Let the sequence $ a_{n\plus{}2} \equal{} a_{n\plus{}1} \plus{} \frac{k}{n}a_n$. For what values of k there is non-zero solution satisfying $ \frac{a_{n\plus{}1}}{a_n} \equal{} \frac{p(n)}{q(n)}$, where $ p$ and $ q$ are polynomials in $ n$.
I have the solution here. I'm gonna reproduce it until the part I didn't understand.
We can rewrite the sequence above this way: $ n(\frac{a_{n\plus{}2}}{a_{n\plus{}1}}\minus{}1) \equal{} k\frac{a_n}{a_{n\plus{}1}}$.
As $ \frac{a_{n\plus{}1}}{a_n} \equal{} \frac{p(n)}{q(n)}$ we can write
$ n(\frac{p(n\plus{}1)}{q(n\plus{}1)}\minus{}1) \equal{} k\frac{q(n)}{p(n)}$ or

$ n(p(n\plus{}1)\minus{}q(n\plus{}1))p(n) \equal{} kq(n\plus{}1)q(n).$
I may suppose WLOG that the polynomials $ p$ and $ q$ have no common roots. Let $ m$ and $ m'$ be the degrees of $ p$ and $ q$ in this order. Let's split the analysis in two cases. $ m\equal{}m'$ and $ m \not\equal{}m'$.
First $ m\equal{}m'$ the degree of the RHS is $ 2m$ where the degree of p(n+1)-q(n+1) must be $ m\minus{}1$. So, the coefficients of the highest degree of $ p$ and $ q$ are the same. [b]As $ p$ and $ q$ do not have common roots, all roots of p(n) in the LHS appear in the RHS as the roots of q(n+1): and so $ p(n)\equal{}q(n\plus{}1)$. [/b] This part in bold I didn't undestand.

[quote="danilorj"]$ n(p(n \plus{} 1) \minus{} q(n \plus{} 1))p(n) \equal{} kq(n \plus{} 1)q(n).$
[/quote]

Notice that all the roots of $ p(n)$ make the LHS equals to $ 0$ however they do not make $ q(n)$ equal to $ 0$ (different roots), so they make $ q(n \plus{} 1)$ equal to zero. Moreover the degree of $ q(n \plus{} 1)$ is also $ m$.

Ok. Let's suppose we have $ m < m'$. So the degree of the left side is $ m \plus{} m' \plus{} 1$ and the degree of the right side is $ 2m'$ and so $ m' \equal{} m \plus{} 1$. And then, none root of $ q(n \plus{} 1)$ can be root of $ p(n \plus{} 1) \minus{} q(n \plus{} 1)$. Why ?


algebra
polynomial
complex analysis
complex analysis unsolved

How many roots does the polynomial $z^4+3z^2+z+1$ have in the right half $z$-plane?

Maybe you can give you answer? I'm intereseted whether my answer is right

Thanks,Nene, I got the same answer: i used argument principle


I am having troubles with this problem:

Let $ A = \{1,2,3,...n\}$ where $n \geq 2 $

a)Find the number of functions $ f:A \rightarrow A $ such that $ f \circ f(1) = 2 $
b)Find the number of functions $ f:A \rightarrow A $ such that $ f $ is onto
c)Find the number of functions $ f:A \rightarrow A $ such that $ f $ is onto and $ f \circ f(1) = 2 $
d)Find the number of functions $ f:A \rightarrow A $ such that $ f $ is onto or $ f \circ f(1) = 2 $

The first problem is set A.  I am unsure what it is defining.  I was thinking that it could be {1,2,3,1,2,3,1...1,2,3,1,2}, but I am unsure.  Any input to help me solve this problem would be greatly appreciated.  Thanks in advance.

Hint 1: The number of all function f : A -> A is n^n
            In general the number of function f : B -> C , is |C|^|B| where |X| is the number of element in X
Hint 2: The number of all function f : A -> A being onto is n!

Edit: P.S The set A is the set of all natural numbers less or equal to n

I'll answer the first question

a)Find the number of functions $ f:A \rightarrow A $ such that $ f \circ f(1) = 2 $

[hide]We want $ f(1)=X $ and $ f(X)=2 $

If $ X=1 ,  $ this is impossible, because $ f(1)=1 => f(X)=1 \neq 2 $

So, we have $n-1$ chances for $X$

Once we have choose $X$, we set $ f(X)=2 $

Now, we have $ f(1) $ and $ f(X) , $ but there are $n-2$ more elements in $ \{1,2,...,n\} $

For that elements, we are free to choose every value we want, so we have $n$ chances for each

Total: $(n-1)n^{n-2}$[/hide]


probability
function
algebra
polynomial
modular arithmetic

Suppose you have a fair three-sided die (don't ask me how :P ) with numbers $0, 1, 2$.  Find the probability that the sum of $n$ throws of the die will add up to a multiple of $3$.

Can you generalize?

Edit:  Just realized that the problem as written is far too easy, and not worthy of Pre-Olympiad difficulty, so here are follow-ups:

1)  Repeat the problem using a tetrahedral die with numbers $0, 1, 2, 3$;

2)  Repeat the problem using a $k$-hedron with numbers $0, 1, 2, 3, 4, ... k-1$.

[hide]
Since the values of the $k$ rolls cover each remainder modulo $k$ exactly once, then given any $k-1$ rolls, there will be a unique roll which will add on the previous $k-1$ to obtain a multiple of $k$. Therefore the probability is $\boxed{\frac 1k}$.
[/hide]

The follow-ups are still asking for multiples of $3$.  :)  (Edit:  Although as far as "Can you generalize?" goes, that would be correct.)

you can do all these with generating functions

I know.  Would you like to post a solution?  The first one is rather simple with or without generating functions :P

can you see if my reasoning is right (for the first part), thanks?

[hide]
Suppose that we throw the dice with number $\{0,1,2 \}$ two times. So, considering 0 multiple of 3, the probability that the sum is multiple of 3 is $\frac{1}{3}$. If we throw another time the dice, we have $3^3$ combinations of results. For each sum that we obtained in the first 2 throws, we have one and only one good result with the third throw (for example if we throw the first 2 times with sum 4, we necessary need a 2). So the good cases are all possible combinations with $n-1$ throws, that are $3^{n-1}$, divided by all possible combinations at throw $n$. So the probability is $\frac{1}{3}$[/hide]

Yes, that's correct.  But it's not so easy when the odds aren't equal ;)

for the first follow up, the generating function is:

$f(x)=(1+x+x^2+x^3)^n$, the sum of the coefficients of $x^{3k}$ is
$\frac{ f(e^{\frac{0\pi*i}{3} })+f(e^{\frac{1\pi*i}{3}})+f(e^{\frac{2\pi*i}{3}})}{3}$
this is easily evaluated...$\frac{4^n+1+1}{3}$, divide this by the total events, or the sum of the coefficients: $f(1)=4^n$, for the answer: $\frac{1}{3}+\frac{2}{3*4^n}$

[quote="felipeochoa"]I don't understand why you get that sum. :blush: Can you please explain...also can you do something similar if instead of 3 you have a regular die?[/quote]

The general gist of the problem is that when you expand a polynomial, such as $P(x) = (x + x^2 + ... + x^6)$, to some $n^{th}$, power, the coefficients of the expansion give the probabilities of a particular sum.

The $P(x)$ used above represents a regular die, with six equal sums ( $1, 2, 3, ... 6$ ) represented by the coefficients.  Raising it to powers is equivalent to rolling multiple dice and adding the results up, and the coefficients give the probability distribution of each sum (given by the powers $x$ is raised to).

When I ask for the sums that are a multiple of $k$, I want the sum of every $k^{th}$ coefficient.  Now, summing every coefficient is easy - substitute $1$.  But how do we sum every $k^{th}$ coefficient?

To simplify the discussion, we will consider the subcase $k = 3$.  Now, let $\omega = e^{ \frac{2i\pi}{3} }$.  The three third roots of unity are $1, \omega, \omega^2$.  Now,

$1 + \omega + \omega^2 = 0$ (this is easily seen from factoring $\omega^3 - 1$).

Suppose $P(x) = ax^n$ and $3$ does not divide $n$.  Then

$P(1) + P(\omega) + P(\omega^2) = a \left( 1 + \omega^n + \omega^{2n} \right) = a \left( 1 + \omega + \omega^2 \right) = 0$

Suppose that $3$ divides $n$.  Then

$P(1) + P(\omega) + P(\omega^2) = a \left(1 + 1 + 1 \right) = 3a$

The reasoning is easily extended to an arbitrary number of terms and an arbitrarily high value of $k$.  Dividing by $k$ gives you the sum of the $k^{th}$ coefficients.  :)

[quote="matt276eagles"][quote="t0rajir0u"]$a \left( 1 + \omega^n + \omega^{2n} \right) = a \left( 1 + \omega + \omega^2 \right)$[/quote]

Could you explain this step?  Is this true for any roots of unity and for any $n$?[/quote]

It's because $n \not\equiv 2n \pmod{3}$ (since $3 \not| n$) so you'll get both $\omega$ and $\omega^2$. In general, if you have

$\omega = e^{\frac{2 \pi}{k}i}$

and $n$ such that $(n, k) = 1$, we can say that

$1+\omega^n+\cdots+\omega^{(k-1)n} = 1+\omega+\cdots+\omega^{k-1}$

because $n, 2n, \ldots, (k-1)n$ are distinct and nonzero modulo $k$.


geometry
analytic geometry
trigonometry
incenter

Let ABC be an equilateral triangle and let D be a point on AB between A and B. Next, let E be a point on AC with DE parallel to BC. FUrther, let F be the midpoint of CD and G the circumcentre of triangle ADE. Determine the angles of triangle BFG.

Let's put the coordinate system such that midpoint of $AB$ is the origin, $AB$ is the $x$-axis and $C$ is, accordingly, on the $y$-axis. Let point $B$ have the coordinates $B(a,0), a>0$ and point $D$ the coordinates $D(x,0), |x|\leq a$. Then we have (prooving all this is easy and I'll supply it if a need be):

$A(-a,0)$
$B(a,0)$
$C(0,a\sqrt{3})$
$D(x,0)$
$E(\frac{x-a}{2},\frac{(x+a)\sqrt{3}}{2})$
$F({x\over 2},\frac{a\sqrt{3}}{2})$
$G(\frac{x-a}{2},\frac{(x+a)\sqrt{3}}{6})$

Therefore

$\overrightarrow{BF}=[{{x-2a}\over 2},\frac{a\sqrt{3}}{2}]$
$\overrightarrow{FG}=[-{a\over 2},\frac{(x-2a)\sqrt{3}}{6}]$

Now

$1^\circ\quad\overrightarrow{BF}\cdot\overrightarrow{FG}=\frac{-ax+2a^2}{4}+\frac{ax-2a^2}{4}=0$

$2^\circ\quad BF^2=\frac{(x-2a)^2}{4}+\frac{3a^2}{4}=3\left(\frac{(x-2a)^2}{12}+\frac{a^2}{4}\right)=3FG^2$

From this we conclude:

$1^\circ\quad\angle BFG=90^\circ$
$2^\circ\quad\tan\angle FGB=\frac{BF}{FG}=\sqrt{3}\implies\angle FGB=60^\circ$
Hence, $\angle FBG=30^\circ$

Hi Farenhajt

can you please post how you got the coordinates of D, E, F, G please

Well, $D$ is just an arbitrary point between $A$ and $B$, hence it's coordinates are $(x,0)$ where $x$ is an arbitrary value between $-a$ and $a$.

By the problem condition, triangle $ADE$ is equilateral. Therefore, $x$-coordinate of $E$ is the midpoint of $AD$, which is calculated as the arithmetic mean of $x$-coordinates of $A$ and $D$, and $y$-coordinate of $E$ is ${\sqrt{3}\over 2}$ times larger than the length of $AD$, which is calculated as the difference between $x$-coordinates of $D$ and $A$.

$F$ is the midpoint of $CD$, hence its coordinates are the arithmetic means of the corresponding coordinates of $C$ and $D$.

$G$ is the centroid of the equilateral triangle $ADE$, hence its coordinates are the arithmetic means of the corresponding coordinates of $A$, $D$ and $E.$

thank you farenhajt  :D your good man!

hey so the circumcentre of a triangle is the arithmetic mean of the coordinates of the three vertices..

$A (a, b) B (c, d) C (e, f) \implies {circumcentre} (\frac{a+c+e}{3}, \frac{b+d+f}{3})$

and i think the incentre is $\frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c}$  where a,b,c are sides opposite x_1, x_2, x_3

am i right? is there a way to find incentre without finding length of sides/

[quote="sen"]
hey so the circumcentre of a triangle is the arithmetic mean of the coordinates of the three vertices. [/quote]

[b]NO![/b] The [i]centroid [/i]is the arithmetic mean of the coordinates. But in this case we use the fact that the triangle is [u]equilateral[/u].

You can also get a nice synthetic solution by rotating the diagram around B by 60 degrees.


[quote="secrets"]Given $ a,b,c \ge 0$ $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ca$ and $ k \ge 3$Prove that:
$ a \plus{} b \plus{} kc \ge 4$[/quote]
because $ kc^2\plus{}(k\minus{}3)(a\plus{}b)c\plus{}(a\minus{}b)^2\geq0$ is true.

I have two similar solutions. Here is the first I found (is not the simplest).
[hide]
Suppose for the sake of contradiction that $ a\plus{}b\le 1$.
Then $ a\plus{}b\le ab\Rightarrow a\plus{}b\le\dfrac{(a\plus{}b)^2}{4}\Rightarrow 4\le a\plus{}b \rightarrow\leftarrow$.
So $ a\plus{}b>1$.

Case 1: If $ ab\plus{}4>2a\plus{}2b$
$ a\plus{}b\plus{}kc\minus{}4\ge a\plus{}b\plus{}3c\minus{}4\equal{}\dfrac{(a\minus{}b)^2\plus{}ab\plus{}4\minus{}2a\minus{}2b}{a\plus{}b\minus{}1}>0$

Case 2: If $ ab\plus{}4\le 2a\plus{}2b$
$ a\plus{}b\plus{}kc\minus{}4\ge a\plus{}b\plus{}3c\minus{}4\equal{}\dfrac{(a\plus{}b\minus{}4)^2\plus{}3(2a\plus{}2b\minus{}ab\minus{}4)}{a\plus{}b\minus{}1}\ge 0$
with equality only if $ a\equal{}b\equal{}2$.
[/hide]

:)

[quote="secrets"][quote="arqady"][quote="secrets"]Given $ a,b,c \ge 0$ $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ca$ and $ k \ge 3$Prove that:
$ a \plus{} b \plus{} kc \ge 4$[/quote]
because $ kc^2 \plus{} (k \minus{} 3)(a \plus{} b)c \plus{} (a \minus{} b)^2\geq0$ is true.[/quote]
I dont' understand,arqady,can you explain? :)[/quote]
$ a \plus{} b \plus{} kc \ge 4\Leftrightarrow(a\plus{}b\plus{}c)(a\plus{}b\plus{}kc)\geq4(ab\plus{}ac\plus{}bc).$ :wink:


Let $ a_1, a_2, a_3\ge0$ and $ a_1^2\plus{}a_2^2\plus{}a_3^2\equal{}1$. Show that at least two of the following four expressions do not exceed $ 1$.

$ |a_1\plus{} a_2\plus{}a_3|$,  $ |a_1\minus{} a_2\plus{}a_3|$,  $ |a_1\plus{} a_2\minus{}a_3|$,  $ |\minus{}a_1\plus{} a_2\plus{}a_3|$.

assume WLOG that $ a_1\leq a_2\leq_3$. then $ \minus{}1\leq 2a_1\minus{}a_3\leq a_1\plus{}a_2\minus{}a_3\leq a_1\leq 1$, so $ |a_1\plus{}a_2\minus{}a_3|\leq 1$

[quote="pluristiq"]assume WLOG that $ a_1\leq a_2\leq_3$. then $ \minus{} 1\leq 2a_1 \minus{} a_3\leq a_1 \plus{} a_2 \minus{} a_3\leq a_1\leq 1$, so $ |a_1 \plus{} a_2 \minus{} a_3|\leq 1$[/quote]
The question is there are at lease two expressions exceed $ 1$.


if you have a problem like the one last year where you needed to give integer solutions to (round 4, problem 2, year 16)
do you have to show the procedure that you did to get an answer, or could you just show that your answer is correct by showing that it satisfies the conditions,
this idea would proceed as such
assume that (x,y,z) is a solution to that equation
show by algerbra that your solution is true
therefore (x,y,z) is indeed a solution
this would not show how you found the solution, but there would be no dispute that it was a correct answer,
would this recieve full points?

Assuming the question doesn't ask you to find all integer solutions, then yes, you would get full credit for such a solution (it's perfectly valid to answer "find an integer solution to x" with "observe that k satisfies the equation").  No number of presented solutions, however, would satisfy a problem that asked for all solutions, since you haven't proven that these are in fact all possible solutions (unless you had a polynomial with a known number of solutions).


this is the 5th(last) problem of the 13th usamo
LINK:http://www.kalva.demon.co.uk/usa/usoln/usol845.html

the answer here is 4...but i have some doubts about that...
plz look at my solution(its not a solution actually...)

Let the polynomial be f(x)

first we prove a lemma: the remainder of f(x) divided by x-a is f(a)

proof: let f(x)=(x-a)q(x)+r

substitute x with a

than we have f(a)=r

hence the remainder is f(a)

f(3n+1)/(3n-1), the remainder should be f(2)=0

if n=4,then f(3n+1)/(3n-1)=730/11, and the remainder is not 0...

i dont know how i got this...but could any1 help me out...maybe tell me where i got wrong? :rotfl:

[quote="bryant124"]this is the 5th(last) problem of the 13th usamo
f(3n+1)/(3n-1), the remainder should be f(2)=0
if n=4,then f(3n+1)/(3n-1)=730/11, and the remainder is not 0...
[/quote]

$f(x)=(x-2)g(x)$ :P

The actual solution is much more complicated than this, involving roots of unity to construct the polynomial. Not sure what in the world you're doing...


geometry
incenter
circumcircle
geometry proposed

In triangle $ ABC,H,I,O$ are orthocenter, incenter and circumcenter, respectively. $ CI$ cuts circumcircle at $ L$. If $ AB\equal{}IL$ and $ AH\equal{}OH$, find angles of triangle $ ABC$.

Please elaborate on how did you get the solution.

It's easy to prove that BCL is equilateral (use lemma - BL=IL=CL). Let AC, BH intersect at P, AC, AB - at Q. We have $ \angle QAC\equal{}\angle PAB\equal{}60^{\circ}$ => $ \angle HCP\equal{}\angle HPQ\equal{}30^{\circ}$ and $ \angle BHC\equal{}60^{\circ}$. Hence HCOB is inscribed quadrilateral, and HO is bisector of CHB. Now using AH=OH and $ \angle OHB\equal{}30^{\circ}$ we get $ \angle HBO\equal{}75^{\circ}$ => $ \angle ABC\equal{}15^{\circ}$ (because <HBQ= <CBO=30).

Angles $\angle LAB$ and $\angle LCB$ share the same arc, so $\angle LAB = \frac{1}{2} \angle C$. Similarly, $\angle LBA = \frac{1}{2} \angle C$. Using the same idea $\angle BLC = \angle A$ and $\angle ALC = \angle B$. Since $\angle LBI = \frac{1}{2} (\angle B + \angle C)$, then $\triangle LBI$ dictates that $\angle LIB = \angle LBI$ and so $IL=AL=BL=AB$. Or you may have already known this result.

Since $\triangle ABL$ is equilateral, $\angle A + \angle B = 60^{\circ}$ and $\angle C = 120^{\circ}$. From observing the triangle formed by $A$, $B$, and the foot of $B$ onto $AH$, $\angle BAH = 90^{\circ} - \angle B$. Likewise, $\angle ABH = \angle 90^{\circ}-\angle A$, so $\angle AHB = \angle A + \angle B = 60^{\circ}$. From $AH=OH$, $\angle AOH = \angle OAH = 120^{\circ} - \angle B$, so $\angle AHO = 2\angle B - 60^{\circ}$. Since $\angle AOB = 2\angle ALB = 120^{\circ}$, $AOBH$ is cyclic, and so $\angle AHO = \angle ABO \Rightarrow 2 \angle B - 60^{\circ} = 30^{\circ} \Rightarrow \angle B = 45^{\circ}, \angle A = 15^{\circ}$.


[hide="only partially solved"]Here is a partial solution that works for functions with continuous derivatives of all orders:

Since $2x+1=x$ when $x=-1$, set $x=-1$ , differentiate, and solve for $f(-1)$. $f(-1)=-\frac{5}{2}$, $f^\prime (-1)=1$, and $f^{(n)} (-1)=0$ for $n>1$.

Then $f(x)=-\frac{5}{2}+(x+1)=x-\frac{3}{2}$.

But this only works if $f$ has continuous derivatives of all orders... now how to prove for ALL continuous $f$? ...[/hide]

[quote="scorpius119"]$f^\prime (-1)=-1$[/quote]

 :huh:  I get $6 f'(-1) = f'(-1) + 5 \implies f'(-1) = 1$.  But other than that, nice proof, even if it's not complete.

[hide="Small extension"]
Consider a continuous, non-differentiable function $g(x)$ such that $3g(2x+1) = g(x)$.  Then $f(x) = -\frac{5}{2} + (x+1) + g(x)$ will satisfy the criterion without being differentiable.  We just need to find all such $g(x)$ now.

We find $3g(-1) = g(-1) \implies g(-1) = 0$, which seems to suggest it might be possible to prove $g(x) = 0$.

Edit:  Engel says the only solution is $f(x) = -\frac{5}{2} + (x+1) = x - \frac{3}{2}$, so there is indeed some kind of a proof that $g(x) = 0$, or there is an alternate derivation.
[/hide]

aha! a way to finish the solution! (which is now fixed...)

[hide]To prove that $g(x)=0$ for all $x$, need to show that $|g(x)|<\epsilon$ for any real $\epsilon>0$.

If $t=2x+1$, then $x=\frac{t-1}{2}$ so $g\left( \frac{t-1}{2} \right)=3g(t)$. Let $h(t)=\frac{t-1}{2}$. and $h^k (t)$ be the result of taking $h$ $k$ times.

Suppose $t>-1$. Then subtracting 1 and dividing by 2, $\frac{t-1}{2}>-1$ so the sequence of $h^k (t)$ is bounded below. Also, adding $t$ and dividing by 2 gives $t>\frac{t-1}{2}$ so that sequence is decreasing. Then that sequence converges.

If $L$ is the value that that sequence converges to, then $L=h(L)\Rightarrow L=-1$.

By continuity of $g(x)$ at $x=-1$, there is some $\delta>0$ such that $|x+1|<\delta\Rightarrow |g(x)|<\epsilon$.

Since the sequence of repeated $h$'s converges to $-1$, there is some positive integer $k$ such that taking the function $k$ times results in $h^k(t)<\delta-1$ so that $|h^k(t)+1|<\delta\Rightarrow |g(h^k (t))|<\epsilon$.

But $g(h^k (t))=3^k g(t)$ so that $|g(t)|<3^{-k}\epsilon<\epsilon$. Then $g(x)=0$ for $x>-1$.

The proof for $x<-1$ is similar; it involves showing that $\frac{t-1}{2}$ is bounded above (by -1) and increasing.

Since $g(x)=0$, $f(x)$ must be $x-\frac{3}{2}$.[/hide]

[quote="towersfreak2006"]Find all continuous functions $f$ satisfying $3f(2x+1)=f(x)+5x$.[/quote]

rewrite as $3( f(2x+1)-(2x+1)+\frac{3}{2})=(f(x)-x+\frac{3}{2})$

define $g(x)=f(x)-x+\frac{3}{2}$, that gives us

$3*g(2x+1)=g(x)$

say $g(x)=\left(\frac{1}{3}\right)^{\log_2(x+1)}*h(x)$, then

$\left(\frac{1}{3}\right)^{\log_2(2x+2)-1}*h(2x+1)=\left(\frac{1}{3}\right)^{\log_2(x+1)}*h(x)$

$h(2x+1)=h(x)$
$h(x)=c$

so our function is

$f(x)=c* \left(\frac{1}{3}\right)^{\log_2(x+1)}+x-\frac{3}{2}$, but this is only defined for positive inputs...


quadratics
analytic geometry
algebra
polynomial
trigonometry
counting
distinguishability

Solve the following system:

[tex]\noindent a+b+c+d=-1 \\ ab+ac+ad+bc+bd+cd=1 \\ abc+abd+acd+bcd=-1 \\ abcd = 1[/tex]

Furthermore, solve the version for [tex]n[/tex] variables.

A word of advice on naming your post -- things that are simple to established problem solvers are rarely so to beginners.

I might suggest that someone wishing to solve this problem start off with the following two systems:

1) Solve for v, w in the complex numbers:
[tex]v + w = -1 \\ vw = 1[/tex]

2) Solve for x, y, z in the complex numbers:
[tex]x + y + z = -1 \\ xy + yz + zx = 1 \\ xyz = -1[/tex]

Name changed. :)

By the way, for the TeX experts, does anyone know why \noindent does not work? Is it because we are already in math mode? How can I avoid the auto-indentation?

I want all the solutions. If you saw them immediately, Rep, leave someone else with less experience to solve it.

I have a kind of cool solution, but there is probably a more obvious solution that I can't see
Define f(x)=(x-a)(x-b)(x-c)(x-d)
f(x)=x^4-(a+b+c+d)x^3+(ab+bc+ac+ad+bd+cd)x^2-(abc+abd+acd+bcd)x+abcd
SUb in each values
[tex]f(x)=x^4+x^3+x^2+x+1[/tex]
we want to find the roots of f(x)
Rewrite the equation:
[tex](x^2+1)^2+(x^2+1)x-x^2=0[/tex]
quadratic formula in x^2+1 we have
[tex]x^2+1=\frac{-x \pm \sqrt {x^2+4x^2}}{2}[/tex]
simplify it we have
[tex]x^2+1=-x(\frac{(1 + \sqrt5)}{2})[/tex]
or
[tex]x^2+1=-x(\frac{(1 - \sqrt5)}{2})[/tex]
solve each quadratic gives us the roots

Thats right for this system Beta - can you solve it for the n-degree version?

The roots are not pretty at all.

[tex]\displaystyle \frac{-1- \sqrt{5} \pm i\sqrt{10-2\sqrt{5}}}{4}[/tex]
and
[tex]\displaystyle \frac{\sqrt{5}-1 \pm i\sqrt{10+2\sqrt{5}}}{4}[/tex]

Rep, how did you solve this?

what's 5th root of unity???

What's are the angles for, and how can it be expanded into complex number???

Sorry for my ignorance in complex numbers/trig, i don't really know much about complex numbers other than the definition.

I m not really sure that I understand your question but I'll try to tell you what I know about the roots of unity.  Roots of unity means the roots of one. That is when you have an equation like [tex]x^4=1[/tex] or [tex]x^8=1[/tex].  The roots of unity are the values of x that satisfy these equations (including complex).  The interesting thing about the roots of unity is that they will space theirself evenly out among the unit circle in the complex plane.  For the equation we are dealing with there are 5 roots i.e. 5 vertices.  Since the unit circle is centered at the origin each root will be 1 unit away from the origin. Since one root is obviously at (1;0 :^o: ).  The others will be spaced evenly so the angles from the origin will be equal throughout.  So the other roots are 360/5=72 degrees apart.  Thus they are (1; 72 :^o: ) (1;144 :^o: )(1;216 :^o:  )(1; 288 :^o: )and we already establishe(1;0 :^o: ) was one.  Oh yeah and I used polar coordinates because they are much easier to work with here.  I hope this helped but you may want someone smarter than me to explain it better.

Here's my solution, which is sort of like Rep123max's solution, but slightly different.

We see the symmetric sums, and think that a, b, c, and d are roots of the polynomial x^4 + x^3 + x^2 + x+ 1 which equals (x^5 - 1)/(x-1).   The solutions to x^5 - 1 are the fifth roots of one.  We only need four of them, since 1 doesn't work.  We see that our solutions are

[tex]
e^{2\pi i/5}
[/tex]

[tex]
e^{4\pi i/5}
[/tex]

[tex]
e^{6\pi i/5}
[/tex]

[tex]
e^{8\pi i/5}
[/tex]

Thanks, I learned A LOT today. Unfortunately I don't have AoPS books, I should get them soon.

LUCKILY, the question asked is a 4th-degree polynomial, which I could solve by using (somewhat cleverly) quadratic formula.  (I tried this problem for almost 2 hours for the n-degree version and got nowhere) :wink:

Honestly, I still don't quite get some of this.

First, what's complex plane? i have heard it once or twice but i can't remember what it is

Second, why are the roots of unity evenly spread on a unit circle?

The complex plane is just like the Cartesian plane(x-y plane) but is extended for complex numbers.   Complex numbers are numbers of the form a+bi.  Note that real numbers fall under the category of complex numbers because they just have b=0.  For example 5=5+0i or :rt2: = :rt2: +0i.  Now to graph points in the complex plane you do it just like the x-y but you use a-b e.g. 4+5i would be graphed like the point (4,5) in the x-y.  I can tell you more if you like but maybe you should look this up on the internet somewhere.  I'm sure they have some great stuff somewhere.

I'm not really sure why the roots of unity are evenly spaced.  Someone with more experience probably could though.

Thanks, and sorry if I asked too many stupid questions.

The complex plane is basically the cartesian coordinate plane, but instead of writing an ordered pair (x,y) we write x+iy. The reason the roots of unity are evenly spaced is because of DeMoivre's Theorem. This theorem says that

[tex]\big(r(\cos\theta+i\sin\theta)\big)^n = r^n\big(\cos n\theta+i\sin n\theta\big)[/tex]

for any real n.

Using this theorem, we can solve x^n=1. Write x in the form [tex]x=r(\cos\theta+i\sin\theta)[/tex]. Then, we need

[tex]r^n(\cos n\theta+i\sin n\theta)=1[/tex]

i.e. r=1 and [tex]n\theta[/tex] is a multiple of [tex]2 \pi[/tex]. So, theta must be a multiple if [tex]2 \pi/n[/tex]. Well, there are n distinguishable multiples of [tex]2\pi/n[/tex], namely

[tex]0\cdot2\pi/n[/tex]
[tex]1\cdot2\pi/n[/tex]
[tex]2\cdot2\pi/n[/tex]
[tex]\vdots[/tex]
[tex](n-1)\cdot2\pi/n[/tex].

The numbers with these angles and each with a magnitude of 1 are equally spaced on the unit circle, i.e. they form a regular polygon.

Thanks to all of you!! I got it now.

Well, something I discovered:

[tex]cos72=\frac{\sqrt{5}-1}{4}[/tex]
[tex]cos36=\frac{\phi}{2}[/tex]


1. Her face was a perfect oval, like a circle that had
its two sides gently compressed by a Thigh Master.

2. His thoughts tumbled in his head, making and breaking
alliances like underpants in a dryer without Cling Free.

3. He spoke with the wisdom that can only come from experience, like a guy who went blind because he looked at a solar eclipse without one of those boxes with a pinhole in it and now goes around the country speaking at high schools about the dangers of looking at a solar eclipse without one of those boxes with a pinhole in it.

4. She grew on him like she was a colony of E.-Coli and he was room-temperature Canadian beef.

5. She had a deep, throaty, genuine laugh, like that
sound a dog makes just before it throws up.

6. Her vocabulary was as bad as, like, whatever.

7. He was as tall as a six-foot-three-inch tree.

8. The revelation that his marriage of 30 years had disintegrated
because of his wife's infidelity came as a rude shock, like a
surcharge at a formerly surcharge-free ATM.

9. The little boat gently drifted across the pond
exactly the way a bowling ball wouldn't.

10. McBride fell 12 stories, hitting the pavement
like a Hefty bag filled with vegetable soup.

11. From the attic came an unearthly howl. The whole scene
had an eerie, surreal quality, like when you're on vacation in
another city and Jeopardy comes on at 7:00 p.m. instead of 7:30.

12. Her hair glistened in the rain like a nose hair after a sneeze.

13. The hailstones leaped from the pavement, just
like maggots when you fry them in hot grease.

14. Long separated by cruel fate, the star-crossed lovers raced
across the grassy field toward each other like two freight trains,
one having left Cleveland at 6:36 p.m. traveling at 55 mph, the
other from Topeka at 4:19 p.m. at a speed of 35 mph.

15. They lived in a typical suburban neighborhood
with picket fences that resembled Nancy Kerrigan's teeth.

16. John and Mary had never met. They were like
two hummingbirds who had also never met.

17. He fell for her like his heart was a mob
informant and she was the East River.

18. Even in his last years, Grand pappy had a mind
like a steel trap, only one that had been left out
so long, it had rusted shut.

19. Shots rang out, as shots are wont to do.

20. The plan was simple, like my brother-in-law Phil.
But unlike Phil, this plan just might work.

21. The young fighter had a hungry look, the kind
you get from not eating for a while.

22. He was as lame as a duck. Not the metaphorical lame duck,
either, but a real duck that was actually lame, maybe from
stepping on a land mine or something.

23. The ballerina rose gracefully en pointe and extended
one slender leg behind her, like a dog at a fire hydrant.

24. It was an American tradition, like fathers
chasing kids around with power tools.

25. He was deeply in love. When she spoke, he thought
he heard bells, as if she were a garbage truck backing up.

26. Her eyes were like limpid pools, only they had
forgotten to put in any PH cleanser.

27. She walked into my office
like a centipede with 98 missing legs.

Some of these could be pretty effective in the right context...For example, sentences like
[quote]The little boat gently drifted across the pond exactly the way a bowling ball wouldn't[/quote]
Are occasionally useful to humor writers. Presumably these students weren't using them right, though.

Another good one to add to that list:

28. It hurt the way your tongue hurts when you accidentally staple it to the wall.

this seems to be stolen from somewhere i know

Psh, yeah right.
And if you were thinking of anything related to you, you are quite wrong.

i was obviously not think of anything related to me in particular how did u get that notion? the sneezing one and the centipede one is funny


function
induction
real analysis
real analysis solved

Let f be a real continuous function such that there exists n with the property that $ f^n$ is bounded. Then either $f$ or $f^2$ is bounded. Here product mean obviously composition.

Let $I_n=f^n(x)$. If $I_1=\mathbb R$, we're done. If we assume $I_1,I_2$ to be unbounded, then they must be intervals unbounded at the same end (either left or right, but not both), s.t. $I_2\subseteq I_1$.

Assume $I_1,I_2,\ldots,I_k$ are unbounded. We have $I_{k+1}=f(I_k)$. If $I_{k+1}$ is bounded, then we write $I_k=f(I_k\cup I_{k-1}\setminus I_k)\subseteq f(\overline{I_k}\cup\overline{I_{k-1}\setminus I_k})$. $f$ is then bounded on both the sets in the last brackets (because we assumed it was bounded on $I_k$, and $\overline{I_{k-1}\setminus I_k}$ is a bounded closed interval), meaning that $I_k$ is, in fact, bounded, contradiction.

We have just shown that $I_{k+1}$ is also unbounded, so we may proceed by induction: $I_1,I_2$ unbounded $\Rightarrow\ I_k$ unbounded, $\forall k\ge 1$.


Αν  p είναι πρώτος της μορφής 4n+1, τότε δείξτε ότι
 $[\sqrt{p}]+[\sqrt{2p}]+...+[\sqrt{np}]= \frac{p^{2}-1}{12}$

To parapanw thema exei sizitithei edw kai einai pragmati ena poli omorfo kai diskolo provlima me mi tetrimmeni kai klassiki lisi.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=2p+sqrt+sqrt&t=5928[/url]

Se sinexeia tou minimatos tou Silouan gia ton prwto arithmo $p$, mou irthe sto mialo kai tha ithela na apodeiksete (kai na exete genika ipopsin sas stous diagwnismous) oti enas prwtos arithmos grafetai san athroisma $2$ tetragwnwn an-n einai tis morfis $4n+1$ (diladi $p\equiv 1 \pmod{4}$)

Alexandros

ωραια λυση. :!: οσο για αυτο που λες Αλεξαντρε ειναι σε καποιο ποστ εδω στο ελληνικο ,αλλα επειδη ειναι πολυ καλο ας το προσπαθησουμε.

Oντως η λυση ειναι ωραια,αλλα λιιιιιιιιιιγο......δυσκολη στο να σκεφτεις απο που να ξεκινησεις........
Αnyway,εγω οταν την πρωτοειδα,επειδη ο p[ηταν της μορφης 4n+1 και ζητουσαμε το ακεραιο μερος αυτου μου ηρθε στο μυαλο ενα λημμα

[b][riza(4n+1)]=[riza(4n+2)]=[riza(4n+3)]=[riza(2n)+riza(2n+1)][/b]

Eπισης απρδειξα οτι [riza(np)]=2n ,[riza((n-1)p)]=2n-1.[riza((n-2)p)]=2n-3
(φραζοντα στα μεταξυ δυο διαδοχικων ακεραιων...κλασσικα!!)αλλα δεν καταφερα να βρω την γενικη μορφη των αριθμων που λειπουν απο αυτο το αθροισμα  (οπως π.χ ο 2n-2).Αν καποιος δουλεψε ετσι την ασκηση και προχωρησε παραπερα,αν μπορει ας γραψει την λυση του...
 :D  :D  :D

[quote="cretanman"]Se sinexeia tou minimatos tou Silouan gia ton prwto arithmo $p$, mou irthe sto mialo kai tha ithela na apodeiksete (kai na exete genika ipopsin sas stous diagwnismous) oti enas prwtos arithmos grafetai san athroisma $2$ tetragwnwn an-n einai tis morfis $4n+1$ (diladi $p\equiv 1 \pmod{4}$) [/quote]

Ωραίο πρόβλημα .
Έστω $p=a^{2}+b^{2}\equiv 0,1,2(mod\ 4)$  .Τα δύο από αυτά υτά απορρίπτονται άρα $p\equiv 1(mod\ 4)$

Τώρα θα δείξουμε ότι αν $p\equiv 1(mod 4)$  τότε υπάρχουν $a,b\in Z$ ώστε $p=a^{2}+b^{2}$ .
Σύμφωνα με το λήμμα Croecker αφού  $p\equiv 1(mod 4)$
υπάρχει $k$  ώστε $p|(k^{2}+1)$  .Άρα ο p διαιρεί τον
$(k+i)(k-i)$ . Αφού τώρα ο $p$ είναι πρώτος θα είναι πρώτος και στο $Z[i]$ άρα ο p θα διαιρεί ή τον k+i  ή τον k-i .
Προφανώς και στις δύο περιπτώσεις έχουμε άτοπο άρα ο $p$ είναι σύνθετος στο $Z[i]$ άρα $p=(a+bi)(a-bi)=a^{2}+b^{2}$
το ζητούμενο .

***Υπάρχει και μια πιο στοιχειώδης λύση που χρησιμοποιεί το πρώτο λήμμα kai to λήμμα Thue


Let $ n$ be an integer.  If the tens digit of $ n^{2}$ is $ 7$, find the units digit.

Well, i could just bash it by knowledge that $ 24^2=576$ but lets look for a better way.

if $ n^2\equiv7x \mod 100$ where x is a digit 0-9, we can just consider the cases. Now we know that $ n^2$ can only end in 0, 1, 4, 5, 6, or 9, so we only have to consider those cases.

if $ n^2\equiv70 \mod 100$ then $ n\equiv0 \mod 10$. but if $ n\equiv 0 \mod 10$, then we know $ n^2\equiv 0 \mod 100$. So thats not right.

if $ n^2\equiv71 \mod 100$ then n is in the form $ 10p\pm1$ for some integer p. squaring it we get $ (10p\pm1)^2=100p^2\pm20p+1$. the 10s digit will be $ 0p^2\pm2p$, which, for all integral p, is even. but our tens digit is 7. so we are wrong again

if $ n^2\equiv 74 \mod 100$ then n is in the form $ 10p\pm2$ for some integer p. squaring it gives us $ 100p^2\pm40p+4$, which doesnt work, because once again, the tens digit will always be even

if $ n^2\equiv 76 \mod 100$ then n is in the form $ 10p\pm4$ which, squared, is $ 100p^2\pm80p+16$, which its 10s digit will always be odd, because it is in the form $ 0p^2\pm8p+1$ which is always odd

if $ n^2\equiv 75 \mod 100$ then we have the same dilemma as we did for $ n^2\equiv 71 \mod 100$.

if $ \n^2\equiv 79 \mod 100$, once again, same dilemma as above

so, our answer is 6

[hide="Much Quicker :) "]
As mz94 noted from base $ 10$ analysis, the only possibilities for the last two digits are $ 70,71,74,75,76,79$.

Since all perfect squares leave remainder $ 0,1\pmod4$ and the $ \pmod4$ residue is determined uniquely by the last two digits, we now also have that the only possibilities for the last two digits are $ 72,73,76,77$.

Clearly, the only intersection of the two sets is $ 7\boxed6$.
[/hide]

oh yea mod 4. forgot about that

*slaps forehead*

[quote="Z-coli"]wow mike you fail.[/quote]

lets put this in perspective james:

who failed PUMaC combinatorics?


number theory
greatest common divisor
number theory theorems

[quote="mao_zai"]Looking for a simple proof.

Let $ p$, $ q$,$ d$ be integers. Prove that $ \gcd(p,d)\gcd(q,d)\ge\gcd(pq,d)$.[/quote]
Let $ (a,b) \equal{} gcd(a,b)$(It is a little shorter :P)
See that $ (pq,d) \equal{} (p,\frac{d}{(q,d)})(q,d) \le (p,d)(q,d)$ QED.


quadratics
number theory
number theory unsolved

Let $ q \equal{} 2p\plus{}1$, $ p, q > 0$ primes. Prove that there exists a multiple of $ q$ whose digits sum in decimal base is positive and at most $ 3$.

Let $ q$ is prime. If $ q\equal{}2$ or $ q\equal{}5$ we find $ N\equal{}10$, suth that $ q|N, S(N)\equal{}1$. If $ q\equal{}3$, then $ q|N\to 3|S(N)$, therefore minimal multiple is N=3 and $ S(N)\equal{}3$.
Let prime $ q>5$. Then $ q|N, S(N)\equal{}2$ if and only if $ \binom{10}{q}\equal{}\minus{}1$, because $ q|10^{(q\minus{}1)/2}\plus{}\binom{10}{q}$.
If $ \binom{10}{q}\equal{}1$, then for your case $ q|10^p\minus{}1$ and $ a_n\equal{}10^n\mod q, n\equal{}0,1,...,p\minus{}1$ are different and belongs $ A\equal{}\{1,2,...,2p\}$.  If $ b\in A$, then $ q\minus{}b\not \in A$.
If $ a\in A, a\plus{}1\not \in A$, then $ a\equal{}10^n\mod q, q\minus{}a\minus{}1\equal{}10^m\mod q$ and $ q|10^n\plus{}10^{m}\plus{}1$. Obviosly exist suth a in A.

What is this symbal
$ \binom{10}{q}\equal{}1$
I don't understand the second line.
 :|

[quote="Hinagiku"]What is this symbal
$ \binom{10}{q} \equal{} 1$
I don't understand the second line.
 :|[/quote]

it's the [url=http://en.wikipedia.org/wiki/Quadratic_reciprocity#Legendre_and_his_symbol]Legendre symbol[/url].

I'm confuse because it doesn't have " _ "
 :)

Since $q=2p+1$, by Fermat's Little Theorem, $10^{2p}\equiv_q1$, therefore $100^p\equiv_q1$, therefore $\mbox{ord}_q100\in\left\{1,p\right\}$.
If $\mbox{ord}_q100=1$, then $100\equiv_q1$, therefore $q|99$, therefore $q\in\left\{3,11\right\}$, and in both cases it's easy to find a multiple of $q$ in the asked conditions.
If $\mbox{ord}_q100=p$, then $100^k$ has exactly $p$ possible values modulo $q$. Since $100^k$ is always a quadratic remainder modulo $q$ and there are exactly $p$ quadratic remainders modulo $q$, $100^k$ may be any quadratic remainder modulo $q$. It's a well known fact that $x^2+y^2+1\equiv_q0$ has solutions $x,y$, therefore there are $a,b$ such that $10^a+10^b+1\equiv_q0$, which solves the problem.

If $q=5$, we can simply take $10$ and so we may assume that $q$ and $10$ are coprime. Consider $ \mbox{ord}_q 10\in \{1, 2, p, 2p\}.$ If $\mbox{ord}_q =1$, we would obtain $q=1$, which is impossible.  If $\mbox{ord}_q =2$, we would have $10^{1} \equiv -1 $ (mod $q$) and so $10^{1} +1$ would work. Similarly, if $\mbox{ord}_q =2p$, we have $10^{p} \equiv -1 $ (mod $q$) and so $10^{p} +1$ works.

The final case is $\mbox{ord}_q =p$. Then the integers $10^{0}, 10^{1}, \ldots, 10^{p-1}$ leave  $p$ distinct remainders modulo $q.$ If we have $10^{a} \equiv p$ (mod $q$) for some $a$, the integer $2 \cdot 10^{a} +1$ would satisfy the conditions. Otherwise there are integers $a,b$ such that $10^{a} +10^{b} \equiv -1$ (mod $q$), because there are only $p-1$ pairs of the form $(1, 2p-1) , (2, 2p-2) , \ldots, (p-1, p+1)$ and then $10^{a} +10^{b} +1$ satisfies all conditions.

$ 10^{2p} \equiv 1 mod(q) \implies ord_q(10) \in \left\{1,2,p,2p\right\}$
the first two cases are easy.
- $ ord_q(10) =2p $:
$ q | 10^p =1 $.
- $ ord_q(10) =p $:
$ \left(\frac{10}{q}\right) =1 \implies 10^i $ is  quadratic remainder modulo $q$ and there is a solution for $ x^2+y^2 +1 \equiv 0 mod q$.


Let $ A_1,A_2,...,A_n$ be $ n$ distinct points in the plane $ (n > 2)$. Let $ a \equal{} max A_iA_j , b \equal{} min A_iA_j (1\leq i< j\leq n)$. Prove that :
$ \frac {a}{b} \geq \frac {\sqrt {3} }{2} (\sqrt {n} \minus{} 1)$

Nobody can solve !
What a difficult problem ! How about using the induction?


1) If $ A\subset B\subset\bar{A}\subset M$, and if $ A$ is connected, show that $ B$ is connected.  In particular, $ \bar{A}$ is connected ($ M$ is a metric space).

[hide="My Thoughts"]Assume that $ B$ is not connected, so $ B\equal{}C\cup D$ with $ C\cap D\equal{}\emptyset$.  Then $ (C\cap A)\cup (D\cap A)\equal{}A$ and $ (C\cap A)\cap (D\cap A)\equal{}\emptyset$.  But $ C\cap A$ is closed in $ M$, as is $ D\cap A$, so from the previous step, $ A$ is disconnected, a contradiction.  I have no idea if this correct.[/hide]

2) If $ f: [a,b]\rightarrow [a,b]$ is continuous, show that $ f$ has a fixed point; that is, show that there is some point $ x$ in $ [a,b]$ with $ f(x)\equal{}x$.

[hide="My Thoughts"]This seems intuitively clear.  However, I can't actually make the algebra work.  I was thinking a proof by contradiction.  Assume that if $ c,d\in [a,b]$, then $ f(c)\equal{}d$ $ \forall c\in [a,b]$ and then somehow conclude that $ f$ is not continuous, but I couldn't get anywhere.[/hide]

For the second, try defining a new function $ g: [a, b] \to \bf R$ by $ g(x) \equal{} x \minus{} f(x)$.

If $ g(a)<0<g(b)$, then $ g(c)\equal{}0$ for some $ c\in [a,b]$, so $ 0\equal{}f(c)\minus{}c\Rightarrow f(c)\equal{}c$ for some $ c$.
But what if it is NOT true that $ g(a)<0<g(b)$?

Do I have the right idea for the first problem?

[quote="JRav"]what if it is NOT true that $ g(a) < 0 < g(b)$?[/quote]

Then we either have $ f(x) < x$ on the interval or $ f(x) > x$ on the interval.  In particular, this is true at the endpoints.  Why is this impossible?

Why don't you actually write down what $ g(a)$ and $ g(b)$ are equal to?

I think I've got it: $ f(b)\leq b$ and $ a\leq f(a)$.  Then $ g(b)\equal{}f(b)\minus{}b\leq 0$ and $ g(a)\equal{}f(a)\minus{}a\geq 0$.  Combining them, $ g(b)\leq 0\leq g(a)$.  Then $ g(x)\equal{}0$ for some $ x\in [a,b]$ so $ 0\equal{}f(x)\minus{}x\Rightarrow f(x)\equal{}x$.

Am I on the right track for the second one?

One interesting question is whether if two continuous mappings of an interval into itself which commute must have a common fixed point.

For the first one, if $ B$ is disconnected, then $ B\equal{}U\cup V$, where $ U$ and $ V$ are disjoint open intervals, where $ U\equal{}E\cup B$ and $ V\equal{}F\cup B$ for open sets $ E$ and $ F$.  I know that's right.  Since $ A\subset B$, $ E\cup A\subset E\cup B\equal{}U$ and $ F\cup A\subset F\cup B\equal{}V$, right?  Since $ U$ and $ V$ are disjoints, subsets of $ U$ and $ V$ are disjoint as well, true?  Then $ (E\cup A)\cap (F\cup A)\equal{}\emptyset$ and $ (E\cup A)\cup (F\cup A)\equal{}A$.  But this last step implies $ A$ is disconnected, a contradiction.  So $ B$ must be connected.  It's the last step I'm not too comfortable with.

[quote="JRav"]if $ B$ is disconnected, then $ B \equal{} U\cup V$, where $ U$ and $ V$ are disjoint open intervals, where $ U \equal{} E\cup B$ and $ V \equal{} F\cup B$ for open sets $ E$ and $ F$.[/quote]  You might want to try this sentence again -- at the moment, it's incomprehensible.  Actually, the whole thing is riddled with either notational ($ \cap$ vs. $ \cup$?) or conceptual errors, e.g., [quote] $ (E\cup A)\cap (F\cup A) \equal{} \emptyset$[/quote] can't possibly be what you were aiming for (it implies $ A \equal{} \emptyset$ immediately).

The problem with your first solution JRav is that you didn't discuss whether or not your disconnection of $ A$ really consists of  two non-empty sets. That's where $ B \subset cl(A)$ comes in.

Why do you keep just guessing on these problems instead of verifying what you did was correct?

For the second problem, if $ f$ has no fixed point then $ \{x | f(x) > x\}$ and $ \{x | f(x) < x\}$ are a disconnection of $ [a,b]$

It suffices to show that $ \bar{A}$ is connected (since if $ A\subset B\subset\bar{A}\subset M$, then $ B\equal{}\text{cl}_{B}(A)$, we can replace $ M$ by $ B$).  Suppose $ \bar{A}\equal{}H\cup K$, with $ H$ and $ K$ disjoint, non-empty open sets in $ \bar{A}$.  Then $ A\equal{}(H\cap A)\cup (K\cap A)$, and the latter are disjoint, non-empty open sets in $ A$.  Then if $ \bar{A}$ is disconnected, so is $ A$.


Murray will obviously win a Grand Slam eventually, unless before that he somehow wins a US Open if Federer fails that.

Um? I never said Murray wouldn't win a Grand Slam. All I said was that Nadal has persistent knee injuries, which suck pretty badly.


I changed my status message to "whats wrong w/ bein a nerd" and this is what came out:

trivialness011 (8:34:47 PM): yes
trivialness011 (8:34:49 PM): if your a nerd
trivialness011 (8:34:51 PM): you love schoo;
trivialness011 (8:34:55 PM): which is an absolute crime
trivialness011 (8:35:02 PM): so you muyst be banished from this world.
tehbeastlyone (8:35:07 PM): no
tehbeastlyone (8:35:09 PM): i dont love school
tehbeastlyone (8:35:09 PM): and
tehbeastlyone (8:35:10 PM): anyway
tehbeastlyone (8:35:12 PM): lots of nerds
tehbeastlyone (8:35:14 PM): dont like school
trivialness011 (8:35:39 PM): No
trivialness011 (8:35:42 PM): nerd = love school
trivialness011 (8:35:44 PM): nerd =
trivialness011 (8:35:45 PM): person
tehbeastlyone (8:35:50 PM): NO
trivialness011 (8:35:51 PM): who fails at real math, good at fake math
tehbeastlyone (8:35:54 PM): nerd=social abonination
trivialness011 (8:35:56 PM): no.
trivialness011 (8:36:02 PM): Nerd = stereotype
trivialness011 (8:36:05 PM): nerd = bad
trivialness011 (8:36:14 PM): nerd = not a spammer, implies has no math skill
trivialness011 (8:36:23 PM): all spammers have math skill, and vice versa.
tehbeastlyone (8:36:25 PM): lol
trivialness011 (8:36:28 PM): nerd = fai
trivialness011 (8:36:33 PM): nerd = phail
trivialness011 (8:36:38 PM): nerd != alex zhai
trivialness011 (8:36:41 PM): nerd is not cool
trivialness011 (8:36:44 PM): because he fails at math
tehbeastlyone (8:36:45 PM): rofl
trivialness011 (8:36:49 PM): therefore
trivialness011 (8:37:03 PM): Nerd = NOT FROM SOVIET RUSSIA = NOT KILLER SHRIMP = BAD
trivialness011 (8:37:33 PM): ok?
tehbeastlyone (8:38:26 PM): k
trivialness011 (8:38:37 PM): nerd = mc person
trivialness011 (8:38:41 PM): nerd != usamo person
trivialness011 (8:38:44 PM): nerd != terrence tao
trivialness011 (8:38:50 PM): nerd != riemann hypothesis
trivialness011 (8:38:53 PM): nerd is not a maht problem
trivialness011 (8:39:00 PM): nerd is not a computer science problem
trivialness011 (8:39:04 PM): nerd is a FAIL problem
trivialness011 (8:39:10 PM): therefore nerd fails.
trivialness011 (8:39:27 PM): qed, ok?

Hmm... depends on how you define a nerd. My definition might be a little different then yours.

Actually, what is seven circles theorem?

Mieh... I just said some random stuff....

LOLLLLLLLLLLLLLLLLLLLL

pythag011, you win. like seriously lol

[quote="pythag011"][quote="andersonw"]By the way, pythag, you've probably seen this already, but recursion could have been used for that 5x5 USACO problem, no stacks needed.[/quote]

But recusion is in effective a stack.

mieh, doesn't really matter...

but recusion is basically a function stack.[/quote]

Ah, okay. You can see that I'm still not very familiar with stacks yet.

A nerd is someone who enjoys and loves learning, especially in a specific academic subject.

I am a nerd because I love math. I dislike school, because I rarely learn anything in it, and I only learn things in boring subjects (bio, english, etc..)

However, if I did not attend school, I couldn't do any math competitions :(, so I go.


inequalities
Cauchy Inequality
inequalities solved

Show  that
1/(x1^n) + 1/(x2^n) + ...+ 1/(xk)^n >=k^(n+1)  for reals  xi  with  sum 1
thanks!

I really doubt they aren't positive. I have a solution for positive:
  Sum 1/(x_i)^n>=k/(x_1*...*x_k)^n/k. But (x_1*...*x_k)^1/k<=(sum x_i)/k=1/k and everything is clear.
  Find another solution using Jensen and the convexity of x^n.

I think the problem must have : x_i are positive real numbers
Solution :
  We have 1/(x_i)^n > 0. The cauchy inequality gives :
 \sum 1/(x_i)^n  \geq k \sqrt (1/(x_1)^n*1/(x_2)^n*...*1/(x_k)^n. (1)
 and : x_1*x_2*...*x_k \leq ((x_1+x_2+...x_k)/k)^k=1/k
 hence 1/(x_1)^n*...*1/(x_k)^n  \geq k^(kn)                                 (2)
From (1) and (2) we have :
  1/(x_1)^n+...+1/(x_k)^n  \geq k^(n+1)


topology Euler Gauss Stanford college MIT geometry

Is it me, or does it seem like there is a lot less respect for applied mathematicians than "theoretical" mathematicians (those who do research in theoretical analysis, algebra, topology, number theory. I get the feeling people think the word "pure" is more concentrated and for smarter people, and applied means just plug-chug. (Not that I do, though.) Think of this: most of the world's most famous mathematicians are best for known for theoretical research, though they may have dabbled in applications: (Cauchy, Euler, Wiles, Gauss, Erdos). The same holds for physicisits: (Einstein, Feynman, Newton, etc). I would especially like to know adults thoughts on this, such as Djiminez, DPatrick, Rruczysc(sp?), MCawford, Ravi, Gyan, tokenadult, Valentin, etc.

Thanks for replying. I am especially witing this because I like applied math more than pure math.

Thank you for asking my direct opinion  :blush: .

First than all, the line separating [i]pure[/i] mathematics from [i]applied[/i] mathematics is very thin, a lot thinner than what most of the people think, specially because basically all the mathematics has proven to be applicable to something in other sciences.

Of course, as you are getting deeper in one of the extremes, there are a lot of differences. In my opinion, if you want to do something too theoretical, the horizon is farther away (I think), mainly because in the last century the applications have grown at an exponential rate, probably faster than the [i]pure[/i] mathematics, therefore, even when you can find hard problems, there are more (a lot more) "easy" problem to solve. While in the more "pure" mathematics, is quite hard to find an "easy" problem. Also, there is more money for applied math than for pure math, and well, mathematicians also need to eat!

Now, probably the idea comes from the fact that there are more people doing "applied" math than "pure" math, and of course, if few people do something, people may think that it is because it is harder to do. I don't know, it is just a few ideas in my head. Let's see what does other people think.

(By the way, my research is in Applied Math :P )

I am sure you can find many famous applied mathematicians

In today's modern world, I do not think there is so much of a schism between applied and pure mathematicians. In fact, I think they work together.

Of course, I am a mere minor, hence I know little of this.

[quote="1234567890"]Is it me, or does it seem like there is a lot less respect for applied mathematicians than "theoretical" mathematicians (those who do research in theoretical analysis, algebra, topology, number theory). I get the feeling people think the word "pure" is more concentrated and for smarter people, and applied means just plug-chug.[/quote]

I think you probably spend a lot of time talking to ("pure") mathematicians.

Hah.  I constantly make fun of applied mathematicians, but that's mostly because I really respect what they do.

Anyways, one of the genuine reasons for a difference is that in a sense, anyone can do applied math (in a non-academic setting).  The basic tools these days are mainly computers, and such proficiency can be trained.  Pure math, on the other hand, is something to which there really aren't many shortcuts, and hence its a bit more.  Of course, applied math in an academic setting is basically equivalent to pure math, but gets tarred with the applied brush just because we feel like it.  :lol:

I just talked to my friend who's a researcher in applied mathematics.  He names Gene Golub (Stanford), Gil Strang (MIT) and Nick Trefethen (Oxford) as three of the most respected applied mathematicians.  He acknowledges that pure mathematicians often don't have respect for applied math.  Applied mathematicians take comfort in the fact that they generally get paid significantly more than pure.

Applied mathematicians work on "real world" problems.  Of course pure math can also lead to solutions of "real" problems.  The use of large primes in cryptography is an example.  The difference is pure math researchers don't usually start with a problem applicable to the real world.  Applied researchers start with a problem in physics, chemistry or biology.  Often this involves solving a PDE.  They look for a unique solution, then a formulation for that solution.  Pure mathematicians would be satisfied at this point but applied mathematicians look further at the computational complexity (how long will it take to compute) and the stability (how accurately can it be computed) of the solution.  The instability that plagues problems like the N-body problem (meaning small perturbations in the input cause huge differences in the output) also plague methods for stable problems.

Applied mathematicians need a broad background.  A solid knowledge of pure math (linear algebra, real and complex analysis, functional analysis, ODEs, PDEs) is required.  Knowledge of computational science (numerical methods in computing) and other sciences (esp. quantum mechanics and solid state physics) is also necessary.

Exciting areas in applied math today include mathematical biology and quantum computation.  A look at the list of MIT math faculty shows that more than a third specialize in applied math.  Lack of respect does not deter them from studying important scientific problems.


find  $ n$ in $ Z$ such that :
$ i$. $ n$ has  $ 10$ divisor in $ Z$.
$ ii$. there are  two prime $ p,q$  numbers such that : $ n\equal{}16\plus{}pq$

A couple of examples -- and I believe the smallest two -- are 162 and 567.

could you show me  how do you find these solutions :oops:


Is it possible to divide a $ 10 \times 10$ board into:

$ (a)$ $ 4$ figures of type $ 1$ and $ 21$ figures of type $ 2$?
$ (b)$ $ 4$ figures of type $ 1$, $ 19$ figures of type $ 2$, and $ 2$ figures of type $ 3$?

a) Divide the board into 2 by 2 squares and color them alternately white and black in a checkerboard pattern. Each type 1 piece covers 2 white and 2 black squares while each type 2 piece covers three of one and one of another. There are an odd number of type 2 pieces; contradiction.

Oddly enough, I can't find a coloring that works for part (b). I'll think about this later.


$\{x,y,z\}\subset R\ ,\ x\ge y\ ,\ x\ge z\ ;\ x+y+z=1\ ,\ xy+yz+zx=\frac{1}{4}$ $\Longrightarrow$ $\sqrt x=\sqrt y+\sqrt z\ .$

$p^{4}+q^{4}+r^{4}-2p^{2}q^{2}-2q^{2}r^{2}-2p^{2}r^{2}=(p+q+r)(p+q-r)(p-q+r)(p-q-r)$

Hence

\begin{eqnarray*}x^{2}+y^{2}+z^{2}-2xy-2yz-2xz=(\sqrt{x}+\sqrt{y}+\sqrt{z})(\sqrt{x}+\sqrt{y}-\sqrt{z})(\sqrt{x}-\sqrt{y}+\sqrt{z})(\sqrt{x}-\sqrt{y}-\sqrt{z})\end{eqnarray*}

But the LHS is equal to $(x+y+z)^{2}-4(xy+yz+xz)=0$, hence either $\sqrt{z}=\sqrt{x}+\sqrt{y}$ or $\sqrt{y}=\sqrt{x}+\sqrt{z}$ or $\sqrt{x}=\sqrt{y}+\sqrt{z}$. However, because of $x\geqslant y\land x\geqslant z$, we have the third option. QED


geometry
combinatorics solved
combinatorics

Six blue, six red and six green points are marked on a plane, no three of them collinear. Prove that the sum of the areas of the triangles with vertexes of the same color is not greater than $1/4$ of the sum of the areas of all the triangles with marked vertexes.

This is Russia 2002. Look here :

http://www.mathlinks.ro/Forum/viewtopic.php?t=3300&highlight=color+colour+triangle

Pierre.


AMC
AIME
AMC 12
AMC 12 B
probability
AoPS Books

After reading and understanding both volumes of the AOPS series, how much did you improve your score on the amc 10/amc 12/aime, what was initial score and then your score after.

After about the first 8 chapters of AoPS 2 I improved about 30 points on the AMC 12. Also, after taking 3 of their courses online my AIME score improved by 12 points from the previous year.

I hit 130s in my AMC, and above 7 in AIME.

I've had a lot of success with the aops books, though i don't work from them exclusively. my freshman and sophomore year i looked through volume 2 mostly, but i'm also very involved with competitive math. i suggest learning from aops, but also learning from your peers. i suppose it depends on what your school is like, but this has got me 132.5 sophomore year on the AMC12B, and a 143(A) and 136(B) on this year's tests. I'm weaker at the AIME but I made a 10 this year.

[quote="codeblue87"]but also learning from your peers. [/quote]

This is a big part of why we built a forum on this site - students learn from each other extremely quickly.

In addition to 'learning from your peers', I'd recommend 'teaching your peers' (or those a little younger).  Nothing teaches you a subject more thoroughly than having to teach it.  I've learned far more math in the last 2 years while preparing to teach than I did when preparing for contests.

Another fun and helpful thing to do is come up with your own problems and share them with your friends.

richard is right, teaching is an extremely good way to solidify what you know. also for a year i almost constantly traded problems with a friend of mine (though they were not our own creations) like probability mentioned.


algebra
polynomial
Vieta
algebra proposed

Prove that for any $ P(x)\in\mathbb{Z}[x]$ there exist polynomials $ Q(x),R(x)\in\mathbb{Z}[x]$, such that
\[ P(x^{2000}) \equal{} R(x)\cdot Q(x)
\]
I don't know why, but I get the feeling that I posted the problem somewhen before.

you have to be more specific...coz you can take $ Q\equal{}1$ any way.

oh, well, I must've mentioned that each of $ Q(x), R(x)$ has a non-zero degree

Sorry, not quite sure what's going wrong here. If $ P(x) \equal{} x\plus{}2$, then $ P(x^{2000}) \equal{} x^{2000} \plus{} 2$ is irreducible by Eisenstein's, so no $ Q,R$ exist?

[quote="Erken"]hmm...I mixed it up, sorry.

For any $ R(x)\in\mathbb{Z}[x]$, you need to prove the existence of $ P(x)$ and $ Q(x)$...hope it is correct now.[/quote]

could somebody please post a solution to the problem.

$ R(x)*R(\minus{}x)\equal{}h(x^2)$.
$ R(x)*R(wx)R(w^2x)R(w^3x)R(w^4x)\equal{}g(x^5)$, where $ w\ne1$ is the root to $ x^5\equal{}1$.
It is not hard to show $ R(wx)R(w^2x)R(w^3x)R(w^4x)\in \mathbb Z[x]$.
Use this iteratively, we can find $ Q(x)$.

[quote="xxp2000"]$ R(x)*R( \minus{} x) \equal{} h(x^2)$.
$ R(x)*R(wx)R(w^2x)R(w^3x)R(w^4x) \equal{} g(x^5)$, where $ w\ne1$ is the root to $ x^5 \equal{} 1$.
It is not hard to show $ R(wx)R(w^2x)R(w^3x)R(w^4x)\in \mathbb Z[x]$.
Use this iteratively, we can find $ Q(x)$.[/quote]

I am not sure what you meant.

We want to show for any $ R$, we can find $ Q$ such that $ P(x^n) \equal{} R(x)Q(x)$, where $ n > 1$.
Let $ r_i,i \equal{} 1,\cdots,n$ be the roots of $ R$, we can write
$ R(x) \equal{} c\prod_{i \equal{} 1}^n(x \minus{} r_i)$.
We let $ A(x) \equal{} \prod_{i \equal{} 1}^n(x^n \minus{} r_i^n)$, obviously $ R(x)|A(x)$.
By Vieta's theorem and Newton-Girard Formula, we can show $ A(x)$ has only rational coefficients,
Now let $ P(x^n) \equal{} M*A(x)$, where $ M$ is a large integer to guarantee $ P$ has only integer coefficients. Done!


algebra
polynomial
induction
Diophantine equation
number theory proposed
number theory

The sequence $ (a_n)$ is defined by: $ a_0\equal{}a_1\equal{}1$ and $ a_{n\plus{}1}\equal{}14a_n\minus{}a_{n\minus{}1}$ for all $ n\ge 1$.
Prove that $ 2a_n\minus{}1$ is a perfect square for any $ n\ge 0$.

The brutal approach works. The characteristic polynomial for the sequence is $ \lambda^2 \minus{} 14\lambda \plus{} 1 \equal{} 0$, with roots $ 7 \pm 4\sqrt {3} \equal{} (2 \pm \sqrt {3})^2$. Therefore, after determining the coefficients (from the first two terms), the general formula is $ a_n \equal{} \frac {1} {4} \left ( (2 \plus{} \sqrt {3})^{2n \minus{} 1} \plus{} (2 \minus{} \sqrt {3})^{2n \minus{} 1} \right )$. It follows $ 2a_n \minus{} 1 \equal{} \frac {1} {2} \left ( \sqrt {(2 \plus{} \sqrt {3})^{2n \minus{} 1}} \minus{} \sqrt {(2 \minus{} \sqrt {3})^{2n \minus{} 1}} \right )^2$, since $ (2 \plus{} \sqrt {3})(2 \minus{} \sqrt {3}) \equal{} 1$. But $ (2 \plus{} \sqrt {3})^{2n \minus{} 1} \equal{} \frac {1} {2^{2n \minus{} 1}} (\sqrt {3} \plus{} 1)^{2(2n \minus{} 1)} \equal{} 2 \left ( \frac {1} {2^{n}} (\sqrt {3} \plus{} 1)^{2n \minus{} 1} \right )^2$ and $ (2 \minus{} \sqrt {3})^{2n \minus{} 1} \equal{} \frac {1} {2^{2n \minus{} 1}} (\sqrt {3} \minus{} 1)^{2(2n \minus{} 1)} \equal{} 2 \left ( \frac {1} {2^{n}} (\sqrt {3} \minus{} 1)^{2n \minus{} 1} \right )^2$. Therefore $ 2a_n \minus{} 1 \equal{} \left ( \frac {(\sqrt {3} \plus{} 1)^{2n \minus{} 1} \minus{} (\sqrt {3} \minus{} 1)^{2n \minus{} 1}} {2^n} \right )^2$.

It remains to check the quantity squared is an integer.
Denote $ A \equal{} \sqrt {3} \plus{} 1$ and $ B \equal{} \sqrt {3} \minus{} 1$. We have $ A^2 \plus{} B^2 \equal{} 8$ and $ A^2B^2 \equal{} 4$. But $ A^{2n \plus{} 1} \minus{} B^{2n \plus{} 1} \equal{} (A^2 \plus{} B^2)(A^{2n \minus{} 1} \minus{} B^{2n \minus{} 1}) \minus{} A^2B^2(A^{2n \minus{} 3} \minus{} B^{2n \minus{} 3})$, hence $ \frac {A^{2n \plus{} 1} \minus{} B^{2n \plus{} 1}} {2^{n \plus{} 1}} \equal{} 4\frac {A^{2n \minus{} 1} \minus{} B^{2n \minus{} 1}} {2^n} \minus{} \frac {A^{2n \minus{} 3} \minus{} B^{2n \minus{} 3}} {2^{n \minus{} 1}}$, so it is proven by induction.

In fact we have proven that $ 2a_n \minus{} 1 \equal{} b_n^2$, with the sequence $ (b_n)_{n\geq 0}$ given by $ b_0 \equal{} \minus{} 1$, $ b_1 \equal{} 1$ and $ b_{n \plus{} 1} \equal{} 4b_n \minus{} b_{n \minus{} 1}$.
(This particular relation is prone to many developments and has been often used; see also TST-Romania 2006).

Thanks for your solution mavropnevma. :first:

Shoki, I would like to see your solution using Pell Equation.

put $ 2a_n\equal{}b_n$
then consider this pell equation : $ x^2\minus{}3y^2\equal{}1$
as u can see we have $ b_1^2\minus{}3(1)^2\equal{}1$ and also $ b_2^2\minus{}3(15)^2\equal{}1$ and also ...
since $ (x_0,y_0)\equal{}(2,1)$ is the fundamental answer of this pell equation,and also the formula which gives all other solutions is
$ x_{n\plus{}1}\equal{}2x_n\plus{}3y_n \ , \ y_{n\plus{}1}\equal{}x_n\plus{}2y_n$
we have $ 2 | x_{2k}$
after finding the general formula for the sequence $ b_n$ and also  the general formula for the sequence $ x_{2k}$ we can easily understand that $ b_n\equal{}x_{2k}$ (maybe u can prove this also by induction!)
so for all $ n$ we have $ b_n^2\minus{}3y_{2k}^2\equal{}1$
which gives
$ (b_n\minus{}1)(b_n\plus{}1)\equal{}3y_{2k}^2$
since $ b_n$ is even so we get $ \gcd(b_n\minus{}1,b_n\plus{}1)\equal{}1$
and so there exists $ (r,s) \in N^2$ such that
$ b_n\minus{}1\equal{}r^2 \ , \ b_n\plus{}1\equal{}3s^2 \ , rs\equal{}y$
and so we get what we wanted.
(note that the case $ b_n\minus{}1\equal{}3r^2 \ , \ b_n\plus{}1\equal{}s^2 \ , rs\equal{}y$ is easily omitted since $ 3r^2\equal{}s^2\minus{}2$ which is a contradiction(modulo $ 3$)) :)

The first few values of $2a_n-1$ for $n\ge 2$ are $5^2,19^2,71^2,265^2$.

Consider the sequence $b_1=1,b_2=5$ and $b_n=4b_{n-1}-b_{n-2}$ for $n\ge 3$. Indeed the first few values are $5,19,71,265$. We will prove by induction that $2a_n-1=b_n^2$, clearly the first few values work.

Not let us suppose we have $2a_i-1=b_i^2$ for all $i\le n$. This means $a_n=\frac{b_n^2+1}{2},a_{n-1}=\frac{b_{n-1}^2+1}{2}$.

Now $a_{n+1}=14a_n-a_{n-1}=14\left(\frac{b_n^2+1}{2}\right)-\frac{1}{2}-\frac{b_{n-1}^2}{2}=7b_n^2-\frac{b_{n-1}^2}{2}+\frac{13}{2}$. So we aim to prove $\frac{b_{n+1}^2+1}{2}=7b_n^2-\frac{b_{n-1}^2}{2}+\frac{13}{2}$.

Now
\begin{align*}\frac{b_{n+1}^2+1}{2} & =7b_n^2-\frac{b_{n-1}^2}{2}+\frac{13}{2}\\ \iff b_{n+1}^2+1 & =14b_n^2-b_{n-1}^2+13\\ \iff (4b_n-b_{n-1})^2 & = 14b_n^2-b_{n-1}^2+12\\ \iff b_n^2+b_{n-1}^2& =4b_nb_{n-1}+6\end{align*}
after simplifying.

To prove $b_n^2+b_{n-1}^2 =4b_nb_{n-1}+6$ we will again use induction. Indeed it is true for the first few values of $n$. Now assuming $b_n^2+b_{n-1}^2 =4b_nb_{n-1}+6$ we see:
\begin{align*}b_n^2+b_{n-1}^2& =4b_nb_{n-1}+6\\ 16b_n^2-8b_nb_{n-1}+b_{n-1}^2+b_n^2 & =16b_n^2-4b_nb_{n-1}+6\\ (4b_n-b_{n-1})^2+b_n^2 & =4(4b_{n}-b_{n-1})b_n+6\\ b_{n+1}^2+b_n^2 & =4b_{n+1}b_n+6\end{align*}
This completes both inductions, so that $2a_n-1=b_n^2$ is a perfect square for all $n\ge 2$. And clearly $2a_0-1=2a_1-1=1$ are squares, so it is true for $n\ge 0$ also.


How many regular n-gons (n odd) can be constructed?

Using Fermat Primes I guess it is:

$\sum_{k=1}^{5}\binom{5}{k}= 31$

Is it right?

Well you could consider the trivial 1-gon to be 32, but doesn't matter really.  Anyway you are correct, assuming that there are indeed only the five Fermat primes, which I beleive is still an open question.

[quote="BaBaK Ghalebi"]but $\sum_{k=1}^{5}\binom{5}{k}=2^{5}=32$...  :huh:
so please edit your post,I think somehow you meant something else...  :maybe:[/quote]

No;
$\sum_{k=1}^{5}\binom{5}{k}=2^{5}=31$
$\sum_{k=0}^{5}\binom{5}{k}=2^{5}=32$

oh sorry I didnt see the fact that $k>0$...  :oops:


Ring Theory
superior algebra
superior algebra unsolved

Let $R$ be a commutative ring and $P_{1}, P_{2}, ..., P_{k}$ be distinct ideals in $R$ such that at most two of them are not prime. Suppose that $I$ is an ideal of $R$ such that $I\subset \bigcup_{i=1}^{n}P_{i}$. Prove that $\exists j, \; 1\leq j\leq n,\; I\subset P_{j}$.

Assume WLOG that the set of ideals $\{P_{i},\ i\in\overline{1,n}\}$ is minimal with respect to covering $I$, in the sense that no proper subset covers $I$. Consequently, for each $i\in\overline{1,n}$ we can find an element $x_{i}\in I$ belonging to $P_{i}$ but not to $P_{j}$ for $i\ne j\in\overline{1,n}$. Also assume that $n\ge 2$ and all $P_{i}$ are prime, except maybe for $P_{1},P_{2}$.

Clearly, $x_{1}+x_{2}$ does not belong to $P_{1}$ or $P_{2}$. By eventually renumbering the $P_{i}$, we may assume that those $P_{i}$ to which $x_{1}+x_{2}$ does not belong are $P_{1},\ldots,P_{m},\ 2\le m\le n$. Now let’s look at the element $x=\prod_{i=1}^{m}x_{i}\cdot\left(\sum_{j=m+1}^{n}x_{m+1}\ldots\widehat{x_{j}}\ldots x_{n}\right)+x_{1}+x_{2}$ of $I$, where $x_{m+1}\ldots\widehat{x_{j}}\ldots x_{n}$ is the product of $x_{j},\ j=\overline{m+1,n}$ except for $x_{j}$. It’s easy to check that $x$ cannot belong to any of the $P_{i}$, contradicting the fact that the $P_{i}$ cover $I$.

Nice solution from grobber (as usual).
(currently : looking for a counterexample in in the case we have only one prime ideal!!)
Any way I don't know whether we have a necessary condition in this thorem or all conditions are somehow useless. :P


Your sketch does not look right; Euler line  appears to pass through the triangle vertex $ A \equiv b \cap c.$ This is possible only when $ b \perp c$ (not the case of your sketch) or when the triangle is isosceles with $ AB \equal{} AC.$ In the 2nd case, Euler line $ e$ must also bisect $ \angle (b, c).$ In both special cases, the solution is trivial but not unique.

Could you describe mutual positions of the sidelines $ b, c$ and the Euler line $ e$ ? For example: $ b, c, e$ are 3 lines in general position (not concurrent and no 2 lines  parallel).

$ b, c$ are 2 triangle sidelines and $ l$ is its Euler line. Label $ A \equiv b \cap c,$ $ E \equiv b \cap l,$ $ F \equiv c \cap l.$ Construct Euler line $ a'$ of $ \triangle AEF,$ say by constructing its orthocenter $ H_a$ and centroid $ G_a.$  Let $ a' \equiv G_aH_a$ cut $ c, b$ at $ B', C',$ respectively. Next, construct Euler line $ l' \parallel l$ of $ \triangle AB'C',$ say by constructing its orthocenter $ H'$ and drawing parallel $ l'$ to $ l$ through $ H'.$ Let $ D' \equiv a' \cap l'$ be intersection of the 2 constructed Euler lines. Let $ AD'$ cut the given Euler line $ l$ at $ D.$ Parallel $ a$ to $ a' \equiv B'C'$ through $ D$ cuts $ c, b$ at $ B, C,$ completing $ \triangle ABC.$

See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=284982[/url] for a proof.


孫子問題
孫子問題記載於中國古代約公元3世紀成書的《孫子算經》內面，是原書卷下第26題:

今有物不知其數，三三數之剩二；五五數之剩三；七七數之剩二，問物幾何?

:blush:  :blush:

太难为情了，我的古文真的学得很烂。 :blush: （居然还能在中国混）
能不能把题目翻译成现代文？

This is probably where the name "Chinese Remainder Theorem" came from. Though in China, they probably call it 孙子定理.

The problem asks for the number which when divided by 3, has remainder 2; when divided by 5, has remainder 3; and when divided by 7, has remainder 2.

[hide]I made a list and got 23.... [/hide]we were suppose to solve it right?

back then the number was a lot larger...

it came from counting the number of soldier by using squares...

faster for the generals...but it is never the less correct.

[quote="paganinio"]孫??題
孫??題記載於中國?代約公元3世紀?書的《孫?算經》內?，是原書?下第26題:

今有物?知其數，三三數之剩二；五五數之剩三；七七數之剩二，?物幾何?[/quote]

Isn't that the "famous" Chinese Remainder Theorem?

我?学的时候就看过的?典的一?。

Try Western encoding or Big-5 untill he comes and repairs his post. ;)

[quote="wenjing"]呵呵，笑啊，
我怎么看?懂啊
什么三三数，五五数啊？
?懂，有没有哪?大侠?解释一下啊？[/quote]
这个对中国人?说应该是著??的, 如果你到一定的年龄都是在中国长大的?...
就是说有一个数, 三个三个数剩下2, (除以3余数是2), ...

这?题在射雕英雄传中出现过,就是黄蓉考瑛姑的三?题中的一?.
瑛姑解题时也是用的"猜想"的方法...
连瑛姑这样的数学工作者都解?出?,说明它还是有一定难度地...... :D

今有物?知其數，三三數之剩二；五五數之剩三；七七數之剩二，?物幾何?

x = 2 (mod3)
x = 3 (mod5)
x = 2 (mod7)

i got x = 23 + 105k ,   where k is integer

[quote="epi14159"]射雕中黄蓉有讲解法[/quote]
是么?我没找到...
还有,瑛姑的答案也是23.

How do you get Chinese characters in here? Could some one tell me how to get the simplified characters....

[quote="isaacchao"]今有物丿知其數，三三數之剩二；五五數之剩三；七七數之剩二，啿物幾何?

x = 2 (mod3)
x = 3 (mod5)
x = 2 (mod7)

i got x = 23 + 105k ,   where k is integer[/quote]
我在一本书上看到口诀：“三人同行七十稀，五树梅花廿一枝，七子团圆正半月，除百零五便得知。”也就是说，3的余数乘以70，5的余数乘以21，7的余数乘以15，相加，除以105，的余数就是最小解。古人就是猛！


I'm having trouble with a problem that was assigned.

How many DISTINCT numbers are there under 1000?

if anyone can help me on ways of solving this it would be greatly appreciated, thanks.

The way you worded the question, the answer is $\infty$, however, the answer to the question replacing numbers with positive integers is 999. This too, is a trivial answer, and I think that you maybe want to reword your question?

My highschool teacher gave us this problem. But I'm pretty sure it's not infinity. Because he told us distinct numbers are numbers wihtout digits repeating, like 605, 604. 101 would not be a distinct number.

Your teacher is using "distinct" in a way different from the accepted definition. Two numbers are distinct if they are not the same. That's why you may have found besstate's reply confusing.

As for your question, a simple counting argument can be used. There are 10 choices for the leftmost digit (0-9). The second digit can be any of the digits that are not the same as the first; hence there are 9 choices. Similarly, there are 8 choices for the units digit. Then there are 10*9*8=720 total numbers less than 1000 without repeating digits.

wow thanks for your help, I did it the slow way and got the same thing.


calculus derivative limit trigonometry function logarithms integration

What is the proof that the derivative of e^x is itself.  I understand that, using the limit definition, one finds that the derivative is:

$ \displaystyle e^x\lim_{h\to 0}\frac {e^h \minus{} 1}{h}$ which we know is equal to e^x and thus to prove that the derivative of e^x is itself we must prove (to be rigorous) that

$ \displaystyle \lim_{h\to 0}\frac {e^h \minus{} 1}{h} \equal{} 1$

Clearly we may not use Taylor Series or L'Hopital's rule as they involve using the derivative of e^x.  The only thing i could think of was using complex substitution.

$ \displaystyle\lim_{h\to 0}\frac {e^h \minus{} 1}{h} \\
 \\
h \equal{} it; h\to 0, t\to 0 \\
 \\
\displaystyle\lim_{t\to 0}\frac {e^{it} \minus{} 1}{it} \equal{} \lim_{t\to 0}\frac {\cos{t} \plus{} i\sin{t} \minus{} 1}{it} \\
 \\
\lim_{t\to 0}\frac {\cos{t} \minus{} 1}{it} \plus{} \frac {i\sin{t}}{it} \equal{} 0 \plus{} \frac {\sin{t}}{t} \equal{} 1$

Although this method proves that the derivative of e^x is itself seemingly rigorously, there is one question raised.  To solve this problem we employed Euler's formula, whose proof using Taylor series I know.  This, however, is circular reasoning still as the Taylor series used to derive Euler's formula employs this very same derivative.  So my question essentially is: is there a proof of Euler's formula that does not use Taylor Series or, in general, the derivative of the exponential function, so that it can be used in evaluating this limit OR is there a another means of either proving the limit to equal 1 or proving that the derivative of the function is itself?

:)

$ y\equal{}e^x$

$ ln(y)\equal{}x$

$ \frac{1}{y} \frac {dy}{dx}\equal{}1$

$ \frac {dy}{dx} \equal{} y \equal{} e^x$

in that proof you used the derivative of ln(x) (which in my "rigorous" formulation I found by doing inverse differentiation through e^x), so let's use straight limits to find the derivative of ln(x) instead.

$ \displaystyle\lim_{h\to 0}\frac {\ln{(x \plus{} h)} \minus{} \ln{x}}{h} \\
 \\
\displaystyle\lim_{h\to 0}\frac {\ln{(1 \plus{} h/x)}}{h} \\
 \\
u \equal{} 1 \plus{} h/x; u\to 1 \\
 \\
\displaystyle\lim_{u\to 1}\frac {\ln{u}}{x(u \minus{} 1)} \equal{} \frac {1}{x}\lim_{u\to 1}\frac {\ln{u}}{u \minus{} 1} \\
 \\
u \equal{} e^t; t\to 0 \\
 \\
\frac {1}{x}\lim_{t\to 0}\frac {t}{e^t \minus{} 1}$
We know that the derivative of ln(x) is 1/x but to know for certain we must prove that that limit equals 1 - which is the same task as proving the limit to equal 1 for the derivative of e^x.  Thus this solution is circular as well.

This is a case in which it matters very much where you start and what your definitions are. Here are the two key questions:

1. How do you define $ a^x$ for $ a>0$ and $ x$ not necessarily rational?

2. How do you define $ e?$

If there is a differentiable function $ f(x)$ that satisfies $ f(x\plus{}y)\equal{}f(x)f(y)$ for all real $ x$ and $ y,$ then we can prove (as you did in your first post) that $ f'(x)\equal{}f'(0)f(x).$ But the previous sentence started with the word "if"; we're begging the question of showing that there is such a function.

One fairly straightforward place to start is to define the logarithm first. Define

$ \ln x\equal{}\int_1^x\frac1t\,dt$ for $ x>0.$

We can then show that $ \ln(ab)\equal{}\ln a\plus{} \ln b$ and that the range of the logarithm function is $ (\minus{}\infty,\infty).$ We define $ e$ to be the unique real number such that $ \ln e\equal{}1.$ We define $ e^x$ as the inverse function to the logarithm. We define $ a^b\equal{}e^{b\ln a}$ for $ a>0.$

There are other logical paths through this, as well. But you do have to decide how you are answering the two questions I started with.

[quote="Dvagner"]What is the proof that the derivative of e^x is itself.  I understand that, using the limit definition, one finds that the derivative is:

$ \displaystyle e^x\lim_{h\to 0}\frac {e^h \minus{} 1}{h}$ which we know is equal to e^x and thus to prove that the derivative of e^x is itself we must prove (to be rigorous) that

$ \displaystyle \lim_{h\to 0}\frac {e^h \minus{} 1}{h} \equal{} 1$

[/quote]

i know
$ e \equal{} \lim_{n\to\infty} \left(1 \plus{} \frac {1}{n}\right)^n$

$ \Rightarrow\quad e \equal{}\lim_{h\to 0} (1 \plus{} h)^{\frac {1}{h} }\quad\Rightarrow\quad \lim_{h\to 0}e^h \equal{} \lim_{h\to 0} (1 \plus{} h)$

so, now
$ \frac {d}{dx} \; e^x \equal{} e^x\lim_{h\to 0}\frac {e^h \minus{} 1}{h} \equal{} e^x\lim_{h\to 0}\frac {(1 \plus{} h) \minus{} 1}{h} \equal{} e^x\lim_{h\to 0}\frac {h}{h} \equal{} e^x$

[quote="Dvagner"]The only thing i could think of was using complex substitution.

$ \displaystyle\lim_{h\to 0}\frac {e^h \minus{} 1}{h} \\
 \\
h \equal{} it; h\to 0, t\to 0 \\
 \\
\displaystyle\lim_{t\to 0}\frac {e^{it} \minus{} 1}{it} \equal{} \lim_{t\to 0}\frac {\cos{t} \plus{} i\sin{t} \minus{} 1}{it} \\
 \\
\lim_{t\to 0}\frac {\cos{t} \minus{} 1}{it} \plus{} \frac {i\sin{t}}{it} \equal{} 0 \plus{} \frac {\sin{t}}{t} \equal{} 1$

Although this method proves that the derivative of e^x is itself seemingly rigorously, there is one question raised.[/quote]

there's no rigor here at all, seemingly or otherwise.  when you take a limit as $ h\to 0$, you are taking the limit through real numbers, whereas in the above you are taking the limit through purely imaginary numbers.

kent's comments below indicate one way that you might prove the result you want, another is to define $ e^x$ by its taylor series; that is, define for all $ x\in\mathbb{R}$,

\[ e^x \triangleq \sum_{n\geq 0} \frac{x^n}{n!}
\]

(by convention $ 0^0\equal{}1$ here), from which it follows by standard results on analytic functions that $ e^x$ is differentiable and its own derivative.  other desired properties, like $ e^xe^y\equal{}e^{x\plus{}y}$ follow without too much work.

maybe this is annoying to you experts, but the idea that you can define e^x or ln(x) in so many different ways and they all lead to the same results bothers me.  I'm so used to a set way to start and then leading up from one idea to the other.  Defining e^x as a series or limit or ln(x) as an integral just doesn't seem right.  Is there a purely algebraic way to define one of the two and then apply it to calculus?

[quote="Dvagner"]in that proof you used the derivative of ln(x) (which in my "rigorous" formulation I found by doing inverse differentiation through e^x), so let's use straight limits to find the derivative of ln(x) instead.

We know that the derivative of ln(x) is 1/x but to know for certain we must prove that that limit equals 1 - which is the same task as proving the limit to equal 1 for the derivative of e^x.  Thus this solution is circular as well.[/quote]

Umm... no I did not use circular reasoning for my "proof" of the derivative of $ e$.
Here is the derivative of $ ln(x)$ without using $ e$ except for a definition of it (no this doesn't ruin or "cicrulate" anything)

$ \lim_{h \to 0} \frac {ln(x \plus{} h) \minus{} ln(x)}{h}$(by definition of the derivative, no argument possible >_<)

$ \equal{} \lim_{h \to 0} \frac {1}{h}ln(1 \plus{} \frac {h}{x})$

$ \equal{} \lim_{h \to 0} ln(1 \plus{} \frac {h}{x})^\frac {1}{h}$       (now let $ a \equal{} \frac {h}{x}$, so $ \frac {1}{h} \equal{} \frac {1}{ax}$)

$ \equal{} \lim_{a \to 0} ln[(1 \plus{} a)^\frac {1}{a}]^\frac {1}{x}$

$ \equal{} \frac {1}{x}ln(e) \equal{} \frac {1}{x}$

There cannot be any argument as to why ln(e) = 1. It is its logarithmic inverse.

So my finding the derivative of $ e^x$ still holds....

[quote="Dvagner"]...  Is there a purely algebraic way to define one of the two and then apply it to calculus?[/quote]
No, there isn't a purely algebraic way, and that is the entire issue.

What do we mean by $ e^{\sqrt{2}}?$ Or to make the question even sharper, what do we mean by $ 2^{\sqrt{2}}?$ We cannot give an algebraic answer to that, in the same way that we can say that $ 2^{5/3}\equal{}\sqrt[3]{2^5}.$

Even to define the meaning of an irrational exponent requires us to use limits, or least upper bounds, or something involving integrals, continuity, and the intermediate value theorem. In other words, we must invoke the machinery of calculus/analysis even to get started.

well, euclid's turtle, i agree with you if you start off with that limit definition of e, but i was not at that stage of thought at that point.

[quote="Dvagner"]maybe this is annoying to you experts, but the idea that you can define e^x or ln(x) in so many different ways and they all lead to the same results bothers me.[/quote]

well ... it's better than the alternative, which is defining things in different ways and having them lead to different results.

[quote="Dvagner"]Defining e^x as a series or limit or ln(x) as an integral just doesn't seem right.  Is there a purely algebraic way to define one of the two and then apply it to calculus?[/quote]

no, and there's no reason to expect there to be, ($ e$ is the unique real number such that $ e^x$ is its own derivative, and derivatives are defined through limiting processes, so why shouldn't $ e$ be defined through some sort of limiting process?)

Mhmm or even so, I just though of using a Differential Equation.
I believe Leibniz began with integration before differentiation. (checks book) Yea he used a trapezoid limit.
Well It sort of uses limits to get the idea but not in the actual formulation (just sort of...discrete :P)
We know $ e^x$  was probably once defined as a function whose rate of change is equal to its derivative.

So.... $ \frac{dy}{dx}\equal{}y$

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

$ ln(y)\equal{}x\plus{}c$

$ y\equal{}e^xe^c$

Using an inital value of $ y(0)\equal{}1$ yields $ c\equal{}0$

So $ y\equal{}e^x$..... come to think of it this doesn't much help without the limits and all as I needed the integral of  $ \frac{1}{a} \equal{} ln(a)$

Finishing so others don't come and think of this nonsense :D


[hide="number 2"]
I got 44 and 5.5 using the formula: $\frac{-b +- \sqrt{b^2 - 4ac}}{2a}$
In this case 4 would be a, 44 would be b and 121 would be c.[/hide]

First one:  U=0
2nd one (perfect sqr 2p-11): P = 5.5 or 11/2
3rd one (sub y sqr w/ another variable): Y = plus or minus 1, plus or minus 9.

[hide="1"]
simplifing gives you: $10u^3=5u^3$
the only number that works for that is $\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{ \boxed{0}}}}}}}$[/hide]
[hide="2"]
$p=\frac{44\pm\sqrt{1936-1936}}{8}$
$p=5.5$[/hide] I don't see how you got 44, math92.  :?
I don't get the third one  :?

of course there are many ways to solve each problem but i think the simplest way to solve the first problem is using zero product property...whoever gave you that problem most likely intended for you to do so seeing as though it works out very simply...thats just my opinion tho  :)

o and when u use zero product property you get the [b]whole[/b] answer which is 0 and .5  most who responded only gave one answer

[hide="#1"]Divide both sides by u^2 and add 5 to both sides. Dividing each side by 10 gets [b]u=1/2[/b].[/hide]
[hide="#2"]Subtract 44 p from both sides and rearrange to get 4p^2-44p+121. This is a perfect square with the root (2p-11). [b]p=11/2[/b].[/hide]
[hide="#3"]This can be factored like a second degree equation: (y^2-1)(y^2-9). [b]y=1 or y=3.[/b][/hide]

btw,  math92, where you messed up is that b is -44, not 44

oh thanks, for telling me that  :)
i was just wondering where i messed up.

[hide="number 1"]
Factor:
$5u^2(2u-1)$=0
So $5u^2=0$ or $2u-1=0$
If $2u-1=0$, then $u=\frac 12$
If $5u^2=0$ then $u=0$
So the solutions are $\{\frac 12,0\}$
[/hide]
[hide="number 2"]
$4p^2-44p+121=0$
That is a perfect square
Factoring,
$(2p-11)^2=0$
Take the square root,
$(2p-11)=0$
So $p=5.5$
[/hide]
[hide="number 3"]
$y^4-10y^2+9=0$
Substitute $x=y^2$
So $x^2-10x+9=0$
Factoring gives $(x-9)(x-1)=0$
So $x=9$ or $x=1$
So $y^2=9$ or $y^2=1$
Taking the square root of both,
$y=\pm 3$
$y=\pm 1$
[/hide]


least common multiple
number theory unsolved
number theory

Let $ n\geq a_1>a_2>\cdots>a_k$ be positive integers such that for all $ i,j$ the least common multiple of
$ a_i$ and  $ a_j$ is less than or equal to $ n$.
Prove that for all $ 1\leq i\leq k$, we have $ ia_i\leq n$.

NO ONE? if this result is incorrect, please give a counterexample.


topology
superior algebra
superior algebra solved

Construct an injective homomorphism from the free group on $ 3$ elements $ a,b,c$ to the free group on two elements $ x,y$.

$ a\mapsto y^1xy^{\minus{}1}, b\mapsto y^2xy^{\minus{}2}, c\mapsto y^3xy^{\minus{}3}$

$ a \mapsto x$
$ b \mapsto xy$
$ c \mapsto xy^2$

@ZetaX:

This is not injective. It is surjective: $ a\mapsto x, a^{\minus{}1}b\mapsto y$. If it would be injective too, it would be an isomorphism $ F_2\cong F_3\Rightarrow 2\equal{}3$.
More explicit: $ ba^{\minus{}1}b \mapsto xy^2$ but $ ba^{\minus{}1}b\neq c$.

[quote="Gockel"]$ a\mapsto y^1xy^{ \minus{} 1}, b\mapsto y^2xy^{ \minus{} 2}, c\mapsto y^3xy^{ \minus{} 3}$[/quote]

Thanks. So, would the prove go like this : denote this homomorphism by $ f$, we want to prove that $ f(a^{e_1}b^{e_2}c^{e_3}a^{e_4}...) \neq 1$ (*) in $ F_2$, where $ e_i \in \{\minus{}1,0,1 \}$. Since $ f(a^{\minus{}1})\equal{}yx^{\minus{}1}y^{\minus{}1} ,f(b^{\minus{}1})\equal{}y^2x^{\minus{}1}y^{\minus{}2}, f(c^{\minus{}1})\equal{}y^3x^{\minus{}1}y^{\minus{}3}$, we have that the product in (*) has the form $ x^{e_1}y^{e_2}x^{e_3}y^{e_4}...$ with $ e_i \in \{\minus{}1,1 \}$ and hence never equal to $ 1$ in $ F_2$.
This is quite clear if you write it down and if i'm not wrong, but maybe there is more rigorous solution ?

Any suggestions for a rigorous proof ?

Thanks.

There is a solution by complete technical overkill: identify $ F_{2/3}$ with the fundamental groups of the graphs $ {\bigcirc}\!\!{\bullet}\!\!{\bigcirc}$ resp. $ {\bigcirc}\!\!{\bullet}\!\!{\bigcirc}\!\!{\bullet}\!\!{\bigcirc}$ and consider the covering map obtained by "twisting" the middle "circle". The required statements can be found in any introductory topology text, e.g., May. owk

By a result of Serre, $ F_2'\equal{}[F_2,F_2]$ is free and a basis is given by commutators $ [x,y]$ with $ x,y\in F_2\setminus\{e\}$.
So just choose 3 of these :D


calculus
integration
trigonometry
real analysis
real analysis solved

In other words :D, Moubinool asks us to prove that
\[\int_{1/(n+1)}^{1/n}\frac{x}{\cos x}dx\leq \frac{\sqrt{(n+1)^2-1}}{n+1}-\frac{\sqrt{n^2-1}}{n}.\]

Since $|\sin x|\leqq |x|$ ,squaring both sides of this,we have $\sin ^2 x\leqq x^2\Longleftrightarrow1-\cos ^2  x\leqq x^2\Longleftrightarrow1-x^2\leqq \cos ^2 x$.

For $-1\leqq x\leqq 1$ we obtain $\sqrt{1-x^2}\leqq |\cos x|$.

Therefore

$\int^{\frac{1}{2004}}_{\frac{1}{2005}} \frac{x}{|\cos x|}dx\leqq \int^{\frac{1}{2004}}_{\frac{1}{2005}} \frac{x}{\sqrt{1-x^2}} dx$, yielding our inequality.

[quote="kunny"]

For $-1\leqq x\leqq 1$ we obtain $\sqrt{1-x^2}\leqq |\cos x|$.

[/quote]

Nice proof kunny, this minoration of $|\cos x|$ is interesting.


Define a domino to be an ordered pair of distinct positive integers.  A proper sequence of dominos is a list of distinct dominos in which the first coordinate of each pair after the first equals the second coordinate of the immediately preceding pair, and in which $(i,j)$ and $(j,i)$ do not both appear for any $i$ and $j.$  Let $D_{40}$ be the set of all dominos whose coordinates are no larger than 40.  Find the length of the longest proper sequence of dominos that can be formed using the dominos of $D_{40}.$

[hide]We can draw a comparison between the domino a set of 40 points in which every point is connected with every other point.

The AOPS book II states that you need to have all even number of segments coming from each point except 0 or 2 which have an odd number of segments coming from the point.  Since there are 39 segments coming from each point it is impossible to touch every segment.

But you can get up to 38 on each segment because you go in to the point then out on a different segment.  Counting going out from the starting and ending at the ending point we have:

(38*40+2)/2

761[/hide]

[quote="joml88"]Define a domino to be an ordered pair of distinct positive integers.  A proper sequence of dominos is a list of distinct dominos in which the first coordinate of each pair after the first equals the second coordinate of the immediately preceding pair, and in which $ (i,j)$ and $ (j,i)$ do not both appear for any $ i$ and $ j.$  Let $ D_{40}$ be the set of all dominos whose coordinates are no larger than 40.  Find the length of the longest proper sequence of dominos that can be formed using the dominos of $ D_{40}.$[/quote]

The wording is very confusing to me, anyone can clarify the problem?

[quote="JBL"]You've made dozens of posts recently whose content is nothing more than "I don't understand."  If you consistently fail to understand what questions mean, you are either not trying hard enough or you are trying to operate at a higher level than you are currently capable of.  If you want to ask not-obnoxious or not-pointless questions, specify precisely what part of the question you find confusing or offer any indication that you've given more than 5 seconds of effort trying to understand, for example by asking whether a particular interpretation is correct.[/quote]

sorry, does the problem ask if (a,b) is one of the dominos then the next domino has to be (c,a), and if (a,b) exist, (b,a) doesn't.  But one thing that i don't understand is how can we do we simulate the porblem into connecting 40 points?

[quote="ehehheehee"]The AOPS book II states that you need to have all even number of segments coming from each point except 0 or 2 which have an odd number of segments coming from the point.  Since there are 39 segments coming from each point it is impossible to touch every segment.[/quote]
I don't have the book but what are 0 and 2?

@cdymdcool: Connecting the points 2 -> 4 -> 5 -> 3 -> 1 is equivalent to playing the domino sequence $ (2, 4), (4, 5), (5, 3), (3, 1)$.

[quote="mathwizarddude"]
I don't have the book but what are 0 and 2?[/quote]

A path that goes through every edge of a graph (an Eulerian path) exists iff all or all but two vertices of the graph have an even number of edges coming out of them. This means 0 or 2 points can have an odd number of edges coming out of them.

[hide="Cleaned solution"] In total, there are at most $\binom {40} {2}=780$ possible "usable dominoes". We create a correspondence to a graph where each distinct number, and two vertices are connected iff they can come in some order. Note the connection, if $a$ is connected to $b$ is connected to $c$, this corresponds to the domino sequence $(a, b), (b, c)$. It is easy to see that this vertex is $K_{40}$. An Eulerian path exists iff there are 0 or 2 vertices of odd degree, so at best we must take 19 edges away, because each evens the degree of at most two vertices. Hence, the answer is 780-19=$761$.  [/hide]


What is the greatest common divisor of all of the members of the set containing all numbers that are the product of four consecutive positive integers?

Four consecutive positive integers have multiples of 3 and 4 and one more even no. Hence GCD will be 24.


function
algebra
polynomial
real analysis
real analysis unsolved

Let $f(x)=(3x^2-4)\left(x-a+\frac{1}{a}\right) (a\neq 0)$.

Find the value of $a$ such that the difference of [b]relative maximum value [/b]and [b]relative minimum value [/b]is minimized.

Are there some bounds for $x$ because on $\Bbb R$ the function is unbounded for every $a$.

Misha

I'm sorry ,I made a mistake in translation.I will correct.

kunny

The derivate of the function is $9x^2-6ax+6\frac{x}{a}-4$. We get the equation: $x^2-\frac{2}{3}ax+\frac{2x}{3a}-\frac{4}{9}=0$ and have to minimize the difference of its roots. This difference is $\sqrt{D}$ where $D$ is the discriminant.
$D=(-\frac{2a}{3}+\frac{2}{3a})^2+\frac{16}{9}$. We have to minimize $(a-\frac{1}{a})^2$. This term assumes its minimum for the same $a$ as $a^2+\frac{1}{a^2}\geq2$. We have equality for $a=\pm 1$.

Misha

I think the diferrence of extrema is described by $k\big|\frac{2a}{3}+\frac{2}{3a}\big|^3$,what value of $k$ did you get?

kunny

Sorry that I misunderstood your problem. I think the solution affords a lot of calculations (putting the roots of the derivate in to the former polynomial and to find the minimum of the difference). Or is there a better way :maybe: ???

Misha

Thank you for your reply!, Misha.
Yes.There is a nice solution.

kunny


For $ A\in M_{n}\mathbb{(C)}$ such that :$ A^{3} = 4.A$ and [b]Tr(A)=2n [/b]. Find[b] A=?[/b]
[b] Love math must come from heart !!![/b]

Eigenvalues of $ A$ can tale next values 2,-2,0. But $ trA\equal{}2n$ so all eigenvalues 2. So $ A\equal{}2E$, $ E$-identity matrix.


A spinner contains eight regions, numbered 1 through 8. The arrow has an equally likely chance of landing on any of the eight regions. If the arrow lands on the line, it is spun again. What is the probability that the arrow lands on an odd number?

Well, you don't really specify how wide the line is... but...
[hide]since odd numbers are 1, 3, 5, 7; there are also 4 even numbers: 2, 4, 6, 8.
Therefore the probability of getting an odd number is 1/2[/hide]

[quote="perfect628"]This isn't HSB either...[/quote]
This is high school basics and some people here (not Interval) are not in advanced classes and are learning basic algebra, like this problem.

Also, this is a [i]basic[/i] skill you should know!

Check out the posting guidelines. This doesn't fit.

Yes, I think interval should pay attention to these guidelines
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32837


Well, I'm off to CCC stage 2.  Should be fun!  Hopefully the problems will be very interesting this year.  I should see Richard Peng there... he's essentially the guy that got me into contest-level programming.  Very interesting stuff.  :D

dangit, what secrets did he tell you?

he never tells me anything.. still dont know how to code C

Yay!  I won a gold medal.  Maybe I get to go to IOI!  We'll see...
Richard won gold too.

Well ok... there are 5 gold medals and 4 people make the IOI team... so given that richard and I both won gold, at least one of us must make it to IOI.  We know Simon won, so he's in it for sure... so that's probably a 2/4 chance that we'll both make it, given even probabilities for the rest of us... hmmm... this will be interesting.

The other winner was a guy named Qiyu.  He was really smart, but he carried a video camera EVERYWHERE!  He kept recording us doing things.  It was a lot of fun!!


calculus
integration
function
real analysis
real analysis unsolved

[quote="arline"]let f,g be continuous functions on interval I=[a,b] such that(riemann)integral  f on I=integral g on I.prove there exists some c in I s.t.f(c)=g(c). :huh:[/quote]

$L(x)=\int_{a}^{x}(f(t)-g(t))dt$ then $L(a)=L(b)=0$ from Rolle we got $L'(c)=0$


invariant
function
geometry
geometric transformation
rotation
abstract algebra
topology

1/Prove that there is no rotationally invariant countably additive measure defined for all subsets of the unit circle of $\mathbb{R}^2$.
2/Prove that one can define a nonnegative finitely additive set function $m(P)$, for all subsets $P$ of the circle, that is rotationally invariant.

I assume these measures are supposed to be finite and positive?

(1)

Let $G$ be the group of rotations around the origin corresponding to rational multiples of $\pi$, and consider a subset $X$ of $\mathbb S^{1}$ containing exactly one representative from each orbit of the action of $G$ on $\mathbb S^{1}$. If $\mu$ is a countably additive rotationally invariant measure on the circle, we have $\mu(\mathbb S^{1})=\mu(\bigcup_{g\in G}gX)=\sum_{n\ge 1}\mu(g_{n}X)$, and since $\mu(gX)=\mu(X),\ \forall g\in G$ ($(g_{n})_{n}$ is an enumeration of $G$), and we find $\mu$ to be trivial: if $\mu(X)>0,\ \mu(\mathbb S^{1})$ would be infinite.

(2)

The group of rotations around the origin of the plane is, of course, abelian. I’ll try to prove the following generalization:

Let $G$ be an abelian group, and let $E$ be the Banach space of bounded functions on $G$ with the supremum norm. There is a linear functional $\Lambda\in E^{*}$ such that

(a) $\Lambda(f)\ge 0$ whenever $f\ge 0$;
(b) $\Lambda({\bf 1})=1$, where ${\bf 1}$ is the constant function equal to $1$;
(c) $\Lambda\circ L_{g}=\Lambda,\ \forall g\in G$, where $L_{g}: E\to E$ is defined by $L_{g}f(h)=f(gh),\ \forall h\in G,\ \forall f\in E$.

(note that it follows from (a) and (b) that $\|\Lambda\|=1$)

Let $K$ be the subset of $E^{*}$ consisting of those functionals which satisfy (a) and (b). $K$ is non-empty, since it contains, for example, all evaluation functionals $\Lambda_{x},\ x\in G$, defined by $\Lambda_{x}(f)=f(x)$. Being a convex closed subset of the unit ball of $E^{*}$, which is weak-$*$ compact, it is a convex compact set in the weak-$*$ topology on $E^{*}$. The group $G$ acts on $E^{*}$ to the left in an obvious way as a group of linear transformations: $g\Lambda=\Lambda\circ L_{g}$. Moreover, $GK=K$, and an element of $K$ which satisfies (c) is just a fixed point of the action of $G$ on $K$. The result we want to prove now follows from:

Let $G$ be an abelian group of continuous linear transformations on a locally convex topological space $X$, and let $K$ be a compact convex subset of $X$ such that $GK\subset K$. There is then a point of $K$ fixed by all elements of $G$.

By a well-known fixed point theorem, for every $g\in G$, the set $\mbox{Fix}(g)=\{p\in K\ |\ gp=p\}$ is non-empty. The sets $\mbox{Fix}(g),\ g\in G$ are convex and, more importantly right now, compact, so in order to prove that they have non-empty intersection, it suffices to prove that every finite subcollection has non-empty intersection. Given $p\in\mbox{Fix}(g)$ and any $h\in G$, we have $g(hp)=h(gp)=hp$, so $hp\in\mbox{Fix}(g)$, i.e. for any $g\in G,\ \mbox{Fix}(g)$ is fixed by $G$ (as a set, not pointwise). Now take $g_{1},\ldots,g_{n}\in G$. $\mbox{Fix}(g_{1})$ is a non-empty compact convex subset of $K$ (convex because $G$ is a group of [i]linear[/i] transformations), and $g_{2}$ acts on it. By the same fixed point theorem that was invoked above, $\mbox{Fix}(g_{2})\cap\mbox{Fix}(g_{1})\ne\emptyset$. Now $g_{3}$ acts on this non-empty set, and continuing like this, we get what we wanted: $\bigcap_{i=1}^{n}\mbox{Fix}(g_{i})\ne\emptyset$.

Edit:

Now prove that there is such a measure which is invariant to all isometries of the circle (i.e. reflections across lines through the origin as well, in addition to rotations).


Να αποδειχθει οτι τα σημεια Steiner,Μ.Ναπολεοντος και το περικεντρο ειναι συνευθειακα..

Sil ειπειδή την έχω προσπαθήσει και δεν την έβγαλα και μου χες πει παλιά ότι έχεις λύση θα ήθελα όποτε βρεις χρόνο έστω και πολύ συνοπτικά να την ποστάρεις. Ή όποιος άλλος την έχει βγάλει.

Δείτε εδώ http://www.mathlinks.ro/Forum/viewtopic.php?t=109092

Το ότι οι ευθείες $ BC,B'C',B''C''$ (το οποίο είναι το κλειδί για τη λύση της άσκησης) μπορεί να προκύψει σχετικά εύκολα με μιγαδικούς αριθμούς

Περιπου αυτη ηταν η λυση μου!!!!νομιζω οτι ειναι πολυ ωραια ασκηση...δεν χρησιμοποιησα ομως μιγαδικους!!! :)


Could someone explain the properties of a surjective, injective bijective function? How you could prove whether or not that a function is surjective, injective bijective?

Thanks :lol:

I might have mixed injective with surjective but...

A surjective function means 'onto.'  A surjective function $f: A\to B$ maps elements of $A$ to all elements of $B$.  Every single element of $B$ is covered by $f$.  I'm not sure how you could prove that a function $f$ is surjective but an idea is to suppose it was not and contradict that using $f^{-1}$.

An injective function means 'one to one correspondence.'  Only one element $a\in A$ goes to only one element $b\in B$.  You can prove that a function $f$ is injective by supposing it was not and showing it is impossible to have $f(a_{1})=f(a_{2})$ for distinct $a_{i}$.

A bijective function is a function both injective and surjective (hence the 'bi' in the name).  To prove that a function is bijective you simply need to prove it is both surjective and injective.