AMC AIME MATHCOUNTS search ARML geometry function

(Mods--I'm not sure if here is the best place to post this; please feel free to move this thread to a more appropriate location.)

Next year, we will try to start a math club at my school, and I would appreciate any suggestions anyone has.

First, I’d like to explain a little about my school and school district. My school is really heavy on humanities/liberal arts and quite lighter on math/science; in fact, the only math competition we offer is basically the AMCs. We’ve only had five AIME qualifications in the history of our school district, and this year, we sent out personal invitations for the AMCs in addition to announcements, but only 7-8 people showed up (some with the motivation of missing class).

My school district does, however, have many strong math students; the middle school MathCounts program is quite successful, for example. Unfortunately, MathCounts ends in 8th grade, and there is no math program to hold up MathCounts alums. Sadly, by sophomore and junior years, student participation in activities differentiates so much, and interest in mathematics (and academic competitions in general) declines substantially.

I strongly believe that, especially in the lower grades, there is interest in mathematics that can be harnessed. For this reason, I'm planning to help spearhead a movement to establish a running math program in the high school. We already have a boost with a supportive administration and substantial funds available from a grant proposal I wrote specifically for this purpose. Additionally, I know probably we’ll need to get an enthusiastic coach (I’m planning to talk to a teacher), and most importantly, members.

I’m wondering about other suggestions/ideas. For example: What is the best way to recruit new members? What kinds of competitions should we participate in? Should they involve lots of travel or be stay-at-home type contests? How many should we do? For a new club, what general structure/schedule would be best to follow? Weekly meetings? How would our meetings be structured, anticipating limited membership and relatively inexperienced problem solvers? What kinds of books/resources should we utilize? Most importantly, how would we best sustain interest in members and develop critical mass? How does math club work at your school?

Thanks in advance. This actually isn’t the first time I’ve tried to help start a math club. We didn’t have a great start last time, so I’m really hoping this try will be successful.  :)

Ah!!! It's a curse when we write up a long thread and no one answers it. Maybe people just like small blocks of text and such... For one thing, I don't know the answer to all your questions and it might be a turn off for some people to try to answer all of them :(

The main thing is to have your team do problem sets regurlarly. Consider going over the problems and/or giving out the solutions afterward. Also consider giving awards and having rankings as incentives for members to study. That's how TJ http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=477000247&t=199100

does it. It motivates a few smart kids to work... meh.

The most important thing is to have a supportive sponsor for your club, and it sounds like you have that.  Having money is also good.  I am shocked that people are giving you money; most clubs have to raise their own money.

Some simple suggestions, based on the fact that you're trying to build math competition culture out of nothing:

Try to compete in whatever local competitions you can.  Even if the quality isn't that great, they tend to be more fun as a social activity than tests that you take at your own school (like AMC).  It also builds camaraderie as well as rivalry with other schools.  After all, you're probably not going to bring a team to ARML and bring down the big boys, but you can certainly try to take on Shelbyville.  Unfortunately, a lot of areas don't have local math competitions.  You might even try non-math competitions just to get people interested in competing.  Stay-at-home contests are also worth doing and require minimal effort, but I don't think they're your key to success.

In general, try to compete as much as possible.  Practice is boring.  Competing is fun.  Most people only practice because it leads to winning.  People have to understand the desirability of winning before they can understand the point of practice.

Try to organize at least one big trip per year, perhaps to a regional math competition.
Everyone likes field trips, especially overnight ones.

Try to convince your math teachers to offer extra credit for math competition participation.  (Note that I said participation rather than performance.)

Instead of inviting everyone to take the AMC, try to convince your math teachers to simply cancel all advanced math classes and just force everyone to take the AMC by default.  Or if you can't do that, at least get them to offer extra credit.

Tell people that being on the math team and winning math awards looks really good on your college application.

I respect what the previous poster wrote, but you should know that what works at TJ is probably not going to work for you.  Maybe if you said where your high school is, others might have more specific advice for you.  I wish you the best of luck.

I'm sure JBL's post would answer all of your questions, but I could offer a little help. If your school system isn't too focused on math/science, take the people you can and ask what afternoons they have free. You can base a schedule off that. Also, check with the parents and print up fliers if they want their child to go to faraway contests or stay-at-home ones.

Entirely based on the people. Like the Articles of Confederation  :P

For those younger students who have no reason or will to join, try doing a separate date for those younger students. If you plan something like Jeopardy, Are you Smarter than a ____ Grader, or a Deal-Or-No-Deal Math game it will certainly attract most people. Even a "lottery" with prizes such as difficult problem solving flash cards could score a few people.

For contests, emphasize the AMC's, and have a couple teams take the iTest. The iTest is free, and it can be found here: http://www.theitest.com/

Once the club has grown in size, you might want to try some bigger competitions, like the ARML Team Selection Test(or equivalent), HMMT, and Princeton. Some people are against driving(especially long distance), so car pooling would be a good idea for the ones that are far away. If you meat weekly at say dinner time, you might want to bring a pizza or two; my Math Circle does that(pizza money is needed, like 10 dollars a month).

My math circle sends out problems on our yahoo group (AMC 12s, AHSMEs, AIMEs, HMMTs, ARMLs, you get it), and every week we meet (at dinner time,  :roll: ), work for an hour, have pizza, and get to working on problems. In the last 45 minutes of the meeting, we usually do either a relay round, a guts round, or work. After three hours are up, people clean up and drive home.

A way to get people to come is to make a member list, and have everyone check their name off when they come. You establish a point system: +3 points for going to a meeting, -2 points for missing a meeting, +100 points for going to a contest, +50 points for competing in the AMCs, +100 points for scoring high/advancing to the next competition, etc.

Also recommend AoPS and it's books, ACoPS, and MAA books.

My MathCounts group has a coach who hands out problems to groups of students, and assistant coaches for each group of students. If you don't know a problem, it can be explained by the coach, or another student. The coaches are people who are in high-school or who have participated in MathCounts competitions for three years.

I strongly second everything that yenlee said. The 1 or 2 hardcore kids might be motivated by your having an ACoPS or AoPS book in the sponsoring teacher's classroom, and they might be motivated by the once-a-year AMC and opportunity to qualify for AIME. But if you really want a good number of people to come, if you want to build an actual program, you need:

- accessible things (reasonably easy contests)
- frequent things (look up Math League for whatever state you're in, Mandelbrot, NAT, other periodic contests)
- travel (people LOVE going to contests together)
- motivation (extra credit for math classes, [i]just for participation[/i], is one of the best ways to do this, as yenlee said)

You could also use your funds to provide some extra motivation, like:

- randomly distributed prizes for people who participate in (say) AMC, or all 6 Math League tests (again participation only)
- keep overall records for people in the club (rankings) and give prizes for that. (this can be a mix of participation and performance)
- provide snacks after the in-school tests (try to make sure that kids who don't take the test can't steal the food)

Thank you so much to everyone who helped. I'll help implement your suggestions next year.

Make it accessible. I HATE WAKING UP IN THE MORNING!!! but you can do this once kids get serious. start rewarding. i know its cliche but it works
take in-school tests and allow teamwork
people love working together

if you want to win though....
practice hard... WITH PIZZA (of course make them pay a little, but not so much that it is a lousy deal)

Our conditions are exactly similar to the OPs conditions except "parents and teachers" despise olympiads and math  :mad:

What Xevarion said is [b]completely[/b] right. My teacher *tried* to start math discussion club out of our school. But all in vain.

Here is our story:
First day, he had around 150 students. But within weeks it crumbled downed to 25. Finally after two months, it stabilized to 2 students. At the regional olympiad, no one turned up except one!(Thats me by the way) and I did not get selected!

So at the end of the day, Sir and I analyzed the reason for failure. The reasons were:

1)No funds. He could not provide good study materials.

2)No schools appreciate learning maths as a hobby! They all turned down offers made by sir. In fact some teachers even advised students not to join the club. They believed the students could invest their time in something more fruitful like "textbook" cramming.

3)When Sir arranged a talk by good mathematicians in the country, only 20 people turned up. Because everyone was afraid they will not understand what they say. Their parents told them it will be a waste of money.

So my Sir is [b]still[/b] fighting this losing battle. I believe if anyone had funded the venture and parents or teachers were supportive, the students would automatically gain interest.

So I believe the most important ingredient in the recipe is support. If you have financial support, you can have good resources, if you have school support, you can get students and if you have parents and faculty support, your students can get good motivation.

Anyway these are my thoughts because of the bad experience I had. It was never meant to discourage you. I really hope and wish that your math club becomes successful

[quote="k8reindeer"]your math club's success will depend on the people in it. not the number of people, the people themselves. focus on getting together your group of people who are interested in math(teachers too)--everything else will follow.[/quote]

It also depends on the number of people to soem extent. You would want a math club to stay alive... with let's say 10 people.

eh, to be honest, i don't participate in math club at all

it's not really interesting, and none of my friends are in it, so it's like "why bother going?"

with that said, there has to be some sort of motivational impetus to get kids interested in it. like i mean if you guys want a core group of really smart kids who get together and do high-tier competitions, that's great. but the annoying truth is that - at least where i live - there aren't that many ridiculously intelligent kids to do those things. so if you want an active math club up and running, you have to get a way of getting more "normal" kids to join, compete, and be motivated. it's definitely not easy, because socially, math has a pretty negative stigma attached to it, and...

basically, good luck.

[color=green]Moderator says: there is [i]never[/i] a good reason to block-quote the entire post preceding your own.[/color]

Disproven by ccy. His club failed worse than cube club

[color=green]Moderator says: there is [i]never[/i] a good reason to block-quote the entire post preceding your own.[/color]

But his club was web design, and about 2 people in our school are interested in that. Contrastingly, we have a whopping 14 people interested in math.

[quote="Xevarion"]- travel (people LOVE going to contests together)[/quote]

Let me emphasize this. Whenever we go to a contest that needs travel(HMMT, Princeton, National MC, you name it), we love playing games. Chess, cards, Contact, and even some "What now?", are great to play as a group.

For card games, Egyptian Rat Screw, Presidents, Mao, and Set are great card games.

[quote="serialk11r"]One good way to attract more math people is bringing food...proven by bookaholic.[/quote]

I would definitely agree with this; even if not all of them are interested in math, it gets more people to at least try it out and experience it before they judge the competitions.

I would add Mafia to that list of games to play, especially if you're going to ARML. Mao is fun (and one of my favorites), but I wouldn't recommend it as a starting game because it tends to annoy a great deal of people, especially those that haven't played it extensively before (I know that the first time I played it at ARML, I was wondering what in the world was going on because people just kept giving me random instructions, and then giving me cards when I didn't follow them  :lol: )

Ummm...to be honest, sponsors, as great as they are, are overrated. If you have a strong math club program, I have found that parents are quite willing to donate and support their kids (and it adds up fairly quickly). It's great that you have a sponsor, and it may give you a lot more options, but focus on the MATH. That's what keeps the core group interested.

Competitionwise, I agree with yenlee that local ones are a lot more fun. If all you're doing are AMC's, a fair number of people are probably going to just get bored and give in. Even if you might not be vying for a top spot, sending a team to ARML would still be a great idea. Although it might not have the toughest math questions (at least compared to USAMO), it's a social competition, and I've found that it tends to be one of the funnest ones, especially if the timing of your plane flights gives you some free time as a team before/after the actual competition.

-piggy *oink* *oink*

yuck...long block of texts...

here's some practical advice:

have food during practice

make it fun, don't just make them do problems, have them work on problems together, or even try to race to see which group can solve it the fastest

go to lots of competitions, especially far off ones, staying in hotel and traveling is awesome.

have food

try to let the parents know about this, they can force their kids to go since i doubt many kids will actually like doing math...

have 1 or 2 meetings per week

hope this helps, i know i'm not very good at giving advise, but meh..i'm doing the best i can.


Let AD,BE,CF be the medians of $\triangle ABC$. Let k be the number for which $a>k|b-c|,b>k|c-a|,c>k|a-b|$ such that $AB\cdot BC\cdot CA>AD\cdot BE\cdot CF$. Determine the minimum value of $k$.

still no reply..... :(
Maybe it'd better for me to post it in the Inequality Forum.


linear algebra
matrix
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be the set of all $ 3 \times 3$ real matrices of determinant $ 1$ of the form
\[ \begin{pmatrix} a & 0 & 0 \\ b & x & y \\ c & z & w \end{pmatrix}.\]Verify that $ G$ is a group. Find a homomorphism from $ G$ onto the group $ GL_2(\mathbb{R})$ of all non-singular $ 2 \times 2$ real matrices, and find its kernel.

G is block lower triangular, so you can multiply block wise.  For the homomorphism just take the bottom-right block.

The problem is basically identical to taking G={[a,0;b,x] : ax=1} and finding a homomorphism to GL(1,R) by sending [a,0;b,x] to x.  In you case "x" is taken from the ring M_2(R) and b from the bi-module R^2, but all of the expressions used when a,b,x are in R still work.


Ring Theory
superior algebra
superior algebra solved

[quote="suedenation"][color=green]In each of the following cases, determine if S is a subring of R.
(a) R = Z[size=75]9[/size], S = { [0], [3], [6] }.
(b) R = Q × Q, S = { (r, s) : r, s in Q, r <= s }.
(c) R = Q, S = { r/s : r, s in Z, s is odd }[/color]

Thanks very much!!!  :D[/quote]

1. Get some sleep.
2. grab your algebra book and check if you know the definitions well.

These steps are meant to rid you of confusion.

3. (a) Make the table for addition and multiplication for $\mathbb Z_9$ and verify that $S$ is an additive subroup stable under multiplication.
4. (b) The operations on $R$ are the component-wise addition and multiplication. Check that $S$ is not an additive subgroup. If $(r,s)\in S$ and $r<s$, then $(-r,-s)\not\in S$.
5. (c) check that this is a ring


algebra
function
domain
geometry
3D geometry

Given that three distinct numbers a,b,c lies in $(0, \infty)$, find the smallest value of a such this equation is true.

$a = b^2 = c^3$

Silverfalcon:  The phrasing of this question suggests that real number solutions are acceptable.   However, both posted answers make the sensible assumption that only integer valued solutions are acceptable.  Should the question be clarified?

And another thing. What do you mean by "Given that three distinct numbers a,b,c lies in (0, :inf: )"?

Of course, a trivial solution would be 1=a=b=c! I think that might be the solution needed, since it is greater than 0 and smaller than (4,8,64).

Billy

To rcv,

Yes, you're right. Perhaps, I should've added it.

To dwx314,

That shows the domain of the answer which is from 0 to any positive numbers, not INCLUDING 0.

[hide]
Since there is an squared and cubed term. we know that we use 2^6.
64=a, b=8, c=4.

[/hide]

[hide="Hint"] 1 doesn't work b/c it asks for distinct integers... and I guess u can assume that the question is asking for an integer or else any numbers would work... It also cannot be negative b/c you have a square... Think of squares and cubes[/hide]
[hide="Solution"]a=64 (Smallest positive cube and square)
b=8 (:sqrt: 64)
c=4 (3 :sqrt: 64)
Answer: a=64
[/hide]
-It's a good time to be a New Englander: GO PATS and SOX!!!
--I can't wait until hockey's back... I'm with the owners on this one...


Hey.

I am looking for a few people who want to have a little competition. The prize for 1st is 200 dollars and it is pretty easy and just for fun. There will be a few questions and it'll be about Groups etc. 1st person to solve all of them wins. It'll be a short list. If you're interested, post here. Contest will be held sometime this coming week. It'll be quick and fun.

I'll give it a try, I'm not doing much this week

Sure, count me in, I just learned some representation theory so this should be interesting.

What are the rules? Also, will you e-mail us to let us know? (My e-mail is jacob.steinhardt@gmail.com).

I may end up being too busy to take the contest, but I still would like to see the problems.

count me in.
I am guessing you mean it is about group theory right?

I'd be [i]very[/i] suspicious that OP is trying to get you all to do his homework for him.

[quote="JSteinhardt"]Who is OP, and do you have any reason to back up this claim?

In response to another post, it is about group theory as well as representation theory, presumably.[/quote]

OP is the original poster (really, isn't it obvious from the context), and I am suspicious since it is near finals time at most north american universities and because he is offering a 200 dollar cash prize.  Really, use some common sense.

[quote="coquitao"][b][color=red]Count me in,
Mr. Duckling!
Gonna be quite a treat,
even if I don't succeed.[/color][/b]

[b]P.S. Could those lines of mine be regarded as haiku?[/b]  :D[/quote]

I thought Haiku only have 3 lines!?

In that case, I think it would be reasonable to ask that the contest not occur until after Christmas. That way we could be sure that finals were over, and thereby minimize any possible ulterior motives without striking down a potentially legitimate contest.


$ x_{1}\equal{} a$
$ 3x_{n}\plus{}\frac{1}{x_{n}}\equal{} x_{n\minus{}1}\plus{}\frac{3}{x_{n\minus{}1}}$
Find a to $ x_{n}$ is integer for all n
In this case find $ lim x_{n}$

$ (3x_{n}\minus{}x_{n\minus{}1})(1\minus{}\frac{1}{x_{n}x_{n\minus{}1}})\equal{}0$, so we have $ a\equal{}\pm 3^{k}$, $ k$ non-negative integer, and $ x_{n}\rightarrow\pm 1$


Princeton college Duke AMC AIME USA(J)MO USAMO

I'm adding the 2007-2008 calendar now, I'll split it off once it gets filled up.  I just copied and pasted last year's schedule but I'll edit it to agree with this year as dates get posted.  So right now almost everything is tentative.

(T = Tentative)

[b][u]November 2008[/u][/b]
[url=http://cgi.math.princeton.edu/mathclub/index.php/Princeton_Math_Competition]Princeton University MathContest[/url]
November 8 -  [url=http://www.math.uga.edu/mathmeet.html/]UGA Math Tournament[/url]
November 13th (Tuesday) - [url=http://www.unl.edu/amc/]AMC 8[/url]
November 17th (Saturday) - [url=http://www.math.duke.edu/dumu/contests/2006/index.html]Duke Math Meet[/url] (T)

[b][u]December 2008[/u][/b]

[b][u]January 2009[/u][/b]
January 19th - Rockdale Mathematics Competition (T)

[b][u]February 2009[/u][/b]
February 6th (Wednesday) - [url=http://math.kennesaw.edu/~ckoppelm/]Kennesaw State Part II[/url] (L)
February 12th (Tuesday) - [url=http://www.unl.edu/amc/]AMC 10/12 A[/url]
February 23rd (Saturday) - [url=http://math.cofc.edu/MATHMEET/]College of Charleston Math Meet[/url]
February 27th (Wednesday) - [url=http://www.unl.edu/amc/]AMC 10/12 B[/url]

[b][u]March 2009[/u][/b]
March 18th - [url=http://www.unl.edu/amc/e-exams/e7-aime/aime.shtml]AIME[/url]

[b][u]April 2009[/u][/b]
April 2nd - [url=http://www.unl.edu/amc/e-exams/e7-aime/aime.shtml]Alternate AIME[/url]
April 19th - Walton MathFest (T)
April 29th - [url=http://www.unl.edu/amc/e-exams/e8-usamo/usamo.shtml]USAMO Day 1[/url]
April 30th - [url=http://www.unl.edu/amc/e-exams/e8-usamo/usamo.shtml]USAMO Day 2[/url]

[b][u]May 2009[/u][/b]
May 30th - [url=http://www.arml.com/]ARML[/url]
May 31st - [url=http://www.arml.com/]ARML[/url]

[b][u]June 2009[/u][/b]

[u][b]Unknowns:[/b][/u]
USC
USC Upstate
Furman
Coastal Carolina
USC Aiken?
Winthrop?

[u][b]Potential Tournaments:[/b][/u]
[url=http://math.kennesaw.edu/~ckoppelm/]Kennesaw State Part I[/url] (L)
[url=http://www.gacs.pvt.k12.ga.us/facstaff/sbaggett/math_tournament.htm]GACS[/url]
[url=http://www.woodward.edu/academics/us/math/mathteam_sjv_tournament.asp]Woodward JV[/url]
[url=http://home.comcast.net/%7Edrgarnerjr/RMC.htm]Rockdale[/url]
[url=http://www2.fultonschools.org/teacher/pricea2/MathVarsity/MVarsityMilTourn.html]Milton Math Tournament[/url]
[url=http://math.kennesaw.edu/~ckoppelm/]Kennesaw State[/url]
[url=http://www.math.gatech.edu/%7Epme/hsmc/]Georgia Tech[/url]

Your friendly neighbor in North Carolina says:
AMC 8 - 11/13/07
AMC 10/12 - 2/12/08, 2/27/08
AIME - 3/18/08, 4/2/08
USAMO - 4/29-30/08

That date for the Duke Math Meet is tentative but probable, I think.

The new AMC dates have been up. They are on the AMC 8 brochures (its a PDF).


algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

Hi
I need some help with this question. $ L$ is a field of characteristic $ 0$ and consider the simultaneous equations over $ L(t)$
$ a^2$= $ b$+$ t$
$ b^2$= $ c$+$ t$
$ c^2$= $ a$+$ t$
I can show that $ a$ either satisfies $ a^2$=$ a$+$ t$ or a certain sextic over $ L(t)$. These are the questions

1) show that the Galois group of the splitting field of the sextic preserves an equivalence relation amongst  it six roots.

2) let $ (x,y,z)$ be an equivalence class of the roots  and let $ f(p,q,r)$ be a symmetric polynomial in three variables. Show that $ f(x,y,z)$ belongs to an extension of $ L(t)$ of degree at most $ 2$

3) Show that this system can be solved by radicals.

I'd really appreciate any help whatsoever.

Regards

1) Galois fell in love with Stéphanie. This caused its loss. If you choose "Galois" as a pseudo, then you can do a small effort to read its life (in wikipedia for instance). Can you do it ?
2) Has somebody a solution ?  I can't show that the equation in $ a$ can be solved by radicals. Necessarily (except for exceptional cases) there is a cubic root (and even $ 2$) in the calculation of the $ 6$ roots.

In fact  a generalization is as follows:
let $ f(x)$ be an irreducible polynomial of degree 2 over $ K$. Then $ f^{(3)}(x)\minus{}x\equal{}(f(x)\minus{}x)p(x)$ where $ degree(p)\equal{}6$. Let $ L$ be the field of decomposition of $ p$. Then $ [L,K]\equal{}18$ and $ p$ is solvable. Why ?
Other example:
let $ f(x)$ be an irreducible polynomial of degree 3 over $ K$. Then $ f^{(2)}(x)\minus{}x\equal{}(f(x)\minus{}x)p(x)$ where $ degree(p)\equal{}6$. Let $ L$ be the field of decomposition of $ p$. Then $ [L,K]\equal{}48$ and $ p$ is solvable. Why ?

The following is a solution of the $ 2$ questions asked in my post#5.
 Asssume that $ p$ is irreducible and let $ a$ be a root of $ p$.
1) Then $ b\equal{}f(a),c\equal{}f^{(2)}(a)$ are also roots of $ p$. Thus $ [K(a,b,c),K]\equal{}6$. Let $ d$ be a fourth root of $ p$.Then $ e\equal{}f(d),g\equal{}f^{(2)}(d)$ are also roots of $ p$. Thus $ L\equal{}K(a,b,c,d,e,g)$ and $ [L,K(a,b,c)]\equal{}[K(a,b,c)(g),K(a,b,c)]\equal{}3$. Finally $ [L,K]\equal{}6.3\equal{}18\equal{}3^2.2$. According to the $ p^nq^m$ Burnside's theorem, $ L$ is solvable over $ K$.
2) Then $ b\equal{}f(a)$ is also a root of $ p$. Thus $ [K(a,b),K]\equal{}6$. Let $ c$ be a third root of $ p$. Then $ d\equal{}f(c)$ is also a root of $ p$. Thus $ [K(a,b,c,d),K(a,b)]\equal{}[K(a,b)(c),K(a,b)]\equal{}4$. Let $ e$ be another root of $ p$. Then $ g\equal{}f(e)$ is also a root of $ p$. Thus $ L\equal{}K(a,b,c,d,e,g)$ and $ [L,K(a,b,c,d)]\equal{}[K(a,b,c,d)(e),K(a,b,c,d)]\equal{}2$. Finally $ [L,K]\equal{}6.4.2\equal{}48\equal{}3.2^4$. According to the $ p^nq^m$ Burnside's theorem, $ L$ is solvable over $ K$.

hey thanks
Oh well I've read about Galois and his life in Stewart's book on Galois theory :)
It's just that i couldn't remember at that instant who Stephanie was ( given the fact that my supervisor's name is the same :D)
Thanks again


superior algebra
superior algebra unsolved

Hello,
I would like to make a small code for Maple and therefore I need a command for a radical of an integer.
The radical of an integer is just the product of its prime factors (each factor appears once in the product):
Examples: $ rad(2^2)\equal{}2$ and $ rad(3*3*5)\equal{}3*5\equal{}15$.

If there is no such command, can anyone help me to make some code to calculate the radical?

with(numtheory):
u:=factorset(n);
> l:=nops(u);
> s:=1:
> for i from 1 to nops(u) do
> s:=s*u[i]:
> od:
> s;

And you may remove that 3rd line ( l:=nops(u); ).


You are given an equation x^2 - 4x + 6k = 0. If the difference of the squares of its roots is 8, find the value of k.

[hide]

solving for the roots, in terms of k, using quadratic formula gives

x=2  :pm:   :sqrt: (4-6k)

squaring each root and subtract smaller from the larger returns

8 :sqrt: (4-6k)

since difference of the squares is 8

8 :sqrt: (4-6k) = 8

:sqrt: (4-6k) =1

4-6k = 1

k=1/2, and the two roots are 3 and 1  [/hide]

Yes, this is correct.

Now try to find another way to do this.

Hint: [hide]Sum of roots and Product of roots[/hide]

I did it that way the first time, but if it's x^2-4x+6k, you know that it is in the form of (x-2)^2 or (x-3)(x-1), and the latter is the only one with diff of squares as 8.

I like the viete-ish method better, though, and I'll let someone else show it

Nice one.

its works if the roots are rational. like if they aren't... but in this problem, they are.

you know that $m+n=4$ (sum of roots) and $m-n=2$ so you just solve the system and get $k$ through the product mn.

Please do not reveal old threads.

I don't mind this happening if the problem wasn't answered by anyone but like in this case, please don't.

Locked.


inequalities
geometric inequality
inequalities proposed

Hello,

Let $x_1,x_2,\cdots,x_n$ for $n\geq 3$ be positive real numbers. Prove that:
\[
{{1}\over{x_2+x_3+\cdots+x_n}}+{{1}\over{x_1+x_3+\cdots+x_n}}
+\cdots+{{1}\over{x_1+x_2+\cdots+x_{n-1}}}<\left({{x_1+x_2+\cdots+x_n}\over{x_1x_2\cdots x_n}}\right)^{{{1}\over{n-1}}}.
\]

Use the arithmetic geometric inequality for the numerators in the left hand side and then the pOWER mean inequality. I do not have a pper, but I think it works.

Well, guess where I have found it: in Komal Contest set of problems for September! So, be careful, rules are very strict!


inequalities logarithms blogs function calculus

Consider the following series:

$S_n = 1+\frac{1}{2}+\cdots+\frac{1}{n}$

$T_n = S_1+S_2+\cdots+S_n$

$U_n = \frac{T_1}{2}+\frac{T_2}{3}+\cdots+\frac{T_n}{n+1}$.

Prove that $T_n+\ln{(n+1)} > U_n+n$.

[hide="hint"]Reading the most recent post of my [url=http://wangsblog.com/jeffrey/]blog[/url] will probably help a lot... [/hide]

Let $F : \mathbb{N} \rightarrow \mathbb{N}$, the function $F(n) = T_n + ln(n+1) > U_n + n$.

[b]Lemma 1.[/b]  $T_n = (n+1)(S_{n+1} - 1)$.  Proof.  In $(n+1)(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n+1})$ we overcount the number of 1's by 1, $\frac{1}{2}$'s by 2, and the number of $\frac{1}{k}$'s by k, so in total we overcounted by 1, $n+1$ times.  Result follows.

--

[b]Lemma 2.[/b] $F(n+1) - F(n) > 0$.  Its equivalent to

$(T_{n+1} - T_n) + (\ln (n+2) - \ln (n+1) ) + (U_n - U_{n+1}) + (n - (n+1)) > 0$.

$\Leftrightarrow S_{n+1} + \ln \frac{n+2}{n+1} > \frac{T_{n+1}}{n+2} + 1$

and using lemma 1, $\Leftrightarrow \ln \frac{n+2}{n+1} > S_{n+2} - S_{n+1}$

or $\Leftrightarrow \ln (\frac{n+2}{n+1})^{n+2}) > 1$, but $(\frac{n+2}{n+1})^{n+2} > e$ where the result follows. (e is the limit when n gets big, and the function is trivially decreasing using calculus)

-----------

We can verify $F(1) > 0$.  Then $F(n+1) > F(n) > \dots > F(1) > 0$, for all $n$.  Result follows.


Can someone tell me what does \null command do? Some example?

\null is a TeX command and is shorthand for \hbox{} which is an empty horizontal box. You'll find the details on this in Donald Knuth's original TeX manual, [url=http://www.amazon.co.uk/TeXbook-Donald-E-Knuth/dp/0201134489/ref=sr_1_1?ie=UTF8&s=books&qid=1240909879&sr=8-1]The TeXbook[/url]. It is not easy reading and is really for those who want to dig deep into the system.
\hbox{} is also mentioned in Victor Eijkhout's freely available book [url=http://www.eijkhout.net/texbytopic/texbytopic.html]TeX by Topic[/url] which you can download from there. I notice that Victor was one of the people who replied to one of your queries on comp.text.tex.


linear algebra
matrix
vector
real analysis
real analysis unsolved

(a) if $ H(p) \equal{} \frac {1}{r}|p|^r$ show $ L(q) \equal{} \frac{1}{s}|q|^s$
is the legendre transform of H when $ \frac {1}{r} \plus{} \frac {1}{s} \equal{} 1$

(b) if $ H(p) \equal{} \frac {1}{2}\sum_{i,j \equal{} 1}^n a_{ij}p_ip_j \plus{} \sum_{i \equal{} 1}^n b_ip_i$ where $ A \equal{} (a_{ij})$ is symm positive def matrix and b is a vector in R^n what is $ L \equal{} H^*$?

(c) define $ \partial H (p) \equal{} \{ q: H(r)\ge H(p) \plus{} q\cdot (r \minus{} p) \forall r\in \mathbb{R} \}$ and show
$ q \in \partial H(p)$ iff $ p \in \partial H^* (q)$ iff $ p\cdot q \equal{} H(p) \plus{} H^*(q)$

i hate you hopf-lax!

I can't understand the text. Could you rewrite it?


can someone give a nive solution to these inequilities please.

1) $ \frac{(x\minus{}2)(x\plus{}3)(x\minus{}4)}{(x\minus{}1)}<0$

2) $ \frac{2}{x\plus{}5}>x\plus{}4$

3) $ \frac{x(x\minus{}2)}{2x\minus{}5}\geq3$

4) $ \frac{x(x\plus{}5)}{x\minus{}4}>\minus{}2$

5) $ \frac{x\plus{}3}{x\minus{}1} \geq \frac{x\minus{}1}{x\plus{}4}$

6) $ \frac{x\minus{}1}{x\minus{}3}<\frac{x\plus{}2}{x\minus{}5}$

i suck at inequilities.

[quote="Elio (n)"]I have never solve inequations of root more than 1. But i think I know.....

1)$ \frac {(x - 2)(x + 3)(x - 4)}{(x - 1)} < 0$ for all $ x\neq{1}$

$ x_1 < 2$
$ x_2 < - 3$
$ x_3 < 4$

2)$ \frac {2}{x + 5} > x + 4$ for all $ x\neq{ - 5}$
$ 2 > (x + 4)(x + 5)\leftrightarrow{2 > x^{2} + 9x + 20}$
$ 0 > x^{2} + 9x + 18\leftarrow{0 > (x + 3)(x + 6)}$

$ x_1 < - 3$
$ x_2 < - 6$[/quote]

Say what?

binomial, have you heard of a "sign-chart", they're used sometimes to solve inequalities with lots of roots which get messy, Try drawing one up for each problem,

If you haven't, post again and I'll draw you up one.

EDIT: Never mind, here is an example of a relatively simple one from my C1 notes

Solve $ (x-4)(x-1) > 0$

\begin{tabular}{|c\parallel{}c|c|c|c|c|}
\hline
& & CV & & CV &  \\ \hline
x & 0 & 1 & 3 & 4 & 5 \\ \hline \hline
x-4 & - & - & - & 0 & + \\ \hline
x-1 & - & 0 & + & + & + \\ \hline \hline
P & + & 0 & - & 0 & +\\
\hline
\end{tabular}

From this it is clear that when the product is positive, it evaluates to a number greater than $ 0$ and therefore, we can say that, $ x<1 \mbox{ or } x>4$

For the record, that was just an excuse to show off some latex which I think look cool

$ 1)$

The roots are at $ \minus{}3, 2, 4$. It is undefined at $ 1$.

Case $ x < \minus{}3$: all results will be greater than 0.
Case $ \minus{}3 < x < 1$: All results will be less than 0.
Case $ 1 < x < 2$: All results will be greater than 0.
Case $ 2 < x < 4$: All results will be less than 0.
Case $ x > 4$: All results will be greater than 0.

Hence the solution is $ \minus{}3 < x < 1 OR 2 < x < 4$.

For the rest of the equalities, solve them similarly. It may help to put all the terms onto one side into one fraction.

Yeah, I agree with Zhero...
But when you move it over, subtract instead of cross-multiplying
Cross-multiplying gives you a wrong answer.
You first find when the term=0
You get x=-3, 1, 2, 4
You just make a number line

<-----|-----------|------|----------|---->
Where you plot -3, 1, 2, and 4

And you can plug in a number from each interval to check...
Just do the same with the rest.
Remember when subtracting to put it all on a common denominator


Show that $\forall x,y,z \geq 0$ we have:
$ \sqrt{xy}+ \sqrt{yz} + \sqrt{zx} \geq 2 \sqrt[4]{xyz(x+y+z)}$.
When does the equality take place?

Put a 2 =yz  b 2 =xz  c 2 =xy  and the inequality becomes

   (a+b+c)^4  \geq 16((ab) 2 +(bc) 2 +(ca) 2 )

sice it is homogeneus you can choose a+b+c=1 and so we have

       (ab) 2 +(bc) 2 +(ca) 2  \leq 1/4 2

this is the well known  HARAZI'S INEQUALITY already discussed many times in this forum.

It seems that in a short period I will be able to write an article with the applications of so called Harazi inequality. It's funny.


[quote="turak"]YAYAYAYAYYAYAYAYAY!
go cynthia!  :D[/quote]

You are next.  :)

wow ty and IT WOULD BE SO AWESOME IF WE ALL MET UP AT MOP NEXT YEAR!!!!!!!!

YAY Cynthia!!!

Sorry, I was in Wyoming. I would have congratulated you sooner.

Maybe if I learn how to solve not-so-challenging geometry problems I'll see you at MOP in 2010. Maybe turak too if she deletes her parenthetical.

you're late :P

turak refuses to delete her parenthetical

turak hates those geometry problems too :(


How many integers $ x$ satisfy the equation
\[ (x^2 \minus{} x \minus{} 1)^{x \plus{} 2} \equal{} 1
\]$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \text{none of these}$

[hide="Solution"]Suppose $ x^2\minus{}x\minus{}1\equal{}1$
We get solutions $ x\equal{}2$ and $ x\equal{}\minus{}1$.
If instead $ x^2\minus{}x\minus{}1\equal{}\minus{}1$, we get $ x\equal{}1$ and $ x\equal{}0$. But $ x\equal{}1$ is not a valid solution because it would make the expression $ \minus{}1$ instead of $ 1$.
Finally, suppose $ x\plus{}2\equal{}0$. We get the solution $ x\equal{}\minus{}2$.
So there are $ \boxed{\textbf{(C)}\ 4}$ solutions, $ x\equal{}\minus{}2,\minus{}1,0,2$.[/hide]


[quote="Hong Quy"][quote="zaizai-hoang"][quote="can_hang2007"]Let $ a,b,c,d$ be positive real numbers such that $ ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd \equal{} 6$. Prove that
\[ \frac {1}{\left(1 \plus{} \frac {1}{a^2}\right)^2} \plus{} \frac {1}{\left(1 \plus{} \frac {1}{b^2}\right)^2} \plus{} \frac {1}{\left(1 \plus{} \frac {1}{c^2}\right)^2} \plus{} \frac {1}{\left(1 \plus{} \frac {1}{d^2}\right)^2} \ge 1
\]
:)[/quote]

Another my nice inequality (but easy)
Let $ a,b,c,d$ be positive real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$. Prove that
\[ 1\ge \frac {1}{\left(1 \plus{} \frac {1}{a^2}\right)^2} \plus{} \frac {1}{\left(1 \plus{} \frac {1}{b^2}\right)^2} \plus{} \frac {1}{\left(1 \plus{} \frac {1}{c^2}\right)^2} \plus{} \frac {1}{\left(1 \plus{} \frac {1}{d^2}\right)^2}
\]
[/quote]
 yes,exactly  it's easy .
$ \sum \frac {1}{((1 \plus{} \frac {1}{a^2})^2} \equal{} \sum \frac {a^4}{(1 \plus{} a^2)^2} \leq \frac {1}{4} \sum 4a^2 \equal{} 1$[/quote]

OMG, I don't know it is very easy like that  :P


Determine the number of positive integral solutions of the equation
a2-7a+b2-7b+2ab=0. Sorry, the 2's are suppose to be squares.

[quote="king_23"]Sorry, the 2's are suppose to be squares.[/quote]

For exponents, you can use the ^ sign. Alternatively, you can also type your expression in LaTeX.

[hide]
$a^2 + 2ab + b^2 - 7(a+b)=(a+b)(a+b) - 7(a+b) = (a+b-7)(a+b)=0$

The factor $(a+b)$ will never be $0$ for positive integer values of $a$ and $b$, so we only look at the factor $(a+b-7)$

The $6$ ordered pair of solutions are then $(6, 1)$, $(5, 2)$, $(4, 3)$ and the pairs where the order is reversed.
[/hide]


i just went to memberlist to look up someone and:

http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=29314

 :P

Something similar happened earlier this week.  It should be fixed soon.

[quote="#H34N1"]But he's a moderator :!:[/quote]

He's not a mod, he just editted his avatar so it says "Moderator."

This is just a glitch with the post counts.

No, he did not change his avatar, that's part of the bug.
But it should be fixed soon, as I don' know if it may be harmful (I think not, but who knows...).

[quote="ZetaX"] as I don' know if it may be harmful (I think not, but who knows...).[/quote]

He shouldn't be, as he's not on the [url=http://www.artofproblemsolving.com/Forum/memberlist.php?level=2]mod list[/url].

He only has 2 posts.  I doubt he is hacking; hackers don't usually contribute to the forum and then hack.  (I think.)
This is odd though.
 :|  :huh:

Moderators don't have any *real* power, anyways. I honestly don't know why people think they do.

[quote="azjps"]That's 1=2, we can just ask him about it?  :huh: (1=2's not a moderator, is he?)[/quote]

Good guess. He is my dad, who is not a hacker.

Should be a glitch. He only has a couple hundred posts, not a couple million.

And he is indeed a moderator.

The memberlist display doesn't affect anything. For some reason, some moderation process puts it off count, however it can always be quickly reevaluated by actually counting one's post so don't worry too much about it for now.


When I rate someone else's post, before I do, the red X which cancels the rating is on the screen, but when I rate, it disappears. Shouldn't it be the other way around? And also, it disappears only for that person.

A few observations:

1) You can't click on the red X if you didn't vote (it looks faded). As to be expected.

2) If you opened the topic without rating a particular post (regardless if you have voted before and retracted, or not), and then vote on that post, the red X will disappear.

3) You can still change you vote while on topic, but the X will not reappear.

4) If you opened a topic having rated a particular post, the red X should appear, not appearing faded. Clicking it retracts the vote, and the X will again look faded. However, this time if you revote, the X will not disappear.

Thanks! So if I want to cancel rating, I should get out of and then get in the topic to do so...


:D Please Help- I'm not sure if this is the right forum- I have a problem that I think that I have worked out and I would like to see the equation for. The question is- assuming that a NC license plate has Three letters, and four numbers, how many possible combinations are there? I have come up with  6,580,454,400. Is this correct? If not, please explain. Thanks- Mike

It depends whether letters and/or numbers can repeat, and whether you have to have a letter or number in a specific spot or not. I seem to get that if both numbers and letters can repeat and you don't have to have a number or letter in specific spot (eg: letter, number, letter, ...) there are $6,151,600,000$ different license plates.

What's the 7 choose 3 for?

EDIT: Ahhh, for the order? But still, why 7 and why 3?
EDIT2: Is that because you'll have $\frac{7!}{3!4!}$ Meaning you have 7 slots, 3 to choose letters, 4 to choose numbers?

[quote="WindSlicer"]EDIT2: Is that because you'll have $\frac{7!}{3!4!}$ Meaning you have 7 slots, 3 to choose letters, 4 to choose numbers?[/quote]Yes.

500th post!! :!:  :D  :D  :thumbup:  :clap:  :clap2:

I did 9P4 and got 3,024
then 26P3 adn got 15,600
then mulitplyed and got 47,174,400

If im wrong tell me why :?

It depends whether the character that looks like a big circle is a 0 (a number) or a O (a letter) or whether there are 2 separate characters or whether that character is omitted.  Also, it matters whether licence plate characters always follow a particular pattern.  (In New York, for example, we always have letter-letter-letter-number-number-number-number.)

[quote="ineedmathhelp"]chances are the question's asking for any license plate with three letters in the first 3 slots and then 4 numbers in the last 4 slots. Then the answer would be:

26^3*10^4=175760000[/quote]

But you also have to multiply that by 7!/(3!x4!) Damn I need some latex....anyway, its because you have 7 slots, 3 letters, 4 numbers.

[quote="Iversonfan2005"][quote="ineedmathhelp"]chances are the question's asking for any license plate with three letters in the first 3 slots and then 4 numbers in the last 4 slots. Then the answer would be:

26^3*10^4=175760000[/quote]

But you also have to multiply that by 7!/(3!x4!) - I need some latex....anyway, its because you have 7 slots, 3 letters, 4 numbers.[/quote]

But Iversofan2005 just said that he is assuming that all license plates in NC follow a general pattern: lllnnnn.


It won't let me log off. BTW I have remember password.

Do you want to give some more detail? Like what happens when you try?

and *please* can you stop using caps (subject of this thread). Its considered extremely rude..

I click logout and it won't let me log out because after i click i go back to the forum which is still signed in. I use Internet Explorer six and this also happened at the library (i hope no body goes to AOPS).


linear algebra
matrix
limit
inequalities
linear algebra unsolved

How do you prove that every matrix can be written as a limit of diagonalizable matrices ?

thanks

How do you write $\begin{pmatrix}0 &1\\ 0 & 0\end{pmatrix}$ as a limit of diagonalizable matrices?

$\left( \begin{matrix}a & 1 \\ 0 &-a \end{matrix}\right)$, where $a \to 0$.

It has been discuseed on the forum before; one needs only to show that the matrices with distinct eigenvalues are dense (via Schur's Theorem).
[Supposing that the matrices are over $\mathbb C$.]

OK, thanks for the reply, I try to prove it for triangular matrices.

The eigenvalues of a triangular matrix are the diagonal elements.
If they are all distinct then the matrix is diagonalizable and we are done. If not there are repeated elements and create the matrix $A_{\epsilon}$ as follows :

$[A_{\epsilon}]_{ij}= [A]_{ij}$ for $i \neq j$ (leave the non diagonal elements unchanged) and add $\epsilon$, $2 \epsilon$... to each of the repeated elements. (for example if an element is repeated three times, instead of it put element + $\epsilon$, element + $2 \epsilon$ and element + $3 \epsilon$ in the three places where this element was).

Now clearly $A_{\epsilon}$ is diagonalisable for $\epsilon$ (except at most a finite number of $\epsilon$) and $\lim_{\epsilon \to 0}A_{\epsilon}= A$

Does this proof work ? I'll try to use Schur to extend it in general


modular arithmetic
inequalities
number theory unsolved
number theory

Let $ p$ be an odd prime. Prove that $ \sum_{k=1}^{p-1}k^{2p-1}\equiv \frac{p(p+1)}{2}(\text{mod}p^{2})$

[hide="My solution (but I think I am wrong)"]

We must prove $ 4\sum_{k=1}^{p-1}k^{2p-1}\equiv 2p \pmod{p^{2}}$

$ k^{2p-1}+(p-k)^{2p-1}\equiv (2p-1)p\cdot k^{2p-2}\pmod{p^{2}}$

so $ (2p-1)(p)(\sum_{k=1}^{p-1}k^{2p-2})\equiv 2p \pmod{p^{2}}\leftrightarrow \sum_{k=1}^{p-1}k^{2p-2}\equiv-2\pmod{p^{2}}$

But I think the last equality is not true..[/hide]

Can anyone help me to find my mistake and solve the problem?
thx..

Sorry if this problem posted before..

Btw why I can not hide my message?

[quote="Yosh..."]Btw why I can not hide my message?[/quote]

i had this problem sometimes before. i think it is because of using the ' symbol in the BBCode:[Hide=My solution (but I think [u]I'm[/u] wrong)]

[Moderator edit: Yes, this is the reason. Edited first post.]

[hide="I hope my help is useful"]

Actually $ 2\sum_{k=1}^{p-1}k^{2p-1}=\sum_{k=1}^{p-1}k^{2p-1}+(p-k)^{2p-1}$ and not $ 4\sum_{k=1}^{p-1}k^{2p-1}=\sum_{k=1}^{p-1}k^{2p-1}+(p-k)^{2p-1}$

So at the end the inequality you get is:

$ \sum_{k=1}^{p-1}k^{2p-2}\equiv-1\pmod{p}$ (and not $ \pmod{p^{2}}$ because you have divided by p)

But this last inequality is true, do you see why?[/hide]

If i am true then it's a problem from Canada MO.


Two vertices of a rectangle are $ (\minus{}3,1)$ and $ (\minus{}3, \minus{}3)$. The line
with equation $ x \equal{} 1$ is a line of symmetry for this rectangle. How many square units are in the area of the rectangle?

The width of the rectangle is the distance from (-3,1) to (-3,-3), which is 4. Since x=1 is the symmetry the length is twice the distance from x=-3 to x=1, or 8. So, the area is 4*8=32


The original statement of Problem 2 of Round 2 did not make explicit that c and d must be positive integers.  The question has been restated to include this information.  Thanks to those of you who pointed this out.

Also, can we use calculators for computation (ex. working out something*sqrt(2)), or does the entire solution have to be by hand? That is, is the purpose of the calculator/computer restriction just to prevent us from writing scroll/test programs, or are we supposed to, if our solution needs it, find sqrt(2) to some number of decimal places and then multiply by hand?

We think the question is unambiguous on this point: "Without using a calculator or computer".

number 4 says the quadrilateral must intersect 4 of the 9 points. does this mean EXACTLY 4 of the 9 points or can it intersect 4 or more points?

maybe I'm missing something.

Read the question carefully; we think it is unambiguous on this point.

Folks, you really *must stop* asking specific questions about these problems!  The rules prohibit *any discussion* of any pending problem.  This includes publicly asking specific questions about any pending problem!

I think Mr. Rusczyk has been kind so far.  But I also think he would be within the rules to disqualify any contestant who publicly asks a specific question.

If you have a question, please ask the contest administrators via PM.  It is exceedingly rare that a USAMTS problem is sufficiently ambiguous to require clarification.  So, chances are high you will be told "the question is clear, as written".  If you still believe the question is unclear or ambiguous, include a statement, with your submission, clearly explaining your reasoning.  Based on e-mails I have exchanged with the previous contest administrators and graders, I have every confidence the current graders will consider your explanation, and you could even earn a "commendation" if you find something the question-writers overlooked.

I am only a coach, so I don't speak for USAMTS.  But please preserve the integrity of the contest.


Find all the pairs of positive integers $ (a,b)$ such that $ a^2 \plus{} b \minus{} 1$ is a power of prime number $ ; a^2 \plus{} b \plus{} 1$ can divide $ b^2 \minus{} a^3 \minus{} 1,$ but it can't divide $ (a \plus{} b \minus{} 1)^2.$

[quote="shoki"]it has been posted, search....(it is China TST but i don't know the year).[/quote]
Exactly it's a problem in China TST 2007 but I don't see any solution. :maybe: This problem seem to be easy, however I can't solve it. :oops:

I'm really sorry :(  but anybody has solution for it? :)

i´ve solved a very similar problem at st. petesburgs olympiad, but i don´t remember the year... :|


If $ x,y,z$ are positive reals such that $ x \plus{} y \plus{} z \plus{} xyz \equal{} 4$, prove that:
$ \frac{x}{\sqrt{y\plus{}z}} \plus{} \frac{y}{\sqrt{z\plus{}x}} \plus{} \frac{z}{\sqrt{x\plus{}y}} \ge \frac{x \plus{} y \plus{} z}{\sqrt{2}}$.

[quote="Raja Oktovin"]If $ x,y,z$ are positive reals such that $ x \plus{} y \plus{} z \plus{} xyz \equal{} 4$, prove that:
$ \frac {x}{\sqrt {y \plus{} z}} \plus{} \frac {y}{\sqrt {z \plus{} x}} \plus{} \frac {z}{\sqrt {x \plus{} y}} \ge \frac {x \plus{} y \plus{} z}{\sqrt {2}}$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127956


its awesome. if u like to play it, check out my blog

I really like to listen to Coltrane recordings of jazz saxophone. That's a great solo instrument.

I don't play the saxophone right now, but I might start to play soon. My brother plays it. :)

I don't play sax at all. I play the violin.

No sax!  Piano.

Rofl, my friend likes a girl whos last name is sax...

I play.

[quote="appillai"]I used to play Alto Sax for about 4 years. Great intrument.

I LOVE KENNY G![/quote]

Ug. No.

I play the bari sax. But in high school, I will switch to the tenor or alto sax. The bari is just too big

[quote="buzzer11"]I play tenor, mainly but started on alto. I think most people (if not all) did at my school.[/quote].

yeah, thats how it is at my school too. btw, very creative with ur signature.


function
analytic geometry
geometry
geometric transformation
reflection
graphing lines
slope

well the only thing that will divide that product is combinations of products of the prime numbers that make it up. (i'm assuming you want a hint, not the solution) i'm not sure if it is clear of what i'm saying either.

[quote="Silentfyre"]Im stuck with this problem:
If p,r, and s are three different prime numbers greater than 2, and n= p x r x s, how many positive factors including 1 and n does n have?[/quote]

I suppose that the x is multiplictation.  It's pretty easy to count the factos.  There's also a formula where you add one two the exponents and multiply them together to get the number of factors (easily provable).  The answer is (1+1)(1+1)(1+1) = 8.

[quote="Silentfyre"]Im stuck with this problem:
If p,r, and s are three different prime numbers greater than 2, and n= p x r x s, how many positive factors including 1 and n does n have?[/quote]

1)n
2)1
3)pr
4)ps
5)rs
6)p
7)r
8)s

prs is the same as n, and order doesn't matter..so 8.

[quote="Silentfyre"]I also have another problem that i need to see how to solve:
h(t)= c - (d - 4t)square

At time t=0, a ball was thrown upward from an initial height of 6 feet. Until the ball hit the ground, its height, in feet, after t seconds are given by the function h above, in which c and d are positive constants, If the ball reached its maximium height of 106 at time t=2.5, what was the height, in feet, of the ball at time t=1?[/quote]

The height was $70$.

[i][edit: Shobhit caught his mistake :P ][/i]

For the second one, from the information given we know that $h(0)=6$ and $h(2.5)=106$, that is:

\begin{eqnarray}
h(0) &=& c-d^2 = 6\\
h(2.5) &=& c-(d-10)^2 = (c-d^2)+20d-100= 106
\end{eqnarray}

If we plug $c-d^2=6$ into $(2)$ we get

\[6+20d-100=106\Rightarrow d=10\]

and thus $c=106$.  So we have $h(t)=106-(10-4t)^2$.  Plug in $t=1$ and we get $h(1)=106-(10-4)^2=\boxed{70.}$

[quote="joml88"][i][edit: Shobhit caught his mistake :P ][/i]

Yes, actually it would be easier if you just plug in numbers. Just say that $n=105$ by letting $p=3,r=5,s=7$ so the factors you have are $105,1,5,3,7,21,35,15$.

Getting stuck with this one:

If k, n, x, and y are positive numbers satisfying x to the -4/3 = k to the -2 and y to the
4/3 = n to the second, what is (xy) to the -2/3 in terms of n and k?

a. 1/nk
b. n/k
c. k/n
d. nk
e. 1

Last few questions:

Let [x] be defined as [x]= x square - x for all values of x. If [a] = [a-2]. what is the value of a?

In the xy-coordinate plane, line m is the reflection of line L about the x-axis. If the slope of line m is -4/5, what is the slope of line L?

I was also wondering how the graph of f would look like if the function f is defined by
f(x) = x square + bx + c, where b and c are constants.

Thanks.

the graph of $ax^2+bx+c$ would be parabola (u-shaped thingie)

and I think line L should have slope of $4/5$

And I'm a little confused on the first but since $[a]=[a-2]$, then $a^2-a=(a-2)^2-(a-2)$, thus $a^2-a=a^2-4a+4-a+2$, so $4a=6$ and $a=3/2$.


Bρειτε το μεγιστο της παραστασης
$ \frac {[(4 \plus{} \sin x)y^4 \plus{} 4y \plus{} 3\sin x \plus{} 12]^2}{(2y^4 \plus{} 4y \plus{} 6)[(\sin x \plus{} 2)y^4 \plus{} 3\sin x \plus{} 6]}$.
Καλο Φθινοπωρο και καλες σπουδες στους επιτυχοντες του forum

Sil φτιαξε το typo του αριθμητη προσθετοντας στην αγκυλη  του αριθμητη το $ \+3sinx$ .Sorry me ta typo..............

Done ! Προσθέτω κάτι ακόμα . Βρείτε και το ελάχιστο της παραπάνω παράστασης καθώς και για ποια ζεύγη $ (x,y)$ επιτυγχάνεται  :wink:
Τέλος δείξτε ότι για κάθε $ y$ υπάρχει ένα $ x$ ώστε να έχουμε ελάχιστο .

Τo διόρθωσα αλλά η αμφιβολία εξακολουθεί να ισχύει

Η άσκηση δεν σου ζητάει να δείξεις ότι για κάθε $ x$ υπάρχει μοναδικό $ y$. Σου ζητάει να δείξεις ότι για κάθε $ y$ υπαρχει $ x$ (σίγουρα όχι μοναδικό).

Ναι πράγματι . Δεν εννούσα μοναδικό . Συγγνώμη για την ελειπή διατύπωση . (Η λύση μου μοιάζει με του κ .Μπόρη)

O.K. Rodolfe .Kanoyme peripoy ta idia me alles synarthseis kai briskv ta akrotata dixvs diaforiko logismo.


integration
real analysis
limit
real analysis unsolved

Suppose that $ f$ is integrable on $ [ \minus{} \pi, \pi]$. Choose $ \epsilon > 0$, can we find a $ \delta$ so that $ |\int_{ \minus{} \delta}^{\delta}f\  dx\ | < \epsilon$ ?

yes, absolute continuity of Lebesgue integral.
We can find $ \delta > 0$ such that for all $ E \subseteq [ \minus{} \pi,\pi]$ with $ m(E) < \delta$ it is $ \int_E |f|dx < \epsilon$

This is a more specific than absolute continuity and has a slightly easier proof than the general case.

Define $ g_{\delta}(x)\equal{}f(x)\mathbf{1}_{[\minus{}\delta,\delta]}(x)\equal{}\begin{cases}f(x)& x\in[\minus{}\delta,\delta]\\0&\text{otherwise}\end{cases}.$

Then $ g_{\delta}(x)\to 0$ [i]a.e.[/i] as $ \delta\to 0.$ Since we also have that $ |g_{\delta}|\le |f|$ everywhere and $ |f|$ is integrable, then the Dominated Convergence Theorem applies and

$ \lim_{\delta\to0}\int d_{\delta}\equal{}\lim_{\delta\to 0}\int_{\minus{}\delta}^{\delta}f\equal{}0.$


Prove whether or not $ n^x \minus{} n$ is divisible by $ x$, for any n

edit: composite number

$ n^x\minus{}n$ is divisible by x, then x and n are relatively prime.

What do you mean?  The simplest example occurs precisely when $ x | n$.

$ n^x\minus{}n\equal{}0 \pmod{n}$ would also be true if x is prime !?
(fermats little theorem)

...
i think this is what Bachukas meant

I seem to have read your problem wrong.  Do you mean the following:

[i]Do there exist composite $ x$ such that $ n^x \equiv n \bmod x$ for all $ n$?[/i]

When $ (n, x) \equal{} 1$, the answer is given by the [url=http://en.wikipedia.org/wiki/Carmichael_number]Carmichael numbers[/url].


inequalities
trigonometry
function
inequalities proposed

Find all solutions $ (x,y)$ , satisfy

$ 20\text{sin} x \minus{} 21 \text{cos} x\equal{}81y^2\minus{}18y\plus{}30$

hello, solving the given equation for $ y$ we get
$ y_{1,2}\equal{}\frac{1}{9}\pm\frac{1}{9}\sqrt{\minus{}29\plus{}20\sin(x)\minus{}21\cos(x)}$ for real solutions it must be
$ \minus{}29\plus{}20\sin(x)\minus{}21\cos(x)\geq0$, but we get $ \minus{}29\plus{}20\sin(x)\minus{}21\cos(x)\equal{}0$  by discussing the function
$ f(x)\equal{}\minus{}29\plus{}20\sin(x)\minus{}21\cos(x)$, from here we get all real solutions as
$ x_1\equal{}\minus{}\arctan(\frac{20}{21})\plus{}\pi\plus{}2k_1\pi$, $ y_1\equal{}\frac{1}{9}$
$ x_2\equal{}\minus{}\arctan(\frac{20}{21})\plus{}\pi\plus{}2k_2\pi$, $ y_2\equal{}\frac{1}{9}$
Sonnhard.

no need to solve anything dr. :wink:
Just observe that the RHS has minimum valu=29 whereas the LHS has maximum value=29=$ \sqrt {20^2 \plus{} 21^2}$.


function algorithm search algebra polynomial logarithms

how about if we all put our ideas to find a formula for prime numbers.i know it is difficult but ,in my opinion it has an exact formula.

      [color=green]1[/color],[color=red]2[/color],[color=red]3[/color],4,[color=red]5[/color],6,[color=red]7[/color],8,9,10,[color=red]11[/color],12,[color=red]13[/color],14,15,16,[color=red]17[/color],18,[color=red]19,[/color]20,21,22,[color=red]23[/color],24,25,26,27,28,[color=red]29[/color],30,[color=red]31[/color],32,33,34,35,36,[color=red]37[/color],38,39,40,[color=red]41[/color],42,[color=red]43[/color],44,45,46,[color=red]47[/color],48,49,50,51,52,[color=red]53[/color],54,55,56,57,58,,,,,,
i think 1 is also a prime number, but another kind of prime i will say why, i am sleepy now,

... Good Luck with that

Try $ f(n) = 6n \pm 1$, although that doesn't get 2 as a prime, and it's not technically a function, but other than that  :lol:

http://mathworld.wolfram.com/PrimeFormulas.html
http://en.wikipedia.org/wiki/Formula_for_primes

Please try to remember that very smart people have been studying number theory for centuries, and if an easily-computable formula for the primes existed, someone would've found it before the creation of this topic.  Stop being so arrogant to think that you don't need to look up any of the previous progress made on a given problem.

And there are a lot of very good number-theoretic reasons [b]not[/b] to consider $ 1$ a prime.  Among them, $ 1$ is a [url=http://mathworld.wolfram.com/Unit.html]unit[/url].

t0rajir0u i know what you are saying, i know that loads of smart people have been studying number theory for centuries,but that doesnt mean we have to stop searching, and lay back. i am not arrogant, and please try not to stop others from working on it,you dont have a 100% proof that primes doesnt have one exact formula do you ?

[color=green]1[/color],[color=red]2[/color],[color=red]3[/color],4,[color=red]5[/color],6,[color=red]7[/color],8,9,10,[color=red]11[/color],12,[color=red]13[/color],14,15,16,[color=red]17[/color],18,[color=red]19,[/color]20,21,22,[color=red]23[/color],24,25,26,27,28,[color=red]29[/color],30,[color=red]31[/color],32,33,34,35,36,[color=red]37[/color],38,39,40,[color=red]41[/color],42,[color=red]43[/color],44,45,46,[color=red]47[/color],48,49,50,51,52,[color=red]53[/color],54,55,56,57,58,,,,,,
ne wayz this is my idea, i might be wrong but..

1+1=[color=red]2[/color]
2+3=[color=red]5[/color]
3+4=[color=red]7[/color]
5+6=[color=red]11[/color]
6+7=[color=red]13[/color]
8+9=[color=red]17[/color]
9+10=[color=red]19[/color]
11+12=[color=red]23[/color]
14+15=[color=red]29[/color]
15+16=[color=red]31[/color]
18+19=[color=red]37[/color]
21+22=[color=red]43[/color]
23+24=[color=red]47[/color]
26+27=[color=red]53[/color]
............
............
............

i think there is a pattern here, every prime is made from adding consecutive numbers before it, except 2, do you notice the pattern?

[quote="mathnerd314"]DiscreetFourierTransform, $ 6n \pm 1$ is not always prime for $ n \in \mathbb{N}$. Try $ n=8$. However, all primes can provably be represented in that form. In other words, IF a number is prime, THEN it can be represented as $ 6n \pm 1$, but not vice versa.

A related result, finding an exact formula for the number of primes less than a given number, is easy... that is, once you've proven the Riemann Hypothesis, one of the seven hardest math problems known to mankind ;)[/quote]

Not quite correct, we can express all numbers as such: 6n+1, 6n+2, 6n+3, 6n+4, 6n+5 6n+6. Now, when testing for primes, we don't choose 6n+2 because that implies divisibility by 2. 6n+3 implies divisibility by 3. 6n+4 implies divisibility by 2 again. And 6n+6 is just 7n, divisibility by 7. The one ones who survived that test are 6n+1 and 6n+5, and the 6n+5 for some integer n, is 6n-1 for some other integer n. That is why 6n +/- 1 works alright. We can easily get better functions with a larger co efficient of n, but it is not very efficient at all.

By the way: Solving the Riemann Hypothesis only IMPLIES that an EFFICIENT algorithm exists, exploiting a result of the RH. Finding an efficient algorithm does not prove RH, and that infact ARE exact results for the Prime Counting Function, but they are not efficient.

[quote="binomial_4eva"]i think there is a pattern here, every prime is made from adding consecutive numbers before it, except 2, do you notice the pattern?[/quote]

Any odd number $ 2k+1$ can be trivially represented in the form $ k+(k+1)$.  It doesn't mean anything.  Instead of searching for a pattern for the primes $ p_{k}$ you're now searching for a pattern for the numbers $ \frac{p_{k}-1}{2}$, which doesn't do anything to change the difficulty of the problem.

Here's another example of what I'm talking about:

$ 5 = \sqrt{24+1}$
$ 7 = \sqrt{24 \cdot 2+1}$
$ 11 = \sqrt{24 \cdot 5+1}$
$ 13 = \sqrt{24 \cdot 7+1}$

Again, instead of searching for a pattern for $ p_{k}$ you might now be tempted to search for a pattern for $ \frac{p_{k}^{2}-1}{24}$.  Doesn't change the difficulty of the problem.  In fact, with these calculations alone you might be tempted to guess that the coefficients are themselves always prime.  But

$ 17 = \sqrt{24 \cdot 12+1}$

[quote="binomial_4eva"]you dont have a 100% proof that primes doesnt have one exact formula do you ?[/quote]

Did you read the links I posted or not?

There [b]are[/b] exact formulas for the primes.  Calculations with any of them are [b]less efficient than searching.[/b]

[quote="Gib Z"]Not quite correct, we can express all numbers as such: 6n+1, 6n+2, 6n+3, 6n+4, 6n+5 6n+6. Now, when testing for primes, we don't choose 6n+2 because that implies divisibility by 2. 6n+3 implies divisibility by 3. 6n+4 implies divisibility by 2 again. And 6n+6 is just 7n, divisibility by 7. The one ones who survived that test are 6n+1 and 6n+5, and the 6n+5 for some integer n, is 6n-1 for some other integer n. That is why 6n +/- 1 works alright. We can easily get better functions with a larger co efficient of n, but it is not very efficient at all.[/quote]

Again, you have merely proven that IF a number is prime, THEN it satisfies that formula. Binomial_4eva's claim is that all numbers satisfying that formula are prime; that is, IF a number satisfies that formula, THEN it is prime. That claim is false. Again, try $ 6 \cdot 8+1$.

And you have a typo: 6n+6 is not 7n, it's 6(n+1).

[quote="binomial_4eva"]t0rajir0u i know what you are saying, i know that loads of smart people have been studying number theory for centuries,but that doesnt mean we have to stop searching, and lay back. i am not arrogant, and please try not to stop others from working on it,you dont have a 100% proof that primes doesnt have one exact formula do you?
...
i think there is a pattern here, every prime is made from adding consecutive numbers before it, except 2, do you notice the pattern?[/quote]

Well, for a start, I notice that you conveniently skipped over 10+11=21 and 12+13=25. So it's obvious that this "pattern" doesn't always work. And by the way, do you know the reason the pattern seemed to work? When you add two consecutive numbers, you get an odd number. And all primes, "except 2", are odd.

By the way, we do have a 100% proof that you can't find a polynomial that always generates primes. Did you read the Wikipedia article?

Oops, t0rajir0u beat me to it.

[quote="mathnerd314"]In other words, IF a number is prime, THEN it can be represented as $ 6n \pm 1$, but not vice versa.[/quote] Unless it's 2 or 3, of course  :wink:

[quote="mathnerd314"]A related result, finding an exact formula for the number of primes less than a given number, is easy... that is, once you've proven the Riemann Hypothesis, one of the seven hardest math problems known to mankind ;)[/quote]
[quote="Gib Z"]By the way: Solving the Riemann Hypothesis only IMPLIES that an EFFICIENT algorithm exists, exploiting a result of the RH. Finding an efficient algorithm does not prove RH, and that infact ARE exact results for the Prime Counting Function, but they are not efficient.[/quote]

So, some more correct information: there is a very good approximate formula for the number of primes less than $ x$ for large values of $ x$.  This is given by the Prime Number Theorem.  However, people are still interested in how large the error for that approximation is.  The Riemann Hypothesis is equivalent to the "best possible" error term.  (Why "best possible"?  Because we're approximating something discrete, the distribution of the primes, with continuous functions -- there always has to be some error.)  Thus, RH implies very strong conditions on the distribution of the primes.

Deterministic, polynomial-time algorithms are known for primality testing.  It was not proven that this was the case until 2002, by Agarwal et al. (working with some undergraduates, I believe).  Their original algorithm ran in time 12th degree in the length of the prime -- this is polynomial, but very slow.  They believed that if they Riemann Hypothesis was true, they could substantially increase the speed of the algorithm.  More generally, any regularity condition on the distribution of the primes is likely to allow more efficient prime testing algorithms.  However, it might be that the speed of the best algorithms for primality testing do not depend on RH to any significant degree.

[quote]Well, for a start, I notice that you conveniently skipped over 10+11=21 and 12+13=25. So it's obvious that this "pattern" doesn't always work. And by the way, do you know the reason the pattern seemed to work? When you add two consecutive numbers, you get an odd number. And all primes, "except 2", are odd.[/quote]

may be the formula has something that skips 10+11=21 and 12+13=25, who knows.

by the way this is good to know that "When you add two consecutive numbers, you get an odd number. And all primes, "except 2", are odd" i didnt know this.

[quote="JBL"]So, some more correct information: there is a very good approximate formula for the number of primes less than $ x$ for large values of $ x$.  This is given by the Prime Number Theorem.  However, people are still interested in how large the error for that approximation is.  The Riemann Hypothesis is equivalent to the "best possible" error term.  (Why "best possible"?  Because we're approximating something discrete, the distribution of the primes, with continuous functions -- there always has to be some error.)  Thus, RH implies very strong conditions on the distribution of the primes.[/quote]

Yes, I was just trying to explain everything in terms that binomial_4eva would understand. (For anyone interested, the prime number theorem gives an estimate of $ \frac{x}{\ln x}$ primes less than or equal to x. A few slightly better approximations exist.)

[quote="JBL"]Deterministic, polynomial-time algorithms are known for primality testing.  It was not proven that this was the case until 2002, by Agarwal et al. (working with some undergraduates, I believe).  Their original algorithm ran in time 12th degree in the length of the prime -- this is polynomial, but very slow.  They believed that if they Riemann Hypothesis was true, they could substantially increase the speed of the algorithm.  More generally, any regularity condition on the distribution of the primes is likely to allow more efficient prime testing algorithms.  However, it might be that the speed of the best algorithms for primality testing do not depend on RH to any significant degree.[/quote]

Interesting point. Does that mean that the Riemann Hypothesis is somehow inextricably related to P=NP?

[quote="binomial_4eva"]may be the formula has something that skips 10+11=21 and 12+13=25, who knows.[/quote]

What's your point? As t0rajir0u stated, using formulas to find prime numbers is pointless. Why exactly would this "formula" skip those numbers? You have to invent something to explain that, and then you'll have to invent something else, ad infinitum.

Oh, and by the way, would somebody please edit binomial_4eva's first post, because it's stretching the forum tables. Thanks.

For those of you who actually want to know a function that generates primes, and are familiar with the floor function, heres one :

$ \pi(m) = \sum_{j=2}^{m}\left\lfloor{(j-1)!+1 \over j}-\left\lfloor{(j-1)! \over j}\right\rfloor \right\rfloor.$

[quote="mathnerd314"][quote="JBL"]Deterministic, polynomial-time algorithms are known for primality testing.  It was not proven that this was the case until 2002, by Agarwal et al. (working with some undergraduates, I believe).  Their original algorithm ran in time 12th degree in the length of the prime -- this is polynomial, but very slow.  They believed that if they Riemann Hypothesis was true, they could substantially increase the speed of the algorithm.  More generally, any regularity condition on the distribution of the primes is likely to allow more efficient prime testing algorithms.  However, it might be that the speed of the best algorithms for primality testing do not depend on RH to any significant degree.[/quote]

Interesting point. Does that mean that the Riemann Hypothesis is somehow inextricably related to P=NP?[/quote]

No.  The problem of primality testing is in P regardless of the status of the RH and P vs. NP.  However, it is possible that the speed of the "best possible" polynomial deterministic algorithm for primality testing depends on the error term in the Prime Number Theorem, and thus on the RH (but this says nothing about NP).

The fate of factoring integers is more closely related to P vs. NP (although not so closely related as NP complete problems), but I don't know if the Riemann Hypothesis is thought to have any implications for easy factoring.

I'm feeling very skeptical about this thread. You're not going to find a polynomial expression that returns a prime value for all integral inputs.

[quote="binomial_4eva"]by the way this is good to know that "When you add two consecutive numbers, you get an odd number. And all primes, "except 2", are odd" i didnt know this.[/quote]

I'm not sure how you'd expect to get anywhere with this problem without knowing that any prime greater than $ 2$ is odd.  Even numbers are, by definition, divisible by $ 2$, so no even number greater than $ 2$ can be prime because it has a divisor that is not $ 1$ or itself.

What kind of formula are you expecting, exactly?  It's known that a polynomial one simply does not exist.  If you're envisioning any kind of closed form in terms of elementary functions, then I suggest you [b]read the links.[/b]

[quote="Gib Z"]My deepest apologies i just realised what I posted before was the formula for the prime counting function. For a function that generates primes, look below:'

$ f(n) = 2+(2n! \,\operatorname{mod}(n+1))$ for all non negative integer n.[/quote]

In the links I posted, as was the other one.  And it doesn't really help, since this function returns $ 2$ for all composite $ n$, so it amounts to a (very slow) prime-checking algorithm.

what about this one:

http://www.primenumbersformula.com/default.htm

Or Ghandi's Formula that i've explained about it in my Blog before:

http://www.mathlinks.ro/weblog_entry.php?t=131537

and wow i didn't know about the formulas in wikipedia [b]t0rajir0u[/b]  :) they are really cool, Prime numbers are really CooOl  :lol:

Several points.

1)I like to point that that there [b]are[/b] prime producing functions.
Some are a consequence of "Wilson's theorem". Notice that this theorem tells us a necessary and sufficient condition for a number to be a prime. However, it is computationally impossible. For example, to see if $ n=101$ is prime we need to consider $ 100!$ which is really big, even for fast computers. So it really isnt efficient.

2)One $ 1$ isn't a prime. One reason why not is that the uniqueness part of the fundamental theorem of arthimetic will be violated.

3)There is [b]no[/b] non-constant polynomial to produces only primes. Try to prove it.

4)Wait a minute! What about $ f(n)=n^{2}+n+41$ for $ n\geq 0$? Doesnt that contradict #3  :rotfl:

[quote="davidyko"]I laughed when I read this:

[quote="binomial_4eva"]by the way this is good to know that "When you add two consecutive numbers, you get an odd number. [/quote]

If you have two consecutive integers, one must be $ 0 \text{ (mod 2)}$, and the other must be $ 1 \text{ (mod 2)}.$ Adding them together yields another $ 1 \text{ (mod 2)}$, which is also known as [b][u]odd[/u][/b].

That was just a fancy way of saying odd + even = odd, if you [u]still[/u] don't understand what I'm saying.[/quote]

Yeah, I typed that thinking that binomial_4eva would recognize it as tautological, but apparently not. I laughed when I read that too :D

$ n^{2}+n+41$... nice :rotfl:. I'm told that $ x^{2}-79x+1601$ is more deceptive :D

[quote="The Zuton Force"]On a side-note, I think it's time that this thread be locked, unless anyone else has any constructive comments on calculating primes.[/quote]

Hmmm, I noticed that nobody has mentioned the Sieve of Eratosthenes (or however it's spelled... :blush:)

[quote="Elemennop"]f(n) = [3n] ??

maybe off by 1 or so[/quote]

:rotfl: I hope you meant that as a joke... because any integer can be expressed as "3n off by 1 or so".

[quote="sylow_theory"]Some are a consequence of "Wilson's theorem". Notice that this theorem tells us a necessary and sufficient condition for a number to be a prime. However, it is computationally impossible. For example, to see if $ n=101$ is prime we need to consider $ 100!$ which is really big, even for fast computers. So it really isnt efficient.[/quote]Actually, one needn't ever calculate numbers larger than 10,000 while applying the Wilson's Theorem-based primality test to 101: since it's a calculation $ \mod 101$, you just reduce after each multiplication and you'll never face numbers larger than $ (101-1)^{2}= 10000$.  (If you use positives as well as negatives, I guess you can actually get this down to $ 50^{2}$.)  The actual problem is the [i]number[/i] of computations necessary, not their size: if $ p$ is, say, 8 (or 200) digits long, the largest number one potentially faces is about 16 (or 400) digits -- not actually that big for a computer (or a very determined elementary schooler with good handwriting) to handle.  The problem is that you need to make $ p$ multiplications -- when $ p = 101$, this is not so bad; when $ p$ is 8 digits, though, you're straining my PC for several minutes at least, and when $ p$ has 200 digits we don't stand a chance.

[i]Post has been edited to remove some faulty information.[/i]

[quote="The Zuton Force"]Ok, I'll try to give some constructive comments.[b] [color=red]I expect you to read this[/color][/b].

First, I see you're interested in math. That's good. Now you need a good place to start. Fortunately, the world is full of [b][color=blue]resources[/color][/b]. One of the things you'll need to realize is that it sometimes helps to start from the basics and build your way up. It's good that you have the interest going, but you won't get too far unless you understand the basics. Pick up a copy of [b][color=blue]AoPS 1[/color][/b]. Read, read, read. Read Wikipedia. [b][color=blue]Wikipedia [/color][/b]is a god of all sorts. Mathworld is great but it's a bit more technical.

The AoPS forum is an excellent resource when you learn how to use it. You need to get a feel of who knows what they're talking about, and [b][color=blue]if they try to help you, take it[/color][/b]. If someone gives you a link to read, it's for your own good to read it and try to understand it. You can start by going all the way back to the beginning of the thread and reading all of t0ra's links.

Next, you need to get a feel of the mathematical method. It's a lot like the scientific method. There are a few keys.
1. [b][color=blue]Relevance, relevance, relevance[/color][/b]. Anything you spend time on in mathematics must have some sort of relevance. It should either connect concepts you know, or work towards a focused goal in terms of a concept you don't know. Questions out of the blue are generally not worth the thought, unless you can think of some sort of relevance. This is by no means something easy; it's something you have to get a feel of. But know that.
2. [b][color=blue]Identify[/color][/b]. When approached with an unknown question, you need to be able to find the root problem. Suppose I ask for the solution to $ (j+456)(x-1)^{19}+(j_{2}^{7}2)(x-1)^{3}+(j_{3}!)(x-1)=42$. What's wrong with that? It's ugly. All the factors with $ j$ in them are different, and there's no reason why you need to have $ x-1$ everywhere. The problem is $ ay^{19}+by^{3}+y=42$. THAT's the problem. If you do this, it will prevent you from asking unnecessary questions and wasting your time.
3. [b][color=blue]Research.[/color][/b] Millions of mathematicians have walked the earth. What we know today is the result of their collective work, and what we don't know today is what none of them could figure out. Their work can help you.
4. [b][color=blue]Build[/color][/b]. Start with easier things, and combine them to form harder ideas. There is a natural flow of things. You need to catch that drift.
5. [b][color=blue]Question[/color][/b]. Is it true? Must it be true? Why is it true? What does it mean if it is true? What kinds of interpretations of it are there? What does it suggest/imply?
6. [b][color=blue]Explore[/color]. [/b]You're already doing this, but without focus. When you want to work towards something, you really should have a goal. Especially in the beginning, when there are huge patches of void, empty space. As these spaces fill up, random work can cause you to run into something. Until then, [b][color=blue]stay focused[/color][/b].

I invite you to get started. Best of luck.

[color=green]On a side-note, I think it's time that this thread be locked, unless anyone else has any [b]constructive[/b] comments on calculating primes.[/color][/quote]

Wow that was really nice :)  i hope it'll help anyone in doing anything in mathematics... :)

OOk the links that i gave are really nice read them, specially i like Ghandi's Formula:

$ p_{k+1}=\left\lfloor 1-\log_{2}\left(-\frac{1}{2}+\sum_{d\: |\: P_{k}}\frac{\mu(d)}{2^{d}-1}\right)\right\rfloor$

where $ p_{k+1}$ is the $ k+1$'th prime and $ P_{k}=p_{1}p_{2}\cdots p_{k}$  :roll:

[quote="K81o7"][quote="JBL"]
Close: his name was Erastothenes.[/quote]

http://en.wikipedia.org/wiki/Eratosthenes

Funny...I thought it was Erasthostenes for some time...[/quote]

Haha, that's what I get for being lazy -- I just put it into my search engine, assuming that if it was wrong it would suggest that I should have used the alternate spelling.  (In fact, both the correct spelling and my (wrong) spelling must be fairly common.)

atleast i can make u all laugh with my stupidity :blush:

It's nice to think about math problems but this problem you are looking at may look deceptively easy, but it is very difficult (at least difficult enough to not belong in this forum :lol: ) ...

In any case, I'm locking this topic.


Prove that if n \geq 5, then the interval [n!, (n + 1)!] contains a multiple of n3.

It was equal to the conclusion that (n+1)! - n! >= n^3
or n*n!>n^3
It can be prove by mathematical induction.

Something similar:

Prove that if n > 9 then there is a perfect cube between n and 3n.

if n = 10,between 10 and 30,27 is a cube.
if n = k,m^3 is between k and 3k
  if (  k+1 != m^3 ){
          m^3 is between k+1 and 3(k+1)
     It is true.
   }
   if( k + 1 == m^3 ){
        ( m+ 1 )^3 > m^3 = k + 1
        ( m + 1 )^3 < 3(k+1)=3m^3 + 3 can be prove easliy
     }


Hello everybody!!!
Can someone help me with pegeons?
What do you know about them?
I feed them every day from my hand, so
Is it safe? how about virus H5N1(ptichij grip)?
They are wild!

Do not keep your pigeon outside!! Ptiji grip is transported by birds wich are sick with Ptiji grip. Pigeon is a good creature. I like pigeons too. In math there is also a kind problem solving method called Pigeonhole Pricinple.


If $ f(x) \equal{} 2$ for all real numbers $ x$, what is the value of $ f(x \plus{} 2)$?

Since $ x$ is real, $ x\plus{}2$ must also be real, meaning $ f(x\plus{}2)\equal{}\boxed{2}.$

bruh i was so confused when i was playing. i was like that must be rly hard. Then I come here and read it carefully  :wallbash:


analytic geometry
geometry
number theory proposed
number theory

Let n>=5 be a positive integer and a_1, b_1, a_2, b_2, . . . , a_n, b_n be integers satisfying the following two conditions :
(i) the pairs (a_i, b_i) are all distinct for i=1, 2,  . . ., n;
(ii) |a_i*b_(i+1) - a_(i+1)*b_i | =1 for i = 1, 2, . . . , n,
where (a_(n+1), b_(n+1)) = (a_1,b_1).
Show that exits i, j with 1<= i, j <=n such that
1<abs(i - j) < n -1 and | a_i*b_ j - a_ j*b_i | = 1

I do not have a solution, but it is a nice problem :

Let M_i be the point of the plane with coordinates (a_i, b_i). Then :
(i) means that the M_i are pairwise distinct integer points.
(ii) means that, for each i, the triangle OM_iM_(i+1) has area 1/2, which means, by Pick's formula, that there is no other integer point than the vertices on the boundary or in the interior of these triangles.

We want to prove that (ii) holds also for an other triangle of the form M_iOM_j where M_i and M_j are not consecutive in the given ordering.

Note that the hypothesis n > 4 is useful since the conclusion does not hold for M_1(1,0), M_2(0,1), M_3(-1,0) and M_4(0,-1).

I hope there are no 10000 subcases to study...

Pierre.


limit
induction
algebra unsolved
algebra

[u][i]Problem [/i][/u]

       [i]     Let $ a  ,b , A  , B$ be positives real numbers and the sequence $ (x_{n})$

                       defined by : $ x_{1} \equal{} a ;  x _{2} \equal{}b$

                    $ x_{n\plus{}1} \equal{} A\sqrt[3]{x_{n}^{2}} \plus{} B\sqrt[3]{x_{n\minus{}1}^{2}}$          $ (n \equal{}2,3,4,......)$

                     Find $ \lim_{n\rightarrow\infty}x_{n}$

     Love An Forever

   Huỳnh Võ Phương An[/i]

The limit is $ (A+B)^3$.

Let $ (y_n)$ be the terms of $ (x_n)$ (in order) that are less than or equal to $ (A+B)^3$, and let $ (z_n)$ be the terms of $ (x_n)$ (in order) that are greater than or equal to $ (A+B)^3$.  It is enough to show that each of these sequences converges to $ (A+B)^3$, if it has any elements.

Evidently if $ (y_n)$ is not finite, it is bounded above by $ (A+B)^3$, and by simple induction, it is bounded below by the increasing sequence given by
\begin{align*}
r_1 &=r_2 = \min(y_1,y_2) \\
r_{n+1} &= (A+B) \sqrt[3]{r_{n-1}^2} , \end{align*}
which evidently converges to $ (A+B)^3$, since all its terms are positive.  Therefore $ (y_n)$ converges to $ (A+B)^3$.

Similarly, $ (z_n)$ is not finite, then is bounded below by $ (A+B)^3$, and bounded above by the decreasing sequence given by
\begin{align*}
s_1 &= s_2 = \max(z_1, z_2) \\
s_{n+1} &= (A+B) \sqrt[3]{s_{n-1}^2}, \end{align*}
which also converges to $ (A+B)^3$.  Hence $ (z_n)$ also converges to $ (A+B)^3$.  Therefore $ (x_n)$ converges to $ (A+B)^3$, as desired.  $ \blacksquare$

[i]    Wrong solution, I afraid

  Please try again :maybe:

    I think  Patrick , Darij  will help me :lol: [/i]

  [i]This problem is from VMO 1990 [/i]

Please show me where my solution is wrong.  Perhaps I am misreading the problem?

I do not see this problem on the [url=http://www.kalva.demon.co.uk/vietnam/viet90.html]1990 Vietnam Math Olympiad[/url] (is this what you meant?).

[i]Sorry

It ' s from Vietnam TST 1990[/i] :blush:

Ah!  I see. :)

Well, other people got the same answer as I did [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=716703#716703]here[/url], and using similar methods.


symmetry
algorithm
combinatorics proposed
combinatorics

What is the simplest way to determine if its correct? If all of the 3x3 grids contains all values from 1-9, is it possible that it is wrong in another way. I was thinking that as long as all the 3x3 grid contains only one value from 1-9, it is nearly impossible that it is wrong in another way.

Is there an easy way to determine if two puzzles are related by the symmetry and even better, is there a normal form (a distinct element for each orbit)?

[quote="LeoYard"]What is the simplest way to determine if its correct?[/quote]

If you solve a Sudoku by hand, then the simplest way is to check each box, line, and column. Of course, if you use a computer program to solve (or to assist you in the solving) then the checking will be instantaneous.

[quote="LeoYard"]I was thinking that as long as all the 3x3 grid contains only one value from 1-9, it is nearly impossible that it is wrong in another way. [/quote]

Actually it is possible to be incorrect, for example put all the 1 of the upper three 3x3 grids in the first row.

Use the algorithm which is used in MathCad.
It is named as "nearest neighbour way to solve".
Thanks by now for replying.


geometry
circumcircle
perpendicular bisector
cyclic quadrilateral
geometry proposed

The triangle $ABC$ is such that $\angle BAC=30^{\circ},\angle ABC=45^{\circ}$. Prove that if $X$ lies on the ray $AC$, $Y$ lies on the ray $BC$ and $OX=BY$, where $O$ is the circumcentre of triangle $ABC$, then $S_{XY}$ passes through a fixed point.

[i]Emil Kolev [/i]

The condition $\angle ABC = 45^\circ$ is irelevant, it is sufficient (and necessary) that the angle $\angle CAB = 30^\circ.$ This makes the triangle $\triangle COB$ equilateral. Let q be a distance greater than or equal to the distance of the circumcenter O from the side CA. Points $X, X' \in CA$ such that OX = OX' = q are intersections of CA with a circle $\mathcal C_1$ centered at O and with radius q, while points $Y, Y' \in BC$ such that BY = BY' = q are intersections of BC with a circle $\mathcal C_2$ centered at B and with the same radius q. Since the circles $\mathcal C_1 \cong \mathcal C_2$ are congruent and since CB = CO, powers of the point C to these 2 circles are equal, $CX \cdot CX' = CY \cdot CY'.$ This means that the quadrilateral XX'YY' is cyclic. The perpendicular bisector of the side YY' is fixed, namely a normal to BC at B, and the perpendicular bisector of the side XX' is also fixed, identical with the perpendicular bisector of CA. Thus the circumcenter P of the cyclic quadrilateral XX'YY' is fixed and the perpendicular bisectors of its remaining sides XY', X'Y and of its diagonals XY, X'Y' all pass through its fixed circumcenter P.

yetti your solution proves that if OX=OX'; the lines pass through a fixed point, but if it is not; your solution doesn't work i think.


algebra
polynomial
vector
superior algebra
superior algebra unsolved

How do you prove that $ \mathbb{Q}(\sqrt [3]{2})$ does not contain in $ \sqrt [3]{3}$?

The standard method of finding an irreducible polynomial with $ \sqrt[3]{2}\plus{}\sqrt[3]{3}$ as a zero looked like it would get very messy.

What else should I try? I am not even sure if it is true, but it seems very intuitive.

It's not terribly messy.  Let $ t \equal{} \sqrt[3]{2} \plus{} \sqrt[3]{3}$; then

$ t^3 \equal{} 5 \plus{} 3 \sqrt[3]{12} \plus{} 3 \sqrt[3]{18} \implies$
$ t^3 \minus{} 5 \equal{} 3 \left( \sqrt[3]{12} \plus{} \sqrt[3]{18} \right) \implies$
$ (t^3 \minus{} 5)^3 \equal{} 27 \left( 30 \plus{} 18 \sqrt[3]{12} \plus{} 18 \sqrt[3]{18} \right) \equal{} 27 \left( 30 \plus{} 6(t^3 \minus{} 5) \right)$.

The first equality on the last line is not true.

EDIT: sorry, it is obviously true

But do you know an easy way of showing that $ t^9\minus{}15t^6\minus{}87t^3\minus{}125$ is irreducible?

This might be useful

$ F({\sqrt[q]{e}}\equal{}F(({\sqrt[q]{f}}) \implies f\equal{}e^tc^q$ where char $ F$= $ 0$ and $ e,f \in F$.

[quote="Abel"]This might be useful

$ F({\sqrt [q]{e}} \equal{} F(({\sqrt [q]{f}}) \implies f \equal{} e^tc^q$ where char $ F$= $ 0$ and $ e,f \in F$.[/quote]
I have never seen that before and I have no idea why that is true. I liked the t0rajir0u's approach of finding a polynomial but I need help proving that it is irreducible.

The minimal polynomials of $ u: = 2^{1/3}, v : = 3^{1/3}$ over $ \mathbb{Q}$ are $ x^3 - 2, x^3 - 3$, respectively and therefore both $ \mathbb{Q}(u), \mathbb{Q}(v)$ are extensions of degree $ 3$ over $ \mathbb{Q}$. If $ \mathbb{Q}(u)$ contained $ v$ then $ \mathbb{Q}(u)$ would be a $ 3$-dimensional $ \mathbb{Q}$-vector space containing the 3-dimensional $ \mathbb{Q}$ vector space $ \mathbb{Q}(v) \Leftrightarrow \mathbb{Q}(u) = \mathbb{Q}(v)$. But then $ u, v$ have to be roots of the same irreducible polynomial over $ \mathbb{Q}$, which they are not.

[quote="puuhikki"]But then $ u, v$ have to be roots of the same irreducible polynomial over $ \mathbb{Q}$[/quote]

Why that?

  darij

[quote="puuhikki"]$ \mathbb{Q}(u) \equal{} \mathbb{Q}(v)$. But then $ u, v$ have to be roots of the same irreducible polynomial over $ \mathbb{Q}$[/quote]
this step is wrong


algebra
polynomial
complex analysis
function
real analysis
real analysis unsolved

Hello,
[quote="kiemkhach"]Find $f:R->R$ satisfies $f(1+f(x))=1+x$ and $f(f(x))=x  \forall x \in R$[/quote]
Are you sure that you are not forgetting some condition. For example, if it is a polynomial, it is very easy (almost trivial to proof) that $f(x)=x$ or $f(x) = -x$. If it has to be continuous, I think that it has to be also $f(x)=\pm x$. Now, if it does not have to be continues (actually, it will just have one discontinuity) there would be a complete class of holomorphic functions that will do it, as $f(x)=\frac1x$. If they don't have to be continuous, well man, you will get a huge family of functions.

Ups,

[quote="jmerry"]$-x$ is not a solution.

A very large family of functions (not continuous) that satisfy the conditions: Let $g(x)$ be an involution of the set $[0,1).$ Define $f(x)$ by the relation $f(x)=g(x)$ if $0\le x<1$ and $f(x+n)=f(x)+n$ for all integers $n.$[/quote]

Yeah, sorry, it seems like in some moment while I was writting the first condition scape of my mind. Jeje, as we would say in my country "Que pelada!" (Translated literally means "what a peeled!", and usually means that you are ashamed for the silly mistake you did!).

Best regards,


Princeton
college
function
calculus
derivative
integration
complex analysis

This is #8 in Ch.1 of Complx Analysis (vol. II of the Princeton Lectures in Analysis):

8. Suppose U and V are open sets in the complex plane. Prove that if f : U → V
and g : V → C are two functions that are differentiable (in the real sense, that is,
as functions of the two real variables x and y), and h = g ◦ f, then
\[ \frac {\partial{h}}{\partial{z}} \equal{} \frac {\partial{g}}{\partial{z}} \frac {\partial{f}}{\partial{z}} \plus{} \frac {\partial{g}}{\partial{\overline{z}}} \frac {\partial{\overline{f}}}{\partial{z}}
\]
and
\[ \frac {\partial{h}}{\partial{\overline{z}}} \equal{} \frac {\partial{g}}{\partial{z}} \frac {\partial{f}}{\partial{\overline{z}}} \plus{} \frac {\partial{g}}{\partial{\overline{z}}} \frac {\partial{\overline{f}}}{\partial{\overline{z}}}
\]

I believe this is only true if you further assume that f and g are differentiable in the complex plane - how else is derivation with respect to z defined?

It's a formalism, sometimes called Wirtinger calculus, the "derivatives" with respect to $ z$ and $ \bar{z}$ defined in terms of partial derivatives in $ x$ and $ y$ so that you get the right result for analytic and anti-analytic functions (so use the C-R equations). It's a very useful notation, because most of the time you can take formulas that are usually stated for functions $ u(x,y)$ and just replace $ x$ with $ z$, $ y$ with $ \bar{z}$, $ \partial x$ with $ \partial z$, $ dx$ with $ dz$ and so on. It works for many changes of variables, Green's formula (but you might need to introduce a constant), etc. Stokes formula in the plane is
$ \int_U \frac {\partial u}{\partial \bar{z}} \, dx \, dy \equal{} \frac {1}{2i} \oint_{\partial U} u(z) \, dz$.

It's not necessary to use it, but it's convenient, and when dealing with functions like for example $ |f|^{2p} \equal{} f^p \bar{f}^p$, $ f$ analytic, thinking in terms of Wirtinger calculus usually simplifies things a lot. It's mentioned in every textbook I've ever seen, but it's never explained why it is a good tool.

As for the given question, I don't see what possible issue you can have with this if you understand the definitions. What is the problem with rigor  :?:

Frankly I can see many "handwavy" solutions to such a problem. But to answer your question, I've done some initial computations (just plugging the said identities in) and the answer doesn't seem to agree...it looks like
\[ \frac {\partial{h}}{\partial{z}} \equal{} \frac {\partial{g}}{\partial{f}} \frac {\partial{f}}{\partial{z}} \plus{} \frac {\partial{g}}{\partial{\overline{f}}} \frac {\partial{\overline{f}}}{\partial{z}}
\]
and its counterpart would fit the bill better, and it is more intuitive.

It is understood that in the original formula that $ \frac {\partial g}{\partial z}$ and $ \frac {\partial g}{\partial \bar{z}}$ are to be evaluated in $ f(z)$. I think you have deduced the correct formula, but you are using incorrect notation in your answer. When one writes $ \frac {\partial g}{\partial f}$, typically $ f$ would have a double meaning as a variable and a function (like one writes y = g(x) or u = h(x,y)). In this case we are not even intending $ g$ to be differentiated against one of its variables in the usual sense, so writing $ \frac {\partial g}{\partial f}$ is very misleading.

So basically I interpreted my $ \frac{\partial{g}}{\partial{f}}$ as $ \frac{\partial{g}}{\partial{(u\plus{}iv)}}$  and then evaluated that in $ u$ and $ v$ using the given equations - is that equivalent to what you're saying it means?

I have no idea what we're talking about anymore. I am not a fan of the accepted differential operator-notation. Maybe I have confused you with poor english. I am saying that the original formula is this:
\[ \frac {\partial{h}}{\partial{z}}(w) \equal{} \frac {\partial{g}}{\partial{z}}(f(w)) \cdot \frac {\partial{f}}{\partial{z}}(w) \plus{} \frac {\partial{g}}{\partial{\overline{z}}}(f(w)) \cdot \frac {\partial{\overline{f}}}{\partial{z}}(w), \quad w \in U
\]
Since for example $ \frac {\partial{g}}{\partial{z}}$ is a function from $ V$ to $ C$, it is understood that in the formula it should be evaluated in $ f(w) \in V$. Note also that when either $ g$ or $ f$ is analytic the formula looks like the usual chain rule in one variable.

I am also saying that $ \frac {\partial{g}}{\partial{z}}(f(w))$ should never be written as $ \frac {\partial{g}}{\partial{f}}$ in this context, because even though we are using the notation $ \frac {\partial}{\partial z}$ this operator is NOT a true derivative - the $ z$ has a special meaning.

I believe Princeton's notation is wrong and prefer the formulation from "Function Theory of One Complex Variable" by Greene and Krantz: Let $ f$ and $ g$ be $ C^1$ functions and assume $ f\circ g$ is defined.  Prove the following version of the chain rule with $ w \equal{} g(z)$:
\[ \frac {\partial}{\partial z} \left(f\circ g\right) \equal{} \equal{} \frac {\partial{f}}{\partial{w}}\frac {\partial{g}}{\partial{z}} \plus{} \frac {\partial{f}}{\partial{\overline{w}}}\frac {\partial{\overline{g}}}{\partial{z}}.
\]
Here, I'm assuming $ f\circ g \equal{} f(g,\overline{g})$.  For example, let $ f \equal{} z^2 \plus{} 2\overline{z}$ and $ g \equal{} 3z \minus{} 4\overline{z}$ then $ \overline{g} \equal{} 3\overline{z} \minus{} 4z$ and $ f\circ g \equal{} (3z \minus{} 4\overline{z})^2 \plus{} 2(3\overline{z} \minus{} 4z)$ and therefore
\[ \frac {\partial}{\partial z}\left(9z^2 \minus{} 24z\overline{z} \plus{} 16\overline{z}^2 \plus{} 6\overline{z} \minus{} 8z\right) \equal{} 18z \minus{} 24\overline{z} \minus{} 8 \equal{} \frac {\partial{f}}{\partial{w}}\frac {\partial{g}}{\partial{z}} \plus{} \frac {\partial{f}}{\partial{\overline{w}}}\frac {\partial{\overline{g}}}{\partial{z}}
\]
where:
\[ \frac {\partial f}{\partial w} \equal{} \frac {\partial f}{\partial z}\Bigg|_{z \equal{} g(z)}
\]

\[ \frac {\partial f}{\partial \overline{w}} \equal{} \frac {\partial f}{\partial \overline{z}}\Bigg|_{\overline{z} \equal{} \overline{g(z)}}
\]
The authors give the suggestion:  Write out everything in real and imaginary parts and use the real variable chain rule from Calculus.  Note if both are analytic then the partial reduces to the ordinary chain rule since by definition $ \frac{\partial f}{\partial \overline{z}}\equal{}0$.