geometry
linear algebra
matrix
vector
geometric transformation
rotation

Have been revising geometry today and have came across some proofs that I can't seem to find in books, but I can't get through either. Any help would be great.

Let A be a 3x3 orthogonal matrix and let x and y be vectors in R^3

a) Show that detA = +/- 1

b) Show that the length of Ax is the same as that of x and that x and y are orthogonal iff Ax and Ay are orthogonal
Suppose further that A represents a rotation through angleθ , with axis of rotation along the unit vector n, show that if m is a unit vector orthogonal to n, then n.m^Am = sinθ

attempt at a)

from defn of orthogonal matrix (Atr.A = I)

det(Atr . A) = det(I) = 1
using standard results such as det(A.B) = detA. detB and det(Atr) = det(A)
we have det(Atr).det(A) = det(A)^2
detA^2=1
therefore det A = +\- 1

b) struggling to start.

Thank you in advance

What you want to do for b) is to prove something even more general: prove that the inner product of $ Ax$ with $ Ay$ is the same as the inner product of $ x$ and $ y.$

Note that the inner product can be defined by matrix multiplication: $ x\cdot y \equal{} y^Tx.$ Use that for all it's worth; use the language of matrix multiplication to guide your steps.

Why would showing that $ (Ax)\cdot(Ay) \equal{} x\cdot y$ give you everything you need? Think about how to compute both the magnitude of a vector and the angle between vectors.

Claim: Ax.Ay = x.y if Ax and Ay are orthogonal
Proof: Ax.Ay = Atr .Ax.y
by use of the property of othogonality this = x.y

Is this the proof that is required?
I still don't really see this :(

[quote="Kent Merryfield"]Why would showing that $ (Ax)\cdot(Ay) \equal{} x\cdot y$ give you everything you need? Think about how to compute both the magnitude of a vector and the angle between vectors.[/quote]  Focus on this line.  Kent is suggesting that you actually do the proof in two steps:
Step 1: Show that if $ A$ is orthogonal then for any vectors $ x$ and $ y$ we have that the dot product of $ Ax$ and $ Ay$ is the same as the dot product of $ x$ and $ y$.  (What you wrote looks like an attempt to prove this, but it's unclear if you had a good grasp on what you were doing.)
Step 2: Use the fact from step 1 to show that the length of $ Ax$ is the same as the length of $ x$ and that $ x$ is perpendicular to $ y$ if and only if $ Ax$ is perpendicular to $ Ay$.  You'll want to use some properties of the dot product here (which is why we've gone through Step 1).


A pyramid with a square base has height $h$ and all of its edge lengths are the sum. Find the volume $V$ of the pyramid

Please ammend the wording of the first sentence. Should it be "are the same"? If it should, then

[hide]$h=\frac{a\sqrt{2}}{2}\implies V=\frac{a^2}{3}\cdot\frac{a\sqrt{2}}{2}=\frac{a^3\sqrt{2}}{6}=\frac{(h\sqrt{2})^3\sqrt{2}}{6}=\frac{2h^3}{3}$[/hide]

Yeah, should be "the same." The answer is correct.

[hide]
If b is the legnth of an edge of the base then
b^2=.5b^2+h^2
b^2=(h^2)*4/3
1/3(h^2*4/3)h
4/9(h^3)
[/hide]


trigonometry
geometry
calculus
analytic geometry
graphing lines
slope
geometric transformation

Given a point  $M$  inside the square $ABCD$  and $N-$ midpoint of  $BC$ such that $\angle{MAD}=\angle{NMD}=\angle{MDC}=x$ .Find the value of $x$.

The triangle $\triangle CDM$ is isosceles with $\measuredangle CDM = \measuredangle CMD = x$. Hence, $CM = CD$ and the point $M$ is on a circle $(C)$ centered at the vertex $C$ and with radius $r = CD$ (side of the square $ABCD$). This circle intersects the line $CD$ at one other point $E$ different from the point $D$. The angle $%Error. "meeasuerdangle" is a bad command.
AMD$ is right, because in the triangle $\triangle AMD$,

$\measuredangle AMD = 180^o - (\measuredangle MAD + \measuredangle MDA) =  180^o - (x + 90^o - x) = 90^o$

Thus the line $AM$ passes through the point $E$. From the right angle triangle $\triangle ADE$ with the legs $DA = r, DE = 2r$,

$\tan x = \frac{DE}{DA} = \frac{2r}{r} = 2$

$x \doteq 63.435^o$

[quote="Mamat"][hide]x=75 :^o:[/hide][/quote]
is this from observation and guessing? i get 75 when i saw this problem for the first time too....

[quote="shobber"][quote="hardsoul"]Given a point  $M$  inside the square $ABCD$  and $N-$ midpoint of  $AB$ such that $\angle{MAD}=\angle{NMD}=\angle{NDM}=x$ .Find the value of $x$.[/quote]
What does $N-$ mean?[/quote]
 I editted  ;) .

[quote="yetti"][quote="hardsoul"]Given a point  $M$  inside the square $ABCD$  and $N-$ midpoint of $BC$ such that $\angle{MAD}=\angle{NMD}=\angle{NDM}=x$. Find the value of $x$.
...
Sorry guys I have editted the problem. :blush: [/quote]
You did not improve things, because now the problem does not have a solution. Since the triangle $\triangle DMN$ is isosceles with $\angle{NMD}=\angle{NDM} = x$, the segments $ND = NM$ are equal and the point $M$ is on a circle $(N, r = ND = NA)$ passing through the square vertices $D, A$. This circle does not pass through any interior points of the square $ABCD$. The angle $\angle DNM = 180^o - (\angle{NMD}+\angle{NDM}) = 180^o - 2x$. The angle $\angle MAD$ spans the major arc $MD$ with the central angle $\angle DNM = 180^o - 2x$, hence

$x = \angle MAD = \frac{\angle DNM}{2} = 90^o - x\ \ \Rightarrow\ \ x = 45^o$

but the point $M$ is not inside the square $ABCD$.[/quote]

Dear yetti I am very ashamed for these errors.Finaly I have editted the problem(hope the last time).

The angle $\angle AMD$ is right, because

$\angle AMD = 180^o - (\angle MAD + \angle MDA) = 180^o - (x + 90^o - x) = 90^o$

Hence, the point $M$ is on a circle $(K)$ with diameter $AD$ centered at the midpoint $K$ of the segment $AD$. The angle $\angle MAD = x$ spans the major arc $MD$ of the circle $(K)$ and it is equal to the angle $\angle NMD = x$. This means that the line $NM$ is a tangent to the circle $(K)$ from the point $N$. To construct the tangent, draw a circle $(O)$ centered at the midpoint $O$ of the segment $NK$ and with radius $ON = OK = \frac{NK}{2}$. The circles $(K), (O)$ intersect at points $M, M'$. Since $\angle NM'D > 90^o$, this point is not acceptable. The circles $(K), (O)$ are congruent, both having diameters equal to the square side. Thus $2MK = NK$ in the right angle triangle $\triangle NMK$, which means that the angle $\angle MKN = 60^o$. But then the angle $\angle AKM = 90^o - \angle MKN = 30^o$ and this is the central angle of the arc $AM$ of the circle $(K)$. Hence, the angle $\angle ADM$ spanning this arc is equal to $\angle ADM = \frac{\angle AKM }{2} = 15^o$ and the desired angle is equal to $x = \angle MDC = 90^o - \angle ADM = 75^o$.

Nice problem.

Regards, Yetti.

What luck! What geomertry program is this thats being used? Ive been looking for one, but cannot find...

The geometry program I am using is called [url=http://www.keypress.com/sketchpad/]The Geometer's Sketchpad[/url] by Key Curriculum Press. It has a lot more capabilities than performing very accurate constructions with compass and straightedge (graphs, analytic geometry, basic calculus and approximation methods, etc.), but I am mainly using this part. It allows you to draw points, lines through 2 points, circles with a given center and passing through a given point, 2 point segments, 3 point arcs, find their intersections or tangency points and perform the basic constructions:

1. find the midpoint of a given segment
2. draw the parallel or perpendicular line to a given line through a given point
3. construct the angle bisector of a given angle

In addition, you can hide unimportant objects (points, lines, circles), find loci of points, lines, circles, etc., perform very accurate measurements (up to 5 decimal places) of segments, angles, slopes, polygon or circle areas, translate, dilate and/or rotate the entire construction, either itself or added as a transformed copy to the current construction, and even animate the construction. It also includes a basic calculator with all trigonometric functions and all the calculated expressions are displayed in the construction area (until deleted).

In principle, you can build custom tools for often performed construction tasks, such as an inversion tool, but I have not built any yet. I use one custom tool provided with the program, that allows to construct conic sections (ellipse or hyperbola from 2 focal points and one point on the curve, any conic section from 5 points, etc.). The conic sections are constructed as loci and the program does not allow to find intersections of lines or circles with loci.

The program can be purchased on-line (click on the above link) and it comes in 3 versions:

1. Student-parent edition for $\$$40.
2. Single user edition for $\$$120.
3. Multiple user edition for $\$$300 (for schools).

The only difference between the single user and student-parent version is the requirement of the student-parent version to have the CD in the drive at the program startup (afterwards, the CD can be removed and the program runs from the hard disk) and that the student-parent edition is not eligible for discounts on future upgrades. The program is very popular, very easy to use and so good that the expense seems to be a mere trifle.

Yetti


trigonometry
Functional Analysis
advanced fields
advanced fields unsolved

Let $ (\mathfrak{X}, \|\cdot\|)$ be a real Banach algebra with unit element 1. Define the linear operations in $ \mathfrak{Y} \equal{} \mathfrak{X} \times \mathfrak{X}$ by:

$ (x_1, x_2) \plus{} (y_1, y_2) \equal{} (x_1 \plus{} y_1, x_2 \plus{} y_2)$
$ (\alpha \plus{} i\beta)(x_1, x_2) \equal{} (\alpha x_1 \minus{} \beta x_2, \alpha x_2 \plus{} \beta x_1)$
$ (x_1, x_2)(y_1, y_2) \equal{} (x_1 y_1 \minus{} x_2 y_2, x_1 y_2 \plus{} x_2 y_1)$.

Then $ \mathfrak{Y}$ is easily seen to be a complex algebra. Now define a norm, which we also denote by $ \|\cdot\|$, on $ \mathfrak{Y}$ as follows:

$ \|(x_1, x_2)\| \equal{} \sup_\theta \|x_1 \cos\theta \plus{} x_2 \sin\theta \|$

With this norm $ \mathfrak{Y}$ is a complex Banach space --- so far so good. What I'm having trouble with is the following: Show that $ \mathfrak{Y}$ is a complex Banach algebra under the norm

$ \|(x_1, x_2)\|' \equal{} \sup \left\{ \|(x_1, x_2)(y_1, y_2)\| \, \|(y_1, y_2)\|^{\minus{}1} \colon (y_1, y_2) \in \mathfrak{Y} \backslash \{0\} \right\}.$

I can show that $ \|\cdot\|'$ is a norm on $ \mathfrak{Y}$ with no problems. However, I can't prove that $ \|\cdot\|'$ is submultiplicative. The best I could do is prove that

$ \|(x_1, x_2) (y_1, y_2)\|' \leq 2 \|(x_1, x_2)\|' \|(y_1, y_2)\|'.$

I can't get rid of the $ 2$. Any help?

Hint: What is the norm of the left multiplication with (x1,x2)? And use $ \|S\circ T\| \le \|S\|\| T\|$ for bounded operators - of course whenever the composition is defined.

Thank you --- can't believe I missed that!


alright i told my club id make a chess tourney and here i have one. it will contain 8 players including me so sign up! anyway when everyone joins the tourney will start. you will be playing others in this tourney but how? its impossible to choose ! [size=200][b]WRONG.[/b][/size]
in the more options section there is an option called confirm manually this means you can choose who you want to play but alot of guests will accept so here are instructions to get the person you want

[hide=" instructions"]1. both of you go on to aops and make a private chatroom
2.go on  to fics with another tab.
3 tell each other your guest names its at the top of the black screen
4. create you control manually game when your correct opponent accepts then type "pending" a list of numbers will show up as well as players who accepted type accept ____ (the number by the game you want) and there youre playing a game for my tourney![/hide]

I join, but I really don't want to use the website, for whatever reason. I'm gonna force the other dude I'm facing to use yahoo.  :P

alright but if no one else is joiining then i am that "other dude"

Uhh I guess I'll join. I'll lose to whoever I play though. :D

il join but all u ppl are owned :D i hope :P

ratings??

I join too.

I am tired being pwn-ed by computer :D

my rating used to b 800. yes i no i suck...  :noo:

we will be using my own methods for this game everyone in the game can play whom the y want but only once untill they have played everyone. winner of this tourney is first to win 3 games. keep track of points and post every game


Asymptote
function
search
algebra
rational function

Find the vertical and horizontal asymptotes of f(x) if there are any

f(x)=

[u]3x/\2 - 9x -30[/u]
x/\3 - 7x/\2 - 18x

[hide]the vertical asymptote is found be setting the denominator equal to zero, so
$x^3-7x^2-18x=0$
$x(x+2)(x-9)=0$
$x=0,-2,9$
these are the vertical asymptotes.[/hide]

Horizontal asymptote: [hide]... is at y=0.
${{(\frac{3x^2 - 9x -30}{x^3 - 7x^2 - 18x})(\frac{\frac{1}{x^2}}{\frac{1}{x^2}}})= \frac{3-\frac{9}{x}-\frac{30}{x^2}}{x-7-\frac{18}{x}}}$

As x approaches infinity, the limit is 0 since the denominator gets vastly greater than the numerator; we have

$\frac{3-(nothing)}{(huge x)-7-(nothing)}$

The huge x in the denominator makes the whole function approach 0.

Thus, in any rational function where the maximum power of x is higher in the denominator, there is a horizontal asymptote at y=0.
[/hide]

Can someone explain how he go the horizontal asymptote

If you don't understand any of the limit stuff yet, don't worry.
There are three situations for the function to take.
1) The degree of the numerator is less than the degree of the denominator.

Ex: $\frac{x+2}{x^2+3}$

The limit or horizontal asymptote in this case is 0.
2) The degree of the numerator is equal to the degree of the denominator.

Ex: $\frac{3x^2+3}{x^2+4}$

The limit in this case is the coeffecient of the leading term in the numerator divided by the coefficient of the leading term in the denominator. Thus, the limit or horizontal asymptote is 3.
3) The degree of the numerator is greater than the degree of the denominator.

Ex: $\frac{x^2+1}{x+3}$

In this case, the asymptote will be what's called a slant asymptote where the asymptote is a slanted line.

If you want to know more, search it up.

I got the horizontal asymptote by multiplying by 1; that is, I multiplied your original function by $\frac{\frac{1}{x^2}}{\frac{1}{x^2}}$, which is 1, since the numerator and denominator are equal.

So consider the new function after the multiplication (look up there). Now, asymptotes describe the behavior of the function at extreme values of x. Note that as x gets huge, every term with x or x^2 in the denominator gets close to zero, and thus is said to have a limit at zero. So discount all the terms with x in the denominator, since they have no value at extremes. We are left with $\frac{3}{x-7}$, and since x is huge, the fraction as a whole approaches 0. Thus, the function approaches 0 at extreme values of x, so the asymptote is 0.


One ticket to a show costs $ \$20$ at full price. Susan buys $ 4$ tickets using a coupon that gives her a $ 25\%$ discount. Pam buys $ 5$ tickets using a coupon that gives her a $ 30\%$ discount. How many more dollars does Pam pay than Susan?

$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 20$

[hide="Solution"] 75% discount off of 20 dollars is 15 dollars (since we are taking a 25% discount).  Therefore, Susan pay 15*4=60 dollars.  The price per ticket Pam pays is 70% of 20 dollars, which is 14 dollars.  Thus, Pam pays 5*14=70 dollars.  70 dollars is 10 dollars more than 60 dollars, so the answer is C.  [/hide]

[quote="4everwise"]One ticket to a show costs $\$20$ at full price. Susan buys $4$ tickets using a coupon that gives her a $25\%$ discount. Pam buys $5$ tickets using a coupon that gives her a $30\%$ discount. How many more dollars does Pam pay than Susan?

$\text{(A)}\ 2 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 20$[/quote]
[hide]So without the discount Susan pays $\$$80. With the discount she pays $\$$60. Pam without the discount pays $\$$100. With the discount she pays $\$$70. So 70-60=10, or C.[/hide]

Susan buys $4$ tickets at $20$ dollars each for a total of $80$ dollars.  Then, after a 25% discount, her final price is $(.75 \cdot 80) = 60$.

Pam buys $5$ tickets at $20$ each for a total of $100$.  Then, after a 30% discount, her final price is $(.70 \cdot 100) = 70$.

$70-60 = \boxed{10}$

[quote="4everwise"]One ticket to a show costs $\$20$ at full price. Susan buys $4$ tickets using a coupon that gives her a $25\%$ discount. Pam buys $5$ tickets using a coupon that gives her a $30\%$ discount. How many more dollars does Pam pay than Susan?

$\text{(A)}\ 2 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 20$[/quote]
[hide="Answer"]20 * 4 = 80. 25% off 80 is 60. 20 * 5 = 100. 30% off 100 is 70. 70 - 60 = [b]10[/b][/hide]


function
combinatorics proposed
combinatorics

Let $ S$ be a sequence that:
\[ \left\{
\begin{array}{cc}
S_0\equal{}0\hfill\\
S_1\equal{}1\hfill\\
S_n\equal{}S_{n\minus{}1}\plus{}S_{n\minus{}2}\plus{}F_n& (n>1)
\end{array}
\right.\]
such that $ F_n$ is Fibonacci sequence such that $ F_1\equal{}F_2\equal{}1$. Find $ S_n$ in terms of Fibonacci numbers.

That's kind of cute:  Let $ S(x) \equal{} \sum_{n \geq 0} S_n x^n$ be the generating function for $ S_n$.  Then $ S(x) \equal{} x \plus{} \sum_{n \geq 2} (S_{n \minus{} 1} \plus{} S_{n \minus{} 2} \plus{} F_n)x^n \equal{} x \plus{} xS(x) \plus{} x^2S(x) \plus{} F(x)$, where $ F(x) \equal{} \frac {x}{1 \minus{} x \minus{} x^2}$ is the generating function for the Fibonacci numbers.  This immediately gives $ S(x) \equal{} \frac {x}{(1 \minus{} x \minus{} x^2)^2}$, so $ xS(x) \equal{} F(x)^2$ and thus $ S_n \equal{} \sum_{k \equal{} 0}^{n \plus{} 1} F_k F_{n \plus{} 1 \minus{} k}$.

Also
\[ S_n \equal{} \frac{nF_{n \plus{} 1} \plus{} 2(n \plus{} 1)F_n}{5}.\]


$ (a;b;c;x;y;z) \geq 0$ and $ x\plus{}y\plus{}z\equal{}1$
prove that:
$ ax\plus{}by\plus{}cz\plus{}2\sqrt{(xy\plus{}yz\plus{}zx)(ab\plus{}bc\plus{}ca)} \leq a\plus{}b\plus{}c$

[quote="peine"]$ (a;b;c;x;y;z) \geq 0$ and $ x \plus{} y \plus{} z \equal{} 1$
prove that:
$ ax \plus{} by \plus{} cz \plus{} 2\sqrt {(xy \plus{} yz \plus{} zx)(ab \plus{} bc \plus{} ca)} \leq a \plus{} b \plus{} c$[/quote]

$ ax\plus{}by\plus{}cz\plus{}2\sqrt{(xy\plus{}yz\plus{}zx)(ab\plus{}bc\plus{}ac)}\equal{}ax\plus{}by\plus{}cz\plus{}\sqrt{(ab\plus{}bc\plus{}ac)(xy\plus{}yz\plus{}zx)}\plus{}\sqrt{(ab\plus{}bc\plus{}ac)(xy\plus{}yz\plus{}zx)}\leq\sqrt{(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}ab\plus{}bc\plus{}ac\plus{}ab\plus{}bc\plus{}ac)(x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}xy\plus{}yz\plus{}zx\plus{}xy\plus{}yz\plus{}zx)}\equal{}\sqrt{(a\plus{}b\plus{}c)^{2}(x\plus{}y\plus{}z)^{2}}\equal{}a\plus{}b\plus{}c$

[quote="mathwizarddude"][quote="zaya_yc"]$ ax \plus{} by \plus{} cz \plus{} \sqrt {(ab \plus{} bc \plus{} ac)(xy \plus{} yz \plus{} zx)} \plus{} \sqrt {(ab \plus{} bc \plus{} ac)(xy \plus{} yz \plus{} zx)}\leq\sqrt {(a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} ab \plus{} bc \plus{} ac \plus{} ab \plus{} bc \plus{} ac)(x^{2} \plus{} y^{2} \plus{} z^{2} \plus{} xy \plus{} yz \plus{} zx \plus{} xy \plus{} yz \plus{} zx)}$[/quote]

Excuse me, but how are you getting the above inequality?[/quote]
basically , it's Cauchy-Schwartz as i can see :D
firstly , you have $ ax \plus{} by \plus{} cz \leq \sqrt{(a^2\plus{}b^2\plus{}c^2)(x^2\plus{}y^2\plus{}z^2)}$

so
 $ ax \plus{} by \plus{} cz \plus{} \sqrt {(ab \plus{} bc \plus{} ac)(xy \plus{} yz \plus{} zx)} \plus{} \sqrt {(ab \plus{} bc \plus{} ac)(xy \plus{} yz \plus{} zx)} \leq \sqrt{(a^2\plus{}b^2\plus{}c^2)(x^2\plus{}y^2\plus{}z^2)} \plus{} \sqrt {(ab \plus{} bc \plus{} ac)(xy \plus{} yz \plus{} zx)} \plus{} \sqrt {(ab \plus{} bc \plus{} ac)(xy \plus{} yz \plus{} zx)}$
and then basically you apply this inequality (which is Cauchy-Schwartz)
$ \sqrt{mp}\plus{}\sqrt{nq}\plus{}\sqrt{st} \leq \sqrt{(m\plus{}n\plus{}s)(p\plus{}q\plus{}y)}$

[quote="mathwizarddude"][quote="zaya_yc"]$ ax \plus{} by \plus{} cz \plus{} \sqrt {(ab \plus{} bc \plus{} ac)(xy \plus{} yz \plus{} zx)} \plus{} \sqrt {(ab \plus{} bc \plus{} ac)(xy \plus{} yz \plus{} zx)}\leq\sqrt {(a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} ab \plus{} bc \plus{} ac \plus{} ab \plus{} bc \plus{} ac)(x^{2} \plus{} y^{2} \plus{} z^{2} \plus{} xy \plus{} yz \plus{} zx \plus{} xy \plus{} yz \plus{} zx)}$[/quote]

Excuse me, but how are you getting the above inequality?[/quote]

$ n\equal{}\sqrt{ab\plus{}bc\plus{}ac}$ and $ m\equal{}\sqrt{xy\plus{}yz\plus{}zx}$ ---> $ ax\plus{}by\plus{}cz\plus{}\sqrt{(ab\plus{}bc\plus{}ac)(xy\plus{}yz\plus{}zx)}\plus{}\sqrt{(ab\plus{}bc\plus{}ac)(xy\plus{}yz\plus{}zx)}\equal{}ax\plus{}by\plus{}cz\plus{}nm\plus{}nm\leq\sqrt{(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}n^{2}\plus{}n^{2})(x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}m^{2}\plus{}m^{2})}\equal{}\sqrt {(a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} ab \plus{} bc \plus{} ac \plus{} ab \plus{} bc \plus{} ac)(x^{2} \plus{} y^{2} \plus{} z^{2} \plus{} xy \plus{} yz \plus{} zx \plus{} xy \plus{} yz \plus{} zx)}$
(Cauchy-Schwartz $ n\equal{}5$)


Weight of basket : 0.2 kg

Tom measured the weigh of baket included one grape.
The measured value was 3.45 kg.
Then how much weigh a grape have?

you must be either kiddin, or forgeting some important parts of the problem ...  :maybe: so which one is it?

Hello, manager!

There are all of conditions where wrote on the text of test.

So, it seems to find out approximat value.

But, are the conditions in text above satisfied with solving problem as approximate value?

If the conditions limited in my text above, isn't the answer of this question is 3.25 kg?

If it has purpose of understanding subtration of approximate value, then is this problem has enough conditions to finding out 3.3 kg?

Please answer me.

first of all, I've never heard of a grape to weight over 3 kilos, so something MUST be wrong with the problem :D or maybe with its translation.

Don't mention any FACTS that we knew before !!

Think it ONLY CONDITIONS for PROBLEM.

Please read again and solve the problem.

There aren't any error in translation.

And.... the CONDITIONS about problem is limited by my problem text.

ONLY those conditions were written in the original text of test.

And.. please solve again....

One more..
If the conditions are limited by my above text, can we know it's about approximate value?

Please think it for GENERAL ideas.

Or.... Is it has enough conditions to know that it's about approximate value?

I think it hasn't enough conditions to know that.

I think the answer is 3.25 kg.

  >>>and..... I SAID the GRAPE is ideal thing.

Excuse me but, why are you so angry? Why are you yelling?
You have no respect for other users, no wonder no one wants to solve your problem.
Valentin: maybe there are 325 grapes each with a weight of 0.01kg. That sounds more realistic and grapes with weight 0.01kg are probably big and juicy, yummy  :cool:  :blush:  ;)


LaTeX
calculus
articles
function
geometry
algebra
floor function

The question was what is infinute over infinite.

[quote="jackdillon"]What is infinete/infinete?(I don't now enough Latex to make infinete signs...)[/quote]

I would guess 1 but im not sure. also u spelled infinite wrong.

Actualy, Infinite over infinite can equal whatever you want.

Infinite isn't a number though, theres no number for infinite.

[quote="jackdillon"]Thats why it can be anything![/quote]

Before you said it could be anything. :maybe:

this is a weird question what do you mean? do you want to know the difference between infinite and infinitesimal? :?:

[quote="Ihatepie"]he wants to know wat infinite/infinite=

as I said b4 i would guess that it is 1 because whenever u divide something from the same thing it = 1[/quote]

that is not true

try $\frac{0}{0}$ on your calculator...i believe it should be 1 but it isn't

[quote="funcia"]well isn't 0/0 undef.? how about 0/8? but if you put infinity over infinity it is just logically 1[/quote]

yes $\frac{0}{0}$ is undef. but i was simply stating to 2 posts aboe that something divided by that same something doesn't always equal 1

$\frac{\infty}{\infty}$ is $\infty$ because no matter how many times you divide it or subtract from it, it is always still infinite.

This has been posted several times in various forums.

$\frac{0}{0}$ does not equal anything. True, $\frac{0}{1}=0$ and $\frac 0{42}=0$ and $\frac 0{12345678987654321}=0$, but that doesn't mean $\frac{0}{0}=0$.

Think of it this way. If $\frac{0}{0}=x$, then it would make sense that $0x=0$ by algebra. But anything times zero is zero, so x can be anything. The expression has no meaning.

That's also why you can't divide by zero. For example, if $\frac{5}{0}=y$, then $0x=5$. But anything times zero is zero, not five.

Now about your original question: Infinity is not a number, so you cannot do normal arithmetic with it. Thus, $\frac{\infty}{\infty}$ has no meaning.

And to whoever said $\infty=1+1+1+...$:
Assume that $\infty=1+1+1+...$.
If you add one to both sides, $\infty+1=1+1+1+...$
So $\infty=\infty+1$ and $1=0$. Clearly false.

There are better definitions when you get into calculus, but I'm not going to go into that here.

You can't treat $\infty$ as a normal number. For example, if you subtract $1$ from it, you still have an infinite number, so there is no choice but to still call it $\infty$. If you subtract a finite number from an infinite number, the infinite number eats up the finite number. Also, some infinite numbers eat up other infinite numbers, like $\epsilon_{0}$ eats up $\omega$.

[quote="jackdillon"][quote="i_like_pie"][quote="SoxFan34"]sorta like how .99999..... forever should equal 1[/quote]
$0.\overline{9}=1$ can easily be proven (just multiplication and division).[/quote]
But it wouldn't realy equal one because you would still have to round up.[/quote]

$0.\overline{9}= x$
$100x= 99.\overline{9}$
$9x=99$

x= 1

so $0.\overline{9}= 1$

[quote="nonie"][/quote][quote="jackdillon"][quote="i_like_pie"][quote="SoxFan34"]sorta like how .99999..... forever should equal 1[/quote]
$0.\overline{9}=1$ can easily be proven (just multiplication and division).[/quote]
But it wouldn't realy equal one because you would still have to round up.[/quote][quote="nonie"]

$0.\overline{9}= x$
$100x= 99.\overline{9}$
$9x=99$

x= 1

so $0.\overline{9}= 1$[/quote]
Yes. This is a fact that some fail to comprehend. Just like the proof of $\sum^{\infty}_{n=1}\frac{1}{2^{n}}=1$
$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=x$
$\frac{1}{2}(1+x)=x$
$\frac{1}{2}=\frac{x}{2}$
So clearly x is 1.

@nonie: some typos in your post....and you could just use 10x

@bpms: not really a rigorous proof....I think you need to prove that that summation actually converges. Without proving that, i could do things like
$1+2+4+\cdots=x$
$1+2(1+2+4+\cdots)=x$
$1+2x=x$
$x=-1$
$1+2+4+\cdots=-1$
which is clearly not true.(just an example)
And also, "the terms get smaller and smaller and eventually closer to 0" doesn't do it. Like consider this:
$1/1+1/2+1/3+1/4+\cdots$ diverges.

@everyone else who says that $\infty/\infty$: read over other ppl's posts again.

@anirudh: you shouldn't really trust a calculator...of course it's right in this case, but it does strange things once in a while....like:
$solve(x^{2}+1=0,x)$
$false$
(just an example...of course you can take care of this problem with cSolve)

And for anyone who's interested, go check out the thread called "Now you can divide by 0!" in Round Table, and go to the link in the first post(bbc article about a british prof who claims he "solved" $0^{0}$. Quite bs imo, but it still gives you a good laugh if you read the comments).

[quote="mathnerd314"][quote="junggi"]@anirudh: you shouldn't really trust a calculator...of course it's right in this case, but it does strange things once in a while....like:
$solve(x^{2}+1=0,x)$[/quote]

Type in log(16)/log(2).
It should give you the answer $4$.
Now take the "greatest integer function" of that answer. This is the greatest integer less than or equal to the answer.
It's $4$, right? No, the calculator thinks it's $3$. :roll:[/quote]
is that on a 89?
b/c my 89 gives the right anwer..... :maybe:

must not be the 89 then!

YOu cant realy trust a calculater, since it is only as smart as you program it to be.  There must some kind of formula used to solve questions with undefined numbers.... Actually, if there isnt one, us Aopsers could try and make one!

[quote="Jiggalog"]Well that may be true but did you ever consider that pi multiplyed by the radius squared is the area of a circle. Take that.  :P [/quote][/code][/quote]What does that have to do with the price of tea in China? :spam:

infinite over infinite is undefined because the infinite on top could be a infinite larger or smaller or equal to the infinite on the bottom due the the reason that infinite is just a sign and not a certain number...  Infinite is a status, not a number.

[quote="nutz_for2.718281828"]$\frac{\infty}{\infty}$[hide]Since $x\infty=\infty$ where x is any number, $\frac{\infty}{\infty}=x$ where x is any number.[/hide][/quote]

As for this your whole proof cancels itself out. To take the infinety away from the x you divide by infinity but how can you infinety on th l.h.s cancel unless infinety/infinety=1 which it doesnt so basicly your sayin that infinety/infinety is indetermanite because   infinety/infinety=1. ???????????? Makes no sence to me  :huh:

[quote="junggi"]maybe this should be locked?(I mean with all the idiots like jiggalog spamming this up, and the question was already answered like um 1000 times)[/quote]

Good idea


here is a nice pb:
a collection E={F1,F2,F3,..} of nonempty subsets of N is called an Antichain if no subset Fi in E is strictly contained in another subset Fj. The question is:
does there exist any uncountable Antichain.

Am I wrong or have you already proposed it on 'Maths et Dlires'? (a french forum)  ;)

Pierre.

I think the answer is 'yes'. It might be wrong, but right now it seems Ok to me:

Take all possibilities of including in our set exactly one of $(n,n+1)$ with $n$ odd, i.e. take precisely one of $1,2$, precisely one of $3,4$, and so on.

very nice. I remember that my example was quite complicate and ugly. Perhaps Pierre remembers that solution i posted a long time ago on a french forum: i do not remember  :D

Look here Alekk to refresh your memory :
http://boumbo.salle-s.org/mathetdelire/forum/forum.php?view=271

Pierre.


modular arithmetic
function
algorithm
number theory
prime numbers
geometric sequence
relatively prime

Can someone post some mod questions??

tks.

[b]First Problem[/b]
Let $ k \equal{} 1 \plus{} 2 \plus{} 3 \plus{} ... \plus{} 100$. Find the remainder when $ k^2$ was divided by $ 7$.
(by using mods, of course)

Is the answer 2?
\[ \mbox{5050 is congruent to 3 in mod 7. So, 3^2 \div 7} \mbox{ has a remainder of 2.}
\]
I don't know if the solution is correct...It looks correct??

Yes that's correct,

Here are couple problems, that raise in difficulty:
1) If today is Thursday, April 16, 2009 then what day of the week will it be on April 16, 2010?
[hide="hint"]
Are you sure you need a hint? Keep trying
[hide="Real hint"]
Note that it's 365 days from now.
[/hide]
[/hide]
2) Show that $ 2222^{5555}\plus{}5555^{2222}$ is a multiple of $ 7$
[hide="hint"]
Calculate the remainder of $ 2222^{5555}$ when divided by $ 7$.
[hide="hint"]
Calculate the remainder of $ 2222$ when divided by $ 7$.
[/hide]
[/hide]
3) Show that if $ p$ and $ q$ are prime numbers larger than $ 3$ then, $ p^2\plus{}q^2$ is never a multiple of $ 3$.
[hide="hint"]
What are the two possible remainders of $ x^2$ when divided by $ 3$?
[/hide]

4) Prove that $ 21n^2\plus{}12n\plus{}6$ is never a perfect square.

5) Show that the sum of the first $ 10,000$ terms in the sequence $ 1,2,4,8,16,32,64,\cdots ,2^{n\minus{}1}$, is a multiple of $ 5$. Where the nth term of the sequence is $ 2^{n\minus{}1}$.

6) Let $ P(n)\equal{}(2\plus{}1)(2^2\plus{}1)(2^4\plus{}1)\cdots (2^{2^{n}}\plus{}1)$. What is the remainder when $ 2(P(200)\plus{}P(201)\plus{}P(202)\plus{}\cdots \plus{}P(300))\plus{}1$ is divided by $ 10$?

[quote="tonypr"]Yes that's correct,

Here are couple problems, that raise in difficulty:
1) If today is Thursday, April 16, 2009 then what day of the week will it be on April 16, 2010?
[hide="hint"]
Are you sure you need a hint? Keep trying
[hide="Real hint"]
Note that it's 365 days from now.
[/hide]
[/hide]
2) Show that $ 2222^{5555} \plus{} 5555^{2222}$ is a multiple of $ 7$
[hide="hint"]
Calculate the remainder of $ 2222^{5555}$ when divided by $ 7$.
[hide="hint"]
Calculate the remainder of $ 2222$ when divided by $ 7$.
[/hide]
[/hide]
3) Show that if $ p$ and $ q$ are prime numbers larger than $ 3$ then, $ p^2 \plus{} q^2$ is never a multiple of $ 3$.
[hide="hint"]
What are the two possible remainders of $ x^2$ when divided by $ 3$?
[/hide]

4) Prove that $ 21n^2 \plus{} 12n \plus{} 6$ is never a perfect square.

5) Show that the sum of the first $ 10,000$ terms in the sequence $ 1,2,4,8,16,32,64,\cdots ,2^{n \minus{} 1}$, is a multiple of $ 5$. Where the nth term of the sequence is $ 2^{n \minus{} 1}$.

6) Let $ P(n) \equal{} (2 \plus{} 1)(2^2 \plus{} 1)(2^4 \plus{} 1)\cdots (2^{2^{n}} \plus{} 1)$. What is the remainder when $ 2(P(200) \plus{} P(201) \plus{} P(202) \plus{} \cdots \plus{} P(300)) \plus{} 1$ is divided by $ 10$?[/quote]

[hide="Answers"]My Answers. Please do tell if there is a better way to do them, because I have only recently been learning about number theory.

1. $ 365 \equiv 1 \pmod{7}$
so we have that it will be one day later in the week. Friday.

2. I found this to be a longer problem, is there a better way of doing it (also without assuming there's a pattern?)

We want to work in mod 7 for this problem. We notice that (we are modding out I believe?):

$ 2222 \equiv 3 \pmod{7}$

By properties of mods, we can raise both sides to any given power. We want to find a pattern, so we try a few values of powers of 3 in mod 7. This will give us:

$ 3, 2, 6, 4, 5, 1, 3, 2, 6... \pmod{7}$

We see that the pattern repeats itself every 6 terms. Therefore, the power of 3 which will work will be:

$ 2222^{5555} \equiv 3^{5555} \equiv 3^5 \equiv 5 \pmod{7}$

Using the same method for finding $ 5555^{2222}$ in mod 7, we find that its 2 mod 7.

$ 5 \plus{} 2 \equiv 0 \pmod{7}$

Proved? Its under the assumption that there is indeed a pattern.

3. We want to work in mod 3. Looking at some primes above three, we see that they can only be congruent to 1 or 2 mod 3.

We have three cases, where both 1 mod 3, where exactly one is 1 mod 3, and when both are 2 mod 3

Checking these all, we will indeed find that in all three cases, $ p^2 \plus{} q^2$ is never a multiple of 3.

4. This is a slightly brute force/guess maneuver at it.

Lets work in mod 4 (for the heck of it, and because it works)

Again, looking at mod 4s of perfect squares, we find that a perfect square can only be 0 or 1 mod 4.

Now, we can look at our cases. We will have four, where n is 0 mod 4, 1 mod 4, 2 mod 4, or 3 mod 4 (the cases by which any and all positive integers must abide by in mod 4)

Case 1: When n is 0 mod 4, squaring it will still mean its 0 mod 4 as will multiplying it by 21.
Multiplying 12 by any number n will ensure that its 0 mod 4.
Adding $ 6 \equiv 2 \pmod{4}$, we find that this number will be 2 mod 4. This cannot be a perfect square.

Case 2: When n is 1 mod 4, squaring it will still give 1 mod 4. Multiplying it by 21, we get:
$ 21 \equiv 1 \pmod{4}$
We needn't worry about the second term, as stated above.
By adding 6, we have $ 1 \plus{} 6 \equiv 3 \pmod{4}$ It cannot be a perfect square.

Continue with the other two cases in the same fashion. We will find that they cannot be perfect squares either.

5. We can see a pattern (AGAIN). Please tell me if this is an incorrect way of doing the problems.
Every four terms will be repeated in mod 5.

$ 1 \plus{} 2 \plus{} 4 \plus{} 8 \equiv 15 \equiv 0 \pmod{5}$

The sum of every four terms is divisible by 5, as are the sums of each of those four term sequences.

Therefore, it is divisible by 5 because 10000 can be split up equally into 2500 groups of 4 terms.

6. This was fun.

$ P(n) \equal{} 3 * 5 * 7 * 5 * 7 * 5..... \pmod{10}$

We see that this product is 5 mod 10, by noting that the product of any number of 5's and only odd digits will result in a final digit of 5, which is 5 mod 10.

$ 2(P(200) \plus{} P(201) \plus{} P(202) \plus{} \cdots \plus{} P(300)) \plus{} 1$

There are 101 values of P that are in there, all congruent to 5 mod 10.

$ 5 * 101 \equiv 5 \pmod{10}$

multiplying by 2 and adding by one:

$ 5 * 2 \plus{} 1 \equiv 11 \equiv 1 \pmod{10}$

Thus, the remainder is 1.

Whao, I would imagine this to be incredibly boring to anyone that reads it, but please do! And please do tell if Im doing anything wrong.

[/hide]

[quote="veezbo"][quote="tonypr"]Yes that's correct,

Here are couple problems, that raise in difficulty:
1) If today is Thursday, April 16, 2009 then what day of the week will it be on April 16, 2010?
[hide="hint"]
Are you sure you need a hint? Keep trying
[hide="Real hint"]
Note that it's 365 days from now.
[/hide]
[/hide]
2) Show that $ 2222^{5555} \plus{} 5555^{2222}$ is a multiple of $ 7$
[hide="hint"]
Calculate the remainder of $ 2222^{5555}$ when divided by $ 7$.
[hide="hint"]
Calculate the remainder of $ 2222$ when divided by $ 7$.
[/hide]
[/hide]
3) Show that if $ p$ and $ q$ are prime numbers larger than $ 3$ then, $ p^2 \plus{} q^2$ is never a multiple of $ 3$.
[hide="hint"]
What are the two possible remainders of $ x^2$ when divided by $ 3$?
[/hide]

4) Prove that $ 21n^2 \plus{} 12n \plus{} 6$ is never a perfect square.

5) Show that the sum of the first $ 10,000$ terms in the sequence $ 1,2,4,8,16,32,64,\cdots ,2^{n \minus{} 1}$, is a multiple of $ 5$. Where the nth term of the sequence is $ 2^{n \minus{} 1}$.

6) Let $ P(n) \equal{} (2 \plus{} 1)(2^2 \plus{} 1)(2^4 \plus{} 1)\cdots (2^{2^{n}} \plus{} 1)$. What is the remainder when $ 2(P(200) \plus{} P(201) \plus{} P(202) \plus{} \cdots \plus{} P(300)) \plus{} 1$ is divided by $ 10$?[/quote]

[hide="Answers"]My Answers. Please do tell if there is a better way to do them, because I have only recently been learning about number theory.

1. $ 365 \equiv 1 \pmod{7}$
so we have that it will be one day later in the week. Friday.

2. I found this to be a longer problem, is there a better way of doing it (also without assuming there's a pattern?)

We want to work in mod 7 for this problem. We notice that (we are modding out I believe?):

$ 2222 \equiv 3 \pmod{7}$

By properties of mods, we can raise both sides to any given power. We want to find a pattern, so we try a few values of powers of 3 in mod 7. This will give us:

$ 3, 2, 6, 4, 5, 1, 3, 2, 6... \pmod{7}$

We see that the pattern repeats itself every 6 terms. Therefore, the power of 3 which will work will be:

$ 2222^{5555} \equiv 3^{5555} \equiv 3^5 \equiv 5 \pmod{7}$

Using the same method for finding $ 5555^{2222}$ in mod 7, we find that its 2 mod 7.

$ 5 \plus{} 2 \equiv 0 \pmod{7}$

Proved? Its under the assumption that there is indeed a pattern.

3. We want to work in mod 3. Looking at some primes above three, we see that they can only be congruent to 1 or 2 mod 3.

We have three cases, where both 1 mod 3, where exactly one is 1 mod 3, and when both are 2 mod 3

Checking these all, we will indeed find that in all three cases, $ p^2 \plus{} q^2$ is never a multiple of 3.

4. This is a slightly brute force/guess maneuver at it.

Lets work in mod 4 (for the heck of it, and because it works)

Again, looking at mod 4s of perfect squares, we find that a perfect square can only be 1 or 2 mod 4.

Now, we can look at our cases. We will have four, where n is 0 mod 4, 1 mod 4, 2 mod 4, or 3 mod 4 (the cases by which any and all positive integers must abide by in mod 4)

Case 1: When n is 0 mod 4, squaring it will still mean its 0 mod 4 as will multiplying it by 21.
Multiplying 12 by any number n will ensure that its 0 mod 4.
Adding $ 6 \equiv 2 \pmod{4}$, we find that this number will be 2 mod 4. This cannot be a perfect square.

Case 2: When n is 1 mod 4, squaring it will still give 1 mod 4. Multiplying it by 21, we get:
$ 21 \equiv 1 \pmod{4}$
We needn't worry about the second term, as stated above.
By adding 6, we have $ 1 \plus{} 6 \equiv 3 \pmod{4}$ It cannot be a perfect square.

Continue with the other two cases in the same fashion. We will find that they cannot be perfect squares either.

5. We can see a pattern (AGAIN). Please tell me if this is an incorrect way of doing the problems.
Every four terms will be repeated in mod 5.

$ 1 \plus{} 2 \plus{} 4 \plus{} 8 \equiv 15 \equiv 0 \pmod{5}$

The sum of every four terms is divisible by 5, as are the sums of each of those four term sequences.

Therefore, it is divisible by 5 because 10000 can be split up equally into 2500 groups of 4 terms.

6. This was fun.

$ P(n) \equal{} 3 * 5 * 7 * 5 * 7 * 5..... \pmod{10}$

We see that this product is 5 mod 10, by noting that the product of any number of 5's and only odd digits will result in a final digit of 5, which is 5 mod 10.

$ 2(P(200) \plus{} P(201) \plus{} P(202) \plus{} \cdots \plus{} P(300)) \plus{} 1$

There are 101 values of P that are in there, all congruent to 5 mod 10.

$ 5 * 101 \equiv 5 \pmod{10}$

multiplying by 2 and adding by one:

$ 5 * 2 \plus{} 1 \equiv 11 \equiv 1 \pmod{10}$

Thus, the remainder is 1.

Whao, I would imagine this to be incredibly boring to anyone that reads it, but please do! And please do tell if Im doing anything wrong.

[/hide][/quote]

[hide]1. clean and simple

2. I guess it's ok but it can be done more precisely with Fermat's little theorem

$ 2222^{5555} \equiv 2222^{925 \cdot 6} \cdot 2222^5 \equiv 2222^5 \equiv 3^5 \equiv 5 \pmod{7}$

$ 5555^{2222} \equiv 5555^{370\cdot 6} \cdot 5555^2 \equiv 5555^2 \equiv 4^2 \equiv 2 \pmod{7}$

summing gives the result

3. Can be done quicker with Fermat's little theorem

$ p^2 \equiv 1 \pmod{3}$

$ q^2 \equiv 1 \pmod{3}$

summing gives the result

4. It might be quicker to prove that $ 7n^2\plus{}4n\plus{}2$ is not multiple of $ 3$. There are only 3 cases which are easy to check and prove.

5. This is geometric sequence which sums to $ 2^n\minus{}1$. Putting in $ n\equal{}10000$ we have:

$ 2^{10000} \minus{}1 \equiv 2^{4\cdot 2500} \minus{} 1 \equiv 1 \minus{} 1 \equiv 0 \pmod{5}$

6. Since second factor equals $ 5 \implies 5|P(n) \; \forall \;n  \ge 2$
$ \implies 5|(P(200) \plus{} P(201) \plus{} P(202) \plus{} \cdots \plus{} P(300))$
$ \implies 10|2(P(200) \plus{} P(201) \plus{} P(202) \plus{} \cdots \plus{} P(300))$
$ \implies 2(P(200) \plus{} P(201) \plus{} P(202) \plus{} \cdots \plus{} P(300)) \plus{} 1 \equiv 1 \pmod{10}$[/hide]

fryeggs, you might want to check [url=http://en.wikipedia.org/wiki/Fermat%27s_little_theorem]Fermat's little theorem[/url], and it's generalizations, [url=http://en.wikipedia.org/wiki/Euler%27s_theorem]Euler's theorem[/url] and [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael's theorem[/url].

When you solve one, post another one up.

Hey flame,
How did you get from
\[ 2222^{5555} \mbox{ to } 2222^{925 \minus{} 6} \cdot 2222^5
\]
Thanks

crap...sorry i didn't see the
\[ 925 \times 6 \mbox{ and not subtract 6}
\]
Flame, can you explain number 2 to me more thoroughly...thanks   :huh:

Find fermat's little theorem and you'll see why flame proceeded as he did.

Considering number 2 I used the fact that any number relatively prime to $ 7$ when raised to power of $ 6$ is congruent to $ 1$ when divided by $ 7$.

This follows from Fermat's little theorem which states that if $ M(a,p)\equal{}1$

\[ a^{p\minus{}1} \equiv 1 \pmod p\]

Since $ 2222$ is relatively prime to $ 7 \implies 2222^6 \equiv 1 \pmod 7 \implies 2222^{6\cdot 925} \equiv 1 \pmod 7$

Some new exercises:

1. If for $ p>3$ numbers $ p$ and $ 2p\plus{}1$ are prime, prove that $ 4p\plus{}1$ is a composite number.

2. If $ a$ is not divisible by $ 5$, than one and only one of the numbers $ a^2\minus{}1$ and $ a^2\plus{}1$ is divisible by $ 5$. Prove it.

3. Prove that numbers $ p$ and $ 8p^2\plus{}1$ can be both prime only if $ p\equal{}3$.

4. If $ M(n,6)\equal{}1$ than $ n^2 \equiv 1 \pmod {24}$. Prove.

5. Prove that all the prime divisors except $ p\equal{}2$ of number $ x^2\plus{}1$ where $ x$ is a positive integer, can be written as $ 4k\plus{}1$.

The only one I think I can do is #2.

Think of first what the squares are in $ \mod{5} : 0,1,4$. Since a is not divisible by 5, we can discard 0. So a is $ \equiv 1 \mod 5$ or $ \equiv 4 \mod5$.

Case 1:

$ a \equiv 1 \mod 5$

$ a^2 \minus{} 1 \equiv 0 \mod5$
$ a^2 \plus{} 1 \equiv 2 \mod5$

Only $ a^2\minus{}1$ is divisible by 5.

Case 2:

$ a \equiv 4 \mod5$
$ a^2 \minus{} 1 \equiv 0 \mod5$
$ a^2 \plus{} 1 \equiv 2 \mod5$

Again, only $ a^2 \minus{} 1$ is divisible by 5.

Thanks Flame.
How do you solve linear congruences?
For example,
\[ 4x\plus{}3 \equiv 1 \pmod{3}\]

[quote="Flame"]
Some new exercises:

1. If for $ p > 3$ numbers $ p$ and $ 2p \plus{} 1$ are prime, prove that $ 4p \plus{} 1$ is a composite number.

2. If $ a$ is not divisible by $ 5$, than one and only one of the numbers $ a^2 \minus{} 1$ and $ a^2 \plus{} 1$ is divisible by $ 5$. Prove it.

3. Prove that numbers $ p$ and $ 8p^2 \plus{} 1$ can be both prime only if $ p \equal{} 3$.

4. If $ M(n,6) \equal{} 1$ than $ n^2 \equiv 1 \pmod {24}$. Prove.

5. Prove that all the prime divisors except $ p \equal{} 2$ of number $ x^2 \plus{} 1$ where $ x$ is a positive integer, can be written as $ 4k \plus{} 1$.[/quote]

Solutions:
[hide="1"] If $ p \equal{} 1(mod 3)$, then $ 2p \plus{} 1 \equal{} 0(mod 3)$ is prime, impossible. So $ 4p \plus{} 1 \equal{} 4(2) \plus{} 1 \equal{} 0(mod 3)$[/hide]
[hide="2"] $ (a^2 \minus{} 1)(a^2 \plus{} 1) \equal{} a^4 \minus{} 1 \equal{} 0(mod 5)$, also $ a^2 \plus{} 1 \neq a^2 \minus{} 1(mod 5)$, so done.[/hide]
[hide="3"] For $ 3 \nmid p$, $ 8p^2 \plus{} 1 \equal{} 8(1) \plus{} 1 \equal{} 0(mod 3)$ is prime, impossible. For $ p \equal{} 3$, $ 8p^2 \plus{} 1 \equal{} 73$ is a prime.[/hide]
[hide="4"] $ n$ is odd $ \Leftrightarrow n^2 \equal{} 1(mod 8)$,  $ 3 \nmid n \Leftrightarrow n^2 \equal{} 1 (mod 3)$ , combining these two gives the desired result.[/hide]
[hide="5"] We have $ x^2 \equal{} \minus{} 1(mod p) \Rightarrow x^4 \equal{} 1(mod p) \Rightarrow 4|p \minus{} 1$ for $ x \neq 1(mod p)$(,i.e. $ p \neq 2$), which is the desired result.[/hide]

[quote="fryeggs"]Thanks Flame.
How do you solve linear congruences?
For example,
\[ 4x \plus{} 3 \equiv 1 \pmod{3}
\]
[/quote]

For small mod like mod 3, we usually simplify it and then by trial. For example, $ 4x \plus{} 3 \equal{} 4x \equal{} (3x) \plus{} x \equal{} x(mod 3)$, so $ x \equal{} 1(mod 3)$, that is , removing any thing divisible by $ 3$ from $ L.H.S.$ because it does not affect the result after you mod it.

If the mod is much larger (for example, mod 123456789), then to express it as continued fraction is a method. (But no this kind of question will be asked, so forget it :P )

[quote="fryeggs"]Thanks Flame.
How do you solve linear congruences?
For example,
\[ 4x \plus{} 3 \equiv 1 \pmod{3}
\]
[/quote]

$ 4x\plus{}3 \equiv x \equiv 1 \mod{3}$. So with this, $ x\equal{}3t\plus{}1$.

Can you  please explain it a bit more thoroughly? I'm a noob at mods  :huh:

Well, $ 3 \equiv 0 \mod 3$ and $ 4 \equiv 1 \mod 3$. Substituting this, $ (1)x\plus{}(0) \equiv x \mod 3$. Since $ x \equiv 1 \mod 3$, it means that x is equal to a multiple of 3 PLUS 1. In other words, $ x\equal{}3t\plus{}1$ for a whole number $ t$.

$ 4x\plus{}3 \equiv 1 \pmod 3$

$ 4x\plus{}2 \equiv 0 \pmod 3$

$ 2x\plus{}1 \equiv 0 \pmod 3$

$ 2x \equiv \minus{}1 \equiv 2 \pmod 3$

$ x \equiv 1 \pmod 3\implies x\equal{}3k\plus{}1$

or

$ 4x\plus{}3\equal{}3y\plus{}1$

$ 4x\minus{}3y\equal{}\minus{}2$

and you have linear diophantine equation

theorem: Given linear diophantine equation $ ax\plus{}by\equal{}c,$ all integer solutions are:

\[ x\equal{}\minus{}bk\plus{}x_1, y\equal{}ak\plus{}y_1, k \in Z\]
where pair $ (x_1, y_1)$ is one solution for given equation.

For our diophantine equation it's easy to guess one pair $ (1,2)$ and using the theorem we have $ x\equal{}3k\plus{}1$.

Sometimes it won't be easy to determine ordered pair just by guessing. Then we use [url=http://mathworld.wolfram.com/EuclideanAlgorithm.html]Euclidean Algorithm[/url].


Εστω $d=(x,y)$ ο μεγιστος κοινος διαιρετης των x,y
Να προσδιορισετε ολα τα ζευγη $(x,y)$ για τα οποια ισχυει

$x+y^{2}+d^{3}=xyd$

 :D  :D  :D

Υ.Γ:Δεν βαζω αλλη ασκηση αν δεν λυθουν και οι 2 προηγουμενες...

A,ξεχασα made in bulgary....(αν αυτο λεει τπτ...)

 :D  :D  :D

Grafw tin lisi i opoia odigei sto apotelesma alla meta apo arketes periptwseis (tis opoies kai den tha analisw pera apo mia endeiktika. Nomizw oti den exei noima apo kei kai pera) ektos ean kapoios exei na proteinei mia omorfoteri (apo apopsi periptwsewn) lisi.

Loipon :

Afou $(x,y)=d$ ara $x=dx_{1}$ kai $y=dy_{1}$ me $(x_{1},y_{1})=1$
Ara i eksiswsi ginetai $x_{1}+dy_{1}^{2}+d^{2}=d^{2}x_{1}y_{1}$ ara $x_{1}=d(dx_{1}y_{1}-y_{1}^{2}-d)$ ap'opou $d|x_{1}$ ki etsi $x_{1}=dx_{2}$ sinepws i teleutaia ginetai $x_{2}=d^{2}x_{2}y_{1}-y_{1}^{2}-d$ diladi ean to thewrisoume ws triwnimo ws pros $d$ exoume: $x_{2}y_{1}d^{2}-d-y_{1}^{2}-x_{2}=0$ kai to opoio exei diakrinousa
$\Delta=1+4x_{2}y_{1}^{3}+4x_{2}^{2}y_{1}$ i opoia prepei na einai isi me miden (pou amesws aporriptetai) 'h teleio tetragwno $r^{2}$.

Ara $4x_{2}y_{1}(y_{1}^{2}+x_{2})=(r-1)(r+1)$ kai epeidi $(r-1,r+1)=2$ ara $\big(\frac{r-1}{2},\frac{r+1}{2}\big)=1$ ki etsi i teleutaia ginetai $x_{2}y_{1}(y_{1}^{2}+x_{2})=\frac{r-1}{2}\cdot\frac{r+1}{2}\ \ (1).$

Omws kathws $(x_{1},y_{1})=1$ ara $(x_{2},y_{1})=1$ ki etsi eukola prokiptei oti $(x_{2}y_{1},y_{1}^{2}+x_{2})=1$ [dioti ean ipirxe prwtos $p$ tetoios wste $p|x_{2}y_{1}$ kai $p|y_{1}^{2}+x_{2}$ tote $p|y_{1}^{3}$ (pollaplasiazw tin deuteuri epi $y_{1}$ kai afairw apo tin prwti) kai $p|x_{2}^{3}$ (pollaplasiazw tin deuteri epi $x_{2}^{2}$kai xrisimopoiw apo tin prwti oti afou $p|x_{2}y_{1}$ ara $p|x_{2}^{2}y_{1}^{2}$). Ara $p|(x_{2},y_{1})^{3}$ ap'opou $p|1$, atopo.]. Etsi prokiptoun kai oi antistoixes periptwseis oi opoies einai para polles.

Endeiktika meletw mia periptwsi gia na ginei katanoito oti einai thema kathara praksewn kai katholou skepsis: An $\frac{r-1}{2}|x_{2}$ kai $\frac{r+1}{2}|y_{1}(y_{1}^{2}+x_{2})$ tote eite $\Big( \frac{r-1}{2}|x_{2}$ kai $\frac{r+1}{2}|y_{1}\Big)$ eite $\Big( \frac{r-1}{2}|x_{2}$ kai $\frac{r+1}{2}|y_{1}^{2}+x_{2}\Big)$

An loipon $\frac{r-1}{2}|x_{2}$ kai $\frac{r+1}{2}|y_{1}$ tote iparxoun akeraioi $k_{1}$ kai $k_{2}$ tetoioi wste

$x_{2}=\frac{r-1}{2}k_{1}$ kai $y_{1}=\frac{r+1}{2}k_{2}$ opote i sxesi $(1)$ ginetai
$k_{1}k_{2}\big(\frac{(r+1)^{2}}{4}k_{2}^{2}+\frac{r-1}{2}k_{1}\big)=1$ ap'opou pairnontas pali periptwseis (p.x. apo tin teleutaia prokiptei oti $k_{1}=k_{2}=\frac{(r+1)^{2}}{4}k_{2}^{2}+\frac{r-1}{2}k_{1}=1$ (den ksexname kai tin periptwsi kapoia apo auta na einai isa me $-1$) ap'opou pairnoume $r=1$ 'h $r=-5$ kai sinexizoume vriskontas me eukolia tis liseis.

Omoia sinexizoume kai gia tis ipoloipes periptwseis.

Alexandros

Y.G. Wraia askisi mexri tin sxesi $(1)$. Apo ekei kai pera einai ligo (ews poli) epipono na vreis tis liseis. Yparxei alli protasi?

αρχικά
made in Bulgaria μην προσβάλλουμε την χώρα
έπειτα πολύ ωραία ασκηση κι βλέπω ο Αλέξανδρος να με καλύπτει απόλυτα

Σωστη φαινεται cretanman.... :lol:
Eχει πολυ καιρο που την ειχα δει(την βρηκα ξεφυλιζοντας κατι παλιες ασκησεις..)και δεν θυμαμαι ακριβως πως την ειχα δουλεψει...Απλως θυμαμαι οτι με ειχε δυσκολεψει....Το μονο που εχω σημειωμενο στην ασκηση ειναι οι λυσεις...
Για αυτο αν μπορεις,γραψε ολες τις λυσεις που προκυπτουν τελικα,γιατι η μεθοδος φαινεται σωστη...

 :D  :D  :D


I hope india wins the third test at Feroz shah Kotla ground in New delhi. But do you think Amit Mishra should be retained instead of Anil Kumble. :lol:

i just hope aussie win....but India doesnt lose :( is it possible??

but i think the ground shud favour mishra :P

so according to ritu,it should be a draw
 :P  :P  :P

but according to me,aussies should win(neverthless they are strong team)

but i hope against it  :D

FSK is fav of kumble,so he should be retained


Funny story: I spent the first 30 minutes trying to figure out which question was easier to do... (all three seemed impossible at the time) and spent next 20 minutes or so coding up

elevator    *  *  *  *  *  *  *  *  *  t  t  (30 lines)
yogfac      *  *  *  *  *  *  *  *  *         (20 lines)

and then eventually giving up on problem 1.

Wow both of your programs are incredibly small.  :D

I did the same questions as you did but I managed to get perfect for elevator.  My program was barely fast enough (worst case around 0.25s)  :P

#1 was very weird though...  Can someone explain how to do it? :?

lol. good job on keeping your programs short.

All three of my programs are extra-long with lots of empty lines :) The 1st problem in the gold division was very similar to the first problem in the February contest. Both are solved using net-flow. However, for this one, you needed to setup some sort of a bipartite mapping. I still have to figure out how to speed up my net-flow:

elevator    *  *  *  *  *  *  *  *  *  *  *
ombro       *  x  *  t  t  t  t  t  t  t  *  *  *  *  *  *
yogfac      *  *  *  *  *  *  *  *  *

I did the silver division and got

alibi       *  *  *  *  x  *  x  x  x  x  *  *  x
outofhay    *  *  *  *  *  x  *  *  *  *  *
yogurt      *  *  t  t  *  *  *  *  t

It's not really that great, but it's better than my other scores.

yeah, i thought ombro was network flow... oh wellz, i'll learn it sooner or later.

My results in the Gold division:

elevator    *  *  *  *  *  *  *  *  *  *  *
ombro       *  *  x  x  x  x  x  x  x  x  *  *  x  *  *  x
yogfac      *  *  *  *  *  *  *  *  *

By the time I realized that my alternative to doing the network flow on ombro was wrong, I had only 15 minutes left, which ended up being not enough time for me to do an actual solution.

I participated in Silver and got this:
alibi       *  *  *  *  *  *  *  *  *  *  *  *  *
outofhay    *  x  *  x  *  x  *  *  *  *  x
yogurt      *  *  *  *  *  *  *  *  *

hehe, in the outofhay prob, my completely wrong algorithm fetcched me 7/11 cases! :lol:
What was the point in keeping a SilverDiv prob (yogurt) in Gold Division??
Wasnt these an easy contest in all three divs:Gold,Silver,Bronze, what do u ppl think?

Hey and also, Is there any permanent IRCchannel/Site for USACO discussions? If not, why not start 1? :) :)

I agree about the chatroom.  I also think we need a forum, but in order for it to have a decent turnout, we probably need to get the coaches to mention it in the e-mails or something.

[quote]And just one more thing, I was wondering if you could have
	a message board type thing where people can talk about the
	training pages and the contests, instead of us just having
	to ask you guys for help all the time.

Yep -- soon.  Of course, they'll be turned off during contests :) .[/quote]

[quote="Aidin Kashigar"]Who are you quoting, h_s_potter2002?[/quote]

An e-mail I got from Rob Kolstad

[quote="countersixte"]I still don't know if this is good or bad, but gold has been a lot easier than it has been recently.  I guess I'll find out soon enough.

elevator    *  *  *  *  *  *  *  *  *  *  t
ombro       *  x  x  x  x  x  x  x  x  x  *  *  x  *  *  x
yogfac      *  *  t  t  *  *  *  *  t

I was afraid there was going to be some DP solution to elevator, but there seemed to be too many variables to keep track of.  Apparently they only chose very easy test cases for that problem.

I guess it makes sense that ombro is network flow.  In the training pages, I'm stuck on that very topic.

Everybody probably did well enough for them to use the USACO scoring system instead of the IOI scoring system this time.[/quote]

for elevator, you didnt use DP and got everything but the last test case?  It ran faster than my DP solution!!!  and there isnt so much things to keep track of  b/c as you can see, I did it in 30 lines (including #includes/IO and other junk).

Maybe he did DP and didn't realize it.  Paladin8 did Prim's in silver and didn't realize it.  I remember I didn't understand the dynamic programming tutorial in the training pages when I first read it (actually I still prefer other sites for my dynamic programming education).  So I got stuck on they money problem for weeks and eventually created the best program ever for it! Only to find out that it was the simplest DP problem ever and anyone that knew any DP could solve it in minutes.  It pays to know your algorithms (or "programming tequnique" in DP's case).

[quote="bugzpodder"]lol I'd give him some credit on knowning whether he did DP or not.[/quote]
Thanks,  bugzpodder.  :lol:

Yeah, I double-checked it in case I really did DP, but I didn't.  I did a straight recursive function (no memoization or anything).  The only thing I did was start at the smaller of the total height reachable and the point at which all blocks disintegrate and worked down from there.

Ombro can be "translated" to:

"How many cows can move to the shelters
if we have only paths shorter than or equal to X. If number of cows
(the flow) is equal to the total number of cows, then our answer is
X!"

Very diffucult task... I've got a working solution, using Floyd Warshall's all-pair shortest path and Maximum Flow with binary search... My solution is 700-800 rows of code (400 of them is prewritten though...). :D

Well, does you 800 lines of code get perfect in the analysis mode in USACO?

If so, can you attach it here.

It gets nearly perfect... About 10% to slow for the test case 10 (probably a optimization issue...). And, of cource, timeout on test case 2 (it's something wrong with that testcase... it runs in "0" time on my computer :P).

lol. 0 time. anyways, can you post your solution then. it would be a great help

usaco has put an analysis to the ombro problem online:
http://ace.delos.com/ombro.htm

thanks. btw how did you find out about this?

there is a thread about this problem in topcoder forums:
http://www.topcoder.com/rtables/viewThread.jsp?forum=7167&thread=489907&message=489906

I did the silver division.  I spent prolly 15 minutes coding and then 45 trying to figure out why it couldn't find the input file on my computer and then I went to bed cause I didn't get to start until late on Monday night.  I got most of the test cases for yogurt though which was the only one i tried, timed out on a couple.


Hey, really quick question. My textbook states that a direct consequence of Gauss' Law is that the change in $ E$, the electric field, from one side of a layer of charge to another is always equal to $ 4\pi q$. My teacher gave me this proof when I asked him:

$ \frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}=4\pi \rho$

Looking at a one dimensional case, where there is a vertical flat sheet of charge intersecting the path, this becomes:

$ \frac{\partial E_x}{\partial x}=4 \pi \rho$

This part gets fuzzier...which makes me think my teacher made some sort of mistake, like maybe leaving out an integral sign or something. I'm just going to write exactly what my teacher did on the board:

$ \int^\text{right}_\text{left} \frac{\partial E_x}{\partial x} \,ds= \int 4 \pi \rho \, ds$

which gives:

$ E_x_\text{right}-E_x_\text{left}=4\pi\rho \, ds$

which gives:

$ E_x_\text{right}-E_x_\text{left}=4\pi\sigma$

My questions are:

What is the physical significance of the gradient of the electric field, other than the direction in which it changes most rapidly? What are the units of $ \rho$ here? What does $ \rho$ even mean? Where did the first identity come from? And why does the integral of $ 4\pi\rho$ become $ 4\pi\sigma$? I would assume that $ \sigma$ is the charge density of the surface.

It's one of Maxwell's equation, the one that can be recognized as the differential form of Gauss Law:

\[ div \; \vec E  \equal{} \nabla . \vec E \equal{} {\rho \over \epsilon}\;,\]

where  $ \rho$ is volumetric charge density and  $ \epsilon$ the permittivity of the medium.

Besides others things, the volume integral comes out kind of  messed up...

1. It's not the gradient of E but the divergence of E.
The gradient's significance is as you said but does not apply here.
A nonzero divergence in a point shows that the flux that exits the (infinitely small) volume "around the point" is different than the flux that enters (there is a net flux).

2. p (ro ) is the charge density.

3. The idea is that you have a net flux only if there is some charge at that point.


AwesomeMath
summer program
MATHCOUNTS
Columbia

Have you ever been away from home?  If so, and you're still here to talk to us, bring that up.  If not, then it becomes more complicated.  You are going to be on a college campus with people whose job is to make sure that you are having fun, learning, and being safe.  If you are a responsible child, then the fear that you cannot take care of yourself is unfounded.  The fear that you will physically not be safe there is equally unfounded.  How to convince here?  I don't see what could happen that couldn't happen at home (i.e., falling down and breaking a bone or something).

[quote="bpms"]My mom is concerned as to my safety at the summer camp. Any suggestions as to how to convince her that I shall be safe?[/quote]
Bring a hunting knife, buy a shirt from the  [url=http://www.nra.org]NRA[/url], and leave an empty gun holster, leather jacket, and motorcycle helmet so that they can barely be seen when you leave your door cracked open.

Honestly, it's what I'd do. The easiest way to win a fight is to never fight one at all.

[quote="krustyteklown"]There are also better paths to never fighting.[/quote]
But this path has the advantage that when he's in the ghetto and someone not-so-politely asks him for his wallet, he can stab them between the ribs. Minimal bleeding and instantaneous death if he punctures the heart.

It's not my fault she's uber paranoid :D
I'm a dude, so I don't need to worry about number 8.

[quote="bpms"]
I'm a dude, so I don't need to worry about number 8.[/quote]
We won't judge you, it's ok.

[quote="bpms"]
I'm a dude, so I don't need to worry about number 8.[/quote]

I'm afraid I don't understand you. I am a dude as well. Why do you not need to worry about it?

[quote="Max300"]
I'm afraid I don't understand you. I am a dude as well. Why do you not need to worry about it?[/quote]

I guess that he doesn't understand how hot the guys must be. :P

Our students' safety is, of course, our top priority.

Members of the camp's staff are on duty around the clock, we have a mentor or assistant in every student apartment, and the campus is constantly patrolled by its own police force.

Furthermore, we take all necessary precautions before the camp begins to ensure that should an unexpected situation arise we are equipped and empowered to handle it.

If there are any additional specific questions or concerns, please don't hesitate to contact us directly (contact information is available on the website).

We do everything possible to make sure Zuming's accuracy with dry-erase markers is the only dangerous part of the experience.

Zac Cox
Program Coordinator, AwesomeMath
zac@awesomemath.org

Another safety hazard involves setting things on fire (thank you for your amazing beginning-of-camp speech, Zac)

To Max300... Zuming owning you is a given anyway, so it doesn't really matter ;)

but hot guys? like wtf man

also, don't hurt yourself playing DDR at camp, it's quite dangerous (soccer is too, kinda)

And then there is the risk that you will make an incredibly stupid mistake in some math problem and expire from shame :yankchain: .

[quote="Max300"]dude your mom is right, there are definitely some safety issues you need to consider before attending awesomemath

i attended camp last year and these are a few of the numerous safety hazards at camp

1. so much math in a day can permanently mess with your brain
2. the kids there are so beastly at ultimate that a properly thrown frisbee could seriously injure you
3. you might have to go naked the rest of the camp, making your body susceptible to other dangers, if you lose your clothes in a game of poker
4. red bull is bad for your health, and unfortunately, is necessary to stay up all night and and still pay attention in class the next day
5. if you fall asleep during class zuming will throw a marker at you. he has deadly accuracy. trust me
6. you might fall out of a roller coaster during the six flags outing and die
7. some of the apartments, especially the girls, are soo dirty and messy by the end of camp that they are classified as biological hazards
8. some of the guys at camp are drop-dead hot, emphasis on drop-dead
9. math camp is just a very dangerous place in general, with all the math gangstas around and stuff
10. zuming will own you
11. zuming will own you
12. zuming will own you
13. "     "
14. "     "
15. "     "
...[/quote]

haha MARKERS *daesun*!! Or well I was in Red but still... haha and I was one of those frisbee people... ;) Haha we played three hours a day and THEN went into the pool. Another hazard: possibly running over zac while buying ice cream.

Oh and another one: Short azns. Nuff said.

By the way HI ZAC!!! GOD I miss AMP.

Man, I didn't even go to AMP and I miss it! (OK, I guess you call this looking forward to it)

[quote="Max300"]Just in case some poor little kid doesn't get to go to camp because his/her parents saw this and took me seriously....I WAS KIDDING (except about the Zuming throwing markers at you part, but seriously, you have to really piss him off *cough* Daesun *cough* to do that)

AMSP is a lot of fun and very safe and as long as you don't put aluminum foil in the oven (hey it wasn't me okay) or mention Mathcounts in front of Zuming you'll be okay. I am hoping I can go back this summer, but I don't know if it will work out.

I have a question Zac. As of a few weeks ago, according to the AwesomeMath site, Pascual was coming back this summer as a mentor. However, his name is no longer there and I think a name that was not listed before is now there. Does this mean he's not going to be back this year?

Also, now that they are changing the way that the camp is being taught from last summer (from learning all subjects to focusing on two), are the classes still going to be split up into different levels?[/quote]

Wait, what's wrong with MATHCOUNTS :D?

Hmm it doesn't seem like Zuming gets pissed at Mathcounts, as he gave away over 6000 dollars in scholarships for his new program in Lexington, MA to the winners at our state competition this year...

what is this new program of Zuming's?  more abuse of Daesun? (oh wait he lives in california - nvm)

seriously, though, the worst you could get injured there is maybe fall and break a bone, but the hospital is nearby, and everybody there will try to help.  you can only get killed if you try to stop one of Leonardo's or Pascual's PKs in soccer.  wonder why felipe's not coming back, eh?

just kidding about Pascual/Leonardo/Felipe.  They won't kill you, and Felipe is still a happy kid in Colombia (?) from what I know.

tell your mom to stop being so over protective..."I know you care about me, and for my safety, but if you care about me and my mathematical interest, you will let me go to the camp" etc...


Who do you think is going to win the World Cup 2006 in Germany?

Heres my take on each of the teams on the ballot:

Germany - Not so much.... Even though they are the host (and recent hosts such as France and South Korea took advantage) they arent the Germany I used to know. Even their peers say that.

Brazil - One I should have voted for. Only factor that might shake them up is the upsets that tend to occur sometimes to a team with such great offensive weapons as this team.

England - The greatest English team to ever appear on the World Cup. Too bad, same goes for Brazil and Argentina. But I voted for them cuz I like England.

Argentina - I say second best to Brazil, only because they sucked last time.

France - Not much even though they god Zidane back. Same as Germany

Holland - Great chance and actually one of few teams that can stop Brazil.

Czech Republic - I dunno why they are on here. They barely made it.

Portugal - I heard the coach complain about the ranking system, but the team wont back up the coach.

USA - Why are they on the poll???? Wheres Italy and Spain? US is not gonna do as well because its in Europe and all the European countries will do better.

here is my order:

BRAZIL

England

Italy

Argentina

and maybe spain and portugal, but most likely not

hrm it's either

Brazil, germany or argentina can't decide...

[quote="xxreddevilzxx"]

USA - Why are they on the poll???? Wheres Italy and Spain? US is not gonna do as well because its in Europe and all the European countries will do better.[/quote]

I don't know  :D  :D

[quote="shobber"]USA is likely to get into 16.[/quote]

don't think so, look at the group usa is in

The world cup result is extremely difficult to predict  :D

I want England to win for these reasons:
1.  I was born there
2. Half my family lives there
3.  David Beckham is the best

Oh well, I'm Brazilian and I'm also aware we have a very strong team. So my vote obviously goes to Brazil.

Argentina has a great team too, but they are in a fairly difficult group.

And I don't know about England, but Liverpool will probably beat São Paulo in the World Championship next Sunday.

Argentina.  Because of old Batistuta, though retired from the National team.

My theory about why USA is on there is because a large portion of the Aopsers are from USA, so its there for consolation?

Czechs are rated no. 2 in the world for some reason.

I vote Brazil, for no other reason than they won the last one.

I don't know anything about football, all i do is follow the path of quarter,semi and really finales

last time belgium got kicked out by bresil
Not a big shame to get kicked out by the winning team, but if I recall correctly, a Belgian goal was disputed by the referee although later images really disproved that

Oh well, nobody got hurt

[quote="Dave Vather"]Go!!! Mexico!!!!!! :first: (yeah, sure )
EDIT Why on the world is USA and Czech Republic on there and Mexico isn´t?
I think Mexico has better chances than these two[/quote]
iran is in the same group with Mexico (Iran will win :) )

i think Brazil has highest chance of winning the world cup,
Germany is not a strong team, the only thing is that they are the host, other wise i don't think they have much chance...

[quote="log0"]My theory about why USA is on there is because a large portion of the Aopsers are from USA, so its there for consolation?

Czechs are rated no. 2 in the world for some reason.

[/quote]

Not really. USA reached the quarters last time out, only to lose to the Scottish referee.

And USA are rated no.8 in the world for some reason  :)

[quote="fredbel6"]seriously, there was something not right with the referees last time right?[/quote]

Yeah, a lot.

Go Saudi arabia :lol:  :lol: They have good playing tactics if you ever watched them play,that's why I have the genes of playing awesome football.

England should do something,this team of their's is a one in a lifetime awesome team.Brazil gonna win though bcos of ronaldinho.

P.S-Zidane is the best footballer alive.

Maybe:

Brazil

Germany

Argentina

 :stretcher:

[quote="warut_suk"][quote="shadysaysurspammed"],that's why I have the genes of playing awesome football.

[/quote]

 :rotfl:  :rotfl:[/quote]

oii,you challenging me?

[quote]iran is in the same group with Mexico (Iran will win  )
[/quote]
I would bet anything to Mexico in that match.
Nobody has answered why did the poll has some teams that surelly won't win (i.e. Czech Republic (sorry czech people)) And there aren't some others that might have better chances
EDIT I can't believe that more people thinks USA will win the World Cop than Germany.

[quote="Dave Vather"]
EDIT I can't believe that more people thinks USA will win the World Cop than Germany.[/quote]

i totally agree. I mean USA ? get real

hey everyone's putting czech down , she isnt soo bad :?   pity nedved went off but there's heinz.

GERMANY forever! :first:

I think the order might be: :winner_first:
Brazil
England
Argentina
France
South Korea
Germany
....
so on
I do not think usa has even a 5% chance of winning... and no offense to iran, czech, mexico, whatever, but I think they most likely won't even make it in the top 10. ;)

Im sorry to said but, you order is imposible.
France, Switzerland and South Korea are in the same group. It's a sure thing that one of them will not get in the round of 16.
Also, some countries that i think should be on the poll and aren't
Spain
Italy
Mexico

[quote="eminem"]

hey everyone's putting czech down [/quote]

the czechs are actually quite good... i would bet anything that the ones that get eliminated in their group will be USA and Ghana...

[quote="Dave Vather"]I didn't know that, but it seems that Czec Republic is really a good team, Ranked number 2 by FIFA (but we all know that that dosen't mean anything) So, it seems that they probably have better chances than i thought. Gruop E isn't a good place to be[/quote]

Indeed...

exactly what i said , i mean look in Euro 2004 , other than a semifinal shock defeat by Greece they looked ready to beat anybody. Pity no Nedved anymore :(


Harvard
college
MIT
induction
number theory
prime factorization
algebra

Yep, that's right!  The [url=http://www.mandelbrot.org/]Mandelbrot[/url] website has been updated and the individual scores for round 2 have been posted.

According to the website, team tests will be graded on January 10th and school standings should be up by January 14th.

Only 5 perfects in the west, whereas 23 in the east?  That's a big difference.  (15 at national level.)

[quote="mathfanatic"]Yep, that's right!  The [url=http://www.mandelbrot.org/]Mandelbrot[/url] website has been updated and the individual scores for round 2 have been posted.[/quote]

Thanks for posting the link. We haven't tried Mandelbrot yet in our local math team. The leaderboard sure included a lot of names that are familiar from the AMC Web site.

oooh, i believe thats Mr. Hon's name up there :P

Who the heck is Mr. Hon?  Or wait, don't answer that one.  I really don't care who he is.  He must be some loser math nerd or something.   :P

No math nerd is a loser.  Besides, mathfanatic is a math nerd, so he must be a loser too, then.

[quote="mathfanatic"]Yep, that's right!  The [url=http://www.mandelbrot.org/]Mandelbrot[/url] website has been updated and the individual scores for round 2 have been posted.  [/quote]

Hey, look at that!  I'm up there.  :shock: :shock:  :P

Alright, I'm done.

[quote="ElectricSnowman"]Hey, look at that! I'm up there.

Alright, I'm done.[/quote]

Quit spamming, you two.

Now, on to more important business.  Grading day for the team tests is today; according to the site, updated team scores so far will be posted on the 14th.

Fourmoredays.  Fourmoredays.

[quote="JBL"]Team tests all graded ... just wait until Sam puts them online :)[/quote]

You were one of the graders?

I am.  I dunno how any teams actually did, though, because they all have 4-letter codes.

I'm a freshman at Harvard.  All of the graders are Harvard or MIT students, along with Sam Vandervelde (who runs the whole contest).

JBL, do you know/have any classes taught by noam d. elkies? he is the mathematician i most respect right now.

Nope.  He was actually a Stuyvesant math team captain, if you care.  I do have a good friend (John Sasmor, for anyone from NYC Math Team) who has just finished a seminar that he taught on Chess and Mathematics.

Oy, I really don't remember, sorry.  There were I think two perfect scores in the national division.

If you're in the national division and you didn't like your score on question 2, you can blame me.

are team completely diff questions in nationsl and regional

No, although this time they were very different.  Usually, the national team includes basically everything in the regional.

ZERO OUT OF FOUR on #2! lol i didn't do this problem on the team but wow... the person thought it was correct. well JBL as you said, I am blaming you.

that's pretty good 2 days earlier than scheduled.

On the memo they posted an alternate solution to the last question on the national individual round.  Could someone post what that question is?

If you tell me what he did, I'll tell you why he was wrong.  If he tried to use prime factorization, it was wrong.  (Only one team that tried to use prime factorization even came remotely close to what they needed -- it's an awful way to go about it, so unless he put down everything that he would have needed, including an obvious inductive structure, I didn't give any points for it.  Doing a bad way slopily (spelling?) is not a big point-earner.)

Oh, and good job Stuy!  Now if only AAST would stop getting perfect scores, you might be able to catch them!

Note:  Yep, he tried to do it with prime factors.  Tsk, tsk, tsk.  All you needed was a straight-forward strong induction.

Lol I keep panicking on these contests. Like I saw exactly what needed to be done after doing #3/#4 and reading #5, but I kept making stupid errors on the induction in #5, and after numerous crossouts, I finally got it correct, up to the part before converting the expression using the Binomial Theorem (which I didn't have time for, panicked). (At least 2/5 on that part)

Err, but #2... ech, still stuck at tied for 10th. Hey at least better than previous years, where we didn't even place in first tier... LOL

"Good job, Stuy"

Haha, Joel, if only you could see us during the contest:

"Is part v done?"
"Yeah, yeah."

--1 min left--

"Wait, where's part v?"
"What?"

... Hence our awe-inspiring score of 0 on that part.  8-)

I was wondering about that -- I thought you 4 could generally be relied upon to write [i]something[/i] down :).  You're still in range of IMSA if you can improve your team round a little bit.

are there awards for team scores?  If so, how are those decided?  First tier?  Top 20 etc.?

I don't think anyone ever actually mentioned it, but it was implied by fairyofwind knowing his team's score on #2.  Anyway, the team scores are up now.

[url]http://www.mandelbrot.org/[/url]

Or for those of you who don't want the sidebar, [url]http://www.mandelbrot.org/scores.html[/url]

There are awards for team scores. They give the teacher a T-shirt or a frisbee I think if your school is in the top 14.


Find all n and k such that $ n^3\plus{}n^2\plus{}n\plus{}1 \equal{} k^2$

[quote="mathemonster"]the left side factors: $ (n \plus{} 1)(n^2 \plus{} 1)$[/quote]

And then $ (n\plus{}1,n^2\plus{}1)\le 2$


[b]Prove that if the sides of trapezoid ABCD (AB || CD) satisfy the condition $ AB^2\plus{}BD^2\equal{}(AB\plus{}CD)^2$, then ABCD is orthodiagonal.[/b]

i couldn't follow the solution they gave in the book.
thanks!

[hide="hint"]If $ a, b, c$, and $ d$ are the sides of a quadrangle, then this quadrangle is orthodiagonal iff $ a^2 \plus{} c^2 \equal{} b^2 \plus{} d^2$.[/hide]


function
real analysis
real analysis unsolved

This is the Lambert W function.  The series can be found directly by using the (little-used) Lagrange Inversion formula.

See  [url]http://en.wikipedia.org/wiki/Lagrange_inversion_theorem[/url].

Does anyone have an elementary proof? I've proved that it is equivalent to:
$ \forall_{n}$
\[ \sum_{k=1}^{n-1}\left({n \atop k}\right)k^{k}(n-k)^{n-k-1}=(n-1)n^{n-1}\]
http://www.mathlinks.ro/viewtopic.php?t=156681


is Fundamentals of Physics by Halliday, Resnick, Walker, appropriate for the prelim (f=ma) exam?

Any introductory university-level physics textbook is sufficient-it's more about your ability to apply the material.  I personally used the book by Halliday, Resnick, and Krane, which had a very nice collection of interesting problems (I think the one by Krane is a little harder).


Let $ (x_n)$ be a sequence such that $ x_0$ and $ x_{n \plus{} 1} \equal{} \frac {2x_n \plus{} 1}{x_n \plus{} 2}$ forall $ n \equal{} 0,1.2...$

determine $ [\sum_{k \equal{} 1}^n x_k]$ where $ [x]$ denote the lagest integer not exceeding $ x$.

It is not hard to find $ x_n\equal{}1\plus{}\frac{2x_0\minus{}2}{(3^n\minus{}1)x_0\plus{}3^n\plus{}1}\equal{}\frac{chna x_0\plus{}shna}{shna x_0\plus{}chna},$ were $ a\equal{}\frac{ln 3}{2}$.
Or $ x_n\equal{}cth(na\plus{}b)$, were $ b\equal{}arcth x_0$.

This is a problem on M&Y in Vietnam. Pleases lock it!


quadratics
geometry
3D geometry
algebra
quadratic formula
algebra unsolved

Find all the integers $x$ and $y$ such that:

$x^{6}+x^{3}y=y^{3}+2y^{2}$

So no solutions by infinite descent, except $(x,y)=(0,0)$

You can subtract $y^{3}+2y^{2}$ from each side to get a quadratic in $x^{3}$. In order for the quadratic to have solutions, $y=n^{2}-n-2$ where n is an integer. After plugging this into the quadratic formula, finding that $x^{3}=-n(n+1)(n-2)$ or $x^{3}=(n-1)(n+1)(n-2)$. None of the RHSs for the above equations will ever be perfect cubes except when the RHSs are 0, so x is always 0 and, after solving the cubic equation that remains, the solutions are $\boxed{(0,0),(0,-2)}$.


calculus
inequalities proposed
inequalities

If $ s\geq r$ and $ 0<m\leq x \leq M$ Prove that:
\[ f(x)\geq 0\]
where
\[ f(x)\equal{}r\left(x^r\minus{}\frac{M^r\minus{}m^r}{M^s\minus{}m^s}x^s\minus{}\frac{M^sm^r\minus{}M^rm^s}{M^s\minus{}m^s}\right)\]

[quote="Sunjee"]If $ s\geq r$ and $ 0 < m\leq x \leq M$ Prove that:
\[ f(x)\geq 0\]
where
\[ f(x) \equal{} r\left(x^r \minus{} \frac {M^r \minus{} m^r}{M^s \minus{} m^s}x^s \minus{} \frac {M^sm^r \minus{} M^rm^s}{M^s \minus{} m^s}\right)\]
[/quote]very easy with some basic calculus  :blush:

First note $ f(m)\equal{}f(M)\equal{}0$.
Suppose first $ r>0$. ($ r<0$ likewise, replacing $ m,M,x$ by $ \frac1m$ etc)
Showing $ f'(m)\ge0$ boils down to showing $ ln(y)\ge1\minus{}\frac1y$. We can assume wlog $ r\equal{}1$ and see that $ f$ has only one extremum. The result follows.  :)

[quote="Sunjee"]Thank u very much. Can u explain me why we must check that
\[ f'(m)\geq 0\sim ?\]
[/quote] oh yes you are right, it's even not necessary to check that, as the coeff of $ x^s$ is negative and $ s\ge r$. So we are assured that $ f(x)$ is negative for big $ x$, and that's all we need. After all, a really simple problem. :P

Even easier: $ f(0)\le 0$  :blush:


So I've watched this movie on metacafe http://www.metacafe.com/watch/384611/can_you_tell_me_how_it_is_possible/

And am wondering the logical solution to that trick. I mean there will be always a 0,(0)1 that will be needed to make the whole 1. Is this thing true, or did he just messed sth up ?

Of course $1=3{1\over 3}=3\cdot 0.\bar{3}=0.\bar{9}$.
It's just 2 ways to write the same number. (like ${1\over 1+\sqrt{2}}=\sqrt{2}-1$)
And $0.\bar{0}1$ doesn't exist. How can you put $1$ after infintely many $0$s?


$\mathcal{F}$ is the family of all $n$-tuples $(A_1,A_2,\ldots,A_n)$, where $A_i$  is a subset of a set containing $1998$ elements. We can easily find that $|\mathcal{F}|=2^{1998n}$.
$\mathcal{P}$ is the family of all $n$-tuples $(A_1,A_2,\ldots,A_n)$, where $A_i$ is a subset of a set containing $1997$ elements. We also have $|\mathcal{P}|= 2^{1997n}$.
For each $x \in X=\{1,2,\ldots,1998\}$, the number of $n$-tuples $(A_1,A_2,\ldots,A_n)$, where $A_i$ is a subset of $X$, and with $x\in A_1 \cup A_2 \cup \cdots \cup A_n$ is $|\mathcal{F}|-|\mathcal{P}|$.
So, the sum we have to determine is $1998(|\mathcal{F}|-|\mathcal{P}|)$.

Let $n, N\in \mathbb{N}^*$, and let $\mathcal{F}$ be the family of all $n$-tuples $(A_1, A_2, \ldots, A_n)$ such that each $A_{i}$ is a subset of $\{1, 2, \ldots, N\}$.  Let $|A|$ denote the number of elements of the set $A$.  Find
\[ \sum_{(A_1, A_2, \ldots, A_n)\in \mathcal{F}} |A_1\cup A_2\cup \cdots \cup A_n| .\]

There are $\binom {N} {k}$ subsets $A \subseteq \{1,2,\ldots,N\}$ with $|A| = k$, $0\leq k \leq N$. For each such $A$, the number of elements $(A_1, A_2, \ldots, A_n) \in \mathcal{F}$ with $A_1\cup A_2\cup \cdots \cup A_n = A$ is obviously $(2^{n} - 1)^k$. Therefore the sum we seek is
\[ \sum_{k=1}^N \binom {N} {k} k(2^{n} - 1)^k = N(2^n-1)\sum_{k=1}^{N} \binom {N-1} {k-1} (2^{n} - 1)^{k-1} = N(2^n-1)2^{n(N-1)} .\]

Most likely there must exist a proof by generating functions.

[quote="mavropnevma"]
Most likely there must exist a proof by generating functions.[/quote]

Define $F_{m,n}(t,x):=\left(1+t\left((1+x)^n-1\right)\right)^m$.  The coefficient of $t^k x^\ell$ is the number of $n$-tuples $\left(A_1,A_2,\ldots,A_n\right)$ of subsets of $\{1,2,\ldots,m\}$ such that $\bigcup_{i=1}^n A_i$ has $k$ elements and that $\sum_{i=1}^n \left|A_i\right|=\ell$.  We want to associate $t^k$ with $k$, and this can be done via differentiating $F_{m,n}(t,1)$ with respect to $t$ and then replacing $t$ by $1$.  The sum of $\left|\bigcup_{i=1}^n A_i\right|$ is then \[\left.\frac{\partial}{\partial t} F_{m,n}(t,1) \right|_{t:=1} =  m \left(2^n-1\right)2^{(m-1)n}\,.\]

This solution is not particularly elegant, but if one is asked to determine the sum of $\left(\sum_{i=1}^n \left|A_i\right|\right) \left|\bigcup_{i=1}^n A_i\right|$, then one can differentiate $F_{m,n}(t,x)$ with respect to each of its variables:  \begin{align*}\left.\frac{\partial^2}{\partial x \partial t} F_{m,n}(t,x) \right|_{t:=1,x:=1} &=  \left.\frac{\partial}{\partial x} m \left((1+x)^n-1\right)(1+x)^{(m-1)n}\right|_{x=1} \\
&=mn2^{(m-1)n-1}\left(m2^{n}-(m-1)\right)\,.\end{align*}  And this is exactly the value of the sum.

[quote=shobber]Let $F$ be the set of all $n$-tuples $(A_1, \ldots, A_n)$ such that each $A_{i}$ is a subset of $\{1, 2, \ldots, 1998\}$.  Let $|A|$ denote the number of elements of the set $A$.  Find
\[ \sum_{(A_1, \ldots, A_n)\in F} |A_1\cup A_2\cup \cdots \cup A_n|  \][/quote]

For an element $x \in \{1,2,\cdots,1998\}$, the probability that $x \in |A_1\cup A_2\cup \cdots \cup A_n| $ is clearly $1 - \frac{1}{2^n}$. Thus the expected cardinality of $|A_1\cup A_2\cup \cdots \cup A_n| $ is $1998 \left ( 1 - \frac{1}{2^n} \right ) $. The summation evaluates to

$$1998 \left ( 1 - \frac{1}{2^n} \right ) \cdot |F| = 1998 \left ( 1 - \frac{1}{2^n} \right ) \cdot 2^{1998n}$$

For each element, there is a $\frac 12$ chance it is not in $A_i$. Thus the probability that a given element is in at least one set is $1-\frac{1}{2^n}$, so the expected value of
\[| A_1 \cup A_2 \cup \ldots \cup A_n | = 1998 \left(1 - \frac{1}{2^n}\right).\]

As there are $2^{1998n}$ $n$-tuples in $F$, our total sum is $\boxed{2^{1998n} \cdot 1998 \cdot \left(1 - \frac{1}{2^n}\right)}$.

Let $\mathbb{I}\{E\}$ be the indicator of the proposition $E$, i.e. $\mathbb{I}\{E\}\in\{0,1\}$ and $\mathbb{I}\{E\}=1$ iff $E$ is true. Set $N\triangleq 1998$ and observe that
\begin{align*}
&\sum_{(A_1,\dots,A_n)\in F} \bigl|A_1\cup A_2\cup \cdots \cup A_n\bigr| \\
& = \sum_{(A_1,\dots,A_n)\in F} \sum_{k=1}^N \mathbb{I}\bigl\{k\in A_1\cup \cdots \cup A_n\bigr\} \\
&=\sum_{k=1}^N \sum_{(A_1,\dots,A_n)\in F} \mathbb{I}\bigl\{k\in A_1\cup \cdots \cup A_n\bigr\} \\
&= N\sum_{(A_1,\dots,A_n)\in F} \mathbb{I}\bigl\{k\in A_1\cup \cdots \cup A_n\bigr\},
\end{align*}
from symmetry. Now, there are $(2^N)^n =2^{Nn}$ $n$-tuples $(A_1,\dots,A_n)\in F$. Of all those, $(2^{N-1})^n$ do not contain $k$. So, the number of those $n$-tuples containing $k$ is
\[
2^{Nn}-2^{(N-1)n} = 2^{(N-1)n}(2^n-1),
\]
yielding the desired answer as $N2^{(N-1)n}(2^n-1)$.


Μια απορια που εχω:

Εστω μια συναρτηση $ f$ παραγωγισιμη στο $ [a,b]$.
Ειναι σιγουρο οτι η $ f'$ ειναι φραγμενη?

[quote="AlexandrosG"]Εστω μια συναρτηση $ f$ παραγωγισιμη στο $ [a,b]$.
Ειναι σιγουρο οτι η $ f'$ ειναι φραγμενη?[/quote]

Ναι αν η $ f'$ είναι συνεχής.

Αλλιώς, σκέψου την εξής κατασκευή (νομίζω ότι δουλεύει αλλά τη σκάρωσα πολύ πρόχειρα οπότε παίζει να έχω κάνει λάθος):

1. Πάρε το διάστημα $ I_n \equal{} [1/n,1/(n \minus{} 1)]$

2. Κατασκεύασε μία παραγωγίσιμη συνάρτηση $ f_n$ ώστε:
2a. $ 0\leq f_n \leq 1/n^2$ σε όλο το $ I_n$ και $ f_n\equiv 0$ εκτός του $ I_n$
2b. $ f_n(1/n) \equal{} 0 \equal{} f_n(1/(n \minus{} 1))$ (redundant εξαιτίας του 2a αλλά δεν πειράζει ;))
2c. για κάποιο $ \xi_n \in I_n$ να έχουμε $ f_n'(\xi_n) \equal{} 1/n$

3. Ένωσε όλες αυτές τις συναρτήσεις, δηλ. πάρε την $ f \equal{} \sum_n f_n$.

Η τελική συνάρτηση είναι παραγωγίσιμη στο $ [0,1]$ (γιατί?) αλλά δεν έχει φραγμένη παράγωγο.

Cheerio,

Durandal 1707

Ευχαριστω για την απαντηση.
Ημουν σιγουρος οτι θα ηταν αρνητικη(αλλιως θα το ηξερα ως θεωρημα :) )
αλλα διαισθητικα δεν μπορουσα να πιστεψω οτι γινεται.

'Ενα απλο παράδειγμα μιας τέτοιας συνάρτησης είναι $ f(x)\equal{}x^2\sin(1/x^2)$ αν $ x\in (0,1]$ και $ f(0)\equal{}0$.

[quote="Jondrette"]'Ενα απλο παράδειγμα μιας τέτοιας συνάρτησης είναι $ f(x) \equal{} x^2\sin(1/x^2)$ αν $ x\in (0,1]$ και $ f(0) \equal{} 0$.[/quote]

Hahaha :D Ήμουν σίγουρος ότι κάποιο παθολογικό ημίτονο θα δούλευε αλλά βαριόμουν να κάνω την παραγώγιση για να σιγουρευτώ :rotfl:

Cheerio,

Durandal 1707


How do I determine algebraically whether or not two polygons are intersecting if I know all of the co-ordinates (x, y) of both polygons. I know that I can check if all of the sides are intersecting, but if it's two octagons that's 64 combinations to check! (I am designing a program, lol, I wouldn't do this problem with a pen and paper! :wink: ).

I think I can use projection, maybe someone knows that method. If there are any methods, I appreciate anyone willing to explain them. I wouldn't mind checking if any of the points of polygon A are inside of polygon B, if anyone knows how to determine if a point is in a polygon that would be good enough!

Note : These could be convex or concave polygons!

This might be better off in the informatics forum, but the best way I can think of right now is just checking if two sides intersect. If you're writing a program, it's an $O(n^{2})$ algorithm, so unless you're using huge inputs shouldn't take too long.

Determining whether a point is inside a polygon does not seem like a faster way (since it would probably involving checking that point with each side anyway).


I  am   using a Windows machine, and  I     download and install the  two programs :  MiKTeX and TeXnicCenter  [url=http://www.mathlinks.ro/Wiki/index.php/LaTeX:Downloads]here[/url]

Now  I  am  trying  to  produc   slides  with   Prosper.   Here is  the  code:

[code]\documentclass[pdf,gyom,slideColor,colorBG]{prosper}

\begin{document}
\begin{slide}{Gauss-Bonnet in 20th Century}
In this talk I will consider the Gauss-Bonnet-Chern formula on
 some open Riemannian manifolds.
\item
(Gauss-Bonnet)Let $\Sigma$ be a smooth closed oriented surface in $R^3$, then
$$\int_{\Sigma}k\mbox{d}s = 2\pi\chi (\Sigma).$$
\item
Hopf, 1925: For a hypersurface $M^n$ in $R^{n+1}$($n$ even), one has
$$\int_{M}k\mbox{d}v =\frac{1}{2}\mbox{vol}(S^n)\chi (M),$$
where $k$ is the Gauss-Kronecker curvature.

\end{slide}

\end{document}
[/code]

The  code  is  right.  How  to  get  the pdf?
 the bar near the top of the TeXnicCenter window reads "LaTeX => PDF"  or  "LaTeX => dvi" ,  which is right?
Many  thanks!

The code is not rght as it doesn't include \begin{itemize} or \end{itemize} so the \item lines cannot be compiled.
The prosper manual tells you that you must use Postscript with LaTeX=>DVI=>PS=>PDF.
To do this you must have Ghostscript installed and use the LaTeX=>PS=>PDF profile. If you can't see this profile then download from [url=http://www.latex-community.org/forum/viewtopic.php?f=31&t=302]here[/url].

Finally, Beamer and pdfscreen produce much nicer slides. See examples of  [url=http://latex-beamer.sourceforge.net/]beamer[/url] and [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=pdfscreen]Manual, for screen viewing[/url]

Thank you very much!

If  the code is right,  make  sure  that the bar near the top of the TeXnicCenter window reads  LaTeX=>PS=>PDF ,
then

1. Select "Build" -> "Current File" -> "Build" from the top menu bar.
2. Select "Build -> "View Output" from the top menu bar.

finally, the  pdf  shows up.

Is  it?

Is  it  suitable  for   Beamer and pdfscreen?