I'm not sure if this should go in basics or intermediate, so feel free to move it if its too easy.

If a basketball player can make 90% of his shots (make one shot 90% of the time) how many shots would he have to take before he had a 50% chance of making 15 shots in a row?

I don't get it. Isn't the probability of him making the shot each time independent of whether or not he made it the previous time?

I'm not sure where the problem came from - my teacher gave it to me.

Also I'm really bad at probability but here is what I thought of:

Basically he would have a $ .9^{15} \equal{} \sim 21$% chance of making 15 shots in a row on his first try.   However, as you pointed out the next set of shots seem to be independent so I don't know of anything you can do from there- perhaps this is a misworded question.

Well, if you shot 100 shots, then there would be higher chances of shooting 15 in a row within those 100 shots, than if you just did 15 shots. I think?

Here's a pre-K* version:
If the basketball player makes each shot with independent probability 90%, how many shots does he have to take before he has a 50% chance of making one shot?

Here's a kindergarten version:
Under the same assumption, what's the probability that the basketball player has made two shots in a row among his first three shots?

Here's an elementary school version:
Under the same assumption, how many shots does the basketball player need to make to have at least a 50% chance of making six in a row?  (Hint: compute $ .9^6$.)

Here's a middle school version version:
Under the same assumption, how many shots does the basketball player need to make to have at least a 50% chance of making seven in a row?  How about eight?

Now here's the high school version:
Under the same assumption,  what's the probability that he has not made three shots in a row by the time he takes eight shots?

Finally, the grown-up version:
Suppose that the basketball player hits each shot independently with probability $ p$.  Fix an integer $ k$.  What is the probability that after taking $ n$ shots, the player has not made $ k$ in a row?  (Key word here is [hide]state diagram[/hide].)

And if you can do that, the given problem is just a horrible computation away from being solved.

This certainly is too hard for high school basics -- I probably would have placed it in Pre-Olympiad, myself.

* Yes, yes, these are jokes.


How'd you guys do at chapter? I got 29 in sprint, 14 in target and I was second overall. First place beat me by a point! :wallbash:
Just wondering, how do you do #16 in the sprint round?

We had 3 perfect scores in our chapter (including me).  4th place had 45 points  :o

We had to use a tiebreaker round for the first time ever in our chapter. and I won!  :lol:

What was number 16?

Samath what chapter are you in?

edit: Like we did last year, do not talk about specific questions until MATHCOUNTS posts the tests. There may still be people taking the test (snow delay etc)

It's in Washington.  We have one of the best chapters, we had 2 people (including me and both current 8th graders) on last year's team, sowe'll probably be back.

I got a 39
I got a 25 on the sprint.
I got a 14 on the target.
Our team got a 20 on the team round.

Our team got a perfect score (on the team round). YAY! : :10:

[hide] 600th POST!!  :wow: [/hide]

I forgot to mention:

7 of our 8 competitors made it to state.
4 on the team.
3 indivisuals.

[quote="samath"]When can we discuss them? Aren't all the competitions supposed to be over by now?[/quote]We can discuss them when the problems are posted on the mathcounts website just to be sure. Think about what controversy it would create if we discussed certain problems when someone whos chapter is tomorrow say them.

I got 28 on the sprint and 16 on the target, but I still tied with 2 other people.  After the tiebraker, :winner_first:

wow. noone realizes that they shouldn't talk about specific problems

I got 26 sprint 14 target. (26th in Peninsula Chapter)

 :dry: Yeah...that's why I didn't deserve to make states.

I also got 26 sprint 14 target...so mad.. :mad: got 12th or something...1 pt away from top 10 or something... :mad:

I would just do guess and check since you have three minutes

I did do guess and check.  I ran out of time around the time I got to 25, so I put down 31 for my answer.

well you should realize that you can't get 29...
i would have put that

[quote="kyyuanmathcount"]I'll tell you there IS a better, slicker solution than total guess/check.

Also, I WILL torture you by never telling you, until the day I die XD.[/quote]
If you won't tell biffandoc, will you tell me?(please!)

Great, first my coach and now you.
I shall BUG YOU FOREVER! muahaha :P

:huuh: There is a slick way to do that. WOAH. Can someone please tell us how to do it. I used up 5:30 mins on that problem.  AND I STILL got it wrong. Please can someone explain it to me. :omighty: However congratulations to kyyuanmathcount for solving it. Now if he will just tell us how. :(

I had 15 seconds left and my answer was between 23, 25, 27, and 29. Why didnt I guess 27???

Actually, I managed to brute force/ guess/check that in about 4 minutes, and thats how I did it during the competition.  But since then, I figured a better method, that you would have to bug me exactly 1 year and 2.1 months to get it from me (thats when my MATHCOUNTS ends).

[quote="kyyuanmathcount"]I WILL torture you by never telling you, until the day I die XD.[/quote]

[quote="Naonao"]I found a simpler way to do it....... but do i fell like tell you
Still kinda upset bout not making state[/quote]

This Forum is to help each other.  If you do not feel this Forum has helped you, you should probably move to another Forum.  If you do feel this Forum has helped you, then you owe a debt to AoPS and to this Forum.

The above quotes contain no useful content, so they are considered spam.  Please avoid spamming the Forum.

Thank you, Ravi, for posting a nice solution.

[quote="Ravi B"][quote="khwain"]Does anyone know a fast way to do the last target round problem? (dart board)  Only 2 people on my team got it right.[/quote]If you choose the five darts to be 2, 4, or 6, then you can get every even score from 10 to 30.  If you choose one dart to be 1 and the other four darts to be 2, 4, or 6, then you can get every odd score from 9 to 25.  It's easy to get the low scores of 6, 7, and 8.  Thus the first score we can't get is 27.[/quote]

Yes, but I would first figure out that when the problem sayed "5 darts," anything you could make with 4,3,2, or 1 dart would be makeable with 5, by converting 4 to 2+2, 2 to 1+1, 6 to 2+4.  *points up* that would be my method, finding more of what you CAN get instead of what you CANNOT get.

Just as side note: The facts are: I really didn't know the good method, though i knew there WAS a good method.  I said I knew because I wanted people to try harder to find that solution.


geometry
inequalities
trigonometry
geometric transformation
number theory

A similar topic was created last yr for COMC 2005. I think it's a good topic.
So what do you guys think the last problem for COMC will be?

Here is the past history of COMC problems:
[b]2005[/b]: Geometry
[b]2004[/b]: Number Theory (and a bit of Combinatorics and
                  Inequalities)
[b]2003[/b]: Algebra
[b]2002[/b]: Geometry
[b]2001[/b]: Geometry
[b]2000[/b]: Combinatorics
[b]1999[/b]: Geometry
[b]1998[/b]: Geometry
[b]1997[/b]: Algebra
[b]1996[/b]: Algebra

Statistics:
Geometry: $\frac{5}{10}$ :o
Algebra: $\frac{3}{10}$
Number Theory: $\frac{1}{10}$
Combinatorics: $\frac{1}{10}$

Looking at the percentages and past history, the expected topic of the problem will be [b]Combinatorics[/b].

However I think we will have [b]geometry[/b] for the last problem. The reason is, geometry problems tend to come in pairs of consecutive years. Also, a geometry problem is barely ever solvable for someone doing only high school math and therefore lets the organizors easily distinguish between ppl at olympiad level and ppl not at olympiad level. However this is just my opinion and I am probably wrong...
So vote and post! :)

In particular, circle geometry has been a heavy favorite for last problems (what about triangles?).   But I think we will have a combinatoris last problem this year.  They can seem approachable to the non-olympian.  Also, last CMO was stuffed with combinatorics.  And I don't think the cms's combinatorics fever has worn off.  Just make it something I solve :D

[quote="Leao"]I hope it's geometry rather than combinatorics (or any other type); I am far stronger in geometry than combinatorics  :lol:[/quote]
That is very true for me too. :)

I have a good feeling it's going to be Geometry. ("Good" as opposed to algebra, which is even worse.) But if you think about it, last year's question, for instance, was technically a geometry problem, but required a lot of algebraic manipulation and use of various basic theorems (or even more algebra if you used trig), but no translations or any of that jazz. Or we might see the classic two-part geometry-combinatorics combo where if you get the first part wrong you're screwed for the second part even though they're unrelated.

Yeah it could be similar to Euclid 10, which has geo, combinatorics, algebra , and nt altogether. :)

[quote="rem"][quote="boxedexe"]I'm hoping for either Combinatorics or Number theory. Also, does anyone have the floor for COMC in oder to make CMO for the past several years?[/quote]
Last yr it was 67 and the year before it was 67, too.[/quote]
It was 67 last year? Someone in our school got 67 and didn't qualify for CMO  :maybe: . The same thing happened the year before, where two people from our school got 67 but did not make CMO. If the cutoff was actually 67, I'll be so mad, I got 32 on part A because of stupid mistakes which brought me to 63.

[quote="rem"]I dunno maybe it was 68 last yr. I am not quite sure. But in 2004 a guy from our school with 67 made it to CMO...[/quote]

Will Ma?  That's because he had cred.  The CMS qualifies anyone with cred; people who have done (well perhaps) the CMO in years past or people who just seem awesome.

Will Ma had 69.
I was the one who had 68. :blush: lol

Interesting score...

Well, you guys did go to Waterloo Collegiate; perhaps the school as a whole has cred?

For the last problem of a contest like COMC , probably the best choice is Combinatorics(a typical Russian style, ie leningrad or ToT).
Solving those problems just needs Creativity , So they give a chance to the inexperienced participants.

I believe it would be a geometry problem since as alexander said, it allows the markers to distinguish between the people at the olympiad level and average problem solvers.

[offtopic]Just out of curiousity, what is your Persian name lomos_lupin[/offtopic]

P.S. God, this forum is really dead.

Masoud Zargar


function
calculus
integration
search
derivative
real analysis
real analysis unsolved

[quote="stergiu"]Find all contineous functions on $\mathbb{R}$ with :

  $\int_{a}^{b}f(x)dx+\int_{c}^{d}f(x)dx =\int_{a+c}^{b+d}f(x)dx , \forall a,b,c,d \in \mathbb{R}$[/quote]

[b]
 Another solution ![/b]

After some effort I found that puting $a=0 , b=y , c=d=x$  we take:

    $\int_{0}^{y}f(t)dt = \int_{x}^{y+x}f(t)dt , \forall x,y \in \mathbb{R}$

 taking derivative with variable $y$  we find $f(x) = f(x+y) , \forall x,y \in \mathbb{R}$. Now we put $x=0$ and we are done.

[u] Babis[/u]


Given a totally multiplicitave function f, that $ f(m) > f(n)$ if $ m > n$, that $ f(n)$ is an integer for an integer n, and that $ f(2) \equal{} 2$, prove that $ f(k) \equal{} k$ for integral k.

Do you mean $ f :\mathbb{Z}\to\mathbb{Z}$?

Well, even if not, we need only prove it for integers.
[hide]We have $ f(1)f(2) \equal{} f(1*2) \equal{} f(2)$ so f(1) = 1. Also, $ 2f(0)\equal{}f(0)f(2)\equal{}f(0)$, so $ f(0)\equal{}0$. Furthermore, $ f(\minus{}1)f(\minus{}1)\equal{}f(1)$ and $ f(\minus{}1)<f(0)\equal{}0$, so $ f(\minus{}1)\equal{}\minus{}1$. Because $ f(\minus{}1)f(k)\equal{}f(\minus{}k)$, $ f(\minus{}k)\equal{}\minus{}f(k)$. Therefore, we need only prove it for positive integers.

We see that it is sufficient to prove the statement only for primes because $ f$ is multiplicative. All numbers $ k$ that are products of those primes will satisfy $ f(k)\equal{}k$. We show a simple lemma:

[b]Lemma.[/b] For all prime $ p>2$, $ p\plus{}1$ and $ p\minus{}1$ must be divisible only by primes less than $ p$.
This is obvious, as $ (p,p\plus{}1)$ and $ (p,p\minus{}1)$ = 1, and all primes bigger than $ p$ are at least $ p\plus{}2$.

Now, we induct on the primes.

Base case: $ f(2)\equal{}2$. Because $ f(2)\equal{}2$ and $ f(4) \equal{} f(2)f(2) \equal{} 4$, $ f(3)\equal{}3$.
Inductive step:Because $ p\plus{}1$ and $ p\minus{}1$ are strictly divisible by primes < $ p$, we know that $ f(p\plus{}1)\equal{}p\plus{}1$ and $ f(p\minus{}1)\equal{}p\minus{}1$. Since $ f$ is strictly increasing and takes integers to integers, we have $ f(p)\equal{}p$ and we are done.[/hide]

[quote="not_trig"]Well, even if not, we need only prove it for integers.[/quote]

I mean as opposed to $ f :\mathbb{R}\to\mathbb{R}$, which would make the proof a lot shorter :)

The phrase "totally multiplicative" is meaningless on $ \bf R$.


search
real analysis
real analysis unsolved

f:R -> R, both f^2, f^3 inf-tely differentiable => f too?

There was lot of discussion about this problem on the forum. Please search for it, if I'm not wrong then it's a post by sam-n. I don't remember if the problem was solved or not, but the only (correct) solution I saw spanned over 6 pages, so I'm definitely not going to write it here.

Let's put this into an abstract form: we have a smooth map $F\colon \mathbb R\to\mathbb R^{2}$, namely $t\mapsto (f^{2}(t),f^{3}(t))$, whose image is contained in the cusped curve $C=\{(x,y)\colon x^{3}=y^{2}\}$. The curve $C$ is covered by $\pi\colon \mathbb R\to C$, where $\pi(s)=(s^{2},s^{3})$. The question is: does $F$ lift to a smooth map $\tilde F\colon \mathbb R\to\mathbb R$ such that $\pi\circ \tilde F=F$?

It surely does in the continuous category. What are the obstructions to lifting maps in the smooth category?  :?:


Define a sequence $ {a_i}$ by $ a_1 \equal{} 3$ and $ a_{i\plus{}1} \equal{} 3^{a_i}$ for $ i \geq 1$. Which integers between 00 and 99 occur as the last two digits in the decimal expansion of infinitely many $ a_i$?

$ \lambda(100) \equal{} 20, \lambda(20) \equal{} 4, \lambda(4) \equal{} 2$.  After $ a_4$ the value of $ a_k \bmod 100$ is fixed; namely,

$ 3 \equiv 1 \bmod 2 \implies$
$ 3^3 \equiv 3 \bmod 4 \implies$
$ 3^{3^{3}} \equiv 7 \bmod 20 \implies$
$ 3^{3^{3^{3}}} \equiv \boxed{87} \bmod 100$

by Carmichael's theorem.

I've never heard of Carmichael's Theorem. Is there a way to do it with just Euler's Theorem?

Of course. It just takes more layers if you use $ \phi$ instead of the actual exponent of the group.

$ \phi(100)\equal{}40$
$ \phi(40)\equal{}16$
$ \phi(16)\equal{}8$
$ \phi(8)\equal{}4$
$ \phi(4)\equal{}2$
$ \phi(2)\equal{}1$

$ a_n\equal{} 1\mod 2$
$ a_{n\plus{}1}\equal{} 3^1\equal{}3\mod 4$
$ a_{n\plus{}2}\equal{}3^3\equal{}3\mod 8$
$ a_{n\plus{}3}\equal{}3^3\equal{}11\mod 16$
$ a_{n\plus{}4}\equal{}3^{11}\equal{}27\mod 40$
$ a_{n\plus{}5}\equal{}3^{27}\equal{}87\mod 100$
and $ a_m\equiv 87\mod 100$ for all $ m\ge 6$.

Unless I am missing something, you actually had to do some work to calculate 3^11 and 3^87 mod 40 and mod 100, right (I'm not complaining--I'm just asking)?

EDIT: Let me try to answer my own question. So, I guess you use Euler's Theorem in two ways in this problem: first to show that $ a_{n} \equiv 3^{a_{n \minus{} 1}\mod{\phi(100)}}\mod 100$ and also to explicitly calculate 3^11 and ^27

EDIT 2: disregard the first edit. I still do not see how you got 3^11 and 3^27.
Also, does the value of a_1 matter?

$ 3^{8}\equal{}(3^4)^2\equal{} 1^2\equal{}1\mod 40$, so $ 3^{11}\equal{} 3^3\equal{}27\mod 40$

I didn't actually calculate $ 3^{27}\mod 100$ myself; I trusted t0rajir0u for that.
Euler's theorem doesn't help for doing the calculation; we just have to multiply a bunch of numbers. An efficient method:
$ 27\equal{}16\plus{}8\plus{}2\plus{}1$
$ 3^1\equal{}3$
$ 3^2\equal{}9$
$ 3^4\equal{}9^2\equal{}81$
$ 3^8\equal{}81^2\equal{}61$
$ 3^{16}\equal{}61^2\equal{}21$
$ 3^{27}\equal{}3\cdot 9\cdot 61\cdot 21\equal{}27\cdot 81\equal{}87$
(All calculations mod $ 100$ for this example. This gets you any exponent up to $ 2^m$ with at most $ 2m$ multiplications mod $ n$, plus the cost of getting the binary expansion of $ m$)

The value of $ a_1$ doesn't matter. I could have gone back one more step and started at $ a_{n\minus{}1}\equal{}0\mod 1$- which is always true for any starting value, and works in all possible versions of the problem.

[quote="FibonacciFan"]I still do not see how you got 3^11 and 3^27.[/quote]

Using only the Totient Theorem, we must proceed as jmerry has described.  Fortunately, the Carmichael Theorem allows us to use $ 3^7$ rather than $ 3^{27}$, which is much simpler :)

[quote="FibonacciFan"]Also, does the value of a_1 matter?[/quote]

Nope.  If $ a_1$ is odd then $ a_1 \equiv 1 \bmod 2$; if $ a_1$ is even then $ a_2 \equiv 1 \bmod 2$.  It's the base of the exponent in the recursion that matters.


let $( {u_{n})}$ be a sequence of positive integers definied by
$u_{0}=1, u_{n+1}=au_{n}+b$
where $a,b$ are natural numbers.  Prove that for any choice of $a$ and $b$ , the sequence $( u_{n} )_ {{n} \geq 0}$ contains infintely many composite numbers

If a=1 $u_n=1+nb$. Let prime $p|u_{n_0}$ then $p|u_{n_0+kp}$ for all k.
If $a\not=1$ sequence $u_n=au_{n-1}+b=a^2u_{n-2}+ab+b=\dots=a^nu_0+b(a^{n-1}+\dots+1)$. It  give $u_n=a^n+\frac{b(1-a^n)}{1-a}$.
If prime $p|u_{n_0}$ then $p|u_{n_0+k|1-a|(p-1)}$ for all k.


I had to do a system restore on my computer recently and I'm trying to get TeXnicCenter working again.  When I open it the Configuration Wizard pops up.  When it says: "Enter the full path of the directory where the executables (latex, tex, etc.) of your TeX-distribution are located:"

What do I do here?  I've tried many things but nothing works.  Also, I have MiKTeX still so that shouldn't be the problem...

Any help?

The path should be something like C:\texmf\MiKTeX\bin\ which should contain programs such as latex.exe, yap.exe, pdflatex.exe etc etc. Look for these files on your system.

Generally speaking LaTeX programs prefers files in folders that have no spaces in their names, so it's not wise to install MiKTeX in Program Files, even if it appears to work.

Thanks, that works.  But now I can't figure out the paths for the DVI, PS and PDF compiler...

It's the same path for the compilers. Do you mean the viewers?
DVI: Uses yap.exe in C:\texmf\MiKTeX\bin\
PDF: uses Adobe Reader, depending on where you installed it
PS: uses Ghostscript or Ghostgum

Use the Wizard which should automatically find MiKTeX. Alt+F7, Wizard


function
algebra
polynomial
arithmetic sequence
4th edition

umm ... i thought q2 was nice (but a bit annoying) but the typical infinite descent problem. my sol for q3 was a stupid co-ord bash (which doesn't really involve too much computations though  :) ).
i have no idea whatsoever how to go about solving q1. it looks totally crazy to me. i think the answer is all functions of form f(x)=x^k where a is a perfect k-th power, and some linear functions. maybe i'm missing some obvious ones tho.
could someone plz post a sol for q1? and maybe a synthetic one for q3?

When you guys will stop telling me that #2 was trivial? I don't see any point why it should be. Anyway, the weaker problem to show that there doesn't exist an infinite arithmetic progression of squares is in Engel's "Problem-Solving Strategies", and it's really a simple infinite descent, but...

#1 was definitiely unattractive to me... I hate those problems where one first has to think for ten minutes what the formula means.

#3 was very nice - and yes, I have a quite synthetic solution  :)  See the attached ZIP file. (To the mods: please [b]don't[/b] delete the TeX, DVI and WMF files; they are rather small and some people could use them.)

  Darij

Q3 can be done using mostly Pythagoras' theorem and Apollonius': just get everything in terms of $r_1$, $r_2$ and the length of the common chord (I think). My attempt at Q1 was as follows: first, all the coeffs have to be rational; now convert into normal form, and look at the integers that produce $a^{(kd)^2}$ for each natural $k$ where $d$ is the degree of the polynomial. Now for sufficiently large $k$, these have to be multiples of each other (the polynomial basically becomes its first term). You can show this by showing that the next integer up is too big, and the next one down too small. This forces the polynomial to be a power of a linear expression, and its pretty much done from there.

How did your infinite descent argument go? I tried for ages and made little progress for Q2.

For Fiachra:
For Q1 i dont see how you deduce from the beggining that all coefficients must be rationals..... But i think the rest of the aproach is ok.

For Darij:
I agree that Q2 wasnt trivial. It was hard, but it is a well known result in the web. I found a solution wich is also a descend, but it was not easy.

And i think Q3 was very easy.

Thanks

For Q1, to show that all the coefficients are rational, note that we have infinitely many pairs $m,n$ of integers such that $m=f(n)$, and each of these yields a 1st-degree equation for the coefficients. When we solve these simultaneously, we must get rational solutions.

About Q2, it can be found in Titu Andreescu, "An Introduction to Diophantine Equations". It is shown that there are no 4 squares in an arithmetic progression.
Use the Lemma : The equation \[x^4-x^2y^2+y^4=z^2\] has only trivial solutions.

It is very ruefully that problem from ML Contest can be such easily found.

[quote="DusT"]About Q2, it can be found in Titu Andreescu, [b][Dorin Andrica,][/b] "An Introduction to Diophantine Equations". It is shown that there are no 4 squares in an arithmetic progression.[/quote]

Curses! That book lies on my bookshelf! Now as you say it, I see that the problem was example 3 in paragraph 2.3.1. But I haven't yet managed to systematically read more than the first chapter...

  Darij

[quote="darij grinberg"]

Curses! That book lies on my bookshelf! Now as you say it, I see that the problem was example 3 in paragraph 2.3.1. But I haven't yet managed to systematically read more than the first chapter...

  Darij[/quote]

ML contests seems to be designed to have a look into new literature.  :D

I cannot possibly know every other book that had this problem solved. Good for you readers :P :)

@Myth: it's hard to create 21 (3x7) problems hard and completely original, that cannot be found in any book :P

Problem no. 3 was my creation, and Problem no. 1 was Harazi's creation.

[quote="Valentin Vornicu"]@Myth: it's hard to create 21 (3x7) problems hard and completely original, that cannot be found in any book[/quote]
I think threre are many people whose opinion you may ask.

[quote]Problem no. 1 was Harazi's creation.[/quote]
It was clear. Who can also invent [b]such[/b] problem?  ;)

I am sorry, Myth if you didn't like it.  :blush:  :blush:  :blush:  :blush:  :blush:.

Well, Q2 is not due to Titu Andreescu but to Fermat himself...

Pierre.

[quote="al.M.V."]Yeah, I agree with Valentin, it is not difficult to come up with a problem of your own, but it's VERY difficult to be perfectly original. I think regardless of the origins of the problems and their difficulties, this contest does serve (at the very least) to bring new problems, and it does help a lot (maybe just for me, but...) Nevertheless, I don't think the problem sets deserve too much criticism. (that's, of course, just my trash... :lol: )[/quote]

I agree with you. I just think that we all should appreciate that Valentin takes his time to create an interesting contest, check solutions and make up rankings. Recently Valentin did not have the time to compile solutions for the contest. That's why I think some forum members as Pierre, Darij, myth etc. could take the source files and generate solutions for it. To share the work might be best. What do you think ?  :)

Exactly orl!!!  :lol: I could not have said that any better!  :P

Thanks for understanding. :cool:

I know it's a very stupid question, but anyway can somebody help me to rephrase Q1... So, as I understood, A is a set of integers of given form and it's a subset of B. Now what does that vertical line mean? I only know that a statement a divides b is denoted by $a|b$, but I think it's something else here...

Ok, I know the statement of the problem is not the happiest one. I prefer the following one: for a given integer $a>1$ find all polynomials with real coefficients f such that the image of the set 1,2,3,... by f contains all the numbers of the form $a^{n^2}$ with n=1,2,3,...

Usually in the definition of a set $A = \{ x \in \mathbb{R} \mid x^2 \leq 1 \}$ for example, means that the set $A$ is formed with all the reals $x$ such that (so the | sign stands for "such that") $x^2 \leq 1$.


Should the IMO add a commemoration slumber party to the official list of events?

Absolutely!  That was the most fun evening at any IMO ever!!  AWESOME!  Next time they should add some big speakers and a good DJ too :D

Heh, but keep it quiet enough for conversation too.

I think this would be great fun, anyway. I'm all for it.

Emily was my favourite weather event ever! I loved it, although when I went to bed at like 4 o'clock, I had no bed to sleep in since it was already token by other persons. Fortunately I'm used to sleeping on the ground.


inequalities
function
induction
inequalities proposed

Solution of (1) :

$ a^3 \plus{} b^3 \plus{} c^4 \leq 9 \Longleftrightarrow a^2 \plus{} b^2 \plus{} \sqrt {9 \minus{} a^3 \minus{} b^3} \geq 3$

We can assume that $ min(a,b) \equal{} a$ without loss of generality.

$ a^2 \plus{} b^2 \leq 3$ and $ a \leq \frac {\sqrt {6}}{2}$

$ f(a) \equal{} a^2 \plus{} b^2 \plus{} \sqrt {9 \minus{} a^3 \minus{} b^3}$

$ f'(a) \equal{} 2a \minus{} \frac {3a^2}{2\sqrt {9 \minus{} a^3 \minus{} b^3}}$

$ f'(a) > 0 \Longleftrightarrow 9a^2 \plus{} 16a^3 \plus{} 16b^3 < 144$

Since $ a^2 \plus{} b^2 \leq 3$

$ 16b^3 \leq 16(3 \minus{} a^2)^{\frac {3}{2}}$

$ 9a^2 \plus{} 16a^3 \plus{} 16b^3 \leq 9a^2 \plus{} 16a^3 \plus{} 16(3 \minus{} a^2)^{\frac {3}{2}} \leq$

$ \frac {27}{2} \plus{} 12\sqrt {6} \plus{} 48\sqrt {3} \approx 126.0323 < 144$

Hence $ f(a)$ is an increasing function and $ f(a) \geq f(0) \equal{} b^2 \plus{} \sqrt {9 \minus{} b^3}$

Now if we prove that $ b^2 \plus{} \sqrt {9 \minus{} b^3} \geq 3$ then the solution is done.

$ b^2 \plus{} \sqrt {9 \minus{} b^3} \geq 3$ $ \Longleftrightarrow$

$ b^2(b^2 \plus{} b \minus{} 6) \equal{} b^2(b \plus{} 3)(b \minus{} 2) \leq 0$ which is obviously true since $ 0 \leq b \leq \sqrt {3} < 2$  :wink:

Equality holds if and only if $ a\equal{}b\equal{}0$ , $ c\equal{} \sqrt{3}$

[quote="crazyfehmy"]Solution of (1) :

$ a^3 \plus{} b^3 \plus{} c^4 \leq 9 \Longleftrightarrow a^2 \plus{} b^2 \plus{} \sqrt {9 \minus{} a^3 \minus{} b^3} \geq 3$.....

[/quote]

thanks. very clear answer.

[quote="4865550150"]Let a^2+b^2+c^2 = 3, a,b,c >=0, prove or disprove the followings:

(1)       $ a^3 \plus{} b^3 \plus{} c^4 \leq 9$,

(2)       $ a^n \plus{} b^n \plus{} c^{n \plus{} 1} \leq (\sqrt {3})^{n \plus{} 1}$, for all positive integers n and n>=2.[/quote]

(1), easy to prove that
$ a^3\plus{}b^3 \le \sqrt{(a^2\plus{}b^2)^3}\equal{}\sqrt{(3\minus{}c^2)^3}$
so we just need to prove that
$ (9\minus{}c^4)^2 \ge (3\minus{}c^2)^3$
$ \Leftrightarrow \left(c^2\minus{}3\right)^2 \left(c^2\plus{}1\right) \left(c^2\plus{}6\right) \ge 0$
clear true.

for (2), I guess, it will holds for any [b]reals[/b] n and $ n \ge 2$.

in the same way, just need to prove that

$ \left(3^{\frac{n\plus{}1}{2}}\minus{}c^{n\plus{}1}\right)^2 \ge (3\minus{}c^2)^n, \quad c \in [0,\sqrt{3}] \qquad (*)$

but I always can't prove it... :oops:

[b]Remark[/b]:
$ n\equal{}2$, (*) will be $ ({c}^{4}\plus{}2{c}^{3}\sqrt {3}\plus{}8{c}^{2}\plus{}4c\sqrt {3}\plus{}6)(c\minus{}\sqrt{3})^2 \ge 0$;
$ n\equal{}4$, (*) will be $ (c^8  \plus{} 2\sqrt 3 c^7  \plus{} 8c^6  \plus{} 10\sqrt 3 c^5  \plus{} 48c^4  \plus{} 48\sqrt 3 c^3  \plus{} 90c^2  \plus{} 36\sqrt 3 c \plus{} 54)(c\minus{}\sqrt{3})^2 \ge 0$;
$ n\equal{}5$, (*) will be $ (c^2\plus{}6)(c^6\plus{}c^4\plus{}12c^2\plus{}9)(c^2\minus{}3)^2 \ge 0$;
...

Solution of Kuing's inequality(but while $ n$ is an integer)  which is

$ \left(3^{\frac {n \plus{} 1}{2}} \minus{} c^{n \plus{} 1}\right)^2 \geq (3 \minus{} c^2)^{n}$ where $ n \geq 2$ is an integer and $ 0 \leq c \leq \sqrt {3}$ is a real.

We will prove by induction. The inequality is true for $ n \equal{} 2$ as Kuing said.

If we prove that $ \left(\frac {3^{\frac {n \plus{} 2}{2}} \minus{} c^{n \plus{} 2}}{3^{\frac {n \plus{} 1}{2}} \minus{} c^{n \plus{} 1}}\right)^2 \geq 3 \minus{} c^2$ then the induction is done.

$ \left(\frac {3^{\frac {n \plus{} 2}{2}} \minus{} c^{n \plus{} 2}}{3^{\frac {n \plus{} 1}{2}} \minus{} c^{n \plus{} 1}}\right)^2 \geq 3 \minus{} c^2$ $ \Longleftrightarrow$

$ 2.c^{2n \plus{} 2} \plus{} 2.3^{\frac {n \plus{} 3}{2}}.c^{n \minus{} 1} \plus{} 3^{n \plus{} 1} \geq 3.c^{2n} \plus{} 2.3^{\frac {n \plus{} 2}{2}}.c^n \plus{} 2.3^{\frac {n \plus{} 1}{2}}.c^{n \plus{} 1}$

If we divide both sides by $ c^{2n}$ , inequality is equivalent to

$ \left[c \minus{} \sqrt {3}.\left(\frac {\sqrt {3}}{c}\right)^n \right]^2 \plus{} 2\sqrt {3}(\sqrt {3} \minus{} c).\left(\frac {\sqrt {3}}{c}\right)^{n \plus{} 1} \geq 3 \minus{} c^2$

$ \Longleftrightarrow$ $ (\lambda^2 \minus{} 2\lambda \plus{} 2)c^2 \minus{} 2\sqrt {3}\lambda c \plus{} 6\lambda \minus{} 3 \geq 0$ where $ \lambda \equal{} \left(\frac {\sqrt {3}}{c}\right)^{n \plus{} 1}$

The last inequality is equivalent to

$ (\lambda \minus{} 1)^2c^2 \plus{} (\sqrt {3} \minus{} c)(2\sqrt {3}\lambda \minus{} c \minus{} \sqrt {3}) \geq 0$

Since $ c\leq \sqrt {3}$ and $ \lambda \geq 1$

$ \sqrt {3} \minus{} c \geq 0$ and $ 2\sqrt {3}\lambda \minus{} c \minus{} \sqrt {3} \equal{} 2\sqrt {3}(\lambda \minus{} 1) \plus{} (\sqrt {3} \minus{} c) \geq 0$ . Induction is done.  :wink:

[quote="kuing"]

(1), easy to prove that
$ a^3 \plus{} b^3 \le \sqrt {(a^2 \plus{} b^2)^3} \equal{} \sqrt {(3 \minus{} c^2)^3}$.....[/quote]

[quote="crazyfehmy"]Solution of Kuing's inequality(but while $ n$ is an integer)  which is

$ \left(3^{\frac {n \plus{} 1}{2}} \minus{} c^{n \plus{} 1}\right)^2 \geq (3 \minus{} c^2)^{n}$ where $ n \geq 2$ is an integer and $ 0 \leq c \leq \sqrt {3}$ is a real.

We will prove by induction. The inequality is true for $ n \equal{} 2$ as Kuing said........
[/quote]

My guess is true :lol: . thx again~~~ :)


Visto l'OliSfratto conviene postare anche in questa sezione l'annuncio del Winter Camp. Spargete la voce.

==============================================

Winter Camp (Italiano) - Informazioni Generali

Il Winter Camp 2007 si svolgera' a Pisa. L'arrivo dei partecipanti e' previsto entro la sera di martedi' 23 Gennaio 2007, la partenza nel pomeriggio di domenica 28 Gennaio 2007.

Lo stage e' pensato per studenti che hanno gia' una buona conoscenza del programma olimpico (vedi Schede Olimpiche e lezioni dello stage senior) e sono ad un buon punto della preparazione sugli esercizi. Lo stage consistera' in sessioni di lavoro su problemi di livello internazionale.

Il Winter Camp si concludera' con una prova finale nelle mattinate di sabato e domenica. [b]Tale prova finale, opportunamente pesata con i risultati di settembre, varra' come prova di selezione per la partecipazione alle "Balkan Mathematical Olympiads" nel caso in cui l'Italia venisse invitata a partecipare[/b] (cosa che finora *non* e' accaduta: l'organizzazione sembra piuttosto in ritardo, al punto che non sono ancora nemmeno note le date ufficiali della manifestazione).

Come gia' accaduto a partire dallo stage Senior 2005, anche al Winter Camp si potra' partecipare come "spesati" (spesati dall'organizzazione) o come "volontari". Gli aspiranti volontari dovranno fare domanda come indicato nell'apposito paragrafo.

Per tutti, invitati o volontari, la partecipazione e' da considerare "sub iudice", nel senso che tutti devono far pervenire le soluzioni scritte di 8 esercizi, con modalita' indicate nell'ultimo paragrafo. Un'apposita commissione valutera' la qualita' del lavoro svolto e sulla base di questo confermera' o meno la possibilita' di partecipare.

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Elenco degli "spesati"

CIMETTA Leone
COLOMBO Maria
CONTI Andrea
DALUISO	Roberto
DE CAPUA Antonio
FOGARI Andrea
GALEOTTI Mattia
KUZMIN Kirill
LAZZARINO Leslie Lamberto
LO BIANCO Federico
LIN Francesco
LOMBARDO Davide
NARDIN Denis
PATIMO Leonardo
SANTI Elia
TRAGER Matthew
VERTECHI Pietro

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Informazioni per i "volontari"

Come fare domanda? Gli interessati dovranno inviare la propria domanda [b]via fax allo 050-2213802 (att. Massimo Gobbino) entro il 6 gennaio 2007[/b].
Per la domanda basta riadattare con i propri dati il fac simile sottostante.

Tutte le domande saranno accolte? Non e' detto.

Come verranno scelti i partecipanti? Sulla base degli esercizi svolti, dell'eta' (con preferenza ai piu' giovani), e del curriculum olimpico.

Quanto costa? La partecipazione alle attivita' dello stage (lezioni, sessioni di esercizi, eventuale prova finale) e' gratuita. Tutto il resto, ed in particolare il viaggio, il vitto e l'alloggio, e' a carico del partecipante, sia come spese, sia come organizzazione (ricerca di treni, alberghi, ...). I partecipanti si devono impegnare a seguire tutto lo stage.

Dove posso alloggiare/mangiare? L'organizzazione provvedera' a prenotare un po' di camere in piu' all'Hotel Di Stefano (lo stesso in cui risiederanno gli spesati). I volontari che vorranno potranno usufruirne, ovviamente a pagamento (a settembre 2006 il costo di una sistemazione in camera multipla e' stato di circa 35 euro a testa per notte). I volontari che lo vorranno potranno mangiare insieme agli spesati alla mensa della Scuola Normale Superiore (a settembre 2006 il costo era di 10.00 euro a pasto).

A cosa ho diritto? A seguire le attivita' dello stage e la prova finale (e' chiaro pero' che i volontari che non hanno partecipato allo stage Senior 2006 saranno svantaggiati nel calcolo del punteggio ai fini dell'eventuale Balkan) alla pari degli altri invitati, ad un attestato di partecipazione.

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

FAC SIMILE DOMANDA DI PARTECIPAZIONE COME VOLONTARIO

Sono <Nome COGNOME>, nato a <Luogo Nascita (Sigla Provincia)> il <Data Nascita>. Quest'anno (2006/2007) sto frequentando il ... anno di corso (nota bene: II liceo classico = quarto anno di corso) del <indicare tipo e nome della scuola> di <citta' della scuola>.
Il mio indirizzo e' <indicare indirizzo completo, con numero di telefono fisso, cellulare ed e-mail>.

Vorrei seguire le attivita' del Winter Camp Pisa 2007.

Le mie precedenti esperienze olimpiche sono: <indicare eventuali stage ai quali si e' partecipato, nonche' gare provinciali e/o nazionali con relativi piazzamenti>.

Se la mia domanda viene accolta, mi impegno a partecipare per tutta la durata della manifestazione.

Provvedero' in maniera autonoma ad organizzare il viaggio. Mi piacerebbe  alloggiare all'Hotel Di Stefano (a mie spese) e/o mangiare alla mensa SNS (a mie spese) (oppure: provvedero' in maniera autonoma ad organizzare il soggiorno, ed a sostenere le relative spese).

<Luogo e data>

Firma dello studente

Firma di un genitore (se lo studente e' minorenne)

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Spedizione esercizi

Modalita' di spedizione: gli svolgimenti dovranno essere spediti in uno dei seguenti tre modi

- per via cartacea a Massimo Gobbino, Dipartimento di Matematica Applicata, via F. Buonarroti 1c, 56127 Pisa;

- per fax allo 050-2213802 all'attenzione di Massimo Gobbino;

- per posta elettronica ([color=blue]un *unico* file in formato pdf (e dimensioni ragionevoli) per ogni partecipante, con nome WC07 seguito dalle prime 4 lettere del cognome senza spazi[/color]) a m.gobbino@dma.unipi.it.

Scadenze: gli svolgimenti dovranno pervenire entro il [b]6 gennaio 2007[/b]. La conferma dell'invito avverra' a quel punto il prima possibile.

Quali sono gli esercizi da svolgere: [b]8 esercizi[/b] scelti tra quelli dell'[url=http://www.mathlinks.ro/Forum/resources.php?c=80&cid=101&year=2006]Iran National Mathematical Olympiad (3rd round 2006)[/url], di cui 2 nel gruppo Algebra, 2 nel gruppo Combinatoria, 2 in Geometria e 2 in Teoria dei Numeri (con esclusione degli esercizi C3, G3 e N1).

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Qualche piccola aggiunta personale all'annuncio del Winter Camp.
[list][*]Attenzione al fatto che ci sono tre problemi esclusi.
[*]Attenzione al fatto che alcuni problemi richiedono, anche solo per capire il testo, nozioni al di fuori del programma olimpico solitamente inteso in Italia. In ogni caso in ogni sezione sono almeno 2 i problemi del tutto aderenti al programma standard.
[*]Attenzione al fatto che i testi di alcuni problemi contengono evidenti errori di stampa (ad esempio l'N5), spesso evidenziati nelle discussioni.  [/list]Spero vivamente che i partecipanti al Winter Camp avessero già familiarità con quei problemi, essendo in vetrina su questo sito da almeno 3 mesi... :wink:

Faccio un elenco degli errori trovati fino adesso:
- A1. Chiaramente $m,n$ devono essere positivi.
- G3. (anche se non è da fare). Dimostrare che vale quella formula se e soltanto se il triangolo è acutangolo o rettangolo.
- N5. Bisogna dimostrare che $\phi(n) \mid L(n)T(n)$
- N6. Le frasi dovrebbero iniziare con "a) $P(x),R(x)$are polynomials..." e "b) $P,R$ are polynomials..."

Sgrunt! :bomb: E' chiaro che non verranno al winter Camp[list]
[*]quelli che hanno spedito gli esercizi per raccomandata (nessuno ha voglia di fare ore di fila all'ufficio postale per ritirarla!);
[*]quelli che hanno spedito file con gli esercizi senza il nome (prova tu a riconoscerli dopo averli stampati);
[*]quelli che hanno spedito più volte o in più modi (vai a capire quale è la versione buona);
[*]quelli che hanno spedito file con scannerizzazioni illeggibili;
[*]quelli che hanno chiamato il file in modo diverso da quello raccomandato (il loro file non compare quando cerco con quel criterio);
[*]quelli che hanno spedito più esercizi del necessario.[/list]

[quote="Xamog"]Sgrunt! :bomb: E' chiaro che non verranno al winter Camp[list]
[*]quelli che hanno spedito più esercizi del necessario.[/list][/quote]

Grazie!!
E se vogliamo essere precisi, sull'e-mail che a me è arrivata da Gobbino c'è scritto:
[quote]
Quali sono gli esercizi da svolgere: 8 esercizi scelti tra quelli dell'Iran National Mathematical Olympiad (3rd round 2006), di cui 2 nel gruppo Algebra, 2 nel gruppo Combinatoria, 2 in Geometria e 2 in Teoria dei Numeri (con esclusione degli esercizi C3, G3 ed N1).
[/quote]
Io 8 esercizi li ho mandati. Non [b]solo[/b] 8 esercizi, purtroppo.
Se poi ho aggiunto qualche altro esercizio dell'Iran, che mi piaceva / mi era venuto / mi andava di scrivere, non potreste semplicemente ignorarlo?
Ero completamente all'oscuro dell'ultima clausola che hai aggiunto!

[quote="edriv"]non potreste semplicemente ignorarlo?[/quote]
Cercheremo un punizione adeguata. :lol:

Una correzione particolarmente larga di maniche :(, e una grossa pazienza di fronte ai disguidi di cui si è parlato in un post precedete :wallbash_red: , ha portato alla seguente lista dei convocati per il WC:

(S) CIMETTA Leone
(S) COLOMBO Maria
(S) CONTI Andrea
(V) COSTA Simone
(S) DALUISO Roberto
(S) DE CAPUA Antonio
(S) FOGARI Andrea
(S) GALEOTTI Mattia
(S) KUZMIN Kirill
(S) LAZZARINO Leslie Lamberto
(V) LEONETTI Paolo
(S) LO BIANCO Federico
(S) LIN Francesco
(S) LOMBARDO Davide
(V) MATTE BON Nicolas
(S) NARDIN Denis
(S) PATIMO Leonardo
(S) SANTI Elia
(S) TRAGER Matthew
(S) VERTECHI Pietro


Let be given f from C^1[0,1] and f(0) = 0, f(1) = 1, 0 <= f(x) <=1 for all x from [0, 1]. Show that there exist a, b from (0, 1), a <> b such that f'(a).f'(b) = 1.

Namdung

Hmmm... Maybe I'm hasty, but this seems pretty easy:

From Lagrange thm there is a c in (0,1) s.t. f'(c)=1. If f'(x)=1 for all x then we just choose a and b to be 2 different reals from (0,1).

Let's assume, then, that f'(x) isn't constant. The set M={f'(x) | x in (0,1)} is an interval because the function f'(x) has Darboux' property. If f'(x)>=1 for all x then f(x)-x is an increasing function, and since f(0)-0=f(1)-1 we get f(x)-x=0 for all x, so f(x)=x, so f'(x)=1 for all x, and we've already discussed this. Something similar happens if f'(x)<=1 for all x, so if f'(x) isn't always 1 then there exist 2 numbers u and v in M s.t. u<1<v, and since M is an interval (u,v) is included in M, and it's now obvious that we can find x=/=y in (u,v) s.t. xy=1.

Think this is right?


inequalities
function
inequalities theorems

I was recently reading an article in "Kvant" about the Jensen's inequality. With use of it the author of an article proves the Cauchy-Shwartz inequality for positive reals by applying Jensen to a convex function [tex]f(x)=x^2[/tex] and substituting [tex]\alpha_i[/tex] by [tex]\frac{m_i}{m_1+m_2+...+m_n}[/tex] in the statement of Jensen's inequality. He multiplies the obtained equality by [tex](m_1+m_2+...+m_n)^2[/tex] obtains:
[tex](m_1+m_2+...+m_n)(m_1\cdot x_1^2+m_2\cdot x_2^2+...+m_n\cdot x_n^2)\geq m_1\cdot x_1+m_2\cdot x_2+...+m_n\cdot x_n[/tex]
By subsituting [tex]m_i=b_i^2[/tex] and [tex]x_i=\frac{a_i}{b_i}[/tex] author gets the Cauchy-Schwartz inequality with its orginal notations.

After this proof there is an exercise in which the reader is asked to prove the Cauchy-Schwartz's inequality for all real numbers [tex]a_i,b_i[/tex]. Can't we apply the already written proof for this case, as none of [tex]\alpha_i[/tex] is negative (we may assume none of them is equal to 0, as the corresponding [tex]b_i[/tex]'s subtracts from both sides of the inequality)? Thank you, for your answer.

[quote="Xixas"]After this proof there is an exercise in which the reader is asked to prove the Cauchy-Schwartz's inequality for all real numbers [tex]a_i,b_i[/tex]. Can't we apply the already written proof for this case, as none of [tex]\alpha_i[/tex] is negative (we may assume none of them is equal to 0, as the corresponding [tex]b_i[/tex]'s subtracts from both sides of the inequality)?[/quote]

Yes, we can. It's strange why the author of the paper considers the case of negative numbers as a case to be considered separately.

  Darij

Another question on using Jensen's inequality:

If [tex]f''(x)>0[/tex], does the Jensen's inequality for a convex function [tex]f(x)[/tex] become strict too?

No.  (To see why, try $f(x)=x^2$)

What are necessary and sufficient conditions for equality in Jensen's inequality?

If the function $f$ is stricly convex then the equality possible $\iff$ $x_1=...=x_n$.
Otherwise we can find distinct $x_1,..,x_n$ s.t. the equality will hold.

What do you mean when you saying "the inequality becomes strict"? We can always attain the eqaulity be putting $x_1=...=x_n$.  :?


geometry
rectangle
circumcircle
analytic geometry
graphing lines
slope
conics

$ABCD$ is a rectangle. $A$, $B$ are on $y=2x+m$, $C$, $D$ are on $y^2=4x$. The equation of the circumcircle of $ABCD$ is $x^2+y^2-x-4y-t=0$.

Find $m$ and $t$.

$AB: y=2x+m$
$P: y^2=4x$
The equation of the circle can be written as
$(x-\frac{1}{4})^2 + (y-2)^2 = t+\frac{17}{4}$

So, if $(K,R)$ is the circle, then $K=(\frac{1}{2},2)$ and $R^2 = t+\frac{17}{4}$

$C(y_c,y_c)$
$D(y_d,y_d)$

$AB \parallel CD$ so the line $CD$ has also slope $2$
so we have $x_c \neq x_d$ and $y_c \neq y_d$ (we will use this fact later)

$(y_c-y_d) = 2(x_c-x_d)$

$C,D \in P \Rightarrow$

$y^2_c=4x_c, \ \ \ y^2_d=4x_d \Rightarrow$

$y^2_c -y^2_d = 4x_c - 4x_d$
$(y_c+y_d)(y_c-y_d) = 4(x_c-x_d)$
$(y_c+y_d)(y_c-y_d) = 2(y_c-y_d)$
$y_c+y_d=2$
because $y_c \neq y_d$

so, if $M(x_m,y_m)$ is the midpoint of $CD$ then $y_m = \frac{y_c+y_d}{2} = 1$

The perpendicular bisector of the chord $CD$ is the line $KM$, and has a slope $-\frac{1}{2}\Rightarrow$

$KM: y-2 = -\frac{1}{2}(x-\frac{1}{2})$

Applying $y_m=1$ we get $1-2 = -\frac{1}{2}(x_m-\frac{1}{2}) \Rightarrow$

$x_m-\frac{1}{2} = 2 \Rightarrow x_m=\frac{5}{2}$

So $M(\frac{5}{2},1)$

The line $CD$ has slope $2$ and passes through $M$. So its equation is

$CD: y-1 = 2(x-\frac{5}{2}) \Rightarrow  y=1+2x-5 \Rightarrow$

$y=2x-4=2(x-2)$

$C$ is on this line and also on the parabola

$y_c = 2(x_c-2) \Rightarrow  y^2_c = 4(x_c-2)^2$

$y^2_c = 4x_c \Rightarrow$
$4(x_c-2)^2 = 4x_c \Rightarrow$
$(x_c-2)^2 = x_c \Rightarrow$
$x^2_c -4x_c+4 = x_c \Rightarrow$
$x^2_c -5x_c+4 = x_c$

This equation has the roots $4$ and $1$.
The one is for $C$ and the other for $D$

Let $x_c=4 \Rightarrow  y_c = 2 \cdot (4-2) = 4\Rightarrow$
$C(4,4)$

applying this values to the initial equation of the circle $x^2+y^2-x-4y=t$ we get:

$t = 4^2+4^2-4-16 \Rightarrow  t = 12$

Since $AB \perp BC$ so $AC$ is a diameter of the circle $\Rightarrow  \overline{CA} = 2\overline{CK}$

$C(4,4)$
${K(\frac{1}{2},2})$

$\Rightarrow A(-3,0)$

$A \in y=2x+m \Rightarrow$

$0=2(-3)+m \Rightarrow  m=6$


calculus
derivative
factorial
calculus computations

The question asks me to find a formula of $f^{(n)}(x)$

I'm going to do a few here, then I will write the question of the one I do not understand and show my work of what I've done so far. Numbers 1 through 4, I've solved...it is number 5 I do not know.

1. $f(x) = x^n$

To solve it, I look at the pattern by figuring out the first couple terms
$f'(x) = nx^{n-1}$
$f''(x) = (n)(n-1)x^{n-2}$

So $f^{(n)}(x)=(n!)x^{n-n}$
$=n!$

2. $f(x) = \dfrac{1}{5x-1}$

$f'(x) = -5(x-1)^{-2}$
$f''(x) = 10(x-1)^{-3}$
$f'''(x) = -150(x-1)^{-4}$

so $f^{(n)}(x) = (-1)^n (5^{n-1})(n!)(5x-1)^{-(n+1)}$

There may be an easier to write this, but I know that this works....

3. $f(x) = e^{2x}$

This one it's easy to see that $f^{(n)}(x) = 2^n e^{2x}$

I wrote the first few out when I solved it, but I don't think it's needed here....

4. $f(x) = \frac1{3x^3}$

$f'(x) = -x^{-4}$
$f''(x) = 4x^{-5}$
$f'''(x) = -20x^{-6}$

I found this one to be
$f^{(n)}(x) = (-1)^n \dfrac{(n+2)!}{6}x^{-(n+3)}$

The sign alternates, that's why there is $(-1)^n$, the 6 in the denominator is caused by 3!...
There may be a cleaner way to write it, but again, I found this to work

Now for the one I could not solve...
5. $f(x) = \sqrt{x}$

It looks like it should be so simple, and I am close, but I have not got it....

$f'(x) = \frac12x^{-\frac12}$
$f''(x) = -\frac14x^{-\frac32}$
$f'''(x) = \frac38x^{-\frac52}$
$f^{(4)}(x) = -\dfrac{15}{16}x^{-\frac72}$
$f^{(5)}(x) = \dfrac{105}{32}x^{-\dfrac{11}{2}}$

I have solved for everything, but what really throws me off is the numerator values for the coefficient of x.
The pattern is 1, 1, 3, 15, 105, 945

I understand that we are multiplying by consecutive odd numbers, but I cannot figure out how to write the general form of that...

So far, I've solved for everything but that:

$f^{(n)}(x) = (-1)^{n+1} (\frac12)^n ??? x^{-\dfrac{2n-1}{2}}$

The ??? represents the product of the first n-1 odd numbers....
I feel that it is something really elementary that I am missing....and I feel kinda dumb for not knowing the pattern of the product of odd numbers...I've tried using factorials, but I can't seem to isolate the odd ones...

Any help with this part is really appreciated....

_ankur

[quote="ankur87"]I don't see how that works....
Could you explain that to me....

If I stick in $n=2$...I get $(2(2)-1)!!$
$=3!! = 6! = 720$

Where does 720 fit into the equation...I am not understanding....[/quote]
You misunderstand $n!!$. $n!! = n (n-2) (n-4) \cdots$
$3!! = 3\cdot 1 = 3$
$4!! = 4\cdot 2 = 8$
$5!! = 5\cdot 3 = 15$

Thank you both for your help, but I still am not understanding 100% of the concept of $n!!$

Is the definition $n!! = n (n-2) (n-4) \cdots$?

If so, then I've never seen that before, but it definitely works

Thanks shyong, your answer works also, I am just figuring out the thought process behind it

Again, thanks for the help, I really appreciate it

[quote="mathisfun1"]Let $\{x_n \} = 1, \ 1 \cdot 3 \cdot 5, \ 1 \cdot 3 \cdot 5 \cdot 7, \ ... \ 1 \cdot 3 \cdot 5 \cdot ...(2n+1)$ where $n =1, \ 2, \ 3, \ ...$.

Note that $1 \cdot 3 \cdot 5 = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}{2 \cdot 4 \cdot 6}$ $= \frac{6!}{2^3 3!}$, and that  in general, \[ 1 \cdot 3 \cdot 5 \cdot ... \cdot (2n+1) = \frac{(2n+2)!}{2^n n!}  \][/quote]

Assuming that I want to do $9 \cdot 7 \cdot 5 \cdot 3 \cdot 1$

That would be $2n+1 = 9 \Rightarrow n=4$

I know that $9 \cdot 7 \cdot 5 \cdot 3 \cdot 1 = 945$

Plugging $n=4$ into your equation, I get

$\frac{(2(4)+2)!}{2^4 4!}$

$=\frac{10!}{16 \cdot 4!}$

$=\frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5}{16}$

$=\frac{151200}{16}$

$=9450$

I think the formula is supposed to be $\frac{(2n+1)!}{2^n n!}$

Thanks for the explanation, that really cleared things up...It really helped me understand shyong's answer as well

I appreciate the help, everyone

[quote="mathisfun1"]I apologize for the error;  $\frac{(2n+2)!}{2^{n+1} (n+1)!}$ is the correct formula.[/quote]

Thanks for clearing that up...Soon after that post I realized what I wrote didnt work for other values of n....this one you have is working
So for all n, this formula will give you the product of odd numbers up to (2n+1)....

Thanks again


Determine the sum of absolute values for the complex roots of $ 20 x^8 \plus{} 7i x^7 \minus{}7ix \plus{} 20.$

$ 20 x^8 \plus{} 7i x^7\minus{}7ix \plus{} 20 \equal{} 20x^4(x^4\plus{}1/x^4)\plus{}7x^4(x^3\minus{}1/x^3)$
Now first substitute $ y\equal{}ix$ and one gets
$ 20y^4(y^4\plus{}1/y^4)\plus{}7y^4(y^3\plus{}1/y^3)$
Now a second substitution with $ z\equal{}y\plus{}\frac{1}{y}$ leads to
$ y^4(20z^4\plus{}7z^3\minus{}80z^2\minus{}21z\plus{}40)$
Now we only have to consider the solutions to the equation
$ (20z^4\plus{}7z^3\minus{}80z^2\minus{}21z\plus{}40)$

By putting z = -2, -1, 0, 1, 2 one sees that this equation has real solutions.
Now looking at the relation between y = a + bi and z
$ z\equal{}a\plus{}bi\plus{}\frac{1}{a\plus{}bi} \equal{} a(1\plus{}\frac{1}{a^2\plus{}b^2})\plus{}bi(1\minus{}\frac{1}{a^2\plus{}b^2})$

As the solutions for z are real we know that
$ 1\minus{}\frac{1}{a^2\plus{}b^2}\equal{}0 \implies a^2\plus{}b^2\equal{}1$
So we now know the absolute value of every solution y of the original equation without even knowing the solutions, looks nice and hopefully correct to me.

Yes I missed an i and I should have written the final answer so thanks to both of you ;-)
Do you have another solution for this problem? Would be nice to see another approach.


[quote][Unparseable or potentially dangerous latex formula.
Error 6 ][/quote]

does this have something to do with the problem?

[quote="math92"]two friends want alex to help them mow lawns.cindy says that she gets 80 dollars for the lawns she wants alex to cut and will give him 60% of this if he works for her. Banto says that he gets 60 dollars for the lawns he wants alex to cut and will give alex 80% of this.  Which friend should alex work for?[/quote]

[hide]Either friend would give alex 48 dollars. [/hide]

i am really sorry.  The dollar signs triggered latex.

[hide]It doesn't matter. They both give the same. If alex wanted he could work for both and get EXTRA MONEY!! :D[/hide]


Find $ f: A\rightarrow A$ such that $ f(x^2\minus{}y^2)\plus{}f(2xy)\equal{}f(x^2\plus{}y^2)$, for $ A\equal{}N,Z,Q,R$

For $ A \equal{} \mathbb{Z}$ at least, I don't think it's that simple: $ f(x) \equal{} x^2$ for $ |x| \neq 1$, $ f(1) \equal{} f( \minus{} 1) \equal{} \text{anything}$ seems to work.

I thought he was asking for an example of such a function  :lol:


probability
percent
articles
geometry
geometric transformation
reflection

I'm sure this there's 99% probability of this thread running into many controversies. For the sake of Pardesi (pardesi is studying so seriously, so lets not disturb him to come and delete posts and move threads), please dont use strong words. Let us discuss the verdict in general and its implications..

The Supreme Court is going to deliver its verdict on the OBC Quota Issue in premier institutions like IIT's, IIM's, etc.
http://ibnlive.com/news/judgement-day-sc-to-deliver-obc-verdict-today/63020-3.html

[hide]On a personal note, I hope they don't implement it this year.[/hide]

I hope they introduce quota for General category....

Breaking News::

SC Decides to uphold 27% quota in premier institutions http://ibnlive.com/news/sc-upholds-obc-quota-in-educational-institutes/63029-3.html.

Might even be implemented this year, but excluding the creamy layer. God knows how they will identify the creamy layer though.

[hide]Maybe I should convert to OBC[/hide]

For all practical purposes,I checked the CNNIBN and it said something about those having annual income of above 2.5 lacs and children of government level 1 and 2 officers as being part of definition of creamy layer.This means even my friends who are bc,st,sc,obc etc wont be able to avail the quota!!(u need atleast an income of more than 2.5 lacs to survive decently in chennai!!!).Well actually this reservation should come as no suprise seeing that the elections are drawing near.... :rotfl:

Iam only unhappy that the judgement has come before april 13 seeing that it is now going to have psycholigical impact on those writing JEE this year:(
Hopefully something can be done

well actually it hardly matters because now the not so useful IIT's get termed as useless IIT's :P  :rotfl:  :rotfl:

i feel that now it is the general category who really needs reservation in these institutions do you all know one thing in AIIMS only 32 seats are available for forward caste all others go the OBC but still our school PS Senior was able to get one student in Anand venkatraman who got a whooping AIR3 in it!!!! :D  :o

[quote="Ram"]

Iam only unhappy that the judgement has come before april 13 seeing that it is now going to have psycholigical impact on those writing JEE this year:(
Hopefully something can be done[/quote]

Definitely there will be a psychological impact. :( I wish the verdict had come after JEE. But I doubt any of us should be worried....Lets just do our best. But the cutoffs will increase for sure....

I think 2.5 lacs/annum implies that you can afford good education for your child (school level that is). The quota's purpose was to reach the poor penniless backward people.

But the big question is: Will it be implemented fully this year or like a 3tier system of 9+9+9.5 percent per year.. And Mr.A.Singh has promised to increase number of seats which means IIT's std may be diluted. because only what is rare, is precious.

[quote="phoenix76"]obc reservations wont matter to genius ones
who r assured of a top rank
it will matter to all those who r at the frontiers of qualifying and not qualifying
who basically aim for qualifying not for a rank :(
so i get to lose my seat if i do some crazy mistakes :o[/quote]

Your last two lines I guess, refer to the likes of me. :( Anyway, the reservation will affect your course. Its gonna be really difficult to get into the popular groups like CSC or EEE. If there are 40 seats in IIT-X , then only 20 will be available to the general public.

lets not get into who is on the frontier line, and who is not shall we??
i feel the people who come here, should be able to clear the exam with ease, but thats not the problem. the minimum rank to not compromise on both college and stream, will go up due the reservation (that is u will get a stream not of ur choice, in an IIT not of ur choice, if u r below that rank). and that should be our main concern, but not now, only after the exam!!
so all the best for JEE 2008!!

[quote="phoenix76"]the bottom 2 lines were meant for me

 :(
if u dont qualify i am sure i wouldnt  either
i just aim to get into an iit
i m not after a branch :)
its for people like shreyas,chapplis,soumyas,pardesis,madness,rakeshs and varuns :([/quote]

So you won't even mind Paper Pulp Engineering or Naval Engineering or Cotton Textile Eng?? Nothing wrong in taking those groups but still.....

[quote="phoenix76"]no i wont mind at all
i wont mind going to iit guwahati also[/quote]
you are a true iitian!!!!! :thumbup:  :lol:

Theres also a proposal for incr no of IITs :mad:
Increasing no. of IITs will definitely bring down d standards, :mad:
few weeks bac an old IIT M dean had blasted off d govt for proposing dis in his article in economic times.
[hide]he said theres a craze into politics so y dont u just incr d size of the parliament
so dt people like soumya get 2 bcom MPS :rotfl:  [/hide]

[quote="phoenix76"]i m pretty realistic
i know where i stand
so i dont expect things which r beyond my reach :)[/quote]
but i think u underestimate yourself too low u have got great marks in fiitjee compared to people here and why do you need to worry about rank!!!! :)

[quote="hash_include"]*long gurgling sound coming from the bathroom flush*

"What was that?! The toilet??"

"No, not really. It's just the Indian Education system going down the drain"

 :(  :mad:[/quote]

a perfect begining for the wrong ending!!!! god knows where this gonna doom to!!!! :mad:  :o

Just Concentrate and do ur best......  If u get a top 100 hundred rank ,, I think nobody can stop u .................

[quote="rituraj007"]quotas doesn't matters 4 AOPS at all :D  :roll:  :winner_first:[/quote]

quotas matters,even for big guns
last year closing rank for iit mumbai computers was 47(general)
don't you think it matters
even in medicals (AIIMS)there are very few seats for general
most probably ,creamy layer is going to be revised (income level is going to be increased,as can be deduced from reactions of politicians,politics is the only justification of quotas)
reservations should be only on the basis of income, not on caste,it segregates society
after all, the standard of iit is going to fall due to it
what is method of implementation (gradually increasing the quota or single slaughter)?
it is a question for most others present here,till my time full fledged quota would have been be implemented :(  :mad:

For all you know, the next thing to happen is a new qouta-general category quota-5% reservation

[quote="SOUMYASHANT NAYAK"]This year, we will probably have a new IIT in Andhra...

8 more IITs coming up and one of them in Orissa....

People will no longer go to private engineering colleges ..Everybody will be an IITian....

A good method to achieve Communist equality...

Down with Communism.... :rotfl:  :rotfl:[/quote]

Ofcourse, thats what the government wants. You see IIT'ians contribute a lot to the country's economy. So now more IIT'ians means more money.!! Nice logic eh?  :rotfl:  :rotfl:

Hail Equality, Hail Socialism, Hail Communism.

and ofc, the unsaid words are " Hail madness " :D
Personally, i tried to refrain from commenting from this topic as i didnt want to swear :D

There is huge caste-based discrimination in India, there's no denying that. It's just that reservation may not be the best solution to solving the problem that is social inequality. But real solutions like major improvement in the primary education seem very difficult and impractical. So I don't see a better way than reservation to make some progress in the direction of social reform. You cannot deny that there has been tremendous growth over the last fifty years due to the implementation of these policies.
Anyway, those are just my two cents. I guess as a student aspiring for the IIT reservation does make a difference. But - look at it this way - if the student who got through in your place was in your socio-economic position and you were in his/hers would he/she have made the most out of it? More than what you have? That's a hard question to answer. But the fact is that not everybody has the good fortune of being born into a good socio-economic situation. It's those who have done exceptionally well considering their socio-economic context who deserve to get a good education despite having average scores when compared to the competition.
Anyway, that being said - I sincerely hope none of you on AoPS will fall prey to that system, but remember that it's not all that senseless.
Ashwath

it never is senseless but there remains this fact:
People who get formal iit training shouldnt get quota benefits.
If people who dont have formal education get quota benefits, most of them join the iitss and fail there and in some cases commit suicide, so is the govt sending them their flight ticket to hell? :huh:

[quote="ashwath.rabindranath"]There is huge caste-based discrimination in India, there's no denying that. It's just that reservation may not be the best solution to solving the problem that is social inequality. But real solutions like major improvement in the primary education seem very difficult and impractical. So I don't see a better way than reservation to make some progress in the direction of social reform. [b]You cannot deny that there has been tremendous growth over the last fifty years due to the implementation of these policies. [/b]
Anyway, those are just my two cents. I guess as a student aspiring for the IIT reservation does make a difference. But - look at it this way - if the student who got through in your place was in your socio-economic position and you were in his/hers would he/she have made the most out of it? More than what you have? That's a hard question to answer. But the fact is that not everybody has the good fortune of being born into a good socio-economic situation. It's those who have done exceptionally well considering their socio-economic context who deserve to get a good education despite having average scores when compared to the competition.
Anyway, that being said - I sincerely hope none of you on AoPS will fall prey to that system, but remember that it's not all that senseless.
Ashwath[/quote]

well written, but i must say that i tend to disagree with you.
do you mean to say ppl belonging to the "Backward classes" have not at all done well till now?
Or do you mean to say that all of those who [b]Have[/b] achieved and who do belong to the aforesaid class are all untalented and depend solely on reservations.
I am sure you dont.
Frankly i think in such competitive exams, if a student wants to get through, He will come what may. This kind of compulsory reservation sometimes tends to bring in more people into iit's in the beginning, though a number of them to survive, there are many others who tend to drop out.This is not what i am saying myself but one of my professors in mathematics who has taught in iit bombay for 40 years and has himself been the part of the jee paper setting comitte severally many times.
Now to the lines i have highlighted:

Just because some development has been made do we become complacent and convince ourselves that yes Reservations are doing some work and we donot have to look for an alternative.
I think you are speaking ignorant of the fact that such decisions regarding quota etc. by whomsoever is indeed to have a good votebank in the following elections
No more, no less.
So We must all remember that reservation is certainly not a wise tool to bring the country forward.
Now a little message for others:
Please do not let such issues perturb you, just do your best in the exams and believe in yourselves.  :D

Well written rags! I have only one thing to add. I personally feel that our leaders must concentrate more on primary education, build better schools in villages and offer higher salaries to entice teachers to teach in Villagers. Those in the city belonging to middle class can afford decent education, but without primary education, many are suffering. That, according to me is the chief issue which must be addressed by the great leaders if they truly care about social development, but if political gimmicks is what people want, thats what they'll get. :(

[quote="Euclidean Geometer"]

well written, but i must say that i tend to disagree with you.
do you mean to say ppl belonging to the "Backward classes" have not at all done well till now?
Or do you mean to say that all of those who [b]Have[/b] achieved and who do belong to the aforesaid class are all untalented and depend solely on reservations.
I am sure you dont.
[b]Frankly i think in such competitive exams, if a student wants to get through, He will come what may.[/b] This kind of compulsory reservation sometimes tends to bring in more people into iit's in the beginning, though a number of them to survive, there are many others who tend to drop out.This is not what i am saying myself but one of my professors in mathematics who has taught in iit bombay for 40 years and has himself been the part of the jee paper setting comitte severally many times.
Now to the lines i have highlighted:

Just because some development has been made do we become complacent and convince ourselves that yes Reservations are doing some work and we donot have to look for an alternative.
I think you are speaking ignorant of the fact that such decisions regarding quota etc. by whomsoever is indeed to have a good votebank in the following elections
No more, no less.
So We must all remember that reservation is certainly not a wise tool to bring the country forward.
Now a little message for others:
Please do not let such issues perturb you, just do your best in the exams and believe in yourselves.  :D[/quote]

If the politicians want to use this to enhance the votebanks - by all means let them do so. As long as the reservations can do some good - I couldn't care less.
And yes - i think that without primary education reservation can only serve a very temporary purpose.
Your highlighted comment reflects your lack of knowledge of how bad circumstances can get. We certainly should try to look for an alternative - primary education being a very important one among them, but I think that equal opportunity for all is vital. Anyway, let's not have ideological debates here - that sort of thing can be reserved for a separate topic.

Oh - when I mean reservation I mean socio-economic reservation. There's no question about that.

[quote="Chronoz"][quote="ashwath.rabindranath"]

If the politicians want to use this to enhance the votebanks - by all means let them do so. As long as the reservations can do some good - I couldn't care less.
And yes - i think that without primary education reservation can only serve a very temporary purpose.
Your highlighted comment reflects your lack of knowledge of how bad circumstances can get. We certainly should try to look for an alternative - primary education being a very important one among them, but I think that equal opportunity for all is vital. Anyway, let's not have ideological debates here - that sort of thing can be reserved for a separate topic.[/quote]

If you had read some of our replies you might have realized that none of us here are against Social Equality. Do you really feel reservation will do good to more than 10 people? Obviously, the creamy layer will fail and again it will reach only those who can afford a good seat in a good college. When have your schemes ever reached the poor?

Well, I think discrimination today is only based on money: The haves and the havenots. There are so many millions who can't afford 2 square meals a day. Why not provide reservation to them? Irrespective of whether they belong to GE or SC or OBC. And many people misuse OBC Quota. By law you are no longer an OBC if you have converted to other religions but I personally know contradictions.[/quote]

Shre is absolutely right. I ve forgotten to mention that.
These reservations are ultimately misused by the ones who belong to this class but can afford to get themselves a "costlier" seat. it really fails to help the people who really need it.
by providing more and more of reservations, i think we are indirectly saying that they are hopeless and can not do without it.
I have quite a few friends belonging to the "OBC" but feel sternly that they are no less than others and needn't reservation beyond a limit.
Another important thing to be remembered is that quality must not be compromised in lieu of the bulk or quantity of ppl that get educated.
Primary education must be strong. it must provide every1 the mental courage that they donot need any reservation.
@ ashwath: you said its impractical improving primary education, easier a solution is reservation.
This i certainly fail to understand.
[hide] a day is surely to come when the so called "Forward class" or "general category" is gonna start desperately needing reservation, or atleast their fair share [/hide]
No offence is meant to any1.

By the way, the IIT's have decided to implement the reservation policy in phases http://sify.com/news/fullstory.php?id=14650981

I matters nothing for me
till 2010 full reservations would be implemented(i would be among first batch of students giving iit in full 50%reservation)

but their is one point i'd ie to make
to give reservations to economically poor is completely right,but is it right to give reservations to well off students(most of reservation beneficials would be among them)
i feel benefit of quota should be given only once to a person and his family not again and again
then only people who really deserve to get quotas would get them and their condition will improve, but most importantly schooling should be improved as many students who went to iit through quotas failed repeatedly and finally had to leave iit.They have been sufferers and they were not benefitted from quotas,in fact they would have better taken admission in NIT and got their degrees.

hey the system is getting down
see social equality is ok.........
but then the cream layer gets redefined and u know this creamy layer affects the education of india.it might not matter to ghenuiseslike u all but it does to us who barely hav a chance a get into iits

think bout it guys and give me ur further thoughts
 :lol:

i also think along the same lines as you

but it would be better if we don't take this discussion ahead
it may hurt others(now hardly any1 comes here)

if you have an examination with a question like'Who has done maximum harm to indian education??????'
the obvious answer would be ARJUN SINGH
in all ways,depreciation in standard of these institutions,creating havoc for students,not giving good students their legitimate share,etc

 :wallbash_red:  :diablo:  :furious:  :starwars:

hey skand i ought to be in the same thinking perspective as u
that s cool bro!

"


fifty-one small insects are placed inside a square of side 1.Prove that at any moment there are at least three insects which can be covered by a single circular disk of radius 1/7

this is very well-known..

[hide]divide the square into 25 squares, each of size $ \frac15\times\frac15$. by php, at least three insects is contained inside one of the unit squares. but the unit square can be covered by a disk of radius 1/7. q.e.d.[/hide]


Could anyone please help solving the following system of non-linear equations with 2 variables?..
(I've tried something with [i]Newton's method[/i], but I'm not sure of it. Any other ways of solving it, please?)
[b]
$ x^{2} \plus{} y^{2} \equal{} 41$
$ x^{3} \plus{} y^{3} \equal{} 189$
[/b]
(now, it may be obvious that a possible solution is  $ (x,y) \equal{} (4,5) \equal{} (5,4)$, but how do I [i]solve[/i] it?)

To prove that is a solution just substitute it back in

$ 4^{2} \plus{} 5^{2} \equal{} 41$
$ 4^{3} \plus{} 5^{3} \equal{} 189$

So $ (4, 5) , (5, 4)$ satisfy the system.

As far as solving I'd say brute force finds the solution the quickest.

The most common method I've seen for these is to find the largest and smallest possible $ x$ or $ y$ values.

Assuming $ x, y \in \mathbb{N}$ then, $ x^{2} \plus{} y^{2} \equal{} 41 \rightarrow 0 < x, y < 7$
and from the second $ x^{3} \plus{} y^{3} \equal{} 189 \rightarrow 0 < x, y < 6$. This helps you know what numbers to try and brute force with which may make the process quicker. Sorry if my wording is confusing.

Brute force doesn't work if you remove the restriction that $ x, y$ are integers.  rofler has given the general method for solving symmetric systems (where $ x, y$ can be interchanged to give the same system); the change of variables he uses is standard.

The most general method is to solve for one variable in terms of the other and plug in, but that's hardly ever nice.  Usually nonlinear systems involve a variety of algebraic tricks such as other kinds of changes of variables.

thank you all for your time and effort, especially [b]rofler[/b]. I really appreciate your immediate response :)

I understand the method [i]rofler[/i] has written, but I do have a question though:
[quote="rofler"]
$ p^3 \minus{} 123p \plus{} 378 \equal{} 0$

guess the root p=9, (honestly no other way other than cardano cubic formula) and we have:

$ (p \minus{} 9)(p^2 \plus{} 9p \minus{} 42)$
[/quote]
Indeed, [b]p=9[/b] is a root. But [i]how[/i] did you extract it?
Also, do you apply [i]Cardano cubic formula[/i], that you mentioned?
I searched at Wikipedia [url]http://en.wikipedia.org/wiki/Cubic_equation#Cardano.27s_method[/url] about [i]Cardano's method[/i] of extracting the roots of a cubic function, but unfortunately I can't figure it out whether you apply [i]Cardano's method[/i] or not.

If you have a nasty wasty polynomial such as we have, it's good to guess roots. $ 378\equal{}2\cdot 3^3\cdot 7$. We guess 2, 3, 6, and 7, but those don't work. We guess 9 and it works. Yay.

We can find that root using the [url=http://en.wikipedia.org/wiki/Rational_root_theorem]rational root theorem[/url].

[quote="rofler"][hide="I want you to fill in the remainder for practice (method that gives all 3 pairs of solutions)"]
Let $ p \equal{} x \plus{} y$ and $ q \equal{} xy$

$ p^2 \minus{} 2q \equal{} x^2 \plus{} y^2 \equal{} 41$

$ p^3 \minus{} 3pq \equal{} x^3 \plus{} y^3 \equal{} 189$

$ p^3 \minus{} 3pq \equal{} p(p^2 \minus{} 2q) \minus{} pq \equal{} 41p \minus{} pq \equal{} p(41 \minus{} q) \equal{} 189$

We know that p is not equal to 0 (since then $ p^3 \minus{} 3pq \equal{} 0$ ) so we have

$ q \equal{} 41 \minus{} \frac {189}{p}$

Plugging back into the first eq. we get

$ p^2 \minus{} 2(41 \minus{} \frac {189}{p}) \equal{} 41$

$ p^2 \plus{} 2\frac {189}{p} \equal{} 123$

$ p^3 \minus{} 123p \plus{} 378 \equal{} 0$

guess the root p=9, (honestly no other way other than cardano cubic formula) and we have:

$ (p \minus{} 9)(p^2 \plus{} 9p \minus{} 42)$

Which has the other 2 roots $ \frac { \minus{} 9\pm\sqrt {81 \plus{} 4*42}}{2} \equal{} \frac { \minus{} 9\pm\sqrt {249}}{2}$

Now, once you have p, you find q with one of the first 2 eq. (just plug in).

Then vieta's formula give 3 pairs of solutions (6 if you count them permuted), 2 extraodinarily ugly, one being (4,5)

P.S. when you get p=9, try that first, as it will probably yield easy solutions[/hide][/quote]

yes , this method works effectively for the problem which is recognized as when you switch x and y by each other , the system remain the same :)

and by the way , i often use computer or calculation to find root of cubic :P


calculus
integration
calculus computations

How about expanding the integrand as a geometric sum?
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112388]Similar problem[/url]


function
complex analysis
real analysis
real analysis unsolved

This is a very difficult result, but knowing the abilities of some members of the forum, I decided to post it here, hoping someone will come up with a relatively easy solution:
  Find all subseries of the series $ \sum_{n\geq 1}{\frac{x^n}{n!}}$ that are bounded for $ x<0$. That is, for which subsets $ A$ of the positive integers is the function $ \sum_{n\in A}{\frac{x^n}{n!}}$ bounded for $ x<0$?

Is the answer $ A\equal{}\mathbb N$ or $ A\equal{}\{4n\plus{}k,4n\plus{}k\plus{}1\}_{n\ge 0}$, $ k\equal{}0,1,2,3$? If so, I'll try to make my half-baked ideas into a rigorous proof; if not, I'll have to think a bit more. My approach is based on Borel transform and Fabry theorem, so it is complex analysis rather than real. Does the known solution follow the same lines?

Hi, Fedja, this is the answer! To my shame, I do not know what Fabry theorem is, but what they use is Laplace transform and a theorem of Szego posted by Vess some time ago and which remained unanswered, of course (I have a solution, but it's far too technical to write it on mathlinks): it's about analytic continuation of power series having only a finite number of distinct coefficients.


vector
Functional Analysis
linear algebra

Let L and K be the liniear subspaces of $R^n$ Euclidean space and dimL<dimK.
Prove that there exists vector in K that is orthogonal to any vector from L

[quote="DVO"]Have you tried using the Gram-Schmidt process?[/quote]

Use it for what? :?

Take an orthogonal basis of $\mathbb{R}^n$ .

Then some of the basis vectors are basis of $K$ and some basis of $L$ .

But $\dim K > \dim L$ hence there is a basis vector in $K$ that it is not

in $L$ and this will be orthogonal to any vector in $L$ .

Oh,thanx a lot.
It was so easy and i didn't solve it  :blush:

Well actually I have a question about the second line of that proof.

[quote]Then some of the basis vectors are basis of $K$ and some basis of $L$.[/quote]

If you start by picking an arbitrary orthogonal basis of $\mathbb{R}^n$, then it may be true that none of those basis vectors are in $L$.  That's a possibility.  So I'm not sure what it means to say that "some of the basis vectors are basis of $K$ and some basis of $L$."

Here's another idea.

Consider the case where $L$ is a one-dimensional subspace of $\mathbb{R}^3$ and $K$ is a two-dimensional subspace of $\mathbb{R}^3$.  Consider $L^{\bot}$, the subspace of $\mathbb{R}^3$ that is perpendicular to $L$.  $L^{\bot}$ has dimension 2.  Since $L^{\bot}$ and $K$ are both two-dimensional subspaces of $\mathbb{R}^3$, they must have a non-trivial intersection.  This implies that there is a non-zero vector in $K$ which is orthogonal to every vector in $L$.

I believe this idea can be used to make a good proof for the general case.

Turns out I'm not using Gram-Schmidt after all.

I think that one needs use Gram-Schmidt. But the idea is basically the same.

[b]Problem.[/b] [color=blue]Let $L$ and $K$ be subspaces of $\mathbb R^{n}$ such that $\textrm{dim}\, L < \textrm{dim}\, K$.

Prove that there is a non-zero vector $v$ in $K$ that is orthogonal to any vector from $L$.[/color]

[i]Solution.[/i]

Let $\ell = \textrm{dim}\, L$ and $k = \textrm{dim}\, K$. By Gram-Schmidt construct a basis $\left\{ e_{1},\ldots,e_\ell \right\}$ for $L$. Next, extend it to a basis $\left\{ e_{1},\ldots,e_\ell,e_{\ell+1},\ldots,e_{n}\right\}$ for $\mathbb R^{n}$. It's clear that $\left\{ e_{\ell+1},\ldots,e_{n}\right\}$ span the subspace orthogonal to $L$. Let this subspace be $L^\prime$.

Then,
\begin{eqnarray*}\textrm{dim}\, \left( L^\prime \cap K \right) &=& \textrm{dim}\, \left( L^\prime \right)+\textrm{dim}\, \left( K \right)-\textrm{dim}\, \left( L^\prime+K \right) \\ \ &=& \left( n-\ell \right)+k-\textrm{dim}\, \left( L^\prime+K \right) \\ \ &=& \left( n-\textrm{dim}\, \left( L^\prime+K \right) \right)+\left( k-\ell \right) \\ \ &\geqslant& 0+\left( k-\ell \right) > 0. \end{eqnarray*}

Thus, $L^\prime \cap K$ has dimension $\geqslant 1$, and the conclusion follows. $\; \; \; \; \; \blacksquare$

The statement is also true in infinite dimensions: If $L$ and $K$ are closed subspaces of a separable Hilbert space and $\dim L<\dim K$ then there exists $u\in K$ with $u\in L^\bot$.


For example, how do you convert 2x^2+3y^2=5 to parametric mode

x=r cos(theta)
y=r sin (theta)
x squared plus y squared = r squared
tangent(theta) = y/x
that should be everything you need to know to solve it

what you just said was about how Convert Cartesian equations to polar equations, not parametric equations.
Parametric equation is like x=at+b y=ct+d...

wly3298456, polar equations are  a kind of parametric equations.  Anything of the form:
$x=f(t)$
$y=g(t)$, counts as a parametric equation.

Plug in $x=at+b$ in your equation.  You cannot find a y of the same form generally.


Find all positive integers $x;y$ greater than 1 so that $2xy-1$ divisible by $(x-1)(y-1)$


In $ \triangle ABC$ with no obtuse angle.Prove that ineq:
                 $ \frac {(1 \minus{} cos2A)(1 \minus{} cos2B)}{1 \minus{} cos2C}$ + $ \frac {(1 \minus{} cos2C)(1 \minus{} cos2A)}{1 \minus{} cos2B}$ + $ \frac {(1 \minus{} cos2B)(1 \minus{} cos2C)}{1 \minus{} cos2A}$ $ \geq$  $ \frac{9}{2}$


For any real numbers, x_0, x_1, ..., x_n, find with proof, the minimum value of f[x]=\sum_{k=1}^{n} |x-x_k|.

Of course we can monotonize so that x_0<=x_1<=...<=x_n, and it is easily seen that the minimum lies within the interval [x_0,x_n]. After experimenting, I've come up with the conjecture that f[x] attains its minimum if and only if x=x_{n/2}+1 (n even) and on the interval [x_{Floor[n/2]},x_{Ceiling[n/2]}] (n odd), but I am unable to prove this.

This question was from Kiran Kedlaya's MOSP 1995 Inequality notes.

I am afraid I am not sure I'm reading what you wrote correctly:  in the case n even, did you mean x = x_(n/2 + 1)  (more likely)  or
x = x_(n/2) + 1  (which is what you wrote, I believe, but seems less likely)?

Assume that the minimum value is not taken at any of the points x=x_0, x=x_1, .. x=x_n, but is taken at x=a.

Now all you have to do is see that [hide]if we consider the point a+b for an arbitrarily small b, then the new value of the function is b times (number of x_i less than a - number of x_i more than a) more than the original. Similarly a-b is (number more - number less)*b more. But one of these must give a new minimum, contradiction, or the number more=number less where we can shift over to the nearest x_i and the minimum is taken there, also contradiction.

So the minimum value is taken at one of the x_i, and its pretty easy to see it has to be the middle one. And as before, if there are an even number of x_i, it will be in that range.[/hide]


algebra
difference of squares
special factorizations

Hi,

I need some help for solve in N^3 this equation : 1 + 4ab = c^2

Thanks a lot.

[hide="Solution"]Taking this mod 4, we see that $ c^2 \equiv 1 \mbox{ mod 4}$, so $ c$ must be odd. Letting $ c \equal{} 2d \plus{} 1$ for some nonnegative integer $ d$, we see that $ 4d^2 \plus{} 4d \plus{} 1 \equal{} 1 \plus{} 4ab$, i.e., $ d(d \plus{} 1) \equal{} ab$.

In addition, all numbers $ a, b$ that satisfy $ ab \equal{} d(d \plus{} 1)$ with $ c \equal{} 2d \plus{} 1$ for some nonnegative integer $ d$ also satisfy the equation, i.e., $ \{(a, b, c) : \exists d \in \mathbb{Z}^\plus{} (ab \equal{} d(d \plus{} 1) \mbox{ and } c \equal{} 2d \plus{} 1\}$ is the solution set (I cannot see any way to simplify the condition that $ ab \equal{} d(d \plus{} 1)$.)
[/hide]

Another way to achieve the same result is to subtract 1, factor using difference of squares, and then use the fact that (c-1)(c+1) cannot be even if c is even.

Overkill is much more fun than parity.

[quote="Temperal"]yet another way is to note that a multiple of four is even and therefore one more than a multiple is odd?[/quote]
Yeah, on hindsight, I didn't even need mods...


Let $ n\in\mathbb{N}$ and $ p,q$ irrational such that $ a\plus{}\sqrt{ab\plus{}n},b\plus{}\sqrt{ab\plus{}n}\in\mathbb{Q}$
Prove that $ n$ perfect square and $ a\plus{}\sqrt{ab\plus{}n}\plus{}b\plus{}\sqrt{ab\plus{}n}\equal{}0$

The desired equality is equivalent to $ a \minus{} b \equal{} 2 \sqrt {n}$, and we know already that $ a \minus{} b$ is rational.

Let $ a \minus{} b \equal{} 2r$ for some rational $ r$ and let $ q \equal{} a \plus{} \sqrt {a(a \minus{} 2r) \plus{} n} \in \mathbb{Q}$.  Then

$ q \minus{} a \equal{} \sqrt {a(a \minus{} 2r) \plus{} n} \implies$
$ a^2 \minus{} 2aq \plus{} q^2 \equal{} a^2 \minus{} 2ar \plus{} n \implies$
$ 2a(r \minus{} q) \plus{} (q^2 \minus{} n) \equal{} 0$

which is linear in $ a$.  But since $ a$ is irrational, it follows that the coefficients must be zero, which gives the result.

Edit:  This is a cute problem.  Source?


Let be $ a,b\in \mathbb{R}$ . Prove that :

$ \left(\frac {3a \plus{} b}{2}\right)^2 \plus{} \frac {13b^2 \plus{} 41}{4}\ge \frac {13a \plus{} 3b}{2}$

$ Mx^2\plus{}Nx\plus{}P \ge 0$ and $ N^2\minus{}4MP \le 0$


[quote="laughinghead505"][quote="davidli"][hide]$(a-1)a(a+1)=8(a-1+a+a+1)$
$(a^{2}-1)a=24a$
$a^{2}=25$

$16+25+36=77$[/hide][/quote]
[hide] When you divided by $a$, you ruled out the possibility of $a=0$ You have to be careful not to do that. :)

An alternative answer would be:

$(-1)^{2}+0^{2}+1^{2}=2$[/hide][/quote]

Oh yeah, that's right. I always forget zero.

When I saw this problem I thought that negative numbers would work but then I saw that it was squares so negative doesn't matter.
When we learned about quadratic functions in school, our teacher kept telling us "DON'T FORGET ZERO!!!" And it worked! :P

It kind of surprises me that people forget zero a lot because it is usually the obvious answer.

[quote="now a ranger"]The product of three consecutive integers is 8 times their sum. what is the sum of their squares?[/quote]
[hide]
$24a=(a-1)(a)(a+1)$
$24a=(a^{2}-a)(a+1)$
$24a=(a^{3}-a^{2}+a^{2}-a)$
$24a=a^{3}-a$
$a^{3}-25a=0$
$a(a^{2}-25)=0$
$a(a-5)(a+5)=0$
$a=0, 5, or-5$
sum of squares if a = 0: 2
sum of squares if a = 5 or -5: $16+25+36=77$
so the answer is either 2 or 77

[/hide]

[quote="bpms"][quote="jhredsox"][hide]

$abc=8(a+b+c)$

it is obviously 4,5,6

so, the answer is $16+25+36=77$

[/hide]

davidli, please tell me this isn't like urs (didn't look to make sure)

-jorian[/quote]
How is that obvious?[/quote]

I bet he just looked at the answer and then wrote that.

omg, why do you think that? no seriously, i got it myself (not that hard)

oh and it's obvious because idk. i thought it was something else that was obvious, then it took me 10 more secs cause it wasn't, but i was too lazy to remove the wording

-jorian

Okay, jorian, your answers are fine. It was just that last time that your post was annoyingly similar to my post. You don't have to say things like you didn't look at mine.

Please post the source next time.

I think this was AMC 10/12 2002B.

[quote="Rafaelloaa"]i think it's [hide]77[/hide]

but i might be miles off[/quote]
You should show your work just in case you made a mistake or something.
BTW, there is one other pair...... (other people have made solutions)


if a,b,c>0 ,and $ab+bc+ca=1$,
please find the minimum of $1/(a+b)+1/(a+c)+1/(c+b)$

thank u everyone :)

[quote="little boy"]if a,b,c>0 ,and $ab+bc+ca=1$,
please find the minimum of $1/(a+b)+1/(a+c)+1/(c+b)$
[/quote]
$\inf\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=2.5$ but my solution is very ugly.

[quote="little boy"]it doesn't matter
can u show me some steps :wink:[/quote]
OK. $\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\geq2.5\Leftrightarrow\frac{a^{2}+b^{2}+c^{2}+3}{a+b+c-abc}\geq2.5\Leftrightarrow$
$\Leftrightarrow4(a^{2}+b^{2}+c^{2}+3(ab+ac+bc))^{2}(ab+ac+bc)\geq$
$\geq25((a+b+c)(ab+ac+bc)-abc)^{2}\Leftrightarrow$
$\Leftrightarrow4\cdot\sum_{sym}(a^{5}b-a^{4}b^{2})+3(a-b)^{2}(a-c)^{2}(b-c)^{2}+$
$+8\cdot\sum_{cyc}(a^{4}bc+a^{3}b^{2}c+a^{3}c^{2}b+\frac{7}{3}\cdot a^{2}b^{2}c^{2})\geq0.$ :)

It was a pleasant surprise to learn of this method
but i want to know how can u think of it
it's too hard for me   :(

How did you come up with the number right away? do you use some kind of method? Because you solve nearly every inequality here...can you tell me that method?Please, please... :help:

There are many receptions how to receive the sum of squares. In our problem I used that $(a-b)^{2}(a-c)^{2}(b-c)^{2}=$
$=\sum_{cyc}(a^{4}b^{2}+a^{4}c^{2}-2a^{4}bc-2a^{3}b^{3}+2a^{3}b^{2}c+2a^{3}c^{2}b-2a^{2}b^{2}c^{2}).$

This is very useful equality.

It's a tough problem which was posted before (ask Soarer). You can find it in Hojoo's notes .

upstairs... do u know a nicer solution about this question?
please write it down ... thank u

please search. Besides arqady's miraculous method (really amazing!!), we can use mixing variables to tackle it.

Yes, mixing variables works. But my proof is still ugly:
Let $f(a,b,c)=\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}-2.5$ and $g(a,b,c)=ab+ac+bc.$
Hence, $g(a,b,c)=g(a,\sqrt{(a+b)(a+c)}-a,\sqrt{(a+b)(a+c)}-a).$
Let $a=\min\{a,b,c\}.$ Then
$f(a,b,c)-f(a,\sqrt{(a+b)(a+c)}-a,\sqrt{(a+b)(a+c)}-a)=$
$=\frac{1}{a+b}-\frac{2}{\sqrt{(a+b)(a+c)}}+\frac{1}{a+c}-\left(\frac{1}{2\sqrt{(a+b)(a+c)}-2a}-\frac{1}{b+c}\right)=$
$=\left(\frac{1}{\sqrt{a+b}}-\frac{1}{\sqrt{a+c}}\right)^{2}-\frac{(\sqrt{a+b}-\sqrt{a+c})^{2}}{2(b+c)(\sqrt{(a+b)(a+c)-a)}}=$
$=\frac{(b-c)^{2}(2(b+c)\sqrt{(a+b)(a+c)}-2a(b+c)-(a+b)(a+c))}{2(a+b)(a+c)(b+c)(\sqrt{a+b}+\sqrt{a+c})^{2}(\sqrt{(a+b)(a+c)}-a)}\geq0,$
which true since, $(b+c)\sqrt{(a+b)(a+c)}-2a(b+c)\geq0$
and $(b+c)\sqrt{(a+b)(a+c)}-(a+b)(a+c)\geq0.$
Id est, remain to prove that $f(a,b,b)\geq0$, where $2ab+b^{2}=1.$ I think it is not hard.

I remembered Soarer posted another way using also mixing variables (from a Chinese book?) .

Yes, I also solved it with mixing variables. But what made you choose a such teribble approach?
[quote="arqady"]Let $f(a,b,c)=\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}-2.5$ and $g(a,b,c)=ab+ac+bc.$
Hence, $g(a,b,c)=g(a,\sqrt{(a+b)(a+c)}-a,\sqrt{(a+b)(a+c)}-a).$
[/quote]

[quote="Lovasz"]
Yes, I also solved it with mixing variables. But what made you choose a such teribble approach?
[quote="arqady"]
Let $g(a,b,c)=ab+ac+bc.$
Hence, $g(a,b,c)=g(a,\sqrt{(a+b)(a+c)}-a,\sqrt{(a+b)(a+c)}-a).$
[/quote][/quote]
Only desire that $g$ did not vary.  :)

Sorry, I meant this one,
[quote="arqady"]
$f(a,b,c)-f(a,\sqrt{(a+b)(a+c)}-a,\sqrt{(a+b)(a+c)}-a) \ge 0.$
[/quote]

[quote="Lovasz"] I meant this one,
[quote="arqady"]
$f(a,b,c)-f(a,\sqrt{(a+b)(a+c)}-a,\sqrt{(a+b)(a+c)}-a) \ge 0.$
[/quote][/quote]
[quote="Lovasz"] But what made you choose a such teribble approach?
[/quote]
Such terrible approach?
Probably I on another am not able. :maybe: :)

thank u !!thank u !!
i love u ,arqady :P

To Lovasc : The mixing variables method Arqady used is to keep $ab+bc+ca$ unchange . This's the most natural method using mixing variables .

[quote="little boy"]:|  could u hlep me to find the connection  if u don't mind[/quote]
sorry, i can't find it now.

here it is:

http://www.mathlinks.ro/Forum/viewtopic.php?t=31377

 :)

[quote="romano"]To Lovasz: The mixing variables method Arqady used is to keep $ab+bc+ca$ unchange . This's the most natural method using mixing variables .[/quote]
Thanks.
[hide=" :!: "]

[quote] Given $a,b,c \ge 0$ satisfying $ab+bc+ca=1$. Prove that: \[{1 \over a+b}+{1 \over b+c}+{1 \over c+a}\ge{5 \over 2}.\] [/quote]

[b]Solution:[/b] Let $f(a,b,c)={1 \over a+b}+{1 \over b+c}+{1 \over c+a}$. Then: \begin{eqnarray*}f(a,b,c)-f\left(a,\frac{b+c}{2},\frac{b+c}{2}\right)&=&{1 \over a+b}+{1 \over a+c}-{2 \over a+{b+c \over 2}}\\ &=&{1 \over a+b}+{1 \over a+c}-{4 \over 2a+b+c}\end{eqnarray*} It’s obviously true since ${1 \over x}+{1 \over y}\ge{4 \over x+y}$.
So that: \[f(a,b,c) \ge f(a,t,t) \quad{\rm where}\quad t^{2}+2ta=1.\] It remains to prove that: \[{2 \over a+t}+{1 \over 2t}\ge{5 \over 2}\\ \iff (t-1)(5t^{2}-4t+1) \le 0.\] Thus... with $t \le 1$.
Done.
[/hide]

I can use Iran 96 to solve this problem but not nice! :blush:

See also here
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=449840&p=3617601#p3617601


Let $ f: R\rightarrow Z$ and:
$ i)f(x+1)=f(x)+1;$
$ ii)$ For all $ n \in N: f(x)=f(\frac{f(nx)}{n});$
Prove that for all $ x \in Q: f(x)=[x]$ or $ f(x)=[x]+1.$

Since f(x+1)=f(x)+1, f(x+2)=f(x+1)+1=f(x)+2, f(x+3)=f(x)+3...
So, for any natural k, f(x+k)=f(x)+k   (1).
If k and x are both natural, then f(k+x)=f(k)+x   (2).
By (1) and (2), we write: f(x)-f(k)=x-k. So, for any natural pair a and b, f(a)+f(b)=a+b, If a and b are reals, then f(a+b) equal to the most natural, not more then x. So, f(x)=x or f(x)=x+1. By putting it in
the second given thing, we will make sure, that these functions are what we were looking for.
Q.E.D.


function
real analysis
real analysis unsolved

Find a differentiable function $f: \mathbb{R}\to\mathbb{R}$ such that $|f|$ is not differentiable anywhere.

$|f|$ is differentiable on the set $\{x\colon f(x)\ne 0\}$.


Solve the equation:
$ \Bigg(\sin^2{x}\minus{}\frac{1}{\sin^2{x}}\Bigg)^2\plus{}\cos^2{x}\minus{}\frac{1}{\cos^2{x}}\equal{}\frac{3}{4}$

[quote="tdl"]Solve the equation:
$ \Bigg(\sin^2{x} \minus{} \frac {1}{\sin^2{x}}\Bigg)^2 \plus{} \cos^2{x} \minus{} \frac {1}{\cos^2{x}} \equal{} \frac {3}{4}$[/quote]
If you put $ x \equal{} \frac{\pi}{4}$, the equation is satisfied.


logarithms
calculus
limit
algebra unsolved
algebra

Determine the number of solutions of the simultaneous equations $ x^2 \plus{} y^3 \equal{} 29$ and $ \log_3 x \cdot \log_2 y \equal{} 1.$

[quote="Mathias_DK"][quote="orl"]Determine the number of solutions of simultaneous equations to $ x^2 \plus{} y^3 \equal{} 29$ and $ log_3 x \cdot log_2 y \equal{} 1.$[/quote]
Let $ x \equal{} 3^k$ and $ y \equal{} 2^j$ we then have $ kj \equal{} 1 \iff k \equal{} \frac {1}{j}$

So $ f(j) \equal{} \sqrt [j]{9} \plus{} 8^j \equal{} 29$. How do you find $ f'(j)$?

For $ j \in ( \minus{} \infty;0]$ there is obviously no solution....
For $ j \in [1;\infty)$ there is only one solution, since $ f(j)$ is increasing.
For $ j \in (0;1)$ there is only one solution cause $ f(j)$ is decreasing.,

I have no idea how to find the solutions.. I'll post, if I get one.

Edit: $ f'(j) \equal{} ln(8) \cdot 8^j \minus{} \frac {ln(9) \cdot \sqrt [j]{9}}{j^2}$[/quote]
Ok,good,i think you will solve it,The result is $ 2$pairs $ (x,y)$
But i cant find clearly $ x\equal{}...,y\equal{}...$ who can?

[quote="Allnames"][quote="Mathias_DK"][quote="orl"]Determine the number of solutions of simultaneous equations to $ x^2 \plus{} y^3 \equal{} 29$ and $ log_3 x \cdot log_2 y \equal{} 1.$[/quote]
Let $ x \equal{} 3^k$ and $ y \equal{} 2^j$ we then have $ kj \equal{} 1 \iff k \equal{} \frac {1}{j}$

So $ f(j) \equal{} \sqrt [j]{9} \plus{} 8^j \equal{} 29$. How do you find $ f'(j)$?

For $ j \in ( \minus{} \infty;0]$ there is obviously no solution....
For $ j \in [1;\infty)$ there is only one solution, since $ f(j)$ is increasing.
For $ j \in (0;1)$ there is only one solution cause $ f(j)$ is decreasing.,

I have no idea how to find the solutions.. I'll post, if I get one.

Edit: $ f'(j) \equal{} ln(8) \cdot 8^j \minus{} \frac {ln(9) \cdot \sqrt [j]{9}}{j^2}$[/quote]
Ok,good,i think you will solve it,The result is $ 2$pairs $ (x,y)$
But i cant find clearly $ x \equal{} ...,y \equal{} ...$ who can?[/quote]
I think it's almost impossible to do without the computer(and very easy with). It does'nt even give a "nice" expression like $ \frac{ln(3) \plus{} ln(2)}{4*4.21^4}$, when i run it through.........

hello, from the second equation we get
$ y=e^{\frac{\ln(3)\ln(2)}{\ln(x)}}$ for $ x>0$,by inserting this in the first equation we have $ x^2+\left(e^{\frac{\ln(3)\ln(2)}{\ln(x)}\right)^3=29}$. I see here no chance to solve this equation explicit with respect to $ x$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, from the second equation we get
$ y = e^{\frac {\ln(3)\ln(2)}{\ln(x)}}$ for $ x > 0$,by inserting this in the first equation we have $ x^2 + \left(e^{\frac {\ln(3)\ln(2)}{\ln(x)}\right)^3 = 29}$. I see here no chance to solve this equation explicit with respect to $ x$.
Sonnhard.[/quote]
But finding $ x$ is very hard (without  computer) Can you?

Mathematica says "The equations appear to involve the variables to be solved for in an essentially non-algebraic way."

Was there some non-calculus solution offered? This one is sort of ugly, and calculus generally isn't required...

claim x^3=a,y^3=b,
therefore loga.logb=6log2.log3=k(say)
and see that a+b=29 and loga.logb=k(some constant)
hence loga.log(29-a)=k
there are only two values possible for a
and the answer follows two solutions
(is it rigorous enough :wink:  :wink: )

i have done a calculation mistake but the idea is the same
correction:
                take x^2=a instead of x^3=a
                now the answer follows in the same way

$x^2 + y^3 = 29$ and $log_3 x \cdot log_2 y = 1$ then from $log_3 x \cdot log_2 y = 1$, we get $x^y=2^3$ , from $x^2 + y^3 = 29$ we get $x=+\sqrt{29-y^3}$, equating we obtain $(29-y^3)^y=2^6$
Let $f(y)=(29-y^3)^y$ ,then $\lim_{y\rightarrow{0}}f(y)=1$ , $f(2)=21^2>2^6$ , and there exist a $y_0$ for which $f(y_0)=0$.
$f'(y)=(29-y^3)^y[log(29-y^3)-\frac{3y^3}{29-y^3}]$ ,then $f'(1)<0$ and $f'(y_0)\approxeq{0}$, also there is a $y_1$ in $(1,y_0)$ for which $f'(y_1)=0$
So $f(y)$ attains two times $2^6$(since $f(2)>2^6$), also after $y>y_0$ the eq. $f(y)<0$
Hence there is two solutions of these equations.

[quote="orl"]Determine the number of solutions of simultaneous equations to $ x^2 \plus{} y^3 \equal{} 29$ and $ log_3 x \cdot log_2 y \equal{} 1.$[/quote]
[quote]Determine the number of solutions of simultaneous equations to $ x^2 \plus{} y^3 \equal{} a (a \geq 17)$ and $ log_3 x \cdot log_2 y \equal{} 1.$
[/quote]

We can find that $y^3+3^{\log_y{2}}=29$. It is apparent upon graphing the equations (disregard variables, the following graphs are on the $xy$ plane)
\[
x^3=y
\]
and
\[
29-3^{\log_x{2}} = y
\]
that there are exactly 2 points of intersection, so the answer is thus $\boxed{2}$.


advanced fields
advanced fields unsolved

We denote $ C(T)$ and $ D(T)$ to be the center set and centroid set respectively of a tree $ T$. For each $ n \geq 2$, construct a tree $ T$ in which $ \text{dist}_T(C(T), D(T)) \geq n$. I don't see how to construct these trees.

Start with a left-right path made of $ n\plus{}1$ vertices, then vertex $ z$, then $ n$ more vertices, last being $ w$. Now add enough leaves, let's say $ 3n$ to the end $ w$.
Vertex $ z$ is the center, at distance $ \leq n\plus{}1$ from any other vertex, while any other vertex is at distance $ \geq n\plus{}2$ from some vertex.
Vertex $ w$ is the centroid, with all subtrees of size $ \leq 2n\plus{}1$, while any other vertex has some larger subtree (for example $ z$ has a subtree of size $ 4n$).
The distance between $ z$ and $ w$ is $ n$.


[quote="GameBot"]The lengths in feet of three pieces of timber are 48, 72, and 40. The sawmill operator needs to cut the timber into logs of equal length with no waste. How many feet long is the greatest possible length she can cut?[/quote]

We need the GCF, which is 8. Thus, the greatest length she can cut is 64.

[quote="nikeballa96"]wait...am i misunderstanding or something? why wouldn't it just be 8? i think i'm failing, but i don't know why...[/quote]

The largest piece is 72. She can cut 64 off of that to get a piece of length 8.


I think it should be $f(x^3+y^3+z^3)=f(x)^3+f(y)^3+f(z)^3$, right¿