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hello, i have just finished my first undergraduate year studying math and i am considering if i want a research career in math/physics. i have some general questions about mathematical research that have been itching me, please help me answer them:

- i've heard people on this forum talk about choosing a mathematical field/discipline, how many are there, and how large are these relative to how much a typical researcher masters. are we talking about fields like algebra, number theory, geometry or their subfields?
- how interconnected are these mathematical disciplines.
- how specialized is your typical researcher
- if you had to classify mathematical researchers according to some standard how would you do this? (amount of publications, quality of publications..) im trying to get a picture of how the math community is divided into levels (or to what extent it is divided)
- how much mathematical knowledge so we have as a whole and how fast is it growing? (like how much mathematics is published every year, in absolute terms and relative terms) perhaps make an analogy to some kind of tree..branching, growth spikes..

i guess if i had to summarize these questions id ask you to give some sort of notion of:
1. the structure of mathematical knowledge in terms of a size comparison and how it branches and interconnects.
2. compare how different mathematical researchers stand relative to each other in terms of ability and achievements. (do mathematicians naturally get classified according to level like sportsmen do?)

i would just like some semi-vague understanding of these questions, perhaps a paragraph or two? thxs

Your questions can't really be answered because they are too vague.  More importantly, I can't imagine how or why the answers would matter.  Nonetheless, I'll try my best to answer.

Mathematics is a large and diverse subject, with many fields or areas of specialization and subfields of specializations, and maybe even subsubfields.  There is no "convention" on what counts as a field or area or whatever.  However, if you want a rough idea of how mathematics breaks apart, take a look at the official AMS classification http://www.ams.org/msc, keeping in mind that even this official classification is inadequate and many research papers do not neatly fit into a particular MSC number and most papers will list multiple MSC numbers.  A coarser classification is given by the math arxiv at http://front.math.ucdavis.edu/math, which has the added benefit that you can see how many papers get posted in each area.  Another coarse classification would be how the NSF breaks down the Division of Mathematical Sciences into its constituent "Programs."

Every area of math (no matter what you mean by "area of math") is highly interconnected with several other areas so as to form an amorphous web of connections.

Some mathematicians work in very narrow fields.  Some work in a wide variety of fields.  Most mathematicians spend the bulk of their careers working on various problems that are at least connected to each other in some small way.

There is not any legitimate sort of ranking method for mathematicians, but the fact is that some researchers are better than others.  In fact, some are MUCH better than others, who in turn are MUCH better than others (just like in your calculus classes).  But there are no "levels."  Again, it's more like an amorphous cloud.  Certainly, a person's publication record is one aspect of their overall reputation.

Mathematics is always growing rapidly.  Most likely, there are more pages of mathematics written each year than in each previous year.

Classifying mathematicians is nothing like classifying sportsmen.  For one thing, one would like to think that mathematicians are in some sense all working together toward a common goal rather than competing against each other.  A better analogy would be musicians, but even that analogy is lousy.

I hope that answers your question.

thank you yenlee, that's the response i was looking for, but im not quite satisfied...

if people specialize, but mathematical branches are highly interconnected, don't they miss many connections by being so specialized?

when i referred to sportsmen and levels i meant that sportsmen are easily classified into levels, (local, district, city,state, national teams.. and finer classifications, like top team in ladder, top players in a team...etc) anyways i would like to find out if a mathematician can and would judge anther's ability or 'level' in order to see if they should collaborate on some research. is there some kind of obvious criterion like the research topic they undertake and/or amount of publications?
is there an upper echelon of top mathematicians that stand high above the rest and only collaborate among themselves? (looking down at the rest of the bunch)

well anyway the heart of the matter is this.. 'if i were a mathematician, where would i stand..' i know i would not be at the top, and i hope i would not be in the bottom :maybe: i have a fear that i might be stuck doing boring research that would be some useless and trivial stuff for the big guns out there... is this true of many mathematicians, is their work boring and quite useless to a top level mathematician... don't be afraid to tell the truth!

Yes, mathematicians undoubtedly miss interconnections because of specialization.  This is one of the reasons why communication and collaboration is so important.

There are many different kinds of collaborations.  Sometimes it's very important for collaborators be "on the same level," but sometimes it is not.  For example, sometimes each collaborator brings a different kind of expertise.  In general, you collaborate with someone if you believe there is some value in doing so.  There are many ways that this value can arise.  Top mathematicians do collaborate with "lesser" mathematicians all the time, for a variety of reasons.

If you ever think that your own research is boring or useless or trivial to others, then you shouldn't be doing it (and trust me, if that were the case, you wouldn't be doing it for long).  In any case, that is rare.  Mathematics is huge, and the "big guns" don't have the time and resources to do everything.  So even if Terry Tao can do something in a week that takes you a year, that doesn't mean that it's not worth doing.  In fact, Tao might even find your paper to be interesting, or he might even have been the one suggesting the problem to you in the first place.

ok that reassures me a bit, but [quote]So even if Terry Tao can do something in a week that takes you a year, that doesn't mean that it's not worth doing.[/quote] sounds pretty depressing...
aside from that im still curious about some general statistics about mathematicians... how many mathematicians are there? how many are at a level like Tao? at what age is a mathematician most productive... many questions to help me get a better feel for the community and scope of mathematics. i want to know as much as i can so i can be confident with my choice of learning more math and physics.

Just to throw in my two cents :).

There is only one Terence Tao, but there was only one Thoger Bang as well. The former one has already published 120+ papers, got the Fields medal, etc.; the second one published only 16 research papers in his life and got no formal awards I'm aware of, but, in my humble opinion, was no less fine mathematician than Tao. Can one directly compete with people like Tao in problem solving? Yes, one can. Rudelson's work on singular values of random matrices is superior to the results of Tao and Vu (yet, have you ever heard of Mark Rudelson?). The point is that there is enough room and opportunities for everybody and nobody should really be afraid to end up doing the "boring stuff".

As to the mathematical hierarchy, it exists but is very flexible. Nobody is prohibited from going to the top but, certainly, not everybody can make it there. Collaboration is mostly completely open. If you have something interesting to offer and just lack certain knowledge or cannot figure out some technical details, most people will give you a hand if they can (there are exceptions, of course, but either they are rare or I was extremely lucky so far). So, if you want to collaborate with somebody "above your level", just make sure that you have an interesting idea that can lead somewhere or something else of value that you can offer. For the last 10 or so years, I haven't published a single paper without co-authors. In all cases, our relative standing in the hierarchy was completely unimportant; all that mattered was that we all had something interesting to say to each other.

Specialization is two-fold. It is like looking at the things in microscope. Your field of vision is greatly reduced but you see many new interesting things that weren't noticeable from the broader perspective. In general, you should be much concerned with your lack of deep knowledge in other fields. All you really need is the ability to adequately restate your problem in other people's language if you want their help or advice.

Where will you stand? Who knows? It all depends on you. If you want not to get lost in the crowd, just learn and think slightly more than you are comfortable with every day and let the time work for you. Be "omnivorous" trying every question coming your way that you can understand, no matter where it came from. Pay no attention to various self-proclaimed gurus telling you their strong opinions about what mathematical problems and fields are important and good and what aren't. The only criterion is whether [b]you[/b] can learn something new from the problem or not and whether you like it. In other words, be youself, follow your taste, take all you can from everybody and do the best you can for everybody and you'll notice quite soon that you are becoming rather unique. There will always be a "bigger fish" somewhere but who cares? It is not a survival game where you can either eat or be eaten or the race where there is only one person who comes to the finish line first. It is rather like constructing a big building where you can add your bricks to the walls or develop blueprints for the next wing.

Are there a lot of people who publish junk even from my perspective? Yes, there are, but they chose to go this way themselves. What they usually do is a blind application of somebody else's idea without real understanding of it to the situation that differs by $ \varepsilon$ from the problem considered in the original work. This is about the worst you can possibly do but even this is somewhat useful: it attracts more attention to the original article, at least.

As to numbers (how many...), google is your best friend for that but do they really matter? When I drive on the highway, I do not count other cars. The fact that there are many people going in the same direction (and equally many going in the opposite one) doesn't diminish or increase my desire to get from $ A$ to $ B$ in the slightest. The same with mathematics: you either want to figure something out or you don't. How many other people are there doing something similar is completely irrelevant. The only ones that count are those who can give you a lift or to whom you can give a lift. In other words, the mathematical community is like any other community: you have some vague feeling of the total size and strength but you really know and interact with your neighbors and friends only and it is entirely up to you how many of those you make. I would say I know a couple of dozen people rather well mathematically and can freely talk to about 100 (in the sense that they would know who I am and would not be surprised by my sending them an e-mail) but my numbers are by no means binding for you: you may prefer to stay completely alone or to communicate to the entire world.

In general, I know this question of "am I any good?" quite well. The short answer to this question is "you'll never know until you seriously try". Another short answer is that "no matter how good you are, it is never good enough". Both are trivialities, of course, but what did you expect?

thank you fedja, your response answered many of my doubts and helped re-affirm my resolve to study mathematics.

[quote]It is rather like constructing a big building where you can add your bricks to the walls or develop blueprints for the next wing.[/quote]
i liked this analogy in particular as iv'e wanted to become an architect ever since i was young. only recently have i been contemplating changing my career path.

a few more questions while i'm posting..
who is Thoger Bang?
once you are engaged in research, how do you spend your time? read papers a lot? think in your office? converse on the phone with colleagues?
is it hard to find papers that are important to your research? is there some detailed list somewhere that classifies what papers are relevant to what subject areas, and which are more fundamental.

oh and 1 more thing
[quote]In general, I know this question of "am I any good?" quite well. The short answer to this question is "you'll never know until you seriously try". Another short answer is that "no matter how good you are, it is never good enough". Both are trivialities, of course, but what did you expect?[/quote]

yes i agree, but there is always a way to compare yourself to others. how good an indicator is olympiad problem solving ability and how does it correlate with age? a thing that bothers me is that i have only suddenly (past year or so) developed a strong interest in math, and as a result im far behind in my knowledge and abilities compared to the general olympiad type people on these forums... so i would like to think that im sub-par beacuse i never really took mathematics to be something 'extra-circular' and now that i am serious i can catch up with my problem solving skills and missing knowledge. so if i can achieve a certain level in olympiad problem solving standards, is this a good indication to my mathematical aptitude?

If you can find a copy of it online, I think the table of contents of [url=http://www.amazon.com/Princeton-Companion-Mathematics-Timothy-Gowers/dp/0691118809]the Princeton Companion to Mathematics[/url] is a good overview of (most of?) the major fields of modern mathematics, and the Companion itself is amazing; for a minimum of background, you can get a pretty good sense of what mathematicians are up to these days.


induction
graph theory
combinatorics unsolved
combinatorics

[quote="modeler"]Prove if there exist a graph satisfies the followings
1. no two edges link to the same pair of vertexes
2. no two edges cross each other
3. each vertex has at least 6 edges linking to it[/quote]

Hmm.. we can prove a stronger claim:
The largest possible minimal vertex degree of a simple planar graph is 5, and for a minimal vertex degree of $2 \leq d \leq 5$ the graph must have at least $\frac{12}{6-d}$ vertices.

Proof:

The constructions for $d=2,3$ are trivial
For $d=4,5$ see attached figures

Denote a graph with $V$ vertices and $E$ edges as $(V,E)$

We first prove that in a simple planar graph $(V,E)$ where $V \geq 3$, $E \leq 3V-6$
The graph $(3,3)$ is a planar complete graph, so for any graph $(3,x)$, $x \leq 3(3)-6$
If for any graph $(n,x)$, $x \leq 3n-6$,
Consider any graph $G=(n+1,y)$
Consider any vertex in $G$ and the "polygon" that encloses it
Let the polygon have $p$ vertices
If the vertex is removed then $p-3$ non-intersecting diagonals can be drawn in the polygonal region containing the vertex, and these do not intersect any edge in the new graph
Thus $y \leq (3n-6)-(p-3)+(p) = 3(n+1)-6$
So by induction $E \leq 3V-6$ for all simple planar graphs $(V,E)$ where $V \geq 3$

If $d \geq 6$ then $6V \leq Vd \leq 2E \leq 6V-12$ $\Rightarrow\Leftarrow$

If $2 \leq d \leq 5$,
$V \geq 3$
$\Rightarrow$ $Vd \leq 2E \leq 6V-12$
$\Rightarrow$ $V \geq \frac{12}{6-d}$


limit
real analysis
real analysis unsolved

Consider the iterated sequence
(1) $ x_0,x_1 \equal{} f(x_0),\dots,x_{n \plus{} 1} \equal{} f(x_n),\dots$,
where $ f(x) \equal{} 4x \minus{} x^2$. Determine the points $ x_0$ of $ [0,1]$ for which (1) converges and find the limit of (1).

[hide]
It's similiar to fixed point methods that is well known in Numerical Analysis. ANd the serie converges to the root of :
$ 4x\minus{}x^2\equal{}x$
$ 3x\equal{}x^2$
$ x\equal{}0$ or $ x\equal{}3$
[/hide]

to ensure convergence we need

$ f([a,b]) \subset [a,b]$ for the numerical concept to hold true

f is clearly monotone increasing on [0,1], $ f (x) > 0, x\in (0,1)$

$ f(0)\equal{}0 \to a\equal{}0$
$ b\le 1 \le 3 \to b \ge b(4\minus{}b) \equal{}f(b)$

thus  $ x_0 \in (0,1] \to \lim_{n\to \infty} x_n \equal{} 3$


group theory
abstract algebra
linear algebra
matrix
superior algebra
superior algebra unsolved

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=21942]here[/url] for a discussion of the simple group of order 168. The two standard models: $ PSL_2(F_7)$ and $ PSL_3(F_2)$.

The elements of order 7 are fairly easy to count in the first model; they belong to equivalence classes with one representative each of the form $ I\plus{}N$ with $ N$ nilpotent, and the subgroup elements are $ I\plus{}kN$ for integers $ k$ mod 7.
A count of nonzero nilpotents: there are six of the form $ \begin{bmatrix}0&a\\0&0\end{bmatrix}$, six of the form $ \begin{bmatrix}0&0\\a&0\end{bmatrix}$, and $ 36$ of the form $ \begin{bmatrix}a&b\\\minus{}a^2b^{\minus{}1}&\minus{}a\end{bmatrix}$, for a total of $ 48$.

Possible numbers of elements of order $ 7$ for any finite group: $ 0,6,48,90,132,\dots$.

without the exact knowledge of the simple group of order 168 solve as follows:

Since $ 168\equal{}7\cdot 24$ holds, all Sylow-7-subgroups are cyclic of order 7. Let $ U\in Syl_7(G)$, then we must have $ 1\equiv |G: N_G(U)|\bmod{7}$ and $ |G: N_G(U)|\in\{1,2,3,4,6,8,12,24\}$. Since $ G$ is simple, this yields $ |G: N_G(U)|\equal{}8$ as the only solution, i.e. we have 8 different Sylow-7-subgroups and $ 48\equal{}8\cdot (7\minus{}1)$ different elements of order 7.

[quote="spider_boy"]How many elements of order $ 7$ are there in a [b]simple[/b] group of order $ 168$?[/quote]

By Sylow's theorem there are exactly $ 8$ $ 7$-Sylow subgroup (we know that the group is simple and, hence does not contain any normal subgroups). Each of them contains $ 6$ elements of order $ 7$ and they don't intersect since there intersection subgroups would have an order, which divides $ 7$. This produces $ 48$ elements of order $ 7$, and there are no others since any element of order $ 7$ must lie in a subgroup of order $ 7$.


Ring Theory
algebra
function
domain
superior algebra
superior algebra solved

Let $ R$ be a domain, and let $ S\subseteq R$ be a multiplicative set.

Prove that: If $ R$ is a PID, then $ S^{ \minus{} 1} R$ is a PID.

You just have to prove: the ideals $I$ of $S^{-1}R$ are all of type $S^{-1}J$ with $J$ an ideal of $R$.


logarithms
trigonometry
Euler
algebra proposed
algebra

Hello Everyone; another question here.

if e^x can be expressed as a power series; does that not mean by the fundamental theorem of algebra that it can be expressed as an infinite product with infinite "roots"?

*please dont kill me; I know  that e^x does not have any roots that I know of, or that I've ever heard of;    :(

so can someone explain to me why it cannot have infinite product expression; or if it does can someone show me what it is?

A power series is [b]not[/b] the same thing as a polynomial.  The relevant generalization is the [url=http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem]Weierstrass factorization theorem[/url].  The factorization of $ e^x$ is, unfortunately, $ e^x$.

You are correct that $ e^x$ has no roots, and this is true even over the complex numbers.  That's what should clue you in that something is wrong.  Incidentally, that kind of logic [b]can[/b] be applied to $ \sin x$, and (although this is very difficult to prove) gives

$ \frac {\sin x}{x} \equal{} \left( 1 \minus{} \frac {x^2}{\pi^2} \right) \left( 1 \minus{} \frac {x^2}{4 \pi^2} \right) \left( 1 \minus{} \frac {x^2}{9 \pi^2} \right) ...$.

Euler used this identity to make some rather amazing conclusions.

I talked a little bit about this here: http://www.mathlinks.ro/Forum/viewtopic.php?t=205007

It's good to be naive though, I never even thought of the possibility of factoring power series as products until I learned about in a class.


\[ \frac{12}{20}\equal{}\frac{x}{13} \Rightarrow x\equal{}\boxed{7.8}\]


articles
geometry
3D geometry
sphere
calculus
algebra
polynomial

1. In the article on rings, I didn't understand what this meant:

http://i519.photobucket.com/albums/u353/dzythegreatest/temp.gif

2. In the article about Hypercubes, I didn't understand any of the highlighted words   :maybe: . (and the related articles on them) I don't really understand the concept of a cube that spans dimensions past the 3rd dimension. Is the 4th dimension supposed to be time in this case?

Please help a confused little boy...   :blush:

1.  The expression given is the [b]definition[/b] of the set $ \mathbb{Z}\left[\sqrt [3]{2}\right]$ -- the set of all numbers that can be written as a sum of integer multiples of $ 1$, $ \sqrt [3]{2}$, and $ {\sqrt [3]{4}}$

2.  In mathematics, dimension is often used in an abstract sense.; it need not be tied to any fixed physical interpretation.  In geometry when we speak of an $ n$-dimensional object, typically we mean a set of points each having $ n$ coordinates.  For example, a four-dimensional unit 'cube' is described by the boundary of the set $ \{(a,b,c,d): 0\leq a,b,c,d\leq 1\}$.  A six-dimensional unit sphere is described by the set $ \{(a,b,c,d,e,f): a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2 \plus{} f^2 \equal{} 1\}$.

Thanks a lot for you reply!

However, I still don't really understand how the ring is defined; Why does it have to be the integers divisible by $ 1, \sqrt[3]{2},$ and $ \sqrt[3]{4}$? Can you tell this just from the LHS?

I now understand the hypercube better now, but how did Mathworld provide a picture of a 4th dimensional cube?

Thanks again!

this is just my opinion but i say that before you do abstract algebra do calculus BC and don't try to learn it from a website like mathworld it just gives a brief introduction buy a book or get a tutor.

Mathworld is a very technical website, meant more as a reference than as a learning tool. You'll find that [url=http://en.wikipedia.org/wiki/Hypercube]Wikipedia[/url] often has more understandable explanations for new concepts.

[quote="AnonAmoz"]I now understand the hypercube better now, but how did Mathworld provide a picture of a 4th dimensional cube?[/quote]

The same way it can provide a picture of a cube on a 2D screen. In a basic cube drawing, you see two squares, you see the edges connecting them (to complete the graph), and they are offset and scaled so that perspective can be used to represent a third dimension.

Here, it does the same thing with two cubes -- it tries to show both the third and fourth dimensions using perspective.

[quote="AnonAmoz"]I still don't really understand how the ring is defined; Why does it have to be the integers divisible by $ 1, \sqrt [3]{2},$ and $ \sqrt [3]{4}$? Can you tell this just from the LHS?[/quote]
It might help first to have a more abstract understanding of polynomials.  We say that a set $ S$ is [b]closed[/b] under an operation $ *$ if for any $ a, b \in S$ we have $ a * b \in S$.  For example, the integers are closed under addition and multiplication.

One way to define the ring of integer polynomials $ \mathbb{Z}[x]$ (read: $ \mathbb{Z}$ [b]adjoin[/b] $ x$) is as the "smallest" ring that contains $ \mathbb{Z}$ and $ x$.  Part of the definition of a ring implies that it must be closed under addition and multiplication, which means that if I take any two elements of this ring, their sum and their product must also be in it.  This means we can add:

$ 1 \plus{} x \in \mathbb{Z}[x]$

and multiply:

$ 2 \plus{} 3 \cdot x \plus{} 4 \cdot x \cdot x \in \mathbb{Z}[x]$.

In other words, [i]polynomials are the result of repeated addition and multiplication.[/i]  (I usually see them defined in terms of monomials.)

The set $ \mathbb{Z}[ \sqrt[3]{2} ]$ (read: $ \mathbb{Z}$ adjoin $ \sqrt[3]{2}$) should be understood in the same way - it is the "smallest" ring that contains both $ \mathbb{Z}$ and $ \sqrt[3]{2}$.  This means we can add:

$ 2 \plus{} \sqrt[3]{2} \in \mathbb{Z}[ \sqrt[3]{2} ]$

and multiply:

$ 1 \plus{} 2 \sqrt[3]{2} \plus{} 7 \sqrt[3]{2} \cdot \sqrt[3]{2} \in \mathbb{Z}[ \sqrt[3]{2} ]$.

In other words, one way to think of this ring is as the ring of "polynomials in $ \sqrt[3]{2}$."  Of course, these polynomials are restricted in degree because once we cube we get an integer back, so this ring can be represented as the set of numbers of the form $ a \plus{} b \sqrt[3]{2} \plus{} c \sqrt[3]{4}$ where $ a, b, c$ are integers.  This set is clearly closed under addition; verify that it is also closed under multiplication as an exercise.

@t0rajir0u: Thanks a lot for your post! I think I understand rings much better now.
Proof for closed under multiplication:

$ (a\plus{}b\sqrt[3]{2}\plus{}c\sqrt[3]{4}) * (d\plus{}e\sqrt[3]{2}\plus{}\sqrt[3]{4})\equal{} ([ad\plus{}2bf\plus{}2ce]\plus{}[ae\plus{}db\plus{}2ef]\sqrt[3]{2}\plus{}[af\plus{}be\plus{}dc]\sqrt[3]{4})$...right?

About hypercubes:

I think I have a better understanding of it now, but it's not really possible to fully imagine the whole tesseract right? Because it has a fourth dimension that humans aren't able to imagine? So we can only see the 3-d "shadows" of it?

More or less.  It's a little melodramatic when you put it that way - we do the same thing with 3D objects (looking at their 2D "shadows") all the time, and you'd be surprised how easy it is to do mathematics with something you "can't" visualize.

In convex geometry (polytopes, etc) we often look at what's known as the 1-skeleton. It's a graph of all of the edges and vertices of the polytope. In three dimensions, this graph is planar (it can be drawn with no crossings) because we can extrude all vertices along unit vectors until they lie on a unit sphere, and then we "poke" a hole in the sphere, which is then unfolded. One reason it's easier to imagine these three dimensional polytopes is that their f-vector contains relatively few elements. An f-vector is a combinatorial object which counts the number of n-faces (with a zero face being a vertex, a one face being an edge, a two face is a normal "face", a three face being cells, etc..) Because we have f-vectors with few elements, it is easier to visualize these.

Another way of looking at polytopes is by looking at their face lattice, or a diagram of subsets of the polytope, partially ordered by inclusion. We look at all of the zero faces, then see what one faces they are connected to, etc. It's a pretty easy way to visualize the "layout" of the faces in more than three dimensions. This is also called a Hasse Diagram.


How many pages are in a book if $ 852$ digits are used to number the pages, starting with page $ 1$?

There are $ 9$ numbers which each has 1 digit: $ 9$ digits
There are $ 90$ numbers which each has 2 digit: $ 180$ digits
There are: $ 852 \minus{} 180 \minus{} 9 \equal{} 663$ digits for the numbers which each has $ 3$ digits
So there are: $ 663: 3 \equal{} 221$ numbers wich each has 3 digits
Therefore, there are: $ 9 \plus{} 90 \plus{} 221 \equal{} 320$ pages


probability
integration
logarithms
limit
function
probability and stats

after reading [url=http://www.mathlinks.ro/viewtopic.php?t=299344]this[/url] post, let's try the following:

Take $ n$ a large integer, $ 0<\alpha< \beta<1$ two real numbers and $ \epsilon>0$.
You generate $ n\minus{}1$ uniformly distributed random numbers between $ 0$ and $ 1$ and place them on that interval. You now have $ n$ segments. Sort these $ n$ segments in increasing lengths and call these lengths $ L_1, L_2, \ldots, L_n$.

If $ P_n(\alpha, \beta, \epsilon)$ is the probability that $ | \sum_{[\alpha n] < i < [\beta n]} L_i \minus{} \int_{\alpha}^{\beta} \ln(\frac{1}{1\minus{}u}) \, d_u| < \epsilon$, could you show that $ \lim_{n} P_n(\alpha, \beta, \epsilon) \to 0$. The heuristic is a simple Poisson approximation, but I cannot make this very rigorous (computer simulations seem to agree with that result)

Hi Alekk
Unfortunately I'm far from having an aswer to your orignal post but as I was thinking about it I tried to find a way to simulate n segments of decreasing length over [0,1] directly without having to sort them after for example drawing n points uniformaly in [0,1].

My guess to this (though far rom your question) problem would be the following algo :
1- Draw from uniformly distributed variable in [0,5;1] one point, let's call it $ a_1$
2- Then draw a second point $ a_2$ in another point in $ [a_1\plus{}1/2(1\minus{}a_1);1]$
.
.
.

n- repeat with the draws with $ a_n$ in $ [a_{n\minus{}1}\plus{}1/2(1\minus{}a_{n\minus{}1});1]$

Then the segments that you obtain are nicely sorted and I think that they have the right corresponding joint law of the "uniformly then sorted" segments.

Alekk, do you think that this is correct ?

Anyway I don't see how this might help to answer your original question but if it works well then it seems to be a trick that could save time during simulation because sorting takes a long time generally

regards

I don't know: could you explain why these points $ a_k$ follow the right distribution. (btw, I did some simulations with $ 10^3$ and $ 10^4$ breaks, and that takes no more than a few seconds)

Here is a plot of the function $ f_n$ that goes through the points $ (x\equal{}\frac{k}{n}, y\equal{}n L_k)$ (normalisation so that $ \int_{[0;1]} f_n(x) \, dx \equal{} 1$) , with $ n\equal{}10^2$ and $ n\equal{}10^3$, compared to the conjectured limit $ f(u) \equal{} \ln( \frac{1}{1\minus{}u} )$

[quote="alekk"] could you show that $ \lim_{n} P_n(\alpha, \beta, \epsilon) \to 0$[/quote]
Of course, you meant $ \lim_{n} P_n(\alpha, \beta, \epsilon) \to 1$.

And, of course, your conjecture is true.

I'll give only rough ideas, which pretty easily can be transformed into rigorous argument.

Step 0. It is enough to prove the statement for $ \alpha \equal{} 0$.

Step 1. Recall that the lengths $ S_1,\dots, S_n$ of the segments have the same joint distribution as  $ E_1/E, \dots, E_n/E$,
where $ E_i$ are i.i.d. with Exp(1) distribution, $ E \equal{} \sum_{i \equal{} 1}^n E_i$. See e.g. Devroye - Non-Uniform Random Variate Generation, p. 208.

Let $ E_{(1)}\le E_{(2)}\le \dots\le E_{(n)}$ be the order statistics of $ \{E_i\}$ (or, if you prefer, the increasing sorting).

We have $ L_i \equal{} E_{(i)}/E$, whence
\[ \sum_{i\le \beta n} L_i \equal{} \frac {\sum_{i\le \beta n} E_{(i)}}{\sum_{i \equal{} 1}^n E_i} \equal{} \frac {\frac1n\sum_{i\le \beta n} E_{(i)} }{\frac1n\sum_{i \equal{} 1}^n E_i}\sim \frac1n\sum_{i\le \beta n} E_{(i)},\]
where the last is by the strong law of large numbers.

Step 2. Let $ \beta \equal{} 1 \minus{} e^{ \minus{} \gamma}$, $ \gamma > 0$. We want to show that
\[ \frac1n\sum_{i\le \beta n} E_{(i)}\sim \frac1n\sum_{i \equal{} 1}^n E_i 1_{E_i\le\gamma}\]
Then we are done, because the last converges to $ E[E_1 1_{E_1\le \gamma}]$ again by the strong law of large numbers.

We have
\[ \gathered \Big|\frac1n\sum_{i\le \beta n} E_{(i)} \minus{} \frac1n\sum_{i \equal{} 1}^n E_i 1_{E_i\le\gamma}\Big| \le E_{(n)} \Big|\beta \minus{} \frac1n \# \{i: E_i\le \gamma \} \Big| \\
\equal{} \frac {E_{(n)}}{\sqrt {n}} \cdot \sqrt {n}\Big|\beta \minus{} \frac1n\# \{i: E_i\le \gamma \} \Big| \equal{} : \frac {E_{(n)}}{\sqrt {n}}\cdot |Z_n|. \endgathered\]
By the central limit theorem $ Z_n$ converges to some normal law (hence bounded in probability). But $ E_{(n)}\sim \log n$ by the well-known property of exponential order statistics (see Devroye, p. 211). So we have  $ \frac {E_{(n)}}{\sqrt {n}}\cdot |Z_n|\to 0, n\to\infty$ a.s.

And we are done :)

Just ask if something is not clear.

[quote="zhoraster"] Step 1. Recall that the lengths $ S_1,\dots, S_n$ of the segments have the same joint distribution as  $ E_1/E, \dots, E_n/E$,
where $ E_i$ are i.i.d. with Exp(1) distribution, $ E \equal{} \sum_{i \equal{} 1}^n E_i$. See e.g. Devroye - Non-Uniform Random Variate Generation, p. 208.
[/quote]

Thanks zhoraster: that was the point I was missing!

Since I did not know that, I simply assumed that the points were the jump times of the usual Poisson process(intensity $ \lambda \equal{} n$) conditioned to be equal to $ n$ at time $ t \equal{} 1$. And because this process satisfies $ X_1 \approx n$, one can forget the conditioning (this was the point to be made rigorous). This is why the length of the segments are approximately iid exponentially distributed (with mean $ \frac {1}{n}$). The rest is clear.


Let $a,b,c$ be three positive [b]integers[/b] such that $a^3+b^3+c^2=3$. Prove that \[a^4b^4+b^4c^4+c^4a^4\leq3\]

I'm sure the word "integers" should be replaced by "reals" or "numbers", otherwise....well, I think you see what will happen.

I think this inequality was posted many times and was solved by
Harazi and Vasc. I just rewrite it.

Use AM-GM : $a^4b^4 \leq a^3b^3 \cdot \frac{a^3+b^3+1}{3}$

Put $x=a^3$, $y=b^3$, $z=c^3$ .

$ x+y+z = 3 $ and we need $ \sum 3xy(x+y+1) \leq 27$

$ \sum 4xy(x^{2}+y^{2}) +3xyz \leq \sum x^{3} + \sum 3xy(x^{2}+y^{2}) +6xyz $

or $ \sum x^{3} + 3xyz \geq \sum xy(x^{2}+y^{2})$.

Can we already post solutions for Mathlinks Contest?

[quote="Myth"]I think, prowler just pointed that it is meaningless to solve that problem on contest, since it has been solved on forum.[/quote]Nevertheless, you [b]SHOULD NOT EVER[/b] post solutions to current problem in the forum  :mad: I already specified this clearly. This is a warning for all users, especially Prowler: please do not do this again. Next time there will be no warning ...  :no:

Please announce me if such disturbance exists. Indeed I searched for the problem on the forum but I could not find it, maybe I was too tired from work.

Due to this I will change the problem right now. The problem this is being replaced with is the following:

If $n>2$ is an integer and $x_1, \ldots ,x_n$ are positive reals such that

\[ \frac 1{x_1} + \frac 1{x_2} + \cdots + \frac 1{x_n} = n \] then the following inequality takes place
\[ \frac{x_2^2+\cdots+x_n^2}{n-1}\cdot \frac {x_1^2+x_3^2+\cdots +x_n^2} {n-1}  \cdots  \frac{x_1^2+\cdots+x_{n-1}^2}{n-1}\geq \left(\frac{x_1^2+...+x_n^2}{n}\right)^{n-1}. \]

[quote="Myth"]Do you mean multiplication sign instead of +?[/quote]Yes, small typo :D Thanks for noticing this .. I am still dizzy :)

I am very sorry, I didn't mentioned that this is round 6 (this session).
I will try not to post solutions on prohibited problems.
Anyway why you post them for discussion so early ?

Again I wanted to say this problem was discussed
in topics
   1) Ineq from city MO
   2) not an easy maxima (last posts are deleted from 21.11.04)

So I agree with you Valentin Vornicu
that it hard to find 3x7 problems almost unknown.

[quote="prowler"]Anyway why you post them for discussion so early ?
[/quote]What do you mean post them "so early" ?!

Megacrazywang was asking about a typo in the first release (i.e. integers instead of reals). No one was asking for a solution.

[quote="Valentin Vornicu"]
Megacrazywang was asking about a typo in the first release (i.e. integers instead of reals). No one was asking for a solution.[/quote]
Valentin, my name is not Megacrazywang...... ;)
Actually the "me" of my name just refers to myself, "crazy" simply means I'm crazy :P  and w[b]o[/b]ng is my surname, which differs from the spelling of china:they use wang instead.

[quote="mecrazywong"]Valentin, my name is not Megacrazywang...... ;)
Actually the "me" of my name just refers to myself, "crazy" simply means I'm crazy :P  and w[b]o[/b]ng is my surname, which differs from the spelling of china:they use wang instead.[/quote]Sorry  :blush: I was typing fast without looking up. The name I made up sounded better to me :P ... sorry again, didn't mean to offend you.

In problem 2, are the sides of a polygon defined to be just the segments that make it up, or the full lines?

[quote="Remike"]I have a question about problem 2. Is it true that P is the set of [b]all[/b] points of the plane?[/quote]Yes all the points in the plane.

What type of inequality is given in round 6 problem 3?

First one starting from x2, second one x1, x3.., last one x1. Is it typo? If not then what are the series?

[quote="mahbub"]What type of inequality is given in round 6 problem 3?
[/quote]Cyclic inequality.

CYCLIC??? :?

but how?

x2, x4....

x1,x3,....

x2,x4...

No it should not be! So then? I am really wondering! :?

in the first one the $x_1^2$ term is missing, in the second factor the $x_2^2$ term is missing, in the last factor the $x_n^2$ term is missing. It is pretty clear I think.

[quote="mahbub"]Oh! NOW I SEE! :blush:

You know i am not so advanced and experienced! :blush:[/quote]There is no problem. We all were like this in the beginning ;) Good luck solving those problems.

You can post it in the Inequalities forum, but write the complete source, and also post here a link of the topic :)

[quote="manlio"]Valentin, can I post this ineq in Inequality forum, for a brief discussion, or it is better to discuss it in this forum ?

$\displaystyle \frac{x_2^2+\cdots+x_n^2}{n-1}\cdot \frac {x_1^2+x_3^2+\cdots +x_n^2} {n-1} \cdots \frac{x_1^2+\cdots+x_{n-1}^2}{n-1}\geq \left(\frac{x_1^2+...+x_n^2}{n}\right)^{n-1}$[/quote]

isn't it just Jensen ?  :)

The link is

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=163636#p163636

What are you seeing , Myth ?

You look frightened :)


Is trans-1-bromo-3-chlorocyclobutane a chiral molecule?

No, it's not chiral. The cyclobutane ring should wiggle around some, around an average flat configuration. On average, the bromine, the chlorine, and the 1 and 3 carbons should lie in a plane, and that plane is a mirror plane for the molecule. The same would be true of the cis-1-bromo-3-chloro isomer.

However, either isomer of 1-bromo-2-chlorocyclobutane would be a different story.

Yes, that's right. However, it is not necessary the ring to wiggle around some average configuration for the molecule to be achiral, since there is already a vertical plane of simmetry.


a) Let there $ m,n \in \mathbb{N} , m > n$ and $ a_k\equal{} \sum_{i\equal{}k}^{m}{\binom{i}{k}} , k \equal{} \overline{0,n}$ . Compute the sum :

\[ \sum_{k\equal{}0}^{n}{k \cdot a_k \cdot \binom{n}{k}}  ;\]

b) Let $ n,k \in \mathbb{N}$ , such that $ 0 \leq k < n$ . Note that :
$ R_{n}^k\equal{}1^k \left( \binom{n}{1} \right)^2 \plus{} 2^k \left( \binom{n}{2} \right)^2 \plus{} \dots \plus{} n^{k} \cdot \left( \binom{n}{n} \right)^2$.
Prove that : \[ \frac{R_n^3}{R_n^2}\equal{} \frac{n\plus{}1}{2} ;\]

OBS : $ \binom{n}{k}\equal{} \frac{n!}{k! (n\minus{}k)!}$ .

The solved problem can be found in the attachement.
I used some theorems that you could find in any combinatorics textbook.


If $ 725x \plus{} 727y \equal{} 1500$ and $ 729x\plus{} 731y \equal{} 1508$, what is the value of $ x \minus{} y$ ?

$ \begin{array}{crcr}
& 729x\plus{}731y & \equal{} & 1508 \\
\minus{} & 725x\plus{}727y & \equal{} & 1500 \\ \hline
& 4x\plus{}4y & \equal{} & 8
\end{array}$
Therefore, $ x\plus{}y\equal{}2$.
Thus, $ x\minus{}y \equal{} 730(x\plus{}y)\minus{}(729x\plus{}731y) \equal{} 1460 \minus{} 1508 \equal{} \boxed{\minus{}48}$.

[quote="Kouichi Nakagawa"]$ \begin{array}{crcr} & 729x \plus{} 731y & \equal{} & 1508 \\
\minus{} & 725x \plus{} 727y & \equal{} & 1500 \\
\hline & 4x \plus{} 4y & \equal{} & 8 \end{array}$
Therefore, $ x \plus{} y \equal{} 2$.
Thus, $ x \minus{} y \equal{} 730(x \plus{} y) \minus{} (729x \plus{} 731y) \equal{} 1460 \minus{} 1508 \equal{} \boxed{ \minus{} 48}$.[/quote]

[/hide]Can you explain more about your solution. How did you get $ x \minus{} y \equal{} 730(x \plus{} y) \minus{} (729x \plus{} 731y) \equal{} 1460 \minus{} 1508 \equal{} \boxed{ \minus{} 48}$.

He knows $ x \plus{} y$, so he's looking for an easy way to express $ x \minus{} y$ as some multiple of $ x \plus{} y$ plus or minus one of the given equations. It just so happens that $ 730(x \plus{} y) \minus{} (729x \plus{} 731y) \equal{} x \minus{} y$. (It makes the computation easier and doesn't require that we know $ x$ and $ y$).

Isn't it 2?? if you subtract 725x+727y=1500 from the other equation you get 4x-4y=8 therefore x-y=2. (Subtracting changes +727 to -727!!) It has to be 2.

Silly Mistakes kill me  :oops_sign: :wallbash_red:  :wallbash_red: , I did this a long time ago, dark times...


calculus
derivative
logarithms
calculus computations

[quote="kunny"]Formula: $ \boxed{\frac {d}{dx} a^x \equal{} a^x\ln a\ (a > 0\ and\ a\neq1)}$

$ \frac {d}{dx}5^{3x} \equal{} 5^{3x}\ln 3\cdot (3x)' \equal{} 5^{3x}\ln 125$.[/quote]

It was wrong.

$ \frac {d}{dx}5^{3x} \equal{} 5^{3x}\ln 5\cdot (3x)' \equal{} 5^{3x}\ln 125$.


quadratics
Euler
number theory unsolved
number theory

The question is:

Show that if p == 1 mod 4, then (a/p) = (-a/p).
(Note that == means congruent).

I know that if X^2==a mod p (p is a prime) is solvable then a is a quadratic residue of p.

For an example, I let p = 5 since 5==1 mod 4. Then, I let X = 2 and 4 just to check the equation. So:

2^2==-1 mod 5...........a=-1 is quadratic residue.
4^2==1 mod 5............a=1 is quadratic residue.

I know that the legendre symbol (a/p) is 1 if a is a quadratic residue mod p and -1 if a is a quadratic non-residue.

From my example, (-1/5)=1 and (1/5)=1, so I have found an example that shows (a/p) = (-a/p) but I'm having trouble proving it in general.

Thanks!

Do you know that the Legendre symbol is multiplicative, i.e. that $\left( \frac{a}{p}\right) \left( \frac{a}{p}\right) = \left( \frac{ab}p \right)$ ¿

Then you just need to find out $\left( \frac{-1}p \right) =1$ for your primes.

So let's say we have 2 quadratic residues.... (a/p) and (-a/p).

Since the Legendre symbol is multiplicative, does
(a/p)(-a/p)=(-a^2/p)?

Yes, but it would be easier to use it with $b=-1$.


For prime numbers $ p$ and $ q$, $ p\plus{}q \equal{} 102$ and $ p>q.$ What is the least possible value of $ p\minus{}q$?

[quote="mathwizarddude"]For prime numbers $ p$ and $ q$, $ p \plus{} q \equal{} 102$ and $ p > q.$ What is the least possible value of $ p \minus{} q$?[/quote]

start from p=q=51 and keep increasing p and decreasing q untill they are both primes.  p=59, q=43 so the difference is 16

[hide="Moo..."]We are looking for primes $ p, q$ such that $ p \plus{} q \equal{} 102$, and $ p \minus{} q$ is minimum. Dividing $ 102$ by $ 2$ gives us $ 51$, so we're looking for odd primes $ 51 \plus{} n , 51 \minus{} n$, such that $ n$ is minimum. Now, we know that the difference between two odd numbers is even, so $ n$ is necessarily even. Eliminating the numbers $ 49, 55, 45$ (which are an even distance from $ 51$, but are not primes), we find $ p \equal{} 59, q \equal{} 43$, as [b]jli[/b] mentioned.

$ \min (p \minus{} q) \equal{} 16$[/hide]


1. Let ABC be a triangle with AB = 15, BC = 14, and CA = 13. Construct the squares $ S_1$ and
$ S_2$ externally on sides AB and AC, and let their respective centers be $ P_1$ and $ P_2$. Suppose the
circle with diameter $ P_1P_2$ intersects BC at two points, X and Y . Find XY .

2. A circle with center O has an area $ 28\pi$. Points A and B lie within the circle so that OA = 4,
OB = 1, and $ \angle AOB \equal{} 120$. Let P be the point on the circle for which $ \angle APB$ is maximized.
Find AP + BP

[hide="Solution to 1"]
I couldn't think of a less computationally intensive way to do this so I used complex numbers. I expressed everything in terms of the complex numbers $ AC$ and $ AB$.

I found the length of $ P_1P_2 \equal{} \frac{\sqrt{36730}}{10}$ and hence the radius $ r$ of the circle (with center $ O$). Drop a perpendicular from $ O$ to meet $ CB$ at a point $ G$, $ OG$ has length $ \frac{\sqrt{365}}{2}$. $ XG^2 \equal{} r^2 \minus{} OG^2 \implies XY \equal{} 2XG \equal{} \frac{\sqrt{230}}{10}$.

You'll eventually need to work out $ \overline{AC} \cdot AB \equal{} 3(33 \plus{} 56i)$ which can be done using the cosine rule on $ \triangle ABC$.
[/hide]

From [url]http://www.mathlinks.ro/viewtopic.php?t=286278[/url] we know $ P_1P_2X$ is a right isoceles triangle, where X is the midpoint of BC.


In a school of 100 students, 90 study English, 75 Spanish, and 42 French. Every student must study at least one subject. What is the least possible number of students who could be studying all three languages

Can someone please explain

Thanks

[hide]# the students from 1-100. 1-90 study english. 26-100 study spanish. which makes it an intersection of 65 (26-90).French is 1-25, 91-100 which makes 35. 42-35 has a leftover of 7. So the answer is [u][b]7[/b][/u][/hide]

[hide]
There are 207 people enrolled in a language class. At most, one person can "count" as 2 people because he is taking 2 different languages. That eliminates 200 of them leaving 7 that must "count" as 3 people as they are taking 3 languages.
[/hide]


Let $ a,b,c$ be positive reals such that $ \sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}\equal{}1$. Prove that

$ \frac{a(ab\plus{}1)}{b(a\plus{}b)}\plus{}\frac{b(bc\plus{}1)}{c(b\plus{}c)}\plus{}\frac{c(ca\plus{}1)}{a(c\plus{}a)}\geq\frac{3}{2}$

[quote="Ligouras"]Let $ a,b,c$ be positive reals such that $ \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \equal{} 1$. Prove that

$ \frac {a(ab \plus{} 1)}{b(a \plus{} b)} \plus{} \frac {b(bc \plus{} 1)}{c(b \plus{} c)} \plus{} \frac {c(ca \plus{} 1)}{a(c \plus{} a)}\geq\frac {3}{2}$[/quote]

substitution   $ a\equal{}x^2,b\equal{}y^2,c\equal{}z^2,(x,y,z>0)$

The following inequality is equivalent to

$ \frac {x^2(x^2y^2 \plus{} 1)}{y^2(x^2\plus{} y^2)} \plus{} \frac {y^2(y^2z^2 \plus{} 1)}{z^2(y^2 \plus{} z^2)} \plus{} \frac {z^2(z^2x^2 \plus{} 1)}{x^2(z^2 \plus{} x^2)}\geq\frac {3}{2}$

substitution   $ x\equal{}\frac{t}{1\plus{}zt\plus{}z^2},y\equal{}\frac{1}{z\plus{}t},z\equal{}z,(t>0)$

Rearranging

The coefficient is positive.


if $x\leftarrow-x$ we have $-\frac{1}{x}f(x)+f(-\frac{1}{x})=-x$ $\rightarrow$ $f(-\frac{1}{x})=-x+\frac{1}{x}f(x),(1)$
But if $x\leftarrow \frac{1}{x}$ we also have that $xf(-\frac{1}{x})+f(x)=\frac{1}{x}$ $\rightarrow$ (using (1) ) $-x^{2}+f(x)+f(x)=\frac{1}{x}$ $\rightarrow$ $f(x)=\frac{1}{2}\left ( \frac{1}{x}+x^{2}\right)$

[quote="santosguzella"]Find all function $f: R^{*}\rightarrow R$ such $\frac{1}{x}f(-x)+f\Big(\frac{1}{x}\Big)=x$.[/quote]

Let $x=-\frac{1}{x}$ and we have:

$-xf\left(\frac{1}{x}\right)+f(-x)=-\frac{1}{x}$

From given equation we have that: $f\left(\frac{1}{x}\right)=x-\frac{1}{x}f(-x)$

Put it in the second and we have:
$-x\left(x-\frac{1}{x}f(-x)\right)+f(-x)=-\frac{1}{x}$
$f(-x)=\frac{x^{3}-1}{2x}$
$f(x)=\frac{x^{3}+1}{2x}$


geometry
3D geometry
algebra
polynomial
inequalities
inequalities unsolved

1. Let $x,y,z > 0$ with $xyz = 1$. Prove that $x+y+z \leq x^2+y^2+z^2$

2. the real numbers $x_1, \cdots x_n$ belong to the interval $[-1,1]$ and the sum of their cubes is zero. Prove that their sum does not exceed $n/3$.

3. let $x_1, \cdots x_5$ be positive reals such that
$(x_{i+1} ^2 -x_{i+3}x_{i+5})(x_{i+2} ^2 -x_{i+3}x_{i+5}) \leq 0$
for $i=1, \cdots 5$ where $x_{n+5}=x_n$ for all $n$. Prove that $x_1=x_2=\cdots=x_5$.

4. let $x,y,z \geq 0$ with $x+y+z=1$. Prove that
$x^3+y^3+z^3+6xyz \geq \frac{1}{4}$

5. Let $P(x)=1+a_1 x + \cdots + a_{n-1}x^{n-1}+x^n$ be a polynomial with complex coefficient. Suppose the roots of $P(x)$ are $\alpha_1, \cdots$ with
$|\alpha_i|>1$ for i=1 to j and $|\alpha_i| \leq 1$ for i=j+1 to n.

prove that
$\prod {|\alpha_i|} \leq \sqrt{|a_0|^2+|a_1|^2+\cdots}$

6. Prove that for any real number $x,y,z$
$3(x^2-x+1)(y^2-y+1)(z^2-z+1) \geq (xyz)^2+xyz+1$

7. (a) prove that any polynomial $P(x)$ such that $P(x) \geq 0$ for all real $x$ can be written as the sum of the squares of two polynomials.

(b) Prove that the polynomial
$z^2(x^2-y^2)(x^2-1)+ y^2(y^2-x^2)(y^2-1)+(1-x^2)(1-y^2)$
is every where nonnegetive, but cannot be written as the sum of squares of any number of polynomials.

By the way, I have solved 1 and 2. 1 by RMS-AM, and then AM-GM. And 2 by Trigonometric substitution with $sin$.

I would be glad if anyone of you can give solution to the rest of the problems.

Solution 4:
use coushy-shwraz $(x+y+z)(x^3+y^3+z^3)\geq (x^2+y^2+z^2)^2$ and we know when $x+y+z=1\Longrightarrow xyz\leq\frac {1}{27}$.Then we most prove the new inequality
in this form  $(x^2+y^2+z^2)^2+6xyz\geq \frac 14\Longrightarrow (x^2+y^2+z^2)^2\geq \frac 14-6xyz\geq \frac 14-\frac 29=\frac{1}{36}$ then we most prove $(x^2+y^2+z^2)\geq\frac 16$ and we know $x^2+y^2+z^2\geq\frac {x+y+z}{3}=\frac 13$. :lol:

[quote="ehsan2004"]Then we most prove the new inequality
in this form  $(x^2+y^2+z^2)^2+6xyz\geq \frac 14\Longrightarrow (x^2+y^2+z^2)^2\geq \frac 14-6xyz\geq \frac 14-\frac 29=\frac{1}{36}$ then we most prove $(x^2+y^2+z^2)\geq\frac 16$ [/quote]
Actually, you replaced the product $xyz$ with the larger value $1/27$, and this is not correct. :P
On the other hand, the following stronger inequality holds:
$x^3+y^3+z^3+\frac{15}{4}xyz \ge \frac{1}{4}$.

Yes, this one is Senderov's inequality, published in Kvant magazine. Standard computations show that in fact it is exactly Schur's inequality. ;)

I know it very well that for harazi and vasc those 7 inequalities are like toy of a baby. So can you please provide the solutions?

Here is a solution for $6$

The inequality

$x^2-x+1\geq \sqrt[3]{\frac {x^6+x^3+1}{3}}$ is equivalent to

$\frac {(x-1)^4(2x^2-x+2)}{3}\geq 0$, which is clearly true.

So we have

$3\prod (x^2-x+1)\geq \prod \sqrt[3]{x^6+x^3+1}\geq (xyz)^2+xyz+1$

which is true by Holder's inequality.

7 is going to use the identity $(a^2+b^2)(c^2+d^2) = (ac+bd)^2 + (ad-bc)^2$ extensively.

I would like to see a solution to $2$.  I believe that it can be handled without much difficulty by smoothing, but I think that there must be a nicer solution.

Mahbub [b]DO NOT[/b] post 7 problems in the same topic. Repost each of them [b]separately[/b]. This topic is closed.


[quote="slamchess"]How would you rank the the national and state competitions easiness and hardness wise?[/quote]

Well last year seemed easy...never had a national test in my hands though  :lol:

[quote="mathgeniuse^ln(x)"]Yeah, I posted a topic like this a long time ago.

Anyways, all of them are easy if you have done them at least once.[/quote]

is it just me or as you progress from year to year it seems to get harder?

[quote="Spectacular"][quote="13375P34K43V312"]Threads like this have been posted in the recent past. Use the search button.[/quote]

he might not want to revive such old posts;  more MC competitions are coming up. Making a new thread like this surely wouldn't hurt.[/quote]
Well our discussion was several pages long, so it was definitely thorough.
Also, since we can't discuss 2007 chapter, it is pointless to start a new thread.
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=difficulty&t=98871
For those who don't feel like searching


[quote="pooyan_seva"]find all function $f: \mathbb{Z} \longrightarrow\mathbb{Z}$ such that $f(x+y)+f(xy)=f(x)f(y)+1$
It was aTST of some where I'm not sure about!!! :roll:  :?[/quote]
$(f(x)-1)(f(0)-1)=0$
$f(0)=1$ then $f(1)=1$ or $f(-1)=0$
+, $f(1)=1 \to f(x)=1$
+, $f(-1)=0 \to f(x)=x+1$
So $f(x)=1$ or $f(x)=x+1$


Let $A,C,E$ lie on a straight line, and $B,D,F$ on another straight line. Show that the midpoints of $AB,CD,EF$ also lie on a straight line.


this is the long answer pactice problem from the ASHIME site.

$W^3+X^3=Y^3+Z^3$

 Let (W, X, Y, Z) be a set of integers that satisfies this equation. Find the minimum value of the product W*X*Y*Z.

How would you go about solving this thing is a strategic way, that doesn't include guess and check.  I mean guess and check is possible because we are dealing with small numbers but still it can lead to a lot of tedious work.  I was thinking that we could use

$(a^3+b^3)=(a+b)(a^2-ab+b^2)$

but i still don't see any simplifications possible

please note its the ASHIMC not the ASHIME, the website is [url]http://www.ahsimc.com[/url]

When you mean by integers do you mean positive distinct integers or any integers because I think an infinitely negative product is possible if the latter is the case.


When I view the images for my Introduction to Number Theory class (from the class transcripts), I get a 404 not found error.  Were the images deleted for some reason?  It's kinda annoying to look at a transcript and not know what the problem was originally.


If  A is shorter than B, than B is not shorter than A, and it takes  BOTH A&B to shake hands...

That's exactly right.

Let's make this case in three people. Abe, Bob, and Carl.

Heights are: Abe>Bob>Carl

Now, Abe can handshake with Bob and Carl but Bob can't handshake with Abe because Abe's taller than Bob. So, Abe and Bob has no handshake. This occurs same for Bob and Carl. Also between Carl and Abe. So, here is what Gyan said if you didn't understand:

Shortest person, Carl, wouldn't handshake with anyone because everybody would be taller than him! And Bob wouldn't handhshake with Abe because Abe's taller and Carl wouldn't handshake with him, so no handshake on Bob! And Abe would have no one to handshake so he would have no handshake also!

So, no handshake.

Wasn't this MC problem? I seem to recognize this.

None, unless these women can magically shrink 0.0 .

ok, thanks i get it. yeah it is a mc problem, but a similar question was in my mathbook.

Yea, 0 because none of the women will shake hands with anyone taller than them so they won't shake hands with anyone.
Otherwise you'd use the handshake formula which I think is
$\frac{n(n-1)}{2}$

Because each woman wouldn't shake hands with herself.

haha, yea, we had that on a math count homework once, didn't think anyone in seventh grade got it... she had to demonstrate and like we had to shake hands and stuff... anyways, the answer is zero because well, if it takes 2 people to shake hands, then one of them has got to be shorter and one taller, so they can't shake hands with each other. even if they are the same height, one is not shorter than the other... but of course, there's always the possiblity of stupid people shaking hands with themselves... jk  :D  :D  :D

the shortest person shakes hands with nobody

[quote="Silverfalcon"]That's exactly right.[/quote]
Hah... Thats a pretty tricky problem. I need to remember stuff like this so i can do better in future competitions. :P

but didnt it say that there are eight different sized woman!

even if they are the same height, they still can't shake hands with each other because if A=B then A is not shorter than B

Part of the key to recognizing this trick question -- or working many regular problems, is recognizing the symmetry of a relationship between two people.  Part of learning mathematics is learning to categorize real world situations in ways that make them easier to think about.

This also helps in computer programming.

Okay the point is you have to shake hands with a person SHORTER than you. That means that you are TALLER than the other person, so she can't shake hands with you.

[quote="mathprincess09"]k in the problem it said that they all have different heights!
rite?
k
so lets say:
1- 5 foot
2-6 foot
3- 4 foot
4- 5ft 10 in
5-6ft 2 in
6- 7ft
7-5ft 4 in
8-3 ft
so they r going to shake hands rite
like one willlll shake hands with 3, and 8.
can some1 plz explain me!
how this thing works![/quote]

yea, but at the same time, 1 can't shake hands with 3 and 8 because 3 and 8 is NOT shorter than 1

ya i thought about it before i went to sleep and i realized that
even if one if smaller than the other than the other is bigger than other
so they cant shake hands.
 :D

[quote=BarbarianKing][quote=ajai]can someone help me with this problem. i don't quite get it.

Eight women of different heights are at a party. Each woman decides to only
participate in a handshake with women shorter than herself. How many handshakes
take place?

[hide]the answer is 0[/hide][/quote]

It's very interesting....[/quote]

Yeah, don't revive 10-year old threads to say "it's very interesting". Like, math92 ALREADY SAID THAT (10 years ago).

[quote=SuperMathWiz]@BarbarianKing@jack74 Pls dont revive old/dead threads.[/quote]

jack74 didn't do anything :P


integration
calculus
calculus computations

One last step: apply an initial condition to fix the constant of integration. In this case, let $ x \equal{} 0.$ Since $ \arctan 0\equal{} 0,$ it will follow that $ C \equal{} 0$ in what you wrote above.

---

From the perspective of the way I'm teaching my students: we did this argument once and for all (with no 3's in it) when we calculated the series centered at $ 0$ for $ \arctan x.$ We memorize that series (along with its radius of convergence, $ 1.$) We then obtain the series for $ \arctan \frac {x}3$ by substitution in the memorized series. And this series converges for $ \left|\frac {x}{3}\right|\le 1,$ or $ |x|\le 3.$

[quote="Kent Merryfield"]One last step: apply an initial condition to fix the constant of integration. In this case, let $ x \equal{} 0.$ Since $ \arctan \equal{} 0,$ it will follow that $ C \equal{} 0$ in what you wrote above.

---

From the perspective of the way I'm teaching my students: we did this argument once and for all (with no 3's in it) when we calculated the series centered at $ 0$ for $ \arctan x.$ We memorize that series (along with its radius of convergence, $ 1.$) We then obtain the series for $ \arctan \frac {x}3$ by substitution in the memorized series. And this series converges for $ \left|\frac {x}{3}\right|\le 1,$ or $ |x|\le 3.$[/quote]Thanks, I appreciate your help!


Let z be a complex number satisifying the following: $|z|=2$, $\arg z=\frac{\pi}6$, $z^3+b=0$. Find the value of $b$.

[quote="kunny"]The answer is $b=-8.$[/quote]
Your answer is wrong. Moreover, please don't just give the answer but hide the solution.

[hide="solution"]by dropping a perpendicular to the real number axis and drawing the distance it is from the origin, we create a 30-60-90 triangle (since $\arg{z}=\frac{\pi}{6}=30^{\circ}$). Thus z=$\sqrt3+i$. Cubing this, we get $8i$. Thus b must be $-8i$. [/hide]

[quote="kunny"]You can use de Moivre's theorem. :D[/quote]

oh yeah  :blush:  I just learned this last night (how surprising!) so I'm not yet familiar with these things yet.

Does De Moivre's Theorem work for just the cubing part?

$(\cos \theta +i\sin \theta )^n=\cos n\theta+i\sin n\theta$ :)


Let $a,b,c,d$ be nonegative real numbers such that $a+b+c+d=1$. Prove that
\[\sqrt{a+b+c^{2}}+\sqrt{b+c+d^{2}}+\sqrt{c+d+a^{2}}+\sqrt{d+a+b^{2}}\geq3.\]
PS: This is an inspiration from an old problem.

[quote="pvthuan"]Let $ a,b,c,d$ be nonegative real numbers such that $ a\plus{}b\plus{}c\plus{}d \equal{} 1$. Prove that
\[ \sqrt{a\plus{}b\plus{}c^{2}}\plus{}\sqrt{b\plus{}c\plus{}d^{2}}\plus{}\sqrt{c\plus{}d\plus{}a^{2}}\plus{}\sqrt{d\plus{}a\plus{}b^{2}}\geq3.\]
PS: This is an inspiration from an old problem.[/quote]

Very nice inequality!

$ \sum\sqrt{a\plus{}b\plus{}c^{2}}\geq 3$
$ \Leftrightarrow\sum\sqrt{(a\plus{}b)(a\plus{}b\plus{}c\plus{}d)\plus{}c^{2}}\geq 3(a\plus{}b\plus{}c\plus{}d)$
$ \Leftrightarrow\sum\sqrt{P_{2}\plus{}Q_{1}}\geq\sum\sqrt{P_{1}\plus{}Q_{1}}$
where $ P_{1}\equal{} b^{2}\plus{}c^{2}\plus{}d^{2}\plus{}bc\plus{}cd\plus{}db$ and $ Q_{1}\equal{} bc\plus{}cd\plus{}db$.
Note that $ P_{i}$'s and $ Q_{i}$'s have an opposite order to the order of $ a,b,c,d$; that is, $ a\geq b$ is equivalent to $ Q_{1}\leq Q_{2}$ and $ P_{1}\leq P_{2}$.

[b]Claim[/b] $ P_{1}\plus{}Q_{1},P_{2}\plus{}Q_{2},P_{3}\plus{}Q_{3},P_{4}\plus{}Q_{4}$ majorizes $ P_{2}\plus{}Q_{1},P_{3}\plus{}Q_{2},P_{4}\plus{}Q_{3},P_{1}\plus{}Q_{4}$.

This can be proved easily, since $ P_{i}$'s and $ Q_{i}$'s have the same order. Note that the fact that elements of $ A$ majorize elements of $ B$ is equivalent to the fact that the maximum sum of $ k$ elements in $ A$ is not smaller than the maximum sum of $ k$ elements in $ B$, for all $ k$. Trivially, the sum of the four numbers are equal for the two groups. The maximum sum of $ k$ numbers among $ P_{1}\plus{}Q_{1},P_{2}\plus{}Q_{2},P_{3}\plus{}Q_{3},P_{4}\plus{}Q_{4}$ is always not smaller than any form of $ P_{a_{1}}\plus{}\cdots\plus{}P_{a_{k}}\plus{}Q_{b_{1}}\plus{}\cdots\plus{}Q_{b_{k}}$, since the maximum is the sum of the biggest $ k$ $ P_{i}$'s and the biggest $ k$ $ Q_{i}$'s. This completes the proof of the claim.

Therefore, by the majorization inequality, we are done, since $ f(x) \equal{}\sqrt{x}$ is concave. Equality holds iff $ (a,b,c,d) \equal{} (\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}), (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)$.

And by the same reasoning, we can obtain another inequality. For $ a\plus{}b\plus{}c\plus{}d \equal{} 1$, we get
\[ \sqrt{a\plus{}b^{2}\plus{}c}\plus{}\sqrt{b\plus{}c^{2}\plus{}d}\plus{}\sqrt{c\plus{}d^{2}\plus{}a}\plus{}\sqrt{d\plus{}a^{2}\plus{}b}\geq 3.\]

For another applications, see http://www.mathlinks.ro/Forum/viewtopic.php?t=163090 .


O is a point inside the acute-angled triangle ABC.The altitudes of the triangle are AD, BE and CF.If the feet of the perpendiculars from O to AD,BE,CF are P,Q,R respectively ,prove triangle PQR is similar to triangle ABC

Let H be the triangle orthocenter. Since the angles $\angle OPH = \angle OQH = \angle ORH = 90^o$ are all right, the pentagon OPHQR is cyclic, its circumcenter being the midpoint M of the segment OH. Depending on the position of the arbitrary point O with respect to the orthocenter H, we have 3 mutually exclusive possibilities: Either the altitude AD intersects the [b]segment[/b] QR, or the altitude BE intersects the [b]segment[/b] RP, or the altitude CF intersects the [b]segment[/b] PQ. WLOG, assume that the altitude CF intersects the segment PQ. The angles $\angle RPQ = \angle RHQ \equiv \angle BHF$ spanning the same arc RQ of the circle (M) are equal and the angles $\angle PQR = \angle PHR \equiv \angle AHF$ spanning the same arc PR of the circle (M) are also equal. But obviously, $\angle BHF = \angle BAC \equiv \angle A$ and $\angle AHF = \angle ABC \equiv \angle B$. Thus the triangles $\triangle PQR \sim \triangle ABC$ are similar, having 2 angles equal.

The triangle $\triangle ABC$ does not have to be acute-angled and the arbitrary point O does not have to be inside of the triangle $\triangle ABC$. The point O can be almost anywhere, even coincident with one of the triangle vertices. It must not coincide with the triangle orthocenter H, because then the triangle $\triangle PQR$ would degenerate into the point H.


I see these very often.  How are they supposed to be solved?  Could someone give me a general method on how to solve them?

I know when the problem is something like, there are $ x$ people, and each of them shake hands with everyone else, its $ \dfrac{x(x\plus{}1)}{2}$.

What if the problem is there are $ x$ people, and each of them shake hands with those shorter than them?  What is a general method on how to solve it?

It's the same! in any pair, one is taller, so shakes hands with shorty.

EDIT. (towards next post) If people are imposed to shake hands with shorter ones, the shorter must obey, thus adding handshakes to their tally! Now we can move towards Ionesco's theater of the absurde ...

Actually, the answer is always 0. In a pair, one person is either shorter, or the same, both of which can't happen.


$ z \plus{} \frac {1}{1!}(n \minus{} 1)a_1 \plus{} \frac {1}{2!}(n \minus{} 1)(n \minus{} 2)a_2 \plus{} \cdots \plus{} \frac {1}{t!}(n \minus{} 1)\cdots(n \minus{} t)a_t$

(1)::: write this formula in binomial form (simplify).

(2)::: what is this formula used for ?

$ z\plus{}\frac{a_1}{n}{n\choose 0}\plus{}\frac{a_2}{n}{n\choose 1}\plus{}\frac{a_3}{n}{n \choose 2}\plus{}\cdots\plus{}\frac{a_t}{n}{n \choose t\minus{}1}$

is this true ?

$ \frac{a_1}{n}{n\choose 0}\equal{}\frac{a_1}{n}\left(\frac{n!}{0!(n\minus{}0)!}\right)\equal{}\frac{a_1}{n}\left(\frac{n!}{n!}\right)\equal{}\frac{a_1}{n}\ne (n\minus{}1)a_1$.

Try using summations to make the notation easier.

ok i found it now its

$ z\plus{}{{n\minus{}1}\choose1}a_1\plus{}{{n\minus{}1}\choose2}a_2\plus{}\cdots\plus{}{{n\minus{}1}\choose t}a_t$

am i wrong ? i dont think its wrong but if it is someone point out plz.

can someone write this in summation form ?

thnkx Temperal
anyone knows what this formula is used for ?

It kind of looks like the Binomial Theorem


k so the question is:

What mass of NH4CL is requried for 5 L of a 0.2 M NH3 solution to make a buffer with a pH of 10.5? Kb = 1.8 x 10^-5

Welcome to this forum Faulkner56.

In a chemistry problem, always begin by writing the relevant chemical equations. In this case,

$NH_{4}Cl(s)+aq \longrightarrow NH_{4}^{+}(aq)+Cl^{-}(aq)$
$NH_{4}^{+}(aq)+H{2}O(l) \rightleftharpoons NH_{3}aq)+H_{3}O^{+}(aq)$.

Now write the equilibrium constant for this reaction and solve for [NH4+]:

$K_{a}= \frac{K_{w}}{K_{b}}= \frac{[NH_{3}][H_{3}O^{+}]}{[NH_{4}^{+}]}\Rightarrow  [NH_{4}^{+}] = \frac{K_{b}[NH_{3}][H_{3}O^{+}]}{K_{w}}$.

Now substitute by the given values to obtain the final equilibrium concentration of ammonium ion:

$[NH_{4}^{+}] = \frac{1.8 \times 10^{-5}\cdot 0.2 \cdot 10^{-10.5}}{1 \times 10^{-14}}= 0.011 M$.

Thus, the buffer capacity of the solution will be low, because $\frac{[NH_{3}]}{[NH_{4}^{+}]}> 10$. Anyway, as the final volume is very closely 5 L, then the required number of moles of NH4+ (and so of NH4Cl) will be

$n = C \cdot V = 0.011 \cdot 5 = 0.055 mol$.

Now use the molar mass of ammonium chloride, M = 53.49 g/mol, to finally obtain the required mass of this salt:

$m = M \cdot n = 53.49 \cdot 0.055 = 2.94 g$.


Let $ a$ be a positive real number and let ${ |z+\frac{1}{z}|=a}$.Find the maximum and minimum value of $ |z|$,where $ z$ belongs to the set of non zero complex numbers.

[hide="my try"]
$ r\equal{}|z|$

$ z\equal{}r(\cos\varphi\plus{}i\sin\varphi)$

$ \frac{1}{z}\equal{}\frac{1}{r}(\cos\varphi\minus{}i\sin\varphi)$

$ |z\plus{}\frac{1}{z}|\equal{}a$

$ |r(\cos\varphi\plus{}i\sin\varphi)\plus{}\frac{1}{r}(\cos\varphi\minus{}i\sin\varphi)|\equal{}a$

$ |r^2(\cos\varphi\plus{}i\sin\varphi)\plus{}(\cos\varphi\minus{}i\sin\varphi)|\equal{}ar$

$ \sqrt{(r^2\plus{}1)^2\cos^2\varphi\plus{}(r^2\minus{}1)sin^2\varphi}\equal{}ar$

$ \sqrt{r^4\plus{}2r^2\cos2\varphi \plus{}1}\equal{}ar$

$ r^4\plus{}2r^2\cos 2\varphi \plus{}1\equal{}a^2r^2$

Derivating with respect to $ \varphi$:

$ 4r^3r'\plus{}4rr'\cos 2\varphi \plus{} 2r^2(\minus{}2\sin 2\varphi)\equal{}2a^2rr'$

Letting $ r'\equal{}0$ for extremes:

$ \minus{}4r^2\sin 2\varphi\equal{}0 \implies \sin 2\varphi\equal{}0 \implies \varphi\equal{}\frac{k\pi}{2}$

[u]Case 1:[/u]

$ k\equal{}0, \varphi\equal{}0$

$ z\equal{}r$

$ |r\plus{}\frac{1}{r}|\equal{}a$

$ r\plus{}\frac{1}{r}\equal{}a$

$ r_{1,2}\equal{}\frac{a\pm \sqrt{a^2\minus{}4}}{2}$

[u]Case 2:[/u]

$ k\equal{}1, \varphi\equal{}\frac{\pi}{2}$

$ z\equal{}ri$

$ |ri\plus{}\frac{1}{ri}|\equal{}a$

$ |(r\minus{}\frac{1}{r})i|\equal{}a$

a) $ r>1$

$ r\minus{}\frac{1}{r}\equal{}a$

$ r_3\equal{}\frac{a\plus{}\sqrt{a^2\plus{}4}}{2}$ (only one solution since $ r$ can't be negative)

b) $ 0<r<1$

$ \frac{1}{r}\minus{}r\equal{}a$

$ r_4\equal{}\frac{\minus{}a\plus{}\sqrt{a^2\plus{}4}}{2}$ (again only one solution since $ r$ can't be negative)

[u]Case 3:[/u]

equivalent to case 1

[u]Case 4:[/u]

equivalent to case 2

Now it's easy to check that $ \frac{a\plus{}\sqrt{a^2\plus{}4}}{2}>\frac{a\plus{} \sqrt{a^2\minus{}4}}{2}>\frac{a\minus{} \sqrt{a^2\minus{}4}}{2}>\frac{\minus{}a\plus{}\sqrt{a^2\plus{}4}}{2}$

Therefore $ \boxed{max\ |z|\equal{}\frac{a\plus{}\sqrt{a^2\plus{}4}}{2}}\ , \ \boxed{min\ |z|\equal{}\frac{\minus{}a\plus{}\sqrt{a^2\plus{}4}}{2}}$[/hide]

[hide="My Approach"]I squared the both sides $ a^{2} \equal{} |z \plus{} \frac {1}{z}|^{2} \equal{} (z \plus{} \frac {1}{z})(\overline{z} \plus{} \frac {1}{\overline{z}})$

Which equals $ \frac {|z|^{4} \plus{} (z \plus{} \overline{z})^{2} \minus{} 2|z|^{2} \plus{} 1}{|z|^{2}}$
Hence $ |z|^{4} \minus{} |z|^{2}(a^{2} \plus{} 2) \plus{} 1 \equal{} \minus{} (z \plus{} \overline{z})^{2}$

[/hide]

[quote="RAGHAV GROVER"]Hence $ |z|^{4} \minus{} |z|^{2}(a^{2} \plus{} 1) \plus{} 1 \equal{} \minus{} (z \plus{} \overline{z})^{2}$
[/hide][/quote]
I think you meant $ |z|^{4} \minus{} |z|^{2}(a^{2} \plus{} 2) \plus{} 1 \equal{} \minus{} (z \plus{} \overline{z})^{2}$

[quote="RAGHAV GROVER"]Let $ a$ be a positive real number and let $ \left|z \plus{} \frac {1}{z}\right| \equal{} a$ . Find the extreme values of $ |z|$ , where $ z\in \mathbb C^*$ . [/quote]

[b][u]Proof[/u].[/b] $ \left|z^2 \plus{} 1\right| \equal{} a|z|$ $ \Longleftrightarrow$ $ \left(z^2 \plus{} 1\right)\left(\overline z^2 \plus{} 1\right) \equal{} a^2z\overline z$ $ \Longleftrightarrow$ $ |z|^4 \plus{} (z \plus{} \overline z)^2 \minus{} 2|z|^2 \plus{} 1 \minus{} a^2|z|^2 \equal{} 0\ \Longleftrightarrow$

$ \left(|z|^2 \minus{} \frac {a^2 \plus{} 2}{2}\right)^2 \plus{} (z \plus{} \overline z)^2 \equal{} \left(\frac {a^2 \plus{} 2}{2}\right)^2 \minus{}$ $ 1\ \stackrel{*!*}{\Longrightarrow}\ \left(|z|^2 \minus{} \frac {a^2 \plus{} 2}{2}\right)^2\le \left(\frac {a^2 \plus{} 2}{2}\right)^2 \minus{} 1\ \Longleftrightarrow$

$ \left| |z|^2 \minus{} \frac {a^2 \plus{} 2}{2}\right|\le$ $ \sqrt {\left(\frac {a^2 \plus{} 2}{2}\right)^2 \minus{} 1}\ \Longleftrightarrow\ \frac {a^2 \plus{} 2 \minus{} a\sqrt {a^2 \plus{} 4}}{2}\le$ $ |z|^2\le\frac {a^2 \plus{} 2 \plus{} a\sqrt {a^2 \plus{} 4}}{2}\ \Longleftrightarrow$

$ \boxed {\ m\equiv\frac { \minus{} a \plus{} \sqrt {a^2 \plus{} 4}}{2}\le |z|\le\frac {a \plus{} \sqrt {a^2 \plus{} 4}}{2}\equiv M\ }$ . Show easily when we have equality left or right !

[b][u]Similar exercises[/u].[/b] Find the extreme values of $ |z|$ , where $ z\in \mathbb C^*$ :

$ 1\blacktriangleright\ \ \left|z^2 \plus{} 1\right| \equal{} |z \minus{} 1|$ .

$ 2\blacktriangleright\ \ \left|z^2 \plus{} 1\right| \equal{} |4z \plus{} 3|$ .

$ 3\blacktriangleright\ \ \left|z^2 \plus{} 1\right| \equal{} 2|z \plus{} 1|$ .

$ 4\blacktriangleright\ \ a\in\mathbb R\ \ \wedge\ \ \left|z^2 \plus{} 1\right| \equal{} |az \plus{} 2|\ \implies\ |z|\ \le\ \sqrt {a^2 \plus{} 3}$ .

$ 5\blacktriangleright\ \ \left|z^2 \plus{} c\right| \equal{} |az \plus{} b|$ , where $ \{a,b,c\}\subset \mathbb R\ ,\ c > 0$ are given.

[hide="Answer."]
$ \boxed {\ \sqrt {\frac {\left(a^2 \plus{} 2c\right)\sqrt c \minus{} \sqrt {\left( (b^2 \plus{} a^2c\right)\left(a^2 \plus{} 4c\right)}}{2\sqrt c}}\ \le\ |z|\ \le\ \sqrt {\frac {\left(a^2 \plus{} 2c\right)\sqrt c \plus{} \sqrt {\left((b^2 \plus{} a^2c\right)\left(a^2 \plus{} 4c\right)}}{2\sqrt c}}\ }$ .[/hide]


algebra
polynomial
modular arithmetic
Ring Theory

Let $ f(x)$ be a polynomial with integer coefficients and let $ p$ be a prime. Denote by $ z_1,...,z_{p\minus{}1}$ the $ (p\minus{}1)$th complex roots of unity. Prove that
$ f(z_1)\cdots f(z_{p\minus{}1})\equiv f(1)\cdots f(p\minus{}1) \pmod{p}$.

The result is obviously valid for $ f(x) \equal{} x$ and for $ f(x) \equal{} a$ with $ a$ integer (by Wilson and little Fermat).
It is also inmediate that if the result is valid for polinomials $ f(x)$, $ g(x)$ then it is also valid for:

1) $ af(x)$ with $ a$ an arbitrary integer.
2) $ xf(x)$
3) $ f(x) \plus{} a$ with $ a$ an arbitrary integer.
3) $ f(x)g(x)$

DONE!

[quote="Albanian Eagle"]why is 3) true?[/quote]

Do i missing something or it is trivial?

I don't get it.
If $ x_1x_2\cdots x_n\equiv y_1y_2\cdots y_n\pmod{p}$
it is not necessarily true that  $ (x_1\plus{}a)(x_2\plus{}a)\cdots (x_n\plus{}a)\equiv (y_1\plus{}a)(y_2\plus{}a)\cdots (y_n\plus{}a)\pmod{p}$

Consider $Q(Z_1)$ and let A be the ring of it's algebraic integers consider I a maximal prime ideal lying over pA.
Hece A modulo I is a field. and inside this field.  $(x-1)...(x-(p-1))=x^{p-1}-1=(x-z_1)....(x-z_{p-1}) $
so consider  y is a primitive root in $F_p$, $y=z_i$ (inside this field).
Because  I interescted with $Z$  is $pZ$. the conclusion follows.

@above Fixed:
This is a rewriting of the answer by paul1703 in the language of $p$-adic numbers.

By Hensel's lemma, the field $\Bbb Q_p$ contains all $(p - 1)$-st roots of unity $z_1, \dots, z_{p - 1}$.
Furthermore, we have $z_i \equiv i\pmod p$, up to reordering.

This justifies the identity $f(z_1)\dots f(z_{p - 1}) \equiv f(1)\dots f(p - 1)\pmod p$ in $\Bbb Z_p$, and hence in $\Bbb Z$.


Most objects made of metal are conductors. Water is also a conductor.

Yeah pure distilled water is definitely not a conductor.  To be a conductor you need free moving ions as I believe.  Normal water doesn't break up into ions very much (although the water molecules do react to form hydronium and hydroxide ions).  Well that is what I was taught about water anyway.

For the second one I'm not sure about it too much either.  If that is multiple choich then I would say b because I believe the question basically means if you have one end( of something probably a conductor) that has a lot of electricity and another that is low which way does it tend to move.  So it would move from a lot to a little.  Someone correct me if I'm wrong.

Potential energy wants to be reduced. Take it from there.

High electrical potential energy = lots of electrons in one place.

Hmmm ... the question (part 2) says "charges" - not "electrons" - so they could be positive charges.

I think we can be sure that they move in one direction or the other - but we don't know which way without more information.

and it also says electrical potential <i>energy</i>, not electrical potential.

It still goes from high potential energy to low potential energy.

I looked up number 2 and found that it is normally from high to low....The book didn't explain it as far as I read though.

Hmm... I know water isn't a conductor, but If I put it into the electrical outlet, wouldn't I die?  (Runs to the outlet with a glass of water)

Well, tap water will probably conduct, because it's got lots of stuff in it.  Purified water wouldn't, though.

[quote="JBL"]Listen -- basic idea behind conductors: if you stick a conductor into an electric socket, you will die.  So, imagine sticking each of those various objects into a socket.  If you think you would die, then it's probably a conductor.  But if you stick, say, wood into an electric socket, worst-case scenario is it catches on fire, which is unlikely to result in your death.  So you would classify wood as an insulator.

[quote="beta"]Water is also a conductor.[/quote]

Well, I would disagree about this -- water is a good conductor of heat, but it's pretty darn bad at conducting electricity.  (I don't know if you've done this, but since water is a polar molecule, it doesn't let electrons just flow freely, which is what something needs to do to conduct electricity.)  I would classify pure water as an insulator.  However, if the water has things like salts or other ionic compounds dissolved in it, it will conduct electricity.

*Props to my chem-major girlfriend for checking that I do in fact know what I'm talking about, sometimes.[/quote]

Don't those saftey buttons on those sockets save you? Isn't electricity like physics?


linear algebra
matrix
linear algebra unsolved

$S \in M_n(\mathbb{R})$ a matrix with positive entries and such that the sum of each column is $1$. Suppose that for every $1 \leq i,j \leq n$ there exists some $m$ such that $S_{i,j}^m>0$, then show that $1$ is a simple eigenvalue of $S$.

And if $S$ is an infinite matrix with the same property. (sum of each column is $1$), what can you say ?

I am extremely interested and perhaps I can provide a (partial) solution as I have seen this before, but please explain :

simple eigenvalue : algebraic multiplicity one?
positive a :  $a\geq 0$ or $a>0$ ?

don't you mean, there is an m such that for all i, j $S_{i,j}^m>0$ ?

because consider : $A=\left[\begin{matrix}0&1\\1&0\end{matrix}\right]$
it satisfies your condition, but it doesn't satisfy mine
it obviously has  as multiple eigenvalue (algebraically and geometrically)

yes, I mean $dim(Ker(S-Id))$, positive means $\geq 0$, and $m$ depends of $(i,j)$. But I don't understand your counter-example. $1$ is a simple eigenvalue of $A$ isn't it ?


calculus
combinatorics unsolved
combinatorics

We have a string with $ n$ characters.How many subsequences can i produce with this string?

is this a permutation?

For one thing, this is not a Calculus question, but Combinatorics.
For seconds, it is not a permutation, for sure.

Let the sequence be $ (x_k)_{k\equal{}1,2,\ldots,n}$. If by subsequence you mean (as you should) a sequence $ (x_{k_m})_{m\equal{}1,2,\ldots,s}$, then there are as many subsequences as non-empty parts of $ \{1,2,\ldots,n\}$, hence $ 2^n \minus{} 1$. If you mean subsequences made by consecutive elements, there are $ n\plus{}(n\minus{}1)\plus{}\cdots\plus{}2\plus{}1 \equal{} \frac {1} {2} n(n\plus{}1)$.

[quote="mavropnevma"]For one thing, this is not a Calculus question, but Combinatorics.
[/quote]

Sorry  :oops: , i was also looking something about calculus and i confused
(someone move my topic)

By subsequence i mean: A subsequence arises from a sequence after removing one or more characters

[quote="mavropnevma"]Then it's $ 2^n \minus{} 2$ (we should not count the empty subsequence, nor the full sequence).[/quote]
yes but why?
thanks!

[quote]... there are as many subsequences as non-empty parts of $ \{1,2,\ldots,n\}$, hence $ 2^n \minus{} 1$ ...[/quote]
... with the added proviso we must also ignore the full part, hence $ 2^n \minus{} 2$.

The indices of the terms in a subsequence can be any such part of $ \{1,2,\ldots,n\}$.

why not count the empty subsequence?
if we remove all characters we have the null sequence


trigonometry function quadratics geometry circumcircle incenter algebra

find the exact value of
$\csc{20^\circ}-\cot{40^\circ}$

please post entire solution, ive been working this for like 2 hours and i can't find a solution. heres the answer if u want [hide]$\sqrt{3}$[/hide]

$\csc \theta=\frac{1}{\sin \theta} \Rightarrow \csc 20^\circ = \frac{1}{\sin 20^\circ}$

$\cot \theta=\frac{\cos \theta}{\sin \theta} \Rightarrow \cot 40^\circ = \frac{\cos 40^\circ}{\sin 40^\circ}$

$\cos 2(20)^\circ=(\cos 20^\circ)^2-(\sin 20^\circ)^2$

$\sin 2(20)^\circ=2\sin 20^\circ \cos 20^\circ$

$\csc 20^\circ - \cot 40^\circ=\frac{1}{\sin 20^\circ}-\frac{(\cos 20^\circ)^2-(\sin 20^\circ)^2}{2\sin 20^\circ \cos 20^\circ}$

$\frac{2\cos 20^\circ-(\cos 20^\circ)^2+(\sin 20^\circ)^2}{2\sin 20^\circ \cos 20^\circ}$

Umm... doesn't look like it will simplify to anything pretty. Another possiblity is to use $\cos 2(20)^\circ=2(\cos 20^\circ)^2-1$, but I don't think that having $-(2(\cos 20^\circ)^2-2\cos 20^\circ-1)$ as the numerator of the final function will help anything...

Can you use a calculator? :P

[b]EDIT:[/b] Hmm... I checked it on my calculator, and it's approximately $1.732$, which is not exactly equal to $\sqrt{3}$...

Wait a second, I might have done something wrong, I switched to degrees instead of radians, and I now get $\approx 1.99$, which is still not $\sqrt{3}$...

[quote="JesusFreak197"]

[b]EDIT:[/b] Hmm... I checked it on my calculator, and it's approximately $1.732$, which is not exactly equal to $\sqrt{3}$...

..[/quote]

$\sqrt{3}=1.73205$ :huuh:

In equilateral triangle draw a cevian at 20 deg  to the side and circumcircle the
   two triangles. Their orthocenters are on the circles.  The circles are equal.
   The incenter of the larger one  is also on the circumcircle of the smaller.
   From here it should not be difficult  to calculate 1/sin20 - cos40/sin40.

  (I would do it myself, but trigo is not my cup of tea).

   Maj.  Pestich

[quote="mkkool64"]that is a surprisingly long solution to a very simple looking problem. :?[/quote]
All right, let me try again:

$\csc{20^o} - \cot{40^o} = \frac{1}{\sin{20^o}} - \frac{\cos{40^o}}{\sin{40^o}} = \frac{2\cos{20^o} - \cos{40^o}}{\sin{40^o}} =$

Using $\cos{(x - y)} - \cos{(x + y)} = 2 \sin x \sin y$ with $x = 30^o,\ y = 10^o$:

$= \frac{\cos{20^o} + (\cos{20^o} - \cos{40^o})}{\sin{40^o}} = \frac{\cos{20^o} + 2 \sin{30^o} \sin{10^o}}{\sin{40^o}} =$

Substituting $\sin{30^o} = \frac 1 2,\ \cos{20^o} = \sin{70^o}$ and using $\sin{(x + y)} + \sin{(x - y)} = 2 \sin x \cos y$ with $x = 40^o,\ y = 30^o$:

$= \frac{\sin{70^o} + \sin{10^o}}{\sin{40^o}} = \frac{2\sin{40^o} \cos{30^o}}{\sin{40^o}} = 2 \cos{30^o} = \sqrt 3$

That's better, isn't it?

This is how I solved this problem......I was shocked when I see Yetti's first solution...


MATHCOUNTS
videos
geometry
function
summer program
MathPath
ARML

It does not play.  :maybe:
Errors display on the page.

hmm I wish I was on the news.. :(
Oh well I got on my local news at chapter :D

Were either of the kids AoPSers? Just wondering

lol they made it seem like it was easy...especially their sample question...

[quote="Sample on-air question"]Carla spent two-fifths of her savings on a bracelet and she had $\$9$ left over.  How many dollars did Carla's bracelet cost?[/quote]
Obviously, this was not a National-level question.  But, they were using the theme "Are you smarter than a math-whiz".  This question was much harder than the typical question on Jeff Foxworthy's "Are you smarter than a fifth grader?"

In general, any publicity for MATHCOUNTS is good publicity.  And two North Carolina Mathletes got some national airtime, along with other shots that showed several other Mathletes.  All in all, I think it was a nice segment.

I know one of the two people interviewed from MathPath. I don't think he's an AoPSer.

Eh, I think that news report made it seem that MathCounts was really easy.

ya seriously... -_-
that sample question was like chapter sprint...

more like chapter practice countdown

but why did they have the school teacher writing those large numbers and adding them???

jorian

[quote="LawOfSigns"]That was...disappointing.  Why didn't they put a much tougher problem like #25 state sprint (canceling middle digit in a fraction) on the news to demonstrate the difficulty of state/nat'l level Mathcounts questions?[/quote]

Because they wanted to give average Americans a chance at solving the problems so they would be engaged and interested and not go HAHAHA YOU'RE DUMB and hold that over their heads?

I also notice there were many other inaccuracies such as saying it was 8th graders only and such.

OH DUDE
DUDE
DUDE

THEY SHOWED ME IN MY PINK SHIRT FOR LIKE
3 SECONDS
BUT I LOOKED PRETTY STUPID
WITH A COWBOY HAT OVER A RED SOX HAT

I was in the background in the same scene as 1337...  With my yellow shirt lol.  I'm in for a split second...  kstan and the other kid on my team i think were blocking me  :(  :mad:  all you could see was my head for like .01 of a second...  I hate being short.

wow - that's you?

when i was thinking about what i should bring when i thought i was gonna go...i was gonna bring this shirt like wrapped in toilet paper, with matching hats and short shorts (and *bright* red shoes)

did anyone else have a weird outfit.

does anyone know if any other news teams did a small story on this event?
if so, speak up!

i wish i can be on the news, did they do it live? if it was live, it would have been pretty fun...... :D

the sample question was just stupid

don't they play mathcounts on espn?
seems like you can see everything on espn this days, anyone saw that rock paper scissor contest where the winner got like 500 thousand dollars.

[quote="alkjash"]The video was so bad... ughhh... adding a bunch of numbers and finding the ones that worked???
Finding the area with a bunch of weird dimensions?
They make math sound like a circus act.  :mad:[/quote]
I'm going to have to agree. They should have said that it seemed like an ugly problem with adding a bunch of numbers, but that they suspected that an elegant smooth solution involving generating functions existed. But still, TV pressure is immense, they didn't have much time to think these things I guess.

Well actually, I did see them talking for like half an hour, and I was like watching them too, before they went on the air. I wasn't eavesdropping or anything, but I do assume they were planning for something. I really should have asked Gray right after (I knew him from MathPath) or like Jenny at ARML (I think she was sitting next to me during the relays). Well I actually wanted to like stick my hand into the camera's view, but that's beside the point. But yeah, they probably got told not to say anything that sounded really complicated, except the use of the word "dimensions" made me fall out of my chair laughing. Also I was like DUDE that problem was pretty easy if you use areas but w/e that's beside the point as well.

[quote="alkjash"]The video was so bad... ughhh... adding a bunch of numbers and finding the ones that worked???
Finding the area with a bunch of weird dimensions?
They make math sound like a circus act.  [/quote]
I know, I didn't get it either. They make it sound so easy, but I didn't solve half the problems at the competition.  :mad: Besides, what did he mean by "weird dimensions"?

Yeah, instead of "rectangle with weird dimensions" they should have said something like "finding the area of a non-regular quadrilateral with irrational sides and trancedential angles" :)


find all sets of 2005 positive integers such that the sum of the fourth potence of any 4 elements of the set is divisible by their product

Probably this can help: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Vietnam+2002&t=37960

And then just take $a=x^2$, $b=y^2$, $c=z^2$, $d=t^2$ and I believe it simplifies the problem...

the equation i ask to solve is the following one:

  $a^4 + b^4 + c^4 + d^4 = nabcd$ i really dont see the relation with problem because the exponents vary can you fill me in?

Ops sorry, it was miss-read (I thought that the powers were 2, not 4). Sorry for posting trash.


superior algebra
superior algebra unsolved

Show that any automorphism of a subfield of $ \bf{C}$ can be extended to an automorphism of $ \bf{C}$.

Did someone made any progress on this one?

Hint: Use a transcendence base and split the field extension of the given field K -> C into (1) a strictly transcendent extension and (2) an algebraic extension. Now extend the given automorphism for (1), then for (2). For (2) use that C is algebraically closed.


Problem:Given a odd integer $ k$($ k>1$).Show that there is an integer m such that $ n^k\plus{}m$ is not a perfect square for every positive integer $ n$.

PS:When $ k\equal{}3$, we can point out a value of $ m$ ($ m\equal{}7$)
     When $ k\equal{}5$, $ m\equal{}31$ satisfies the problem.
     When $ k\equal{}7$, $ m\equal{}7$ satisfies the problem.
     .............................................................

For any $ k>3$, one can take $ m\equal{}1$ and apply Mihăilescu's theorem (ex-Catalan conjecture) to prove that $ n^k \plus{} 1$ is never a square.

[quote="maxal"]For any $ k > 3$, one can take $ m \equal{} 1$ and apply Mihăilescu's theorem (ex-Catalan conjecture) to prove that $ n^k \plus{} 1$ is never a square.[/quote]
What is Mihăilescu's theorem?Is it hard to prove?
I don't think $ n^k\plus{}1$ is never a square for any odd integer $ k$.

[quote="SUPERMAN2"]Is it hard to prove?[/quote]
For me the it is hard to prove but I don't know how talent you are. You can read for example the book "Catalan's Conjecture" by René Schoof. It contains a proof of the theorem but you should know some algebraic number theory. There are some irritating typos in the edition I have.


geometry
circumcircle
cyclic quadrilateral
projective geometry
geometry solved

1. ...
2. Does anybody knows a simple solution of the theorem:
    Let $ABCD$ be a cyclic quadriteral and let $M= AB\cap CD$,
    $N=AD\cap BC$ and $L=AC\cap BD$. Then $MN$ is a polar of $L$

[color=blue][b](2)[/b][/color] Construct tangents to the circumcircle $(I)$ of the cyclic quadrilateral $ABCD$. Let the tangents at $A, B$ meet at a point $P$, the tangents at $B, C$ at a point $Q$, the tangents at $C, D$ at at a point $R$ and the tangents at $D, A$ at a point $S$. The intersection points $P, Q, R, S$ form a tangential quadrilateral $PQRS$ with the diagonals $PR, QS$, the incircle $(I)$ and the contact quadrilateral $ABCD$. The points $P, Q, R, S$ are poles of the chords $AB, BC, CD, DA$ and conversely, the chords $AB, BC, CD, DA$ are polars of the points $P, Q, R, S$. Consequently, the intersections $M \equiv AB \cap CD$, $N \equiv BC \cap DA$ are poles of the diagonals $PR, QS$ and the line $MN$ connecting the poles $M, N$ is a polar of their intersection $K \equiv PR \cap QS$ and conversely, $K$ is a pole of the line $MN$.

Using Pascal's theorem for the cyclic quadrilateral $ABCD$ considered as a degenerate hexagon $ABBCCD$, the intersections $M, L$ of the chord pairs $AB \cap CD$, $AC, \cap BD$ and the intersection $Q$ of the tangents $PQ \cap QR$ at the points $B, C$ are collinear.

Using Pascal's theorem for the cyclic quadrilateral $ABCD$ considered as a degenerate hexagon $AABCDD$, the intersections $M, L$ of the chord pairs $AB \cap CD$, $AC, \cap BD$ and the intersection $S$ of the tangents $RS \cap SP$ at the points $D, A$ are collinear.

Hence, the 4 points $L, M, Q, S$ are all collinear.

Using Pascal's theorem for the cyclic quadrilateral $ABCD$ considered as a degenerate hexagon $AABBCD$, the intersections $N, L$ of the chord pairs $BC \cap DA$, $AC, \cap BD$ and the intersection $P$ of the tangents $SP \cap PQ$ at the points $A, B$ are collinear.

Using Pascal's theorem for the cyclic quadrilateral $ABCD$ considered as a degenerate hexagon $ABCCDD$, the intersections $N, L$ of the chord pairs $AB \cap CD$, $AC, \cap BD$ and the intersection $R$ of the tangents $QR \cap RS$ at the points $C, D$ are collinear.

Hence, the 4 points $L, N, P, R$ are all collinear.

Since the point $L$ is on both the diagonals $PR, QS$ of the tangential quadrilateral $PQRS$, it is identical with their intersection $K$, i.e., $K \equiv L$. But it was shown earlier that the point $K$ is a pole of the line $MN$.

Hmm. Yes, well I mean shorter one and more elementary.
Anyway, thank you Yetti.

let $P \equiv NL\cap DC,Q \equiv NL\cap AB$

in triangle NDC use ceva and menalus you can get:

$\frac{NB}{BC}.\frac{CP}{PD}.\frac{DA}{AN}=1$ (1)

$\frac{NB}{BC}.\frac{CM}{MD}.\frac{DA}{AN}=1$ (2)

from (1) and (2) you can get:$(MP,DC)=-1$

and therefore $N(MQ,AB)=-1$

I think it's shorter and easier.

I didn't understand first problem $Number$ $1$, can U explain more?

Thanks yetti for your brilliant and very interesting solution!!!
But I do not understand how you taught of the applications of the Pascal's them 4 times considering  the degenerate hexagon with the given cyclic quadrilateral ABCD??
Please share it.


Hi everyone,

Could anybody help me with these problems.

1.A man placed a ladder against the wall such that it would just reach the window.It is inclined at an angle BETA.Later he placed the ladder such that it would reach the parapet.The foot moved a distance of 'a' meters and is inclined at an angle 'ALPHA'.Prove that height of the parapet from
the window is [a*(cot(ALPHA+BETA)/2)]

2.The angle of elevation of a tower from a point is ALPHA and the angle of elevation of a man standing on the tower is BETA.The height of the man is 'a' meters.Prove that height of the tower is
[a*sin(ALPHA)*cos(ALPHA+BETA)]/[sin(BETA)]

Thanks in advance,
Alfy.

[quote="kunny"]Can you give us a figure?[/quote]

For problem 1 -

[quote="kunny"]Can you give us a figure?[/quote]
For Problem 2

[quote="asebastian"]
1.A man placed a ladder against the wall such that it would just reach the window.It is inclined at an angle BETA.Later he placed the ladder such that it would reach the parapet.The foot moved a distance of 'a' meters and is inclined at an angle 'ALPHA'.Prove that height of the parapet from
the window is [a*(cot(ALPHA+BETA)/2)][/quote]

I haven't solved this problem.I think this statement has something to lack of condition. :maybe:

Here is my approach.

Let $d$ be the distance to the window, we have $h_{w}=d\tan \beta,\ h_{p}=(d-a)\tan \alpha .$ Elliminating $d,$ we have $h_{p}=\left(\frac{h_{w}}{\tan \beta}-a\right)\tan \alpha .$ Thus the desired hight is $h_{p}-h_{w}=\frac{\tan \alpha-\tan \beta}{\tan \beta}h_{w}-a\tan \beta .$

[quote="kunny"][quote="asebastian"]
2.The angle of elevation of a tower from a point is ALPHA and the angle of elevation of a man standing on the tower is BETA.The height of the man is 'a' meters.Prove that height of the tower is
[a*sin(ALPHA)*cos(ALPHA+BETA)]/[sin(BETA)][/quote]

This is easy to solve more than the problem 1.

Let $d$ be the distance to the tower, we have $h=d\tan \alpha ,\ a+h=d\tan \beta.$ Elliminating $d,$ we have $h=\frac{a\tan \alpha}{\tan \beta-\tan \alpha}.$[/quote]

Can we arrive at a*sin(ALPHA)*cos(ALPHA+BETA)]/[sin(BETA) from the equation u have mentioned above?

[quote="kunny"]Yes, you can. Rewtite $tan \alpha =\frac{\sin \alpha}{\cos \alpha}$

Use the formula $\cos (x+y)=......$[/quote]
Hi,

I am not sure I got that..Could you please explain a little further..I was not able to come at the equation using that..

Thanks for your help

[quote="kunny"]All right.

$h=\frac{a\tan \alpha}{\tan \beta-\tan \alpha}=\frac{a\cdot \frac{\sin \alpha}{\cos \alpha}}{\frac{\sin \beta}{\cos \beta}-\frac{\sin \alpha}{\cos \alpha}}$

$=\frac{a\sin \alpha \sin \beta}{\sin \beta \cos \alpha-\cos \beta \sin \alpha}$

$=\frac{a\sin \alpha \sin \beta}{\sin (\beta-\alpha)}$[/quote]

Thanks for your efforts
But isn't it different from [a*sin(ALPHA)*cos(ALPHA+BETA)]/[sin(BETA)]?

This seems unattainable for me  :(


Anyone have the 2007 NEAML and State meet problems from this year?

I could get the NEAML ones from my coach...

Bump, I am also looking for these problems (particularly the team round).

[quote="hedgehog12"]I don't have it, but you could google it.[/quote]

The only place that hosts NEAML problem sets is this page http://www.matharchive.org/neaml/


function
calculus
real analysis
topology
algebra
functional equation
real analysis theorems

I am reading a book called "The Calculus Gallery" where it states "It is fair to say that any function ever considered prior to 1900 belonged to the family of Lebesgue-measurable functions." It doesnt seem wrong to me, but can anyone think of any counterexamples to the claim?

Thanks!

Los

I believe this post should go in one of the forums under "College Playground," probably in the "Real Analysis" forum.

The claim seems fair to me. How could you possibly produce a counterexample? Not by accident.

In the early 19th century, Fourier caught flack for talking about piecewise-defined functions, even functions that were "pretty nice," as in piecewise $ C^1.$ But they violated expectations that were rooted in what we now recognize as analyticity. The modern definition of function didn't really exist until sometime in the first half of the 1800s.

As late as the 1870s or 1880s, one can find quotes decrying such examples as a nowhere differentiable continuous function as being abominations and not what mathematics should be about - and we can write such things very explicitly as uniformly convergent series. There's no doubt about measurability.

I can't imagine talking about anything like the "standard construction" or a nonmeasurable set, or about additive (but not linear) functions on the reals, before the general (and controversial) notions about sets and cardinality as advanced by Cantor in the 1870s and 1880s, and Cantor's work itself did not go far enough by itself to produce anything nonmeasurable.

In looking up Zermelo's Axiom of Choice, Zorn's Lemma, and the Hausdorff Maximal principle, I see dates quoted that range from 1904 to 1922. Given that, I find the "before 1900" claim as quite believable.

OK, someone before then might have wondered about solutions of the functional equation $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ - but without some Axiom of Choice equivalent, how are you supposed to show that there are any monsters among the solutions to that?

The claim isn't so shocking since measure theory itself was born around the turn of the century.  It's sort of like saying non-Hausdorff topological spaces were never considered before topology was invented.

But the claim depends on what it means for a "function" to have been "considered."  Our modern set-theoretic concept of a function essentially comes from Cantor (or at least he is the one most responsible for its wide adoption), and since Cantor's concept of a function does include non-measurable functions, one might say that such functions were "considered," even though no *specific* examples of such a functions were contemplated.

To me the more interesting historical statement is that non-measurable functions probably weren't even sensible mathematical objects in the pre-Cantor world (just as Prof. Merryfield said).  From a modern point of view, it's quaint how the function concept evolved over time.  Calculus proved the usefulness of  the function concept, but for a very long time a function had to be something given by a formula.

And, indeed, one can still [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=253408]find people[/url] who believe that every function must have a formula.   :wink:


Hi folks.

I would like to have all my equations \noindent. Or flushed left. Is there any ways of doing this? Tried using \mathindent, but that didnt work (dont know which usepackage).

Sincerly,
Kristian

Use the fleqn option as in \documentclass[12pt,fleqn]{article} or whatever class you are using. Then all equations will be flushed left with an indent given by \mathindent (which only works with fleqn) so to have no indent at all you put \setlength{\mathindent}{0cm}.

Thanks for the help! Worked the way it should. Im new to this website and really impressed by the quick answer!