Let $ n$ denote the minimum number of squares that must be removed from a 4 by 4 checkerboard so that there does not remain a 1 by 3 configuration of squares. What is $ n$?

I think it's one. If you take one from one of the middle four squares... I'm not sure. I mean like you take one from the middle and then it's four, two, one, four, four. I mean by the rows. Then by columns, it's basically the same thing. So there is no there by one configuration.

[geogebra]5535a01293428801f0dcb28a119d8e0e6097b7b0[/geogebra]

The ones with the dots are the ones taken out. From that, we see that we must take out $ \boxed{5}$ squares.

Can't we just do this? Or do they mean like ANYWHERE there can't be a 1 by3 configuration? If that's the case then you can just do this... sorry if I reposted an answer...  :|

hm let's see...
you're picture has $ \boxed{12}$ 1 x 3 configurations ;)

Can't we do

 [geogebra]c54adb2f5919a10ead7a605793353848e5480642[/geogebra]

First we do the square (a2,2)--(a2,4)--(a4,2)--(a4,4) and then take (a1--3) and (a3--1). That's 3 squares? Or can't we even just take the whole board?

Well if we can do it YOUR way, then we can just take one square- the whole entire board. I doesn't state that they have to be a certain size either...

[quote="BOGTRO"][..] Or can't we even just take the whole board? [..][/quote]

I said that. So I guess the answer would be 1?


what are the four quantum numbers for iodine?

I'm not sure what you mean. Quantum numbers, assuming we are talking about the same thing, are used to specify where electrons are in an atom (i.e. $ n, l ,m_l,$ and of course spin).

That's right. As stated, that question does not make any sense. Remember that quantum numbers are a product of quantum theory, and are used to describe orbital's energy, form, and spatial orientation, and electrons. Let's recall briefly here those concepts.

According to quantum theory, an orbital is a region around the nucleous of an atom where there is a high probability of finding one or two electrons. How are orbitals described? By three numbers that we call "quantum numbers". The principal quantum number, [i]n[/i], takes on natural values (1,2,3, ...) and is related with the orbital's energy. The bigger [i]n[/i] is, more energetic will be the orbital. Observe that by "orbital's energy" we really mean "the energy of an electron when it "lies" on that orbital". What energy is this? Well, an electron is negatively charged and the nucleous is positively charged, so there is a potential energy "between" them. As you know, this electric potential energy is inversely proportional to the distance between charges, so an orbital further apart from nucleous will have an increasing energy. That's the significance of [i]n[/i]: the greater it is, bigger will be the energy of an electron in that orbital because of the greater distance from nucleous.
The second quantum number (or angular momentum quantum number), [i]l[/i], describes the tridimensional form of the orbital, and for a given [i]n[/i] the only possible values of [i]l[/i] are 0, 1, 2, ..., n-1. Each value of [i]l[/i] corresponds to a given type of orbitals: orbitals with [i]l[/i] = 0 are named s orbitals, and have spherical simmetry; orbitals with [i]l[/i]= 1 are the p orbitals; those with [i]l[/i] = 2 are the d orbitals, and those with [i]l[/i] = 3 are the f orbitals.
The third quantum number (or magnetic quantum number), $ m_l$, describes the spatial orientation of orbitals. For a given [i]l[/i], the possible values of $ m_l$ are -l, -l+1, ..., 0, 1, ..., l-1, l.
Those three quantum numbers are needed to describe completely a given orbital. However, to describe a given electron on a given orbital, a fourth quantum number is needed: the spin quantum number $ m_s$, whose only possible values are $ \pm \frac12$. These are related with the fact that an electron behaves as if it was rotating about an axis that passes through its center (just like the Earth does) - it spins -, and there are only two possible directions of rotation.

Thanks for the explanation! I meant to ask about the valence electron of the highest energy in the ground-state Iodine atom.

Iodine is in block p, period 5, group 17, and the "previous" inert gas is Kr. Therefore, the electron configuration of iodine is $ [Kr]5s^25p^5$, with the 5p electrons being the most energetic. Therefore, a possible set of quantum numbers for one of such electrons is $ \left(5,1,\minus{}1, \frac12\right)$.

so there is no set rule for whether it is $ \pm \frac 12$? and for the $ m_l$ quantum number, does it go from left to right on the periodic table from positive to negative?

I don't think there is any fixed rule about $ m_s$ being $ \plus{}\dfrac{1}{2}$ or  $ \minus{}\dfrac{1}{2}$.
But I have read that an electron with $ m_s \equal{} \plus{}\dfrac{1}{2}$ (clockwise spin) has a lower energy, and thus greater stability than an electron with $ m_s \equal{} \minus{}\dfrac{1}{2}$ (counter-clockwise spin).

So I [i][b]think[/b][/i] the convention is to use $ \plus{}\dfrac{1}{2}$.

so, for iodine, according to stability, it should be $ \minus{}\frac 12$ right? since its not as stable as a full octet.

I am not sure, but i think it's $ \plus{} \dfrac{1}{2}$.

The atom as such is not as stable as a noble gas. But that doesn't say much abt the quantum no. of the electron.

Usually, an electron (or any particle for that matter) will always prefer to be in a state of lower energy(greater stability) rather than in a state of greater energy(lesser stability).

When filling the orbitals first (using Hund's rule) with electrons of parallel spins, i guess u need to use clockwise spins first ($ \plus{} \dfrac{1}{2}$). Only when u are trying to pair the spins(Pauli's exclusion principle), u need to use a counterclockwise spin ( $ \minus{} \dfrac{1}{2}$).

Since, in an Iodine atom, the last p-electron is a lone electron (unpaired), i think u should use  $ \plus{} \dfrac{1}{2}$. Atom's stablity as a whole ned not be considered, because that is essentially because of a lack of fully filled configuration and has nothing to do with the spin :maybe:

I might be wrong though

@ hurdler:

I am really sorry.   :oops_sign:

What I have been saying all this while is true only if the system is placed in an external magnetic field. Only then does the electron with $ m_s \equal{} \plus{}\dfrac{1}{2}$  (clockwise spin) have a lower energy than an electron with $ m_s \equal{} \minus{}\dfrac{1}{2}$(counter clockwise spin).  :oops:

I apologise for my mistake....


abstract algebra
group theory
superior algebra
superior algebra unsolved

Show that if G is a finitely generated group, then every quotient group of G is finitely generated.

Show that if $1 \rightarrow N \rightarrow G \rightarrow Q \rightarrow 1$ is a short exact sequence of groups, and if both N and Q are finitely generated (resp. finitely presented), then so is G.

The proof of the first statement is a one-liner: the images of the generators generate the quotient.

First I assume that the groups are abelian. This is because I borrow the definition of finite presentation from modules, and the $\mathbb{Z}$-modules
are exactly the abelian groups.
We have the following diagram in which the rows and columns are exact:
\[ \begin{array}{ccccccccc} \ &\ &\mathbb{Z}^{n_2}&\ &\ &\ &\mathbb{Z}^{q_2}&\ &\ \\ \ &\ &d_2\downarrow &\ &\ &\ &\downarrow\partial_2&\ &\ \\ \ &\ &\mathbb{Z}^{n_1}&\ &\ &\ &\mathbb{Z}^{q_1}&\ &\ \\ \ &\ &d_1\downarrow &\ &\ &\ &\downarrow\partial_1&\ &\ \\ 1&\to &N&\stackrel{i}{\to}&G&\stackrel{p}{\to}&Q&\to&1\\ \ &\ &\downarrow &\ &\ &\ &\downarrow &\ &\ \\ \ &\ &1 &\ &\ &\ &1&\ &\ \\ \end{array}  \]
This just comes from the definitions.
We upgrade it to:
\[ \begin{array}{ccccccccc} 1&\to&\mathbb{Z}^{n_2}&\stackrel{i_{n_2}}{\to} &\mathbb{Z}^{n_2}\oplus\mathbb{Z}^{q_2}&\stackrel{p_{q_2}}{\to} & \mathbb{Z}^{q_2}&\to&1 \\ \ &\ &d_2\downarrow &\searrow^ {i_{n_1}\circ d_2}&\downarrow\rho_2 &\swarrow f_2&\downarrow\partial_2&\ &\ \\ 1&\to&\mathbb{Z}^{n_1}&\stackrel{i_{n_1}}{\to} &\mathbb{Z}^{n_1}\oplus\mathbb{Z}^{q_1}&\stackrel{p_{q_1}}{\to} &\mathbb{Z}^{q_1}&\to&1 \\ \ &\ &d_1\downarrow &\searrow^ {i\circ d_1}&\downarrow\rho_1 &\swarrow f_1&\downarrow\partial_1&\ &\ \\ 1&\to &N&\stackrel{i}{\to}&G&\stackrel{p}{\to}&Q&\to&1\\ \ &\ &\downarrow &\ &\downarrow &\ &\downarrow &\ &\ \\ \ &\ &1 &\ &1 &\ &1&\ &\ \\ \end{array}  \]
This diagram is constructed as follows:
\par The maps on the first and second row are natural injections and projections. With these construction it is obvious that now all the rows are short
exact sequences. Notice that on the first two rows we also have reverse maps because of the direct sum property.
\par We use the following lemma to construct $f_1$ and $f_2$:
\par\textbf{Lemma:} Let $p: G\to Q$ be a surjective group homomorphism. Assume there exists $n\in\mathbb{N}$ and $\pi: \mathbb{Z}^n\to Q$ a surjective
group homomorphism. Then there exists a group homomorphism $\bar\pi: \mathbb{Z}^n\to G$ such that $p\circ\bar\pi=\pi$.
\par\textbf{Proof of lemma:} Just choose $\bar\pi(e_i)\in f^{-1}(\pi(e_i))\ \forall i=\overline{1,n}$, and extend by
linearity.\eop\\
\par We apply the universality property of the direct sum to find maps $\rho_1: \mathbb{Z}^{n_1}\oplus\mathbb{Z}^{q_1}\to G$ and
$\rho_2: \mathbb{Z}^{n_2}\oplus\mathbb{Z}^{q_2}\to\mathbb{Z}^{n_1}\oplus\mathbb{Z}^{q_1}$ such that every square in the diagram is commutative.
\par It follows from the snake lemma that $\rho_1$ is surjective.
\par From the commutativity of the diagram, $\rho_1\circ\rho_2=i\circ d_1\circ d_2\circ p_{n_2}=i\circ 1\circ p_{n_2}=1$.
\par Also from the snake lemma it follows that there is an exact sequence
$coker d_2\stackrel{\bar i}{\to} coker\rho_2\stackrel{\bar p}{\to} coker\partial_2\to 1$. By exactness of the first and third column,
$coker d_2=N$ and $coker\partial_2=Q$. The maps $\bar i$ and $\bar p$ are maps induced by $i_{n_1}$ and $p_{q_1}$ i.e. $\bar i(\hat a)= \widehat{i_{n_1}(a)}$ etc. The identifications $coker d_2=N$ and $coker\partial_2=Q$ allow us to define $\bar i: N\to coker\rho_2$ by
$\bar i(d_1(a))=\widehat{i_{n_1}(a)}\ \forall a\in\mathbb{Z}^{n_1}$ and $\bar p: coker\rho_2\to Q$ by
$\bar p(\hat b)=\partial_1(p_{q_1}(b))\ \forall b\in\mathbb{Z}^{n_1}\oplus\mathbb{Z}^{q_1}$.
\par Let $A=\mathbb{Z}^{n_1}\oplus\mathbb{Z}^{q_1}$. We have a surjection $\rho_1: A\to G$ and a canonical projection that I will call
$\alpha: A\to coker\rho_2$. It is clear that $\ker\alpha=Im\rho_2$. $\rho_1\circ\rho_2=1\Rightarrow \ker\rho_1\supseteq Im\rho_2$. By the universality
property of the factor group, there is a group homomorphism $\beta: coker\rho_2\to G$. The snake lemma again proves that $\beta$ is an isomorphism.
This proves $G\simeq A/Im\rho_2$. But $\rho_1$ is surjective so $A/\ker\rho_1\simeq G$. This and the inclusion $Im\rho_2\subseteq\ker\rho_1$ prove
$Im\rho_2=\ker\rho_1$ i.e. the exactness of the middle column.
-----------------------------------------------------------------------------------------------------------------------

The non-abelian case is a bit unclear to me.
It works very much the same up to a point by replacing $\mathbb{Z}^n$ by the free group $\mathbb F_n$, and the direct sum of abelian groups by the free product of groups.
I'm not sure that the definition $G$ is finitely presented if and only if there exists an exact sequence $\mathbb F^n\to \mathbb F^m\to G\to 1$ is equivalent to $G$ is a given by a finite set of generators and relations, but it seens plausible.

The problem that appears is the existence of $coker$ in the snake lemma.


Is there some specific equation or formula that could find the equation of the Euler Line of a triangle, given all coordinates each vertex on the triangle?


Let $ 0\leq a,b,c\leq \frac{\pi}{2}$ be such that

$ \sin a+\sin b+\sin c=1$ and $ \sin a\cos 2a+\sin b\cos 2b+\sin c\cos 2c=-1$.

Find all possible values of $ \sin^{2}a+\sin^{2}b+\sin^{2}c$.

I denote $ x=sina$ $ y=sinb$ $ z=sinc$
Using the formula $ cos(2\alpha)=1-2sin^{2}(\alpha)$ the second relation is equivalent to:
$ \sum x(1-2x^{2})=-1$ $ \Leftrightarrow$ $ \sum x-2\sum x^{3}=-1$ $ \Leftrightarrow$ $ \sum x^{3}=1$ ,because $ \sum x=1$ so we have the system:
$ x+y+z=1$
$ x^{3}+y^{3}+z^{3}=1$

From $ (x+y+z)^{3}=x^{3}+y^{3}+z^{3}+3(x+y)(y+z)(x+z)$ $ \Rightarrow$ $ y+z=0$ $ \Rightarrow$ $ x=1$
Moreover because $ y\geq 0$ ,$ z\geq 0$ and $ y+z=0$ $ \Rightarrow$ $ y=z=0$ ,therefore $ x^{2}+y^{2}+z^{2}=1$


algebra polynomial function calculus derivative trigonometry algorithm

Hi everyone,

I'm having some difficulty grasping the need for determining Taylor's polynomials when expanding functions by the Taylor's series. I realise that the formula foir evaluating Taylor's polynomials can be extended to generate the Taylor's series for that particular function. In that case, why evaluate Taylor's polynomials ? What is their significance ?

The reasoning usually adopted to determine the polynomial involves determining a tangent to the curve f(x), where f(x) is your function (for which you require the Taylor's series). The tangent equation is said to provide the best "local approximation" of the required function. I don't see how this is so ? I scoured the internet for some information and I came across this statement "The fundamental idea in differential calculus is that a function can be 'locally' approximated by its tangent line." Now, this gets more and more confusing. Fundamental idea in differential calculus ????

I'd appreciate and be grateful for some help on this topic.

Thanks, Siddharth

Taylor polynomials are useful because they (some of them, at least) are useful in approximating other differentiable functions, up to some degree of accuracy.  For example,
$ e^{x}\equal{}1\plus{}x\plus{}\frac{x^{2}}{2!}\plus{}\cdots\plus{}\frac{x^{n}}{n!}\plus{}R_{n}$
where $ R_{n}$ is the remainder, or error term.  By taking derivatives over and over (assuming we can, there is some deep analysis behind the validity of differentiation), we can approximate a function more and more accurately.

Your calculator computes values of sines and cosines using Taylor polynomials.  Using [url=http://en.wikipedia.org/wiki/Taylor%27s_theorem]Taylor's theorem[/url], it is possible to compute exactly what degree of Taylor polynomial you need to achieve $ 13$-digit accuracy and then stop there.

More generally, studying the sequence of Taylor polynomials is a useful way to get some grip on their limit - the Taylor series itself.  We cannot really examine the Taylor series [i]except[/i] as a limit of the Taylor polynomials, and Taylor series are very useful when they have the nice properties we want them to have (i.e. convergence).

JRav and t0rajir0u have been speaking from a point of view that assumes that the function you're interested in is real analytic - that is, infinitely differentiable and given locally by its Taylor series. But what if the function you're interested in is not analytic, or you don't know that it is? What if all you know is that it has three derivatives, say, or four derivatives? Then the Taylor polynomial captures our knowledge.

The field of numerical analysis has many algorithms for approximating things: roots of functions, integrals, solutions of differential equations. Always, we want to know both whether the method converges and how fast it converges. And all of those estimates usually have some hypothesis of differentiability - $ C^1,C^2,C^4,\dots$ - and an estimate of the size of some derivative is contained in the error estimate. All of those error estimating theorems use either Taylor polynomials or interpolating polynomials.

If I want to discuss finding and classifying the extrema of a smooth function of several variables, I can best explain what is going on by looking at the second order Taylor polynomial about the point in question.

There are certain "bump" functions such that the derivatives always exist, but when you calculate the Taylor series, you actually get that the series is 0, despite the fact that the function is clearly not the 0-function.

Hmmm, I've got a better understanding of Taylor's series now. I do have a more fundamental question. In the exponential series, if I were to substitue 'jΘ' in place of 'x', by Euler's formula I would be able to get infinite series for the cos Θ and sin Θ terms respectively. An interesting fact over here is that all the even powers correspond to the cosine series, while the odd powers move to the sine series. I think these are Taylor's seriesfor the cosine and sine functions respectively (real-analytic functions) and I wanted to know how the interesting split of terms between the cosine and sine series was done.

I was reading up some litreature on SOS Math, when I came up with doubts on Taylor's polynomials.

Coming to my questions:
a) Am I right in assuming the two infinte series generated are Taylor's series ?
b) If so, could someone point as to why only even (or odd) powers of Θ remain in the series ?

P.S: My office has blocked out certain features of Mathlinks. So, I cannot put up better images of theta,etc. Bloody *****.[/code]

[quote="Siddharth Rao"]In the exponential series, if I were to substitue 'jΘ' in place of 'x', by Euler's formula I would be able to get infinite series for the cos Θ and sin Θ terms respectively. An interesting fact over here is that all the even powers correspond to the cosine series, while the odd powers move to the sine series. I think these are Taylor's seriesfor the cosine and sine functions respectively (real-analytic functions) and I wanted to know how the interesting split of terms between the cosine and sine series was done.[/quote]
You mean [url=http://en.wikipedia.org/wiki/Euler%27s_formula]Euler's formula[/url]?  That is one of several "proofs."  I prefer the interpretation in terms of the [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=216343]matrix exponential[/url].

The infinite series generated are indeed Taylor series, but you must justify the re-arrangement of terms in your series.  That is, you can't just move all of the even powers to one and side and the odd powers to the other side for any series.  The series must have a certain property (which is...?).

To your second question, what do you mean "only even (or odd) powers" remain?  Both even and odd powers of $ \theta$ remain in the series, or else you could not get both sin and cosine.

Just like with real numbers, you can think of a complex number as a point on a circle.  It's "length" is determined by the radius of the circle, and it's precise real and imaginary coordinates are determined by $ \theta$.  So any complex number $ z$ can be written $ z\equal{}r(\cos{\theta}\plus{}i\sin{\theta})$.

While many high school students glibly work with sin x, most mathematicians feel more comfortable having its Taylor series. As far as Taylor polynomials go, they are a never ending source of computational pain to college multivariable calculus students and allegedly useful (and extremely painful) when it comes to finding extrema of functions.

In short it is nice for theoretical interest, but, as with most other things you learn in regular math classes, mathematica can do the work for you without you ever knowing the nuts and bolts. I've given up trying to memorize the different forms of the remainder, for example; I've decided that's what reference books are for.

I still think Weierstrass' Approximation Theorems are pretty useful.  I would much rather work with a polynomial than some crazy function I can't fathom.

For theoretical manipulations of Taylor series, this characterization is nice:
Suppose $ f$ is $ n$ times differentiable at $ a$. The Taylor polynomial $ p$ of degree $ n$ is the unique polynomial of degree $ \le n$ satisfying $ \lim_{x\to a}\frac{f(x)\minus{}p(x)}{x^n}\equal{}0$.

This allows us to easily prove that a polynomial which is a good approximation is a Taylor polynomial, as might appear with $ e^{\minus{}x^2}\approx 1\minus{}x^2\plus{}\frac{x^4}{2}\plus{}\cdots$ and similar substitutions.

If you can actually calculate the derivatives easily, this isn't worth it. $ \sin$ and $ \cos$ are the absolute best cases for the error terms in Taylor's theorem.


inequalities
function
inequalities unsolved

If $ 0<x\leq 1$ and $ o<y\leq 1$ prove or disprove that:

$ {\rm{(1 \minus{} x)}}^{1 \minus{} y}  \plus{} (1 \minus{} y)^{1 \minus{} x}  \ge 2 \plus{} 2xy \minus{} 2x \minus{} 2y$

[quote="delegat"]If $ 0 < x\leq 1$ and $ o < y\leq 1$ prove or disprove that:

$ {\rm{(1 \minus{} x)}}^{1 \minus{} y} \plus{} (1 \minus{} y)^{1 \minus{} x} \ge 2 \plus{} 2xy \minus{} 2x \minus{} 2y$[/quote]
Set $ a\equal{}1\minus{}x,b\equal{}1\minus{}y, 0 \le a,b <1$
The inequality becomes:
$ a^b\plus{}b^a \ge 2ab$
Because the function $ f(t)\equal{}t^m, m\in R$ is decreasing on $ [0,1)$
So $ a^b\plus{}b^a \ge a^1\plus{}b^1 \ge a^2\plus{}b^2 \ge 2ab$


inequalities
logarithms
inequalities proposed

[quote="manlio"]For $ a,b,c,x,y,z$ positive reals

$ a^2 \plus{} b^2 \plus{} c^2 \geq \sqrt {\frac {2x}{x \plus{} y}}ab \plus{} \sqrt {\frac {2y}{y \plus{} z}}bc \plus{} \sqrt {\frac {2z}{z \plus{} x}}ca$[/quote]
By buffalo's way we need to prove that $ \sum_{cyc}(x^4z^2\minus{}x^2y^2z^2)\geq0,$ which is obvious.

[quote="FelixD"]@arqady: what's buffalo's method?[/quote]
If $ a > 0$ and $ b^2 \minus{} 4ac\leq0$ then $ ax^2 \plus{} bx \plus{} c\geq0$ for all value of $ x.$ :wink:

[quote="arqady"][quote="manlio"]For $ a,b,c,x,y,z$ positive reals

$ a^2 \plus{} b^2 \plus{} c^2 \geq \sqrt {\frac {2x}{x \plus{} y}}ab \plus{} \sqrt {\frac {2y}{y \plus{} z}}bc \plus{} \sqrt {\frac {2z}{z \plus{} x}}ca$[/quote]
By buffalo's way we need to prove that $ \sum_{cyc}(x^4z^2 \minus{} x^2y^2z^2)\geq0,$ which is obvious.[/quote]

Dear arqady,

nice and neat solution, but can you please explain better your computation? Thank you very much. :)

Here is my proof:
$ a^2 \plus{} b^2 \plus{} c^2 \geq \sqrt {\frac {2x}{x \plus{} y}}ab \plus{} \sqrt {\frac {2y}{y \plus{} z}}bc \plus{} \sqrt {\frac {2z}{z \plus{} x}}ca\Leftrightarrow$
$ \Leftrightarrow a^2 \minus{} \left(b\sqrt {\frac {2x}{x \plus{} y}} \plus{} c\sqrt {\frac {2z}{z \plus{} x}}\right)a \plus{} b^2 \plus{} c^2 \minus{} bc\sqrt {\frac {2y}{y \plus{} z}}\geq0.$
Hence, it remains to prove that
$ \left(b\sqrt {\frac {2x}{x \plus{} y}} \plus{} c\sqrt {\frac {2z}{z \plus{} x}}\right)^2 \minus{} 4b^2 \minus{} 4c^2 \plus{} 4bc\sqrt {\frac {2y}{y \plus{} z}}\leq0,$
 which is equivalent to
$ \frac {(x \plus{} 2y)b^2}{x \plus{} y} \plus{} \frac {(z \plus{} 2x)c^2}{x \plus{} z} \minus{} 2bc\left(\sqrt {\frac {2y}{y \plus{} z}} \plus{} \sqrt {\frac {xz}{(x \plus{} y)(x \plus{} z)}}\right)\geq0,$
for which we need to prove that
$ \left(\sqrt {\frac {2y}{y \plus{} z}} \plus{} \sqrt {\frac {xz}{(x \plus{} y)(x \plus{} z)}}\right)^2 \minus{} \frac {(x \plus{} 2y)(z \plus{} 2x)}{(x \plus{} y)(x \plus{} z)}\leq0.$
But $ \left(\sqrt {\frac {2y}{y \plus{} z}} \plus{} \sqrt {\frac {xz}{(x \plus{} y)(x \plus{} z)}}\right)^2 \minus{} \frac {(x \plus{} 2y)(z \plus{} 2x)}{(x \plus{} y)(x \plus{} z)}\leq0\Leftrightarrow$
$ \Leftrightarrow x^2z \plus{} y^2x \plus{} z^2y \plus{} xyz\geq\sqrt {2xyz(x \plus{} y)(x \plus{} z)(y \plus{} z)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^4z^2 \minus{} x^2y^2z^2)\geq0.$

[quote="manlio"]Thank you very much, Arqady. :)  Your proof is very nice.

I tried to find a proof by AM-GM and Cauchy-Schwarz, but I failed. :([/quote]
You can use AM-GM in that way: (But I think it's the same with [b]arqady[/b])
Use AM-GM inequalitiy we have
$ \sqrt {\frac{{2x}}{{x \plus{} y}}} ab \plus{} \sqrt {\frac{{2z}}{{z \plus{} x}}} ca \equal{} a\left( {\sqrt {\frac{{2x}}{{x \plus{} y}}} b \plus{} \sqrt {\frac{{2z}}{{z \plus{} x}}} c} \right)$
                 $ \le{a^2} \plus{} \frac{1}{4}{\left( {\sqrt {\frac{{2x}}{{x \plus{} y}}} b \plus{} \sqrt {\frac{{2z}}{{z \plus{} x}}} c} \right)^2}$
So it's suffficies to show that
$ {b^2} \plus{} {c^2} \ge \frac{1}{4}{\left( {\sqrt {\frac{{2x}}{{x \plus{} y}}} b \plus{} \sqrt {\frac{{2z}}{{z \plus{} x}}} c} \right)^2} \plus{} \sqrt {\frac{{2y}}{{y \plus{} z}}} bc$
$ \Leftrightarrow {b^2}\left( {4 \minus{} \frac{{2x}}{{x \plus{} y}}} \right) \plus{} {c^2}\left( {4 \minus{} \frac{{2y}}{{y \plus{} z}}} \right) \ge 4bc\left( {\sqrt {\frac{{2y}}{{y \plus{} z}}}  \plus{} \sqrt {\frac{{xz}}{{\left( {x \plus{} y} \right)\left( {z \plus{} x} \right)}}} } \right)$
Notice that $ 4\minus{}\frac{2x}{x\plus{}y}>0$ and $ 4\minus{}\frac{2z}{z\plus{}x}>0$ then we can use AM-GM ineq in here, so we just have to prove that
$ \left( {4 \minus{} \frac{{2x}}{{x \plus{} y}}} \right)\left( {4 \minus{} \frac{{2y}}{{y \plus{} z}}} \right) \ge 4{\left( {\sqrt {\frac{{2y}}{{y \plus{} z}}}  \plus{} \sqrt {\frac{{xz}}{{\left( {x \plus{} y} \right)\left( {z \plus{} x} \right)}}} } \right)^2}$
This inequality is equivalent to
$ \frac{x}{{x \plus{} y}} \plus{} \frac{y}{{y \plus{} z}} \plus{} \frac{z}{{z \plus{} x}} \plus{} \sqrt {\frac{{2xyz}}{{\left( {x \plus{} y} \right)\left( {y \plus{} z} \right)\left( {z \plus{} x} \right)}}}  \le 2$
$ \Leftrightarrow \frac{y}{{y \plus{} x}} \plus{} \frac{z}{{z \plus{} y}} \plus{} \frac{x}{{x \plus{} z}} \ge 1 \plus{} \sqrt {\frac{{2xyz}}{{\left( {x \plus{} y} \right)\left( {y \plus{} z} \right)\left( {z \plus{} x} \right)}}}$ [b](I)[/b]
This inequality is a well-known.
P/s: To prove[b] (I)[/b] you can see here (in page 19)

[quote="arqady"][quote="leedt26"]
$ \frac {y}{{y \plus{} x}} \plus{} \frac {z}{{z \plus{} y}} \plus{} \frac {x}{{x \plus{} z}} \ge 1 \plus{} \sqrt {\frac {{2xyz}}{{\left( {x \plus{} y} \right)\left( {y \plus{} z} \right)\left( {z \plus{} x} \right)}}}$ [b](I)[/b]
This inequality is a well-known.
P/s: To prove[b] (I)[/b] you can see here (in page 19)[/quote]
Problem 3 (from the download)
Show that for any positive real numbers $ a,$ $ b$ and $ c$ we have
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc\geq\sqrt [3]{abc}(a \plus{} b \plus{} c)^2\]
My proof:
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$
Hence, $ a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc\geq\sqrt [3]{abc}(a \plus{} b \plus{} c)^2\Leftrightarrow f(v^2)\geq0,$ where
$ f(v^2) \equal{} \minus{} 3uv^2 \plus{} 3u^3 \minus{} u^2w \plus{} w^3.$
But $ f$ gets a minimal value when $ v^2$ gets a maximal value, which happens  when two numbers from $ \{a,b,c\}$  are equal ($ a,$ $ b$ and $ c$ are real roots of the equation $ x^3 \minus{} 3ux^2 \plus{} 3v^2x \minus{} w^3 \equal{} 0$ :wink: ).
Thus, for the proof enough to check only one case: $ b \equal{} c \equal{} 1.$
In this case the original inequality is equivalent to $ g(a)\geq0,$ where
$ g(a) \equal{} \ln(a^3 \plus{} 6a \plus{} 2) \minus{} \frac {1}{3}\ln a \minus{} 2\ln(a \plus{} 2).$
But $ g'(a) \equal{} \frac {2(a \minus{} 1)(a^3 \plus{} 9a^2 \minus{} 3a \plus{} 2)}{3a(a^3 \plus{} 6a \plus{} 2)(a \plus{} 2)},$
which gives $ a_{min} \equal{} 1$ and our inequality is proven.[/quote]
we can prove the stronger $ 8(a^3\plus{}b^3\plus{}c^3)\plus{} 54abc \ge 9\sqrt [3]{abc}.(a \plus{} b \plus{} c)^2$
By shur,we have $ 8(a^3 \plus{} b^3 \plus{} c^3) \plus{} 30abc \ge 2(a \plus{} b \plus{} c)^3$
and from AM-GM
$ (a \plus{} b \plus{} c)^3 \plus{} (a \plus{} b \plus{} c)^3 \plus{} 27abc \ge 9\sqrt [3]{abc}.(a \plus{} b \plus{} c)^2$
From it,we get
$ 8(a^3 \plus{} b^3 \plus{} c^3) \plus{} 57abc \ge 9\sqrt [3]{abc}.(a \plus{} b \plus{} c)^2$
This problem post by tranquocluat in maths.vn

In fact, the stronger inequality holds:

If $ a,$ $ b,$ $ c$ are positive real numbers such that $ abc\equal{}1,$ then
$ 3(a^3\plus{}b^3\plus{}c^3)\plus{}27 \ge 4(a\plus{}b\plus{}c)^2.$

[quote="can_hang2007"]In fact, the stronger inequality holds:

If $ a,$ $ b,$ $ c$ are positive real numbers such that $ abc \equal{} 1,$ then
$ 3(a^3 \plus{} b^3 \plus{} c^3) \plus{} 27 \ge 4(a \plus{} b \plus{} c)^2.$[/quote]
I think the maximal $ k$ for which the inequality
$ a^3\plus{}b^3\plus{}c^3\plus{}(9k\minus{}3)abc\geq k\sqrt[3]{abc}(a\plus{}b\plus{}c)^2$
holds for all non-negative $ a,$ $ b$ and $ c$ is
$ k\equal{}\min_{a>0}\frac{1}{18}\left(a^2\plus{}9a\plus{}15\plus{}\frac{2}{a}\right)\equal{}1.316...$

[quote="arqady"][quote="can_hang2007"]In fact, the stronger inequality holds:

If $ a,$ $ b,$ $ c$ are positive real numbers such that $ abc \equal{} 1,$ then
$ 3(a^3 \plus{} b^3 \plus{} c^3) \plus{} 27 \ge 4(a \plus{} b \plus{} c)^2.$[/quote]
I think the maximal $ k$ for which the inequality
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} (9k \minus{} 3)abc\geq k\sqrt [3]{abc}(a \plus{} b \plus{} c)^2$
holds for all non-negative $ a,$ $ b$ and $ c$ is
$ k \equal{} \min_{a > 0}\frac {1}{18}\left(a^2 \plus{} 9a \plus{} 15 \plus{} \frac {2}{a}\right) \equal{} 1.316...$[/quote]
I only prove in the case $ k \ge \frac {3}{2}$
Let abc=1.
$ f(a,b,c) \equal{} a^3 \plus{} b^3 \plus{} c^3 \minus{} k(a \plus{} b \plus{} c)^2$
$ f(a,b,c) \minus{} f(a,\sqrt {bc},\sqrt {bc}) \equal{} (\sqrt {b} \minus{} \sqrt {c})^2[(b \plus{} c \plus{} \sqrt {bc})^2 \minus{} 2ka \minus{} k(\sqrt {b} \plus{} \sqrt {c})^2]$
$ a \equal{} min(a,b,c)$.It is easy to show that $ bc \ge 1$
We have
$ (b \plus{} c \plus{} \sqrt {bc})^2 \minus{} 2ka \minus{} k(\sqrt {b} \plus{} \sqrt {c})^2 \ge \frac {9(\sqrt {b} \plus{} \sqrt {c})^4)}{16} \minus{} k(\sqrt {b} \plus{} \sqrt {c})^2 \minus{} 2k \ge 4.(\frac {9}{4} \minus{} k) \minus{} 2k \equal{} 9 \minus{} 6k \ge 0$
Finally,we only need to show that in the case b=c,which is easy.

[quote="arqady"]
I think the maximal $ k$ for which the inequality
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} (9k \minus{} 3)abc\geq k\sqrt [3]{abc}(a \plus{} b \plus{} c)^2$
holds for all non-negative $ a,$ $ b$ and $ c$ is
$ k \equal{} \min_{a > 0}\frac {1}{18}\left(a^2 \plus{} 9a \plus{} 15 \plus{} \frac {2}{a}\right) \equal{} 1.316...$[/quote]
Now I see that it's wrong.
We can say only that the maximal value of $ k$ grater than $ 1.316...$
Indeed, your inequality is true, Can.

[quote="manlio"]For $ a,b,c,x,y,z$ positive reals

$ a^2 \plus{} b^2 \plus{} c^2 \geq \sqrt {\frac {2x}{x \plus{} y}}ab \plus{} \sqrt {\frac {2y}{y \plus{} z}}bc \plus{} \sqrt {\frac {2z}{z \plus{} x}}ca$[/quote]
nice inequality,it generalize an old and hard Vasc's one.
it's easy to show that the inequality is equivalent with,
$ a^2 \plus{} b^2 \plus{} c^2 \geq \sqrt {\frac {2}{1 \plus{} x^2}}ab \plus{} \sqrt {\frac {2}{1 \plus{} y^2}}bc \plus{} \sqrt {\frac {2}{1 \plus{} z^2}}ca$
where $ x,y,z$ are non negative real numbers such that $ xyz \equal{} 1$. this inequality is symetric for $ x,y,z$ assume then that $ z \equal{} max\{x,y,z\}$ then $ xy \equal{} min\{xy,yz,zx\}\leq 1$
using Cauchy Schwarz we have,
$ \sqrt {\frac {2}{1 \plus{} x^2}}ab \plus{} \sqrt {\frac {2}{1 \plus{} y^2}}bc \equal{} b\left(a\sqrt {\frac {2}{1 \plus{} x^2}} \plus{} c\sqrt {\frac {2}{1 \plus{} y^2}}\right)\leq b\sqrt {2(a^2 \plus{} c^2)}.\sqrt {\frac {1}{1 \plus{} x^2} \plus{} \frac {1}{1 \plus{} y^2}}$
$ \equal{} b\sqrt {2(a^2 \plus{} c^2)}.\sqrt {1 \plus{} \frac {1 \minus{} x^2y^2}{(1 \plus{} x^2)(1 \plus{} y^2)}}\leq b\sqrt {a^2 \plus{} c^2}.\sqrt {1 \plus{} \frac {1 \minus{} xy}{1 \plus{} xy}}$
$ \equal{} b.X\sqrt {\frac {2z}{1 \plus{} z}}$ where $ X \equal{} \sqrt {2(b^2 \plus{} c^2)}$, then it's suffice to prove that,
$ b.X\sqrt {\frac {2z}{1 \plus{} z}} \plus{} ac\sqrt {\frac {2}{1 \plus{} z^2}} \leq a^2 \plus{} b^2 \plus{} c^2$
using Cauchy-Schwarz we get $ \sqrt {\frac {2}{1 \plus{} z^2}} \leq \frac {2}{1 \plus{} z}$ then it's suffice to prove that:
$ b^2 \minus{} bX\sqrt {\frac {2z}{1 \plus{} z}} \plus{} a^2 \plus{} c^2 \minus{} \frac {2ac}{1 \plus{} z}\geq 0$
$ \Delta \equal{} \frac {2zX^2}{z \plus{} 1} \minus{} 4\left(a^2 \plus{} c^2 \minus{} \frac {2ac}{z \plus{} 1}\right)$
$ \equal{} \frac {4z(a^2 \plus{} c^2)}{z \plus{} 1} \plus{} \frac {8ac}{z \plus{} 1} \minus{} \frac {4(a^2 \plus{} c^2)(z \plus{} 1)}{z \plus{} 1}$
$ \equal{} \minus{} \frac {4(a \minus{} c)^2}{z \plus{} 1}$ if $ a\neq c$ then $ \Delta < 0$ and then we deduct our inequality, therefore it's suffice to check the case of $ a \equal{} c$ in that case, the inequality is equivalent with,
$ 2ab\sqrt {\frac {2z}{z \plus{} 1}} \plus{} \frac {2a^2}{1 \plus{} z}\leq b^2 \plus{} 2a^2$
$ \Leftrightarrow b^2 \plus{} \frac {2a^2z}{z \plus{} 1}\geq 2ab\sqrt {\frac {2z}{z \plus{} 1}}$
which is juste an application of Am-Gm inequality, that complete the proof.
equality holds if and only if $ x \equal{} y \equal{} z$

[quote="manlio"]For $ a,b,c,x,y,z$ positive reals

$ a^2 \plus{} b^2 \plus{} c^2 \geq \sqrt {\frac {2x}{x \plus{} y}}ab \plus{} \sqrt {\frac {2y}{y \plus{} z}}bc \plus{} \sqrt {\frac {2z}{z \plus{} x}}ca$[/quote]
By AM-GM: $ \frac{1}{2}\sqrt{\frac{2x}{x\plus{}y}}(a^2\plus{}b^2) \ge \sqrt{\frac{2x}{x\plus{}y}}ab$
By Cauchy-Schwarz: $ c\sqrt{\frac{2y}{y\plus{}z} \plus{} \frac{2z}{z\plus{}x}}\sqrt{b^2\plus{}a^2} \ge \sqrt {\frac {2y}{y \plus{} z}}bc \plus{} \sqrt {\frac {2z}{z \plus{} x}}ca$
Let $ 2x^2 \equal{} a^2\plus{}b^2$. Then it is enough to prove:
$ 2x^2\plus{}c^2 \ge \sqrt{\frac{2x}{x\plus{}y}}x^2 \plus{} \sqrt{\frac{4y}{y\plus{}z} \plus{} \frac{4z}{z\plus{}x}}cx \iff$
$ x^2 \left ( 2 \minus{} \sqrt{\frac{2x}{x\plus{}y}} \right ) \plus{} c^2 \ge \sqrt{\frac{4y}{y\plus{}z} \plus{} \frac{4z}{z\plus{}x}}cx$
And therefore it is enough to prove:
$ 4\left ( 2 \minus{} \sqrt{\frac{2x}{x\plus{}y}} \right ) \ge \frac{4y}{y\plus{}z} \plus{} \frac{4z}{z\plus{}x} \iff$
$ 1 \minus{} \frac{y}{y\plus{}z} \minus{} \frac{z}{z\plus{}x} \ge \sqrt{\frac{2x}{x\plus{}y}} \minus{} 1 \iff$
$ \frac{z(x\minus{}y)}{(x\plus{}z)(y\plus{}z)} \ge \frac{x\minus{}y}{\sqrt{2x(x\plus{}y)} \plus{} x\plus{}y} \iff$
$ (x\minus{}y)(z\sqrt{2x(x\plus{}y)} \minus{} z^2\minus{}xy ) \ge 0 \iff$
$ (x\minus{}y)(2z^2x^2 \minus{} z^4 \minus{}x^2y^2) \ge 0$
Since the original inequality was cyclic in $ x,y,z$ it is enough to prove the two cases $ x \ge z \ge y$ and $ y \ge z \ge x$.

Case $ x \ge z \ge y$ gives $ x\minus{}y \ge 0$ and $ x^2z^2 \ge z^4, x^2z^2 \ge x^2y^2$ and we are done.
Case $ y \ge z \ge x$ gives $ x\minus{}y \le 0$ and $ z^2x^2 \le z^4,z^2x^2 \le y^2x^2$ and we are done.

So the proof is finished :)

[quote="peine"]Then inequality is equivalent with,
$ a^2 \minus{} a\left(b\sqrt {\frac {2x}{x \plus{} y}} \plus{} c\sqrt {\frac {2z}{z \plus{} x}}\right) \plus{} b^2 \plus{} c^2\geq 0$
[/quote]
I think it should be
$ a^2 \minus{} a\left(b\sqrt {\frac {2x}{x \plus{} y}} \plus{} c\sqrt {\frac {2z}{z \plus{} x}}\right) \plus{} b^2 \plus{} c^2 \minus{} bc \sqrt{\frac{2y}{y\plus{}z}} \ge 0$

[quote="Mathias_DK"][quote="peine"]Then inequality is equivalent with,
$ a^2 \minus{} a\left(b\sqrt {\frac {2x}{x \plus{} y}} \plus{} c\sqrt {\frac {2z}{z \plus{} x}}\right) \plus{} b^2 \plus{} c^2\geq 0$
[/quote]
I think it should be
$ a^2 \minus{} a\left(b\sqrt {\frac {2x}{x \plus{} y}} \plus{} c\sqrt {\frac {2z}{z \plus{} x}}\right) \plus{} b^2 \plus{} c^2 \minus{} bc \sqrt {\frac {2y}{y \plus{} z}} \ge 0$[/quote]
thank's a lot for your remark,   :!:  . I edited my message, and I posted a new solution, which is too nice, hope it's true.


Numbers $p,q,r$ lies in the interval $(\frac{2}{5},\frac{5}{2})$ nad satisfy $pqr=1$. Prove that there exist two triangles of the same area, one with the sides $a,b,c$ and the other with the sides $pa,qb,rc$.

We know that if triangle $ABC$ has length of sides are $a,b,c$ then its area equal to $\frac{2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2}-a^{4}-b^{4}-c^{4}}{4}$. And if $a,b,c$ are positive reals such that $2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2}-a^{4}-b^{4}-c^{4}>0$ then $a,b,c$ are lengths of sides of a triangle.

Now, WLOG we can assume that $a=1$. We need prove that : Exist positive reals $b,c$ such that $0<2b^{2}+2b^{2}c^{2}+2c^{2}-1-b^{4}-c^{4}=2p^{2}q^{2}b^{2}+2q^{2}r^{2}b^{2}c^{2}+2p^{2}r^{2}c^{2}-p^{4}-q^{4}b^{4}-r^{4}c^{4}.$

At this time, that is all what i know.

Are you kidding me?
The problem does not require $p, q $ and $ r$ to be distinct. So let $p = q = r = 1$ and $1$ lies in the interval $ (\frac{2}{5},\frac{5}{2}) $. The two triangles are congruent because $pa = a, qb = b, rc = c$ and have the same area.  Done and totally acceptable!

On the other hand, if $p, q $ and $ r$ are distinct, arbitrarily let $r$ = 1 and $ \frac{1}{p} = q$, the two triangles to be $ABC$ and $A'B'C'$ with $AB$ = $a$, $BC$ = $c$, $CA$ = $b$,  and $A'B'$ = $pa$, $B'C'$ = $rc$ = $c$, $C'A'$ = $qb$. We can also let $ \angle BAC$ =  $ \angle B'A'C'$ and $Sx$ and $Sy$ be the areas of $ABC$ and $A'B'C'$, respectively.

We have $Sx$ = $ \frac {1}{2}ab sin\angle BAC$ = $ \frac {1}{2}paqb sin\angle B'A'C'$ = $Sy$.
Actually, the interval range in the problem is not really important; to be nice you can use two numbers that are inverses of each other.


I thought you want a integer solution... :lol:
Let x=ab, y=a+b, the equation becomes that
$(x+y+20)^2=20xy$
Since 20 is a divisor of a perfect square, the square must have the divisor 100,
So 5|xy and 10|x+y+20 that is 10|x+y, and 5|x or 5|y
Suppose that 5|x,it's easy to get 5|y. It's clear that 5|a and 5|b.
Let $x=5x_1, y=5y_1$. We get $20x_1y_1=(x_1+y_1+4)^2$
Similarly $5|x_1y_1, 10|x_1+y_1+4$
By5|a, 5|b we have 25|x, that is $5|x_1, x_1=5x_2$, and we get $y_1\equiv 1(mod 5), y_1=5y_2+1$
so $4x_2(5y_2+1)=(x_2+y_2+1)^2 \Rightarrow 4x_2\equiv(x_2+1)^2(mod y_2)$
and $(y_2+1)^2\equiv 0(mod x_2)$ these are $x_2|(y_2+1)^2, y_2|(x_2-1)^2$
what's more, $x_2\pm y_2$ is odd.
Let $tx_2=(y_2+1)^2$ put it into the other equation
we get $y_2|2-2t, 2t\equiv 2(mod y_2)$ we can get $x_2\equiv 2(mod y_2), x_2-2|y_2$
similarly $x_2|2+2r, 2r\equiv 2(mod x_2)$ we can get $y_2\equiv -2(mod x_2), y_2+2|x_2$
so $y_2+2|x_2-y_2-2, x_2-2|y_2-x_2+2$
From first we know $x_2-y_2-2$ is odd,but one of the 2 divisors must be even.Contradiction.

[quote="Ghang Hwan"][quote="Skyward_Sea"]I thought you want a integer solution... :lol:
Let x=ab, y=a+b, the equation becomes that
$(x+y+20)^2=20xy$

I don't think so.

it will be $(x+5y+25)^2=20x(y+5)$[/quote][/quote]

Sorry I made such a stupid mistake... :P
A new method here!
You equation is equal to $(ab-5a-5b-25)^2=0$
so ab-5a-5b-25=0 that is (a-5)(b-5)=50
the steps below are easy now, I think .

Yeah, $(ab-5a-5b-25)^2=0$ seems to be the best rearrangement.

Perhaps a more natural variation on your initial substitution would have been the best way to get it - rather than $y=a+b$, let $y=a+b+5$, and you get $(x+5y)^2=20xy$, which rearranges quite neatly into $(x-5y)^2=0$.


set theory
advanced fields
advanced fields theorems

Does anyone know what a large capital E means in set theory papers written in the 1930s? This is a large italic E not a sigma, and it is not backwards like an existential symbol.

Just a guess, but it might be an old existential quantifier.

Or in might mean union (E for "ensemble.") Or maybe it's not safe to guess.

I also saw an old paper where it was used for $\in$ once. If you give us a few examples of usage, I think everything will become clear at once. Without that, any of our guesses seems to be as good (bad) as any other.

It's something like that:

$E_{x}(x>0)$?

It's old notation for $\{x| x>0\}$.


This of Ductrung is so nice that I think it must have its own topic

\[ ab\plus{}bc\plus{}ca\le\frac{a^{3}(b\plus{}c)}{a^{2}\plus{}bc}\plus{}\frac{b^{3}(c\plus{}a)}{b^{2}\plus{}ca}\plus{}\frac{c^{3}(a\plus{}b)}{c^{2}\plus{}ab}\le a^{2}\plus{}b^{2}\plus{}c^{2}\]

Hello manlio,

I used Vornicu-Schur for both inequalities in my first solution. Since then I also found Cauchy-Schwarz-only solutions (for both) but I still think the Vornicu-Schur based are the most "direct" and natural.

Note that
\[ \sum_{cycl}\frac{a^{3}(b\plus{}c)}{a^{2}\plus{}bc}\plus{}\sum_{cycl}\frac{abc(b\plus{}c)}{a^{2}\plus{}bc}\equal{}\sum_{cycl}\frac{a(b\plus{}c)(a^{2}\plus{}bc)}{a^{2}\plus{}bc}\equal{}2(ab\plus{}bc\plus{}ca)\]
The given inequality is therefore equivalent to
\[ \sum_{cycl}\frac{abc(b\plus{}c)}{a^{2}\plus{}bc}\le ab\plus{}bc\plus{}ca\le\frac{1}{2}\sum_{cycl}\left( a^{2}\plus{}\frac{abc(b\plus{}c)}{a^{2}\plus{}bc}\right)\ \ (*)\]
The left-hand side is a well-known inequality by Hoo-joo Lee [Crux 2580]; the right-hand side is equivalent to
\[ 0\le\sum_{cycl}\left( a^{2}\plus{}\frac{abc(b\plus{}c)}{a^{2}\plus{}bc}\minus{}a(b\plus{}c)\right)\equal{}\sum_{cycl}\frac{a^{2}(a\minus{}b)(a\minus{}c)}{a^{2}\plus{}bc}\]
which is true by Vornicu-Schur lemma.

BTW: the left-hand side of (*) can also be obtained as a special case of (which I posted at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=159311[/url]):
For any real number $ n\ge 1$ and any positive real numbers $ a,b$ and $ c$,
\[ \frac{a^{2}(b^{n}\plus{}c^{n})}{a^{2}\plus{}bc}\plus{}\frac{b^{2}(c^{n}\plus{}a^{n})}{b^{2}\plus{}ca}\plus{}\frac{c^{2}(a^{n}\plus{}b^{n})}{c^{2}\plus{}ab}\le a^{n}\plus{}b^{n}\plus{}c^{n}\]
(by the change of variables $ (a,b,c)\rightarrow (a^{\minus{}1},b^{\minus{}1},c^{\minus{}1})$) or of:
If $ k\ge 0$, $ p\ge 1$ and $ a,b,c$ are positive real numbers, then
\[ \frac{b\plus{}c}{a^{k}(a^{2}\plus{}pbc)}\plus{}\frac{c\plus{}a}{b^{k}(b^{2}\plus{}pca)}\plus{}\frac{a\plus{}b}{c^{k}(c^{2}\plus{}pab)}\le\frac{2}{p\plus{}1}\left(\frac{1}{a^{k\plus{}1}}\plus{}\frac{1}{b^{k\plus{}1}}\plus{}\frac{1}{c^{k\plus{}1}}\right)\]

Dear Ductrung, youd said it is possible to prove

$ \sum\frac{a^{3}(b\plus{}c)}{a^{2}\plus{}bc}\leq\sum a^{2}$

using only Cauchy, but I don't see that. I tried it in two ways but I failed. :oops:

$ 1st)$

Using your trick I rewrote the inequality in an equivalent form

$ \sum ab\le\frac{1}{2}(\sum(a^{2}\plus{}\frac{abc(b\plus{}c)}{a^{2}\plus{}bc})$

Then I used Cauchy in this form

$ \sum\frac{(b\plus{}c)^{2}}{(a^{2}\plus{}bc)(b\plus{}c)}\geq\frac{2(\sum a)^{2}}{\sum ab(a\plus{}b)}$

and putting in our inequality after some calculations we have

$ \sum ab(a\plus{}b)(a\minus{}b)^{2}\geq abc\sum (a\minus{}b)^{2}$

but now we have to use SOS criterio to prove it (so we need not only Cauchy)

$ 2nd)$

$ \sum\frac{a^{3}(b\plus{}c)}{a^{2}\plus{}bc}\leq\sum a^{2}$

I rewrote

$ \sum\frac{a^{3}(b\plus{}c)}{a^{2}\plus{}bc}\equal{}\sum\frac{a^{4}(b\plus{}c)^{2}}{(a^{2}\plus{}bc)a(b\plus{}c)}$

But $ (a^{2}\plus{}bc)a(b\plus{}c)\equal{}ab(a^{2}\plus{}c^{2})\plus{}ac(a^{2}\plus{}b^{2})$ so by Cauchy we have

$ \frac{a^{4}(b\plus{}c)^{2}}{(a^{2}\plus{}bc)a(b\plus{}c)}\equal{}\frac{a^{4}(b\plus{}c)^{2}}{ab(a^{2}\plus{}c^{2})\plus{}ac(a^{2}\plus{}b^{2})}\leq\frac{a^{3}b}{a^{2}\plus{}c^{2}}\plus{}\frac{a^{3}c}{a^{2}\plus{}b^{2}}$

Finally we remain with

$ \sum\frac{(a^{3}\plus{}b^{3})c}{a^{2}\plus{}b^{2}}\leq\sum a^{2}$

which I cannot say if it is false or true. It is only easy to prove that

$ \sum\frac{(a^{3}\plus{}b^{3})c}{a^{2}\plus{}b^{2}}\geq\sum ab$

Ductrung, can you please give me a  hint for a only Cauchy solution? :)

Thank you very much.

[quote="manlio"]I solved them by SOS method and Vornicu-Shur. Have you a more direct solution?

Thank you very much.[/quote]
Dear manlio,could you post your sos solution???I am a newbie who is learning sos.
Thanks. :)

Hi Manlio,

[quote="manlio"]
$ \sum ab(a\plus{}b)(a\minus{}b)^{2}\geq abc\sum (a\minus{}b)^{2}$
but now we have to use SOS criterio to prove it (so we need not only Cauchy)
[/quote]
the above can be proved without full expansion, without appealing to any theorem but purely algebraic manipulation using some known identity.

[quote="manlio"]
$ \sum\frac{(a^{3}\plus{}b^{3})c}{a^{2}\plus{}b^{2}}\leq\sum a^{2}$
[/quote]
The above is not true!

For stephen38:

$ \sum (ab(a\plus{}b)\minus{}abc)(a\minus{}b)^{2}\geq 0$

$ \sum C(a\minus{}b)^{2}\geq 0$

where $ C\equal{} ab(a\plus{}b)\minus{}abc$ ,$ B\equal{} ca(c\plus{}a)\minus{}abc$,$ A\equal{} bc(b\plus{}c)\minus{}abc$

Then it is very easy to prove (WLOG assume $ a\geq b\geq c$) that

$ B\geq 0$ , $ A\geq 0$ , $ B\plus{}C\geq 0$ which is third criterium of SOS method.

Ductrung, can you please give me the known identity used to avoid SOS method.

Thank you very much.

[quote="manlio"]
$ \sum ab(a\plus{}b\minus{}c)(a\minus{}b)^{2}\geq 0\ \ (*)$
[/quote]
We know the transform between SOS form and Vornicu-Schur form:
$ \sum_{cycl}x(b\minus{}c)^{2}\equal{}\sum_{cycl}(x\plus{}y)(c\minus{}a)(c\minus{}b)$
Thus (*) is equivalent to
$ 0\le\sum_{cycl}[ c^{2}(a\plus{}b)\plus{}c(a\minus{}b)^{2}](c\minus{}a)(c\minus{}b) \equal{}\sum_{cycl}c^{2}(a\plus{}b)(c\minus{}a)(c\minus{}b)\ \ (**)$
The last equality is due to the fact that the sum
\[ \sum_{cycl}c(a\minus{}b)^{2}(c\minus{}a)(c\minus{}b)\]
is vanished. To prove (**), assume $ a\ge b\ge c$. Then
$ c^{2}(a\plus{}b)(c\minus{}a)(c\minus{}b)\ge 0$
Further,
$ a^{2}(b\plus{}c)(a\minus{}b)(a\minus{}c)\plus{}b^{2}(c\plus{}a)(b\minus{}a)(b\minus{}c) \equal{} (a\minus{}b)\left[ a^{2}(b\plus{}c)(a\minus{}c)\minus{}b^{2}(c\plus{}a)(b\minus{}c)\right]$
$ \ge (a\minus{}b)\left[ a^{2}(b\plus{}c)(b\minus{}c)\minus{}b^{2}(c\plus{}a)(b\minus{}c)\right]$
$ \equal{} (a\minus{}b)\left[ (b\minus{}c)[ a^{2}(b\plus{}c)\minus{}b^{2}(c\plus{}a) ]\right]$
$ \equal{} (a\minus{}b)\left[ (b\minus{}c)[ ab(a\minus{}b)\plus{}c(a^{2}\minus{}b^{2}) ]\right]$
$ \equal{} (a\minus{}b)^{2}\left[ (b\minus{}c)(ab\plus{}bc\plus{}ca)\right]\ge 0$
and we are done. OK, I must confess that it is faster to invoke Vornicu-Schur lemma on (**).
But what about the following alternative: To prove
\[ ab\plus{}bc\plus{}ca\le\frac{1}{2}\sum_{cycl}\left( a^{2}\plus{}\frac{abc(b\plus{}c)}{a^{2}\plus{}bc}\right)\]
as you noted, by Cauchy-Schwarz inequality, it suffices to show that
\[ 2(ab\plus{}bc\plus{}ca)\le a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}\frac{2abc(a\plus{}b\plus{}c)^{2}}{a(b^{2}\plus{}c^{2})\plus{}b(c^{2}\plus{}a^{2})\plus{}c(a^{2}\plus{}b^{2})}\]
Let $ X \equal{} a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}2(ab\plus{}bc\plus{}ca)$. The above then can be rewritten as
$ 0\le X\sum_{cycl}a(b^{2}\plus{}c^{2})\plus{}2abc(a\plus{}b\plus{}c)^{2}$
$ \equal{} X\sum_{cycl}a(b\minus{}c)^{2}\plus{}2abc[ (a\plus{}b\plus{}c)^{2}\plus{}3X ]$
$ \equal{} X\sum_{cycl}a(b\minus{}c)^{2}\plus{}2abc[ 4(a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}ab\minus{}bc\minus{}ca) ]$
$ \equal{} X\sum_{cycl}a(b\minus{}c)^{2}\plus{}4abc\sum_{cycl}(b\minus{}c)^{2}$
$ \equal{}\sum_{cycl}a(b\minus{}c)^{2}[ X\plus{}4bc ]$
$ \equal{}\sum_{cycl}a(b\minus{}c)^{2}(b\plus{}c\minus{}a)^{2}$
which is obviously true.

Thank you very much, Ductrung.
Your reasoning is very interesting. :)


Whenever I make a small equation and compile it in $ \text{\LaTeX}$, the result I get is a page with my formula and a lot of blank space. Is there a way I can resize the page so that I don't have any of that blank space (like the formulas you see in this website)?

LaTeX is for writing documents so if you only want one equation, that's what you'll get on a page. This site extracts an image of the equation and puts it in a post. If you want a document then there is no reason why you can't put lots more on the page but if you want an image on its own then use the TeXer, where you can copy the image.


What is the smallest integer value of $ n$ such that $ 2^{\minus{}n} < \frac{1}{1000}$?

hello, taking the logarithm of both sides your inequality we have
$ \minus{}n\ln(2)<\ln\left(\frac{1}{1000}\right)\equal{}\ln(1)\minus{}\ln(1000)\equal{}\ln(1000)$
Then we get
$ \minus{}n<\frac{\ln(1000)}{\ln(2)}$
By multiplying with minus one
$ n>\frac{\ln(1000)}{\ln(2)} \approx 9,965784...$
So we have finally $ n\geq10$.
Sonnhard.

This can be rewritten as $ \frac {1}{2^n} < \frac {1}{1000}$
Both numbers are positive, so we can cross multiply to get

$ 1000 < 2^n$

I'm betting you guys probably know your powers of 2 pretty well, and that $ 2^{10}$=1024 (the infamous kilobyte).

9 is too small, hence $ \boxed{10}$

While logs are relevant, it's probably overkill for this one.


Given real numbers $a,\alpha,\beta, \sigma \ and \ \varrho$ s.t. $\sigma, \varrho > 0$ and $\sigma \varrho = \frac{1}{16}$, prove that there exist integers $x$ and $y$ s.t.

\[ - \sigma \leq (x+\alpha_(ax + y + \beta ) \leq \varrho  \]

whats the complete question
are they asking the min of them or max or is the comma actually a bracket


dudes, MIT is over.. Im so happy. no more physics nor chem for 1 month. though i'll miss the math i have to admitt

yeaaa

Are you going to be there for IAP?  If you are you should do the mystery hunt with the Simmons Hall team :)

Also, I'm going to get more events going with our lounge next semester assuming everyone signs up again.  Hopefully I can get a weekly (or every two weeks, or something) where we do a couple practice problems and, of course, get free food.

I love the idea of IAP.  It better be as purely awesome as it sounds.

im not staying for iap man..
hey i havent met u. u live in simmons right/ who are ya? do i know u

Yeah, I live in Simmons.  But we haven't met yet. :(

aw ok i'll look for u in Feb.
i live in 7c.. the best floor ever

joml88, are tu taking 100 B next semester?
i think i am taking 100 C instead. one of those 2... you should take it too

I'm going take two of 18.03, 18.06, and 18.100.  Not sure yet.  Maybe 18.03 and 18.100C...

if u take 18.03 and 18.06 you will get bored ! take 100b.. i think ill take 100b, because 100c is too much work and i can clear out the CI-M requirement with seminars that are easier ( no psets, no tests) and many of them look interesting

is 18.06 linear algebra?
with strang?
because i need lots of help for that.
will someone explain vectors and graphing because i am really simple-minded

Postnikov is teaching it in the Spring.  I suggest asking about vectors in the pre-olympiad forum.

THANKS!  i never thought about that. is Prof. Strang staying at MIT?

I think vectors are a little too advanced for a middle schooler, although I have done physics vectors stuff and linear algebra in the summer before in class.

I don't think Strang's going anywhere(?).  Professors don't necessarily teach the same classes each semester.

[quote="funcia"]I know but I thought Strang was THE LINEAR ALGEBRA GUY. Partially because he wrote the book I use.

AND HOW DO YOU KNOW I AM A MIDDLE SCHOOLER MAYBE I AM SOME HACK WHO LIKES TO PRETEND THEY ARE!!!!

Ok I am so wtf I gave too much info on the net.[/quote]

I meant I AM A MIDDLE SCHOOLER  :lol:

phew too many people know everything aboubt me *ggrrrr*** well too bad i revealed(or did i) that i am a middle schooler

postnikov is a good friend of mine!

hes a combinatorialist...

i dont know why he is teaching  a basic math class.. hes a great mathematician.. but he is hard to follow.. i mean, its wierd for me thinking of him giving a computational class like 18.06, that so many non-math people take.

a paradox would be that he teaches basic math class to have time to youthfully study other more complicated activities in more spare time.  :P


I have seen this SO many times.
[hide] Write the ratios, with g = number of girls, b = number of boys
$\frac{g}{b}=\frac{2}{3}$, $\frac{b-44}{g}=\frac{2}{5}$
Multiply the two equations:
$\frac{b-44}{b}=\frac{4}{15} \Longrightarrow 15b-660=4b \Longrightarrow 11b=660 \Longrightarrow b=60$
$\frac{g}{60}=\frac{2}{3} \Longrightarrow g=40$
$g+b-44=40+60-44=\boxed{56}$[/hide]

[quote="ch1n353ch3s54a1l"]After one-fifth of the girls left a school dance, the ratio of girls to boys was 2:3. Then, 44 boys left and the ratio of boys to girls was 2:5. How many students remained at the dance??[/quote]
[hide]First equation
$.8g/b=2/3 2.4g=2b b=1.2g$
next equation
$(.8g)/(b-44)=5/2 1.6g=5b-220.$
Substitution yields
$4.4g=220. g=50 b=60.$
The answer is [b]56[/b][/hide]

A simpler solution :P  :D
[hide]
Flip both fractions so that they are boys over girls. Find a common denominator since the number of girls doesn't change (in the last part of the problem). The two fractions are 15/10, 4/10. 15-4=11 <- change in boys. 44/11=4 so there are 4 people per unit in the fractions. 4*4+10*4=56.
[/hide]

Huge bump but surely this is better than starting a new thread when this question has been posted so many times.
So I get it that we can find the ratios of boys to girls before and after. Wouldn't you set $\frac{11}{10}=\frac{44}{x}$? I doubt that is the right proportion and I get so lost after here..

Call the number of girls $g$ and boys $b$.

Then we set the proportion $\frac{0.8g}{b}=\frac{2}{3}$. From this we get $6g=5b$.

Now, $\frac{0.8g}{b-44}=\frac{5}{2}$. From this we get $1.6g=5b-220$. But we know from our previous result that $5b=6g$, hence the substitution $\rightarrow 4.4g=220$ and $g=50, b=60$.

However, only 16 boys and 40 girls stayed $\rightarrow \boxed{56}$.


This is for people who went to nationals last year. Which state do you think had the coolest pins? Also, which had the stupidest? (i think i know that answer)
My opinion, best- Virgin Islands (also a strange story to go along w/ it)
Worst- my home, connecticut

Nah Illinois was probably the worst.  Even though Illinois got 2nd two years ago, our pin was simply a plastic one in the shape of illinois (probably with "illinois" written on it)... very creative...

No wonder no one liked ours, but I suppose the previous (or previous previous, however you want to look at it) finish paid off, as people still were willing to trade for it :-O

Maybe we AoPS people can arrange a place to meet and trade pins or something...

at least yours didn't say something stupid, like "Connecticut, we're full of surprises"

If I recall correctly, the worst pin was Mississippi.  It was a plastic pin that said "Mississippi".  No image of the state, no duck (a la Minnesota), no flag, etc.  Just the word "Mississippi".

Incidently, I ended up with about 40 Mississippi pins because I traded my ESPN Zone card to some guy from Mississippi who didn't care about pin trading.  I was gonna use those pins as extra firepower to get all the states, but some states didn't bring pins.  So now I have two almost-complete pin collections from '02 and '03.

[quote="probability1.01"]I think pin-trading is too tiring, scrambling about trying to find those possessing the rarest pins.  I propose some strategies:

1.  Work as a team so that no extra pins will be leaking out, ensuring the value of your team's pins.

2.  Don't trade until later to raise demand.

3.  Forget pin-trading... I'd rather eat my meal peacefully and perhaps get a few exchanges while I'm at it.[/quote]
lol, what i did last year was just the opposite. i traded away 4 of my state's pins at once, and getting many extras for other states. of course, then my team got really pissed at me and stole from me. You know who you are....... :evil:

I believe that the purpose of pin-trading is to get to know the other guys.  It's not like we actually hold value in the pins (although knowing that we're math geeks, maybe we do...).  That's why this year I'm not even going to try (ok, I won't be able to resist temptation to do it a little).  I'll mostly chat w/ the other people around.


Nice problem, my friend  :) . Using the lenma: $ \frac {1}{x^2} \plus{} \frac {1}{y^2} \plus{} \frac {1}{z^2} \geq\ \frac {27}{(x \plus{} y \plus{} z)^2}$
And I think the stronger is not trues  :maybe: (but I haven't checked it yet)

[quote="nguoivn"] Using the lenma: $ \frac {1}{x^2} \plus{} \frac {1}{y^2} \plus{} \frac {1}{z^2} \geq\ \frac {27}{(x \plus{} y \plus{} z)^2}$[/quote]

I would like to see how you've proceeded then.

My proof is based on the fact that : $ \frac {2}{3(a \plus{} b)^2} \plus{} \frac {1}{3(b \plus{} c)^2} \geq \frac {9}{(c \plus{} 3b \plus{} 2a)^2}$ ; we sum cyclically and we get the desired result.

[quote="nguoivn"]And I think the stronger is not trues  :maybe: (but I haven't checked it yet)[/quote]
However,I can't find a counterexample and It seems true :maybe:

[quote="Evariste-Galois"]hello " rachid" I think that this lemme can be used like that:

$ \frac {1}{x^2} \plus{} \frac {1}{y^2} \plus{} \frac {1}{y^2}\geq\frac {27}{(x \plus{} 2y)^2}$
and we sum cyclically and we get
$ \sum\frac {1}{x^2}\geq\sum{\frac {9}{(x \plus{} 2y)^2}}$
and let $ a \plus{} b \equal{} x$ and $ b \plus{} c \equal{} y$ and  $ c \plus{} a \equal{} z$
we get the result
thank you for the lemma[/quote]
I don't think so my friend,I guess that nguoivn's Idea is different than mine,since what you've done is also my Idea posted before  :wink:  :)

Now let's try to prove/disprove the stronger:

[quote="rachid"]I think that the following inequality is also true:

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

for $ a,b,c \geq 0$[/quote]

[quote="iqtimo"][quote="rachid"]Let $ a,b,c\geq 0$.Prove that:

$ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \geq \frac {9}{(a \plus{} 3b \plus{} 2c)^2} \plus{} \frac {9}{(b \plus{} 3c \plus{} 2a)^2} \plus{} \frac {9}{(c \plus{} 3a \plus{} 2b)^2}$

[hide]
I think that the following inequality is also true:

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

for $ a,b,c \geq 0$[/hide][/quote]
using :
$ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(b \plus{} c)^2} \geq \frac {27}{(a \plus{} 3b \plus{} 2c)^2}$[/quote]
Yes, that's thing I want to mention  :)

The second inequality is false . Try $ a\equal{}b\equal{}c\rightarrow 0$ .

[quote="alex2008"]The second inequality is false . Try $ a \equal{} b \equal{} c\rightarrow 0$ .[/quote]
You're wrong alex2008  :) .[b]However[/b],remind that no two of the variables are equal to zero.

[quote="FelixD"]But they are not!![/quote]
I know,it's just for precision.

Just see that for $ a \equal{} b \equal{} c$ we get the equality case,wich contradicts what alex2008 says.

Do you have any counterexample ?

[quote="Evariste-Galois"]Big mistake "Rachid"  :wink:  :wink:[/quote]
Be sure of what you write before posting anything  :wink: :D

:rotfl:  :rotfl:  what do you say !!!!!!

I sorry for starting this "quarell" :( . It is my bad for miscalculating something  :D  . When $ a\equal{}b\equal{}c$ we have equality case , rachid was right . And when i said $ \rightarrow 0$ i didn't reffered one of the variables is zhero , i just reffered it is very close to zhero , but even so the inequality is true . Now , i think that rachid's inequality is true  :)

[quote="rachid"]
[quote="rachid"]I think that the following inequality is also true:

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

for $ a,b,c \geq 0$[/quote][/quote]
Yes, it's trues and very nice  :lol: . My solution is only Am-Gm and it's based on: $ 3(a \plus{} b \plus{} c)^2 \geq\ 2\sum\ a^2 \plus{} 7\sum\ ab$
 @rachid: Can you give me your name? Indeed, I really want to have this problem in our book, And we want to write exactly the author's name. Thank you very much  :)

BTW,I've proved that the following inequality is true :

$ \frac {1}{(a \plus{} 3b \plus{} 2c)^2} \plus{} \frac {1}{(b \plus{} 3c \plus{} 2a)^2} \plus{} \frac {1}{(c \plus{} 3a \plus{} 2b)^2}$    $ \geq$  $ \frac {9}{8(a^2\plus{}b^2\plus{}c^2) \plus{} 28(ab\plus{}bc\plus{}ca)}$

for $ a,b,c \geq 0$

[quote="nguoivn"]
[quote="rachid"]I think that the following inequality is also true:

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

for $ a,b,c \geq 0$[/quote]
Yes, it's trues and very nice  :lol: . My solution is only Am-Gm and it's based on: $ 3(a \plus{} b \plus{} c)^2 \geq\ 2\sum\ a^2 \plus{} 7\sum\ ab$[/quote]
I would like to see your AM-GM's proof  :)

@nguoivn : see your PM :)

Yes, I sent it to you. How do you feel with my proof?  :)

Very nice and elementary proof,nguoivn  :lol: .Thank you !

I'll respect your want to not reveal the proof for now.

Now, I can post it ^^!

The first, consider the case $ a \le b \le c$, using Am-Gm Inequalily with the form $ \dfrac{1}{(x \plus{} y)^2} \le \dfrac{1}{4xy},$ we have $ \frac {1}{{{{(3a \plus{} 2b \plus{} c)}^2}}} \le \frac {1}{{4(a \plus{} 2b)(c \plus{} 2a)}}.$
So, we only need to prove that $ \frac {1}{{(a \plus{} 2b)(c \plus{} 2a)}} \plus{} \frac {1}{{(b \plus{} 2c)(a \plus{} 2b)}} \plus{} \frac {1}{{(c \plus{} 2a)(b \plus{} 2c)}} \le \frac {1}{{ab \plus{} bc \plus{} ca}},$

Or $ (a \plus{} 2b)(b \plus{} 2c)(c \plus{} 2a) \ge 3(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca).$
Equivalent to  $ (a \minus{} b)(b \minus{} c)(c \minus{} a) \ge 0$ (obviously trues with $ a \le b \le c$).
Consider the last case is $ a \ge b \ge c$.
We also use Am-Gm but with another hint like the following $ \dfrac{1}{(3a \plus{} 2b \plus{} c)^2} \le \dfrac{1}{4(a \plus{} b \plus{} c)(2a \plus{} b)}$.
With this estimation, easily too see that the proof'll complete if we can show that $ \frac {1}{{a \plus{} b \plus{} c}}\left( {\frac {1}{{2a \plus{} b}} \plus{} \frac {1}{{2b \plus{} c}} \plus{} \frac {1}{{2c \plus{} a}}} \right) \le \frac {1}{{ab \plus{} bc \plus{} ca}},$

Equivalent to $ (2a \plus{} b)(2b \plus{} c)(2c \plus{} a)(a \plus{} b \plus{} c) \ge (ab \plus{} bc \plus{} ca)[2({a^2} \plus{} {b^2} \plus{} {c^2}) \plus{} 7(ab \plus{} bc \plus{} ca)].$

Because $ a \ge b \ge c$, so $ (2a \plus{} b)(2b \plus{} c)(2c \plus{} a) \ge 3(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca),$
And we bring the problem to prove the stronger $ 3{(a \plus{} b \plus{} c)^2} \ge 2({a^2} \plus{} {b^2} \plus{} {c^2}) \plus{} 7(ab \plus{} bc \plus{} ca).$
But this last ineq is obviously trues by Am-Gm again. We have done!
Equality occurs if and only $ a \equal{} b \equal{} c.$
PS: It's also a nice problem in our first book "Bat dang thuc va nhung loi giai hay" (V.Q.B.Can - T.Q.Anh, Ha Noi Publishing House 2009)  :)
Thanks again Rachid Benchlikha - the author of this very nice problem  :)


function
limit
integration
real analysis
real analysis unsolved

[quote="Slizzel"]Let $n\in\mathbb{N}, \ n\geq 2$  and the function $f_{n}(x)=\sum_{k=0}^{n}{|x-k|}$ and $F_{n}: \mathbb{R}\to{R}$ a primitive of $f_{n}$. Compute:

(a) $\lim_{n\to\infty}\frac{F_{n}(1)-F_{n}(0)}{n^{2}}$

(b) $\lim_{x\to\infty}\frac{F_{n}(x)-F_{n}(0)}{x^{2}}$

Since $f_{n}$ is continous using Leiniz-Newton formula we have (a) $\lim_{n\to\infty}\frac{F_{n}(1)-F_{n}(0)}{n^{2}}=\lim_{n\to\infty}\int_{0}^{1}f(x)dx$ right? and for (b) same thing... Well i don't know if this helps  :blush:[/quote]
For a):
Use Stolz's Theorem twice:
$\lim_{n\to\infty}\frac{F_{n}(1)-F_{n}(0)}{n^{2}}=\lim_{n\to\infty}\frac{\int_{0}^{1}f_{n}(x)dx}{n^{2}}=\lim_{n\to\infty}\frac{\int_{0}^{1}f_{n}(x)-f_{n-1}(x)dx}{2n-1}$
$=\lim_{n\to\infty}\frac{\int_{0}^{1}|x-n|dx}{2n-1}$
$=\lim_{n\to\infty}\frac{\int_{0}^{1}|x-n|-|x-n+1|dx}{2}=\frac{1}{2}$

For b),Use L'hosptial's Rule:
$\lim_{x\to\infty}\frac{F_{n}(x)-F_{n}(0)}{x^{2}}=\lim_{x\to\infty}\frac{\int_{0}^{x}f_{n}(t)dt}{x^{2}}=\lim_{x\to\infty}\frac{f_{n}(x)}{2x}$
$=\lim_{x\to \infty}\frac{\sum_{k=0}^{n}{|x-k|}}{2x}=\frac{n+1}{2}$
For why we can use the two theorems is obvious,I think.


geometry
conics
parabola
calculus
integration
algebra proposed
algebra

There is a certain parabola, which opens up, that has no unit lattice points lying on it, and it is increasing
on 1 < x < 5.  Seven unit lattice points, grand total, lie (below it and above the x-axis) for 1 < x < 5.  These seven internal points lie on the lines x = 2, x = 3, and x = 4.  Three of the seven points are vertically below the parabola a distance of 0.1 unit apiece.

Find the area, to the nearest half of a square unit, for the region bounded by the curve/lines:  the certain parabola, the x-axis, x = 1, and x = 5.

Here's a solution without integral calculus.

Under the conditions, the seven points would have to be these:

(2,1), (3,1), (3,2), (4,1), (4,2), (4,3), (4,4)

Then the three points that can be 0.1 unit below (apiece) from the parabola are (2,1), (3,2), and (4,4), making (2,1.1), (3,2.1),
and (4,4.1) known points on the parabola.   A QuadReg selection on a TI series graphics model calculator (for instance)
applied to these gives an equation of y = 0.5x^2 - 1.5x + 2.1

For the endpoints, when x = 1, y = 1.1, and when x = 5, y = 7.1

I will use Pick's Theorem:  (1/2)b + i - 1, where b is the # of
boundary points and i is the # of interior points of a polygon, where
it is on a grid of unit lattice points (in this case).

Pick's Theorem is meant for use on polygons, so I am using the section of the parabola where it is relatively close to unit lattice points, together with x = 1, x = 5, and y = 0 for the polygon.

All the 16 boundary points are:  (1,0), (2,0), (3,0), (4,0), (5,0), (1,1), (2,1), (3,2), (4,4), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (5,7)

All of the 4 interior points are:  (3,1), (4,1), (4,2), (4,3)

(1/2)16 + 4 - 1 = 11

11 square units........Answer

If I were to check by integrating  (0.5x^2 - 1.5x + 2.1)dx from
x = 1 to x = 5, I would get

approximately  11.067


The "Inverse" function is defined recursively as follows:
The nth number is the inverse of the sum of the previous n-1 terms.

Does the sum of the elements in this infinite sequence converge?  If so, what to?

Please explicate what you mean by "the inverse of the previous n-1 terms"

Sorry.  I meant inverse of the sum.  I changed it.

[hide="Solution"]We will prove that the series diverges.
Let the terms of the sequence be $a_{1},a_{2},\ldots$ and $s_{n}: =a_{1}+\ldots+a_{n}$. We have $a_{n}= \frac{1}{s_{n-1}}$, so $s_{n}= s_{n-1}+a_{n}= s_{n-1}+\frac{1}{s_{n-1}}$. Let us assume that $\left\{s_{n}\right\}_{n=1}^{\infty}$ converges and $s: =\lim_{n\to\infty}s_{n}$. Then $s=s+\frac{1}{s}$, so $\frac{1}{s}=0$ - a contradiction. Thus the sequence $\left\{s_{n}\right\}_{n=1}^{\infty}$ diverges and the series $\sum_{n=1}^{\infty}a_{n}$ also diverges.[/hide]

The Inverse-Inverse function is defined as follows: the nth term is the inverse of the sum of the inverses of the previous terms in the sequence.  Does this sum converge?  If so, what to?

[hide="Solution"]
Consider the "Inverse-Inverse-Inverse" function or the sequence whose terms ore the reciprocals of the corresponding term in this sequence.  The nth term of that sequence is the sum of the previous n-1 terms.  Therefore, the sum of the first n elements is half the sum of the first n+1 elements.  If the sum doubles each term, each term after the 2nd one must be twice the previous one.  This means the sequence goes 1, 1, 2, 4, 8, 16....  The inverse of this is $1, 1, \frac{1}{2}, \frac{1}{4}...$ which as everybody knows, converges to 3.
[/hide]


sin2x = tgx - cos2x

What's the value of six-cosx?
Please help me

$ \sin 2x \equal{} \mathrm{tg} x \minus{} \cos 2x\Leftrightarrow$ $ \sin x \minus{} \cos x \plus{} 2\sin^2 x\cos x \minus{} 2\sin x\cos^2 x \equal{} 0\Leftrightarrow$ $ (\sin x \minus{} \cos x)(1 \plus{} 2\sin x \cos x)\equal{}0\Leftrightarrow \sin x \minus{} \cos x \equal{} 0\vee 1 \plus{} 2\sin x \cos x \equal{} 0$

$ 1 \plus{} 2\sin x \cos x \equal{} 0\Leftrightarrow 2\sin x\cos x \equal{} \minus{} 1\Leftrightarrow$ $ \cos^2 x \minus{} 2\sin x \cos x \plus{} \sin^2 x \equal{} 2\Leftrightarrow |\sin x \minus{} \cos x| \equal{} \sqrt {2}\Leftrightarrow$ $ \sin x \minus{} \cos x\in \{ \minus{} \sqrt {2}\ ,\ 0\ ,\ \sqrt {2}\}$


If $ a_1,a_2,\cdots,a_n$ are real positive numbers , prove that:

\[ \sqrt{\frac{a^2_1\plus{}a^2_2\plus{}\cdots\plus{}a^2_n}{n}} \ge \frac{a_1\plus{}a_2\plus{}\cdots\plus{}a_n}{n}\]

[quote="Obel1x"]If $ a_1,a_2,\cdots,a_n$ are real positive numbers , prove that:
\[ \sqrt {\frac {a^2_1 \plus{} a^2_2 \plus{} \cdots \plus{} a^2_n}{n}} \ge \frac {a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n}{n}\]
[/quote]

this ineq holds for any [b]reals[/b] $ a_i$.

by CS, we have

$ n(a^2_1 \plus{} a^2_2 \plus{} \cdots \plus{} a^2_n)$

$ \equal{}(1\plus{}1\plus{}...\plus{}1)(a^2_1 \plus{} a^2_2 \plus{} \cdots \plus{} a^2_n)$

$ \ge (a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)^2$

so we get

$ \sqrt {\frac {a^2_1 \plus{} a^2_2 \plus{} \cdots \plus{} a^2_n}{n}}$

$ \ge \frac {\left| a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \right|}{n}$

$ \ge \frac {a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n}{n}.$

Function $ f(k)=\left(\frac{a_1 ^k +...+a_n ^k}{n}^\right)^{\frac{1}{k}}$ is increasing since $ f'(k)\geq 0$ ,then the conclusion follows .


SMO 2004 Open #24

Find the number of ways to tile a $5 \times 2$ grid with blue $1 \times 1$ tiles, red $2 \times 1$ tiles and green $2 \times 2$ tiles.

Hint: Use recursion.

[hide="Probably Wrong"]
Call the number of ways to tile an $n\times 2$ rectangle $T_{n}$.  There are six ways to cover the $n$th column: two blue tiles, a red tile, half of two red tiles, half a green tile, and half a red tile and a blue tile in two different configurations.

The first two can be obtained from an $n-1\times 2$ rectangle tiling, and the last four from an $n-2\times 2$ tiling.  Thus $T_{n}=2T_{n-1}+4T_{n-2}$.

We can easily see that $T_{1}=2$ and $T_{2}=8$.  Thus $T_{3}=24$, $T_{4}=80$, and $T_{5}=256$.
[/hide]

It's greater because for the last two configurations, there's another space which could be filled by a blue tile (which it looks like you assume), or a red tile.
[hide="if I did not mess up..."]
Let $a_{n}$ be the number of ways to tile an $n\times 2$ grid, and $b_{n}$ be the number of ways to tile an $n\times 2$ grid with a corner removed.

For a recursion for $b_{n}$, note that green tiles can't fill in the corner adjacent to the removed one. If it is filled blue, then $a_{n-1}$ ways to fill in the rest, and if filled red, $b_{n-1}$ ways to fill in the rest. So we get
\[b_{n}=a_{n-1}+b_{n-1}\]
Now for a recursion for $a_{n}$, if we align the grid such that there are 2 rows, consider what space fills in the upper right corner. If blue, there are $b_{n}$ ways. If green, there are $a_{n-2}$ ways. If red, either the red tile is horizontal or vertical. If vertical, there are $a_{n-1}$ ways. If horizontal, the lower right corner can be blue for $b_{n-1}$ ways, or red for $a_{n-2}$ ways. This means
\[a_{n}=b_{n}+a_{n-1}+b_{n-1}+2a_{n-2}=2a_{n-1}+2a_{n-2}+2b_{n-1}\]
Then $b_{n-1}=\frac{a_{n}}{2}-a_{n-1}-a_{n-2}$, so $b_{n-2}=\frac{a_{n-1}}{2}-a_{n-2}-a_{n-3}$. Subtracting,
\[a_{n-2}=\frac{a_{n}}{2}-\frac{3a_{n-1}}{2}+a_{n-3}\]
\[\Rightarrow a_{n}=3a_{n-1}+2a_{n-2}-2a_{n-3}\]
We find $a_{1}=2$ and $a_{2}=8$. Also, $b_{1}=1$, so for purposes of applying the recursive formula, $a_{0}=1$. Now we just compute:
\[a_{3}=26,\ a_{4}=90,\ a_{5}=\boxed{306}\]
[/hide]

I think Scorpius119 has it  :)  Very good consideration about the corner being removed there.


induction
number theory proposed
number theory

Let $A$ be a set of positive integers such that
a) if $a \in A$, the all the positive divisors of $a$are also in $A$;
b) if $a,b \in A$, with $1<a<b$, then $1+ab \in A$.
Prove that if $A$ has at least 3 elements, then $A$ is the set of all positive integers.

hi Moroccan citizen how re u?
u re from ??
i m from rabat! i ve passed the IMOs USA2001 and UK2002
so about your problem:
BY INDUCTION :roll:
if we assume that,giving a n>=5, for all k between 1 and n : k is in A!!!!
{
if n is  odd
let's show that n+2 is in A
n(1+k)+1=k(n+2)
with k=(n+1)/2#n (coz n>3)
(k+1) in A
and n in A
so k(n+2) in A and finally n+2 in A
n(n+2)+1=(n+1)² in A so n+1 in A
}
else n is even=2k
and n+1=2k+1 in A if k>2 (n>=5)
Now!!
for the start values:
1 in A (divisor of an element in A)
2 in A
3 in A
so 7=(2*3)+1 in A
2*7+1=15 in A
and 5 in A
(3*5)+1=16 so 4 in A
we may start with {1,2,3,4,5}

It is the same problem with Romanian Junior Balkan TSTs 2004, Test 5, problem 3, posted here: http://www.mathlinks.ro/Forum/viewtopic.php?t=5892
Continue discussion there.


integration
logarithms
Putnam
limit
trigonometry
real analysis
LaTeX

[color=blue][i]Στο χώρο αυτό οι φοιτητές μας έχουν τη δυνατότητα να ανταλάσσουν ιδέες , προβλήματα  ή και να στέλνουν

τις απορίες ή τις λύσεις τους σε προτεινόμενες ασκήσεις άλλων φοιτητών.

 Με αυτό τον τρόπο , όλα τα θέματα θα είναι συγκεντρωμένα και θα είναι εύκολη η πρόσβαση, χωρίς χρονοτριβή.[/i][/color]

[b] Καλές εμπνεύσεις !!![/b]

[color=red][b]ΑΣΚΗΣΗ 1[/b][/color]

Να υπολογιστεί το γενικευμένο $ \displaystyle\int^1_0 \frac {lnx}{1 \plus{} x}\,dx$
A) με μιγαδική ανάλυση β) με σειρές Fourier γ) με τίποτε από τα δυό προηγούμενα . Αν υπάρξει ενδιαφέρον  θα δώσω λύση γισ την γ) περίπτωση σε λίγες μέρες

r_boris, μπορεις να γραψεις την λυση τωρα?  :roll:

Mηηηη, όχι λύση, ακόμα την παλεύουμε εδώ πέρα!!!

Καλά, προφανώς αστειεύομαι, έτσι? :) :) :) Πάντως, αν είναι να μπει η λύση από τώρα, γίνεται να μπει σε spoiler ή σε attachment για να μην φαίνεται με το που μπαίνει κάποιος στο topic?

Προφανώς κάνω κάτι λάθος γιατί με έναν τρόπο βγάζω $ \minus{}\frac{\pi^2}{2}$ και με άλλον τρόπο βγάζω $ \minus{}\frac{\pi^2}{12}$ (που το εμπιστεύομαι πιο πολύ γιατί έχει λιγότερες πράξεις).

Cheerio,

Durandal 1707

[quote="Durandal"] και με άλλον τρόπο βγάζω $ \minus{} \frac {\pi^2}{12}$ (που το εμπιστεύομαι πιο πολύ γιατί έχει λιγότερες πράξεις).
[/quote]

Καλά κάνεις και το εμπιστεύεσαι  :wink:

Σωστό είναι το $ \minus{}\frac{\pi^2}{12}$ αλλά πως? γιατί πιστεύω πως είναι θαυμάσια απόδειξη

r_boris, μπορεις να γραψεις την λυση σου και να την βαλεις σε spoiler οπως ειπε και ο Durandal. Δηλαδη αναμεσα σε tags  [hide][/hide]

Heh, Silouan thanks για την ψήφο εμπιστοσύνης :) Εσύ πώς το έλυσες και ήσουν σίγουρος για το αποτέλεσμα?

Οι δύο δικοί μου τρόποι είναι οι εξής:

i. Brute force μιγαδική ανάλυση (r_boris, ο τρόπος μου είναι μάλλον άκομψος οπότε, αν μπορείς, γράφεις και τη δική σου μιγαδική λύση στο τέλος?)
[hide]$ \int_0^1 \frac {\log x}{x \plus{} 1} \textrm{d}x \equal{} \minus{} \int_0^\infty \frac {s}{1 \plus{} e^s} \textrm{d}s \equal{} \minus{} \int_{ \minus{} \infty}^\infty \frac {e^u}{1 \plus{} e^{e^u}}\textrm{d}u$
Στο τελευταίο ολοκλήρωμα έχουμε απλό πόλο στο $ \log\pi \plus{} i\frac {\pi}{2}$, υπολογίζουμε το υπόλοιπο, παίρνουμε την κλασική ημικυκλική διαδρομή, αγκομαχάμε λίγο για να βγάλουμε τα φράγματα στο ημικύκλιο, κάνουμε πολλάααααααα λάθη στις πράξεις, και τελικά ελπίζουμε ότι έχουμε βγάλει σωστά $ \minus{} \frac {\pi^2}{12}$.

Αν τώρα κάπου έχω κάνει λάθος στις πράξεις και απλώς οδηγούσα το χέρι μου εκεί υποσυνείδητα, apologies... οι πράξεις κι εγώ είμαστε σε διάσταση τα τελευταία 15 χρόνια ;)[/hide]

ii. Με διπλή ολοκλήρωση
[hide]Ξεκινάμε με την παρατήρηση: $ \int_0^1 \frac {\log x}{x \plus{} 1} \textrm{d}x \equal{} \log x\cdot \log(x \plus{} 1)\Big|_0^1 \minus{} \int_0^1 \frac {\log(x \plus{} 1)}{x} \textrm{d}x$.
Τώρα, έχουμε επίσης $ \frac {\log(x \plus{} 1)}{x} \equal{} \int_0^1 \frac {1}{1 \plus{} xy} \textrm{d}y$, οπότε το δεύτερο ολοκλήρωμα θα γράφεται: $ \int_0^1\int_0^1 \frac {1}{1 \plus{} xy} \textrm{d}y\,\textrm{d}x$. :!:  Αυτό είναι ωραία μορφή επειδή $ \frac {1}{1 \plus{} xy} \equal{} 1 \minus{} xy \plus{} (xy)^2 \plus{} \cdots$ και, κατά συνέπεια:
$ \int_0^1\int_0^1\frac {1}{1 \plus{} xy} \textrm{d}x\,\textrm{d}y \equal{} \sum_{n \equal{} 0}^\infty\int_0^1\int_0^1 ( \minus{} 1)^n(xy)^n \textrm{d}x\,\textrm{d}y \equal{} \minus{} \sum_{n \equal{} 1}^\infty \frac {( \minus{} 1)^n}{n^2}$.

Όμως $ \sum_{n \equal{} 1}^\infty \frac {1}{n^2} \equal{} \frac {\pi^2}{6}$ οπότε λίγα rearrangements δίνουν $ \sum_{n \equal{} 1}^\infty \frac {( \minus{} 1)^n}{n^2} \equal{} \minus{} \frac {\pi^2}{12}$

Ta daaah...    :)

r_boris, απλώς πανέμορφη άσκηση! Αν δεν είχα δει παλιά έναν υπολογισμό του $ \zeta(2)$ με ένα διπλό ολοκλήρωμα, δε θα σκεφτόμουν ποτέ των ποτών τα παραπάνω!
[/hide]

Υπέροχη άσκηση!

Cheerio,

Durandal 1707

Οντως τελεια ασκηση και τελεια λυση του Durandal  :)

Να ρωτισω ομως κατι. Γιατι εχεις το logx και οχι lnx? Και πως προκυπτει οτι $ \frac{log(x\plus{}1)}{x}\equal{}\int_0^1 \frac{1}{1\plus{}xy}dy$? Επισης το αρχικο $ logxlog(x\plus{}1)$ που πηγε?

[quote="mailo"]Να ρωτισω ομως κατι. Γιατι εχεις το logx και οχι lnx? Και πως προκυπτει οτι $ \frac {log(x \plus{} 1)}{x} \equal{} \int_0^1 \frac {1}{1 \plus{} xy}dy$? Επισης το αρχικο $ logxlog(x \plus{} 1)$ που πηγε?[/quote]

$ \log x, \ln x$, θέμα συνήθειας ;)

Το ολοκλήρωμα που ρωτάς, είναι απλώς ένας υπολογισμός: θεώρησε το $ x$ σαν σταθερά (αφού ολοκληρώνεις ως προς $ y$) και βγαίνει.

το $ \log x \log(x\plus{}1)$ είναι $ 0$ για $ x\equal{}1$ και τείνει στο $ 0$ καθώς $ x\to 0$.

Cheerio,

Durandal 1707

[quote="Durandal"]

Το ολοκλήρωμα που ρωτάς, είναι απλώς ένας υπολογισμός: θεώρησε το $ x$ σαν σταθερά (αφού ολοκληρώνεις ως προς $ y$) και βγαίνει.
[/quote]

Μου βγαινει $ \int_0^1 \frac {1}{1 \plus{} xy}dy\equal{}ln(x\plus{}1)$  :roll:

[size=150][b][color=red]ΑΣΚΗΣΗ 2[/color][/b][/size]

Να υπολογιστεί το: $ \int_{ \minus{} \infty}^\infty\cdots\int_{ \minus{} \infty}^\infty \frac {\textrm{d}x_1\cdots \textrm{d}x_n}{(1 \plus{} x_1^2 \plus{} \cdots \plus{} x_n^2)^\frac {n \plus{} 1}{2}}$

Cheerio,

Durandal 1707

PS: Mailo, πρόσεχε όταν παραγωγίζεις $ \frac{d}{dy}\log(1 \plus{} xy) \equal{} \frac {x}{1 \plus{} xy}$.

Η παράγουσα του  $ \frac {1}{1 \plus{} xy}$  είναι η $ \frac {1}{x}\ln(1 \plus{} xy)$ εφόσον παραγωγίζουμε ως προς y  και το χ θεωρείται  σταθερά.

Παναγιώτη με πρόλαβες ... :lol:

Να κανω μια ερωτηση. Οταν εχουμε μια ρητη συναρτηση $ F(x)\equal{}\frac{f(x)}{g(x)}$ της οποια ο βαθμος του παρονομαστης $ g(x)$ ειναι μεγαλυτερος του βαθμου του αριθμητη $ f(x)$ και ειναι δευτερου βαθμου με αρνητικη διακρινουσα, τοτε πως αυτη η συναρτηση αναλυεται σε αθροισμα απλων κλασματων?

Αν εχουμε θετικη διακρινουσα τοτε ειναι απλα τα πραγματα, βρισκουμε τις ριζες $ r_1,r_2$ του $ g(x)$ και ετσι η $ F(x)$ γινεται $ F(x)\equal{}\frac{f(x)}{g(x)}\equal{}\frac{f(x)}{a(x\minus{}r_1)(x\minus{}r_2)}\equal{}\frac{A}{x\minus{}r_1}\plus{}\frac{B}{x\minus{}r_2}$

Ummm.. μάλλον elementary μου φαίνεται.

[hide]Θέτεις $ t\equal{}e^{\minus{}u^2}\minus{}1$ στο δεύτερο ολοκλήρωμα και εναλλάσσεις τη σειρά της ολοκλήρωσης στο διπλό ολοκλήρωμα που προκύπτει.

Αν δεν κάνω λάθος, το τελικό αποτέλεσμα είναι $ \minus{}\frac{1}{2e}$? Ή κάτι τέτοιο?[/hide]

Χάνω κάτι?

Cheerio,

Durandal 1707

[quote="AlexandrosG"][b]Ασκηση 4[/b]

Αν $ F_n$ ειναι η ακολουθια Fibonacci με $ F_0 \equal{} 1$ , $ F_1 \equal{} 1$ , $ F_{n \plus{} 1} \equal{} F_n \plus{} F_{n \minus{} 1}$ να αποδειξετε οτι

$ F_{m \plus{} n \plus{} 1} \equal{} F_{m \plus{} 1}\cdot F_{n \plus{} 1} \plus{} F_m\cdot F_n$  m,n φυσικοι

Ενδιαφερον θα ειχε μια λυση με επαγωγη ως προς m και n. Η λυση που ειχε εκει που βρηκα το προβλημα το κανει καταλληλο για το topic και ειναι αρκετα ομορφη και συντομη.

Εκτος απο αυτη ειναι μερες αλυτες και οι ασκησεις 2 και 3. Οποιος μπορει ας βαλει τις λυσεις τους.   :)[/quote]

Βγαίνει και με επαγωγή στο $ n$. Για $ n\equal{}1$ το δοσμένο εύκολα ισχύει για κάθε $ m$. Εστω ότι ισχύει γαι $ n$, θα αποδείξουμε ότι ισχύει και για $ n\plus{}1$.

$ F_{m\plus{}n\plus{}1\plus{}1}\equal{}F_{m\plus{}1\plus{}n\plus{}1}\equal{}F_{m\plus{}2}\cdot F_{n\plus{}1}\plus{}F_{m\plus{}1}\cdot F_n\equal{}$

$ \equal{}(F_m\plus{}F_{m\plus{}1})F_{n\plus{}1}\plus{}F_{m\plus{}1}(F_{n\plus{}2}\minus{}F_{n\plus{}1})\equal{}$

$ \equal{}F_{m\plus{}1}\cdot F_{n\plus{}2}\plus{}F_m\cdot F_{n\plus{}1}$

Q.E.D.

[quote="mostel"]Nαι, elementary είναι ο τρόπος λύσης, απλώς νόμιζα πως ο τρόπος με τον όποιο το κατασκεύασα είναι καλά κρυμμένος (την τύφλα μου :P)

Η ιδέα μου ήταν ότι:

Αρχικά ισχύει:

$ \int_{0}^{x}f(t)dt \plus{} \int_{0}^{f(x)}f^{ \minus{} 1}(t)dt \equal{} x\cdot f(x)$

Όλα αυτά θεωρώντας ως συνάρτηση την :

$ f(x) \equal{} e^{ \minus{} x^2} \minus{} 1$

όπου guess ποια είναι η αντίστροφη και πως απλουστεύει αυτό τη λύση του προβλήματος!

Μάλλον φαιά ουσία, μιας και μου πήρε αρκετή ώρα να τη φτιάξω.

Στέλιος[/quote]
Μια μικρη διόρθωση στον τύπο $ \int_{0}^{x}f(t)dt \plus{} \int_{f(0)}^{f(x)}f^{ \minus{} 1}(t)dt \equal{} x\cdot f(x)$ Εδώ βέβαια  είναι $ f(0)\equal{}0$ και πιο γενικά $ \int_{y}^{x}f(t)dt \plus{} \int_{f(y)}^{f(x)}f^{ \minus{} 1}(t)dt \equal{} x\cdot f(x)\minus{}y\cdot f(y)$

Βαζω τη λυση που ελεγα

Εστω M={{1,1},{1,0}}

Με επαγωγη δειχνουμε οτι $ M^n$={{$ F_{n \plus{} 1}$,$ F_n$},{$ F_n$,$ F_{n \minus{} 1}$}}
Τωρα απο την ιδιοτητα $ M^{m \plus{} n}$=$ M^n*M^m \equal{} M^m*M^n$ γραμμενη σε μορφη πινακων και απο το πανω αριστερα στοιχειο
παιρνουμε την  $ F_{m \plus{} n \plus{} 1} \equal{} F_{m \plus{} 1}*F_{n \plus{} 1} \plus{} F_m*F_n$

εστω α άρητος και το σύνολο Α={ma+n|m,n ανήκουν στο Z}.
νδο το Α είναι πυκνό στο R

[b][color=red][size=150]ΑΣΚΗΣΗ 5[/size][/color][/b] (Dimitris T)

Αν $ a$ άρρητος, νδο το σύνολο $ A \equal{} \{ma \plus{} n: m,n\in\mathbb{Z}\}$ είναι πυκνό στο $ \mathbb{R}$.

Κλείνω την απόδειξη σε spoiler για να μην τη χαλάσω σε όσους τη δουλεύουν ακόμα :)

[hide]Κατ'αρχάς, θεωρώντας μόνο το δεκαδικό μέρος, αρκεί να δείξουμε ότι το σύνολο $ \{m a: m\in\mathbb{Z}\}$ είναι πυκνό στο $ [0,1]$. Ή, πιο κομψά, δουλεύοντας $ \mod \mathbb{Z}$, αρκεί να δείξουμε ότι το σύνολο $ C \equal{} \{e^{2\pi i m a}: m\in\mathbb{Z}\}$ είναι πυκνό στον μοναδιαίο κύκλο $ S^1 \equal{} \mathbb{R}/\mathbb{Z}$.

Προφανώς, εφόσον ο $ a$ είναι άρρητος, το σύνολο $ C$ θα είναι άπειρο και άρα θα έχει κάποιο οριακό σημείο $ \xi$ στο συμπαγή κύκλο $ C$. Τότε, αν $ \epsilon > 0$ θα υπάρχουν $ n_{1,2}$ με $ n_1\neq n_2$ ώστε $ |e^{2\pi i n_{1(2)} a} \minus{} \xi| < \frac {\epsilon}{2}\implies 0 < |e^{2\pi i n_1 a} \minus{} e^{2\pi i n_2 a}| < \epsilon$ (από εφαρμογή τριγωνικής ανισότητας; θυμηθείτε και ότι ο $ a$ είναι άρρητος). Αυτό όμως δείχνει ότι για $ k \equal{} n_2 \minus{} n_1$, θα έχουμε $ 0 < |e^{2\pi i k a} \minus{} 1| < \epsilon$ (δηλαδή και το $ 1$ είναι οριακό σημείο του $ C$).

Tώρα, έστω $ \theta \equal{} 2k\pi a$ οπότε $ 0 < |e^{i\theta} \minus{} 1| < \epsilon$. Έτσι, αρκεί να δείξουμε ότι για κάθε $ w\in S^1$ υπάρχει κάποιο $ n\in\mathbb{Z}$ ώστε $ |e^{i n\theta} \minus{} w| < \epsilon$. Αυτό όμως είναι απλό να το δει κανείς: οι αριθμοί $ e^{i\theta}, e^{2i\theta}\ldots e^{i n\theta}\ldots$ προχωράνε καλύπτοντας σιγά-σιγά τον κύκλο με βήματα $ < \epsilon$. Άρα, για κάποιο $ n$, θα βρεθούν σε απόσταση το πολύ $ \epsilon$ από το $ w$, QED.

Note: Το παραπάνω είναι για να καταλαβαίνουμε τι γίνεται. Για όσους θέλουν μία πιο "αποστειρωμένη" διατύπωση, αυτό που λέω είναι ότι οι περιοχές $ \{z: |z \minus{} e^{in\theta}| < \epsilon\}, n\in\mathbb{Z}$ αποτελούν μία ανοιχτή κάλυψη της συμπαγούς $ S^1$ και τα υπόλοιπα κυλάνε ομαλά :) [/hide]

Ωραίο πρόβλημα :) Επιτέλους, τώρα σιγά-σιγά ξαναβρίσκω λίγο χρόνο για το forum...

Cheerio,

Durandal 1707

Poli oraia,
Ti lete gia ena bima parapera :

An $ a$ arritos kai $ P(x)\equal{}a\cdot x^n \plus{}a_{n\minus{}1}\cdot x^{n\minus{}1}\plus{}\ldots\plus{}a_0$ einai pragmatiko polionimo,  n.d.o to sinolo $ A\equal{}\{P(m)\plus{}n: m,n \in Z\}$  einai pikno sto $ R$.

Stergios

[b][color=red]Τα νέα προβλήματα και τις  λύσεις  προτείνω να τα γράφετε όχι πια στο στικυ , αλλά στον κανονικό χώρο. Με αυτόν τον τρόπο θα είναι πιο εύκολο να βλέπετε τις νέες ασκήσεις και τις λύσεις .Αν χρειαστείτε κάποιο στικυ για άλλη χρήση ,εκτός ασκήσεων ,  να μου το ζητήσετε .[/color][/b].

 Εννοείται πως η βοήθειά σας [u]και[/u] στους χώρους των ολυμπιάδων ή των σχολικών ασκήσεων είναι πολύτιμη.

[u] Μπάμπης[/u]

[quote="silouan"]Να υπολογιστεί το ολοκλίρωμα

$ \int\frac {x \plus{} 1}{x(e^x \plus{} 1)}$[/quote]

Silouan θα ήθελα αν μπορείς να βάλεις σε hide την λύση γιατί την παλεύω εδώ και λίγες μέρες την άσκηση και δεν έχω καταφέρει κάτι....
Είναι καιρό άλυτη οπότε δεν πιστεύω να διαφωνήσει κάποιος..

Katarxas sorry gia ta greeklish alla grafw apo kinhto... Exw ena provlhmataki se mia askhsh pithanothtwn:
Se ena kentro minhmatwn ftanoun minhmata ipshlhs mesaias kai xamhlhs proteraiothtas me sixnothta emfanishs 50% , 30% kai 20% anristoixa. Ta minhmata ipshlhs proteraiothtas kostizoun 5 monades, ta mesaias 3 monades kai ta xamhlhs 1 monada... Poia h pithanothta mia seiras 100 minhmatwn na exei kostos ligotero apo 200 monades?

To exw palepsei poli alla den mporw na vrw akrh :(

ok. Lithike... Skeftomouna pio poliplokes diadikasies alla den htan kati telika

[quote="modulo"]Καλημέρα. :) Μήπως γνωρίζει κανείς πώς λέγεται στα Αγγλικά η μέθοδος του κατοπτρισμού; :) Έχω κολλήσει σε ένα προβληματάκι προσδιορισμού ενός διανυσματικού πεδίου και έχω δοκιμάσει ό,τι έχω σκεφτεί σε mirror ή reflection για να βρω πληροφορίες στο google αλλά μάταια. :/

(Θεώρησα περιττό ν' ανοίξω νέο topic γι' αυτό το πραγματάκι, ελπίζω να το 'γραψα στο σωστό μέρος)[/quote]

Δε νομίζω να είναι και πολύ χρήσιμο μετά από 2 μήνες :( αλλά μάλλον θέλεις τη μέθοδο των   [url=http://en.wikipedia.org/wiki/Method_of_image_charges]image charges[/url]...

Cheerio,

Durandal 1707

Θα μπορούσατε να με βοηθήσετε με τον υπολογισμό του ολοκληρώματος του {(1+χ^1/2)^1/3}*χ^-1

Ευχαριστώ εκ των προτέρων


geometry
MATHCOUNTS
number theory
modular arithmetic

I feel that one area that I am weak in is modular arithmetic. Are there any good places where I could learn it? I know I could buy the Introduction to Number Theory, and it would be in it, but are there any free ones? Also, I am not looking for a definition, like wikipedia, but rather an explanation.

I'm afraid it's too complicated to explain in one post. I suppose the only option is to buy Introduction to Number Theory, or read certain chapters in the classics.

Well, I just ordered the classics, and in the table of contents, I only see one chapter about Number Theory in Volume 2. Also, I understand most of the Number Theory book, it is just the last few chapters that are about modular arithmetic. Correct me if I am wrong.

I was looking for links to some good website resources, like how coolmath.com has lessons for algebra.

Stick with AoPS volume 1. It's got a whole chapter dedicated to number theory.(chapter 5)

Other resources include UCSMP math in addition to AoPS

If you wanted, you could also have a look at this:
[url]http://www-history.mcs.st-and.ac.uk/~edmund/lnotes/node5.html[/url]

In AoPS volume 1, one of the first few chapters explains basic modular arithmetic very well. In AoPS volue 2, it get a LOT more in depth, so read the chapter in volume 1 before reading the chapter in volume 2.

This link is also decent, though I'd recommend either buying Intro to Number Theory or borrowing it from a library/teacher/friend.
[url]http://www.math.rutgers.edu/~erowland/modulararithmetic.html[/url]

Introduction to Number Theory from AoPS is better than the free links here (though obviously it's not free).

Santos's dispences are the best in my hopinion

Look at this one http://www.openmathtext.org/lecture_notes/number_theory_book.pdf  :wink:

I personally feel that intro. to number theory explains the best, AOPS vol.1, doesn't have that many chapters dedicated to that topic.
What helped me was the advanced MATHCOUNTS/AMC 8 course. I learned modular arithmetic, base numbers much better for Mathcounts preprartion. I suggest you to try that!


calculus
integration
calculus computations

Hi everyone,

Could someone please explain to me how to solve the integral x^3-x^2-x+1/x^+9

I know that I have to do long divition first.

After doing this I got x-1-10x+10/x^2+9, but the correct answer for this step is x-1-8x-1/x^2+9. Does anyone see where I went wrong?

After that I am supposed to use a table and the correct answer is

1/2x^2-x-8(1/2ln(x^2+9))+8(1/3tan^-1(x/3))+c

Can someone explain to me how they got this as their final answer, please?

Thank you very much

That's incomprehensible. Use parentheses!

Do you mean $ \int\frac{x^{3}\plus{}x^{2}\plus{}x\plus{}1}{x^{2}\plus{}9}\,dx$?

Yes

except it's -x^2 and -x

x^3-x^2-x+1/(x^2+9)

I then did long division and got x-1-(10x+10)/(x^2+9)

I said that x^2+9 goes into x^3-x^2 X times. Then, after subtracting the first row from the top row I got -x^2-10x+1 and said that that went into the original equation -1 times. I then got -10x+10 as a remainder. Do you see the error?

Thank you

You're still not using enough parentheses- but I think it's actually meant to be $ \int\frac{x^{3}\minus{}x^{2}\plus{}x\minus{}1}{x^{2}\plus{}9}\,dx.$
That would account for the difference between the 8 and the 10.

Well, to kill the remaining, make a long division, split the integral and use the following properties

$ \int\frac{f'(x)}{f(x)}\,dx\equal{}\ln|f(x)|\plus{}k$ & $ \int\frac{1}{a^{2}\plus{}u^{2}}\,du\equal{}\frac{1}a\arctan\frac{u}{a}\plus{}k$

Of course, find them the constraints.


Let $ S$ be the set of all five-digit numbers such that the sum of their digits is 43. What is the probability that a number randomly selected from set $ S$ will be divisible by 11? Express your answer as a common fraction.

What is meant by the first set of numbers and the second set of numbers in the solution?

All the five-digit numbers with the sum of the digits 43 are either 99997 or 99988 or a permutation of those.  there are 3 ways to arrange them so that ther'e divisible by 11.  15 total, so 3/15=1/5


function
integration
probability and stats

Let $ Y$ be a real random variable and its characteristic function is $ \varphi_Y(t) \equal{} e^{\minus{}|t|^p}$, where $ p \in (0, 2]$. Find all $ p$ such that $ \mathbb{E}|Y| < \infty$.

If $ E(|Y|)$ were finite (that is, $ Y\in L^1$), then $ \phi_Y$ would be in $ C^1$ - continuously differentiable. For these functions, the only question about differentiability would be at $ t\equal{}0.$ Examining that, we see that $ e^{\minus{}|t|^p}\in C^1$ if and only if $ p>1.$

That's a necessary condition: is it sufficient? In this case, yes. Since $ \phi_Y(t)$ decays rapidly at $ \infty,$ the density $ f(y)$ is very smooth. That means the only issue will be the rate of decay as $ y\to infty,$ and that should look like a power of $ y$ that depends in a predictable fashion on $ p.$

Note that whenever the mean exists at all, we have $ E(Y)\equal{}0.$

The borderline case is a failure. If $ p\equal{}1,$ then $ \phi_Y(t)\equal{}e^{\minus{}|t|}$ and the density of $ Y$ is $ \frac1{\pi(1\plus{}y^2)}$ on $ (\minus{}\infty,\infty)$ and $ E(|Y|)\equal{}\infty.$

Thanks for your reply and an idea how to deal with the case $ p \in (1, 2]$ (the fact that it is necessay condition is clear). However, I can't do it. I only know that the density is (since $ \varphi_Y$ is integrable) $ g(x) \equal{} \frac{1}{2\pi}\int_\mathbb{R}e^{\minus{}itx}e^{\minus{}|t|^p}dt$. But how to prove that $ E|Y| \equal{} \int_\mathbb{R} |y|g(y)dy < \infty$. Could you post more hints?


inequalities
algebra
polynomial
inequalities proposed

Let $ x$, $ y$ and $ z$ be real numbers such that $ x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}1$. Prove that:\[ 7(x^{4}\plus{}y^{4}\plus{}z^{4})\plus{}8(x\plus{}y\plus{}z)(xyz)\ge 4(xy\plus{}yz\plus{}zx)\plus{}1\]

Ok . Nice and easy problem .  :)
The inequality is equaivalent to $ 7\sum x^{4}\plus{}8xyz(x\plus{}y\plus{}z)\minus{}4\sum xy\cdot\sum x^{2}\minus{}(\sum x^{2})^{2}\geq 0$
or $ 6(\sum x^{4})\plus{}4xyz(x\plus{}y\plus{}z)\geq 4\sum (xy(x^{2}\plus{}y^{2}))\plus{}2\sum x^{2}y^{2}$ which true since

$ \sum x^{4}\plus{}xyz(x\plus{}y\plus{}z)\geq\sum xy(x^{2}\plus{}y^{2})$ from schur and the trivial $ x^{4}\plus{}y^{4}\plus{}z^{4}\geq x^{2}y^{2}\plus{}y^{2}z^{2}\plus{}z^{2}x^{2}$

Shoot...I don't like when problems get 'schur-ed'...My solution was that the inequality was equivalent to:

$ \sum (x^{2}\plus{}y^{2}\minus{}xy\minus{}z^{2})^{2}\ge 0$

I wonder if it is true that any symmetric inequality that is equivalent a sum of a bunch of squares of polynomials can be proven by schur/muirhead.


[quote="t0rajir0u"]It seems to me that

$(a^{2}+2bc)(b^{2}+2ca)(c^{2}+2ab) \ge (ab+2bc)(bc+2ca)(ca+2ab)$

Follows from the well-known $a^{2}+b^{2}+c^{2}\ge ab+bc+ca$ and a few variations (in other words, from Rearrangement.)[/quote]Very Good!

[quote="The soul of rock"]Prove that for all $a,b,c>0$ we have
$abc(a+2c)(b+2a)(c+2b)\le(a^{2}+2bc)(b^{2}+2ca)(c^{2}+2ab)$[/quote]
Here is my solution:
Let $x=\frac{a}{c},y=\frac{c}{b},z=\frac{b}{a}$, then $x,y,z>0,xyz=1$. The inequality becomes
\[(x+2)(y+2)(z+2) \le \left( \frac{2x}{y}+1 \right) \left( \frac{2y}{z}+1 \right) \left( \frac{2z}{x}+1 \right) \]

\[\Leftrightarrow 4\sum{x}+2\sum{xy}\le 4\sum\frac{x}{z}+2\sum\frac{x}{y}\]
This inequality follows from these inequalities
\[\sum\frac{x}{y}\ge \sum{x}\tag{1}\\ \sum\frac{x}{z}\ge \sum{xy}\tag{2}\]
You only need to use AM - GM Inequality to prove.

We are done!

Have fun! :)

Wow, a nine proof. But, please send your solution for two inequalities

[quote="me@home"]Don't quite see how it is rearrangement  :|
Are you mistaking something with $a+b+c\geq x+y+z\implies abc\geq xyz$? thats not true either...[/quote]
It occurs when $(x,y,z)\succeq(a,b,c)$ and $x,y,z,a,b,c>0$ which follow by Karamata's inequality
@toanhocmuonmau: Nice proof

[quote="me@home"]Don't quite see how it is rearrangement  :| [/quote]

Nearly all inequalities of this form can be solved by expanding them out and using Rearrangement  :rotfl:


Find all permutations of 2, 3, 4, ... , 102 in which the nth number is divisible by n for each n.

[hide][b]Claim:[/b] the only such permutation is $ 2,3,...,102$, i.e., the identity.
Let $ \sigma$ be a permutation on $ 2,3,...,102$. Denote the number of prime factors, counting multiplicity, of $ n$ as $ s(n)$. We prove our claim by strong induction on $ s(n)$.

[u]Base:[/u] $ s(n) \equal{} 1$. Since $ \sigma(n)|n$, $ \sigma(n) \equal{} 1, n$. But $ 1$ is not an element. Therefore $ \sigma(n) \equal{} n$.
[u]Inductive Step:[/u] suppose it holds for $ s(n) \equal{} 1,...,k \minus{} 1$. If $ s(n) \equal{} k$, for any $ d|n$ with $ s(d) < k$ and $ d\neq 1$, we have $ \sigma(d) \equal{} d < n$. Since $ \sigma(n)|n$, $ \sigma(n) \equal{} 1$ or $ s(\sigma(n)) \equal{} k$. But $ 1$ is not an element. Therefore $ s(\sigma(n)) \equal{} k$, $ \sigma(n) \equal{} n$. QED
[/hide]

Edited: I add my claim to clarify my intent. Correct?

Nevermind. The above is remained as a testament for my stupidity.

I don't understand what you are proving. :maybe:  The problem asks you to find all permutations, not only to find the number of permutations.

[b]@ timwu:[/b] The numbers are intentionally offset by 1...

[hide]Partition the set into two subsets: divisors of 102 and nondivisors of 102. Non-divisors must be in the position of their own number. Now figure out what to do with the divisors. For example, 102, 2, 3, 4, ... 101 works. You can also switch 102 with any of its divisors, and it'll still work. Say you switched it with 51. Then you can in turn switch 51 with one of its divisors and it will still work, etc. Basically, it's the number of ways to arrange the divisors.

A useful approach is to start from the 101st position and work backwards. Each time, there are only two numbers greater than or equal to the position number of the position you want to fill: that number, and 102. We can obviously only use 102 in a position that is a divisor of 102. Whichever number we displace goes into the first position because any number is divisible by 1.[/hide]

by the way, the original formulation of this problem asks you to find the number of permutations. :D

ps: i'm sorry that i've posted this twice. i forgot that i've posted this. :oops:

[hide="Solution"]There are a total of 101 numbers. The 52nd through 101st numbers are 52-101, respectively. So we have positions 1-51, with numbers 2-51, and 102.

102 can only go into spots which are divisors of 102, and those are 1, 2, 3, 6, 17, 34, and 51. Likewise, the divisors of 102 can only take up those spots. Now we take the first spot that isn't taken up by a divisor of 102: 50. The only number that can go there is 50. The next spot can only be filled with 49, etcetera, and we skip all divisors of 102. Thus the $ n$th spot is filled with $ n$ for all numbers that are not divisors of 102. Thus the only permutation happens in spaces 1, 2, 3, 6, 17, 34, and 51.

Now the number of the space that 102 fills is either 1, a prime, or a product of two distinct primes. If 102 fills the first spot, then 51 fills spot 51, 34 fills spot 34, and so on. That is one case. If 102 fills a prime spot, the prime must go in spot 1. That is three more cases. If 102 fills a product of primes, the product of primes can only go in one of the two prime slots, and the prime-numbered slot that it goes in must be in the first slot. That is 6 cases.

1+3+6=10[/hide]

Forgive any un-rigor, this is my first combinatorics solution in the pre-olympiad section.


function
induction
algebra unsolved
algebra

Let $f$ be a function from the set of positive integers to the set of non-negative integers such that $f(1)=0$ and for all $n \geq 2$ : $f(n)= \max \{f(i)+f(n-i)+i\}$
Determine $f(2005)$

$f(n)=\frac{n(n-1)}{2}$ by induction:

$f(1) = 0 = \frac{1 \cdot 0}{2}$ is obviously true.

Now assume the above formula is true for all $i \in {1, 2, ..., n-1}$. Then

\begin{eqnarray*}
{max \left(f(i) + f(n-i) + i\right)&=&max \left(\frac{i(i-1)}{2} + \frac{(n-i)(n-i-1)}{2} + i\right)\\
&=&max \left(\left(i-\frac{n-1}{2}\right)^2-\left(\frac{n-1}{2}\right)^2+\frac{n^2-n}{2}\right) = \\
&=&max \left(\left(n-1-\frac{n-1}{2}\right)^2-\left(\frac{n-1}{2}\right)^2+\frac{n^2-n}{2}\right) = \frac{n(n-1)}{2}
}\end{eqnarray*}.

I think that the real problem comes from Taiwan 2000, and the condition is max for $i$ between $1$ and $n/2$. In this case the answer is much more cool, and the problem is less trivial than it looked without it.