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Assume that a function $f(x)=\frac{3x+a}{x^2+3}$ has maximal value  $\frac{3}{2}.$

(1) Find the value of constant number $a$.
(2) Find the range of the value of $b$ for which the equation for $x,\ f(x)=b$ has two distinct real roots.

[hide] I don't know LATEX so I am just going to describe the process I used.  First, differentiate using the quotient rule.  Set the derivative equal to zero and then solve for "a".  Substitute this back into the original and set the original equal to 3/2.  This will give you the x-coordinate of the maximum.  Substitute this value into the original equation that still contains the "a" value and solve for "a".  I have not yet tried the second half of your problem...I will try to post soon.[/hide]

That was way too much to Latex...
[hide][img]http://img.photobucket.com/albums/v194/mukmasterxps/solnsc2.png[/img][/hide]

PART A: Your answer is right.
Comment: Let $p$ be the value such that $f'(p)=0,$ maximal value is $\frac{3}{2p}.$

PART B: Is $bx^2-3x+3(b-1)=0$ always quadratic equation? :(

kunny

[quote="The Zuton Force"]Since b is a constant, wouldn't it be?  :?  :?[/quote]

Yes, $b$ is costant, but in case of $b=0,$ you can't use discriminant.


I practice solving Olympiad problems, and when I tried this German problem I downloaded it from internet, I failed. Can anyone tell me how to solve this problem?

Question:
Let x be a non-zero real number satisfying the equation $ ax^2+bx+c=0$. Furthermore, let a, b, and c be integers satisfying $ |a|+|b|+|c|>1$. Prove that

$ |x| \ge \displaystyle \frac{1}{|a|+|b|+|c|-1} $

Just try dividing equation with x. And something else :)


Not yet  :wallbash:
Some little hint ? :oops:

i don't know how to do it either...but if i find out,i'll post it. Lets see who solves it first. :lol:

[quote="Altheman"]does this help?

n^2< c/a<(n+1)^2

2n<-b/a<2n+2[/quote]

That' s all I've got too ! :mad:

[quote="Tinu"]Let a,b,c be integers, a>0.If the quadratic $ax^{2}+bx+c=0$ has 2 distinct roots in $(n;n+1)$,where " n" is an integer,prove that \[a\geq5\]  \[c\geq5n^{2}+1\] [/quote]

uh, are you sure $a \ge 5$? a quadratic with distinct roots $5,6$ is just $x^{2}-11x+30$.

The problem says "in $(n; n+1)$" and not "in $[n;n+1]$" - meaning, you can have $n<x_{1}<x_{2}<n+1$ and not $n\leqslant x_{1}<x_{2}\leqslant n+1$.

I didn't find a good solution but if you want, you can see the solution on advanced section, algebra, unsolved problems. it was posted on Fri, 13 Oct and the title is also strange quadratic. I didn't quite understand that one because its wasn't written in Latex.Maybe u will. I'll also try to find a solution because i think i'm very close.

[color=green]Part 1[/color][hide]We will prove that if $f(x)=ax^{2}+bx+c$ has real roots in $(0;1)$ then $a \ge 5$
It's easy to show that the minimum of f(x) is $f(-\frac{b}{2a})=-\frac{b^{2}-4ac}{4a}$. Also $f(0) \ge 1$ and $f(1) \ge 1$.
Consider the function $g(x)=f(x)+\frac{b^{2}-4ac}{4a}$. We have $g(0)>1$ and $g(1)>1$
[color=red]Case 1:[/color] $-\frac{b}{2a}\le \frac12$
 then $g(0)=c+\frac{b^{2}-4ac}{4a}=\frac{b}{4a}=a*(\frac{-b}{2a})^{2}\le \frac{a}4>1$ so $a>4$ or $a \ge 5$.
[color=red]Case 2:[/color]$-\frac{b}{2a}>\frac12$ so $1-(\frac{-b}{2a})=1+\frac{b}{2a}< \frac12$
then $g(1)=a+b+c+\frac{b^{2}-4ac}{4a}=\frac{4a^{2}+4ab+b^{2}}{4a}=a*(1+\ac{b}{2a})^{2}<\frac{a}{4}>1$ and again $a \ge 5$ :D[color=green][/hide]Part 2[/color]
[hide]Let $f(x)=ax^{2}+bx+c$ has real roots in $(n;n+1)$ then $g(x)=f(x+n)$ has real roots in $(0;1)$ and obviosly $g(x)=ax^{2}+...$ so for any $(n,n+1)$ $a \ge 5$[color=green][/hide]

Part 3[/color][hide]$f(x)=ax^{2}+bx+c$ has real roots in $(n;n+1)$ then $n<-\frac{b}{2a}<n+1$ so $-\frac{b}{2a}=n+\frac{m}{2n}$ where $0<m<2a$ or $b=-2an-m$.
$f(n)=an^{2}-(2an+m)n+c=-an^{2}-mn+c \ge 1$. So $c \ge an^{2}+mn+1$.
When $n \ge 0$ this means $c \ge an^{2}+1$
[color=darkred]Remark 1:[/color] As $m>0$ this means that $c = an^{2}+1$ only when $n=0$ else $c \ge an^{2}+n+1$
[color=darkred]Remark 2:[/color] when $n<0$ the limits of c are not the same. For  examlple $f(x)=6x^{2}+5x+1=0$ has real roots $x=-\frac12$ and $x=-\frac13$ but $1=c<6(-1)^{2}+1$.
For negative n: $n \ge a(n+1)^{2}+1$
All the ideas of this proof was given to me from a friend  :) [size=150][b]Rosnik[/b][/size][/hide]


What is the product of all possible digits $ x$ such that the six-digit number $ 341,4x7$ is divisible by 3?

Add the digits up (not including x).

You should get 19.

Then there are three possible digits that result in a number divisible by 3.

Those numbers are 2, 5, and 8.

Multiply, and you should clearly see that the answer is [b]80[/b].


The remainder, when [i]f[/i]([i]x[/i]) is divided by ([i]x[/i] - [i]a[/i])([i]x[/i] - [i]b[/i]), is written in the form [i]A[/i]([i]x[/i] - [i]a[/i]) + [i]B[/i]([i]x[/i] - [i]b[/i]); prove that

$ A\equal{}\frac{f(b)}{b\minus{}a}, B\equal{}\frac{f(a)}{a\minus{}b}$

Write $ f(x) \equal{} (x \minus{} a)(x \minus{} b) g(x) \plus{} A(x \minus{} a) \plus{} B(x \minus{} b)$ and substitute the two obvious values.


When we start writing in cotniuing Arabic numbers, up to n.If totally we use 3289 Arabic numericals to complete , then what is the number of n?

$ \mathsf{12345678910111213.....}$

[hide]
Numbers with less than or equal to one digit $ \equal{} (d \le 1) \equal{} 9$

$ d \le 2 \equal{} 99 \minus{} 9 \equal{} 90$
$ d \le 3 \equal{} 999 \minus{} 99 \equal{} 900$...

$ \Rightarrow S_n \equal{} 1(9) \plus{} 2(90) \plus{} 3(900) \plus{} 4(9000)...$

$ S_n \le 3289 \Rightarrow S_n \equal{} 1(9) \plus{} 2(90) \plus{} 3(900) \equal{} 9(321) \equal{} 2889$

$ \Rightarrow 3289 \minus{} 2889 \equal{} 400$

There are $ 400$ digiths worth of $ 4$ digit numbers $ \Rightarrow \dfrac{400}{4} \plus{} 999 \equal{} 1099$

Therefore $ n \equal{} 1099$[/hide]


function inequalities parameterization trigonometry limit logarithms calculus

This theorem addresses the behavior of a meromorphic function $ f(z)$ in a deleted neighborhood surrounding an essential singularity, $ a$.  In essence it states that for any positive $ \epsilon$, the inequality:

$ |f(z) \minus{} c| < \epsilon$ holds for any $ c$ an infinite number of times in $ 0 < |z \minus{} a| < \delta$

Answer the following:

(1) For a given $ c$, can I find an infinite set of concentric rings around the singularity $ 0 < . . . r_3 < r_2 < r_1 < k$ for some $ k$ say $ k \equal{} 1$ such that between succeeding rings, I can always find only one point $ z_n$ satisfying $ |f(z_n) \minus{} c| < \epsilon$ for $ r_{n \plus{} 1}\leq|z \minus{} a|\leq r_n$

(2) If (1) is not true no matter how small I make the difference between successive rings, then how do the multiple points accumulate?  What are their distribution and is this distribution dependent on the type of essential singularity?

(3) Would the distribution of points depend on the function or would it be a general property of essential singularities?

(4) Is the qualitative distribution independent of the value $ c$ and $ \epsilon$?

(5) If (4) is not true, how would the distribution depend on these parameters?

Since the set $ \{z\colon |f(z)\minus{}c|<\epsilon\}$ is open, you should rethink your approach to the problem.

Great . . . I don't entirely understand that which means of course I don't understand the theorem. I think you mean in any small neighborhood of a point $ z_n$ satisfying the inequality, there is an open set of points surrounding it that also satisfies the inequality.  I'll continue to look into it.

Perhaps I should change the inequality to "equality" since that is what the theorem is saying:  ". . . the value c will be taken at some point $ z_1\in N_{\delta_1}(a)$ where $ N_{\delta_1}(a)\equal{}\{z : 0<|z\minus{}a|<\delta_1$".  So now, let $ \delta_2<|z_1|$ then c will be taken at some point $ z_2\in N_{\delta_2}(a)$.  Continue in this manner with $ \delta_n<|z_{n\minus{}1}|$

I'll end up with the sequence $ \{\delta_n\}$.  Now, can this set be used to characterize essential singularities?

More precisely, any point $ z_0$ for which $ |f(z_0)\minus{}c|<\epsilon$ has a neighborhood in which [b]all[/b] points satisfy $ |f(z)\minus{}c|<\epsilon$. So there is no way how you can have just one such point in a circular ring.

It's convenient to assume that the point of essential singularity is $ \infty$. Thus one is led to study the behavior of [url=http://en.wikipedia.org/wiki/Entire_function]entire functions[/url] at infinity. Things like order, number of asymptotic values, Nevanlinna theory, etc. There are many thousands of pages on this topic.

Finally, there is not that much difference between $ \exp$ and $ \sin$ in the complex plane: $ \sin z\equal{}(\exp(iz)\minus{}\exp(\minus{}iz)/(2i)$.

[quote="mlok"]More precisely, any point $ z_0$ for which $ |f(z_0) \minus{} c| < \epsilon$ has a neighborhood in which [b]all[/b] points satisfy $ |f(z) \minus{} c| < \epsilon$.
[/quote]

So then I have a "region" which satisfies the inequality within an annulus $ \{z: r_2\leq |z \minus{} a|\leq r_1\}$.  Then between the annulus $ \{z: r_3\leq |z \minus{} a|\leq r_2\}$ I'd expect to find another set of regions since by the theorem ". . . c will be taken at some point $ z_2\in N_{\delta_2}(a)$ with $ N_{\delta_2}(a) \equal{} \{z: 0 < |z \minus{} a| < r_2\}$.  Continuing in this manner, I'd expect then to find a "smattering" of regions in each annulus as I head towards the singularity.  I wonder then how these regions are distributed throughout the neighborhood  of $ a$ . . . thousands of pages, huh?

That makes more sense. Still, you should be aware of situations like $ f(z)\equal{}e^{1/z}$ and $ c\equal{}0$ (i.e., $ c$ is an omitted value of $ f$), in which case $ \{z\colon |f(z)\minus{}c|<\epsilon\}$ is just one region (namely, an open disk whose boundary contains $ 0$.)

Thought this was interesting:  In the plots below I've plotted the Casorati-Weierstrass regions for the sets $ \{z: |f(z) \minus{} (1 \plus{} i)| < 1\}$ for the two functions $ f(z) \equal{} e^{1/z}$ and $ f(z) \equal{} Sin(1/z)$.  The plots  give you the impression of "scaling" suggesting the trends continue ad infinitum towards the singularity like a fractal.  However this may be due to the limitations of the graphics or code below since the functions are  oscillating with increasing amplitude and frequency as they approaches the singularity.

[img]http://img253.imageshack.us/img253/8021/cwregionsiv1.jpg[/img]

For reference, here's my Mathematica code for the first plot:

[code]f1[z_] := Exp[1/z];
z0 = 1 + I;
zmax = 0.2;
dz = 0.0011;
plist = Flatten[Table[{x, y,
           Abs[f1[z] - z0] /. z -> x + I*y},
         {x, -zmax, zmax, dz}, {y, -zmax, zmax,
           dz}], 1];
sel = Select[plist, #1[[3]] < 1 & ];
points = Point /@ ({#1[[1]], #1[[2]]} & ) /@
         sel;
p1 = Show[Graphics[points], Axes -> True,
     PlotRange -> {{-zmax, zmax},
         {-zmax, zmax}}][/code]

It's not really a fractal (self-similar) structure: it's a periodic pattern (both $ e^z$ and $ \sin z$ are periodic functions) after Mobius transformation $ z\mapsto 1/z$.

Note on names- it's [url=http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Casorati.html]Casorati[/url] with an a.

Great . . . seems to me I could have accomplished the same result above by simply using the "multi-inverse-functions" defined by:

$ f^{ \minus{} 1}(w) \equal{} \frac {1}{ln(r_w) \plus{} i(\Theta_w \plus{} 2n\pi)}$

$ f^{ \minus{} 1}(w) \equal{} \frac {1}{ \minus{} iLog[iw \plus{} (1 \minus{} w^2)^{1/2}]}$

Granted questions of analyticity come into play with the inverse functions.  Suppose we ignore that for now and consider: Can a function have an essential singularity without it's multi-inverse involving in some way, the multi-valued Log function?  Why do I think the answer to that question is no.

If so, then it just seems more natural then to re-cast Casorati-Weierstrass in terms of such multi-inverses.  Here's a start:

If $ f^{ \minus{} 1}(w)$ contains an infinitely-valued multi-function in some particular way that I currently can't define, then $ f(z)$ will contain an essential singularity at each of the branch points of $ f^{ \minus{} 1}(w)$ with $ f(z)$ "taking" all values of the inverse multi-function in neighborhoods surrounding those branch points.  No doubt there are some problems with this though.

(I corrected the spelling above, thanks.  I'm particular about that too).

[quote]Can a function have an essential singularity without it's multi-inverse involving in some way, the multi-valued Log function?[/quote]
Of course it can. Choose any sequence of nonzero complex numbers $ c_n$ ($ n\ge 0$) such that $ \lim_{n\to \infty}|c_n|^{1/n} \equal{} 0$. This sequence defines an entire function $ f(z) \equal{} \sum_{n \equal{} 0}^{\infty}{c_nz^n}$ which has an essential singularity at infinity. There is no reason for a random power series like that to be related to logarithms. You can consider $ f(1/z)$ if you prefer to have a singularity at $ 0$.

The reason logarithms come up in the most obvious examples is that $ \exp$ and its relatives $ \sin$ and $ \cos$ are the first transcendental entire functions that come to mind.

Ok.  Figured I was stretching my understanding of all this.  Actually this stems from my wish to understand the proof of Picard's Great Theorem.  Others wishing to review the proof can find it in "Complex Analysis" by M. Gonzalez (vol 2),  and "The Theory of Functions" by Titchmarsh.  It looks accessible although can't say I follow it yet but I'll spend time with it as well as this theory.  I still like the idea of the multi-inverse function and will also play around with those in Mathematica.  For me, Mathematica is an integral part of understanding mathematics.  Just my own personal style, that's all. :)


hi! I want to customizing the style of a document. In my cls file I wrote

\newcommand{\name}[1]{\center{\bf{#1}}}

my problem is that in the .tex file when I do \name{blabla} it centers not only blabla but the rest of the text that I wrote after blabla. Any ideas??

And I also would like to know if there's any difference between writing

\newcommand{\name}[1]{\center{\bf{#1}}} and \newcommand{\name}[1]{\center\bf{#1}}

because I've seen that not always it's used the { between \

thanks in advance!!!

Centred text uses an environment rather than a command. Try
[code]\newcommand{\name}[1]{\begin{center} \bf{#1} \end{center}}[/code]


Today was my geometry regent, and there was one question where we had to construct a perpindicular line through a point OFF the line.

I made 2 arcs {like constructing a perpindicular bisector} and made the arcs intercept the point. Then I joined the points. Will I be given full credit for this...?  :(

It depends on the person grading it.  :P

I didn't understand the question. Is it like this?:

 [geogebra]b45c6ebddfc5646e395be20069394cad4b9395f7[/geogebra]

If so, what information do you have?

I would think...
 [geogebra]fd5ea0d23316e3dc2e2985c0837b07e06a165490[/geogebra]
(The final line should be perpendicular, if I wasn't using an online drawing tool)

The diagram is exactly what you gave pi, but the point is moved over past the midpoint. I drew 2 arcs, {like when you construct a perpindicular bisector} and made them intersect that point, then I made a line. Is that valid?

Use geogebra to show what you did, then I'll tell you whether or not it is valid.

http://www.handprint.com/HP/WCL/IMG/LPR/perp.gif It looks like this, however the point is NOT on the line, AND its shifted a little to the right of the midpoint. Same concept tho.

I am pretty sure it is correct, but you have to make sure the two blue circles have the same radius (very important). A faster way is to just find half the distance between the two center points of the blue circles. The circle way works too because half the diameter is the radius. Nice technique though!

[hide="what I did"]
I drew two arcs intersecting the line from the point above the line. Setting any radius with the compass such that it reaches farther than half the distance from the two points on the line, draw an arc below the line. Keeping the radius, place the pointy end on the other point on the line and draw an arc that intersects the previous arc. Now just connect the intersection and the point above the line.
(sorry if this is a little hard to follow)[/hide]

A fast method:

 [geogebra]9b6dbb650c61c06438eaae55f03d3bf524056b32[/geogebra]

Where the intersection point by the right angle mark is the midpoint of the line formed by the points of intersection of the circle and the original line.

Actually, that wouldn't be any faster. Assuming you're only given a compass and straightedge, you would still have to construct the midpoint, which does not take less time than constructing an intersection of two arcs.


Ok, I've been trying to solve this question

A line is a tangent to a circle at B.  Points A and C are on the line on opposite sides of B.  A chord MN is parallel to the tangent. Prove that triangle MBN is isosceles.

 I'm suppose to use the Tangent-Chord Theorem, but i must prove it first in order to use it. I spent the past hour and i still can't figure out the Tangent Chord Theorem so I can use it in my proof :( i've tried using the alternate angle theorem to prove that <CBN = <BNM and <ABM = <BMN now I'm just really stuck because I can't prove <BMN = <BNM.  If I did, than I'll know it's a isosceles triangle

if the center is $O$, draw $BO$, $MO$, and $NO$, we know that $BO$ is perpendicular to $AC$ (why?)

let acr $MB=x$, so $\angle MNB=\frac{x}{2}$

by alternate interior angles, $\angle NBC=\frac{x}{2}$

since $\angle OBC=90$, subracting gives us $\angle OBN=90-\frac{x}{2}$

since $\triangle OBN$ is isosceles (radii are the congruent sides) and there are 180 degrees in a triangle, we have that

$\angle NOB +90 -\frac{x}{2}+90-\frac{x}{2}=180$ -> $\angle NOB=x$

since it is a central angle, arc $NB=x$, since arc $NB$ and arc $MB$ are equal, the chords $MB$ and $NB$ are equal, making $\triangle MNB$ isosceles

I understand the steps and process you went through, but I'm unsure how you got the angle MNB = x/2

Also, is there a way to use the tangent chord theorem to prove it?

an inscribed angle is half the arc it intercepts, the proof of this is not that hard, but you have to consider multiple cases


$A,\ B,\ C,\ D,\ E$ and $F$ lie (in that order) on the circumference of a circle. The chords $AD,\ BE$ and $CF$ are concurrent. $P,\ Q$ and $R$ are the midpoints of $AD,\ BE$ and $CF$ respectively. Two further chords $AG \parallel BE$ and $AH \parallel CF$ are drawn. Show that $PQR$ is similar to $DGH$.

JUST ANGLE CHASING
let the point of concurrent be $X$,its easy to see that$PQRX$ is cyclic quadrilateralm so:

$\angle HDG =\angle HAG=\angle QXC=\angle QPR \ ,\ \angle PQR=\angle PXR=\angle PAH=\angle DGH$

It is just angle chasing after the quadrilateral PQRX is known to be cyclic, but I could not see it (not that it is cyclic, but why).

Let X be a fixed point inside a circle (O) of radius R at the distance d = OX from the circle center. Let AB be an arbitrary chord passing through the point X and M the midpoint of this chord. The locus of midpoints M is a circle with the diameter OX.

Denote a = XA, b = XB, d = XO. WLOG, assume $a \ge b$. Denote $\theta = \angle AXO$. Then $\angle BXO = 180^\circ - \theta$. Using the cosine theorem for the triangles $\triangle AXO, \triangle BXO$,

$XA^2 + XO^2 - 2\ XA \cdot XO \cos \angle AXO = OA^2$

$XB^2 + XO^2 - 2\ XB \cdot XO \cos \angle BXO = OB^2$

Rewriting:

$a^2 + d^2 - 2ad \cos \theta = R^2$

$b^2 + d^2 + 2bd \cos \theta = R^2$

Subtracting these 2 equations

$a^2 - b^2 - 2d(a + b) \cos \theta = 0$

Reducing by $a + b \neq 0$:

$d \cos \theta = \dfrac{a - b}{2} = \dfrac{a + b}{2} - b$

$XO \cos \theta = \dfrac{AB}{2} - XB = MB - XB = XM$

The proposition follows.


Suppose you have two cars, A and B, approaching the finish line of a race.

The front of A crosses over the finish line first, but then it either just slows, comes to a complete stop, or goes backward by whatever combination of
forces applied inside to its controls and/or to the outside of it.  At any rate
it turns out that the entirety of car B makes it past the finish line before car A
does, or it is the case (although car A initially crossed the finish line, car A never does continue moving across).

I don't know the answer to this question:

Which car has won the race, the first to get the front part of its body across the finish line, or the first car to get its entire body past the finish line (and presumably stay there, and not move backwards past the finish line)?

It depends what the rules of the particular race are.  But I suppose you mean what would they do if no rules were defined beforehand...

In that case, I believe this would belong in the Round Table...

Personally, I think that

if you have to get "across" the finish line, you have to completely pass it.
If you have to get "to" the finish line, you have to reach it.
if no rules are defined, you have to get your center of mass to cross the finish line first.


What is the smallest positive perfect cube that can be written as
the sum of three consecutive integers?

If a number is the sum of 3 consecutive cubes, it must be div 3. The smallest cube div 3 is 3^3=27


A isosceles trapezoid has a longer base, $ a$ and a shorter base, $ b$.  If $ a$ is also equal to the diagonal of the trapezoid and $ b$ the height of the trapezoid, then what is the ratio $ \frac{b}{a}$


Prove that $ x^{2}+y^{2}=133703$ has no integer solutions for $ (x,y)$.

$ k^{2}\equiv0,1\mod4$

Therefore $ a^{2}+b^{2}$ cannot be equivalent to $ 3\mod4$. However, $ 133703\equiv3\mod4$


This thread is for the Team Chess Tournament Round 1: Team Beta (White) vs. Team Eta (Black). Play begins on December 7th. Any questions may be asked [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164632]here[/url]. All information/FAQ is being updated [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Chess/Chess_Tournament_Info]here[/url].

[hide="Team Beta"]
   1. Temperal
   2. CircleSquared
   3. abacadaea
   4. miyomiyo
   5. davidyko
[/hide]

[hide="Team Eta"]
   1. dbkarp
   2. rem
   3. ZeroFive1
   4. moogra
   5. Patterns_34
[/hide]

Each team will have 3 days for each move.

Good luck!

[b]Move 1 (White): d4[/b]

[img]http://i2.tinypic.com/85bepug.jpg[/img]

Yes that was a mistake: it should've been [b]f[/b]5, not [b]d[/b]5.
Thank you for the catch 1=2.  :)

[b][Edit][/b] I think my method of entering the list must've been misinterpreted. I was planning on creating the list of moves given so far and building on it as the game progresses. Example (this isn't the real game, just illustrating an example):
[b]Black's 3rd move:[/b]
[list]1. e4 c5
2. Nf3 Nc6
3. d4 $ \boxed{\text{cxd4}}$[/list]
And so on.
I'll re-edit it and from now on enter the moves in as just the move one side made so it isn't misinterpreted.
I apologize for the inconvenience that temporarily occurred today.

[quote="1=2"]I know that I am not in the game, but...

http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Chess/Chess_Tournament_Info/Round_1/Game_2

Same game, different info. Is this a mistake?[/quote]

Thanks for the notification.  :) I'll correct that.

[quote]Any error in the Wiki page will be amended by Temperal  :)  so do not stress about that.[/quote]

That's right.  :D

[quote="Patterns_34"][quote="Temperal"][quote="mz94"]ok, like, do we win by forfeit? idk how much time they have but its been at least a week or 2[/quote]

Team [color=red]Eta[/color] now has a warning for exceeding the time limit.

You can win by any normal chess methods, or if the game is taking too long, the TDs will decide.[/quote]
Referring to us or Team Zeta?
We apologize for not posting our move sooner; we thought the time limit was $ 3$ days instead of $ 2$ for a moment (more technically, we were $ 2$ hours away from reaching the 3 day limit).

Regardless, this is our move: [b]3. ... e6[/b].
Again, apologies for the delay.[/quote]

i think its Zeta. its probably a typo on Temperal's behalf. if he wanted to give you a warning, he probably wouldve posted it here

This topic is dead... :(  I wanted to play...

Patterns sent me a message saying how his entire team is inactive due to school work (finals, I'm sure). As the current active team in this game, you are allowed to accept their temporary vacation or penalize them.

[hide="Psst"]It's best in everyone's interest to accept the vacation; be a good sport =P. If no one responds by February it is an automatic forfeit and Team Beta gets a win while Team Eta gets the draw.[/hide]

I apologize that we're not responding.

At this moment, most of us are still having finals, for this [i][u]is[/u][/i] the week some of us are taking it (I will be available to respond a move for the team by Monday or so after finishing up my finals AND taking the SAT this Saturday).

As a person who is still active for the team, I ask those of Team Beta to be patient for just a little longer, for they will get a respond and a move from us by at most Monday.  :wink:

I apologize I didn't post a move on the date I promised.
On behalf of my inactive team:

[b]7. ... Nd7[/b]

I guess this is what my team wanted: :maybe:

[b]8. Qd3[/b]

[quote="Patterns_34"]Team $ \beta$ has a replacement?
If so, it would be helpful if the opposing team (like us) was also aware.  :)

In addition, just to let the moderators know, the [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Chess/Chess_Tournament_Info/Round_1/Game_2]AoPSWiki page[/url] of our game hasn't been amended yet.[/quote]

Temperal is taking a few weeks off. I don't think he needs a replacement, we have four people left. I guess I will amend the Wiki page.

I certainly don't need a replacement. I'm just preparing for state MC and AMC 12...

I apologize we responded late. As for our move:
[b]8. ... a6[/b]

*nudge*

[size=9]random text here to make the message longer[/size]

[b]9. ... b5[/b]
I greatly apologize for the delay.  :(


If $ x \plus{} y \equal{} 5$ , $ x \plus{} z \equal{} 8$ and $ y \plus{} z \equal{} 1$ , what is the value of
$ x \plus{} y \plus{} z$ ?


Ειπα στην αρχη να μην ανοιξω αυτο το τοπικ και να βαλω το ερωτημα μου στο τοπικ που εχει ανοιξει ο Δημητρης με τιτλο Θεωρια Γραφηματων. Ειδα ομως οτι δεν εχει δωθει αποδειξη στο προηγουμενο θεωρημα που εχει βαλει, οποτε βαζω την απορια μου εδω και αν κρινουν οι μοντς σωστο το μετακινουνε αργοτερα. Οριστε αυτο που αναρωτιομουν...

Υπαρχει ευθεια αποδειξη στο θεωρημα του Ντιρακ (οχι απαγωγη σε ατοπο, επαγωγη κτλ.) που λεει οτι αν σε ενα γραφημα με ν κορυφες και καθε κορυφη να ειναι ενωμενη με ν/2 κορυφες τουλαχιστον, τοτε υπαρχει hamiltonian circuit.

Ααα, και χωρις χωρισμο περιπτωσεων, χρηση αλλων θεωρηματων ή λημματων...

Κάπως βιαστικά:

Η συνηθισμένη απόδειξη του Dirac μπορεί να γραφτεί χωρίς άτοπο: Από κάθε μονοπάτι σε λιγότερο από ν κορυφές μπορούμε να δημιουργήσουμε ένα μεγαλύτερο μονοπάτι (και η απόδειξη μας εξηγεί πως να το κάνουμε αν θέλουμε) και τέλος, κάθε μονοπάτι σε ν κορυφες μπορούμε να το μετατρέψουμε σε κύκλο Hamilton (και πάλι η απόδειξη μας λέει πως να το κανουμε).

Aν μπορεις γραψε μου την αποδειξη ή δως' μου ενα link γιατι θελω να τη μαθω και δεν βρισκω τιποτα...

Κεφάλαιο 4.2 [url=http://www.ecp6.jussieu.fr/pageperso/bondy/books/gtwa/gtwa.html]εδώ[/url].

Η απόδειξη είναι με άτοπο αλλά μπορεί να αποφευχθεί όπως εξηγώ στο προηγούμενο post.

Ευχαριστω, θα τη διαβασω.


calculus
derivative
real analysis
real analysis unsolved

when performing partial differentiation with respect to one variable, all the others are treated as constant. But what about numbers unrelated to variables.

for example 1 + x y^-1

does the 1 disappear as usual when diff w.r.t y?

its been so long since i've done this!

thanks

Yes.  It's constant in $y$, so it vanishes under differentiation.


floor function
number theory
relatively prime
rational numbers

Prove that there is no positive rational number $x$ such that \[x^{\lfloor x\rfloor }=\frac{9}{2}.\]

Put $ x\equal{}\frac{p}{q}$ with $ p,q$ relatively prime, and $ \lfloor x\rfloor\equal{}n$. Then we want that $ 2p^{n}\equal{}9q^{n}$, so $ 2|q$, for $ n>1$ this means the RHS has more than one factor $ 2$ so $ 2|p$, contradiction with relative prime.

So $ n\equal{}1$ (since $ n\equal{}0$ is impossible), so $ 2p\equal{}9q$, so $ x\equal{}\frac{9}2$, but then $ n>1$, contradiction.

Another way (though not as good):

Suppose $x$ is a positive rational. If $x<1 $ then $x^{\lfloor x\rfloor } = x^0 = 1 \neq 9/2 $. If $1\leq x < 2 $, $x^{\lfloor x\rfloor } = x^1= x \neq 9/2 $ since $9/2 > 2$. If $ 2\leq x < 3 $ then $x^{\lfloor x\rfloor } = x^2 \neq 9/2 $ since $3/\sqrt{2}$ is irrational. If $ 3\leq x $ then $ x^{\lfloor x\rfloor } \geq 3^3 = 27 > 9/2 $. So there are no positive rational solutions to $x^{\lfloor x\rfloor }= \frac{9}{2}$.

$ My $ $ solution $

$ (a,b)=1 $
$ x=\frac{a}{b} $
$ \lfloor x\rfloor=c $
$$ x^{\lfloor x\rfloor}=\frac{a^c}{b^c}=\frac{9}{2}\implies a^c=9,b^c=2\implies c=1,a=9,b=2$$
$ Then $
$ \lfloor x\rfloor=1,x=\frac{9}{2} $  No solution


geometry
3D geometry
tetrahedron
circumcircle
geometry unsolved

Let ABCD be a tetrahedron whose 4 altitudes are concurrent and let R be the radius of its circumcircle (sphere). Let $ h_1;h_2;h_3;h_4$ be the lengths of the altitudes corresponding A,B,C,D and $ R_1;R_2;R_3;R_4$ be the circumcircle's radius of the opposite faces of A,B,C,D. Prove that
$ \frac{1}{h_1\plus{}2\sqrt{2}R_1}\plus{}\frac{1}{h_2\plus{}2\sqrt{2}R_2}\plus{}\frac{1}{h_3\plus{}2\sqrt{2}R_3}\plus{}\frac{1}{h_4\plus{}2\sqrt{2}R_4} \geq \frac{1}{R}$.
When does equality occurs?

What's a [b]Tetrahedron[/b]? I don't know this definition!

4 points in a space define a Tetrahedron!:D


[i]A circle is placed within another circle, not necessarily the centre.  Find the area of all the circles that can be placed between the smaller and the larger circle, while touching both circles.[/i]

Masoud Zargar

The problem isn't clear. If what's wanted is the SUM of the areas of all such circles, then it's obviously infinite, since in the special case of concentric circles we have infinitely many circles with constant area (general case can be solved via integral calculus, but the resulting integral also diverges).

What's exactly wanted in the problem?

Well, let's imagine we have a circle $C_1$ placed somewhere in $C_2$.  Between the two circles, there is a gap with a certain area.  Now, between the area, we construct as many circles at possible that are tangent to both circles.  All the surrounding circles are also linked.  What is the area of these small tangent circles?

Masoud Zargar

Well then, it should've been said that each circle ALSO touches the two adjacent ones, besides the two given circles :)

Well, I usually lose lots of marks on math tests for communication.  :blush: :).  Now, let's begin solving for I do not have a solution.

But there's also another problem: such construction may not be possible if we want the added circles to be connected throughout. It can be easily shown in the case of concentric circles: if the center of the given circles is $O$, their radii $R$ and $r$, and the two adjacent new circles have centers $C_1$ and $C_2$, then (proof will be supplied if needed) $\angle C_1OC_2=2\arcsin \frac{R-r}{R+r}$, and $\frac{\pi}{\arcsin \frac{R-r}{R+r}}$ may not be integer.


calculus
integration
trigonometry
complex analysis
complex analysis unsolved

How to solve the integral $ \int_{0}^{\plus{}\infty}\frac{sen(x)}{x^p}dx$ for $ 0 < p < 2$?

 Please, I need some help!!!!!

Maple says
\[ \int _{0}^{\infty }\!{\frac {\sin \left( x \right) }{{x}^{p}}}{dx}\equal{}{
\frac {\sqrt {\pi }{2}^{\minus{}p}\Gamma  \left( 1\minus{}1/2\,p \right) }{\Gamma
 \left( 1/2\plus{}1/2\,p \right) }}\]


A. 342. Prove that for any p prime of the form 4k+1
[sqrt(p)]+[sqrt(2p)]+..[sqrt((p-1)p)]=(p-1)(2p-1)/3.

I know a similar problems was published here, and pbornstein gave a solution, but I can't remember it and can't find it on the forum.

Let $LPR(x)$ denote the least positive residue of $x$ modulo $p$.

First, observe that for all primes $p == 1 (mod 4)$, $(\frac{-x}{p}) = (-1)^{\frac{p-1}{2}} (\frac{x}{p}) = (\frac{x}{p})$.  Therefore, $\{LPR(x^2)\}_{x=1}^{\frac{p-1}{2}}$ can be paritioned into $\frac{p-1}{4}$ pairs of residues of the form $\{k, p-k\}$.  It follows that ${\sum_{1}^{p-1}\{\frac{x^2}{p}\} = \frac{2}{p} \sum_{1}^{\frac12 (p-1)} LPR(x^2)} = \frac{2}{p} * p(\frac{p-1}{4}) = \frac{p-1}{2}$.  Therefore,
$S = \sum_{1}^{p-1} [pk]$
$=(p-1)^2 - \sum_{1}^{p-1} [\frac{y^2}{p}]$ (because both equal the number of lattice points under the graph $y=\sqrt{xp}, 1 \leq x \leq p-1$)
$= (p-1)^2 - \sum_{1}^{p-1}\frac{y^2}{p} + \sum_{1}^{p-1}\{\frac{y^2}{p}\}$
$=(p-1)^2 - \frac{(p-1)(2p-1)}{6} + \frac{p-1}{2}$, by our above remarks,
$= \frac{(p-1)(2p-1)}{3}$, after simplification, which proves the required.

Neat problem.


there exist positive $ x_{i}$ for $ i=1,2,...,n$ with sum $ S$ such that $ a_{i}=\frac{S-x_{i}}{x_{i}}$
now $ \left(\sum \sqrt{\frac{S-x_{i}}{x_{i}}}\right)^{2}\geq n(n-1)^{2}+\sum \frac{S-x_{i}}{x_{i}}\geq (n-1)^{2}\left(n+\sum \frac{x_{i}}{S-x_{i}}\right)$
thus we are left with showing
$ n\geq \sum \sqrt{\frac{x_{i}x_{j}}{(S-x_{i})(S-x_{j})}}$  :wink:


inequalities
quadratics
geometry
inequalities unsolved

I see that the inequality resembles discriminat of a quadratic, but experimenting, I couldn't solve it.

I also wonder if Schur (generalized perhaps) could solve it,

or perhaps just multiplying it all out? (I tried to use brute force, but I couldn't get it) :blush:

I find the inequlaity very hard and unfortunately cannot really solve it.
Now the fact that the inequlaity resembles a quadratic means that if you can find at least one real solution for $ x$ in the equation:
$ (a_{1}a_{2}\plus{}a_{1}a_{3}\plus{}a_{2}a_{3})x^{2}\plus{}(a_{1}b_{2}\plus{}a_{1}b_{3}\plus{}a_{2}b_{1}\plus{}a_{2}b_{3}\plus{}a_{3}b_{1}\plus{}a_{3}b_{2})x\plus{}(b_{1}b_{2}\plus{}b_{1}b_{3}\plus{}b_{2}b_{3}) \equal{} 0$
you will be finished since that would mean the discriminant(which is  the LS of ineq) is non-negative...
However I don't know how to prove that yet...
where is this from?

[quote="rem"]Now the fact that the inequlaity resembles a quadratic means that if you can find at least one real solution for $ x$ in the equation:
$ (a_{1}a_{2}\plus{}a_{1}a_{3}\plus{}a_{2}a_{3})x^{2}\plus{}(a_{1}b_{2}\plus{}a_{1}b_{3}\plus{}a_{2}b_{1}\plus{}a_{2}b_{3}\plus{}a_{3}b_{1}\plus{}a_{3}b_{2})x\plus{}(b_{1}b_{2}\plus{}b_{1}b_{3}\plus{}b_{2}b_{3}) \equal{} 0$
you will be finished since that would mean the discriminant(which is  the LS of ineq) is non-negative...
However I don't know how to prove that yet...
where is this from?[/quote]

It is from IMO Longlist 1987 #4, proposed by Australia. All the other Australia problems were easy one-liners that year, so I wonder why this one is so hard.

Anyway, I tried letting $ a_{1}\equal{} b_{1}k_{1}$, $ a_{2}\equal{} b_{2}k_{2}$, $ a_{3}\equal{} b_{3}k_{3}$, and substituting $ k_{1}$, $ k_{2}$, $ k_{3}$ into the quadratic, and I couldn't get anything very good, but problably I was to tired. However, I did get something that ended up looking alot like generalized Schur....

Then I tried to multiply out the entire L.H.S. and R. H. S. and see if I could do anything to it. I got something else that looked like Schur....

I couldn't combine the two methods together, or make them work against each other to solve the problem so.... anyway, I was probably too tired

This is, indeed, a "one-liner" (in the sense of: one idea and it's trivial) assuming as known some facts about geometrical inequalities. The proof goes as follows:

I rewrite the problem using simpler notations:

[color=blue][b]Problem.[/b] Let a, b, c, x, y, z be nonnegative reals. Prove the inequality

$ \left(ay \plus{} az \plus{} bz \plus{} bx \plus{} cx \plus{} cy\right)^{2}\geq 4\left(bc \plus{} ca \plus{} ab\right)\left(yz \plus{} zx \plus{} xy\right)$,

with equality if and only if $ a: x \equal{} b: y \equal{} c: z$.[/color]

[i]Solution.[/i] According to the Conway substitution theorem ([url=http://www.mathlinks.ro/Forum/viewtopic.php?p=204393#204393]http://www.mathlinks.ro/Forum/viewtopic.php?t=2958 post #3[/url]), since a, b, c are nonnegative reals, there exists a triangle ABC with area $ S \equal{} \frac12\sqrt {bc \plus{} ca \plus{} ab}$ and with its angles A, B, C satisfying $ \cot A \equal{} \frac {a}{2S}$, $ \cot B \equal{} \frac {b}{2S}$, $ \cot C \equal{} \frac {c}{2S}$ (note that we [b]cannot[/b] denote the sidelengths of triangle ABC by a, b, c here, since a, b, c already stand for something different). Similarly, since x, y, z are nonnegative reals, there exists a triangle XYZ with area $ T \equal{} \frac12\sqrt {yz \plus{} zx \plus{} xy}$ and with its angles X, Y, Z satisfying $ \cot X \equal{} \frac {x}{2T}$, $ \cot Y \equal{} \frac {y}{2T}$, $ \cot Z \equal{} \frac {z}{2T}$. Now, by [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=204393#204393]http://www.mathlinks.ro/Forum/viewtopic.php?t=2958 post #3[/url] Theorem [b](c)[/b], we have

$ \cot A\cot Y \plus{} \cot A\cot Z \plus{} \cot B\cot Z \plus{} \cot B\cot X \plus{} \cot C\cot X \plus{} \cot C\cot Y$
$ \geq 2$,

what rewrites as

$ \frac {a}{2S}\cdot\frac {y}{2T} \plus{} \frac {a}{2S}\cdot\frac {z}{2T} \plus{} \frac {b}{2S}\cdot\frac {z}{2T} \plus{} \frac {b}{2S}\cdot\frac {x}{2T} \plus{} \frac {c}{2S}\cdot\frac {x}{2T} \plus{} \frac {c}{2T}\cdot\frac {y}{2T}\geq 2$,

what simplifies to

$ ay \plus{} az \plus{} bz \plus{} bx \plus{} cx \plus{} cy\geq 2\cdot 2S\cdot 2T$
$ \equal{} 2\cdot 2\cdot\frac12\sqrt {bc \plus{} ca \plus{} ab}\cdot 2\cdot\frac12\sqrt {yz \plus{} zx \plus{} xy} \equal{} 2\sqrt {\left(bc \plus{} ca \plus{} ab\right)\left(yz \plus{} zx \plus{} xy\right)}$.

Upon squaring, this becomes

$ \left(ay \plus{} az \plus{} bz \plus{} bx \plus{} cx \plus{} cy\right)^{2}\geq 4\left(bc \plus{} ca \plus{} ab\right)\left(yz \plus{} zx \plus{} xy\right)$,

qed. (the equality case is left to the reader).

[b]EDIT:[/b] The same problem has been discussed at http://www.mathlinks.ro/viewtopic.php?t=151682 .

  Darij

Thanks for everyone sharing their crazy smartness!  :omighty:

Before I came back to check on this post, I was playing around with the discriminant stuff, and I think I have a solution based on that theme.

So we must find a value for $t$ such that the quadratic (using Darij's notation),

$f(t)=(ab+ac+bc)t^{2}+(ay+az+bx+bz+cx+cy)t+(xy+xz+yz)$ is negative (or non-positive), since very big $t$ will result in a positive value for the quadratic.

This is accomplished by:

if $\frac{x}{a}=m$, $\frac{y}{b}=n$, $\frac{z}{c}=p$, and (without loss of generality) $m \geq n \geq p$, then $t=n$ suffices. The proof is just an easy matter of cancelling terms, and reduces to $0 \geq ac(n-m)(n-p)$

Sunchips :)

P.S. Darij-how did you learn all that crazy math, what is your secret? :wow:


geometry
logarithms
integration
calculus
calculus computations

Find the area bounded by $ e^{ \minus{} |x| } \minus{} \frac {1}{2}$ and $ |x| \plus{} |y| \equal{} \ln2$.
please help me to find the solution.

You might want to specify which area you want to find.

$ |x| \plus{} |y| \equal{} \ln 2$ itself encloses a region and the equation $ y_1 \equal{} e^{ \minus{} |x|} \minus{} \frac {1}{2}$ slices through that closed region. The question is which piece are we interested in: the portion above or below $ y_1$?

Looking at it graphically, there are two possible interpretations of this, the first of which is more likely:

The first is equal to:

$ 2\left (\int^0_{ \minus{} \ln 2} e^x \minus{} \frac {1}{2}dx \plus{} \frac {(\ln 2)^2}{2}\right )$

The second is equal to $ 2 (\ln 2)^2$ subtract the first.

can somebody tell me what are the points of intersections ??

They intersect at two points: $ (\ln 2,0),( \minus{} \ln 2, 0)$

There are two domain. The graph of $ y \equal{} e^{ \minus{} |x|} \minus{} \frac {1}{2}$ is like tractrix,

The small area $ S_1\equal{}2*\left(\frac {1}{2}(\ln 2))^2 \minus{} \int_0^{\ln 2} \left(e^{ \minus{} x} \minus{} \frac {1}{2}\right)\ dx\right) \equal{} \boxed{(\ln 2)^2 \plus{} \ln 2 \minus{} 1}$

The large area $ S_2\equal{}(\sqrt {2}\ln 2)^2 \minus{} S_1 \equal{} \boxed{(\ln 2)^2 \minus{} \ln 2 \plus{} 1}$.

@ eastymorton.

I agree that they intersect at $ (\ln 2,0),( \minus{} \ln 2, 0)$ but i feel that that the point $ \ \minus{} ln2 , 0$ is satisfied only when $ x,y < 0$ ie it is in the third quadrant with $ \minus{} y \equal{} e^x \plus{} 1/2$   and   $ x \plus{} y \equal{}\minus{} \  ln2$ . the curve $ \minus{} y \equal{} e^x \plus{} 1/2$ intersects the y baxis only at $ \minus{} 1/2$ .

so  i fell the the other curve which is now in the 2nd quadrant should be there in the 3rd quadrant,but now there is no area enclosed .

so how do we find it ..please help ansd pardon me if  i am wrong


What is the units digit of $ 1! \plus{} 3! \plus{} 5! \plus{} 7! \plus{} 9! \plus{} 11!$ ?

Every factorial after $ 5!$ ends in 0, so we just have $ 1 \plus{} 6 \equal{} 7$.

$ \ 1!$ ends in 1
3 ends in 6
5 ends in 0
7 ends in 0
and so on for 9, and 11.
so 6+1=7
Answer is 7
EDIT: Beaten(why am i always beaten :( )


If f(0)=2, f(1)=5, f(n+1)=f(n)-f(n-1) for all integers n such that n$\geq$ 1 find the value of f(7)

[quote="yeppyyep"]ok answer is really 5, you missed one more but that's ok. is it possible to write explicit formula for f(n)?[/quote]

[hide="Y Not?"]$f(n) = \left( 1-\frac{4i\sqrt{3}}{3}\right)\left( \frac{1+i\sqrt{3}}{2}\right)^{n}+\left( 1+\frac{4i\sqrt{3}}{3}\right)\left( \frac{1-i\sqrt{3}}{2}\right)^{n}$

where $i=\sqrt{-1}$[/hide]

[url]http://www.cs.cmu.edu/~adamchik/21-127/lectures/recursion_4_print.pdf[/url]

But that's definitely way more trouble then it's worth in this case.


calculus
derivative
trigonometry
function
calculus computations

Um...For what values of $ x$ does the graph of $ f(x) \equal{} x\plus{}2\sin{x}$ have a horizontal tangent?

So, the derivative of that function is $ 1\plus{}2\cos{x}$, so $ \cos{x}\equal{}\frac{\minus{}1}{2}$. Solving for $ x$, I get $ \frac{2\pi}{3}$.

At this point, I understand that there are infinite solutions, seperated by some interval related to $ \pi$ in some way, but I don't remember how to express them the correct way. Help please?

$ \cos (x)\equal{}\minus{}\frac{1}{2}\equal{}\cos\left(x\plus{}2\pi c_{1}\right)$

$ x\equal{}2\pi c_{1}\minus{}\frac{2\pi }{3}\lor x\equal{}2\pi c_{1}\plus{}\frac{2\pi }{3}$

For periodic functions, just find trivials solutions and add a integer multiple of the period.

My book says the solution is $ (2n\plus{}1)\pi\pm\frac{\pi}{3}$ That's the same as your solution, right?


floor function
number theory proposed
number theory

Let $ a_i$ be a non-negative integer sequence such that
\[ a_{k \plus{} 1} \equal{} \lfloor a_k(1 \minus{} \frac {1}{p_k})\rfloor\ \forall k\in \mathbb N
\]
Suppose $ n\in \mathbb N,a_1 \equal{} p_n \minus{} 2$, then prove that $ a_n \equal{} 0$

[hide="Terminology"]1) $ p_m$ denotes the $ m$-th prime number.
2) $ \lfloor x\rfloor$ denotes the maximum integer $ \le x$[/hide]
[hide="Examples"]
$ a_1\equal{}p_3\minus{}2\equal{}3,a_2\equal{}1,a_3\equal{}0$

$ a_1 \equal{} p_9 \minus{} 2 \equal{} 21,10,6,4,3,2,1,0,0 \equal{} a_9$

$ a_1 \equal{} p_{13} \minus{} 2 \equal{} 39,19,12,9,7,6,5,4,3,2,1,0,0 \equal{} a_{13}$

$ a_1 \equal{} p_{25} \minus{} 2 \equal{} 95,48,32,25,21,19,17,16,15,14,13,\cdots ,3,2,1,0,0 \equal{} a_{25}$[/hide]

No one interested? :maybe:
I can tell you that it's not a random conjecture..., I have some ideas, but I want to know your solutions, opinion.

O.K. I am joining! :lol:

I found a recursion like expression, probably you have already found it....
But it is hard to continue with it because of the greatest integer function. I think it will be better for me to think over it if you post the idea from which you generated this extraordinary problem.

[hide="click here"]P/s: Tahole dekhbe besh kichu lok agrohi hoye ei topic e join koreche.....tumi ki etar opor kono seminar e bolbe? Tahole beshi na janano ekdike valo...kintu ekdom kono idea nei to....fole kotha theke suru korbo kichui bujhte parchina, atleast tomar kivabe mone holo seta bolo...[/hide]


[quote="shadysaysurspammed"]Find $n$ such that....
$-5^4+5^5+5^n$ is a perfect square.[/quote]

N=5.

-5^4 + 5^5 + 5^5 = 75^2

[hide]$-5^{4}+5^{5}+5^{n}=x^{2}$, for some integer $x$.  Clearly $n$ must be a nonnegative integer, or the expression would not be an integer.

$5^{4}(-1+5)+5^{n}=x^{2}$
$50^{2}+5^{n}=x^{2}$
$x^{2}-50^{2}=5^{n}$
$(x+50)(x-50)=5^{n}$

Searching for two powers of 5 with a difference of 100, we find that 25 and 125 are the only ones.  Therefore, $5^{n}=(25)(125)\Rightarrow n=5$, and there are no other solutions.[/hide]

i did this the same way as matt276eagles.
another problem like that is,
find $n$ such that $2^4+2^7+2^n$ is a perfect square. (i found this and the previously posted problem in the art and craft of problem solving.)

[quote="maokid7"]
find $n$ such that $2^4+2^7+2^n$ is a perfect square. [/quote]

Again...
$2^n=(m+12)(m-12)$
8 and 32 are 2 powers of 2 which differ by 24....so
$2^n=2^3*2^5$
$n=8$


analytic geometry
graphing lines
slope
geometry
3D geometry
trigonometry
calculus

Can someone post the solutions to the following problems?
1.  A graphing calculator has on its display pixels that are 0.01 apart horizontally.  Find the value of n such that the graph of sin(nx) will be graphed as a horizontal line: $n=\frac{100}{\pi}$, $100$, $100\pi$, $\frac{200}{\pi}$, $200\pi$.

2.  $\displaystyle N=\{(a,b) | b < \frac{1}{1+a^2}\}$ is a subset of $R^2$.  Given any line contained in N, what is its maximum slope?

3.  John is happily making a right circular cone of volume 10 cubic feet, with a base of radius $r$ and a perpendicular height $h$.  What is the approximate dimension (r,h) of the cone which would minimize the construction costs if the material cost for the side of the cone is 34 cents per square foor, and for the bottom the cost is 44 cents per square foot.

[hide="1."]
If it appears as a horizontal line, then $\sin{(nx)}$ has a period of $.01$, or $\sin{(nx)}=\sin{(n(x+0.01))}$ so $nx+2\pi=nx+0.01n \Rightarrow n=200\pi$.[/hide]

Oh, good solution.  Any solutions for the other two?

For #2, does it have to be a whole line (extending to infinity) or can it be a line segment?

3)
[hide]
I'm not sure if having different costs for material will affect anything.  This may be completely wrong but:
$V = \pi r^2h = 10$
$h = 10/(\pi r^2)$

$A = 2\pi r^2 + 2r\pi h$
$A = 2\pi r^2 + 2r\pi(10/(\pi r^2))$
From here I think you'd need calculus or a graphing calculator to solve the problem.
Here's a calculus solution just for fun.  I hid it since we're not suppose to post calculus in this forum:
[hide]
$\frac{dA}{dx} = 4\pi r - 20r^{-2}$
You want to find critical numbers:
$0 = 4\pi r - 20r^{-2}$
$0 = 4r(\pi - 5r^{-3})$
$4r = 0$
$r = 0$
$\pi - 5r^{-3} = 0$
$\pi = 5r^{-3}$
$\frac{\pi}{5} = \frac{1}{r^3}$
$\sqrt[3]{(\frac{\pi}{5})} = \frac{1}{r}$
$\frac{1}{\sqrt[3]{(\frac{\pi}{5})}} = r$
The function decreases on the interval from 0 to r and then increases from r to infinity so this must be a minimum.
$h = 10/(\pi * r^2)$
$h = 2.335$
$(r,h) = (1.168, 2.335)$
[/hide]
[/hide]

Thanks, SnowStorm; however, since it's a cone, it should be V=(1/3)(pi*r^2*h)=10.

Does anyone have a solution to #2?  The answer, by the way, is:
[hide]0[/hide]

Also, here's another problem I need a solution to :) :
How many numbers are there that satisfy the following equation?
$\displaystyle n+n^2+\frac{1}{n}+\sqrt{n}=10$
(these are just numbers, not integers or postive, etc.)

I got 2, but i'm not sure how to prove it without using a quasi-calculus solution.

[hide="2."]
Consider the graph $b = \frac{1}{1+a^2}$. It has a horizontal asymptote at $b=0$. If $b < \frac{1}{1+a^2}$, the line must be under the horizontal asymptote for any $a$. However, if a line has a non-zero slope, its range is all reals and it won't satisfy the inequality. Thus the only slope of a line contained in $N$ is $0$.[/hide]

so, yea, i turned it into a 6th order polinomial (your last question). From there, i decided to let my calculator solve it for me....both of these are estimates:

n= 1.62476
n= 1.98750

complex:
n= -1.22952+1.56676i
n= -1.22952-1.56676i
n= -1.57665+1.95906i
n= -1.57665-1.95906i

im sure i didnt help at all...but...im gonna post anyway


geometry
symmetry
incenter
angle bisector

The quadrilateral $ ABCD$ inscribed in a circle wich has diameter $ BD$. Let $ A',B'$ are symmetric to $ A,B$ with respect to the line $ BD$ and $ AC$ respectively. If $ A'C \cap BD \equal{} P$ and $ AC\cap B'D \equal{} Q$ then prove that $ PQ \perp AC$

call $ E: AB \cap CD$ so $ \angle BPC \equal{} \angle BCA' \minus{} \angle CBD \equal{} \angle ADB \minus{} \angle CBD \equal{}$
$ \equal{} 90 \minus{} \angle ABD \minus{} \angle CBD \equal{} 90 \minus{} \angle ABC \equal{} \angle BEC$.
Then BCPE is cyclic and $ EB \perp BD$.

Then the problem is equal to this:

H is the orthocenter of ABC and AH,BH,CH meet respectively BC,CA,AB on D,E,F. Call H' the symmetric of H w.r.t. ED and ED meet AF on G and AH' on K. Then $ KF \perp ED$.

We can prove it in this way: F is the harmonic cojugate of G w.r.t. AH, but G on $ \triangle AKH$ is the feet of the internal angle bisector of $ \angle AKH$, so F is the feet of the external angle bisector of $ \angle AKH$ and then $ FK \perp KG$.

Denote AC meet BD at G,  B.G.D.P is harmonic division GC is bisector of angle BQC

not a hard one
Hint : $AC \cap BD=T$ and $BB' \cap AC=L$ . now use Menelaus theorem for$\bigtriangleup BB'D$ and point $Q$ and use sin law for $\bigtriangleup BCD$ and point $P$ then you will see that $\frac {PB}{PD}=\frac {QB'}{QD}=\frac {BT}{TD}$ so done.

By symmetry, $\Delta AB'Q=\Delta ABQ\implies \angle BQT=\angle DQT$, so we need to show that $\frac{BP}{PD}=\frac{BT}{TD}$, where $T=AC\cap BD$.
We see easily that $D$ is the incenter of $\Delta ACP$, so $D, B$ are harmonic conjugate w.r.t. $\overline{PT}$, hence $\frac{BT}{BP}=frac{TD}{PD}\iff\frac{PD}{PB}=\frac{TD}{BT}$; since $TQ$ is internal angle bisector of $\angle BQD$, it follows that $PQ$ is the external bisector of $\angle BQD$ and $PQ\bot AC$.

Best regards,
sunken rock

Let $AC\cap BD= E $ and $BB'\cap AC = M $. Also, let $BB'\cap PQ = X $.
It is easy to see that $A'$ lies on $\odot (ABCD) $. As, $BA = BA'$, so, $BC $ bisects  $\angle PCQ $. Thus, $(-1) = (P,E;B,D)\stackrel{Q}{\doublebarwedge}(P,E;B,D)\stackrel{Q}{\doublebarwedge}(X,M;B,B')$. But, $M$ is the midpoint of $BB'$. Hence, $X $ is a point at infinity and so, $BB' || PQ $. This gives $\angle PQC = 90^{\circ} $.

Let $\{L\}$=$BA'\cap CD$, $\{T\}$=$AC\cap BD$, $\{Q\}$=$BC \cap A'D$.
Simple angle chasing gives $T,Q,L$ colliniar and $LQ \perp BD$.

By a well, known lemma, we get that $QT$ is the polar of $P$, so $P$ is on the polar of $T$ which gives $(P,T,B,D)=-1$.

Now, all we have to prove is that $\frac{BT}{TD}=\frac{QB'}{B'D}$, which is Menelaus' theorem in triangle $BB'D$ with line $QT$ $\blacksquare$

Well though the computations took a while
They were pretty straightforward and simple
And ended nicely
I blindly used complex numbers and solved it.


A random pole:

Do you usually type in the BBCode or do you click the icons?

Its becoming more popular these days to let people type their own HTML in their messages, (of course, drastically limited/cleaned etc). I quite like that

I use both. I type the bold tags and the easy ones. When I use hide, I click it. To date I've only used bold, underline, and hide on this forum (as far as I remember). I've used quote but only quoting from the thread

I got so confused once while typing my biology lab in Microsoft Word I started to type "[ b ]" instead of Ctrl B.  :D


limit
real analysis
real analysis solved

Denote by p_n te sequence of prime numbers.
Calculate lim n->oo from [product from k=1 to n from (n+k)^1/p_(n+k)]

Let Q(n) = prod_(k=1 to n) (n+k)^(1/p_(n+k))
x(n) = lnQ(n) =  \sum _(k= 1 to n) ln(n+k)/p_(n+k)
x(n) = n.(x(n)/n)

I am going to get an equivalent of x(n)/n with a mixture
equivalent technic and Cesaro-Stolz technic that I learned on
mathlinks  :D

(x(n) - x(n-1))/(n - (n-1)) = ln(2n)/p_(2n) but the n-th prime p_n is equivalent when n tends to infinity to n.lnn

(x(n) - x(n-1))/(n - (n-1)) ~ ln(2n)/(2n.ln(2n)) ~ 1/(2n)

x(n) =  n.(x(n)/n) ~ n(x(n) - x(n-1))/(n - (n-1)) ~ n/(2n)

and in 6th grade I learned  n/(2n) = 1/2 (if n=/=0)

 \lim x(n) = 1/2
 \lim Q(n) =   \sqrt e


integration
trigonometry
function
calculus
derivative
LaTeX
absolute value

$ \int_0^\infty\mid e^{\minus{}x}\sin x\mid\,dx$

Split it into the intervals $ [n \pi, (n \plus{} 1)\pi]$.  The function has a nice enough anti-derivative between points where the sign changes, so you can compute on each interval with no problem.  Then sum the resulting series (for which it will help to know that $ \sin x \equal{} \frac {e^{ix} \minus{} e^{ \minus{} ix}}{2i}$).

For purposes of calculus, absolute value is just another piecewise linear function, and piecewise linear functions are best dealt with by splitting at the joints and working separately on each piece.

I have

$ \int_0^\infty\mid e^{ - x} \sin x \mid \, dx$

$ = \int_0^\pi e^{ - x} \sin x\, dx - \int_\pi^{2\pi} e^{ - x} \sin x\, dx + \int_{2\pi}^{3\pi} e^{ - x} \sin x\, dx - ...$

$ = \sum_{n = 0}^\infty (\int_{2n\pi}^{(2n + 1)\pi} e^{ - x} \sin x\, dx - \int_{(2n + 1)\pi}^{(2n + 2)\pi} e^{ - x} \sin x\, dx)$

$ = \sum_{n = 0}^\infty (\frac { - e^{ - x}}{2} (\cos x + \sin x)]_{2n\pi}^{(2n + 1)\pi} - (\frac { - e^{ - x}}{2} (\cos x + \sin x)]_{(2n + 1)\pi}^{(2n + 2)\pi})$

$ = \sum_{n = 0}^\infty (\frac {e ^{ - 2n\pi}}{2} - \frac {e ^{ - (2n + 2)\pi}}{2})$

$ = \frac {1}{2}( 1 - e ^{ - 2\pi} + e ^{ - 2\pi} - ...)$

$ = \frac {1}{2}.$

Some intermediate steps not written as it took about three minutes to work out (and is probably wrong) and over twelve minutes to $ \text{\LaTeX}$ not very well, during which time (I've just spotted) JBL posted. I must learn to $ \text{\LaTeX}$.

Well, actually the summing part is simpler than I suggested (I somehow didn't notice that the values of sine and cosine are all at integer multiples of pi  :| )  Somewhere you lost the sign you introduced, I think, but I'm not sure where -- $ \frac{1}{2}$ is the value of the integral [i]without[/i] the absolute value.

Thanks, Immanuel.

I got a sign wrong and now make it

$ \frac {(1 \plus{} e^{ \minus{} \pi})^{2}}{2} \sum_{n \equal{} 0}^\infty e^{ \minus{} 2n\pi}$,

(that is, half your answer)

$ \equal{} \frac {(1 \plus{} e^{ \minus{} \pi})^{2}}{2}(\frac {1}{1 \minus{} e ^{ \minus{} 2\pi}})$

$ \equal{} \frac {(1 \plus{} e^{\minus{}\pi})}{2(1 \minus{} e^{ \minus{} \pi})}$

which is approximately $ 0.545$.

Have I still got it wrong?

You're welcome. Sorry forgetting the half...


Is there any place to get the National Competition videos online?

Thanks.

The 2006 competition wasn't broadcasted but you can find a short 2005 clip [url=http://www.mathcounts.org/webarticles/anmviewer.asp?a=765&z=71]here[/url]

I have videos of myself in countdown somewhere...

But doesn t that only have the final round and interviews?

its the only one published

http://www.mathcounts.org/webarticles/anmviewer.asp?a=765&z=71

[quote="Klebian"]http://www.msnbc.msn.com/id/12772703/

Click on the picture for the video, that's all we've got, really...[/quote]

isnt it the same video

no, it s not.

There s another video.

[url]http://www.wusatv9.com/video/player.aspx?aid=31628[/url]

I found another one:
http://www.raytheon.com/about/contributions/mathmovesu/

Also, I tried Google Video and YouTube, but neither have MATHCOUNTS Videos.

Thank You very much.  We appreciate the videos you guys put.

I only saw the video online at MathCounts, and that was pretty short.

[quote="Ignite168"]The one on the MathCounts site is for 2005 and the one on the MathMovesU site is for 2004.[/quote]

yes. i should have mentioned earlier. :oops:

Is there a Mathcounts Movie? Mr. Rusczyk gave a link to something on idmb that said he starred in the Mathcounts Movie 2004 adn 2005...

It's the 60 minute show that was shown on ESPN with highlights of the National Competition in 2004 and 2005 not a movie.


It's almost here.

Who do you think/believe will win? who are you cheering for?

Me: cheering for Agassi(if he plays), Safin, Nadal, Sharapova, Dementieva, Vaidisova and Clijsters

Kim Clijsters injured her foot, don't know if she will play or not.

this is going to be the boringest Aus open ever.

[quote="manuel"]this is going to be the most boring aus open ever...
nadal, agassi and safin are out...

thats what i heard

federer can win it playing with his left hand[/quote]

you never know...federer has lost already once this year. and a sleeper might come out and beat federer. it's possible (hopefully andy roddick :D )

i love these threads...
i want roddick to win, or hewitt, but i think federer will win

by the way, this is a little off topic, but any seniors here gonna play college tennis?

im going to play in mit yayyyy... practiced with the team last month..

I heard dementieva went as well......less star studded aus open this year,sad burt it'll be awesome if ivan ljubicic wins,my fav out of the playing lot.Safin isn't playin :( He's so cool.

How many say federe will win?

Dissappointing, but not terribly suprising. I mean, she's played like what, 3 matches since Wimbledon?

who wud have saw that coming?
L. Hewitt beat by Chena? Ouch! :stretcher:

Roddick gone  :(

I think american men's tennis era is probably over, women too.

Oh yeah!!!I like the guy who beat roddick.Seen him for the first time.

ok , fed's won already.  :(  all gone , hewitt ,roddick..... maybe ljubicic can do something :huh:

tribute to haas , he had fed really screwed up, too baaadd fed won :(

nice goin haas  :coolspeak:

[b]Alright guys..BIG TIME UPDATE!!![/b]
BAGDATIS is in the FInal!!!Wooohoooooo...he beat nalbandian.Brought him down to earth bagdatis.Federer's in the final.

And in the [i]ladies[/i]..Our new champ....Amelie.. .(Finally she got a title).mauresmo!!!!!

at the moment i'm seeing paes/damm vs buggin brothers..oops bryan bros...first set paes/damm second bryans and third its 2 games all.

I feel bad for Mauresmo, not being able to win it, to have that moment. But great sportsmanship when Amelie went up to sit by Justine to find out how she was feeling. That was nice to see.

:ewpu: sounds confusinggg, smelly fart sayin mauresmo wins the title , while Master says "not being able to win it." .....hey guys , so who won? i missed the match and now  i cant find score on TV anywhere :(

Federer won at last :first:

However it was some real good tennis from Bagdatis :winner_second:

5-7:7-5:6-0:6-2
Wow!!!

[quote="The Master"]Sorry, maybe I didn't include as many details as I should have.

Basically, Henin-Hardenne retired because she was not feeling well. Henin said it was a stomach-ache. That's why I said I feel sorry for Mauresmo, not having that moment, not being able to 'win' it, but it kind of being handed to her.

Many people believe that Henin should have just 'sucked it up', and played through the pain, at least finish the match.[/quote]

I'm disgusted by the opinions that the pundits are throwing out.  Before this match, anyone who had watched even a tiny amount of women's tennis in the last three years would call her the grittiest player in the womens' game, but she retires from a big match after playing a pair of tight three setters the previous couple of matches and everyone throws her under the bus.

[quote="blahblahblah"]I'm disgusted by the opinions that the pundits are throwing out.  Before this match, anyone who had watched even a tiny amount of women's tennis in the last three years would call her the grittiest player in the womens' game, but she retires from a big match after playing a pair of tight three setters the previous couple of matches and everyone throws her under the bus.[/quote]Ya, sometimes people are too quick to judge without thinking. I don't think any professional athlete would retire from a match like that without very good reason. It is not in their nature.


number theory
greatest common divisor
algebra unsolved
algebra

let $ a_n \equal{} n^2 \plus{} 20$ and $ d_n$ denotes the greatest common divisor of $ a_n$ and $ a_{n \plus{} 1}$
 then prove that $ d_n$ divides $ 81$.


geometry
circumcircle
geometric transformation
reflection
trigonometry
geometry proposed

Since the reflection $ D$ of $ H$ in $ B_{0}$ is the second intersection of $ BO$ with the circumcircle, the lines $ HB_{0}$ and $ PQ$ are parallel if and only if the pencil $ (PQ, PH, PB_{0}, PD)$ is harmonic. The conclusion follows from the fact that $ BH \cap (PQ, PH, PB_{0}, PD) \equal{} (B, Q, H, B_{1}) \equal{} \minus{} 1$, where $ B_{1}$ is the foot of the $ B$-altitude in triangle $ ABC$.

Just a question to clarify the brilliant solution of Pohoatza: is point D the second intersection of BO with the circumcircle? Thanks

My solution is metric. If B_1 is the foot of the altitude from B, we can consider triangles B_1B_0H and B_1PQ. Then it suffices to prove the equality B_1Q/B_1H = B_1P/B_1B_0 . Both quotients are equal to 2sinAsinC/cos(A-C).

If $ \{T\}\equal{}HB_0\cap BO$ then $ \frac{BH}{BT}\equal{}\frac{2R\cos{B}}{2R}\equal{}\cos{B}$.
In addition, $ \triangle{BC_1Q}\sim \triangle{BCP}$, thus $ \frac{BQ}{BP}\equal{}\frac{BC_1}{BC}\equal{}\cos{B}$.
Thus, $ \frac{BH}{BT}\equal{}\frac{BQ}{BP} \Longleftrightarrow PQ\parallel{}HB_0$.

Construct the diameter $ BOT$. In fact that the quadrilateral $ AHCT$ is a parrallelogram. We have :
$ \frac{BP}{PH}\equal{}\frac{[BMN]}{[HMN]}\equal{}\frac{BM.BN}{HM.HN}$
$ \frac{BQ}{BT}\equal{}\frac{[BAC]}{[TAC]}\equal{}\frac{BA.BC}{HA.HC}$
Easy to proof that $ \frac{BM.BN}{HM.HN}\equal{}\frac{BA.BC}{HA.HC}$
Hence, by Thales theorem we get that $ PQ // HT  \Rightarrow PQ // HB_{0}$

Let BT be a diameter of (ABC); it’s easy to see that $ \triangle BAT \sim  \triangle BA_{1}H$ and P and Q are homologues points, hence the conclusion follows.

Best regards,
sunken rock

[quote]In a scalene triangle $ ABC$ the altitudes $ AA_{1}$ and $ CC_{1}$ intersect at $ H, O$ is the circumcenter, and $ B_{0}$ the midpoint of side $ AC$. The line $ BO$ intersects side $ AC$ at $ P$, while the lines $ BH$ and $ A_{1}C_{1}$ meet at $ Q$. Prove that the lines $ HB_{0}$ and $ PQ$ are parallel.[/quote]
[b]Solution:[/b] Let $BP \cap HB_0=X$. Let $\ell || HB_0$ through $P$ intersects $BB_1$ at $Y$
$$-1=(H, ~ X ; ~B_0, ~ \infty_{HX}) \overset{P}{=} (H, ~ B ; ~ B_1, ~Y)$$
But, $$-1=(A , ~ C ; ~ A_1C_1 ~ \cap ~ AC, ~B_1)  \overset{B}{=} (C_1, ~ A_1 ; ~A_1C_1 ~ \cap ~ AC , ~ Q)  \overset{C}{=}   (H , ~ B ; ~B_1, ~Q)$$
Hence, $Y \equiv Q$

Just compute everything!

Let $D$ be the foot of the $A$-altitude, and let $A'$ be the $A$-antipode; it suffices to show $\frac{AQ}{AP} = \frac{AH}{AA'}$. Notice that $\angle AQB_1 = 90^\circ - B + C$, so $$\frac{AQ}{AB_1} = \frac{\sin B}{\cos(B-C)} \implies AQ = \frac{2R \cos A \sin B\sin C}{\cos(B-C)}.$$ Also, $$AP = \frac{AD}{\cos(B-C)} = \frac{2R\sin B\sin C}{\cos(B-C)}.$$ It is evident now that $\frac{AQ}{AP} = \cos A = \frac{AH}{AA'}$, as needed.

[color=#9a00ff][b]Remark.[/b][/color] It turns out that $\overline{AQ}$ and $\overline{AP}$ are similar parts in the inverse similarity $\triangle AB_1C_1 \sim \triangle ABC$, which explains a lot.

huh
[hide = solution]
Consider the map $\tau$ which reflects a point in the bisector of $\angle ABC$ and dilates with a factor of $\cos B$ wrt to $B$. Let $B_1$ denote the $B$-antipode of $\triangle BAC$. Remark that under $\tau$ the segment $BB_1$ maps to $BH$ since triangle $\triangle BAC$ goes to $\triangle BA_1C_1$. In particular $P = BB_1 \cap AC$ goes to $Q = BH \cap A_1C_1$. Thus $BP/BB_1 = BQ/BH$ from where it follows that $PQ \parallel HB_1$. Since $H, B_0, B_1$ are collinear this finishes.
[/hide]


I have a problem getting set up with latex i dont know what file to chose for the Distrabution Directory could someone please help?

On top of your screen is a button leading you to an explanation site :

[url]http://www.mathlinks.ro/LaTeX/AoPS_L_About.php[/url]

part of the forum is the latex forum where questions like this belong :
[url]http://www.mathlinks.ro/Forum/index.php?f=123[/url]

Sorry didnt see that part of the forum. And ive looked at the site and it dosent help
Thank though


2. Al gets the disease algebritis and must take one green pill and one pink pill each
day for two weeks. A green pill costs $ \$1$ more than a pink pill, and Al's pills
cost a total of $ \$546$ for the two weeks. How much does one green pill cost?

$ (A) \$7 \qquad (B) \$14 \qquad (C) \$19 \qquad (D) \$20 \qquad (E) \$39$

[hide]If it costs 546 dollars for 14 days, then it should cost 39 dollars per day. This means that a green pill is 20 dollars, while a pink pill is 19. [b]A green pill costs 20 dollars. [size=150]D[/size][/b][/hide]

Let $ x$ be the cost of a green pill; then $ x \plus{} (x\minus{}1) \equal{} \frac{546}{2 \times 14} \equal{} 39 \Longrightarrow x \equal{} 20$.

I actually got this question wrong when taking a practice test by misreading.  :oops:


function
algebra
domain
complex analysis

Let $ f$ be analytic in a neighborhood of the set $ Im(z)\geq 0$ and assume $ f(i)=0$
and that $ f(z)\leq 1$ when $ Im(z)=0$. Show that
\[ |f(z)| \leq \left|\frac{ (z-i) }{ (z+i) }\right| \]
for all $ z$ with $ Im(z)\geq 0$

Basically you just want to use Schwarz's Lemma, but to put it into the correct form you need Cayley's transform.

So note, that $ M(z)=\frac{z-i}{z+i}$ (Cayley's transform) maps the upper half-plane conformally onto the unit disk.  Then let $ g(z)=f(M^{-1}(z))$ and note that $ g: D \rightarrow D$.  That is actually a little tricky since you were only given $ f(z)\leq1$ when $ \Im z=0$, but you can deduce it by realizing that conformal maps map boundaries into boundaries and knowing that $ f(i)=0$ maps an interior point into an interior point.

Now we have $ |g(z)| \leq 1$ and $ g(0)=0$.  Therefore, $ |g(z)| \leq |z|$ by Schwarz's Lemma.  That is equivalent to:  $ |f(M^{-1}(z))| \leq |z|$.  Replacing $ z$ by $ M(z)$ gives $ |f(z)| \leq \left| \frac{z-i}{z+i}\right|$ for $ \Im z \geq 0$.

[quote]That is actually a little tricky[/quote]
That part is so tricky that there is a counterexample: $ f(z)=\frac{e^{-i z}-e}{1+e}$.

Thanks Mlok for the counterexample.

So it looks like if we change Color's condition that $ |f(z)| \leq 1$ when $ \Im z =0$ to $ |f(z)| \leq 1$ when $ \Im z \geq 0$, then the proof I provided will work.

This does raise a question of my own though.  For a function $ f$ satisfying the original conditions of the problem, I mentally imagined the boundary curve (the real axis) being conformally mapped by $ f$ to a simple closed curve contained in the unit disk or possibly being the boundary of the unit disk itself.  Since I knew one interior point of the upper half-plane mapped into the interior of this closed curve, I thought all of the points would as well via a Jordan curve argument.

So my question is, under what conditions can we be assured that will be true?  Or do I need something like my boundary curve not containing the point at $ \infty$?

[quote]Since I knew one interior point of the upper half-plane mapped into the interior of this closed curve[/quote]
But we don't know that. In my counterexample the image of the real axis is indeed a simple closed curve [namely a circle] which lies in the unit disk -- but $ 0$ is outside of the curve.

For unbounded domains the maximum principle is replaced by the [url=http://en.wikipedia.org/wiki/Phragm%C3%A9n-Lindel%C3%B6f_principle]Phragmén-Lindelöf principle[/url]. The wikipedia page gives a rather sketchy special case. When working with the sector $ \{z\colon |\arg z|<\beta\}$, you need $ |f(z)|$ to be majorized by $ \exp(c|z|^{\alpha})$ with $ \alpha<\frac{\pi}{2\beta}$.

An accessible reference: [i]Potential Theory in the Complex Plane[/i] by T. Ransford.


The room rate at the Detroit Marriott hotel for Guests at MATHCOUNTS Nationals is 129 dollars/night + 15% tax.  How much money could your Dad/Uncle/Grandpa save if they share a room with me?  How much if two people share a room with me?

OK, so this isn't a very interesting math question, but I'm talking real money.  I'm the father of a nationals competitor and I've reserved a room for four nights (arriving May 4th). I am looking for someone (adult male) to share a room and hotel expenses for some or all of those nights.  So, if you have any relatives who are coming and want to ease the cost, PM or e-mail me please.

Thanks, and good luck to you all at nationals!

David

[quote="dwx314"]So, first of all, this 129 figure, is this for sharing, or a separate room?[/quote]

That's the room rate.   One extra person is no extra charge.   I didn't ask about beyond double occupancy.  The room has two beds, and is nonsmoking.  So I'm looking at roughly 600 dollars for the single occupancy room for four nights.  That's a lot more than the airfare, so I'd love to share expenses.

[quote="dwx314"]but don't you have to room with a fellow teammate though?[/quote]

It should be apparent I'm a parent.  (See first post, and sorry for the pun)  Competitors all room with other competitors, and there is no charge for them of course.  But guests (parents, etc.) pay for rooms.

So, a night =159.85
4 nights = 639.4
Wow. Thats a lot to save.

Sorry for confusing you with a competitor. (entropy2?! how obvious could it get?)

Good luck to your son/daughter. Too bad I didn't go. (can you tell I'm a student)

[hide="the answer"]593.40 is the total cost for 4 nights, so divide by 2. . . 296.70![/hide]

Thanks for wishing me luck, dwx.  I stole my dad's username (meaning he uses entropywins other places) so he had to use a new one.  We were in the mathcounts class together btw  :D


I need really good help on painting cubes here.

I have like really few senses on cubes and whenever this appears on AMC-10, it's like dead killer to me.

Can anyone post some introduction and then some intermediate and so forth so I can increase my cubic ability?

Thank you very much!!!

Well.. here are the VERY basics.. (this is all I know)

For a painted cube divided into n^3 smaller cubes...
Number of smaller cubes with [b]NO [/b]paint on them = (n-2)^3.
(Imagine a smaller cube shielded from paint by the cubes on the surface)

Number of smaller cubes with paint on [b]one [/b]side = 6(n-2)^2
These cubes are on the faces of the large one,
but we can't count the cubes on the corners or edges,
so imagine a smaller square with side length n-2 for each of the 6 faces of the original cube.

Number of smaller cubes with paint on [b]two[/b] sides = 12(n-2)
A cube has 12 edges, and each edge has n smaller cubes, but we're not counting the corners.

Number of smaller cubes with paint on [b]three[/b] sides = 8 (on the corners)

Oh, and total surface area of n^3 unit cubes = 6n^3
There are n^3 cubes, each with surface area 6.

Tip: whenever you see such a problem, draw a diagram. Make a big cube and put lines on it to show where the unit cubes are. If the problem doesn't give a specific number of unit cubes, arbitrarily set the side length of the big cube to whatever you think would be most helpful in visualizing it.

That is SOOOOOOOO useful!

To be honest, I didn't know any of them. :(

Hmm.. Thanks! :D

Most number of cubes visible at once:

n:^3: - (n - 1):^3:


Find all finite set $S\subset {\mathbb Z}_+$ which satisfying both the following conditions:
1) There is any power of two in $S$, ie... there is $k\in{\mathbb Z}_+$ which $2^k\in S$.
2) Any $x,y\in S$ which $x\neq y$ then $\frac{x+y}{\gcd(x,y)}\in S$.

Enjoy~  :)

Assume that there is a set $S$ satisfying (1) and (2) with more than one element.

It's not clear from the notation whether the power of two that's in the set $S$ could be $1 = 2^0$.  If so, then putting $y=1$ we have $x \in S \Rightarrow x+1 \in S$ and so $S$ is infinite.

If now there's a power of 2 other than 1 in $S$ then there is an even number in $S$.  If $x$ is odd and $y$ is a power of 2  then $\mathrm{hcf}\{x,y\} = 1$, so $x \in S \Rightarrow x+y \in S$ and again $S$ is infinite.

Now suppose that all the elements of $S$ are even.  Let $x$ and $y$ be the first two elements of $S$ in order.  Then $\mathrm{hcf}\{x,y\} \ge 2$ and so $z = \frac{x+y}{\mathrm{hcf}\{x,y\}} \le \frac{x+y}{2} < y$, a contradiction.

So the only possibility is that $S = \{ 2^k \}$ for some $k$.


inequalities
trigonometry
function
triangle inequality
inequalities unsolved

[quote="ityaa"]let ABC be a triangle,and x,y,z>0 the following inequality is true or not?
$ xtan \frac A2 + ytan \frac B2 +ztan \frac C2 \geq \sqrt {xy+yz+zx} $[/quote]

at least, this is not true if z is very big (compared with x and y), and the triangle ABC is a degenerate pseudo-triangle with angles A = 90°, B = 90°, C = 0°. who knows, maybe it will hold if the angles A, B, C of the triangle ABC and the numbers x, y, z are equally sorted, so, for instance, $A\geq B\geq C$ and $x\geq y\geq z$. the problem is, i have no time to check right now.

greetings from saarbrücken this time (tomorrow will be the first demo exam)

darij

thanks,But How about the follow one?
$ (y+z)tan \frac A2 + (z+x)tan \frac B2 +(x+y)tan \frac C2 \geq \ 2 \sqrt {xy+yz+zx} $

Your new conjecture is right. Also, the old conjecture is right for the case when the number arrays (x; y; z) and (A; B; C) are equally sorted. I will prove this below.

At first, I change the notations:

[color=blue][b]Problem.[/b] Let $\alpha$, $\beta$, $\gamma$ be the three angles of a triangle, and let x, y, z be three nonnegative real numbers.

[b](a)[/b] Prove the inequality

$\left(y+z\right)\tan\frac{\alpha}{2}+\left(z+x\right)\tan\frac{\beta}{2}+\left(x+y\right)\tan\frac{\gamma}{2}\geq 2\sqrt{yz+zx+xy}$.

[b](b)[/b] If, additionally, the number arrays (x; y; z) and $\left(\alpha;\;\beta;\;\gamma\right)$ are equally sorted, then prove the inequality

$x\tan\frac{\alpha}{2}+y\tan\frac{\beta}{2}+z\tan\frac{\gamma}{2}\geq\sqrt{yz+zx+xy}$.[/color]

[i]Solution.[/i] Let $A=90^{\circ}-\frac{\alpha}{2}$, $B=90^{\circ}-\frac{\beta}{2}$, $C=90^{\circ}-\frac{\gamma}{2}$. Since $\alpha$, $\beta$, $\gamma$ are the three angles of a triangle, we have, for instance $\alpha<180^{\circ}$, what yields $A=90^{\circ}-\frac{\alpha}{2}>90^{\circ}-\frac{180^{\circ}}{2}=0^{\circ}$, and similarly we get B > 0° and C > 0°, and we also get $\alpha+\beta+\gamma=180^{\circ}$ by the sum of the angles in the triangle, what yields

$A+B+C=\left(90^{\circ}-\frac{\alpha}{2}\right)+\left(90^{\circ}-\frac{\beta}{2}\right)+\left(90^{\circ}-\frac{\gamma}{2}\right)=270^{\circ}-\frac{\alpha+\beta+\gamma}{2}$
$=270^{\circ}-\frac{180^{\circ}}{2}=180^{\circ}$.

Thus, the angles A, B, C are all positive, and add up to 180°. Hence, A, B, C are the angles of a triangle. Since $A=90^{\circ}-\frac{\alpha}{2}$, we have $\cot A=\tan\frac{\alpha}{2}$, and similarly $\cot B=\tan\frac{\beta}{2}$ and $\cot C=\tan\frac{\gamma}{2}$. Also, since $A=90^{\circ}-\frac{\alpha}{2}$, $B=90^{\circ}-\frac{\beta}{2}$, $C=90^{\circ}-\frac{\gamma}{2}$, the number arrays $\left(\alpha;\;\beta;\;\gamma\right)$ and (A; B; C) are oppositely sorted; but since the function f(t) = cot t is monotonically decreasing on the interval ]0°; 180°[, the number arrays (A; B; C) and (cot A; cot B; cot C) are oppositely sorted. Hence, the number arrays $\left(\alpha;\;\beta;\;\gamma\right)$ and (cot A; cot B; cot C) are equally sorted. Thus, requiring that the number arrays (x; y; z) and $\left(\alpha;\;\beta;\;\gamma\right)$ are equally sorted is equivalent to requiring that the number arrays (x; y; z) and (cot A; cot B; cot C) are equally sorted. Hence, the above problem follows from the following theorem:

[color=blue][b]Theorem 1.[/b] Let A, B, C be the three angles of a triangle ABC, and let x, y, z be three nonnegative real numbers.

[b](a)[/b] We have the inequality

$\left(y+z\right)\cot A+\left(z+x\right)\cot B+\left(x+y\right)\cot C\geq 2\sqrt{yz+zx+xy}$.

[b](b)[/b] If, additionally, the number arrays (x; y; z) and (cot A; cot B; cot C) are equally sorted, then we have the inequality

$x\cot A+y\cot B+z\cot C\geq\sqrt{yz+zx+xy}$.[/color]

And the [i]proof of Theorem 1[/i] can be done as follows:

First we will prove Theorem 1 [b](a)[/b]. The inequality in question,

$\left(y+z\right)\cot A+\left(z+x\right)\cot B+\left(x+y\right)\cot C\geq 2\sqrt{yz+zx+xy}$,

is homogeneous in the three variables x, y, z (in fact, these three variables occur in the power 1 on the left and on the right hand side of this inequality). Thus, we can apply dehomogenization by WLOG assuming that yz + zx + xy = 1.

Since the numbers x, y, z are nonnegative, there exist angles X, Y, Z in the interval ]0°; 90°] satisfying x = cot X, y = cot Y, z = cot Z.

The equation yz + zx + xy = 1 is equivalent to zx + xy = 1 - yz, what rewrites as x (y + z) = 1 - yz. Thus, $x=\frac{1-yz}{y+z}$. Since x = cot X, y = cot Y, z = cot Z, we thus have $\cot X=\frac{1-\cot Y\cot Z}{\cot Y+\cot Z}$. But by the addition formula for cotangents, $\frac{\cot Y\cot Z-1}{\cot Y+\cot Z}=\cot\left(Y+Z\right)$. Thus, $\cot X=\frac{1-\cot Y\cot Z}{\cot Y+\cot Z}=-\frac{\cot Y\cot Z-1}{\cot Y+\cot Z}=-\cot\left(Y+Z\right)=\cot\left(180^{\circ}-\left(Y+Z\right)\right)$. Since all three angles X, Y, Z lie in the interval ]0°; 90°], this yields X = 180° - (Y + Z). Thus, X + Y + Z = 180°. Hence, the angles X, Y, Z lie in the interval ]0°; 90°] and add up to 180°. Thus, these angles X, Y, Z are the angles of some triangle XYZ.

Now, applying http://www.mathlinks.ro/Forum/viewtopic.php?t=2958 post #3 Theorem [b](c)[/b] to the triangles ABC and XYZ, we get the inequality

$\cot A\cot Y+\cot A\cot Z+\cot B\cot Z+\cot B\cot X+\cot C\cot X+\cot C\cot Y\geq 2$.

Since x = cot X, y = cot Y, z = cot Z, this becomes

$\cot A\cdot y+\cot A\cdot z+\cot B\cdot z+\cot B\cdot x+\cot C\cdot x+\cot C\cdot y\geq 2$.

In other words,

$\left(y+z\right)\cot A+\left(z+x\right)\cot B+\left(x+y\right)\cot C\geq 2$.

Since yz + zx + xy = 1, this becomes

$\left(y+z\right)\cot A+\left(z+x\right)\cot B+\left(x+y\right)\cot C\geq 2\sqrt{yz+zx+xy}$,

and Theorem 1 [b](a)[/b] is proven.

The proof of Theorem 1 [b](b)[/b] is substantially simpler: Since the number arrays (x; y; z) and (cot A; cot B; cot C) are equally sorted, the Chebyshev inequality yields

$\frac{x\cot A+y\cot B+z\cot C}{3}\geq\frac{x+y+z}{3}\cdot\frac{\cot A+\cot B+\cot C}{3}$.

Upon multiplication with 3, this becomes

$x\cot A+y\cot B+z\cot C\geq\left(x+y+z\right)\cdot\frac{\cot A+\cot B+\cot C}{3}$.

Now, since $\left(x+y+z\right)^2-3\left(yz+zx+xy\right)=\frac12\left(\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2\right)\geq 0$, we have $\left(x+y+z\right)^2\geq 3\left(yz+zx+xy\right)$, so that $x+y+z\geq\sqrt{3\left(yz+zx+xy\right)}$. Also, by a well-known triangle inequality, $\cot A+\cot B+\cot C\geq\sqrt3$. Thus,

$x\cot A+y\cot B+z\cot C\geq\left(x+y+z\right)\cdot\frac{\cot A+\cot B+\cot C}{3}$
$\geq\sqrt{3\left(yz+zx+xy\right)}\cdot\frac{\sqrt3}{3}=\sqrt{yz+zx+xy}$,

and Theorem 1 [b](b)[/b] is proven.

  darij

darij grinberg,thank you for your nice solution.
you are a kind and careful man

thanks very much for your solution.nice.


Each of the ten volumes of the collected works of Theodore Sturgeon is available in paperback for $ \$$15 or in hardcover for $ \$$25. Theresa buys a copy of each of the ten volumes for a total of $ \$$220. How many hardcover volumes did she buy?

Assume she bought all paperback. This would be 15*10=150 dollars. Each of the hardbacks is 10 dollars more expensive. She paid 70 dollars more than 150 dollars. 70/10= $ \boxed{7}$ hardbacks


geometry rectangle parallelogram rhombus

Each angle of a rectangle is trisected. The intersections of the pairs of trisectors adjacent to the same side always form:
$ \textbf{(A)}\ \text{a square} \qquad\textbf{(B)}\ \text{a rectangle} \qquad\textbf{(C)}\ \text{a parallelogram with unequal sides} \\
\textbf{(D)}\ \text{a rhombus} \qquad\textbf{(E)}\ \text{a quadrilateral with no special properties}$

[hide="Click for solution"]
The diagonals of the quadrilateral formed lie along the two perpendicular lines joining the midpoints of the opposite sides of the rectangle.  These diagonals are of different lengths, and are perpendicular bisectors of each other.  Therefore, the figure is a rhombus, or $ \boxed{\textbf{(D)}}$.[/hide]

But a kite also has diagonals of different length that are perpendicular


Does anyone else participate in UIL math/number sense?

Hmmm...I do wish I still did. (Un)fortunately, I don't live in Texas anymore?

(Un)fortunately, I have never lived in TX except at a Holiday Inn Hotel...

i did district uil, but i didn't do state.

I do.
Anyone going to state?
It'd be cool to see some people who do real math.
I'm going in Math/Number Sense/Calculator/Science/Computer Science.
So PM me if you are going in any of those and perhaps we'll meet up.


When I press back I always see this
Fatal error: Allowed memory size of 8388608 bytes exhausted (tried to allocate 40 bytes) in /data/home/mathlinks/www/includes/functions_cache.php on line 339

wt's wrong?

btw, can we discuss apmo questions now?

did the apmo 2004 take place yet?

that is because sometimes the memory is full. it's a strange phenomena, because I have extended the memory from 8Mb to 16Mb :?

We're not supposed to discuss the APMO questions until they appear on the CMS site (which should be in a week's time or so, I suppose).  I think it will be safe to discuss APMO questions after Friday or so, because we certainly don't want this year's marks to all be voided like it happened in 2001 and 2002.

could you please be more explicit about this rule? what's its purpose? didn't you guys already took the APMO ?! :?

Well, the APMO takes place at different times for participants from different countries.  Before 2003 it could be done over a week, but due to much question-leaking in 2001 and 2002 the results of the APMO were voided in those 2 years.

As a result above all the APMO question sheets there are instructions which specifically say "Do not reveal the contents of this exam to anyone, especially over the Internet, until the exam is posted on the official APMO website."  It does take a bit long for the questions to get uploaded though, so the general rule is that you wait 3 or 4 days (and in that time, the questions have already been marked :D ) then it should be safe.

well, I do not understand why not all people take apmo in the same time ... is that really that hard? :S I mean I am confused about the "week" difference ...

Well, there are people in USA and Latin America, as well as Hong Kong, Taiwan, Singapore etc. taking APMO.  Clearly it's not possible to have it all at the same time, so the Americas takes it on the afternoon of 15 March and the Western Pacific Rim takes it on the morning of 16th March.

I guess in the past countries just had more flexibility about which day it would be on.

Well, a week is still a bit too much, isn't it?


1) Going across a period on the periodic table, what is the relationship between electron shielding (from the nucleus's pull) and first ionization energy?

2) What is exactly meant by "base strength" and "acid strength"?

3) What structural features must a compound have to be able to undergo autoionization?

4) If $HClO_{4}$ and $H_{2}NNH_{3}^{+}$ react, what would the products be?

2) Isn't that how completely the acid/base dissociates in water? Like it would be a weak acid/base if only a small fraction of it dissociates and strong acid/base if all or most of it dissociates:
[url]http://en.wikipedia.org/wiki/Strong_acid[/url]
[url]http://en.wikipedia.org/wiki/Strong_base[/url]


let Pk denonate the nth prime
prove that $lim_{\infty}\prod\left(1 - \frac{1}{p_n}\right) = 0$
-edit- latex added in

Is that equality correct?

If it is so, then one of the terms of the product must be zero, and I can't see which one this might be.

True, but I don't think that's what the problem is attempting to ask.

I think the questions asks us to prove that \[\lim_{\infty}\prod\left(1 - \frac{1}{p_n}\right) = 0.\]

[quote="shyong"]obviously the series converge to $0$ if the product is to infinity ....[/quote]
not necessarily. (1-1/4)(1-1/9)(1-1/16)... converges to 1/2, though the product is to infinity

It suffices to prove that $\prod\frac{1}{1-p_i}$ goes to $\infty$.  But because $\frac{1}{1-p}=1+p+p^2+...$ we have that this product like the harmonic series.

[quote="AntonioMainenti"][quote="Magnara"]It suffices to prove that $\prod\frac{1}{1-p_i}$ goes to $\infty$.[/quote]How do you know this? I would understand if you said $\prod\frac{p_k}{p_k-1}$, but then again I have almost no familiarity with infinite products. [quote="Magnara"]But because $\frac{1}{1-p}=1+p+p^2+...$ we have that this product like the harmonic series.[/quote]I thought that $\frac{1}{1-x}=1+x+x^2+\ldots$ was only valid for something like $|x|<1$.[/quote]
every individual p is less than 1, so his proof still holds

er...primes are rarely smaller than one; perhaps y'all are talking about this infinite product: $\prod_{i=1}^{\infty} \frac{1}{1-\frac{1}{p_i}} = \prod_{i=1}^{\infty} \sum_{k=0}^{\infty} \frac{1}{p^k} = \sum_{i=1}^{\infty} \frac{1}{i}$, which is indeed the harmonic series, eh?

[quote="MysticTerminator"]$\prod_{i=1}^{\infty} \frac{1}{1-\frac{1}{p_i}} = \prod_{i=1}^{\infty} \sum_{k=0}^{\infty} \frac{1}{p_i^k}$.[/quote]How'd you get that?

Yep, that's a well - known equality, it has a name but I can't remember it right now.

[quote="MysticTerminator"]sum of geometric series, right?[/quote]Obviously..   :blush: I was thinking it was some kind of sophisticated computation.

So basically to show that $\lim\prod_i\left(1-\frac{1}{p_i}\right)=0$, you show that $\lim\prod_i\left(1-\frac{1}{p_i}\right)^{-1}=\lim\prod_i\frac{1}{1-\frac{1}{p_i}}=\lim\prod_i\sum_k\frac{1}{p_i^k}\ge\lim\sum_i\frac{1}{i}$, which is infinite. I think I'll have remember that one; very nice.

This fact is a corrolary of Euler's proof of the divergence of [tex]\sum_{p \in \mathbb{P}}^\infty \frac{1}{p} = \infty[/tex] where [tex]\mathbb{P}[/tex] is the set of prime numbers.

[quote="Singular"]How does one get $\lim\prod_i\sum_k\frac{1}{p_i^k}\ge\lim\sum_i\frac{1}{i}$?  I had to use holder.[/quote]The left hand side is \[\left(1+\frac{1}{2}+\frac{1}{2^2}+\ldots\right)\left(1+\frac{1}{3}+\frac{1}{3^2}+\ldots\right)\left(1+\frac{1}{5}+\frac{1}{5^2}+\ldots\right)\cdots\]When this is expanded (theoretically..) you'll get every $\frac{1}{n}$ term for $n\ge1$. At first I thought you would get repeats but I think I was wrong, so just $\lim\prod_i\sum_k\frac{1}{p_i^k}=\lim\sum_i\frac{1}{i}$.

[quote="blahblahblah"]Not getting repeats is exactly the fundamental theorem of arithmetic![/quote]That's precisely what made me change my mind.  ;)  :D


limit
trigonometry
function
logarithms
inequalities
continued fraction
number theory

Hi,
I have a question about a trigonometric limit which is the following one:
Does the following limit exists
$\lim_{n\rightarrow \infty}n(1-\sin n)$,
and in case it does is it $\infty$?
Didilica

The function is unbounded but is also zero infinitely often.  The limit doesn't exist.

(Useful) [b]Theorem:[/b] Let a function $f$ be defined on a neighbourhood $J$ of a point $a,$ except possibly at the point $a$ itself. Then, \[\lim_{x \to a}f(x) = L\] if and only if \[\lim_{n \to \infty}f(x_{n}) = L\] for every sequence $x_{1},\ x_{2},\ x_{3}, \ldots$ of points in $J$ such that $x_{n}\not= a$ for any positive integer $n$ and $x_{n}\to a$ as $n \to \infty.$
---------------------------------------------
Coming to our problem, make the substitution $x = \frac1{n}.$
So, $\lim_{n \to \infty}n(1-\sin n) = \lim_{x \to 0}\frac{(1-\sin \frac1{x})}{x}$

Now, if we pick two sequences $\{a_{n}\}$ and $\{b_{n}\}$ such that $a_{n}= \frac1{2\pi n}$ and $b_{n}= \frac1{(2n+\frac12)\pi},$ then we note $a_{n}\to 0$ and $b_{n}\to 0.$ Also, $a_{n},\ b_{n}\not= 0$ for any $n.$

But, $\lim_{n \to \infty}\frac{1-\sin 2\pi n}{\frac1{2\pi n}}= \infty,$ whereas

$\lim_{n \to \infty}\frac{1-\sin (2n+\frac12)\pi }{\frac1{(2n+\frac12)\pi }}= 0.$

Hence,  $\lim_{n\rightarrow \infty}n(1-\sin n)$ doesn't exist.

[quote="didilica"]FieryHydra,
What you have shown is that $\lim_{x\rightarrow 0}\frac{1-\sin\frac{1}{x}}{x}$ does not exist. Next you plugged instead of $x$ the sequence $\frac{1}{n}$ and have concluded that $\lim_{n \rightarrow \infty}n(1-\sin(n))$ does not exist.
Let`s apply your argument to the following problem.
It is known that $\lim_{x\rightarrow \infty}\sin x$ does not exist. Next by using your ideea we take $x_{n}=2\pi n$ which is a sequence which converges to ${\infty}$ and we should conclude that $\lim_{n\rightarrow \infty}\sin(2\pi n)=0$ does not exist, which is false.
I do not think that your argument is correct. To prove that the limit does not exits by using the theorem you quoted you have to pick two sequences of natural numbers and show that the function evaluated at these two sequences converge to two different limits....
Didi[/quote]
didilica, if you read what I wrote earlier carefully, you will see that the theorem assumes that $f$ is defined on some neighbourhood of a point $a.$ In your example, $a = \infty$ is not a valid point because you cannot construct a neighbourhood around $\infty.$ We must take care of that first.

So, we write $\lim_{x\rightarrow \infty}\sin x = \lim_{x \to 0}\sin \frac1{x}.$

Now, we consider two sequences $a_{n}$ and $b_{n}$ such that $a_{n},\ b_{n}\to 0,$ and show that $\lim_{n \to \infty}\sin \frac1{a_{n}}\not = \lim_{n \to \infty}\sin \frac1{b_{n}}.$ In this case, we can choose $a_{n}= \frac1{2n\pi}$ and $b_{n}= \frac1{(2n+\frac12)\pi}.$

Lastly, I can't see how my argument implied in anyway that for $x_{n}= 2n\pi,\ \lim_{n\rightarrow \infty}\sin(x_{n})$ does not exist.

FieryHydra, you make the substitution $x = \frac1{n}$ and then use the sequence $a_{n}= \frac1{2 \pi n}$, whose (multiplicative) inverse is $2 \pi n$, which is not an integer!
Look one more time at the theorem you stated: It says that $\lim f(x) = L$ iff $\lim f(x_{n}) = L$ happens for all sequences, not that $\lim f(x)$ does not exist iff $\lim f(x_{n})$ does not exist for all sequences.

[i]Remark.[/i] The neighbourhoods of $\infty$ contain intervals of the form $\left( a, \infty \right)$, in which case your theorem does hold :!:

Are we clear now?

As for a solution to the problem, I was thinking of using the fact the $\left\{ 1,2,\ldots \right\}$ is dense on the trigonometric circle and some estimates on how big is the $n$ which is very close to $\frac{\pi}2$. But I got stuck... I'll let somebody else :)

You want to show that there is a constant $c>0$ such that there are infinitely many pairs $m,n$ with $|n-m\pi|<\frac{c}{n}$. (then $n|\sin n|=n|\sin(n-m\pi)|<c$)
Hint: continued fractions.

I'm not sure about this but I don't think we really need continued fractions for this although as you mentioned, I think you can just straight up use continued fraction by noting that $n < 4m$ or something reasonable like that.

Anyways:

Clearly there are infinitely many $n,m \in \mathbb{Z}^{+}$ such that $|\frac{n}{m}-\pi| < \frac{1}{2m}$  Multiplying through by $m$ gives us that $|n-m\pi| < 1/2$, but now we know that since $|n-m\pi| < |\sin{n-m\pi}|$ we have $n|\sin{n}| < \frac{n}{2}$ for arbitrarily large $n$.  And we get that $n-n\sin{n}> \frac{n}{2}$ for arbitrarily large $n$, and so no limit exists.

That's not good enough; you've shown that our sequence doesn't have a finite limit, but we'd rather show that the sequence takes small values infinitely often and doesn't tend to $\infty$.

I was misreading the question slightly; my statement works well for showing that $n\sin n$ takes small values, but we need the altered version below for this sequence:
Show that there are infinitely many pairs $m,n$ and some constant $c$ such that $|n-(2m+\frac12)\pi|<\frac{c}{n}$.

Then $\sin n>1-\frac{c^{2}}{2n^{2}}$ and $n(1-\sin n)<\frac{c^{2}}{2n}$; we have a subsequence tending to zero.

Oh I see.  So we want to show that there must be some M such that there is no N s.t. all n greater than N must have n(1-sin n) > M?  So we have to actually show a bounded sequence.  hmmm,

Well, continuing from your arguement, all we really need to show is  that we can find  $|n-(2m+\frac12)\pi|<\frac{c}{n}$ for infinitely many $n,m$.   Why don't we instead find the equivalent:
$|\frac{2n}{4m+1}-\pi| < \frac{2c}{n(4m+1)}$.
It is well known that if you take rational approximations to $\pi$ given to us by continued fractions then the sequence of approximations $\frac{p_{n}}{q_{n}}$ differ by less than $\frac{1}{q_{n}^{2}}$ from $\pi$.  So if we knew that we could set our fraction $\frac{2n}{4m+1}$ to be one of these approximations we would have that:
$|\frac{2n}{4m+1}-\pi| < \frac{1}{(4m+1)^{2}}$.

But the bound is not quite what we want, we want an $n$ in there, but notice that since this is supposed to be a good approximation to $2n < D(4m+1)$, where I'm too lazy to check what D might be, probably 4 or something.  So we have that certainly:

 $|\frac{2n}{4m+1}-\pi| < \frac{1}{(4m+1)^{2}}< \frac{D}{n(4m+1)}$

Now the issue is what if $P_{n}$ is not even, and $Q_{n}$ is not 1 mod 4.  Well, first of all, we know the two are relatively prime, so we need only worry about what $Q_{n}$.  First we make sure that is odd infinitely often.  This is true because in fact: $(Q_{n-1},Q_{n}) =1$.  So we know that no two consecutive denominators are even.  Given any of these odd denominators, what if we get unlucky and we get only things that are 3 mod 4.  That's okay, why then just set: $2n = 3P_{n}$ and $4m+1 = 3Q_{n}$.  In this case both $n$ and $4m+1$ go up by a factor of three, so just in case we need to do this why don't we just add a factor of 10 into D.  So we have our infinite pairs.

That's it. See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=90103]here[/url] for another proof of this result.

Hi,
Here is (in my opinion ) a very nice connection:

Do you remember that the following problem was open:

Study the convergency of the following series: $\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{2+\sin n}{3}\right)^{n}$.

Claim that $\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{2+\sin n}{3}\right)^{n}=\infty$.
Proof: By way of contradiction we assume that the sum converges. Then, by a well known theorem of Olivier (since the series has positive terms) we get that $\lim_{n\rightarrow \infty}n\frac{1}{n}\left(\frac{2+\sin n}{3}\right)^{n}=\lim \left(\frac{2+\sin n}{3}\right)^{n}=0$. This implies that $\lim \left(\frac{3}{2+\sin n}\right)^{n}=\infty$, or equivalently
$\lim n\ln \frac{3}{2+\sin n}=\infty$.
Now we have, via the inequality $\ln (1+x)\leq x$, $\forall x \geq 0$ that
$\lim n \frac{1-\sin n}{2+\sin n}=\infty$.
On the other hand since $n \frac{1-\sin n}{2+\sin n}\leq n \left(1-\sin n\right)$ , we get that $\lim_{n \rightarrow \infty}n\left(1-\sin n\right)=\infty$, which is a contradiction.
Thus the series diverges.
What do you think?
I think that the same statement holds for the following series
$\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{2+\sin(\beta n)}{3}\right)^{n}=\infty$, where $beta$ is any irrational number.

That's not a proof of anything; the "theorem of Olivier" you quote is not true without additional hypotheses. Some terms of the series are very close to $\frac1n$, but most are far smaller.

See this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22093]old thread[/url] for a correct proof that the series converges.

you are right, the terms should be nonincreasing to apply that theorem, I missed that.

[quote="zdyzhj"]this question and other similar questions  can easily be solved if use the fact that sinn is dense in [-1,1][/quote]
It's not that easy.

Consider this example: define the sequence $ x_n$ defined by

$ \frac12,\frac34,\frac14,\frac78,\frac58,\frac38,\frac18 ,\frac{15}{16},\dots$

We do have that $ x_n$ is dense in $ [0,1].$ Does it follow that $ \liminf nx_n\equal{}0?$

Well, no. The best we can do is the following subsequence:

$ 1\cdot x_1\equal{}\frac12.$

$ 3\cdot x_3\equal{}3\cdot 14\equal{}\frac34.$

$ 7\cdot x_7\equal{}7\cdot18\equal{}\frac78.$

In general, $ (2^k\minus{}1)x_{(2^k\minus{}1)}\equal{}\frac{2^k\minus{}1}{2^k}$ and this doesn't tend to zero. In fact, $ \liminf nx_n\equal{}1.$


function
complex analysis
complex analysis unsolved

Show that the automorphisms of the region complex plane minus 2 points are all mobius transformation.

I'm looking forward to see the answer to this.

This is not true. Say, if $R$ is a simply connected proper subset of $\mathbb C$, then there exist exactly two conformal automorphisms of $R \setminus \{x, y\}$ (where $x$ and $y$ are two distinct points in $R$). One is the identity map. The other is the automorphism of $R$ that exchanges $x$ and $y$. Clearly, the latter map doesn't have to be a Möbius transformation.

Update: as I understand the question, it asks about automorphisms of the set that equals a region (of the complex plane) with two points removed

Yeah, but now we are talking about the whole plane.

First of all note that our automorphism cannot have essential singularities at infinity or any of the two removed points, since that would contradict bijectivity. So, removing 3 singular parts we get a costant function by Liouville's theorem, and hence the automorphism is a rational function. Now just check which ones work.


What is the 13th number in the arithmetic sequence:
$ 7,10,13,\ldots$ ?

$ a_n \equal{} 7\plus{}(n\minus{}1)3 \equal{} 3n\plus{}4$.
Thus, $ a_{13} \equal{} 3 \cdot 13 \plus{} 4 \equal{} \boxed{43}$.