mathleague.org
ARML
email
geometry
analytic geometry
Stanford
college

Any info on how ARML tryouts work this year and where and when the CA mathleague contests will be would be much appreciated. Tim's website isn't being much help, so I'm asking here.

Thanks!

The website has some of the contest dates and locations posted, including the first one next Saturday (9/26) in San Jose.

As for ARML tryouts, the extent of my knowledge is from Tim and I bouncing ideas around after the bus ride last spring.  I think the basic format is going to remain the same, but you'd have to ask Tim if you want any kind of official word.

we're pushing back the date of the first one to October 10, but things this year will generally be the same as last year.  the website isn't much help yet because details of how to register haven't been released yet.  i will post here and send an email out to the sfba-arml mailing list (it's fixed now but i don't think it's taking new registrations yet) as soon as i have everything confirmed for this year including sites and times.  all the dates are up though (http://mathleague.org/hsevents.php) and have been cross-checked with other bay area contests.  the december one conflicts with something in the south bay that is only open to 12th graders living in Santa Clara County, so if you're one of the few people eligible for that one you'll want to plan your ARML tryouts around that event.  we're also on the same day as the Miller Math Marathon, so junior high students will need to plan their tryouts accordingly as well.  any other schedule conflicts are a direct result of someone else's failure to respond to my attempts to coordinate scheduling; hopefully we'll avoid that this year but there are always a few rogue contest hosts who refuse to coordinate dates..  :)

real quick note - i got confirmation that we're on at SJSU for the morning of october 10; today i'll be working on getting the website set up to accept registrations and getting the official invitation letter out.  this letter will be mailed to schools i have contact info for (which includes every school in the Bay Area that participated in the AMC 10/12 last year), as well as to the SFBA-ARML email list.  if you don't receive this email by Thursday morning please email operations@mathleague.org and ask to be put on the list.  feel free to start registering as soon as the event shows up with an active registration link on our website.  i'll post more info as it becomes available..

http://mathleague.org/hsevents.php

[Just sent this out to the mailing list, but it'll take a couple of hours to reach everyone.  If you don't end up getting it, send me an email so i can add you to the list:]

mathleague.org is pleased to announce that the second annual Northern California championship meet for high school math students will take place this year on April 24 at Stanford University.  Students and teams may qualify to attend the championship meet by scoring at least 50% on any event at one of our local qualifying rounds or through an optional mail-in round available to mathleague.org member schools.  Although these contests are designed for high school students, junior high students may compete at both the qualifying meets and the championship meet. Be sure to go to http://mathleague.org/california for more information about the qualifying rounds (including locations, schedules, and test format), how to register your students for these contests or for the mail-in contest, and how to obtain practice materials.

Qualifying rounds this year have been scheduled for the following dates:
•	October 10, San Jose State University
•	November 14, University of California, Berkeley
•	December 5, University of California, Davis
•	December 12, Amador Valley High School
•	January 9, Stanford University
•	February 13, University of California, Berkeley
•	March 6, [location TBA]

Further details for the contests we currently have scheduled can be found on our website; if none of the listed locations is convenient for you, please contact us at mathleague@mathleague.org to discuss how your school may host a qualifying meet for schools in your area.  Please note that students attaining 50% correct in any event at any qualifying meet are eligible to take all tests at the championship meet.

mathleague.org is also using these contests to recruit students to represent our region at the national ARML competition. We will be competing at the American Regions Mathematics League national contest in Las Vegas this June. This year's ARML meet is scheduled for June 4 and 5 at the campus of the University of Nevada - Las Vegas.  For more information about ARML, feel free to check out our team webpage aat http://mathleague.org/armlsfba.php.

Looking toward next year, mathleague.org will be setting the dates for our 2010-2011 contests in May.  We always strive to avoid scheduling conflicts with other contests; as such, any requests for specific dates, or information about other contests we should coordinate scheduling with, should be emailed by May 1 to mathleague@mathleague.org.  If your school is interested in hosting a qualifying round for next year, feel free to contact us anytime for more information.

In other news...

Contest Prep Classes:  This year, mathleague.org is also organizing a series of math contest preparation classes to help students prepare for AMC and MATHCOUNTS competitions; please visit us on the web at http://mathleague.org/classes.php for more details or to sign up for these classes.  mathleague.org is happy to offer a referral credit to anyone who refers us to students who end up registering for one of our classes; see our website for more details.

Junior High Contests:  mathleague.org is once again providing tests for a number of junior high math contests  this year, including a contest on October 31 and several in the spring.  Please visit
http://mathleague.org/jhevents.php for more information; listings for next semester's contests will be posted as details are confirmed.

Other Contests:  In our eternal quest to coordinate with other contests to avoid scheduling conflicts, we received the following replies from organizers of various contests, each of whom was kind enough to include information about their contest -
"The Mathematics Department at Fresno State will hold its next (annual) Math Field Day, a competition for middle and high school students, on Saturday, April 17, 2010. The competition will take place on our campus.  http://csufresno.edu/math/news_and_events/field_day/index.shtml"
"BAMO is scheduled, as usual, for the last Tuesday of February, Feb 23 2010. The awards ceremony is on Sunday, March 7."
"the Santa Clara Valley Math Olympiad (sponsored by SCVMA for high school seniors) is always scheduled for the second Saturday in December (this year, Dec. 12)."
"SMT will probably be in February."
You should be able to find information about each of these contests online; if that fails, feel free to send us an email and we can put you in touch with the organizers.  Please note that any contest not listed above represents a failure of  the contest organizer to respond to our inquiries.  Unfortunately mathleague.org cannot accept any responsibit]lity for conflicts that arise as a result of contest organizers ignoring our efforts to coordinate scheduling.

Looking forward to another productive math contest season,

Tim Sanders
President, mathleague.org
Head Coach, SFBA ARML

tim, where is the october 10 contest in SJSU?  Do you know yet? Is it Macquarrie or somewhere else?

did you register for the contest?  that information will be emailed to registered students in a day or two..  :)

I know its a day late, but is there any way possible we (Lynbrook) could still get in?

Thanks,
Timothy Chu

yeah actually i'm going to be making photocopies thursday afternoon, so i can still accommodate anyone who registers by tomorrow at noon.  also, the location is SCI 142 at San Jose State.  a campus map is at http://www.sjsu.edu/about_sjsu/docs/SJSU_campus_map.pdf if you'd like to look it up..

BUT...even though you all know the location now, don't show up without registering because i would hate not to have enough tests to go around..

a lot of people who told me they'd be registering for the october 10 contest haven't yet.  i'm making a few extra tests but not many, so don't expect there to be tests available on saturday if you haven't registered.  if you want to sign up at this point, i'll put you on a waiting list for tests and will distribute tests saturday based on the order in which people signed up.  remember, that sign up location is http://mathleague.org/register.php..


trigonometry
inequalities
algebra
system of equations
algebra proposed

Solve the system of equations:
$\frac{b(x+y)}{x+y+cxy}+\frac{c(z+x)}{z+x+bzx}=a$;
$\frac{c(y+z)}{y+z+ayz}+\frac{a(x+y)}{x+y+cxy}=b$;
$\frac{a(z+x)}{z+x+bzx}+\frac{b(y+z)}{y+z+ayz}=c$.

[quote="Farenhajt"]Substitution

\begin{eqnarray*}u &=& \frac{x+y}{x+y+cxy}\\ v &=& \frac{y+z}{y+z+ayz}\\ t &=& \frac{z+x}{z+x+bzx}\end{eqnarray*}

yields

\begin{eqnarray*} bu+ct &=& a\\ cv+au &=& b\\ at+bv &=& c\end{eqnarray*}

which is easily solved (for example by Cramer's rule), and we get

\begin{eqnarray*}u &=& \frac{a^2+b^2-c^2}{2ab}\\ v &=& \frac{b^2+c^2-a^2}{2bc}\\ t &=& \frac{a^2+c^2-b^2}{2ac}\end{eqnarray*}

(Inasmuch this resembles Cosine Law, we can't assume that $a,b,c$ satisfy the triangle inequalities since such a condition isn't given in the problem.)

Now

\begin{eqnarray*} u=\frac{x+y}{x+y+cxy} &\implies& \frac{1}{u}=1+\frac{cxy}{x+y}\\ &\implies& \frac{xy}{x+y}=\frac{1-u}{cu}\\ &\implies& \frac{1}{x}+\frac{1}{y}=\frac{cu}{1-u}\end{eqnarray*}

Next

\begin{eqnarray*}\frac{cu}{1-u} &=& \frac{c(a^2+b^2-c^2)}{2ab-a^2-b^2+c^2}\\ &=& \frac{c(a^2-2ab+b^2-c^2+2ab)}{2ab-a^2-b^2+c^2}\\ &=& -c+\frac{2abc}{(-a+b+c)(a-b+c)}\\ &=& -c+\frac{2abc(a+b-c)}{P}\end{eqnarray*}

where $P=(-a+b+c)(a-b+c)(a+b-c)$

Hence

$\frac{1}{x}+\frac{1}{y}=\frac{2abc(a+b-c)}{P}-c\qquad\qquad (1)$

In the same manner we find

$\frac{1}{y}+\frac{1}{z}=\frac{2abc(b+c-a)}{P}-a\qquad\qquad (2)$
$\frac{1}{z}+\frac{1}{x}=\frac{2abc(c+a-b)}{P}-b\qquad\qquad (3)$

This system is easily solved: $(1)+(3)-(2)$ gives

$\frac{2}{x}=\frac{2abc(3a-b-c)+P(a-b-c)}{P}\implies x=\frac{2P}{2abc(3a-b-c)+P(a-b-c)}$

Similarly

$y=\frac{2P}{2abc(3b-c-a)+P(b-c-a)}$
$z=\frac{2P}{2abc(3c-a-b)+P(c-a-b)}$[/quote]
You have just my solution and similar answer.

Please don't quote such big things when it isn't necessary.


$SO_{2}$ gas is bubbled into a solution of $MnO_{4}^{-}$. Write a balanced equation to show this reaction.

Is $SO_{2}$ called Sulfur Dioxide? I know about Sulfite and Sulfate ($SO_{3}, SO_{4}$)
Is $MnO_{4}$ called Manganate?

$SO_{2}$: Sulphur Dioxide.
$SO_{3}$: Sulphur Trioxide.
$SO_{3}^{2-}$: Sulphite Ion.
$SO_{4}^{2-}$: Sulphate Ion.

$Mn^{2+}$: Manganous Ion.
$MnO_{4}^{2-}$: Manganate Ion.
$MnO_{4}^{-}$: Permanganate Ion.

I think SO2 will become SO3 as it's oxidised by MnO4-.
I think the permanganate ions may come from Potassium Permanganate which is a oxidising agent

Hello,
  The reaction is:
    2KMnO4 + 2H2O + 5SO2 ----> K2SO4 + 2MnSO4 + 2H2SO4.
(Kindly read appropriately the chemical symbols as using I have not mastered Latex yet).
 Here,sulphur dioxide is oxidised to sulphuric acid by potassium permanganate.True,it is first oxidised to sulphur trioxide which dissolves in the aqueous medium to yield sulphuric acid.
Sathej

Oh, so it's further reacts w/ the water present in the aqueous solution, isn't it?

If it's so, will be there any acid mist?
As I know that SO3 react vigorously with water to become sulphuric acid

Hello,
   No,there will not be any visibly mentionable 'acid mist'.Such a mist occurs only if SO3 is directly bubbled into water.In this case,SO3 is only thought of as a possible intermediate.
Sathej

The suggested answer is in conflict with the question which specified MnO4- not K2MnO4, which becomes 2K+ and MnO4-2.

While one normally doesn't see an ion specified as a reactant, perhaps it has a mixture of cations, none of which react.

After using the proper reactants, determine which products will occur (redox tables), finally balance the equation.

Hello,
  The given answer is correct.MnO4- is the permanganate ion.KMnO4 is written as K+ and MnO4-.I have taken potassium permanganate KMnO4 and not potassium manganate K2MnO4.Anyway writing it as an ionic equation is a simple consequence of the given equation.The reactant can very well be given as an ion  and here MnO4- can stand for KMnO4 or NaMnO4 or any such compound, not a mixture of cations.
Sathej


I can't find any simpler way to do this than by actually doing the whole thing out.

$ 88_9 \implies 8 \times 9^1 \plus{} 8 \times 9^0 \implies 72 \plus{} 8 \implies 80_{10}$

$ 555_6 \implies 5 \times 6^2 \plus{} 5 \times 6 \plus{} 5 \times 6^0 \implies 5(6^2 \plus{} 6 \plus{} 1) \implies 5(43) \implies 215_{10}$

$ 2222_3 \implies 2 \times 3^3 \plus{} 2 \times 3^2 \plus{} 2 \times 3 \plus{} 2 \times 2 \times 2^0 \implies 2(3^3 \plus{} 3^2 \plus{} 3 \plus{} 1) \implies 80_{10}$

$ 80 \plus{} 215 \minus{} 80 \equal{} 215$

$ 215 \equal{} (11^2)a \plus{} 11b \plus{} c$, where $ \overline{abc}$ is our number in base $ 11$, and $ a,b,c < A$ ($ A$ is the notation for $ 11$ in a specific digit place.)

$ 215 \equal{} 121(1) \plus{} 11b \plus{} c$

$ 94 \equal{} 11b \plus{} c$

$ 94 \equal{} 11(8) \plus{} c$

$ 6 \equal{} c$

So our answer is $ \boxed{186_{11}}$.
_________________________

I understand part of my solution is somewhat vague.  Do you need any help understanding it?


MiKTeX cannot be downloaded anymore.  When it asks you to choose a site to download from, it shows you an error.  For those who have already downloaded it, count yourself luky.

Uh... I can download it fine...

http://prdownloads.sourceforge.net/miktex/basic-miktex-2.5.2421.exe?download


Factor it into $(x+1)(x-1)*(x+2)(x-1)*(x+1)(x-5)*(x+2)(x-5)$, so the answer is $\boxed{A}$.

That took me less than a minute to do.

Yeah, I just factored them out in my head, didn't take too much more than a half a minute.

To increase your proficiency in figuring out how to factor quadratics quickly, just practice practice practice, and soon you'll be able to factor quadratics rather quickly (If they are factorable). But, yeah, that's the quickest way I can think of.

yeah, I think this should be moved to the middle school section because that was just way to easy.

it was easy.  I just recognized x+1 and x-1 in x 2 -1
then i tried to get x-5 so i looked at the last one.  and i looked at the x 2 -3x-10 and div. and got x+2
so it's x-2

:lol: I could tell from looking at it. A  :lol:

factoring should be a basic skill. Clearly C and D are not possible from the first parenthesis.

Everyone here know synthetic division?
Find the zeroes of the function
x^2 + 2x + 1
using synthetic division.

[quote="karthik_k"]Q.5 : Which of the following is not a factor of $(x^2-1)(x^2+x-2)(x^2-4x-5)(x^2-3x-10)$
(A)$x-2$
(B)$x+2$
(C)$x+1$
(D)$x-1$
(E)$x-5$

Any shortcuts ?? I know you can do the stuff the loooooooooong way , but then , just thought there might be a different way of solving it.[/quote]

[hide="Hint"]

The key is to recognize that each factor is itself factorable. Write out the complete factorization and then go through and check off the true factors one at a time. The only one left will be your answer.
[/hide]


logarithms
floor function
number theory
prime factorization

[hide]The number of 0s at the end of a number is the smallest of the values of the exponents of 2 and 5 in the prime factorization. In this case, the prime factorization is 2^32 5^25, so there are 25 0s at the end. 2^7=128, which has 3 digits, so there are 3+25=28=D digits.[/hide]

[quote="Silverfalcon"]The number of digits in $4^{16} 5^{25}$ (when written in the usual base 10 form) is

A. 31
B. 30
C. 29
D. 28
E. 27[/quote]

[hide]$4^{16}5^{25}=10^{25}2^7$

$10^{25}$ has $26$ digits with $25$ $0$s, and $2^7=128$, which has three digits.

So the total number of digits is $25+3=28$

$\boxed{D}$[/hide]

I did the calculator way

[hide]$4^{16}5^{25}$
so $16 \cdot \log{4}=10$ and
$25 \cdot \log{5}=18$

Therefore, $10+18=\boxed{28}$.[/hide]

You don't need a calculator to do it that way, as long as you have $\log 2$ and $\log 3$ memorized. :) It's still simpler to do it the other way, though.

Calculator was forbidden in 80's AMC so unless you memorized $\log 2$ and $\log 3$, you'll have to do what first few posts did.

Yes, I have $\log 2$ and $\log 3$ memorized; they aren't too hard to get down, and you can get almost every other log from that. Only the primes are a little difficult. But, in the case of this problem, the other method is easier, anyway.

Here's a method that you may enjoy...not considering whether or not you need a calculator.

You say that $4^{16}5^{25}=5^{13.5+25}=5^{39.5}\approx 5^{39}$.
You then say that $\lfloor 39\log 5+1\rfloor =28$.
$\therefore$ 28 is the answer.

[hide="What immediately popped into my mind"]I thought about pairing 5's and 2's!  :)

$4^{16}5^{25}=2^{32}5^{25}=10^{25}2^7$, which has 28 digits.

D is the answer.[/hide]

[quote="YooSam"][quote="apusey"]Unless you have a new way to do it or a contradiction you don't need to post the same solution over and over and over again[/quote]

What do you mean by that?  :?:

Sometimes people don't look at others' posts because they want to do that on their own.[/quote]

I don't understand that.

Ok if you don't want to look at others posts and try to find a solution. But i think it's absolutly nonsense posting same thing several times, this just makes people waste their time. So i think all of us can do the right thing.
:spam:


calculus
integration
vector
function
derivative
topology
calculus computations

Vector Field  F(x,y)=P(x,y)i + Q(x,y)j, where P(x,y)= - y / (x²+y²)    Q(x,y)= x / (x²+y²)
a) Show that there is a gradient field f so that F= ∇ f
b) Calculate the work of vector field F along (x²/4) + (y²/9) = 1

I think that was a question to get you to check whether it was conservative or not, not to look it up.  :wink:

There are several equivalent definitions for conservative.

Your calc book should say somewhere that a vector field is conservative if $ \frac{\partial P}{\partial y}\equal{}\frac{\partial Q}{\partial x}$.

It is a conservative field. θP/θy = θQ/θx <=> rotF=0
I'm just trying to have a solution of the problem in order to study it, and solve others like it for an exam coming up. Thanks you

Since it is conservative, we know that $ f$ exists. I'm not sure if (a) wants us to find it, or just justify why it exists.

For (b), note that $ C: \frac{x^2}{4}\plus{}\frac{y^2}{9}\equal{}1$ is a closed curve and the work is given by $ \int_C \vec{F}\cdot d\vec{r}$. You should have a theorem that allows you to calculate this without actually doing any calculation (sometimes called the "fundamental theorem of line integrals," or sometimes it is the way people define "conservative field").

This isn't a conservative vector field. A vector field $ \vec{F}$ on an open set U $ \subseteq$ $ \mathbb R^{2}$ is conservative $ \Longleftrightarrow$ $ \oint{\vec{F}\cdot ds} \equal{} 0$ for all closed curves C in U. And, as previously proposed, if a vector field F = (P,Q) is conservative $ \Longrightarrow$ $ \frac {\partial P}{\partial y} \equal{} \frac {\partial Q}{\partial x}$

BUT $ \frac {\partial P}{\partial y} \equal{} \frac {\partial Q}{\partial x}$ $ \not\Longrightarrow$ F is conservative (this implication is only true if U is convex, but fails to hold for more general open sets).

[hide]The potential function, f, you're looking for is $ f \equal{} tan(\frac {y}{x})$, which isn't well defined when x = 0 (if you dont believe me that this is the correct function, differentiate and see for yourself). So it's not a function on the whole of $ \mathbb{R}^2\setminus\{(0,0)\}$ and thus, not a potential function. Integrating around any closed curve will give you a value of 2$ \pi$[/hide]

You're right, this isn't conservative, but I think your reasoning isn't quite right. $ U$ doesn't have to be convex, it can be any open simply connected region (which $ \mathbb{R}^2$ satisfies both of these anyway), the condition that was missing is that $ \frac{\partial P}{\partial y}\equal{}\frac{\partial Q}{\partial x}$ AND all first-order partial derivatives of $ P$ and $ Q$ are continuous (which is not satisfied in this case) $ \Rightarrow$ $ F$ conservative.

Oh right, my mistake. Reading off of half smudged notes from last term will do that to me -_- Thanks for the correction!


I don't really like AoPS too much, and I am too retarded for math.

Just delete me from the AoPS, so my username JWHooper will not exist in this forum. Thanks mods (or whoever is responsible for this)

[quote="JWHooper"]I don't really like AoPS too much and I don't need AoPS anymore[/quote]Who cares? Accounts are not deleted if you post stuff so I can't help you. Just don't use it :)


calculus
derivative
inequalities
inequalities proposed

Let $ a,b,c$ be nonnegative numbers, and let $ x\equal{}\sqrt \frac {2a}{b\plus{}c}$, $ y\equal{}\sqrt \frac {2b}{c\plus{}a}$, $ z\equal{}\sqrt \frac {2c}{a\plus{}b}$. Prove that

$ \frac 1{x}\plus{}\frac 1{y}\plus{}\frac 1{z}\minus{}3 \ge \sqrt{2}(x\plus{}y\plus{}z\minus{}3)$.

[quote="Vasc"]Let $ a,b,c$ be nonnegative numbers, and let $ x \equal{} \sqrt \frac {2a}{b \plus{} c}$, $ y \equal{} \sqrt \frac {2b}{c \plus{} a}$, $ z \equal{} \sqrt \frac {2c}{a \plus{} b}$. Prove that

$ \frac 1{x} \plus{} \frac 1{y} \plus{} \frac 1{z} \minus{} 3 \ge \sqrt {2}(x \plus{} y \plus{} z \minus{} 3)$.[/quote]
[hide]Note that $ x^2y^2\plus{}y^2z^2\plus{}z^2x^2\plus{}x^2y^2z^2\equal{}4$, from now we can use EV method of VasC.

PS: $ \sqrt{2}$ on the right hand is also the best constant for this inequality. :)[/hide]

[quote="Vasc"][quote="can_hang2007"] Note that $ x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2 \plus{} x^2y^2z^2 \equal{} 4$, from now we can use EV method of VasC.
[/quote]
How can you use EV method ?[/quote]
I am sorry, VasC.
I have a big mistake in calculaton the derivative, but I found this inequality can by solved by the LCF-RCF method of you.  :)

[quote="can_hang2007"]I found this inequality can by solved by the LCF-RCF method of you.  :)[/quote]
Indeed, this is my solution, too.
We can generalize the inequality to $ n$ variables, as follows:

$ \sum \frac 1{x_1} \minus{}n \ge \sqrt{n\minus{}1}\ (\sum x_1\minus{}n)$,

where $ x_1\equal{}\sqrt \frac{(n\minus{}1)a_1}{a_2\plus{}...\plus{}a_n}$.

[quote="Javkhaa"][quote="Vasc"][quote="can_hang2007"]I found this inequality can by solved by the LCF-RCF method of you.  :)[/quote]
Indeed, this is my solution, too.
We can generalize the inequality to $ n$ variables, as follows:

$ \sum \frac 1{x_1} \minus{} n \ge \sqrt {n \minus{} 1}\ (\sum x_1 \minus{} n)$,

where $ x_1 \equal{} \sqrt \frac {(n \minus{} 1)a_1}{a_2 \plus{} ... \plus{} a_n}$.[/quote][/quote]
Download files :) :


all you physics thalais, let physics go to hell and solve this, and i want an easy solution, and i am sure there is one as this was a jee problem.

Freshly prepared al(oh)3 and mg(oh)3 are stirred vigorously in a buffer solution containing 0.25 mol/L of nh4cl and 0.05 mol/L of nh4oh. Calc the conc. Of al3+ and mg2+ ions in solution. Kb(Nh4Oh)=1.8 X 10^-5
Ksp(Mg(OH)2)=8.9X10^-12 and Ksp(Al(OH)3= 6 X 10^-32.

first latexify that u can use
[code]
$ \ce{NH2} $
[/code]
to get $ \ce{NH2}$  :)

Rohit I don't use Chem Package, I use only Latex for even chemistry, Its pretty convenient actually considering you can use subscript, superscript anyway.

[hide="Soln"]
I think this should be the answer. However I am not very satisfied with my reasoning on why this should be the answer and this is a beautiful example of what Mr.Ravi would call Blind Problem Solving though evidently, it is much more common for (the general public) in physics than in chem.

Here's the algo

Anyway, since this is a buffer
$ P_{OH}\equal{} PK_b \plus{} log \frac{Salt}{Base}$
Using $ P_{OH}$, we find $ [OH]^\minus{}$

Now obviously since we know KSp's of the given hydroxides, divide it by the square and cube of OH conc to get the conc of cation. However I am not really sure if this method is right, so better verify the numerical value.

[/hide]

Rohit, considering you have posted when he says Physics thalai.
QED.

that's the only different thing da in the chem package and its better  :D

[quote="Euclidean Geometer"]i think when ever there is such a problem on several equilibria act together,
the best and easiest method is the charge balance method.
i guess the solution cms off in a SINGLE LINE![/quote]

Would you mind throwing some light on this.

[quote="Cyber-Shreyas"]Rohit I don't use Chem Package, I use only Latex for even chemistry, Its pretty convenient actually considering you can use subscript, superscript anyway.

[hide="Soln"]
I think this should be the answer. However I am not very satisfied with my reasoning on why this should be the answer and this is a beautiful example of what Mr.Ravi would call Blind Problem Solving though evidently, it is much more common for (the general public) in physics than in chem.

Here's the algo

Anyway, since this is a buffer
$ P_{OH} \equal{} PK_b \plus{} log \frac {Salt}{Base}$
Using $ P_{OH}$, we find $ [OH]^ \minus{}$

Now obviously since we know KSp's of the given hydroxides, divide it by the square and cube of OH conc to get the conc of cation. However I am not really sure if this method is right, so better verify the numerical value.

[/hide]

Rohit, considering you have posted when he says Physics thalai.
QED.[/quote]
unfortunately this method doesnt seem to work. :noo:
[quote="euclidean geometer"]
i think when ever there is such a problem on several equilibria act together,
the best and easiest method is the charge balance method.
i guess the solution cms off in a SINGLE LINE![/quote]
actually this makes more complicated, coz you have so many ions in solution.

@varun: Why doesn't it work?? I don't find anything conceptually wrong in my method though I am not very satisfied with it.

If it is a recent IIT-JEE question, then I am sure you will find the soln somewhere. Which year is it?

i dunno why it doesnt work dats why i posted it here. i have the soln but its really long and invloves solving a cubic eqn. dats why i was asking fer a nice and easy answer.


Here's question.

Prove that among 10 consecutive integers that there is at least one that is not divisible by 2,3,5,7.

Here's another version of the problem:

Prove that among 14 consecutive integers that there is at least one that is not divisible by 2,3,5,7,11.

Finally:

Find thirteen consecutive numbers such that every number in the set is divisible by 2,3,5,7,11 or 13.

I've been able to prove the first part(im not quite sure if it is right) in spoiler:

[hide]First, we have the 10 integers. 5 are odd. 5 are even.  We are looking to maximize the number of integers divisible by either 2,3, 5, or 7.  So. We have the 10 integers.  5 such numbers are divisible by 2.  This leaves all the odd numbers.  For the odd numbers, the maximum number of numbers that are divisible by 3 in the set of five numbers is 2. 5-2=3 numbers left.  This leaves the divisors 5 and 7, which can only possibly divide two of the remaining numbers.  Thus, the answers is 3-2=1[/hide]

This approach, however doesn't get the job done for the second part.

There was a typo near the end of your solution.  When you said "this leaves the divisors 7 and 11", I think you meant "this leaves the divisors 5 and 7".

I didn't understand your very last sentence.  Did you mean this approach does work for the second part, or does not work?

The second part was posed already in another thread (the one about relocating to Wyoming, of all things).  Anyway, I will repeat my hint for it below.

Hint in spoiler: [hide]Prove that among 7 consecutive odd integers, there are at least three neither divisible by 3 nor 5.[/hide]

Solution to second part:

[hide]We know that 7 of these are even. Cross them out. We are left with seven consecutive odds, say ABCDEFG. Thus, either A,D,G are divisible by 3, or B,E or C,F are divisible by three. (The last two cases are symmetric).

Case 1: A,D,G divisible by 3. At most one is divisible by 5, so we have three numbers left over and 7 and 11 to deal with. So one must be divisible by none of the integers.

Case 2: B,E divisible by 3. We can have two integers divisible by 5, leaving three to be reckoned with. Again, this case produces one number not divisible by any of the numbers listed.

[/hide]

Solution to the last part:

[hide]Playing off my solution earlier, we have 7 evens, 6 odds (ABCDEF). B,E must be divisible by 3 (G is invalid, it is the last term), so A, F are divisible by 5. This leaves C to be divisible by 7 and E to be divisible by 11. So what is the smallest solution to 11x - 7y = 2? x = 4 and y = 6 works, but x and y cannot be even, so x = 11 and y = 17. This gives us integers 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126.

[/hide]

cool, someone else who took maml2

how do you think you scored?? ( i think i got roughly 75)

i got all 3 parts of that question during the test... however my solution to part 2 was the complete opposite of elegant, but i'll still get credit for it as there was no BS involved.


geometry
vector
trigonometry
analytic geometry
graphing lines
slope

2007 Girls Math Olympiad #5
a geometry problem

We can draw in M, the midpoint of AC, draw DM and angle chase to find that we need that A, D, E are collinear. There are also a bunch of similar triangles...maybe that will help.

Let's not be synthetic...

[hide="vectors"]

Scale the figure so that $ CD \equal{} AD \equal{} 1$

Let $ D$ be the origin.  Let $ BD \equal{} x$.  Let $ \angle CDB \equal{}\theta$.

$ E \equal{} (B\plus{}C)/2$ while $ F \equal{}\frac{2C\plus{}A}{3}$

$ E\cdot (E\minus{}F) \equal{}\frac{B\plus{}C}{2}\cdot (\frac{B\plus{}C}{2}\minus{}\frac{2C\plus{}A}{3})$

To prove this is zero, it suffices to consider $ (B\plus{}C)\cdot (3(B\plus{}C)\minus{}2(2C\plus{}A)) \equal{} (B\plus{}C)\cdot (3B\minus{}C\minus{}2A)$

This is $ 3 |B|^{2}\minus{}|C|^{2}\plus{}2B\cdot C\minus{}2C\cdot A\minus{}2B\cdot A$

In terms of $ x$ and $ \theta$, we get $ 3x^{2}\minus{}1\plus{}2x\cos(\theta)\minus{}2\cos(120)\minus{}2x\cos(240\minus{}\theta)$

Simplifying we get $ (x)(3x\plus{}2(\cos(\theta)\minus{}\cos(240\minus{}\theta)))$

So we only need to show that $ 3x\plus{}2(\cos(\theta)\minus{}\cos(240\minus{}\theta))$ is equal to $ 0$.

By the sine law on $ \triangle ABD$, $ x \equal{}\frac{\sin(\theta\minus{}120)}{\sin(60)}$

If we plug this into the expression above, we get everything in terms of $ \sin(\theta)$ and $ \cos(\theta)$.  After some simplification, everything cancels easily and we get $ 0$.
[/hide]

This shouldn't be too hard with regular coordinates...we can assign coordinates to A, B, C, then it's trivial to find E, F, M, and we can use slopes and 30-60-90 triangles to get D.


limit
trigonometry
logarithms
real analysis
real analysis unsolved

Let $ a_n$ be a bounded sequence of real numbers and let $ \displaystyle\lim_{n\to\infty}(a_n\minus{}2a_{n\plus{}1}\plus{}a_{n\plus{}2})\equal{}0$.
Show that  $ \displaystyle\lim_{n\to\infty}(a_n\minus{}a_{n\plus{}1})\equal{}0$.

Let $ a'_n = a_{n+1} - a_n$ and $ a''_{n} = a'_{n+1} - a'_n = a_{n+2} - 2a_{n+1} + a_{n}$.

Suppose $ a'_n$ does not converge to $ 0$. Then there is $ \epsilon > 0$ such that for all $ N$, there is $ n$ such that $ |a'_n| \geq \epsilon$. Without loss of generality, assume there are infinitely many $ n$ such that $ a'_n \geq \epsilon$. Enumerate all such $ n$ in increasing order and denote them by $ n(k)$.

If $ a'_n$ contains only finite negative terms, then there is $ N$ such that $ a'_n \geq 0$ for all $ n \geq N$. This indicates that

\begin{eqnarray*}
a_{n(N+k)+1} & = & a_{n(N+1)} + \{ a_{n(N+1)} + a'_{n(N+1)+1} + \cdots + a'_{n(N+k)-1} + a'_{n(N+k)} \}\\
& \geq & a_{n(N+1)} + \{ a'_{n(N+1)} + a'_{n(N+2)} + \cdots + a'_{n(N+k-1)} + a'_{n(N+k)} \}\\
& \geq & a_{n(N+1)} + k\epsilon,
\end{eqnarray*}

so $ a_n$ cannot be bounded. This is a contradiction! Therefore $ a'_n$ contains infinitely many negative terms.

Temporaily fix a positive integer $ m$, then there is $ N$ such that $ |a'_{n+1} - a'_{n}| = |a''_n| < \epsilon / m$ for all $ n \geq N$.

Let $ k_0$ be the smallest $ k > n(N)$ such that $ a'_k < 0$. Then by our choice of $ N$, there is $ n(N) \leq k_m < k_{m-1} < \cdots < k_1 < k_0$ such that $ (i-1)\epsilon / m \leq a'_{k_i} \leq i \epsilon / m$ for each $ i = 1, 2, \cdots, m$. Then

\begin{eqnarray*}
a_{k_0}
& = & a_{n(N)} + \{ a'_{n(N)} + a'_{n(N)+1} + \cdots + a'_{k_0 - 1} \}\\
& \geq & a_{n(N)} + \{ a'_{k_m} + \cdots + a'_{k_1} \}\\
& \geq & a_{n(N)} + (\epsilon / m )\{ (m-1) + \cdots + 0 \} \\
& = & a_{n(N)} + (m-1) \epsilon / 2.
\end{eqnarray*}

This shows that $ a_n$ cannot be bounded by choosing sufficiently large $ m$. So $ a'_n$ must converge to $ 0$.

Thanks!!!
Can we say that the series $ \displaystyle\sum_{i\equal{}1}^{\infty}a'_n$ is convergent.

No, since  for $ a'_{n}\equal{}\frac{1}{n}$ for $ n\equal{}1,...$ serie diverges

If $ a'_n\equal{}\frac{1}{n}$, then the sequence $ a_n$ isn't bounded.So $ a'_n$ can't be $ \frac{1}{n}$

$ a_n \equal{} \sin \log n$ gives a counter-example. $ \frac{d}{dx} \sin \log x \equal{} \frac{\cos \log x}{x} \to 0$ as $ x \to \infty$. Then by MVT, $ a'_n \to 0$ as $ n \to \infty$.


There are one-way flights between $100$ cities of a country. It is possible to fly  starting from the capital city and visiting all other $99$ cities and returning again to the capital city. Let $ N$ be the smallest number of flights inorder to form such a flight combination. Among all flight combinations (satisfying previous condtions), $ N$  can be at most ?

$\textbf{(A)}\ 1850  \qquad\textbf{(B)}\ 2100  \qquad\textbf{(C)}\ 2550  \qquad\textbf{(D)}\ 3060  \qquad\textbf{(E)}\ \text{None}$

To decrease the lost in translation, i will give a small example (in fact the original text is also complex to understand, this is what i understood):

Consider a country with $ 1\Rightarrow 2 \Rightarrow 3 \Rightarrow 1$. In this situation the only $ N\equal{}3$.

Consider a country with $ 1 \Leftrightarrow 2 \Leftrightarrow 3$ and $ 1\Rightarrow 3$. $ 1\Rightarrow 2 \Rightarrow 1 \Rightarrow 3 \Rightarrow 2 \Rightarrow 1$ is a possible path with $ N\equal{}6$. But it is not the smallest because there is a path satisfying above condition such as $ 1\Rightarrow 2 \Rightarrow 3 \Rightarrow2 \Rightarrow1$. This yields for this country-path combination $ N\equal{}5$.

We dont think all combinations yet but current $ max(N)\equal{} max(3,min(6,5)) \equal{} 5$.

We want to find the country-path combination which makes $ N$ the biggest.


Factorize   the     polynomial     in    x

$(t^2+1)x^3-(t^2-1)x^2-(t^2+1)x+(t^2-1)$

please solve this sum .i have been trying this a lot of days.thanks


integration
abstract algebra
probability
function
probability and stats

prove that any true random walk in $ \mathbb{Z}^3$ (true in the sense that the random walk does not always stay in the same hyperplane) is transient.

Hi Alekk,

By embedding the $ \Z^3$ random walk in a 3-dimensionnal Brownian Motion,  is it not the same as proving that the 3-dim brownian motion is transient ?

you mean by using something like the Donsker invariance principle ? How would you make this more precise ?

Well I'm sorry Alekk,

I was in fact referring to something I've read in "Rogers and Williams" 's Book (Vol 1) some time ago

I can remember that they use an embedding trick to prove various results by finding the right successive stopping times of a Brownian Motion to get the distributions of any random walk ( or may be it's the increments) and yes I think that there's a link with Donsker's Invariance Principle, but as I do not presently have this reference at hand I just mention it hoping this might help

see ya

Otherwise there's a proof based on hte following fact :
if $ S_n$ is random walk over $ \mathbb{Z}^d$ of law $ \nu$ the points of $ \mathbb{Z}^d$ are recurent iff
$ lim_{r\to1}\int_{[0,2\pi[^d}\mathcal{R}e[\frac {1}{1 \minus{} r\hat{\nu}(t)}]dt \equal{} \plus{} \infty$
Then it's only a matter of proving that if $ d > 2$ then the previous limit is finished.

You can find the proof page 107 in the pdf document "Cours de probabilités Approffondies 2005-2006" at Université Pierre et Maris Curie by Lacroix and Priouret in the Appendix as I know that you can read French (sorry for the others non-francophones readers)

[quote="fedja"]One quick question: do I understand it right that by a random walk you mean just any process with i.i.d. increments?[/quote]
yes Fedja.

Hi Alekk

I took a closer look to the embedding I was refering to and it happens to be disappointing. It's only Skohokhod Embedding and it does not bring anything new to the problem, and the possible link to DIP was only that the Skohokhod Embedding can be used during its proof.

Nevertheless I would have found it nice to use transience of 3-dim BM to get the proof, I simply don't know how to acheive this.

see ya

PS:

Is there something like scaling and speed measure for random walks ?

Let $ P$ denote the kernel of the walk.  We can find $ \varepsilon > 0$ such that $ P \equal{} \varepsilon P' \plus{} (1 \minus{} \varepsilon)Q'$ such that $ P'$ has mean $ 0$ and is finitely supported.  By the LCLT you can show that the probability of returning to $ 0$ after $ n$ steps under $ P'$ (a well behaved kernel!), which we denote by $ P'_n(0)$, is at most $ c/n^{3/2}$.  It's fairly straightforward to show that the same bound (possibly with a different constant) holds for $ P_n(0)$.  But by the usual criterion, this is enough to show transience of SRW on $ \mathbb{Z}^3$, since $ \sum_n cn^{ \minus{} 3/2} < \infty$.

In fact, one can probably argue directly, since if $ P$ does not have mean $ 0$ or finite variance, it should be transient.  Then perhaps one can use some cruder version of the CLT to get a similar bound to the above.  But fooling around with the LCLT or characteristic functions is tricky and it's easier in practice to look at some sort of book which has fairly sharp estimates for these things :)


For a positive integer $ n$, denote $ \frac {10^n \minus{} 1}{9} \equal{} \overbrace{111\cdots 111}^{\text{n's 1}}$ by $ \boxed{n}$. For example $ \boxed{1} \equal{} 1,\ \boxed{2} \equal{} 11,\ \boxed{3} \equal{} 111$.

(1) Let $ m$ be a non negative integer. Prove that $ \boxed{3^m}$ is divisible by $ 3^m$ and indivisible by $ 3^{m \plus{} 1}$.

(2) Prove that $ n$ is divisible by 27 is the necessary and sufficient condition for which $ \boxed{n}$ is divisible by 27.

[b]Generalization (lifting the exponent):[/b]  Let $ a, k, n$ be positive integers with $ n > 1$ and let $ l$ be a non-negative integer.  Let $ p$ be a prime.  Suppose that $ p^k \parallel a \minus{} 1$ and $ p^l \parallel n$.  Then $ p^{k\plus{}l} \parallel a^n \minus{} 1$ unless both $ p \equal{} 2$ and $ k \equal{} 1$.

The desired result follows by setting $ p \equal{} 3, k \equal{} 2, l \equal{} 3$.

[quote="kunny"]
(1) Let $ m$ be a non negative integer. Prove that $ \boxed{3^m}$ is divisible by $ 3^m$ and indivisible by $ 3^{m \plus{} 1}$.
.[/quote]
It seems too obvious,isn't it? Are you mistaken in something,kunny? :wink:


problem:

((1/3)*t^3)+t=s

find t in terms of s?

answer should just be t in terms of just s.

Don't post the same problem twice.

http://planetmath.org/encyclopedia/CubicFormula.html


probability
conditional probability
probability and stats

Let X0, X1... be a finite irreducible aperiodic Markov Chain started from its stationary distribution, let x in the States space. Let N_n (x) = the summation of 1(Xk=x) from k=1 to n. SHow that [N_n (x)]/n converges to pi(x) in probability.

where pi(x) is its stationary distribution.

I was trying to do this question by chebyskev's inequality.
I showed that E([N_n(x)]/n) converges to pi(x). But I'm not sure how to show that the Var([N_n(x)]/n) converges to 0.

May someone helps?

Looks we're taking the same class.... :|

for this problem, u can expand the square thing and take expectation one by one. most terms are independent by markov chain

I think it's more complicated than that.  The events $ X_i \equal{} x$ and $ X_j \equal{} x$ aren't independent in a Markov chain, since the transition probabilities aren't all equal to each other.  Instead what happens is that $ X_i \equal{} x$ and $ X_j \equal{} x$ are "nearly independent" if $ i$ and $ j$ are very far apart.  The hard part is to quantify this near independence and use it to estimate the variance.

More precisely, let $ Y_k \equal{} P(X_{m \plus{} k} \equal{} x | X_m \equal{} x)$ (this is well defined since we're in a Markov chain).  $ Y_k$ converges to $ \pi(x)$, and we have
\[ E(N_n^2) \equal{} \sum_{i \equal{} 1}^n \sum_{j \equal{} 1}^n P(X_j \equal{} x \wedge X_i \equal{} x)
\]
We now split the sum up into diagonal terms ($ i \equal{} j$) and off diagonal terms.
\[ \equal{} \sum_{i \equal{} 1}^n P(X_i \equal{} x) \plus{} 2\sum_{i \equal{} 1}^n \sum_{j \equal{} i \plus{} 1}^{n} P(X_j \equal{} X_i \equal{} x)
\]
Using the definition of $ Y_k$ and conditional probability, this is
\[ \sum_{i \equal{} 1}^n P(X_i \equal{} x) \plus{} 2\sum_{i \equal{} 1}^n \sum_{k \equal{} 1}^{n \minus{} i} P(X_j \equal{} x)Y_k
\]
To control the double sum, we use a standard trick from analysis.   Fix $ \epsilon > 0$.  We know there's some $ k_0$ for which $ k > k_0$ implies that $ Y_k$ is within $ \epsilon$ of $ \pi(x)$.  Now split the sum up into two parts.

(1) Those terms with $ k < k_0$.  You should argue that the sum of these terms is much less than $ n^2$.

(2) Those terms with $ k \geq k_0$.  You should argue that those terms have sum approaching $ \pi(x)^2 n^2 \plus{} error$, where $ \frac {error}{n^2}$ goes to 0 as $ \epsilon$ goes to 0.


$\triangle{ABC}$ can cover a convex polygon $M$.Prove that there exsit a triangle which is congruent to $\triangle{ABC}$ such that it can also cover $M$ and has one side line paralel to or superpose one side line of $M$.

Move the sides of the triangle $ABC$ closer to the polygon, keeping them parallel to themselves, until they all touch the border of the polygon.

If one of them intersects the border along a non-degenerate segment, we're done: that side was parallel to a side of the polygon to begin with. Otherwise, the new sides $BC,CA,AB$ meet the border of the polygon $\mathcal M$ in the three points $X,Y,Z$ respectively.

Now start rotating the lines $BC,CA,AB$ at the same time, with equal angular velocities, around $X,Y,Z$ respectively. Eventually, we'll reach a position in which one of the lines will meet the border of the polygon along a non-degenerate segment. The triangle we've obtained now is similar to the initial one, but possibly smaller. Enlarge it (move its sides away from the polygon, keeping them parallel to themselves) until it reaches the initial size. This last triangle has the desired properties.

You need to find a point to rotate,and the point is the intersection point of two of the vertical lines from $X,Y,Z$


function
probability
trigonometry
real analysis
real analysis unsolved

Let $ f$ be continuous function on $ [a,b]$,Prove or disprove the measure of the following set is $ 0$:
$ D\equal{}\{x|f(x)\equal{}0,\forall \delta>0,\inf \{f(y)|y\in (x\minus{}\delta,x\plus{}\delta)\}<0\}$

Let $ K$ be a [url=http://en.wikipedia.org/wiki/Smith-Volterra-Cantor_set]Cantor-type set of positive measure[/url], and $ f(x) \equal{} \minus{} \mathrm{dist}(x,K)$.

Additional problem,Can such $ f$ cross the $ 0$ point in a positive measure?
(I suppose you know the meaning by "cross")
It is like an imc problem.

Depends a lot of what you mean by "cross". Alas, I don't know what this words means in this particular context (or rather I see too many possible meanings that lead do different answers).

Or I am sorry,
I mean $ f$ cross the  $ 0$ at $ x$ by $ x$ such that:
(1)$ f(x) \equal{} 0$
(2)$ \forall \delta > 0,\inf \{f(y)|y\in (x \minus{} \delta,x)\} < 0,\sup \{f(y)|y\in (x,x \plus{} \delta)\} > 0$ or
$ \forall \delta > 0,\inf \{f(y)|y\in (x \minus{} \delta,x)\} > 0,\sup \{f(y)|y\in (x,x \plus{} \delta)\} < 0$
Do you think the defination of "cross" reasonable?

With this definition (I assume that $ \inf$ and $ \sup$ in the last line should be exchanged) all you need is to modify mlok's example a little bit introducing signs on complementary intervals to the Cantor-type set $ K$ of positive measure (let's say, randomly). Then all but countably many points of $ K$ are accumulation points of those intervals from both sides, so, with probability $ 1$, almost each point of $ K$ is a crossing point. An explicit construction is also not very hard.

Sorry,I should exchange the $ \inf$ and $ \sup$ in the last line.
Thank you,fedja,does your inplicit example use the measure of the function space?I knew few about it.
Can you show me an explicit example?

An explicit example could be constructed by replacing each triangle in the graph of $ \mathrm{dist}(x,K)$ by one period of a sinusoidal wave of the same height.

But How can show the height can have different sign in the consecutive interval?

It will even have different signs within the same interval.

Oh,Yes,I'm sorry.But how to see such construction is reasonable?
Just define the function(sinusoidal wave of the same height) on the open interval which has been dropped by the construcion of cantor type set consecutively? If like this,I think all the crossing point in the same in the same interval is still countable.
Am I wrong :blush: ?

With your definition of crossing, every point of $ K$ will be a in the cross set now, because in any neighborhood you can find a full sine wave, which has both positive and negative parts.

Oh,Yes.You are right,Thank you.I Have just thought the cross point in the interval maybe have a positvie measure but haven't thought the point in $ K$. :blush:
So let's disscus the problem of your definition.

I think that with my definition the set of crossing points is at most countable. You can try to associate an open interval to every crossing point so that all such intervals are disjoint.

[quote="mlok"]Another definition of "crossing": for any $ x$ such that $ f(x) \equal{} 0$ there exists $ \delta$ such that $ f(x \plus{} \delta_1)f(x \minus{} \delta_2)\le 0$ for all $ \delta_1,\delta_2\in [0,\delta]$. We still assume that the set $ f^{ \minus{} 1}(0)$ has empty interior. Can it have positive measure?[/quote]
Ok,Do you mean if $ x$ is a crossing point,then no other crossing point in the interval $ (x \minus{} \delta,x \plus{} \delta)$?
It's quite nice,and not hard to prove :)


Ring Theory
number theory
relatively prime
superior algebra
superior algebra unsolved

a)Prove that $\mathbb{Z}[\sqrt{-5}]$ has class number $2$.
b)Using a), show that $x^{2}+5=y^{3}$ has no solutions in rational integers.

a) $ (1\plus{}\sqrt{\minus{}5})(1\minus{}\sqrt{\minus{}5}) \equal{} 2\cdot 3$ shows that it is not UFD, thus not PID, thus class number $ \geq 2$. The Minkowski bound gives class number $ \leq\frac{2!}{2^{2}}\cdot\frac {4}\pi\sqrt{20}\leq 2.85$, the result follows.

b) Let $ x,y$ be solutions, then $ (x\plus{}\sqrt{\minus{}5})(x\minus{}\sqrt{\minus{}5}) \equal{} (y)^{3}$ as elements and as ideals. The ones on the left are coprime (take some modulo considerations on the equation to get this, don't want to do this now), thus $ (x\plus{}\sqrt{\minus{}5}) \equal{}\mathfrak a^{3}$ in ideals.
But $ \mathfrak a$ is principle ideal, since $ 0 \equal{} (x\plus{}\sqrt{\minus{}5}) \equal{}\mathfrak a^{3}\equal{}\mathfrak a$ in the class group (seen as $ \mathbb{Z}/(2)$).
Thus $ \mathfrak a \equal{} (a) \equal{} (b\plus{}c\sqrt{\minus{}5})$, giving (all units are $ \pm 1$, the $ \minus{}1$ case can be directly reduced to the other one) $ x\plus{}\sqrt{\minus{}5}\equal{} (b\plus{}c\sqrt{\minus{}5})^{3}\equal{} (b^{3}\minus{}15bc^{2})\plus{}(3b^{2}c\minus{}5c^{3})\sqrt{\minus{}5}$, yielding $ c(3b^{3}\minus{}5c^{2}) \equal{} 3b^{2}c\minus{}5c^{3}\equal{} 1$, thus $ 3b^{3}\minus{}5 \equal{}\pm 1$, impossible.

PS: Here an elementary proof:

We easily see that $ y\equiv 1\mod 4$.
Now $ x^{2}\plus{}4$ has no prime factors $ p\equiv 3\mod 4$ (otherwise $ \minus{}1$ would be a square $ \mod p$) and if we consider the factoring $ x^{2}\plus{}4 \equal{} y^{3}\minus{}1 \equal{} (y\minus{}1)(y^{2}\plus{}y\plus{}1)$, we get that $ y^{2}\plus{}y\plus{}1$ has also no such factors, thus is sum of two squares (we need only that primes $ \equiv 1\mod 4$ are of that type and the known identity $ (a^{2}\plus{}b^{2})(c^{2}\plus{}d^{2}) \equal{} (ac\minus{}bd)^{2}\plus{}(ad\plus{}bc)^{2}$, so we can avoid some parts of the full two-squares-theorem's proof), contradicting $ y\equiv 1\mod 4$ (which gives $ y^{2}\plus{}y\plus{}1\equiv 3\not\equiv a^{2}\plus{}b^{2}\mod 4$).

I didn't understand why $ y\equiv 1 \bmod{4}$.
"$ x^2 \plus{} 4$ has no prime factors" where did this come from?

Look at the equation $ \mod 4$ and consider all cases for $ x$.
And the fact that $ x^2 \plus{} 4$ has no prime factors $ p \equiv \minus{} 1 \mod 4$ comes from the simple fact that this would mean $ x^2 \equiv \minus{} 4 \mod p$, thus $ \left( \frac x2 \right)^2 \equiv \minus{} 1 \mod p$. And the latter is impossible for $ p \equiv \minus{} 1 \mod 4$.

By the way, we don't need to prove that $ y^2 \plus{} y \plus{} 1$ is a sum of two squares, it suffices to prove that a product of $ 2$'s and primes $ \equiv 1 \mod 4$ is never $ \equiv \minus{} 1 \mod 4$, being much more easy.

Thats a square root of $ \minus{}5$  :wink:

And you may want to read the "PS:" of the second post...

[quote="ZetaX"]thus $ \left( \frac x2 \right)^2 \equiv \minus{} 1 \mod p$. And the latter is impossible for $ p \equiv \minus{} 1 \mod 4$.
.[/quote]

why, please?

I can't prove this statement, some one can prove it?

$ p \in P$, $ p \equal{} 4k \minus{} 1$ Then $ p$ doesn't divide$ x^2 \plus{} 4$

That's because $ \minus{}1$ is not a square mod $ p$.

[quote="berti"]That's because $ \minus{} 1$ is not a square mod $ p$.[/quote]

yes that was my question

why a square can't be -1 mod p???

with p=-1 mod 4

Assume that $ x^2 \equiv \minus{}1 \mod p$. Then $ \minus{}1 \equal{} (\minus{}1)^\frac{p\minus{}1}{2} \equiv (x^2)^\frac{p\minus{}1}{2} \equal{} x^{p\minus{}1} \equiv 1 \mod p$, the first equation by $ p \equiv 3 \mod 4$, the last one by little Fermat. But $ \minus{}1 \equiv 1 \mod p$ is contradictory.

[quote="ZetaX"] $ x^{p \minus{} 1} \equiv 1 \mod p$, the first equation by $ p \equiv 3 \mod 4$, the last one by little Fermat. But $ \minus{} 1 \equiv 1 \mod p$ is contradictory.[/quote]

$ x^{p \minus{} 1} \equiv 1 \mod p$

we need x and p are relatively prime!

[quote="HilbertThm90"]Uh, $ x$ and $ p$ are relatively prime because $ p$ is a prime.[/quote]

p might divide x !!!!

$ x$ and $ p$ are relatively prime because $ p | x^2 \plus{} 1 \implies \gcd(p, x^2) \equal{} 1$.


Let $ m$ and $ n$ denote the greatest and least positive three-digit
multiples of 7, respectively. What is the value of $ m \plus{} n$ ?

The greatest is $ m\equal{}994\equal{}142(7)$ and the least is $ n\equal{}105\equal{}7(15)$ and $ 994\plus{}105\equal{}\boxed{1099}$.


What is that largest positive integer $n$ for which $n^{3}+100$ is divisible by $n+10$?


[b][hide="3"][/b]in $\mod {12}$ we have $2+11=1$

so, maybe in a clock?[/hide]

[hide="2"]
English?
[/hide]

Were these from a school planner thing?  I have the same brainteasers in my school planner.

[quote="dogseatcheese"]1.It's impossible to survive longer than one week without drinking water; yet Maia managed a ten-day desert crossing without finding water or taking any with her-how? [hide="BAD ANSWERS"]She drank juice, a coconut etc.[/hide]

2."Therein lies the secret!" proclaimed Sherlock. "THere is a 7-letter word in Englsih that contains 10 words without rearranging any of its letters." What is the word?

3. When does 2 plus 11 equal 1?[/quote]

[hide]
1. Maia is camel or something? not sure
2. Therein (there, in, here, ere, rein, the, her)
3. In a weird world? [/hide]

[quote="dogseatcheese"]Yes I did get these from my planner. :D
I guess that 1 and 3 are Camel and Clock. However, I am not sure about 2.[/quote]

What does being a camel have anything to do with it?  You stated that it's impossible without drinking water, and Maia wasn't allowed to carry water...

If something may not carry water, it may not have any water in it...so I was thinking along the lines of an inanimate object, such as a rock, or speck of dust...

[quote="dogseatcheese"]1.It's impossible to survive longer than one week without drinking water; [/quote]Also, this refers to the human, I think

1.[hide]was maia a camel?[/hide]

and did u get this from a Discover planner/agenda?
i no that that #3 was there, as i saw it the other day.

Read the rest of the topic, both of your points have been made/discussed...

Maybe Maia didn't survive the crosing, but still made it across.

Maybe she kept water in her mouth, and drank it after 5 days (see? she had water once every five days).

Maybe she crossed the desert quicker than a normal person!

Maybe it randomly rained?

Maybe Maia means "ocean" or something...

[quote="dogseatcheese"]1.It's impossible to survive longer than one week without drinking water; yet Maia managed a ten-day desert crossing without finding water or taking any with her-how?[/quote]

Just a few clarifications: Is Maia a human? Is "drinking" used as a participle? Is "crossing" used as a noun? Was Maia alive after crossing the desert? Did she cross the desert in ten Earth days? Although she didn't find water or take any with her, did she somehow obtain it in the desert?

Thanks!

[quote="dogseatcheese"]I just wrote what ever was on the Brainteaser in my DISCOVER planner. However, I believe that the 7 days without water=death was meant for humans. I think that Maia was a camel and she survived without bringing or drinking water. (Water in the literal sense as in WATER and not water like fat or a camel's hump). Yea.......however.....number 2 is very difficult......and well....it is easy to understand. :)[/quote]

2 is difficult??? I got it as soon as I read the first half of the question. I just had to check to see if my answer was correct.

Wait...so the answer to 2 is Therein? but i can't find 10 words....

okok. I made a few riddles for you.

What 5 letter word becomes shorter when you add 2 letters to it?
Is it possible to put 2 things into a box and only have one thing in it?
When does 1+1=0?
Can you unfold a piece of paper 5 times and keep the area of the paper the same?

[hide="2"]Yes, if one is a hole.[/hide]
[hide="4"]Yes, it doesn't matter how it's folded, the area is the same.[/hide]
[hide="10 words"]therein, herein, here, in, there, the, he, her, rein, ere[/hide]

got only 2 wrong.
I put [b]Nothingness[/b] in it!

for the area, isnt the area of a paper the same unless u cut it


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a finite group of order $n$.

As a consequence of Cayley's Theorem, it is well know that $G$ is isomorphic to a subgroup of $\mathrm{GL}_n(\mathbb{Z})$.

Here a two open questions:

[list] 1. Is $G$ is isomorphic to a subgroup of $\mathrm{SL}_n(\mathbb{Z})$? (I think the answer is yes, but how to prove it?)

2. Is there an integer $m<n$ such that $G$ is isomorphic to a subgroup of $\mathrm{GL}_m(\mathbb{Z})$ (same for $\mathrm{SL}_m(\mathbb{Z})$)?

[/list]
Thanks,
JC

Your first question is ok, I had it on ens exam this year (shame on me, I gave a wrong solution during the examination). The idea is that $S_{n-1}$ can be seen as subgroup of $GL_n(Z)$: make any permutation act on the basis $(e_1-e_2,...,e_1-e_n)$ on the hyperplane of equation $x_1+...+x_n=0$. After that, everything is clear.

Thanks for your answer harazi.

I understand that $\mathfrak{S}_{n-1}$ can be viewed as a subgroup of $\mathrm{GL}_n(\mathbb{Z})$.

But I don't understand how you show that $G$ is isomorphic to a subgroup of $\mathrm{SL}_n(\mathbb{Z})$.

Could you or someone who understands elaborate please?

You can find here ftp://ftp.drimm.u-bordeaux1.fr/UPS/files/corriges-2006/pdf/m06lm1e.pdf this exam and you question about embedding in $SL_n(\mathbb{Z})$ is #16 in this paper. Here ftp://ftp.drimm.u-bordeaux1.fr/UPS/files/corriges-2006/pdf/m06lm1ca.pdf are the solutions to the exam questions.

Thanks eugene. It's in French but I got it  ;)

Any idea about my second question?

Your second question seems really difficult and it's highly improbable to have a nice answer. It depends a lot on the group itself. Just take a cyclic group for example: then of course one can see it as a subgroup of $GL_m(Z)$ with $m<n$.

I agree with you harazi.

If $G$ is supposed commutative we might have a "nice" answer :?:


category theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ (C_*, \partial)$ be a chain complex of free abelian groups such that $ H_n(C_*) \equal{}0$ for all $ n>0$.   Let $ G\equal{}H_0(C_*)$.   Suppose there is a functor $ F:$[b]Ab[/b] $ \rightarrow$ [b]Ab[/b] which satisfies that $ F($coker $ f)\equal{}$coker$ F(f)$, for all morphism $ f$.  Then

[b]1.[/b]  Show that $ (F(C_*),F(\partial))$ is a chain complex.
[b]2.[/b]  Define $ F_n(G)\equal{}H_n(F(C_*))$.   Show that $ F_n(G)$ does not depend on the choise of $ C_*$.
[b]3.[/b]  Show that every short exact sequence of abelian groups

\[ 0 \rightarrow G \rightarrow G' \rightarrow G'' \rightarrow 0\]

induces an exact sequence

\[ 0 \rightarrow F_1(G) \rightarrow F_1(G') \rightarrow F_1(G'') \rightarrow G \rightarrow G' \rightarrow G'' \rightarrow 0\]

Thank you.

Can it be derived from hypothesis that the functor is additive so that we can say $ F(0)\equal{}0$ and thus we would have the first part?

Second part:

It makes sense to construct augmented chain complexes to get reduced homology, so we build

$ \dots \to C_n \to C_{n\minus{}1} \to \dots \to C_0 \to G \to 0,$

so that all the homology groups are 0. This is a "free resolution" of $ G$. If you check
Hatcher, it's a standard result than any two free resolutions of a group are equivalent
(chain isomorphic up to chain homotopy). Then the chain complexes

$ \dots \to F(C_n) \to F(C_{n\minus{}1}) \to \dots \to F(C_0) \to 0$ are
chain homotopic as well (if $ F$ is additive; you'd either need to check that it is
or use something slightly weaker), and so $ F_n(G)$ is independent of $ C_*$; $ F_n$
is called a derived functor of $ F$, I believe.

[quote="Nukular"]For the first part:

Claim: If $ f : A \to B$ is a zero morphism, then $ F(f) : F(A) \to F(B)$ is too.
Proof:
f is a zero morphism => coker f = B.
 coker(F(f)) = F(coker f) = F(B) => F(f) = 0.

Therefore, since $ \partial^2 \equal{} 0$, $ F(\partial)^2 \equal{} F(\partial^2) \equal{} 0$.[/quote]

If $ A$ is finitely generated, is $ F(A)$ as well ?.  If not, we can't claim that $ Coker(F(f)) \equal{} F(coker f) \equal{} F(B) \Rightarrow F(f) \equal{} 0.$

[quote="Nukular"]Second part:

It makes sense to construct augmented chain complexes to get reduced homology, so we build

$ \dots \to C_n \to C_{n \minus{} 1} \to \dots \to C_0 \to G \to 0,$

so that all the homology groups are 0. This is a "free resolution" of $ G$. If you check
Hatcher, it's a standard result than any two free resolutions of a group are equivalent
(chain isomorphic up to chain homotopy). Then the chain complexes

$ \dots \to F(C_n) \to F(C_{n \minus{} 1}) \to \dots \to F(C_0) \to 0$ are
chain homotopic as well (if $ F$ is additive; you'd either need to check that it is
or use something slightly weaker), and so $ F_n(G)$ is independent of $ C_*$; $ F_n$
is called a derived functor of $ F$, I believe.[/quote]

I have two question about that:

1. What is the map from $ C_0$ to $ G$ ?

2. Also, this only gives us the independence of $ F_n$ when $ n > 0$ , so what about the negative ones?

Thank you

In fact, if $ \partial_0$ is identically zero then I believe than there is a surjection between $ C_0$ and $ G$.


How to type fraction? Should I type something like \frac {1}{2}?

Just finish it with $ \$$ signs. Basically, you do this:
[list]$ \$$\frac{1}{2}$ \$$, which yields $ \frac {1}{2}$[/list]

You can read more on how to type some [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Commands]commands[/url] and [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols]symbols[/url] from the following links provided.  :wink:


MATHCOUNTS
AMC
USA(J)MO
USAMO
ARML
SuMAC
summer program

[quote="dts"]You might be an AoPSer if...

...more than half of the shirts in your drawer came from math competitions.[/quote]

I have around 16 math t-shirts out of 30 I think :D How many do you guys have?

i have about 6 or so math t-shirts (which are probably gone if they didn't shrink to fit a 6-year old :P ) with about 3 from MATHCOUNTS. One is from chapter and one is from state and the last is one that our coaches/teachers bought for us. How many MATHCOUNTS t-shirts do y'all have?

I only have one maths related shirt.  :(  It's an old navy one lol that isred and says 'I *calculator* math'...so like i calculate math or i love math or something...i haven't figured it out yet. :)

I have 6 NYC Math Team shirts alone.  One from Mandelbrot, one from USAMO, one from MOP, several from NYSML and ARML.  So, a dozen plus.  My NYC shirts are getting to be in bad shape, like the one from 1999.  On the other hand, one of the coaches always wears his 1991 shirt to ARML, and it's in worse shape than mine.

ha. i've got a couple, one is the most tasteless color yellow you can imagine and spells out my school's name in formulas. it way rad. my parents hate it because of the color, though.

[quote="h_s_potter2002"]I don't have any math shirts.  My parents don't believe in buying shirts with words on them.  :([/quote]Math shirts usually are not bought but given to you in different competitions.

Everybody needs [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=154752#p154752]some[/url] inspiration for making a new poll.  :lol:

[quote="JBL"]...one from USAMO...[/quote]

How'd you get one from USAMO...?  Did they give out USAMO shirts in the past, and then just stop? :?

I have five - two from mathcounts, and three from mathcamp.

seven or eight I think... 2 from mathcounts chapters, (maybe 1 from states), 1 from nationals; 1 from arml, 1 from sumac, 2 from wpi

too bad they are so gigantic that all except 1 are in my dad's wardrobe :lol:

plus a hat :D

I've got like 5 of them. I actually kinda like them big...

I think I have about six shirts.... I don't know about other people but I really like math shirts...lol :D

I have 2 math t-shirts, both from mathcounts nationals.  (Chapter/state didn't give any out).  4 if you count collared shirts - in '02 and '03 they gave out those as well at nats.

I think I must have over a dozen.  Almost all of them are from Mathcounts.

[quote="eryaman"]can you write out the equation here?[/quote]

Yea sure....*gets shirt*

Here it is...mind you, my latex skills are crappy

$(4 cos \theta, 4 sin \theta)
(\pm 1.5 + 0.25 cos \theta, \pm 1.5 + 0.25 sin \theta)
(2 cos (\theta/2), -0.25 + 2 sin (-\theta/2))$

Graph all: $0 \leq \theta \leq 2\pi$

So the first item in each parentheses is the x, and the second is the y. But you probably already knew that.

I had trouble deciding whether to choose 0 or "all my t-shirts are math shirts". The second choice was related to 0/0, though, so I guessed it was safer to go with plain old 0. Says a lot about me.

[quote="Mariag"]i have 6

3 from NYC team at NYSML and ARML
2 NYSML team
1 NYSML individual
1 HCSSiM shirt
1 AoPS (will get it soon, for USAMTS)[/quote]
Wow a USAMTS winner can't add?
 ;)

I don't have any.  I want one :D

My adress is...  :lol:

I have like 3... and hopefully 4 by the end of this month :)

I got 1 chapter MC shirt and 1 state MC shirt. I got a shirt with  :pi: on it. And it's not just the symbol-the symbol is made up of really tiny numbers! (around like 6,000 digets of pi) I'm hoping to get a shirt that says "there are 10 types of people in the world, those who can read binary and those who can't" :lol:

[quote="JStaver"]I'm hoping to get a shirt that says "there are 10 types of people in the world, those who can read binary and those who can't" :lol:[/quote]
i got that one but it's a little different. it has 11111010101 followed by "there are 10 types of ppl, those who understand this shirt and those who don't" i wish it said binary instead of understand, but it's still cool. got it at college of charleston math meet.
and my number of shirts is continuing to grow. i now have 8 and i'll get another one at arml. in fact, i'm wearing one now. :P

[quote="Slizzel"]I saw a very nice confusing math t-shirt. On his front was written: [size=134][b]2+2=5[/b][/size] (for large values of 2)  :D That's funny !!!  :)[/quote]

I've seen that shirt... I wish I had one  :P

http://www.thinkgeek.com/tshirts/science/60f5/zoom/

Ask and ye shall receive.

Not only is it not free, it's expensive! And there's probably shipping, too! I don't know if it's worth it... :P

of my 6 or 7 math tshirts, my favorite is one with this on it: http://www.toothpastefordinner.com/022703/gang-sines.gif


In the classroom, there is an interactive mandelbrot set if you have java. What does it doo? Can anyone help me on what it does? Please help. :?

I really want to know what to do with the thing.
I still need help........................ :(

I think he means what does the Mandlebrot set do within math. To tell the truth...i'm not sure either!

[quote="Klebian"]Basically, it's just there to tell you if you have java or not.
If you don't see it, then you need to get java.
If you see it and want to know what to do with it, you can select like a section of it and it'll zoom in.
The only real use of it, though (at lest in the classroom), is to tell you if your computer has java.[/quote]

oh. okay... I think. But I still dont really get it.


function
analytic geometry
combinatorial geometry
advanced fields
advanced fields unsolved

[url=http://en.wikipedia.org/wiki/Knaster-Kuratowski-Mazurkiewicz_lemma][b]Knaster-Kuratowski-Mazurkiewicz Theorem[/b] (also known as Knaster-Kuratowski-Mazurkiewicz Lemma)[/url][b].[/b]

Let $ X$ be a Banach space.
Let $ x_1,\ldots,x_n$ be a finite set of points in $ X$. Let $ F_1,\ldots,F_n$ be closed subsets of $ X$.
For any set $ M\subset X$, denote by $ co[M]$ the convex hull of $ M$.
Assume that for any $ 0 < k\leq n$ and any $ 0\leq i_1 < i_2 < \ldots < i_k\leq n$, we have $ co[x_{i_1},\ldots,x_{i_k}]\subset\cup_{j \equal{} 1}^kF_{i_j}$.

Prove that $ \cap_{i \equal{} 1}^nF_i\neq\emptyset$.

It's an easy consequence of [url=http://planetmath.org/encyclopedia/SpernersLemma.html]Sperner's lemma[/url], but here's a proof using the [url=http://en.wikipedia.org/wiki/Brouwer_fixed_point_theorem]Brouwer fixed point theorem[/url].

Let $ S \equal{} (s^1,\ldots,s^n)$ be the $ n \minus{} 1$-dimensional simplex, and let $ f: S\to\mbox{co}(x_i)$ be the function which sends $ \sum\lambda_is^i$ onto $ \sum\lambda_ix_i$, where $ \lambda_i\in[0,1],\sum\lambda_i \equal{} 1$. We'll denote the preimages of the $ F_i$'s by the same names, in order to keep things simple. From now on we forget about the $ x$'s, and work only with $ S$.

Assume there's no point of $ S$ belonging to all sets $ F_i$, and let $ \varphi: S\to S$ be the application which sends $ x\in S$ onto $ \frac 1{\sum d(x,F_i)}(d(x,F_1),\ldots,d(x,F_n))\ (*)$. $ \varphi$ is, by our assumption, well-defined, since there's no $ x$ for which the denominator of the fraction in $ (*)$ is zero. Since $ \varphi: S\to S$ is continuous, it has a fixed point $ x_0$. If $ i_1 < \ldots < i_k$ are the indices of the positive coordinates of $ x_0$ when written as a convex combination of the $ s^i$'s, then the fact that $ x_0$ is a fixed point of $ \varphi$ says that $ d(x_0,F_{i_1}),\ldots,d(x_0,F_{i_k}) > 0$, so $ x_0\in\mbox{co}(s^{i_1},\ldots,s^{i_k})\setminus\bigcup_{j \equal{} 1}^kF_{i_j}$, and this contradicts the hypothesis.


geometry
inequalities
quadratics
geometry proposed

Let $ ABC$ be a triangle with three sides $ a;b;c$ and areas $ S$.Put $ p\equal{}\frac{a\plus{}b\plus{}c}{2}$.Prove that:
$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c} <\frac{p}{S}$

[quote="chien than"]Let $ ABC$ be a triangle with three sides $ a;b;c$ and areas $ S$.Put $ p \equal{} \frac {a \plus{} b \plus{} c}{2}$.Prove that:
$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} < \frac {p}{S}$[/quote]

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} < \frac {p}{S}\Leftrightarrow{h_a \plus{} h_b \plus{} h_c}<{a \plus{} b \plus{} c}$

$ {h_a \plus{} h_b \plus{} h_c}\le{m_a\plus{}m_b\plus{}m_c}<{\frac{b\plus{}c}{2}\plus{}\frac{a\plus{}b}{2}\plus{}\frac{a\plus{}c}{2}}\equal{}a\plus{}b\plus{}c$

I will prove stronger result: $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\leq \sqrt {\frac {3}{4}}\frac {p}{S}$

Take standard substitution for sides of triangle :$ a \equal{} y \plus{} z$ , $ b \equal{} x \plus{} z$ and $ c \equal{} x \plus{} y$.

Then it is easy to find that:$ \sqrt {\frac {3}{4}}\frac {p}{S} \equal{} \sqrt {\frac {3(x \plus{} y \plus{} z)}{4xyz}}$
and our inequality is equivalent to:
$ \frac {1}{x \plus{} y} \plus{} \frac {1}{y \plus{} z} \plus{} \frac {1}{z \plus{} x}\leq \sqrt {\frac {3(x \plus{} y \plus{} z)}{4xyz}}$

But notice that:
$ \frac {1}{x \plus{} y}\leq \frac {\sqrt {z}}{2\sqrt {xyz}}$

After summing smilar inequalities our inequality reduces  to its equivalent:
$ \sqrt {x} \plus{} \sqrt {y} \plus{} \sqrt {z}\leq\sqrt {3(x \plus{} y \plus{} z)}$ which is just Power mean inequality (or inequality between arithmetic and quadratic mean)

Equality holds when $ a \equal{} b \equal{} c$


calculus
integration
trigonometry
calculus computations

Supposed that $ f(x)$ has $ f'(x)$ and for any real numbers $ x,\ y$, $ \int_y^{x \plus{} y} f(t)\ dt \equal{} \int_0^x \{f(y)\cos t \plus{} f(t)\cos y\}\ dt$ holds.

(1) Express $ f(x \plus{} y)$ in terms of $ f(x),\ f(y).$
(2) Find $ f'\left(x \plus{} \frac {\pi}{2}\right)$. Note that $ f\left(\frac {\pi}{2}\right) \equal{} 2.$
(3) Find $ f(x).$

(1)f(x+y)=f(y)cosx+f(x)cosy
(2)ans=-2
(3)f(x+pi/2)=f(pi/2)cosx,next easy.


There are 1150 members of a gym. 1/5 of the males and 1/4 of the females do not swim. A total of 250 members do not swim. How many males do swim?

I'm stuck. Can someone help?

[hide="Hint"]

Try writing two equations for this and solving.

[/hide]

[hide="Solution"]
We can write two equations for this.

Let $ x$ be the number of males in the gym, and $ y$ be the number of females in the gym.

$ x \plus{} y \equal{} 1150$

$ \frac{1}{5}x \plus{} \frac{1}{4}y \equal{} 250$

To solve this, let's use substitution. Take the first equation and write $ y$ in terms of $ x$.

$ x \plus{} y \equal{} 1150$

$ y \equal{} 1150 \minus{} x$

Now we can substitute this value into our second equation. But first, let's clean up that second equation - fractions are messy, annoying, easy to mess up.

$ 20(\frac{1}{5}x \plus{} \frac{1}{4}y) \equal{} 250(20)$

$ 4x \plus{} 5y \equal{} 5000$

Finally we can substitute.

$ 4x \plus{} 5(1150 \minus{} x) \equal{} 5000$

$ 4x \plus{} 5750 \minus{} 5x \equal{} 5000$

$ 5750 \minus{} x \equal{} 5000$

$ x \equal{} 750$

There are $ 750$ males in the gym, and $ \frac{1}{5}$ can't swim. Thus, $ \boxed{600}$ males can swim.[/hide]

[quote="nick42"]yeah...what isabella said. if you still don't get it, guess and check  :D

$ a^2 \plus{} b^2 \equal{} c^2$[/quote]

it would take forever to guess and check unless you were really lucky


Euler function algebra domain logarithms trigonometry calculus

There's a famous equation made by Euler: e^(pi*i)=-1
so:
e^(pi*i)=i^2

Take both sides to the i power

e^(pi*i)i=i^2i

simplifies to

e^(pi*-1)=i^2i

1/e^pi=i^2i

Take the square root of both sides

1/sqr(e^pi)=i^i

Is this right?

Theres actually an infinite number of values for i^i. Very confusing.

All the solutions are e^(2pi*n)*e^-(pi/2).

Speaking of complex numbers, did anyone solve my ``paradox'' about e^(2*i*pi*n)?

There is actually only one value of i^i that's consistent with our normal definition of ln z. It's sort of like the Square Root function. Although x^2 - 4 = 0 has two solutions, -2 and +2; when looked at as a function, :sqrt:4 only has one value, 2. That's because functions, by definition, have to be single valued. So we make a choice. And by convention, we make the square root function positive on the positive reals.

Likewise, we make a choice with the ln function. ln z could equal w + (2pi*n)*i, for any integer n. But so as to make ln z real valued on the positive real numbers we make a choice to let n = 0. This choice is called a Branch Cut.

So to remain consistent with our choice for ln z, i*i can only have one value, e^-(pi/2). But for each other possible choice of ln z, there corresponds a consistent value for i^i, e^-(n+1/2)*pi.

How to handle functions like this, which are naturally multivalued is dealth with by Riemann Surfaces. Basically you change your Domain from the Complex Plane to some other surface which covers the complex plane multiple times, sort of like layers of tape. So two points appear to be in the same place but really they're on different "layers" of the surface. Hence the function is single-valued on the surface. What you're doing when you make a choice in one of these functions is you're "cutting it" so that you restrict the domain to just one of these layers.

Yes, but ln(x) isn't the unique inverse of e^x. Anyway, ln(x) is only defined when x is real. A better inverse of e^z is Log(z), where - :pi:<Arg(z)<:pi:. We can just as easily choose our branch of log(z) to be Log(z)+2:pi:i. Thus all the values of i^i are valid.

It depends on how you're using your notation. Where I come from ln z usually refers to the natural logarithm and log z refers to the common log. ln z by definition is the inverse function of e^z, but ln z and log z are just multiples of each other so they have the same standard branch cut.

It isn't that the other values of i^i are invalid, it's just that they are not consistent with the standard branch of the ln function on the complex plane. Once that decision is made, the value of i^i is determined - it can only have one value. Furthermore, within any cut you can still define the ln of negative and even complex numbers in a consistent way. For example, from i^i = e^-(pi/2) you can compute what the value of ln (i) should be.

Just something weird: i^i = e^ (ln (i^i)) = e^ (i ln i) = cis (ln i). Since jeffrey has already shown this is equal to 1/sqrt(e^Pi), this means cos (ln i) = 1/sqrt (e^Pi) and sin (ln i) =0, i think. Isn't that weird?

Of course, it doesn't really mean anything (as it has been pointed out, ln only works w/ reals, but I still think it's cool.

No that's not right. cis z = e^(iz) = cos z + i sin z, but that does [i]not[/i] mean cos z = Real (cis z) and sin z = Im (cis z) in all circumstances. That rule only applies when z is Real!

Because z = ln (i) is not a real number that rule does not apply. The correct rule is cos z = (e^(iz) + e^(-iz))/2, and sin z = (e^(iz) - e^(-iz))/(2i). When z is pure imaginary, z = ix, that means cos (ix) = cosh (x) and sin (ix) = -i*sinh(x), where cosh and sinh are the hyperbolic cosine and hyperbolic sine functions.

Oh darn sorry.  :oops:

Is there a course where you learn this sort of stuff dealing with complex numbers?

sounds cool. is it college level, bc i'm pretty sure there's nothing remotely like that at the high school.

What year are you, and what high school do you go to?

You can take complex analysis through EPGY, as I did. [url]http://epgy.stanford.edu[/url]. Otherwise you'll probably have to wait until college unless you go to a math magnet school.

Zeta you might enjoy a book called Fourier Series and Integrals by H. Dym and H.P. McKean. It has lots of very interesting topics in it.

I am going to be a freshman this year, and I go to a public school. From what you've said so far, I guess I'll have to wait till college.....unless you could tell me what EPGY is?

It stands for Education Program for Gifted Youth. It allows people to take classes that they wouldn't usually be allowed to take for a few more years in school so that they can get ahead. It's run by Stanford University, but it's a correspondence course, so you don't have to be local to take their classes. They offer classes in many fields, but they have mostly math classes. Check it out. It's a good program.

Hey Mysitic, have you ever thought about the NC School for Science and Math? They offer a course in complex analysis there. It's probably not as complete or in depth as you would get in a college course (and even if you took it there, you might still want to take it again in college). But they offer some interesting classes for those who exhaust the typical high school ciriculum.

I went to the SC School and I loved my experience there. You could also take a class at a local university (if there are any nearby) after you complete BC Calculus.

Well, my school system has always sent kids over to UNC once they finish the high school classes, which is why I never really thought about going to S&M. I thought the only other advantage to going to S&M besides the advanced topics would be a slightly better math team, but they really aren't doing that well recently. So I figured it probably wouldn't be worth it.[/i]

You could also start studying on your own.

sorry nothing connected to this topic as such.
I use [url=http://www.mathlinks.ro/Forum/files/complex_941.rar]this[/url] little DOS program to check for all complex number proofs etc... I like it a lot


inequalities
geometry
geometric transformation
reflection
function
algebra

[quote="can_hang2007"][quote="hxy09"]$ a,b,c,d\ge 0$ and $ abcd \equal{} 1$
Prove that $ \sum \frac {1}{1 \plus{} a \plus{} a^2 \plus{} a^3}\ge 1$[/quote]
This inequality will appear with a nice solution in the up-coming book...[/quote]

I am waiting for it :lol:

[quote="hxy09"]$ a,b,c,d\ge 0$ and $ abcd \equal{} 1$
Prove that $ \sum \frac {1}{1 \plus{} a \plus{} a^2 \plus{} a^3}\ge 1$[/quote]

for $ a,b,c,d>0,abcd\equal{}1$, setting $ a\equal{}e^x, b\equal{}e^y, c\equal{}e^z, d\equal{}e^w$ and $ f(x)\equal{}\frac {1}{1 \plus{} e^x \plus{} e^{2x} \plus{} e^{3x}}$, ineq becomes

$ x,y,z,w \in \mathbb{R}, x\plus{}y\plus{}z\plus{}w\equal{}0$ then

$ f(x)\plus{}f(y)\plus{}f(z)\plus{}f(w) \ge 1.$

we can prove that $ f(x)$ is "half concave and half convex" in $ \mathbb{R}$, so, using "Half concave and half convex theorem", to prove $ f(x)\plus{}f(y)\plus{}f(z)\plus{}f(w) \ge 1$ just need to prove that

$ 3f(x)\plus{}f(\minus{}3x) \ge 1,$

that's equivalent to

$ \frac {3}{1 \plus{} k \plus{} k^2 \plus{} k^3}\plus{}\frac {k^9}{k^9 \plus{} k^6 \plus{} k^3 \plus{} 1} \ge 1$

where $ k\equal{}e^x>0.$

but

$ \frac {3}{1 \plus{} k \plus{} k^2 \plus{} k^3}\plus{}\frac {k^9}{k^9 \plus{} k^6 \plus{} k^3 \plus{} 1} \minus{} 1$

$ \equal{}\frac{(k\minus{}1)^2 (2 k^4\plus{}k^3\plus{}k\plus{}2)}{(k\plus{}1)(k^2\plus{}1)(k^2\minus{}k\plus{}1)(k^4\minus{}k^2\plus{}1)}$

$ \ge 0,$

so ineq holds.

[quote="hxy09"]$ a,b,c,d\ge 0$ and $ abcd \equal{} 1$
Prove that $ \sum \frac {1}{1 \plus{} a \plus{} a^2 \plus{} a^3}\ge 1$[/quote]

I've a very simple proof, is it correct?
By AM-GM, $ a\plus{}a^2\le 1\plus{}a^3$. Thus $ \sum \frac {1}{1 \plus{} a \plus{} a^2 \plus{} a^3}\ge \frac12\sum \frac {1}{1 \plus{} a^3}\ge\frac {4}{1 \plus{} \sqrt{\sqrt{(abcd)^3}} }\equal{}1$. The last step uses twice the well known $ \frac {1}{1 \plus{} a }\plus{}\frac {1}{1 \plus{} b }\ge\frac {2}{1 \plus{} \sqrt{ab} }$.
   :roll:

$ \frac{1}{1\plus{}a} \plus{} \frac{1}{1\plus{}b} \ge \frac{2}{1\plus{}\sqrt{ab}}$ if and only if $ ab \ge 1$.

[quote="spanferkel"][quote="hxy09"]$ a,b,c,d\ge 0$ and $ abcd \equal{} 1$
Prove that $ \sum \frac {1}{1 \plus{} a \plus{} a^2 \plus{} a^3}\ge 1$[/quote]

I've a very simple proof, is it correct?
By AM-GM, $ a \plus{} a^2\le 1 \plus{} a^3$. Thus $ \sum \frac {1}{1 \plus{} a \plus{} a^2 \plus{} a^3}\ge \frac12\sum \frac {1}{1 \plus{} a^3}\ge\frac {4}{1 \plus{} \sqrt {\sqrt {(abcd)^3}} } \equal{} 1$. The last step uses twice the well known $ \frac {1}{1 \plus{} a } \plus{} \frac {1}{1 \plus{} b }\ge\frac {2}{1 \plus{} \sqrt {ab} }$.
   :roll:[/quote]

ineq:
$ \frac {1}{2}\sum \frac {1}{1 \plus{} a^3}\ge\frac {4}{1 \plus{} \sqrt {\sqrt {(abcd)^3}} } \equal{} 1$
dose not hold...
you can try $ a\equal{}b\equal{}c\equal{}3,d\equal{}\frac{1}{27}$.

[quote="hxy09"]$ a,b,c,d\ge 0$ and $ abcd \equal{} 1$
Prove that $ \sum \frac {1}{1 \plus{} a \plus{} a^2 \plus{} a^3}\ge 1$[/quote]
We have also

$ \sum \frac {1}{1 \plus{} \alpha a \plus{} \beta a^2 \plus{} \gamma a^3}\ge 1$

for any nonnegative real numbers $ \alpha,\beta,\gamma$ such that $ \alpha\plus{}\beta\plus{}\gamma\equal{}3$.

[quote="Vasc"][quote="hxy09"]$ a,b,c,d\ge 0$ and $ abcd \equal{} 1$
Prove that $ \sum \frac {1}{1 \plus{} a \plus{} a^2 \plus{} a^3}\ge 1$[/quote]
We have also

$ \sum \frac {1}{1 \plus{} \alpha a \plus{} \beta a^2 \plus{} \gamma a^3}\ge 1$

for any nonnegative real numbers $ \alpha,\beta,\gamma$ such that $ \alpha \plus{} \beta \plus{} \gamma \equal{} 3$.[/quote]

Oh god,how amazing :!:
Could you please post your solution ,my dear friend:lol:

[quote="hxy09"] Oh god,how amazing :!:
Could you please post your solution ,my dear friend:lol:[/quote]
Recently, I sent to Mathematical Reflections an article with two applications of SCF-Theorem (Side Convex Function-Theorem). One of these applications is the following.

Let $ a_1,a_2,...,a_n$ ($ n\ge 4$) be positive real numbers such that $ a_1a_2...a_n\equal{}1$.
If $ \alpha,\beta,\gamma$ are nonnegative real numbers satisfying $ \alpha\plus{}\beta\plus{}\gamma\ge n\minus{}1$, then

$ \sum_{i\equal{}1}^n \frac 1{1\plus{}\alpha a_i\plus{}\beta a_i^2\plus{}\gamma a_i^3} \ge \frac n{1\plus{}\alpha\plus{}\beta\plus{}\gamma}$.

A solution given in the book written by Yang Xuezhi :lol:
The main idea is to prove $ \frac{1}{1\plus{}a^2\plus{}a^4\plus{}a^6}\plus{}\frac{1}{1\plus{}b^2\plus{}b^4\plus{}b^6}\ge\frac{1}{1\plus{}x^3y^3}$
Can anyone create a nicer proof for this two-variable inequality? :)

[quote="hxy09"]A solution given in the book written by Yang Xuezhi :lol:
The main idea is to prove $ \frac {1}{1 \plus{} a^2 \plus{} a^4 \plus{} a^6} \plus{} \frac {1}{1 \plus{} b^2 \plus{} b^4 \plus{} b^6}\ge\frac {1}{1 \plus{} a^3b^3}$
Can anyone create a nicer proof for this two-variable inequality? :)[/quote]
Very nice stuff! You may find the proof is not really 'nice', but it is quite ingenious anyway.

Dear Vasc, can your "Side Convex Function Theorem" also prove that one? And what about the next one?

If $ a,b,c\ge0$, prove $ \boxed{\sum\frac {1}{1 \plus{} a^3 \plus{} a^6 \plus{} a^9 \plus{} a^{12} \plus{} a^{15}}\ge\frac {1}{1 \plus{} a^5b^5c^5}}$ :?:

[hide="Remark."]The polynomial $ P(a) \equal{}1 \plus{} a^2 \plus{} a^4 \plus{} a^6$ in the LHS denominator with its equally distributed powers seems to be the "biggest" one (in a sense yet to be defined :maybe: ) such that $ \frac {1}{P(a)} \plus{} \frac {1}{P(b)}\ge\frac {1}{1 \plus{} a^3b^3}$. e.g. , the ineq is false for the bigger $ P(a) \equal{} 1 \plus{} a \plus{} a^5 \plus{} a^6$ or $ P(a) \equal{} 2 \plus{} 2a^6$ as above in post#7 (:blush:). As if some deeper result were hidden behind...

BTW It looks like kuing's method can be applied to prove generally $ \sum\frac {1 \minus{} a_i}{1 \minus{} a_i^n }\ge1$ (for $ n$ non-negative variables $ a_i$ with product $ 1$).[/hide]

[quote="hxy09"]$ a,b,c,d\ge 0$ and $ abcd \equal{} 1$
Prove that $ \sum \frac {1}{1 \plus{} a \plus{} a^2 \plus{} a^3}\ge 1$[/quote]

This inequality is [b]Problem 6.33[/b] (page 310) of new book:

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://canhang2007.wordpress.com/2009/12/08/new-book-inequalities-with-beautiful-solutions-v-cirtoaje-v-q-b-can-t-q-anh/][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009.

[quote="kuing"][quote="hxy09"]$ a,b,c,d\ge 0$ and $ abcd \equal{} 1$
Prove that $ \sum \frac {1}{1 \plus{} a \plus{} a^2 \plus{} a^3}\ge 1$[/quote]

for $ a,b,c,d > 0,abcd \equal{} 1$, setting $ a \equal{} e^x, b \equal{} e^y, c \equal{} e^z, d \equal{} e^w$ and $ f(x) \equal{} \frac {1}{1 \plus{} e^x \plus{} e^{2x} \plus{} e^{3x}}$, ineq becomes

$ x,y,z,w \in \mathbb{R}, x \plus{} y \plus{} z \plus{} w \equal{} 0$ then

$ f(x) \plus{} f(y) \plus{} f(z) \plus{} f(w) \ge 1.$

we can prove that $ f(x)$ is "half concave and half convex" in $ \mathbb{R}$, so, using "Half concave and half convex theorem", to prove $ f(x) \plus{} f(y) \plus{} f(z) \plus{} f(w) \ge 1$ just need to prove that

$ 3f(x) \plus{} f( \minus{} 3x) \ge 1,$

that's equivalent to

$ \frac {3}{1 \plus{} k \plus{} k^2 \plus{} k^3} \plus{} \frac {k^9}{k^9 \plus{} k^6 \plus{} k^3 \plus{} 1} \ge 1$

where $ k \equal{} e^x > 0.$

[/quote]

to prove $ f(x) \plus{} f(y) \plus{} f(z) \plus{} f(w) \ge 1$ just need to prove that

$ 3f(x) \plus{} f( \minus{} 3x) \ge 1,$       ？？？？？？？

$ f(x) \plus{} f(y) \plus{} f(z) \plus{} f(w) \ge 3f(x) \plus{} f( \minus{} 3x)$   ？？？？？？

$ f( \minus{} 1) \plus{} f( \minus{} 1) \plus{} f(1) \plus{} f(1) \equal{} 1.351945727...$

but

$ 3f( \minus{} 1) \plus{} f(3) \equal{} 1.931860047....$

$ 1.351945727 > 1.931860047$    ???

I think  to prove $ f(x) \plus{} f(y) \plus{} f(z) \plus{} f(w) \ge 1$ just need to prove that

$ f(x) \plus{}3 f(\minus{}\frac{x}{3}) \ge 1$

that's equivalent to

$ \frac{(2t^4\plus{}t^3\plus{}t\plus{}2)(t\minus{}1)^2t^3(t^2\minus{}t\plus{}1)(t^4\minus{}t^2\plus{}1)}{t^3\plus{}t^2\plus{}t\plus{}1}\geq 0$

where $ t \equal{} e^{\minus{}\frac{1}{3}} > 0.$

[quote="hedeng123"][quote="kuing"][quote="hxy09"]$ a,b,c,d\ge 0$ and $ abcd \equal{} 1$
Prove that $ \sum \frac {1}{1 \plus{} a \plus{} a^2 \plus{} a^3}\ge 1$[/quote]

for $ a,b,c,d > 0,abcd \equal{} 1$, setting $ a \equal{} e^x, b \equal{} e^y, c \equal{} e^z, d \equal{} e^w$ and $ f(x) \equal{} \frac {1}{1 \plus{} e^x \plus{} e^{2x} \plus{} e^{3x}}$, ineq becomes

$ x,y,z,w \in \mathbb{R}, x \plus{} y \plus{} z \plus{} w \equal{} 0$ then

$ f(x) \plus{} f(y) \plus{} f(z) \plus{} f(w) \ge 1.$

we can prove that $ f(x)$ is "half concave and half convex" in $ \mathbb{R}$, so, using "Half concave and half convex theorem", to prove $ f(x) \plus{} f(y) \plus{} f(z) \plus{} f(w) \ge 1$ just need to prove that

$ 3f(x) \plus{} f( \minus{} 3x) \ge 1,$

that's equivalent to

$ \frac {3}{1 \plus{} k \plus{} k^2 \plus{} k^3} \plus{} \frac {k^9}{k^9 \plus{} k^6 \plus{} k^3 \plus{} 1} \ge 1$

where $ k \equal{} e^x > 0.$

[/quote]

to prove $ f(x) \plus{} f(y) \plus{} f(z) \plus{} f(w) \ge 1$ just need to prove that

$ 3f(x) \plus{} f( \minus{} 3x) \ge 1,$       ？？？？？？？

$ f(x) \plus{} f(y) \plus{} f(z) \plus{} f(w) \ge 3f(x) \plus{} f( \minus{} 3x)$   ？？？？？？

$ f( \minus{} 1) \plus{} f( \minus{} 1) \plus{} f(1) \plus{} f(1) \equal{} 1.351945727...$

but

$ 3f( \minus{} 1) \plus{} f(3) \equal{} 1.931860047....$

$ 1.351945727 > 1.931860047$    ???

I think  to prove $ f(x) \plus{} f(y) \plus{} f(z) \plus{} f(w) \ge 1$ just need to prove that

$ f(x) \plus{} 3 f( \minus{} \frac {x}{3}) \ge 1$

that's equivalent to

$ \frac {(2t^4 \plus{} t^3 \plus{} t \plus{} 2)(t \minus{} 1)^2t^3(t^2 \minus{} t \plus{} 1)(t^4 \minus{} t^2 \plus{} 1)}{t^3 \plus{} t^2 \plus{} t \plus{} 1}\geq 0$

where $ t \equal{} e^{ \minus{} \frac {1}{3}} > 0.$[/quote]

yeah, you are right, I'm make mistake... :oops:
thanks, hedeng.

[quote="hxy09"]A solution given in the book written by Yang Xuezhi :lol:
The main idea is to prove $ \frac {1}{1 \plus{} a^2 \plus{} a^4 \plus{} a^6} \plus{} \frac {1}{1 \plus{} b^2 \plus{} b^4 \plus{} b^6}\ge\frac {1}{1 \plus{} x^3y^3}$
Can anyone create a nicer proof for this two-variable inequality? :)[/quote]
See also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=18857]here[/url]

[quote="hxy09"]$ a,b,c,d\ge 0$ and $ abcd \equal{} 1$
Prove that $ \sum \frac {1}{1 \plus{} a \plus{} a^2 \plus{} a^3}\ge 1$[/quote]
An expanding gives:
 $\sum_{sym}(2a^2+4a^3+6a^3b^2+3a^3b^3+6a^3b-3ab-4abc-6a^2bc-2a^2b^2c^2-6a^2b^2c)\geq0$,
which is true by Muirhead.


logarithms articles geometry parameterization limit LaTeX

For example, I have this piece of code:

\begin{proof} The result is trivial if $g = 1$. If $g\ne1$, noting that as $\chi$ runs through all precharacters so does $\chi\chi_g$, we get
\begin{center}$\displaystyle\sum_\chi\chi(g) = \sum_\chi\chi\chi_g(g) = \chi_g(g)\sum_\chi\chi(g) = 0$, since $\chi_g(g)\ne1$.
\end{center}
\end{proof}

What I want, really, is to have the 'end of proof box' on the same line as the calculations. If not, it's a waste of space.

Cheers.

Is there a reason why you are not using display mode with \ [...\ ] which automatically centres the mathematics and uses displaystyle?
[code]\begin{proof} The result is trivial if $g = 1$. If $g\ne1$, noting that as $\chi$ runs through all precharacters so does $\chi\chi_g$, we get
\[
\sum_\chi\chi(g) = \sum_\chi\chi\chi_g(g) = \chi_g(g)\sum_\chi\chi(g) = 0, \text{ since } \chi_g(g)\ne1. \qedhere
\]
\end{proof}[/code]
Note the use of \qedhere to put the end of proof symbol exactly where you want it.

The reason I'm not using
\[
and
\]
is that I have text in between... Is it still possible to use
\[
and
\]
and not have the text in between in italics and without spacings?

also, I have a similar problem with the following code

\begin{lem}\label{chi0lem2} $\displaystyle P(s)=\sum_p \frac{1}{p^s}\rightarrow\infty$ as $s\rightarrow 1^+$.
\end{lem}
\begin{proof} Since $\displaystyle\frac{1}{1-r}\le1+2r\le e^{2r}$ for $\displaystyle0\le r\le\frac{1}{2}$ and $\displaystyle0\le p^{-s}\le \frac{1}{2}$ we have
\begin{center} $\displaystyle\zeta(s)=\sum_{n\ge1}\frac{1}{n^s}=\prod_p\frac{1}{1-p^{-s}}\le\prod_pe^{2p^{-s}}=e^{2P(s)}\rightarrow\infty$ as $s\rightarrow1^+$,
\end{center}
by lemma \ref{chi0lem1}. Taking $\log$ on both sides yields the required result.
\end{proof}

because there is a new line after 'proof', but there is no need for a new line there

Before I do anything:
1. you will need to supply a complete example since lem is not a standard environment so I don't know how you've defined it;
2. replace \begin{center} $ \$$\displaystyle ... $ \$$ \end{center} with \ [ ... \] which is the correct way to do things. If you want text inside an equation then use \text.

Ok, half the problem is solved and I've deleted all the centers and displaystyles. Now, in the following example (I'm taken the lem away), there is a line after 'proof', which I don't want.

\begin{proof}\begin{align*}\sum_\chi \overline{\chi}(b)P(s,\chi)
&= \sum_\chi \overline{\chi}(b)\sum_p \chi(p)\frac{1}{p^s} & \text{(definition}\\
&= \sum_p \frac{1}{p^s} \sum_\chi \overline{\chi}(b)\chi(p) & \text{(absolute convergence)}\\
&= |G|\sum_{p \equiv b \mod a} \frac{1}{p^s} & \text{(proposition)} & \qedhere
\end{align*}
\end{proof}

[quote="justin"] there is a line after 'proof', which I don't want.[/quote]I don't see that here but you need to give a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal working example[/url]to be sure.

Sorry about that, here is a small working example:

\documentclass[12pt]{article}
%
\oddsidemargin -8pt
\topmargin -24pt
%
\textheight 8.5 in
\textwidth 6.8 in
%
\usepackage{amsmath, amsthm, amssymb}
%
\newtheorem{thm}{Theorem}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{deff}[thm]{Definition}
\newtheorem{prop}[thm]{Proposition}
\theoremstyle{remark}
\newtheorem*{rem}{Remark}
\newtheorem*{nota}{Notation}
%
\begin{document}
%
\begin{proof}\begin{align*}\sum_\chi \overline{\chi}(b)P(s,\chi)
&= \sum_\chi \overline{\chi}(b)\sum_p \chi(p)\frac{1}{p^s} & \text{(definition})\\
&= \sum_p \frac{1}{p^s} \sum_\chi \overline{\chi}(b)\chi(p) & \text{(absolute convergence)}\\
&= |G|\sum_{p \equiv b \mod a} \frac{1}{p^s} & \text{(proposition)} & \qedhere
\end{align*}
\end{proof}
%
\end{document}

Aha! I thought you meant that there was a line after the proof, whereas you mean a line after the word 'proof'.
But of course this is bound to happen since displayed equations [i]always[/i] start on a new line wherever they occur otherwise how could they be centered? This is normal behaviour. You could put some words after 'proof' such as 'We have'.

I'm not sure why you have % in lots of places as they don't do anything. And please use the geometry package which sets the margins etc correctly and adjusts lots of hidden parameters accordingly.

Sorry to bring up an old..ish topic, but I'm having problems with the following code:
[code]\mbox{(e)} \lim_{h\to0}{\frac{(2+h)^3-8}{h}} = \lim_{h\to0}{\frac{h\cdot((2+h)^2+2(2+h)+2^2)}{h}} = (2+0)^2 + 2(2+0) + 4 = \boxed{12}[/code]
which renders as:
\[ \mbox{(e)} \lim_{h\to0}{\frac{(2\plus{}h)^3\minus{}8}{h}} \equal{} \lim_{h\to0}{\frac{h\cdot((2\plus{}h)^2\plus{}2(2\plus{}h)\plus{}2^2)}{h}} \equal{} (2\plus{}0)^2 \plus{} 2(2\plus{}0) \plus{} 4 \equal{} \boxed{12}\]
How could I make it so that everything's on one line? Thanks.

[quote="undefined117"]How could I make it so that everything's on one line? Thanks.[/quote]
Put the whole lot inside an \mbox
\[ \mbox{\text{(e)} \lim\limits_{h\to0}{\cfrac{(2+h)^3-8}{h}} = \lim\limits_{h\to0}{\cfrac{h\cdot((2+h)^2+2(2+h)+2^2)}{h}} = (2+0)^2 + 2(2+0) + 4 = \boxed{12}}\]
Also:
1. replace \mbox{(e)} by \text{(e)}
2. to ensure that the h \to 0 is below \lim use \limits
3. for larger fractions use \cfrac


geometry
circumcircle
inequalities
trigonometry
geometry unsolved

Let $ ABC$ be an acute triangle and denote by $ Q$ the center of its Euler's circle. The circumcircle of $ ABC$, which has radius $ R$, meets $ AQ,BQ,CQ$ respectively at $ M.N,P$.
Prove that
$ \frac{1}{QM}\plus{}\frac{1}{QN}\plus{}\frac{1}{QP}\geq\frac{3}{R}$

[color=darkred][b]I think that $ \frac {1}{QM} \plus{} \frac {1}{QN} \plus{} \frac {1}{QP}\le\frac {3}{R}\ (*)$ is correctly (see $ a \equal{} 5$ , $ b \equal{} 4$ , $ c \equal{} 3$ ) ! [/b][/color] [color=darkblue]Observe that

$ \sum\frac {1}{QM} \equal{} \sum\frac {QA}{QA\cdot QM} \equal{} \frac {1}{p_w(Q)}\cdot \sum QA$ , where $ p_w(Q)$ is the power of the point $ Q$ w.r.t. the ciecumcircle $ w$ of $ \triangle ABC$ . Prove easily that

$ 4\cdot QA^2 \equal{} R^2 \plus{} b^2 \plus{} c^2 \minus{} a^2$ and is well-known the relation $ GO^2 \equal{} R^2 \minus{} \frac 19\cdot\sum a^2$ . Thus, the inequality $ (*)$ is equivalently with

$ \sum QA\le\frac {3p_w(Q)}{R} \equal{} \frac 3R\cdot \left(R^2 \minus{} QO^2\right)$ $ \Longleftrightarrow$ $ \frac 12\cdot\sum\sqrt {R^2 \plus{} b^2 \plus{} c^2 \minus{} a^2}\le \frac 3R\cdot\left(R^2 \minus{} \frac 94\cdot GO^2\right) \equal{}$ $ \frac 3R\cdot\left[R^2 \minus{} \frac 14\cdot\left(9R^2 \minus{} \sum a^2\right)\right] \equal{}$

$ \frac {3}{4R}\cdot\left( \minus{} 5R^2 \plus{} \sum a^2\right)$ . In conclusion, $ \boxed {\ \frac {1}{QM} \plus{} \frac {1}{QN} \plus{} \frac {1}{QP}\le\frac {3}{R}\Longleftrightarrow 15R^2 \plus{} 2R\sum\sqrt {R^2 \plus{} b^2 \plus{} c^2 \minus{} a^2}\le 3\cdot\sum a^2\ }$ .

[b][u]Remark.[/u][/b] $ \left\{\begin{array}{c} R^2 \plus{} b^2 \plus{} c^2 \minus{} a^2 \equal{} R^2\cdot\left\{\left[2\cos A \plus{} \cos (B \minus{} C)\right]^2 \plus{} \sin^2(B \minus{} C)\right\}\ge 0 \\
 \\
\sqrt {R^2 \plus{} b^2 \plus{} c^2 \minus{} a^2}\ge R\left[2\cos A \plus{} \cos (B \minus{} C)\right] \\
 \\
\sum\cos (B \minus{} C)\ge \frac {3(4r \minus{} R)}{R}\ ;\ \sum\cos A \equal{} 1 \plus{} \frac rR\end{array}\right\|$ $ \implies$ $ \sum\sqrt {R^2 \plus{} b^2 \plus{} c^2 \minus{} a^2}\ge 14r \minus{} R$ .

$ \sum\frac {1}{QM} \equal{}$ $ \sum\frac {QA}{p_w(Q)}\ge \frac {1}{R^2}\cdot \sum QA \equal{}$ $ \frac {1}{2R^2}\cdot\sum\sqrt {R^2 \plus{} b^2 \plus{} c^2 \minus{} a^2}\ge$ $ \frac {14r \minus{} R}{2R^2}$ , i.e. $ \sum\frac {1}{QM}\ge \frac {14r \minus{} R}{2R^2}$ .

In conclusion, $ \boxed {\ \frac {14r \minus{} R}{2R^2}\ \le\ \sum\frac {1}{QM}\stackrel {?}{\ \le\ }\frac 3R\ }$ . [b]Now yet I don't know to prove the right inequality ![/b]

[b][u]Remark.[/u][/b] $ b^2 \plus{} c^2 \minus{} a^2 \equal{} 8R^2\cos A\sin B\sin C$ in any triangle $ ABC$ .

[b][u]Remark.[/u][/b] $ \sum\frac {1}{QM} \equal{} \sum\frac {QA}{p_w(Q)}\ge\frac {1}{R^2}\cdot\sum QA$ . Now I"ll apply the [b]Erdos - Mordell[/b]'s inequality. Therefore, if denote

[b]the distance $ \delta_d(X)$ of $ X$ to the line $ d$[/b], then $ \sum\frac {1}{QM}\ge \frac {2}{R^2}\cdot \sum\delta_{BC}(Q) \equal{} \frac {1}{R^2}\cdot\sum\left[\delta_{BC}(H) \plus{} \delta_{BC}(O)\right] \equal{}$

$ \frac {1}{R^2}\cdot \left[2R\cdot\sum\cos B\cos C \plus{} (R \plus{} r)\right] \equal{}$ $ \frac {1}{R^2}\cdot \left[2R\cdot\frac {p^2 \plus{} r^2 \minus{} 4R^2}{4R^2} \plus{} (R \plus{} r)\right] \equal{} \frac {p^2 \plus{} r^2 \plus{} 2Rr \minus{} 2R^2}{2R^3}$ .

In conclusion,$ \boxed {\ \sum\frac {1}{QM}\ge \frac {p^2 \plus{} r^2 \plus{} 2Rr \minus{} 2R^2}{2R^3}\ }$ [/color]

[quote="Virgil Nicula"] In conclusion, ${ \boxed 15R^2 + 2R\sum\sqrt {R^2 + b^2 + c^2 - a^2}\le 3\cdot\sum a^2\ }$ .
[/quote]
This Inequality right?? :maybe:  :maybe:


geometric series
real analysis
real analysis unsolved

Prove : if a power series converges pointwise on the set A $ \subseteq$R,
then it converges uniformly on any compact set K $ \subseteq$ A.

THX  :lol:

If that were strict inclusion $ K\subset A$, this would be an utterly natural consequence of the existence of a radius of convergence. Inside a smaller radius, comparison to a geometric series gives uniform convergence.

As asked, this involves tricky stuff on the boundary.

can u give a solution pls maybe? I'm really stuck on this and it is not hwk question
but something I do for practice.
I don't really no how to use the information "K is a compact set" in my proof unless I
preassume all power series are continuous...but in the hint it says "the theorem leads
to the fact that a power series is continuous at every point at which it converges."

and I think K $ \subseteq$ A preassumes A doesn't equal to R.

The issue is not whether $ A$ is the whole line. Actually, strict inclusion isn't really what we need for the easy version- we need strict inclusion at all boundary points, best ensured by making $ A$ open.

Here's the version in which $ A$ is open:
Let $ a$ be the center of our power series$ \sum c_n(x\minus{}a)^n$, and let $ R\equal{}\sup_{x\in A}|x\minus{}a|$. Given a compact set $ K\subset A$, let $ r\equal{}\max_{x\in K}|x\minus{}a|$. We have $ r<R$. There is some $ y\in A$ with $ r<|y\minus{}a|<R$; define $ s\equal{}|y\minus{}a|$. Since the series converges at $ s$, the absolute terms $ |c_n|s^n$ are bounded by some $ C$. Now, for all $ x$ with $ |x\minus{}a|\le r$, we get $ |c_n(x\minus{}a)^n|\le C\frac{r^n}{s^n}$, and the series converges uniformly by the M-test comparing to a geometric series.

This argument works more generally than just in $ \mathbb{R}$.

At the boundary in your original question, this is connected to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244472]your previous thread[/url].

But isn't there another abel's theorem which allows for uniform convergence on closed disks contained in the interior of the disk of convergence? It is also proved by Weierstrass M-Test and comparison to geometric series

If it's what I just proved in that paragraph, I don't think it's worthy of a name.