In the Upstate NY area(a little north of Albany), there is a whole team of home-schooled MathCounts competitors, and a couple of them nearly won Nationals. And it was made up entirely of home-schooled students who were interested in mathematics and were from the local area.

I would recommend starting one of your own if you aren't from that area, or aren't from an area that doesn't have a team.

If you can't join other homeschoolers in your area to form a team, find out who your local coordinator is.  You can enter the chapter competition as an individual.  The registration deadline is in early December, but you should apply sooner so your parents can get the study materials for you.  It might also be worth trying to get your hands on old chapter, state, and national exams, even if you have to buy them.

My son who is homeschooled did this in the 6th grade last year, finished second in his chapter, went on to finish first in the state, and went with the state team to nationals.   Advancement from one level to the next is only based on the individual scores and not the team score.

Granted we're not from New York and the competition here is not at the same level -- but it's absolutely possible for an individual homeschooler to compete in MathCounts.  Without teammates you won't be able to do the team competitions, at least not officially.  My son did the team round unofficially with 3 individuals from a local school at chapters.  He could not do that round at state.  And of course he was on a team at Nationals.

(Technically by MathCounts rules each homeschool is its own school.  There's a provision for combining teams from small schools in the same or neighboring chapters to compete as one team.  So if you have a way to get in touch with homeschoolers in your chapter and neighboring chapters within the same state, you just might be able to come up with a team.  Even three can compete as a team, but the sum of your individual scores gets divided by four.  So you're not likely to win but still get the experience of doing the competition.)

The MathCounts website should help you if any of this isn't clear --

Hope this helps,

Blue Morpho

The selective advantage to being a homeschooler is that you usually have a lot more time than students who attend a private or public school. Use this to your benefit and you'll see how much farther you progressed!  :)

[quote="Math46"]HI I don't have anywhere to participate in MATHCOUNTS! So I don't do it.  But I'm homeschooled in 6th grade and very accelerated in Math. :)[/quote]
I am in 7th grade, and I was quite better at math last year. I've done Intro to Number thoery, algebra, and Geometry, none quite actually finished for one reason or another,
If you want to be in MATHCOUNTS, you can set up a home-schooling group, or you can sign up as an indivisual. I was at mathcounts last year and I signed up for a KIND-OF school I go to, there was a class but I was the only one there. So I had to do the team section alone.
I ended up scoring 52nd place, with about 70 people there.
I probably would have gotten higher if I had actually studied, but lower if I hadn't taken AOPS classes.
So, just a hint, MATHCOUNTS math is realatively easy if you actually do it, but if you don't study, no matter how smart you are, you probably won't get that high.

Well, I'm home schooled and I have never done MathCounts but there is a MathCounts homeschool team around my area and there might be one around yours.

you could always go to britian and participate in the British Math Olympiad...
[url]http://www.bmoc.maths.org/home/bmo.shtml[/url]

[hide="Letter from the queen"]MESSAGE FROM, HER ROYAL HIGHNESS, THE QUEEN

To the citizens of the United States of America from Her Sovereign
Majesty Queen Elizabeth  II

In light of your failure in recent years to nominate competent
candidates for President of the USA and thus to govern yourselves, we
hereby give notice of the revocation of your independence, effective
immediately. (You should look up 'revocation' in the Oxford English
Dictionary.)

Her Sovereign Majesty Queen Elizabeth II will resume monarchical
duties over all states, commonwealths, and territories (except Kansas
, which she does not fancy).

Your new Prime Minister, Gordon Brown, will appoint a Governor for
America without the need for further elections.
Congress and the Senate will be disbanded.  A questionnaire may be
circulated next year to determine whether any of you noticed.
To aid in the transition to a British Crown dependency, the following
rules are introduced with immediate effect:
-----------------------
1. The letter 'U' will be reinstated in words such as 'colour,'
'favour,' 'labour' and 'neighbour.'  Likewise, you will learn to spell
'doughnut' without skipping half the letters,  and the suffix '-ize'
will be replaced by the suffix '-ise.'  Generally, you will be
expected to raise your vocabulary to acceptable levels.  (look up
'vocabulary').
------------------------
2. Using the same twenty-seven words interspersed with filler noises
such as ''like' and 'you know' is an unacceptable and inefficient form
of communication. There is no such thing as U.S.English.. We will let
Microsoft know on your behalf.  The Microsoft spell-checker will be
adjusted to take into account the reinstated letter 'u'' and the
elimination of  '-ize.'
      -------------------
3. July 4th will no longer be celebrated as a holiday.
-----------------
4. You will learn to resolve personal issues without using guns,
lawyers, or therapists.  The fact that you need so many lawyers and
therapists shows that you're not quite ready to be independent.  Guns
should only be used for shooting grouse.  If you can't sort things out
without suing someone or speaking to a therapist,then you're not ready
to shoot grouse..
----------------------
5. Therefore, you will no longer be allowed to own or carry anything
more dangerous than a vegetable peeler.  Although a permit will be
required if you wish to carry a vegetable peeler in public.
----------------------
6. All intersections will be replaced with roundabouts, and you will
start driving on the left side with immediate effect.  At the same
time, you will go metric with immediate effect and without the benefit
of conversion tables.  Both roundabouts and metrication will help you
understand the British sense of humour.
--------------------
7. The former USA will adopt UK prices on petrol (which you have been
calling gasoline) of roughly $ \$$10/US gallon.  Get used to it.
-------------------
8. You will learn to make real chips.  Those things you call French
fries are not real chips, and those things you insist on calling
potato chips are properly called crisps.  Real chips are thick cut,
fried in animal fat, and dressed not with catsup but with vinegar.
-------------------
9. The cold, tasteless stuff you insist on calling beer is not
actually beer at all.  Henceforth, only proper British Bitter will be
referred to as beer, and European brews of  known and accepted
provenance will be referred to as Lager.  Australian beer is also
acceptable, as they are pound for pound the greatest sporting nation
on earth and it can only be due to the beer.  They are also part of
the British Commonwealth - see what it did for them.  American brands
will be referred to as Near-Frozen Gnat's Urine, so that all can be
sold without risk of further confusion.
---------------------
10. Hollywood will be required occasionally to cast English actors as
good guys.    Hollywood will also be required to cast English actors
to play English characters.  Watching Andie Macdowell attempt English
dialogue in Four Weddings and a Funeral was an experience akin to
having one's ears removed  with a cheese grater.
---------------------
11. You will cease playing American football.  There is only one kind
of proper football; you call it soccer.  Those of you brave enough
will, in time, be allowed to play rugby (which has some similarities
to American football, but does not involve stopping for a rest every
twenty seconds or wearing full kevlar body armour like a bunch of
nancies)..
---------------------
12. Further, you will stop playing baseball.  It is not reasonable to
host an event called the World Series for a game which is not played
outside of America .  Since only 2.1% of you are aware there is a
world beyond your borders, your error is understandable.  You will
learn cricket, and we will let you face the Australians first to take
the sting out of their deliveries.
--------------------
13. You must tell us who killed JFK.  It's been driving us mad.
-----------------
14. An internal revenue agent (i.e. tax collector) from Her Majesty's
Government will be with you shortly to ensure the acquisition of all
monies due (backdated to 1776).
---------------
15. Daily Tea Time begins promptly at 4 p.m. with proper cups, with
saucers, and never mugs, with high quality biscuits (cookies) and
cakes; plus strawberries (with cream)  when in season.

God Save the Queen!

Originally from
[url]http://www.funnycoolstuff.com/2008/10/24/the-letter-from-the-queen-of-england/[/url][/hide]


search function AMC USA(J)MO USAMO MATHCOUNTS AIME

Or go to the Missouri Forum.

Also, Alan Chou no longer exists, if it's the Alan Chou I think you're talking about.

Alan Chou isn't in MC...? Unless I'm missing something.

I looked here: [url]http://mathcounts.org/Document.Doc?id=341[/url]
and searched for Alan Chou (using the search function of course), and there is no such person going to nationals.

I know! I've heard a lot about alan chou cause someone said "Are you alan chou's bro or something?" I was like...... Ok.. And I googled alan chou and it seemed as if he's a smarticle. But I don't see much activity with him. I think he's gone.  :|

[quote="dnkywin"]Predict what this year's nats will be, like this:
[hide]Champion: Bobby Shen
Runner-up: Darryl Wu
Semifinalists: Calvin Deng, Alan Chou
Quarterfinalists: Jason Hyun, Runpeng Liu, Steven Chen, Sam Zbarsky
Qualifiers: David Yang, Mark Selke, Scott Wu, Albert Chu[/hide][/quote]
woah...egotism here by voting for yourself????? :huh:

Let's not make predictions for ourselves or our teams, cuz that's like bragging.

But I do predict that Bobby would do better than Darryl this year.
And Texas would win.

COUNTDOWN (those who'd surely make it): Bobby Shen, Darryl Wu, Sam Zbarsky, Jason Hyun, Alan Zhou.

That's all I'm putting down for now. We can add more when time goes by, closer to Nats.

And no offense to him or anything, but why Albert Chu? WA's state CD was official, and Albert started at only 10th at states for written but beat everyone to win cd and the state title...

By making all these predictions all you are doing is adding a lot of pressure to all these kids.  They are all very good and any of them could win if they have a good day.  The national champion(s) is not always the top kid or team.  It's the kid or team that was the best on THAT ONE DAY.  If they had another competition the next week the results would be close but I would bet that there would be several differences.

[quote="gauss1181"]

Let's not make predictions for ourselves or our teams, cuz that's like bragging.

[/quote]

It's much better than showing fake modesty, which is seriously very annoying. Also, it can depress others when they [b]know[/b] that a person is better than them, but that person keeps saying that they fail.

Also, I agree with JBoyd; if you have a competition several different times in a week, it is pretty likely that you will have several different champions.

I also agree with something that math154 said to me last night: "meh speculation is so pointless...but addicting"  :P .

I bet there will be some random people in the cd round. You  guys may not no some people who are silent, but good. For example, there is this 7th grader from VA that got a perfect paper in state, and she owned everyone in cd round winning 3-0 on most of the rounds and 2-1 in one. Or it may just be that someone who you don't think will do good might have been practicing really hard and will have a good day at nats. So really it all depends...

I completely agree with Mr. Boyd. We should probably stop talking about predictions.

[quote="BOGTRO"][quote="pythag011"]No; Alan Chou changed his last name to Zhou, so now he's Alan Zhou.[/quote]

lol. Add a [/sarcasm] next time.[/quote]

I'm serious.

If creativity is real math, what are formulas?

Uh... formulas are not real math. Creativity is.

6-7th graders don't have creativity? Whatcha talking about? We have plenty!

[quote="ReseesPieces"]umm i guess the more expereinced u are u know more formulas....if thats what u mean....andno, 6th graders and 7th graders dont have more creativity...[/quote]

This is somewhat funny. And has no truth to it at all.

My opinion is that whoever works the hardest has the best chance of winning. Well, I suppose whoever works the hardest and the smartest has the best chance, but that's really there is to it. Of course there's some luck involved, but my point still stands.

i think we are talking about insight, not creativity: creativity is what you need to write and draw; hence the name "creativity"
when you find a elegant solution/ alternative solution to a tough olympiad problem, it's called insight: it wasn't knowing the math, because elegant= less math, but it was knowing how to approach it

[quote="CubeX"]The thing is, the little people(6th or 7th) can actually solve the problems that use the tiniest bit of creativity really good, and the big(8th) people know all the formulas and such, so they will do good on problems based on real math.

Don't put pressure on people going to NATS. :!:[/quote]

Creativity really isn't used in mc problems; all of the mc probs I've seen are straightfoward (I'm not saying that its a simple application of one theorem, but the solution is extremely intuitive....)

I agree with xpmath. (Though in my opinion luck is a huge part of mc.)

Hmm... advanced formulas can be extremely helpful if you know how to use them well (I've used Lagrange Interpolation version of Chinese Remainder Theorem three times on mc probs...) but generally the quickest solution to mc problems uses only simple theorems..

I've also noticed that quite a few mc probs are simple applications of standard tricks...

Hm. I should make problems when I grow up. I think izzy should too since those Problem of the Days are stumping me...  :wink:

[quote="jonathanchou711"]Yeah! It all depends! And just cause you're an 8th grader or something doesn't mean you have more abilities than us! Darryl Wu had skillz and he was in 6th grade![/quote]

Not to mention being one of the three six-grader USAMO-ists.

(which includes pythag011 and [sarcasm]this really random guy called Lewis Chen[/sarcasm])

And two of them are on the California national team!

(in case anyone missed the sarcasm I am Lewis Chen... lol)

anyways, in terms of so-called "creativity", we have 2 2008 USAMO-ists, and the only other team rivaling that is Maryland.

[quote="BOGTRO"][quote="ReseesPieces"]umm i guess the more expereinced u are u know more formulas....if thats what u mean....andno, 6th graders and 7th graders dont have more creativity...[/quote]

I find this very offensive. I'm willing to bet that I have more creativity then you do.[/quote]

Indeed, you have 7 Creativity Units whereas ReseesPieces has only 6.5, a fact which was absolutely necessary to mention, extremely important, and not pathetic bragging at all!

[quote]
Not to mention being one of the three six-grader USAMO-ists.

(which includes pythag011 and [sarcasm]this really random guy called Lewis Chen[/sarcasm])[/quote]

Firstly, pythag is a seventh grader. Secondly, "insight" and "creativity" are relative, meaningless terms, and one could argue that making USAMO requires far more in the way of experience than insight. Thirdly, the point of sarcasm is subtlety, which you have utterly disgraced. Never attempt to use sarcasm again.

james4l was talking about how Darryl was one of the three 6th grade USAMO qualifiers last year, along with pythag and himself.

Pythag was in seventh grade last year, too, but he failed it.

[size=59]okay fine i was wrong[/size]

[quote="ReseesPieces"]w/e...i feel stupid right now hey maybe its cause i am stupid

sigh*

 back on topic.....i geuss... i think CA will produce atleast 3 finalits, Texas willl produce 3, VA atleast 3, Washington maybe 1 or 2 and the last couple from Michigan, Illinois, New Jersey, or Ohio....[/quote]

What's a finalist?

[quote="ReseesPieces"]w/e...i feel stupid right now hey maybe its cause i am stupid

sigh*

 back on topic.....i geuss... i think CA will produce atleast 3 finalits, Texas willl produce 3, VA atleast 3, Washington maybe 1 or 2 and the last couple from Michigan, Illinois, New Jersey, or [b]Ohio[/b]....[/quote]

As if.

You forgot Maryland anyway. I don't think that all three of CA, TX, and VA will produce three each.
There's also Missouri, North Carolina, etc... Also how can you forget Alan Zhou from Massachusetts? Sigh.

Man I love reading all these comments from Nat competitors....since it means your aren't practicing and will be less likely to out do my kids :-)  And that was sarcasm!

Seriously guys( and gals) all this talk is pointless!!!  Especially since I've been to 8 Nats and only twice times have 3 kids from one state made top 12.  I would be really surprised to see it again this year.

If you are on my team and reading this......get back to work!!  See you at practice in the morning :lol:

Best of luck to everyone at Nats!!

[quote="frost13"]Man I love reading all these comments from Nat competitors....since it means your aren't practicing and will be less likely to out do my kids :-)  And that was sarcasm![/quote]

I do not have a very strong will and thus cannot resist the temptation to speculate about this. Also I get a free pin if I predict well enough, which clearly is much more important than winning.  :ninja:

[quote="frost13"]If you are on my team and reading this......get back to work!!  See you at practice in the morning :lol:

[/quote]

I can assure you that one of your team members just saw your post. Possibly because I gave him the link.

Hmm...

I'm hoping certain things about nationals (Though I would love it if my wish that all mc probs were usamo #1-level came true, this is clearly unrealistic.) After an easy chapter round and a very easy state round, will nationals be easier or harder?

[quote="tinytim"]Obviously Mathcounts saved all of its hard problems and loaded them into the National Test... (example: 2005)[/quote]

I'd like Nats to be mid-AIME level / trivial Olympiad #0 probs

Killer, but not really killer.

Specifically to HidN428 who wrote in part "I bet there will be some random people in the cd round"... -- in part I agree with the "gist" of what you wrote, but I have to comment on where you wrote the above "random people" comment.  How do you define "random"? If someone analyzed all past competitions including but not limited to MATHCOUNTS (ie: including AMC results, AIME results, other math competitions, etc.) and some student who wasn't involved or didn't score high in any of those suddenly "won" - then maybe some would consider that "random" --, but I'd suggest that people outside your picks are only "random" to you because you are not familiar with their prior accomplishments or current abilities - would you agree with that?

Personally, I'd like Mathcounts to be Mathcounts level, which appears to be what it does best.

[quote="JBoyd"]I agree with Frost13.  The kids that are serious about nationals are working hard and not spending a lot of time in forums talking about nationals.  There is an old saying that 'talk is cheap, show me something'.  I actually told my kids that I did not want them on AoPs for the next month unless they were helping on a problem or playing FTW.

All these conjectures mean absolutely nothing.  What everyone has done at chapter and state up to this point means nothing.  What means a lot is how much qualilty time you are putting into practicing your individual skills and getting to know your team mates.  Even then what means the most is how your team does on the ONE DAY at Nationals.  You could be the best team in the country hands down but if you have a bad day, it's all over.  What the teams are doing now will help to minimize the possibility of having a bad day.

With that said....Good luck to everyone that is working hard![/quote]
Good words..... :omighty:

[quote="JBoyd"]I agree with Frost13.  The kids that are serious about nationals are working hard and not spending a lot of time in forums talking about nationals.  There is an old saying that 'talk is cheap, show me something'.  I actually told my kids that I did not want them on AoPs for the next month unless they were helping on a problem or playing FTW.

All these conjectures mean absolutely nothing.  What everyone has done at chapter and state up to this point means nothing.  What means a lot is how much qualilty time you are putting into practicing your individual skills and getting to know your team mates.  Even then what means the most is how your team does on the ONE DAY at Nationals.  You could be the best team in the country hands down but if you have a bad day, it's all over.  What the teams are doing now will help to minimize the possibility of having a bad day.

With that said....Good luck to everyone that is working hard![/quote]

Man. That makes me think of the song "Bad Day". I think that at least one kid will have a bad day with a deadly cough and a runny nose.


[size=150]Let $ a,b > 0$,$ t,m$ are real numbers ,prove that:

$ \frac {a^{t \plus{} m}}{b^t} \geq \frac {a^t}{b^{t \minus{} m}} \plus{} a^m \minus{} b^m,$
        if $ tm \geq 0.$

$ \frac {a^{t \plus{} m}}{b^t} \le \frac {a^t}{b^{t \minus{} m}} \plus{} a^m \minus{} b^m,$
        if $ tm< 0.$

BQ[/size]

[size=150]If the above is 2-variables ,Now discuss 3-variables:

Let $ a,b,c > 0$,and $ m,t$ are real numbers,prove that:

$ \frac {a^{t \plus{} m}}{b^t} \plus{} \frac {b^{t \plus{} m}}{c^t}\geq \frac {a^t}{b^{t \minus{} m}} \plus{} \frac {b^t}{c^{t \minus{} m}} \plus{} a^m \minus{} c^m.$

   if $ tm \geq 0.$

$ \frac {a^{t \plus{} m}}{b^t} \plus{} \frac {b^{t \plus{} m}}{c^t}\le \frac {a^t}{b^{t \minus{} m}} \plus{} \frac {b^t}{c^{t \minus{} m}} \plus{} a^m \minus{} c^m.$

   if $ tm<0.$

BQ[/size]

4-variables:

$ \frac {a^{t \plus{} m}}{b^t} \plus{} \frac {b^{t \plus{} m}}{c^t} \plus{} \frac {c^{t \plus{} m}}{d^t} \geq \frac {a^t}{b^{t \minus{} m}} \plus{} \frac {b^t}{c^{t \minus{} m}} \plus{} \frac {c^t}{d^{t \minus{} m}} \plus{} a^m \minus{} d^m.$
             if $ tm\geq 0$.

$ \frac {a^{t \plus{} m}}{b^t} \plus{} \frac {b^{t \plus{} m}}{c^t} \plus{} \frac {c^{t \plus{} m}}{d^t} \le \frac {a^t}{b^{t \minus{} m}} \plus{} \frac {b^t}{c^{t \minus{} m}} \plus{} \frac {c^t}{d^{t \minus{} m}} \plus{} a^m \minus{} d^m.$
             if $ tm<0$.

Come on, guys:
       Congratulations to the generation of this great inequality！
     Pythagoras is killing more than 100 head of cow?
      So let me cry a few sound ...I am sorry，You've got to keep sleep。

BQ

[size=150]thank limes123,dear limes123,
This inequality has always been your,from you....
BQ[/size]

well,My job is done,That you have shown，solution....

BQ

[quote="xzlbq"]n-variables:

[size=150]Let $ x_i > 0,i \equal{} 1,...,n,n \geq 2,$,prove that:

$ \sum_{i \equal{} 1}^{n \minus{} 1}\frac {x_i^{t \plus{} m}}{x_{i \plus{} 1}^t}\geq \sum_{i \equal{} 1}^{n \minus{} 1}\frac {x_i^t}{x_{i \plus{} 1}^{t \minus{} m}} \plus{} x_1^m \minus{} x_n^m.$
    if $ tm \geq 0$.

$ \sum_{i \equal{} 1}^{n \minus{} 1}\frac {x_i^{t \plus{} m}}{x_{i \plus{} 1}^t} \le \sum_{i \equal{} 1}^{n \minus{} 1}\frac {x_i^t}{x_{i \plus{} 1}^{t \minus{} m}} \plus{} x_1^m \minus{} x_n^m.$
    if $ tm < 0.$

BQ[/size][/quote]

$ n \equal{} 2, 3$     see here

[url]http://www.irgoc.org/viewtopic.php?f=25&t=252&sid=ff42ade211ee93c04d13ac35714f41e8[/url]

or $ n \equal{} 3$


probability
Support
combinatorics proposed
combinatorics

A and B each has a pair of dice . A will win if he throws sum 7 and B will win if he throws sum 4 and they throw simultaneously untill one of them wins, equal throws being disregarded.Find P(A)

the chance of throwing a 7 is 1/6 and the chance of throwing a 4 is 1/12. so the chance of both is 1/72 and the chance of neither is 55/72. so there is a 1-55/72-1/72=2/9 chance of one of them winning in any given round, and there is a 1/6*11/12=11/72 chance of A winning in any given round o the probability A wins is (11/72)/(2/9)=[b]11/16[/b]

Is this really an Olympiad problem, seems too easy to me

actually here the throwing is repeated like ABAB.....
so it would include a gp or such things. not really hard for olympiad level but harder than [i]latios315[/i] wrote. :wink:

what do you mean? I solved as if they threw until someone won.

game can end in 9 ways 6 support to A and 3 to B.Hence P(A)=6/9=2/3

First of all i would like to comment that the language of the problem is confusing and bad.  :wink: i mean even $ P(A)$ is not defined even if we leave aside the fact that the following portion :[quote="makar"]A and B each has a pair of dice . A will win if he throws sum 7 and B will win if he throws sum 4 and [u][i][b]they throw simultaneously untill one of them wins,[/b][/i][/u] equal throws being disregarded.Find P(A)[/quote] does not make sense grammatically. :huh:
also equal throws are disregarded according to [i]makar[/i]. So there are actually 2 ways(not 3) in which B can win.
for the moment we leave that condition(equal throw or else the 3/36 will be replaced by 2/36)
now for the problem. it is not clear if the throwing occurs like ABABAB... or only for one round.(for the word [b]"until"[/b] in the question).
If the process is repeated. then probability for A to win is
$ P(A) = [(6/36)(1 - 3/36)] + \\
[(1 - 6/36)(1 - 3/36)(6/36)(1 - 3/36)] + \\
[((1 - 6/36)(1 - 3/36))^2(6/36)(1 - 3/36)] + ... \\
= (1/6)(11/12)[1 + 55/72 + (55/72)^2 + ...] \\
= (1/6)(11/12)*72/17 \\
= 11/17.$

if the process ends in one single round:
then $ P(A) = (6/36)(1 - 3/36) = 11/72$

if the dice is thrown by some person $ C$ and if the problem was as follows:
A will win if the sum is 7 and B will win if the sum is 4 and dices are thrown simultaneously untill one of them wins.Find P(A).

and if the game was to end in a single round,
then only the game can [u][b]end[/b][/u] in 9 ways but they are not all the ways.obviously the game might not end in a  single round.

See the place you are getting confused is to think that each round is seperate and independent which is not the case. please see example2 on page 387 in Hall&Knight if you are not convinced. Also the problem in the book asked for Expectation, not probability which is a bit different.
I will try to clarify the thing:
what if A throws 7 and B throws 4 simultaneously?this case was not considered in your solution. So in any round A can win iff A throws 7 and[b] B doesn't throw 4.[/b]
according to you probability of winning for A in second round is 2/3. But actually probability for A to win in second round is equal to:
(probability of not winning for A in 1st round)*(probability of not winning for B in 1st round)*(probability of  winning for A in 2nd round)*(probability of not winning for B in 2nd round) :wink:  :)

[quote="makar"]First of all I want to say sorry if language of the problem is not clear.some one asked me the same problem.I found the similler in Higher Algebra By Hall and Knight (Examples XXXII c problem 15). Problem goes like this:

"A makes a bet with B of Rs. 15 to Rs. 6 that in a single throw with two dice he will throw seven before B throws four. Each has a pair of dice and they throw simultaneously untill one of them wins, equal throws being disregarded; find B's expectation."

solution:
In any round they are throwing simultaneously
In each round it is 2 to 1 in favour of A
hence if chance of winning the game by B is X then for A it will be 2X
X +2X =1,
X=1/3
P(A)=2X=2/3

ALTERNATIVELY:

Moreover when i said game can end in 9 ways by this I means whenever it will end there can be 9 ways only out of which 6 favours to A and 3 to B.
Hence P(A)=6/9=2/3[/quote]

I assume that the question is really supposed to say the following:

A and B play a game with pairs of dice.  A has a "good" throw if the sum of As dice is 7, while B has a "good" throw if the sum of Bs dice is 4.  In each round, they throw simultaneously; if one of them has a good throw and the other does not, that player is declared the winner; otherwise, they throw again.

Assuming this is a correct reading of the problem, the only person who's posted what appears to be a correct solution is latios315.

@ithpower: you appear to not understand the method that latios used.  (Or perhaps just don't understand the notion of simultaneous throws?  In this case, the rest of my comments to you are off-point.  The phrase you highlighted as ungrammatical actually makes perfect sense, although "equal throws being disregarded" does not.)  If the probability that A wins a given round is $ p$ and the probability that B wins a given round is $ q$ ($ 0 < p \plus{} q \leq 1$) then the probability that A wins in the end is $ \frac {p}{p \plus{} q}$, a result that one can (but need not) derive using the series approach.  The series that you wrote down is not correct: the probability that A wins in the second round is $ \left(\frac {6}{36} \cdot \frac {3}{36} \plus{} \left(1 \minus{} \frac {6}{36}\right) \cdot \left(1 \minus{} \frac {3}{36}\right)\right) \cdot \frac {6}{36} \cdot \left(1 \minus{} \frac {3}{36}\right)$, not what you wrote.  Had you done it correctly, you would have ended up with latios' answer.

@makar: in the first computation, you've treated every throw of 7 by A as a victory, but this is not the case (because B might throw a 4).  The 9 possible ending positions are not equally likely, either.

First I want to say thanks to JBL, actualy when i first solved the problem i got the same answer by both series way as well as by p/p+q as that of latios315 but when i come across this hall and knight problem, Expectation of B was not maching. My one friend suggested me the same solution which i posted over here to get the opinions of experts i also objected the solution but answer is coming out from P(B)=1/3
as expectation of B = (1/3)*15-(2/3)*6=1 which is the answer in the book.
usually answers in hall and knight are correct
JBL can u give a try to Hall and knight problem which i am mentioning

"A makes a bet with B of Rs. 15 to Rs. 6 that in a single throw with two dice he will throw seven before B throws four. Each has a pair of dice and they throw simultaneously untill one of them wins, equal throws being disregarded; find B's expectation."

Thank you JBL for clearing the doubt and sorry for the computation error. :blush:
By the way sorry to latios315 because i didnot read the post carefully.

Hi makar,

That answer is certainly inconsistent with my understanding of the problem (I think it should be $ 15 \cdot \frac{5}{16} \minus{} 6 \cdot \frac{11}{16} \equal{} \frac{9}{16}$) -- most likely this means that there is some sort of typo or error in the text (something which happens to nearly every textbook).


Is there a website or something where I can download it?

I believe that you would have to buy AutoCAD before you can download it.

There are however, [url=http://www.osalt.com/autocad]open source alternatives[/url] which are free.

You can download a free student version of AutoCAD (and free student versions of many other Autodesk products) at [url=http://students.autodesk.com/]http://students.autodesk.com/[/url]. I think the free versions have the same features as the professional versions.

@Ollie10: AutoCAD is a 3d design software from Autodesk.


I applied arrangement inequaltiy.
First the inequality is equivalent to:
$\frac{4a-1}{1-a^{2}}+\frac{4b-1}{1-b^{2}}+\frac{4c-1}{1-d^{2}}+\frac{4d-1}{1-d^{2}}\ge 0$
But WLOG:$a \ge b \ge c \ge d$
we have:
$4a-1 \ge 4b-1 \ge 4c-1 \ge 4d-1$,Also:
$\frac{1}{1-a^{2}}\ge \frac{ 1}{1-b^{2}}\ge \frac{1}{1-c^{2}}\ge \frac{1}{1-d^{2}}\ge 0$
So by the arrangement inequality.
It seems so easy,So I doubt if my solution is ture?

[quote="arqady"][quote="zhaobin"]
So by the arrangement inequality.[/quote]
Are you mean by Chebishov? :wink:
[quote="zhaobin"]
It seems so easy,So I doubt if my solution is ture?[/quote]
I think, it is very nice proof. :)[/quote]
Thank you :P
Yes I mean by  Chebyshev inequality.
In fact Chebyshev inequality can be proved by arrangement inequality so I just wrote that. :)
And also because I think I may spell "Chebyshev" in a wrong way then may mislead others. :P
See,Now you didn't spell it in a right way  :wink:

[quote="zhaobin"]
And also because I think I may spell "Chebyshev" in a wrong way then may mislead others. :P
See,Now you didn't spell it in a right way  :wink:[/quote]
I think you are wrong, zhaobin. :P  :)
When he (Chebyshov) was alive (150 years ago), in Russian it was written Чебыш[b]ё[/b]в (Chebysh[b]o[/b]v).
But as it is often not accepted to write the letter to Russia [b]ё[/b],
the mistake has turned out: Чебыш[b]е[/b]в (Chebyshev).
In occasion of the letter [b]y[/b] (Cheb[b]y[/b]shov) it I agree with you.
Id est, in Russian: Чебышёв=Чебышев.
But to say in both cases it is necessary [b]o[/b] (if we wish to respect to P.L.Chebyshov).
In English: Chebyshov only. :)
The gift for you, zhaobin:
Let $a>0,$ $b>0$ and $c>0.$ Prove that
$\frac{\sqrt[3]a}{2a+b}+\frac{\sqrt[3]b}{2b+c}+\frac{\sqrt[3]c}{2c+a}\leq\frac{\sqrt[3]a}{2b+c}+\frac{\sqrt[3]b}{2c+a}+\frac{\sqrt[3]c}{2a+b}.$

[quote="arqady"]
The gift for you, zhaobin:
Let $a>0,$ $b>0$ and $c>0.$ Prove that
$\frac{\sqrt[3]a}{2a+b}+\frac{\sqrt[3]b}{2b+c}+\frac{\sqrt[3]c}{2c+a}\leq\frac{\sqrt[3]a}{2b+c}+\frac{\sqrt[3]b}{2c+a}+\frac{\sqrt[3]c}{2a+b}.$[/quote]
Thank you,Arqady.But I think you are old... :P You have post this inequality on the forum,and I rememeber the inequality is from your student who take part in IMO.
Any way I think I give a solution,and you said the solution is nice. :lol:

[quote="zhaobin"]I rememeber the inequality is from your student who take part in IMO.
[/quote]
Yes! The idea from him, but  the forming is mine. :)


[quote="Alexander Khrabrov"]For positive numbers $a_k$, $b_k$ and $c_k$ prove an inequality
\[ \sum (a_k+b_k+c_k) \sum \frac{a_k b_k + b_k c_k + c_k a_k}{a_k+b_k+c_k}
   \sum \frac{a_k b_k c_k}{a_k b_k + b_k c_k + c_k a_k} \leq
   \sum a_k \sum b_k \sum c_k .\][/quote]
let $t$=right-left
$\sum a_k \sum b_k \sum c_k =\sum a_i b_j c_k $
$ \sum (a_k+b_k+c_k) \sum \frac{a_k b_k + b_k c_k + c_k a_k}{a_k+b_k+c_k}
   \sum \frac{a_k b_k c_k}{a_k b_k + b_k c_k + c_k a_k}$=$ \sum (a_i+b_i+c_i) \frac{a_j b_j + b_j c_j + c_j a_j}{a_j+b_j+c_j} \frac{a_k b_k c_k}{a_k b_k + b_k c_k + c_k a_k}$
chang $i ,j, k$poistion
I think it can be proved,but I havn't prove it now,I will post again when I prove it


A round pencil has length 8 when unsharpened, and diamter 1/4. It is shrpened perfectly so that it remains 8 inches long, with a 7-inch section still cylindrical and the remaining 1 inch giving a conical tip. What is the volume?

[hide="Solution"]

We have a cylinder with height $ 7$ and radius $ \frac{1}{8}$ and on top of that we have a cone whose height is $ 1$ and whose radius is $ \frac{1}{8}$.

So the volume of the sharpened pencil is

$ \pi (\frac{1}{8})^{2}\times 7 \plus{} \frac{1}{3} \pi (\frac{1}{8})^{2} \times 1$

$ \equal{} \frac{7\pi}{64} \plus{} \frac{\pi}{192}$

$ \equal{} \frac{22 \pi}{192}$

$ \equal{} \frac{11 \pi}{96}$.

[/hide]


[quote="risinglegend"]How to do this sum? $(-2x^3+3x^2+7x+9) + (3x^3-5x^2-11)$[/quote]

Add like terms together to get $x^3-2x^2+7x-2$.

Note that when you add the polynomials, the exponents do not change, and you can only add like exponents (for example, you can do 3x^2 + 4x^2 = 7x^2, but not 3x^2 + 3x = 6x^3)

Yeah, just add the coefficients of like polynomials


Prove that

$ 2\minus{}\frac{r}{R}\leq\frac{1}{2}(\frac{h_a}{r_a}\plus{}\frac{h_b}{r_b}\plus{}\frac{h_c}{r_c})\leq\frac{R}{r}\plus{}\frac{r}{R}\minus{}1$

[quote="Ligouras"]Prove that

$ 2 \minus{} \frac {r}{R}\leq\frac {1}{2}(\frac {h_a}{r_a} \plus{} \frac {h_b}{r_b} \plus{} \frac {h_c}{r_c})\leq\frac {R}{r} \plus{} \frac {r}{R} \minus{} 1$[/quote]

$ \frac {1}{2}\sum {\frac {{h_a }}{{r_a }}} \equal{} \frac {1}{2}\sum {\frac {{h_a (s \minus{} a)}}{S}} \equal{} \sum {\frac {{s \minus{} a}}{a}} \equal{} \frac {{s(ab \plus{} bc \plus{} ca)}}{{abc}} \minus{} 3$

$ \equal{} \frac {{s^2 \plus{} 4Rr \plus{} r^2 }}{{4Rr}} \minus{} 3$

so we have
\[ 2 \minus{} \frac {r}{R} \le \frac {1}{2}(\frac {{h_a }}{{r_a }} \plus{} \frac {{h_b }}{{r_b }} \plus{} \frac {{h_c }}{{r_c }}) \le \frac {R}{r} \plus{} \frac {r}{R} \minus{} 1\]
\[ \Leftrightarrow 2 \minus{} \frac {r}{R} \le \frac {{s^2 \plus{} 4Rr \plus{} r^2 }}{{4Rr}} \minus{} 3 \le \frac {R}{r} \plus{} \frac {r}{R} \minus{} 1\]
\[ \Leftrightarrow 16Rr \minus{} 5r^2 \le s^2 \le 4R^2 \plus{} 4Rr \plus{} 3r^2 ,\]
which is Gerretsen Inequality...so ineq holds.


Prove that

$ \frac{a_1^2}{1\minus{}a_1} \plus{} \frac{a_2^2}{a_1\minus{}a_2} \plus{} \frac{a_3^2}{a_2\minus{}a_3} \plus{} \ldots \plus{} \frac{a_n^2}{a_{n\minus{}1}\minus{}a_n} > \frac12(a_1\plus{}2a_2\plus{}3a_3\plus{}\ldots\plus{}na_n)\minus{}1$

whenever $ 1>a_1>a_2>\ldots>a_n>0$.

[quote="MTM"]Prove that

$ \frac {a_1^2}{1 \minus{} a_1} \plus{} \frac {a_2^2}{a_1 \minus{} a_2} \plus{} \frac {a_3^2}{a_2 \minus{} a_3} \plus{} \ldots \plus{} \frac {a_n^2}{a_{n \minus{} 1} \minus{} a_n} > \frac12(a_1 \plus{} 2a_2 \plus{} 3a_3 \plus{} \ldots \plus{} na_n) \minus{} 1$

whenever $ 1 > a_1 > a_2 > \ldots > a_n > 0$.[/quote]

Are there any solutions?


modular arithmetic
combinatorics proposed
combinatorics

Let $ A\equal{}\{1,2,\ldots, 2008\}$. We will say that set $ X$ is an $ r$-set if $ \emptyset \neq X \subset A$, and $ \sum_{x\in X} x \equiv r \pmod 3$. Let $ X_r$, $ r\in\{0,1,2\}$ be the set of $ r$-sets.

Find which one of $ X_r$ has the most elements.


calculus
integration
logarithms
calculus computations

Find $ \int_{0}^{\infty}\ln^{n}x~e^{\minus{}kx}dx$, where $ n\in\mathbb{N},k > 0$.

$ \int_{0}^{\infty}t^{x\minus{}1}e^{\minus{}kt}dt\equal{}\Gamma(x)\, k^{\minus{}x}$

Differentiate $ n$ times with respect to $ x$ and set $ x\equal{}1$ .

$ \int_{0}^{\infty}\ln^{n}(t)\, e^{\minus{}kt}dt\equal{}D^{n}\Big(\Gamma(x)\, k^{\minus{}x}\Big)\Big|_{x\equal{}1}$

The last term is after Leibniz rule

$ \sum_{j\equal{}0}^{n}{n\choose j}\Gamma^{(j)}(1)\frac{(\minus{}\ln k)^{n\minus{}j}}{k}$ whereas $ \Gamma^{(j)}(1)\equal{}\left(e^{\ln\Gamma}\right)^{(j)}(1)$

after Faá di Bruno's formula can be written as

$ \sum_{\ell_{1}\plus{}2\ell_{2}\plus{}...\plus{}j\ell_{j}\equal{}j}\frac{j!}{\ell_{1}!\cdots\ell_{j}!}\, (\minus{}\gamma)^{\ell_{1}}\prod_{m\equal{}2}^{j}\left((\minus{}1)^{m}\frac{\zeta(m)}{m}\right)^{\ell_{m}}$


Hey smiley monster (LOL)/suma/SM:D did you find out if you made state MC or not???

I REALLY WANNA DIE TO SEE YOU THERE :D  :D

I hope your coach knows what place you got...or hoping that you'd get a successful partition if your chapter only sends top school... :maybe:

So there's this thing called the Private Message button. ;)

* returns to being outside of Illinois *

Lol well this seems more like a illinois issue just to see if there's consistency between chapters...cuz last year there seemed to be a debate over one of them... :maybe:

[quote="Aryth"]Hmm, gauss, Josh and I were talking about you today.  :jump:[/quote]
lol really? what was it about?

Something about you trying to conceal your name until you make Nats? And also how you're going to be at ARML tryouts.

Eh, I'm not sure yet if I'm gonna be at ARML tryouts even though I really wanna be...My mom (who is an idiot) isn't so sure...

Hmmm, lol, what's with the name concealing until Nats thingy?

[quote="suma_milli"]lol.

Oh yeah and I made state for sure,  found out this morning :D.  Read my blog for more details![/quote]
GOOD JOB!!! YOU DID IT!!!
I hope we'd have a funny entertaining supercoolawesome time together :D  :D


geometry
LaTeX
inradius
ratio
perimeter
similar triangles
geometric sequence

i don't really know if this problem should be in this forum, but here it is
right triangle ABC with right angle at B has an incircle of radius 25
points X and Y are drawn on AB and AC respectively such that
XY is perpendicular to AB and is tangent to the incircle of ABC
if the incircle of AXY has radius 20, what is the area of triangle ABC?

[hide="hint"]Similar triangles and $rs=A$ and tangents from a point are equal[/hide]

I have an answer
I'll try to get it in Latex
but I'm not that great with it
answers would be appreciated so I can check mine

[hide="my solution, not in latex so it looks pretty ugly"]
the incircle of ABC has radius 25, so it has diameter 50
I believe this makes BX = 50
the incircle of AXY has radius 20, so diameter 40
put in the tangent perpendicular to AB through that circle
call it MN
triangle AMN should have inradius 16, diameter 32
so an infinite series of incircles with diamteric ratio 4/5 gives us AB = 50 + 40 + 32 + 128/5 + 512/25 + ........ = 250
this is where it gets ugly
A = rs
so 2A = rp (p is perimeter)
so 2A/p = r
twice the area can be expressed as AB*BC
call BC x
2A = 250x
since this is a right triangle
the perimeter can be expressed as (x+250+(sqrt(62500 + x^2)))
(x+250+(sqrt(62500 + x^2)))  = p
(250x)/p = 25
so (250x)/25p = 1
so (10x)/p = 1
so 10x = p
10x =  x+250+(sqrt(62500 + x^2))
9x - 250 = (sqrt(62500 + x^2)
81x^2 - 2250x + 62500 = 62500 + x^2
80x^2 - 2250x = 0
80x^2 = 2250x
x = 225/8
(1/2)*250*(225/8) = [b]3515.625[/b]
:lol:
[/hide]

my bad
i forgot to cancel out the twos so i solved for half of x
double my answer to get my new answer
which i believe is correct
[hide="the real answer"]7031.25[/hide]

[hide]

Use similar triangles, use the easily proved formula $r = \frac{a+b-c}{2}$. I got side lengths of 150, 62.5 and 162.5. Giving an area of 4687.5

[/hide]

one problem with your solution
you have the longest leg of the triangle being 150
but the first 5 terms of the geometric sequence
s = 50+40+32+128/5+512/25
exceeds 150
by the way i drew out my triangle of leg lengths 250 and 56.25
and the inscribed circle had a radius of 25
i truncated the triangle tangent to its incircle
new triangle had inradius 20

a little bit easier method to get AB:

Let the ratios of semi-perimeters S[AXY]/S[ABC] = r.
Then ratio of areas K[AXY]/K[ABC] = r^2 (1)

Then 1/S[AXY]/ 1/S[ABC] = 1/r.  (2)

Multiply 1 and 2, and you get ratio of inradii to be r. So r=4/5.
Then it is trivial to show AB=250, given AX=50.

From there, 'sapphyre's' solution is straightforward


[hide]
I did $\sqrt{1}*\sqrt{2}*\sqrt{3} \cdots *\sqrt{15}$

that turned out to be 1143535.906

$\sqrt{20*21*22*23*24*25}=11292.12115$

$\frac{1143535.906}{11292.12115}=\boxed{101.2684766}$

While i was typing this i realized that the 10. 11, and 12 on the top could cancel out with the 20, 22, and 24 on the bottom.
So you would have

$\boxed{\boxed{\frac{1}{2100}}}$

I don't know which one is right or if they are both wrong.
[/hide]

A simpler and more straightforward way of doing this would be to prime factorize all the values under the square root sign.

Then you can cancel out the values that are the same in both the numerator and denominator to make things simpler.

[b]Then,[/b] you can evaluate.

Which would be $\framebox{36\sqrt{\frac{182}{23}}}$ in simplest radical form.

The decimal approximation of $36\sqrt{\frac{182}{23}}$ is
$101.26847657502351781155318805262338546949849801877586209204...$ (I obviously used the computer for the computing).

So $101.268476575023517811553188052623385469...\approx{\framebox{101.3}}$


function
calculus
integration
Gauss
geometry
ratio
3D geometry

A point charge of charge +Q is placed a distance H above an infinite conducting plane. If you follow a field line that starts at the point charge and is parallel to the conducting plane at that point, where does it intersect the plane?

I don't understand, how do I identify a single line. There are infinitely many lines all ending at different spots.

You are asked to follow a field line that starts out parallel to the conducting plane.

It's easy to find the equipotentials in rectangular without simplifying them, but I cannot get them in a form that would allow an easy enough transformation into the field lines :(

So it's clear that the equipotentials are:

$\displaystyle \Phi(x,y) = \frac{1}{\sqrt{(y-H)^2+x^2}}-\frac{1}{\sqrt{(y+H)^2+x^2}}$

Thus it remains to calculate $\displaystyle F(x,y) = \frac{\Phi_y}{\Phi_x}$. I can't find a simplified form so that I can finish, any hint? Is this the wrong way to approach the problem?

OK grr... $\Phi(x,y)$ is obviously conservative, therefore there must exist some function $\Psi$ such that $\frac{d}{dx}\left(\Psi(x,y)\right) = \Phi_y + \Phi_x\frac{dy}{dx} = 0$. This is just a matter of getting this math right. I will keep trying.

There is a very neat solution using Gauss's law.  :D
The best way to follow field lines is to consider enclosed flux in one way or another. For example, this is how you show that when you accelerate an electric charge, there is a tangential component to its electric field.

Yes, you really just want the appropriate surface integral for the Gauss style attack.

A key point to note is that the horizontal distance from the 'normal' (perpendicular) line dropped from the charge to the plane will be measured in units of the vertical distance from the plane.  It is a geometry type solution (a ratio) that has no units.  You can also express the answer in the form of an angle between a straight line drawn from the charge to the intersection and the 'normal'.

I do not see how Guass' Law will help, but I am eager to see your solution.

I am now fairly convinced that the general solution to these field lines is non elementary, and when you think about the shape of the lines, it actually makes sense as they are very funky :) especially at large y.

Could you post your solution hidden so that people who don't want to see it don't have to look?

sheesh.. thats a nice problem..

no idea here

[quote="skylarth"]sheesh.. thats a nice problem..
[/quote]Yes.  Hardly a high-school calculus problem.
You will need to do some contour integrals as well, if you want to reduce the complexity of the problem.

The solution. As you can see, it does only use high school calculus. :)

The key observation is this:
Consider the surface formed by the flux lines that start out parallel to the conducting plane. [b]Half of the flux from charge Q lies above this surface, and half of the flux lies below this surface.[/b] Intuitively, this is because in the immediate vicinity of a point charge, the field is spherically symmetric. For a more rigorous argument, consider a small sphere around the point charge, and divide it into an upper hemisphere and a lower hemisphere...you can fill in the details.

Now the problem is reduced to this one: find R such that a circle of radius R, centered below Q, contains half the flux from Q. All that is left is a simple integration.

Using an image charge (see diagram 1), we see that the field at a radius r (see diagram 2) is
$\frac{2kQ}{(r^2+h^2)}\cos\theta=\frac{1}{\pi\epsilon_0}\frac{Q}{r^2+h^2}\frac{h}{(r^2+h^2)^\frac{1}{2}}=\frac{1}{\pi\epsilon_0}\frac{Qh}{(r^2+h^2)^\frac{3}{2}}$
Integrate this from 0 to R, we get
$\int^R_0 2\pi r\frac{1}{\pi\epsilon_0}\frac{Qh}{(r^2+h^2)^\frac{3}{2}} dr=\frac{2hQ}{\epsilon_0}\int^R_0\frac{r}{(r^2+h^2)^{\frac{3}{2}}}dr=(\frac{2hQ}{\epsilon_0})\frac{1}{h}-\frac{1}{(R^2+h^2)^\frac{1}{2}}$
Since this integral must equal half the flux from Q,
$\frac{Q}{2\epsilon_0}=(\frac{2hQ}{\epsilon_0})(\frac{1}{h}-\frac{1}{(R^2+h^2)^\frac{1}{2}})$
$\frac{3}{4}=\frac{1}{\sqrt{\frac{R^2}{h^2}+1}}$
$\frac{R}{h}=\frac{1}{\sqrt{3}}$

[quote="Rogue Physicist"]You will need to do some contour integrals as well, if you want to reduce the complexity of the problem.[/quote]
So post your solution.

[quote="schulmannerism"]Using an image charge (see diagram 1), we see that the field at a radius r (see diagram 2) is
$\frac{2kQ}{(r^2+h^2)}\cos\theta=\frac{1}{\pi\epsilon_0}\frac{Q}{r^2+h^2}\frac{h}{(r^2+h^2)^\frac{1}{2}}=\frac{1}{\pi\epsilon_0}\frac{Qh}{(r^2+h^2)^\frac{3}{2}}$[/quote]

You lost a 2 in the denominator of $k$. It should be $\displaystyle \frac{1}{2\pi\epsilon_0}\frac{Qh}{(r^2+h^2)^\frac{3}{2}}$

Disregarding that 2, the rest of it is fine until you get here:
[quote="schulmannerism"]$\frac{3}{4}=\frac{1}{\sqrt{\frac{R^2}{h^2}+1}}$
$\frac{R}{h}=\frac{1}{\sqrt{3}}$[/quote]

Given $\frac{3}{4}=\frac{1}{\sqrt{\frac{R^2}{h^2}+1}}$, clearly $R={\frac{\sqrt{7}}{3}h}$.

Now if you had added in the 2 back in the beginning, you would have ended up with $R={\sqrt{3}h}$.

[quote="Spoon"]You lost a 2 in the denominator of $k$.
[/quote]
No, there is a factor of two in the numerator (because there is the original charge and its image) and this cancels out the 2 in the denominator of k.

You're right about the second mistake though, so the answer is $\sqrt{7}/3$

[quote="Spoon"]So post your solution.[/quote]You've done some excellent and serious work here.
Why should I spoil your victory?  You deserve the win.
Cheers!

But you two should sort out the trivial corrections, so people can follow the trail.
Well done.

I did nothing of the sort. That is not my solution, it's his. All I did was point out that he left off a 1/2 in the first step.

You claim to have a different solution, a more complicated one... If you truly do, I'd like to see it. I would be much more interested in your solution than this one, even if this is simple and to the point.

[quote="schulmannerism"]Yeah you're right Spoon, I should get my fundamental constants sorted out.  :P[/quote] No sweat.

Let $Q$ denote the location of the charge, $\bar{Q}$ its image charge, and P an arbitrary point.

Using the idea from my original solution, it is easy to find a numerical relationship between R (distance from line $Q\bar{Q}$, Y (height above a point above plane) and H ($\frac{Q\bar{Q}}{2}$ ), which is satisfied by any point $P$ on the field line that starts out parallel to the plane. The relation turns out to be this:
$\cos\angle\bar{Q}QP+\cos\angle Q\bar{Q}P=1$

This generalizes:
$\cos\angle\bar{Q}QP+\cos\angle Q\bar{Q}P$
is constant for every flux line!

This shows that the field lines actually do satisfy a simple relationship, just not in rectanglular coordinates. Since $\angle\bar{Q}QP$ and $\angle Q\bar{Q}P$ uniquely specify a point, there are probably some mathematical solutions using different coordinate systems.

Elegant!  Coordinate-free is the way to go.[quote]You claim to have a different solution, a more complicated one... If you truly do, I'd like to see it. [/quote]You are mistaken.  I merely suggested a less intelligent approach, after glancing at the problem.  I would have muddled through and taken a lot longer.

When you've seen the best, why mess with the rest?


That's what we use . . .   Pretty straightforward install.  Haven't had any problems with it.

After you install the whole deal, the application you'll be wanting to use is pdflatex.

Pick whichever one is closest to you. Actually it probably doesn't matter at all.

[quote="Rep123max"]What do i do when i gives those 4 choices for like VA, NC, MN and Ireland?[/quote]

I assuming you're just refering to the download location. Just pick the one that is closest to where you live. It doesn't really matter.

It will explain to you exactly how it should be configured. Just follow the instructions, and you should be fine.

[quote="bugzpodder"]i tried miketex, but after installation i had no idea what to do with it!!!  anyone care to tell me what is suppose to happen??

now i am using mathType + acrobat + ms word to make pdf files...[/quote]

you make your .tex file, and then you go to a command prompt, and change folders to wherever your .tex file is, and then type [b]pdflatex[/b] filename.tex

and that will (if there are no errors in the .tex file) create your pdf.


We have 120 unit squares, arbitarily arranged in a 20 x 25 rectangle (both position and orientation are arbitary). Prove that it is always possible to place a circle of unit diameter inside the rectangle without intersecting any of the squares.

[hide] We can partition the side of length 20 by 13 lines with the distance between them >V2 (so that no square can cut more than one line). Then these lines are cut by the squares a total of 120 times.

By pigeonhole, we know that one of these lines is cut by at most 9 lines. If there is no space on any of these lines for the diameter of a circle with diameter 1, then 10 + 9V2 >= 25.  However, since 10+9V2 < 25, there must be an uncut segment on that line of length >1.

We can place the diameter of that circle on that segment. Then the circle will not be cut by any of the squares on that line, or by those on the neighboring lines...is this sufficient?

Where is this problem from?

[/hide]

This is an old sovjet problem :

[hide]http://www.kalva.demon.co.uk/soviet/ssoln/ssol6112.html[/hide]


geometry
inequalities
inequalities unsolved

[quote="tranthanhnam"]${a^2 +b^2 +c^2 }$ + ${4\sqrt{3}S} $ ${\geq}$ $\frac{6(a^2.b^2+b^2+c^2 +c^2a^2)}{a^2+b^2 +c^2}$[/quote]
where a,.b,c are lenght three triangle  and S is its area

In fact, this holds with reversed sign:

[color=blue][b]Theorem.[/b] If ABC is a triangle with area $\Delta$, then

$a^2+b^2+c^2+4\sqrt3\Delta\leq\frac{6\left(b^2c^2+c^2a^2+a^2b^2\right)}{a^2+b^2+c^2}$.[/color]

[i]Proof.[/i] Denote $U=a^2+b^2+c^2$ and $V=b^2c^2+c^2a^2+a^2b^2$. Then, by the Heron formula,

$\Delta=\frac14\sqrt{\left(a+b+c\right)\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}$
$=\frac14\sqrt{4\left(b^2c^2+c^2a^2+a^2b^2\right)-\left(a^2+b^2+c^2\right)^2}=\frac14\sqrt{4V-U^2}$,

so the inequality in question becomes

$U+4\sqrt3\cdot\frac14\sqrt{4V-U^2}\leq\frac{6V}{U}$.

This simplifies to $U+\sqrt{3\left(4V-U^2\right)}\leq\frac{6V}{U}$. After subtracting U, this becomes $\sqrt{3\left(4V-U^2\right)}\leq\frac{6V}{U}-U$. Now, the left hand side of this inequality is trivially $\geq 0$ (as a square root), and the right hand side of this inequality is also $\geq 0$ (since $\Delta=\frac14\sqrt{4V-U^2}$ yields $4V-U^2\geq 0$, so that $4V\geq U^2$, and thus $6V>4V\geq U^2$, so that $\frac{6V}{U}>U$ and thus $\frac{6V}{U}-U>0$). Hence, we can square this inequality, and obtain $3\left(4V-U^2\right)\leq\left(\frac{6V}{U}-U\right)^2$. But this is trivially true, since $\left(\frac{6V}{U}-U\right)^2-3\left(4V-U^2\right)=4\left(\frac{3V}{U}-U\right)^2\geq 0$. Thus, the theorem is proven.

  Darij


algebra
polynomial
integration
LaTeX
IMC
college contests

How many nonzero coefficients can a polynomial $ P(x)$ have if its coefficients are integers and $ |P(z)| \le 2$ for any complex number $ z$ of unit length?

Let's give a try. Take $ P$ such a polynomial and observe that Parseval's formula already shows that the sum of the squares of the polynomial is at most $ 4$, thus there are at most $ 4$ nonzero coefficients. If there are $ 4$, the sum of the squares must be $ 4$ and thus by Parseval $ \int_{0}^{2\pi}{|P(e^{it})|^{2}dt}=4\cdot 2\pi$ and by hypothesis this implies $ |P(z)|=2$ for all $ |z|=1$. From here we deduce that $ P(z)P(1/z)=4$ for $ |z|=1$ and thus for all $ z$. Thus the only zero of $ P$ is $ 0$ and so $ P(X)=aX^{n}$ for some integer $ a$ and some $ n$. But then we have contradiction.
  Thus, there are at most $ 3$ nonzero coefficients. Let's suppose we have $ 3$, then they are all $ 1,-1$ and since we can throw out any factor of the form $ z^{s}$ of $ P$, we may assume that $ P(z)=1+az^{k}+bz^{l}$ with $ k<l$ and $ a,b$ equal to $ 1$ or $ -1$. Let $ z_{j}$ be the roots of the equation $ bz^{l}=-1$. Then $ |2+az_{j}^{k}|\leq 2$ and because the sum of $ 2+az_{j}^{k}$ equals $ 2l$, we must have everywhere equality and also $ 2+az_{j}^{k}=2$, which is impossible. Therefore our dear polynomial has at most $ 2$ nonzero coefficients, which is obviously possible.

Hi,
I'm sorry but I don't know how to type math expressions here :blush:
Just produce some pdf file.
Here is my solution.(maybe wrong).
Apreciated any observation.Thank you all!

I don`t have practice in latex but I got (some copy & paste  :oops: ) anywhere
to avoid the download and to follow better the proof looking at the previous message.

$ \begin{gathered}{\text{Suppose that }}\exists\left|{z_{0}}\right| \equal{} 1{\text{ with}}\left|{P(z_{0})}\right| < 2.\hfill\\ {\text{Than for a suitable neighborhood [a}}{\text{,b]}}\subset{\text{[0}}{\text{,2}}\pi{\text{] (content strictly) of }}z_{0}{\text{ we have: }}\hfill\\ \frac{1}{{2\pi }}\int_{0}^{2\pi }{\left|{P(e^{i\theta })}\right|d\theta }\equal{}\frac{1}{{2\pi }}(\int_{0}^{a}{\left|{P(e^{i\theta })}\right|d\theta }\plus{}\int_{a}^{b}{\left|{P(e^{i\theta })}\right|d\theta }\plus{}\int_{b}^{2\pi }{\left|{P(e^{i\theta })}\right|d\theta }) <\hfill\\ <\frac{1}{{2\pi }}(2(a\minus{}0)\plus{}2(b\minus{}a)\plus{}2(2\pi\minus{}b)) \equal{} 2 < 3 \equal{}\frac{1}{{2\pi }}\int_{0}^{2\pi }{\left|{P(e^{i\theta })}\right|^{2}d\theta }.\hfill\\ {\text{So }}\frac{1}{{2\pi }}\int_{0}^{2\pi }{\left|{P(e^{i\theta })}\right|d\theta }<\frac{1}{{2\pi }}\int_{0}^{2\pi }{\left|{P(e^{i\theta })}\right|^{2}d\theta }{\text{ and then }}\left|{P(z)}\right| > 1{\text{ }}{\text{,}}\forall\left| z\right| \equal{} 1.\hfill\\ {\text{Now}}{\text{, evaluating for z \equal{} 1}}{\text{, we must have }1 < }\left|{P(1)}\right| \equal{}\left|{a\plus{}b\plus{}1}\right|\leqslant 2\hfill\\ {\text{leaving }}\left|{a\plus{}b\plus{}1}\right| \equal{} 2.Contradiction!\hfill\\ {\text{So }}\left|{P(z)}\right| \equal{} 2{\text{ }}{\text{,}}\forall\left| z\right| \equal{} 1{\text{ and again P(z) \equal{} cz}}^{\text{n}}{\text{ for suitable c and n}}{\text{.}}\hfill\\ \end{gathered}$

I'm very very sorry,but this solution is WRONG!!!
Apologize me!
 :blush:  :blush:


The MathZoom Summer Camp 2008 will be held at the University of California, Los Angeles, from July 13 through August 2, 2008.

Please visit the [url=http://www.artofproblemsolving.com/Forum/index.php?f=453]MathZoom[/url] forum or [url]http://www.mathzoom.org[/url] for details.

I'd just like to say, I've never seen the local forums so new post-filled.


About how much time should we give ourselves to take both the pre and post tests?

This varies greatly from class to class.  I wouldn't so much worry about time spent as about whether or not you can solve the problems.

Well, if you study for the test right before you take it, obviously you'll do well.

But I think what you want to test is whether you [i]know[/i] the information well enough to avoid taking another test. If you study for it and don't retain it, then you just inflated your "score" on the post test, and you might actually need to take it.

My advice: just take the test, if you know the stuff you should be able to do it or at least know basicly what to do on most of the problems. But if you find yourself lost or getting the wrong answers on a good amount of problems, then you might need to take the class.

[quote="Calculus"]Also, should I prepare for the tests or just take it? I believe if I spend a few days preparing I could do a lot better on it.[/quote]

If you can study for a few days, then ace the post test, I would say you didn't need the course since your study can supplant it.  It's your call if you'd like to approach it this way.  Personally, I'd just take the test, then make a personal call on whether I need the class if I got 60-70%.  If I got less than 60%, definitely take the class.  80%+, definitely don't.  In between, your call.


$ \int \left[ \left(\frac{x}{e} \right)^{x}\plus{} \left(\frac{e}{x} \right)^{x} \right] log_ex dx$

delete the $ \ln x$ after your last integral sign.. :wink:  nice solution btw

[quote="milin"]${ \left(\frac {x}{e}\right)^x + \left(\frac {e}{x}\right)^x = e^{x\left(\ln x - 1\right)}} + e^{ - x\left(\ln x - 1\right)} = 2 \cosh\left(x\left(\ln x - 1\right)\right)$

Now we have:

$ \int \left(\left(\frac {x}{e}\right)^x + \left(\frac {e}{x}\right)^x \right) \ln x\,dx = 2\int \cosh\left(x\left(\ln x - 1\right)\right) \ln x \,dx$

Use $ u = x\left(\ln x - 1\right)$ and you get:

$ 2\int \cosh\left(x\left(\ln x - 1\right)\right) \ln x \,dx = 2\int \cosh\left(u\right)\,du = 2 \sinh\left(u\right) =$

$ = 2 \sinh\left(x\left(\ln x - 1\right)\right) = \left(\frac {x}{e}\right)^x - \left(\frac {e}{x}\right)^x$

[EDIT:] thanks yeahnigga[/quote]

Thanks for the solution!


Find the minimum value of $ E\equal{}\frac{1}{a\plus{}1}\cdot \frac{b^2\minus{}1}{\minus{}b}\plus{}\frac{1}{b\plus{}1}\cdot \frac{a^2\minus{}1}{\minus{}a}$ , where $ a,b\in (0,1)\ ,\ a\plus{}b\equal{}1$ .


vector
probability
combinatorics unsolved
combinatorics

[b]GAME: [/b]There are $n$ players. Each write a positive integer on piece of paper. Then all papers are taken and winner is who written smallest number, so that noone else has written it (for example if $n=7$ and numbers are 1,1,1,2,3,3,4 the winner is the one who wrote number 2).

I was trying to find a strategy, but I found that for such game, where are $n$ players and they can do whatever they want, there's hard to find something. BTW: Can we, in games like this one, find strategy? Or a way how to play optimal? I'm curious about it....

in games like the one you have described(which by the way is not a deterministic game at all because if you let the state of a game be the vector of the numbers they have all chosen,then this set of possible states of teh game is infinite) there cannot be strategies-simply because all the players are at the same footing to start with.so if there is a strategy for one,it is also  a strategy for everyone else.
even in a probabilistic sense i don't think that you can better you chances to more than $\frac{1}{n}$ since no one has any more information than the others.

Thanx. But I know one game that is very simalar and there's 'strategy' like: with propability blabla do blablabla, with propability blabla do blablabla, ... and at the end you know that with propability blablabla you'll win with other players playing anyhow. Is something like this for all games? For this game?

well i can give you an example v concretely.
suppose there are 2 players.the first writes down two positive numbers both between $0$ and $1$. he shows one of these to the second player who has to guess if this is the smaller number.if his guess is right,he wins else the first player wins.(if both numbers are equal then the second player wins regardless of what he says so the first player will obviously not choose both numbers equal!).
in this game if the second player randomly decides whether the number seen is the smaller one or larger then his chances are 50%. but he can do better than that.
in the game you propose, i however don't see a better way to improve too since there is no information available at any stage and moreover each player is in exactly the same state.so whatever is applicable to one of them is also applicable to the other.


$AB$ is the diametre of a circle. $P$ is any point on $AB$. $QPR$ is a chord such that $\angle QPA= 45^{\circ}$. Prove that $AB^{2}=2PQ^{2}+2PR^{2}$.

[hide]Let $O$ be the midpoint of $AB$, $M$ the midpoint of $QR$,$d=OM$, $R=AO$. WLOG assume $P$ is on $OB$. As $OM=PM$ ($\angle QPR =45^{o}$)
$QP=QM+MP=\sqrt{R^{2}-d^{2}}+d$ and $PR=RM-PM=\sqrt{R^{2}-d^{2}}-d$
$QP^{2}+PR^{2}=2R^{2}=\frac{AB^{2}}{2}$.[/hide]


Prove that in any gathering of $ 6$ people, there are either $ 3$ who are mutual acquaintances or three who are strangers to each other.

[hide="Hint"]Draw out 6 points representing people. A red line segment connecting would mean acquaintance, and blue is stranger.[/hide]

Hmm are acquaintances necessarily mutual?

[hide=" Solution"]

We can call the $ 6$ people $ 1, 2, 3, 4, 5$ and $ 6$ and write $ 12A$ if $ 1$ and $ 2$ are acquainted and $ 12S$ if $ 1$ and $ 2$ are strangers.

Then of the pairs $ 12, 13, 14, 15, 16$ at least three are $ A$ or at least three are $ S$. Suppose that $ 12, 13$ and $ 14$ are $ A$ and consider $ 23, 24$ and $ 34$: if none of these is $ A$ then $ 234$ are $ S$; if one of these (say $ 23$) is $ A$ then $ 123$ are $ A$.

We can repeat tihs argument with $ A$ and $ S$ interchanged if we begin with three $ S$ pairs.

[/hide]

[hide="A graph-theory solution."]Consider a 6-node graph with one node separated from the rest. Let these nodes represent the people. We draw a red line if two people are acquaintances and a blue line if they are strangers. This means that a blue triangle represents 3 mutual acquaintances and a red triangles represetns three complete strangers.

Consider the 5 lines from the separated node to each of the others. By the pigeonhole principle there must be at least three lines of the same colour. Assume WLOG that there are three blue lines. Now consider the lines joining the three nodes who these blue lines were drawn to. If any of the lines between these three were blue then there would be a blue triangle. If there are no blue lines between these three then they are all red and these lines form a red triangle. Thus there is either a red triangle or a blue triangle.

Hope this is clear enough.[/hide]

For the record, this is the first non-trivial Ramsey number $ R(3;3)$ (check on Google keyword "Ramsey Theory").


[quote="GameBot"]Mohan is selling cookies at the economics fair. As he decides how to package the cookies, he finds that when he bags them in groups of 4, he has 3 left over. When he bags them in groups of 5, he has 2 left over. When he bags them in groups of 7, he has 4 left over. What is the least number of cookies that Mohan could have?[/quote]

$ n \equiv 3 \pmod 4$
$ n \equiv 2 \pmod 5$
$ n \equiv 4 \pmod 7$
$ n \equiv \minus{}13 \pmod 4$
$ n \equiv \minus{}13 \pmod 5$
$ n \equiv \minus{}13 \equiv 7 \pmod 20$
$ n \equiv \minus{}73 \pmod 7$
$ n \equiv \minus{}73 \pmod 20$
$ n \equiv \minus{}73 \equiv \boxed{67} \pmod 140$.

I don't get it. Where did -13 come into the picture?
Can you please create a full detailed explanation please?
I'm not very good with mods. Can you also explain how to go about similar problems like this?

We have $n \equiv 3 \pmod 4$, $n \equiv 2 \pmod 5$, and $n \equiv 4 \pmod 7$. Thus, $n \equiv -13 \pmod 4$, $n \equiv -13 \pmod 5$, so $n \equiv -13 \equiv -73 \pmod {20}$. Since $n \equiv 4 \pmod 7$, $n \equiv -73 \pmod 7$, and thus $n \equiv -73 \equiv \boxed{67} \pmod {140}$.


vector
topology
geometry
calculus
calculus open

Is it true that to disconnect an n-dimensional space at an n-1 dimensional space is required.Prove or disprove it.What bout R-infinity or nay arbitrary topological space.Waiting 4 ur reply.

The question is actually very interesting, but it depends on your definition of 'space' (the word itself is very vague in math).

If you are talking about vector space over $\mathbb{R}$, the answer is YES, and the proof is very simple. Suppose that it is not true, then, for some value of $n$ there is a closed set $C$ strictly contained in a vector subspace $V\subset\mathbb{R}^n$, where $\mbox{dim}V=n-1$ such that $C$ disconects $\mathbb{R}^n$, then, the strategy is very simple, I am just going to tell it informally. Take two of the disconnected pieces of $\mathbb{R}^n$. If both contain points in the complement of $V$, say $x$ in one of them and $y$ in the other, then, those points have to be in different sides of $V$. Consider $w\perp V$, and consider $v\in V$ such that $v\notin C$, then $v-w$ and $v+w$ are in different sides of $V$, and we can join $x$ to one of them by a strait segment not touching $V$, and $y$ to the other, contradiction, because is more than conected, is actually arcconected. If one of the connected parts contains the complement of $V$  and all the others are strictly contained in $V$, well, at least in the usual topology they cannot be open (the one contained in $V$), condition necessary to disconnect, but on the other hand, you can thing about it in this way: Just pick one of the points in one of the disconnected parts inside V, callit $u$, then you can connect $v$ to $v+w$ by a straight segment without touching any other point in $V$ (and therefore in $C$).

On the other hand, with a different definition of space, even a metric space, it is not true. For example, $S^n/S^{n-1}\equiv S^n \lor S^n$ constructed as a CW complex can be defined as a $n$-dimensional space that can be disconnected by taking just one point (a zero dimensional space). Actually, you can construct many spaces using algebraic geometry, define a metric space over it, define a dimension appropiately, and with the topology indiced by the metric you defined, be able to disconnect the space with things with dimension smaller than $n-1$.

Which of the two parts answer your question?

[quote="the game"]Yup you are right in the 1st part I was talking about R-n.
but I dont think it answered my question about disconnecting any arbitrary topological space..Thanks for the solution.[/quote]

Read the second part. In general is hard to define the dimension in a general topological space (what would be the dimension of a $\mathbb{R}^n$ under the undiscrete or the discrete topologies?). Any way, for CW-complexes there is a well studied and defined theory of dimension, and I am giving you a topological space of dimension $n-1$ (sorry, I didn't notice that) that is separable by only taking a 0-dimensional set (a single point).

Best,

Yup you are right, i guess the definition of space is vague as you have put it.Now you have confused me.
I ll think about it..


Let $ a,b,c>0$ prove that
\[ \frac{1}{5(a^2\plus{}b^2)\minus{}ab}\plus{}\frac{1}{5(a^2\plus{}b^2)\minus{}ab}\plus{}\frac{1}{5(a^2\plus{}b^2)\minus{}ab}\ge \frac{1}{a^2\plus{}b^2\plus{}c^2}\]
Good problem for $ Cauchy$ inequality.

[quote="M.A"]Let $ a,b,c > 0$ prove that
\[ \frac {1}{5(a^2 + b^2) - ab} + \frac {1}{5(a^2 + b^2) - ab} + \frac {1}{5(a^2 + b^2) - ab}\ge \frac {1}{a^2 + b^2 + c^2}
\]
Good problem for $ Cauchy$ inequality.[/quote]
It's a very nice Vasc's inequality!
$ \sum_{cyc}\frac{1}{5a^2+5b^2-ab}=\sum_{cyc}\frac{(a+b+2c)^2}{(a+b+2c)^2(5a^2+5b^2-ab)}\geq$
$ \geq\frac{16(a+b+c)^2}{\sum(a+b+2c)^2(5a^2+5b^2-ab)}.$
Thus, it remains to prove that:
${ 16(a+b+c)^2(a^2+b^2+c^2)\geq\sum_{cyc}(a+b+2c)^2(5a^2+5b^2-ab)},$ which is equivalent to
$ \sum_{sym}(3a^4+3a^3b-8a^2b^2+2a^2bc)\geq0,$ which obvious.

Thank you, M.A!
The following Vasc's inequality is true too:
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ \frac {1}{4a^2 \minus{} ab \plus{} 4b^2} \plus{} \frac {1}{4a^2 \minus{} ac \plus{} 4c^2} \plus{} \frac {1}{4b^2 \minus{} bc \plus{} 4c^2}\geq\frac {9}{7(a^2 \plus{} b^2 \plus{} c^2)}
\]

Are "Vasc's inequalities" a type of problem or a collection of single problems proposed by "Vasc"?


Let $ a,b,c>0$ . Prove that :

\[ \frac{1}{a^2\plus{}2}\plus{}\frac{1}{b^2\plus{}2}\plus{}\frac{1}{c^2\plus{}2}\le \frac{\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}}{abc\sqrt{2}}\]

[quote="Pain rinnegan"]Let $ a,b,c > 0$ . Prove that :
\[ \frac {1}{a^2 \plus{} 2} \plus{} \frac {1}{b^2 \plus{} 2} \plus{} \frac {1}{c^2 \plus{} 2}\le \frac {\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}}{abc\sqrt {2}}\]
[/quote]
It's wrong. Try $ a\equal{}b\equal{}c\equal{}2.$ :wink:

I think it's the best choice of a,b,c to show that this is wrong.Great counterexample since it cancels out $ \sqrt{2}$.
Most of the times though I assume that the person who's posting is putting a correct ineq so i was actually trying to prove it while i saw your message..lol.Anyways,I guess i need more practise solving ineq so I will be able to figure out faster if some ineq is correct or wrong.

Peace

~Kostas

I'm sorry , it must be :

\[ \frac {1}{a^2 \plus{} 2} \plus{} \frac {1}{b^2 \plus{} 2} \plus{} \frac {1}{c^2 \plus{} 2}\le \frac {\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}}{\sqrt{2abc}}\]

Okay,I am really sorry but i think that your new inequality is still wrong :(
Anyways,let me show you what i did and tell me your opinion:

We have: $ a^2 + 2 = (a^2 + 1) + 1 \geq 2a + 1 \geq 2\sqrt {2a}$.
Hence the left hand side of your desired equation is less or equal to:

$ \frac {1}{2 \sqrt {2a} + \frac {1}{2 \sqrt {2b} + \frac {1}{2 \sqrt {2c}$ = $ \frac {\sqrt {ab} + \sqrt {bc} + \sqrt {ca}}{2\sqrt {2abc}}$.

But the inequality(I think): $ \frac {\sqrt {ab} + \sqrt {bc} + \sqrt {ca}}{2} \leq \sqrt {a} + \sqrt {b} + \sqrt {c}$ is not necessarily correct for $ a,b,c > 0$.

~Kostas

[quote="new_member"]Okay,I am really sorry but i think that your new inequality is still wrong :(
Anyways,let me show you what i did and tell me your opinion:

We have: $ a^2 \plus{} 2 \equal{} (a^2 \plus{} 1) \plus{} 1 \geq 2a \plus{} 1 \geq 2\sqrt {2a}$.
Hence the left hand side of your desired equation is less or equal to:

$ \frac {1}{2 \sqrt {2a}} \plus{} \frac {1}{2 \sqrt {2b}} \plus{} \frac {1}{2 \sqrt {2c}}$ = $ \frac {\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}}{2\sqrt {2abc}}$.

But the inequality(I think): $ \frac {\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}}{2} \leq \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}$ is not necessarily correct for $ a,b,c > 0$.

~Kostas[/quote]

I think you are wrong .  Think again . :wink:

[quote="new_member"]I don't think i can find anything else.Could you please post a solution?You could hide it if you want in case other people wanna try.Thanks[/quote]

I can't solve it . That's why i posted it in the unsolved section .

Lol,hehe i am sorry i just noticed it's actually in the unsolved section.I just saw your problem from the recent posts.
Anyways,i guess we''ll just have to wait till someone gives a solution.But yeah,I don't think i can help more than that.

[quote="Pain rinnegan"]I'm sorry , it must be :
\[ \frac {1}{a^2 \plus{} 2} \plus{} \frac {1}{b^2 \plus{} 2} \plus{} \frac {1}{c^2 \plus{} 2}\le \frac {\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}}{\sqrt {2abc}}\]
[/quote]
Let $ a \equal{} x^2\sqrt2,$ $ b \equal{} y^2\sqrt2$ and $ c \equal{} z^2\sqrt2,$ where $ x,$ $ y$ and $ z$ are positive numbers.
Hence, your inequality is equivalent to the following
$ \sum_{cyc}\frac {1}{x^4 \plus{} 1}\leq\frac {x \plus{} y \plus{} z}{xyz}.$
Now let $ y \equal{} 3$ and $ z \equal{} 1.$
We obtain $ \frac {1}{x^4 \plus{} 1} \plus{} \frac {1}{82} \plus{} \frac {1}{2}\leq\frac {x \plus{} 4}{3x},$ which is wrong for $ x\rightarrow \plus{} \infty.$ :wink:


Prove that the limit of the sequence $ s_n \equal{} \sqrt {4n^2 \plus{} n} \minus{} 2n$ is $ \frac {1}{4}$ (formal epsilon proof).

$ 2n \plus{} \frac {1}{4} \minus{} \frac {1}{64n}\leq \sqrt {4n^2 \plus{} n}\leq 2n \plus{} \frac {1}{4}$ for $ n\geq 1$ (readily proven just by squaring both sides), so
\[ \left \vert \sqrt {4n^2 \plus{} n} \minus{} 2n \minus{} \frac {1}{4}\right \vert \leq \frac {1}{64n}
\]
which, as a bonus, gives a good estimate on the rate of convergence.


[hide] Rewrite as $(x-1)^{7}= 7x^{6}+35x^{4}+21x^{2}+1$. Noting that the RHS is the even-powered terms of $(x+1)^{7}$, so we use the standard trick to write it as $\frac{(x+1)^{7}+(-x+1)^{7}}{2}$. Then The equation becomes

$(x-1)^{7}= \frac{(x+1)^{7}-(x-1)^{7}}{2}\Rightarrow 3(x-1)^{7}= (x+1)^{7}$.

Since we are looking for real solutions only, we simply take the $7$th root of both sides to get

$\sqrt[7]{3}x-\sqrt[7]{3}= x+1 \Rightarrow x = \frac{\sqrt[7]{3}+1}{\sqrt[7]{3}-1}$,

which we can verify to be the same as the given result through an expansion of a finite geometric series. [/hide]


[b] Problem. [/b] $ m, n, a_1, a_2, \ldots, a_n$ are positive integers such that the following inequalities hold:

$ a_1 \le 1^m$

$ a_1 \plus{} a_2 \le 1^m \plus{} 2^m$

$ a_1 \plus{} a_2 \plus{} a_3 \le 1^m \plus{} 2^m \plus{} 3^m$

$ \vdots \vdots \ddots$

$ a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n \le 1^m \plus{} 2^m \plus{} \ldots \plus{} n^m$

Prove that $ \sqrt[m]{a_1} \plus{} \sqrt[m]{a_2} \plus{} \ldots \plus{} \sqrt[m]{a_n} \le 1 \plus{} 2 \plus{} \ldots \plus{} n$.

Let $ s=1+\cdots+n$. $ f(x)=x^{1/m}$ is concave so by Jensen's
\[ \frac1sa_1^{\frac1m}+\frac2s\left(\frac{a_2}{2^m}\right)^{\frac1m}+\cdots+\frac ns\left(\frac{a_n}{n^m}\right)^{\frac1m}\le\left(\frac{a_1}{s}+\frac{a_2}{2^{m-1}s}+\cdots+\frac{a_n}{n^{m-1}s}\right)^{\frac1m},\]so
\begin{align*}
a_1^{\frac1m}+\cdots+a_n^{\frac1m}&\le s\left(\frac1s\right)^{\frac1m}\left(\frac1{n^{m-1}}(1^m+\cdots+n^m)+\left(\frac1{(n-1)^{m-1}}-\frac1{n^{m-1}}\right)(1^m+\cdots+(n-1)^m)+\cdots\right)^{\frac1m}\\
&=s\left(\frac1s\right)^{\frac1m}s^{\frac1m}\\
&=1+2+\cdots+n.
\end{align*}

Yeah, that's pretty much what I did too. Yay for Jensen? :\


You mean, the area of $|x|+|y| \le 9$ for $-5 \le x \le 4$?

That's it. Sorry, but I don't know much LaTeX

[quote="vamathletes"]What is the number of square units in the region of
absolute value of x + absolute value of y is less than or equal to 9

-5 is less than or equal to x which is less than or equal to 4[/quote]

[hide]The graph of $|x|+|y| \le 9$ is interior of a square with vertices at $(0,9),$ $(9,0),$ $(0,-9),$ and $(-9,0).$

Then this is cut in three by lines $x=-5$ and $x=4$.

We find the area of regions in between that has shape of a hexagon, which can be calulated by adding up two trapezoids (divided by $y$-axis):

(left): bases 8 and 18 [median=13], height 5
(right): bases 18 and 10 [median=14], height 4

Adding, the area is $13 \cdot 5+14 \cdot 4=\boxed{121}$ (sq. units).[/hide]


modular arithmetic
number theory unsolved
number theory

Solve in positive integers:

$ (x\plus{}y)(xy\plus{}1)\equal{}2^z$

I have reached to the equation $ 2^{n\minus{}2}(2^{2m \minus{} n \minus{} 2} \minus{} 1) \equal{} k(k\minus{}1)$ for some integer k, but I cannot continue from here...


Gauss
number theory unsolved
number theory

We now that $p$ is prime number greater than $n+1$. Prove that $n+1|\left(\begin{array}{cc}n+p\\\ n\end{array}\right)$

Note that $(n+1+p)C_{n+p}^p = (n+1)C_{n+1+p}^p$.
If $q$ is a prime which divides both $n+1$ and $n+1+p$ then $q$ divides $p$, so that $q=p$ since $p$ is a prime. It follows that $p$ divides $n+1$ which contradicts that $p>n+1.$ Thus, $n+1$ and $n+1+p$ are coprime and, from Gauss' theorem, we deduce that $n+1$ divides $C_{n+p}^p.$

Pierre.


inequalities
geometry
inradius
trigonometry
inequalities proposed

[quote="Sergic Primazon"]Let $ a$ and $ b$ sides ,  $ r$ - inradius of $ \triangle ABC$ . Find minimum : $ \boxed{\frac {ab}{r^2}}$ and then does minimum holdes?[/quote]
For the area we have $ \frac {ab}2\sin\gamma \equal{} rs$, where $ s \equal{} \frac {a \plus{} b \plus{} c}2$, so we have to minimize $ {\frac {ab}{r^2}} \equal{} \frac {2s}{r\sin\gamma}$. Suppose $ r$ and $ \gamma$ fixed, then it is geometrically clear that $ s$ is minimal when $ a \equal{} b$. Now, $ \frac ar \equal{} \cot\frac\beta2 \plus{} \cot\frac\gamma2 \equal{} \cot\frac\beta2 \plus{} \tan\beta$.
This becomes minimal for $ \cot\frac\beta2 \equal{} \sqrt {\phi^3} \equal{} \sqrt {2 \plus{} \sqrt {5}}$, equivalently $ \tan\beta \equal{} \sqrt {\phi} \equal{} \sqrt {\frac {1 \plus{} \sqrt {5}}2}$ or

$ \boxed{\cos\alpha \equal{} \cos\beta \equal{} \phi^{ \minus{} 1} \equal{} {\frac {\sqrt {5} \minus{} 1}2}}$ and $ \boxed{\cos\gamma \equal{} \phi^{ \minus{} 3} \equal{} \sqrt {5} \minus{} 2}$, and we get $ \boxed{\min\frac {ab}{r^2} \equal{} \phi^5 \equal{} \frac {11 \plus{} 5\sqrt {5}}2}$.
:)


Probably a trick question, but is it [hide]A[/hide] ?

TheDon

Can you show how you got it?

P.S. Your answer may or may not be right. So, if you start explaining, you will confirm your answer or finds it's wrong.

Here's what I got.

Answer:
[hide]C[/hide]

Work:

[hide] .8t + 18 = 1.5t + 15
.7t = 3
t = 30/7
t :approx: 4.29
[/hide]

Careful ASquare
[hide]
t is indeed 30/7 or 4 2/7
This means the time is 4 + 6 = 10:something
2/7 is the number of minutes
60 minutes in an hour, so we do 2/7 * 60 = 120/7 = 17.1 or rounded is 17 minutes
so answer is 10:17 or B
[/hide]

Hmmm..

[hide="Answer"]Solve equation 0.8t +18 = 1.5t + 15 to get t = 30/7
Keep in mind this is the number of hours. This equals approximately 257 minutes, so the answer is B, 10:17 AM.[/hide][/hide]

[quote="Silverfalcon"]#6

At a campground near Lake Michigan, the temperature outside a cabin is 0.8t+18 degrees Celsius, and the temperature inside is 1.5t+15 degrees Celsisus, where t is the number of hours after 6 am. To the nearest minute, at what time will the temperatures inside and outside be the same?

A) 6:04 am
B) 10:17 am
C) 10:29 am
D) it is always warmer outside
E) None of these

For the last choice that doesn't have to do with this problem but for fun:

[hide="F"]Impossible to figure it out since I don't have thermometer with me  :P [/hide][/quote]

[hide="solution"]$.8t+18=1.5t+15$

$.7t=3$

$t=\frac{30}{7}=4$ hours $\frac{2}{7}60$ minutes

t=4 hours and 17 mins

10:17

(B)[/hide]


[hide="Think more elementary"] Telescoping sum. ;) [/hide]


compute the two possible positive integral values of x such that
x^2+X+39 is a perfect square.

[hide]I don't know how to compute that, but I made a simple program and the answers are 6 and 38.[/hide]

Set $x^{2}+x+39=m^{2}$ and multiply by $4$ .

[hide="Explanation"]$4x^{2}+4x+156=4m^{2}\iff (2x+1)^{2}+155=(2m)^{2}\iff\\ \iff(2m+2x+1)(2m-2x-1)=155$
Factorize RHS and find the possible values for $x$ and $m$.[/hide]

Meaning 155=5*31, let a,b, be the two factors (note a>b). always a+b=4m, a+b=156 or 36, m=39 or 9, so x is 6 or 38


geometry
rectangle
symmetry
angle bisector
geometry solved

In rectangle ABCD, the angle bisector of angle B intersects diagonal AC and
side AD in E and F, respectively. A line through E parallel to AB intersects
diagonal BD in K. Prove that line FK is perpendicular to AC.

Let $T=AK\cap BE$. Because of the symmetry if the figure, $AT=BT$, and since $\angle ABT=45^{\circ}$, it means that $\angle ATB=90^{\circ}\Rightarrow AK\perp BF=EF$. We thus have $AK\perp EF,\ EK\perp AF$, meaning that $K$ is the orthocenter of $AEF$, so $FK\perp AE$, Q.E.D.


Prove that if $ 2^k\minus{}1$ is prime, then $ 2^{k\minus{}1}(2^k \minus{} 1)$ is a perfect number.
A perfect number is a number where the sum of its divisors is twice itsef. For example 6 is a perfect number because 1 + 2 + 3 + 6 = 12 = 2(6).

that's a std question
See http://en.wikipedia.org/wiki/Mersenne_prime  :)


La un moment dat aveam impresia ca monologhez pe FORUM.
Ultimele observatii ale dl. prof Virgil Nicula, ma fac insa sa perseverez in a
pune pe forum, probleme care mie mi s-au parut a fi frumoase!

In attachment veti gasi o frumoasa problema, care va va aduvce aminte de
problema lui Gh. Titeica.

Aceasta problema se regaseste si in culegerea lui PRASOLOV:
"PROBLEMS in PLANE and SOLID GEOMETRY", este vorba despre
problema 3.26 de la pagina 61-62. In aceasta culegere gasiti o alta
solutie a acestei frumoase probleme, care difera in mod esential de
solutia pe care am dat-o eu!

Pe internet se gasesc atat versiunea in lb. rusa cat si cea in lb. engleza
si pot fi descarcate GRATUIT.
In lb. rusa o gasiti la adresa:
            http://www.mccme.ru/free-books/
la pozitia 33, a paginii care se deschide!
(cea in engleza nu mai stiu unde este).


[quote="silouan"]If $xy+yz+zx=1$ Prove that
$(x+y+z)(\frac{2x}{2x+y+z}+\frac{2y}{2y+x+z}+\frac{2z}{2z+x+y})\geq\frac{3\sqrt 3}{2}$ :)[/quote]
There are no conditions on $x,y,z$? Are they >0 ?

Homogenalize the inequality,\[(x+y+z)(\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y})\geq\frac{3\sqrt 3}{4}\sqrt{\sum xy}\]
Dehomogenalize with x+y+z=1, we have to prove\[\sum\frac{x}{1+x}\ge\frac{3\sqrt 3}{4}\sqrt{\sum xy}\]
By Cauchy Schwarz,\[\sum\frac{x}{1+x}\ge\frac{1}{\sum x(1+x)}=\frac{1}{1+\sum x^2}\]The result follows from $\sum xy(1+\sum x^2)^2=2(2\sum xy)(1-\sum xy)^2\le2(\frac{2\sum xy+1-\sum xy+1-\sum xy}{3})^3=\frac{16}{27}$.

You cannot say $x+y+z=1$ because we have $x+y+z=xy+yz+zx=1$ which is impossible. Your solution is wrong I think

He is allowed to do that, since he "removed" the condition $xy + yz + zx = 1$ first.

he told this [quote="mecrazywong"] we have to prove\[\sum\frac{x}{1+x}\ge\frac{3\sqrt 3}{4}\sqrt{\sum xy}\]
[/quote]
but now $xy+yz+zx\neq 1$ So itis not enough to prove this

[quote="mecrazywong"]Homogenalize the inequality,\[(x+y+z)(\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y})\geq\frac{3\sqrt 3}{4}\sqrt{\sum xy}\][/quote]

It is enough to prove this inequality, right?
(Since then we CAN take $xy + yz + zx = 1$ and we get the inequality you wanted)

Well, since this is a homogenous inequality, it is sufficient to prove it for $x + y + z = 1$.
Then, from that "special" case, you can conclude that it holds for all positive reals.
So, then we are done.

Do you get it?

I first homogenalise the inequality so the condition xy+yz+zx can be trashed away. Therefore I can dehomogenalise with another condition.
If you are still not persuaded by this method, write $(x+y+z)(\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y})\geq\frac{3\sqrt 3}{4}\sqrt{\sum xy}$ as $\sum\frac{x}{x+y+z}\sum\frac{\frac{x}{x+y+z}}{1+\frac{x}{x+y+z}}\ge\frac{3\sqrt 3}{4}\sqrt{\sum(\frac{x}{x+y+z}\frac{y}{x+y+z})}$.

Edit: Thanks Arne for explaining this :)

Thank you I get it but could someone post another solution because this solution it seems to me correct but not so good . Please try it .

Well, if you are not satisfied, why don't you look for more solutions yourself?

In my opinion, mecrazywongs solution is very short and very neat. Excellent!


[quote="chien than"]Prove that
$ (\sum_{sym} a^2b)^2 \geq \sum ab . \sum a^2b^2$
where $ a;b;c$ be positive real numbers.[/quote]
Are you mean that $ \left(\sum_{cyc}(a^2b\plus{}a^2c)\right)^2\geq4\sum_{cyc}ab\cdot\sum_{cyc}a^2b^2$ $ ?$ If so
$ \left(\sum_{cyc}(a^2b\plus{}a^2c)\right)^2\geq4\sum_{cyc}ab\cdot\sum_{cyc}a^2b^2\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}a^2b^2(a\minus{}b)^2\plus{}2abc\sum_{cyc}(a^3\minus{}a^2b\minus{}a^2c\plus{}abc)\geq0.$ :wink:

[quote="arqady"]$ \left(\sum_{cyc}(a^2b \plus{} a^2c)\right)^2\geq4\sum_{cyc}ab\cdot\sum_{cyc}a^2b^2$[/quote]

I can just repeat: http://www.mathlinks.ro/Forum/viewtopic.php?t=105871 ;)
Note that $ a$, $ b$, $ c$ may be arbitrary reals. This is part of an article I have submitted for the Romanian journal "Math Time".

  darij

I have missed it. It's beautiful! Thank you, Darij!


who can tell me something about the names above the stars??
For example, what is the "Riemann Hypothesis"?

millenium problems
more info : http://www.claymath.org/millennium/

[quote="shobber"]
Can you provide a link??[/quote]

Please do a search, I (and others) talked about it somewhere. :wacko:

[quote="EUCLA"][quote="shobber"][quote="EUCLA"]you can also read THE MILLENIUM PROBLEMS author: KEITH DEVLIN[/quote]
Can you provide a link??[/quote]
that's a book  :D[/quote]

That is a good one!  :D  ;)

[quote="djimenez"]You know, in about two months that I have been here, I have seen the same question over and over again, about once a week I think. May be it would be wise to put a sticky post in the Round Table and or the F&G forum about this, to see if such questions stop. I mean, probably people would stop asking something that has been already answered over and over again :D . (I don't critizice the people who ask because I think I did the same question :blush: )[/quote]
that's a good idea.

[quote="djimenez"][/quote][quote="EUCLA"][quote="shobber"][quote="EUCLA"]you can also read THE MILLENIUM PROBLEMS author: KEITH DEVLIN[/quote]
Can you provide a link??[/quote]
that's a book  :D[/quote][quote="djimenez"]

That is a good one!  :D  ;)[/quote]
it makes me understood lots of things :D


A uniform heavy chain is placed on a table with a part of it hanging over the edge.I just begoins to slide when this part is one third of its length.What is the coefficvent of friction between the table and chain?

[color=blue]The coefficvent of friction between the table and chain is $\frac{1}{3}$, right?[/color]

shadysaysurspammed, is that the correct answer? I'm asking because I get something totally different.

Yeah, I thought that the condition should be $\frac{mg}3 = \frac{2mg}3 \cdot \mu$, but I'm not that sure, though.

we observe the "limes"(lol) case when $F_{friction}=F_{g}$ (friction between table and whole chain and gravitational force of the third that hangs)

\begin{eqnarray*}F_{friction}&=& F_{g}\\ m\cdot g\cdot \mu &=& \frac{1}{3}\cdot m\cdot g\\ \mu &=& \frac{1}{3}\end{eqnarray*}

(friction is not defined by the area that presses the surface but by the force that presses the surface)

NOTE: Im puzzled... Is the force that causes friction in this problem equal to $mg$ or $\frac{2mg}{3}$ ? Could someone care to explain in detail?

Assume a small displacement dx after equilibrium is attained.
Work energy theorem gives -
$\frac{mg dx}{3}$ = $\frac{\mu{2mg dx}}{3}$

Solving ,$\mu$ = $0.5$

here the normal force is indeed $\frac{2mg}{3}$ since only the part on the table contributes to the normal force..Here we have not considered the kinetic energy term since a small displacement immediaely after equilibrium gives zero K.E.

[quote="ashwinkrishnan"]since only the part on the table contributes to the normal force..[/quote]
Could you explain this in detail?

Just before the instant the chain is about to slip over the table, it is in a state of equilibrium. After the equilibrium is attained ,even a slight displacementmakes the chain to slide down the table.
After the first elemental displacement, there is a change in the potential energy of the overhanging part and this is compensated by the work done against friction.Since we have considered the first displacement,the kinetic energy doesn't come into the picture.

I understand the thing ur talking about. I [i]dont[/i] uderstand why is the normal force on the surface equal to $\frac{2mg}3$!

Obviously,the normal force is caused due to omly the portion of the chain left on the table and not the overhanging part.


[quote]Given a polynomial $x^6+x^4-115x^3+x^2+x+1=0$, find the sum of the real roots.[/quote]

Anybody know a solution?

[quote="joml88"]The symmetry suggests that we divide by $x^3$ and make a variable substitution:

\[x^3+\frac 1{x^3}+x +\frac 1x-115=0.\]

So let $x+\frac 1x =w.$  Then $x^3+\frac 1{x^3}=w^3-3w.$

\[w^3-2w-115=0.\]

By inspection, $w=5$ is a zero.  It then factors into $(w-5)(w^2+5w+23).$  The latter product procuces no real roots.  The former produces:

\[x+\frac 1x=5\Rightarrow x^2-5x+1=0 \Rightarrow r_1+r_2=5.\]

Thus the sum of the real roots is $5$.[/quote]

All this could have been correct, if there was no 'x' term in the given equation.

because of the presence of 'x' in the equation, there will be a term containing $\frac 1{x^2}$,  which is missing in your proof.  Please check it.

Shoot, that's what I get for trying to do that problem before going to school and rushing.

You know the sum of all the roots is 0. So if we can find the imaginary roots, its trivial, actually just the real parrt of the imaginary terms


limit
logarithms
function
complex analysis
complex analysis unsolved

Let $ p(k)$ denote the $ k$th prime number, with $ p(1) \equal{} 2$.  For a fixed $ z\in\mathbb{C}$, define the sequence
\[ s_k \equal{} s_{k \minus{} 1} \plus{} \frac {1}{p(k)^z}\left(1 \minus{} s_{k \minus{} 1}\right)
\]
with $ s_0\equal{}0$.  Also, define
\[ u(z) \equal{} \lim_{k\rightarrow\infty} s_k
\]
For what part of the complex plane is $ u(z)$ well defined, and what is it equal to?

case when $ s_{k} \to \infty$ is not interesting Roughly speeking this is hint:
say $ P(k) \approx k \ln k$ and consider $ s'(k) \approx s(k) \minus{} s(k \minus{} 1) \equal{} \frac {1}{P^{z}(k)}(1 \minus{} s(k \minus{} 1))\approx (1 \minus{} s_{k})\frac {1}{k^{z}\ln^{z}k}$ e.t.c

Well an estimate of that sort will tell you that it converges for, say, $ \left\{z\mid\Re(z) >1\right\}$, but I don't think it gets you much closer to figuring out the actual value.
[hide="hint"]
You can get $ s_k$ to telescope, and the Zeta function rears its head.
[/hide]


function
topology
vector
Ring Theory
real analysis
real analysis solved

here an interesting question (i think ..):
find all continuous morphism of ring from C([0;1],R) to R.
(C([0;1],R) is the set of all continuous functions from the compact interval [0;1] to R)

any idea ..

Alekk, could you please give a definition of continuous morphism? I did not meet this till now. Thanks

I think they are morphisms which are also continuous functions, but one should define the topological spaces we're working with. For example, is the topology defined on $C([0,1],\mathbb R)$ the one induced by the norm $\sup_{x\in[0,1]}|f(x)|$?

I think I am close to find all morphisms, not necessarily continuous. Or am I wrong? Please, someone check the following solution:
  First of all, we exclude the trivial zzero morphism. Thus, we may assume that $ f(1)=1$ ( here 1 is the constant function). Since the restriction of the morphism (which will be denoted by f) to the set of continuous constant functions behaves exactly like a field morphism of the reals,  we must have $ f(r)=r$ for the constant function r. I will prove that any morphism has the following form: there exists $s$ in [0,1] such that for any function tau we have $ f(tau)=tau(s)$. Suppose we have a morphism f which is not of this form. Thus, for any $0\leq u\leq 1$ there is a function $g_u$ such that $ g_u (u)$ is different from $ f(g_u)$. Thus there is an open set which contains u, call it $ V_u$ and an $eps_u>0$ such that for all x in $V_u$ we have $ (g_u (x)-f(g_u))^2>eps_u$. Now, the compact [0,1] is covered by a family of open sets, thus we can find a finite couver. So, there exists $ V{}_u{}_1,..., V{}_u{}_n$ which cover [0,1]. Then for all x in [0,1] we have $\sum_{i=1,n}{(g{}_u{}_i (x)-f(g{}_u{}_i))^2}>eps$, where $eps=min(eps{}_u{}_1,...,eps{}_u{}_n)$. Consider the function $ h(x)=\sum_{i=1,n}{(g{}_u{}_i (x)-f(g{}_u{}_i))^2}$. Using the aditivity and multiplicativity of f as well as the fact that the restriction of f to the constant functions is the identity we find that $ f(h)=0$. But we also have $ h>eps$, thus we can write $h=eps+t^2$ with t a function. Then $ 0=f(h)=f(eps+t^2)=eps+f^2(t)$, which is clearly impossible.
    Now, the fundamental question: where is the mistake?

hi Harazi,

in fact I think you're right and i've been idiot
to impose continuity to the problem. Yesterday I arrived
to the same conclusion as you but someone claimed on another
maths forum that there were far much morphisms and it was foolish to
find all of them:so i restricted to the continuous one. So now i'm sure
he was totally wrong ;-)

I posted this question because it's related to the following problem:
find all the maximal ideal of the ring C([0;1],R). Now we know that
all morphisms from C([0;1],R) to R are surjective and of the form f->f(s),
it's quite straightforward to conclude ..

in fact I misunderstood what I was said yesterday
on another forum. These are the morphisms of group
from C([0;1],R) to R that are far too numerous to
be clearly described... sorry..

Alekk, I simply do not understand what you mean in your last post? Thus, my solution is wrong? Are there really many starnge such morphisms or what?

harazi was right; point evaluations are the only [b]ring[/b] homomorphisms, even without the requirement of continuity. For something like vector space homomorphisms, there are too many to characterize simply.

Point evaluation is also continuous in the infinity-norm, which is the standard metric anyway.

excuse my english Harazi. I wanted to say that you were right.
But for morphisms of group, as said Jmerry, there are far too many morphisms: it suffices to take an arbitrary value for each element of a basis of R which is a vector space over Q ..