function
algebra
domain
LaTeX
calculus
derivative
complex analysis

I am a mathematical hobbyist teaching myself complex analysis from Gamelin's book. I have already struck a problem. Would appreciate help.

On page 16 of Gamelin, "Complex Analysis" we find the following text:

"Now we turn to the problem of finding an inverse function for w = . Every point w not equal to 0 is hit by exactly two values of z, the two square roots . In order to define an inverse function, we must restrict the domain in the z-plane so the values of w are hit by only one z. There are many ways of doing this, and we proceed rather arbitrarily as follows.

Note that as the rays sweep out the open right half of the z-plane, with the angle of the ray increasing from  to , the image rays under w = sweep out the entire w-plane except for the negative axis, with the angle of the ray increasing from  to  . This leads us to draw a slit, or branch cut, in the w-plane along the negative axis from  to 0 and to define the inverse function on the slit plane C\(,0]."

Now Gamelin does say this is somewhat arbitrary - but he does not motivate this "branch cut" idea well. Why are we doing this - presumably because when arg(z) =  it repeats the same values as when arg(z) =  - and so we are getting rid of repeated values to achieve a single valued inverse function. Is this the reason for doing this?

Presumably we could cut the plane at another ray? Is this right?

Another question is why are we removing the point 0 - the function and the inverse seem to be defined there and there is no question of repeated or multiple values. Is the reason something I will discover [possibly related to continuity or differentiability?] later in the book

You have some missing symbols there - you need to find some way to include them. The best way is use $ \text{\LaTeX},$ which is enabled on this site. On the other hand, in most cases, it's possible to guess from context what you actually meant.

The branch cuts were arbitrary, and could have been done differently - Gamelin specifically mentions that they could have been done differently. You probably don't really want to use cute swirly spirals, and it would probably offend your sensibilities to have a square root of a positive real to be anything other than a positive real - but you still have a lot of choices.

As for the situation at zero: there is an extra consideration here. We're interested not just in finding an inverse function, but in finding an inverse function which is itself analytic. If we have $ w=f(z),$ where $ f$ is an analytic function, and $ f'(z_0)\ne 0,$ then there is a neighborhood of $ z_0$ on which $ f$ is one-to-one, and we can choose an analytic inverse $ f^{-1}(w)$ on a neighborhood of $ w_0=f(z_0).$  And we have that $ (f^{-1})'(w_0)=\frac1{f'(z_0)}.$

But if $ f'(z_0)=0,$ then none of this works, and in particular we cannot find a function that has any of the properties of an inverse and still has a complex derivative at that point. (Note from the paragraph above the relationship between asking about that derivative and dividing by zero.)


integration
logarithms
trigonometry
calculus
calculus computations

Define $ I_{n}$ by $ I_{n}\equal{} \int_{0}^{1} \frac{x^{n}}{\sqrt{1\plus{}x^{2}}}$,  $ n \ge 0$. Evaluate $ I_{0}$ and $ I_{1}$.

Establish the reduction formula $ I_{n} \equal{} \frac{\sqrt{2}\minus{}(n\minus{}1)I_{n\minus{}2}}{n}$ for suitable values of $ n$, which you should state.

Hence calculate $ I_{5}$ and $ I_{6}$.

$ I_0\equal{}[\sinh^{\minus{}1}x]_{0}^{1}\equal{}[\ln\left(x\plus{}\sqrt{1\plus{}x^2}\right)]_{0}^{1}\equal{}\ln\left(1\plus{}\sqrt{2}\right)$,
$ I_1\equal{}[\sqrt{1\plus{}x^2}]_{0}^{1}\equal{}\sqrt{2}\minus{}1$.

To find $ I_n$, trig substitution $ x\equal{}\tan{\theta}$ and integral by part will help. I have to go now...


MATHCOUNTS ARML geometry AMC AIME trigonometry

If you are, tell what state :D... and kevin, force you know who to come!!

[quote="indianamath"]uh i'm going. but i suck.[/quote]

ok.  thats nice to know.

I'll be there with Wisconsin Red (A)...I hope

ARML is a really good competition for 8th-12th graders... some people will argue (cough richard cough) that it makes you slower, but if you think about it, doing two of the harder problems within 10 min is extremely difficult, and requires good intuition.  Anyways... its fun :D

It depends on where you live.  For ARML, regions send all-star teams, kind of like Mathcounts, except that the regions can be anything (within certain limits).  In the midwest, most of the teams are state teams.  On the east coast and in California, a lot of teams come from smaller regions.  Examples would be New York City, Western Massachusetts, San Fran Bay Area, San Diego, Montgomery County (MD), AAST and TJ (high schools that have their own ARML team), etc.  Any region can send teams if someone from the region organizes them, and if the territory they claim doesn't overlap with anyone else's territory.

Anyway, each region's organizers are free to pick the team however they choose.  Most have some sort of qualifying process, usually based on local competitions as well as the AMC's.  A few (Bay Area being one, I don't know what the others are) allow any student to join, and then hold tryouts for the better teams.

Oh, and it's not just for 8th to 12th graders.  It's for anyone who joins a team--in some regions, it's near impossible to make the team in middle school.  The Bay Area team, by contrast, sends lots of middle school students, even going down to 6th graders in some cases.

The Southern California ARML teams will have about 5 8th graders this year out of about 45 total students (3 teams). There are no 7th graders this year, but some of this year's 8th graders participated last year as 7th graders.

There were some qualifying standards this year, since more than 50 wanted to come (and there's only one bus). The standards weren't all that tough.

Is it the same qualifying test for everyone? Or are there variations?

ARML is also for people who get forced by their nationals teammate to join because SOMEBODY has something going on with her.

Mathcrazed is a bit crazed about other... subjects, as well  :wink:

huh what are you talking about dude? :wink: ... some states don't require qualifying tests... i think its just based on the coaches decision anyways... based on your performance

I'm going. Despite being repeatedly told I suck and everyone else has gotten better than me.

hehehe... im not that mean am I?... besides... you just need to get back from post-mathcounts depression syndrome :D... PM[size=59]d[/size]S... anyways... last year, some random kid from zionsville won ARML also... Deja Vu anyone?? lol

ehh your dad is saying it too lol. get on gmail later.

I'm going, Florida B, Penn.

along with ckck and ragnarok.

Then again, it depends if xxazurewrathxx (chen) accepts my payment of fake pesos i sent today...

lol fake pesos??? lmao lol ^^>..<.. i would hope that SOMEONE's mom would let her go... but uh -_- not happening.. or is it?

SAT's?? u can reschedule those? or like graduation or something else.

I'm missing my graduation party for ARML /cry.

Oh, and lyndon, u have to serve us at the graduation reception. It's 2:00 - 4:00 so u can still make it by 6:00.

Back on topic:

How comparable are mathcounts and ARML? Or how about ARML vs. AMC10/12/AIME?

So to qualify/invited, you just do well on any of the Math League, AMC, AIME, or anything else right?

Depends on your team.  Some teams have specific tryout contests that you have to come to; others just go through the high score lists and pick people.  There's a nice "How to Join an ARML Team" thread in the ARML forum where you can ask about your local team.

im in md, and i live in montgomery county.  the way our coach does it is give practice sets consisting of nyciml contests.  then he takes the top 60 high schoolers and 15 middle schoolers.  btw, which location are you guys going to.  im penn state

(me) Kevin Tian, TX (hopefully at least silver =P)
Kevin Li, TX
Kevin Chen, TX
Jeff Feng, TX
Bobby? (maybe) Generally we don't have 6th graders, but bobby is just like pwnage...

Dude, just because he's good at mathcounts doesn't mean he'll be good at ARML (Talking about Bobby Shen).  ARML is comparable to harder problems on the AMC 10/12, except harder... Sometimes, #8 will involve high level concepts, but usually, they are on the level of AIME problems from 1-10, which are actually pretty simple.  ARML problems, though, can be solved much faster.

You can't really compare them because the problems are quite easy and are like one-step but you get very little time (especially the team round) which makes it hard.

Only like four or five out of the 18 problems on the individual and team rounds are actually difficult.

hmmm...
on individual rounds i average about 5-7 (i always miss no. 8) minus careless calculation errors...
so i think bobby can get at least a 5 probably

[quote="Sly Si"]The thing with ARML problems is that they're often just one step, but it can be quite difficult to find that step.  I think this is generally more true on the individuals--team round problems are a bit more normal.[/quote]

yeah that was the point I was attempting to make, as you can see I am not great at putting things into words...

who are you jhredsox?? are you going to be in 9th grade...? it depends on what state you're in :D

you will need to be able to get about 4 on AIME.(I got 3 and i'm still doing it though.)

no you don't... don't listen to him -_-... if you want to be on Team A, you have to know most high school math and how to apply them.  Practice is the most important though, and again, always try to find several solutions to problems.

(it's a little rude just to rate simple questions and answers as spam)

well, i really don't know how to explain what i know/don't know, so it's kind of hard. i'm not exactly sure what "high school stuff" implies, so if someone could provide an answer to that, that'd be great.

till then

lol... people just love rating my posts as spam.. :D... but w/e. Ummm high school math= everything up to calc. So, just knowing algebra or geometry will sometimes actually be useful, but its usually algebra II and some more advanced geometry.  Sometimes, Trig will make it a lot simpler, but it almost never requires it.

[quote="jhredsox"]do you think i could be good enough to think about trying out for arml next year?
thanks[/quote]

That's not a very well phrased question. What does good mean? Why do you have to be good to try out? Isn't the whole point of trying out to pick the optimal team?

well, i wanna know if i'd even have a chance of making the connecticut a team...

thanks

Actually I recall Ignite168 saying that it was near impossible for sophomores and below to make the A team, because apparently they have a weird selection process. I would presume it's too late to try out by now, you should ask him.

I realize since we select 52 people, and cause a lot of people to put it nicely, are not as gifted and about 90% of the ones who are attend prep schools which typically do not go to ARML that it is relatively easy for any sophomore and under with skills to make the team. But yes, the selection process makes it near impossible for a sophomore with mediocre skills to get on the team as opposed to a senior with mediocre skills.

I really can't say whether you have a good shot at making A team or not seeing as we don't find out until we actually arrive at Penn State and I don't know your strengths at all. I could give you some runoff questions and stuff though if you message me but I have no idea what the qualifying score was for the team but as far as I know, I think I probably got over double the qualifying score.


On the equation ax+b=0, what does a, x  and b stand for?

What is their defenitions?

a and b are constants, they have a fixed value. x is a variable, depending on a,b.

$ x\equal{}\minus{}\frac{b}{a}$.

x is the variable, a is the slope of the line, and b is the y-intercept (value when x=0)


prove in triangle ABC
$ (s\minus{}b)^2/(s\minus{}c)^2\equal{}sqr root(c/b)$
s is semi perimeter

[quote="hinhhoc273"]'sqrroot(c/b)'. What does it mean? :huh:[/quote]Maybe he means $ \sqrt{\frac{c}b}$.

The equation does not hold for many cases. Take, for instance, $ a\equal{}3,b\equal{}4,c\equal{}5$. But it holds for equilateral triangles, which is trivial.  :D


Let $ a$, $ b$, $ c$, $ d$, $ S$ be five positive constants.
Find the minimum value of the expression
$ A \equal{} \frac {a}{x} \plus{} \frac {b}{y} \plus{} \frac {c}{z} \plus{} \frac {d}{t}$, where $ x$, $ y$, $ z$, $ t$ are four positive variables satisfying $ ax \plus{} by \plus{} cz \plus{} dt \equal{} S$.

This is a one line solution...I will wait till post because this might be from some journal...

Generally, we can also solve

Given that: $ \sum x_i a_i\equal{}k$ where $ x_i\ge 0$ are variable, $ a_i\ge 0$ constant, find the minimum value of:

$ \sum\frac{a_i}{x_i}$

[quote="nmtvn"]Find the minimum value
a,b,c,d,S is 5 positive constants.
$ ax \plus{} by \plus{} cz \plus{} dt \equal{} S$
Find the minimum value of the expression
$ A \equal{} \frac {a}{x} \plus{} \frac {b}{y} \plus{} \frac {c}{z} \plus{} \frac {d}{t}$, with $ x,y,z,t$ is 4 positive variables.[/quote]
We have:
$ A \equal{} \frac {a}{x} \plus{} \frac {b}{y} \plus{} \frac {c}{z} \plus{} \frac {d}{t} \geq \frac {(a \plus{} b \plus{} c \plus{} d)^2}{ax \plus{} by \plus{} cz \plus{} dt} \equal{} \frac {(a \plus{} b \plus{} c \plus{} d)^2}{S}$
(Note that  $ \frac {(a \plus{} b \plus{} c \plus{} d)^2}{S}$ is constant)
Equality occurs when $ x\equal{}y\equal{}z\equal{}t$.

[Moderator edit: Fixed a mistake.]

The problem was apparently posted as an attempt to cheat at he M&Y quadrilateral problem ( http://www.mathlinks.ro/Forum/viewtopic.php?t=164083 ; however, the attempt is pretty useless for the quadrilateral problem, since the equality case $ x\equal{}y\equal{}z\equal{}t$ cannot be fulfilled in general :) ). It can be now discussed again.

  Darij

See the problem of this topic , I remember an inequality that i studied in my high school :
" Let $ a, b, c, d, x, y, z, t$ be positive real numbers such that  $ ax \plus{} by \plus{}cz \plus{} dt \equal{} xyzt$. Prove that
$ x \plus{} y \plus{} z \plus{} t > \frac{4}{3}.(\sqrt{1\plus{}\sqrt{a\plus{}b}\plus{}\sqrt{a\plus{}c}\plus{}\sqrt{a\plus{}d}\plus{}\sqrt{b\plus{}c}\plus{}\sqrt{b\plus{}d}\plus{}\sqrt{c\plus{}d}} \minus{} 1)$ "


geometry
geometric transformation
reflection
circumcircle
geometry proposed

Given an acute triangle $ ABC$, with altitudes $ AA_1, BB_1, CC_1$, with $ A_1, B_1, C_1$ lying on $ BC,CA,AB$ respectively, consider the reflections of $ A_1, B_1, C_1$ over the midpoints of $ BC,CA,AB$ respectively as $ A_2, B_2, C_2$. Consider the intersection of the circumcircles of  $ AB_2C_2$ and $ ABC$, $ BC_2A_2$ and $ ABC$ and $ CA_2B_2$ and $ ABC$ as $ A_3, B_3, C_3$ respectively. Prove that $ A_1A_3, B_1B_3, C_1C_3$ are concurrent.

It is [b]Vietnam TST 2009-problem 1[/b].

let $ AA'$ parallel to $ BC$. we'll prove $ A'$ $ \equiv$ $ A3$ or $ AA'B2C2$ is cyclic. let $ A4$ lie on $ CA'$ such that $ AC2$=$ A'A4$. we have $ AC1$.$ AB$=$ AB1$.$ AC$=> $ CB2$.$ CA$=$ CA4$.$ CA'$=>$ AA'B4B2$ is cyclic so $ B2$ lie on $ (AA'A4). Moever$C2$ lie on$(AA'A4) => (QED).
       It's easy to see $ AA3$ through Centroid $ G$ of $ ABC$. Similarly BB3,CC3 through G=> AA3,BB3,CC3 concurent  :)
 Soory if I'm wrong  :)


Prove that the numbers $ \sqrt {2} , \sqrt {3}$ and $ \sqrt {7}$ can't be the terms of an arithmetic progression .

If they were, then $ a\equal{}\sqrt{2}$, $ a\plus{}d\equal{}\sqrt{3}$ and $ a\plus{}2d\equal{} \sqrt{7}$.

$ d\equal{} \sqrt{3}\minus{}\sqrt{2} \equal{} \sqrt{7}\minus{}\sqrt{3}$.

Square both sides and then square again and we get $ \sqrt{126}$ is rational, a contradiction.

Hope this is OK; I'm in a rush again.

Then $ \frac {\sqrt{7} \minus{} \sqrt{3}} {\sqrt{3} \minus{} \sqrt{2}}$ is rational. Perform the same convenient squarings.

[hide]
Let $ m(\sqrt 7-\sqrt 3)=n(\sqrt 3-\sqrt 2)$ where $ m,n\in \mathbb N$ with $ \gcd(m,n)=1$. Then:
\begin{align*}
m\sqrt 7-m\sqrt 3&=n\sqrt 3-n\sqrt 2\\
m\sqrt 7+n\sqrt 2&=(m+n)\sqrt 3.\end{align*}

Squaring both sides yields\[ 7m^2+2n^2+mn\sqrt{14}=3(m+n)^2,\]or $ \sqrt{14}=\frac{3(m+n)^2-7m^2-2n^2}{mn}$, implying that $ \sqrt{14}$ is rational, contradiction.[/hide]


a four-digit number ABCD is called snakelike if A<B>C<D.

how many four-digit snakelike numbers with distinct digits exist?

The answer is larger than the number of times this has been posted before on the forum:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63321

Do a search!  (Especially when posting questions from popular exams like the AIME.  This is 2004 number 6.)


nope just the fact that I feel REALLY bad for the kid....  :(

that's about it...

I don't think it would really happen.

I think the kid may feel a shock as he jumps in, but probably not enough to kill him.

However, that would definitely not short the circuit.

[quote="gregx007"]In the movie Syriana, a light in a swimming pool is cracked, exposing its circuitry to the pool.  When a child jumps in, he is electrocuted, and the lights all go out as presumably a fuse is blown.  Does anyone know if Hollywood has portrayed something which could actually happen?

It seems to me that given the size of the pool, the amount of current that would flow through the child would be miniscule, especially as it could only flow through him and back into the water, which should only be through a small potential difference.  Also, I am not sure why his jumping into the pool would cause a fuse to blow, as he does not add anything to the short circuit.  Any thoughts?[/quote]
In terms of the Thevenin equivalent circuit, the pool light represents a resistive load, which becomes an open circuit when the pool light malfunctions. The pool light is essentially replaced by the pool itself, a conductive bath, driving the current in the circuit to its maximum. If anything in the circuit were to short out, it would have done so at this point -- if the circuit were fused, the fuse would blow, no more current would flow, and the boy would have been perfectly safe.

As it was, the boy jumped into the pool and was electrocuted, as he certainly would have been, as he then represented a resistive load himself through which all the current in the pool would have passed. Would it have been enough to kill him? Hard to say. His heart was certainly part of the current path, and it takes precious little current to stop one's heart - but in any case, the explosion and sparks and what not was pure theatre.

I saw on TV some time ago a report about someone who got electrocuted in a swimming pool and then he (she) wasn't able to move, but was found on time. I don't remember the exact circumstances, but I don't think a light caused it.

But then again, it's well-known that one shouldn't trust what one sees on TV.


An urn contains 2009!+1 black balls and 2009!+1 red balls . In every step we take out 2009 balls and discard all of them if they are all of the same color or keep the red ones and discard the black ones if they are of different colors . Given that 1 ball remains at the end after a finite number of steps find the probability that it is a red ball .

Let the number of red balls in the urn at any time be R.
Then (R)Modulo(2009) remains constant.
Since initially (R)Modulo(2009) is 1 , the number of red balls at the end has to be unity.
Thus the probability is 1. :lol:


this seemed nice:

1) A hunter shoots 3 birds out of a stole of 10 which lied on a fence. How many birds are left after the shooting? :D

Answer: none, because the 3 dead birds are dead and dropped off the fence, and the other 7 flew away.

2) On a lake there are 10 ducks. A hunter shoots 3 of them. How many ducks are there after the shooting? :D

Answer: 3, all of the dead ducks, the other have obviously flyed away.

3) In a cage there are 10 birds. A hunter shoots 3 of them. How many birds are there after the shooting? :D

Answer: 10, all of them are, because they cannot exit the cage :D

4) A hunter shoots 3 birds that are flying in the sky in a stole of 10. How many birds are there after the shooting? :D

Answer: Finally it's 10-3=7! birds left (flying in the air)! .... huuugh - really tough this mathematics can get :D:D

I actually knew the answer because I have heard a variation of the first question from my dad when I was a little kiddie. Here is another one so easy but it still tricks a lot of kids without patience for second thoughts:

Q) If you got a rectangular table and you cut off one corner, how many corners are 'left'?

Gladly to say, it worked out on most kids when I was in elementary school :D

5) A hunter shoots a bird on the tree in a stole of 3. How many birds are there after the shooting?

Answer: 1. The dead one dropped, one other flied away, the rest one was stunned there  :D

6) A hunter shoots a bird on the tree in a stole of 3. How many birds are there after the shooting?

Answer: -1. The hunter accidently shot god who came down as a bird for the fun of it and broke the universe  :D


group theory
abstract algebra
number theory
greatest common divisor
superior algebra
superior algebra unsolved

Prove that:

1)If a group $ G$ contains a subgroup ($ \not\equal{} G$) of finite index, it contains a normal subgroup ($ \not\equal{} G$) of finite index.

2)If a normal subgroup $ N$ of order $ p$ ($ p$ prime) is contained in a group $ G$ of order $ p^n$, then $ N$ is in the center of $ G$.

1. If $ H$ is a subgroup of finite index, pick the normal subgroup $ \cap_{g \in G} gHg^{\minus{}1}$
2. Consider the natural map (conjugation) $ \sigma: G \rightarrow Aut(N)$, i.e. $ g$ is sent to the automorphism $ T_g$, which is conjugation by $ g$. The kernel of this map is the centralizer of $ N$. Now, $ H \equal{} \mathrm{Im} \sigma$ is a subgroup of $ Aut N$ of order $ \varphi(p)$ and is the homomorphic image of $ G$ which is of order $ p^n$. By Lagrange's theorem $ |H|$ divides $ \mathrm{gcd}(\varphi(p), p^n) \equal{} 1$, i.e. the image is 1 which means that G commutes with N.

[quote="Soarer"]1. If $ H$ is a subgroup of finite index, pick the normal subgroup $ \cap_{g \in G} gHg^{ \minus{} 1}$[/quote]
Why does this have finite index?  (I'm aware that the finite intersection of subgroups of finite index has finite index, but it seems like $ H$ may have infinitely many conjugates)

The number of conjugates is the index of the normalizer, so if $ H$ has finite index it only has finitely many conjugates.

$ G$ operates on the set of cosets $ G/H$ by left-multiplication.
This induces a group homo $ G\to S_{|G: H|}$ with kernel $ K\equal{}\bigcap_{g\in G}gHg^{\minus{}1}$.
So $ |G: K|$ divides $ |G: H|!<\infty$.

Dear olorin, can you explain why kernel $ K\equal{}\bigcap_{g\in G} gHg^{\minus{}1}$ ?

$ \forall g : xgH \equal{} gH \Leftrightarrow$
$ \forall g : g^{\minus{}1}xg \in H \Leftrightarrow$
$ \forall g : x \in gHg^{\minus{}1}$.

[quote="olorin"]$ G$ operates on the set of cosets $ G/H$ by left-multiplication.
This induces a group homo $ G\to S_{|G: H|}$ with kernel $ K \equal{} \bigcap_{g\in G}gHg^{ \minus{} 1}$.
So $ |G: K|$ divides $ |G: H|! < \infty$.[/quote]

How do you know that |G:K| divides |G:H|?

$ |G: K|$ does not necessarily divide $ |G: H|$, but it divides $ |S_{|G: H|}| \equal{} |G: H|!$, since, as olorin showed, $ G/K$ is isomorphic to a subgroup of $ S_{|G: H|}$.


\neq  for example $ a \neq b$ (not equal). Have you downloaded the file in the topic "How to learn $\LaTeX$"? It contains most of the important symbols.

If you want to know how to include \neq in files outside mathlinks you must use the amssymb package.
I have had some problems trying to fit it in the amsmath and amsfonts. No success in that department  :D .

\ne also works on this site. It does exactly the same thing as \neq. Also, \ge is the same as \geq and \le is the same as \leq.

\ne : $\ne$ . \neq : $\neq$ .
\le : $\le$ . \leq : $\leq$ .
\ge : $\ge$ . \geq : $\geq$ .


Have a go. :D

A nonlinear oscillator has potential given by
$ U(x)\equal{}\frac{kx^{2}}{2}\minus{}\frac{m\lambda x^{3}}{3}$
with $ \lambda$ small.  Find the solution to the equation of motion to first order in $ \lambda$ assuming $ x\equal{}0$ and $ t\equal{}0$.

Can we simply assume that the object undergoes small oscillations about its equilibrium points?

This is all the information I have on the problem.  If you'd like to assume that, then fine.

Seems $ \; \lambda \;$ is a constant, so what means "first order in  $ \; \lambda \;$?

Could you tell us, please, where from comes this problem?

It comes from a book of old exam problems.  In this case, Princeton.

[quote="Immanuel Bonfils"]Seems $ \; \lambda \;$ is a constant, so what means "first order in  $ \; \lambda \;$?[/quote]

It means that $ \lambda$ is small enough (consider it $ \epsilon$) such that you can neglect terms in the final expression that involve powers of $ \lambda$ greater than 1.

The process of solving this problem should be rather straightforward, but you end up with a differential equation that I don't know how to solve.  We have:

\[ F(x) \equal{} m\lambda x^2 \minus{} kx \implies \frac{d^2 x}{dt^2} \equal{} \lambda x^2 \minus{} \frac{k}{m}x\]

Then substituting $ v \equal{} \frac{dx}{dt}$, we get

\[ \frac{dv}{dt} \equal{} v\frac{dv}{dx} \equal{} \lambda x^2 \minus{} \frac{k}{m}x \implies\]
\[ vdv \equal{} (\lambda x^2 \minus{} \frac{k}{m}x)dx \implies\]
\[ v \equal{} \frac{dx}{dt} \equal{} \sqrt{\frac{2\lambda x^3}{3} \minus{} \frac{k}{m}x^2 \plus{} C}\]

That last equation I don't know how to solve.  Or is there a smarter way?

this problem falls quickly to a perturbative approach.  details are ugly, but this is a routine calculation.

wikipedia explains the general idea if you are unfamiliar: http://en.wikipedia.org/wiki/Perturbation_theory

in this question, the anharmonic term is a perturbation to the rest of the Hamiltonian (which you know how to solve, etc.)

Cool, thanks for the info.  Equipped with knowledge gained from a cursory reading of the perturbative method, I believe I can finish the problem.

I will replace $ \lambda$ with $ \epsilon$ to emphasize the fact that it's small.  Starting from $ \displaystyle \frac {d^2 x}{dt^2} \equal{} \epsilon x^2 \minus{} \frac {k}{m}x$, we can let
\[ x(t) \equal{} x_0 \plus{} \epsilon x_1 \plus{} \epsilon^2 x_2 \plus{} \cdots \mbox{ where } x_i(0) \equal{} 0.
\]
Since we want the answer to first order in $ \epsilon$, after plugging this expression into the differential equation, we get
\[ x_0'' \plus{} \epsilon x_1'' \equal{} \epsilon x_0^2 \minus{} \frac {k}{m}(x_0 \plus{} \epsilon x_1).
\]
Now equating coefficients of powers of $ \epsilon$, we get two equations:
\[ x_0'' \equal{} \minus{} \frac {k}{m}x_0,
\]

\[ x_1'' \equal{} x_0^2 \minus{} \frac {k}{m}x_1
\]
The first one gives $ \displaystyle x_0(t) \equal{} A\sin\left(\sqrt {\frac {k}{m}}t\right)$, the regular harmonic solution ($ A$ is your amplitude), reducing the second equation to $ \displaystyle x_1'' \plus{} \frac {k}{m}x_1 \equal{} A^2\sin^2\left(\sqrt {\frac {k}{m}}t\right)$.  Abbreviate $ \displaystyle \sqrt {\frac {k}{m}}$ as $ \omega$.  This last equation can be pwned via undetermined coefficients: guess the solution $ x_1(t) \equal{} a\sin^2\omega t \plus{} b\cos^2\omega t$, whereupon
\[ x_1'' \equal{} 2(a \minus{} b)\omega^2\cos 2\omega t \equal{} 2(a \minus{} b)\omega^2(\cos^2\omega t \minus{} \sin^2\omega t).
\]
Plugging this into the equation gives
\[ x_1'' \plus{} \omega^2 x_1 \equal{} (2a \minus{} b)\omega^2\cos^2\omega t \plus{} (2b \minus{} a)\omega^2\sin^2\omega t \equal{} A^2\sin^2\omega t,
\]
and solving the equations
\[ 2a \minus{} b \equal{} 0
\]

\[ 2b \minus{} a \equal{} \frac {A^2}{\omega^2}
\]
gives $ a \equal{} A^2/3\omega^2 \equal{} mA^2/3k$ and $ b \equal{} 2mA^2/3k$.  Considering the solution of the homogeneous equation (for SHM) associated with this last equation, our general solution for $ x_1$ is
\[ x_1(t) \equal{} \frac {mA^2}{3k}\sin^2\left(\sqrt {\frac {k}{m}}t\right) \plus{} \frac {2mA^2}{3k}\cos^2\left(\sqrt {\frac {k}{m}}t\right) \plus{} A^2\sin\left(\sqrt {\frac {k}{m}}t\right).
\]
Finally (if you prefer $ \lambda$),
\[ x(t) \equal{} x_0(t) \plus{} \lambda x_1(t) \equal{}
\]

\[ (A\plus{}\lambda A^2)\sin\left(\sqrt {\frac {k}{m}}t\right) \plus{} \lambda \frac {mA^2}{3k}\left[\sin^2\left(\sqrt {\frac {k}{m}}t\right) \plus{} 2\cos^2\left(\sqrt {\frac {k}{m}}t\right)\right].
\]
Note that this solution satisfies the initial conditions only approximately.  I hope this is right.

So this problem is more interesting than I initially thought it would be, maybe because I've never seen a perturbative method of solving differential equations until today.  Time to learn more about ODEs...

genericme

You can use energy balance equation but it comes out to the same integral, and I doubt it has an exact solution; that is why the problem asks for a solution to first order in $ \;\lambda\;$ .

[hide] \[ x \equal{} (A\plus{} \lambda B) sin \omega_o t \plus{} {{A^2 \lambda} \over {6 \omega_o^2}} ( 3 \plus{} cos 2 \omega_ot \minus{} 4 cos \omega_o t)\]
where  $ \; \omega_o \equal{} \sqrt{{k \over m}}$[/hide]

Hmm, could you elaborate on your solution?  What does $ B$ represent in your equation?

Ah, and your solution does satisfy the initial conditions exactly.  Actually, I'm starting to feel that most of my solution was bunk.

I guess you guess want an answer?  Here we go:

[hide]The equation of motion of the nonlinear oscillator is
$ \frac{md^{2}x}{dt^{2}}=-\frac{dU(x)}{dx}=-kx+m\lambda x^{2}$

Neglecting the $ m\lambda x^{2}$ term, we obtain the zero-order solution of the equation
$ x_{(0)}=A\sin(\omega t+\varphi)$

where $ \omega=\sqrt{{k}/{m}}$ and $ A$ is an arbitrary constant.  As $ x=0$ at $ t=0$, $ \varphi=0$ and we have
$ x_{(0)}=A\sin(\omega t)$

Suppose the first order solution has the form $ x_{(1)}=x_{(0)}+\lambda x_{1}$.  Substituting it into the equation of motion and neglecting terms of orders higher than $ \lambda$, we have
$ \ddot{x_{1}}+\omega^{2}x_{1}=x^{2}_{(0)}=\frac{A^{2}}{2}\left[1-\cos(2\omega t)\right]$

To solve this equation, try a particular integral
$ x_{1}=B+C\cos(2\omega t)$

Substitution gives
$ -3\omega^{2}C\cos(2\omega t)+\omega^{2}B=\frac{A^{2}}{2}-\frac{A^{2}}{2}\cos(2\omega t)$

Comparison of coefficients gives
$ B=\frac{A^{2}}{2\omega ^{2}}\hspace{1cm} C=\frac{A^{2}}{6\omega^{2}}$

The homogeneous equation
$ \ddot{x}_{1}+\omega^{2}x_{1}=0$

has solution
$ x_{1}=D_{1}\sin{\omega t)+D_{2}\cos(\omega t)}$

Hence we have the complete solution
$ x_{(1)}=(A+\lambda D_{1})\sin(\omega t)+\lambda\left[\frac{A^{2}}{2\omega ^{2}}+D_{2}\cos(\omega t)+\frac{A^{2}}{6\omega^{2}}\cos(2\omega t)\right]$

The initial condition $ x=0$ at $ t=0$ then gives
$ D=-\frac{2A^{2}}{3\omega^{2}}$

and
$ x_{(1)}=A'\sin(\omega t)+\frac{\lambda A^{2}}{\omega^{2}}\left[\frac{1}{2}-\frac{2}{3}\cos(\omega t)+\frac{1}{6}\cos(2\omega t)\right]$

where $ A'$ is an arbitrary constant.[/hide]

OK, I see that I made a silly mistake, but my method was the same.

Not so bad indeed.
But your difficulty to match with the initial condition is why your solution is not a [b]general solution [/b]; try it.

By the way, that is the problem with my [b] B [/b]: if we put the initial velocity , as it comes out from $ \;x_o\;\;, A \omega_o  \;$ we'll  have B = A.

As a matter of fact this problem is much more math. rather phys. : it's OK as it's MathLinks!
But there is no sense in the physical dimensions:  $ \; \lambda \;$ should be adimensional...

Thanks for attempting.  I'll post some more problems when I get around to it.  I'll give you some time to debate a solution and then I'll post the correct solution some time later.

Your answer is the same as mine , replacing $ \; A \plus{} \lambda B\;\;$ by A'

But, kind of echoing genericme, two arbitrary constants? A and A'?

Thanks.

[quote="Immanuel Bonfils"]
But your difficulty to match with the initial condition is why your solution is not a [b]general solution [/b]; try it.
[/quote]

Yeah, I realized that I didn't even consider initial conditions when solving that last equation with undetermined coefficients - the trick was to express $ \sin^2\omega t$ as $ \frac {1}{2}(1 \minus{} \cos 2\omega t)$, as JRav did, which would have allowed me to guess a more general solution.


Consider the acute-angled triangle $ ABC$ , altitude$ AD$  and point $ E$ - intersection of $ BC$ with diameter from $ A$ of circumcircle. Let $ M, N$ be symmetric points of $ D$ with respect to the lines $ AC$ and $ BC$ respectively.
           Prove that  angle EMC = angle BNE
  It is RMO but I do not find the sulotion for it
I think the problem was posted .Could you give me the sulotion or the link

It is a problem of Iran TST Junior 2008, and its solution was posted.

http://www.mathlinks.ro/viewtopic.php?t=209259
I don't now about Iran TST, but it was also given at Romanian Junior TST 2008.


Find all solutions of the equation $ (U\equal{}\mathbb{C})$

[list][list]$ 3x^2\minus{}4x\plus{}\sqrt{3x^2\minus{}4x\minus{}6}\equal{}18$[/list][/list]

Let $ y\equal{}3x^2\minus{}4x\minus{}6$ so that our equation is $ y\plus{}6\plus{}\sqrt y\equal{}18\implies\sqrt y\equal{}12\minus{}y\implies y\equal{}144\minus{}24y\plus{}y^2$

$ \implies y^2\minus{}25y\plus{}144\equal{}(y\minus{}9)(y\minus{}16)\equal{}0\implies y\equal{}\{9,16\}$.

However, we see that $ y\equal{}16$ does not satisfy $ \sqrt y\equal{}12\minus{}y$, so $ y\equal{}9$ is the only solution.

Thus, we have $ 3x^2\minus{}4x\minus{}6\equal{}9\implies3x^2\minus{}4x\minus{}15\equal{}(3x\plus{}5)(x\minus{}3)\equal{}0\implies x\equal{}\boxed{\left\{\minus{}\frac53,3\right\}}$.


geometry
3D geometry
modular arithmetic
number theory proposed
number theory

Prove that none of the numbers $2^{2^{n}}+1, n = 0, 1, 2, \cdots$ is a perfect cube.

[hide="Solution"]

Assume that for some $k \in \mathbb{N},$ $2^{2^{n}}+1 = k^{3}.$ We then rearrange to get:

$2^{2^{n}}= k^{3}-1 = (k-1)(k^{2}+k+1).$ But $k^{2}+k+1 = k(k+1)+1 \equiv 1 \bmod{2}.$ This is a contradiction, as the left hand has no odd divisors. Thus, $2^{2^{n}}+1$ is never a perfect cube for $n = 0, 1, 2, \cdots.$ $\blacksquare$

[/hide]

[quote="Karth"]Prove that none of the numbers $2^{2^{n}}+1, n = 0, 1, 2, \ldots$ is a perfect cube.[/quote]
More generally: "[i]there is no triple $(n,a,k) \in \mathbb{N}^{3}$ such that $k \ge 2$ and $2^{2^{n}}+1 = a^{k}$ (i.e., no Fermat numbers are perfect powers, apart from triviality - I mean the case when $k = 1$).[/i]"

Proof: Mihailescu's theorem kills it! But I see this way would make you all discontent. :wink: Then for future reference let $F_{n}= 2^{2^{n}}+1$, for $n = 0, 1, 2, \ldots$ Obviously $F_{0}= 3$ is O.K. with the claim. So admit $n \ge 1$ and by way of contradiction suppose there is any triple $(n,a,k) \in \mathbb{N}^{3}$, with $k \ge 2$, such that $F_{n}= a^{k}$. Then $a^{k}\equiv 2 \bmod 3$, entailing both $k$ and $a$ are odd [in particular, $k \ge 3$ and $a \ge 5$, because $\gcd(F_{m}, F_{n}) = 1$, whenever $m, n \in \mathbb{N}$ are distinct - which makes $3 \nmid F_{n}$]. Thus there are $b,c \in \mathbb{N}$, with $1 \le b < c$ and $b+c = 2^{n}$, such that $a-1 = 2^{b}$ and $\sum_{i=0}^{k-1}a^{i}= 2^{c}$, i.e. $\sum_{i=0}^{k-1}(2^{b}+1)^{i}= 2^{c}$. Anyway this is absurd, as $\mbox{LHS}\equiv k \equiv 1 \bmod 2$, whereas $\mbox{RHS}\equiv 0 \bmod 2$.

Could you cite Fermat's Last Theorem for this? I was considering it, but that would sort of be cheating...  :P not to mention that you probably couldn't have used it in 1990.

Fermat's Last theorem should be rather useless here...

[hide="another proof of general problem"]
Let $2^{n}=m$ and suppose $2^{m}+1=a^{k}$ where $a,k>1$. Let $p$ be largest integer such that $a-1$ is divisible by $2^{p}$. Then $k>1$ implies $m>p$, so reduce modulo $2^{p+1}$. We then have
\[1\equiv 2^{m}+1\equiv a^{k}\equiv (2^{p}+1)^{k}\equiv k\cdot 2^{p}+1\pmod{2^{p+1}}\]
so that $k$ even. Let $k=2j$. We now have
\[2^{m}=(a^{j}+1)(a^{j}-1)\]
Now $a^{j}+1$ and $a^{j}-1$ are powers of 2 that differ by 2, so they must be 4 and 2. Then $a^{j}=3$, so $a=3,j=1,m=3$. But $2^{n}=3$ gives a contradiction!
[/hide]


topology
Ring Theory
superior algebra
superior algebra unsolved

Let $ A,B$ be commutative rings with unity and $ f: A\to B$ a morphism of rings.
Let $ \mathfrak{p}$ be a prime ideal of $ A$. Show that $ f^*^{-1}(\mathfrak{p})$ is naturally homeomorphic to $ Spec(B_{\mathfrak{p}}/pB_p)=Spec(k(p)\otimes_A B)$, where $ k(p)$ is the residue field of $ A_p$.

Since k(p) can be reached after the two step procedure of firstly localizing A at p, then reducing wrt the maximal ideal in the localisation, it is sufficient to show:

(i) If $ I$ is an ideal in $ A$ then $ Spec(B\otimes A/I)$ is naturally homeomorphic to $ f^*^{-1}(V(I))$;
(ii) if $ S$ is a multiplicative subset of $ A$, then $ Spec(S^{-1}B)$ is naturally homeomorphic to $ f^*^{-1}(Spec(S^{-1}A))$.
Note that there is a natural isomorphism $ S^{-1}B=B\otimes_AS^{-1}A$ of $ S^{-1}A$-algebras and that $ Spec(S^{-1}A))$ is naturally identified with the [u]homeomorphic[/u] image in $ Spec(A)$ via the inclusion $ Spec(S^{-1}A))\to Spec(A)$ (coming from $ A\to S^{-1}A$).

for every morphism $ f : X \to S$ of schemes, we have $ f^{\minus{}1}(s) \equal{} Spec(\kappa(s)) \otimes_S X$.

This is certainly true, but I am just wondering what definition of $ f^{\minus{}1}(s)$ you use to understand what your category is you naturally identify the fiber and the product $ Spec(k(s))\times_SX$ with each other: category of topological spaces or schemes?  :wink:

you know the answer, right? first we consider the fiber just as a topological space, but this homeomorphism above yields a scheme structure.

Yes I do. But then how do you establish the homeomorphism between the fiber and the underlying topological space of the fiber product? I cannot see any proof without going back to affine charts (which is clear form a point of view of categories). I mean that we do need to proof the exercise in this thread, right?

But maybe you mentioned the result as a generalization of the exercise ...  :)


prove that this numbers is not a prime number:
11...111(this number has 1991 digit)

[hide="Solution"]1991 is a composite number thus 111...11 is composite too[/hide]

[b][size=200]HIDE[/size][/b]

[b][size=200]YOUR[/size][/b]

[b][size=200]SOLUTIONS[/size][/b]

[b][size=200]!!!!!!!!!!!!!!!!!!!!!!![/size][/b]

[size=0]please...[/size]

This is the most annoying thing about the intermediate topics. People do not hide their solutions and have no consideration to the people who would WANT to solve these problems. So from now on, hide your solutions like this:

[code]
[hide=LIKE THIS!!!]you see???[/hide][/code]

[hide="LIKE THIS!!!"]you see???[/hide]

You can't use apostrophes in the hide text.


Hi folks

I'm a newbie who was a math fan back in the early/mid 1990s.  I attended Woburn C.I. (Toronto, Canada) and had Larry Rice and Peter Crippen for calculus and algebra.

I'm now writing a paper on astigmatic correction.  I have attached an image from a very old classic textbook in the field of ophthalmology (Duke Elder's System of Ophthalmology) that references "Euler's Theorem Concerning Curved Surfaces".  I have googled the term but have not been able to find what this theorem is (if it even exists).  I was hoping someone could point the way to what the author might be referring to.

Thanks in advance.

http://img383.imageshack.us/img383/435/3copyio5.jpg
[url=http://www.imagehosting.com/][img]http://img383.imageshack.us/img383/435/3copyio5.jpg[/img][/url]

Isnt this Diffrential Geometry??
Then maybe you are refering to Prinicipal Curvature theorem.
That'll be this:
[url]http://en.wikipedia.org/wiki/Euler%27s_theorem_(differential_geometry)[/url]


Let f be a continuous function from [a,b] to itself. Define the sequence {xn} by x0 in [a,b] and for all n

  x_(n+1)= f(xn)

Prove that {xn} converges if x_(n+1) -x_n tends to zero

i try to find $ f$ such that:
i) $ x_{f(n)}$ converges
ii) $ \exists M\in\mathbb{N}^*,\forall n\in\mathbb{N}: \ 0<f(n\plus{}1)\minus{}f(n)\le M$

if i find it, the rest is easy:
because $ \forall k\in\{0,1,2,....,M\}\: x_{f(n)\plus{}k}\equal{}x_{f(n)}\plus{}\sum_{j\equal{}0}^{k\minus{}1}x_{f(n)\plus{}j\plus{}1}\minus{}x_{f(n)\plus{}j}$ converges.
and $ \mathbb{N}\minus{}\bigcup_{k\equal{}0}^{M}\{f(n)\plus{}k: \ n\in\mathbb{N}\}$ is finite

Maybe u could find it. But my solution uses another idea, it's not really elementary

  Let 's prove that if  a<b are two limit points of the sequence {xn} ( i.e there exists a subsequence converges to it ). Then for all c  in [a,b] , c is limit point ( under hypothese that x_(n+1) - x_n tends to zero ). It's trivial to show that if c is a limit point of {xn}, then

 f(c)=c.

 So if {xn} is not convergent, there exist two limit points a<b. We must have f(x)=x for all x on [a,b], that leads to x_n=x_(n+1)=... for n large enough


Rezolvati in nr reale ecuatia:
[x+y]=[x]*[y]+1
[a] partea intreaga a nr real a.

Asa cum e scrisa ecuatia nu se intelege daca ar trebui sa citim \[[x+y]=[x]\cdot[y]+1\] sau \[[x+y]=[x]^{[y]}+1\,.\] Care este varianta dorita?

Prima este varianta corecta.Cred ca o solutie este $\ x=y=1$, dar inca nu am o demonstratie. :(

[hide="solutie"] fie $[x]=k$, $[y]=p$, $a=x-k$, $b=y-p$.
ecuatia devine $k+p+[a+b]=kp+1$.
$[a+b]$ poate fi $0$ sau $1$ si de aici doua cazuri principale

A. $k+p=kp+1$, sau $(k-1)(p-1)=0$.
B. $k+p=kp$, sau $(k-1)(p-1)=1$. ($k$ si $p$ sunt intregi)

etc...
[/hide]


function
induction
algebra unsolved
algebra

Functions $f,g : [2,4] \rightarrow [2,4]$ satisfy conditions :
\[f(g(x))=x\]
\[g(f(x))=x\]
\[f(x)g(x)=x^{2}\]
Show, that for all $x\in [2,4]$ ,$ f(x)=g(x)=x$

The first two conditions give that $f$ and $g$ are bijective and that $f^{-1} = g.$

Let $x \in [2;4].$
Define $f^n(x) = [f(x)]^n $ and $f^{(n)}(x) = f(f^{(n-1)} (x))$ where $f^{(1)} = f.$
Let $a = \frac {f(x)} x.$
From the third given relation, we deduce that $f^{(2)}(x))x = f(f(x))g(f(x))=f^2(x).$
Thus $f^{(2)}(x) = a \cdot f(x).$ (1)

From (1) and with an easy induction, we may prove that for all $n \geq 1$, we have $f^{(n)}(x) = a^{n-1} \cdot f(x).$

It follows that for all $n$, we have $2 \leq a^{n-1} \cdot f(x) \leq 4 .$ Taking limit, it leads to $a=1$, that is $f(x)=x.$ Thus $g(x)=x$ too, and we are done.

Pierre.


$k$ is a real number. Prove that $k \in \mathbb{Q}$ if and only if there are three different terms in the sequence
\[k,\ k+1,\ k+2,\ k+3,...\]
that form a geometric sequence.

Let $a_{n}=k+n$

$\Leftarrow$
Assume, that $s<r<t$ and $(a_{s},a_{r},a_{t})$ is a geometric sequence. It is equivalent to : $a_{r}^{2}=a_{s}\cdot a_{t}\iff k(2r-t-s)=st-r^{2}$. It's easy to prove, that $2r-t-s \not=0$, so $k=\frac{st-r^{2}}{2r-t-s}\in\mathbb{Q}$

$\Rightarrow$
$k=\frac{p}{q}$, then $(a_{s},a_{r},a_{t})$ is a geometric sequence with ratio $q+1$ for:

$s=p$
$r=p(q+2)$
$t=p(q^{2}+3q+3)$


Let $ 0 < a < 1$ and we define a sequence $ \{x_n\}_{n \equal{} 0}^{\infty}$ as $ x_n \equal{} 3^na \minus{} \lfloor 3^na\rfloor$,which satisfies

$ 0 < x_{3m} < \frac {1}{3},\frac {1}{3} < x_{3m \plus{} 1} < \frac {2}{3},\frac {2}{3} < x_{3m \plus{} 2} < 1$, for all nonnegative integer $ m$.Determine $ a$.


group theory
abstract algebra
topology
advanced fields
advanced fields unsolved

I thought about this a few days ago, and was baffled.  I'm not sure if this question is trivial or meaningless, but I thought I'd throw it out there.

So with some groups, it seems like the direct product operation applied to their subgroups results in something like a free idempotent commutative monoid.  For example, in the finite cyclic group order $ 6$, or the symmetric group $ S_{3}$.  It also seems, with such groups, that if $ A/B \equal{} C$, then $ A \equal{} C$x$ B$.  This seemed to me to be sufficient structure to ask if it applies in general.  However, it doesn't seem so simple with topological spaces.  For example, $ R^{2}/Z^{2}\equal{} S^{1}$x$ S^{1}$, the torus, but $ R^{2}$ and $ S^{1}$x$ S^{1}$x$ Z^{2}$ are pretty different spaces in terms of open sets, etc.  Then, I wondered if $ R^{2}$ could be expressed as the direct product of $ Z^{2}$ and another space.  "More or less," $ R^{2}\equal{} [0,1)$x$ [0,1)$x$ Z^{2}$.  However, if we use the subspace topology for $ [0,1)$ and $ Z^{2}$, $ [0,1)$x$ [0,1)$x$ 0$ is an open set, so $ [0,1)$x$ [0,1)$x$ Z^{2}$ is strictly finer than $ R^{2}$.  I just wondered why so many more things fail when we move away from the discrete "purely algebraic" groups, and whether there's a way to define things so that they don't fail?

No, you can't write $ \mathbb{R}^{2}$ as a Cartesian product that way, even if you ignore the group structure. The connected components of $ A\times B$ are the products of the connected components of $ A$ and the connected components of $ B$; if $ A$ is the discrete space $ \mathbb{Z}^{2}$, there are infinitely many connected components in the product.
As a group ignoring the topology, you can't write $ \mathbb{R}^{2}$ as a direct product unless both factors are divisible groups. That rules out $ \mathbb{Z}^{2}$ and the like.

Also, this fails most of the time for finite groups already. That $ \mathbb{S}_{3}$ you mentioned is already a counterexample. You don't even have to go nonabelian- look at $ \mathbb{Z}/4\mathbb{Z}$.

How are $ S_{3}$ and $ \mathbb{Z}/4\mathbb{Z}$ counter-examples?

Neither is a direct product in any nontrivial way. They've got subgroups and quotients, but they don't combine as direct products.

Oh, sorry - to clarify.  My choice of wording was horrible.  Indeed, wrong.  Instead of direct product, I meant to say something more like "product of subgroups."  As in, take one element from each subgroup and multiply them, and do this in all possible ways.  Actually, I guess that clarifies the problem, because now the $ \mathbb{R}^{2}$ example works.  In that case, I wonder if it generalizes to all groups...

Given a group $ G$, with $ H$ a subgroup of $ G$, can all elements of $ G$ be expressed as a product of an element in $ H$ and a primitive representative of an equivalence class in $ G/H$?

Ok, now it's trivial.  Thanks jmerry, and sorry for posting a one-sided conversation!


I am looking to see if anyone has the team round for 2002 for national that someone could share.

Thanks,

I have it in JPG format so I'll post it here soon. Could someone convert it to pdf and repost it here?

I think this round was in Biffanddoc's nats thingie.

Hmmmm. You're right. Don't use mine then, his is better quality.

Here I am attaching the pdf document for 2002 team round

*sigh*

this team round along with the sprint target and solutions for 2002 and 2001 are on http://www.freewebs.com/tzpublicsite

The PDF is better too


Solve the equation
$ (x\plus{}4)^4\equal{}2(2x\plus{}13)^3\plus{}50(2x\plus{}13)$
Can you take me a simple solution? :|
[/hide]

The substitution isn't necessary, one can expand as is and then factor the resulting quartic to $ (x^2\minus{}12x\minus{}114)(x^2\plus{}12x\plus{}42)\equal{}0$ but that requires a little more effort.

hello, bute when it is factorized, then the equation is easy to solve, one get
$ 6\plus{}5\sqrt{6}, 6\minus{}5\sqrt{6},\minus{}6\plus{}i\sqrt{6},\minus{}6\minus{}i\sqrt{6}$
Sonnhard.


geometry
3D geometry
number theory
prime factorization

Determine the number of positive odd and even factor of $ 5^6\minus{}1$.

[quote="wangsacl"]Determine the number of positive odd and even factor of $ 5^6 \minus{} 1$.[/quote]

Through the difference of 2 squares and the sum/difference of 2 cubes we get

$ 5^6 \minus{} 1 \equal{} 2^3 \times 3^2 \times 7^1 \times 31^1$

If the question is asking for how many odd and even factors[b] all together[/b], then the answer is

$ 4 \times 3 \times 2 \times 2 \equal{} 48$ (note this includes $ 5^6 \minus{} 1$ as a factor [if not then $ 47$ proper divisors])

If it asks for how many even, and how many odd.

Note that for even numbers, there must be a power of 2 where the power is greater than 0. Such  divisors are in the form

$ 2^a \times 3^b \times 7^c \times 31^d$

$ a$ can be any three values (3, 2 or 1) and the rest can be as was done above, hence there are

$ 3 \times 3 \times 2 \times 2 \equal{} 36$

The odd factors must only have $ 2^0$ in its prime factorization, so there are

$ 1 \times 3 \times 2 \times 2 \equal{} 12$

In conclusion there are 36 even factors (or if you dont count $ (5^6 \minus{} 1)$ as a factor then there are 35) and there are 12 odd factors.


Harvard college MIT Princeton calculus derivative integration

Right now, my top three college choices are Harvard, MIT, and Princeton.  I want to apply to one of these early, but am having a hard time deciding which I like the most.

To help me out, could you guys say which you would pick if you had the chance to go to any and why?

Well, I'm at Princeton, but I don't really have a convincing reason because I didn't really consider anywhere else.

Since they're all great universities in terms of education, one thing you can consider is the environment. Harvard and MIT are in Boston, a good-sized city. Princeton is in, well, Princeton, which is a pretty small and quiet place. You can go to New York City or Philadelphia by train from Princeton. You could visit. Also, if you have any specific questions, there are people here you can ask. As I said, I go to Princeton. JBL goes to Harvard. Many people here go to MIT.

Hmm okay.  I did visit all of them, but I was able to look around Harvard and MIT more extensively because I stayed in Cambridge for two days.  When I went to Princeton, all I really had time for was a tour and information session.

I liked the city atmosphere at MIT and Harvard, but I don't know how much of a difference this would really make in my enjoyment of college.  Xevarion, what kind of things are there to do in Princeton (the town or at school)?  Is there a lot besides that one big street right outside of the school (this is all I saw when i visited)?  Also, could you comment on the math classes you've take...some things like how many other prospective math majors there are and the general math class sizes and quality.

I have some other questions for anyone knowledgeable about the mathematics at Princeton, MIT, and Harvard.  One of the attracting points of Harvard is Math 55.  My first question is: does Princeton or MIT offer an equivalent type of course?

Also, I know that you can take courses at Harvard if you go to MIT, and vice-versa.  So I could take Math 55 even if I was at MIT.  But after this freshman course, does MIT offer theoretical math courses that rival Harvard's in terms of difficulty and rigor?  Otherwise, I would be taking most of my math classes at Harvard, which doesn't make much sense.

My last question(s) concerns the sizes of the math departments of each school.  First, what exactly do people mean when they compare the size of a department for these colleges--number of professors, courses, or what?  How much does the size affect undergraduate education?  Furthermore, according to whatever your definition may be, how do the sizes of the math departments of the three universities compare?  Specifically, how do their theoretical math departments compare?  I ask this because although I've heard that MIT's department is larger than Harvard's, this does not necessarily mean their pure math department (the one in which I am interested) is.

I don't know anything about Harvard or MIT. I dont' even know what "Math 55" is. Here are some things about Princeton:

Last year, I went places other than Nassau Street (the street you were talking about) very few times. There's a shopping center with a great dim sum restaurant; if you need to buy anything or crave authentic Chinese food, you can go there. There's a movie theater that's pretty far away. As I said before, you can take a train to New York City. Some people go to NYC often. I don't think people go far from campus for any other reason, really.

There are a lot of student performance groups, and some of them are very good. Also, from time to time professional performers will come to Princeton. There are plays, classical concerts, and modern music (the latter is usually on campus, and paid for by a student group). I recall last year Andras Schiff and Lang Lang both had concerts in Pton.
A common pastime is going to "The Street" (which is Prospect Avenue, not Nassau Street) in the evening to party. Prospect Avenue houses all of the eating clubs, which I'm sure you've heard of. At night, they party. Sometimes they have cool stuff (once I saw a comedian, Elliot something-stereotypically-asian. He was funny.). They have copious booze. I'm not much of a drinker, so I don't go to the Street much. I can refer you to an expert if you want though.

Regarding math classes: Princeton, Harvard, and MIT are among the best universities for mathematics in the world. You will get a great education at any of them. I've been at Princeton for one year, and the classes I took were:
Math 215: Analysis in a Single Variable. [url=http://en.wikipedia.org/wiki/John_Horton_Conway]The Prof[/url]. This class was basically rigorous real analysis. Sequences and series, limits, continuity, derivatives, Riemann integrals, uniform continuity and convergence. When Conway finished the curriculum (about halfway through the semester) he went on to teach us whatever he felt like talking about. We learned basic point-set topology, Alexander's Horned Sphere, how to make magic squares, how to get the Taylor series for trig functions using the Bernoulli numbers, and how the Bernoulli numbers are related to the Riemann zeta function. Conway was awesome. This class made me fall in love with math. Conway doesn't use a textbook, by the way.
Math 217: Honors Linear Algebra. [url=http://www.amazon.com/s/103-7985618-3328648?ie=UTF8&index=books&rank=-relevance%2C%2Bavailability%2C-daterank&field-author-exact=R.%20C.%20Gunning]The Prof[/url]. This was a rigorous introduction to linear algebra. We learned all the basic stuff you'd expect: matrices, Gaussian elimination, determinants, eigenvalues and eigenvectors, rational canonical form, Jordan form, self-adjoint and normal matrices, some stuff about finite-dimensional Hilbert spaces. One shortcoming of this class is that we really didn't talk about trace at all, so I don't know anything about it still. We also didn't learn the Frobenius norm (whatever that is). There was almost no application in this class. I found this refreshing. Basic linear algebra is sort of boring, but I don't blame this on Princeton. Gunning is old but nice and sometimes pretty funny. We used the Hoffman and Kunze book.
Math 330: Fourier Series and Partial Differential Equations. [url=http://en.wikipedia.org/wiki/Yakov_G._Sinai]The Prof.[/url] This was a rigorous introduction to the theory of Fourier series and the Fourier transform. We proved a lot of convergence results (about the series) and the standard theorems about the transform. After learning all of the theory, we did some PDE problems, mostly with Laplace's equation, the heat equation, and the wave equation. In the end we discussed finite Fourier analysis and the Fast Fourier Transform. This class was wonderful. Sinai (being Russian) had an accent, but after the first day I could understand him easily. We use Stein's book, but Sinai doesn't teach right from it, he just uses it as a resource.
Math 332: Integration Theory and Hilbert Spaces. [url=http://www.agnesscott.edu/Lriddle/women/chang.htm]The Prof[/url]. In my opinion most of the material in this class was EXTREMELY cool, but unfortunately I didn't really like the professor's lecturing style. She taught from transparencies and only used the board occasionally for proofs, and everything was straight from the book. This was Stein's book also (he wrote a series of four), so fortunately it was a great book. This is just evidence that no matter where you go, you won't like every single professor. Anyway, this class covered measure theory, the Lebesgue integral and differentiation theory, infinite dimensional Hilbert spaces, including some examples (we discussed the Fourier transform, some complex analysis stuff I didn't understand that well because it assumed a prior course in it, and weak solutions to partial differential equations). We finished with Hausdorff measure, which was cool.

Some further comments: The 300-level classes are not usually intended for freshman. However, Princeton lets you do pretty much whatever you want. If you go against their advice, and you do okay, great. If you go against their advice and get screwed, they won't be sympathetic. (They didn't advise me not to take those classes, by the way. But my second semester was pretty hard.)
The math classes are extremely difficult. I spent far more time second semester on problem sets for Math 330 and 332 than I did with anything else. But I never really felt bored or that I was doing busy work. Some of my elders seem to think I've been taught too much modern math too quickly, and that I need a stronger foundation in the basics, but again, this isn't entirely Princeton's fault, as I did go ahead.
At the summer program I just completed, there were some Texas A&M graduate students, and I was noticably more knowledgeable about analysis than most of them. (Not the two guys who were studying SCV though.)

Finally, regarding the size of the department: I don't know about MIT or Harvard. There are 15-25 majors in math each year at Princeton. (You don't declare your major until the end of sophomore year, unless you're an engineer.) This means that there are more math professors than undergraduates in the major. I do not know how many graduate students there are; I think it is around the same amount per year though. I think it's really nice -- I can easily talk to any professor, and when I am doing my independent work junior/senior year, most likely my advisor will only be dividing his attention between me and one or two other students, if he has any others at all!

I hope I answered most of your questions regarding Princeton. I'm afraid I can't help with MIT or Harvard. Feel free to PM me or post your questions here again if there's anything else you want to know. :)

Yeah, I have Sinai again next semester for probability theory (I'll wager that's what the research you refer to was about). Maybe I can impress him with a reference. :D :D

[quote="orl"][quote="Elemennop"][quote="Xevarion"]
Math 330: Fourier Series and Partial Differential Equations. [url=http://en.wikipedia.org/wiki/Yakov_G._Sinai]The Prof.[/url] [/quote]

Woahhhhhh

My dad has done alot of research work with that guy[/quote]

So this leaves the question who your father is, isn't it?  :D[/quote]

Chernov. He's mainly done research with him related to billiards and statistical models with billiards. And actually, now that I think about it, he was actually my dad's advisor for his PhD/masters/whatever it was when he was a student at Moscow University (I'm pretty sure this is right, since my dad mentioned Kalmagorov was his grand-advisor, and Sinai was a student of Kalmagorov's).

I would guess Sinai would introduce alot of billiards in his Probability Theory class, Xevarion.

Haha that's what he's famous for, eh? You know, Egorov was Kolmogorov's advisor, I think. So your dad is in a chain of pretty awesome mathematicians.

Well, he must be pretty lucky in his acquaintances.

Maybe not all of you are familiar to how possibly the previous posters in this thread retrieved the information who was who's Ph.D. advisor, then I suggest checking the Mathematics Genealogy Project for Yakov Grigor'evich Sinai [url=http://www.genealogy.ams.org/html/id.phtml?id=10481]here.[/url]

I was wrong! Lusin was Kolmogorov's advisor. (The reason I know about this is we had a discussion about it in class one time. I think I used Lusin's theorem [about the Lebesgue integral] in some proof.) But Sinai said that Egorov couldn't have been Lusin's advisor; they were about the same age. (Maybe he was wrong? From Google I find Egorov was 14 years older.)

I have a similar question as to which school I should apply early to.  However, my debate is not over which school I like the best.  Rather, I'm curious as to which of the top schools has the most generous early application program, and where I am most likely to be accepted.

The word on the street is that MIT does not give an advantage to early applicants. It appears that ALL other schools do. So the strategic thing to do, I suppose, if you like MIT and a few other schools, is to apply to your favorite of the other schools early, and to apply to MIT and all the others during the regular round. But if your first-choice school is not a single-choice early action school, or an early decision school, you could apply early to all the schools you like that permit you to do so on a nonexclusive basis.

Making decisions about which college to attend is simply difficult when financial considerations are involved. I visited a state mu alpha theta (MAO) conference a few months ago and talked to two then seniors (whom I did not know that well, but through chatting, I knew that their achievements are similar). Both had placed in top 20 in state MAO in their grade and both were national merit finalists. One was accepted and decided to go to MIT on a free ride since his family income was low. The other made the decision to stay in honor’s program in a state university, where he would get a free ride based on his achievements and state residency. However, if the later decided to go to an ivy league school or MIT (which I imagine that he should not have any problems as long as he applied for more than 3 of them), his family would have to pay the ticket price (during the conversation, he told me his dad is a cardiologist). I felt very sorry for him. In my mind, going to a top tier school isn’t really for education, but for networking and meeting exceptional people. Unfortunately, some of the well qualified students do not feel that they want their family to pay for their college (even their families have no problems in paying and willing to). Those students may lose the opportunity to go to Harvard, MIT…

tokenadult and Gyan, do you have any comments and suggestions :?:

To Gyan, was your son (the one at Duke) on Duke’s scholarship (the AB scholarship or other scholarship)? If a student got accepted by Harvard (has to pay ticket price) and Duke (awarded AB scholarship), which one should he/she go? Duke is certainly great, but Duke is not Harvard (at least in my mind). Gyan, I hope your son did not make that decision based on Duke's scolarship.

qvc - Yes my son was on AB Duke scholarship.  There were a few kids in his batch (and in later batches too) who were also accepted in Harvard (virtually every AB duke finalist had other offers from Harvard/Princeton etc) and virtually all (though, obviously,  a few exceptions )   chose Duke.   Within the people I know  well (and these guys are financially so well off  that   tuition etc  was not a factor at all in the selection process)  all three out of three  chose Duke over Harvard  .  Actually once they visited Duke (and other schools)  the decision was not even that difificult or close.  (Duke recruiting process is very nice  and AB duke scholarship  offers things like -  summer program in Oxford etc - Heck they  pay your tuition, if you desire to go  take a few semesters  in other places (including Harvard :) ) .  Not to mention, AB Duke scholarship  is very prestigious.

Gyan, thank you for your reply. You have all the reasons to be proud of your sons. I do understand how prestigious AB Duke Scholarship is. Getting AB Duke Scholarship is generally much more difficult than merely getting accepted by Harvard. My older daughter (just started her junior year) will apply to Duke and a few others. She likes MIT. But she told me on multiple occasions that if she is awarded AB Duke Scholarship, she is likely to go to Duke. Obviously, my husband and I want her to go to Harvard.

Hi, qvc, I think your observation is correct that the merit scholarship programs have a stronger pull for families with high enough incomes that they won't receive the HUGE financial aid awards that are possible at need-based-only financial aid schools like Harvard and MIT. Then, the issue looks like spending serious money for one thing (Harvard or MIT) versus getting something else (Duke, Caltech, etc.) for free. One AoPS parent who private-messaged me a year or two ago about his son, a prominent math competitor, was wondering whether he should pay out of pocket (maybe not full list price, but a serious amount of money) for MIT or have his son attend his state's state university. He finally decided to spend the money to have his son go to MIT. That's what I would do in the same circumstance--but I am NOT in the same circumstance. We (our family) is likely to qualify for huge need-based financial aid, so the merit aid offers don't look as appealing to us.

So you see Gyan and I might look at the same choice and urge our children to do different things. I can't say it's wrong to decide to take a merit aid offer from Duke--Duke has a fine math program--but what I can say is that you should urge your child to apply to several interesting programs and gather as much information as possible, and then make a decision based on the best information all of you have about what fits your family's situation.

Good luck in the applications.

Token Adult, qvc -IMO  criteria is more than just financial consideration.  For example,  When it comes to AB Duke scholarship, (or other honors program)  it's , I think,  little to do with financial considerations.  Whats make it attractive is the program (and prestige too) which goes with it.

I know of a few cases,  where Harvard "matched" the   Duke's AB scholarship's  financial part (Full tuition)  after learning that the candidate was offered ABD scholarship.  (I was  told that it is routinely done , though I don't see  any "merit scholarship"  on Harvard's Web site - They do have Presidential scholarship,  and the financial assistance associated with is quite 'flexible' or so I was told ). These candidates, not surprisingly, still chose Duke.

About 4 years ago, when I was doing all the college visits, virtually all the professors (including the one who  was  at MIT- who told me, if it was his son his first choice,  - taking into factor like his   interest in math and classic languages,  - would be  Duke) talked highly of Duke's math program.

qvc - Best wishes  for your daughter.

Tokenadult and Gyan, thanks for your inputs. They are very helpful.

My daughter still has a year before applying for college. She is not at imo level. She did qualify for usamo in her freshman and sophomore years without much effort. Any additional advancement beyond that level would need some serious effort, which is very difficult for her schedule (fully loaded in IB program with 7 different subjects, plus 3 online AP courses, piano and violin lessons and weekend Chinese class).

We do not qualify for any need-based-only financial aid (We are willing to pay for prestigious college). That “Harvard matched the Duke's AB scholarship's financial part (Full tuition) after learning that the candidate was offered ABD scholarship” is certainly an interesting piece of information. We have signed up to visit MIT and Duke’s recruiting shows “college in your town” in a month. Duke’s show is actually a combination of Duke, Harvard, U of Penn and Georgetown. I will keep you posted about my daughter’s progress.

[quote]We do not qualify for any need-based-only financial aid[/quote]

There are certainly families of which this is true, but do be sure to check what income range is ABSOLUTELY beyond the range for financial aid eligibility, because of course when colleges cost more than 40 thousand dollars per year, incomes above 100 thousand dollars per year can still be eligible for financial aid. It doesn't hurt to apply if there is any chance at all of receiving need-based aid.

[quote="tokenadult"]Hi, qvc, I think your observation is correct that the merit scholarship programs have a stronger pull for families with high enough incomes that they won't receive the HUGE financial aid awards that are possible at need-based-only financial aid schools like Harvard and MIT. Then, the issue looks like spending serious money for one thing (Harvard or MIT) versus getting something else (Duke, Caltech, etc.) for free.
...
[/quote]

I can tell you that Caltech's financial aid is top-notch.  They routinely meet 99-100% of need as determined by FAFSA.  One of the major factors influencing my decision to go there is the fact that they gave me ~ 4x as much financial aid as any other school (outside of UMD's full-ride), and was slightly cheaper, too.  There are some changes going on in the whole administration right now that may affect financial aid, though.  The current students aren't too happy about it and letting everyone know about it, so I'm wondering how it will all play out.

I heard that the ivy league (or some of its colleges) have ended their early decision program, very recently. :|

This year the early round application programs are working just as they did last year. Next year (2007-2008 school year, for class of 2008 high school students applying to college class of 2012), Harvard, Princeton, and the U of VA will no longer have an early round of admission application decisions. It's hard to say how many other colleges will follow their example.


I took a course in problem solving in elementary school that used two books by a lady author who used journaling and encouraged groups of children of all ages to solve problems their own way and then present their solutions to the others so each group would be exposed to the others method of solution.  I can't remember her name as I let the books "get away" temporarily.  Does anyone know her name?

Also:  How do you make the text of these messages larger?  I can hardly read  the text.  nancyelizabeth

The small text is probably your browser.  Try holding Alt and moving your mouse scroll ball (if you have one).  Or if that doesn't work go up to the pull down menus at the top of the screen and look for the text size one.

Perhaps the author you are asking about is Marilyn Burns?

Regarding the text size, besides the suggestion below, you can click on the Forum link on the left side of the page, and there should be a line near the top of the Forum page that lets you select "Font Size".


Find the number of even numbers from 10000 to 79999 which also have no same digits .

Case 1:The last digit is 2,4, or 6
In this case, there are 3 possibilities for the last digit, then 6 for the first digit, then 7,6, and 5 for the other digits for a total of 3780
Case 2: The last digit is 0 or 8
In this case, there are 2 possibilities for the last digit, then 7 for the first digit, then 7,6, and 5 for the other digits for a total of 2940.

There are 6720 in total.


geometry
geometric transformation
rotation

How many ways are there to sit 8 boys and 13 girls that no boys sit next to each other and each girl sit next to at least one boy?

Is there a better way to do this other than listing?

Are rotations different?
[hide]Insert 1-2 girls in the 8 slots between each boy. So find the number of strings of 8 1s and 2s so the sum of the numbers is 13. Clearly, there are 5 2s and 3 1s. Since the table is circular, we just need to find how many ways we can place the three ones in the eight slots. This is $ {8 \choose 3}$ if rotations are different.[/hide]

Rotations are considered the same.
you still need to permute those persons right?


induction
real analysis
real analysis solved

Let p be a natural number p>=2, fixed and (x_n) defined like so:
x_0>0 and x_(n+1)+1=(px_n+1)^(1/p), for any n natural. Compute lim(n-->oo) [sum(k=0,n)x_k]/ln(n)

"Dan Barbilian" contest 6 dec 2003

Use AM-GM to obtain that x_n+1<=x_n. Also, by induction we obtain x_n>=0. So x_n is convergent and it's limit verifies (1+l)^p=1+pl, from where l=0. It suffices to find lim n*x_n since by Cesaro-Stolz theorem
lim (x_0+x_1+...+x_n)/lnn=lim x_n+1/(ln(1+1/n))=lim n*x_n. We calculate lim n*x_n using Cesaro_Stolz: lim n*x_n=lim n/(1/x_n)=lim 1/(1/x_n+1-1/x_n)=lim 1/(1/((1+px_n)^1/p-1)-1/x_n). Put 1+px_n=(1+y_n)^p. It follows that lim y_n=0 and we have to find lim 1/(1/y_n-1/((1+y_n)^p-1)), which is trivial.

can someone compute the last limit, it's not trivial for me! :)


function
calculus
integration
real analysis
real analysis unsolved

No solution has been proposed for this problem ([b]U19[/b]) so far in the online publication.

Let $f_{0}$ be a real valued function which is continuous on the interval $[0,1]$. For each non-negative integer $n$, let $f_{n+1}(x) = \int_{0}^{x}f_{n}(t) \, dt$. Suppose there is a positive integer $k$ such that $f_{k}(1) = \frac{1}{(k+1)!}$. Prove that there exists $x_{0}$ such that $f_{0}(x_{0}) = x_{0}$.

if not, $f_{0}(x)<x$ for all $x$, or $f_{0}(x)>x$ for all $x$. Integrate and conclude that $f_{k}(x) < \frac{x^{k+1}}{(k+1)!}$ for all $x$, or $f_{k}(x) > \frac{x^{k+1}}{(k+1)!}$ for all $x$


The sum to infinity of $\frac{1}{7} + \frac{2}{7^2} + \frac{1}{7^3} + \frac{2}{7^4} + \cdots$ is:
\[ \text{(A)}\ \frac{1}{5} \qquad \text{(B)}\ \frac{1}{24} \qquad \text{(C)}\ \frac{5}{48} \qquad \text{(D)}\ \frac{1}{16} \qquad \text{(E) none of these} \]

[quote="amirhtlusa"]write it as the sum of two geometric sequences
you get $\frac{7}{48}+\frac {1}{24}$
so the final asnwer is $\frac {9}{48}$[/quote]
Yeah. I get the same thing, writing them as two infinite series:
\[\displaystyle t_{n}=\frac{1}{7}\cdot\left(\frac{1}{7^2}\right)^{n-1}\]
\[\displaystyle t_{n}=\frac{2}{7^2}\cdot\left(\frac{1}{7^2}\right)^{n-1}\]
These are the two sequences. So their infinite sums, respectively, are:
\[\displaystyle \frac{\frac{1}{7}}{1-\frac{1}{7^2}}+ \frac{\frac{2}{7^2}}{1-\frac{1}{7^2}}\]
Simplifying this, we get:
\[\displaystyle \frac{\frac{7}{49} + \frac{2}{49}}{1-\frac{1}{49}}\]
\[\displaystyle = \frac{\frac{9}{49}}{\frac{48}{49}}\]
\[\displaystyle = \frac{9}{48}\]
\[\displaystyle = \frac{3}{16}\]

So that would be $\displaystyle (E)$ Are we right? :D

[quote="joml88"]Hmm...why is the title of this "[i]Alternating[/i] Series Sum"?[/quote]
because of $1$ and $2$.(alternating doesn't always mean it is going positive, and negative)

O, lol...i didn't see those 2's... I need better glasses...or i should use my contacts...


algebra
polynomial
system of equations
algebra unsolved

We are given the following polynomial : \[ P(x)=x^4-6x^3+23x^2-34x+26 \]

Considering that $1$ is solution to the system of equations :\[ -4x^4+12x^3-34x+26=0 \ \text{and} \ -12x^3+46x^2-34x=0 \] show that $z=1-i$ is a complex root of $P$

Thanks for your ideas!
Mark.

There is no mystery:

$P(x)=(x^2-2x+2)(x^2-4x+13)$, and $1+i, 1-i$ are the roots of $x^2-2x+2$.

But apparently the question is how to spot such a factorisation thanks to this system of equations?


geometry
3D geometry
invariant
conics
ellipse
geometric transformation
rotation

Let $p, (p\ge 2),$ be a fixed integer. If ${\mathbf x}: =(\alpha,\beta,\gamma)\in{\mathbb R}^{3}$ and $\alpha^{p}+\beta^{p}=\gamma^{p}$let us write  ${\mathbf x}\in{\mathcal P}\; .$ [b] [color=green]
 Find all linear transformations $A:{\mathbb R}^{3}\to{\mathbb R}^{3}$ which preserve the class ${\mathcal P}$ in the sense that  $A\left({\mathcal P}\right)\subseteq{\mathcal P}\; .$ [/color] [/b]

[size=75] [b] Remark:[/b]  Such non-trivial mapping exists. For instance,in case $p=2$ let $A({\mathbf x})=M{\mathbf x}^{T}$ where $M=\left(\begin{array}{ccc}2&1&2\\ 1&2&2\\ 2&2&3 \end{array}\right) \; .$ [/size]

Consider the case $p=2$. $A$ leaves the cone $P: =\{\mathbf{x}\colon x_{1}^{2}+x_{2}^{2}=x_{3}^{2}\}$ invariant precisely when the circle $C=\{(x_{1},x_{2},1)\colon x_{1}^{2}+x_{2}^{2}=1\}$ is mapped into the cone. Since $C\subset N: =\{\mathbf{x}\colon \langle \mathbf{x},e_{3}\rangle=1\}$, the image of $C$ lies in the plane $A(N)=\{\mathbf{x}\colon \langle \mathbf{x},\mathbf{v}\rangle\}=1$ with $\mathbf{v}=(A^{*})^{-1}e_{3}$. Since $A(N)$ must intersect the cone along an ellipse, we have $v_{3}^{2}>v_{1}^{2}+v_{2}^{2}$. Of course, this is only one necessary condition. The other condition is that the ellipse of intersection coincides with $A(C)$, but I could not get this in a clean form.

If $p\ge 3$, then $\mathbf{v}$ must be collinear with $e_{3}$ (otherwise the intersection of $A(N)$ with $P$ is not an affine image of $N\cap P$), and $A$ must be a rotation (+reflection) about the $x_{3}$ axis multiplied by a scalar... or so I think.  :maybe:


[quote="sharkman"]In his will, a man leaves 1/2 of his money to his wife, 1/2 of what is left to his older child and 1/2 of the remainder to his youngest child.  His two cousins divide the rest equally, each receiving 2000 dollars.  What was the total amount of money BEQUEATHED in the man's will?

Also, what does the word BEQUEATHED mean as used in the word problem?[/quote]
"Bequeathed" means given.
[hide] Say that man has $x$ dollars.  $\frac{x}{2}$ goes to his wife.  This leaves him with $\frac{x}{2}$ of his dollars left.  Now, half of this goes to his older child.  Thus is $(\frac{x}{2})(\frac{1}{2})=\frac{x}{4}$.  So $\frac{x}{4}$ of his dollars goes to his older son.  Now, there is still $\frac{x}{4}$ dollars remaining.  Half of this goes to his younger son.  Thus, $(\frac{x}{4})(\frac{1}{2})=\frac{x}{8}$ dollars goes to his younger son.  Thus, we have that there is $\frac{x}{8}$ dollars left.  Half of this goes to one cousin, and the other half goes to the other.  Thus, each cousin gets half of $\frac{x}{8}$, which is $\frac{x}{16}$.  Now, since each cousin gets 2000 dollars, we know that $\frac{x}{16}=2000\Rightarrow x=2000(16)=\boxed{32000}$ [/hide]

Why all of these problems?  Are they homework or something?

Bequeathed means, in the sense of the problem, the total sum of money available to give to people, that the will dictates.

[hide]$\frac{x}{2^{4}}=2000$

$x=32000$

[/hide]


Prove that for an obtuse triangle ABC with $ A \geq 90$ we have

$ b\plus{}c \leq \sqrt{2}a$

where $ a,b,c$ are sidelengths of triangle ABC opposite at angles $ A,B,C$

[hide="Solution"]

In obtuse triangle $ a^2>b^2\plus{}c^2$, and as $ b^2\plus{}c^2\ge2bc$, the proof is complete.
In right angled triangle is the equality case, for $ b\equal{}c$
[/hide]


[hide]

We want $n^2 \equiv 10 \ldots{} 19$ (mod 20)

Note that $n^2 \equiv (n+20)^2$ (mod 20), so we only need to find how many $n$ in $\{1,2,3,\ldots{},20\}$ have the tens digit of $n^2$ odd and multiply it by 5.

These numbers are 4,6,14,16. So the answer is (B).

[/hide]


I have 14.  levans has confirmed that there are no more than 14.

[quote="PhireKaLk6781"]I currently have nine badges, one for each of the three categories.[/quote]

I sadly have only 13. :)

How come there is a maximum of 5 badges for Wins and Rating but not for Games?

[quote="PhireKaLk6781"]I currently have nine badges, one for each of the three categories.[/quote]
wait so in ftw $ 1\cdot 3\equal{}9$? no wonder.
i have 13 :(

i have ten badges
4-games
5-wins
1-rating

w8 does that make any sense? how can you win more games then you play?

who has the most???
probably Jongy or AIME15
right?

@cgyao15: Probably the win standards are lower? Perhaps...

@shadow123blast: I know Jongy and AIME15 both have 14, I think the other >2000 rating ones also probably have 14.

there's a leaderboard.
in all-time, everyone over 2000 has 14(most likely)

Isn't it impossible to have 1 badge? When you play 1 game, you get 1 game and rating badge.


Let $ a$, $ b$, $ c$ be the sides of a triangle, and $ S$ its area. Prove:
\[ a^{2} \plus{} b^{2} \plus{} c^{2}\geq 4S \sqrt {3} \plus{} \sqrt {ab}(a\minus{}b)^2\plus{}\sqrt {bc}(b\minus{}c)^2\plus{}\sqrt {ca}(c\minus{}a)^2
\]
In what case does equality hold?

[quote="Shinraehee"]Let $ a$, $ b$, $ c$ be the sides of a triangle, and $ S$ its area. Prove:
\[ a^{2} \plus{} b^{2} \plus{} c^{2}\geq 4S \sqrt {3} \plus{} \sqrt {ab}(a \minus{} b)^2 \plus{} \sqrt {bc}(b \minus{} c)^2 \plus{} \sqrt {ca}(c \minus{} a)^2
\]
In what case does equality hold?[/quote]
Are you sure?


These are some problems from Loren Larsen's "Problem Solving Through Problems" Section 7.1 that I need help with. Please just give a hint if you can:

1) Suppose that a,b,c are positive numbers. Prove that:

(a) $ a^2 b^2 \plus{} b^2 c^2 \plus{} c^2 a^2 \geq abc(a \plus{} b \plus{} c)$

(b) If $ a \plus{} b \plus{} c \equal{} 1$, then $ ab \plus{} bc \plus{} ca \leq 1/3$

2) Prove that

$ \frac {1}{2}\cdot \frac {3}{4}\cdots \frac {999999}{1000000} < \frac {1}{1000}$

(Hint: Square each side and "give a little" to create a "telescoping product)

3) If $ \frac {a \plus{} b}{b \plus{} c} \equal{} \frac {c \plus{} d}{d \plus{} a}$ prove that either $ a \equal{} c$ or $ a \plus{} b \plus{} c \plus{} d \equal{} 0$.

Thanks!

1) a) Pick any two terms on the left, and use AM-GM.

b) If you homogenize the right hand size (multiply by $ a\plus{}b\plus{}c$ until both sides are the same degree) and multiply out, you'll have an inequality that can be solved in the same way as #1.

2) $ \frac{1}{2} < \frac{2}{3}$ and so forth.  The hint you posted really says it all, if it doesn't make sense then experiment with a smaller number of terms.

3) Multiply out, move everything to one side, and try factoring $ a\minus{}c$ out of the expression.

[quote="LydianRain"]b) If you homogenize the right hand size (multiply by $ a \plus{} b \plus{} c$ until both sides are the same degree) and multiply out, you'll have an inequality that can be solved in the same way as #1.
[/quote]

I can homogenize both sides:

$ ab\plus{}bc\plus{}ca \leq 1/3 \equal{} (a\plus{}b\plus{}c)^2/3 \equal{} \frac{1}{3}\left(a^2\plus{}b^2\plus{}c^2\plus{}2(ab\plus{}bc\plus{}ac)\right)$

But I don't see how you can apply AM-GM to that.

[quote="FibonacciFan"]If $ \frac {a \plus{} b}{b \plus{} c} \equal{} \frac {c \plus{} d}{d \plus{} a}$ ...[/quote]

If $ \frac{p}{q} \equal{} \frac{r}{s}$ then either $ p \equal{} \minus{}r, q \equal{} \minus{}s$ or $ \frac{p}{q} \equal{} \frac{r}{s} \equal{} \frac{p \plus{} r}{q \plus{} s}$.

[quote="FibonacciFan"]
I can homogenize both sides:

$ ab \plus{} bc \plus{} ca \leq 1/3 \equal{} (a \plus{} b \plus{} c)^2/3 \equal{} \frac {1}{3}\left(a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ac)\right)$

But I don't see how you can apply AM-GM to that.[/quote]

First rearrange the inequality to

$ 0\le \frac{1}{3} (a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)$.

Here we can either apply AM-GM or note that $ a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca\equal{}\frac{1}{2}((a\minus{}b)^2\plus{}(a\minus{}c)^2\plus{}(b\minus{}c)^2)$.

[quote="FibonacciFan"][quote="Alfred"]
$ 0\le \frac {1}{3} (a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)$.

Here we can either apply AM-GM or ...[/quote]

How can you apply AM-GM?[/quote]

$ \frac{a^2 \plus{} b^2}{2} \geq ab$

$ \frac{b^2 \plus{} c^2}{2} \geq bc$

$ \frac{c^2 \plus{} a^2}{2} \geq ca$

Add 'em up! :)  This is also a good strategy for 1) a).


Will it include mechanical waves and fluid mechancs?

From what I saw on last year's exam, no.

Is last year's exam available online somewhere?

I doubt it's online.  I think you can order old exams from the site, though.

well last year's test had one fluid mechanics question at the very end...


Jo is thinking of a positive integer less than 100. It is one less than a multiple of 8, and it is three less than a multiple of 7. What is the greatest possible integer Jo could be thinking of?

The largest number under the first criteria is $ 95$.  Then you keep subtracting $ 8$ until your number meets the second criteria.  However, since $ 95$ meets both criteria, the answer is in fact $ \boxed{95}$.


Let $ a$, $ b$, and $ c$ be non-negative real numbers and let $ x$, $ y$, and $ z$ be positive real numbers such that $ a\plus{}b\plus{}c\equal{}x\plus{}y\plus{}z$. Prove that
\[ \dfrac{a^3}{x^2}\plus{}\dfrac{b^3}{y^2}\plus{}\dfrac{c^3}{z^2} \ge a\plus{}b\plus{}c.\]
[i]Hery Susanto, Malang[/i]

It's obvious by Holder's inequality.
By Holder's inequality
[hide]$ (x \plus{} y \plus{} z)(x \plus{} y \plus{} z) (\frac {a^{3}}{x^{2}} \plus{} \frac {b^{3}}{y^{2}} \plus{} \frac {c^{3}}{z^{2}}) \ge (a \plus{} b \plus{} c)^3 \implies \frac {a^{3}}{x^{2}} \plus{} \frac {b^{3}}{y^{2}} \plus{} \frac {c^{3}}{z^{2}} \ge a \plus{} b \plus{} c$[/hide]

[quote="Raja Oktovin"]Let $ a$, $ b$, and $ c$ be non-negative real numbers and let $ x$, $ y$, and $ z$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} x \plus{} y \plus{} z$. Prove that
\[ \dfrac{a^3}{x^2} \plus{} \dfrac{b^3}{y^2} \plus{} \dfrac{c^3}{z^2} \ge a \plus{} b \plus{} c.\]
[i]Hery Susanto, Malang[/i][/quote]
$ \frac{a^3}{x^2}\plus{}x\plus{}x \geq 3a$

By Holder inequality ,we have

$ (x\plus{}y\plus{}z)(x\plus{}y\plus{}z)\left(\frac{a^3}{x^2} \plus{} \frac{b^3}{y^2} \plus{} \frac{c^3}{z^2}\right) \ge (a\plus{}b\plus{}c)^3$

$ \frac{a^3}{x^2} \plus{} \frac{b^3}{y^2} \plus{} \frac{c^3}{z^2}\ge\frac{(a\plus{}b\plus{}c)^3}{(x\plus{}y\plus{}z)^2}$

since $ x\plus{}y\plus{}z\equal{}a\plus{}b\plus{}c$ so

$ \frac{a^3}{x^2} \plus{} \frac{b^3}{y^2} \plus{} \frac{c^3}{z^2}\ge\frac{(a\plus{}b\plus{}c)^3}{(x\plus{}y\plus{}z)^2}\equal{}a\plus{}b\plus{}c$

EDITED : Sorry I didn't open your quoting (dgreenb801) before  :oops:

I think the problem without the condition $ \sum a \equal{}\sum x$ was asked in the Belarus Olympiad 2000....Nevertheless, a useful tool for solving tougher inequalities...

This problem is an imediate result of the following inequality:

If: $ a_i\geq0,b_i>0,i\equal{}1,2,\dots,k,$ and $ k,n\in N,k\geq1,$ then:

$ \frac{{a_1}^{n\plus{}1}}{b_1^n}\plus{}\frac{a_2^{n\plus{}1}}{b_2^n}\plus{}\cdots\plus{}\frac{{a_k}^{n\plus{}1}}{b_k^n}\geq\frac{(a_1\plus{}\cdots\plus{}a_k)^{n\plus{}1}}{(b_1\plus{}\cdots\plus{}b_k)^n}$

You can notice that I've proved this inequality, using my lemma, at http://www.mathlinks.ro/viewtopic.php?t=310721 , as an application of it.
Of course, it could also be considered as a result of Holder's inequality.

For $ a_1\plus{}\cdots\plus{}a_k\equal{}b_1\plus{}\cdots\plus{}b_k$ we obtain:

$ \frac{{a_1}^{n\plus{}1}}{b_1^n}\plus{}\frac{a_2^{n\plus{}1}}{b_2^n}\plus{}\cdots\plus{}\frac{{a_k}^{n\plus{}1}}{b_k^n} \geq a_1\plus{}\cdots\plus{}a_k$

For $ k\equal{}3, n\equal{}2$ we get to the initial problem.

I don't know what Holder ineq is,but I can second-kill it by using Carlson's ineq:
by Carlson's ineq
$(\sum\frac{a^3}{x^2})(\sum x)(\sum x)\ge (\sum a)^3$
what follows is trivial.

[quote="marin.bancos"]This problem is an imediate result of the following inequality:

If: $ a_i\geq0,b_i>0,i\equal{}1,2,\dots,k,$ and $ k,n\in N,k\geq1,$ then:

$ \frac{{a_1}^{n\plus{}1}}{b_1^n}\plus{}\frac{a_2^{n\plus{}1}}{b_2^n}\plus{}\cdots\plus{}\frac{{a_k}^{n\plus{}1}}{b_k^n}\geq\frac{(a_1\plus{}\cdots\plus{}a_k)^{n\plus{}1}}{(b_1\plus{}\cdots\plus{}b_k)^n}$
[/quote]

This inequality is right for n any positive real !

It was also in district olympiad of Kazakhstan in 2010 january for 9 grades. Then this problem was again in city (regional) olympiad in march of 2010 again for 9 grades .  It was interesting to see one problem twice  in two consecutive rounds. :lol:

We have : $\frac{a^3}{x^2}+x+x\geq3a $ (by Cauchy )
Similar : $\frac{b^3}{y^2}+y+y\geq 3b  $ and $\frac{c^3}{z^2}+z+z\geq 3c  $
=> $\frac{a^3}{x^2}+\frac{b^3}{y^2}+\frac{c^3}{z^2}\geq 3(a+b+c)-2(x+y+z)=a+b+c$ ( too easy for a TST problem ) .

Expanding problem : $\frac{x^{n+1}}{a^n}+\frac{b^{n+1}}{b^n}+\frac{z^{n+1}}{c^n}\geq a+b+c$ .

[quote=achen29]One step by Titu's Lemma[/quote]

Thanks, for your advice about this trivial inequality. Of course, there were no any analogous solutions before. THANK YOU!!!


algebra
polynomial
linear algebra
matrix
geometry
3D geometry
number theory unsolved

Let $ P(x)$ be a polynomial that sends integers to integers.  Let $ \alpha_1, \alpha_2, ... \alpha_n$ be the complete list of conjugates of some algebraic integer.  Prove or disprove that $ P(\alpha_1) \plus{} ... \plus{} P(\alpha_n)$ is always an integer.

[hide="Remark"] Of course it suffices to show this for $ P(x) \equal{} {x \choose k}$.  I have some ideas about how this proof should go but I haven't worked it out yet. [/hide]

What I see is a trace. If $ A$ is an integer matrix with the $ \alpha_i$ and only them as eigenvalues (always possible with a companion matrices), we want to show that the trace of $ P(A)$ is always an integer, or find some counterexample.

For a quick counterexample, let $ \omega,\omega^2$ be the non-real cube roots of $ 1$. Let $ A\equal{}\begin{bmatrix}0&1\\\minus{}1&\minus{}1\end{bmatrix}$; we have $ \text{tr}(A^n)\equal{}\begin{cases}2&3|n\\\minus{}1&\text{otherwise}\end{cases}$. We get $ \text{tr}\binom{A}{3}\equal{}\frac16\left[\text{tr}(A^3)\minus{}3\cdot \text{tr}(A^2)\plus{}2\cdot \text{tr}(A)\right]\equal{}\frac16[2\plus{}3\minus{}2]\equal{}\frac12$. In other words, $ \binom{\omega}{3}\plus{}\binom{\omega^2}{3}\equal{}\frac12$.

It's false.


The card (figure on the left) will be placed on a table with unit cubes stacked on top of  it to create a solid such that the number in each square indicates the number of cubes to be stacked in the column above the square ( the figure on the right shows two of the eight stacks already in place).  Including the 8 sq un of the bottom face of the solid, what is the surface area of the entire solid formed once all the cubes are in place?
card:
3        4

2        3

5        2

1        3
(I can solve this and you can too, but how do you find a quicker method to do it without counting everything up or multiplying for each rectangle or something?)

[hide="general method"]First the top, bottom, and the sides. Now, add in the differences between adjacent numbers, This will give you the surface area (make sure you see why).[/hide]


The vertices of the graph $ O_k$ are the k-element subsets of $ \{1,2,..,2k\plus{}1\}$. Two vertices are adjacent if and only if they are disjoint sets. Prove that the girth (the minimum circle) of $ O_k$ is $ 6$ if $ k\ge 3$.

Given any two disjoint subsets of size $ k$ from a $ 2k\plus{}1$ element set, there is one element left over. We can use this element to "label" each edge.

Lemma: Starting with a particular set, $ S$, there are exactly $ k\plus{}1$ edges coming from it. For each element, $ x$, not in $ S$ but in $ 1,...,2k\plus{}1$, there is a unique edge coming from $ S$ labeled $ x$.
Proof: This is obvious.

So without loss of generality, we can start with the set $ 1,2,...,k$. Now there are exactly $ k\plus{}1$ edges coming from this set and they are labeled $ k\plus{}1,...2k\plus{}1$. Lets continue going out another edge. The next set is $ k\plus{}1,...,m\minus{}1,m\plus{}1,...,2k\plus{}1$. So the edges connected to this are labeled $ 1,2,...,k,m$. But if we go along the edge labeled $ m$, we traversed an edge and went back. Since we are looking at cycles, we cannot do this. So the options for the second labeled edge are $ n\equal{}1,2,...,k$.

So if we start with $ 1,2,...,k$ and are looking at a cycle, the next edge is $ m>k$, then that is connected to by an edge labeled $ n\le k$. Now observe:

\[ 1,2,....,k \rightarrow m \rightarrow k\plus{}1,k\plus{}2,...,m\minus{}1,m\plus{}1,...,2k\plus{}1\rightarrow n\rightarrow 1,2,...,n\minus{}1,n\plus{}1,..., k, m\]

So the effect edges $ (m,n)$ was to replace $ n$ by $ m$.

Using this observation it is easy to construct a $ 6$ cycle. We just take the set $ 1,2,...,k$ and replace $ k$ by $ k\plus{}1$, then replace $ k\plus{}1$ by $ k\plus{}2$ and replace $ k\plus{}2$ by $ k$.

So, in particular, we get the sequence
[hide]$ 1,2,...,k$
- $ k\plus{}1$ ->
$ k\plus{}2,...,2k\plus{}1$
- $ k$ ->
$ 1,2,...,k\minus{}1,k\plus{}1$
- $ k\plus{}2$ ->
$ k,k\plus{}3,...,2k\plus{}1$
- $ k\plus{}1$ ->
$ 1,2,...,k\minus{}1,k\plus{}2$
- $ k$ ->
$ k\plus{}1,k\plus{}3,...,2k\plus{}1$
- $ k\plus{}2$ ->
$ 1,2,...,k$

and just to note for what I wrote above to make sense, we need in the 5th set written there to have $ k\plus{}3\le 2k\plus{}1\iff k\ge 2$ (and if you check, that $ 6$ cycle is valid for $ k\equal{}2$ [/hide]

Now we must show that there is no shorter cycle. Clearly there are no one or two cycles.

Now lets look at $ 3$ and $ 5$ cycles. A 5 cycle is represented by edges $ m_1,n_1,p,n_2,m_2$ where $ m_1\ne n_1$, $ m_2\ne n_2$, $ m_1,m_2\ge k\plus{}1$, $ n_1,n_2\le k$ (note by looking at the cycle going the other way we get $ m_2,n_2$).

Now since $ k\ge 3$, there is an $ l\le k$ that is not equal to $ n_1,n_2$. It is easy to see that after one edge, $ l$ is not in the set, and after two edges, $ l$ is in the set (since $ l\ne n_1$), then after $ 3$ edges, $ l$ is not in the set which contradicts the $ 3$ cycle case. Then continuing, since $ n_2\ne l$, after $ 4$ edges $ l$ is in the set, $ 5$ edges not in the set which is a contradiction to the $ 5$ edge cycle.

Now we just need to look at $ 4$ cycles. Indeed, this is represented by $ m_1,n_1,n_2,m_2$ which satisfy the same conditions as above. But we look at the middle set. We must arrive at the same set by replacing $ n_1$ with $ m_1$ as we would with replacing $ n_2$ with $ m_2$. This obviously forces $ m_1\equal{}m_2$ and $ n_1\equal{}n_2$. But all that means is that we are going back and forth along two edges which is not a cycle.


analytic geometry
conics
ellipse
probability
parabola
LaTeX

:arrow:    an ellipse is inscribed in a circle and a point within the circle is chosen at random.if the probability that this point lies outside the ellipse is 2/3  ,then the eccentricity of the ellipse is_____

 :arrow:   a variable parabola of latus rectum l ,touches a fixed parabola ,then axis of the two curves being parallel .the locus of the vertex of the moving parabola is a parabola whose latus rectum is _____

Is the eccentricity of ellipse :-  (2/3) * sqrt(2)

PS:- Sorry don't know how to use latex!!!!!

probability for point inside ellipse = 1 - 2/3 = 1/3

so, 1/3  =  (pi*a*b)/(pi*a*a) or b = a/3 ....(i)

now , b^2 = a^2(1-e^2)

Replacing with (i)

e = (2/3) * sqrt(2)

Is it correct/ forgive me again for not using latex!!!!

nice solution.

please try the second one also.

That was my solution also but I think it is wrong as it is not necessary that the center of the ellipse be the same as the center of the circle.

Than how can we inscribe the ellipse :maybe:


Let be $ a,b,c\ge0$. For $ 0\le k\le7$, prove that :

$ \sum_{cyc}a^k(b \minus{} c)^2 (b^3 \minus{} c^3)^2 (a \minus{} b)(a \minus{} c)\le 0$.

[quote="fjwxcsl"]Let $ {a,b,c}\geq0,{k,n}\in{N},$prove

$ \sum_{cyc}a^{k}(b \minus{} c)^{2}(b^{n} \minus{} c^{n})^{2}(a \minus{} b)(a \minus{} c)\le 0$

hords  iff  $ {0}\le{n}\le{2k \plus{} 1}.$[/quote]

  Sorry!   The inequality hords  iff  $ {0}\le{k}\le{2n \plus{} 1}.$


[quote="ragingbullrocky"]Know anybody who does every single problem in a textbook?[/quote]

This should go into G&FF...

and yes, many AoPSers would do all the problems in their math book =)

Depending on what textbook it is. :rotfl:
[if it's my school calculus textbook then i would be wasting time]

I do. It's good for practice. Even if some are easy, I do them.

Good luck trying to do this with college level math textbooks. I'll bet if someone did every single problem in Dummit & Foote's algebra book, grad school quals in algebra would be a breeze. I get so exhausted just trying to understand the proofs of the theorems in the books I read, I often have little motivation to do problems at the end.

But doing problems helps a lot, especially when you're learning techniques rather than concepts.

Never, if it's for school. It's unbearably dull...
AoPS Vol. II is a completely different matter, however.

[quote="#H34N1"]Waste of time for textbooks. AoPS books are a different story though.[/quote]

I find this false. The problems especially the later ones in the textbooks are generally what I would find is taught in the actual lessons of AoPS which is the process and steps of deriving it. This may come easily to people on AoPS (which is why it's in the main section) but it's much more helpful for students to know where it came from and a homework reinforcing that with a proof helps a lot more than applying it over and over.

I try to do as many as possible- but sometimes it can get tedious! Depends on the topic matter I suppose, and how lazy I am!

I do the ones I find interesting, which accounts for around 1/10 of all the problems in the book

i do as many as possible (if its an interesting book)
in my opinion, if you are working in a book where you can easily do all the problems, you're wasting your time :)

[quote="rnwang2"][quote="daermon"]in my opinion, if you are working in a book where you can easily do all the problems, you're wasting your time[/quote]
This is so true...yet for some reason I feel like I must go through AoPS 1 and do every single problem...:|[/quote]

One word: don't.  :ninja:

Ahem ... a bit of wiser opinion.

Depends on what you aim at. Let's get this straight: if you are aiming for a 42 in IMO or something nearabout, [i]go for every single problem in every single textbook you have[/i]. I don't actually think that the problem of wasting time is serious: just take a pencil in hand, look at a sum for ten seconds, think of a solution, put a tick and go to the next sum. That way you can go through a fat yet easy textbook in a week.

However I concur a bit with Derek'e opinion - except that I put in a clause of my own: I do the ones I [i]can[/i] and find interesting. I have somehow never felt a strong urge to solve problems that are too tough. If you want to enjoy life [i]and[/i] enjoy maths, this is the option for you. Thos who do all they can don't enjoy maths, those who do all they find interesting don't enjoy life, and those who do ALL are best left unspoken of. :P

My school calculus textbook? I never do any homework ever.

USAMO 1972-1986 by Murray Klamkin? That's a different story. Modern Higher Algebra by E. Artin? Most of them, though I often skip trivial exercises merely for lack of time (like "Multiply everything out and verify it's a group etc.). I do, however, now and then try to prove a theorem before I read their prove.