function
limit
trigonometry
probability
floor function
real analysis
real analysis unsolved

I cannot solve the following problem. Calculate
\[ \mathop{\lim }_{n\to\plus{}\infty }{\text{cos}}^{n}\left({2\pi n!x}\right)\quad ,\quad x\notin\mathbb{Q}.\]
Please help me.

I can say that zero is a limit point for almost all $ x$, but I also believe the sequence diverges for almost all $ x$. It certainly depends on the exact value; we have convergence to 1 if $ x\equal{}e$.

I think there's a very good chance you're asking the wrong question.

I believe that, if we observe your limit for $ x\in\mathbb{R}$, the you limit equals $ 1_{\mathbb{Q}}$, but I'll get back with explanation later, but it should go like this, let $ x \equal{}\frac{p}{q}\in\mathbb{Q}$, then there after some $ n\in\mathbb{N}$, we would have $ n!\frac{p}{q}\in\mathbb{Z}$, so we get $ 2\pi\cdot z,z\in\mathbb{Z}$, so you limit then equals $ 1$, (so that is in line with $ 1_{\mathbb{Q}}$) again if $ x\in\mathbb{R}\backslash\mathbb{Q}$, then $ \cos{(something)}< 1$ so you limit is $ 0$, what is again in line with $ 1_{\mathbb{Q}}$.
Not exactly the best explanation but as I said I'll get back with a better explanation.

A family of examples which shows the variety:

Let $ x \equal{} a_{0}\plus{}\sum_{n \equal{} 1}^{\infty}\frac{a_{n}}{n!}$, where $ \minus{}\frac{n}{2}\le a_{n}\le\frac{n}{2}$ for all $ n > 0$, and the $ a_{n}$ are integers. The fractional part of $ (n\minus{}1)!x$ is $ \frac{a_{n}}{n}\plus{}O(\frac{1}{n})$.
The limit in this case is $ \exp\left(\minus{}\frac{1}{2}\lim_{n\to\infty}\frac{a_{n}^{2}}{n}\right)$, which can be anything in $ [0,1]$, and fails to exist if that limit doesn't exist.

Every real number can be represented in this form; just use the sequence of closest approximations by numbers of the form $ \frac{m}{n!}$.

This can also be used to prove the assertion in my first post. If we choose a real number at random from $ [0,1]$ uniformly, the $ a_{n}$ are distributed independently and almost uniformly on their possible ranges (if $ \frac{n}{2}$ is an integer, $ \pm\frac{n}{2}$ get half weight).
If the sequence converges, there is some $ m$ for which either $ \frac{a_{n}^{2}}{n}>1$ for all $ n>m$ or $ \frac{a_{n}^{2}}{n}<2$ for all $ n>m$. For any particular $ m$, the probability of the first is $ \prod_{n>m}1\minus{}\frac{2\lfloor\sqrt{n}\rfloor\plus{}1}{n}\equal{}0$ and the probability of the second is $ \prod_{n>m}\frac{2\lfloor 2\sqrt{n}\rfloor\plus{}1}{n}\equal{}0$. These events happen with zero probability; by countable (sub)additivity, the probability of either happening for any $ m$ is still zero, and thus the sequence diverges with probability 1.


geometry
geometric transformation
homothety
geometry proposed

Let $\Gamma$ be a circle, and let $ABCD$ be a square lying inside the circle $\Gamma$. Let $\mathcal{C}_a$ be a circle tangent interiorly  to $\Gamma$, and also tangent to the sides $AB$ and $AD$ of the square, and also lying inside the opposite angle of $\angle BAD$. Let $A'$ be the tangency point of the two circles. Define similarly the circles $\mathcal{C}_b$, $\mathcal{C}_c$, $\mathcal{C}_d$ and the points $B',C',D'$ respectively.

Prove that the lines $AA'$, $BB'$, $CC'$ and $DD'$ are concurrent.

I think I solved it, but I used some projective transformations. Here's a hint (and, at the same time, an extension of the text of the problem):

Try to consider just two circles ($C_a$ and $C_b$, for example), and try to show that the center of the large circle, the intersection of $AA'$ and $BB'$ and the center of the considered square are collinear.

This would solve the problem, and it would say something more about the intersection pt of $AA'$ and $BB'$ (and $CC'$ and $DD'$, of course). It really is a nice problem! It's just the kind of problem which makes me feel sorry for not being a contestant in the TST :).

dear Grobber,could you post your solution (I 'm not good at projective transformation)? :blush:

Let's introduce some notations:

Let $O_a$ and $O_b$ be the centers of $C_a$ and $C_b$ respectively. Assume these two circles touch $AB$ at $A_1$ and $B_1$. Let $O$ be the center of the large circle and let $P$ be the center of $ABCD$. Assume $OA'$ and $OB'$ cut $AB$ at $A_2$ and $B_2$ respectively. Let $X$ be $A'B' \cap AB$.

It's easy to see that $X$ is the external homothety center of $C_a$ and $C_b$, so it also lies on $O_aO_b$. This means that the lines $A'B'$, $O_aO_b$, $A_2B_2$ and $OX$ are concurrent in $X$, so $(A',A_2;O_a,O)=(B',B_2;O_b,O) \ (*)$. Consider the homographic transformation from the pencil of lines through $A$ to the pencil of lines through $B$ which maps $AA'\rightarrow BB'$, $AO_a\rightarrow BO_b$ and $AO\rightarrow BO$. From $(*)$ we find that it also maps $AB=AA_2\rightarrow BB_2=AB$, so it has a fixed element, and it's well-known result that such transformations are perspectivities (at least that's what they're called in Romanian :)), i.e. the locus of the intersection points of corresponding lines is a line, so $O=AO\cap BO$, $P=AO_a\cap BO_b$, $AA'\cap BB'$ are collinear, and that's exactly what we wanted.

I know that there are some much simpler solutions (a contestant showed me something using Menelaus, for example), but I like this idea.

Well, I think my solution is more beautiful.
You see M and N and P and Q are the points on the circle that tangents from them to circle are respectively parallel to AB,BC,CD,DA.
The tangents from Q and M to intersect at A".
The tangents from M and N to intersect at B".
The tangents from N and P to intersect at C".
The tangents from P and Q to intersect at D".
Homothety from A' takes A to A".So AA' goes from A".
So we must prove AA",BB",CC" and DD" are concurrent.
But this is obvious because sides of ABCD and A"B"C"D" are parallel.
And so theses lines are concurrent on a line connecting center of square and circle.
(Center of A"B"C"D" is center of circle)
____________________________________________
No pain ,no gain.

I heard of a solution given by a contestant which used homothety and this might be it. It is indeed nicer than mine :).

Hi guys  :)
In my solution I havn't used that $ABCD$ is a square :blush: ,this problem is true when $ABCD$ is just a circumscribed quadrilateral ;) .
So try this problem now,I thing now it is more interesting :) .

[quote="Tiks"]...this problem is true when $ABCD$ is just a circumscribed quadrilateral...[/quote]
This is like telling the solution. Let $(I)$ be the incircle of the quadrilateral $ABCD.$ $A$ is the internal homothety center of the circles $\mathcal C_a, (I)$ and $A'$ is the external homothety center of the circles $\mathcal C_a, \Gamma.$ Therefore, the internal homothety center $H$ of the circles $\Gamma, (I)$ lies on $AA'.$ Similarly, $H$ lies on $BB', CC', DD',$ so that it is the concurrency point.

It suffices to prove that $AA', BB', CC'$ are concurrent. By Desargues' Theorem, we just need to show that $AB \cap A'B', BC \cap B'C', CA \cap C'A'$ are collinear. By Monge's Theorem, $AB \cap A'B'$ is the exsimilicenter of $\mathcal{C}_a, \mathcal{C}_b$ and $BC \cap B'C'$ is the exsimilicenter of $\mathcal{C}_b, \mathcal{C}_c.$ Let $\omega$ be the incircle of $ABCD.$ By Monge's Theorem on $\omega, \mathcal{C}_a, \mathcal{C}_c$, we know that $AC$ goes through the exsimilicenter of $\mathcal{C}_a, \mathcal{C}_c.$ By Monge's Theorem on $\Gamma, \mathcal{C}_a, \mathcal{C}_c$, we know that $A'C'$ also does. Hence, $AC \cap A'C'$ is the exsimilicenter of $\mathcal{C}_a, \mathcal{C}_c$, and Monge's Theorem on $\mathcal{C}_a, \mathcal{C}_b, \mathcal{C}_c$ finishes.

$\square$


A farmer has previously fenced a field in the shape of a trapezoid as shown.  The farmer divided this field into exactly four congruent trapezoids.  How many additional meters of fencing are needed? Express you answer to the nearest whole number.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=519[/img]

[hide] Because the triangle has an area of 2400, each individual triangle must have an area of 600. Because the dimensions of the trapezoid include 40's and and 80, it'd be safe to assume that the bases of each trapezoid are 20 and 40. That would also make the height 20. Now it's simply a matter of how to arrange the trapezoids. It's obvious that the trapezoids are in the shape of the original big trapezoid. It's also obvious that the diagonal for two of the trapezoids must land on the diagonal of the bigger trapezoid. The new trapezoid looks like the figure below. You get 80 + 20*sqrt.2 and that adds up to about [b]108[/b] [/hide]


Two candles have different lengths and thicknesses. The long one can burn $ 3\frac{1}{2}$ hours; the short one, 5 hours.
After 2 hours burning, the candles are equal in  length. Two hours ago, what fraction of the long candle's height gave the short candle's height ?

[size=75][i]from book "The Moscow puzzles" by Boris A. Kordemsky

Please hide Your solution . Write inside Hide ... /Hide[/i]
[/size]
 :moose:

Let x and y be the length of the long and short candles respectively. After 2 hours, $ 2\div3\frac{1}{3}\equal{}\frac{4}{7}x$ has burned, and $ \frac{2}{5}y$ has burned. This leaves $ \frac{3}{7}x\equal{}\frac{3}{5}y$. Thus the short candle had a height that was $ \boxed{\frac{5}{7}}$ of the long one.

[quote="$ LaTeX$"]Let x and y be the length of the long and short candles respectively. After 2 hours, $ 2\div3\frac {1}{3} \equal{} \frac {4}{7}x$ has burned, and $ \frac {2}{5}y$ has burned. This leaves $ \frac {3}{7}x \equal{} \frac {3}{5}y$. Thus the short candle had a height that was $ \boxed{\frac {5}{7}}$ of the long one.[/quote]

Typel $ 2\div3\frac {1}{3} \equal{} \frac {4}{7}x$ . Should be $ 2\div3\frac {1}{2} \equal{} \frac {4}{7}x$
 :maybe:


Find the ordered pair of rational numbers $(x,y)$ for which $4^{\log_{16} 27}=2^x3^y$.

[hide="Hint"]Don't make the same careless mistake I made at first by taking the $\log_4$ of both sides and messing up.  ;) [/hide]

[quote="4everwise"]Don't make the same careless mistake I made at first by taking the $\log_4$ of both sides and messing up.  ;) [/quote]
Why not? :)
$4^{\log_{16} 27}=2^x3^y$
$\log_4 4^{\log_{16} 27}=\log_4 2^x3^y$
$\log_{16} 27=\log_4 2^x3^y$
Since $\log_a b=\log_{a^n} b^n$, we have $\log_4 2^x3^y=\log_{16} 2^{2x}3^{2y}$
Therefore $27=2^{2x}3^{2y}$ and so $(x,y)=(0,\frac{3}{2})$


abstract algebra
algebra
polynomial
quadratics
superior algebra
superior algebra unsolved

Let y be an integer, which is not a square of another integer.

How do I prove that Q[sqrt(y)] = {a + bsqrt(y) : a,b in Q} is isomorphic to Q[x]/(x^2 - y) where Q is the set of rational numbers?

Thanks for your help.

Do you know the first isomorphism theorem?  What is the kernel of the surjective homomorphism $ f: \mathbb{Q}[x] \to \mathbb{Q}[\sqrt{y}]$ defined by $ f(p(x)) \equal{} p(\sqrt{y})$?

Okay, I get this part. How do I use this to prove that Q[sqrt(2)] is not isomorphic to Q[sqrt(3)]? What property of a ring fails?

HINT  If $ \alpha, \alpha' \notin \mathbb{Q}$ are conjugate then their images under a hom remain so,
because the hom preserves their minimal polynomial $ (x \minus{} \alpha)(x \minus{} \alpha') \in \mathbb{Q}[x]$.
But $ 2|(x \minus{} x')^2$ in $ \mathbb{Z}[\sqrt {2}]$ versus $ 3|(x \minus{} x')^2$ in $ \mathbb{Z}[\sqrt {3}]$.

Hence the name "discriminant", i.e. it discriminates among quadratic number field
isomorphism classes, as you will learn later in your study of algebraic number theory.

Alternately, the polynomial $ x^2 \minus{} 2$ splits in $ \mathbb{Q}[ \sqrt{2} ]$ but is irreducible in $ \mathbb{Q}[ \sqrt{3} ]$.

Or the prime $ 7$ splits in the first, but not in the second ;)
Well, overkill solutions apart, the principal idea is that $ 2$ is a square in one, but not the other; and this can be shown diretly using calculation and irrationality of $ \sqrt 3$.


quadratics algebra polynomial calculus integration

Since quadratic integers of the form $a+b\sqrt{c}$ occur in quadratic equations in conjugates, is it impossible for an integer cubic equation to have 3 quadratic integers as roots?

In general, do roots of the form $a+b\sqrt{c}$ have to occur in conjugate pairs in integer polynomials?

[quote="breez"]quadratic integers of the form $a+b\sqrt{c}$[/quote]

You mean quadratic roots, right?

Anyway, it is impossible for a polynomial with integral coefficients to have 2 roots of the form $a+b\sqrt{c}$, because they do have to occur in conjugates.

But irrationals don't necessarily have to appear in conjugate pairs right? Just quadratic integer roots in integer polynomials.

[quote="breez"]In general, do roots of the form $a+b\sqrt{c}$ have to occur in conjugate pairs in integer polynomials?[/quote]

Yes.  We can prove this with a very simple lemma.  Let $\overline{a+b \sqrt{c}}= a-b \sqrt{c}$.

[b]Lemma:[/b]   Given a polynomial $P(x)$ with integer coefficients, $P( \bar{x}) = \overline{ P(x) }$ for any $x \in \mathbb{Z}[ \sqrt{c}]$.

This is certainly not true of cubic roots, for example.

Edit:  Alternately, we can introduce the concept of a "minimal polynomial."

[b]Definition:[/b]  The minimal polynomial $m(x)$ of some member $r$ of a field extension of $\mathbb{Q}$ is the monic rational polynomial of smallest degree such that $m(r) = 0$.

Then we make two claims.

[b]Theorem:[/b]  Let $r$ be a root of some rational polynomial $p(x)$.  Then its minimal polynomial divides $p$.

[b]Theorem:[/b]  The minimal polynomial of $a+b \sqrt{c}$ is $P(x) = (x-(a+b \sqrt{c}))(x-(a-b \sqrt{c}))$.


I wanted to go back over all the puzzles with my dad.  Are there any readily accessible solutions?   :oops: I forgot everything.

Yup, the solutions are up on the website along with the puzzles:

http://www.mathcamp.org/hunt/2005/

Dave


geometry
perimeter
combinatorics proposed
combinatorics

Unit square $ABCD$ is divided into $10^{12}$ smaller squares (not necessarily equal). Prove that the sum of perimeters of all the smaller squares having common points with diagonal $AC$ does not exceed 1500.

[i]Proposed by A. Kanel-Belov[/i]

I believe the follow process of division of $ ABCD $ into $ n$  ($ {n}\equiv1\bmod 3 $) smaller squares, will give, for each $ n $, the maximum sum of the perimeters of smaller squares  that intersect $ AC $:

1) Divide $ ABCD $ into 4 equal squares
2) Divide the smaller square which has $ A $ as vertex into 4 equal squares
3) Divide the smaller square which has $ C $ as vertex into 4 equal squares
4) etc

In each process we will divide one of the smaller squares, whose main diagonal belongs to $ AC $ into 4 equal squares. Actually this division will be done in the direction $ A $ to $ C $ and after dividing the smaller square which has $ C $ as one vertex, we will  start the process again, dividing the smaller square which has $ A $ as a vertex into 4 equal squares.

Note that in this way all squares intersect $ AC $ in at least one point.

Does anyone know if the process above gives the maximum sum and how to prove this?

In the positive case the bound 1500 can be improved.


conics
ellipse
geometry
analytic geometry
graphing lines
slope
geometry unsolved

Lines $k$ and $l$ (intersecting in $A$) are tangent to ellipse $\xi$ in points $B$ and $C$. Let $F$ and $G$ be focuses of $\xi$. Let $D$ be the projection of $A$ on $FG$. Prove that
\[\angle BDA=\angle CDA !!! \]

Let YZ be the ellipse major axis and P the midpoint of YZ. Let (P) be a circle with center P and diameter YZ. Let normals to YZ from B, C meet (P) at B', C' on the same side of YZ as B, C. WLOG, assume that Y, C', B' Z follow on (P) in this order. Let YC', ZB' meet at X. Since YZ is a diameter of (P), $YB' \perp ZB' \equiv ZX,$ $ZC' \perp YC' \equiv YX,$ and YB', ZC' are the Y-, Z-altitudes of the $\triangle XYZ$ intersecting at its orthocenter H. $XH \perp YZ$ is the X-altitude, let $D \in YZ$ be its foot. Let (N) be the 9-point circle of the $\triangle XYZ$ passing through D. Let (N) meet XD again at A'. Since $P \in (N)$ and $A'D \perp YZ \equiv PD,$ PA' is a diameter of (N). Since $B',\ C' \in (N)$ and PA' is a diameter of (N), $PB' \perp B'A',\ PC' \perp C'A'.$ Since PB', PC' are radii of (P), B'A', C'A' are tangents of (P). $\triangle DB'C'$ is the orthic triangle of the $\triangle XYZ,$ hence $DA' \equiv DX$ bisects the angle $\angle B'DC'.$ Using a parallel projection, project the circle (P) into the ellipse $\xi.$ The line $XY \equiv FG$ and the point $D \in XY$ stay in place. The tangents B'A', C'A' of (P) go into the tangents BA, CA of $\xi.$ The line $DX \equiv DA' \perp XY$ is projected to itself, so that DA remains perpendicular to $XY \equiv FG.$ Since DA' bisects the angle $\angle B'DC',$ DB', DC' are symmetrical with respect to DA', and it follows that DB, DC are symmetrical with respect to DA, hence DA bisects the angle $\angle BDC.$

[quote="derezins"][color=darkred]Lines $k$ and $l$ (intersecting in $A$) are tangent to ellipse $\xi$ in points $B$ and $C$. Let $F$ and $G$ be focuses of $\xi$. Let $D$ be the projection of $A$ on $FG$. Prove that $\widehat{ADB}\equiv\widehat{ADC}\ .$[/color][/quote]
[color=darkblue][b][u]An analytical proof.[/u][/b] Suppose w.l.o.g. that the equation of the ellipse $\xi$ is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\ .$
Let $A,B$ be two mobile points which belong to the given ellipse, i.e.

$\{\begin{array}{ccc}B(am,bn) & , & \boxed{m^{2}+n^{2}=1}\\\\ C(ap,br) & , & \boxed{p^{2}+r^{2}=1}\end{array}\|\Longrightarrow\{\begin{array}{ccc}\mathrm{the\ tangent\ }BB & : & bmx+any-ab=0\\\\ \mathrm{the\ tangent\ }CC & : & bpx+ary-ab=0\end{array}\|\Longrightarrow$

$A[\frac{a(r-n)}{mr-np}\ ,\ \frac{b(m-p)}{mr-np}]\Longrightarrow\{\begin{array}{c}\mathrm{\ the\ slope\ of\ }DB\mathrm{\ is\ }s(DB)=\frac{b}{a}\cdot (mr-np)\cdot \frac{n}{m(mr-np)+n-r}\\\\ \mathrm{the\ slope\ of\ }DC\mathrm{\ is\ }s(DC)=\frac{b}{a}\cdot (mr-np)\cdot \frac{r}{p(mr-np)+n-r}\end{array}\ .$

Therefore, $s(DB)+s(DC)=$ $\boxed * \cdot [(mr-np)(mr+np)+(n-r)(n+r)]=$

$\boxed * \cdot [r^{2}(m^{2}-1)+n^{2}(1-p^{2})]=$ $\boxed * \cdot (-r^{2}n^{2}+r^{2}n^{2})=0\ ,$

where $\boxed * =\frac{b\cdot(mr-np)}{a\cdot [p(mr-np)+n-r]\cdot [m(mr-np)+n-r]}\ .$

In conclusion, $s(DB)+s(DC)=0\ ,$ i.e. $\widehat{ADB}\equiv\widehat{ADC}\ .$

[b][u]Remark.[/u][/b] You can prove easily if the ellipse $\xi$ is a circle $c=C(O)$ and

the point $D$ is the projection of the mobile point $A$ to the fixed diameter $[XY]\ .$

Indeed, the points $B,C,D$ belong to the circle with the diameter $OA$ and

$\widehat{ABC}\equiv\widehat{ACB}\ ,\ \widehat{ABC}\equiv\widehat{ADC}\ ,\ \widehat{ACB}\equiv\widehat{ADB}$ $\Longrightarrow$ $\widehat{ADB}\equiv\widehat{ADC}\ .$

Then you can project on an any plane with the common line $XY$ ([u]the nice Yetti's proof[/u]).[/color]


A digital watch displays hours and minutes with $ \text c{AM}$ and $ \text c{PM}$. What is the largest possible sum of the digits in the display?

$ \textbf{(A) } 17\qquad \textbf{(B) } 19\qquad \textbf{(C) } 21\qquad \textbf{(D) } 22\qquad \textbf{(E) } 23$

yeah RFT has been bumping a lot of these threads, I've seen 7 or 8 by now :)

[quote=rohan.sp]Hi, RFT can bump these posts as long as he posts a solution.

Thanks,
rohan.sp[/quote]

I mean yeah his solutions are also really good tho

but it clutters the forum and all of the 'newer' posts get cluttered at the bottom

[quote=pith0n]How was this an AMC 12?[/quote]

It's AMC 12A 2006 Problem 4, normally, the earlier problems of earlier years are easier.

In military time it would actually be 19:59  for an answer of 24 but this question asks for normal 12 hour time...


[img]http://www.part.lt/img/3d3ecacd0e147b142cc1617b599b43a65.JPG[/img]

$ \angle{XX_1X_2} \equal{} \angle{XX_1X_3}$ also $ \angle{XX_2X_1} \equal{} \angle{XX_2X_3}$ also $ \angle{XX_3X_1} \equal{} \angle{XX_3X_2}$

Prove that point X is the ortocenter of triangle ABC.

X is center of inscribed circle of $ X_1X_2X_3$.
Let's call angles at $ X_1,X_2,X_3$, 2x,2y and 2z, and angles at A,B,C we call a,b,c
We find that x+z=b,x+y=c,z+y=a
Solving we get z=90-c
Quadrilateral $ AX_2XX_3$ is cyclic, so $ \angle X_2AX\equal{}\angle X_2X_3X\equal{}z$
Let AX intersect BC at M. Then angles in ACM are: z,c,$ \angle AMC$. But c+z=90, so $ AX\bot BC$. Analogous we get X is the ortocenter.

[quote="Bugi"]Quadrilateral $ AX_2XX_3$ is cyclic, so $ \angle X_2AX \equal{} \angle X_2X_3X \equal{} z$[/quote]

Wow  :)

I wasn't clever enough to find  $ \angle{X_2AX}$

Let $ \angle X_2AX\equal{}\angle X_1BX\equal{}x;\angle X_2CX\equal{}\angle X_3BX\equal{}y;\angle X_3AX\equal{}\angle X_1CX\equal{}z$
Since then we have $ x\plus{}y\equal{}B,y\plus{}z\equal{}C,z\plus{}x\equal{}A$
Sum up to get $ x\plus{}y\plus{}z\equal{}90$ or $ z\plus{}B\equal{}90$
Which implies $ AX\perp BC$.
Similarly, we also have $ BX\perp AC$ and $ CY\perp AB$.
Therefore X is the ortocenter of triangle ABC.


function
algebra
domain
functional equation
algebra unsolved

Find all function $ f: \mathbb{Z}\rightarrow\mathbb{Z}$ such that $ f(x\plus{}f(y))\equal{}f(x)\minus{}y$ for all $ x,y\in\mathbb{Z}$

There is no such function.
See pb #25 here :
http://www.animath.fr/cours/eqfonc.pdf

Pierre.

Here is my solution:

1. Let $ f(0) = c$. From $ x = y = 0$ in our equation we have $ f(c) = c$.

From $ x = c,y = 0$ we have $ f(2c) = c$.

From $ x = y = c$ we have $ f(2c) = 0$.

Hence $ c = 0\implies f(0) = 0$. Thus from $ x = 0,y = t$ we get $ \boxed{f(f(t)) = - t\,\forall t\in\mathbb Z}$.

2. Let $ f(a) = f(b)$. Then
\begin{eqnarray*} & & f(a + f(a)) = f(a + f(b)) \\
& \implies & f(a) - b = f(a) - a \\
& \implies & a = b\end{eqnarray*}
Thus $ f$ is injective.

3. Assume that $ f(m) = n$. Then
\begin{eqnarray*} & & f(m) - n = 0 \\
& \implies & f(m + f(n)) = 0 = f(0) \\
& \implies & m + f(n) = 0 \\
& \implies & m = - f(n)\end{eqnarray*}
Thus $ f$ is surjective.

4. Let $ x = f(a),y = b$. Then we have
\begin{eqnarray*} & & f(x + f(y)) = f(y) - x \\
& \implies & f(f(a) + f(b)) = f(f(a)) - b = - (a + b) = f(f(a + b)) \\
& \implies & f(a + b) = f(a) + f(b)\,\forall a,b\in\mathbb Z\end{eqnarray*}
This is Cauchy's equation with solution $ f(x) = mx$. Substituting into our equation we get $ m(x + my) = mx - y\implies m^2y = - y\implies m = i\implies f(x) = ix$, which is impossible as $ f: \mathbb Z\to\mathbb Z$.

Thus no such function exists.

we take $ a \equal{} \minus{} f(0)$:we have $ y \equal{} 0$ gives $ f(x \plus{} a) \equal{} f(x)$ then $ f(x \plus{} f(y)) \equal{} f(x \plus{} f(y \plus{} a))$ ==> $ f(x) \minus{} y \equal{} f(x) \minus{} y \minus{} a$ ==> $ f(0) \equal{} \minus{} a \equal{} 0$
$ x \equal{} 0$ gives $ fof(y) \equal{} \minus{} y$ then $ f(x \minus{} y) \equal{} f(x \plus{} fof(y)) \equal{} f(x) \minus{} f(y)$ so $ y \equal{} \minus{} 1$ gives $ f(x \plus{} 1) \equal{} f(x) \minus{} f( \minus{} 1)$
then $ \forall x\in\mathbb{Z}: \ f(x) \equal{} xf( \minus{} 1)$ and $ \minus{} 1 \equal{} fof(1) \equal{} f( \minus{} 1)^2\ impossible$

Actually, it's been posted on this forum.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150758

I think this problem generalizes to real numbers too, right?

I have a question: If it was given that $ f: \mathbb C\to\mathbb C$, then wouldn't $ f(x) \equal{} ix$ be the only solution? Actually I never handled a functional equation in complex domain before, so I'm a bit confused.. :blush:

I haven't done any complex ones either, but I can't think of any special nuances to them. Anyways... from $ f(0) \equal{} 0, f(f(x)) \equal{} \minus{}x$:

Notice that $ f(\minus{}x) \equal{} f(f(f(x))) \equal{} \minus{}f(x)$, so $ f$ is odd.

Put $ y \equal{} \minus{}f(y)$, and we get $ f(x\minus{}f(f(y))) \equal{} f(x) \minus{} (\minus{}f(y))$, or $ f(x\plus{}y) \equal{} f(x) \plus{} f(y)$.

For $ \mathbb{R}$, there's been a topic about the solutions to $ f(f(x)) \equal{} \minus{}x$ (all of which aren't continuous), and I don't know if any of them are also solution to the Cauchy Equation.

For $ \mathbb{C}$, I don't know how Cauchy's equation generalizes to complex numbers.


logarithms
calculus
derivative
induction
algebra unsolved
algebra

Let $c_1,c_2,...,c_n$ be real numbers such that $c_1^k+c_2^k+...+c_n^k > 0$ for all $k=1,2,...$

Let's put $f(x)=\frac{1}{(1-c_1x)...(1-c_nx)}$

Show that $f^{(k)}(0)>0$ for all $k=1,2,...$

ps. Unfortunatelly, there is no olimpyc analysis section, so I place this post here.

http://www.artofproblemsolving.com/Forum/index.php?f=84

Here the place for this contest

Here is the solution I got from theyre site:
Put [tex]g ( x ) = \log f ( x )[/tex]. Then

[tex]g ( x ) = - \sum_{j = 1}^n l \tmop{og} ( 1 - c_j x ) \text{\quad and \quad}
g^{( k )} ( x ) = \sum_{j = 1}^n \frac{c_j^k}{( 1 - c_j x )^k}[/tex]

The assumption [tex]( P )[/tex] gives

[tex]g^k ( 0 ) = c_1^k + c_2^k + \cdots + c_n^k > 0[/tex]

Now observe that if all the derivatives of [tex]g[/tex] at the origin are positive,
then [tex]e^g[/tex] has the same property. For this show by the induction that

[tex]\left( e^g \right)^{( k )} = e^g \cdot S[/tex]

and [tex]S[/tex] is a finite sum of terms of the form

[tex]a \left( g^{( l_1 )} \right)^{m_1} \left( g^{( l_2 )} \right)^{m_2} \cdots
\left( g^{( l_r )} \right)^{m_r}[/tex]

where [tex]a, l_j[/tex] and [tex]m_j[/tex] are positive integers. For instance:

[tex]( e^{g ( x )})' = e^{g ( x )} g' ( x )[/tex]

[tex]( e^{g ( x )})'' = e^{g ( x )}( g' ( x )^2 + g'' ( x ))[/tex]

[tex]( e^{g ( x )})''' = e^{g ( x )} ( g' ( x )^3 + 3 g' ( x ) g''( x ) + g''' ( x ))[/tex]

[tex]( e^{g ( x )})^{( 4 )}=e^{g ( x )} ( g' ( x )^4 + 6g' ( x )^2 g'' ( x ) + 3 g'' ( x )^2 + 4 g' ( x ) g''' ( x ) + g^{( 4 )} ( x ))[/tex]

Wow, I didn't think it would be that easy, thanks for it cyberjoe.


12. Hydrochloric acid 0.1 moles per dm$^{3}$ is added bit by bit into a flask that contains 20.0 cm$^{3}$ sodium hydroxide solution 0.1 moles per dm$^{3}$ and 10.0 cm$^{3}$ of water. What is the total volume of the mixture in the flask when the solution becomes neutral?

A. 30.0 cm$^{3}$
B. 40.0 cm$^{3}$
C. 50.0 cm$^{3}$
D. 60.0 cm$^{3}$

13. When a piece of iron plated with metal X is scratched, the iron does not rust, but when a piece of iron plated with metal Y is scratched, the iron rusts quickly. From the above observations, we can conclude that the arrangement of the metals X, Y and iron in order of decreasing electropositivity is:

A. X, iron, Y
B. iron, X, Y
C. Y, X, iron
D. Y, iron, X

15. When a white solid Q is heated, a brown gas is released. When distilled water is added to white solid Q, an aqueous solution of Q is formed. When ammonia solution is added to the aqueous solution of Q, a white precipitate, soluble in excess ammonia solution, is formed.  Q is most likely to be:

A. calcium nitrate
B. zinc nitrate
C. aluminium nitrate
D. lead (II) nitrate

[hide="12"]There is 20$cm^{3}$ of NaOH; and exactly that amount of HCl is needed to neutralize the solution since they both are 0.1 mol per cubic $dm^{3}$. So when the solution is neutralized, there will be 30+20=50 mL of mixture?  [/hide]

[hide="13"]Decreasing EP sounds like increasing EN, so wouldn't it be X, iron, Y? (A)[/hide]

[hide]For 15 I'm gonna take a stab and say A, since CaCO3 is calcium carbonate, and it's white... :maybe: [/hide]


Suppose $z=a+bi$ is a solution of the polynomial equation $c_4z^4+ic_3z^3+c_2z^2+ic_1z+c_0=0$, where $c_0$, $c_1$, $c_2$, $c_3$, $a$, and $b$ are real constants and $i^2=-1$. Which of the following must also be a solution?

$\textbf{(A) } -a-bi\qquad \textbf{(B) } a-bi\qquad \textbf{(C) } -a+bi\qquad \textbf{(D) }b+ai \qquad \textbf{(E) } \text{none of these}$

[hide]taking conjugates of both sides, and using the distribution properties, if $z$ is a root, then $-\overline{Z}$ is a root so $\boxed{C}$[/hide]

Hello, the correct answer is actually choice $B$.

The answer is B, because by Viete's, the product of the roots is a real number. The only answer choice that satisfies this condition is the conjugate of the given root.

Nonono guys the coefficients of the odd-degree terms are complex so the whole "conjugate must be a root" thing doesn't exactly work out

[hide="Solution"]
As Altheman suggests, note that taking the conjugate of both sides and noting that $\overline{i}=-i$ yields \begin{align*}\overline{c_4z^4+ic_3z^3+c_2z^2+ic_1z+c_0}&=0\\c_4\overline{z^4}+c_3\overline{iz^3}+c_2\overline{z^2}+c_1\overline{iz}+c_0&=0\\c_4\overline{z}^4-c_3i\overline{z}^3+c_2\overline{z}^2-c_1i\overline{z}+c_0&=0\\c_4(-\overline{z})^4+ic_3(-\overline{z})^3+c_2(-\overline{z})^2+ic_1(-\overline{z})+c_0&=0.\end{align*} Thus if $z=a+bi$ is a root, then $-\overline{z}=-(a-bi)=-a+bi$ is also a root.  The correct answer choice is $\boxed{\textbf{(C)}}$.[/hide]

What is the motivation for taking the conjurgate of both sides? Also, how would you notice that $c_4\overline{z}^4-c_3i\overline{z}^3+c_2\overline{z}^2-c_1i\overline{z}+c_0=c_4(-\overline{z})^4+ic_3(-\overline{z})^3+c_2(-\overline{z})^2+ic_1(-\overline{z})+c_0$?

A good amount of the motivation came from the fact that I knew what the answer was :P

I guess some other motivation can come from the fact that all the answer choices have something to do with negatives and/or conjugates.  It thus makes sense to try to experiment and see if these types of manipulations work.  (This technique is pretty common; in fact, it's the easiest way to prove that if all the coefficients of a polynomial $P(x)$ are real, then $P(a+bi)=0\Longleftrightarrow P(a-bi)=0$.)  Then once you get to the last line, it's all about fudging with the expression to write it in exactly the right form.

[hide=Alternate Method]Let $f(x)=c_4z^4+ic_3z^3+c_2z^2+ic_1z+c_0$ and let $g(k)=f(ki)$ for some $k$. This means that if $a+bi$ is a solution to $f(x)$ then $\frac{a+bi}{i}$ is a solution to $g(k)$. Since $g(k)$ has no imaginary coefficients, we have that the conjugate rule works here so we first rationalize the denominator to get that $-ai+b$ is a solution to $g(k)$. By the conjugate rule $ai+b$ is also a root. Plugging this into the equation $g(k)=f(ki)$ we see that this implies that $(ai+b)*i=bi-a$ is a solution to $f(x)$. Thus our answer is $(C)$$-a+bi$.[/hide]

[hide=Intuitive Idea]In your standard polynomial with complex roots but real coefficients you have roots in the form $a+bi$ and $a-bi$. The roots sum to a real number and the product is integer. In this one you still need integer products so you still think think of the ide behind cojugates where you use $(a-b)(a+b)=a^2-b^2$. But you want the real parts to die out so it makes sense that the other root is $-a+bi$. [/hide]


the squence of functions $f_{0}$,$f_{1}$.... í given by the conditions
$f_{0}(x)$=/$x$/  :forall $x$ :in R
$f_{n+1}(x)$=/$f_{n}(x)-2$/ for $n$=0,1,2.... and all x  :in R
For each positive integer $n$ ,solve the equation $f_{n}(x)$=1


linear algebra
matrix
symmetry
algebra
polynomial
vector
topology

Let $ A$ be $ n\times n$ nilpotent matrix of nilpotency $ n$. Let $ B$ be $ n\times n$ positive definte and symmetric matrix. What can we say about the matrix $ C\equal{}AB$? Is it also nilpotent?

Thanks,
bhupala.

What do you expect? Positive definiteness obviously can't help (the algebraic closure of positive definite symmetric matrices are all symmetric matrices), and symmetry doesn't make $ AB$ nilpotent either: $ \left(
\begin{array}
[c]{cc}
0 & 1\\
0 & 0
\end{array}
\right)  \left(
\begin{array}
[c]{cc}
0 & 1\\
1 & 0
\end{array}
\right)  \equal{}\left(
\begin{array}
[c]{cc}
1 & 0\\
0 & 0
\end{array}
\right)$ is idempotent rather than nilpotent.

  darij

darij, in your example, $ B$ wasn't positive definite.

But the larger point still holds - this proposition stands no chance of being true.
\[ AB\equal{}C\]
\[ \begin{bmatrix}1&\minus{}1\\1&\minus{}1\end{bmatrix}\begin{bmatrix}5&2\\2&1\end{bmatrix}\equal{}\begin{bmatrix}3&1\\3&1\end{bmatrix}\]
The example was picked with no care at all; the presumption is that in "most" cases (in some sense I don't want to specify), you'd get a counterexample. Here $ A$ is nilpotent, $ B$ is positive definite, and $ C,$ which has nonzero trace, cannot possibly be nilpotent. (The eigenvalues of $ C$ are $ 0$ and $ 3.$)

As I said, positive definiteness is a red herring. If some polynomial identity - like $ \left(AB\right)^n \equal{} 0$ in our case - holds for every symmetric positive definite matrix $ B$ (note that we consider $ A$ fixed here, and that "polynomial identity" means "polynomial identity in the entries of the matrix $ B$", not necessarily "polynomial identity in the matrix $ B$"), then it also holds for every symmetric (not necessarily positive definite) matrix $ B$. In fact, the positive definite symmetric matrices are an open subset of the vector space of all symmetric matrices, and if a polynomial identity holds on an open set, then it holds everywhere. Of course, this only works if our assertion is really a polynomial identity - so, for instance, if our assertion would be "the matrix $ AB$ is diagonalizable over $ \mathbb{R}$" or "the matrix $ AB$ is diagonalizable over $ \mathbb{C}$", then we couldn't kill it that easily. (Of course, it would then be wrong for a different trivial reason.)

  darij


Prove for any x given integer that [tex]1-\frac{1}{x}<a_{x}<1[/tex] where
 [tex]a_{0}=\frac{1}{2} [/tex] and  [tex]a_{k}=a_{k-1}+\frac{a_{k-1}^{2}}{x} [/tex] for k=1,2,..,x.

rewrite the equation as

[tex]\frac{1}{a_{k-1}}-\frac{1}{a_k}=\frac{1}{n+a_{k-1}}[/tex]

use telescopic sum to get

[tex]\frac{1}{a_0}-\frac{1}{a_n}=\frac{1}{n+a_0}+...+\frac{1}{n+a_{n-1}}[/tex]

since [tex]a_k>0[/tex] for all [tex]k[/tex], therefore

[tex]\frac{1}{a_0}-\frac{1}{a_n}<\frac{1}{n}+...+\frac{1}{n}=1[/tex]

it follows that [tex]a_n<1[/tex]

furthermore we have

[tex]\frac{1}{a_0}-\frac{1}{a_n}>\frac{1}{n+1}+...+\frac{1}{n+1}=\frac{n}{n+1}[/tex]

after rearranging the terms, we have

[tex]a_n>1-\frac{1}{n+2}>1-\frac{1}{n}[/tex]


1. A parallelogram has one side of length 20, and 2 diagonals of length 24 and 10. What is its area? (Does anyone know how to do this without Heron?)

2. A circle is inscribed in a quadrilateral of lengths 5, 7, 8, and x. Find x.

(I might have messed up some of these; my geometry teacher asked me about them earlier, but I can't remember the real problems.)

I don't think those lengths are possible for #1. I could be wrong though. It won't be the first time.

I'll ask my teacher for the real numbers tomorrow. Sorry.

Number 1 is impossible, as Simon said.  (Hey, have you gotten a lot of "Simon says" jokes in the past?)

Number 2 is a good question, with a minor limitation:  you have to know what order the sides of the quadrilateral are in.

Over the years, I have. It started to get old when I was about six years old. Stephen subtly mentioned it once a few months ago.

2) Assuming lengths are right,[hide]6.5, I think.[/hide]  I did it werd, so maybe not.

Nope, although numerically close.  What was your thinking?

Ummmmm...I actually drew it out, not very clean, but it worked.  I know there has to be an easier way, I just didn't want to bother finding it.

I don't see how you could have drawn it out -- how did you know where the circle went relative to the sides?

Honestly, I don't know how I did it either.


I think that we should have different tournaments for people in different states (ex. everyone in North Carolina play in a big new game).  What do you guys think?

We can if you want..but it may be competetive for states who have a lot of people

usually that's what happens in the evening, though :huh:

[quote="KingSmasher3"]I don't think enough people from a single state will be online at one time to play a big, competitive game.  :huh:[/quote]

Some states will though, like NC and probably California.

I don't think that it is a bad idea, but (like KingSmasher said) it might be hard for some states, and would exclude quite a few people.

[quote="MC10"]I don't think this is such a good idea. What if only 1 person comes here and plays FTW from a state? He/she would have no one to play. Maybe state vs. state? (First person in state to answer produces the state's answer)[/quote]

That is why there is something called organizing.  :)

how about each state holds "tryouts" (essentially a game with only the people from that state)
the winners of those will represent the state in the state vs. state competition

if only one person from a state shows up at the tryout, he/she represents

if no people show up from a state to the tryout, the state is not represented

that is a great idea vallon :D  yeah tryouts

well...i'm in arizona, so....

Nobody else on AoPS is from arizona that I know except for 2 people.

State tryouts is a good idea.

But some states are going to have a lot of people making it really hard. I think you might need to split up some states like north and south of a state.

yeah NC has loads of members and you would have to have a huge instate tournament just to find one representative   for the state. and how would you get everyone on at the same time because of time zones. it is a good idea though :thumbup:

And in other places, there are too little amount of people..so it might not be that good of an idea, if you really think about it

What if we had different states against each other?  (We could have tryouts, and the top 2 people could play each other  :D  I'm sure theres at least two people in each state)

I was thinking about holding a "mega-tournament" along these lines where people could join, and I would split them up into groups, winners play (if nessacary in more groups and winners play) round-robin to win, but I figured it would take too much time and would probably die.

A somewhat different, albeit not nessacarily better idea that I had was to have a "Grand Prix" cycle, where a tournament would start every week (or 2, or w/e) awarding points and then the person with the most points at the end of the period (couple months? 6 months? year?) would win, although that would also probably die after the uniqueness of it wears off after a few tournaments.

I like the idea vallon. im from wi we have a couple people

just get some random hacker that would make all the computers say "TOURNAMENT FOR EVERYONE ON FTW!" as soon as they go on aops. That would only cost... several thousand, millions, maybe. :ninja:

Judging from the history of this forum, it is for posts with a seriousness content of at least 0.1%.

Even though it's not a revival, it almost is. Don't do such things.

Moderator, please lock.

Also, note that if you were only going to say one of those three (now these four) things, you should simply Private-Message, and only post like this when a significant amount of stuff is to be said.


USA is a free country. You can practice any religion you like and USA doesn't have an official religion. But, why then, do they read out of the Bible when the President elected? Why then, does the President have to say "So help me god" or whatever crap on the inauguration? Why then, do public schools celebrate Christmas but not Diwali or hannukah? Why then, do kids have to say the Pledge of Allegiance which contains the lines "one nation under  [b]god[/b][b]? Why then, do they ask you your relgion (hindu, muslim, christian etc) for any kind of assistance from the government (Medicaid/Medicare for example)? In my opinion, the USA is ambiguously christian. It says it is a free country and it may have been so in the past not so much in the current. Express your opinions. [/b]

Most of the things you addressed are traditional.  Second, no one HAS to say the pledge of allegiance in school.  Many schools do celebrate other religions' holidays.  For example, I almost positive that the SUNY university system in New York grants days off for Jewish holidays, as do many public high schools in NY.  When you are asked about your religion on government forms, they are optional.  You are not required to give your religion on anything.

The fact that the US is a free country means that you can practice whatever religion you want.  The country considers itself a secular one, as there is no official religion.  When a county is 78% Christian (or whatever the number is), then yes, most people are going to be Christian.  But the government does not employ missionaries to convert people to Christianity.  I am an atheist and I would love to remove all of the references to God and whatnot, bit it's not going to happen, at least not for a while.  About 16% of the country considers itself atheist/agnostic/no religion, which is actually more than any religion aside from Christianity, but as you can tell, not nearly enough.

Right, a big part of the population is christian, but do not pretend that this is an issue separate from freedom. A lot of christians to some extent want everyone to live under their particular moral rules (all laws are moral I guess, but I mean laws which do not have to do with aggression, fraud, theft etc). Since the US has democracy, or at least a strong opinion can change laws, this allows for the introduction of plenty of laws that are in contradiction to freedom. Of course, this is really just an argument that a democracy without an extremely strong constitution is (almost always) unfree.

I don't see anything in itself unfree about being able to elect a christian president, it's just that the powers of the government/people are way too strong.

[quote="Ihatepie"] As JBL said, kids are not required to state the Pledge.[/quote]
I'm not JBL, as he hasn't posted in this thread. :)

Public schools tend to honor holidays from multiple religions but private schools can do as they choose.  Just as a Catholic school is not going to celebrate Yom Kippur or Ramadan, a Yeshiva is not going to celebrate Christmas.

[quote="Poincare"] In my opinion, the USA is ambiguously christian.  [/quote]

We aren't ambiguously Christian, we are tacitly Christian.  When the majority are a single religion, it carries over into the spirit of the nation and thus the nation's institutions.  I wouldn't worry about it if I were you, as the current trend is to remove this institutionalized Christianity.  I mean, really, how can you cite Christmas Break as a continuing problem when most of the time we refer to it as Winter Break and associate it with Santa Clause and rein deer bringing presents.

To me, your position is overblown because evidence overwhelmingly points to the fact that Christianity is losing influence in America.  People like you lived a generation ago and already set the trend of persecuting majority groups / removing the privileges of majority groups.  So, I can't understand why you are saying it has gotten worse.

http://www.newsweek.com/id/192583 this says that there are 10% fewer Christians now than 20 years ago, so why would Christian influence in society be on the rise?

[quote="Ihatepie"][quote="Poincare"]USA is a free country. You can practice any religion you like and USA doesn't have an official religion. But, why then, do they read out of the Bible when the President elected?[/quote]It's not required. If I remember correctly, John Quincy Adams used a law book while Theodore Roosevelt didn't use a Bible either.

[quote] Why then, does the President have to say "So help me god" or whatever crap on the inauguration?[/quote] It's not required either. In fact, afaik the majority of Presidents have not used it. However, due to the nature of modern politics, andy politician who does not state those words would have their approval ratings plummet.

[quote] Why then, do public schools celebrate Christmas but not Diwali or hannukah?[/quote] I would assume this is because of the huge number of Christians in this country. If it was not celebrated, the masses would probably be very angry.

[quote]Why then, do kids have to say the Pledge of Allegiance which contains the lines "one nation under  [b]god[/b][b]?[/quote] As JBL said, kids are not required to state the Pledge. In addition, the phrase "under God" (which was added in 1954) has been widely contested as a violation of the separation of church and state.[/quote]

I agree with everything accept for the third one. Also, In my post I also just meant to bold god, not the whole paragraph.

[quote="Poincare"]I agree with everything accept for the third one. Also, In my post I also just meant to bold god, not the whole paragraph.[/quote]
Please be careful while composing a message/post! I think, in the present context, the word should be "except" and not "accept".

About the pledge, it is supposed to represent the ideals and freedoms of the USA.  'under god' was thrown in there in 1954 to show that we were a strong country under god while in fact doing so showed our weakness of hypocrisy.  I have heard people say that under god is in the pledge because the founding fathers were Christian and formed our country under god's name but the founders are dead and who cares what religion they were.

"Under God" was actually added to distinguish the US from the officially atheist Communists.

hi every body
In the dollar bill,it is written that ''in god we trust",
along with this, it seems that the general atmosphere in america ( prosperous countries) is secuirity of living, hopes and dreams are  strongly linked to finance. So the correlation leads to ''in money we trust" so then money is god in america. And this can explain why the government can autorise himself to kill a country to save the god. But since tht god is virtual and has no existence by himself, he will swallow the existence of his worshiper.

[quote="Poincare"]Why then, do they ask you your relgion (hindu, muslim, christian etc) for any kind of assistance from the government (Medicaid/Medicare for example)?[/quote]
I don't know why you think this is malicious; it's almost certainly for statistical purposes.

Look, there's no contradiction between the following two statements:

1) The United States population consists primarily of Christians, and many of its traditions are Christian traditions.

2) Non-Christians in the United States are free to practice their religion.

Traditions, whether Christian or otherwise, are hard to give up.  This isn't a unique phenomenon.

Asking for religion basically has to be for statistical purposes because I think it is illegal to make religious information mandatory (I could be mistaken, this is what I was told).
[quote="t0rajir0u"]Traditions, whether Christian or otherwise, are hard to give up.[/quote]
Unfortunately, it has gotten worse in the last several decades rather than better.  America has become an increasingly spiritual nation while Western Europe has become more secular.  It is well-known that several of the Founding Fathers, including Franklin and Jefferson, despised religion.  So I don't understand where people are coming from when they say this country was built on Christianity.  Ten commandments maybe?  Those were around (if you believe the Bible) long before Jesus, and I'm pretty sure that the citizens of most countries would realize "Hey, perhaps I shouldn't steal my neighbor's stuff and then kill him," without the Bible.

[quote="Poincare"]U Why then, do public schools celebrate Christmas but not Diwali or hannukah? Why then, do kids have to say the Pledge of Allegiance which contains the lines "one nation under  [b]god[/b][b]? [/b][/quote]
Our school does celebrate other holidays. For example, this year we did not have school on Rash Hashanah. I think that the school does not celebrate many holidays because they are not as common, and if they did celebrate all the holidays, there would be very little school. Also, the Pledge of Allegiance is [i]not[/i] required- the parent can submit a form saying that they do not want their child to participate. For example, a kid in my class is a Jehova's Witness and she does not say it.

[url=http://www.nyc.gov/html/dot/html/motorist/scrintro.shtml]Alternate side parking[/url] was suspended on Diwali in New York City last year (or maybe the year before?) for the first time.  (Other religious holidays on which it is suspended: Asian Lunar New Year, a wide assortment of Jewish holidays, a variety of Christian holidays (including some on the Orthodox calendar), and two Islamic holidays: Idul-Fitr and Idul-Adha.)

[quote="JRav"]So I don't understand where people are coming from when they say this country was built on Christianity.[/quote]
Then you didn't pay enough attention in your history classes.  America didn't elect a non-Protestant President until 1960.

On the other hand, if you're talking about the structure of government, it's more accurate to credit the Enlightenment.  But it's hardly as if all Enlightenment philosophers were atheist, either.

I'm saying that America was not established as a nation to be governed by members of the clergy, who take their orders from Christianity.  Citizens are not to elect officials because they are or aren't Christian, and politicians are not supposed to make decisions based on what their church would do.

And as far as the 1960's comment, Lincoln  did not belong to any particular sect of the Church, Johnson did not have any denominations, and Jefferson said, "Millions of innocent men, women and children, since the introduction of Christianity, have been burned, tortured, fined and imprisoned. What has been the effect of this coercion? To make half the world fools and half hypocrites; to support roguery and error all over the world..."  Did he have religious affiliations?  Yes, but he was hardly a proponent of running the country based on religion.

I would classify the Enlightenment philosophers as deists, as were many of the founders.  Certainly isn't atheist, but it isn't full-fledged Christianity either.

[quote="t0rajir0u"]America didn't elect a non-Protestant President until 1960.[/quote]  This is true, but I'm not quite sure of its relevance -- the US must have been something like 75% Protestant at that time, and that number has been monotonically decreasing.  The (correct and reasonable) distinction JRav was drawing was between a Christian nation and a nation of Christians.  The US has been the latter for all of its history (although decreasingly so), but it has never been the former, and the distinction is really important.  It's why, for example, it's illegal to have school-mandated prayer.  Engel v. Vitale -- it's actually quite impressive how flimsy and generic the prayer struck down was: [quote="NYS Board of Regents"]Almighty God, we acknowledge our dependence upon Thee, and we beg Thy blessings upon us, our parents, our teachers and our country. Amen.[/quote]

[quote="NYS Board of Regents"]Almighty God, we acknowledge our dependence upon Thee, and we beg Thy blessings upon us, our parents, our teachers and our country. Amen.[/quote]
Prayer fail.

I think that the U.S.A. allows all peaceful practice of religion as long as nothing completely inane is done (protesting naked, overturning cars, etc.), but I'm not completely sure. If the U.S.A. has enacted a law restricting peaceful and un-inane (exactly how do I spell this word?) practice of religion, then it should be repealed.

Also I've never thought about theistic specie/paper money/oaths. In my opinion I don't think that one should be forced to say the theistic parts of oaths (so help me god, under god, etc.), but I honestly don't know about currency. I think it's good as it is, to be honest.

Also I'm agnostic.

Celebrating only Christian holidays is an obvious 'favor' towards Christianity, but if one decides to counter that by celebrating holidays of other religions, then it's likely for the 'major' (most followed) religions come first (hence the celebrations of Buddhist, Jewish, and Islamic holidays at some schools), which, in turn, is likely to cause an uproar from the representatives of not-so-widespread religions -- "Do you consider our religions to be less important so that you don't celebrate [b]our[/b] holidays?!" On the other hand, celebrating holidays of all religions (I can't even imagine this happening) will probably cut the school year to a week or two of classes at best -- the rest of the time will be spent on satisfying the representatives of different religions (and since there is probably no official separation between 'real' and 'not real' religion, you will probably have pastafarians and like demanding their own holidays). Would it not be convenient/logical not to celebrate religious holidays at all, limiting all the school holidays to every, say, quarter, or use other time-spaced breaks? Sorry for the off-topic.

[quote="3100"]pastafarians [/quote]

lol, I found term funny, at first I thought you misspelled Rastafarians, but I then realized...  :lol:

Anyways, I just wanted to suggest to you that it is not a great policy to give no one cake because there isn't enough cake for all.

winternights1, is there really a fair alternative though? I would rather treat all of the religions fairly, than give some of them an advantage simply because they have more followers/are more known in the U.S.

Great, why don't I start my own religion, then?

The purpose of schools giving holidays is not to endorse or celebrate those holidays. Rather, they aim to best serve the students and the community.

- If a large fraction of the student body feels compelled to skip school on a given day, then it may not be useful to hold school on that day, since many students would fall behind.

- On the other hand, if only a small fraction of the student body feels compelled to skip school on a given day, then it is definitely not the right decision to cancel school for everyone, thus either eliminating a school day for many students or pushing everyone's school year end date one day ahead.

The fairest thing to do is to give an excused absence to students who wish to take a day off for a less popular (legitimate) religious holiday, and for teachers to offer to help them catch up on material.

Also, I doubt anyone comes to the US with the intention of having school boards nationwide call their holidays school holidays. I'm a Hindu, and frankly, I don't care that they don't call off, say, Diwali. I've talked to Muslims on this matter, and they tell me that they'd rather not have the whole school shut down so that they can go home and pray -- they like that to be their own thing. And I totally understand that feeling.

TZF is right on.  Again, it depends on the region of the country.  In Long Island, and in NY in general (as TZF and JBL can attest to), there is a large Jewish population.  Therefore, many public schools, including universities in the SUNY system acknowledge Jewish holidays in addition to Christian ones.  The particular town I come from has a very large proportion of Jews.  Now I would say that a vast majority are not very religious but nevertheless, I'm sure many parents would hold their kids out of school (and the kids would jump at the chance!) on  Jewish holidays, just as Christian parents would hold their kids out of school on Easter or Christmas, if school actually was in session.

http://en.wikipedia.org/wiki/Utilitarianism

Well, these people think it's fair to pursue the greatest good to the greatest number.  I think a happy medium exists between what this philosophy dictates and what you would have us do.

First off, "Fair" can mean a lot of things.  You use it to mean equal and unbiased in the context of your sentence.  You can treat all of your slaves equally and still be a cruel person.

What would I suggest? I would prefer a charter school system, in which parents give their tax money to the school their kid attends.  It allows for private / religious schools to have more affordable tuition.  That way the dominant faiths could organize their own schools.  But since this isn't going to happen, the way things are now is not so bad.  I guess you can make some small adjustments and concessions to certain groups.  I've never actually heard any complaining about stuff like this where I live... I didn't even know anyone cared.  We have spring break tied in with Easter, thereby secularizing it, and Winter break is equated with Christmas break, thereby secularizing it.  These are the only 2 big Christian holidays celebrated, so where's the problem?

edit: also, what the above 2 posters said

[quote="winternights1"]That way the dominant faiths could organize their own schools.[/quote]
That's a huge step backward as "Intelligent Design" shows.

In general: Majority's decisions doen't make things right. In fact, it's an essential part of democracy to protect some common values and minorities.

That is a good point.  However, the dominant faiths already have organized their own schools, so if intelligent design is being taught then, it will be because it is taught now.  Nothing will change between now and the implementation of what I said, except that there would be more religious schools with greater enrollment for less tuition a person.  The main goal of my idea is actually to make the quality of each person's education superior, as private prep schools are usually better than inefficient, standardized test obsessed, drop out plagued, public schools.

In any case, most religious groups are willing to accept evolution as true.  All three of the private, religious schools I've been to have taught evolution only, and I've been to schools of 3 different denominations.

Hmmm... my idea might be a bit too flimsy to work.  But, really, I don't want to argue the details about some school system that will never exist.  I gave it as a general suggestion because the guy asked for an alternative.

I don't get where you are going with the majority thing, though.  I'm pretty sure a minority of people support intelligent design.

edit: I have half a star rating!  Does that mean I am new or does it mean that everyone has rated my posts at between 1 and 0?   :huh:

[quote="JRav"]Asking for religion basically has to be for statistical purposes because I think it is illegal to make religious information mandatory (I could be mistaken, this is what I was told).[/quote]
Hmm... when my parents fill out those forms, its almost always necessary to mark your religion.

1) You were responding to only the first tiny bit of what you quoted.  Including the rest of the quote makes it harder to understand, and clutters up the page.  I've fixed it for you, but in the future you should do it yourself.

2) What you said is not in contradiction with what JRav said.  The existence of a spot to state your religion is not the same as that being mandatory.

3) What forms precisely are you talking about?

Perhaps the one that excuses a student from saying the Pledge of Allegiance.

I don't think (at least at my school) it would be a major problem if you left out the "one nation under god" part out of the pledge.
Secondly my mom who is a teacher and its in her list of stuff to teach is to briefly touch on religious holidays, such as Kwanzaa, Hanukkah and Christmas, because they influence culture or something like that.

Also don't revive 1 year old topics.

[quote=Joe__2020]Why not revive 10 year old topics?[/quote]

1. Because
2. There is no #2

And when the US references God, although it is the Christian God, you can apply this to any god.  As long as the government does not specify which religion it is referring to, it's not a violation of the 1st amendment.

[quote=Iamnobody][quote=samrocksnature][quote=Joe__2020]What if you are Buddhist or follow another religion that doesn't have a god?[/quote]

Well then too bad, live in a country like China then.[/quote]

This seems a little harsh...

But America was founded on Christian ideals. They're not forcing you to have a certain religion. Think of "God" as the god you believe in.[/quote]

I'm saying that because China has the highest population of Buddhists. I'm saying if you live in a country, then you have to live with it's traditions and culture.

[quote=Joe__2020]What if you are Buddhist or follow another religion that doesn't have a god?[/quote]

Or if you are polytheistic.

[quote=happycupcake]America, the land of the free*

*Except for a bunch of stuff and some other things and yeah.[/quote]

It more like:

[size=1000][color=#f00]America, the land of the free[/color] [size=50][color=#ccc]*[/color][/size]
[size=50][color=#ccc]*Except for a bunch of stuff and some other things and yeah. Also if you found this text dont tell your friends bc we lik propaganda and nationalism. <3[/color][/size]

Eh more like: "America, the land of the free (except if you disagree with the general populace)."

Edit:
Going back on track though could someone give an explanation why exactly we need to be forcing people to agree with "the traditions and culture" when:
1. The culture has changed over time.
2. There isn't really any one real culture.
3. Religious freedom is part of the first amendment, and certain religions have "traditions and culture" that may differ.


integration
calculus
limit
algebra
linear equation
calculus computations

I'm having a hard time separating out y from t in this problem

8(dy/dt)+y=56t

I was able to get as far as dy/dt+(y/8)=7t but then I can't get the ydy=tdt set up. Thanks for any insight.

I vote that you look for a particular and homogeneous solution.

A very simple classification scheme for first order differential equations: they can be linear, separable, both linear and separable, or neither.

Suppose you rewrite it as $\frac{dy}{dt}=g(t,y).$

If you can factor $g(t,y)=a(t)b(y),$ then it is separable.

If $g(t,y)=p(t)y+q(t),$ then it is linear.

With this classification scheme come two methods: separable equations can be solved by separation of variables, and linear equations can be solved by the method of multiplying factors.

This equation is linear. It is not separable. You appear to be trying to solve it by separation of variable. That won't work - it's the wrong method for this problem.

The right method is multiplying factors. First write this in standard form for a linear equation:

$\frac{dy}{dt}+\frac 18y=7t.$

The multiplying factor is $\mu(t)=e^{\int\frac18\,dt}=e^{t/8}.$

$e^{t/8}\frac{dy}{dt}+\frac18e^{t/8}y=7te^{t/8}$

$\frac{d}{dt}\left(e^{t/8}y\right)=7te^{t/8}$

$e^{t/8}y=\int7te^{t/8}$

$e^{t/8}y=56te^{t/8}-448e^{t/8}+C$ (integrate by parts)

$y=56t-448+Ce^{-t/8}.$

yep that was staring me in the face. So when I boil it down I get y=56t
I'm given Y(0)=8

However my solution appears to be wrong.

Your last post makes no sense to me.

The initial condition $y(0)=8$ would result in $C=456$ in my solution:

$y(t)=56t-448+456e^{-t/8}.$

Now, let me ask a calculus problem:

The initial condition is positive, but the solution starts out decreasing. You can actually see this from the differential equation itself. If $t=0$ and $y=8,$ then

$\frac{dy}{dt}+\frac18(8)=0$ or $\frac{dy}{dt}=-1.$

On the other hand, $\lim_{t\to\infty}y(t)=\infty$ since $56t$ is the dominant term for large $t.$

Question: For $t\ge 0,$ do we ever have $y(t)<0?$ Where does $y(t)$ reach its minimum?

----

The main point of my first post: we classify differential equations so that we know how to solve them. For first order equations, the two major classification are separable and linear These classifications are not mutually exclusive. They're not exhaustive either. If the equation is neither linear nor separable, there are some other classifications it might (or might not) belong to: homgeneous, exact, Bernoulli, Riccati, a few others. None of these minor classifications is anywhere near as important as linear and separable, and they come nowhere close to exhausting the possibilities. There are many differential equations without closed form elementary solutions - even equations of the form $y'(t)=g(t).$

Thanks Kent you clarified a big whole in my knowledge classifying the different types of problems I'll come across. I think I'll run through that line up each time i try a new problem now. Sorry about the confusion I posted a reply to JBL before I saw yours.

JBL's recommendation was right on (the same as yours but approched from a different angle). I only did half of the problem (albeit with crappy algebra) the particular solution should be Yp=56t-448

Then you solve for the homogeneous equation 8(dy/dt)+y=0

You add the two together and plug in your initial conditions
Y(0)=8 to solve for C. You end up with the exact same answer you came up with!


Find all $f: R\to  R$ so that for all real $x, y$ we have $f(x+f(y)) =f(x).$

Case 1: $f(x) = 0$, $\forall x \in \mathbb{R}$.
This is trivial.

Case 2:
Let $f(\mathbb{R}) = \{f(x) | x \in \mathbb{R}\}$.

Then, let $z \in f(\mathbb{R})$, such that $z$ is nonzero.
$f(x+z) = f(x)$, $\forall x \in \mathbb{R}$.

From this, $f(f(x+z)) = f(f(x)) = f(f(x)+0) = f(0)$, $\forall x \in \mathbb{R}$.

Therefore, $f(f(\mathbb{R})) = \{f(0) \}$.

So, over the subset $f(\mathbb{R})$, $f$ is a constant function.

However, there are many such functions over all of $\mathbb{R}$ which satisfy this.

For example, the function $1_{\mathbb{Q}}$ which is $1$ if $x$ is rational, and $0$ if $x$ is irrational.

To PTynan89:
If$f(0) =0$ then  how will you solve ???


function
geometry
analytic geometry
calculus
integration
topology
advanced fields

I have this problem in measure thoery. It was discussed in the class but I could not follow that. ANy help would be greatly appreciated .. thanks..
[/code]

I dont think the solution exists. But I would like to hear a second opinion

$U$ is an open halfspace, and $x_{0}\in\partial U$. Define $\delta(x)=\mathrm{dist}(x,U)^{-1}$. The question is: for which $p>0$ is the function $f^{p}$ is integrable on the ball $B(x_{0},1)$?

Hmm... what is $f$?  :maybe:

Upps!!!.. I used two different notation. I mean $f(x)=\delta(x)$ actually.. I am attaching the correct question.

Are $x_{0}$ and $x_{\text{fixed}}$ also the same? If so, then the function is infinite on half of the ball.

Given $x_{0}, N\in R^{n}$ consider the open set $U=\{x\in R^{n}|~~ <x-x_{o},N>~ >0\}$
and let $\delta(x)=\frac{1}{dist(x,U)}$, where ${dist(x,U)}$ is the Euclidean distance of $x$ from $U$. Determine the range of $p>0$ for which $|\delta(x)|^{p}\in L^{1}(U\cap B(x_{0},1))$, where $B(x_{0},1)=\{x\in R^{n}~| ~~\|x-x_{0}\|<1$. We can simplify the geometry of the problem to solve this.

Still incorrect. It does not make much sense to ask when $|\delta(x)|^{p}\in L^{1}(U\cap B(x_{0},1))$ because $\delta(x)=\infty$ for all $x\in U$. If you wanted to know when $|\delta(x)|^{p}\in L^{1}(B(x_{0},1)\setminus U)$, that would be a meaningful question. And the answer is: $0<p<1$. Choose the system of coordinates so that $N=\langle 1,0,0,\dots,0\rangle$ and use the Fubini-Tonelli theorem, integrating with respect to $x_{1}$ first.

I think in that case $0<p<n$. Don't you think so?

No I don't. Let $I\subset \mathbb R^{n}\setminus U$ be a line segment that is perpendicular to $\partial U$ and has an endpoint on $\partial U$. The integral of $\delta(x)^{p}$ over the segment is convergent if and only if $\int_{0}^{1}\frac{1}{x^{p}}dx$ is convergent, that is, $p<1$. Fubini-Tonelli reduces the integral over $B\setminus U$ to integration over such segments.


How i can find  problems of JBKMO team selection test 2000 up to 2005 without 2004!!!? ;)

1.: I hope with 2004 is also possible  :lol:
2.: JBK = junior balkan ¿
3.: this is not a very clear question! (when 2. is right: there is more than one country in JBK)

I mean romanian selection test for JBKMO.
 :blush:


geometry
circumcircle
analytic geometry
graphing lines
slope
function
algebra

Probably well known:

We have $n$ fixed points $z_1 , ... , z_n$ on the circle - with it's circumcenter in $(0,0)$ and radius $1$.
Prove that if $|z-z_1| |z-z_2| ... |z-z_n| \leq 2$ for all $z$ on the circle, then
$z_1 , ..., z_n$  are vertices of a regular  $n$-gon.

in the complex formulation of the problem we know that $z_1,...,z_n$ are actually
roots of the equation $z^n=1$ and so $z_1,...,z_n$ form the regulated n-gon.
this is what i remember of the problem formulated in complex number...

problem reformulation in elementary terms
consider distance from every P(x1,y1),Q(x2,y2) points in euclidean plane
 $d(P,Q)=\sqrt{(x1-x2)^+(y1-y2)^2}$
consider n points on the circle (z1,...,zn) so that
taking any point of the circle of radius one wich would mean
some
$T(x,y)$ with $x^2+y^2=1$
$\prod_{i=1}^n {d(z_i,T)}\leq 2$
and now proove that
proove that
$d(z_1,z_2)=d(z_2,z_3)=...=d(z_{n-1},z_n)=d(z_n,z_1)$

Nope, it's the other way around. (Hopefully)

could we consider a small case like n=6
see what happens ?

let $W(z)=arctan(\frac{Re(z)}{Im(z)})$
if that is true then $W(z)$ is actually the slope of the segment [OZ]
expressed in radians (where O is the center of the circle and Z a point on it
corresponding to its affix z).
if we want to find the angle between 2 segments we can construct this function
$J(z1,z2)=W(z1)-W(z2)$
so at last we find that whatever the ordered set {${z1,...,zn}$}  (*)
might be we have
$2\pi=\sum_{i=1}^{n-1}J(z_i,z_{i+1})$
this is what all sets of the form * have in common

this proof (which I started) needs some correction

maybe the problem posted by Rafal can be approached through
Measure Theory
there is this problem similar to this one but much harder
http://www.cecm.sfu.ca/~mjm/Lehmer/lc.html

Here's an incomplete thought, which I haven't gotten to pay off yet.

Consider the polynomial $f(z)=(z-z_1)(z-z_2)\cdots (z-z_n).$

Write it as $f(z)=a_0+a_1z+\cdots+a_{n-1}x^{n-1}+z^n.$

Note that $|a_0|=1$ since it is a product of points on the circle.

By the Maximum Modulus Principle, $|f(z)|$ assumes its maximum value somewhere on the unit circle, and we are assuming that that maximum value is $\le2.$

If we could somehow use this to prove that $a_1=a_2=\cdots=a_{n-1}=0,$ we would be done.

(I have not yet done that. That's why I called this an incomplete thought.)

Here is the classical proof of the inequality $\max_{|z|=1} |f(z)|\ge 2$ (in Kent's notation). Consider $f(z)-2a_0$. It is still a polynomial with the leading coefficient $1$ and the free term $-a_0$ that is $1$ in absolute value. Hence, it has a root $\zeta$ with $|\zeta|\le 1$. Thus, by the maximum principle, $\max_{|z|=1} |f(z)|\ge |f(\zeta)|=2$. Try to proceed from here ;).

ok so the maximum of the function is at least 2.
so we have still to proove any ngon not regular has the product >2.
wait ! i'm confused
doesnt the problem say that we have to proove $|f(z)| \leq 2 \ \forall \ |z|=1$ ?
you just prooved $max(f)\geq 2$ but how do we propose to proceed ?

[quote="spx2"]
wait ! i'm confused
doesnt the problem say that we have to proove $|f(z)| \leq 2 \ \forall \ |z|=1$ ?
you just prooved $max(f)\geq 2$ but how do we propose to proceed ?[/quote]
No, the problem's [b]assumption[/b] is that $\max |f|\le 2$. By now you know that it means that it actually equals $2$. So, every inequality in the proof I cited is actually an equality. In particular, all roots of $f(z)-2a_0$ lie on the unit circle and so on. Try to squeeze as much information from this as possible and use it wisely.

You do not need any complex analysis beyond this point. Just high school algebra and the identity $\frac 1z=\bar z$ for $|z|=1$. If I say any more, I'll reveal the entire solution. Try to make the last step yourself :).

why do you say here
in post #11
that $-a_0$ is 1 in absolute valuea ?

i remember there was a physics problem with vectors with some points arranged
in a polygon shape(those vectors were forces) and all added gived something
nice.
does it resemble to this ?

excuse my stupid question but how does one interpret geometrically a multiplication of
n complex numbers wich are all on a circle?

[edit : oh yes http://www2.edc.org/makingmath/mathtools/complex/complex.asp
here it says that:
When you multiply complex numbers, you multiply moduli and add arguments.
so now what Kent said is obvious :) ]


Hi Chinese MLinkers ,
 I saw a problem in a recent Chinese magazine but no solution was given . Could you please help me to search for me a solution ?
  Let $ABC$ be a triangle . $AB=c, BC=a , AC=b$ . $x,y,z$ are three positive real numbers . $M$ is inside $\triangle ABC$ . Prove that :
 $x.MB.MC +y.MA.MC +z.MA.MB \geq min(xbc, yac,zab)$
  Thank you .


Find all positive integer solutions x,y,z,p, with p a prime, of the equation:
 x^p + y^p = p^z.

I might have made some mistakes, but even if I have, I think the solution can still be 'saved':

It's not that hard to get the solutions for p=2: (x,y,z,p)=2^k,2^k,2k+1,2).

Now let's assume p>2, so p>=3. x=-y(mod p). We can also assume p does not divide x and y, because if it does we just divide everything with p^p and we get another equation. p^z=(x+y)(x^(p-1)-x^(p-2)*y+..-x*y^9p-2)+y^(p-1)), so p^a=x+y and p^b=(x^(p-1)-x^(p-2)*y+..-x*y^9p-2)+y^(p-1)).

It's easy to see that for p>=5 x^(p-1)-x^(p-2)*y+..-x*y^9p-2)+y^(p-1)>x+y (*). The largest power of p which divides x^(p-1)-x^(p-2)*y+..-x*y^9p-2)+y^(p-1) is p because x^(p-1)-x^(p-2)*y+..-x*y^9p-2)+y^(p-1) = px^(p-1) (mod p^2)<>0(mod p^2), so x^(p-1)-x^(p-2)*y+..-x*y^9p-2)+y^(p-1)=p and x+y<p, so x+y=p^a<p, so x+y=1 and we get some contradictions. This means that p=3. I think (*) holds for p=3 and both x and y >1 so we get the same contradictions as above, so x=1 or y=1. This means 3^z-1=x^3. We tinker with (*) a bit more (we can see that it always holds with >= instead of > etc) and we can easily find x=2. So another type of solution is (x,y,z,p)=(2,1,2,3). All other solutions are obtained from this one by interchanging x and y or by multiplying x and y with p^k for a certain k.

I didn't prove everything completely, but, like I said, I think this is the main idea.


topology algebra function domain Asymptote

Let $ A$ a bounded subset of $ \mathbb{R}^2$, homeomorfic with the open disk. Then boundary of  $ A$, contains a contiunous curve(not const)??

Not necessarily. Unfortunately, the counterexample I have is too ugly to post. I'll see if I can simplify it to something readable before posting or find some reference (I have no doubt that the question has been discussed in the literature).

what do you mean that the boundary of $ A$ contains a continuous non constant curve?

Sorry it took so long. Here goes the (sketch of a) proof (it is better than what I had originally but still a bit long).

Let $ \varphi(r)\equal{}\begin{cases}0, &0\le r\le 1
\\
\frac15(r\minus{}1)^2,& 1\le r\le 2
\\
1\minus{}\frac 8{5r},&r\ge 2
\end{cases}$. Then $ T(z)\equal{}z\varphi(|z|)e^{i\psi(|z|)}$ is a non-expanding map provided that $ \psi\equal{}0$ on $ [\tfrac 32,\plus{}\infty)$ and
$ \psi'(r)$ is much less than $ (r\minus{}1)^{\minus{}2}$ on $ (1,\tfrac 32]$. This still allows for $ \psi(r)\to\infty$ as $ r\to 1$, which is the last condition we shall impose. Note also that $ |T(z)\minus{}z|\le 10$ for all $ z\in\mathbb C$.

Now, for a compact set $ K\subset\mathbb C$, define $ S(K)$ to be $ T^{\minus{}1}(K)$ if $ 0\notin K$ and the union of $ T^{\minus{}1}(K\setminus\{0\})$ and the unit circumference $ \{|z|\equal{}1\}$ if $ 0\in K$. Note that $ S(K)$ is connected if $ K$ is connected and $ T(S(K))\equal{}K$. Assume that $ 0\in K$ and there exists a smooth curve ending at $ 0$ and not intersecting $ K$ at any other point. Then $ S(K)$ does not intersect a curve that winds around the unit circle infinitely many times:
[asy]size(250);

D(D((-8,0))..(-7,0.3)..(-6,-0.2)..(-5,0));
MC("S", D((-4,0)--(-2.3,0),linewidth(0.5)+black,EndArrow), 2*N);

real r=2,t=0;
D(CR((0,0),0.8),red);
path P=(3,0)--(2,0);
for (int n=100;n<300;++n)
{
P=P--(r*dir(t));
t+=10;r-=1/n;
}
D(P);[/asy]
This means that if $ C\subset S(K)$ is a continuous curve, than either $ C$ is a subset of the unit circle, or $ T(C)$ does not pass through $ 0$.

Now comes the main construction. We shall call a compact set $ K$ [i]good[/i] if it is connected, separates $ 0$ from $ \infty$, there is a countable set $ D$ of points of $ K$ such that any non-trivial continuous curve contained in $ K$ passes through infinitely many points in $ D$, and every point of $ K$ is an endpoint of a smooth curve whose all other points do not belong to $ K$. Clearly, the unit circumference is a good set.

Now, for $ w\in \mathbb C$ and $ \rho>0$, consider the transformation $ T_{w,\rho}(z)\equal{}w\plus{}\rho T(\tfrac{z\minus{}w}{\rho})$ and $ S_{w,\rho}$, which is the similarly rescaled and shifted $ S$. The key point is that if $ K$ is good, $ w\in K$, and $ 10\rho<\text{dist\,}(0,K)$, then $ S_{w.\rho}(K)$ is also good. Moreover, the countable set $ D$ for $ S(K)$ can be obtained by taking the union of $ T_{w,\rho}^{\minus{}1}(D\setminus\{w\})$ and some countable open set on the circumference $ \{|z\minus{}w|\equal{}\rho\}$ where $ D$ is the countable set for $ K$.

Now, choose some sequence $ \rho_k>0$ such that $ \sum_k\rho_k<0.01$ and take $ K_0$ to be the unit circumference. Also choose some countable set $ D_0$ for $ K_0$. Now take $ w_1\in D_0$ and consider $ K_1\equal{}S_{w_1,\rho_1}(K_0)$. It is also a good set and we can choose $ D_1$ to be the image of $ D_0$ united with some dense subset $ D_1'$ of the circumference $ \{|z\minus{}w_1|\equal{}\rho_1\}$.

Now recall the old game when two people take turns and at each turn the first person adds a countable number of new balls to the pile while the second one removes one ball from the pile. Recall how the second person can leave the pile empty after they both do counatbly many moves. Play the same game with our dense sets: at each stage we may take one point in $ D$ and blow it into a circle at the cost of adding countably many new points on this circle. We can organise the process so that every point we ever had or added will be blown into a circle. I would hate to introduce the corresponding formal enumeration and my language here is a bit imprecise (namely, I do not distinguish between the points of the images of $ D_0$ and the points of $ D_0$), but I hope it is clear what I mean.

In this way, we get a sequence of mappings $ T_j\equal{}T_{w_j,\rho_j}$, their "inverse" mappings $ S_j$, and good compact sets $ K_j\equal{}S_j(K_{j\minus{}1})$. We want to show that the latter have some "limit" $ K$ and $ K$ still separates $ 0$ and $ \plus{}\infty$ and is connected but has no non-constant continuous curves in it. Once it is done, to get the domain the problem asks for, just take the connected component of $ \mathbb C\setminus K$ containing the origin. Since $ K$ is connected, it is simply connected, so it is homeomorphic to the disk. On the other hand, its boundary is a subset of $ K$.

The formal definition of $ K$ is as follows. A point $ x$ on the plane belongs to $ K$ is and only if for every $ \varepsilon>0$ the ball $ B(x,\varepsilon)$ intersects all but finitely many $ K_j$. We shall also need the limiting mappings $ T^{(n)}\equal{}T_{n\plus{}1}\circ T_{n\plus{}2}\circ T_{n\plus{}3}\circ\dots$. Their existense can be shown using the fact that $ T_j$ moves any point by at most $ 10\rho_j$ and the fact that each $ T_j$ is non-expanding. Note that $ T^{(n)}$ maps $ K$ to $ K_n$ (one more careful passage to the limit is needed to justify that). Showing that $ K$ is connected and separates $ 0$ from $ \infty$ is a fairly straightforward exercise. Now, assume that $ K$ contains a non-constant continuous curve $ \Gamma$. Then $ T^{(n)}\Gamma$ is a continuous curve in $ K_n$ for all $ n$ and it is non-constant for all sufficiently large $ n$. Taking one such $ n$, we see that this curve must pass through infinitely many points of $ D_n$ but each of these points (one is enough) is blown into a circle for some $ m>n$, so $ T^{(m)}\Gamma$ cannot be continuous.

This post is an (uneasy) compromise between trying to be short and trying to be clear. Feel free to ask as many questions as you need to understand everything but don't expect me to answer immediately at this time :).

Thank You!!! First questions is, what kind of program did you use for making this figure?

Have you been on AoPS during the last year at all, csabafarkas? :? :P. [[Asymptote]], of course!


Oh wow, your classroom math must be really hard.

[hide] What's $\log_{2}9\sqrt3$? [/hide]

[quote="dragonx"]What is the solution to $\log_{3}\sqrt{27}+\log_{2}3\sqrt{27}$

Pay attention to the bases of the logs![/quote]

$\log_{3}\sqrt{27}+\log_{2}3\sqrt{27}$

$\frac{1}{2}\log_{3}27+\log_{2}9+\frac{1}{2}\log_{2}3$

$\left(\frac{1}{2}\times 3 \right)+\log_{2}9+\frac{1}{2}\log_{2}3$

$\frac{3}{2}+3.962406252$

$\approx 5.462406252$.

by use of calculator of course  :wink:

edit: oh yea mods this needs to be in HSB.

[quote="dragonx"][quote="xpmath"]You don't by any chance go to Boulan Park Middle School. Wait... I think I recognize that username. Could you perhaps be Aishvar?
Anyway we had that for homework! Then our teacher said to change the 2 to a 3 in the 2nd term.[/quote]

yes it is me but if it was in base 2 what would be the answer is my question[/quote]

what do you mean base 2?  You want the answer in base 2 form?  I dont think thats what you are implying though because that answer woulld be more complicated(hard to simplify anywhere)

[quote="dragonx"]What is the solution to $\log_{3}\sqrt{27}+\log_{2}3\sqrt{27}$

Pay attention to the bases of the logs![/quote]
[hide]
$\log_{3}\sqrt{27}= \log_{3}3^{\frac{3}{2}}= \frac{3}{2}\\\\\log_{2}3\sqrt{27}= log_{2}3^{\frac{5}{2}}\\\\\text{simplifying we get...}\\\\5.46$[/hide]

[quote="Art of Owna"]

what do you mean base 2?  You want the answer in base 2 form?  I dont think thats what you are implying though because that answer woulld be more complicated(hard to simplify anywhere)[/quote]

He meant log base 2.

Anyways, for $\log_{3}{\sqrt{27}}+\log_{2}{3\sqrt{27}}$, you can mess around with the properties of logs to get

$\frac{1}{2}\log_{3}{27}+\log_{2}{3}+\frac{1}{2}\log_{2}{27}= \frac{3}{2}+\log_{2}{3}+\frac{1}{2}\log_{2}{27}$

$\approx 5.46241$

For the base 2 logs, use the [url=http://en.wikipedia.org/wiki/Logarithm#Change_of_base]change in base theorem[/url], which you may or may not learn in the near future if you haven't already, which makes it possible to plug into your calculator.


algebra
polynomial
linear algebra
matrix
linear algebra theorems

You have this theorem:

[quote="Known Theorem"]Let $A,B \in \mathcal M_2 ( \mathbb C )$. If the minimal polynomial of $A$ is its characteristic polynomial and $AB=BA$, then $B=\alpha A + \beta I_2$, $\alpha,\beta \in \mathbb C$.[/quote]

This can be proven by direct computations.

What about the generalization of this theorem? How do we prove that? Computations can't work, I think.

i don`t think it can be generalised.......

Let's at least state the proposed generalization in a way that makes sense:

Suppose that $A,B\in\mathcal{M}_n(\mathbb{C})$ and that the minimal polynomial of $A$ is its characteristic polynomial. If $AB=BA,$ then $B=P(A)$ where $P(x)$ is a polynomial of degree $<n.$

Now, is that true or false?

My teacher told me that for all $n$, $B$ can be written as a combination of the powers of $A$. Kent Merryfield, you say that's true or untrue?

Here's what I found:

$B_1=I_5$;

$B_2=A+A^3-A^4$;

$B_3=A^2+2A^3-2A^4$;

$B_4=I_5-2A^3+2A^4$;

$B_5=A-A^3+A^4$;

$B_6=A^2$.

Also, you can directly notice that $B_3+B_4-B_6-B_1=0$, so that space's dimension is $\leq 5$.

I don't know whether this "generalization" is true, but your counter-example does not seem to work. (perhaps there is a trivial counter-example to it)

I really hope that I didn't make any mistakes in my computations.

Silly me :). I don't know what the heck I was thinking about.

Anyway, your teacher's assertion is true, I think. I'll post an outline of the proof later.

Edit:

You may as well assume that $B$ has Jordan form, i.e. it's a block diagonal matrix, with diagonal blocks of the type $\begin{pmatrix}\lambda&1&0&\cdots&0\\0&\lambda&1&\cdots&0\\0&0&\lambda&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&\lambda\end{pmatrix}$
with only one block for each eigenvalue $\lambda$, and with the block $B_\lambda$ corresponding to the eigenvalue $\lambda$ having size equal to $m_\lambda$, the multiplicity of $\lambda$ in the characteristic polynomial $p_{B}$ of $B$.

Now notice that if $AB=BA$, then for every $\lambda$ we have $A(B-\lambda I)^{m_\lambda}=(B-\lambda I)^{m_\lambda}A$, so $A$ fixes $\ker (B-\lambda I)^{m_\lambda}$. This translates to the fact that $A$ also has a block diagonal structure, with a block $A_\lambda$ in the same position as the corresponding block $B_\lambda$ of $B$. Next you find the commuting space of a block $B_\lambda$. Check what happens for small dimensions. It should be easy enough to prove that this space consists of those upper-triangular matrices which which have equal entries on each diagonal parallel to the main one. In other words, the space generated by the powers of $\begin{pmatrix}0&1&0&\cdots&0\\0&0&1&\cdots&0\\0&0&0&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&0\end{pmatrix}$. This means that the space of matrices which commute with $B$ consists of block diagonal matrices which have diagonal blocks of that form. Its dimension is the sum of the dimensions of the blocks, i.e. exactly $n$. The linear combinations of powers of $B$ are easily seen to form a subspace of this space, but the dimension of this subspace is also $n$, so we're done: the two spaces coincide, i.e. every matrix which commutes with $B$ is a linear combination of powers of $B$, with exponents $0,1,\ldots,n-1$.

The lemma given by Kent Merryfield is true if A has distinct eigenvalues ,regarding the exercice now the condition relating to the minimal polynomial means in this case that I and A are no-linear ,we have 2 cases here :
-whether A has 2 distinct eigenvalues then we deduce that A and B are similar to two diagonal matrixes in the same base the answer to the answer then is an easy conclusion.
-otherwise A is similar to a triangular matrix (for instance lower triangular) then we prove that B is similar to a lower triangular matrix in the same base and by computation we reach the answer of the exercice again :).

[quote="spix"]$A$ and $B$ should be different from $I_n$.[/quote]

I don't understand what you are saying  :?

If $AB=BA$ then $B=f(A)$ . Take $A=I_n$ then it would be that $B$ is of diagonal form. false  :|

We consider $I_n$ to be $B^0$. And saying that $B\ne I_n$ is redundant, since our hypothesis tells us that the minimal polynomial of $B$ is the same as its characteristic polynomial, and this clearly doesn't hold for $I_n$. In other wods, those conditions $A,B\ne I_n$ have no place here :).

Yes  :blush:  I realised that during my nap. Sorry.

[quote="grobber"]You may as well assume that $B$ has Jordan form[/quote]

Why can you assume this?

 :?

Because any matrice over $C$ can be reduced to this form using the Jordan diagonalisation process,i.e $M=P^{-1}DP$,where $D$ is a matrice with only Jordan blocks on the diagonal. It's not a difficult thing to prove it but you should know some reduction theory.

All matrices that commute with $M$ also commute with $D$? This is what I was asking.

Read what Grobber wrote in his post.He explained what are you asking right now:"This means that the space of matrices which commute with $B$ consists of block..... ".
So the answer is yes.


Does anyone know any simple solutions to the difficult questions? Thanks! (I think I got a 92(?) :blush: but I'm only in 7th grade.)

Which difficult questions are you refering to? Check all the other threads - the ones where people are discussing the problems - they might have stuff there too.

there was an AMC-10 Math Jam a week ago...u can check out the transcript. It went through all the last five questions. Check it out at the Math Jam sections.


Two teams play a game a of soccer, which team A wins 10-8. Assuming any order of the 18 goals is equally likely with another, what is the probability the score was never tied after the first goal?

This is [url=http://en.wikipedia.org/wiki/Bertrand%27s_ballot_theorem]Bertrand's ballot theorem[/url].

Oh, thanks man. Might be handy on a future AIME.


calculus
integration
logarithms
trigonometry
calculus computations

I have asked a lot of questions. Now I think it is time I contributed to the fun by asking a question not for help but for the enjoyment of others   .

Find

$ L\equal{}\int_0^1\frac{\ln^3\left(\frac{1\plus{}x}{1\minus{}x}\right)}{x}\,dx$

Have fun.  :D

Well, JRav, I'm pretty sure it exists. I don't really have time for calculations, but the answer is

[hide="Answer"][hide="Pretty useless, but just showing that it exists in closed form"]$ \frac {\pi^4}{8}$[/hide][/hide]

Maybe I'll post a solution later.

I meant in terms of an antiderivative.  Or does one have to do weird expansions and substitutions and the like.

Simple(?) substitution $ x \equal{} \tan (\tfrac{u}{2})$ shows that the given integral is equal to $ \int_{0}^{\infty} \frac{u^3}{\sinh u} \, du \equal{} \frac{\pi^4}{8}$.

But well... I don't think $ u^3 \text{csch} \, u$ has an elementary antiderivative.

[quote="JRav"]Are you sure this integral indeed exists in a closed form?[/quote]

Well, I liked one of professor Merryfield's comments on this subject:

[quote="Kent Merryfield"]I'm not familiar enough with the esoteric algebraic topic of elementary representation of integrals to make any definitive pronouncements, but I will say this: there is no good reason to assume that it can be represented in elementary terms.

["Elementary terms" means starting with powers, trig and inverse trig functions, exponentials and logarithms, and building the function in question in a finite number of arithmetic or function-composition steps. A power series involves an infinite number of steps. Of course, benorin is right; this has an easily found power series which is useful for many - but not all - purposes.]

And that's the usual response to many similar questions: there's no good reason to assume that it can. One should really get in the habit of thinking of elementary representation of integrals as the exception rather than the rule.[/quote]

[quote="sos440"]Simple(?) substitution $ x \equal{} \tan (\tfrac{u}{2})$ shows that the given integral is equal to $ \bf{\color{red}{\int_{0}^{\infty} \frac {u^3}{\sinh u} \, du \equal{} \frac {\pi^4}{8}}}$.

[/quote]

Yeah, but prove it   :D

[quote="sos440"][quote="Mathstud28"]Yeah, but prove it   :D[/quote]

Oops... Here it is:

\begin{eqnarray*} \int_{0}^{\infty} \frac {x^3}{\sinh x} \, dx & = & 2 \int_{0}^{\infty} \frac {x^3 e^{ - x}}{1 - e^{ - 2x}} \, dx \\
& = & 2 \int_{0}^{\infty} x^3 e^{ - x} \sum_{n = 0}^{\infty} e^{ - 2nx} \, dx \\
& = & 2 \sum_{n = 0}^{\infty} \int_{0}^{\infty} x^3 e^{ - (2n + 1)x} \, dx \\
& = & 2 \sum_{n = 0}^{\infty} \frac {\Gamma(4)}{(2n + 1)^4} \\
& = & 2 \Gamma(4) \cdot \left( 1 - \frac {1}{2^4} \right) \zeta(4) \\
& = & \frac {\pi^4}{8} \end{eqnarray*}

The change of order between integral and sum is justified by Tolleni's theorem.[/quote]

That is very similar to my solution! Very good!  :D

[quote="sos440"]The change of order between integral and sum is justified by Tolleni's theorem.[/quote]

I couldnt find this theorem on google. What is it please?

[quote="skylarth"]I couldnt find this theorem on google. What is it please?[/quote]

I'm sorry I swapped 'n' and 'll'. [url=http://en.wikipedia.org/wiki/Fubini%27s_theorem#Tonelli.27s_theorem]'Tonelli's theorem'[/url] is the right name for that theorem.


[quote="pkothari13"]The question posed was

Find the integer closest to $1000\displaystyle\sum\limits_{n=3}^{10000}\dfrac{1}{n^2-4}$

In their solution, they use a telescoping series but say it reduces to

$250(1 + 1/2 + 1/3 + 1/4 - 1/9999 - 1/10000 - 1/10001 - 1/10002)$

My question is, don't the $1/3$, $1/4$, $1/9999$, and $1/10000$ all telescope out as well? Am I just overlooking something or what?[/quote]
No - the first and last four terms don't cancel.  Because the sequence telescopes as follows

$S = 10^3 \displaystyle\sum\limits^{10^4}_{n=3} \dfrac{1}{n^2 - 4} = 250 \left(\displaystyle\sum\limits^{10^4}_{k = 3} \dfrac{1}{n -2} - \dfrac{1}{n + 2} \right)= 250 \left(\displaystyle\sum\limits^{10^4}_{n=3}\dfrac{1}{n-2} - \displaystyle\sum\limits^{10^4 + 4}_{n=7} \dfrac{1}{n - 2}\right)$

you can see that the first terms $n = 3,4,5,6$ and the last terms $n = 10^4 + 1, 10^4 + 2, 10^4 + 3, 10^4 + 4$ do not cancel.  I hope this makes sense.

Thank you very much zanttrang! I now see my error.  :)


Heya,
Ive been wanting to write a computer program for quite a while (like 7 years) but this one thing always stumps me, and so i always drop it after a bit.  I figured now i would ask for help  :)  I had no idea where else to post this, so you folks get it :)

You have two spaceships with known starting locations.  Assume instant acceleration.  Ship A starts out at a particular velocity and in a particular direction in 3d space from its starting point.  Ship B has a know starting point, and a known maximum velocity (which it will always use and accelerate to immediately.  Nice technology eh?).

What i need to determine is:
1. Can ship B overtake ship A (easiest part)
2. Where does it overtake ship A (point in space)
3. How long does it take.
4. What is the angle of pursuit that ship B must take to intercept ship A

I could do this brute force in a computer program in 2d, but adding the third dimension added a whole new layer of complexity.  Also, my brute force method wasnt the most elegant way of determining the answer, and i was hoping for somehting a little cleaner.

I think you should post this in another section of the forum. Maybe try [url=http://www.artofproblemsolving.com/Forum/index.php?f=331]Computer Science and Informatics[/url].

I think you can do it yourself in 2D. You won't need a computer. And in 3D the concept of overtaking is unclear.


$ \bigstar$ Let $ a,b,c,d\ge 0$ such that $ a\plus{}b\plus{}c\plus{}d\equal{}4$. Prove that:
\[ abcd\plus{}abc\plus{}bcd\plus{}cda\plus{}dab\le 5\]

Setting $ f\left( {a,b,c,d} \right) \equal{} abcd \plus{} abc \plus{} bcd \plus{} cda \plus{} dab$ then we have

$ f\left( {a,b,c,d} \right) \le f\left( {\frac{{a \plus{} b}}{2},\frac{{a \plus{} b}}{2},c,d} \right)$

without any condition so we will also have

$ f\left( {\frac{{a \plus{} b}}{2},\frac{{a \plus{} b}}{2},c,d} \right) \le f\left( {\frac{{a \plus{} b}}{2},\frac{{a \plus{} b}}{2},\frac{{c \plus{} d}}{2},\frac{{c \plus{} d}}{2}} \right)$

$ f\left( {\frac{{a \plus{} b}}{2},\frac{{a \plus{} b}}{2},\frac{{c \plus{} d}}{2},\frac{{c \plus{} d}}{2}} \right) \le f\left( {\frac{{a \plus{} b \plus{} c \plus{} d}}{4},\frac{{a \plus{} b \plus{} c \plus{} d}}{4},\frac{{a \plus{} b \plus{} c \plus{} d}}{4},\frac{{a \plus{} b \plus{} c \plus{} d}}{4}} \right)$

AND

$ f\left( {\frac{{a \plus{} b \plus{} c \plus{} d}}{4},\frac{{a \plus{} b \plus{} c \plus{} d}}{4},\frac{{a \plus{} b \plus{} c \plus{} d}}{4},\frac{{a \plus{} b \plus{} c \plus{} d}}{4}} \right) \equal{} 5$

So we have done and the equality hold for $ a \equal{} b \equal{} c \equal{} d \equal{} 1$

We have a nice solution (and easy):Apply AM_GM inequality we have:
$ 16(abc\plus{}bcd\plus{}cda\plus{}dab)\equal{}16ab(c\plus{}d)\plus{}16cd(a\plus{}b)\le 4(a\plus{}b)^2(c\plus{}d)\plus{}4(c\plus{}d)^2(a\plus{}b)\equal{}4(a\plus{}b)(c\plus{}d)(a\plus{}b\plus{}c\plus{}d)\le (a\plus{}b\plus{}c\plus{}d)^3\equal{}64$
$ abcd\le (\frac {a\plus{}b\plus{}c\plus{}d}{4})^4\equal{}1$
So the inequality is proven. :D

very nice proof, I don't think my problems is easy like that  :blush:

[quote="zaizai-hoang"]your solution don't show anything  :ninja: Can you post clearly? Using Mixing Variable method is also my proof for this one  :wink:[/quote]

Because of   $ \left( {\frac {{a \plus{} b}}{2}} \right)^2 \ge ab$    zaizai-hoang. We have

$ f\left( {a,b,c,d} \right) \equal{} ab\left( {cd \plus{} c \plus{} d} \right) \plus{} cd\left( {a \plus{} b} \right)$

$ f\left( {\frac {{a \plus{} b}}{2},\frac {{a \plus{} b}}{2},c,d} \right) \equal{} \left( {\frac {{a \plus{} b}}{2}} \right)^2 \left( {cd \plus{} c \plus{} d} \right) \plus{} cd\left( {a \plus{} b} \right)$

So $ f\left( {a,b,c,d} \right) \le f\left( {\frac {{a \plus{} b}}{2},\frac {{a \plus{} b}}{2},c,d} \right)$
without any condition