calculus
integration
derivative
linear algebra
matrix
algebra
system of equations

Hello,

This is my first post, but I've browsed and searched these forums for help and suggestions for a good while now.

I'm not sure if this portion of this problem belongs in this forum, but as I get further into the problem integration will play a role and I'm sure I'll have questions about that so I assume it fits here.

Anyway, I'm given a chart of time and velocity of an object:

t =  0,  10,  20,  30,  40,  50,  60   (respectively)
v = 0,   5,   21,  42,  62,  78,  83

Now I'm asked to solve a system of equations to find a model in the form: v = at^3 + bt^2 + ct + d

I know the solution to this is basic, but I'm drawing a complete blank here.  I'm not looking for the answer specifically, but more of which direction to take this.

Thanks!

The solution won't involve integrals. Depending on the methods you use, it might involve derivatives, but I've decided to move it here, to the linear algebra section.

The first thing to note: this problem is [b]overdetermined[/b]. You have seven points of data. Any four of those points will suffice to determine a cubic equation. But that means that if you take the first four points, you will (probably - I haven't checked) determine a different cubic than the one you determine with the last four points.

What do we do with that? One approach is to argue that because no such equation will go through all seven points, then the equation has no solution - and walk away. And before we make it out the door, the experimental scientist who collected the data is yelling for us to come back - we can't get away that easily. To the experimentalist, having seven points is [i]better[/i] data than having a mere four points would be - and he's right. We need to find the "best compromise" solution among our possibilities - as soon as we define what "best" means.

How familiar are you with the subject of [b]least-squares curve fitting[/b]?

Least-squares curve fitting can be done using a matrix [url=http://en.wikipedia.org/wiki/Pseudoinverse]pseudoinverse[/url], preferably with Matlab. The following code
[quote]t=0:10:60
v=[0,5,21,42,62,78,83]
A=[(t.^3)',(t.^2)',t',ones(size(t))']
coeff=pinv(A)*v'
approx=coeff(1)*t.^3+coeff(2)*t.^2+coeff(3)*t+coeff(4)
[/quote]
returns four coefficients and also the cubic approximation to $v$:
$\textrm{approx }=-0.0952,\ 5.2619,\ 20.9048,\ 41.6667, \ 62.3810,\ 77.8810, \ 83.0000$
Accurate within $0.4$.  :)

[quote="Kent Merryfield"]How familiar are you with the subject of [b]least-squares curve fitting[/b]?[/quote]

Not famaliar at all; but if I recall, that would be using a straight line to approximate an equation for the set of data?  Maybe?

I didn't think it would be quite this complicated.  I was assuming you could just grab a few simple equations using the points given to derive the velocity equation.  I'm just stumped where these equations would come from.

Thanks

Well, a line fit - a regression line - is one particular kind of least squares curve fitting. However, in this instance, we were asked to fit a cubic equation, not a linear equation.

This can be expressed as a max/min problem. You have given data $(t_{i},v_{i})$ for $1\le i\le 7.$ Define

$g(a,b,c,d)=\sum_{i=1}^{7}(at_{i}^{3}+bt_{i}^{2}+ct_{i}+d-v_{i})^{2}$

Now, find $a,b,c,d$ that minimize this function.

[The answer to this problem should be identical to the one mlok produced, although he chose not to report $a,b,c,d$.]

Alright, a few problems here:

First, this is all [i]wayyyy[/i] over my head.

Second, I have to solve this in [i]Mathematica[/i], and am very new to this software as well and am not even a little bit sure how that translates to Mathematica.

Also, is a summation like the one stated considered a system of equations?  If I'm missing something obvious, you'll have to forgive me as this is all a bit past my level of understand currently.

Thanks

The system of equation to solve is
\[\begin{cases}\frac{\partial g}{\partial a}(a,b,c,d)&=0\\ \frac{\partial g}{\partial b}(a,b,c,d)&=0\\ \frac{\partial g}{\partial c}(a,b,c,d)&=0\\ \frac{\partial g}{\partial d}(a,b,c,d)&=0\\ \end{cases}\]
because at the point of minimum (which is unique) all partial derivatives are equal to zero. For example,
\[\frac{\partial g}{\partial a}=\sum_{i=1}^{7}2(at_{i}^{3}+bt_{i}^{2}+ct_{i}+d-v_{i})t_{i}^{3}\]
by the Chain Rule. You get four equations with four unknowns.

Four [i]linear[/i] equations in four unknowns - there's a reason why I moved this into the matrix section, and why mlok used a matrix command in MATLAB for his solution.


Dude, I am willing to bet you $\$ 100$ right now that you make it.

i hear its easiest to make MOSP when you're a freshman or junior. why is that?

and it's hardest when you're a senior because you have to be a USAMO winner? what exactly does that mean?

Note: these numbers are not exact, but only rough estimates.

Well, when you're a freshman, you can make Red MOSP, which is a group of about 30 freshmen, who scored the highest on USAMO.
Since fewer people are eligible, those that are have a better chance.  The cutoff here is usually ~7.

All grades, with the exception of seniors, are eligible for Blue MOSP, which has about 20 or so participants.  Thus, as a junior you will be the most experienced you can possibly be while still being eligible.  The cutoff here is generally 15-20.

Everyone is eligible for Black MOSP (and is the ONLY thing that seniors are eligible for).  Only 12 students are selected, the top 12 scorers on the USAMO, also called the USAMO winners (Blue MOSP is then the next 20 or so non-seniors after that).  The cutoff to make black is generally between 25-30, although it was 35 in 2002 (easy USAMO that year, though).

Black MOSPers then take the Team Selection Test, after which 6 of them will be selected for the US IMO team.

[quote="flamesong522"]i got 216.5 as usamo index

what is my chance to getting in

last year it was 217

i am so worried about it

p.s i'm junior[/quote]
The AIME this year was harder than the AIME last year, AND the USAMO is being expanded more, so I'd be extremely surprised if 216.5 doesn't make it. Start studying for USAMO.

I'd be more looking at "is it going to be <205 or not?" not "will I get in with a 216.5?" So don't worry  :wink:

As people said, there are many many factors that will make the cutoff much lower than last year's.

Please try to refrain from individual threads like this. Instead, post in the bigger threads about USAMO indices.

Locked.


linear algebra
matrix
limit
absolute value
linear algebra unsolved

Find necessary and sufficient condition  for matrix $ A$ such that the limit $ \displaystyle\lim_{n\to\infty}A^n$ exists.

Let $ A$ be a square complex matrix, the limit matrix will only exist if all the eigenvalues are either one or have absolute value less than one. (In fact $ \lim_{n\to \infty}A^n\equal{}0$ if and only if all the eigenvalues of $ A$ have absolute value less than one.) Just triangularize the given matrix to prove these assertions.


Έστω $ P(x)$ πολυώνυμο με πραγματικούς συντελεστές τέτοιο ώστε $ P(x)\geq 0$ για κάθε πραγματικό αριθμό $ x$. Αποδείξτε ότι υπάρχουν πολυώνυμα $ Q_{1}(x),Q_{2}(x)$ με πραγματκούς συντελεστές τέτοια ώστε $ P(x)\equal{}Q^{2}_{1}(x)\plus{}Q^{2}_{2}(x)$ για όλους τους αριθμούς $ x$.

Ως μη αρνητικο το πολυωνυμο εχει ριζες πραγματικες αρτιας πολλαπλοτητας και συζυγεις μιγαδικες .Επισης τα πολυωνυμα $ P(x) \plus{} iQ(x)$ kai  $ P(x) \minus{} iQ(x)$ εχουν ριζες συζυγεις (ευκολα αποδ.)
Οποτε

$ P(x) \equal{} [(x \minus{} r_1)...(x \minus{} r_k)(x \minus{} z_1)...(x \minus{} z_m)][(x \minus{} r_1)...(x \minus{} r_k)(x \minus{} \overline{z_1})...(x \minus{} \overline{z_m})]\equal{}
(P(x) \plus{} iQ(x))(P(x) \minus{} iQ(x))\equal{}P(x)^2 \plus{} Q(x)^2$

Με εντυπωσίασε η λύση σου. Δεν ξέρω εαν η δική μου είναι σωστή γι'αυτό θα ήθελα να την κοιτάξεις...

Το πολυώνυμο $ P(x)$ είναι της μορφής $ P(x) \equal{} c\cdot \prod^{n}_{k \equal{} 1}(x^{2} \minus{} p_{k}x \plus{} q_{k})$ με $ c,p_{k},q_{k}\in R$, $ c\geq 0$ και $ p^{2}_{k} \minus{} 4q_{k}\leq 0$ για $ k \equal{} 1,2,...,n$.

$ x^{2} \plus{} p_{k}x \plus{} q_{k} \equal{} \left(x \plus{} \frac {p_{k}}{2}\right)^{2} \plus{} \left(\frac {\sqrt {4q_{k} \minus{} p^{2}_{k}}}{2}\right)^{2} \equal{} A^{2}_{k}(x) \plus{} B^{2}_{k}(x)$

Άρα αρκεί να αποδείξουμε ότι το γινόμενο $ c\cdot \prod^{n}_{k \equal{} 1}(A^{2}_{k}(x) \plus{} B^{2}_{k}(x))$ γράφεται ως άθροισμα 2 άλλων πολυωνύμων και το οποίο ισχύει απο την ταυτότητα Lagrange, δηλαδή  $ c\prod^{n}_{k \equal{} 1}(A^{2}_{k}(x) \plus{} B^{2}_{k}(x)) \equal{}c\cdot (Q^{2}_{1}(x) \plus{} Q^{2}_{2}(x))$.

Σωστος ...  συμληρωσε μονο στο τελος οτι τα πολυωνυμα ειναι επι $ (\sqrt{c})^2$


There are $n$ points on a sphere of radius $1$. If the $n$ points are spread apart so that the surface is uniformly covered, what is the distance between two adjacent points?

spread out equally. the distance between any two adjacent points is invariant.

How do you know there always exists a way to "uniformly distribute" n points on a sphere?

Once you establish that there is, how do you know that there's not [u]more[/u] than one way to "uniformly distribute" n points on a sphere, each with different distances between the points?

Exactly. Take three points. Any "equilateral curved triangle" satisfies your criterion of uniform distribution, and the "side" of such "triangle" can be anywhere from $0$ to ${2\pi\over 3}$

Here's the statement which is more workable and is much closer to what I think you have in mind: $N$ electrically charged particles, all positive and with equal amount of charge, are located on the surface of a unit sphere and can move freely over the surface. What is the distance between any two adjacent particles in the final state of equilibrium?

Yes, but where the "charged particles" would move may depend on how the force varies with distance.  I think that there might be more problems.
The best way to word the problem is:
There are $N$ points on a unit sphere.  The least distance between two points is $d$.  What is the maximum possible value of $d$?

(This is a packing problem, and I highly doubt you will find a gerneral solution.)

[quote="draperp"]Yes, but where the "charged particles" would move may depend on how the force varies with distance.  I think that there might be more problems.[/quote]

Because of the repelling forces, the particles will tend to distance from each other as much as possible, and they will also tend to position themselves so that the potential energy of the system is minimized. There are easy physics formulae which describe both the distance and the potential energy of such a system, though the resulting equations admittedly can be very cumbersome.

[quote="Potato Theory"]Well, I'm not sure if this works or not, but:

$A = 4\pi r^{2}$

Only the values of $n = 4, 6, 8, 12, 20$ can be placed equidistant on a sphere, so $\frac{4\pi r^{2}}{n}$ would be the area each point occupies.  Hmm... not sure where to go from there though...[/quote]

that's what i'm thinking. then i thought about drawing a circle centered at each point, passing through all the adjacent points. then considering overlap maybe you could find an expression and set it equal to the surface area. but i'm not sure about finding the surface area of a circular cap.

Taking a swing at this:

Since distances between points are invariant, there are three solution types:

1. Two points must be on opposite sides of the sphere. Distance is 2.

2. A regular polygon inscribed in any circular cross section containing the center of the sphere. They are all equidistant, and if confined to the sphere, "charged particles" in those places could not go anywhere. There are infinite such solutions. The distance between adjacent points is  $d=\sqrt{2-2cos(\frac{360^{o}}{n})}$

3. A regular polyhedron, inscribed within the sphere, where n is the number of verteces. Since there are only five regular polyhedrons, each with its own number of vertices, there are only five values of n that can be solved in this way, and the corresponding distances can be calculated case-wise.


[url=http://en.wikipedia.org/wiki/Laugh_track]wikipedia[/url]
at the risk of making a post similar to the 'which show you like best' topic, I'd like to ask something about laugh tracks
Maybe it should be in games and fun, but my other post was moved to this place so therefore i try it here..

I catched a glimpse of a rerun of 'O Doctor Beeching' and 'Step by step' and I noticed something.  I wasn't laughing .  Then who was?  The laugh track! :ninja:

Being a huge tv and movie fan, I wondered if I am only the only one opposing laugh tracks?

I think laugh tracks stands for cheap, predictable humour that isn't funny but they're telling you to do so.

The 'new' wave of shows I listed like Family Guy, Futurama, Simpsons... don't use laugh tracks.  And what happens?  They actually are funny!!
It also allows being funny all the time with hidden jokes, without 22 minutes of incessant laugh tracks.

I personally think TV is going down worldwide to horrible evils, among them reality tv, we must not give up! :cleaning:

I dislike laugh tracks.  If just Mr. Bean had not used laugh tracks...it would be that much funnier...
And yes, shows without laugh tracks are generally far better.


How could you denote a combination using the parenthesis...

I have attempted to do so using and array within the larger parenthesis... but it comes out looking huge!
 :oops:


A sequence $(a_{n})$ is deﬁned by $a_{1}= 1$ and $a_{n}= a_{1}a_{2}...a_{n-1}+1$ for $n \geq 2.$ Find the smallest real number $M$ such that $\sum_{n=1}^{m}\frac{1}{a_{n}}<M\; \forall m\in\mathbb{N}$.

Yes, because $\lim_{n\to\infty}a_{n}=\infty$ therefore $\lim_{m\to\infty}(2-\frac{1}{a_{m+1}-1})=2$.


I'm not sure if i did the math right and i need someone to check it over please~

The question is:

Suppose a mountain being uplifted at 2m per 1000 years. It is also being eroded at 2cm per year.

-Is it rising faster then eroded?
-explain why
-after 100 years what will be the change in the height of this mountain.

So
2m=1000years(rise)
2cm=1 year(eroded)

I converted cm to meter

1cm=0.01
2cm=0.02

So to figure out by how much it is being eroded every years and the rate of its uplift i did

2m=1000years(rise)(2/1000=0.002m every year.)
2cm=1 year(eroded) 0.02m every year.

So the mountain is being eroded faster then the rise? I don't think thats right... unless they picked random numbers in this problem as the uplift and eroding. Mountains should rise faster then it eroding.


Can anyone suggest astrophysics presentation topics for a school physics workshop?(grade 12)

How broad is the term "astrophysics"?  I would suggest a topic in cosmology, perhaps dark energy and the expanding universe.

I chose "Black holes:A thermodynamical Aspect"
I hope to include Hawking Radiation, how the laws of thermodynamics can be applied to black holes and the concept of a White hole(the time reversal of a black hole) and how it is entropically indistinguishable from a black hole

Please tell what u think I can do to make it more intersting
and give ur suggestions.

Thank u,
Awaiting ur reply

I don't really have much interest in astrophysics (I prefer cosmology) but it seems interesting.  Make sure you convey what makes black holes fascinating.


Let trangle ABC such that $ sinA\equal{}2sinB.sinC$
Prove: $ B\equal{}C\equal{}45^0$

[hide]
$ \sin(180^\circ \minus{} (B\plus{}C)) \equal{} 2\sin{B}\sin{C}$

$ \sin((180^\circ\minus{}B) \minus{} C) \equal{} 2\sin{B}\sin{C}$

$ \sin(180^\circ \minus{} B)\cos(C) \minus{} \sin(C)\cos(180^\circ\minus{}B) \equal{} 2\sin{B}\sin{C}$

$ \sin{B}\cos{C} \plus{} \sin{C}\cos{B} \equal{} 2\sin{B}\sin{C}$

$ \sin{C}\cos{B} \equal{} \sin{B}\sin{C}$

$ \tan{C} \equal{} \tan{B}$

since B and C are both angles in the same triangle, $ B\equal{}C$

back to the original equation

$ \sin(180^\circ \minus{} 2B) \equal{} 2\sin^2{B}$

$ \sin(2B) \equal{} 2\sin^2{B}$

$ 2\sin{B}\cos{B} \equal{} 2\sin^2{B}$

$ \tan{B} \equal{} 1$

so $ B \equal{} 45^\circ$ (because B=C, it cannot be 135)

whew, that was rather longer than I expected
[/hide]

[quote="nayel"][quote="chickendude"]$ \sin{B}\cos{C} \plus{} \sin{C}\cos{B} \equal{} 2\sin{B}\sin{C}$

$ \sin{C}\cos{B} \equal{} \sin{B}\sin{C}$[/quote]

I think something's wrong in this part.[/quote]

Yeah, $ \sin{B}\cos{C}\plus{}\sin{C}\cos{B}\equal{}\sin{(B\plus{}C)}$

$ \sinh = 2\sinh \sinh$, and $ A + B + C = 180^0$
$ \sin(B + C) = 2 \sinh \sinh$
$ \sinh\cosh + \sinh\cosh = 2\sinh\sinh$, because $ \sinh \sinh \neq 0$
$ \coth + \coth = 2$

This doesn't imply they must be equal. Try $ \arctan2$ and $ \arctan\frac{2}{3}$.

P.S. What's wrong with the display?

When you used \sinB, apparently sinB was treated as one thing, and it doesn't exist, so nothing showed up


Here's a question I've been having a lot of trouble with. If I could have some help, that'd be great.

According to a police report, a car was sitting at rest at a stoplight when it was hit from the rear by a truck weighing 1.5 times as much as the car. At the time of the collision both vehicles had their four-wheel brakes on, and skid marks showed that they skidded together 7.8 m in the original direction the truck was traveling before coming to rest. Assuming a coefficient of friction of 0.8, what must have been the approximate speed of the truck just before the collision?

This is from my homework assignment, and I just really have no clue about how to approach it.  Help?

I think the braking force is factored into the coefficient of friction, so use work-energy principle to find kinetic energy of the car + truck just after the collision; then, use conservation of momentum to find the original velocity of the truck.

Okay, so I tried this problem to.  I am a bit confused.  So if you do work-energy theorem you get 1.25mv^2 = the KE just after the collision, right?  But we are unable to solve for velocity or KE, so when you go to the next step, the conservation of momentum, we have no values to put in for the velocity.  Plus, the cars are moving together so there is not a opposite reaction.  So, you can't use the mvo = -mvf  (vf = final velocity and vo=original velocity)  right?   Some help would be greatly appreciated.

What was the change in energy? What was the work done due to external forces? Those are equal.

[color=blue]Here is my opinion
Consider $M$ and $m$ are the mass of the truck and the car.
The conservation of momentum give us
$\large Mv_0=(M+m)v \Longleftrightarrow v=\frac{M}{M+m}v_0$
The conservation of enery give us
$\large\frac{1}{2}(M+m)v^2=k(M+m)gS\Longleftrightarrow \frac{1}{2}\frac{M^2}{M+m}v_0^2=k(M+m)gS$
$\large\Longrightarrow v_0=\frac{M+m}{M}\sqrt{2kgS}$
If I'm wrong, please correct.[/color]


induction
combinatorics proposed
combinatorics

there are 8 different symbols designed on n diffrent T-shirt,where n>=2,it's known that each shirt contains at least one symbol,and for any two shirts, the symbols on them are not all the same, also, for any k symbols,1=<k<=7, the number of shirts containing at least one of the k symbols is even,find the value of n.

Dear omid I know the sol but I'm searching for alternative one :D

No reason to leave this one unsolved for so long :).

We can consider the T-shirts to be subsests of a set $S$ with $8$ elements. Assume $n$ is even. That statement about parity can also be translated as follows: for each subset $S'$ of $S$ with $1\le k\le 7$ elements, the number of subsets of $S'$ among our sets has the same parity as $n$. Since $n$ is even, this number always has to be even.

Let $1\le k\le 7$ be the first positive integer for which we have a set $S'$ with $k$ elements. The number of subsest of $S'$ must be even, but it's odd, because $S'$ is the only subset of $S'$ which appears on a T-shirt, so we have a contradiction. This shows that $n$ is odd (just like Omid said :)).

We can now easily prove that all the non-void subsets of $S$ are part of a T-shirt by induction on the number of elements, going from $1$ to $8$ and keeping in mind that for each $\emptyset\ne S'\subseteq S$, the number of subsest of $S'$ which appear on a T-shirt has the same parity as $n$, so it's odd.


I thought this movie was particularly funny, especially the physics classroom scene.  I mean, what physics professor would ask for a computation of eigenvalues in the middle of class?  And this was the funniest part:

[i]Professor[/i]: Can't anyone here tell me the eigenvalues of this map?  Somebody, please!

Peter?

[i]Peter[/i]: 0.23 ... electron volts.

[i]Professor[/i]: Excellent work.

--------------

Eigenvalues don't have units!!!!

I liked Dr. Oct more than the green goblin, which was why I liked this movie more than the first one. I hope the next one doesn't involve the green goblin, but I am pretty sure that he is coming back.

Actually, it seems likely that the filmmakers woud wait another film before attempting to resolve the issue of the Green Goblin and his potential successor.  Harry's pursuit of Spiderman was a driving factor in <I>Spiderman 2</I>, and it might be prudent to wait until the fourth film or later to advance the issue any further.


Dear everyone.

I got a very interesting problem just right now.
Because of my ugly solution, I post it on forum. :blush:

Please read the following.
For two triangle ABC and DEF, their areas are equal.
AB>DE and DF>AC.
Do there exist two points G, H (on BC, EF respectively) satisfying that AG=DH, AGC=DHF (their area).

May I ask you proofs of art ?  :)
Thank you for reading my shoddy post.

Hello!

[...] - the area of the triangle ...
We know that $ \left[ABC\right]\equal{}\left[DEF\right]$

Construct the triangle $ \Delta AMN$ such that $ \Delta AMN\cong\Delta DEF$, $ BM\parallel{}CN$ and
$ \left(MN\right)\cap\left(BC\right)\equal{}\left\{G\right\}\Rightarrow\frac{GB}{GC}\equal{}\frac{GM}{GN}\Rightarrow\frac{BC}{GC}\equal{}\frac{MN}{GN}$   (1)
Construct the point $ H\in\left(EF\right)$ such that $ HF\equal{}GN$   (2)
$ \Delta AGN\cong\Delta DHF\Rightarrow AG\equal{}DH$   (*)

(1),(2) $ \Rightarrow\frac{BC}{GC}\equal{}\frac{EF}{HF}\Rightarrow \frac{\left[ABC\right]}{\left[AGC\right]}\equal{}\frac{\left[DEF\right]}{\left[DHF\right]}\Rightarrow\left[AGC\right]\equal{}\left[DHF\right]$   (**)

(*),(**) $ \Rightarrow$ we are done!

Best regards, Petrisor Neagoe

@Petry: Do you know a neat way of proving such construction is possible? I think that is the center of the result.

I have some proofs of the problem  but I can't avoid using intermediate value theorem to prove that my constructions (which follow the same idea as your proof) are possible.

 :)


abstract algebra
symmetry
real analysis
calculus
group theory
algebra
function

I will be taking Abstract Algebra next semester, and the professor is trying to figure out which book go with.  We currently use the book by Dummit and Foote, and the professor is deciding between that book and Nicholson's.  Does anyone have any suggestions?  Which books worked well, which didn't?  Strengths, weaknesses?

Just a note: don't suggest a graduate-level book that already assumes knowledge since this is a first course.

Could you explain the contents of the course a bit more? It's a first course; is it 1st year, 2nd year undergraduate? What material will be covered? Only groups, or also rings, modules, fields...?

Personally I like Dummit and Foote; I just bought it to supplement the [url=http://www.amazon.com/Groups-Symmetry-Undergraduate-Texts-Mathematics/dp/0387966757/ref=sr_1_1?ie=UTF8&s=books&qid=1236540434&sr=8-1]horrible book by Armstrong[/url]in my Group Theory course. Dummit & Foote has clear explanantions, with lots of examples, but does have a 'theorem-proof'-format. In fact, I heard it is a widely used graduate-level book, but in my opinion it is perfectly suitable for 2nd year undergraduate.
It's not clear to me whether you already have a copy of Dummit & Foote?

Other well known books are Algebra by Artin, Topics in Algebra by Herstein.

PS: I think this topic should be in the Superior Algebra section instead of Advanced Fields.

I'd advise against Artin if only because Artin himself is currently working on a second edition and I don't know whether your professor will be able to get it in time.  There are some substantial revisions, at least as far as the exercises go.  The first edition is certainly a great reference if you already have a copy.

[quote="BlueVelvet"]Could you explain the contents of the course a bit more? It's a first course; is it 1st year, 2nd year undergraduate? What material will be covered? Only groups, or also rings, modules, fields...?[/quote]
The course assumes you know linear algebra so it is a first course in algebra.  As for what year level, that's more difficult to quantify.  I know several 2nd year students taking algebra now and I certainly could have taken it this year but I opted for analysis.  Similarly, there are certainly upper level students who for some reason or another haven't taken it yet.  I'll be taking it 3rd year but again, only because I opted to take analysis this year.

I don't currently own any of the books because I don't want to shell out $ \$150$ for a book only to discover that we're using a different one.

Since you already have the book, I'll tell you what sections of the book are currently being covered and you could figure out exactly what stuff it is:
1.1-1.7
2.1-2.5
3.1-3.5
4.1-4.6
7.1-7.6
8.1-8.3
"Selected topics of chpt 9"

Again, I'm not really sure exactly what material that constitutes.
[quote="BlueVelvet"]Dummit & Foote has clear explanantions, with lots of examples, but does have a 'theorem-proof'-format.[/quote]
I was under the impression that most upper level books tend to be theorem-proof.  Is this any different from, say, Rudin?

Rudin is a horrible example of a good math textbook.  While it is terse, complete, and provides a variety of good exercises, Rudin does a notoriously bad job of actually explaining the motivation for any of the definitions or discussing the historical development of the main ideas of real analysis.  For that reason, several people prefer Spivak's [i]Calculus[/i], which in its own words
[quote]combines leisurely [b]explanations[/b], a profusion of examples, a wide range of exercises and plenty of illustrations in an easy-going approach that enlightens difficult concepts and rewards effort.[/quote]
I'm under the impression that many math textbooks of the Rudin variety are primarily intended as references for practicing mathematicians when they need to look up the proof of a certain result.  This is not the same thing as a textbook that is good to learn from.

[quote="JRav"]Since you already have the book, I'll tell you what sections of the book are currently being covered and you could figure out exactly what stuff it is:
(...)
Again, I'm not really sure exactly what material that constitutes.[/quote]
Chapter 1-4 is all about groups (Ch.1: Intro to Groups, Ch.2: Subgroups, Ch.3: Quotient Groups and Homomorphisms, Ch.4: Group Actions (incl. Sylow)). We covered this stuff in one semester. Ch. 7 and 8 are about Rings (Intro to Rings, Euclidean Domains, Principal Ideal Domains and Unique Factorization Domains), Ch. 9 is about Polynomial Rings.

Dummit & Foote actually cover way more, there are 19 chapters (over 900 pages), but that's probably the case with every Abstract Algebra book.

[quote]I was under the impression that most upper level books tend to be theorem-proof.  Is this any different from, say, Rudin?[/quote]
Yes, all upper level books are theorem-proof. What I meant was that, for example, Artin tends to have some discussion with a proof of a theorem in it, and then states something like "we have proved the following:". Or he doesn't state a theorem marked as a theorem, but it's 'hidden' in a story.
Armstrong does this even more (he often has definitions and proofs hidden in examples); personally I don't like it because it makes it hard to look things up without having to reread long discussions. I like it when definitions, theorems and proofs are clearly stated and emphasised. But that's a matter of taste.

You might also want to take a look at Gallian's [url=http://www.amazon.com/Contemporary-Abstract-Algebra-Joseph-Gallian/dp/0618514716/ref=pd_cp_b_0?pf_rd_p=413864201&pf_rd_s=center-41&pf_rd_t=201&pf_rd_i=0547165099&pf_rd_m=ATVPDKIKX0DER&pf_rd_r=1W69FP6373GYK3M9FKXG]Contemporary Abstract Algebra[/url], which is much more introductory, with sometimes longwinded (inelegant) proofs, many examples, famous quotes, and even film suggestions.

Another introductory text is Fraleighs [url=http://www.amazon.com/First-Course-Abstract-Algebra-7th/dp/0201763907/ref=cm_cr_pr_product_top]A First Course in Abstract Algebra[/url]. Personally I would go directly for the real deal, i.e. the Dummit & Foote / Artin level. That's probably also the course' level (but you might want to have one introductory text to fall back to, in order to see things explained in a simple way).

Probably the best thing is to look at these in your library and see which one suits you the best.

Yes, I'm completely aware that Dummit & Foote is much longer-what I listed is the material covered by the Algebra I course
[quote="t0rajir0u"]For that reason, several people prefer Spivak's Calculus...[/quote]
I already own Rudin and Spivak, and I don't really need an analysis book.  Are you familiar with Carothers' book "Real Analysis"?  We use this book for analysis and it seems to provide motivation for its concepts but perhaps we still have a different idea of "theorem-proof".

I really like Dummit and Foote, but it's a bit advanced for a first-time look at algebra.  (It was already noted that it's meaty enough for a graduate course.)  At the easier end of the spectrum, I like what I've seen of Gallian, which looked very friendly.

I don't understand the relevance of the new edition of Artin.  Is it because there might be some nether-region when the both the 1st and 2nd eds are unavailable?  I think I probably learned stuff from Artin, but my memory of that book has somehow been erased.  No one has mentioned Lang, who I think has an undergrad book as well as a (more common) grad book.

Neither here nor there:  Baby Rudin is not really a reference for the practicing mathematician.  A practicing mathematician ought to have internalized all of the useful content of the book (or be one of those mathematicians who never use real numbers).  Also, there's a lot in there on Riemann integrals, which are never mentioned in real math.  (I won't even start on the good old Stieljies (sp?) integral.)  The time when Rudin really earns its pricetag is probably in whatever analysis course you take *after* taking Rudin.  It really is a beautiful book (except for the ugly measure theory chapter).  Although I think I learned nothing from it the first time, I really enjoyed it the second time.  I can see how Rudin could also be a good companion for a course with good lectures.

Thank you for your input.

To correct you, it's Stieltjes  :P.  Rudin's book isn't my required text for analysis but I find that it's a good supplement to my other book, since it provides a different angle on some of the proofs.

[quote="yenlee"]
Neither here nor there:  Baby Rudin is not really a reference for the practicing mathematician.[/quote]

Rudin also doesn't 'name' theorems which have a well-known name, which is another reason that it's not a particularly good reference.  That being said, I liked it a lot as a textbook, and the maturity that you gain from having to do a lot of the work of understanding the material yourself (since Rudin doesn't really hold your hand) is quite valuable.  On the other hand, this could just as easily be taken as a negative for the book.

I think he names them, he just doesn't write
[b]Theorem 1.1[/b] (Fundamental Theorem of Algebra),
he tends to give the name in the paragraph above/below or name it in the section title.  However, I do agree that the lack of motivation for many theorems is somewhat frustrating.

Some of them are not named at all.  One that jumps out in my mind is [url=http://en.wikipedia.org/wiki/Dini's_theorem]Dini's Theorem[/url].

In several of the books I've seen, Dini's theorem is never actually presented in the text, but rather given as an exercise.  And then after the exercise description it will usually have in parentheses something like "this result is sometimes called Dini's theorem."


Anyone use the Gaim (and aMSN) plugin TeXIM?

I can't seem to get it to work. *prods computer*

Yeah, I've just discovered it a couple of days ago and tried, but I'm not sure how to get it to work either.  When I enable the plugin, it gives an error to specify paths, but you have to enable it to be able to do that.... :huh:

This is the webpage:
http://www.student.cs.uwaterloo.ca/~kzorin/texim.php

Ask DPopov.  He posted about this in the $\LaTeX$ forum and said he had troubles also but now knows how to get it to work.

The author of the plugin has told me that you should try redownloading the DLL file from http://www.student.cs.uwaterloo.ca/~kzorin/texim.php and you should be fine now...

And also make sure to install the GhostScript properly using the link on the author's page.
He also told me to download the new one where the error msg was removed, but now you might not know if you have correctly installed the necessary programs, because once i reinstalled GhostScript, the error msg went away.

can you send latex pictures then to normal msn users as well?

Hey, I share the same problem with all you guys. I read through the details from the webpage   http://www.student.cs.uwaterloo.ca/~kzorin/texim.php
and download all the required softwares, which is quite a long list from different locations including the amsn and the others. At some stage I was required to download the "Imagemagic" as well and I didn;t know how to open and execute. Can anyone of the "successors" here show us a list of procedure to get it installed?  :(

Well guys I was succesfully able to install it and use it. Just a problem...I think normal msn users can't view them. Well I installed it on amsn on a linux box (gentoo) and oh well it works perfectly. you just type [code] /tex /Sigma [/code] and you get the famous Sigma. and so forth. You know just tex on. I am going to try it on windows at school. I am a peruvian user btw.

Well if any of you has it installed please let me know. IS THERE A WAY THAT NORMAL MSN USERS CAN WATCH MY TEX THINGIES...IT IS KINDA ABSURD TO TALK ALONE.

[quote="aproan"]IS THERE A WAY THAT NORMAL MSN USERS CAN WATCH MY TEX THINGIES...IT IS KINDA ABSURD TO TALK ALONE.[/quote]that's what I wondered too...

you know, as msn lil custom emoticons are shown to anyone perhaps there is a way (or in the code itself) to make latex into an image instantly and send it as an emoticon. and thus being able to talk.

If a solution does not exist we should try to program one.

I've downloaded  "ImageMagick-6.2.6-4.zip"  . Is that the one you're supposed to download? After that, I don't know where to go. How do you open and execute Image Magick?


Determine all the $ k$ that exists a set of $ 2008$ consecutive positive integers and you can partition it in $ k$ sets with the same sum.

You forgot to add something on the desired  partition unless there is no real question. :wink:
Maybe are you asking for the same sum in each set of the partition?

Pierre.

Conditional is there exist n to : $ k|1004(2n\plus{}2007)\equal{}4.251(2n\plus{}7)$ and $ 0<k<2009$


Ring Theory
superior algebra
superior algebra unsolved

Let $D$ be a division ring, $C$ its center and $S$ be a division subring of $D$ which is stabilized by every map $x\mapsto dxd^{-1},d\neq0$ in $D$. Show that either $S=D$ or $S\subset C$

Any answer?

People this is a challenging problem Solve it

Suppose that $S\not\subset C$. We have to prove that $D=S$. Suppose that it is not true take $a\in D$ which is not in $S$. And let $x\in S$ be a non-central element. Put $b=a-1$. We have:

$a(a^{-1}xa- b^{-1}xb) = x(b+1)-(b+1)b^{-1}xb=x-b^{-1}xb$
where $b=a-1$. Take $A=a^{-1}xa-b^{-1}xb$ and $B=x-b^{-1}xb$.
We obtain $a=BA^{-1}\in S$. Contradiction.


Currently, I am in the mathcounts problem series. I want to join the intermediate counting and probability class though. I am comfortably being able to do the Are you ready test or the pretest, but I can't do the Do you need this(post test). Is it okay if i join?

It's perfect for you then (it should be), because you only need the stuff in the classes.

236factorial is correct.  The pretest is meant to test that you know what you need to know in order to begin the class.  If you were able to pass the post test, you wouldn't need the class.  In that regard, it's exactly the level of class we would recommend to you.

I plan on doing Intermediate Counting and Probability this Fall.  I can do most of the "Are You Ready" quiz problems but not with comfort.  I was wondering if this was the best class for me, or if I could perhaps see some of the material for the Introduction class so that I can be sure I can do well in the class.

That's hard to say without opening up the curriculum further than the pre- and post-tests.

Perhaps you or someone you know will pick up a copy of the Intro Counting/Probability book we'll be publishing soon.  From that book you would get anything you were missing and a large number of good practice problems.  Then perhaps you would feel more comfortable jumping straight into the Intermediate Class.

Speaking of which, when will you be finished publishing all these subject books?

I intend on purchasing the full set.

Final dates are unpredictable, but we hope to have the Counting/Probability book out very soon.  Unless something crazy happens, it will be this semester.


geometry
3D geometry
sphere
pyramid
geometry proposed

A sphere is inscribed into a quadrangular pyramid. The point of contact of the sphere with the base of the pyramid is projected to the edges of the base. Prove that these projections are concyclic.

Let $ABCD$ be  the base of this pyramid and $ E$ its opposite vertex. Let $ X,Y,Z,W$ be the tangency points of its insphere $\omega$ with the faces $ EAB,$ $EBC,$  $ECD,$ $EDA$ and $ P$ its tangency point with the base $ABCD.$ It is easy to figure out that $ \triangle EXB \cong \triangle EYB,$ because of $ BX \equal{} BY$ and $ EX \equal{} EY.$ [i]Mutatis mutandis[/i] $ \triangle EYC \cong \triangle EZC,$ $\triangle EZC \cong \triangle EWD$ and $\triangle EWA \cong \triangle EXA.$ Then we get

$ \angle CYB \equal{} 360^{\circ} \minus{} (\angle EZC \plus{} \angle EXB)$

$ \angle AWD \equal{} 360^{\circ} \minus{} (\angle EZD \plus{} \angle EXA)$

$ \angle CYB \plus{} \angle AWD \equal{} 720^{\circ} \minus{} (\angle EZC \plus{} \angle EXB \plus{} \angle EZD \plus{} \angle EXA)$

$\angle CYB \plus{} \angle AWD \equal{} \angle CZD \plus{} \angle AXB$

Since $ AX \equal{} AW \equal{} AP,$ $BX \equal{} BY \equal{} BP,$ $CY \equal{} CZ \equal{} CP$ and $DZ \equal{} DW \equal{} DP$ we get

$ \angle AXB \equal{} \angle APB , \ \angle BPC \equal{} \angle BYC$

$\angle CZD \equal{} \angle CPD, \ \angle AWD \equal{} \angle APD$

$\Longrightarrow  \angle APD \plus{} \angle BPC \equal{} \angle CPD \plus{} \angle APB$

$ \Longrightarrow \angle APD \plus{} \angle BPC \equal{} 180^{\circ} , \ (*)$

Let  $ M,N,L,K$ be the orthogonal projections of $ P$ onto the edges $ AB,BC,CD,DA.$ Since the quadrilaterals $ PMBN,$ $PNCL,$ $PLDK,$ $PKAM$ are all cyclic, it follows that

$ \angle APD \equal{} \angle KMA \plus{} \angle KLD , \ \angle BPC \equal{} \angle NMB \plus{} \angle NLC$

Combining with $ (*)$ yields $ \angle KLM \plus{} \angle NMK \equal{} 180^{\circ}$ $\Longrightarrow$ $ MNLK$ is cyclic.


Find all functions $ f: \mathbb R^\plus{}\rightarrow\mathbb R^\plus{}$ satisfying $ (f(x))^2\ge f(x\plus{}y)f(f(x)\plus{}y)$ for every $ x,y>0$

Sorry, I posted incorrently.  :oops:
It must be:
Find all functions $ f: \mathbb R^+\rightarrow \mathbb R^+$ satisfying $ (f(x))^2\ge f(x+y)(f(x)+y)$ foe every $ x,y\in%Error. "mathh" is a bad command.
R^+$

[b]Assumption:[/b]
There exists some $ a,b\in \mathbb R^ \plus{}$ such that $ f(a \plus{} b) \plus{} b \minus{} f(a)$ is positive.

[b]Disproof by contradiction:[/b]
For any positive real $ l$ there exists a positive integer $ k$ (however large) satisfying $ b \minus{} \frac {bf(a \plus{} b)}{f(a \plus{} b) \plus{} \frac {b}{k}} < l$
because that is equivalent to: $ k > \frac {b(b \minus{} l)}{lf(a \plus{} b)}$

Let $ l \equal{} f(a \plus{} b) \plus{} b \minus{} f(a)$, so there exists $ k$.

The given functional inequation is equivalent to: $ f(x \plus{} y)\le f(x) \minus{} \frac {yf(x)}{f(x) \plus{} y}\ \forall x,y\in \mathbb R^ \plus{}$
So:
$ f(a \plus{} b)\le f(a) \minus{} \sum_{i \equal{} 1}^{k} \frac {\frac {b}{k}f(a \plus{} (k \minus{} i)\frac {b}{k})}{f(a \plus{} (k \minus{} i)\frac {b}{k}) \plus{} \frac {b}{k}}$
[hide="because"]$ f(a \plus{} b) \equal{} f(a \plus{} k\frac {b}{k})$
$ \le f(a \plus{} (k \minus{} 1)\frac {b}{k}) \minus{} \frac {\frac {b}{k}f(a \plus{} (k \minus{} 1)\frac {b}{k})}{f(a \plus{} (k \minus{} 1)\frac {b}{k}) \plus{} \frac {b}{k}}$
$ \le f(a \plus{} (k \minus{} 2)\frac {b}{k}) \minus{} \frac {\frac {b}{k}f(a \plus{} (k \minus{} 2)\frac {b}{k})}{f(a \plus{} (k \minus{} 2)\frac {b}{k}) \plus{} \frac {b}{k}} \minus{} \frac {\frac {b}{k}f(a \plus{} (k \minus{} 1)\frac {b}{k})}{f(a \plus{} (k \minus{} 1)\frac {b}{k}) \plus{} \frac {b}{k}}$
$ \cdots$[/hide]

It is clear from the given inequation that $ f$ is decreasing. So:
$ \sum_{i \equal{} 1}^{k} \frac {\frac {b}{k}f(a \plus{} (k \minus{} i)\frac {b}{k})}{f(a \plus{} (k \minus{} i)\frac {b}{k}) \plus{} \frac {b}{k}}\ge \sum_{i \equal{} 1}^{k} \frac {\frac {b}{k}f(a \plus{} b)}{f(a \plus{} b) \plus{} \frac {b}{k}} \equal{} \frac {bf(a \plus{} b)}{f(a \plus{} b) \plus{} \frac {b}{k}} > b \minus{} l$

$ f(a \plus{} b) < f(a) \minus{} b \plus{} l$
But $ f(a \plus{} b) \equal{} f(a) \minus{} b \plus{} l$
Hence contradiction.

-------------------------------------

So for all $ x,y\in \mathbb R^ \plus{}$: $ f(x \plus{} y) \plus{} y \minus{} f(x)$ is non-positive.
So $ f(x)\ge f(x \plus{} y) \plus{} y\ \forall \ x,y\in \mathbb R^ \plus{}$
Now it is clear that there exists no such function satisfying the given inequation.


algebra
polynomial
induction
superior algebra
superior algebra unsolved

Show that if $ \lambda$ is an eigenvalue of $ \tau$, then $ p(\lambda )$ is an eigenvalue of $ p(\tau ),$ for any polynomial $ p(x).$ Also, if $ \lambda \ne 0,$ then $ {\lambda}^{\minus{}1}$ is an eigenvalue for $ {\tau}^{\minus{}1}$.

Thanks in advance for assisting me.

Its quite simple, it only involves plugging $ \tau$ into your polynomial $ p$ and turning all factors of $ \tau v$ into $ \lambda v$. Here's how to do it in detail:

By definition, $ \lambda$ is an eigenvalue of $ \tau$ if $ \tau v \equal{} \lambda v$ for all $ v$. So let $ p(\tau)\equal{}a_n\tau^n \plus{} a_{n\minus{}1}\tau^{n\minus{}1}\plus{}... a_1\tau \plus{} a_0$. Apply this operator to $ v$ getting $ p(\tau)v \equal{} a_n\tau^n v \plus{} a_{n\minus{}1}\tau^{n\minus{}1} v\plus{}... a_1\tau v \plus{} a_0 v$. Now, we know $ \tau^{n}v \equal{} \tau^{n\minus{}1}\tau v$ so by induction, we get $ \tau^n v \equal{} \lambda ^ n v$. Thus we get $ p(\tau) v \equal{} a_n\lambda^n v \plus{} a_{n\minus{}1}\lambda^{n\minus{}1} v \plus{}... a_1 \lambda v \plus{} a_0 v$. Now, we can pull out $ v$ gain to get $ p(\tau) v \equal{} (a_n\lambda^n\plus{}...a_1\lambda \plus{} a_0) v \equal{} p(\lambda) v$

For the second part:
$ \tau v \equal{} \lambda v$. Left multiply by $ \tau^{1}$ to give $ v \equal{} \tau^{\minus{}1}\lambda v$ But $ \lambda$ is a scalar so we can pull it out front to get $ v \equal{} \lambda\tau^{\minus{}1} v$. Move $ \lambda$ to other side to get $ \lambda^{\minus{}1} v \equal{} \tau^{\minus{}1} v$


algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

could you please write a solution to the following one

find the Galois group for $ x^3\minus{}x\minus{}1\in\mathbb{Q}[X]$

thanks in advance :)

the polynomial is irreducible and the discriminant $ \minus{}4(\minus{}1)^3\minus{}27(\minus{}1)^2\equal{}\minus{}23$ has no root in $ \mathbb{Q}$, thus the galois group is $ S_3$.


modular arithmetic
number theory unsolved
number theory

Let $ p$ is a prime and $ n\equal{}k\cdot p^2\plus{}1$ for some $ k<p$. Suppose that $ 2^k\neq 1(\mod n)$ and $ 2^{n\minus{}1}\equal{}1 (\mod n)$. Prove that $ n$ is prime.

The case $ p\equal{}2$ can be disposed of easily, so assume $ p$ odd. There exists a prime $ q$ and positive integer $ j$ such that $ q^j$ divides $ n$ and $ 2^k\not\equiv 1\pmod{q^j}$; otherwise, $ 2^k$ would be congruent to 1 modulo every prime-power divisor of $ n$, therefore congruent to 1 modulo $ n$.

In this case, the order of 2 modulo $ q^j$ divides $ n\minus{}1$ but not $ k$, so must be divisible by $ p$. In particular, $ p$ divides $ q^{j\minus{}1}(q\minus{}1)$. It's easy to see that $ p\not| q$, so we must have $ p|(q\minus{}1)$. As $ 2^{n\minus{}1}\equiv 1\pmod{n}$, $ n$ is necessarily odd: so is $ q$, and we can find nonnegative integers $ a,b$ such that
\[ q\equal{}2ap\plus{}1,\frac{n}{q}\equal{}2bp\plus{}1\Rightarrow n\equal{}1\plus{}2p(a\plus{}b)\plus{}4p^2ab\]
As $ p$ odd, we must have $ a\plus{}b$ divisible by $ p$. If $ b$ were positive (and of course $ a$'s positive), then we would require $ a\plus{}b\geq p$ and $ ab\geq p\minus{}1$ implying
\[ k\equal{}\frac{n\minus{}1}{p^2}\geq 2\plus{}4(p\minus{}1)\equal{}4p\minus{}2\geq p\]
a contradiction. So $ b\equal{}0$ and $ n\equal{}q$, a prime.


function trigonometry geometry rectangle limit integration Euler

Two suggestions:

1. Consider the function $f(z)=\frac1{z^2\sin(\pi z)}$

If $n\in\mathbb{Z}\setminus\{0\}$ then $f$ has a simple pole at $n$ with residue $\frac{(-1)^n}{\pi n^2}.$

It has a pole of order $3$ at $0.$ Your next task will be to compute the residue there.

Let $C_N$ be the closed contour bounding a rectangle whose boundaries are $x=\pm \left(N+\frac12\right)$ and $y=\pm N.$

We can argue that $\lim_{N\to\infty}\int_{C_n}f(z)\,dz=0.$

Part of that argument is that $|\sin(x+iy)|$ tends rapidly to $\infty$ as $|y|\to\infty.$

So the sum of all of the residues for this functions is zero. We can algebraically solve the resulting equation to find $\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}.$

Given this, we can do some simple algebra to derive the sum we want.

2. Do something similar, but with $f(z)=\frac{\cot(\pi z)}{z^2}.$

Both methods can be easily adapted to the calculation of $\zeta(2k)=\sum_{n=1}^{\infty}\frac1{n^{2k}}.$ The larger $k$ is, the higher the order of the pole at $0$ and thus the harder you have to work to find the residue, but there are no other complications.

i really dont understand your idea Kent , but you can use complexe considering the polynome (X+1)^2n+(X-1)^2n to fund the some of cotan²((2k+1)pi/2n) k=0,...,n and deduice from this some the some of 1/sin²(((2k+1)pi/2n) k=0,...,n , then we need only the analytic formula : for each x in ]0,pi/2[ we have sin(x)<x<cotan(x) and then by encadring the some 1/k² you found pi²/6
i hope that Kent give more explanitions because his idea seems to be very intersted :roll:

I'll admit that I don't understand the method that yassinus was proposing.

I was using the residue theorem: $\int_Cf(z)\,dz=2\pi i\sum_{w\in\Omega}\text{Res}(f,w)$ where the curve $C$ appropriately bounds the region $\Omega.$

Following the first version above, in which $f(z)=\frac1{z^2\sin \pi z}$:

The sequence of contours $C_N$ splits halfway between the poles. From that, the $\frac1{z^2}$ factor causes the contribution of the vertical sides of the box to shrink as $N\to\infty$ and the contribution of the horizontal sides goes rapidly to zero. Hence:

$\lim_{N\to\infty}\sum_{|w|\le N}\text{Res}(f,w)=2\pi i\lim_{N\to\infty}\int_{C_N}f(z)\,dz=0$

As stated above, $f$ has a simple pole at $\pm n$ with residue $\frac{(-1)^n}{\pi n^2}.$

We get the residue at zero by long division:

$\frac1{z^2\sin \pi z}=\frac1{\pi z^3-\frac{\pi ^3z^5}6+O(z^7)}=\frac1{\pi}z^{-3}+\frac{\pi}6z^{-1}+ O(z^{-3})$

Hence the residue at zero is $\frac{\pi}6$ (the coefficient of the $z^{-1}$ term.)

This gives us that $\lim_{N\to\infty}\frac{\pi}6+\sum_{n=1}^N\frac{(-1)^n}{\pi n^2} +\sum_{n=-N}^{-1}\frac{(-1)^n}{\pi n^2}=0.$

That is, $2\sum_{n=1}^{\infty}\frac{(-1)^n}{\pi n^2}+\frac{\pi}6=0$ or $\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}=\frac{-\pi^2}{12}.$

Now suppose $\sum_{n=1}^{\infty}\frac1{n^2}=A.$ Then $\sum_{k=1}^{\infty}\frac 1{(2k)^2}=\frac A4.$

$2\sum_{k=1}^{\infty}\frac 1{(2k)^2}-\sum_{n=1}^{\infty}\frac1{n^2}$

$=2\left(\frac14+\frac1{16}+ \cdots\right)-\left(1+\frac14+\frac19+\frac1{16}+\cdots\right)$

$=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}=\frac{2A}4-A=-\frac A2.$

So $-\frac A2=-\frac{\pi^2}{12}$ and hence $A=\frac{\pi^2}6.$

think you Kent , a friend from college told me yesterday about the residu theorem,i'm studing at preparatory classes (you can told Alekk about it,we are studing for 2 years ,...,it's for engineery) so i have no idea about homolorph functions(entires functions ) and i'm so angry beeing like that  :mad:
About my suggestion , the roots of the polynome (X+1)^2n+(X-1)^2n are :
x_k=-i.cotan((2k+1)pi/4n) , k in {0,...,2n-1} and we have x_k=x_(2n-1)-k since
(X+1)^2n+(X-1)^2n =P(x²) where P is a polynom of dgree n that we know the coefficients (wich are a_k=2nC2k ) hence we have Sum(x²_k,k=0,...,n-1)=n(2n-1)  and we coclude that Sum(1/sin²((2k+1)pi/4n),k=0,...,n-1)=2n², now using that for every x in ]0,pi/2[ :cotan²(x)<1/x²<1/sin²(x) we have :
 Sum(cotan²((2k+1)pi/4n), k=0,...,n-1)<Sum((4n/(2k+1)pi)²,k=0,...,n-1)<...
then when we tend n to oo we get that Sum(1/(2n+1)²)=pi²/8 and we finish it like you have do in your proof .
i want to tell you that you dont need to clarify how to passe from Sum((-1)^n/n²) to Sum(1/n²) , : i'm not so stupid  :D

Euler calculated $\sum_n\frac1{n^k}$ for even $k$ up to 26. Here is his method:

Step 1: $\frac{\sin \pi x}{\pi x}=\prod_{n=1}^\infty\left(1-\frac{x^2}{n^2}\right)$
This formula is correct, but it needs justification that Euler didn't give, involving serious complex analysis.

Step 2: Apply Viete's relations to calculate the coefficients of this "polynomial" in terms of its roots, and solve for the desired quantities one by one.
Again, no justification given here. This is also some very intense algebra.

[quote="ifai"]How to prove $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ by using the tools of complex analysis?[/quote]

you can also prove with Parceval identity

[quote="jianuovidiu"]you can also prove with Parceval identity[/quote]
I'd call that Fourier analysis, as something distinct from complex analysis - which is why I went with the residue theorem approach above.

The subject came up in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=51675]this thread.[/url] Take particular note of the reference given by blahblahblah in post #10 of that other topic.

If you follow that reference, you'll find 14 different calculations of $\zeta(2).$ Of particular note:

Proof 4 uses Parseval's identity.

Proof 7 is Euler's original argument, which jmerry referenced in post #6, above.

Proof 8 is a residue theorem calculation, using the cotangent, rather than my use of the cosecant, above.

Proof 9 is essentially the argument that yassinus gives above in #5.

I like proof 9 (post #5) because it's elementary.

Have a look there : [url]http://www.maths.ex.ac.uk/~rjc/etc/zeta2.pdf[/url] : there are 14 proofs, some involving complex analysis

Edit : oops, it's the same reference as blahblahblah's sorry for useless post

And if you knew that there is an $elementary$ proof for this one. ;)

Could you show us your [i]elementary[/i] proof Cezar? :)

More general: can someone explain me the borders of elementary¿ Everything is elementary for one who knows the subject good enough, so...

No, this elementary proof only uses the well-known inequality: \[ \sin x < x < tg x  \] ;)

I think this elementary inequality is quite well-known. Isn't it?  ;)  It is based on some computations with $\sum {ctg^2{\frac{k\pi}{2m+1}}}$. By the way, Cezar, it may look cool and fancy and so on to you, but at least once a month you could post a solution, you know...

Exactly, you are right here. This is the main idea.  ;)

Well, anyway.. can you post this proof Cezar please? :)

I'm assuming Cezar is referring to proof #9 in the file referenced in tµtµ's post #10, above.

Ah yeah, you're right Kent! :)

Anyway, I don't think it's [i]that[/i] elementary..  :blush:


inequalities
LaTeX
search
inequalities unsolved

It's In Math& Young in current Use. Please stop do this and try yourself... As I see, It's very easy.

[quote="babylearnmath294"]I think It's very easy.And I think it's an old inequality[/quote]
Yes, that is a well known inequality. Squaring $A$ and then using AM-GM solve this.

Posted again and the right solution is Nickolas one.

http://www.mathlinks.ro/Forum/viewtopic.php?t=44295

sorry, but after squaring, we get $\frac{x^{2}y^{2}}{z^{2}}+\frac{y^{2}z^{2}}{x^{2}}+\frac{x^{2}z^{2}}{y^{2}}+2$, how do we apply AM-GM? i got something really messy and couldn't get the answer. hope you clarify thanks :)

Let me use Latex

[quote="babylearnmath294"]Oh,after squaring,we have
                    $\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{x^2z^2}{y^2}+2$

    I think it's very easy to prove :
  $\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{x^2z^2}{y^2}\geq x^2+y^2 +z^2=1$[/quote]

as we know that generally $a^2+b^2+c^2\geq ab+bc+ca$

[quote="tranthanhnam"]let $x,y,z$be three  numbers positive  :
      find the  maximal value of expression :
      $A=\frac{xy}{z} +\frac{zy}{x} +\frac{xz}{y}$
 where $x^2+y^2+z^2=1$[/quote]
IF $x\rightarrow0+,y=z\rightarrow\frac{1}{\sqrt2}-$ then $A\rightarrow+\infty.$ :lol:

I think it is a typo. We are not search for the maximum but for the minimum

let$\frac{xy}{z}=a$ ,.....
         we have : by cauchy inequality
$( a+b+c)^2 \geq 3(ab+bc+ca) =3$


The solution excluded the 1 cm wide top surface of the rim of the box, with an additional area of 16cm*24cm-14cm*22cm = 76 sq. cm.  (It also lost 60sq.cm. when incorrectly multiplying 16*24.)  The answer is 2216 sq. cm.

A little time can be saved by combining the top rim with the inside floor and the outside bottom (top and bottom view areas):
  2* 16cm * 24cm = 8 * 96 cm^2 = 768 cm^2
add the outside of the walls
  10cm * 2 * (16cm + 24cm) = 800 cm^2
add the inside of the walls
  9cm * 2 * (14cm + 22cm) = 9 * 72 cm^2 = 648 cm^2

The sum is 8 * (96+100+81) cm^2 = 8*277 cm^2 = 2216 cm^2


summer program
Mathcamp
Ross Mathematics Program

Hi

I happen to be a very good chess player. Anybody interested in late night blitz ?
I am wondering whether to bring my chess clock or not ( Boards and coins must be there).
Any rated players coming there ? I have a FIDE elo rating of 2113 ( though my playing level would be somewhere around 2175 ) ?

Also, Table Tennis and Basketball anyone ( though any american should beat me hollow in bball ) ?

Akash

hmm, that's too bad. I didn't apply for Mathcamp but I am accepted in the Ross program and I am waiting for the HCSSIM...
I got a Fide elo 2050 but it is not my best rating.
With my dragon and my closed frencf, you would have some trouble in blitz games...

Well, I specially love playing against Dragon ( I play 9. 0-0-0 ). However Carokann is an opening I feel extremely uncomfortable with. Against 1.... e5, I play Ruy Lopez.
With black for 1.d4,c4,Nf3 I play Kings Indian Defence kind of setups, though I am planning to change to nimzo. Against Carokann, I play Smyslov ( e4 c6 d4 d5 nc3 )

Chess and other board games are always very big at camp. Once can never have too many chesssets, though there never seems to be a shortage either.

Table tennis is also very popular; but group sports, with the possible exception of soccer, havn't been all that big in the years i've gone to camp. Of course, that doesn't mean things cant change  ;)

Usually ultimate frisbee has a following, though.

I wouldn't worry about bringing boards and pieces. The people that run mathcamp seem to bring a lot of games. (Like Ricochet Robots)

Too bad I wasn't able to be the next Dan, though.

Wow, you guys have such high ratings; isn't 2100 almost at masters level? I like chess too but my rating is only around 1100.


[u][b]Simple:[/b][/u] Can you find four different positive integers $ a$, $ b$, $ c$ and $ d$ such that their reciprocals add up to $ 1$?

[u][b]Not so simple:[/b][/u] Find four different positive integers $ a$, $ b$, $ c$ and $ d$ such that their reciprocals add up not to $ 1$, but as close to $ 1$ as possible.

[hide="Simple"]
Consider $ (a,b,c,d) \equal{} (2,3,8,24)$
[/hide]

[hide="Not So Simple"]
Let $ (a,b,c) \equal{} (2,3,6)$ and $ d$ be arbitrarily large.
[/hide]

[b][u]Even less simple:[/u][/b] [i]Find 4 distinct natural numbers such that the sum of their reciprocals is [u]less than[/u] 1 but as close to 1 as possible.[/i]

[quote="The Zuton Force"] [b][u]Even less simple:[/u][/b] [i]Find 4 distinct natural numbers such that the sum of their reciprocals is [u]less than[/u] 1 but as close to 1 as possible.[/quote]

I think the best case here (below 1) is $ \frac{1}{2}\plus{}\frac{1}{3}\plus{}\frac{1}{7}\plus{}\frac{1}{43}\equal{}1\minus{}\frac{1}{1806}$

But I've not demonstrated that's the best.


probability
combinatorics unsolved
combinatorics

There are 2N prisoners and a room. In the room there are 2N envelops numbered 1 to 2N, and inside envelops there are cards numbered also 1 to 2N.  Each envelop contains one card, but the number on the envelop has nothing to do with the number on the card (basically randomly allocated).  Now the prisoners are also numbered and taken into the room one by one, and allowed to open N envelops each to attempt to find the card with their own number, and if all of them succeed they will be released.

They are allowed to agree on a certain strategy beforehand but certainty no information exchange is allowed once the game starts. how will they maximise their hope and what's the probability?

[quote]
Are they allowed to rearrange the envelops? Then p=1/2[/quote]
Now, they may not rearrange.

Was discussed [url=http://www.mathlinks.ro/viewtopic.php?t=191186]here[/url].
I can add that if allocation is non-random (in particular, if it contains a huge cycle), then it is also possible to survive with the same probability.

[quote="FelixD"]I guess they have to close every envelop they've opened? Moreover, if they found the card with their own number in one of the envelops, do they have to put it back into the envelop and close the envelop afterwards?[/quote]
Your guess is correct.


Based on previous years, can anybody tell me about how long it takes the USAMO graders to finish? I forget how long it took last year.

Thanks.

I seem to remember that MOP calls would start going out May 7, but not to ask about your score until May 11.

Usually, they post all the scores at once on the website.  Last year, I got my score before I got my MOP invitation.

Yeah, I think the scores will probably come up pretty soon after the May 6-7 weekend.

The teacher's manual says they grade the tests May 4-6 and release the scores on May 8.

so, apparently, according to the USAMO manual page four: grading goes on May 4,5,6 and they post results on May 8

however, according to the USAMO manual page 13: the grading takes place on May 6,7.  MOSP invites start coming out on May 7.  you get your copy of the USAMO back on May 9 (with solutions), and you shouldn't ask them about results until May 11.

hmm...reminds me of the 'May 10rd' errors they made earlier (although those may have been fixed :?: )

EDIT: so, everybody's correct, depending on which page they remembered.

[quote="Teki-Teki"]hmm...reminds me of the 'May 10rd' errors they made earlier (although those may have been fixed :?: )
[/quote]

Speaking of that, did anyone notice that the date on the first day of the test was wrong?  I think it said April 26.  :)

Hm, well, that's better than the GA State Varsity tournament results, which claim it's still 200[b]5[/b]...

Anyway, well, if you don't check the website till say May 13, I can _guarantee_ your score will be there.

As it says in the Teachers' Manual, page 4, Section III, grading is on May 4-5-6, with results posted May 8.  As it says on page 13 in the Teachers' Manual in the Sample S-I letter which has already been mailed and recevied by you:  USAMO winners and MOSP invitations will be telephoned on May 6-7.

We do not return your USAMO papers to you.  We will send a copy of the USAMO and the USAMO solutions to your Exam Manager along with your total score on May 9.  We do not report individual problem scores, only the total.  Of course, you can already obtain a copy of the USAMO problems and solutions on our website.

All your questions are addressed in the Teachers' Manual.  Do not inquire about anything until May 11, we are too busy before then to answer your questions.

Steve Dunbar
Director, American Mathematics Competitions


Suppose that $p(x)\ne 0$ is a polynomial, define $q(x)=p(x)*p(-x)$, prove that $q(x)$ is a polynomial whose nonzero terms have even degree

q(x) is a product of polynomials, so it is a polynomial

q(x) is an even polynomial, so it has no non-zero terms with odd degree (if it did, consider the difference q(k)-q(-k) for all positive k; it should equal zero always, but if it had odd degree terms than it wouldn't be zero.)

[hide]
Let the polynomial $P$ be
\[a_{0}+a_{1}x+a_{2}x^{2}+...+a_{n}x^{n}\] where $n$ is the degree of $P$.

Let $\lfloor\frac{x}{2}\rfloor=j$ and $\lfloor\frac{x-1}{2}\rfloor=k$ (notice that $j = k$ or $k+1$.
Then $P(x)P(-x)$ = $[(a_{0}+a_{2}x^{2}+\cdots+a_{2j}x^{2j})+(a_{1}x+a_{3}x^{3}+\cdots+a_{2k}x^{2k})][(a_{0}+a_{2}x^{2}+\cdots+a_{2j}x^{2j})-(a_{1}x+a_{3}x^{3}+\cdots+a_{2k}x^{2k})]$ = (polynomial with only even coefficients)^2 - (polynomial with only odd coefficients)^2, which has only even coefficients.
[/hide]

new problem: let p(x) be a polynomial with real coefficients, let $p(x)p(x\omega)p(x\omega^{2})=q(x)$ where $\omega^{2}+\omega+1=0$, prove that $q(x)$ is a polynomial with real coefficients whose terms of degree divisible by 3

More generally, if $p$ any positive integer and $\omega$ is primitive $p$th root of 1, then for any polynomial $f$ with real coefficients, let
\[g(x)=\prod_{k=0}^{p-1}f(\omega^{k}x) \]
Then $g$ is a polynomial with real coefficients and nonzero coefficients of degree divisible by $p$.
[hide]
Since $f$ has real coefficients, $\overline{f(z)}=f(\overline{z})$ for all complex $z$. This means that for real $x$,
\[\overline{f(\omega^{k}x)}=f(\omega^{p-k}x) \]
Then
\[\overline{g(x)}=\overline{f(x)f(\omega x)f(\omega^{2}x)\cdots f(\omega^{p-1}x)}\]

\[=f(x)f(\omega^{p-1}x)f(\omega^{p-2}x)\cdots f(\omega x)=g(x) \]

\[\Rightarrow g(x)=\frac{g(x)+h(x)}{2}\]
for all real $x$, where
\[g(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n}x^{n}\Rightarrow h(x)=\overline{a_{0}}+\overline{a_{1}}x+\overline{a_{2}}x^{2}+\cdots+\overline{a_{n}}x^{n}\]
Since it holds for infinitely many $x$, the two polynomials above are equal. However, $\frac{g(x)+h(x)}{2}$ clearly has real coefficients, so $g(x)$ has real coefficients.

To show that all nonzero coefficients are of degree divisible by $p$, observe that $g(x)=g(\omega x)$. Then we can let $a_{k}$ be the coefficient of the $x^{k}$ term: then $a_{k}(\omega^{k}-1)=0$. If $a_{k}\neq 0$, then $k$ is divisible by $p$.
[/hide]

i was going for:

p(x) polynomial, $q(x)=\prod_{k=0}^{n-1}p( a_{k}x)$ where $a_{k}$, are solutions to $x^{n}=1$, then $q(x)$ has only terms with degree divisible by $n$

proof: let $p(x)=a\prod (x-r)$

then $q(x)=a^{n}\prod (a_{k}*x-r)=a^{n}\prod(x^{n}-r^{n})$, the result follows

clearly, your result is pretty much the same...primitive root implies all roots covered when taking to powers, etc


trigonometry
geometry
perpendicular bisector
geometry proposed

[color=darkblue]Let $O$ be a point inside of a $\triangle ABC$ such that $\measuredangle\ OAB=\measuredangle\ OAC=20^\circ,\ \measuredangle\ OCA=10^\circ$ and $\measuredangle\ OCB=30^\circ$, then prove that $\measuredangle\ OBC=80^\circ$[/color]

Another short trigonometric proof  :)
Denote $\angle{OBC}=x$, thus $\angle{BOC}=150-x, \angle{AOC}=70+x, \angle{OAC}=100-x$. From the law of sin we have $\frac{OB}{\frac{1}{2}}=\frac{OC}{\sin{x}}$ and $\frac{OA}{\sin{10}}=\frac{OC}{\sin{(100-x)}}=\frac{OC}{\cos{(x-10)}}$. By dividing them , and because $OA=OB$, we have $\frac{1}{2}\cos{(x-10)}=\sin{10}\sin{x}=\frac{\cos{(x-10}-\cos{(x+10)}}{2}$, thus $\cos{(x+10)}=0$, so $x=80^\circ$.

¬¬ thanks anyway but looking for no trig.  :whistling:  :whistling:

Greetings  :surf:

$AO$ meets $BC$ at $D$.
$E$ is a point that triangle $EAB$ is aquivalent
we have $\angle EBC=\angle ABC-\angle ABE=100-60=40$
and $\angle DAE=\angle BAE-\angle BAD=60-20=40$ so $A,B,D,E$ are cyclic
Thus $\angle ADE=\angle ABE=60$.
Otherwise $\angle ADC=180-\angle ADB=180-\angle AEB=120$
so $DE$ is the bisector of angle $\angle ODC$
But $\angle DOC=\angle OAC+\angle OCA=20+10=30=\angle OCD$ so triangle $ODC$ is isosceles
Thus $DE$ is the perpendicular bisector of $OC$
So $\angle EOC=\angle ECO=\angle ECB-\angle OCB=40=\angle EBC$
so $B,O,E,C$ are cyclic. So $\angle OBE=\angle OCE=40$
we have $\angle OBC=\angle OBE+\angle EBC=80$ $q.e.d$

Thanks Huyền Vũ  :)

Now the problem is solved  :ninja:


Square ABCD and equilateral triangle AED are coplanar and share AD, as shown. What is the measure, in degrees, of angle BAE?
[asy]size(3cm);
pair A,B,C,D,E;
A = (0,0);
B = (0,1);
C = (1,1);
D = (1,0);
E = dir(60);
draw(A--B--C--D--E--A--D);
label("$A$",A,dir(180));
label("$B$",B,dir(180));
label("$C$",C,dir(0));
label("$D$",D,dir(0));
label("$E$",E,dir(0));[/asy]

$ \angle BAE\equal{}\angle BAD\minus{}\angle EAD\equal{}30^\circ$


Find the minimum value of the function:
$ f(x)\equal{}(32x^5\minus{}40x^3\plus{}10x\minus{}1)^{2006}\plus{}(16x^3\minus{}12x\plus{}\sqrt{5}\minus{}1)^{2008}$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202139]Another problem you posted[/url] comes from an ongoing contest, so I suspect that this one does as well, especially since you have posted six problems.


pigeonhole principle
ratio
number theory unsolved
number theory

Let a,b,c,d be integers.  Show that the product of the differences b-a, c-a, d-a, c-b, d-b, d-c is divisible by 12.

this is trivial so here goes: we have 4 numbers so by pigeonhole two of them have the same residue mod 3 hence their diference is divisible by 3. now to prove divisibility by 4 we have two kases:
1.- if two of them are equal mod 4 we are done.
2.- if not then all residues mod 4 appear that is (a,b,c,d) mod 4 is (0,1,2,3) hence 3-1=2 is even and 2-0=2 is even so two of the diferences are even and hence the product is divisble by 4. so now the product is divisible by 3 and 4 and hence by 12. hooray my fistr correct post¡¡¡
too bad the problem is easy :(

I think the idea is on the correct path but what if all the factors of the product are for example mod 1.  then, it would not be divisible by 4 correct?

[quote]I think the idea is on the correct path but what if all the factors of the product are for example mod 1. [/quote]

See above  Marko said "1.- if two of them are equal mod 4 we are done."

  IOW if a and b both are  1 mod 4, (a-b) would be multiple of 4.

How about extensions to the problem?  For example, if we have 5 integers, and consider the product of all their differences, what is the best statement we can make about its divisibility?  What if we start with n integers?

Well i mean for every prime less than or equal to n we have by pigeonhole that one of the factors is a multiple of p. i think that is as much as we can go to generalize this. i mean the next thing we could ask ourselves is how many times does each prime divide the product. but for that we would have to know the ratio n/p but we cant really work with that. so i think we could  do no more to generalize this. but clearly if we take any partikular number than we could definitly work with it. but to generalize more than what i already did i think is impossible. but thats my humble opinion maybe i dont know something that i should. :blush:


trigonometry
quadratics
geometry proposed
geometry

BC=a, angle BAC=α, AD=h
Then find the value of AB+CA :?:

Thanks!

[color=darkblue]We have

$ S\equal{}\dfrac{1}{2}ah$

$ bc\equal{}\dfrac{2S}{\sin\alpha}\equal{}\dfrac{ah}{\sin \alpha}$

Denote $ b\plus{}c\equal{}x$ and $ bc\equal{}y$. Then $ b^2\plus{}c^2\equal{}x^2\minus{}2y\equal{}z$

Also

$ 16S^2\minus{}a^4\equal{}b^4\plus{}c^4\minus{}b^2c^2\minus{}a^2(b^2\plus{}c^2)\equal{}z^2\minus{}3y^2\minus{}a^2z$

We got quadratic with respect to $ z$.

You can find $ z$ and then $ x$, cause $ y$ is known, but these are "simple" computations.[/color]


Three hemispheres, each of radius $ r$ lie [b]fixed[/b] on a horizontal surface packed together with their flat sides on the horizontal surface. A [b]sphere[/b] of radius $ r$ and mass $ m$ is placed on top of these hemispheres. Find the [b]Normal Reaction Force[/b] exerted by an individual hemisphere.

[hide="Read Me:"]This is an interesting question a friend gave me. A common mistake (which I too made and which irritated me until I corrected myself) observed while solving this question is that of the resolution of angles of the triangles concerned with the setup.
[/hide]

N sin 60 = mg/3

N = 2mg/3root(3)

But is it obvious the other component of normal forces cancel?

Seems Conjurer made same mistake Rajiv referred at...

[hide] $ N = {{mg} \over {\sqrt {6}}}$[/hide]

[quote="Immanuel Bonfils"]Seems Conjurer made same mistake Rajiv referred at...

[/quote]

Exactly!!!

[quote="Immanuel Bonfils"]$ N = {{mg} \over {\sqrt {6}}}$[/quote]

This is the right answer!

Find $ BE$ and accordingly $ \angle EBD$. Read the names of the points concerned carefully. Refer the diagram for help.


Constellations are the patterns and figures of groups of nearby stars named after various things.
There is a list from Wikipedia [url=http://en.wikipedia.org/wiki/List_of_constellations]here[/url].

Have a constellation you like in particular?
Which ones can you readily recognize?

I do not really have [i]one[/i] particular one that I like but, I do rate them according to the objects they have in them.  Some being Lyra (M57), Andromeda (M31), Pegasus (M15), Hercules (M13, M92), Ursa Major (M81/82, M101), Gemini (M35, M36, M37), Orion (M42).
Those M-number are objects from Messier's catalog.
[url=http://en.wikipedia.org/wiki/List_of_Messier_objects]Wikipedia - Messier Objects[/url]
As for recognizing, the North Hemisphere constellations for the most part.

I like Ursa major, and Orion. I can recognize Orion, Usra Major, Ursa Minor, Cancer, Andromeda, and The dragon one.

People living in the north are missing the biggest treasures. Centaurus (NGC 5139 and some galaxy), Scorpius-Sagittarius-Serpens Cauda(M4,6,7,8,20,16,17...). I like these better than the northern ones (UMa, Her, Leo, CmB...)  Not to miss Carina($ \eta$), Ophiucus (globulars)...

:( The observors in the North Hemisphere also miss the Large and Small Magellic Clouds, those must be fantastic.

Oh, I miss them too. I am still north of the equator you see. My limits are perhaps Crux/Carina :?:


How many distinct positive factors does 32 have?

There's a really neat formula you can use for these kinds of problems.

First, you need to find the prime factorization. Let's use 180 as an example.

The prime factorization is: $ 2^2 \cdot 3^2 \cdot 5^1$

Now, each factor has to have each of these primes as a factor in its own prime factorization, and no other primes. For example, the factor 45 has a prime factorization of $ 2^0 \cdot 3^2 \cdot 5^1$. For a 2, there are 3 possible exponents, 0, 1, and 2. For a 3, there are three possible exponents as well, 0, 1, and 2. For the 5, there are only 2, 0 and 1. Do you see why?

Thus we can see as a general formula that the number of factors for a number with prime factorization $ p_1^a \cdot p_2^b$ is $ (a \plus{} 1)(b \plus{} 1)$. (Note that there can be more than just 2 primes.)

So the prime factorization of 32 is just $ 2^5$. So $ 5 \plus{} 1 \equal{} 6$, thus our answer is $ \boxed{6}$.


MATHCOUNTS percent LaTeX probability factorial function quadratics

The rules are simple, I'll post a problem or two, and people have a predetermined amount of days from the time I post them to PM me the answers. After a day, the qualifiers will move on, and the problems will get a bit more difficult, until we have narrowed the field down to a number of competitors(to be determined by me) and will hold a mock countdown round at a time determined by all of the qualifiers. Round 2ers are up

Round 2 is over.

I'll TRY....  :huh: But how long do we have to do each problem? I know the first round is till Sunday night; what about the others?

And I'll be away for July, so maybe not.  :(

It will depend on the problem, and however the time frame may get smaller, and most likely will for each problem, because this is the introductory round it is longer than it normally will be, and there is a fair chance that this competition will reach its culmination by July, so hold out hope. Also, on this problem I'm not requiring justification, but on others solutions are needed.

I'll do these Mathcounts competitons untill I start school.
Wow! I have a 1/8 chance of getting it right from a guess.

And we're eliminated if we miss a problem here?  :o

At this first round, if you miss the problem, you are out. In further rounds, the rules might be different. As this isn't a very hard problem if you explore it for 5-10 minutes, I think the rule is fair :) Especially considering you have four days. (well, three at this point)

[quote="bob123"]I'll do it

I'm assuming the letters in this problem have to be different, right? (cuz it doesn't say so)[/quote]

Yeah, the letters have to be different.

ATTENTION: A small amount of people have sent me their answers.

YOU NOW HAVE ONE AND A HALF DAYS TO SEND ME YOUR ANSWER. If you can't get it by Sunday night, and you have one last option. Guess!

A couple people who said they were participating haven't sent me their answers yet. Be sure you get them in.

[quote="sirorange"][quote="236factorial"]Wait, the competitors answered all that in one day?!  :o[/quote]

Yes, they truly are amazing  :D[/quote]

I don't know, at first sight, I don't think i can do any of them. I'll look into them soon.

W00T
 :winner_second:

Now for the dumb question:
[hide]Are you going to post how many questions each of us got right? [/hide]

My answers:
[hide]Spending Spree
bowl=$55$ ; candlestick=$19$ ; plate=$26$

White On Red
$6$ white socks ; $4$ red socks

'Round Goes the Gosip
$9$ phone calls

Lots of Confusion
$18$ lots

Ticket to Ride
Originally-$5$ stations ; New-$4$ stations

Tennis Anyone
28 sets would be won, and each player gets $x^{2}$ dollars where $x=$ the number of matches the player won.
The largest sum of money that could be payed would be $1^2+2^2+3^2+4^2+5^2+6^2+7^2=140$ dollars. $\frac{140}{8}=17.50$ dollars. this is why the club took $17.50$ dollars each.

One, Two, Three
$192;384;576$
$219;438;657$
$273;546;819$
$327;654;981$

No Help Here
$\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$

Proof:

$\phi=\frac{1+\sqrt{5}}{2}$

$\phi^{3}=2+\sqrt{5}$

$\frac{1}{\phi^{3}}=-2+\sqrt{5}$

So:
$\sqrt[3]{\phi^{3}}+\sqrt[3]{-\frac{1}{\phi^{3}}}=\phi-\frac{1}{\phi}$

Since:
$\phi-1=\frac{1}{\phi}$

$\phi-\frac{1}{\phi}=\phi-\phi+1=1$

Which means:
$\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$

$Q.E.D.$

$w00t!$

Blue on Blue
$9$ balls

Heads or Tails
$P=\frac{1}{2}$[/hide]

[quote="nat mc"]:first:  YAY!!    Congrats to Xantos as well, with a :clap:.   Phew, that was xtreme[/quote]

Lol, that [i]was[/i] xtreme. You should do another one, sirorange, this one was nicely run.

[quote="sirorange"]nat mc missed 5 and 8 (you misinterpreted the question), while Xantos C. Guin missed 3 and 9.[/quote]
wait, I got the same answer as him for #8!

Is #9 asking for the number of balls total, or just the blue ones?

Also, what was nat mc's proof on number No Help Here? Wasn't a proof required?

Number 9 was asking for the total, and he did provide a proof for no help here, he just didn't post it.

[quote="sirorange"]No, you assumed that you get 5 dollars for each game won.[/quote]
But thats number 6 you're talking about! I am confused. So I missed 5 and 6.  Cause I got #8 right, my proof showed it.

I don't get how I missed #5, I guess I misunderstood it.  If each sends it to another, and my first number was 3, then there were 3*2/2=3 sent first, and then 52 more are printed, and 3+52=55=11*10/2.  Doesn't mine work?

My solution actually proved it intead of assuming that it 3 and 11 would work

[quote="sirorange"]Number 9 was asking for the total....[/quote]

Darn. I could have won.

Congratulations nat mc and good try Xantos C. Guin!!!


Hi, I am a 2nd year student in Economics. I have some difficulties with probability. Here is a problem that I cannot solve and I need some help.

It is supposed that only 50% of the students of a sample will go up to the level master. One chooses by chance 18 students.
Which is the probability so that at least 10 go up to the level master.

I want just to know in this exercise, which is the law of probability that one must apply.
Thank you for your assistance, and I'm so sorry for my bad English.

$1 = \text{Probability that at least 10 go up}+\text{Probability that 9 go up}+\text{Probability that at most 8 go up}$

Now, given a situation where at least $10$ go up, we can assign a corresponding situation where at most $8$ go up (by reversing the "status" of each student), so the two probabilities are equal.

The probability that exactly $9$ students go up is

$\frac{{18 \choose 9}}{2^{18}}\approx 18.55\%$

(Because out of $18$ students, we choose $9$ of them, where there are $2^{18}$ possible arrangements of students going up or not going up.)  Hence our probability $P$ that at least $10$ students go up can be solved for:

$1 \approx P+0.1855+P$
$P \approx 0.4073 = 40.73\%$

Thanks you very much for the answer. It helps me a lot !
 :)


Ένας γυμναστής παρατήρησε ότι με τα 23 αγόρια που προπονεί , μπορεί να κάνει το εξής :
  α) Να ορίζει ως  διαιτητή ένα οποιοδήποτε από τα 23 παιδιά .
  β) Τα υπόλοιπα 22 να τα  χωρίζει σε δύο ομάδες των 11 ατόμων , έτσι ώστε κάθε ομάδα να έχει το  ίδιο συνολικό βάρος.

 Αν το βάρος κάθε παιδιού είναι θετικός ακέραιος , να αποδειχθεί ότι και τα 23 παιδιά έχουν το ίδιο βάρος !

[b] Σχόλιο[/b]

 Είναι από το βιβλίο A.Soifer : Colorando math Olympiad ....και είναι πιστεύω πολύ βασική για αυτή την κατηγορία ασκήσεων. Έχουμε και τη λύση της .

πολύ ωραίο πρόβλημα κ. Στεργίου!
 Εχει όμως ξανασυζητηθεί εδώ:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=253162[/url]


geometry perimeter calculus integration inradius

[hide="Solution"]
From $A = rs$ we conclude that we desire all integer right triangles such that $r = 2$.  All integer right triangles have an integer inradius, and because $r = 2$ we can assume that the triangle is either primitive or double a primitive triangle with $r = 1$.  The only such triangle is $(3, 4, 5)$, hence, the only other possible solution is $(6, 8, 10)$.  (Note that $(5, 12, 13)$ is the unique primitive triangle with $r = 2$.  Note also that for larger values of $r$ this is not necessarily true.)
[/hide]

[quote="amcavoy"]The $\left(5,12,13\right)$ right triangle has area and perimeter both equal to $30$.  Find all right triangles $T$ with integral sides such that the area of $T$ equals the perimeter of $T$.

[hide="Partial Solution"]
$x+y+\sqrt{x^{2}+y^{2}}=\frac{1}{2}xy\implies y=\frac{4\left(x-2\right)}{x-4}$

?
[/hide][/quote]

i just follow your step: y= (4x-16+8)/x-4 = 4+8/(x-4)
since x and y are positive integers , so x can be 5,6,8,12 so y=12,8,6,5 as correspond, so the solution is (6,8,10)


Diophantine equation
number theory proposed
number theory

Let $ k\in N$
Prove that exist a infinite  $ (m,n)\in N$ such that
$ \frac{m\plus{}k}{n}\plus{}\frac{n\plus{}k}{m}$ is an natural.

You are wrong. We can prove it only for $ k\equal{}1$ (work $ (m,n)\equal{}(ak,bk)$, were $ (a,b)$ solution for $ k\equal{}1$).
Let $ k\equal{}1, s\equal{}m\plus{}n,t\equal{}m\minus{}n$, then $ \frac{m\plus{}1}{n}\plus{}\frac{n\plus{}1}{m}\equal{}\frac{4s(s\plus{}1)}{s^2\minus{}t^2}\minus{}2$. If $ d\equal{}(s,t)$ we get $ d|4$, therefore $ \frac{16(s\plus{}1)}{s^2\minus{}t^2}$ must be integer. It had only finite solutions.

[quote="Rust"]You are wrong. We can prove it only for $ k \equal{} 1$ (work $ (m,n) \equal{} (ak,bk)$, were $ (a,b)$ solution for $ k \equal{} 1$).
Let $ k \equal{} 1, s \equal{} m \plus{} n,t \equal{} m \minus{} n$, then $ \frac {m \plus{} 1}{n} \plus{} \frac {n \plus{} 1}{m} \equal{} \frac {4s(s \plus{} 1)}{s^2 \minus{} t^2} \minus{} 2$. If $ d \equal{} (s,t)$ we get $ d|4$, therefore $ \frac {16(s \plus{} 1)}{s^2 \minus{} t^2}$ must be integer. It had only finite solutions.[/quote]
You are wrong?
Do you want an example?

[quote="Rust"]You are wrong. We can prove it only for $ k \equal{} 1$  [...].[/quote]

@Rust:
If the case $ k\equal{}1$ hasinfinitely many solution pairs $ (m_i,n_i)$, then consider $ (k*m_i,k*n_i)$ for general $ k$.

[quote="test20"][quote="Rust"]You are wrong. We can prove it only for $ k \equal{} 1$  [...].[/quote]

@Rust:
If the case $ k \equal{} 1$ hasinfinitely many solution pairs $ (m_i,n_i)$, then consider $ (k*m_i,k*n_i)$ for general $ k$.[/quote]
Oh yes .
$ k\in N$
Other Prove that exist infinite $ (m,n)\in N$ such that
$ mn|m^2 \plus{} n^2 \plus{} k$
This will help us solve my problem.

Prove that if the Pell equation
$ ax^2\minus{}by^2\equal{}c$ where $ a,b\in N$ and $ ab$ is not a square.
has a solution then it has infinite solution.
Use to prove that
For $ a,k\in N$
Prove that exist infinite $ (m,n)\in N$ such that
$ \frac{an\plus{}k}{m}\plus{}\frac{bn\plus{}k}{m}$ is an integer.

[quote="TTsphn"]Prove that if the Pell equation
$ ax^2 \minus{} by^2 \equal{} c$ where $ a,b\in N$ and $ ab$ is not a square.
has a solution then it has infinite solution.
Use to prove that
For $ a,k\in N$
Prove that exist infinite $ (m,n)\in N$ such that
$ \frac {an \plus{} k}{m} \plus{} \frac {bn \plus{} k}{m}$ is an integer.[/quote]
I show, that first problem was not correct.
I think these problem will be correct if
$ \frac{am\plus{}k}{n}\plus{}\frac{bn\plus{}k}{m}$ is an integer.

I will prove for you
$ d \equal{} \gcd(m,n)$
$ m \equal{} da,n \equal{} db$
Then
$ d^2ab|d^2(a^2 \plus{} b^2) \plus{} kd(a \plus{} b)$
$ \Leftrightarrow dab|(a \plus{} b)(d(a \plus{} b) \plus{} k)$
Chose $ d \equal{} a \plus{} b$ then
$ ab|(a \plus{} b)^2 \plus{} k \Leftrightarrow ab|a^2 \plus{} b^2 \plus{} k$ (1)
I prove that exist infinite $ (a,b,k)$ by Pell equation.

Prove that exist infinite $ (x,y)\in N$ and $ \gcd(x,y)\equal{}1$ and
$ x|y^2\minus{}5$
$ x|y^2\minus{}5$
More $ k\in Z$ and k is different from 0
Prove that exist infinite $ (x,y)\in N$
$ x|y^2\plus{}k$
$ y|x^2\plus{}k$

[quote="TTsphn"]Prove that exist infinite $ (x,y)\in N$ and $ \gcd(x,y) \equal{} 1$ and
$ x|y^2 \minus{} 5$
$ x|y^2 \minus{} 5$
More $ k\in Z$ and k is different from 0
Prove that exist infinite $ (x,y)\in N$
$ x|y^2 \plus{} k$
$ y|x^2 \plus{} k$[/quote]
TTsphn, why you post one problem many times?
Obviosly these problem followed from $ xy|(x^2 \plus{} y^2 \plus{} k)$.

The first is $ k\in N$ but the general $ k\in Z$


geometry
rectangle
quadratics
Gauss
number theory unsolved
number theory

Just draw a rectangle with vertices $(0,0),(a,0),(0,b),(a,b)$ and count, using this large rectangle's diagonal, the points inside or on the boundary of the smaller rectangle with vertices $(0,0),\left(\frac{a-1}2,0\right),\left(0,\frac{b-1}2\right),\left(\frac{a-1}2,\frac{b-1}2\right)$.

No, some standard theorem used by some proves of quadratic reciprocity, I think.

I'm pretty sure it was in a past MO, but in a slightly differenbt form.

Bomb

The solution to this problem is partially one of several proofs of the Gauss quadratic reciprocity theorem ;).

Grobbers post is a part of Eisenstein's approach and solution (however he did for $a,b$ primes, but when $(a,b)=1$ then it's a proof of the Jaboci's version of that theorem).


Maria sees these first four figures of a sequence in her textbook.  The first figure is made from 6 toothpicks.  Maria has 41 toothpicks.  What is the Figure number of the largest figure of this sequence Maria can create if the sequence were to continue forever?
[metapost=d33b49fb5ddf96d3861b7a4f46968644]save u, hexagon; path hexagon;

 % Set the value of u to scale the entire image
 u:=.5cm;

 hexagon:=(u*right)--(u*dir(60))--(u*dir(120))--(u*left)--(u*dir(-120))--(u*dir(-60))--cycle;

 % Figure 1
 draw hexagon;
 label.bot(btex Figure $ 1$ etex,u*sqrt(3)/2*down);

 currentpicture:=currentpicture shifted (3.5u*left);

 % Figure 2

 draw hexagon shifted (u*left);
 draw hexagon shifted (u*dir(60));

 label.bot(btex Figure $ 2$ etex,u*sqrt(3)/2*down);

 currentpicture:=currentpicture shifted (5u*left);

 % Figure 3

 draw hexagon shifted (u*1.5*left);
 draw hexagon shifted (u*sqrt(3)/2*up);
 draw hexagon shifted (u*1.5*right);
 label.bot(btex Figure $ 3$ etex,u*sqrt(3)/2*down);

 currentpicture:=currentpicture shifted (6.3u*left);

 % Figure 4

 draw hexagon shifted (2u*left);
 draw hexagon shifted (u*dir(60)+u*left);
 draw hexagon shifted (u*right);
 draw hexagon shifted (2u*right+u*dir(60));
 label.bot(btex Figure $ 4$ etex,u*sqrt(3)/2*down);

 currentpicture:=currentpicture shifted (4u*left);

 % ...

 draw (0,0) withpen pencircle scaled 3;
 draw (u,0) withpen pencircle scaled 3;
 draw (2u,0) withpen pencircle scaled 3;
[/metapost]


Let x, y, z be real numbers in the interval (0; 1). Prove that
$ \frac{1}{x(1\minus{}y)}\plus{}\frac{1}{y(1\minus{}Z)}\plus{}\frac{1}{z(1\minus{}x)}\geq \frac{3}{xyz \plus{} (1\minus{}x)(1\minus{}y)(1\minus{}z)}$

[quote="lnnvnu"]Let x, y, z be real numbers in the interval (0; 1). Prove that
$ \frac {1}{x(1 \minus{} y)} \plus{} \frac {1}{y(1 \minus{} Z)} \plus{} \frac {1}{z(1 \minus{} x)}\geq \frac {3}{xyz \plus{} (1 \minus{} x)(1 \minus{} y)(1 \minus{} z)}$[/quote]

Let $ x\equal{}\frac{1}{1\plus{}a},y\equal{}\frac{1}{1\plus{}b},z\equal{}\frac{1}{1\plus{}c},a,b,c>0$, inequality becomes
\[ \frac{1}{{b\left( {1 \plus{} c} \right)}} \plus{} \frac{1}{{c\left( {1 \plus{} a} \right)}} \plus{} \frac{1}{{a\left( {1 \plus{} b} \right)}} \ge \frac{3}{{1 \plus{} abc}},\]
\[ \Longleftrightarrow \sum_{cyc} {ab\left( {1 \plus{} b} \right)\left( {ca \minus{} 1} \right)^2 }  \ge 0,\]
obvious true, so inequality holds.

[quote="lnnvnu"]Let x, y, z be real numbers in the interval (0; 1). Prove that
$ \frac {1}{x(1 \minus{} y)} \plus{} \frac {1}{y(1 \minus{} Z)} \plus{} \frac {1}{z(1 \minus{} x)}\geq \frac {3}{xyz \plus{} (1 \minus{} x)(1 \minus{} y)(1 \minus{} z)}$[/quote]

[hide]Chắc bạn là người Việt.Không nên post ở đây vì nó ở trong THTT số này! khi chưa hết hạn nộp bài.[/hide]
Close topic  :)