[hide="hint"]
Any sequence of A's and B's of length equal to $ n$ or any sequence of A's and B's of length less than $ n$ and containing at least one $ B$ uniquely determines a sequence of the desired form.[/hide]

I hope this is correct...
[hide]
Let $ A_{n}$, $ B_{n}$, and $ C_{n}$ be the number of sequence of length n that end in $ A$, $ B$, and $ C$, respctively. Therefore:
$ A_{n\plus{}1}\equal{}A_{n}\plus{}B_{n}\plus{}C_{n}$
$ B_{n\plus{}1}\equal{}A_{n}\plus{}B_{n}\plus{}C_{n}$
$ C_{n\plus{}1}\equal{}B_{n}\plus{}C_{n}$

$ A_{n\plus{}1}\plus{}B_{n\plus{}1}\plus{}C_{n\plus{}1}\equal{}3(A_{n}\plus{}B_{n}\plus{}C_{n})\minus{}A_{n}$
$ S_{n\plus{}1}\equal{}3S_{n}\minus{}A_{n}$
$ S_{n\plus{}1}\equal{}3S_{n}\minus{}S_{n\minus{}1}$

(I tried to find the explicit formula for this recursion, but it seems too ugly to type out)

So making a table listing out from $ S_{1}$ to $ S_{10}$, one can see that
$ \boxed{S_{10}\equal{}10,946}$
[/hide]

[quote="MellowMelon"]I don't believe your recursion specifies that the $ C$s all occur in one block. :maybe:

[hide="Solution with K81o7s hint"]
The number of sequences of $ A$s and $ B$s with length $ n$ is $ 2^{n}$. The number of sequence of $ A$s and $ B$s with length $ n$ and at least one $ B$ is $ 2^{n}\minus{}1$.

Then the number of sequences of length $ n$ containing $ A,B,C$ that satisfy the conditions is:
$ 2^{n}\plus{}\sum_{i \equal{} 1}^{n\minus{}1}{2^{i}\minus{}1}$
$ 2^{n}\plus{}\sum_{i \equal{} 0}^{n\minus{}2}{2^{i\plus{}1}}\minus{}(n\minus{}1)$
$ 2^{n}\plus{}2\cdot (2^{n\minus{}1}\minus{}1)\minus{}(n\minus{}1)$
$ 2^{n}\plus{}2^{n}\minus{}2\minus{}n\plus{}1$
$ 2^{n\plus{}1}\minus{}n\minus{}1$
This gets the answer $ 2037$ for $ n \equal{} 10$.
[/hide][/quote]
You can't put an apostrophe in a hide tag.

[quote="MellowMelon"]I don't believe your recursion specifies that the $ C$s all occur in one block. :maybe:

[hide="Solution with K81o7s hint"]
The number of sequences of $ A$s and $ B$s with length $ n$ is $ 2^{n}$. The number of sequence of $ A$s and $ B$s with length $ n$ and at least one $ B$ is $ 2^{n} \minus{} 1$.

Then the number of sequences of length $ n$ containing $ A,B,C$ that satisfy the conditions is:
$ 2^{n} \plus{} \sum_{i \equal{} 1}^{n \minus{} 1}{2^{i} \minus{} 1}$
$ 2^{n} \plus{} \sum_{i \equal{} 0}^{n \minus{} 2}{2^{i \plus{} 1}} \minus{} (n \minus{} 1)$
$ 2^{n} \plus{} 2\cdot (2^{n \minus{} 1} \minus{} 1) \minus{} (n \minus{} 1)$
$ 2^{n} \plus{} 2^{n} \minus{} 2 \minus{} n \plus{} 1$
$ 2^{n \plus{} 1} \minus{} n \minus{} 1$
This gets the answer $ 2037$ for $ n \equal{} 10$.
[/hide][/quote]

I don't understand why the number of sequence of $ A$s and $ B$s with length $ n$ and at least one $ B$ is $ 2^{n} \minus{} 1$. I don't get what you mean by at least one $ B$. You said that the sequence has both A's and B's.

These 4 lines are confusing:
$ 2^{n} \plus{} \sum_{i \equal{} 1}^{n \minus{} 1}{2^{i} \minus{} 1}$
$ 2^{n} \plus{} \sum_{i \equal{} 0}^{n \minus{} 2}{2^{i \plus{} 1}} \minus{} (n \minus{} 1)$
$ 2^{n} \plus{} 2\cdot (2^{n \minus{} 1} \minus{} 1) \minus{} (n \minus{} 1)$
$ 2^{n} \plus{} 2^{n} \minus{} 2 \minus{} n \plus{} 1$
$ 2^{n \plus{} 1} \minus{} n \minus{} 1$

someone please explain

[quote="bigman"]I don't understand why the number of sequence of $ A$s and $ B$s with length $ n$ and at least one $ B$ is $ 2^{n} \minus{} 1$. I don't get what you mean by at least one $ B$. You said that the sequence has both A's and B's.[/quote]
By a sequence containing $ A$s and $ B$s I mean a sequence all of whose elements are either $ A$ or $ B$, and nothing else. Therefore $ AAA$, $ BBB$, $ ABA$, are all valid, because all of their elements are $ A$ or $ B$.

When I say at least one $ B$, I mean that a sequence containing only $ A$s is not allowed. To see why these are equivalent, not containing one $ B$ means containing no $ B$s, which means all $ A$s. If the length of the sequence is specified, the only sequence that gets removed from the set in the previous paragraph is $ AA \ldots A$. So we subtract $ 1$ from the size of the set in the previous paragraph ($ 2^n$).

[quote="bigman"]These 4 lines are confusing:
$ 2^{n} \plus{} \sum_{i \equal{} 1}^{n \minus{} 1}{2^{i} \minus{} 1}$
$ 2^{n} \plus{} \sum_{i \equal{} 0}^{n \minus{} 2}{2^{i \plus{} 1}} \minus{} (n \minus{} 1)$
$ 2^{n} \plus{} 2\cdot (2^{n \minus{} 1} \minus{} 1) \minus{} (n \minus{} 1)$
$ 2^{n} \plus{} 2^{n} \minus{} 2 \minus{} n \plus{} 1$
$ 2^{n \plus{} 1} \minus{} n \minus{} 1$[/quote]
They are ambiguous; you'll have to forgive me since that post is a year old. Here is a hopefully better exposition.

$ 2^{n} \plus{} \sum_{i \equal{} 1}^{n \minus{} 1}{\left( 2^{i} \minus{} 1 \right) }$
Summations are linear, so separate the two terms inside.

$ 2^{n} \plus{} \sum_{i \equal{} 1}^{n \minus{} 1}{2^{i}} \minus{} \sum_{i \equal{} 1}^{n \minus{} 1}{1}$
Shift the index by $ 1$ in the first summation, and evaluate the second.

$ 2^{n} \plus{} \sum_{i \equal{} 0}^{n \minus{} 2}{2^{i \plus{} 1}} \minus{} (n \minus{} 1)$
You can factor out a $ 2$ ($ 2^{i\plus{}1} \equal{} 2 \cdot 2^i$), then use the geometric series formula.

$ 2^{n} \plus{} 2 \cdot (2^{n\minus{}1} \minus{} 1) \minus{} (n \minus{} 1)$
And now it's a bunch of arithmetic to get to the answer.

[quote="mdk"]Let S(n) denote the number of sequences of length n formed by the three
letters A,B,C with the restriction that the C’s (if any) all occur in a single
block immediately following the first B (if any). For example ABCCAA,
AAABAA, and ABCCCC are counted in, but ACACCB and CAAAAA are
not. Derive a simple formula for S(n) and use it to calculate S(10).[/quote]
To make sure that I interpret it correctly, can you answer me whether this following case is acceptable: ABCCCBCC?Sorry if it sounds stupid but I don't want to waste  my time doing it in the wrong interpretation.

why the n-1 at the top of the summation. i don't see how we have a length of n-1...

someone? anyone? this is a bit confusing to me

comeon guys, this should be a peace of cake

[quote="K81o7"]
Any sequence of A's and B's of length equal to $ n$ or any sequence of A's and B's of length less than $ n$ and containing at least one $ B$ uniquely determines a sequence of the desired form.[/quote]

There are $ 2^n$ strings that don't contain a 'C'. Now, we consider the strings that do have a 'C'; then, deleting the 'C's, we have a set of shorter strings of 'A's and 'B's, but each string of 'A's and 'B's also uniquely determines the string including the 'C's. Of course, after deleting the 'C's, the maximal length of the string is $ n \minus{} 1$.


from Leibnitz that point is the projection of the centroid on $ \triangle ABC$ on $ r$.


probability
modular arithmetic
LaTeX
function

King Arthur and n knights sit around the round table and play a game.

King Arthur starts with a dice, and they start taking turns rolling it.
If, the dice is a:
1 or 2, the roller wins,
3 or 4, the dice is passed to the right,
5 or 6, the dice is passed to the left.

What is probability that King Arthur wins?

how many knights? :?:
if its just king arthur, of course he'll win.

The idea is to create an expression involving N so that you have a probability for all numbers of knights.

Fix $n$, and number the players modulo $n+1$ with King Arthur numbered $0 = n+1 =-n-1 = \ldots$.  Note that the probability of a player winning at a given point in the game depends only on who happens to hold the die, not on the history of the game up until this point.  So, let $P(i)$ be the probability that King Arthur wins, given that the die is currently held by player $i$.  We seek to calculate $P(0)$ (since at the beginning of the game, Arthur holds the die).  Note that $P(0) = \frac{1}{3}(1+P(1)+P(-1))$ and that for $i \not\equiv 0 \pmod{n+1}$ we have $P(i) = \frac{1}{3}(P(i+1)+P(i-1))$.  This gives you a system of $n+1$ linear equations in $n+1$ variables, which you can in principle solve.

It's not clear to me what bloodysloth is suggesting the answer is (math looks better in LaTeX), but doesn't appear to be correct for even $n = 0$.  The next few values are 5/11, 9/20 and 13/29.  I don't immediately see a pattern here.

well... i figured it out.  method somewhat similar to the above poster's suggestion.  just a different setup.

[hide]
W(n) = root(5)/5 + 2 * root(5)/5 * [1.5-root(1.25)]^n / {1 - [1.5-root(1.25)]^n}
[/hide]

Does anyone know why, ∑ from 0 to ∞ of [ (2n)!/(n!n!) * (1/3)^(2n+1)] = 1/root(5)?

This was used in part of my solution.

Yeah, I don't know how to use Latex or whatever... someone show me...

Let's turn this thread into a new problem:

Prove that
∑ from 0 to ∞ of [ (2n)!/(n!n!) * (1/3)^(2n+1)] = 1/root(5)

I have no clue if this is possible

Use $\sum_{n=0}^\infty \binom{2n}{n}x^{n}= \frac{1}{\sqrt{1-4x}}$.  (This is a wellish-know generating function/taylor series result.  When/if you learn about generating functions and Tayler series, you should learn about this, hopefully.)  There is a LaTeX forum here, and there is a LaTeX guide (in the AoPS skin, click the big link on the left that says "LaTeX.")  You don't need any of the part about downloading stuff in order to use LaTeX on the forums (only if you want to create your own documents), just learn the codes and then slap them between two dollar signs.


The ratio of teachers to students in a particular school is 1 to 11. The ratio of female students to the total number of students is 4 to 9. If there are 396 female students, how many teachers are there?

First, we set up a proportion for the students using the number of femal students.

$ \frac49\equal{}\frac{396}{x}\implies x\equal{}891$.

Now, do so for the teachers.

$ \frac{1}{11}\equal{}\frac{y}{891} \implies y\equal{}\boxed{81}$.

[quote="DanielParkJMC"]im practicing CD rounds for MC[/quote]

Ok wrong page.

This place might be better.

[url]http://www.artofproblemsolving.com/Forum/index.php?f=132[/url]


Who attended the competitions?

I attended the Physics Challenge (2nd) and the Physics Bowl (1st). I didn't see any teams from Howard County, which I found a little surprising.

I was there.  I only heard about it yesterday though, so I couldn't scrape together a full team from Blair.


Find the largest even number which cannot be expressed as the sum of two odd composite numbers.

Davron

Using Cheby's Chef Theorem (that one on primes, not that from the ugliest mathematics ever) with the upper bound $\pi(n) \cdot \log(n) \leq \log(4) \cdot n$ gives that for $n>64$ there are less than $\frac{n}{4}$ primes $\leq n$.
So assume that a even number $2n$ is not the sum of two composite odd numbers. Then we can pair up $k,2n-k$ for all odd $k$, and at least one has to be prime (or $1$) then. There will be $n$ pairs, thus at least $\frac{n}{2}$ primes below $2n$ (note that $1$ is not prime but is substituted by $2$ in the counting). But by the above this means $2n \leq 64$.
Thus just $32$ cases to check  ;)

Without this theorem, you could use that $\frac{\varphi(210)}{210} \leq \frac{1}{4}$ and use a similar reasoning, but with higher bound.


superior algebra
superior algebra unsolved

[quote="wiosna"]Prove:
 The sum of all nil left ideals in any ring R coicides with the sum of all nil right ideals.[/quote]

Assuming $ R$ is unitary: It's clear from $ (ab)^n\equal{}a(ba)^{n\minus{}1}b$ that, for any $ x \in R,$ the left ideal $ Rx$ is nil iff the right ideal $ xR$ is nil. let $ I$ be any nil left ideal of $ R.$ Then $ \sum_{x \in I} Rx \equal{}I \subseteq \sum_{x \in I} xR.$


There will be a
[size=200]HUGE PARTY[/size]
happening on Saturday, September 23rd, 2006.

This party lasts pretty much all day and should be quite a blast.  I encourage all of you to come out.  There will be beats, bands, food, fire-spinning, console gaming, juggling, and more.  Admission is free for everyone.

Once the sun goes down, it transforms into a full-fledged dance party, and I will be one of your DJs there.  If you guys thought the parties during frosh week were fun, wait until you see this!

For more info, see [url]http://b2green.undergrad.ca/[/url].  There is some sample music up there that you can download, although last time I checked, some of the links were dead.

I think your poll needs revamping.  Consider adding the option of: [i]I don't go to Waterloo :( BUT I'M COMING ANYWAYYYYYY!!!!!!!![/i]

I think that counts as YES!!!!!!

But... all the waterloo guys are too busy with their orientation to answer right now... lol.

Can't make it! Big event in Toronto that weekend, can't miss it. To bad, I wanted to meet a few of you irl.


me and my friend in prague (on the boat)

[quote="funcia"]although i have some bad eye sight from reading in the dark i'd say you all look related :)[/quote]

 :lol:

the boy behind me (looks like he is thinking about something) is my brother. the boy in front of me is a friend. the boy behind my stearing brother is ... i don't know who is he.


geometry
circumcircle
power of a point
radical axis
geometry unsolved

[color=darkblue] [b]Jayme[/b], Here is again a similar nice and easy problem with a right triangle. I hope you like it. [/color]

[color=darkred]Let $ ABC$ be a triangle with $ A$ - excircle $ w_a \equal{} C(I_a,r_a)$ and $ B$ - excircle $ w_b \equal{} C(I_b,r_b)$ . Denote

the midpoint $ M$ of $ [AC]$ and  the tangent points $ X$ , $ Y$ of $ w_a$ , $ w_b$ with the sideline $ AC$ . The circumcircle

of $ \triangle BMX$ cuts again the circumcircle of $ \triangle ABC$ in the point $ Z$ . Prove that $ B\in YZ$ $ \Longleftrightarrow$ $ A \equal{} 90^{\circ}$ .[/color]

[hide="Solution"]
The problem is equivalent to showing that $ Y$ lies on the radical axis between the circumcircles of $ \triangle BMX$ and $ \triangle BAC$ if and only if $ \angle BAC\equal{}90$, which is true if and only if $ AY\cdot YC\equal{}MY\cdot YX$ if and only if $ \angle BAC\equal{}90$.  Now, letting $ AB\equal{}c$, $ BC\equal{}a$, $ AC\equal{}b$, and the semiperimeter be $ s$, we have that $ s\minus{}c\equal{}AY$, $ YC\equal{}s\minus{}a$, $ MY\equal{}AY\minus{}AM\equal{}s\minus{}c\minus{}\frac{b}{2}\equal{}\frac{a\minus{}c}{2}$, and $ YX\equal{}YC\plus{}CX\equal{}s\minus{}a\plus{}s\minus{}b\equal{}c$.  Thus it suffices to show that \[ 2(s\minus{}c)(s\minus{}a)\equal{}(a\minus{}c)c\iff \angle BAC\equal{}90\] \[ (a\plus{}b\minus{}c)(b\plus{}c\minus{}a)\equal{}2ac\minus{}2c^2\iff \angle BAC\equal{}90\] \[ b^2\minus{}(c\minus{}a)^2\equal{}2ac\minus{}2c^2\iff \angle BAC\equal{}90\] \[ b^2\minus{}c^2\minus{}a^2\plus{}2ac\equal{}2ac\minus{}2c^2\iff \angle BAC\equal{}90\] \[ b^2\plus{}c^2\equal{}a^2\iff \angle BAC\equal{}90\] which is true, so we are done.  [/hide]

Indeed, it is easily. I remark only that the division $ (A,Y,C,X)$ is harmonically

iff $ A \equal{} 90^{\circ}$ because $ (A,Y,C,X)$ is harmonically $ \Longleftrightarrow$ $ \frac {YA}{YC} \equal{} \frac {XA}{XC}$ $ \Longleftrightarrow$

$ \frac {p \minus{} c}{p \minus{} a} \equal{} \frac {p}{p \minus{} b}$ $ \Longleftrightarrow$ $ A \equal{} 90^{\circ}$ $ \Longleftrightarrow$ $ YM\cdot YZ \equal{} YA\cdot YC$ a.s.o.

The initial conclusion in my problem was " If $ D\in BC$ so that $ AD\perp BC$ , then $ \widehat {CDY}\equiv\widehat {CDX}$ " .


inequalities
number theory unsolved
number theory

Find all pairs of nonnegative integers ($x$, $y$) such that \[x^{3}+8x^{2}-6x+8 = y^{3}\]

I got that [tex](0,2)[/tex] is the only solution but I'm afraid of posting my vodoo-mathematics proof :)

Well. Does that prove that voodo is wrong? :)
(9,11)

Anyways, the proof is more important. Anyone?

well, LHS = (x+2)^3 + 2x(x-9). So when x>9, y>=x+3, but that's too big (easily verifiable). So only solutions are x=0 and x=9.  :D

Could you add more detail? (because I still do not understand)

How is $y \geq x+3$? What does that prove?

oops, i had a mistake (+ sign instead of -) in my previous post, now it should be okay... :D

Because of x,y are positive integers it is enough to prove that x+1<y<x+3 to make a conclusion y=x+2
      First we will prove x+1<y (1)
       we get after cubing x^3+3x^2+3x+1<x^3+8x^2-6x+8 which is equal to
          5x^2-9x+7>0 , 5x^2-5x-4x+4+3>0
          5x*(x-1)-4*(x-1)+3>0 , (x-1)*(5x-4)+3>0 (*)
          last inequality is correct for nonegative integers
     On second hand we will prove y<x+3
         after cubing x^3+8x^2-6x+8<x^3+9x^2+27x+27
                   and we get x^2+33x+19>0 which is corecct so we make a conlusion y=x+2
       x^3+8x^2-6x+8=(x+2)^3
       x^3+8x^2-6x+8=x^3+6x^2+12x+8
          2x^2-18x=0
          2x*(x-9)=0
         only solutions x=0 y=2 and x=9 y=11


Given rectangle ABCD with a random point X, with

    AX = j
    BX = k
    CX = n
    DX = m

    What is the value of j?

    * m
    * k^2 + n^2 - m^2
    * k^2 - m^2
    * sqrt (k - n)
    * none of the above


I need help in finding the equations of two parabola's. One has a turning point at (60,10). The second crosses through the origin (0,0). They both need to join smoothly and have the points (24,5). Differentiation needs to be used to ensure that the gradient of the curve meet smoothly. The graph look something like this to give you the  whole picture. The graph needs to be considered as two parabolas.

[url=http://img468.imageshack.us/my.php?image=graphwl8.png][img]http://img468.imageshack.us/img468/8946/graphwl8.th.png[/img][/url]

Is anyone able to help me please. Try and show the basic steps of how you came to solve it.

Thanks

P.S. [b]DiscreetFourierTransform[/b] you were a legend and helped with my last problems. Thanks heaps and I hope you are able to help with this one as well. Your working out was tops!

Differentiation $\subset$ Calculus.  Calculus belongs in the calculus forum.


articles calculus integration geometry MATHCOUNTS inequalities college

In all the math competitions I have participated in, and in all my math classes, I have seen very few blacks (or African-Americans, although I do not like that term: Am I European-American?). I have read articles about the race-gap, but I would like to hear opinions of math people. Any comments welcome.

Sorry if I am not referring to a wide audience. Once I wrote this, I realized that I was forgetting all the mathematicians in African countries. I was mainly referring to blacks as a minority.

Well, I suppose that my opinion is more of the same, here we go again. I live in Atlanta, Georgia, and here there is a big population of black people* (See note) and I have seen that most of the low income population in the city is black, and most of the black population has low income. I don't say that it is impossible, but it is hard to do well on school when your family has no money. Also, low income population is more prone to drop out of school, and that is not a race problem (that kind of thinking is, in my opinion, BS!!), but because life is harder. Also, there is this thought that if you are a mathematician you are going to starve to death because there is no job you can do (few things can be farther away from reality, but that is what most people think), and well, some people that have had a hard life in the economical sence has a tendence to go for "more productive" or "more estable" carriers, if they handle to get into college.

Now, Calderbank made me realize something: Every single non-white mathematician that I know, is also non-American (understanding American as from the US). Nevertheless, I do know a few good mathematicians who hapen to have a lot more melamine on their skin than me.

* Note: I do not consider the term "black people" politically incorrect, as well as "white people". I also think that African American is out of order. Would you say that I am European-Central American because I can track that four generation ago, most than half of my family was native European? C'mon, most of the black population on the US has at least five or six generations to be in America (I mean, the biggest time when black people came into the US was before 1863, isn't it?, and well, 140 years, about at least five generation). Many of them really care about their ancestral origins, but most of them do not. Think about this, Do white people identify themselves with the Europeans? Most of them, at most ten generations ago, comes almost completely from somewhere there (Even I partially do, I also have native central american blood, what is kind of obvius when you see my skin color!) Why is it bad to say that someone is black if he has black skin, but it is not bad to say that someone is white if he has white skin? I also think that this kind of double moral is somehow, discrimination! My skin is brown, so what!, does it make me superior to you? No it doesn't. Does it make me inferior to you? No, it doesn't! I can not do any more than quote

[quote="Mr. Martin Luther King"]I have a dream that my four children will one day live in a nation where they will not be judged by the color of their skin but by the content of their character.[/quote]

In Belgium, we have a huge amount of nationalities. Mainly people from Italy, Marocco, Turkey, etc. Where I live, you mainly have moroccan families. (i suppose 1,200 of the :pm: 3,500 inhabitants are from another country). I don't bother, because most of them are actually pretty nice people. Of course some of them cause troubles but not more than 'white people'.
In primary school, there were several allochthones in my class every year. They weren't exactly more stupid or more smart than other students. Especially not in maths... But that changes when I went to secondary school. Most of those 'black people' then went to a technical school, while 'white people' went to all sorts of schools. I think their parents just don't want them do study any longer than necessary, and they probably want their kids to be able to start working from the moment they leave school. Maybe the kids feel like that too, i dunno for sure  :?
I've also noticed how only few 'black people' go to maths contest  :(  I don't know the reason for that, although i suppose technical schools don't encourage their students to participate, while general secondary schools do...

I think the situation is different than in the USA, because we have a better social safety net. If people have to emigrate, and they come to Belgium, they receive -as an unemployed person- an amount of money which is almost as much as the Belgian minimum salary. It sounds stupid, but it's just like that. Therefore, you have to cause financial problems yourself if you really want to be 'poor'. But most 'black people' have a job, and they probably work as good as a 'white person'. :)

With respect to the term African-American, it does provide a very useful distinction, you all just aren't thinking about it from the correct point of view.  My neighborhood is approximately 85% "black."  However, the population is very dissimilar from the black population of Atlanta -- most of the residents of my neighborhood are of relatively recent Carribean descent (or themselves immigrants).  To label them and Atlantans both with the term black destroys a lot of significant information about their cultural and historical backgrounds.

With respect to racial minorities* and math competitions, this is true of the entire mathematics world -- in America, nearly all tenured mathematicians are white men.  This is a problem on its own, but it is also tied in to exceedingly complicated societal issues.  I think it's probably safe to say that a major reason the U.S. [i]doesn't[/i] have a social safety net like Belgium (or most of the rest of Western Europe) is that America had a significant black population at the time when those safety nets were being established, and Europe didn't.

It's also been a perpetual problem that in those few places where we have any meaningful racial integration, tracking almost invariably is a force of separation.

[quote="mcalderbank"]In all the math competitions I have participated in, and in all my math classes, I have seen very few blacks (or African-Americans).[/quote]

same here. on my math team in the 3 years i have been on it, there has been no black people and only 1 girl (the teacher's daugher). in my ap statistic and ap euro classes, there are no black poeple.
and i to dislike the african-american term as i am not a english-irish-french-german-canadian-american (phew...that's a mouthful).

The biggest reason why there are so few African-Americans in mathematics (both in contests and professionally) has to do with social stigma and the lack of prominent role models in these areas. People subconsciously pattern their self-identity after those whom they are close to or those whom they can emotionally identify with. This tends to be people from their own "race" or culture. So having few African-Americans involved in mathematics in one generation leads to a cyclical system of inspiring few of them to go into mathematics in the next generation.

There are also some unfortunate social stigmas involved which have to do with mathematics being perceived by African-Americans as being not very practical (either financially or professionally). So in a culture which has been struggling to make tangible and practical advances - mathematics has not been very high on the radar for many Black-American intellectuals in the last 40 years.

Fields such as law, medicine, politics, art and literature and even engineering have been much more popular. (To this day my parents still wonder why I don't become a doctor - as if that's the only successful thing that you can be if you're Black and good at science or math :roll:).

In regard to why the term African-American is used - I believe it gained popularity in the mid-20th century to replace the terms Negro and Colored, which were seen to have negative connotations at the time. It was also used to re-emphasize the heritage and history of Black-Americans at a time it had been purposefully forgotten, neglected, or denigrated by many.

So, in short, there are several reasons why Black-Americans would choose to be called African-American which don't generalize well to other American ethnicities. It all has to do with historical context.

[quote="JBL"]There is an interesting article available  [url=http://www.nytimes.com/2005/03/20/magazine/20HARVARD.html?ei=5090&en=e9727ddcbbbd4431&ex=1268974800&partner=rssuserland&pagewanted=print&position=]here[/url] that is somewhat related to the topic at hand, I think.  (It's from the New York Times Magazine -- you may need a subscription to nytimes.com in order to view it.)[/quote]

Thanks for the link Joel, that was a fascinating article.

Yes, causation is complicated when you look at social issues like this. Basically, you look at the life history of millions of individual human beings, aggregate them into approximate categories (both the African-American group in the United States population and the European-American group have some overlap in membership, and some debatability about the boundaries of the group) and then turn wild speculation mode on. Most explanations for group differences need to be treated very tentatively.

I also saw the New York Times Sunday magazine story, which relates one of the latest attempts to gain valid information on this long-standing issue. The man profiled in the article is, of course, an African-American man with considerable mathematical competence (greater competence than my own, for sure).

one of my friends from the Bay Area who's african-american got a 25 on the AMC 8 and a pretty high score on the AMC 10 this year. he was the smartest person at our middle school, and one of only...20 black kids out of 350 in the 8th grade, when most of the minority were asians, including myself. pretty impressive :D

I live in Belgium too.  I have  only seen two non fully Belgians in the finales of physics or mathematics olympiad, and both were chinese.  It's not a racial issue.  It's just that immigrants in our country tend to do badly, they often don't speak the language well etc...

two things
what about recent immigrants from africa, who may or may not be black (i'm egyptian, thus african, but usually put down ethnicity as caucasian/white), and there are also other people who might be from subsaharan africa but for whom the "recent immigrants" hypothesis of strict parents wanting kids to excel stereotype might apply, or is that not true?

i rememebr more african americans involved in math team in junior high than in high school, though there were still a good number in honors classes

Now that the issue has come into focus, I recal hearing something describing how the majority of African-American's ( who possess the oppurtunity ) choose to go into fields like Law opposed to Academia.  It's easy to see how such trends can create, "intellectual" stereo-types, for Black students aspiring to enter University, and ultimately dictate what endeavours they set for themselves ( like math-competitions ) during their high-school years.

Who is here on AOPS who's black as well?

[quote="blahblahblah"]There is more AA practiced in law school admissions than in graduate school admissions, which doubtless provides impetus for that direction.[/quote]

I'm not so sure this is true. Of course there's the famous case of Michigan Law school that comes to everyone's mind, but that doesn't really count as evidence.

I think there may actually be more AA in a lot of areas of graduate school admission than in law school admissions, simply because the need for more blacks to go to graduate school is so much greater -- and that's what AA is supposed to remedy.

And even if it were true that AA was practiced more in law school admissions, I really don't think it would give [i]that[/i] strong of an impetus for people to choose law school over graduate school. I'm not concinced that there are thousands and thousands of blacks going to law school simply because they feel they could get into Law School with the help of AA, but couldn't get into graduate school without it.

just my 2 cents..

I blame GOP discrimination. Mind you, this is not a blanket statement, as not all Republicans are racist, and not all racists are Republicans, but I actually feel that in areas were African Americans are more heavily present, education funding is not what it should be.

I think the main reason that you don't see that many black in math-related careers is because many of them live in the inner-city, and come from single-parent homes, and aren't encouraged as much to do well in academics and stuff.

Proportionally less white academics come from the inner city as well, so that doesn't really mean anything, as poverty causes essentially the same problems independent of which ethnic group it is visited upon.

[quote="LordoftheMorons"][quote="blahblahblah"]Your opinion is for all intents and purposes irrelevant if you can't spell 'racist' correctly.  I don't usually nitpick on grammar or spelling, but come on...[/quote]

...How does a typo constitute an opinion being disregarded?[/quote]

  His typo abused the underlying foundation of this topic.  Either that, or blahblahblah is simply being a pest ( a role he plays most wonderfully, if I might add  :lol:  ).

In January 2003, there was a session at the Joint Mathematics Meetings on recruiting more females and ethnic minorities into contest participation.  Some of the talks were about what the barriers are to contest participation for African-American students in particular.  The reason most cited was among those listed by gauss202:

[quote="gauss202"]
There are also some unfortunate social stigmas involved which have to do with mathematics being perceived by African-Americans as being not very practical (either financially or professionally). So in a culture which has been struggling to make tangible and practical advances - mathematics has not been very high on the radar for many Black-American intellectuals in the last 40 years.

Fields such as law, medicine, politics, art and literature and even engineering have been much more popular. [/quote]

Also, there's a set of databases of black mathematicians at
http://www.math.buffalo.edu/mad/00.INDEXmad.html
which includes profiles of lots and lots of black mathematicians.  Definitely worth checking out.

At my school there is a pretty intresting stiuation, it is located in a "black" (not to be racist or how ever u spell that word  ;) but for a lack of better word since a couple of my "black" friends don't like the term African American because they really grew up here and have roots here too) neighborhood, and is an IB school. So an intresting situation pops up, (Due to the reasons listed above) as it turns out most of the "black" people are in the general class. Yet, our math organiztion (Mu Alpha Theta, Mathcounts, Math Olympiad) is headed by two black teachers one of which is going to Harvard. Soo.. I think my school is an exception and an example to your rule of there are not a lot of "black" people doing advanced math and other higer classes. ;)

Hello everyone. I agree with the replies thus far. I'm Black and math has been my favorite subject all of my life. My mother even used math to punish me in the past (No Math this week, just your regular math homework) lol. That punishment got to me the worst. I would say that Blacks don't like math mostly because they have to actually think harder for math than other subjects. From my experience, Blacks dread everything that dealt with math. It was annoying to hear people question the teacher as to (why I gotta be doing this, it hard for me, etc). There may be other factors to contribute to this, I don't know.

[b]Race differences in average IQ are largely genetic[/b]

Medical Research News
Published: Tuesday, 26-Apr-2005

[b]A 60-page review of the scientific evidence, some based on state-of-the-art magnetic resonance imaging (MRI) of brain size, has concluded that race differences in average IQ are largely genetic.[/b]

The lead article in the June 2005 issue of Psychology, Public Policy and Law, a journal of the American Psychological Association, examined 10 categories of research evidence from around the world to contrast "a hereditarian model (50% genetic-50% cultural) and a culture-only model (0% genetic-100% cultural)."

The paper, "Thirty Years of Research on Race Differences in Cognitive Ability," by J. Philippe Rushton of the University of Western Ontario and Arthur R. Jensen of the University of California at Berkeley, appeared with a positive commentary by Linda Gottfredson of the University of Delaware, three critical ones (by Robert Sternberg of Yale University, Richard Nisbett of the University of Michigan, and Lisa Suzuki & Joshua Aronson of New York University), and the authors' reply.

[ . . . ]

Complete text at http://www.news-medical.net/?id=9530

[quote="smbelcas"]The article cited above seems to have been sponsored by or posted by someone at (but is definitely linked to) http://www.charlesdarwinresearch.org/, which openly declares its bias.  Therefore I don't see any reason to trust that the summary at news-medical.net is accurate.  (I was immediately suspicious when I noted that the authors claim that on average sub-Saharan Africans are borderline severely and profoundly mentally retarded; my suspicion grew when they used the big-brain claim.)[/quote]

Since you don't trust the summary, here is the original APA research paper: http://www.udel.edu/educ/gottfredson/30years/

This is a good website to go to to find out your own biases.

https://implicit.harvard.edu/implicit/demo/selectatest.jsp

One of the two authors (Rushton) is the contact for the eugenics website that you linked to, smbelcas.  (Giving Darwin a bad name, I might add.)  The other seems to have been up to this sort of nonsense since at least 1969.  The fact that they cite [i]The Bell Curve[/i] in a generally approving manner makes me feel free to discard the paper without thinking further about it.

The three rebuttals are fun reading, though.  I like the second one (by Nisbett).

[quote="Asian-American"]
Since you don't trust the summary, here is the original APA research paper: http://www.udel.edu/educ/gottfredson/30years/[/quote]

Thank you for the link.  After reading through some of the papers in that issue (on each side), my conclusion is that none of the arguments in favor of the hereditarian hypothesis seem to have any evidence which points to a genetic component to IQ/g/etc.  Everything cited is statistical/correlative in nature (rather than causal).  The evidence only says that either there is a nonenvironmental component (perhaps physiological/developmental rather than genetic) to IQ/g/etc. or that we have not yet figured out how to consistently neutralize environments.  So, it seems to me that the system is still too complex for us to understand.

And, JBL, I agree that the website linked from news-medical.net does indeed give Darwin a bad name.  Very sad.  Perhaps it's this sort of thing that causes some 60% of the populace to discredit evolution?  (Yes, that's just wishful thinking on my part.)

[quote="smbelcas"][quote="Asian-American"]
Since you don't trust the summary, here is the original APA research paper: http://www.udel.edu/educ/gottfredson/30years/[/quote]
Thank you for the link.  After reading through some of the papers in that issue (on each side), my conclusion is that none of the arguments in favor of the hereditarian hypothesis seem to have any evidence which points to a genetic component to IQ/g/etc.  Everything cited is statistical/correlative in nature (rather than causal).  The evidence only says that either there is a nonenvironmental component (perhaps physiological/developmental rather than genetic) to IQ/g/etc. or that we have not yet figured out how to consistently neutralize environments.  So, it seems to me that the system is still too complex for us to understand.[/quote]
While the mountain of genetic data that will arrive over the next few decades will still leave us far short of a complete understanding of human development, it will resolve the question being discussed in this thread.


A positive real number a and an angle with vertex A are given. Show that all the circles which pass through A and intersect the sides of the given angle at points B,C with AB+AC=a have a second common point in addition to A.

Let $X,Y$ be points on the sides of the angle s.t. $AX=AY=\frac a2$. Now let the perpendiculars from $X$ to $AX$ and from $Y$ to $AY$ intersect at $P$. Given $B,C$ as above, it's clear that $PB=PC,\angle BPC=\angle XPY$, and from here we have $PB\cdot AC+PC\cdot AB=BC\cdot PA$, meaning that $ABPC$ is cyclic.

I did the same in reverse way.

Anyway, why is it in Open Questions?


inequalities
logarithms
inequalities unsolved

$ (a)$ Find the minimum value of $ x^x$ for $ x$ a positive real number.
$ (b)$ If $ x$ and $ y$ are positive real numbers, show that $ x^x\plus{}y^y>1.$

$ (x^x)' \equal{} (e^{x \ln{x} })' \equal{} e^{x \ln{x} } (x \ln{x} )' \equal{} e^{x \ln{x} } (x' \ln{x} \plus{} x (\ln{x})' ) \equal{} x^x ( \ln{x} \plus{} 1) \equal{} 0$
$ \Leftrightarrow x \equal{} \frac{1}{e}$
$ \min{x^x} \equal{} \left(\frac{1}{e}\right)^{\frac{1}{e}} \equal{} \frac{1}{\sqrt[e]{e}} > \frac{1}{2}$ as $ \sqrt[e]{e} < 2 \Leftrightarrow e < e^2$


LaTeX
combinatorics unsolved
combinatorics

[quote="soccer022483"]Note: $f_n$ counts the number of ways to tile a 1 by n-1 board with dominoes and monominoes (a square). use this fact to give a combinatorial proof that:  $1+f_1+f_3+...+f_{2n-1}=f_{2n}$[/quote]

Well, at first, the formula should be $f_1+f_3+...+f_{2n-1}=f_{2n}$.

For the proof, we will use double counting. Consider a $1\times\left(2n-1\right)$ board. Label its cells with $c_1$, $c_2$, ..., $c_{2n-1}$. Each of these cells is either covered by a monominoe or covered by a dominoe; a monominoe covers a single cell, while a dominoe covers two neighboring cells.

We will count the number of all possible coverings of the $1\times\left(2n-1\right)$ board. On the one hand, by the condition of the problem, this number is $f_{2n}$. If we succeed to find another way of counting this number, which shows that this number is $f_1+f_3+...+f_{2n-1}$, then it will follow that $f_1+f_3+...+f_{2n-1}=f_{2n}$, and the problem will be solved.

Thus, it remains to show that the number of all possible coverings of the $1\times\left(2n-1\right)$ board is $f_1+f_3+...+f_{2n-1}$.

Well, for an arbitrary covering of the $1\times\left(2n-1\right)$ board, let $k$ be the number of the first cell covered by a monominoe; in other words, let $k$ be the smallest positive integer such that the cell $c_k$ is covered by a monominoe. (For instance, if the very first cell $c_1$ is covered by a monominoe, then $k = 1$; if the first $2$ cells are covered by dominoes and the third cell $c_3$ is covered by a monominoe, then $k = 3$; if the first $6$ cells are covered by dominoes and the seventh cell $c_7$ is covered by a monominoe, then $k = 7$.)

This number $k$ exists for all coverings; this is because, no matter how we cover our $1\times\left(2n-1\right)$ board, some cell must be covered by a monominoe (else, all cells would be covered by dominoes, so they would be grouped into pairs of neighboring cells, and thus, their number would be even; but their number is $2n - 1$ and thus odd).

For every covering of the $1\times\left(2n-1\right)$ board, the number $k$ is odd (in fact, all cells with numbers smaller than $k$ are covered by dominoes, and thus, they are grouped into pairs of neighboring cells, so that their number must be even, and therefore, $k$ is odd). And, clearly, we have $1\leq k\leq 2n-1$ (else, speaking of the "cell $c_k$" would be nonsense).

Hence, every covering of the $1\times\left(2n-1\right)$ board yields an odd integer $k$ such that $1\leq k\leq 2n-1$.

Conversely, for every odd integer $s$ with $1\leq s\leq 2n-1$, there exist exactly $f_{2n-s}$ coverings of the $1\times\left(2n-1\right)$ board for which we have $k = s$. In fact, requiring that $k = s$ is equivalent to requiring that the cell $c_s$ is the first cell covered by a monominoe; hence, all the previous cells $c_1$, $c_2$, ..., $c_{s-1}$ must be covered by dominoes, and thus, the only freedom we have is to choose the covering of the remaining cells $c_{s+1}$, $c_{s+2}$, ..., $c_{2n-1}$; these are exactly $\left(2n - 1\right) - s = 2n - s - 1$ cells, and, of course, the number of different ways of covering them equals to the number of ways to cover a $1\times\left(2n-s-1\right)$ board, and, as we know, this number is $f_{2n-s}$.

Altogether, every covering of the $1\times\left(2n-1\right)$ board yields an odd integer $k$ such that $1\leq k\leq 2n-1$, and for every odd integer $s$ with $1\leq s\leq 2n-1$, there exist exactly $f_{2n-s}$ coverings of the $1\times\left(2n-1\right)$ board for which we have $k = s$. Thus, the number of all coverings of the $1\times\left(2n-1\right)$ board is the sum of $f_{2n-s}$ over all odd integers $s$ satisfying $1\leq s\leq 2n-1$. Hence, this number is

$f_{2n-1}+f_{2n-3}+...+f_{2n-\left(2n-3\right)}+f_{2n-\left(2n-1\right)}=f_{2n-1}+f_{2n-3}+...+f_3+f_1$
$=f_1+f_3+...+f_{2n-1}$.

And this completes our proof.

Was this halfways understandable?

Of course, the numbers $f_1$, $f_2$, ... are the famous Fibonacci numbers. The combinatoric interpretation via the $1\times\left(n-1\right)$ board is easily seen to be isomorphic to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14984]the gappy set interpretation[/url]. As an exercise, I leave it to find a similar proof of the formula $1+f_2+f_4+...+f_{2n}=f_{2n+1}$. In fact, we have to consider a $1\times 2n$ board, but this time, there exists one case when $k$ is not defined (this is the case when the whole board is covered by dominoes)...

  Darij

Uh... Actually, isn't there a '+1' at the end? The original formula is correct.
There is the case where all the tiles are dominoes. That would be your f_0=1.

That's a very formal proof. Certainly understandable! If I were to do it, I'd let k be the number of leading dominoes. Then I don't need to talk about odd integers. Ha ha. Just nitpicking. So after establishing the bijection, I'd have sum of f_(2n-(2k+1)) for k=0 to (n-1). Plus 1.

Pardon my lack of LaTeX knowledge.

[quote="Yossarian"]Uh... Actually, isn't there a '+1' at the end?[/quote]

Thanks, that was a typo. I've corrected it now.

[quote="Yossarian"]The original formula is correct.
There is the case where all the tiles are dominoes. That would be your f_0=1.[/quote]

No, not for a $1\times\left(2n-1\right)$ board. In fact, $f_n$ was defined as the number of ways to cover a $1\times\left(n-1\right)$ board with dominoes and monominoes; not a $1\times n$ board.

  Darij

Oh, oops. I thought f_n was for a 1xn board. (In which case there'd be an f_0) Misread the question. Heh. I'm always making careless mistakes like this. I hope it's just that I haven't done much math for a couple of years.


Solve equation
$ \left\{  \begin{array}{l}    x\plus{}\sqrt{y^2\minus{}2y\plus{}2}\equal{}3^{y\minus{}1}\plus{}1 \\    y\plus{}\sqrt{x^2\minus{}2x\plus{}2}\equal{}3^{x\minus{}1}\plus{}1   \end{array}  \right.$

Let $ z\equal{}x\minus{}1,t\equal{}y\minus{}1$, then
(1) $ z\equal{}f(t),t\equal{}f(z)$, were $ f(x)\equal{}3^x\minus{}\sqrt{x^2\plus{}1}$.
It is not hard to chek, that $ f(0)\equal{}0,f(x)>x>0,f(x)<x<0$. Therefore (1) had only solution $ z\equal{}t\equal{}0$ or $ x\equal{}y\equal{}1$.

[quote="Rust"]$ f(0) \equal{} 0,f(x) > x > 0,f(x) < x < 0$. .[/quote]
Why? Can you post complete solution


Can anyone tell me how to find the 5th last digit from the right in the expansion of

             5 raised to the power 5 to the power 5 to the power 5 to the power 5 to the power 5...sorry don't know how to use LateX.

Source: Mathematics Olympiad challenges , Titu Andreescu.

I believe the period is $ 1,7,9,5,6,2,4,0$ for the fifth digit from the right of $ 5^n$.


calculus
integration
trigonometry
logarithms
function
calculus computations

Evaluate the integral

$\int^{2\pi}_{0}\frac{x^{2}\sin x}{8+\sin^{2}x}dx$

I try to do substitution like $x=\pi-t$ , $x=\frac{\pi}{2}-t$ , $\sin x=t$ and also tried Integration by parts but still cannot do it , someone please enlighten me  :maybe:

[hide="Here is the my solution."]
[color=darkblue]I'll try the substitution $\underline{t=\phi (x)=\pi-x}\ !$

$\boxed{I}\equiv\int_{0}^{2\pi}\frac{x^{2}\sin x}{8+\sin^{2}x}\ dx=$ $\int_{-\pi}^{\pi}\frac{(\pi-x)^{2}\sin x}{8+\sin^{2}x}\ dx=$ $\int_{0}^{\pi}\left[\frac{(\pi-x)^{2}\sin x}{8+\sin^{2}x}+\frac{-(\pi+x)^{2}\sin x}{8+\sin^{2}x}\right]\ dx$ $\Longrightarrow$ $\boxed{I=-4\pi\cdot J}\ ,$ where

$J\equiv\int_{0}^{\pi}\frac{x\sin x}{8+\sin^{2}x}\ dx=$ $\int_{0}^{\pi}\frac{(\pi-x)\sin x}{8+\sin^{2}x}\ dx=\pi\cdot\int_{0}^{\pi}\frac{\sin x}{8+\sin^{2}x}\ dx-J$ $\Longrightarrow$ $\boxed{J=\frac{\pi}{2}\cdot K}\ ,$ where

$K=\int_{0}^{\pi}\frac{\sin x}{8+\sin^{2}x}\ dx=$ $\int_{0}^{\pi}\frac{(\cos x)'}{cos^{2}x-9}\ dx=$ $-\int_{-1}^{1}\frac{1}{x^{2}-9}\ dx=$ $-2\cdot\int_{0}^{1}\frac{1}{x^{2}-9}\ dx=$ $\frac{1}{3}\cdot \ln\left|\frac{3+x}{3-x}\right |_{0}^{1}=\frac{1}{3}\ln 2\Longrightarrow \boxed{\ K=\frac{1}{3}\ln 2\ }\ .$

Therefore, $I=-4\pi\cdot\frac{\pi}{2}\cdot\frac{1}{3}\ln 2$ $\Longrightarrow$ $\boxed{\ I==-\frac{2\pi^{2}}{3}\cdot\ln 2\ }\ .$[/color][/hide]

Thank you very much Virgil Nicula , the solution is very nice  :D


What is a [b] "vector " [/b] ?
What is a [b] " force " [/b] in Physics  ?
What is a [b] " force " [/b] in  Mechanics ?
What is a [b]" free=vector" [/b] in Mechanics . It's a  vector with some properties, or it's a set or an equivalence class ?
What is the  [b] " acceleration " [/b] ?
Please  rigurous definitions. I am not specialist in Physics !

[quote="perfect_radio"]Wait a minute. Does "vector" has a different definition in physics than in mathematics?[/quote]No, I don't think so.

The definition of force is this:

force = mass times acceleration (also with vectors).

That's the most rigorous definition I can imagine, really.

well, technically, force is is the derivative of the momentum vector with respect to time..

Also see http://scienceworld.wolfram.com/physics/Force.html and http://en.wikipedia.org/wiki/Force

For vectors try http://mathworld.wolfram.com/Vector.html

You can find many math- and physics-related stuff on the wolfram website.

a vector is a representation of an amount and a direction, like you could have a vector of velocity.  like if a car was driving 40 kilometers per hour north east, you could have a vector 4 units long pointing 45 degrees, it would somehow have to be communicated that north is up and 1 unit is 10 kilometers per hour.  it could be for acceleration or anything that has a direction to it.

most commonly i've used vectors with forces.  just say you have a mass $m$ and a change of acceleration of $a$, and you have a force $f$ (in Newtons) that equals $ma$.  what the force means is it takes $f$ newtons to accelerate the amount of mass $m$ by the amount $a$.

force is the same in physics as mechanics.

could you go into what you mean by "free=vector" im not sure what you're referring to.

acceleration is the change of velocity for a given amount of time.  $m/s^2$ can be read as "meters per second per second", which is the change of velocity per second.

As said above, the definition of acceleration is $\overrightarrow{a} = \frac{d\overrightarrow{v}}{dt}$.

In the 9th grade (seems ages ago :) ) I learned that a vector is an ordered pair of points. We called this "vector legat" and used the notation $\overrightarrow{AB}$. The free vector was introduced indeed as an equivalence class: a free vector represents all equivalent vectors ("vectori legati"), i.e. which only differ in the application point and used the notation $\overline{AB}$. Sorry if this is inaccurate.


According to X telephone company, the number of text messages being sent in one month wen from 930 million to 1.2 billion from June 2002 to June 2003. Assuming this trend continues, what would be the amount sent in June 2004.?

Well it's not a pretty number:
$ 1,548,387,096$    :roll:

Numbers arent always pretty

$ \boxed{\boxed{\boxed{\boxed{1548387097}}}}$


[quote="hainam-sp"]Find the maximum value of $P$: $P=3yz+3zx+3xy-xyz$ when $x,y,z>0$  and $x^3+y^3+z^3=3$[/quote]

  The answer is 8. The equality occurs when $x=y=z=1$
 From $3=x^3+y^3+z^3 \geq 3xyz$ so $xyz \leq 1$
  From schur, we have:
   $3+ x^3+y^3+z^3+3xyz \geq \sum (x^2y+y^2x+1) \geq 3(xy+yz+xz)$
$\leftrightarrow$ $6+3xyz \geq 3 (xy+yz+xz)$
$\leftrightarrow$ $3(xy+yz+xz)-xyz \leq 6+2xyz \leq 8$


calculus
integration
function
trigonometry
calculus computations

OK, I have three integrals, and I know that they require a substitution to transform them to rational functions before I can solve them, but I'm not sure how to do this:

1) int_0^4 [(1 - root(x)) / (1 + root(x))]

2) int [dx / (root(x) + cuberoot(x)]

3) int_(pi/3)^(pi/2) [(sin(x)dx) / (cos^2(x) + cos(x) -2)]   --> x is actually given as theta, but I assume that it doesn't matter which is used

Thanks in advance for your help

All of these will be "forced" substitutions rather than "natural" substitutions. The first hint about forcing a substitution: write it backwards; let $x$ depend on $u$ rather than the other way around.

1. $\int_0^4\frac{1-\sqrt{x}}{1+\sqrt{x}}\,dx$

Let $u=\sqrt{x}.$ Taking my "write it backwards" hint, $x=u^2,\,dx=2u\,du.$ The integral becomes

$\int_0^2\frac{1-u}{1+u}\,2u\,du.$

2. $\int\frac{dx}{\sqrt{x}+\sqrt[3]{x}}$

The common denominator for $\frac12$ and $\frac13$ is $\frac16.$ Try $u=x^{1/6},\,x=u^6,\,dx=6u^5\,du.$ The integral becomes

$\int\frac{6u^5\,du}{u^3+u^2}=6\int\frac{u^3\,du}{u+1}$

3. $\int_{\frac{\pi}3}^{\frac{\pi}2}\frac{\sin \theta\,d\theta}{\cos^2\theta+\cos \theta -2}.$

This time let $u=\cos\theta.$ This is actually what I call a "natural" substitution; it doesn't have to be forced.

Hi Kent, thanks so much for your help... I'm just not sure about what you mean by a "forced" and "natural" substitution. Is a forced substitution simply any "u" substitution, or is it a substitution used to covert the integral to another type to solve?

The issue is mostly whether when you write $u=g(x),\,du=g'(x)\,dx,$ there's a nice convenient $g'(x)\,dx$ (or a constant multiple of it) sitting there waiting to be used, and no leftover pieces. It's a "forced" substitution when that's not true but you want to make the substitution anyway.

One does anticpate with a forced substitution that the substitution itself is unlikely to be the last step and there's probably more work to be done.

Thanks a lot for clarifying that... another question... for 3), when subbing u=cosx, do I need to change the integral of integration at any point in the exercise?

You have a choice: either change the limits of integration at the same time you make the substitution, and work everything in terms of the new variable, or don't. If you don't then in effect you're working an antiderivative problem, and you've committed yourself to returning to the original variable.

Here are examples of what I mean, for a very simple integral:

$\int_0^{\frac{\pi}2}\cos^3x\sin x\,dx =$

Let $u=\cos x,\,du=-\sin x\,dx$ and when $x=0,u=1;\,x=\frac{\pi}2,u=0.$

$=\int_1^0u^3(-du)=\int_0^1u^3\,du=\left.\frac14u^4\right|_0^1 =\frac14(1-0)=\frac14.$

Versus the following:

$\int_0^{\frac{\pi}2}\cos^3x\sin x\,dx=\int_?^?-u^3\,du= \left.-\frac14u^4\right|_?^?=$

${=\left.-\frac14\cos^4x\right|_0^{\frac{\pi}2}}= -\frac14(0-1)=\frac14.$

When I teach, I do some problems one way and some problems the other way, and insist that my students be comfortable with both. I also object (frequently!) to students working it the second way but writing limits of $0$ and $\frac{\pi}2$ for $du$ integrals. That's not right, because those aren't the limits that $u$ goes between. I'm OK with those question marks, which basically stand for "the correct limits of integration for this new variable, which I haven't bothered to find because I don't plan to use them."


$\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}\ge \frac{2}{R}$
We all know what is $R,h_a,h_b,h_c$

Justi rewrite $h_a,h_b,h_c$ as $\frac{2P}a, \frac{2P}b, \frac{2P}c$ and $R$ as $\frac{abc}{4P}$. B this substitution inequality is $abc\geq (a+b-c)(a-b+c)(-a+b+c)$ which is realy well known.

[quote="Beat"]Prove this well known inequality.[/quote]
assume $a \ge b \ge c$
if $b+c-a <0$,then it is obvious.

if not,we have $a+b-c,b+c-a,c+a-b \ge 0$
there are too many method.
solution1:let $x=b+c-a,y=....$

solution2:$(a+b-c)(a+c-b)=a^2-(b-c)^2 \le a^2$ :)

Hmm, very nice solution. Is this inequality discovered? Becouse I invented it when i was solving another one (I`m talking about the inq in the 1st post)


Let $\varphi : [0,\frac{\pi }{2}{\rm ]}\to R$ a continuous function such that $\varphi (\frac{\pi }{2}) \ne 0$ ,
show that $\int_{0}^{\frac{\pi }{2}}{\varphi (t)\sin^{2n}}(t)dt$ is an equivalent to $\varphi (\frac{\pi }{2})\int_{0}^{\frac{\pi }{2}}{\sin^{2n}(t)dt}$ when n tends to infinity .

[quote="maskman"]no one can help?[/quote]
Posted a mere hour and ten minutes after you posted the problem. Please show more patience than that!

Note that $\sin^{2}(t)$ is continuous, nonnegative, always $\le 1$ and equal to $1$ at exactly one place: $\frac{\pi}2.$ These will be the key ingredients of the solution.

Perhaps Mean value theorem for integrals may help :wink:


geometry
geometric transformation
vector
analytic geometry
graphing lines
slope
geometry solved

What is the area of the largest regular hexagon H that can be inscribed in (or covered by) a unit square S?

I think i solved the problem... The result would be l=(sqrt(3)-1)/ sqrt(2); this is the case where the hexagon has two adiacent vertices (M and N) that rest on adiacent sides (AB and BC) of the square and BD is bisector for the segment MN.

The only combinatorial part of my proof (but even this is relative) is to prove that the hexagon has 2 adiacent vertices that lay on 2 adiacent sides of the square(this is simple). The rest is denoting BM and N by x and y and proving that the best position for the hexagon is when x=y. This is done by some computation and the observation that only none, one or two of the verices opposite to M and N also lay on the square's sides.

I don't want to write to complete solution to the problem because i think there could be a more elegant one. This one ... stinks

I think I got a nicer solution, even though it still involves some computation. Indeed, the maximum length of the haxagon's side is  (\sqrt 3-1)/ \sqrt 2. It can be attained, as kappa showed.

Now, let's prove that if a regular hexagon H is inscribed in the square then the hexagon G is also inscribed in the square, where G is obtained from H by translating it so its center would be the center of the square.

Proof: let H' be the symetrical of H around the center of the square. Then H' is also the translation of H of vector OO' (where O is the center of H and O' is the center of H'). Now this is easily proved because a regular hexagon is centrally symetric. So since the translation H' of H of vector OO' is in the square, then so would be the translation G of vector OO'/2, since the interior of the square is a convex set.

So, suppose that there is a regular hexagon of side length L>(\sqrt 3-1)/ \sqrt 2 inscribed in the square of verices (1,0), (-1,0), (0,1), (0,-1). We can assume (because of the lemma above) that the hexagon has the center (0,0). And each of the vertices of the hexagon would have the coordonates x and y such that |x|+|y|<=1 (because it is inscribed)

Now, let m1, m2, ..., m6 be the slopes of the vectors OV1, ...OV6, where O is the origin and V1, V2, ..., V6 are the vertices of the hexagon. Because  it is a regular hexagon, one of the angles corresponding to the slopes is between 60 degrees and 120 degrees, so that slope is greater, in modulus, than  \sqrt 3. So, let m be that slope and x and y be the coordinates of the corresponding vertex (y>0 because of how we picked the slope).

So keep in mind that y=m|x|, x 2 +y 2 =L 2 >2- \sqrt 3 and that |x|+y<=1 and you will get a contadiction.


Is this true?

All the cubic polynomial has at least one real roots. If yes can we prove it.

Also can the same be considered that every odd degree polynomial with real co-efficient has at least one real root?

Can you please show the proof?

What happens as $ x$ goes to positive infinity?  As $ x$ goes to negative infinity?  (The answer to your second question is yes.)

First idea: Consider $ x\to\minus{}\infty$ and $ x\to\infty$. As the sign of $ f$ cannot be the same in both cases, we can use the continuity of $ f$ to show that there is a solution of $ f(x)\equal{}0$.

Second idea: Since $ f(z)\equal{}0\iff f(\bar{z})\equal{}0$ all non-real zeros of $ f$ come in pairs; factoring out a complex number and its conjugate decreases the polynomial's degree by 2, so in the end we're left with a polynomial of degree 1 and real coefficients, which trivially always has a real root.

quite trivial.

wlog assume that the leading coefficient is positive. consider the minimum and maximum value of $ x$ such that $ f'(x)\equal{}0$ (that is, the turning point), let $ a$ be the minimum and $ b$ the maximum. If $ x<a$, clearly $ f(x)$ is increasing (by our assumption that the leading coefficient is positive and that polynomial's highest degree is odd), which implies that $ f(x)$ goes to $ \minus{}\infty$ as $ x$ goes to $ \minus{}\infty$. If $ x>b$, $ f(x)$ is also increasing, so $ f(x)$ goes to $ \infty$ as $ x$ goes to $ \infty$. Since the polynomial must be continuous, there must be a real value of $ x$ such that $ f(x)\equal{}0$.

I think that solution overcomplicates things to the point of being not quite right. There are plenty of cubic polynomials $ f$ with no solutions to $ f^\prime (x) \equal{} 0$. What do you do for them?

You shouldn't need derivatives to argue $ f \rightarrow \pm \infty$ as $ x \rightarrow \pm \infty$.

[quote="hsbhatt"]Doesnt this argument suffice?:  The complex roots, if any, come in conjugate pairs (as the coefficients are given to be real). Since the total number of roots is odd, there has to be at least one real root[/quote]

yeah that is fine.

Personally, I'm inclined to the proof that uses the Intermediate Value Theorem alone, since proving that polynomials are continuous is much easier than proving the Fundamental Theorem of Algebra.  Of course, both proofs are illuminating.

Isn't there a proof that there is an even number of complex roots of a polynomial???

1=2, yes - assuming the coefficients are real.  This is equivalent to the complex roots coming in conjugate pairs.

The proof is simple: $ a\plus{}bi$ is a root of $ P(x)$, a polynomial with real coefficients.  Let $ D(x) \equal{} (x\minus{}(a\plus{}bi))(x\minus{}(a\minus{}bi)) \equal{} x^2\minus{}2ax\plus{}a^2\plus{}b^2$.  Divide the $ P$ by $ D$, let the quotient be $ Q(x)$ and the remainder be $ R(x)$. $ R(x)$ can have at most a degree of 1, and must have real coefficients.  If $ x\equal{}a\plus{}bi$, $ D(a\plus{}bi) \equal{} 0$ so $ P(a\plus{}bi) \equal{} D(a\plus{}bi)Q(a\plus{}bi) \plus{} R(a\plus{}bi) \equal{} R(a\plus{}bi)$, but $ P(a\plus{}bi) \equal{} 0$, so $ R(a\plus{}bi) \equal{} 0$.  Since $ R$ has real coefficients and at most a degree of $ 1$, this is only possible if $ R(x) \equal{} 0$ for all $ x$, which then implies that $ a\minus{}bi$ is a root of $ P$


Simplify: $ \sqrt{50} \plus{} \sqrt{18}$ . Express your answer in simplest radical
form.

$ \sqrt{50}\equal{}\sqrt{25\cdot 2}\equal{}5\sqrt{2}$
$ \sqrt{18}\equal{}\sqrt{9\cdot 2}\equal{}3\sqrt{2}$

$ \sqrt{50}\plus{}\sqrt{18}\equal{}8\sqrt{2}$

sqrt 50 is 5sqrt2, sqrt 18 is 3sqrt2, so their sum is 8sqrt2


calculus
algebra
polynomial
india
inequalities proposed
Inequality

nice cezar lupu  :D

but what about $x^3$$y^3$($x^3$ + $y^3$) ≤ 2  ;)

It is very easy . Just use AM-GM
  Use $a^3+b^3=(a+b)(a^2-ab+b^2)=2(4-3ab)$
Then apply AM-GM for  $(4-3ab) ,ab,ab,ab$ . Finish!

Thi can also be solved by factorising and then substituting the given values.
 We then get $t^3(4-3t)\geq 0$, which is proved by applying A.M-G.M to 4-3t ,t,t,t
  and we are done. :D
That is essentially what that guy did but as I did solve it on my own I decided to post it.
  Ashwath

Ganitham Namaha

[quote="Together"][quote="Fermat -Euler"]$x$, $y$ are positive reals such that $x + y = 2$. Show that $x^3$$y^3$($x^3$ + $y^3$) ≤ 2.[/quote]
We have:  $x^3y^3(x^2+y^2-xy) \leq (x+y)^8 = 1$[/quote]
Sorry  :oops:
I want tho write: $x^3y^3(x^2+y^2-xy) \leq (\frac{((x+y)^2}{4})^4) = 1$

[quote="cezar lupu"]Guys, I have proved the following inequality: if $x,y\geq 0$ such that $x+y=2$ then \[ x^{2}y^{2}(x^{2}+y^{2})\leq 2.  \] :D

 Actually, just consider $x=1-t, y=1+t$ with $t\in [0,1].$ Then $x^{2}y^{2}(x^{2}+y^{2})=2(1-t^{2})(1-t^{4})\leq 2$ ;)[/quote]

Irish olympiad 2000.

[url]http://www.kalva.demon.co.uk/irish/soln/sol006.html[/url]

That's EXACTLY the same solution...  :huh: :?

I have never checked Kalva ,so I didn't know that their solution was the same. But I'll try another solution, because I don't want to be suspected of something else. ;)

Another solution: replace $(x+y)$ with $2$ and $xy$ with $m$.Then the required ineq is equivalent to:
$2(4-3m)m^3 \le 2$
equivalent $(4-3m)m^3 \le 1$
equivalent $3m^4- 4m^3+ 1 \ge 0$
equivalent $(m-1)(3m^3-m^2-m-1) \ge 0$
equivalent $(m-1)^2(3m^2+2m+1) \ge 0$ (*)
Becasue discriminant of $(3m^2+2m+1)$ is negative then $(3m^2+2m+1) >0$ and so (*) is true. The result follows.

This question appeared in the term-end exam of the math class in a Japanese University
(where high-school calculus of polynomials is reviewed for students in the faculty of Business Administration)
http://www.setsunan.ac.jp/~t-azuma/2009ko/termendRM092Dweb.pdf
Although the material is in Japanese I hope you can at least recognize
mathematical formula, where we prove 0 =< t^3(2^3-3*2*t) =<2
(t=xy and 0=<t=<1) using differentiation, rather than AM-GM technique.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2941769#p2941769]Irish 2000:[/url] Let $ x; y \geq 0$ with$ x + y = 2$. Prove that  $x^2y^2(x^2 \plus{} y^2) \leq 2.$

See also here
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=542177

hello, the given inequality is equivalent to
$x^2y^2(4-2xy)\le 2$
or
$(a-1)(a^2-a-1)\geq 0$
with $a=xy$
this is true since we have
$\frac{x^2+y^2}{2}\geq xy$
this is eqivalent to
$xy\le 1$
Sonnhard.

The inequality is analogous to proving $x^3y^3(4-3xy)\leq 1$ as we are given $x+y=2$ with $x,y$ being positive reals.
Note that because $x+y=2$, then by A.M.-G.M. we have $xy\leq 1$.
So $x^3y^3\leq xy\leq 1$.
Now $x^3y^3(4-3xy)\leq xy(4-3xy)=-3(xy-\dfrac{2}{3})^2+\dfrac{4}{3}\leq 1$. And we are done!
The last inequality follows from the constraint that $0<xy\leq 1$.

Use jensens function is x3(2-x)6 by applying jensens this simplifies in one line to less than or equal to 2

If true, we will discover a new interesting ineq.

[quote=Grotex]Is $x^ny^n(x^n+y^n)\leq 2$ true for any integer $n$ when $x,y\geq 0$ and $x+y=2?$[/quote]
http://www.artofproblemsolving.com/community/c6h542177p3128531

Yes the general case can be proved by again applying jensen's inequality with the function as x^2n(2-x)^n

See also here:
http://www.artofproblemsolving.com/community/c6h176919
http://www.artofproblemsolving.com/community/c6h219238

[quote=Fermat -Euler]If $x$, $y$ are positive reals such that $x + y = 2$ show that $x^3y^3(x^3+ y^3) \leq 2$.[/quote]
If $x$, $y$ and $z$ are positive reals such that $x + y +z = 3$ . [url=http://www.artofproblemsolving.com/community/c6h439750p2478234]show that[/url] $x^3y^3z^3(x^3+ y^3 +z^3) \leq 3 .$

I have known one generalization from Dr. Tran Nam Dung. But first you can do some to guess the generalization:
Let $n$ is a positive integer and $x, y$ are positive reals such that $x + y = 2$. Prove that:
a) $(xy)(x^2 + y^2) \le 2$
b) $(xy)^{3}(x^3 + y^3) \le 2$
c) $(xy)^{6}(x^4 + y^4) \le 2$
d') Can you guess now?
d) $(xy)^{10}(x^5 + y^5) \le 2$

If $x$, $y$ are positive reals such that $x^3y^3(x^3+ y^3) =2$ show that $$x + y \geq2$$
$$\iff$$
Let $a,b>0 $ and $ab(a+b)=2.$ Prove that $$\sqrt[3]a+\sqrt[3]b\geq 2$$
Solution of DAVROS:
 Let $2x=\sqrt[3]{a}+\sqrt[3]{b}, y=\sqrt[3]{ab}$ then $ab(a+b)=2 \iff y^3(8x^3-6yx)=2 \iff 4x^3y^3=3.xy^4+1$

Then AM-GM yields $4x^3y^3 \ge 4\sqrt[4]{x^3y^{12}} = 4x^{\frac34}y^3 \implies x\ge1$
[url=http://www.artofproblemsolving.com/community/c6h542249p3132090]h[/url] [url=http://www.artofproblemsolving.com/community/c6h241650p1329763]h[/url] [url=https://artofproblemsolving.com/community/c752559h1768870p13546719]h[/url] [url=https://artofproblemsolving.com/community/c4h2821329p25891293]h[/url]

Solution: We rewrite the $x^3y^3(x^3+ y^3) \leq 2$ as $x^3y^3(x+ y)(x^2-xy+y^2) \leq 2$ so we need to prove that $x^3y^3(x^2-xy+y^2) \leq 1$. But,
$x^3y^3(x^2-xy+y^2) \leq 1$ is equivalent to $x^3y^3((x+y)^2-3xy) \leq 1$, then $x^3y^3(4-3xy) \leq 1$ so we need to prove that $4x^3y^3 \leq 1+3x^4y^4$ which is AM-GM on numbers $x^4y^4,x^4y^4,x^4y^4, 1$

Yes we can prove that in general x^Sn-1Y^Sn-1(X^N+Y^n)<=2 with the condition that X+Y=2 by jensen's equality set function as A^Sn (2-A)^Sn-1 this is less than equal to 2F(1) which is precisely 2! Note that Sk denotes the sum of first k natural numbers

This inequality may turn out be useful we can in general prove that If X+Y=k for some nk in N atleast the fact that XY^Sn-1 (X^n+Y^n)<=2((n/2)^(2Sn-1+n))

And I used Lagrange Multipliyers....too easy?

This is my solution:
$x+y$=$2\geq2({xy})^{1/2}$ which is equivalent to $1\ge xy$ or $1\ge x^{3}y^{3}$...(1)
Now on doing cube of $x+y$ and solving we get $x^3+y^3 = 8-6xy$, thus we nee to prove $x^3y^3(4-3xy)\leq 2$.
Now $3\geq 3xy$ on subtracting by 4 we get $1\geq (4-3xy)$...(2)
On multiplying (1) &(2) the result follows

[hide=Solution]
Let \(p=xy\). Then,
\begin{align*}
x^3y^3(x^2 - xy + y^2) &\leq 1 \\
\iff (1-p)(p^2 + p + 1 - 3p^3) &\geq 0
\end{align*}
which is clear.
[/hide]

[quote=Fermat -Euler]If $x$, $y$ are positive reals such that $x + y = 2$ show that $x^3y^3(x^3+ y^3) \leq 2$.[/quote]

[Hide=one-liner]
$t=xy$
$$t^3(4-3t)\leq 1\Leftrightarrow (t-1)^2(2t^2+(t+1)^2)\geq 0$$[/hide]
On a second thought, is the condition of positive reals necessary? Because even if they are allowed to be negative, the inequality will still be true

Sorry for reviving old thread but my [hide = solution]
First of all note that $8 = (x+y)^3 = x^3 + y^3 + 3xy(x+y) = x^3 + y^3 + 6xy$
Applying AM-GM on the terms $x^3 + y^3,2xy,2xy,2xy$ we get $$2 = \frac{(x^3 + y^3) + 2xy + 2xy + 2xy}{4} \geq \sqrt[4]{8x^3y^3(x^3+y^3)}$$
This simplifies to $$\boxed{x^3y^3(x^3 + y^3) \leq 2}$$

[/hide]

Let $a_1,a_2,\cdots,a_n>0$ and $a_1+a_2+\cdots+a_n=n$$ (n\ge 2) .$ Show that $$a_1a_2\cdots  a_n(a_1^2+a_2^2+\cdots+a_n^2) \leq n$$
$$a^3_1a^3_2\cdots  a^3_n(a_1^3+a_2^3+\cdots+a_n^3) \leq n$$
Let $a,b,c,d >0$ and $a+b+c+d=4.$ Show that $$abcd(a^2+b^2+c^2+d^2) \leq 4$$
$$a^3b^3c^3d^3(a^3+b^3+c^3+d^3) \leq 4$$
[url=https://artofproblemsolving.com/community/c6h360304p12685646]h[/url] [url=https://artofproblemsolving.com/community/c6h542514p3129761]h[/url] [url=https://artofproblemsolving.com/community/c6h542177p3125665]h[/url]


trigonometry
calculus
derivative
calculus computations

\[ \begin{gathered}
  f\left( x \right) \equal{} \sin \left( {\frac{{\cos x}}
{x}} \right) \hfill \\
  f'\left( x \right)? \hfill \\
\end{gathered}\]

Solve
Im solved it, but solution its too long
halp pls

Let, $ f(x)=\sin (g(x))$, $ g(x)=\frac{\cos x}{x}$.
So,
\begin{align*}
f'(x)
&= \frac{df}{du} \ \frac{du}{dx} \\
&= \cos u \ \frac{-\sin x - x \cos x}{x^2} \\
&= \cos \left(\frac{\cos x}{x}\right) \ \frac{-\sin x - x \cos x}{x^2} \\
&= -\frac{1}{x^2}\cos \left(\frac{\cos x}{x}\right) (\sin x + x \cos x) \\
\end{align*}

I'm not sure why so many people write "derivate;" it's not a word.  The correct term is "differentiate."  (I'm not targeting  you, I've seen it too many times).

Use the chain rule.  It's not pretty but it works:
Pull out the derivative of $ \frac{\cos{x}}{x}$, which itself requires the quotient rule:
$ \frac{\minus{}x\sin{x}\minus{}\cos{x}}{x^{2}}$.
Therefore, continuing with the chain rule,
$ f'(x)\equal{}\minus{}\frac{x\sin{x}\plus{}\cos{x}}{x^{2}}\cos\left(\frac{\cos{x}}{x}\right)$.

Kouichi, check your quotient rule.

Sorry.
I'm typo.

$ \left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}$.
So,
\begin{align*}
f'(x)
&= \frac{df}{du}\ \frac{du}{dx} \\
&= \cos u\ \frac{-x\sin x-\cos x}{x^{2}} \\
&= \cos\left(\frac{\cos x}{x}\right)\ \frac{-x\sin x-\cos x}{x^{2}} \\
&= -\frac{1}{x^{2}}\cos\left(\frac{\cos x}{x}\right) (x\sin x+\cos x)
\end{align*}


Euler
limit
number theory open
number theory

Wow! how ignorant one can be... or how missinformed. I had been always told that $\gamma$ was for sure, not only irrational but transcendental. Nevertheless, doing a small research along the web, it seems like both constants are not known to be irrational, those are open problem!!! Does any one wants to give it a try?

In the book "Concrete Mathematics" by Graham, Knuth, Patashnik, they say that calculations by Richard Brent have shown that, if Euler's constant is rational, its denominator must be more thatn 10,000 decimal digits long.  Therefore nobody believes it is rational, but nobody so far has been able to prove that it is irrational.


4*5=20
4+5=9
20-9=11
I M SURE ITS
NOT WRONG
UNTIL I
SUBTRACT IT
WRONG! :blush:

[quote="MCrawford"]What is the positive difference between the product of the two numbers 4 and 5 and their sum?[/quote]
[hide="solution"]
Product = 20
Sum = 9
20 - 9 = [b]11[/b]
[/hide]

[quote="MCrawford"]What is the positive difference between the product of the two numbers 4 and 5 and their sum?[/quote]

[hide]4 times 5=20

4 plus 5=9

20 minus 9 equals [b]11[/b][/hide]

lol, these seem like the problems you come up with when u literally cant think of anything else

[hide](4*5) - (4 + 5)=   20-9  = [b]11[/b][/hide]

[quote="math92"]oops!  forgot to use hide/spoiler tags!!!  SORRY!!!!!!!!!!!!!![/quote]

it doesn't really matter because youre like the 10th post.


a square with a quarter circle inside...

that is a right (iscoseles) triangle to the top right

what is the hypotenus of that triangle given that the side of the square is $4+4\sqrt{2}$?

[hide]Draw diagonal from center of quarter circle to opposite vertex. Length of that is:

$\sqrt{2}*(4+4\sqrt{2})=8+4\sqrt{2}$

Length of the segment without the radius of the quarter circle:

$8+4\sqrt{2}-4-4\sqrt{2}=4$

The length of the hypotenuse is twice that, so $8$.

[/hide]

[hide]The area of a parabolic segment is $4/3$ times the area of the triangle with the same base and height. Therefore, $(4+4\sqrt{2})^2-\pi (4+4\sqrt{2})^2/4=2/3x^2$. Now we find

$(4+4\sqrt{2})^2-\pi (4+4\sqrt{2})^2/4=2/3x^2$
$16(1+\sqrt{2})^2(1-\pi/4)=2/3x^2$
$16(1+\sqrt{2})^2\pi=8x^2$
$\sqrt{2\pi}(1+\sqrt{2})=x$

So $x\sqrt{2}=2\sqrt{\pi}(1+\sqrt{2})$

Nvm, [color=red]THIS IS WRONG![/color] :blush: [/hide]

[hide="solution"]
from the centre of the circle to the tangent point the line segment measures $4+4\sqrt{2}$
if we drop a perpendicular to the side below the perpendicular=$4+2\sqrt{2}$
from the point of tangancy if we drop a perpendicular to the neareast side(left side) the perpendicular=$2\sqrt{2}$
now it becomes obvious that $\frac{1}{2}$ the hypotenose is $2\sqrt{2}*\sqrt{2}=4$
So, the hypotenuse is $8$
[/hide]


how do i find the asymptote of the following (if it exists):

$ y \equal{} 1 \minus{} e^x$

You can find about asymptotes [url=http://en.wikipedia.org/wiki/Asymptote]here[/url].

This particular curve has asymptote $ y\equal{}1$ since it converges to $ 1$ when $ x \rightarrow \minus{}\infty$

We know that the graph of $ y\equal{}e^x$ has an asymptote at the x-axis.

It follows from this, that the graph of $ y\equal{}\minus{}e^x$ also has an asymptote at the x-axis.

It then follows that $ y\equal{}\minus{}e^x\plus{}1$ has an asymptote at the line $ y\equal{}1$.

In general, $ y\equal{}a^x$ has a horizonal asymptote at $ y\equal{}0$, for all $ a$.


HMMT geometry calculus quadratics algebra polynomial MATHCOUNTS

Who's going?

I'm taking geometry and combinatorics =]
How about you guys?

I'll be taking the Algebra and Combinatorics.  Geometry's too hard  :P

I'll be there. I'm taking Algebra and Combinatorics as well. There is no number theory for some reason...

Yes, algebra and geometry are easy for me.

algebra + calculus

a lot of algebra + combinatorics here.  :|

wait what is algebra like? is general test easy

[quote="kyyuanmathcount"][quote="worthawholebean"]Calculus and Geometry.

The most exciting subjects.[/quote]

I contend that geometry is exciting only for those good at it :( hence, me doing geo is like staring at the paper hoping it will solve the problem for me.

Darn, with so many people doing Algebra and Combo, the weightings are going to be bad.[/quote]

Well, maybe some people should take Calc then  :lol: I heard that is the lowest number.

[quote="worthawholebean"]I'm not particularly good at geometry, but I like it.[/quote]
This statement is my favorite statement of all time, especially when it's true.

I'm going, and I'm taking the General test. I'm going to take the General test for the rest of my life and mock Daesun Yim. :P

[quote="andersonw"]I'm taking algebra and calculus, because I can't do geometry for my life, and I can't count for my life either.
Come to think of it, I can't really do algebra or calculus too, but at least I'll have some idea what to do on those tests. :P[/quote]

Similar to my rationale to taking those two tests, as I found that 80% of the easy problems in the gut round that I cannot do involve geo or counting.

egh I'm pretty good at geometry; it's def. the best in how you don't need to know anything to get a problem right

and combinatorics is the second most fun  :P

I suck at algebra-my knowledge ends at the quadratic formula -_-

alg + geo? maybe I should like, learn something about polynomials before I go

or I could learn calculus but that's pretty unlikely

[quote="t0rajir0u"]Everyone should come to HMMT :)  I plan on giving a lunch lecture on the Fibonacci numbers.[/quote]

This also sounds like a good idea!
I would love to meet you.
Where will you be hosting your lunch lecture?

I still suggest that aops people meet somewhere... (see the other post)

[quote="t0rajir0u"]Everyone should come to HMMT :)  I plan on giving a lunch lecture on the Fibonacci numbers.[/quote]

Oh my goodness, lunch [i]and[/i] Fibonacci?! Please tell me when/where.

Geometry / Combinatorics for me

I fail at solving polynomial equations under the crunch of time. I would do Calculus but I also don't know series well enough.

[quote="t0rajir0u"]Whoops, I meant a mini-event lecture, I guess; it'll be right after lunch.  The details should be on the HMMT website at some point.[/quote]
awesome i would love to meet such a prolific poster haha

Calculus is actually one of the better tests to take, because if you're fairly competent with the majority of calculus, you will get a score within a stone's throw of the top 10 or even in the top 10. The new scoring system also helps, because I think the calculus scores are artificially deflated, since you have a bunch of kids who are not extremely serious competitors in standard school tracks taking the least objectionable options (to them): Algebra and Calculus.

Me? I'm taking the general because I'm ok at everything, but far from good at anything.  Better to fail moderately at everything then to fail miserably at one thing, I guess.  Anyway, we at AAMC have found that the more people take the general test over the subjects, the higher our team score will be, since everyone can get at least some problems right.

t0rajir0u, your autograph, my forehead, saturday.

I think that most (non-AoPSer) people would take Algebra and Geometry?

I'm taking Algebra + Combinatorics. HMMT's gonna be fun. Especially meeting 'famous people' :lol:

yeah thats what im taking, alg and geo, famous, i need autograph sir !!

i wish i could get people to sign my aops shirt but looks like ill have to live with them signing my mathcounts shirt

Are there anything special that will be sold tomorrow?

I'm goin tomorrow. =D

I'm on the same team as NittanyLion.

Hope I'll see you there guys.

Algebra/Calculus.

[quote="t0rajir0u"] I plan on giving a lunch lecture on the Fibonacci numbers.[/quote]

I was familiar with all of the proofs of Binet that you presented, but I liked the last part of the talk that synthesized them. Thanks for the lecture!

i wanted to go to the fibonacci lecture but i got lost and went to the rubiks cube thing instead is there anyway i can obtain a written form of the lecture or something?

The slides from the lecture are available on my blog (link in signature).  My apologies for the room-change; it was very last-minute, but the room I was originally assigned didn't have a computer projector.


Here's an inequality I made. I hope it's original... please tell me if it isn't.

Given that [tex]a,b,c[/tex] are positive reals, prove that:

[tex]
\\
\displaystyle\frac{a^2}{3^3}+\displaystyle\frac{b^2}{4^3}+\displaystyle\frac{c^2}{5^3} \geq \displaystyle\frac{(a+b+c)^2}{6^3}
[/tex]

Hm. Cauchy anyone?

[tex]\displaystyle \left(\frac{a^2}{3^3}+\frac{b^2}{4^3}+\frac{c^2}{5^3}\right)\left(3^3+4^3+5^3\right) \ge(a+b+c)^2[/tex].

Since [tex]3^3+4^3+5^3=6^3[/tex], dividing by [tex]6^3[/tex] finishes up.

wow, that was a lot easier than my solution...

holder on the LHS, [tex]3^3+4^3+5^3[/tex], [tex]a+b+c[/tex], then cubing and dividing by [tex]6^3(a+b+c)[/tex].  Well ok... my solution wasn't much longer  :)

it is trivial we have by cauchy's inequality
[tex](3^3+4^3+5^3)(a^2+b^2+c^2)>=(a+b+c)^2[/tex](alt+x)
but [tex] 3^3+4^3+5^3=6^3[/tex](alt+x)
and we are done


With him getting in the office again soon I thought it would be funny to take out good 'ol [url=http://www.dubyaspeak.com]dubyaspeak[/url] for some laughs. Since that website gonna have four more years of speechs.  :D

lol he said "liberty" 14 times in his speech. not that i counted. dan rather did.

That site sucks.  I wasted an entire night laughing at the audio instead of doing my homework, now I've got a lot of catching up to do.

[quote="Valentin Vornicu"][quote="theone853"]tsk tsk.... we've gone to blaming the president for us not doing our homework....[/quote]how far can you get? :) I blamed the martians in 3rd grade, so at this point i've gona only about 80,000,000 km or so with the blame :D[/quote]

sorry, what do you mean? I do not understand.

oh right, now I get it.

yeah, is liberty like the new word everybody emphasizes? liberty this, liberty that even Newark Airport is now Liberty Airport

Liberty, democracy, freedom, and all of those words that the American political culture loves so very much.


It takes Jesse 4 days longer to prepare an order of pies that it takes James. Together they baked pies for 2 days and then James left. Jesse took 7 additional days to complete the order of pies. Working alone, how long would it have taken James to prepare the pies?

Please help me :(

Kindly show the representation, working equation and the solution.

THANKS :D

Over and out.

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=254862[/url]


I need a good (free, online) tutorial for Visual Basic 6.  Know where I can find one?  Also, I think it would be good to have a programming subforum.  Do we already have something like that?

Well, i think this section is fine! but if u want to be in contact of more programmers then u can go to acm.uva.es. there u will find the world greatest programmers :D  And u will enjoy practicing there (I hope) BTW, there they deal with contest programming like ACM or IOI. OK?

Mahbub

You can try [url]http://www.vbtutor.net/vbtutor.html[/url]

For practice and examples, you can go to [url]http://www.vb-helper.com[/url]  ;)

Is there a classroom (non-"internet") program that offers VB or programming lessons?  I know Duke TIP does something like that, but I don't think I would pay $3,000.00!

yes,
This is one example
[url]http://www.downseek.com/download/11404.asp[/url]
it's a demo, but I believe you can find the full version elsewhere---such as using Music Planet or BitComet ;)

Thanks, I'll check that out.  But what I meant was a summer course, such as a class from ... MathPath  Know of anything like that?

Thanks

If you've got the time and the patience , try using MSDN...It isn't tutorial type , but then it's got resources that you . well , more than need.....


search
combinatorics unsolved
combinatorics

The square 0 ≤ x ≤ n, 0 ≤ y ≤ n has (n+1)2 lattice points. How many ways can each of these points be colored red or blue, so that each unit square has exactly two red vertices?

Things that would have improved the quality of this post:
1) Having done a search for the problem: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=985187362&t=156356
2) Putting the problem source somewhere in the post.
3) Using LaTeX.

I also posted a follow-up problem there that no one has answered:
how many ways are there to color the vertices with 4 colors so that each unit square has one vertex of each color?


algebra
polynomial
function
search
algebra proposed

Find the polynomial satisfy $ f(x^2\plus{}1)\equal{}f(x)^2\plus{}1$

The solution set consists of all polynomials of type $ f(x)\equal{}x\plus{}b$. Clearly, all polynomials of this type fulfill the equation. Conversely, let $ f$ be a solution. Then $ f\not\equal{}0$, therefore write $ f(x)\equal{}\sum_{v\equal{}0}^na_vx^v$ with $ n\equal{}deg(f)$ and $ a_n\not\equal{}0$. If $ n>1$ we calculate $ 1\equal{}f(x^2\plus{}1)\minus{}f(x^2)\equal{}na_nx^{2(n\minus{}1)}\plus{}q(x)$ with $ deg(q)<2(n\minus{}1)$, a contradiction. Therefore, we can write $ f(x)\equal{}ax\plus{}b$, which immediately yields $ a\equal{}1$.

Unfortunately,

$ f(x^2 \plus{} 1) \equal{} x^2 \plus{} (b\plus{}1)$

while

$ f(x)^2 \plus{} 1 \equal{} x^2 \plus{} 2bx \plus{} (b\plus{}1)$

so actually the only such function that works is $ f(x) \equal{} x$.

[quote="hjbrasch"]The solution set consists of all polynomials of type $ f(x) \equal{} x \plus{} b$. Clearly, all polynomials of this type fulfill the equation. Conversely, let $ f$ be a solution. Then $ f\not \equal{} 0$, therefore write $ f(x) \equal{} \sum_{v \equal{} 0}^na_vx^v$ with $ n \equal{} deg(f)$ and $ a_n\not \equal{} 0$. If $ n > 1$ we calculate $ 1 \equal{} f(x^2 \plus{} 1) \minus{} f(x^2) \equal{} na_nx^{2(n \minus{} 1)} \plus{} q(x)$ with $ deg(q) < 2(n \minus{} 1)$, a contradiction. Therefore, we can write $ f(x) \equal{} ax \plus{} b$, which immediately yields $ a \equal{} 1$.[/quote]

I think you made an error : you must compute $ 1 \equal{} f(x^2 \plus{} 1) \minus{} f(x)^2$ and not $ 1 \equal{} f(x^2 \plus{} 1) \minus{} f(x^2)$

And by the way, also $ p(x)\equal{}x^2\plus{}1$ works...

A sequence of polynomial satisfy is $ f_0(x)\equal{}x,f_{n\plus{}1}(x)\equal{}f_n(x)^2\plus{}1$

[quote="tdl"]Find the polynomial satisfy $ f(x^2 \plus{} 1) \equal{} f(x)^2 \plus{} 1$[/quote]

I remember a general problem on this forum a couple of months ago
however I could not find it with the search function..  :blush:

It stated something like:
Find all $ P$ polynomial such $ P(Q(x)) \equal{} Q(P(x))$ (in this case $ Q(x) \equal{} x^2 \plus{} 1$ ) for $ Q$ another polynomial.

Sorry, I misread the problem as $ 1\equal{}f(x^2\plus{}1)\minus{}f(x^2)$


Help me to find Atbin Doroodchi!
send me message if you know about him. :wink:

What?!

This is almost certainly not the right forum to post this.

Actually, I think this belongs in the Alabama forum.

[quote="b-flat"]I'll just have to take your word for it that this belongs in the Alabama forum, because I have no idea.

Moved to Alabama.[/quote]

Well, I just googled. There were only 2 results, one was an Alabama Math Contest, the other I dunno.

To "despammatize" this topic:

Atbin Doroodchi is hiding somewhere on an infinite number line. He starts out at some given number and is moving at some constant integer speed in some direction. (For example, starting out at the number 5 and moving two numbers to the left per minute.) You can only check one place per minute. Find a strategy to find him in a finite amount of time.

:D

Now seriously, though, who is he?

Um...I think the whole "help me find him" idea is basically the problem

Hey, If U R Yashar Kiarashi, I will b happy to cantact with u, because i have really missed My old school :(

U m these topics are from 2006 why do you feel the need to dig them up?


$1)$ Prove the quadratic formula.

$2)$ Factor: $49x^{2}-316x+132=0$

[hide="#1"]
take our standard form of the quadratic equation and complete the square.
$ax^{2}+bx+c = 0$
$x^{2}+\frac{bx}{a}=-\frac{c}{a}$
$(x+\frac{b}{2a})^{2}= \frac{b^{2}-4ac}{4a^{2}}$
$x+\frac{b}{2a}= \frac{\sqrt{b^{2}-4ac}}{2a}$

$\boxed{x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}}$
[/hide]

[hide="#2"]$(49x-22)(x-6)$

Note: This is my thought process as I am factoring the quadratic.

$49 = 7^{2}$
$132 = 2^{2}\cdot 3 \cdot 11$

We have thus the prime factors $2^{2}\cdot 3 \cdot 7^{2}\cdot 11$ to split into two numbers that add to $316$. Both numbers cannot be odd, so both must be even. 7 does not divide 316 so both factors of 7 must be in one of the numbers. A bit of trial and error reveals that the two numbers are $2 \cdot 3 \cdot 7^{2}= 294$ and $2 \cdot 11 = 22$. With this information we can get the factoring above.[/hide]


linear algebra
matrix
linear algebra unsolved

Suppose $ A_{N \times N}$ and $ B_{N+1 \times N+1}$ are two symmetric positive definte matrix such that: $ B= \begin{bmatrix}& & & & b \\ & A & & & b \\ & & & &b \\ b& b& ...&b& b^{'}\end{bmatrix}$\\
Let use define operator $ SUM$ for a matrix $ G_{k \times k}$ such that:
$ SUM(G)=\Sigma_{i=1}^{k}\Sigma_{j=1}^{k}G_{ij}$ then:
$ SUM(B^{-1})=\frac{SUM(A^{-1})+b^{'}-2bb^{'}SUM(A^{-1})}{1-b^{2}b^{'}SUM(A^{-1})}$
wher$ G^{-1}$ means inverse of the matrix $ G$

Hi apahliani.
Your formula is false for $ b=0$.


function
number theory unsolved
number theory

[quote="Akashnil"][quote="Euclid's Turtle"]Can't just by observation say $ f(x) \equal{} x$ ?[/quote]Then most functional equations wouldn't qualify to be in competitions.[/quote]
For completeness, one should write "Determine [i]all[/i] functions...."

Yea tahts what I was getting at. He didn't mention that we should find $ all$ of them.

This equation seems to be in fact "classical subject" (for 9th graders, obviously).
Check, for example, [url=http://projectpen.files.wordpress.com/2008/01/pen01s.pdf]this article[/url], page 2.

[quote="kihe_freety5"]determine f:N → N with
a) f(mn)=f(m)f(n) with (m,n)=1
b) f(m)>f(n) with m>n
c) f(2)=2[/quote]
In Russian forum we give solution for
1) exist N, suth that $ f(m,n)\equal{}f(m)f(n)$ if $ (m,n)\equal{}1, m>N,n>N$.
2) $ f(m)\ge f(n)$.
then exist k: $ f(x)\equal{}x^k$.

I am sorry :(
It is " Determine all functions"

Determine all function f:N → N with
a) f(mn)=f(m)f(n) with (m,n)=1
b) f(m)>f(n) with m>n
c) f(2)=2


inequalities
geometry
inequalities proposed

Let a,b,c be sidelengths, S be area and $ r_{a},r_{b},r_{c}$ be the radius of escribed circles of an triangle ABC
Prove that
$ 2S(ar_{a} \plus{} br_{b} \plus{} cr_{c}) \plus{} (abr_{a}r_{b} \plus{} bcr_{b}r_{c} \plus{} car_{c}r_{a}) \geq 24S^2$

hello, using
$ r_a\equal{}\frac{S}{s\minus{}a}$
$ r_b\equal{}\frac{S}{s\minus{}b}$
$ r_c\equal{}\frac{S}{s\minus{}c}$
we get after simplifying and factorizing
$ \frac{4 \left(7 a^3\minus{}6 a^2 b\minus{}6 a^2 c\minus{}6 a b^2\plus{}15 a b c\minus{}6 a c^2\plus{}7 b^3\minus{}6 b^2 c\minus{}6 b c^2\plus{}7 c^3\right)}{(\minus{}a\plus{}b\plus{}c) (a\plus{}b\minus{}c) (a\minus{}b\plus{}c)}\geq0$
setting
$ a\equal{}y\plus{}z,b\equal{}x\plus{}z,c\equal{}x\plus{}y$ in the inequality above we get
$ \frac{x^3\plus{}3 x^2 y\plus{}3 x^2 z\plus{}3 x y^2\minus{}21 x y z\plus{}3 x z^2\plus{}y^3\plus{}3 y^2 z\plus{}3 y z^2\plus{}z^3}{x y z}\geq0$
and this is true by AM-GM.
Sonnhard.


geometry
trapezoid
rhombus
geometry unsolved

What are the quadrilaterals which have  3 equal sides and 2 equal angles??

No ideas?
Must be a trapesoid or a square.

What about a rhombus?
Answers: rhombuses, isosceles trapezoids.


abstract algebra
linear algebra
matrix
number theory
greatest common divisor
superior algebra
superior algebra unsolved

Let $A$ be the matrix of a homomorphism $\phi: \mathbb{Z}^{n}\to\mathbb{Z}^{m}$ between free modules.  Prove that $\phi$ is surjective if and only if the gcd of the determinants of the $m\times m$ minors of $A$ is 1.

The few ideas I've had have led me to dead ends.  Can anyone give me some insight?  Is one direction of the implication easier than the other?  Thanks!

It's a particular case of http://www.mathlinks.ro/Forum/viewtopic.php?t=124137 problem [b]c)[/b], but another question is whether I am going to have time to post my proof... You could try to figure it out from the hints by Perfect_radio and Harazi...

Or, what is probably more useful, you could check out some good algebra books, as I am most likely not the first to come up with the result...

  darij


Βάζω μια εξαιρετικά δύσκολη αλλά και υπέροχη άσκηση ......
Έστω $P$  ένα πολυώνυμο με πραγματικούς συντελεστές .Βρείτε όλες τις συναρτήσεις από το R->R έτσι ώστε να υπάρχει πραγματικός $t$  έτσι ώστε

$f(x+t)-f(x)=P(x)$ για κάθε πραγματικό αριθμό $x$

Σιλουανέ  η άσκηση  είναι  αρκετά  καλή   ;)  ;) .......Μου  φαίνεται  οτι πιο  έυκολο  είναι  να  ζητηθεί  το  t   ωστε  να  προσδιοριστεί  το  P  παρα  να  βρεθεί  η  f .Για  παράδειγμα  αν    $P(x) = 2x+ 1$  τότε  προφανώς   $f(x) = x^2 + x$  και   t =1  . Μπορούμε  λοιπόν  γενικότερα  να  δείξουμε  οτι  για  κάθε   πολυώνυμο  δευτέρου  βαθμού  υπάρχει  ενα  τρίτου  καθώς  και  ένα  t  ώστε  να  ισχύει  δεδ0μένη  σχέση $P(x) = f(x+t) - f(x)$
    Τώρα  όσο  για  την  απόδειξη  .............Στην  αρχή  νόμιζα  οτι  την  βρήκα ........... τώρα  την  έχω  αποδείξει  μονάχα  για  ειδικές  περιπτώσεις   όπως  έγραψα  παραπάνω . Τώρα  για  πολυώνυμο  ν  βαθμού  μπορεί  να  ισχύει  κάτι  αντίστοιχο

........Για  ότι   νεότερο  θα  γράψω .........

Έχω την εντ'υπωση ότι δουλεύεις αρκετά καλά και η ψιλοεπαγωγική λύση που πας να κάνεις μπορεί να σε οδηγήσει σε καλό δρόμο ......  :D
Απλά πρέπει να προσέξεις κάτι για το $t$ . ;)

Μου είπαν αρκετοί ότι την προσπάθησαν χωρίς αποτέλεσμα .Βάζω μια λύση και την αφιερώνω στο Μέγιστο Γιώργο   :winner_first:
Προφανώς αν το $P=0$  τότε  για $t=0$  και για κάθε συνάρτηση από το R->R  ισχύει το ζητούμενο .έστω  $P(x)=\sum_{k=0}^na_kx^k$  με $a_n\neq 0$  και  $t\in R^{*}$.Και ψάχνουμε για ένα πολυώνυμο $Q_t(x)=\sum_{k=1}^{n+1}b_kx^k$
τέτοιο ώστε $Q_t(x+t)-Q_t(x)=P(x)$  με x πραγματικό.
Κάνοντας εξίσωση συντελεστών έχουμε :
$\binom{n+1}{1}tb_{n+1}=a_n$
$\binom{n+1}{2}t^2b_{n+1}+\binom{n}{1}tb_n=a_{n-1}$
......

$\binom{n+1}{n+1}t^{n+1}b_{n+1}+......+\binom{1}{1}tb_1=a_0$
Παρατηρούμε ότι το σύστημα έχει μοναδική λύση , άρα υπάρχει μοναδικό πολυώνυμο
$Q_t$ ώστε $Q_t(x+t)-Q_t(x)=P(x)$  με  $Q_t(0)=0$  και
$deg\ Q_t=1+deq\ P$
Tέλος θέτουμε $g(x)=f(x)-Q_t(x)$ .
Τότε η $f$  ικανοποιεί το ζητούμενο  αν και μόνον αν $g(x+t)-g(x)=0$  για κάθε  πραγματικό x δηλαδή η $g$  να είναι περιοδική με περίοδο  $t$.
Άρα  $f(x)=Q_t(x)+g(x)$  όπου $g$  μια οποιαδήποτε περιοδική συνάρτηση με περίοδο $t$

 :lol: . Ελπίζω να σας άρεσε............


[quote="2424"]Let $ a,b,c$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} \frac {4}{3}$ Prove that
  $ \frac {(a \minus{} 1)^2(b \minus{} 1)^2}{(a \plus{} b \minus{} 1)^2} \plus{} \frac {(b \minus{} 1)^2(c \minus{} 1)^2}{(b \plus{} c \minus{} 1)^2} \plus{} \frac {(c \minus{} 1)^2(a \minus{} 1)^2}{(c \plus{} a \minus{} 1)^2} \geq 1/3$[/quote]

Okay.It's easy
Let:$ p\equal{}a\plus{}b\plus{}c;q\equal{}ab\plus{}bc\plus{}ca$
We have:$ p^2\minus{}2q\equal{}\frac{4}{3}$
By cauchy-swarchz
$ LHS \ge \frac{(q\minus{}2p\plus{}3)^2}{2q\minus{}4p\plus{}\frac{17}{3}}$
We need prove that:
$ \frac{(q\minus{}2p\plus{}3)^2}{2q\minus{}4p\plus{}\frac{17}{3}} \ge \frac{1}{3}$
Because:$ q\equal{}\frac{p^2\minus{}\frac{4}{3}}{2}$ So:
Ineq
$ <\equal{}>(3p^2\minus{}12p\plus{}14)^2 \ge 12p^2\minus{}38p\plus{}52$
$ <\equal{}>[3(p\minus{}2)^2\plus{}2]^2 \ge 12(p\minus{}2)^2\plus{}4$
$ <\equal{}>9(p\minus{}2)^4 \ge 0$
It's always true :)

We have $ a^2\plus{}b^2\plus{} \frac{(a\minus{}1)^2(b\minus{}1)^2}{(a\plus{}b\minus{}1)^2} \geq 1$ :P

[quote="2424"][quote="WiZ@Rd"][quote="2424"]We have $ a^2 \plus{} b^2 \plus{} \frac {(a \minus{} 1)^2(b \minus{} 1)^2}{(a \plus{} b \minus{} 1)^2} \geq 1$ :P[/quote]

Can you prove that, please?[/quote]
Let  $ x \equal{} \frac {(a \minus{} 1)(b \minus{} 1)}{(a \plus{} b \minus{} 1)}$
$ \equal{} > ab \plus{} bx \plus{} xa \equal{} a \plus{} b \plus{} x \minus{} 1$
$ \equal{} > a^2 \plus{} b^2 \plus{} x^2 \equal{} (a \plus{} b \plus{} x \minus{} 1)^2 \plus{} 1 \geq 1$  :P[/quote]

you made small mistake.
you wrote
ab + bx + xa = a + b + x - 1
a^2 + b^2 + x^2 = (a + b + x - 1)^2 + 1
but it's
ab-ax-bx=a+b-x-1
a^2 + b^2 + x^2=(a+b-x-1)^2+1

[quote="amature fermat"][/quote][quote="2424"][quote="WiZ@Rd"][quote="2424"]We have $ a^2 + b^2 + \frac {(a - 1)^2(b - 1)^2}{(a + b - 1)^2} \geq 1$ :P[/quote]

Can you prove that, please?[/quote]
Let  $ x = \frac {(a - 1)(b - 1)}{(a + b - 1)}$
$ = > ab + bx + xa = a + b + x - 1$
$ = > a^2 + b^2 + x^2 = (a + b + x - 1)^2 + 1 \geq 1$  :P[/quote][quote="amature fermat"]

you made small mistake.
you wrote
ab + bx + xa = a + b + x - 1
a^2 + b^2 + x^2 = (a + b + x - 1)^2 + 1
but it's
ab-ax-bx=a+b-x-1
a^2 + b^2 + x^2=(a+b-x-1)^2+1[/quote]
iam sorry!
$ x= \frac{(a-1)(b-1)}{1-a-b}$


I don't quite understand the solution on the AMC Packet.

Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters?

A) 250
B) 300
C) 350
D) 400
E) 500

PUT THE EXPLANATIONS!!! I don't want the post saying "A," B," or something like that. I need a good explanation. ;)

[hide="Hint"]Remember they are running in opposite directions.  At the first meeting, what is the total distance run by Brenda and Sally?  At the second meeting, what is the total distance run by Brenda and Sally?[/hide]

Brenda ran 100 meters before the first meet.

Sally run 150 meters between the first and second meet.

(Those are given.) ;)

Hmm...The answer key said that the sum of two people's distance before the first meet = the half of the total distance. But I have no clue on why or how that's like that...hm.m.m.

Diametrically opposite points divide the track in half. So the combined distance they travel before the first point is half the track's length.
The distance they travel between the first and second points is the total distance.
So Brenda runs 200 meters between the first and second points.
200+150=350.

[hide]since the start diametrically opposite each other... the total distance they run between meeting is: half the length of the track... so Sally runs 150 meters and brenda runs 200 meters.... so the track is 350 meters long.
[/hide]


[quote="mdk"]The Morse code consists of points anddashes. How many letters can be in the Morse code if no letter contains more than four signs, but all must have at least one?[/quote]
what do you mean by letter?

and cant there be inifite, how many different individual sings are possible?

[hide="Solution"]
For the one symbol case: two letters.
For the two symbol case: fix the first letter (2 choices) and fix the second letter (2 choices) = 4 choices.
For the three symbol case: fix the first letter (2 choices), fix the second letter (2 choices), fix the third: 8 total letters.

For the four symbol case: it's just 2^4.

The sum is $ 2+2^{2}+2^{3}+2^{4}= 30$ letters, right?
[/hide]

you could also say it's $ 2^{n+1}-2$, where n is the number of signs.


floor function
number theory unsolved
number theory

for $0<a<1$ assume $A=\{\lfloor \frac{n}{a}\rfloor\}_{n\in\mathbb{N}}$
and  $B=\{\lfloor \frac{n}{1-a}\rfloor\}_{n\in\mathbb{N}}$

prove the disjoing sets $A$ and $B$ toghether will form every number in $\mathbb{N}$

This is obviously not true when $a = 1/2$. In fact the conclution is true if and only if the number $a$ is irrational. Indeed, assuming $a$ is irrational this statement is equivalent  to a beautyful theorem by Beatty which states that if $\alpha$ and $\beta$ are positive irrational numbers such that $1/\alpha + 1/\beta = 1$, then the sequences $\{a_n\}$ and $\{b_n\}$, where $a_n = [\alpha n]$ and $b_n = [\beta n]$, form a partion of the set of natural numbers.

[quote="Solar Plexsus"] Indeed, assuming $a$ is irrational this statement is equivalent  to a beautyful theorem by Beatty which states that if $\alpha$ and $\beta$ are positive irrational numbers such that $1/\alpha + 1/\beta = 1$, then the sequences $\{a_n\}$ and $\{b_n\}$, where $a_n = [\alpha n]$ and $b_n = [\beta n]$, form a partion of the set of natural numbers.[/quote]

where can i get the proof??!?

For a proof of Beatty's theorem, check the following link:

[url]http://planetmath.org/encyclopedia/ProofOfBeattysTheorem.html[/url]


What is the value of $ x$?
\[ \frac{20}{50} \equal{} \sqrt{\frac{20}{x}}\]

$ \dfrac{20}{50}$ can be represented as $ \dfrac{2}{5}$.

square the both side.

you get $ \dfrac{4}{25}\equal{}\dfrac{20}{x}$.

so 4x=500.

so x=125.

answer : 125.