inequalities
linear algebra
inequalities unsolved

Right in this forum, mathlinkers describe their proposals as beautiful, nice, ugly, or wonderful,....

It is quite common for us to have different sense of the beauty of inequalities in particular. And sometimes, people have bitter argument about what is supposed to be nice.

Now, HAVE YOUR SAY here. Please let us know your opinion. Give the inequality that you think is the most beautiful posted in 2006 on Mathlinks.

Thank you.

Maybe this topic will make more argument increase, Thuan. I think firstly we should ask that,
Do you like to use "superior mathematics" (although its idea) to solve inequalities?
There are two tendacies of that fact. One part is to agree but one part is not. I my seft don't agree, because if that, almost problems are lost their essential beauties.

The answer is clear. Most people here appreciate solutions with minimum or no use of the so-called superior mathematics. High school mathematics requires a lot of techniques and tricks.

If you talk to some professional mathematicians at universities and insist their view on what you are after. You will have your answer for yourself.

You also don't understand me, Thuan, I suppose. I love elementary knowledge and hate anything from superior mathematics which uses for elementary inequalities. I'm extreme that, but I love inequalities, so I want to protect it, at least its pristine beauty.

[quote="harazi"]I promised never posting something on this forum, but I will make an exception now. For our dear elementary master hungktn, I would like to express my total disgust for this completely childish and arogant way of seeing mathematics. Those who do not know the unity of mathematics are born with this special feature of making publicity of their books on mathlinks. I just could not imagine that this forum will become a market of selling vietnamese books with "elementary" and sacred mathematics and of endless discussions which are completely void. Hungktn, look at those superiour mathematics and you will see the real power and beauty of mathematics. If you are incapable of doing anything else than endless and completely useless computations, so be it, but do not force the others sharing your sick conceptions of mathematics. You don't like mathematics and sincerely, mathematics doesn't like you either. Please, continue these void discussions as long as you like. It's a pity we are so unfortunate on mathlinks to be obliged to assist as such disgusting ideas and conversations. I apologise for this hard and probably impolite words, but not to you, rather to the other users.[/quote]

Dear Harazi, thank u for ragard, but maybe u dont understand my opinion. I love maths, not only elementary but also superior. But especially in inequalities, I hate to use "superior mathematics" to solve them, because if that, almost of them are insignificant. Exept of inequalities, I still learn superior mathematics. But... oh, only in elementary inequalities, for me, at least for me, it's not acceptable to use superior mathematics to solve.

About the book, I never have idea to market it in Mathlinks. Untill I post enough 1000 posts, I will leave and train the superior mathematics and E, also say good bye to everyone.

Do you understand me? Harazi?

Hello! What an interesting and informative discussion! Actually, I didn't plan to reply here, as finding a place to live in Munich where I am starting to study next week is a somewhat more urgent thing than proving inequalities (I am not saying it is a more important thing in general, but the one week left yields a certain situation at the moment), but I got somewhat annoyed by the fact that each day I wake up, click on "View posts since last visit" and see about 5 topics in the inequalities forum containing insults between Hungkhtn and Pvthuan (I hope I have spelt both equally incorrect now so nobody feels favoured). It seems that assertions of the kind "this is my last post on this forum, have fun" have become quite the fashion in recent times, and based on the above I can understand the deeper motives behind them besides having a possibly round post count. On the other hand, judging by the recent posts of some people, I am wondering if it wouldn't better if they would have already quit some time before.

Some words to Harazi: As you have been out of this forum for very long, you are misunderstanding the whole matter. It's not about the means of solutions being elementary or not. If it is about mathematics at all, then it is about solutions being nice or ugly (in fact, most of the non-elementary methods referred to - the methods generally referred to by three-letter abbreviations - tend to become pretty ugly when applied to particular inequalities, but that's not a general rule, and Muirhead, being elementary, is ugly in its applications as well). But actually it is much more about two Vietnamese groups publishing books about inequalities simultaneously and trying to denounce each other's work. I don't know who of the two started that but I wish the discussion to return to mathematical grounds as soon as possible. I am just a contributor to Pvthuan's project, providing some problems and solutions, and I don't know much about the situation in Vietnam, so please excuse me if I am misjudging something.

Finally let's try out something new and contribute to the topic. As 2006 is not over yet, here is my proposal:

For any three positive reals a, b, c satisfying abc = 1, we have

$a+b+c\geq 3\left(\frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2a}\right)$.

I have an beautiful proof using WTF. Of course, it is trivial using OMG, but this is advanced mathematics and thus the proof stinks. Don't ask me what this all means, it's top secret and will be published on my gravestone.

[Jokes aside, I don't have a nice proof of the above inequality. I just know by computer algebra (completely elementary and absolutely disgusting) that it is correct. I look forward to a constructive discussion about it.]

  darij

I don't believe that everything now is becoming so bad as that. There's something not good happening after the time my book is published in Vietnam. Yes, afer my book which I wrote a long, there are some groups of people "producing inequalities" and sell, I'm very sad because the book cause a bad antecedent as that, because sometime there's some people say that "Ah, I have a very strong method, method and method..., I dont regard anybody :( ". A bad fact :(.  I'm really sad because of that. I want to quite inequality, but reviewing postes in Mathlinks and also other website in Vietnam, I can not suffer. The tendancy of inequalities in Vietnam now is very bad, very bad... I feel very blameful about it. This is the reason why I become aggressive with almost Vietnamese members who producing book in this forum. I hate them, who are organizing groups producing inequalities use maths for marketing and... money :(. So bad!!!

Thuan and I have some argument, because I hate to solve problems of him, he creates it then leaving, I dont think that before posting his problems, he had spend some minutes about it, for me that's not responsible. That's not relating to books. So many problems're false. It's all the reason why I think solving his problem is tired. In real life, he and I are good friends, at least I think that. So it's better if we dont tell anymore about book and book here, Thuan? Dont make abroad people think Vietnamese not good? Okie?

Some word to Dariji. Hi my friend, for a longtime I dont see you in inequality forum, so It's pittiful because we haven't any oppotunity to discuss about maths. I see you in IMO, but maybe you dont know  :blush: . Dont think bad about me, I'm only the person who is faithful in inequality, or exactly, its beauty. Now, I will leave here and maybe I wont return in the future, good luck and take care.

Bye.  :(

Hungkhtn, there is nothing I have against you and your book, and the same holds for Thuan, but I was just upset that in the last time, the two of you turned a number topics here in the Inequalities forum into pretty pointless discussions about beauty and elementariness. I don't dismiss everything which is not mathematics, but these discussions were really of no use or interest to anyone, they always followed the same pattern (one of you calling either the inequalities or the proofs of the other bad because ugly or non-elementary).

I am also faithful about the beauty of inequalities, and, though I am mostly working on linear algebra right now, I am taking some time to find new ideas and ways to prove inequalities. However, I don't tend to criticize others' proofs because I don't like them - a proof is always better than no proof. I hope you won't do it as well, and some other people, particularly on the Inequalities forum either (unfortunately, the Inequalities section has become home of spammers insulting others for ugly problems or long proofs and not posting anything constructive on their own; remember this was one of the reasons Harazi left).

As I said, there is nothing personal in my posts here. On the contrary, I would be glad to see your posts here, and same holds for Thuan, I only wanted people to end these unnecessary discussions.

  darij

[quote="hungkhtn"]I love elementary knowledge and hate anything from superior mathematics which uses for elementary inequalities.
.[/quote]
It is also my view. It seems that you did not understand my English. Sorry for my bad English.

Forget all the rude words. We all love inequalities and share your views of what make an inequality beautiful.

I think the following ineq is also nice[and strong]: \[ 2\cdot\sum_{cyc}\left(\frac{b}{a+b}\right)^{2}\geq\sum_{cyc}\frac{a}{a+b}\] ,for positives a,b,c

[quote="andyciup"]I think the following ineq is also nice[and strong]: \[2\cdot\sum_{cyc}\left(\frac{b}{a+b}\right)^{2}\geq\sum_{cyc}\frac{a}{a+b}\] ,for positives a,b,c[/quote]

After we expand, it remains us to apply a simple Schur for $ab,bc,ca$ and then $AM-GM$, i will tell you more on ym.

You can discuss the problem[and write the solution] in the problem
[hide="thread"]http://www.mathlinks.ro/Forum/viewtopic.php?t=108710[\hide][/hide]
Also,I think it can`t be done by Schur,as you say,since it is not symetric :P


Can someone provide a list of math competitions where one can send his/her solutions through via e-mail or by post? Thanks very much!

What about problems in gazette/journal where one could send solutions in?


algorithm
geometric series
real analysis
real analysis unsolved

[size=150]given any real number x, is there a subsequence of the harmonic sequence such that ∑an converges to x[/size]

i'm not even sure where to start since there are both rational (like Geometric Series) and irrational (like e) numbers for which there is a subsequence whose series converges to x... could someone answer in yes/ no and/or give hints on where to start

Greedy algorithm.

Of course, you can't get any negative sums.

sorry, this doesn't help :blush:  (deleted)

This is true of precisely the positive reals (or the non-negative reals if you allow the empty subsequence).  The algorithm jmerry is suggesting is as follows:

1.  Start with an empty subsequence $ S$ and look at the first term.  If it's less than $ x$, add it to $ S$; otherwise, go to the next term.
2.  If the current term $ t$ satisfies $ \sum_{s \in S} s \plus{} t \le x$, add it to $ S$.  If equality is achieved, we are done; otherwise, go to the next term.
3.  Repeat.

Since the harmonic series is unbounded and since the terms in the harmonic series go to zero, any positive $ x$ can be reached and approximated to arbitrary precision.

is it rigorous?
how do we prove for every e>0 there's an N such that n>N --> |x - sn| < e?
i tried to prove it by contradiction, but doesn't seem to work, could someone else do it?

The key point is that divergence of a series is "tail" behavior - it's unchanged by throwing out finitely many terms.  Since you can make new terms arbitrarily small, the only way for the algorithm to fail is if the remaining terms aren't big enough to reach $ x$, which never occurs because $ \sum_{k > N} \frac{1}{k}$ diverges for every $ N$.

That's how we get close.

Once we get past the initial stretch of including every $ \frac1k$, each term covers most of what's left, and the error decreases (faster than) geometrically.


algebra
linear equation
linear algebra
superior algebra
superior algebra unsolved

For $p$ an odd prime number,  let $m_a$ the number of solutions of the linear equation $x_1+x_2+x_3+x_4=a$, where $x_1,\dots, x_4\in\mathbb F_p^{\times}$ (the set of non-zero (invertible) elements of $\mathbb F_p=\mathbb Z/p\mathbb Z$).

Prove that $m_0\geq m_1$.

eum... if it is linear, why isn't this in linear algebra?
[quote="cimabue"]Let $m_a$ the number of solutions of the linear equation $x_1+x_2+x_3+x_4=a$, where $x_1,\dots, x_4\in\mathbb F_p^{\times}$. Prove that $m_0\geq m_1$.[/quote]
And isn't $m_1=m_0=p^3$? Or am I misgetting something?  $\mathbb F_p^{\times}=\{0,1,...,p-1\}$ right?

I think that's usually written $\mathbb{F}_p^*$, which is the set of nonzero elements of the field. That makes this problem slightly nontrivial.


Hello!

I'm writing a paper on logarithmic spirals and am not a math major.
I need to write about graphing logarithmic spirals on a graphing calculator
and need a few examples of realistic values for a and k, given the formula
for a logarithmic spiral:  r=ae^k*theta

Can any of you math geniuses help me out? I need to express the formulas
 in the regular form (as above) and in the form of a natural log.

Thanks in advance!

ann-marie

http://mathworld.wolfram.com/LogarithmicSpiral.html

--------------------------------------------------------------------

Mice and logarithmic spiral

http://mathworld.wolfram.com/MiceProblem.html

:)

HELLO!

THANKS SO MUCH FOR THE LINKS RE: LOG SPIRALS. THEY
WERE VERY HELPFUL.

ANNMARIE


Let $n$ be positive integers.Find the value of $n$ for which $n,n+2$ and $n+4$ are all prime numbers.

Very easy.

If n = 3, then 3, 5, 7 are primes.
If n > 3,
(1) n = 3k, then n is not prime.
(2) n = 3k + 1, then n + 2 = 3k + 3 is not prime.
(3) n = 3k + 2, then n + 4 = 3k + 6 is not prime.

So the only solution is n = 3.


induction
function
limit
algebra unsolved
algebra

[quote="LideR_M"][hide="hint"]Induction and grapf of $ f(x)=\frac{x^{4}+9}{10x}$[/hide][/quote]
Easy to say. So, post your solution.  :wink:

[quote="Yuriy Solovyov"]Let $ x_{n+1}=\frac{x_{n}^{4}+9}{10x_{n}}, x_{1}=2.$
Prove, that $ \frac{465}{511}<x_{n}\leq{\frac{5}{4}}$ for all $ n>1.$[/quote]
I know this isn't a 'cute' solution, but I havent got idea for better.
[hide="Solution"]
Let 1st prove the left inequality.
We've got $ x_{1}>\frac{465}{511}$, and now:
$ x_{n+1}=\frac{x_{n}^{4}+9}{10x_{n}}$
And now we consider the function $ f(x)=\frac{x^{3}}{10}+\frac{9}{10x}$, hence $ f'(x)=\frac{3}{10}x^{2}-\frac{9}{10}\frac{1}{x^{2}}$, hence $ f'(x)=0$ for $ x=\pm\sqrt[4]{3}$ and $ f'(x)>0$ for $ x>\sqrt[4]{3}$ and $ f'(x)<0$ for $ x<\sqrt[4]{3}$. We have $ \lim_{x\to 0^{+}}f(x)=+\infty$ and we can show that $ f\left(\sqrt[4]{3}\right)>\frac{465}{511}$ (with calculator... i havent gor idea for show that quick in other way), so for $ x>\frac{465}{511}$ we have $ f(x)>\frac{465}{511}$, hence $ x_{n+1}>\frac{465}{511}$

And now the right inequality... this same way... and we should show that $ f\left(\frac{465}{511}\right)\leq\frac{5}{4}$ and $ \sqrt[4]{3}>\frac{5}{4}$[/hide]


[quote="arqady"][quote="M.A"]hello. here is the inequality
 prove that if a,b,c>0  and a+b+c=1 then
 ab/(3+b+2c) + bc/(3+c+2a) +ca/(3+a+2b) <=1/12
 :lol:[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=89774[/quote]

thank you arqady
and i have a question where can i find a lot of inequalities by solutions  :huh:

[quote="arqady"][quote="M.A"]
and i have a question where can i find a lot of inequalities by solutions  :huh:[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100399
By my opinion it's a very good book. :wink:[/quote]

 thanks for your suggestion. i really want to buy this book but it isnt in my country :(


calculus
integration
trigonometry
inequalities
calculus computations

show that $ \int_{1}^{\infty}x^{ \minus{} 1}|\sin{x}|dx$ diverges

my idea:  I think you have to show that there is a constant, $ c > 0$ such that $ \int_{n\pi}^{(n \plus{} 1)\pi}x^{ \minus{} 1}|\sin{x}|dx > c\int_{n\pi}^{(n \plus{} 1)\pi}x^{ \minus{} 1}dx$ for all $ n\geq 1$. but i dont know how to show this :maybe:

That could work, but it's probably simpler to switch the roles: $ \int_{n\pi}^{(n\plus{}1)\pi}\frac1x|\sin x|\,dx\ge \frac1{(n\plus{}1)\pi}\left|\int_{n\pi}^{(n\plus{}1)\pi}\sin x\,dx\right|\equal{}\frac{2}{(n\plus{}1)\pi}$.
The comparisons are easier when we're estimating the decreasing functions. The maximum possible value of your $ c$ would come from the $ n\equal{}1$ case, and it's not something that can be calculated exactly.

The sum of those terms is a constant multiple of the harmonic series, and thus it diverges.

could  you explain briefly how you got that inequality?

Another way to show this is to show $ \int_1^\infty \frac{\cos 2x}{x}\ dx$ converges (integration by parts helps here) and conclude that $ \int_1^\infty \frac{\sin^2 x}{x}\ dx$ diverges.

$ \frac1x$ is decreasing, so it's greater than its value at the right endpoint.

[img]http://www.math.tamu.edu/~lkovalev/files/maple-sin.jpg[/img]

This has been a public shaming of Maple 11. Thank you.

Does it do any better if you use definite integral notation and ask it to plot

$ \int_0^x\left|\frac{\sin t}{t}\right|\,dt\ ?$

Yes, it does - which is why I did not use that approach  :)
It's just that it thinks that $ \mathrm{sign}(f)\int^x f$ is a symbolic antiderivative of $ |f|$. Which is true locally, but not globally.


If $ a,b,c$ are positive numbers and $ a \plus{} b \plus{} c \equal{} 3$, prove that
$ \sum_{cyc}\frac {a \plus{} b}{9 \minus{} c^{2}}\ge\frac {3}{4}$

it's easy,
  inequality  is  equavilent to :
$ \frac{1}{3\plus{}a}\plus{}\frac{1}{3\plus{}b} \plus{} \frac{1}{3\plus{}c} \geq \frac{3}{4} \\
 \Leftrightarrow 5(ab\plus{}bc\plus{}ca) \plus{} 3abc  \leq 18$
On the  other hand , we have :  $ ab\plus{} bc\plus{}ca  \leq \frac{(a\plus{}b\plus{}c)^2}{3}\equal{}3$and $ abc  \leq \frac{(a\plus{}b\plus{}c)^3}{27} \equal{}1$

You can handle it even easier.

The inequality is equivalent to
$ \frac{1}{3\plus{}a}\plus{}\frac{1}{3\plus{}b} \plus{} \frac{1}{3\plus{}c} \geq \frac{3}{4}$ and by AM-HM

$ \frac{1}{3\plus{}a}\plus{}\frac{1}{3\plus{}b} \plus{} \frac{1}{3\plus{}c} \geq \frac{9}{3\plus{}a\plus{}3\plus{}b\plus{}3\plus{}c}\equal{}\frac{9}{12}\equal{}\frac{3}{4}$.

[quote="SAPOSTO"]Nice solutions [b]Hong Quy[/b] and [b]mornik[/b]! The inequality is really very easy but not so much if you don't know Caushy-Schwarz, you will need to do something like in [b]Hong Quy[/b]'s solution.[/quote]

I don't see where either of us used Cauchy-Schwarz (I used AM-HM and Hong Quy, as I see, used AM-GM).  :maybe: Although I see my solution would be equivalent to using Cauchy in Engel form.

What is this "Engel form"? I talk about Caushy-Schwarz used in the following form: $ \sum_{i \equal{} 1}^{n}\frac {a_{i}^{2}}{b_{i}}\ge \frac {(\sum_{i\equal{}1}^{n}a_{i})^{2}}{\sum_{i \equal{} 1}^{n}b_{i}}$. Hm, maybe this is the "Engel form" because I see you are using what I wrote and you call this "Engel form". Sorry, I didn't know it is called so.  :blush:

Yes, this is Cauchy in Engel form, however, that was not what I had in mind.
I just used

$ \frac{r_1\plus{}r_2\plus{}r_3}{3}\geq \frac{3}{\frac{1}{r_1}\plus{}\frac{1}{r_2}\plus{}\frac{1}{r_3}}$. This is AM-HM (arithmetic mean - harmonic mean) inequality which is equivalent to Cauchy in Engel form for $ a_i\equal{}1$ ($ r_1\equal{}\frac{1}{3\plus{}a}$, $ b_1\equal{}3\plus{}a$ and so on).

OK! I didn't know that $ AM\ge HM$ comes from Caushy-Schwarz.

We only use expansion for other inequalities, not in this case. I think Hong Qui has used Cauchy in his solution and Cauchy- Schwartz is a consequence of Cauchy so there is no argument here.


Let $a$, $b$ and $c$ be positive reals. Show that \[ \left(\frac{2a}{b + c}\right)^{2/3} + \left(\frac{2b}{c + a}\right)^{2/3} + \left(\frac{2c}{a + b}\right)^{2/3} \geq 3. \]

This is an improvement of Nesbitt pretty known on ML. See

http://www.mathlinks.ro/Forum/viewtopic.php?t=12763
http://www.mathlinks.ro/Forum/viewtopic.php?t=73225
http://www.mathlinks.ro/Forum/viewtopic.php?t=30733
http://www.mathlinks.ro/Forum/viewtopic.php?t=69692

  dg


inequalities
function
algebra
polynomial
inequalities proposed

Find the largest m such that for all a,b,c,d>0 with sum 1 we have:
  m<(a^3+1)(b^3+1)(c^3+1)(d^3+1).

just work it out and you'll see that the highest possibility is with a=b=c=d

so (1+(.25)^3)^4

being 1,063980162143707275390625

but please note that you should write m \leq ... because m CAN be attained

Perhaps this easiest way (but not the smartest way) is to see that ln (1+x^3) is convexe on 0..1 and then to apply Jensen

For any function f which is convex on an interval I,
for any positive reals a1, a2, ..., an with sum 1,
for any positive reals x1, x2, ..., xn in I,

\[\sum a_if(x_i) \geq f(\sum a_ix_i).\]

This is my favourite inequality, it's very powerful:
most well known inequalities are an immediate consequence of Jensen.

except for am-gm, which comes from applying jensen to log function, what other inequalities can be derived from jensen? how would you prove maclaurin?

Find also the smallest M such that for all a,b,c,d>0 with sum 1 we have (a^3+1)(b^3+1)(c^3+1)(d^3+1)<M.

Unsolved! Come on guys, this can't be that hard? :D

$(a^3+1)(b^3+1)\le(a+b)^3+1\Leftrightarrow3(a+b)\ge a^2b^2$. Note that $a+b\le1$, thus $3(a+b)\ge3(a+b)^4\ge48a^2b^2$. Therefore the max. is attained when one of a,b,c,d equals 1, and other others 0.


[color=brown]I made up this riddle when it happened to me.
The [u]answer[/u] to (1) surprised me, because it is very simple.
I found a [u]solution[/u], but I search for a solution, more elegant (the trivial way)[/color]

[u]The Problem[/u]
John is driving a car from city A to city B. At the beginning of his trip, the fuel tank is completely full. He should deliver the car at the city B, with a full tank.

Fortunately, there is a filling station in every point in $[A,B]$.
Unfortunately, the price of the petrol is increasing: It is given by the formula $P(x) = P_{0}-bx$, where $x$ is the distance from the city B.

The fuel consumption of the car is constant and it is $a\ lt/km$.

Solve one of the following problems:
a) If only $n$ fillings are allowed, which are the optimal points (of the filling)?
b) Prove that, it we increase $n$, then we can achieve a better (lower) cost
c) Find the marginal cost (infinite fillings) in terms of $a,b,P_{0}$ and the distance $AB$

[color=green][u]Remarks[/u]
- We assume that, the petrol that the fuel tank can hold, is enough to cover the distance from city A to city B
- Of course, once he stop at a filling station, it is better to [u]fill up[/u] (the petrol is more expensive at the following filling stations)
- Certainly, he should fill up at the city B. If $n=1$, it means that he cannot fill anywhere else
- $P(x)>0$, but it could be zero at $A$ - that would not be a problem
- Sorry for my English[/color]

[color=blue][u]Explanation[/u]
Of course, John wants to fill up at the very first stations, because they have cheap petrol. The problem is that he has not enough free space in the fuel tank.[/color]

[hide="Hint"]
Start with $n=1,2,...$

[hide="Hint 2"]Try to give some numbers to the problem, for example $AB=100km$, $P(x)=1-0.01x$, $a=0.1\ lt/ km$[/hide][/hide]

[hide] when he gets to the first gas station, he puts his remaining gas into a seperate tank, then he fills his tank up from the station, when he is at city B, he puts the gas from the tank into the car, and it should be just right.

Visualize it as a graph... then the solution should become obvious. Basically, to chart the money you pay, you have a figure that looks like n steps that must be above some line that looks like x+y = c, for some c. Then obviously you must pay the area below the line, but you then have n isosceles triangles whose bases sum to a fixed length. The problem then basically becomes to optimize $\sum x^{2}$ given fixed $\sum x$, which is easy.

So, you mean that $x_{1}=x_{2}=\cdots=x_{n}=\frac{d}{n}$, where $d$ is the distance $AB$.

That was my answer too. Could you please post more details?

By the way, in my solution I used recursion on $n$


Use this thread to discuss problems regarding Thermodynamics, including those posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185862]here[/url].

[hide="SORRY CARCUL"]a
the work done will be $ \int_{V_{1}}^{V_{2}} PdV \equal{} nRT \ln \frac {V_{2}}{V_{1}}$
b)now in adiabatic process final pressure $ P_{ad} \equal{} (\frac {V_{1}}{V_{2}})^{\gamma}P_{i}$ since
also for isothermal $ P_{is} \equal{} \frac {P_{1}V_{1}}{V_{2}}$ since $ V_{1} < V_{2}$ and $ \gamma > 1!$ we have $ P_{ad} < P_{is}$[/hide]

[hide="Problem 2"]a) because $ \Delta U \equal{} nC_{v}\Delta T$ and $ \Delta H \equal{} nC_{p}\Delta T$
b)because $ C_{p} \equal{} \frac {\gamma R}{\gamma \minus{} 1}$[/hide]

[hide="problem 3"]$ \begin{array}{ccccccc} \mathrm{No.} & \mathrm{q} & \mathrm{w} & \mathrm{\Delta U} & \mathrm{\Delta H} & \mathrm{\Delta S} &{ \mathrm\Delta S_{\mathrm{univ}}} \\
a) & + & + & + & + & - & 0 \\
b) & + & + & + & + & - & 0 \\
c) & 0 & - & - & - & 0 & 0 \\
d) & + & - & 0 & 0 & + & - \end{array}$ will post rest later(if no one posts)[/hide]

Pardesi, how about respect the others and hide your answers? Wouldn't that be nice?

Also, your answer for 1. (a) and for 2 are wrong.

[hide]18. The first two are essentially definitions of specific heats, namely $ \frac {1}{T} C_P, \frac {1}{T} C_V$. Next, $ dH \equal{} T dS \plus{} V dP$ and $ dS \equal{} \frac {1}{T} C_P dT \minus{} \alpha V dP$, so $ dH \equal{} \frac {1}{T} C_P T dT \minus{} \alpha VT dP \plus{} V dP$ and hence the desired expression is $ V(1 \minus{} \alpha T)$.

19. We start with $ dV \equal{} \minus{} \kappa V dP \plus{} \alpha V dT$, so $ dU \equal{} T dS \minus{} P dV \equal{}$ $ T (\frac {1}{T} C_P dT \minus{} \alpha V dP) \minus{} PV (\alpha dT \minus{} \kappa dP)$ $ \equal{}$ $ (C_P \minus{} PV \alpha) dT \plus{} (\kappa PV \minus{} \alpha TV) dP$, which yields the desired relation.

20. This also follows from the above.

21. $ dH \equal{} T dS \plus{} V dP \equal{} T (\frac {1}{T} C_V dT \plus{} \Big(\frac {\partial P}{\partial T}\Big)_V dV ) \plus{} V ( \Big(\frac {\partial P}{\partial T}\Big)_V dT \plus{} \Big(\frac {\partial P}{\partial V}\Big)_T dV )$. Hence the desired expression is $ T \Big(\frac {\partial P}{\partial T}\Big)_V \plus{} V \Big(\frac {\partial P}{\partial V}\Big)_T$. We simplify this by applying the Euler chain rule for the first term and the coefficients of volume change: $ T \alpha V \frac {1}{\kappa V} \minus{} V \frac {1}{\kappa V} \equal{} \frac {\alpha T \minus{} 1}{\kappa}$.

22. $ dU \equal{} T dS \minus{} P dV \equal{} T (\frac {1}{T} C_V dT \plus{} \Big(\frac {\partial P}{\partial T}\Big)_V dV) \minus{} P dV \equal{} C_V dT \plus{} \Big(\frac {\alpha T}{\kappa} \minus{} P \Big) dV$, so $ \Big(\frac {\partial T}{\partial V}\Big)_U \equal{} \minus{} \Big(\frac {\partial T}{\partial U}\Big)_V \Big(\frac {\partial U}{\partial V}\Big)_T \equal{} \frac {1}{C_V} \Big( P \minus{} \frac {\alpha T}{\kappa} \Big)$

23. $ dH \equal{} T dS \plus{} V dP \equal{} T (\frac {1}{T} C_P dT \minus{} \alpha V dP) \plus{} V dP$ so $ \Big(\frac {\partial T}{\partial P}\Big)_H \equal{} \minus{} \Big(\frac {\partial T}{\partial H}\Big)_P \Big(\frac {\partial H}{\partial P}\Big)_T \equal{} \frac {V(\alpha T \minus{} 1)}{C_P}$.

24. $ d(G/T) \equal{} \frac {1}{T^2} (T dG \minus{} G dT) \equal{} \frac {1}{T^2} (T ( \minus{} S dT \plus{} V dP) \minus{} G dT)$, so the desired value is $ \frac { \minus{} G \minus{} TS}{T^2} \equal{} \minus{} \frac {H}{T^2}$.

25. Note $ dV \equal{} \alpha V dT \minus{} \kappa V dP$ so $ \frac {1}{T} C_P dT \plus{} \Big(\frac {\partial S}{\partial P}\Big)_T dP \equal{} dS$ $ \equal{}$ $ \frac {1}{T} C_V dT \plus{} \Big(\frac {\partial S}{\partial V}\Big)_T dV$, so $ \minus{} \frac {1}{T} (C_P \minus{} C_V) dT \equal{} \Big(\frac {\partial S}{\partial P}\Big)_T dP \minus{} \Big(\frac {\partial S}{\partial V}\Big)_T dV$ $ \equal{}$ $ \Big(\frac {\partial S}{\partial P}\Big)_T dP \minus{} \Big(\frac {\partial S}{\partial V}\Big)_T (\alpha V dT \minus{} \kappa V dP)$ and $ \frac {1}{T} (C_P \minus{} C_V) \equal{} \minus{} \Big(\frac {\partial S}{\partial V}\Big)_T \alpha V \equal{} \minus{} \Big(\frac {\partial P}{\partial T}\Big)_V \alpha V \equal{} \minus{} \frac {\alpha^2 V}{\kappa}$.

26. $ \Big(\frac {\partial}{\partial P}\Big)_T \Big(\frac {\partial S}{\partial T}\Big)_P \equal{} \Big(\frac {\partial}{\partial T}\Big)_P \Big(\frac {\partial S}{\partial P}\Big)_T \equal{} \minus{} \Big(\frac {\partial}{\partial T}\Big)_P \Big(\frac {\partial V}{\partial T}\Big)_P \equal{} \minus{} \Big(\frac {\partial^2 V}{\partial T^2}\Big)_P$; the $ T$ may be moved outside the first differentiation.[/hide]

At first sight those seem to be right: after, they were mainly mathematical problems. Also, hide your answers please.

[hide]27. same as 26

30. In an analogous manner to 25, we write $ \Big(\frac{\partial P}{\partial V}\Big)_S \equal{} \Big(\frac{\partial P}{\partial V}\Big)_T \plus{} \Big(\frac{\partial P}{\partial T}\Big)_V\Big(\frac{\partial T}{\partial V}\Big)_S$ so applying the Maxwell relation and subtracting one from both sides of the equation, it suffices to prove $ \kappa V \Big(\frac{\partial P}{\partial T}\Big)_V \Big(\frac{\partial P}{\partial S}\Big)_V \equal{} \frac{1}{C_V} \frac{TV\alpha^2}{\kappa}$ which is true as $ \frac{1}{T} C_V \Big(\frac{\partial P}{\partial S}\Big)_V \equal{} \Big(\frac{\partial P}{\partial T}\Big)_V$ which is in turn equal to $ \frac{\alpha}{\kappa}$.

31. Hm having trouble differentiating the specific heats with respect to these other variables. Need to think about that.

36. [/hide]


AMC
USA(J)MO
USAMO
function
ratio
algebra
functional equation

I'm wondering if you all can help me with USAMO-style functional equations (things like find all functions f which map the integers to the integers such that f(m+f(n))=f(m)+n). I'm currently at MOP because I'm a freshman and we had a class on them. I understood the solutions but I don't think I would be able to come up with them. So please post general strategies for solving them or some sample problems. Thanks.

[quote]Dont mind me asking, but what is the interest?[/quote]
I'm sure he meant integers.

In general, experiment with functional equations. Look for simple solutions, and always look for what happens when you substitute simple values like 0 and 1.

Using your problem f(m+f(n))=f(m)+n as an exxample:
Try m=0: f(f(n)) = f(0) + n     (1)
Try n=0: f(m+f(0)) = f(m)     (2)
The equation (2) suggests that f(0) = 0, which would also simplify (1).
Trying another example, f(m+f(1)) = f(m) + 1    (3).
By (2) applied repeatedly, f(m + k*f(0)) = f(m)
By (3) applied repeatedly, f(m + k*f(1)) = f(m) + k.
Therefore, 0 = f(m + f(1)*f(0)) - f(m) = f(0)
Now (1) becomes f(f(n)) = n,
so f(m + n) = f(m + f(f(n))) = f(m) + f(n)
and f must be linear: f(n) = n*f(1), and f(f(n)) = n*f(1)*f(1)
This forces f(1) =  :pm: 1 and either
f(n) = n   or    f(n) = -n

[quote="jpark"]i'm just wondering..
You said that f(m+f(0)) = f(m) implies that f(0)=0, but is there a chance that f(x) is a periodic function?[/quote]

He said that f(m+f(0)) = f(m) suggests that f(0) = 0, he didn't say "implies". Then he proved several lines later.

It's a nice solution!

Thanks for all this.

Sorry about "interest". I edited that.

Do any of you know of any problems similar in difficulty to the one I posted so I could try it? They gave me a problem set here but the difficulty of the problems in it seems to increase exponentially :-).

How about this problem?

Find all functions f that satisfy the functional equation
[tex]\displaymath
f(x) + f(\frac{1}{1-x}) = x
[/tex]
for every x not equal to 0 or 1.

Notes on Ravi's problem:

Hint 1:[hide] The mapping that sends x to 1/(1-x) is periodic of period 3.[/hide]

Hint 2:[hide] f(2) = 7/4; f(-1) = 1/4; f(1/2) = -5/4.[/hide]

Hint 3:[hide] You can find out what has to happen to the three points of an "orbit" under the mapping in Hint 1 by solving three linear equations in three unknowns. That's where the numbers in Hint 2 are coming from.[/hide]

Hint 4:[hide] The answer is a rational function. (That is, a ratio of polynomials.) Unsurprisingly, the denominator of the fraction has both x and 1-x as factors.[/hide]

Answer:[hide] f(x) = (-x^3 + x -1)/(2x(1-x))[/hide]

Let [tex]\mathbb{R}^{+}[/tex] be the set of all positive real numbers. Find all functions [tex] f :\mathbb{R}^{+}\rightarrow \mathbb{R}^{+}[/tex] such that [tex]f(x)f(yf(x)) = f(x + y)[/tex] for all [tex]x, y \in \mathbb{R}^{+}[/tex].

for fuzzylogic's problem:

[hide]1/(1+kx) works for all positive k. (The right side with the nested functions seems to demand something reciprocal-ish.)[/hide]

Not sure where to go from there.

[quote="Arne"]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8606[/quote]

Wow! A lot of interesting problems. Here is one not in the list (if I read your list carefully):

Let [tex]\mathbb{R}^{+}[/tex] be the set of all positive real numbers. Find all functions [tex] f :\mathbb{R}^{+}\rightarrow \mathbb{R}^{+}[/tex] such that [tex]f(f(x)+y) = xf(1+xy)[/tex] for all [tex]x, y \in \mathbb{R}^{+}[/tex].


calculus
integration
limit
logarithms
real analysis
real analysis unsolved

[quote="xirti"]Find

\[ \lim\limits _{n\rightarrow \infty}\frac{1}{n^2}\cdot \int_1^e\ln (1+x^n)\cdot \ln(1+2x^n)dx  \][/quote]
Since $\ln(1+kx^n) = n \left(\ln x + n^{-1}\ln(k+x^{-n}) \right)$, the limit becomes
$\lim_{n \to \infty} \int_1^e \left(\ln x + n^{-1}\ln(1+x^{-n})\right)\left( \ln x + n^{-1} \ln(2+x^{-n})\right) dx$
which is $\int_1^e \left( \ln x\right)^2 dx = e-2$.

It seems that it was not hrad:D Thank you!


Hello!

Let $ EFG$ be a triangle such that $ \angle EFG>90^\circ$, $ EG\equal{}r>0$ and $ FG\equal{}p>0$. Let $ I\in [EG]$ such that $ \angle IEF\equal{}\angle IFG$. Let $ H\in FI$ on the opposite side of $ EG$ from $ F$ such that $ EH\equal{}q>0$ where $ q^2\equal{}r^2\minus{}p^2$. Prove that $ \angle EHI\equal{}\angle HGI$. :)

[quote="csbonny"]I think EFG < 90 is ok[/quote] No! $ 90^\circ<\angle EFG<180^\circ$.

[quote="yuyang"]No ideas  :o[/quote][color=purple][b]
Probable the your proposed problem is very easily ![/b][/color]

[quote="Virgil Nicula"][color=darkred][b][u]An nice equivalent enunciation.[/u][/b] Let $ ABCD$ be a convex quadrilateral.

Prove that if two among the lower relations are truly, then and the another relation is truly :

$ \|\begin{array}{cc} 1\blacktriangleright & \widehat {ABD}\equiv\widehat {CAD} \\
 \\
2\blacktriangleright & \widehat {BCD}\equiv\widehat {CDB} \\
 \\
3\blacktriangleright & BD^2 = BC^2 + AD^2\end{array}$.[/color][/quote]
[color=darkblue][b]Proof.[/b] Denote the intersection $ I\in AC\cap BD$. Therefore,

$ \{\begin{array}{ccc} \widehat {ABD}\equiv\widehat {CAD} & \Longleftrightarrow & ADI\sim BDA\Longleftrightarrow AD^2 = ID\cdot BD\ . \\
 \\
\widehat {BCD}\equiv\widehat {CDB} & \Longleftrightarrow & BCI\sim BDC\Longleftrightarrow BC^2 = BI\cdot BD\ . \\
 \\
BD^2 = BC^2 + AD^2 & \Longleftrightarrow & BI\cdot BD + ID\cdot BD = AD^2 + BC^2\ .\end{array}$

[b]Proposed problem.[/b] Find the convex quad. $ ABCD$ for which $ \{\begin{array}{c} AC^2 = AB^2 + CD^2 \\
\ BD^2 = AD^2 + BC^2\end{array}$[/color]


probability
advanced fields
advanced fields unsolved

Let $(X,B,m)$ be a probability space, $T$ is a measure-preserving transformation on $X$ and $f:X\to\mathbb{R}$ is a measurable function. Prove that $\limsup \frac{1}{n}f(T^nx)=0$ or $+\infty$ almost surely.

Comments:  This one is much harder than it may seem, I think so. If $f$ is integrable, then we can use the Birkhoff Ergodic Theorem. But unfortunately, it is not.

[i]Edited by Myth[/i]

[quote="Kent Merryfield"]Shall we assume $f\ge0$ as a hypothesis?[/quote]

 Of course, in fact the $f\le0$ part is insignificant.

Sorry, Myth. No proof yet; no counterexample either.

It is a very strange problem. I thought on it for two hours three days ago...


Hölder  :P

I've just copied it from Deutsch forum  :lol:

But how to write it in LaTeX??


real analysis
limit
logarithms
real analysis unsolved

let $ a_i > 0\ ,\ i \equal{} 1,2,3,\dots$ :
$ a)$ prove $ \lim \sup _{n\rightarrow \infty} (\frac {a_1 \plus{} a_{n \plus{} 1}}{a_n})\ge e$
$ b)$ for every $ a\ge e$ find sequence $ a_n > 0$ that :$ \lim \sup _{n\rightarrow \infty} (\frac {a_1 \plus{} a_{n \plus{} 1}}{a_n}) \equal{} a$

(a) Assume the contrary. So that for $ n > N$,
\[ \frac {a_1 \plus{} a_{n \plus{} 1}}{\sqrt [n]{e} a_n} < 1,
\]
hence $ a_{n \plus{} 1} < e^{1/n} a_n \minus{} a_1$, Let $ H_n \equal{} \sum_{k \equal{} 1}^n{1/k}$, multiply $ e^{ \minus{} H_n}$ on both side
\[ e^{ \minus{} H_n} a_{n \plus{} 1} < e^{ \minus{} H_{n \minus{} 1}} a_n \minus{} e^{ \minus{} H_n} a_1 < e^{ \minus{} H_{n \minus{} 1}} a_n \minus{} \frac {1}{n \plus{} 1}a_1.
\]
Hence $ e^{ \minus{} H_n} a_{n \plus{} 1} < e^{ \minus{} H_{N \minus{} 1}} a_N \minus{} \sum_{k \equal{} N \plus{} 1}^{n \plus{} 1}{1/k}$
Since $ \sum{\frac {1}{k}} \equal{} \plus{} \infty$, left side tends to $ \minus{} \infty$ as $ n\to\infty$, since $ e^x > 0$, $ a_n < 0$, contradiction.

(b) Let $ k \equal{} \ln a$, $ a_1 \equal{} 1$, $ a_n \equal{} n / (k \plus{} 1/n \minus{} 1)$.


Two angles are complementary. The difference in their measures is $ 40^{\circ}$. What is the number of degrees in the larger angle?

hello, let $ x$ and $ y$ be the both angles with $ x \geq y$, then we get
$ x\minus{}y\equal{}40^{\circ}$
$ x\plus{}y\equal{}90^{\circ}$
solving this system we have $ x\equal{}65^{\circ}$ and $ y\equal{}25^{\circ}$.
Sonnhard.


Hello, I have a few questions about
\[ \\\limsup_{n\rightarrow\infty}E_n=\bigcap_{n=1}^{\infty}\left[\bigcup_{k=n}^{\infty}E_k\right]\\ \liminf_{n\rightarrow\infty}E_n=\bigcup_{n=1}^{\infty}\left[\bigcap_{k=n}^{\infty}E_k\right]  \]
Could you please define them, and also give some real examples from set theory so I understand the meaning better?

Thanks in advance.

Masoud Zargar

Well, this [i]is[/i] a definition. You can rewrite it as

\[ \limsup_{n\rightarrow\infty}E_n=\{x : \forall n \in \mathbb{N} ~ \exists k \in \mathbb{N}_{\geq n} : x \in E_k\} \\\\ \liminf_{n\rightarrow\infty}E_n=\{x : \exists n \in \mathbb{N} ~ \forall k \in \mathbb{N}_{\geq n} : x \in E_k\} \]
Can't you just consider an example by yourself? This is a good exercise for [i]you[/i]. See also [url=http://en.wikipedia.org/wiki/Limit_inferior#Sequences_of_sets]Wikipedia[/url].


inequalities
inequalities unsolved
algebra

Let a, b, and c denote the sides of a triangle. Show that
$a^{3}+b^{3}+c^{3}-3abc \geq 2\sqrt{9+6\sqrt{3}}(a-b)(b-c)(c-a)$.

[quote="arqady"][quote="ideahitme"]Let $a, b$ and $c$ denote the sides of a triangle. Show that
$a^{3}+b^{3}+c^{3}-3abc \geq 2\sqrt{9+6\sqrt{3}}(a-b)(b-c)(c-a)$.[/quote]
I think, $a^{3}+b^{3}+c^{3}-3abc \geq \sqrt{9+6\sqrt{3}}(a-b)(b-c)(c-a)$ is true for all non-negative $a,$ $b$ and $c.$
And the estimation $\sqrt{9+6\sqrt{3}}$ is best. :wink:
From this that is necessary follows.[/quote]

Yeah, you're right. In fact, it was the original version.
The obvious question is how to kill the general version!

[quote="ideahitme"]
The obvious question is how to kill the general version![/quote]
Very easy:
Determine the most real number M such that the inequality
$a^{3}+b^{3}+c^{3}-3abc\geq M(a-b)(b-c)(c-a)$
holds for all real non-negative numbers a, b and c. :D
Solution:
Let $M>0$ and $a=\min\{a,b,c\},$ $c\geq b$
and $b=a+x,$ $c=a+y$ for $y\geq x\geq0.$
Hence, $a^{3}+b^{3}+c^{3}-3abc\geq M(a-b)(b-c)(c-a)\Leftrightarrow$
$\Leftrightarrow(a+b+c)((a-b)^{2}+(a-c)^{2}+(b-c)^{2})\geq2M(-x)(x-y)y\Leftrightarrow$
$\Leftrightarrow(3a+x+y)(x^{2}-xy+y^{2})\geq Mxy(y-x).$
Let $y\neq0.$
If $a=0$ then $f(t)=t^{3}+Mt^{2}-Mt+1\geq0,$ where $t=\frac{x}{y}$ and $0\leq t\leq1.$
$f'(t)=3t^{2}+2Mt-M.$ $t_{min}=\frac{-M+\sqrt{M^{2}+3M}}{3}.$
$f(t_{min})\geq0\Leftrightarrow2M^{3}+9M^{2}+27\geq(2M^{2}+6M)\sqrt{M^{2}+3M}\Leftrightarrow$
$\Leftrightarrow M^{4}-18M^{2}-27\leq0\Leftrightarrow M\leq\sqrt{9+6\sqrt{3}}.$
Id est, $a^{3}+b^{3}+c^{3}-3abc\geq\sqrt{9+6\sqrt{3}}(a-b)(b-c)(c-a)$
for all non-negative $a,$ $b$ and $c.$
Equality holds for $a=0,$ $b=\frac{-M+\sqrt{M^{2}+3M}}{3}$ and $c=1,$
where $M=\sqrt{9+6\sqrt{3}}.$ :)

Using Mixing Variable completely by considering that when we reduce 3 variables above by the same quantity and assume that this quantity is the smallest(eg: c=min(a,b,c)>=0 then considering that f(a-c,b-c,0) is reduced, but M(a-b)(b-c)(c-a) is still not changed!Therefore, prove when c=0 and we just work with 2 variable a and b). We also can assume (a,b,c) belongs to the interval (a,a+1)(a>0) since they are three sides of a triangle!

[quote="ideahitme"]Let a, b, and c denote the sides of a triangle. Show that
$ a^{3} \plus{} b^{3} \plus{} c^{3} \minus{} 3abc \geq 2\sqrt {9 \plus{} 6\sqrt {3}}(a \minus{} b)(b \minus{} c)(c \minus{} a)$.[/quote]

wlog $ a\equal{}$max{$ a,b,c$}
if $ a\geq b \geq c$ then its obvious since the left side is positive and the right is negative

so we need to prove it only in case $ a\geq c \geq b$

Let $a,b,c>0$ . [url=https://artofproblemsolving.com/community/c6h625144p7487740]Show that[/url]
$$a^3+b^3+c^3 \ge 3abc+\sqrt{9+6\sqrt3}|(a-b)(b-c)(c-a)|$$


Find all functions $ f,g: \mathbb{N} \to \mathbb{N}$ such that for any $ n \in \mathbb{N}$:
1) $ f^{g(n)}(n)\equal{}n$ ($ f^m$ stands for combining the function $ f$, $ m$ times with itself)
2) $ f(n)g(n)\equal{}n^2$

substitute f(n)=n^2/g(n) to the first equation. Divide into two cases: g(n) is even or g(n) is odd. Both leads to f(n)=g(n)=n for all n.

Edit: the above solution is wrong. New solution (outline): f is injective. Take the smallest a for which f(a) is not a. So f(a)>a, find a contradiction.

Good idea! But how to get the contradiction?

as i think about it again, it turned out that my solution was wrong. how do you solve this pmvald?

[quote="pmvald"]Find all functions $ f,g: \mathbb{N} \to \mathbb{N}$ such that for any $ n \in \mathbb{N}$:
1) $ f^{g(n)}(n) \equal{} n$ ($ f^m$ stands for combining the function $ f$, $ m$ times with itself)
2) $ f(n)g(n) \equal{} n^2$[/quote]

I would be very interested too in your general solution, pmvald, since I found infinitely many solutions but have some difficulty to give them a general form. For example :

1) identity
$ f(n) \equal{} g(n) \equal{} n$

2) nearly identity
$ f(n) \equal{} g(n) \equal{} n$ $ \forall n\notin\{6,18\}$
$ f(6) \equal{} 18$ and $ g(6) \equal{} 2$,
$ f(18) \equal{} 6$ and $ g(18) \equal{} 54$

3) previous case extended :
$ \forall a$ not perfect square :
$ f(2a) \equal{} 2a^2$ and $ g(2a) \equal{} 2$
$ f(2a^2) \equal{} 2a$ and $ g(2a^2) \equal{} 2a^3$

$ f(n) \equal{} g(n) \equal{} n$ for all other values.

4) with some odd $ g(n)\ne n$ :
$ f(n) \equal{} g(n) \equal{} n$ $ \forall n\notin\{45,135,405\}$
$ f(45) \equal{} 135$ and $ g(45) \equal{} 15$
$ f(135) \equal{} 405$ and $ g(135) \equal{} 45$
$ f(405) \equal{} 45$ and $ g(405) \equal{} 3645$

5) ...
It's possible to build infinitely many other different solutions.

[b][u]Thanks for giving us your solution, pmvald[/u][/b]

Thanks for your attentions ... in fact this problem was supposed to be easier than this (to fit into an Olympiad problem), but it turned out to be a rather research problem. The main reason is the way I formulated expression (1). What I had in my mind was that $ g(n)$ must be the order of $ f$ over $ n$, i.e., the SMALLEST integer $ k$ such that $ f^k(n)\equal{}n$. But expression (1) means $ g(n)$ is a multiple of $ k$ and this way I screwed up the problem, but as we will see this screwed-up version is not free from implications and it leads to some other olympiad-style problems. The original problem is not hard at all, the underlying idea for this problem (and some other combinatorial function equation) is to consider $ f$ over a special set. For example in the above problem, the special set would be what I call an $ f$-circuit: a ordered set $ (a_0,a_1,\cdots,a_{k\minus{}1})$ where $ f(a_i)\equal{}f(a_{i\plus{}1})$ for $ i\equal{}0,1,\cdots,k\minus{}1$ (index calculations are mode $ k$). Expression (1) implies that the set $ \mathbb{N}$ is partitioned into $ f$-circuits. In the original problem, one can easily see that any $ f$-circuit must be of length $ 1$ which implies $ f(n)\equal{}n$. In fact let $ (a_0,a_1,\cdots,a_{k\minus{}1})$ be an $ f$ circuit of length $ k > 1$. Then for any $ i\equal{}0,1,\cdots,k\minus{}1$ we have $ g(a_i)\equal{}k$ and (by applying expression (2)) $ a_{i\plus{}1}\equal{}a_i^2/k$. These relations lead to the conclusion that $ k\equal{}n$ which is contraction. So $ f(n)\equal{}n$ is the only answer when $ g(n)$ is defined to be the order of $ f$ over $ n$. But my mistake in formulating it (in expression (1)) led to the fact that $ g(n)$ is but a multiple of the order of $ f$ over $ n$. Lets see where it takes us. Similar to the previous case, any such function $ f$ induces a partitioning of $ \mathbb{N}$ into $ f$-circuits. So the first logical step would be to identify $ f$-circuits. In the previous case we had only one $ f$-circuit, namely the trivial (i.e., length $ 1$) one. But it is different in this case. pco's post contains some $ f$ circuits of length greater than $ 2$, namely $ (6,18)$ (case (2)), $ (2a,2a^2)$ (case 3), $ (45,135,405)$ (case (4)). This suggests that $ f$-circuits are no longer trivial. A necessary and sufficient condition for the ordered set $ (a_0,a_1,\cdots,a_{k\minus{}1})$ to be an $ f$-circuit is that $ k\cdot a_{i\plus{}1}|a_i$ for all $ i\equal{}0,1,\cdots,k\minus{}1$. So the function $ f$ is defined uniquely with a partitioning of $ \mathbb{N}$ into $ f$-circuits. This condition along with expression (2) implies that $ k | a_i$ for any $ 1\equal{}0,2,\cdots,k\minus{}1$. So we can write $ a_i\equal{}k\cdot b_i$. Then for the ordered set $ (b_0,b_1,\cdots,b_{k\minus{}1})$ we get the following property $ b_{i\plus{}1}|b_i^2$. I called any ordered set with this property $ 1$-circuit. At this point one can see we have many diverse $ 1$-circuits. As a special case let $ b_i \equal{}p^{c_i}$ where $ p$ is an arbitrary prime number. Then $ (b_0,b_1,\cdots,b_{k\minus{}1})$ is a $ 1$-circuit if and only if $ c_{i\plus{}1} \leq 2c_i$. Obviously there are many such integers $ c_i$. So the general structure is $ 1$-circuits is not restricted enough to be easily formulated in math language, so is the structure of $ f$-circuits (which can be obtained from $ 1$ circuits by multiplying each member by the length of the circuit). When I'm dealing with a hard problem like this, instead of trying to solve it, I try to recognize the sub-problems it suggests. This way I may come up with some more user-friendly problems than the main problem. So at this point, even though we are stuck with the task of determining the general structure of $ f$, but instead there are some interesting sub-problems to deal with. The concepts of $ f$-circuit, $ 1$-circuit and partitioning $ \mathbb{N}$ into $ f$-circuits are all sources of good problems. pco introduced $ f$ circuits of length 2 and 3, so I asked myself what lengths an $ f$-circuit or (equivalently) a $ 1$-circuit can have. A simple argument shows that for any integers $ n>1$ and $ k\geq 1$ there is a $ 1$ circuit  of length $ k$ containing $ n$. A generalized version of this problem may replace the single integer $ n$ with a set of integers. It leads to the following Olympiad-style problem:

Given a set $ A \subset \mathbb{N}$. Prove that there exists a set $ B\equal{}\{b_0,b_1,\cdots,b_{k\minus{}1}\}$ with $ A \subset B$ and $ b_{i\plus{}1}|b_i^2$ for all $ i\equal{}0,1,\cdots,k\minus{}1$ (index calculations are mod $ k$) if and only if all members of $ A$ have the same set of prime factors.

As for the partitioning of $ \mathbb{N}$ into $ f$-circuits, some pco's examples for $ f$ induce one non-trivial $ f$-circuit (e.g.,$ (6,18)$,...) along with trivial $ f$-circuits for all the rest of integers which are not in the non-trivial $ f$-circuit. This partitioning of $ \mathbb{N}$ is very close to the trivial partitioning consisting of all trivial $ f$-circuits (except for the members of the non-trivial $ f$-circuit). I was looking for a more radical partitioning, i.e., avoiding trivial $ f$-circuits as much as possible. There are some obligations, for example expression (2) forces us to have $ f(p)\equal{}p$ for any prime number $ p$. So for any partitioning induced by a function $ f$ we must have trivial $ f$-circuits $ (p)$. Can they be the only trivial $ f$-circuits for some $ f$? I proved the following result which is a fit problem for Olympiad too:

Let $ P$ be the set of all prime numbers. Prove that $ \mathbb{N}\minus{}P$ can be partitioned into non-trivial $ f$-circuits.

many other questions can be asked. For example the above problem states that we need to discard the set $ P$ before we can partition the rest into $ f$-circuits of length $ \geq 2$. Now replacing $ 2$ with a general constant $ c \in \mathbb{N}$, we may ask which set we need to discard before partitioning the rest into $ f$-circuits of length $ \geq c$ ... I'll post more problems of this kind.

[quote="pco"][quote="pmvald"]Thanks for your attentions ... [ snip ] ......[/quote]

So the problem was wrong and you did not have the solution for it.
And we wasted our time looking for a solution.

[quote="pmvald"] I'll post more problems of this kind.[/quote]
 :mad:[/quote]You should read the whole post, pco. What he meant was problems like this:

[quote="pmvald"]Given a set $ A \subset \mathbb{N}$. Prove that there exists a set $ B \equal{} \{b_0,b_1,\cdots,b_{k \minus{} 1}\}$ with $ A \subset B$ and $ b_{i \plus{} 1}|b_i^2$ for all $ i \equal{} 0,1,\cdots,k \minus{} 1$ (index calculations are mod $ k$) if and only if all members of $ A$ have the same set of prime factors.[/quote]

and this:

[quote="pmvald"]Let $ P$ be the set of all prime numbers. Prove that $ \mathbb{N} \minus{} P$ can be partitioned into non-trivial $ f$-circuits.[/quote]

:D


greatest common divisor
number theory unsolved
number theory

For a natural number n>=4. Find the minimal k positive integer(k>1), such that every subset S from Dn = {1,2,3,...,n} of k elements is that the greatest common divisor of all the elements exist in S or is equal 1.