[quote="ragnarok23"]Bah, I wish there was another way, but- you really must be outgoing. Go talk to people and don't wait for other to talk to you. That's a start.
I am impossibly shy, and thus, I've got no friends. :([/quote]

could u plz let me be ure friend? ;)

I have some friends.  I got them by just hanging out with them and eventually, they just started talking to me and thought of me as a friend.

Cheer up, everyone! I think the average number of  friends an Aopser has is less than 1!
It's okay to be antisocial! :blush: (just kidding...or maybe not?)

Anyways, just find somebody with similar interests with you and it will work out fine.  ;)

Yes, but that is easy to say. I have no idea how to become like everyone else. The thing is, a very large portion of the time, I am home, reading something on an academic subject, I have not even been to the tons of palces most people my age go, like I have never been in bars or nightclubs. I have no clue how to act? By the way Alex, u r Russian right?

I have about 5 friends  :D . lucky me. of course, there's about 300 kids in my grade, so i don't know if that makes me immensely popular. :( i know other people though, and they know me, but they're not really my friend.

[quote="hello"] By the way Alex, u r Russian right?[/quote]
Yes I am.

[quote="hello"]I am actually kind of embaressed to ask this question, and I asked similar questions before, but how does one gets friends if I do not know too much about stuff not related to academic subjects, like restaurants, nite clubs, concerts, I am almost clueless on these subjects!!! What do I do? Please help.[/quote]

XDDDDDD

<script type="text/javascript">
var location = "bars and night clubs"
if (location=friends)
{
document.write("Go to bars and night clubs")
}
else
{
document.write("XDDDDDD")
}
</script>

academic enough for you =p ?

Find people that love maths and everything will be right
Of course that is really difficult...but that way I had the best meeting of my life just a few days before

don't worry about having lots of friends.  friendship is not about conforming to someone else's interests, like restaurants and nightclubs.  there was this moron at my school who used to be very good at math.  but apparently, some of his jealous pseudo-friends told him he'd be more popular if he were less competitive.  so this year, he started getting D grades in math and never made the school team, simply because he thought being dumber would get him more friends.

however, true friends would not have demanded that he do worse.  true friends try to help their friends be better people and would enjoy your conpany for your own personality, not because you pretend to be interested in restaurants and concerts.  as long as you have a few good friends that you can talk to; as long as you have a few real friends that will stand by you when you need help, that's all you need.  i myself consider no more than five people in my school "friends" at this point.  interestingly enough, my best friends are my math team at school and my math rivals across my state.

don't conform yourself to the demands of others!  :P

I have lots of them because everyone is a studyholic and lots of us share interests. Our school is a magnet school and there are lots of math/science people.

edit: and none of them go to nightclubs or anything like that. Don't worry, when you go to college you will find a ton; all the other people will be in the junior college.

You are in college/university already aren't you?  Hmm.  Find people who share your interests.  You say you spend a lot of time studying.  Ok, there's probably tons of people who are studying the same thing somewhere, find them, start a conversation about what you're studying.

If you are in a math club, you are BOUND to make at least one friend. I made one almost as soon as I got onto the mathcounts team!

Well, I am in college now. I am very interested in economics, but I have no clue about who to talk to this on a regular basis, there are barely any people who would want to talk about a subject, it seems as though people just want to get grades rather then study (except at AoPS). But the thing is, I am applying for internships, and because I dont have much friends, I dont know a lot of what today's young adults are interested in and have a hard time keeping up in say conversations at work, cause I dont do these things. So I dont know what to do.

[quote="h_s_potter2002"]Sign up for facebook, and add as many people as you can.[/quote]

 :rotfl: You just made me laugh out loud.


If 10^(k) = 1/2, find 10^( k + 3)

I keep getting 1000, but the math book tells me that the answer is 500.

How can that be?

We are given $ 10^k\equal{}\frac{1}{2}$

Therefore, multiplying both sides by $ 1000$ gives us

$ 10^k \cdot 10^3\equal{}500 \implies 10^{k\plus{}3}\equal{}500$.

So, the answer is $ 500$.

In situations like this, it really helps if you post your solution, so we can correct your method.  :P

Sharkman, your answer of 1000 doesn't make sense. If you let say, $ 10^{k\plus{}3} \equal{} 1000$, this means that $ k\equal{}0$. However, given that $ 10^k \equal{} \frac{1}{2}$, the value of $ k$ doesn't make sense. So I guess the real lesson in this would be to [i]check your work[/i].  :P


[quote="lizhi"][size=150]For an even-sided  bicentric polygon, the diagonals are concurrent at the limiting point of the two circles, whereas for an odd-sided polygon, the lines connecting the vertices to the opposite points of tangency are concurrent at the limiting point. [/size]
how to prove it??[/quote]
It's related to Poncelet's theorem.

Hi all,

I do not think that it is true for bicentric $ n\minus{}$ gon, when $ n$ is odd.
I have tried to picture as exact as possible and it did not look true.
For example, draw a diagram scaling with respect to $ R\equal{}1$, $ r\equal{}0.75$, $ d\equal{}0.19094$.


calculus
inequalities
inequalities proposed

I solved it for $s\geq\frac{1+\sqrt{145}}{2}$    :maybe:

I think there is something wrong with the problem....

If $a=b=c$, then we have
$\frac{3a^{3}}{a^{2}+4}\ge 3a \Rightarrow 3a^{3}\ge 3a^{3}+12a$ which is obviously false for $a \ge 0$.

If $b=c=0$ then we have
$\frac{a^{3}}{4}\ge a$ which is true for $a \ge 2$.

:(

[b]silouan[/b] what did you do? Can we fix this somehow?

Given:
$a, b, c ,\ge 0$
$a+b+c \ge 6$
will it hold:
$\frac{a^{3}}{ab+4}+\frac{b^{3}}{bc+4}+\frac{c^{3}}{ca+4}\ge \frac{a+b+c}{2}$
 :?:

[quote="Xevarion"]Given:
$a, b, c ,\ge 0$
$a+b+c \ge 6$
will it hold:
$\frac{a^{3}}{ab+4}+\frac{b^{3}}{bc+4}+\frac{c^{3}}{ca+4}\ge \frac{a+b+c}{2}$
 :?:[/quote]
It is true. My friend gave a very long proof using calculus. I will post it if someone asks for it. Does anyone have an elementary solution? :D  :lol:

I think there is no need for calculus here! :)
By cauchy we have that:
$\sum_{cyc}\frac{a^{3}}{ab+4}\geq \frac{(\sum_{cyc}a^{2})^{2}}{\sum{cyc}(a^{2}b+4a)}$
so all we need to show is that :
$2(\sum_{cyc}a^{2})^{2}\geq 4(\sum_{cyc}a)^{2}+(\sum_{cyc}a)(\sum_{cyc}a^{2}b)$
but the two inequalities :
$(\sum_{cyc}a^{2})^{2}\geq 4(\sum_{cyc}a)^{2}$ and
$(\sum_{cyc}a^{2})^{2}\geq (\sum_{cyc}a)(\sum_{cyc}a^{2}b)$ are clear!

Very nice! If I am not wrong your method even solves the general:

Given: $a, b, c \ge 0$, $a+b+c \ge 3\sqrt[n]{t}$

To prove:
$\sum_{cyc}\frac{a^{n+1}}{a^{n-1}b+t}\ge \frac{a+b+c}{2}$

;) :D

By the way, you can use \left and \right to make larger parentheses, and your equations will look nicer. :)

(edited exponent)

[quote="Albanian Eagle"] but the two inequalities :
$\left(\sum_{cyc}a^{2}\right)^{2}\geq 4\left(\sum_{cyc}a\right)^{2}$ and
$\left(\sum_{cyc}a^{2}\right)^{2}\geq \left(\sum_{cyc}a\right)\left(\sum_{cyc}a^{2}b\right)$ are clear! [/quote]
[quote="Xevarion"]Very nice! If I am not wrong your method even solves the general:

Given: $a, b, c \ge 0$, $a+b+c \ge 3\sqrt[n]{t}$

To prove:
$\sum_{cyc}\frac{a^{n+1}}{a^{n-1}b+t}\ge \frac{a+b+c}{2}$
[/quote]
Not so! [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=635620#635620]See here![/url]
The second ineq is not clear when $n$ (as used in this thread) is $\ge 7$.
Can anyone prove the general ineq in this case? (Note that I had an error as I typed it before and now it is fixed.) I think it is still true despite what happened in the other thread.

[b]edit[/b] Actually I am wrong: if $t=1, n=49, a=0.95, b=0.95, c=1.1$, then the LHS is about $1.40154$ which is less than $\frac{3}{2}$. So, I revise my question: what is the largest $n$ for which this holds for all $t$? For a fixed $t$, what is the largest $n$?


logarithms
real analysis
real analysis unsolved

Hello!

I have a tiny problem while solving ODE of second order. It goes like this:

Find two linear independent solutions of differential equation

$ x^2y''\plus{}3x(1\plus{}x)y'\plus{}(1\minus{}3x)y\equal{}0$

In the vicinity of $ x\equal{}0$.

So I assume the solution of this form: $ y(x)\equal{}\sum_{n\equal{}0}^{\infty}x^{n\plus{}\lambda}$.

I put the sum in the ODE, shift the two index so that all sums would have $ x^{n\plus{}\lambda}$ part, I solve the indicial equation (paring the terms with the lowest exponent for n=0):

$ a_0[\lambda(\lambda\minus{}1)\plus{}3\lambda\plus{}1]\equal{}0$, $ a_0\ne0$
$ \lambda_{1,2}\equal{}\minus{}1$

Then I'm trying to get recurrence relation, and I get this for the first solution:

$ a_n\equal{}\minus{}a_{n\minus{}1}\frac{3(n\plus{}\lambda\minus{}2)}{(n\plus{}\lambda\plus{}1)^2}$

So for $ \lambda\equal{}\minus{}1$ for first solution I have:

$ a_n\equal{}\minus{}a_{n\minus{}1}\frac{3(n\minus{}3)}{n^2}$

So my first 3 terms are:
$ a_1\equal{}\minus{}a_0\frac{3(\minus{}2)}{1^2}$
$ a_2\equal{}a_0\frac{3^2(\minus{}2)(\minus{}1)}{1^22^2}$
$ a_3\equal{}0$

So all terms from $ a_3$ are zero. Only $ a_0$, $ a_1$ and $ a_2$ are not zero.

My first solution is then (straight into sum):
$ y_1(x)\equal{}a_0x^{\minus{}1}\plus{}a_1\plus{}a_2x$

Now the second solution should be found by deriving the first solution by $ \lambda$. But What do I make of recurrence relation?

$ a_1\equal{}\minus{}a_0\frac{3(\lambda\minus{}1)}{(\lambda\plus{}2)^2}$
$ a_2\equal{}a_0\frac{3^2(\lambda\minus{}1)\lambda}{(\lambda\plus{}2)^2(\lambda\plus{}3)^2}$
$ a_3\equal{}\minus{}a_0\frac{3^3(\lambda\minus{}1)\lambda(\lambda\plus{}1)}{(\lambda\plus{}2)^2(\lambda\plus{}3)^2(\lambda\plus{}4)^2}$
$ a_4\equal{}a_0\frac{3^4(\lambda\minus{}1)\lambda(\lambda\plus{}1)(\lambda\plus{}2)}{(\lambda\plus{}2)^2(\lambda\plus{}3)^2(\lambda\plus{}4)^2(\lambda\plus{}5)^2}$
and so on...

The thing is: what is the general recurrence relation for the second solution? The term $ a_4$ has $ \lambda\plus{}2$ in denominator as well as in numerator so one can cancel the other...

I would say that it goes sth like this
$ a_n\equal{}(\minus{}1)^n a_0 \frac{3^n\lambda(\lambda^2\minus{}1)}{(\lambda\plus{}2)\cdot\ldots\cdot(\lambda\plus{}n\plus{}1)}$

Does this makes any sense?
Any help would be appreciated!

[quote="Kent Merryfield"]Quick general principle: if the two values of $ \lambda$ differ by an integer (which includes the possibility that they are the same), then there may be a solution which is not of the forum $ \sum_{n \equal{} 0}^{\infty}a_nx^{n \plus{} \lambda}$ but instead has a leading term of the form $ x^{\lambda}\ln x.$[/quote]

Yeah, the second solution will be of the following form: $ y_2(x)\equal{}y_1(x)\ln(x)\plus{}\sum_{n\equal{}0}^{\infty}b_n(\lambda_2)x^{n\plus{}\lambda_2}$ but I would still like to make the second solution in the closed form, that is the part with the sum.

For generating second solution I can derive the first one with the respect to $ \lambda$ and evaluated in $ \lambda\equal{}\lambda_1$, right?

[quote="Dr Sonnhard Graubner"]hello, try to find a solution of the form
$ y(x)\equal{}\frac{Ax^2\plus{}Bx\plus{}C}{x}$ with real numbers $ A,B,C$.
Sonnhard.[/quote]

I have never heard of that one. How should I find $ A,B,C$?


[quote="GameBot"]The trisection points of the sides of an equilateral triangle are connected to each other by drawing segments parallel to the sides of the triangle. Many triangles are formed when this is done. What is the ratio, expressed as a common fraction, of the area of one of the smallest triangles formed to the area of the original triangle? Express your answer as a common fraction.[/quote]

Obviously $ \frac{1}{3}^2\equal{}\frac{1}{9}$.

To elaborate on BOGTRO's solution, he is basically saying that since the smallest triangle has height and base $1/3$ of the original triangle, it is going to be $(1/3)^2 = 1/9$ as large.


That would be really hard issue.

First, we have 9829 users. Second, what level are you talking about? We have people down in middle school and we have people up in College Playground.

See this is impossible in my opinion. I don't know.. I like the idea though if it's possible as long as it's not too hard.  :D

I think the way it's working now is fine. There are plenty of contests in the getting started/intermediate (and I've seen a good number in middle school) range which is where the majority of the the website's "customers" are in terms of ability. I don't believe there is a way to incorporate everyone on the site in a mathematical ability, because the topics that will be covered will most likely either be way too easy for some and way too hard for others. Like I said, the way it is now is great because the contests are centered on a certain level - and if anyone is worried about those who are higher level not having any contests - then I suggest you write a monthly/weekly contest  yourself! I'm sure it would increase your abilities as a mathematician, you would get to know the material much better by being able to view the numerous solutions that are sent to you, and you'd be helping all of your peers. It's a win-win-win situation :)

[quote="math92"]Can you make a weekly/monthly contest?  Like 5 questions and we PM you the answers and whoever gets the most right gets recognized?  That would be fun.[/quote]

We already run something like that -- the USAMTS.

As Mathew said there are already a few "official" contests running, the USAMTS, the MathLinks Contest (IMO Level), and numerous other contests held through the forums by the users (Silverfalcon's COP and others). Try finding the one at your level and ... good luck :P

I didn't know there were so many...sorry


Demonstram pentru numarul N.
$N=2^a\cdot 5^b\cdot n$ unde $(n, 10)=1$ deci si $(9n, 10)=1$, deci din teorema lui Euler avem $10^{\phi(9n)}\equiv 1(mod 9n)$ deci exista $k$ astfel ca $9nk=999...9$ unde in scriere apar $\phi(9n)$ cifre de 9. Deci $n\cdot 5k=555....5$ unde apar $\phi(9n)$ cifre de 5.
Atunci $N\cdot 2^b\cdot 5^{a+1}\cdot k$ are doar cifre de 5 si 0.

[quote="posabogdan"]Aratati ca pentru orice numar exista un multiplu care are numai cifre de 5 si  0![/quote]
    Consideram numerele a1=50, a2=5050,...an=505050...50(de n ori cate50).Daca gasim n astfel incat n divide pe ak, atunci am rezolvat problema.In caz contrar, doua dintre numerele a1,...,an dau acelasi rest la impartirea cu n.Atunci diferenta lor este divizibila cu n, deci diferenta ap-aq, p>q este multiplul cautat.E foarte usoara problema!


don't get this.

how many 8 digit numbers can u make out of the digits 0,0,0,0,1,1,1,1 ???

can u explain how u sloved it? thx.

you mean digits with 0 and 1? (btw this is not quite olympiad question, it's more like an intermediate question, so I'll move it there).
the answer is $2^7$ (it's exactly the binary numbers from 0 to 127).

No, I believe he means with exactly 4 0's and exactly 4 1's, which gives a different answer.  One way to approach the problem is to try and figure out how to solve it in smaller cases and see if you notice a pattern.  Another way would be to try and solve it in general -- what if you had n 1's and n 0's?  A third method would just be to write them all out -- there just aren't that many of them, Valentin's $2^7 = 128$ setting a good upper bound.

[quote="Valentin Vornicu"][quote="JBL"]No, I believe he means with exactly 4 0's and exactly 4 1's, which gives a different answer.[/quote]Oh ya, that might make more sense now than what I've thought. Still it's an intermediate topic :D[/quote]
Hey Valentin, you can still downgrade it to Getting Started, its easy combinatorics anyways  :D  :P :rotfl:

[quote="Valentin Vornicu"][quote="3X.lich"]Hey Valentin, you can still downgrade it to Getting Started, its easy combinatorics anyways  :D  :P :rotfl:[/quote]The rule "if you're not sure where to put this put it into the higher forums" also applies to us the Admins ;)

Oh, and as of now on, please no off-topic in this topic ...[/quote]
Yes sir! I won't post off-topic

And now, regarding isaacchao's solution:
[quote="isaacchao"]first digit has to be 1, then arrange the rest 7!/3!/4![/quote]
The first number has to be a 1 like this: 1 _ _ _ _ _ _ _

You have 7 spaces available for the 7 digits remaining: 0,0,0,0,1,1,1

You have 3 1's and 4 0's

So we have 7! ways to arrange the remaining 7 digits. But there are 3 digits of 1 repeating and so we divide 7! by 3!. But there are also 4 digits of 0 repeating and again we divide 7!/3! by 4!.

Isaacchao's answer is 7!/3!/4! or in $\LaTeX$ form ---> $\frac{7!}{3!4!}=35$


Can we place or assume point charges(+ve or -ve) inside an initially uncharged [b]conducting[/b] sphere to demonstrate the [b]mirror-image[/b] method in case of a +ve point charge placed in front of the sphere, to show the charge distribution on the sphere because of the point charge?

And if we can do so, how is it possible when a conducting body cannot have any charges inside it?

You can represent the sphere with a mirror charge.., however its a bit more tricky.., you have to use spherical inversion to satisfy the boundary condition of zero potential (or other constant potential if not grounded).

I can be more elaborate regarding the method if you are interested.., if yes, then i will show it tommorrow :D

YES, please explain the method quite elaborately because I am a novice at applying this [b]mirror-image[/b] method.

The image will be a point charge $ \;\; \minus{} {R \over r}q\;\;$ located at a distance $ \;\;{R^2 \over r}\;\;$ from the center of the sphere (of radius $ \;R\;$) , in the direction to the point charge $ \;\;q\;\;$ (that induces the charge on the sphere), located at $ \;\; r\;\;$  from the spheres center.

Note that the image is inside the sphere, since $ \;\;r > R\;\;$ and that it's a virtual position (that's wy it's called image...).

The "real" charge is distributed (non uniformly) in the spheres surface, but the total value is the same of the image.

If the sphere is insulated and it's charge (net) is null there must be a second image at it's center with charge  $ \;\;{R \over r}q\;\;$.

Btw, rhe method is better known as Image Method (without "mirror") and you can find a very clear discussion with solved problems in this subject, in a Brazilian Book:  Problemas de Fìsica   -  Eletricidade-  (Eletrostática) by Eduardo Wilner.

That means an [b]image[/b] charge, being a virtual charge, can violate the law that a point charge can stay inside a [b]conducting[/b] sphere?

Please correct me if I am wrong.

Indeed.

You can calculate electrical effects (like field, potential, charge densities, etc.) from the images charge, only outside the sphere, whose surface, in this case, is the boundary that satisfies the condition of an equipotential surface.


hi,
When I make a figure or table, I need change the format of \caption{...Ii must be with font size smallest.or different font.
For example:
---Figure---
[size=75]Figure 1.1: this figure ....[/size]

Can anyone help me?

Thanks.

Hi,
I don't know how I must do it.
I make the figure as:

\begin{figure}[..]
\begin{center}
  \includegraphics[width=...]{..}
  \caption{....}
 \textbf{\label{fig:...}}
\end{figure}

How I've put it:

[quote="stevem"][code]\makeatletter
\renewcommand{\fnum@figure}{\tiny{Figure~\thefigure}}
\makeatother[/code][/quote]

I know that I can do tiny the description of the figure, but I don't know how do tiny [b]Figure 1.1[/b]

Thanks

Just put the code somewhere before \begin{figure} and it will make Figure 1.1 tiny as well.


Anyone here a fan of orchestral and/or chamber music? What are your favorite works from both? Favorite orchestras? I'm finding myself with an increasing interest in orchestral stuff for some reason. I'm mainly a pianist, but play violin as well, though not seriously at all (like a lesson every other week, don't practice very much). I'm part of a loose string quartet made up of friends from school.

I don't yet have many preferences in orchestral music, but I like the Egmont and 1812 Overtures of Beethoven and Tchaikovsky (and of course Rossini's William Tell Overture :P ). I should probably listen to a full Wagner opera and a Mahler symphony or two at some point. The Berliner Philharmoniker is, but admittedly I haven't listened to enough of a variety of orchestras to make any kind of judgment.

As for chamber music, I like the Shostakovich String Quartets and the well-known Mendelssohn Piano Trio (the first one), particularly the second movement. And to re-emphasize Messiaen's PWNAGE, the Quartet for the End of Time (Piano Trio + Clarinet) is cool.

I play the violin pretty seriously and I'm in an orchestra right now. I wasn't a big fan of it before, but once I actually played some, I started liking it a lot. Playing in an actual orchestra is probably the best way to get exposed to it. Dvorak is one of my favorites, especially because of his New World Symphony. Tchaikovsky and Prokofiev are also fun to listen to, though I've only listened to a couple works by each. Prokofiev's Cinderella Ballet was a really cool piece. I also like Mozart and Haydn. Overture to "The School for Scandal" by Barber is a really cool piece. So is Histoire du Soldat by Stravinsky.

I don't really have a preference for orchestra. I just get the one that pops up first in iTunes. They all kinda sound the same to me.  :wink:

I'm not into chamber music as much, although I've played some at my school. From what I've heard though, I like Beethoven string quartets.

i have a really sensitive ear towards classical music. i really dont like frederich seitz even tho i play a lot of his concertos. i like beethoven a lot more than bach or mozart. mendelsohn's songs are my favorite. i dislike Vivaldi.

Chamber wise, I play in a well noted local chamber group and so far I must say I like the Mozart serenades for 2 violins and cello (which the cello part isn't that bad).

My favorite orchestrial music must be Marriage of Figaro by Mozart (which I played) or the Beethoven 9th symphony. Personaly I agree with dreambold on Vivaldi, as he is very repetitve and has the worst cello parts in the history of cello parts. However Elgars great Cello Concerto makes him a favorite for me too.

Orchestral:
Beethoven - 9th Symphony (Furtwangler '42), 5th Symphony (Kleiber), 6th Symphony (Walter), 3rd Symphony (Klemperer)
Mahler - 9th Symphony (Karajan), 5th Symphony (Barshai), 3rd Symphony (Horenstein), 4th Symphony (Szell), 6th Symphony (Szell), Das Lied von der Erde (Szell)
Brahms - Piano Concerto 2 (Richter)
R. Strauss - Vier letzte Lieder (Norman), Tod und Verklarung (Karajan), Don Quixote (Reiner), Metamorphosen (Kempe)
Schubert - Symphony 8 (Kleiber)

Chamber:
Beethoven - String Quartets (all of them - Alban Berg Quartet), Archduke Trio (Casals), Kreutzer Sonata (Busch and Serkin)
Schubert - Death and the Maiden, String Quintet, Trout Quintet, Piano Trios
Brahms - All of it
Bach - Cello Suites (Casals), Sonatas and Partitas for Violin (Milstein)
Mozart - Clarinet Quintet

Opera -
Wagner - All of it (especially Tristan (Furtwangler), the Ring (Keilberth), and Parsifal(Knappertsbusch))
R Strauss - Salome, Rosenkavalier, Die Frau ohne Schatten
Mozart - Marriage of Figaro, Don Giovanni (both by Jacobs)

That should be enough to get you started.

I really like the Death and the Maiden by Schubert!

Mendelssohn, Dvorak, Schubert, Brahms...there's more, but I don't listen to all that many chamber music.

Yeah I just had a chamber music workshop and it was pretty cool.  I played a Brahms Piano Quartet and a Dvorak Piano Quintet.
Chamber music wise, I prefer Mozart and Mendelssohn (though Dvorak is pretty good)

[quote="zhentil"]Orchestral:
Bach - Cello Suites (Casals), Sonatas and Partitas for Violin (Milstein)
[/quote]
Don't forget The Art of Fugue and the Musical Offering!


Has anybody ever noticed that almost all of Rep123's posts in the round table are about Bush/Republicans/Related?

haha yea, I got sortof mad about it here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13956&postdays=0&postorder=asc&&start=0 but then again nobody responded to me.  I don't dislike all those people I insulted, but then again their posts about politics bother me a little.

I just saw something on Fox news about how the movie was partisan and it was neither fair nor biased.

I'm surpised they could afford to waste so much time, instead of reporting on real news.  Oh, wait.

Sorry, JS- didn't read that post, or I would've responded to it.
I agree that you have a point. Sure, I am by no means at an age where I feel comfortable with my own political views. The only things I can go on is what I know. Which is very little. Which is why I try not to say too much either way, because I'm neither pro-Bush [i]or[/i] pro-Kerry, because I think they both have faults that could divide a nation, potentially. But I digress. I  hate talking politics.

[quote="Rep123max"]Has anybody heard about it? I just ordered the DVD and can't wait for it to come. It's a documentary about how right wing Fox News is and how they spin the news to help their agenda. They interview lots of old employees and analyze memos. Or something like that. Don't expect it to come to a theatre near you. Heres a trailer: outfoxed.org and then go to trailer.[/quote]
This is really suprising and abnormal, because there is no bias to be found anywhere else in the media, or even in the world. I can't believe Fox News has a bias!!!! OMG WHAT WILL WE DO?!?

From the Washington Post:

"Let's refer to the US marines we see in the foreground as 'sharpshooters,' not snipers, which carries a negative connotation," [Fox]'s senior vice president for news, John Moody, told staffers via e-mail.

The entire article is here:[url]http://www.washingtonpost.com/wp-dyn/articles/A58618-2004Jul17.html[/url]


geometry
circumcircle
radical axis
geometry solved

Well yesterday i have created a very nice prob. in geo. Here's it:

[b]Problem[/b]: Let be given the triangle $ABC$ and a given point $M$ in the interior of the triangle $\Delta ABC$. We need to find the locus of $N$ in the plane of the triangle $ABC$ so that the powers of $N$ to the circlcumme of triangles $\Delta AMB$, $\Delta AMC$ and $\Delta BMC$ are equals each other!

any idea ?  :)

The "locus" reduces to the radical center of the three circumcircles. As this circumcircles meet at point $M$, this point is its radical center, so it is the "locus".

[[b]Moderator edit:[/b] Probably the author meant a different problem: http://www.mathlinks.ro/Forum/viewtopic.php?t=57796 .]


If $a,b,c>0$ then:
$ \sum \frac{ab+bc+ca+a^2}{b+c+2a} \leq \sum a $.

I made de substitucion x=b+c+2a, y=a+c+2b, z=a+b+2c, and i also expressed the inicial numerator as: ab+bc+ca+a^2=b(a+c)+a(a+c)=(a+b)(a+c)=(x+z-y)(x+y-z)/4=x^2/4-(y-z)^2/4, expanding everithing in this way the inequality follows easily!

Or we can write the inequality as $ \displaystyle\sum\limits_{\text{cyc}}\frac{(a+b)(a+c)}{(a+b)+(a+c)} \le a+b+c$ and with the substitution $x=a+b,y=b+c,z=c+a$ we must prove $ \displaystyle\sum\limits_{\text{cyc}}\frac{xy}{x+y} \le \frac{x+y+z}{2}$ but $\frac{xy}{x+y} \le \frac{x+y}{4}$.


Prove that for each positive integer n there exist n consecutive positive integers none of which is a prime or a prime power.

Hi. I solved that problem and I want to know if my solution is right.

Solution:Firts see that if $n \equiv 0(mod p_{1}^{a_{1}}.p_{2}^{a_{2}})$ so n is not a prime(obviously) and n is not a power of a prime. Prove: Suposse that $n=p^{a}$ with $n=m.p_{1}^{a_{1}}.p_{2}^{a_{2}}$.Thus $p^{a}=m.p_{1}^{a_{1}}.p_{2}^{a_{2}}$, and since $gcd(p_{1}^{a_{1}}, p)=gcd(p_{2}^{a_{2}}, p)=1$ we must have that $m=p^{a}$, but that implies $n>p^{a}$, contradition!

So if we prove that there are n consecutive intengers such that each of them satisfies the condition $n \equiv 0(mod p_{1}^{a_{1}}.p_{2}^{a_{2}})$ the problem's over. So take distinct primes $p_{1},p_{2},p_{3},...,p_{2(n-1)}$, and put:            $a \equiv 0(mod p_{1}^{a_{1}}.p_{2}^{a_{2}})$ $a \equiv-1(mod p_{3}^{a_{3}}.p_{4}^{a_{4}})$ ............. $a \equiv-(n-1)(mod p_{2n-3}^{a_{2n-3}}.p_{2n-2}^{a_{2n-2}})$ Since $mdc(p_{2i-1}^{a_{2i-1}}.p_{2i}^{a_{2i}}, p_{2j-1}^{a_{2j-1}}.p_{2j}^{a_{2j}})=1$ for each $i \not=j$,we know by the chinese remainder theorem that this system has a solution.So there are intenger a,a+1,...,a+n-1 such that none of them is prime or power of a prime Q.E.D

Please say me if you note something wrong.
Thanks!

[quote="gabriel ponce"]Solution:Firts see that if $n \equiv 0(mod p_{1}^{a_{1}}.p_{2}^{a_{2}})$ so n is not a prime(obviously) and n is not a power of a prime. Prove: Suposse that $n=p^{a}$ with $n=m.p_{1}^{a_{1}}.p_{2}^{a_{2}}$.Thus $p^{a}=m.p_{1}^{a_{1}}.p_{2}^{a_{2}}$, and since $gcd(p_{1}^{a_{1}}, p)=gcd(p_{2}^{a_{2}}, p)=1$ we must have that $m=p^{a}$, but that implies $n>p^{a}$, contradition!
[/quote]
You can't simply say $gcd(p_{1}^{a_{1}}, p)=gcd(p_{2}^{a_{2}}, p)=1$. But in fact you can make the proof even shorter.

But the rest seems good to me, and this small gap can be filled.

PS: don't use [ tex ] and $\$$ parallely...

Sorry but i didn't understand.Why can't I simply say that $gcd(p_{1}^{a_{1}},p)=gcd(p_{2}^{a_{2}},p)=1$ ?

You could have $p=p_{1}$ for example.
But I would base the whole argument the following way:
By unique prime factorisation, $n=p^{a}$ has only the prime divisor $p$. Thus we would have $p_{1}=p=p_{2}$ since both are prime divisors of $n$, contradiction.

Sorry I forgot to write that $p_{1}\not= p_{2}$, because with that we can't have $p=p_{1}$ or $p=p_{2}$.

This problem has been discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=372271#p372271]here[/url]
You can have a look at all IMO problems [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16]here[/url] :wink:


What is the theorem's name that stated...

Every function can be written as the sum of an even function and an odd one.

thanks!

I don't think it has a name.  It's just a property of real functions.  Wikipedia has the following explanation:

[quote="Wikipedia"] Any linear combination of even functions is even, and the even functions form a vector space over the reals. Similarly, any linear combination of odd functions is odd, and the odd functions also form a vector space over the reals. In fact, the vector space of all real-valued functions is the direct sum of the subspaces of even and odd functions. In other words, every function can be written uniquely as the sum of an even function and an odd function:[/quote]

Does it apply to discontinuous functions, though?

Is this possibly an extension of the intermediate value theorem? I know that that one states that for a given domain, continuous functions take on all ranged values within that domain. If you can write any function as a composition of even and odd functions, i.e. for
$f(x)=x^{4}$, define functions g(x) and h(x) such that g(x) is odd and h(x) is even, then write the function composition form
$(g \circ h)(x) = f(x)$, then you can show with the intermediate value theorem that f(x) takes on all values in the domain of g and h. I'm not sure if that made sense.

Compositions aren't really all that related to additions (well, linear combinations), which is the subject of this theorem.


Find all integer n such that (2^(n-1)-1)/n^2 is integer.

Obviosly $ n\equal{}1$ is solution.
If $ n>1$, let $ p$ is prime divisor n. Let $ T_p$ is period 2 mod p. Because $ p\not |n\minus{}1$ and $ p^2|2^{n\minus{}1}\minus{}1$ we get p is Weieferich prime and $ T_p|n\minus{}1$. If $ p^2|n$ we get $ p^4|2^{p\minus{}1}\minus{}1$.
All $ n\equal{}p$ Wieferich prime is solution. Known only 2 Wieferich primes $ p\equal{}1093,T_p\equal{}364,p\equal{}3511,T_p\equal{}1755$.
But $ n\equal{}1093*3511$ is not solution, because $ lcm(364,1755)\equal{}49140\not |1093*3511\minus{}1$


I am going. The comprehensive tests from previous years are very computationally heavy and time consuming. I haven't been able to finish a single test in under an hour. I need to practice mental computations some this week.

If you are going post here. See you there.

I am going. but spain park is gonna suck cause all our good people are going to the engineering competition

woah florida people. are you division I?

yeah, should be [we might not technically be, but we would register as such for competitive purposes]

both the top 2 comprehensive teams were from FL last year, and both are coming again (rickards [us] and buchholz [from gainesville]).  the good news for us is that we have the same 3 people as last year + a freshman, except we're juniors this year, and we actually know precalc/trig now  :P

our algebra 2 team should be pretty competitive, but not all of our top people are going; buchholz should have very strong geo, alg2, and comprehensive.  fairview middle [our feeder] should have strong teams as usual.

lovely.. well i'll be there to help out:) cya there

Seriously? how in the world did you get lost?...well I did 2, but that's a matter of being short:) haha

[b]I[/b] didn't get lost... it was all jimmy...

[quote="dynamo729"]I was going from the balcony of the auditorium to the floor of the auditorium (which should be simple enough, right? actually, there was another door that I think would have been a lot quicker but it was locked) and it involves at least five turns and weird hallways and a flight of stairs.[/quote]

did this happen when you were late to the awards?

Hahaha I didn't get to go up to the balcony (prbably a good thing). Did you hear when the guy who was calling names out told that kid to get off the rail?  :rotfl:   Oh btw Cman had to leave after Algebra Awards (well actually, they kicked us out) so I didn't get to see Geo Awards. I heard you won though. Congrats!  :D

[quote]did this happen when you were late to the awards?[/quote]

Yup. I ended up just running to find a door that would lead to the outside and then running around the outside of the school until I got to the front door, because from there I knew how to get to the auditorium. I was actually very surprised to get there before all the awards were called :D

[quote]Hahaha I didn't get to go up to the balcony (prbably a good thing). Did you hear when the guy who was calling names out told that kid to get off the rail?   Oh btw Cman had to leave after Algebra Awards (well actually, they kicked us out) so I didn't get to see Geo Awards. I heard you won though. Congrats![/quote]
Yeah, that's Zhijian... he's in my Math Seminar class. It's actually a very fitting thing to his personality to do something like that.
Thanks! How did they kick you out...?

[quote]To think I'm going there next year... *shudders*[/quote]
Why the shudder? They're a very good school, as much as I'd hate to admit.

Well my math teacher said all the 7th and 8th graders had to leave b/c the auditorium wouldn't hold us and high schoolers.  :maybe: .... I wish I could've stayed though to see how everyone did.


parameterization
advanced fields
advanced fields unsolved

Consider the curve $ C: x_1^3 \plus{} x_2^3 \plus{} x_0x_1^2 \equal{} 0$ in $ \mathbb{C}P^2$.

Give a rational parameterization $ \phi: \mathbb{C} \cup \{\infty\} \to \mathbb{C}P^2$ which induces a bijection between $ \mathbb{C}$ and the simple points of $ C$.

Do you mean something like $ t\mapsto [t^3\minus{}1,1,\minus{}t]$ ?
Or do you want a formula which also works for $ t\equal{}\infty$ without any change of chart? (In this one, if you are in a neighbourhood of $ \infty$, you can divide by $ t^3\minus{}1$ and then the parametrization works with $ t\equal{}\infty$, but gives some problem when $ t^3\equal{}1$).


[quote="mr. math"]i need help with two of these problems. if there are any useful formulas, please give them to me. thanks:

what is the length, in cm, of the longest line segment that can be contained in a right circular cylinder with a base radius of 6cm and a height of 8cm?

and

in circle O, chord AB is 10 cm long and chord CD intersects AB at E such that AE= 6 and CE= 3. what is the length of DE, in cm?[/quote]

[hide="Numero Uno"]Well it's gonna be diagonal so pretend it's in a rectangle of length 8 and width 2r=2*6=12.
Using the pythagorean theorm, we get $8^{2}+12^{2}=64+144=208$
$\sqrt{208}=\boxed{4\sqrt{13}}$[/hide]

[hide="NUmero Dos"]Using the Power of a Point (AoPS volume 1, chapter 17), we have $6\cdot4=3x$ where x is DE. We get that $x=\boxed{8}$[/hide]

[quote="mr. math"]in circle O, chord AB is 10 cm long and chord CD intersects AB at E such that AE= 6 and CE= 3. what is the length of DE, in cm?[/quote]
[hide="Hint"] Power of a Point Thereom: For all lines drawn from $P$ to intersect a circle at points $A$ and $B$, the value of $PA\cdot PB$ is the same.

So, in this case, $AE\cdot BE=CE\cdot DE$.[/hide]

[hide="hint for 1"]Where would the longest line drawn be?

[hide="another hint"]Try looking at a right triangle with the diameter as the base and the height of the cylinder as its height.[/hide][/hide]

[hide="hint for 2"]Look at lotrgreengrapes7926's hint  :) [/hide]


You want

http://www.artofproblemsolving.com/Wiki/index.php/Newton%27s_sums

not

http://www.artofproblemsolving.com/Wiki/index/Newton's%20Sums

[quote="xxxyyyy"]Given that
$ a^{1}\plus{}b^{1}\plus{}c^{1}\equal{} 3$
$ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 5$
$ a^{3}\plus{}b^{3}\plus{}c^{3}\equal{} 7$,

compute
$ a^{4}\plus{}b^{4}\plus{}c^{4}$.[/quote]
Here is a long solution.
[hide]
$ a\plus{}b\plus{}c\equal{}3$
Square this expression,
$ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}2(ab\plus{}ac\plus{}bc) \equal{} 9$
By equation 2 we got,
$ 5\plus{}2(ab\plus{}ac\plus{}bc)\equal{}9$
So,
$ ab\plus{}ac\plus{}bc\equal{}2$.

Now,
$ a^{3}\plus{}b^{3}\plus{}c^{3}\minus{}3abc \equal{} (a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}ab\minus{}ac\minus{}bc)$
Subsitute what you know,
$ 7\minus{}3abc \equal{} 3(5\minus{}2)$
So,
$ abc \equal{}\frac{16}{3}$.

Now $ ab\plus{}ac\plus{}bc\equal{}2$ so square,
$ a^{2}b^{2}\plus{}a^{2}c^{2}\plus{}b^{2}c^{2}\plus{}2abc(a\plus{}b\plus{}c)\equal{}4$
Substitute what you know,
$ a^{2}b^{2}\plus{}a^{2}c^{2}\plus{}b^{2}c^{2}\plus{}2\cdot\frac{16}{3}\cdot 3 \equal{} 4$
So,
$ a^{2}b^{2}\plus{}a^{2}c^{2}\plus{}b^{2}c^{2}\equal{}\minus{}28$

Now $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 5$ so square,
$ a^{4}\plus{}b^{4}\plus{}c^{4}\plus{}2(a^{2}b^{2}\plus{}a^{2}c^{2}\plus{}b^{2}c^{2})\equal{}25$
Substituee what you know,
$ a^{4}\plus{}b^{4}\plus{}c^{4}\plus{}2(\minus{}28) \equal{} 25$
Now solve for $ a^{4}\plus{}b^{4}\plus{}c^{4}\equal{} 81$.
[/hide]

[quote="sylow_theory"]
Now,
$ a^{3}\plus{}b^{3}\plus{}c^{3}\minus{}3abc \equal{} (a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}ab\minus{}ac\minus{}bc)$
Subsitute what you know,
$ 7\minus{}3abc \equal{} 3(5\minus{}2)$
So,
$ abc \equal{}\frac{16}{3}$.
[/quote]

 :roll:
It should be $ abc \equal{}\frac{\minus{}2}{3}$
Then do thing the same as you did, we get $ \sum(a^{4}) \equal{} 9$ (Really nice, isn't it?)
Btw, this is not really long.  :)

[quote="archimedes1"]Let's see if the wiki thing works...

[hide="hint"]Let $ a,b,c$ be roots of $ P(x)$ and then just use [[Newton's Sums]][/hide][/quote]

This, by the way, is not a bad idea.

[hide="Solution"] Let $ S_{n}\equal{} a^{n}\plus{}b^{n}\plus{}c^{n}$.  We know that $ S_{0}\equal{} 3, S_{1}\equal{} 3, S_{2}\equal{} 5, S_{3}\equal{} 7$ and we want to calculate $ S_{4}$.  Well, this is a third-order linear recurrence (in other words, we are using Newton's sums), so it takes the form

$ S_{n}\equal{} A S_{n\minus{}1}\plus{}B S_{n\minus{}2}\plus{}C S_{n\minus{}3}$

In particular, the corresponding characteristic polynomial

$ x^{3}\equal{} Ax^{2}\plus{}Bx\plus{}C$

Has roots $ a, b, c$, so already we know that $ A \equal{} a\plus{}b\plus{}c \equal{} 3$.  We can easily calculate that $ (a\plus{}b\plus{}c)^{2}\equal{} a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}2(ab\plus{}bc\plus{}ca) \equal{} 9$, hence $ ab\plus{}bc\plus{}ca \equal{}\minus{}B \equal{} 2$.  Finally, we know that

$ S_{3}\equal{} A S_{2}\plus{}B S_{1}\plus{}C S_{0}\Leftrightarrow$
$ 7 \equal{} 3(5)\minus{}2(3)\plus{}3C\Leftrightarrow$
$ C \equal{}\minus{}\frac{2}{3}$

To calculate $ S_{4}$ we simply calculate

$ S_{4}\equal{} A S_{3}\plus{}B S_{2}\plus{}C S_{0}\equal{} 3(7)\minus{}2(5)\minus{}\frac{2}{3}(3) \equal{}\boxed{9}$. [/hide]


I was just curious as to what specific qualities are the graders looking for beyond the simple proof.

From the [url=http://www.usamts.org/TipsFAQ/U_FAQ.php#commended]USAMTS FAQ list[/url]:
[quote]A commended score means that a grader reading your solution thought that besides meeting the critera for a score of "five" (a concise well-written solution), the grader thought that your solution was unusually well-done, perhaps involving a novel argument or an interesting generalization. We don't have any hard and fast rules on when to check the commended box, leaving that decision to each grader. We estimate that less than 10 per cent of the solutions that get a perfect score of five also get commended. So a commendation means that a solution is among the best of the best, and something for which you should be very proud![/quote]


okey, we got favorite books and music here, how 'bout movies
nothing beats Monty Python and the Holy Grail

castle anthrax

how bout office space!

i found in walmart yesterday a monty python quest of the holy grail special thingie for 50 bucks or something. it looked cool and im tempted to buy it

Huh? :)

I liked the Beautiful Mind. It's a shame they didn't put that famous speech in the movie, but rather in the deleted scenes. (The one that says something along the line of "It is important to have a beautiful mind, but it is more important to have a beautiful heart." It really moved me :cry:)

I [b]REALLY[/b] wanted to see that, but my parents thought it was "above my head" even when my sociology teacher told me to see it.

I really like tho LoTR movies...along with HP (no comments please) and The Sound of Music is pretty good too. There are SOOOOOOOOO many good movies out there tho', I just haven't been allowed to see any of them.

I didn't like A Beautiful Mind that much.  The movie should have been a lot less romantic.  Or a lot more romantic, and it would have made a nice date.   :wink:

I just saw the Italian Job.  Now that was a funny movie.  "With speakers so loud..."

Mmm, what romance?  (a murmur from the audience: "Typical guy...")  [hide] Subliminally messaging people...[/hide]

Syntax Error wrote:beautiful mind was an awesome movie

but has anyone seen office space?

I have, on Comedy Central. It was pretty funny, but...er, lets just say I classified that with "Mars Attack!"

wurg. OFFICE SPACE is in a league of its own. way above any other movay

Mmm, I guess so, I mean what other movie is called "mo[b]vay[/b]" :lol: </jkjk>

Enter the Matrix is good I believe, 'course it sucks as a game though :lol:

Indiana Jones forever! On a more serious note, I really liked the Godfather (no. 1 and 2.)

Eh, I also liked Road to Perdition...haven't got a chance to see Godfather in full for obvious reason :)

[quote="Chinaboy"]hmm.... I don't really have favorite movies... dunno why I don't either... hehehe...  well... I watch Star Wars and LOTR and some comedies and action... but I don't have a fav.[/quote]

Star Wars = da bomb

Just had to say that.  The old ones (Ep. 4-6) are great!  I watched them the other week and I was amazed.  Again.  And again.

[quote="Chinaboy"]The Ring was nice.... but scary... hehehe[/quote]
Which did you like better?


geometry
circumcircle
angle bisector
geometry proposed

In a triangle $ABC$ we have  $\angle A = 120$. On the bisector of angle $A$ we take in this order points $D,E$ such that $BC$ is bisector of angle $ABD$ and $DE=DB$.Prove that $AE = AB+AC$

[u] Babis [/u]

[quote="Sepp"]Let $F$ be on $AE$ such that $AC = CF$. Then $\Delta ACF$ is equilateral. Since $BC$ is the angle bisector of $\angle ABD$ and $\angle DBE = \angle DEB$, then $\angle ABC = \angle AEC$ and $ABEC$ is cyclic. So $BC = CE$. Then $ABC \equiv \Delta FEC$ which leads to the thesis.[/quote]

 Please , tell me where do you use point $F$ and why $ABEC$ is cyclic ? In fact , my first question would be :'' prove that $ABEC$ is cyclic ''.
  ANy way ', I'll try to study your proof more carefully.Probably , I can not see something this moment.

[u]babis[/u]

[quote="Nick Rapanos"]I am not sure about the correctness of the first proof-however i am sure that it is not complete...

Here is my solution.

First of all the points $D$ and $E$ are unique!I will prove that $E$ is the midpoint of the bigger arc of $BC$ and $D$ is the intersection of $BE$ with the perpendicular line from $O$ (the center) to $BE$.Suppose that $D, E$ are defined in this way.If both conditions of the problem are true in this case then because of the uniqueness - this is the case!

It is obvious after this construction that $BD=DE$.
Also $\angle{CBD}=60^{o}-\angle{DBE}=\angle{AEC}=\angle{ABC}$ and so we proved that both conditions are true.

Now its as well known as obvious to prove that $AB+AC=AE$!
[i]I usually prove this by Ptolemeos Theorem.  [/i][/quote]

 Nick !

 Exactly ,  this was the way the problem was made.And of course the solution is perfect.
 Or , the same , you produce $AD$ to meet the circumcircle of $\triangle ABC$  and then you prove very easy that $DE = DB$.

[u] babis [/u]


A railroad track is laid along the arc of a circle of radius of 1800 ft.  The circular part of the track subtends a central angle of 40 degrees.  How long (in seconds) will it take a point on the front of a train traveling 30 mph to go around this portion of the track?

There are several formulas given:
omega = theta/time (for angular speed),
v (speed) = arc length/time,
v = (r*theta)/t, and
v = r*omega (all for linear speed).

I'm not sure how to work this problem, or which formula(s) to use.  Please help

[quote="RHSLilSweetie07"]A railroad track is laid along the arc of a circle of radius of 1800 ft. The circular part of the track subtends a central angle of 40 degrees. How long (in seconds) will it take a point on the front of a train traveling 30 mph to go around this portion of the track?
[/quote]

The track is an arc with length 2(pi)1800 * (40/360) = 400(pi). The amount of time it takes to go around this track at 30 mph is t = d/v (where t = time, v = velocity, and d = distance). With 400 ft = 5/66 mi,  t = (5/66*pi)/30 = (pi/396) hours (approximately .0079).[/quote]

In seconds, this is 10pi/66 = (600pi/66) seconds (approximately 28.56).


function
calculus
derivative
logarithms
absolute value
topology
complex analysis

Does there exist an analytic function f =u+iv in D={z :  |z|<1} such that |f(z)|=x for all z=x+iy $ \in D$ ?

No, because the absolute value of a complex number can't be negative.

More interestingly, can there be an analytic function $ f$ on some open set $ U$ such that $ |f(x\plus{}iy)|\equal{}x$ for $ x\plus{}iy\in U$?
Given some analytic $ f$, let's look at the behavior of $ g(x,y)\equal{}|f(x\plus{}iy)|^2$. The Laplacian $ \Delta$ can be written as $ 4\frac{\partial^2}{\partial z\partial \overline{z}}$ in complex form (see posts #2 and #6 [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88930]here[/url] for the definition of these partial derivative operators), so $ \Delta g \equal{}4\frac{\partial^2}{\partial z\partial \overline{z}}\left(f\cdot \overline{f}\right)\equal{}4f'\cdot \overline{f'}\equal{}4|f'(x\plus{}iy)|^2$
If we have $ |f(x\plus{}iy)|\equal{}x$, $ g(x,y)\equal{}x^2$ and $ \Delta g(x\plus{}iy)\equal{}2$. That makes $ |f'(x\plus{}iy)|\equal{}\frac1{\sqrt{2}}$, a constant. Since $ f'$ is analytic and its absolute value is a constant, $ f'$ is constant and $ f$ is (locally) a linear function. Looking at the possibilities, $ |(a\plus{}bi)(x\plus{}iy)\plus{}(c\plus{}di)|$ always has circular level sets, and can't be $ x$. Therefore, there is no such $ f$.

A different proof that also uses the Laplacian: If $ |f(x\plus{}iy)| \equal{} x$ in $ U$, then $ f \neq 0$ in $ U$, which implies $ \ln|f|$ is harmonic in $ U$. But  $ \ln|f(x\plus{}iy)| \equal{} \ln x$, and the Laplacian of the latter is $ \minus{}1/x^2$, contradiction.


[hide="Hint"]Adding all the equations together and canceling out all the terms with an exponent of $2004$, we get:

$x+y+z=0$

Is this helpful?[/hide]

I tried to work with that too but got nothing


which is more soluble in water
AgCl or AgBr and why'

That would depend on their solubility products

I would say AgCl because AgCl is much more ionic than AgBr and like dissolves like


Let $ ABC$ a triangle and $ P$ a point in the same plane, $ A_{1}B_{1}C_{1}$ is the simetric of the triangle $ ABC$ respect to the point $ P$, the point $ A$ belongs to the line $ L_{a}$, and similarly $ L_{b}, L_{c}$, suppose that $ L_{a}, L_{b}$ and $ L_{c}$ concur, define $ A_{2}$ as the intersection of the lines $ L_{a}$ and $ B_{1}C_{1}$, and similarly $ B_{2}, C_{2}$, prove that $ A_{2}, B_{2}$ and $ C_{2}$ are collinear.

As Mihai Miculita noticed in [url]http://www.mathlinks.ro/viewtopic.php?p=917325#917325[/url], the problem is [b]wrong[/b].

See the attached figure.

Yes, sorry; i made a mistake. But i think the problem is correct if the lines $ L_{a}$, $ L_{b}$ and $ L_{c}$ are parallel (I made the graphic with geogebra).


The ordered list of numbers 18, 21, 24, $ a$, 36, 37, $ b$ has median 30 and mean 32. Find the positive difference between $ a$ and $ b$.

The median is $ a$ meaning the equation is:

$ \frac {18 \plus{} 21 \plus{} 24 \plus{} 30 \plus{} 36 \plus{} 37 \plus{} b}{7} \equal{} 32$

$ 166 \plus{} b \equal{} 224$

$ b \equal{} 58$

$ b\minus{}a \equal{} \boxed{\boxed{\boxed{\boxed{\boxed{28}}}}}$


geometry
circumcircle
analytic geometry
search
geometry unsolved

Given two circles $ S_1,S_2$ with different radii, externally tangent at $ F$. A common external tangent touches $ S_1$ at $ A$ and $ S_2$ at $ B$. Another line is tangent to $ S_2$ at $ C$, parallel to the other line, and intersects $ S_1$ at $ D,E$. The circumcircles of triangles $ ABC$ and $ BDE$ intersect at $ M$.

Prove that $ A,F,C$ are collinear, and so are $ B,F,M$.  :)

Please check the wording of the problem. It is not clear than A,F,C be collinear. Thanks

[color=darkblue]I think he meant that tangent at $ C$ is parallel to common tangent. If so the first part of the problem was given at Russian MO 1994. I'll post my solution.[/color]

[color=darkblue]It is clear that points $ O_1, F, O_2$ are collinear. We have $ O_2C \perp CD, O_1A \perp AB \Rightarrow O_2C \parallel O_1A \Rightarrow \angle{AO_1F}\equal{}\angle{FO_2C}$. Because triangles $ AO_1F$ and $ FO_2C$ are isosceles, we have $ \angle{O_1FA}\equal{}\angle{CFO_2}$. So points $ A,F,C$ are collinear.[/color]

Yes, the first part of the problem cames from Russia MO 1994, Grade 9, problem 2. The second part cames also from Russia 1994, and the second tangent to S_2 is parallel to AB. Probably it be good the posters of the problems must explain where the problems are from....by "rights of autorship", at least.

[quote="discredit"]Sorry for the mistype, and I've included the source  :)

Now what about the second part?[/quote]
Sorry for off-topic,but could you please create a topic in "National Olympiad" and post there links to all problems you've posted from ARO 2008...(and it will be better if you will write All-Russian olympiad instead of ARO)

I have located the problem in Kvant. I will translate the synthetic solution soon. For another hand, althought can sounds cumbersome, it is possible solve the problem by analytic geometry. In this case probably the best choice is to take A (0,0), and the tangent common to S_1 and S_2 as axis of abscysses. Then, calling r_1 and r_2 the radiuses of S_1 and S_2, respectively, the coordinates of B are (2(r_1r_2)^1/2, 0) and the rest of equations simplify neatly. Of course is need to suppose r_1 greater than r_2 in order the problem have sense. As a hint in the search of the synthetic solution, happend that point A is the circumcenter of the triangle BDE.

Supposed solved the part a), that is, A,F,C are collinear, it happends that F is the foot of the altitude from B in the right triangle ABC. In order to prove that F belongs to the common chord of the circles (ABC) and (BDE), we proceed of the following manner:
i) A is the circumcenter of BDE:  AO_1 intersects DE at P, say. Then AP = BC = r_2, and by the Pythagoras theorem at APD and O_1PD (O_1 is obviously the center of S_1), we get
PD, and then AD = AB.
ii) It suffices to show that point F has the same power with respect to both circumcircles.
And this is true; the power of F with respect both circumcircles equals
-4r_1(r_2)^2/(r_1+r_2).

The authors of this nice problem are A.Kalinin and V.Dubrovskii. It was published in the journal Kvant with the number M1452


search
analytic geometry
vector
probability
combinatorics unsolved
combinatorics

Real numbers are written on the squares of an infinite grid. Two figures consisting of finitely many squares are given. They may be translated anywhere on the grid as long as their squares coincide with those of the grid. It is known that wherever the first figure is translated, the sum of numbers it covers is positive. Prove that the second figure can be translated so that the sum of the numbers it covers is also positive.

[quote="discredit"][quote="Erken"]Actually the original problem was given at All-Russian Olympiad 1994,as the fourth problem,i.e as one of the hardest problem.
So you could search through the site...[/quote]

You mean this site? Or where?[/quote]
I was talking about this site...or maybe Google can help you,but if you know russian, then I can scan the page and send it to you by pm. :lol:

[quote="discredit"][quote="Erken"]
I was talking about this site...or maybe Google can help you,but if you know russian, then I can scan the page and send it to you by pm. :lol:[/quote]

The problem is I don't know Russian...

Maybe you can post at least the outline of the solution? Or the full solution would be good.[/quote]
Well I'll do it later...if not today,then tomorrow for sure. :wink: Or maybe someone else will post a solution

It does not seem to me that hard, although perhaps a little awkward to explain. If I didn't miss something, you should be able to assign coordinates to the grid squares (taking some arbitrary square to be (0, 0)), and then consider the sets A and B of coordinates covered by the two figures (placed arbitrarily, as long as they are oriented correctly). Then look at the multiset $ \{ a \plus{} b | a \in A, b \in B \}$, where we use vector addition to add $ a$ and $ b$.

To [b]discredit[/b]:
I guess now there is no reason for me to post the outline of a solution,because [b]probability 1.01[/b] already done it... :wink:


ok if u don't want a proof by well ordering try it by induction..though both are equivalent
fix $ n$ we will prove that for every $ m$ and any $ n$ there exist $ q$ and $ 0\leq r < n$ such that $ m \equal{} nq \plus{} r$
for $ m \equal{} 1$ this is trivial set $ q \equal{} 0$ and $ r \equal{} 1$
or else if $ n \equal{} 1$ set $ r \equal{} 0$ and $ q \equal{} 1$
now let this be true for $ m \equal{} k$ for some natural number $ k$ then
there exists $ q'$ and $ r'$ such that
$ k \equal{} nq' \plus{} r'$ $ 0\leq r' < n$
now if $ 0 \leq r' < n \minus{} 1$
$ k \plus{} 1 \equal{} nq' \plus{} r' \plus{} 1 \equal{} nq \plus{} r$ where $ q' \equal{} q$ and $ r' \plus{} 1 \equal{} r < n$
if $ r' \equal{} n \minus{} 1$
then $ k \plus{} 1 \equal{} n(q' \plus{} 1) \equal{} nq \plus{} 0$
so we are done

wth da fool. i fully typed a proof by WOP and then submitted and found it was locked  :mad: . Stupid moderator  :mad:

Nice... :D Can you pm me the full proof rohit? Someone wanted it..  Thats why I posted this question in the first place, sorry about its trivial nature oh master of mathematics!

sorry da...when i read the thread there was nothing there u can post it here if u want...

dei dumbo ******* i typed full of connotations and things and u say coolly?  :mad:  :mad:
@shruthi : tell me who wanted it and i shall chose my discretion to tell the place where the proof is  :)

dei avasarekudeke ....u shud not have posted  this
[quote="madness"]Wth go mod n? :huh:
A remainder can only be less than or equal to its divisor. You want me to prove that  :huh: ? . Dont infuriate with such problems  :ninja:
@pardesi: WOP is well ordering principle
plz look at ur pm[/quote]
this implied u were disinterested and infuriated :D

Well, a person called Arjun Ramesh, whom you would probably not know..

posts can be misleading  :D  And i was infuriated bcos of sth else due to the author which made me show my anger here da  :|

Who is Arjun Ramesh? Where does he study? I need to know these things to tell the source And probably i know his source too  :D

Aiyo.... He used to study in CV.. Now he is working ok? Internship.. And he was bugging me with random problems, and I got reallllly bugged.. So I just posted the first problem I couldnt take here.. Now can you please?

I have a book whose first page is this WOP  :D . Std proof. I dont have the patience to type it all again  :) . So ask pardesi to type as it involves a lot of tedious symbols  :P . i have to go now. have to study for chem pracs

dei rohit ore problem ke solution kudeke yen da evelave alatikare :huh:

dei naye i have board chem pracs da and as i said the proof involves much of tedious latex da so i cant have to go neways.  :D
@shreyas if u see this: Have asked pardesi wat u told me to ask.  :D

seri let me do it it isn't long
Consider the set $ S\equal{}[m\minus{}nq>0: q\in \mathbb{N}]$
that is the set of numbers of the form $ m\minus{}nq$ is positive and  where $ q$ ranges over  integers..
by WOP there exists a $ q'$ for which $ m\minus{}nq'$ is minimum say that value is $ r$ now we prove $ m\minus{}nq'\equal{}r <n$
because if $ m\minus{}nq'\equal{}r>n$ then $ m\minus{}q'r>m\minus{}(q'\plus{}1)n\equal{}r\minus{}n\equal{}r'>0$ so a contradiction since we assumed that
$ r$ is the minimum [b]Positive[/b] value of $ m\minus{}nq$ .
So done :)

Lol.. Thanks anyway.. All the best for chem Rohit...

I didnt have chem pracs today.. had bio.. Went great!

chem pracs over and i dont wanna speak much abt it as i am already up in flames  :mad:  :mad:  :mad:

The proof goes as follows:
Consider $ S = [b - ak|b - ak \geq 0,k \in \mathbb{Z}]$
$ \implies b + |ab| \in S$.
$ \implies$ $ |S| \neq 0$.
By WOP, $ S$ has a least element say $ b - aq = r$ .
If $ r \geq |a|$ then $ 0 \leq r - |a| < r \in S$
a contradiction.
$ \implies 0\leq r < |a|$ .

[color=blue]MOD EDIT:I $ \text{\LaTeX}$ ified it [/color]

Thanks Rohit :D And why up in flames about the practs da?

[quote="madness"]viva ques went to my neighbours while i was simply sitting and nodding my head whether they where right as if i was the external examiner  :D   :huh:  Actually i should be paid the fees for the day  :P[/quote]

I know. Same here. I answered most that I got but the fact remains that I got very less. I am very worried. WTH Gandu got the sappest of the quest and he faltered. Wrong NV.

same here da!!!  :D  And i got only 2 ques da  :mad:

@madness
everything is gud enuf without $ \text{\LaTeX}$ da so no need :)

i want that belongs to in latex da plz  :D  :D then only it will look like a professional proof  :D


Prove that every odd integer $ n\neq1,-1$ Has infinitly many multiples cosisting, in base 10, only of the digits 1and 0, with not 2 consecutive digits 1. ($ n$ not a multiple of 5)

 :)

[hide="a generalization"] For any positive integers $ k$ and $ n$ such that $ (n, 10)=1$, we have that there are infinitely many multiples of $ n$ which are just several versions of $ k$ in a row (for example, if $ k$ were $ 10$, of the form $ 10101010...010$.)
[/hide]

[hide="solution"] Suppose $ k$ has $ d$ digits.  Let $ k_{i}$ just be $ i$ $ k$'s in a row (ie, $ k+10^{d}k+10^{2d}k+...+10^{d(i-1)}k=k\sum_{m=0}^{i=1}10^{md}$).  Then consider the residues modulo $ n$ of
\[ k, k_{2}, k_{3}, ... \]
Clearly, two of these must be congruent modulo $ n$.  Then $ n$ divides their difference.  We then have something like
\[ n|10^{dj}k_{i}\]
But $ (n, 10)=1$, so $ n|k_{i}$.  It then follows that $ n$ divides two of $ k_{i}$ in a row, three in a row, etc...[/hide]

I don't think that you are done...

You showed that there exists $ i$ and $ j$ such that $ n|(k_{i}-k_{j})$...but then you deduce that $ n|k$.

Like if you think about the way that you proved this, you essentially showed that for any $ k$, $ n|k$.

Looking at your proof, this would be the different step. If wlog $ i>j$, then $ n|k*10^{j}*\frac{1}{9}*(10^{i-j}-1)$.

Basically you proved that there exists $ q$ such that if $ (n,10)=1$, then $ n|(10^{q}-1)$.

The core of the problem is still not solved...Of course once you find one $ k$, you can trivially construct more.

[quote="Altheman"]I don't think that you are done...

You showed that there exists $ i$ and $ j$ such that $ n|(k_{i}-k_{j})$...but then you deduce that $ n|k$.
[/quote]

This wasn't the argument.  Here is what the logical progression was...

$ k_{i+j}\equiv k_{j}\mod{n}$
$ n|(k_{i+j}-k_{j})$, $ k_{i+j}-k_{j}=10^{dj}k_{i}$
$ \implies n|10^{dj}k_{i}\implies n|k_{i}$

I did not say $ n|k$: rather, I said $ n$ divides [b]some[/b] $ k_{i}$.

Haha okay I missed a huge step. I didn't see how you collapsed those terms.

Essentially you showed that given a positive integer, $ d>1$, there exists $ q$ such that $ n|\frac{10^{qd}-1}{10^{d}-1}$ where $ (n,10)=1$.

This uses like a pidgeonhole argument on partities mod $ n$ and then subtracting, etc.

That lemma essentially finishes the problem...

I was thinking of doing like $ 1000...0001$ where you extend the number of $ 0$'s, but I didn't think of extending the number of blocks of $ 100000...00$.

Good solution.

[quote="ElChapin"]:huh:

I understand all the prove, but I can't get how does it apply to the problem.

???[/quote]

So okay...take $ k$ to be any number formed by only 1's and 0's which doesn't have two consecutive 1's in a row and doesn't both begin and end in a 1.  $ k=10$, would probably be the simplest example.

mmm... I see...
As you don't even mentioned it, I supose it is trivial to find such $ k$ for all $ n$.
Well nice solution then.
__________________________________________________________________

Mine goes like this:

[b]Lemma 1:[/b]

All integers $ n$ relativly prime with $ 10$, have a multiple with only ones (a repunit).
[find $ x$ such that $ 9n|10^{x}\minus{}1$]

[b]Lemma 2:[/b]

$ 1\plus{}x\plus{}x^{2}\plus{}...\plus{}x^{n\minus{}1}\equal{}p|q\equal{}x^{a_{1}}\plus{}x^{a_{2}}\plus{}...\plus{}x^{a_{n}}$ if
$ \{1, 2,\cdots,n\}\equiv\{a_{1}, a_{2},\cdots, a_{n}\}\pmod{n}$
[$ n$ roots of the unit]

 :)

[hide="Solution"]
Let $ n|p\equal{}1\plus{}10\plus{}10^{2}\plus{}...\plus{}10^{k\minus{}1}$ with $ k>1$ an  integer, and $ q$ as in lemma 2.
The problem is reduced then to prove there are infinitely many sets A such that
$ A\equal{}\{a_{1}, a_{2},\cdots, a_{n}\}\equiv\{1, 2,\cdots,n\}\pmod{n}$ in order to have $ q$ with no two consecutive 1's, wich is plain.
For example:
$ a_{i}\equal{}b_{i}n\plus{}i$ for $ i\equal{}1,2,\cdots,n$  and $ a_{i\plus{}1}>a_{i}\plus{}1$ for $ i\equal{}1,2,\cdots,n\minus{}1$

 :)
[/hide]

I guess here is how I would write up a proof.

Let $ S$ be the set of numbers with the desired property. Define $ f(x)\equal{}\frac{100^{x}\minus{}1}{99}$. Consider the $ n\plus{}1$ integers: $ f(1)$, $ f(2)$,..., $ f(n\plus{}1)$ modulo $ n$. Clearly two must be equivalent modulo $ n$ by the pidgeonhole principle. Suppose that $ f(i)\equiv f(j)\mod n$ where $ i>j$. Then $ n|[f(i)\minus{}f(j)]\equal{}10^{j}*f(i\minus{}j)$. Since $ (n,10)\equal{}1$, $ n|f(i\minus{}j)$. Hence $ S$ is nonempty.

Suppose for sake of contradiction that $ S$ had a finite number of elements. Since $ S$ is a set of positive integers with a finite number of elements, it has a maximal element. Call it $ x$. If $ x$ has $ d$ digits, then consider $ x*10^{d\plus{}1}\plus{}1$. This number is in $ S$ and larger than $ x$, which is a contradiction. Hence $ S$ has an infinite number of elements.


conics
algorithm
parameterization
algebra
polynomial
number theory
Diophantine equation

I don't understand how to do them very well. Linear ones are easy, but once they get more advanced I get lost. Could somebody post guidelines for solving them and/or some practice problems?

Thanks in advance.

One thing that bites a lot of people is the fact that $ x^2$ represents both a positive and negative solution. In some equations, like with square roots, one must check for extraneous solutions. Also notice that in conics, the constant is also the square of a certain value, not the value itself. That caused me a lot of trouble, and made my Conics test grade ([b][color=red]C-[/color][/b]) by far the worst of my math grades.

I don't see the relevance of the post above  :huh: What do conics have to do with Diophantine equations?

Linear Diophantine equations, as you already know, have a standard algorithm for obtaining the solutions. Thanks to Yuri Matiyasevich, we know today that such algorithm doesn't exist for all Diophantine equations (there is a standardized way of solving second-order D. e., and the algorithm for third order is yet to be found).

What should you use for olympiad D.es? Well, the toolbox could include:

-modular arguments (very helpful when it comes to squares and their residues), which can lead to use of CRT etc.
-factoring and cases-solving
-elliptic curves
-constructing solution families (when asked to prove that an equation has infinitely many solutions)

http://www.mathlinks.ro/index.php?f=466

PEN Diophantine equations might be too hard, but you could see the application of common techniques, and then return to easier problems on Intermediate and Pre-Olympiad and try to apply the techniques there.

/just my $ \frac{1}{50}\$$/

[quote="hsiljak"]I don't see the relevance of the post above  :huh: What do conics have to do with Diophantine equations?[/quote]
Well, conics are the next simplest form of Diophantine equation to look at after linear ones.  There are a few overlapping perspectives that allow you to solve several classes of them, including the notion of a [url=http://en.wikipedia.org/wiki/Pell%27s_equation]Pell's equation[/url] and some standard factorization arguments (over the integers) allowing you to compute Pythagorean triplets and the solutions to other related equations, as well as more elaborate factorization arguments over other rings.  In general, it is known that every conic in two variables has a rational parameterization, hence the solutions to every homogeneous conic in three variables over the integers has a polynomial parameterization.

Thank you, Qiaochu: conics were always $ \mathbb{R}\to \mathbb{R}$ in my (algebraic) perspective, since I always related them to my Algebra courses (which never implied searching for its integer solutions). In other words, conics in this context are what I called 'second-order Diophantine equations', right?


geometry
incenter
circumcircle
projective geometry
geometry unsolved

$ I$ is the incenter of $ ABC$ triangle and $ ABC$ is not isosceles. $ A_1 , B_1 , C_1$ are the points of intersection of incircle to  $ BC , AC , AB$ respectively.prove that circumcenters of $ AIA_1 , BIB_1 , CIC_1$ are collinear

$ \mbox{Let } O_a, O_b, O_c \mbox{circumcenters of }\triangle{IAA_1}, \triangle{IBB_1} \mbox{ and } \triangle{ICC_1}. \\
\mbox{Let }
A_2\in BC, AA_1\perp AA_2; B_2\in AC, BB_1\perp BB_2\\
 \mbox{ and } C_2\in AB, CC_1\perp CC_1\perp CC_2. \\
\mbox{The points } A_2, B_2, C_2 \mbox{ are on the trilinear polar line of the point I and the points } \\
O_a, O_b \mbox{ and } O_c \mbox{ is midpoints of } [IA_2], [IB_2], [IC_2] \mbox{ respectively}\Rightarrow \\
\Rightarrow \mbox{ the points } O_a, O_b, O_c \mbox{ are colinear.}$

your solution is nice
but is there any other solution?


sorry i'm a senior in hs and haven't applied myself in math until recently. Could you please state the theorem or rule that you use to find log(10/2)? it would "make sense" and i believe you than mean that 10-2 is idk 3log(2)???

Could you understand the meaning of $ \log 2?$

if your implying thats theres some generalization that states log(a)=(2a/2) which seems obvious but not with logs then i guess yes; yet i dont know where to go from there. To be honest i don't know how this math is so difficult for me. Maybe i tried to go through aops v1 too fast.

[quote="adrenalinefix36"]if log(2)=.3010 and log(3)=.4771 the value of x when 3^(x+3)=135 is approximately
A)5
B)1.47
C)1.67
D)1.78
E)1.63

no clue how to solve it since i constantly get a log(5) is there any relationship between the log2 the log3 and the log5? I guess this isn't an intermediate problem but please stoop.[/quote]

[hide="Hint 1"]$ 135 \equal{} 27\cdot5 \equal{} 3^3\cdot5$[/hide]

[hide="Hint 2"]We want base 10 here, so note that $ 5\cdot2 \equal{} 10$.[/hide]

$ \log 2$ is a number $ x$ such that $ 10^x \equal{} 2.$ Let $ 10^y \equal{} 5$ we have $ 10^{x \plus{} y} \equal{} 2*5.$ Thus $ x \plus{} y \equal{} \log 2*5\Longleftrightarrow \log 2*5 \equal{} \log 2 \plus{} \log 5.$
Similarly $ 10^{x \minus{} y} \equal{} \frac {2}{5}\Longleftrightarrow x \minus{} y \equal{} \log \frac {2}{5}\Longleftrightarrow \log 2 \minus{} \log 5 \equal{} \log \frac {2}{5}$.

i'm sorry sorry for my stupidity. After coming up with the two hints, which i had done previously, where do we go? PLease show steps as if i was an alien entirely ignorant of any rule. lol.thanks

yes 100%! please show calculation of answer

Before showing the answer, complete the following expressions.

Let $ a,\ b,\ c,\ p,\ M$ are positive numbers not equal to one.

1. $ \log_a M \plus{} \log_a N \equal{} \boxed{A}.$

2. $ \log_a M \minus{} \log_a N \equal{} \boxed{B}.$

3. $ \log_a M^p \equal{} \boxed{C}.$

4. $ \frac {\log_c b}{\log_c a} \equal{} \boxed{D}.$

Sorry my first post was useless.  :oops: Now you can solve your problem in using formula 4.

Good luck! :)

kunny

I apologize once more for my ignorance i simply do not understand logs. I guess i understand that some log operations like log(5)=(log(10)/log(2)) but i don't know that value(log(10)). It is it 3(log(2))? by subtraction or am i supposed to know what log(10)?(which i don't understand since im only given log(2) and log(3).

yea confusing in reality wouldn't it be log(2)/log(2) but in this case its log(2)-log(2) which is 0? since a-a=0

[quote="kunny"]What is the meaning of $ \log_2 2$?[/quote]

$ \log_2 2$ is the number $ x$ such that $ 2^x\equal{}2$. Thus $ \log_a a\equal{}1$ for $ a>0$ and $ a\neq 1.$

[quote="kunny"]$ \log_2 2$ is the number $ x$ such that $ 2^x \equal{} 2$. [/quote]

This is fundamental.  Most times I find that the difficulty students have with logarithms is that they do not understand the basic [b]definition[/b] of a logarithm, so that even if they memorize all of the relevant properties they are constantly confused as to basic calculations like this.  Here are some examples, and I will write out the base of the logarithm to avoid all possible confusion.

$ \log_{10} 1 \equal{} 0$ because $ 10^0 \equal{} 1$.
$ \log_{10} 10 \equal{} 1$ because $ 10^1 \equal{} 10$.
$ \log_{10} 2 \equal{} 0.3010...$ because $ 10^{0.3010...} \equal{} 2$.
$ \log_{10} 3 \equal{} 0.4771...$ because $ 10^{0.4771...} \equal{} 3$.
$ \log_{10} 1000 \equal{} 3$ because $ 10^{3} \equal{} 1000$.

If you understand exponents and you understand the definition of a logarithm, then you understand logarithms.  Therefore, ask yourself:

- [i]Do you understand exponents?[/i]
- [i]Do you understand the definition of a logarithm?[/i]

What kunny has been trying to communicate to you is that

$ \log_{10} 5 \equal{} \log_{10} \left( \frac{10}{2} \right) \equal{} \log_{10} 10 \minus{} \log_{10} 2 \equal{} 1 \minus{} \log_{10} 2 \equal{} 1 \minus{} (0.3010) \equal{} .6990$.

The first equality is arithmetic.  The second equality is the logarithm property

$ \log_a \frac{b}{c} \equal{} \log_a b \minus{} \log_a c$

which follows from the corresponding exponent property

$ a^{x \minus{} y} \equal{} \frac{a^x}{a^y}$

where $ x \equal{} \log_a b, y \equal{} \log_a c$.  The third equality follows from kunny's discussion that $ \log_a a \equal{} 1$.

Now i understand so this is process. First you take the equation with exponents and simplify it from 3^(x+3)=135 to 3^x=5. Then seeing the exponents, we convert to logs. We see that log3(5)=x=(log(5))/(log(3)). From there we simplify logs, the log(5) following the common concept of numerical division. Hence, log(10/2)/log(2). The important thing to then realize, is that given the fact that all logs unless noted otherwise are of base 10, and the fundamental relationship of loga(N)/loga(N)=1 by the law of exponents we have (1-log(2))/log(3). From there just use substitution of known quantities. Thanks, i'll try to make it easier next time by learning latex.


limit
integration
trigonometry
logarithms
calculus
derivative
calculus computations

$ \lim_{x\to + \infty}{(\frac {1}{lnx}\int_{0}^{x}{\frac {dt}{\sqrt [5]{1 + t^{5}}}})} = \lim_{x\to\infty} \frac {1}{\sqrt [5]{1 + x^{5}}} \cdot x = 1$ by Lopitale rule
$ \lim_{x\to + 0}(\frac {1}{x^{2}}\int_{0}^{x}{t^{1 + t}dt}) = \lim_{x\to 0} \frac {x^{1 + x}}{2x} = \lim_{x\to 0} \frac12 x^x = \frac12$ . the same idea
since $ \sqrt [n]{x}sin(x) \to \sin x,\ n\to \infty$ then $ \lim_{n\to + \infty}{\int_{0}^{\pi}{\sqrt [n]{x}sin(x) dx}} \to \int_0^{\pi} \sin x\ dx = 2$ by Lebesque convergence theorem
since $ sin^{n}{x} \to 0\ (mod \lambda)$ then $ \lim_{n\to + \infty}{\int_{0}^{\frac {\pi}{2}}{\frac {sin^{n}{x}}{\sqrt {1 + x}}dx} = \int_{0}^{\pi/2} 0 \,dx = 0}$ by Lebesque convergence theorem

$ \forall x\in[n,n + 1]: \ \frac {n}{n^5 + 1} \ge \frac {x}{x^5 + 1} \ge \frac {n + 1}{(n + 1)^5 + 1}$ so $ \frac {n^5}{n^5 + 1} = n^{4}\int_{n}^{n + 1}{\frac {n\ dx}{n^{5} + 1} \ge n^{4}\int_{n}^{n + 1}{\frac {xdx}{x^{5} + 1} \ge n^{4}\int_{n}^{n + 1}{\frac {n + 1\ dx}{(n + 1)^{5} + 1} = \frac {n^5 + n^4}{(n + 1)^5 + 1}}}}$. RHS and LHS tends to 1 when $ n\to\infty$ so answer is 1

$ \lim_{n\to + \infty}{n^{3}\int_{n}^{2n}{\frac {xdx}{x^{5} + 1}} = \lim_{n\to + \infty} \frac {\frac {2\cdot 2n}{(2n)^{5} + 1} - \frac {n}{n^5 + 1} }{ - 3n^{ - 4} } = \frac13\frac {n^4(32n^6 - 4n^6 + ....)}{32n^{10} + ...} = \frac {28}{3\cdot 32}=\frac{7}{24}}$ by Lopitale rule

and 1)
$ \lim_{n\to + \infty}{\frac {\sqrt {n^{3n}}}{n!}{\prod_{k = 1}^{n}{sin(\frac {k}{n\sqrt {n}})}}} = \lim exp( \sum \ln\left(\frac {\sin\frac {k}{n\sqrt {n}}}{\frac {k}{n\sqrt {n}}}\right))$
now

 $ \lim \sum \ln\left(\frac {\sin\frac {k}{n\sqrt {n}}}{\frac {k}{n\sqrt {n}}}\right) = \lim \sum \ln\left(\frac {\frac {k}{n\sqrt {n}} - \frac16 \frac {k^3}{n^4\sqrt {n}} + o(\frac {k^4}{n^6})}{\frac {k}{n\sqrt {n}}}\right) = \lim \sum \ln\left(1 - \frac16\frac {k^2}{n^3} + o(\frac {1}{n\sqrt {n}})\right) = \lim - \frac16 \sum \frac {k^2}{n^3} + o(\frac {1}{n\sqrt {n}}) = - \frac16 \frac {n(n + 1)(2n + 1)}{6n^3} + o(\frac {1}{\sqrt {n}}) \ to - 1/36$. one can shown that all $ o(..)$ here is uniformly bounded.. so
answer is $ e^{ - 1/36}$

[quote="anik_andrew"]
$ \lim_{n\to + \infty}{n^{3}\int_{n}^{2n}{\frac {xdx}{x^{5} + 1}} = \lim_{n\to + \infty} \frac {\frac {2n}{(2n)^{5} + 1} - \frac {n}{n^5 + 1} }{ - 3n^{ - 4} } = \frac13\frac {n^4(32n^6 - 2n^6 + ....)}{32n^{10} + ...} = \frac {10}{32}}$ by Lopitale rule
[/quote]

This can't be correct.  As $ x \to + \infty$ we see that $ f(x) = \frac {x}{x^5 + 1} \sim \frac {1}{x^4}$, so

$ \int_{x = n}^{2n} \frac {x \, dx}{x^5 + 1} \sim \int_{x = n}^{2n} \frac {dx}{x^4} = \frac {7}{24n^3}$.

Thus $ \lim_{n \to + \infty} n^3 \int_{x = n}^{2n} \frac {x \, dx}{x^5 + 1} = \frac {7}{24}$.

The error is that you computed the derivative of the integral with respect to $ n$ incorrectly.  It should be

$ \frac {d}{dn}\left[\int_{x = n}^{2n} \frac {x \, dx}{x^5 + 1} \right] = \frac {d}{dn}[2n]\frac {2n}{(2n)^5 + 1} - \frac {n}{n^5 + 1}$.

hello, yes for 7) i have also got
$ \lim_{n \to \infty}n^3\int_{n}^{2n}\frac{x}{1\plus{}x^5}\,dx\equal{}\frac{7}{24}$.
Sonnhard.

atomicwedgie, thanks )) i forgot this derivirtate ) i have edited )  :blush:

and what do you mean
[quote="atomicwedgie"]

 we see that $ f(x) \equal{} \frac {x}{x^5 \plus{} 1} \sim \frac {1}{x^4}$, so

$ \int_{x \equal{} n}^{2n} \frac {x \, dx}{x^5 \plus{} 1} \sim \int_{x \equal{} n}^{2n} \frac {dx}{x^4}$.

[/quote] ?

hello, for 6)
$ \lim_{n \to \infty}n^4\int_{n}^{n\plus{}1}\frac{x}{x^5\plus{}1}\,dx$ i have got $ 1$ as the limit.
Sonnhard.


geometry
incenter
circumcircle
projective geometry
power of a point
radical axis
geometry solved

First of  all we observe that $A'I$ is tangent to the circle $IBC$ (by simple angle chasing)

Denote center of the circle $IBC$ as $O_1$, its radius - $R_1$ and $M$ -midpoint of $BC$.

We will prove that $A'O^2- A'I^2 =R^2$ , same for $B,C$
(which will give us that this all this points are collinear and $OI\perp A',B',C'$)

$A'O^2 = A'M^2+OM^2= A'O_1^2 -O_1M^2+OM^2$
$A'I^2 = A'O_1^2-R_1^2$

$A'O^2-A'I^2 = R_1^2+OM^2-O_1M^2$

But we can see that $OM = Rcos(a), R_1 = 2Rsin(a/2), O_1M = 2Rsin^2(a/2)$

And $A'O^2-A'I^2 = R^2cos^2(a)+4R^2sin^2(a/2)- 4R^2sin^4(a/2)= R^2$ c.t.d.

Construct polar line of the incenter. Line OI intersects it at X.
     Three circles with centers at A',B',C'   passing thru I and X  are coaxal
      with A'-B'-C' the axis - mid-perpendicular to IX,  or midline in the
     corresponding triangles with vertex at I.

      Maj.  Pestich

I am sorry pestich, but could you explain why $A'X = A'I$ ,
(I mean $A'$ is center of the circle which passes through $I,X$)

[quote="jhaussmann5"]Let I be the incentre of triangle ABC.  Let the line passing through I perpendicular to AI intersect side BC at A'.  Points B', C' are defined similarly.  Prove that A', B', C' lie on a line perpendicular to IO, where O is the circumcenter of triangle ABC.[/quote]

My proof of this uses more or less the same idea as Prowler's one, but is a bit more general:

Since the point I is the incenter of triangle ABC, it lies on the angle bisectors of the angles CAB, ABC and BCA. Thus, $\measuredangle IAB=\frac{A}{2}$, $\measuredangle IBA=\frac{B}{2}$ and $\measuredangle ICB=\frac{C}{2}$.

WLOG assume that the point A' lies on the extension of the segment BC beyound the point B. Then, < A'IB = < AIB - < AIA'. But since $IA^{\prime}\perp AI$, we have < AIA' = 90°, and thus < A'IB = < AIB - 90°. But by the sum of the angles in triangle AIB, we have $\measuredangle AIB=180^{\circ}-\measuredangle IAB-\measuredangle IBA=180^{\circ}-\frac{A}{2}-\frac{B}{2}$. Hence,

$\measuredangle A^{\prime}IB=\measuredangle AIB-90^{\circ}=\left(180^{\circ}-\frac{A}{2}-\frac{B}{2}\right)-90^{\circ}=90^{\circ}-\frac{A}{2}-\frac{B}{2}$
$=90^{\circ}-\frac{A+B}{2}=90^{\circ}-\frac{180^{\circ}-C}{2}$       (since A + B = 180° - C by the sum of the angles of triangle ABC)
$=90^{\circ}-\left(90^{\circ}-\frac{C}{2}\right)=\frac{C}{2}=\measuredangle ICB$.

Hence, by the converse of the tangent-chordal angle theorem, it follows that the line IA' is the tangent to the circumcircle of triangle BIC at the point I. Thus, the point A' is the point where the tangent to the circumcircle of triangle BIC at the point I intersects the line BC. Similarly, the points B' and C' are the points where the tangents to the circumcircles of triangles CIA and AIB at the point I intersect the lines CA and AB, respectively. Now, our assertion that the points A', B', C' lie on one line perpendicular to the line IO becomes a particular case of the following general theorem:

[color=blue][b]Theorem 1.[/b] Let P be an arbitrary point in the plane of a triangle ABC, and let the tangents to the circumcircles of triangles BPC, CPA, APB at the point P intersect the lines BC, CA, AB at the points A', B', C', respectively. Then, these points A', B', C' lie on one line perpendicular to the line PO, where O is the circumcenter of triangle ABC.[/color]

[i]Proof of Theorem 1.[/i] Let k be the circumcircle of triangle ABC, and let p be the "zero circle" with center P (this is the circle with center P and radius 0). Then, the power of the point A' with respect to the circle k equals $A^{\prime}B\cdot A^{\prime}C$, and the power of the point A' with respect to the circle p equals $A^{\prime}P^2$. But by the intersecting secant and tangent theorem, we have $A^{\prime}B\cdot A^{\prime}C=A^{\prime}P^2$ (since the point A' lies on the tangent to the circumcircle of triangle BPC at the point P). Hence, the point A' has equal powers with respect to the circles k and p; thus, the point A' lies on the radical axis of the circles k and p. Similarly, the points B' and C' also lie on this radical axis. Hence, all three points A', B', C' lie on one line, namely on the radical axis of the circles k and p. This radical axis is clearly perpendicular to the line PO, since O and P are the respective centers of the circles k and p, and the radical axis of two circles is always perpendicular to the line joining their centers. This proves Theorem 1.

  Darij


Given$ a,b,c>0$.Prove that:
$ \frac{20ab}{(a\plus{}c)(b\plus{}c)}\plus{}\frac{15bc}{(b\plus{}a)(c\plus{}a)}\plus{}\frac{12ca}{(c\plus{}b)(a\plus{}b)} \ge 11$ :)

[quote="nsato"][hide]Expanding and simplifying, this becomes
\[ 9a^2 b + 9ab^2 + a^2 c + ac^2 + 4b^2 c + 4bc^2 \ge 22abc.
\]
But by weighted AM-GM,
\begin{align*} & 9a^2 b + 9ab^2 + a^2 c + ac^2 + 4b^2 c + 4bc^2 \\
& = 36 \cdot \frac {a^2 b}{4} + 18 \cdot \frac {ab^2}{2} + 12 \cdot \frac {a^2 c}{12} + 18 \cdot \frac {ac^2}{18} + 12 \cdot \frac {b^2 c}{3} + 36 \cdot \frac {bc^2}{9} \\
& \ge 132 \left( \frac {a^2 b}{4} \right)^{36/132} \left( \frac {ab^2}{2} \right)^{18/132} \left( \frac {a^2 c}{12} \right)^{12/132} \left( \frac {ac^2}{18} \right)^{18/132} \left( \frac {b^2 c}{3} \right)^{12/132} \left( \frac {bc^2}{9} \right)^{36/132} \\
& = 22abc. \end{align*}
Equality occurs if $ a: b: c = 2: 1: 3$.[/hide][/quote]
Nice solution :)

[quote="quykhtn-qa1"]Given$ a,b,c > 0$.Prove that:
$ \frac {20ab}{(a \plus{} c)(b \plus{} c)} \plus{} \frac {15bc}{(b \plus{} a)(c \plus{} a)} \plus{} \frac {12ca}{(c \plus{} b)(a \plus{} b)} \ge 11$ :)[/quote]
After expanding we need to prove that
$ (9b\plus{}c)a^2\plus{}(9b^2\minus{}22bc\plus{}c^2)a\plus{}4b^2c\plus{}4bc^2\geq0.$
If $ 9b^2\minus{}22bc\plus{}c^2\geq0$ the inequality is true.
But for $ 9b^2\minus{}22bc\plus{}c^2<0$ enough to prove that
$ (9b^2\minus{}22bc\plus{}c^2)^2\minus{}16bc(b\plus{}c)(9b\plus{}c)\leq0,$ which is equivalent to
$ (3b\minus{}c)^2(9b^2\minus{}54bc\plus{}c^2)\leq0,$ which is true.

[quote="arqady"][quote="quykhtn-qa1"]Given$ a,b,c > 0$.Prove that:
$ \frac {20ab}{(a \plus{} c)(b \plus{} c)} \plus{} \frac {15bc}{(b \plus{} a)(c \plus{} a)} \plus{} \frac {12ca}{(c \plus{} b)(a \plus{} b)} \ge 11$ :)[/quote]
After expanding we need to prove that
$ (9b \plus{} c)a^2 \plus{} (9b^2 \minus{} 22bc \plus{} c^2)a \plus{} 4b^2c \plus{} 4bc^2\geq0.$
If $ 9b^2 \minus{} 22bc \plus{} c^2\geq0$ the inequality is true.
But for $ 9b^2 \minus{} 22bc \plus{} c^2 < 0$ enough to prove that
$ (9b^2 \minus{} 22bc \plus{} c^2)^2 \minus{} 16bc(b \plus{} c)(9b \plus{} c)\leq0,$ which is equivalent to
$ (3b \minus{} c)^2(9b^2 \minus{} 54bc \plus{} c^2)\leq0,$ which is true.[/quote]
Nice :)

I am very sorry for my doubts..... :?:
I am thinking, that if we replace $ a, b, c$ by $ ka, kb, kc$ respectively then the inequality remains unchanged, then can we assume that : $ abc\equal{}1$?? :huh:

Can anyone help me to clear my doubts??
PS: I think I was looking for a proof without Expanding.

[quote="Potla"]I am very sorry for my doubts..... :?:
I am thinking, that if we replace $ a, b, c$ by $ ka, kb, kc$ respectively then the inequality remains unchanged, then can we assume that : $ abc \equal{} 1$?? :huh:

Can anyone help me to clear my doubts??
PS: I think I was looking for a proof without Expanding.[/quote]
Yes, you can assume any condition,my friend :)


assuming that the inert gas configuration of atoms is their most stable  state it follows that it (in. gas. config.)must have the min. energy.(for eg unstable nucleii gain stability by releasing enormous amt.s of energy and are respectively converted to their less energy states which are stable) . It follows therefore that there must be inherent tendency in each particle to release energy and get converted into its most stable state.
Now let us look towards the ionization of metals.All metals of the s block are stabilized by the release of electrons.they are converted into a stabler state and hence energy must be released in the process .WE can say that  DEL(H)<0.also there is an increase in the randomness of the system ,thus DEL(S)>0(from a bound electron we get a free one and the no. of particles increases too).thus the ionization process ought to be feasible at all temperatures and  must be accompanied with a simultaneous release of energy.
What we see in the world is just the reverse.We have to provide heat energy ,light photons etc to the metals for overcoming their ionization energy , ionizatiion potential etc.Why is this so? :?:

The ionization process has an activation energy, doesn't it?  The [b]net[/b] enthalpy and entropy make it favorable, but an initial energy is necessary to begin the reaction.

why at all??
is this activation energy is used for overcoming the coulombic  barrier?
 on the other hand,an electron in the outermost shell is nearer to the electron cloud and must experience a greater repulsive force.:-(electrostatic force is  inversely proportional to the square of the  distance.)
(the nucleus may be assumed to be the size of a tip of a pin paced inside a well of 1 meter  diameter :the electro cloud)
there are reactions which have high act.energy but eventually turn out to be exothermic. Why is the ionization proces not one of them??


calculus integration logarithms limit inequalities LaTeX induction

Let $ (a_n)$ be a sequence of positive real numbers, decreasing and the sum of whose terms is infinite. Prove that the series whose general term is $ min(a_n,\frac{1}{n}) $ is also divergent. Not a difficult, but a very beautiful problem.

The problem came up in the 6th German TST this year and was already posted by Darij.

Misha

I'm not really suprised becaused nearly all problems of the German TSTs this year
were from the ISL 2003.

Let's start with a simpler problem: Supose $x_n>0$ and $\sum x_n=\infty$. Then we can prove that $\sum \min(1,x_n)=\infty$. This is clear as follows. If $x_n>1$ infinitely many times, then $\min(1,x_n)=1$ infinitely many times, and the series diverges because its terms don't tend to zero. On the other hand, if $x_n>1$ only finitely many times, then $\min(1,x_n)=x_n$ for all sufficiently large $n$ and $\sum \min(1,x_n)$ diverges by comparison to $\sum x_n$.

Now we employ the Cauchy condensation test for convergence: if $(b_n)$ is a nonincreasing sequence of positive numbers, then $\sum b_n$ converges if and only if $\sum 2^kb_{2^k}$ converges. (This is merely an extension of Oresme's 14th-century proof of the divergence of the harmonic series).

The minimum of two monotone sequences is monotone, so $\min\left(a_n,\frac1n\right)$ is nonincreasing. By the condensation test, we know that $\sum 2^ka_{2^k}$ diverges. Applying the condensation test again, we see that $\sum \min\left(a_n,\frac1n\right)$ is comparable to $\sum 2^k\min\left(a_{2^k},\frac1{2^k}\right)=\sum\min\left(2^ka_{2^k},1\right)$. The result in the first paragraph shows that this diverges.

Your proof is the same as the best proof in the TST. At that time I was not so good in Calculus and I even didn't know the condensation test by Cauchy.

Misha

All tests for the convergence of positive term series are at heart comparison tests. Out of all of those tests, are there any for which "$a_n$ is decreasing" is a key hypothesis? There are two: the integral test and the condensation test. Unsurprisingly, these two tests tend to be well suited to the same specific examples, such as $\sum \frac1{n(\ln n)^2}$. It was the hypothesis of decreasing terms that caused me to think of the condensation test.

Sorry to revive such an old thread, but here's something related to this. The following problem is sort of a converse to the one posted here:

Given a decreasing sequence $(b_n)_n$ of positive reals with the property that for every decreasing sequence of positive reals $(a_n)_n$ with $\sum a_n=\infty$ we have $\sum\min(a_n,b_n)=\infty$, prove that $\inf_n nb_n>0$.

It appeared in some AMM from a couple of years ago, I think.

That should be $\liminf$; the long-term behavior is what matters.

I'm pretty sure it works with $\inf$ though.

The problem is that it's vacuous with $\inf$. The $\inf$ is always positive.

Why? Isn't it possible to find large $n$'s for which $nb_n$ approach zero? I mean, no, it's not possible, since that's what the problem states, but why do you find it obvious?

i made a nice proof:suppose that Sum(b_n) <+00 where (b_n)=(inf(a_n,1/n))
since (a_n) and (1/n) are bouth positifs and decreasing it's to for (b_n) , and so
0<=n.b_2n<=Sum(b_k,k=n,...,2n)=o(1)  and because (b_n) is decreasing we conclude that  b_n=o(1/n) , so for sufficient large n :b_n=a_n witch gaves a contradiction. :P

Sorry yassinus,

I am probably too stupid, but still can't see why $\sum_{k=n}^{2n}b_k = o(1)$ (1). For example $\lim_{n\to \infty}\sum_{k=n}^{2n}\frac{1}{k} =\ln 2$, but I assume you use a different argument to get (1)... or do you mean

$\sum_{k=n}^{2n}b_k = O(1)$?

Many thanks!

[i][b]Edit[/b][/i]   To Kent Merryfield: I don't know the proof that you mean of the so-called ``condensation criterion'', but isn't it just a consequence of the inequalities

$\sum_{k=3}^{\infty}b_k\leq \sum_{k=1}^{\infty}2^k b_{2^k} \leq 2 \sum_{k=1}^{\infty}b_k$?

Many thanks!

jifko777: Yes, that's the proof. Think of it as a generalization inspired by Oresme's ~800 year old proof of the divergence of the harmonic series.

Kent, I must confess that that's the easiest proof I have seen so far, which makes it sort of brilliant, doesn't it.

Cheers.

but we know that sum(bk,k=n,+00) ->0 when n->+00 because i suppose the series convergent ( your exemple is not a counter-exemple because harmonic  sum is not convergent as we know)
 0<=sum(bk,k=n...2n) <=sum(bk,k=n,+00)  and we can conclud that sum(bk,k=n...2n)=o(1)  :huh:

Thanks [b]yassinus[/b], very nice proof indeed! I presume I wasn't concentrated enough when reading your posting.

Actually the same technique may be used to prove that the harmonic series diverges.

Just a minor request to you: Could you please latex your postings so that they can be read more easily. If you still haven't done latex at all, you can start by holding the mouse pointer on a latex typed formula, then you can see that some symbols appear in a yellow (Windows) box. You can produce the same formula by copying the symbols from the yellow box and putting the dollar sign at the beginning and at the end of your formula.

Once again thanks very much for your reply!

Does Cauchy condensation test hold for $ \sum 3^n a_{3^n}$

[quote="Kent Merryfield"]Let's start with a simpler problem: Supose $ x_n>0$ and $ \sum x_n \equal{} \infty$. Then we can prove that $ \sum \min(1,x_n) \equal{} \infty$. This is clear as follows. If $ x_n>1$ infinitely many times, then $ \min(1,x_n) \equal{} 1$ infinitely many times, and the series diverges because its terms don't tend to zero. On the other hand, if $ x_n>1$ only finitely many times, then $ \min(1,x_n) \equal{} x_n$ for all sufficiently large $ n$ and $ \sum \min(1,x_n)$ diverges by comparison to $ \sum x_n$.

Now we employ the Cauchy condensation test for convergence: if $ (b_n)$ is a nonincreasing sequence of positive numbers, then $ \sum b_n$ converges if and only if $ \sum 2^kb_{2^k}$ converges. (This is merely an extension of Oresme's 14th-century proof of the divergence of the harmonic series).

The minimum of two monotone sequences is monotone, so $ \min\left(a_n,\frac1n\right)$ is nonincreasing. By the condensation test, we know that $ \sum 2^ka_{2^k}$ diverges. Applying the condensation test again, we see that $ \sum \min\left(a_n,\frac1n\right)$ is comparable to $ \sum 2^k\min\left(a_{2^k},\frac1{2^k}\right) \equal{} \sum\min\left(2^ka_{2^k},1\right)$. The result in the first paragraph shows that this diverges.[/quote]

The minimum of two monotone sequences is monotone?

They have to be monotone in the same direction for that to be true, but yes. (And in this situation, they were monotone in the same direction.)

Suppose $ (a_n)$ and $ (b_n)$ are nonincreasing sequences and let $ c_n\equal{}\min(a_n,b_n).$ Then $ c_n$ is nonincreasing.

Let $ m<n.$ Suppose WLOG that $ c_m\equal{}a_m.$ Then $ c_n\le a_n\le a_m\equal{}c_m.$

Ok,I find a more general conclusion is still true:
Suppose  $ (a_n)$ and $ (b_n)$ are nonincreasing positive sequences and let $ c_n\equal{}\min \{a_n,b_n \}$
Then $ \sum c_n$ diverges,if and only if $ \sum a_n$ and $ \sum b_n$ are diverges.

And if there are more than two sequences,then take the minimum,the conclusion will still be true.

zhaobin, that is false. One can find an example in which $ \sum a_n \equal{} \infty,$ $ \sum b_n \equal{} \infty,$ but $ \sum c_n < \infty.$

In fact, when I posed this question in one of the closed portions of this site, I got this response:

[quote="DrKevin"][quote="Kent Merryfield"]
(b) (Very difficult.) Find nonincreasing sequences $ a_n$ and $ b_n$ such that $ \sum_{n \equal{} 1}^{\infty}a_n \equal{} \infty$ and $ \sum_{n \equal{} 1}^{\infty}b_n \equal{} \infty$ but $ \sum_{n \equal{} 1}^{\infty}\min(a_n,b_n)$ converges.
[/quote]

[hide="My shot at 3b"]
Since $ \sum_{n \equal{} 1}^{\infty}\frac {1}{2^n}$ converges, the goal is to construct $ a_n$ and $ b_n$ so that $ \min(a_n, b_n) \equal{} \frac {1}{2^n}$.  We also want both $ \sum a_n$ and $ \sum b_n$ to diverge, so neither $ a_n$ nor $ b_n$ can be the minimum all the time.  Therefore they have to alternate: each assumes the smaller value (i.e. $ \frac {1}{2^n}$) for a certain number of times, while the other stays the same.  The length of each such "run" should be such that the larger value is repeated enough times to add up to 1.  Then, in each of the two sums $ \sum a_n$ and $ \sum b_n$, there will be infinitely many segments with sum 1, therefore they both diverge to $ \infty$, and yet the series of the minimums is convergent.

Thus we can set up the following:
\[ a_1 \equal{} a_2 \equal{} \frac {1}{2}, b_1 \equal{} \frac {1}{2}, b_2 \equal{} \frac {1}{4} \\
a_n \equal{} \frac {1}{2^n}, b_n \equal{} \frac {1}{8}, \quad n \equal{} 3,\ldots,10 \\
b_n \equal{} \frac {1}{2^n}, a_n \equal{} \frac {1}{2^{11}}, \quad n \equal{} 11,\ldots,2^{11} \plus{} 10 \\
a_n \equal{} \frac {1}{2^n}, b_n \equal{} \frac {1}{2^{2^{11} \plus{} 11}}, \quad n \equal{} 2^{11} \plus{} 11,\ldots,2^{2^{11} \plus{} 11} \plus{} 2^{11} \plus{} 10 \\
 \\
\cdots \cdots \cdots
\]
You can see the length of each "run" increases rapidly.  Obviously the solution is not unique.  In fact, we can make $ \{a_n\}$ and $ \{b_n\}$ strictly decreasing ...
[/hide][/quote]

:blush: Sorry,there is something wrong in my "proof".Thank you.
And by your solution and the construction above,I find follow theorems which is obvious.

(1)Suppose  $ (a_n)$ are nonincreasing positive sequences,and $ \sum a_n$ diverges,then $ \sum \min \{a_n,\frac {1}{n \ln n}\}$ diverges.or $ \sum \min \{a_n,\frac {1}{n \ln n \ln \ln n}\}$ diverges.

(2)for any nonincreasing positive sequence $ c_n$ such that $ \sum c_n$ converges,we can find two sequence $ a_n,b_n$ such that $ \sum a_n,\sum b_n$ diverge.And $ c_n \equal{} \min \{a_n,b_n \}$

The theorems that pointed out by zhaobin will make the problem more clear.
But I don't know how the theorem(1) be proved .
A new one or ask for help from the conclusion in the  front.

Sorry,The theorem(1) may be wrong. :blush:

[quote="zhaobin"]Sorry,The theorem(1) may be wrong.[/quote]
Take the first part. We're looking at $ \sum \min\left(a_n,\frac1{n\ln n}\right).$

Use the condensation test. We compare to $ \sum \min\left(2^ka_{2^k},\frac1k\right)$

We'd now like to quote the original theorem of this topic. But to do that, we'd have to have that $ 2^ka_{2^k}$ is nonincreasing - but we don't know that. This is a hole in the proposed proof. I don't know yet whether it is a fatal hole - that would require the construction of a counterexample, which I haven't done. But it's a hint about how one might approach a possible counterexample.

Yes,you are right,I haven't noticed the fatal hole :blush: .

And in fact,grobber have pointed  out the conclusion above,but it is hard to construct a counterexample. :(
[quote="grobber"]
Given a decreasing sequence $ (b_n)_n$ of positive reals with the property that for every decreasing sequence of positive reals $ (a_n)_n$ with $ \sum a_n \equal{} \infty$ we have $ \sum\min(a_n,b_n) \equal{} \infty$, prove that $ \inf_n nb_n > 0$.

It appeared in some AMM from a couple of years ago, I think.[/quote]

mlok has done a very close thing in one of his posts. If $ \liminf nb_n\equal{}0$, then, for every $ \varepsilon>0$, we can find arbitrarily large $ N$ with $ b_n<\varepsilon/N$ for $ n\ge N$. Now we want to construct $ a_n$. Do it by induction choosing $ \varepsilon_k\equal{}2^{\minus{}2k}$ and, taking $ n_{k\plus{}1}>2^k n_k$ satisfying the above property for $ \varepsilon_k$ and putting $ a_n\equal{}2^{\minus{}k}/n_k$ for $ 2^{k\minus{}1} n_{k\minus{}1}<n\le 2^{k}n_{k}$. This sequence $ a_k$ is obviously non-increasing and the series $ \sum_k a_k\equal{}\plus{}\infty$ because $ \sum_{n\equal{}2^{k\minus{}1}n_{k\minus{}1}}^{2^k n_k}a_n\ge \frac 12$. On the other hand, $ \min(a_n,b_n)\le 2^{\minus{}k}/n_k$ for $ 2^{k\minus{}1}n_{k\minus{}1}<n\le n_k$ and $ \le 2^{\minus{}2k}/n_k$ for $ n_k<n<2^k n_k$, so $ \sum_{n\equal{}2^{k\minus{}1}n_{k\minus{}1}}^{2^k n_k}\min(a_n,b_n)\le 2^{\minus{}k}$, whence $ \sum_n\min(a_n,b_n)<\plus{}\infty$.


[b](1) Find all Integers which are measures, in degrees, of the interior angles of the regular polygons.[/b]

well, the interior angles of a regular polygon of $ n$ sides measure $ \frac {180(n \minus{} 2)}{n} \equal{} 180 \minus{} \frac {360}{n}$, so the $ n$ that work are the divisors of 360 greater than or equal to 3.


$ x,y$ are real numbers , such that : \[ 1\leq x^2\minus{}xy\plus{}y^2\leq 2\]
Prove that : \[ \frac{2}{9} \leq x^4\plus{}y^4 \leq 8\]

solution:
le prove the first:x°4+y°4<= 8,we have x² +y²<=2+xy <=>x°4 +y°4 -8 <= -(x²y²-4xy+4)=-(xy-4)²<= 0.
the second is easy. :wink: