
g(x) = 1/{log(1+x)} - 1/x

In 2) is your question this-

$ g(x) \equal{} \frac{1}{log(1\plus{}x)} \minus{} \frac{1}{x}$

or is the log in fractional part?

no it is not fractional part..
the question is same as u mentioned

[b]Q1.[/b] The function $ y \equal{} f(x)$ is given by the parametric equations (with the parameter $ t$)
\[ x(t): \equal{} \frac {1}{1 \plus{} t^2}\qquad\text{and}\qquad y(t): \equal{} \frac {1}{t(1 \plus{} t^2)}.\]
It is easy to see that $ t\neq 0$. Moreover, $ x(t)\in (0,1)$ iff $ t\in\mathbb{R}$. Hence, we consider $ t\in\mathbb{R}\setminus\{ 0\}$ and split the proof in two parts:
[b]Part 1.[/b] Assume that $ t > 0$. Then $ x(t)$ is decreasing for such $ t$'s. Choose (WLOG) $ t_1,t_2$ with $ 0 < t_1 < t_2$. Then $ x(t_1) > x(t_2)$. Now we shall prove that also $ y(t_1) > y(t_2)$ which follows evidently in virtue of the fact that $ (t_2^3 \minus{} t_1^3) \plus{} (t_2 \minus{} t_1) > 0$ as the sum of two positive numbers.

[b]Part 2.[/b] Analogously ...


factorial
algebra
polynomial
combinatorics proposed
combinatorics

evaluate the sum:
$\sum_{m=1}^{N}m^n$ in terms of factorials, Stirling numbers and binomial coefficients. Show that it is a polynomial of degree $n+1$ in $N$.

Hi.

In my opinion, this is a beautiful identity! It's also discussed in "Problem Solving Strategies", so that's the reason why I know it ;)

Here's the solution:

We'd like to count all n-tuples [tex](a_1,a_2,...,a_{n+1})[/tex], where [tex]a_i\leq N+1[/tex] and [tex]a_i<a_{n+1}, i=1,2,...,n[/tex]. On the other hand, we get [tex]1^n+2^n+...+N^n[/tex] since there are [tex]m^n[/tex] such tuples if [tex]a_{n+1}=m+1[/tex]. On the other hand, we can separate the tuples by counting their different numbers ocurring; if there are [tex]k[/tex] such different numbers, then there are [tex]N+1\choose k[/tex] possibilities to choose these numbers. Then, we have to partition the set [tex]\{1,2,...,n\}[/tex] into [tex]k-1[/tex] subsets to decide which of the [tex]a_i,i=1,2,...,n[/tex] get the same number in the tuple we're constructing; by means of the Stirling Numbers, there are [tex]S(n,k-1)[/tex] ways to do such a partitioning; now, we have to decide, which "group" of [tex]a_i[/tex] gets which of the [tex]k-1[/tex] numbers - to do this, we have [tex](k-1)![/tex] possibilities. So, putting together all these things and summing up for [tex]k=2,...,n+1[/tex], we get [tex]1^n+2^n+...+N^n=\sum_{k=2}^{n+1} {N+1\choose k} S(n,k-1) (k-1)!=\sum_{k=1}^{n} {N+1\choose k+1} S(n,k) k![/tex] and we're done.

Hanno


Yeah, this is probably a newb question but here goes. I downloaded MikTeX 2.5, TeXnicCenter, Ghostview, and Ghostscript for writing my thesis.

I picked up a templete from a former math professor but there's a problem. TeXnic seems to be unable to find the .cls file assoicated with the .tex files involved even though it's right there in the same folder. Any idea how to fix this?

There may be other files referred to in the cls file that it can't find. Make a minimal document such as:
\documentclass{whatever}
\begin{document}
Testing
\end{document}
and then post the log containing the errors.

This is pdfeTeX, Version 3.141592-1.30.6-2.2 (MiKTeX 2.5) (preloaded format=latex 2007.6.28)  2 JUL 2007 00:44
entering extended mode
**Test.tex
(Test.tex
LaTeX2e <2005/12/01>
Babel <v3.8g> and hyphenation patterns for english, dumylang, nohyphenation, ge
rman, ngerman, french, loaded.

! LaTeX Error: File `kcmath2e.cls' not found.

Type X to quit or <RETURN> to proceed,
or enter new name. (Default extension: cls)

Enter file name:
! Emergency stop.
<read *>

l.2 \begin
          {document}

*** (cannot \read from terminal in nonstop modes)

Here is how much of TeX's memory you used:
 12 strings out of 95420
 154 string characters out of 1187358
 44877 words of memory out of 1048577
 3259 multiletter control sequences out of 60000
 3640 words of font info for 14 fonts, out of 1000000 for 2000
 14 hyphenation exceptions out of 8191
 10i,0n,7p,45b,8s stack positions out of 5000i,500n,10000p,200000b,32768s
No pages of output.

Figure the bottom part is probably unimportant but it's best to be thourough. (Spelling?)

I assume your test document starts with the line
\documentclass{kcmath2e}

Ensure that the directory where your latex and kcmath2e.cls are does not have any spaces in the file name. Spaces can cause problems with latex programs.

If changing the directory doesn't help, move kcmath2e.cls into wherever your tex files are. If they are in c:\miktex (for example) then make a new subdirectory and put kcmath2e.cls in c:\miktex\tex\latex\kcmath2e\. Then you must tell MikTeX that you've done this by going to MiKTeX->Settings->Refresh FNDB
Then try compiling again.

This is rather embarrising but I can't find the Refresh FNDB option. Could I get a walk though for that as well?

In Windows Start -> All Programs find MiKTeX 2.5, then in the submenu click on Settings. Then it's the top button on the right.

You ever spend days looking for something then someone points out what is literally the most obvious place for it and that's exactly where it is?

Yeah. Thank you!


parameterization
parametric equation
real analysis
real analysis unsolved

Um i would just like to know how to get the parametric equation for the equation $x-1 = \frac{y}{2}-\frac{z}{3}$.
I think somethings wrong with the given but if you could suggest a way to so. It would be greatly appreciated.

Using $y$ and $z$ as parameters:
\[\begin{cases}x=&1+y/2-z/3 \\ y=&y \\ z=&z\end{cases}\]


Hello everyone,

For the last week, I've been working on my last project - creating a website.
Its purpose will be to gather people interested in different areas - economics, geography, english literature, physics, chemistry, history,etc.
It will be a place where people can discuss different topics related to large number of sciences.
The website will consist of forum; index and a few pages that you will see when the page is finished.
News (scientific areas) will be posted; topics and discoveries will also be placed somewhere on the website.

I believe I will be ready with it in the next 1-2 weeks.
I am looking for people who can be Moderators, Administrators, and Articles Uploaders in different areas.
If you are interested/have any questions, please PM or write in this topic :)
Thank you!
:)

I know there are prominent history, literature, chemistry and physics forums out there. What I haven't seen is a good, prominent economics forum. So, if you want to focus your efforts, that would probably be a good one (and I'd be interested in checking it out).

In any case, I don't think someone who goes to your forum for one topic would be very likely to post significantly in other topics, so if you were to make some multi-topic forum like that, it would basically be like bashing several totally different forums into a single website.

My 2¢

My first idea was to be about economics :)
So yes, most probably I will make it only about economics, but I will help because I am not that qualified to support the whole website on my own.
Everyone who is interested in helping me to create a [b]economics website + forum[/b] PM or write here :)
Today I will think of a new name, if you have any suggestions - please share.

Thank you again,

S.


combinatorial geometry
combinatorics unsolved
combinatorics

Let $T$ be a set of $2005$ coplanar points with no three collinear. Show that, for any point of the $2005$ points, the number of triangles it lies strictly within, whose vertices are points in $T$, is even.

Reformulate it as:
Let $n+1$ (with $n$ even) points in general position be given in the plane. Take one of them ($P$). Prove that $P$ lies inside of an even number of triangels.

Prove:
When $P$ 'moves' around, the number of triangels it is contained in remains the same as long as $P$ doesn't cross the line between any two points.
Since the statement is true when $P$ lies outside of the convex hull of all $n$ other points, it suffices to show that the parity of that number remains the same when $P$ crosses a border.
Let $AB$ the border $P$ crosses, and $a$ be the number of points on the same side of $AB$ as $P$, different from $P$.
Well, when $P$ is near enough to $AB$, then $P$ is contained in all $a$ triangles formed with $AB$ on the same side as $P$, and not in those having $A,B$ and of the $n-a-2$ points of the other side as vertices. When now $P$ moves across the border, this reverses, so $P$ will lie inside of the $n-a-2$ triangles formed with $A,B$and one the points that are now on the same side as $P$. Since the 'being inside' property of all triangles not having $AB$ as a side will remain unchanged, the number of triangels $P$ lies in changes by $n-2(a+1)$, so parity remains the same iff $n$ is even, and it is here.


limit
probability
calculus
calculus computations

$ \lim_{n\infty}  e^{\minus{}n} \sum_{k\equal{}0}^{n} \frac{n^k}{k!}$

hello, WolframAlpha gives for $ e^{\minus{}n}\sum_{k\equal{}0}^{n}\frac{n^k}{n!}\equal{}\frac{\Gamma(n\plus{}1,n)}{n\Gamma(n)}$.But how we can compute the searched limit?
Sonnhard.

Let $ X_n$ be a random variable with a Poisson distribution with mean $ n$. Then

$ e^{\minus{}n}\sum_{k\equal{}0}^n\frac{n^k}{n!}\equal{}P(X_n\le n).$

$ X_n$ can be considered to be the sum of $ n$ independent identical Poisson variables of mean $ 1.$ Hence, the Central Limit Theorem applies to it.

Let $ Z_n\equal{}\frac{X_n\minus{}n}{\sqrt{n}}.$ Then $ Z$ has mean zero and variance one, and by the CLT, $ Z_n$ converges in distribution to a standard normal distribution.

$ P(X_n\le n)\equal{}P(Z_n\le 0).$ As $ n$ tends to infinity, this tends to the probability that a standard normal random variable is $ \le 0,$ and that probability is $ \frac12.$

So the answer is $ \frac12.$

(No, I don't really know how to prove that without invoking probability, or at least following a plan strongly motivated by probability.)


Hi everyone,

I want to annouce you of the upcoming start of the [b]4th edition of the MathLinks Contest[/b], on October 11th, 2004. The questions will be online at 22,00 GMT time under the contest tab (circle).

Please choose to be notified of changes in this topic, to keep up with upcoming MathLinks Contest announcements.

This contest features [b]hard, IMO-like, olympiad-level[/b] problems which are to be solved without any exterior help.

The full-regulations of the contest can be found in this forum http://www.mathlinks.ro/Forum/viewforum.php?f=10

Previous contests, plus the contest main page is located at
http://www.mathlinks.ro/Forum/contest.php

A notice: the final results of the 3rd edition of the MathLinks Contest will be published on the 12th of October, 2004. I am really sorry for the (huge) delay.

We have a new round starting now, round 2. Please download the questions from the contest page, located at http://www.mathlinks.ro/Forum/contest.php

Hi folks,

Round 3 questions were just posted.

Good luck!

Hi guys,

I just finished correcting round 2, and the results are the following:
(they can also be found from now on, on the contest pages)

This is just a remainder that the questions of Round 5 are posted in the following thread

http://www.mathlinks.ro/Forum/viewtopic.php?t=20550

There has been a change in the problem list for the Round 6. As for this moment (20 December 2004 -  4:30 PM GMT Time) the problem 3 of the 4th Edition - Round 6 is replaced by the following problem:

If $n>2$ is an integer and $x_1, \ldots ,x_n$ are positive reals such that

\[ \frac 1{x_1} + \frac 1{x_2} + \cdots + \frac 1{x_n} = n \] then the following inequality takes place
\[ \frac{x_2^2+\cdots+x_n^2}{n-1}\cdot \frac {x_1^2+x_3^2+\cdots +x_n^2} {n-1}  \cdots  \frac{x_1^2+\cdots+x_{n-1}^2}{n-1}\geq \left(\frac{x_1^2+...+x_n^2}{n}\right)^{n-1}. \]

Regards,
Valentin Vornicu.

Last round in 4th Edition has been posted in the [url=http://www.mathlinks.ro/Forum/contest.php]Contest[/url] section. Have fun!

A small (and rather obvious) mistake in the statement of problem no. 3 appeared (was found originally by Myth, but for some reason the modification didn't appear in the final .pdf file which I posted, so billzhao announced me about it): instead of proving that $P_n(X)$ is irreducible for all positive integers $n$, we must prove that it is irreducible for all positive integers $n$, with $n\geq 3$, because for $n=2$, we have $P_2(X)=(X+1)^2$.

Please choose to be notified of the following topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=31524 to receive announcements about the 5th Edition of the MathLinks Contest :)


In forward stepwise regression, what is the advantage of using a relatively small alpha-to-enter value for adding variables?  Also, what is the advantage in using a large alpha-to-enter value?

smaller alpha (p-value) will allow fewer covariates to enter the model.
larger alpha will allow more variables into the model.

So...if i specify an 0.05 entry criterium, then only variables at the 0.05 significance level can enter the model. If I specify an 0.25 significance level then more variables will enter the model. A true stepwise regression model will have an entry and stay criteria.


In a basketball tournament, there are 6 teams in a group and each team plays each other team exactly once. Only the top 2 teams will qualify for the next round. What is the minimum number of games a team must win to be sure that they qualify for the next round?

[hide]Obviously, if a team wins only once or twice, they're most likely not going to make the top $ 2$, so we can eliminate those pretty quickly.

If all $ 6$ teams were equally competitive and had close records, they would look like this:

A- 2-3
B- 3-2
C- 2-3
D- 3-2
E- 2-3
F- 3-2

Since there are $ 3$ teams that had the best records and only $ 2$ of them can make the top $ 2$, the least amount of wins a team has to have would be $ 4$.[/hide]

It shuld be 5. I also thought that 4 would be enough.

A/B/C- 4-1
D/E/F - 1-4 is possible. Just have A/B/C win every game  except once to one of the other A/B/C (forms a "circle") and D/E/F lose every game except once to the other D/E/F (also a circle). So the minimum must then be 5 wins.

it would be 4

5,4,3,3,3,3

the team with 4 gets in

please post the question in the appropriate forum this is really easy for high school basics

...and so easy that you got it wrong... :wink:

cygao you did not even read the question right it is asking the least number of games

4 would be the right answer

because two teams get into the playoffs

this is so easy that you got it wrong yourself......................

1)i'm not cygao
2)read the above posts, then you notice that they are [b]5[/b] not 4.

It says the lowest you can get to be [b]sure[/b]. So yes, you are wrong. The answer is 5 as I already showed in my post.

[quote="calculatorwiz"]you havent been listening to what I  have been writeing[/quote]
Hahaha, I'm sorry, but this made me laugh  :rotfl:

[quote="calculatorwiz"]just because the posts above say it is 5 doesnt mean it is right[/quote]
And please look at the previous posts before posting. You do this so many times in other threads where you revive a dead topic and offer nothing new to say. And in this case, could you at least point out where other posts are incorrect?

oh sure 4 is lower than three if i understand your post...

first off, let me say what is wrong with this thread
i see flaming, accusing, sarcasm (which i didn't understand, BTW), and making fun of something someone wrote (although i think you didn't mean it that way)
i also predict that more flaming will continue

so, clean up this thread, act responsibly, and let's continue this discussion in a slightly nicer manner

[quote="calculatorwiz"]just because the posts above say it is 5 doesnt mean it is right

clearly the lowest is 4 because it is asking the lowest out of all possibilities

5,4,3,3,3,3

you havent been listening to what I  have been writeing[/quote]

i'm assuming the numbers you listed are the team's wins
if it is, let me just say that it is impossible to have that many wins

there are 6 teams, meaning there will be 6C2=15 matches
15 matches means 15 wins
the number of wins you have listed total 21, which is definitely not equal to 15

therefore, there must be something wrong with your solution, and thus you answer may be wrong

if you could post a better solution, then we could consider your answer and debate whether the correct answer is 5 or 4


modular arithmetic
algebra unsolved
algebra

Let $ x,y,z,\in N$ .Solve the equation .
$ x(x\minus{}y)\plus{}y(y\minus{}z)\plus{}z(z\minus{}x)\equal{}2010^{n}\minus{}1$

__________________________________________
Azerbaijan Land of the Fire  :ninja:

[quote="Pirkuliyev Rovsen"]Let $ x,y,z,\in N$ .Solve the equation .
$ x(x \minus{} y) \plus{} y(y \minus{} z) \plus{} z(z \minus{} x) \equal{} 2010^{n} \minus{} 1$ [/quote]

So $ (x\plus{}y\plus{}z)^2\minus{}3(xy\plus{}yz\plus{}zx)\equal{}2010^n\minus{}1$ but $ LHS\equal{}0\pmod 3$ or $ LHS\equal{}1\pmod 3$ and so $ n\equal{}0$, else $ RHS\equal{}2\pmod 3$

So $ x(x \minus{} y) \plus{} y(y \minus{} z) \plus{} z(z \minus{} x) \equal{} 0$ which may be written $ (x\minus{}\frac{y\plus{}z}2)^2\plus{}3(\frac{y\minus{}z}2)^2\equal{}0$ and so $ x\equal{}y\equal{}z$

Hence the solutions : $ \boxed{(x,y,z,n)\equal{}(u,u,u,0)}$ for any $ u\in\mathbb N$ (and this suppose $ n\equal{}0$ is allowed).


What's the ideal score for the individuals of ARML? I scored 5 this year and I, along with a team mate, got the medals. I answered numbers 1,2,3,4 and 7 correctly...Is it too low? or just average?

The standards vary with the test's difficulty, but 5 is normally a very good score.
For 2007, a 5 would be typical for an average member of a Division A team, or for a leader of a Division B team.

There's a lot of analysis in the various results threads to look at.

I've been an ARML coach for 6 years, from 2002 through 2007. We had one team in 2002 which qualified for Division A; in subsequent years, we had a multi-team organization in which the top team competed in Division A. If I look at the scores of the top team, I get a sample of 90 person-years of a team that finished in the top 20 teams overall in every one of those years. Here's the score distribution:

8: 1
7: 6
6: 13
5: 12
4: 20
3: 17
2: 10
1: 6
0: 5

The 8 was from 2002, but all of the 7's and most of the 6's were from 2006 and 2007. (In the two most recent years, it was harder than usual to score 8 but easier than usual to score 7.) For the 2003 team, the two top scores were 4's - that was a very difficult year.

[quote="Kent Merryfield"]Here's the score distribution:

8: 1
7: 6
6: 13
5: 12
4: 20
3: 17
2: 10
1: 6
0: 5
[/quote]

I'm going to take that data and say the following things:
median is 4
mode is 4
mean is about 3.83


If $a,b$, and $c$ are the roots of the equation $x^{3}-x^{2}-x-1=0$,
(a) show that $a,b$ and $c$ are distince
(b) Show that
\[\frac{a^{1000}-b^{1000}}{a-b}+\frac{c^{1000}-a^{1000}}{c-a}+\frac{b^{1000}-c^{1000}}{b-c}\]
is an integer.

[hide="a:  cheap solution"] The first derivative is $3x^{2}-2x-1$ which does not share roots with the original equation (or else the original equation would have a linear factor), hence there are no repeated roots. [/hide]


How many integers from 100 to 199 inclusive have no
consecutive identical digits? (For example, 121 has no
consecutive identical digits, but 322 does.)

The first digit is obviously 1. the second can be any of the otehr 9 digits. The 3rd can be any digit (including 1) thats not the 2nd digit. tehrefoe tere are 1*9*9=81 numbers that fit the criteria

[quote="vallon22"]just a little latex

The first digit is obviously $ 1$. the second can be any of the other $ 9$ digits. The 3rd can be any digit (including $ 1$) thats not the 2nd digit. tehrefoe tere are $ 1 \times 9 \times 9 \equal{} 81$ numbers that fit the criteria[/quote]

It's ok, vallon.  You don't have to latex-ify everyone of mz's posts. :rotfl:

Yes; ernie is right. If I find that the post is difficult to read and understand, I'll edit it myself. :wink:

Thanks for the effort though!

Oh, and this is getting off-topic, so locked.


geometry
geometric transformation
reflection
circumcircle
vector
geometry proposed

Suppose $ABC$ be a triangle inscribed in a circle $\left(O\right)$. Prove that there exists a triangle $A'B'C'$ such that for any circumcevian triangle $A_{1}B_{1}C_{1}$ of triangle $ABC$, one of the products $A_{1}A'\cdot B_{1}C_{1}$, $B_{1}B'\cdot C_{1}A_{1}$, $C_{1}C'\cdot A_{1}B_{1}$ is equal to the sum of the other two.

This is a wonderful problem, Treegoner.
Let A', B', C' be the reflections of the orthocenter H wrt the sidelines BC, CA, AB respectively (we know that A', B', C' lie on the circumcircle)
Suppose that A1B1C1 is the circumcevian triangle of a point P.
We have A1A' = 2.R.sin<A1AA' = 2.R|cos(<BC,AP)| = 2.R.|(BCxAP)|/a/AP (where BCxAP is the dotproduct of the vectors BC and AP)
Now B1C1 = |u| a/BP/CP where u is the power of P wrt the circumcircle.
Hence A1A'.B1C1 = 2.R.|u|.|(BCxAP)|/(AP.BP.CP).
Same way, B1B'.C1A1 = 2.R.|u|.|(CAxBP)|/(AP.BP.CP) and C1C'.A1B1 = 2.R.|u|.|(ABxCP)|/(AP.BP.CP).
The result follows from the fact that (BCxAP)+(CAxBP)+(ABxCP) = 0

To jpe,
       Your solution is definitely beautiful too. My triangle A'B'C' is the same as yours.  I have another solution but I want to keep it for a few days.
       I 'd like to know how you can "sense" the circumcevian triangle of the orthocenter H wrt to ABC. BTW, do you think there is another triangle A'B'C' satisfying this property ?
   Khoa.

Darij on his website in his paper on isogonal conjugates of medians

   proved a wonderful theorem about a chord made by

   traces on circumcircle by just two isogonal cevians from one vertex.  He proved

   that this chord is parallel to the triangle side opposite to the vertex (where the

  isogonal cevians originate).  We have to use this theorem because  it makes it so

  much  easier to multiply chords (that's what  is required in this problem)  when

  they are parallel to each other (at least I find this task to be more manageable when

  the   angle between then is 0).

  Orthocenter and circumcenter are two very popular isogonal conjugate centers

  and this  way one rather quickly  can zero in on the solution - the orthocenter of the

  referrence  triangle. Along the way a pleasant surprise is  waiting : circumcevian

  triangle of circumcenter turns out to have sides parallel to the referrence

  triangle.

 And then just follow JPE's wonderful proof.

  Thank you.

  M.T.

Treegoner wrote :
" I 'd like to know how you can "sense" the circumcevian triangle of the orthocenter H wrt to ABC"

In fact, i've supposed that A'B'C' was the circumcevian triangle of a point Q and seen that necessarily Q = H.
Proof : B1C1.A1A' = 2.R.a.|u|.sin(<QAP)/(BP.CP) with u = power of P wrt the circumcircle.
If we take P on the line AQ, we must have
b.sin(<QBP)/CP = c.sin(QCP)/BP and now, when P->A,
sin(<QBA) = sin(<QCA); same way sin(<QCB) = sin(<QAB) and sin(<QAC = sin(<QBC)
It follows easily that Q=H.

I don't know if the circumcevian triangle of H is the only solution (I mean, if we don't suppose that A'B'C' is a circumcevian triangle)


calculus
integration
limit
trigonometry
real analysis
logarithms
function

Evaluate

\[\lim_{n\to\infty}\int^1_0\frac{(1+x-\cos nx)^2}{1+x^2}dx\]

I would rewrite it as

$\int_0^1\frac{(1+x)^2}{1+x^2}\,dx- 2\lim_{n\to\infty}\int_0^1\left(\frac{1+x}{1+x^2}\right) \cos(nx)\,dx + \lim_{n\to\infty}\int_0^1\frac{\cos^2(nx)}{1+x^2}\,dx$

And in the last integral, write $\cos^2(nx)=\frac{1+\cos(2nx)}2.$

Based on some other topics we've had recently, it should be clear how to proceed from there.

We need only a very, very weak version of Riemann-Lebesgue here. (No $L^1$ norms, no approximations, no appeal to deep theorems of analysis.)

Suppose that $f\in C^1[a,b].$ Then

$\int_a^bf(x)\cos(nx)\,dx=\frac{f(x)\sin(nx)}n|_a^b- \frac1n\int_a^bf'(x)\sin(nx)\,dx.$

Taking limits on the right hand side, and noting that both $|f|$ and the integral of $|f'|$ must be bounded, we see that

$\lim_{n\to\infty}\int_a^bf(x)\cos(nx)\,dx=0.$

If $f:[a,b]\rightarrow R$ continuous then  $\lim_{n\rightarrow +\infty}\int_{a}^{b}f(x)e^{inx}dx=0$

The theorem is true not just for continuous functions but also for $f\in L^1(a,b).$ The route to proving it lies in approximating $f$ in the $L^1$ norm by a "nice enough" function - $C^1$ would do as one "nice enough" possibility. Since the function in kunny's problem was already $C^1$ I chose to quote an easier version.

[quote="Kent Merryfield"]Somewhere in all of this I suppose that someone should answer kunny's question.

$\int_0^1\frac{(1+x)^2+\frac12}{1+x^2}\,dx= \int_0^1 1+\frac12\cdot\frac1{1+x^2} +\frac{2x}{1+x^2}\,dx$

$=1+\frac{\pi}8+\ln 2.$[/quote]

You are right.


The median connects one vertex of the triangle and the middle of the opposing side.

[[Median (geometry)]]

A median is one of the most common types of cevians of a triangle. Each triangle has three medians, one from each vertex to the midpoint of the opposite side. The three medians are concurrent at a point called the centroid (which is the "balancing point", which I forget the technical term for, of the triangle), which is a trivial result of Ceva's theorem.

The median to side $ BC$ of triangle $ \triangle ABC$ has length
$ \frac 12 \sqrt{2AB^2\plus{}2AC^2\minus{}BC^2}$
and by symmetry, the other two sides have similar formulas.

[quote="abacadaea"]i know what it is, but i dont get it in a problem like [url=http://www.artofproblemsolving.com/Wiki/index.php/2005_AIME_I_Problems/Problem_10]this[/url]

what line is the median?[/quote]
AM and well, the unnamed point and M, but we don't need that.

Whoops, didn't read the part about the AIME problem.

I think what he doesn't understand is
[quote]The line containing the median to side BC has slope -5.[/quote]

That simply means the line that's formed by extending the median of the triangle.


geometry
ratio
analytic geometry
conics
parabola
number theory
prime numbers

1) The equal sides of an isosceles triangle have lengths of $5x-30$ and $3x+10$. What is the greatest possible integer value of the third side?

2) After 10 tests, Nilda's average test grade was 84. Her teacher decided to drop each student's lowest and highest test grade. This brought Nilda's average up to 85. If her lowest test score (of the original 10 tests) was 24 points lower than her highest test, what was her highest test grade.

3) Jake left his house at 3 PM, and determined that he would just make it to his appointment by 4:30 PM the same day if he averaged 60 mi / hr  for the trip. After 10 minutes, his car was caught in a major traffic jam until 3:25 PM, during which his car did not move at all. What did he need to average (to the nearest mi/hr) for the rest of thr trip to just make his appointment?

4) 144 + 142 - 140 - 138 + 136 + 134 - 132 - 130 + ... + 8 + 6 - 4 - 2 = ?

5) A pattern of x's has 1 x in the first row, 2 x's in the second, ... , 50 in the 50th row, 49 in the 51st row, 48 in the 52nd row, ... , and 1 in the last row. How many x's are in the pattern?

6) What digit is in the 2004th place after the decimal point in the decimal expansion onf 1/81.

7) A rectangular solid has 2 parallel square faces, a surface area of 162 and integer dimensions. Determine all possible volumes for this solid. Disregard units.

8) The difference of the squares of 2 prime numbers is 10197. What is the sum of the squares of the 2 numbers?

9) The hypotenuse of a right triangle has length $10^{2004} + 1$ and one of its legs hsa length $10^{2004} - 1$. If the other leg has length $m(10^n)$, where m and n are integers, what is the ordered pair $(m,n)$?

10) Points D and E are on sides BA and BC, respectively, of right triangle BAC (right angle at A). DEB is a right angle, and the area of triangle DEB is 1/9 the area of triangle BAC. If the ratio of BD to BE is 3:2, what is the ratio of BD to DA?

11) Consider 100 externally tangent congruent circles with all centers collinear (on a horizontal line). The total area of the 100 circles is $25\pi$. If P is the center of the leftmost circle and Q is the center of the rightmost circle, and points A and B are on the first and last circle respectively such that AP and BQ are perpendicular to PQ, and A and B are on opposite sides of PQ,  what is the length of AB?

12) Determine the coordinates of all points on the parabola $y = \frac{1}{24}(x-3)^2 - 4$ which are at a distance of 16 units from the point $P(3,2)$.

13) Evaluate $\sum_{n=0}^{\infty}2^{2-4n}-2^{-4n}$

14) For how many positive integers n is [n/9] = [n/10], where [x] is the greatest integer  <=  x?

15) A deck of 16 cards contains 4 'A' cards numbered from 1 through 4, 4 'B' cards, 4 'C' cards and 4 'D' cards similarly numbered. A 'hand' consists of any 4 of these 16 cards, arranged in any order. A hand is said to have exactly one matching pair if it consists of exactly 2 cards with teh same letter, for example B2, B1, D2, C3 (note this is the same hand as B1, B2, C3, D2). How many different such hands are possible in this deck?

Does anybody have the solutions to these problems?


In square ABCD, E is on AB (A-B-E) such that BC=EO where O is the center of the sqaure. Find the angle measure of BEO.

Not sure but I think that if we let the length of the side of the square be $ 2$ and drop a perpendicular from $ O$ to $ AB$, meeting $ AB$ at its midpoint $ M$, then triangle $ OME$ is right-angled at $ M$, $ OM\equal{}1, OE\equal{}2$ and so $ \angle MEO \equal{} \angle BEO \equal{} 30^{\circ}$.


Nobody is interested in this? I have tried the same method used for USA 2003, but I failed. Homogenous method does not work for this because we can not turn each of the three expression on the left side into an expression of a single variable. Say, $a+b+c=3$, ....

Maybe brute force can help but I am too lazy to do this. Or if anyone can end this topics by giving a counterexample.

[quote="pvthuan"]Prove that if $ a,b,c > 0$ then
\[ \frac {(4a \plus{} b \minus{} c)^2}{2a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {(4b \plus{} c \minus{} a)^2}{2b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {(4c \plus{} a \minus{} b)^2}{2c^2 \plus{} (a \plus{} b)^2}\geq8.
\]
[/quote]
By Cauchy Schwarz Inequality, we get
\[ \sum \frac {(4a \plus{} b \minus{} c)^2}{2a^2 \plus{} (b \plus{} c)^2} \ge \sum \frac {(4a \plus{} b \minus{} c)^2}{2(a^2 \plus{} b^2 \plus{} c^2)}
\]
It suffices to prove that
\[ \sum (4a \plus{} b \minus{} c)^2 \ge 16\sum a^2
\]
which is obviously true because
\[ \sum (4a \plus{} b \minus{} c)^2 \equal{} \sum (16a^2 \plus{} 8 a(b \minus{} c) \plus{} (b \minus{} c)^2) \equal{} 16\sum a^2 \plus{} \sum (b \minus{} c)^2 \ge 16\sum a^2
\]
We have done.
:)


The area of a square inscribed in a semicircle is to the area of the square inscribed in the entire circle as:

$ \textbf{(A)}\ 1: 2 \qquad\textbf{(B)}\ 2: 3 \qquad\textbf{(C)}\ 2: 5 \qquad\textbf{(D)}\ 3: 4 \qquad\textbf{(E)}\ 3: 5$

[hide]

Let the radius of the circle be $ r$. Also, let the side length of the square inscribed in the semicircle be $ s_1$. If we draw a line connecting the center of the circle to one of the points the square touches the circumference, we get a right triangle with legs $ s_1/2$, $ s_1$ and hypotenuse $ r$. So, $ s^2_1\equal{}\frac{4r^2}{5}$.

Let the side length of the square inscribed in the whole circle be $ s_2$. If we draw the diagonal of the square, we have $ 2r\equal{}s_2\sqrt2$, so $ s_2^2\equal{}2r^2$. Hence, the ratio of of the squares is $ \frac{\frac{4r^2}{5}}{2r^2}\equal{}\frac{2}{5}$, or $ \mathrm{ (C) }$

[/hide]

Believe it or not, this question appeared again in its exact form* in the 2007 MATHCOUNTS State Sprint Round.

* Obviously, not word-for-word. It included pictures as well.


If $ 2^{3x} \equal{} k$, find $ 8^{x\plus{}1}$ in terms of $ k$.


O παίκτης $A$ και ο παίκτης $B$ παίζουν το εξής παιχνίδι:
Στην αρχική θέση του παιχιδιού υπάρχουν τρεις στείλες με κάποιο αριθμό σπίρτων η κάθεμια. Η κίνηση του κάθε παίκτη είναι να πάρει όσα σπίρτα θέλει από μία μόνο στήλη (αν θέλει και όλα τα σπίρτα της στήλης) αλλά είναι υποχρεωτικό να πάρει τουλάχιστον ένα σπίρτο. Οι παίκτες παίζουν εναλλάξ (κάνει κίνηση ο ένας και στη συνέχεια είναι η σειρά του άλλου να παίξει). Νικητής του παιχνιδιού είναι αυτός που θα πάρει και το τελευταίο σπίρτο με αποτέλεσμα ο άλλος παίκτης να μην έχει κίνηση.
Ο παίκτης $A$ έχει το πλεονέκτημα να επιλέξει πόσα σπίρτα θα περιέχει η κάθε στήλη αρχικά (θα πρέπει να υπάρχει τουλάχιστον ένα σπίρτο σε κάποια στήλη) και στη συνέχεια ο παίκτης $B$ έχει το πλεονέκτημα να επιλέξει ποιος παίκτης θα παίξει πρώτος.

Να βρεθεί αν υπάρχει στρατηγική νίκης για κάποιον από τους δύο παίκτες και αν ναι, ποιά είναι αυτή και για ποιόν παίκτη;

[b]Ένα παράδειγμα:[/b] Έστω ότι στην αρχική θέση ο $A$ επιλέγει να υπάρχει $1$ σπίρτο στην πρώτη στήλη, $2$ στη δεύτερη και $3$ στην τρίτη. Ο παίκτης $B$ στη συνέχεια επιλέγει  να παίξει πρώτος και αφαιρεί $2$ σπίρτα από την τρίτη στήλη. Ύστερα ο παίκτης $A$ αφαιρεί όλα τα σπίρτα από τη δεύτερη στήλη. Τώρα ο $B$ αφαιρεί $1$ σπίρτο από οποιαδήποτε από τις εναπομείναντες στήλες (πρώτη και τρίτη) και ο $A$ κερδίζει το παιχνίδι γιατί παίζει τελευταίος παίρνοντας και το τελευταίο σπίρτο.

Stelios...

Δεν ενδιαφέρθηκε κανείς για αυτό το πρόβλημα ή είναι τόσο δύσκολο;  :maybe:
Μήπως να δώσω ένα hint;
[hide="hint"]Για ευκολότερη αντιμετώπιση του προβλήματος μετατρέψτε τον αριθμό των σπίρτων στην κάθε στήλη σε δυαδική μορφή. [/hide]

 :play_ball:

Είναι κλασσικό πρόβλημα.

[hide]... δηλαδή γενικεύεται: Game of Nim[/hide]

Πράγματι έψαξα και το βρήκα στο internet και ως Nim, αλλά δεν βρήκα κάποια στοιχειώδη λύση έστω και για την απλή περίπτωση που έβαλα. Το μόνο που βρήκα είναι η στρατηγική νίκης χωρίς απόδειξη.
Αν κάποιος ξέρει μια στοιχειώδη λύση και μπορεί ας την γράψει.  :idea:

Γεια σε όλους!
Πολύ ωραίο παιχνίδι!  :)

Στέλιο, δε φαντάζομαι να υπάρχει λυση χωρίς τη χρήση της XOR
Το έψαξα κι εγώ λίγο και βρήκα απόδειξη στο internet  για τη στρατηγική

[hide]

Στρατηγική έχει πάντα ο παίκτης Β.
Αν το XOR άθροισμα των τριών ποσοτήτων είναι ίσο με μηδέν, τότε επιλέγει να παίξει δεύτερος. Αν δεν είναι μηδέν, τότε επιλέγει να παίξει πρώτος. Και στις δύο περιπτώσεις κερδίζει αν ακολουθήσει τους κανόνες που αναφέρονται [url=http://www.cut-the-knot.org/nim_theory.shtml]εδώ[/url], όπου υπάρχει και μία καλή απόδειξη.

Επίσης, [url=http://www.cut-the-knot.org/nim_st.shtml]εδώ[/url] μπορεί κάποιος να κάνει πρακτική εξάσκηση.

Έχοντας κατά νου αυτά που αναφέρονται στο  cut-the-knot μπορούμε να παίξουμε το παιχνίδι χωρίς να χρειαστεί χαρτί για τους υπολογισμούς. Απλώς μπορούμε νοερά να απλοποιούμε λίγο τα νούμερα (όποτε είναι δυνατόν).

Η στρατηγική που περιγράφεται στο cut-the-knot σε απλά λόγια μεταφράζεται ως εξής:
Σε κάθε στήλη χωρίζουμε τα σπίρτα κατά το δυαδικό σύστημα, δηλαδή σε 1άδες, 2άδες, 4άδες, 8άδες κλπ.
Σκοπός μας είναι να μετασχηματίσουμε τις στήλες έτσι ώστε να έχουμε:
- άρτιο πλήθος 1άδων
- άρτιο πλήθος 2άδων
- άρτιο πλήθος 4άδων
κλπ
Και αυτό να συμβαίνει σε κάθε μας κίνηση

Θεωρούμε ότι και οι δύο παίκτες παίζουν με το βέλτιστο τρόπο.
Έστω (χ,ψ,ω) η τριάδα που δηλώνει το πλήθος των σπίρτων στην κάθε στήλη

Ας δούμε πρώτα μία υπερ-απλουστευμένη περίπτωση.
Αν καταφέρουμε να "δώσουμε" στον αντίπαλο μία τριάδα της μορφής (χ,χ,0), τότε κερδίζουμε το παιχνίδι.
Πράγματι, απο κει και μετά αρκεί να κάνουμε τις ίδιες κινήσεις με τον αντίπαλο (αν πχ πάρει 1 σπίρτο από τη μία, τότε εμείς παίρνουμε 1 σπίρτο από την άλλη, κ.ο.κ.)

Δηλαδή όταν του δίνουμε δύο χ-άδες, κερδίζουμε.
Ε, κάπως έτσι είναι και με το άρτιο πλήθος των 2άδων, 4άδων κλπ.

Δηλαδή, όπως στην πρώτη περίπτωση φροντίζουμε σε κάθε μας κίνηση να του δίνουμε δύο στήλες με το ίδιο πλήθος σπίρτων, έτσι και εδώ φροντίζουμε να του δίνουμε δύο σύνολα Χ,Υ με το  ίδιο πλήθος 1άδων, 2άδων, 4άδων κλπ

Πιο συγκεκριμένα, σε κάθε κίνηση φροντίζουμε να αφαιρούμε τα σπίρτα που "περισσεύουν", αφού κάνουμε νοερά κάποιες απαλοιφές.

Αν θελήσει κάποιος, μπορώ να δώσω αριθμητικό παράδειγμα.
[/hide]

Πολύ ενδιαφέροντα links!
Μια τέτοιου είδους απόδειξη σαν αυτή που υπάρχει στο πρώτο [url=http://www.cut-the-knot.org/nim_theory.shtml]link[/url] έψαχνα να βρω...! Ευχαριστώ!  :)


[quote="MIA6"]In 2006, Linkup, a phone service provider, has more than 50 million customers worldwide and is growing at an amazing rate. One estimate from an analyst predicts linkup could have up to 245 million customers by the end of 2008. Write Linkup's worldwide number of customers in 2006 in scientific notation. What percent of increase does Linkup expect to attain by the year 2008?  I wrote: (245-50)/50, but i am not sure should i divide by 50 or 245? hope you can help, thanks.[/quote]
divide by 50

Yes, divide by 50.  The formula for percent increase is \[\text{\% increase}=\frac{\text{New}-\text{Old}}{\text{Old}}\times 100\%\]


for a given natural number $ n$, what is the size of the smallest poset in which every poset of size $ n$ can be embedded?


Two different integers are randomly selected from the set of integers greater than 0 and less than 9. What is the probability that they have no common prime factor? Express your answer as a common fraction.

The sets for each possible prime factor $ 2,3,5,7$ are $ \{2,4,6,8\}$, $ \{3,6\}$, $ \{5\}$, and $ \{7\}$, respectively. The way to get a common prime factor is to choose two numbers from one of these sets. Here, this means $ \binom42\equal{}6$ ways from the $ 2$ set or $ \binom22\equal{}1$ way from the $ 3$ set, so there are $ 7$ bad cases. The answer is $ 1\minus{}\frac7{\binom82}\equal{}1\minus{}\frac7{28}\equal{}1\minus{}\frac14\equal{}\boxed{\frac34}$.


1. Describe the relationship between speed and velocity.

2. Define force in your own words and provide an example.

3. You and a friend are pulling on a desk. How can you adjust the angle between you, the desk, and your friend to make your efforts most efficient?

[Please help me on these. I just want to clear some definitions up before I start my AP Physics class this year.]

ill leave (2) and (3) to some other posters as they will probably have better responses than me.

(1) speed is the magnitude of velocity.  speed is a scalar quantity, meaning that it has a magnitude or a numerical representation.  velocity is a vector, which has a magnitude (its speed) and a direction.  if you say something is going at 10 meters per second, that is the speed of the object.  if you say that the object is going north, that is the direction of the object.  velocity is the combined direction and magnitude of the object.

Does anyone have any ideas about the second and third questions?


If April 4 is the first day of a new project, and June 6 is the last day, for how many days did the project run?

Is this problem inclusive or exclusive?

[hide="Inclusive"]There are 30 days in April. So, the project runs for 27 days in April.  There are then 31 days in May for which the project runs.  There are then 6 days of the project in June.  So, the answer is $ \boxed{64}$days.
[/hide]

[hide="Exclusive"]
The project would then run for 26 days in April, 31 days in May, and 5 days in June.  This is $ \boxed{62}$days.
[/hide]

It would be inclusive, because those days are the first and last days of the project. I think it would only be ambiguous if it used the word "between".


i dont know if this is mathcounts level but anyways

a rocket ship is traveling through space at an initial speed of 60mph on january 1st 2000. it gains 2 mph a day.

1) what day will it reach 5000mph

2) after what year will it have reached 1 million miles

i belive number 1 is

October 6,2006.  You have to note that 2000 and 2004 are a leap year (366 days) then use arithmetic formula fof 60+2(k-1)=5000, so k is the 1971st day, if jaunary 1, 2000 is the 1st day.

second is august 28, 2002 using same idea but you have to use summation to find it instead of just the kth term.  you have to sum up to the kth term.

sorry, i think i missed something. do not take into account leap years for number 1. number 2 i dont think it matters

huh Im confused?

What do you mean dont take account leap years, then still both of my answers are only 2 days and 1 day off respectively

number 2 yes leap years matter because 2000 is a leap year.

He means don't add one extra day every 4 years, and to Abacedea, it does matter, i think for problem 2.

Yea did you read what I just said nexmus>

If you dont want to count leap years, then my orignial answers are 2 and 1 days off repectively (which i already said)...

And i already said number 2 leap years matter.

Ooops, srry i didn't see the last part of your second post.  :blush:  Srry about that Firecricket.

[quote="firecricket91"]Eiter did I sorry, but then are my answers correct?  (just take day off for number 2..)[/quote]

finish these solutions. im too lazy to figure these out:

1. easy-(5000-60)/2 days later whenever that is

2. $ 2x(x+1)/2+60x=1,000,000$. its basically 60+(60+2)+(60+4) so thats where i get the 2x(x+1)/2. they cancel each other out(the 2s, sorry im being lazy). get $ x^{2}+61x-1,000,000=0$ do the quadratic equation or completing the square. once again im too lazy to do it.


probability
combinatorics unsolved
combinatorics

Can anybody help me with the following problems:

1. A baised coin is tossed 2n times. Show that the probability that the number of heads is the same as the number of tails is $\binom{2n}{n}{(pq)}^{n}$ Find the limit of this as ${n\rightarrow \infty }$.

2. Show that

$\sum_{k=0}^{n}(-)^{k}\binom{n}{k})=0$

$\sum_{k=0}^{n}\binom{n}{k})=2^{n}$

$\sum_{k=0}^{n/2}\binom{n}{2k})={2}^{n-1}$ If n is even;

$\sum_{k=0}^{n/2}\binom{n}{k})={2}^{n-1}$ If n is even.

[quote="223rem"]

$\sum_{k=0}^{n}(-)^{k}\binom{n}{k})=0$(1)
[/quote]
Because $(1-1)^{n}=0$!!

[quote]

$\sum_{k=0}^{n}\binom{n}{k})=2^{n}$(2)
[/quote]
Because $2^{n}=(1+1)^{n}$ !

[quote]

$\sum_{k=0}^{n/2}\binom{n}{2k})={2}^{n-1}$ If n is even;
[/quote]
From (1) and (2) we have this statement!

[quote]

$\sum_{k=0}^{n/2}\binom{n}{k})={2}^{n-1}$ If n is even.[/quote]
From (2) and $\binom{n}{k}=\binom{n}{n-k}$ we have this statement!


calculus
integration
inequalities
function
calculus computations

If the function $f\colon [1,2]\to R$ is such that $\int_{1}^{2}f(x) dx=\frac{73}{24}$, then show that there exists a $x_{0}\in (1;2)$ such that
\[x_{0}^{2}<f(x_{0})<x_{0}^{3}\]

[Edit: $f$ is continuous]

[quote="olorin"]$f(x)=\{\begin{array}{cl}8&\text{for }x\in [1,{3\over 2}]\\-{23\over 12}&\text{for }x\in ({3\over 2},2] \end{array}$
Then $\int_{1}^{2}f(x)dx={73\over 24}$ with $f(x)>x^{3}$ for $x\in [1,{3\over 2}]$ and $f(x)<x^{2}$ for $x\in ({3\over 2},2]$.[/quote]

how did u get the function and further it is given in the question to prove that there exists a $x_{0}$ which satisy the conditions and in ur case that diifers for the left hand side identity nand thje right hand side

[quote="pardesi"]how did u get the function and further it is given in the question to prove that there exists a $x_{0}$ which satisy the conditions and in ur case that diifers for the left hand side identity nand thje right hand side[/quote]
I simply searched $c,d\in\mathbb{R}$, $c$ large enough, $d$ small enough
with $f(x)=\{\begin{array}{cl}c&\text{for }x\in [1,{3\over 2}]\\d&\text{for }x\in ({3\over 2},2] \end{array}$,
such that $\int_{1}^{2}f(x)dx={c+d\over 2}={73\over 24}$.
I therefore proved the problem incorrect.

Adding continuity as an hypothesis fixes it, I think.


Gauss
number theory unsolved
number theory

Show that the sum of all primitive roots modulo $ n$ is congruent to $ \mu(n\minus{}1)$ modulo n.

[quote="Payman Morteza"][quote="Mathias_DK"][quote="Payman Morteza"]Show that the sum of all primitive roots modulo $ n$ is congruent to $ \mu(n \minus{} 1)$ modulo n.[/quote]
What is $ \mu$?[/quote]
$ \mu$ is mobius function.Look here :

http://mathworld.wolfram.com/MoebiusFunction.html[/quote]
Hmm.. Then if $ n \equal{} 5$ we get $ \mu(\minus{}1) \equal{} 1$ And the primitive roots $ \bmod 5$ is $ 2,3$ so $ 2\plus{}3  \equiv \minus{}1(5\minus{}1) \bmod 5$. But that is wrong. Hm... What am I missing?

$ \mu (4) \equal{}0$, so no problem.
But for the problem itself, take a look at the PEN section, it was solved there.

[quote="ZetaX"]$ \mu (4) \equal{} 0$, so no problem.
But for the problem itself, take a look at the PEN section, it was solved there.[/quote]
Ohhhh.... Now I understand! :)
If thought: $ \mu \equal{} \minus{}1$ and then we have $ \mu(5\minus{}1) \equal{} \minus{}1 * 4$. But of course $ \mu(5\minus{}1) \equal{} \mu(4) \equal{} 0$. Thx! :)

This is a very old result due to Gauss, so I'm not sure why it's appearing on a contest.

Those that have nothing better to do (  :wink:  ) could try to evaluate $ n_2 : \equal{} \sum_{a , b \in P, \ a \leq b} ab \mod p$ (or alternatively $ s_2 : \equal{} \sum_{a,b \in P, \ a \neq b} ab \mod p$), where $ P$ denotes the primitive roots $ \mod p$.


任庆春 天津耀华中学 高三
邓煜 深圳高级中学 高二
沈才立 浙江镇海中学 高二
柳智宇 华中师大一附中 高三
甘文颖 湖北武钢三中 高三
金龙 东北师大附中 高三

:( It is said that the 7th and the 8th are Shanghaiese

[quote="xuclever"]Who are the 7th and 8th?
Just curious?

BTW, why Jiayin Kang is not in the team?
I am surprised about it.[/quote]

Because Jiayin Kang didn't attend the exam.He hasn't learnt the high school lessons because of IMO,
so he wants to studies them this year.

And it is not bad news.You see,Jiayin Kang's exit has made room for other students. :lol:

[quote="xuclever"]Who are the 7th and 8th?
Just curious?

BTW, why Jiayin Kang is not in the team?
I am surprised about it.[/quote]

It is said that the 8th is Zili Zhang and the 7th may be Zilin Jiang.
Zili Zhang is in Grade 3 and Zilin Jiang is in Grade 2.

[quote="nirvanatear"]......
牛A<康嘉引<邓煜<牛C.......[/quote]
What does it mean?I think that 邓煜 is not better than 康嘉引 now.(Maybe he will be better than 康嘉引 next year).
And 牛A and 牛C mean what?

[quote="jin"]Really? I think Zhang Anru from Hunan maybe the 7th or 8th he's been the first three before the last two problem.And Zhang Anru,Zhangzili,Jiangzilin were in the same room with me.[/quote]
Maybe……
Zhang Zili said that he was 8th.Jiang Zilin said that his teacher told him that his score was highest in Shanghai team.
And they are all in Shanghai team…………

[quote="jin"]Really? I think Zhang Anru from Hunan maybe the 7th or 8th he's been the first three before the last two problem.And Zhang Anru,Zhangzili,Jiangzilin were in the same room with me.[/quote]
Congratulationgs, Jin!

[quote="jin"]Thank you![/quote]
Hi Jin,
which university will you enter?
what about the other 5 guys?

Congratulations, Jin!

Good luck in IMO.

I remember that all of us will go to pku

wow...
congrats!!!
it's the chinese national team man...
best team theres ever been on this planet!

好像集训队的很多人都去北大了

各位好强
祝贺Jin  本人最近考TOEFL  好久没来这里了
在Slovenia 好好表现 好好玩  祝福你

[quote="jin"]好像集训队的很多人都去北大了[/quote]

PKU.... ;)

也有几个人去了清华,念基科班,不过好像基科班现在不是很好.并且主要还是学物理的占优势.

大家好!

我是英国队的一个学生。

下个星期两三天我去北京看你们怎么练习。我陪两个拍电视的人，他们在做一个电影为了IMO。

我发现这里有中国队的几位，所以介绍我自己一点。

又开心下个星期可以看到你们那么聪明的人！

我的中文是在上海学的，这里有我的照片。。。

http://imo2006.dmfa.si/participants/GBR.html   - contestant 4. :)

[quote="Skyward_Sea"][quote="shobber"][quote="Skyward_Sea"]本人最近考TOEFL[/quote]
How was your score?  是不是还要等六周?[/quote]
成绩出来了 考得很好 我可能会忙出国而不是竞赛了[/quote]
多少分啊？

[quote="shobber"][/quote][quote="Skyward_Sea"][quote="shobber"][quote="Skyward_Sea"]本人最近考TOEFL[/quote]
How was your score?  是不是还要等六周?[/quote]
成绩出来了 考得很好 我可能会忙出国而不是竞赛了[/quote][quote="shobber"]
多少分啊？[/quote]
653+4.5

wow! I can't imagine people from the chinese IMO people accutally post here!
I'm soo proud to be a Chinese becuase we always win the math things (except i'm pretty bad at it)
The only math contest i took in China when i still lived there was the school wide contests and Hua Lou Gen

北大还是个很好的地方，很多学长都去了

wodediannaodabuchuzhongwen... buguonimendouhaolihaia! woyecuanshishanghairenlexixi. Yep, it's pretty hard 2 read. But I just said you guys were smart and that I'm sort of Shanghainese too. But I'm not in high school yet. :(


algebra
system of equations
algebra proposed

If $ A$ is nonzero integer, solve the following system of equations:

$ x\plus{}y^{2}\plus{}z^{3}\equal{}A$

$ \frac{1}{x}\plus{}\frac{1}{y^{2}}\plus{}\frac{1}{z^{3}}\equal{}\frac{1}{A}$

$ xy^{2}z^{3}\equal{}A^{2}$.

If we let $ x\equal{}a, y^{2}\equal{}b, z^{3}\equal{}c$ then the system becomes:
$ a\plus{}b\plus{}c\equal{}A$ (*)
$ abc\equal{}A^{2}$ (**)
$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{}\frac{1}{A}$ (***)
Clearing the denominators of (***) we get:
$ ab\plus{}bc\plus{}ca\equal{}\frac{abc}{A}\equal{}\frac{A^{2}}{A}\equal{}A$ (substituting from (**)).
Therefore, $ a\plus{}b\plus{}c\equal{}A, ab\plus{}bc\plus{}ca\equal{}A, abc\equal{}A^{2}$ so $ a,b,c$ are the roots of the following cubic equation:
$ m^{3}\minus{}Am^{2}\plus{}Am\minus{}A^{2}\equal{}0$. we can factor it to get:
$ (m\minus{}A)(m^{2}\plus{}A)\equal{}0$
Then the roots of the equation are $ A, i\sqrt{A},\minus{}i\sqrt{A}$ where $ i^{2}\equal{}\minus{}1$
Then, we just take any permutation of $ \{x,y^{2},z^{3}\}$, set it equal to $ \{A,i\sqrt{A},\minus{}i\sqrt{A}\}$ and for each of the $ 6$ permutations get a solution $ (x,y,z)$.