I was just thinking of something, when this suddenly cropped. Its easy, but still its my own.;)

When initially scientists started analyzing charge, they found that there are 2 types of charge. They found that some charges attracted each other and some repelled each other. So they decided that all charges which attract must be of the same type and all those which repel must be of opposite type. So the named the charges Type A and Type B.

Are their actions and decisions correct? Explain with proper assumption.

Yes yes its by convention, one from cats fur or sth and another from a glass rod?

They named it as +ve and negatice bcos: $ a \plus{} ( \minus{} a) \equal{} 0,  a \in \mathbb{R}$

but how did they know that like charges will repel and unllike charges will attract each other??? :maybe:

no no .......how were they aware of the property that like charges repel and unlike charges attract?? what experiments did they do to find out?? :maybe:

How was i able to post within 10 minutes???? :maybe:

Because you've been promoted as mod  :rotfl:  :rotfl:

[hide]@Madness: No mathematics please. Actually how we handle charges today is extremely mathematical. What is great about the scientists was not the discovery of charge, but the notations they associated with it. Many of us do not realize how abstract a concept charge or for that matter mass is. The plus and minus are just symbols and refer to nothing. To analyze charge is difficult. But the simple ways we use it in mathematical notations clearly illustrate their forward thinking[/hide]

Okay, anand Leave that fact.... See none of us can see charge, nor visualize it. It is an intrinsic property. What we can do is that.... take different charges and bring them closer and see whether the forces are attractive or repulsive in nature. and thereby we say since 2 kinds of interactions exist, there have to be ATLEAST two types of charge. We do not know anything about positive or negative, they are just notations.

Now a smart alec (me?) makes a claim that all charges which attract have to be of the same type, and all those which repel have to be of a different type.

Now remember, I do not know any of the underlying mathematics related to the charge...

What if i oppose and say that it is unlike charges that repel and not like......what will you say?? :D  :D

edit: what have you written shreyas? :maybe:
"Now a smart alec (me?) makes a claim that all charges which attract have to be of the same type, and all those which repel have to be of a different type."

i thought all charges which attract have to be of different type (i mean unlike) :)

it actually depends upon how you define charges.

it was confirmed that there were 2 kinds of charges. then it was seen, that when 2 objects are rubbed with each other, then they attract. since charges should be conserved, the 2 bodies must acquire unlike charges(so that they cancel each other). hence it was deduced that unlike charges attract.

also, one other factor that could have compelled them to think that the body acquires opposite charges is a loose transalation of Newtons 3LOM (also the concept of duality, for eg. in Angels and Demons by Dan Brown, Vetra creates matter, and hence, he deduces that he must have created anti matter too!!!)

now the only flaw i can find in the above argument is:
transport urself back to the time when these experiments were being done. what compels u to think that charges should be conserved??

[quote="anandghegde"]
i thought all charges which attract have to be of different type (i mean unlike) :)[/quote]

Thats exactly what I'm trying to oppose. From Class VI we learn that Like Charges Repel and unlike charges attract. Have any of us taken two attracting bodies and seen that they are of the opposite type. Definitely not. Because none of us can see charge. Its just an intrinsic property. Now I make a claim that what we've been learning all till today is wrong. Bodies which attract are the same type and those which repel are of opposite type?

Its a logical question come on.... Pardesi, you wanna try? ;)

[hide]P.S: There is a flaw in the argument.... so spot it if you can. [/hide]

@achu: What conservation of charge?

the flaw in my argument was that what compelled the scientists to think that charges cannot be created from nowhere, and if charge of one type is generated, then charge of another type should also be created.

if u ignore this flaw, then it is pretty simple. when u rub 2 objects, if u consider them to develop opposite charges, and it is seen that they attract, this leads us to the conclusion that unlike charges attract, and logically like charges repel.
if u say that like charges attract, then it means that while rubbing, like charges must have developed on the 2 bodies, and this is not possible by conservation of charge.

@Achu

what does your signature mean?? :D  :maybe:

btw ...see here......very serious discussion...abt free lunches
http://www.goiit.com/posts/list/physical-chemistry-lunch-4027.htm
 :D

[quote="slogger"]dunno if it is correct
but maybe it has to do with stability ie lowering of energy
Each system tries to lower its energy
now potential energy of a system is given by q1q2/r
so if they are like charges then r has to be increased to minimise potential energy
while if they are opposite charges then the numerator is negative and r has to be decreased to minimise the potential energy
thus opposite charges tend to come closer to each other(attract)and vice versa for like charges[/quote]
Anchit,
the timeline shreyas is talking abt is jut wen charges were discovered
They wuldn have even defined electric potential, leave alone the expression for potential. :wink:

Achu is kind of right... But I didn't get the exact answer I expected (somebody did give me this, but not on AOPS)..

Simple contradiction.. Let us assume the scientist/smart alec is correct. (Like Charges Attract, unlike repel)

Now take two bodies (Body 1 and Body 2) which attract. They must be of same type. So let them be of Type 1. Take another body which repels the Body 1. It must be Type 2. Now according to our notations Body 2 and Body 3 must also repel because they are opposite type, but it was found that they attract. And hence the contradiction was how scientists accepted like charges repel and unlike attract.

In reality:
Body 1: +ve
Body 2: -ve
Body 3: +ve


function
inequalities
vector
geometry
3D geometry
binomial coefficients
inequalities open

Let $f$ be a convex function on an interval $I$ and take $n$ variables $x_1,\ldots,x_n\in I$.
For $k=1,\ldots,n$ define the mean $s_k$ by
\[ s_k={n\choose k}^{-1}\sum_{i_1<\cdots<i_k}f\left(\frac{x_{i_1}+\cdots+ x_{i_k}}{k}\right). \]
[hide="Note"] (Note that from the convexity of $f$ it follows immediately that $s_1\ge\cdots\ge s_n$.)
This agrees with the intuitive idea that $s_{k+1}$ is more “averaged” than $s_k$ and thus should be smaller than $s_k$, considering the U-shape of the convex function $f$.[/hide]
For all $1\le k< \ell< m\le n$, denote by $P_n (k, \ell ,m)$ the following statement : \[ P_n (k, \ell ,m)\Leftrightarrow\boxed{ (m-\ell)s_k+(\ell-k)s_m\ge (m-k)s_\ell\text{ holds }\forall x_1,\ldots,x_n\in I }. \]
In the case $n=m$, we’ll omit the index, writing $P (k, m,n)$ for $P_n (k, m,n).$ (We always suppose $k<m<n$)
Note that $P(1,2,3)$ is the same as the standard Popoviciu inequality and more generally $P(1, m,n)$ is the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21786]Popoviciu-Cirtoaje-Zhao inequality[/url] (from about post #8 onwards). It is straightforward to derive from it that $P(k, m k ,nk)$ is true for all $k\in\mathbb N$.
[b]The main question is, how to caracterize the triples $k< m< n$ such that $P (k, m,n)$ is true ? [/b]
This problem seems to me like having hit a beehive... Let’s consider it first for $k+m=n$. This already seems sufficiently tough.
By majorization it is not hard to show that for $\lambda \in\mathbb N$, the vector $x=v(\lambda) : =[\lambda_{n-\lambda } (\lambda-n)_\lambda]$ (here indices mean multiplicities) yields a counterexample if $\lambda k\le n$ and ${n-1\choose k-1}\ge{n-\lambda \choose k}$. This can only occur for $\lambda\in\{2,3\}.$ (I haven’t proven that thoroughly, but it should not be too hard with some calculus.)

Thus, $\lambda=2$ yields that $P(k,m,k+m)$ is false for $k<m\le\frac{\sqrt{5}+1}2 k +\frac{5-\sqrt{5}}{10}$ (this is just an explicit way of writing ${k+m-1\choose k-1}\le{k+m-2\choose k}$, it is easy to see that both are equivalent for $k,m\in\mathbb N$ ).
$\lambda=3$ yields that $P(k,m,k+m)$ is false for $m=2k+1,k\ge3$ ; $m=2k+2,k\ge10$ ; $m=2k+3,k\ge17$ etc., which is[i] roughly[/i] equivalent to $2k+1\le m\le\frac{58k+15}{27}.$

In the table below, the rows correspond to $k$ and the colums to $m$. ‘$x$’ means $P(k,m,k+m)$ is known to be true (i.e. $m$ is a multiple of $k$). ‘$2$’ and ‘$3$’ mean it is false, $v(2)$ or $v(3)$ yielding a counterexample.

$.\ \ \hskip .08in2\hskip .095in3\hskip .095in4\hskip .095in5\hskip .095in6\hskip .095in7\hskip .095in8\hskip .09in9\hskip .045in 10\hskip .045in 11\hskip .045in 12\hskip .045in 13\hskip .045in 14\hskip .04in 15\hskip .04in 16\hskip .04in 17\hskip .04in 18\hskip .04in 19\hskip .04in 20\hskip .04in 21\hskip .04in 22\hskip .04in 23\hskip .04in 24$
$1\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx\hskip .093inx$
$2\hskip .089in\hskip .089in\hskip .089in2\hskip .089inx\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089inx$
$3\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in2\hskip .089in2\hskip .089inx\hskip .089in3\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx$
$4\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in2\hskip .089in2\hskip .089in\hskip .089in\hskip .089inx\hskip .089in3\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx$
$5\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in\hskip .089in\hskip .089inx\hskip .089in3\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx$
$6\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx\hskip .089in3\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx$
$7\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx\hskip .089in3\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx\hskip .089in\hskip .089in$
$8\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx\hskip .089in3\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx$
$9\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx\hskip .089in3$
$10\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx\hskip .089in3\hskip .089in3$
$11\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in2\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089in\hskip .089inx\hskip .089in3\hskip .089in3$

From dabbling around, I have come to conjecture a sort of $(n-\lambda)EQ$ theorem for $\lambda\in\{2,3\}$, i.e.
[b]Conjecture : $P(k,m,k+m)$ is true everywhere else (above the diagonal), so all the holes should become ‘$x$’.[/b]
Note that if this is so, it would imply for instance that $P(6,10,16)$ holds in spite of $P(3,5,8)$ being false :wallbash: .
[hide="Remarks"] concerning the case $k+m\neq n$:
I would guess that $P (k, m,n)$ is always false for $k+m>n$. But counterexamples may be far from straightforward (i.e. highly asymmetrical), as the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88806]P(2,3,4) case[/url] shows. :blush:

For $k+m\le n$, one could think that $P(k, m,n)$ implies $P(k, m ,n+1)$. It seems for instance that $P(k, mk,n)$ is true for all $n\ge (m+1)k$ (recall it’s true for all those $n$ that are as well multiples of $k$), BUT in general there must be some more conditions for that. e.g. $P(2,5,n)$ is false for all $n\ge9$, but I have a hard time to find counterexamples for $n=7$ and $n=8$. Similarly, $P(4,7,n)$ is false for all $n\ge12$, but may be true for $n=11$. :wallbash:
A bit more generally, if my ugly calculations :blush: are true, $P(k,2k-1,tk+1)$ is false for $k\ge2$ and $t\ge3$, by virtue of $v(t)$. It should be possible from there to disprove $P(k,2k-1,n)$ for all $n\ge 3k$ by ‘interpolating’ the $v(t)$ in some way.  Lots of ugly calculations with multiple products of binomial coefficients  :oops:
By now, I have just proven some special cases like $P(2,3,n)$ is false for all $n\ge4$, and so is $P(3,5,n)$ for all $n\ge6$.
Might it be possible that “$P(k, m,n)$ is false”  implies “$P(k, m ,n+1)$ is false”?? This would be very strong. [/hide]

It seems that we can add another $ x$ to the table, though it is rather an $ x$ for "$ x$-tra ugly". The assertion $ P\left( 2,5,7\right)$ (that is, $ P_{7}\left( 2,5,7\right)$) is true. I am going to sketch the proof below; feel free to ask questions about the steps you don't understand. And feel free to report typos, as I am not going to proofread this in the nearest future.

It is less the proof of this fact than the idea of the proof that makes me post this, because the proof is not quite a stroke of genius, but the idea [i]may[/i] be interesting. If I am not mistaken, using this idea you can prove or disprove $ P_s\left(2,s,2 + s\right)$ by checking finitely many cases, and the number of cases grows quadratically with $ s$.

The proof of $ P\left( 2,5,7\right)$ below is an elaboration of the tactics I introduced(?) with [url=http://www.mathlinks.ro/viewtopic.php?p=140145#140145]http://www.mathlinks.ro/viewtopic.php?t=19097 post #11[/url] Lemma 3. To say it in one sentence: Inequalities like Popoviciu are linear optimization problems.

Before we come to the proof, let's first formulate $ P\left( 2,5,7\right)$ (note that I am calling the function $ \varphi$ in order to prevent confusion of $ f$'s):

[color=blue][b]Popoviciu-$ P\left( 2,5,7\right)$ inequality.[/b] Let $ \varphi$ be a convex function from an interval $ I\subseteq\mathbb{R}$ to $ \mathbb{R}$. Let $ a$, $ b$, $ c$, $ d$, $ e$, $ f$, $ g$ be seven points from $ I$. Then,

$ 2\sum\limits^{21}\varphi\left( \frac {a + b}{2}\right) + 9\cdot7\varphi\left( \frac {a + b + c + d + e + f + g}{7}\right)$
$ \geq5\sum\limits^{21}\varphi\left( \frac {a + b + c + d + e}{5}\right)$.

Here, $ \sum\limits^{21}$ means "sum over all $ 21$ different permutations"; for instance,

$ \sum\limits^{21}ab = ab + ac + ad + ae + af + ag + bc + bd + be$
$ + bf + bg + cd + ce + cf + cg + de + df + dg + ef + eg + fg$.[/color]

[i]Proof (sketched).[/i] The inequality in question will follow from Karamata once we have shown the majorization

$ \left( \underbrace{\text{all }21\text{ different permutations of }\frac {a + b}{2}}_{\text{each of them }2\text{ times}},\right.$
$ \left. \ \ \ \ \ \ \ \ \ \ \underbrace{\frac {a + b + c + d + e + f + g}{7}}_{9\cdot7\text{ times}}\right)$
$ \succ\left( \underbrace{\text{all }21\text{ different permutations of }\frac {a + b + c + d + e}{5}}_{\text{each of them }5\text{ times}}\right)$.

According to the symmetric majorization criterion ([url=http://www.mathlinks.ro/viewtopic.php?p=140145#140145]http://www.mathlinks.ro/viewtopic.php?t=19097 post #11[/url] Lemma 1), this majorization will be proven once we show that for every real $ t$, we have

$ 2\sum\limits^{21}\left\vert \frac {a + b}{2} - t\right\vert + 9\cdot7\left\vert\frac {a + b + c + d + e + f + g}{7} - t\right\vert$
$ \geq5\sum\limits^{21}\left\vert \frac {a + b + c + d + e}{5} - t\right\vert$.

If we substitute $ a - t$, $ b - t$, $ c - t$, $ d - t$, $ e - t$, $ f - t$, $ g - t$ for $ a$, $ b$, $ c$, $ d$, $ e$, $ f$, $ g$, then this simplifies to

$ 2\sum\limits^{21}\left\vert \frac {a + b}{2}\right\vert + 9\cdot7\left\vert\frac {a + b + c + d + e + f + g}{7}\right\vert$
$ \geq5\sum\limits^{21}\left\vert \frac {a + b + c + d + e}{5}\right\vert$,

i. e. to

[color=green][b](1)[/b][/color] $ \sum\limits^{21}\left\vert a + b\right\vert + 9\left\vert a + b + c + d + e + f + g\right\vert \geq\sum\limits^{21}\left\vert a + b + c + d + e\right\vert$.

All that remains is now to prove this inequality [color=green][b](1)[/b][/color] for arbitrary reals $ a$, $ b$, $ c$, $ d$, $ e$, $ f$, $ g$.

Consider the function $ F: \mathbb{R}^{7}\rightarrow\mathbb{R}$ defined by

$ F\left( a,b,c,d,e,f,g\right)$
$ = \sum\limits^{21}\left\vert a + b\right\vert + 9\left\vert a + b + c + d + e + f + g\right\vert - \sum\limits^{21}\left\vert a + b + c + d + e\right\vert$.

Then, proving [color=green][b](1)[/b][/color] is equivalent to showing that $ F\left( a,b,c,d,e,f,g\right) \geq0$ for all reals $ a$, $ b$, $ c$, $ d$, $ e$, $ f$, $ g$.

Now let me first show an attempt at proving this which is not "good enough" but explains it better how I come to the "right" idea:

Consider the following $ 22$ linear "inequalities":

$ a + b\ \maltese\ 0$ and all $ 21$ different permutations of this inequality;
$ a + b + c + d + e + f + g\ \maltese\ 0$.

If you replace the $ \maltese$ sign (a templar cross) by a $ \leq$ or a $ \geq$ sign in each of these $ 22$ "inequalities", you get $ 22$ linear inequalities. Let $ V_{i}$ be the set consisting of all points $ \left( a,b,c,d,e,f,g\right) \in\mathbb{R}^{7}$ which satisfy all of these $ 22$ inequalities. Here, $ i$ is just an index which says which of the $ \maltese$ signs are to be replaced by $ \leq$ signs and which ones are to be replaced by $ \geq$ signs. There are $ 2^{22}$ ways to replace the $ \maltese$'s by $ \leq$'s and $ \geq$'s, so there are $ 2^{22}$ different indices $ i$ and $ 2^{22}$ sets $ V_{i}$. The union of these sets $ V_{i}$ is the whole $ \mathbb{R}^{7}$.

Each of the sets $ V_{i}$ is obviously a linear polytope in $ \mathbb{R}^{7}$ (under a "linear polytope", I understand a subset of $ \mathbb{R}^{n}$ which is defined by a system of finitely many linear inequalities). The function $ F$ is concave on each of these polytopes $ V_{i}$. In fact, the "positive part" of $ F$ (that is, the term $ \sum\limits^{21}\left\vert a + b\right\vert + 9\left\vert a + b + c + d + e + f + g\right\vert$) is a linear function on $ V_{i}$ (since $ a + b$ and its $ 21$ different permutations, and $ a + b + c + d + e + f + g$ never change their signs on $ V_{i}$), and the "negative part" of $ F$ (that is, the term $ \sum\limits^{21}\left\vert a + b + c + d + e\right\vert$) is a convex function (as a sum of modulus functions), so that $ F$ is a convex function subtracted from a linear function, and thus a concave function.

We want to show that $ F\left( a,b,c,d,e,f,g\right) \geq0$ for all reals $ a$, $ b$, $ c$, $ d$, $ e$, $ f$, $ g$. An obvious approach to this is to find the minima of $ F$ and show that $ F$ is $ \geq0$ at these minima. Nicely enough, $ F$ is concave on each polytope $ V_{i}$, and the minimum of a concave function on a polytope is attained at a vertex of the polytope...

... almost. The snag is that, with our above polytopes $ V_{i}$, the minimum of $ F$ on $ V_{i}$ need not exist, because the polytopes $ V_{i}$ are unbounded. A very obvious fix for this is to artificially bound them by intersecting them with a big enough cube.

This is what we are going to do now. Since the function $ F$ satisfies $ F\left( ta,tb,tc,td,te,tf,tg\right) = tF\left( a,b,c,d,e,f,g\right)$ for nonnegative $ t$, it is clear that in order to verify $ F\left( a,b,c,d,e,f,g\right) \geq0$ for all $ \left( a,b,c,d,e,f,g\right) \in\mathbb{R}^{7}$, it will be enough to verify $ F\left( a,b,c,d,e,f,g\right) \geq0$ for all $ \left( a,b,c,d,e,f,g\right) \in\mathbb{R}^{7}$ that lie inside the cube $ \left[ - 1,1\right] ^{7}$ (id est, that satisfy $ - 1\leq a\leq1$, $ - 1\leq b\leq1$, ..., $ - 1\leq g\leq1$).

Let $ U_{i} = V_{i}\cap\left[ - 1,1\right] ^{7}$. Then, $ \bigcup_{i} V_{i} = \mathbb{R}^{7}$ yields $ \bigcup_{i}U_{i} = \left[ - 1,1\right] ^{7}$. Hence, in order to show that $ F\left( a,b,c,d,e,f,g\right) \geq0$ for all $ \left( a,b,c,d,e,f,g \right) \in\mathbb{R}^{7}$ that lie inside the cube $ \left[ - 1,1\right] ^{7}$, it will be enough to prove that for every $ i$, we have $ F\left( a,b,c,d,e,f,g\right) \geq0$ for all $ \left( a,b,c,d,e,f,g\right) \in U_{i}$. Now, $ U_{i}$ is a bounded polytope, and the function $ F$ is concave on $ U_{i}$ (since it is concave on $ V_{i}$, and $ U_{i}\subseteq V_{i}$). Hence, the function $ F$ assumes its minimum on $ U_{i}$ in a vertex of $ U_{i}$ (because the minimum of a concave function on a polytope is assumed at a vertex of the polytope and exists when the polytope is bounded).

Hence, instead of proving that $ F\left( a,b,c,d,e,f,g\right) \geq0$ for all $ \left( a,b,c,d,e,f,g\right) \in U_{i}$, it will be enough to prove that $ F\left( a,b,c,d,e,f,g\right) \geq0$ whenever $ \left( a,b,c,d,e,f,g\right)$ is a vertex of the polytope $ U_{i}$.

Now, how to characterize a vertex of $ U_{i}$ ? We should not hope to explicitely find the vertices of such an implicitely given polytope, but it is a good idea to assume that $ \left( a,b,c,d,e,f,g\right)$ is a vertex of the polytope $ U_{i}$ and to try to conclude as much as possible about $ \left( a,b,c,d,e,f,g\right)$ from this assumption.

If we have a polytope $ \left\{ x\in\mathbb{R}^{n}\mid w_{u}x\leq k_{u}\text{ for every }u\in U\right\}$, where $ w_{u}$ is a constant covector (that is, a linear transformation from $ \mathbb{R}^{n}$ to $ \mathbb{R}$) and $ k_{u}$ is a real constant for every $ u\in U$, then every vertex $ x_{0}$ of this polytope satisfies $ w_{u}x = k_{u}$ for at least $ n$ different $ u\in U$, and these $ u$'s can be chosen in such a way that the corresponding $ w_{u}$ are linearly independent. In other words, if a polytope is characterized by some linear inequalities, then at every vertex of the polytope, at least $ n$ of these inequalities become equalities, and these $ n$ inequalities can be chosen in a way that their homogeneous parts are linearly independent.

The polytope $ V_{i}$ is characterized by the following $ 22$ inequalities:

$ a + b\ \maltese\ 0$ and all $ 21$ different permutations of this inequality;
$ a + b + c + d + e + f + g\ \maltese\ 0$;

here, the $ \maltese$ signs are to be understood as $ \leq$'s or $ \geq$'s according to the choice of $ i$.

The polytope $ U_{i}$, being the intersection of $ V_{i}$ with $ \left[ - 1,1\right] ^{7}$, is therefore characterized by the following $ 22 + 7 + 7 = 36$ inequalities:

$ a + b\ \maltese\ 0$ and all $ 21$ different permutations of this inequality;
$ a + b + c + d + e + f + g\ \maltese\ 0$;
$ a\leq1$, $ b\leq1$, ..., $ g\leq1$;
$ - a\leq1$, $ - b\leq1$, ..., $ - g\leq1$.

Now, if we let $ \left( a,b,c,d,e,f,g\right)$ be a vertex of this polytope $ U_{i}$, then at least $ 7$ of these $ 36$ inequalities become equalities at $ \left( a,b,c,d,e,f,g\right)$, and these inequalities can be chosen in a way that their homogeneous parts are linearly independent. In other words, (at least) $ 7$ of the following $ 36$ equalities are fulfilled:

$ a + b = 0$ and all $ 21$ different permutations of this equality;
$ a + b + c + d + e + f + g = 0$;
$ a = 1$, $ b = 1$, ..., $ g = 1$;
$ - a = 1$, $ - b = 1$, ..., $ - g = 1$.

And these $ 7$ equalities have linearly independent homogeneous parts.

If one of these $ 7$ fulfilled equalities happens to be $ a + b + c + d + e + f + g = 0$, then we are done, because $ a + b + c + d + e + f + g = 0$ entails $ F\left( a,b,c,d,e,f,g\right) = 0$ (as can be easily seen from the definition of $ F$), and thus $ F$ is $ \geq0$ in this vertex $ \left( a,b,c,d,e,f,g\right)$. So assume that $ a + b + c + d + e + f + g = 0$ is not one of the $ 7$ fulfilled equalities. Hence, (at least) $ 7$ of the following $ 35$ equalities are fulfilled:

$ a + b = 0$ and all $ 21$ different permutations of this equality;
$ a = 1$, $ b = 1$, ..., $ g = 1$;
$ - a = 1$, $ - b = 1$, ..., $ - g = 1$.

We will refer to the $ 21$ permutations of the equality $ a + b = 0$ as "indirect equalities", and to the equalities $ a = 1$, $ b = 1$, ..., $ g = 1$ and $ - a = 1$, $ - b = 1$, ..., $ - g = 1$ as "direct equalities". If one and the same variable appears both in a direct equality and in an indirect equality (for instance, the variable $ a$ in $ a = - 1$ and $ a + b = 0$), then we can replace the indirect equality by a direct equality (in this case, we replace $ a + b = 0$ by $ b = 1$). This replacement does not change the linear independency of the homogeneous parts of our $ 7$ equalities, but it decreases the number of indirect equalities. Perform such replacements until there remains no variable that appears both in a direct equality and in an indirect equality.

After this is done, we have our $ 7$ variables $ a$, $ b$, $ c$, $ d$, $ e$, $ f$, $ g$. Some of them - say, $ t$ of them - appear in direct equalities and only in them, and the other $ 7 - t$ variables appear in indirect equalities only (or nowhere at all). Let the [i]direct variables[/i] be the $ t$ variables that appear in direct equalities, and the [i]indirect variables[/i] be the $ 7 - t$ other variables.

We must have $ t$ direct equalities (in fact, we have $ t$ variables that appear in direct equalities, so we must have at least $ t$ direct equalities, and we cannot have more than $ t$ direct equalities because two direct equalities for one and the same variable would have linearly dependent homogeneous parts, while our $ 7$ equalities must have linearly independent homogeneous parts). Hence, we have $ 7 - t$ indirect equalities.

These $ 7 - t$ indirect equalities contain only the $ 7 - t$ indirect variables (as we have seen above), and are all of the form $ \left( \text{sum of two indirect variables}\right) = 0$. Hence, the indirect equalities are $ 7 - t$ homogeneous linear equations for the $ 7 - t$ indirect variables. And these $ 7 - t$ equations must have linearly independent homogeneous parts, i. e. be linearly independent. But a system of $ k$ linearly independent homogeneous linear equations for $ k$ variables can have only one solution: $ \left( 0,0,...,0\right)$.

Hence, all the $ 7 - t$ indirect variables must equal $ 0$ if $ \left( a,b,c,d,e,f,g\right)$ satisfies our $ 7$ equations. The direct variables all lie in the set $ \left\{ - 1,1\right\}$ (because they satisfy the direct equations, which are of the kind $ a = 1$, $ b = 1$, ..., $ g = 1$ and $ - a = 1$, $ - b = 1$, ..., $ - g = 1$).

Hence, if $ \left( a,b,c,d,e,f,g\right)$ is a vertex of the polytope $ U_{i}$, then every of the reals $ a$, $ b$, $ c$, $ d$, $ e$, $ f$, $ g$ is an element of the set $ \left\{ - 1,1,0\right\}$ (namely, it is $ 1$ or $ - 1$ if it is a direct variable and $ 0$ if it is an indirect variable).

Hence, in order to show that $ F\left( a,b,c,d,e,f,g\right) \geq0$ for every vertex $ \left( a,b,c,d,e,f,g\right)$ of the polytope $ U_{i}$, it remains to prove that $ F\left( a,b,c,d,e,f,g\right) \geq0$ for every $ \left( a,b,c,d,e,f,g\right) \in\left\{ - 1,1,0\right\} ^{7}$.

Thus we have reduced the proof of [color=green][b](1)[/b][/color] to a check of finitely many - namely, $ 3^{7}$ - cases. Fortunately, the function $ F$ is symmetric, so it is enough to consider the cases $ \left( a,b,c,d,e,f,g\right) = \left( 1,1,...,1, - 1, - 1,..., - 1,0,0,...,0\right)$. These are still quite many cases to check by hand.

Here is an attempt to be somewhat smarter: Let $ \left( a,b,c,d,e,f,g\right) \in\left\{ - 1,1,0\right\} ^{7}$. Assume that among the reals $ a$, $ b$, $ c$, $ d$, $ e$, $ f$, $ g$, the number $ 1$ appears $ u$ times, the number $ - 1$ appears $ v$ times, and the number $ 0$ appears $ w = 7 - \left( u + v\right)$ times. Here, of course, $ u$ and $ v$ are nonnegative integers satisfying $ u + v\leq7$. Then, it is easily seen that

$ F\left( a,b,c,d,e,f,g\right) = 2\binom{u}{2} + 2\binom{v}{2} + uw + vw + 9\left\vert u - v\right\vert$
$ - \binom{u}{2}\left\vert \left( u - v\right) - 2\right\vert - \binom{v} {2}\left\vert \left( u - v\right) + 2\right\vert - \binom{w}{2}\left\vert u - v\right\vert$
$ - uv\left\vert u - v\right\vert - uw\left\vert \left( u - v\right) - 1\right\vert - vw\left\vert \left( u - v\right) + 1\right\vert$.

We can WLOG assume that $ u > v$ (in fact, if $ u < v$, then we replace $ \left( a,b,c,d,e,f,g\right)$ by $ \left( - a, - b, - c, - d, - e, - f, - g\right)$ to get $ u > v$, and if $ u = v$, then we have $ a + b + c + d + e + f + g = 0$ and thus $ F\left( a,b,c,d,e,f,g\right) = 0$ so we are done), so that $ u - v\geq1$. Then, this simplifies to

$ F\left( a,b,c,d,e,f,g\right) = 2\binom{u}{2} + 2\binom{v}{2} + uw + vw + 9\left( u - v\right)$
$ - \binom{u}{2}\left\vert \left( u - v\right) - 2\right\vert - \binom{v}{2}\left( \left( u - v\right) + 2\right) - \binom{w}{2}\left( u - v\right)$
$ - uv\left( u - v\right) - uw\left( \left( u - v\right) - 1\right) - vw\left( \left( u - v\right) + 1\right)$.

Unfortunately, we have to split into two more cases $ u - v\geq2$ and $ u - v = 1$.

If $ u - v\geq2$, then this becomes

$ F\left( a,b,c,d,e,f,g\right) = 2\binom{u}{2} + 2\binom{v}{2} + uw + vw + 9\left( u - v\right)$
$ - \binom{u}{2}\left( \left( u - v\right) - 2\right) - \binom{v}{2}\left(\left( u - v\right) + 2\right) - \binom{w}{2}\left( u - v\right)$
$ - uv\left( u - v\right) - uw\left( \left( u - v\right) - 1\right) - vw\left( \left( u - v\right) + 1\right) = 2v\left( 6 - u\right)$

(where we have used that $ w = 7 - \left( u + v\right)$), so that $ F\left( a,b,c,d,e,f,g\right) \geq0$ because either $ u\leq6$ and thus $ 6 - u\geq0$, or $ u = 7$ and thus $ v = 0$ (because $ u + v\leq7$).

If $ u - v = 1$, then the above becomes

$ F\left( a,b,c,d,e,f,g\right) = 2\binom{u}{2} + 2\binom{v}{2} + uw + vw + 9\cdot1$
$ - \binom{u}{2}\cdot1 - \binom{v}{2}\left( 1 + 2\right) - \binom{w}{2}\cdot1$
$ - uv\cdot1 - uw\left( 1 - 1\right) - vw\left( 1 + 1\right) = 3\left( u - 1\right) \left( 4 - u\right)$

(where we have used that $ w = 7 - \left( u + v\right)$ and that $ u - v = 1$), so that $ F\left( a,b,c,d,e,f,g\right) \geq0$ because $ u - 1\geq0$ and $ 4 - u\geq0$ (since $ u - v = 1$ yields $ v = u - 1$, so that $ u + v\in\left[ 0,7\right]$ becomes $ u + \left( u - 1\right) \in\left[ 0,7\right]$, so that $ u\in\left[ 1,4\right]$ and thus $ 1\leq u\leq4$).

In both cases, $ F\left( a,b,c,d,e,f,g\right) \geq0$, and we are done!!!!!

[b]EDIT:[/b] I said the method generalizes, and here is what one obtains by generalizing it: For every $ s\geq 4$, the inequality $ P_{s + 2}\left(2,s,s + 2\right)$ is true. The corresponding analogue of [color=green][b](1)[/b][/color] is

$ \sum_{i < j}\left\vert a_i + a_j\right\vert + \frac {\left(s - 2\right)\left(s + 1\right)}{2}\left\vert a_1 + a_2 + ... + a_{s + 2}\right\vert$
$ \geq\sum_{i < j}\left\vert\left(a_1 + a_2 + ... + a_{s + 2}\right) - a_i - a_j\right\vert$

for any $ s + 2$ reals $ a_1$, $ a_2$, ..., $ a_{s + 2}$.

  Darij

EDIT: the very last thing (about reducing to cyclic sums) is nonsense. :blush: As I can't edit the post anymore,  :mad:  please ignore that.

Spanferkel: I have written a (rather poor) SML code to check your inequality on reals from some given set. As I showed in my post #2, in order to prove the inequality $ P\left(2,s\right)$ (that is, $ P_{s\plus{}2}\left(2,s,s\plus{}2\right)$), it is already enough to verify it for $ f$ being the modulus function and the reals $ x_1$, $ x_2$, ..., $ x_n$ all coming from the set $ \left\{1,0,\minus{}1\right\}$ - what is, indeed, easy to do by a computer. Unfortunately, this is not sufficient anymore for $ P\left(k,m\right)$ with $ k>2$; for instance, $ P\left(3,4\right)$ holds for all reals $ x_1$, $ x_2$, ..., $ x_n$ from the set $ \left\{1,0,\minus{}1\right\}$ but does not hold in general.

For this reason, a program of this kind can only provide counterexamples, but not give proofs for $ P\left(k,m\right)$ with $ k>2$. Unfortunately, even checking the case of $ x_i\in\left\{1,0,\minus{}1\right\}$ requires a rather long time for my program for "big" $ k$ and $ m$. The only new result I have found at the moment is that:

$ P\left(4,7\right)$ is wrong for $ x_1\equal{}x_2\equal{}x_3\equal{}x_4\equal{}x_5\equal{}x_6\equal{}1$, $ x_7\equal{}x_8\equal{}x_9\equal{}x_{10}\equal{}x_{11}\equal{}\minus{}1$ and $ f\left(x\right)\equal{}\left|x\right|$.

Looking at this one more time, I am wondering why nobody noticed it before. ;)

Here are some more results I got by my program, but all of which are known to you:

$ P\left(2,3\right)$ is wrong for $ x_1\equal{}x_2\equal{}x_3\equal{}1$, $ x_4\equal{}x_5\equal{}\minus{}1$ and $ f\left(x\right)\equal{}\left|x\right|$.
$ P\left(3,4\right)$ is wrong for $ x_1\equal{}x_2\equal{}x_3\equal{}x_4\equal{}x_5\equal{}1$, $ x_6\equal{}x_7\equal{}\minus{}2$ and $ f\left(x\right)\equal{}\left|x\right|$.
$ P\left(3,5\right)$ is wrong for $ x_1\equal{}x_2\equal{}x_3\equal{}x_4\equal{}x_5\equal{}1$, $ x_6\equal{}x_7\equal{}\minus{}2$, $ x_8\equal{}0$ and $ f\left(x\right)\equal{}\left|x\right|$.
$ P\left(3,7\right)$ is wrong for $ x_1\equal{}x_2\equal{}x_3\equal{}x_4\equal{}x_5\equal{}x_6\equal{}x_7\equal{}1$, $ x_8\equal{}x_9\equal{}x_{10}\equal{}\minus{}2$ and $ f\left(x\right)\equal{}\left|x\right|$.
$ P\left(4,5\right)$ is wrong for $ x_1\equal{}x_2\equal{}x_3\equal{}x_4\equal{}x_5\equal{}x_6\equal{}1$, $ x_7\equal{}x_8\equal{}x_9\equal{}\minus{}1$ and $ f\left(x\right)\equal{}\left|x\right|$.

A "positive" result: $ P\left(3,8\right)$ holds if $ f\left(x\right)\equal{}\left|x\right|$ and $ x_i\in\left\{\minus{}3,\minus{}2,\minus{}1,0,1,2,3\right\}$. However, I fear that this is not quite informative.

Here is the SML code, in case anyone can optimize it (though, I guess, true "optimization" means rewriting it in a different language):

[hide="Some really boring code"][code](* SOME SML COMMANDS FOR MAJORIZATION THEORY. *)

(* first some commands for handling lists: *)

(* maplist : function f; list [x_1, x_2, ..., x_n] => list [f(x_1), f(x_2), ..., f(x_n)] *)

val rec maplist = fn f => fn nil => nil | h :: t => f(h) :: maplist f t;

(* leftappendlist : list a; list [x_1, x_2, ..., x_n] => list [a @ x_1, a @ x_2, ..., a @ x_n] (hereby, x_1, x_2, ..., x_n must be lists) *)

val leftappendlist = fn a => (let val f = fn x => a @ x in
                              fn l => maplist f l
                              end);

(* sumlist : list l => sum of elements of l *)

val rec sumlist = fn nil => 0 | h :: t => h + sumlist(t);

(* the following operations on lists are analogous to some standard operations on sets in mathematics. *)

(* reducetohold : mapping f that transforms objects into boolean values; list l => sublist of l that consists of all elements x of l for which f(x) holds *)

val rec reducetohold = fn f => fn nil => nil | h :: t => if f h then h :: reducetohold f t else reducetohold f t;

(* binomiallist : number n; list l => the list of all sub-lists of l of length n *)

val rec binomiallist = fn n => (fn nil => if n = 0 then nil :: nil else nil | h :: t => (leftappendlist (h :: nil) (binomiallist (n-1) t)) @ (binomiallist n t));

(* COMMANDS FOR POPOVICIU-LIKE INEQUALITIES. *)

(* partitionlist : number n; list l => list of all lists of the form (a,a,...,a,b,b,...,b,...,...,k,k,...,k) with exactly n elements such that a, b, ..., k occur in the list l in this order *)

val rec partitionlist = fn 0 => (fn l => [nil]) | n => (fn nil => nil | h :: t => (leftappendlist [h] (partitionlist (n-1) (h :: t))) @ (partitionlist n t));

(* hundredten : number n => list of all lists of the kind [1,1,...,1,-1,-1,...,-1,0,0,...,0] with n elements altogether *)

val rec hundredten = fn n => partitionlist n [1,~1,0];

(* abssums : number k; list l => list of the absolute values of the sums of all k-element sublists of the list l *)

val abssums = fn k => fn l => maplist (fn x => abs (sumlist x)) (binomiallist k l);

(* binomial : number n; number k => binomial coefficient n! / (k! (n-k)!) *)

val rec binomial = fn 0 => (fn k => if k = 0 then 1 else 0) | n => (fn k => binomial (n-1) (k-1) + binomial (n-1) k);

(* spanferkel : number k; number m => list of all (k+m)-tuples consisting of 1's, -1's and 0's only which do NOT satisfy spanferkel's P(k,m,k+m) popoviciu inequality for f(x) being the modulus function *)

val spanferkel = fn k => fn m => reducetohold (fn l => sumlist (abssums k l) + ((m-k) * binomial (m+k) m) div (m+k) * abs(sumlist l) - sumlist (abssums m l) < 0) (hundredten (k+m));

(* spannferkel : the same but with tuples consisting of 1's, -1's, 2's, -2's and 0's *)

val spannferkel = fn k => fn m => reducetohold (fn l => sumlist (abssums k l) + ((m-k) * binomial (m+k) m) div (m+k) * abs(sumlist l) - sumlist (abssums m l) < 0) (partitionlist (k+m) [0,1,~1,2,~2]);

(* spannnferkel : the same but with tuples consisting of 1's, -1's, 2's, -2's, 3's, -3's and 0's *)

val spannnferkel = fn k => fn m => reducetohold (fn l => sumlist (abssums k l) + ((m-k) * binomial (m+k) m) div (m+k) * abs(sumlist l) - sumlist (abssums m l) < 0) (partitionlist (k+m) [0,1,~1,2,~2,3,~3]);

[/code][/hide]

  darij

[quote="darij grinberg"]
$ P\left(4,7\right)$ is wrong for $ x_1 \equal{} x_2 \equal{} x_3 \equal{} x_4 \equal{} x_5 \equal{} x_6 \equal{} 1$, $ x_7 \equal{} x_8 \equal{} x_9 \equal{} x_{10} \equal{} x_{11} \equal{} \minus{} 1$ and $ f\left(x\right) \equal{} \left|x\right|$.

Looking at this one more time, I am wondering why nobody noticed it before. ;) [/quote]
Well, my statement [quote="spanferkel"] the vector $ x\equal{}v(\lambda)$  yields a counterexample if $ \lambda k< n$ and
$ {n\minus{}1\choose k\minus{}1}>{n\minus{}\lambda \choose k}$[/quote]
is of course just one sufficient condition. Your choice of $ \vec x\equal{}[1_6(\minus{}1)_5]$ is equivalent to plugging $ v(5)$ in the original $ P(4,7)$  inequality. I’ve found that the same way $ P(5,9)$ is disproved by $ \lambda\equal{}6$ :) but this series doesn’t continue: $ P(6,11)$ is not disproved by $ \lambda\equal{}7$, nor the subsequent ones.  I would call those cases ‘sporadic’, in contrast to the patterns for $ v(2)$ and $ v(3)$. And I'm not at all sure that there are other ‘sporadic’ counterexamples using $ v(\lambda)$ 's.

So we have two more entries now:

$ \texttt{.\ \ 2\ \ 3\ \ 4\ \ 5\ \ 6\ \ 7\ \ 8\ \ 9\hskip.1in 10\ 11\ 12\ 13\ 14\ 15\ 16\ 17\ 18\ 19\ 20\ 21\ 22\ 23\ 24\ \ }$
$ \texttt{1\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x}$
$ \texttt{2\ \ \ \ \ 2\ \ x\ \ d\ \ x\ \ d\ \ x\ \ d\ \ x\ \ d\ \ x\ \ d\ \ x\ \ d\ \ x\ \ d\ \ x\ \ d\ \ x\ \ d\ \ x\ \ d\ \ x}$
$ \texttt{3\ \ \ \ \ \ \ \ 2\ \ 2\ \ x\ \ 3\ \ \ \ \ x\ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ x}$
$ \texttt{4\ \ \ \ \ \ \ \ \ \ \ 2\ \ 2\ \ 5 \ \ x\ \ 3\ \ d\ \ \ \ \ x\ \ \ \ \ d\ \ \ \ \ x\ \ \ \ \ d\ \ \ \ \ x\ \ \ \ \ d\ \ \ \ \ x}$
$ \texttt{5\ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\ \ 2\ \ 2\ \ 6 \ \ x\ \ 3\ \ \ \ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ \ \ \ }$
$ \texttt{6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\ \ 2\ \ 2\ \ \ \ \ \ \ \ x\ \ 3\ \ \ \ \ d\ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ d\ \ \ \ \ \ \ \ x}$
$ \texttt{7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\ \ 2\ \ 2\ \ 2\ \ \ \ \ \ \ \ x\ \ 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ }$
$ \texttt{8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\ \ 2\ \ 2\ \ 2\ \ 2\ \ \ \ \ \ \ \ x\ \ 3\ \ \ \ \ \ \ \ d\ \ \ \ \ \ \ \ \ \ \ x}$
$ \texttt{9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\ \ 2\ \ 2\ \ 2\ \ 2\ \ \ \ \ \ \ \ \ \ \ x\ \ 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ }$
$ \texttt{10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\ \ 2\ \ 2\ \ 2\ \ 2\ \ 2\ \ \ \ \ \ \ \ \ \ \ x\ \ 3\ \ 3\ \ \ \ \ }$
$ \texttt{11\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\ \ 2\ \ 2\ \ 2\ \ 2\ \ 2\ \ 2\ \ \ \ \ \ \ \ \ \ \ x\ \ 3\ \ 3}$

Now you say
[quote="darij grinberg"]  for instance, $ P\left(3,4\right)$ holds for all reals $ x_1$, $ x_2$, ..., $ x_n$ from the set $ \left\{1,0, \minus{} 1\right\}$ but does not hold in general. [/quote]
I disagree. $ P(3,4)$ is disproved by plugging $ v(2)$ in the original inequality, which is equivalent to putting $ \vec x\equal{}[1_5(\minus{}1)_2]$ in [color=green](1)[/color]. The former yields $ [6_{30}0_{35}(\minus{}1) _{60}(\minus{}8) _{15}]\prec [8_{20}1_{80}(\minus{}6)_{40}]$ which is wrong since $ [6_{30}]\not\prec [8_{20}1_{10}]$.

By the way, I can’t see why your argumentation with the indirect variables etc. shouldn’t be valid also for $ k>2$. :?:
If it is, and if my conjecture $ G(u,\lambda)\ge G(n \minus{} \lambda,\lambda)$ for $ u>\lambda$ is true (now, this is really iffy!), we would only need to check $ v(\lambda)$ for $ 2\le \lambda<\frac n2$ in order to prove or disprove $ P(k,m)$.

Just to add something new:

$ P\left(6,10\right)$ is wrong, confuted by $ x_1 \equal{} x_2 \equal{} ... \equal{} x_{11} \equal{} 1$ and $ x_{12} \equal{} x_{13} \equal{} ... \equal{} x_{16} \equal{} \minus{} 2$ and $ f\left(x\right) \equal{} \left|x\right|$. [This $ 16$-tuple and the one obtained by replacing all $ x_i$ by $ \minus{} x_i$ in it are the only two $ 16$-tuples of reals $ x_1,x_2,...,x_{16}\in\left\{ \minus{} 2, \minus{} 1,0,1,2\right\}$ that don't satisfy $ P\left(6,10\right)$ for $ f\left(x\right) \equal{} \left|x\right|$.]

$ P\left(6,11\right)$ is wrong, confuted by $ x_1 \equal{} x_2 \equal{} ... \equal{} x_{11} \equal{} 1$ and $ x_{12} \equal{} x_{13} \equal{} ... \equal{} x_{17} \equal{} \minus{} 2$ and $ f\left(x\right) \equal{} \left|x\right|$. [This $ 17$-tuple and the one obtained by replacing all $ x_i$ by $ \minus{} x_i$ in it are the only two $ 17$-tuples of reals $ x_1,x_2,...,x_{17}\in\left\{ \minus{} 2, \minus{} 1,0,1,2\right\}$ that don't satisfy $ P\left(6,11\right)$ for $ f\left(x\right) \equal{} \left|x\right|$.]

Is this much of a surprise, knowing that $ P\left(5,11\right)$ and $ P\left(7,11\right)$ are both wrong? Who knows. What is more surprising, in my opinion, is the correct $ P\left(11,22\right)$ under the wrong $ P\left(10,22\right)$ (maybe you could check your computation again? - I just find it surprising), and probably above something else that is also wrong (I don't say "exactly one line above").

[quote="spanferkel"][quote="darij grinberg"]  for instance, $ P\left(3,4\right)$ holds for all reals $ x_1$, $ x_2$, ..., $ x_n$ from the set $ \left\{1,0, \minus{} 1\right\}$ but does not hold in general. [/quote]
I disagree. $ P(3,4)$ is disproved by plugging $ v(2)$ in the original inequality, which is equivalent to putting $ \vec x \equal{} [1_5( \minus{} 1)_2]$ in [color=green](1)[/color]. The former yields $ [6_{30}0_{35}( \minus{} 1) _{60}( \minus{} 8) _{15}]\prec [8_{20}1_{80}( \minus{} 6)_{40}]$ which is wrong since $ [6_{30}]\not\prec [8_{20}1_{10}]$.[/quote]

I was imprecise. When I said "$ P\left(3,4\right)$ holds for all reals $ x_1$, $ x_2$, ..., $ x_n$ from the set $ \left\{1,0, \minus{} 1\right\}$", I meant it holds for all such reals under the condition $ f\left(x\right) \equal{} \left|x\right|$. This is all that my program can check. It cannot check majorization - though I am thinking about rewriting it so that it can.

[quote="spanferkel"]I’ve found that the same way $ P(5,9)$ is disproved by $ \lambda \equal{} 6$ :)[/quote]

Though, under the condition $ f\left(x\right) \equal{} \left|x\right|$, the counterexample is harder to find. This makes me think about rewriting the code.

[quote="spanferkel"]By the way, I can’t see why your argumentation with the indirect variables etc. shouldn’t be valid also for $ k > 2$. :?:[/quote]

Look here:

[quote="darij grinberg"]If one and the same variable appears both in a direct equality and in an indirect equality (for instance, the variable $ a$ in $ a \equal{} \minus{} 1$ and $ a \plus{} b \equal{} 0$), then we can replace the indirect equality by a direct equality (in this case, we replace $ a \plus{} b \equal{} 0$ by $ b \equal{} 1$).[/quote]

In the case of $ k \equal{} 3$, you can have, for instance, the equations

$ a \equal{} \minus{} 1$, $ b \equal{} \minus{} 1$, $ c \equal{} \minus{} 1$, $ a \plus{} e \plus{} f \equal{} 0$, $ b \plus{} f \plus{} d \equal{} 0$, $ c \plus{} d \plus{} e \equal{} 0$,

which lead to $ a \equal{} \minus{} 1$, $ b \equal{} \minus{} 1$, $ c \equal{} \minus{} 1$, $ d \equal{} \frac12$, $ e \equal{} \frac12$; $ f \equal{} \frac12$. So your variables don't lie in the set $ \left\{ \minus{} 1,0,1\right\}$ anymore. They are still rational and can assume only finitely many different values, but it should be rather hopeless to check these cases on a computer.

  darij

[quote="darij grinberg"]

In the case of $ k \equal{} 3$, you can have, for instance, the equations

$ a \equal{} \minus{} 1$, $ b \equal{} \minus{} 1$, $ c \equal{} \minus{} 1$, $ a \plus{} e \plus{} f \equal{} 0$, $ b \plus{} f \plus{} d \equal{} 0$, $ c \plus{} d \plus{} e \equal{} 0$,

which lead to $ a \equal{} \minus{} 1$, $ b \equal{} \minus{} 1$, $ c \equal{} \minus{} 1$, $ d \equal{} \frac12$, $ e \equal{} \frac12$; $ f \equal{} \frac12$. So your variables don't lie in the set $ \left\{ \minus{} 1,0,1\right\}$ anymore. They are still rational and can assume only finitely many different values, but it should be rather hopeless to check these cases on a computer.

  darij[/quote]Yes you are right  :blush:

I've checked some more cases in which way the majorization can be violated and found a much simpler pattern:[b]
In addition to the positions already mentioned, $ v(3)$ also yields a counter-example for all $ k < m < 2k$[/b].  :)  :blush:

So the updated table (overlaying the 2's etc.) is simply

$ \texttt{.\hskip.02in \ \ 2\ \ 3\ \ 4\ \ 5\ \ 6\ \ 7\ \ 8\ \ 9\hskip.1in 10\ 11\ 12\ 13\ 14\ 15\ 16\ 17\ 18\ 19\ 20\ 21\ 22\ 23\ 24\ \ }$
$ \texttt{1\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x\ \ x}$
$ \texttt{2\ \ \ \ \ 3\ \ x\ \ d\ \ x\ \ d\ \ x\ \ d\ \ x\ \ d\ \ x\ \ d\ \ x\ \ d\ \ x\ \ d\ \ x\ \ d\ \ x\ \ d\ \ x\ \ d\ \ x}$
$ \texttt{3\ \ \ \ \ \ \ \ 3\ \ 3\ \ x\ \ 3\ \ \ \ \ x\ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ x}$
$ \texttt{4\ \ \ \ \ \ \ \ \ \ \ 3\ \ 3\ \ 3\ \ x\ \ 3\ \ d\ \ \ \ \ x\ \ \ \ \ d\ \ \ \ \ x\ \ \ \ \ d\ \ \ \ \ x\ \ \ \ \ d\ \ \ \ \ x}$
$ \texttt{5\ \ \ \ \ \ \ \ \ \ \ \ \ \ 3\ \ 3\ \ 3\ \ 3\ \ x\ \ 3\ \ \ \ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ \ \ \ }$
$ \texttt{6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ x\ \ 3\ \ \ \ \ d\ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ d\ \ \ \ \ \ \ \ x}$
$ \texttt{7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ x\ \ 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ }$
$ \texttt{8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ x\ \ 3\ \ \ \ \ \ \ \ d\ \ \ \ \ \ \ \ \ \ \ x}$
$ \texttt{9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ x\ \ 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ }$
$ \texttt{10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ x\ \ 3\ \ 3\ \ \ \ \ }$
$ \texttt{11\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ 3\ \ x\ \ 3\ \ 3}$

So the most interesting/challenging open cases are now $ P(3,8)$ and $ P(5,12)$. And $ P(9,20)$ where it may come very very close.
And note there is still that sort of possibility that $ P(6,14)$ may be true even though $ P(3,7)$ is false.

EDIT: [quote="darij grinberg"]Just to add something new:

$ P\left(6,10\right)$ is wrong, confuted by $ x_1 \equal{} x_2 \equal{} ... \equal{} x_{11} \equal{} 1$ and $ x_{12} \equal{} x_{13} \equal{} ... \equal{} x_{16} \equal{} \minus{} 2$ and $ f\left(x\right) \equal{} \left|x\right|$. [This $ 16$-tuple and the one obtained by replacing all $ x_i$ by $ \minus{} x_i$ in it are the only two $ 16$-tuples of reals $ x_1,x_2,...,x_{16}\in\left\{ \minus{} 2, \minus{} 1,0,1,2\right\}$ that don't satisfy $ P\left(6,10\right)$ for $ f\left(x\right) \equal{} \left|x\right|$.]

$ P\left(6,11\right)$ is wrong, confuted by $ x_1 \equal{} x_2 \equal{} ... \equal{} x_{11} \equal{} 1$ and $ x_{12} \equal{} x_{13} \equal{} ... \equal{} x_{17} \equal{} \minus{} 2$ and $ f\left(x\right) \equal{} \left|x\right|$. [This $ 17$-tuple and the one obtained by replacing all $ x_i$ by $ \minus{} x_i$ in it are the only two $ 17$-tuples of reals $ x_1,x_2,...,x_{17}\in\left\{ \minus{} 2, \minus{} 1,0,1,2\right\}$ that don't satisfy $ P\left(6,11\right)$ for $ f\left(x\right) \equal{} \left|x\right|$.]

Is this much of a surprise, knowing that $ P\left(5,11\right)$ and $ P\left(7,11\right)$ are both wrong? Who knows. What is more surprising, in my opinion, is the correct $ P\left(11,22\right)$ under the wrong $ P\left(10,22\right)$ (maybe you could check your computation again? - I just find it surprising), and probably above something else that is also wrong (I don't say "exactly one line above").

  darij[/quote]I think it is more straightforward to look at the table by rows and not by columns. :)
Especially now, in the light of the 'new' $ v(3)$ thing.

EDIT: Looking at it again... each row has its 'Popoviciu island' (=the entry for $ P(k,2k)$). And the first columns probably don't. But if you look at $ P(10,22)$ - no error! This is systematical for $ k\ge10$, as the number of $ 3$'s to the right of the island increases roughly like $ \frac {4k \plus{} 15}{27}$. :D


geometry
rectangle
real analysis
real analysis unsolved

Let $L$ be a continuous (make it $C^1879$ if you want) curve in $R^2$. Let $A(L)$ the set of rectangles $R$ of $R^2$ such that $L$ is contained in the interiour of $R$. What is the $sup$ value over all curves $L$ of length 1 of the $inf$ value over $R$ in $A(L)$ of the area of $R$ (sorry for the very bad explanation).

The answer is $\frac{1}{4}$. The worst curve is the broken  line consisting of 2 perpendicular intervals of length $\frac{1}{2}$.
Here is a [hide="hint."] To show that every curve of length $1$ is contained in a rectangle of area $\frac{1}{4}$, consider the least rectangle containing our curve with one side parallel to the line containing the endpoints of the curve. Then the projection of the curve to the perpendicular side of the rectangle covers it at least twice. This is all you need to make the estimate.
[/hide]


algebra
polynomial
USAMTS
algebra unsolved

Consider a polynomial with integer coefficients f(x).such that f(200)=f(7)=2007.
Again f(0) lies between 0 and 2007.proof that f(0) doesnot depend on the choice of the polynomial and find f(0).

You probably forgot something, because if $ f$ satisfies the 1st condition, then $ f(x)\minus{}f(0), f(x)\minus{}f(0)\plus{}1, \cdots, f(x)\minus{}f(0)\plus{}2007$ all satisfy both conditions.

[quote="murgi"]Consider a polynomial with integer coefficients f(x).such that f(200)=f(7)=f(2007).
Again f(0) lies between 0 and 2007.proof that f(0) doesnot depend on the choice of the polynomial and find f(0).[/quote]
So consider $ f(x) \equal{} (x\minus{}7)(x\minus{}200)(x\minus{}2007) \plus{} 1 \plus{} 7 \cdot 200 \cdot 2007$ and $ g(x) \equal{} (x\minus{}7)(x\minus{}200)(x\minus{}2007) \plus{} 2 \plus{} 7 \cdot 200 \cdot 2007$.
We get $ f(200)\equal{}f(7)\equal{}f(2007)$ and $ g(200)\equal{}g(7)\equal{}g(2007)$. But $ f(0)\equal{}1$ and $ g(0)\equal{}2$... So it does depend on the choice of polynomial............

I believe that the original problem [url=http://www.usamts.org/Tests/USAMTSProblems_19_3.pdf]USAMTS Year 19 Round 3 #3[/url] had $ f(200) \equal{} f(7) \equal{} 2007$ NOT $ f(2007)$.

But, this might be a strikingly similar yet different problem.

[quote="murgi"]I am Sorry It will Be f(200)=f(7)=2007.[/quote]

The condition $ \Leftrightarrow f(x)\minus{}2007\equal{}(x\minus{}7)(x\minus{}200)Q(x)$, where $ Q(x)$ is a polynomial with integer coefficient.

So $ f(0)\minus{}2007\equal{}1400 Q(x)$. As $ 0<f(0)<2007 \Leftrightarrow \minus{}2007<f(0)\minus{}2007<0$,
$ f(0)\minus{}2007\equal{}\minus{}1400, f(0)\equal{}607$


modular arithmetic
number theory unsolved
number theory

Use Fermat's Little Theorem: If $ p$ is prime and $ \gcd(a,p)\equal{}1$, then $ a^{p\minus{}1}\equiv 1 \pmod{p}$.

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=317660[/url]


Don't try forcing him to do anything.  Give him guidance but let him do what he wants.

[quote="bubala"]My brother (in 4th grade--I'm in 8th) isn't that interested in math because he wants to be his own person and doesn't want to follow in my footsteps.  How do I make him love math?  Obviously, he is interested in math because he's smart and he realizes that math is actually cool, but he doesn't really pursue that much.

I know this isn't the norm.  In many families, every child loves math.  HELP ME![/quote]

The question is...did you love math in 4th grade? Give him some time untill he starts to understand stuff better

[quote="bubala"]My brother (in 4th grade--I'm in 8th) isn't that interested in math because he wants to be his own person and doesn't want to follow in my footsteps.  How do I make him love math?  Obviously, he is interested in math because he's smart and he realizes that math is actually cool, but he doesn't really pursue that much.

I know this isn't the norm.  In many families, every child loves math.  HELP ME![/quote]

Why do you want to force your brother to like math?

Maybe he's just not into it.

It must be hard for him to follow a big brother's footsteps, so it IS the norm, anthropologically speaking for younger siblings to rebel because they don't want to be in the shadow of an older sibling.

Reason I began to like math was because of Mathcounts.  Give him some Mathcounts stuff, and tell him it would be really cool if he did well as a 6th grader.  I also liked math because of some interesting things you can learn.  You should teach him random math whenever you have time, focusing on the particularly clever and useful aspects.

My brother (who is in 9th grade and i am in 12th) is very  rebelious. he is much more adapted to the american society and sees himself as "cool" and me as a nerd because i like math. this is his way to express himself since he doesn't want to just follow my footsteps. he's very smart, but he intentionally likes the things i don't like (or maybe it just happened that way?) but i think it's pretty prevalent for siblings to not want to do what their older brothers/sisters like.

Try giving him some interesting logical thinking problems and make them mathematical. Or, you could also get [u]The Joy of Mathematics[/u], which has a lot of interesting things in it.

Or, if all else fails, use the force.  :starwars:

Don't force him. What are you gonna do, make him do extra math problems at age 10? I don't care how many other people here did that, but I think it's a screwy idea. If he really likes math, you won't have to make him love it.

To end this, I'll quote the comic strip Peanuts.

Linus: It bugs me to see Sally crawling around all the time. When will she finally be able to walk?
Lucy: Don't rush her. She's got all the time in the world to learn. And once you learn how to walk, you're committed for life.

I have some excellent advice. I used to hate math in 4th grade as well. Last year in 7th grade, i skipped 2 years of math because I started enjoying it and got adept at it.

First, people enjoy things they are good at. Introduce him to some interesting problems and help him improve at mathematics. Also, peoples' minds mature more as they grow older. He may have a much different attitude as he grows up.

 :roll:

Kinda same situation here

My twin sister ( I'm fatarnel twin ) doesn't like math, and my dad thinks she is lacking at the knowledge of math.
and I've been trying to teach her, but gave up in a month

all I could teach you is use your parents' authority to do math

Hey bubala, I'm also in the same situation.  My lil brother is only 9 though.  I was thinking before that oh, when I was this age I learnt to do this and this, and could do that, but after a while, you kinda realize that he'll do what he likes and wants.  You can let him know every once in a while how cool it is, like, show him patterns in pine cones, give him a cool puzzle.

But don't force it on him.  He'll come to see it himself, hopefully.  If it is forced on him, he'll start to hate it and pull all the "I'm not like my brother/sister!!"

In the meantime just have fun with math.  Don't show him anything unless he wants to see it, but give him questions to think about that are fun.

I also have an young brother in grade 7(I'm in grade 11). He doesn't like math at all. :(
I have a question. Why would you want to get your younger siblings to love math? Even if you love math, it doesn't mean that your younger siblings have to love math.

Many people don't like math. I personally blame on education. Schools teach kids calculation for first 6 years or so(or is it just my younger brother?). They never get to see beauty of problem solving in school.

What if I sprang a book like "The Joy of Modern Art" on you and started asking you to start making you paint or copy various pieces of art?

Depends. What are you holding in the other hand? A pistol? A bag of money? Skittles?

Personally, I don't see any reason to stress siblings into math, or any other subject for that matter. I only really discovered my interest and aptitude for mathematics three years ago, in high-school. There's certainly no need to start a serious study earlier, I'd say.

I'd say definitely make sure they are aware of the tremendous fun of math but don't force it.

Yea, and as evident by myself, you don't have to be good at math to love it....

Sometimes, people only like things they are good at....

why would it be that for 2 siblings- or even fraternal twins- one would like something alot and the other hate it with a passion??? it seems to be the norm, but why

I don't really like math as much as I like thinking, and math just happens to be a clear and percise way of thinking about interesting things.

Don't try to convince your brother to like math. If your brother happens to be someone who enjoys critical thinking, logic, or debate, then you can kind of introduce him to some simple but interesting math question that do not require knowledge. That's how I became interested anyways.

[quote="liyi"]Let him read some books. "The Joy of Mathematics" is a good one...I was not interested in math until read such books. Anyway, you don't need to except a 4th grader interested in math...because the math course is not math, it is just arithmetic... Who was interested in that?[/quote]
Can you introduce sth about it??

i don't know how much this would apply to other people, but my liking for math started because i had a good deal of my education in egypt before i moved to the US. there, not as much time was spent teaching people how to add (i was learning high school algebra in 6th grade). when i moved here, i wanted more than what the normal school class was offering me because i already knew that stuff, so i sought math team to console me and found a place to do cool math in it. on the other hand, my brother came right after learning multiplication table and spent another year learning it here, he didn't experience "more" so he didn't have the same urge to find new stuff. if there is an age difference between the two siblings of a newly immigrated family, this situation might be the same.
he was forced into math team, however, and he's slowly beginning to like it :)


Each of the sides of a square $ S_1$ with area $ 16$ is bisected, and a smaller square $ S_2$ is constructed using the bisection points as vertices. The same process is carried out on $ S_2$ to construct an even smaller square $ S_3$. What is the area of $ S_3$?

$ \textbf{(A)}\ \frac {1}{2} \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

[quote="worthawholebean"]Each of the sides of a square $ S_1$ with area $ 16$ is bisected, and a smaller square $ S_2$ is constructed using the bisection points as vertices. The some process is carried out on $ S_2$ to construct an even smaller square $ S_3$. What is the area of $ S_3$?

$ \textbf{(A)}\ \frac {1}{2} \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$[/quote]

[hide]After each process, the square is decreased in size by a factor of 1/2. Thus, since this is done twice, we get: $ 16 \cdot 1/2 \cdot 1/2 \equal{} \boxed{4, \text{ E}}$[/hide]


Hello, I was just reading over the Latex tutorial and for the life of me, can't seem to space out my paragraphs.  The tutorial explicitly stated Latex only creates spaces between paragraphs when an extra space was added, as "supposedly" demonstrated by the "American civil-war speech".  When I typed this into the TeXnix centre, however, there were no gaps between the paragraphs.  Would someone mind explaining this to me.....and how to space out my paragrpahs.....
  Thanks.

  EDIT: O' and here's what I typed into TeXnix Centre:

  [i]\documentclass[11pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{array}
\usepackage{xy}

\pdfpagewidth 8.5in
\pdfpageheight 11in
\topmargin 0in
\headheight 0in
\headsep 0in
\textheight 7.7in
\textwidth 6.5in
\oddsidemargin 0in
\evensidemargin 0in
\headheight 77pt
\headsep 0.25in

\begin{document}

Fourscore and seven years ago our fathers brought forth on this continent a new nation, conceived in liberty and dedicated to the proposition that all men are created equal.

Now we are engaged in a great civil war, testing whether that nation or any nation so conceived and so dedicated can long endure. We are met on a great battlefield of that war. We have come to dedicate a portion of it as a final resting place for those who died here that the nation might live. This we may, in all propriety do. But in a larger sense, we cannot dedicate, we cannot consecrate, we cannot hallow this ground. The brave men, living and dead who struggled here have hallowed it far above our poor power to add or detract. The world will little note nor long remember what we say here, but it can never forget what they did here.

It is rather for us the living, we here be dedicated to the great task remaining before us--that from these honored dead we take increased devotion to that cause for which they here gave the last full measure of devotion--that we here highly resolve that these dead shall not have died in vain, that this nation shall have a new birth of freedom, and that government of the people, by the people, for the people shall not perish from the earth.[/i]

\end{document}

The default settings in LaTeX put no space in the paragraph skip.  To have a line skip, do this:

\setlength{\parskip}{length1 plus length2 minus length3}

where length1, 2, 3 are .....length!

length1 is the basic skip for your paragraph
length2 is the length you allow latex to add to ajust the layout of your page
length3 is the length you allow latex to remove to ajust the layout of your page

It is not a good thing to use both a pararagraph skip and an indent.  To remove the indent, do this:

\setlength{\parindent}{0ex}

And what would be the optimal length setting for someone who is using the packages mentioned above and uses Latex solely for document-writing?

personnally, I use 3ex plus 1ex minus 0ex but with one and half spacing.  Try and see what you like.


function
analytic geometry
algebra
domain
topology
real analysis
real analysis unsolved

Does there exist a function $ f$: $ R$->$ R$ such that for each real $ y$, $ f^{\minus{}1}(y)$ is an uncountable set?

Yes, because $ \mathbb{R}\times\mathbb{R}$ has the same cardinality as $ \mathbb{R}$. Compose projection on the first coordinate with a surjective map from $ \mathbb{R}$ to $ \mathbb{R}^2$.

Can you explain a little bit further, or give an example?

I'm pretty sure an example is precisely what he gave you.

It's been discussed multiply on these forums, the existence of a surjection f:[0,1]->[0,1] such that the image under f of any open set in the domain is [0,1].

That's slightly different- the example I've used has countably infinite preimages.


$ f(x)\equal{}x\plus{}x^x$ is a increasing function, so the only answer is 5.
We can check that for negative numbers and for 0<x<1 there are no solutions, as then $ x\plus{}x^x$ is less than 2.
For x>1, it's increasing, so it's 5

I agree with the general idea of your solution Bugi, but there are a few problems in the details. First of all, $ 5 \plus{} 5^5 \neq 30$ (maybe you meant to say $ x \equal{} 3$).

And how do you support some of the claims in your explanation? Some of them are not so obvious. Like how do you know there are no solutions for negative numbers? This function behaves very strangely for negative rational numbers.

hello, no, the solution is indeed $ x\equal{}3$, because $ 3\plus{}3^3\equal{}3\plus{}27\equal{}30$.
Sonnhard.

Bugi, it's not $ x\plus{}x^2$ but $ x\plus{}x^x$. $ 5\plus{}5^5 \neq 30$.
Still, I don't think it's enough to just look at positive integers as others said. You need to check all rational and irrational numbers as well.

[quote="Dr Sonnhard Graubner"]hello, no, the solution is indeed $ x \equal{} 3$, because $ 3 \plus{} 3^3 \equal{} 3 \plus{} 27 \equal{} 30$.
Sonnhard.[/quote]

How do you know that's the only solution?

This problem can be tackled by noticing the increasing rate of $ x^x$  and establishing a limit for x, but these problems should probably be solved by virtue of graph.

hello, let $ f(x)\equal{}x\plus{}x^x$, then $ f(x)$ has a local minimum at $ x\approx0.103818$ and $ y\approx0.8942625235$ and further we have $ \lim_{x\to 0\plus{}}f(x)\equal{}1$, $ \lim_{x \to \plus{}\infty}f(x)\equal{}\plus{}\infty$ so we have only one positive solution of our equation $ x\plus{}x^x\equal{}30$.
Sonnhard.


trigonometry
inequalities
inequalities proposed

i found this problem on a "shortlist" from the ibero 96 held in costa rica. i couldn't solve the problem and when i checked the solution it was like 3 pages long (as far as i remember). anyway i consider it a nice problem... enjoy it!

Let $\cos\theta_1+\cos\theta_2+...+\cos\theta_{2n}=0$.
Prove that $|\cos\theta_1+2\cos\theta_2+3\cos\theta_3+...+2n\cos\theta_{2n}|\leq n^2$

i'm so sorry for the mistake... i've edited it now  :blush:

From the condition $\cos\theta_1+\cos\theta_2+\cos\theta_3+...+\cos\theta_{2n}=0$ and the inequality $|\cos x|\leq 1, \forall x\in R,$ we obtain:
$\cos\theta_2+\cos\theta_3+...+\cos\theta_{2n}=-\cos\theta_1\in [-1;1]$
$\cos\theta_3+\cos\theta_4+...+\cos\theta_{2n}=-\cos\theta_1-\cos\theta_2\in [-2;2]$
...
$\cos\theta_{2n-1}+\cos\theta_{2n}\in[-2;2]$
${\cos\theta_{2n}\in [-1;1]}$
Hence
\[ |\cos\theta_1+2\cos\theta_2+3\cos\theta_3+...+2n\cos\theta_{2n}|\leq  \] \[ \leq |\cos\theta_1+\cos\theta_2+\cos\theta_3+...+\cos\theta_{2n}|+  \] \[ +|\cos\theta_2+\cos\theta_3+...+\cos\theta_{2n}|+...+|\cos\theta_{2n-1}+\cos\theta_{2n}|+|\cos\theta_{2n}|\leq  \]
\[ \leq 0+1+2+...+(n-1)+n+(n-1)+...+2+1=n^2.  \]


My biology book (in the biochem section) depicts a nonionized phosphate group with both hydroxyls attached to the phosphate. The ionized obviously has the negative oxygen instead of the hydroxyl.

My question would be, if only one hydroxyl was removed (and a negative oxygen in it's place), would the molecule be considered ionized? Or partial ionized?

it wud still be called ionised an atom/molecule cannot be partially ionised

I think this depends on the terminology used by a given author. Sometimes, we can say that a molecule is "partially ionized" when it has the capacity to donate more than one proton (for example). On the other hand, partially ionized is also often used in relation with weak acids/bases: if an acid/base is weak, then when in the presence of water (or other solvent) it will, when equilibrium is attained, be only partially ionized.

So a phosphate group with one hydroxyl ionized [i]is[/i] possible?


Why is it that there is no polyhedron with 7 vertices, 11 edges and 6 faces, when in fact it satisfies the Euler's formula: vertices - edges + faces = 2?

Is there even a polyhedron with 11 sides?

How do you know there isn't? Maybe you just don't know how to construct it... (not saying it must exist, but just a possibility).

There is such a polyhedron. Why did you think there isn't one?


$a,b,c$ are positive such that
$(a^2+b^2+c^2)^2 \ge 2(a^4+b^4+c^4)$ prove that
$a,b,c,$ are the sides of triangle.

It is very well known inequality related to Heron's formulae.
It is the same as $(a+b+c)(a+b-c)(a-b+c)(-a+b+c)\geq 0$.


geometry
trigonometry
ratio
projective geometry
trig identities
Law of Sines
geometry proposed

Points $ A_1$, $ B_1$, $ C_1$ are chosen on the sides $ BC$, $ CA$, $ AB$, respectively of a triangle $ ABC$. Denote by $ G$, $ G_a$, $ G_b$, $ G_c$ are the centroids of triangles $ ABC$, $ AB_1C_1$, $ BC_1C_1$, $ CA_1B_1$, respectively. The centroids of triangles $ A_1B_1C_1$ and $ G_aG_bG_c$ denoted by $ G_1$, $ G_2$, respectively. Prove that:
$ a.$ Points $ G$, $ G_1$, $ G_2$ are collinear.
$ b.$ Lines $ AG_a$, $ BG_b$, $ CG_c$ are concurrent if and only if lines $ AA_1$, $ BB_1$, $ CC_1$ are concurrent.

[hide="Solution"]
For the first part, we proceed with complex numbers.  Let point $ X$ be $ x$ in the complex plane.  Thus, we have that $ G_a \equal{} \frac {a \plus{} b_1 \plus{} c_1}{3}$, $ G_b \equal{} \frac {b \plus{} c_1 \plus{} a_1}{3}$, and $ G_c \equal{} \frac {c \plus{} a_1 \plus{} b_1}{3}$.  Thus, the centroid of $ \triangle G_aG_bG_c$, which is $ G_2$, is $ \frac {a \plus{} b \plus{} c \plus{} 2a_1 \plus{} 2b_1 \plus{} 2c_1}{9}$.  Also, the centroids of $ \triangle ABC$, which is $ G$, and $ \triangle A_1B_1C_1$, which is $ G_1$, are $ \frac {a \plus{} b \plus{} c}{3}$ and $ \frac {a_1 \plus{} b_1 \plus{} c_1}{3}$ respectively.  Therefore, $ G_1$, $ G_2$, and $ G$ are collinear if and only if $ \frac {g \minus{} g_1}{g_2 \minus{} g_1} \equal{} \frac {\frac {a \plus{} b \plus{} c}{3} \minus{} (\frac {a_1 \plus{} b_1 \plus{} c_1}{3})}{\frac {a \plus{} b \plus{} c \plus{} 2a_1 \plus{} 2b_1 \plus{} 2c_1}{9} \minus{} (\frac {a_1 \plus{} b_1 \plus{} c_1}{3})}$ is real.  Yet, this is just $ 3$ after simplifying, so $ G$, $ G_1$, and $ G_2$ are collinear.

For the second part, I will present two solutions.

[u]Solution 1:[/u]
Let $ M$ be the midpoint of $ B_1C_1$.  Then, applying the Law of Sines on $ \triangle AMC_1$ and $ \triangle AMB_1$ yields that $ \frac {\sin \angle C_1AM}{\sin \angle AMC_1} \equal{} \frac {C_1M}{AC_1}$ and $ \frac {\sin \angle MAB_1}{\sin \angle AB_1M} \equal{} \frac {MB_1}{AB_1}$.  Dividing yields that $ \frac {\sin \angle C_1AM}{\sin \angle AMC_1} \equal{} \frac {AB_1}{AC_1}$.  Yet, line $ AM$ is the same as line $ G_aA$ since $ G_a$ is the centroid of $ \triangle AC_1B_1$.  Doing the same for the rest and multiplying yields that
\[ \frac {\sin \angle C_1AG_a}{\sin \angle G_aAB_1}\cdot \frac {\sin \angle B_1CG_c}{\sin \angle A_1CG_c}\cdot \frac {\sin \angle A_1BG_b}{\sin \angle C_1BG_b} \equal{} \frac {AB_1}{AC_1}\cdot \frac {A_1C}{BC_1}\cdot \frac {BC_1}{A_1B}
\]
Thus, one of these is equal to one if and only if the other one is equal to one as well.  By the converse of Sine Ceva and regular Ceva, we see that $ AA_1$, $ BB_1$, and $ CC_1$ concur if and only if $ AG_a$, $ BG_b$, and $ CG_c$ concur.

EDIT:

[u]Solution 2:[/u]
Let $ M$, $ N$, and $ P$ be the respective midpoints of $ B_1C_1$, $ A_1C_1$, and $ A_1B_1$.  Let $ AM$, $ CP$, and $ BN$ meet $ BC$, $ AB$, and $ AC$ at $ X$, $ Z$, and $ Y$ respectively.  Notice that $ \frac {BX}{XC} \equal{} \frac {[AXB]}{[AXC]} \equal{} \frac {\frac {AB}{AC_1}\cdot [AXC_1]}{\frac {AC}{AB_1}\cdot [AXB_1]}$.  Yet, since the diagonal $ AX$ bisects the diagonal $ B_1C_1$ of quadrilateral $ AC_1XB_1$, we have that $ [AC_1X] \equal{} [AB_1X]$, so $ \frac {BX}{XC} \equal{} \frac {AB}{AC}\cdot \frac {AB_1}{AC_1}$.  We do the same with the rest, so $ \frac {CY}{AY} \equal{} \frac {BC}{AB}\cdot \frac {BC_1}{BA_1}$ and $ \frac {AZ}{ZB} \equal{} \frac {CA}{BC}\cdot \frac {CA_1}{BC_1}$.  After multiplying everything together and noting that $ \frac {BC}{AB}\cdot \frac {AC}{BC}\cdot \frac {AB}{AC} \equal{} 1$, we have that
\[ \frac {CY}{AY}\cdot \frac {AZ}{ZB}\cdot \frac {BX}{XC} \equal{} \frac {BC_1}{BA_1}\cdot \frac {CA_1}{B_1C}\cdot \frac {AB_1}{AC_1}
\]
Thus, the left hand side of this equation equals one if and only if the right hand side of this equation equals one.  By Ceva's Theorem, we see that $ AX$, $ BY$, and $ CZ$ concur if and only if $ AA_1$, $ BB_1$, and $ CC_1$ concur.  Yet, line $ AX$ is the same as line $ AM$ and since $ G_a$ is the centroid of $ \triangle AB_1C_1$, we see that line $ AX$ is the same as line $ AG_a$.  Thus, $ AG_a$, $ BG_b$, and $ CG_c$ concur if and only if $ AA_1$, $ BB_1$, and $ CC_1$ concur. [/hide]

Dear April,

this problem is indeed quite nice, but it's not as hard as it looks if you know some useful properties of the centroid. Because of the fact that we are dealing with centroids who can be treated as centers of mass, part a) can be seen quickly.

[u][b]Part a):[/b][/u] This follows from Theorem 5 in http://forumgeom.fau.edu/FG2008volume8/FG200816index.html
It states: [i]The centroid of a set of points $ \mathcal{S}$ is also the centroid of the system of centroids of the k-subsystems of $ \mathcal{S}$.[/i]
We can even calculate the ratio $ \frac{|G_1G|}{|GG_2|}$ if we want to, using similar theorems.

[u][b]Part b):[/b][/u] For the names of points, look at the picture.
Now, $ \Delta ABC$ and $ \Delta G_aG_bG_c$ are concurrent if and only if $ \Delta ABC$ and $ \Delta A_2B_2C_2$ are.
By Desargues' theorem, this is true if and only if $ X_2,Y_2,Z_2$ are collinear.

Now, we note that $  C_1X_1| = 2 |B_2X_2|$, so we have that $ \frac{X_2B_2}{X_2C_2} = \frac{X_1C_1}{X_1B_1}$. Now Meneloas' theorem shows that $ X_2,Y_2,Z_2$ are collinear if and only if $ X_1,Y_1,Z_1$ are collinear. This last statement is, using Desargues again, clearly equivalent to saying that $ \Delta ABC$ and $ \Delta A_1B_1C_1$ are perspective. $ \Box$

[u][b]Picture:[/b][/u]

[img]http://img528.imageshack.us/img528/5505/aprilcentroidssa8.png[/img]

[b][size=100]Part (a)[/size][/b]. - It is easy to prove, as a neat application of the [b][size=100]ERIQ theorem[/size][/b] ( [b]E[/b]qual [b]R[/b]atios [b]I[/b]n [b]Q[/b]uadrilateral ). Also, it is true that $ GG_{2} \equal{} 2\cdot G_{2}G_{1}.$

[b][size=100]Part (b)[/size][/b]. - Also it is easy to prove, as a direct application of the [b][size=100]Cevian Nest theorem[/size][/b].

I will come back soon with more details.

Kostas vittas.

Part a) just assign weights p,q,r to A,B,C so that $ pA \plus{} qB \equal{} (p \plus{} q)C'$ and cyclically.

Then mass point geometry just destroyed the problem.

b) These very same mass points I do believe pwn the problem, anyone want to verify?

[quote="vittasko"][b][size=100]Part (a)[/size][/b]. - It is easy to prove, as a neat application of the [b][size=100]ERIQ theorem[/size][/b] ( [b]E[/b]qual [b]R[/b]atios [b]I[/b]n [b]Q[/b]uadrilateral ). Also, it is true that $ GG_{2} \equal{} 2\cdot G_{2}G_{1}.$

[b][size=100]Part (b)[/size][/b]. - Also it is easy to prove, as a direct application of the [b][size=100]Cevian Nest theorem[/size][/b].[/quote]
Let $ M$ be, the midpoint of $ BC$ and also let $ N_{1},\ N_{2},\ N_{3}$ be, the midpoints of the segments $ B_{1}C_{1},\ C_{1}A_{1},\ A_{1}B_{1},$ respectively.

We denote as $ P,\ Q,$ the midpoints of the segments, $ G_{b}G_{c},\ N_{2}N_{3}$ respectively and in the non-convex quadrilateral $ BCN_{3}N_{2},$ based on the [b][size=100]ERIQ theorem[/size][/b],

already has been posted in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=97493]Collinearity of orthocentres[/url] of [b][size=100]shobber[/size][/b], we conclude that the points $ M,\ P,\ Q,$ are collinear and also that $ MP \equal{} 2\cdot PQ$ $ ,(1)$

We consider now, the non-convex quadrilateral $ AN_{1}QM$ and because $ \frac {AG_{a}}{G_{a}N_{1}} \equal{} \frac {MP}{PQ} \equal{} 2$ and also because of $ \frac {AG}{GM} \equal{} \frac {G_{a}G_{2}}{G_{2}P} \equal{} \frac {N_{1}G_{1}}{G_{1}Q} \equal{} 2,$

aplying again the [b][size=100]ERIQ theorem[/size][/b], we conclude that the points $ G,\ G_{2},\ G_{1},$ are collinear and also that $ GG_{2} \equal{} 2\cdot G_{2}G_{1}.$

So, it is completed the solution of the [b][size=100]part (a)[/size][/b], of the proposed problem.

$ \bullet$ As for the [b][size=100]part (b)[/size][/b], we can say that when the lines segments $ AA_{1},\ BB_{1},\ CC_{1}$ are concurrent, the result of the concurrency of the line segments

$ AG_{a}\equiv AN_{1},\ BG_{b}\equiv BN_{2},\ CG_{c}\equiv CN_{3}$ it follows, because of the line segments $ A_{1}N_{1},\ B_{1}N_{2},\ C_{1}N_{3},$ are concurrent at the centroid $ G_{1}$ of $ \bigtriangleup A_{1}B_{1}C_{1}.$

It is the well known [b][size=100]Cevian Nest theorem[/size][/b], posted by [b][size=100]darij grinberg[/size][/b] somewhere in this forum and I will post here later the link, because I can’t find it now.

Kostas Vittas.

PS. I temporaryly named as [b][size=100]ERIQ theorem[/size][/b] this power theorem, but I would like to find another better name. So I am looking forward in your oppinions and suggestions about his name, or if there is a reference about it.

[quote="vittasko"]It is the well known [b][size=100]Cevian Nests theorem[/size][/b], posted by [b][size=100]darij grinberg[/size][/b] somewhere in this forum and I will post here later the link, because I can’t find it now.[/quote]
See at the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=6579]Concurrency (useful!): Cevian Nests Theorem (or part of it)[/url], of [b][size=100]Arne[/size][/b].

Kostas Vittas.


integration
real analysis
real analysis solved

The infinum is $\frac1e$ (I think).

I'll post my argument a little later.

Let $g(t)=e^{-t}f(t).$ Then $g'(t)=e^{-t}(f'(t)-f(t))$ and $g\in E'$ where

$E'=\{g\in C^1[0,1]:g(0)=0,g(1)=e^{-1}\}.$

The infinum we are looking for becomes

$\inf_{g\in E'}\int_0^1e^t|g'(t)|dt$

But $\int_0^1e^t|g'(t)|dt> \int_0^1|g'(t)|dt\ge\left|\int_0^1g'(t)dt\right|=e^{-1}.$

We can't equal this but we can approach it by having $g(t)$ rise quickly from 0 to 1 very near $t=0$ and then remain 1 for the rest of the interval.


We have the following pattern:

3 can be decomposed as    3 , 1+2 , 1+1+1
          # of ways = 3
4 can be decomposed as    4 , 1+3 , 1+1+2 , 1+1+1+1
          # of ways = 4
5 can be decomposed as    5 , 1+4 , 1+2+2 , 1+1+3 , 1+1+1+2 ,
                                     1+1+1+1+1
          # of ways = 6
6 can be decomposed as    6 , 1+5 , 1+1+4 , 1+1+2+2 ,
                                     1+1+1+3 , 1+1+1+1+2 , 3+3
                                     1+1+1+1+1+1+1 , 2+2+2 , 1+2+3 ,
          # of ways =10

Is there a general formula for the number of ways to decompose a given integer?

http://en.wikipedia.org/wiki/Integer_partition

If this topic interests you, you may want to have a look at "additive number theory."


sorry if it's dumb.....

i havent downloaded latex yet, so i have written x to the power 5 as x^5,ok?

prove that the expression......
x^5 -  b(x^3) + c(x^2)+dx - e will be the product of a complete square and a complete cube if.....
12b/5 = 9d/b = 5e/c = d^2/c^2

Don't have to download anything... the forum has built in code. :)[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5261]Link[/url]

i believe you have to input the values in terms of any term $a,b,c,d,e$ and then factorise the thing and get the result but im too lazy to do all that.
 :D


linear algebra
matrix
linear algebra unsolved

$ A = \left(\begin{array}{ccc}-1 & 1 & 1\\
1 & -1 & 1\\1 & 1 & -1\end{array}\right)\$
Calculate $ A^n$
Something more generality....If $ B \equal{} \left(\begin{array}{ccc}a & b & b\\
b & a & b\\b & b & a\end{array}\right)\
Calculate$$ B^n$

We are to use Null potent, then binominal theorem, I think.

Let $ J$ be the matrix of all ones; $ J^2\equal{}mJ$, where $ m$ is the size of the matrix. Now $ (J\minus{}2I)^n\equal{}a_nJ\plus{}b_nI$; $ a_{n\plus{}1}\equal{}ma_n\minus{}2a_n\plus{}b_n$ and $ b_{n\plus{}1}\equal{}\minus{}2b_n$. Solve these linear recurrences (with initial conditions $ a_0\equal{}0,b_0\equal{}1$: $ b_n\equal{}(\minus{}2)^n$ and $ a_n\equal{}\frac{1}{m}(m\minus{}2)^n\minus{}\frac{1}{m}(\minus{}2)^n$.

The diagonal entries are $ \frac1m\left((m\minus{}1)(\minus{}2)^n\plus{}(m\minus{}2)^n\right)$ and the off-diagonal entries are $ \frac1m\left((m\minus{}2)^n\minus{}(\minus{}2)^n\right)$.

You're looking at the special case $ m\equal{}3$. The generalization is similar.


algebra
polynomial
calculus
derivative
algebra unsolved

Find all polynomials $P(x,y)$ with real coefficients such that:
$P(x,y)=P(x+1,y)=P(x,y+1)=P(x+1,y+1)$

Maybe I'll do something stupid but:

$P(x,y)=P(x+1,y)\implies \frac{\partial P}{\partial x}=0\implies P(x,y)=f(y)$

$P(x,y)=P(x,y+1)\implies\frac{\partial P}{\partial y}=0\implies \frac{\text{d}}{\text{d}y}f(y)=0\implies f(y)=C$

Hence $P(x,y)=C$

Or, without resorting to partial derivatives:

The only solution of the polynomial equation $Q(t)=Q(t+1)$ is $Q(t)=C$. Therefore if $P(x,y)=P(x+1,y)$, then $P(x,y)$ doesn't depend on $x$, hence $P(x,y)=R(y)$. Then from $P(x,y)=P(x,y+1)$ we find $R(y)=R(y+1)\implies R(y)=C$, which gives $P(x,y)=C$

nice solution. :lol:
seems I have to change the topic's name :maybe: , it wasn't that much tough


Let $ a,b,c\in [0,1].$Prove that
$ \sum\frac{a}{1\plus{}bc}\plus{}abc\leq\frac{5}{2}$

[quote="DCTPKTCSPHN"]Let $ a,b,c\in [0,1].$Prove that
$ C = \sum\frac {a}{1 + bc} + abc\leq\frac {5}{2}$[/quote]
Assume $ a \geq b \geq c$
$ = > C \leq \frac {a}{1 + bc} + \frac {b + c}{1 + bc} + abc \leq \frac {a}{1 + bc} + abc + 1 = V$
$ V = \frac {a(bc^2 + bc + 1)}{1+bc} + 1 \leq 5/2$ (Because ${ a,b,c\in[0,1]}$ :lol:  :P


While competing in a school mini-triathlon, Ray ran 4 miles in 45 minutes, swam 1 mile in 20 minutes, and biked 15 miles in 50 minutes. What was his average rate for the race?

Total distance: 20 miles.
Total time: 115 minutes.

Rate: $ \frac{20}{115}$=$ \boxed{\frac{4}{23}}$

[hide="solution"]First we find how far he travelled (4 miles running + 1 mile swimming + 15 miles biking = 20 miles total)
Then we divide that total by his total time (45 minutes running + 20 minutes swimming + 50 minutes biking)

So, our distance / time is 20 miles/1:55.
The problem asks for his average speed, so we will convert our time into minutes (115 minutes).. 20/115=.174 miles/minute.  Multiply by 60 to find his average mph and you get [b]10.435 mph[/b][/hide]


Let $p(x)=x^{6}+x^{5}+\cdots+1.$ Find the remainder when $p(x^{7})$ is divided by $p(x).$


[quote="fallinlovewithmaths"]Is there a covering f:R to R function such that for different values of x and y , f(x+y) - f(x) - f(y) can be exactly 0 and 1.[/quote]

Could you be more precise in your problem ?

If you are looking for functions such that $ f(x\plus{}y) \minus{} f(x) \minus{} f(y)\equal{}0$ $ \forall x,y\in\mathbb{R}$, choose $ f(x)\equal{}ax$, for example

If you are looking for functions such that $ f(x\plus{}y) \minus{} f(x) \minus{} f(y)\equal{}1$ $ \forall x,y\in\mathbb{R}$, choose $ f(x)\equal{}ax\minus{}1$, for example


How many functions $f: \{ a,b,c \} \to \{0,1,2,3,4\}$ have the property that $f(a)+f(b)+f(c)=11$?

:huh:
At least two numbers of $f(a)$, $f(b)$ and $f(c)$ are 4, so remaining number is 3. Therefore the answer is 3.