Exactly what the topic says, signups and the such for mafia moderators.
Signing up for games should be [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=534&t=490400]here[/url].

[u][b]Queue[/b][/u]
[quote="Mini Game Queue"]
[list=1]
RTG - Level 2
summitwei- Level 2
abean077 - level 0 (not sure large or mini - please specify)
etr217 - level 2[/list]
[/quote]
[quote="Large Game Queue"]
[list=1]summitwei - Level 0
relay400 - level 3
statisticslol + Kagebaka - level 3
SFGiantsFan - level 0
whatshisbucket - level 3
fdas - level 4
benjaminsz - level 3
math_turtle37 - level 1
hdgehrke - level 0
jeffisepic - level 0
pandabear10 - level 0
fdas - level 3
[/list]
[/quote]
[quote="Newbie Queue"]
Just do newbies whenever you want if you can tbh

No formal queue is really necessary but here it is
GeneralCobra19
[/quote]
[quote="Co-Mods"]
[list=1][/list][/quote]
[hide="Levels"]Level 0 - request for game modding
Level 1 - vague theme ready
Level 2 - basic gameplay and mechanics ready
Level 3 - specific role distribution and other details ready
Level 4 - reviewed and balanced setup
Level 5 - fully polished setup ready
[hide="Levels in Detail"]
Level 0- Sign up
Level 1- Have a basic idea of flavor (or no flavor)
Level 2- Be ready with basic mechanics, end situations, etc.
Level 3- Start finishing up most of the stuff, making sure everything works. At this point, you should have a setup ready to review
Level 4- Once you have your setup reviewed, you're at this stage. Submit proof (by PM) to a mod to get your setup put to this level
Level 5- At this point, you're just waiting. If you're doing advanced mechanics, you'll be doing it between stage 4 and 5 (Not recommended for new mods).
[/hide][/hide]

[u]Notes to Mods[/u]
Rules for playing are in the other thread. This is more for the mods of games.
Setups should be balanced. All to-be game mods must PM some experienced player with their setup before the beginning of the game, including specifics (such as whether it is an [url=http://www.mafiascum.net/wiki/index.php?title=Open]open[/url] game, whether specific actions such as quoting PMs are allowed, among others). Ask them whether they have interest on the game before giving the setup though; if you have a closed setup and they want to play, obviously you can't give the setup.
Balancing is important. One of the most important things to remember is have lots of townies for the scum to hide in.
If you feel inexperienced, feel free to have a backup mod. It is best to keep the front post of the game updated with who is dead and alive, and the roles of the dead (and generally when they died).

[i]Here are some details that I think should be addressed: [/i]
[list][*]Quoting role PMs. This is usually not to be allowed, but feel free to experiment.
[*]Posting the townie (citizen) PM in the first or second post of the thread. This way the scum have at least something to go by.
[*]Giving scum fakeclaims. This means give them plausible rolenames (esp if it is a theme game) that are not given to other players, and give them plausible [i]roles[/i] that are not given to other players.
[*]How to reveal roles of the dead. In general, all you need to do is post the rolename of the player, the role of the player, and his alignment. That is all that is usually necessary.
[*]A standard set of rules. I want all mods, after creating their topic, to put the rules (together with story, player list, etc) in the first post. If you're unsure of what to put in the rules, check a few games.
[*]Open games vs. Closed games. The first 20+ games on this site were open- that is, all the roles were known to the public before the game.
[*]Millers. (Not so much an issue anymore) It is pretty standard that if a miller (a townie that turns up guilty to cops) is going to be included in a game, he should be aware that he is a miller. No one else need be aware that he is a miller.
[*]Roles in general. See the mafiascum wiki for a variety of standard and non-standard roles.
[*]Dealing with inactivity. Use the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=534&t=218829]Replacement Thread[/url] for finding replacements.[/list]

Added. Also, if you're looking at this thread, please read the rules. They're good.

[quote="Klebian"][url=http://mikeburnfire.sitesled.com/mscumB.swf]This[/url] Flash thing is pretty awesome.[/quote]

Agreed.(but it has some off color [a tad bit] animations, so I would rate this PG-13, and 12 and under should look at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=177754]this[/url] version of the rules)

That link you gave is just rules- my link was for ROLES. Maybe a bit off color but then so is the essential concept of mafia- killing people daily and nightly?

PG 13 is a bit extreme.

1=2, I doubt there are many (if any) under-13-year-olds here besides us.

does it matter, lots of people at school have watched at least 1 R rated movie

I don't sign up, and the tutorial is pretty good

That flash, unfit for people below 13? Rofl. I know few people who hadn't seen porn/violence/whatever at age 13.

Especially people on a site like this (who are a bit more mature than the age average) should have no problem. The violence isn't even animated there... I think some people suffer from paranoia.

[quote="Peter"]Especially people on a site like this (who are a bit more mature than the age average) should have no problem.[/quote]I'd say people here have seen less porn/violence/whatever than average.

Do any of you guys have themes for game 30 yet?

If don't, would it be weird if I suggest one?

I have this cool theme...

put me back on the mod list
I'll make it less complicated this time

27's over town win

I might mod...
Nah I'm not responsible enough.
Maybe later.

[quote=scannose]lol aops default emojis[/quote]

fr

[quote=scannose][quote=awesomeming327.]ok but i wrote "uwu" so i get automatic precendence[/quote]

qwq > uwu[/quote]

what even is a qwq

makes no sense
unpronouncable, unvisualizable, hasn't gathered any connotation uwu has

[img]https://cdn.discordapp.com/attachments/836327783357087754/1111054115066351666/image.png[/img]

look it lacks a clear identity

[quote=ad333][quote=scannose][quote=awesomeming327.]ok but i wrote "uwu" so i get automatic precendence[/quote]

qwq > uwu[/quote]

what even is a qwq

makes no sense
unpronouncable, unvisualizable, hasn't gathered any connotation uwu has[/quote]

ok but

[quote=scannose][quote=awesomeming327.]ok but i wrote "uwu" so i get automatic precendence[/quote]

qwq > uwu[/quote]

you're looking for TwT

hey snoozing wanna ghostmod for me again <3

[quote=scannose][quote=awesomeming327.]ok but i wrote "uwu" so i get automatic precendence[/quote]

qwq > uwu[/quote]

wait qwq is supposed to be crying

wow i can't believe you're encouraging emotional maniipulation like that

im not sure but qwq
is like crying but
not quite crying

[quote=ad333][quote=WisteriaV]if you’re gonna do crying
use T_T[/quote]

wait is ToT also crying

or no[/quote]

looks like screaming + crying
or some other crying with your mouth open

don’t take my word for it

edit: google says it’s just a crying emoji

T_T is like dying inside
ToT is like the sobbing emoji


Find the number of real roots of the equation: $ 2^x\equal{}x^2\plus{}1$.

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

$ f(0) \equal{} 0$

$ f(1) \equal{} 0$

$ f(2) \equal{} \minus{} 1$

$ f(3) \equal{} \minus{} 2$

$ f(4) \equal{} \minus{} 1$

$ f(5) \equal{} 6$

Therefore ,$ f(x) \equal{} 0$ has odd number of roots and hence atleast one root between $ 4$ and $ 5$ and hence $ f(x) \equal{} 0$ has atleast $ 3$ roots.To prove that $ f(x) \equal{} 0$ has exactly $ 3$ roots it needs to be proved that $ f'(x)$ is positive for $ x\geq 5$.

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

is clearly positive for $ x\geq 5$.

What is left now is to prove that there exists exactly one root between $ x \equal{} 4$ and $ x \equal{} 5$.Let the root between $ x \equal{} 4$ and $ x \equal{} 5$ be $ x \equal{} c(f(c) \equal{} 0)$.
Consider the interval $ [4,c]$.By [b]LMVT[/b] one can see that $ f'(c_{1}) \equal{} \frac {f(c) \minus{} f(4)}{c \minus{} 4} \equal{} \frac {1}{c \minus{} 4}\geq 0$
for $ c_{1}\in(4,c)$.

Similarly,Consider the interval $ [c,5]$.By [b]LMVT[/b] one can see that $ f'(c_{2}) \equal{} \frac {f(5) \minus{} f(c)}{5 \minus{} c} \equal{} \frac {6}{5 \minus{} c}\geq 0$
for $ c_{2}\in(c,5)$.

Further we can assume two roots between $ x\equal{}4$ and $ x\equal{}5$ ,say $ d_{1}$ and $ d_{2}$.Thus by Rolle's Theorem there must exist some $ x\equal{}e$ for which $ f'(e)\equal{}0$ for some $ e\in(d_{1},d_{2})$ which we see is not possible since none of $ f'(c_{1}),f'(c_{2})$ is zero,rather they turn out to be positive quantities.

Therefore between $ x \equal{} 4$ and $ x \equal{} 5$ we have $ f'(x)\geq 0$ and hence since $ f(x)$ is an increasing function on $ [4,5]$ there can exist only [u]one[/u] root.

Therefore,in totality $ f(x) \equal{} 0$ has exactly [b]3[/b] roots.

Let me know whether the answer and solution are correct or not

Find the number of real roots of the equation:
$ 2^x\equal{}x^2\plus{}n$, where $ n\in N$.

[quote="soruz"]Find the number of real roots of the equation:
$ 2^x \equal{} x^2 \plus{} n$, where $ n\in N$.[/quote]
But first tell me whether I am right or wrong

[quote="soruz"]Find the number of real roots of the equation:
$ 2^x \equal{} x^2 \plus{} n$, where $ n\in N$.[/quote]

$ f(x)\equal{}2^x\minus{}x^2\minus{}1$
Let us assume that there are $ 4$ roots of $ f(x)\equal{}0$.
Then $ f'(x)\equal{}0$ must have atleast $ 3$ roots.
Then $ f''(x)\equal{}0$ must have atleast $ 2$ roots.
Then $ f'''(x)\equal{}0$ must have atleast $ 1$ root i.e $ 2^x(log_{e}2)^3\equal{}0$ must have atleast $ 1$ root which is clearly a contradiction.
Therefore,the equation $ f(x)\equal{}0$ has exactly $ 3$ roots.

If $ f(x)\equal{}2^x\minus{}x^2\minus{}n$
then the equation $ f(x)\equal{}0$ can have at most $ 3$ roots if we go by the same logic demonstrated above.

[quote="pankajsinha"]Let $ f(x) \equal{} 2^x \minus{} x^2 \minus{} 1$
What is left now is to prove that there exists exactly one root between $ x \equal{} 4$ and $ x \equal{} 5$.Let the root between $ x \equal{} 4$ and $ x \equal{} 5$ be $ x \equal{} c(f(c) \equal{} 0)$.
Consider the interval $ [4,c]$.By [b]LMVT[/b] one can see that $ f'(c_{1}) \equal{} \frac {f(c) \minus{} f(4)}{c \minus{} 4} \equal{} \frac {1}{c \minus{} 4}\geq 0$
for $ c_{1}\in(4,c)$.

Similarly,Consider the interval $ [c,5]$.By [b]LMVT[/b] one can see that $ f'(c_{2}) \equal{} \frac {f(5) \minus{} f(c)}{5 \minus{} c} \equal{} \frac {6}{5 \minus{} c}\geq 0$
for $ c_{2}\in(c,5)$.

Further we can assume two roots between $ x \equal{} 4$ and $ x \equal{} 5$ ,say $ d_{1}$ and $ d_{2}$.Thus by Rolle's Theorem there must exist some $ x \equal{} e$ for which $ f'(e) \equal{} 0$ for some $ e\in(d_{1},d_{2})$ which we see is not possible since none of $ f'(c_{1}),f'(c_{2})$ is zero,rather they turn out to be positive quantities.

Therefore between $ x \equal{} 4$ and $ x \equal{} 5$ we have $ f'(x)\geq 0$ and hence since $ f(x)$ is an increasing function on $ [4,5]$ there can exist only [u]one[/u] root.

Therefore,in totality $ f(x) \equal{} 0$ has exactly [b]3[/b] roots.

Let me know whether the answer and solution are correct or not[/quote]

I don't believe it is right. The Lagrange's mean value theorem that you have applied speaks only at x=c1. Your c1 is fixed, but then you consider it is arbitrary. There lies your mistake.

Yes,I realized my mistake and therefore i have posted another solution after this .


Hi, how should I centre a table ?
I tried \begin{center} ... \end{center}

and \centralize

but it didn't worked...

ee, if those don't work....is your table really large?
do you have a "bad box"?

and just to make sure, you want to center the table in the page, not center the text in the cells?

glad I could help for the \subitem!

ledurt

Hi,

is your table really large?-> no , 5*3 cels
do you have a "bad box"? ->no

and just to make sure, you want to center the table in the page, not center the text in the cells?->yes, center the table in the page

oops..
just found it:
I typed:

\centering
\begin{table}
\begin{tabular}

which didn't worked.
It had to be:

\begin{table}
\centering
\begin{tabular}

thx.

well, good!

glad I could help... ;)

ledurt


Hi!
Which is the main idea of black when he plays at the sicilian e6? (can you give me any links?)

Well, I never play it, but it does have its charm: standard 3. Bc4 is a ? move here, leaving the initiative to the Black. Because of its resemblance to French defense, (and possible transfer in its positions in as the game continues), this line is called French Variation, ECO B40-B49, see http://www.silex.hr/asp/openings/openings.asp for detailed variants.

See more:

http://www.chessgames.com/perl/explorer?node=38154&move=3&moves=e4.c5.Nf3.e6&nodes=21720.32033.32034.38154

http://www.experiencefestival.com/a/Sicilian_Defence_-_Open_Sicilian/id/5462788

as i play against the e6 variation with white, and i play it with black, i  consider it to be a slightly trickier line than the d6 lines. of course, black can just play d6 and transpose into a scheveningen, but sometimes black threatens bb4(although most of the time it is bad, considering that an exchange of that bishop for the night severely weakens d6), pinning the white night on c3 and putting pressure on the e4 pawn with the f6 night. Also, if black desires a scheveningen formation, the e6 variation aviods certain side lines, such as 1.e4 c5 2.nf3 d6 3.Bb5+(which is by no means bad for black). However, sometimes if white delays the pawn break d4, black is left with an awkward position with weaknesses in the dark squares, especially if black heads for a fianchetto of the kings bishop, which is typical in the closed sicilian. for example: e4 c5 nf3 e6 nc3 nc6 be2 g6? d4! with a clear advantage for white.

the sicilian is actually trying to pressure one part of the side of the opponent and leaving the queen and the rooks to pressure the other side. I usually play the dragon which white usually plays the yugoslavian. It's actually a good strategy, but I have no idea how to end...


We have $P_k(x)=\frac{x^k-1}{x-1}$, so using Newton's binom we obtain \[LHS=\frac{(x+1)^n-2^n}{x-1}=2^{n-1}\frac{\left(\frac{x+1}{2}\right)^n-1}{\frac{x+1}{2}-1}=2^{n-1}P_n\left(\frac{x+1}{2}\right)=RHS.\]


geometry
3D geometry
sphere
trapezoid
geometric transformation
ratio
conics

Let $ \epsilon$ be a plane and $ k_1,k_2,k_3$ be spheres on the same side of $ \epsilon.$ The spheres $ k_1,k_2,k_3$ touch the plane at points $ T_1,T_2,T_3,$ respectively, and $ k_2$ touches $ k_1$ at $ S_1$ and $ k_3$ at $ S_3$. Prove that the lines $ S_1 T_1$ and $ S_3 T_3$ intersect on the sphere $ k_2$. Describe the locus of the intersection point.

Consider the diametral section $ \alpha$ of the spheres $k_2,k_3$ containing their centers $ O_2,O_3$ and the right trapezoid $ O_3O_2T_2T_3.$ Since $ S_3$ is the insimilicenter of $ k_3 \sim k_2,$ then $ T_3S_3$ goes through the antipode $ P$ of $ T_2$ on the sphere $ k_2.$ Similarly, by considering the  diametral section $ \beta$ of the sphere $ k_1,k_2$ orthogonal to $ e,$ then $ T_1S_1$ goes through $ P.$ Since

$ \frac {(T_2T_1)^2}{(T_2T_3)^2} \equal{} \frac {4r_1r_2}{4r_2r_3} \equal{} \frac {r_1}{r_3} \equal{} \text{const}$

It follows that $ T_2 \in e$ moves on  the Apollonian circle $ \mathcal{C}$ of the segment $ T_1T_3$ with ratio $ \sqrt {\frac {_{r_1}}{^{r_3}}}.$ As a result, $ P$ moves on the surface of the right cylinder with base $ \mathcal{C}$ on the plane $e.$ On the other hand, in the right trapezoid  $ PO_3T_3T_2,$ we have:

$ {PO_3}^2 \equal{} 4r_2r_3 \plus{} (2r_2 \minus{} r _ 3)^2 \equal{} 4{r_2}^2 \plus{} {r_3}^2.$  Similarly, $ {PO_1}^2 \equal{} 4{r_2}^2 \plus{} {r_1}^2$

$\Longrightarrow {PO_3}^2 \minus{} {PO_1}^2 \equal{} {r_3}^2 \minus{} {r_1}^2 \equal{} \text{const}$ $\Longrightarrow$ $ P$  lies on the radical plane $ \pi$ of  $ k_2,k_3.$ Thus, locus of $ P$ is a Ellipse $ \mathcal{E},$ i.e. intersection of the right cylinder $ \mathcal{C}$ with the plane $ \pi.$


What score is usually required to make cd at nationals?

i don't remember the 2007 score, but in 2008 i think it was 38(maybe 36?)

one of my teammates(who doesnt have an account here) got a 34 but placed in the low twenties, if that helps.

[quote="Yongyi781"]Quite obviously, 12 people make cd every year. So the answer is yes. :)[/quote]
I think he means anyone who made CD, who has an account here.
 :rotfl: Silly kid.  :rotfl:

7 people had accounts if I am not mistaken...


Let a, b, c be positive numbers. Prove that

$ (a\plus{}b\plus{}c)^2\plus{}6abc(a\plus{}b\plus{}c)\geq8(ab\plus{}bc\plus{}ca)\minus{}3$

With

$ a^2\plus{}b^2\plus{}c^2\equal{}1$

--------------------------
 :wink: Ligouras

[quote="Ligouras"]Let a, b, c be positive numbers. Prove that

$ (a \plus{} b \plus{} c)^2 \plus{} 6abc(a \plus{} b \plus{} c)\geq8(ab \plus{} bc \plus{} ca) \minus{} 3$ (*)

With

$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$

--------------------------
 :wink: Ligouras[/quote]

$ p\equal{}a\plus{}b\plus{}c;q\equal{}ab\plus{}bc\plus{}ca;r\equal{}abc$

$ (*) <\equal{}> 7\minus{}3p^2\plus{}6pr \geq 0$

By Schur we have : $ r \geq max \{ 0; \frac{p(p^2\minus{}2)}{9} \}$

* $ 0< p \leq \sqrt{2}$

We have : $ r \geq 0$

We have : $ 7\minus{}3p^2\plus{}6pr \geq 7\minus{}3p^2 >0$

* $ \sqrt{2} \leq p \leq \sqrt{3}$

We have :  $ r \geq \frac{p(p^2\minus{}2)}{9}$

And $ 7\minus{}3p^2\plus{}6pr \geq 7\minus{}3p^2\plus{}\frac{2p^2(p^2\minus{}2)}{3} \geq 0 <\equal{}> (p^2\minus{}3)(2p^2\minus{}7) \geq 0$

Done ! :)

Could you explain how you got $ r\geq max\{ 0;\frac{p(p^{2}\minus{}2)}{9}\}$ from $ 7\minus{}3p^2\plus{}6pr\ge0$ by Schur's?

Thanks. :blush:


analytic geometry
geometry
projective geometry
geometry proposed

Two tangent lines are drawn to the circle centered at the origin with radius 1 from the point $ P(x_{0},\ y_{0})$ outside the circle.

(1) Let $ Q$ be the mid point of the two tangent points, express the coordinate of $ Q(x_{1},\ y_{1})$ in terms of that of $ P(x_{0},\ y_{0})$, then prove that $ OP\cdot OQ=1$.

(2) When the point $ P$ moves on the line $ x+y=2$, find the locus of the point $ Q$.

You are just constructing the inversive image of $ P$.

Let $ H$ be the foot of the perpendicular from $ (0,0)$ to $ x+y=2$. Let $ H'$ be the inversive image of $ H$ in the circle. The locus is a circle diameter $ OH'$ minus the point $ O$ [that last bit is arguable. I am pretty sure that we are not accepting conventions of projective geometry in this context, so I stand with the exclusion of $ O$ from the locus].


I think that LaTeX is a really nice program and I am using when doing math papers. But I just have one thing which is irritating and that is the lack of picture and graphs in my papers and that is beacuse I do not know how to import a picture or a graph, and by the way can you recommend a program in which one can draw skecthes of things.

Hope you got it all and that you can help me!

hello, maybe you should try to read the not so short introduction to latex first...

[url=http://people.ee.ethz.ch/~oetiker/lshort/lshort.pdf]people.ee.ethz.ch/~oetiker/lshort/lshort.pdf[/url]

you can find most of the information inthere and otherwise just search trought the forum first...

greetz,
A

I have already got that guide, which I think is horrible long :)

But I hoping that someone would advice me how to import images and stuff. And do you know any mathematical free drawing program for computer?

Have a look at the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=53217]Ultimate guide to inserting Pictures in LaTeX[/url] discussion on this forum

[quote="Diddy"]and by the way can you recommend a program in which one can draw skecthes of things.[/quote]
I use [i]SmartDraw[/i], but think [i]Canvas[/i] is another good one.

Ok I will have a look at that.

And these 2 programs that you mention can I download them for free, and if yes where can I find them?

Trial versions only

[url=http://www.smartdraw.com/specials/smartdraw.asp?type=35545&id=35545]Smartdraw[/url] and [url=http://www.acdamerica.com/support-canvas/canvas-software-downloads/canvas-downloads.html]Canvas[/url]

Ok I see,

but the problem is that, this did not get me pretty far, in fact I have just got even more confussed, so if someone would please advice me on how to put graphs and images in LaTeX in an easy way. It would be most appreciated!

Come on some of you must this stuff, so please help.


limit
integration
real analysis
topology
real analysis unsolved

Not difficult

$f$:$[0;1] \rightarrow R$ continous

$g \in C^{1}([0;1];[0;1])$ such that $|g'|<1$ on $[0;1]$

Prove that there exist a unique $b$ independant of $f$ such that

$\lim_{n\rightarrow +\infty}\int_{0}^{1}f(w_n(x))=f(b)$ where

$w_n=g\circ g\circ\dots\circ g$ n-times

I am realy interested of a solution of this problem  :?

1. Any continuous mapping from $[0,1]$ to $[0,1]$ has at least one fixed point. This (you can think of it as the easy one-dimensional case of the Brouwer fixed point theorem) follows from the Intermediate Value Theorem.

2. $g$ cannot have two fixed points, since for $x\ne y,\,|g(x)-g(y)|=|g'(\xi)|\cdot|x-y|<|x-y|.$ Therefore, there is a unique $b$ such that $g(b)=b.$

3. Let $I_n=w_n([0,1]).$ That is, $I_0=[0,1]$ and $I_{n+1}=g(I_n).$ Since the continuous image of a closed bounded interval is a closed bounded interval, we have that $I_n$ is a closed bounded interval. Furthermore, $I_0\supset I_1\supset I_2\supset\cdots.$

4. Let $J=\bigcap_{n=0}^{\infty}I_n.$ $J$ is a closed bounded interval that contains $b.$ We must have $g(J)=J.$ If $y\in J$ then for each $n,$ $y\in I_{n+1}$ which implies that $y=g(x)$ for $x\in I_n.$ Since this is true for all $n$, $x\in J$ and $y\in g(J).$

5. But if $\text{diam}(J)>0$ then we cannot have $g(J)=J,$ since by the argument in #2, for $x\ne y,\,|g(x)-g(y)|<|x-y|.$ This forces the diameter of $g(J)$ to be strictly less than the diameter of $J,$ unless that diameter is zero.

6. Hence $J=\{b\}.$ It follows that $\lim_{n\to\infty}\text{diam}(I_n)=0.$

7. For all $x$ and all $n,$ $|w_n(x)-b|\le\text{diam}(I_n).$ Given any $\epsilon>0$ there exists $\delta >0$ such that $|y-b|<\delta$ implies that $|f(y)-f(b)|<\epsilon.$ Hence for sufficiently large $n,$ $|f(w_n(x))-f(b)|<\epsilon$ for all $x.$

8. We have shown that $f(w_n(x))$ converges uniformly to $f(b).$ Hence,

$\int_0^1f(w_n(x))\,dx\to\int_0^1f(b)\,dx=f(b).$

---

That was on the overcomplete and overly pedantic side - I was just making sure that all my students in this semester's course in which such subjects as uniform converges were covered would be able to read it.

I think that it is quite straightforward with dominated convergence, isn't it ?

I didn't want to use dominated convergence because it wasn't necessary. That is, the whole problem can be settled at a pre-Lebesgue level. Note that to use dominated convergence, you would still have to prove pointwise convergence, and that takes most of the argument above.

Note that my #1 was redundant and unnecessary; if you follow the rest of it, I prove the existence of a unique fixed point.

In fact, we have this result:

Suppose $K$ is a compact metric space, and suppose $f : K\to K$ satisfies the following weak contraction property: $\forall\,x,y\in K,\,d(f(x),f(y))<d(x,y).$ Then there exists a unique $b\in K$ such that $f(b)=b.$

We do need compactness as a hypothesis here, as shown by the example of $f(x)=x+\frac1x$ on $[1,\infty).$


For a triangle $\triangle ABC , F$is area and $a,b,c$ is three side Prove that:
$a^2+b^2+c^2 \geq 4\sqrt{3}F +\frac{(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2}{a^2+b^2+c^2}$
have fun  :)

hello, rewriting your equation in the following form
$ a^2\plus{}b^2\plus{}c^2\minus{}\frac{(a^2\minus{}b^2)^2\plus{}(b^2\minus{}c^2)^2\plus{}(c^2\minus{}a^2)^2}{a^2\plus{}b^2\plus{}c^2}\geq 4\sqrt{3}F$
setting $ F\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$ with $ s\equal{}\frac{a\plus{}b\plus{}c}{2}$ squaring both sides, subtracting from each other and factoring we get
$ \frac{(a^4\plus{}b^4\plus{}c^4\minus{}a^2b^2\minus{}a^2c^2\minus{}b^2c^2)^2}{(a^2\plus{}b^2\plus{}c^2)^2}\geq0$
and this is true, the equal sign holds if $ a\equal{}b\equal{}c$.
Sonnhard.


calculus
integration
trigonometry
calculus computations


trigonometry
analytic geometry
vector
complex numbers

Calculate :

$ S\equal{}1\plus{}\cos \ \frac{2\pi}{5}\plus{}\cos \ \frac{4\pi}{5}\plus{}\cos \ \frac{6\pi}{5}\plus{}\cos \ \frac{8\pi}{5}$

[quote="Yongyi781"][hide="Brute force"]
$ \cos \frac {2\pi}5 \equal{} \frac { \minus{} 1 \plus{} \sqrt 5}4$
$ \cos \frac {4\pi}5 \equal{} \frac { \minus{} 1 \minus{} \sqrt 5}4$
$ \cos \frac {6\pi}5 \equal{} \frac { \minus{} 1 \minus{} \sqrt 5}4$
$ \cos \frac {8\pi}5 \equal{} \frac { \minus{} 1 \plus{} \sqrt 5}4$

$ 1 \plus{} \cos\frac {2\pi}5 \plus{} \cos\frac {4\pi}5 \plus{} \cos\frac {6\pi}5 \plus{} \cos\frac {8\pi}5 \equal{} 1 \plus{} 2\frac { \minus{} 1 \plus{} \sqrt 5}4 \plus{} 2\frac { \minus{} 1 \minus{} \sqrt 5}4 \equal{} 1 \minus{} 1 \equal{} 0$.[/hide]

[hide="Complex numbers"]
Note that this sum is the sum of the x-coordinates of a regular pentagon in the complex plane, given by $ z_j \equal{} \text{cis}\,\frac {2\pi j}{5}$. But it is known that the sum of $ n$ complex numbers that form a regular $ n$-gon is 0, so $ \sum_{j \equal{} 0}^5 \text{cis}\,\frac {2\pi j}{5} \equal{} 0$, so both the real and imaginary parts of the sum is 0. The result follows directly.[/hide][/quote]

No nice solution , except complex numbers ? :(

Why are people so afraid of complex numbers?  Yongyi's solution can be phrased in vector terms alone:  "the center of gravity of a regular $ n$-gon inscribed in the unit circle is the origin."  That's intuitive enough, right?

In some countries the complex numbers are learned in X grade . Why using complex numbers , when you can solve a problem more easy such that (every) student to understand ?

[hide="Without complex numbers"]Let $ x \equal{} \frac {2\pi}{5}$ and multiply by $ 2\sin \frac {x}{2}$

$ 2\sin \frac {x}{2}(S \minus{} 1) \equal{} 2\sin \frac {x}{2} \cos \ x \plus{} 2\sin \frac {x}{2}\cos \ 2x \plus{} 2\sin \frac {x}{2}\cos \ 3x \plus{} 2\sin \frac {x}{2}\cos \ 4x$

Using prod formulas , we get :

$ S \minus{} 1 \equal{} \frac {2\sin \frac {4x}{2}\cdot \cos \frac {5x}{2}}{2\sin \frac {x}{2}} \equal{} \minus{} \frac {\sin \frac {4\pi}{5}}{\sin \frac {\pi}{5}} \equal{} \minus{} 1$[/hide]

However, that solution is more "bashy" than using a fact of roots of unity.

By the way, these things aren't learned even in "X" grade in America... they are not taught in regular school curricula except for five minutes saying "roots of unity are solutions to $ x^n\equal{}1$, there is nothing more, end of lesson".

Consider $ P(z)\equal{}z^5\minus{}1$.  Clearly, the roots are the complex numbers $ cis \frac{2n\pi}{5}$ for $ n\equal{}0,1,2,3,4$
The sum of the roots is $ 0$ or in other words the sum of the real parts of the roots is $ 0$, i.e.  $ 1\plus{}\cos\ \frac{2\pi}{5}\plus{}\cos\ \frac{4\pi}{5}\plus{}\cos\ \frac{6\pi}{5}\plus{}\cos\ \frac{8\pi}{5} \equal{} 0$

 :lol:

so what if certain topics like complex numbers arent thought or thoought only in X ??? i think i person does these stuff out of interest and when there is interest i dont see the limitation? why is it wrong to use complex numbers instead of trig?? there can always be elementary solutions however "advanced" the theory used is.
It is time people started embracing higher stuff otherwise i am afraif they arent spending time on maths only [b]contests[/b].I request to be forgiven because my accomplishments are lesser than nothing so i might have no right but this is what i feel.


limit
logarithms
calculus
integration
real analysis
real analysis unsolved

Calculate
\[ \lim_{x\to\infty}{e^{\frac {x^3}{9} - \frac {x}{12}}\prod_{k = 1}^{x}\left(\frac {k}{x}\right)^{k^2}
}\]
[hide] :P  :D [/hide]

hello, i have not a complete proof yet, but i think the searched limit is zero.
Sonnhard.

I/m not sure about exact calculation of this limit , as a number , but
try to check my old result :
\[ \sum\limits_{n \equal{} 1}^{N}n^{2}\ln n \equal{} (\frac {1}{3}N^{3} \plus{} \frac {1}{2}N^{2} \plus{} \frac {1}{6}N)\ln N \plus{} \frac {N}{12} \minus{} \frac {N^{3}}{9} \plus{} c_{1} \plus{} o(1)
\]
your limit must be $ e^{c_{1}}$

[hide="Mathematica seems to think"]
it's $ \boxed{e^{\frac {\zeta(3)}{4\pi^2}}}$.  Extremal's approach (though I think the lhs should be negative) together with some contour integration seems like a promising way to get there, since Mathematica also suggests
\[ \sum_{n \equal{} 1}^N{n^2 \log n} \equal{} \frac {\zeta(3)}{4\pi^2} \plus{} \zeta^{(1,0)}( \minus{} 2,N \plus{} 1)
\]
where $ \zeta$ refers to the generalized Riemann zeta function.[/hide]

This limit seems to converge rapidly, as my numerical approach may it seem - but i cann't calculate further than 27 (using software) without getting the same answer, over and over. The limit converges around the value 1.0308107031825,
but i cann't determine the actual value

[quote="Hidden Scofield"][/quote]

[hide] :mad: [hide] :P [hide] :D [/hide][/hide][/hide]

[hide] :rotfl:

like u dude

:wink: [/hide]


baba pashin biain begin chand ta hal kardin degh kardim

yek soale:tedad halate soale 1 chand ta mishe?yani ba davaran va ya bedoone davaran bayad mishmordim?(n*2^(n-5) ya ...)

sunset fekr konam manzoor ba davaran boode chon az oonayi ke man porsidam 90% eshoon ba davaran dar nazar gerefte boodan .
dar morede kafam fekr konam kaf hoodoodaye 23,24 nomre bashe .
too soale 3 fekr mikonin be rahe halle ba adade mokhtalet nomre bedan ya na ?

agah ye javunmard peyda mishe soalaye emtehano bege?harki harkodumesho ke hale type  dare bege hadaghal! :oops:

5)mikhahim adade 10 raghami besazim ke dar anha faghat az arghame 1,2,3,...,9 estefade shode bashad.midanim ke har 2 adad ya dar bish az 1 ragham ekhtelaf darand ya dar yek ragham bish az 1 vahed ekhtelaf darand.maximuim tedade shomareha chand tast?entekhabe in max be chand tarigh momken ast.

4-sabet konid tanha adadi ke be ezaye har $ n$ e tabiyi $ 4(a^{n} \plus{} 1)$ moka'abe kamel ast , $ a\equal{}1$ mibashad.
6-dar mosalase $ ABC$ ertefae $ AH$ ra rasm karde im . az $ H$ be do zel'e $ AB$ va $ AC$ amood mikonim va paye do amood ra be tartib $ T$ va $ T'$ minamim . $ O$ markaze dayereye mohitie mosalase $ ABC$ ast . neshan dahid agar
$ 2OT \equal{} AC$ angah $ 2OT' \equal{} AB$

javabe soale 5 mishe : $ \frac {9^{10} \plus{} 1}{2}$ va javabe ghesmate $ B$ ham mishe 1 .

hi.harki chandta va kodoom soalaro hal karde?!!!
be nazare shoma rooze dovvom [color=red]sakhttar[/color] bood?!!! :?:[/img]

man rooze aval 2 ta hal kardam soale 1 am joob zadam vali shayad 3 in begiram rooze 2vom ham 4 va 6 hal kardam fek konam kaf to khafantarin halatesh 25 bashe..!kheliam sakht bood

baba biain begin chandta hal kardin dg eyy baba

sala man rooze avval 3 ta hal kardam
rooze dovvom faghat soale 6 ro nesfe va shayad bishtar (va shayad kamtar!) hal kardam  :(  :(  :(  omidam be come inam javabe soale 3  :
avval rahnamayiish ro sabet kon  :D
bad ba esteghra sabet $ f^{1387}$ ham homographice (yani ghiafash to hamoon mayehas)
sepas az in nokte estefade kon ke $ f^{1387}(f(x)) \equal{} f(x) , f^{1387}(f(f(x)) \equal{} f(f(x))$ bad hala sabet kon ke $ f(f(x)) \ne x$ intori ke migi ba borhane kholf agar $ f(f(x)) \equal{} x$ angah $ f^{1387}(x) \equal{} f(x)$ pas $ f(x) \equal{} x$ ke khalafe farze masalast .

Man rooze avval 2,3 ro neveshtam , rooze dovvomam har 3 ta ro neveshtam . albatte soale 3 ro ba mokhtalet neveshtam nemidoonam nomre midan ya na , dar har soorat nomram beyne 28 ta 35 hast .

[quote="sepehr ahmadzadeh"]toxic u tehrani?
kodoom madreseeii?? ye amar begirim bebinim kia chandta hal kardan har ki bege ta kafo hads bezanim dg[/quote]
ey baba hal dige maro yadet nemiad ?!?! man az Sabzevaram

khashi toro khodam midoonam baghie begiin baWwWW

[quote="Ashegh"]sallam azizanam,yeki mitoone hendese haro inja benvise?

mamnoon[/quote]

2-dar mosalase $ ABC$ , $ I_{a}$ markaze dayereye mohatie khareji nazire rase $ A$ mibashad . In dayere bar emtedade $ AB$ va $ AC$ be tartib dar $ B'$ va $ C'$ momas ast . agar $ P$ va $ Q$ mahale barkhorde $ BI_{a}$ va $ CI_{a}$ ba $ B'C'$ bashand ( be tartib ) va mahale taghatoe $ BQ$ ba $ CP$ , $ M$ bashad angah neshan dahid andazeye amoode vared az $ M$ bar zel'e $ BC$ barabar ba $ r$ mibashad . (shoae dayereye mohatie dakhelie mosalas)
soale 6am balatar neveshtam .

mamnoon azizam,baraye man ke modatha az hendese door boodam,soale2,1min bishtar tool nakeshid,

nemidoonam,be nazaram solae sade E hastesh,ishalla hame hallesh kardan dige?

soale 6 ro ham alan didam,roosh fekr mikonam :wink:

ishalla hame  be arezoohatoon beresin, :wink:


Hi,
My question is when I apply for a college I send them a transcript.  However, my current high school transcript only has the contest awards I got in High School on it.  I have heard that it looks good to colleges that you have participated in MATHCOUNTS.  Do I add MATHCOUNTS (obviously done in middle school) in my highschool transcript or do I tell the college about it in some other way?

PS. I have never filled out a college application before, so if there is a section that I could list my awards and achievements that doesn't involve my high school transcript, PLEASE tell me about it and how it works.

THANKS SOOOO MUCH!

Your transcript has awards?  That's unusual.  There's usually a separate portion of the application for awards and achievements, but I think MATHCOUNTS is a little too far back to make much of a difference.

There are different applications for each college (many are commonapp) but most have a section specifically set aside for awards. I wouldn't mention a middle school award though unless it's really significant (e.g. winning mathcounts nats). But in those cases, you'll generally have high school awards which can further outweigh the middle school awards.

You could offhandedly mention it in an essay (maybe it made you more interested in higher-level math? I know it did for me). Try not to be obnoxious when doing so, though.

I think there is a word limit or preference for the application.

Therefore, you should mention the AMC/AIME/USAMO/etc.s more, because that is a high school contest and more prestigious than a middle school competition like Mathcounts.


geometry
parallelogram
rotation
algebra
system of equations

Given the following system of equations:
$y=-\frac{3}{4}x+3$, $x=0$, and $x=4$,
find either of the two equations that will complete
the system so that the region bounded by the
graphs of the equations in the final system
is a parallelogram with an area of 12 square
units. Express the equation in the form
$ax+by=c$.

[hide]If we rotate the graph by 90 degrees, it's obvious that the height of the parallelogram is $4$ units, so the base of the parallelogram is $12/4=3$ units.
Also, its not hard to know $y=-\frac{3}{4}x+3$ has the y-intercept of 3. So the other line of the parallelogram must be parallel to $y=-\frac{3}{4}x+3$ and its 3 units away from $y=-\frac{3}{4}x+3$.

By shifting $y=-\frac{3}{4}x+3$ up or down 3 units, you get

$\boxed{y=-\frac{3}{4}x}$ or $\boxed{y=-\frac{3}{4}x+6}$[/hide]

[quote="Allan Z"]If we rotate the graph by 90 degrees, it's obvious that the height of the parallelogram is $4$ units, so the base of the parallelogram is $12/4=3$ units.
Also, its not hard to know $y=-\frac{3}{4}x+3$ has the y-intercept of 3. So the other line of the parallelogram must be parallel to $y=-\frac{3}{4}x+3$ and its 3 units away from $y=-\frac{3}{4}x+3$.

By shifting $y=-\frac{3}{4}x+3$ up or down 3 units, you get

$\boxed{y=-\frac{3}{4}x}$ or $\boxed{y=-\frac{3}{4}x+6}$[/quote]
The answer should be in ax+by=c form... Also, hide your solution.

[hide]with allan's reasoning, you get y=-3/4 x and y = -3/4 x + 6. in ax + by = c form, the answer is 3/4 x + y = 0 and 3/4 x + y = 6.[/hide]

[quote="bpms"]How can there be two answers in a chapter target[/quote]
It says find either of the two solutions.


[quote="kunny"]Let $c$ be a positive constant number.
Find the maximum value of $\sqrt{c+x}-\sqrt{c^{2}-x^{2}}+\sqrt{c-x}$ for $|x|\leq c.$[/quote]
[hide]Let $a=\sqrt{c+x}$ and $b=\sqrt{c-x}$.

So $f(x)=a+b-ab=1-(1-a)(1-b)$ and $a^{2}+b^{2}=2c$.

$f$ is convex in $a,b$ so the maximum is attained at the endpoints $|x|=\sqrt{c}$

$f_{max}=\sqrt{2c}$[/hide]

[quote="kunny"]Your view point is good, we need to calssify several cases for $c.$[/quote]You are right, f is convex for some positive values of c. I'll have to work on it.


inequalities
geometry
email
inequalities proposed

:? Let ABCD be a convex quadrilateral. When and where there is thisalways inequality:
\[3\sqrt S\ \le s+\sqrt [4] {abcd}\]
where S is the area of ABCD. I don't know if this inequality is true or false. :?:  :?:

[quote="levi"]:? Let ABCD be a convex quadrilateral. Then there is the inequality:
\[3\sqrt S\ \le s+\sqrt [4] {abcd}\]
where S is the area of ABCD. I don't know if this inequality is true or false. :?:  :?:[/quote]

It is false. See http://www.mathlinks.ro/Forum/viewtopic.php?t=1427 .

  darij

I knew that there are some quadrilaterals for which this inegality is false.
But, in my opinion, to solve this conjecture, in the offing, means to find the largest class of quadrilaterals  for which the inequality is true. Do you know when is true, i.e. in what class of triangles is it true ? This is the question ! I read the mentioned message and I have not the conviction that this conjecture was solved. For d$\rightarrow0$ inegality becomes (for certain triangles) $s\ge 9r$ and there is such triangles. [u]For example[/u], a=24,b=c=13  ==> s=25, S=60, r=12/5 and $3\sqrt 60\le 25\ (\ i.e.\ 25\ge 9\cdot\frac{12}{5})$ is true.
 :arrow: This conjecture is a case "zebra crossing" !
[u]Another example[/u]. Let ABCD a inscribed quadrilateral for which [b]a+c=b+d=s[/b] (I don't know what is named in English). We have S=$\sqrt{(s-a)(s-b)(s-c)(s-d)}=\sqrt{abcd}=sr$. The inequality becames $s\ge 2\sqrt S,\ i.e\ s^2\ge 4S\ i.e.\ s\ge 4r.$   (*) It is true.
 Please, can you to explain me more clear ? Thank you !
[b]Note.[/b] In any convex quadrilateral there is the inequality:
$4S\le {\min\{(a+c)((b+d),(a+b)(c+d),(a+d)(b+d)\}}$
which in our case becames $4S\le s^2$, i.e  (*).
[b]Remark[/b]. I launched this conjecture in my book "Geometrie plana. Culegere de probleme. Editura GIL, 2002", from which sir "xxxxtt" taked over (he confirmed me in a recent Email through Forum) and launched at http://www.mathlinks.ro/Forum/viewtopic.php?t=1427 at 24 oct. 2003. [u]Do you know[/u] this "conjecture" were known long before ? If yes, [u]can you give me a adrress[/u] ? Thanks.


A charity sells 140 benefit tickets for a total of $\$$2001. Some tickets sell for full price (a whole dollar amount), and the rest sells for half price. How much money is raised by the full-price tickets?

\[\bfseries{(A)}\ \$782 \qquad \bfseries{(B)}\ \$986 \qquad \bfseries{(C)}\ \$1158 \qquad \bfseries{(D)}\ \$1219 \qquad \bfseries{(E)}\ \$1449\]

Problem Solving Strategy:

On this test, I got the other 24, and then couldn't get this one with 3 minutes left.  So I did this:

I realized that AMC would probably put the full-price ticket total and the half-price ticket total as answer choices to confuse people.  So I realized that just from that, A and D were the only choices that added up to 2001.  I guessed A, and got it right.

Here's the real solution though:

[hide="Solution"]
Let n be the number of tickets (full-price) and let p be the price of each one.  So:

$np+(140-n)\cdot \frac{p}{2}= 2001$

So

$p(n+140) = 4002$

Now we see that $n+140$ must be a factor of 4002.  Listing out all 8 factor pairs, we see the only possible value is $n+140=174$.  So $n=34$.  Plugging back in, $p=23$ and the answer is $34 \cdot 23 = \boxed{782}$
[/hide]

[quote="mysmartmouth"]I got the other 24, and then couldn't get this one [/quote]

 :huh:

This was definitely one of the easier questions on this test, if I remember correctly  :P

That was the same problem I was facing. I got most of the others but not this one because I couldn't figure out how to approach it :wallbash:

Hm, well I was going at it the wrong way.  I  didn't think AMC would be *** enough to make me do trial and error wtf.

I did end up getting it after the test and I was like WTF what a stupid problem.

21 on this test did give me some problems.  It took up like 10 minutes, and I ended up with a unfactorable cubic polynomial.  Eventually, I just plugged in the answers as roots, and I got lucky when the second one worked.


function
real analysis
real analysis unsolved

(a) Give example of a function $f: \Re\to\Re$ with lateral limits in every point such that $f(\Re)=Q$.
(b) Does there exist $f: [0,1]\to \Re$ with lateral limits in every point such that $f([0,1])=[0,1]\cap Q$?
(c) Is it posible that the function from (a) to be non-constant on every interval?

for the first one, if $\phi: \mathbb{N}\to \mathbb{Q}$ is a bijection, then one can take $f(x)=\phi([x])$.


[quote="spiders_hate_chestnut"]1. Alex parked his car at Leggett Airport on Monday at 7 a.m. He removed his car from the parking lot on Friday of the same week at 10 a.m. If he is charged one dollar per hour, but cannot be charged over 7 dollars in a 24 hour period, how much would he have to pay when he picks up his car?

(I'm having problem with "cannot be charged over 7 dollars in a 24 hr period")

2. Let B(n) denote the sum of the digits of the binary (base 2) representation of the variable n. Let T(n) denote the sum of the digits of the ternary (base 3) representation of n. For example, B(9) = B(1001 sub 2) = 2 and T(9) = T(100 sub 3) = 1. What is the smallest positive integer n greater than 1 such that       B(n)=T(n)?

(what are binary and ternary?)

also, the computer is not letting me use the dollar sign, saying that something is happening with latex, so if someone could help me with that too, it would be very helpful.[/quote]

For #1, determine how many full 24 hour periods there are from 7 am monday to 10 am friday.

For #2, binary is base 2 and ternary is base 3.

For dollar sign, you would do this
[code]
$\$$
[/code]
Which would produce $ \$$

[quote="spiders_hate_chestnut"]1. Alex parked his car at Leggett Airport on Monday at 7 a.m. He removed his car from the parking lot on Friday of the same week at 10 a.m. If he is charged one dollar per hour, but cannot be charged over 7 dollars in a 24 hour period, how much would he have to pay when he picks up his car?

(I'm having problem with "cannot be charged over 7 dollars in a 24 hr period")

2. Let B(n) denote the sum of the digits of the binary (base 2) representation of the variable n. Let T(n) denote the sum of the digits of the ternary (base 3) representation of n. For example, B(9) = B(1001 sub 2) = 2 and T(9) = T(100 sub 3) = 1. What is the smallest positive integer n greater than 1 such that       B(n)=T(n)?

(what are binary and ternary?)

also, the computer is not letting me use the dollar sign, saying that something is happening with latex, so if someone could help me with that too, it would be very helpful.[/quote]

Ok so umm..
[hide="The Solutions"]
1) So from Monday 7:00am to Friday 7:00am, five 24 hour would have elapsed, meaning he would be charged $ 5\times7 \equal{} 35\$$. However, from 7:00~10:00, 3 more hours have elapsed, so the answer would be $ 35 \plus{} 3 \equal{} 38\$$
2) Binary and Ternary are base 2 and base 3. So I simply did guess and check on this one.
Binary goes like this:
0,1,10,11,100,101,110,111,1000,1001, etc.
Ternary goes like this:
0,1,2,10,11,12,20,21,22,100, etc.
So....
First would be 110 in Binary and 20 in Ternary.
Which, converted to base 10, would be 6.
[/hide]


How would I write R, for the reals in mathmode? Furthermore, does the method work for F(field of anything), Q, and C?
Just want to make sure.

\mathbb{stuff}

$ \mathbb{N,\,R,\,Q,\,C}$

\mathbf{stuff}

$ \mathbf{N,\,R,\,Q,\,C}$

Note: The "stuff" should be in capital letters; otherwise you get symbols instead of letters.

You can avoid the "{}", when you're writing in mathbb mode (when it's just a single letter). So put for example \mathbb R and you're done.


Μήπως ξέρει κανένας site με e-books??
Κυρίως για μαθηματικά αλλά και γενικά..

Η πιο κάτω σελίδα έχει αρκετά βιβλία κυρίως μαθηματικής θεματολογίας
http://math.stuff.gr/gr/

Bouzoukman και demetris επειδή είστε νέα μέλη στο φόρουμ, σας παραπέμπω σε ένα αντίστοιχο topic που είχε ανοιχτεί πριν μερικούς μήνες.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=134325

 :D

Ρε παιδιά το stfu δε σας καλύπτει??? :lol:  :wink:

[quote="bouzoukman"]Βασικά φίλε protonios εγώ το άνοιξα αυτό το topic για πει κανένας άμα ξέρει που θα βρω κανα dictionary με ορολογία γιατί δεν έχω βρει κάτι αξιόλογο σε κανά βιβλιοπωλείο...
Και σκέφτηκα να το γενικεύσω...[/quote]

 [b]Τι είδους  dictionary  ψάχνεις για να δω αν μπορώ να βοηθήσω ; Ελληνικό , ξένο και τι να έχει ;Μόνο όρους ή και αναπτυγμένη κάπως θεωρία;[/b]

    Μπάμπης

[quote]Βασικά φίλε protonios εγώ το άνοιξα αυτό το topic για πει κανένας άμα ξέρει που θα βρω κανα dictionary με ορολογία γιατί δεν έχω βρει κάτι αξιόλογο σε κανά βιβλιοπωλείο...
Και σκέφτηκα να το γενικεύσω...[/quote]

Πω ρε και μας είχαν προτείνει ένα στο σχολείο, δεν το θυμάμαι. Θα το αναζητήσω πάντως αν μπορώ να σε βοηθήσω... :wink:

Βρήκα ένα εδώ [color=red][i]http://math.stuff.gr/common/e-rafi.php?id=170[/i][/color]

[quote="bouzoukman"]Βασικά κ.Στεργίου το μόνο που μου λείπει είναι ένα ελληνοαγγλικό.Αγγλοελληνικό και αγγλικό με θεωρία έχω βρει κανά δυο...[/quote]

Έχω δει λεξικά μαθηματικών όρων στο Σαββάλα.Υπάρχουν αρκετά. Είχα και εγώ όταν ήμουν φοιτητής , αλλά τώρα έχουν εκδοθεί πολύ καλύτερα(αγγλοελληνικά κλπ) και για μαθηματικά και για φυσική. Ρίξε μια ματιά ή κάνε ένα τηλέφωνο σε κάποιο μεγάλο βιβλιοπωλείο. Καλύτερα όμως να το δεις μόνος σου πριν το παραγγείλλεις.

 Μπάμπης


[quote="kunny"]By observing, $x=3.$[/quote]
Please post your solution! Concrete!


algebra
polynomial
geometric series
complex numbers
algebra solved

Let $ n$ be a positive integer. Prove that the polynomial $ p\left(x\right) \equal{} \left(x^2 \plus{} 1^2\right)\left(x^2 \plus{} 2^2\right)...\left(x^2 \plus{} n^2\right) \plus{} 1$ is irreducible in $ \mathbb{Z}\left[x\right]$.

[i]Alternative formulation:[/i] Let $ n$ be a positive integer. Prove that the polynomial $ p\left(x\right) \equal{} \displaystyle\prod^{n}_{k \equal{} 1}\left(x^2 \plus{} k^2\right) \plus{} 1$ is irreducible over $ \mathbb{Z}$.

For $n=1$, the result is obvious; $p$ has degree 2 and no real roots.

For $n>1$, we prove the more general fact that $p$ is irreducible over the extended ring $\mathbb{Z}[i]$ (and its quotient field).
Suppose $p(x)=f(x)g(x)$, where the coefficients of $f$ and $g$ are in $\mathbb{Z}[i]$. We have $p(j \cdot i)=1$ for $j= \pm 1, \pm 2, \dots, \pm n$ by the given construction of $p$. Therefore, $f(j \cdot i)$ is a unit of  $\mathbb{Z}[i]$ for the same values of $j$. The only units in this ring are $1,-1,i,-i$, so each of the $2n$ values of $f$ is one of these.
Now, since $x-y|f(x)-f(y)$ by the geometric series formula, $(n+k)i|f(n\cdot i)-f(-k\cdot i)$ for $1 \le k \le n$. Since $n+k\ge n+1 \ge 3$, this forces $f(n\cdot i)=f(-k\cdot i)$. Similarly, $f(-n\cdot i)=f(k\cdot i)$ and $f$ takes the same value at all $2n$ points. The polynomial $h(x)=f(x)-f(i)$ is zero at these $2n$ points, so it is either identically zero or has degree at least $2n$. In either case, the factorization is trivial (either $f$ is a unit or $g$ is a unit), and $p$ is irreducible.

There is a proof without gaussian integers.
A problem from Romanian TST $2003$ states that if $f(X)$ is an irreducible polinomial with $|f(0)|$ not a perfect square then $f(X^2)$ is also irreducible.
Hence it suffices to prove the irreducibility of $f(X)=\prod_{i=1}^n (x+k^2)+1$
Suppose $f=gh$ with $deg(g)\leq \frac n2$.
Then as $f(-k^2)=1, 1\leq k \leq n$ we get $g(-k^2)=h(-k^2)=\pm1$.
Thus the monic polynomial $g^2-1$ has $n$ roots and degree at most $n$ and is not identically $0$ hence $g^2-1=(X+1^2)\cdots(X+n^2)$ thus
$g^2(X)=(X+1^2)\cdots(X+n^2)+1$ which cannot be possible (put $X=0$)
QED

[Edit: edited according to RobertuX's observation below]

I think it's a cute problem. We will prove our statement by contradiction, using complex numbers as arguments of our polynomials.
Assume there exist two polynomials $Q$ and $R$ in $Z[x]$ such that $P(x)=Q(x)R(x)$ Then for $1\leq k \leq n$ we would have $Q(ki)R(ki)=1$ Which means that $|Q(ki)|^2|R(ki)|^2=|1|^2=1$ But as $Q,R \in Z[x]$, we have $|Q(ki)|^2 \in Z$ and $|R(ki)|^2 \in Z$ , so $|Q(ki)|^2=|R(ki)|^2=1$ this means that $Re(Q(ki))^2+Im(Q(ki))^2=1$ and in analogical way for $R$. But we have also $Re(Q(ki)) \in Z$ and $Im(Q(ki)) \in Z$ So we obtain two possibilities   for each $k$:either $Q(ki)=1$ or $Q(ki)=i$ and the same for $R$ But we need $R(ki)Q(ki)=1$ so we get that for every $1 \leq k \leq n$ $R(ki)=1$ and $Q(ki)=1$ this means that $Q$ has a form \[ Q(x)=Q_1(x)(x-i)(x-2i)...(x-ni)+1 \] and for $R$ :  \[ R(x)=R_1(x)(x-i)(x-2i)...(x-ni)+1 \] for some polynomials $R_1, Q_1$. As $P=QR$, we may get \[ 2n=deg(P)=deg(Q)+deg(R)=n+n+deg(Q_1)+deg(R_1) \] which means that  both $Q_1$ and $R_1$ are constant. They also have to be integer. Consider coefficent of $x^{2n}$ . In $P$ it will be $1$ but from the other side this coefficent is equal to $Q_1P_1$, so $Q_1=P_1$. We've obtained then \[ R(x)=Q(x) \] This means that $P(x)=Q(x)^2$ so for every $m$ being integer, $P(m)$ would be a perfect square. Here we get contradiction cos' \[ P(0)=1^22^23^2...n^2+1=(n!)^2+1 \] which is not a perfect square.


Find an explicit formula for
$a_{0}=j$
$a_{n}=\frac{1}{4n+3}(8n(a_{n-1})+3)$, where $n>0$.

Not sure if there is an easy way to do this.  :maybe:

Generating functions might help, but it might just get too complicated beyond a certain point.


hello folks  :D

can you sum :

$ \text{(i)}\quad \sum_{k\equal{}0}^\infty\;\dfrac{1}{k^3\plus{}4}$

and

$ \text{(ii)}\quad \sum_{k\equal{}0}^\infty\;\dfrac{1}{k^4\plus{}5}$

I think it's simple exercises on complex analysis, I believe that your  students Know well this part of math  :lol:


1)ABC dik açısı A köşesinde olan bir dik üçgen ve D noktası A dan BC hipotenüsüne inilen yükseklik ayağı olsun. ABC ABH ve AHC üçgenlerinin iç teğet çemberlerinin yarıçapları toplamının AH uzunluğuna eşit olduğunu ispatlayınız.

2) Bir dışbükey ABCD dörtgeninde AD ve BC kenarlarının orta dikmeleri dörtgenin içindeki bir P noktasında; AB ve CD nin orta dikmeleri ise yine dörtgenin içindeki bir Q noktasında kesişiyor.
m(APD)=m(BPC) ise m(AQB) = m(CQD) dir, gösteriniz.

Evde çalışırken bayağı uğraştırdı... paylaşmak istedim. Kolay gelsin

ilk probleme bir çözüm verelim.ikinci problem için zambak'ın meraklısına geometri kitabında AC=BD köşegen eşitliği ispatlanarak bir çözüm vardı.iyi çalışmalar...


function
integration
real analysis
real analysis unsolved

Let $ \mu$ a locally finite Borel measure in $ \mathbb{R}^n$,and $ f$
non-negative $ \mu$ measurable functions $ \int f
d\mu\equal{}\int_{0}^{\infty}\mu({x: f(x)>t})dt$.

Solution:
Let $ A\equal{}\{(x,t): f(x) \geq t\}$.
Then $ \int_0^{\infty}\mu(\{x\in X: f(x) \geq t\})dt\equal{}\int_0^{\infty}\mu(\{x: (x,t)\in A\})dt$=$ \int \mathcal{L}^1(\{t \in [0,\infty]: (x,t) \in A\})d\mu x$=$ \int \mathcal{L}^1([0,f(x)])d\mu x\equal{}\int f d\mu x$


[size=200]then $ m \geq 0$ ,or $ m \le 0$ hold.
BQ[/size]

eg1:
let $ f(x,y,z) \equal{} x$,then
$ det(M) \equal{} x^2zy \minus{} y^2x^2 \minus{} y^2z^2 \plus{} xz^2y \plus{} xzy^2 \minus{} x^2z^2\le 0,$
hold.

[size=150]eg2:
let $ f(x,y,z) \equal{} yz$,then
$ det(M) \equal{} \minus{}x^4y^2z^2\plus{}x^3y^3z^2\plus{}x^2y^3z^3\minus{}x^2y^2z^4\minus{}x^2y^4z^2\plus{}x^3y^2z^3\le 0$
hold.

eg3:
let $ f(x,y,z) \equal{} x(y\plus{}z)$,then
$ det(M) \equal{} x^4y^2z^2\minus{}x^4y^4\minus{}z^4y^4\plus{}x^2y^2z^4\plus{}x^2y^4z^2\minus{}x^4z^4\le 0$
hold.

eg4:
let $ f(x,y,z) \equal{} (y\minus{}z)^2$,then
$ det(M) \equal{} 5x^4y^4\plus{}5z^4y^4\plus{}5x^4z^4\minus{}60x^4y^2z^2\minus{}60x^2y^2z^4\minus{}60x^2y^4z^2\minus{}z^8\plus{}40x^3y^3z^2\plus{}40x^2y^3z^3\plus{}40x^3y^2z^3\minus{}10xy^3z^4\minus{}10xy^4z^3\minus{}10x^3y^4z\minus{}10x^3z^4y\minus{}10x^4z^3y\minus{}10zx^4y^3\plus{}30z^5xy^2\minus{}20z^6xy\plus{}30z^5x^2y\plus{}30zy^5x^2\plus{}30z^2y^5x\minus{}20y^6xz\plus{}30x^5y^2z\minus{}20x^6zy\plus{}30x^5z^2y\plus{}4y^7z\plus{}4y^7x\minus{}2z^5y^3\minus{}4z^6x^2\minus{}2z^5x^3\plus{}4z^7x\plus{}4z^7y\minus{}2z^3y^5\minus{}4z^6y^2\minus{}2y^5x^3\minus{}4y^6z^2\minus{}4y^6x^2\minus{}2x^5y^3\minus{}4x^6z^2\minus{}2x^5z^3\plus{}4x^7z\plus{}4x^7y\minus{}4y^2x^6\minus{}x^8\minus{}y^8\le 0$
hold.

eg5:
let $ f(x,y,z) \equal{} x(y\minus{}z)^2$,hold.

eg6:
let $ f(x,y,z) \equal{} y(y\minus{}z)^2$,does not hold!
why?
Because f(x,y,z) is  not equal f(x,z,y).

eg7:
let $ f(x,y,z) \equal{} y\plus{}z\minus{}tx$,hold.

BQ[/size]


I have seen a solution to this problem before which uses invariance but I can't remember it, so I am looking for a solution to it again. An invariance solution would be preferred, but any welcome. Thanks in advance.

The numbers $ 1,1,2,2,3,3...,n,n$ are to be arranged in such a way that there are exactly $ k \minus{} 1$ numbers between the two occurrences of the number $ k$. So the two $ 1$'s must be adjacent, the two $ 2$'s must have one number between them, the two $ 3$'s must have two numbers between them etc. For example an arrangement for $ n \equal{} 4$ might look like this:

$ 4,1,1,3,4,2,3,2$

Prove that it is impossible to find such an arrangement for all $ n \equiv 2$ or $ 3 \bmod{4}$.

See Hamster1800's post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1568413#1568413]here[/url] for one proof method.  An equivalent phrasing is as follows: $ 1 \plus{} 2 \plus{} \ldots \plus{} n$ is the sum over $ i$ of the distances from the first copy of $ i$ to the second copy of $ i$.  Adding this to $ 1 \plus{} 2 \plus{} \ldots \plus{} 2n$ gives a sum over $ i$ of twice the position of the second copy of $ i$.  Thus $ (1 \plus{} 2 \plus{} \ldots \plus{} n) \plus{} (1 \plus{} 2 \plus{} \ldots \plus{} 2n) \equal{} \frac {n(n \plus{} 1) \plus{} 2n(2n \plus{} 1)}{2} \equal{} \frac {n(5n \plus{} 3)}{2}$ must be even.  Now conclude.

I would not be shocked if there were a somewhat more elegant or transparent version of this argument.

Ahh,that's very clever.

So for $ \frac{n(5n\plus{}3)}{2}$ to be even, $ n(5n\plus{}3) \equiv 0 \bmod{4}$

This means that either at least one of $ n$, $ 5n\plus{}3$ are congruent to 0 mod 4, or they are both 2 mod 4. The latter is impossible since if $ n \equiv 2 \bmod{4}$ then $ 5n\plus{}3 \equiv 1 \bmod{4}$. Thus our only possibility is that one of them is 0 mod 4.

Thus we are left to solve two equations.

$ n \equiv 0 \bmod{4}$ gives $ n \equiv 0 \bmod{4}$  :P

$ 5n\plus{}3 \equiv 0 \bmod{4}$ gives $ n \equiv 1 \bmod{4}$

Thus if the arrangement is to have any chance of existing, n must be congruent to $ 0$ or $ 1 \bmod{4}$.


Points $ A$ and $ B$ lie on a circle centered at $ O$, and $ \angle AOB\equal{}60^\circ$. A second circle is internally tangent to the first and tangent to both $ \overline{OA}$ and $ \overline{OB}$. What is the ratio of the area of the smaller circle to that of the larger circle?

$ \textbf{(A)}\ \frac{1}{16} \qquad
\textbf{(B)}\ \frac{1}{9} \qquad
\textbf{(C)}\ \frac{1}{8} \qquad
\textbf{(D)}\ \frac{1}{6} \qquad
\textbf{(E)}\ \frac{1}{4}$

[hide]let $ P$ be the center of the small circle with radius $ r$, and let $ Q$ be the point where the small circle is tangent to $ OA$, and finally, let $ C$ be the point where the small circle is tangent to the big circle with radius $ R$.  Then $ PQO$ is a right triangle, and a 30-60-90 triangle at that.  So, $ OP\equal{}2PQ$.  Since $ OP\equal{}OC\minus{}PC\equal{}OC\minus{}r \equal{} R\minus{}r$, we have $ R \minus{} r \equal{} 2PQ$, or $ R\minus{}r \equal{} 2r$, or $ \frac{1}{3} \equal{} \frac{r}{R}$.  Then the ratio of areas will be $ \frac{1}{3}$ squared, or $ \frac{1}{9}$.[/hide]

DUDE WE DID THIS PROBLEM IN GEOMETRY CLASS THE DAY BEFORE HAHAHA
AND I HAD IT ON A PRACTICE ARML RELAY
APPARENTLY THEY DISPLAYED THIS PROBLEM AT THE ANNUAL NCTM MEETING

[hide]Extend OA and OB to A' and B' so that A'B' is tangent to the little circle and OA'B' is equilateral. Then note that the little circle is the incircle of equilateral OA'B'. Thus its center O' is the centroid. Then let OO' hit the big circle at P. Thus OO'/O'P=2, O'P/OP=1/3. square it. yay.[/hide]

[hide="Alternate approach"]
[asy]draw((-3,0)--(3,0));
draw(3*dir(60)--3*dir(240));
draw(3*dir(120)--3*dir(300));
draw(Circle((0,0),3));
draw(Circle((0,2),1));
draw(Circle((0,0),1));
draw(Circle((0,-2),1));
draw(Circle((-sqrt(3),1),1));
draw(Circle((-sqrt(3),-1),1));
draw(Circle((sqrt(3),1),1));
draw(Circle((sqrt(3),-1),1));[/asy]
Each central angle has a measure of $ 60^\circ$.  It's obvious that the smaller circle has a radius $ \frac{1}{3}$ the size of the larger one's radius, so the area ratio is $ \boxed{\frac{1}{9}}$.[/hide]


Find all functions $f,g,h: \mathbb{R}\to\mathbb{R}$ such that \[\forall x,y,z\in\mathbb{R}\; : f(h(g(x))+y)+g(z+f(y))=h(y)+g(y+f(z))+x.\]

(*) $f(h(g(x))+y)+g(z+f(y))=h(y)+g(y+f(z))+x$
$y=z$ gives
(**) $f(h(g(x))+y)=h(y)+x$
and $y=0$ gives
(***) $f(h(g(x)))=x+h(0)$.
So $g$ is injective, and (*) and (**) give
$g(z+f(y))=g(y+f(z))$ and $z+f(y)=y+f(z)$ for all $y,z$.
$z=0$ gives $f(y)=y+c$ for all $y$ with $c: =f(0)$.
Then (**) gives
(****) $h(g(x))+y+c=h(y)+x$
and $x=0$ gives $h(y)=y+d$ for all $y$ with $d: =h(g(0))+c$.
Then (****) gives
$g(x)+d+y+c=y+d+x$ and $g(x)=x-c$ for all $x$.

So solutions are $f(x)=x+c$, $g(x)=x-c$ and $h(x)=x+d$ for all $c,d\in\mathbb{R}$.


Let $ V$ be a finite-dimensional real inner product space, and let $ U$ and $ T$ be self-adjoint linear operators on $ V$ such that $ UT \equal{} TU$. Prove that there exists an orthonormal basis for $ V$ consisting of vectors that are eigenvectors of both $ U$ and $ T$.

Use this to show that if $ A$ and $ B$ are symmetric $ n\times n$ matrices such that $ AB \equal{} BA$, there exists an orthogonal matrix $ P$ such that $ P^TAP$ and $ P^TBP$ are both diagonal matrices.

Any help? There are a few hints: For any eigenspace $ W\equal{}E_{\lambda}$ of $ T$, $ W$ is both $ T$- and $ U$-invariant. Also, $ W^{\perp}$ is both $ T$- and $ U$-invariant.

We have the well-known result: let $ T$ be a linear operator on a finite-dimensional real inner product space $ V$. Then $ T$ is self-adjoint if and only if there exists an orthonormal basis $ \beta$ for $ V$ consisting of eigenvectors of $ T$.

The following result may also help: let $ W$ be a finite-dimensional subspace of an inner product space $ V$, and let $ y\in V$. Then there exist unique vectors $ u\in W$ and $ z\in W^{\perp}$ such that $ y\equal{}u\plus{}z$.

All your hints give the result.

The second hint suggests that $ U$ and $ T$ are diagonalizable. With a well-known result, we know that they are simultaneously diagonalizable.

Choose 1 vector $ v$ that simultaneously diagonalises $ U$ and $ T$ with the same eigen-value $ \lambda$. Note that $ U \vline_{span(v)}$ and $ T\vline_{span(v)}$ are also self-adjoint linear operators on this subspace and $ Tv \equal{} Uv \equal{} \lambda v$. Then, we just have to normalise to get a vector $ v' \equal{} \alpha v$ such that $ (v',v') \equal{} 1$.

Then induction on the space $ span(v)^{\perp}$ to get the desire basis. Here we make use of the third hints: $ V \equal{} span(v) \oplus span(v)^{\perp}$.


vector
geometry
cyclic quadrilateral
geometry solved

Let $ ABC$ be a triangle and $ H$ its orthocenter. Take $ M,N$ on $ CA$ such $ MN \equal{} AC$. $ D,E$ is projection of $ M,N$ on $ BC,AB$. Prove that B,D,H,E is cyclic

Translate by the vector $ AM$ or vector $ CN$ we have: $ CH\rightarrow ND,AH\rightarrow ME$
Then $ H \equal{} AH\cap CH\rightarrow ME\cap ND \equal{} F\Rightarrow HF//AC\Rightarrow\angle BHF \equal{} 90^{0}$
Then $ \angle BDF \equal{}\angle BHF \equal{}\angle BEF\Rightarrow B,D,H,E$ cyclic

We denote as $ H',$ the intersection point of $ MD,$ $ NE$.

Through $ M,$ $ N,$ we draw two lines and parallel to $ AB,$ $ BC$ respectively and let $ L$ be, their intersection point.

Also, we denote the points $ D'\equiv LN\cap MD$ and $ E'\equiv ML\cap NE.$

Because of $ ML\parallel AB$ and $ LN\parallel BC$ and $ MN = AC,$ we conclude that the triangles $ \bigtriangleup MLN,$ $ \bigtriangleup ABC,$ are congruent and so,

we have $ MH'\parallel = AH,$ where $ H',$ is the orthocenter of $ \bigtriangleup MLN.$

We conclude now, that $ HH'\parallel AC$ and so, it is easy to show that the quadrilateral $ BEHH',$ is cyclic because of

$ \angle EH'H =\angle ENA =\angle HCA =\angle EBH.$

Because of now, the cyclic quadrilateral $ BEH'D,$ we conclude that the points $ B,$ $ D,$ $ H',$ $ H,$ $ E,$ are cyclic and the proof is completed.

Kostas Vittas.

[quote="darij grinberg"]also http://www.mathlinks.ro/Forum/viewtopic.php?t=5789 .

darij[/quote]

...."Nothing is new, under the geometry sun".... ([b]Francois Rideau[/b])

Kostas Vittas.


Given M and N (middle points of AB and AC- triangle ABC) and P the middle point of the BC arch (in which A isn't) in the circunscript circumference. BAC acute

Build the ABC triangle.

Please, help me!
I need it for Sunday evening.

Regards.  :lol:

it has been also posted here:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=168418[/url]


Putnam
number theory
number theory unsolved

Suppose $ a,b,c,d$ are integers tiwh $ 0\le a\le b\le 99$, $ 0\le c\le d\le 99$. For any integer $ i$ let $ n_{i}=101i+1002^{i}$. Show that if $ n_{a}+n_{b}$ is congruent to $ n_{c}+n_{d}$ mod 10100, then $ a=c$ and $ b =d$.

Any pointers on how to start? I dont think there is anything special about the numbers.
so i tried replacing them with smaller numbers so see if i noticed anything. but sadly this didnt help much.

[quote="maokid7"]Suppose $ a,b,c,d$ are integers tiwh $ 0\le a\le b\le 99$, $ 0\le c\le d\le 99$. For any integer $ i$ let $ n_{i}=101i+1002^{i}$. Show that if $ n_{a}+n_{b}$ is congruent to $ n_{c}+n_{d}$ mod 10100, then $ a=c$ and $ b =d$.[/quote]

Actually, this is Putnam 1994 and the problem statement is different. In fact, $ n_{i}=101i+100\cdot 2^{i}$. That should make it easier to work with, at least. It's still a hard problem.

[hide="Hint"]Look at it mod 101 and mod 100. Use Fermat's Little Theorem[/hide]

alright. thanks, ill give it a shot and hopefully after a few hours ill have it.  :oops:

[quote="maokid7"]question. is it true that if $ (a,n)=1$ and
$ a^{k}\equiv a^{y}\bmod n$ then $ k+y\equiv 0\bmod \phi(n)$?[/quote]
No, see at $ k=y=1$ :P

[b]maokid7[/b], the following is true:
If $ (a,n)=1$ and $ a^{x}\equiv a^{y}(mod n)$
then $ x\equiv y (mod d)$ where d is the smallest natural number such that $ a^{d}\equiv 1 (mod n)$

thanks for correctly stating the problem for me and even providing the hint about fermat's litthe theorem. however, i am still unable to solve the problem. sorry. could someone please show a partial solution or give an even bigger hint.
thanks.


Let a, b, c > 0. x, y, z be positive reals satisfying  xyz = ax + by + cz. Show that
         x + y + z > sqrt(a+b) + sqrt(b+c) + sqrt(c+a)

This inequality is beautiful but very hard for me.
Can you please post your solution?

Thank you very much.

Indeed, it isn't trivial at all. Only now I had a good idea. I tried to solve it two days ago, but I failed. Now it's her turn to dye:
   Let a=myz, b=nzx, c=pxy. Then m+n+p=1 and we have to prove that
  sqrt(z)*sqrt(my+nx)+sqrt(x)*sqrt(nz+py)+sqrt(y)*sqrt(px+mz)<x+y+z. Apply Cauchy for the LHS and use the fact that my+nx+nz+py+px+mz<1.

I cannot understand the inequality

                       my+nx+nz+py+px+mz<1

in Harazi's solution.
As Harazi has left for three days , can anyone else be so kind to explain to me that inequality?

Thank you very much.

You are right, Manlio, Harazi had a misprint, it shoudl be

my+nx+nz+py+px+mz < x+y+z

which exactly the a/z + a/y + b/x + b/z + c/y + c/x < x + y + z.

Sorry for my inattention for first reading solution. It could be found before.

Have fun,

Namdung

[quote]My solution is not so far from Harazi's

xyz = ax + by + cz => x = a/yz + b/z + c/y > b/z + c/y. Similarly
y > a/z + c/x, z > a/y + b/x. Sum this inequality and add x+y+z to both side, we have
2(x+y+z) > x+y+z + (b+c)/x + (c+a)/y + (a+b)/z
Now, everything is clear.

I have some interesting discovery about the question: "What is the minimum value of x+y+z". I will come back with this one soon. Before that, let think about it.

Namdung
[/quote]
i don't understand  :mad:

Oh; i'm sorry this is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=127807]stronger[/url]

Please concious about your behavior mathboy_cntHDVN. It maybe annoyed many people, included me.  :wink:

The following inequality holds for $ax+by+cz=xyz$:

         $\frac{x+y+z}{2}\geq \sqrt{a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}+\frac 3{4}\sqrt[3]{abc}}$.

[quote="Vasc"]The following inequality holds for $ ax + by + cz = xyz$:

         $ \frac {x + y + z}{2}\geq \sqrt {a + b + c + \sqrt {ab} + \sqrt {bc} + \sqrt {ca} + \frac 3{4}\sqrt [3]{abc}}$.[/quote]
For non-negative $ a,$ $ b,$ $ c,$ $ x,$ $ y$ and $ z$ we need to prove that
$ (x+y+z)^2(ax+by+cz)\geq\sum_{cyc}(4a+4\sqrt{ab}+\sqrt[3]{abc})xyz.$
But $ (x+y+z)^2(ax+by+cz)\geq\sum_{cyc}(4a+4\sqrt{ab}+\sqrt[3]{abc})xyz\Leftrightarrow$
$ \Leftrightarrow a(x^3+y^2x+z^2x+2x^2y+2x^2z)+b(y^3+x^2y+z^2y+2y^2x+2y^2z)+$
$ +c(z^3+x^2z+y^2z+2z^2x+2z^2y)\geq\sum_{cyc}(2a+4\sqrt{ab}+\sqrt[3]{abc})xyz,$
which is true because by AM-GM we obtain: $ ax^3+by^3+cz^3\geq3\sqrt[3]{abc}xyz,$
$ a(y^2x+z^2x)+b(x^2y+z^2y)+c(x^2z+y^2z)\geq2(a+b+c)xyz$ and
$ a(2x^2y+2x^2z)+b(2y^2x+2y^2z)+c(2z^2x+2z^2y)=$
$ =(2ax^2z+2by^2z)+(2ax^2y+2cz^2y)+(2by^2x+2cz^2x)\geq$
$ \geq4(\sqrt{ab}+\sqrt{ac}+\sqrt{bc})xyz.$
Done!


function
complex analysis
complex analysis unsolved

Good evening.

Let $g$ be a function from $D_r=(z \in \mathbb{C} | |z|\le r)$ such that for any $z \in \mathring{D_r}$ there exists $b \in \mathbb{C}$ , a neighbourhood V of 0 such than $V+z \subset \mathring{D_r}$ and a function $\varepsilon$ from $V$ to $\mathbb{C}$ such that $g(z+h)=g(z)+bh+h\varepsilon (h)$ for any $h \in V$
Let $K$ be a nonempty compact of $\mathring{D_r}$ and $M=max(|f(z)| ,z \in K)$.
Prove the existence of $z$ in $\partial K$ such that $|f(z)|=M$.

My teacher told me this was known as the principle of maximum. Yet this exam does not require any complex analysis knowledge. Therefore I would appreciate having an elementary proof, or a short explanation of the complex analysis tools involved.
Thanks.

Maybe you know the proof for analytic function ? Then such a function is in fact an analytic function, this is a classical theorem of complex analysis. You will find the proof in any book about the subject.


modular arithmetic
combinatorics unsolved
combinatorics

A phone company starts a new type of service. A new customer can choose $k$ phone numbers in this network which are call-free, whether that number is calling or is being called. A group of $n$ students want to use the service.

(a) If $n\geq 2k+2$, show that there exist 2 students who will be charged when speaking.

(b) It $n=2k+1$, show that there is a way to arrange the free calls so that everybody can speak free to anybody else in the group.

[i]Valentin Vornicu[/i]

(a) Since each student can choose at most $k$ numbers, there are at most $nk$ free connections. However, in total, there are $\frac{n(n-1)}2\ge\frac{n(2k+1)}2>nk$ connections among the students, so one of them will not be free.

(b) Label the students $0,1,\ldots,2k$. $i$ chooses to speak for free with $j$ iff $j-i\in\{1,2,\ldots,k\}\pmod{2k+1}$. This arrangement works.

In graph-theoretical terminology
a) No digraph on $ n \geq 2k\plus{}2$ vertices and maximal outdegree at most $ k$ can have $ K_n$ as its subjacent unoriented graph;
b) Among all digraphs on $ n \equal{} 2k\plus{}1$ vertices and maximal outdegree at most $ k$ there exists one having $ K_n$ as its subjacent unoriented graph.

Notice in the proof above that in case a) not just one, but at least $ \frac {n} {2}$ connections will not be free. The model presented for case b) can be prolonged for $ n\equal{}2k\plus{}2$ to ensure that no more than $ \frac {n} {2} \equal{} k\plus{}1$ connections are not free.


How to prove the Maclaurin's Inequality?
http://planetmath.org/encyclopedia/MacLaurinsInequality.html

See here for example :) :
http://www.mathlinks.ro/Forum/viewtopic.php?t=12379
http://www.mathlinks.ro/Forum/viewtopic.php?t=20919


calculus
integration
function
trigonometry
complex analysis
complex analysis unsolved

I am confused on how to integrate this. This problem is asking me to find the integral using the definition of the integral is complex analysis. The function is

$ \int_C \sin z dz$ where $ C$ is the contour from $ \minus{}2i$ to $ 1\minus{}2i$. I don't see how to do this at all.  How do I rewrite this in order to do this problem?  There is probably a simple way to do this one, but I am not seeing it right now for some reason. Thanks in advance for an idea of where to go with this.

$ \sin$ has an antiderivative. Fundamental theorem.


geometry
topology
calculus
integration
real analysis
real analysis theorems

i am presently studying stokes theorem.
can anyone suggest a book ?
thanks in advance :lol:

for online resources, this migh help u:

[url]http://tutorial.math.lamar.edu/Classes/CalcIII/StokesTheorem.aspx[/url]

for hard copy,  any good bookon differential geometry will do

If you really want to understand Stokes' theorem, look for something with "Manifolds" in the title. The theorem about line and surface integrals is one case of a theorem that unifies the Fundamental Theorem of Calculus across many kinds of integrals, and the general theorem also carries Stokes' name.

Once you look at it from that perspective, the hard part is defining all of those integrals. Stokes itself is easy on a box, and everything looks like a nonoverlapping union of boxes.


limit
logarithms
trigonometry
real analysis
real analysis unsolved

We define a sequence in the fowllowing way:

$x_{2n}=\ln(x_{2n-1})$ and $x_{2n-1}=\cos(x_{2n})$

Also, $x_{0}$ is chosen in a way so that for ever natural number like $k$, $x_{k}$ exists. Prove that:

$\lim_{n\to\infty}x_{2n}=0$ and $\lim_{n\to\infty}x_{2n-1}=1$

Let $a_n=x_{2n}$ and $b_n=x_{2n+1}$. Then $a_n=\ln(\cos(a_{n-1}))\le a_{n-1}$ So, $a_n$ is convergent to $\ell_1$. $b_n=\cos(a_{n-1})\to \ell_2=cos(\ell_1)$. Solving the system of ecuations $\cos(\ell_1)=\ell_2$ and $\ln(\ell_2)=\ell_1$ we get the conclusion.

[quote="xirti"]Let $a_n=x_{2n}$ and $b_n=x_{2n+1}$. Then $a_n=\ln(\cos(a_{n-1}))\le a_{n-1}$ [/quote]
And why is $\ln(\cos x)\le x$? If anything, I would claim the opposite (note that $a_n$ are negative for $n\ge 1$) but even then it holds only if $x$ is small enough. Meanwhile, we only know that all $x_k$ exist but are given no information about their size...  :roll:

Some ideas:
  \[ \Big( \ln(\cos(x)) \Big)' = - \tan(x)  \]

Let $x \in [-0.5, 0[$. Then it is $-\tan(x) \in ]0,1[$.

There is a $\lambda \in [-0.5, 0[$ with
  \[ - \tan(\lambda) = \frac{\ln(\cos(x)) - \ln(\cos(0))}{x - 0} \iff - x \tan(\lambda) = \ln(\cos(x))  \]

So it is $x < \ln(\cos(x)) < 0$, but, of course, only for $x \in [-0.5, 0[$.

The next crucial step to solve the problem would be:
* Show, if $x_{2n}$ exists for every $n$, then there is a $n_0$ with $x_{2n_0} \in [-0.5, 0]$.


vector
function
calculus
derivative
calculus computations

The solution to the following is explained in my book, but the it is a bit unclear, so if anyone could explain how it is done step by step, it would be greatly appreciated!

Suppose the temperature at point (x, y, z) in space is given by the function:

f(x,y,z) = 1/(1 + x^2 + y^2 + z^2)

A bee is located at the point P = (1,1,1) in space.

Find a direction (as a three-dimensional vector) in which the initial rate of change in the temperature is 0.

1) The rate of change in a certain direction is called a directional derivative.
2) When the direction is represented by a unit vector $ \vec u$, the directional derivative can be found by the formula $ D_{\vec u}f=\nabla f\cdot \vec u$.
3) You are given $ f$, so you can find its gradient $ \nabla f$ and evaluate it at the point $ (1,1,1)$. What's next? You don't have $ \vec u$ -- in fact, you are asked to find it.
4) You must find a vector $ \vec u$ such that $ \nabla f(1,1,1)\cdot \vec u=0$. Recall that the dot product of two vectors is zero precisely when they are perpendicular. So, you want a vector $ \vec u$ that is perpendicular to $ \nabla f(1,1,1)$.
5) This is where you can use a standard property of the cross product: $ \vec a\times \vec b$ is perpendicular to both $ \vec a$ and $ \vec b$. Just take the cross product of $ \nabla f(1,1,1)$ with an arbitrary vector (like $ \vec i$) and you are done.


[quote="Kent Merryfield"]27 Enero:

Si las soluciones de la ecuación $x^2+4x-5=0$ son soluciones de la ecuación $2x^3+9x^2-6x-5=0,$ ¿cuál es la tercera raíz de la segunda ecuación?[/quote]
Vamos muchachos, esta es bien facil y nadie la ha intentado. Alguien se anima?

[quote="Severius"]$x^2+4x-5=(x+5)(x-1)=0$

Así que nos dicen que $1, -5$ son raíces de $2x^3+9x^2-6x-5=0$. Ahora dividimos eso por los factores dados, y obtenemos $2x^3+9x^2-6x-5=(x+5)(x-1)(2x+1)=0$. La tercera raíz es $-\frac{1}{2}$...[/quote] :first:


inequalities
function
real analysis
real analysis unsolved

what do you think about this? find best constant $C$ and say when hold equality:
$Cf(\sum{a_{i}x_{i}})\leq \sum{a_{i}f(x_{i})}$ where $f''(x)<0$ and
$a\leq f(x)\leq b$ is it know? or is it possible?

[quote="mlok"]You assume $\sum a_{i}=1$, right? And $f''(x)<0$ might just as well be $f''(x)\le 0$: all you achieve by imposing a strict inequality is non-existence of extrema, with no better bound.
$C=1$, of course.[/quote]
I mean for  convex function.  Later I shall formulate more exact conditions
. but I think there exist some const. :)

I don't know what you think you're going to find but there's no better constant than 1.

First, if $f$ is actually a linear function (in the elementary precalculus version of what you mean by that), then

$f\left(\sum a_{i}x_{i}\right)=\sum a_{i}f(x_{i}).$

Now, suppose the graph of $f$ has a tangent line. That means that in a neighborhood of that point, that tangent line is the first approximation to the function itself. ("A nearsighted flea thinks it's a straight line.") Let all of the $x_{i}$ be in that neighborhood - then you'll have a near-equality. You're always going to be able to do that, to produce a near-equality. And that means you can't beat 1 as a constant.


I wondered why I have so many posts per day until I noticed that my registration date from the AOPS site was used. But I registered at ML before. How do you choose the entry date ? By random or is there an algorithm ?  :D

Not sure how it happened in your case - I don't really remember how we ended up assigning the entry date for those with overlapping accounts (we used a phpbb tool for part of the merge, but were mostly concerned with getting all the accounts properly merged in other places - i.e. your post total, and assigning the right posts to the right users)

Well, I became a ML members in July after the IMO 2003.  :?  And I discovered AOPS far later (December 2003).

[quote="rrusczyk"]i.e. your post total[/quote]

:roll:

Well I don't think the post total is really correct for most members now...


calculus
integration
logarithms
trigonometry
function
complex analysis
real analysis

hello, for 3) we get
$ \int_{0}^{\infty}\frac{\ln(x)}{1\plus{}e^x}\,dx\equal{}\minus{}\frac{1}{2}\ln(2)^2$.
Sonnhard.

As for the third integral, if you have access to the JStore archive, you may find it here

H.F.Sandham; Ernest Trost, Amer.Math.Monthly
Vol.58, No.10 (Dec., 1951), pp.705--706, Problem n.4394,

or

M.S.Klamkin; H.F.Sandham; M.R.Spiegel,  Amer.Math.Monthly
Vol.62, No.8 (Oct., 1955), pp.588--590, Problem n.4592

Also let me tell you that oo is $ \plus{}\infty$

For the third integral, we use the equality

$ \int_{0}^{\infty} \frac{x^{s-1}}{1+e^x} \, dx = \eta(s)\Gamma(s),$

where $ \eta(s)$ is the Dirichlet eta function, defined for $ \text{Re}s > 0$ by $ \eta(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s}$.

Differentiating both side by s and plugging $ s = 1$, we have

$ \int_{0}^{\infty} \frac{\log x}{1+e^x} \, dx = \eta'(1) - \gamma \log 2.$

Now it suffices to show that $ \eta'(1) = \gamma \log 2 - \frac{1}{2} \log^2 2$, and we can show this by elementary means.

It is not hard to see that the defining series of $ \eta(s)$ converges uniformly for any $ 0 < \delta \leq \text{Re} s$ by Dirichlet test. So

$ \eta'(1) = \sum_{n=1}^{\infty} \frac{(-1)^n \log n}{n}.$

Let $ H_n =  \sum_{k=1}^{n} \frac{1}{k}$ be the harmonic number. Then

\begin{eqnarray*} \sum_{k=1}^{2n} \frac{(-1)^k \log k}{k}
& = & 2\sum_{k=1}^{n} \frac{\log 2k}{2k} - \sum_{k=1}^{2n} \frac{\log k}{k} \\
& = & H_n \log 2 - \sum_{k=n+1}^{2n} \frac{\log k}{k} \\
& = & H_n \log 2 - \sum_{k=1}^{n} \frac{\log (n+k)}{1+\frac{k}{n}}\frac{1}{n} \\
& = & (H_n - \log n) \log 2 + \log n \left[ \log 2 - \sum_{k=1}^{n} \frac{1}{1+\frac{k}{n}}\frac{1}{n} \right] - \sum_{k=1}^{n} \frac{\log \left(1 + \frac{k}{n} \right)}{1+\frac{k}{n}} \frac{1}{n}
\end{eqnarray*}

But it is easy to see that, if $ f$ is a $ C^1$ function defined on $ [0, 1]$, then $ \sum_{k=1}^{n} f\left( \frac{k}{n} \right) \frac{1}{n} = \int_{0}^{1} f(x) \, dx + O \left( \frac{1}{n} \right)$, by mean value theorem. So taking $ n\to\infty$, we have

$ \eta'(1) = \eta \log2 - \int_{0}^{1} \frac{\log (1+x)}{1+x} \, dx = \gamma \log 2 - \frac{1}{2} \log^2 2$

as desired. This method can be extended to evaluate $ I_n = \int_{0}^{\infty} \frac{\log^n x}{e^x + 1} \, dx$ for any $ n$, though it requires bunch of painful calculations. For example, you can show that

$ I_2 = \frac{\pi^2}{6} \log 2 - \gamma^2 \log 2 + \frac{1}{3} \log^3 2 - 2\gamma_1 \log2$

where $ \gamma_k$ are [url=http://en.wikipedia.org/wiki/Stieltjes_constants]Stieltjes constants[/url].

Also, note that

$ \int_{0}^{\infty} \frac{x^{s-1}}{e^x + 1} \, dx = (1 - 2^{1-s}) \zeta(s) \Gamma(s).$

and by definition of Stieltjes constants, the Laurant series for $ \zeta(s)$ at $ s =1$ is given by

$ \zeta(s) = \frac{1}{s-1} + \sum_{n=0}^{\infty} \frac{(-1)^n \gamma_n}{n!}(s-1)^n.$

Combining two results gives more efficient way to calculate $ I_n$.

it looks  integrals from complex analysis than real analysis. So I think it's  not so easy by solving via method from real analyisis. for example
in the 2 integral if you consider function  $ f(z) \equal{} \frac {1}{ze^{z}(\ln z \minus{} \pi i )}$ and integrate over the contour

[url=http://images.google.ru/imglanding?q=contour%20integration&imgurl=http://upload.wikimedia.org/wikipedia/commons/thumb/8/8d/Keyhole_contour.svg/180px-Keyhole_contour.svg.png&imgrefurl=http://en.wikipedia.org/wiki/Methods_of_contour_integration&usg=__6vso8Mea8gwXSy4O_eu4KD3GNcc=&h=180&w=180&sz=5&hl=ru&sig2=6-N-_t0PWnc4bTctJ9EIFQ&um=1&tbnid=RER1xu1pk4JOaM:&tbnh=101&tbnw=101&prev=/images%3Fq%3Dcontour%2Bintegration%26ndsp%3D18%26hl%3Dru%26lr%3D%26sa%3DN%26start%3D72%26um%3D1%26newwindow%3D1&ei=N6c0S47yD4ah-Ab82ZWvCg&ndsp=18&lr=&sa=N&start=79&um=1&newwindow=1#tbnid=RER1xu1pk4JOaM&start=83]contour[/url]

you will get after taking Im parts' that your integral is equal to 1/2. your firts and third problem I left you as exercises try to find needed function and contour. :)


[b]Problem[/b]: It's well-known that the equation
\[3^x+4^y=5^z\]
has only one sollution in positive integers $(2,2,2)$. The question is that : does there exist an infinitely many triple $(a,b,c)$  positive integers in which $a^2+b^2=c^2$ such that the equation $a^x+b^y=c^z$ has unique solution in positive integers is $(x,y,z)=(2,2,2)$

[quote="bomb"]a=5,b=12, c=13 is another such example.
I'm still working on general case.

Bomb[/quote]

Thanks and good luck friend  :D

[quote="bomb"]a=5,b=12, c=13 is another such example.
I'm still working on general case.

Bomb[/quote]

Working for a year? ejejejej. Any solution?

[quote="spx2"]Jose...bomb seems to be very ambitious don't you think  :lol:[/quote]

Yes!


Hi, this is a problem in the log chapter in AOPS V. 2.  Can anyone give me hints on how to approach it, I don't want to look at the soltuions yet.

If one uses only the information $10^{3}=1000$, $10^{4}=10000$, $2^{10}=1024$, $2^{11}=2048$, $2^{12}=4096$, $2^{13}=8192$, what are the largest $a$ and smallest $b$ such that one can prove $a<\log_{10}2<b$?

[hide="hint"]
$10^a < 2 < 10^b$
Try comparing the values you are given. Notice that $1000$ is close to $1024$ and $10000$ is close (well sort of) to $8192$.[/hide]

[quote="xxreddevilzxx"]does a and b have to be integers?[/quote]

No, just use the given info and find the tightest bound.


[quote="Lovasz"]Let $x_i,i=\overline{1,n}$ be integers satisfying: $x_1^3+x_2^3+...+x_n^3=2006$ , then find the greatest value of:
$E=x_1+x_2+...+x_n$
 :D  :roll:  :)[/quote]

That is easy. By power mean, we have:
$\left (\frac{2006}{n} \right)^{1/3} \geq \frac{E}{n}$ and then after multiplying by n and using integer part we have desired.

[quote="shyong"]Im guessing $x_i=1$ for all $i$ by inspection so the greatest sum $E\leq 2006$ .
 :D[/quote]
No, I don't think so, shyong, how could you make sure that $x_i=1$.

There wont be any maximum if ALL integer all allowed ( i.e positive and negative ) . Easy reason ,

suppose now there are 2006's 1 which cube sum is 2006 . if we remove a 1 and replace it with a negative integer $-n^3$ , we wil have to compensate it with + $n^3+1$'s 1 to make it 2006 . But when we do so , the sum is increased by $n^3+1-n-1=n(n+1)(n-1)>0$  . So you can see that if we keep doing this process , the sum would be infinity large . Similar argument can also prove that there are no minimum for it .

I think that you have to restrict $x_i$ to positve integer , then we can use cauchy to prove

$(\sum x_i)(\sum x^3_i)\geq \frac{1}{n^2}(\sum x_i)^4$ $\implies 2006n^2\geq (\sum x_i)^3$ . But for postive integer $x_i$ , the maximum of $n$ is $2006$ , hence the max of $E$ is 2006 when all $x_i=1$

[quote="Jan"]I don't think your solution is correct. Power mean can only be used on positive reals as far as I know, and not on integers, which can be negative, like we are dealing with here.  :)[/quote]

Actually, my solution is correct. It is easy to see, that in the text of the given example is written, the given is $n$, not the n integers. And the sum is maximum for all nonnegative integers. That's easy to see. And than my previous post tells the right one solution.

(why necessary nonnegative and not only positive integers? because for too many great values of $n$)..

Have a nice day

[quote]Actually, my solution is correct.[/quote]
That's possible, all I was saying is that Power mean is not true for negative numbers. And since you didn't prove that the max will occur when all numbers are positive, your proof isn't valid.

[quote]And the sum is maximum for all nonnegative integers. That's easy to see. [/quote]
You'll have to prove this first. It's not trivial.
If you prove it, you CAN use Power Mean and then your solution is correct  :)

Let me provide a counterexample for conceive's conjecture.

$1^3+1^3+(k+1)^3+(k-1)^3-k^3-k^3+(6p)^3 = 6k+2+(6p)^3$. For every p there's a k satisfying $6k+2+(6p)^3=2006$.

$1+1+(k+1)+(k-1)-k-k+(6p) = 2+6p$. The sum can get arbitrarily high, while perfectly being cancelled by k in the sum of third powers.

So for $n\ge 7$, there is no maximum for sure. I'll leave you the pleasure of exploring the case $n\le 6$.  [edit: yes, pleasure on the caseworking was ironic]


linear algebra
matrix
function
linear algebra unsolved

$ f: M_{2}(C) \minus{}> M_{2}(C)$ , $ f(x)\equal{}x^m$ .
If $ f$ surjective prove that $ m\equal{}1$

$ \begin{pmatrix}
 0&1\\
0 &0 \end{pmatrix}$ cannot be written $ A^m$ with $ m>1$.

What about $ f\colon M_n(\mathbb{C})\to M_n(\mathbb{C})$, $ f(x)\equal{}x^m$?
Essentially the same example works if $ m\ge n$.

For $ m>1.$ Just take JC_math's $ 2\times 2$ example as the upper left corner and pad it out with zeros.

Related, but more evolved surjective function following this link
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126962


[quote="Mîndrilă Claudiu"][i]Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$. Prove that $ \frac {a^{10}}{b \plus{} c} \plus{} \frac {b^{10}}{c \plus{} a} \plus{} \frac {c^{10}}{a \plus{} b} \geq \frac {a^7}{b^7 \plus{} c^7} \plus{} \frac {b^7}{c^7 \plus{} a^7} \plus{} \frac {c^7}{a^7 \plus{} b^7}$.[/i]
[b]Claudiu Mîndrilă[/b][/quote]
$ LHS \equal{} \frac {a^{10}}{a^3b^3c^3(b \plus{} c)} \plus{} \frac {b^{10}}{a^3b^3c^3(c \plus{} a)} \plus{} \frac {c^{10}}{a^3b^3c^3(a \plus{} b)}$

$ \equal{}\frac {a^7}{b^3c^3(b \plus{} c)} \plus{} \frac {b^7}{a^3c^3(c \plus{} a)} \plus{} \frac {c^7}{a^3b^3(a \plus{} b)}$

$ \rightarrow LHS \ge\ \frac {a^7}{b^7 \plus{} c^7} \plus{} \frac {b^7}{c^7 \plus{} a^7} \plus{} \frac {c^7}{a^7 \plus{} b^7}$

Because $ a^7 \plus{} b^7 \ge\ a^3b^3(a\plus{}b)$ Which is obvious


Define a square matrix A by $ a_{ij}\equal{} (x_{i}\plus{}y_{j})^{n}$for $ 0\le i,j\le n$

Show that $ \det A \equal{}\prod_{k \equal{} 1}^{n}{\left(\begin{array}{l}{\rm n}\\ {\rm k}\\ \end{array}\right)}\prod_{i > j}{(x_{i}\minus{}x_{j})(y_{j}\minus{}y_{i})}$

[hide="hint"]rewrite it as the product of two Vandermonde Determinants  :wink: [/hide]


Let $ X$ be a metric space.

Let $ E$ be a nonempty subset of $ X$. Define $ E$ be a nowhere dense set if $ X\backslash \bar{E}$ is a dense set (where $ \bar{E}$ denotes the closure of $ E$ in $ X$).  Define $ E$ to be of first category if $ E$ is the union of a countable collection of nowhere dense sets.

Let $ E$ be a subset of first category in $ X$.

Can $ E$ be dense in $ X$ if $ X$ is complete? Can $ E$ be dense in any $ X$?


superior algebra
superior algebra unsolved

i have  \[ \forall a,b\hspace{5pt}ab\equal{}ba\Rightarrow a\equal{}b\]
how can i prove from this :

that aaa=a  ?

 :(

Write down the semigroup axioms.  Which ones look like they might help?

there is a isomorphism that:

\[ (x,y).(z,t).(x,y)=(x,t).(x,y)=(x,y)\Rightarrow
(x,y)^{3}=(x,y)\Leftrightarrow
a^3=a\]

Thanks

I doubt I'm alone in having absolutely no idea what $ x$, $ y$, $ z$, $ t$, $ (\minus{},\minus{})$, or $ .$ are.

Key problem-solving technique: working backwards.  You wish to conclude $ aaa\equal{}a$.  You are given a theorem that allows you to conclude that $ b\equal{}a$.  What might be a convenient choice for $ b$? :)


search
algebra
polynomial
factorial
calculus
linear algebra
matrix

This would probably be much easier if there would be a good way to search for two-dimensional sequences in Sloane...

Let $ n\in\mathbb{N}$, where $ \mathbb{N}$ means $ \left\{0,1,2,...\right\}$.

[b](a)[/b] Prove that there exists one and only one family $ \left(a_{n;i,j}\right)_{\left(i,j\right)\in\mathbb{N}^2}\in\mathbb{R}^{\mathbb{N}^2}$ of reals such that

$ \binom{XY}{n} \equal{} \sum_{i \equal{} 0}^{\infty}\sum_{j \equal{} 0}^{\infty}a_{n;i,j}\binom{X}{i}\binom{Y}{j}$ in the polynomial ring $ \mathbb{R}\left[X,Y\right]$.

[b](b)[/b] Prove that $ a_{n;i,j}$ is an integer for every $ n\in\mathbb{N}$, $ i\in\mathbb{N}$ and $ j\in\mathbb{N}$.

So far the part I have solved. Now the interesting part:

[b](c)[/b] Prove or disprove that $ a_{n;i,j}$ is a nonnegative integer for every $ n\in\mathbb{N}$, $ i\in\mathbb{N}$ and $ j\in\mathbb{N}$.

I have checked this for $ n\leq 6$. If this is correct for all $ n$, the next question would probably be the combinatorial/representation-theoretic meaning of the $ a_{n;i,j}$.

  darij

This is really interesting. The first two parts are easy. For (c) would looking at the duplication formula of the Sheffer sequence of the falling factorial help?
It goes something like
\[ \binom{XY}{n} \equal{} \sum_{k \equal{} 0}^n\binom{X}{k}\frac {\langle Y\left((t \plus{} 1)^Y \minus{} 1\right)^k|X^n\rangle}{n!}\]
:maybe:
I don't have much time now, but I will try to work on this soon.

EDIT:
Never mind Umbral calculus for a moment, I think it's easy to show that $ a_{n; i,j}$ counts the number of ways of placing $ n$ ones in an $ i\times j$ matrix so that there is at least a $ 1$ in every row and every column  :wink:
In particular one has $ a_{n;i,j}\neq 0$ iff $ \max \{i,j\}\le n\le ij$,
 $ a_{n;n,n} \equal{} n!$ and $ a_{n;n,n \minus{} 1} \equal{} a_{n;n \minus{} 1,n} \equal{} \frac {n!(n \minus{} 1)}{2}$ etc..

A question that I came up with inspired by the above (a generalization),
if $ \binom{x}{\lambda} \equal{} \prod_{i}\binom{x}{\lambda_i}$ for any partition $ \lambda\in \mathbb{Y}$. If we have the variables $ X_j,1\le j\le n$ and denote $ \binom{\vec{X}}{\lambda} \equal{} \sum_{\sigma\in S_n}\prod_{i}\binom{X_{\sigma(i)}}{\lambda_i}$ we will have an identity
\[ \binom{\prod_{i}X_i}{\lambda} \equal{} \sum_{l(\mu) \equal{} n}a_{\lambda \mu}\binom{\vec{X}}{\mu}\]
What can we say about the $ a_{\lambda \mu}$?
Note that such $ a_{\lambda \mu}$ exist because the $ \binom{\vec{X}}{\lambda}$ are a basis for $ \Lambda_n$

EDIT:
[hide="partial answer"]when $ l(\lambda) \equal{} 1$ the $ a_{\lambda \mu}$'s are non negative integers, they count the number of ways of placing $ 1$'s in the entries of an $ n$-dimensional rectangle whose dimensions are given by the rows of $ \mu$ so that projected along each dimension they cover all possible coordinates.[/hide]

EDIT2:
[hide="continued partial answer"]actually all $ a_{\lambda \mu}$ are non negative integers. So it would be interesting to find combinatorial interpretations of the cases when $ \lambda$ has more than 1 row, or their representation-theoretical meaning in general.[/hide]

Really sorry for triple posting, but can not edit the posts above anymore.
$ a_{\lambda \mu}$ counts the number of ways of placing the squares of $ \lambda$ on an $ l(\mu)$ dimensional rectangle whose dimensions are given by the row lengths of $ \mu$ so that projecting along each dimension covers all coordinates. The squares in the same row of $ \lambda$ are indistinguishable while ones in different rows aren't.
But I still feel like there should be something more about these coefficients..

Now looking back maybe this is not the only way to think of generalizing this. What about the following:
Let $ X \equal{} \{X_i\}_{i \equal{} 1}^n,Y \equal{} \{Y_i\}_{i \equal{} 1}^n$ and $ XY \equal{} \{X_iY_i\}_{i \equal{} 1}^n$, with notation as above
\[ \binom{\overrightarrow{XY}}{\lambda} \equal{} \sum_{\mu \nu \in \mathbb{Y}}c_{\mu \nu}^{\lambda}\binom{\vec{X}}{\mu}\binom{\vec{Y}}{\nu}\]
What can we say about $ c_{\mu \nu}^{\lambda}$?

Wow, your solution is great. It reminds me of Stanley's work on nonnegative-integer stochastic matrices, though I don't see a direct connection. Maybe you see one...

I thought of extending this into another direction (actually motivated by $ \lambda$-rings): For every $ n\in\mathbb{N}$ and $ m\in\mathbb{N}$, there exists a unique sequence of reals $ \left(b_{n,m;i}\right)_{i\in\mathbb{N}}\in\mathbb{R}^{\mathbb{N}}$ such that $ \binom{\binom{X}{n}}{m}\equal{}\sum_{i\equal{}0}^{\infty} b_{n,m;i}\binom{X}{i}$ in $ \mathbb{Z}\left[X\right]$. These $ b_{n,m;i}$ are nonnegative integers.

[hide="Combinatoric description"]
Actually, $ b_{n,m;i}\equal{}\left|\left\{S\subseteq\mathcal{P}_n\left(\left\{1,2,...,i\right\}\right)\mid \left|S\right|\equal{}m\text{ and }\bigcup_{s\in S}s\equal{}\left\{1,2,...,i\right\}\right\}\right|$, where $ \mathcal{P}_n\left(T\right)$ means $ \left\{Q\subseteq T\mid \left|Q\right|\equal{}n\right\}$ for every set $ T$ and every $ n\in\mathbb{N}$.
[/hide]

As for your last generalization, I would start by considering $ \binom{\vec{XY}}{n}$ (since $ \binom{\vec{XY}}{\lambda}$ is just a product of these, and products have obvious combinatorial meaning), but I cannot concentrate well enough at the moment to go any further.

  darij

That's a nice result!
Btw, I'm not familiar with Stanley's work.. Do you have a reference?
Oh, note that I defined $ \binom{\vec{X}}{n}$ as a symmetric sum. Given $ X = \{X_1,X_2,\dots,X_n\}$ And $ \lambda = \{\lambda_1,\lambda_2,\dots,\lambda_n\}$
\[ \binom{\vec{X}}{\lambda} = \text{Per}\{\binom{X_i}{\lambda_j}\}\]
I guess the question becomes is there any nice combinatorial interpretation of binomial permanents?

Just for fun of course one can also define $ ^{B}s_{\lambda}(X) = \frac {\det(\binom{X_i}{\lambda_j + n - j})}{\det(\binom{X_i}{n - j})}$ (stands for binomial schur polynomial  :P, I doubt the term actually exists) and ask similar questions, and right now I cant tell which one is harder.

for Schur polynomials there is an identity
\[ s_{\lambda}(XY) = \sum_{\mu,\nu}\gamma_{\mu,\nu}^{\lambda}s_{\mu}(X)\s_{\nu}(Y)\]
where the $ \gamma_{\mu,\nu}^{\lambda}$ are given by ${{ \gamma_{\mu,\nu}^{\lambda} = \langle \chi^{\lambda}, \chi^{\mu}\chi^{\nu}}\rangle_{S_n} = \frac {1}{n!}\sum_{w\in S_n}\chi^{\lambda}(w) \chi^{\mu}(w)\chi^{\nu}}(w)$ which is why it's interesting to look at the questions above. Also looking at the binomial analogues of the Littlewood-Richardson coefficients might be interesting too but I want to look at this first, they are closely related anyway.
Ideas? Darij, t0rajir0u?

[quote="Albanian Eagle"]I guess the question becomes is there any nice combinatorial interpretation of binomial permanents?[/quote]
Certainly; binomial permanents count the number of $ k$-tuples of lattice paths from a certain configuration of sources to a certain configuration of sinks in $ \mathbb{Z}^2$ where only moves to the right and up are allowed and each source is matched to one sink.  Incrementing the $ x_i$ while keeping the $ \lambda_i$ fixed corresponds to moving, say, the sinks only to the right.

This is a very interesting discussion; might I suggest someone investigate what happens to other polynomial sequences of binomial type?

[hide="Some stupid thoughts."]:) I was afraid you were going to say that.
I'll try to see if that will help saying something about $ c_{\mu \nu}^{\lambda}$. But the problem is that even though that is a nice way of combinatorially interpreting permaments it still isn't that easy to handle (not to me at least, right now  :blush: ). I'm still thinking about [url=http://www.mathlinks.ro/viewtopic.php?search_id=979262680&t=290277]this[/url] where the natural way is to use Lindstrom's interpretation of binomial determinants as non-intersecting paths, but never thought of how to get rid of that extra $ b^{\text{something}}$ factor by a "clean" combinatorial procedure, which I suspect has something to do with roughly scaling the sources by a factor of $ b$. Here we are scaling by a factor of $ Y$ and I'm facing the same "wall"  :oops:
Btw, If I remember correctly Macdonald does talk about these symmetric binomial polynomials at some point, it just doesn't really say anything about their properties.[/hide]


$ a,b,c>0$ . Prove that :
$ \frac{1}{\sqrt{2a^2\plus{}3ab\plus{}2b^2}}\plus{}\frac{1}{\sqrt{2b^2\plus{}3bc\plus{}2c^2}}\plus{}\frac{1}{\sqrt{2c^2\plus{}3ca\plus{}2a^2}} \ge \frac{9}{\sqrt{7}(a\plus{}b\plus{}c)}$

$ (2a^2 \plus{} 3ab \plus{} 2b^2)(2b^2 \plus{} 3bc \plus{} 2c^2)(2c^2 \plus{} 3ca \plus{} 2a^2) \leq\ 343$
with $ a \plus{} b \plus{} c \equal{} 3$
it's not hard but rather long  :wink:

I want nice solution for this .
Your ineq will kill by MV but I''dont like that .

[quote="Lanyes"]By Holder inequality, the problem will be easy[/quote]
$ \left(\sum_{cyc}\frac {1}{\sqrt {2a^2 \plus{} 3ab \plus{} 2b^2}}\right)^2\sum_{cyc}c^3(2a^2 \plus{} 3ab \plus{} 2b^2)\geq(a \plus{} b \plus{} c)^3.$
Hence, it remains to prove that
$ 7(a \plus{} b \plus{} c)^5\geq81\sum_{cyc}(2a^3b^2 \plus{} 2a^3c^2 \plus{} 3a^3bc),$ which is wrong. :wink:
What did you mean? Thank you!


trigonometry
inequalities proposed
inequalities

Guys maybe this problem isn't interesting :( ,but try it please :)

[quote="Tiks"]$A,B$ and $C$ is angle of triangle.
Prove that $\sum \cos^4A\sin^2A\geq \frac 13\sin^2A\sin^2B\sin^2C$
 See this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=65597]www.mathlinks.ro/Forum/viewtopic.php?t=65597[/url][/quote]
Wher is the solution????????????????????? :?  :roll:

[quote="Tiks"]$A,B$ and $C$ is angle of triangle.
Prove that $\sum \cos^4A\sin^2A\geq \frac 13\sin^2A\sin^2B\sin^2C$
 See this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=65597]www.mathlinks.ro/Forum/viewtopic.php?t=65597[/url][/quote]
Nobody,why?
This problem isn't hard!


Triangle $ABC$ and $ABD$ are isosceles with $AB=AC=BD$, and $\overline{BD}$ intersects $\overline{AC}$ at $E$. If $\overline{BD}$ and $\overline{AC}$ are perpincicular, find $\angle{C}+\angle{D}$.

[hide]$\angle BAD = \angle D = \angle BAC+90^{\circ}-\angle D = 180^{\circ}-2\angle C+90^{\circ}-\angle D$.
Thus $\angle C+\angle D = 135^{\circ}$.[/hide]


geometry
parallelogram
vector
IMO Shortlist
IMO Longlist

Let $P$ and $P^{\prime }$ be two parallelograms with equal area, and let their sidelengths be $a,$ $b$ and $a^{\prime },$ $b^{\prime }.$ Assume that $a^{\prime }\leq a\leq b\leq b^{\prime },$ and moreover, it is possible to place the segment $b^{\prime }$ such that it completely lies in the interior of the parallelogram $P.$

Show that the parallelogram $P$ can be partitioned into four polygons such that these four polygons can be composed again to form the parallelogram $%
P^{\prime }.$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

If ABCD whith AB=b and BD=a is one of the two parallelograms, let E on BD a point so that AE=b'. This is possible since we know that b'>b but b' < AD (the biggest of the two diagonals). Then if C' is the midpoint of ED, let's take a point D' on AD s.t. AD'=C'D' in this way we have bisected the parallegram whith a segmente long b'.
Now we translate D'DDC' of the vector DA, s.t. D'-->A" and C'-->B".

After this we take a point A' on A"B" s.t. D'A'=a' this is possible since, as the two parallelograms are equivalent, a' is greater than the altitude wrt D'C' of the parallelogram A"D'C'B" and a' < a = D'A".
If G=A'D'^AB, let D"F the translation of A'G of the vector AD.
If one translate A"D'A' of the vector D'C' we obtain the parallegram A'B'C'D' equivalent to ABCD and having sides a', b'.

The parallelogram ABCD have  been partitioned into four polygons and these four polygons have been composed again to form the parallelogram A'B'C'D', as requested.

Perhaps the picture attached, make more clear the concept.