algebra
polynomial
search
induction
number theory proposed
number theory

Given a polynomial $ P(x)$ with integer coefficients. For each natural number $ n$, $ P(n)>n$. Consider the sequence $ x_1\equal{}1,x_2\equal{}P(x_1),...,x_n\equal{}P(x_{n\minus{}1}),...$. Given that for each natural number $ n$, there exists a member of the sequence which is divisible by $ n$. Prove that $ P(x)\equal{}x\plus{}1$.

Umm... I have no idea how to search this. Please help. Or at least give the outline of solution.

It 's not very tough!

Put Pm(x)=P(P(...(x))..) ( m times)

Assume that there exists an index m such that  d= Pm+1(1) - Pm(1) > 1

Remark that

Pm+2(1) - Pm+1(1) is divisible by Pm+1(1) - Pm(1), that yields d| (Pm+2(1) - Pm+1(1))

Hence Pm+2(1) - Pm(1) is divisible by d

By induction we could prove that  d| Pn(1) - Pm(1) for all n>m  (*)

Now we can take N= d^k , k is selected so that   d^k is not divisible by Pm(1) and  d^k > max{ 1,P1(1),...., Pm(1)}  (**)

According to the hypothese, we could find an index j such that  N|Pj(1)

Because of (**) , we must have j> m. Combining with (*) we must have d| Pm(1)

Hence Pm(1) is not less than Pm+1(1) - Pm(1)

If there exist an infinite number of such index m, there exist infinite x such that   P(P(x)) does not exceed  2.P(x)  --> deg P<2

It 's not hard to continue.....

[quote="mto"]k is selected so that   Pm(1) is not divisible by Pm(1) and  d^k > max{ 1,P1(1),...., Pm(1)}  (**)

[/quote]

What do you mean?

As d>1 , we can always select k so that d^k is not divisible by Pm(1) and  d^k > max{1,P1(1),...,Pm-1(1) }

I just mistyped something


parameterization
trigonometry
algebra
system of equations

[hide]Here's the question that I,again, don't get the true result provided by the book!
An inclined plane making an angle of 25 degree with the horizontal has a pulley at its top. A 20kg block on the plane is connected to a freely hanging 10kg block by means of a cord passing over the pulley. Compute the distance the 10kg block will fall in 2s, starting from the rest. Neglect friction.
My solution:
We resolve the gravitation of the 20kg block into 2 components:x-axis which is equivalent to the tension of a cord and y-axis.
To compute the tension of the cord, we have: 20.9,8.sin25=N_x=T
Now, since the 10kg block falls down, the acceleration also goes down, so we have the equation:
10.9,8-T=10.a(a is the acceleration of the 10kg block), so we get: a=1.52(rounded result)
Applying the formula: s=v_o.t+1/2.a.t^2, we get: s=0+1.52.2=3.04m, which is different from the book's result:1.0m :(
PS:I'll upload my figure later![/hide]

Hi!!

your mistake is that, the block on the inclined plane is moving too, since its connected to the falling block via the pulley, so you cant compute the tension just as you did.
Your case would be if the block on the plane would be in equilibrium, but its not since its accelerating.. :D
So just make the correction in the equation of motion for the block on the plane, and together with the equation of the falling block you get a system of equations, with the unknowns as the tension and the acceleration of the blocks. Eliminate the tension and you get the acceleration.

Oh, and just a hint, try using parameters all the way, and plug the numerical data at the very end of the problem, because then the solution is much more clear.. :idea:

Because the question asks for the distance the 10kg block falls, we can assume that the system is moving in the direction of the 10kg block. By isolating each block and analyzing the forces on each, we can find a system of equations that will allow us to solve for $ a$.

If we call the movement in the direction of the 10kg block to be positive, and we let $ m_1 \equal{} 10$ and $ m_2 \equal{} 20$, then we get the following system of equations:

By Newton's second law:
$ T \minus{} m_2g\sin\theta \equal{} m_2a$
$ m_1g \minus{} T \equal{} m_1a$

Add these equations to eliminate the T term:
$ m_1g \minus{} m_2g\sin\theta \equal{} (m_1 \plus{} m_2)a$

$ a \equal{} \frac{g(m_1 \minus{} m_2\sin\theta)}{m_1\plus{}m_2}$

I also agree with Thaakisfox, plug in the numbers after :)

After this, using $ s \equal{} v_0t \plus{} \frac{1}{2}at^2$ will give you the correct result as shown by the book.


What about rewriting it as
$\displaystyle \sum \frac{1}{\frac{1}{a}+1+b} \leq 1$
and then use Jensen's with $\displaystyle f(x)=\frac{1}{x}$.  This reduces to proving $\displaystyle 6 \leq \sum (a+\frac{1}{a})$ which is obviously true

[quote="blahblahblah"]It expands into $(abc-1)^2\geq 0$, which is obviously true.  I don't have a more insightful proof, because equality holds whenever $abc=1$.[/quote]
Yes that's how I did it. It was a extremely ugly when expand it. So I wanna know if there is a better proof therefore I put it here.

The problem is that unless we introduce some auxilliary variable, almost all of the usual inequalities we employ will be useless against it.


It was on the 2008 School Sprint Round, and just happened to be one of the problems I got wrong...

I have gotten to this part...and I cannot solve it.

(1 + 3 + 5 + . . . + 2009) - (2 + 4 + 6 + . . . + 2008)

My answer was 4017. I do not believe that is correct.

...Or you could group the terms in a way so that it will be easy to evaluate.
\begin{align*}(1 + 3 + 5 + \dots + 2007 + 2009) - (2 + 4 + 6 + \dots + 2006 + 2008) & = (1 - 2) + (3 - 4) + \dots + (2007 - 2008) + 2009 \\
& = - 1004 + 2009 \\
& = \boxed{1005}.\end{align*}
Perhaps if you could explain how you got the answer 4017, I could point out where you went wrong. (If you want.)

Since its a school round, problem you should think easier.

What number was this anyway. I don't focus my attention on school rounds anymore.

Solution  [hide] I grouped the numbers (1-2)+(2-3)+(3-4)+.....+(2007-2008)+2009.
There are 2008/2=1004 pairs of numbers from 1 to 2008 inclusive.
So, the answer would be -1(1004)+2009= -1004+2009=1005[/hide]


salam
migam chetore ta ghabl az shorue dore ye maraton rah bendazim ke soalash mesle marhale 3 bashe:
[size=150]  [color=cyan] problem 1 [/color][/size]
farz konid $ABC$ yek mosalase hadezaviye bashad ke $BC<CA$ va $O$ markaze dayereye mohitiye an bashad.va $H$ markaze ertefayi va$F$ paye artefae $CH$ bashad.amoode bar $OF$ dar $F$ zele$AC$ ra dar $P$ ghat mikonad.sabet konid:
$\angle FHP =\angle BAC$
 ;) masaleye ghashangiye.

eival  ;)  dar vaghe bayad ye gharine nesbat be noghteyef bezanim va bad kheili chiza moshakhas mishe:
pas masaleye bad:
[size=150][color=cyan]problem 2 [/color][/size]
yek morabae $(n-1)(n-1)$ be $(n-1)^2$ morabae vahed be raveshe mamuli taghsim shode ast.
be chand tarighe mokhtalef mitavan $n^2$ rase hameye moraba haye vahed ra ba do range ghermez va abi rang kard ke har moraba vahed daghighan 2 rase ghermez dashte bashad :?:
inam soale ghashangiye. ;)
amir ba in jomle kheili narahat shodam:
$\text {good bye mathlinks forever}$ :(
bi khial in khaste khoda boode ke intori beshe;joone man ziad narahat nasho manam delam migire. :(

$(n-1)^2$ moraba
doroste? ya man eshyebah mikonam

agha sharmande.
yekbar nashod man ye masale badam va kamelan dorost bashe. :oops:

ebteda hokm ra taamim midahim:
tedade rang amizihaye matloob baraye shabakeye $m*n$ az noghate safhe barabar ast ba $2^m+2^n-2$
hokme bala ra ba esteghra rooye $m+n$ sabet mikonim.
baraye $m+n=4$(paye)yani $m=2$ va $n=2$ ya $m=3$ va $n=1$(ya baraks) ke hokm be rahati ghabele barresi va dorost ast.
hal farz konim hokm be ezaye $m+n=k$ dorost bashad. dorostie an ra be ezaye $k+1$ sabet mikonim. shabake ra bedoone sotoone $n$ om dar nazar migirim.farz konim tedade rangamizihay
shabake be tori ke dar sotoone $n-1$ om hadeaghal do noghteye motavalie yekrang vojood dashte bashad $A$ va tedade rangamizihai ke dar sotoone $n-1$om hich do noghteye motavalie yekrang nabashad $B$ bashad.baraye halate $A$ khaneye $(1,n)$ ra ba yek rang(mokhalefe khaneye $(1,n-1)$) mitavan rang kard zira dar gheire in soorat rangamizi kamel nemishavad.ba in kar baghieye shabake bedoone moshkel va yekta rang mishavad.
baraye halate $B$ khaneye $(1,n)$ ra ba do rang mitavan rang kard va sotoone akhar dar har do soorat bedoone moshkel rang mishavad.
pas tedade kole rangamizihaye shabake $A+2B$ ast amma $A+B$mosavist ba tedade rangamizihaye shabakeye $m*(n-1)$ ke tebghe farze esteghra barabar ast ba $2^m+2^{n-1}+2$ pas baraye esbate hokm kafist sabet konim $B=2^{n-1}$. dar haghighat tedade ranamizihai az shabakeye $(n-1)*(n-1)$ ke sotoone akhar hich do noghteye motavalie yekrang nadarad ra mikhahim.
baraye in kar ebteda do khaneye $(1,1)$ va $(2,1)$ ra be do tarigh sepas $(1,2)$ va $(2,2)$
ra be do tarigh va ..... ta do satre avval kamel shavad be tori ke dar har sotoon do rang bashad.
vazeh ast ke baghieye shabake be tore yekta por mishavad.
pas hokm sabet shod.
baraye halte $n*n$ javab barabar ast ba $2^{n+1}-2$.

afarin ;)
soale badi ham khodet post kon. :)

For a positive integer $n$ let $d(n)$ be the sum of its digits.when written in demical form.Prove that for each positive integer $k$ there exists a positive integer $m$ such that the equation    $x+d(x)=m$ has exactly $k$ positive integer solutions. :huh:
masaleye sakhtie.albatte kheili ham ghashange.

[quote="mohamadhosein"]demical form manzoor mabnaye dah dahiye?
ta farda hal mishe. ;)[/quote]
are mabnaye dahe

[quote="John-Nash"][quote]ta farda hal mishe.[/quote]

Omran. :D[/quote]
omran?

ei baba dige az ma gozashte , man dige pir shodam.
mage mano nemishnasi? man john nasham.

baba khafan to ke nobel prizo ostadesh kardi dige olampiad chiye? ;)

[quote="amirhtlusa"][quote="samanfarhat"]For a positive integer $n$ let $d(n)$ be the sum of its digits.when written in demical form.Prove that for each positive integer $k$ there exists a positive integer $m$ such that the equation    $x+d(x)=m$ has exactly $k$ positive integer solutions. :huh:
.[/quote]

fekr konam male TST e Romania bashe...(2002 ya 2003 motmaen nistam).

vali soal e sadei hast.
ba esteghra be rahati hal misheh.

agar $x+d(x)=m_k$ daghigan $k$ javab dashteh bashad,
aangah $\exists$ $0\leq t\leq 8$ s.t. $2+m_k+2t=9l$
$l=\frac{2+m_k+2t}{9}$
gharar bedahid: $m_{k+1}=m_k+10^l+1$
angah $x+d(x)=m_{k+1}$ daghighan  $k+1$ javab khahad dasht. QED

yek nafar e dige mitooneh soal e baadi ro post koneh.[/quote]
soale ghashangie dar zemn az oon soalaye emtehanie khoobe ke dar eine sadegi vaghte ziadi migire :)

kasi soale badi ro nemikhad bede?
vagarna man ye soal bedam ke topic baste beshe. :lol:


[hide]If you pick $2$ keys your chances are doubled, so the probability is $\frac{2}{12}=\frac{1}{6}$.

Edit: To generalize, say there are $n$ things to choose from and only one "works."

Then the probability of picking the good one if you choose $m$ of them is $P=\frac{\text{number of ways to pick the good one}}{\text{total number of possible selections}}=\frac{\binom{n-1}{m-1}}{\binom{n}{m}}=\frac{m}{n}$, as expected.[/hide]

[quote="allrighty"]The key to a locked door is one fo the twelve keys in a cabinet. If a person selects two keys at random from the cabinet and takes them to the door, what is the probability that he can open the door?[/quote]

[hide]the probability for a key to open the cabunet is 1/12

since there are 2 keys, it is 2/12=1/6[/hide]

[quote="jli"][hide]the probability for a key to open the cabunet is 1/12

since there are 2 keys, it is 2/12=1/6[/hide][/quote]hence my post..[quote="AntonioMainenti"][hide]If you pick $2$ keys your chances are doubled, so the probability is $\frac{2}{12}=\frac{1}{6}$.[/hide][/quote]

People are encouraged to post their solutions without necessarily reading those of other people -- that's what the hidden text is for.

$\displaystyle\frac16$?
Is that the correct answer?
I knows nothing about probability so can anyone tell me where i can get some basic knowledge about the stuff online?
Thank you!


I know the basic PIE formula such as finding the total number of elements in at least one of the sets. But, is there a formula to calculate how many things are in at least k (k>1) out of the n sets? How about the number of things that are in exactly k out of n sets?


trigonometry
geometry
AMC
AIME
calculus
integration
modular arithmetic

Find the number of ordered pairs of real numbers $ (a,b)$ such that $ (a \plus{} bi)^{2002} \equal{} a \minus{} bi$.

$ \textbf{(A)}\ 1001\qquad \textbf{(B)}\ 1002\qquad \textbf{(C)}\ 2001\qquad \textbf{(D)}\ 2002\qquad \textbf{(E)}\ 2004$

Since $ z \cdot \bar z = |z|^{2},$ we have:
\[ z^{2002}= \bar z = \frac{|z|^{2}}{z}\iff z^{2003}= |z|^{2}\]
Since $ |z|$ is a real number, there are $ 2003$ values for $ (a,b)$ that satisfy this equation.

[quote="indianamath"][quote="Karth"]$ z \cdot \bar z = |z|,$[/quote]

how did you get that>?[/quote]

It's supposed to be $ z \cdot \bar z = |z|^{2},$ actually. If we let $ z = a+bi,$ we then get $ z = a-bi.$ Since $ |z|^{2}= a^{2}+b^{2},$ it follows.

[quote="LeFromage"][hide]
$ z^{2002}=\bar z\implies |z|^{2002}=|z|$.

Then, $ |z|=0$ or $ |z|=1$. The first case has 1 solution. For the second case, $ z\bar z=1\implies \bar z =\frac{1}{z}$.

Substituting into the original equation, $ z^{2003}=1$, so there are 2003 solutions.

There are a total of 2004 solutions.[/hide][/quote]

Ah, you're right. I forgot the $ a = b = 0$ case...  :blush:

My Solution:
We have $ (a + bi)^{2003} = a^2 + b^2 = (a^2 + b^2)^{2003/2} (\text{cis} 2003 \theta)$
so $ (a^2 + b^2)^{2001/2}(\cis 2003 \theta) = 1$ (for nonzero a, b)
It follows that $ \sin 2003 \theta = 0\implies 2003\theta = 2n\pi, n\in \mathbb{Z}$ while keeping $ \cos 2003\theta$ positive.
So $ \theta = (2/2003)n\pi = \arctan(a/b)$ where $ a^2 + b^2 = 1$.  So there are 2003 different values for $ \arctan(a/b)$ and since $ \arctan$ is bijective, there are 2003 different values for a and b.
Also, we must include the case when $ a = b = 0$ which gives a total of $ 2004$ possible ordered pairs.

We are given $(a+bi)^{2002}=a-bi$ so multiplying through by $a+bi$ we get $(a+bi)^{2003}=(a+bi)(a-bi)=a^2+b^2$
The third equality is just putting this into trigonometric form. We want to make it into $rcis(\theta)$ where $r$ is the distance to the origin, which is just $\sqrt{a^2+b^2}$, so by de moivre's theorem, $(rcis(\theta))^n = r^n * cis(n\theta)$ so $\sqrt{a^2+b^2}^{2003}=(a^2+b^2)^{2003/2}$ and we have some value of $\theta$ in the equation.

Thanks Abe27342, can you explain rest of mathwizarddude's solution for me? Thanks. Sorry for having you explain everything.

Yeah, so we know
$(a^2+b^2)^{2003/2}(cis2003\theta)=a^2+b^2$
dividing through by a^2+b^2 we get
$(a^2+b^2)^{2001/2}(cis2003\theta)=1$
so he concludes $sin2003\theta=0$ because $cis2003\theta=cos2003\theta+isin2003\theta$ and obviously $a^2+b^2$ is real, so $cis2003\theta$ must also be real. $cos2003\theta$ must be positive because $a^2+b^2$ will always be positive and the product of the two must be 1.
Then by geometry we know $\theta=arctan(b/a)$ (it's actually b/a, although it doesn't matter) as $tan\theta=b/a$ (look on the complex plane to see this) so we conclude
$\frac{2n\pi}{2003}=arctan(b/a)$ and $a^2+b^2=1$
There are 2003 different values because n can belong to any of the numbers between 0 and 2002 as by excluding any solution where sin(x)=0 but cos(x)<0 (He did this by only including $2n\pi$ and not $n\pi$, as all solutions for $2n\pi$ have cosx>0)
That being said, There are 2003 solutions to that part.
Then you need to also include if the number is 0, which gives 2004 solutions

$ \text {By the way, what is the source for this problem? Is it AIME? } $

[quote="AlphaMath1"]Multiple choice and topic (imaginary numbers $\in$ Alg II/Pre-Cal) suggests it is AMC 12.[/quote]
Ah! I see. I never realized that AIME did not have multiple choice.

[quote="w squared parabola"]Thanks, Abe27342. If you don't mind one more question, there's one part in your solution I do not understand. In $ \frac{2n\pi}{2003}=arctan(b/a) $, why $ n $ could not be anything greater than 2003?[/quote]When $n=2003$, we get
$\frac{2n\pi}{2003}=\frac{2\cdot2003\pi}{2003}=2\pi.$

However, this is the same angle as that obtained when $n=0$. Similarly, when $n=2004$ we get the same angle as when $n=1$. By restricting $n$ from $0$ to $2002$, he is ensuring that every angle is counted exactly once.

For those interested, this is the [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44&year=2002&sid=6e7e38d7efd07c9031803d332b7e98ff#pp899529]2002 AMC 12A #24[/url].

MathWise, but look since $ sin2003\theta=0 $, then $ \frac{2n\pi}{2003}=\theta $ because $ sin2n\pi=0 $. So wouldn't any n will work because the result will always be $ sin2003\theta=0 $? So wouldn't there will not be any repeating involved like in your explanation?

Every integral value of $n$ works. However, plugging in n=0 and n=2003 we see that we get the same number. Same for n=1 and n=2004. It doesn't matter that every n works, it matters that there are exactly 2003 numbers that get RETURNED across all values of n. Have you learned about roots of unity? It's essentially the same concept.

Ok, I see. Thanks for being very patient with me. I was looking for a way to solve the problem using DeMoivre's Theorem and this is what I am looking for.

Can somebody please tell me what is wrong with the following? I have a tendency to make mistakes with DeMoivre's but I cannot figure out why. Thanks.

[hide="Incorrect solution"]$a + bi = cos(x) + isin(x) = e^{ix}$

$(a + bi)^{2002} = e^{2002ix} = cos(2002x) + isin(2002x)$

$(a + bi)^{2002} = cos(2002x) + isin(2002x) = a - bi$

$cos(2002x) + isin(2002x) = cos(x) - isin(x)$

$cos(2002x) = cos(x)$ and $sin(2002x) = -sin(x)$

It seems to follow that;

$2002x \equiv 360 - x \bmod{360}$

$2003x \equiv 0 \bmod{360}$

$x \equiv 0 \bmod{360}$

$a + bi = cos(x) + isin(x) = cos(0) + isin(0) = 1$. This is obviously wrong, though.[/hide]

Well, there isn't anything WRONG there, it's just that what you have arrived at does not help in solving the problem. What you've done by comparing the real and imaginary parts and then looking $\mod{360}$ is correct. By the way \[\cos 2002x = \cos x \implies 2002x \equiv x \pmod{360} \iff 2001x \equiv 0 \pmod{360},\] but since we also have to consider the $2003x \equiv 0 \pmod{360}$ condition, this does not change anything. Ultimately, we get that $x \equiv 0 \pmod{360}$ as you have done above. This means that $\cos x + i \sin x = 1 = \cos x - i \sin x = a + bi = a - bi$. This not wrong at all, since $1^{2002} = 1$. However the information we have found simply does not allow us to take any further steps in solving the problem, and thus it is just the wrong approach.

(If you're wondering, the reason it does not help is because it is never explicitly stated that $x$ must be an integer)

I think it does help, actually. If we have;

$2003x \equiv 0 \bmod{360}$

Then we must be able to express $x$ in the following form;

$x = \frac{360a}{2003}$

$a$ can have any integer value. But the above form demonstrates that $x$ can only have $2003$ unique values- when $a$ ranges from $0$ to $2002$. Once $a$ exceeds $2002$, you'll get repeats for values of $x$.

If $a_m \equiv a_n \bmod{2003}$

then $x_m \equiv x_n \bmod{360}$.

[quote="LeFromage"][hide]
$ z^{2002}=\bar z\implies |z|^{2002}=|z|$.

Then, $ |z|=0$ or $ |z|=1$. The first case has 1 solution. For the second case, $ z\bar z=1\implies \bar z =\frac{1}{z}$.

Substituting into the original equation, $ z^{2003}=1$, so there are 2003 solutions.

There are a total of 2004 solutions.[/hide][/quote]

Can someone explain the following:

1. How does $ z^{2002}=\bar z\implies |z|^{2002}=|z|$?

2. What is the motivation behind looking at the magnitudes?

1. Take the magnitude of both sides and use the fact that magnitude map is compatible with multiplication.
2. You're supposed to see this and realize that 1 has to be a possible magnitude.

Thanks, also one more question, where did $z^{2003}=1$ come from?

It is well known that if $|z| = 1$, then $z\bar{z} = 1$ (this is very important).

Or in general, $|z|^2=z\overline{z}$ for any complex number $z$.  (This is because $(a-bi)(a+bi)=a^2+b^2$.)

This question involves solution one in the wiki:

"Let $s=\sqrt{a^2+b^2}$ be the magnitude of $a+bi$. Then the magnitude of $(a+bi)^{2002}$ is $s^{2002}$, while the magnitude of $a-bi$ is $s$. We get that $s^{2002}=s$, hence either $s=0$ or $s=1$.

For $s=0$ we get a single solution $(a,b)=(0,0)$.

Let's now assume that $s=1$. Multiply both sides by $a+bi$. The left hand side becomes $(a+bi)^{2003}$, the right hand side becomes $(a-bi)(a+bi)=a^2 + b^2 = 1$. Hence the solutions for this case are precisely all the $2003$rd complex roots of unity, and there are $2003$ of those.

The total number of solutions is therefore $1+2003 = \boxed{2004}$.

Why is the equation $s^2002=s$ true? While I get that both $a+bi$ and $a-bi$ have the same magnitude, why are we able to replace them with $s=|a+-bi|$? Thanks

If two complex numbers are equal, then their magnitudes must be equal in particular.

[b]Solution:[/b] Let $a+bi = rcis\theta$. Then we have that $(rcis\theta)^{2002} = rcis(-\theta)$. Ignoring the case where $r=0$ for now, and dividing, we have that $r^{2001}cis(2003\theta) = 1$. Clearly, $r=1$ in this case and that leaves us with $2003$ possible distinct complex numbers. However, including the case with $r=0$ (which results in $(a,b) = (0,0)$), we have one additional possibility. Hence the answer is $2003 + 1 = 2004$, or E.

MAA was really nice not to put 2003 as an answer choice


Let $\omega=\frac{z^2-3z+6}{z+1}$ and $z=1+i$. Find $|\omega|$ and $\arg\omega$.

Have a look at: http://scholar.hw.ac.uk/site/maths/topic11.asp?outline= :)

$arg$ is the angle made by the line segment from the origin to $(a, b)$ of the complex number $a + bi$ off of the real axis in an argand plane.


trigonometry
calculus
integration
calculus computations

Prove that the integral of sinmx*sinnx dx on the interval -pi to pi = {0 if m does not = n, and pi if m = n

This one has me more than a little puzzled....


geometry
circumcircle
incenter
geometric transformation
rotation
conics
ellipse

Let $ ABC$ be a triangle and denote by $ (O)$, $ (I)$ its circumcircle and incircle, respectively. Let $ A'$ be a point on $ (O)$ and let $ B'$, $ C'$ be the intersections of the circumcircle $ (O)$ with the tangents from $ A'$ to $ (I)$. Prove that the locus of the symmedian point of triangle $ A'B'C'$ (as $ A'$ moves on $ (O)$) is an ellipse.

Let $K,U$ be the symmedian points of $\triangle ABC$ and $\triangle A'B'C'.$ If $R$ denotes the circumradius of $\triangle ABC,$ by Lagrange theorem for the symmedian point $K$ and incenter $I$ of $\triangle ABC,$ we get the expression:

$\frac{a^2 \cdot IA^2+b^2 \cdot  IB^2+c^2 \cdot IC^2}{a^2+b^2+c^2}=KI^2+R^2-KO^2$

Then, substituting the following identities

$IA=\sqrt{\frac{bc(s-a)}{s}} \ , \ IB=\sqrt{\frac{ac(s-b)}{s}} \ , \ IC=\sqrt{\frac{ab(s-c)}{s}}$

$\Longrightarrow \ \frac{a^2bc(s-a)+b^2ac(s-b)+c^2ab(s-c)}{s(a^2+b^2+c^2)}=KI^2+R^2-KO^2$

After  simplifying terms and using the identity $a^2+b^2+c^2=2s^2-2r^2-8Rr,$ we obtain:

$4Rr \left ( \frac{r^2+4Rr}{s^2-r^2-4Rr} \right) =KI^2-KO^2+R^2 \ \ (1)$

On the other hand, the power of $K$ with respect to $(O)$ is given by:

$R^2-KO^2=\frac{3a^2b^2c^2}{(a^2+b^2+c^2)^2}=3 \left (\frac{2Rrs}{s^2-r^2-4Rr} \right)^2 \ \ (2)$

In order to have the locus of $U$ we  need to eliminate $s$ from $(1)$ $\cap$ $(2)$ which is not "poristically" fixed. Plugging $s$ from $(1)$ into $(2),$ after simplifying terms we obtain:

$\frac{_1}{^3}(R^2-KO^2)(r^2+4Rr)-Rr(KI^2-KO^2+R^2)=\frac{_1}{^4}(KI^2-KO^2+R^2)$

$\frac{_1}{^3}(R^2-UO^2)(r^2+4Rr)-Rr(UI^2-UO^2+R^2)=\frac{_1}{^4}(UI^2-UO^2+R^2)$

It remains to choose an appropriate projective reference to identify the locus of $U,$ say a cartesian reference $x,y$ where $O \equiv (0,0)$ and $I \equiv (d,0)$ for $d= \overline{IO}.$ Therefore:

$UO^2=x^2+y^2 \ , \ UI^2-UO^2=(x-d)^2+y^2-x^2-y^2=d^2-2dx$

$\frac{_1}{^3}(R^2-x^2-y^2)(r^2+4Rr)-Rr(d^2-2xd+R^2)-\frac{_1}{^4}(d^2-2dx+R^2)^2=0$

This is a second degree equation where the $x^2,y^2$ coefficientes are different and have equal sign. Furthermore, there's no xy- term that indicates a rotation of main axes or linear y-term. Thus, the locus of $U$ is an ellipse $\mathcal{K}$ that has $IO$ as one of its axes. $\square$
_________________________________

Interestingly, the locus of the three ex-symmedian points $U_a,U_b,U_c$ of $\triangle A'B'C'$ is also a single conic, but not necessarily ellipse, i.e. the vertices of the tangential triangles of $\triangle ABC$ and $\triangle A'B'C'$ lie on a conic $\mathcal{K}',$ inverse image in the circumcircle of the Limaçon $\mathcal{L}$ embracing the incircle $(I)$ with cusp $O.$

Locus of the othocenter of $\triangle A'B'C'$ is a circumference $\omega$ whose center is the reflection $O'$ of $O$ about $I.$ Let $N$ be the 9-point center of $\triangle ABC.$ Since 9-point cirle $(N)$ and incircle $(I)$ are internally tangent $\Longrightarrow$ $NI=\frac{_1}{^2}R-r$ $\Longrightarrow$ $O'H=2NI=R-2r$ is a fixed distance. Consequently, the locus of the centroid of $\triangle A'B'C'$ is obviously the circumference $\omega'$ homothetic to $\omega$ through the homothety $(O,\frac{_1}{^3}).$

I'll finish some minor Details left by @above
[quote]$\frac{a^2 \cdot IA^2+b^2 \cdot IB^2+c^2 \cdot IC^2}{a^2+b^2+c^2}=KI^2+R^2-KO^2$[/quote]

We know by lagrange on $K,I$ with $K \equiv ([KBC]:[KCA]:[KCA])$ that $\frac{IA^2\cdot [KBC]+IB^2\cdot [KCA]+IC^2\cdot [KAB]}{[ABC]}=KI^2+p(K,\mathcal{O})=KI^2+R^2-OK^2$
Now use  barycentric cordinates wrt $ABC$.Hence $[KBC]=\begin{vmatrix*} 0 & 0 & 1 \\ 0 & 1 & 0 \\ \dfrac{a^2}{a^2+b^2+c^2} & \dfrac{b^2}{a^2+b^2+c^2} & \dfrac{c^2}{a^2+b^2+c^2} \end{vmatrix*} \cdot [ABC] \Rightarrow [KBC]=\dfrac{a^2}{a^2+b^2+c^2}\cdot [ABC]$.Similar variants give the conclusion.$\square$

[quote]$a^2+b^2+c^2=2s^2-2r^2-8Rr,$[/quote]

Note that $r^2=\dfrac{\Delta^2}{s^2}=\dfrac{(s-a)(s-b)(s-c)}{s}$ and $4Rr=\dfrac{abc}{s}$.Now substitute and simplify.

[quote]$R^2-KO^2=\frac{3a^2b^2c^2}{(a^2+b^2+c^2)^2}$[/quote]

[We again use barycentric co-ordinates wrt $ABC$.Clearly $K=\left(\dfrac{a^2}{a^2+b^2+c^2},\dfrac{b^2}{a^2+b^2+c^2},\dfrac{c^2}{a^2+b^2+c^2}\right)$.Note that the power of ay point $P$ wrt $\odot{ABC}$ is $-a^2yz-b^2zx-c^2xy \Rightarrow OK^2-R^2=p(K,\mathcal{O})=\sum a^2\cdot \dfrac{b^2c^2}{(a^2+b^2+c^2)^2}=-\dfrac{3a^2b^2c^2}{(a^2+b^2+c^2)^2}$.Hence the conlusion.$\square$


(1) Evaluate
\[ \log_{2}\sqrt{2} \plus{} \log_{2}\sqrt[4]{4}\plus{}\log_{2}\sqrt[8]{8}\plus{}\cdots\plus{}\log_{2}\sqrt[2^n]{2^n}\plus{}\cdots.
\]

(2) Find a closed form for the series
\[ 1 \plus{} 2x \plus{} 3x^2 \plus{} \cdots z\plus{} nx^{n\minus{}1}
\]
for $ x\neq 1$.

Thanks in advance!

[quote="math154"][hide="1"]
$ \log_{2}\sqrt {2} \plus{} \log_{2}\sqrt [4]{4} \plus{} \log_{2}\sqrt [8]{8} \plus{} \cdots$

$ \equal{} \log_{2}2^{\frac12} \plus{} \log_{2}2^{\frac24} \plus{} \log_{2}2^{\frac38} \plus{} \cdots$

$ \equal{} \frac12 \plus{} \frac24 \plus{} \frac38 \plus{} \cdots$

$ \equal{} \left(\frac12 \plus{} \frac14 \plus{} \frac18 \plus{} \cdots\right) \plus{} \left(\frac14 \plus{} \frac18 \plus{} \frac1{16} \plus{} \cdots\right) \plus{} \cdots$

$ \equal{} 1 \plus{} \frac12\cdot1 \plus{} \frac14\cdot1 \plus{} \cdots$

$ \equal{} \boxed2$
[/hide]
[hide="2"]
$ S \equal{} 1 \plus{} 2x \plus{} 3x^2 \plus{} \cdots \plus{} nx^{n \minus{} 1}$

$ \implies Sx \equal{} x \plus{} 2x^2 \plus{} \cdots \plus{} nx^n$

$ \implies S(1 \minus{} x) \equal{} 1 \plus{} x \plus{} x^2 \plus{} \cdots \plus{} x^{n \minus{} 1} \minus{} nx^n$

$ \equal{} \frac {1 \minus{} x^n \plus{} nx^{n \plus{} 1} \minus{} nx^n}{1 \minus{} x}$

$ \implies S \equal{} \boxed{\frac {nx^{n \plus{} 1} \minus{} (n \plus{} 1)x^n \plus{} 1}{(1 \minus{} x)^2}}$
[/hide][/quote]

How did you get from $ \frac {1 \minus{} x^n \plus{} nx^{n \plus{} 1} \minus{} nx^n}{1 \minus{} x}$ to $ S \equal{} \boxed{\frac {nx^{n \plus{} 1} \minus{} (n \plus{} 1)x^n \plus{} 1}{(1 \minus{} x)^2}}$?  Thanks!

$ S(1\minus{}x)\equal{}\frac {1 \minus{} x^n \plus{} nx^{n \plus{} 1} \minus{} nx^n}{1 \minus{} x}$, so dividing both sides by $ 1\minus{}x$ gives

$ S\equal{}\frac {nx^{n\plus{}1}\minus{} x^n\minus{}nx^n \plus{} 1}{(1 \minus{} x)^2}\equal{}\frac{nx^{n\plus{}1}\minus{}(n\plus{}1)x^n\plus{}1}{(1\minus{}x)^2}$.

another solution:
let $ f_n(x) \equal{} x \plus{} x^2 \plus{} \cdots \plus{} x^n \equal{} \frac{x(1\minus{}x^n)}{1\minus{}x}$
$ f_n'(x)\equal{}1\plus{}2x\plus{}3x^2 \plus{} \cdots \plus{} nx^{n\minus{}1}\equal{}..$

[hide="Another solution for #2."]
$ 1 \plus{} 2x \plus{} 3x^2 \plus{} ... \plus{} nx^{n \minus{} 1} \equal{} z$
$ 1 \plus{} x \plus{} x^2 \plus{} ... \plus{} x^{n \minus{} 1} \equal{} \frac {1}{1 \minus{} x}$
$ z \minus{} \frac {1}{1 \minus{} x} \equal{} x \plus{} 2x^2 \plus{} 3x^3 \plus{} ... \plus{} nx^n \equal{} z \cdot x$
$ z(x \minus{} 1) \equal{} \minus{} \frac {1}{1 \minus{} x}$
$ z \equal{} \frac {1}{(1 \minus{} x)^2}$

[/hide]

Edit: oops, misread the question.

That's not correct: $ 1\plus{}x\plus{}x^2\plus{}...\plus{}x^{n\minus{}1}\equal{}\frac{1\minus{}x^n}{1\minus{}x}$--however, $ \lim_{n\to\infty}1\plus{}x\plus{}x^2\plus{}...\plus{}x^{n\minus{}1}\equal{}\frac1{1\minus{}x}$.


Find all positive integers $x,y,z$ such that
a)$\frac{x-y\sqrt{2006}}{y-z\sqrt{2006}}\in\mathbb{Q}$ , and
b)$x^{2}+y^{2}+z^{2}$ is a prime.

Let $p,q$ be integers such that $\frac{x-y\sqrt{2006}}{y-z\sqrt{2006}}= \frac{p}{q}$. Then $(xq-yp)+(pz-yq)\sqrt{2006}= 0$. As $x,y,z,p,q$ are integers, this means that $xq = yp$ and $pz = yq$. Therefore, $\frac{x}{y}= \frac{p}{q}= \frac{y}{z}$, so $xz = y^{2}$.
Now we want $x^{2}+y^{2}+z^{2}= x^{2}+xz+z^{2}= (x+z)^{2}-xz = (x+z)^{2}-y^{2}= (x+y+z)(x-y+z)$ to be prime. As $x,y,z$ are positive integers, $x+y+z > x-y+z$, so we should have $x-y+z = 1$. But then $x^{2}+y^{2}+z^{2}= x+y+z$, i.e. $x(x-1)+y(y-1)+z(z-1) = 0$. Each term at the RHS is non-negative, so every term should be equal to 0, so $x = y = z = 1$.


function
factorial
limit
real analysis
real analysis unsolved

Can any of you solve this?  :blush:

Firstly, some notation:

Let $ \Pi(x) \equal{} \Gamma(x\plus{}1)$ where $ \Gamma(x)$ is the usual gamma function i.e. an extension of the factorial to the complex numbers.

Let $ log^{n} (x) \equal{} log( log( \cdots log( x ) ) )$ where  $ log$ s applied [b]n[/b] times to[b] x[/b] e.g.

$ log^{4} (x) \equal{} log( log( log( log( x ) ) ) )$

Similarly, let $ {\Pi}^{n} (x) \equal{} \Pi( \Pi( \cdots \Pi( x ) ) )$ where $ \Pi$  is applied [b]n[/b] times to [b]x[/b].

So The Question IS
Let  $ a_{n} \equal{} log^{n} ( {\Pi}^{n} (3) )$
Evaluate, using a computer of otherwise,  $ \lim_{n \rightarrow \infty}{ a_{n} }$  to five decimal places.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180173
I'm pretty sure the limit to five places is 2.1164, but would be interested in seeing a rigorous upper estimate (if it's not too messy). Failing that, I'd like to know the value of $ a_5$.  :)


calculus
derivative
algebra
binomial theorem
calculus computations
poll

I was very curious about this topic. But, I have feeling that second choice would be the answer. What do you guys think?

\[ 5.6^5 \equal{} \left(5 \plus{} 0.6\right)^5 \equal{} 3125 \plus{} 5(625)(0.6) \plus{} 10(125)(.36) \plus{} \cdots \plus{} .07776 \equal{} 5507.31776
\]
Hmm, the ninth derivative one might take a little bit.

[hide=":P"]
Just kidding, just had to do that.  [/hide]

[quote="azjps"]\[ 5.6^5 \equal{} \left(5 \plus{} 0.6\right)^5 \equal{} 3125 \plus{} 5(625)(0.6) \plus{} 10(125)(.36) \plus{} \cdots \plus{} .07776 \equal{} 5507.31776
\][/hide][/quote]

 :) I used this method to find $ 11^7$ in calculus yesterday!  Similarly, binomial theorem can be used to find that $ (x\minus{}2y)^7\equal{}\sum_{i\equal{}0}^7 \binom{7}{i} x^{7\minus{}i}(\minus{}2y)^i$.

As for your four term expansion, multinomial theorem is the approach I would use if I needed do do something like take a 5 digit number to the twenty-third power. (but there better be a GOOD reason first.)

This reminds me of in kindergarden where people were obsessing about how far they could "count" the integers. I claimed that I could count to any number, but no one would accept this because they felt I couldn't just stand there and count to a million

Yes. The ends justify the means in some ways, especially in mathematics.
If someone can do that, then that person can usually make further inferences on other topics.

I'm assuming that your problems are examples of all mathematical problems and that by saying "can do" it implies the person can do all problems of that kind.

I think most of us CAN do them. But most of us WOULD NOT do it since calculating them does not benefit us to understand math deeply.
Nevertheless, a mathematician should have patience to do a lot of boring things.

I don't think being able to do those types of problems makes one a math genius because they each can be solved with a simple algorithmic procedure (albeit repititive and boring) that produces rather uninteresting results.

However, that's not to say that being able to do things like that aren't impressive.


email
algebra
polynomial
algebra unsolved

could you guys please solve this factoring problem for me...
i need it by sunday, it's plus 10 points for us in class if we get this problem, that is why i need help. txs.

x4yz + xy4z + xyz4 - y3z3 - x3z3 - x3y3

btw, the numbers are all exponents, i just don't know how to put the numbers up.Ü

please email me the solution and answer at mabel_costas@hotmail.com

i really need ur help on this, hope you guys can help me.

txs.

MaBeL :)

[quote="lordWings"]It's an interesting problem.
$P(x,y,z)=x^{4}yz+xy^{4}z+xyz^{4}-y^{3}z^{3}-x^{3}z^{3}-x^{3}y^{3}$.

If we write it as a polynomial on $x$, we have
$P_{yz}(x)=(yz)x^{4}-(y^{3}+z^{3})x^{3}+(y^{3}+z^{3})yzx-(yz)^{3}$,
it's easier to identify the root $x^{2}=yz$.

Since $x,y,z$ are entirely symmetrical, we'll have
$P(x,y,z)=k(x^{2}-yz)(y^{2}-xz)(z^{2}-xy)$,
and finally it can be checked that $k=1$.[/quote]

Thanks!Ü


Compute x{n} if x{0}=p; x{n+1}=x{n}^2-a*x{n}+b. n natural.

xn+1=(x_n-a/2) 2+c', where c'=b-a 2 /4.
Now take y_n=x n -a/2. It checks yn+1=y_n 2 +c, where c=c'-a/2=b-a 2 /4-a/2.

This final recurence is the best you can get from the problem.
It leads you smth like y_n=(...(p' 2+c) 2 +c) 2+c)...) 2 +c.

Hope you don't mind me moving this topic to solved.


calculus
integration
function
real analysis
real analysis unsolved

Here is an infinite series:\[ \sum^\infty_{n=1} \frac{(n )!}{n^n} = 1.87985386... \]

What does this series equal? (ie. what known numbers, integrals, functions or series can it be expressed as?). Or at least, what are any (non trivial) properties about this series?

Well, one property is the rate of convergence. One way to express that: the power series

$\sum\frac{n!}{n^n}x^n$

has radius of convergence $e.$ Saying that is a weak form of Stirling's approximation.

But is the sum above a "nice" number? A number with a name that we can roll around in our hands and manipulate as we see fit?

The cardinality of the real numbers is uncountable. What is the cardinality of the "nice" numbers? Answering that question would require us to make a precise definition of a squishy concept, but I'm inclined to think the cardinality is finite. (For instance, it wouldn't include all rational numbers, because some rational numbers have numerators or denominators too big for us to write down.)

So pick a number given to us by some limit process - some infinite sum, some integral, a fixed point or root of some function. If we know nothing else about the process that produced this number, we'd have to say that it's pretty unlikely that the number is "nice." There could be something here - but there's no particularly good reason to suspect that there is.


[quote="simon_math"]A uniform $ 2 kg$ cylinder rests on a laboratory cart. The coefficient of static friction between the cylinder and the cart is $ 0.5$. If the cylinder is $ 4 cm$ in diameter and $ 10 cm$ in height, which of the following is closest to the minimum acceleration of the cart needed to cause the cylinder to tip over? (in m/s^2)

(A) 2     (B) 4     (C) 5     (D) 6

The answer is (B), but I can't see why. Anyone wanna help me?[/quote]

Draw the FBD of the cylinder.
The centre of mass lies at the geometric centre of the cylinder. At the critical moment, when the cylinder is about to topple about an edge, the Normal reaction passes through an edge. Resolve the torques with the bottom edge as your reference point. The friction passes through this point, hence you needn't bother about it.
Now,
$ \begin{align*} mg(2 \times 10^{ - 2}) = m(a_{max})(5 \times 10^{ - 2}) \\
& a_{max} = g\left(\frac {2}{5}\right) \\
& = 10 \times 0.4 \\
& = 4 \, m/s^2$

Hence your answer is Option $ B)$.


Let $ a,b,c \in (0,1)$ and $ x,y,z \in \mathbb{R}$ so that $ a^x\equal{}bc,b^y\equal{}ca,c^z\equal{}ab.$ Prove that $ x\plus{}y\plus{}z \ge 6.$ When does the equality holds?

[quote="moldovan"]Let $ a,b,c \in (0,1)$ and $ x,y,z \in \mathbb{R}$ so that $ a^x \equal{} bc,b^y \equal{} ca,c^z \equal{} ab.$ Prove that $ x \plus{} y \plus{} z \ge 6.$ When does the equality holds?[/quote]

we have $ x\equal{} \frac{ ln(bc)}{ln(a)}\equal{}  \frac{ ln(b)\plus{}ln(c)}{ln(a)}  \equal{} \frac{ \minus{}ln(b)\minus{}ln(c)}{\minus{}ln(a)}$ etc...  and the result follows by am-gm


This one is not strong. The following very strong inequality holds

if $abc=1$,$a,b,c>0$
prove $\frac{1}{(1+a)^3}+\frac{1}{(1+b)^3}+\frac{1}{(1+c)^3} \geq \frac 3 8$

The last is equivalent to Vietnam 2005 TST ineq.

Sorry about my mistake, but can you tell me where you found this question? I'm curious, and if that's in Chinese, I can read it too.

Frank Morgan

you are wrong...silouan's problem is right,it is Vietnam 2005 TST ineq.
and my problem is from a friend of mine

But there should be nothing wrong about this: $\ \frac{1}{a} + a \geq 2$
then $\ \ 2+a+b+c + \frac{1}{a} +\frac{1}{b} + \frac{1}{c} \geq 8$
$\ \frac{1}{(1+a)^3}+\frac{1}{(1+b)^3}+\frac{1}{(1+c)^3} \geq \frac 3 8$

Sorry, I was wrong. The probabal reason is because I was solving the problem while concentrating on programming class teacher. :D. I'm really sorry.

[quote="zhaobin"][quote="keira_khtn"]Here is my approach for both two above inequals:
According a China TST :
Let $abcd=1$.Prove that:
$\sum\frac{1}{(1+a)^2}\geq 1$
Proof of this:Using the result:
$\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}\geq \frac{1}{1+ab}$ for all reals $a,b$
and the anologous ineq for $c,d$ ,combine with the original relation we have done...
For TST VN 2005,take $d=1$ and using Cauchy ineq we have result.
And for zhaobin's ineq let's take $d=\frac18$![/quote]
I'm sorry.I can't understand.in my inequality,it is $3$ in $\sum \frac{1}{(1+a)^3}$[/quote]

  Use Power-mean ineq to deal with it !

[quote="nttu"]It's my mistake !
 In Chinese ineq , the equal holds for $a=b=c=d =1$ , so when you put $d=8$ , the equal doesn't hold .
   The ineq for exponent 2 is wrong (for example, a=2,b=3,c=1/6 )[/quote]
yes.but thanks alll the same :)
continue,man :lol:

Sorry, applying amgm in denominator reverses the sign.

Oh,it's not as easy as i think.Indeed,the generalization of this was posted by Harazi in this forum:
http://www.mathlinks.ro/Forum/viewtopic.php?t=6017&sid=346b3eb9b70ad1519382e983b7494dc4

This question's generalization should be a special case of :
$\ \frac{1}{(1+kx_1)^2}+\frac{1}{(1+kx_2)^2}+...+\frac{1}{(1+kx_n)^2} \geq \frac{n}{(1+k)^2}$(Is here a "$\ 2$" missing on the right side for $\ \frac{n}{2(1+k)^2}$?)
in which is the case of $\ k=1$
Let $\ a_1,a_2,...a_n (n \in N, n \geq2)$ be all positivereal numbers such that $\ a_1a_2a_3...a_n = 1$
Find the min(or maybe, infimum) value of ${\ \sum{\frac{1}{(1+a_i)^2}}}$(Which should be $\ \frac{n}{2(n-1)^2}$)

[quote="keira_khtn"][quote="zhaobin"]I think they are not compeltely same. :)
and somebody have a solution?[/quote]
Your problem is easy case :n=3.By mixxing variables method, we have:
$\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}\geq 1$
with $abc=1$
Proof:Assume c=min{a,b,c}
Using the fact:$\frac{1}{x+1}+\frac{1}{y+1}\geq\frac{2}{\sqrt{xy}+1}$
holds when $xy\geq 1$
Substituting $x=2a,y=2b$ s.t.:$ab\geq 1\geq \frac14$(because c=min)
$\frac{1}{1+2a}+\frac{1}{1+2b}\geq \frac{2}{1+2\sqrt{ab}}$
$\to f(a,b,c)\geq f(c,\sqrt{ab},\sqrt{ab})$
We remain to prove that:$f(t,t,\frac{1}{t^2})\geq 1$
$\to \frac{2}{2t+1}+\frac{1}{1+\frac{2}{t^2}}\geq 1$
$\to 1+t^2\geq 2t$ which is obvios..
Finally using Power mean inequal,we have done...[/quote]
I don't know what things is wrong.
in my problem there is $a+2$,not $2a+1$
and we only have $\sum \frac 1 {a+2} \leq 1$
with $abc=1$
have you notice that?? :)  :)


linear algebra
matrix
linear algebra unsolved

Find all $ A,B,C \in M_n{\mathbb{(C)}}$ are Hermitian matrices such that :
          $ A^{6}\plus{}B^{6}\plus{}C^{6}\plus{}6(A^{4}\plus{}B^{4}\plus{}C^{4})\plus{}10(A^{2}\plus{}B^{2}\plus{}C^{2})\minus{}2(A^3B\plus{}B^{3}C\plus{}C^{3}A)\minus{}6(AB\plus{}
BC\plus{}CA)\equal{}0$

Add hard result :
  Find $ (A,B,C)$ Hermitian matrix numbers such that :
   $ A^{6} \plus{} B^{6} \plus{} C^{6} \minus{} 6(A^{4} \plus{} B^{4} \plus{} C^{4}) \plus{} 10(A^{2} \plus{} B^{2} \plus{} C^{2}) \minus{} 2(A^{3}B \plus{} B^{3}C \plus{} C^{3}A) \plus{} 6(AB \plus{} BC \plus{} CA) \equal{} 0$


function
real analysis
real analysis unsolved

Let $ f: \mathbb{R}^n\rightarrow\mathbb{R}$ be a measurable function. If for any Lebesgue integrable function $ g: \mathbb{R}^n\rightarrow\mathbb{R}$, $ fg$ is also Lebesgue integrable show that $ f$ is essentially bounded, i.e. $ \exists C > 0$ s.t. $ |f|\leq_{L}C$

If $ f$ is not essentially bounded, then there exists a sequence of disjoint sets $ A_n$ of finite positive measure such that $ |f| > 2^n$ on $ A_n$. Let $ g \equal{}\mathrm{sign}\,(f) \sum_n \frac {2^{ \minus{} n}}{|A_n|}\chi_{A_n}$.


Search "AwesomeMath 2009" on Facebook and join! See you there!  :D


geometry
perimeter
circumcircle
trigonometry
geometry proposed

If the feet of the altitudes of the triangle $ ABC$ are the vertices of a second triangle $ PQR$, such that the primeter of $ PQR$ equals the diameter of the circumcircle of $ ABC$, prove that $ \sin A\cdot \sin B\cdot\sin C=\frac{1}{2}$ and $ \sin^{3}A\cdot\cos (B-C)+\sin^{3}B\cdot\cos (C-A)+\sin^{3}C\cdot\cos (A-B)=\frac{3}{2}$.

[hide="Part (a)"] Denote $ H,\,O$ be the orthocentre and circumcentre of $ \triangle ABC$
We have $ QR=AH\cdot\sin A=2R\cdot\cos A\sin A=a\cdot\cos A$
$ \implies PQ+QR+RP=\sum a\cdot\cos A=\frac{\sum a\cdot R\cdot\cos A}{R}=\frac{\sum S_{OBC}}{R}=\frac{S_{ABC}}{R}$
On the other hand, $ \sum QR=R\implies \frac{S_{ABC}}{R}=R\implies\frac{abc}{4\cdot R^{3}}=1\implies\prod\sin A=\frac{1}{2}$[/hide]

I think that, we have  $ \sin^{3}A\cdot\cos (B-C)+\sin^{3}B\cdot\cos (C-A)+\sin^{3}C\cdot\cos (A-B)=3 \sin A\cdot \sin B\cdot\sin C$ for all triangle $ ABC$. Do you agree?  :maybe:


function
irrational number
real analysis
real analysis unsolved

Do you also want it to be continuous at every irrational number or just discontinuous at rational numbers?

$ f(x) \equal{} \begin{cases}0 \,\,\, \mbox{if} \,\,\, x\not\in \mathbb{Q}\cap[0,1] \\
x\,\,\, \mbox{if} \,\,\,x\in\mathbb{Q}\cap [0,1] \end{cases}$

is discontinuous at every rational number.

Jrav,
your function is NOT non-deareasing funtion

[quote="t0rajir0u"]Let $ q_n$ be an enumeration of the rationals on $ [0, 1]$ and let $ f(x)$ be the sum of the elements of the set $ \{ 2^{ \minus{} n} : q_n \in [0, x] \}$.[/quote]

t0rajir0u, I got you. Thanks.

[quote="dn324307"]Hi dear,

I am looking for a monotone function ([b]i.e non-decreasing function[/b]) on $ [0,1]$ that is discontinuous at every rational number.  I start with the following function .It is monotone BUT discontinuous every where

$ f(x) \equal{} \begin{cases}e^{x} \,\,\, \mbox{if} \,\,\, x\not\in \mathbb{Q}\cap[0,1] \\
x \,\,\, \mbox{if} \,\,\,x\in\mathbb{Q}\cap [0,1] \end{cases}$

How can I modify it to get a function that monotone function on $ [0,1]$ that is discontinuous at every rational number.

thanks so much[/quote]

this function isn't monotone as it looks like $ 1$ for small irrational numbers near and like $ 0$ for small rational numbers.  Pick $ p<x<q$ to be small with $ p,q$ rational and $ x$ irrational.


function
number theory open
number theory

Does exist an integer-valued function $f:N^* -> N^*$ , such that:
$f(f(n)) = p_n$ , where $p_n$ is the $n$ - [i]th[/i] prime.
There is any explicit definition?
Can you prove or disprove this?

What if we extend to:
$f:N^* -> N^*$
$f^k (n) = p_n$ , where $p_n$ is the $n$ - [i]th[/i] prime and $k \geq 2$.
$f^k$ means $f$ composed $k$ times.

[quote="heartwork"]Does exist an integer-valued function $f:N^* -> N^*$ , such that:
$f(f(n)) = p_n$ , where $p_n$ is the $n$ - [i]th[/i] prime.[/quote]

There is such a function!
Take $a(m,1)$ as the $m-th$ composite and $a(m,n+1)$ the $a(m,n)-th$ prime.
For each positive integer $N$ there exists $m,n$ such that $N=a(m,n)$.
For example define $f$ as:
$f(a(2m-1,n))=a(2m,n)$
$f(a(2m,n))=a(2m-1,n+1)$.


When an amount of calcium reacts with carbon to product $ 64$ grams of solid $ \text{CaC}_2$, the heat that released was $ 59.8$ kJ.
On producting $ 2.8$ grams of solid $ \text{CaO}$, the heat that released was $ 31.76$ kJ.
The released heat in combusting $ 11.2$ liters of acetylene $ (\text{C}_2\text{H}_2)$ (STP) was $ 650$ kJ.
It is known that the enthalpy change of the following reaction $ \text{CaO}_{(s)} \plus{} \text{H}_2\text{O}_{(l)} \to \text{Ca(OH)}_{2(aq)}$ was $ \minus{} 631,1$ kJ, and the standard enthalpy of formation of $ \text{CO}_{2(g)}$ was $ \minus{} 393.5$ kJ.

Calculate the enthalpy change of the following reaction
$ \text{CaC}_{2(s)} \plus{} 2\text{H}_2\text{O}_{(l)} \to \text{Ca(OH)}_{2(aq)} \plus{} \text{C}_2\text{H}_{2(g)}$ when $ 3.2$ grams of $ \text{CaC}_2$ was reacted.

[b]Atomic weights[/b] Ar(Ca) = 40, Ar(C) = 12, Ar(O) = 16

Standard application of Hess Law. Write down the reaction corresponding to each data given, and then think in a way of algebraically combine them in order to obtain the desired reaction.


Is there a way I can partition more space from D drive to C drive?

Sorry...I'm pretty stupid when it comes to hardware stuff...*shrugs*

[url=http://www.partitionmagic.com/partitionmagic/]PartitionMagic[/url] ought to do it.

Be careful not to delete all the contents of both drives when you do that.


aga NA va NB ham W2  ro too X,Y  ghat kone be tori ke x,c dar yek taraf va Y,D dar tarafe digare 2 dayere bashand      darim

con M,N paad motojanes and va MC , NX rooye mehvar asli hamresand ⇒ X,C ham paad motojanes and   va be hamin shekl D,Y ham paad motojanes and

chon M markaz tajanose W1 , W ast ⇒  momas bar W1 too C ba momas too W dar B movazie.      be hamon shekl Momas too X bar W2 ba momas too C dar W1 movazian⇒  C,X   va B,Y   mahale momas moshtarake do dayereye W1 ,W2  e     ⇒    CD bar W1 momase .......

chi migi? :?:  :?:

fekr konam mikhasti rahe hale in soal ro bedi.
http://www.mathlinks.ro/Forum/viewtopic.php?t=99054


This one belongs to a middle school --> the basics section...

Anyways , years is the right answer....

[quote="momtonate"]In the following equation, determine the units on t if S is measured in dollars and p is measured in dollars per year
S=pt[/quote]

momtonate,

Evidently you've gotten a number of very smug replies, from those who already know their 1st year Algebra quite well.  Let's see if I can actually answer your question though.

The quantities we use in everyday arithmetic each have some type of "unit" associated with them that describes what they are. Dollars, yards, milligrams, days, volts, gallons, etc. are all examples.

Your problem then is one of [b]dimensional analysis[/b], a fancy way of saying "what kind of units are you using?"

Fortunately units can be treated just like variables in an algebra equation therefore:

$S =pt$ where $S=\text{dollars}$ and $p=\frac{\text{dollars}}{\text{year}}$ becomes

$\text{dollars} =\frac{\text{dollars}}{\text{year}}t$

If we now divide both sides by $\text{dollars}$ then

$\frac{\text{dollars}}{\text{dollars}}=\frac{t}{\text{year}}$

The $\text{dollars}$ cancel out on the left hand side and then

$1=\frac{t}{\text{year}}$

If we multiply both sides by $\text{year}$ then

$\text{year}=t$

That is $t$ has units of year(s).

Typically the singular versus plural from the standpoint of English grammar makes no difference in the math.

Thanks to Dr. No for providing a good explanation to a serious request for help.

[color=darkblue][Moderator's note:  Dimensional Analysis is a difficult concept for many people -- including some high school and college algebra students.  Even some college Physics and Engineering students have difficulty mastering the concept.   This was not a challenge.  This was a request for help.  Please, let's make the forums inviting to new members.  [size=75]Various smug remarks have been removed.][/size][/color]


vector
linear algebra
linear algebra unsolved

Suppose that $V$ is a vector space over the zero characteristic field $\mathbb F$ and $T$ is an operator on $V$ which $\forall k>0\ \ \ \mbox{Tr}(T^k)=0$. Prove that $T$ is nilpotent.

Suppose the eigenvalues of T are $a_1,...,a_n$ and suppose $S_k = a_1^k + ... + a_n^k$. It follows that $S_1=...=S_n = 0$, which means $a_1 = ... = a_n = 0$


abstract algebra
group theory
superior algebra
superior algebra unsolved

How many additive abelian groups of order 16 have the property that $ x\plus{}x\plus{}x\plus{}x\equal{}0$?
Why?

I assume you're not familiar with (or prefer not to use) the structure theorem for finitely-generated abelian groups, since using that theorem makes the problem completely routine (list all the abelian groups of order 16, take the ones that have exponent $ 2$ or $ 4$).

If the question is how to do this "with bare hands", I suppose you could do something like this: start by classifying the groups of order $ 4$. There are only two of them, and they are both abelian: $ C_4$ and $ C_2 \times C_2$ where $ C_m$ denotes a cyclic group of order $ m$. This is easy to see: either there's an element of order $ 4$ (in which case the group is cyclic), or all non-trivial elements have order $ 2$ (in which case you can explicitly construct the multiplication table).

So back to our abelian group $ A$ of order $ 16$, either it has an element of order $ 4$ or it has all elements of order $ 2$. In the latter case the group is obviously isomorphic to $ C_2 \times C_2 \times C_2 \times C_2$. In the former case $ A$ has a cyclic subgroup $ K$ of order $ 4$, and $ A/K$ also has order $ 4$ so it's either cyclic or $ C_2 \times C_2$.

If $ A/K$ is cyclic of order $ 4$, there's an element $ x \in A$ such that $ x, xK, x^2K, x^3K$ are all distinct, which means that $ x$ itself must have order $ 4$ in $ A$. Setting $ H \equal{} \langle x \rangle$ we see that $ H \cap K \equal{} \{1\}$ so $ A \equal{} H \times K \cong C_4 \times C_4$.

A similar analysis of the other case yields $ A \cong C_4 \times C_2 \times C_2$.

Overall, there are 3 abelian groups of order $ 16$ and exponent $ 4$.


topology
function
vector
advanced fields
advanced fields unsolved

Let $X$ and $Y$ be topological spaces and $f$, $g$ be two continuous functions from $X$ to $Y$. Prove or disprove that $A$ is closed. $A$ is the set of all points in $X$ s.t $f(x)=g(x)$.

I've proved it when $X$ is a metric space using sequences.

It is certainly true for $Y$ vector space endowed with Hausdorff topology because $\{0\}$ is closed and $A=(f-g)^{-1}(0)$ is hence closed as well but is not true in general. For example endow $Y$ with topology in which $\{0\}$ is open then $A$ is also open.

There's a flaw in that argument, Vladimir. We are only told that $Y$ is a topological space, not that it is a vector space (or that is has any algebraic properties.) Hence, $(f-g)$ is not defined.

Let $F$ map $X$ to $Y\times Y$ by $F(x)=(f(x),g(x))$. $F$ is continuous, and $A=F^{-1}(D)$ where $D$ is the diagonal $\{(a,a): a\in Y\}$. If $D$ is closed in $Y\times Y$. we have the result.

Lemma: $D$ is closed in $Y\times Y$ if and only if $Y$ is Hausdorff.
Proof:
The basic open sets in $Y\times Y$ are of the form $U\times V$, when $U$ and $V$ are open. $U\times V$ intersects $D$ if and only if $U$ and $V$ intersect.
If $Y$ is Hausdorff, we can find disjoint open sets $U,V$ containing $p,q$ respectively in $Y$ for any ordered pair $(p,q)$ of distinct points. The union of these $U\times V$ is $Y\times Y\setminus D$, so $D$ is closed.
Conversely, if $D$ is closed, we can find a basic open set $U\times V\subset Y\times Y\setminus D$ containing $(p,q)$ for any $(p,q)\not\in D$. These $U$ and $V$ are disjoint open sets containing $p,q$ respectively, and $Y$ is Hausdorff.

It's pretty easy to construct counterexamples when $Y$ is not Hausdorff (as long as $X$ isn't discrete). This includes cases in which $X$ has a metric topology.

[Edit: Typo fixed]

My purpose was to show that $A$ can be both closed and open. So I gave example and counterexample. The case when $Y$ is vector space is the most easy for it. :)

jmerry, you're wrong, I'm afraid. As I said, I've proved that $A$ is closed when $X$ is metric(or even first-countable) regardless of the topology of $Y$.You have said $A=F^{-1}(D)$(a typo... ) and if $D$ is closed, we have the result. What if $D$ is not closed?

All right- here's an example.

Let $X=\mathbb{R}$ and $Y=\{0,1\}$ with the indiscrete topology. Every function into $Y$ is continuous.
Let $f(x)=0$ for all $x$, and let $g(x)=1$ when $x$ is rational, 0 when $x$ is irrational.

That's what can happen if $D$ is not closed. We might get lucky and have $F^{-1}(D)$ closed anyway, but it's likely we won't.

If you used distances to define continuity in the metric case, you probably implicitly assumed that $Y$ is metric (and Hausdorff) as well.


algebra
polynomial
superior algebra
superior algebra unsolved

Hello,Everyone,When I doing the problem [url]http://www.mathlinks.ro/viewtopic.php?t=153576[/url].I came out the following problem.

Problem:Let $f(x)\in\mathbb Q[x]$ be an irreducible polynomial of odd degree, and $\rho\in\mathbb C$ be a root of $f$, and $K=\mathbb Q(\rho)$.then for every $u_{1},u_{2},\cdots,u_{n}\in Q(\rho)$ is linear independent over $\mathbb{Q}$ if and only if $u^{2}_{1},u^{2}_{2},\cdots,u^{2}_{n}\in Q(\rho)$  is linear independent over $\mathbb{Q}$.

[quote="zhaobin"]
Problem:Let $f(x)\in\mathbb Q[x]$ be an irreducible polynomial of odd degree, and $\rho\in\mathbb C$ be a root of $f$, and $K=\mathbb Q(\rho)$.then for every $u_{1},u_{2},\cdots,u_{n}\in Q(\rho)$ is linear independent over $\mathbb{Q}$ if and only if $u^{2}_{1},u^{2}_{2},\cdots,u^{2}_{n}\in Q(\rho)$  is linear independent over $\mathbb{Q}$.[/quote]

I was playing around with some field extensions on paper and I think I have a way to make it stronger, but inevitably more complicated.

[b]Problem:[/b] Let $f(x) \in \mathbb{Q}[x]$ be an irreducible polynomial of Kummer degree... blah, blah, blah (same conclusion).

And by "Kummer" I mean the degree is not a power of two. For example, 6 or 10. Though it can still be even. (I just made that term up to make my conjecture more readable).

But sorry, I cant help. I can prove $n=2$  :rotfl:

EDIT: I am wrong.

I don't think you conjecture is right,consider $f(x)=x^{6}-2$ and $\sqrt{2},1$...

For the original problem, take $\sqrt[3]2,\sqrt[3]3,\sqrt[3]2+\sqrt[3]3$. owk

Erm... What is the polynomial for such a choice? :?

It does not matter, it has to be the minimal polynomial of a primitive element of an extension of $\mathbb Q$ having odd degree, "containing" $\sqrt[3]2,\sqrt[3]3$. Take $x^{9}-3\cdot x^{6}+165\cdot x^{3}-1$ if you like. owk

Thank you,owk.
but what about the converse?
If $u_{1},u_{2},\cdots,u_{n}\in Q(\rho)$ is linear independent over $\mathbb{Q}$,then $u^{2}_{1},u^{2}_{2},\cdots,u^{2}_{n}\in Q(\rho)$  is linear independent over $\mathbb{Q}$


Am atasat un material legat de inegalitati. El poate fi util atat celor care vor sa se initieze in acest domeniu, cat si celor avansati. Sper sa va placa si sa va fie de folos.

Anexez inca citeva materiale legate de inegalitati culese de pe net.
Mihai Miculita


calculus
integration
parameterization
induction
calculus computations

Help me to consider the solution is correct or wrong?

$ *\,\,I\left( y \right) \equal{} \int\limits_{0}^{a}{y.{{e}^{x.y}}.dx} \equal{} \int\limits_{0}^{a}{{{e}^{x.y}}.d\left( xy \right)} \equal{} \left[ {{e}^{xy}} \right]_{x \equal{} 0}^{x \equal{} a} \equal{} \left( {{e}^{ay}} \minus{} {{e}^{0}} \right)$

$ \Rightarrow {{I}^{'}}\left( y \right) \equal{} \int\limits_{0}^{a}{xy.{{e}^{x.y}}.dx} \equal{} a.{{e}^{ay}},\,\,\,\,\Rightarrow {{I}^{\left( n \right)}}\left( y \right) \equal{} \int\limits_{0}^{a}{y.{{x}^{n}}.{{e}^{x.y}}.dx} \equal{} {{a}^{n}}.{{e}^{ay}}$

$ \Rightarrow {{I}^{\left( n \right)}}\left( 1 \right) \equal{} \int\limits_{0}^{a}{{{x}^{n}}.{{e}^{x}}.dx} \equal{} {{a}^{n}}.{{e}^{a}}$

$ but\,\,{{I}^{''}}\left( 1 \right) \equal{} \int\limits_{0}^{a}{{{x}^{2}}.{{e}^{x}}.dx}\ne {{a}^{2}}.{{e}^{a}}$

where am I wrong?

You are totally mixing up differentiating and integrating with respect with $ x$ or $ y$ (see how your $ dx$ moves to $ dy$). Try it again paying attention to which is the variable (for differentiating and/or integrating), and which is the  constant for that operation!

hello, we have
$ I'(y)\equal{}ae^{ay}$
$ I{''}(y)\equal{}a^2e^{ay}$
and by induction
$ I^{(n)}(y)\equal{}a^ne^{ay}$.
Sonnhard.

The mistakes start when you write
[quote]$ {I}^{'}\left( y \right) \equal{} \int\limits_{0}^{a}xy.{{e}^{x.y}}.dx$[/quote]
As you differentiate with respect to $ y$, and $ {I}\left( y \right) \equal{} \int\limits_{0}^{a}y{{e}^{xy}}dx$, it follows that $ {I}^{'}\left( y \right) \equal{} \int\limits_{0}^{a}(e^{xy} \plus{} xy{{e}^{x.y}})dx$
(since $ \frac {d} {dy} ye^{xy} \equal{} \left ( \frac {d} {dy} y \right ) e^{xy} \plus{} y\left ( \frac {d} {dy} e^{xy} \right ) \equal{} e^{xy} \plus{} yxe^{xy}$).


linear algebra
matrix
limit
inequalities
vector
linear algebra unsolved

Let $ A \in \mathbb{R}^{n \times n}$ and $ \parallel{}\cdot\parallel{}$ be an arbitrary Matrixnorm. Let $ \rho(A)$ be the spectral radius of $ A$.
Show $ \rho(A) \equal{} \lim_{n \rightarrow \infty} \parallel{}A^n\parallel{}^{\frac {1}{n}}$ with $ \rho(A)$ being the spectral radius.

This is Gelfand's formula. To prove it, solve the following exercises by the order indicated:

1) Show that, $ \forall k \in \mathbb{N}$, for every $ A \in \mathbb{C} ^{n \times n}$, and for every matrix norm $ \parallel{} \cdot \parallel{}$ we have $ \rho(A) \leq \parallel{}A^k\parallel{}^{1/k}$.

2) Let $ A \in \mathbb{C} ^{n \times n}$. For each $ \delta > 0$ define $ B_{\delta} \equal{} \frac {1}{\rho(A) \plus{} \delta} A$. Show that $ \lim_{k \to \infty} B_{\delta}^k \equal{} O_{n \times n}$.

3) Prove Gelfand's formula.

Thx for your help!

1) Let $ (v, \lambda)$ be an eigenvector-eigenvalue pair (obviously $ v \neq 0$) for the matrix $ A \in \mathbb{C}^{n \times n}$
$ |\lambda|^k \parallel{}v\parallel{} \equal{} \parallel{}\lambda^k v\parallel{} \equal{} \parallel{}A^k v\parallel{} \leq \parallel{}A^k\parallel{} \cdot \parallel{}v\parallel{}$
$ \Leftrightarrow |\lambda|^k \leq \parallel{}A^k\parallel{}$
$ \Leftrightarrow \rho(A) \leq \parallel{}A^k\parallel{}^{\frac{1}{k}}$

2) I'm not sure what $ O_{n \times n}$ means, but I tried to prove it this way:
$ B_{\delta} \equal{} \frac{1}{\rho(A) \plus{} \delta}A$, then $ \rho(B_{\delta}) \equal{} \frac{\rho(A)}{\rho(A)\plus{}\delta} < 1$
so it is $ \lim_{k \rightarrow \infty} B_{\delta}^k \equal{} 0$.
This means, a natural Number $ N \in \mathbb{N}$ exists such that $ \forall k \geq N \Rightarrow \parallel{}B^k\parallel{} < 1$, which means:
$ \parallel{}A^k\parallel{} < (\rho(A) \plus{} \delta)^k$ $ \Leftrightarrow$ $ \parallel{}A^k\parallel{}^{\frac{1}{k}} < (\rho(A) \plus{} \delta)$.

3) Combining 1) and 2) we get: $ \rho(A) \leq \parallel{}A^k\parallel{}^{\frac{1}{k}} < (\rho(A) \plus{} \delta)$, which by definition is:

$ \lim_{k \rightarrow \infty} \parallel{}A^k\parallel{}^{\frac{1}{k}} \equal{} \rho(A)$

[quote]$ \parallel A^{k}v\parallel \leq \parallel A^{k}\parallel \cdot \parallel v\parallel$[/quote]

From where does that inequality come from? In the first place, you are using the same notation for both vector norms and matrix norms. Second, that is only true for compatible (vector and matrix) norms. You may want to try another approach.

[quote]I'm not sure what $ O_{n \times n}$ means[/quote]

It's the square zero matrix of order n.

Sorry, i did forget to mention that we're talking about consistent norms. These norms are sub-multiplicative norms, this way i can use this inequality.

Thx Carcul

But we don't have to restrict ourselves to such norms, as Gelfand's formula is valid for [i]any[/i] matrix norm. Why don't you try to produce a more general proof to 1)?  :)

:) you're right, well, let me try.

I can do it the same way 2) was shown:

Let's consider the matrix $ C_{\delta} \equal{} \frac{1}{\rho(A) \minus{} \delta}A$, then $ \rho(C_\delta) \equal{} \frac{\rho(A)}{\rho(A) \minus{} \delta} > 1$.
This means a natural number $ N_1 \in \mathbb{N}$ exists such that $ \forall k \geq N_1 \Rightarrow \Vert C^k \Vert > 1$, this means
$ \forall k \geq N_1 \Rightarrow \Vert A^k \Vert > (\rho(A) \minus{} \delta)^k$ or $ \forall k \geq N_1 {\Rightarrow \Vert A^k \Vert}^{\frac{1}{k}} > (\rho(A) \minus{} \delta)$.
But this way I showed $ \rho(A) \minus{} \delta < {\Vert A^k\Vert}^{\frac{1}{k}}$ and not $ \rho(A) \leq {\Vert A^k\Vert}^{\frac{1}{k}}$

Hint: start by showing that, for any matrix norm $ \parallel{} \cdot \parallel{}$ and $ A \in \mathbb{C}^{n \times n}$, we have $ \rho(A) \leq \parallel{}A\parallel{}$; the conclusion then follows from the fact that $ \rho(A^k) \leq \parallel{}A^k\parallel{}$ and $ \rho(A^k) \equal{} [\rho(A)]^k$.

Next try:

$ \lambda$ is eigenvalue for $ A$.

Let $ x \neq 0$ so it is $ Ax \equal{} \lambda x$. $ \Leftrightarrow$ $ \Vert A \Vert \geq \frac{\Vert Ax \Vert}{\Vert x \Vert} \equal{} \vert \lambda \vert$ $ \Rightarrow$ $ \Vert A \Vert \geq \rho(A)$.
So it is $ \rho(A^k) \equal{} \left[\rho(A)\right]^k \leq \Vert A^k\Vert$.

Your are again restricting yourself to consistent norms. Another hint: for $ Ax \equal{} \lambda x$ define the matrix $ A_x \equal{} [x \,\, 0\,\, ... \,\, 0]$ (first column is vector $ x$ and the remaining columns are zero).

So I use $ A_x$ instead of $ x$. This way I'm just using matrices and no vector. The proof is the same i posted before just with $ A_x$ instead of $ x$, right?

Thx Joe

Yes, it's similar. More exactly:

Let $ \lambda$ be an an eigenvalue of $ A$ such that $ |\lambda| \equal{} \rho(A)$, and let $ x$ be a corresponding eigenvector. Then, $ Ax \equal{} \lambda x$, and so $ AA_x \equal{} \lambda A_x$. Therefore,

$ \parallel{}\lambda A_x\parallel{} \equal{} |\lambda| \parallel{}A_x\parallel{} \equal{} \parallel{}AA_x\parallel{} \leq \parallel{}A\parallel{} \parallel{}A_x\parallel{}$  (1).

Now, since $ A_x \neq O_{n \times n}$, we have $ \parallel{}A_x\parallel{} \neq 0$, and so from (1) we get

$ \parallel{}A\parallel{} \geq |\lambda| \equal{} \rho(A)$. $ \Diamond$

The norms on $ \mathcal{M}_n$ are equivalent. Then it is sufficient to prove the result using a particular norm.


Petya and Vasya are given equal sets of $ N$ weights, in which the masses of any two weights are in ratio at most $ 1.25$. Petya succeeded to divide his set into $ 10$ groups of equal masses, while Vasya succeeded to divide his set into $ 11$ groups of equal masses. Find the smallest possible $ N$.

I guess the answer is $ 50$.
Solution:
Let $ W \equal{} \{a_1\leq a_2\leq\dots\leq a_N\}$ be set of masses of our weights.WLOG suppose $ a_1 \equal{} 4$,it follows that $ a_N\leq 5$.
Suppose that $ 41\leq N\leq 49$,then there are $ 10$ groups,name them $ A_1,A_2,\dots,A_{10}\in W$,of equal masses.
Let $ S(A_i)$ means sum of masses of weights from $ A_i$.
If there exist $ |A_i|\leq 3$,then $ S(A_i)\leq 15$,but there exist $ A_j$,such that $ |A_j|\geq 5$,hence $ S(A_j)\geq 20$.Contradiction.
Now suppose that there exist $ |A_i|\geq 6$,then $ S(A_i)\geq 24$.It follows that $ |A_j|\geq 5$,for all $ 1\leq j\leq 10$,but then $ N\geq 51$.Contradiction.
Hence there are only two types of $ A_i$,with $ 5$ or $ 4$ weights.
Obviously if there exist $ |A_i| \equal{} 4$,then $ A_i \equal{} \{5,5,5,5\}$,it follows that if $ |A_j| \equal{} 5$,then $ A_j \equal{} \{4,4,4,4,4\}$.Hence there are only masses  equal to $ 4$ or $ 5$.
Suppose that there are $ 4k$ of number $ 5$ and $ 5l$ number $ 4$.We conclude that $ 20|N$,contradiction.
If there exist $ |A_i| \equal{} 3$,then $ |A_j| \equal{} 3$,for all $ 1\leq j\leq 10$,hence $ N \equal{} 30$.Similarly we conclude that $ N$ can be equal only two $ 20,30,40$,and they are consists from groups with number of elements equal to $ \frac {N}{10}$.
Let $ N \equal{} 40$,and now let $ B_1,B_2\dots,B_{11}$ be eleven groups with equal sum of masses.
Then there exist $ B_i$ and $ B_j$,with $ |B_i|\leq 3$ and $ |B_j|\geq 4$,but then $ S(B_i)\leq 15$ and $ S(B_j)\geq 16$,contradiction.
Similarly we can obtain contradictions for cases $ N \equal{} 20$ and $ N \equal{} 30$.
The last step is to show example for $ N \equal{} 50$:
Let $ W \equal{} \{4,4,\dots,4,5,\dots,5\}$,consists only by $ 4$ and $ 5$,and $ 4$ repeats $ 30$ times,$ 5$ repeats $ 20$ times.
Also let's take $ A_i \equal{} \{4,4,4,5,5\}$,for all $ 1\leq i\leq 10$
And $ B_j \equal{} \{4,4,4,4,4\}$,for $ 1\leq i\leq 6$, and $ B_j \equal{} \{5,5,5,5\}$ for $ 7\leq j\leq 11$.
We are done.

This is pretty equivalent I believe, but worded differently.

[hide="Solution with v_Enhance"]
We claim the answer is $50$.

$50$ can be achieved as above with a set of 30 elements weighing 4, 20 weighing 5.

We show nothing less than $50$ works.

Consider Vasya's groupings:

Obviously $[0,10]$ fail.

Similarly, $[12,21]$, $[23,32]$ and $[34,43]$ fail (since the masses of the sets with less elements are too small compared to the sets with one more element).

$11,22,33$ all similarly fail when you consider Petya's 10-grouping.

For $[44,49]$, let $N=40+n$.

Then considering the 10-grouping, we necessarily have $n$ 5-sets and $10-n$ 4-sets (otherwise its obviously impossible).

Since for some $a$, $5a \leq \text{mass of 5-set} = \text{mass of 4-set} \leq 5a$, and we have equality only when each member of a 5-set weighs $a$ and each member of a 4-set weighs $1.25a$, there must be $5n$ weights weighing $a$ and $40-4n$ weighing $1.25a$. But then the total mass is $200a$, and each set of the 11-grouping is composed of weights weighing $a$ and $1.25a$, which cannot be combined to form a total of $\frac{200a}{11}$, so $[44,49]$ is impossible as well.

Thus, the minimum $N$ is 50.[/hide]

The answer is $50$.Now,let the mass of the smallest weight be $4$.Now,example for $50$ is just take $20$ weights of mass $5$ and $30$ weights of mass $4$.The grouping here is obvious(for $10$,just take groups like this:${4,4,4,5,5}$ and $10$ of them and for $11$ just take $5$ groups like ${5,5,5,5}$ and $6$ groups like ${4,4,4,4,4})$.Now,lets prove that it is impossible for n less than $50$.If $n$ is less than $44$ and not divisible with $11$ we will have in partition of $11$ groups that this is impossible since we have $2$ groups whose number of elements differ by $1$.Also,if it is divisible with $11$ than it is not with $10$ so similary we get a contradiction.Now,if it is $44$ in partition of $10$ groups we conclude that the weights are only $4$ or $5$ and then easily get a contradiction(this is cinda obvious so I won't post it) and if is bigger than $44$ and less than $50$ we get that the sum of elements in groups in both groups are equal so contradiction,so we are done.


Can anyone help with determining the structures of A - K?

[img]
http://classes.yale.edu/chem225/probsets/PS08spr/PS8-S08/PS8-S08-str1.gif [/img]

The first clue I used is that ozonolysis of 2 butene gives one product K, so the 2 butene must be the trans isomer. B/c A is in the center and reacts with Lindlar catalyst, it must be a alkyne. However, I think A must be a terminal alkyne b/c It reacts with NaNH2 to form I. However, if A is a terminal alkyne, then it must be propyne, and B must be propane. In which case the heat of combustion would not equal 1317 kcal/mol. I first had it where A was 2-octyne, but then 2 octyne is an internal alkyne and does not react with NaNH2.

Can anyone help. Something is not working out here.

I'm getting the same discrepancy as you between n-octane and propane...

maybe there's an error?[/hide]

it is 2 octyne. i dunno how you deduced that 2 butene is the trans isomer as even the cis form will give the same products(in this case product). and A need not be a terminal alkyne. even internal alkynes react with nanh2/ at high trmp(in paraffin, i think) to give a terminal alkyne. all the other data matches with A being 2 octyne. check this out for heat of combustion-
http://home.fuse.net/clymer/rq/index.html


In 1983, Norman Johnson set a record by slicing each inch of a 12-inch cucumber into 22 pieces. It took him 13.4 seconds. When Norman finished, into how many pieces was the cucumber divided?

Since it is $ 12$ inches, and each was divided into $ 22$ pieces, we have $ 12\times{22} \equal{} \boxed{264}$.


What is the area of a triangle with altitudes 12, 15, and 20?
I know the answer.

[quote="vibgyor"][hide]These altitudes work as a right triangle. 15-20-25.[/hide][/quote]
How do you know?

[quote="vibgyor"][hide]These altitudes work as a right triangle. 15-20-25.[/hide][/quote]
I already knew it was a right triangle ;). How would you solve it if it wasn't necessarily one?

[quote="Mathalete"][quote="inom"]These altitudes work as a right triangle. 15-20-25.
Because
$15^{2}+20^{2}=25^{2}$[/quote]

That doesn't neccesarily make it a right triangle.[/quote]
Um yes. It does...
EDIT: 15, 20, 25 are the sides...

oh.  if they are the sides then its trivial [hide]$150$[/hide]

[quote="ProtestanT"][hide="solution"]
this is just aan arithmetic problem. we can write 6 equations to solve for 6 variables
Let $A$=the area
Let $a,b,c,d,e,f$ = sections of sides (2 per side), with $(a+b)*12=A$, $(c+d)*20=A$, $(e+f)*15=A$
$a$ is adjacent to $e+f$, $b$ is adjacent to $c+d$
$c$ is adjacent to $e+f$, $d$ is adjacent to $a+b$
$e$ is adjacent to $c+d$, $f$ is adjacent to $a+b$
Using the pythagorean theorem, these six equations are true.
$a^{2}+12^{2}=(f+e)^{2}$
$b^{2}+12^{2}=(c+d)^{2}$
$c^{2}+20^{2}=(e+f)^{2}$
$d^{2}+20^{2}=(a+b)^{2}$
$e^{2}+15^{2}=(c+d)^{2}$
$f^{2}+15^{2}=(a+b)^{2}$
Now someone else do all this hard solving stuff that i'm too lazy to do, or come up with an easier way.
[/hide][/quote]
I got that too I think it can be simplified further b/c altitudes are geometric means of two sections of the side (i.e ab=144)
im pretty sure thats true but I'm not positive

anybody have a solution for the altitudes?

[quote="bpms"][quote="daermon"]oh.  if they are the sides then its trivial [hide]$150$[/hide][/quote]
Yes, but the problem doesn't tell you that ;). How would you solve this only knowing that the altitudes are 12, 15, and 20?[/quote]
[hide]Let $a,b,$ and $c$ be the side lengths of the triangle ($\triangle ABC$). We have \begin{eqnarray}12a=15b=20c.\end{eqnarray} Now there are two things you could do after this:
We have \[12=\frac{[ABC]}{a}=\frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{16a},\] and then substitute values for $b$ and $c$ in terms of $a$ and then solve the equation.

However, I like this (cheap  :P ) method better:
Choose any $a,b,c$ such that $(1)$ is satisfied. For example, $3,4,5$ works fine. Now calculate the height of the triangle to the side of length $3$. It turns out to be $4$. However, the actual triangle we are looking for has a height of $20$ to the shortest side. This means our numbers $3,4,5$ are too small by a ratio of $\frac15$. So just multiply them by $5$ to get the actual side lengths, $15,20,$ and $25$.  :D [/hide]


AMC
USA(J)MO
USAMO
MATHCOUNTS
greatest common divisor
geometry
circumcircle

dude. We're like cursed to tie each other on AMC stuff.

[quote="uldivad9"]I got a 10.
Wait a minute, what happened to my at least two problems?[/quote]

I told you that you overestimated your score. You're not from MA, shoo.

I got 13, as predicted. I am quite good at predicting my score, evidently.

I hope you know that the highest possible score is 42.

[quote="uldivad9"]I got a 10.
Wait a minute, what happened to my at least two problems?[/quote]

Hey david, I talked to you at nats!  :D
You beat me at everything.

[quote="Phelpedo"]Do you know what David got?[/quote]

David got 13.

i get 12  :P

beast all of y'alls middle school scores

[quote="uldivad9"]I got a 10.
Wait a minute, what happened to my at least two problems?[/quote]

agh. i hate you.

I think I got like 4/1/0/4/0/0. which would suck.  I forgot to state that $ GCF(\mbox{odd number}, 2^x\plus{}\mbox{odd number})\equal{}GCF(2^x,\mbox{odd number})\equal{}1$ I think...

Just said that since the difference is a power of two and both are odd, they're relatively prime. -_-.  Something like that. Ugh.

T.T screwed up

There's no way you got two 4's. You most likely got 7/0/0/2/0/0.

I heard they gave out a lot of 3's on #4. Maybe 4/2/0/3/0/0?

I got a 9... so I'm thinking either 7/0/0/2/0/0 or 7/0/1/1/0/0. Probably the former.

you have no way of hearing that. it was far more likely 5/0/0/5/0/0

I am pretty sure the graders do not release individual results. And as Carl said, getting a 4 or 3 is very unlikely...

[quote="hello.world"][quote="calc rulz"]How did Lexington do overall?[/quote]

 :whistling:

personally i failed it. not too sure about everyone else's scores, so i'll leave it at that.[/quote]

Hopefully everyone is ok with me posting this.

me=13
noah=11
sway=10 (i hear they lowered the cgmo cutoff...but i haven't talked to her yet)
stephen=8
ted=6

[quote="Fanatic"][quote]
you have no way of hearing that. [/quote]

Actually I do in fact have ears.[/quote]

Unless you heard this from a grader, it's most likely completely useless information you're getting. However, if he did indeed get a 3 on #4, then the score distribution almost definitely would have been 6/0/0/3/0/0.

I'm pretty sure I made nontrivial progress on number 2 though... So that leaves 8 points left.  Maybe 6 and 2?  But I'm also sure that my proof for #4 was pretty good, probably a few minor holes.  So maybe 5 and 3?

I doubt what you did on #2 constituted nontrivial progress. It seems like the type of problem that making a legitimately helpful first step is quite difficult. What exactly did you do?

6 and 1 and 2 or 7 and 2 are far more likely that 5 and 1 and 3 ;)

Circumcenter of the triangle is G.  AG is a radius of the larger circle and a diameter of the circumcircle of APN (something like that).  With some difficulty, you can prove that FGBC are cyclic, and FG is parallel to BC and in turn parallel to PN.  Trying to think what else because I don't have my actual written solutions.

So I guess I got 16, David got 13, and the other people probably don't want their scores disclosed... Well, we have our first (albeit Red) Mopper in quite some time.

At least six (!) from MA have been confirmed to have made MOP.

Kid from RL which is not perfect628 (Red)
Me (Red, 13)
David (Red, 13)
Elizabeth (CGMO, 8)
Sway (CGMO, 10)-I actually haven't talked to her because she's at Physics Camp at the moment but I assume she didn't turn down the invitation
Sergei (Blue, unless he opted not to go as he was on the HM list)

Phelpedo, did the other freshman from Andover make it? Also I recall Hamster saying Vlad got 3 problems, but I don't have any idea whether or not he made blue.

[quote="evil person"]Kid from RL which is not perfect628 (Red) [/quote]
Yeah yeah yeah...I still don't see how I failed so badly...

Andrew got a 10 btw.

[quote="CatalystOfNostalgia"]FOBC is indeed cyclic (if you actually proved this, it may get a point), but I don't think that FO || BC. Another key step is proving that <AFC=<AFB but I forget whether or not you need that to get that FOBC is cyclic.[/quote]

Ya I actually proved FOBC is cylic. Parallel thing was accident, not true.  I didn't write that on my proof either.

[quote="CatalystOfNostalgia"]At least six (!) from MA have been confirmed to have made MOP.

[b]Kid from RL which is not perfect628 (Red)[/b]
Me (Red, 13)
David (Red, 13)
Elizabeth (CGMO, 8)
Sway (CGMO, 10)-I actually haven't talked to her because she's at Physics Camp at the moment but I assume she didn't turn down the invitation
Sergei (Blue, unless he opted not to go as he was on the HM list)

Phelpedo, did the other freshman from Andover make it? Also I recall Hamster saying Vlad got 3 problems, but I don't have any idea whether or not he made blue.[/quote]

Wow. I feel honored to be identified like this.  :|


geometry
trigonometry
circumcircle
trig identities
Law of Sines
perpendicular bisector

Let AD be a median of triangle ABC. Let point E lie on AD (extended if necessary) such that CE is perpendicular to AD. Suppose that angle ACE equals angle ABC.

Prove that either AB =AC or angle BAC is a right angle.

[b]A metrical solution.[/b]

I suppose w.l.o.g. $E\in (AD)$, i.e. $C>B$.
Thus, $AD=b\sin B+\frac a2\sin (C-B)\Longleftrightarrow$
$2m_a=2b\sin B+(a\cos B)\sin C-(a\sin B)\cos C=$
$2b\sin B+(c-b\cos A)\sin C-(b\sin A)\cos C=$
$2b\sin B+c\sin C-b(\sin C\cos A+\cos C\sin A)=$
$2b\sin B+c\sin C-b\sin B=b\sin B+c\sin C\Longleftrightarrow$
$2m_a=b\sin B+c\sin C\Longleftrightarrow \boxed {\ 4Rm_a=b^2+c^2\ }\ \ (1)\ .$

[u]Remark.[/u] $2m_a=b\sin \widehat {CAD}+c\sin \widehat {BAD}\ \ .$

From $(1)$ and the well-known relation $\boxed {\ (b^2-c^2)^2+16S^2=4a^2m^2_a\ }$ results :
$4R^2(b^2-c^2)^2+64R^2S^2=a^2(b^2+c^2)^2\Longleftrightarrow$
$4R^2(b^2-c^2)^2+4a^2b^2c^2=a^2(b^2+c^2)^2\Longleftrightarrow$
$4R^2(b^2-c^2)^2=a^2(b^2-c^2)^2\Longleftrightarrow$
$b=c\ \vee\ a=2R\Longleftrightarrow b=c\ \vee\ A=90^{\circ}\ .$

Therefore, $\boxed {\ 4Rm_a=b^2+c^2\Longleftrightarrow b=c\ \vee\ A=90^{\circ}\ }\ .$

Another sort of solution...

Let $\angle ABC = \angle ACE = \theta$ and $\angle BCE = \phi$.

We find that $\angle BAD = 90+\phi-\theta$ and $\angle CAD = 90-\theta$. By Law of Sines on both $\triangle ABD$ and $\triangle ACD$, we obtain

$\frac{\sin{\theta}}{\sin{(90+\phi-\theta)}} = \frac{\sin{(\theta-\phi)}}{\sin{(90-\theta)}}$, or

$\sin{\theta}\cos{\theta} = \sin{(\theta-\phi)}\cos{(\theta-\phi)}$. This becomes $\sin{2\theta} = \sin{(2\theta-2\phi)}$.

From this we either have $2\theta = 2\theta-2\phi \Rightarrow \phi = 0 \Rightarrow \angle ABC = \angle ACB \Rightarrow AB = AC$, or

$180-2\theta = 2\theta-2\phi \Rightarrow 2\theta-\phi = 90$, but we also had

$\angle BAC = \angle BAD+\angle CAD = (90+\phi-\theta)+(90-\theta) = 180-(2\theta-\phi) = 90$.

Therefore one of the two must be true.

Let k be the circumcircle  of ABC, and CE intersect k in K. As <AKC = <ABC triangle AKC is equilateral so AE is a perpendicular bisector of CK so O - the circumcenter of AKC and ABC lies on AE. If O is different from D ABC is equilateral and if O = D  <A=90.


Each side of an equilateral triangle is divided into $ 1997$ equal parts, and each vertex of the triangle is joined by a segment to each of the division points on the opposite side. All the intersection points of these segments inside the triangle are marked. How many marked points are there?

hi
Count number of paralellograms and triangles and use it.. it is the key to solve


geometry
circumcircle
geometric transformation
reflection
perpendicular bisector
geometry proposed

Let $ ABCDE$ be a convex pentagon with radius of circumcircle is $ R$.
Suppose $ r_1;r_2;r_3;r_4$ are radius of incircle of triangle $ ABC;AED;ABD;AEC$,respectively.
Prove that if $ r_1\equal{}r_2;r_3\equal{}r_4$ then triangle $ ABC$ and triangle $ AED$ are similar.

Let $ (O)$ be the circumcircle of $ ABCDE.$ Let $ r_5, r_6$ be inradii of the $ \triangle BCD, \triangle CDE.$ The sum of triangle inradii does not depend on the triangulation of a cyclic polygon, hence

$ r_1 (\triangle CAB) \plus{} r_4 (\triangle CEA) \plus{} r_6 (\triangle CDE) \equal{}  r_2 (\triangle DEA) \plus{} r_3 (\triangle DAB) \plus{} r_5 (\triangle DBC)$

Since $ r_1 \plus{} r_4 \equal{} r_2 \plus{} r_3,$ it folows that $ r_5 \equal{} r_6.$ The $ \triangle DCB, \triangle CDE$ have a common side $ CD$ and equal inradii, hence $ I_5I_6 \parallel CD.$ They also have the common circumcircle $ (O),$ therefore $ OI_5 \equal{} OI_6.$ Since they are not identical, they are oppositely congruent $ \triangle BCD, \cong \triangle EDC$, one is reflection of the other in the perpendicular bisector of $ CD,$ and $ BC \equal{} ED.$ The $ \triangle ABC, \triangle DEA$ have equal  inradii $ r_1 \equal{} r_2$ and common circumcircle $ (O),$ therefore $ OI_1 \equal{} OI_2.$ Since they have equal sides $ BC \equal{} ED,$ they are congruent, and since they have the common vertex $ A,$ the congruence is opposite, $ \triangle ABC \cong \triangle AED,$ one is reflection of the other in the perpendicular bisector of $ CD.$


A prime number $ p$ is given. On the left side of the blackboard numbers $ 1,2,...,p\minus{}1$ are written, on the right just one $ 0$. We consider a sequence of $ p\minus{}1$ moves, each of them as follows: we choose a number from the left side, add it to all he numbers on the blackboard and erase it. Determine for which values of $ p$ one can choose such a sequence of moves so the number remaining on the blackboard after all of them will be divisible by $ p$.

I am really sorry by what follows.

Lemma 1: for every $ k > 1$ such that $ k|p - 1$, there is a number $ a \not\equiv 1 \text{ mod } p$ such that $ k = ord_pa$.

Proof: take $ a = g^{\frac {p - 1}{k}}$, where $ g$ is a primitive root modulo $ p$. Clearly $ a^k \equiv 1 \text{ mod }{p}$. Moreover, $ a^d \equiv 1 \text{ mod } p \Rightarrow g^{\frac {(p - 1)d}{k}} \equiv 1 \text{ mod } p$. Since $ g$ is a primitive root, ${ p - 1|\frac {(p - 1)d}{k}} \Rightarrow k|d \Rightarrow k \le d$, then $ k$ is the order.

Lemma 2: suppose $ k = ord_pa$. There is a partition $ \{1, 2, ..., p - 1\} = A_1 \bigcup A_2\bigcup...$ such that $ A_i = \{t, at, a^2t, ..., a^{k - 1}t\}$ (all numbers are seen $ \text{ mod } p$) for some $ t$, for each $ i$.

Proof: every number is in one of the sets $ S_t = \{t, at, a^2t, ..., a^{k - 1}t\}$, because $ t \in S_t$. Now we just need to prove that $ S_u \bigcap S_t$ is empty if $ S_u \not = S_t$ (in general, given a family of sets, if any two are equal or disjoint and their union is $ A$, we have a partition of $ A$). Indeed, if $ S_u \bigcap S_t \not = \emptyset$, we have $ a^mu \equiv a^nt$, therefore $ u \equiv a^{n - m}t$, therefore $ S_u = \{a^{n - m}t, a^{n - m + 1}t, ..., a^{n - m + k - 1}t\} = S_t$, q.e.d.

Suppose $ p = 5$ and the numbers $ a, b, c, d$ are erased in this order ($ \{a, b, c, d\} = \{1, 2, 3, 4\}$). First the numbers become

Left ......... Right
$ a + b$........ $ a$
$ a + c$
$ a + d$

Then they become

$ 2a + b + c$ ........$ 2a + b$
$ 2a + b + d$

Then they become

$ 4a + 2b + c + d$ ........$ 4a + 2b + c$

The remaining number is $ 8a + 4b + 2c + d$. This happens in general: the last number is of the form $ a_1 + 2a_2 + ... + 2^{p - 2}a_{p - 1}$, where $ \{a_1, a_2, ..., a_{p - 1}\} = {1, 2, ..., p - 1}$. So we need to prove that there is a permutation for which this ugly sum is a multiple of $ p$. We are going to show that this works for $ p > 3$. First, let us assume $ p$ is odd. Let $ d = ord_p2$. We have

$ a_1 + 2a_2 + 4a_3 + ... + 2^{p - 2}a_{p - 1} \equiv (a_1 + a_{d + 1} + ...) + 2(a_2 + a_{d + 2} + ...) + 4(a_3 + a_{d + 3} + ...) + ... + 2^{d - 1}(a_d + a_{2d} + ...) \text{ mod } p$

Let $ k = \frac {p - 1}{d}$ and let $ c$ be a number such that $ k = ord_pc$. Case $ d < p - 1$, in our lemma 2, we can take $ A_i = \{a_i, a_{d + i}, ...\}$ (observe that we would really have $ |A_i| = \frac {p - 1}{d} = k$, as needed). To conclude this sum is divisible by $ p$, we just need to show that $ \sum_{a \in A_i}a$ is, for a fixed $ i$. But $ t + ct + c^2t + ... + c^{k - 1}t = t\frac {c^k - 1}{c - 1}$. The numerator of the fraction is divisible by $ p$, while the denominator is not. This ends the proof.

Case $ d = p - 1$, we just take $ a_i \equiv 2^i \text{ mod } p$ (since $ 2$ is a primitive root modulo $ p$ ($ d = p - 1$), the $ a_i$'s are going to be distinct). We need to prove that

$ \sum_{i = 1}^{p - 1}a_i^2 \equiv \sum_{i = 1}^{p - 1}2^ia_i \equiv 0 \text{ mod } p$.

We can observe that $ \sum_{i = 1}^{p - 1}a_i^2 = \sum_{i = 1}^{p - 1} i^2 = \frac {(p - 1)p(2p - 1)}{6}$, that is $ 0 \text{ mod }p$ for $ p > 3$. Hence we can attain the goal for all $ p > 3$. It is easy to check the goal is impossible for $ p = 2$ or $ p = 3$.

Was I clear? Is something wrong? If somebody understood and think it is correct, please tell me... I am not very sure of it.


[quote]Uh...I'm confused by the notation.[/quote]

Another way to write it:

$b^{\frac{1}{a}} \times a^{\frac{1}{b}}=4\sqrt{2}^\sqrt{2}$


trigonometry
calculus
integration
calculus computations

Three possible strategies:

1. Remember (or re-derive) the identity which says that $ \cos A\cos B\equal{}$ a sum or difference of sines or cosines of something like $ A\plus{}B$ and $ A\minus{}B$ or maybe $ \frac{A\plus{}B}2$ and $ \frac{A\minus{}B}2.$ How good is your memory for identities like this?

2. Give up on remembering that formula and just write $ \cos 4x\equal{}\frac12(e^{4ix}\plus{}e^{\minus{}4ix})$ and $ \cos 5x\equal{}\frac12(e^{5ix}\plus{}e^{\minus{}5ix}),$ multiply it out, integrate, then see if you can put it back together as cosines or sines.

3. Integrate by parts, twice, and solve for algebraically for the integral. This would be particularly useful if this were a definite integral over some nice interval consisting of whole quadrants.

$ \int cos4x \ cos 5x \ dx$

using: $ cos A \ cos B \equal{} \frac {1}{2}(cos (A \plus{} B) \plus{} cos (A \minus{} B))$

$ \equal{} \int \frac {1}{2}(cos (9x) \plus{} cos (x))$

$ \equal{} \frac {1}{18}sin 9x\ \plus{} \ \frac {1}{2}sin x$

hello, you meant $ \cos(A)\cos(B)\equal{}\frac{1}{2}\left(\cos(A\minus{}B)\plus{}\cos(A\plus{}B)\right)$.
Sonnhard.


algebra
polynomial
function
real analysis
real analysis solved

Show that the polynomial $ P_n (x) = x^n + x - 1 $ has exactly one positive root $k_n$. Show that $k_n$ tends to $1$  as $n$ tends to $ \infty $ . (this part follows immediately)
The problem is how fast it tends? (unsolved part)
I suppose that we have to find some "elementary" function $f(n)$ such that exists fixed reals $a,b > 0$ and:
$ a f(n) \leq k_n \leq b f(n) $ ?

the easy part of the problem is trivial. The function has the derivate nx^(n-1)+1 so it is strictly increasing for positive x. Since f(0)<0 it has exactly 1 positive root.
From x^n+x=1 follows x^(n-1)(x+1)=1 The left side tends to x^n and the root to 1.

Misha

Is it something like
 $k_n=1-y_n$ with $\frac{ln n}{2n}\leq y_n \leq \frac{2ln n}{n}$

I did it in the same way:
$1- \frac{ln n}{n} \leq k_n \leq 1- \frac{ln n}{2n}$ ,
regarding the behaviour of $x^n+x$ , since:
$(1- \frac{ln n}{n})^n + 1- \frac{ln n}{n} \leq 1 \leq (1- \frac{ln n}{2n})^n + 1- \frac{ln n}{2n}$ ,
so $k_n = 1 - O(\frac{ln n}{n})$ .


function
floor function
ceiling function
combinatorics proposed
combinatorics

A network is a simple directed graph such that each edge $ e$ has two intger lower and upper capacities $ 0\leq c_l(e)\leq c_u(e)$. A circular flow on this graph is a function such that:
1) For each edge $ e$, $ c_l(e)\leq f(e)\leq c_u(e)$.
2) For each vertex $ v$: \[ \sum_{e\in v^\plus{}}f(e)\equal{}\sum_{e\in v^\minus{}}f(e)\]

a) Prove that this graph has a circular flow, if and only if for each partition $ X,Y$ of vertices of the network we have:
\[ \sum_{\begin{array}{c}{e\equal{}xy}\\{x\in X,y\in Y}\end{array}} c_l(e)\leq \sum_{\begin{array}{c}{e\equal{}yx}\\{y\in Y,x\in X}\end{array}} c_l(e)\]

b) Suppose that $ f$ is a circular flow in this network. Prove that there exists a circular flow $ g$ in this network such that  $ g(e)\equal{}\lfloor f(e)\rfloor$ or $ g(e)\equal{}\lceil f(e)\rceil$ for each edge $ e$.


algebra
polynomial
geometry
algebra open

Can anyone help me and give some references for the Newton polygons regarding irreductible polynomials? A net site or some basic theory about them would be just perfect :)

Ok, I see it's useless to say every single day that this is not the right area for  such questions. Anyway, why do I bother, after all?


Find all $m$ and $n$ such that $19^{19}=m^4+n^3$


AMC
AIME
AMC 10
AMC 12
Support
MATHCOUNTS
USA(J)MO

[url]http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2007-aa/07-AIMEstats.shtml[/url]

Stats are up!

So about a hundred and fifty people were above floor yet below index. I find that mildly comforting.

So they haven't separated AIME I and AIME II results as yet?  Did they even count them separately?  (I heard the AIME II was easier...)

Oooh, a real-life troll? Where does he live? I want to go observe what a troll looks like [from a careful distance ;)]. Is he a cave-troll or a mountain-troll?

[Yes, I shouldn't have fed the troll.]

It still makes for quite a few girls who pwned him. I am making the perhaps over-generous conversion of AMC10 - 20 pts. to get an approximate AMC12 score.

[quote="Ignite168"]Judging by the fact that females naturally suck at math, I think they mixed up the male and female columns here: http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2007-aa/2007--AIMEindex.pdf[/quote]

Thanks for the compliment :roll:

[quote="diophantient"][quote="solafidefarms"][quote="Ignite168"]I think they mixed up the male and female columns here: http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2007-aa/2007--AIMEindex.pdf[/quote]

Just because you believe that free speech is good doesn't mean you have to take advantage of it in insulting ways. That's why certain people believe that we have too much free speech.

But I think you are right that they mixed up the males and females. Given that the top female scored a 14 on the AIME and the top female score on the AMC 12 was a 145.5, no female could have gotten a 296 index.[/quote]
No one in existance could've gotten a 296 index[/quote]
even though ignite is most likely correct in this case
it is possible they incorrectly stated the top female scores(in the other direction)

[quote="xpmath"]@kyyuanmathcount
I don't think you took the Amc 10, but if you did, you probably could have gotten a perfect. It wasn't really hard[/quote]

Actually not...both of this year's AMC10 #25's were on the 12.  I got the 12A one wrong and left the 12B one blank ^_^.  Thus, I got killed by him.

Females will one day be better.  Some day.  <-- I'm allen, and I support this message.

Yeah but ur forgetting that u probably would have more time if u had taken the amc10 to look over the prob. 25s.
The amc10 problems are quicker to finish, so u would have gotten at least the 10a #25 right, given that kind of extra time.

[quote="worthawholebean"][quote="solafidefarms"][quote="Ignite168"]I think they mixed up the male and female columns here: http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2007-aa/2007--AIMEindex.pdf[/quote]

Just because you believe that free speech is good doesn't mean you have to take advantage of it in insulting ways. That's why certain people believe that we have too much free speech.[/quote]
It's Ignite, quite possibly one of the biggest trolls ever. He's probably the only person I know who is a troll in real life. Don't take the bait.[/quote]

I feel offended.

[quote="kyyuanmathcount"]Females will one day be better.  Some day.  <-- I'm allen, and I support this message.[/quote]

LOL

I think I should have taken 12A/10B instead of 10A/12B, I would most likely have ended up with 135-144 instead of 150-120

and I should have played the AIME II card, because evidently it's possible to use Mathcounts 3 days before as an excuse...I would have had 9 instead of 7...

and I actually would have qualified for index, by alot, because my index would be 225 instead of my lame 190.

They posted the state stats, but i think some of them might be sort of messed up.  For instance in my state, New Mexico, there were two freshman who qualified for USAMO, yet it shows the scores for freshman as one 7, one 3 and one 2, so like wtf?

(Also, i was pretty sure i got a 10 but it shows the highest score as an 8, though maybe I switched two bubbles or something)

[quote="Ignite168"][quote="samath"]WOOHOO

top 9th grade score on the AIME!  :D[/quote]

Didn't you say you got a 13? I didn't see a 13 or a 12?[/quote]

He thought he got a 12, but he got #11 wrong.

Well, nobody's answered, so I'll ask again...
Are they going to post stats for the AIME I and AIME II separately?  I'm curious to know if anyone actually got a 15 on the I.

[quote="calc rulz"]None of the males got a 250 index and arnav had 250 right? hmm I guess that tells us something about him...?[/quote]

yeah maybe it's something like I don't need to write down that I'm male on every single form

since my awesome masculinity is already so obvious for all the world to see

it is common knowledge by this point that I get all the girls

[quote="Ignite168"][quote="samath"]WOOHOO

top 9th grade score on the AIME!  :D[/quote]

Didn't you say you got a 13? I didn't see a 13 or a 12?[/quote]

I said I got a 12 at first before I saw the answers, but I got one wrong.
I got an 11.

[quote="davidyko"]lol @ 97.5/13.  :huh:[/quote]

I'd actually like to meet that person.

Did anyone else notice that if we believe the official stats, the total number of people taking either AIME dropped over 60% from last year? 22,764 in 2006, 8472 in 2007...that can't be right.

That can be right. Please note that the year that's out of line is 2006, not 2007. The typical number of AIME qualifiers is 10,000 to 15,000, and not all of them follow through and take the AIME. The number was much higher than usual in 2006.

This year seemed to have the lowest number of AIME takers since 2000 at the latest, though.

And the new scoring... last year I'm pretty sure the cutoff did not drop below 100 and probably would have been much higher if they looked at percentages only.

[quote="kyyuanmathcount"][quote="Zmastr"]Wow.  There were both a 139.5/0, and a 126/150[/quote]

Well, I think 139.5/0 just didn't take the AIME.  126/15 probably means that you are really good, just you had a semi-off day on the AMC12.

Haha, apparently, somebody smart enough to get 10 on the AIME wasn't smart enough to mark his/her gender.

 :clap2:  three perfect scores, all 12th graders.

Hmmm...an 8th grader tied for 11 on AIME full-scored the AMC10, thus owning me in index.[/quote]

I think that is Kevin Chen from Texas


How would you rate the difficulty of the  2005 AMC 12B compared to past AMC 12 tests?

not much different from last year's
but it was just that i was worse last year than this year ( i am still a sophomore)

same as usual: 1st 20 are doable, 21,22,23 are challenging, and OH MY GOSH, i didnt know what to do with 24 or 25. espeically 25 becuase it was just so messy. I just didnt know where to start.

Me thoughts:

Easier to qualify for AIME, harder to get a perfect score, because of the dastardly difficult 24 and 25.

I still do not fully understand those problems. I solved all the other problems I missed afterwards in about an extra half-hour, so my real problem is speed.

I got a 110.

I didn't think it was TOO bad. If I could read correctly, I could've got 143. But no, I'm illiterate. So I got 131. However, I have to say that those last two problems were like... not cool. At all.

How would you rate the chances of the cutoff score being 96 or below?

slim to none...the first 15 were rather easy so I think the cutoff will stay at 100

first 21 were easy, except for that number 18, i totally screwed up, i forgot to inscribe the circle into the trapezoid, so i kept getting 90.

not too bad....1-21 is not to bad,24,25 is hard,
which make a lot of people in 130's,which isn't  bad but still hard enough to get to AIME


Let $ F$ be a subset of a metric space $ (X,d)$. Then the following are equivalent:
  a) $ F$ is closed
  b) $ A\equal{}\{x: d(x,F)\equal{}0\}\subset F$

Use  the fact that $ x\in \overline{F}$ in the metric space $ (X,d)$ iff  there is a sequence $ (x_{n})\subset F$ st $ x_{n}\to x$. This is proved as following :

if $ x\in\overline{F}$ then for any $ n>0$ the ball $ B(x,\frac{1}{n})\cap F\neq\emptyset$ so we can choose $ x_{n}\in F$ and $ d(x,x_{n})\leq \frac{1}{n}$. Let $ n\in \mathbb{N}$ then we can find a sequence $ (x_{n})\in F$ st : $ d(x,x_{n})\leq \frac{1}{n}$

The other direction : use the fact that for any $ \epsilon >0$ , we can find $ N>0$ st : $ x_{n}\in B(x, \epsilon)$ for $ n>N$ so $ x\in\overline{F}$


Find x:

when x^x=10.

Where can I begin?

[hide="Solution"] $x \ln x = \ln 10$

Let $u = \ln x$.  Then

$ue^{u}= \ln 10$
$u = W(\ln 10)$

Where $W$ is the Lambert $W$-function:  http://mathworld.wolfram.com/LambertW-Function.html

Hence

$\ln x = W(\ln 10)$
$x = \boxed{ e^{ W(\ln 10) }}$ [/hide]

wow. Is this the simplest solution? Because I'm pretty sure this wasn't covered in class. =/

Well, I miswrote it slightly, but as far as a purely algebraic solution goes, yes.  I think.  Basically, $x^{x}$ is an ugly and difficult function.

yeah, I caught on to the log omission.

thanks for your help! The reason I was questioning was because my teacher's notorious for giving out difficult problems that can be solved using simple methods...I just usually don't see them and choose the ugly way to solve them.

I guess it's unavoidable in this case. =/


I am looking for a book that provides a nice transition into intermediate/ olympiad number theory for someone who has mastered the material equivalent to AoPS Intro number theory,  and has some knowledge on intermediate number theory, but not at all experienced with it. Is there such a book that thouroughly covers intermediate level work, while diving (deeply) into the realms of Olympiad level problems? I am working the AoPS Volume 2 Chapter on Number Theory now.

Niven and Zuckerman is the old classic introduction to NT; two good recent books are Rosen and Dence & Dence.

They are all "elementary" number theory books suitable for self study, because of their completeness; and for the most parts, they don't require any Calculus or abstract algebra.

A few Dover books (cheap) are also good (Andrews, Giola, Leveque), if you don't care to spend $100 for a book, but the Dover books are faster pace and require greater math preparations.

The textbooks mentioned here wouldn't include Olympiad-level problems, just the techniques you need for solving them. You'll have to find the problems elsewhere.