geometry
perimeter
trigonometry
geometry solved

ABCD = a quadrilateral.
|CD|=|DA|=s

(let vX denote the interior angle at X)
vA=40 DEG
vB=80
vC=80
vD=160

Show that also |BC|=s

Thank you alot !

Does |CD| mean the length of CD?

Also, does s = (a + b + c + d)/2?

I'm guessing |CD| is length, but s is just some constant (not the semiperimeter - a nondegenerate polygon can't have 1 side equal in length to the semiperimeter, much less 2 sides equal in length to the semiperimeter (make sure you see why)).

Yeah, if one side was the semiperimeter, it would have to be one half the perimeter of the shape, and that can only happen with a squished down quadrilateral like this:

[img]http://img.photobucket.com/albums/v259/drspyder/quadrilateral.gif[/img]

Probably, the reason why this problem has been unsolved for such a long time is that it was posted in the wrong section - the Olympiad Section is the only appropriate one. Anyway, before we solve the problem, let's restate it without the misleading notation s:

[color=blue][b]Problem.[/b] Let ABCD be a quadrilateral such that CD = DA, < DAB = 40°, < ABC = 80°, < BCD = 80° and < CDA = 160°. Prove that BC = CD = DA.[/color]

My [i]solution[/i] goes as follows:

We will work with non-directed angles.

Let W be a point on the ray CB such that < CDW = 50°. We will prove that W = B; in order to prove this, we will show that the lines CB, DW and AB are concurrent.

In fact, consider the triangle CDA. Since CD = DA, this triangle is isosceles, and since the angle at its apex equals < CDA = 160°, its base angle is

$\measuredangle DAC=\measuredangle ACD=\frac{180^{\circ}-\measuredangle CDA}{2}=\frac{180^{\circ}-160^{\circ}}{2}=\frac{20^{\circ}}{2}=10^{\circ}$.

From < DAB = 40° and < DAC = 10°, it follows that < BAC = < DAB - < DAC = 40° - 10° = 30°. From < BCD = 80° and < ACD = 10°, it follows that < ACB = < BCD - < ACD = 80° - 10° = 70°.

Finally, < CDW = 50° and < CDA = 160° lead to < WDA = < CDA - < CDW = 160° - 50° = 110°.

Now, by the trigonometric version of the Ceva theorem, applied to triangle CDA, the lines CB, DW and AB are concurrent if and only if the equation

$\frac{\sin\measuredangle CDW}{\sin\measuredangle WDA}\cdot\left(-\frac{\sin\measuredangle DAB}{\sin\measuredangle BAC}\right)\cdot\left(-\frac{\sin\measuredangle ACB}{\sin\measuredangle BCD}\right)=1$

holds (the minus signs appear here since the lines AB and CB divide the angles CAD and DCA of triangle CDA externally). But since < CDW = 50°, < WDA = 110°, < DAB = 40°, < BAC = 30°, < ACB = 70° and < BCD = 80°, we have

$\frac{\sin\measuredangle CDW}{\sin\measuredangle WDA}\cdot\left(-\frac{\sin\measuredangle DAB}{\sin\measuredangle BAC}\right)\cdot\left(-\frac{\sin\measuredangle ACB}{\sin\measuredangle BCD}\right)$
$=\frac{\sin 50^{\circ}}{\sin 110^{\circ}}\cdot\left(-\frac{\sin 40^{\circ}}{\sin 30^{\circ}}\right)\cdot\left(-\frac{\sin 70^{\circ}}{\sin 80^{\circ}}\right)=\frac{\sin 50^{\circ}}{\sin 110^{\circ}}\cdot\frac{\sin 40^{\circ}}{\sin 30^{\circ}}\cdot\frac{\sin 70^{\circ}}{\sin 80^{\circ}}$.

Now, since 110° = 180° - 70°, we have sin 110° = sin 70°; also, since 50° = 90° - 40°, we have sin 50° = cos 40°. Moreover, $\sin 80^{\circ}=\sin\left(2\cdot 40^{\circ}\right)=2\sin 40^{\circ}\cos 40^{\circ}$. Finally, it is well-known that $\sin 30^{\circ}=\frac12$. Thus,

$\frac{\sin\measuredangle CDW}{\sin\measuredangle WDA}\cdot\left(-\frac{\sin\measuredangle DAB}{\sin\measuredangle BAC}\right)\cdot\left(-\frac{\sin\measuredangle ACB}{\sin\measuredangle BCD}\right)$
$=\frac{\sin 50^{\circ}}{\sin 110^{\circ}}\cdot\frac{\sin 40^{\circ}}{\sin 30^{\circ}}\cdot\frac{\sin 70^{\circ}}{\sin 80^{\circ}}$
$=\frac{\cos 40^{\circ}}{\sin 70^{\circ}}\cdot\frac{\sin 40^{\circ}}{\left(\frac12\right)}\cdot\frac{\sin 70^{\circ}}{2\sin 40^{\circ}\cos 40^{\circ}}=1$.

Thus, the equation

$\frac{\sin\measuredangle CDW}{\sin\measuredangle WDA}\cdot\left(-\frac{\sin\measuredangle DAB}{\sin\measuredangle BAC}\right)\cdot\left(-\frac{\sin\measuredangle ACB}{\sin\measuredangle BCD}\right)=1$

is proven, and we can conclude that the lines CB, DW and AB are concurrent. In other words, the line DW passes through the point of intersection of the lines CB and AB. But the point of intersection of the lines CB and AB is clearly the point B, and thus, the line DW passes through the point B. In other words, the point W lies on the line DB. On the other hand, we also know that the point W lies on the line CB. Hence, the point W is the point of intersection of the lines DB and CB. In other words, W = B. Hence, the equation < CDW = 50° becomes < CDB = 50°. Combining this with < BCD = 80° and using the sum of angles in triangle BCD, we get

  < DBC = 180° - < CDB - < BCD = 180° - 50° - 80° = 50°,

so that < DBC = < CDB. Thus, the triangle BCD is isosceles, and we have BC = CD. Since we were given that CD = DA, we can thus conclude that BC = CD = DA. And the problem is solved.

  Darij

Perhaps an easier  elementry method:

Let K be the point on line AB such that  <KDA=100 degreesl so (<DKA = 40 degrees)
  and since <DAK= <DKA (=40 degrees) ===>    KD =  AD
Also  <CDK = 60 degrees, and CD=DK (Because CD=DA and DA = DK)
Hence  triangle  CDK is equilateral  and CD=CK and angle  <CKD = 60 degrees.
Hence  angle  <CKB = 80 degrees = <CBK
Hence CB=CK which is equal to  CD or DA. :)
  [hide="Curiously"]if   CD and AB meet at X,  AX is also equal to AD=DK=KC=BC !!! - a source of many similar  problems :)) (Also as Darji notices  <CBD = 50 degrees ; and <BCA = 70 degrees - All come out very nice!)  [/hide]

Thanks for the simple solution! Still I find it funny that this problem can be solved in the same way as two other adventitious angles problems: http://www.mathlinks.ro/Forum/viewtopic.php?p=169022 and http://www.mathlinks.ro/Forum/viewtopic.php?t=16296 ...

  Darij


Ring Theory
group theory
abstract algebra
calculus
derivative
number theory unsolved
number theory

Prove that :
 The equation :
 $ x^3+y^3=2z^3 $ has no possitive integers solution ($ x $ is different from $ y $ )

The main idea is to use an infinite discent.Try to find out that x,y,z are even.

I am not very sure about it. I will preffer doing something like: $(x+y)(x+yw)(x-yw)=2z^3$ and working in $Z+Zw$ where $w$ is such that $w^3=1$ and $w^2+w+1=0$ it is a UFD and i think analizing some cases, we may obtain the result. At least in this way, i showed up $x^3+y^3=z^3$ ahd no solution.

Pascual: Are you sure you obtain results this way? working in $\mathbb{Z} [i\sqrt{3}]$ will surely solve $x^3+y^3=3z^3$ (using for example that $(1-w)^2$ is associate of $3$ but I dont have idea for this problem. I'll gladly see the solution in $\mathbb{Z} [i\sqrt{3}]$

Megus I am going to post a solution I hope to find tonight tomorrow, jaja!

Hey Pascual, so what about your solution - can you post it ?

sorry megus, all my aproaches have failled in this one!

I'll try to solve this problem next week (after VMO05)  :lol:

Here is the nice and modern way to think about this equation.

First, it is clear that rational solutions are sufficient. Then, we can get rid of $z=0$ easily. Dehomogenize by dividing by $z^3$ to obtain the equation $x^3+y^3=2$. Via the regular maps $x\mapsto 24/(x+y)$ and $y\mapsto 72(y-x)/(y+x)$ we get the isomorphic equation (when I say isomorphic, I mean the variety described by one equation is isomorphic to the variety described by the other equation)
$y^2=x^3-4\cdot 432$.

This is an elliptic curve (and the fact that it is an elliptic curve comes from the fact that the original equation $x^3+y^3=2z^3$ had the solution $(1,1,1)$ to begin with) and in general tables (such as http://www.maths.nott.ac.uk/personal/jec/ftp/data/INDEX.html) this curve is written simply as $[0,0,0,0,-4\cdot 432]$. Moreover, this curve is isomorphic to the elliptic curve
$y^2=x^3-27$. This curve has conductor 36, has torsion subgroup isomorphic to $\mathbb Z/2\mathbb Z$ with the nontrivial element of $\mathbb Z/2\mathbb Z$ corresponding to the pair $(3,0)$.

General elliptic curve theory says that the rational solutions to the equation form a finitely generated group. We already know the subgroup of torsion points (i.e., points such that a multiple is equal to 0) so all we need is the rank of the curve. There are two way to do this: either we compute generators and bound the rank from below in this way, or if the rank is 0 or 1 we could use really hard theorems (Coates-Wiles or Kolyvagin) to bound the rank from above by computing derivatives of $L$-functions.

In this case, since the conductor is 36 (very small), Cremona (http://www.maths.nott.ac.uk/personal/jec/ftp/data/INDEX.html) has already computed the rank to be 0. So there are two rational solutions to the curve: the identity point and $(0,3)$. It is easy to translate back into the original case, where $(1,1,1)$ is the identity, and $(1,-1,0)$ is the other point.

For more information, see http://arxiv.org/PS_cache/math/pdf/0403/0403116.pdf

Andrei

[quote="ajorza"]

For more information, see http://arxiv.org/PS_cache/math/pdf/0403/0403116.pdf

Andrei[/quote]

But the link doesn't work,Can you give another? :)

for a more elementary solution, see chapter 2.14 of "Sierpinski-Elementary Theory of Numbers".(too long to copy it)


I'm probably not a good problem solver, but I think that for the harder questions on AHSME, it's not unusual to take 30 minutes. Some AHSME questions can take me an hour.

[quote="Archimedes27"]I must admit with shame that it takes me somtimes about half an hour to solve a AHSME problem.Maybe more. How long should a good problem solver take to solve a problem? What is a good time?[/quote]

It all depends on the problem. Some problems can take a lot of time to solve if you don't know one rule or shortcut or something.

[quote="JSRosen3"]If you're taking the AHSME for practice, then it's good to spend as much time as it takes to solve the problem.

However, on the actual AMC exam, if you see a problem that you think will take you an unreasonable amount of time, then you should skip it--2.5 free points![/quote]

Heh, my math teacher had all his students take the AMC 10 this year, and gave extra credit points on tests based on what we got on the AMC 10. One kid just did 3 problems and took a 73 and got 7 points extra credit on his next test.

[quote="yif man12"] Heh, my math teacher had all his students take the AMC 10 this year, and gave extra credit points on tests based on what we got on the AMC 10. One kid just did 3 problems and took a 73 and got 7 points extra credit on his next test.[/quote]

How many points will your teacher give you if you don't answer any questions at all  :?, free 62.5 points?

Honestly, it depends how well you are trying to do.  If you want to move on to AIME, that's one question.  If you want to move on to USAMO, that's another.  If you want a 150, you're probably overly driven, but that would require another level all together.  If you are just aiming to make AIME, my strong suggestion is that you figure out exactly how many questions you need, allowing for one or two wrong answers, in order to pass.  This turns out to be something like 16 or 17 questions answered or so.  This is a really worthwhile excercize.  Then just aim to answer that many questions.  Check over those questions.   And then only answer additional questions if you are supremely confident of your answer.  This (along with intelligent question picking -- that is, don't just assume number 15 is easier than number 18 because it comes first) will allow you to spend much more time on each problem, so brute speed becomes less important.

If you want to move on to the USAMO, the equation is greatly complicated by how well you think you can do on the AIME.

If you really want a 150, then you should be able to go through the entire contest twice in the time alotted.  That works out to about 1-2 minutes a question, on average.  Of course, that is bound to be slanted:  if you're spending 30 minutes apiece on questions 1-5, you're in much bigger trouble than if you're spending that much time on questions 20-25.  I would say that a really top-level AMCer should be able to do the middle five questions of an average-difficulty AMC in, say, 15 to 20 minutes including a thorough checking.  But you should bear in mind that it's not at all necessary to be a top-level AMCer in order to succeed at the AMC, AIME, USAMO, or life.


this is one of tiplers problems i don't know why whatever i do doesn't give the right answer...

A Helium balloon is used to lift a load of $110$  $N$. The weight of the balloon's skin is $50$  $N$ and the volume of the balloon when fully inflated is $32$ $m^{3}$ . The temperature of the air is $0^{ o}C$ and the atmospheric pressure is $1$  $atm$.The balloon is inflated with sufficient Helium gas so that the net upward force on the balloon and its load is $30$  $N$. Neglect changes of temperature with altitude.

[b]Find the number of moles of Helium gas in the balloon.[/b]

and find the max altitude that the balloon can go...

$V_{B}= 32\ \text m^{3}$ is the balloon volume, $W_{L}= 110\ \text N$ is the load weight, $W_{S}= 50\ \text N$ is the balloon skin weight, $W_{H}$ is the helium gas weight, $F_{L}= 30\ \text N$ is the balloon lift, $F_{B0}$ the buoyancy force at the ground level.

$F_{B0}-(W_{L}+W_{S}+W_{H}) = F_{L},\ \ \ W_{H}= F_{B}-(W_{L}+W_{S}+F_{L})$

By Archimedes law, $F_{B0}= g\rho_{A0}V_{B},$ where $g = 9.81\ \text{m/sec}^{2}$ the gravitational acceleration, $\rho_{A0}= \frac{M_{A}}{v}= \frac{p_{0}M_{A}}{RT}$ is the air density at the ground level, $M_{A}$ the air molar mass, $p_{0}= 1\ \text{atm}= 101.325\ \text{kPa}$ the air pressure, $T = 0^\circ\text{C}= 273.15^\circ\text{K}$ the air temperature, and $R = 8314.5\ \text{J}/(^\circ\text{K}\cdot \text{kmol})$ the gas constant. Neglecting water vapor in the air and assuming air composition as 78% of $\text N_{2},$ 21% of $\text O_{2},$ and 1% of $\text{Ar},$

$M_{A}= 0.78 \cdot 28+0.21 \cdot 32+0.01 \cdot 40 \doteq 29.0\ \text{kg/kmol}$

$\rho_{A0}= \frac{p_{0}M_{A}}{RT}\doteq 1.29\ \text{kg/m}^{3}$

$F_{B0}= g \rho_{A0}V_{B}= \frac{gp_{0}M_{A}V_{B}}{RT}\doteq 406\ \text N$

$W_{H}= F_{B0}-(W_{L}+W_{S}+F_{L}) \doteq 216\ \text N$

$n_{H}= \frac{m_{H}}{M_{H}}= \frac{W_{H}}{gM_{H}}\doteq 5.5\ \text{kmol}$

where $M_{H}= 4\ \text{kg/kmol}$ is the molar mass of $\text{He}.$ The hydrostatic air pressure decreases with increasing altitude h as

$\text dp =-g\rho_{A}(h)\ \text dh =-\frac{gpM_{A}}{RT}\ \text dh,\ \ \ \frac{\text dp}{p}=-\frac{gM_{A}}{RT}\ \text dh$

$\ln p =-\frac{gM_{A}}{RT}\ h+\text{const},\ \ \ p = p_{0}\exp\left(-\frac{gM_{A}}{RT}\ h\right)$

Consequently, the air density and the buoyancy force decrease as

$\rho_{A}(h) = \frac{p(h) M_{A}}{RT}= \rho_{A0}\exp\left(-\frac{gM_{A}}{RT}\ h\right)$

$F_{B}(h) = g \rho_{A}(h) V_{B}= F_{B0}\exp\left(-\frac{gM_{A}}{RT}\ h\right)$

At the maximum altitude the balloon can reach, the buoyancy force is balanced by the combined weights of the load, baloon skin, and the helium gas,

$F_{B}(h_{max}) = F_{B0}\exp\left(-\frac{gM_{A}}{RT}\ h_{max}\right) = W_{L}+W_{S}+W_{H}= 376\ \text N$

$h_{max}= \frac{RT}{gM_{A}}\ \ln \left(\frac{F_{B0}}{W_{L}+W_{S}+W_{H}}\right) \doteq 613\ \text m$

seems to be a nice solution thanx :) ...but i have a probelm here:

[quote="yetti"]Neglecting water vapor in the air and assuming air composition as 78% of $\text N_{2},$ 21% of $\text O_{2},$ and 1% of $\text{Ar},$

$M_{A}= 0.78 \cdot 28+0.21 \cdot 32+0.01 \cdot 40 \doteq 29.0\ \text{kg/kmol}$
[/quote]

how did you find the $M_{A}$ the air molar mass?!!!!!

One kmol of any gas (Avogadro number of molecules) takes the same volume $v = \frac{RT}{p}$ at the same temperature T and pressure p ($v \doteq 22. 4\ \text m^{3}$ at $T = 0^\circ\text C \doteq 273^\circ\text K$ and $p = 1\ \text{atm}\doteq 101\ \text{kPa}$). The relative abundance of oxygen, nitrogen, argon in air [b]by volume[/b] are 78%, 21%, 1%. From the periodic table, the atomic masses of the nitrogen, oxygen, and argon atoms are 14 AMU, 16 AMU, 40 AMU, respectively. (AMU is the atomic mass unit.) Oxygen and nitrogen form molecules $\text N_{2},\ \text O_{2},$ argon does not. The molecular masses of $\text N_{2},\ \text O_{2},\ \text{Ar}$ are then 28 AMU, 32 AMU, 40 AMU and their molar masses (Avogadro number of molecules) $28\ \text{kg/kmol},\ 32\ \text{kg/kmol},\ 40\ \text{kg/kmol}.$ Combine this with the relative abundance of nitrogen, oxygen, argon in air by volume.


What does it mean when a function is odd, even, or neither?

For $ y\equal{}f(x)$, if $ f(\minus{}x)\equal{}f(x)$, then $ f(x)$ is called even function and if $ f(\minus{}x)\equal{}\minus{}f(x)$, then called odd function.


limit logarithms calculus integration function quadratics trigonometry

today was my final exam of Calculus1, here are the problems, i solved all of them but i solved number $1$ very badly...

$1.$ Solve the equation:

$z^{4}+7(1+i)z^{2}+25i=0$

$2.$ Find the following limit:

$\lim_{x\rightarrow0}\left(\frac{1}{\ln\left(x+\sqrt{1+x^{2}}\right)}-\frac{1}{\ln\left(1+x\right)}\right)=?$

$3.$ For which real numbers $\alpha$ and $\beta$ the following integral is convergent:

$\int^{1}_{0}\frac{(\ln x)^{2}}{x^{\alpha}(1-x)^{\beta}}dx$

$4.$  Let ${a_{n}=\frac{1\times3\times5\times\cdots\times(2n-1)}{2\times4\times6\times\cdots\times(2n)}}$,

$a)$ For which real numbers $s$ the series $\sum^{\infty}_{n=1}a^{s}_{n}$ is convergent?

$b)$ Find the interval of convergence of the series $\sum^{\infty}_{n=1}a^{2}_{n}\left(2x^{2}-1\right)^{n}$.

$5.$Find the Taylor series of function $f(x)=\frac{2-x}{x^{2}-x+1}$ in $x_{0}=1$ and find its radius of convergence.

Wow, you had a challenging Calc I class...

The way I'd do $1$, is to say let $u=z^{2}$, then you have a quadratic and can apply the quadratic formula: $u^{2}+7(1+i)u+25i=0$, so $u=\frac{-7(1+i)\pm\sqrt{49(2i)-4(25i)}}{2}= \frac{-7(1+i)\pm\sqrt{-2i}}{2}$. So your roots are $z=\pm\sqrt{u}= \pm\sqrt{\frac{-7(1+i)\pm\sqrt{-2i}}{2}}$. I'm sure this can be simplified, but that would be my process.

this was my solution for $1$:(thats why i said i solved it badly)

$u=\frac{-7(1+i)\pm\sqrt{-2i}}{2}$
and as $\sqrt{i}=\sqrt{e^{i\frac{\pi}{2}}}=e^{i\frac{\pi}{4}}$ or $e^{i\frac{5\pi}{4}}$

I just used $\sqrt{i}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$

so
$u=\frac{-7\sqrt{2}\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)\pm\sqrt{2}i\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)}{2}$

$=\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)\left(\frac{-7\sqrt{2}}{2}\pm\frac{\sqrt{2}}{2}i\right)$
$=\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)\times5\left(\frac{-7\sqrt{2}}{10}\pm\frac{\sqrt{2}}{10}i\right)$

let $\theta=\arctan\frac{\frac{\sqrt{2}}{10}}{\frac{-7\sqrt{2}}{10}}$

so
$u=5e^{i\frac{\pi}{4}}e^{\pm i\theta}=5e^{i\left(\frac{\pi}{4}\pm\theta\right)}$

and so $z=\sqrt{5}\left(\cos\left(\frac{\frac{\pi}{4}\pm\theta+2k\pi}{2}\right)+i\sin\left(\frac{\frac{\pi}{4}\pm\theta+2k\pi}{2}\right)\right)\ \ \ \ \ \ \ \ \ (k=0,1)$

but after the exam when discussing with friends I found out how easy the problem was beacuse you can assume that $u=x+iy$ and the problem will be solved easily....

the other problems are easy.....

the answers i got for each of them was:

[hide]2. $L=\frac{-1}{2}$

3.$\alpha<1$ and $\beta<3$

4.$a)\ \ \ \ \ s>2$,     $b)\ \ \ \ \ (-1,1)$

5.$\frac{2-x}{x^{2}-x+1}=\sum^{\infty}_{n=0}\left(x-1\right)^{3n+2}-\sum^{\infty}_{n=0}2\left(x-1\right)^{3n+1}+\sum^{\infty}_{n=0}\left(x-1\right)^{3n}$ and the radius of convergence is $1$.[/hide]

well i got 96 marks in the final exam i was the second in University.. :lol: . i'm really happy guys. :D .. thank you very much to everyone...i learned a lot here about calculus...

that was a calc 1 final? wow looks kind of hard for just calc 1, but still way to go.

we call it here  math1, i don't know maybe i'm calling it wrong...maybe it's not calculus1... :maybe:

HI
we Can say :an=C(n,2n)/2.^2n .We Use Inequalities and  obtain:an<1/√2n+1 and I think we can say: an<1/√π*n. .
Then by Cauchy Inequality we can say that: an>=1/n+1.

Then:(1/n+1)<=an<1/√2n+1. we see that ∑(an).^s.   is divergent for  s<=1.
and by left inequality we see for :s>2 our serie is  Convergent.(Because ∑n.^(-P)  Is Convergent for p>1).
For 1<s<=2  We use:Rabeh test and it say for this interval our serie is divergent .for Rabeh  test see ere:


Stanford
college
geometry
geometric transformation
reflection
ARML
MATHCOUNTS

Thanks to everyone who came out to the contest at Stanford yesterday.  Even with short notice we still managed to have a reasonable turnout and a very competitive contest.  i announced yesterday that the December 14 contest will be held at SJSU - i'll update the website shortly to reflect this change, but i wanted to get this information out to you right away so you can begin making plans..

i know a lot of you are probably interested in attending these contests to post tryout scores for ARML, but i would really like to encourage you to bring full teams of six from your schools whenever possible, because the more teams we have the more interesting the competition will be for the first ever Northern California championship.  We had a few really strong teams yesterday, and it was clear that many of the other schools would have been competitive with full teams..

Remember, information about the contests can be found at http://mathleague.org/california and there is a link at the bottom of the page to register for the contest..

I was unable to make it to the contest :(

Which team / group did the best?

I had to leave early...

Are the results going up?

You were there? I was. I forgot the scores already.

Hmmm,

first place got 195, second/third/fourth place got 185, and I (7th place) got a 175.  :(

That's all I remember

chenhsi you were there?

WHOA!  195!?

Hmm...that's out of 200 right?  That means 29/30 on sprint, and 8/8 target.

If I'm lucky, I got 90

:blush:

Yes, its TargetScore*10+SprintCorrect*4-SprintWrong [max 200]

I would like to point out that the sprint scoring system is proportionally equivalent to the SAT scoring system (+1, -0.25) Raw Score.

Eh, yeah, I missed 3 on sprint, 1 on target (note to self: 6+8 does not equal 16)

Schools were calculated by
SumOfTopSixPeopleIndividualScore/6+RelayTeam1+RelayTeam2+20*TeamScore
For a maximum of 400

I think...

(Relay: 3minute = 8, 6minute = 4)

Our "unofficial team score" would be

(175+61)/6+0+120=159.333...

Hmmm, that is surprisingly pathetic. Then again:
1) my teammate (leafiness0) is only a MATHCOUNTS guy
2) my team has 2 people, thereby invalidating our "relay" score
3) the Team Round was a one-man-operation...

Top school was Lynbrook, then Paly.

Wait, james, where were you sitting?

If you started at the top, and walked all the way down the middle aisle, down to the first row, turn left, and I'm the first guy.

[quote]Top school was Lynbrook, then Paly.[/quote]

HoldonasecWHAT?! I was told by the people I sent from Paly that they completely failed (as they should have, with only four people)...

it was actually Gunn that was second place.  Paly only had 4 people so could not be competitive for top school.  anyone reading this from Paly, please send more people to the next one so you can show us what you can do..  :)

Sadly I was at Stanford but couldn't take the contest due to a MUN conference.  I guess it's alright though as I probably achieved higher honors at the conference than I could have at the math contest :D

[quote="james4l"]Then you're Eugene Chen...

I was furthest right, then I switched between 2nd and 3rd row[/quote]

HA! I guessed your name right, AIME15.

[quote="AIME15"]Do you have a seating chart then?  Hmm...

I have absolutely no idea who was furthest right and switched between rows. Gotta wait till IDEA MATH.  :P[/quote]
I assumed that the Cross-Country shirt person was you. Also, you were playing Puzz Pack... Your shirt's back had a star (like most sport shirts) and the name next to the star was Eugene Chen.


Taylor series proves right inequality; for the left, I assume it should be $ n$ not $ x$. Expanding formally the left-hand expression gives

\[ 1 \plus{} \frac{\binom{n}{1}}{n} \plus{} \frac{\binom{n}{2}}{n^2} \plus{} \ldots \le 1 \plus{} \frac{1}{1} \plus{} \frac{(n\minus{}1)/n}{2!} \plus{} \frac{(n\minus{}1)(n\minus{}2)/n^2}{3!} \le 1 \plus{} \frac{1}{1} \plus{} \frac{1}{2!} \plus{} \frac{1}{3!} \plus{} \ldots\]

(where = holds if $ n<2$)

Taking a limit and the Binomial theorem does the trick.

Another easy one is to show that $ e$ is not commensurable!


What do you guys think of the "new" NCSSM Online? Is it a good substitution for going to NCSSM? Is it better or worse than NCVPS?

As an NCSSM student / NOT an NCSSM online student, it seems to me as if the courses on NCSSM Online and the [some? the corresponding ones] real NCSSM courses are becoming quite similar.

Web Physics is a painful class for all -- tons of work.

My NCSSM friend taking Computational Chem thinks it is too easy/slow-paced (but he does research in computational chem and got Siemens semifinalist for it... so... yeah.)

The rest, I have no idea.


Find a polynomial function that passes through the following points:

A(−2,0), B(−1,1), C(0,0), D(1,1), E(2,0), and F(3,1).

Any help or a starting point would be greatly appreciated. =)

Source: enrichment hmk.

[hide="General approach"]Since there are $6$ points, the polynomial must be of the fifth degree. Hence write $P(x)=ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f$, plug in the given pairs of values and solve the system with 6 variables.[/hide]

[hide="Particular approach"]Six points determine the polynomial of the fifth degree. Since $P(-2)=P(0)=P(2)=0$, we know that $P(x)=(x+2)x(x-2)(ax^{2}+bx+c)$. Plugging in the values $x=-1, x=1, x=3$, we get the system with 3 variables. Answer: $P(x)={1\over 15}x(x^{2}-4)(2x^{2}-5x-2)$[/hide]


I got about 20 errors because of this:
[code] $$\begin{eqnarray*} -103 &=& y^{2}-zx
&=& y^{2}-(y+5)(20-2y)
&=& 3y^{2}-10y-100 \end{eqnarray*} $$ [/code]
What's up with that?

1. Remove the dollar signs. Eqnarray converts from text to math mode by itself so shouldn't be used in math mode and automatically uses display mode.
2. You've forgotten \\ at the end of each equation
3. Please don't use eqnarray and use its modern replacement align. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ[/url]. In this case use [code]\begin{align*}
 -103 &= y^{2}-zx \\
 &= y^{2}-(y+5)(20-2y) \\
 &= 3y^{2}-10y-100
\end{align*}[/code]


function
algebra
polynomial
calculus
Euler
geometry
number theory

1. limit`existence problem

A right-side limit of f(x)=x at x=0 is 0.         //P1     //Thus,  lim from { x -> 0_+ } x=0
A limit of f(x)=5x-2 at x=2 is 8.                 //P2     //Thus,  lim from { x -> 2 } 5x-2=8
If t satisfies 0≤t<ε for every ε>0, then t=0.  //Q

Here, P1 and Q contradict each other.
P2 and Q too.

Proof.

Q is as follows.
There not exists t to satisfy 0<t<ε for every ε>0.  //Q`
Q` is as follows.
If 0<t is, then t<ε is impossible for every ε>0.    //Q``

P1 is as follows.   //ε-δ method
Given every ε>0 there exists a δ>0 such that |f(x)-A|<ε is possible for every x satisfying 0<x-a<δ. //provided A=a=0
It is as follows.
Given every ε>0 there exists a δ>0 such that x<ε is possible for every x satisfying 0<x<δ.  //P1`     //∵f(x)=x>0, A=0

Here, 0<x   //∵right-side limit
∴ if Q`` is true, then x<ε is impossible.
Because x is the same as t of Q``.
∴ if Q`` is true, then P1` is false.
∴ Q`` and P1` contradict each other.
∴ Q and P1 too.  //Q.E.D of P1

P2 is as follows.   //ε-δ method
Given every ε>0 there exists a δ>0 such that |f(x)-8|<ε is possible for every x satisfying 0<|x-2|<δ.  //p2`

Here, f(x) is an injective function.
∴ if x≠2, then f(x)≠8.
∴ 0<|f(x)-8| for x satisfying 0<|x-2|<δ.
∴ if Q`` is true, then |f(x)-8|<ε is impossible.
Because |f(x)-8| is the same as t of Q``.
∴ if Q`` is true, then P2` is false.
∴ Q`` and P2` contradict each other.
∴ Q and P2 too.  //Q.E.D of P2  //Q.E.D all

If Q is true, then each of 0, 8 is not a limit of each of P1, P2
and further, it can be proved that there exists no limit at all.

It is not a problem of special function. It is general problem. I will show you it.

Existence condition of limit A of f(x) at x=a is as follows.   //ε-δmethod  //Thus,  lim from { x -> a } f(x)=A
Given every ε>0 there exists a δ>0 such that |f(x)-A|<ε is possible for every x satisfying 0<|x-a|<δ.

Here, if Q is true, and if f(x) is not constant function in 0<|x-a|<δ, then there not exists limit A.

proof.

Here, f(x) is not constant function in 0<|x-a|<δ.
∴f(x) value can be different according to x in 0<|x-a|<δ.
∴f(x)≠A for some x in 0<|x-a|<δ
∴0<|f(x)-A| for the x.
∴ If Q`` is true, then |f(x)-A|<ε is impossible for the x.
Because |f(x)-A| is the same as t of Q``.
∴ if Q`` is true, then A can not be a limit of f(x) at x=a.
∴ if Q`` is true, there not exists limit A.
∴ if Q is true, there not exists limit A.   //Q.E.D

The result above means that if Q is true, then the following funtions have no limit at every x.
.All polynomial functions except for constent function. For example, 8x+3, x^2, x^5 -3x,...
.exp(x), log(x)
.sinx, cosx, tanx
...

It means that if Q is true, our almost common functions have no limit in almost domains.

Of course, it is not consistent with our common sense in mathematics.
We think these functions have limit in almost domains.
It means that unconsciously, we are denying Q.

If we will accept Q, then we should renounce Analysis.
In no limit existing, how can we set up Analysis?

Of course, Q is necessary to get limit`uniqueness.    //It is written in every book about Analysis.
But if Q is accepted, then limit can not exist at all.

Limit`uniqueness and limit`existence contradict each other.
We should throw away one of these.

What will you throw away in these?
I will do limit`uniqueness.

And I can prove limit`uniqueness false.
The known proof is false.
There is a false logic.

Analysis was drowned in fantasy.
Analysis committed an error.
An unbelievable thing happened in mathematics.
But it`s real.

Reference :
t above is a real number.
I wish you to think that every undefined character is a real number.

Reference :
Existence condition of limit A of function f(x) at x=a is as follows. //It was said above.
Given every ε>0 there exists aδ>0 such that |f(x)-A|<ε is possible for every x satisfying 0<|x-a|<δ.

Here, attend to "for every x".
It means that if |f(x)-A|<ε is impossible "for some x" satisfying 0<|x-a|<δ, then A is not a limit.
The proof above was from it.

Reference :
Negation of Q`` is as follows.
Though t>0 is, t<ε is possible for every ε>0.
And it is as follows.
There exists t satisfying 0<t<ε for every ε>0.
And it is as follows.
t satisfying 0≤t<ε for every ε>0 is not only 0.

p.s :
I regret my poor English.
I worry about my expression to be understood to you.

I`m from South Korea.
I want to unify world languages.
Further, I want to unify world itself.
For I do not want to waste my time, my life.
It`s sad to every one that every one is isolated.
I deplore myself.
I want to talk to every one in the world.
But I can`t.
I`m angry at my being in this situation.
...

p.s : Continuously, the problems of Analysis and the answer will be brought in with the following contents.
1. limit`existence problem
2. real number`density problem
3. limit`duplicity problem
4. the history of h-problem
5. a lucky misunderstanding
6. the problem of limit`duplicity in special case in pure mathematics
7. the problem of limit`duplicity in special case in pure mathematics(2)
8. cluster point problem
9. 0＜1÷n＜ε problem
10. problem related to natural number law
11. separation theorem
12. every positive number problem
13. separation theorem(2)
14. continuity problem
15. limit`duplicity in sequence
16. postface

I wish you to suspend judgement on this paper until you read it wholly.

If you will know the whole contents, you go to the following site.
http://cafe.nate.com/selftel/bbs/b5/list.html

my E-mail : monotel@daum.net

There is no contradiction. You simply don't understand the quantifiers in the definition of a limit. There is no universal $ \delta$, just a $ \delta(\epsilon)$ for each choice of $ \epsilon$.

The indication from you is not related to the problem.
I wish you not to have a prejudice.
So, would you like to read all, esp chapter 3, 12, 13? //3. limit`duplicity problem //12. every positive number problem //13. separation theorem(2).
It`s short. It not takes your time much.
http://cafe.nate.com/selftel/bbs/b5/list.html

**What do you think of the following?

2. real number`density problem

If t satisfies 0≤t<ε for every ε>0, then t=0.  //Q

Here, real number`density and Q contradict each other.

proof.

Q is as follows.  //I said above.
There not exists t to satisfy 0<t<ε for every ε>0.  //Q`

But t should exist according to real number`density, because ε≠0.
It is that ε is not 0 at all, so there should exist another number between ε and 0.

Therefore, real number`density and Q` contradict each other.    //Q.E.D

So, what is false?
It is Q.              //It will be proved in chapter 11~13.
Real number`density is true.

If I knew what you meant by "real number density", I could tell you where you went wrong, and why there is in fact no contradiction. It's not a standard English term. I'm willing to hazard a guess anyway: that you aren't reading the "for every" quantifier correctly.

The real number`density means the following.
The whole set of real numbers is dense.
Thus, if c,d∈R and c<d, then there exists x∈R satisfying c<x<d.

And I`d like to show you a part of chater 11.             //11. separation theorem  //The separation theorem means the negation of Q
It is as follows.

proof5 : the method : There exists b∈R` satisfying b>a for every a∈R`  //R` : whole set of positive real numbers   //the proof of separation theorem
Here, b>a ∴ 1/b<1/a  //∵b>a>0
And 0<1/b ∴ 0<1/b<1/a
Here, we can take 1/b=t, 1/a=ε ∴ We have 0<t<ε.
Here, a is every positive number, so 1/a=ε is every positive number too.
∴There exists t to satisfy 0<t<ε for every ε>0.   //Q.E.D

Reference : Intuitive understanding separation theorem with number line

0---------t--------ε-----------t`-----------------   //0<t<ε<t` for every ε>0

ε is not 0, no matter how ε is "every" positive number.
ε is merely positive number, no matter how ε is "every" positive number.

There exists no neighboring number to 0 such that there not exists any other number between 0 and the number.
No matter how a number exists very close to 0, there must exist a crack between 0 and the number.
And there exists at least a number in the crack.
This is just number`density.
However we take ε close to 0, ε should be separated from 0.
Therefore, however we take ε close to 0, there exists t more close to 0.
Therefore, there can exist t to satisfy 0<t<ε for every ε>0.
This is just separation theorem.

There exists t`∈R` to satisfy t`>ε for every ε>0, though ε is very very big.
Equally, there exists t∈R` to satisfy t<ε for every ε>0, though ε is very very small.
It was that proof5 showed just it mathematically.

Whole set of positive numbers is an infinite set.
Because there exists a positive number bigger than every positive number.
Whole set of positive numbers is an infinite set too.
Because there exists a positive number smaller than every positive number.

There exists a positive number smaller than every positive number?
It looks queer.
But it is just the number`density and the property of the infinite set.

**

There exists b∈R` satisfying b>a for every a∈R`  //R` : whole set of positive real numbers
This is the root of mathematics.
I think that you knew it already.

[quote]There exists a positive number smaller than every positive number? [/quote]

No, that's not the assertion.

FOR EVERY positive real number $ x$, THERE EXISTS a positive real number $ y$ such that $ 0 < y < x$.

This does not imply that you can find ONE $ y$ which 'works' for all $ x \in \mathbb{R}_\plus{}$.

jmerry's comment on the quantifiers is exactly this.

$ \forall x \in \mathbb{R}_\plus{} : \exists y \in \mathbb{R}_\plus{} : y < x$ is NOT the same as $ \exists y \in \mathbb{R}_\plus{} : \forall x \in \mathbb{R}_\plus{} : y < x$.

You said below.
"FOR EVERY positive real number , THERE EXISTS a positive real number  such that .
This does not imply that you can find ONE  which 'works' for all . "

I did not say "imply".
I did say "prove".
You see the proof5 above.
And see below.   //It`s another proof.

"There exists x>0 satisfying x<1/x."
It is as follows.
"It is for some x>0 that x<1/x is possible."
Here, one sample x>0 is sufficient to prove it.
Because it is the proof for some x.
For example, x=1/2

"There exists x>0 satisfying conditional p(x)."
It is as follows.
"It is for some x>0 that conditional p(x)  is possible."
Here, one sample x>0 is sufficient to prove it.
Because it is the proof for some x.

"There exists x>0 satisfying x<y for every y>0."  //W
It is as follows.
"It is for some x>0 that x<y  is possible for every y>0."  //W`
Here, one sample x>0 is sufficient to prove it.
Because it is the proof for some x.
For example, x=y/2
Therefore, W` is true.
Therefore, W is true.

Attend to W.
W is the negation of Q`

Do you agree above?

I wish you to see chapter 12, 13.   //12. every positive number problem  13. separation theorem(2)
http://cafe.nate.com/selftel/bbs/b5/list.html

p.s : What do you think of my other proof above? For example, real number`density problem

***********************************************

3. limit`duplicity problem

It is impossible that a limit exists uniquely.

proof.

If a limit of f(x) at x=a is A, then it is as follows. //ε-δ method
Given every ε>0 there exists aδ>0 such that |f(x)-A|<ε is possible for every x satisfying 0<|x-a|<δ.

And if real number`density is real, then it is as follows.
There exists B satisfying 0<|A-B|<ζ for every ζ>0.

Here, |f(x)-B|=|f(x)-A+A-B|≤|f(x)-A|+|A-B|<ε+ζ
∴ |f(x)-B|<ε+ζ
Here, each of ε,ζ is every positive number, so ε+ζ is every positive number too.
∴ B is a limit of f(x) at x=a too.
And 0<|A-B| ∴ B≠A

∴ If there exists a limit, then there exists another the one.  //Q.E.D

It`s a tragedy.
But it`s a little tragedy.
Because limit`s duplicity is a microscopic problem.
B≠A is possible but |A-B|<ζ for every ζ>0.
It means that the difference between A and B is infinitesimal at most.
For example, if 1 is a limit, then -2, 1/2, 2/3, 100/101, 0.9999999999, 101/100, 2, 100 are not limits.
So to speak, limit`duplicity is just a problem of infinitesimal size.
Don`t interpret the problem extensively.

Though limit is not unique, it is not fatal in Analysis.
Though limit is not unique, Infinitesimal Calculus is efficient as formerly, except for special case in pure mathematics.
So to speak, Euler, Langrange, Laplace`Infinitesimal Calculus is not inflicted with limit`s duplicity at all.
Limit`s duplicity may be ignored in most cases, in most people.

The known Analysis says that there not exists the difference between limits even infinitesimally.
The modified Analysis says that there exists the difference between limits infinitesimally.
This is the only difference between the known Analysis and the modified Analysis.

Reference :
0<|A-B|<ζ is as follows.
A-ζ<B<A+ζ  //B≠A
Here, ζ>0 ∴ A-ζ≠A+ζ
Therefore, according to real number`density, there exist innumerable real numbers between A-ζ and A+ζ.
There exists a real number that is not A.
Therefore, there exists B satisfying 0<|A-B|<ζ for every ζ>0.

Reference : Limit`duplicity can be induced more intuitionally.
Existence condition of limit A of f(x) at x=a is as follows.   //ε-δmethod
Given every ε>0 there exists aδ>0 such that |f(x)-A|<ε is possible for every x satisfying 0<|x-a|<δ.

Here, what means f(y) for 0<|y-a|<δ?

|f(x)-A|<ε is possible for every x satisfying 0<|x-a|<δ.
∴|f(y)-A|<ε is possible for every y satisfying 0<|y-a|<δ.

Here, |f(x)-f(y)|=|f(x)-A+A-f(y)|≤|f(x)-A|+|A-f(y)|<ε+ε=2ε
∴ |f(x)-f(y)|<2ε
∴ f(y) is a limit of f(x) at x=a too.

Here, y satisfying 0<|y-a|<δ is not unique.   //real number`density.
∴ If f(x) is not a constant function in 0<|x-a|<δ, then f(y) is not unique.

Essentially, limit can not be unique.
It is that all f(x) for 0<|x-a|<δ are limits.
This is similar to the concept of classical limit.

Reference : Here, I expect the following counterargument against my proof.
"Let B≠A be.
Here, ε is every positive number.
∴ We can take ε=|A-B|/2.
∴ We can induce a contradiction.
∴ We can deny B≠A.
∴ We can induce limit`s uniqueness."

It is wrong.
There is a false logic.
The reason is seen at chapter12.  //12. every positive number problem

I pointed out the problem in ε-part.      //; "|f(x)-A|<ε"
In fact, δ-part has the same problem too.  //; "0<|x-a|<δ"
Because δ is decided according to ε.  //It`s that jmerry mentioned above too.
∴δ becomes every positive number, because ε is every positive number.
∴ If Q` is true, then there not exists x satisfying 0<|x-a|<δ.
Because |x-a| is the same as t of Q` and δ is the same as ε of Q`.
In conclusion, Q causes the problem in both ε-part and δ-part.

reference :
A right-side limit of f(x)=x at x=0 is 0.         //P1
Here, 0<x<ε   //∵right-side limit and ε-δ method  //I said above.
Here, you should accept at least the following Y or Z.
x not exists. //Y
x exist innumerably. //Z
Y is from Q`, and Z is from real number`density.
If you choose Y, then the limit not exists.
If you choose Z, then the limits exist innumerably.
It is not acceptable that only one x exists.
If x exists, then x exist innumerably.
What do you choose out of Y or Z?

reference :
I don`t think that I`m the first person to raise an objection to Analysis.
Because this problem is not difficult at all.
The error that I pointed out in Analysis is the obvious error that everyone can easily discover too,
only if not suppressed under the authority of mathematics.

We should not forget that mathematicians are people and mathematics is made with the people`hands.

The recognition and the solution of this problem is not from ability but from spirit.

Some of the characteristic things that mathematical cranks do:

1. They write at great length. Their writings are often very verbal.

2. The invent their own terminology, using their own non-standard phrasing rather than standard terminology.

3. Grandiosity. They tend to claim that their insight will solve great unsolved problems, or that it will overturn some large segment of established mathematics when it is properly understood.

4. They often work in areas that are well-settled, and ignore proofs to the contrary.

5. When confronted over some flaw in their chain of reasoning - often a flaw so basic that it renders the entire long-winded structure inoperative - they never, ever admit error but continue to defend their original statements with essentially no change in argument.

6. They tend to claim that they are persecuted, or that they are lone voices in the wilderness standing up to the dissent-suppressing oligarchy of mainstream mathematics.

It turns out that analysis cranks aren't very common, not when compared to geometry cranks (circle-squarers, angle trisectors) or number theory cranks (Fermatists), but monotel fits enough of these descriptions to be labeled a crank. Note in particular the claim of suppression in his last post. A few months ago on another subforum of this site, there was a very long-running thread in which members of this community attempted to engage a different crank (a number theory crank). As dialogue, it was fruitless. As a general rule, dialogue with cranks is fruitless; they don't change their minds.

This thread is now locked.


[hide]Hi[/hide]

Try clicking the "hide" above? Does it work, as in do you see the text inside? It's certainly not working for me... It worked until yesterday...
My browser is Firefox 3.5.6 with Javascript enabled (if that has anything to do with it)
If it's just me, what can I do to fix this?

It works fine for me. I'm using Firefox 3.5.6 as well, so I don't know what's wrong.

I think something went wrong with the cache files or sth. Idk, but it started working after clearing cookies, history, and all that crap with ccleaner. So I guess it was just me :)

ya, it has to be you, because it works for me too?
I'm using firefox :)


Let $ a,b,c>0$ such that: $ a^2\plus{}b^2\plus{}c^2\plus{}abc\equal{}4$. Prove that:
$ (\sqrt{2\minus{}a}\plus{}\sqrt{2\minus{}b}\plus{}\sqrt{2\minus{}c})(\sqrt{\frac{2\minus{}a}{2\plus{}a}}\plus{}\sqrt{\frac{2\minus{}b}{2\plus{}b}}\plus{}\sqrt{\frac{2\minus{}c}{2\plus{}c}}) \ge 3\sqrt{3}$  :)

[quote="quykhtn-qa1"]Let $ a,b,c > 0$ such that: $ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \equal{} 4$. Prove that:
$ (\sqrt {2 \minus{} a} \plus{} \sqrt {2 \minus{} b} \plus{} \sqrt {2 \minus{} c})(\sqrt {\frac {2 \minus{} a}{2 \plus{} a}} \plus{} \sqrt {\frac {2 \minus{} b}{2 \plus{} b}} \plus{} \sqrt {\frac {2 \minus{} c}{2 \plus{} c}}) \ge 3\sqrt {3}$  :)[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48926


algebra
polynomial
geometry
circumcircle
analytic geometry
rectangle
6th edition

Can anyone write here conditions of problems of Round 7. My computer can't read it.
Beforehand thanks!

Problem 1.

Write the following polynomial as a product of irreducible polynomials in $\mathbb{Z}[X]$
\[ f(X) = X^{2005} - 2005 X + 2004 . \] Justify your answer.

Problem 2.

Let $ABCD$ be a cyclic quadrilateral. Let $M, N$ be the midpoints of the diagonals $AC$ and $BD$ and let $P$ be the midpoint of $MN$. Let $A',B',C',D'$ be the intersections of the rays $AP$, $BP$, $CP$ and $DP$ respectively with the circumcircle of the quadrilateral $ABCD$.

Find, with proof, the value of the sum
\[ \sigma = \frac{ AP}{PA'} + \frac{BP}{PB'} + \frac{CP}{PC'} + \frac{DP}{PD'} . \]

Problem 3.

A lattice point in the Carthesian plane is a point with both coordinates integers. A [i]rectangle minor[/i] (respectively a [i]square minor[/i]) is a set of lattice points lying inside or on the boundaries of a [i]rectangle[/i] (respectively [i]square[/i]) with vertices lattice points. We assign to each lattice point a real number, such that the sum of all the numbers in any square minor is less than 1 in absolute value.

Prove that the sum of all the numbers in any rectangle minor is less than 4 in absolute value.

[quote="Simo_the_Wolf"]I'm wrong if i say for n°3 that a rectangle minor is also a square minor?[/quote]Of course. A square minor is also a rectangle minor, but obviously not the other way around.

[quote="Valentin Vornicu"]A [i]rectangle minor[/i]  is a set of lattice points lying inside or on the boundaries of a [i]rectangle[/i] [/quote]

And what do you mean with this sentence, that i choose a rectangle with vertices lattice point and then i can choose some points inside and some points on the boundaries or i can choose ALL points inside or ALL points on the boundaries... Explain, please  :P

I think it's pretty clear what you have to prove ;) (just think that you have to find an upper bound!).

[quote="toumaf"]Well... Simo_the_wolf is right. Every rectangle is contained in some square, so every subset of the set R of [b]all[/b] points in the rectangle or on its boundary is a subset of S the set of [b]all[/b] points inside the square.[/quote]You guys are not reading the whole problem I guess. Sure every rectangle minor is [geographically] contained within a square minor, but that doesn't help you with the problem (not in the way you think it does anyway).

What I read is "[i]a rectangle minor[/i] (respectively a [i]square minor[/i]) is [size=167][b]a[/b][/size] set of lattice points lying inside or on the boundaries of a [i]rectangle[/i] (respectively a square)".

In this definition we get that any set of points with bounded coordinates is both a rectangle and a square minor.

What? :huh:
I think it's very clear: a square minor is also a rectangle minor, while not all rectangle minors are square minors (just like not all rectangles are squares!).

Post edit: hmm, ok I think I understand where the confusion arrives from. I have modified the statement of the problem as follows

[quote="Problem 3"]
A lattice point in the Carthesian plane is a point with both coordinates integers. A [i]rectangle minor[/i] (respectively a [i]square minor[/i]) is a set of lattice points formed by [i]all[/i] the lattice points lying inside or on the boundaries of a [i]rectangle[/i] (respectively [i]square[/i]) with vertices lattice points. We assign to each lattice point a real number, such that the sum of all the numbers in any square minor is less than 1 in absolute value.

Prove that the sum of all the numbers in any rectangle minor is less than 4 in absolute value.[/quote]

I think a better formulation would be:

If $R$ is any [i]rectangle[/i] (respectively [i]square[/i]) with vertices lattice points, then the set of all lattice points lying inside or on the boundaries of $R$ is called a [i]rectangle minor[/i] (respectively a [i]square minor[/i]).

Am I right, Valentin?

[b]EDIT:[/b] sorry, didn't note you already updated the wording in your post before.

  darij

Yes Darij, that is a nicer formulation indeed :)

:( can anyone tell me if squares and rectangles like (1,1);(1,-1);(-1,1);(-1,-1)  :huh:

wow an highly unusual place to start with...there were surely more obvious forums to post this... :)
anyways what do u want?...post that [url=http://www.mathlinks.ro/index.php?f=149]here[/url]

i am asking whether to these the same rule applies

I think he is asking whether rectangles in problem 3 are assumed to have their sides parallel to the axes of the coordinate system or not. So in case anyone still remembers this ML Contest...

  darij


3 equations with 3 variables as side lengths

[quote="abacadaea"]1)a rectangular prism has faces with areas of 21, 70, and 30. what is the volume of the rectangular prism?

variation: if the areas were x, y, and z, then what would the volume be?[/quote]
Ok let's call the length, width, and height l, w, and h.
Therefore,
$ x\equal{}lw$
$ y\equal{}lh$
$ z\equal{}wh$
Now the volume would be $ lwh$
So if we did $ xyz$, it would be equal to $ lw \times lh \times wh\equal{}l^2w^2h^2$
$ \sqrt{l^2w^2h^2}\equal{}lwh$
So we just take the square root of the product of the three faces.
$ V\equal{}\sqrt{xyz}$
Now just plug in the values and you get that the volume of the prism is 210.

Draw a picture.  For area 21, the sides have length 3 and 7.  Then to get an area of 70, pair the side of 7 with a side of length 10.  Then the final area will be the product of 3 and 10, or 30 as the question asks.  Then evaluate $ 3\times7\times 10\equal{}210$.

[quote="abacadaea"]1)a rectangular prism has faces with areas of 21, 70, and 30. what is the volume of the rectangular prism?

variation: if the areas were x, y, and z, then what would the volume be?[/quote]

Just use logical thinking to figure out the sides.

For example, $ 3,7$ are going to be essential. And $ 3\times 10\equal{}30$ So $ 10$ is the other side.

On variation, I totally agree with ZhangPeijin.

[quote="abacadaea"]what if you have something like 55, 11, and 10. you cant really figure out the sides then. :).

(there was something like that on a countdown round)[/quote]
Following the formula I made up it would be $ 55\sqrt{2}$

[quote="abacadaea"]1)a rectangular prism has faces with areas of 21, 70, and 30. what is the volume of the rectangular prism?[/quote]
The dimensions would be 3, 7, and 10, because 3*7=21, 7*10=70, and 3*10=30. So the volume of the prism is 3*7*10=210 cubic units.

[quote="ZhangPeijin"][quote="abacadaea"]what if you have something like 55, 11, and 10. you cant really figure out the sides then. :).

(there was something like that on a countdown round)[/quote]
Following the formula I made up it would be $ 55\sqrt {2}$[/quote]
I don't think you made the formula up. It's in the AoPS Book 1 Chaper 18 or whatever chapter that covers the 3D Geometry.

Well, basically, you use what ZhangPeijin had posted.

Whoa...
Are you serious?
 :|
Ok so it's not my formula anymore.
 :(
But anyways, I was telling the truth, I made that formula up on the spot.

[quote="alanchou"]really?  Then you're screwed.

Yeah, if the areas of the faces were like, say $ 12, 18, and 36$, try finding side lengths then.  :P[/quote]
Following the formula that was in AoPS volume 1 or something like that... (Whatever DK said)
It would be $ 36\sqrt {6}$
Ok sidelengths...
Wait I'll figure it out...
Though it could be a number of different combinations.
Wait I got one.
$ 2\sqrt{6},3\sqrt{6},\sqrt{6}$
 :|

i dont even had vol 1 :(. i just got it off some countdown round we did. only AoPS book i have is intro to geo. i got it as a "pre christmas" gift

Ahh I see.
I got volume one and two + solutions...
Right before Mr. Frost started offering us a 10% discount for them...
So yeah, stupid decision...
$ \$10$ lost.
 :(

I lost my Volume 1 :(
But, if you look in Ch. 13 of Vol. 2, even though it specifically doesn't say rectangles, it tells you right there the technique to solving these.

[quote="ZhangPeijin"][quote="DonkeyKong"][quote="alanchou"]really?  Then you're screwed.

Yeah, if the areas of the faces were like, say $ 12, 18, and 36$, try finding side lengths then.  :P[/quote]

Me think that the sides are going to be radical. Do you know how to find those without trial and error?[/quote]
There's a way to do that?
I just made educated guesses based on the factors.
:P[/quote]
Oh, I didn't see that you had an answer. I am sorry.

Well, I guess that $ 6$ is going to be crucial factor to find the lengths, because sides are all divisible by $ 6$.

Then I guess it's obvious that sides aren't going to be integers, so it has to( or may be) radicals.

I think I see a way to find those by using the prime factoring method, but I am probably wrong.


AMC
AIME
USA(J)MO
USAMO
AIME I
AMC 10
algebra

I would never be able to avoid careless mistakes on the AIME if I weren't extremely verbose with my work.  The danger is that I will add something wrong and miss a problem because of it.  Also, if you did last year's AIME I, I missed number 1 because I didn't write anything down.

[quote="SnowEverywhere"]When you write the AIME and AMCs, do you guys generally do all of the non-algebra work in your head, do you write partial solutions or do you write out complete solutions? I am just wondering because I tend to do questions in my head because of the difference in time in certain cases; however sometimes it causes me to miss something. What do you guys recommend?[/quote]

Writing Solutions:
Not on AMC 8.
Not on AMC 10
Not on AMC 12
Even not on AIME.

You will on USAMO, a [b]complete solution[/b] Grading is very strict.

And DONT DO STUFF IN YOUR HEAD on AIME!! Work it out. The AIME problems can get very hard, and I'm pretty sure most people don't want to ruin hard work by making a careless error

I think what Snow meant, AT, was that how much work do you do on scratch paper. If I'm correct in this interpretation, I generally just scribble down numbers and equations, enough so that I can tell what I did and easily check for mistakes, but not so much as to waste too much time; my only problem with this is that I tend to write small and cramped so as to maximize the amount of content that can fit on piece of paper. :P I'm also not very organized and thus my work on several problems will run into one blob of numbers, so I try to keep everything in its own problem.

I am very prone to making careless mistakes (though thankfully, much less so recently). So for me, I usually write down a lot of my computation.

Also, for harder problems such as last 5 AIME, and olympiad problems, I write out every observation that I've made to try and make it easier for me to solve the problem. I also try and figure out a plan before I attack the problem.

AMCs: I generally just write down what i'm thinking. For example if I was supposed to solve a system of two linear equations with two variables, I would probably write down the equations, then write down a similar system of equations that I can use elimination on, I add/subtract in my head, I write a linear equation with one variable down, and I solve for that variable.
I then plug it in (usually in my head if it's not too complicated) to get the other variable.

AIMES: Full solutions. Heck, you've got 3 whole hours!

USAMO: You're SUPPOSED to write full solutions.

[quote]AIMES: Full solutions. Heck, you've got 3 whole hours![/quote]

3 hours is not a long time. At all.

Writing full solutions on an AIME is not a very good idea... writing outlines takes a much shorter amount of time.

So you probably have to write a lot of work on scratch paper on AIME, and 3 hours is not a lot of time. 18 minutes per problem. Don't let that scare you.

You've still got time.

Do some tests without time first, don't worry about how much you take. (I took 9(!)hours once)
Then after you are confident you are capable of solving the problems, you can start with time.

Actually it's 12 minutes. There are 15 problems.


Pythagorean Theorem
geometry
geometry unsolved

Show how to construct right angled triangle with legs of lengths $ n$ and $ \sqrt{2n\plus{}1}$

Draw a line segment $ AB$ of length $ n \plus{} 1$. Construct the circle $ w$ with diameter $ n \plus{} 1$. Draw the circle $ m$ with centre $ A$, radius $ n$. Let circles $ w$ and $ m$ intersect at $ C,C_1$. Then traingles $ ABC$ and $ ABC_{1}$ are the required triangles, because $ AB \equal{} n \plus{} 1$; $ \angle{ACB} \equal{} \angle{AC_{1}} \equal{} 90^o$ as $ AB$ is diameter of $ w$; also $ AC \equal{} AC_{1} \equal{} n$ because circle $ m$ has radius $ n$. Then, bu the Pythagorean theorem, $ BC \equal{} BC_{1} \equal{} \sqrt {AB^2 \minus{} AC^2} \equal{} \sqrt {(n \plus{} 1)^2 \minus{} n^2} \equal{} \sqrt {2n \plus{} 1}$ which is what we want.

We always use word [b]leg[/b] when we talk about right angled triangles.
Leg is used for 2 shorter sides of right angled triangle (and hypotenuse is third and greatest side).

[b]rem[/b] thank you for your solution.


We can prove inductively that $a_k=a_{k-3}$ $\forall k\geq3,$  $k \in N$

2004 is a multiple of 3 so using the above relation
$a_{2004}$ will finally be equal to $a_0$

that is $a_{2004}=a_0=100$

I think I'm right!

 :)


$ Let a, b, c \ge0 such    that        ab \plus{} bc \plus{} ca \equal{} 3.      Prove that a^3 \plus{} b^3 \plus{} c^3 \plus{} 7abc \ge10.$

[quote="ABRORBEK"]Let $ a, b, c \ge0$ such that $ ab \plus{} bc \plus{} ca \equal{} 3.$ Prove that $ a^3 \plus{} b^3 \plus{} c^3 \plus{} 7abc \ge10.$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185868


Hello.  I have done the following problem two ways.  Please tell me which one is correct - Solution 1, or Solution 2:

Question: Find the length L from point A to the top of the pole (please see attachment):

Solution 1: Since we can make it into a 30-60-90 triangle with one leg 28, across from angle A = 30 since angle B = 60.  Thus AB = 2 * 28 = 56 feet

Solution 2: First, assume that angle C is a 90° angle. Set the equation as sin A = BC/AB (remember that the sin of an angle equals the opposite side divided the hypotenuse). Therefore, sin (60°) = 28/L (line AB = L, line BC = 28). Rearranging the equation to solve for L gives the following equation: L = 28/sin (60°). The solution gives the following result:

L=28/0.866025

L=32.33 ft

Or, 32 ft 4 in

[quote="Mark1"]
Solution 1: Since we can make it into a 30-60-90 triangle with one leg 28, across from angle A = 30 since angle B = 60.  Thus AB = 2 * 28 = 56 feet
[/quote]

well, that solution doesnt work

in a 30-60-90 right triangle, the side opposite the 30 degree angle is $\frac{1}{2}$ the side opposite the 90 degree angle.

the side opposite the 60 degree angle is the side opposite the 30 degree angle times $\sqrt{3}$

what you can do is use law of sines.

$\frac{sinA}{a} = \frac{sinB}{b} = \frac{sinC}{c}$

so in this case, that's

$\frac{sin60}{28} = \frac{sin30}{b} = \frac{sin90}{L}$

so what we have is,

$\frac{sin60}{28} = \frac{sin90}{L}$

$sin60$ is $\frac{\sqrt{3}}{2}$ and $sin90$ is $1$

so all you need to do is solve for $L$.

$L = \frac{28}{\frac{\sqrt{3}}{2}}$

$L = \frac{56\sqrt{3}}{3}$

Hi kee,

Excellent use of the law of sines.  Thank you for your time and excellent work.

 :)


Take an equilateral triangle $ABC$ with side length $1$. Take points $D$, $E$ and $F$ on sides $BC$, $CA$ and $AB$ respectively such that $BD=x$, $CE=y$ and $AF=z$. Then by area ratios, we can show that \[\sum y(1-z) = \frac{\triangle{AEF}+\triangle{BDF}+\triangle{CDE}}{\triangle{ABC}}< 1\]


modular arithmetic
vector
number theory
prime factorization

For a prime $ p$ an a positive integer $ n$, denote by $ \nu_p(n)$ the exponent of $ p$ in the prime factorization of $ n!$. Given a positive integer $ d$ and a finite set $ \{p_1,p_2,\ldots, p_k\}$ of primes, show that there are infinitely many positive integers $ n$ such that $ \nu_{p_i}(n) \equiv 0 \pmod d$, for all $ 1\leq i \leq k$.

[b]Lemma[/b] $ a$, $ b$ and $ k$ is positive integer and $ p$ is a prime number satisfying $ p^{k}>b$. Then $ v_{p}(ap^{k}\plus{}b)\equal{}v_{p}(ap^{k})\plus{}v_{p}(b)$.
[b]Proof[/b] $ v_{p}(ap^{k}\plus{}b)\minus{}v_{p}(ap^{k})$ is equal to $ p$-exponent of $ (ap^{k}\plus{}1)(\cdots)(ap^{k}\plus{}b)$.
Because $ p$-exponent of $ b$ is less than $ k$, $ ap^{k}\plus{}c$ and $ c(\le b)$ have same $ p$-exponent.
Therefore $ v_{p}(ap^{k}\plus{}b)\minus{}v_{p}(ap^{k})\equal{}v_{p}(b)$.

Let's make a infinite sequence $ \{x_{i}\equal{}p_{1}^{e_{i,1}}\cdots p_{k}^{e_{i,k}}\}$ such that $ x_{i}< p_{j}^{e_{i\plus{}1,j}}$ for all $ 1\le j\le k$. and $ x_{1}\plus{}\cdots\plus{}x_{m}\equal{}y_{m}$
There are only $ d^{k}$ possible case of $ (v_{p_{1}}(y_{m}),\cdots,v_{p_{k}}(y_{m}))\pmod d$
So there is a infinite subsequence of $ y_{m}$s $ \{y_{i_{m}}\}_{m}$ such that above vector is same.
Then by Lemma, $ v_{p_{j}}(y_{i_{a}}\minus{}y_{i_{b}}) \equiv 0 \pmod d$  $ (1\le j\le k, a>b)$
These $ y_{i_{a}}\minus{}y_{i_{b}}$ are all can be a solution.


Trying to solve this limit, I was looking at the solutions and something here doesnt make sense to me on the third phase.

The exercise was that the limit Exists and finite and you need to find a and b, and the limit.

$\lim_{x\to 0}\frac{ \frac{1+ax^{2}}{(1+bx^{2})}-cosx}{x^{6}}$

$\lim_{x\to 0}\frac{ 1+ax^{2}-(1+bx^{2})cosx}{(1+bx^{2})x^{6}}$

Third phase:

$\lim_{x\to 0}\frac{ 1+ax^{2}-(1+bx^{2})cosx}{x^{6}}$

Where does the $(1+bx^{2})$ in the denominator dissapears?

Im sure its not a mistake because the answers comes out right after a mclauren expansion of cosx and taking a and b.

This one has been bugging me for ages!

Any help would be appriciated.

Thanks.

what is done is that ${\lim_{x\to 0}}{1+bx^{2}}$ is equal to  1
so to cancle out the power's of x easiy they have applied the limit to only that part because

${\lim_{x\to 0}a \over b }$ ${=}$  ${\frac{{\lim_{x\to 0}a }}{\lim_{x\to 0}b}}$


$1340$ numbers are chosen from the set ${1,2,3,\ldots,2007}$. What is the probability that at least three of the chosen numbers are consecutive?

[hide] the number of possibilites to choose $x$ consecutive numbers from $n$ consecutive is $n-x+1$ and in this case $2007-3+1=2005$ we are to choose $1337$ numbers from $2004$ numbers, so  $2005\cdot\binom{2004}{1337}$ ops I forgot to compute the number of not consecutive numbers...it's jut all the possibilites -what I got, isn't it?[/hide]

keep thinking about it. there's a subtlety

[hide="Hint"]Contradiction. Find an example with no consecutive triplet.[/hide]

[hide]
Break up the $2007$ numbers into $669$ ascending sets of $3$ ex: $(1, 2, 3), (4, 5, 6) ... (2005, 2006, 2007)$. Dividing the $1340$ numbers chosen among the $669$ sets, $2$ slots out of each set are filled using $1338$ of the numbers chosen. We have $2$ numbers left to choose, which means that at least $2$ sets of $3$ will be filled, so we are guaranteed to have $3$ consecutive integers in any set we choose. Probability = 1.
[/hide]

[quote="Chocoboom"]OK theres only one way to get a set without at least 3 consecutive numbers. (1,2,4,5,7,8,10,11....2006,2007)
so the answer is
$1-\frac{1}{\displaystyle{2007\choose 1340}}$[/quote]

We take out all multiples of 3 right? So $2007$ would have to be taken out because $2005$ is included 2007 - 2007/3 = 1338 so there is no way to not have 3 consecutive.

your mistake is that you haven't considered other possibilities like four,five,etc. nos. to be consecutive[/quote]

amm I just thought now that my mistake is that I count the same combination more than one time


Using the 6 points on the circle for vertices, how many triangles are possible that have point $ P$ as one vertex?
[asy]draw((0,1)..(1,0)..(0,-1)..(-1,0)..cycle);
dot((1,0));
dot((-1,0));
dot((0,-1));
dot((-.8,-.6));
dot((-.707,.707));
dot((.8,.6));
label("$P$",(-.707,.707),NW);[/asy]

Since you require P as a vertex, then there are $ \binom{5}{2}$ or $ \boxed{10}$ ways to make a triangle for the other two vertices.


function
real analysis
real analysis unsolved

$ f: R \minus{} > R$ is defined by $ f(x) \equal{} x$ if $ x$ is rational, and $ f(x) \equal{} 0$ is irrational.

I need to show f does not have a limit at c if c is not equal to 0.  I've shown this true if c is rational, but I'm having a problem with if it's irrational, any help?

Let me also mention, I've been trying to do this with sequences.


algebra
function
domain
advanced fields
advanced fields unsolved

For two operators $ A,B$ (we suppose they're both defined in not trivial domains where the operations we are doing are possible) which both commute with their commutator, we know that $ \displaystyle e^{A}e^{B}\equal{}e^{A\plus{}B}e^{\frac{1}{2}[A,B]}$; is there a simple way to generalize it to two general (non-commuting) operators?

it's not a simple nor a "nice" generalization, but here you have
http://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula
where the only requirement you should need is that the set of brackets involving just $ A$'s and $ B$'s is finite dimensional (e.g. they live in the image of the representation of a finite-dimensional lie algebra).

does anyone know a more general result?


calculus
integration
trigonometry
function
real analysis
probability
floor function

establish the convergence / divergence of these 4 improper integrals:

$( \;a\; ) \;\int_{-\infty}^{\infty}\; \frac{\sin{(x^{2007})}}{x^{2006}}\;dx\;$

$( \;b\; ) \;\int_{-\infty}^{\infty}\; \frac{\sin{(x^{2007})}}{x^{2008}}\;dx\;$

$( \;c\; ) \;\int_{-\infty}^{\infty}\; \frac{\sin^{2007}{(x)}}{x^{2006}}\;dx\;$

$( \;d\; ) \;\int_{-\infty}^{\infty}\; \frac{\sin^{2007}{(x)}}{x^{2008}}\;dx\;$

(hint: [b]not [/b]as easy as it seems...  :D  need to look at
$\; \int_{-\infty}^{-1}, \int_{-1}^{0}, \int_{0}^{1}, \;\;and\;\;\int_{1}^{\infty}\;$ )

[ PS: trig functions such as $\;\sin{(x)}\;$ are not good
when it comes to looking at what's happening at $\;\pm\infty\;$ ;
therefore, limit tests will [u]not [/u]be appropriate ! ]

In all four cases, the integral of the absolute value over the set $|x|\ge 1$ is bounded by $\int_{|x|\ge 1}\frac1{|x|^{m}}\,dx$ where $m\ge 2006.$ Those all converge absolutely, hence converge. That means that in all four cases the convergence of the integral depends only on the convergence of the integral from $-1$ to $1.$ We can address this by finding the appropriate asymptotic expression for the function as $x\to 0.$

In case (a) and case (c), the integrand is asymptotic to $x$ as $x\to0.$ Hence the integral from $-1$ to $-1$ is not even truly improper, and certainly converges.

In case (b) and case (d), the integrand is asymptotic to $\frac1x$ as $x\to 0.$ Hence, in both cases, the integral diverges. (In both cases, a Cauchy principal value will converge and be equal to zero.)

Nice integrals. However, since

$\int_{-\infty}^{\infty}\frac{\sin^{2007}x}{x^{2007}}dx$ converges, it is natural to ask the following question:

Calculate

$\int_{0}^{\infty}\frac{\sin^{2007}x}{x^{2007}}dx=?$.

[quote="didilica"]Calculate

$\int_{0}^{\infty}\frac{\sin^{2007}x}{x^{2007}}dx=?$.[/quote]

computing it's [b]exact [/b]value, well, that i couldn't do...  :maybe:

in general, for
$\int_{0}^{\infty}\;\left( \frac{\sin x}{x}\right)^{m}\;dx\;\;\;\quad where \;\;\;m=1,2,3,4,5\cdots\;\;\;\cdots$

i could only go as far as $m=7$ and couldn't really spot a pattern
to be able to speculate (or compute) the integral for $m=2007$ !   :|

I can get an asymptotic equivalent for that. The first thing to to is to get it to a form friendlier to my preferred definition of the Fourier transform, with $2\pi$ in the exponent of the exponential.

$\int_{0}^{\infty}\left(\frac{\sin x}{x}\right)^{n}\,dx=\frac{\pi}2\int_{-\infty}^{\infty}\left(\frac{\sin\pi x}{\pi x}\right)^{n}\,dx$

The Fourier transform of $\frac{\sin \pi x}{\pi x}$ is the box function, $B(\xi)=\begin{cases}1,&|\xi|<\frac12\\0,&\text{otherwise}\end{cases}.$

Hence, the Fourier transform of $\left(\frac{\sin\pi x}{\pi x}\right)^{n}$ is the $n$-fold convolution of $B$ with itself, and the integral over the whole line of the original function is the value at zero of that $n$-fold convolution. It would be possible to express that with a sum formula, but I'll choose to approximate it instead. The box function is the density of a probability distribution with mean zero and variance $\frac{1}{12}.$ The $n$-fold convolution of the box function with itself is the density of a probability distribution of mean zero and variance $\frac{n}{12},$ or $\sigma=\sqrt{\frac{n}{12}}.$ By the Central Limit Theorem, this tends as $n\to\infty$ to a normal distribution, and the density at the mean of a normal distribution of variance $\sigma^{2}$ is $\frac{1}{\sqrt{2\pi}\,\sigma}.$ Hence we get an approximate value for the $n$-fold convolution at zero of $\sqrt{\frac{6}{\pi n}}.$ Multiply that by ${\frac{\pi}2}$ to get:

As $n\to\infty,$ $\int_{0}^{\infty}\left(\frac{\sin x}{x}\right)^{n}\,dx$ is asymptotic to $\sqrt{\frac{3\pi}{2n}}.$

In retrospect, we could arrive at the same place by noting that $\left(\frac{\sin x}{x}\right)^{n}\sim e^{-nx^{2}/6}$ for large $n.$

Here's the summation formula I alluded to above. The general value of the $n$-fold convolution of $B$ with itself is
\[\frac1{(n-1)!}\sum_{k=0}^{n}\binom{n}{k}(-1)^{k}H\left(\xi+\frac{n}2-k\right)\left(\xi+\frac{n}{2}-k\right)^{n-1}\]
where $H(x)$ is the Heaviside unit step function - $1$ for positive numbers, $0$ for negative $x.$

Evaluate this at zero to get
\[\frac1{(n-1)!}\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{k}(-1)^{k}\left(\frac{n}{2}-k\right)^{n-1}\]
Multiply that by $\frac{\pi}2$ to get the integral we're interested in.

Nice solution Kent,

Another method, avoiding Fourier techniques,  for calculating integrals of the form $\int_{0}^{\infty}\frac{\sin^{a}x}{x^{b}}dx$ is given by the  following two nice integral formulae:

1) $\int_{0}^{\infty}\frac{\sin^{2p}x}{x^{m}}dx=\frac{(2p)!}{\Gamma(m)}\int_{0}^{\infty}\frac{t^{m-2}}{(t^{2}+4)(t^{2}+4^{2})\cdots (t^{2}+4p^{2})}dt$, valid for $1<m<2p+1$.

2) $\int_{0}^{\infty}\frac{\sin^{2p+1}x}{x^{m}}dx=\frac{(2p+1)!}{\Gamma(m)}\int_{0}^{\infty}\frac{t^{m-1}}{(t^{2}+1)(t^{2}+3^{2})\cdots (t^{2}+(2p+1)^{2})}dt$, valid for $0<m<2p+2$.

Now in view of these formulas the exact calculation of improper integrals of the form $\int_{0}^{\infty}\frac{\sin^{a}x}{x^{b}}dx$ is solved.


algebra
polynomial
function
floor function

find an $ n_{th}$ formula for a sequence from this list of numbers
you are allowed to only choose one number from each row at a time.starting from the first row going down.and the list made only from four rows, which is coloured.

1)$ \color{red}1,3,7,9,13,15$
2)$ \color{blue}4,6,14,17,20,23$
3)$ \color{yellow}10,12,16,18,22,24$
4)$ 4,6,14,17,20,23$
5)$ 1,3,7,9,13,15$
6)$ 10,12,16,18,22,24$
7)$ \color{green}2,5,8,11,19,21$
8)$ 10,12,16,18,22,24$
9)$ 1,3,7,9,13,15$
10$ 2,5,8,11,19,21$
11)$ 4,6,14,17,20,23$

if the list was to be continued could we still make atleast one $ n_{th}$ formula

why does no one help me ?please i need to find the answer for that.

I'm sorry -- could you rephrase your question?

ok i will rephrase
basically this is a list made from those four rows that are coloured.each of the rows are one of those rows that are coloured.
1)$ \color{red}1,3,7,9,13,15$
2)$ \color{blue}4,6,14,17,20,23$
3)$ \color{yellow}10,12,16,18,22,24$
4)$ 4,6,14,17,20,23$
5)$ 1,3,7,9,13,15$
6)$ 10,12,16,18,22,24$
7)$ \color{green}2,5,8,11,19,21$
8)$ 10,12,16,18,22,24$
9)$ 1,3,7,9,13,15$
10$ 2,5,8,11,19,21$
11)$ 4,6,14,17,20,23$

ok the question is this:
is it possible to make a formula or $ n_th$ formula (by $ n_{th}$ formula i mean something like $ n_{th}\equal{}{1,4,9,16,25}\equal{}n^2$).if we only choose one element from each row once.

i mean if we choose one element from row like $ 1,4,10,4,1,10,2,10,1,2,4$ does that have a formula?
that is only an example, but can we find a sequence from that list if we choose one element from each row that has an $ n_{th}$ formula ?

Well, every finite sequence of numbers can be expressed as a polynomial, but this is often not the most appropriate formula because it often does not capture the behavior of whatever the sequence is measuring. Did these numbers come from a particular situation?

I remember you posted a sequence a while back, and when you posted the motivation, it made sense to people. Sequences can come from all kinds of places, and it is not always easy to see the motivation behind such a sequence from just a few numbers.

If it helps...
I found a function that helps repeat certain values in a polynomial function...
one sets u(x) = x - a*int( x/a ), where int() is the greatest integer function (but any step function works) and then one just makes the compostion of
P(u(x)) for some polynomial P()

see if you can find a pattern to define each polynomial in terms of those numbers listed and good luck - I hope I helped :)

you mean $ U(x)=x-a\left\lfloor{\frac{x}{a}\right\rfloor}$

can you explain more please , how do you use step function for that ?please explain more


function
advanced fields
advanced fields unsolved

I have a question about a problem which I couldn't solve completely.
For an integer $ k > \equal{} 2$ we call a positive integer $ n$ a $ k$-th power free if there is no prime number $ p$ such that
$ p^k$ divides $ n$. Define $ a_k (n) \equal{} 1$ if $ n$ is $ k$-th power-free and $ a_ k(n) \equal{} 0$ if $ n$ is not $ k$-th power free.

Define $ S: \equal{} \sum_{n \equal{} 1} ^{\infty} a_k (n)n^{ \minus{} s}$. Prove that $ S \equal{} \frac {\zeta (s)}{\zeta (ks) }$ if Re $ s > 1$.

Where $ \zeta (s) \equal{} \sum_{n \equal{} 1}^{\infty} n^{ \minus{} s} \equal{} \prod_{p} \frac {1}{1 \minus{} p^{ \minus{} s}}$ is the zeta-function.

To do this. There is a hint giving: write the LHS of the first equation as a product over the primes $ \prod_{p}(...)$.

I started as fellow:
I showed that $ S$ converges absolutely on $ \{s \in \mathbb{C} |\text{Re }s > 1 \}$ and that $ a_k(n)$ is an arithmetical function. By a theorem from the book we use we have $ S \equal{} \prod_{p} (\sum_{j \equal{} 0}^{\infty} a_k (p^j)p^{ \minus{} js})$. Since $ a_k(p^m) \equal{} 0$ for all $ m$ s.t $ k|m$ we can write the sum  inside the product as a difference of two sums:
$ \sum_{j \equal{} 0}^{\infty} a_k (p^j)p^{ \minus{} js} \equal{} \sum_{j \equal{} 0}^{\infty}p^{ \minus{} s} \minus{} \sum_{i \equal{} 1}^{\infty}p^{ \minus{} iks}$.

This gives: $ \sum_{j \equal{} 0}^{\infty} a_k (p^j)p^{ \minus{} js} \equal{} \frac {1}{1 \minus{} p^{ \minus{} s}} \plus{} \frac { \minus{} 1}{1 \minus{} p^{ \minus{} ks}} \plus{} 1$=
$ \frac {1}{1 \minus{} p^{ \minus{} s}} \plus{} \frac { \minus{} p^{ \minus{} ks}}{1 \minus{} p^{ \minus{} ks}}$.
Now I have my goal is to write :

$ S \equal{} \prod_{p} \frac {1 \minus{} p^{ \minus{} ks}}{1 \minus{} p^{ \minus{} s}} \equal{} \frac {\zeta (s)}{\zeta (ks) }$

but for some reason I don't succeed.
Can anyone help me?

Just a small interpretation mistake I think: $ a_k(p^{\alpha})\equal{}1$ if $ \alpha < k$, $ 0$ otherwise.

[quote="bambaman"]Just a small interpretation mistake I think: $ a_k(p^{\alpha}) \equal{} 1$ if $ \alpha < k$, $ 0$ otherwise.[/quote]
yes....
for example $ a_3(p^3)\equal{}a_3(p^4)\equal{}0$ but $ a_3(p)\equal{}1$.

Which means that this step is wrong:
[quote="badida"]By a theorem from the book we use we have $ S \equal{} \prod_{p} (\sum_{j \equal{} 0}^{\infty} a_k (p^j)p^{ \minus{} js})$. Since $ a_k(p^m) \equal{} 0$ for all $ m$ s.t $ k|m$ we can write the sum  inside the product as a difference of two sums:
$ \sum_{j \equal{} 0}^{\infty} a_k (p^j)p^{ \minus{} js} \equal{} \sum_{j \equal{} 0}^{\infty}p^{ \minus{} s} \minus{} \sum_{i \equal{} 1}^{\infty}p^{ \minus{} iks}$.[/quote]
You later cancel this error out by falsely deciding that $ \frac {1}{1 \minus{} p^{ \minus{} s}} \plus{} \frac { \minus{} p^{ \minus{} ks}}{1 \minus{} p^{ \minus{} ks}} \equal{} \frac {1 \minus{} p^{ \minus{} ks}}{1 \minus{} p^{ \minus{} s}}$, which happens to be the expression that you would have ended up with if you hadn't made the first error.  For the final step, just note that $ \prod \frac{x}{y} \equal{} \frac{\prod \frac{1}{y}}{\prod \frac{1}{x}}$ and you're done.

Ok! Now I see what I was doing wrong....!!

By the way I didn't  'falsely decide' what you have written but I wanted to get that expression I just didn't succed because of my previous mistakes.

Thanks a lot.


Find the least $ n\in \mathbb{N}$ such that

$ \sum_{k\equal{}1}^n\frac{k}{(k\plus{}1)!}\equal{}\frac{1}{2!}\plus{}\frac{2}{3!}\plus{}\cdots \plus{}\frac{n}{(n\plus{}1)!}\ge 0.999$

[hide="Answer"]$ \frac {k}{(k \plus{} 1)!} \equal{} \frac {k \plus{} 1 \minus{} 1}{(k \plus{} 1)!} \equal{} \frac {1}{k!} \minus{} \frac {1}{(k \plus{} 1)!}$

$ \therefore \sum_{k \equal{} 1}^n\frac {k}{(k \plus{} 1)!} \equal{} \sum_{k \equal{} 1}^n\left(\frac {1}{k!} \minus{} \frac {1}{(k \plus{} 1)!}\right) \equal{} 1 \minus{} \frac {1}{(n \plus{} 1)!}$

$ 1 \minus{} \frac {1}{(n \plus{} 1)!}\ge 0.999\equal{}\frac{999}{1000} \implies \frac {1}{(n \plus{} 1)!}\le \frac {1}{1000}$

$ \implies (n \plus{} 1)!\ge 1000$

Notice that $ 6! \equal{} 720$ and $ 7! \equal{} 5040$, so the least $ n\in \mathbb{N}$ such that the sum is greater than $ 0.999$ (or $ \frac {999}{1000}$) is $ \boxed{n \equal{} 6}$.[/hide]

[quote="Patterns_34"][hide="Answer"]
$ \sum_{k \equal{} 1}^n\left(\frac {1}{k!} \minus{} \frac {1}{(k \plus{} 1)!}\right) \equal{} 1 \minus{} \frac {1}{(n \plus{} 1)!}$
[/hide][/quote]

How did you get that closed form?

Expanding, we get [hide="this."]$ \sum_{k \equal{} 1}^n\left(\frac {1}{k!} \minus{} \frac {1}{(k \plus{} 1)!}\right) \equal{} \sum_{k \equal{} 1}^n\frac {1}{k!} \minus{} \sum_{k \equal{} 1}^n\frac {1}{(k \plus{} 1)!}$

$ \equal{} \left(1 \plus{} \frac {1}{2!} \plus{} \frac {1}{3!} \plus{} ... \plus{} \frac {1}{n!}\right) \minus{} \left(\frac {1}{2!} \plus{} \frac {1}{3!} \plus{} ... \plus{} \frac {1}{n!} \plus{} \frac {1}{(n \plus{} 1)!}\right)$
Can you finish it?[/hide]


[color=indigo]Discussion of censorship issues.  Please feel free to disscuss your views on censorship, however do not directly accuse people as we would like no hurt feelings, this is meant to be a friendly disscussion.

This is also the place to sign up for patrols.  Once 5 people sign up, we will provide more info, if you don't agree with patrols, please don't be mean about it, try to share your views in an appropriet and even tempered way.  [b][i][u]DON'T CONFRONT ANY USERS DIRECTLY!!![/u][/i][/b]  If you have found a disturbing item, please e-mail the AoPS main site, url will be posted shortly, or talk to a mod.  The mods will do whatever they want with the information you give them, this is just a little help to keep AoPS a safe place (I paraphrased nicholaskramer here).

Thank you,
lyra[/color]

I recal reading about a member who displayed interests in "girls/sex", as he put forth in his introductory post, but quite frankly, it is absurd to fear individuals on such a weak basis, especially on an internet forum where ( I assume ) you do not post any of your personal information.  In fact, I'm sure some individuals would be offended by your post, arguing a healthy interest in sex is not reasonable evidence to deem someone a "sexual predator".

I am not implying in anyway, that sex is bad or unhealthy, I am implying rather, that this is a private interest and, due to the number of young people on AoPS it could be quite dangerous/disturbing.

If you don't know what sex is, seeing the word won't do anything. If you do know, seeing the word won't do anything. On the other hand, this post is surely a very positive influence to those young innocent children. Remember kids, Big Brother is watching!

Also, someone who really is a sexual predator is not going to put anything suggesting that on his/her profile.

regardless of whether you know what it means or what it will do, it's inappropriate and does not belong here. If it was intended as a joke, (most likely, because if it were true why would it be posted) it really offends me.

Censorship is a dangerous thing.  I suggest you ignore it - if the user posts sexually themed posts he will face action from the administrators, and otherwise unless you click on his or her profile you have no reason to read it.

[quote="lyra"]In response to the issues some AoPS users are raising.  I believe that:

*It is inapropriate and wrong for sexuall things to be posted on AoPS, AoPS is a forum for math and learning to pranks.
*The administrators can not possibly keep track of 16944 and counting users.
*I am a elementry/middleschool student, and many of the students in the AoPS class I am taking are around the same age and are interested.
*What may seem like humor to highschool/college students should not be protrayed due to the wide age range and religious beliefs of AoPS users.

~lyra[/quote]

Criticism of the government is also a very dangerous thing, and we should assign a task force of one out of every 50 people to make sure that their neighbors are in line. After all, can you imagine the damage they could do?

[quote="blahblahblah"]There are currently about 45 moderators or administrators on the site, many of whom are legally adults.  Generally when inappropriate content is posted (it does happen occasionally), it is dealt with quickly.[/quote]Actually there are more than 60 mods and admins all together :) I think there is no need for "patrols" ;)

Doesn't this seem more like a "round table" discussion?

My view is that sex, alone, as an interest could mean their a reproductive health therapist, as a job, and mathematician as an interest.  I feel it unfair to, just because of that word, ban a user.  Modern-day society tends to look down at certain human activities...

I'm not going to take a position in this, but this does raise an interesting question. Let's say someone tells a joke to a group of friends in a public area. Everyone that person tells it to is fine with it and laughs. Someone who overhears it and is offended sues in court. Who do you think should win?

[hide]The person who told the joke! Because it's likely to be someone I know[/hide]

[quote="lyra"]Thank you, I had only put out the idea of a patrol with the "ok" of rrusczyk:

"just email us when you find something inappropriate. Do understand that we may not always agree with what is deemed inappropriate, but we will appreciate your input."

Just trying to help,
lyra[/quote]

Most people would not read that as approval, you know...

A: I didn't copy all of the e-mail since it wasn't public, and B: I am not saying ban the user, just maybe a litttle heads up that there are people of different ages on here.  I don't know what the admins do, but I know that a certain users interest was removed.

I believe that the Mods and Admins can see if there are any explicit things. THey search EVERYwhere in the forums. OMG they even go into the Bay Area forum....It is like the most empty forum there is....

[quote="lyra"]Who says everyone on here is a mathamatician?[/quote]

Yea...I agree some people use AoPS just for......the Other Topics Forum. ;)

I see the sides of people who are saying that it is a safe place and that the mods can take care of it but a little extra input from the users certainly wont hurt! regardless what we think is "appropriate" and what isnt if Lyra sends it to a mod,it is their decision what to do with it;basically a tip "Hey mod,this so and so happened....." if you disagree with the "patrol" you don't have to do it!!If you want to help make the fourm a safer place....great!! In all what I'm saying is... (A)it doesn't hurt anything and (b) we don't decide what to do with it,a mod does. So don't argue with us about "proper punishment"...argue with them!!

To Lyra:I personally support the "patrol" idea if you didn't pick that up already  :P

[quote="yif man12"][quote="kyyuanmathcount"]As a moderator of my forum, I make sure I check every category at least once, sometimes twice a day, I feel it is quite safe.  After all, how many mathematicians do dirty things?[/quote]

I'm pretty sure mathematicians have sex just like the rest of us. If not, it's time to look at different careers.[/quote]

Err... I meant, as in talk about that, and have it as an interest more than the average joe.

I'm not sure what this discussion is actually stemming off of, but I assure you that moderators constantly go through the threads in their forums. The whole point of having multiple moderators assigned to certain forums is that they're large enough to where it's likely one person will be unable to read or at least glance through all the content being posted. Similarly, the point of having moderators is so that admin don't have to poke through all the content themselves. :-) Any user that finds inappropriate content should send a link to the most active moderator on that forum. That's probably the best way for people to help.

As far as censorship itself goes, there's usually not much to censor except spam. Sexually explicit material RARELY lands on AoPS and when it does, we usually locate it. When we don't, the users do.

if it's in the "games and fun" forum, and it's kept to a fairly "innocent" level (like a link to a story with some sexual jokes, or something), it's perfectly fine. I mean, we're (mostly) all kids here and we do find these things funny. just no porn or anything.

[quote="yif man12"] have sex just like the rest of us. [/quote]

  As odd as it may seem, I've yet to meet an adolescent who has engaged in sexual intercourse.  Please, when you make comments that refer to your personal history, do not use terms like "us" in the process of explaining so.

  Ironically enough, never before has the topic of "sex" been more talked about than with the arrival of this "anti-sexual-explicit" post.  Congrats.

  Finally, I would like to volunteer myself as the very first "sexually offensive post" patroller.  On my agenda today:  Sam Sullivan, Rudyard Kilping, Chuck Norris, and the ever infamous Celine Dion.

i volenteer as the second!!as some people argued above...it could be meant as a "joke" but i ask....for the sake of our users...please keep it on other fourms(if they are ok with your joke that is)...i will try to help later....im a bit busy this week...

[quote="h_s_potter2002"][quote="G-UNIT"]As odd as it may seem, I've yet to meet an adolescent who has engaged in sexual intercourse.[/quote]

You probably have, but you just didn't know it.[/quote]

  I've yet to meet one in the sense I do not know him personally, nor has he made it known to me.  So perhaps that is a possibility.

and more signups? we curently have :lyra,g-unit,and me!!!! so keep argueing and signing up! :) we get...um....a safer fourm as our reward ;) lets go for 6 including lyra!


In the newest AoPs, does the solution to Problem 246 consider both odd and even cases?

I think I've found one that can skip the casework  :D :

We write $k * {n \choose k}$ as ${n \choose k}{k \choose 1}$ and use the identity ${n \choose k}{k \choose m} = {n \choose m}{n-m \choose k-m}$ to get $k * {n \choose k}={n \choose 1}{n-1 \choose k-1}$. We can now factor out ${n \choose 1}=n$ from all the terms and get $n * \left( {n-1 \choose 0} + {n-1 \choose 2} + {n-1 \choose 4} + ... \right)$ which is $ n* \left( (1+1)^{n-1} - (1-1)^{n-1}\right)/2 = n*2^{n-2}$

I'm sorry if this is the wrong place to post these kinds of dumb comments.

Well, it [i]is[/i] the other problem solving topics section. The art of problem solving book does consider both cases.

However, I think your solution is better.  :D

Well, if you say so... but, could you at least post the actual problem?

The problem is:

Evaluate ${n\choose1}+3{n\choose3}+5{n\choose5}+...$ in closed form.


search
topology
real analysis
real analysis unsolved

Let $ \{ K_\alpha \}_{\alpha \in I}$ be a family of compact sets in a metric space $ (X,d)$. Suppose that any finite intersection of sets in this family is nonempty. Prove that the intersection of all sets in this family is nonempty.


I have been scanning google for software that can help with algebra but they look very basic.

I was wondering if you people here at the math forum use any software to breakdown algebra into programming code.

My algebra is a bit weak but I am learning again, I still would like to use any good software to check results.

Basically I would like to enter in a formula and have a step by step breakdown so I can code it.

Thanks

Tinkerz

[quote="Tinkerz"]
I was wondering if you people here at the math forum use any software to breakdown algebra into programming code.[/quote]

Breakdown algebra into programming code?  :huh: Could you expand on what exactly you mean by that?

He wants a program that will solve equations for him and tell him how they were solved.

You're going to run into a compromise here. There are a few step-by-step solvers out there, but they are mostly for linear equations, quadratic equations, and other very simple forms, to my knowledge.

Programs that can crank through rough equations often use algorithms that might not be appropriate for you to use even if you knew what they were and how they worked, especially if you are learning.

[b]Could you clarify [i]what[/i] kind of equations you want to solve?[/b]

I suggest posting problems you find difficult here on AoPS. I'm sure many members would be glad to help you.

I strongly agree with igiul. You are presently using the 'software' for which you are searching: AOPS. This is the site for any mathematical or scientific question. As far as programmable software, Mathematica is what I use, and there is a free trial, but it isn't very easy to obtain nor is it permanent. And although it can show how it arrives at answers, the full version is expensive and isn't--currently--useful for learning; it's more of a mathematical tool. Moral of the story: stick with AOPS!


calculus
algebra
polynomial
function
derivative
calculus computations

Hello all, im new to the forums and im here to learn from you guys which i hope you can help me with this. I just started taking calculus and i kinda forget some stuff sometimes and since i dot have a book for it i dont have anyway to verify how something is done.
so yea here is my problem

In F(x)=ax^3+bx^2+cx +d, determine the values of the constants a,b,c,d so the polynomial function has a relative minimum at (0,0) ad a relative maximum of (2,2).

i know this might seem simple but i just cant seem to find a way around it >.> Ive tried a couple of approaches but cant seem to get anywhere.

Min = $ (0,0)$ and Max = $ (2,2)$ then $ a <0$. Since $ (0,0)$ belongs to the graph of $ f$ then $ d\equal{}0$.
\[ \begin{gathered}
  f\left( x \right) \equal{} ax^3  \plus{} bx^2  \plus{} cx \plus{} d \hfill \\
  f'\left( x \right) \equal{} 3ax^2  \plus{} 2bx \plus{} c. \hfill \\
\end{gathered}
\]Then $ 0$ and $ 2$ are roots of $ f'(x)\equal{}0$ which yields
\[ \begin{gathered}
  c \equal{} 0 \hfill \\
  3a \plus{} b \equal{} 0 .\hfill \\
\end{gathered}
\]Since $ (2,2)$ belongs to the graph of $ f$ then
\[ 8a \plus{} 4b \equal{} 2.
\]Thus
\[ a \equal{} \frac{{ \minus{} 1}}
{2},b \equal{} \frac{3}
{2},c \equal{} 0,d \equal{} 0.
\]

wow, fast response, ty very much, yet one comment, how did you get 8a+4b=2? actually, if you dont mind could you do it step by step so i know how to do it? because this is my first time doing a problem like this

Since $ c\equal{}d\equal{}0$ then $ f(x)\equal{}a x^3\plus{}b x^2$. Plug in the point $ (2,2)$ and you get $ f(2)\equal{}2\equal{}2^3a\plus{}2^2b\,\rightarrow\, 8a\plus{}4b\equal{}2$.

So therefore he did the same thing to get 3a+b=0. Or am i wrong?[/code]

No. He used the fact that $ f^{'}(2)\equal{}0$ so $ 12a\plus{}4b\equal{}0 \rightarrow 3a\plus{}b\equal{}0$

Thank you so much for your fast responses and the great help:D

if you dont mind would you like to tell me how to et the roots of the derivative?


Suppose $a \not= \frac{k\pi}{2}$ for any integer $k$ and

\[ T = \frac{\sin{a} + \tan{a}}{\cos{a} + \cot{a}}. \]
Which of the following statements s true?
(a) $T$ takes negative values only.
(b) $T$ takes nonnegative values only.
(c) $T$ takes positive values only.
(d) $T$ takes both positive and negative values.

[hide]For $a \not = \frac{k\pi}{2}$
$|\tan{a}|>|\sin{a}|$ and $|\cot{a}|>|\cos{a}|$ so the numerator's parity is always the same as that of $\tan{a}$ and the denominator's parity is always the same as $\cot{a}$, but $\tan{a}$ and $\cot{a}$ are always the same parity which means $T$ is always positive. $T$ also cannot be $0$ so the answer is $C$.

Would $B$ also be considered correct since positive is included in nonnegative?[/hide]


[quote="_NightWish_"]Sa se determine f:[0,infinit) - R  continue astfel incat

    F(2^x)-f(2^(x-1)=4^(x-1)  oricare ar fi x real[/quote]

fie t=2^(x-1), ecuatia devine f(2t) - f(t)=t^2, pentru orice t pozitiv sau aplicand de n ori avem:

f(2^n t) - f(t) = t^2 (4^n - 1) / 3, pentru orice n pozitiv. notam u=2^n t si avem:

f(u) - f(u/2^n) = [ u^2 (4^n - 1) ] / [ 4^n 3 ]. trecand la limita n->inf si tinand cont de continuitatea lui f avem:

f(u) - f(0) = u^2 / 3, adica f(u) = u^2/3 + c, cu c pozitiv.

tibi socaciu


calculus
derivative
function
analytic geometry
conics
parabola
real analysis

My text says that a sufficient set of conditions for differentiability is that one of the partials must be continuous at the point in question and all others must exist and be finite at the point.

Why is this not a counterexample $ f(x,y)\equal{}\sqrt{|xy|}$? It is definitely not differentiable at $ (0,0)$, but both partials exist and are continuous at $ (0,0)$, since $ f_x\equal{}0$ and $ f_y\equal{}0$.

In your example, neither partial derivative is defined in a neighborhood of $ (0,0)$. Being continuous is commonly (albeit not always) understood so that the function must be defined in a neighborhood of the point where it is continuous.

OK, that makes sense.

Here is something that is bothering me then. How is it possible that a function can have directional derivatives in every direction at a point, yet still not be differentiable at that point? What is it about the definition that gives it that extra strength that the two aren't equivalent.

I know that $ f(x,y) \equal{} \begin{cases} \frac {x^2y}{x^4 \plus{} y^2} ~\text{if}~ (x,y)\neq (0,0) \\
0 ~\text{if}~ (0,0) \equal{} 0 \end{cases}$ gives you the counterexample. But can't you take the Jacobian (which exists since the partials exist) to be the linear transformation that works? And then all you need if for a single one of the partials to be continuous to get the whole thing to work. I understand that in this case lack of continuity is the reason, but you can construct one that is continuous with this property too.

Crap, I should have thought these things through before the night before my final.

The essential point here is not really about derivatives - it's that the existence and equality of all directional limits does not give you the existence of a limit. Draw the coordinate plane with half-parabolas $ y\equal{}x^2$ and $ y\equal{}2x^2$, where $ x\ge 0$. Let $ G$ be the region between them. If $ f$ is a function that is equal to $ 0$ everywhere outside of $ G$, then all of its directional limits (and directional derivatives too) at $ (0,0)$ are equal to zero. This is true [u]no matter what $ f$ does inside $ G$.[/u]


function
parameterization
probability
probability and stats

I'm not really sure what steps to take here so any help would be appreciated!

Find the generating function for the negative binomial distribution below:

P(x=k) = (k-1 choose r-1) p^r * (1-p)^(k-r)

$ X \equal{} X_1 \plus{} X_2 \plus{} ... \plus{} X_r$,
where $ X_i$ are independent variables with geometric distribution of parameter $ p$.
Hence, the probability generating function is
$ G_X(t) \equal{} G_{X_1 \plus{} X_2 \plus{} ... \plus{} X_r}(t) \equal{} (G_{X_1} (t))^r$,
$ G_{X_1} (t) \equal{} \sum_{n \equal{} 1}^{ \plus{} \infty}pq^{n \minus{} 1}t^n \equal{} \sum_{n \equal{} 1}^{ \plus{} \infty}pq^{ \minus{} 1}(qt)^{n} \equal{} \frac {pt}{1 \minus{} qt}$.

Thanks, that makes sense.

I was wondering - the expectation of the negative binomial distribution is r(1-p) divided by p.

And the variance of the negative binomial distribution is r(1-p) divided by p^2.

How can one tell this by looking at the generating function:  (sp/(1-(p-1)s))^r

[quote="cornzxo"]
the expectation of the negative binomial distribution is r(1-p) divided by p.[/quote]
the expectation of the negative binomial distribution is r divided by p


Let 2*1994 boxes be arranged on a line. Can one distribute the numbers 1,1,2,2,...,1994,1994 in the boxes such that every box contains one number and for each i among the boxes with the numbers i there exactly i-1 other boxes, i = 1,2,...,1994.

Answer: it's impossible

Soln: Let's suppose we have a sequence with 2n positions formed with 1, 1, 2, 2, .., n, n s.t. between i and i there are i-1 numbers (if there are i numbers bet. i and i this is called a Langford sequence). Let the first occurence of i be a_i, so the second occurence of i must be a_i+i, so 1+2+..+2n=a_1+..+a_n+(a_1+1)+(a_2+2)+..+(a_n+n), so             2n(2n+1)/2=2(a_1+..+a_n)+n(n+1)/2, so a_1+..+a_n=(4n^2+2n-n^2-n)/4, so 4|3n^2+n=n(3n+1) iff n=0 (mod 4) or n=1 (mod 4). Unfortunately ( :D ) our n is 1994 = 2 (mod 4), so it's impossible.

I saw this soln adapted for the Langford sequences (on a website) to prove that they exist only if n=0 or 3 (mod 4).


sallam azizaye man,mikham bedoonam ageh kasi maghale ya tarhi jadid

dar riazi ebda koneh, aya jayi hast keh betooneh oonja erae bedeh, ta

benamesh sabt beshe?

aya sazmane jahani vasash vojood dareh?

too iran chetor?

ageh etelae darin mamnoon misham begin. :wink:

[quote="mohamadhosein"]ashegh man fekr konam bayad aval tooye yek journale maroof chap she.[/quote]

mohamad hoseine aziz,man aval mikham be taeed berese,

keh bad az oon beram va erae bedam, akhe ageh ghalat bashe,bad mishe

khodam keh moshkeli toosh nemibinam, vali in dalil nemishe keh kamelan

dorost va bedoone naghs bashe. :wink: