I believe that this is an old problem but can't find it searching:

$f(x)$ is a polynomial with real coefficents and no non-negative real roots.
Prove that for some positive integer $n$:

$(1+x)^{n}\cdot f(x)$ has all positive real coefficents

x^2+x+1 has no positive real roots since the discriminent is -3

and -x^2-x-1 has the same roots as x^2+x+1 because multiplying 0 by -1 still creates a value of 0.

Since all the coefficients are negative and (1+x)^n has all positive coefficients each term when expanded is all negative and when combining like terms, they are still all negative

so then it does not have all positive real coefficents which means there is probably something missing

I have found the missing condition:

$f(x) > 0$ for all $x \geq 0$

It is a problem from the 1997 IMO shortlist.

There is a generalization of problem posted but I can't find it :
$f(x_{1},x_{2},\dots,x_{n})$ is a homogeneous polynomial that $f(x_{1},\dots,x_{n})>0$ if $x_{1},x_{2},\dots,x_{n}\geq0$ then there is $N$ that all coefficients of $(x_{1}+x_{2}+\dots+x_{n})^{N}f$ are positive.


function
limit
real analysis
real analysis unsolved

[size=150]Is  there existing a function $ f(x) \in C^1(\mathbb{R})$, satisfy $ f(x)>0$ and $ f'(x)\equal{}f(f(x))$, for $ \forall x\in \mathbb{R}$

prove or disprove[/size]

Note that, if $ f$ is a continuous increasing function which is bounded below, then $ \lim_{x\to\minus{}\infty} f(x)$ exists. This simple observation will lead us to a contradiction if we assume such $ f$ exists.

Let's assume that such $ f$ exists. Since $ f$ is positive and increasing, there exists $ 0 \leq l \equal{} \lim_{x\to\minus{}\infty} f(x)$. Also, since $ f'$ is positive and increasing, we find that $ f'(x)$ converges as $ x \to \minus{}\infty$. Moreover, if the limit is not zero, then we know $ f(x)$ cannot converge to some finite value as $ x \to \minus{}\infty$. So $ \lim_{x\to\minus{}\infty} f'(x) \equal{} 0$. Now we have a contradiction that

$ 0 \equal{} \lim_{x\to\minus{}\infty} f'(x) \equal{} \lim_{x\to\minus{}\infty} f(f(x)) \equal{} f(l) > 0.$


[hide]C.) 46

30*16

separate the polygon into 2 rectangles.[/hide]

You realize that this is for the resources section, so:

[quote="worthawholebean"]
If you want to get your solutions posted:
1. Again, avoid extraneous comments; these can be posted in a separate reply.
2. (As always) hide your solution; the parser automatically strips out the hide tag, so it isn't an issue. The parser won't have trouble with unhidden solutions, so you should hide just for common courtesy.
3. Use LaTeX.
4. For diagrams, use Asymptote or LaTeX, for the same reasons as above.
5. Be clear and thorough.
[/quote]

5*6+4*4=46 square units. And in your attachment, how do you hide the answer?

Type

[code]
[hide]
Hello, world!
[/hide]
[/code]

That comes out as

[hide]
Hello, world!
[/hide]

You need [/hide] for the last one, not [hide].

simple way: click the hide button, type whatever you want, then click the hide button again.

[hide]LALALA[/hide]

in order to have something like this:

[hide="HEY WORLD"]LALALA[/hide]

use,
[code][hide=HEY WORLD]LALALA[/hide]
[/code]

[hide="Back on topic, heres my solution"]
The area is greater than $ 6 \times 5\equal{}30$ and less than $ 6 \times 9\equal{}54$, so $ \boxed{\textbf{(C)}}$ must be correct.
[/hide]

[hide="Another solution"]
Extending $ \overline{FE}$ partitions the polygon into a rectangle and a square whose areas are $ 30$ and $ 16$ respectively, so they sum to $ 46$, or choice $ \boxed{\textbf{(C)}}$.
[/hide]

[hide="Yet another solution"]
Extending $ \overline{AF}$ and $ \overline{DC}$ to form a large rectangle shows that the area is $ (6 \times 9)\minus{}(4 \times 2)\equal{}54\minus{}8\equal{}46$, or answer choice $ \boxed{\textbf{(C)}}$.
[/hide]

[hide]

One square of side 4*4 and one rectangle of side 5*6.So that equals 46.

[/hide]

^OR
[hide="Alternate Solution"]
$6 \cdot 9 - (6-4) \cdot (9-5) = 6 \cdot 9 - 2 \cdot 4 = 54 - 8 = \boxed{\text{(C)} 46}$
[/hide]

You can also break it up into two rectangles and just add the areas. There are a lot of good ways to do this. This is a good problem.

I did 6*5 + 4*4


superior algebra
superior algebra solved

Let $R$ be a ring with unity and let $J$ be an ideal of $R$. Show that there is a bijection between the prime ideals of $R$ containing $J$ and the prime ideals of $R/J$.

This is standard; it's done every time you see this in a course. It doesn't take much thought to put it together- just do what the system lets you do.

All you can use is the projection $R \rightarrow R/J$, which gives the required bijection.


I saw many of you here are interested in religion and science, but is anyone here intersted in philosophy? Specifically I am interested in aesthetics, philosophy of art and music... and obviously philosphy of mathematics..

Yeah, I think philosophy of math is interesting. What can you say about the philosophy of math?

Well I don't know that much about this topic, but i read godel's proof, and have read other papers and notes on logic, but I would like to know the direction that the contemporary researchers in this area are taking...

I'm actually very interested in philosophy because my dad majored philosophy and actually teaches me at home

what area of philosophy you and your dad are interested?

Hello.... Right now i'm taking IB and it's compulsory to take philosophy. :(
i have to write an essay about philosophy of math in a few weeks time.
If you are interested in it, can you help me out to write an essay?

The question i have to answer to is:

[color=blue][size=150]Is Knowledge in mathematics and other Areas of Knowledge (ethics, science, history, art) dependent on culture to the same degree in the same way?[/size][/color]

Thank you :)


let$a,b,c$ be negative real numbers  .
 Prove that :
$(a^{2}+b^{2}+c^{2}) \leq \sum{\sqrt{a^{2}-ab+b^{2}}\sqrt{b^{2}-bc+c^{2}}}$

I am probably doing something wrong, I get:

$\sum{\sqrt{a^{2}-ab+b^{2}}\sqrt{b^{2}-bc+c^{2}}}\geq \sum (a-b)(a-c)=a^{2}+b^{2}+c^{2}-\sum ab$
but ab is positive so the ineq follows
this makes no sense it is way too easy..

[quote="me@home"]I am probably doing something wrong, I get:

$\sum{\sqrt{a^{2}-ab+b^{2}}\sqrt{b^{2}-bc+c^{2}}}\geq \sum (a-b)(a-c)=a^{2}+b^{2}+c^{2}-\sum ab$[/quote]

You are doing something wrong (first, you are proving a much weaker inequality; then, you can't multiply two inequalities $A\geq B$ and $C\geq D$ if some of A, B, C, D are negative).

Tranthanhnam is doing something ever wronger: the problem is from a Vietnamese competition with deadline November 6, 2006. Moderators please LOCK this topic.

  darij


what are the differences between the mean and the standard deviation

The mean is a measure of central tendency. It's the average; it's the point about which the probabilities balance. The standard deviation is a measure of dispersion, of how broad the distribution it.

Here are two possible games. In the first game, roll one fair six-sided die, and receive a payment of $ \$2$ times the number showing on the die. In the second game, roll two such dice, add together the numbers on the two dice, and receive a payment of $ \$1$ times that sum.

These games both have the same expected payoff - in other word, the mean. In both cases, it's $ \$7.$ But the standard deviation of the two games is different. The first game, which has more "spread-out" results, has a standard deviation of about $ \$3.42,$ while the second game, in which the payoffs cluster more closely around the mean, has a standard deviation of about $ \$2.42.$

Hi,

This page will show you the differences in how you calculate standard deviation and mean:

[url=http://standard-deviation.appspot.com/]Standard Deviation Example[/url]

B.


I sent my Entry/Permission forms by Snail Mail the same day as I sent my Round 1 solutions. It's been a month now, and my Profile page still says that I haven't submitted my forms.  :mad: (I don't think even Snail Mail would take a month.)

It is possible that we may have misfiled a group of entry/permission forms.  We'll be going through our files this week and hopefully we can get it straightened out.

Update: we did indeed find a whole pile of entry/permission forms that we thought had been entered into the database, but in fact were not.   :oops_sign:

We're in the process of entering them now.  They should hopefully all be in the system in the next 48 hours.

Note that some people sent in entry or permission forms sent us stuff that was either illegible or unsigned.  So if you sent us a form, and it's not listed by the end of the week, that may be the reason.

Update: We now believe that we have filed all of the forms that we have received.

If you sent a form, and it's not listed on the USAMTS website, one reason may be that it was lacking a necessary signature.  In this case, please resend the form with your Round 2 submissions, and make sure that all the necessary/optional signatures (both yours and your parent's) are there.  Also, please make sure that the information on the form is legible: we had to exclude a couple of forms because we could not decipher the handwritten ID#.

I have sent mine along with the second solution set, but it seems to not have been marked as received on "My Usamts" page. Should I resend it again?


modular arithmetic
number theory proposed
number theory

Find all positive integers $ n,m$ such that

$ 3^m\plus{}2\equal{}5^n$

$ m\equal{}n\equal{}1$ is a solution. Let $ m>1, \ n>1$

$ 5^n \equiv 2 \pmod{9} \Longrightarrow n \equiv 5 \pmod{6}$

$ 3^m \equiv 3 \pmod{5} \Longrightarrow m \equiv 1 \pmod{4}$

$ n \equiv 5 \pmod{6} \Longrightarrow 5^n \equiv 3 \pmod{7} \Longrightarrow$

$ 3^m \equiv 1 \pmod{7} \Longrightarrow m \equiv 0 \pmod{6}$ but $ m$ is odd.

$ m\equal{}n\equal{}1$.  :wink:

Yes...that's what I tried to do as well but I made an arithmetic error. Thanks


search MATHCOUNTS Putnam MIT college function

I was wondering, has anyone here ever read the book [i]Countdown[/i]? Its about the 6 members of the IMO team in 2001 (?). I read it recently and found the book extremely interesting, and to anyone who did read the book, I would love to hear your thoughts.

Yes I have read it, I think that many people on this site have read that book also.  It is very good.  It even inspired me a little bit.

I read it, too. A very interesting read, in fact - I really liked how they explained the problems; makes some sounds almost... easy! :P Anyway, everyone should read it because it is inspiring and it shows what some IMO competitors are like, in case you have any stereotypes about them (maybe not so common here, but you know what I mean).

yeah--and the author discusses many interesting issues like women in mathematics, the correlation between math and music, etc.

I fully agree with bubala's post there, it discusses so much about math/problem solving in general.  As for the math/music correlation... search this forum or read the book.

[quote="noneoftheabove"]yes press alt f4 i have and plz bring advance wars to camp :D[/quote]

that's no cool....so i was curious and pressed alt f4 and *poof*! away went my screen.  :lol:

I have read it, right before I went to IMO2004. It's a good book.

[quote]yes press alt f4 i have and plz bring advance wars to camp [/quote]
noneoftheabove? mark? I guess it must be.

Anyway, I found it very inspiring as well and many of the problems they explained sounded simple after just one key insight or extra line. Oh and does anyone no any other storys about math teams and such like this one?

Another good book is [u]In Code: A Mathematical Journey[/u] by Sarah Flannery, although its not about a math [i]team[/i]. Its about an individual student and how she developed a math project for the Intel Science Fair.

[quote="noneoftheabove"]err thats his user name[/quote]

ok, so guess who feels like a total idiot right now  :blush:

In my opinion (which nobody ever listens to), [i]Countdown[/i] is a very interesting book.

It is an illuminating book for those of us who have not witnessed math competitions at the level of the olympiads.  In most sports, including lower levels of math competition, the coaches have no involvement in creating or scoring the test, so it was interesting to read how different is the system at the math olympiad.

The book also has a good discussion of the preponderance of Asians, especially males, at the highest level of math competition in the US.  Steve Olson points out that students who were born overseas, or who have at least one parent born overseas, do disproportionately well in math competition in this country.  After watching mathcounts on ESPN today, this is my hypothesis too, and it also seems to apply, although to a lesser extent, to children of European immigrants.  Feng Hsiung Hsu, the brain behind the chess-playing computer Deep Blue, has another theory about Chinese intellectual dominance: the age old Chinese system of meritocracy, whereby the best civil service jobs were filled through open competition and exams, which meant that anyone who was smart enough and studied hard enough would gain entry to the top jobs.  He is from Taiwan and describes in his autobiography how he trained himself to excel academically.

It would be interesting to find out in 5 or 10 years time what Reid Barton is doing - he is clearly an extremely gifted mathematician.

err why would i be mark press alt f4( u will never guess)

[quote="xfactor"] It would be interesting to find out in 5 or 10 years time what Reid Barton is doing - he is clearly an extremely gifted mathematician.[/quote]

And Gabriel Carroll, too. Both have been Putnam fellows 4 times. Both are supposedly graduating /or already graduated in May this year; Barton from MIT, Carroll from Harvard.

Did read and very much enjoyed the book. Got me interested in problem solving in general. Bought some of the references mentioned in Countdown and have struggled with the following problem. We worked out that the solution is imaginary, but that there is a solution. We don't see any elegant (algebraic) approach to the problem. Any hints or ideas?

sin(sin(sin(sin(x)))) = cos(cos(cos(cos(x))))

[quote="MathFiend"]that's no cool....so i was curious and pressed alt f4 and *poof*! away went my screen.  :lol:[/quote]
i did the same thing and also feel stupid except my screen didn't disapear.  :D

I would guess that there are infinitely many solutions, and I don't mean like something + n*2*pi... but I'm not very good with the whole trig/complex numbers thing.

You should go look in the olympiad forums for the answers to math questions.  There is not real point at which those functions are equal; I'm sure there are complex values, but I haven't any idea how to go about finding them, and I'd rather that people not try to do it here.

[quote="Tunio"]Is there really a correlation between math and music? Enlighten me :P[/quote]

  Numerous studies have been conducted on this matter, and many point out that pupils who play musical instruments perform better on tests concerning mathematical depth.  This pattern, most likely occurs because pupils  who play musical instruments will have longer attention spans (because playing a musical instrument takes hours and hours of rigorous practice) and greater aptitude for studying.  Meaning they have the span to study mathematics for longer periods of time when compared to other pupils.  Playing a musical instrument, to my knowledge, dosent strengthen ones mathematical mind in any direct manner.  But it goes to say, a student who simply studied lots and lots of mathematics would perform just as well as a student who was musicaly talented.


calculus
analytic geometry
geometry
rectangle
function
integration
calculus computations

I'm working on Linear Differential Equation and here's the two problems that I got stuck on from the book called:" Multivariable Calculus (Tom Apolos)". Any detailed solutions would be greatly appreciated, since I have no idea how to set up the equation from the given information. Anyway here's those 2 problems:

1. If a strain of bacteria grows at a rate proportional to the amount present and if the population doubles in one hour, by how much will it increase at the end of two hours?
2. A curve with Cartesian equation $ y \equal{} f\left(x\right)$ passes through the origin. Lines drawn parallel to
the coordinate axes through an arbitary point of the curve form a rectangle with two sides on
the axes. The curve divides every such rectangle into two regions $ A$ and $ B$, one of which has an
area equal to $ n$ times the other. Find the function $ f$

1. Let $ P$ be the bacteria population. Read what problem gives you:

'a strain of bacteria grows at a rate proportional to the amount present' gives you $ \frac {dP}{dt} \equal{} kP$, where $ k$ is the proportionality constant;

'the population doubles in one hour' gives you $ P(1) \equal{} 2P(0)$.

The rest is straightforward

2. Draw the rectangle and the given curve passing through two opposite corners of it. Label one of the corners as the origin and label regions $ A$ and $ B$. Label the other corner with an arbitrary point $ (x_0,y_0)$ on the curve. Start translating the conditions of the problem:

'one of which has an area equal to $ n$ times the other': What's the area of $ A$? Set up an integral, say $ A \equal{} \int_0^{x_0}f(x)\, dx$. So $ B \equal{} x_0y_0 \minus{} A$ (remember the area of the rectangle is $ x_0y_0$.) Now try to incorporate the condition using $ n$.

'The curve divides every such rectangle': That means you can 'drop' the subscripts on the arbitrary point and find the function $ f$.

[quote="ghjk"]Tom Apolos[/quote]  As a general rule, if you have no idea how to set up the question, you should go reread the chapter in the text that explains how to do problems of this sort, paying special attention to the worked examples.  Maybe in this case you should start with the front cover where the author's name is written :P

(On content, 10000th User gives some very good suggestions for these two problems.)


calculus
integration
real analysis
topology
real analysis unsolved

(Maybe this should be in advanced topics)

In Stein's analysis book, he develops a general theory of integration on abstract measure spaces, pretty much in the same way one normally constructs Lebesgue integration. However, he assumes (and uses, in a critical step) the fact that his measure space is $\sigma$-finite.

My question is: Can we develop a similar theory of integration on a measure space that is not $\sigma$-finite? Or do we just not care because such measure spaces are annoying?

In Folland's [i]Real Analysis[/i] much of the measure theory is developed for semifinite measures. A measure $m$ is semifinite if for any measurable set $E$ with $m(E)=\infty$ there exists a measurable set $F\subset E$ such that $0<m(F)<\infty$. It is clear that $\sigma$-finite $\implies$ semifinite.

An important example of non-$\sigma$-finite measure is the Hausdorff measure $\mathcal H^{s}$ on $\mathbb R^{n}$, where $s<n$. I think it may be semifinite, but can't prove it right now.


The rhizomes of ginger ([i]Zingiber officinale[/i]) are well known for their medicinal and flavoring properties, and in Ayurveda (the traditional system of medicine in India), different formulations of ginger are used for the treatment of gastrointestinal problems, common cold and other ailments. Several compounds are responsible for the pungent taste of ginger, but the principal one is Zingerone ($ C_{11}H_{14}O_{3}$). This organic compound, when pure, exists as a crystalline solid that is sparingly soluble in water, but soluble in ether. Zingerone gives positive $ FeCl_{3}$, 2,4–DNP (2,4–dinitrophenylhydrazine), and iodoform tests - however, it does not react with Tollen’s reagent. On reaction with bromine water Zingerone gives only one monobrominated product. The IR spectrum of Zingerone indicates the presence of a weak intramolecular hydrogen bond, which is still present even after Clemmensen reduction. The 1H-NMR spectrum of Zingerone shows the following peaks: 2.04 (singlet, 3H), 2.69 (triplet, 2H), 2.71 (triplet, 2H), 3.81 (singlet, 3H), 5.90 (broad singlet, D2O exchangeable, 1H), 6.4 - 6.8 (two doublets and one singlet, 3H). Deduce the structure of Zingerone.

i think this is a   question from ICho,2001 held at India :maybe:

Yes it is, from problem 6, but I adapted it slightly.


function
inequalities
inequalities unsolved

let a;b;c >0 and t>1
prove that:
[tex]\frac{a^t}{b^t+c^t}+\frac{b^t}{c^t+a^t}+\frac{c^t}{a^t+b^t}>=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}[/tex]

Oh,the function:
$f(t)=\frac{a^t}{b^t+c^t}+\frac{b^t}{c^t+a^t}+\frac{c^t}{a^t+b^t}$
is a stricly incresing.
Is it your solution?


$ a,b,c in R^{\plus{}}$

We have：$ Sigma sqrt((a^{2}\plus{}ab\plus{}b^{2})/(c^{2}\plus{}ab))\geq (3sqrt(6))/2$

Can an adminstrator lock this?  Not only has this problem been posted before, but it is still on the first page of this very section!  No solution has been posted yet, but hopefully one will be soon.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=160854


calculus
integration
LaTeX
function
real analysis
real analysis unsolved

Prove that:
$\int_{-1}^1f^2(x)dx\ge \frac 1 2 (\int_{-1}^1f(x)dx)^2 +\frac 3 2(\int_{-1}^1xf(x)dx)^2$

Please give a proof without using even and odd functions. (the oficial proof uses those and seems to be un-natural) :D

[quote="xirti"]Prove that:
$\int{-1}^1f^2(x)dx\dots$
[/quote]
What is $-1^1$ doing here? Is it just $-1$? Or you tried to write $\int_{-1}^1$ but made a mistake in $\LaTeX$?

Yes sorry for the mistake. I have edited it.
Please help me with this problem

In the inner product space $L^2[-1,1]$, the elements $\sqrt{\frac12}$ and $\sqrt{\frac32}x$ both have unit length, and are orthogonal. Therefore, $\sqrt{\frac12}\int_{-1}^1 f(x)\,dx$ is the component of $f$ in 1, and $\sqrt{\frac32}\int_{-1}^1 xf(x)\,dx$ is the component of $f$ in $x$.

So, this amounts to saying that the norm of $f$ is at least as big as the sum of the squares of some of its components. The even and odd functions are just a convenient way of splitting into two orthogonal subspaces with direct sum equal to everything.


Let $ a_n \equal{} \frac{1}{\sqrt{n^2\plus{}8n\plus{}1}}\plus{}\frac{1}{\sqrt{n^2\plus{}16n\plus{}1}}\plus{}\frac{1}{\sqrt{n^2\plus{}24n\plus{}1}}\plus{}...\plus{}\frac{1}{\sqrt{9n^2\plus{}1}}$ for all $ n\in N$
Find the value of
$ lim_{n\rightarrow \infty } a_n$

[quote="Bloody-Joker"]Let $ a_n \equal{} \frac {1}{\sqrt {n^2 \plus{} 8n \plus{} 1}} \plus{} \frac {1}{\sqrt {n^2 \plus{} 16n \plus{} 1}} \plus{} \frac {1}{\sqrt {n^2 \plus{} 24n \plus{} 1}} \plus{} ... \plus{} \frac {1}{\sqrt {9n^2 \plus{} 1}}$ for all $ n\in N$
Find the value of
$ lim_{n\rightarrow \infty } a_n$[/quote]
  i think so,
$ L\equal{} \int\limits_{0}^{1} \frac{2\sqrt{2}}{\sqrt{1\plus{}8x}} dx$

Note that for $ k \equal{} 1..n$:
$ \frac {1}{\sqrt {n^2 \plus{} 8kn}} \minus{} \frac {1}{\sqrt {n^2 \plus{} 8kn \plus{} 1}} < \frac {1}{n^3}$, so:
$ (\frac {1}{\sqrt {n^2 \plus{} 8n}} \plus{} ... \plus{} \frac {1}{\sqrt {9n^2}}) \minus{} (\frac {1}{\sqrt {n^2 \plus{} 8n \plus{} 1}} \plus{} ... \plus{} \frac {1}{\sqrt {9n^2 \plus{} 1}}) < \frac {1}{n^2}\rightarrow 0$, if $ n$ moves to infinity.

$ lim_{n \minus{} > \infty}a_n \equal{} lim_{n \minus{} > \infty}\sum_{k \equal{} 1}^{n}\frac {1}{\sqrt {n^2 \plus{} 8kn}} \equal{} lim_{n \minus{} > \infty}\sum_{k \equal{} 1}^{n}\frac {1}{n}\frac {1}{\sqrt {1 \plus{} 8\frac {k}{n}}} \equal{} \int\limits_{0}^{1} \frac {1}{\sqrt {1 \plus{} 8x}} dx \equal{} 0.5$


In this problem you are asked to convert the general standard form of a quadratic polynomial into the completed square form. Suppose

$ f(x)= Ax^{2}+Bx+C$

where $A$, $B$, and $C$ are the coefficients of the quadratic polynomial, and $A\neq 0$. Then $f$ can be written as

$ f(x) = a(x-h)^{2}+k$

where
$a=$ ,
$h=$ , and
$k=$ .
Enter your answers as algebraic expressions in $A$, $B$, and $C$. Remember that mathematics is case sensitive so in this case your variables are capital letters, and you must enter them as such.


geometry
rhombus
circumcircle
3D geometry
trigonometry

[b]1.[/b]Three equal circles have a common point H,and they are intersectating two by two in the points $ A,B,C$.Prove that the circumscribe circle of $ \Delta ABC$ is equal with the three circles.

First, draw a diagram and construct all the radii (the obvious ones at least)

[hide]Then construct the circumradii of our triangle ($ \overline{AH}$, $ \overline{BH}$,$ \overline{CH}$)

Given this information, we may be led to hypothesis that $ \overline{BH} \parallel \overline{AO_2}$ (where $ O_2$ is one of the centres of a circle which is tangent at $ B$

Then we might try and prove the stronger result, which is equivalent to the proof

$ AHBO_2$ is a rhombus

[/hide]

This is the main idea of this problem....but we can have another beautifull proof:
See the attachment for the figure: $ A_1,B_1,C_1$ are on the Euler's circle of the triangle ABC.But $ A_1,B_1,C_1$ are on the Euler's circle of the triangle $ O_1O_2O_3$ from where $ C(ABC)\equal{}C(O_1O_2O_3)$ and therefore $ C(O_1)\equal{}C(ABC)\equal{}C(O_2)\equal{}C(O_3)$

Another way...

Notice that $ m\angle{ABC}\equal{}\frac{1}{2}(m\widehat{AH}\plus{}m\widehat{HC})$.  Both possible $ m\widehat{AH}$ are equal in measure, because the circles all have the same radius: the same holds for $ \widehat{HC}$.  Hence, $ m\angle{ABC}\equal{}\frac{1}{2}m\widehat{AC}$
Therefore, $ m\widehat{AC}$ has the same measure in the circumcircle of $ AHC$ and in the circumcircle of $ ABC$, and so the two circles have the same radius.

Project a cube onto a plane ;) (parallel projection).

This is the most beautifull way to solve this problem Altheman...This problem has some extensions:We will note the initial problem with ***:
[b]Extension 1[/b] [b][color=darkblue]There are given three equal equilateral triangles having a common point O.Prove that the circumscribe circles of the equilateral triangles are intersectating two by two in three points such as the circumscribe circle for the triangle formed by those three points  is equal to each triangle given.[/color][/b]
Let $ O_1,O_2,O_3$ the centers of the circles $ \Rightarrow$ $ C(O_1O_2O_3) = C(O_1) = C(O_2) = C(O_3)$.But relating to *** we will have $ C(O_1O_2O_3) = C(XYZ)$ where $ X,Y,Z$ are the intersection points of the circles $ C{O_1),C(O_2),C(O_3)}$.

[b]Extension 2[/b] [b][color=darkblue]There are given three equal equilateral triangles $ OA_1B_1,OA_2B_2$ and $ OA_3B_3$. Prove that the circumscribe circles of the triangles $ OA_1B_1$,$ OA_2B_2$,$ OA_3B_3$ are intersectating two by two in three points that form an equilateral triangle equal with those considered.The same property takes place for the circumscribe circles of the triangles $ OA_1B_3,OA_2B_1,OA_3B_2$[/color][/b]
Let $ R$ be the raze of the circle with center $ O$ who passes thru the other points ot the equilateral triangles given.Let $ A",B",C"$ be the centers of the circumscribe circles of the triangles $ OA_3B_2$,$ OA_1B_3$,$ OA_2B_1$ and $ A,B,C$ the intersections points for these circles .Note with $ D,E,F$ the midles of the segments$ A_1B_1 , A_2B_2,A_3B_3$,with $ A',B',C'$ the midles of the segments $ A_3B_2$,$ A_1B_3$ ,$ B_1A_2$,with $ M,N,P,Q$ the midles of the segments $ OC,OB,OA_3,OB_3$, with $ 2\alpha = \angle A_2OB_2$,$ 2\beta = \angle A_1OB_3$, $ 2\gamma = \angle A_2OB_1$.
$ B'$ is the center of the cricumscribe cifcle for the triangle built on $ DF$ as a side.The same propertyes have $ A'$ and $ C'$.It is known that the centers of the circumscribe circles for the equilateral triangles built in the exterior ,on the sides of a triangle form an equilateral triangle $ \Rightarrow$ $ \Delta A'B'C'$ is equilateral.
The quadrilaterals $ A'A_3PA"$ and $ B'B_3QB"$ are inscriptible and therefore we can write that
\[ OA' \cdot OA" = OB' \cdot OB" = \frac {R^2}{2}
\]
$ Rightarrow$ $ A'B'$ and $ A"B"$ are antiparallels in $ \Delta OA'B'$.Therefore $ \angle OB"A" = \angle OA'B'$ and similar $ \angle OC"A" = \angle OA'C'$ $ \Rightarrow$ $ \angle OB"A" + \angle OC"A" = \frac {\pi}{3}$
In $ \Delta A"B"C"$, $ m(\angle A") = m(\angle B"OC") - m(\angle OB"A") - m(\angle OC"A") = \beta + \gamma$
Consider the circle with $ OA"$ diameter with the observation that $ M,N,P$ are on this circle.
\[ \angle MA"N = \angle PA"O
\]
$ \Rightarrow$ $ MN = OP = \frac {R}{2}$ $ \Rightarrow$ $ BC = 2MN = R$.In the same way $ AB = AC = R$.Therefore $ \Delta ABC$ is equilateral and equal with the three equilaterals trangle given.The same for rest.

[quote="Vikernes"]And what about the sine theorem?

Let $ R$ the radii of the three circles and let $ T$ their common intersection. We define $ |AT| \equal{} m, |BT| \equal{} n, |CT| \equal{} l$.
By the sine theorem in the triangle $ ACT$ we have $ m \equal{} 2R\sin{\angle{ACT}}$, but by the sine theorem in the triangle $ ABT$ we deduce that $ m \equal{} 2R\sin{\angle{ABT}}$, so $ \angle{ACT} \equal{} \angle{ABT} \equal{} \alpha$. By the same way $ \angle{BAT} \equal{} \angle{BCT} \equal{} \beta, \angle{CAT} \equal{} \angle{CBT} \equal{} \gamma$

Now let $ S\equal{} AT\cap BC$. $ \angle{STB} \equal{} \alpha \plus{} \beta$ hence $ \angle{ATB} \equal{} \pi \minus{} (\alpha \plus{} \beta)$
If $ R'$ is the radii of the circumcircle of the triangle $ ABC$, we have by the sine theorem in $ ABC$
$ |AB| \equal{} 2R'\sin{(\alpha \plus{} \beta)}$ but the sine theorem in $ ATB$ give us $ |AB| \equal{} 2R\sin{(\pi \minus{} (\alpha \plus{} \beta))}$. Finaly $ R \equal{} R'$[/quote]
More simple, the problem follows from Johnson's Theorem :D.
See Johnson's theorem in [url]http://agutie.homestead.com/files/johnsont1.html[/url] (you need Flash Player).


function
complex analysis
complex analysis unsolved

Let $ f: \mathbb{C}\rightarrow\mathbb{C}$ be an analytic non-constant function. Without making use of Picard's theorem prove that there exists a point $ z$ and a natural number $ n$ such that $ f(z)^n\equal{}1$
[Picard's theorem asserts that te image of an entire nonconstant function ommits at most one point on the complex plane]

By the open mapping theorem, the image of $ f$ cannot contain any point of the unit circle, as then it must contain an open ball about that point and some point within this ball will be a root of unity. Hence the image of $ f$ is either contained entirely within or entirely outside the unit circle; in the latter case, consider instead $ \frac{1}{f}$ so that one obtains a bounded analytic nonconstant function, which is impossible by Liouville.

correct
My solution is almost the same. Using Liouville's theorem we find $ M > 1$ such that $ |f(z)| > M$ and we can also find a $ m < 1$ such that $ |f(z)| < m$ using Louville for $ 1/f$. therefore, by the continuity of $ |f(z)|$ there exists w such that |f(w)|=1. Using the open mapping theorem we find $ f[D(w,e)]$ open so the latter intersects an arc of the unit circle. The set of the roots of unity is dense in the unit circle, so we find a root of unity in $ f[D(w,e)]$, which is the case.


E frumos acum, dupa un an de mathsup, caci probleme de tipul asta ies direct: A,B sunt diagonalizabile (privite ca matrici complexe) si comuta, deci exista o baza comuna de diagonalizare si atunci totul se reduce la primalitatea a doua polinoame, care numai dificila nu e.

I think there is a solution without diagonalisation theory, because

in 10th,11th,12th grade in romania you don't see diagonalisation theory

Moubi, I know that solution, but let me show I have learnt something in France.  :D

I am interested a solution without diagonalisation theory
thx

Ok, I did not learn that in France. "Merde"! Ok, here is the romanian solution:
  If it weren't, then there is a non-zero column vector x such that $ Ax=-(I+B)x$ Then $ x=-(I+B)^{1977}x$ ( just make the 1977 power of the equality). Thus, we are left with the primality of the polynomials $ x^{1978}-1$ and $1+(1+x)^{1977}$ which is immediate.

I find it is interesting to have different solution of a problem
thx

Sincerely, I find the second solution much more direct. You are right, if there is something interesting, then finding multiple solutions must be included in it.

[quote="silviu"]Fie $A,B \in M_n(R)$.Daca $AB=BA,A^{1977}=B^{1978}=I_n$
Aratati ca matricea $A+B+I_n$ este inversabila.[/quote]

Moubinool,

yes, there is a solution without diagonalization, because one can look at the universal example of an algebra generated by such $A,B$.  If the statement of this problem is true, then it is true where $n$ is the dimension of this (finite dimensional) algebra, therefore it is true in the universal example, therefore it is true purely as a result of algebraic identities satisfied by such $A, B$.

So if the problem is correct, then there is an element of $\mathbb{Q}[A,B]$ inverse to $1+A+B$.   We can see that the problem is correct by considering eigenvalues; no 1997-th root + 1998-th root of unity can have a sum of -1.

The question is then how to find explicitly the inverse, without writing down matrices of size 1977x1978.


[quote="balan razvan"]hi there, i'm  a biginner, and i want to solve more interesting problems with complex numbers and functions.i'll apreciate your help. :)[/quote]

First welcome to mathlinks. Secondly, what texactly do u want? Suggestions for books, with compex numbers?

Yep, I would like sugestions for books with complex numbers...mor exactly  ebooks :D:D .


MATHCOUNTS conics parabola geometry algebra polynomial

Hey guys,

I forgot to post this thread up for COPSD. I'm still busy so I won't be able to grade or give any scores away till this Friday (which is two days from today.. guess how many minutes that is :D ).

Even you didn't do this contest, do the problems for fun. It's a good Mathcounts level problem and should give you a good exercise.

Problems can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=330324#p330324]here[/url].

As for the result, 236factorial was only one who sent me the answer.  :( I was hoping for more people to compete but oh well. Anyway, he got the perfect score of getting all 10 questions right.

Let's look at the solutions.

[hide="Solution 1"]
Since the coefficient of $x^2$ is positive, we know that this is a parabola that opens upward. Hence, we can find the minimum. To do this, we complete the square:

$y = x^2-12x+2 = (x-6)^2-34$. The minimum occurs at when $x = 6$ and that is $\fbox{-34}$. [/hide]

[hide="Solution 2"]
When we add all three equations, we get $4a+4b+4c = 36$. Since we're looking for just $a+b+c$, we divide by 4 to get the answer $\fbox{9}$. [/hide]

[hide="Solution 3"]
For each question, we can either answer TRUE or FALSE. Hence, there are two options for each 10 qustions. $2^{10} = \fbox{1024}$. [/hide]

[hide="Solution 4"]
This type of problem is something that you might want to learn how to do because it appears very often in lower level (middle school). I even saw couple in high school level ones. There is even a chapter in AoPS volume II. Anyway:

$\sqrt{4+\sqrt{4+\sqrt{4 + \cdots}}}$ is what we have. Let's say the whole thing equals $x$. Note that after first term, second term and the rest looks exactly like the $x$. So:

$\sqrt{4+\sqrt{4+\sqrt{4+\cdots}}} = x$ can be written as:

$\sqrt{4+x} = x$. Solving this equation gives two values (one positive and one negative) but we only want positive ones. Then, $x = \dfrac{1+\sqrt{17}}{2}$ and $a = 1$, $b = 17$. [/hide]

[hide="Solution 5"]
In my opinion, this problem is perhaps the TOUGHEST one among this COPSD. First, you must see that the second triangle $A'B'C'$ has one-fourth the area of the triangle $ABC$. $A''B''C''$ has $\Big(\dfrac{1}{4}\Big)^2$ area of triangle $ABC$. Since $ABC$ has area of $25 \sqrt 3$ (it's just regular equilateral triangle with side length 10), all equilateral triangles will have area of:

$25 \sqrt{3} + \dfrac{1}{4} \cdot 25 \sqrt 3 + \Big(\dfrac{1}{4}\Big)^2 \cdot 25 \sqrt{3} + \Big(\dfrac{1}{4}\Big)^3 \cdot 25 \sqrt 3 + \cdots$

This is a geometric series. To find the sum, we use formula $\dfrac{a_1}{1-r}$ where $a_1$ is the first term and $r$ is the constant rate. The final answer is $\boxed{\dfrac{100 \sqrt 3}{3}}$. [/hide]

[hide="Solution 6"]
Let's say $x$ is the EXACT value. Then if we square the equation, we get:

$x^2 = \Big(\sqrt{4 + 2 \sqrt 3} + \sqrt{4 - 2 \sqrt 3}\Big)^2\\ x^2 = 4+2 \sqrt 3 +4 - 2 \sqrt 3 + 2 \sqrt{(4+2 \sqrt 3)(4 - 2 \sqrt 3)}\\ x^2 = 12\\ x = \boxed{2 \sqrt 3}$ [/hide]

[hide="Solution 7"]
This problem is pretty complicated. First, the other root is $-4i$. To find the rest, we note that $(x - 4i)(x+4i) = x^2+16$. So, dividing $x^2+16$ from original polynomial $x^4-5x^3+22x^2-80x+96$, we find that roots are $2,3$. Thus, the final answer is $\boxed{-4i,2,3}$.

Just for sidenote, the polynomial with integer coefficient has roots in form of $a+bi$ and $a-bi$. Otherwise, it would have $i$, where $i = \sqrt{-1}$, in one of the terms. [/hide]

[hide="Solution 8"]
Note that the expression $\dfrac{1}{r^2} + \dfrac{1}{s^2} = \dfrac{r^2+s^2}{(rs)^2}$. We can find what $r$ and $s$ but think this way.

Squaring $r+s$ gives $r^2+2rs+s^2 = 25$. Subtracting $2rs$ which is 20, we get $r^2+s^2 = 5$. Since we know what $rs$ is, we get $(rs)^2 = 10^2 = 100$.

Thus, $\dfrac{r^2+s^2}{(rs)^2} = \dfrac{5}{100} = \boxed{\dfrac{1}{20}}$. [/hide]

[hide="Solution 9"]
First, note this. Any number that has sum of its digit of 18 will contain the at least one square as its positive factor. Why? That's because we know that if the number has sum of 9 or its multiple, it's divisible by 9. Listing all possible choices, we find that there are only 189 and 378 work. So, the answer is $\fbox{2}$ [/hide]

[hide="Solution 10"]

I like to brute force in this small-type problems. Only $(1,3),(1,5),(2,4),(3,5)$ work so the answer is $\fbox{4}$. The easy way to see this is that the numbers either have to be all even or all odd. [/hide]


inequalities
algebra
polynomial
inequalities unsolved

1. Prove that for any reals x, y, z, which satisfy condition x 2 + y 2 + z 2 = 2 we have
x+y+z)  \leq xyz + 2
(Poland, 1991)

2. If all the roots of the polynomial x3 + ax2 + bx + c are real, show that 12ab + 27c  \leq 6a3 + 10(a2 - 2b)3/2. When does equality hold?
(Vietnam 2002)

These inequalities are really twin brothers. Do they?

Namdung

Hi all,

The first one was solved in this forum. Can anyone solve the second one? One told me that in that year, only three students solved it (by three totally different way).

Namdung

I have the official solution. Should I post it?

Arne, can you please post the solution?

Thank you very much.

We want to prove that -6a(a2 - 2b) \leq -27c + 10(a2 - 2b)3/2.
Let the real roots of the equation be x, y, z. We want to prove
6(x + y + z)(x2 + y2 + z2) \leq 27xyz + 10(x2 + y2 + z2)3/2.
Wlog, |x| \leq |y| \leq z and x2 + y2 + z2 = 9.
So the inequality is equivalent to 2(x + y + z) - xyz \leq 10.
Now, we recognise the first inequality - it is the same type.
And exactly this inequality was proved here:
[url]http://www.mathlinks.ro/phpBB/viewtopic.php?t=52&highlight=[/url]
So we are done.

[quote="Namdung"][b]2[/b]. If all the roots of the polynomial $ x^3 + ax^2 + bx + c$ are real, show that

$ 12ab + 27c \leq 6a^3 + 10\left(a^2 - 2b\right)^{3/2}$.

When does equality hold? [/quote][quote="Arne"]We want to prove that

$ - 6a\left(a^2 - 2b\right) \leq - 27c + 10\left(a^2 - 2b\right)^{3/2}$.

Let the real roots of the equation be $ x, y, z$. We want to prove

$ 6(x + y + z)\left(x^2 + y^2 + z^2\right) \leq 27xyz + 10\left(x^2 + y^2 + z^2\right)^{3/2}$.[/quote]
$ 100\left(x^2 + y^2 + z^2\right)^3 - \left[6(x + y + z)\left(x^2 + y^2 + z^2\right) - 27xyz\right]^2$

$ = 100\left(a^2 - 2b\right)^3 - 9(9c + 4ab - 2a^3)^2$

$ = 64a^6 - 456a^4b + 1056a^2b^2 + 324a^3c - 800b^3 - 648abc - 729c^2$

$ = \frac {1}{4}\left(16a^3 - 57ab + 108c\right)^2 - 135\left(4a^3c - a^2b^2 - 18abc + 4b^3 + 27c^2\right)$

$ + \frac {5}{4}b^2\left(87a^2 - 208b\right)$

$ = \frac {1}{4}\left[16\left(x^3 + y^3 + z^3\right) - 9(x^2y + x^2z + y^2z + y^2x + z^2x + z^2y) + 33xyz\right]^2$

$ + 135(y - z)^2(z - x)^2(x - y)^2$

$ + \frac {5}{4}(yz + zx + xy)^2\left[87\left(x^2 + y^2 + z^2\right) - 34(yz + zx + xy)\right]\geq0\Longrightarrow$

$ 10\sqrt {\left(x^2 + y^2 + z^2\right)^3}\geq 6(x + y + z)(x^2 + y^2 + z^2) - 27xyz$,

with equality if and only if $ y = z = - 2x\geq 0$ or any cyclic permutation, i.e. $ 4a^3 = - 27c\leq0 = b$.

By the way, the following inequality

$ \frac {\sqrt {\left(x^2 + y^2 + z^2\right)^3}}{6}\left[\sqrt {\frac {(k + 2)\left(k^2 + 4k + 40\right) + 6\sqrt {3}\left(k^2 + 4k + 8\right)}{3k}} + \sqrt {\frac {(k + 2)\left(k^2 + 4k + 40\right) - 6\sqrt {3}\left(k^2 + 4k + 8\right)}{3k}}\right]$

$ \geq (x + y + z)\left(x^2 + y^2 + z^2\right) - kxyz$

holds if and only if $ k\geq 8\left(1 - \sqrt {\frac {2}{3}}\right) = 1.4680\cdots$; with equalty if $ \left(k + \sqrt {k^2 + 4k - 8}\right)x = 2(2 - k)y\wedge y = z\geq 0$.

For $ k = 2$, we have

${ (x + y + z)\left(x^2 + y^2 + z^2\right) - 2xyz\leq \sqrt {2\left(x^2 + y^2 + z^2\right)^3}}$,

with equality if $ x = 0,y = z = 1$, see here: http://www.mathlinks.ro/Forum/viewtopic.php?t=72157

[hide="Other special cases."]${ (x + y + z)\left(x^2 + y^2 + z^2\right) - \frac {3}{2}xyz\leq \frac {13}{9}\sqrt {\left(x^2 + y^2 + z^2\right)^3}}$,

with equality if $ x = 1,y = z = 2$;

${ (x + y + z)\left(x^2 + y^2 + z^2\right) - \frac {9}{2}xyz\leq \frac {5}{3}\sqrt {\left(x^2 + y^2 + z^2\right)^3}}$,

with equality if $ x = - 1,y = z = 2$;

${ (x + y + z)\left(x^2 + y^2 + z^2\right) - \frac {19}{12}xyz\leq \frac {25}{4}\sqrt {\frac {\left(x^2 + y^2 + z^2\right)^3}{19}}}$,

with equality if $ x = 1,y = z = 3$;

${ (x + y + z)\left(x^2 + y^2 + z^2\right) - \frac {19}{6}xyz\leq \frac {13}{2}\sqrt {\frac {\left(x^2 + y^2 + z^2\right)^3}{19}}}$,

with equality if $ x = - 1,y = z = 3$;

${ (x + y + z)\left(x^2 + y^2 + z^2\right) - \frac {22}{3}xyz\leq 5\sqrt {\frac {2\left(x^2 + y^2 + z^2\right)^3}{11}}}$,

with equality if $ x = - 2,y = z = 3$;

${{ (x + y + z)\left(x^2 + y^2 + z^2\right) - 8\left(1 - \sqrt {\frac {2}{3}}\right)xyz\leq \frac {8\sqrt {2} + \sqrt {3}}{9}\sqrt {\left(x^2 + y^2 + z^2\right)^3}}}$

with equality if $ x = y = z = 1$.[/hide]

Hello, very good work Jichen:

I am thinking about this. Could you help?

\[ A(a\plus{}b\plus{}c\plus{}d)(a^2\plus{}b^2\plus{}c^2\plus{}d^2)\leq B(abc\plus{}bcd\plus{}cda\plus{}dab)\plus{}C(a^2\plus{}b^2\plus{}c^2\plus{}d^2)^{3/2}.\]
I want to find all possible $ A, B, C$ such that the inequality is true for all $ a,b,c,d$.

[quote="Namdung"]1. Prove that for any reals x, y, z, which satisfy condition x 2 + y 2 + z 2 = 2 we have
x+y+z)  \leq xyz + 2
(Poland, 1991)
Namdung[/quote]
We have :
$ 2\equal{}x^2\plus{}y^2\plus{}z^2 \ge 2yz$ or $ yz \leq 1$
By Bunhiacopsky we have :
$ [x(1\minus{}yz)\plus{}y\plus{}z]^2 \leq [x^2\plus{}(y\plus{}z)^2][(1\minus{}yz)^2\plus{}1] \leq (2\plus{}2yz)(y^2z^2\minus{}2yz\plus{}1) \leq 4$
(because $ yz \leq 1$)
There fore:
$ |x\plus{}y\plus{}z\minus{}xyz| \leq 2$
or $ x\plus{}y\plus{}z \leq 2\plus{}xyz$


Do we live in the 3rd dimension? (Just kidding)

Can a 2-dimentional figure exist in the 3rd dimentional world?

nope. theres a big difference between 2-dimensional objects and very thin objects. 2 dimentional objects have no thickness.

We probably live in an $8\frac{\log(2)}{\log(3)}$-dimensional space.

[quote="Gyan"][quote]Can a 2-dimentional figure exist in the 3rd dimentional world?[/quote]
Try to do an experiment and find out the result. If  you (generic you) can't do that experiment, there is no point asking that question . - Theory ofQ uantum physics .[/quote]
ha... hahaha....

Atoms have height, weight, length, and width..

What about empty space (vacuum)???

You can't just get that anywhere, you know.

[quote]Atoms have height, [b]weight[/b], length, and width..
[/quote]
  Huh ?  Mass certainly but   weight?  Weight  is force due to gravity .. and BTW they (I mean atoms ) do not even have height or length or width.. they  just have wave functions ....

Yes, cuz if u draw something on a piece of paper and look at it, it exists,(the drawing itself, i dont care about the pencil lead's thickness and weight, or the papers thickness and weight)

i dont think thats rite...

If you account for quantum uncertainty in the object's thickness, then no.  There is a probability at any given time that it has some thickness.

[quote="Athena"][quote="Ion"]Wouldn't shadows be 2 dimensional?[/quote]

good point. but im not really sure.

and for the drawing thing, that wouldnt count because u have to count the thickness of the paper lead no matter what.[/quote]

Shadows include the triangle up to the thing that causes it.

what about String Theory and 26 dimensions that we live in  :D ?

Depends how far said world extends, since planes are infinite.

Don't mind me, I'm just babbling what could very easily be nonsense.

Well, if the string theorists are right and we live in 11 or 26 dimensions, depending on which view you take, then the 2-D figures could live here but the third dimension would have to be kind of rolled up and put away, so to speak.  Like the poem:

Steve Weinberg, returning from Texas
Brings dimensions galore to perplex us
But the extra ones all
Are rolled up in a ball
So tiny it never affects us.

well just think about it.  if we were 2D. then we'd be as thin as a sheet of paper. (i think.  btw, my computer got fixed :lol: )

[quote="cardwacko9999"]string theory: the theory that the world is made up of tiny vibrating strings (like quarks, maybe?)- says that many dimensions

This is sort of theoretical, but, cant a plane (not airplane) exist in a 2D world?[/quote]
in case u dont get why that explains many dimensions, we could have equatiosn set up using string theoretical equations and energy would go out.  By conservation of energy, energy cant do that.  SO it must have leaked into another dimension
and if u say
thats just a theory, then they have proved that electrons exibit wave-like behavior , though they are clearly particles.

and for the question, it could

[quote="gameworld7"]actually it could happen in the next 1 million years[/quote]

ahem... sound familiar?... u hypocrite!!!! (j/k)

what do you mean? it could happen! just pray to God and He might do that.

true... but u just said

[quote="gameworld7"]what do you mean "it could"? it just can't happen. imposible. nada. cero. no. get it?[/quote]

i don't know. i'm contradicting myself all the time. i don't know what i'm saying.

look who doesnt have a brain NOW

oh wait... u never had one o_O

j/k

for two dimensionalthingy, i take two blocks
i put them together
the area which they touch are points
therefore the 1st dimension exists here.
and the second dimension follows

random point: cardwacko please change ur sign. its bugging me.

you EXISTING is bugging me...

scrambled- how does 2-d follow 1-d?

(1-d's a line, rite?)

why do people hate me so much? i'm just trying to make the world a better place...

See?  More spamming.

2-d follows 1-d because a plane is a line squared.

For those of you wondering if shadows are 2-d well I'm going to tell you about it.

  Shadows don't have dimensions. Light doesn't have a dimension. In fact shadows aren't even light, they are only a lack of light because something is blocking tha light.

true. and i dont get the string theory or watever but heres sum logic that i dont think anyone mentioned yet even tho its very obvious.  2-d is width and lenght, or area. cause its 2- DIMENSIONAL. means it only has 2 dimentions which are length and width. 3-dimensional is 3 dimensions ( length, width, height) 0-d is empty space (i think) but for 4-d.....i dont think that its weight. usually its something thats visible so im not sure about that.

[quote=/cardwacko]... you dont have one either, sunnyray, athena took (aka stole) it from you years ago[/quote]

huh? there was nothing to steal in the first place

0-D are points (no length, no width, no height, no depth, no ....)


$sgn(\sigma\tau)=sgn(\sigma)\cdots gn(\tau)$
the proof is as they say
$sgn(\sigma\tau)=\prod_{1 \leq i,j \leq n} \frac{(\sigma\tau)(j)-(\sigma\tau)(i)}{j-i}= \prod_{1 \leq i,j \leq n} \frac{\sigma(\tau(j))-\sigma(\tau(i))}{\tau(j)-\tau(i)}\cdot \prod_{1 \leq i,j \leq n} \frac{\tau(j)-\tau(i)}{j-i}$
No problem until this...very nice...
now they say that
$\prod_{1 \leq i,j \leq n} \frac{\sigma(\tau(j))-\sigma(\tau(i))}{\tau(j)-\tau(i)}=sgn(\tau)$
wich they say is evident .I disagree.I cant get it

[quote="spx2"]$\prod_{1 \leq i,j \leq n} \frac{\sigma(\tau(j))-\sigma(\tau(i))}{\tau(j)-\tau(i)}=sgn(\tau)$[/quote]
I think it should be (the $<$ is missing in all your products!):
$\prod_{1 \leq i<j \leq n} \frac{\sigma(\tau(j))-\sigma(\tau(i))}{\tau(j)-\tau(i)}=sgn(\sigma)$

To prove this, see that $\prod_{1 \leq i<j \leq n} \frac{\sigma(j)-\sigma(i)}{j-i}=sgn(\sigma)$ does not depend on anything special on $i,j$ in the sum:
- neither $i<j$ is really necessary
- nor the ordering in which you multiply the numbers
as long as the product goes over all pairs $\{i,j\}, i \neq j$ exactly once.
SO to permutate $i,j$ with any permutation like $\tau$ before multiplying doesn't change the product away from being $sgn(\sigma)$.


Why is that topic locked?

Anyway, nobody worked out how to make the calculation easier. Rather than say x^3 - y^3 = (x-y)(x^2 + xy + y^2) = x^2 + xy + y^2 where x-y=1, which involves 3 multiplication, or (x+y)^2 - xy, which still involves a lot of calculation.. it equals (x-y)^2 + 3xy = 1+3xy. Only really involves one multiplication.

oops sorry meant to split it into the actual topic and another topic about calculators and whatnot. didn't notice that it didn't split. sorry. well anyway, this can be the thingie about the actual topic and the other one you can have your nice little discussion about calcs.

or instead of all this factoring...(x+1)^3-x^3=3x^2+3x+1=3x(x+1)+1. that's why i don't like these consecutive number problems it's always so much easier simply to plug in.


Is the USAMTS a descendant of the old Wisconsin Math
Talent Search?  Are there any published compilations of
the old Wisconsin problems?  I stumbled on some
photocopies of the latter (from 60-70s) back in high
school and got good mileage out of them.

Concerning the LaTex issue, some politically incorrect
editorial comments:

- Plain text (ASCII as in old-fashioned Unix terminal email)
    remains the only universal format

- LaTeX comes in many flavors and isn't standardized.
     Sending "my" LaTex doesn't mean it will compile
     on "your" machine.

- LaTeX may be obsolete fairly soon.  If anything one
     could argue for learning the HTML/MathML/XML-type
     markup notation, particularly for a web-based contest.

- It is just as educational, and more *mathematically*
    educational compared to learning LaTeX, to develop
    the ability to express mathematics in text email, without
    recourse to TeX or fancy diagrams, and with a minimum
    of notation.  It makes you think about how to concisely
    summarize a diagrammatic or algorithmic construct, as
    well as the overall idea being expressed.
   The mental exercise of reading such an email is comparable
    to reading TeX and also, in my opinion, at least as educational.
    (e.g. visualizing a diagram, or drawing it from a written
    specification).

- For problem-solving by correspondence it is definitely the
    case that, if not a huge INCENTIVE, then it is at least
    an important part of the CULTURE, to have solvers recognized
    and brilliant solutions published (and likewise for problem
    proposers).  This was the case with the old Wisconsin
    talent search by mail, as well as the Arbelos (which Richard
    may remember), and the problem solving columns of
    Mathematics Magazine, the American Mathematical Monthly,
    the published USAMO/IMO/Putnam solutions ("special
    prize solution by so-and-so"), Crux Mathematicorum,
    Mathematical Gazette, and so on.  Paul Erdos got started
    this way as one of the recognized repeat solvers in the
    Komal high school journal, as did many other Hungarian
    mathematicians; even today although only a few make
    the Hungary IMO team, dozens or hundreds compete in the
    annual problem solving contest by mail.

- Completely agree with rcv that the contest shouldn't
   involve extraneous requirements -- just mathematics and
   written correspondence, be it by mail or e-mail.
   In an article on "how to get through math grad school",
   Ilan Vardi made some point to the effect that one should
   avoid spending time to learn Unix, programming, Tex
   beyond minimal needs, and other math-extraneous stuff,
   and a similar principle may apply at high school level.
   If LaTeX specifically were useful for other purposes
   (such as computer programming as a job credential)
   it might be a different story.

- Concerning accessibility and entry fees, the MAA (and Prof.
   Walter Mientka specifically, godfather of the modern
   AHSME/AIME/USAMO, who printed the AHSME out
   of his garage to keep the fee under $1.00 per student)
   bent over backwards to make it accessible.  The involvement
   of a school was to promote the integrity of the contest;
   I believe that one could arrange to take the exam at other
   schools if not offered locally.  This is less of an issue for
   an untimed correspondence contest on the honor system
   such as the MTS, so to compare accessibility in this sense
   is apples-to-oranges.

- I do think some nod should be made to the open-access
  principle exemplified by Profs. Mientka and Berszenyi,
  if nothing else then as a continuation of what they did.
  Of course, the inconvenience here would not be borne
  by me so it may be glib to suggest this.  Nevertheless,
  making the student adapt to technological limitations or
  preferences of the contest organizers seems contrary
  to the original open spirit.

[quote="fleeting_guest"]Concerning the LaTex issue, some politically incorrect
editorial comments:

- Plain text (ASCII as in old-fashioned Unix terminal email)
    remains the only universal format
[/quote]
I find a good magic marker will write on almost anything ... oh, wait, you mean universal [i]electronic[/i] format.

[quote="fleeting_guest"]- LaTeX comes in many flavors and isn't standardized.
     Sending "my" LaTex doesn't mean it will compile
     on "your" machine. [/quote]
I had to grade a test with a mismatched font set, which was tough to read. But that was a PDF file.

A minimal version of LaTeX2e, without any of the packages except graphics and fancyhdr, is standardized.

[quote="fleeting_guest"]- LaTeX may be obsolete fairly soon.  If anything one
     could argue for learning the HTML/MathML/XML-type
     markup notation, particularly for a web-based contest.[/quote]
HTML is not up to handling the mathematical notation. I am not familiar with MathML.

[quote="fleeting_guest"]- It is just as educational, and more *mathematically*
    educational compared to learning LaTeX, to develop
    the ability to express mathematics in text email, without
    recourse to TeX or fancy diagrams, and with a minimum
    of notation.  It makes you think about how to concisely
    summarize a diagrammatic or algorithmic construct, as
    well as the overall idea being expressed.
   The mental exercise of reading such an email is comparable
    to reading TeX and also, in my opinion, at least as educational.
    (e.g. visualizing a diagram, or drawing it from a written
    specification). [/quote]
If you want to send in your solutions that way, it is fine with me. It is a pleasure to grade well-written prose. The only requirement for LaTeX in the USAMTS is that only solutions in LaTeX will be posted as official solutions. Use of other formats has no effect on the score of a USAMTS submission.

The main purpose of the official solutions is so that new participants can read through them and see our standards and goals for the USAMTS and learn from the techniques that other participants used in the past. Reading mathematics that does not use special fonts and diagrams requires more skill than reading mathematics that uses the fonts and diagrams. We want to have the official solutions as readable as possible.

[quote="fleeting_guest"]- For problem-solving by correspondence it is definitely the
    case that, if not a huge INCENTIVE, then it is at least
    an important part of the CULTURE, to have solvers recognized
    and brilliant solutions published (and likewise for problem
    proposers).  This was the case with the old Wisconsin
    talent search by mail, as well as the Arbelos (which Richard
    may remember), and the problem solving columns of
    Mathematics Magazine, the American Mathematical Monthly,
    the published USAMO/IMO/Putnam solutions ("special
    prize solution by so-and-so"), Crux Mathematicorum,
    Mathematical Gazette, and so on.  Paul Erdos got started
    this way as one of the recognized repeat solvers in the
    Komal high school journal, as did many other Hungarian
    mathematicians; even today although only a few make
    the Hungary IMO team, dozens or hundreds compete in the
    annual problem solving contest by mail. [/quote]
Unfortunately, a lot more USAMTS solutions are of the quality that they could be official solutions than anyone would want to read through. We limited ourselves to a single good representation of any common method of solution (At least that was my standard in Years 14 and 15, not that anyone has seen the Year 15 official solutions yet).  There was a heavy random element to the selection of the official solutions in the years that LaTeX was not required. Adding this year's requirement for LaTeX has probably reduced the randomness.

[quote="fleeting_guest"]- Completely agree with rcv that the contest shouldn't
   involve extraneous requirements -- just mathematics and
   written correspondence, be it by mail or e-mail.
   In an article on "how to get through math grad school",
   Ilan Vardi made some point to the effect that one should
   avoid spending time to learn Unix, programming, Tex
   beyond minimal needs, and other math-extraneous stuff,
   and a similar principle may apply at high school level.
   If LaTeX specifically were useful for other purposes
   (such as computer programming as a job credential)
   it might be a different story. [/quote]

I repeat, LaTeX is not a requirement to participate in the USAMTS.

I disagree with the sentiment that a mathematics student should focus solely on mathematics. Some non-mathematical skills such as programming are heavily used by mathematicians and it does help to master those skills.  A mathematics students should learn not just mathematics but also the tools of mathematics.

Of course, many participants in the USAMTS will choose some other career than mathematics. I admit that LaTeX is not as useful to non-mathematicians. But likewise HTML and other electronic formats are not of much use in many professions.

I don't know where to find rcv's comments to respond to them.

[quote="fleeting_guest"]- I do think some nod should be made to the open-access
  principle exemplified by Profs. Mientka and Berszenyi,
  if nothing else then as a continuation of what they did.
  Of course, the inconvenience here would not be borne
  by me so it may be glib to suggest this.  Nevertheless,
  making the student adapt to technological limitations or
  preferences of the contest organizers seems contrary
  to the original open spirit.[/quote]

Originally, Prof. Berzsenyi published the USAMTS solutions in CONSORTIUM magazine, were he was limited to about a page, so he could not print the full solutions, let alone student-written solutions. The ability to post student-written solutions is an outgrown of the Internet. And until this year, it was limited by the ability of the USAMTS's solutions editor to transcribe printed solutions into electronic form, which took about 2 hours per solution (this was with document typesetting software, not with LaTeX). Using standard electronic formats increases our ability to post student-written solutions as official solutions. For now, the AoPS people are restricting themselves to LaTeX for the solutions they will post. I am not part of the AoPS, so I don't know their future plans, but I believe that if some other electronic format becomes as good and as widely used in the mathematics community as LaTeX, they will use that format too.

Erin Schram
USAMTS grader and former solutions editor

To the best of my knowledge, the Wisconsin contest still exists, as does a Southern California offshoot of it run by Pomona College.

[url]http://www.math.pomona.edu/talentsearch.html[/url]

These are independent of USAMTS. They are similar to USAMTS in that they are mail-in, proof-oriented contests. They differ in some minor stylistic ways - the mix of questions is a little different, the amount of partial credit awarded in grading is a little different, the Pomona contest allows students to enter as cooperating groups. These efforts complement each other more than they compete with each other; there's no reason a student couldn't do both.

The UW math talent search is at math.wisc.edu/~talent. I think it uses the same problems as the California talent search.

[quote="Erin J. Schram"]I don't know where to find rcv's comments to respond to them.[/quote]

I think fleeting_guest was referring to the fact that, some time ago (in [url=http://www.artofproblemsolving.com/Forum/topic-12736.html]this[/url] thread), rcv argued that $\LaTeX$ shouldn't be required if you want your solutions published.

[quote="fleeting_guest"]Is the USAMTS a descendant of the old Wisconsin Math
Talent Search?  Are there any published compilations of
the old Wisconsin problems?  I stumbled on some
photocopies of the latter (from 60-70s) back in high
school and got good mileage out of them.[/quote]

I do believe the Wisconsin Talent Search was an inspiration for the USAMTS.  I do not know how to get back problems for the Wisconsin/Pomona program.  I am a fan of the Wisconsin/Pomona program and have used the latter to find students for our Math Circle here in San Diego.

[quote="fleeting_guest"]
Concerning the LaTex issue, some politically incorrect
editorial comments:

- Plain text (ASCII as in old-fashioned Unix terminal email)
    remains the only universal format

- LaTeX comes in many flavors and isn't standardized.
     Sending "my" LaTex doesn't mean it will compile
     on "your" machine.
[/quote]

We ask the students to send in an output file, not just plain LaTeX.  We have not yet had any compatability problems with the solutions we have chosen to publish.

[quote="fleeting_guest"]
- LaTeX may be obsolete fairly soon.  If anything one
     could argue for learning the HTML/MathML/XML-type
     markup notation, particularly for a web-based contest.
[/quote]

I haven't seen a whole lot of evidence of LaTeX being obsolete.  Moreover, I haven't seen any system for rendering mathematics on the web that is as easy to use and as flexible and transferrable to other media as LaTeX.

[quote="fleeting_guest"]
- It is just as educational, and more *mathematically*
    educational compared to learning LaTeX, to develop
    the ability to express mathematics in text email, without
    recourse to TeX or fancy diagrams, and with a minimum
    of notation.  It makes you think about how to concisely
    summarize a diagrammatic or algorithmic construct, as
    well as the overall idea being expressed.
   The mental exercise of reading such an email is comparable
    to reading TeX and also, in my opinion, at least as educational.
    (e.g. visualizing a diagram, or drawing it from a written
    specification).
[/quote]

This I pretty strongly disagree with - mathematics written in plain text is a nightmare to read and understand.  This is why notation was developed.  Students need to learn how to communicate their ideas most effectively.  Many ideas simply cannot be communicated as effectively with plain text as they can with notation.  Look at the very old (pre-TeX and pre-math smileys) posts in this forum and you'll see.  The addition of LaTeX to this board has been a tremendous boon, as information can be conveyed much more effectively.

Having to think about how to concisely explain your solution is not mitigated by recourse to typesetting tools - if anything, you have to think about it more because you have more options.

[quote="fleeting_guest"]
- For problem-solving by correspondence it is definitely the
    case that, if not a huge INCENTIVE, then it is at least
    an important part of the CULTURE, to have solvers recognized
    and brilliant solutions published (and likewise for problem
    proposers).  This was the case with the old Wisconsin
    talent search by mail, as well as the Arbelos (which Richard
    may remember), and the problem solving columns of
    Mathematics Magazine, the American Mathematical Monthly,
    the published USAMO/IMO/Putnam solutions ("special
    prize solution by so-and-so"), Crux Mathematicorum,
    Mathematical Gazette, and so on.  Paul Erdos got started
    this way as one of the recognized repeat solvers in the
    Komal high school journal, as did many other Hungarian
    mathematicians; even today although only a few make
    the Hungary IMO team, dozens or hundreds compete in the
    annual problem solving contest by mail.
[/quote]

Our top solvers are recognized even if they don't use LaTeX - if they choose, their names are listed on the Leaders page, and top solutions are commended (and these students are listed as well).

We reward professionally written solutions by publishing those written in LaTeX (which are usually the most well-written anyway, as students who take care to write in LaTeX tend to take care in writing their solution).  We have some nice non-LaTeX solutions, but typically 8 of the top 10 are written in LaTeX.

[quote="fleeting_guest"]
- Completely agree with rcv that the contest shouldn't
   involve extraneous requirements -- just mathematics and
   written correspondence, be it by mail or e-mail.
   In an article on "how to get through math grad school",
   Ilan Vardi made some point to the effect that one should
   avoid spending time to learn Unix, programming, Tex
   beyond minimal needs, and other math-extraneous stuff,
   and a similar principle may apply at high school level.
   If LaTeX specifically were useful for other purposes
   (such as computer programming as a job credential)
   it might be a different story.
[/quote]

You don't need to use LaTeX to submit your solutions or to win prizes, only to be published.

[quote="fleeting_guest"]
- Concerning accessibility and entry fees, the MAA (and Prof.
   Walter Mientka specifically, godfather of the modern
   AHSME/AIME/USAMO, who printed the AHSME out
   of his garage to keep the fee under $1.00 per student)
   bent over backwards to make it accessible.  The involvement
   of a school was to promote the integrity of the contest;
   I believe that one could arrange to take the exam at other
   schools if not offered locally.  This is less of an issue for
   an untimed correspondence contest on the honor system
   such as the MTS, so to compare accessibility in this sense
   is apples-to-oranges.
[/quote]

I'm not sure what this is in reference to, but for US citizens and residents, it's tough to get more accessible than our contest.  We are trying to raise funds to open the contest to international students as well.

[quote="fleeting_guest"]
- I do think some nod should be made to the open-access
  principle exemplified by Profs. Mientka and Berszenyi,
  if nothing else then as a continuation of what they did.
  Of course, the inconvenience here would not be borne
  by me so it may be glib to suggest this.  Nevertheless,
  making the student adapt to technological limitations or
  preferences of the contest organizers seems contrary
  to the original open spirit.[/quote]

Again, I disagree that we have violated anything here by rewarding students who take the time to present their work more professionally than others do. Students who haven't the time or inclination to learn LaTeX still can participate in every other way.

I think I can also point to this message board as an example that students really like to be able to convey their mathematics cleanly and professionally - we've even had some teachers write us and thank us for the LaTeX tutorial and incentives we built for students to learn it because the students are doing their *homework* in TeX (as many of my friends in grad school do).

Furthermore, there is a real value in learning how to fight with a computer to make it do what you want it to do - the people who master this skill are far more likely to be effective in their professional lives than those who don't.  In this vein, I'll expound on the earlier comment by Erin Schram by noting that our goal here is not just to teach mathematics or develop mathematicians.  Educators in general, and those in mathematics specifically, have a much broader calling - to help students learn how to think, how to solve problems, and how to be effective in whatever endeavor they choose.

I definitely agree that LaTeX is a helpful tool and skill.  I never would have found the time on my own to learn it, but the USAMTS provided a reason and some motivation (something specific to try typesetting).  I handwrote solutions for the first two rounds and then used LaTeX for the third - I love how much neater and communicable my math writing is now (and without much of a learning curve!)  It's definitely a transferrable skill, too, at least within the science-math-programming-enginneering family.  So I just want to back up Mr. Rusczyk's comments as one of the grateful students he described :)

[quote="fleeting_guest"]
- It is just as educational, and more *mathematically*
    educational compared to learning LaTeX, to develop
    the ability to express mathematics in text email, without
    recourse to TeX or fancy diagrams, and with a minimum
    of notation.  It makes you think about how to concisely
    summarize a diagrammatic or algorithmic construct, as
    well as the overall idea being expressed.
   The mental exercise of reading such an email is comparable
    to reading TeX and also, in my opinion, at least as educational.
    (e.g. visualizing a diagram, or drawing it from a written
    specification). [/quote]
Okay, if you think it's more educational, then type it up.  Anything you can do with LaTeX can be done by hand.  Are you attempting to say that it is better to put it into ASCII format than to write it by hand?  And I don't see how you could put a paper like [url=http://www.google.com]this[/url] into ASCII.  How, in any way, could you resort to typing in ASCII a large, complex, problem involving, for example, a clebsh diagonal cubic without a picture?  Sometimes a picture is necessary, unless the sole purpose of your paper was to make your grader or reader do a lot of work.  These graders don't want to do great mental excercises -- they have a lot of papers to grade.  Secondly, I could write a program in machine code only, to show my ability to express code in 1's and 0's, and it's a great mental excercise to translate it into anything meaningful.

[quote="fleeting_guest"]
- For problem-solving by correspondence it is definitely the
    case that, if not a huge INCENTIVE, then it is at least
    an important part of the CULTURE, to have solvers recognized
    and brilliant solutions published (and likewise for problem
    proposers).  This was the case with the old Wisconsin
    talent search by mail, as well as the Arbelos (which Richard
    may remember), and the problem solving columns of
    Mathematics Magazine, the American Mathematical Monthly,
    the published USAMO/IMO/Putnam solutions ("special
    prize solution by so-and-so"), Crux Mathematicorum,
    Mathematical Gazette, and so on.  Paul Erdos got started
    this way as one of the recognized repeat solvers in the
    Komal high school journal, as did many other Hungarian
    mathematicians; even today although only a few make
    the Hungary IMO team, dozens or hundreds compete in the
    annual problem solving contest by mail. [/quote]
People want to ge able to read your solutions without having to strain their eyes.  I remember there was a case wher there was a hand-written proof written so well that they just scanned and posted it.  And certainly, a "good" proof is one that is easy to understand -- in which LaTeX is certainly better than ASCII.

[quote="fleeting_guest"]
- Completely agree with rcv that the contest shouldn't
   involve extraneous requirements -- just mathematics and
   written correspondence, be it by mail or e-mail.
   In an article on "how to get through math grad school",
   Ilan Vardi made some point to the effect that one should
   avoid spending time to learn Unix, programming, Tex
   beyond minimal needs, and other math-extraneous stuff,
   and a similar principle may apply at high school level.
   If LaTeX specifically were useful for other purposes
   (such as computer programming as a job credential)
   it might be a different story. [/quote]
The contest does not have extraneous requirements.  LaTeX is but just one of the many options available to people writing proofs.  And according that article, we can learn Tex "up to minimal needs." which is what's needed for the USAMTS.

[quote="fleeting_guest"]
- I do think some nod should be made to the open-access
  principle exemplified by Profs. Mientka and Berszenyi,
  if nothing else then as a continuation of what they did.
  Of course, the inconvenience here would not be borne
  by me so it may be glib to suggest this.  Nevertheless,
  making the student adapt to technological limitations or
  preferences of the contest organizers seems contrary
  to the original open spirit.[/quote]
The preferences of the contest organizers is that they should be able to grade it without spending enormous quantities of time, which is entirely a reasonable request.  That is what TeX is for.


hi people,

Can you help me to find the unknow r3 in the equation??

my boyfriend got 12.7 but I think it is not right.

 [url=http://img127.imagevenue.com/img.php?loc=loc208&image=46173_math.JPG][img]http://img127.imagevenue.com/loc208/th_46173_math.JPG[/img][/url]

Is that some kind of statistics probelm?

Or are you supposed to solve it in terms of $\chi$


Let a,b and c be real numbers such that a>=b>=c>=0.Determine all pairs of non-zero real numbers (m,n) such that
(a^m-b^m)*(b^n-c^n)>=(a^n-b^n)*(b^m-c^m)

Using Cauchy theorem we have $m \ge n$
Isn't it?


Let AC be a diameter of a circle K. Chords AB and CD intersect in a point P. Does the center of circle PBD lies on K? If yes, why?

let O be the circumcenter of PBD and let $ \angle DPA \equal{} x$ clearly x<90. Let a point X on the arch BD that not contain A, then $ \angle BOD \equal{} 2x$ and $ \angle BXD \equal{} 90 \plus{} x$ so $ \angle BXD > \angle BOD$ so O is external to the circle. Then O lies on K iff P is on K.


vector
topology
real analysis
real analysis unsolved

One of the ways of defining compactness on a metric space in introductory real analysis books is by introducing open coverings.

Do we have an equivalent formulation of compactness using closed coverings ? or semi open (semi closed coverings) ?

Closed coverings are totally different from open coverings.  In a metric space, each point can be covered by itself, so every infinite set has a closed covering not admitting a finite subcovering.  Of course one can replace every set involved in the usual definition by its complement, but this is probably not what you mean.

You should think more about what the open cover definition really means.

The other way to define a compact metric space is using sequences. In our topic, I agree with t0rajir0u. If we replace the open sets with close, we won't have a compact space either we use semi-open or clopen sets.

What we could do is applying duality, i.e. taking complements in this case. Then a space is compact iff every family of closed sets with empty intersection has a finite subfamily with empty intersection by taking complements in the covering axiom.

I agree with the others that we won't have any characterization of compactness in terms of coverings by closed sets. The most we can expect are assertions about families of closed sets with empty intersection. All this applies to topological spaces in general which includes metric spaces and normed vector spaces.

Not germane to coverings but compactness in that space can be discussed in terms of the finite intersection property of a class of closed sets


If $ a,b,c$ are pozitiv real numbers such that $ abc\equal{}1$ then
 $ \frac {ab}{a^{5}\plus{}b^{5}\plus{}ab} \plus{}\frac {bc}{b^{5}\plus{}c^{5}\plus{}bc} \plus{} \frac {ca}{c^{5}\plus{}a^{5}\plus{}ca} <1$

[quote="enndb0x"]If $ a,b,c$ are pozitiv real numbers such that $ abc \equal{} 1$ then
 $ \frac {ab}{a^{5} \plus{} b^{5} \plus{} ab} \plus{} \frac {bc}{b^{5} \plus{} c^{5} \plus{} bc} \plus{} \frac {ca}{c^{5} \plus{} a^{5} \plus{} ca} < 1$[/quote]
It follows from the basic
$ a^5\plus{}b^5\geq a^2b^2(a\plus{}b)$
So we have that
$ \displaystyle\sum \frac{1}{ab(a\plus{}b)\plus{}abc}\geq\displaystyle\sum \frac{ab}{a^5\plus{}b^5\plus{}ab}$
So we have to prove that
$ 1\geq\displaystyle\sum\frac{1}{ab(a\plus{}b)\plus{}abc}$
But $ \displaystyle\sum \frac{1}{ab(a\plus{}b)\plus{}abc}\equal{}\displaystyle\sum \frac {abc}{ab(a\plus{}b\plus{}c)}\equal{}\displaystyle\sum \frac{c}{a\plus{}b\plus{}c}\equal{}1$  :)

This was A1 on the 1996 IMO shortlist. Please confine discussion to that topic: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=37207#37207

Additionally, the inequality is not strict.

what do you mean the inequality is not strict?

it means that the expression could be $ \le 1$,
instead of lesser than 1


Why?
[hide]
$ \left\lceil\frac{37.50}{200}\right\rceil\equal{}\lceil.1875\rceil \equal{}\$.19$
[/hide]

yes correct!!!! that does seem like a lot of money for me to pledge myself still....it doesn't get any better with this nice blue background.

[quote="xpmath"]Basically, we need to find how much money she gets per lap. Just divide.[/quote]

Yeah, I think the probelm's just trying to confuse you by putting extra stuffs in there.

[quote="funcia"]Ok, Funcia is swimming a swim-a-thon. Funcia wants to raise$ \$37.50$. Funcia swims 200 laps. If people will donate per lap that Funcia does(and we know she is definitely doing 200 of them! such a hardcore swimmer :rotfl: ) how much monies(l :rotfl: ) would Funcia need to trick a person to give her if she plans to only raise the money using one sponsor?

Oh and example:
Funcia tricks, um, "GoogleBot" into giving her 10 cents per lap.  She gets 20 dollars.[/quote]
so its just 37.50/200?

[hide="solution"]funcia needs 3750 dollars and swims 200 laps. $ \frac{\$37.50} {200}$[/hide]
how can you be too "lazy" to do this

well, first of all, you know i wasn't too lazy to be on my computer...but i'm always on my computer, so that doesn't count. i don't have my calculator...but i have ten or so on my computer...so why not make it really confusing it post it on aops...

 :rotfl:

yeah so like, obviously, this problem has been solved, i guess the only useful additions to it would be if anyone had a really easy way to divide it out.

[hide=":rotfl:"]
$ \frac{37.50}{200}$
or, $ \frac{3750 * \frac{1}{100}}{2 * 100}$ so if we "invert and multiply" in this case,
we see that we will divide by 2(taking the 2 out of the denominator) to get 1875(we assume you can do this)
we get the 100 out of the denominator in a similar way,
and have, $ 1875 * 1/10^4$ which is easy to see that it is $ .1875$  [/hide]

[quote="hunter34"]Why?
[hide]
$ \left\lceil\frac {37.50}{200}\right\rceil \equal{} \lceil.1875\rceil \equal{} \$.19$
[/hide][/quote]

Heh heh, not entirely correct. Technically  $ \lceil.1875\rceil\equal{}1$. So it's $ \frac{\left\lceil\frac {37.50}{200}\cdot 100\right\rceil}{100}$.


When I tried to look at my badges...

[quote="Alcumus Server"]Fatal error: Cannot use string offset as an array in /var/www/Alcumus/Badges.php on line 76[/quote]


Hello, tayyaba

[hide]

ST is the hypotenuse of right triangle SPT.

ST2 = SP2 + PT2 = 62 + 72 = 85

Therefore: ST = sqrt{85}

Angle RSV = angle STP.

Hence, right triangle SVR is similar to right triangle SPT.

Then RV/SR = SP/ST ---> RV/9 = 6/sqrt{85}

Therefore: RV = 54/sqrt{85}

[/hide]


inequalities
induction
inequalities unsolved

[quote="hedeng123"]
Guess  $ n\geq 12$

$ \displaystyle\sum_{i \equal{} 1}^n \dfrac{1}{ih_i} > 2$[/quote]

Why guess? It's clearly wrong $ \forall n \geq 12$...

Yeah, it's just what I said...because $ \sum_{i \equal{} 1}^{n}\frac{1}{ih_{i}}\geq\sum_{i \equal{} 1}^{12}\frac{1}{ih_{i}}> 2 \quad \forall n\geq 12$...
but if you want to keep calculating, excel did it for me until $ n\equal{}10000$, where we have $ \sum_{i \equal{} 1}^{10000}\frac{1}{ih_{i}}\equal{}3,16823590860162...$^^

[quote="Martin N."]Yeah, it's just what I said...because $ \sum_{i \equal{} 1}^{n}\frac {1}{ih_{i}}\geq\sum_{i \equal{} 1}^{12}\frac {1}{ih_{i}} > 2 \quad \forall n\geq 12$...
but if you want to keep calculating, excel did it for me until $ n \equal{} 10000$, where we have $ \sum_{i \equal{} 1}^{10000}\frac {1}{ih_{i}} \equal{} 3,16823590860162...$^^[/quote]

sorry    I did not understand what you mean before

you are right

:lol:

but

$ n \equal{} 10000$, my answer is $ \sum_{i \equal{} 1}^{10000}\frac {1}{ih_{i}} \equal{} 1.1682359086016323245...$

You made a small mistake :lol:

[quote="xin007"]Let $ h_n \equal{} \displaystyle\sum_{i \equal{} 1}^n\dfrac{1}{i}$

Prove $ \displaystyle\sum_{i \equal{} 1}^n \dfrac{1}{ih_i} < 2$[/quote]

I'm sorry, just  inequalities is:
 Let $ h_n \equal{} \displaystyle\sum_{i \equal{} 1}^n\dfrac{1}{i}$

Prove $ \displaystyle\sum_{i \equal{} 1}^n \dfrac{1}{ih_i^2} < 2$

[quote="xin007"]
I'm sorry, just  inequalities is:
 Let $ h_n \equal{} \displaystyle\sum_{i \equal{} 1}^n\dfrac{1}{i}$

Prove $ \displaystyle\sum_{i \equal{} 1}^n \dfrac{1}{ih_i^2} < 2$[/quote]

I'll prove
\[ \sum_{i \equal{} 1}^n \dfrac{1}{ih_i^2} \leq 2 \minus{} \frac{1}{h_n} \qquad (1)\]
by induction on $ n$.

To start, taking $ n\equal{}1$, we have equality in $ (1)$; assuming that $ (1)$ holds for some $ n \in \mathbb{N}$ we want to prove
\[ \sum_{i \equal{} 1}^{n\plus{}1} \dfrac{1}{ih_i^2} \leq 2 \minus{} \frac{1}{h_{n\plus{}1}}\mbox{.} \qquad (2)\]

Subtracting $ (1)$ from $ (2)$, we get
\[ \dfrac{1}{(n\plus{}1)h_{n\plus{}1}^2}\leq \frac{1}{h_n}\plus{}\frac{1}{h_{n\plus{}1}}\equal{}\frac{h_{n\plus{}1}\minus{}h_n}{h_nh_{n\plus{}1}}\equal{}\frac{1}{(n\plus{}1)h_nh_{n\plus{}1}}\]
\[ \Leftrightarrow h_{n\plus{}1}\geq h_n\mbox{,}\]
which is obvious and completes the induction.  :)

Furthermore, as the sequence $ h_i$, $ i\equal{}1,2,\ldots$ diverges, $ 2$ is the best constant such that the proposed inequality holds.


There are exactly 3 integers $ x$ satisfying the inequality
$ x^{2}+bx+2\leq 0$.
How many integer values of $ b$ are possible?

This is an easy problem!
$ x^{2}+bx+2\leq 0 (*)$
$ x_{1,2}=\frac{-b \pm \sqrt{b^{2}-8}}{2}$ asume $ x_{1}<x_{2}$
as there are exactly 3 integers  satisfying the inequality $ (*)$
then $ x-1<x_{1}\leq x$ and $ x+2<x_{1}\leq x+3$  so
$ 2\leq x_{2}-x_{1}\leq 4$,but $ x_{2}-x_{1}=\sqrt{b^{2}-8}\Rightarrow 4\leq b\leq 4 \Rightarrow b=4$
which is a solution!


Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \sqrt{\frac{a^2\plus{}bc}{b^2\plus{}bc\plus{}c^2}}\plus{}\sqrt{\frac{b^2\plus{}ca}{c^2\plus{}ca\plus{}a^2}}\plus{}\sqrt{\frac{c^2\plus{}ab}{a^2\plus{}ab\plus{}b^2}} \ge \sqrt{6}\]
:)
I inspired it from the old one of VasC
\[ \frac{a^2\plus{}bc}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{b^2\plus{}ca}{c^2\plus{}ca\plus{}a^2}\plus{}\frac{c^2\plus{}ab}{a^2\plus{}ab\plus{}b^2} \ge 2\]
:)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \sqrt {\frac {a^2 \plus{} bc}{b^2 \plus{} bc \plus{} c^2}} \plus{} \sqrt {\frac {b^2 \plus{} ca}{c^2 \plus{} ca \plus{} a^2}} \plus{} \sqrt {\frac {c^2 \plus{} ab}{a^2 \plus{} ab \plus{} b^2}} \ge \sqrt {6}
\]
:)
I inspired it from the old one of VasC
\[ \frac {a^2 \plus{} bc}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b^2 \plus{} ca}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {c^2 \plus{} ab}{a^2 \plus{} ab \plus{} b^2} \ge 2
\]
:)[/quote]
\[ \sum {\sqrt {\frac{{a^2  \plus{} bc}}{{b^2  \plus{} bc \plus{} c^2 }}} }  \le \sqrt {3\sum {\frac{{a^2  \plus{} bc}}{{b^2  \plus{} bc \plus{} c^2 }}} }
\]
and problem1 ,i can kill it by Cauchy -Schwarz or Holder :)

[quote="tronghieu"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \sqrt {\frac {a^2 \plus{} bc}{b^2 \plus{} bc \plus{} c^2}} \plus{} \sqrt {\frac {b^2 \plus{} ca}{c^2 \plus{} ca \plus{} a^2}} \plus{} \sqrt {\frac {c^2 \plus{} ab}{a^2 \plus{} ab \plus{} b^2}} \ge \sqrt {6}
\]
:)
I inspired it from the old one of VasC
\[ \frac {a^2 \plus{} bc}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b^2 \plus{} ca}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {c^2 \plus{} ab}{a^2 \plus{} ab \plus{} b^2} \ge 2
\]
:)[/quote]
\[ \sum {\sqrt {\frac {{a^2 \plus{} bc}}{{b^2 \plus{} bc \plus{} c^2 }}} } \le \sqrt {3\sum {\frac {{a^2 \plus{} bc}}{{b^2 \plus{} bc \plus{} c^2 }}} }
\]
and problem1 ,i can kill it by Cauchy -Schwarz or Holder :)[/quote]
Can you show your solution?

I think it's very cool!, stronger than the Vasc one.  :)

[quote="tronghieu"]\[ \sum {\sqrt {\frac {{a^2 \plus{} bc}}{{b^2 \plus{} bc \plus{} c^2 }}} } \le \sqrt {3\sum {\frac {{a^2 \plus{} bc}}{{b^2 \plus{} bc \plus{} c^2 }}} }
\]
and problem1 ,i can kill it by Cauchy -Schwarz or Holder :)[/quote]
Dear tronghieu, I don't like spam. I think every word we say must be included with our action, that makes our words becomes more believable.

[quote="can_hang2007"][quote="tronghieu"]
\[ \sum {\sqrt {\frac {{a^2 \plus{} bc}}{{b^2 \plus{} bc \plus{} c^2 }}} } \le \sqrt {3\sum {\frac {{a^2 \plus{} bc}}{{b^2 \plus{} bc \plus{} c^2 }}} }
\]
and problem1 ,i can kill it by Cauchy -Schwarz or Holder :)[/quote]
Dear tronghieu, I don't like spam. I think every word we say must be included with our action, that makes our words becomes more believable.[/quote]that is my solution
\[ \sum {\sqrt {\frac{{a^2  \plus{} bc}}{{b^2  \plus{} bc \plus{} c^2 }}} } )^2 ((a^2  \plus{} bc)^2 (b^2  \plus{} bc \plus{} c^2 )) \ge (\sum {a^{^2 } }  \plus{} \sum {ab} )^3
\]
\[ (\sum {a^{^2 } }  \plus{} \sum {ab} )^3  \ge 12\sum {a^2 b^2 (a^2  \plus{} b^2 )}  \plus{} 6\sum {a^3 b^3 }  \plus{} 6abc\sum {a^3 }  \plus{} 12abc\sum {ab(a \plus{} b)}  \plus{} 36a^2 b^2 c^2
\]
because f ''(abc)=0.We need to prove in 2 case:c=0 anh b=c :)

[quote="tronghieu"][quote="can_hang2007"][quote="tronghieu"]
\[ \sum {\sqrt {\frac {{a^2 \plus{} bc}}{{b^2 \plus{} bc \plus{} c^2 }}} } \le \sqrt {3\sum {\frac {{a^2 \plus{} bc}}{{b^2 \plus{} bc \plus{} c^2 }}} }
\]
and problem1 ,i can kill it by Cauchy -Schwarz or Holder :)[/quote]
Dear tronghieu, I don't like spam. I think every word we say must be included with our action, that makes our words becomes more believable.[/quote]that is my solution
\[ \sum {\sqrt {\frac {{a^2 \plus{} bc}}{{b^2 \plus{} bc \plus{} c^2 }}} } )^2 ((a^2 \plus{} bc)^2 (b^2 \plus{} bc \plus{} c^2 )) \ge (\sum {a^{^2 } } \plus{} \sum {ab} )^3
\]

\[ (\sum {a^{^2 } } \plus{} \sum {ab} )^3 \ge 12\sum {a^2 b^2 (a^2 \plus{} b^2 )} \plus{} 6\sum {a^3 b^3 } \plus{} 6abc\sum {a^3 } \plus{} 12abc\sum {ab(a \plus{} b)} \plus{} 36a^2 b^2 c^2
\]
because f ''(abc)=0.We need to prove in 2 case:c=0 anh b=c :)[/quote]
But unfortanetly, the inequality
\[ (\sum {a^{^2 } } \plus{} \sum {ab} )^3 \ge 12\sum {a^2 b^2 (a^2 \plus{} b^2 )} \plus{} 6\sum {a^3 b^3 } \plus{} 6abc\sum {a^3 } \plus{} 12abc\sum {ab(a \plus{} b)} \plus{} 36a^2 b^2 c^2
\]
is not true.
You should check it carefully. ;)

[quote="Ji Chen"]
it remains to be proven

$ \left[\sum{(2x \plus{} y \plus{} z)\left(x^2 \plus{} yz\right)}\right]^3\geq 6\sum{(2x \plus{} y \plus{} z)^3\left(x^2 \plus{} yz\right)^2\left(y^2 \plus{} yz \plus{} z^2\right)}$.
[/quote]
It's obviously true after expanding:
$\sum_{sym}(2x^9+12x^8y-3x^7y^2-8x^6y^3+6x^5y^4+18x^7yz+24x^6y^2z-12x^5y^3z-$
$-12x^4y^4z+18x^5y^2z^2-39x^4y^3z^2-6x^3y^3z^3)\geq0$, which is true by Muirhead.

[quote=can_hang2007]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \sqrt{\frac{a^2\plus{}bc}{b^2\plus{}bc\plus{}c^2}}\plus{}\sqrt{\frac{b^2\plus{}ca}{c^2\plus{}ca\plus{}a^2}}\plus{}\sqrt{\frac{c^2\plus{}ab}{a^2\plus{}ab\plus{}b^2}} \ge \sqrt{6}\]
[/quote]
Let  $a, b, c$ are  non-negative real numbers .Prove that$$ \frac{\sqrt{a^2+bc}}{b+c}+\frac{\sqrt{b^2+ca}}{c+a}+\frac{\sqrt{c^2+ab}}{a+b}\geq\frac{3\sqrt{2}}{2}.$$
[url=http://www.artofproblemsolving.com/community/c6h626396p3756981]Scret..[/url]
[url=http://www.artofproblemsolving.com/community/c6h1173743p5650118]here[/url]

for 2.62
http://www.artofproblemsolving.com/community/c6h97690

for 2.63
http://artofproblemsolving.com/community/q1h553983p3218845

[quote=arqady]2.64
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$\frac{a\sqrt{a^2+3bc}}{b+c}+\frac{b\sqrt{b^2+3ac}}{c+a}+\frac{c\sqrt{c^2+3ab}}{a+b}\geq a+b+c$$
Proof.
By AM-GM $\sum_{cyc}\frac{a\sqrt{a^2+3bc}}{b+c}\geq\sum_{cyc}\frac{2a(a^2+3bc)}{(b+c)^2+a^2+3bc}$.
Hence, it remains to prove that $\sum_{cyc}\frac{2a^3+6abc}{a^2+b^2+c^2+5bc}\geq a+b+c$, which is obvious.[/quote]
Very nice.
Thank arqady.

[quote=sqing]Vasile Cîrtoaje,Symmetric Rational And Nonrational Inequalities:[/quote]

65
[img]http://s13.sinaimg.cn/middle/0018zOAxgy70f4RIRfSdc&690[/img]


Squaring both sides of the equation x+y=1, we have

x^2+2xy+y^2=1
x^2+-8+y^2=1
x^2+y^2=9

QED

[quote="chess64"]Ch3cK 0U+ +His 1:

If the two sides of right triangle $\triangle ABC$ are $x$ and $y$, and the hypotenuse is $\sqrt{2987}$, find $x^2+y^2$. :gleam:[/quote]
$x^2+y^2=(\sqrt{2987})^2$
$x^2+y^2=2987$

[quote="chess64"]Ch3cK 0U+ +His 1:

If the two sides of right triangle $\triangle ABC$ are $x$ and $y$, and the hypotenuse is $\sqrt{2987}$, find $x^2+y^2$. :gleam:[/quote]

$x^2$+$y^2$=$\sqrt{2987}^2$
because $x$ and $y$ are the legs of the triangle and
$\sqrt{2987}$ is the hypotenuse.

btw you're all right the answer is 9.

[quote="chess64"]Ch3cK 0U+ +His 1:

If the two sides of right triangle $\triangle ABC$ are $x$ and $y$, and the [b]hypotenuse[/b] is $\sqrt{2987}$, find $x^2+y^2$. :gleam:[/quote]

No hypotenuse?  :roll:

Unless you mean the first problem.

[quote="anirudh"]what hypotenuse? There is no hypotenuse in thi s problem.[/quote]
I was answering chess64's problem...and I think that there is a hypotenuse in it! :roll:

[quote="anirudh"]did you copy me  ;)[/quote]
did you copy PenguinIntegral?

Sorry. I didn't. Sorry abouut my accusations.


[hide]Every day, Tom drives Nebula $\frac13$ crazy, and Alex drives Nebula $\frac16$ crazy.  Thus, working together, each day, Nebula gets $\frac16+\frac13=\frac12$ crazy.

$\therefore2\mbox{ days}$[/hide]

[hide]$\frac{x}{3}+\frac{x}{6}= 1$
Solve for x and
$x = \boxed{2}$[/hide]

[hide]The answer is the solution to

$\frac{x}{6}+\frac{x}{3}=1$.

This is $x=2$

$2$[/hide]

[hide]Alex: 1/6 per day;Tom:1/3 per day.
Therefore together it takes them 2 days[/hide]

[hide]since Alex has half of the driving crazy power of Tom, the power is one and a half times Tom's amount is 3 days, so 1.5x=3.
x=2

2 days[/hide]

[quote="nealth"]Alex can drive Nebula crazy in six days, Tom can drive Nebula crazy in three days. How many days does it take Alex and Tom together to drive Nebula crazy?[/quote]

[hide]$\text{Alex works at a rate of}$ $\frac{1}{6}$ $\text{and Tom works at a rate of}$ $\frac{1}{3}$
$\frac{1}{6}+\frac{1}{3}= \boxed{\frac{1}{2}}$ [/hide]

... before he gets banned, could someone please tell me who imaretard is?

Thanks.

Because he knows my real name (well, at least he can mispell it)

[hide="my answer"]$\frac{1}{6}+\frac{1}{3}=\frac{1}{2}$ in 1 day so we need 2 days total to drive Nebula crazy.  [/hide]


Let $x,y,z>0$ such that $xy+yz+zx+xyz=4$ Prove that

      $1+x+y+z \le xyz+\frac 1x+\frac 1y+\frac 1z$

Let p,q,r = x+y+z, xy+yz+zx, xyz    respectively.

given $q+r = 4 \Rightarrow q = 4-r$

need $r + pr \leq r^2 + q$

or $r^2 + 1 + 3 \geq (2+p)r$

where it remains to show $3 \geq pr$.

---

ill finish later

You can perhaps begin like that :

$xy+xz+yz+xyz=4$ implies that $1/x+1/y+1/z+1=4/xyz$

We have thus :

$xyz+1/x+1/y+1/z=xyz+4/xyz-1$ and you have to find the minimum but I don't know later....

This is the my first proposed problem (June 22, 2005) on http://www.mathlinks.ro and not even now is solved !

[quote="keira_khtn"]Yes.As far as I remember I solved it 3 months ago.I've posted it and the main idea is trigonometry.[/quote]

how??
 :?:

[quote="fuzzylogic"][quote="socrates"]Nobody explains me??  :([/quote]

See here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=trig&t=41928[/quote]

Thanks!  :)


logarithms
limit
function
inequalities
calculus
calculus computations

Determine whether the series $ \sum_{n=2}^{\infty}a_n$ converges absolutely,converges conditionally or diverges.

$ a_n=\frac{(-1)^n}{\sqrt[n]{ln(n)}}$

My attempt:

$ a_n=\frac{(-1)^n}{e^{\frac{1}{n}ln(ln(n))}}=\frac{(-1)^n}{e^{\frac{1}{n}ln((n-1)+O(n^2))}}
=\frac{(-1)^n}{e^{\frac{1}{n}((n-2)+O(n^2))}}=\frac{(-1)^n}{e^{1-\frac{2}{n}+O(n)}}=(-1)^n(\frac{2}{n}+O(n))$
so I think that this limit diverges.

[quote="azatkgz"]

$ a_n = \frac {( - 1)^n}{e^{\frac {1}{n}ln(ln(n))}} = \frac {( - 1)^n}{e^{\frac {1}{n}ln((n - 1) + O(n^2))}} = \frac {( - 1)^n}{e^{\frac {1}{n}((n - 2) + O(n^2))}} = \frac {( - 1)^n}{e^{1 - \frac {2}{n} + O(n)}} = ( - 1)^n(\frac {2}{n} + O(n))$
so I think that this limit diverges.[/quote]

I don't agree with this.
I think it's not true
we can't say that
$ \ln n =n-1+O(n^{2})$ for big $ n$
when you use formula
$ \ln 1+x = x-x^{2}/2 ...$
it's true when $ -1< x \leq 1$
and when hyou wrote
${ \ln n = \ln 1+ (n-1} =  (n-1)+ O(n^{2})$
you can't do this because (n-1 ) is not less than $ 1$ !!!
serie divergent because lim $ \frac {\ln \ln n}{n}=0$

Like this?if $ b_n\equal{}(\minus{}1)^n$ is bounded and $ c_n\equal{}\frac{1}{\sqrt[n]{ln(n)}}$ does not monotonically converg to zero,then from ABel's Test series $ a_n$ does not converge.

It's true that the terms don't go to zero- but you didn't calculate their approximate size accurately (They're actually much closer to 1). That's what Extremal is complaining about.

Could not understand why it diverges? :(

OK. Look :
I want to prove that your serie is like this $ \sum (\minus{}1)^{n}$ which divergent (it's wellknown ) or generally :
serie  $ \sum x_{n}$ diverge , if $ x_{n}$ don't go to $ 0$ when $ n\minus{}>\plus{}oo$
now let's consider it for your serie :

you have $ \frac{1}{(\ln n )^{\frac{1}{n}}}$
in your case $ x_{n}\equal{} \frac{1}{(\ln n )^{\frac{1}{n}}}$
at first prove that limit $ \lim _{n\minus{}>\plus{}oo}\frac{1}{(\ln n )^{\frac{1}{n}}}$ is equal to $ 1$. try to prove this if you can't then I will post solution

$ \lim_{n\rightarrow \plus{}\infty}\frac{1}{(lnn)^{\frac{1}{n}}}\equal{}\lim_{n\rightarrow \plus{}\infty}\frac{1}{e^{\frac{1}{n}ln(ln(n))}}\equal{}\frac{1}{e^{ \lim_{n\rightarrow \plus{}\infty} \frac{1}{n}ln(ln(n))}}\equal{}1$

$ \lim_{n\rightarrow \plus{}\infty}\frac{1}{n}ln(ln(n))\equal{}\lim_{n\rightarrow \plus{}\infty}\frac{1}{nlnn}\equal{}0$from I'Hopital Rule

[quote="azatkgz"]$ \lim_{n\rightarrow \plus{} \infty}\frac {1}{(lnn)^{\frac {1}{n}}} \equal{} \lim_{n\rightarrow \plus{} \infty}\frac {1}{e^{\frac {1}{n}ln(ln(n))}} \equal{} \frac {1}{e^{\lim_{n\rightarrow \plus{} \infty} \frac {1}{n}ln(ln(n))}} \equal{} 1$

$ \lim_{n\rightarrow \plus{} \infty}\frac {1}{n}ln(ln(n)) \equal{} \lim_{n\rightarrow \plus{} \infty}\frac {1}{nlnn} \equal{} 0$from I'Hopital Rule[/quote]
yes you are right !!! :) remember function $ n^{a}$ is more increasing than the function  $ (C\ln n )^{b}$ where $ a,b>0$
therefore limit :
$ \frac{(C\ln n )^{b}}{n^{a}} \equal{} 0$

Limit comparison is a test for absolute convergence- that doesn't apply here. What we're looking at is the general statement Extremal gave: a convergent series must have its terms tend to zero, and the contrapositive (If the terms don't go to zero, the series diverges).

Your latest post is an accurate proof that the terms of this series don't go to zero.

I have a nicer way to prove that $ \sqrt[n]{\ln n}\to 1$.
It's well known that $ \sqrt[n]{n}\to 1$ (easy proof with using AM-GM inequality) and then we have for $ n>2$

$ \leftarrow 1\leq\sqrt[n]{\ln n}\leq\sqrt[n]{n}\rightarrow 1$, hence:
$ \sqrt[n]{\ln n}\to 1$

The actual asymptotic $ a_n \equal{} (\minus{}1)^n \plus{} O(\frac{\log \log n}{n})$ . So indeed the terms don't go to zero, the series does not converge.