the question is something like...

How many numbers less than 1000 can be expressed as the sum of 2 positive odd primes?

If there is some shortcut for this plz post it! :lol:

No, there isn't really any shortcut. It doesn't seem like you said the problem correctly, because there are lots of prime numbers less than 1000. There could be some kind of formula for this kind of problem, but I doubt it.

there would be 499. golbach's conjecture, which is not proven, says every even number greater than two can be written as the sum of two primes. so, there would be (1000/2)-1 even numbers other than two that can be written as the sum of two primes.

If golbach's conjecture is true, than the answer is 498, not 499.
4 cannot be expressed as a sum of 2 ODD primes.

I think that Goldbach's conjecture has been "proven" to work on the  interval of numbers up to a number certainly greater than a 1000.  This was accomplished simply by virtue of super computers.  Although Goldbach's conjecture itself isn't proven, the answer can be ascertained by the computers that have tried disproving it.  (Goldbach's conjecture will need a large counter-example, like Polya's conjecture.)

So yes, I would safely say the answer is 498. (4=2+2)

I'm so stupid, I forgot all about Goldbach! Thanks everyone! Your help was highly appreciated!

[quote="batteredbutnotdefeated"]what if the number of primes were...say...about 100? Wouldn't Goldbach's conjecture be incorrect? I did all the work and I got 16! I'm very confused now...[/quote]
100 primes added up!?
Well I guess the best way to approach that would be list the first 100 primes, and see their sum. Now if you have an upper bound, I guess you could see the sum of the primes that would bring you closest to the upper bound and then see how many lesser integers could be made from that.  That is just a guess\outline though.

[quote="batteredbutnotdefeated"]No i did not mean the sum of the primes! I simply meant how many numbers under 100 could be expressed as the sum of two odd primes![/quote]
Oh. Golbach's conjecture still works then. Just remember to forget 4.

what if the question was the sum of two DISTINCT positive odd primes...?

Then subtract the compliment from the answer to the original question.

a lot of even numbers, especially as they get larger, can be written in more than one way as the sum of two primes. So, like for 10, although it can be written as 5+5, it can also be written as 3+7, which satisfies your requirements. Just keep that in mind when doing the distinct primes question. Also, one good way would be doubling all primes less than 500, and seeing if that number can be written as two distinct primes, if not, subtract one from your total.

I got that all even numbers greater than 6 can be expressed as the sum of two distinct odd prime numbers.

If i said positive, would it make a difference? :ninja:  :rotfl:  :D

[quote="batteredbutnotdefeated"]I still am a bit puzzled... Could someone please work out the distinct primes question? I wrote them all out for the numbers under a 100 and got 16. I also verified it to check if it was correct and it was. All i want to know is the quick way to do these kinds of [i]distinct [/i]problems.

The original question was how many numbers under 100 can be expressed as the sum of 2 distinct odd primes?[/quote]

First you figure out how many numbers can be expressed as the sum of two odd primes regardless of if they are different or not. In this case it would be 47. Then we subtract the number of numbers that are the sum of two equal primes. In this case there are 14 (you can just count the number of primes less than fifty apart from 2. The answer turns out to be 33.

aaaaah! Oooops... I posted the wrong question all this time... Here is the right one.

How many numbers under a 1000 can be expressed as the sum of two positive odd primes?


Just to clarify here are the steps to register right?
1) make a username and write down your I.D. (check)
2)Send your solutions (im handwriting) on USAMTS paper in USAMTS format in the mail by October 14, with the entry form.


Let  a,b,c \geq 0 . Show  that
a^3+b^3+c^3 +6abc \geq1/4(a+b+c)^3
Thanks

Take a+b+c=1 and use that sum a^3=3abc+1-2sum ab, Schur inequality ( sum ab<=(9abc+1)/4) and probably the fact that abc<=1/27.


This is for discussion on the March Tournament, open from now until the end.  Try to keep all discussion here.

Since this would just clog up the registration thread, I'll put it here:

I need two people for my team.  :ninja:

Ok...
[list][*]Should I make a list of people who need a team?
[*]Should someone else, because that seems painful?
[*]Should I shorten this tournament to 8 teams, no more no less, due to the fact that many people might be taking Spring break?[/list]

RETURNING TEAM!

Boyz of Darkness

verizonwires
vallon22
emton

Okay guys, registration is now available at the website at http://aopsftw.freehostia.com/home/

The two teams that have already signed up here do not need to register.

Oh, CircleSquared.  I found it(obviously).

Wait, is the March tournament over already?! :huh:

No, are you confused with the first tournament run by #H34N1?  That one was canceled but this one is still on!
:)

I guess you are not confused. ;)
Too hard to explain.  It was a complete mess.


AMC USA(J)MO USAMO AIME geometry inequalities induction \/closed

Would it be possible to hold a USAMO Problem Seminar?

The material covered on the USAMO is very broad and deep.  It often takes 30 minutes to an hour to even discuss one problem and we do not feel that a few hours of seminar could possibly help students immediately fill their gaps of knowledge the way that an AIME seminar could.

We do however feel that our higher level courses are great preparation for the USAMO.  I think this is partially evidenced by the fact that about a dozen students in each the Geometry and Inequalities classes made the USAMO and that list of students includes veterans of MOsP who wouldn't need to take subject classes that were not Olympiad level.  This is part of the reason we have begun labeling our classes using the names of the famous math competitions -- it gives students a good marker of identification for the difficulty level of the material.

One of the courses that we will be adding in the future is a class on general problem solving techniques that will include ideas such as induction, parity, bijections, isomorphisms, and a host of other ideas that are essential to Olympiad level problem solving.  This is a class I would recommend to any student who hasn't already achieved a top 10 score on the USAMO (and maybe some who have).

Mr. Crawford,

As a student very much engaged in math competitions but never having taking the USAMO, or any contest that required proofs, I think what may help students is to offer a seminar (over the span of a couple days, like the AIME one, as opposed to a full-scale class) that teaches them how to write proper proofs.

Personally, I rarely see formal proofs (ones that would be worth 7 points) for olympiad problems online, except for the official USAMO/IMO/Putnam solutions for the past few years that are posted on the AMC site. The "solutions" I usually see to problems online are usually just steps of logic that make up the bulk of a formal solution. For example, one mistake that students often make when proving inequalities is that they write their proof backwards: they start with the given inequality and transform it with algebraic manipulations into a true statement.

Personally, I would just hate to figure out how to solve a problem on the USAMO and then end up getting a 5/7 because I didn't write the proof very clearly or properly.

[quote="MCrawford"]I myself once found learning to write a formal proof frustrating for precisely the reasons you mentioned.

While we do not plan to run a seminar on proof-writing, we are gradually integrating proof-writing lessons into our subject courses.  I plan to talk a bit about formal inequality proofs in my current inequalities class.  We also plan to add material to the site that relates directly to formal proof writing.  Unfortunately time is limited for us and the best features will not be ready before this year's USAMO, but students taking our higher level classes will get better and better doses of formal proof writing soon.

Now that we have TeX on the site, explaining a well-written proof will be much easier, but this feature is only hours old.  Bear with us, we are doing our best to add good problem solving curriculum when we can.[/quote]

Are you studying for a doctorate?

[quote="churchilljrhigh"]
Are you studying for a doctorate?[/quote]

No.

[quote="Fierytycoon"]As a student very much engaged in math competitions but never having taking the USAMO, or any contest that required proofs, I think what may help students is to offer a seminar (over the span of a couple days, like the AIME one, as opposed to a full-scale class) that teaches them how to write proper proofs.[/quote]

Hmm. . .   Good idea.  What we're going to do about teaching how to write proofs and how to approach Olympiad-style contests is start with a Math Jam next Friday.  Sometime this spring or summer, we'll write a long article (or articles) about how to write a proof, using a lot of the discussion in the Math Jam to guide what questions we'll try to address in the article.


geometry
circumcircle
vector
geometric transformation
reflection
cyclic quadrilateral
power of a point

if A, midpoint of AB, midpoint of AC, and centroid are concyclic, find all such triangles.

i think only equilateral triangles work (i've got a tedius proof using ptolemy's theorem).  but what is interesting that if i only used power of point to arrive at 2aa=bb+cc, which does not characterise the triangle as equilateral - it only shows cosA>=1/2 or angle A is less than or equal to 60 degrees.  but power of point is necessary and at the same time sufficient for four points to be concyclic, why doesnt it work this time?

if you have a better proof, by all means post it.

my proof
-------------------
median length:
mb=(1/4)(2aa+2cc-bb)
mc=(1/4)(2aa+2bb-cc)

power of a point:
(c/2)c=(2(mb)/3)(mb)
3cc/4=(1/4)(2aa+2cc-bb)
3cc=2aa+2cc-bb
cc=2aa-bb

similarly, bb=2aa-cc
same equation :S

2aa=bb+cc
aa=bb+cc-2bccosA
hence 4bccosA=bb+cc=2aa
but bb+cc=4bccosA>=2bc
hence 4cosA>=2
cosA>=1/2
A<=60 (i see thats what you got)

bb+cc=2
2bccosA=1
bb+cc+2bc=2+1/cosA
(b+c)^2=2+1/cosA
(b-c)^2=2-1/cosA

and by ptolemys theorem,

2a*ma=b*mb+c*mc
hence
2(2bb+2cc-1)=b(2bb+2-c)+c(2cc+2-b)
6=2bbb+2ccc+2b+2c-2bc
note that ccc+bbb=(c+b)(cc+bb)-cb(c+b)
hence
2((c+b)(cc+bb)-cb(c+b)+(b+c)-bc)=6
6(b+c)-2(b+c)bc-2bc=6
this can be transformed into
6sqrt(2+1/cosA)-sqrt(2+1/cosA)/cosA-1/cosA=6
this is a cubic in terms of 1/cosA=t

6sqrt(2+t)-sqrt(2+t)t-t=6
sqrt(2+t)=(6+t)/(6-t)
(2+t)(6-t)(6-t)=(6+t)(6+t)

which factors into (t-2)(tt-9t+18)=0
hence we have t=2,t=-1.68 or t=10.68
but the last two does not satisfy the condition that cosA>=1/2 hence
t=1/cosA<=2
and A should be acute because if A is obtuse then if the centroid is M,
and midpoint of AB is P, midpoint of AC is P, then 180-is obtuse, but in a cyclic quadrilateral opposite angles sums up to pi.
but pi/2+pi/2=pi
hence t=2 is the only solution
bb+cc=2
2bccosA=1
bc=1

bb-2bc+cc=(b-c)^2=0
hence b=c
so equilateral

Let O be the circumcenter, G the centroid, M midpoint of BC, G1 the point of intersection of the median AG with the circumcircle and A1 the point of intersection of OA with the circumcircle.. We can quickly see that MG=MG1, so G is the middle of the segm AG1. Because angle A1G1A=90 deg angle OGA is also equal to 90 degrees, so R^2=OG^2+GA^2 (*). By using the formula (I think it belongs to Leibniz) GA^2+GB^2+GC^2+3OG^2=3R^2 this is actually a particular case of Leibniz's formula) and also the equality (*), we get 2GA^2=GB^2+GC^2. If we also put the condition that GA+GB+GC=0 (these are vectors, not segments), we get cosx=-(GB^2+GC^2)/4*GB*GC (**), where x is the angle between GB and GC. I think condition (**) is enough and (**) does not imply the fact that ABC is equilateral, so I believe that all triangles satisfying (**) are solutions, not just the equilateral ones. Observe that for GB=GC we get x=120 degrees, which corresponds to the equilateral triangle.

Let $ M,N,P$ be the midpoint of $ BC,CA,AB$ respectively and $ G$ the centroid.

The lines $ BPA,BGN$ are secants to the circle, so $ BP \cdot BA = BG \cdot BN \Rightarrow \frac{1}{2}c^{2}= \frac{2}{3}BN^{2}\Rightarrow$

$ \boxed{3 c^{2}= 4 BN^{2}}\ \ (1)$ and similarly $ \boxed{3 b^{2}= 4 CP^{2}}\ \ (2)$

From the median formula we have
$ c^{2}+a^{2}= 2BN^{2}+\frac{b^{2}}{2}$

$ b^{2}+a^{2}= 2CP^{2}+\frac{c^{2}}{2}$
$ \hline$
adding $ \Rightarrow$

$ \frac{c^{2}}{2}+\frac{b^{2}}{2}+2a^{2}= 2BN^{2}+2CP^{2}$, now from $ (1)$ and $ (2)\Rightarrow$

$ \frac{c^{2}}{2}+\frac{b^{2}}{2}+2a^{2}= 3\frac{c^{2}}{2}+3\frac{b^{2}}{2}\Rightarrow \boxed{2a^{2}= b^{2}+c^{2}}$

Again from the median formula $ b^{2}+c^{2}= 2AM^{2}+\frac{a^{2}}{2}$

So $ 2a^{2}= 2AM^{2}+\frac{a^{2}}{2}$ which finally gives $ AM^{2}= 3\cdot\frac{a^{2}}{4}\Rightarrow \boxed{AM = \frac{a\sqrt 3}{2}}$

[quote="Bugzpodder"][color=darkred]In $ \triangle ABC$ denote the midpoint $ P$ of the side $ [AB]$, the midpoint $ N$ of the side $ [AC]$ and the its centroid $ G$.
Prove that the quadrilateral $ APGN$ is concyclically if and only if there is the relation $ 2a^{2}=b^{2}+c^{2}$.[/color][/quote]

[color=darkblue][b]Proof I.[/b] Denote the midpoint $ M$ of the side $ [BC]$ and the reflection $ R$ of the point $ G$ w.r.t. the point $ M$, i.e. the reflection of the point $ A$ w.r.t. the point $ G$. Observe that $ 3\cdot MR=AM=m_{a}$. Therefore, the quadrilateral $ APGN$ is cyclically $ \Longleftrightarrow$ the quadrilateral $ ABRC$ is cyclically $ \Longleftrightarrow$ $ MB\cdot MC=MA\cdot MR$ $ \Longleftrightarrow$ $ \frac{a^{2}}{4}=\frac{m_{a}^{2}}{3}$ $ \Longleftrightarrow$ $ 4m_{a}^{2}=3a^{2}$ $ \Longleftrightarrow$ $ 2(b^{2}+c^{2})-a^{2}=3a^{2}$ $ \Longleftrightarrow$ $ \boxed{b^{2}+c^{2}=2a^{2}}$.

[b]Proof II.[/b] The quadrilateral $ APGN$ is cyclically $ \Longleftrightarrow$ $ CG\cdot CP=CN\cdot CA$ $ \Longleftrightarrow$ $ 4m_{c}^{2}=3b^{2}$ $ \Longleftrightarrow$ $ 2(a^{2}+b^{2})-c^{2}=3b^{2}$ $ \Longleftrightarrow$ $ \boxed{b^{2}+c^{2}=2a^{2}}$.[/color]

[quote="Virgil Nicula"][color=darkred][b]An easy extension.[/b] Let $ X(\alpha ,\beta ,\gamma )$ be an interior point of the triangle $ ABC$ (with barycentrical coordinates w.r.t. $ \triangle ABC$).
Prove that the quadrilateral $ APXN$ is cyclically $ \Longleftrightarrow$ there is the relation $ \boxed{\frac{a^{2}}{\alpha }=\frac{b^{2}}{\alpha+\beta }+\frac{c^{2}}{\alpha+\gamma }}$.

[b]Particular cases.[/b] The quadrilateral $ APXN$ is cyclically $ \Longleftrightarrow$ $ \{\begin{array}{cccc}1\blacktriangleright & X: =G & \mathrm{\ (centroid)\ }& 2a^{2}=b^{2}+c^{2}\\\\ 2\blacktriangleright & X: =I & \mathrm{\ (incenter)\ }& A=60^{\circ}\\\\ 3\blacktriangleright & X: =N & \mathrm{\ (Nagel's\ point)\ }& a^{2}=(p-a)(b+c)\\\\ 4\blacktriangleright & X: =S & \mathrm{\ (Lemoine's\ point)\ }& a^{2}=bc\\\\...... & ......... & .................... & ..........................\end{array}$[/color][/quote]


i) Let $a,b,c\geq0$, prove that \[\Bigl(\frac a{a+b}\Bigl)^{2}+\Bigl(\frac b{b+c}\Bigl)^{2}+\Bigl(\frac c{c+a}\Bigl)^{2}+\frac{ab+bc+ca}{(a+b+c)^{2}}\geq1.\] ii) Let $a,b,c\geq0$, prove that \[\frac a{a+b}+\frac b{b+c}+\frac c{c+a}+\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}}\leq\frac52.\]

ii)<=>6sum[a^2(b+c)(a-b)(a-c)]+
                6sum[abc(a-b)(a-c)]+
                2sum[(ab^2+bc^2)(a-c)^2)>=0,
        the inequality holds  !


[quote="doicu"]hallo,
find all functions $f: R \to R$ such that
$f(xf(x)+f(y))=y+f(x)^{2}$. :)[/quote]

If we put $x=0$ we have : $f(f(y))=f(0)^{2}+y$      $(\ast)$.
now it's clear that for every $x_{1},x_{2}\in R$ that $f(x_{1})=f(x_{2})$ , we must have $x_{1}=x_{2}$ and so $f^{-1}(x)$ exists.
now in $(\ast)$ we put $y=f^{-1}(x)$ and we have $f(x)-f^{-1}(x)=f(0)^{2}$         $(\oplus)$. this means that $f(x)-f^{-1}(x)$ is constant for every $x$.
now if in $f(xf(x)+f(y))=y+f(x)^{2}$ we put $x=y$ we have: $f(xf(x)+f(x))=x+f(x)^{2}$
and if we put $f^{-1}(x)$ in place of $x$ and we have: $f(x(1+f^{-1}(x)))=f^{-1}(x)+x^{2}$
and now if we put $x=0$ we have: $f(0)-f^{-1}(0)=0$ and by $(\oplus)$ we have $f(0)^{2}=0$ or for every $x$ we have $f(x)=f^{-1}(x)$ and so the only answer is $f(x)=x$.

The solution posted is incorrect since $ f(x) \equal{} x$ and $ f(x) \equal{} \minus{}x$ are both solutions.  To prove a function is a bijection, one has to show that it is both 1-1 and onto. The solution posted only shows that the function is 1-1. But it is easy to show that the function is a bijection.

$ f(xf(x) \plus{} f(y)) \equal{} y \plus{} f(x)^2$-----------(*)

Putting $ x \equal{} 0$, we get $ f(f(y)) \equal{} y \plus{} f(0)^2$. Since $ y$ can range over the set of reals, $ f$ is onto.
Suppose $ f(u) \equal{} f(v) \implies f(f(u)) \equal{} f(f(v)) \implies u \equal{} v$. Hence, $ f$ is 1-1.
We can conclude that $ f$ is a bijection.

Since $ f$ is a bijection, there exists a $ x_0$ such that $ f(x_0) \equal{} 0$.
Putting $ x\equal{} x_0$  in (*), we get $ f(f(y)) \equal{} y$. This implies $ f(0)^2 \equal{} 0 \implies f(0) \equal{} 0.$

Putting $ y \equal{} 0$ in (*), we get $ f(xf(x)) \equal{} f(x)^2$.
Replacing $ x$ by $ f(x)$, we get
$ f(f(x)f(f(x)))  \equal{} f(f(x))^2 \implies f(xf(x)) \equal{} x^2 \implies f(x)^2 \equal{} x^2$.

It is easy to verify that $ f(x) \equal{} x$ and $ f(x) \equal{} \minus{}x$ are both solutions to the problem.
Suppose there exists a solution such that $ f(x_1) \equal{} \minus{}x_1$ and $ f(y_1) \equal{} y_1$ for some $ x_1 \neq 0$ and $ y _1\neq 0$.
Putting $ x \equal{} x_1$ and $ y \equal{} y_1$ in (*), we get $ f(\minus{}x_1 ^2 \plus{} y_1) \equal{} y_1 \plus{} x_1^2$.
Squaring both sides, we get $ f(\minus{}x_1 ^2 \plus{} y_1)^2 \equal{} (y_1 \plus{} x_1^2)^2 \implies (\minus{}x_1 ^2 \plus{} y_1)^2 \equal{} (y_1 \plus{} x_1^2)^2 \implies 4x_1^2y_1 \equal{} 0$.
This is not possible since $ x_1 \neq 0$ and $ y_1 \neq 0$. Hence, the only solutions to the problem are $ f(x) \equal{} x$ and $ f(x) \equal{} \minus{}x$.


How many regular hexagons appear in the figure below?

I can only vaguely describe this, but...
[hide]
Any regular hexagon here must contain at least 6 congruent equilateral triangles.

We define 1 unit to be a side length of the smallest possible triangle, and a Size $ n$ hexagon to be a hexagon with side length $ n$ units. We can only use integral $ n$, because we are only given segments as such.

Then we can identify a Size 1 hexagon by its top triangle, which points downwards. Thus there are $ 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} 6 \equal{} 21$ Size 1 hexagons that may "fit" using the top triangle as a reference. Similarly, there are $ 1 \plus{} 2 \plus{} 3\equal{} 6$ "fitting" Size 2 hexagons (using top 2 triangles). No size 3 or higher hexagons exist, so the answer is $ 21 \plus{} 6\equal{} \fbox{27}$[/hide]

Another way of looking at it is if you have a hexagon, then you can extend the sides to form an equilateral triangle whose number of sides is a multiple of three (we can say that we extend the side that is horizontal and on the bottom, etc so we don't overcount).

Can't you just count possible points for centers?

Care to elaborate? Two different hexagons can have the same center...

Number of centers possible for a radius 1 hexagon: the (9-3)th = 6th triangle number = 21
Number of centers possible for a radius 1 hexagon: the (9-2*3)th = 3th triangle number = 6
Number of centers possible for a radius 1 hexagon: the (9-3*3)th = 0th triangle number = 0


abstract algebra
group theory
superior algebra
superior algebra unsolved

if G is a finite group and 3 is not a facter of cardinal G, if P:G---G,given by p(x)=xxx is a homomorphism ,then G  is abelian. who can prove it,thanks!

P is a homomorphism means $x^3y^3 = (xy)^3 \Rightarrow x^2y^2 = (yx)^2$ for all $x,y$ in G.
Lemma $x^3 = e \Rightarrow x = e$
Proof this is true because there is no element of order 3.
Lemma $x^3 = y^3 \Rightarrow x = y$
Proof this is equivalent to $(xy^{-1})^3 \Rightarrow xy^{-1} = e$

$(xyx^{-1})^3 = xy^3 x^{-1}$
On the other hand, $(xyx^{-1})^3 = x^3y^3x^{-3} = (x^3yx^{-3})^3$
Thus by lemma, $xyx^{-1} = x^3yx^{-3}$
$y = x^2yx^{-2}$
i.e. $x^2$ and $y$ commute.

Since $x^2y^2 = (yx)^2$
$xxyy = yxyx$
$yxxy = yxyx$ (since $x^2$ and $y$ commute)
$xy = yx$

@o : Are you sure? I can prove this only, when $n$ is odd, or with the assumption $gcd(n,|G|)=1$.

[quote="-oo-"]@Soarer: What do you mean here by special cases?[/quote]
I mean how to prove your statement? (It is a special case of o's statement)

Oh, I've made a mistake ;). In $G=S_3$, $x \mapsto x^7 = x$ is an automorphism and $gcd(7,|G|)=1$.

@o: How did you prove your generalization, when $n$ is even? :)

[quote="o"]i think it forther.if G in finite.h(x)=x^n is a homomorphismof groups.where n>2 and for all k=3,4,...,n,k is not divisor of IGI ,then G is abelian![/quote]

Hi everyone.

I found a proof for this one, but it's long. I will use the fact, that the map $x\mapsto x^n$ is bijective if and only if |G| and n are relative prime.
Furthermore I will use, that $x\mapsto x^{n-1}$ is an antihomomorphism iff $x\mapsto x^n$ ist an homomorphism:
$x^n\cdot y^n=(xy)^n \Leftrightarrow x^{n-1}\cdot y^{n-1}=(yx)^{n-1}$

First case: |G| is odd, then every non-trivial common divisor of n and |G| would be odd and therefore in the set {3,...,n}. so gcd(|G|,n)=1. Analogue n-1 and |G| are relative prime. So $x\mapsto x^{n-1}$ and $x\mapsto x^n$ are injective.
So we have for arbitrary x and y:
$\begin{math} xy^n x^{-1}=(xyx^{-1})^n=x^n y^n x^{-n}\\ \Rightarrow  y^n=x^{n-1} y^n x^{-(n-1)}=(x^{n-1} y x^{-(n-1)})^n\\ \Rightarrow  y=x^{n-1} y x^{-(n-1)}\\ \Rightarrow  x^{n-1}=y^{-1} x^{n-1} y=(y^{-1} xy)^{n-1}\\ \Rightarrow  x=y^{-1} xy\\ \Rightarrow  \text{G is abelian} \end{math}$

Special case: n=3.
Since 3 doesn't divide |G|, n and |G| are relative prime, so $x\mapsto\ x^n$ is an automorphism. Again we have:
$\begin{math}xy^3 x^{-1}=(xyx^{-1})^3=x^3 y^3 x^{-3}\\ \Rightarrow  y^3=x^2 y^3 x^{-2}=(x^2 y x^{-2})^3\\ \Rightarrow  y=x^2 y x^{-2} \Rightarrow  y x^2=x^2 y\\ \Rightarrow  x^2 y^2=y^2 x^2=(xy)^2\\ \Rightarrow  xy=yx\\ \end{math}$

Third case: |G| is even, n > 4 and n odd
This is a tricky one: 4 doesn't divide |G| and so |G|=2u for an odd number u. Like in case 1 we have that $x\mapsto x^n$ is an injective map, but this time $x\mapsto x^{n-1}$ isn't injective because n-1 is an even number.
Let x,y again be arbitrary elements from G, then analogue to case 1 the identity $y=x^{n-1} y x^{-(n-1)}$ holds. $\Rightarrow  x^{n-1} y^{n-1}=y^{n-1} x^{n-1}=(xy)^{n-1}$ So the map $x\mapsto x^{n-1}$ is a homomorphism $G\rightarrow Z(G)$ where Z(G) denotes the centre of G. Let now a be an element of the kernel of this homomorphism: $a^{n-1}=1$. Then (because there are no odd common divisors of n-1 and |G|) its order has to be a power of two and since |G|=2u, it has to be 1 or 2. So a lies in one and therefore in all 2-sylowgroups.
In fact there is an element of order 2 in the kernel, because $x\mapsto\ x^{n-1}$ isn't injective. So the kernel is the (only) 2-sylowgroup of G. Now G/Z(G) is isomorphic to a subgroup of $\mathbb{Z}_2$ and therefore cyclic. So G is abelian.

Fourth case: |G| is even, n>4 and n is even
This is similar to the third case: Now n-1 is odd and therefore relative prime to |G| and so the antihomomorphism $x\mapsto x^{n-1}$ is injective.
$\begin{math} xy^{n-1}x^{-1}=(xyx^{-1})^{n-1}=x^{-(n-1)} y^{n-1} x^{n-1} \Rightarrow  y^{n-1}=x^{-n} y^{n-1} x^n = (x^{-n} y x^n)^{n-1} \Rightarrow  y=x^{-n} y x^n \end{math}$
Now the argument is the same: $x\mapsto x^n$ is now a homomorphism $G\rightarrow Z(G)$, its kernel is the only 2-sylowgroup, so the quotientgroup G/Z(G) is cyclic and G is abelian.

Finally: QED :)


calculus
analytic geometry
derivative
graphing lines
slope
Euler
Pythagorean Theorem

How do you solve this problem????

A right triangle has legs of length 3 and 4. What is its hypotenuse?

I've tried every single method but I just can figure it out!!! :(  I even used calculus!!! :o

I have an idea. What if you make a triangle with coordinates $ (3,0),(0,4),(0,0)$, and then use the distance formula to find the distance fromula? Oh wait. That won't work. Maybe if you find the derivative of $ 4x\plus{}3y\equal{}12$, that will help!!!

Isn't there some kind of theorem that solves these kinds of problems? Some weird greek dude invented it...(i think)... :maybe:

I have no idea. What I do know is that the derivative of $ y\equal{}\minus{}\frac43x\plus{}4$ is $ \minus{}\frac43$, so the slope is $ \minus{}\frac43$. Now, since we have the slope, and two points on the line, we can find the equation of the line!!! Using point-slope, we have $ y\minus{}4\equal{}\minus{}\frac43x$, and the equation of the line is $ y\equal{}\minus{}\frac43\plus{}4$!!! Now we have an equation. Maybe solve and use the distance formula?

:Prompt $ A,B$
:Disp $ \sqrt {A^2 \plus{} B^2}$

Is something wrong with this? I let $ A$ be the hypotenuse. Then, if $ \sqrt{A^2\plus{}B^2}\equal{}A$, $ A^2\equal{}A^2\plus{}B^2$ and $ B\equal{}0$!!!

Yes!!! I finally found it!!!! Apparently, using the Pythagorean Theorem, we can solve this problem!!. Anyone know who came up with the Pythagorean Theorem????? Beats me.... :maybe:

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

Wikipedia says "There is much evidence that Babylonian mathematicians understood the principle, if not the mathematical significance".

So, I conclude that the Babys found it first.

Wait, isn't Pythagoras that person who discovered $ \pi$????

No, I think first Euler made it up while studying trigonometry. He got bored with degrees so while eating dessert, he made up radians, and effectively, $ \Pi$.

What????? But I thought Euler was the guy that came up with Fermat's Last Theorem....

No you guys are all total noobs.

Fermat's last theorem was thought up by....Andrew Wiles!

But Andrew Wiles proved that there are finitely many prime numbers!!!

But I thought Archimedes proved that.... :huh:

I think he's referring to "that" in your sentence, which refers to Andrew Wiles's finite existence. I didn't know Wiles was immortal. :maybe:  :huh:

So let me get his straight:

Pythagoras invented $ \pi$,

Euler invented Fermat's Last Theroem,

Archimedes proved that there is a finite number of primes,

and,

Andrew Wiles is immortal?

If you flip the W over and rearrange the letters, Wiles becomes Smile. Does immortality make [b]you[/b] smile?

As of now it does.
So now who discovered pythagorean theorem?

No he said the Babylonians did.
There is a big difference.

Yes but all Babylonians were babies once. So former babies invented it. Former babies invented everything. Except CNRhK. Chuck Norris was never a baby. He was born beastly.

All of you are quite wrong. The Flying Spaghetti Monster was the founding of the universe, and all of its stuff like Ragu and Pythagorean Theorem


Let $ a$, $ b$, and $ c$ be non-zero integers satisfying \[ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{}\frac{1}{a\plus{}b\plus{}c}.\] Find all integer $ n$ such that \[ \frac{1}{a^n}\plus{}\frac{1}{b^n}\plus{}\frac{1}{c^n}\equal{}\frac{1}{a^n\plus{}b^n\plus{}c^n}.\]

if $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{}\frac{1}{a\plus{}b\plus{}c}$ then $ (a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}0$.
and $ \frac{1}{a^n}\plus{}\frac{1}{b^n}\plus{}\frac{1}{c^n}\equal{}\frac{1}{a^n\plus{}b^n\plus{}c^n}$ iff $ (a^n\plus{}b^n)(c^n\plus{}b^n)(b^n\plus{}c^n)\equal{}0$.
now if $ n$ is even ,the last equality can't be realised , so $ n$ must be odd and recall the identity $ x^n\plus{}y^n\equal{}(x\plus{}y)(x^{n\minus{}1}\plus{}.....)$ we get the result.
so $ n$ is odd.


Not very many of us, but post a hi or something.

Hi.

Um...I don't actually live in Kansas now, but I've lived in Kansas City for about three months last year before summer, so I guess I'm still eligible to post here, considering the inactiveness of this forum. :) To avoid being spamming, I'll comment some on my stay there: a great place for enjoying life, but not that good for studying math. For sure this is not the reason that I moved, though...

Well, I confess that I was just being bored somehow...


I wonder if he meant  to put the x there.

You seem to be confusing two ways to write composite functions.  I believe you just mean $f(g(x)) = g(f(x)) = x$, or in the other notation, $(f \circ g) (x) = (g \circ f) (x) = x$.

[hide="1"] $f(g(x)) = f \left( \frac{1}{3}(4-x) \right) = 4-3 \left( \frac{4-x}{3}\right) = 4-(4-x) = x$

$g(f(x)) = \frac{4-(4-3x)}{3}= \frac{3x}{3}= x$ [/hide]
[hide="2"] $f(g(x)) = \frac{1}{ \frac{1}{x}}= x$

In this example $f(x) = g(x)$ so of course $g(f(x)) = f(g(x))$. [/hide]

Good question.

What does the circle mean?

$(f \circ g)(x)$ is the same as saying $f(g(x))$, but it is another notation

Unfortunately, some people use $(f \circ g)(x)$ to mean $g(f(x))$, which is why I don't bother with that notation.

The circle means the operation composition, i.e. composition of functions.


An awesome number is a number that is the product of two consecutive primes, for example, 77 is awesome because 7 x 11 = 77. What is the least common multiple of the 3 smallest awesome numbers?

The 3 least awesome numbers are 11x13, 13x17 and 17x19 whose LCM is 11x13x17=2431 of course

I am a big idiot. I didn't read the question properly. I thought three smallest 3-digit awesome numbers.lol

[quote="darkdieuguerre"][hide="Solution"]The 3 smallest awesome numbers are $ 2\times 3$, $ 3\times 5$, and $ 5\times 7$, or $ 6$, $ 15$, and $ 35$.

Their LCM is $ 2\times 3\times 5\times 7 \equal{} \boxed{210}$.[/hide][/quote]

We don't need to calculate the actual awesome numbers, just their prime factorizations.

[quote="gouthamphilomath"]Yeah, we need not find the exact numbers. Just find the lcm as I did in my first post[/quote]

Again, you worked out the process for the l.c.m. incorrectly in your first post.

[quote]
11x13, 13x17 and 17x19 whose LCM is 11x13x17=2431 of course
[/quote]

11x13,  13x17, 17x19 whose LCM is 11x13x17x19 = 46189
That's what it would have been.

But darkdieuguerre showed the smaller factors with the correct product.

uh oh, Idiot I am. I get the idea but mess it up in calculations. Is there a cure for that? Thanks


Anybody who would like to help: just pm me your email address (since i have to forward it to you). (I promise i won't spam, threaten, or do anything else to you)

This poem was written by a terminally ill young girl in a New York Hospital.

It was sent by a medical doctor - Make sure to read what is in the closing statement AFTER THE POEM.

SLOW DANCE
Have you ever watched kids
On a merry-go-round?
Or listened to the rain
Slapping on the ground?
Ever followed a butterfly's erratic flight?
Or gazed at the sun into the fading night?
You better slow down.
Don't dance so fast.
Time is short.
The music won't last.
Do you run through each day
On the fly?
When you ask How are you?
Do you hear the reply?
When the day is done
Do you lie in your bed
With the next hundred chores
Running through your head?
You'd better slow down
Don' t dance so fast.
Time is short.
The music won't last.
Ever told your child,
We'll do it tomorrow?
And in your haste,
Not see his sorrow?
Ever lost touch,
Let a good friendship die
Cause you never had time
To call and say,"hi"
You'd better slow down.
Don't dance so fast
Time is short.
The music won't last.
When you run so fast to get somewhere
You miss half the fun of getting there.
When you worry and hurry through your day,
It is like an unopened gift....
Thrown away.
Life is not a race.
Do take it slower
Hear the music
Before the song is over.

PLEASE FORWARD THIS TO HELP THIS LITTLE GIRL.
ALL FORWARDED E-MAILS ARE TRACKED TO OBTAIN THE TOTAL COUNT.

Dear All:

PLEASE pass this mail on to everybody you know.
It is the request of a special little girl who
will soon leave this world as she has cancer.
Please send this to everyone you know or don't know.
This little girl has 6 months left to live, and as her dying wish,
She wanted to send a letter telling everyone to live
their life to the fullest, since she never will.
She'll never make it to prom, graduate from high
school, or get married and have a family of her own.
By you sending this to as many people as possible,
you can give her and her family a little hope,
because with every name that this is sent to,
The American Cancer Society
will donate 3 cents per name to her treatment and recovery plan.
One guy sent this to 500 people! So I know that we can
send it to at least 5 or 6.
Just think it could be you one day. It's not even
your money, just your time!

PLEASE PASS ON AS A LAST REQUEST

Dr. Dennis Shields, Professor
Department of Developmental and Molecular Biology

The poem and the entire thing is really touching, but I believe they have proved that all chain letters like this one where some organization will donate some money to a kid for every time it's forwarded, is fake. :dry:

But, the poem is really nice and touching...  :|

[quote] It's not even
your money, just your time! [/quote]

  Are you kidding me?  Time [i]is[/i] money.

  Beyond this, a mediocre peom, nothing truly original.  It is giving us two motives simutaneosly, as to why to pass this letter on.  One of pity, the other of profit.  An effective combination for getting people to cash out the bucks, and a common trait of "scam" emails.  Bah, if we were truly interested in helping cancer patients we'd donate money oursleves.  As for "enjoying life to the fullest", this is one of many phrases we never quite stop hearing enough of, perhaps its time we started learning from the consequences of our actions.

i am not sure but it seems to me like this is a chain letter because it sounds like one because i got one a long time ago but nothing happend

I've gotten so many of these types of chain things that I'm no longer sure whether it's true of not...

XD

I've got around ten of these mails this month.

Always look up such things on snopes.com to avoid embarrassing yourself. :blush:

Now for my contribution to the cause of dupes:

I think people should really read this. And no, this
email so does not contain a
virus. I sent it myself! It's a chain of copying and
pasting, not forwarding. :)

IF YOU DON'T SIGN THIS THEN YOU ARE SICK!
by the way, highlight it all and copy (ctrl + c) and
paste (ctrl + v)
and go to compose a new message, do not forward.
This is so terrible. A site that we were able to shut
last year has
returned We have to try to shut it down again! (
http://www.bonsaikitten.com) A
Japanese man in New York breeds and sells kittens that
are called
BONSAI CATS. That would sound cute, if it weren't
kittens that were
put into little bottles after being given a muscle
relaxant and then
locked up for the rest of their lives!! The cats are
fed through a straw and
have a small tube for their faces. The skeleton of the
cat will take on the
form of the bottle as the kitten grows. The cats never
get the opportunity
to move. They are used as original and exclusive
souvenirs. These are the
latest trends in New York, China, Indonesia and New
Zealand. If you
think you can handle it, go to this site
http://www.bonsaikitten.com ! and have a look
at the methods being used to put these little kittens
into bottles. This
petition needs 500 names, so please put your one name
on it!!! Copy the
text into a new email and put your name on the bottom,
then send it to
everyone you know!

Unfortunately, this email had 446 people at the bottom.

 :rotfl:

Billy

Hey I've gotten that one. It's sick...if its true

Yeah, I guessed that(I'm still not sure)

i dont see the point of chain mail there is nothing good if the whole world sees it

[quote="mathclass"][quote="nonie"]Hey I've gotten that one. It's sick :mad: ...[/quote]

dude. you do know that bonsai kittens isn't for real right? It's trying to show how we can be hypocritical. like, how we can shut up cows for their entire lives and kill them and leave them in horrible conditions, and when the cute little kitties are faced with the same thing, everyone protests! but who cares about the ugly cows or pigs?[/quote]

The poor cows in feedlots are taken away from their mothers, shipped far away on train cars stuffed together with lots of other cows, and put in a feedlot. There they live in manure a foot deep, with caked manure weighing 10 pounds on their tail, for the rest of their life, without enough room to lie down. Every week or so they are sprayed with a toxic drench to keep down the flies. If they get sick, they may either get antibiotics to keep them alive till they can be slaughtered, or they may be taken straight to the butcherhouse. At the end of their miserable existance in feedlots, they are shipped again to the butcherhouse., where if a cow is sick but can still walk(even if it has to be prodded!), it can be killed and made into people food. And they wonder why meat is becoming less popular.

Note that I am in no way advocating not killing animals, nor being vegetarian. I know that the only humane way to raise a cow or sheep or other ruminant is on what God made them to eat - grass. I'm just saying that feedlots should not exist.

Oh, yes, Nonie, search snopes.com for "bonsai kitten" and hear all about it! It's actually quite humorous that people would believe that: what business sends people to it's website and doesn't let you order one?

Billy

This ones halarious:

http://www.savetoby.com/

i went on snopes.com

Claim:   A 7-year-old girl named Amy Bruce is dying of lung cancer and a brain tumor.

Status:   False.

Example:   [Collected on the Internet, 1999]

Sorry.  I'm an idiot.   :roll:

[quote="noneoftheabove"]what is snopes.com what does it do what is the point of it and how does it know if it is true or not[/quote]

you know, you could GO THERE YOURSELF....

anyways, i think the bonsai kittie thing was a bunch of kids from MIT. honestly, too much time on their hands!

holy crap...that is sick and twisted
but funny in a dark way lol

[quote="zeb"]Here's another sad story:

I was rushing around trying to get some last minute shopping done. It was dark, cold, and wet in the car park as I loaded my car up with the gifts I felt obligated to buy. I noticed that I was missing the receipt which I would need to get out of the car park, so mumbling under my breath, I retraced my steps. As I was searching the wet pavement for the lost receipt, I heard a quiet sobbing.

The crying was coming from a poorly dressed boy of about 10 years old. He was short and thin. He had no coat. He was just wearing a ragged flannel shirt to protect him from the cold evenings chill. Oddly enough, he was holding two fifty pound notes in his hand. Thinking that he had got lost from his parents, I asked him what was wrong and he told me his sad story.

He said that he came from a large family. He had three brothers and four sisters. His father had died when he was seven years old. His mother was poorly educated and worked two full time jobs to make ends met. Nevertheless, she had managed to scrimp and save two hundred pounds to buy her children Christmas presents. The young boy had been dropped off, by his mother, on the way to her second job. He was to use the money to buy presents for all his brothers and sisters and save just enough to take the bus home. He had not even entered the shopping centre, when an older boy grabbed two of his fifty pound notes and disappeared into the night.

"Why didn't you scream for help?" I asked.
The boy said, "I did."
"And nobody came to help you?"
The boy stared at the ground and sadly shook his head.
"How loud did you scream?" I enquired.
The soft-spoken boy looked up and meekly whispered, "Help me!"
I realised that absolutely no one could have heard that poor boy cry for help.

So I grabbed his other two fifty pound notes and ran off.  :ninja:

 :rotfl: (from boreme.com).[/quote]

LOL! that was funny. it is a joke right? if it were true i'd feel so bad for the little boy! :(

How about Happy stories :D? Does anybody wants to tell one :P?


function
complex analysis
complex analysis unsolved

Let $ f$ be an entire function. Prove that if $ f$ is injective then $ f(z) \equal{} az \plus{} b$, for some $ a,b \in \mathbb{C}$, $ a\neq 0$.

Can I show this elementary? (not using Picard, or working in the extended complex plane)

$ g(z) \equal{} f(\frac {1}{z})$ defined on $ \mathbb C\setminus\{0\}$ is injective, hence has a non-essential singularity at $ z_0 \equal{} 0$, so $ f$ is a polynomial. Since $ f$ is injective, it is affine.

Hope this is elementary enough.  :)

That was my idea too... I only used the terminology "$ f$ has a singularity at $ \infty$ if $ f(1/z)$ has a singularity at $ 0$". That's what I meant with "working in the extended complex plane". I think this is the most straightforward approach though.

hmm ... how does one show that the singularity at $ 0$ is nonessential? This follows from big Picard but is there something simpler you're using ?

[quote="yannick"]Can't we use Casorati-Weierstrass for that?[/quote]Yes, this is what I had in mind.


[quote="darij grinberg"][quote="chien than"]Let $ ABC$ be a triangle with incircle $ (I)$.$ (I)$ touch $ BC;CA;AB$ at $ D;E;F$,respectively.$ AD$ cuts circle $ (I)$ at $ X$.$ BX;CX$ cut $ (I)$ at $ Y;Z$,respectively.Prove that $ AX;BZ;CY$ are concurrent.[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=2619

  darij[/quote]
$ AY$ cuts circle $ (I)$ at $ P$;$ AZ$ cuts $ (I)$ at $ Q$
Prove that $ AD;FP;EQ$ are concurrent


Αν $p(p+2k)$ διαιρει το $(p-1)!+1$ και $k<p(p-1)/2$ τοτε $p$ διαιρει το $k!-1$ ($p,k$ θετ ακεραιοι και $p$ πρωτος.)

Τοσες μερες αλυτη!Αν εχει κανεις καμια ιδεα ας την προτινει. :wink:

Το ότι ο $p$ είναι πρώτος δεν προκύπτει από το θεώρημα Wilson?

epishs kai o $p+2k$ einai prwtos  :ninja:

Συγνωμη παιδια γιατι χαρις την συνομιλια που ειχα με τον Κωστα για την παρουσα ασκηση ανακαλυψα οτι εχω κανει λαθος.Παντος ο galeos βρηκε αυτο που ηθελα και σε αυτο ειχα σκοπο να επικεντρωθειτε.Για να το διορθωσω αποτι βλεπω αντι για $p$ διαιρει το $k!-1$ επρεπε να γραψω $p+2k$ διαιρει το
$(2k)!-1$. :wink: Μπραβo galeo. :)

panw pou etoimazomoun na grapsw programma gia na dw an isxuei! :roll:
to neo problhma bgainei "grhgora" xrhsimopoiwntas oti $\binom{p+2k-1}{p-1}\equiv 1\mod (p+2k)$. (apo thn idiothta $\binom{p-1}{k}\equiv(-1)^{k}\mod p$ an $p$ prwtos kai $k<p$)
alliws mporoume na poume oti $p+2k/(p+2k-1)!+1$, $p+2k/ (p-1)!+1$ apo tis opoies vgazoume oti $p+2k/ p(p+1)\cdots(p+2k-1)-1$. To zhtoumeno akolouthei an parathrhsoume oti $p\equiv-2k\mod (p+2k),\ldots,p+2k-1\equiv-1\mod (p+2k)$. :!:


ratio
geometry
circumcircle
analytic geometry
vector
similar triangles
geometric transformation

Let two fixed points $A, B$ . $A \in Ox$ and $B \in Oy$ . Let $M , N$ move on $Ax , By$  respectively satisfying : $BN =k.AM$ , $k>0 , k=const$ . Prove that : The circumcircle of $\triangle OMN$ passes through a fixed point

$A(a,0)$, $B(0,b)$, $M(m,0)$, $N(0,n)\ ;\ M\in [Ax\Longrightarrow n\ge a;\ N\in [By\Longrightarrow n\ge b\ .$
$BN=k\cdot AM\ (k>0)\Longrightarrow n-b=k(m-a)\Longrightarrow \boxed {\ n=km+b-ka\ }\ .$
The equation of the circumcircle of the triangle $OMN$ is
$\left (x-\frac m2\right)^2+\left(y-\frac n2\right)^2=\frac 14 (m^2+n^2)$, i.e.
$x^2+y^2=mx+ny\Longrightarrow x^2+y^2=mx+(km+b-ka)y\Longrightarrow$
$m(x+ky)=x^2+y^2+(ka-b)y$. The coordinates of the fixed point $F$ (if it exists)
are the solution of the following system: $\boxed {\ x+ky=0,\ x^2+y^2+(ka-b)y=0\ }\ ,$
i.e.$\boxed {\ x_F=\frac{k(ka-b)}{k^2+1},\ y_F=\frac{b-ka}{k^2+1}\ }\ .$ Now you can easily find a syntetical solution for this problem.

A slightly different formulation:

[color=blue][b]Theorem 1.[/b] Let x and y be two lines intersecting at a point C, and X and Y two points on these lines x and y, respectively. Furthermore, let k be a real number. Then, for any two points A and B on the lines x and y such that $YB=k\cdot XA$, the circumcircle of triangle CAB passes through a fixed point.[/color]

Here, the segments are directed.

[i]Proof of Theorem 1.[/i] Let X' be any point on the line x, and Y' be the point on the line y such that $YY^{\prime}=k\cdot XX^{\prime}$. Since we also have $YB=k\cdot XA$, we thus get $\frac{XA}{XX^{\prime}}=\frac{YB}{YY^{\prime}}$.

Let R be the common point of the circumcircles of triangles CXY and CX'Y' different from C. Then, using directed angles modulo 180°, we have < RXC = < RYC, what rewrites as < RXX' = < RYY', and similarly we get < RX'X = < RY'Y. Thus, the triangles RXX' and RYY' are directly similar. Now, since $\frac{XA}{XX^{\prime}}=\frac{YB}{YY^{\prime}}$, the points A and B are corresponding points in these two similar triangles (in fact, they divide the corresponding sides XX' and YY' in the same ratio), and since corresponding points in directly similar triangles form equal angles, this yields < RAX = < RBY, what rewrites as < RAC = < RBC and thus shows that the circumcircle of triangle CAB passes through the point R. Since R is a fixed point, this proves Theorem 1.

  Darij

Circles CXY and CAB cut in the center O of the spiral similarity that carries AB to XY with coefficient k. And for any two points
A',B' on x,y such that XA'/YB'=k and vectors XA',YB' preserve the same direction  the circumference CA'B' goes through the  same center O of the spiral similarity, this fixed point is O.


inequalities
number theory
relatively prime
algebra proposed
algebra

Let n be an integer which is greater than 1, not divisible by 1997.
Let $ a_m\equal{}m\plus{}\frac{mn}{1997}$ for all m=1,2,..,1996
$ b_m\equal{}m\plus{}\frac{1997m}{n}$ for all m=1,2,..,n-1
We arrange the terms of two sequence $ (a_i), (b_j)$ in the ascending order to form a new sequence $ c_1\le c_2\le ...\le c_{1995\plus{}n}$
Prove that $ c_{k\plus{}1}\minus{}c_k<2$ for all k=1,2,...,1994+n

[quote="mr.danh"]Let n be an integer which is greater than 1, not divisible by 1997.
Let $ a_m \equal{} m \plus{} \frac {mn}{1997}$ for all m=1,2,..,1996
$ b_m \equal{} m \plus{} \frac {1997m}{n}$ for all m=1,2,..,n-1
We arrange the terms of two sequence $ (a_i), (b_j)$ in the ascending order to form a new sequence $ c_1\le c_2\le ...\le c_{1995 \plus{} n}$
Prove that $ c_{k \plus{} 1} \minus{} c_k < 2$ for all k=1,2,...,1994+n[/quote]
Have anyone tried to solve this problem :| ? If u have, plz solve it for me, cuz I have'nt known the solution ...~.~

We can generalize this problem as follows:

Given two relatively prime integers $A,B$, and
$a_m = m(1+ \frac{A}{B}), m = 1,..., B-1$
$b_m = m(1 + \frac{B}{A}), m = 1,..., A-1$
And assemble $c_i$ as described in the original problem, prove that the difference between 2 neighboring $c_i$ is less than 2.

It suffices to prove that for each $a_k$, we can find another $a_i$ or $b_j$ that lies between $a_k$ and $a_k+2$, and vice versa, for each $b_l$, we can find another $a_i$ or $b_j$ that lies between $b_l$ and $b_l + 2$.
Without loss of generality, we may assume that $A > B$.
The distance between two consecutive $b_j$s are $b_{j+1} - b_j = 1 + B/A < 2$, so for each $b_l$, we're guaranteed that $b_l < b_{l+1} < b_l+2$.

Now we're left to consider $a_k$.
Fix $k$, and now take the smallest $l$ such that $b_l > a_k$. This is always possible because:
\[a_k \leq a_{B-1} = (B-1)(1+A/B) < (A-1)(1+B/A) = b_{A-1}\]
(the middle inequality is true because $A > B$.)

If $l$ is the smallest such $l$, that means
\[b_{l-1} \leq a_k \iff (l-1)(1+B/A) \leq k(1+A/B) \iff l/A - k/B \leq 1/A\]

So
\[b_l - a_k = l(1+B/A) - k(1+A/B) = (A+B)(l/A - k/B) \leq (A+B)/A < 2\]
which completes the proof


If $a,b,c$ are positive numbers prove that :
$(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c})\geq \frac{9}{1+abc}$

[quote=harazi]Use the inequality $\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\geq \frac{3}{1+abc}$ and the rearangement inequality.  I remember Vasc conjecturing some time ago that a general inequality in n bariables should be true. Does anyone have a solution?[/quote]
http://www.artofproblemsolving.com/community/c6h1061354p4599534


inequalities
function
integration
probability
real analysis
real analysis unsolved

Let $ f$ be a Schwarz function on $ \mathbb{R}^{n}$. Prove that the following inequality holds:
\[ \parallel{}f\parallel{}^{2}_{L^{2}(\mathbb{R}^{n})}\leq\frac {4\pi}{n}\inf_{y\in\mathbb{R}^{n}}\left[\int_{\mathbb{R}^{n}}|x \minus{} y|^{2}|f(x)|^{2}dx\right]^{1/2}\cdot \inf_{z\in\mathbb{R}^{n}}\left[\int_{\mathbb{R}^{n}}|\xi \minus{} z|^{2}|\hat f(\xi)|^{2}dx\right]^{1/2}.
\]

[quote="Cezar Lupu"]Let  ....
\[ \leq\frac {4\pi}{n}\inf_{y\in\mathbb{R}^{n}}\left[\int_{\mathbb{R}^{n}}|x \minus{} y|^{2}|f(x)|^{2}dx\right]^{1/2}\cdot \inf_{z\in\mathbb{R}^{n}}\left[\int_{\mathbb{R}^{n}}|\xi \minus{} z|^{2}|\hat f(\xi)|^{2}dx\right]^{1/2}
\]
[/quote]
 and
$ \int_{\mathbb{R}^{n}}|\xi \minus{} z|^{2}|\hat f(\xi)|^{2}dx \equal{} \infty$ for nonzero $ f$

 :)

http://www.mathnet.or.kr/mathnet/kms_tex/99938.pdf   theorem 3.4

It would help if you gave the definition of Fourier transform that you're using - in particular, where the $ 2\pi$ is in your definition.

Note that if you consider $ |f(x)|^2$ and $ |\widehat{f}(\xi)|^2$ to be (constant multiples of) probability densities, the infinums on the the right hand side are achieved by setting $ y$ equal to the mean (center of mass) for $ x$ and $ z$ equal to the mean (center of mass) for $ \xi.$ Under that interpretation, the whole right hand side is the product of two standard deviations.

I put the proof for $ n\equal{}1$ as Chapter 14 in [url=http://www.csulb.edu/~kmerry/]my course notes.[/url]


i am writing in WinEdt. I just notice that the page numbering starts  with 1 with each chapter. How do i change that i counts continuously to the end? Thx.

i found my error - i was using \usepackage{chappg}. Now i wonder how i can number my pages continuously in the main part and start with page 1 in each chapter in the appendix (A-1, B-1, ...) Any ideas about that?? Thx.


[quote="filomen"]If $ a,b,c > 0$, then

$ \frac {5c^2 \minus{} 2a^2 \minus{} 2b^2}{c(a \plus{} b)} \plus{} \frac {5a^2 \minus{} 2b^2 \minus{} 2c^2}{a(b \plus{} c)} \plus{} \frac {5b^2 \minus{} 2c^2 \minus{} 2a^2}{b(c \plus{} a)}\leq\frac {3}{2}$[/quote]

$ \Longleftrightarrow 4(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \plus{} 8\sum(b^2c^2(a \minus{}$ $ b)(a \minus{} c)) \plus{} 6abc\sum(a(a \minus{} b)(a \minus{} c)) \plus{} 7abc\sum((b \plus{} c)(a \minus{} b)(a \minus{} c))\geq 0$

 :lol:

[quote="filomen"]If $ a,b,c > 0$, then

$ \frac {5c^2 \minus{} 2a^2 \minus{} 2b^2}{c(a \plus{} b)} \plus{} \frac {5a^2 \minus{} 2b^2 \minus{} 2c^2}{a(b \plus{} c)} \plus{} \frac {5b^2 \minus{} 2c^2 \minus{} 2a^2}{b(c \plus{} a)}\leq\frac {3}{2}$[/quote]

More strong:

$ \frac {5c^2 \minus{} 2a^2 \minus{} 2b^2}{c}\frac{a\plus{}b}{4ab} \plus{} \frac {5a^2 \minus{} 2b^2 \minus{} 2c^2}{a}\frac{b\plus{}c}{4bc} \plus{} \frac {5b^2 \minus{} 2c^2 \minus{} 2a^2}{b}\frac{c\plus{}a}{4ca}\leq\frac {3}{2}$
<=>
$ 6cab\minus{}3a^2b\minus{}3ca^2\plus{}4b^3\minus{}3b^2c\minus{}3bc^2\plus{}4c^3\minus{}3ab^2\minus{}3c^2a\plus{}4a^3 \geq0$
<=>
$ (\minus{}a\plus{}2b\plus{}2c)(b\minus{}c)^2\plus{}(\minus{}b\plus{}2c\plus{}2a)(c\minus{}a)^2\plus{}(\minus{}c\plus{}2a\plus{}2b)(a\minus{}b)^2\geq0$
<=>
$ \frac{1}{2}\sum{(a\plus{}b\plus{}c)(b\minus{}c)^2}\plus{}3\sum{a(a\minus{}b)(a\minus{}c)} \geq 0$
BQ


The product of three integers is 36 and their sum is 11. What is the sum of their squares?

The three integers are $ 2$, $ 3$, and $ 6$, and the sum of their squares is $ 2^2\plus{}3^2\plus{}6^2\equal{}13\plus{}6^2\equal{}7^2\equal{}\fbox{49}$.

Any way to solve this without guess and check?

I don't think so, you just need to think through it really quickly.


Two circles of respective radii 7 and 5 are externally tangent to each other. How long is their common external tangent?


We have the maximum of:

$ ( {x_{i_1 } - x_{i_2 } } )^2 + ( {x_{i_2 } - x_{i_3 } } )^2 + ( {x_{i_3 } - x_{i_4 } } )^2 + ......... + ( {x_{i_{n - 1} } - x_{i_n } } )^2 + ( {x_{i_n } - x_{i_1 } } )^2$

When $ A=x_{i_1 }.x_{i_2 }+x_{i_2 }.x_{i_3 }+...+x_{i_n }x_{i_1 } \hspace{1cm}$ is minimum.

We will prove by induction that we have the minimum of $ A$ when:
$ A=x_nx_1+x_1x_{n-1}+x_{n-1}x_3+.....+x_{n-4}x_4+x_4x_{n-2}+x_{n-2}x_2+x_2x_n$

For $ n=3$ is obvious.
For $ n=4$ is easy too.

Suppose that it is true for $ n=k$

Let see $ A$ like a string. We choose the position of $ x_1$ and $ x_{k+2}$ in any string with the numbers $ x_2,x_3,...,x_k,x_{k+1}$ such that $ A$ is minimum.

Suppose that the string after we choose the position of $ x_1$ and $ x_{k+2}$ is $ ....x_j,x_1,x_p....x_h,x_{k+2},x_l.....$

Then when we add $ x_1$ to the string the value of $ A$ increase by:
$ B=x_jx_1+x_1x_p-x_jx_p$
$ B=-(x_1-x_j)(x_1-x_p)+x_1^2$
We have the minimum of $ B$ when:
$ C=(x_1-x_j)(x_1-x_p)$is maximum
Then we have the maximum of C when $ x_j=x_{k+1}$ and $ x_p=x_{k+2}$
(in this form we can prove that it is true for $ n=4$).

On the same way the best position of $ x_{k+2}$ is when $ x_h=x_1$ and $ x_l=x_2$
By the supposition, the minimum of $ A$ with the numbers $ x_2,x_3,...,x_k,x_{k+1}$ is:
$ x_{k+1}x_2+x_2x_{k}+x_{k}x_4+.....+x_{k-3}x_5+x_5x_{k-1}+x_{k-1}x_3+x_3x_n$

And choosing the best position of $ x_1$ and $ x_{k+2}$:
$ A=x_{k+1}x_1+x_1x_{k+2}+x_{k+2}x_2+x_2x_{k}+x_{k}x_4+.....+x_{k-3}x_5+x_5x_{k-1}+x_{k-1}x_3+x_3x_n$
Then it is true for $ n=k+2$ and by the induction is true for all $ n$.

Then the rearrange is:

${ ( {x_{n } - x_{1 } } )^2 + ( {x_{1 } - x_{n-1 } } )^2 + ( {x_{n-1 } - x_{3 } } )^2 + ......... + ( {x_{n-2} } - x_{2 } } )^2 + ( {x_{2 } - x_{n } } )^2$


i can't find the solution to this equation:
16(x^5)=5(y^4)+10(y^2)+1
x,y are integers

[quote="redsmartie"]Where did you find this problem?
Way to tough for me![/quote]

what grade math are you taking?

and you can simplify that equation to

$16x^{5}= 5y^{4}+10y^{2}+1$

[quote="soccerstar123"]:P  :P  :P[/quote]

that is spam...please do not post unless you are posting an answer or helping the OP to their original question

16x^5=5y^4+10y^2+1

(1,1) jumps out at me right away...  i dont know if there are any other solutions :lol:

Using modular arithmetic, you can find that because 16 divides both sides, y must be 1 greater or less than a multiple of 4 and x must be 1 greater than a multiple of 5. I'm not sure what to do with this from here.

ok, here is something that might be useful...
it's obvious that y must be odd(sorry, but i don't know how to use latex), so we put y=2k-1.
we calculate all those powers, divide by 16(left side is divisible by 16 for every odd y,it's easy too see that), and then we get:

x^5=5k^4-10k^3+10k^2-5k+1.

this can be written as:

x^5=k^5-(k-1)^5 or x^5+(k-1)^5=k^5, so by fermat's theorem there aren't any solutions except k=1,x=1, and that gives us x,y=1 as the only solution...

 :D


Εδώ θα μαζέψουμε ασκήσεις από τη συνδυαστική Γεωμετρία !  Είναι ίσως το δυσκολότερο μέρος του βιβλίου , αλλά σίγουρα θα μαζέψουμε καλά προβλήματα. ΝΑ γράφουμε αναλυτικές λύσεις , διότι παρουσιάζουν ιδιομορφία και δυσολία στην επιχειρηματολογία !
  Αρκούν [color=red]20 Ασκήσεις [/color]. Ας μαζέψουμε όμως αυτές και ξαναβλέπουμε.

 [color=red] Καλή Δύναμη !!![/color]

ΑΣΚΗΣΗ 1- (Γιάννης Παναγέας)
Δίνεται μία σκακιέρα (2n+1)*(2n+1) γεμισμένη με 2 σύμβολα το + και το - (κάθε κουτί έχει 1 σύμβολο).
Μπορείς να κάνεις "αλλαγή" των + και - ως εξής:

Επιλέγεις 2n+1 κουτάκια που ανά 2 δεν βρίσκονται στην ίδια γραμμή και στήλη και αλλάζεις το πρόσημο στο κάθε
κουτί από αυτά που επέλεξες( αν είναι - το κάνεις + και αν είναι + το κάνεις - ). Να δείξεις ότι μπορείς μετά από συγκεκριμμένες "αλλαγές" να αφήσεις στην σκακιέρα το πολύ 2n + και τα υπόλοιπα να είναι - (ανεξάρτητα από την αρχική κατάσταση της σκακιέρας)

ΛΥΣΗ-(Γιάννης Παναγέας)
Πρέπει να βρεις μία αναλοίωτη. Αρχικά μετράω τα + και αν είναι περιττά σε αριθμό αλλάζω την κύρια διαγώνιο ώστε να γίνουν άρτια σε αριθμό(ανα δύο δεν είναι στην ίδια γραμμή και στήλη) Αν είναι άρτια τα αφήνω!
Αρχική σκέψη είναι ότι μπορείς να αλλάξεις τα σύμβολα των κορυφών ενός οποιουδήποτε ορθογωνίου
χωρίς να αλλάξεις ουσιαστικά κανένα άλλο σύμβολο πάνω στην σκακιέρα ( αλλάζοντας αρχικά δύο από τις απέναντι κορυφές και κάποια άλλα 2n-1 και στη δεύρερη αλλαγή αλλάζοντας τις άλλες δύο που μένουν και τα ίδια 2n-1 που επέλεξες στην πρώτη αλλαγή)
Επίσης να παρατηρήσεις ότι με αυτο τον τρόπο δεν αλλάζει το πλήθος των + mod 2 . Με αυτή τη διαδικασία μπορείς να πετύχεις
κάθε φορά τα τετράγωνα $i_{1},\ i_{2},\ i_{3},.....,\ i_{i},\ i-1_{i},\ i-2_{i}\ ,......,\ 1_{i}$ να έχουν το πολύ 1 + και έτσι προχωράς από i=2n+1 μέχρι i=2.
Άρα στο τέλος θα μείνουν το πολύ 2n+1 +. Όμως δεν γίνεται να είναι περιττού αριθμού τα +. άρα θα έχουν μείνει το πολύ 2n +.

[size=150][color=red]Άσκηση 2[/color][/size]

Φανταστείτε ένα τετραγωνικό πλέγμα με απόσταση ανάμεσα στους κόμβους του (σημεία τομής των κάθετων και οριζόντιων γραμμών του) ίση με 1. Επίσης, έστω σχήμα εμβαδού μικρότερου από 1, κατά τα άλλα αυθαίρετο. Δείξτε ότι είναι δυνατόν να τοποθετήσουμε αυτό το σχήμα πάνω στο πλέγμα κατά τρόπο ώστε να μην πέφτει πάνω σε κανέναν κόμβο του πλέγματος.

(Η λύση στο pdf...)

[color=red][size=150]Άσκηση 3[/size][/color]

Υποθέτουμε ένα παιχνίδι κυβικής 8 x 8 x 8 τρίλιζας στον τρισδιάστατο χώρο. Πόσες διαφορετικές νικηφόρες ευθείες γραμμές οκτάδων (από ‘X’ ή ‘Ο’) είναι δυνατές?

(Η λύση στο pdf...)

[b]ΑΣΚΗΣΗ 4 [/b]

Στο επίπεδο δίνεται πεπερασμένο σύνολο πολυγώνων τα οποία ανά δύο έχουν κοινό σημείο.Αποδείξτε ότι υπάρχει ευθεία που τέμνει όλα αυτά τα πολύγωνα.

[i]Ε.Σ.Σ.Δ 1976[/i]

[b]Λύση[/b]

Προβάλλουμε όλα τα πολύγωνα σε κάποια ευθεία (ε).Η προβολή κάθε πολυγώνου στην (ε) είναι κάποιο ευθύγραμμο τμήμα.Σύμφωνα με την υπόθεση , οποιαδήποτε δύο τέτοια τμήματα έχουν κοινό σημείο.Από αυτό προκύπτει ότι υπάρχει σημείο $P$ που
ανήκει σε όλα τα τμήματα (για να αποδειχτεί αυτό , αρκεί να θεωρηθεί η (ε) ως άξονας συντεταγμένων και να σημειωθεί το ελάχιστο από τα δεξιά πέρατα αυτών των τμημάτων).Η ευθεία που είναι κάθετη στην (ε) και διέρχεται από το σημείο $P$ τέμνει
όλα τα πολύγωνα.

Επειδή βλέπω ότι το topic αυτό όσο πάει και δυσκολεύει :wink: να και μία απλή άσκηση χρωματισμών από τη Μεσογειάδα του 2003.

Χρωματίζουμε όλα τα σημεία του επιπέδου μπλε ή κόκκινα ώστε να υπάρχει τουλάχιστον ένα μπλε και ένα κόκκινο. Γίνεται:
1) κάθε κύκλος ακτίνας 1 να έχει στην περιφέρειά του ακριβώς ένα μπλε σημείο?
2) ακριβώς δύο μπλε σημεία?

Για το 1:
Έστω ότι γίνεται η υπόθεση. Επιλέγω ένα μπλε σημείο Μ1 και γράφω κύκλο ακτίνας 1. Σημειώνω το μοναδικό μπλε σημείο στην περιφέρειά του Μ2. Θεωρώ ένα σημείο Κ στην περιφέρεια του κύκλου τέτοιο ώστε <Μ2Μ1Κ=60. Το Κ είναι κόκκινο αλλιώς η υπόθεση δεν ισχύει. Επειδή το τρίγωνο που σχηματίζεται είναι προφανώς ισόπλευρο, αν διαγράψουμε τον κύκλο με κέντρο Κ και ακτίνα 1 θα έχει τα Μ1 Μ2 στην περιφέρειά του. Άτοπο αφού είναι και τα δύο μπλε. Άρα δε γίνεται.

Για το 2: (νομίζω)
Γίνεται αρκεί να θεωρήσουμε παράλληλες ευθείες σε όλο το επίπεδο που να απέχουν μεταξύ τους 2. Βάφουμε μπλε τα σημεία τους και κόκκινο το υπόλοιπο επίπεδο και έχουμε ότι κάθε κύκλος ακτίνας 1 θα εφάπτεται δύο των ευθειών ή θα τέμνει μία σε δύο σημεία.

Έχω κάνει κάτι λάθος? Υπάρχει μεγάλο πρόβλημα με αυτά τα Μεσογειακά γιατί δε βγάζουν στο διαδίκτυο τις λύσεις. :mad:


Since when could guests post and participate in Math Jams and stuff?

They can't - I hadn't set up the permissions on that new forum correctly.  They're fixed now.


How many real numbers sanisfy the following?
$3^{2x+2}-3^{x+3}-3^x-3=0$

[hide]Let $a=3^x$.

Then $9a^2-27a-a-3=0$
$9a^2-28a-3=0$
which has one positive and one negative root.  Since $3^x$ can't be negative, I'd say there is one solution.[/hide]


vector
induction
linear algebra
linear algebra unsolved

Let $K$ be an infinite field and $V_{1},V_{2},...,V_{n}$ are vector sub-space of $V$ (all these spaces are over $K$). Prove that $\cup_{i=1}^{n}V_{i}$ is sub-space of $V$ iff exists $i$ such that $V_{j}\subset V_{i}$ for all $j$. What about if $K$ is finite?

If $V$ is a finite dimensional vector space over a finite field $K,$ then there are only finitely many elements of $V.$ We can enumerate all of the nonzero elements of $V$ as $x_{i}$ and let $V_{i}=\text{span}(x_{i}).$ Then the union of all of the $V_{i}$ is $V$ itself, but (assuming the dimension of $V$ is at least 2,) no one of the $V_{i}$ contains the others.

Ok, but remain first part of problem :)
[quote="N.T.TUAN"]Let $K$ be an infinite field and $V_{1},V_{2},...,V_{n}$ are vector sub-space of $V$ (all these spaces are over $K$). Prove that $\cup_{i=1}^{n}V_{i}$ is sub-space of $V$ iff exists $i$ such that $V_{j}\subset V_{i}$ for all $j$. [/quote]

There must exist $V_{1}$ and $V_{2}$ such that no $V_{j}$ contains both $V_{1}$ and $V_{2}.$

[color=blue][I've tried several things to repair this portion of the argument - I don't have a whole argument yet. I think perhaps we need to proceed by induction on the number of subsets. Think of this as a partial argument, needing repairs.] [/color]

Choose $x\in V_{1}\setminus V_{2}$ and $y\in V_{2}\setminus V_{1}.$

Consider the set $A=\{(1-t)x+ty: t\in K\}.$ We claim that for any $j,\ V_{j}\cap A$ consists of at most one point.

To prove the claim, suppose there are two points, $x_{u}=(1-u)x+uy$ and $x_{v}=(1-v)x+vy.$ Then for any $t\in K,$ we can let $r=\frac{t-u}{v-u}.$ Then $(1-r)x_{u}+rx_{v}=(1-t)x+ty.$ In other words, we can represent anything in $A$ as a linear combination of $x_{u}$ and $x_{v}.$ Thus any subspace that contains two points in $A$ contains all of $A.$ In particular, it contains both $x$ and $y.$

[color=blue][I was trying to claim that this shows it would contain both $V_{1}$ and $V_{2}$ - I can't really claim that, yet.][/color]

But if we had that, then if each other subspace could cover only one element [color=blue][or finitely many elements][/color] of $A,$ then finitely many subspaces couldn't cover all of $A.$

Suppose $u \in V_{1}\cup \cdots \cup V_{n-1}-V_{n}$, $v \in V_{n}-V_{1}\cup \cdots \cup V_{n-1}$ and consider $u+kv$, where $k \in K$

Obviously for any $k$, $u+kv$ is not in $V_{n}$. Thus $u+kv \in V_{1}\cup \cdots \cup V_{n-1}$. Since $u+kv$ are all distinct for different k, there are infinitely many of them spread in V_1 up to V_{n-1}. WLOG assume $V_{1}$ contains infinitely many of $u+kv$. This would then mean $v \in V_{1}$, contradiction.


For natural numbers rab = scd is valid. Prove that r(a^2+b^2) + s(c^2+d^2) is no prime.

We can safely assume r, s, a, b, c, d>0. If (r, s)>1 then we're done, so we can further assume that (r, s)=1. This means that ab=ks and cd=kr with k=(ab, cd).

We multiply r(a2+b2)+s(c2+d2) with k and we get cd(a2+b2)+ab(c2+d2)=(ac+bd)(ad+bc), so what we must prove is that N=(ac+bd)(ad+bc)/(ab, cd) is not a prime.

I think this can be done by ordering a, b, c and d. If we assume N is prime then ac+bd | (ab, cd) or ad+bc | (ab, cd). We may assume WLOG a is the greatest among the 4 numbers. If b isn't the second greatest then ac>=ab or ad>=ab and we're done, so a>=b>=c, d. Now we have bc>=cd so ad+bc>cd so ad+bc can't divide cd so we have a contradiction.

I hope no mistakes wormed their way into this.

Thanks grobber , it's correct!
But the last step of u is long . Just goes as:
   Assume that a=max{a,b,c,d} this implies min{ac,ad} >= cd so
min{ac+bd,ad+bc} > cd >= gcd(ab,cd)
This contracts!


Find all natural number n such that:

$a=\frac{1000^{n}-1}{999}$ is a prime

never.
$a=\frac{1000^{n}-1}{999}$ is composite if n>1 (a=1 for n=1).

http://www.cs.utexas.edu/users/EWD/ewd13xx/EWD1307.PDF

I was browsing [url=http://www.cs.utexas.edu/users/EWD/indexEWDnums.html]this page[/url] and I chose a random paper to read. Incredibly, I ran over this problem. What are the odds of that?


Hello,

I wanted to make a table like the one on page 45 (Table 2.9) here (http://tobi.oetiker.ch/lshort/lshort.pdf).

I tried to do it but I couldn't and when I create a normal table and add a long text it just continue to the left last of the page and doesn't make a new line (I hope you get what I mean)

Thanks in advance

In \begin{tabular}{...} instead of r or c use p{[i]width[/i]} where [i]width[/i] is whatever width you want for that column eg 3cm

Sorry but seems I miss understood something, see this example:

\begin{tabular}{|c|c|}\hline{\bf Name}&{\bf Name}\\ \hline Name & Name \\ Name & Name \\ \hline \end{tabular}

\begin{tabular}{p{10cm}}\hline{\bf Name}&{\bf Name}\\ \hline Name & Name \\ Name & Name \\ \hline\end{tabular}

p{width} makes one column of that width; that is, it replaces exactly one instance of c (or r, or whatever).  If you want two columns, use p{width} p{width}.

$ \begin{tabular}{|p{5cm}|p{5cm}|}\hline{\bf Name} & {\bf Name} \\
\hline Name & Name \\
Name & Name \\
\hline\end{tabular}$

There are two columns so you must have two entries in {...} after \begin{tabular}
$ \begin{tabular}{|c|p{5cm}|}
\hline
\textbf{Name}& \textbf{Name}\\
\hline
Name & this is a long line;  notice how it is wrapped into a new line \\
Name & another long line; you can force a new line here \newline using\ \verb=\newline=\\
\hline
\end{tabular}$
By the way {\bf Name} is obsolete - it should be \textbf{Name}

Thanks a lot for clarification  :) .

I still don't know how to remove the outside borders like the site I gave in my first post

Thank you

Just remove \hline at start and end and the |
$ \begin{tabular}{cp{5cm}}
\textbf{Name}&\textbf{Name}\\ \hline
Name & this is a long line; notice how it is wrapped into a new line\\ Name & another long line; you can force a new line here\newline using\ \verb\equal{}\newline\equal{}\\
\end{tabular}$


calculus
integration
calculus computations

[color=darkred]Study the sequencies (i.e. monotonicity, boundedness and if has limit, find it) :

$ a_n \equal{} \int_0^1\frac {1}{x^n \minus{} x \plus{} 1}\ \mathrm {dx}$ and $ b_n \equal{} \int_0^1\frac {1}{x^n \plus{} x \plus{} 1}\ \mathrm {dx}$, where $ n\in \mathcal N^*\ .$[/color]

You should have said monotonicity, boundedness,  and find the limit of..., i.e.,

Study the monotonicity, the boundedness, and find the limit of the following two sequences:

$ a_{n} \equal{}$

and

$ b_{n} \equal{}$.

Thank you, Didi ! I think that now I wrote correctly.

I think it would make no difference if we consider the following two sequences, we modify a bit Virgil's problem:

Let $ k\geq 0$ be a nonnegative integer and let

$ a_{n}\equal{}\int\limits_{0}^{\infty}\frac{x^{k}}{1\minus{}x\plus{}x^{n}}dx$,

and

$ b_{n}\equal{}\int\limits_{0}^{\infty}\frac{x^{k}}{1\plus{}x\plus{}x^{n}}dx$.

Prove that the first sequence diverges and the second one converges, in which case find its limit.


Let Circle $ (A)$ and Circle $ (B)$ be the inversion images of each other with respect to Circle $ (P)$，show that the three circles $ (A),(B)$and$ (P)$ are coaxial.  Further，denote the pole of the above-mentioned common axis with respect to Circle $ (P)$ as $ Q$，show that $ P$ and $ Q$ are the two homothetic centers of Circle $ (A)$ and Circle $ (B)$．

These are basic properties of the inversion, and they are refered in all documents about inversion which can easily found on Internet.


inequalities
LaTeX
real analysis
real analysis unsolved

[img]http://img340.imageshack.us/img340/9832/11393784.jpg[/img]

I don't have good idea ,thanks~

First of all, you can use $ \text{\LaTeX}$ on this forum directly, no need to cut and paste images (especially those that run beyond the normal window width).

Now, about the question. Let $ 0\le x<y=x+h$. We have 2 estimates for $ |f(x)-f(y)|$: it is bounded by $ 2y^\alpha$ (trivial) and by $ 3x^{\alpha-2}h$ (Mean value theorem). Now just show that the minimum of these two bounds is $ \le Kh^{\alpha/2}$ for all $ x$.


Let $ z_1,z_2,z_3\in \mathbb{C}^*$ distincts such that $ |z_1|\equal{}|z_2|\equal{}|z_3|$ and $ z_1\plus{}z_2z_3\ ,\ z_2\plus{}z_3z_1\ ,\ z_3\plus{}z_1z_2\in \mathbb{R}$ . Prove that $ z_1z_2z_3\equal{}1$ .

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Note: All angles are treated mod $ 2\pi$.

We may write $ z_1 \equal{} r e^{i \alpha}$, $ z_2 \equal{} r e^{i \beta}$, $ z_2 \equal{} r e^{i \gamma}$ where $ \alpha, \beta, \gamma \in \mathbb{R}$ are distinct and $ \mathbb{R} \ni r > 0$. ($ r \equal{} 0$ would imply $ z_1 \equal{} z_2 \equal{} z_3 \equal{} 0$.)

This way the conditions become (by Euler's formula)

$ r\sin(\alpha) \plus{} r^2\sin(\beta \plus{} \gamma) \equal{} 0$
$ r\sin(\beta) \plus{} r^2\sin(\gamma \plus{} \alpha) \equal{} 0$
$ r\sin(\gamma) \plus{} r^2\sin(\alpha \plus{} \beta) \equal{} 0$

Suppose first that one of the angles (WLOG $ \alpha$) equals $ 0$ or $ \pi$.
Then the first equation becomes $ r\sin(\beta \plus{} \gamma) \equal{} 0$, which means that either $ \beta \plus{} \gamma \equal{} 0$ or $ \beta \plus{} \gamma \equal{} \pi$.

1) $ \alpha \equal{} 0$, $ \beta \plus{} \gamma \equal{} 0$. The second equation becomes $ \sin(\beta) \minus{} r\sin(\beta) \equal{} 0$ from which we get $ r\equal{}1$ (Note that $ \beta \equal{} 0$ or $ \beta \equal{} \pi$ is not possible.). Thus $ z_1z_2z_3 \equal{} 1^3 e^{i(\alpha \plus{} \beta \plus{} \gamma)} \equal{} 1$.
2) $ \alpha \equal{} 0$, $ \beta \plus{} \gamma \equal{} \pi$. The second equation becomes $ \sin(\beta) \plus{} r\sin(\beta)$ from which we get $ r\equal{}\minus{}1$, which is a contradiction.
3) $ \alpha \equal{} \pi$, $ \beta \plus{} \gamma \equal{} 0$. The second equation becomes $ \sin(\beta) \plus{} r\sin(\beta)$ from which we get $ r\equal{}\minus{}1$, which is a contradiction.
4) $ \alpha \equal{} \pi$, $ \beta \plus{} \gamma \equal{} \pi$. The second equation becomes $ \sin(\beta) \minus{} r\sin(\beta)$ which is the same situation as in the case (1).

Now suppose all of $ \sin \alpha$, $ \sin \beta$ and $ \sin \gamma$ are non-zero. Then

$ r\sin(\alpha) \plus{} r^2\sin(\beta \plus{} \gamma) \equal{} 0 \Leftrightarrow$
$ \sin(\alpha) \plus{} r \sin(\alpha \plus{} \beta \plus{} \gamma \minus{} \alpha) \equal{} 0 \Leftrightarrow$
$ \sin(\alpha) \plus{} r \sin(\alpha \plus{} \beta \plus{} \gamma) \cos(\alpha) \minus{} r \cos(\alpha \plus{} \beta \plus{} \gamma) \sin(\alpha) \equal{} 0 \Leftrightarrow$
$ 1 \plus{} r \sin(\alpha \plus{} \beta \plus{} \gamma) \cot(\alpha) \minus{} r \cos(\alpha \plus{} \beta \plus{} \gamma) \equal{} 0$

Do the same for other two equations and say $ X \equal{} \cos(\alpha \plus{} \beta \plus{} \gamma)$, $ Y \equal{} \sin(\alpha \plus{} \beta \plus{} \gamma)$. We get

$ 1 \plus{} r Y \cot(\alpha) \minus{} r X \equal{} 0$
$ 1 \plus{} r Y \cot(\beta) \minus{} r X \equal{} 0$
$ 1 \plus{} r Y \cot(\gamma) \minus{} r X \equal{} 0$

Suppose now that $ Y \neq 0$ then

$ \cot(\alpha) \equal{} \cot(\beta) \equal{} \cot(\gamma) \equal{} \frac{rX \minus{} 1}{rY}$

But the equation $ \cot(\alpha) \equal{} c$ has two solutions, since we have 3 equal cotangents we must have 2 of the angles $ \alpha, \beta, \gamma$ equal, which is a contradiction. Thus $ Y \equal{} 0$.

$ \sin(\alpha \plus{} \beta \plus{} \gamma) \equal{} 0$

Therefore we know $ \alpha \plus{} \beta \plus{} \gamma$ equals $ 0$ or $ \pi$. If $ \alpha \plus{} \beta \plus{} \gamma \equal{} 0$ then
$ X \equal{} 1$ and we get $ 1 \minus{} r \equal{} 0$ thus $ r\equal{}1$ and $ z_1z_2z_3 \equal{} 1$. If $ \alpha \plus{} \beta \plus{} \gamma \equal{} \pi$ then
$ X \equal{} \minus{}1$ and we get $ 1 \plus{} r \equal{} 0$ leading to a contradiction.

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Take $ z_1 \equal{} re^{i\alpha}$, $ z_2 \equal{} re^{i\beta}$, $ z_3 \equal{} re^{i\gamma}$, with distinct $ \alpha,\beta,\gamma \in [0, 2\pi)$, $ r \in \mathbb{R}_\plus{}^*$, and take $ \sigma \equal{} \alpha\plus{}\beta\plus{}\gamma$. Then relation $ z_1\plus{}z_2z_3 \in \mathbb{R}$ implies $ \sin \alpha \plus{} r\sin(\beta\plus{}\gamma)\equal{}0$; similarly the other cyclic two. This means that $ \alpha,\beta,\gamma$ are (all three) roots of the equation $ f(x) \equal{} \sin x \minus{} r\sin(x \minus{} \sigma) \equal{} 0$. But $ \sin x \minus{} r\sin(x \minus{} \sigma) \equal{} (1\minus{} r\cos\sigma)\sin x \plus{} (r\sin\sigma)\cos x$.

Computing $ (1\minus{}r\cos\sigma)^2 \plus{} (r\sin\sigma)^2$, this value is zero if and only if $ \sigma \equal{} 2k\pi$ and $ r\equal{}1$, whence $ \boxed{z_1z_2z_3 \equal{} 1}$. Otherwise, let's take $ \theta \equal{} \arctan \frac {1\minus{}r\cos\sigma} {r\sin\sigma}$; then $ f(x) \equal{} \sqrt{(1\minus{}r\cos\sigma)^2 \plus{} (r\sin\sigma)^2}\cos (x \minus{} \theta)$, and cannot have three distinct roots in $ [0, 2\pi)$, therefore two of the angles must be equal, absurd.


what is problem's solution
$ x^{20}\plus{}\sqrt{x\plus{}1}\minus{}x^2\equal{}1$

hello, should $ x$ be a real number?
Sonnhard.

We need $ x\geq \minus{}1$; but there LHS is $ 0$. We can write $ x^2(1\minus{}x^{18}) \equal{} \frac {x} {1\plus{}\sqrt{x\plus{}1}}$, with obvious solution $ x\equal{}0$. For $ x\neq 0$ we can cancel $ x$ to get $ f(x)\equal{}x(1\minus{}x^{18}) \equal{} \frac {1} {1 \plus{} \sqrt{x\plus{}1}} \equal{} g(x)$.
Solutions can exist only for $ x \in (0, 1)$, where $ g$ is decreasing from $ 1/2$ to $ \sqrt{2} \minus{}1$, while $ f$ looks like an inverted "parabola", with value $ 0$ in $ 0$ and $ 1$. There are therefore two more solutions in $ (0,1)$. Wolfram alpha gives as approximate values $ 0.453398$ and $ 0.969324$.


Just wondering, would things like PAI, PSS, and PCA be 'universally' accepted as abbreviations in proofs on things like USAMTS?

I cant help you , cuz i dont know what they mean...
Interpret my answer

I think it would be safe to leave them as the full form, because I can't figure them out either.

(PAI= Parallel Alternate Interior?  PCA= Parallel Corresponding angles? )

PAI=If two parallel lines are cut by a transversal, than Alternate Interior Angles are congruent.
PSS='           '                       '          '       ', then Same-Side Interior Angles are supplementaryl
PCA='          '                '                   '   ', then Corresponding Angles are congruent.
 [size=75]I got so used to using them in my math class. *sigh*[/size]:)

I had a feeling PSS mean something about same side...

I personally wouldn't recommend using those abbreviations. I've seen books just say something like:

"Since lines AB and CD are parallel, then angle [something] = angle [something else]."

But books are books...

I would say write the entire thing out, as a lot of people are not familiar with what they stand for.


Can anyone verify this?

If a polynomial $f(x) = \sum_{i=0}^{n}a_{i}x^{n-i}$ has roots $r_{1}$, $r_{2}$, $r_{3}$, ..., $r_{n}$ with $a_{0}=1$, then $0 = \sum_{k=0}^{n} \left(a_{k} \sum_{i=1}^{n}r_{i}^{m+1-k} \right)$ for some positive integer $m$.

I got it from those whatchamacallems... Newton's Sums or something? For this pattern:

$(a+b)(a^m+b^m)=a^{m+1}+b^{m+1}+(a^{m-1}+b^{m-1})ab$
$(a+b+c)(a^m+b^m+c^m)=a^{m+1}+b^{m+1}+c^{m+1}+(a^{m-1}+b^{m-1}+c^{m-1})(ab+ac+bc)-(a^{m-2}+b^{m-2}+c^{m-2})abc$
$(a+b+c+d)(a^m+b^m+c^m+d^m)=a^{m+1}+b^{m+1}+c^{m+1}+d^{m+1}+(a^{m-1}+b^{m-1}+c^{m-1}+d^{m-1})(ab+ac+ad+bc+bd+cd)-(a^{m-2}+b^{m-2}+c^{m-2}+d^{m-2})(abc+abd+acd+bcd)+(a^{m-3}+b^{m-3}+c^{m-3}+d^{m-3})abcd$

etc, etc, continuing the pattern with more and more variables, but instead replace the $a,b,c,d,...$ with roots of a polynomial and move everything to the right side.

Or have I done something terribly wrong?

No, that looks about right. The $k$th power sum is equal to the alternating sum of the product of corresponding lower power sums and symmetric sums of the roots.

Can you prove it plz??

I think I don't understand the oroginal statement??? :(  :roll:


Machine $ A$ produces $ 50$ calculator batteries per hour and costs $ \$10$ per hour to operate. Machine $ B$ produces $ 60$ calculator batteries per hour and costs $ \$15$ per hour to operate. Production demands are so great that at least $ 580$ calculators must be produced in an 8 hour day. Of all the possible ways to meet production demands, how many hours should each machine be run to minimize costs? Write your answer as an ordered pair (Machine $ A$, Machine $ B$).

Let $ a \equal{}$ number of hours machine $ A$ is run, $ b \equal{}$ number of hours machine $ B$ is run.
For an eight hour day, $ a \leq 8$ and $ b \leq 8$.
Since at least $ 580$ calculators must be made, $ 50a \plus{} 60b \geq 580$.
To get the minimum cost, one of the machines must run for $ 8$ hours.
When Machine $ A$ is ran for eight hours, we have $ 50(8) \plus{} 60b \equal{} 580$, so $ b \equal{} 3$.
When Machine $ B$ is ran for eight hours, we have $ 50a \plus{} 60(8) \equal{} 580$, so $ a \equal{} 2$.
The equation for the cost is $ C \equal{} 10a \plus{} 15b$.
Plugging in the solutions for $ a$ and $ b$:
If Machine $ A$ is ran for eight hours:  $ C \equal{} 10(8) \plus{} 15(3) \equal{} 125$.
If Machine $ B$ is ran for eight hours:  $ C \equal{} 10(2) \plus{} 15(8) \equal{} 140$.
The cost is lower to run Machine $ A$ for eight hours,
so the answer is $ (8,3)$.


Find the limit as

lim (n-->infinity and h-->0) (sinh+sin2h+sin3h+.........+sin((n-1)h))

[quote="kenn4000"]this limit is poorly defined, are n and h related?[/quote]

no....

Sorry, we can't understand the problem, or what should we find?


vector
linear algebra
combinatorial geometry
linear algebra unsolved

These are two suggested exercises from my second year linear algebra course. Could someone please show me what the proof of these would look like? I can see the answer intuitively, but I have no idea how to express it mathematically.

Let V be a vector space and A be any subset of V. There exist linear subspaces of V that contain A (for example, V). Let T={N: N is a linear subspace of V that contains A}. Show that T contains a smallest element (here the book suggests trying the intersection of all the N's in T).

I can see how this makes sense, but I have no idea how to prove that the intersection is the smallest element of T.

And the next question:

Let V be a vector space and let K be any set of linear subspaces M of V. Let U be the union of all the subspaces in K., ie

A={x in V: x in M for some for M in K}

Apply the previous question to conclude that there exists a smallest linear subspace on V that contains every M in K.

For this one, I think the answer would logically be the sum of all the M's in K, where the sum is defined as

if M,N are linear subspaces of V, M+N={m+n|m in M, n in N}

But again, how do I express that mathematically, and how do I show that a subspace is the "smallest" or "biggest"?

Thanks in advance:)

Both these exercises rely on the hereditary property of linearity with respect to intersection, and on the fact that the intersection of a family of sets is the largest set included in all. To wit:

Let $ \mathcal{F}$ be a non-empty family of subspaces of a linear space $ V$. Then $ \cap\mathcal{F} \equal{} \bigcap_{S \in \mathcal{F}} S$ is easily checked to be a linear subspace of $ V$ (just check the defining conditions).

Take now a set $ A \subseteq V$, and the family $ \mathcal{F}$ of all subspaces $ S$ such that $ A \subseteq S$ (not empty, since $ V$ is there). Then $ A \subseteq \cap\mathcal{F}$ (in fact this subspace itself belongs to the family). It is the smallest containing $ A$ since for any subspace $ T$ containing $ A$, $ T$ will belong to the family above, therefore it will also contain $ \cap\mathcal{F}$.

The same procedure is used to define the convex hull of a set, since convexity is also hereditary.


function
logarithms
complex numbers
complex analysis
complex analysis unsolved

Let $ \mathbb D : \equal{} \{z \in \mathbb C: |z|<1\}$ and $ f : \mathbb D \to \mathbb C$ a bounded analytic function such that $ f(0) \ne 0$. Suppose $ \{ z_n\}_{n\equal{}1}^\infty \subset \mathbb D$ is a sequence of complex numbers such that $ z_n \ne z_m$ whenever $ n \ne m$ and $ f(z_k)\equal{}0$ for each $ k$. Show that the series of real numbers
\[ \sum_{n\equal{}1}^\infty {\ln |z_n|}\]
converges.

Thank you  :( .

Take a look what  Blashke product is  :wink:
P.S friendly advise- don't post so many problems in one time. Just try to understand all hints and solutions for the problems,you have posted before.Then you will  solve next problems by yourself :)

[quote="Kalle"]The problem is obviously set up to be a direct application of Jensen's formula.[/quote]

Thank you very much, indeed by Jensen's formula we obtain the following
\[ \ln \left| {f\left( 0 \right)} \right| \minus{} \ln M \leqslant \sum\limits_{n \geqslant 1} {\ln \left| {{z_n}} \right|} \leqslant 0\]
where $ M$ is nothing but an upper bound of $ f$.


A spaceship of mass $ m_{0}$ moves in the absence of external forces with a constant velocity $ v_{0}$. To change the motion direction, a jet engine is switched on. It starts ejecting a gas jet with velocity $ u$ which is constant relative to the spaceship and directed at right angles to the spaceship motion. The engine is shut down when the mass of the spaceship decreases to $ m$. Through what angle $ \alpha$ did the motion direction of the spaceship deviate due to the jet engine operation?

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What I've set out to do, is this:

The standard 'rocket equation' goes like this, $ m\vec{w}\equal{}\minus{}\frac{dm}{dt}\vec{u}$. $ \vec{u}$ is always perpendicular to the motion of the spaceship, and hence so is $ \vec{w}$. That makes $ \vec{w}$ the centripetal acceleration that causes the spaceship to deviate from the straight path tracing out a circular path with constant velocity $ v_{0}$ (since acceleration is at right angles to it).

So now what? Does the angle that the circular path make at its centre equal the deviation angle $ \alpha$? (If so, how?  :huh: ) How do you find that angle any way?.. Or am I off the mark totally?

This problem comes from a [i]famous[/i] Russian book... Any suggestions people?

[/hide]

[hide="name of the book "]Problems in General Physics by  I.E.Irodov  :D [/hide]

[hide="solution"]
yes u are right upto the point that it moves in a circle and the general law governing such a motion that is
$ m\vec{w}=-\frac{dm}{dt}\vec{u}$
now let's have a general derivaton that helps u in other similar problems let the radius of curvature at time t be $ \rho(t)$ now clearly since there is only one normal accelaration and the body moves in a plane
we have $ \frac{mv_{0}^{2}}{\rho(t)}=\frac{-\lambda u}{m(t)}$ where $ \lambda =\frac{dm}{dt}$
now this gives $ \rho(t) =\frac{v^{2}}{-\lambda u}$
also that $ v_{0}=\omega(t)\rho(t)$
note that body need not be moving in a circle for these equaions to hold true
plunging the value of $ \rho(t)$
we get
$ \frac{-u\lambda}{v m(t)}=\omega(t)$
this gives
${ \frac{u}{v_{0}}}\int_{m_{0}}^{m}\frac{dm}{m}=\int_{0}^{\alpha}d\theta$
this gives $ {}alpha=\frac{u}{v_{0}}\ln\frac{m_{0}{m}$ [/hide]


search
inequalities unsolved
inequalities

Let $ a,b,c\geq 0$ and $ abc\equal{}1$. Prove that
$ \frac{1}{2\plus{}a}\plus{}\frac{1}{2\plus{}b}\plus{}\frac{1}{2\plus{}c}\leq\frac{1}{1\plus{}a\plus{}b}\plus{}\frac{1}{1\plus{}b\plus{}c}\plus{}\frac{1}{1\plus{}c\plus{}a}$
[color=red]I need a nice solution[/color]

I think It must be      $ \frac{1}{1\plus{}b\plus{}c}\plus{}\frac{1}{1\plus{}c\plus{}a}\plus{}\frac{1}{1\plus{}a\plus{}b}\leq\frac{1}{2\plus{}a}\plus{}\frac{1}{2\plus{}b}\plus{}\frac{1}{2\plus{}c}$
It's bulgaria 1997
See here : http://www.mathlinks.ro/viewtopic.php?search_id=1147236415&t=243  :wink:


Hello, can anyone help me with this problem. I knew I got this wrong at the chapter mathcounts.

An ammonia and water mixture fills a five-gallon container. Eighty percent of the mixture is ammonia but some of the mixture will be drained and replaced with pure water. If a five-gallon mixture of fifty percent ammonia is desired, how many quarts of the mixture need to be drained before the water is added? (A quart is one-fourth of a gallon.) Express your answer as a decimal to the nearest tenth.

[quote="Construe"][hide="Solution"]At the start, there are 5*80%=4 gallons of ammonia and 5-4=1 gallon of water. At the end (after adding the water), there are 5*50%=2.5 gallons of ammonia and the same amount of water. There will still be 2.5 gallons of ammonia just before the water is added at the end, because only water is added.

So, the initial mixture must be drained until there are 2.5 gallons of ammonia remaining, that is, 1.5 gallons of ammonia must be removed. However, removing ammonia also implies removing water, since it is a mixture of both that is being drained. Since 1.5/4 = 3/8 of the ammonia is removed, 3/8 of the water = 3/8 gallons of water will be removed.

Overall, 1.5+3/8=12/8+3/8=15/8 gallons need to be drained. The question asked for the answer to be in terms of quarts, so:

15/8*4=15/2=$ \boxed{7.5}$ quarts

I hope I explained that clearly.[/hide][/quote]
I'm sorry, why would you do $\frac{1.5}{4}$? There are 2 elements, that's my confusion.

How many elements there are has nothing to do with it. To find the fraction of ammonia being removed, you divide the amount of ammonia being removed by the total amount of ammonia at the beginning.


If $ x \diamondsuit y \equal{} \frac{\sqrt{xy}}{|x\minus{}y|}$, find $ 4 \diamondsuit (9 \diamondsuit 1)$ to the nearest tenth.

9 diamond 1 is just sqrt(9)/8 or 3/8.

Then we have sqrt(3/2) divided by 29/8, or in other words 8sqrt(3/2) divided by 29 = sqrt(3*32)/29 = sqrt(96)/29. We estimate this is roughly sqrt(100)/30 = 1/3 which rounds to 0.3

Sidenote: The estimation is off by about 0.0045.

[quote=LLin1]Since $ x \diamondsuit y \equal{} \frac {\sqrt {xy}}{|x \minus{} y|}$, simply plug the numbers into the
equation and you have your answer.

$ \frac {\sqrt {9\times1}}{|9 \minus{} 1|}$

$ \frac {3}{8}$

Then...

$ \frac {\sqrt {4\times\frac {3}{8}}}{|4 \minus{} \frac {3}{8}|}$

$ \frac {\sqrt {\frac {3}{2}}}{\frac {29}{8}}$

$ \sqrt {\frac {3}{2}}\times\frac {8}{29}$

$ \frac {8\sqrt {3}}{29\sqrt {2}}$

Multiply the fraction by $ \sqrt {\frac {2}{2}}$....

$ \frac {8\sqrt {6}}{58}$

$ \frac {4\sqrt {6}}{29}$

That is approximately $ \boxed{0.3}$[/quote]

How do you calculate 4sqrt6/29 without a calc in under 45 seconds

[quote=knightime1010][quote=Possible]How do you calculate 4sqrt6/29 without a calc in under 45 seconds[/quote]

We have $\sqrt{2} \approx 1.4$ and $\sqrt{3} \approx 1.7$, so $\sqrt{6} \approx 2.38$. Multiply this by $\dfrac{4}{29} \approx \dfrac{1}{7}$ to get $0.34 \approx \boxed{0.3}$.[/quote]

Maybe you can do that but I cant in 45 seconds.


HClO, HClO2, HClO3, and HClO4. List in order of increasing acidity, and give reason.

$ HClO < HClO_{2}< HClO_{3}< HClO_{4}$. In order to see this you have to draw their Lewis (Electron-Dot) Diagram. The $ O$ atom has a very high electrognegativity and has the tendency to draw electron density towards itself and away from the $ O\minus{}H$ bond and therefore weakening it. Hence the $ H^{\plus{}}$ is more easily ionised as the number of oxygens increase and this accounts for the increasing acidity.

For a given resonance hybrid, why does the oxidation number of each atom stay the same for each atom in each resonance structure while the formal charge of some atoms may differ in different resonance structures?

Ressonance structures are different Lewis structures that can be drawn for a given species. All ressonance structures must have the arrangement of atoms but differ in the distribution of electrons (although the number of unpaired electrons must be the same, as the total charge).
Formal charge and oxidation number are two different approaches that can be used to give a charge for each atom in a species. Formal charge is based in a pure covalent model, where each atom is given all its lone pairs and half the shared electrons: in essence, the formal charge is the difference between the number of valence electrons of the atom in the free state and in a particular species. As different ressonance structures have different electron distributions around each atom, then the formal charges will differ for each atom.
Oxidation number is an alternative way of giving a charge to an atom by considering an ionic model, where the electrons in covalent bonds are supposed to "belong" to the most electronegative atom. So, whichever ressonance structure you consider the number of electrons belonging to each atom will always be the same, and its oxidation number will not vary.


calculus
integration
trigonometry
logarithms
calculus computations

Compute the integral:

$\displaystyle \int_{0}^{1} \frac{ \ln(1-x^2)}{1+x^2} dx$.

thank you!

Then how to compute
$\int_0^{\tan 1}\ln (1-\tan^2 t)dt?$

first of all u have
$\displaystyle \int_0^{\ pi /4}\ln (1-\tan^2 t)dt? $


Let $ a,b,c$ are positive real numbers, and $ abc \equal{} 1$.
Show $ a \plus{} \frac {1}{a} \plus{} b \plus{} \frac {1}{b} \plus{} c \plus{} \frac {1}{c}\geq 4(\frac {bc}{b \plus{} 1} \plus{} \frac {ac}{c \plus{} 1} \plus{} \frac {ab}{a \plus{} 1})$.
If you have got a solution, please let me know
miwalin@163.com

By AM-HM,

$ b \plus{} \frac {1}{c} \geq \frac {4}{\frac {1}{b} \plus{} c} \equal{} \frac {4ab}{a \plus{} abc} \equal{} \frac {4ab}{a \plus{} 1}$

$ a \plus{} \frac {1}{b} \geq \frac {4}{\frac {1}{a} \plus{} b} \equal{} \frac {4ac}{c \plus{} abc} \equal{} \frac {4ac}{c \plus{} 1}$

$ c \plus{} \frac {1}{a} \geq \frac {4}{\frac {1}{c} \plus{} a} \equal{} \frac {4bc}{b \plus{} abc} \equal{} \frac {4bc}{b \plus{} 1}$

Adding those up gives the inequality that you want.

We have:

\[ a \plus{} \frac {1}{a} \plus{} b \plus{} \frac {1}{b} \plus{} c \plus{} \frac {1}{c}\geq 4(\frac {bc}{b \plus{} c} \plus{} \frac {ca}{c \plus{} a} \plus{} \frac {ab}{a \plus{} b})\].

[quote="fjwxcsl"]We have:
\[ a \plus{} \frac {1}{a} \plus{} b \plus{} \frac {1}{b} \plus{} c \plus{} \frac {1}{c}\geq 4(\frac {bc}{b \plus{} c} \plus{} \frac {ca}{c \plus{} a} \plus{} \frac {ab}{a \plus{} b})
\]
.[/quote]
An expanding gives
$ 2(a\plus{}b\plus{}c)^2\minus{}18\plus{}\sum_{cyc}(a^3b^2\plus{}a^3c^2\minus{}2a)\plus{}\sum_{cyc}(a^3b\plus{}a^3c\minus{}2a^2b^2)\geq0,$
which is obvious.


I think this problem will help: Solve in real numbers : $2^{x}+5^{x}=3^{x}+4^{x}$.
P.S. MVT helps here. :)