algebra
polynomial
quadratics
product of roots
algebra theorems

I am new to this forum, so I don't know if a question about thoerem is supposed to be put here.

When we have the sum and the product of roots, for example, a quadratic equation.
ax^2+bx+c=0
We can create a sequence,
aS_n+bS_n-1+cS_n-2=0
But why, may anyone show me a proof?

It's obvious that we can define the sequence.  Just give it two initial values, and the other values can be calculated uniquely.  I think I know what you're really asking, though.

I'll approach this problem from the other direction.

Suppose we have a sequence $s_{n}= au^{n}+bv^{n}$ for some $a, u, b, v \in \mathbb{C}$.  Verify for yourself the following identity:

$s_{n+1}= (u+v) s_{n}-uv s_{n-1}$

Let $A = u+v, B =-uv$.  We can write this as

$s_{n+1}= A s_{n}+B s_{n-1}$

A second-order linear recurrence.  Note that the quadratic equation

$x^{2}= Ax+B$

Has roots $u, v$.  I think that was the object of your question?

Thanks t0rajir0u
I think I am clear for the original question.
Does C mean Constant?
so we can also use the sequence to find $2u^{5}+v^{5}$ if they are the roots of a quadratic equation with the value of product and sum known
and I would expect that there at least two values, right?

[quote="Command"]Does C mean Constant?
so we can also use the sequence to find $2u^{5}+v^{5}$ if they are the roots of a quadratic equation with the value of product and sum known
and I would expect that there at least two values, right?[/quote]

$\mathbb{C}$ means Complex.

Yes, you can use the recursion to find arbitrary sums.  For example, start $a_{0}= 2+1, a_{1}= 2u+v$, and apply the recursion.


show that the equation
x^2+y^2+z^2=(x-y)(y-z)(z-x) has infinitely many solutions for integers x,y,z.

I think not. The LHS is then 9 times, and RHS 27 times original for x,y,z

It suffices to find $ \langle x',y',z' \rangle$ satisfying $ \left[(x')^2 \plus{} (y')^2 \plus{} (z')^2\right] \equal{} k[(x' \minus{} y')(y' \minus{} z')(z' \minus{} x')]$ where $ k$ is any integer, and then take $ \langle x,y,z \rangle \equal{} \langle kx',ky',kz' \rangle$.

Such an example would be $ \langle x',y',z'\rangle \equal{} \langle 2n \minus{} 1,2n,2n \plus{} 1\rangle$, where the product of the pairwise differences is $ 2$ and the sum of squares is $ 12n^2 \plus{} 2$, so $ \langle x,y,z \rangle$ $ \equal{} \langle(6n^2 \plus{} 1)(2n \minus{} 1), (6n^2 \plus{} 1)(2n), (6n^2 \plus{} 1)(2n \plus{} 1)\rangle$ for all integers $ n$ is a set of infinite solutions.


Given$ a,b,c>0$ such that:$ abc\equal{}1$.Prove that:
$ \frac{ab\plus{}a\plus{}1}{c^9(1\plus{}a)^2}\plus{}\frac{bc\plus{}b\plus{}1}{a^9(1\plus{}b)^2}\plus{}\frac{ca\plus{}c\plus{}1}{b^9(1\plus{}c)^2} \geq\ \frac{9}{4}$ :)

[quote="quykhtn-qa1"]Given$ a,b,c > 0$ such that:$ abc \equal{} 1$.Prove that:
$ \frac {ab \plus{} a \plus{} 1}{c^9(1 \plus{} a)^2} \plus{} \frac {bc \plus{} b \plus{} 1}{a^9(1 \plus{} b)^2} \plus{} \frac {ca \plus{} c \plus{} 1}{b^9(1 \plus{} c)^2} \geq\ \frac {9}{4}$ :)[/quote]
Use the equality $ \frac {1}{ac \plus{} c \plus{} 1} \plus{} \frac {1}{ab \plus{} b \plus{} 1} \plus{} \frac {1}{bc \plus{} b \plus{} 1} \equal{} 1$ and Cauchy-Schwartz, we prove
$ \frac {1}{c^5(1 \plus{} a)} \plus{} \frac {1}{b^5(1 \plus{} c)} \plus{} \frac {1}{a^5(1 \plus{} b)}\geq \frac {3}{2}$
Let $ a \equal{} \frac {y}{x},b \equal{} \frac {z}{y},c \equal{} \frac {x}{z}$, this inequality equivalents to:
$ \sum\frac {x^{10}}{x^5y^5 \plus{} y^4zx^5}\geq \frac {3}{2}$
Easy if we use Cauchy-Schwartz  :D


[quote="admire9898"]Let $ n\geq 1$ be an integer. Find all real numbers $ x$ such that $ (cos x)^{n} \minus{} (sin x)^{n} \equal{} 1$.[/quote]

[color=darkred][b]You just need to solve this when n is an odd number.and $ n\geq 3$
Anybody can give me the solution?[/b][/color]

[quote="admire9898"]Let$ n\geq 1$ be an integer. Find all real numbers $ x$ such that $ (cos x)^{n} \minus{} (sin x)^{n} \equal{} 1$.[/quote]
Use that $ (\cos{x})^n\leq (\cos{x})^2$ and $ \minus{} (\sin{x})^n\leq (\sin{x})^2$, so $ (cos x)^{n} \minus{} (sin x)^{n} \equal{} 1$ holds if and only if $ (\cos{x})^n \equal{} (\cos{x})^2$ and $ \minus{} (\sin{x})^n \equal{} (\sin{x})^2$.

I think Problem 1 has difficulty
 :)  I change all $ x_{i}$ to $ x_{1}$ by the conditions,But It doesn't work. :wink:

[b]Who can hadle problem1
details needed.Thanks！ :P [/b]


calculus
derivative
integration
calculus computations

I have seen many times that when solving a problem many users perform something along the lines of, $I(t)=\int_{a}^{b}f(tx)dx$ so
 $\frac{d}{dt}I(t)=\frac{d}{dt}\int_{a}^{b}f(tx)dx=\int_{a}^{b}\left(\frac{d}{dt}f(tx)\right)dx$

my question is, how would one justify doing this if they were to use when writing a formal solution.

It's easiest to justify if we phrase it using integration rather than differentiation; if $f(t,x)=\int_{c}^{t}g(s,x)\,ds$, $I(t)=\int_{a}^{b}\int_{c}^{t}g(s,x)\,ds\,dx=\int_{c}^{t}\int_{a}^{b}g(s,x)\,dx\,ds$ by Fubini's theorem (as long as $g$ is nise enough, of course). My $g$ is your $\frac{\partial f}{\partial t}$.


Given that the digits 1, 2, 3, 4, 5 and 6 are placed in the boxes shown, what is the greatest possible positive difference that can be obtained?

$$\begin{array}{r}
\Box \Box \Box\\
- \Box \Box \Box\\
\hline
\end{array}$$

[quote="AceOfDiamonds"]You want the $ 6$ to be minused, and as big as possible.
Since it can't be in hundreds(or else negative), place in tens.
Now similarly $ 5$ put in ones
Now you know that the difference in only the hundreds digit is $ 1$ so it's smallest
so you get $ 412\minus{}365$   (1 and 2 similar to 6 and 5)
and calculating that you get $ 47$[/quote]
FTW wants the greatest difference, AceOfDiamonds
[hide="The Solution"]
Make the first number as big as possible and the difference as small as possible. $654 - 123 = \boxed{531}$
[/hide]


I know i seem kinda clueless, but what exactly [i]is[/i] spam on this message board? I know what email spam is, but i hardly think it would be the same. and same with censored. i remember reading something about it when i signed up, but i forgot. ya, u can probably tell i'm new here. :blush:

Whatever you said just know is spam, because it is not related to the topic at all.

But yeah, I am not from Indiana but I know a lot of people from Indiana who own the Math Contests

[quote="mathcountsfreak"]I just want to say that no one is using the IN forum (this one), so I made this topic so people from IN could say what they or their school have accomplished. It doesn't have to be about math, it could be sports, science, geography, etc. So please post if you want to keep this forum alive.
And please, no  :spam: or  :censored: .[/quote]

i am from IN. im a west sider

[quote="cognos599"]

i am from IN. im a west sider[/quote]

Clarification: "west sider" = from West Lafayette Jr/Sr High School. I know because i'm from there, too.

what state are you from [b]indiana[/b]math :wink:

[quote="roadnottaken"]I'm from Ohio (right next door!), and we have a fairly active forum.  I don't think anyone would object if people from Indiana joined it :lol: .[/quote]

I think it should be the other way around... :wink:  .


i made a blog called Problems and later deleted it
when I tried to make another one, i can't.
also, can you make more than 1 blog?

No you can't have more than one blog. I've restored your previous one. I will take a look at the blogging system in more detail starting next week.

also, how do u change the name of the blog
and what are the color codes?

This should be all under blog options.

And color codes are like:

000000- which is black

or

999999- which is white i think

They can also contain letters, like FFFFFF. (which I have no clue what it is)

Just pick up any HTML/CSS book. I'm sure there's a good explanation of color codes. Or just use Google to find a good color chart. :)

also, how do u do a blog template and how does the 'timeframe' thingy work?


vector
geometry
topology
Euler
invariant
advanced fields
advanced fields theorems

it strikes me that you use the vanishing homology of euclidean space a good deal in electromagnetism. This is perhaps most obvious when finding potentials, as you're directly using that closed differential forms implies exactness, i.e. vanishing of the de Rham cohomology. It also seems like you also use it implicitly in a lot of other contexts too, though. For example, when talking about the vector area (such as when defining magnetic dipole moment), you consider a loop and then integrate something over the surface that the loop bounds, but the existence of this surface, i.e. the fact that any $ S^1$ is a boundary, is again topological information. so how much does electromagnetism change when working over a topologically nontrivial space?

originally posted in advanced physics without getting responses; I'm trying again here

I actually jotted notes about this down one day in class precisely because I forgot that we were working on $ \mathbb{R}^3$. I got furious when the teacher made that comment, and so I started constructing a counterexample. It was only when I got back to my room that day did I realize my mistake that I was in a physics class and not a math class.

I didn't really pursue it after that, though. I think that this is trivial to get around, but I was always concerned with the fact that when you set a charge in space, you have to ignore that point when talking about potentials since it blows up there. It seems to me that we are working with punctured space in that case, in which the de Rham cohomology is no longer the trivial case.

There is a bit on this in the introduction to Bott and Tu, but it's barely enough to whet your appetite and there are no references.

Some of what you are looking for may be explained in:  A Course in Mathematics for Students of Physics: Volume 2, by Bamberg and Sternberg, but I've never actually read it, so I don't know what's really in there.  One could also trying to talk to Shlomo himself.

If you want to go deeply into the relationship between EM and topology, you want to read about what the physicists call U(1) gauge theory.  The Maxwell's equations page on Wikipedia says something about this, but it looks poorly written.  The basic idea is that Maxwell's equations make perfect sense on a Lorentzian manifold M (and mathematically they make sense on any Riemannian manifold, too).  The electromagnetic field is described by some 2-form F, and Maxwell's equations are linear PDEs on F, one of which says that F is closed.  As you pointed out, closed is not the the same as exact, so there need not be a global vector potential.

However, an important observation is that the vector potential isn't really a "vector" anyway (in the sense of differential geometry).  It is actually something called a connection (on a complex line bundle over M), and F is the curvature of this connection.  The gauge theory point of view is to treat the connection as the fundamental mathematical object, while F is just some object that is derived from it.  When viewed this way, Maxwell's equations are just the Euler-Lagrange equations for some functional defined on the space of connections.

The physicists Yang and Mills generalized this construction to more general connections (for physics-y reasons), and many years later Donaldson realized that one could use this theory to study the topology of manifolds.  Specifically, focusing attention on the Riemannian case (not Lorentzian), the vaccuum case of Maxwell's equations simply says that the Hodge Laplacian of F vanishes, and the Hodge Theorem then implies that the solutions of Maxwell's equations are in 1-1 correspondence with the elements of H^2(M,R).  In other words, if no one had ever invented H^2, it could theoretically have been discovered by observing that the dimension of the space of solutions of the (Riemannian vaccuum) Maxwell's equations is a topological invariant.  Donaldson generalized this idea: He studied the space of solutions (called a moduli space) of the "Yang-Mills equations" (which generalize Maxwell) and showed that this space had some topologically invariant properties.  He also proved some bad-a** theorems using this new machinery.

Okay, so I have no idea how much of what I just wrote makes sense to you, but once I got going I couldn't stop.  I'm not an expert on anything I just wrote, so take it all with a grain of salt.


trigonometry
function
geometry
geometric transformation
reflection
IMO

I've noticed that back in the 1960's there was almost a trig question with basic functions sin and cos on every single tests. Also, the questions seem more down to the point rather than all wordy like the current IMO's. Is this suggestive of an underscoring trend in mathematical teaching that occurred over the years?

No, it just means that the IMOs have gotten harder.

IMO does not really reflect the overall education system. It reflects the training of the elite mathletes.

Finally, IMO isn't wordy, especially when compared to those programming olympiads :D


linear algebra
matrix
algebra
polynomial
vector
linear algebra unsolved

Hi! I just found here! It's so cooooool! I wanna propose a problem:

[color=blue][The following approximation of Fred 1998's original problem has been re-added by moderator. As blahblahblah says below - please don't delete them. It makes it hard to read the rest of the topics.][/color]

If $A$ is a real matrix, show that all of the eigenvalues of $A-A^{T}$ must be pure imaginary.

[quote="Kent Merryfield"]Let $B=A-A^{T}.$ Then $B$ is antisymmetric; that is, $B^{T}=-B.$

Let $C=B^{T}B.$ Then $C$ is symmetric and positive semidefinite. All of the eigenvalues of a symmetric matrix are real. As for the positive, consider this:

Suppose $Cx=\lambda x,$ that is, $B^{T}Bx=\lambda x.$ Multiply on the left by $x^{T}.$ Then $x^{T}B^{T}Bx=\lambda x^{T}x$ which becomes $\|Bx\|^{2}=\lambda\|x\|^{2}.$ This forces $\lambda>0.$

So all of the eigenvalues of $C$ are nonnegative. But $C=-B^{2},$ so all of the eigenvalues of $B^{2}$ are nonpositive.

If $\lambda$ is an eigenvalue of $B,$ then $\lambda^{2}$ is an eigenvalue of $B^{2}.$ But then $\lambda^{2}\le0,$ hence $\lambda$ is pure imaginary.[/quote]

Thanks. Why if  lambda is an eigenvalue of B, then lambda ^ 2 is an eigenvalue of B^ 2? [/quote]

It's a standard argument.

If $Bx=\lambda x$ then

$B^{2}x=B(Bx)=B(\lambda x)=\lambda(Bx)=\lambda^{2}x.$

More generally, if $P$ is any polynomial, then if $\lambda$ is an eigenvalue of $B,$ it follows that $P(\lambda)$ is an eigenvalue of $P(B).$

From that we deduce that if $B$ satisfies some polynomial equation, then all of the eigenvalues of $B$ must satisfy the same polynomial equation.

[quote="Kent Merryfield"]Let $B=A-A^{T}.$ Then $B$ is antisymmetric; that is, $B^{T}=-B.$

Let $C=B^{T}B.$ Then $C$ is symmetric and positive semidefinite. All of the eigenvalues of a symmetric matrix are real. As for the positive, consider this:

Suppose $Cx=\lambda x,$ that is, $B^{T}Bx=\lambda x.$ Multiply on the left by $x^{T}.$ Then $x^{T}B^{T}Bx=\lambda x^{T}x$ which becomes $\|Bx\|^{2}=\lambda\|x\|^{2}.$ This forces $\lambda>0.$

So all of the eigenvalues of $C$ are nonnegative. But $C=-B^{2},$ so all of the eigenvalues of $B^{2}$ are nonpositive.

If $\lambda$ is an eigenvalue of $B,$ then $\lambda^{2}$ is an eigenvalue of $B^{2}.$ But then $\lambda^{2}\le0,$ hence $\lambda$ is pure imaginary.[/quote]

$x^{T}B^{T}Bx=\lambda x^{T}x$ which becomes $\|Bx\|^{2}=\lambda\|x\|^{2}$

How?

Think of your vectors as being written as $n\times 1$ (one-column) matrices. Then $y^{T}x$ is a $1\times 1$ matrix - that is, a real number. In fact,

$y^{T}x=\sum_{i=1}^{n}x_{i}y_{i}$ is the inner product of $x$ and $y.$

Hence, $x^{T}x$ is the inner product of $x$ with itself, which can also be written as $\|x\|^{2},$ as long as we are using the usual Euclidean norm.

This is an extremely useful way to write norms and inner products; I highly recommend it.

In the particular case above, $\|Bx\|^{2}=(Bx)^{T}(Bx)=x^{T}B^{T}Bx.$


calculus
integration
trigonometry
calculus computations

Using $\sin x = \frac{2\tan (x/2)}{1+\tan^{2}(x/2)}$, $\cos x = \frac{1-\tan^{2}(x/2)}{1+\tan^{2}(x/2)}$ and setting $\tan(x/2) = u$ the integral can be reduced to $\frac{2^{n}}{(a+b)^{n}}\int_{0}^{\infty}\frac{u^{n-1}}{(1+{\lambda}^{2}u^{2})^{n}}du$ where $\lambda^{2}= \frac{a-b}{a+b}$. Putting $1+\lambda^{2}u^{2}= v$ and simplifying the resulting integral yields $\frac{2^{n-1}}{(a^{2}-b^{2})^{n/2}}\int_{1}^{\infty}\frac{(1-\frac{1}{v})^{\frac{n}{2}-1}}{v^{\frac{n}{2}+1}}dv$. Set $\frac{1}{v}= z$ in the last integral to get the desired result.


Triangle $ ABC (\sphericalangle ACB > 90$) is inscribed in circle $ S$. $ D$ is a point where $ CS$ cuts $ AB$. $ |AC| \plus{} |BC| \equal{} 2|CS|$. Proof that circles inscribed in triangles $ ADC$ and $ BDC$ have the same radiuses.

It seems to be really difficult.. I thought that I had a solution, but I was wrong:/ I hope someone will solve it couse I'm really interested:)

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=307888[/url]


Hi everyone. I had a couple of questions pertaining to EMF. I am not 100% sure if my solutions are correct and I would like to know if someone could please confirm how to solve the following questions.

Thank you very much.

[b]1. Suppose the emf of the battery in circuit shown in Fig. 1 varies with time t so that the current is given by i(t) = 3.1 + 7.0t, where i is in amperes and t is in seconds. Take R = 4.4 Ω and L = 5.1 H. For t = 0, what is (a) the battery emf and (b) its rate of change? [/b]

[i]I know that I need to use the loop rule but I am a little confused with respect to how exactly to apply it to this particular problem.[/i]

[b]2. For the circuit of Fig. 1, assume that ε = 14.0 V, R = 7.26 Ω, and L = 8.11 H. The ideal battery is connected at time t = 0. (a) How much energy is delivered by the battery during the first 2.54 s? (b) How much of this energy is stored in the magnetic field of the inductor? (c) How much of this energy is dissipated in the resistor?[/b]

[i]I know that the energy stored in the magnetic field of a coil at any time depends on the current through the coil at that time (U = Li^2), so part (b) isn't really a problem. I am just confused as for how I go about solving (a) and (c).[/i]

[b]3. The magnetic field of a cylindrical magnet that has a pole-face diameter of 4.4 cm can be varied sinusoidally between 42.9 T and 44 T at a frequency of 20 Hz. At a radial distance of 2.1 cm, what is the amplitude of the electric field induced by the variation?[/b]

[i]I know that since R>r, then amp = (r/2)(dB/dt)[/i]

Thank you very much, in advance. I appreciate the help.

[color=blue][b](1)[/b][/color]
$\epsilon(t) = Ri(t) + L\ \frac{\text di}{\text dt}$

$\frac{\text d\epsilon}{\text dt} = R\ \frac{\text di}{\text dt} + L\ \frac{\text d^2i}{\text dt^2}$

$i(t) = 3.1 + 7t$ is known, $\frac{\text di}{\text dt} = 7$ and $\frac{\text d^2i}{\text dt^2} = 0.$ Substitute this and the R, L values into the equations for $\epsilon(t),\ \frac{\text d\epsilon}{\text dt}.$

[color=blue][b](2)[/b][/color]
$Ri(t) + L\ \frac{\text di}{\text dt} = \epsilon$

$L\ \frac{\text di}{\text dt} + \frac R L\ i(t) = \frac \epsilon L$

This differential equation has a general solution

$i(t) = C \exp \left(-\frac R L\ t \right) + \frac \epsilon R$

where C is an arbitrary constant. At t = 0, $i(0) = 0$, hence, $0 = C + \frac \epsilon R,\ C = -\frac \epsilon R$ and

$i(t) = \frac \epsilon R \left[1 - \exp \left(-\frac R L\ t \right)\right]$

Power delivered by the battery at time t is $P(t) = \epsilon i(t)$ and energy delivered from t = 0 to $t = \tau = 2.54\ \text{sec}$ is

$E = \int_0^{\tau} P(t)\ \text dt = \epsilon \int_0^{\tau} i(t)\ \text dt = \frac{\epsilon^2}{R} \left[t + \frac L R \exp\left(-\frac R L\ t \right)\right]_0^{\tau} = \frac{\epsilon^2}{R} \left\{\tau + \frac L R \left[\exp\left(-\frac R L\ \tau \right) - 1\right]\right\}$

On the other hand, since $\epsilon = Ri(t) + L\ \frac{\text di}{\text dt}$,

$E = R \int_0^{\tau} i^2(t)\ \text dt + L \int_0^{\tau} i(t) \frac{\text di}{\text dt}\ \text dt = E_R + L \int_0^{i(\tau)} i\ \text di = E_R + \frac L 2 i^2(\tau)$

The first term is energy dissipated in the resistor from t = 0 to $t = \tau$ and the second one energy stored in the coil. Since both E and $E_L = \frac L 2 i^2(\tau)$ have been calculated, $E_R = E - E_L$. It is certainly possible to calculate $E_R$ the hard way.

[color=blue][b](3)[/b][/color]
$\epsilon = -\frac{\text d\Phi}{\text dt} = -\pi r^2 \frac{\text dB}{\text dt}$

$B = \overline B + B_0 \sin(\omega t)$

where $\overline B = \frac{42.9 + 44}{4} = 43.45\ \text T,\ B_0 = \frac{44 - 42.9}{2} = 0.55\ \text T,$ and $\omega = 2 \pi f,\ f = 20\ \text{Hz}$. Therefore,

$\epsilon = -\frac{\text d\Phi}{\text dt} = -\pi r^2 \omega B_0 \cos (\omega t)$

That's an electromotive force (voltage) induced in a circular loop of radius r inside the magnet. The electric field along this loop is

$E = -\frac{\epsilon}{2 \pi r} = \frac{r \omega B_0}{2} \cos (\omega t) = E_0 \cos (\omega t)$

so that its amplitude is $E_0 = \frac{r \omega B_0}{2} = \pi r f B_0$.

Thank you for your help. I perfectly understand your reasoning for questions (1) and (3), but I am slightly confused on question (2), parts (a) and (c).

I understand that the energy stored in the magnetic field = E (sub L) = 12.1348 J in my case. What would be the values for the energy is delivered by the battery during the first 2.54 s and the energy dissipated in the resistor? I realize that E (sub R) is the energy dissipated in the resistor, yet [u]I am having trouble finding a numerical value for the energy delivered by the battery during the first 2.54 s aka part (a). [/u]

It is obvious to me that the difference between the energy delivered by the battery during the first 2.54 s and the energy stored in the magnetic field is equal to the energy dispitated in the resistor.

Thank you very much yetti. Have a good day.

Thanks - that makes MUCH more sense now. I really appreciate your help. Like you, I also got caught up in my algebra and make a careless mistake.

I found that $L$ reffers to $\frac{\Phi}{i}$. So there's some magnetic flux involved in there. How do you practically do that?

Also, would the situation have been different if the arrow $\epsilon_L$ pointed in the opposite dirrection? Would any sign flip?

I don't think you actually need to work with $L = \frac{\Phi_{\text{B}}}{I}$; just know that if you substitute this into Faraday's Law of Induction, this gives you that $\mathcal{EMF} = -L \frac{dI}{dt}$. From this, we have $dE = V dQ = ( L \frac{dI}{dt}) (I dt) = L I dI$, so integrating, we find that $E = \frac{1}{2} LI^2$.


geometry
circumcircle
Euler
ratio
incenter
angle bisector
geometry proposed

For a triangle [b]ABC[/b], its incircle ([b]I[/b]) touches [b]BC[/b],[b]CA[/b],[b]AB[/b] at [b]D[/b],[b]E[/b],[b]F[/b] respectively.
Let its circumcircle be ([b]O[/b]) and  the circumcircles of  triangles [b]AEF[/b], [b]BDF[/b], [b]CDE[/b] meet ([b]O[/b]) at [b]P[/b], [b]Q[/b], [b]R[/b] respectively.

(1) [b]PD[/b], [b]QE[/b], [b]RF[/b] are concurrent (let the point of concurrence be [b]S[/b])
(2) [b]S[/b] is on the line [b]OI[/b]

Thank you for reading my post, in advance. :)

[b]Lemma:[/b] $ H_a,H_b,H_c$ are the feet of the altitudes of $ \triangle ABC$ and $ P$ is any point on its plane. Then the circles $\odot (PAH_a),$ $\odot (PBH_b)$ and $\odot(PCH_c)$ meet again on $PH.$

Inversion WRT $ (I)$ takes $\odot( AFE) , \odot( CED), \odot(BDF)$ into the lines $FE,ED,DF$ and the circumcircle $ (O)$ of $\triangle ABC$ into the 9-point circle $(O')$ of $ \triangle DEF$ $\Longrightarrow$ $ P,Q,R$ are taken into the altitude feet  $ X,Y,Z$ of $ \triangle DEF.$ Lines $ PD,FR,QE$ are taken into the circles $ \odot(DIX),\odot(EIY)$ and $\odot( FIZ).$ From the previous lemma, these latter circles meet on the Euler line  $IO$ of $ \triangle DEF.$ Therefore, their inverse lines $ PD,$$ FR$ and $ QE$ under the referred inversion concur on $ IO$ as well.

1.triangle ARE and BRD are similar  so  RF bisects angleARB

2.denote U,V,W  midpoints o fthe minor arc BC,CA,AB
  then we have angle DEF and UVW are similar with corresponding sides parallel and the ratio is ID/OU=r/R
  S lies on OI

[b]Here is my solution (Using inversion :D )[/b]
Consider the inversion through pole $ I$, power $ r^2$, where $ r$ is the radius of the incircle wrt  $ \triangle ABC$. Let $ (C_1)$, $ (C_2)$, $ (C_3)$ be the circumcircles of $ \triangle AEF$, $ \triangle BFD$, $ \triangle CDE$, respectively. It is obviously that $ I(I,r): A\mapsto A_0$, $ B\mapsto B_0$, $ C\mapsto C_0$, where $ A_0,B_0,C_0$ are the midpoints of $ EF,DF,DE$, respectively, $ D\mapsto D$, $ E\mapsto E$, $ F \mapsto F$. Hence, $ I(I,r): (C_1)\mapsto EF$, $ (C_2)\mapsto FD$, $ (C_3)\mapsto ED$, $ (O)\mapsto (A_0B_0C_0)$. Since, $ P$ is the intersection of $ (C_1)$ and $ (O)$, therefore if $ I(I,r): P\mapsto P'$, therefore, $ P'$ is the intersection of $ (A_0B_0C_0)$ and $ EF$, arguing the same for $ Q$ and $ R$. Now, it is easy to notice that $ (A_0B_0C_0)$ is the Euler circle wrt $ \triangle DEF$, therefore, $ DP', EQ', CR'$ are concurent, further, they are also the altitudes of $ \triangle DEF$. Since the fact that $ E,F,Q',R'$ are concyclic, hence, $ E,F,Q,R$ are also concyclic (easy to notice this through the $ I(I,r)$). Now, let $ X$ be the intersection of $ EF$ and $ RQ$, hence, $ \overline {XE}. \overline {XF} \equal{} \overline {XQ}.\overline {XR}$, which implies that $ \mathcal {P}(X,(I)) \equal{} \mathcal {P}(X,(O))$. With the same arguement, we also get that $ (P,F,D,R)$ and $ (E,D,Q,P)$ are sets of concyclic points. Let $ Y$, $ Z$ define the same as $ X$, respectively, wrt $ (P,F,D,R)$ and $ (E,D,Q,P)$. It is now easy to see that $ X$, $ Y$, $ Z$, respectively, have the same power wrt $ (I)$ and $ (O)$. Therefore, $ X,Y,Z$ are collinear. Hence, according to [b]Desgaurte theorem[/b], we get that $ PD,QE,RF$ are concurrent.
$ S$ is the concurrent point of these lines. It is easy to notice that $ I(I,r): H_i\mapsto S$, where $ H_i$ is the orthocenter wrt $ \triangle DEF$. Therefore, $ I,S,H_i$ are collinear. We need to prove that $ I,O,H_i$ are collinear. But this is obviously true, you can see at this following link: [color=blue][b][u]http://www.mathlinks.ro/viewtopic.php?t=152601[/u][/b][/color]. Therefore, we will get the result of the whole problem
Which ends our proof :lol:

[b]@: Sorry if my solution is the same to any posts[/b]

Posted before: http://www.mathlinks.ro/viewtopic.php?t=159997  :D

First I will restate the problem:

[b]Problem:[/b]
Let $ (O)$ and $ (I)$ in that order be the circumcenter and incenter of $ \triangle ABC$. $ (I)$ tangents to $ BC,CA,AB$ at $ A_0,B_0,C_0$. $ (AB_0C_0), (BC_0A_0),(CA_0B_0)$ intersects $ (O)$ again at $ A_1,B_1,C_1$ respectively. Prove that $ A_0A_1,B_0B_1,C_0C_1$ concur at a point on the Euler's line of $ \triangle A_0B_0C_0$.

[b]Solution:[/b] (inversion combined with angle chasing)
Denote by $ A_3$ and $ A_2$ the intersection of $ IA$ and $ IA_1$ with $ B_0C_0$, we have $ \widehat{AA_1A_2} \equal{} \widehat{AA_3A_2} \equal{} 90^o$ so $ A,A_1,A_2,A_3$ are concyclic, which implies that $ \overline{IA}.\overline{IA_3} \equal{} \overline{IA_1}.\overline{IA_2} \equal{} r^2$, where $ r$ is the in-radius of $ \triangle ABC$. Moreover: $ I$ is the midpoint of arc $ B_0C_0$ of circle $ (AB_0C_0)$ so $ A_1A_2$ is the angle bisector of $ \widehat{B_0A_1C_0}$, but $ \triangle A_1C_0B_0 \sim \triangle A_1BC$ so $ A_1A_0$ is also the angle bisector of $ \widehat{BA_1C_1}$. By this, $ \widehat{A_1A_2C_0} \equal{} \widehat{A_1A_0B}$, which implies that $ A_2$ is the foot of the altitude from $ A_0$ of $ \triangle A_0B_0C_0$.
Similarly, the images of $ B_1,C_1$ through the inversion with center $ I$ and power $ r^2$ are the feet of altitudes of $ \triangle A_0B_0C_0$.
But in $ \triangle A_0B_0C_0$, three circles $ (IA_0A_2),(IB_0B_2),(IC_0C_2)$ have a common point on its Euler's line, and so do $ A_0A_1,B_0B_1,C_0C_1$, which ends the proof. :roll:


geometry
incenter
circumcircle
geometric transformation
reflection
ratio
dilation

Let $I$ be the incenter of the triangle $ABC$, et let $A',B',C'$ be the symmetric of $I$ with respect to the lines $BC,CA,AB$ respectively. It is known that $B$ belongs to the circumcircle of $A'B'C'$.
Find $\widehat {ABC}$.

Pierre.

Hope this makes sense.

Let the tangent points of the incircle of ABC and on the sides BC, AC, and AB be D, E, and F.  Also let the intersection of the incircle of ABC and BI to be X.

Now since DI=DA', IE=EB', ID=DC', and IDA', IEB', and IDC' are straight lines all by the definition of a reflection, then A'B' // DE, A'C' // DF, and B'C' // EF.  In other words, DEF is similar to A'B'C', and is in the ratio 1:2.

So, a dilation taking D to A', E to B', and F to C' will also take the circumcircle of DEF to the circumcircle of A'B'C', taking X to B.  So, looking at triangle ID, we have DI=r  (r is the radius of the circumcircle of DEF)  and BI=2r, with <BDI=90.  So, we have a 30-60-90 triangle, with <IBD=30.

Similarly, <FBI=30, so angle ABC=60

isnt the solution just angle chasing?

kennyworm

there is another solution.
it very palpable that IA'=IB'=IC'
 <C'IB'=180-<A
<A'IB'=180-<C
<A'BC'=2<B
so <A'B'C'=1/2 (<A+<C)
since A',B',C', B are on the same circle
<C'BA'+<A'B'C'=180
2<B+1/2(<A+<C)=180
3/2<B+1/2(<A+<B+<C)=180
3/2<B=90

<B=60

Why not simply: I is the circumcentre of A'B'C' (Erratum: was ABC). Thus IB = 2r. Thus IBC = 30deg... and so ABC = 60 deg.

Yes , you are right [b]kueh[/b].It is so simple.
By the way you have typo in your post : I is the circumcentre of A'B'C'.


geometry
trigonometry
3D geometry
prism
inequalities

A parallelepiped has surface area 216 and volume 216. Show that it is a cube.

Let the sides of one face be $ a$, $ b$ and $ A$ be the area of that face. Thus $ A\equal{}ab\sin(\theta)\le ab$. Let $ h$ be the distance from this face to the opposite face. Then we have $ Ah\equal{}216$. Project this face onto the plane of the opposite face. This rectangular prism cannot have more area than the parallelepiped, so $ 2A\plus{}h(2a\plus{}2b)\le 216\Rightarrow A\plus{}h(a\plus{}b)\le 108$ Therefore $ 108\ge A\plus{}h(a\plus{}b)\ge A\plus{}h(2\sqrt{A})\equal{}A\plus{}2\frac{216}{\sqrt{A}}\ge 108$. From this, you must have equality in all inequalities, so it's a cube.

I haven't worked out the details but shouldn't this be doable with AM-GM or something?

I don't think that works until you can establish that it's a rectangular prism. Of course, once you establish that, then it's just:

$ abc\equal{}216 \implies \frac{a^{\frac{1}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}}}{6}\equal{}1$

$ 2(ab\plus{}bc\plus{}ca)\equal{}abc \iff \frac{a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{1}{3}}\plus{}a^{\frac{4}{3}}b^{\frac{1}{3}}c^{\frac{4}{3}}\plus{}a^{\frac{1}{3}}b^{\frac{4}{3}}c^{\frac{4}{3}}}{3}\equal{}abc$

$ \implies a\equal{}b\equal{}c$

Let $a, b, c$ to be the sides of the parallelepiped. We have that $abc = 216 (I)$ and that $2(ab + bc + ca) = 216 (II)$.

From $(I)$ and $(II)$ we get $\sum\dfrac{1}{a} = \dfrac{1}{2}$. Squaring both sides, we get $2\sum \dfrac{1}{ab} = \dfrac{1}{4} - \sum \dfrac{1}{a^2}$ $(*)$.

By Cauchy, $2\sum \dfrac{1}{ab} \ge \dfrac{9}{54} \Leftrightarrow \dfrac{1}{4} - \sum \dfrac{1}{a^2} \ge \dfrac{9}{54}$ $\Leftrightarrow \dfrac{9}{108} \ge \sum \dfrac{1}{a^2} \ge \sum \dfrac{1}{ab} \ge \dfrac{9}{108}$.

So we get equality in AM-GM, what implies $a=b=c$.

I think this is the easiest way to solve the system :
$$abc=216$$
$$ab+bc+ca=108$$

Apply Newton's inequality and C.B.S. one time to get:

$$108^2=(ab+bc+ca)^2 \ge 3abc(a+b+c) \ge 3abc \sqrt{3(ab+bc+ca)} =3 \cdot 216 \cdot 3 \cdot 6 = 54 \cdot 216 = 108^2$$

Hence we have equality everywhere and so $a=b=c$,                  $$\odot$$


Three people are standing in the plane at points (0,0), (3,0), and (0,3) respectively. Only one person can move at a time, and that person can move any distance in a line parallel to the line joining the other two people.

Can the three people end up at the points (-1,1), (-1,3) and (2,5)?

Need they end up there in that order?  Or could any one of them be in each of those positions?  (It might matter.)

By the way, where does this question come from?  I know I've seen it before, but I'm not sure where.

i havent worked on it yet, but do they have to move in order?  person 1 moves, then 2, then 3, then 1, etc. ?

They can move in any order, and the people aren't distinct :D so yeah, they don't need to end up there in any particular order.

Hmm, I remember seeing this with them having to end up there in order, but I might be imagining it.  You didn't say where you found the problem, though.

I found it in the 'anything' forum of a non-math forum. Weird, that.

could you please not answer this question? it is from this year's mathcamp qualifying quiz. nr. 5 i believe.

Oh, so that's where I saw it.  You might go tell the people on the non-math forum not to help cheaters :).

Oh ok, sorry. But I had also seen this problem a few years ago, so its not original or anything. Somebody can lock this topic then.

oh and by the way, I doubt anyone was cheating since the guy who posted it wasn't american, and it was posted as a challenge (we're onto the 27th page where the person who solves it posts anotehr puzzle :D)


analytic geometry
real analysis
real analysis unsolved

hi,
i'm french and this is my first message, so be gentle ;)
okay... so i have a surface and i can calculate each point's mean curvature, and each directions for each principal curvature.
i want to move each point so that mean curvature becomes the same everywhere(a specific value i fix).
so, consider only one point.... i want to move that point so that the curvature of the surface at this point is a given real(that i choose)
How can i do ... do u know equations that could give me the new point ?
thanks

Mike

Sorry, I can't understand it.

Probably, the moderators should move it to College Playground.

Sorry, perhaps i didn't explain good enough my problem (i'm not a super english speaker)...
my surface is a NURBS surface.....
i want to move a surface point in order to give this surfacepoint a given mean curvature(which is the half of the sum of the principal curvatures). so i need to find the new coordinates for the point.
(and after i will need to find the coordinates of the control points to give the pointOn SUrface the wanted coordinates.)

Thanks

Mike

You're posting in a forum that is for intermediate high school students.  The problem you are posting is one of differential geometry.  Try looking in the College Playground section of this site for a forum that is more appropriate for this question where people will be able to answer it.


Calculate: $ \frac{\frac34\plus{}(\frac56 \div \frac23)}{\frac34\minus{}(\frac56 \div \frac23)}$


MATHCOUNTS
email
geometry
geometric transformation
reflection
symmetry
Pythagorean Theorem

Can you please help me in this one, I am more confused on this???

Georgia lives three miles from the bank of a straight river. Her friend Caleigh lives two miles from the river, on the same side, but 12 miles upstream. Georgia must get some water from the river and then go to Caleigh’s house. If Georgia jogs from her house to the river bank and then to Caleigh’s house at a rate of five miles per hour, what is the least amount of time required? Express your answer as a mixed number.

This is my way of doing it. It is really hard for me to explain without drawing a picture(and the pic is quite simple :P ), so if you can't quite understand it, I'm very very sorry.

[hide]Ok, so if you are looking for the least amount of time required, the distances she travels to go to the river and to her friend's house must be the same. If we draw the path out, with the river horizontally across, along with 2 perpendicular lines that show the distance from the 2 houses to the river, we get 2 right triangles with their bases along the river.

Some terms I will use...
Georgia's triangle=The "triangle" shape that appears, marked by her path to the river, the distance along river from the point at which she turns, to the line perpendicular to the river through her house, and the line perpendicular to the river through her house.
Caleigh's triangle=same idea, with Caleigh's side of the river.
____Side of the river= The side containing ____'s triangle.

Let's assume that the base of Georgia's triangle is $ x$. This means the base of Caleigh's triangle is $ 12 \minus{} x$.

Now we can apply Pythagorean theorem to the situation! Our "hypotenuse" is the distance she travels to the river, or the distance from the river to Caleigh's house. Since they are the same, we can set the square of bases equal to each other. Georgia's triangle's hypotenuse is $ 3^2 \plus{} x^2$, while Caleigh's triangle's hypotenuse is $ 2^2 \plus{} (12 \minus{} x)^2$.

So, $ 3^2 \plus{} x^2 \equal{} 2^2 \plus{} (12 \minus{} x)^2$. We will rearrange to make $ 24x \equal{} 135$ and $ x \equal{} \frac {45}{8}$, which is Georgia's base.

This means the distance of the "hypotenuse", or the path from her house to the river, is $ \sqrt {3^2 \plus{} (\frac {45}{8})^2} \equal{} 6.375$.

However, she has to travel twice that distance, cause after going to the river, she goes the same distance to her friend's house. $ 6.375*2 \equal{} 12.75$

Now it's a simple $ d \equal{} rt$ problem. $ 12.75/5 \equal{} 51/20 \equal{} \boxed{2\frac {11}{20}}$

I think that's the correct answer. I know the solution looks confusing, with all the triangles and such, but if only I could draw a picture.... :P   [/hide]

btw, do you have an email? I can email you the picture that'll help explain everything...if I can draw it.

oops, taht's not right. $ x\equal{}139/24$ I messed up in my solving...hypotenuse is about 9 miles so her time would be around...$ 18/5\equal{}\boxed{3\frac{3}{5}}$...I think.

[hide="The intended solution"]
To minimize the amount of time, we need to minimize the distance Georgia needs to walk.
Let Georgia's house be point $ A$, let Caleigh's house be point $ B$, let the point where Georgia stops at the river be point $ P$, and let the river be line $ m$.
Consider the shortest route from point $ A$ to $ P$ to $ B$.
We can reflect point $ B$ across line $ m$ to point $ B'$.
By symmetry, $ AP\plus{}PB \equal{} AP\plus{}PB'$, however $ AP\plus{}PB'$ is minimized when $ A, P, B'$ are all collinear.
Then $ AP\plus{}PB' \equal{} AB' \equal{} \sqrt{12^2\plus{}5^2} \equal{} 13$.
The least amount of time required is $ \frac{13}{5} \equal{} \boxed{2\frac{3}{5}}$.[/hide]

Ok, first she runs to the river. That's $ 3$ miles. Then she runs to Caleighs house which is $ \sqrt{4\plus{}144}$ away from that point on the river.

She runs $ 3\plus{}2\sqrt{37}$ miles at 5 mph, so the answer is $ \frac{3\plus{}2\sqrt{37}}{5}$.


The problem is really nice.
I have almost got a solution. But I am stuck somewhere


inequalities AMC USA(J)MO USAMO geometry calculus derivative

So I've been told by a number of people that research math, by and large, doesn't require the ingenuity that hard contest problems requre. Is there any merit to this statement? Or do many research problems require more ingenuity than hard contest problems?

Thanks,
-Arachnotron

Both definitely require ingenuity, each is just a type of "mathematical problem".  The only difference is that most research problems are too difficult, require too much technical knowledge, or take too much time to put on a contest.  Research questions are generally much more open ended, like "Is it possible to do something like this?" or "What interesting facts are there here?"  Research is largely about asking the right questions, and determining what you actually want to prove/what's interesting.  Hence it takes more time: it's more open ended.

Many olympiad problems are motivated by research, so you can't necessarily separate the two completely.  Just remember that most research problems are boiled down to something more concrete and specific, something more like "prove this" rather than "what is it possible to prove" if they're going to be made into olympiad problems.

Another difference I should mention is that contest problems tend to have the property that figuring out their motivation can be a big help towards solving them, since it gives you an idea of what the intended solution is.  On the other hand, for research problems you will ideally already know the motivation and there is no intended solution.  This makes contest math more of a puzzle or psychological game on occasion.

[quote="blackbelt14253"]I'm not sure how much ingenuity is required in research math, but I can tell you that research math is definitely a lot more bashy than olympiad math.[/quote]
You say that like it's a bad thing!  I'm sure you're thinking of examples like Lagrange multipliers.  My response is that research mathematics is concerned with general principles, and Lagrange multipliers are a great general principle, as are the underlying principles of convexity behind inequalities like Muirhead and Karamata.  Personally, I'm a big fan of generality.  Proving inequalities using AM-GM and Cauchy alone is an interesting recreational activity, but it's not serious mathematics.  (Proving something about the set of [b]all[/b] inequalities that can be proven using AM-GM and Cauchy alone, on the other hand, is an interesting research question.)

Point to ponder: do you think that any fields medallist would be able to score a 42 on IMO? Is there a disconnect between professional math achievement and olympiad strength?

I would think that many would not be used to the time constraints, but then again, there are people who have done well on IMO like Perelman (perfect score) and Tao, that have gone on to become fields medalists.

Well I mean without the time limits: do you think they could solve all of the problems?

There are certainly Fields Medalists who can do well on IMO/Putnam type stuff. There are also ones who probably couldn't solve one of the problems in a day ("no time limit" is a funny concept). In my limited experience, I'd say that the skills required to perform well in olympiad type math are very different from those required to be a good researcher. Many people excel at both and many excel at one but not the other. This is because research is about asking the right questions and making the right definitions, whereas contest problems are usually about finding some clever trick. To their credit, these tricks often seem to be analogues of techniques in research (c.f. last year's USAMO#6), except possibly in geometry.

So basically IMO/Putnam stuff are more like puzzles that try to test quick, insightful thinking, but research is more about trying to discover mathematical truths and adding to the sum of knowledge (or in the case of applied math, using it to solve "real world" problems)?

The skills required for both types of activity are complementary; it's useful for a mathematician to have both.  It can help to know some high-level stuff to see the motivation behind certain problems (remember that contest problems tend to be [b]written[/b] by research mathematicians!), and conversely if in the middle of your research you need to prove a technical lemma it's useful to have the training to be able to do this quickly.

Please don't try to separate the two types of activities using short slogans.  Math is math.

math is math, but solving olympiad problems under timed constraints is significanly different than doing mathematical research. The differences are worth pointing out for several reasons. One reason is that there are quite a few young people, most of whom all they know about mathematics is from contets, who mistakenly view Math Olympiads as a measurement of their ability to do mathematical research.

There are some great math researchers who were good at Mathematical Olympiads - there are many more who were not.

The set of successful math researchers (however you define successful) have a diverse and broad array of talents and abilities. Some are good problem solvers, some are good "synthesizers", some are good expositors, some are good at making essential insights...

I think its important for students to realize that it is great to train at Olympiad mathematics because it strengthens your overall problem solving and strategy development skills -- which can be useful in many areas, not just mathematics. But that the skills for successful in mathematical research aren't completely derivative of those skills for used in math competitions.

Hm, I found this stunningly relevant speech by Andrew Wiles from IMO 2001:

http://www.claymath.org/Popular_Lectures/Andrew_Wiles/

[quote] What then distinguishes the mathematics we professional mathematicians do from the mathematical problems you have faced in the last week? The two principal differences I believe are of scale and novelty. [/quote]

Of course, keep in mind that many great mathematicians aren't necessarily good at "contest" math.  With serious math, they can be the greatest in the world, but with contest math some can't.

Example: John Nash was able to prove Fermat's Little Theorem (out of [u]Men of Mathematics[/u]) with virtually no number theory background when he was around 16 years old-which is, to say, a great accomplishment.  He never did well on the Putnam Competition, however, although in his first year of graduate school at Princeton (after graduating from Carnegie Mellon in 3 years) came up with his Equilibrium.

I just don't understand [i]why[/i] that is though? I mean, if serious math consists of unsolved problems and take months/years to solve, then why do people who can do that have trouble with problems engineered to be solved in 90 minutes by high school students? I could understand them not being able to perform well on contests because of time constraints, but would they really have trouble just working on those problems as take-home problem sets?

[quote="Arachnotron"]would they really have trouble just working on those problems as take-home problem sets?[/quote]
Personally, I think the time constraint is everything.  Another issue is training: strong Olympians are very familiar with common proof techniques in Olympiad problems, whereas serious mathematicians may be more specialized in an esoteric subject and unfamiliar with elementary but useful results outside of their subject.  And yet a third is that some serious mathematicians think in such abstract terms that they may simply be very rusty at solving concrete problems.

Here's an example of what I mean.  Alexandre Grothendieck, one of the greatest "theory-builders" of the modern age, was once asked by a colleague to provide an example of some statement he was making about primes.  He said, "all right - take $ 57$."

Regardless of how good serious mathematicians are at [i]solving[/i] difficult Olympiad problems, I think a serious mathematician is much more likely to [i]generalize[/i] difficult problems and [i]relate[/i] them to other mathematical areas, as well as pick out the [i]underlying concepts[/i] more readily.

Speaking to anyone in general (t0rajir0u?) with experience, did you find strictly olympiad competition math more fun, or is it the research math is overall more enthralling? Can this olympiad math be viewed as a concrete basis for the extended topics in a career?

One aspect could be perhaps that as a career, a mathematician can work in the field that they find most interesting. For olympians, they have to be adept in many different fields, though certainly not to the same extent.

I always find it amusing that John Nash is used in an example of someone who did poorly on the Putnam.  I think it's fair to say that he only did poorly by his own standards.

Heh, John Nash didn't get top 5, but he was like in top 10 or something like that. But I too would be interested in seeing an answer to veezbo's question.

[quote="veezbo"]Can this olympiad math be viewed as a concrete basis for the extended topics in a career?[/quote]

For the most part no, but anything is possible. Some Olympiad problems come from ideas that were found in the course of doing mathematical research. Sometimes the problems are just one part of a much larger problem, or are simplified or boiled down to concrete instances of some more general principle. Sometimes they are made harder by leaving out some (known) bits of information. But for the most part, the type of math found in olympiad problems is not modern enough to be a basis for a career in math.

Of course, if a problem can be generalized or extended into a new direction - maybe it could be the basis for interesting research. But it is unlikely.

[quote="veezbo"]did you find strictly olympiad competition math more fun, or is it the research math is overall more enthralling?[/quote]
My favorite Olympiad problems are the ones related to the ideas in "serious" mathematics I find the most interesting.  So I would say that I hate taking Olympiads but that I like dissecting them afterwards.

For me, it's strange...I agree with t0r in that the problems which generalize and have connections to larger math concepts are among the most interesting, but there's also something about very elementary, but general, powerful ideas which come up in olympiad problems (in particular, in some olympiad combinatorics) which appeals to me.  Of course, my favorite part of olympiads is, as t0r says, dissecting the problems afterwards.  In fact, I think most people who did olympiads can agree that that's the best part: the nice math problems.

One difference...

IMO problems are known in advance to have solutions that can be written in a reasonable amount of space.  Research problems are not.

The three volumes on the Putnam Competition list the Putnam Fellows for every year, together with their current positions if known.  Most ended up at major research universities.

Has anyone noticed that problems in graph theory/combinatorics/number theory tend to resemble olympiad problems? I've looked at the open problem garden and problems in those fields really resemble olympiad problems in their statements...

Like I said, Olympiad problems are written by mathematicians; it's not surprising that the questions are superficially similar.  What's surprising is that questions can sound similar in their wording and be at incredibly different levels of difficulty.

Indeed. t0r, if mathematicians write these questions, how come most textbooks (even grad level) don't come with problems that require that level of thought? I think it'd be great for education in general if people were exposed to harder problems. There's no reason to only allow a handful of students to see these "gems" of problems...

Once I started doing AMC/AIME and eventually USAMO stuff, I could *feel* myself getting smarter -- and it's ridiculously satisfying!

The simplest answer is that it's really hard to come up with good problems.  The next simplest answer is that there are, in fact, textbooks with really hard problems in them; I'm thinking in particular of the exercises in [url=http://math.mit.edu/~rstan/ec/]EC1 and 2[/url], some of which are open problems.

The long answer is that Olympiad problems and textbook exercises have very different goals, so it doesn't make much sense to compare the two.  The goal of an Olympiad problem is to challenge test-takers as fairly as possible; the problem should not be well-known or have a trivial solution if you happen to know a particular theorem.  It should be solvable by cleverness alone, but it also shouldn't hinge on only a single observation; that is, it should require significant effort.  The goal of a textbook exercise, on the other hand, is to familiarize yourself with the material you've just learned and apply it.

If all you want is a book of problems, then those exist, too.

[quote="Arachnotron"]So if the majority of textbook exercises are designed to test understanding, how do (future) mathematicians work on developing problem solving skills needed to tackle significant open problems? Are olympiads/contests the only way or are there other avenues to significant creative mathematical thought?[/quote]

Problem solving skills aren't necessarily the most important skill needed for tackling open problems. Most importantly, you need to have a good knowledge of the history of work done on various problems so you're not rehashing ground that's already been travelled. You also need a deep (almost philosophical) understanding of the connection the problem has to other areas of mathematics (or physics, or biology, or anything that can point to a new line of attack). This foundation is laid in undergrad, but it's mainly cultivated in graduate school. Going to talks by other researchers, reading up on current developments in a field, and starting to prove things on your own is how you build up the skills necessary to do original research.

What a strange question to ask.  The AMCs have been around for maybe fifty years.  How long have we been doing mathematics?

This may be a really stupid question but...isn't math all about solving problems? Even if you're building theories, isn't that still solving problems?

EDIT: Also, how do the olympiads fit into the context of research math if undergrad builds a foundation and graduate builds research/problem solving skills? Is it just meant for talent identification and fun?

[quote="Arachnotron"]This may be a really stupid question but...isn't math all about solving problems? Even if you're building theories, isn't that still solving problems?[/quote]
Good question.  A standard answer is this: there are (very roughly speaking) [url=http://www.dpmms.cam.ac.uk/~wtg10/2cultures.pdf]two types[/url] of mathematicians:  those who think that the goal of building theories is to solve problems, and those who think that the goal of solving problems is to build theories.  Which one you see as the end result is a matter of preference.

[quote="Arachnotron"]EDIT: Also, how do the olympiads fit into the context of research math if undergrad builds a foundation and graduate builds research/problem solving skills? Is it just meant for talent identification and fun?[/quote]
I think Olympiads do a great deal to help students discover interesting mathematics to which they would not otherwise have been exposed; in other words, I think it helps to create mathematicians more than simply identify them.  Again, as Kedlaya put it, if you discover that you prefer to solve a problem in an hour than a minute, perhaps you also prefer to solve a problem in a year than an hour.

For me, the main difference is "exploration" vs "reaching the goal." (reminds me of that xkcd...)

Olympiad problems usually involve some kind of cute trick, combined with some computation. The goal is well-defined. But when I do Olympiad problems, I sometimes begin to wander off, looking at some other interesting facets of the structure being described in the problem. Then, I have to tell myself to snap out of it and actually solve the problem at hand.

In research, though (and I can say I don't have that much experience with this, just RSI), one has the freedom (and opportunity) to bumble around and run into new ideas, new ways of doing things. (Incidentally, this is how I tend to do olympiad problems - I screw around until I get something I know. I'm pretty sure most of the better problem-solvers go through some set of techniques first [ok, maybe I have a few?] and these are often sufficient.) I get to explore a subject, and try to understand it fully, or even completely. I am not zeroing in on a certain aspect; I try to build a broad base of knowledge, including not only the subject matter but related areas.

Another difference is that of "knowledge." Olympiad problems are generally "elementary." I feel comfortable here; it's not TOO abstract, TOO difficult for me to understand. Research, however, (excepting, somewhat, combinatorics) [wow that's a lot of commas] tends to require (perhaps not quite require, but close) knowledge of more advanced, abstract fields. And then I get owned. But this "abstractness" is a matter of mathematical maturity (which I don't yet have, judging from my completely futile attempts at learning higher math), and problem-solving is a valuable method of gaining this maturity. Just don't get too hung-up on Olympiads - they're not the end of the road!

Hmm, so what is the problem solving process for a hard research problem that does take a year? What is most of that time spent doing?

[quote="Arachnotron"]...What is most of that time spent doing?[/quote]

Reading, talking to people, experimenting, and (if necessary) doing examples to help you understand general principles (which does often involve problem solving skills).

And then trying to prove things that you think might be true.

[quote="Arachnotron"]So I'm curious: what makes unsolved problems like the Hadwiger conjecture (graph theoretical one or combinatorial geometry) or Riemann so...tough to crack? Especially for people who have no trouble solving contest stuff? Is it because nobody's had the "bright idea" to tackle it or are there paths that people can sort of see but the technology (theory?) isn't quite there to really solve them?[/quote]

Most likely, it is because there is a lot more math that can be applied to different problems so the ones that are harder to solve take longer because there are so many different paths.

Thus, talking to others for ideas and playing around is essential to finding out a solution.

Look, you really have to keep in mind that contest problems are written with a reasonable solution in mind.  The search space of possible solutions is therefore extremely small.  There's absolutely [b]no comparison[/b] to a legitimate open problem where nobody knows where the solution might come from.

For example, a problem in combinatorics might be solved by methods in topology.  This might seem farfetched, except that [url=http://en.wikipedia.org/wiki/Necklace_splitting_problem]it's already happened[/url].  And of course Wiles' proof of Fermat's Last Theorem is another great example:  the problem's been around for 350 years but the tools Wiles used didn't exist until a few decades ago (if that).

For particular cases, I recommend Terence Tao's explanation of [url=http://terrytao.wordpress.com/2007/03/18/why-global-regularity-for-navier-stokes-is-hard/]why Navier-Stokes is hard[/url] as a great example.  There are also, for example, several clearly-identified reasons why P=NP is hard; a good reference is [url=http://www.scottaaronson.com/papers/algstoc.pdf]this paper of Scott Aaronson's[/url].

Yeah, I'll probably stop posting on this thread considering I'm just a 10th grader who has nearly no math knowledge and I am not incredibly talented in it either.

Hmm, I guess (wait, what grade are you in again?), if you're aiming to be good at competitive math, practice for that.

If you're aiming to have a good time, idk. Do whatever you want. I'm sure if you do some searching, you can get yourself associated with some form of research if you want to try that out (provided you have a pretty strong mathematical background).

Well I'm a first year college student. I did pretty well in the AMC series (didn't quite make USAMO though), so I was inspired to actually learn problem solving, which made me want to become a mathematician. I'm asking primarily so that I know what to expect in the transition from the contest problems I do for fun and the real world.

If you want to know how people solve problems in research mathematics, and how to gain the skills to do this, an excellent reference is Terrence Tao's blog. You can also view the polymath blogs and wikis to see, among others, some of the world's top mathematicians (Tao, Gowers, Graham, etc.) trying to solve open problems.

Here are the relevant links:

Acquiring mathematical skills:
http://terrytao.wordpress.com/career-advice/

In particular,
http://terrytao.wordpress.com/career-advice/solving-mathematical-problems/
http://terrytao.wordpress.com/career-advice/does-one-have-to-be-a-genius-to-do-maths/

Polymath:
http://gowers.wordpress.com/2009/01/27/is-massively-collaborative-mathematics-possible/
http://michaelnielsen.org/polymath1/index.php?title=Main_Page

with a recent development discussed somewhat at
http://terrytao.wordpress.com/2009/08/08/polymath4-finding-primes-now-officially-active/

with roots at
http://polymathprojects.org/2009/07/28/deterministic-way-to-find-primes-discussion-thread/

and a wiki at
http://michaelnielsen.org/polymath1/index.php?title=Finding_primes

EDIT: Also, while you can get much more informed opinions at the links I gave above, my personal opinion is that Olympiad math gives you good training for technical skills related to mathematical research. However, as Qiaochu has noted, technical skills are by no means the most important part of research mathematics.

This is definitely a simplification of my thoughts on Olympiad math, but I'll stop here since I mostly agree with what Terry Tao and Qiaochu have to say (also because I am still not entirely sure what I think about Olympiad math).

[quote="JSteinhardt"]technical skills are by no means the most important part of research mathematics.[/quote]
Right.  Some mathematicians seem to be very good at guessing what [i]should[/i] be true based on a very high-level and heuristic view of their subject, and they leave it to other mathematicians to actually prove these things.  A great example is the [url=http://en.wikipedia.org/wiki/Borsuk%E2%80%93Ulam_theorem]Borsuk-Ulam theorem[/url], where the story goes that Ulam simply intuited the statement of the theorem in a flash and it took Borsuk all of his technical expertise to actually prove it.  (I'm not sure how much I believe this story, but given what else I've heard about Ulam it's within the realm of possibility.)

So I e-mailed Dr. Vandervelde (author of 2009 USAMO #3) about this topic and here's his response:

My question:
[quote] Now my question is: do many research problems require the same (or greater) ingenuity and problem solving ability that a difficult USAMO problem (like #3) requires? [/quote]

Response:
[quote]The short answer to your question is yes and no.  Yes, in the sense that a lot of research projects require much more time and mental energy to develop and thoroughly explore.  But no, "research" math is not necessarily harder than something like #3 on the USAMO.  (Which is one of the more difficult problems that I've worked on in the past year or two!)[/quote]

Are contest problems sometimes results/lemmas from research papers?


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There didn't really seem to be an appropriate place to put this, so I just decided to post it here.

Suppose [b]ω=N={0,1,2,3,...}[/b] is the natural numbers ordered by [b]ε[/b].

then what is [b](ω+1)*ω[/b] in normal form?

My answer: [b](ω+1)*ω=sup{(ω+1)*α|α<ω}=(ω+1)+(ω+1)+(ω+1)+(ω+1)+...=ω+(1+ω)+(1+ω)+(1+ω)+...=ω+ω+ω+ω+...=sup{ω*α|α<ω}=ω²[/b]

Is this correct?

nvm I guess I am correct according to this thread http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1396363779&t=191817 for nearly the same reason.. cool... but was my answer rigorous enough?


geometry
vector
3D geometry
sphere
inequalities
puzzles

I will start by posting two "paradox" math problems or math problems with unbelievable answers.  If you can think of a reasonable explanation of the paradox, post it and give another paradox.  This should work so that there are two problems to be solved at all times.  When you can explain one of the two given paradoxes, copy the remaining one into your post so that it is easier to see what problems have not been yet solved.  Do not answer your own paradox.  If you think a paradox is literally [i]impossible[/i], you may put up a new post as if it has been solved and say that the paradox is impossible.  And one more thing: have fun! :D

Paradox # 1:
What is the area of a circle with radius i?  How is this possible?  How would you draw this?

Paradox # 2:
If [i]z[/i] is one divided by zero, what is 0 times [i]z[/i]?  Is it one or is it indeterminate?  (It is certainly not undefined.)

[hide="1"] $i$ is not an acceptable value of a metric, which must be a non-negative real. [/hide]

[quote="nateharman1234"]for number 2 I say it's zero,  1/0 is undetermined, but anything times zero is zero so it doesnt matter if it is undetermined[/quote]

$z$ is undefined, therefore $0 \times z$ is also undefined.

$z=\frac{1}{0}$

The flaw comes in saying this.  Equality is something we only can assign to certain mathematical objects, ie numbers, vectors, matrices, etc.  This mathematical object does not fall under any of thoe categories, so we can't define equailty for it.

[quote]I say for #1 we use the complex plane so it would be a normal circle with r = 1.[/quote]

I do not think we can use the complex plane because if we let's say... calculate the radius vertically, the answer would of course be $i$... but if we calculate it horizontally, the radius would be $1$

We know that for any circle or sphere all radii must be equal... so there we have a contradiction since $1 \not=i$

I'm not sure if this could be a valid proof so guys fell free to correct it.............

Radius is defined as the [b]distance[/b] between the center of the circle and the points that make it up.

$\text{ Distance is a non-negative real, even in the complex plane.}$

I'm moving this to the Puzzles & Brainteaseers forum.

If we imagine we can create a "minusland" or a geometrical area where no positive values are allowed (i.e., a square with area 4 has side length -2), we can place the i circle into minusland.  Thus the area of the circle is negative pi.

We know we must be in minusland because of the negative area, but only negative values and 0 can be drawn in minusland; therefore i is less than or equal to 0.  However, -i can also be drawn in minusland, making -i less than or equal to zero.  Multiply both sides of that inequality by -1 and we know that not only is i less than or equal to 0, it is also greater than or equal to 0.  If you see my conclusion, i = 0. :o

Where is my incorrect step? :?

[quote="miyomiyo"][b]If[/b] we imagine we can create a "minusland" or a geometrical area where no positive values are allowed[/quote]

This is a very bad "if."  There is no guarantee that such an axiomatic system is self-consistent.  This would be your incorrect step.

Edit:  The other problem is that you then apply "plusland" logic to minusland.  "Greater than" and "less than" are concepts that [b]do not exist with complex numbers. [/b]

Edit:  This whole discussion is nonsense.  You're arbitrarily changing axioms so that questions with no answers suddenly have answers.  That isn't math.

The definition of the circle is:

[quote="MathWorld"]the set of points in a plane that are equidistant from a given point O.[/quote]

Therefore, the area of any circle is 0. :rotfl:

But, seriously, this whole thing does not make sense. These things are not even paradoxes, more like pseudomath.

[quote="miyomiyo"]The paradox contest was supposed to continue with people adding more paradoxes.  Here's another:

What is infinity to the zeroth power?

And here's a logic paradox for all logic people ;) :
This statement is false.[/quote]

$\infty^{0}$ is an indeterminate form.

The second one is the first actual example of a paradox.  It has no resolution.  Look up Godel's Incompleteness Theorem.

Well, most here is nonsense and "pseudomath" as called above, but when you define $x^2+y^2=r^2$ to be a circle with radius $r^2$ in the plane, then setting $r=i$ and (this is important) working in $\mathbb{C}^2$, a circle with radius $i^2=-1$ makes totally sense: the set is nontrivial then (and I wouldn't draw it as long as I can't influence four-dimensional spaces in total freedom ;) ). Only problem: the hypervolume would not be finite, and you loose some other properties.
Such things are (well: very simple) parts of algebraic geometry.


I think the following is true:
The following colours are defined: red, green, blue, yellow, cyan, magenta, black, yellow. But you can extend this a long way by using the RGB system which is the way you mix the primary colours Red, Green, Blue.
The syntax is:

[b]\color[rgb]{x,y,z} where ${ 0 \leq x,y,z \leq 1}$ and each represents the proportion of red, green and blue in the colour.[/b]

Thus
\color[rgb]{1,0,0} is $ \color[rgb]{1,0,0}{red}$
\color[rgb]{0,1,0} is $ \color[rgb]{0,1,0}{green}$
\color[rgb]{0,0,1} is $ \color[rgb]{0,0,1}{blue}$
and then you mix them so
\color[rgb]{0.5,0.5,0.5} is $ \color[rgb]{0.5,0.5,0.5}{grey}$
\color[rgb]{0.82,0.41,0.12} is $ \color[rgb]{0.82,0.41,0.12}{chocolate}$

Then for fun:   $ \text{\Large\color[rgb]{ 0.2000 , 0.3922 , 0.7647}G \color[rgb]{ 0.9529 , 0.0980 , 0.0118}o \color[rgb]{ 0.9686 , 0.8431 , 0.1529}o \color[rgb]{ 0.2000 , 0.3922 , 0.7647}g \color[rgb]{ 0.2667 , 0.7686 , 0.0235}l \color[rgb]{ 0.9529 , 0.0980 , 0.0118}e}$
(taken from [url=http://www.stanford.edu/~dgleich/notebook/2007/12/a_fun_latex_command.html]David Gleich's Notebook[/url])

You will need to experiment. You can use [url=http://web.njit.edu/~kevin/rgb.txt.html]RGB to Color Name Mapping[/url] to help, [i]provided[/i] you use their numbers as a proportion of 255. Thus on that page [i]chocolate[/i] is 210,105,30. Change that to $ \frac {210}{255},\frac {105}{255},\frac {30}{255} \approx 0.82,0.41,0.12$ to get $ {\color[rgb]{0.82,0.41,0.12}\int_{0}^{1}\frac {x^{4}\left( 1 - x\right) ^{4}}{1 + x^{2}}\,dx} = \frac {22}{7} - \pi$

You can apply colours to selected parts of the expression by
1) using {} as in {\color{red}x}y $ {\color{red}x}y$
[i]or[/i]
2) using a new colour \color{red}x\color{black}y  $ \color{red}x\color{black}y$
[i]or[/i]
3) using \normalcolor \color{red}x\normalcolor y $ \color{red}x\normalcolor y$

[b][color=red]PLEASE only use LaTeX colours for maths - for text like this, use the forum's font colour drop-down box[/color][/b]

(I have moved this thread into the LaTeX forum as others may want to refer to it).

[quote="stevem"][img]http://www.math.harvard.edu/computing/latex/color_files/colf8.gif[/img]
Source: [url=http://www.math.harvard.edu/computing/latex/color.html#SECTION001119000000000000000000]LaTeXe's Colour Extensions: Shades of colours[/url][/quote]This is very complicated. I cannot get it.

You need LaTex when entering math.

$\color[rgb]{0.2835,0.746754,0.92573}a^2+b^2=c^2$

If you test it:
[hide="[color=#004000]hi[/color]"]BLAH[/hide]
The underline is still blue.

[quote="Letti"][quote="yetti"]Testing $ \text{\LaTeX}$ colors:

[code]$\color{red} \int_a^b f(x)\ \text dx = \color{green} F(b) - F(a)$[/code]

$ \color{red} \int_a^b f(x)\ \text dx = \color{green} F(b) - F(a)$
$ \color{blue} \int_a^b f(x)\ \text dx = \color{cyan} F(b) - F(a)$
$ \color{magenta} \int_a^b f(x)\ \text dx = \color{yellow} F(b) - F(a)$

Basic colors (red, green, blue, cyan, magenta, yellow) work fine, but I am unable to guess names of other colors. It would be nice to know names of basic colors with less luminosity (dark red, dark green, dark blue, ...). Can dark colors be used for $ \text{\LaTeX}$ on Mathlinks and if yes, what are their names ?

Thank you.[/quote]also, you do not have to use [color=orange]LaTeX[/color][/quote]
That is bump number one!
[quote="Subul"]Can colors be used in Hidden Text? When you want to hide something, can the color be something else rather than blue?[/quote]
Bump number two!
Stop please guys!
Just make a revival thread titled Latex colors revival.
This should also go in the Latex test forum

$ \color{red} Red? $
$ \color{green} Green? $
$ \color{blue} Blue? $
$ \color{blue} Purple? $
$ \color{blue} Gray? $
$ \color{blue} Chocolate? $
$ \color{yellow} Yellow? $
$ \color{blue} Orange? $
$ \color{blue} Pink? $
$ \color{blue} Dark red? $
$ \color{blue} Dark blue? $
$ \color{blue} Dark green? $

Does every color work?

$\color[rgb]{1,0.3333,0}\text{\LaTeX}$

Use the color slider to make any color imaginable.

[color=#408080]You don't need to use $ L^AT_EX.$ You can just use the thing that says font color right next to the font size.[/color]

Not if what you want to do is use colors in LaTeX  :roll:

$\color[rgb]{1,0.3333,0}\text{\LaTeX}$

\color[rgb]{r,g,b}, r, g, b being amounts of red green blue respectively, ranges from 0 to 1.

[quote="Kent Merryfield"]$ \color[rgb]{.8,.75,0}\text{Yellow you can see?}$

$ \color[rgb]{1,0,0}r\color[rgb]{.9,0,0}e\color[rgb]{.8,0,0}d\ \color[rgb]{.7,0,0}t\color[rgb]{.6,0,0}o\ \color[rgb]{.5,0,0}d\color[rgb]{.4,0,0}a\color[rgb]{.3,0,0}r\color[rgb]{.2,0,0}k\color[rgb]{.1,0,0} \minus{}$

$ \color[rgb]{0,0,1}b\color[rgb]{0,0,.9}l\color[rgb]{0,0,.8}u\color[rgb]{0,0,.7}e\ \color[rgb]{0,0,.6}t\color[rgb]{0,0,.5}o\ \color[rgb]{0,0,.4}d\color[rgb]{0,0,.3}a\color[rgb]{0,0,.2}r\color[rgb]{0,0,.1}k$[/quote]
it's very interesting for me.

[quote="Subul"]If nothing goes right, go left.[/quote]
How far right is all right? I have little problems with a german, who has rassistic ideas. So far as I can see, there is nothing of his ideas, what he turns in practice. He is a son of a professor. And he is a very unpractic man. He is already somewhat elder and he does not find a woman as "friend". This he takes as occasion for his drifting away from the "Golden Mean". Perhaps in the deep of his heart he is much more left than right. Do I can have hope, that he might become better, when I tell him:

"Wenn dir alles nicht recht ist, dann muss es dir links sein." oder: "Wenn nichts recht klappen will, dann versuch´s doch lieber links." or: "The heart in the body beats on the left side." / "Das Herz im Körper, das schlägt auf der linken Seite."?

The described man often says: "Das führt doch zu nichts." (This though leads to nothing.)

Black is a so-called uncolor like the gray of the cats by night. -  And brown?

I tried your "ThisColorIsAwesome!" and wanted to compile it to be able to see it in better quality as pdf, but I have got the error-message:[code]! Package color Error: Argument `4' not in range [0,1].[/code]

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1541609#p1541609]post 2[/url] in this thread.

Thanks for the reply. The rgb-concept was already clear for me.
My question was more explicitely formulated: Why does this code here work?[code]$ \color[rgb]{4,10.9,5}{ThisColorIsAwesome!} $[/code]$ \color[rgb]{4,10.9,5}{ThisColorIsAwesome!} $

Probably because the code is changed before the image is produced.

When the code is changed, by whom? And why? This code like I have cited it works, but only here on AoPS. I have tried it myself.

The code that you type is parsed by the AoPS system before it is compiled as a TeX document. For example, \displaystyle is added to the beginning of the formula even when using dollar signs. A check is also made to ensure that there are an even number of dollar signs. There are other changes that are made, many of which I've forgotten about as well as others that I don't know about (some of these may be for security reasons).
It's trivial to use PHP code to change \color[rgb]{4,10.9,5} to something else that works, but I'm not privy to what it is.

Testing $\text{\LaTeX}$:

$\color[rgb]{0.9, 0.1, 0.97}{Pink}$
$\color[rgb]{5, 1, 2}{Light Green}$

$ \color[rgb]{66.99,77.99,99.99}{ThisColorIsAwesome!}\[ \color[rgb]{66.99,77.99,99.99}{ThisColorIsAwesome!}\] \color[rgb]{66.99,77.99,99.99}{ThisColorIsAwesome!}$ $ \color[rgb]{66.99,77.99,99.99}{ThisColorIsAwesome!}$

oops... I think I did something wrong...

$ \color[rgb]{66.99,77.99,99.99}{ThisColorIsAwesome!}$    $ \color[rgb]{66.99,77.99,99.99}{ThisColorIsAwesome!}$    $ \color[rgb]{66.99,77.99,99.99}{ThisColorIsAwesome!}$

[quote="PythonNut"]You guys really should test this in the [url=http://www.artofproblemsolving.com/Forum/viewforum.php?f=224]Test Foum[/url]. (not here)[/quote]

$ \color[rgb]{1,10.1,99.00001}{using intense limits plz quote if you want to see those intense limits}$


[b]Does anyone know who was the youngest candidate this year? and how old?[/b]

I know there was a very young contestant in the Irish team (Galin Ganchev is his name, I think, he's 14 or something), and maybe some Japanese contestant is still younger, but I don't really know.

They're discussioning about that [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=45310]here[/url]. :)


Λοιπόν προτείνω να συζητήσουμε λίγο για την ΙΜΟ 2006 .Τι προβλήματα λέτε να μπουν ??Εύκολα δύσκολα ??Ποια θα είναι η δυσκολία τους σε σχέση με πέρσυ ??Πού θα είναι τα medal cutoffs??Και άλλα πολλά  :D
Λοιπόν πιστεύω ότι θα μπουν δύο γεωμετρίες κλασσικά 1ο και 5ο θέμα.......
Μία θεωρία αριθμών για 2ο θέμα . Για τέταρτο δεν έχω ιδεα....Για έκτο συνδιαστικό πρόβλημα ή ......συναρτησιακή εξίσωση!!!!!!!(έχει χρόνια να μπει ;) ) .Για ανισότητα δεν ξέρω .......Και επίσης πολυώνυμο έχει να μπει καιρό ;)
Για τη δυσκολία πιστεύω οτι πολλοί θα επιζητήσουν δύσκολα........αλλά θα δείξει
Περιμένω απόψεις..........

σε ποια χωρα θα γινει η ΔΜΟ?

[quote="kostas"]σε ποια χωρα θα γινει η ΔΜΟ?[/quote]
Σλοβενία ρε Κώστα ;)

ναι το ειδα μετα.παντως εξαρταται και απο τη δυσκολια της περυσινης.πως σας φανηκαν τα περισινα? :)

Ενταξει δεν ηταν και απο τις πιο δυσκολες.Εχω και μια προαισθηση οτι ερχεται το χρυσο!Για να δουμε,υπομονη.Αληθεια ξερει κανεις πια χρονια ειχαμε παει καλυτερα?

Λοιπόν την καλύτερη θέση την πήραμε στην πρώτη μας εμφάνηση το 1975 .
Οι χώρες ήταν 17 και πήραμε την 15η θέση .Τη χειρότερη την πήραμε το 2000 . Πήραμε την 56η θέση σε 82 χώρες . Πιστεύω όμως ότι η καλύτερη εμφάνισή μας έγινε το 2004 εδώ στην Αθήνα .Πήραμε την 26η θέση σε 85 χώρες..........
Τότε τα πράγματα είχαν ως εξής
Αχιλλέας Πορφυριάδης     Εύφημη μνεία
Στέργιος Αντωνακούδης  6       5      2      7     4       0  σύνολο    24   ασημένιο
Στέφανος Αρετάκης          7      2      0      7      5      0    σύνολο   21  χάλκινο
Φάνης Παπαμιχάλης          7      3      0      7      3      0     σύνολο   20  χαλκινο
Μάριος Παπαμιχάλης         7      2      0      7      7      0     σύνολο   23   χάλκινο
Μηνάς  Παγωνάκης           7      7      0      7      6      0     σύνολο   27   ασημένιο

Την πρώτη θέση το 2004 μοιράστηκαν οι
1      CAN       Jacob Tsimerman  42   Gold Medal
1      HUN       Bela Andras Racz  42   Gold Medal
1      RUS       Andrey Badzyan  42   Gold Medal
1      RUS       Mikhail Dubashinskiy  42   Gold Medal

Όσο για πέρσυ κάποιος μπορούσε να πάρει μετάλλιο μέχρι και με 8 !!!!!!!!!
Όμως κάποιος για να πάρει χρυσό έπρεπε να γράψει τουλάχιστον 35 !!!!!!!

νομιζω οτι την μεγαλυτερη επιτυχια την ειχαμε το 1996! πολυ καλα και το 2003. το λεω γιατι το 2004 τα θεματα ηταν(για ΔΜΟ παντα) ευκολα σχετικα.

απο μεταλλια? θα παρει κανεις χρυσο? ποσα αργυρα και χαλκινα βλεπετε? :first:

ποτε φευγει η αποστολη της Ελλαδας?

λοιπον, βλεπω οτι η συζητηση εχει φουντωση κανονικα :D so θα ηθελα να ευχηθω καλη επιτυχια στα παιδια που θα πανε ΔΜΟ.  ΚΑΛΗ ΕΠΙΤΥΧΙΑ ΠΑΙΔΙΑ !!!!!!! :thumbup:

άκουσα ότι φεύγουν αυτή τη Δευτέρα...
καλή επιτυχία σε όλους αλλα και ιδίως σε αυτούς που δε θα έχουν δεύτερη ευκαιρία

Καλη επιτυχια και απο εμενα σε ολους τους διαγωνιζομενους!

 :)

Καλη επιτυχια παιδια :) Σας ευχομαι αυτη η ολυμπιαδα να σας μινει αξεχαστη με τις καλυτερες εντυπωσεις.Χρυσο  για την Ελλαδαρα μας. :wink:

Χαιρετώ όλους τους Έλληνες mathlinkers !
 Πώς τα πήγε η Εθνική μας ομάδα ?  Γνωρίζει κανείς ?

   Ω.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=574686#574686[/url]

Προτελευταία λίστα στο post.

Οι παύλες ( - ) σημαίνουν ότι δεν έχει διορθωθεί το συγκεκριμένο πρόβλημα στο συγκεκριμένο γραπτό.

EDIT: Για να μην τρέχετε:

[quote]
Greece
GRC1 0 1 0 2 0 0    3
GRC2 7 0 1 7 0 0    15
GRC3 7 1 1 7 0 0    16
GRC4 7 1 0 4 0 0    12
GRC5 7 1 0 6 0 0    14
GRC6 7 1 0 1 0 0    9
[/quote]

Συγχαρητήρια σε όλους και σε εσένα cmad για το μετάλλιο στην πληροφορική  :)
EDIT :
Gr 1 -> Nikos Bountas
Gr 2 ->Ralis Karamichalis
Gr 3 ->Kostas Psychas
Gr 4 ->Nick Rapanos
Gr 5-> Nikos Romanidis
Gr 6->Eleni Stauropoulou

EDIT : Από ότι φαίνεται θα έχουμε τέσσερα χάλκινα και μία εύφημη μνεία   :lol:

Πώς τα βλέπεις cmad τα πράγματα θα πάρουμε 4 μετάλλια ?
 Που έκτιμάς τη βάση , μιας και είσαι ειδικός σε αυτά τα θέματα .

 Ω.

Εγώ πιστεύω οριακά και θα το πάρει το μετάλλιο....... :lol:

:(  :(  :( .Δυστυχώς τα πράγματα πήγαν λίγο χειρότερα......
Τελικός απολογισμός δύο χάλκινα (Ειδικά  ο Ράλλης πήρε το μετάλλιο με τον ένα πόντο από την ανισότητα  :) ) και τρεις ευφημες μνείες .(Να πω την αλήθεια περίμενα ότι το medal cut-off για το χάκινο θα ήταν 12 σε αυτή την περίπτωση θα πέρναμε τέσσερα χάλκινα) .Αλλά δεν έχει σημασία συγχαρητήρια σε όλους ..... :lol:

ΑΝΑΝΕΩΣΗ ΤΟΥ ΠΟΣΤ :
Καταταγήκαμε συνολικά στην 56η θέση ανάμεσα σε 90 χώρες με 69 βαθμούς .
Πρώτη η Κίνα με 6!!!!!!! χρυσά δεύτερη η Ρωσία , τρίτη η Κορέα , τέταρτη η Γερμανία , 5η η αμερική , έκτη η Ρουμανία , η Κύπρος 81η και τελευταία η Μοζαμβίκη με συνολικό σκορ  0     :lol:

Συγχαρητηρια σε ολα τα παιδια. :)

Γεια!!!!
Και μια ανασκοπηση για την ΙΜΟ.....
Οντως θα μπορουσαμε να ειχαμε γυρισει με περισσοτερα μεταλλια η με λιγοτερα....αλλα τελικα γυρισαμε μονο με 2 και στην 56η θεση.....
Η αληθεια ειναι οτι η κουραση του ταξιδιου επηρεαζει σημαντικα την αποδοση των διαγωνιζομενων (υπαρχουν βεβαια και οι <ντοπαρισμενοι> κινεζοι) :)
ερχεται και η παραξενη βαθμολογηση  αν ξεφυγει κανεις εξω απο το grating scheme,και μερικες φορες συμβαινουν λιγο παραξενα πραγματα.....
Για παραδειγμα ο GRC5 παρολο που γυρισε με honorable mention θα μπορουσε να γυρισει μεχρι και με αργυρο μεταλλιο και σιγουρα πρωτος απο την ελληνικη ομαδα.Επισης ο GRC1 θα μπορουσε να παρει mention απο το 4ο θεμα αλλα τελικα πηρε μονο 2 βαθμους μονο και μονο γιατι δεν εξετασε την περιπτωση x<0 ,ενω αν το ιδιο γραπτο διορθωνοτανε απο ελληνες εξεταστες(π.χ στον αρχιμηδη)θα μπορουσε να παρει μεχρι και 6 ποντους.Επισης η GRC6 ενω τα εγραφε ωραια :)  στο  4ο θεμα  πηρε μονο μια μοναδα...Τελος ο GRC4 απο βλακεια του εχασε το 7 στο 4ο θεμα και κατα συνεπεια το χαλκινο μεταλλιο,ενω ο GRC2 και GRC3 δεν πηρανε ουτε 1 μοναδα απο το 6ο θεμα παρολο που αποδειξανε αρκετες ειδικες περιπτωσεις ,αλλα δυστηχως με 1 βαθμο ανταμοιβοταν καποιος μονο για την αποδειξη του ζητουμενου στην περιπτωση του πολυγωνου το οποιο εχει κεντρο συμμετριας και σε καμια αλλη περιπτωση... :(
Αυτα.... για να ξερουν ολοι χοντρικα πως προεκυψαν οι τελικες βαθμολογιες....

ΡΑΛΛΗΣ :D

(ΔΕΝ ΑΡΕΣΑΝ ΣΕ ΚΑΝΕΝΑΝ ΤΑ ΘΕΜΑΤΑ ΤΗΣ Ε.Μ.Ο=ΕΞΩΓΗΙΝΗ ΜΑΘΗΜΑΤΙΚΗ ΟΛΥΜΠΙΑΔΑ= ΕΧΤRATERRESTRIAL MATHEMATICAL OLYMPIAD???? ΠΕΡΙΜΕΝΩ ΑΠΑΝΤΗΣΕΙΣ ΓΙΑ ΟΛΑ ΤΑ ΘΕΜΑΤΑ....)

Ο φίλος μας ο Ράλλης πάντα υπερβολικός όσο πατάει η γάτα για την ντόπα των κινέζων Ράλλη σε χαιρετώ κι συγχαρητήρια για το μετάλλιο σου πάντα να μας τιμάς ως Ελληνες


A particle of mass m moves from rest on a rough horizontal table under an attraction to a fixed point of the table proportional to its distance from the point.Show

(a) there are two positions of limiting equilibrium on the line of motion

(b )that if the initial displacement is large enough the particle describes half-cycles of simple harmonic motion about these two positions alternately

(c) that the particle completes a whole number of half-cycles before coming to rest.

Hm, what is a position of limiting equilibrium? Thank you :).

Message for bus

The particle is in a position of limiting equilibrium when it is instantaneously at rest, that is the velocity is zero and its direction is changing.
In this question static and kinetic coefficients of friction are assumed to be the same usually denoted by the  letter of the Greek alphabet mu (I cant type Greek letters using the standard keyboard).THerefore you can assume in this question that F=(mu)N at alll stages of the motion.

Example suppose a particle is placed on a rough inclined plane and that the force normal to the plane is N and the friction force acting up the plane is F Then

  F/N<(mu) when the particle is in equilibrium
  F/N=(mu) when the particle is in limiting equilibrium ,that is the particle is on the point of slipping down the plane
  F/N=(mu) when the particle is in motion down the plane.

I hope this answers your question.Good luck in solving it !

Yes, that helped, thank you. By the way, you can put greek letters and all kinds of math formulas in your posts using LaTeX tags - try typing $\mu$ .

We know that in SHM we have two limiting equillibrium positions. Do we mean to prove the same in this problem?

Do you want a mathematical answer, or an intuitive one?

Because the force of attraction is proportional to the distance, you can model it with Hooke's law, and therefor a particle connected to a spring(with unstretched position at the point of no attraction)

And then you see that the sum of forces is $kx - friction$

and... yeah.  It all follows.

Answer to Varun; Yes,you must prove the existence of two  points on the line of motion where speed is zero.Hint :The square of the speed is a quadratic function of distance from the point of attraction.

Answer to Daveed : I was hoping for a full mathematical answer!


For every positive integer $k$ let $a(k)$ be the largest integer such that $2^{a(k)}$ divides $k$. For every positive integer $n$, determine $a(1)+a(2)+a(3)+...+a(2^{n})$.

[hide="solution"]
We are trying to find $\sum_{i=1}^{2^{n}}{a(i)}$.
Find a first estimate by counting all even terms as "1": total value=$2^{n-1}$
Now add a 1 for every fourth term, to make up for the undercount: we add another $2^{n-2}$.
Now you keep correcting, and you will get: $\sum_{i=0}^{n-1}{2^{i}}=\boxed{2^{n}-1}$.
And that's it...
[/hide]


geometry
function
3D geometry
sphere
geometric transformation
invariant
analytic geometry

Here's a problem I like a lot (despite the fact that some people will probably solve it immediately) :).

Let ${\bf S}$ be the set of all polygonal surfaces in the plane (a polygonal surface is the interior together with the boundary of a non-self-intersecting polygon; the polygons do not have to be convex). Show that we can find a function $f:{\bf S}\to (0,1)$ s.t. if $S_1,S_2,S_1\cup S_2\in{\bf S}$ and the interiors of $S_1,S_2$ are disjoint, then $f(S_1\cup S_2)=f(S_1)+f(S_2)$.

It's even assumed their interiors are disjoint. I forgot to add that, sorry. As the title suggests, this function behaves exactly like the area function.

I remembered this problem today, and had to solve it again, since my initial proof was wrong :). Anyway, I thought it was a good chance to bump it up. If it still doesn’t look appealing to anybody, I guess we’ll just have to let it die :).

Just pull it to a sphere by stereographic projection, then project that back on to the plane using some projection with bounded area. Now let the function be the ordinary (measure) area function on the plane in the resulting image, scaled to total measure 1. [Edit: the Mercator projection has unbounded image, so it doesn't work]

Any translation-invariant area function must of course be unbounded.

Here's a completely different example: Put coordinates on the plane. Enumerate the rational points $P_n$. Give the $n$-th point weight $2^{-n}$. Then define $f(S)=\sum_{P_n\in S}2^{-n}($angle around $P_n$ contained in $S$/$2\pi$)

Any continuous $f(x,y) > 0$ with $\int f = 1$ will do, such as $f = g(x)g(y)$ with $\int g(t) dt = 1$.

Or, take area measure on disc (or square, triangle etc) and apply a homeomorphism of this finite figure with the entire plane.

[quote="jmerry"]Just pull it to a sphere by stereographic projection, then project that back on to the plane using some projection with bounded area. Now let the function be the ordinary (measure) area function on the plane in the resulting image, scaled to total measure 1. [Edit: the Mercator projection has unbounded image, so it doesn't work] [/quote]

It seems to work --- if I understand correctly, you mean to assign to any polygon in the plane the surface area of its stereographic projection onto a sphere of area 1.  The example with rational points is very nice, by the way.


limit
calculus
integration
Euler
calculus computations

It looks simple at the first sight, but...
Determine the following limit:
$lim_{n\to \infty} e^{-n} \sum_{i=0}^n \frac{n^i}{i!}$

Let $S=\sum_{i=0}^{n}\frac{n^i}{i!}$, $T=\sum_{i=n+1}^{\infty}\frac{n^i}{i!}$, we have $S+T=e^n$, further more, it' easy to prove that $S-T=e^n*o\left(1\right)$, thus $\lim_{n\to+\infty}e^{-n}S=\frac{1}{2}$. In fact, $\lim_{n\to+\infty}\sqrt{n}\left(e^{-n}S-\frac{1}{2}\right)=\frac{2}{3\sqrt{2\pi}}$

[quote="Buffalo"] In fact, $\lim_{n\to+\infty}\sqrt{n}\left(e^{-n}S-\frac{1}{2}\right)=\frac{2}{3\sqrt{2\pi}}$[/quote]

How did you manage for the last one?
Can you detail it please!

Write  $S(n)=\sum_{i=0}^{n}\frac{n^i}{i!}$ with integral and use Laplace method
for an equivalent

To go to the next order, Euler-Maclaurin formula $\sum_{i=1}^{n}f\left(i\right)=\int_{1}^{n}f\left(x\right)dx+\frac{1}{2}\left(f\left(n\right)-f\left(1\right)\right)+\sum_{k=1}^{\infty}\frac{B_{2k}}{\left(2k\right)!}\left(f^{(2k-1)}\left(n\right)-f^{(2k-1)}\left(1\right)\right)$ is needed. The integral mainly comes from the "tiny" region that x is "close" to n.


Is it true that a numerable intersection of compact connected sets is connected? What if the sets are decreasing for inclusion?

In the first case the answer is 'no' even if there are two sets (consider two closed arcs of a circle intersecting at both ends), and in the second case it's 'yes', and it's been discussed before.


calculus
integration
logarithms
calculus computations

How to find the integral

$\int x^2\arctan x \,dx$    :oops:

Integration by part gives us.
\begin{eqnarray*}\int x^2 \arctan x \, dx &=& \frac{1}{3} \int \arctan x \, d(x^3) \\ &=& \frac{1}{3} \left( x^3 \arctan x - \int \frac{x^3}{x^2+1} \, dx \right) \end{eqnarray*}

To calculate $\int \frac{x^3}{x^2+1} \, dx$.
]
\begin{eqnarray*}\int \frac{x^3}{x^2+1} \, dx &=& \int x - \frac{x}{x^2+1} \, dx \\ &=& \frac{x^2}{2}+C_1 - \frac{1}{2} \int \frac{1}{x^2+1} \, d(x^2) \\ &=& \frac{x^2}{2} - \frac{1}{2} \ln (x^2+1) + C \end{eqnarray*}

So $\int \frac{x^3}{x^2+1} \, dx = \frac{x^2}{2} - \frac{1}{2} \ln (x^2+1) + C$.
and \[ \int x^2 \arctan x \, dx = \frac{1}{3} \left( x^3 \arctan x - \frac{x^2}{2} + \frac{1}{2} \ln (x^2+1) + C \right) \]


$k$ is a non-negative integer and the integers $a_{1}, a_{2}, ... , a_{n}$ give at least $2k$ different remainders on division by $n+k$. Prove that among the $a_{i}$ there are some whose sum is divisible by $n+k$.

Perhaps I am wrong, but I think it works with $k+1$ instead of $2k$ (and even $k$ when $k \geq 2$ and $n+k$ is odd). Suppose the conclusion is false. WLOG $a_{1},a_{2},...,a_{k+1}$ give different remainders; $S$ is their sum. Take the sequence $a_{1},a_{2},...,a_{k+1},S-a_{1},S-a_{2},...,S-a_{k+1},S,S+a_{k+2},S+a_{k+2}+a_{k+3},...$. There are $n+k+2$ numbers in the sequence. $a_{j}\equiv S-a_{j}(\bmod n+k)$ occurs at most twice. Therefore, since none of these numbers are multiples of $n+k$, there must be two whose difference is a multiple of $n+k$ (and which are not $a_{j},S-a_{j}$). Their difference is, anyway, also a "sum", impossible.


limit
logarithms
integration
real analysis
real analysis unsolved

Ok, I know I'm about to say something stupid, but I don't get this... :(

We have $\frac 1{\sqrt[n]k}>\int^{k+1}_k x^{-\frac 1n}dx$. Apply this to $k=1,2,\ldots,n$ and sum everything up to get $\frac 1{\sqrt[n]1}+\frac 1{\sqrt[n]2}+\ldots+\frac 1{\sqrt[n]n}>\int^{n+1}_1 x^{-\frac 1n}dx=\frac{x^{1-\frac 1n}}{1-\frac 1n}\big|^{n+1}_1$. This last expression has value $\frac{(n+1)^{1-\frac 1n}-1}{1-\frac 1n}$.

That is equivalent to $n$, but we must show that a sequence larger than this is equivalent to $\frac n{\ln n}$. :?

We need to look at that last expression more closely:

$(n+1)^{1-\frac1n}=(n+1)\cdot e^{-\frac{\ln (n+1)}{n}}$
$=\left(n+1\right)\left(1-\frac{\ln (n+1)}{n}+o(\frac1n)\right)=n+1-\ln (n+1)+o(1)$

The whole expression is then $\frac{n-\ln(n+1)+o(1)}{1-\frac1n}=n-\ln n+1+o(1)$

This isn't $\frac{n}{\ln n}$, so grobber seems to be right.

For an estimate on the other side, $\frac1{\sqrt[n]{k}} < \int_{k-1}^k x^{\frac1n}\, dx$, so
$\frac 1{\sqrt[n]1}+\frac 1{\sqrt[n]2}+\cdots+\frac 1{\sqrt[n]n}<1+\int^{n}_1 x^{-\frac 1n}dx=1+\frac{x^{1-\frac 1n}}{1-\frac 1n}\big|^n_1$
This last expression becomes $1+\frac{n^{1-\frac1n}-1}{1-\frac1n}$, which is $1+n-\ln n +o(1)$ as $n\to \infty$

The result we actually get from these two limits is that $\lim_{n\to \infty}\left(\frac 1{\sqrt[n]1}+\frac 1{\sqrt[n]2}+\cdots+\frac 1{\sqrt[n]n}-n+\ln n\right)=1$

The original problem was wrong.

Actually, the fact that $\frac 1{\sqrt[n]1}+\ldots+\frac 1{\sqrt[n]n}$ is equivalent to $1$ is pretty clear intuitively, since when $n$ becomes large, all the terms of the sum are awfully close to $1$.

It depends on how close you're talking about. They are the same multiplicatively, but the additive difference from $n$ still goes to infinity.

I guess the original problem is right.but I can not tell the reason

Nene, could you please fix the problem ... or confirm that it's correct! I'm pretty sure the limit you are giving is not one - I think it's infinity.

Many thanks!
jivko777

OK, as Nene doesn't reply, I would like to give my reasons about the incorrectness of his/her problem:

$\frac{1}{\sqrt[n]{k}} = e^{-\frac{1}{n}\ln k}$, $k = 1, 2, \ldots, n$.

For $f(x) = e^{-x}$ we have the following Taylor expansion:

$f(x) = f(0) + x f'(0) + \frac{x^2}{2}f''(\xi) = 1 - x + \frac{x^2}{2} e^{-\xi},$ where $\xi \in [0, x]$.

Now for $x_{k,n} = \frac{1}{n} \ln k,\, k = 1, 2, \ldots, n$

$f(x_{k,n}) = 1 - \frac{1}{n} \ln k +\frac{1}{2} \left(\frac{\ln k}{n}\right)^2 e^{- \xi_{k,n}} ,$

where $\xi_{k,n} \in \left[0, \frac{\ln k}{n}\right]$ for $k = 1, 2, \ldots, n$. Therefore

$\sum_{k = 1}^{n} f(x_{k,n}) = \sum_{k=1}^{n} \left(1 - \frac{1}{n}\ln k\right) + \frac{1}{2}\sum_{k=1}^{n}\frac{\ln^2 k}{n^2} e^{- \xi_{k,n}}$.

Obviously

$0 \leq\sum_{k=1}^{n}\frac{\ln^2 k}{n^2} e^{- \xi_{k,n}} \leq \sum_{k=1}^{n}\frac{\ln^2 k}{n^2} \leq \frac{1}{n}\sum_{k=1}^n \frac{\ln^2 k}{k} \stackrel{n \rightarrow\infty}{\longrightarrow} 0$

since $\lim_{k\rightarrow \infty}\frac{\ln^2 k}{k} = 0$.

Therefore $\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{k=1}^{n}f(x_{k,n}) = \lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n} \left(1 - \frac{1}{n}\ln k\right) = \lim_{n\rightarrow\infty} \left(1 - \frac{1}{n^2}\ln(n!)\right) = 1$

since $\lim_{n\rightarrow\infty} \frac{1}{n^2}\ln(n!) = 0$.

Note: I think I could have considered $f(x) = f(0) + x f'(\xi), \, \xi\in [0,x]$ instead of using second order expansion and it seems to me that the same arguments work, but saw it few seconds ago and don't want to edit my message now  :).  Sorry for the inconvenience.

If you spot any mistakes, please let me know. Many thanks!

Cheers,
jivko777

Actually there is a slight advantage of considering second order Taylor expansion instead of only first and it is that now it's easy to show that

$\lim_{n\rightarrow\infty}\left(n - \sum_{k=1}^{n}f(x_{k,n})\right) = \lim_{n\rightarrow\infty}\left(n - \sum_{k=1}^{n}\left(1 - \frac{1}{n}\ln k\right)\right) \\\\
= \lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n}\ln k = \infty$.

But the same result could have been shown by only noting that

$0 = \frac{\ln 1}{n} \leq \frac{\ln 2}{n} \leq \ldots \leq \frac{ln n}{n} \stackrel{n\rightarrow\infty}{\longrightarrow} 0$

and therefore $e^{-\xi_{k,n}} > \frac{1}{2}$ (for example) for $n$ big enough... and thus we could have done the same stuff with expanding only up to first order ... Sorry!

So ... still no real advantages of using the second order expansion ...

jivko777

And finally, one can easily show that

$\lim_{n\rightarrow\infty}\frac{1}{\ln(n)}\left(n - \left(1 + \frac{1}{\sqrt[n]{2}}+\ldots+\frac{1}{\sqrt[n]{n}}\right)\right) = 1$

... by (again) considering only first order Taylor expansion.

jivko777

Alternately, you could have just read the posts by grobber and me. The final result I stated is even stronger than yours.

That's a good point, jmerry. Thanks!  :lol:

I guess after reading grobber's posting I went straight to the last one, by zhaobin, and skipped yours. Sorry! Therefore I left the topic with the impression that the question of correctness of Nene's problem is still unclear. Now I read your posting and liked your solution. Well done!

BTW you are quite right that your result is stronger than mine, but it's also true that I can get your result quite easily with my solution:

$\lim_{n\rightarrow\infty}\left(n - \sum_{k=1}^n \frac{1}{\sqrt[n]{k}} - \ln(n)\right) = \lim_{n\rightarrow\infty}\left(\frac{1}{n}\sum_{k=1}^n \ln(k) - \ln(n)\right) = \lim_{n\rightarrow\infty}\ln\left(\frac{\sqrt[n]{n!}}{n}\right) = 1.$

Cheers,
jivko777

Sorry. I have been out of town for more than three weeks and then I did not follow the discussion about the limit I proposed sometimes ago. I agree that the  limit is wrong in the formulation in which "N.Bourbaki" shows it for the reason that are been told.

[quote="Nené"]Finally I find the error in the limit with the help of "Bourbaki". The limit is:

\[\displaystyle \mathop {\lim }\limits_{n \to  + \infty } \frac{{\ln n}}{n}\left( {1 + \frac{1}{{\sqrt[n]{1!}}} + \frac{1}{{\sqrt[n]{2!}}} + ...\frac{1}{{\sqrt[n]{n!}}}} \right) = 1
\]
Infact : \[{ \frac{1}{{\sqrt[n]{1!}}} + \frac{1}{{\sqrt[n]{2!}}} + ...\frac{1}{{\sqrt[n]{n!}}}} \sim \frac n {\ln n}\qquad \textrm{if}\quad n\to +\infty.\][/quote]
are you sure?
but I find ${\frac{1}{{\sqrt[n]{1!}}} + \frac{1}{{\sqrt[n]{2!}}} + ...\frac{1}{{\sqrt[n]{n!}}}} \sim e \ln n \qquad \textrm{if}\quad n\to +\infty.$

Hint of "Bourbaki ": confront every term $ (p!)^{-1/n}$  with  $n^{-p/n}$

can you give a strict solution?
I still think I'm right :D  :D

[quote="zhaobin"]
... I find ${\frac{1}{{\sqrt[n]{1!}}} + \frac{1}{{\sqrt[n]{2!}}} + ...\frac{1}{{\sqrt[n]{n!}}}} \sim e \ln n \qquad \textrm{if}\quad n\to +\infty.$[/quote]

I also found  $e\ln n$

After careful cosideration (I hope!), I agree with Zhaobin and Moubi and think that" Bourbaki" was wrong.


What is the correlation between vapor pressure and molecular weight, and why?  Thanks!

For a start, you must specify in what family of compounds you want to study such a correlation, for the obvious reasons (I think).

well, ideally, it shouldn't matter (the ideal gas law PV=nRT that is pretty correct for reasonable pressures)

but the van der waals equation says

$ P\equal{}\frac{nRT}{V\minus{}nb}\minus{}\frac{n^2a}{V^2}$

The two corrections from the ideal gas law are:

V-nb: correction for the actual volume in which the gas molecules are free to move

$ \frac{n^2a}{V^2}$ : since attractions between molecules will decrease the number of and speed of gas molecule-container wall collisions, (proportional to the number of atoms divided by the volume, squared, since this gives a measure of the "crowdedness" of the container)

a and b are constants that are different for each gas, with a being some constant and b being the actual volume of a mole of atoms.

So, my point in mentioning this is that as molecules tend to get bigger, V-nb tends to get smaller, and a tends to get bigger (by not-noticeable amounts at pressures like 1 atm)

So to answer your question, it depends on the magnitudes of a and b here, as a rule there shouldn't be an  important correlation unless you're dealing with very, very high pressures where the number of molecules is very very high