Does anyone know of any math (math- not science) programs in the NYC metro area? I am very interested in Math, am a junior, and am a girl. I applied for the columbia summer program "A Survey for Modern Mathematics" but that is only a month (july), and I would like to do more- during the schoolyear and August, perhaps.

Any ideas?
thanks for any help

Hi there.  I'm a Stuyvesant graduate, now a sophomore in college, and I was very involved in mathematics in the city while I was in high school.  Where are you in school?

One program I can suggest is the CCNY Summer Scholars program, which I've been involved in for the last 4 summers as a student and a teaching assistant.  It runs through July and August, so you wouldn't be able to attend both it and the Columbia program, unfortunately, but if you aren't accepted at the Columbia program, it would be a good choice.

Also, during the year, there are a variety of programs offered through different colleges (including, but not limited to, NYU and CCNY) that allow high school students to take college-level classes free or at low cost.

I'm not really sure what kind of thing you're interested in -- do these sound like the right type of program?  The New York City Math Team, which you can qualify for via various competitions, is also a wonderful group that is worth trying to find a place on, if competitions are something you like.

If you have any questions about the programs I've mentioned, I'd be happy to give you as much information about them as I have.

I HIGHLY recomend going to a program such as HCSSIM.

http://www.hcssim.org

instead of wasting your money away at a program such as Columbia's Summer School.

I went to both programs and, in retrospect, feel that both were important in my development in math problem solving skills and in my personality as well. I would say that the CCNY program is more introductory while HCSSiM was more intense. Of course, my view is biased because I went to CCNY after freshman year and did not have enough experience for the more advanced classes.
Both are in July though, and the HCSSiM application is out.


Given an equilateral tringle $ ABC$ how can we inscribe a [color=red][b]not regular [/b][/color]hexagon $ DEFGKL$ (there is not unique )  with equal sides, as in the figure ?

[b]NOTE[/b]

This problem arises with the IMO - 2005

Babis

( This problem has been discussed in a greek math forum)

It is easier to do the opposite: to construct a hexagon with equal sides and correct directions of lines containing three sides. You can do three adjacent sides two of which have prescribed directions in any way you want (within reason), after which the problem reduces to finding 2 points $ A$ and $ B$ on two given circles with the prescribed vector $ \overline{AB}$, which is achieved by translating one of the circles by this vector and finding the intersection point with another circle:
[asy]size(200);
pair A=(0,0), B=A+dir(60), C=B+dir(20), D=C+dir(-60);
D(A--B--C--D, brown+linewidth(1));
pair E=IP(D(shift((1,0))*D(CR(A,1)),red),D(CR(D,1)),1),
F=E-(1,0);
D(D--E--F--A, green+linewidth(1));
D("A",E); D("B",F);[/asy]

Happy Holidays Season to all my friends !

  M.T. (Ret. Maj.)


[quote="sri340"]f(x) = 3− 2x , find f −1(5).

A. 3    B. – 1      C. 2     D. 1    E. none of these[/quote]

Do you mean $ f^{\minus{}1}(5)$?

First, compute $ f^{\minus{}1} (x)$.

$ x \equal{} 3 \minus{} 2 f^{\minus{}1}(x)$

$ x \minus{} 3 \equal{} \minus{}2 f^{\minus{}1}(x)$

$ f^{\minus{}1}(x) \equal{} \frac{3 \minus{} x}{2}$

$ \therefore f^{\minus{}1}(5) \equal{} \frac{3 \minus{} 5}{2} \equal{} \boxed{\minus{}1}$.

So the answer is [b]B[/b].

This way is easier:
Substitute $ 5$ for $ f(x)$ to get $ 5\equal{}3\minus{}2x$.

You could also solve it rather quickly in your head by noting that $ 3\minus{}2x$ goes through the $ y$ axis at $ 3$ and slopes downward, which means a value of $ 5$ (which is greater than $ 3$) must occur before the line $ x\equal{}0$. Clearly, the only feasible answer is the negative one.


function
integration
calculus
trigonometry
calculus computations

Find the function $ f (x)$ if it satisfies  the condition  $ f(x) \equal{} x \plus{} \int_{0}^{\pi/2}\sin(x \plus{} y)f(y)\,dy$

Differentiate twice under the integral sign to get that $ f''(x)+f(x)=0.$ That means that we can write $ f(x)=A\cos x+B\sin x$ for some constants $ A$ and $ B.$

Expand out $ \sin(x+y)(A\cos y+B\sin y)=(\sin x\cos y+\cos x\sin y)(A\cos y+B\sin y)$ and integrate, getting
\begin{align*}A&=\frac12A+\frac{\pi}4B\\B&=\frac{\pi}4A+\frac12B\end{align*}
The only solutions of this are $ A=B=0,$ so the only solution to the original problem is that $ f$ is the zero function.

sorry kent i jus typed the question wrong.please have a look at it now in the 1st post.Sorry :wallbash:

So now the differential equation is $ f''(x)+f(x)=x,$ which has the general solution $ f(x)=x+A\cos x+B\sin x.$

We can compute that $ \int_0^{\frac{\pi}2}y\sin(x+y)\,dy=\cos x+\left(\frac{\pi}2-1\right)\sin x.$ Combine that with my previous work to get

$ x+A\cos x+B\sin x=x+\left(\frac12A+\frac{\pi}4B+1\right)\cos x+\left(\frac{\pi}4A+\frac12B+\frac{\pi}2-1\right)\sin x$

This leads to the system
\begin{align*}\frac12A-\frac{\pi}4B&=1\\-\frac{\pi}4A+\frac12B&=\frac{\pi}2-1\end{align*}
If I solved that correctly (not a sure thing), I get $ A=\frac{-8+4\pi-2\pi^2}{\pi^2-4},\ B=\frac{8-8\pi}{\pi^2-4}$ so that

$ f(x)=x+\frac{(-8+4\pi-2\pi^2)\cos x+(8-8\pi)\sin x}{\pi^2-4}.$


Calculate: $ {n \choose 1}^2\plus{}2.{n \choose 2}^2\plus{}3.{n \choose 3}^3\plus{}...\plus{}n.{n \choose n}^2$

I dont get it .does all the terms have power 2 or other power ?

[hide]
First, we show that $ \displaystyle\sum_{k\equal{}0}^n \binom{n}{k}^2\equal{}\binom{2n}{n}$:

Consider $ \binom{n}{k}^2$.  This is equal to $ \binom{n}{k}\binom{n}{n\minus{}k}$.  Now suppose we have $ n$ boys and $ n$ girls and we want to make a group of $ n$ boys and girls.  Then $ \binom{n}{k}\binom{n}{n\minus{}k}$ is the number of ways to choose $ k$ boys and $ n\minus{}k$ girls.  If we sum this over all $ 0\le k\le n$ then we have the number of ways to choose $ n$ people out of the total $ 2n$ without regards to the number of boys or girls.  Thus, $ \displaystyle\sum_{k\equal{}0}^n \binom{n}{k}^2\equal{}\binom{2n}{n}$.

Now, let

$ S\equal{}\binom{n}{1}^2\plus{}2\binom{n}{2}^2\plus{}\cdots\plus{}(n\minus{}1)\binom{n}{n\minus{}1}^2\plus{}n\binom{n}{n}^2$

$ \equal{}\binom{n}{n\minus{}1}^2\plus{}2\binom{n}{n\minus{}2}^2\plus{}\cdots\plus{}(n\minus{}1)\binom{n}{1}^2\plus{}n\binom{n}{0}^2$

Then using these to expressions of $ S$, we find $ 2S\equal{}S\plus{}S\equal{}n\binom{n}{0}^2\plus{}n\binom{n}{1}^2\plus{}n\binom{n}{2}^2\plus{}\cdots\plus{}n\binom{n}{n\minus{}1}^2\plus{}n\binom{n}{n}^2$.  Factoring out $ n$ we see an expression we have seen already.  Using $ \displaystyle\sum_{k\equal{}0}^n \binom{n}{k}^2\equal{}\binom{2n}{n}$ we find that

$ 2S\equal{}n\binom{2n}{n}$

Thus, $ S\equal{}\frac{1}{2}n\binom{2n}{n}$[/hide]


If in a triangle $ ABC$ with fixed base $ BC$,the vertex $ A$ move such that $ \cos{B}\plus{}\cos{C}\equal{}4\sin^{2} {\frac{A}{2}}$ then prove that $ b\plus{}c\equal{}2a$

[hide="Solution"]A bit too lazy to find a geometric solution right now... recall that
\[ \cos{B} = \frac {c^2 + a^2 - b^2}{2ca},\quad\cos{C} = \frac {a^2 + b^2 - c^2}{2ab},\quad\sin\frac {A}{2} = \sqrt {\frac {(s - b)(s - c)}{bc}},\]so\begin{align*}\cos{B} + \cos{C} & = 4\sin^2\frac {A}{2} \\
\frac {bc^2 + a^2b - b^3}{2abc} + \frac {a^2c + b^2c - c^3}{2abc} & = 4\frac {a(a + b - c)(c + a - b)}{4abc} \\
(a + b - c)(c + a - b)(b + c) & = 2a(a + b - c)(c + a - b) . \\
\end{align*}The result follows from the Triangle Inequality.[/hide]

[quote="Maths Mechanic"]If in a triangle $ ABC$ with fixed base $ BC$,the vertex $ A$ move such that $ \cos{B} \plus{} \cos{C} \equal{} 4\sin^{2} {\frac {A}{2}}$ then prove that $ b \plus{} c \equal{} 2a$[/quote]

[hide="my solution"]$ cosB \plus{} cosC \equal{} 2(1 \minus{} cosA)$

$ \Leftrightarrow \frac {a^2 \plus{} c^2 \minus{} b^2}{2ac} \plus{} \frac {a^2 \plus{} b^2 \minus{} c^2}{2ab} \equal{} 2(1 \minus{} \frac {b^2 \plus{} c^2 \minus{} a^2}{2bc})$

$ \Leftrightarrow \frac {a^2(b \plus{} c) \plus{} bc(b \plus{} c) \plus{} (b \plus{} c)( \minus{} b^2 \minus{} c^2 \plus{} bc)}{2abc} \equal{} 2.\frac {a^2 \minus{} (b \minus{} c)^2}{2bc}$

$ \Leftrightarrow \frac {(b \plus{} c)[a^2 \minus{} (b \minus{} c)^2]}{2abc} \equal{} \frac {a^2 \minus{} (b \minus{} c)^2}{bc}$

$ \Leftrightarrow b \plus{} c \equal{} a$[/hide]


inegalitatea este urmatoarea:

(1+x1)(1+x2)....(1+Xn) mai mare sau egal decat (1+ radical de ordinul n din x1*x2*...*xn) ^n; xk pozitive k=1,n

cum se demonstreaza?

[color=brown]Vezi (la pagina 467) exercitiile 104 si 107 ([u]inegalitatea Huygens[/u]) in cartea
"[b][u]Analiza matematica pentru clasa a XI - a, Editura Teora, 1999, Bucuresti[/u][/b]".[/color]

[color=darkblue]Nici nu mai exista in librarii si nici in stocul Editurii Teora. Am cautat-o si eu. Daca totusi o gasesti, cumpara si pentru mine cel putin un exemplar (exista m.p. pe MLS). Vezi la biblioteca sau la ... un bun prieten si/sau coleg.[/color]

ideea ce imi vine la prima vedere ar fi asta...
[hide="solutie"]
intai demonstrez pentru $n=2$. demonstrez e mult spus, ca se verifica imediat.
apoi, prin inductie dupa $k$, arat pentru $n=2^{k}$. (folosesc ipoteza de inductie si inegalitatea pentru $n=2$).
in final, apelez la "trucul" de a lua $k$ astfel incat $2^{k}\geq n>2^{k-1}$, si, folosind punctul precedent, iau variabilele $x_{1},x_{2},\ldots,x_{n}$ si apoi completez cu $x_{n+1}=x_{n+2}=\ldots=x_{2^{k}}=\sqrt[n]{x_{1}x_{2}\ldots x_{n}}$. aplicand punctul anterior, o sa mi se reduca exact atatea paranteze cate trebuie.
sper sa fie bine ce am zis
[/hide]

Are si o forma mai generala:
[quote="Old and New Inequalities, page 123"]
Daca $a_{1}, \ldots, a_{n}, \, b_{1}, \ldots, b_{n}, \, p_{1}, \ldots, p_{n}> 0$ si $p_{1}+\ldots+p_{n}= 1$, atunci $\prod_{i=1}^{n}\left( a_{i}+b_{i}\right)^{p_{i}}\geq \prod_{i=1}^{n}a_{i}^{p_{i}}+\prod_{i=1}^{n}b_{i}^{p_{i}}$.
[/quote]
Ideea de rezolvare e una foarte frumoasa:

Daca inlocuim $\left( a_{1}, b_{1}\right)$ cu $\left( k a_{1}, k b_{1}\right)$, atunci obtinem o inegalitate echivalenta, deci putem presupune "WLOG" ca $a_{1}+b_{1}= 1$. De asemenea, vom lua $a_{2}+b_{2}= 1, \ldots, a_{n}+b_{n}= 1$.
Inegalitatea pe care trebuie s-o demonstram devine $1 \geq \prod_{i=1}^{n}a_{i}^{p_{i}}+\prod_{i=1}^{n}b_{i}^{p_{i}}= S$.
Dar din medii avem $\prod_{i=1}^{n}a_{i}^{p_{i}}\leq \sum_{i=1}^{n}p_{i}a_{i}$ si $\prod_{i=1}^{n}b_{i}^{p_{i}}\leq \sum_{i=1}^{n}p_{i}b_{i}$, deci $S \leq \sum_{i=1}^{n}p_{i}\left( a_{i}+b_{i}\right) = \sum_{i=1}^{n}p_{i}= 1$.

o alta solutie, nu e mare diferenta, luam $\sum{\frac{a_{i}}{a_{i}+b_{i}}}\geq \frac{\sqrt[n]{a_{1}...a_{n}}}{\sqrt[n]{\prod{(a_{i}+b_{i})}}}$ din medii.

acum analog $\sum{\frac{b_{i}}{a_{i}+b_{i}}}\geq \frac{\sqrt[n]{b_{1}...b_{n}}}{\sqrt[n]{\prod{(a_{i}+b_{i})}}}$, si adunandu-le rezulta chiar inegalitatea.

EDIT: de fapt e la fel.

Dar inegalitatea asta se rezolva direct din Holder..(Nu o sa arate a solutie :) ):
\[\left(1+a_{1}\right)\left(1+a_{2}\right)\cdots\left(1{}_{a}{}_{n}\right)\geq\left(1+\sqrt[n]{a_{1}\cdot a_{2}\cdots a_{n}}\right)^{n}\]


:ninja: ok im a noob in math  :blush:  i want to be able to answer the questions we do in math enrichment faster lols so could anyone post some choosing and permutation probs...and any random problems lol tx :lol:

check around, there are literally hundreds of problems here :)

Heres one that have:
Bob job is to put house numbers up on people's doors.  Bob has the numbers 3, 4, and 9.  How many house numbers can he make if each house number is three digits long and he can reuse house numbers?


inequalities
quadratics
conics
parabola
calculus
algebra
inequalities unsolved

Hey everyone, I have an EXTREMELY important question, and it's driving me mad. the 2007 exams are available here, http://www.studyjapan.go.jp/en/toj/toj0307e.html to check out and download/print to practice on. My question relates to the ~Math(A)~ Exam.

Question 1 of the math exams states this:

The solution of the inequality 2x^2 - 3x -2 [is less than or equal to] 0 is:

____ [is less than or equal to] "X" [which is less than or equal to] _______

Okay, so my question comes in here. According to the answer key, the answer should be this:
1/2 [is less than or equal to] "X" [which is less than or equal to] 2.

However, no matter what I do, I cannot get this under any circumstances. This is a straightforward, fairly easy inequality. Why then am I getting NEGATIVE 1/2 <= X <= 2?

Solving it by factoring gives me this as solutions:

2x^2-3x-2=0
(2x+1)(x-2)=0.
(2x+1)=0 ; (x-2)=0
2x=-1 ; x=2
x=-1/2

So, x=-1/2, and x=2.

Solving by the quadratic equation gives me this as a solution:

x= -b [pus or minus] sqr(b^2 - 4ac)
all over 2a.

x= 3 [plus or minus] sqr(9 - 4*2*-2)
all over 2*2

x= 3 [plus or minus] sqr(25)
all over 4.

x= 3 [plus or minus] 5
all over 4

x=3+5 ; x=3-5
over 4 ; over 4

x=8/4 ; x=-2/4
-----> { x=2 ; x=-1/2 } <-----

So, as you can see, both of those ways work out to solve for x as -1/2 and 2. not 1/2 and 2, but [negative!] 1/2 and 2.

I even took out my calculator to triple check this equation, and on the graph that came up (used a TI-83) The points where the parabola crossed the x-axis were at NEGATIVE 1/2 and 2.

So, To quadrouple check, I went in to my calculus teacher, and asked her what she got as a solution. My CALCULUS teacher also got NEGATIVE 1/2 and 2 as solutions, and drew a number line to check it as well. No matter what I do, I am not getting the POSITIVE 1/2 and 2 that they are. Maybe it was a mistype on the answer key, I can't think of much else it could be.

PLEASE WRITE BACK, MAYBE GIVE ME SOME INSPIRATION BY TELLING ME THERE WAS SOMETHING ALL OF US MISSED THAT THE MATH(A) EXAM WANTED US TO FIND!!!

You are correct; it was probably a typo.


do you mean that like ab=2? then it would be solvable.

look again, i edited it.  thanks for pointing that out though.  now try again.

[hide="hint"]$you need to find (a+b)^{2}.$[/hide]

[hide="further hint"]Then factoring $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$.[/hide]

ummmm....that would be 1

so is the answer $(a+b)^{2}+(a^{2}+b^{2})$?

you need to find ab with $(a+b)^{2}-(a^{2}+b^{2})$

just tell me the answer and then show how you got it.

ummm... how do you multiply 3 binomials together. :oops:  :blush:

[quote="rd5493"]How do you solve problems like this?

If $a+b=1$ and $a^{2}+b^{2}=2$ what is the value of $a^{3}+b^{3}$?  Express your answer as a common fraction.[/quote]
[hide]So you have $a+b=1$. Square this equation to get $a^{2}+2ab+b^{2}=1$. Since $a^{2}+b^{2}=2$, substitute it back in to get that $2ab+2=1$. So $2ab=-1$, and $ab=-\frac{1}{2}$. Then factor out $a^{3}+b^{3}$ to $(a+b)(a^{2}-ab+b^{2}$. The $(a+b)$ part equals 1 so we can forget about that. Substituting everything in we get that $a^{2}-ab+b^{2}=2+\frac{1}{2}$. So our final answer is $\frac{5}{2}$. [/hide]

[quote="superwizard188"][hide]lol, I just took a class on this so I should've known it. xD My way is dumber, but only slightly. :P

(a+b)^3 = a^3 + b^3 + 3a^2b + 3b^2a = 1
a^3 + b^3 = 1 - 3ab(a+b)
We can solve for ab = -1/2 so...

a^3 + b^3 = 1 - 3(-1/2)(1) = 1 + 3/2 = 5/2 :D Same thing, sorta.[/hide][/quote]
Lol, your way is the same way as everyone else's.
Edit: Nvm, I didn't see your first step.


The points $ A(1,4)$, $ B(\minus{}1,4)$ and $ C(\minus{}1,\minus{}4)$ lie in a two-dimensional plane. What is the sum of the coordinates of point $ D$ given that $ ABCD$ is a rectangle?

Not quiet sure how to explain this, but the coordinate is $ D(1,\minus{}4)$, so our sum is $ \boxed{\minus{}3}$.

Either by visualization or a quick diagram, one can realize that $ A$ is in Quadrant I, $ B$ is in Quadrant II, and $ C$ is in Quadrant III, so $ D$ is in Quadrant IV. Now, since it is a rectangle, $ AB$ and $ CD$ are parallel and equal. So since the distance from $ A$ to $ B$ is $ 2$, the distance from $ C$ to $ D$ is also $ 2$. This means that $ D$ is the point $ (1,\minus{}4)$, and hence oursum is $ 1\plus{}(\minus{}4)\equal{}\boxed{\minus{}3}$.


The arithmetic mean of $ x$, 50, 53, 21 and 75 is greater than 55 and less than 70. How many possible integer values are there for $ x$?

since it doesnt ask for the actual numbers, just for the number of numbers, we can just perform:

70-55   =   15*5   =   75-2   =   73
highest # - lowest # * 5 (for 70, 70.2, 70.4, etc.) - 2 (for 70 and 55) = 73

meh.
i dont wanna use LATEX.
u gotta problem?  :bomb:

yea, thats what i thought :D


How would you go about finding the point $P$ within a cyclical quadrilateral $ABCD$, such that $AP+BP+CP+DP$ is minimized?

[b]Example:[/b] $AB=2, BC=3, CD=5, DA=4$. If $AP+BP+CP+DP \geq k$, find $k$.

(edit: Ok, so my conjecture doesn't make sense.  :P  )

woa man...it might just be the intersection of the diagonals...WHAT A THOUGHT

this is meh for an arml problem...minimize the sum of the distances from the verticies of a quadrilateral is old...but i guess it is fine

Well if we try to make an analog similar to triangles, the circumcenter of a triangle doesn't minimize $AP+BP+CP$ because this point can be outside of the triangle... The same happens for cyclic quads.

EDIT: Try using vectors and see if you find that the minimum occurs when $P$ is the centroid of the cyclic quad. Just a guess...


probability
complementary counting
conditional probability

In this new kind of poker, people are dealt cards that have numbers from 0 to 9 inclusive.  However, the cards are completely independent of each other.  Also, there are no suits.  Cards go only by their number.  People are dealt five cards each.  You are just as likely to get each card.
EDIT: There are an infinite number of cards.

Compute the probability of a person getting:
1. a pair (like 6-6-2-9-5)
2. two pairs (like 7-7-3-3-8)
3. a three of a kind (like 0-0-0-5-2)
4. a straight (like 3-4-5-6-7 or 5-2-6-3-4)*
5. a full house (a three of a kind and a pair, like 4-4-4-1-1)
6. four of a kind (like 8-8-8-8-4)
7. five of a kind (like 2-2-2-2-2)

*In a straight, the cards need not be in order, but you must be able to rearrange them so that they are in order.  Zeroes can be high or low, but not both in the same hand.  So 0-1-2-3-4 and 6-7-8-9-0 are both straights.  However, 8-9-0-1-2 is not.

So we are assuming there are infinite cards?

[hide="Number Six"]
The probablility of JUST getting the four you need, not five:
The P of getting the four you need is $\frac{1}{10^{4}}$ The probability of the last card not being the same as the first four is $\frac{9}{10}$ The conditional probability that both happen is $\frac{9}{10^{5}}$
Is this correct?
[/hide]

I'm assuming for example a pair is 6-6-1-2-3 and not 6-6-6-1-2 or 6-6-1-1-2.

[hide="1"]
$\binom{5}{2}\cdot 10\cdot 9\cdot 8\cdot 7$
[/hide]

[hide="2"]
$\frac{\binom{5}{2}\binom{3}{2}}{2!}\cdot 10\cdot 9\cdot 8$
[/hide]

All out of $10^{5}$

I got something else for 6:
[hide="6"]Look at A-A-A-A-B, where A and B are distinct.  There are 10 possibilities for A and 9 possibilities for B.  Then we can rearrange that in 5 ways.  So there are $5\cdot10\cdot\9$, or $450$, ways.  So the answer is $\frac{450}{100000}$, or $\frac{9}{2000}$.[/hide]
For 1, I agree with JavaMan's method.

I missed the rearrangement part. I forgot that the order of the four of a kind doesn't matter.

[quote="mathnerd314"]#1 is pretty easy too
[hide]using complementary counting:

Ways to get no pairs: $10 \times 9 \times 8 \times 7 \times 6$
Total number of ways: $10^{5}$

Our answer is $69760$[/hide][/quote]

Incorrect.

eh let's give it a try. Here are my sorutions. *

[hide="1"]
$\binom{5}{2}\cdot 9 \cdot 8 \cdot 7 \Rightarrow \frac{63}{1250}$[/hide]
[hide="2"]
$\binom{5}{1}\binom{4}{2}\cdot 9 \cdot 8 \Rightarrow \frac{27}{1250}$
[/hide]
[hide="3"]
$\binom{5}{3}\cdot 9 \cdot 8 \Rightarrow \frac{9}{1250}$
[/hide]
[hide="5"]
$\binom{5}{3}\cdot 10 \cdot 9 \Rightarrow \frac{9}{1000}$
[/hide]
[hide="6"]
$\binom{5}{4}\cdot 10 \cdot 9 \Rightarrow \frac{9}{2000}$
[/hide]
[hide="7"]
$10$ (probably the hardest one out of all of them)
[/hide]

and here's 4, i kinda couldn't get it back in the numbering for some reason
[hide="4"]
$6 \cdot 5! \Rightarrow \frac{9}{1250}$[/hide]

I think your solutions for 2, 5 and 6 are wrong.

I think that #4 is wrong, too.  I think you forgot that 0 can be high or low.

[hide="4"]There are $10^{5}$ total ways to get the 5 cards.  The sequence can start with 0, 1, 2, 3, 4, 5, or 6, for 7 total sequences.  There are $5!=120$ ways to rearrange the cards for order.

Thus, the probability is $\frac{840}{10^{5}}=\frac{21}{2500}$[/hide]

[hide="#7"]
There are 10000 possible values for 5 cards and 10 possible ways to get a five of a kind, for a probability of $\frac{1}{1000}$[/hide]

[quote="moogra"]Wait, in poker you are allowed to discard up to 5 cards and redraw, is that allowed? I noticed that's not in the rules.[/quote]

That's only one type of poker and since it isn't in the rules, no.

[quote="madskillz1"]so what are the answers?[/quote]
Their posted numerous times(I think). I also know you're new but in the future don't revive old topics.

[quote="madskillz1"]The ANSwers are not RIGHT! and they are differnet[/quote]
Lots of people got the same answer for some of the questions.


A lot of this forum is dedicated to "what college admissions officers care about." So, I have some MORE questions, regarding this, that I'd like to present for discussion.
1. "Do college admission officers care" about your weighted GPA? That is to say, really, if you have a high B in an AP class, is this any better or about equal to having a high A in a regular class? On a weighted GPA, they will come out exactly the same (as a 4.0) but unweighted? Not a pretty picture.
2. "Do college admission officers care" about your job? That is, do they care about what kind of job you've had, and how long you've held it, etc? For example, if one has held a job in the field which one desires to enter as a career, is this "more desirable" than holding a job for the strict purpose of acquiring money? Can a job, in essence, work as an "extracurricular" because one is using it as a learning tool?
3. "Do college admission officers care" about random volunteering? I mean to say, if, in addition to regular volunteer work, one also does small "side jobs" when one is bored, is this seen as even mentionable? How does one cite this on a resume?

Ok, that should get us started.

The brief answer to your question: yes, probably.

The slightly longer answer:  Imagine you were a college admissions officer.  Consider what you are looking for in a student.  (If, at this stage, you have an image of yourself in mind, you should start over.)  There are lots and lots of qualities that make an individual stand out, and only some of them are academic.  Each college probably has its own system for interpreting GPAs, weighted and unweighted.  Your jobs and volunteering are iteresting because they give a picture of what you will be like as a member of a university community, beyond the time you spend in the library studying.

I think what you, and most other people asking such questions, really mean is, "How much are my accomplishments really worth?" since it should be obvious that they are worth [i]something[/i].  And that's the question that is incredibly hard to answer: two different colleges, or even two different officers at the same college, might value various activities and academics and so on very differently.  You might as well put down every last thing you've done in high school on your applications, not because one of them will trigger an instant "admit this student" response in admissions offices around the country, but instead because they will help build up a larger positive image of you that will tend to increase your chance of being viewed as a good person to accept.


function
analytic geometry
calculus
derivative
topology
real analysis
real analysis theorems

Why do books state that the notion of an inverse function is lost as the number of variables exceed one? Wouldn't the inverse of a multivariable function be a coordinate point with the same number of ordered points as the number of variables?
BTW, does anyone know ways to find inverses of single variable functions through a numerical approach? Comparable to how certain functions that can't be integrated through elemental functions are integrated by their Taylor expansion.
Thank you for your time, these are questions I cant seem to find the answer through the internet =E

*feel free to throw me any topic I'm afraid I've started math too late and i want to learn =D

[quote="ManuelS"]Why do books state that the notion of an inverse function is lost as the number of variables exceed one?...[/quote]

Which books say this? Generally speaking, there are one-to-one functions $ f: U \subseteq \mathbb{R}^2 \to \mathbb{R}$. But if $ U$ is open (or contains an open set), they are not continuous. Such one-to-one functions do, however, have inverse functions from $ V \subseteq \mathbb{R} \to \mathbb{R}^2$.

[quote]BTW, does anyone know ways to find inverses of single variable functions through a numerical approach?...[/quote]

One way to "find" the inverse of a function, $ f$, numerically is to numerically solve the equations $ f(x) \equal{} c$ for a bunch of different values of $ c$. This will give you a sampling of (approx) values of $ f^{ \minus{} 1}$.

What i meant is that there is no functional inverse for a function of two or more variables - it was in some text introducing multivariable functions, i can't recall the name
[b]What I am wondering, is whether there are "partial inverses" for function of more than one variable, comparable to partial derivatives[/b].
Let's say that we have a function of two vars z=f(x,y), and let's assume that it can be put in the form x=g(z,y) and we plug in g into f. So we have z=f(g(z,y),y) and the function is only expressed in two variables. if we solve again for "z" we would have a function defined only by "y". Wouldn't the inverse of z(y) be a partial inverse, in terms of "y"?
idk am just wondering...it would suck if it canceled into an identity at "z=f(g(z,y),y)"

oh, and that approach you told me to "finding" inverses I kind of figured, I was wondering whether there was something more elegant...i guess if you had some points and used some interpolation curve it would be nice..=D

[quote="ManuelS"]What i meant is that there is no functional inverse for a function of two or more variables - it was in some text introducing multivariable functions, i can't recall the name[/quote]

I don't know what this means. Any function which is one-to-one, has an inverse function. There are one-to-one multivariable functions, so it's possible for multivariable functions to have inverse functions. Maybe they said something about continuous functions from $ \mathbb{R}^2$ to $ \mathbb{R}$ not having inverse functions?

[quote]What I am wondering, is whether there are "partial inverses" for function of more than one variable, comparable to partial derivatives.[/quote]

If you take a multivariable function, say $ f(x,y)$, and look at it for fixed $ y$ - it's just like a single variable function of $ x$. So suppose you define $ f|_y(x) : \equal{} f(x,y)$ for fixed y. Then if $ f|_y: x \mapsto z$ is one-to-one, it will have an inverse function $ g|_y: z \mapsto x$.

[quote]Let's say that we have a function of two vars z=f(x,y), and let's assume that it can be put in the form x=g(z,y) and we plug in g into f. So we have z=f(g(z,y),y) and the function is only expressed in two variables. if we solve again for "z" we would have a function defined only by "y". Wouldn't the inverse of z(y) be a partial inverse, in terms of "y"?
idk am just wondering...it would suck if it canceled into an identity at "z=f(g(z,y),y)"[/quote]

By the definition of what an inverse is, you will get an identity, namely z=z.

[quote="ManuelS"]
[b]What I am wondering, is whether there are "partial inverses" for function of more than one variable, comparable to partial derivatives[/b].
[/quote]
Try to google "implicit function theorem"
[quote]
Let's say that we have a function of two vars z=f(x,y), and let's assume that it can be put in the form x=g(z,y) and we plug in g into f.
[/quote]
Apparently, you'll get a simple identity (try yourself with $ f(x,y)\equal{}x\plus{}y$) :P

yeah  :blush: it did cancel out.. it was just an idea...=E well you guys are the bomb, thanx for all the help!! XD


calculus
inequalities proposed
inequalities

If $ a$ and $ b$ are positive real numbers, without using calculus, prove that $ (a\plus{}b\plus{}3)^{3} > 8a^{2}(b\plus{}2)$

Let $ b\plus{}2\equal{}c$
We have to prove that:
$ (a\plus{}c\plus{}1)^3>8a^2c$
Let $ a\equal{}d^3$ and $ c\equal{}e^3$

$ d^3\plus{}e^3\plus{}1>2d^2e$
Let $ d\equal{}6$; $ e\equal{}4$
 $ d^3\plus{}e^3\plus{}1\equal{}216\plus{}64\plus{}1\equal{}281<288\equal{}2d^2e$

Thank you for this counter example Most probably I missed some condition ( this is problem from my math teacher so I will ask him). Sorry for this mistake.


function
integration
inequalities
real analysis
real analysis unsolved

$f \in L^{p}(\mathbb{T})$ where $1<p \leq 2$. Prove that there exists a constant $C_{p}$ (that depends only on $p$) that satisfies:
$\sum_{k \neq 0}\frac{|\hat{f}(k)|^{p}}{|k|^{p}}\leq C_{p}\| f \|_{p}^{p}$

Isn't it rather $\parallel f\parallel _{p}^{p}$? It's not homogeneous as stated. In this case, I think (but I'm not sure) it's useful to note that the fourier coefficients of $\int_{0}^{x}{f(t)dt}$ are closely related to $\frac{c_{k}(f)-c_{k}(0)}{k}$.

Ok, my method does not seem to work...  :blush:  :(

Assume that $1<p<2.$ The $p=2$ case is easier.

Let $q$ be the dual exponent to $p.$ That is, $\frac1p+\frac1q=1.$ Note that $p<q.$

By the Hausdorf-Young inequality, $\|\widehat{f}\|_{q}\le\|f\|_{p}.$

Now just use Hölder's inequality:

$\sum\frac{|\widehat{f}(k)|^{p}}{|k|^{p}}\le \left(\sum|\widehat{f}(k)|^{q}\right)^{\frac{p}{q}}\left(\sum\frac1{|k|^{pq/(q-p)}}\right)^{1-\frac{p}{q}}$

Check the exponent of $|k|$ in the denominator of that last sum.

$\frac{pq}{q-p}=\frac{p\cdot\frac p{p-1}}{\frac p{p-1}-p}= \frac p{2-p}$

But $\frac p{2-p}>1$ for all $1<p<2,$ so that sum converges to a number that depends only on $p.$

And the Hausdorf-Young inequality shows that

$\left(\sum|\widehat{f}(k)|^{q}\right)^{\frac{p}{q}}= \|\widehat{f}\|_{q}^{p}\le\|f\|_{p}^{p}.$

And we are done.

P.S. Harazi - I also tried looking at the antiderivative of $f$ (assuming WLOG that $\widehat{f}(0)=0).$ That idea didn't lead anywhere for me, either.


search
function
number theory unsolved
number theory

Let $a$ and $b$ be two distinct positive integers, such that their
 Arithmetic Mean, Geometric Mean , Harmonic Mean are all integers.
 find the minimum possible positive value of $\displaystyle\ a - b$ .

Probably search function ;)
(For example "arithmetic geometric harmonic integer" works)

Well, first, I remembered I had seen it.
Second, I searched for it and used [i]exactly[/i] ZetaX's keywords. :)


A polynomial $P$ satisfies $P(n)\in{\mathbb{Z}}$  $\forall n\in{\mathbb{Z}}$. Prove that $P$ has rational coefficients.

Even more can be proved: Every such polynomial is of the form $\sum_{k=0}^n a_k \binom{x}{k}$ with integers $a_k$.

For the idea of the prove see http://www.mathlinks.ro/Forum/viewtopic.php?t=40407 .

For sure, but here Lagrange interpolating polynomials provides a easy and direct answer.


function integration algebra polynomial induction calculus derivative

it is well known that $exp(-\frac{t^2}{2})$ has itself as fouriertransform

but what about other eigenfunctions (allow complex valued functions, with complex eigenvalues?)

is any result known??

my guess is that that gaussian I just gave is about it?

Let's assume you set your constants of $2\pi$ up in appropriate ways as to give yourself a unitary operator. That is, if $\mathcal{F}(f)$ is the Fourier transform of $f,$ then $\|\mathcal{F}(f)\|_2^2=\|f\|_2^2.$ (Call that the Plancherel theorem.)

Then $\mathcal{F}^4$ is the identity operator.

From that we see that the possible eigenvalues of $\mathcal{F}$ are the fourth roots of 1, namely $\pm1,\pm i.$

Each of these is an eigenvalue, and each eigenspace is infinite dimensional. Since we have infinite dimensional eigenspaces, we have free choice about a basis.

A particularly nice basis would be the Hermite functions. Let $f_k(x)=H_k(x)e^{-\frac{x^2}2},$ where $H_k$ is the $k$th order Hermite polynomial.

Then $\mathcal{F}(f_k)=i^kf_k.$

Thank you very much Kent Merryfield but could you please elaborate a bit further
First of all this is my definition of the fourier transform of a function f

$\frac{1}{\sqrt{2 \pi}}\int f(t) e^{\omega t i} dt$

Now you make use of a 2-norm, is this allowed?  Or does existence of transform automatically imply square integrability?

Also could you give me some help (like a hint) how I could prove that those constructions with hermite polynomials leads to eigenfunctions
I assumed this definition was the most appropriate : $f_{n}= \frac{d^{n}( e^{-x^2})}{dx^{n}} e^{-\frac{x^2}{2}}$
but what standard technique to use now (there a lot of fourier tricks that pop up in my mind but none leads to a nice proof, no proof actually, even with induction)

And when you have found your four eigenvalues, how do prove that this really is an infinite basis? How do we know there aren't any others?  Or is this is exceedingly difficult to prove?

That $\frac1{\sqrt{2\pi}}$ out front does make that definition a unitary operator. However, the fact that you have $+i\omega t$ in your exponent instead of $-i\omega t$ should mean that the eigenvalues for the Hermite functions are the conjugates of what I said.

Remember that if you take derivatives of a function, you multiply the Fourier transform by a polynomial; multiplying a function by a polynomial amounts to taking a differential operator of the Fourier transform.

You might also investigate the differential equation that the Hermite functions satisfy. (Same as the one-dimensional time-independent Schrödinger equation for a harmonic oscillator.)

thanks mister merryfield for the hints ( I am very focussed on this problem as our professor asked us to work on these things, to get things straight, it's technically not homework as it is goes completely ungraded)
with

 $f_{n}= \frac{d^{n}( e^{-x^2})}{dx^{n}} e^{-\frac{x^2}{2}}$
I 've managed to find that it satisfies this differential equation :

$f_{n}^{(2)}-x^2 f_{n}+(2 n+1) f_{n}=0$

which is indeed very much like harmonic oscillator equation

now applying the rules with the polynomials (I still used the + with the omega in my fourier definition) I found that my fourier transform also had to satisfy this differential equation

but how does this lead to full determination of my fourier transform, this second order differential equation leads to a twodimensional linear space of equations when i understand correct?

The Fourier transform of a smooth function that decays rapidly at infinity must be a smooth function that decays rapidly at infinity. Imposing the condition of tending to zero at infinity on the solutions of your differential equations should restrict you to a one-dimensional subspace of the solution space. Then there are several possible ways of finding the right normalization within that one dimensional space.


two balls collide elastically on a pool table.  Ball A is moving at 2m/s to the left while ball B is moving at 1 m/s to the right before they collide.  If Ball A is twice as massive as ball B.  What is the velocity of Ball A and Ball B after the collision?

[quote="aadil"]apply momentum and kinetic energy conservation[/quote]

The given link leads to a formula which is derived from the COM and COME.
However, note that the given formula can be used only in perfectly elastic collisions.

I don't know what is COM & COME either, but the velocity just invert their direction, after collision.

[quote="Immanuel Bonfils"]I don't know what is COM & COME[/quote]

Conservation of Momentum and Conservation of Mech Eng.


inequalities
LaTeX
inequalities unsolved

This inequality is supposed to be easy but I could not manage to solve it...

If a, b, c are positive numbers such that abc = 1, prove the inequality

1/sqrt(b + 1/a + 1/2) + 1/sqrt(c + 1/b + 1/2) + 1/sqrt(a + 1/c + 1/2) >= sqrt(2)

( sqrt = square root, sorry but these happens when u dont know lattex)

[quote="erbed"]If $a, b, c$ are positive numbers such that $abc = 1$, prove the inequality

$\frac{1}{\sqrt{b+\frac{1}{a}+\frac{1}{2}}}+\frac{1}{\sqrt{c+\frac{1}{b}+\frac{1}{2}}}+\frac{1}{\sqrt{a+\frac{1}{c}+\frac{1}{2}}}\ge \sqrt{2}$
[/quote]
It is okay?

Nice one,but realy easy :) .

Let that $x/y=a,y/z=b,z/x=c$,then we need to prove that
$\sum{\sqrt{\frac{x}{2y+2z+x}}}\geq 1$
but as you can easy see $\sqrt{\frac{x}{2y+2z+x}}\geq \frac{x}{x+y+z}$=>$\sum{\sqrt{\frac{x}{2y+2z+x}}}\geq \sum{\frac{x}{x+y+z}}=1$
equality holds when $x;y=0$ and $z=1$ :wink:

[quote="SpongeBob"]How can $a=\frac{x}{y}$ when $x,y=0$ ;)[/quote]
When I said about equality case,I forget how I set up x,y,z.In that case there is no equality case and the initial inq is strict! :P

[quote="pohoatza"]simple ineq, but i have an other soln :)

so we have to prove that if $abc=1$ the following ineq holds:

$\sum{\frac{1}{\sqrt{b+\frac{1}{a}+\frac{1}{2}}}}\geq \sqrt{2}$

$\Rightarrow \sum{\frac{1}{\sqrt{\frac{2ab+2+a}{2a}}}}\geq \sqrt{2}$

$\Rightarrow \sum{\frac{1}{\sqrt{\frac{2ab+2+a}{a}}}}\geq 2$

but, from $AM-GM$ we have:
$\sqrt{\frac{2ab+2+a}{a}\cdot 1}\leq \frac{\frac{2ab+2+a}{a}+\frac{a}{a}}{2}$

$\Rightarrow \sqrt{\frac{2ab+2+a}{a}}\leq \frac{ab+a+1}{2a}$

$\Rightarrow \frac{1}{\sqrt{\frac{2ab+2+a}{a}}}\geq \frac{2a}{ab+a+1}$

So it remains us to prove that :

$\sum{\frac{2a}{ab+a+1}}= 2$ , which is trivialy true(amplify the first term with $c$ , and the second with $ac$)

PS: sorry if i made any mistake , i`m really tired.[/quote]

I'm afraid it' s the same solution :wink: ,you just don't let that $a=\frac{x}{y}$ :) ,but this makes your work easier!


Let $ {x_n}$ such that
$ x_0\equal{}2,x_{n\plus{}1}\equal{}\frac{2x_n\plus{}1}{x_n\plus{}2}$
Find $ [\sum^n_{k\equal{}1}x_k]$ where $ [x]$ denote the largest integer not exceeding $ x$

we prove $ x_n>1$
that $ x_n<2\Rightarrow [x_n]\equal{}1$

$ x_n \equal{} \tan \left(\theta _n \plus{} \frac {\pi}{4}\right)$ with $ \tan \theta _n \equal{} \frac {1}{3^n}\ (n \equal{} 0,\ 1,\ 2,\ \cdots)$.

Set $ y_n\equal{}\frac{x_n\minus{}1}{x_n\plus{}1}$ to find the relation of $ y_n$ and $ y_{n\plus{}1}$.


If $ x \equal{} \frac{2}{3}$ and $ y \equal{} \frac{3}{4}$, what is the value of $ (x \plus{} y)(\frac{x}{y})$ expressed as a common fraction?

\begin{align*}
\left(\frac23 +\frac34 \right) \left(\frac{\frac23 }{\frac34 }\right) &= \left(\frac{17}{12}\right) \left(\frac{8}{9}\right) \\
&= \boxed{\frac{34}{27}}
\end{align*}


I'm organizing a group of people to help out the Community Dining Room in Branford, CT. It's a soup kitchen and you need to be free Saturdays. If you're interested, reply here and I'll get to you with more information.

I've already got a couple of people who're going to come.

1st and 2nd Saturdays of each month, we'll pick a time.

I was looking on the event page and I just realized I don't take civics nor am I in the '10 class...  :roll:

You don't have to be.

Jenny Wang, Demi Chen, and David Chen all might be coming and they go to Guilford High..

Also, it's easier to explain everything on AIM.

Alternatives:

Downtown Evening Soup Kitchen
Colombus House

Both in New Haven

The lady has not emailed me back yet.

Saturday October 21st
Saturday December 16th
Saturday December 30th.
9am to 2:00pm.

Do you want to carpool?

The 21st is also Homecoming and the dance is at 7 but you won't be able to go to any of the games.

i might be able to go sometime (i live near hartford) and help out (i also need service hours for scouts)
-jorian

do u have more info?
thx


vector
linear algebra
matrix
geometry
parallelogram
linear algebra unsolved

Everyone knows the geometric meaning of $\det(A)$ is the volume of parallelepipe formed by the $n-$vectors
${A_{i}=\left[\begin{array}{c}A_{1,i}\\ A_{2,i}\\ \vdots \\ A_{n,i}\end{array}\right]}$ where $i\in\{1,\ldots,n\}$

How about the geometric meaning of ${\rm Tr}(A)$  :?:

If $A$ is a projection, that is, $A^{2}=A$, then ${\rm tr}A$ is the dimension of the image of $A$. I don't know about arbitrary matrices.

I was trying to piece together something for which the ingredients possibly included direction fields, differential equations, infinitesimal generators, and the exponential map. But it wouldn't have been simple, and calling it "geometric" would have been a stretch.

:idea:
I usually think of a square matrix as a way of representing a map from $\mathbb R^{n}$ to itself: $\mathbf x\mapsto A\mathbf x$. But sometimes it is better to think of it as a vector field: for every point $\mathbf x$ there is a vector $A\mathbf x$ attached to it. (The difference is that $A\mathbf x$ is interpreted as an element not of $\mathbb R^{n}$, but of its tanget space.) With this interpretation, $\mathrm{tr}A$ is the divergence of the vector field $\mathbf v(\mathbf x)=A\mathbf x$. Does the [url=http://en.wikipedia.org/wiki/Divergence]divergence[/url] have a geometric meaning? I think so.

Take 3 by 3 matrices for example $A=\left[\begin{array}{ccc}A_{1,1}&A_{1,2}&A_{1,3}\\ A_{2,1}&A_{2,2}&A_{2,3}\\ A_{3,1}&A_{3,2}&A_{3,3}\end{array}\right]$ which could be understood as a number of 3 3-vectors
$A_{1}=\left[\begin{array}{c}A_{1,1}\\ A_{1,2}\\ A_{1,3}\end{array}\right]~,~ A_{2}=\left[\begin{array}{c}A_{2,1}\\ A_{2,2}\\ A_{2,3}\end{array}\right]~,~ A_{3}=\left[\begin{array}{c}A_{3,1}\\ A_{3,2}\\ A_{3,3}\end{array}\right]$

(1)  ${\rm Tr}(A)=A_{1,1}+A_{2,2}+A_{3,3}$ is the total sum of the length of the projections of $A_{1},A_{2},A_{3}$ onto the 1st, 2nd and 3rd axis

(2)  $\det\left[\begin{array}{cc}A_{1,1}&A_{1,2}\\ A_{2,1}&A_{2,2}\end{array}\right]$ is the area of the parallelogram bounded by the projection of the vectors $A_{1},A_{2}$ onto the plane between the 1st and 2nd axis
$\det\left[\begin{array}{cc}A_{2,2}&A_{2,3}\\ A_{3,2}&A_{3,3}\end{array}\right]$ is the area of the parallelogram bounded by the projection of the vectors $A_{2},A_{3}$ onto the plane between the 2nd and 3rd axis
$\det\left[\begin{array}{cc}A_{3,3}&A_{3,1}\\ A_{1,3}&A_{1,1}\end{array}\right]$ is the area of the parallelogram bounded by the projection of the vectors $A_{3},A_{1}$ onto the plane between the 1st and 3rd axis

(3)  $\det (A)$ is the volume of the parallelogram formed by the three vectors $A_{1},A_{2},A_{3}$ projected onto the volume bounded by the 1st, 2nd and 3rd axis

and the same geometrical meaning goes to all other dimensions  :)


How many size n combinations of birthdays are possible for n random people?

How many of these combinations contain at least 2 of the same birthdays?

[hide="Part 1"]
Each person can have one of $365$ distinct birthdays, so the total number of combinations is $365^n$.
[/hide]

[hide="Part 2"]
Consider the combinations with all distinct birthdays.  This is simply $365 \times 364 \times ... \times (365 - (n-1)) = 365 P n$ (where $nPr$ denotes the number of permutations of $r$-sized subsets of $n$, or $\frac{n!}{(n-r)!}$), so the number of combinations that have at least two birthdays the same are $365^n - 365 P n$.
[/hide]

The Birthday paradox, huh?   :rotfl:


Hey all,

Could someone find the pattern to this pyramid of numbers (see link)?

http://img501.imageshack.us/img501/4522/photo1mp1.jpg

Thanks! Any help is appreciated.

[hide]This is an old and popular problem.

Read each row out loud, and interpret what you hear as a description of the previous row.[/hide]

where'd you get it from?
or is the whole problem just recognizing a pattern?

oh that was from the book 'Empire of the Ants' by Bernard Werber

i think

This is a common problem. If you haven't heard it before, it is hard to figure out, as this is neither arithmetic nor geometric progression.

Challenge: Explain why 4 can never appear.

[hide="Solution for the extension"]
To get $ 4 \ \ 1$ you need $ 1 \ \ 1 \ \ 1 \ \ 1$ in the previous row but to get that you need $ 1 \ \ 1$ in the row before that.  However, that would have been written $ 2 \ \ 1$ not $ 1 \ \ 1 \ \ 1 \ \ 1$.
Continues for the other numbers.
[/hide]


Consider numbers a1 < a2 < a3 < a4 and b1 < b2 < b3 < b4. Suppose also that a2<b1 and a4<b4. If a1+a2+a3+a4>=b1+b2+b3+b4. In other words say, which the smallest, the second smallest, the third smallest etc

Trivial.

We have $a_1<b_1$ (since $a_1<a_2<b_1$), $a_2<b_2$ (since $a_2<b_1<b_2$) and $a_4<b_4$. If we would have $a_3\leq b_3$, then it would follow that $a_1+a_2+a_3+a_4<b_1+b_2+b_3+b_4$, contradicting the condition $a_1+a_2+a_3+a_4\geq b_1+b_2+b_3+b_4$. Thus, we must have $a_3>b_3$. Hence,

$a_1<a_2<b_1<b_2<b_3<a_3<a_4<b_4$.

This solves the problem.

  darij


[quote="mathsss"]$ a;b;c \ge 0;a \plus{} b \plus{} c \equal{} 1$ prove that
$ \frac {3(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}{(a^2 \plus{} b^2 \plus{} c^2)^3} \plus{} 2(ab \plus{} bc \plus{} ca \minus{} 1)^2 \ge 1$[/quote]
Let $ a\plus{}b\plus{}c\equal{}3u,$ $ ab\plus{}ac\plus{}bc\equal{}3v^2$ and $ abc\equal{}w^3.$
Hence, your inequality is equivalent to $ f(w^3)\leq0,$ where $ f$ is a convex function.
Thus, $ f$ gets a maximal value, when $ w^3$ gets an extremal value, which happens when two numbers from $ \{a,b,c\}$ are equal or when maybe $ w^3\equal{}0,$ which easy to check.


Nice:

Prove that if m and n are positive integers, then 5m + 5n can be written as the sum of two perfect squares if and only if m - n is even.

I think I got it

[hide]

=>

5^a+5^b modulo 8 is always 6 if a-b is odd.

But the sum if 2 squares modulo 8 is never 6.

<=

(a+b)(c+d) = (ac-bd) + (ad+bc) so is enough to prove that 5^n + 5^(n+2k) can be written as the product of the sum of 2 squares.

this is clear : 5^n + 5^(n+2k) = 5^n (1 + (5^k))

now 5 = 1 + 2 so 5^n can be written as the product of the sum of two square (by induction).[/hide]


This is a topic I have always been curious about. so many people have so many different opinions about this particluar subject. Most people believe that media does cause violence, but I care to differ slightly. I think that media can induce violent behavior only if the kid is willing to lets him/herself be affected by the nature of the media program. A person like me for example, wouldn't care if a the Texas Chainsaw dude came and killed someone on T.V.(as long as it doesn't happen in real life - but people somehow think that is "cool" and they want to kill - what I call immature reasoning though it can't be expected of most people), because I really don't care(I'm not interested). However, T.V. violence could start to affect the minds of people who have been watching too much television by itself. Advertisements don't help either.

If America is trying to be a crime free country, shouldn't everyone work together in one effort. If companies advertise guns and beer while promoting peace, it has no effect. They always say the welfare of "our" children are above most things in this country, but I guess money has struck victory again.(this entire paragraph is one side of the debate )

What is your opinion?

I believe that anything that a kid does, watches, reads, or plays could affect on how he/she sees violence. Therfore this leads me straight to the point that nowadays anything could lead to the violence that occurs in childrens minds. You cant deny it everyone has violent thoughts every once in a while, it just comes with the territory of being human. With all the war and murder that goes on in America or with America kids will always find out whats happening no matter what it is through that being movies or books. I think there will always be violence in this world as long as knowledge is present. Thats what causes it, the thing that we use in good ways. Have you ever thought about the people in prison....well of course youve thought of them but have you ever associated them with being smart. The truth is some of the smartest people in the world lurk in America's prison systems. We dont think about that because most dont want to realize that. But its the truth they have plenty of time to think in there considering they are isolated from all of the world except those who are their cell mates if any.

i believe the person chooses to expose himself to the tv.  It is his own fault if he chooses to get violent after exposure.  After all, he exposed himself to the tv.  Anyone with half a brain would know that stuff was either fake (movies) of terrible(news).  Illogical reasoning states
a)that man killed on tv is real.  The movies make money.  Ill make money.
b)the part where the guy gets shot is cool.  Ill do it too, and ill be cool.
c) killing isnt that bad (thinks man on the set is real).  Thats why this movie hasnt been banned from tv.  Ill kill too.

I think that what a kid watches especially th yonger ones has a direct connection to what they think and do.  They also should though realize that the things on T.V. are fake except on the news.   I think it would probally be better if we just didn't watch T.V. since it rots our brains anyway (I know I can't loose any moreI am bad enough at Math already).   That is my opinion though.

There was a similar discussion (focusing on video games, but also including movies and television) [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8482&start=0]here[/url].

My two cents:

It is (in most cases) unlikely that television or movie violence lead directly to actual violence, although there certainly are documented cases of this.  A 17-year-old boy just got out of prison after 3 years in Florida -- he killed a (smaller, younger) playmate of his while immitating moves he'd seen in professional wrestling.
Usually, thought, people do realize the difference between the fictional violence of television and real-life violence.  Movies and television wouldn't necessarily have to "make people do violent things" in order to cause violence, though.  Everyone has in us a certain capacity for violence.  I'm certainly capable of knocking my little brother around a little bit without thinking about it very much.  Thousands of people volunteer for armies all around the world.  Violence is something that humans are very, very good at.  However, most of us have a fairly limited extent to which we realize that capability for violence -- I would recoil at the thought of ever, say, breaking my brothers bones or hitting him badly enough to bruise, even if I push him around.  Similarly, even violent people are fairly reluctant to engage in an unknown world of violence such as shooting or stabbing other people.
But now imagine someone who has watched thousands of murders on television -- the act of murder, of taking another human's life, would seem far less abnormal.  In that circumstance, it would (might) be much easier for that person to actually commit such an act himself.

Mmm, my opinion: I think it really do depend on the age and the maturity of the child. If they are, say 7, then they can and might immitate the TV without thinking. A person who don't know the consequences of the action may also repeat that action seen. However you can't really prove that if a child sees violence on TV and does something violent, then the violence on TV caused it. I mean if we never see violence then would we necessarily be peace-loving good people? How do we even know what is peace if we have never seen violence?
Also is it possible to prevent a child from seeing any violence? Will that actually be good for the child in the long run? (assuming he/she won't go out hurting people) I mean think about it, if a person that naive goes out and sees someone say get punched, they'll get confused, scared, or even repeat it just to understand what that action does to people. Course i doubt this will ever happen but supposing the people working to ban violence actually do ban all violence from TV, then imagine the world afterwards...

My advice to the parents: Talk about violence with your kids when they are old enough to understand before they get miseducated by TV. It's virtually impossible to protect your child from violence, instead try to help them understand and overcome the violence instead.

...*reads over thread* Interesting how none of the people who's discussing this is an adult :lol:

Under most legal measures, I'm an adult.

according to the al franken and other democrats fox is a biased conservative tv. station i always watch nbc with tom brockaw and i only watch the cartoons (simpsons,shaman king) on fox.

Here's an interesting question re: the Superbowl halftime show:

Why are people upset about a woman's breast being shown on TV, when they aren't upset about the act of a man forcefully ripping off a piece of a woman's clothing?

The woman allowed the man to do that. I read in newsweek that when timberlake ripped off the shirt, jackson was supposed to be wearing a red bra.

hmm scrambled, her shirt is apiece of clothing/....

and hwat was that star like thing on her breast?

i don't get what all the fuss is about. they have nudity everywhere especially in movies (from pg-13 to nc-17)

[quote="churchilljrhigh"]i don't get what all the fuss is about. they have nudity everywhere especially in movies (from pg-13 to nc-17)[/quote] except when they have naked women, they dont show the boobs or their reproductal organ.  Besides,  13>kids  watch the superbowl.

i think that violence on tv does affect anyone to a certain extent. as a christian i try to limit the violence i see on tv. i think it affects little kids the most because their minds are so impressionable. it is kind of like cussing around a three year old. they are automatically going to say what you just said regardless of if you just told them not to. the older we get the more we accept things as being "normal" it is normal to hear about murders and suicide on the news. why make up stories about it and put them on tv when we have enough tragedy right here in our own world that really happens. i watched the passion of the christ a couple of months ago for the first time. it is amazing what mel gibson had to do to that movie to get it to reach people. we are so numb to violence that seeing a true event like that doesnt phase us in the least. if we can watch a man being beat with whips with bones on the ends, see a crown of thorns shoved on his head, and watch him be hung on a cross without being moved, this world has gone too far.

I think violence does influence teens.  The way I see it, all the violence exposure gives teens the impression that violence is normal, and socially acceptable.  Althought it probably doesn't always [i]directly[/i] cause violence, I think it does at least have some impact, especially for very young kids (kindergarten age) and teens.

Did you never heard about a kid who jumped from the roof of his home because he wanted to fly as Superman does? well, that is a very extreme case, and I don't think that anyone over, may be, eith years old would try it (and even eight seem to old to try such stupidicity), but in that case is because the kid cannot conscientiously distinguish between fantasy and reality.

After certain age, you can conscientiously distinguish reality from fiction, nevertheless, the eye is very powerful, and in the subconscient it turns to be not that easy to do such diferenciation. Hence, if you see some situation over and over again in TV, and eventually you are in a similar situation, you would probably have a similar reaction, because it is in the back of your head already... and many of such reactions are very violent!

[quote="djimenez"]Hence, if you see some situation over and over again in TV, and eventually you are in a similar situation, you would probably have a similar reaction, because it is in the back of your head already... [/quote]I've seen tons and tons of Steven Segal movies, but I still can't do any of this Aikido moves :? :D :D

[quote="Valentin Vornicu"]I've seen tons and tons of Steven Segal movies, but I still can't do any of this Aikido moves :? :D :D[/quote]

I do ... but well, I have trained 10 hours a week for over a year  :D  :P  ;)


summer program
Mathcamp
PROMYS
Ross Mathematics Program
AwesomeMath
AMC
USA(J)MO

In your opinion, which math camp is the best for starting proof writing and easier olympiad problems and why do you think it is that mathcamp?

I understand that promys and ross are pretty long and focus alot on number theory so they would be very useful, but I'm wondering if awesomeMath is still better for starting olympiad problems because Dr. Titu Andreescu started the program. I also know that canada/usa is also excellent and slightly longer than awesome math but the program sometimes goes into some topics only tangentially related to math competitions.

So what do you think?  :)

amp is about the only well known (well known as in i heard of it) math camp that [i]specializes[/i] in olympiad problems
oh and of course mop, but making mop requires slight trickiness
oh and LOL THERE IS MATH ZOOM but no one cares about that the last time i checked
canada/usa is more geared toward "higher" mathematics (they still do offer classes in olympiad problem solving, i think, but i am inclined to believe that this class is probably inferior to the amp program, just based on # of hours) and promys and ross towards research style math, iirc

gogo awesomemath :O

I agree with everything you said, but something still confuses me.

Alot of my school's math team captains have gone to canada/usa or Ross but I don't know of any who went to awesome math. I'm not sure if more of them went to canada/usa or Ross because they already had alot of experience with contests (and would have done well in contests regardless of where they went) and they wanted to research, or if canada/usa and Ross are just very helpful for the higher level math in contests.

Any thoughts on that?

Canada/USA Mathcamp is not geared toward contests.  It does have classes in problem solving (which are usually very good, although certainly not as intensive as AMP), and there are two weekly competitions.  But theoretical math that has very little application to contests is the focus, and you probably shouldn't go if you are [i]only[/i] interested in problem solving.  If you are interested in problem solving, and also want to learn cool advanced math, then you definitely should go.  :)

Also note that 11 of the top 24 finishers on the USAMO last year have gone to Mathcamp at least once.  You would have an excellent peer group if you want to ask questions about problem solving.


Could someone explain, Does Ohm's Law always hold? If not, when does this law hold?
Thanks in advance.

no ohm's law doesn't always hold infact it holds for some special 'conductors' called the ohmic conductors.

why? that can be known once you go through  the "proof" of ohm's law .
the basic assumption behind ohm's law derivation are-
1.the electrons are virtually only affected by the external electric field you supply
2.there is uniform distribution of speed
3.if an electron belonging to a cross section $A-{1}$ reaches another $A_{2}$ then every electron in the previous cross section reaches there(basically the same as that for ideal liquid flow)
4,.there exists something called relaxation time which on an average is constant

when these conditiona and some other swhich i dont remember(most are here) :maybe:  are need e for a conductor to be "ohmic"

when these are not obeyed then the conductor becomes non ohmic

For example semiconductors aren't ohmic.


Incircle of $\triangle{ABC}$ tangents sides $BC,AC,AB$ at $P,Q,R$ respectively. $AA_{0}$ , $BB_{0}$ $CC_{0}$ are perpendiculars with $BC$ ,$AC$, $AB$ .And $P_{0}, Q_{0}, R_{0}$ are midpoints of  $AA_{0}$ , $BB_{0}$ $CC_{0}$ respectively. Prove that $PP_{0}, QQ_{0}, RR_{0}$ are concurrent


Find av function f(x) so that f(t^2+t+1)=t for all positive real numbers of t.

[quote="mna851"]I don't think there is one.[/quote]
But there is

I'm sorry, I didn't read the part about positive real $t$.

[quote="mna851"]The function[hide] $f(x)= \frac{\sqrt{4x-3}-1}{2}$[/hide] works.[/quote]
Thats right

[hide]

1- By varying t it is clear that f is surjective over positive reals

2- t^2+t+1 is increasing over positive reals, and thus is not equal for any value of t in that domain. => f is injective and thus bijective

3- bijectivity=>inverse g(y) exists

g(y)=y^2+y+1
setting g(y)=x and solving for y gives f(x)

note the negative solution to the quadratic can be rejected as t is positive.
=> f(x) = (-1+sqrt(4x-3))/2
[/hide]


Putnam
topology
probability
algorithm
complex analysis
real analysis

I recently stumbled onto the career section on this site, and I was wondering how a quantitative trading job is like. After reading this section, I am really interested in DE Shaw. I am about to be a sophomore in a Top 25 college in the East Coast, but it is not an Ivy or top tech university. I'd rather not tell my exact college, for privacy reasons. I will be graduating a year early, so I will be applying for a job and/or Ph.d./Masters programs pretty soon. I know a lot about the latter, but I am clueless about the first, so please excuse my seemingly dumb questions. Thanks!
1. How do I apply to DE Shaw? When? How do I get an interview?
Does job experience matter?
2. How hard is it to get accepted from a Top 25, but not elite, school?
3. Do I need to take linear programming courses now? They're important, but at my university, they are easy, to be frank. Do firms not like to hire people with lots of theoretical courses?
3. Here is some background info on me so I can let the wonderful posters here help me to decide if I would fit at such firms.....
Background info: I was homeschooled, and never particpated in any high school math contests. However, I did manage to take several college classes in high school. But I did take the Putnam this year and got a score of 32, which I have heard is pretty good. I've done a research paper with my advisor in the field of abstract algebra, too, this summer. I am a math/physics major. My relevant classes I've taken are listed below. I managed to keep a 4.0.
Freshmen Year (3 trimesters)-
MTH- GRAD General Topology
MTH- GRAD Algebraic Topology
MTH- Abstract Algebra 1
MTH- Abstract Algebra 2
MTH- GRAD Complex Analysis
MTH- GRAD Real Analysis
PHYS- Newtonian Mechanics
PHYS- Quantum Mechanics
ECON- Intermediate Microeconomics
ECON- Intermediate Macroeconomics
EECS- Data Structures

What I scheduled this year (3 trimesters)
MTH- GRAD Probability Theory
MTH- GRAD Stochastic Processes
MTH- Algebra 3
MTH- Algebra 4
MTH- Game Theory
MTH- GRAD Combinatorics
PHYS- GRAD E&M
PHYS- GRAD Thermal Physics
ECON- Econometrics
EECS- Algorithms
EECS- Foundations of Computer Science

Thanks for any feedback!

as far as what i've learned about getting a job as a quantitative trader, the only thing you're missing is computer programming courses.  if you can delve into that and get good at it (maybe as a minor or something) then as long as your interview goes well, i can't see any reason why they wouldn't want to hire you.


Here are some of the problems from this round:

1. There is an odd number of persons on a field and each one throws a pie to its closest neighboor. Prove one of the persons doesnt get hit by a pie.

3. Alice and Bob play a game. Alice writes down $n$ positive integers on a board and then Bob erases some of them and write a $+$ or a $-$ before each of the remaining number, and then computes their sum. Bob wins if he gets a multiple of $2005$, otherwise Alice wins, for wich $n$ does Alice wins? for wich $n$ does Bob?

I will post the remaining problems tomorrow

The first one is easy, and it appeared in the french correspondance program in 2002-2003 : http://www.animath.fr/tutorat/dossier_02032.html
(problem 3)

problem 1 is well known,

it also appears in Engel's Problem Solving Strategies

As for the second one, I think the answer is: for $n\le 10$ Alice wins, otherwise, Bob wins.

Assume $n\ge 11$. There are $\ge 2^{11}-1=2047>2005$ non-empty subsets of our set, so there must be two which are equal modulo $2005$, and from here it's clear. If $n\le 10$, Alice can take the numbers to be $2^0,2^1,\ldots,2^{n-1}$. I hope it's Ok. I've been posting all sorts of stupid things today :).

well Grobber you are right, This problems are really easy, I hope to post the remaining ones tomorrow, i think in the 11th and 10th grades paper there where only 2 original problems, one of them was the one Grobber solved (but it is still easy)


ARML
AMC
AIME
number theory
prime numbers

So I know AMC doesn't allow calculators anymore, ARML doesn't either, and AIME originally didn't(I think).

Anyways, so I am wondering, should I use calculators when solving books like Introduction to Number Theory?
I find it as practice to not use a calculator, but when you are asked to find the 5 smallest prime numbers greater than 1000, even if you narrow it down to a few numbers, it takes a lot of time checking if these are prime or not.

I am not being lazy here; I try not to use my calculator when I am doing school homework, unless it is absolutely necessary. However, it seems to take too much time and hinder my progress. So I am wondering, what do other AoPS'ers do here?

Do you use a calculator or not, when preparing?

<Sorry if this is the wrong section of the forum to ask these questions>

No, you shouldn't. There are some problems that are ment to be solved without a calculator, although they may be ugly. Plus, there are some really nice solutions to problems that may seem to require a calculator. A really basic example to is 1+2+3+...+n. You wont get as much out of practice if you use it.

On a completely random note, shouldn't this topic be moved?

If you practice with a calculator, youre going to get used to it. When the test comes and youre not allowed to use a calculator, youre going to be completely lost (I know this from previous experience :P)

On a (decent) math contest you're not going to be asked to compute huge arithmetic unless there's a nice trick.  There isn't really a "nice trick" to finding the 5 smallest primes over 1000.


number theory
prime numbers
number theory unsolved

Prove that : for each n is integer  let v(n) is the number distinct prime divisor of n.then
         $\sum_{d|n}{\mu(d).v(n/d)}=0 or 1$

Let $\varepsilon_p = \left\{\begin{array}{l} 1, p \mbox{ prime}\\ 0, \mbox{ otherwise} \end{array}\right.$, and $\omega ( n )$ be the number of prime divisors of $n$.
Then
\begin{eqnarray*} \sum_{d|n} \varepsilon_d & = & \omega ( n ) \end{eqnarray*}
Therefore by the Moebius inversion formula,
\begin{eqnarray*} \sum_{d|n} \mu ( d ) \cdot \omega ( n / d ) & = & \varepsilon_n \end{eqnarray*}
Hence the result.

First of all let's factor $n$ as the product of primes: $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}q_1q_2\cdots q_l$. Here $p_1,p_2,\ldots ,p_k,q_1,q_2,\ldots ,q_l$ are distinct prime numbers and $\alpha_i>1,\ \forall i=\overline{1,k}$. Note that none $p_i^2$ can divide $d$ for $d$ to contribute something potentialy non-zero in the sum (we get this from the definition of the Mobius' function). Then it is easy to conceive that the desired sum equals $\sum_{m=0}^{k+l}\sum_{j=0}^m {l\choose j}{k\choose {m-j}}(k+l-j)(-1)^m$. However using Vandermonde's identity we get the following chain of equalities:
\[{ \sum_{m=0}^{k+l}\sum_{j=0}^m {l\choose j}{k\choose {m-j}}(k+l-j)(-1)^m=\\ \sum_{m=0}^{k+l}((-1)^m(k+l)\sum_{j=0}^m {l\choose j}{k\choose {m-j}}+(-1)^{m+1}\sum_{j=0}^m j{l\choose j}{k\choose {m-j}})=\\ \sum_{m=0}^{k+l}((-1)^m(k+l){{k+l}\choose m}+(-1)^{m+1}\sum_{j=1}^m j{l\choose j}{k\choose {m-j}})=\\ (k+l)\sum_{m=0}^{k+l}(-1)^m{{k+l}\choose m}+\sum_{m=0}^{k+l}(-1)^{m-1}l\sum_{j=1}^m {{l-1}\choose {j-1}}{k\choose {m-1-(j-1)}}=\\ (k+l)(1-1)^{k+l}+l\sum_{m=0}^{k+l}(-1)^{m-1}\sum_{j=0}^{m-1} {{l-1}\choose j}}{k\choose {m-1-j}}=\\ l\sum_{m=0}^{k+l}(-1)^{m-1}{k+l-1\choose {m-1}}=\\ l(1-1)^{k+l-1}=0  \]
Note that those equalities are well-defined whenever $k+l>1$ (because then no $0^0$ appears). However checking cases $n=1$ ($k=l=0$) and $n=p_1^{\beta_1}$ ($k=1$ or $l=1$) gives that the sum we are computing then equals $1$ or $0$ as well.


find asymptotic  when $ t\to 1-0$ of this sum
$ \sum t^{n!}$

There is probably something wrong with this:

For positive $ t<1$ we have $ \sum t^{n!}\leq \sum t^{n}=\frac{1}{1-t}$, we have $ \lim_{t\to 0^{+}}\sum t^{n!}\leq 1$.

For negative $ t>-1$ we have $ \sum t^{n!}\geq \sum t^{n}=\frac{1}{1-t}$, so $ \lim_{t\to 0^{-}}\sum t^{n!}\geq 1$. Therefore $ \lim_{t\to 0}\sum t^{n!}=1$.

I probably shouldn't post this with the knowledge that something is off, it seems like the answer should be $ 0$.


Να και μια τριγωνομετρική άσκηση:
Αν σε τρίγωνο ABC είναι $ (\frac{sin(B)}{sin(A)})^{2007} \equal{} (\frac{a}{b})^{2007}$ να βρείτε το είδος του τριγώνου. (Έχω αλλάξει κάπως τα νούμερα, αλλά δε παίζει σημασία.)

Θα δώσω τη λύση κρυφή. Με την άσκηση αυτή αξίζει να ασχοληθεί κανείς.
[hide]
$ (\frac{sin(B)}{sin(A)})^{2007} \equal{} (\frac{a}{b})^{2007}(I)$
από το νόμο των ημιτόνων έχουμε $ \frac{a}{sin(A)} \equal{} \frac{b}{sin(B)} < \equal{} > (\frac{sin(B)}{sin(A)})^{2007} \equal{} (\frac{b}{a})^{2007} (II)$
$ (I), (II) \equal{} > (\frac{a}{b})^{2007} \equal{} (\frac{b}{a})^{2007} < \equal{} > a \equal{} b$ άρα το τρίγωνο είναι ισοσκελές.

Η επίσημη λύση είναι με περιπτώσεις και άτοπο αν $ a < b$ ή $ a > b$, οπότε μένει το $ a \equal{} b$ που ισχύει. Προσωπικά, θεωρώ ότι αυτός ο τρόπος είναι πολύ μεγάλος.

[/hide]

Μανιέ οπως σου πα πρεπει να μαθω Latex για να δώσω λύση αλλά νομίζω οτι εάν εφαρμόσουμε νόμο  συνημιτόνων θα βγεί εύκολα.

λαθος ημιτόνων ήθελα να πω μπερδεψα το  sin με το cos

Βγαίνει γελοία. Την έβαλα απλώς γιατί η επίσημη λύση παίρνει περιπτώσεις της σύγκρισης των ημιτόνων, και δε μου άρεσε.


Let $ a,b,c$ be positive numbers such that $ abc\equal{}1$.

Prove that

$ (a\plus{}b\plus{}c)^2\minus{}(a\plus{}b)(b\plus{}c)(c\plus{}a)\ge1$

if and only if at most one of $ a,b,c$ is greater than 1.

[quote="Mageling"]Let $ a,b,c$ be positive numbers such that $ abc \equal{} 1$.

Prove that

$ (a \plus{} b \plus{} c)^2 \minus{} (a \plus{} b)(b \plus{} c)(c \plus{} a)\ge1$

if and only if at most one of $ a,b,c$ is greater than 1.[/quote]
$ (a \plus{} b)(b \plus{} c)(c \plus{} a) \equal{} (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} abc \equal{} (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} 1$ so it is equivalent to:
$ a \plus{} b \plus{} c \ge ab \plus{} bc \plus{} ca$.
Let $ a \equal{} \frac {x}{y}, b \equal{} \frac {y}{z}, c\equal{}\frac {z}{x}$, and wlog assume $ x \ge y,z$.
After multiplying with $ xyz$ it becomes:
$ \sum_{cyc}xy^2 \ge \sum_{cyc}x^2y \iff$
$ (x \minus{} y)(y \minus{} z)(z \minus{} x) \ge 0$.
So it holds if and only if $ x \ge z \ge y$, and then $ a \ge 1, b \le 1, c \le 1$, that is: At most one of $ a,b,c$ are greater than $ 1$. QED :)

[quote="Mathias_DK"][quote="Mageling"]Let $ a,b,c$ be positive numbers such that $ abc \equal{} 1$.

Prove that

$ (a \plus{} b \plus{} c)^2 \minus{} (a \plus{} b)(b \plus{} c)(c \plus{} a)\ge1$

if and only if at most one of $ a,b,c$ is greater than 1.[/quote]
$ (a \plus{} b)(b \plus{} c)(c \plus{} a) \equal{} (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} abc \equal{} (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} 1$ so it is equivalent to:
$ a \plus{} b \plus{} c \ge ab \plus{} bc \plus{} ca$.
Let $ a \equal{} \frac {x}{y}, b \equal{} \frac {y}{z}, c \equal{} \frac {z}{x}$, and wlog assume $ x \ge y,z$.
After multiplying with $ xyz$ it becomes:
$ \sum_{cyc}xy^2 \ge \sum_{cyc}x^2y \iff$
$ (x \minus{} y)(y \minus{} z)(z \minus{} x) \ge 0$.
So it holds if and only if $ x \ge z \ge y$, and then $ a \ge 1, b \le 1, c \le 1$, that is: At most one of $ a,b,c$ are greater than $ 1$. QED :)[/quote]

x=1/a , y=1/b, z=1/c
=> xyz=1
and  x+y+z < xy+yz+Zx ! Why ? :maybe:
x,y,z is one case of a,b,c => absurd !

Assume, without loss of generality, $ a\ge b\ge c$. Hence, $ 1\ge c^2 \Rightarrow 1\ge c \Rightarrow ab\ge1\Rightarrow a\ge 1$.
Note that $ (a \plus{} b)(b \plus{} c)(c \plus{} a) \equal{} (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} abc \equal{} (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} 1$
Thus the inequality is equivalent to
$ a \plus{} b \plus{} c\ge ab \plus{} bc \plus{} ca\Rightarrow a \plus{} b \plus{} c\ge\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$
Substituting $ c \equal{} \frac {1}{ab}$ gives
$ a \plus{} b \plus{} \frac{1}{ab}\ge\frac {1}{a} \plus{} \frac {1}{b}\plus{}ab \Rightarrow\frac {(a \minus{} 1)(1 \minus{} b)(ab \minus{} 1)}{ab}\ge 0$
This is true iff $ b\le 1\Rightarrow a\ge 1\ge b\ge c$.
Hence, at most one of $ a,b,c$ is greater than 1. $ \blacksquare$

[quote="Elixir10"][quote="Mathias_DK"][quote="iqtimo"]x=1/a , y=1/b, z=1/c
=> xyz=1
and  x+y+z < xy+yz+Zx ! Why ? :maybe:
x,y,z is one case of a,b,c => absurd ![/quote]
 :huh:
What are you talking about? :what?:[/quote]He may've just misread your post.

Your solution, though, looks correct.[/quote]
i think inequality false ! because can't value a+b+c and ab+bc+ca
if x=1/a , y=1/b, z=1/c
suppose a+b+c $ \geq$ ab+bc+ca then
xy+yz+zx $ \geq$ x+y+z

[quote="Elixir10"]Assume, without loss of generality, $ a\ge b\ge c$. Hence, $ 1\ge c^2 \Rightarrow 1\ge c \Rightarrow ab\ge1\Rightarrow a\ge 1$.
Note that $ (a \plus{} b)(b \plus{} c)(c \plus{} a) \equal{} (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} abc \equal{} (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} 1$
Thus the inequality is equivalent to
$ a \plus{} b \plus{} c\ge ab \plus{} bc \plus{} ca\Rightarrow a \plus{} b \plus{} c\ge\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$
Substituting $ c \equal{} \frac {1}{ab}$ gives
$ a \plus{} b \plus{} \frac {1}{ab}\ge\frac {1}{a} \plus{} \frac {1}{b} \plus{} ab \Rightarrow\frac {(a \minus{} 1)(1 \minus{} b)(ab \minus{} 1)}{ab}\ge 0$
This is true iff $ b\le 1\Rightarrow a\ge 1\ge b\ge c$.
Hence, at most one of $ a,b,c$ is greater than 1. $ \blacksquare$[/quote]
if exist a,b,c (abc=1 ) serve inequality a+b+c $ \geq$ ab+bc+ca then exist x,y,z ( 1/a,1/b,1/c ) don't serve inequality because xy+yz+zx $ \geq$ x+y+z ! :maybe:
Why ? :blush:

[quote="Mathias_DK"]@Elixir10: I was not adressing you. Your proof looks totally fine :)

@iqtimo: If $ a \ge 1 \ge b,c$ and $ x,y,z \equal{} \frac {1}{a},\frac {1}{b},\frac {1}{c}$, then $ y,z \ge 1 \ge z$. So of course $ a \plus{} b \plus{} c \ge ab \plus{} bc \plus{} ca$ and $ x \plus{} y \plus{} z \le xy \plus{} yz \plus{} zx$. I don't really know where you are going...[/quote]
suppose $ a \plus{} b \plus{} c \ge ab \plus{} bc \plus{} ca$ with all abc=1  (1).
a',b',c' serve $ a \plus{} b \plus{} c \ge ab \plus{} bc \plus{} ca$
we choose x=1/a' , y=1/b' ,z=1/c' => xyz=1
but xy+yz+zx  $ \ge$ x + y+ z => xyz=1 but x,y,z don't serve x+y+z $ \ge$ xy+yz+zx
from(1) => absurd  ! :blush:

[quote="iqtimo"][quote="Mathias_DK"]@Elixir10: I was not adressing you. Your proof looks totally fine :)

@iqtimo: If $ a \ge 1 \ge b,c$ and $ x,y,z \equal{} \frac {1}{a},\frac {1}{b},\frac {1}{c}$, then $ y,z \ge 1 \ge z$. So of course $ a \plus{} b \plus{} c \ge ab \plus{} bc \plus{} ca$ and $ x \plus{} y \plus{} z \le xy \plus{} yz \plus{} zx$. I don't really know where you are going...[/quote]
suppose $ a \plus{} b \plus{} c \ge ab \plus{} bc \plus{} ca$ with all abc=1 .
a',b',c' serve $ a \plus{} b \plus{} c \ge ab \plus{} bc \plus{} ca$
we choose x=1/a' , y=1/b' ,z=1/c' => xyz=1
but xy+yz+zx  $ \ge$ x + y+ z
absurd ! :blush:[/quote]Read the necessary condition for the inequality. At most 1 can be greater than 1.

If $ a,b,c$ satisfy this, you get $ z\ge y\ge 1\ge x$, meaning either

a) $ y \equal{} 1$; in this case, equality holds
b) $ y > 1$; in this case, both $ y$ and $ z$ are greater than 1, violating the condition $ \Rightarrow$ inverse of inequality holds.

@Mathias: I know, though using the $ x,y,z$ transformation works well too. And, thanks.

[quote="Elixir10"][quote="iqtimo"][quote="Mathias_DK"]@Elixir10: I was not adressing you. Your proof looks totally fine :)

@iqtimo: If $ a \ge 1 \ge b,c$ and $ x,y,z \equal{} \frac {1}{a},\frac {1}{b},\frac {1}{c}$, then $ y,z \ge 1 \ge z$. So of course $ a \plus{} b \plus{} c \ge ab \plus{} bc \plus{} ca$ and $ x \plus{} y \plus{} z \le xy \plus{} yz \plus{} zx$. I don't really know where you are going...[/quote]
suppose $ a \plus{} b \plus{} c \ge ab \plus{} bc \plus{} ca$ with all abc=1 .
a',b',c' serve $ a \plus{} b \plus{} c \ge ab \plus{} bc \plus{} ca$
we choose x=1/a' , y=1/b' ,z=1/c' => xyz=1
but xy+yz+zx  $ \ge$ x + y+ z
absurd ! :blush:[/quote]Read the necessary condition for the inequality. At most 1 can be greater than 1.

If $ a,b,c$ satisfy this, you get $ z\ge y\ge 1\ge x$, meaning either

a) $ y \equal{} 1$; in this case, equality holds
b) $ y > 1$; in this case, both $ y$ and $ z$ are greater than 1, violating the condition $ Rightarrow$ inverse of inequality holds.

@Mathias: I know, though using the $ x,y,z$ transformation works well too. And, thanks.[/quote]
you don't explain which i read ! true or fales  :?:  :huh:

[quote="Mageling"]Let $ a,b,c$ be positive numbers such that $ abc \equal{} 1$.

Prove that

$ (a \plus{} b \plus{} c)^2 \minus{} (a \plus{} b)(b \plus{} c)(c \plus{} a)\ge1$

if and only if at most one of $ a,b,c$ is greater than 1.[/quote]

Very nice.
We have:
$ (a\minus{}1)(b\minus{}1)(c\minus{}1) \ge 0$
So:$ a\plus{}b\plus{}c \ge ab\plus{}bc\plus{}ca$
Inq
$ <\equal{}>(a\plus{}b\plus{}c)^2\minus{}(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca) \ge 0$
It's true  :P

[quote="caubetoanhoc94"][quote="Mageling"]Let $ a,b,c$ be positive numbers such that $ abc \equal{} 1$.

Prove that

$ (a \plus{} b \plus{} c)^2 \minus{} (a \plus{} b)(b \plus{} c)(c \plus{} a)\ge1$

if and only if at most one of $ a,b,c$ is greater than 1.[/quote]

Very nice.
We have:
$ (a \minus{} 1)(b \minus{} 1)(c \minus{} 1) \ge 0$
So:$ a \plus{} b \plus{} c \ge ab \plus{} bc \plus{} ca$
Inq
$ < \equal{} > (a \plus{} b \plus{} c)^2 \minus{} (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \ge 0$
It's true  :P[/quote]
if a=2 , b=3 ,c=1/6 , abc=1
a+b+c $ \ge$ ab+bc+ca is not true !

[quote="caubetoanhoc94"]You should note:
if and only if at most one of $ a,b,c$ is greater than 1  :P[/quote]
i'm sorry . i don't read  careful it  :oops:   !


calculus
integration
logarithms
limit
function
trigonometry
real analysis

Calculate: $ \int_{0}^{\infty}{\frac {\ln^{2}(t + 1)}{(t + 1)\sqrt {t}}\textbf dt}$

[hide]My solution:
$ \textbf I=\int_{0}^{\infty}{\frac{\ln^{2}(t+1)}{\sqrt{t}}\textbf d(\ln (t+1))=\frac{\ln^{3}(t+1)}{\sqrt{t}}|_{t=0}^{t=\infty}-\textbf J=-\textbf J}$ where $ \textbf J=2\int_{0}^{\infty}{\frac{\ln^{2}(t+1)}{(t+1)\sqrt{t}}\textbf dt}-\frac{1}{2}\int_{0}^{\infty}{\frac{\ln^{3}(t+1)}{t^{3/2}}\textbf dt}$, we get:
$ \textbf I=-2\textbf I+\frac{1}{2}\int_{0}^{\infty}{\frac{\ln^{3}(t+1)}{t^{3/2}}\textbf dt}$ or $ \textbf I=\frac{1}{6}\int_{0}^{\infty}{\frac{\ln^{3}(t+1)}{t^{3/2}}\textbf dt}=\frac{1}{6}\textbf G$ where $ \textbf G=\int_{0}^{\infty}{\frac{\ln^{3}(t+1)}{t^{3/2}}\textbf dt}$. now again partial integration:
$ \textbf G=\int_{0}^{\infty}{\ln^{3}(t+1)\textbf d\left(-\frac{1}{2}t^{-1/2}\right)}=-\frac{1}{2}t^{-1/2}{\ln^{3}(t+1)|_{t=0}^{t=\infty}+\frac{3}{2}\int_{0}^{\infty}{\frac{\ln^{2}(t+1)}{(t+1)\sqrt{t}}\textbf dt=\frac{3}{2}\textbf I}}$ or $ \textbf I=0$  :maybe: this is strange.
[/hide]

[quote="Yuriy Solovyov"]$ \textbf I = \int_{0}^{\infty}{\frac {\ln^{2}(t + 1)}{\sqrt {t}}\textbf d(\ln (t + 1)) = \frac {\ln^{3}(t + 1)}{\sqrt {t}}|_{t = 0}^{t}}$[/quote]

$ \sqrt{t}$ is not a constant ;)

you joking with me?  :D
$ \lim_{t\to\infty}\frac {\ln^{3}(t \plus{} 1)}{\sqrt {t}} \equal{} 0$
$ \lim_{t\to 0}\frac {\ln^{3}(t \plus{} 1)}{\sqrt {t}} \equal{} 0$
so....

My point is that the substitution $ u \equal{} \ln (t \plus{} 1)$ gives $ \int_{0}^{\infty} \frac{u^2}{ \sqrt{ e^u \minus{} 1} } \, du$, the indefinite integral of which is clearly not $ \frac{u^3}{ \sqrt{e^u \minus{} 1} }$.

maybe you forgot that i wrote $ \textbf J$ after... :) i don't understand :(

:blush:

I seem to have read that as an equal sign for some reason.

It is strange for me because function $ \frac {u^2}{\sqrt {e^u \minus{} 1}} > 0$, but integral of it is $ 0$. maybe i've done a mistake, but i don't see it now :(

[quote="Yuriy Solovyov"]Calculate: $ \int_{0}^{\infty}{\frac {\ln^{2}(t + 1)}{(t + 1)\sqrt {t}}\textbf dt}$
[/quote]

$ t\rightarrow x^2$
$ x \rightarrow \tan\,\theta$
$ \theta\rightarrow\cos^{-1}y$

we have $ I=2\int_0^\infty\dfrac{\ln^2(1+x^2)}{1+x^2}\;\textbf dx=8\int_0^{\dfrac{\pi}{2}}\ln^2\cos\,\theta\;\textbf d\theta$
$ =8\int_0^1\dfrac{\ln^2y}{\sqrt{\; 1-y^2\;}}\;\textbf dz=\boxed{2\pi\left[2\ln^2 2+\zeta(2)\right]}$

this was shown on the forum before.

i don't think it is strange at all. you   just managed to go round and essentially came back to where you started.

$ \textbf I = \int_{0}^{\infty}{\frac {\ln^{2}(t + 1)}{\sqrt {t}}\textbf d(\ln (t + 1)) = \underbrace{\frac {\ln^{3}(t + 1)}{3\sqrt {t}}\bigg|_0^{\infty} }_{ = \;0} + \frac {1}{6}\int_{0}^{\infty}{\frac {\ln^{3}(t + 1)}{t^{3/2}}\textbf dt}}$

now again partial integration:
$ \textbf I = \frac {1}{6}\ \int_{0}^{\infty}{\ln^{3}(t + 1)\textbf d\left( - 2t^{ - 1/2}\right)}$

$ = \frac {1}{6} \left[ - 2t^{ - 1/2}{\ln^{3}(t + 1)\bigg|_0^{\infty} + 6\int_{0}^{\infty}{\frac {\ln^{2}(t + 1)}{(t + 1)\sqrt {t}} \;\textbf dt \right] = \frac {1}{6}(0 + 6 \; \textbf I) = \textbf I}}$

 you have a few mistakes in your calculations. first off, i don't know why you had to go through a lot of stuff just to arrive at $ \textbf G$. what you have for $ \textbf G$ later on is actually obvious when you did integration by parts the   first time, like i have shown above. also, the expression you have for $ \textbf G$ at the end is not correct, it should say $ \textbf G = \textbf I$, implying that you have come back to square one.

Yes misan, thanks, i see where i'm wrong  :blush: $ \frac{1}{t^{3/2}}\textbf dt\equal{}\textbf d\left(\minus{}2t^{\minus{}1/2}\right)$ but not $ \textbf d\left(\minus{}\frac{1}{2}t^{\minus{}1/2}\right)$ stupid mistake :(


In trapezoid $ ABCD$, $ \overline{AB}$ and $ \overline{CD}$ are perpendicular to $ \overline{AD}$, with $ AB\plus{}CD\equal{}BC$, $ AB<CD$, and $ AD\equal{}7$. What is $ AB\cdot CD$?

$ \textbf{(A)}\ 12 \qquad
\textbf{(B)}\ 12.25 \qquad
\textbf{(C)}\ 12.5 \qquad
\textbf{(D)}\ 12.75 \qquad
\textbf{(E)}\ 13$

[geogebra]5f3a179ae9599bd34d70e307b0bb740265b16faa[/geogebra]
if AB=x and CD=y, we have BC=x+y.

by pythagorean theorem, we have $ (x\plus{}y)^2\equal{}(y\minus{}x)^2\plus{}49$

solving the equation, we have $ 4xy\equal{}49 \implies xy\equal{}12.25$

answer: B

Sorry for 12 year bump (wow) but [hide=Solution]Since $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD},$ we have that $ABCD$ is a rectangle with $AB=CD=3.5$ and $BC=AD=7,$ which gives the answer $3.5\times3.5=\boxed{\textbf{(B)}~12.25}$ and we're done.

Let $H$ be the foot from $A$ to $CD$. Call $AB$ as $x$ and $HD$ as $Y$. We just need to find $x(x+y)$. By the Pythagorean Theorem, we have $BC=AH=\sqrt{y^2+49}$, so we have $$2x+y=\sqrt{y^2+49} \rightarrow 4x^2+4xy=49 \rightarrow 4x(x+y)=49 \rightarrow x(x+y)=\boxed{12.25}.$$


This is just a problem I came across...Try to solve this! It fits the season!  ;)

Swimming pools around the country opened this past weekend for the summer.  At Tate’s neighborhood pool, it is estimated that the number of people at the pool on opening day is 253% of the average daily number of people who go to the pool throughout the year.  If 432 people came to the pool on opening day, what is the average number of people per day throughout the summer?  Express your answer to the nearest whole number.

I saw the solution online, but I think their explanation was wrong... I think it is
[hide]177[/hide].

Someone please tell me if I am right.

[hide]I divided 432 by 2.53 to get 170.75 which rounds to 171[/hide]

[hide]The answer is

$\frac{432}{2.53}$  which approximately is 171

$171$.

Was that the answer to the question that you found online arvindsn?[/hide]

[hide] You turn it into a proportion and get x = 43200/253. Were you supposed to use a calculator on this, because I used one thinking that it was a target round question. [/hide]

[quote="mathgeniuse^ln(x)"]I assume you could since the author never said that you couldn't use a calculator.[/quote]

No, when I saw it on the Mathcounts site.

Okay, I guess you can't then.

I guess you divide it the way you learned in 4th Grade.

[quote="Arvind_sn"]yeeah thats how I got it wrong in the first place... AAGH! I made a silly mistake!!! :furious:[/quote]

Well, good job on getting the question right.

Nice emoticon.


inequalities
algebra
polynomial
algebra proposed

Sure about that 54? Even if the highest difference was 54 it'd have been above 990...

To have existence possible, I guess 54 must be replaced with a number around 4.9441323248

Unless it's the 990 needing replacement of course.

it's not possible, but just because $54>\sqrt{2200}$, which is a much worse bound than peters. how did you get yours?
for mine: sum up $(x_i-x_j)>54^2$ to get $9\sum x_i^2-2(\sum_{1\leq i<j\leq n}x_ix_j)\geq 45\cdot 54^2$, but we have$0=(\sum x_i)^2=\sum x_i^2+2(\sum_{1\leq i<j\leq n}x_ix_j)$, thus $\sum x_i^2\geq \frac {45\cdot 54^2}{10}>990$.

Peter

Here is a slightly more interesting variant of this problem:

==================
Determine all real numbers $ x_1,\ldots,x_7 $ that satisfy

* $ \sum_{k=1}^7 x_k = 0 $

* $ \sum_{k=1}^7 x^2_k \le 7$

* $|x_j-x_k| \ge 1/2 $ for all $j\ne k$

==================

Admin Edit: [tex ] tags are valid only with BBCode enabled. Please change your default profile setting.

Peter, I proved that even with 5 as limit you had a 1017+ minimum, so for the exact value I used excel (just to make it clear that it's not a typo in that number)

I rather think he will have meant the difference between $x_i^2$ and $x_j^2$ or anything alike? :)

Also, I'll make test20's post readable:

[quote]Determine all real numbers $x_1,\ldots,x_7$ that satisfy
*$\sum_{k=1}^7 x_k = 0$
*$\sum_{k=1}^7 x^2_k \le 7$
*$|x_j-x_k| \ge 1/2$ for all $j\ne k$[/quote]

[b]Solution(s) and generalization.[/b]

Suppose that $ X=\{x_1,x_2,...,x_n\}\in {\mathbb R}^n  $ are such that
\[ \sum\limits_{k=1}^n x_k = A \; \; \; ,\; \;
 \sum\limits_{k=1}^{n} x_k^2= B\; .\]
 Let $ \mu = \displaystyle \frac{A}{n}\; , \; $ $ m(X):= \displaystyle \min\limits_{1\le i,j\le n}|x_i-x_j| \;
 $ and  \[\Delta:= nB-A^2= n\sum\limits_{k=1}^n(x_k-
 \mu)^2=\sum\limits_{1\le i\le j\le n}|x_i-x_j|^2\; . \]
 [b] THEOREM[/b][i] The following inequality holds
 \[  m(X)\le \frac{2}{n}\sqrt{\frac{3\Delta}{n^2-1}}\; \; ,\]
 the equality case being valid if and only if
 $ \; X=\displaystyle \{y_1,y_2,...,y_n\}\; $ where \[
  (*)\; \; \; \; \;  y_k:=\mu +\frac{n-2k+1}{n}\sqrt{\frac{3\Delta}{n^2-1}}\; \; ,\;
   \; \; k\in \{1,2,...,n\}.
 \]
[/i]
 [b]Proof.[/b]   Observe that $ A, B, \Delta $ are symmetric in $x_1,x_2,...,x_n. $  Let us assume $x_1\ge x_2\ge \ldots \ge x_n .$ Because $ \displaystyle \sum\limits_{k=1}^{n}(x_k-\mu)= 0 , $ there exists a natural number $ k_0\; ,\; 1\le k_0\le n-1,$ such that
\[
 x_1-{\bf \mu }\ge x_2-{\bf \mu }\ge \, ...
\ge x_{k_{0}}-{\bf \mu }\ge 0>x_{k_{0}+1}-{\bf \mu }\ge \, ... \ge
x_n-{\bf \mu }\; .
\]
We shall denote \[
s=\left\{
\begin{array}{ccc}
\ds k_0 & \mbox{if}\;  & \ds |x_{k_{0}}-{\bf \mu }|\le |x_{k_{0}+1}-{\bf \mu }|\\
&&\\
\ds 1+k_0\; &\; \mbox{if}\;  &\ds
 |x_{k_{0}}-{\bf \mu }| > |x_{k_{0}+1}-{\bf \mu }|
\end{array}\right. .\]
First , it may be observed that when  $ i\le s $ we have
\[x_i-{\bf \mu }=\sum\limits_{j=i}^{s-1} (x_j-x_{j+1})+(x_s-{\bf
\mu })\ge (s-i)m(X)+(x_s-{\bf \mu }) .\] Because for $ i>s $
\[x_i-{\bf \mu }=(x_s-{\bf \mu })-\sum\limits_{j=s}^{i-1}
(x_j-x_{j+1})+(x_s-{\bf \mu })\le (x_s-{\bf \mu })+(s-i)m(X)\; ,\]
it follows \[\frac{{\bf \Delta}}{n}=\sum\limits_{i=1}^s (x_i-{\bf
\mu })^2+\sum\limits_{i=s+1}^n (x_i-{\bf \mu })^2\ge \] \[\ge
\sum\limits_{i=1}^s ((s-i)m(X)+(x_s-{\bf \mu }))^2+
\sum\limits_{i=s+1}^n ((x_s-{\bf \mu })+(s-i)m(X))^2\; .\] This may be written as
\[(**)\; \;
 \frac{{\bf \Delta}}{n^2}\ge \ds \frac{6n^2-6s(n+1)+(n+1)(2n+1)}{6}
 m^2(f)+\]
\[
\ds +(2s-n-1)(x_s-{\bf \mu })m(X)+(x_s-{\bf \mu })^2
\]
with equality iff \[ x_j-x_{j+1}=m(X)\; \;  ,\; \; \;   (j\in \{1,
\, ..., n-1\}) . \] Taking into account (**) we find \[\frac{{\bf
\Delta}}{n^2}\ge \frac{(n^2-1)}{12} m^2(f)\; \; \; \;
\textrm{or}\; \; \;
 m(X)\le \frac{2}{n} \sqrt{\frac{3{\bf \Delta}}{n^2-1}}\; ,\]
the equality case being valid if and only if $ x_1, \, ... x_n $ satisfy \[x_s-{\bf \mu }=\frac{n-2s+1}{2}m(X)\hsp ,\hsp  (s\in \{1, \, ..., n\})\; .\]

[b]PARTICULAR CASES:[/b]

[b] CASE 1:[/b] [i]  When  $ n=7\; ,\;  A=0\; ,\; B=7\; $[/i] we have  $ \Delta= 49 \; ,\; \mu=0,\; $ and $ m(X)\le\displaystyle \frac{1}{2}\; . $ On the other hand, in proposed Problem by [b]Peter VDD[/b] it is required to have  $ \; m(X)\ge\displaystyle  \frac{1}{2}$. Therefore , if  $\{x_1,...,x_7\} $ exists we must have  $ \; m(X)=\displaystyle
\frac{1}{2} =\displaystyle \max\limits_{1\le i,j\le n}|x_i-x_j|\;
,\; $ that is (see (*)) \[ X=\{y_1,y_2,...,y_n\} \; \; \mbox{\rm with}\; \; \; y_k:=2-\frac{k}{2}\;  \; ,\; \; \; k\in \{1,2,...,n\}.
\]

 [b] CASE 2:[/b]  [i] Let  $ n=10 \; ,\; \; A=0\; ,\; B=990\; .$ [/i] Then $\Delta=\displaystyle 100\times 99  \; ,\; \mu=0 $ ,  and from Theorem we conclude  that does not exists real numbers  $\{x_1,x_2,...,x_{10}\}$  as required in  the question proposed by [b]flip2004[/b] .

Possible [b]references[/b]:

[1]A.L., [i]Problem 246[/i] Mat.Vesnik 8(23) (1971).

[2]A.L., [i]Problem 372[/i] Mat.Vesnik 13(28) (1976) 237-238.

[3]A.L., [i]Inequalities for the roots of a class of polynomials.[/i] Univ.Beograd Publ.Elektrotehn.Fak.Ser.Mat.Fiz. No.577-598 (1977) 79-85.

[4] G.V. Milovanovic , D.S. Mitrinovic, Th.M.Rassias , [i] Topics in polynomials: Extreal problems, Inequalities, Zeros.[/i] World Scientific , Singapore,London, 1994, (see pages 101-106, Theorem 1.3.7).

Regards, Alex


Find all functions f: R --> R satisfying :
 $ f(x+yf(x)) = f(x) + xf(y) $ , $ \forall x,y \in R $

$f(x+f(x)y)=f(x)+xf(y)$
Put $y=0$ to get $f(0)=0$
Now suppose $f(k)=0$ putting $x=k$ we get $k=0$
Put $x=y=-1$and obtain $f(-1-f(-1))=0$ hence $f(-1)=-1$
Suppose that $f(k)=-1$.Put $x=y=k$ and get $k=-1$
Put $y=1$,$x=-1$ we get $f(-2)+f(-1)=-1$
Put $x=1$,$y=-2$ and get $f(-2f(1)-1)=f(1)+f(-2)=-1$ hence $f(1)=1$
Now in original equation putting $x=1$ we get $f(1+y)=1+f(y)$
Now by an easy induction we obtain
(i)$f(x+y)=x+f(y)$,$x$-is an integer and $y$-is real.
(ii)$f(nx)=nf(x)$,$x$-real,$n$-natural.
Puting $x=-1$ in original equation we get $f(-1)+f(-y)=f(-1-y)=-1-f(y)$hence $f(-x)=-f(x)$ for all real $x$'s.
using (ii) we get $f(r)=r$ for all rational $r$.
I don't know how to proceed further to real numbers.

Yes probably I missed this function when I put $f(k)=0$ putting $x=k$ we get $kf(y)=0$ for all $y$ so $f(x)=0$ also satisfies this equation.


AMC AIME algebra polynomial quadratics function geometry

Here's an idea:

[hide]Well you want to find some combination of the first three equations that will give you the final equation, yes? So lets see if we can find three numbers, a, b, and c, so that a times the first equation plus b times the second equation plus c times the third gives us the final equation.

But rather than working in lots numbers, you want to use the pattern of the coefficents to help you find what a, b, and c should be. You noticed that they were squares - good. Now how could you use that to figure out what a, b, and c need to be?[/hide]

A hint that gives away the whole enchilada:

[hide]In each variable, the combination of its coefficients should give: a*n2 + b*(n+1)2 + c*(n+2)2 = (n+3)2. (Although we only need this to be true for n = 1,...,7, if we can find a, b, and c that make it true for all n then that will work too!)[/hide]

No. Your answer is correct. Someone might ask though, how did you know to do those steps?

In my solution, the key is to notice that difference of squares of 2 consecutive numer is an odd number.

I just thought of a polynomial in the form of

c :^2: (x_1) + (c+1) :^2: (x_2) + ... (c+6) :^2: (x_7), where x_1, x_2, ... x_7 are the coefficients and are constant.

So for the first equation,
c :^2: (x_1) + (c+1) :^2: (x_2) + ... (c+6) :^2: (x_7) = 1 as c = 1

For the second equation,
c :^2: (x_1) + (c+1) :^2: (x_2) + ... (c+6) :^2: (x_7) = 12 when c = 2

And last
c :^2: (x_1) + (c+1) :^2: (x_2) + ... (c+6) :^2: (x_7) = 123 when c = 3.

The polynomial c :^2: (x_1) + (c+1) :^2: (x_2) + ... (c+6) :^2: (x_7) is going to be a quadratic equation with respect to c, see for yourself, it should be clear from looking at the above.

(1,1), (2,12), and (3,123) are three pairs that we know on the quadratic and we're trying to find what the polynomial would equal when c = 4 so instead of trying to find the equation, we do this:

In a quadratic function, instead of a linear function where it's an arithmetic series with the y-values, it's an arithmetic sequence with the [i]differences [/i]in consecutive y-values.
1 + [b]11 [/b]= 12
12 + [b]111 [/b]= 123
123 + [b]211 [/b]= 334

When c = 4, c :^2: (x_1) + (c+1) :^2: (x_2) + ... (c+6) :^2: (x_7) = [b]334[/b]  Slick isn't it?

wow.  that's amazing.  i think that's the method i was thinking of.

From some familiarity with recurrence relations, I know that any quadratic sequence satisfies the recurrence relation

[tex]a_n = 3a_{n-1}-3a_{n-2}+a_{n-3}[/tex].

Specifically, I know that

[tex]3(9)-3(4)+1(1)=16[/tex],
[tex]3(16)-3(9)+1(4) = 25[/tex],

and so on. So, 3(third expression)-3(second expression)+1(first expression) = fourth expression, and the answer is 3(123)-3(12)+1(1)=334.

[quote="Nosepuppy"]i know of someone who completed this problem in ~15 seconds after writing it down.  he explained to me how he solved it, but it was unclear to me.  anyhow, it was quite elegant, and only took 15 seconds worth of aritehmatic.  i'll ask him when he gets back from mop.[/quote]

I think that's probably [b]zableman[/b]'s method or pretty close.

By the way, [b]beta[/b]'s method is very close to the differences method. Since the coefficients are quadratic, the 3rd differences [tex]\Delta^3=0[/tex].

Let the value we want to find be a, then

The 1st differences [tex]\Delta[/tex] are: 12-1=11, 123-12=111, a-123

The 2nd differences [tex]\Delta^2[/tex] are: 111-11=100, (a-123)-111=a-234

The 3rd differences [tex]\Delta^3[/tex] are: (a-234)-100=a-334

As I said the 3rd differences is identically 0, so a-334=0, that gives us a=334.

[quote="fuzzylogic"][quote="zabelman"]From some familiarity with recurrence relations, I know that any quadratic sequence satisfies the recurrence relation

[tex]a_n = 3a_{n-1}-3a_{n-2}+a_{n-3}[/tex].
[/quote]

Hmm..., this is pretty cool by noticing the recurrence relation!

So if all the coefficients are cubes (or cubic polynomials) and add one more equation as conditions. Then you can find the value of the 5th one by using the recurrence relation:

[tex]a_n - 4a_{n-1}+6a_{n-2}-4a_{n-3}+a_{n-4}=0[/tex].

Am I right? I am learning something everyday. :-)[/quote]

Yes, that's right. Now, can you prove it? i.e. can you prove that any cubic sequence satisfies that recurrence relation? Furthermore, can you prove that the ONLY sequences that satisfy that relation are polynomials with degree 3 (or less)? Generalize?!

[quote="zabelman"]Yes, that's right. Now, can you prove it? i.e. can you prove that any cubic sequence satisfies that recurrence relation? Furthermore, can you prove that the ONLY sequences that satisfy that relation are polynomials with degree 3 (or less)? Generalize?![/quote]

This is pretty standard material about linear recursive sequences. The way you applied linear recursive sequences to that particular problem is pretty cool in my opinion.

For people who are interested in learning linear recursive sequences. I refer you to a Berkeley Math Circle handout:

http://mathcircle.berkeley.edu/BMC6/ps/linear.pdf

[quote="WarpedKlown1335"]I just thought of a polynomial in the form of

c :^2: (x_1) + (c+1) :^2: (x_2) + ... (c+6) :^2: (x_7), where x_1, x_2, ... x_7 are the coefficients and are constant.

(1,1), (2,12), and (3,123) are three pairs that we know on the quadratic and we're trying to find what the polynomial would equal when c = 4 so instead of trying to find the equation, we do this:

Slick isn't it?[/quote]

Wow, Nice observation!

This works for [b]not equally spaced c values[/b] as well.

If we knew (1,1), (3,123) and (4,334) are three pairs on the quadratic,  we can use Lagrange Interpolating Polynomial to find other points such as (2, y).

http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html

[quote="WarpedKlown1335"]It probably was :-), but Ravi B almost beat me to it, he was hinting to a way similar to mine.  Who was the kid you're talking about that showed you the elegant solution?[/quote]

mildorf.  he has some killer trig problems in the advanced forums.

It's not nearly as neat and tidy (and also much easier to make arithmetic mistakes), but the classic ARML solution is to assume that x4 through x7 are all zero and solve the resulting system.


For a positive integer number $r$, the sequence $a_{1},a_{2},...$ defined by $a_{1}=1$ and $a_{n+1}=\frac{na_{n}+2(n+1)^{2r}}{n+2}\forall n\geq 1$. Prove that each $a_{n}$ is positive integer number, and find $n's$ for which $a_{n}$ is even.

We can prove $a_n={2\cdot\sum_{i=1}^ni^{2r+1}\over n(n+1)}$ by simple induction. Then it reduces to prove, that $n(n+1)$ divides $2\cdot\sum_{i=1}^ni^{2r+1}$. But that's simple work with modulo, if you realize $i^{2r+1}+(a-i)^{2r+1}\equiv0[a]$, where $a$ is $n$ or/and $n+1$.

Radar, there seems to be a part missing from your solution.

Let $r\ge 1,\rho=2r+1$ and $a_n:={2\over{n(n+1)}}\sum_1^n k^\rho$. Then $a_n$ is even iff
$n\equiv 0,3(4)$. Proof:
The terms of $k^{\rho}$ modulo 2 for $k\ge 1$ are $1,0,\cdots$, hence for $n\equiv 1,2(4)$
we have $\sum_1^n k^\rho \equiv 1(2)$ and $a_n$ is odd. Now take $n=4s$.
Then modulo $4s$ we have \[\sum_1^{4s}k^\rho \equiv \sum_1^{4s-1}k^\rho =
(2s)^\rho+\sum_1^{2s-1} (k^\rho+(4s-k)^\rho)\equiv 0\] Hence $a_n$ is even.
Also we have $(n+2)a_{n+1}-na_n=2(n+1)^{2r}$ as in the exercise, hence
for odd $n$ we have $a_{n+1}\equiv a_n(2)$. The result for $n\equiv 3(4)$ follows.