Twelve congruent disks are placed on a circle $C$ of radius 1 in such a way that the twelve disks cover $C$, no two of the disks overlap, and so that each of the twelve disks is tangent to its two neighbors. The resulting arrangement of disks is shown in the figure below.  The sum of the areas of the twelve disks can be written in the from $\pi(a-b\sqrt{c})$, where $a,b,c$ are positive integers and $c$ is not divisible by the square of any prime. Find $a+b+c$.

[asy]
real r=2-sqrt(3);
draw(Circle(origin, 1));
int i;
for(i=0; i<12; i=i+1) {
draw(Circle(dir(30*i), r));
dot(dir(30*i));
}
draw(origin--(1,0)--dir(30)--cycle);
label("1", (0.5,0), S);[/asy]

[hide]

We wish to find the radius of one circle, so that we can find the total area.

Notice that for them to contain the entire circle, each pair of circles must be tangent on the larger circle. Now consider two adjacent smaller circles. This means that the line connecting the radii is a segment of length $2r$ that is tangent to the larger circle at the midpoint of the two centers. Thus, we have essentially have a regular dodecagon whose vertices are the centers of the smaller triangles circumscribed about a circle of radius 1.

We thus know that the apothem of the dodecagon is equal to 1. To find the side length, we make a triangle consisting of a vertex, the midpoint of a side, and the center of the dodecagon, which we denote A, M, and O respectively. Notice that OM=1, and that $\triangle OMA$ is a right triangle with hypotenuse $OA$ and $m \angle MOA = 15^\circ$. Thus $AM = (1) \tan{15^\circ}= 2-\sqrt{3}$, which is the radius of one of the circles. The area of one circle is thus $\pi(2-\sqrt{3})^{2}= \pi (7-4 \sqrt{3})$, so the area of all 12 circles is $\pi (84-48 \sqrt{3})$, giving an answer of $84+48+3 = \boxed{135}$.

[/hide]

A bit wordy, but that solves it.

[quote="calc rulz"][hide]

Notice that for them to contain the entire circle, each pair of circles must be tangent on the larger circle.

[/hide][/quote]

[hide]Wait, so the point of tangency is on the larger circle C?  I seem to get a contradiction when that happens....tell me what I did wrong.

So, if I connect the center of the large circle to the point of tangency of one of the smaller circles, I get the radius of the circle. However, if I connect the center of the large circle to the center of one of the smaller circles, I also get the radius of the circle.  The radius of the smaller circle to the point of tangency is supposed to be perpendicular to the tangent line, but how it be perpendicular to the radius if it forms an isoceles triangle...?[/hide]

[quote="Zellex"]if I connect the center of the large circle to the center of one of the smaller circles, I also get the radius of the circle.[/quote]

The answer to your paradox is that the above statement is false.

Oh, I see, so the center of the disks are not on the circle?

Sorry, I only looked at the diagram instead of reading the whole problem  :blush: .

The radius should be $ \sin 15$...and the total area is $ 12\sin^2 15^\circ \pi\equal{}(6\minus{}3\sqrt{3})\pi$, so $ a\plus{}b\plus{}c\equal{}6\plus{}3\plus{}3\equal{}\boxed{012}$.

[quote="Altheman"]The radius should be $ \sin 15$...and the total area is $ 12\sin^2 15^\circ \pi \equal{} (6 \minus{} 3\sqrt {3})\pi$, so $ a \plus{} b \plus{} c \equal{} 6 \plus{} 3 \plus{} 3 \equal{} \boxed{012}$.[/quote]

The picture is very bad, the center of the disks is not suppose to be on the circle, so r is tan15.

Altheman, Ah, you made the same mistake I did in Mar 03 2007 :P.  The radius is actually $ \tan 15$.

Ha...that picture is confusing...I didn't really read the problem...and assumed that the centers lied on the circle like the diagram shows...That is obviously not the case if you actually read the problem :D.

I'm confused... Is the answer 12 or 135?  Are the centers of the small circles on the large circle?  I don't think the question said they weren't...

Thanks

The problem precisely describes a situation.  As was indicated several times in the thread, in this situation we have that the centers of the small circles are not on the large circle (proof: this leads to a contradiction, as in Zellex's post).

Here is a more accurate diagram. (Asymptoted from the diagram at [url=http://www.artofproblemsolving.com/Wiki/index.php/1991_AIME_Problems/Problem_11]the AoPS Wiki[/url])

[hide]
[asy]
unitsize(100);
draw(Circle((0,0),1));
dot((0,0));
draw((0,0)--(1,0));
label("$1$", (0.5,0), S);

for (int i=0; i<12; ++i)
{
dot((cos(i*pi/6), sin(i*pi/6)));
}

for (int a=1; a<24; a+=2)
{
dot(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12)));
draw(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12))--((1/cos(pi/12))*cos((a+2)*pi/12), (1/cos(pi/12))*sin((a+2)*pi/12)));
draw(Circle(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12)), tan(pi/12)));
}

[/asy][/hide]

Is it possible to replace the diagram on the AoPS Resources AIME with the new diagram (of 1-1 is 3)?

[quote="calc rulz"]
Notice that for them to contain the entire circle, each pair of circles must be tangent on the larger circle.
[/quote]
Can someone explain why this is true? Why couldn't the smaller circles' point of tangency lie outside(or inside) of the bigger circle?

If the disks were tangent outside or inside $C$, there would be space in between disks and they would not cover $C$. However, the problem statement mentions that they do cover $C$. (To clarify for the future, this means that the radius of each of the disks is $\tan{15^{\circ}} = 2 - \sqrt{3}$; thus, the total area of the $12$ disks is $\pi(84 - 48\sqrt{3})$, which gives the answer of $\boxed{135}$.)
-----
Indeed, the diagram is quite confusing; the centers of the disks cannot lie on $C$ because they would be tangent inside the circle, which is not possible as per the problem statement.

Is it possible for a moderator to replace the original post with the corrected diagram given in post #12 for future users? Thanks! :)


If three cubes of sizes 3 by 3, 2 by 2, and 1 by 1 are arranged in some manner, what is the least possible surface area of the combined figure?

Your best chance is always to stack them onto each other. 3*3*3 first 2*2*2 second and last 1*1*1. So you get a surface area of: 74!

:D

[quote]Your best chance is always to stack them onto each other. 3*3*3 first 2*2*2 second and last 1*1*1. So you get a surface area of: 74!

[/quote] Nah that's not right if you stack the 2 by 2 and the 1 by 1 on top of the 3 by 3 you get 72

Yep. 72's right.

It was two Olympiads ago this year. I figured I was officially allowed to spill the beans


floor function
Pascal\u0027s Triangle
algebra
binomial theorem

I need help!

1)Find a formula for the greatest number in the nth row of a pascal's triangle.

2)prove that the numbers  in any row of pascal's triangle can always be divided into 2 sets with the same sum.

thanks

[hide="1"]I'm not sure if this is correct, but I get $ \binom{n}{\lfloor\frac{n}{2}\rfloor}$[/hide]
[hide="2"]For rows with an even number of numbers, simply split the row in half. For rows with an odd number of numbers, take every other number into one set, and the remaining numbers into a second set. The two sets will always have an equal sum. Obviously you can't split the first row into more than one set.[/hide]

Answer

I assume that the first row begins 1, 1.

Lets take the nth row.

If n is even, then there is an odd number of terms in the row. The middle term, $ x$ is the largest, and is given by the following binomial coefficient:

$ x \equal{}\binom{n}{\frac{n}{2}}$

If n is odd, then there are an even number of terms in the row. Each of the two middle terms, $ x$ is the largest, and is given by the following binomial coefficient:

$ x \equal{}\binom{n}{\frac{n}{2}\minus{}\frac{1}{2}}$

In general:

 $ x \equal{}\binom{n}{\left[\frac{n}{2}\right]}\equal{}\frac{n!}{\left[\frac{n}{2}\right]!\left[\frac{n}{2}\right]!}\equal{}\frac{n!}{\left(\left[\frac{n}{2}\right]!\right)^{2}}$

Answer

Lets consider the nth row of pascal's triangle.

If n is odd, then there are an even number of terms in this row. You can split the row down the middle, and obtain exactly the same terms on both sides, and hence there are two sets with the same sum.

This is because $ \binom{n}{r}\equal{}\binom{n}{n\minus{}r}$

If n is even, then there are are odd number of terms in this row. From the binomial theorem (we assume $ n\geq 2$):

$ (1\plus{}x)^{n}\equal{}\binom{n}{0}x^{n}\plus{}\binom{n}{1}x^{n\minus{}1}\plus{}\binom{n}{2}x^{n\minus{}2}\plus{}\cdots\plus{}\binom{n}{n}x^{0}$

Let $ x \equal{} 1$:

$ \Rightarrow 2^{n}\equal{}\binom{n}{0}\plus{}\binom{n}{1}\plus{}\binom{n}{2}\plus{}\cdots\plus{}\binom{n}{n}$

$ \Rightarrow 2^{n\minus{}1}\equal{}\binom{n\minus{}1}{0}\plus{}\binom{n\minus{}1}{1}\plus{}\binom{n\minus{}1}{2}\plus{}\cdots\plus{}\binom{n\minus{}1}{n\minus{}1}$

Using pascal's identity: $ \binom{n}{r}\plus{}\binom{n}{r\plus{}1}\equal{}\binom{n\plus{}1}{r\plus{}1}$

$ 2^{n\minus{}1}\equal{}\binom{n}{1}\plus{}\binom{n}{3}\plus{}\cdots\plus{}\binom{n}{n\minus{}1}$

But these are all terms in the nth row, and they sum to half of $ 2^{n}$, which is the sum of all the terms on the nth row.

Hence we have found a way to split the terms into two equal sets, the other set containing all the terms in the nth row not belonging to this first set.


let m,n,p are positive real
prove: sqrt( m / (m+p))+sqrt( p / (n+p))+sqrt( n / (m+n)) <=3/sqrt(2)

See http://www.mathlinks.ro/Forum/viewtopic.php?t=17549 . Myth gave a really great proof in post #11.

  Darij


I use firefox for pretty much everything but CSS as it doesn't render as well in fire fox.
-The other flaw about fire fox is amount of memory it takes compared to other browsers, but i really like its add on ability.

My Internet Explorer caused my computer to freeze...so I had to switch to Netscape...which really sucks.

I switched to FIREFOX like 3 months ago. Used to use IE.

FIREFOX FTW!!!!! :D  :D

lol my dad calls IE 'internet exploder'  :bomb: no offense to ppls that like IE

LOL!!! :P

I spam spam spam tastes HORRIBLE.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=178953]here[/url].[quote="jmerry"]What about the multibrowser option? I use Safari and Firefox. On the Windows machine, that's Firefox for everything, while it's Safari for almost everything (except AOPS) on my Mac. [/quote]

I'm sorry but you guys need to discover Opera if you already haven't as it eats the others for breakfast.

Mmm.... my friends Wii Internet runs on Opera if I am not mistaken.

hmm.... most of the time firefox, just to use chatzilla... but IE7 could get irritating some times

I use both firefox and internet explorer...

use Internet Explorer on my laptop, Safari on my iPod touch  :D

On my dad's desktop, I use Firefox.

On my laptop, I use Internet :bomb:.

 :rotfl:

[quote="leoxnlin"]99% of the time Firefox. Although for someone reason I can't get java apps to work. So I have to switch to IE for that.[/quote]

I think there's an add-on called IE Tab for firefox so you can open a new tab in "Internet explorer" or something

Firefox only becuase of all-in-one-gesture and the more user-friendly tools/options

[quote="leoxnlin"]99% of the time Firefox. Although for someone reason I can't get java apps to work. So I have to switch to IE for that.[/quote]
Can you give an example of a site that doesn't work for you in FF? Also (this is rather stupid to ask, but still) do you have java enabled in your preferences?

internet explorer :D i know it's not very common but it works for all the websites i've ever heard of and isnt that slow. so i'll be stickin with it :lol:

[quote="herefishyfishy1"]I use Firefox. I also don't know of any reasons to use Internet Explorer. Did you know that a single line of JavaScript code can crash most versions of IE?

Also, visit http://www.ie7.com/ for information about Internet Explorer 7.[/quote]

Wow nice firefox propaganda.
I don't really see the difference, though I like the all-in-one gesture add-on, add-ons in general, and the spell check.

At school I use Safari even though Firefox is available. I just do it out of tradition.
At home I use IE because it is the only one available.
I have not had many problems with either one.


Write an equation of the line (or lines) that satisfies the condition stated below.

All points equidistant from the lines y = 3x - 1
and y = 3x + 9.

This question is about LOCUS.

[hide]y=3x+4

When a line is equidistant from another line, it is parallel. In order for a point to be equidistant from two lines, it must be in the middle of them, 4 is in the middle of -1 and 9.[/hide]

[quote="Interval"]In other words, you added, 9 and -1 (addition of BOTH y-intercepts) and then divided by 2 to find 4 as the new y-intercept, right?[/quote]

That is correct.


[quote="dmath328"]I want to know my error (I misunderstand mixing varible I think..)

When we meet  next problem,

Pblm)  $ x,y,z > 0$.Prove that $ f(x,y,z) \ge M$  (when $ g(x,y,z) \equal{} 0$)

Mixing variable) $ f(x,y,z) \ge f(\frac {x \plus{} y}{2},\frac {x \plus{} y}{2},z)$for all $ x,y,z > 0$,$ g(x,y,z) \equal{} 0$

Then it's enough to show $ f(a,a,b) \ge M$ for all $ a,b > 0$, $ g(a,a,b) \equal{} 0$

This method doesn't have any logical error??[/quote]

If it doesn't have any error, let's see one problem.

Pblm) $ x,y,z>0$,$ xyz\equal{}1$
Prove that $ \frac{1}{(x\plus{}1)^2}\plus{}\frac{1}{(y\plus{}1)^2}\plus{}\frac{1}{(z\plus{}1)^2}\plus{}\frac{1}{x\plus{}y\plus{}z\plus{}1} \ge 1$

Sol) let $ g(x,y,z)\equal{}xyz\minus{}1$,$ f(x,y,z)\equal{}\frac{1}{(x\plus{}1)^2}\plus{}\frac{1}{(y\plus{}1)^2}\plus{}\frac{1}{(z\plus{}1)^2}\plus{}\frac{1}{x\plus{}y\plus{}z\plus{}1}$

Then it's easy to prove $ f(x,y,z) \ge f(\frac{x\plus{}y\plus{}z}{3},\frac{x\plus{}y\plus{}z}{3},\frac{x\plus{}y\plus{}z}{3})$ by Jensen

Then, it's enough to show $ f(a,a,a) \ge 1$ when $ a>0$,$ g(a,a,a)\equal{}0$

So a=1. We are done.

But I think it's not correct solution. Please give me logical error please.

[quote="dmath328"]
 Please give me logical error please.[/quote]
It's wrong because $ g(x,y,z)\neq g\left(\frac{x\plus{}y\plus{}z}{3},\frac{x\plus{}y\plus{}z}{3},\frac{x\plus{}y\plus{}z}{3}\right).$ :wink:

[quote="arqady"][quote="dmath328"]I want to know my error (I misunderstand mixing varible I think..)

When we meet  next problem,

Pblm)  $ x,y,z > 0$.Prove that $ f(x,y,z) \ge M$  (when $ g(x,y,z) \equal{} 0$)

Mixing variable) $ f(x,y,z) \ge f(\frac {x \plus{} y}{2},\frac {x \plus{} y}{2},z)$for all $ x,y,z > 0$,$ g(x,y,z) \equal{} 0$

Then it's enough to show $ f(a,a,b) \ge M$ for all $ a,b > 0$, $ g(a,a,b) \equal{} 0$

This method is right???[/quote]
You are right  if $ g(x,y,z) \equal{} g\left(x,\frac {y \plus{} z}{2},\frac {y \plus{} z}{2}\right)$ is true. :wink:[/quote]

Thank you very much, arqady!

Now I know what is my error!


modular arithmetic
geometry
incenter
circumcircle
trigonometry

Dunno if this counts:
$(r,\theta)^{p/q}=(r^{p/q},\frac{p{\theta}}{q}+\frac{2\pi{k}}{q})$
DeMoivre's...it's pretty basic.

[quote="1=2"]It's just an extension of Euler's Toient Theorem.[/quote]
It is not an extension; it is a trivial consequence.

[quote="1=2"]List all of the intermediate formulas you know here! I'll compile a list of the formulas as the days go by. Here's one to start:

$a^{(\phi(m)+x)}\equiv a^{x}\pmod m$

where a and m are relatively prime. It's just an extension of Euler's Toient Theorem.[/quote]
1. you spelled totient wrong ;)
2. This isn't an extension, just divide both sides by $a^{x}$ and you have it. It's like saying if a=b then ac=bc.

A couple of simple geometry formulas/theorems:

(1)  [u] Ptolemy's [/u]
In a quadrilateral $ABCD$ we have: [\ AB \cdot CD + BC \cdot AD \geq AC \cdot BD \] with equality iff $ABCD$ is cyclic.

(2) If triangle $\triangle ABC$ has points $P,Q,R$ on the respective sides $AB, BC, CA$,
\[\frac{AP}{PB}= \frac{BQ}{QC}= \frac{CR}{RA}= n \implies \frac{ [PQR] }{[ABC]}= 1-\frac{3n}{(n+1)^{2}}\]
The pedal triangle (P,Q,R are midpoints) is a special case of this, where $n=1$ and thus its area is $\frac{1}{4}$ that of the initial triangle.

(3) Similarly, in a quadrilateral $ABCD$ with points $P,Q,R,S$ placed on each side as in (2),
\[\[ \frac{AP}{PB}= \frac{BQ}{QC}= \frac{CR}{RD}= \frac{DS}{SA}= m \implies \frac{ [PQRS] }{[ABCD]}= 1-\frac{2m}{ (m+1)^{2}}\]
(4) [u] Euler's Formula [/u]
If the distance between the incenter and the circumcenter of a triangle is $d$, then:
\[d^{2}= R^{2}-2Rr \]
with the usual notation.

how many times are you going to see 3,4...lol...those are problem specific...

Also, there are waaaay too many Euler's formulas.
This one, trig, e+2=f+v, etc.

Lagrange interpolation in Intermediate?  :huh:

[quote="Altheman"][quote="cincodemayo5590"]

The pedal triangle (P,Q,R are midpoints) is a special case of this, where $n=1$ and thus its area is $\frac{1}{4}$ that of the initial triangle.

[/quote]

midpoints is the medial triangle

pedial is something else...i forget exactly...but it is in geo revisited...it might be triangle formed by perpendiculars to the sides...[/quote]

Pedal triangle is just if you take any point $P$ in the interior of $ABC$ and drop perpendiculars to the sides.

[quote="paladin8"]

Pedal triangle is just if you take any point $P$ in the interior of $ABC$ and drop perpendiculars to the sides.[/quote]

Oh yea...ok...and there was some cool result, like if you itterate that process 3 times, you get a triangle similar to the first

[quote="Altheman"][quote="cincodemayo5590"]

The pedal triangle (P,Q,R are midpoints) is a special case of this, where $n=1$ and thus its area is $\frac{1}{4}$ that of the initial triangle.

[/quote]

midpoints is the medial triangle

pedial is something else...i forget exactly...but it is in geo revisited...it might be triangle formed by perpendiculars to the sides...[/quote]

no, the altitudes is orthic triangle, i think

[quote="Altheman"][quote="cincodemayo5590"]

The pedal triangle (P,Q,R are midpoints) is a special case of this, where $n=1$ and thus its area is $\frac{1}{4}$ that of the initial triangle.

[/quote]

midpoints is the medial triangle

pedial is something else...i forget exactly...but it is in geo revisited...it might be triangle formed by perpendiculars to the sides...[/quote]

Oh, sorry. Wrong triangle... but I can't edit my post anymore. Rawr.


c varies inversley as d.  If c=.6 when d=20, then find d when c=0.25

[hide]Since it's an inverse relationship,

6 * 20 = 0.25 * d
120 = .25d
d = 480 [/hide]


[code]
\begin{document}
\maketitle

\section{Solutions to Problem Set 2}
\label{sec:Solutions to Problem Set 2}

\begin{enumerate}
\item[1.3.1.]

\item[(a)]Using $\pi r^{2}$ you find that the area of the circle is $81\pi\approx254.34$ khet$^{2}$.\\

\item[(b)]You first find the radius which is 4.5 khet. Now solve the equation:
\begin{eqnarray*}
4.5^{2}x&=&64\\
20.25x&=&64\\
x&=&\displaystyle\frac{64}{20.25}\\
x&\approx&\boxed{3.16}
\end{eqnarray*}

\item[1.4.1.]

\item[(a)]

The line segments $\overline{AB}$ and $\overline{BC}$ are of same length. If you connect $\overline{AB}$ and $\overline{BC}$ at point $B$, such that $\overline{AC}$ is a straight line, then the length of $\overline{AC}=\overline{AB}+\overline{BC}$

\ifx\JPicScale\undefined\def\JPicScale{1}\fi
\unitlength \JPicScale mm
\begin{picture}(60,80)(0,0)
\linethickness{0.3mm}
\put(10,80){\line(1,0){20}}
\linethickness{0.3mm}
\put(40,80){\line(1,0){20}}
\put(10,80){\makebox(0,0)[cc]{A}}

\put(10,80){\makebox(0,0)[cc]{A}}

\put(30,80){\makebox(0,0)[cc]{B}}

\put(40,80){\makebox(0,0)[cc]{B}}

\put(60,80){\makebox(0,0)[cc]{}}

\put(60,80){\makebox(0,0)[cc]{C}}

\put(20,60){\makebox(0,0)[cc]{}}

\linethickness{0.3mm}
\put(10,60){\line(1,0){40}}
\put(10,60){\makebox(0,0)[cc]{A}}

\put(30,60){\makebox(0,0)[cc]{B}}

\put(50,60){\makebox(0,0)[cc]{C}}

\end{picture}
\item[(b)]

\item[1.5.2]
\item[(a)-(i)]
See attached documents.\\

\item[1.5.3]
\item[(a)-(d)]
See attached documents.\\

\item[1.5.4]
\item[(a)-(d)]
See Attached documents.

\end{enumerate}
\end{document}
[/code]

The output produces a large space the picture, then more large space. Is there a way to correct this.

I used jpicedt for the picture.

The space before the picture is because [code]\unitlength \JPicScale mm[/code] should be [code]\setlength{\unitlength}{\JPicScale mm}[/code] The space after the box is because you have a picture of height 80mm which is too large for your drawing, so reduce the height and maybe the width.
Please don't se eqnarray, it is obsolete. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ 5[/url].

What would be the recommended length, and width?

I just tried height (the second coordinate) and replaced the all 80mm's with 10mm and all the 60mm's with 2mm and it looks OK but only you can judge what looks right. It's really a matter of trial and error to get what you want.


Suppose, $ k(p)\equal{}\frac{p^3\plus{}6p^2\plus{}5p}{p^2\plus{}p}$ represents the number of peaches Katey can consume in $ p$ hours. While Kyle here can consume $ k(k(p))$ peaches in $ p$ hours. If they both started consuming at the same time, then, how many peaches did Katey Consume if Kyle finished consuming $ 13$ peaches at the end of time??

Write $ k(p)\equal{}\frac{p(p^{2}\plus{}6p\plus{}5)}{p(p\plus{}1)}\equal{}\frac{(p\plus{}1)(p\plus{}5)}{p\plus{}1}\equal{}p\plus{}5$.  That should help simplify your calculations a lot.


analytic geometry
Putnam
number theory
prime numbers
number theory proposed

A point $ P$ of integer coordinates in the Cartesian plane is said [i]visible[/i] if the segment $ OP$ does not contain any other points with integer coordinates (except its ends). Prove that for any $ n\in\mathbb{N}^*$ there exists a visible point $ P_{n}$, at distance larger than $ n$ from any other visible point.

[b]Dan Schwarz[/b]

I have a solution, but it uses Dirichlet's Theorem.

Can anybody find one without it?

[quote="usa"][hide="Hint"]Chinese Remainder Theorem[/hide][/quote]

Yes, using it we can guarantee that the points within distance $ n$ are not visible. But what about the point itself?

It is clear that $ P(x, y)$, with $ x, y\in\mathbb{N}^*$, is visible if and only if $ \gcd(x, y) \equal{} 1$. The idea is to place $ P_{n}(a, b)$
at the center of a $ (2n\plus{}1)\times (2n\plus{}1)$ square having all lattice points in its interior or on its sides not visible.

To this purpose, consider a set of $ (2n\plus{}1)^{2}\minus{}1$ distinct prime numbers $ p_{i_{j}}$ with $ \minus{}n\leq i, j\leq n$ (but $ i, j$ not simultaneously null),
all larger than $ n$. The systems of congruences $ a\equiv \minus{}i (\mod p_{i_{j}})$ and $ b\equiv \minus{}j (\mod p_{i_{j}}),$ for all $ i, j,$
have positive integer solutions $ a, b > n$, according to the [color=red]Chinese Remainder Theorem[/color].

Therefore all points $ (a \plus{} i, b \plus{} j)$ (except maybe $ P_{n}(a, b)$) are [u]not[/u] visible, since $ p_{i_{j}} | \gcd(a \plus{} i, b \plus{} j).$
Hence the distance from $ P_{n}(a, b)$ to any other visible point is larger than $ n.$

The original part comes now. If $ \gcd(a, b) \equal{} 1,$ then $ P_{n}$ is visible, andwe are done. If not, denote by $ \pi \equal{}\prod p_{i_{j}},$
$ \delta \equal{}$ the product of all the primes dividing both $ a$ and $ b,$ and $ \rho\equal{}$ the product of all the other primes that divide $ a.$

Then $ \gcd(a, b,\pi) \equal{} 1$, since $ p_{i_{j}}$ cannot divide both $ i$ and $ j,$ being larger inabsolute value, and $ i, j$ not being simultaneously null.
Take $ b^{\prime} \equal{} b\plus{}(\delta\plus{}1)\rho\pi.$ Clearly we can substitute $ b^{\prime}$ for $ b$, since they differ by a multiple of $ \pi.$
Now, for a prime $ d | \delta,$ we have $ d | a$ and $ d | b^{\prime}$, since $ d | b$ but $ d | (\delta \plus{} 1)\rho\pi.$
For $ a$ prime $ r |\rho$, we have $ r | a$ and $ r | b^{\prime},$ since $ r | (\delta \plus{} 1)\rho\pi$ but $ r  | b.$

Since any prime dividing $ a$ is necessarily of one of the two types above, it follows that $ \gcd(a, b^{\prime}) \equal{} 1,$ so
this new point $ P_{n}(a,b^{\prime})$ is [u]visible[/u].

I foud that problem in: [b]Putnam and beyond[/b] (Titu Andreescu and Razvan Gelca) as an application of [u]Chinese Remainder Theorem[/u] (see [b]problem 793[/b]).

Proposed by [b]A.A. Mullin[/b] in [i]American Mathematical Monthly, 1977[/i].


arithmetic sequence
Extremal combinatorics
Ramsey Theory
Arithmetic Progression
IMO
IMO 1983
combinatorics

Is it possible to choose $1983$ distinct positive integers, all less than or equal to $10^5$, no three of which are consecutive terms of an arithmetic progression?

Hmm... strangely, I get a yes answer.

Consider all numbers that consist only of $0$ and $1$, $\text{base 3}$. Due to not being equal, every 2 numbers must differ in at least one digit, so in a possible arithmetic sequence the third number would have a $2$ in that place. And since $1983<(12.10^{10}) _\text{base 2}$ while $100000>(12.10^{10})_\text{base 3}$, the first $1983$ such numbers are a valid sequence?

Let $T_1$ be the set containing $\{ 1, 2, 4, 5 \} $ and if $T_n=\{ a_1, ..., a_k \}$ that $T_{n+1}=\{ a_1, ..., a_k, 2 \cdot a_k+a_1, 2 \cdot a_k +a_2, ..., 3a_k \}$. $T_{n+1}$ clearly doesn't contain three numbers forming arithmetic sequence. Let $f(n)$ be the greatest number in $T_n$. We see that ${|T_{n+1}|=2|T_n|}$ and $f(n+1)=3f(n)$.
$|T_1|=4$ and $f(1)=5$ so $|T_{10}|=2048>1983$ and $f(7)=3645 \Rightarrow f(8)<11000 \Rightarrow f(10)<99000<10^5$ so we're done.

I think this should work?

I claim that it is possible.

We start off with this solution using base $3$.

[b][u]Claim:[/u][/b] If we put any three consecutive terms of an arithmetic sequence with [b]positive difference[/b] in base $3$, there will be at least $1$ occurrence of each digit $0$, $1$, and $2$, in all of these numbers.

To clarify: Let the first term be $1$ (mod $3$) and the common difference be $2$ (mod $3$), then the final digit of the $3$ numbers will be $1$ -> $0$ ->$2$ (in base $3$).

The only case that contradicts this claim is if the difference is $0$, but these $3$ numbers will not be distinct.

[b]Thus, the minimal case, is if all the numbers in this sequence when written in base $3$, doesn't include a $2$.[/b]

Now we want to know the $1983$rd digit of this sequence. Because each number in this sequence only contains $0$s and $1$s (in base $3$), we will want to know $1983$ in base $2$.

$1983_{10}= 11110111111_{2}$

Lastly, the largest number in this sequence is the base $10$ representation of $11110111111_{3}$.

This is just $87844$.

So in the minimal case, $87844$ is the largest number in the sequence. Thus, the answer is $\boxed{Yes}$.

[hide=Remark] The sequence will go $1, 3, 4, 9, 10, 12, 13, 27, 28, 31$, ...[/hide]

[url=https://dgrozev.wordpress.com/2021/04/26/progression-free-sets/]Some more optimal constructions[/url].

[quote=solver2]2,1,4,3,6,5,8,7,......
the changes are -1,+3,-1,+3,...  which ensures that an ap is not made[/quote]

$\ 1,3,5$ is AP


analytic geometry
geometry
trigonometry
circumcircle

In acute triangle ABC, $BD$ and $CE$ are altitudes. Construct $BF \perp ED$ at $F$ and $CG \perp DE$ at $G$.
Prove: $EF=DG$

That's not hard.

[hide]$BFE$ and $BDC$ are similar, so $EF/BE = CD/BC$.
$CDG$ and $CBE$ are similar, so $DG/CD = BE/BC$.
Hence $EF = DG$[/hide]

I have a very ugly coordinate proof.   :P

Okay here's an actual geometry proof:

BCDE is cyclic because of the two 90 degree angles.

angle BDE=90-angle B=DCG

DG=CDsinGCD=CB cosBcosC.

Using a similar argument for EF, we find that DG=EF

My proof is exactly the same as Arne's.

By the way, there is an geometrical way to prove it by the cyclic points fourierseries mentioned.
Since $B, E, D, C$ are cyclic and $\angle{BEC}=90^{o}$, so the diameter of the circumcircle of $BEDC$ is $BC$.
So let $O$ be the midpoint of $BC$, so $O$ is also the center of the circle. Therefore if we construct $OM \perp DE$, we will have $EM=DM$.

Also because we have BF // OM // CG, and BO=OC, hence FM=MG. Therefore:
\[FE= FM - EM = MG - DM = DG\].


9.99g of $ \ce{CaCl2}$ are dissolved in 162g of $ \ce{H2O}$. Find the boiling point of the solution obtained [K(water) = 0.51 K kg/mol].

Since $ CaCl_2$ is an electrolyte, and it dissociates into it one $ Ca^{2\plus{}}$ ion and two $ Cl^\minus{}$ ions, the Van't Hoff factor $ i \equal{} 3$

Now, since you know molar mass of $ CaCl_2$, its mass and the mass of the solvent (water), find molality.

We know that elevation in boiling point is given by :

$ \Delta T_b \equal{} K.m$ where m=molality, K is given.

Once you find $ \Delta T_b$, the boiling point of the solution = B = $ 100 \plus{} \Delta T_b$.

And why are all my latex formulae [i]italicized[/i]???

Try coming here [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179608[/url] and reading item #2 in the first post.

hope that helps!


Game over. Innocents win.

Starting roles list: [list][*][b]3 mafia:[/b] standard mafia role[*][b]1 inspector:[/b] standard inspector role[*][b]1 angel:[/b] standard angel role, but saves against both mafia and werewolf kill.[*][b]1 healer:[/b] same as angel. But beware: if a person gets saved by both angel and healer, he dies from overdose.[*][b]0 werewolves:[/b] standard werewolf role[*][b]1 werewolf mother:[/b] each night, the werewolf mother decides if she want to go out or stay home. If she goes out, she can bite one person at night. That person loses his role and becomes a werewolf instead. The only exception is the mafia, who will suicide when bitten instead, to prevent all mafia being known. If she stays at home, she cannot be killed during the night by any means. When the werewolf mother dies, all werewolves die and the werewolf team is eliminated.[*][b]1 werewolf slayer:[/b] the werewolf slayer is immune to the werewolf mother bites and cannot be killed at night by werewolves. He can pick a target each night, if the target is a werewolf or a werewolf mother that is going out, then that target dies. If the werewolf slayer targets a werewolf mother at home, he will not notice any difference with a normal citizen.[*][b]3 network masons:[/b] network masons are like normal masons except that they don't know eachother at the start of the game. Each night, each network mason can target another player, and if another network mason is targetted, they are informed on eachothers identity and can talk the next nights.[*][b]5 citizens:[/b] standard citizen role.[/list]

I will tell every night who died by what. I will not tell how or what if the kill was prevented. The role of the peron who died will be revealed to the killers only. (i.e. mafia kills to mafia only, ww kills to ww's only, slayer kills to slayer only - except for the mother)

Edit: Yes, this means you don't get to know every role that died.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38004]standard rules can be found here.[/url]

[hide="questions dts"][quote="dts"]What is the policy on votes?  Are we allowed to switch our votes at any time, or not at all, or only once, or what?[/quote]switch without abuse. No hop-around without reason.

[quote="dts"]Should the Network Masons also suicide when caught by the Werewolf Mother, in line with the Mafia suiciding so the other Mafia don't have their identities blown?[/quote]no. if that means the werewolves know all three masons then too bad. (not that they'll first kill the masons anyway...)

[quote="dts"]Can the Werewolves make an attack on the same night they're bitten?[/quote]no. until end of night, they keep their own role. At end of night, they die and re-emerge as werewolf. Their normal ability for that night still goes on.

[quote="dts"]Are the Citizens made aware who is killed by what (i.e. you say, for instance, 142857 was killed by Mafia, Adunakhor was killed by Werewolves, bubka was killed by Werewolf Slayer)?[/quote]I'm still thinking on that. I'll go for the one with most tactical options probably.

[quote="dts"]In the same vein, will we be told if a Mafia suicides to avoid the clutches of the Werewolves?[/quote]The answer will probably be the same as above.

[quote="dts"]If the Werewolf Mother goes outside to bite someone and is killed by either the Werewolf Slayer or the Mafia, will the bite be successful (i.e., does the person the Werewolf Mother targets get killed)?[/quote]Yes.

[quote="dts"]Finally, if the Werewolf Mother stays inside, can it still be killed by a joint Angel/Healer overdose, or does that fall under "cannot be killed during the night by any means"?[/quote] Indeed, she cannot be killed in that way either. She is far away in her hole outside the city and cannot die from things happening in the city. :)

[quote="dts"]OK, that's enough questions, I guess.  This looks like it'll be an interesting game, can't wait for it to start![/quote]it's started, go ahead! ;)[/hide][hide="questions fredbel6"][quote="fredbel6"]Peter could you just add everything in the initial post about who can know what and is notified in what manner of what?[/quote]
Of course...

[quote="fredbel6"]Am I correct there is only 4 bad guys?[/quote]for starters, yes. There will become more werewolves of course.

[quote="fredbel6"]Is time of great essence here?  I mean when I played mafia with just a detective nobody really had to hurry, there was a plan for everything and one thing came after the other.
For instance, the werewolf mother, does she immediately  have to inform the moderator when night starts about her decision to go out or not?  Are all other reports ignored by the moderator until he gets such a pm?  And how many times can the slayer hit someone at night?  And what if
- night starts
- slayers pm's , says he wants to hit B
- B, who is the mother, pm's moderator to say she is going out
Is this a hit or not?[/quote]I don't understand your problem... at the end of the night, I check my PM box and do everything I have in it. You just send in something at any time during the night... time of sending does not matter at all.

[quote="fredbel6"]Does a protect by angel or healer imply also a protect from mother werewolf bites?[/quote]Nope. No antidote for mother bites exists.

[quote="fredbel6"]Are we gonna get the full description of events after the game?[/quote]Yes, this has happened in most games so I don't see why I won't do this.

[quote="fredbel6"]Where does the allegiance lie of a civilian, inspector, angel...
Does he care for himself or ALL the nonbaddies?
I mean, is sacrifice tactically interesting?
IF his own survival is top priority, it might actually be interesting for him to get bitten, if he thinks the werewolves are taking over, to become one[/quote]You win if your team wins. So a citizen wins if mafia and werewolves are eliminated. As for "interesting to get bitten", that would be stupid, since there's no way you can influe the fact of getting bitten. (a smart ww mother will not bite a person asking for it, since that's probably the slayer or a person that knows he/she'll die anyhow.

[quote="fredbel6"]And what if one does still act in the interest of his civilian group and sabotages the werewolf attempt?[/quote]after biting? that's stupid, since they win with the werewolf team after being bitten.[/hide][hide="questions klebian"][quote="klebian"]If the werewolf mother goes outside and is saved by both angel and healer, then will she die?[/quote] yes.

[quote="klebian"]So if werewolf slayer targets a werewolf, we're announced that a werewolf dies or not?[/quote] I think so. I will elaborate on the announcinf part in a minute.

[quote="klebian"]If werewolf slayer targets werewolf mother and there are already 3 other werewolves, are we told that these 4 people died and that they were werewolves?[/quote] yes. Therefor it is usually wise for the WW mother not to go out at night if there are already some werewolves.

[quote="klebian"]Are we told when someone (without naming who) has been bit by werewolf mother?[/quote]of course not, then you'd know a werewolf. :P

[quote="klebian"]Can angel or healer save from being bit by werewolf mother?[/quote] nope.

[quote="klebian"]Can angel or healer save from being killed by werewolf slayer?[/quote]No, he cannot save werewolves from being slain.

[quote="klebian"]If a network mason targets a player and finds out that he is also a mason, the next night, they'll be able to talk and try 2 different people to see if either is mason right?[/quote] Yes.

[quote="klebian"]Can werewolf slayer die by mafia?[/quote] Of course.

[quote="klebian"]Is werewolf slayer innocent?[/quote]Of course. (that's quite obvious, right?)

[quote="klebian"]Can werewolf and werewolf mother communicate at night?[/quote] Good question. Yes, they can.

[quote="klebian"]Important question: if werewolf mother bites inspector/healer/angel, then we won't know that this power role was bitten, right? [/quote]Indeed, you will not know.

[quote="klebian"]So essentially all our power roles could be werewolves? Then we wouldn't know how close we were to winning/losing?[/quote]In theory, that's possible, but it's highly unlikely.[/hide][hide="questions bubka"][quote="bubka"]
1) [b]Least serious[/b]: Do werewolves continue their original roles? [/quote]no. as it says...

[quote="bubka"]2) [b]Somewhat more serious[/b]: A mafia vs. werewolves game ends when one party is eliminated. Does that mean slaying the werewolf mother ends the game?[/quote]no, 2 parties have to be eliminated. As always.

[quote="bubka"]3) [b]MOST SERIOUS[/b]: [i]Is the werewolf mother female[/i]?[/quote]How many male mother do you know? ;)
The ww mother is female, but that implies nothing on the person playing it.[/hide]

so we just start making random accusations now?
On IMO, you could just tell from their faces almost immediately.

Wouldn't it be nice if we could wipe out the werewolf mother immediately?

Edit : other ideas I had were calling for a temporary truce between civilians and mafia to wipe out the werewolf mother?

And how about somehow making a chart to collect information : if a person dies from suicide we immediately know he was mafia, that there are only two others left, and that he wasn't the werewolf mother.
If a person dies from overdose, we know for sure that he was not angel or healer.

[quote="fredbel6"]If a person dies from overdose, we know for sure that he was not angel or healer.[/quote]Just a remark: this is not true: angel/healer can save himself and accidently being saved by the other party too.

None whatsoever. But your name is similar to somebody else's on this forum. I won't mention whose though.

But I don't know how to make random accusations, and moreover have no desire to be called trigger-happy (that expression is a cliche if you ask me). So I must wait and see ...

Are you refering to perfectnumber628? I swear it was completely coincidental, and I'd never seen that name before.

Just to get the ball rolling: [hide="I"]am not sure what were doing, but I think that we should just chat, and accuse the most suspicious, so I accuse no one yet.[/hide]

[quote="nutz_for2.718281828"]I'm confirming, but the citezen is still a boring role. Why didn't you make an interesting citezen role?(normal citezens)

oh, the game has begun[/quote]
Are you saying you are citizen?  :huh:

No, just that the citezens still have boring roles.

i figure for all the questions, petervdd will settle intricacies at his discretion?

so to begin with, does anyone have any better info than a random guess?

[quote="dts"]Well, it's pretty much either a random guess or an attempt at psychoanalysing past posts at this point.

I haven't really wanted to post much here, since I'd feel obliged to get the game moving, and that would involve making an accusation, and it seems almost like the first to accuse in any game is the first to be lynched.  However, it looks rather like everyone else has the same idea, so...

I [color=red]ACCUSE[/color] 142857.  His only question concerned both the Werewolf Mother and the Mafia, so... maybe he's one of them?

Yes, quite random, I know.  But someone's got to do it eventually.[/quote]
Umm...almost half these questions are concerning the werewolves....

That is the hard  :maybe:  part even the poorest actor can hide his mafianess or whatever when he is behind the screen.

Okay, everyone post a copy of your pm you got from PeterVDD on a to be determined exact same time! !!!! :D :roll:

... I realize myself that makes no sense but what other way is there??

[quote="fredbel6"][b]That is the hard  :maybe:  part even the poorest actor can hide his mafianess or whatever when he is behind the screen.[/b]

Okay, everyone post a copy of your pm you got from PeterVDD on a to be determined exact same time! !!!! :D :roll:

... I realize myself that makes no sense but what other way is there??[/quote]
That makes it all the more challenging and fun. :D

uhhhh.... greetings from heaven i guess

u guys killed me, go to <censored>

this game finally finished!!!!!  congratulations everyone!!!!! especially fredbel for his plan!!!!!  yay!!!!! :lol:

I finally notice that I die.

I'm the first mafia to post here after the end of the game. Let me tell you our plan. Our first night kill is bubka, the most skillful player in the game. We have thought of killing Klebain at that time, and the reason is:[quote="142857"]O.K. If everyone think that we should roleclaim, I'll roleclaim.
I'm a
[hide]ww slayer. But I'm just a citizen right now as the only ww mother is dead. I get the idea that Klebian is the ww mother from the accusation on me by dts. He accused me as I have ask a question about mafia and ww mother. Then, I think that I may find out the ww mother by reading all questions by all the people and I found that Klebain has ask 12 questions with 11 questions about ww and 1 silly question about the mason:
[quote="Klebian"]If a network mason targets a player and finds out that he is also a mason, the next night, they'll be able to talk and try 2 different people to see if either is mason right?[/quote]
So, I think that Klebian may be a ww mother and slay him at the following night. Luckily, he is the one and we have much less worry about the guilties.[/hide][/quote]
And then we continues to kill the most skillful people left. At last, we kill the masons, angel or healer. Also, we plan to make a lots of post suggesting who the angel or healer should protect in order to confuse them. It is what actually happen in day 4 (page27-28).

I think we should have won the game, before ragnarok roleclaim to be inspector and said a citizen is a mafia. Luckily, no one believe Nutz or Tim that my reason on ragnarok's help and lynch me. Maybe it is also our plan: be very very quiet when someone is suspicious about you. It may not work when there is some very good mafia-player in the game, but it will certainly work when there is only mafia new player (online).

I actually want to be inactive in the beginning of the game, but dts accuse me in the very beginning and force me to become active. Thanks dts to help us kill the ww mother in the beginning of the game.

One last thing to ask Peter: why would a people get overduse if he is target by the angel, healer and mafia. I think that the mafia's kill should have balanced one angel's heal and the effect on that person is a healer's heal only.

Good job innocent :(

[quote="Ignite168"][quote="dts"]I didn't know, really... I just had a feeling.[/quote]

May it's because Klebian has been guilty in games 15, 18, 19, and now 20?

He wasn't playing in 16 and was modding 17.

That's 4 games in a row.

If he's guilty in 21 and 22, that'd make 6 :)[/quote]

i was citizen in 16.

[quote="142857"]I'm the first mafia to post here after the end of the game. Let me tell you our plan. Our first night kill is bubka, the most skillful player in the game.[/quote]

OMG thanks a lot for that compliment.  :blush: But you chose to spare the second most skilful player in the game - fredbel6, who, although a bit slow, did finally prove to be your nemesis!  :wink:

[quote="142857"]Our first night kill is bubka, the most skillful player in the game.[/quote]Just to be curious, what makes you think this? I never noticed much input from bubka actually (in any game), maybe I've just not been paying attention (I'd put dts as best player by far).

[quote="142857"]One last thing to ask Peter: why would a people get overduse if he is target by the angel, healer and mafia. I think that the mafia's kill should have balanced one angel's heal and the effect on that person is a healer's heal only.[/quote]Mainly a tactical thing: if for example someone claims inspector, it's quite difficult to save him without revealing either angel or doctor. Usually it's 2 doctors who each give a dose of morphium, but as angel is the way the role is called here, I used that name too. It's quite common practice to do this with 2 doctors. And it allows some nice tricks, as the one you've been killed with.

[quote="dts"]Why, thank you... :D  although I fear my lifespan in Game 23 may have just taken a dive, unless of course I'm Mafia. :maybe:

I must be a fast learner, though, as I've never played Mafia except the half-dozen or so games I've been in on AoPS.[/quote]

Not even making it past the first day is quite a dive.

I'm still confused  :maybe:  about 142857, who was amazingly active and convincing yet mafia, even though it is his first game. :!:

[quote="142857"][quote="tim1234133"][quote="nutz_for2.718281828"]Dd you notice that ragnarok said that 142857 AND I(now tim) were innocent when he claimed inspector? That was to make suspicion in two people: me(now tim) and 142857.[/quote]

Why would ragnorak want to incite suspicion in 142857? Weren't they mafia together?[/quote]
Yes, and this is the main reason why we(mafia) lose.[/quote]

Nope. Ragnarok made a good move by making sure he threw in a surefire innocent in with you. You would've lost either way, since you had like no way of killing angel/healer.

The thing with ragnarok isn't the suspicion on 142857, the thing is losing their precious inspector role and a mafia's life at once, just for a citizen's life (which even reduced their hiding possibility).


function
induction
algebra proposed
algebra

Find all functions $f: \mathbb N \to \mathbb N$ such that for all positive integers $n$,
\[ f(f(n) - n) = 2n.\]

[quote="MJ GEO"]find all function $ f$ from $ N$ to $ N$ and $ f(f(n) \minus{} n) \equal{} 2n$[/quote]

Since $ f(n): \mathbb N\to\mathbb N$, and in order $ f(f(n)\minus{}n)$ be defined, we need $ f(n)>n$

then $ f(f(n)\minus{}n)>f(n)\minus{}n$ and so $ 2n>f(n)\minus{}n$ and $ f(n)<3n$ and so $ 3n>f(n)>n$

Consider now that we got $ an>f(n)>bn$. This implies $ a(f(n)\minus{}n)>2n>b(f(n)\minus{}n)$ and so $ \frac{b\plus{}2}bn>f(n)>\frac{a\plus{}2}an$

So we can define the two sequences :
$ a_1\equal{}3$ and $ b_1\equal{}1$ and $ a_{n\plus{}1}\equal{}\frac{b_n\plus{}2}{b_n}$ and $ b_{n\plus{}1}\equal{}\frac{a_n\plus{}2}{a_n}$

And we get $ a_{2p\plus{}1}n>f(n)>b_{2p\plus{}1}n$ and $ b_{2p}n>f(n)>a_{2p}n$

And it is easy to show that these two sequences have the same limit $ 2$ and so $ \boxed{f(n)\equal{}2n}$ which, indeed, is a solution.

[quote="MJ GEO"]find all function $ f$ from $ N$ to $ N$ and $ f(f(n) \minus{} n) \equal{} 2n$[/quote]
$ f: \mathbb{N} \to \mathbb{N}$
Let $ g(n) \equal{} f(n) \minus{} n$, then $ g: \mathbb{N} \to \mathbb{N}$. And $ f(g(n)) \equal{} 2n \iff g(n) \plus{} g(g(n)) \equal{} 2n$.
If $ g(a) \equal{} g(b)$ then $ g(g(a)) \plus{} g(a) \equal{} g(g(b)) \plus{} g(b) \iff a\equal{}b$, so $ g$ is injective. I will prove by induction that $ g(n) \equal{} n$.
$ g(1) \plus{} g(g(1)) \equal{} 2$ gives $ g(1)\equal{}1$ since $ g(1),g(g(1)) \ge 1$.
Assume that $ g(1)\equal{}1,g(2)\equal{}2, \cdots, g(n) \equal{} n$. Then $ g(n\plus{}1) \ge n\plus{}1$ and $ g(g(n\plus{}1)) \ge n\plus{}1$, since $ g$ is injective. And $ g(n\plus{}1)\plus{}g(g(n\plus{}1)) \equal{} 2(n\plus{}1)$ gives $ g(n\plus{}1)\equal{}n\plus{}1$. So $ g(n) \equal{} n$ and hence $ f(n)\equal{}2n$, which is easily seen to be a solution.

[quote="Mathias_DK"][quote="MJ GEO"]find all function $ f$ from $ N$ to $ N$ and $ f(f(n) \minus{} n) \equal{} 2n$[/quote]
$ f: \mathbb{N} \to \mathbb{N}$
Let $ g(n) \equal{} f(n) \minus{} n$, then $ g: \mathbb{N} \to \mathbb{N}$. And $ f(g(n)) \equal{} 2n \iff g(n) \plus{} g(g(n)) \equal{} 2n$.
If $ g(a) \equal{} g(b)$ then $ g(g(a)) \plus{} g(a) \equal{} g(g(b)) \plus{} g(b) \iff a \equal{} b$, so $ g$ is injective. I will prove by induction that $ g(n) \equal{} n$.
$ g(1) \plus{} g(g(1)) \equal{} 2$ gives $ g(1) \equal{} 1$ since $ g(1),g(g(1)) \ge 1$.
Assume that $ g(1) \equal{} 1,g(2) \equal{} 2, \cdots, g(n) \equal{} n$. Then $ g(n \plus{} 1) \ge n \plus{} 1$ and $ g(g(n \plus{} 1)) \ge n \plus{} 1$, since $ g$ is injective. And $ g(n \plus{} 1) \plus{} g(g(n \plus{} 1)) \equal{} 2(n \plus{} 1)$ gives $ g(n \plus{} 1) \equal{} n \plus{} 1$. So $ g(n) \equal{} n$ and hence $ f(n) \equal{} 2n$, which is easily seen to be a solution.[/quote]

Nice, and prettier than mine  :blush:


Let x,y,z be positive real numbers.If
$ M\equal{}\frac{x}{y\plus{}2004z}\plus{}\frac{y}{z\plus{}2004x}\plus{}\frac{z}{x\plus{}2004y}$
find the min value of M.

[hide]We have
$ \sum_{cyc}^{x,y,z}\frac{x}{y\plus{}2004z}\geq\frac{(x\plus{}y\plus{}z)^{2}}{2005(xy\plus{}yz\plus{}zx)}\geq\frac{3}{2005}$
When $ x\equal{}y\equal{}z\equal{}1$, $ M\equal{}\frac{3}{2005}$ [/hide]


Let $A,B,C,D$ be 4 points (no 3 collinear) and $M,N$ be the midpoints of $AD,BC$. Also let $S \in AB$ si $T \in DC$, $k \in \mathbb{R}$ s.t. $\frac{\overline{SA}}{\overline{SB}}=\frac{\overline{TD}}{\overline{TC}}=k$, $\{ L \} = ST \cap MN$. Prove that:

[b](a) $k = \frac{\overline{LM}}{\overline{LN}}$[/b]

[b](b) $L$ is the midpoint of $ST$[/b]

I already posted a vector solution here ( see http://www.mathlinks.ro/Forum/viewtopic.php?p=319742#319742 - post #5 ), and I'm awaiting for a sintetic solution. (I guess it should be very easy, but I can't figure it out by myself)

( The notation $\overline{**}$ reffers to orientated segments and it means that we should take the products or ratios with a sign )

Note that [MNS] is a weighted average of [MNA] and [MNB], while [MNT] the same weighted average of [MNC] and [MND].  [MNC] = [MNA], [MND] = [MNB].  Hence SL = TL.  Now apply the same concept to compare [STN] to [STM].


trigonometry
limit
calculus
calculus computations

Can anyone prove to me that lim x--> infinity, sinx / x = infinity? wait, is that equal to infinity or other answers?  I tried to, I differentiate both, so i got cos x / 1, but what is cos infinity? I don't know, on the graph, y seems to go between -1 and 1, so I really have no idea. Will it be the same to find out sin 2/x / 2/x ? Hope u can help.

since $ \minus{}1\leq \sin x\leq 1$, we  have $ \minus{}\frac{1}{x}\leq \frac{\sin x}{x}\leq \frac{1}{x}$ for $ x>0$. Thus by squeezing theorem, we obtain $ \lim_{x\rightarrow \infty} \frac{\sin x}{x}\equal{}0$.

[quote="MIA6"]I tried to, I differentiate both[/quote]

l'Hopital's rule is only valid if the limit in question is an indeterminate form, which this isn't.


Estaria bien si alguien tuviera el libro de [b]Coxeter "Geometry Revisited"[/b] para que todos podamos tenerlo please  :(

Yo lo tengo en inglés... si alguien lo quiere que me pase su dirección de e-mail o que simplemente alguien me enseñe a poder adjuntarlo acá...

M4RIO, el problema es que la extensión no está autorizada. Mandame un PM con tu mail y yo te lo mando.

[quote="M4RI0"]podrias adjuntar el archivo jose??[/quote]

Yo colgue el libro ace un tiempo pero me dijeron que era ilgela, pideselo a jose en privado

suerte

Ese libro lo pueden encontrar en formato DJVU (bajarse el programita, no es dificil encontrarlo) en la siguiente direccion

http://lib.org.by/_djvu/M_Mathematics/MSch_School-level/

Saludos


function
number theory
relatively prime
number theory proposed

Let $\varphi_k(n)$ denote the sum of the $k$th powers of the numbers $\leq n$ and relatively prime to $n$. Prove that
$\sum_{d|n}\frac{\varphi_k(d)}{d^k}=\frac{1^k+\hdots+n^k}{n^k}$ where the sum is extended over the positive divisors of $n$.

Let $f(\cdot)$ be a function defined for any rational $x\in [0, 1]$ and put $F(n) = \sum_{t=1}^n f\!\left(\frac{t}{n}\right)$ and $F^*\!\left(n\right) = \sum_{t=1}^n^{\prime} f\!\left(\frac{t}{n}\right)$, where the prime refers the RHS summation is taken over any integer $k = 1, 2, \ldots, n$ such that $\gcd(n,k) = 1$, for any $n\in\mathbb{Z}^+$. Then it's well-known $F^* = \mu \star F$, viz $F = \mu^{-1} \star F^*$, where $\star$ denotes Dirichlet's convolution. So fix $k \in \mathbb{R}^+$ and let $f(x) = \frac{x^k}{n^k}$, for $0 \leq x \leq 1$. You get $F(n) = \frac{1^k + 2^k + \ldots + n^k}{n^k} = ( \mu^{-1} \star F)(n)$, for any $n \in \mathbb{Z}^+$. Anyway $\mu^{-1} = u$, the identity function. Hence $\frac{1^k + 2^k + \ldots + n^k}{n^k} = (u \star F^*)(n) = \sum_{d \mid n}^n u(d) \cdot F^*\!\left(\frac{n}{d}\right)$. Now it's only counts...


When I complete the square of 3x^2 - 4x + 2 = 0, I get a totally different answer than when solving it using the quadratic formula.

Can anyone here solve this using completing the square and show me step by step?

The answer is:

x = (2 + sqrt{2}i)/3

and

x = (2 - sqrt{2}i)/3

Also, does it matter if I place the i before the square root symbol or after?  In other words, should it be (sqrt{2}i) or i(sqrt{2}) for the question above?

The $i$ acts like a coefficiant so it would go before the radical.

As for me, I like the form of $a+bi\ (a,\ b\in \mathbb{R})$.
Which do you like better, $\sqrt{2}$ times $a$, $\sqrt{2}a$ or $a\sqrt{2}$?

This is interesting, as the quadratic equation [i]is[/i] completing the square.

[quote="sharkman"]Also, does it matter if I place the i before the square root symbol or after?  In other words, should it be (sqrt{2}i) or i(sqrt{2}) for the question above?[/quote]

...remind me again...isnt there a special thread for sharkman/interval to post his questions, or did that die...?  :|

but anyways.

mathematically, $i\sqrt{2}=\sqrt{2}i$, as long as you make sure you dont accidentally put the $i$ under the radical.  im not sure if there is a "technical way" to write it, but it seems to be most commonly written, and easily understandable, as $i\sqrt{n}$.

but seeing as you still havent learned LaTeX, im sure nothing you write will be easily understandable.  :rotfl:

okay, sorry that was kind of mean, but PLEASE, learn LaTeX, or at least post in your thread.  :wink:

EDIT: HAZAA. i found it.  enjoy sharkman.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=153452[/url]


limit
integration
trigonometry
calculus
function
real analysis
real analysis unsolved

Compute $ \lim_{x \minus{} > \infty}\frac {\int_0^x\frac {dt}{4 \plus{} cost}}{x}$

Note that L'Hôpital's rule will not work here.

Let $ A\equal{}\frac1{2\pi}\int_0^{2\pi}\frac{dt}{4\plus{}\cos t}$

(Since this is the integral of a continuous function over a closed interval, there's no question that $ A$ exists.)

We claim that the desired limit is $ A.$ For proof:

Suppose $ 2\pi N\le x<2\pi(N\plus{}1).$ Then

$ \frac1x\int_0^x\frac{dt}{4\plus{}\cos t}\equal{}\frac{2\pi NA\plus{}\int_{2\pi N}^x\frac{dt}{4\plus{}\cos t}}{2\pi N\plus{} (x\minus{}2\pi N)}\equal{}\frac{A\plus{}\frac1{2\pi N}\int_{2\pi N}^x\frac{dt}{4\plus{}\cos t}}{1\plus{}\frac{x\minus{}2\pi N}{2\pi N}}$

Then $ \left|\frac1{2\pi N}\int_{2\pi N}^x\frac{dt}{4\plus{}\cos t}\right|\le\frac1{2\pi N}\cdot\frac13(x\minus{}2\pi N)\le \frac1{3N}$

(where we used that $ 0<\frac1{4\plus{}\cos t}\le\frac13$ )

and $ 0<\frac{x\minus{}2\pi N}{2\pi N}<\frac1N$

So as $ x\to\infty,$ we also have $ N\to\infty.$ The numerator of the fraction tends to $ A$ and the denominator tends to $ 1,$ so the overall limit is $ A.$

As for the specific computation of $ A:$ that's a calculus problem (quite doable) that I leave to someone else.

Use the substitution $ t\equal{}\tan\frac{x}{2}\Rightarrow\cos{x}\equal{}\frac{1\minus{}t^{2}}{1\plus{}t^{2}}$.

[hide="Short answer"]$ A \equal{} \frac1{\sqrt {15}}$.

Checked numerically. Note that it has to be between $ \frac15$ and $ \frac13.$ It turns out to be the geometric mean of those numbers.[/hide]


Let $ a,b,c$ be positive real numbers such that $ ab\plus{}bc\plus{}ca\ge a\plus{}b\plus{}c.$ Prove that
\[ \frac1a\plus{}\frac1b\plus{}\frac1c \leqslant\ 3\]

That's clearly wrong. Just take $ a\to 0$, $ b \equal{} 2$, $ c \equal{} 3$. Then the given condition $ ab \plus{} bc \plus{} ca \ge a \plus{} b \plus{} c$ is equivalent to $ 6\ge 5$ and your inequality becomes $ \infty \le 3$. :P

it's true that:
if $ a \plus{} b \plus{} c \ge ab \plus{} bc \plus{} ca$
then
$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\ge3$

because
$ ( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})(ab \plus{} bc \plus{} ca) \ge 3(a \plus{} b \plus{} c)$

No, that's also not true. Take again the values $ a \to 0$, $ b\equal{}c\equal{}1$. Then the condition becomes $ 2 \ge 1$, which is true and the inequality becomes $ \infty \le 3$. :wink:

[quote="itiselizabeth"]Let $ a,b,c$ be positive real numbers such that $ ab \plus{} bc \plus{} ca\ge a \plus{} b \plus{} c.$ Prove that
\[ \frac1a \plus{} \frac1b \plus{} \frac1c \leqslant\ 3\]
[/quote]
I think that this problem need condition other  :wink:
Please itiselizabeth check problem

I edited post #3
maybe that's this problem

I was actually trying to prove that $ \frac1{1\plus{}a\plus{}b}\plus{}\frac1{1\plus{}b\plus{}c}\plus{}\frac1{1\plus{}c\plus{}a}\le1.$ I thought I could apply AM–HM to each term of the LHS and then sum up – but it’s obvious now that I can’t do that. :oops:

I think Dumel's problem is true
$ \sum a\geq\sum ab\Leftrightarrow\sum a\sum ab\geq(\sum ab)^2\geq 3abc\sum a$


1) You forget to mention the arrival obey poisson dist.
i)I think the prob should be P(X=5)^5
'coz "during any period of hour between 7 p.m. and 11 p.m., exactly 5 people will arrive" means each hour has 5 arrival.
ii)use the property of memoryless, easy
iii)easy, directly use poisson dist.

2)
the quality = mean + Zb*var
where F(Zb)=0.98
F(.) is the C.D.F of the dist. you can easily compute it from pdf.

3)
I think you are right.

[quote="dadida"]1)Between 7p.m. and 11p.m., arrivals of patients at the casualty department of a hospital occur at random at an average rate of 6 per hour.
i)Find the probability that, during any period of hour between 7 p.m. and 11 p.m., exactly 5 people will arrive.
ii)A patient arrives at exactly 10.15 p.m. Find the probability that at least one more patient arrives before 10.35 p.m.
iii)Use a suitable approximation to estimate the probability that fewer than 20 patients arrive at the casualty department between 7 p.m. and 11 p.m. on any particular night.

I can't do part ii and iii, Can anyone help me on part ii and iii? And, can someone check whether my answer for part i is correct or not, thanks.

i)P(X=5) = (e-6 . 6^5) / 5! = 0.1606
Since it says any period of hour between of 7 p.m. and 11 p.m., I think we should use binomial distribution whivh gives
P(X=5) = 4C1 * (0.1606) * (0.8394)^3 = 0.3799

---------

2)A factory is supplied with grain at the beginning of each week. The weekly demand, X thousand tonnes, for grain from this factory is a continuous random variable having the probability density function given by
f(x) = 2(1-x) , 0<=x<=1
= 0 , otherwise
Find
iii)The quantity of grain in tonnes that the factory should have in stock at the beginning of a week, in order to be 98% certain that demand in that week will be met.

I can't do part iii, I've no idea of doing it. (I've done part i and part ii, which they ask to find mean and variance)

----------

3))Eggs sold in a supermarket are packed in boxes of 12. For each egg, the probability that it is cracked is 0.05, indepently of all other eggs. A random sample of n boxes is selected and the variance of the sample mean number of cracked eggs in a box is 0.019. Find the value of n.

var=np(1-p)
0.019=n0.05(1-0.05)
0.019=n0.0475
n=0.019/0.0475
n=0.4

but I think it's wrong? Can anyone help? Thanks.[/quote]


function
floor function
number theory unsolved
number theory

Let $ \alpha$ be a positive irrational number. Prove that the two sequences,

$ \lfloor 1\plus{}\alpha \rfloor, \lfloor 2\plus{}2\alpha \rfloor, ... , \lfloor n\plus{}n\alpha \rfloor$

and

$ \lfloor 1\plus{}\alpha^{\minus{}1} \rfloor, \lfloor 2\plus{}2\alpha^{\minus{}1} \rfloor, ... , \lfloor n\plus{}n\alpha^{\minus{}1} \rfloor$

contain every positive integer exactly once. Prove that this is false if $ \alpha$ is rational.

Note that $ \frac{1}{1 \plus{} \alpha} \plus{} \frac{1}{1 \plus{} \frac{1}{\alpha} } \equal{} 1$ and then the result follows from the lemma cited [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=226535]here[/url].


function
inequalities
algebra solved
algebra

I suppose that you mean p=<a,b,c,d,e<=q
Well, the expression suma* sum1/a is a convex function with respect to each of these variables and therefore it reaches the maximum at the endpoints of the interval. n of them equal p and the other 5-n equal q.
it is simple to verify the inequality at this point,since the only variable is 5=>n>=0.

I have solution in my book. This solution spans 1.5 pages and it is ugly.

Myth said : "I have a solution in my book" What book is that?

"Foreign mathematical olympiads", 1987. It is very "advanced" book.

i think it can be easily proved by noticing that the function
f(a,b,c,d,e)=(a+b+c+d+e)(1/a+1/b+1/c+1/d+1/e) is convex in each variable a,b,c,d,e ... so the maximum is achieved for a 5-uplet of the form (g1,g2,g3,g4,g5) with gi=p or gi=q and it's quite straightforward to finish ..
am I wrong ??


Given real numbers $ a,b,c,x,y$ and $ z$ with
$ (a\plus{}b\plus{}c)(x\plus{}y\plus{}z)\equal{}3$ and $ (a^2\plus{}b^2\plus{}c^2)(x^2\plus{}y^2\plus{}z^2)\equal{}4$  Prove that
$ ax\plus{}by\plus{}cz \ge 0$

[quote="Mathpro"]Given real numbers $ a,b,c,x,y$ and $ z$ with
$ (a \plus{} b \plus{} c)(x \plus{} y \plus{} z) \equal{} 3$ and $ (a^2 \plus{} b^2 \plus{} c^2)(x^2 \plus{} y^2 \plus{} z^2) \equal{} 4$  Prove that
$ ax \plus{} by \plus{} cz \ge 0$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70625


algebra
polynomial
calculus
calculus computations

n S k=1 (k+2)(k-2)

not sure how to write the sigma so it's represented as S
k is the summation index

so my question is, can I solve this simply by pluging in the numbers?
ie. = -3 + 8 + 5 + 12.....

or do I solve using algabraic rules?
like: n Sk k=1   =   n(n+1) / 2

if that is the case, then how sould I go about solving my question?[/hide][/code][/quote]

Try the following:

To begin or end an equation, use a dollar sign ($\$$).

To make a subscript, use _{stuff}, so x_{yz} is $x_{yz}$.

To make a superscript, use ^{stuff}, so x^{yz} is $x^{yz}$.

To make a sum, use \sum, so \sum_{n = 1}^{10} x_{y}^{z} gives $\sum_{n = 1}^{10}x_{y}^{z}$.

For fractions, use \frac{numerator}{denominator}, so \frac{13}{2} gives $\frac{13}{2}$.

Is that enough for what you're trying to write?

Also, general rule: "plugging in numbers" is equivalent to "checking that something isn't always wrong," except in cases where particular nice theorems apply. This is not equivalent to proving anything (again, with certain exceptions: it's possible to prove things about polynomials of known degree by plugging in sufficiently many values).  So, you probably should be doing algebraic manipulations.

$\sum_{k=1}^{n}(k-2)(k+2)$ is equal to $\sum_{k=1}^{n}(k^{2}-4)$, which is the same as $\left(\sum_{k=1}^{n}k^{2}\right)-4n$. The answer can be found by watching flash animations on the left margin of this page.  :)

you guys are wicked!!
that was SUCH a fast reply!!
thanks!

watching the animation now...


In a convex quadrilateral $ ABCD$, $ \angle BAC \equal{} \angle CAD$, $ \angle ABC \equal{} \angle ACD$, the extensions of $ AD$ and $ BC$ meet at $ E$, and the extensions of $ AB$ and $ DC$ meet at $ F$. Determine the value of
\[ \frac {AB \cdot DE}{BC \cdot CE}.
\]

What is the purpose of $ F$? [hide]Let $ \angle BAC \equal{} x$, $ \angle ABC \equal{} y$, and $ \angle ACB \equal{} z$.  Then $ x \plus{} y \plus{} z \equal{} 180$.

By AA similarity, $ \triangle ABC \sim \triangle ACD$.  So $ \frac {AB}{BC} \equal{} \frac {AC}{DC}$.

Next, we have that $ \angle DCE \equal{} x$, because $ BCE \equal{} y \plus{} z \plus{} \angle DCE$ is a straight line.  Hence, $ \triangle ACE \sim \triangle CDE$.  Thus, $ \frac {AC}{CE} \equal{} \frac {DC}{DE}$.

Finally, $ \frac {AB \cdot DE}{BC \cdot CE} \equal{} \frac {AC \cdot DE}{DC \cdot CE} \equal{} \frac {DC \cdot DE}{DE \cdot DC} \equal{} 1$.
[/hide]

$ F$ might be a red herring, I thought so during the contest.  :)


Given $ a, b, c > 0$. Prove that $ \frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}}\plus{}\frac{2(a\plus{}b\plus{}c)^4}{27(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)} \ge 5$  :)


solve in R

$ \left \lfloor x^2\right \rfloor\equal{}2x^2\minus{}7x\plus{}6$

[quote="tony123"]solve in R

$ \left \lfloor x^2\right \rfloor \equal{} 2x^2 \minus{} 7x \plus{} 6$[/quote]
$ \left \lfloor x^2\right \rfloor\le x^2$ then $ (x\minus{}1)(x\minus{}6)\le 0$
Now we only need check $ x\equal{}1,2,3,4,5,6$ to find the result

[quote="Allnames"][quote="tony123"]solve in R

$ \left \lfloor x^2\right \rfloor \equal{} 2x^2 \minus{} 7x \plus{} 6$[/quote]
$ \left \lfloor x^2\right \rfloor\le x^2$ then $ (x \minus{} 1)(x \minus{} 6)\le 0$
Now we only need check $ x \equal{} 1,2,3,4,5,6$ to find the result[/quote]
don't spam,check your solution carefully, $ 1 \le x \le 6$ not mean $ x\equal{}1,2,3,4,5,6$


[quote="nonie"][quote="ch1n353ch3s54a1l"][quote="nonie"]A wheel of Marci's bike makes 1056 revolutions in 1 mile. WHat is the number of inches in the diameter of the wheel? (To the nearest whole number.)[/quote]
[hide]
The wheel makes 1056 revolutions in 5280 feet. 5280/1056=5, so the wheel goes 5 feet in 1 revolution. This also means the circumference is 5 feet, and the diameter is $\frac{5}{\pi}$ feet, which, to the nearest integer, is 2. [/hide][/quote]

The answer needs to be expressed in inches. That should change your answer significantly.[/quote]

><
[hide]$\frac{5\times12}{\pi}\approx19$[/hide]


Y'all have probably played this before, but I can't seem to find a thread here.

Sooo...

[quote="Some guy"]The purpose of this game is to find a (real) word that contains a given sequence of three letters.

An (fallacious) example is:

yst
OMGTEHDA[u][b]YST[/b][/u]ARITBURNSSSSS

A few more examples are:

mpl
example

nce
balance

Rules:
1) Must be a real word (preferably from the English language, or a common loan word.) Must not be an abbreviation. Must not be a proper noun.
2) When a word is figured out, the person who figured it out must pose the next set.
3) You must have figured out for the next set of letters before you post. No random zzz posts.
4) Abuse of the rules is encouraged.
5) If you don't think that there is a word that fits rule (1) with a given three-letter grouping, you may call BS on it.
6) Do NOT use electronic or print aids. That just ruins the fun. (Unless everyone is calling BS on a three-letter grouping.)[/quote]

I'll start with a relatively easy one:
[b]
xho[/b]

egocentric

uac

(anyone remembers the old game called words with letters in it?)

zigzag

what's a guacamole? I had something else in mind for uac. I forgot waht. Somebody else can try.

vab

guacamole is a dip made with smushed avacados.

lovable

hak

Hey - shouldn't I get to do the new sequence (seeing that I BS'd it)?

If so, [b]bno[/b]

Nobody's going to attempt [b]aou[/b]?

If not, I'll reveal the word I had in mind...

...and the word is

[hide][b]giaour[/b], which went through Persian, Turkish, the Italian dialect Venetian, and finally French, and means "one outside the Islamic faith"

This word was used in the 2006 national spelling bee under "World Religions", in the "Advanced" section.[/hide]

Extremely obscure words are fun, aren't they? :D

Wrong. Ri[u]jst[/u]tafel is the name of an Indonesian rice dish. Take the cards! :D Horribly obscure words are soooo much fun! :harhar:

Maybe I'm being [i]too[/i] obscure. [b]lch[/b]

Whoa whoa one at a time

[quote="i_like_pie"]a[u]lch[/u]emy

[b]tle[/b][/quote]

boo[b]tle[/b]g

BS on [b]okk[/b]

Bookkeeping -- the work or skill of keeping account books or systematic records of money transactions


Let $ \{a_{n}\}_{n\ge 1}$ be a sequence in the field $ \mathbb{K}$,let $ v_{k}\equal{}(a_{k},a_{k\plus{}1},\cdots,a_{k\plus{}n\minus{}1})^{t}$
Prove that:
$ \dim\{span(v_{1},v_{2},\cdots)\}\equal{}\max_{k\ge 1}\dim span(v_{k},v_{k\plus{}1},\cdots,v_{k\plus{}n\minus{}1})$

I am not sure for it is right.

yes zhaobin it's a very nice problem. The problem is true for $ n\equal{}2$ I think, but it seems hard to check for the general case.


ARML mathleague.org AMC AIME USA(J)MO USAMO email

How was the championship meet? (I didn't get to go b/c of a band thing). Who won? Who's on A team? Did they announce B team?

Top three teams were Lynbrook, Amador Valley, and Cupertino (I think; might have been MSJ).

The A team list will be sent out to the ARML mailing list soon; if you're not on the mailing list, sign up for ARML (at mathleague.org) and you'll be put on it.  B team was not announced; I think that's in large part because the B team is likely to change quite a bit between now and the contest as people decide not to go to ARML and others get bumped up in their place.

I can, however, tell you the formula for team selection:
Let A be the sum of your top three scores on qualifying contests (200 each, 600 total possible).
Let B be your AIME score.
Let C be the number of times you have previously been a USAMO winner.
Then your selection index is $ A \plus{} 20B \plus{} 600C$.

Hmmm...that's an interesting selection criteria.  I really hope that email is sent out soon, as I left before awards!

C is basically just for Evan O'Dorney.

Blehhh my horrible AIME score is affecting everything this year, including ARML, where I am at the B team level compared to A team last year.  However, it looks like about 5 people who were on the A team list have said that they will not be able to make it because of issues with timing or pride or other stuff, so that's good news for B team members.

Target round was pretty disgusting/computational this time, so everyone's scores were 30-40 points lower than usual.  That's tough for those for whom this is the third tryout...

wait was the email sent lol? I didn't get it... @archimedes: what is your "sfba arml index"?

LOL "pride".  :)  yeah i hope you all found the tests today to be tougher than usual (except power of course), and congrats to Lynbrook, the Northern California champions.  BTW Lynbrook you got pwned by two schools in Iowa of all places, so now you have to go to Kansas City to defend your honor against them..  :)

[quote]Target round was pretty disgusting/computational this time, so everyone's scores were 30-40 points lower than usual. That's tough for those for whom this is the third tryout... [/quote]

^Yeah, this was my third tryout and my score was at least a good 30 points lower than my average (I got 3-4 less than my average target score today), so I'm probably going to end up dropping from the A team last year to the B or C team this year =(

Does anyone know what "ARML index" was necessary to make the A team?

Argh the Target today was amazingly computational, but not that difficult...

Did anyone else here feel that Sprint was easier than normal?

And my AIME arithmetic errors are going to kill my ARML index...hoping to make B team.

Amador got 2nd?  WOOT!

is it confirmed that the emails will be sent out today?
what are everybody's arml indices?

[quote="Sly Si"]
Let C be the number of times you have previously been a USAMO winner.[/quote]
LOL GOOD BYE A TEAM

I also feel like the target round relied on too much case-checking this time, and was harder than the previous target rounds - unfortunately, I was relying on this score as my 3rd tryout score.

3rd place team was Cupertino, I remember.

Go AMADOR VALLEY!

Seeing as how no email has been sent by mathleague yet, does anyone remember the current A team?

Mr. Sanders, will the email with the list of A team qualifiers be sent out tonight for those of us that are anxious to know if we made the team?

Sorry, I guess being on the border line creates unnecessary anxiety?

AoPS Wiki implies that Blue MOP consists of:

Top non-senior USAMO finishers: In addition to the winners, the next 18 or so non-senior non-Canadian finishers are invited to attend MOP. This group is viewed as potential IMO team members for future years, although in extreme circumstances (including 2006) IMO team members for that year have been drawn from this pool.

fascinating.  okay, so listen everyone, don't stay up tonight waiting for ARML information because i'm working late.  look for it tomorrow morning though..  :)

All ARML information will be mailed to those on the mailing list, right?

Question that kinda gets back on topic:

Can middle-schoolers compete in the National meet?

I was like, A + 20B + 600C, how in the world do I get into A team with 100 weighted AIME score+550-ish mathleague score until I reread the multiplier as 20 instead of 10. So, ARML A for me? Hmmm.

Blargh lol and i didn't even make B team.  Shame on me.  Guess those arithmetic errors on AIME cost me more than making usamo, eh?

I can't go to ARML anyways, I have another commitment.

[quote="farmertim"]i'm doing what i can to get that out ASAP.  please be patient.  :)  BTW i'm trying to find some information online that will enable me to determine the stability and consistency of the blue mop selection process.  obviously i can't be giving out any free lifetime A team memberships to people who fall into a category that isn't well-defined or that isn't the same from year to year.  black mop is of course well-defined so no problem.  anyone care to help?[/quote]

Actually, if you really want "well-defined", it's probably better to take Winners and Honorable Mentions. Black MOP isn't necessarily the top 12, because sometimes people don't go and so they take some non-winners into black. I think HM is the top 12 after Winners, so it is also well defined unlike Blue MOP.

wait
uh
blue mop is like
the top 20ish nonseniors after the winners
and
the seniors
kind of graduate
so the next year we dont care about them for arml
?

Well, blue MOP is still not completely "well-defined" because some people who make it don't go.

[quote="DataBox"]Cuz RSI > Blue MOP =)[/quote]

darn this has a positive probability of being true :/

@taylor: feel the wrath of the berserkergang?


How do you all view the national debt?  The media always seems to criticize budgets that aren't balanced, but I've heard economics professors say that it isn't that big of a deal.  After all, our deficit is usually around 2-5%(?) of our GDP and firms run at much higher borrowing rates than that.  On the other hand, government bonds must eventually be paid back--with interest.  Are politicians just passing frivolous debt to future generations?  Is anyone here for a balanced budget amendment or is that a stupid idea?  Just wondering your thoughts...

Carrying a debt is fine, so long as the investment is worth it.

I've never understood the argument that debt is [i]de facto[/i] a good idea.  Should any institution spend more just for the sake of issuing bonds?

Debt allows bond holders a degree of control over government that makes me uncomfortable at times.  It bothers me that China controls an enormous amount of power of the U.S. government in the form of financing (bonds).

I'm also wary of the Keynesian argument for using debt to control interest rates.  Counter-cyclical demand theory seems dead to me.  If liquidity of U.S. markets ever got bad enough for a little money present in the form of interest rate setting become a lot of money -- we'd know something really bad was happening in the world.

I would argue that much of government spending is wasteful and that it's not clear that a debt is necessary for that reason.  Of course, it will take years to pay back what the federal government already owes.

All that said, I think a balanced budget amendment is an absurd idea.  It keeps government from doing its job when we need it most.

A better idea would simply be to educate citizens well enough to understand, demand, and enforce efficient fiscal management.

I just wrote a long paper on the National Debt for my political science class!! Not sure I understood everything and all but at least it was good information gathering etc. It appears that there are several different views and everything about this. There is for example, the Ricardian Equivalence theory proposed by Professor Robert Barro that it should not matter to most people whether the government finances its money through borrowing by issuing bonds or making changes to taxes or spending. Yet, there is a lot of empirical evidence against this theory. I am attaching my paper to this post.

[quote="Watson"]I have to follow Heinlein on this. What a national debt does is it forces the goverment to pay for making money, and makes overproduction worse as taxes aren't spent consuming things but to investments. Play an economic simulation, and watch what happens as production rises beyond the money supply.[/quote]

That might be generally true when we look at examples around the world, but I do think debt can play a positive role in an economic enterprise, be it a company or a nation.  However, that role is almost always played when the entity is young, or invests in a completely new plan.  For a nation to carry debt through the best of times hints to the problem you mention, as well as others, such as corruption or other fiscal mismanagement.

[quote="Watson"]Why can't a nation just print more money instead of borrow? Both increase the monetary supply. And nations have many kinds of debt. Borrowing forigen monies to do trade is a good idea, but borrowin from banks to finance programs is quite expensive.[/quote]

The costs in printing and borrowing money are different, and should be weighed against each other.  For certain, they are not the same.

Printing currency devalues existing currency.  That implicitly taxes holders of cash to the degree that they hold it.  That also makes it an unnatural disincentive to hold cash, whereas there is no doubt we don't need new factors skewing the incentives of capital and investment.

Selling bonds however doesn't create new paper money, so it doesn't have the same effect.  Taxes must be levied to pay interest (or profits payed away from businesses), but those taxes come from general tax funds, spread throughout the whole base of taxpayers (and profits from owners, and therefore their workers as well).  Sure, there are also imbalances in the system as a result, but they are more incremental and are already systematic (with cultural acceptance) to the degree that they are law.

Printing paper money fast has a history of causing currency devaluations, which in turn affects international commerce.  I'd be inclined to avoid it.

I didn't do a very good job of summarizing the costs or making them clear, but unfortunately I don't have time to do a better job just now.

Fractional reserve banking also increases the money supply. Let's say I am a bank in a small town, total money 100,000. I have half of it in my bank. Someone needs a loan of 50,000. I loan all of it to him and remain solvent, despite having that much exactly. How?
Simple. I let him make checks instead of having him walk out with my cash. When people deposit the checks I am safe. The problem is if my clients want to close their accounts simultainiously. But let's look closer at what happens next:
The person making the loan pays me back eventually. But for 30-days the amount of cash in the town is 1.5 of normal. And the person pays for the privillage. Now we see the goverment pays bankers for coining currency, which the goverment can do by itself.

Fractional reserve doesn't actually increase the total amount of money wealth.  It just leverages existing wealth at the cost of an increased probability of local systemic collapse (not that I'm knocking the practice -- there is always a value to volatility that we can measure via the Sharpe ration, and such exchanges of risk are positive if controlled well).

The overall difference is one of providing liquidity versus trying to conjure new wealth out of thin air (which must be paid for by capital holders by the devaluation of their dollars).


is the answer $ \sqrt{2}$........

because coefficeint of restiution is the ratio of relative spped of separation by relative spped of approach........
by momentum conservation and by symmetry teh 2 balls move at 45 degrees to the line of approach of the 3rd ball and using the definition we get the answer to be $ \sqrt{2}$


Find the easiest prove of the fact that $lim_{n\rightarrow \infty}\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdot\cdot\cdot +\frac{1}{n^2}=\frac{\pi^2}{6}$

I hate to say that, but it's the nonsense of a question .:roll:

You could look at http://www.mathlinks.ro/Forum/viewtopic.php?t=52359 and the links given there.

Maybe he is asking about some sort of "elementar proof"?
I'm pretty sure there isn't such  (althought questionable is what should be classified under the term "elementar" )

Every proof must use calculus, and with calculus there are some easy ones.
@Beat: what exactly you want¿


In triangle ABC which AB=BC draw line BD to AC,draw a line AE to BC intersecting BD at P and AP=AD .given angles ACB=50,CAE=20,ABD=50
  Find angle AED=? :)

To: [b]Virgil Nicula[/b]

we know answer.

Give HINT please!!! or the solution!!!

I can give you HINT
  Draw a circle around  triangle

[quote="Maruf"][color=darkred]In triangle $ABC$ which $AB=BC$ draw line $BD$ to $AC$,draw a line $AE$ to $BC$ . Given angles $ACB=50^{\circ}$ , $CAE=20^{\circ}$ , $ABD=50^{\circ}$ . Find angle $AED\ .$[/color][/quote]
[color=darkblue][b]Proof[/b] ([u]for beginners[/u]).
\[\underline{\overline{ \|\begin{array}{c}\\ AB=BC\Longrightarrow |\begin{array}{c}m(\widehat{BAE})=m(\widehat{EBD})=30^{\circ}\\ \\ m(\widehat{ADB})=80^{\circ}\ ,\ m(\widehat{AEB})=70^{\circ}\end{array}\\ \\ m(\widehat{AED})=x\Longrightarrow m(\widehat{BDE})=80^{\circ}-x\\ \\ \boxed{\ \sin\widehat{ABD}\cdot\sin\widehat{DAE}\cdot\sin\widehat{EDB}\cdot\sin\widehat{BEA}=\sin\widehat{DBE}\cdot\sin\widehat{EAB}\cdot\sin\widehat{BDA}\cdot\sin\widehat{AED}\ }\\ \\ \sin 50\sin 20\sin (80-x)\sin 70=\sin^{2}30\sin 80\sin x\\ \\ 4\sin 50\sin 20\cos (10+x)\cos 20=\cos 10\sin x\\ \\ 2\sin 50\sin 40\cos (10+x)=\cos 10\sin x\\ \\ (\cos 10-\cos 90)\cos (10+x)=\cos 10\sin x\\ \\ \cos 10\cos (10+x)=\cos 10\cos (90-x)\\ \\ \cos (10+x)=\cos (90-x)\\ \\ 10+x=90-x\\ \\ \boxed{\ x=40\ }\ .\\ \\ \end{array}\|}}\]
[/color]

:thumbup: Good solution, I got it!!  :thumbup:

Could you solve it without using trigonometry, please! :roll:


256 people play a game where the players wander around a room while a music is playing.

when the music stops, they have to have a 'rock-paper-scissors' game with the person nearest to them.
the loser of the 'rock-paper-scissors' game follows the winner around the room when the music starts again.

that repeats until there is 1 winner.

if poohead :D  is playing, what is the probability that he will win?

thats correct. there is a 1/2 chance poo head will survive each round, and there will be 8 rounds (2^8=256). therefore, the probability is 1/256.

Or everyone has an equal chance of winning... 1/256.


1) $ \frac{z^4}{8}\plus{}i\sqrt{3}\equal{}\minus{}1\Longleftrightarrow z^4\equal{}16\left(\minus{}\frac{1}{2}\minus{}i\frac{\sqrt{3}}{2}\right)\equal{}16e^{i\pi\frac{7}{6}}\Longleftrightarrow z\equal{} 2\cdot e^{i\pi\frac{7\plus{}12k}{24}}$ for some $ k\equal{}0,1,2,3$
2) is unspecified

[quote="Borat"]1.  z^4/8+i√3=-1

2.  (2+i√12)^5

 Thanks for the help :D[/quote]
you did not give but let's say it's $ h$ so we then have $ z^{4} \equal{} \minus{}1 \minus{} i\sqrt{3}$
$ z^{4} \equal{} 2(\frac{\minus{}1}{2} \minus{} \frac{i\sqrt{3}}{2}) \equal{} 2e^{i\pi\frac{7}{6}}$ so $ z \equal{} \sqrt[4]2e^{i\pi\frac{7}{24}}$


Arnold and Boris are given a sufficiently long paper strip, one with letter $A$ written, and the other with letter $B$. Every minute, one of them (not necessarily one after the other) writes the word written on the other strip on his own strip, either to the left or to the right of the word already present. Prove that the next day, one will be able to cut the word on Arnold's strip into two words and exchange their places, obtaining a palindromic word.

What if Arnold writes $B$ next to his $A$. Then he gets $AB$. I can't see any palindromes...

The problem says that there's a step "every minute" and then the end is "next day", so maybe you have to wait $24\times 60$ steps.  :maybe:

but that's still wrong, assume arnold write B next to A and get AB. Then Boris do all the remaining steps (arnold not doing anything).
arnold will still have that AB even after a year..........

Any Russian know where is the problem :?:


I sure hope not :rotfl:

The competition isn't about how good your handwriting is, it is about how good  your math is. If your handwriting is legible, I think that it would be fine.

But, you'd have to see his handwriting to actually know how bad it is. 1337...'s isn't that bad. MINE is SO much worse.

Mine is pretty horrible too, unless I make a special effort, which I did. Now it is legible, but barely.

[quote="13375P34K43V312"]If we have a really good solution, but it's somewhat illegible, are you gonna take off? Not like illegible to the point where the graders give up on trying to read it, just hard to read.[/quote]The graders have the discretion to deduct 1 point for an especially poorly-written solution.  They are instructed only to do this in extreme cases.

Part of mathematics is effectively communicating your work.  Thus it is legitimate to consider this as part of our scoring criteria.

Uh oh :rotfl: . That doesn't look good for me.....

Last year, I hand-wrote all my solutions, and I've had teachers tell me to type up inclass essays so they could read them.

I got 100 last year, so I'm figuring they don't take off unless the writign is really bad.  Plus, good math students tend to not have great handwriting (I had to hand soemthing in to my gym teacher last year.  He looked down and said "I'll bet you're good at math" =)

JB

[quote="jackb1115"]Good math students tend to not have great handwriting[/quote]
This is not exactly true.  Have you ever seen a solution handwritten by Anders Kaseorg?  His handwriting is probably the neatest I have ever seen.

[quote="13375P34K43V312"]Who is Anders Kaseorg?[/quote]

IMO gold medalist, Davidson fellow.  All-around smart guy.

yea... anders's handwriting is quite... amazing

If I'm not mistaken, he writes his a's like they appear in this forum (not the convention way of "handwriting" an a ("a" versus "$a$") - if you're a guy that is  :P )

my handwriting is legible and not messy, but it isn't neat either. Do you think math students tend to not have as good handwriting? mine is only average, but then again I just really joined a *math team* this year.


integration
Functional Analysis
complex numbers
advanced fields
advanced fields unsolved

$H$ a separable Hilbert space. An operator $T : H \to  H$ is called a Hilbert-Schmidt operator if there exists a Hilbert base (e_1,e_2,..,e_n,..) of $H$ satisfying : $\sum_n |Te_n|^2 < \infty$.
Show that such an operator is compact. Is the converse true ?

For a counterexample to the converse, consider a separable Hilbert space with orthonormal basis $e_0,e_1,\ldots,e_n,\ldots$, and take the diagonal operator $T$ defined by $Te_n=\frac 1{\sqrt n}e_n$. It's compact because a diagonal operator with diagonal entries $\alpha_n$ is compact iff $\alpha_n\to 0$, and this is certainly the case here, but it's not Hilbert Schmidt because $\sum\|Te_n\|^2=\sum\frac 1n=\infty$.

As for the problem itself, we can treat it in a slightly more general setting. Let our Hilbert space be some $\mathcal H=L^2(\Omega)$, where $(\Omega,\mu)$ is a measure space, and consider the operator $T$ defined by $Tf(x)=\int_{\Omega}K(x,y)f(y)\mbox d\mu(y)$ for some $K\in L^2(\Omega\times\Omega)$. Then $T$ is compact.

In order to prove this, it suffices to show that whenever $f_n\to 0$ weakly form some sequence $(f_n)_n$ contained in the unit ball of $\mathcal H,\ Tf_n\to 0$ strongly. Now, since $\int_\Omega f\bar g\mbox d\mu\to 0,\ \forall g\in\mathcal H$, by taking $g(y)=\overline{K(x,y)}$ for $x\in\Omega$ we get $Tf_n(x)\to 0,\ \forall x\in\Omega$, i.e. $Tf_n$ converges to $0$ simply. We want $\int_\Omega\left|\int_\Omega K(x,y)f_n(y)\mbox d\mu(y)\right|^2\mbox d\mu(x)\to 0$, but this now follows from the fact that the sequence of maps $x\mapsto\left|\int_\Omega K(x,y)f_n(y)\mbox d\mu(y)\right|^2=|Tf_n(x)|^2$ converges to $0$ simply and every term of the sequence is dominated by $x\mapsto \int_\Omega|K(x,y)|^2\mbox d\mu(y)$, which belongs to $L^1(\Omega)$ because $K\in L^2(\Omega\times\Omega)$. The Dominated convergence Theorem applies, and the conclusion follows.

In our case, we can take $\Omega$ to be the set of natural numbers, and $\mu$ to be the counting measure.

One can suppose that $H=L^{2}([0;1])$. Prove that every Hilbert-Schmidt is
of the form
$(Tf)(x)= \int_{[0;1]}K(x,y)f(y)dy$
where $K(.,.) \in L^{2}([0;1] \times [0;1])$

Let $(e_{n})_{n}$ be an orthonormal basis for $\mathcal H$, and check that $K(x,y)=\sum_{n}(T\bar e_{n})(x)e_{n}(y)$ works ($\overline f$ is the complex conjugate of $f$).


For some odd prime p, let $ M_p \equal{} 2^p \minus{}1$

Prove that all prime divisor are $ 2pk\plus{}1$ for some positive integer k.

The order of $ 2 \mod q$ is $ p$ for $ q|2^p \minus{} 1$.Thus $ p|q \minus{} 1$ and $ 2|q \minus{} 1$ is rather trivial.


function
number theory
prime numbers
number theory solved

1. N = 2000 * 2002 *2006* 2008+ n ...
find smallest possible n such that N is a perfect square...

2. The number of digits of the integer 2^101

3. Find the number of pairs of positive integers (x,y) that satisfy the equation

1/x + 1/y = 1/2004

er... Can u solve it then???
Mr Pro??
The solution please??
Action speaks louder than words...

Well since this is SMO I must reply :) (and yes, I am inclined to agree that the questions are easy :( , but once again let me clarify: this was first round, 25 questions in 2.5 hours, full solutions not necessary).

1) $N = 2000 \times 2002 \times 2006 \times 2008 = (2004^2 - 16)(2004^2 - 4) = (2004^2-10)^2 - 36$, so $n = 36$.

2) In [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=21633&highlight=#21633]this[/url] thread Myth gave a pretty solution to the number of digits in $2^{1000}$, which is harder.

Using the same idea, it's obvious that there are at least $1 + 3 \times 10 = 31$ digits because $2^{10} = 1024$, and using $(1 + \frac{1}{n})^n < e$ we have $2^{100} = 10^{30}(1+\frac{3}{125})^{10} < 10^{30} \sqrt{e} < 10^{30} \times 2$, hence there are 31 digits.

3) We can rewrite it as $(x - 2004)(y - 2004) = 2004^2 = 2^4 \times 3^2 \times 167^2$, which has 45 factors.  Hence there are 45 possible pairs $(x,y)$.

(2004^2-16)(2004^2-4) is not equal to 2004^4 - 100
but i figure the solution using your way... Thanks...

as for the second and third one...
2^101
i dun get ur method... can give me more details..

how can u rewrite it as that equation ???
how u get it???

Thanks alot...

another question...
(185^n) + (149^n) k is divisible by 2004 for odd positive n...

Find all prime integers n such that the product of its digits equal to n^2 - 10n - 22...

THanks once again...

The first one is false: take n=1 to get 334

Also, larger odd n are not divisible by 2004 since
[tex]185^n+149^n=(167+18)^n+(167-18)^n[/tex]
[tex]=\displaystyle \left(167^n + 167^{n-1}18^1{n\choose1} + ... + 18^n \right)+ \left(167^n - 167^{n-1}18^1{n\choose1} + ... - 18^n\right) [/tex] for odd n

Summing these gives:
[tex]2 \displaystyle \left(167^n + 167^{n-2}18^2{n\choose2} + ... + 167^118^{n-1}{n\choose{n-1}}\right) [/tex] which is divisble by 334 but not 668 due to the [tex]167^n[/tex] term

Now for the second new one .... note that the roots of [tex]n^2-10n-22[/tex] are [tex]5\pm\sqrt{47}[/tex], so for [tex]n>5[/tex] we have [tex]n^2-10n-22[/tex] increases and for [tex]n<5+\sqrt{47}\approx11.86[/tex] is less than 0.

Also, note that [tex]18^2-10*18-22=122>100[/tex] is larger than the possible product of digits in a 2-digit number.  Therefore the only possible prime candidates are 13 and 17 so we test these:

[tex]13^2-10*13-22=169-130-22=17\not=3[/tex]
Since we know that this function increases for [tex]n>5[/tex] we know that [tex]17^2-10*17-22>17>7[/tex] so this cannot be a solution either.

Now for three digits and larger (call it d), the smallest number, [tex]10^{d-1}[/tex] when plugged into [tex]n^2-10n-22[/tex] gives [tex]10^{2d-2}-10^{d}-22>9^{d}[/tex] and as such there can be no solutions with three or more digits.  Therefore we conclude there are no prime numbers such that the product of its digits equals [tex]n^2-10n-22[/tex]

On a side note, this property does hold for 12, but that is not a prime.

I apologise for writing something so ridiculously stupid yesterday :blush:.  The post has been corrected.

2) $2^{101} > 2^{100} = (2^10)^{10} = 1024^{10} > (10^3)^10 = 10^{30}$
$2^{101} = 2 \times 10^{30}(1+\frac{3}{125})^{10} < 2 \times 10^{30} e^{10 \times 3/125} < 10^{30} \times 4$.

3) I'm sure you can get to that equation.  From then onwards we see that $x-2004$ and $y-2004$ must be a pair of factors of $2004$ which multiply to be $2004$.


calculus
integration
function
real analysis
real analysis unsolved

suppose that $ x_n$ is an equi-distributed sequence in $ [0;1]^d$, and $ f: [0;1]^d \to \mathbb{R}$ a function. If $ f$ is nice enough (continuous, or Riemann integrable, etc ..), one knows that
$ \frac{1}{n} \sum_1^n f(x_k) \to \int_{[0;1]^d} f(x) dx$
But how can one bound the speed of the convergence. I'm looking for something like
$ | \frac{1}{n} \sum_1^n f(x_k) \minus{} \int_{[0;1]^d} f(x) dx | \leq \psi(n)$
where $ \psi$ can depend on $ f$ and $ (x_k)_{\mathbb{N}}$. (but not too much  :lol: ) One can feel that the more $ f$ is regular, and the better $ (x_k)$ is equidistributed, the faster the convergence is. But how can one formalize that ?

If that's random equidistribution (the $ x_k$ are independent), we get a Central Limit Theorem estimate, and the expected error is $ O(n^{\minus{}\frac12})$, as long as $ f$ is at least $ L^2$.

Let's try a more regular distribution: $ x_k\equal{}ky\mod L$, where $ L$ is the integer lattice. This is best handled with Fourier series; let $ f(x)\equal{}\sum_{s\in L}c_se^{2\pi i s\cdot x}$. We get $ \sum_{k\equal{}0}^{n\minus{}1}f(kx)\equal{}nc_0\plus{}\sum_{s\in L,s\neq 0}c_s\frac{1\minus{}e^{2\pi i ns\cdot x}}{1\minus{}e^{2\pi i s\cdot x}}$. Dividing by $ n$, the error is at most $ \frac1n\sum_{s\in L,s\neq 0}|c_s|\left|\frac{1\minus{}e^{2\pi i ns\cdot x}}{1\minus{}e^{2\pi i s\cdot x}}\right|$

If the Fourier coefficients decay fast enough, that sum will be $ O(n^{\minus{}1})$; $ O(n^{\minus{}1})$ for the coefficients that don't have small denominators, and $ O(n^{\minus{}1})$ for the small fraction (about $ \frac1n$) that do.

$ o(n^{\minus{}1})$ is too much to ask for in all cases- since the newest term adds $ O(n^{\minus{}1})$ error, we can only hit that by a lucky coincidence.

[quote="alekk"]thank you very much Jmerry for your interest :-) I was wondering if there are some other results in this way. Nevertheless I would like to have such result for functions not that regular. In your case, one needs something "much" stronger that $ \sum_{\mathbb{Z}^n} |c_k| < \infty$. But I am trying to apply this to functions that are not so regular: maybe continuous, but surely not $ C^1$, which is quite a big problem for the size of the Fourier coefficient. Any new ideas ?[/quote]What you are looking for is known as [b]quasi-Monte Carlo integration[/b]. See e.g. [url=http://www.amazon.com/Generation-Quasi-Monte-CBMS-NSF-Conference-Mathematics/dp/0898712955]here[/url].


Functional Analysis
real analysis
real analysis unsolved

Here is a nice problem that I want to share with mathlinkers. You will enjoy solving it, as I did  :lol:

Prove that no infinite Banach space can have a countable linear dimension.

Hmm, yes you are right, but it is nice of you if you show more detalis, isn't it? ;)

A finite dimensional subspace is closed and has empty interior. Therefore, and subspace with countable linear dimension is of 1st category.

Is there argument which shows that the dimension of a Banach space is at least continuum? (Not using Continuum Hypothesis, of course)