Students at a math contest were asked to try 3 new flavors of ice cream from the local ice cream parlor: Huckleberry, Chokecherry and Apple Dumpling.  Eight people sampled all three.  45 had huckleberry, 56 has chokecherry, and 63 had apple dumpling.  18 sampled both huckleberry and chokecherry, 26 tasted apple dumpling and chokecherry, and 20 had huckleberry and apple dumpling.  How many people had ONLY apple dumpling?

-oops, this is a set theory problem, not a probability problem

Look at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=138148[/url]
or
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=138146[/url]

I would make a ven diagram where each circle represents one flavor.

[quote="Phoenix257"][hide]Don't we need to know how many students there are total?[/hide][/quote]

[hide="Not really"]Let $n(P)$ be the number of elements of set $P$. $\cup$ is for union, and $\cap$ is for intersection.

$n(A \cap H \cap C)$, $n(A)$, $n(A \cap C)$, and $n(A \cap H)$ are all given. You can figure out the value of $n(A-(H \cup C))$ by performing the following operation (preferably in your Venn diagram):

$n(A)-n(A \cap C)-n(A \cap H)+n(A \cap H \cap C)$
$=63-26-20+8=\boxed{25}$.

You can even figure out the total from the information given:
$n(A \cup B \cup C)$
$=n(A)+n(B)+n(C)-n(A \cap B)-n(A \cap C)-n(B \cap C)+n(A \cap B \cap C)$[/hide]

[hide]I used a venn diagram and it was really easy that way. The answer was 25.[/hide]


geometry
3D geometry
sphere
calculus
integration
function
symmetry

[quote="kimby_102"]Source: ACoPS

A spherical, 3D planet has center @ $(0,0,0)$ and radius 20. At any point $(x,y,z)$ on the surface of this planet, the temperature is $T(x,y,z)=(x+y)^2+(y-z)^2$ degrees. What is the avg. temperature of the surface of this planet?[/quote]

since it is a sphere, $x^2+y^2+z^2=400$, then

$T=400+y^2+2xy-2yz$

then for all x and -x are on the sphere, etc so all pairs cancel out giving

$T=400+y^2$

then i think there is some calculus here...we have a temp of y^2 for a circumference of $2\pi\sqrt{r^2-y^2}$

then i believe that it is then just the average value thing:

$\frac{1}{r}\int_0^r2\pi*y^2\sqrt{r^2-y^2}\, dy=\frac{\pi*r^3}{16}$, and add in 400, to get

$\frac{\pi*r^3}{16}+400$

not sure if i made some mistakes...but i think it is something like this...i might have too many r's in my final answer

I think for average value you would find the sum of all $T$ then divide by the surface area, not $r$.  So maybe...............

Adding $T(x,y,z)+T(-x,y,z)+T(x,y,-z)+T(-x,y,-z)$ (all points for a constant $y$), we get $1600 + 4y^2 + 4xy - 4xy -4yz + 4yz = 1600 + 4y^2$
Then we have $4\int_{-20}^{20} 400 + y^2 dy = \frac{256000}3$
However, when $y=-20,20$, we overcount since $x=z=0$, so we have to subtract the three other T's.  At $y=-20,20$, $T=800$, and we take away 6, so the new sum is $\frac{241600}3$
The surface area is $1600\pi$, so the average value is $\frac{151}{3\pi}$

[quote="kimby_102"]Wait, u have to use calculus?

$T(x,y,z)+T(y,z,x)+T(z,x,y)=4(x^2+y^2+z^2)=4*20=80$

I think this is right

so wouldn't the avg. Temp. be $\frac{80}{3}$?[/quote]

I don't know if your method is correct, but doesn't $x^2+y^2+z^2=800$? :?

[quote="kimby_102"]Wait, u have to use calculus?

$T(x,y,z)+T(y,z,x)+T(z,x,y)=4(x^2+y^2+z^2)=4*20=80$

I think this is right

so wouldn't the avg. Temp. be $\frac{80}{3}$?[/quote]

the temperatures are not evenly distributed in that way...

this might not be appropriate for the intermediate topics as calculus is required to get the average function value as the surface integral.

I was wrong before.
Since we have $T=400 + y^2 + 2xy -2zy$, every temp is at least 400.  So we can take out the 400 and just add it to the final average we get.
For each $y$, we have a circle with equation $x^2 + z^2 = 400-y^2$ where $y$ is constant.  For each set of distinct $|x|$ and $|z|$, we have the points $(\pm x, \pm z)$, and adding the T's, the trailing two terms indeed cancel.  Therefore, we only have the $y^2$s left.
Here, I think, what we must do it find how many points there are, which is the circumference, and multiply that by $y^2$
So we have $\int_{-20}^{20} y^2 2 \pi \sqrt{400-y^2} dy = 4 \pi \int_0^{20} y^2 \sqrt{400-y^2}dy$
Then we divide by the surface area which is $1600\pi$ to get
$\frac{ \int_0^{20} y^2 \sqrt{400-y^2} dy}{400} + 400$
I don't know if there is a good way to take that integral... but with calculator, I get $478.539...$  If this is from acops, I don't think this is right since there must be some rational answer and probably without calculus... but, at least to me, my way makes sense :lol:

I saw the hint in the back of the book and it says to imply symmetry by adding $T(x,y,z)+T(y,z,x)+T(z,x,y)$. But I'm not sure whether u have to use calc. to solve....

Hm, well doing that cancels everything out, giving 1600.  However, it overcounts where x=y=z (simple fix though)... it's just a matter of figuring out how many times to apply this.

i reccommend using spherical coordinates and taking a surface integral and then dividing by the entire area.  This should give you the average surface temperature with little work.  In order to do so you must make x,y,z into spherical ordinates, which isn't that bad.  Then take the double integral with r=constant

Wouldn't you not have to worry about the points where x=y=z?
Since there are only two points, you can remove them and the average won't change

I think you can remove any finite number of points and the avareage won't change. Possibly even a countably infinite number of points.


Hola que tal, ojala que a los hispanos les agrade las matematicas discretas sonrealmente hermosas, bueno aqui un problemita un tanto facil

[b]Determine el maximo número de aristas que puede tener un grafo simple de 8 vertices en el cual no hay cuadrilateros.[/b]
[color=red]Observacion:[/color]
[color=darkblue]1 Un grafo simple se considera a aquel grafo en el cual no hay loops y para cada par de vertices hay a lo mas una arista que los une
2 debe entenderse que un cuadrilatero en grafos son aquellas figuras que forman un ciclo de 4.[/color]

Haber amigos hispanos es bacan el problema. :lol:

Que Dios les Bendiga
Bye

ok, voy a poner mi solkucion ya que si es  lindo:

Contemos el  numero $N$ de triplas de vertices $(a,b,c)$ tales que $ab$ es un arco y $bc$ es un arco. (a,b,c distintos)

Para cada pareja $(a,c)$ puede haber maximo un $b$ (o habria cuadrilatero) de donde $N$ es menor o igual que $\binom{8}{2}$ Por otro lado si un vertice $b$ tiene grado $d_b$, hay $\binom{d_b}{2}$ parejas $(a,c)$ tales que $ab$ y $bc$ son arcos.

Se sigue que si los grados de los vertices son $d_1,d_2...,d_8$ entonces $N=\binom{d_1}{2}+\binom{d_2}{2}...+\binom{d_8}{2}$

entonces: $\binom{8}{2}\geq \binom{d_1}{2}+\binom{d_2}{2}...+\binom{d_8}{2}$

De la ultima desigualdad si fiajmos $d_1+d_2..+d_8$ tenemos un minimo para todos los $d_i$ iguales o difiriendo por 1. ahora es facil evaluar el minimo para esta suma. (esto da una cuadratica si no estoy mal)

Mas detalles despues...


hello

I am curious as to know which courses in mathematics would be best to take when working with computer related problemsolving. I know that there are sites where proffesionals give advice, however, im talking about strictly algorithmic nature and problem solving in computer programing.

If you dont mind,
name your top three most important studies in math related to programing.

thanks in advance

well if it had to be a book
i wud say
concrete mathematics by knuth, graham and patashnik
it's a wonderful treatise
try getting ur hands onto it
best regards

thanks sounds good I am really going to look into it, however that cant be all can it, anybody?

I agree, Concrete Mathemathics is a great book.

From me, I can't come up with a specific book, but there is a cool place:
MIT OpenCourseWare : ocw.mit.edu

Try it, there is a "Mathemathics for Computer Science" course.


Να βρείτε τις μη αρνητικές ακέραιες λύσεις της εξίσωσης:
$ 9^x\minus{}3^x\equal{}y^4\plus{}2y^3\plus{}y^2\plus{}2y$

Ενδιαφέρουσα άσκηση. Προτείνω την παρακάτω λύση:

ή,
[hide]$ \leftrightarrow 3^x(3^x \minus{}1) \equal{} y(y^2\plus{}1)(y \plus{} 2)$
2 περιπτωσεις
-αν $ y \equiv 0 (mod2)$ τοτε,
$ y(y\plus{}2)\equal{}3^x \minus{}1$
$ y^2\plus{}1\equal{}3^x$
εξισωσεις που δινουν, $ y\equal{}0,x\equal{}0$
-αν $ y \equiv 1 (mod2)$ τοτε,
$ y^2\plus{}1\equal{}3^x\minus{}1$
$ y(y\plus{}2)\equal{}3^x$
εξισωσεις που δινουν $ y\equal{}1,x\equal{}1$[/hide]


how do you do this problem:

Find the sum of the first sixteen terms in the arithmetic sequence 1,7,13,19, 25.........

include explanation

[hide]you might want a calculator.
i took 1 + 2 + 3...+15 because for every term after the first you just add 6 more. this comes out to be 120, which is how many times 6 is added.  120*6 is 720, plus 16 (1 for each term) and you get 736.[/hide]

[hide]
There is a formula for this.
It is $\text{sum}=\frac{\text{first term + last term}}{2}\cdot \text{number of terms}$
There are 16 terms. The first is 1 and the last is $6(15)+1=91$
Applying the formula it is:
$\frac{92}{2}\cdot16=736$[/hide]

[hide]16-1 is 15
15*6= 90
90+1 = 91
and apply the sum of numbers formula
$((1+91)/2)*16$
$which = 736$[/hide]
Yay, my first math mode and LaTeX equation.

Another neater equation:

$\displaystyle\sum_{k=0}^{15} 6k+1=6\times \displaystyle\sum_{k=0}^{15} k +16=\frac{6(15)(16)}{2}+16=736$

heres what i would do :oops: [hide]

its a process you do 6*0 then 6*1.......6*15 figure out that that is 120 6's do 120*6=720 plus the 16 extra ones we left out which equils 736:)[/hide]


exits or not 2n positive integer numbers $x_{1},x_{2}...x_{n},y_{1},y_{2}...y_{n}$ are distint such that
$\left\{ \begin{array}{c}{1}\sum_{i=1}^{n}x_{i}= \sum_{i=1}^{n}y_{i}\ \sum_{i=1}^{n}x_{i}^{2}= \sum_{i=1}^{n}y_{i}^{2}\end{array}\right.$ with $(n \in N ,n\geq 2 )$

Sorry I'm wrong! $x_{1},x_{2},...,x_{n},y_{1},y_{2},...,y_{n}$ are distint ! I edited my post . Thankyou very much  :) .Can you solve it?? :)


In $ \triangle{ABC}$, $ \angle{A}\equal{}60^0$ and $ \angle{B}\equal{}20^o$. $ D$ is point on segment $ BC$ such that $ \angle{ADC}\equal{}40^o$.

Show that $ \frac{1}{AC}\equal{}\frac{1}{AB}\plus{}\frac{1}{AD}$.

WLOG, assume $ AC\equal{}1$. Note that $ \angle C\equal{}100^{\circ}$. So $ AB/\sin100^{\circ}\equal{}AC/\sin20^{\circ}$, and hence $ AB\equal{}\sin100^{\circ}/\sin20^{\circ}$.

In triangle $ ADB$, $ \angle D\equal{}140^{\circ},\angle B\equal{}20^{\circ}$. So $ AD/\sin20^{\circ}\equal{}AB/\sin140^{\circ}$, and hence $ AD\equal{}\sin100^{\circ}/\sin140^{\circ}$.

Therefore $ \frac1{AD}\plus{}\frac1{AB}\equal{}\frac{\sin20^{\circ}}{\sin100^{\circ}}\plus{}\frac{\sin140^{\circ}}{\sin100^{\circ}}\equal{}\frac{2\sin80^{\circ}\cos60^{\circ}}{\sin100^{\circ}}\equal{}1\equal{}\frac1{AC}$. Q.E.D.


calculus integration logarithms trigonometry quadratics function LaTeX

Nice integral, and I have an equally nice solution! :)

First, note that $\int_{a}^{b}f(x)\, dx = \int_{a}^{b}f(a+b-x)\, dx.$ ... $(*)$

This is quite easy to show: just use the substitution $y = a+b-x$ in the second integral, and you have the result.

Now, $I = \int_{0}^{\infty}\frac{\log x}{x^{2}+2x+4}dx = \int_{0}^{\infty}\frac{\log x}{(x+1)^{2}+3}dx$

We make the substitution $x+1 = \sqrt3 \tan y$ to get

$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{\ln (\sqrt3 \tan y-1)}{\sqrt3}\, dy$

$= \frac1{\sqrt3}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\ln \left( \sqrt3 \tan (\frac{\pi}{2}+\frac{\pi}{6}-y)-1 \right)\, dy$ ... [Using the identity shown in $(*)$.]

$= \frac1{\sqrt3}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\ln \left(-\frac{\sqrt3}{\tan (\frac{\pi}{6}-y)}-1\right)\, dy$

$= \frac1{\sqrt3}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\ln \left(-\frac{\sqrt3 (\sqrt3+\tan y)}{1-\sqrt3 \tan y}-1 \right)\, dy$ ... [ Note $\tan(a-b) = \frac{\tan a-\tan b}{1+\tan a \tan b}$]

$= \frac1{\sqrt3}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\ln (\frac{4}{\sqrt3 \tan y-1}) \, dy$

$= \frac1{\sqrt3}\ln 4 \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\, dy-I$

$\Rightarrow 2I = \frac{2\ln 2}{\sqrt3}(\frac{\pi}{2}-\frac{\pi}{6})$

$\Rightarrow I = \boxed{\frac{\pi}{3\sqrt3}\ln 2}.$

yes, that's my solution, too. I was thinking about another solution using residues. What do you think? :)

[color=darkblue][b]Yes, FieryHydra, you are an equally nice solution ![/b]

 Using the your method, I obtained $\boxed{\ \int_{0}^{\infty}\frac{\ln x}{x^{2}+2ax+b}\ dx=\frac{1}{2\sqrt{b-a^{2}}}\cdot\arctan\frac{\sqrt{b-a^{2}}}{a}\cdot \ln b\ }\ ,\ b>a^{2}\ .$ What happen if $a>0$ and $0<b\le a^{2}\ ?$
An interesting particular case is $\boxed{\ \int_{0}^{\infty}\frac{\ln x}{x^{2}+a^{2}}\ dx=\frac{\pi}{2a}\cdot \ln a\ }$ , where $a>0\ .$ More particularly, $\boxed{\ \int_{0}^{\infty}\frac{\ln x}{x^{2}+1}\ dx=0\ }\ .$[/color]

Virgil Nicula, thanks very much for showing us the generalisation. By the way, I very much enjoy your generalisations on all kinds of problems. :)

Thanks and I. And yet, FieryHydra, what happen if $0<b\le a^{2}$ and $a>0\ ?$

A generalization

Consider the case a>0, b non zero, c>0 and b^2-4ac<=0. Then
int ln(x)/(a*x^2+b*x+c)dx=

=(ln(c/a))/(sqrt(4ac-b^2))arctan((1/b)sqrt(4ac-b^2)),

and I think that the integral can be calculated even if b^2-4ac>0

 the substitution x=sqrt(c/a)t  solves the problem.

The new integral which contains the logarithm is  zero (via a new

substitution of the form t=1/y), and the integral which is left simply contains the reciprocal of a quadratic function which can be calculated by simple fraction decomposition.
If b=0 we see that the integral is 1/(2sqrt(ac))ln(c/a)Pi/2.
Didi

[/hide][/code]

Virgil Nicula, I am sorry I could't find enough time during the day to answer your question. I haven't read didilica's post above, but I hope he has answered the question you posed. It gets harder these days to respond to posts quickly, now that the summer holidays are over!  :(

[quote="didilica"][color=darkred]I think that the integral can be calculated even if b^2-4ac>0. The substitution x=sqrt(c/a)t  solves the problem. The new integral which contains the logarithm is  zero (via a new substitution of the form t=1/y), and the integral which is left simply contains the reciprocal of a quadratic function which can be calculated by simple fraction decomposition.[/color][/quote][color=darkblue]Please, write in LaTeX ! I understood nothing. Can you offer some details ? Sorry, I think you was wrong.[/color]

Dear Virgil Nicula,
Can you explain to me please ( I do know how to use latex) how to post my calculations in latex? Do I have to create a separate file and then to attach it to the forum ?
Didi[/quote]

You can enclose the code in dollar signs thus:
$\$$\int_a^b f(x) dx = F(b) - F(a) $\$$ ==> $\int_{a}^{b}f(x)dx = F(b)-F(a)$
You can click someone else's LaTeX to see what code they used. There's a guide on the site -- see the link on the sidebar. There's also a forum for LaTeX questions and a forum for testing.

Let $a>0$, $b\neq 0$, $c>0$ and $b^{2}-4ac \geq 0$.
$\int_{0}^\infty \frac{\ln x}{ax^{2}+bx+c}dx=\frac{\ln \frac{c}{a}}{\sqrt{4ac-b^{2}}}\arctan \frac{\sqrt{4ac-b^{2}}}{b}$.
The substitution $x=\sqrt{\frac{c}{a}}t$, solves the problem.
Didi

[color=darkblue][b]Very nice, Didilica ![/b] Here is the your idea.

[b]Lemma.[/b] $\boxed{\ a>0\ ,\ b>0\Longrightarrow\ L\equiv\int_{0}^{\infty}\frac{\ln x}{ax^{2}+bx+a}\ dx=0\ }\ .$
[b]Proof.[/b] $t=\phi (x)=\frac{1}{x}$ $\Longrightarrow$ $L=\int_{0}^{\infty}\frac{\ln x}{ax^{2}+bx+a}\cdot \left(-x^{2}\right)\cdot \phi '(x)\ dx=$ $\int_{\infty}^{0}\frac{\ln \frac{1}{t}}{\frac{a}{t^{2}}+\frac{b}{t}+a}\cdot\left(-\frac{1}{t^{2}}\right)\ dt=$ $-\int_{0}^{\infty}\frac{\ln t}{at^{2}+bt+a}\ dt=-L$ $\Longrightarrow$ $L=0\ .$

[b]Remark.[/b] The substitution $x=t\sqrt{\frac{c}{a}}$ in the previous exercise is [b][u]"dreadfully" ![/u][/b][/color]

Thanks for the tip Xevarion.
Let`s play with Cesar`s integral. We make the substitution $x=2t$, and get that
$I=\int_{0}^\infty\frac{\ln x}{x^{2}+2x+4}dx=\frac{\ln 2}{2}\int_{0}^\infty\frac{dt}{t^{2}+t+1}+\frac{1}{2}\int_{0}^\infty\frac{\ln t}{t^{2}+t+1}dt=\frac{\ln 2}{2}\int_{0}^\infty\frac{dt}{t^{2}+t+1}$, since the second integral is $0$ (via the substitution $t=\frac{1}{y}$. The general integral can be calculated by the same ideea, we only need to have that the signs of both $a$ and $c$ are the same so that we can make the substitution $x=\sqrt{\frac{c}{a}}t$.
Didi

[quote="Cezar Lupu"]Calculate the following integral \[\int_{0}^{\infty}\frac{\log x}{x^{2}+2x+4}dx.\] [/quote]

It's a standard exercise on integration with residue, three or four lines really.
Function is $\frac{\log^{2}z}{z^{2}+2z+4}$. You can read details in complex analysis textbooks. I don't know standard course in English, in Russian it's Lavrentiev & Shabat Functions of Complex Variables: Theory and Methods.

[quote="Cezar Lupu"]I know Residues work just fien here, but please, feel free to post solutions. Foe example see http://www.mathlinks.ro/Forum/viewtopic.php?t=111666  :wink:[/quote]

I have no time to typesetting the solution. I have showed function. Contour is a circle with cutting along positive real line. Type-lovers have a good chance to write complete solution.

[quote="perfect_radio"]I think that the solution should be similar with [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=602279#602279]this one[/url].[/quote]

Not quite, because function $x^{2}+2x+4$ is not a even function.


geometry
geometric transformation
rotation

I don't know if this has been posted before but the solution was quite interesting.

Using 2 3s and 2 8s find a way of getting a value of 24, by using no square roots, no decimal points, nothing, except for addition, subtraction, multiplication, division, and parantheses.  And of course the equal sign.

Good Luck.

This is actually quite easy.

[hide="Really really cheap solution"]
Turn one of the threes around so it looks something like an E, and then stick the open end onto the other three, to make an 8. Now you have a total of three 8's, and simply do 8+8+8.
[/hide]
:D :D :D

Okay, I should be more specific.  Only one equal sign.  No rotations of any of the numbers, particularly 3, and if there are any other cheap solutions, I will make add additional rules.

There is one solution that involves some great thinking.

[hide="cheap solution that follows the current rules"] $\\ ( )\quad )\\ ( )\times \; )+8\times3-8\times3.$it uses parentheses :D [/hide]
Edit:
[hide="partially cheap solution"]
$(\frac{8}{8-3-3})!$[/hide]

Oh, yeah, don't use anything that is cheap.  Seriously, don't use other symbols to make them look like numbers.  There is a solution that you can type on your calculator that follows all of these rules and is not cheap.

It just involves some thinking.

[quote="nutz_for2.718281828"][hide="the real, non-cheap solution"]$(\frac{8}{\frac{8}{3}})*8$
or

(8/(8/3))*8[/hide][/quote]

That has THREE eights and only ONE three!

[hide="My solution"]$\boxed{\frac{8}{3-\frac{8}{3}}}$[/hide]

Congratulations, we have a winner.

And yes nutz, please learn to notice the topic, and how to count to 3.

JK

another problem: make 24 using 4, 4, 7, 7.  and no cheap stuff.

[hide="With a decimal point"]$7+7+\frac4{.4}$[/hide]

try without a decimal point.  there's a cool-ish solution- actually i think this one is on the picture on some of the "24 challenge" boxes.

[quote="mathgeniuse^ln(x)"]Are you sure there is a solution to this?

this seems impossible.[/quote]
[hide="it is possible"]
$7\times(4-\frac{4}{7})$[/hide]

Wow, that took me a whole 15 seconds.

$(4-\frac47)7=24$


counting
combinatorics
IMO Shortlist
Integer sequence

Find all finite sequences $(x_0, x_1, \ldots,x_n)$ such that for every $j$, $0 \leq j \leq n$, $x_j$ equals the number of times $j$ appears in the sequence.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[hide="Solution, from IMO Compendium"]Clearly, all terms in the sequence are nonnegative integers. $ x_0\equal{}0$ gives a contradiction, so we have $ x_0>0$.
Now let $ m$ be number of positive terms in $ x_1,x_2,\ldots,x_n$. Since $ x_i$ counts the terms equal to $ i$, $ x_1\plus{}x_2\plus{}\cdots\plus{}x_n$ counts the total number of positive terms in the sequence, which is $ m\plus{}1$. For this to be possible, we must have $ m\minus{}1$ of these terms equal to 1, one of the terms equal to 2, and the rest equal to 0. Thus, only $ x_0$ can be greater than 2, so only one of $ x_3,x_4,\ldots,x_n$ can be positive, with the positive term corresponding to the value of $ x_0$. This implies that $ m \leq 3$.
We now casework the values of $ m$:
$ m\equal{}1$: $ x_1\equal{}2$ is impossible, so $ x_2\equal{}2$. This gives us $ x_0\equal{}2$, leading to the sequence $ (2,0,2,0)$.
$ m\equal{}2$: Either $ x_1\equal{}2$, or $ x_2\equal{}2$. These give $ (1,2,1,0), (2,1,2,0,0)$.
$ m\equal{}3$: We have $ x_k>0$ for some $ k>2$. Then $ x_0\equal{}k$, and $ x_k\equal{}1$. $ x_1\equal{}1$ then is impossible, so $ x_1\equal{}2$, leading to $ x_2\equal{}1$. These are all the positive terms in the sequence, so we end up with $ (k,2,1,0,0,\ldots,0,1,0,0,0)$, where there are $ k\minus{}3$ zeroes in the middle.[/hide]

there must be something wrong with your solution, i haven't read it completely but i have found a counter-example.

Consider this for n = 6

3,2,1,1,0,0,0.

This clearly works, and n is bigger than 3. So I think you MIGHT have a slight mistake somewhere.

Probably you should reread the solution given -- $m$ and $n$ are different letters ;)

Note that
\[
x_0+x_1+x_2+\cdots = 0\cdot x_0 + 1\cdot x_1 + 2\cdot x_2 + \cdots
\]
so
\[
x_0=x_2+2x_3+3x_4+\cdots
\]
Now we do casework on $x_0$.

If $x_0=0$, then there is a $0$, so $x_0\ge 1$, a contradiction.
If $x_0=1$, then it's easy to see that the equation is satisfied only if $x_2=1$ and $x_3=x_4=\cdots = 0$. There must be at least one $2$, which means $x_1=2$. There is one $0$, so $n=3$, and $(1,2,1,0)$ is a valid sequence.
If $x_0=2$, then either $x_2=2$ and $x_3=x_4=\cdots = 0$, or $x_3=1$ and $x_2=x_4=x_5=\cdots = 0$. In the first case, since $x_3=x_4=\cdots = 0$ and there are two $2$s already, $x_1\ge 2$ is impossible. $x_1=0$ and $x_1=1$ yield the valid sequences $(2,0,2,0)$ and $(2,1,2,0,0)$. The second case is impossible
Finally, suppose $x_0=a$ with $a\ge 3$. This implies $x_a\ge 1$, and in fact $x_a=1$ since if $x_a \ge 2$, we get a contradiction in the equation. This implies the equation can only hold if $x_2=1$ and $x_3=\cdots = x_{a-1}=a_{a+1}=\cdots = 0$. Now since $x_2=1$, we need a $2$ somewhere, which can only come from $x_1=2$. Adjusting the length of the sequence with the correct number of $0$s yields the sequence $(a,2,1,0,0,\cdots 0,1,0,0,0)$ with $a$ total $0$s for each $a\ge 3$.

[hide=Answer]
For
       $n=1,2$  there are no such sequences.
       $n=3$       there are $2$ soulutions : $(1,2,1,0)$ and $(2,0,2,0)$.
       $n=4$      there is $1$ solution :    $(2,1,2,0,0)$
       $n=5$  there are no such sequences
       $n>=6$     one solution for every such $n$:
                  $(n-3,2,1,0, ... ,0,1,0,0,0)$. [where $"..."$ implies all entries are $0$ here][/hide]

[b][u]Solution[/u][/b]

[b]First Simple Key Observation[/b] See that \[ \sum_{i = 0}^n x_i = n+1\] This is because every number in the sequence is from {$0,1,..,n$} as each represent number of occurances of a particular $0<=j<=n$ in the sequence so each $x_i<=n+1$ and if one $x_i=n+1$ this can't hold(as there are only n remaining places to fill and $i$ can't be $n+1$). So every number in the sequence is either $0$,$1$ , ..., or $n$. And thus  \[ \sum_{i = 0}^n x_i = n+1\] as $n+1$ is the total number of numbers in the sequence.
[b]Second Simple Key Observation[/b] Let $x_0=k$. Observe that $x_0>0$ as if $x_0=0$ implies $0$ is not present in the sequence and simultaneously it is present once (a contradiction).
[color=#00f][b]Main pivotal reasoning[/b][/color] Let $x_0=k$ [$k>0$]. Now leaving $x_0$ and the $k$ other $x_i$ which have value $0$ there are $(n+1)-(k+1)=n-k$ other terms of the sequence whose sum is $n+1-k=(n-k)+1$.
Each of these takes values $>=1$(else that contradicts definition of $k$). After giving each of them a $1$ there is still a $1$ left which must go with one of them making that a $2$.(argument similar to proving Pigeon-Hole-Principle if you notice :yup: )
So exactly $n-k-1$ values among them are $1$ and only $1$ of them takes value $2$.($(n-k+1)=(n-k-1)*1+1*2$.

"So we see that there are always
[b]i>[/b] $n-k-1$ entries of value $1$
[b]ii>[/b] $1$ entry of value $2$
[b]iii>[/b] $1$ entry of value $k$
[b]iv>[/b] $k$ entries of value $0$".   ([b]Statement I[/b])

Next we consider 3 cases :
[b]Case 1 : $k>=3$[/b]
[b]Case 2 : $k=2$[/b]
[b]Case 3 : $k=1$[/b]

[b][u]Case 1 : $k>=3$[/u][/b] The whole sequence is filled with $0,1,2$ and $k$. So for only these "$i$"s $x_i$ is non-zero. We know there is exactly one $2$, exactly one $k$. So $x_2=1$ and $x_k=1$ and $x_0=k>2$. We know there is a $2$ somewhere and thus $x_1$ must be $2$. This in turn suggests $n-k-1=2$ or , $k=n-3$. As we assumed $k>=3$ so here $n>=6$. Thus for every $n>=6$ there is exactly one sequence $(n-3,2,1,0,...,1,0,0,0)$.$\blacksquare$

[b][u]Case 2 : $k=2$[/u][/b] As $k=2$ put that in the [b]Statement I[/b] we get that there will be exactly $2$ "$2$"s in the sequence, exactly $n-2-1=n-3$ "$1$"s and $2$ zeroes. Again as the sequence is filled with only $0$,$1$ and $2$ only $x_0,x_1$ and $x_2$ can be non-zero and $x_0=x_2=2$. What remains is only $x_1$. $x_1$ can't be $2$ (then $x_2=3$ which is not). So $x_1=0$ or $x_1=1$.
If $x_1=0$ $n-3=0$ or, $n=3$ and the sequence is $(2,0,2,0)$. $\blacksquare$
If $x_1=1$ $n=4$ and the sequence is $(2,1,2,0,0)$. $\blacksquare$

[b][u]Case 3 : $k=1$[/u][/b] Again put $k=1$ in [b]Statement I[/b] and we get that there are exactly $1$ "$2$", exactly $1$ "$0$" and exactly $n-1-1+1=n-1$ "$1$"s in the sequence. As there is one $2$ somewhere in the sequence and except $x_0,x_1$ and $x_2$ all other $x_i=0$ and $x_0=x_2=1<2$ so $x_1$ must be that $2$. This in turn gives $n=3$ and our solution is $(1,2,1,0)$. $\blacksquare$ $\blacksquare$ $\blacksquare$  :play_ball:

We claim the sequences that work are $(1,2,1,0)$, $(2,0,2,0)$, $(2,2,0,1,0)$, and $(n-3,2,1,0,\ldots,0,1,0,0,0)$ for any $n\ge 6$. It is easy to see that these work, so we now show that they are the only ones.

Note that $0\le x_k\le n+1$ as $x_k$ is the size of some subset of $\{0,1,\ldots,n\}$. Furthermore, if $x_k=n+1$, then all the $x_i$s are $k$, so the sequence is $(n+1,\ldots,n+1)$, which doesn't work. Therefore, $0\le x_k\le n$ for all $k$. We have the following key claim.

[b]Claim:[/b]  We have \[n+1=x_0+\cdots+x_n=x_1+2x_2+\cdots+(n-1)x_{n-1}+nx_n.\]

[i]Proof:[/i]  Note that $x_0+\cdots+x_n$ is the sum of the number of times $k$ appears in the sequence over all $k$. Since each term of the sequence is between $0$ and $n$, this means that $x_0+\cdots+x_n$ is counting the number of terms in the sequence, which is $n+1$, so \[x_0+\cdots+x_n=n+1.\] Furthermore, $x_0+\cdots+x_n$ is the sum of the terms of the sequence, which is \[\sum_{k=0}^n k\cdot(\text{number of times }k\text{ appears})=\sum_{k=0}^n kx_k,\] which proves the claim. $\blacksquare$

In particular, we have \[\boxed{x_0=x_2+2x_3+\cdots+(n-1)x_n}.\] We now split into cases based on the value of $x_0$.

[b]Case 1[/b]: Suppose $x_0=0$. However this can't happen, as the sequence then has at least one $0$, so $x_0\ge 1$.

[b]Case 2[/b]: Suppose $x_0=1$. Thus, \[1=x_2+2x_3+\cdots+(n-1)x_n,\] so $x_2=1$ and $x_3=\cdots=x_n=0$. But we have exactly one $0$, so we must have $n\le 3$. Now, \[x_1=(n+1)-(x_0+x_2+\cdots+x_n)=(n+1)-(1+1)=n-1,\] so the sequence is $(1,n-1,1,0)$ or $(1,n-1,1)$. The first gives to $(1,2,1,0)$ and the second gives $(1,1,1)$. We see that $(1,1,1)$ doesn't work, so the only solution from this case is $\boxed{(1,2,1,0)}$.

[b]Case 3[/b]: Suppose $x_0=2$. Thus, we have \[2=x_2+2x_3+3x_4+\cdots+(n-1)x_n,\] so $x_4=\cdots=x_n=0$. We have two subcases. [list]     [*] [b]Case 3.1[/b]: Suppose $x_2=2$ and $x_3=0$. Then, $x_3=\cdots=x_n=0$, so $n\le 4$ as $x_0=2$. Now,     \[x_1=(n+1)-(x_0+x_2+\cdots+x_n)=(n+1)-(2+2)=n-3,\]     so the sequence is $\boxed{(2,0,2,0)}$ or $\boxed{(2,1,2,0,0)}$.     [*] [b]Case 3.2[/b]: Suppose $x_2=0$ and $x_3=1$. Then, $x_2=x_4=\cdots=x_n=0$, so $n\le 4$ as $x_0=2$. Now,     \[x_1=(n+1)-(x_0+x_2+\cdots+x_n)=(n+1)-(2+1)=n-2.\]     Since $x_3=1$, we have $x_1\ge 1$, so $n=3,4$. These give $(2,1,0,1)$ and $(2,2,0,1,0)$. Neither of these work, so no solutions here. [/list]

[b]Case 4[/b]: Suppose $x_0\ge 3$. Let $a=x_0$. Then, we have $x_a\ge 1$, and we have \[a=x_2+2x_3+\cdots+(n-1)x_n\ge (a-1)x_a.\] If $x_a\ge 2$, then we have $a\ge 2(a-1)$, or $a\le 2$, which isn't the case. Therefore, $x_a=1$, so \[\sum_{\substack{k=2\\ k\ne a}}^n (k-1)x_k=1.\] This means $x_2=1$ and $x_k=0$ for $k\in\{3,\ldots,n\}\setminus\{a\}$. Now, $a\ge 3$, so the sequence has exactly two $1$s except for possibly $x_1$. This means that $x_1=2$. But we have \[x_0+\cdots+x_n=n+1,\] so \[a+2+1+1=n+1,\] so $a=n-3$. This gives the sequence $\boxed{(n-3,2,1,0,\ldots,0,1,0,0,0)}$ for any $n\ge 6$ (as $a=n-3\ge 3$).

This completes the proof.

We claim that the only sequences are $(1,2,1,0)$, $(2,0,2,0)$, $(2,1,2,0,0)$, and $(n-3,2,1,0,\ldots,0,1,0,0,0)$ for all $n\ge 6$.

First, note that $x_0+\ldots+x_n=n+1$. As there are $x_0$ $0$s, there are $n-x_0$ positive numbers with positive index. However, they sum to $n+1-x_0$, so they must be comprised of all 1s and a single 2. This means that the only positive numbers in our sequence have indices a subset of $\{0,1,2,x_0\}$. Now, we break into cases based on the value of $x_0$.

If $x_0=1$, note that we only have one $2$, so $x_2=1$. Now, we must have $x_1=2$. No other numbers can be positive, so this only yields $(1,2,1,0)$.

If $x_0=2$, our sequence now has two $2$s, so $x_2=2$. Now, we can have $x_1=0,1$, leading to the solutions $(2,0,2,0)$ and $(2,1,2,0,0)$.

Finally, if $x_0\ge 3$, we must have $x_2=1$ and $x_{x_0}=1$, so $x_1=2$. This yields the aforementioned infinite class of solutions.

Nice problem! Here's my solution: The only possible sequences are $$(2,1,2,0,0),(1,2,1,0),(2,0,2,0) \text{ and } (n-3,2,1,\underbrace{0,0, \dots ,0}_{(n-6) \text{ times}},1,0,0,0) \quad \forall n>5$$

Notice that the sequence consists of only $0,1, \dots ,n$. Consider the sets $A$ and $B$ given by $$A=\{i \text{ such that }\exists k \text{ with } x_k=i\} \text{ and } B=\{i \text{ such that } x_i \neq 0\}$$
Note that the given condition implies that $A=B$. Now, we have $x_0+x_1+ \dots +x_n=n+1$ as the LHS simply represents the total number of times each number $0,1, \dots ,n$ is present in the sequence. Also, $x_1+2x_2+\dots +nx_n$ is the sum of the elements of the sequence. Thus, we get $$x_0+x_1+ \dots+x_n=x_1+2x_2+ \dots +nx_n=n+1$$
$$\Rightarrow x_0=x_2+2x_3+ \dots +(n-1)x_n$$
If there exist $3$ numbers more than $1$ in set $B$, then let $i>j>k \geq 2$ be the three maximal integers in $B$. This gives $$x_0 \geq (i-1)x_i+(j-1)x_j+(k-1)x_k \geq i+j+k-3 \geq i+3+2-3=i+2$$
But then, since $x_0$ lies in set $A$, so $x_0 \in B$ as well, which contradicts the fact that $i$ is the maximal element in $B$. So there are at most $2$ elements in $B$ which are more than $1$. If there are exactly $2$ such elements, then let $i>j \geq 2$ be these numbers. Then $$x_0=(i-1)x_i+(j-1)x_j \geq i+j-2 \geq i$$
As shown above, $x_0 \in B$, which means that we must have $x_0 \leq i$ as well. Thus, equality must hold everywhere, which gives $x_0=i,$ $x_i=x_j=1$ and $j=2$. Then there are $2$ ones in the sequence, and so $x_1=2$. Since $x_0+x_1+x_i+x_j=n+1$, so this gives $x_0=n-3$, and hence $i=n-3$. As $i>2$, so we must also have $n-3>2$, i.e. $n>5$. This gives the infinite class of sequences mentioned in the beginning.

Now $B$ must have atleast one element more than $1$, since otherwise $x_0=0$ (which is not possible as this means that $0 \in A$ but $0 \not \in B$). So suppose $B$ has exactly $1$ element more than $1$, and let this element be $i$. Then $x_0=(i-1)x_i$. If $i>3$, then $x_0 \in B$ gives that some multiple of $i-1$ also lies in $B$, which is contrary to the fact that $i$ is the only element more than $1$ in set $B$. So we have $i=2$, and $x_0=x_2$. Then we get $2x_0+x_1=n+1$. But $x_3,x_4, \dots ,x_n$ are all zeros, and so $x_0 \geq n-2$. Also, $x_2 \leq 2$ since it represents the number of times $2$ appears in the sequence, and we cannot have $x_0=x_1=x_2=2$. Then, as $x_0=x_2$, so we get $2 \leq x_0 \leq n-2$, which gives $n \leq 4$. A simple case bash then yield the remaining answers (in particular, remember that $1 \leq x_0 \leq 2$). $\blacksquare$

Let $S$ be the sum $x_0+x_1+\dots+x_n.$ While $S$ is that, $S$ is also zero times the number of times zero appears in the sequence, plus one times the number of times one appears in the sequence, and so on adding to $x_1+2x_2+3x_3+\dots+9x_9.$ Where $x_i=0$ if $x_i$ does not ever happen in the sequence.
\\$~$\\
Note that $x_{10}=x_{11}=\dots=0$ because numbers $\ge 10$ aren't digits. We have $9\ge x_0=x_2+2x_3+\dots+8x_9.$ When $x_0=0$ this is impossible. When $x_0=1$, then $x_2=1$ and that's it for index greater than one. Thus, $(1,2,1,0)$
\\$~$\\
Note that $x_0=2$ implies that $(x_2,x_3)=(2,0)$ or $(0,1).$ This gives $(2,0,2,0)$ or $(2,1,2,0,0)$ and none for $(0,1)$ because there's a two.
\\$~$\\
For $x_0\ge 3$, note that $(x_0-1)x_{x_0}\ge x_0-1$ so there is one other two and that's it. This gives $(x_0,2,1,0,\dots,0,1,0,0,0).$

The only solutions are $\boxed{(1,2,1,0), (2,0,2,0), (2,1,2,0,0), (x-3,2,1,0,0,\ldots, 0,1,0,0,0)}$ for any $x\ge 6$ (where the entries in $\ldots$ are zero). We can check that these all work.

Notice that \[x_0 + x_1 + \cdots + x_n = 0\cdot x_0 + 1\cdot x_1 + \cdots + n \cdot x_n,\]so
\[x_0 = x_2 + 2x_3 + 3x_4 + \cdots + (n-1)x_n\]

Clearly $x_0>0$.

Case 1: $x_0 = 1$.
Then we must obviously have $x_2 = 1$ and $x_3 = x_4 \cdots = x_n = 0$. Since there is only one zero in the sequence, we have $n=3$. Since $x_0 = x_2 = 1$, we have $x_1  = 2$. This gives $(1,2,1,0)$.

Case 2: $x_0 = 2$.
Then we must have $x_4 = x_5 = \cdots = x_n = 0$ and either $x_3 = 1, x_2 = 0$ or $x_2 = 2, x_3 = 0$.

If $x_3 = 1, x_2 = 0$, then we must have $x_i = 3$ for some $i$, which means $x_1 = 3$, but this is obviously impossible.

Thus, $x_2 = 2, x_3 = x_4 = \cdots = x_n = 0$. Notice that $x_i\ne 1\forall i\ne 1$, so $x_1\in \{0,1\}$. We must have exactly two zeroes, so $x_1 = 0$ implies $n=3$ and $x_1 = 1$ implies $n=4$. This gives $(2,0,2,0)$ and $(2,1,2,0,0)$.

Case 3: $x_0 > 2$.
Let $x_0 = k$. Since $k\le n$, we have $(k-1)x_k \le x_0 = k$, so $x_k$ is at most $1$. Since $x_0 = k$, we have $x_k\ge 1$, so $x_k = 1$. Now, $(k-1)x_k = k-1$, so \[\sum_{i=2, i\ne k}^n (i-1)x_i  = 1,\]so we must have $x_2 = 1$ and $x_3 = \cdots = x_{k-1} = x_{k+1} = \cdots = x_n = 0$. Then $x_1\ne 0$, so there are exactly $n-3$ zeroes in the sequence, which means $x_0 = n-3$ and $n>5$. Now $x_2 = x_k = 1$, so $x_1 = 2$ must hold. This gives $(n-3,2,1,0,0,\ldots, 0, 1,0,0,0)$, as desired.


inequalities
function
complex analysis
linear algebra
linear algebra unsolved

Let $||.||$ be a norm in $M_n(C)$ and $A_1,A_2,...,A_p$ are $n \times n$ matrices in $M_n(C)$.Prove that for any $x>0$ we can take a complex number $z$, s.t. $|z|<x$ and
$|| (I_n-zA_1)^{-1}+(I_n-zA_2)^{-1}+...+(I_n-zA_p)^{-1} || \geq p.$

$f: z \to (Id-zA_1)^{-1}+..+(Id-zA_p)^{-1}$ is defined on a neighboorhood of zero $V$ and is analytic in this neighboorhood. And in the Banach algebra $(M_n({\mathbb{C}}),|.|)$, many results of complex analysis still hold. So one can apply the maximum modulus principle and the fact that $|f(0)|=p$.
PS: I think you meant $>p$ ?

Ooppss, it seems it was more or less (probably more...) trivial. I didn't have this insight... too bad. ANyway, it doesn't work with strict inequality (take all matrices zero). But I have a question: when is it true that the norm of this function is constant? In particular, is it true that in this case all matrices are nilpotent?


[quote="jeez123"]Solve the inequality [b]algebraically[/b]. Support your answer graphically.
$2\cos2x>1$, for $0<x<2\pi$

[hide="What I did"]
$\cos 2x>\frac{1}{2}$
$x>\frac{\pi}{6}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{6}$

This is a "show your work" problem, I guess my solution should have been more algebraic... what steps am I missing? I only got the solutions because there's another similar problem and I get the results (the pies) from there. [/hide]
Thanks![/quote]

to add on you can take $\text{arc}\cos$ of both sides? it does show more work  :D

Yeah, just let $t = 2x$, take the $arccos$ of $t$, and then divide by $2$.

jeez123, set intervals and test the values when it is >1/2 and when it is <1/2. You only found the critical points, now you need to choose the intervals to make your original statement true...


Find all pair of integer numbers $(n,k)$ such that $n$ is not negative and $k$ is greater than $1$, and satisfying that the number:
\[ A=17^{2006n}+4.17^{2n}+7.19^{5n}  \]
can be represented as the product of $k$ consecutive positive integers.

Let $A=17^{2006n}+4.17^{2n}+7.19^{5n},B=A(mod \ 8)$. We have B=4 if n is even, and B=2 if n is odd. Therefore k<4.
1.If k=2 then $A=x(x+1)$ if n>0 then $x>17^{1003n}$. It give contadition. k=2,n=0,A=12,x=3 is solution.
2. k=3 A=(x-1)x(x+1), x is even. $A(mod 5)=2(-1)^n, x(x^2-1)=0,\pm 1.$ Therefore had not solution.


topology
real analysis
real analysis unsolved

This question's been buggin me for a while:

Call a metric space "totally bounded" if, for every $\epsilon > 0$, the metric space is the union of a finite number of closed balls of radius $\epsilon$. Prove that a metric space is totally bounded [b]if and only if [/b]every sequence has a Cauchy subsequence.

Hausdorff's theorem about characterisation of compactness in terms of $\varepsilon$-nets.

for the [b]only if[/b], first for $\epsilon = 1$, the infinite terms of the sequence $\{ x_{n}\}$ contained in a finite set of closed balls of radius $1$. So there must be one ball containing infinite number of terms. Let the term with the smallest index, we shall call $y_{1}$, be the first term in the subsequence.

So for $\epsilon=\frac{1}{2}$, we inside the previous ball of radius $1$. Again. there must be one ball of radius $\frac{1}{2}$ containing infinite number of terms. Again pick the smallest-index one (we shall exclude $y_{1}$) and let it be $y_{2}$.

And so on and so forth. It's not difficult to check that the subsequence obtained is Cauchy.

for the [b]if[/b] part, for any $\epsilon$, we simply pick any one point in the metric space, say $x_{1}$. If the whole space is covered by the ball of radius $\epsilon$, then we are done. Else we pick a point outside the ball, and you carry on checking if the metric space is covered by the union of the two balls. and so on and so forth.

If this process doesnt end, then we've obtained a sequence whereby there is no Cauchy subsequence. which you should be able to check.

yah. there's this concept about $\epsilon$-nets here. but i cant remember exactly what the concept is.

[quote="bythecliff"]Ok, I don't think this problem needs such a complicated theorem. All the concepts we covered so far are closed sets, compactness, connectness, metric spaces, cauchy sequences, etc... of that nature.[/quote]

This problem don't need Hausdorff theorem, this problem is Hausdorff theorem exactly. More precisely it's one of the many reformulation of this theorem.

mao_zai, can you please explain why we can just let our subsequence be $\{y_{n}\}$? I don't understand how we can just pick our subsequence pts to be simply any point contained in a ball with infinite points.

Let $\epsilon =1$. Let $B_{1}$ be the ball of radius 1 containing the infinite number terms from the sequence. Define $n_{1}=\min \{i|x_{i}\in B_{1}\}$. So let $x_{n_{1}}$ be the first term in the subsequence. Define $A_{1}=\{x_{i}|x_{i}\in B_{1}\}\backslash\{ x_{n_{1}}\}$ and note that $A_{1}$ is an infinite set.

Now. the inductive step. Suppose $B_{i}$'s, $A_{i}$'s and $x_{n_{i}}$'s are all chosen. We note that $A_{i}$'s are infinite sets and $B_{i}$ is of radius $\frac{1}{i}$. Moreover,  $A_{1}\supset A_{2}\supset \ldots A_{i}$ and so $n_{1}< n_{2}< \ldots < n_{i}$. For $i+1$, we know that there must be a ball $B_{i+1}$ of radius $\frac{1}{i+1}$ containing an infinite subset of $A_{i}$. We pick $n_{i+1}=\min\{j|x_{j}\in B_{i+1}\cap A_{i}\}$. And let $A_{i+1}=B_{i+1}\cap A_{i}\backslash \{x_{n_{i+1}}\}$.

Hmmm. Seems like i'm making it sound complicated. Errr... You try to picture the metric space as a "net". for radius 1, there must be a hole admitting infinite number of terms. Then within "the hole of radius 1" and looking at "the infinite terms which passed through", there must also be also a hole of smaller radius which admits an infinite number of terms. so, basically, u keep looking closer at each hole (with a smaller radius) and you can see infinite terms inside it. To make a subsequence, just pick one term each time. The rest are just technical details.

mao_zai, can u please explain the other way proof, the one where you start out assuming every sequence of a metric space has a Cauchy sequence, and then going on to show that it has to be totally bounded?

Hmmm. Fix $\epsilon$. Pick any one point in the metric space, say $x_{1}$. If the whole space, $M$, is covered by the ball of radius $\epsilon$ centered at $x_{1}$ (we shall denote as $B_{\epsilon}(x_{1})$), then we are done. Else there exist a point $x_{2}\in M\backslash B_{\epsilon}(x_{1})$. Again, if $M\subset B_{\epsilon}(x_{1}) \cup B_{\epsilon}(x_{2})$, then we are done. Otherwise, we shall pick $x_{3}\in M\backslash B_{\epsilon}(x_{1}) \cup B_{\epsilon}(x_{2})$. So on and so forth.

If this process does not terminate, we have a sequence $x_{1}, x_{2},\ldots$. By the premise, we have a Cauchy subsequence $x_{n_{1}}, x_{n_{2}},\ldots$. So for $\frac{\epsilon}{2}$, there must exist a $N$, such that $d(x_{n_{N}},x_{n_{N+1}})<\frac{\epsilon}{2}$. That is, $x_{n_{N+1}}\in B_{\epsilon}(x_{n_{N}})$, which contradicts the construction.


calculus
derivative
limit
calculus computations

Find the derivative of $ f(x)\equal{}2x^2\minus{}3$ by using the definition of the derivative.

e.g. $ f'(x)\equal{}\lim_{h\rightarrow 0 }{()\equal{}}$ ?

hello
$ f'(x) \equal{} \lim_{h\rightarrow 0 }[ \frac {f(x \plus{} h) \minus{} f(x)}{h}]$
$ \equal{} \lim_{h\rightarrow 0 }[\frac {(2(x \plus{} h)^2 \minus{} 3 \minus{} 2(x)^2 \plus{} 3}{h}]$
$ \equal{}\lim_{h\rightarrow 0 }[2h \plus{} 4x]$
$ \equal{} 4x$
thank u


$ abc = 1$

PT-

$ \frac{1}{1 + a + b} + \frac{1}{1 + b + c} + \frac{1}
{1 + c + a} \leq  \frac{1}{2 + a} + \frac{1}{2 + b} + \frac{1}{2 + c}$

[color=blue]MOD edit:i latexified it [/color]
u can $ \text{\LaTeX}$ [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]here[/url]

look at the two posts next to my solution [url=http://www.mathlinks.ro/viewtopic.php?t=169433]here[/url]

thanks..

i had a different approach of setting a,b,c to be the roots of a trinomial.. it gets long and reduces to sum to squares is positive but that was tedious.

good post some more questions
P.S r u appearing for the INMO-2008

sure about posting :?:
yes i am writing the INMO-2008

ha yeah. i couldve made that clear. yes the posting only. and yes, im writing it this year.


An ABCD square and an any point are given for lying inside this square. Show that the centres of gravity of triangles ABO , BCO, CDO, DAO form the square about permanent field.

I think that the problem is like this,

Let $ O$ be any point inside a square $ ABCD$. Show that, the centroids of $ \triangle ABO, \triangle BCO, \triangle CDO$ and $ \triangle DAO$ forms a square which has an invariant area.

[hide="My solution"]
Its a trivial problem.
Let $ A_1,B_1,C_1,D_1$ be the midpoints of the sides $ AB,BC,CD,DA$ respectively.
So,$ A_{1}B_{1}=\sqrt {2}\frac{AB}{2}$ .Now let the side of the square(that u asked)$ =a$
${ a=\frac{2}{3}A_{1}B_{1}}=\frac{2\sqrt {2}\frac{AB}{2}}{3}=\frac{\sqrt 2}{3}AB$.
So the area of the square only depends on the length $ AB$.

[/hide]
[url=http://img74.imageshack.us/my.php?image=squareqr2.png][img]http://img74.imageshack.us/img74/1226/squareqr2.th.png[/img][/url]

the area is $ \frac{d^{2}}{9}$ by basic proportionality theorem
where d is the diagonal of the square
quite trival :)


hey,

determine all natural such as $ 2^8\plus{}2^{11}\plus{}2^n$ is a perfect square

good luck[/code]

We can check all n up to 8, we find none of them work.
If n is greater than 8, then you can divide the whole number by $ 2^8$ and the result must still be a square.
So $ 9 \plus{} 2^m \equal{} k^2$
$ 2^m \equal{} (k \plus{} 3)(k \minus{} 3)$
Both k+3 and k-3 are powers of 2, and they differ by 6. It is clear that 2 and 8 are the only such numbers. So k=5, m=4. So n=8+4=12.

yes, me i was an other solution

$ 2^8\plus{}2^{11}\plus{}2^n\equal{}m^2$
$ 2^n\equal{}(m\minus{}48)(m\plus{}48)$

$ s$ and $ t$ are two natural such as

$ s\plus{}t\equal{}n$
$ m\plus{}48\equal{}2^s$
$ m\minus{}48\equal{}2^t$

so $ 2^t\plus{}96\equal{}2^s$

so $ 2^t(2^{s\minus{}t}\minus{}1)\equal{}3*2^5$

so $ s\equal{}7$ and $ t\equal{}5$

so $ n\equal{}12$


algebra
polynomial
trigonometry
inequalities
function
complex numbers
rational function

For every positive odd integer $k$, is there a $k$ degree polynomial with integer coefficients where every $tan(\frac{n\pi}{k})$ (where $n$ is an integer) is a root of the polynomial?  If so, what is the general form?

[quote="t0rajir0u"]Sorry, but for $k = 5$ the sum and product of the tangents are incommeasurable.  Note that such a polynomial would probably imply the existence of a polynomial for $\tan n \theta$ in terms of $\tan \theta$, which is not possible.[/quote]

Actually, I think $\prod_{n=1}^{5}\ (x-tan(\frac{n\pi}{5}))=x^{5}-10x^{3}+5x$
So the sum and product of the tangents are zero, which is the coefficient of the $x^{4}$ and $x^{0}$ terms.  But that's still integer coefficients.
And
I don't think my question implies what you're saying it implies.
Although I may be being stupid and wrong lol.
I also edited the phrasing of my question cause it might've been confusing.

[hide="hint"]
there is a result due to Chebyshev which establishes a class of polynomials $T_{k}\in \mathbb{Z}\left[x\right]$ such that $T_{k}\left(\cos{\theta}\right) = \cos{k\theta}$.  One can show this using complex numbers, from which one obtains a general form involving nasty binomials and whatnot.  it's not too difficult of a proof.  Anyway, the important thing that we get from the proof is that $\deg{T_{k}}= k$.  i think there is something similar but a little more complicated for the sine function.  anyway, you could maybe use this to show that such a polynomial $P_{k}$ must be not a polynomial but a rational function. :?:
[/hide]

[quote="Teki-Teki"]i think there is something similar but a little more complicated for the sine function.[/quote]

Such a polynomial exists for odd $k$ only, and I already established the existence of a rational function in this case.

$\prod_{n=1}^{1}\ \left(x-tan(\frac{n\pi}{1})\right)=x^{1}$
$\prod_{n=1}^{3}\ \left(x-tan(\frac{n\pi}{3})\right)=x^{3}-3x$
$\prod_{n=1}^{5}\ \left(x-tan(\frac{n\pi}{5})\right)=x^{5}-10x^{3}+5x$
$\prod_{n=1}^{7}\ \left(x-tan(\frac{n\pi}{7})\right)=x^{7}-21x^{5}+35x^{3}-7x$

This looks [i][b]really [/b][/i]patternish... which is why I brought this problem up.
Pascal, I'm guessing.

Edit:  Wow scorpius gets it like 2 seconds after I post.  Yay.  Although that entire thing sorta went over my head.

[quote="t0rajir0u"]Sorry, I miscalculated.  Here is my issue:

Suppose we had a family of polynomials $P_{k}(x)$ such that

$P_{k}\left( \tan \theta \right) = \tan k \theta$

Then

$P_{k}\left( \tan \frac{n \pi}{k}\right) = \tan \pi = 0$

So this would imply the existence of your desired polynomials; but in fact we have

$P_{k}(x) = \frac{(1+ix)^{k}-(1-ix)^{k}}{i (1+ix)^{k}+(1-ix)^{k}}$

And I can find no simple way of showing that this is a polynomial, if it even is, for odd $k$.[/quote]
$P_{k}$ doesn't have to be a  polynomial! We only need the roots, so we can just take the numerator of the rational function. The numerator will be a polynomial with the desired roots.

And yes, the coefficients are binomial coefficients: you get the coefficient of the $x^{2p+1}$ term to be $(-1)^{p}\binom{k}{2p+1}$.

[quote="bloodysloth"]$\prod_{n=1}^{1}\ \left(x-tan(\frac{n\pi}{1})\right)=x^{1}$
$\prod_{n=1}^{3}\ \left(x-tan(\frac{n\pi}{3})\right)=x^{3}-3x$
$\prod_{n=1}^{5}\ \left(x-tan(\frac{n\pi}{5})\right)=x^{5}-10x^{3}+5x$
$\prod_{n=1}^{7}\ \left(x-tan(\frac{n\pi}{7})\right)=x^{7}-21x^{5}+35x^{3}-7x$

This looks [i][b]really [/b][/i]patternish... which is why I brought this problem up.
Pascal, I'm guessing.[/quote]

The pattern immediately follows once you see the triangular numbers.

[quote="scorpius119"]$P_{k}$ doesn't have to be a  polynomial! We only need the roots, so we can just take the numerator of the rational function. [/quote]

... I figured I was missing something like that.    :oops:


will someone please explain you can apply the trick?

[quote="kyyuanmathcount"]AoPS Wiki has a good explanation and a link to Mr. Rusczyk's video about it.

http://www.artofproblemsolving.com/Movies/simon2.php

for the second video.[/quote]
What about the third?

simon's favorite factoring trick is best when you have an equation with only xy, x, and y as variables.

$xy+x+y+1=(x+1)(y+1)$
$xy-x-y+1=(x-1)(y-1)$
These are just special cases of
$(a+b)(c+d)=ac+ad+bc+bd$.


What is
$\displaystyle \left(1-\displaystyle \left(\frac{1}{2}\right)^2\right)\displaystyle \left(1-\displaystyle \left(\frac{1}{3}\right)^2\right)\displaystyle \left(1-\displaystyle \left(\frac{1}{4}\right)^2\right)...\displaystyle \left(1-\displaystyle \left(\frac{1}{56}\right)^2\right)$

[hide]If you notice that you can break it into sections: $\boxed{\frac34}$, $\frac34 \times \frac89=\boxed{\frac46}$, $\frac46 \times \frac{15}{16}=\boxed{\frac58}...$.After the expression is multiplied by $ (1-(\frac1{n})^2)$, the value is $\frac{1+n}{2n}$.  When n=56, that equals $\frac{57}{112}$.[/hide]

I think.  :P

[hide="Stuff"]
Lol, the problem was meant to be done by the fact that (1 - 1/n^2) can be turned into (n-1)(n+1)/(n^2)... it's basically an advanced telescoping series.

Yeah, when the terms cancel out you're left with (1/2)(57/56), which simplifise to 57/112.
[/hide]


algebra
polynomial
function
trigonometry
calculus
calculus computations

[quote="maokid7"]How would you solve a differential equation in the form

$ay''+by'+cy=d$? Where $a,b,c,d$ are constants.[/quote]

1) y=y_0+d/c, where y_0 - solution of ay''+by'+cy=0.

2) ay''+by'+cy=0.

We find solution as $y=e^{\lambda x}$. We have
$a\lambda^{2}+b\lambda+c=0$. From there

$y=C_{1}e^{\lambda_{1}x}+C_{2}e^{\lambda_{2}x}$, where $\lambda_{1,2}$ are roots of $a\lambda^{2}+b\lambda+c=0$.

If $\lambda_{1}=\lambda_{2}$, then $y=(C_{1}x+C_{2}) e^{\lambda_{1}x}$.

Solution for the general differential equation of the form $a_{n}y^{(n)}+a_{n-1}y^{(n-1)}+...+a_{1}y'+a_{0}y=c$, where $a_{0},a_{1},...,a_{n},c$ are constants, is in the form $y=y_{H}+y_{P}$ where $y_{H}$ is the solution of the homogenous differential equation and $y_{P}$ is the solution of the particular equation. The homogenous equation is solved by substituting $y=e^{zx}$. You get $z^{n}a_{n}e^{x}+z^{n-1}a_{n-1}e^{x}+...+za_{1}e^{x}+a_{0}e^{x}=0$(notice the 0 here, this is why its called the [i]homogenous[/i] equation). Divide thru $e^{x}$ and solve the correspondent polynomial equation. You will get solutions $z_{n},z_{n-1},...,z_{1}$. If all of the solutions are different then $y_{H}=A_{n}e^{z_{n}x}+A_{n-1}e^{z_{n-1}x}+...+A_{1}e^{z_{1}x}$ else if some $z_{1}=z_{2}$ then $y_{H}=A_{n}e^{z_{n}x}+A_{n-1}e^{z_{n-1}x}+...+A_{2}xe^{z_{1}x}+A_{1}e^{z_{1}x}$ else if some $z_{1}=z_{2}=z_{3}$ then $y_{H}=A_{n}e^{z_{n}x}+A_{n-1}e^{z_{n-1}x}+...+A_{3}x^{2}e^{z_{1}x}+A_{2}xe^{z_{1}x}+A_{1}e^{z_{1}x}$ else ... (you get the point, just keep multiplying by x the 'equal' ones) The solution for the particular equation is obtained by substituting $y=b$, where $b$ is some constant, into the original equation. You get $a_{0}b=c$, $y_{H}=\frac{c}{a_{0}}$. And finally, just add $y_{H}$ and $y_{P}$ to get the solution.

Cheers  :P

so if instead of some constant $c$ we had $cx^{2}+dx+e$ we would let $y=f(x)$ where $f(x)$ is a quadradic?

If the right side is a polynomial function the solution should exist. Solve for $y_{H}$ in the same manner and solve the particular equation(note that particular solution is [b]any[/b] that fits) by substituting $y_{P}$ with a polynomial function of the same degree as the right side. Then just plug it in and determine coefficients.

In your example it would go the following way:
$a_{n}y^{(n)}+...+a_{2}y^{''}+a_{1}y^{'}+a_{0}y=cx^{2}+dx+e$
$y_{P}=Ax^{2}+Bx+C$
$y_{P}^{'}=2Ax+B$
$y_{P}^{''}=2A$
$y_{P}^{'''}=0$
$2a_{2}A+2a_{1}Ax+a_{1}B+a_{0}Ax^{2}+a_{0}Bx+a_{0}C=cx^{2}+dx+e$
By grouping coeficients by $x^{2}$, $x$ and $x^{0}$ you get
$a_{0}A=c$ by $x^{2}$
$2a_{1}A+a_{0}B=d$ by $x$
$2a_{2}A+a_{1}B+a_{0}C=e$ by $x^{0}$
Now you got A, B and C. End of story, bye bye  :P

P.S. Sometimes you will have to use euler's formula $e^{ix}=\cos{x}+i\sin{x}$


probability
function
Ross Mathematics Program
quadratics
inequalities
complex analysis
algebra

Can someone please help me understand these problems?

(1) Let P equal the 79th natural number not divisible by 6, and Q equal the 29th natural number not divisible by 2 and 3.  Find Q-P.

(2) Using the digits 0, 1, 3, 5, and 7 w/o repetition, how many 5 digit natural numbers can be formed which are divisible by 5?

--I know that the first number can't be zero and the last number has to be 5 or 0, so I tried using the counting principle, 4 possibilities for the first, 4 for the second, 3 for the third, 2 for the fourth, and 1 for the fifth.  4*4*3*2*1 = 96, which is not the correct answer. I think you need to subtract out the times when 0 and 5 are used up before the 5th digit...?

Thanks!

Good thought for the counting principle, but you've counted the possibilities like 50137.  You're going to have to be a little careful for this one.

A suggestion: [hide]Try different cases for the last place, and then use your same logic for the other 4 digits.[/hide]

(3*3*2*1*1)+(4*3*2*1*1) = 42 So easy, thanks for the suggestion JBL!

Any hints on the other one?

Here's another one:

What is the probability that a three digit natural number will have exactly 2 digits the same?

Here's what I did:

9*10*10 = 900 three digit natural numbers

XXY, XYX, YXX are the three patterns to have 2 digits the same

11Y, 1Y1, Y11 are the patterns for 1, also have this pattern for 2,3,4,5,6,7,8,9.  There also is Y00, but no others w/zero.

3*10 + 1 = 31

31/900 = probability.

However, thats not correct...

I dont get it (hint 444) but it is not the correct answer (I have the solutions).  Can you give me a hint about how to solve it?  Anyway thanks for your time...

It was a hint about what you were doing wrong.  Try re-reading the question, looking over your solution, and thinking about my hint.

OHH I get it!  Like 444 would be a # that I am counting by does not have EXACTLY 2 digits the same.

My work:

What is the probability that a three digit natural number will have exactly 2 digits the same?

Here's what I did:

9*10*10 = 900 three digit natural numbers

XXY, XYX, YXX are the three patterns to have 2 digits the same

11Y, 1Y1, Y11 are the patterns for 1, also have this pattern for 2,3,4,5,6,7,8,9. There also is Y00, but no others w/zero.

3*9 + 1 = 31

31/900 is not right though, not even close....

OKAY im thinking about it......

Okay I think I am close but this is still not correct:

What is the probability that a three digit natural number will have exactly 2 digits the same?

9*10*10 = 900 three digit natural numbers

XXY, XYX, YXX are the three patterns to have 2 digits the same

11Y, 1Y1, Y11 are the patterns for 1, also have this pattern for 2,3,4,5,6,7,8,9. There also is Y00, but no others w/zero.

For 1:

11Y = 1*1*9 = 9 possibilities (b/c Y can equal 0, 2-9)
1Y1 = 1*9*1 = 9 possibilities
Y11 = 8*1*1 = 8 possibilities (b/c Y can equal 2-9 but not 0)

11Y or 1Y1 or Y11 = 9 + 9 + 8 = 26 (I dont really get why Im doing this...)

Now there are 1-9 numbers (X values)

26*9 = 234 +1 (the Y00 case) 235

235/900 = probability

I am SO close, the answer is 27/100 !!

Ohh wait I forgot to add 9 instead of 1 (Y00, Y=1-9). 27% Yea!

Can anyone give me a hint on the original #2 I posted?

[quote="JS1527"]Can anyone give me a hint on the original #2 I posted?[/quote]

Hint:  Use your counting principle, but separately count the case where the units digit is 0 versus the case where the units digit is 5.

[quote="JBL"]He actually meant the #1 he posted.[/quote]

Of course, he did not mean #2.  If I had read your first response, or his counter-response, that would have been perfectly clear.  Sorry.

Unfortunately, that leaves me holding the bag to answer #1.

I contend that this question, as written, is ambiguous.  Although I have a preference, I have not been able to convince myself that one meaning is conclusively superior.

I consulted the "American Heritage Book of English Usage, 1996", online at [url]http://www.bartleby.com/64/C003/021.html[/url].  I also consulted "The Chicago Manual of Style", 13th Edition, and Strunk's "The Elements of Style".

I was unable to find the exact usage, as used in the question, but I was able to find several cases where the guides pointed out ambiguous usage.

Perhaps the closest online example I found is at the link above:

[quote]all in negative sentences.  Be careful with sentences that have an all  not  form.  They may be hazardous to your clarity.  The sentence "All of the departments did not file a report" may mean that some departments did not file or that none did. If you want the first meaning, you can express it unambiguously by saying "Not all of the departments filed a report".  If you want the second meaning, try a paraphrase such as "None of the departments filed a report" or "All of the departments failed to file a report".  Note that the same problem can arise with other universal terms like "every" in negated sentences, as in the ambiguous "Every department did not file a report".[/quote]

In this particular question, the author is using "and" in a negative sentence.  I am wondering how many of the words preceding "2" the question-writer intended to be implicitly duplicated by the word "and"?

I.e., which of the following does the question-writer mean?
[list]- (the 29th natural number not divisible by 2) and (3).
- the 29th natural number not divisible by (2 and 3).
- the 29th natural number not divisible (by 2) and (by 3).
- the 29th natural number not ((divisible by 2) and (divisible by 3))
- the 29th natural number ((not divisible by 2) and (not divisible by 3)).[/list]

I don't believe the question-writer intended any of the former 3 usages.  I don't believe they "make sense" in the context of the problem.

If the reader wonders whether the author of the question meant to use the word "or" in place of the word "and", he/she has a valid question.  The latter 2 usages, above, are equivalent, by de Morgan's Theorem, to the following:
[list]- the 29th natural number ((not divisible by 2) or (not divisible by 3)).
- the 29th natural number not ((divisible by 2) or (divisible by 3)).[/list]

Let's parse "the 29th natural number not ((divisible by 2) and (divisible by 3))":
[list]- "divisible by 2" = {2, 4, 6, 8, 10, 12, ...}
- "divisible by 3" = {3, 6, 9, 12, ...}
- "(divisible by 2) and (divisible by 3)" = {6, 12, ...}
- "not ((divisible by 2) and (divisible by 3))" = {1, 2, 3, 4, 5, 7, 8, 9, 10, 11, ...}[/list]

Let's parse "the 29th natural number ((not divisible by 2) and (not divisible by 3))":
[list]- "not divisible by 2" = {1, 3, 5, 7, 9, 11, ...}
- "not divisible by 3" = {1, 2, 4, 5, 7, 8, 10, 11, ...}
- "(not divisible by 2) and (not divisible by 3)" = {1, 5, 7, 11, ...}[/list]

My preference on parsing this question is to bind as few words by the conjunction as possible.  Thus the phrase I prefer is "not ((divisible by 2) and (divisible by 3))".  As illustrated above, this is equivalent to "not divisible by 6".  This interpretation would also be consistent with a question-writer, who intended a simple, elegant solution, with a simple twist for the student to find.

Let P equal the 79th natural number not divisible by 6.  P=94.  [There are floor(94/6)=15 multiples of 6 from 1 to 94, so there are 94-15=79 non-multiples of 6.]

Let Q equal the 29th natural number not divisible by 6.  Q=34.  [There are floor(34/6)=5 multiples of 6 from 1 to 34, so there are 34-5=29 non-multiples of 6.]

P-Q = 94-34 = 60.

I have no idea where the question originates or the "correct" answer.  Is this correct?

Yes I meant #1 (sorry, my mistake).  I am at school now and will check the answer when I get home...

I am not the author of the problem, so I can't tell you the meaning, and I can't solve the problem (that's why I'm asking for your help), so I don't know which wording the author meant.

My guess would be
P = the 79th natural # not divisible by 6
Q = the 29th natural # divisible by 2 and divisible by 3 (ie divisible by 6, right?)

I think that the "not divisible" goes w/P only...

Anyway the answer is 80.

For your work, can you explain what "floor" means?

Your original problem statement:
[quote="JS1527"](1) Let P equal the  79th natural number not divisible by 6, and Q equal the 29th natural number not divisible by 2 and 3. Find Q-P.[/quote]

[quote="JS1527"]I think that the "not divisible" goes w/P only...

Anyway the answer is 80.

For your work, can you explain what "floor" means?[/quote]

You repeated the phrase "not divisible by" in both the P and the Q clauses of the original problem statement.  Did you paraphrase the original problem or did you misquote it?  Please tell us the source of the problem.

I agree with you, if the problem had asked:
"Let P equal the 79th natural number not divisible by 6, and Q equal the 29th natural number divisible by 2 and 3. Find Q-P."

Then there are no ambiguities.  The problem is simple and straightforward, and the given answer matches my calculations.  P=94 (as shown previously).
Q=29*6=174.  Q-P=80.

To answer your other query, floor(x) gives the greatest integer less than or equal to x.  (I.e., rounded down.)  As I used the function, floor(n/m) gives the number of times a multiple of m occurs among the first n natural numbers.

Ohh Im so sorry, you are correct:

Let P equal the 79th natural number not divisible by 6, and Q equal the 29th natural number divisible by 2 and 3.

SO sorry.

Please forgive my ignorance.  I did not mean to make you do all of that work just b/c I made a typo.  Sorry again!

JS -- instead of posting many messages in a row, you can use the "Edit Post" button on the upper-right-hand corner of each message to add, remove, or change text in any post you make.

JBL -- Thanks for the hint.  I'm a new user (as you can see).

rcv -- If you can forgive me for my copying error I have some questions about your work:

[quote]Let P equal the 79th natural number not divisible by 6. P=94. [There are floor(94/6)=15 multiples of 6 from 1 to 94, so there are 94-15=79 non-multiples of 6.]
[/quote]

I don't understand how you got 94.  It seems like you knew the answer and worked backward to show how it worked.

[quote]Q=29*6=174. [/quote]

I also don't know why you multiplied here.  It seems like since floor(29/6) =4, there are 4 multiples of 6 in the first 29 natural numbers.  The only way would be able to do it was to do a list (6, 12, 18, 24, all the way up to the 29th #).  Your way is better, I just don't get why Q=29*6...

[quote="JS1527"]rcv -- If you can forgive me for my copying error I have some questions about your work:[/quote]

It's OK, JS1527.  Please copy carefully, hereafter.

Let's start with the rule that floor(n/m) returns the number of factors of m between 1 and n.  Let's start with an easy example -- factors of 3.  floor(1/3)=0, floor(2/3)=0, floor(3/3)=1, floor(4/3)=1, floor(5/3)=1, floor(6/3)=2.

Do you see the pattern?  Every 3rd increment in the numerator, n, results in 1 additional factor of 3.  Try it for m=6 and m=7 if you don't see it.

As far as finding the 29th multiple of 6, I simply multiplied 29 by 6.  The first multiple of 6 is 6=6*1.  The second multiple of 6 is 12=6*2.  The third multiple of 6 is 18=6*3.  Thus the 29th multiple of 6 should be 174=6*29.

If you are unsure that this is true, you could check that 174 is a multiple of 6, and you could use floor(174/6) to see how many multiples of 6 exist between 1 and 174.

The other problem (calculating P) did require a little trial and error.  To begin with, I knew that 5/6 of the natural numbers were not divisible by 6.  So I took the goal (the 79th such number) and multiplied by 6/5.  79*6/5=94.8.

Next, I figured the result would be near 94 or 95, so I could have used guess and check to find the correct value.

floor(95/6)=15.  But 95-15=80, so this was not the correct answer.  I need a slightly smaller answer.  Next I tried floor(94/6)=15.  Now 94-15=79, so we have proven that 94 is the 79th natural number not divisible by 6.

I hope this helps.

Thanks for the response.

One more question:

[quote]The other problem (calculating P) did require a little trial and error. To begin with, I knew that 5/6 of the natural numbers were not divisible by 6. So I took the goal (the 79th such number) and multiplied by 6/5. 79*6/5=94.8.
[/quote]

Why do you multiply 79 by 6/5 to approximate the P value?  I don't understand why this works.

Thanks!

1/6 of all numbers are divisible by 6.  Thus 5/6 are not.  Thus, in general, we expect 5/6 of the numbers less than a given number to be not divisible by 6.  Thus, the nth non-multiple of 6 is in general roughly equal to 6n/5.

[quote]Thus, the nth non-multiple of 6 is in general roughly equal to 6n/5.[/quote]

Why is this though?  I understand the first two statements you made, I just don't see why they show this one.

Let us consider the number n, or integers near it.  About 1/6 of the numbers less that this number are divisible by 6.  Thus, about n/6 of the numbers less than n are divisible by 6, so n is approximately the 5n/6th number not divisible by 6.  Letting n = 6m/5, we have that 6m/5 is approximately the mth number not divisible by 6.

[quote]Let us consider the number n, or integers near it. About 1/6 of the numbers less that this number are divisible by 6. Thus, about n/6 of the numbers less than n are divisible by 6, so n is approximately the 5n/6th number not divisible by 6.[/quote]

Gotcha.

[quote]Letting n = 6m/5, we have that 6m/5 is approximately the mth number not divisible by 6.[/quote]

Sorry but I don't get this.

Is there any way that you can explain this to me using dimensional analysis sortof method?  This usually answers my questions about WHY you multiply/divide in other problems.

We have "n is the 5n/6th number not divisible by 6."  We want an expression that looks like "The mth number not divisible by 6 is l," where l is something in terms of m.  So we just have to equate terms, since the letters in the two different sentences represent the same things: m = 5n/6, l = n.  Solving this equation for l, we get l = 6m/5.

Another way of thinking about it is this:  We have a list of numbers of length A which we put through a process.  No matter what A is, our process makes the list A into a list of length 5A/6.  Now, suppose you have the value B that comes out of the process, and you want to know what A went in.  Well, you just plug them into the expression B = 5/6 * A, and what pops out is that A = 6B/5.  So, in our particular example, we know that 79 came out at the end, so 79 = 5/6*x, which gives us x = 6/5*79, or about 94.

(All of this is approximating, of course -- you have to round off decimals and guess a little bit at the end no matter what.)

By the way, did you know that there is a fairly general, exact method for finding the nth positive integer that is not <fill in the blank>?

Let F(1), F(2), ..., be a (strictly) increasing sequence of positive integers.  These are the numbers that we are trying to avoid.  We will try to find a formula for the nth positive integer that is not in the sequence F.

Let f(n) = F(n) - n.

Let g(n) be the number of positive integers m such that f(m) < n.  Note that g is kind of like the inverse function of f.

Let G(n) = g(n) + n.

Then G(n) is the nth positive integer that is not in the original sequence F(1), F(2), ....

Let's try it on the example you were looking at: the 79th positive integer that is not divisible by 6.  We are trying to avoid multiples of 6.  So let F(n) = 6n.  Then f(n) = 5n.  With a little algebra, we find that g(n) = ceiling(n/5) - 1.  Thus G(n) = n + ceiling(n/5) - 1.  In particular, G(79) = 79 + 16 - 1 = 94, which agrees with the previous work.

I learned this method from Ross Honsberger's book, [i]Ingenuity in Mathematics[/i] (Chapter 12 on Complementary Sequences).

Challenge problem:  Find a formula for the nth positive integer that is not a perfect square.

[quote]Challenge problem: Find a formula for the nth positive integer that is not a perfect square.[/quote]

I can't look at this problem right now but I will look at it when I get home later tonight.  Would you guys mind waiting and letting me answer it?  Thanks!

--Julie

Okay, I usually do not take so long to understand a problem, but I think that I get it.  I will have to do more work on Ravi's question though.  Could you guys by can chance post a random question like the one I had so much trouble understanding (just the find P part) and let me answer it, then tell me if its correct?  Sorry to bother you but it would def. help since I need much more practice w/this.

Okay, what is the 100th positive integer that is not a multiple of 4?

1,2,3,__,5,6,7,__...

1/4 natural #s are multiples of 4, 3/4 natural numbers are not multiples of 4.

You have the number n.  n is the 3n/4 natural number that is not a multiple of 4.

The mth natural number that is not a multiple of 4 is L.

L=n  m=3n/4  n=4m/3

The mth natural number that is not a multiple of 4 is 4m/3.

m=100, 4m/3  :approx: 133

floor(n/m) returns the number of multiples of m between 1 and n
floor(100/4) = 25.  There are 25 multiples of 4 between 1 and 100.

133-25 dosent = 100

So the answer is 125.

??????

Ravi B wrote:Challenge problem: Find a formula for the nth positive integer that is not a perfect square.

Ok, here's my algebra (in spolier):

[hide]

First, you know that F(n)=n-floor(sqrt n) or y=x-sqrt(x) (I'll reintroduce the floor in a bit), now we want to solve for the inverse of the statement.

y=x-sqrt(x)

x=y-sqrt(y)

sqrt(y)=(y-x)

y=y^2-2xy+x^2

0=y^2-2xy-y+x^2

0=y^2+y(-2x-1)+x^2

now solve the quadratic

(-(-2x-1)+(-)sqrt((-2x-1)^2-4*1*x^2))/2*1=y

so (2x+1+sqrt(4x+1))/2=y

now floor the entire function

y=floor((2x+1+sqrt(4x+1))/2)

[/hide]

Note: NAS asked me if he could do this problem.  Just so nobody will get mad at him for ignoring my request. =)

[quote="JS1527"]floor(n/m) returns the number of multiples of m between 1 and n
floor(100/4) = 25.  There are 25 multiples of 4 between 1 and 100.[/quote]

You don't want to know the number of multiples of 4 less than 100 ... you want to know the number of multiples of 4 less than 133, to see if it is in fact the number you need or whether it might be 134 or something.

Okay I changed it:

1,2,3,__,5,6,7,__...

1/4 natural #s are multiples of 4, 3/4 natural numbers are not multiples of 4.

You have the number n. n is the 3n/4 natural number that is not a multiple of 4.

The mth natural number that is not a multiple of 4 is L.

L=n m=3n/4 n=4m/3

The mth natural number that is not a multiple of 4 is 4m/3.

m=100, 4m/3  133

floor(n/m) returns the number of multiples of m between 1 and n
floor(133/4) = 33. There are 33 multiples of 4 between 1 and 100.

133-33 = 100

So the answer is 133.

??????

I don't think your formula is quite right.  For example, your formula says that the second positive non-square is 4, when it is actually 3.  Similarly, your formula says that the sixth positive non-square is 9, when it is actually 8.

I think the part where you went astray is by dropping the floor function, only to add it back at the end.  You never justified why that transformation would preserve the answer (it doesn't).

Ok, assuming my origanal formula is right [hide]F(n)=n-floor(sqrt n)[/hide]

how do I find the inverse it while keeping the floor function?

To tell the truth, I would have solved the problem by using the 4-step method I described in my previous post.  Namely, find the function F(n) [different from your function F], then find f(n), then find g(n) [the hardest part], and finally find G(n).  It works out well.

Alternatively, in your method, you have observed that if m is a nonnegative integer, then the number of non-squares between 1 and m (inclusive) is m - floor(:sqrt: m).  Thus, to find the nth positive non-square, you have to find the smallest integer m such that m - floor(:sqrt: m) :ge: n.  But that inequality is equivalent to m - n :ge: floor(:sqrt: m), which is equivalent to m - n > :sqrt:(m) - 1.

In general, a good way to deal with floor(x) is to note that it is the unique integer such that x - 1 < floor(x) :le: x.


function
limit
complex numbers
absolute value

Consider the sequence-based function defined by:

$a_{0}(x)=x$
$a_{n+1}(x)=\sqrt{1+a_{n}(x)}-a_{n}(x)$ for positive integers $n$
$f(x)=\lim_{n\to\infty}a_{n}(x)$

a. With justification, find the values of $x$ for which $f(x)$ converges.

b. With justification, find the values of $x$ for which $f(x)$ converge through only real numbers (i.e., there is no $n$ such that $a_{n}(x)$ is not real)

c. With justification, find a closed form expression for $f(x)$

[hide]If $a_{n}(x)$ converges at all,
x=$\sqrt{1+x}-x$ or $4x^{2}=1+x$ or $4x^{2}-x-1$ so x=$\frac{1\pm\sqrt{17}}{8}$, but the negative root is untrue since $a_{n}(\frac{1-\sqrt{17}}{8})$ is positive.
If x<-1, then $a_{n}(x)$ is imaginary and x>3, $a_{n+1}(x)$ imaginary.  Therefore, if -1<x<3, then $a_{n}{x}$ will be purely real.
To prove convergence, assume $ki+r=a_{n}(x)$.  Now, $a_{n+1}(x)$ will have an imaginary part that has an absolute value less than k, because the sign of the imaginary part under the radical is opposite that of the -ki being subtracted.  This process, when repeated infinitely, removes the imaginary part from consideration.
I can't seem to show that the reals converge.  I think what needs to be proved is that the real part of $a_{n+2}(x)$ is always between the real part of $a_{n}(x)$ and $a_{n+1}(x)$, but I can't seem to do it.[/hide]


[quote="orl"]Recently there has been a thread called: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=13035]Is power inherently evil?[/url] In one of the most recent issues of the [url=http://www.economist.com/]"The Economist"[/url] I read that based on the massive complaints of unmoral behaviour of leading managers, especially MBA graduates, in industry many top business universities, e.g. London School Of Economy, Havard etc., introduce courses on entrepreneurial ethnics. Do you think this will change the situation ? Does money contradict ethnics ?[/quote]

Reallity is that the problem is about your own principles, and your own will. Unfortunatelly, that seems not to be something that many people worry to have. Of course that the human been is greedy by nature, that is what made us build cities, create currency, etc, and well, when you see that you can obtain with ease something if you just ignore a little your own principles, well, what is better, $10K this week and I forget my principles, or I stand with them and the hell with that much money.

The problem, actually is quite complex.

Best,


function
limit
real analysis
real analysis unsolved

If the conditions is not strong enough try to add this:

there exists $ c \in [a,b]$ s.t. $ f(c) \equal{} g(c)$

The added condition actually [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=175804]follows from the preceding ones[/url].

It's easy to see that $ f$ has a fixed point, i.e. $ f(x_0)\equal{}x_0$ for some $ x_0$. Then observe that $ x_1: \equal{}g(x_0)$ is also a fixed point for $ f$, because $ f(g(x_0))\equal{}g(f(x_0))\equal{}g(x_0)$. Furthermore, all elements of the sequence $ x_{n\plus{}1}\equal{}g(x_n)$ are fixed points for $ f$. Choose a convergent subsequence: its limit is a fixed point for both $ f$ and $ g$, as required.

Incidentally, this answers the question I raised in the thread linked above.

Could you explain how does the limit be a fixed point of $ g$? What about $ x_2 \equal{} x_0$?

Sorry, my "proof" is wrong, and I even don't see a way to fix it.  :( Done too much grading recently.

Whatever, inspired by your solution to that problem, I find this proof:

[hide]
Suppose there is no common fixed point for $ f$ and $ g$. Then $ f(a) > a$ or $ g(a) > a$. If it's the former case, let $ s \equal{} \min\{x : f(x) \equal{} x\}$ (since f is continuous). $ f(x) \ge x$ when $ x \in [a,s]$. And $ m \equal{} \min\{f(x): x \in [a,s]\} > a$, so $ f([a,s]) \subseteq [m,b] \subset [a,b]$. Let $ I_1 \equal{} f([a,s]) \cap g([a,s])$, we have $ f(I_1) \subseteq I_1$, $ g(I_1) \subseteq I_1$, $ I_1 \subset [a,b]$. Do the same when $ g(a) \neq a$. This process can be continued, and we obtained a closed interval sequence $ I_1 \supset I_2 \supset \cdots \supset I_k \supset \cdots$. Let $ I \equal{} \bigcap_{k \equal{} 1}^{\infty}{I_k} \equal{} [c,d]$, $ d \ge c$. We are now to prove that $ f(c) \equal{} g(c) \equal{} c$.

Since $ h(x) \equal{} \min\{f(t) : x \le t \le c\}$ is continous. For any $ k \in \mathbb{N}$, $ I_k \equal{} [c_k, d_k]$ and $ h(c_k) \le c_{k \plus{} 1} < c$. Since $ {c_k}$ convergent to $ c$, $ h(c) \equal{} \lim_{k\to\infty}{h(c_k)} \le \lim_{k\to\infty}{c} \equal{} c$, but $ h(c) \equal{} f(c) \ge c$ by $ f(I) \subseteq I$. By the same way we deduced that $ f(c) \equal{} g(c) \equal{} c$, a contradiction.

So there must exists a common fixed point.
[/hide]

sorry, my proof is also wrong.. this problem is from zorich's mathematical analysis chapter * Continuity exercise 4 c).
Our teacher said that there is a counterexample.

[color=green]I'm sorry but i saw a different problem :

Let two continuous functions $ f: [0,1] \longrightarrow [0,1]$ and : $ g: [0,1]\longrightarrow [0,1]$ so that $ f(g(x))\equal{}g(f(x))$ for all $ x\in\;[0,1]$ and $ f$ is a decreasing function .Prove that has at least a number $ c\in\;[0,1]$ so that $ f(c)\equal{}g(c)\equal{}c$ [/color]

[quote="quangpbc"][color=green]I'm sorry but i saw a different problem :

Let two continuous functions $ f: [0,1] \longrightarrow [0,1]$ and : $ g: [0,1]\longrightarrow [0,1]$ so that $ f(g(x)) \equal{} g(f(x))$ for all $ x\in\;[0,1]$ and $ f$ is a decreasing function .Prove that has at least a number $ c\in\;[0,1]$ so that $ f(c) \equal{} g(c) \equal{} c$ [/color][/quote]
If $ f$ is a decreasing function,then $ f$ has only one fixed point.The solution follows from molk's argument obviously.


geometry
trigonometry
rectangle
Pythagorean Theorem

[hide="If we must prove that XCD is equilateral..."]This only works if $ABCD$ is a square.  Obvious counterexample below.

WLOG let the square have side length 2.  Looking at isosceles $\triangle ABX$, the altitude from $X$ is also a median.  Let $M$ be the midpoint of $\overline{AB}$.  We have right $\triangle AMX$ with $AM=1$.  So $MX=\tan\frac{pi}{12}=2-\sqrt{3}$.  Extend $\overline{MX}$ to $N$ on $\overline{CD}$.  It can be shown that $AMND$ is a rectangle by using parallel lines and right angles.  So $MN=AD=2$, therefore $XN=\sqrt{3}$.  Also, $DN=AM=1$.  So $N$ is a midpoint, hence $CN=1$.  By the Pythagorean Theorem, $CX=DX=2$.  So we have $CX=DX=CD=2$, and we are done.[/hide]


Harvard college calculus Yale AMC USA(J)MO USAMO

Hi, Harvard students,

JBL last year kindly shared Web links to the Math 25 and Math 55 courses at Harvard in the 2003-2004 school year:

http://abel.math.harvard.edu/~laurent/math25/math25.html

http://www.math.harvard.edu/~siu/math55b/index.html

Now a new school year has begun, and I think the course Web sites this year are

http://abel.math.harvard.edu/~tomc/math25/

and

http://abel.math.harvard.edu/~schmid/55a/

Is any AoPSer taking either of those courses this year?

There are some Harvard Web pages that provide information about the two courses, for example

http://math.harvard.edu/undergrad/Pamphlets/freshmenguide.html

I thought these courses deserved their own thread on AoPS.

Out of curiosity, what is so special about these two classes?

That characterization is not really correct.  Most colleges have an introductory multivariable calculus course of some sort.  At Harvard, that course is Math 21.  Many schools also have a more theoretical version of such a course, aimed at potential math majors.  Such a course would introduce methods of proof, probably do some linear algebra, and be generally more rigorous in its approaches to its material.  At Harvard, that class is math 23.  Very few other schools have courses intended to be taken by freshmen above this level.  Harvard has 2, math 25 and math 55.  Unlike many courses, they are designed to be taken for a full year, so the material from the two semesters is highly integrated (unlike what you find in math 21, where math 21A is multi and math 21B is lin. alg, and anyone can take both or either of those courses in whatever order they choose).  The two courses assume a very strong mathematical background, in particular with respect to ideas like formal proof and problem solving skills.  They both go in-depth into their material and treat it with a high degree of rigor.  (Last year, the math 55 class began by rigorously defining the real numbers, for example.)  They are, in general, much more like the type of courses that get called "analysis" elsewhere than the types of classes that get called "calculus."  This is unusual for freshmen classes.  Also, they both have quite impressive workloads and are taken only by people who are pretty serious about math.  So, not that math 55 is unpopular, it's just that it's a crazy amount of work so no one who didn't really care would ever take it.  Math 25 is somewhat less intense than math 55.

I do know of one other school with a similar course: the Analysis class taught by Paul Salley (Sally?) at the University of Chicago struck my as being reasonably similar.  It is not limited to freshmen, however, and is taken by both freshmen and sophomores (and maybe even some upperclassmen).  I do not know for a fact that it is at the same level, but it's certainly a very high-quality class that approaches math with the same level of seriousness, if perhaps not the same workload, as math 25 and 55.  Also, Sally (Salley?) is just amazing :).

Anyhow, that's everything I can tell you about math 25 and 55.  I can swear to the reasonable accuracy of my descriptions of the courses, but I don't really know that much about what most other colleges offer along these lines.  My assertations that they are unique courses are hearsay from people who would be more likely to know than I and small amounts of anecdotal evidence I have collected.  For example, a math major friend of mine at Yale took their hardest freshman level class and, upon looking through the math 23 course materials, guessed that it was at a similar but slightly lower level.  In other words, nothing whatsoever like math 55.

Hi, JBL,

Thanks for the further details about various freshman math courses at Harvard.

BTW, the correct spelling is "Paul Sally," which I just checked by Googling. I know some U. of Chicago alumni here, and they are very impressed with him, as I also am by knowing them, his students.

My interest in the Harvard math courses goes back quite a few years, to some occasion when I was browsing the Web and I saw, on Harvard's Web site, a whopping FOUR different courses (as you have described) that all seemed to be titled "multivariable calculus and linear algebra." Of course I guessed that the courses with the higher course numbers are harder, but in those days I didn't see the Web pages that explain the differences among those courses

http://math.harvard.edu/undergrad/Pamphlets/freshmenguide.html

http://math.harvard.edu/undergrad/Pamphlets/which_course.html

and I think the page that simply lists the courses

http://www.math.harvard.edu/teaching/courses.html

didn't used to have such detailed descriptions of each course. So I wondered, for years, what made one course different from the others. Over the years, I've gradually heard various legends about the Math 55 course, which seems to be the specific attraction of Harvard to some applicants, and so I'm interested in getting a reality check on the legends. I am very grateful to the Harvard faculty for putting more and more course syllabuses and homework sets on publicly accessible Web sites, which gives all of us on the outside a better chance to understand how Harvard math courses are conducted.

[quote="MithsApprentice"]I don't know much about Math 25. But apparently Math 55 is a special program that teaches all of undergrad math in, what, 1 year? It's highly accelerated, not very popular, and extremely difficult. :-)[/quote]

I remember that John Scholes once wanted to put up a script/book for an accelerated version of an undergraduate program as he felt that most ones are too basic and slow in his opinion.  :lol:

My two math goals right now are make USAMO and test into Hon. Analysis with Sally next fall (I got in!  Hooray!).  Though, I predict the first will fail and the second will suceed.

[quote="orl"]I remember that John Scholes once wanted to put up a script/book for an accelerated version of an undergraduate program as he felt that most ones are too basic and slow in his opinion.  :lol:[/quote]By the way, orl, do you know whether Scholes is planning to update his kalva site?  Several months ago, I e-mailed him the solutions to 3 olympiad problems that he requested on his site, but I never got a reply. :(

I emailed him awhile back with a new solution to IMO 2000/A2 awhile back and didn't get a reply either.

I have a few more questions about the difficulty of the courseload for Math 23, 25, and 55- How much experience have students had in extracurricular mathematics or college courses before taking one of those classes? I understand that Math 55 is probably the domain of IMO medalists, but what about the other two courses? For example, would an AIME qualifier with only AP math experience be able to perform at a high level in 23 or 25? Do most of the students taking one of those courses already have another multivariable course under their belts before they take 23 or 25? Should a student be at least at the USAMO level before they can handle the rigorous proof-style of 25?

I know this was a rather long list of questions. Thanks for the information.

From what I've seen, 25 is a typical rigorous type analysis/algebra course, with the difference that the problemsets are harder than you would see at most universities (though JBL can surely tell you more)

A student with no analysis background will have difficulties most likely; I believe they use Rudin, and Rudin is legendarily terse.  An AIME qualifier with only AP calculus would find this quite difficult; its amazing how many unwarranted assumptions you make when learning calculus for the first time.

I still think the most important prereq is a familiarity with analysis (not necessarily in metric spaces) and with basic linear algebra.  But I could be wrong.  And of course, problem solving skills translate over.

[quote="Fountainhead"]How much experience have students had in extracurricular mathematics or college courses before taking one of those classes? I understand that Math 55 is probably the domain of IMO medalists, but what about the other two courses?  For example, would an AIME qualifier with only AP math experience be able to perform at a high level in 23 or 25?[/quote]
So, math 23 should be within the range of BC calc students.  It really does not assume any experience with rigorous proof, although if one isn't putting in sufficient work (or I suppose, doesn't have the capacity, although that is more theoretical than likely) to grasp how to prove things, the course could get very difficult very quickly.
Math 55 is largely, but not exclusively the domain of IMO participants -- other students do take it.  (I have several friends who have not been IMO participants who took or are taking 55, including people who didn't ever make USAMO.  However, many of those people had other experience with proofs, for example through college courses.)  The number of people who take it depends on the professor: last year, I believe there were ~20 or 25 students in 55 when Prof. Siu taught.  This year, Prof. Schmid is teaching it and the class size is closer to 10 or less.
So, let me try and give a somewhat revealing anecdote.  My year, math 25 started with 25 students.  In registering for the spring, only 9 of those people re-signed up.  I found the course difficult, but not too difficult: in the fall, I had a roughly 66 (out of 100) average on our exams and problem sets, which was good enough for an A.  However, there were students who were spending 20 hrs. a week on their problem sets and doing appreciably worse than that.  A number of students had test averages of 30 (out of 100) and below.  Most of these students dropped down to 23, where they found it to be not very tough.  I have friends of a perfectly-smart-but-not-particularly-mathy variety who took math 23 and found themselves spending 10-15 hrs. per week on the problem sets.  (These would be people who took BC calc, but probably nothing higher, and likely had never done proofs before.)  Many (30 or more?) students dropped math 23 mid-year.  However, only one person dropped down from math 55 to math 25, and I tend to think of him as a student of physics rather than math.
For students deciding between math 25 and 55 (such as myself), there is a considerable grace period in which to make the decision.  (In general, at Harvard, registration happens after one week of classes have passed.  In this case, they give you at least 2 additional weeks to make the switch with not penalties whatsoever, and the classes are supposed to parallel each other somewhat for this time.)
[quote]Do most of the students taking one of those courses already have another multivariable course under their belts before they take 23 or 25?[/quote]
For 25, almost all of us did.  (There was at least one student in 55 who had never learned integral calculus, so the first definition of an integral he became familiar with was Rudin's "the infimum over all partitions of the sum of the upper rectangles, if and only if it is equal to the supremum over all partitions of the sum of the lower rectangles."  However, people who get 41/42 at IMO should be considered exceptions to the rule :) )  For 23, it certainly isn't necessary but it wouldn't hurt, either.  I have to say that from time to time in 25 I found myself realizing that we were covering something that I had done in multivariable calculus in high school and it just was appearing in entirely different form.  Also, anyone who seriously believes that she will be a math major should take 21 instead of 23 -- if you can't cut 23, there's no way you can find 8 math classes to meet the major requirement.  And I would make the personal suggestion that anyone at least sit in on 25 for a while to see if they feel comfortable with it, although it is by no means a requirement to major in math that one take 25 or 55.
[quote]Should a student be at least at the USAMO level before they can handle the rigorous proof-style of 25?[/quote]
Well ... you might get different answers from asking the 9 of us who stuck with 25 than if you asked the 16 who dropped down (at least 1 of whom [i]was[/i] USAMO quality or thereabouts and dropped because he didn't want to be the worst student left in the class).  The beauty of Harvard's registration system is you actually get a real chance to see if you can deal with the class before you commit to it, and the math department is quite understanding of folks who can't decide between the 2X's.  So, I would say that the best way to answer that question is to try to do a problem set.
[quote]I know this was a rather long list of questions. Thanks for the information.[/quote]
Sure, my pleasure.  I think I'd say one more thing, which is that in all of these classes, many students work in groups.  In the second semester of math 25, we had class that ended at 11 AM Friday and problem sets due at 5 PM, and we just all piled into a room in the library and spend much of those six hours finishing off our problem sets.  And it was really fun!  Much more fun than the first semester had been, when I had worked primarily on my own.  I think finding a good group to work with can really make the difference in how much one enjoys (and whether or not one survives) these classes -- there has to be a relative evenness of level and an actual ability to accomplish something.

Whew!  I suppose one more comment: these classes do change.  The professor who taught me 25 is doing research somewhere in France now and the professor for 55 this year is different from last year's.  I think that the general parameters of difficultly and whatnot should be relatively constant, but there's no way to be sure until one actually sits down in the classroom.

Okay, so now I've lied about being finished twice.  For what blahblahblah said, I don't think he's far off the mark.  My year, we didn't use any textbook at all, so I've never actually used Rudin, but I think it's being used this year.  (I think 55 always uses it.)  I'm not really sure how much analysis background is needed -- I only took the class with my own background, which while not truly outstanding compared to many people at Harvard and on this site, was certainly very strong.  I think problem solving skills make a major difference, probably between a 5-10 and 10-20 hour problem set.  I certainly think I could have done the 23 problem sets twice as quickly as many people in 23 just on the strength of my math team-related knowledge and pieces of multivariable calculus.  I would encourage any student coming to Harvard and looking at majoring in math to sit in on {23, 25} or {25, 55}, depending on their strength.  I think that 23 probably pulls more students down who would be in 25 than students up who would be in 21, if it didn't exist, but there definitely is a class of math-interested students without a strong enough background to get enough out of it for what they'd have to put in.

Incidentally, my reasons for taking 25 rather than 55 were a preference for the 25 professor (for which many of my fellow 25ians have given me odd looks since) and a believe that I would not put sufficiently much extra work into 55 to actually gain a personal benefit over 25.  So I took 25 and had a grand old time.

Now, I'm really done :)

as far as i know there is also another website about 55 taught in 2002 by Professor  Noam D. Elkies
the site is  http://www.math.harvard.edu/~elkies/M55a.02/index.html
               http://www.math.harvard.edu/~elkies/M55b.02/index.html
there are many lecture notes there

by the way , is there anyone who knows that how to get the lecture notes from Prof. Siu and Prof. Schmid or can anyone be so kind to send it to me by email: ltsingeo@gmail.com      thank you very much!

[quote="JBL"][quote="ltsin"]by the way , is there anyone who knows that how to get the lecture notes from Prof. Siu and Prof. Schmid[/quote]

If they aren't available on those professor's websites, they probably aren't publicly available at all.  Sorry.[/quote]

I'm reviving this old thread to check the statement above. For a year or two, it was possible to follow the links in this thread to find the lecture notes for various years of the Harvard Math 55 course, which looks quite interesting. Are there any current Math 55 students here on AoPS? I'm wondering if Harvard has changed its policies about Web access systematically, because it seems that now many course Web pages are password-protected, which didn't used to be the case.

Does anyone have any public links that work anymore for accessing lecture notes for Harvard's Math 55 course (from any year, but especially from this year)? It's regrettable that Harvard hasn't come up with something quite like MIT's [url=http://ocw.mit.edu/index.html]OpenCourseWare[/url] initiative, which although only a partial representation of MIT lecture notes is a very helpful resource for learners of math and many other subjects.

Yes, and Elizabeth Denne is teaching 25.

[quote="blahblahblah"]I think Noam Elkies taught it this year; check his website.[/quote]

Thanks, blahblahblah, that worked exactly. A Google on the professor's name leads to his Math Links page

http://www.math.harvard.edu/~elkies/math.html

which in turn leads to the home page of this year's (so far, last semester's) Math 55 course.

http://www.math.harvard.edu/~elkies/M55a.05/index.html

And that page leads to all the lecture notes, Q.E.F.

Thanks too to JBL; a similar Google process leads to this year's Math 25 notes.

Noam Elkies' website contains material and notes from other courses he's taught in the past; I've found these quite useful on occasion.