$ 1)$ αν $ a,b,c$ μη αρνητικοι αριθμοι με $ a^2\plus{}b^2\plus{}c^2\equal{}1$ νδο
$ a\plus{}b\plus{}c\leq \sqrt{3}$

2)αν $ x,y$ μη αρνητικοι αριθμοι νδο
$ x^2\plus{}y^2\plus{}1 \geq x\sqrt{y^2\plus{}1}\plus{}y\sqrt{x^2\plus{}1}$
(Εσθονία)

[quote="DIMITRIS T"]$ 1)$ αν $ a,b,c$ μη αρνητικοι αριθμοι με $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$ νδο
$ a \plus{} b \plus{} c\leq \sqrt {3}$

2)αν $ x,y$ μη αρνητικοι αριθμοι νδο
$ x^2 \plus{} y^2 \plus{} 1 \geq x\sqrt {y^2 \plus{} 1} \plus{} y\sqrt {x^2 \plus{} 1}$
(Εσθονία)[/quote]

Πολύ ωραία γεωμετρική εικόνα και στις δύο :)

Cheerio,

Durandal 1707

Για την πρωτη θετω $ a\equal{}\sqrt{x} b\equal{}\sqrt{y} c\equal{}\sqrt{w}$.Τοτε ισχυει x+y+w=1.Θεωρω την συναρτηση $ \sqrt{x}$ που ευκολα μπορω να δειξω οτι εχει δευτερη παραγωγο αρνητικη.Οποτε απλη εφαρμογη της ανισοτητας Jensen.Για την δευτερη απλα την διπλασιαζω,ολα στο πρωτο μελος και ειναι αθροισμα τετραγωνων.

Εδώ συμφωνώ με τον Παναγιώτη
Πράγματι πού ωραία γεωμετρική εικόνα
Α) Θεωρείστε τα διανύσματα $ \vec{u}\equal{}(a,b,c) , \vec{v}\equal{}(1,1,1)$  τότε είναι $ |\vec{u}|.|\vec{v}| \ge \vec{u}.\vec{v}$ μεά τις εύκολες πράξεις προκύπτει άμεσα το ζητούμενο
Β)Θεωρείστε το ορθογώνιο παραλληλεπίπεδο  με βάσεις τα ορθογώνια ΟΑDΒ , ΚLΜΝ και ακμές τις ΟΚ,AL,DΜ, ΒΝ
Αν ΟΑ=x, ΟΒ=y ,ΟK=1 τότε στο ορθογώνιο τρίγωνο  OMN  είναι $ OM^2\equal{}x^2\plus{}y^2\plus{}1, MN\equal{}x , ON\equal{}\sqrt{1\plus{}y^2}$  και είναι γνωστό ότι το εμβαδόν του ειναι $ \le$  από το εμβαδόν του ορθογωνίου και ισοσκελούς τρίγωνου με υποτείνουσα την ΟΜ ,,διότι εχει > ύψος, οπότε προκύπτει η σχέση $ \frac{x^2\plus{}y^2\plus{}1}{2} \ge x.\sqrt{1\plus{}y^2}$ και αντίστοιχα από το OBM είναι $ \frac{x^2\plus{}y^2\plus{}1}{2} \ge y.\sqrt{1\plus{}x^2}$ προσθέτοντας κατά μέλη το ζητούμενο
Απο αλγεβρικές λύσεις θα προτιμούσα για το Α) την BCS και για το Β) $ x^2\plus{}(\sqrt{y^2\plus{}1})^2 \ge 2x.\sqrt{1\plus{}y^2}$ και την ανάλογή της

1)

$ 1 = a^2 + b^2 + c^2\geq 3\left(\frac {a + b + c}{3}\right)^2 = \frac {(a + b + c)^2}{3}$

Επισης λυνεται πολυ ευκολα με την μεθοδο των πολλαπλασιαστων του Lagrange.
Δειχνουμε οτι το $ \left(\frac {1}{\sqrt {3}},\frac {1}{\sqrt {3}},\frac {1}{\sqrt {3}}\right)$ ειναι σημειο ολικου μεγιστου της $ f(a,b,c) = a + b + c$ υπο συνθηκη $ a^2 + b^2 + c^2 - 1 = 0$

2)

Απο ΑΜ-ΓΜ

$ x^2 + (y^2 + 1)\geq 2x\sqrt {y^2 + 1}$

και $ (x^2 + 1) + y^2\geq 2x\sqrt {y^2 + 1}$

Προσθετουμε κατα μελη και διαιρουμε με το 2 οποτε

$ x^2 + y^2 + 1 \geq x\sqrt {y^2 + 1} + y\sqrt {x^2 + 1}$


A 2-cup mixture is 1/3 flour and 2/3 cornmeal. If 1 cup of flour is added to the 2-cup mixture, what fraction of the new 3-cup mixture is flour? Express your answer as a common fraction.

Right now, there is $ 2 \cdot \frac{1}{3}\equal{}\frac{2}{3}$ cups of flour. If you add one more cup, then there will be $ \frac{2}{3}\plus{}1\equal{}\frac{5}{3}$ cups. This is $ \frac{\frac{5}{3}}{3}\equal{}\boxed{\frac{5}{9}}$ of the $ 3$ cup mixture.


induction
inequalities proposed
inequalities

Prove that: $ \frac { \sqrt{6}}{5} + \ldots +  \frac{ \sqrt{2n(2n+1)}}{4n+1}< \frac{n}{2}$, $ \forall n \in N^{*}$.

Here is my approach!  Induction:

for $n=1$ we have  $ \frac{ \sqrt{6}}{5} < \frac{1}{2}$  which gives $24<25 $ which is true!!
lets consider it true for $n-->n$ and prove it for $n+1$

$ \sum_{k=1}^{n}  \frac { \sqrt{(2n(2n+1)}}{4n+1)} + \frac { \sqrt{(2n+2)(2n+3)}}{4n+5} < \frac{n}{2}+ \frac { \sqrt{(2n+2)(2n+3)}}{4n+5} < \frac{n+1}{2} $
all that remains is to prove the last ineq:
$ \frac{n}{2} + \frac { \sqrt{(2n+2)(2n+3)}}{4n+5} < \frac{n+1}{2} $ which is obious as we after some computation we get that $16<25$ which is obiuosly true!
Indution works!
Is there another aproach?
cheers! :D :D

Yes, there exist. We have 2n(2n+1) < 4n^2 + 2n + 1/4 = (2n+1/2)^2. So sqrt(2n(2n+1))/(4n+1) < (2n+1/2)/(4n+1) = 1/2. Each summand is less than 1/2 so the sum is less that n/2.

Namdung


probability
conditional probability
probability and stats

This is a classic, yet interesting problem.

A fair coin is flipped repeated. Find the probability that two consecutive heads occur before three consecutive tails.

A bunch of inelegant algebra $ \to$ $ \frac{7}{10}$.

Is there a simple combinatorial way to see this?

Flip the coin twice and see what state we're in. There are four possibilities: HH, HT, TH, and TT. If it's HH, then we're done.

Let $ x$ be the conditional probability of eventually finishing HH given that the current state is HT.

Let $ y$ be the conditional probability of eventually finishing HH given that the current state is TH.

Let $ z$ be the conditional probability of eventually finishing HH given that the current state is TT.

If the current state is HT, then the probability is $ \frac12$ that the next state is TH and $ \frac12$ that the next state is TT. Hence, $ x = \frac12y + \frac12z.$

If the current state is TH, then there is a $ \frac12$ probability that the next state is HT and a $ \frac12$ probability that we finish HH immediately. Hence, $ y = \frac12x + \frac12.$

If the current state is TT, then there is a $ \frac12$ probability that the next state is TH and a $ \frac12$ probability that we finish TTT immediately. Hence, $ z = \frac12y.$

The system
\[ x = \frac12y + \frac12z \\ \\
y = \frac12x + \frac12 \\ \\
z = \frac12y
\]
can be solved quite readily. We get $ x = \frac35,y = \frac45,z = \frac25.$ Each of these states has a $ \frac14$ chance of occurring in the first two flips, along with the $ \frac14$ chance of ending immediately. That gives the probability we want as:
\[ P = \frac14\left(1 + \frac35 + \frac45 + \frac25\right) = \frac7{10}.
\]
This is a fairly typical use of conditional probability/conditional expectation arguments. I could also have called this a Markov chain with absorbing states.


parameterization
calculus
calculus computations

$ x \equal{} 1/(t^2\minus{}1)$
$ y \equal{} t^2 / (t\minus{}1)$

I need help with this, I can't find how to derivate such a thing

$ \frac{dy}{dx}\equal{}\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$ \frac{d^2y}{dx^2}\equal{}\frac{d}{dt}\left(\frac{dy}{dt}\right)\frac{dt}{dx}$

Thx, but I find that even more complicated, could you perhaps explain in words, I live in Belgium and we don't use that notation (yet)

Then How about plugg $ y \minus{} x \equal{} t \plus{} 1\Longleftrightarrow t \equal{} y \minus{} x \minus{} 1$ into $ x \equal{} \frac {1}{t^2 \minus{} 1}$ to elliminate parameter $ t$?

No, that was not useful.

$ x \equal{} \frac {1}{t^2 \minus{} 1}\Longrightarrow \frac {dx}{dt} \equal{} \minus{} \frac {2t}{(t^2 \minus{} 1)^2}$

$ y \equal{} \frac {t^2}{t \minus{} 1} \equal{} t \plus{} 1 \plus{} \frac {1}{t \minus{} 1}\Longrightarrow \frac {dy}{dt} \equal{} 1 \minus{} \frac {1}{(t \minus{} 1)^2}$

$ \therefore \boxed{\frac {dy}{dx} \equal{} \frac {\frac {dy}{dt}}{\frac {dx}{dt}}} \equal{} \minus{} \frac {t(t \minus{} 2)}{(t \minus{} 1)^2}\cdot \frac {(t^2 \minus{} 1)^2}{2t}$

$ \equal{}\minus{}\frac 12(t \minus{} 2)(t \plus{} 1)^2$

$ \therefore \frac {d^2y}{dx^2} \equal{} \frac {d}{dx}\frac {dy}{dx}$

[quote="kunny"]Then How about plugg $ y \minus{} x \equal{} t \plus{} 1\Longleftrightarrow t \equal{} y \minus{} x \minus{} 1$ into $ x \equal{} \frac {1}{t^2 \minus{} 1}$ to elliminate parameter $ t$?

No, that was not useful.

$ x \equal{} \frac {1}{t^2 \minus{} 1}\Longrightarrow \frac {dx}{dt} \equal{} \minus{} \frac {2t}{(t^2 \minus{} 1)^2}$

$ y \equal{} \frac {t^2}{t \minus{} 1} \equal{} t \plus{} 1 \plus{} \frac {1}{t \minus{} 1}\Longrightarrow \frac {dy}{dt} \equal{} 1 \minus{} \frac {1}{(t \minus{} 1)^2}$

$ \therefore \boxed{\frac {dy}{dx} \equal{} \frac {\frac {dy}{dt}}{\frac {dx}{dt}}} \equal{} \minus{} \frac {t(t \minus{} 2)}{(t \minus{} 1)^2}\cdot \frac {(t^2 \minus{} 1)^2}{2t}$

$ \equal{} \minus{} \frac 12(t \minus{} 2)(t \plus{} 1)^2$

$ \therefore \frac {d^2y}{dx^2} \equal{} \frac {d}{dx}\frac {dy}{dx}$[/quote]

Thx once more but I still don't get it (sorry)
I get the first 2 steps because that just a simple derivation.
But the following steps (starting from the one with the box) I can't follow anymore.

This is a formula $ \frac {dy}{dx} \equal{} \frac {\frac {dy}{dt}}{\frac {dx}{dt}}$

Oh ok, last problem:

$ \therefore \frac {d^2y}{dx^2} \equal{} \frac {d}{dx}\frac {dy}{dx}$

WHat is this:
$ \frac {d}{dx}$

What is this:
$ \frac {d}{dx}$

Are you asking the meaning for it?? :)

$ \frac {d}{dx}f(t)$ means differentiate $ f(t)$ with respect to $ x$.

For example, $ \frac {d}{dx}t^2 \equal{} \frac {d}{dt}\cdot t^2\cdot \frac {dt}{dx}$, yielding $ \frac {d}{dx}t^2 \equal{} 2t\frac {dt}{dx}$ by chain rule.


geometry
3D geometry
rotation
combinatorics proposed
combinatorics

Define a R-cube to be a flat block of 2*2 cubes, with another cube on top of the 4 base cubes, so it has 5 cubes in it. Find all n, for which a n*n*n cube can be filled up exactly and precisely with R-cubes. Note that they can be rotated, reflected and stuff.

c'mon any ideas people? its a great problem, please try solving it.


Find all functions $f: \mathbb R \to \mathbb R$ which satisfy $f(0) = 0$ and
\[ f(f(x)\plus{}y)\equal{}2x\plus{}f(f(y)\minus{}x)\]
for all real $x$ and $y$.

[quote]Find all $ f: R\to R$ which satisfy
\[ f(0) \equal{} 0;\ f(f(x) \plus{} y) \equal{} 2x \plus{} f(f(y) \minus{} x)\ (1)\]
[/quote]
Take $ y: \equal{} \minus{} f(x)$ into $ (1)$, we obtain $ \minus{} 2x \equal{} f(f( \minus{} f(x)) \minus{} x)$, $ \forall x\in \mathbb {R}$, implying $ f$ is surjective function.
Denote $ g(x)$ by the function defined on $ \mathbb {R}$ such that $ g(x) \equal{} f(x) \minus{} x$. Then from $ (1)$, we get:
$ g(g(x) \plus{} x \plus{} y) \plus{} g(x) \equal{} g(g(y) \plus{} y \minus{} x) \plus{} g(y)$ $ (2)$, $ \forall x,y\in \mathbb {R}$.
Take $ x: \equal{} 0$ into $ (2)$, we have $ g(g(y) \plus{} y) \equal{} 0$, $ \forall y\in \mathbb {R}$ or $ g(g(x) \plus{} x) \equal{} 0$, $ \forall x\in \mathbb {R}$.
But since $ f$ is surjective, then $ g(x) \equal{} 0$, $ \forall x\in \mathbb {R}$, yielding that $ f(x) \equal{} x$, $ \forall x\in \mathbb {R}$.

[i]Conclusion.[/i] $ f(x) \equal{} x$ is only the solution to our problem.

[quote="pco"][quote="mathVNpro"][quote]Find all $ f: R\to R$ which satisfy
\[ f(0) \equal{} 0;\ f(f(x) \plus{} y) \equal{} 2x \plus{} f(f(y) \minus{} x)\ (1)\]
[/quote]
Take $ y: \equal{} \minus{} f(x)$ into $ (1)$, we obtain $ \minus{} 2x \equal{} f(f( \minus{} f(x)) \minus{} x)$, $ \forall x\in \mathbb {R}$, implying $ f$ is injective function.[/quote]

Why should $ \minus{} 2x \equal{} f(f( \minus{} f(x)) \minus{} x)$ imply $ f$ injective ?  :huh:[/quote]
Sorry, I mean "surjective"  :blush: . I have edited it!

[quote="peter117"]Find all $ f: R\to R$ which satisfy
$ f(0) \equal{} 0$
and $ f(f(x) \plus{} y) \equal{} 2x \plus{} f(f(y) \minus{} x)$[/quote]

Setting $ x\equal{}0$, we get $ f(f(y))\equal{}f(y)$ and so $ f(x)\equal{}x$ $ \forall x\in f(\mathbb R)$

And since mathVNpro proved that $ f(x)$ is surjective, $ f(\mathbb R)\equal{}\mathbb R$ and $ f(x)\equal{}x$ $ \forall x$


[quote="Kent Merryfield"]No, it diverges, which can be seen instantly once you know that $\sum\frac1k$ diverges.

Of course, in this case, you can find an explicit formula for

$\prod_{k=1}^{n}\left(1+\frac1k\right).$

It's a "telescoping" product. Have you looked at that?[/quote]
It becomes $\frac{k+1}{k}$ so when we multiply out, these term will cancel and it only left the first and the last term. Hence , it is $n+1$ right?


[quote="momtonate"]Can anybody help me out with this?

On December 31, 1995, there were an estimated 411 prison inmates per 100,000 U.S. residents.  This number rose to an estimated 476 inmates per 100,000 U.S. residents by December 31, 2002.

a.What percentage of the U.S. population were prison inmates at the end of 1995?
b.What percentage of the U.S. population were prison inmates at the end of 2002?[/quote]

My dear friend, this is not exactly what we consider a [b]Superior Algebra Unsolved Problem[/b]. I am sure that soon one of our efficient moderators will move this problem to another forum. Nevertheless, let's give it a solution.

They are telling you that in 1995, for each 100 000 U.S. Residents, there were 411 prision inmates, then, the ratio between the total of the population and the prision inmates is $\frac{411}{100,000}=0.00411$ therefore the persentage of the population that was prision inmates at the end of 1995 was 0.411%. Now, applying the same idea, you can find that at the end of 2002 the percentage of pupulation in jail was 0.475%. If we assume that the people in jail increases exponentially an try to obtain the rate from the information we have, then, by 2038 we will have that 1 inmate per 100 U.S. residents. Scary, isn't it?

Best,


AMC USA(J)MO USAMO search geometry inequalities AIME

Can't the problem be reduced to showing that 0 and 1 are in S?  This would generate all the integers by (c).  Does this make the problem any easier?

I'm new to AoPS, by the way, and would like some adivce.  I made USAMO last year but didnt this year because I severely messed up on the AIME.  I would like to do some preparation by myself (i.e. not at a camp) for next year's USAMO.  What would you recommend I do?  I already have purchased some books that I think may be useful.

Thanks for any advice.

Hi there, and welcome to the forums!

Your analysis of number 2 is correct.  The difficulty of the problem is showing that 1 is in the set, if I remember correctly.

The question you asked is one that a lot of people are interested in.  You might try a search of the forums (the search button is in the upper-right), as well as looking through old posts in this forum, the Other Problem Solving Topics forum, and the other contest forums to see if you can find any information there.  I personally recommend lots of solving problems as the best way to improve for contests and other people have had more specific suggestions on where good problems can be found or for books targetting specific subject areas (inequalities, geometry, et cetera.)

Good luck!

[quote="cn2_71828182846"]I'm new to AoPS, by the way, and would like some adivce.  I made USAMO last year but didnt this year because I severely messed up on the AIME.  I would like to do some preparation by myself (i.e. not at a camp) for next year's USAMO.  What would you recommend I do?  I already have purchased some books that I think may be useful.

Thanks for any advice.[/quote]

Welcome to the Art of Problem Solving!

For books, many people like the books "The Art and Craft of Problem Solving" by Paul Zeitz, and "Problem-Solving Strategies" by Arthur Engel.  Both are intermediate to advanced problem solving books, which would be useful for AIME/Olympiad preparation.

Also, you can get previous AMC tests, AIMEs, and Olympiads in the Publications section of the AMC website (http://www.unl.edu/amc).

You might look through the Math Jam transcripts for the USAMO Day 1 Math Jam where we walked through that problem.


let $ABC$ be a triangle.and $a,b,c$ be the sidelength of it. prove that area of triangle $ABC$ is $\frac{1}{2} \times a \times b \times sin(C)$.
( i have already post this problem in basics and it is hard for them but i think it is easy for you.) :D

[quote="Altheman"]drop a perpendicular from A to BC, this is the height, and is $\sin{C}*b$, and the base is $a$, so using $\frac{1}{2}bh$, we have $A=\frac{1}{2}ab\sin{C}$[/quote]
exactly Altheman,you solved it nice. :D

[quote="K.W.M.A.N"]or another way to look at this:
if we say $D=A+\vec{AB}+\vec{AC}$ then the area of ABDC equals  $\vec{AB}\cdot\vec{AC}$, this is the double of the area of ABC.[/quote]

that should be $\vec{AB}\times\vec{AC}$, you have to use, the Cross product not the dot product.

the area of triangle is $\frac{1}{2}|\vec{AB}\times\vec{AC}|$which based on the defenition of cross product is :
$|\vec{AB}||\vec{AC}|\sin{\theta}$ where $\theta$ is the angle bettween the vecotrs (sides).


MATHCOUNTS
AMC
AIME
Support
geometry
AMC 10

Well Im multi tasking and playing right now.  It's an awesome 3rd person medeavil rpg.  I was just wondering who else plays (if any).  For people who don't play it was voted best browser game of 04 and is totally free unless you want members ($5 a month).  Just go to http://www.runescape.com

it's an awesome game, but be warned it can be very addictive
I'm not gonna post my username cause I'm paranoid, but feel free to post yours, talk about the game.  And who knows, maybe someone new will find out about the game.

Fr4nkthetank.  Frankthetank wasn't available, so I substituted a 4.

i used to play....for 3 months or so. yea it is really addicting. i stopped on the day of thanksgiving in seventh grade (im in eighth now) on that exact day, it got hacked and the person just took all teh stuff and......yea. its kinda sad....cause i really lyked my account.....and i was doing sumthing important.

awesome... the accounts seemed to get hacked too often.  They have a problem with that, I guess people get so into the game that they are willing the steal other people's stuff.  You just have to be smart with what you do, and you're safe.  My account has 1110 skill total, so i have to be kindof careful about where I give the name out to... you understand.

[quote="Athena"]whats ur combat levels? mine was.....47 or 48 or somewhere around that when i got hacked.[/quote]
84.. yep I play too much I admit it but comsidering some people are 120+ I feel a tad better

Hey fellow runescape members!
My name is "al_funcoot"
I'm a member and my combat lvl is 64 (I trained too much on fletching), but anyways someone contact me if they're a member and they want to play castlewars.

My friend got a membership for Christmas, and he was combat level 50 after about a week.  Yes, there are much crazier people, but this kid's parents are strict and stuff.

My combat level is 40.  Mostly, I have been doing peaceful things.

Does anyone want to give me a job (in Runescape)?  I'm poor  :( .

My Runescape username is Blooblyblobl  :D
I have combat 63, and I don't need to contact jeff since I know him in real life...
Frankthetank, what kind of job would you be thinking of?  To this end, what are your non-combat skills.
I just became a member, and I now lvl up on men to get herbs (unids go for 1k each!!!).  LOL!

Frankthetank, like what entropy said what are your noncombat skill's lvl and what job do u want to do. Oh yah how much do u want to be payed?
Can u mine coal?

I used to play but not for very long. I played for about a week and got bored of it.

Maplestory is a lot less repetitive. I've never played Runescape before (I thought it was run-escape before :P ), but this is what I've heard from my friends.

http://www.mapleglobal.com

Somedude, zezima should make you feel a lot better...  All skills except herblore 99, 96 herblore! :D

BTW athena, I've been hacked twice, scammed out of a mask once, and still enjoy playing.  Just use password recovery, and spend a couple of days getting the most important items (rune pick) back.  It's really sad, but doesn't ruin the entire experience  ;)

I had a pretty good account but I don't paly anymore

[quote="ragnarok23"]I used to play, but now I don't as my parents got really mad whenever I played.
Looking back, I'm wondering why I ever played that game in the first place.  :rotfl:[/quote]
wow, same here...
indian parents think alike i guess

umm... wow I have not been on AoPS for a while now.

I dont play runescape anymore. heh.

Runescape makes you stupid.  :rotfl:

i have a vll 79 with a blue phat, ova 180 mil!! and i bought it for 90mil (it went way up, inflation)

well, i play and i have a decent account...It isn't that addicting...

"...makes you stupid"

ditto! I should get off my computer and start setting timers when I'm posting!

[quote="now a ranger"]FIRST OF ALL... I HATE RUNESCAPE

SECOND OF ALL... I HAD 80 mining 77 smithing.. what an idiot

DO NOT PLAY IT .. IT WILL MAKE YOU a POOPOOHEAD![/quote]
WHY THE - ARE YOU REVIVING 35109235014376504765 OLD TOPICS JUST TO SAY RUNESCAPE SUCKS. THAT'S LIKE REVIVING 10976012650192OLD TOPICS IN THE OLYMPIAD FORUMS TO SAY 1+1=2
why isnt this locked yet

I played for a few years a while back, and If I hadn't ever found out about it i'd probably be in nationals mathcounts by now  :D
it's veeeeery addicting and a big waste of time.  I was really bad too.  only got to lvl 57 or something like that

This is an old topic...

I got to level 93 and quit when I realized I was wasting my time. As daermon said, it's very addicting. :ninja:


Marty wants to paint a box.  He can choose to use either blue, green, yellow, or black paint.  Also, he can style the paint by painting with a brush, a roller, or a sponge.  How many different combinations of color and painting method can Marty choose?

there are 4 colors, and 3 ways to paint it
therefore, there are $ 4\times3$ ways to paint the box, or $ 12$ ways

it's easy:4 colors x 3 techniques= 12 :)   :idea:

It is simply choice 1: 4, Choice: 3 so 4*3=12 :wink:

[quote=Leon11]I got 12[/quote]

elaborate?

[quote=Butler]Ok. Because there are 4 possibilities for paint color, and 3 for type of painting style. 4*3=[u][b]12[/b][/u] possibilities[/quote]

[quote=flash07]I got 12 too.[/quote]

[quote=Leon11]I got 12[/quote]

behold..... the sheer honesty of the aops community


You mean this $\frac{2xy}{x+y} + \sqrt{\frac{x^2 + y^2}{2}} \geq \sqrt{xy} + \frac{x+y}{2}$????

$\frac{2xy}{x+y} + \sqrt{\frac{x^2 + y^2}{2}} \geq \sqrt{xy} + \frac{x+y}{2}\Leftrightarrow\frac{(x-y)^2}{2(\sqrt\frac{x^2+y^2}{2}+\sqrt{xy})}-\frac{(x-y)^2}{2(x+y)}\geq0\Leftrightarrow$
$\Leftrightarrow x+y-\sqrt\frac{x^2+y^2}{2}-\sqrt{xy}\geq0\Leftrightarrow(\sqrt x-\sqrt y)^2-(\sqrt\frac{x^2+y^2}{2}-\sqrt{xy})\geq0\Leftrightarrow$
$\Leftrightarrow(\sqrt x-\sqrt y)^2-\frac{(\sqrt x-\sqrt y)^2(\sqrt x+\sqrt y)^2}{2(\sqrt\frac{x^2+y^2}{2}+\sqrt{xy})}\geq0\Leftrightarrow\sqrt{2(x^2+y^2)}-x-y\geq0\Leftrightarrow$
$\Leftrightarrow(x-y)^2\geq0.$

I saw that somebody else posted a solution while I was typing mine  :oops: . they are basically the same but so I wouldn't have wrote it for nothing I will post it .
$\displaystyle\left(\sqrt{\frac{x^2+y^2}{2}}-\sqrt{xy}\right)^2\geq 0 \Rightarrow  \frac{x^2+y^2}{2}+xy\geq 2\sqrt{xy\frac{x^2+y^2}{2}} \Rightarrow$
$\displaystyle (x+y)^2\geq\frac{x^2+y^2}{2}+xy+2\sqrt{xy\frac{x^2+y^2}{2}}=\left(\sqrt{\frac{x^2+y^2}{2}}+\sqrt{xy}\right)^2 \Rightarrow$
$\displaystyle x+y\geq \sqrt{\frac{x^2+y^2}{2}}+\sqrt{xy}\Rightarrow  \frac{(x-y)^2}{x+y}\leq\frac{(x-y)^2}{\sqrt{\frac{x^2+y^2}{2}}+\sqrt{xy}} \Rightarrow$
$\displaystyle\frac{(x+y)^2-4xy}{2(x+y)}\leq\frac{\frac{x^2+y^2}{2}-xy}{\sqrt{\frac{x^2+y^2}{2}}+\sqrt{xy}} \Rightarrow \frac{x+y}{2}-\frac{2xy}{x+y}\leq\sqrt{\frac{x^2+y^2}{2}}-\sqrt{xy} \Rightarrow$
$\frac{x+y}{2}+\sqrt{xy}\leq\frac{2xy}{x+y}+\sqrt{\frac{x^2+y^2}{2}}$

[quote="Ramanujan"]Well i was writing a solution too but arqady is pretty fast  :)[/quote]
 :yup:

You're welcome mps.
[quote="Ramanujan"][quote="JellyWorld"]You could have just written $HM+QM\ge AM+GM$ :P[/quote]

Is this true for more that two variables  ???  :?[/quote]

I think it is. I have a better question ... If the $\alpha$-mean($M_{\alpha}$) of $x_1,...,x_n$ is $\left(\frac{x_1^{\alpha}+...+x_n^{\alpha}}{n}\right)^{\frac{1}{\alpha}}$ then what conditions must $\alpha>\beta>\gamma>\delta$ satisfy so that $M_{\alpha}+M_{\delta}\geq M_{\beta}+M{\gamma}$, but I also have to get in bed to get some sleep so I'll see what you are up to later  :roll:

It is also my problem I proposed in my book. I put $w=x/y$ to solve it.


function
quadratics
algebra
functional equation
algebra unsolved

Find all functions $f$ which are defined for all $x\in\mathbb{R}$ and, for any $x$, $y$, satisfy
$xf(y)+yf(x)=(x+y)f(x)(y)$

Is this $xf(y)+yf(x)=(x+y)f(x)\times{y}$ or $xf(y)+yf(x)=(x+y)f(xy)$

[quote="abdurashidjon"]Find all functions $f$ which are defined for all $x\in\mathbb{R}$ and, for any $x$, $y$, satisfy
$xf(y)+yf(x)=(x+y)f(x)f(y)$[/quote]

setting $x=0$ we obtain $yf(0)=yf(0)f(y) \implies f(y)=1$ $\forall y \in \mathbb {R}$

[quote="hydro"]setting $x=0$ we obtain $yf(0)=yf(0)f(y) \implies f(y)=1$ $\forall y \in \mathbb {R}$[/quote]
Actually $yf(0)=yf(0)f(y) \implies f(y)=1$ $\forall y \in \mathbb {R^*}$ [b]or[/b] $f(0)=0$.

There are infinity solutions:
1. f(x)=0 for all x.
2. f(x)=1 for all $x\not =0$, f(0) is all.

I have the solution but from book i could not understand the restriction. My solution follows as:

Set $y=x$ then easily we can get that $f(x)(f(x)-1)=0$
$(i) f(x)=0$
$(ii) f(x)=1.$
Set $y=-x$ then $xf(-x)-xf(x)=0\rightarrow f(x)=f(-x)$ so function is even
Set  $y=1$ and take $f(x)=ax^2+c$, where $a$ is either function or any real number,$c$ is constant(we will look for $a$ and $c$). Then we will find $a=0$ and $c=0$ and $1$. It follows that $f(x)=0$ or $1$ as we found before, so where is the restriction?

I don't agree with the solution from your book, because I don't know why it assumes the quadratic form. Maybe it's not $ax^2$ but $a(x^2)$?. Anyway, the restriction is that the first time $f(x)=0$ or $f(x)=1$ for each separate $x$ (any combination of 0's and 1's is still possible), and the second time $f(x)=0$ for every $x$ or $f(x)=1$ for every $x$ (only the constant solutions are possible).
My solution is more similar to Rust's, but I'll check.

[quote="lordWings"]I don't agree with the solution from your book, because I don't know why it assumes the quadratic form. Maybe it's not $ax^2$ but $a(x^2)$?[/quote]

The solution was mine not from book. And I just asked where is the restriction of my solution.
The solution from book as:
$(i)$ and $(ii)$ and next is as in book it shows that function is even for all $x$, $i$, $e$., so function can be continues (later I will post the solution of book)

a) $x=y$ $\to$ $2xf(x)=2xf(x)^2$ $\to$ if $x\neq 0$, $f(x)\in\{0,1\}$.
So, if the set of zeroes of $f$ is $Z=\{z\in\mathbb{R}|f(z)=0\}$, then the solution must be:
$f(x)=0$, $x\in Z$, $f(x)=1$, $x\not\in Z$, except for $f(0)$, which is free.

b) If $f(z)=0$, $x=z$ $\to$ $zf(y)=0$.
This means that there are three possibilities for the set of zeroes : $Z=\emptyset$, $Z=\{0\}$, or $Z=\mathbb{R}$.

Using a)+b), the three possibilities are:
- $f(x)=1$, except for $f(0)$, which is free.
- $f(x)=1$, except for $f(0)=0$ (falls into the first possibility).
- $f(x)=0$

Exactly like Rust.

Lordwings, we also can show function is even and can someone find even part of function
Abdurashid

The problem has already been solved (with only two steps), and you can see that all solutions are even functions (without having had to prove it).
But, if you want,
$y=-x$ $\to$ $x(f(-x)-f(x))=0$ $\to$ $f(x)=f(-x)$, even function.
I don't understand your question about the even part of the function.

The problem is that true for even function such there can be many solutions for  even function, if it's true so ithas to be proven or just need to be said?
Abdurashid

In this case we have proceeded in order to:
a) find restrictions that help us eliminate functions that are not solutions to the functional equation,
b) prove that all the remaining functions are possible solutions.

In this case, proving that the function had to be even was possible, but unnecessary. (With other functional equations, it could have been a very useful information.)


[quote="byuns9334"]Find all f: Q -> Q that satisfies f(x^2 + y) = f(x^2) + f(y) for all x, y ..[/quote]

Put $ x\equal{}0$ we get $ f(0)\equal{}0$

[b]Case 1:[/b] there does not exist $ x_0$ such that $ f(x_0) \neq 0$ . Then $ f(x)\equal{}0 \forall x \in Q$ is the solution.

[b]Case 2:[/b] there exists $ x_0\equal{}\frac{q}{p}$ such that $ f(x_0) \neq 0$ and $ p,q \in Z$

then for any $ y\equal{}\frac{q_1}{p_1}$ , where $ p,q \in Z$

, let $ c\equal{}\frac{1}{pp_1}$ .

$ f(c^2\plus{}y)\equal{}f(c^2)\plus{}f(y) \Rightarrow f(kc^2)\equal{}kf(c^2) \forall k \in Z$

So $ x_0\equal{}qp_1^2c^2 \Rightarrow f(x_0)\equal{}qp_1^2f(c^2)$
$ y\equal{}q_1p^2c^2 \Rightarrow f(y)\equal{}q_1p^2f(c^2)$

So $ \frac{f(y)}{f(x_0)}\equal{}\frac{y}{x_0}$  (because $ f(c^2) \neq 0$ or $ f(x_0)\equal{}0$)
$ \Rightarrow f(y)\equal{}\frac{f(x_0)}{x_0}y \forall y \in Q$
$ \Rightarrow f(x)\equal{}kx \forall x \in Q$ , where $ k$ is a constant $ \in Q$ , which is the solution after checking.


induction
linear algebra
matrix
vector
abstract algebra
linear algebra unsolved

Let $ \mathcal{S} \in \mathcal{M}_n(F)$ be a set with the property that every $ (n\plus{}1)$ members of $ \mathcal{S}$ have a common eigenvector.
Show that there is a common eigenvector for all elements of $ \mathcal{S}$.

Let me try it :). Hepefully I don't make any mistake.
We will use induction on $ k$ that every $ n\plus{}k$ elements of $ S$ has a common eigenvalue.
The statement is clearly true for $ k\equal{}1$. Assume it is true for $ k$.
For $ k\plus{}1$ let $ T\equal{}\{M_1,M_2,\cdots, M_{n\plus{}k\plus{}1}\}\subset S$. Let $ T_i: \equal{}T\backslash \{M_i\}$ and let $ \lambda_i$ be the common eigenvalue of matrices in $ T_i$ (such $ \lambda_i$ exists by the induction hypothesis).
Now $ \lambda_1,\cdots,\lambda_{n\plus{}k}$ are eigenvalues of $ M_{n\plus{}k\plus{}1}$. Since an $ n\times n$ matrix has at most $ n$ eigen values, then $ \lambda_i\equal{}\lambda_j$ for some $ 1\leq i<j\leq n\plus{}k$.
Then $ \lambda_i\equal{}\lambda_j$ is a common eigen value of  $ T_i\cup M_i\equal{}T$.

Are you sure you solved the right problem? ;)

  darij

Thanks Darij, I didnt read carefully. It should be common eigen vector instead of common eigen value :).

The result looks pretty but is not true in general.
Here is a simple counter-example.
In $ F^2$, consider three different $ 1$-dimensional linear subspaces
$ D_1$, $ D_2$ and $ D_3$.
Let $ P_1$ (resp. $ P_2$, resp. $ P_3$) denote
the projection of $ F^2$ onto $ D_2$ with kernel $ D_3$
(resp. onto $ D_3$ with kernel $ D_1$, resp. onto $ D_1$ with kernel $ D_2$).

Consider then the set $ \mathcal{S}$ consisting of the six matrices
$ \begin{pmatrix}
P_1 & 0 \\
0 & 0
\end{pmatrix}$, $ \begin{pmatrix}
P_2 & 0 \\
0 & 0
\end{pmatrix}$, $ \begin{pmatrix}
P_3 & 0 \\
0 & 0
\end{pmatrix}$,
$ \begin{pmatrix}
0 & 0 \\
0 & P_1
\end{pmatrix}$, $ \begin{pmatrix}
0 & 0 \\
0 & P_2
\end{pmatrix}$ and $ \begin{pmatrix}
0 & 0 \\
0 & P_3
\end{pmatrix}$.

Then $ \mathcal{S}$ is a counter-example (its elements have no common eigenvector, although
any subset of $ \mathcal{S}$ of cardinal $ 5$ has a common eigenvector).

dSPaBJ is right.
I looked at the problem again and I omitted an important asumption, sorry  :blush:
I can't edit the original post, but I'd like to add that all element of $ \mathcal{S}$ have a fixed point (i.e. 1 is an eigenvalue).
Sorry again for those of you who searched that problem without that asumption.

[quote="JC_math"]... add that all element of $ \mathcal{S}$ have a fixed point (i.e. 1 is an eigenvalue).[/quote]
@ JC_math: But even this property is fulfilled by all matrices of dSPaBJ's counterexample as they are all (non-trivial) projections!?! Or I am missing something here?

I looked again, is the "counter example" from dSPaBJ made of $ 5\times 5$ matrices?


2t*4 - 4t*2 + 2
* means expoent

Which are the factors?
Should we use Ruffinis way to solve this problem or make t*4 = d*2? :ninja:

what's your point?

do you want it convert it to Binomial?

do you want to convert it to a binomial?

$\displaystyle 2t^4 - 4t^2 + 2 = 2(t^4 - 2t^2 + 1) = 2(t^2 - 1)^2 = 2(t+1)^2(t-1)^2$

So the factors are $\displaystyle 2$, $\displaystyle(t+1)$, and $\displaystyle(t-1)$.

EDIT: What is Ruffinis?????

Your answer is correct but the theres also onother way to solve it that is= dividing it by its zero. ;) that is dividing by (x-zero).
We can do division simply by way that a mathematic called Ruffini created. I still dont have latex system so i cant show you that.
Explore about Ruffini anywhere and youll see what I mean.  :o  :|  :)  ;) I was asking only the easiest way but yours also quite simple.


ratio
superior algebra
superior algebra unsolved

Hi,

I need to prove that the infinite series are convergent using the comparison test and the ratio test.

$ \displaystyle\sum_{n\equal{}1}^{\infty}\frac{2^n \plus{} 7^n}{3^n \plus{} 11^n}$

I have proved it using the comparison test.

Now using the ratio test, I got:

$ \displaystyle\frac{a_{n\plus{}1}}{a_{n}}\equal{}\frac{2^{n\plus{}1} \plus{} 7^{n\plus{}1}}{2^n \plus{} 7^n}*\frac{3^n \plus{} 11^n}{3^{n\plus{}1} \plus{} 11^{n\plus{}1}}$

I am struggling to simplify it further. Would appreciate some help. Thank you.

hello, writing your quotient in the following form
$ \frac{(2\cdot2^n\plus{}7\cdot7^n)(3^n\plus{}11^n)}{(3\cdot3^n\plus{}11\cdot11^n)(2^n\plus{}7^n)}\equal{}$
$ \frac{2\cdot6^n\plus{}7\cdot21^n\plus{}2\cdot22^n\plus{}7\cdot77^n}{3\cdot6^n\plus{}11\cdot22^n\plus{}3\cdot21^n\plus{}11\cdot77^n}\equal{}$
$ \frac{2\cdot\left(\frac{6}{77}\right)^n\plus{}7\cdot\left(\frac{21}{77}\right)^n\plus{}2\cdot\left(\frac{22}{77}\right)^n\plus{}7}{3\cdot\left(\frac{6}{77}\right)^n\plus{}11\cdot\left(\frac{22}{77}\right)^n\plus{}3\cdot\left(\frac{21}{77}\right)^n\plus{}11}$.
I hope this will help you.
Sonnhard.


group theory
superior algebra
superior algebra unsolved

Hey. So during a lesson last week, we were given the following question.

"Let X = {(a,b):b=2a} ⊂ [b]R[/b][sup]2[/sup] and define a map *: A x A --> [b]R[/b][sup]2[/sup] by *((a,b),(c,d)) = (a+c, b+d). Show that * defines a b inary operation on X and that (X,o) forms a group."

Can anyone post up the proof? It's more for totality of knowledge rather than anything else, but it'd still be greatly appreaciated, because I have no clue what I'm doing.

It kind of follows from the definitions. if you know what a [hide="binary operation"]http://en.wikipedia.org/wiki/Binary_operation[/hide] is and what a [hide="group"]http://en.wikipedia.org/wiki/Group_(mathematics)[/hide] is then you'd only have to check your required properties are satisfied. Give it some thought and post any questions if you're stuck with any concept. Seeing some examples in a group theory book just to get the idea wouldn't hurt though  :P

I'll give it a go. The problem is that our notes are extremely basic, fairly void of examples, and access to books is fairly limited :P

*goes and looks at it*


vector
superior algebra
superior algebra unsolved

I have 3 questions :

1. How to identify the additive group $ {\mathbb{F}_4}^ \plus{}$;

2. Prove that every element of $ \mathbb{F}_p$ has exactly one $ p$th root;

3. Let $ p$ be a prime. Describe the integers $ n$ such that there exist a finite field $ K$ of order $ n$ and

an element $ \alpha \in K^{\times}$ whose order in $ K^{\times}$ is $ p$.

Thanks!

1. It's just a vector space over $ \mathbb{F}_2$, so it's the Klein 4-group.

2. If $ x \in \mathbb{F}_p$, what is $ x^p$? This should help.

3. Do you know what the structure of $ \mathbb{F}_n^{\times}$ is?


A sequence $x_0=A$ and $x_1=B$ and $x_{n+2}=x_{n+1} +x_n$ is called a Fibonacci type sequence. Call a number $C$ a repeated value if $x_t=x_s=c$ for $t$ different from $s$.
Prove one can choose $A$ and $B$ to have as many repeated value as one likes but never infinite.

At first it's easy to see that it's impossible to construct a sequence with
infinitely many repeated values .

From reccurence we have : $x_n=\frac{\alpha(1+\sqrt{5})^n-\beta(1-\sqrt{5})^n}{2^n\sqrt{5}}$ ($\alpha,\beta$ are some real values) , so the only possibilities are that sequence
$x_n$ is constant ($x_0=0,x_1=0$) or $x_n->\infty$ or $x_n->-\infty$ ,what
enables us to find an infinite number of repeated values ([Edit] forgott to add about strictly increasing or decreasing ${x_n}$ - this is sufficient).

Suppose we define our sequence backwards , so also mention
$ F_{-1},F_{-2},F_{-3} ... F_{-k} $ . We can take any $k$ we like and that's
our last step. Using Fibbonacis reccurence let's define ... $...,F_{4}=5,F_{3}=3,F_2=2,F_1=1,F_0=1
F_{-1}=0,F_{-2}=1,F_{-3}=-1,F_{-4}=2,F_{-5}=-3,F_{-6}=5,F_{-7}=-8,F_{-8}=13,...$ ... so we see that $F_{-2}=F_{0},F_{-4}=F_{2},F_{-6}=F_{4},F_{-8}=F_{6},...,F_{-2n}=F_{2n-2}$ , so putting $k$ as much as we like we have repeated values-finitely-as much  as we like...

i can't understand the first part. can someone explain? i think it is not that kind of easy. proving that this sequence is monotonic is a wrong way. this is false :)


Να δείξετε ότι για καθε $a,b\in R$
$|a+b|^{r}\leq |a|^{r}+|b|^{r}$ με $r\in[0,1]$

Λοιπόν: Διαίρεσε με $|a|^{r}$ την ζητούμενη και θέλεις ισοδύναμα να δείξεις ότι

$\left|1+\frac{b}{a}\right|^{r}\leq 1+\left|\frac{b}{a}\right|^{r}$

Θεωρούμε τώρα την συνάρτηση $f(x)=1+x^{r}-(1+x)^{r}$ με $r\in[0,1]$ η οποία είναι γνησίως αύξουσα για $x\geq 0$ άρα

$f\left(\left|\frac{b}{a}\right|\right)\geq f(0)=0$

Αρα $\left(1+\left|\frac{b}{a}\right|\right)^{r}\leq 1+\left|\frac{b}{a}\right|^{r}$

Όμως λόγω τριγωνικής ανισότητας (και ότι η συνάρτηση $g(x)=x^{r}, x\geq 0$ και $r\in[0,1]$ είναι γνησίως αύξουσα), έχουμε $\left|1+\frac{b}{a}\right|^{r}\leq \left(1+\left|\frac{b}{a}\right|\right)^{r}$ απ'όπου έπεται το ζητούμενο.

Αλέξανδρος

Ωραία λύση Αλέξανδρε!!!
Η άσκηση είχε και δεύτερο μέλος που έλεγε ν.δ.ο

$|a+b|^{r}\leq c_{r}(|a|^{r}+|b|^{r})$ με $c_{r}=2^{r-1}$ για $r>1$.

Και αυτό λύνεται με τον ίδιο ακριβώς τρόπο. Διαίρεση με το $|a|^{r}$, όλα στο πρωτό μέλος και θεωρούμε τη συνάρτηση

$f(x)=c_{r}+c_{r}x^{r}-(1+x)^{r}, \ x\geq 0, r>1$ η οποία είναι γν. φθίνουσα στο $[0,1]$ και γν. αυξουσα στο $[1,+\infty]$

Άρα για κάθε $x\geq 0$ ισχύει $f(x)\geq f(0)=0$ άρα για $x=\left|\frac{b}{a}\right|$ έχουμε

$f\left(\left|\frac{b}{a}\right|\right)\geq 0$ και λόγω της $\left|1+\frac{b}{a}\right|^{r}\leq \left(1+\left|\frac{b}{a}\right|\right)^{r}$ που δείξαμε στο πρώτο post, έχουμε το ζητούμενο.

Αλέξανδρος

JENSEN  $\frac{f(2a)+f(2b)}{2}\ge f(a+b), f(x)=|x|^{r}=convex$

An dn kano la8os,ine  gnosti anisotita..

$\frac{a+b}{2}\leq (\frac{a^{r}+b^{r}}{2})^{1/r}$ me $a,b>0,r>1$


inequalities
function
inequalities unsolved

let $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}3$ prove that

 \[ \frac{a\plus{}1}{b^2\plus{}1}\plus{}\frac{b\plus{}1}{c^2\plus{}1}\plus{}\frac{c\plus{}1}{a^2\plus{}1}\geq3\]

[quote="mitdac123"]This ineq was posted in "secret in ineq " of hunghktn ;)[/quote]

 is  (secret in inequality ) a book

[quote="M.A"]let $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 3$ prove that
\[ \frac {a \plus{} 1}{b^2 \plus{} 1} \plus{} \frac {b \plus{} 1}{c^2 \plus{} 1} \plus{} \frac {c \plus{} 1}{a^2 \plus{} 1}\geq3
\]
[/quote]

it is very simply:
use Jensen's  ineq for $ f(x) \equal{} \frac{1}{x^2\plus{}1}$
 we  have :  $ (a\plus{}1)f(b) \plus{} (b\plus{}1)f(c) \plus{} (c\plus{}1)f(a) \geq 6f(\frac {3\plus{}ab\plus{}bc\plus{}ca}{6})\geq 6f(1) \equal{} 3$

[quote="Sergic Primazon"]
it is very simply:
use Jensen's  ineq for $ f(x) \equal{} \frac {1}{x^2 \plus{} 1}$
 we  have :  $ (a \plus{} 1)f(b) \plus{} (b \plus{} 1)f(c) \plus{} (c \plus{} 1)f(a) \geq 6f(\frac {3 \plus{} ab \plus{} bc \plus{} ca}{6})\geq 6f(1) \equal{} 3$[/quote]

No, your function is concave only in case $ x \ge \frac{1}{3}$ :wink:

[hide]for $ x\geq0$ we have $ \frac{1}{1\plus{}x^2}\geq 1\minus{}\frac{x}{2}$[/hide]

[quote="FOURRIER"][quote="Sergic Primazon"]
it is very simply:
use Jensen's  ineq for $ f(x) \equal{} \frac {1}{x^2 \plus{} 1}$
 we  have :  $ (a \plus{} 1)f(b) \plus{} (b \plus{} 1)f(c) \plus{} (c \plus{} 1)f(a) \geq 6f(\frac {3 \plus{} ab \plus{} bc \plus{} ca}{6})\geq 6f(1) \equal{} 3$[/quote]

No, your function is concave only in case $ x \ge \frac {1}{3}$ :wink:[/quote]

Yes, my function is concave  in case $ x \ge \frac {1}{\sqrt{3}}$
let $ a\leq b$  and  $ a\leq c ;$
   if $ a < \frac {1}{\sqrt{3}}$  , then for min in this inequality  must be : a = 0 or b = c ;
   if a = 0 then min > 3,...
   if ( b = c )  min will be then a = b = c = 1


For those of you who've taken calculus, or who are taking it now - what book did/do you use, and what do you think of it?

The first calculus book I read was [i]Calculus the Easy Way[/i], which is great for a first book. I read a bunch of others after that, although my criterion for selecting was which books were available at the library. I used Swokowski for my AP calculus class and Stewart for Multivariable Calculus. Stewart is quite a nice book since it has some non-standard (contest-like) problems for interested students.

if you are looking for a concise (900 pages) and full featured calculus book (Calculus 1-3)
go for Calculus 8th Edition by Varberg, Purcell, and Rigdon

adn there is this set of old (my dad's) some like six volume set of like 10000 medium to challengning calculus probelms its in chinese translated from russian. (if i ever find the english name i'll post it)

I use [i]Calculus, Seventh Edition[/i], by Anton, Bivens, and Davis.  It's a good one, as far as I can tell.  (I don't have anything to compare it to, but it's pretty easy to understand.)

Here it is on Amazon: [url]http://www.amazon.com/exec/obidos/tg/detail/-/0471381578/qid=1075879662/sr=1-1/ref=artofproblems-20/104-8634216-0877525?v=glance&s=books[/url]

I hope that worked.  I doctored the link a little to make AoPS the referrer (by adding artofproblems-20 to the appropriate place).

In high school, we used a textbook whose name I have forgotten, and it's just as well.  In our college honors class, we used an unforgettable book -- Spivak's Calculus.  Just for the (challenging) problems alone, it was worth every penny.  There is also a solutions manual available.

Spivak was the only calculus book that Paul Zeitz gave in his list of recommended math books.

I think Spivak is far too theoritical as a first book into Calculus.  If you intend to slowly break into the world of Calclus, a very popular textbook is Stewart's Calculus (with various incarnation of the integral sign / violin on the cover).  Spivak, Apostle and others are quite a bit more theoritical, and serves as a bridge between a calculus course and an analysis course.

I agree, Spivak is a great book, but not for an introduction to Calculus for high school students.

What we are using  is the  Calculus by Stewart.  It 's  a  nice  book. and  I believe  many are  using  that  book.

I think I used that varberg book for Cal BC and I used Stewart for Cal III. I'm using a book by Patrick Fitzpatrick (what a cool name) for real analysis.

I did a search for Stewart's Calculus on google, but most of the results were for his solution guides and such. Can someone please tell me the exact name of his books in order?

Stewart's calculus book was at one time probably the most widely used calculus book in the country. It may now be being overtaken by some of the newer books though. There are several versions of the book that, unfortunately, all look very similar (with the integral sign/violin on the cover, but with different colors).

The different versions that I know of are:

Single Variable Calculus (with CD-ROM), 850 pages
Publisher: Wadsworth Publishing; 5th Edition (December 6, 2002)
ISBN: 0534393667

Multivariable Calculus (with CD-ROM), 630 pages
Publisher: Wadsworth Publishing; 5th Edition (December 20, 2002)
ISBN: 0534393578

Calculus (with CD-ROM), 1300 pages
Publisher: Wadsworth Publishing; 5th Edition (December 20, 2002)
ISBN: 053439339X

Calculus: Early Transcendentals (with CD-ROM), 1320 pages
Publisher: Wadsworth Publishing; 5th Edition (December 20, 2002)
ISBN: 0534393217

I use [url=http://www.amazon.com/exec/obidos/tg/detail/-/1559531177/qid=1076261097//ref=sr_8_xs_ap_i1_xgl14/002-5333374-1363210?v=glance&s=books&n=507846]this book[/url] for Calculus; it's pretty nice.

Gauss, you seem to be quite experienced on this subject (I bet that statement has been said thousands of times sicne 1777).
What book would you recommend for someone who hasnt started calculus at all?


M is the midpoint of the segment AB.The point $ C\in AB$ and $ C\neM$.In the same plane by the segment AB are build two isosceles triangles ACK(AK=CK) and BCL(BL=CL).We also know that K,L,M and C lie on one circle.Prove that KL||AB or $ KA\bot LB$.

May I ask how point $ N$ is defined? Also the rest of the text isn't clear...

Well,you must prove that $ KA \bot LB$ or $ KL\parallel{}AB$.


Show that n^4 + 4 is not a prime number for any integer n greater than 1.

For even integers its straight forward. But what about odd integers?

Posted many primes (omg, I really wrote "primes" instead of "times" here) before. It's the identity of Sophie Germain which kills the problem: $n^{4}+4=(n^{2}+2n+2)(n^{2}-2n+2)$.


If you have been taking a class on a Tuesday and it is ending but you want to take another class that is on a Tuesday as well, and the classes overlap for 1 or 2 weeks, will you be able to access both classrooms for just the two weeks?

Because whenever I try to get into the classroom from two windows, one of them always gets disconnected...

Thanks.

Also, today, when I logged into both the math jam and my regular class, I was disconnected from the math jam.

Thank you!  I did not see it before I posted it though.  Thanks so much.


AMC
USA(J)MO
USAMO
inequalities solved
inequalities

Let a' = a 1/3 , b' = b 1/3  and c' = c 1/3  and you got the well-known USAMO 1997.

Pierre.

It's a tournament of the towns problem too :D


One apple and one orange cost a total of $ 38$ pence.

Three apples and four oranges cost a total of $ 136$ pence.

What is the total cost of three apples and three oranges?

Find the cost of (i) one orange, (ii) one apple.

[hide="Solution"]$ a\equal{}\text{cost of apple}$ and $ o\equal{}\text{cost of orange}$:

$ a\plus{}o\equal{}38$

$ 3a\plus{}4o\equal{}136 \\
3(a\plus{}o)\plus{}o\equal{}136 \\
3 \cdot 38 \plus{} o\equal{}136 \\
o \plus{} 114\equal{}136 \\
\boxed{o\equal{}22}$

$ a\plus{}o\equal{}38 \\
a\plus{}22\equal{}38 \\
\boxed{a\equal{}16}$

$ 3a\plus{}3o\equal{}3(a\plus{}o)\equal{}3 \cdot 38 \equal{} \boxed{114}$ [/hide]


[b]Aici[/b] se pot gasi cateva probleme pentru capacitate, la partea a doua. :D


geometry
3D geometry
number theory
prime factorization

GIven positive integers $a,b,c,d$ such that $a^{3}=b^{2}, c^{3}=d^{2}$, and $c-a =25$, determine $a,b,c,d$.

[hide]

Say that $a^{3}=b^{2}=m$.  All the exponents of the primes in $m$'s prime factorization must be divisible by both 2 and 3, so $m$ is a perfect sixth power.  For $a^{3}$ to be a perfect sixth power, $a$ must be a perfect square.  The same is true for $c$, clearly.  So we must find two perfect squares that differ by 25.  $13^{2}=169$ and $12^{2}=144$ are the only ones that fit the requirements, so $c=169$ and $a=144$.  Thus $b=12^{3}=1728$ and $d=13^{3}=2197$.

[/hide]


Prove the followings:

1) Equation $x^3=y^2+3$ does not have any natural solution.

2) Equation $x^n=y^2+3$ does not have natural solution for $n=4k-1$

For the second we can assume that $p=4n-1$ is prime (in general $4n-1$ has such factor).
Now $\mod 8$ we get $x^p = y^2+3 \equiv 3,4,7 \mod 8$, thus $x \equiv -1 \mod 4$.

Lets solve $x^p+1=y^2+4$.

Now $x^p+1=(x+1)(x^{p-1}-x^{p-2}+-...+1)$ and $\gcd(x+1 , x^{p-1}-x^{p-2}+-...+1 ) | p$.
But $p|(x+1)$ is excluded since that implies $p|y^2+4 \implies (2^{-1}y)^2 \equiv -1 \mod p \implies p \not \equiv -1 \mod 4$.
Thus $\gcd(x+1 , x^{p-1}-x^{p-2}+-...+1 )=1$.

But by known conditions when an integer can be written as sum of $2$ squares (see the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=76610]Formulary[/url]) , this gives that $x^{p-1}-x^{p-2}+-...+1$ is the sum of two squares, contradicting $x^{p-1}-x^{p-2}+-...+1 \equiv (-1)^{p-1}-(-1)^{p-2}+-...+1 = p \equiv -1 \mod 4$.


My question to Aops/ML users is why did you choose to do IMO/MO math?
There are tonnes of other cool things to do in high school, so why would you choose this one?

In my opinion, IMO/MO math is very nice and interesting(even though it might not be that useful in most jobs). Also, I think it is the most fun thing to do in high school if you love Math: you go to camps and then if you get really good, to IMO, which from what I have heard is a lot of fun.
And you get scholarships to universities if you do well on olympiads, espescially IMO.

It's ok if you repeat something that someone else said already.

I think you've anticipated most of the reasons that other people will give.  I'll give you my own answer:

First, let me say that it's not quite right for me to say I've ever really done "IMO math," because I've never gotten even close to qualifying for the IMO.  I only qualified twice for USAMO as a high school student and my best finish in USAMO was top-40 or so (as a senior).

But I still deeply enjoy solving old USAMO problems (and occasionally, IMO problems).  Here's why:

1. It's amazing how many challenging problems you can pose and solve just using the bare-bones material that one learns in high school.  We all tend to take this for granted (nobody seems to worry that we'll eventually "run out" of truly original problems to put in our math olympiads).  And yet it is no easy task to "come up with" an Olympiad problem.  To my mind, the best Olympiad problems are discovered, not manufactured, which makes it that much more of a challenge to put together a great olympiad each year.

2. Unlike problems in current mathematical research, a lot of good Olympiad problems can be posed in ways that most high school students can understand (even though the solutions cannot).  A fairly educated layperson could understand what a lot of these problems ask for (if you gave them a short description of what "mathematical proof" is about).  Because of this, olympiad problems have broad appeal.  Most of the problems I work on in my research would take hours to explain, even to someone with an undergraduate education in math.

3. Unlike most of the problems you encounter in high school, you can't simply grind out an olympiad problem.  A lot of olympiad problems - even some of the hardest ones - are ones where you might come up with the crucial insight as you're about to go to sleep, when you're in the shower, on the bus to school, etc. (assuming you're working on them outside of a competitive context).

4. Yes, there are benefits (e.g. scholarships) available to top performers in math olympiads.  However, I don't think this is that strong a motivation for the people who are really great at this kind of thing.  Most of the people who are really serious about olympiad math do it because they love it.

well my school gives free donuts during the AIME test and it probably does so with the USAMO test as well

I do math because it is fun..... Who needs a better reason than that?

Let's assume that you do something else instead of IMO math. Then we may ask the question, "Why are you doing that instead of all the other cool things out there?" And we're back where we started. It's like asking why people do anything since if you ever pick something to do there are millions of other options...

In any case, I think most people here do math for the joy of it, for fun if you will. And maybe they enjoy competition; the AMCs are a good place to go for that. Coincidentally, if you do well enough, you get to go to the IMO. It's not like you chose specifically to do IMO problems. You just ended up there because you went in the direction of your interests.

I get to miss English once a month if I do the AMC series.

I miss 15 hours of school for the AMC, AIME, USAMO.

Who wouldn't want that?

It is exciting for me to struggle to reach the top, especially on a subject that I have interest in. And skiping class is the extra bonus :D (although sometimes it backfires :( )

Week off from school for every stage I qualify, various bonuses that come from being finalist of Polish MO, spirit of competition multiplied by the fact that I just love math riddles.

And in Iran , If you manage to qualify for the 3-th level (Its a little hard :D ) you will be granted a seat in any university you want.

We have A HARD exam for university exams in iran which you should study at least A YEAR to pass, so you can generally Save a year.


You have 200 sticks. The first is one unit long, the second is two units, etc. You sort them into 50 piles, each pile does not necessarily have the same number of sticks.

Some guy that doesn't understand mathematics comes and says that none of the piles have three sticks that can be arranged into a triangle (in the Euclidean plane). Prove that this guy is wrong and that it is possible to find three sticks within one pile that can make a triangle.

[quote="abacadaea"]is this a trick question or what? i dont need pigeon hole we just have 2,3,4. or do we need to prove it for a random pile?[/quote]
That's not a proof;it's only one example.

What if 2,3, and 4 did not go into the same pile?

The idea is to prove there exists at least ONE pile such that a triangle can made from sticks only in that pile.

Since nobody has attempted this in a long time, I shall post a hint.

[hide="hint"]
To solve the problem, you only need to consider sticks of lengths 100 through 200.
[/hide]

There are 101 sticks of lengths from 100-200. By Pigeonhole, there must be 3 sticks in the same pile (as 101/50>2). Those three sticks from a triangle. The sense of gratification is lost when you read the hint before the actual question.


symmetry
superior algebra
superior algebra unsolved

If R and T are rings. g:R->T is the only homomorphism from R to T , and h:T->R is the only homomorphism from T to R. Why g is an isomorhism?

[i]g[/i] may or may not be an isomorphism. Are you sure you didn't mean $ h \circ g$?

[quote="coquitao"][i]g[/i] may or may not be an isomorphism.[/quote]
Could you give an example of (unitary) rings and theirs homomorphisms satisfying above conditon, such that $ g$ is not isomorphism?

[quote="coquitao"] Are you sure you didn't mean $ h \circ g$?[/quote]
I think if above conditions implies:
$ h\circ g$ - isomorphism
then by symmetry also
$ g\circ h$ - isomorphism
so $ g$ is injective and surjective homomorphism - hence $ g$ is isomorphism.
Am I missing something?

The only homomorphism from $ \mathbb{Z}/7\mathbb{Z}$ to $ \mathbb{Z}/3\mathbb{Z}$ is the zero map. The only homomorphism in the opposite direction is the zero map.

Usually the zero map is not considered a ring homomorphism, because it doesn't map unity to unity.

All right, here's a counterexample which works with unity, on the same principle. Let $ R$ be $ \mathbb{F}_2\times \mathbb{Z}$, with pointwise operations. Let $ T$ be $ \mathbb{Z}$. In mapping $ R$ to $ T$, we must map the identity $ (1,1)$ to $ 1$, and since we map $ 0\equal{}2(1,0)$ to zero, we must map $ (1,0)$ to zero. The only homomorphism $ g$ is thus $ g(x,y)\equal{}y$. In the opposite direction, mapping $ 1$ to the identity $ (1,1)$ determines everything, and the only homomorphism $ h$ has $ h(x)\equal{}(x,x)$.
Neither $ g$ nor $ h$ is an isomorphism, although the composition $ g\circ h$ is. I could break that as well; if I made $ T\equal{}\mathbb{F}_3\times \mathbb{Z}$, neither map ($ g(x,y)\equal{}(y,y),$ $ h(x,y)\equal{}(y,y)$) would be injective or surjective.


modular arithmetic
number theory proposed
number theory

Prove that there exists infinitely many pairs of primes $ (p,q)$ that satisfies $ 2^{p\minus{}1} \equiv 1 \pmod{q}$ and $ 2^{q\minus{}1} \equiv 1 \pmod{p}$.

[hide]I have solution but it is not natural at all. Since first, we have to think of the Fermat's number: $ F_n \equal{} 2^{2^n} \plus{}1$. I hope there will be more natural solutions.[/hide]

This is a problem from Mongolian Olympiad 2008. The (official) solution also makes use of Fermat numbers.

[hide]Let $ F_n\equal{}2^{2^n}\plus{}1$, for all $ n\geq 2$. Consider primes $ p$ and $ q$ such that $ p| F_n$ and $ q| F_{n\plus{}1}$. In the multiplicative group $ (\mathbb Z_p^*,\cdot\,)$ we have $ \widehat{2^{2^n}}\equal{}\minus{}\hat{1}$, so $ \widehat{2^{2^{n\plus{}1}}}\equal{}\hat{1}$, implying that the order of $ \hat{2}$ is $ 2^{n\plus{}1}$. As a consequence, \[ 2^{n\plus{}1}| p\minus{}1. \tag{1}\]

On the other hand, the prime $ p$ is of the form $ 8k\plus{}1$ and $ \left(\frac{2}{p}\right)\equal{}(\minus{}1)^{\frac{p^2\minus{}1}8}\equal{}1$. This implies $ 1\equal{}\left(\frac{2}{p}\right)\equal{}2^{\frac{p\minus{}1}2}$ (mod $ p$), so $ 2^{n\plus{}1}\mid\frac{p\minus{}1}2$, and consequently $ 2^{n\plus{}2}\mid p\minus{}1$.

As $ q| 2^{2^{n\plus{}1}}\plus{}1$, it follows that $ q| 2^{2^{n\plus{}2}}\minus{}1$. Combining this with $ 2^{n\plus{}2}| p\minus{}1$ we get $ q| 2^{p\minus{}1}$, as needed.

From $ (1)$ we have $ 2^{n\plus{}2}| q\minus{}1$. Using $ p| 2^{2^n}\plus{}1$ we obtain $ p| 2^{2^{n\plus{}1}}\minus{}1$ and finally $ p| 2^{q\minus{}1}$.

It remains to show that there exists infinitely many primes as requested. Notice that if a prime $ p$ divides both $ F_n$ and $ F_m$, the order of the element $ \hat{2}$ in the group $ (\mathbb Z_p^*,\cdot)$ equals $ 2^{n\plus{}1}$ and $ 2^{m\plus{}1}$ simultaneously, so $ n\equal{}m$. This shows that $ p\neq q$, and, moreover, there are infinitely many pairs of primes $ (p,q)$.[/hide]


Let $ p_n$ be the n-th prime.
$ S \equal{} \lim_{n \to \plus{} \infty}{\prod_{i \equal{} 2}^n{(1 \plus{} ((p_n \minus{} 1)^k \minus{} 1)^{ \minus{} 1})}}$, where $ k > 1$ is real and fixed.

Does S converge?


A TV screen measures 24 by 16 inches. If each dimension is increased by 20$ \%$, by what percent is the area increased?

the new dimensions are each 1.2 of the previous dimensions, so the area is scaled by a factor of $ 1.2^2\equal{}1.44$, which is a $ \boxed{44\% }$ increase


Why is it that some people have their names in green (like theone(#) and sunnyray) when they are logged on but do not have their name as a moderator next to any of the different sections of the forum?

I'd like to know the answer to this question too.

sunnyray and i actually co-moderate a class forum because sometimes middle school kids get way out of hand when posting online and we will be de-moderated sometime next month when the class ends and the class forum disappears


hello

can you please help me with this question.... thanx

Show that if G is a bipartite simple graph with v vertices and e edges then
e <= v^2 / 4

[hide]
ehh AM-GM it, e <= mn and v = m+n where m and n are the number of vertices on either side.

so by AM-GM squared v^2/4 >= mn >= e[/hide]

i don't understand your thread please explain !!!!

[quote="worthawholebean"]AM-GM is the Arithmetic Mean-Geometric Mean Inequality; for all positive $a$ and $b$, $\frac{a+b}{2}\ge\sqrt{ab}$.[/quote]
Or, more generally,
$\frac{1}{n}\sum^{n}_{k=1}a_{k}\geq \sqrt[n]{ \prod^{n}_{k=1}a_{k}}$
for any whole number n, where $a_{1}$, $a_{2}$, ... $a_{n}$ are all positive integers with equality iff $a_{1}=a_{2}=...=a_{n}$


Given a convex quadrilateral $ ABCD$. Let $ R_a$, $ R_b$, $ R_c$ and $ R_d$ be the circumradii of triangles $ DAB$, $ ABC$, $ BCD$, $ CDA$. Prove that inequality $ R_a < R_b < R_c < R_d$ is equivalent to \[ 180^{\circ}\minus{} \angle CDB < \angle CAB < \angle CDB\]

[i]Author: O.Musin[/i]

Assuming $ 180^o\minus{}\angle CDB<\angle CAB<\angle CDB (*)$ then $ \angle CDB>90^o$ and $ A$ lies out of $ (BCD)$
$ \Rightarrow \angle ADB>\angle ACB \Leftrightarrow R_a\equal{}\frac{AB}{sin \angle ADB}<\frac{AB}{sin \angle ACB}\equal{}R_b$
From $ (*)$ we get $ sin \angle BAC>sin \angle BDC \Leftrightarrow R_b<R_c$
Finally, $ \angle DBC>\angle DAC$ so $ R_c<R_d$
Conversely, we can prove similar.


How many of the 900 three-digit numbers have at least one even digit?

[hide]How many do not?

There are 5^3=125 numbers that have all odd digits. Thus 900-125=775 have at least one even digit. [/hide]


quadratics
number theory proposed
number theory

Let p be a prime with the property that :
For each prime q = 1 mod p, then there exist positive integers x,y such that

q = x^2+py^2.

Determine whether there exists precisely one such prime.

Bomb

$q = x^2 + py^2$ and $q \equiv 1 \mod p$.
So looking $\mod p$ we get $q \equiv x^2 \equiv 1 \mod p$. So $q$ is a quadratic residue of $p$.
For the law of quadratic reciprocity $\left(\frac{p - 1}{2}\right)\left(\frac{q - 1}{2}\right)$ must be even.
Now consider the expression $\mod 4$. $(x^2, y^2)$ can be:
I. $(0, 0)$, so $q \equiv 0 \mod 4$. Contradiction.
II. $(0, 1)$, so $q \equiv p \mod 4$
III. $(1, 0)$, so $q \equiv 1 \mod 4$
IV. $(1, 1)$, so $q \equiv 1 + p \mod 4$.
Now we look for $(p - 1)(q - 1) \equiv 0 \mod 4$.
II. $(p - 1)^2 \equiv 0 \mod 4$. So $p \equiv 2$ or $p \equiv 1$.
III. Always true.
IV. $p(p - 1) \equiv 0 \mod 4$.
So we can take $p \equiv 3 \mod 4$. Hence $p = 4k + 3$ and $p = 3, 7, 11, 19...$.
Take $p = 3$ and $q = 7$. So $7 = x^2 + 3y^2$ which is satisfied.
Now take $p = 7$ and $q = 29$. Then $29 = x^2 + 7y^2$ which is satisfied too.
So there exist more than one prime with that property.
Does it work?  :)


algebra
polynomial
trigonometry
LaTeX
function
real analysis
real analysis unsolved

Consider the trigonometric polynomial
\[ T_n(x) \equal{} a_0 \plus{} \displaystyle\sum_{i \equal{} 1}^n(a_icosix \plus{} b_isinix),\]
where $ a_i^2 \plus{} b_i^2 > 0, i\equal{}1,2,\ldots,n$. Show that the number of roots of the equation $ T_n(x) \equal{} 0$ in $ [0, 2\pi]$ is not more than $ 2n$.

Use Euler's formula. T_n(x) becomes a polynomial in z=exp(ix), of degree 2n. This will have 2n roots. But all may not have magnitude one (as required by z= exp(ix)). Thus, the number of roots is utmost 2n.

I don't see how it's possible to write $ T_n$ as a polynomial in $ e^{ix}$, even allowing for complex coefficients. Notice that $ T_n$ is a real-valued polynomial and that the coefficients of the cosine and sine are different. You could write $ T_n$ as $ \sum_{k \equal{} \minus{} n}^n c_k e^{ikx}$, but that is not exactly a polynomial.

$ \sum_{k \equal{} \minus{} n}^n c_k e^{ikx} \equal{} e^{ \minus{} inx}\sum_{k \equal{} 0}^{2n} c'_k e^{ikx}$.  The latter factor is a polynomial in $ e^{ix}$, and the former is never 0 (so contributes no roots).

bharathhm, the quality of your posts is constantly undermined by the relative difficulty of reading math not typeset in [[LaTeX]].  Please learn a little (see also [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690[/url]).

IOO, LaTeX has built-in commands for most common functions -- use \sin and \cos for $ \sin$ and $ \cos$, for example.


geometry
geometric transformation
reflection

what will an air bubble in water act like ??( type of lens or mirror , concave or convex or something like that)

which surface??? I believe it depends on curvature of mirror and refrative indices of air and water...

can someone tell the answer  along with soln please.....

Dude i've already posted this....see the "optics anyone" topic(maybe on the second page),it has snice optics questions,but sadly,no one replied :(

I think it would clearly act as a diverging lens,simply cuse of the differences in the refractive inices of air and water...

Where are the light rays coming from. Left????

Well, does the direction of light matter if the bubble is spherical?

Yeah Rushil,the bubble is spherical,so it doesnt matter ;)

[quote="shadysaysurspammed"]I think it would clearly act as a diverging lens...[/quote]Right. The bubble is shaped as converging lens but the refractive index of air is lower than that of water so it will be actually diverging lens. In the center of course, on the sides it loses focus rapidly. The focal point seems to be just outside of the bubble.

(n = 1.33)
(blue = reflected rays)

Whoops, there was a little bug in my computations so the blue rays are little off - it's not too bad though. Anyway, here's the focal point:

[quote="eminem"]yup you are right bus....

but what i got confused is that what if total internal reflection takes place?[/quote]Total internal reflection can only happen in the medium with greater refractive index which is in this case water. They just reflect from the surface of the bubble - the blue rays on my picture.

[quote="eminem"]so , i mean.... wont it create a problem , like it wont be a perfect diverging lens as some rays would be reflected also....?[/quote]That's right, it won't be perfect diverging lens. Not even real lens with spherical surface is perfect - it's reasonably good near the center but to be perfect, it needs to be parabolical (I think) and infinitely thin. That's why they make lenses thin too.

[quote="eminem"]what i mean to say is , wont it be a ......ummmmm... a different type of lens ?

an ordinary diverging lens has got its material denser than the outside . however this lens is rarer than the outside!! So isnt there bound to be some difference ? Or am i just acting stupid..... ok forget it...[/quote]

cmon,eminem,dont just give up u may have a point(maybe..)

Now u said that this lens is rarer than the outside...what if the bubble's floatin in air?Of course it aint a perfect divergin lens cause its not uniform throughout,u know.

i 'm saying when it IS in water! When it IS in water . when the bubble is in the water, what will it act like! so there's no question of the bubble being outside is there?

[quote="eminem"]what i mean to say is , wont it be a ......ummmmm... a different type of lens ?

an ordinary diverging lens has got its material denser than the outside . however this lens is rarer than the outside!! So isnt there bound to be some difference ?[/quote]Of course, it probably won't be identical to any ordinary glass lens. If you look through the bubble some distance from the center, you'd see only fuzzy images and who knows what and it won't be any kind of lens at all. Actually, if you looked through the "blue rays" part of it, it would act like a mirror! And on the other hand, in the center it looks just like ordinary lens. There's no special categorization for this kind of object I think - the best you can do is to say that it's diverging lens in the middle and mirror on the sides and then it acts weird in between. Look at the pictures of focal point and the traced rays to see for yourself...

yup thats what i meant to say bus  :coolspeak:  that ya got my thoughts!. that middle lens and from the edges mirror is precfisely the point i wanted to express

ne way guys the show's over . the question's over ,lets go on to morer worthy ones....


Hi ,everyone,i will very appreciate if you give me some advice.

from grade 1 to grade 2,the probability is 1/2;from grade  to grade 3,the probability is 1/2,
else it will down to grade 1.what is the equation from grade1 to grade 3?

[quote="wengewsh"]Hi ,everyone,i will very appreciate if you give me some advice.

from grade 1 to grade 2,the probability is 1/2;from grade 2  to grade 3,the probability is 1/2,
else it will down to grade 1.what is the expectation from grade1 to grade 3?[/quote]


trigonometry
inequalities
inequalities proposed

Triangle ABC is acute angled

Prove 3/2<cosA/(cosB+sinC) +cosB/(sinC+cosA) +sinC/(cosA+cosB) <2

To prove this inequality we use three lemmas
 1)  For x,y,z are three sides of a triangle we have
   x/(y+z) + y/(z+x) + z/(x+y)  < 2

 2)  For all positive x,y,z we have
    x/(y+z) + y/(z+x) + z/(x+y)  >= 3/2

 3)  For all acute triangle ABC:  cosA, cosB , sinC are three sides of a triangle.

 I think these lemmas complet the proof.

The equality is impossible because there is no triangle ABC such that A=B=90-C

Wow, I didn't think it would be so simple... I actually was irritated by the strange asymmetric terms and didn't even manage to write them down without confusing "sin" and "cos"...

Two notes about the lemmas you use:

[quote="tranminhhoang"]2)  For all positive x,y,z we have
    x/(y+z) + y/(z+x) + z/(x+y)  >= 3/2[/quote]

This is actually the so-called [b]Nesbitt inequality[/b].

[quote="tranminhhoang"]3)  For all acute triangle ABC:  cosA, cosB , sinC are three sides of a triangle.[/quote]

And this one was posted and proven in [url=http://mathlinks.ro/Forum/viewtopic.php?t=15019]another thread started by Michael Niland[/url]. Aha, now it's clear how he came up with the above asymmetric inequality!

  Darij


group theory
abstract algebra
least common multiple
superior algebra
superior algebra unsolved

Let G be an abelian group, k a fixed positive integer, and H={a an element of G|  |a| divides k}.  Prove that H is a subgroup of G.  Thanks.

$ \vert e \vert \equal{} 1$, therefore $ e \in H$.  :D

Now, if $ a,b \in H$, let u=ord(a) and v=ord(b).

If (u;v)=1 the result follows from  $ (ab)^{uv} = e$ and the fact that u|k, v|k, (u,v)=1 imply uv|k. [color=red][REMEMBER: $ (ab)^{uv} = e$ means that ord(ab)=uv or that ord(ab) is an integer that divides uv. It is a well-known property of the order...][/color]

Otherwise, let L be the Least Common Multiple of u and v. The result follows here from ${ (ab)^{L}} = e$ and the following fact,

2) u|k, v|k imply that $ L = \frac {uv}{(u;v)}$ | k. [color=red][Remember definition of L.C.M.][/color]

It remains to see what is going on with the inverses... If $ b \in H$ then v=ord(b) divides k. Since
$ b^{v} = e$ implies $ (b^{ - 1})^{v} = e$, we conclude that H is indeed a subgroup of G.


Express $ 0.6\overline{333}$ as a common fraction.


calculus
integration
algebra
function
domain
LaTeX
Ring Theory

(a) Let F(R) = all f: R -->R
For f,g belongs to F(R) define f+g: R -->R by x-->f(x)+g(x)
and f.g:R-->R by x-->f(x)g(x)
Show that F(R) is a commutative ring which is not an integral domain

(b) Define C(R), D(R) to be all f in F(R) which are respectively continuous and differentiable. Show that D(R)=<C(R)=<F(R)
Calculate F(R)* and U(F(R)) where F(R)* is a ring with non-zero elements in it. And U(F(R)) is unit of the ring

Note: I used R to denote for Real number not Ring R because I dont use Latex

I dont know how to do part (b) , can somebody help please?
I intend to use analysis result to prove that D(R)=<C(R)=<F(R) but I forgot how to do it.

Thank you fro any ideas and suggestions

To use latex put the cash symbol around sets of characters. So $ \left( \frac ab \right)$ is just  *cash*\left( \fracab \right) *cash*
You wouldn't need \left \right for the brackets normally but without them used how they were it'd come up as $ (\frac ab)$

a) That $ F(\mathbb{R})$ is a ring is trivial: here is how you demonstrate it: if $ f$ and $ g$ are real valued functions, so is $ f\plus{}g$ so $ f\plus{}g\in F(\mathbb{R})$. Same thing with products. Both operations commute and multiplication distributes. Now, to show it is not integral consider the function $ f(x)\equal{}1$ if $ x<0$ and  $ f(x) \equal{} 0$ everywhere else. Now consider the function $ g(x)\equal{}0$ if $ x<0$ and $ g(x)\equal{}1$ everywhere else. Then $ f(x)g(x)\equal{}0$ identically but neither are identically $ 0$. Thus we do not have an integral domain.

b) Now, a priori to being differentiable or continuous, your function $ f$ has to be a function (real valued in this case) so clearly $ D(\mathbb{R}) \subset F(\mathbb{R})$ and $ C(\mathbb{R}) \subset F(\mathbb{R})$. Now for a function to be differentiable at a point, it must be continuous at that point so $ D(\mathbb{R})\subset C(\mathbb{R})$. Now, an element in the ring $ F(R)$ is the zero element (additive identity) iff it is identically zero, so $ F(R)*$ is the set of all functions excluding $ 0$ (I'm not sure if this is what you're looking for). The unit of the ring is the multiplicative identity which is the function that is everywhere $ 1$.


Figure 7 below illustrates a typical set-up of the Young’s Double Slit experiment.

The distance between the double slits to the screen is 3.0 m.  The slit separation is 0.2 mm, and the wavelength of incident light is 633 nm.

(i)	Determine the angular displacement of the third order minima.

(ii)	Determine the fringe width of the interference pattern observed on the
	screen.
(iii)	Estimate the theoretical maximum number of bright fringes that can be observed.

(iv)	If the widths of the slits are assumed to be infinitely small and y is the distance from the central maxima on the screen, sketch a clearly labelled graph to show the variation of the intensity of the fringe pattern with y.  The intensity of the waves from each slit is I0.  Express the intensity in terms of I0.

(v)	The slit S1 is now covered with a filter such that the light emerging from S1 is reduced by half in amplitude.
Sketch on the same graph for part (iv), the new intensity pattern you would expect to observe.  Label your graphs clearly.

start with providing the forumulaes 1,2 are pretty direct

I may be blabbering here. But for (iii), isn't the width of the screen required?  :maybe:

We actually produce young's double slit and the precision slits are cut with a laser. Our engineering department has some great photos of the slits and optical signatures.

Contact me with any questions you have.


LaTeX
Functional Analysis
advanced fields
advanced fields unsolved

Let $H$ be a Hilbert space and let $T$ be a linear mapping from $H$ to $H$ such that $(Tx,y) = (x,Ty)$ for all $x,y \in H$

Prove that T is a bounded operator.

By the [url=http://mathworld.wolfram.com/ClosedGraphTheorem.html]Closed Graph Theorem[/url]., it suffices to prove that whenever $x_{n}\to x$ and $Tx_{n}\to y$, we have $y=Tx$.

Let $z$ be arbitrary. On the one hand, $(Tx_{n},z)\to(y,z)$, and, on the other hand, $(Tx_{n},z)=(x_{n},Tz)\to(x,Tz)=(Tx,z)$. Hence, for all $z$, we have $(y-Tx,z)=0$. Apply this to $z=y-Tx$ and you get $y=Tx$, as desired.

Edit:

Whenever I try to type \| in $\LaTeX$, in order to get to close vertical parallel lines (sort of like this ||), here's what happens: $\\mid $. It used to work before the server change.

Grobber beat me :).  This is sometimes known as the Hellinger-Toeplitz theorem.


How many sequences $ a_1,a_2,...,a{}_2{}_0{}_0{}_8$ are there such that each of the numbers $ 1,2,...,2008$ occurs once in the sequence, and $ i \in (a_1,a_2,...,a_i)$ for each $ i$ such that $ 2\le i \le2008$?

There are $ 2^{n\minus{}1}$ such permutations.

$ 2$ can be either $ a_1$ or $ a_2$. $ 3$ could have have been $ a_1$ or $ a_2$ or $ a_3$ but since $ 2$ took up one place we have two choices. We can see that for $ 2\leq m\leq n$ we have exactly two choices. One all the $ 2,\dots,n$ are fixed, it would automatically decide $ 1$'s position.


geometry
analytic geometry
geometry unsolved

Let ABC is a triangle, G is the intersection of the medians. (O) is the outcircle. Prove that [b][B,C,O,G is on a same circle][/b] $ \Longleftrightarrow AB^4 \plus{} AC^4 \equal{} BC^4 \plus{} AB^2 \cdot AC^2$

Sorry, (O) is the circumscribed circle of ABC.
No one solves my problem :( .It's really interesting.

We use barycentric coordinates with respect to $ \triangle ABC.$ The equation of the circle that goes through $ O \equiv (a^2S_A: b^2S_B: c^2S_C),$ $ B \equiv (0: 1: 0)$ and $C \equiv  (0: 0: 1)$ is

$ (O_a) \equiv a^2yz + b^2zx + c^2xy - \frac {b^2c^2}{2S_A}x(x + y + z) = 0$

$ G,O,B,C$ are concyclic $\Longleftrightarrow$ $G \equiv (1: 1: 1)$ satisfies the equation of $ (O_a).$ Then

$2S_A(a^2 + b^2 + c^2) - 3b^2c^2 = (b^2 + c^2 - a^2)(a^2 + b^2 + c^2) - 3b^2c^2 = 0$

$a^2b^2 + a^2c^2 - a^4 + b^4 + b^2c^2 - a^2b^2 + b^2c^2 + c^4 - a^2c^2 - 3b^2c^2 = 0$

$b^4 + c^4 = a^4 +b^2c^2.$


What are all real number solutions to the equation $x^{y}= y^{x}$?

Extension: Which solutions are integral and which are rational?

Trivial solutions are pairs $(t,t)$. Put $y=qx, q\neq 1$. Then

\begin{eqnarray*}x^{qx}=(qx)^{x}\iff x^{q}=qx\iff x^{q-1}=q\iff x=q^{1\over q-1}\iff y=qx=x^{q}=q^{q\over q-1}\end{eqnarray*}

Hence all non-trivial solutions are given by $(q^{1\over q-1},q^{q\over q-1}),q\neq 1$

Extensions are now easy to complete.

I beg all Mathlinkers: please stop posting this problem! It has been posted too many times.

http://www.qc.edu.hk/math/Advanced%20Level/y%5Ex_x%5Ey.htm

Really.

take the same base: $t=y\to x^{t}=x\cdot t$, so $x=t^{\frac{1}{t-1}}$


Let the Trianlge ABC with M is the mid point of BC.I is a point in ABC such that AI, BI, CI are the bisectors of angle A,B,C.
H,K are the two points on AB,AC such that $ IH \bot AB,IK \bot AC$. P is a point on HK such that $ MP \bot HK$
Prove that: PBC is an obtuse trangle.

Come on! It is not too hard. My friends :wink: .


Ring Theory
vector
superior algebra
superior algebra unsolved

Let $ (A,\plus{},\cdot)$ be a finite commutative ring and $ S\equal{}\left\{x \in A | x^2\plus{}1\equal{}0\right\}$. Prove that if $ S$ has more than 2 elements, then $ \sum_{a \in S}a\equal{}0$

Hey,what about $ \frac{\mathbb{Z}_2[x]}{(x^2\plus{}1)}$
Then $ S\equal{}\{1,x\}$

[quote="LydianRain"]above:

[quote="Aalexandru"]if $ S$ has more than 2 elements[/quote][/quote]
Sorry.I thought 2 is more than 2... :blush:

If the ring has characteristic $ >2$, you can pair numbers with their negatives (this works cause the solutions are units).
So we are left with rings of characteristic $ 2$. And in that case, we can restrict to the subring $ R$ generated by the solutions to $ x^2\minus{}1\equal{}x^2\plus{}1\equal{}0$. In fact, $ R$ is a $ \mathbb Z /(2)$ vector space generated by the solutions (easy to check), so let $ a_i$, $ i\equal{}1,...,k$, be a basis consisting of these solutions to $ x^2\equal{}1$.
Now all elements of our ring are of type $ s\equal{}\sum_{i \in I} a_i$ , so $ s^2 \equal{} |I|$. Thus the solutions of $ x^2\equal{}1$ are exactly those elements with an odd number of nonzero coordinates. Finally, the solutions are a affine subspace with $ 2^{k\minus{}1}$ elements, which are easy to seen to have sum $ 0$ if $ k>2$ (for $ k \leq 2$, we have at most $ 2$ such solutions).

Hi,nice solution.But I have a stupid question...
If the ring has characteristic 4,How did you  pair numbers with their negatives?(In case also have $ x \neq \minus{}x$?)
For example,$ \mathbb{Z}_4[x]/(4x^2\plus{}1)$,we have $ 2x\equal{}\minus{}2x$.

That's why we need the solutions to $ x^2 \plus{} 1 \equal{} 0$ are units: If $ x \equal{} \minus{} x$, then $ 2x \equal{} 0$; now multiply by $ x$ to get $ \minus{} 2 \equal{} 0$, giving $ 2 \equal{} 0$.
In more generality, if $ x$ is a unit, it isn't a zero divisor, thus $ ax \equal{} 0$ gives $ a \equal{} 0$.

In your example, you have $ 4x^2 \plus{} 1 \equal{} 1$ as $ 4 \equal{} 0$. Thus the ring you gave is the trivial ring $ 0$.

Thank you,you are right.I just use $ 2x\equal{}0$ to get $ 4x^2\equal{}\minus{}4\equal{}0$,so I consider char 4.I am sorry for my stupidness.Anyway,Nice solution.


linear algebra
matrix
inequalities
function
inequalities unsolved

prove that $ \frac {1}{a \plus{} 3b} \plus{} \frac {1}{b \plus{} 3c} \plus{} \frac {1}{c \plus{} 3a}\leqslant\frac {1}{a \plus{} 2b \plus{} c} \plus{} \frac {1}{b \plus{} 2a \plus{} c} \plus{} \frac {1}{a \plus{} 2c \plus{} b},\;a,b,c > 0$

The sign should be reversed. Let $ \mathbf{x}: \equal{} \begin{pmatrix}a \\
b \\
c\end{pmatrix}$, $ \mathbf{P}: \equal{} \begin{pmatrix}3 & 1 & 0 \\
0 & 3 & 1 \\
1 & 0 & 3\end{pmatrix}$, $ \mathbf{Q}: \equal{} \begin{pmatrix}2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2\end{pmatrix}$. The inequality can be written as
$ \sum_{i \equal{} 1}^{3}\frac {1}{\left(\mathbf{P}\mathbf{x}\right)_{i}}\geqslant \sum_{i \equal{} 1}^{3}\frac {1}{\left(\mathbf{Q}\mathbf{x}\right)_{i}}$.
Let $ \mathbf{y}: \equal{} \mathbf{P}\mathbf{x}$. Then $ \mathbf{Q}\mathbf{x} \equal{} \mathbf{Q}\mathbf{P}^{ \minus{} 1}\mathbf{y} \equal{} \mathbf{R}\mathbf{y}$, where $ \mathbf{R}: \equal{} \mathbf{Q}^{ \minus{} 1}\mathbf{P} \equal{} \frac {1}{7}\cdot\begin{pmatrix}4 & 1 & 2 \\
2 & 4 & 1 \\
1 & 2 & 4\end{pmatrix}$. The inequality is equivalent to
$ \sum_{i \equal{} 1}^{3}f\left(\left(\mathbf{y}\right)_{i}\right)\geqslant \sum_{i \equal{} 1}^{3}f\left(\left(\mathbf{R}\mathbf{y}\right)_{i}\right)$,
where $ f\left(t\right): \equal{} \frac {1}{t}$. The function $ f$ is convex and since the matrix $ \mathbf{R}$ is doubly stochastic, we have $ \mathbf{y}\succ\mathbf{R}\mathbf{y}$ and the inequality follows by Karamata.

Let $ x \equal{} a \plus{} 3b; y \equal{} b \plus{} 3c; z \equal{} c \plus{} 3a.$The inequality becomes:
$ \sum_{sym}^{} \frac {1}{x} \geqslant \sum_{cyc}^{} \frac {7}{4x \plus{} 2y \plus{} z}$
But this is true because:
$ \sum_{sym}^{} \frac {1}{x} \equal{} \frac{1}{7}\sum_{cyc}^{} (\frac {4}{x} \plus{} \frac {2}{y} \plus{} \frac {1}{z}) \geqslant \frac{1}{7}\sum_{cyc}^{} \frac {49}{4x \plus{} 2y \plus{} z} \equal{} \sum_{cyc}^{} \frac {7}{4x \plus{} 2y \plus{} z}$

It's deduced from the event: $ \frac{1}{1\plus{}3b}\plus{}\frac{1}{b\plus{}2c\plus{}a}\geq \frac{4}{(a\plus{}3b)\plus{}(b\plus{}2c\plus{}a)}\equal{}\frac{2}{a\plus{}2b\plus{}c}$


calculus
derivative
calculus computations

A curve y=x^2and the point R(-1,0).P(s,0) is a variable point on the x-axis ands is increasing at the rate of 5 units per second . Q is a point on the the curve such that PQ is perpendicular to the x-axis

Calculate the rate of increasing of the length of QR when s=2 units.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26438]Please do not post calculus problems in this forum![/url]

http://www.mathlinks.ro/Forum/viewtopic.php?t=56425

...

You do not need to post your problems more than once!


geometry
3D geometry
algebra
polynomial
linear algebra
matrix
complex numbers

prove that $ \sqrt[3]{2}\plus{}\sqrt[3]{3}\plus{}\sqrt[3]{5}$ is irrational

If you cube it.  You should get an integer, and some radicals.  If $ \sqrt[3]{2}\plus{}\sqrt[3]{3}\plus{}\sqrt[3]{5}$ was rational, so would it's cube, meaning those radicals have to be rational as well, but they're not.

Yes, but showing that the sum of those radicals is irrational isn't any easier than this problem.  Have you tried actually doing that?

One approach to this problem is to use the identity

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

and another is heavily computational:  one can explicitly compute the characteristic polynomial of the linear transformation $ x \mapsto (\sqrt[3]{2} \plus{} \sqrt[3]{3} \plus{} \sqrt[3]{5})x$ in $ \mathbb{Q}[ \sqrt{2}, \sqrt{3}, \sqrt{5} ]$ and show that the corresponding minimal polynomial has to have degree greater than $ 1$.  (We expect it to have degree $ 27$.)

In my mind, solving this particular problem is about as hard as proving much more general facts about algebraic integers, so one might as well become familiar with the general case and then this problem becomes completely trivial.

[b]Definition[/b]: a complex number $ z$ is an [b]algebraic integer[/b] if there are integers $ a_0, a_1, \ldots, a_{n-1}$ such that $ z^n + a_{n-1}z^{n-1} + \ldots + a_1 z + a_0 = 0$.

Two of the most fundamental facts about algebraic integers are the following.

[b]Fact 1[/b] If $ x$ and $ y$ are algebraic integers, then so is $ x+y$.

[b]Fact 2[/b] If $ x$ is an algebraic integer and $ x \in \mathbb{Q}$ then $ x \in \mathbb{Z}$.

Once we prove both facts, showing that $ u=\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{5}$ is irrational becomes easy. Obviously $ \sqrt[3]{m}$ is an algebraic integer for any integer $ m$ since it is the root of $ x^3-m=0$. Therefore $ u$, which is the sum of algebraic integers, is itself an algebraic integer, by Fact 1. However, if it had been rational, then by fact 2 it would be an integer, which clearly it is not (it's easy to estimate $ 4 < u < 5$).

[b] Definition[/b]: Let $ x_1, x_2, \ldots, x_n$ be complex numbers. We say that they [b]span[/b] a number $ x$ if $ x = a_1x_1 + a_2x_2 + \ldots + a_nx_n$ for some (ordinary) integers $ a_i$.

To prove Fact 1 above, we need an alternative characterization of algebraic integers.

[b]Lemma[/b]: Let $ z$ be a complex number. Suppose there are nonzero complex numbers $ x_1, \ldots, x_n$ such that whenever they span a number $ x$, they also span the number $ zx$. Then $ z$ is an algebraic integer.

[b]Proof[/b]: The numbers $ x_1, \ldots x_n$ certainly span $ x_i = 0 \cdot x_1 + \ldots + 1 \cdot x_i + \ldots + 0 \cdot x_n$. Therefore they also span $ zx_i$, which means that

$ z x_i =  a_{i,1}x_i + a_{i,2}x_2 + \ldots + a_{i,n}x_n$

for some integers $ a_{i,j}$. We can consider this as a system of linear equations in the variables $ x_i$, as follows:

$ (a_{1,1}-z)x_1 + a_{1,2}x_2 + \ldots + a_{1,n}x_n = 0$

$ a_{2,1}x_1 + (a_{2,2}-z)x_2 + \ldots + a_{2,n}x_n = 0$

$ \vdots$

$ a_{n,1}x_1 + a_{n,2}x_2 + \ldots + (a_{n,n}-z)x_n = 0$.

From linear algebra, we know that the determinant of the coefficient matrix has to be zero in order for this system to have a nonzero solution (which, by assumption, it does). If we write out that determinant explicitly, we obtain a polynomial in $ z$ with leading coefficient $ 1$ and ordinary integers as coefficients. Thus, $ z$ is an algebraic integer.  QED

[b]Proof of Fact 1[/b]: Suppose $ x$ and $ y$ are algebraic integers, so $ x^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 = 0$ and $ y^m + b_{m-1}y^{m-1} + \ldots + b_1y + b_0 = 0$. Consider the set of numbers $ u_{i,j} = x^i y^j$ for $ 0 \leq i < n$ and $ 0 \leq j < m$. If $ u$ is spanned by the $ u_{i,j}$, then so are $ xu$ and $ yu$. This is very easy to see: just replace every occurrence of $ x^n$ by $ -a_{n-1}x^{n-1} - \ldots - a_1x - a_0$. Therefore $ (x+y)u$ is also spanned by the $ u_{i,j}$: just span $ xu$ and $ yu$, and add. Thus $ x+y$ is an algebraic integer. QED

[b]Proof of Fact 2[/b]: This is quite trivial. Suppose $ x = \frac{a}{b}$ is an algebraic integer, where $ a$ and $ b$ are relatively prime integers. Then for some integers $ c_0, c_1, \ldots, c_{n-1}$ we have

$ (\frac{a}{b})^n + c_{n-1}(\frac{a}{b})^{n-1} + \ldots + c_1\frac{a}{b} + c_0 = 0$.

Multiplying by $ b^n$ this becomes

$ a^n + c_{n-1}a^{n-1}b + \ldots + c_1 a b^{n-1} + c_0b^n = 0$.

But everything on the LHS except $ a^n$ is visibly divisible by $ b$, therefore $ a^n$ is also divisible by $ b$, but since $ a$ and $ b$ are relatively prime we must have $ b = \pm 1$. QED


LaTeX
induction
algebra
polynomial
function
algebra open

It is well known fact that if \[x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1,\] then $xyz=0$. The same holds for the equations \[x+y+z=x^2+y^2+z^2=x^4+y^4+z^4=1\] and \[x + y + z = x^2 + y^2 + z^2 = x^5 + y^5 + z^5 = 1.\] But the problem has this obvious generalization:

Find all integers $n>5$, such that if $x,y$ and $z$ are complex such that \[x+y+z=x^2+y^2+z^2=x^n+y^n+z^n=1\] and $xyz$ is real, then $xyz = 0.$ My computations (made with computer) show that there is no answer for $n < 150$. But I can't prove that there aren't such numbers n at all. May be anyone could help?

[color=red]Moderator edit: I think this is an "open" question. [+ LaTeX][/color]

Hi,
Let S(n)=x^n+y^n+z^n, S(0)=3, S(1)=1, S(2)=1 and P=xyz.
We have (x+y+z)^2=S(2)+2(xy+yz+zx), hence xy+yz+zx=0.

Your problem is : if x,y,z are complex and S(1)=S(2)=S(n)=1 do we have P=0?

Since S(1)=S(2)=1, x,y,z are the roots of the polynom X^3-X^2-P.

Hence, for n>=3, we get S(n)=S(n-1)+PS(n-3).

Thus S(n)=1+nP, for n=3,4,5 and the answer is yes.

But, if n>=6, the answer is no, because, by induction we prove that
S(n)=Sum(a_k(n)P^k, k=0..[n/3]), with a_0(n)=1, a_1(n)=n (also, if n=3p then a_p(n)=3) and a_k(n) is, for 2<k<[n/3], in a positive integer.

So S(n)=1 <=> P*(n+Sum(a_k(n)P^(k-1), k=2..[n/3])=0.

Note that, this proves :
for all n>=3, if x,y,z are positive reals and S(1)=S(2)=S(n)=1 then we have xyz=0.

Friendly,
Georges

Hi, GeorgesZ!
Excuse me, but you did not understand my problem completely! In your terms, I am interested whether the polynomial S(n)-1 (for n>5) has unique real root p=0. The fact that p=0 is a root, is obvious. But is it the only one in R?
Nice try! Keep solving!

Hi Svet,
I don't understand what you say by :
"The fact that p=0 is a root, is obvious. But is it the only one in R?"

If you mean : 1) are some n such that S(1)=S(2)=S(n)=1 and P=xyz real implies some value for P<>0? (Response yes many)
or 2) are some n such that S(1)=S(2)=S(n)=1 and P=xyz real implies some value for P=0 multiple? (Response no)
or 3) are some n such that S(1)=S(2)=S(n)=1 and P=xyz real implies the only value for P in R+ is 0? (Response yes all of them).

You don't need to suppose P real.

My proof gives :
if x,y,z are complex such that S(1)=S(2)=S(n)=1,
then, if n=6, then P is 0 or -2,
if n=7, then P is 0 or -1
etc, since I proved that we have P*(n+a_2(n)P+...+a_([n/3])(n)P^([n/3]-1))=0.

Calculating the a_k(n) with Maple, I observe a curious fact :

If p is any prime integer, then all the coeficient  a_2(p), ..., a_([p/3])(p) (positive integers) are multiple of p (remember that a_1(p)=p , if p is prime or not).

This seems to me difficult to prove.

Friendly,
Georges

Hi Svet,
I understand now your question and it is a difficult one.

You could have said to me that you allready knew what I explained.

Anyway, we set X=xyz, and, for n>=3, T(n)=(S(n)-1)/X and get T(3)=3,T(4)=4,T(5)=5, and, for n>=6,
(*) T(n)=T(n-1)+1+X*T(n-3)

To answer : for n>=6, there are x,y,z complex with X=xyz real<>0 and
S(1)=S(2)=S(n)=1, we must prove that the polynom T(n) has a real negative root.

This is clear if degree(T(n)) is odd.

For a>0, we can try to prove A(n)=T(n)(-a)<0, this sequence is given by, A(3)=3, A(4)=4, A(5)=5, and, for n>=6, by :
(**) A(n)=A(n-1)+1-a*A(n-3).

Can we have an explicit expression for A(n) as a function of a?

Friendly,
Georges

Two counter examples:
the roots of t^3-t^2+2 verify x^6+y^6+z^6=1;
the roots of t^3-t^2+1 verify x^7+y^7+z^7=1.
          Friendly.
         Loup  blanc.


induction
linear algebra
matrix
trigonometry

Let $ A \equal{} \left( \begin{array}{cc} a & b \\
c & d \end{array} \right)$ with $ ad \minus{} bc \equal{} 1$ and $ a,\ b,\ c,\ d\in\mathbb{R}$ for a postive integer $ n$,

define $ x_n,\ y_n$ by $ \left( \begin{array}{c} x_1 \\
y_1 \end{array} \right) \equal{} \left( \begin{array}{c} 1 \\
0 \end{array} \right),\ \left( \begin{array}{c} x_{n \plus{} 1} \\
y_{n \plus{} 1} \end{array} \right) \equal{} A\left( \begin{array}{c} x_n \\
y_n \end{array} \right)$.

If $ x_2^2 \plus{} y_2^2 \equal{} x_3^2 \plus{} y_3^2 \equal{} 1$, then prove that $ x_n^2 \plus{} y_n^2 \equal{} 1$ for any $ n$.

Did you mean $ \ \left( \begin{array}{c} x_{n \plus{} 1} \\
y_{n \plus{} 1} \end{array} \right) \equal{} A\left( \begin{array}{c} x_n \\
y_n \end{array} \right)$?

[hide]We  have
      $ A^2\minus{}Tr(A)A\plus{}\det(A)I_2\equal{}0$
Then $ A^2\left( \begin{array}{c} x_{n} \\ y_{n} \end{array} \right)\minus{}Tr(A)A\left( \begin{array}{c} x_{n} \\ y_{n} \end{array} \right)\plus{}\left( \begin{array}{c} x_{n} \\ y_{n} \end{array} \right)\equal{}0$ (by $ \det(A)\equal{}1$)
Or : $ \left( \begin{array}{c} x_{n\plus{}2} \\ y_{n\plus{}2} \end{array} \right)\minus{}Tr(A)\left( \begin{array}{c} x_{n\plus{}1} \\ y_{n\plus{}1} \end{array} \right) \plus{}\left( \begin{array}{c} x_{n} \\ y_{n} \end{array} \right)\equal{}0$
 Implies : $ x_{n\plus{}2}\equal{}Tr(A)x_{n\plus{}1}\minus{}x_{n}$ ,$ y_{n\plus{}2}\equal{}Tr(A)y_{n\plus{}1}\minus{}y_{n}$
 Easy see that : $ \left( \begin{array}{c} x_{2} \\ y_{2} \end{array} \right)\equal{}\left( \begin{array}{c} a  \\ c  \end{array} \right)$,$ \left( \begin{array}{c} x_{1} \\ y_{1} \end{array} \right)\equal{}\left( \begin{array}{c} 1  \\ 0  \end{array} \right)$
 $ x_3\equal{}Tr(A)a\minus{}1,y_3\equal{}Tr(A)c$ , $ a^2\plus{}c^2\equal{}1$ and $ x^{2}_{3}\plus{}y^{2}_{3}\equal{}Tr^{2}(A)(a^2\plus{}c^2)\minus{}2Tr(A)a\plus{}1\equal{}1$ then $ Tr(A)\equal{}0$ or $ Tr(A)\equal{}2a$
a) If $ Tr(A)\equal{}0$ True ! by $ x_{n\plus{}2}\equal{}\minus{}x_n,y_{n\plus{}2}\equal{}\minus{}y_n$ easy see that $ x_n^2 \plus{} y_n^2 \equal{} 1$
b) If $ Tr(A)\equal{}2a$ or $ \frac{Tr(A)}{2}\equal{}a$
  Use induction
       With n=1,2,3 ir correct
  Suppose correct to $ n \geq 3$
 We have
$ x_{n\plus{}1}^2 \plus{} y_{n\plus{}1}^2 \equal{}(Tr(A)x_n\minus{}x_{n\minus{}1})^2 \plus{}(Tr(A)y_n\minus{}y_{n\minus{}1})^2\equal{}Tr^2(A)(x_n^2\plus{}y_n^2)\plus{}(x_{n\minus{}1}^2\plus{}y_{n\minus{}1}^2)\minus{}2Tr(A)(x_nx_{n\minus{}1}\plus{}y_ny_{n\minus{}1})\equal{}Tr^2(A)\plus{}1\minus{}2Tr(A)(x_nx_{n\minus{}1}\plus{}y_ny_{n\minus{}1})$ [b](1)[/b]
 $ x_nx_{n\minus{}1}\plus{}y_ny_{n\minus{}1}\equal{}Tr(A)(x_{n\minus{}1}^2\plus{}y_{n\minus{}1}^2)\minus{}(x_{n\minus{}1}x_{n\minus{}2}\plus{}y_{n\minus{}1}y_{n\minus{}2})\equal{}Tr(A)\minus{}(x_{n\minus{}1}x_{n\minus{}2}\plus{}y_{n\minus{}1}y_{n\minus{}2})$ and ($ x_2x_1\plus{}y_2y_1\equal{}a$)
  $ x_nx_{n\minus{}1}\plus{}y_ny_{n\minus{}1}\equal{}\frac{Tr(A)}{2}\plus{}(\minus{}1)^{n\minus{}2}(x_2x_1\plus{}y_2y_1\minus{}\frac{Tr(A)}{2})\equal{}\frac{Tr(A)}{2}\plus{}(\minus{}1)^{n\minus{}2}(a\minus{}\frac{Tr(A)}{2})\equal{}\frac{Tr(A)}{2}$
 and from ([b]1[/b]) we have $ x_{n\plus{}1}^2\plus{}y_{n\plus{}1}^2\equal{}Tr^2(A)\plus{}1\minus{}2Tr(A)\frac{Tr(A)}{2}\equal{}1$[/hide]
 [b] Sorry by english of me so bad[/b] ! :blush:

Yeh, we can understand your English. :lol:

I will check your solution later.

This problem is difficult for Japanese university entrance exam, I think.

How do you think about it? :)

ThankS !
[quote]This problem is difficult for Japanese university entrance exam, I think.
[/quote]
  :read:  [b]Yes ,I think ,so. [/b]

$ x_{n\plus{}1}\equal{}ax_n\plus{}by_n$ and $ y_{n\plus{}1}\equal{}cx_n\plus{}dy_n$
Observe that
$ A\equal{}cx_{n\plus{}1}\minus{}ay_{n\plus{}1}\equal{}\minus{}y_n$
and
$ B\equal{}ax_{n\plus{}1}\plus{}cy_{n\plus{}1}\equal{}x_n\plus{}(ab\plus{}cd)y_n$
So $ A^2\plus{}B^2\equal{} x_{n\plus{}1}^2\plus{}y_{n\plus{}1}^2\equal{}x_n^2\plus{}y_n^2\plus{}(ab\plus{}cd)y_n(2x_n\plus{}(ab\plus{}cd)y_n)$---(*)

Taking $ n\equal{}2$ we get
$ (ab\plus{}cd)c(2a\plus{}(ab\plus{}cd)c)\equal{}0$
[b]Case 1:[/b] If $ ab\plus{}cd\equal{}0$ then we are done.
[b]Case 2:[/b] If $ c\equal{}0$ then we get $ a\equal{}\pm1$. Hence  $ x_n\equal{}\pm1$ and $ y_n\equal{}0$ for all $ n$.
[b]Case 3:[/b] If $ ab\plus{}cd$ and $ c$ both are non-zero. Then
$ 2a\plus{}(ab\plus{}cd)c\equal{}0 \Longrightarrow ab\plus{}cd\equal{}\frac{\minus{}2a}{c}$

$ (*) \Longrightarrow x_{n\plus{}1}^2\plus{}y_{n\plus{}1}^2\equal{}x_n^2\plus{}y_n^2\plus{}\frac{2(ab\plus{}cd)}{c}y_n(cx_n\minus{}ay_n)$
$ \Longrightarrow x_{n\plus{}1}^2\plus{}y_{n\plus{}1}^2\equal{}x_n^2\plus{}y_n^2\minus{}\frac{2(ab\plus{}cd)}{c}y_ny_{n\minus{}1}$
Now easy calculation shows that $ a\equal{}\minus{}d \Longrightarrow A^2\equal{}\minus{}I_2$
[hide]$ 2a\plus{}(ab\plus{}cd)c\equal{}0 \Longrightarrow 2a\plus{}abc\plus{}c^2d\equal{}0\Longrightarrow 2a\plus{}a(ad\minus{}1)\plus{}c^2d\equal{}0$
$ \Longrightarrow a\plus{}d(a^2\plus{}c^2)\equal{}0 \Longrightarrow a\plus{}d\equal{}0$
Moreover $ ad\minus{}bc\equal{}1 \Longrightarrow a^2\plus{}bc\plus{}1\equal{}0$
Hence
 $ A^2 \equal{} \left( \begin{array}{cc} a & b \\
c & d \end{array} \right) \left( \begin{array}{cc} a & b \\
c & d \end{array} \right)\equal{} \left( \begin{array}{cc} a & b \\
c & \minus{}a \end{array} \right)\left( \begin{array}{cc} a & b \\
c & \minus{}a\end{array} \right)$
$ A^2\equal{}\left( \begin{array}{cc} a^2\plus{}bc & 0 \\
0 & a^2\plus{}bc \end{array} \right)\equal{}\left( \begin{array}{cc} \minus{}1 & 0 \\
0 & \minus{}1 \end{array} \right)\equal{}\minus{}I_2$
[/hide]
Induction on $ k$ gives $ y_{2k\minus{}1}\equal{}0$
So $ y_ny_{n\minus{}1}\equal{}0 \Longrightarrow x_{n\plus{}1}^2\plus{}y_{n\plus{}1}^2\equal{}x_n^2\plus{}y_n^2\equal{}1$

Let $ u_0\equal{}[1,0]^T,u_n\equal{}A^nu_0$. We know that $ A^2\equal{}tr(A)A\minus{}I,\parallel{}u_0\parallel{}\equal{}\parallel{}u_1\parallel{}\equal{}\parallel{}u_2\parallel{}\equal{}1$ where $ \parallel{}.\parallel{}$ is the euclidean norm. $ \parallel{}u_2\parallel{}^2\equal{}(tr(A))^2\plus{}1\minus{}2tr(A)<u_1,u_0>\equal{}1$;$ <u_1,u_0>\equal{}a$ gives $ tr(A)\equal{}0$ or $ tr(A)\equal{}2a$.
case1: $ tr(A)\equal{}0$  then $ A^2\equal{}\minus{}I$ and it is easy.
case 2: $ tr(A)\equal{}2a$ and $ a\equal{}d$. $ A\equal{}\begin{pmatrix}cos(\theta)&b\\sin(\theta)&cos(\theta)\end{pmatrix}$ with $ cos(\theta)^2\minus{}bsin(\theta)\equal{}1$.
1) The case $ sin(\theta)\equal{}0$ gives $ A\equal{}\begin{pmatrix}\epsilon&b\\0&\epsilon\end{pmatrix}$ and after it is easy (here $ \epsilon\equal{}\pm{1}$).
2) The case $ sin(\theta)\not\equal{}0$ implies that $ A$ is  a rotation.