function
algebra
domain
conics
parabola
analytic geometry

is there anyone who have finished all the problems?
I think there is something wrong with the second last question?

449. Let S = {x : x > −1}. Determine all functions from S to S which both
(a) satisfies the equation f(x + f(y) + xf(y)) = y + f(x) + yf(x) for all x, y 2 S, and
(b) f(x)/x is strictly increasing or strictly decreasing on each of the two intervals {x : −1 < x < 0} and
{x : x > 0}.

I still haven't finished them, didn't get much time. :blush:
What did you find wrong with the problem?

I finished all the others but this one.
Actually, I came up an answer which contradicted to the statement"x<-1".
 :blush:

No, there is a solution. :wink:
it's a nice question.

I solved everything except 448 and 445, but I am not spending much time on these problems. We still have till August 15, that's like 3 weeks.

I don't think we are allowed to tell you the answer, because this is like an individual competition(sort of); however I can explain the problem to you if you want.

I just have a question: i don't get number 448:

448. A criminal having excaped from prison travelled for $10$ hours before his escape was detected. He was then pursued and gained upon at $3$ miles per hour. When his pursuers had been $8$ hours on the way, they met a train going in the opposite direction at the same rate as themselves, which had met the criminal $2$ hours and $24$ minutes earlier. In what time from the beginning of the pursuit will the criminal be overtaken?

I have several questions about this, I would be really happy if someone answered them here:
1. What does the problem mean by "gained upon"?
2. Is the train actually going in the opposite direction to the pursuers?(because that would be a little irrational: why would the criminal want to travel towards the prison...)

[quote="rem"]
I have several questions about this, I would be really happy if someone answered them here:
1. What does the problem mean by "gained upon"?
2. Is the train actually going in the opposite direction to the pursuers?(because that would be a little irrational: why would the criminal want to travel towards the prison...)[/quote]
[hide="1"]By gained upon, they mean catching up with the criminal. So the distance between the criminal and the pursuers was decreased at the rate of 3 miles/hr.[/hide]
[hide="2"]I don't think the criminal hoped on the train. This just means that the criminal met the train 2hrs and 24min earlier. The question doesn't say that the criminal or the pursuers got on the train. Its a ittle hard to explain, but think of it like two marbles( :D ). One's the train, and the other is the criminal. When it says they met, that means, in marble terms, that both marbles passed each other.[/hide]
P.S. I may be wrong on these, but thats how I saw them. :P

[quote="rem"]Ok I get it. Thanks

And we must probably assume the criminal's speed is const. :)[/quote]
yeah, its const. :)

[quote="andy_tok"]I didn't mean to get an answer just to make sure there's an answer.
Anyway, I'll try again. Thanks! :)

question 448 have something seem not clearly stated but straightforward though.[/quote]
Whenever it seems not clearly stated, that means it could be tricky. :P

again  "functions from S to S" means the domain and range  are both S?

Yes. It means both $x$ and $f(x)$ belong to $S$.

A pretty technical point, but I like rem's statement of it better.  "Range" refers to the image of the function, which may not cover all of $S$.

but does that mean not all elements of S are covered?

All elements in $S$ are covered, ie for all $x$ in $S$ there will be a value for $f(x)$.

But not every element in $S$ is necessarily a value of $f(x)$ for some $x$.

Speaking of Olymon, it is now August, but they only have on the cms website, the problems for June...Is it supposed to be behind?

It was behind for all three years that I did it, and since Prof Barbeau has been running Olymon for ages, he probably hasn't changed the way he runs it.  Also, there's sometimes less Olymon sets in the summer - maybe one or two sets for the whole summer.

Yeah it is supposed to be behind.
Btw thanks for reminding me that it's already August, and I still have to send in the solutions.  :!:

I know this is late, but could someone explain 445,
[hide="445"]
Two parabolas have parallel axes and intersect in two points. Prove that their common chord bisects the segments whose endpoints are the points of contact of their common tangent.[/hide]
First of all, whats the common chord? Also, whose common tangent is this of, the parabola or the common tangent?
I think i messed up on this one very bad. :(

[quote="rem"]The common chord is the line segemnt joining the two points of itersection of the two parabolas; the common tangent line is a line that is tangent to both parabolas, teh common tangent is the line segment that has its endpoints the piints of tangency of the common tangent line to the two parabolas.
You want to prove that the line containing the common chord bisects the common tangent. :)[/quote]
Thanks for the fast reply Alex. :)
Now i have to do my solution again. It wasn't like i wrote a lot anyway. :D

[quote="rem"]The common tangent line is a line that is tangent to both parabolas, teh common tangent is the line segment that has its endpoints the points of tangency of the common tangent line to the two parabolas.
You want to prove that the line containing the common chord bisects the common tangent. :)[/quote]
This a little confusing, Could you explain what:
[quote="Olymon"]...the segments whose endpoints are the points of contact of their common tangent.[/quote]
means? Thanks

OK. let's say u have to parabolas, $p_{1}, p_{2}$. They intersect at points $A, B$. Let point $C$ be on $p_{1}$ and point $D$ on $p_{2}$ such that $CD$ is tangent to both $p_{1}$ and $p_{2}$.
You need to prove that $AB$ bisects $CD$.

That was so obvious, i must be still half-asleep. :blush: :sleeping:
Thanks a lot.

[quote="jpark"]Although I didn't do much of the olymon, I can tell you that many students send in their Olymon solutions later than the deadline. You should wait a bit longer before discussing the problems.[/quote]
thats true, we should wait a bit. This year was harder than usual though...probably because Prof. E.J. Barbeau wrote it and selected the questions.

[quote="rem"]I don't think so. Soem previous issues were harder in my opinion...[/quote]
Well, this is my second, and out all the others i looked at...this was tough. Which one did you find harder?

The issue for January.
[url]http://www.math.ca/Competitions/MOCP/2006/prob_jan.pdf[/url]

That is tough. Incidentally, it's also by Mr. Barbeau. :)

yes, Mr. Barbeau always makes harder problems, about one level harder than the problems by other people.

[quote="andy_tok"][quote="lomos_lupin"]Ok its now after the aug 15 so we can discuss the problems :

Who got the parabolas ,445?[/quote]
i just assigned two functions to them and proved that the mid point of the common tangent segment is on the common chord.[/quote]

Elaborate your solution.

this off topic, but this is the first time i have seen the Canadian forum being inactive for more than a few hours. :o  :D

[quote="Kileel"]If we can discuss the solutions now...

For 445, coordinates!  Wolog, the parabolas are y=x^2 and y=ax^2 + bx + c.  Consider one common tangent.  Let it meet the parabolas at x=u and x=v respectively.  Get equations from this (*).  I show that the x-coordinate of the common tangent's midpoint equals the x-coordinate of the intersection between the extended common chord and the common tangent using (*).[/quote]

What is the common tangent equation ?

y=2ux - u^2 and y=(2av + b)x + c-av^2.  Since this is a common tangent, these are equivalent.  Hence 2u=2av+b and -u^2=c-av^2.

Mine is quite similar to Kileel's.
But I set the parabolas to be y=ax^2-amx and y=bx(x-n)
so one of the intersection is (0,0)
and the common chord is abx(m-n)-(a-b)y=0
……

When do you think the solution for the previous month and this month will arrive? I really want to see how they solve the questions. :)


algebra proposed
algebra
number theory
combinatorics
Additive combinatorics

Determine whether there exists an arithimethical progression consisting of 40 terms and each of whose terms can be written in the form $ 2^m \plus{} 3^n$ or not. where $ m,n$ are nonnegative integers.

Restate the problem as follows: Prove that in an arithimethical progression consisting of $ 40$ different positive integers, at least one among them can't be written in the form $ 2^k \plus{} 3^l.$ Where $ k,l$ are nonnegative integers.

This is official proof,I wish to see more proofs here :)

I think it is more like a number theory one than algebraic one :maybe:

That is a kick *** solution. Nice work hxy09.

Strange solution, it seems like 40 can be reduced, even for this proof.

Congratulations hxy09, by the way. When will the Chinese team be announced? - it looks like you are in -

[quote]I heard that no one got this problem during the test. Is that true?[/quote]

Actually, quite to the contrary, a lot of people got this problem. It just takes a bit of time to realize that a combination of modular arithmetic and algebraic estimates work well together. Btw, the longest such arithmetic progression has 6 terms.

Seriously? Wow! Sorry, I just heard from some person that there was a question in the China TST 2009/2010 which no one got and was like algebra/number theory, so I just guessed it was this one.

From what I heard, this inequality had a very low score:
http://www.artofproblemsolving.com/community/c6h265482p1440850

But I think the following is even more difficult (being taken directly from an Erdos paper):
http://www.artofproblemsolving.com/community/c6h366317p2015072

In my year (2008), nobody solved this (Ljunggren result):
http://www.artofproblemsolving.com/community/c6h198296p1090306

Actually that whole test had only 11 problems solved in total by about 40 contestants.

[quote=hxy09]Thank you ahmedMido  :)
One of my friends from Shanghai told me it can be killed by $ mod 23$Can anybody continue with it?[/quote]

Really wonderful and fantastic solution! :winner_first: May I ask how you and your friends come up with such a cool solution? :)

A fact easy to be confirmed is that $ 2^m + 3^n$ will never be the multiple of $23$ ,which doesn't seem to do an effective work without algebra inequalities help to estimate.

"Actually, quite to the contrary, a lot of people got this problem. It just takes a bit of time to realize that a combination of modular arithmetic and algebraic estimates work well together. Btw, the longest such arithmetic progression has 6 terms."

Here is a solution without any number theory or modular arithmetic whatsoever.  Assume that such a progression existed.  Look at the last $20$ terms, so they are all within a factor of $\frac{39}{20}=1.95$.  Then for each term $2^m+3^n$, let $f(2^m+3^n)$ be whichever of $2^m, 3^n$ is bigger.  Clearly Since $2f(x)>x>f(x)$, this means all values of $f$ on these last $20$ are within a factor of $3.9$.  Since $f$ only takes on powers of $2$ and $3$, there exists $A$ and $B$ so that the range of $f$ on these last $20$ terms is contained in $\{2^A, 2^{A+1}, 3^B, 3^{B+1}\}$.  Now, look at the subset of these last $20$ where $f(x)=2^M$ for some fixed $M$.  If there are at least $5$ of them, then $2^M+3^{x_1}<2^M+3^{x_2}<2^M+3^{x_3}<2^M+3^{x_4}<2^M+3^{x_5}$ are in an arithmetic progression of length $20$, so $3^{x_5}-3^{x_1} \le 19(3^{x_2}-3^{x_1})$.  This is impossible, as $\frac{3^{x_5}-3^{x_1}}{3^{x_2}-3^{x_1}} \ge \frac{27(3^{x_2})-3^{x_1}}{3^{x_2}-3^{x_1}} \ge 27$.  Hence, $f$ hits each of $2^A, 2^{A+1}$ at most $4$ times among these last $20$.  But similarly, if $f(x)=3^N$ for $7$ values $2^{x_i}+3^N$ for $1 \le i \le 7$, we get that $19 \ge \frac{2^{x_7}-2^{x_1}}{2^{x_2}-2^{x_1}} \ge 32$, a contradiction.  So of these among last $20$, $f$ hits $2^A, 2^{A+1}$ $4$ times each and $3^B, 3^{B+1}$ $6$ times each.

Among the last $20$ there are some $3^B+2^{a_i}$ for $1 \le i \le 6$, all at most $2(3^B)$, and then some $3^{B+1}+2^{b_i}$ for some $1 \le i \le 6$.  So the latter $6$ must be among the last $14$, and then $2^{b_6}-2^{b_1} \le 13(2^{b_2}-2^{b_1})$, a contradiction.

There is an even more effortless solution to this problem.  Look at the final $30$ terms of the AP, which are off by a factor of at most $\frac{39}{10}$.  We let $f(2^m+3^n)=\max(2^m,3^n)$ again.  Then on these terms, $f$ is off by a factor of at most $\frac{39}{5}<8$, so there exist $A,B$ so that the range of $f$ is contained in $2^A, 2^{A+1}, 2^{A+2}, 3^B, 3^{B+1}$.  But there are at most $6$ points with $f=2^M$ and at most $5$ with $f=3^N$ by a similar argument as before.  As $30>28$ we are done.

[quote=HuangZhen]A fact easy to be confirmed is that $2^m + 3^n$ will never be the multiple of $23$ ,which doesn't seem to do an effective work without algebra inequalities help to estimate.[/quote]
Let me write down the complete mod $23$ solution here: first note that $p=23$ is the smallest prime mod which $2,3$ are quadratic residues while $-1$ is not ($\Leftrightarrow p\equiv-1$ mod $24$). The property guarantees $2^m + 3^n$ is never a multiple of $p$, therefore if at least $p$ such integers form an arithmetic progression they must be congruent to each other mod $p$. On the other hand for any prime $p\equiv-1$ mod $4$ and $k\in(\mathbb{Z}/p)^\times$, from the identity
\[\sum_{x=0}^{p-1}\left(\frac{k-x^2}{p}\right)=1\]
one can deduce the number of ways to write $k$ as an ordered sum of two nonzero quadratic residues are $\frac{p-1}{4}-\frac{1}{2}\left(\frac{k}{p}\right)$. Thus when $2^m + 3^n$ varies in any fixed residue class mod $23$, there are $\leq 6$ possibilities of $(2^m,3^n)$ mod $23\longleftrightarrow(m,n)$ mod $12$. Finally if $3$ different pairs of $(m,n)$ are congruent to each other mod $12$, then the ratio between the largest and smallest values of $2^m+3^n$ is greater than $2^{11}$ so these values surely don't fit into an arithmetic progression of positive integers with $\leq 40$ terms. Therefore an arithmetic progression with terms of the form $2^m+3^n$ and congruent to each other mod $23$ couldn't have $>12$ terms.

There does not even exist such a progression with $13$ terms.

Suppose for the sake of contradiction that such a progression exists. Both $2$ and $3$ are quadratic residues modulo $23,$ and $-1$ is not. So $2^m + 3^n$ is never a multiple of $23,$ and all terms of the progression are congruent to a nonzero residue $\pmod{23}.$

Consider the set $S$ of ordered pairs $(j,k)$ of not-necessarily-distinct nonzero quadratic residues $\pmod{23}.$ It is known that there are $5$ nonzero quadratic residues $k$ such that $k+1$ is also a quadratic residue, and $6$ nonzero quadratic residues where $k+1$ is not a quadratic residue. For any arbitrary nonzero quadratic residue $j,$ we may multiply by $j$ each of the $5$ nonzero quadratic residues $k$ where $k+1$ is also a quadratic residue, yielding $5$ nonzero quadratic residues $j$ less than a quadratic residue. We may also multiply by $j$ each of the $6$ nonzero quadratic residues $k$ where $k+1$ is not a quadratic residue, yielding $6$ nonzero quadratic residues $j$ less than a quadratic nonresidue.

So there are $5 \times 11= 55$ pairs in $S$ summing to a nonzero quadratic residue (two nonzero quadratic residues cannot sum to $0$) and $6 \times 11 = 66$ summing to a quadratic nonresidue.

Given two nonzero quadratic residues $p$ and $q,$ we claim the number of pairs in $S$ summing to $p$ is the same as the number of pairs in $S$ summing to $q.$ This is because each pair of the former type can be multiplied by $\frac{q}{p}$ to get a pair of the latter type, and each pair of the latter type can be multiplied by $\frac{p}{q}$ to get a pair of the former type. So there are exactly $5$ pairs in $S$ summing to every nonzero quadratic residue. Similarly, there are exactly $6$ pairs in $S$ summing to every quadratic nonresidue.

In our sequence of terms of the form $2^m + 3^n,$ there are at most $6$ possibilities for the remainders of $(m,n) \pmod{12}.$ By the Pigeonhole Principle, the remainders of $(m,n) \pmod{12}$ are constant across at least three terms. Either the largest of these terms is at least $2^{12}$ times the smallest, or we have three terms of the form $2^{m} + 3^{n}, 2^{m+12} + 3^{n}, 2^{m+24} + 3^{n},$ or $2^{m} + 3^{n}, 2^{m} + 3^{n+12}, 2^{m} + 3^{n+24}.$ Both possibilities are impossible in a progression of $13$ terms. $\square$


What do you guys think of the difficulty/point distribution at the International Mathematical Olympiad (IMO) 2006?

[b]Question number | Mean score | Standard deviation[/b]

[i]1   |     5.614/7  | 2.497
2   |     1.833/7  | 2.500
3   |     0.659/7  | 1.685
4   |     4.998/7  | 2.480
5   |     1.183/7  | 2.152
6   |     0.187/7  | 0.988[/i]

[b]The following table shows the quotient Mean score(x) / Mean score(y):[/b]

[i]1->2: 3.1
2->3: 2.8
4->5: 4.2
5->6: 6.3[/i]

Selecting the IMO question you first have to select those questions that represent your wanted difficulty level. The wanted difficulty level is kind of optimization problem. On the hand you don't really want anybody to score zero but on the other you don't want to make the exam too easy to loose distinction among the best contestants. I think IMO is an amazing experience to our young mathematicians. Thus to avoid frustration and keeping the motivation for solving challenging problems everybody should have at least some partial success or even solve a problem completely. I think those issues worked out very well for that IMO in 2006. There were sufficiently many easy questions, in particular #1 and to quite some extent question 4. Question six was used to distinguish among gold medal winners and perfect scorers. And #3 was required to be within the higher-scoring gold medal winners. But that selection procedure gives rise to a hard time for those participants who easily can solve questions #1 and #4 but yet have trouble solving #3 and especially #6. Solving three problems, that is cracking #2 or #5 on top, yields a silver medal easily. But those who want to secure those gold medals which are hardly beyond the gold cut-off have to make sure not to risk any simple mistakes in their solutions. One of those candidates was Alexandr Zamorzaev-Orleanschii from Moldova which was not lucky enough to get full credit on problems #4 and #5. I mean even just getting 6 points on any of those questions #1, #2, #4 or #5 requires you to get at least one point for #3 or #6 which turned out to be quite hard, especially for the latter problem.
Looking at the score distribution we see that the difficulty jump is notably quite high from problem #4 to #5 and from problem #5 to #6.

In conclusion I liked that a certain proportion got medals as usual and even a real huge number got a certificate which is quite crucial to feed further motivation of participants. But unfortunately students who are aiming at the lower gold ranks have a hard time making making no easy errors in their solutions or omitting important information which may be hard to compensate for at questions #3 or #6.

Which score distribution do you suggest, prefer etc.?

[b]References:[/b]

[i]- http://www.srcf.ucam.org/~jsm28/imo-scores/2006/
- http://imo2006.dmfa.si/results_itd.html[/i]

My opinion: I agree with you in some points. I think that it's not profitable to all the contestant to get 0 or 42, so there must be a high (but not extreme) level. But to motivate the students to continue solving problems, like you said, there would be nice to have a problem like p1 this year, so that the average of this problem is $>5$. This mean that a lot of contestants have, at least, an honourable mention. Then, p4 has to be more difficult than p1, but not THAT hard, so "good", but not excelent solvers, can go home with a nice 14.
Then, p2 and p5 are separate people in the range 14-28 points, and to establish the cut-off. In a IMO like this one, in 2006, solving 4 problem (p1, p2, p4, p5), or scoring 28, need to significate a gold medal, because both p2 and p5 were difficult (p5 may have to be stronger than p2).
Then, p3 and p6 (p6 more difficult than p3): They have to be to separate the IMOers: Getting p3, for example, means getting 35 (7-7-7-7-7-0), and this mean being very good, and far better than a 28 boy. Then, p6 is to distinguish the best ones: the perfect scorers (that 42 boys who are sooo great). I mean, solving p6 for me in an IMO shas to significate being one of the best high-school students in math around the world, so an average $n$ of $0,1<n<0,2$ would be nice.
Well, post your ideas in this nice topic!!!!!!!!!!


Post your first name and what you were named after/what it means.

I thought this might be fun since AoPS/ML is a diverse international community  :)

I'll start
Tim--"honoring god" in greek (grass in english), named after Christmas Carol character/biblical figure

American name
Taylor(im a guy goddammit)-no idea what it means. my mom thought it was a cool name
korean name
junggi(my sn yea.)-it means something crappy-"basis of the government" or whatever.

Mines Kylie and I have this keyring that says it means "boomerang"...  :?: kinda weird....

Jialiu
It's too hard to explain what it means.

Nathan: gift (Hebrew)
Savir: [I dunno but it's an Israeli name]

Peter:ROCK!!!!

I was named after my great-great grandfather, or great-uncle, or something like that

Alina (Russian) it means something along the lines of being pretty ( :P  :lol: ) I was named that because my mom liked the name and she had a friend or some distant relative or something named that.

I think my mom's name is kinda cool. Her name is Vita wich means life in Italian.

Allen, it's an common asian name.  I don't really have an official Chinese name because my parents never really bothered to give me one, I was born in the US.

[quote="akalra1"]Arvind... means somethin in some Indian language... or something[/quote]

....

 :idea:
[size=0]Intelliegence abounds.[/size]

k ur ze mod tat controls spam wat was tat

YEA I READ IT I POSTED AFTER I READ IT IN FACT BUT I STILL DONT GET IT

akalra you are the smartest person I know and Klebian agrees with me.

[quote="bubka"][quote="Klebian"][quote="akalra1"]Arvind... means somethin in some Indian language... or something[/quote]

....

 :idea:
[size=200]Intelliegence abounds.[/size][/quote]

really it does.

Arvind means:
1) Lotus
2) Greatest
3) Copper[/quote]

copper pwns

Frédéric, awful French spelling, I know, and I can only type in on an Azerty keyboard.

 :mad: It must be pronounced in a Dutch way though, like "Frederik".

"frede" is related to the Dutch word "vrede" and the German word "Friede", both meaning "peace"

"rik" means "powerful"

 :D so "Frédéric"= "peaceful ruler" or "mighty protector" :yoda:

[quote="akalra1"]as in "angel" as in "mafia 20"[/quote]
Actually I was the healer. :maybe:

I may have protected some people (and I actually saved Tim)  but I wasn't peaceful at all in that game (why spare the mafia?) and I wasn't a ruler, I left that to Tim...

[quote="funcia"]my name is rare and that of an opera(sort of)[/quote]

carmen?

that's a weirdass name, it reminds me of that crazy carmen sandiago lady, she used to piss me off so much

no she is really weird especially that puzzle of the world i had to do...i won't say to keep myself rather safe(i told people i was schizophrenic) but it has something to do with my favorite operas which are classical so you'll just have to guess.

[quote="funcia"]schizophrenic[/quote]
OMG that's the second time you use that word. What does it mean? :|

[url]http://en.wikipedia.org/wiki/Schizophrenic[/url]

Actually I mean more of the encarta definition...

Hey, "schizophrenic" apparently describe Opal Koboi in Artemis Fowl. I don't think that's the definition though, nor do I think that helped.

Er..yeah. I have an identity crisis. I forgot my name. The last thing I was called is "Pretty Pink Princess Pony", and I know for a fact that's not my name.

DID YOU GUYS KNOW I LOVE OPAL!!!!? I actually also right almost all my notes in gnommish OMG there's to be another AF!!!! err sorry that i had to tell you all that but i've already ordered it...is it yet out? but actually i think i liked artemis when he was evil and vile but now jon spiro is my favorite

I also suffer from mulitiple personalities so yeah last time i was called a name it was pansy and that is NOT my name

Alex means the "defending man" or something similar.
I don't know why i was named like this  :D

i am daesun

i believe it means something to do with being big in korean, except it's based of hanja, which is what korean people call chinese sometimes

in english
Nicholas = Victory

in chinese
jia= california
lei= thunder
jia + lei = jialei= my chinese name

athanasios=immortal (but not imortal :( )

[quote="azndemon43"][quote="Treething"]isn't jia home or something?[/quote]
yeah, but you know how the chinese language is so confusing
multiple characters for each ping ying word

the intended meaning was the jia in jia zhou, or california, for non-chinese speaking people[/quote]

Oh I thought it was "jia" as in "zhi jia ge" as in CHICAGO.

SRSLY. Who names their kid California??  :rotfl:

[quote="fredbel6"]What's next, a person named Orlando  :D ?[/quote]

rimshot.

[quote="fredbel6"]What's next, a person named Orlando  :D ?[/quote]
I was in a hotel in Orlando and the person at the desk was apparently named Rolando :roll:


firstly thanks for help,

i would like to know if have any nitrats(NO3/NO2) that are soluble.

tks!

i think all of nitrate are unsoluble, and i would like to know if monobasic bismuth nitrate is soluble or how can i know if this is soluble.
thanks!

yes, but i dont have the numbers, please calculated to me!

Bismuth Nitrate is hygroscopic and forms a pentahydrate.  The CRC lists it as "reacts with H2O".

In general, the vast, vast majority of nitrates are soluble.


limit
logarithms
real analysis
real analysis unsolved

try to find sequence $ a_{n}$ such as :
$ a_{n}^{2} \equal{} 1 \plus{} 2na_{n\plus{}1}$

You wanted to say expression for $ a_n$?

yes, but not as the sum or product or... .
Find any  closed form .  I spent whole 2 hours but can't find it.

Oh, I   forgot condition on $ a_{n}$
$ \lim 2^{ \minus{} n} \ln \frac{ 1}{a_{n}^{2}} \to 0$ as $ n \to \infty$


how do u raise a power to a power

i.e. (2^3)^4 except w/o parenthesis?? :ninja:

Curly brackets can be used to enclose anything, including an exponent: 2^{3^4} will give $2^{3^4}$. (And by the way, this notation means 2^(3^4), not (2^3)^4.)


Hi! Here is a nice maths problem.

Three guys pay for a meal at a restaurant 45$. So each pays 15$. They leave and the waiter realisez that the meal is only 40$ so he gives 5$ to his son to give them back to the three guys. On the way to them this fellow realisez that these 3 guys cannot split 5$ so he buys a 2$ ice-cream and gives the 3 men 3$ back. So each pays 14$.
Now a small calculus: 14*3+2=44. So.. where is the other dollar?

where can it be?!
:D :D :D

this one is very well known... do I answer, or do I wait so that the other can think on this "problem"  :D  ?

Hence
3*14 + 3 = 45
and
3*14 - 2 + 5 = 45.

There is no dollar missing.

There is no dollar missing,like Daniel pointed out.Unless our Lagrangia lost his finger in the men's pockets [img]http://community.the-underdogs.org/smiley/happy/jump3.gif[/img]


induction
Pythagorean Theorem
geometry
real analysis
real analysis unsolved

Let $ (a_n)_{n \ge 0}$ be a sequence such that: $ 13^{a_n}\equal{}12^{a_{n\minus{}1}}\plus{}5^{a_{n\minus{}2}}, \forall n \in \mathbb{N}, n \ge 2, a_0\equal{}0, a_1\equal{}1.$

Prove that the sequence $ (a_n)_{n \ge 0}$ is convergent, and find its limit.

$ 13^{a_n} \equal{} 12^{a_{n \minus{} 1}} \plus{} 5^{a_{n \minus{} 2}}, \forall n \in \mathbb{N}, n \ge 2, a_0 \equal{} 0, a_1 \equal{} 1$.

it looks like Pythagorean theorem:D

divide both sides by $ 13^{x}$ where $ x$ is the limit of $ a_{n}$
so $ a_{n}\rightarrow 2$ when $ n\rightarrow\infty$

We show with induction that $ 2\geq a_{n} \geq a_{n\minus{}1}$.

[b]Base of induction[/b]: $ 2\geq a_{3}\geq a_{2} \geq a_{1}$.

[b]Inductive step1:[/b] $ a_{n}\geq a_{n\minus{}1} \geq a_{n\minus{}2} \Rightarrow a_{n\plus{}1}\geq a_{n}$.

$ 13^{a_{n\plus{}1}}\geq  13^{a_{n}}
\Leftrightarrow 12^{a_{n}}\plus{}5^{a_{n\minus{}1}} \geq 12^{a_{n\minus{}1}}\plus{}5^{a_{n\minus{}2}}  \surd$

[b]Inductive step2:[/b] $ 2 \geq a_{n}\geq a_{n\minus{}1} \Rightarrow 2 \geq a_{n\plus{}1}$.

$ 12^{a_{n\minus{}1}}\plus{}5^{a_{n\minus{}2}}\leq  12^{2}\plus{}5^{2}\equal{}13^{2} \Rightarrow a_{n\plus{}1}\leq 2  \surd$.

So $ \{a_{n}\}$ converges.

Now we have to see just that the only solution of $ 12^{L}\plus{}5^{L}\equal{}13^{L}$ is $ L\equal{}2$. One possible argument would be that $ f(x)\equal{}13^{x}$ and $ g(x)\equal{}12^{x}\plus{}5^{x}$ are both strictly convex on $ \mathbb {R}$ and $ f(x)$ increases quicker than $ g(x)$ on $ \mathbb{R^{\plus{}}}$, therefore they can have just one intersection on $ \mathbb{R^{\plus{}}}$.

Thank you both [color=darkblue][b]1234567a[/b][/color] and [color=darkblue][b]tmrfea[/b] [/color]for your posts. The first post shows that there is a relation between this exercise and Pythagorean theorem, and the second one shows a very nice proof for the fact that the sequence is convergent. I spent more than one hour trying to prove that the sequence is convergent, but I didn't succeed. Thank you again.


Is it possible to find the sum of the infinite series:

0!/x + 1!/x^2 + 2!/x^3 + 3!/x^4 + 4!/x^5 + .....

Thanks.

A necessary condition for a sum to converge is that the terms get increasingly closer to $ 0$ for sufficiently large indexes.  In this case, if $ x$ is fixed, what happens to $ \frac {k!}{x^{k \minus{} 1}}$ as $ k$ gets large?

Well, for any x whose absolute value is greater than one, terms do eventually approach zero - so the sum should converge (presuming lxl > 0).

Regardless of how large $ x$ is, $ k! > x^{k \plus{} 1}$ for sufficiently large $ k$, so the terms do not approach $ 0$ as $ k\to \infty$.

Do you perhaps have the fractions in reverse? If you are looking for the sum $ \sum_{k \equal{} 0}^\infty \frac {x^{k \plus{} 1}}{k!}$, that does converge (to $ xe^x$), but it's a topic that belongs in the calculus section of this board.


Hi there,

I've got this Cryptarithm (the title was Squareroot):

GR'OG gives root RL;
-NR = OYOG;
-AAD = OUE

My problem is the notation of the first line.
The solution is 46'14 gives root 67
I just can't bring these numbers together. What does 46'14 mean?

I would be happy if someone could explain this to me.

Thank you,

dexters

$\sqrt{46.14} \approx 6.7$ could be a solution.


A soccer ball is constructed using 32 regular polygons with equal side lengths. Twelve of the polygons are pentagons, and the rest are hexagons. A seam is sewn wherever two edges meet. What is the number of seams in the soccer ball?

There are $ (12 \times 5) \plus{} (20 \times 6) \equal{} 180$ seams in total. We want to find how many meet. Thus, $ 180 \div 2 \equal{} \boxed{90}$.

i was having trouble with that one on ftw....
Thank you


[quote="aktyw19"]Which  logarithm has greatest value
$ log _{3}7$ or $ log _{7}27$[/quote]

$ log _{3}7$ > $ log _{7}27$

I do not understand
Please,  it explain it more


Hi,

Tonight I got two emails, both identified by my antivirus software as virus-infected, that appeared to come from AoPS forum participants. I am aware that usually infected emails spoof return addresses, so that you can't tell who they REALLY come from, but they often do so by reading other addresses on the infected computer. So since I got two emails with the same virus infection that both appeared to come from AoPSers, I figure that someone with a lot of AoPS addresses in his mailbox--perhaps yet another AoPSer--is infected. PLEASE check your computer for virus infection if you haven't done so recently.

If you really want to reach me, try a private message.

Unless you've been posting your e-mail on this forum, it is most likely one of the site administrators.

Thanks for letting me know, though.

By the way, how do you know they came from AoPS participants?

[quote="mathfanatic"]Unless you've been posting your e-mail on this forum, it is most likely one of the site administrators.[/quote]

That's what I was thinking: the only computer that would have all the real email addresses of participants here is the administrator's computer.

[quote="mathfanatic"]By the way, how do you know they came from AoPS participants?[/quote]

The two emails I received both had email return addresses that used AoPS user names with well-known ISP domain names. But I didn't think that either of the individuals spoofed know my real name, and thus my real email address, and that's why it seems most plausible that the infected computer is one carrying the member list for this forum.

Richard and I have been sent a three digit number of virused emails between us since the latest big outbreak.  We have virus detection software scanning incoming email and checking our computers regularly.  I believe we have been safe from all infection, but those emails with unicode characters that are infected have come from many of our students, associates, and people who have ordered the books.


salam, man ghablan in soalo gozashte budam va ye nafar ye rahehal eraye karde amma man nafahmidam, lotfan age kasi rahehalo motevajeh shod uno tozih bede, merc
[b]problem:[/b]N contestants answered an exam with n questions. ai points are awarded for a correct answer to question i and nil for an incorrect answer. After the questions had been marked it was noticed that by a suitable choice of positive numbers ai any desired ranking of the contestants could be achieved. What is the largest possible value of N?

[b]solution:[/b]The largest value of N is n. Obviously we have example with n contestants when everybody has solved only one (different) problem.
It remains to prove that .
Consider the matrix  where  is 0 if the i-th contestant hasn't solved the j-th problem and 1 if he has. We shall prove that the row-vectors are linearly independent. Suppose they are linearly dependent. Then there exhist natural numbers  such that:
 where  are the row vectors. But now if the sum of first l n-s is smaller than the sum of the rest then we cannot make the first l contestant to be the first in the ranking and if the sum of the first l of n-s is bigger we cannot make them to be the last in the ranking. In each case we cannot make any ranking we want.
So the row vectors are linearly independent but we cannot have more than n n-dimensioned linearly independent vectors so we are done.

[quote="exir"]hichkas in raho motevajeh nashode ya idei baraye in soal nadare? :([/quote]

ie lotfi mikoni soalo be farsi begi?

[quote]ie lotfi mikoni soalo be farsi begi?[/quote]
sherkatkonandegan dar azmuni teti bayad be [b]n[/b] soal javab dahand. pasokhe sahih be soale [b]i[/b]om [b]ai[/b] emtiyaz darad ke [b]ai[/b] adadi tabieest (i andise a hast) va be pasokhe ghalat emtiyazi ta'logh nemigirad (emtiyazash 0 ast) nomereye har kas majmue emtiyazatist ke kasb karde, pas az rotebandi motevajeh mishavim ba haman javabhayi ke sherkatkonandegan dadeand mitavan [b]a1,a2,...,an [/b]ra tori taghir dad ke har rotbebandi digari momken bashad. hadeaksare[b] n[/b] chand ast?

bebakhshid rahehali ke aval gozashtam ye ghesmathayish moghe copy paste ja oftade bud ke alan motevajeh shodam va in kamele hamun rah hast: (dar payin manzur az [b]ni[/b] andise i om az n hast)

[b]Solution:[/b]Let the number of contestants=N. The largest value of N is n. Obviously we have example with n contestants when everybody has solved only one (different) problem.
It remains to prove that [b]N=<n[/b]
Consider the matrix[b] A=(aij)[/b]where[b]aij[/b] is 0 if the i-th contestant hasn't solved the j-th problem and 1 if he has. We shall prove that the row-vectors are linearly independent. Suppose they are linearly dependent. Then there exhist natural numbers [b]n1,...,nN[/b] such that:
[b]n1v1 + n2v2 +...+ nlvl = nl+1vl+1...nNvN[/b] where [b]v1..vN[/b]  are the row vectors. But now if the sum of first l n-s is smaller than the sum of the rest then we cannot make the first l contestant to be the first in the ranking and if the sum of the first l of n-s is bigger we cannot make them to be the last in the ranking. In each case we cannot make any ranking we want.
So the row vectors are linearly independent but we cannot have more than n n-dimensioned linearly independent vectors so we are done.

hichkas nemitune in raho tozihe bishtari bede ya rahe digeyi pishnahad kone? :maybe:


geometry
perimeter
cyclic quadrilateral
Pythagorean Theorem
geometry unsolved

Cyclic quadrilateral $ABCD$ has positive integer side lengths $AB$, $BC$, $CA$, $AD$. It is known that $AD=2005$, $\angle{ABC}=\angle{ADC} = 90^o$, and $\max \{ AB,BC,CD \} < 2005$. Determine the maximum and minimum possible values for the perimeter of $ABCD$.

[quote="shobber"]Cyclic quadrilateral $ABCD$ has positive integer side lengths $AB$, $BC$, $CA$, $AD$. It is known that $AD=2005$, $\angle{ABC}=\angle{ADC} = 90^{\circ}$, and $\max \{ AB,BC,CD \} < 2005$. Determine the maximum and minimum possible values for the perimeter of $ABCD$.[/quote] [i]Proof.[/i] Let $\overline{AB}=a$, $\overline{BC}=b$, $\overline{CD}=c$ and $\overline{DA}=d=2005$. Since $\angle{ABC}=\angle{ADC} = 90^{\circ}$, we have by the Pythagorean theorem $a^2+b^2-c^2=2005^2$ $(*)$. Hence if we find a solution $(a$, $b$, $c)$ of $(*)$ satisfying $\max \{a$, $b$, $c\} < 2005$, then $a$, $b$, $c$, $d$ are the sidelengths of a cyclic quadrilateral.

Let us first determine the minimum value of the perimeter of $ABCD$, i.e. the minimum value of $a+b+c$, since $d=2005$ is constant. Let wlog $a \ge b$ and thus $a\ge 1000$. The triple $(2004$, $45$, $4)$ satisfies $(*)$ and gives $a+b+c=2053$. I'll now prove that $2053$ is indeed the minimum. We evidently have $(b+c)^2>(b+c)(b-c) = (2005-a)(2005+a)$ and thus $a+b+c>a+\sqrt{(2005-a)(2005+a)}=:f(a)$. $f$ is convex. Moreover, we have $f(1000)>2053$, $f(2003)>2053$ and $f(2004)<2053$. This, together with the convexity, implies $f(a)>2053$ on $[1000$, $2003]$. Hence the minimum is attained for $a=2004$. A quick computation shows that $b+c \ge 49$ and thus the minimum of $a+b+c+d$ is $4058$.

Next, I'll show that the maximum value of $a+b+c+d$ is $7772$ and thus that the maximum value of $a+b+c$ is $5767$.
[b]Lemma.[/b] If $pq=rs$ for some positive integers $p$, $q$, $r$, $s$ satisfying $r>p>q>s$, then $p+q<r+s$.
[i]Proof of the Lemma.[/i] Assume $p+q \ge r+s$ and thus $pq+q^2 \ge qr+qs \Leftrightarrow rs+q^2\ge qr+qs \Leftrightarrow (r-q)(s-q) \ge 0$, which is clearly a contradiction, since $r>q$ and $q>s$. Therefore, the Lemma is proved.
Back to the problem.  If $b+c>2005+a$, then we have by the lemma that $2b=(b+c)+(b-c) > (2005+a)+(2005-a)=4010$ and thus $b>2005$, clearly a contradiction. Hence $b+c < 2005 +a$ and $b-c>2005-a$. Let $b-c=2005-a+x$ and $b+c=2005+a-y$. Consequently, \[y= \frac{x(2005+a)}{2005-a+x}. \quad \quad (**)\] Since $2005+a \equiv 2005-a \mod 2$, $x$ is even. Assume that $a+b+c=5767$ is not the maximum. Then $2005+2a-y=a+b+c >5767 \Leftrightarrow 2a-3762>y$. Suppose that $x \ne 2$ and thus $x \ge 4$. From $(**)$ we get $y \ge 4(2005+a)/(2009-a)$. Therefore, $(a-1881)(2009-a) >2(2005+a) \Rightarrow 0>(a-63)^2$, which is a contradiction. Therefore, $x=2$. A similar estimation gives $(a-1881)(2007-a) > 2005+a$. Let $a=1881+a'$. Hence $(a-1881)(2007-a)>2005+a \Leftrightarrow a'(126-a') > 3886+a'$ and therefore $881>(2a'-125)^2 \Leftrightarrow 19 > |2a'-125|$. Thus, we get $53\le a' \le 72$ and consequently $1934 \le a \le 1953$. Hence $2007-a \in [54$, $73]$. Since $y$ is an integer, $2007-a | 2(2005+a) \Leftrightarrow 2007-a | 8024=2^3 \cdot 17 \cdot 59$. Thus, $2007-a=59$ or $2007-a=68$. Both values of $a$ lead to $a+b+c=5767$ and therefore the maximum.

Thus, the minimum value of the perimeter is $4058$, the maximum value is $7772$.

I think there is a slight mistake in the above for the minimum value, since I believe $(2004, 45, 4)$ doesn't work. The below solution should give the correct value of the minimum.

I think that the minimum value of the perimeter of $ABCD$ is $4160$, attained at $(1999, 155, 1).$ We will now show that this is minimal. Let $a, b, c$ be as $\mathbf{FelixD}$ defined above, also assuming that $a \ge b$. As $\mathbf{FelixD}$ did above, we can observe that $a+b+c > 4160$ whenever $a \le 1998$ (plug them into $f$), and so it hence only suffices to find the minimal values of $a+b+c$ when $a \in \{1999, 2000, 2001, 2002, 2003, 2004\}$. Since we have that $2005^2 - a^2 = b^2 - c^2$, from $b+c | 2005^2-a^2$ and $b+c > b-c$, we can easily finish with case-checking when noting that $2005^2-2004^2 = 19 \cdot 211, 2005^2 - 2003^2 = 48 \cdot 167, 2005^2-2002^2 = 3 \cdot 4007, 2005^2 - 2001^2 = 8 \cdot 2003, 2005^2 - 2000^2 = 89 \cdot 225,$ and $2005^2 - 1999^2 = 154 \cdot 156,$ where each factor pair is the one which minimizes the larger factor. Since all of them other than $156$ are greater than $160$, and $a \ge 2000$ except for the $156$, it's readily seen that $4160$ is the minimum here.


quadratics
number theory unsolved
number theory

p is a prime of the form (2^n)+1 , n>1 prove 3 is a primitive root

now, p is a prime of the form 4q+1, where q is an odd prime, prove 2 is a primitive root.

unknown source

For the first problem, it suffices to show that $3$ is not a quadratic residue since $\varphi(\varphi(p))=\frac{p-1}2$, thus there are the same numbers of quadratic nonresidues and of primitive roots [and primitive roots are always quadratic nonresidues]. But now quadratic reciprocity finishes it.

For the second,  $\varphi(\varphi(p))=2(q-1) = \frac{p-1}2-2$. Since $q$ is odd, we get $p \equiv 5 \mod 8$. Thus $2$ is a quadratic nonresidue $\mod p$. By the same reasoning as for the first problem, all we have to do is to find two quadratic nonresidues $\mod p$ that are $\neq 2$ and not primitive roots. Here $\pm i \mod p$, the solutions of $x^{2}\equiv-1 \mod p$, work:
Clearly they are not primitive roots since their $4$-th powers are $1$. They are quadratic nonresidues because if $x^{2}\equiv i \mod p$, then $\mathrm{ord}_{p}(x)=8$, implying $8|p-1$ by the order lemma, contradicting $p \equiv 5 \mod 8$.

[quote="Keno20"]

now, p is a prime of the form 4q+1, where q is an odd prime, prove 2 is a primitive root.
[/quote]

We can actually make this statement stronger!

If p an q are "Germain primes" that is p=2q+1 (and therefore of the form 4q+1) then, 2 is a primitive root for p.

[quote="sylow_theory"](and therefore of the form 4q+1)[/quote]
That $q$ would not necessarily be prime, and even more $2q+1 \equiv-1 \mod 4$ for all odd $q$, especially prime ones.

The statements is wrong if $2$ is a quadratic residue $\mod p$ [exactly happening if $q \equiv 3 \mod 4$], as we can see for $q=3, p=7$. But if $q \equiv 1 \mod 4$, the same proof as before works (this time we need one nonsquare not being a primitive root, so we take $-1$). In fact, if $p=2^{k}q+1$ with $p,q$ prime, we always have that the primitive roots are exactle that nonsquares not being a $2^{k}$-th root of $1$.

hmmm..., out of curiosity, is there anyway to get away with this without using  a complex number?

There is no complex number used. Do you fear the letter $i$¿ ;)

Sorry for the spam, but this is just gorgeous!  :rotfl:  :rotfl:

:blush:  sorry. how stupid. i was half asleep.  got it, thanks guys.


Is $ x^4\plus{}2x^3y\minus{}3x^2y^2\plus{}2xy^3\plus{}y^4$ factorable?

I wouldn't think so (at least not in a nice manner, though). Neither does Mathematica 6.

i assume you mean over the rationals, that means the result of the factoring can only have rational coefficients.
can the factored result have irrational coeffecients?
anyway, a way to factor this is to treat y as a constant and factor the quartic in x

solution:
so i dont have a rigorous solution but you realize this is almost a perfect square ($ x^4\plus{}2x^3y\plus{}3x^2y^2\plus{}2xy^3\plus{}y^4$) and the polynomial we are trying so factor is only $ 6x^2y^2$ less than this. therefore we can rewrite our polynomial as
$ (x^4\plus{}2x^3y\plus{}3x^2y^2\plus{}2xy^3\plus{}y^4)\minus{}6x^2y^2$

$ (x^2\plus{}xy\plus{}y^2)^2\minus{}(\sqrt6xy)^2$
using diff. of squares

$ (x^2\plus{}(\sqrt6\plus{}1)xy\plus{}y^2)(x^2\minus{}(\sqrt6\minus{}1)xy\plus{}y^2)$

and these are easily (maybe not the arithmetic) factored by the quadratic formula...
well it gets super messy so i wont do it here... this is definitely NOT classroom math


I don't know where to put this one, and let it be here
hm this is a inernational math forum, so it's not easy to find here people from such little countries as for example Georgia. I'm from Georgia and I want to find kids from my countrey who like math. There is no georgian math forum at all because too many kids are not interested in math here. I just want to have new friends who like math.

If anybody is here from Georgia, tell me please.

send me a PM or contact me on the sky pe
sky pe: anika_555


Circle O is the inscribed circle of triangle ABC. The points of tangency on BC, CA, AB are D, E, F respectively. line DO intersects EF at A', similarly we can find points B' and C'. Prove: AA', BB', CC' are concurrent.

thxx!

Using Ceva's theorem in angle form, it is suffice to show that $ \frac {sin\angle A'AB}{sin\angle A'AC}*\frac {sin\angle B'BA}{sin\angle B'BC}*\frac {sin\angle C'CB}{sin\angle C'CA}\equal{}1$

since $ sin\angle A'AB \equal{} \frac {A'F*sin\angle AFE}{A'A}$

do this for all six sin values and multiply them, we get $ \frac {A'F}{A'E}*\frac {B'E}{B'D}*\frac {C'D}{C'F} \equal{} 1$, which is true by Ceva's law

Yeah..more generally we have

$ ABC$ is a triangle, $ AD$, $ BE$, $ CF$ are concurrent cevians in $ \triangle{ABC}$.  $ DA'$, $ EB'$, $ FC'$ are concurrent cevians in $ \triangle{DEF}$.  Then $ AA', BB',$ and $ CC'$ concur.  Just Trig Ceva.

Let $BC\cap AA'=A"$, a.s.o.
By Ceva, from first two triangles we get: $\frac{AF}{BF}\cdot\frac{BD}{CD}\cdot\frac{CE}{AE}=1\ (\ 1\ )$ and $\frac{EA'}{FA'}\cdot\frac{FB'}{B'D}\cdot\frac{DC'}{C'E}=1\ (\ 2\ )$.
Next, draw $FX\parallel EY\parallel BC, \{X,Y\}\in AA"$.
Now: $\frac{AF}{AB}=\frac{FX}{BA"}, \frac{CA"}{EY}=\frac{AC}{AE}$; multiply the last 2 equalities to get
$\frac{AF}{AB}\cdot\frac{CA}{AE}=\frac{FX}{BA"}\cdot\frac{CA"}{EY}=\frac{FX}{EY}\cdot\frac{BA"}{CA"}=\frac{FA'}{EA'}\cdot\frac{BA"}{CA"}$ $\iff\frac{BA"}{CA"}=\frac{AF}{AB}\cdot\frac{CA}{AE}\cdot\frac{EA'}{FA'}\ (\ *\ )$.

Likewise we get other two equalities for $FD, ED$ respectively which, multiplied get the desired $\frac{BA"}{CA"}\cdot\frac{CB"}{AB"}\cdot\frac{AC"}{BC"}=1$.

Best regards,
sunken rock

Dear Mathlinkers,
with a brutal force we can apply the cevian nests theorem...
Sincerely
Jean-Louis

Dear Mathlinkers,
with more finess, AA' is the A-median of ABC (http://www.artofproblemsolving.com/Foru ... &t=479138).

Sincerely
Jean-Louis


I was working on a problem and came across a derivative where my answer just didn't seem to match the intended answer.

How would you people differentiate x^ln(x) ?

What I would do is [hide]let y = x^ ln x and then take logs and differentiate implicity. I get x^(ln x - 1) * 2 ln x.[/hide]

How were you trying to do it?

I plead the fifth on how I was trying to solve it to begin with.  The log then implicit differentiation worked.  You're answer is correct, but in a different form.

[hide]

y = x^ln x

ln(y) = (ln x) :^2:

Differentiate to get:

(1/y)dy/dx = 2(ln x)(1/x)

dy/dx = y(2/x)(ln x)

Since y = x^(ln x)

dy/dx = (2/x)(ln x)(x^ln x)

[/hide]

An other way :

x^ln(x) = e^(ln(x)^2) then use (e^u)' = u' e^u


calculus
integration
real analysis
real analysis unsolved

Could someone help with the following integration please?

$ \int{\frac{e^{\minus{}x^2}}{R^2\plus{}(W\minus{}x)^2}dx}\equal{}\frac{1}{2R}\left(\int{\frac{e^{\minus{}x^2}}{R\plus{}(W\minus{}x)\cdot i}dx}\plus{}\int{\frac{e^{\minus{}x^2}}{R\minus{}(W\minus{}x)\cdot i}dx}\right)\equal{}$
$ \equal{}\frac{1}{2R}\left(\int{\frac{e^{\minus{}x^2}}{R\plus{}W\cdot i\minus{}x\cdot i}dx}\plus{}\int{\frac{e^{\minus{}x^2}}{R\minus{}W\cdot i\plus{}x\cdot i}dx}\right)\equal{}$
$ \equal{}\minus{}\frac{i}{2R}\left(\int{\frac{e^{\minus{}x^2}}{\minus{}R\cdot i\plus{}W\minus{}x}dx}\plus{}\int{\frac{e^{\minus{}x^2}}{\minus{}R\cdot i \minus{} W\plus{}x}dx}\right)$

So, we should evaluate the integral of the type: $ \int{\frac{e^{\minus{}x^2}}{a\plus{}x}dx}$
This integral could not be evaluated.

Hi Yuriy,

Thanks for your help.  I was wondering how did you arrive at the first expression?  What properties did you use?
Thanks!

We know that $ a^2\plus{}b^2\equal{}(a\plus{}b\cdot i)\cdot (a\minus{}b\cdot i)$ and we will have:
$ \frac{1}{R^2\plus{}(W\minus{}x)^2}\equal{}\frac{1}{(R\plus{}(W\minus{}x)\cdot i)\cdot(R\minus{}(W\minus{}x)\cdot i)}\equal{}$
$ \equal{}\frac{1}{2R}\cdot \left(\frac{1}{R\plus{}(W\minus{}x)\cdot i}\plus{}\frac{1}{R\minus{}(W\minus{}x)\cdot i}\right)$
Then we use that $ \frac{1}{i}\equal{}\minus{}i$  :)


geometry
3D geometry
tetrahedron
geometry unsolved

Let $ S$ be the surface area of a tetrahedron, show that the sum of areas of the six angle-bisectors of this tetrahedron doesn't exceed $ \frac {\sqrt{6}}{2}S$

[quote="thesunflower"]Let $ S$ be the surface area of a tetrahedron, show that the sum of areas of the six angle-bisectors of this tetrahedron doesn't exceed $ \frac {\sqrt {6}}{2}S$[/quote]
To Mods: Please lock this topic, this is a problem from M&Y, issue 3,,2008, and we can't discuss it now


AMC
USA(J)MO
USAMO
geometry
geometric transformation
complex numbers
geometry open

So...is the following true? If so, how to prove it? By construction?

given two points A and B, constant $k$ and angle $\alpha$, there is exactly one point $O$ such that the spiral similarity centered at $O$ with constant $k$ and angle $\alpha$ carries $A$ to $B$.

I'm pretty sure the following is true, but kinda shaky on my arguments:

given points $A$, $B$, $A'$, $B'$, there is exactly one point $O$ such that the spiral similarity centered at $O$ carries $A$ to $A'$ and $B$ to $B'$.

Once the second one is proven I have a lightning solution to USAMO 2006 number 6 :D

Complex numbers do this job in an easy ways:
$w(A-O)+O = B$
$wA-wO+O = B$
$O(1-w) = B-wA$
$O = \frac{B-wA}{1-w}$

 :)

A spiral similarity with centre (0,0) is multiplication my a complex number $w$.


abstract algebra
geometry
trigonometry
search

I'm thinking of accelerating (i.e. to be exempted from/skip) this module at school next year and have to take a test to check whether I have what it takes to accelerate.

[quote]
This  advance (sic)  course  in  trigonometry  focuses  on  the
concepts  of  trigonometric  functions,  identities  and
equations.  Topics  include  addition  theorem,  double  angle
formula,  factor  formula and R-formula. The unit of circular
measure is also discussed here.
[/quote]

How should I study this topic (in-depth, olympiad style though)? [I am taking the AMCs next year] The test will be held in July next year. My strength is algebra and I'm getting better at geometry but I don't know much about trigonometry.

Any recommendations? Also, where can I get problems? Recommend any books/e-books if possible. Thanks!
[/quote]

titu's book is good. it starts from basic level to tougher concepts.{103 trignometry problems }
u can find more books at esnips.com and 4shared.com

I managed to get Titu's book here: http://www.esnips.com/doc/b1103866-afa8-4f5c-a038-13b0d90d8a57/Zumung-Feng---103-Trigonometry-Problems. Is this the correct version?

yes, this is it. if u need a pdf book rather than reading it online, then i can give u if u wish
and yes this is a good book but if u want more u can search on those websites.

Ok. What's the difference between a PDF book and the one I found?

EDIT: Do you know where to download Titu's 101 Problems in Algebra

well, one that u found, i think cant be downloaded so i was saying a ebook that u can read even while offline.if it can be downloaded then its fine.


limit
logarithms
calculus
integration
real analysis
real analysis unsolved

Hello!

Can you help me with the computation of the following limit?

$ \lim_{n\to\infty}n\cdot\left (1\minus{}\prod_{j\equal{}n\plus{}1}^{\infty}\frac{j^2}{j^2\plus{}1}\right ).$

I think, the value of the limit above equals [b]1[/b], but I'm not able to prove it.
Thanks!

hello, i think your product is given by $ \prod_{j\equal{}n\plus{}1}^\infty\frac{j^2}{j^2\plus{}1}\equal{}\frac{\Gamma(1\minus{}i\plus{}n)\Gamma(1\plus{}i\plus{}n)}{\Gamma(n\plus{}1)^2}$.
Sonnhard.

[code]In[1]:= Product[j^2/(j^2 + 1), {j, n + 1, Infinity}]

Out[1]= (Gamma[(1 - I) + n] Gamma[(1 + I) + n])/Gamma[1 + n]^2[/code]

It looks suspiciously like you just copied and pasted from Mathematica, so unless you're going to show us how you derived that expression, there is no reason for you to be posting this.

It doesn't even really simplify the computation of the limit.

Perhaps it is worth studying the following problem:

Let $ k\geq 1$ be a natural number. Prove that

$ \lim\limits_{n\rightarrow \infty}n^{k}\left(1\minus{}\prod\limits_{j\equal{}n\plus{}1}^{\infty}\frac{j^{k\plus{}1}}{j^{k\plus{}1}\plus{}1}\right)\equal{}\frac{1}{k}$.

Back to the original question:

$ \ln\left(\prod_{j\equal{}n\plus{}1}^{\infty}\frac{j^2}{j^2\plus{}1}\right)\equal{}\sum_{j\equal{}n\plus{}1}^{\infty}\ln\left(1\minus{}\frac1{j^2\plus{}1}\right)$

$ \equal{}\sum_{j\equal{}n\plus{}1}^{\infty}\minus{}\frac1{j^2}\plus{}O(j^{\minus{}4})\equal{}\minus{}\frac1n\plus{}O(n^{\minus{}2}).$

Exponentiate that to get that the product is $ 1\minus{}\frac1n\plus{}O(n^{\minus{}2})$

Hence, the original limit is $ 1,$ as claimed.

didilica's generalization comes down to the following assertion, which can be shown by the integral test for the convergence of series:

$ \sum_{j\equal{}n\plus{}1}^{\infty}\frac1{j^{k\plus{}1}}\equal{}\frac1{kn^k}\plus{}O(n^{\minus{}k\minus{}1}).$

And we don't need for $ k$ to be a natural number to claim that: $ k$ a positive real number will suffice.


[Table]
----------------------------------
Year     Host       Games   Average
----------------------------------
1930    Uruguay   18        3.89
1934    Italy        17        4.12
193     France      18        4.67
1950   Brazil         22       4.00
1954   Switzerland 26      5.38
1958   Sweden     35       3.60
1962   Chile         32       2.78
1966   England     32       2.78
1970   Mexico      32       2.97
1974   Germany    38       2.55
1978   Argentina   38       2.68
1982   Spain        52       2.81
1986   Mexico       52      2.54
1990   Italy          52      2.21
1994   USA           52     2.71
1998   France       64      2.67
2002  Korea, Japan64      2.52
----------------------------------
Table 1 shows the average goals per game in each World Cup

Questions:
1)Base on the data in Table 1, calculate the mean of the goals per game for the World Cup from 1930 to 2002

2)Base on the data in Table 1,calculate the standard deviation of the goals per game for the World Cup from 1930 to 2002

3)Base on answer in 1(a), the number of goals per game may assumed to be a Poisson variable, find the probability that

a)exactly 3 goals in the final match.
b)more than 2 goals in both the semi-finals matches.
c)exactly 2 goals in two out of four quarter-finals matches.
d)exactly 4 goals in the final match or a total of 7 goals in the
 semifinal matches.

[hide="1"]i got 2.23 on my calculator.[/hide]
[hide="2"]i got 3.15[/hide]

[hide="#1"]Actually, I got 3.228 on my calculator. Did you divide by the wrong number? You should note that some of the years from the 1940s are missing.[/hide]

By the way, what's the formula for standard deviation again? I forgot it  :P .

so may i know...for the question 1...izit correct if i use the
 'total sum of average in the table/ total no.of years?

or use the formula 'Efx/Ef' where 'f' is no of games and 'x' is average?


How do we predict whether we will have $ 2s\minus{}2p$ mixing or not?

[quote="BanishedTraitor"]How do we draw the Molecular Orbitals for Heteronuclear Molecules? (e.g. $ \ce{CO}$). How different is it from Homonuclear diatomic molecules?[/quote]

Well you do it as normally you do for homonuclear molecules  :huh:   :maybe:

I think s and p mixing depends on energy difference between s and p orbitals


Could someone please tell me how to get Asymptote to work on TeXnicCenter?
I downloaded everything, made a tool from the AoPSWiki page etc.
So when I press my "hotkey," I see the asymptote command prompt come up for half a second and the poof, nothing happened.
Do I have to write special code in it?  I am just playing around with asymptote.  Nothing "serious."
Thanks in advance!
minicon
 :lol:

Hi stevem,
I still have questions, sorry!  :oops:
On the [[Asymptote:_Help]] page, I have problems 1, 2, and 3.
How should I start?
_______________

Also, for the Asymptote Doc in TeXnicCenter, do you just type in the code
[code]draw((5,5)--(10,10))[/code]
or something else?

Thanks so much for helping me!
 :omighty:

Have you followed the instructions exactly in [[Asymptote:_Getting_Started/Windows/Interactive_Mode#Using_TeXnicCenter_for_Asymptote]] including the example there? Have you followed the solutions to the problems to Problems 1,2,3 particularly taking care the file name ends in .asy not .asy.tex?

You can just type in the code draw((5,5)--(10,10)) provided you add the vital ; at the end so that you get draw((5,5)--(10,10));

I think it would take some time to fix.  I have another way around it.  Thanks for helping stevem!
:lol: :)


In how many ways can you distribute 9 identical candy bars to 3 people if each person does not have to recieve any candy bars?

I'm not sure if this is right, but....

[hide]Case 1: 1 person receives candy bars - 1 possible way per person, but 3 people, so 3 ways

Case 2: 2 people receive candy bars - $\dbinom{3}{2}= 3$ ways of choosing 2 people, and 8 ways of distributing candy bars to each of them, so $2*8=16$ ways

Case 3: All 3 people receive candy bars - so $7+6+5+4+3+2+1 = 28$ possible ways

So, total is: $3+16+28 = 47$ possible ways total... :lol:  :lol: [/hide]

- Nawahd Werdna

Well that can't be, since:
[hide]
$\binom{11}{2}= \frac{11\cdot 10}{2}= 55$
[/hide]
Too lazy to explain...well,
[hide]
It is equivalent to the number of solutions to
$x+y+z=9$
which can be solved by balls-in-urns formula, for those of you who knew
[/hide]

[quote="Nawahd Werdna"]@pkerichang:

I don't think that 55 is right, because you are counting the cases that were counted in case 1 and case 2. Basically, you are including (incorrectly) the 8 cases where 1 or more people have 0, which we have already counted... :maybe:  :maybe:[/quote]
actually I get pkerichang's answer.  The easiest method is balls and urns.  I looked over what you did and for the case wehre two people get a candy bar, that is not 16, its [b]3[/b]*8=24 (you did 2*8), its 3*8 because its 3C2.  So just add 8 to your 47 taht you got and you get 55.

To formalize it:
[hide]
This problem is equivalent to the number of non-negative x, y, and z solutions to:

$x+y+z=9$

So let each candy bar be a #. In some order, we have 9 #s separated by 2 +s, for instance:

###++######

is a case that works, and is equivalent to (x, y, z)=(3,0,6). They key is we now have 11 total objects, 2 of type + and 9 of type #.

As such, the problem reduces to: How many different ways can we place the two +s in the 11 objects? This is merely 11 choose 2:

${11\choose{2}}=\boxed{55}$

And we're done.
[/hide]

Follow-up: Solve the same problem, except with the stipulation that each person has to receive at least 1 candy bar.

Hint:
[hide]
Take into account the 1 for each x, y, and z in the system x+y+z=9
[/hide]

[hide]Then let's say $(x-1)+(y-1)+(z-1)=6$ and $\binom{8}{2}= \boxed{28}$.[/hide]
Same problem (different numbers) was also in MC forum.

Ahhh, I see what I did wrong....  :blush:  :blush:

for reference, the balls and urns formula says that the number of ways of placing $b$ indistinguishable balls into $u$ urns is given by

${b+u-1 \choose u-1}$

How is balls and urns "different" then combinations?  :oops:


I was reading a chapter on proving trig identities when it came down to this:

$\sin^{2}\theta-\sin^{2}\theta \cos^{2}\theta$

and the factored version was

$\sin^{2}\theta (1-\cos^{2}\theta)$

is this right? i don't think iti s...but then again i just started trig  :wink:

Of course that's right. It is just normal factoring.

lol omg im soo stupid wasn't thinking straight sorry!

of course, I would simplify that to $\sin^{4}\theta$...

yeah but u needed it to be

$\sin^{2}\theta (\sin^{2}\theta)$ for some fraction work


If n is natural and gcd(n,3)=1 prove that there exist a number $ m$ for each $ n$ such that $ 3^{m}\minus{}1$ is divsible by n.

According to Euler's Theorem,let m=$ \phi (n)$


Find the value of the expresion:

$E=[(sec(x)+csc(x))^{2}+(tan(x)+cot(x))^{2}]sin(2x)$


inequalities
LaTeX
inequalities proposed

If $ a\geq 2$, $ b\geq 2$, $ c\geq 2$ are reals, then prove that
$ 8\left(a^3\plus{}b\right)\left(b^3\plus{}c\right)\left(c^3\plus{}a\right)\geq 125\left(a\plus{}b\right)\left(b\plus{}c\right)\left(c\plus{}a\right)$.

Lets write $LHS$ as $8*(a^3b^3c^3+abc+a^4b^3+b^4c^3+c^4a^3+a^4c+b^4a+c^4b)$.
From the Muirheads inequality we have that $a^4b^3+b^4c^3+c^4a^3\geq \sum a^3b^2c^2=a^2b^2c^2(a+b+c)$ and $a^4c+b^4a+c^4b\geq \sum a^3bc=abc(a^2+b^2+c^2)$.$(**)$
Now lets see that $a^2b^2c^2=(ab)(bc)(ca)\geq(a+b)(b+c)(c+a)$(This is easy to prove).
From the $(**)$ we get that
$LHS\geq8*(a^3b^3c^3+abc+a^2b^2c^2(a+b+c)+abc(a^2+b^2+c^2))
=8*a^2b^2c^2(abc+\frac{1}{abc}+a+b+c+\frac{a^2+b^2+c^2}{abc})$ so we have to prove that :
$abc+\frac{1}{abc}+a+b+c+\frac{a^2+b^2+c^2}{abc}\geq\frac{125}{8}$ or
$8a^2b^2c^2+8+8abc(a+b+c)+8(a^2+b^2+c^2)\geq125abc$
WLOG let $a \geq b \geq c$
Now let $abc=P$ we will make that following change $a\rightarrow\frac{a}{\epsilon}$ and $b\rightarrow b\epsilon$ where $\epsilon\geq1$ and $a\geqq\epsilon$. The $RHS$ doesnt change. In the $LHS$ the first part also doesn't change.
$a+b\geq\frac{a}{\epsilon}+b\epsilon$ equivalent to $(\epsilon-1)(a-b\epsilon)$ which is true. Also we get that
$a^2+b^2\geq\frac{a^2}{\epsilon^2}+b^2\epsilon^2$. So as we get the numbers closer to each other the $LHS$ decreases while the $RHS$ remains the same so it is enough to prove the inequality for the case $a=b=c$ which is equivalent to :
$8a^6+8+24a^4+24a^2\geq125a^3$ Which is pretty easy. I hope I didn't make mechanical mistakes or missused Muirheads.(Sorry for so many edits :? )

Thank you ,Myth .You are so kind  :)
I think my post is not so difficult to read .

@ RaduSorici:

[quote="RaduSorici"]Lets write $LHS$ as $8*(a^3b^3c^3+abc+a^4b^3+b^4c^3+c^4a^3+a^4c+b^4a+c^4b)$.
From the Muirheads inequality we have that $a^4b^3+b^4c^3+c^4a^3\geq \sum a^3b^2c^2=a^2b^2c^2(a+b+c)$ and $a^4c+b^4a+c^4b\geq \sum a^3bc=abc(a^2+b^2+c^2)$.$(**)$
Now lets see that $a^2b^2c^2=(ab)(bc)(ca)\geq(a+b)(b+c)(c+a)$(This is easy to prove).
From the $(**)$ we get that
$LHS\geq8*(a^3b^3c^3+abc+a^2b^2c^2(a+b+c)+abc(a^2+b^2+c^2))
=8*a^2b^2c^2(abc+\frac{1}{abc}+a+b+c+\frac{a^2+b^2+c^2}{abc})$ so we have to prove that :
$abc+\frac{1}{abc}+a+b+c+\frac{a^2+b^2+c^2}{abc}\geq\frac{125}{8}$ or
$8a^2b^2c^2+8+8abc(a+b+c)+8(a^2+b^2+c^2)\geq125abc$
WLOG let $a \geq b \geq c$
Now let $abc=P$ we will make that following change $a\rightarrow\frac{a}{\epsilon}$ and $b\rightarrow b\epsilon$ where $\epsilon\geq1$ and $a\geqq\epsilon$. The $RHS$ doesnt change. In the $LHS$ the first part also doesn't change.
$a+b\geq\frac{a}{\epsilon}+b\epsilon$ equivalent to $(\epsilon-1)(a-b\epsilon)$ which is true. Also we get that
$a^2+b^2\geq\frac{a^2}{\epsilon^2}+b^2\epsilon^2$. So as we get the numbers closer to each other the $LHS$ decreases while the $RHS$ remains the same so it is enough to prove the inequality for the case $a=b=c$ which is equivalent to :
$8a^6+8+24a^4+24a^2\geq125a^3$ Which is pretty easy.[/quote]

RaduSorici, your solution is essentially correct, but you made a little argumentation mistake: The two inequalities

$a^4b^3+b^4c^3+c^4a^3\geq a^3b^2c^2+b^3c^2a^2+c^3a^2b^2$;
$a^4c+b^4a+c^4b\geq a^3bc+b^3ca+c^3ab$

do [b]not[/b] follow from Muirhead (since Muirhead can be only used for [i]symmetric[/i] inequalities, and these two inequalities are not symmetric). But still these two inequalities can be easily proved using weighted AM-GM:

For the first inequality, note that weighted AM-GM yields

$\frac{6}{13}a^4b^3+\frac{2}{13}b^4c^3+\frac{5}{13}c^4a^3\geq \left(a^4b^3\right)^{\frac{6}{13}}\cdot\left(b^4c^3\right)^{\frac{2}{13}}\cdot\left(c^4a^3\right)^{\frac{5}{13}}=a^3b^2c^2$,

and similarly

$\frac{6}{13}b^4c^3+\frac{2}{13}c^4a^3+\frac{5}{13}a^4b^3\geq b^3c^2a^2$;
$\frac{6}{13}c^4a^3+\frac{2}{13}a^4b^3+\frac{5}{13}b^4c^3\geq c^3a^2b^2$.

Summing up these three inequalities, we get exactly $a^4b^3+b^4c^3+c^4a^3\geq a^3b^2c^2+b^3c^2a^2+c^3a^2b^2$ (since $\frac{6}{13}+\frac{2}{13}+\frac{5}{13}=1$).

Similarly, we can prove the second inequality: By weighted AM-GM,

$\frac{9}{13}a^4c+\frac{3}{13}b^4a+\frac{1}{13}c^4b\geq \left(a^4c\right)^{\frac{9}{13}}\cdot\left(b^4a\right)^{\frac{3}{13}}\cdot\left(c^4b\right)^{\frac{1}{13}}=a^3bc$,

and similarly

$\frac{9}{13}b^4a+\frac{3}{13}c^4b+\frac{1}{13}a^4c\geq b^3ca$;
$\frac{9}{13}c^4b+\frac{3}{13}a^4c+\frac{1}{13}b^4a\geq c^3ab$.

Summing up these three inequalities, we get exactly $a^4c+b^4a+c^4b\geq a^3bc+b^3ca+c^3ab$ (thanks to $\frac{9}{13}+\frac{3}{13}+\frac{1}{13}=1$).

So both of your inequalities are proven.

@ namanhams:

I don't understand your solution. Could you please explain what you call Cauchy and what you call Bunhiacopski? For me, it has always been one and the same thing. Maybe you actually used Hölder rather than Cauchy-Schwarz-Buniakowski?

[EDIT: Now I understand your solution. "Bunhiacopski" meant Hölder. Thanks Myth.]

@ problem:

I have a different solution of the original problem, based on an idea used by RaduSorici:

We have to prove the inequality

$8\left(a^3+b\right)\left(b^3+c\right)\left(c^3+a\right)\geq 125\left(a+b\right)\left(b+c\right)\left(c+a\right)$

for any reals a, b, c which are all $\geq 2$.

Well, since $a\geq 2$, we have $a-2\geq 0$ and also $2a^2-a-2>0$ (the latter is because $a\geq 2$ implies $a-\frac14\geq 2-\frac14=\frac74$, so that, as both sides of this inequality are positive, squaring yields $\left(a-\frac14\right)^2\geq\left(\frac74\right)^2=\frac{49}{16}>\frac{17}{16}$, and thus $\left(a-\frac14\right)^2-\frac{17}{16}>0$; but we can easily see that $\left(a-\frac14\right)^2-\frac{17}{16}=\frac12\cdot\left(2a^2-a-2\right)$, and thus we conclude that $2a^2-a-2>0$).

Multiplying the inequalities $a-2\geq 0$ and $2a^2-a-2>0$, we get $\left(a-2\right)\left(2a^2-a-2\right)\geq 0$. But $\left(a-2\right)\left(2a^2-a-2\right)=2\left(a^3+2\right)-5a^2$. Thus, $2\left(a^3+2\right)-5a^2\geq 0$, so that $2\left(a^3+2\right)\geq 5a^2$. Since $b\geq 2$, we conclude that $2\left(a^3+b\right)\geq 2\left(a^3+2\right)\geq 5a^2$. Similarly, $2\left(b^3+c\right)\geq 5b^2$ and $2\left(c^3+a\right)\geq 5c^2$. Multiplying these three inequalities together, we get

$2\left(a^3+b\right)\cdot 2\left(b^3+c\right)\cdot 2\left(c^3+a\right)\geq 5a^2\cdot 5b^2\cdot 5c^2$;

in other words,

$8\left(a^3+b\right)\left(b^3+c\right)\left(c^3+a\right)\geq 125a^2b^2c^2$.

But $a\geq 2$ and $b\geq 2$ yield $a-2\geq 0$ and $b-2\geq 0$, and thus $ab-\left(a+b\right)=\left(a-2\right)+\left(b-2\right)+\left(a-2\right)\left(b-2\right)\geq 0$. Hence, $ab\geq a+b$. Similarly, $bc\geq b+c$ and $ca\geq c+a$. Altogether, we conclude

$8\left(a^3+b\right)\left(b^3+c\right)\left(c^3+a\right)\geq 125a^2b^2c^2=125\cdot ab\cdot bc\cdot ca$
$\geq 125\cdot\left(a+b\right)\cdot\left(b+c\right)\cdot\left(c+a\right)$.

And the inequality is proven.

[b]PS.[/b] See http://www.mathlinks.ro/Forum/viewtopic.php?t=1617 for a similar inequality.

  Darij

Namanhams's solution is great. I am very glad there are people who care about readers and about elegancy.

Indeed, he slightly messed with names:
Bunhiacopski=general Holder
Cauchy=AM-GM (actually, it is Cauchy in russian literature, at least)


Entrance to ISI and CMI will be based on an examination which is of RMO to INMO standard.
So if u prepare urself for the olympiads, u can easily get admission into them.  :)

[quote="anorithdialga"]:D Can anyone in this forum tell me if there are any suitable coaching institutions in KOLKATA providing training for the ISI and CMI Entrance Exams?

I know of one such institution called RAMANUJAN SCHOOL OF MATHEMATICS located near ISI, Kolkata .Does anyone know if it is good? :maybe:

Waiting eagerly for replies...... :D[/quote]

i have heard about this institution . the thing is they have a good reputation for selected candidates in RMO, ISI,CMI etc but very few of their students get through INMO (none from last year)

Then are there any institutions training for INMO ?

@Aravind: I know that the CMI & ISI entrance exams  questions have a standard higher than RMO but lower than INMO . But the problem is the number of questions is much higher . They have a lot of MCQ's and also some subjective ones. Doesn't this make those exams a whole lot tougher?  :huh:

[quote="Akashnil"][quote="apratimdefermat"]none from last year[/quote]
No. [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=38552]Ratul Saha[/url] was a [url=http://www.isical.ac.in/~rmo/inmolist_08.pdf]INMO 08 awardee[/url].
By January this year I did 4-5 special classes on INMO this year in Ramanujan. But I can't give you an assessment because I think that is not enough to tell how it really is.
P.S. Ratul was with us in those 4-5 classes.[/quote]

wat do u mean by special classes?
i mean to say do u take coaching in Ramanujan.
can u tell what exactly do they teach there in those classe ?

[quote="Akashnil"]this year before INMO they offered some class for those going to write INMO. That was not regular class. Among the 7 ppl who attended it most were their regular students, but I was not. I was in class IX. But they normally teach XI and XII. As there was only 4-5 classes, they didn't teach much theory. We did problems. They taught some techniques for problem-solving like recurrence, infinite decent while discussing problems.[/quote]

I wanna know if they solve ur personal queries .
I have been thinking about joining them do u think it wud be a good idea?

They answered personal questions in our class.


conics
parabola
analytic geometry
graphing lines
slope
quadratics
calculus

Hello! I'm new here   :D

This problem was proposed over a year ago, and I've got to a point where I have given up.  The problem goes as follows:

"Find the equation in Y= form for a curve that is both equidistant and perpendicular to the parabola $y=x^2$ and the X-axis"

Here’s a [size=75](poorly drawn) [/size]illustration:
[img]http://www.quigibo.brickfilms.com/images/parab.GIF[/img]
The blue is the parabola and the green is the curve I need.  As you can see, the final result should have a plus-minus sign because there are actually 2 curves that satisfy the condition.

Yes… It may [i]seem[/i] simple… but it’s [b]not[/b]!

I’ve written a 3 page step by step process on how to solve it and I eventually got to a point where I needed to solve for $x$ in the quartic equation $x^4+ux-u^2=0$.  But I can’t seem to find the quartic equation anywhere.  Anyway, that’s where I stopped.  For those of you interested in how I got that far, I’ll try to sum it up in the next few lines.  It may be a spoiler to those who want to solve it themselves

1) The curve relative to the parabola would have coordinates:
$(x+a,x^2-b)$
2) But because the slope of a parabola is $2x$, the coordinates are actually:
$(x+2xb,x^2-b)$
3) The distance from the parabola is equal to the height so:
$(2xb)^2+(b)^2=(x^2-b)^2$
4) Which simplifies to:
$b=\frac{-1\pm\sqrt(1+4x^2)}{4}$
5) Now we know b in terms of x and to get Y= we set:
$u=x+2bx$
6) Which becomes:
$x^4+ux-u^2=0$
I’m stuck!!!  All that’s left is to solve for x and to get $y=x^2-b$ in terms of $u$.  Even if I did use the quartic formula, I still wouldn’t know which of the quartic's 4 roots or which of the 2 roots in step 4 (I used the quadratic formula) to use.

So… that’s about it.  Can someone please try this?  For the sake of my sanity? :P

Take a particular point on the graph, $P(x_1, y_1)$.  Draw the perpendiculars to $y = x^2$ and to $y = 0$.  Call the point where the perpendicular intersects the parabola $A(x_2, x_2^2)$, and call the point where the perpendicular intersects the axis $B(x_1, 0)$.  Draw the tangent lines to $A, B$, which intersect at a point $C(x_3, 0)$.  Draw the triangles $\triangle ACP, \triangle BCP$.  These triangles are congruent and right.

Let's calculate $x_3$.  Calculus tells us that the slope of $AC$ is $2x_2$, so write $2x_2 = \frac{x_2^2 - 0}{x_2 - x_3} \implies 2x_2 (x_2 - x_3) = x_2^2 \implies x_2 - x_3 = \frac{x_2}{2} \implies x_3 = \frac{x_2}{2}$.

Note that we can also write the slope of $AC$ as $2x_2 = 4x_3 = \tan (\angle ACP + \angle BCP) = \tan (2 \angle BCP)$ (remember, similar triangles - these angles are equal).  Similarly, we can write the slope of $PC$ as $\frac{y_1 - 0}{x_1 - x_3} = \tan (\angle BCP)$.  Let's break out the tangent half-angle formula.

Unfortunately, the standard tangent half-angle formula does not give $\tan (\frac{\theta}{2})$ in terms of $\tan \theta$, which is what we would like, so instead we work backwards from the double-angle formula $\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$, substituting $t = \tan \theta$.

$4x_3 (1 - t^2) = 2t$
$4x_3 t^2 + 2t - 4x_3 = 0$
$t = \frac{-2 \pm \sqrt{4 + 64x_3^2} }{2} = -1 \pm \sqrt{1 + 16x_3^2}$.

We take the positive solution, so we have $\sqrt{1 + 16x_3^2} - 1 = \frac{y_1}{x_1 - x_3}$.  Almost there!

We need to express $x_3$ in terms of $x_1$.  Let's use the fact that $\triangle ACP \cong \triangle BCP$.

$AP = \sqrt{ (y_1 - x_2^2)^2 + (x_1 - x_2)^2 } = \sqrt{ (y_1 - 4x_3^2)^2 + (x_1 - 2x_3)^2 }$
$BP = y_1$
$AP = BP$
$y_1^2 = (y_1 - 4x_3)^2 + (x_1 - 2x_3)^2$
$8x_3 y_1 = 16x_3^2 + (x_1 - 2x_3)^2$
$y_1 = 2x_3 + \frac{ (x_1 - 2x_3)^2 }{8x_3}$

And from our above work we have

$y_1 = (x_1 - x_3) ( \sqrt{1 + 16x_3^2} - 1)$

Hmm... let's do some more equivalences.

$AC = \sqrt{ (x_2^2 - 0)^2 + (x_2 - x_3)^2 } = \sqrt{16x_3^4 + x_3^2} = x_3 \sqrt{1 + 16x_3^2}$
$BC = x_1 - x_3$
$AC = BC$

Well, I'm stuck again.

Edit:  For the parametric equation, call $t = x_3$ and we can write

$y_1 = t \sqrt{1 + 16t^2} ( \sqrt{1 + 16t^2} - 1)$
$x_1 = t + t \sqrt{1 + 16t^2}$

This may be very n00bish of me, but I finally understand how those math images were so simple to make! They were BBcode  :blush: .  Well, I edited my first post.  It should be easier to read now.

Here's my idea, and i'll leave the details to you.

1. Take an arbitrary point of the parabola $(c, c^2)$;
2. Find the tangent and the normal to the parabola;
3. Find the vertex of the angle between the tangent and the $x$-axis;
4. Find the bisector of the angle between the tangent and the $x$-axis: one point of the bisector is the vertex from Step 3, and the slope can be computed from the relationship $k_{tangent}={{2k_{bisector}} \over {1-k_{bisector}^2}}$ (easily derived from the fact that the slope of a line is the tangent of the angle between the line and the $x$-axis). Since this is a quadratic equation, you'll have two solutions, for the obtuse and the acute angle;
5. Find the intersection of the normal and each bisector in turn - these are the desired points of the two curves, i.e. their parametric equations.

I have already done this, and derived the equation in terms of a parameter $x_3$ which is the x-coordinate of the x-intercept of the tangent lines - however, the problem asks for $y = f(x)$ form, which requires solving a general quartic by this method  :(

Does anyone [i][b]know[/b][/i] the quartic formula?  Mathworld's page on it is way too confusing for me and it doesn't actually give the formula, just how to dirive it. :|

The actual formula is too much of a beast to write out explicitly (even writing the cubic formula explicitly is... just  :huh: ) so you must go through the entire derivation in order to keep the calculations sane.  That's why I'm convinced there should be an easier way to do this problem, if it's doable with high school mathematics at all.

Oh! I found it on wiki!  :lol:

[quote="Wikipedia"]Given the quartic equation
$A x^4 + B x^3 + C x^2 + D x + E = 0$
its solution can be found by means of the following calculations:
$\alpha = - {3 B^2 \over 8 A^2} + {C \over A}$
$\beta = {B^3 \over 8 A^3} - {B C \over 2 A^2} + {D \over A},$
$\gamma = {-3 B^4 \over 256 A^4} + {C B^2 \over 16 A^3} - {B D \over 4 A^2} + {E \over A},$
if $\beta=0$ solve $u^4+\alpha u^2 + \gamma = 0$ and substitute $x=u-{B\over 4A}$ finding the roots
$x=-{B\over 4A}\pm_s\sqrt{-\alpha\pm_t\sqrt{\alpha^2-4\gamma}\over 2},\qquad\beta=0$.
$P = - {\alpha^2 \over 12} - \gamma,$
$Q = - {\alpha^3 \over 108} + {\alpha \gamma \over 3} - {\beta^2 \over 8},$
$R = {Q\over 2} \pm \sqrt{{Q^{2}\over 4}+{P^{3}\over 27}}$, (either sign of the square root will do, as long as $R$ does not disappear unnecessarily; in the case of $P=0$ we want $R=Q$)
$U = \sqrt[3]{R}$, (there are 3 complex roots, any one of them will do)
$y = - {5 \over 6} \alpha -U + \begin{cases}U=0 &\to 0\\U\ne 0 &\to {P\over 3U}\end{cases},$
$x = - {B \over 4 A} + { \pm_s \sqrt{ \alpha + 2 y} \pm_t \sqrt{-\left(3\alpha + 2 y \pm_s {2\beta\over\sqrt{\alpha +2y}} \right) }\over 2 }.$
The two $\pm_s$ must have the same sign, the $\pm_t$ is independent. To get all roots, find $x$ for $\pm_s$,$\pm_t$ = +,+ and for +,- and for -,+ and for -,-. Double roots will be given twice, triple roots 3 times and quadruple roots would be given 4 times (although then $\beta = 0$, which is a special case). The order of the roots depends on which cubic root $U$ is chosen.[/quote]

Though I may still need a little help with this... $(A=1,B=0,C=0,D=u,E=-u^2)$

[quote]I'm convinced there should be an easier way to do this problem, if it's doable with high school mathematics at all.[/quote]
I'd rather just get an answer with this method than spend too much time looking for an alternitive way right now.  And besides, it may not be doable with high school level mathematics.

It's not really clear to me why the proponent of the problem asks for the explicit form of the curve - as if parametric equations are not valid...

If quartic is unaviodable (and it seems it is), then this surely doesn't belong to high school (perhaps not even to college - there's just too much handiwork involved).


Ethel made a list of all the whole numbers between 1 to 100. How many times did she write the number 2?

[quote="mathmanman"]Hmm, well, here is my reasoning :
[hide]once between 1 and 10 (2)
twice between 10 and 20 (12 and 20)
ten times between 20 and 30 (21,22,23,24,25,26,27,28,29)
and, then, once between 30/40, 40/50 etc..
So, summing it up, we get : 20 :)
So, there are 20 2's in her list[/hide][/quote]

Yep. There's only 1 between 10 and 20, as 'between' is not including 20, and you counted '20' in the 'ten between 20 and 30'. So, it's [hide]19.[/hide]

No, mathmanman is right.

There are 10 2's written in the tens digit and 10 2's written in the units digit.

There is a different question that this can be confused with, which is, [b]how many of the integers from 1 to 100 are written with a 2?[/b]

The answer to that question is 19 and we can find it by multiplying our count of all the integers without a 2 digit and subtracting the total from 100:
\[ 100 - (10 - 1)(10 - 1) = 100 - 81 = 19. \]

Think about this proportionality method and why it works.

[hide]
I listed all of the numbers with a 2 in them.
2
12
20
21
22
23
24
25
26
27
28
29
32
42
52
62
72
82
92
There are 19 numbers, but in 22 there are two 2's, so the answer is 20.[/hide]

i think that MCrawford's way is easier than pascal_1623's.

But we don't learn the nice methods without significant exploration, such as pascal_1623 was doing.

The best way to become better at math is to solve problems, and then look back at what you've done and see if you can wrap it up into a nicer package.  My method is just the condensed version of exactly what pascal_1623 was doing and it's important not to overlook that fact.

Much of better mathematics is a matter of applying the basics [i]more efficiently[/i].

[hide]ten 2s in the ones place
ten 2s in the tens place

20 all together.
[/hide]

[hide]
There is 20, 21, 22, 23, 24, 25, 26, 27, 28, 29

2 is used 10 times as the tens digit.

2, 12, 22, 32, 42, 52, 62, 72, 82, 92

2 is used 10 times in the units digit.

So 2 was used 20 times.
[/hide]

[quote="236factorial"]For this type of problem, first I see how many numbers have the units digit 2. There are 10.

Now, every number between 20 and 29 have a 2, that means there are 11 2's (22 has two 2's). But when we counted the units digit, we also counted a two in 22, so we have a overlap. We need to take that overlap away.

so 10+11-2=20 2's.

My answer disagrees with the administrators?  :D[/quote]

You only need to take away 1 from the overlap -- you still need to count the other to single-count everything.
\[ 10 + 11 - 1 = 20. \]
Still, I'm not sure this is the best way to look at the problem.  It's not as nice an organization of the counting process.  Need we count in such a way that we need to consider overcounting?

I think your original way was correct and the best one. Ask the question, how many numbers from 1-100 don't have a 2? That's 9*9=81.
So there are 19 numbers with 2's and add the one from the extra in 22 to get 20. This way, there's no overcounting involved.

But consider this problem, a much harder one probably harder than even mathcounts (split if necessary):

How many numbers from 1 - 1000 have at least one 2 or at least one 5, but not both digits?

Can you do this without the overcounting methods?


Basicity order is :
p-Methoxy Aniline>Aniline>p-Nitro Aniline
en donating group increases en density on N of aniline and en withdrawing grp decr. the en density
hence the order .

With amines as bases, the conjugate acid will have a positive charge, as the best base will be the one that can stabilize more readily that positive charge. So we are looking for anilines with electron donating substituents, and correct answer is the one given by Madness.

[quote="Pardesi"]yes that's what i thought but my book says the opposite order[/quote]

I strongly recommend you to chose another book for study Organic Chemistry. If the problems you have been posting are from that book, then it is not clearly the appropriate book for you.

Actually i am reading as u know from Morrison But have this book for doing Problems strongly recommended in our country for IIT-JEE so i am helpless( but u r there  :) )


limit
logarithms
calculus
calculus computations

$\gamma$
[hide=";)"]No sensefull question $\implies$ no sensefull answer[/hide]


In this game, all you have to do is solve math problems. In teams, one team gives another team a problem which the opposing team has to solve. No calculus, no open problems. I think sign up is pointless, but for this specific game, its necessary

People that have joined:
1. Poincare (no duh)
2. FantasyLover
3. Brut3Forc3
4. 1=2
5. mewto555555

Waiting for people to sign up!

People that have joined:
1. Poincare (no duh)
2. FantasyLover
3. Brut3Forc3
4. 1=2
5. mewto55555
6. AIME15

Jeez...you forgot me.  The third post.

How will be divided into teams?

[quote="dragon96"]/in

How many people?
What lvl problems (MC, AMC, AIME, etc)?
How is it played in the forums?[/quote]
idk, probably hard AMC to hard AIME/beginning USAMO. And, if someone proves that the solution that is the official solution to a problem is wrong, well, the first team loses. If we have more people, i may introduce points which would make it more fun
and why \out 1=2?

i'll join. should we start making teams. i hope you change your mind of the beginning USAMO. Do school mathcounts questions.

SECOND CAPTAIN!

although i dont think were doing captains.

heh mew, 1=2 already dropped out.

[b]PICK: moogra[/b]

On the edge of being out...

Pick: (undecided)

egglv99! its not gonna be too hard! i was talking about like the end of the part when the lvl was that high! if you guys want to, we can lower the level!

OK. do the problems have to be of original design?

Either way, I join.

Mew hasn't posted for 23 hours, so I get another pick!

[b]PICK: pythag011[/b]

i think so...we're still picking teams :P

mewto55555's team: nobody
AIME15's team: moogra, pythag011

Locked for the same reason as explained in http://www.mathlinks.ro/viewtopic.php?t=233458 .
Go to the tournament threads ;)

PS: prove this (not open, not calculus!):
Let $ K,L$ be number fields such that $ \mathrm{Gal}(\bar K | K) \cong \mathrm{Gal}(\bar L | L)$ as profinite groups. Show that $ K \cong L$.


I'd like advice on efficient ways to solve this problem given the time constraints of a contest situation:

In a right triangle, all the sides have lengths that are integers.  The length of one of the legs is 60.  Find the sum of all distinct values for the length of hypotenuse.

The answer is 2791 -- there are 13 triples to generate.  It seems to me that a calculator program using Maseres Rule might be helpful here if one needs to obtain the solution quickly??


Let E be a point inside nonisosceles acute triangle ABC
lying on median AD, and drop perpendicular EF to line BC. Let
M be an arbitrary point on EF, and let N and P be the orthogonal
projections of M onto lines AC and AB, respectively. Prove that the
angle bisectors of angles PMN and PEN are parallel.

[b]HINT:[/b] [i]use the theorem[/i]:
$ \mbox{In quadrilateral ABCD the bisectors of angles ABC and ADC are parallel }\\
\Leftrightarrow \angle BAD\equal{}\angle BCD.$

Here is my pf:
join $ {BE , CE}$;draw $ {EX}$⊥$ {AC}$;$ {EY}$⊥$ {AB}$;$ {EW}$⊥$ {NM}$;$ {EZ}$⊥$ {MP}$;altitude $ {AH}$ on $ {BC}$; WE just need $ {\triangle {EXN}}$ ∽ $ {\triangle{EYP}}$;so first :
 Because  $ {D}$ is the midpoint of side $ {BC}$ ; then we get $ {S\triangle {ACE}=S\triangle {ABE}}$; $ {\longrightarrow}$ $ {EX*AC=EY*AB}$ $ {\longrightarrow}$ $ {\frac {EX}{EY}=\frac {AB}{AC}}$...................(1).
Easy to get that :$ {\triangle{EWM}}$∽$ {\triangle {ACH}}$; AND $ {\triangle {EZM}}$∽$ {\triangle {AHB}}$ $ {\longrightarrow}$ $ {\frac {EW}{AH}=\frac {EM}{AC}}$and$ {\frac {EZ}{AH}=\frac {EM}{AB}}$ $ {\longrightarrow}$ $ {EW*AC=AH*EM=EZ*AB }$ $ {\longrightarrow}$  $ {  \boxed{\frac {EW}{EZ}}=\frac {AB}{AC}=\frac {EX}{EY}}$ ..................(2)
Notice that $ {\boxed{{EW=CN} and  {EZ=YP}}}$ so we get :${ {\frac {CN}{YP}}=\frac {EX}{EY}}$; $ {\rightarrow}$ $ {\triangle {EXN}}$ ∽ $ {\triangle{EYP}}$; $ {\longrightarrow}$ By  mihai miculita ' s lemma we have done !
 :ewpu: ugly solution  I think


suppose $ n \in N$ has written on a black board.

we clear this number and write the number $ n\plus{}2^b\minus{}2^a$, where $ a$ is the greatest number such that $ 2^a \mid n$ and $ b$ is the least number such that $ n<2^b$.

if we start with $ n\equal{}42$,what is the written number on the board after $ 113$ times doing this property?

$ a_0 \equal{} 42 \equal{} 2\times 21$, and checking the first values you get:

$ a_1 \equal{} 42 \plus{} 2^6 \minus{} 2^1 \equal{} 104 \equal{} 2^3\times 13$

$ a_2 \equal{} 104 \plus{} 2^7 \minus{} 2^3 \equal{} 224 \equal{} 2^8 \minus{} 2^5 \equal{} 2^5 (2^3 \minus{} 1)$

$ a_3 \equal{} 224 \plus{} 2^8 \minus{} 2^5 \equal{} 448 \equal{} 2 (2^8 \minus{} 2^5) \equal{} 2 a_2 \equal{} 2^9 \minus{} 2^6 \equal{} 2^6(2^3 \minus{} 1)$

$ a_4 \equal{} 2a_3 \equal{} 2^2a_2$

$ \cdots$ by induction

$ a_{113} \equal{} 2^{111}a_2 \equal{} 2^{111}(2^8 \minus{} 2^5)$

$ \boxed{a_{113} \equal{} 2^{119} \minus{} 2^{116}}$

For the general case here come some ideas:
1) first of all write each $ a_i$ on base two. Example: $ 42\equal{}101010_2$
2) the operation basically eliminates the rightmost one and adds one digit one to the left. Example: $ 101010_2 \Rightarrow 101000_2 \Rightarrow 1101000_2$
3) once you have only one chain of consecutive ones the number can be written like $ 2^b\minus{}2^a$ and start doubling itself.


integration
calculus
logarithms
trigonometry
real analysis
real analysis unsolved

(1) Show that
$\int_{x}^{\infty}e^{-t^{2}}\, dt < \frac{1}{2x}e^{-x^{2}}$
for all $x > 0$.

(2)  Discuss the convergence of $\int_{0}^{\infty}x \sin(e^{x}) \, dx$.

*On (1), is there a good solution that is not as artificial as using that $(\frac{1}{2t^{2}}+1)e^{-t^{2}}> e^{-t^{2}}$.

An approach for number 1) is this:

Let $F(x)=2x\int_{0}^{\infty}e^{-t^{2}}dt-2x\int_{0}^{x}e^{-t^{2}}dt-e^{-x^{2}}$.

Then $F'(x)=2\int_{x}^{\infty}e^{-t^{2}}dt>0$ hence $F$ increases. It follows that $F(x)\leq F(\infty)=0$ and the result follows.

2) The substitution $e^{x}=y$ transforms your integral to

$\int_{1}^{\infty}\frac{\ln y}{y}\sin y\, dy$ which converges by an application of the Dirichlet's criteria.

Integration by parts is a powerful tool.

First problem:
$\int_{x}^\infty \frac1{2t}\cdot 2te^{-t^{2}}\,dt=\left[-\frac1{2t}e^{-t^{2}}\right]_{x}^{\infty}-\int_{x}^{\infty}\frac1{2t^{2}}e^{-t^{2}}\,dt=\frac1xe^{-x^{2}}-\int_{x}^{\infty}\frac1{2t^{2}}e^{-t^{2}}\,dt$
This process could be repeated to generate an asymptotic series for the integral $\int_{x}^{\infty}e^{-t^{2}}\,dt= e^{-x^{2}}\left(\frac1{2x}-\frac1{4x^{3}}+\frac3{8x^{5}}-\frac{15}{16x^{7}}+\cdots\right)$. The terms strictly alternate, and the original integral is alternately smaller or larger than the sum, with error less than the next term. Unfortunately, the full series diverges for all $x$.
I consider this very natural.

Second problem: $\int_{0}^{\infty}xe^{-x}\cdot e^{x}\sin(e^{x})\,dx=\left[-xe^{-x}\cos(e^{x})\right]_{0}^{\infty}+\int_{0}^{\infty}(e^{-x}-xe^{-x})\cos(e^{x})\,dx$
$= \int_{0}^{\infty}(1-x)e^{-x}\cos(e^{x})\,dx$, which converges by its exponential decay.

Note: "evaluation" at $\infty$ is treated as a limit here, just as in the definition of improper integrals.

[Edit: At least our nearly simultaneous posts aren't entirely redundant...]

I would approach #1 by integration by parts: write $e^{-t^{2}}$ as $\frac1{2t}\cdot 2te^{-t^{2}},$ then integrate the $2te^{-t^{2}}$ while differentiating the $\frac1{2t}.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107053]This topic[/url] has all of this and more.


function
real analysis
real analysis unsolved

Does there exist a function $f: [0,1]\to \Re$ such that $f$ is verticaly asimptotic in every rational point of $[0,1]$?
[hide="answer"] no [/hide]

Consider $F_n=\{x: f(x)<n\}$,and let $\bar{F_n}$ be the closure of $F_n$. We have that $\cup_1^\infty \bar{F_n}=[0,1]$. By baire th we have that there exists $[a,b]\subset \bar{F_n}$ for some $n$. In interval $[a,b]$ all values of $f$ ore less than $n$, so $f$ cannot be asimptotic to $\infty$ in any point of the interval, hence it can't be in any rational point in this interval.

I hope it works  :lol:

[quote="spix"]My solution is about building a set of intervals $I_n$ with the property $|f(x)|>n$ , then choose $x_n$ rational in $I_n$ and $V$ a neighborhood of $x_n$ such that $|f(x)|>n+1$ . I chose $I_{n+1}$ as a inteval in the intersection of $I_n$ and $V$. notice that $I_{n+1}$ is included in $I_n$ thus there exist a point $a$ such that a belongs to all the intervals considered so $|f(a)|>n ,\forall n\in \mathcal l N$ contradiction![/quote]

It works but you must pay attention to the fact that $f(x)\ge n+1,\forall x\in V\setminus \{x_n\}$. then you may chose $I_{n+1}\subset (V\setminus \{x_n\})\cap I_n$. And it exists since $V$ have nonempty interior even we take one point  from it.

I got an example that may interest you. First take $V$ a Vitali set, which is formed by a system of representants in $[0,1]$ of $\mathbb{R}/\mathbb{Q}$. Now take biyection $h: [-1,1]\cap\mathbb{Q}\longrightarrow\mathbb{N}$ and a proyection $\pi: [0,1]\longrightarrow V$ st for each $x\in[0,1]$, $x-\pi(x)\in\mathbb{Q}$. Finaly consider $f: [0,1]\longrightarrow\mathbb{N}$
\[ f(x)=h(x-\pi(x))  \]
This $f$ is pretty weird,  for each $a\in[0,1]$, there are sequences $(x_n)\subseteq[0,1]$ st $f(x_n)\longrightarrow\infty$. Well, it's existence depends on the axiom of choice since Vitali's set does.

Another examples are non-continupus solutions of Cauchy exquations. Their graph is dense in $\mathbb{R}^2$ :!:


analytic geometry
graphing lines
slope
geometry
rectangle
ratio
algebra unsolved

Hi,

It seems intuitive that a particle bouncing around a box will follow a path that eventually returns to its starting point. If we increase the number of particles and allow them to bounce off each other in totally elastic collisions will each particle return to its original starting point despite the collisions?

My hunch is that they will, but is it possible to prove that each particle will return to its starting point in a finite time or number of bounces? Is there any formula for the mean free closed path of n particles in a box of a given area. 2D is fine, if it helps make the problem solvable.

Thanks for any help :)

[EDIT] I suspect this question is not in the right section. Can anyone recommend which section it should be in?

[u]Here's a simple counterexample.[/u]

Consider a box with dimensions $ 1 \times \pi$, and a particle fired from a corner at a $ 45^{\circ}$ angle to the edges, and suppose that it's speed is $ \sqrt {2}$ (this choice just makes the numbers easier). Then it's speed component along each edge is $ 1$.

Thus, it will reach one of the edges of length $ \pi$ whenever $ t \equal{} 1 \cdot n \equal{} n$ for any integer $ n$, and will reach an edge of length one whenever $ t \equal{} \pi \cdot m$ for any integer $ m$.

If it returns to its starting point at some time $ t$, then it must be at a length-$ 1$ edge as well as a length $ \pi$ edge, since any corner is the intersection of two such edges. Thus, we have $ t \equal{} n \equal{} \pi m$ for some integers $ m, n$.

However, this implies $ \pi \equal{} \frac {n}{m}$ for some integers $ m,n$, which is a clear contradiction, since $ \pi$ is irrational.

Therefore, this particle will never return to its starting point.

Thanks Zuton,

Nice counterexample. Seems I came to the right place after all ;)

[hide="A single particle"] The motion of a single particle is straightforward.  Given dimensions $ m \times n$, consider the lattice $ L : \{ (ma, nb) | a, b \in \mathbb{Z} \}$ and suppose that our particle starts at the origin traveling on a line with slope $ s$.  Extend the path of the particle until it hits a point in $ L$.

$ L$ divides up the real plane into rectangles.  Note that every time the side of a rectangle is hit, the particle behaves as if it had bounced off the side of a box of dimensions $ m \times n$ (except that its motion continues in another rectangle).  There is therefore an exact correspondence between the path of the particle through this lattice and the path of the particle bouncing around a box.

When will the particle hit a point in $ L$, if ever?  Well, the particle has coordinates $ (x, sx)$ at any time, so we want to know when it is possible that

$ (x, sx) \equal{} (ma, nb) \Leftrightarrow$
$ s \equal{} \frac{nb}{ma} \equal{} \frac{n}{m} \cdot \frac{b}{a}$.

In other words, the ratio of $ s$ and $ \frac{n}{m}$ needs to be rational.  This constraint is called [b]commensurability.[/b]  In the example The Zuton Force gave, $ s \equal{} 1, \frac{n}{m} \equal{} \pi$, which are not commensurable.  If $ m, n$ are integers, the condition is that $ s$ needs to be rational. [/hide]
I think we can best phrase your question as such (for simplicity):

[i]Given $ n$ particles of equal (negligible) mass starting from a single point in the corner of a box with integer dimensions and (initially) traveling along lines of rational slope with the same velocity, will each particle eventually return to the initial corner?[/i]

Note that I have added two constraints:  equal mass and equal velocity.  The problem is hard enough as is, in my opinion; in particular, it's no longer possible to predict the path of any particular particle.


superior algebra
superior algebra solved

Give an example of a nonabelian group of order 16 whose center is not cyclic.

That's very easy: $Q_{8}\times\mathbb{Z}_{2}$ is such a group, because the center is $\{\pm 1\}\times\mathbb{Z}_{2}\cong V_{4}$